[
{
"id": 5001,
"subject": "General Science",
"question": "Let y = y(x) be the solution of the differential equation $$\\left( {(x + 2){e^{\\left( {{{y + 1} \\over {x + 2}}} \\right)}} + (y + 1)} \\right)dx = (x + 2)dy$$, y(1) = 1. If the domain of y = y(x) is an open interval ($$\\alpha$$, $$\\beta$$), then | $$\\alpha$$ + $$\\beta$$| is equal to ______________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nLet y + 1 = Y and x + 2 = X
$$ \\Rightarrow y = 12$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5004,
"subject": "General Science",
"question": "Let y = y(x) be the solution of the differential equation dy = e$$\\alpha$$x + y dx; $$\\alpha$$ $$\\in$$ N. If y(loge2) = loge2 and y(0) = loge$$\\left( {{1 \\over 2}} \\right)$$, then the value of $$\\alpha$$ is equal to _____________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$\\int {{e^{ - y}}} dy = \\int {{e^{\\alpha x}}} dx$$
If y = y(x) is the solution of the differential equation $$\\left( {1 + {e^{2x}}} \\right){{dy} \\over {dx}} + 2\\left( {1 + {y^2}} \\right){e^x} = 0$$ and y (0) = 0, then $$6\\left( {y'(0) + {{\\left( {y\\left( {{{\\log }_e}\\sqrt 3 } \\right)} \\right)}^2}} \\right)$$ is equal to
\nHence option D is correct.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5020,
"subject": "General Science",
"question": "
If the solution curve $$y = y(x)$$ of the differential equation $${y^2}dx + ({x^2} - xy + {y^2})dy = 0$$, which passes through the point (1, 1) and intersects the line $$y = \\sqrt 3 x$$ at the point $$(\\alpha ,\\sqrt 3 \\alpha )$$, then value of $${\\log _e}(\\sqrt 3 \\alpha )$$ is equal to :
Let the solution curve of the differential equation $$x \\mathrm{~d} y=\\left(\\sqrt{x^{2}+y^{2}}+y\\right) \\mathrm{d} x, x>0$$, intersect the line $$x=1$$ at $$y=0$$ and the line $$x=2$$ at $$y=\\alpha$$. Then the value of $$\\alpha$$ is :
If the solution curve of the differential equation $$\\frac{d y}{d x}=\\frac{x+y-2}{x-y}$$ passes through the points $$(2,1)$$ and $$(\\mathrm{k}+1,2), \\mathrm{k}>0$$, then
Let $$y=y(x)$$ be the solution curve of the differential equation $$ \\frac{d y}{d x}+\\left(\\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\\right) y=\\frac{(x+3)}{x+1}, x>-1$$, which passes through the point $$(0,1)$$. Then $$y(1)$$ is equal to :
$$\n\\begin{aligned}\n&y \\cdot \\frac{(x+1)^{2}(x+2)}{x+3}=\\int(x+1)(x+2) d x \\\\\\\\\n&y \\frac{(x+1)^{2}(x+2)}{x+3}=\\frac{x^{3}}{3}+\\frac{3 x^{2}}{2}+2 x+c\n\\end{aligned}\n$$\n\n
Curve passes through $(0,1)$\n\n
$$\n\\begin{gathered}\n1 \\times \\frac{1 \\times 2}{3}=0+c \\Rightarrow c=\\frac{2}{3} \\\\\\\\\n\\text { So, } y(1)=\\frac{\\frac{1}{3}+\\frac{3}{2}+2+\\frac{2}{3}}{\\frac{\\left(2^{2} \\times 3\\right)}{4}}=\\frac{3}{2}\n\\end{gathered}\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5027,
"subject": "General Science",
"question": "Let $y=y(x)$ be the solution of the differential equation \n
$\\left(3 y^{2}-5 x^{2}\\right) y \\mathrm{~d} x+2 x\\left(x^{2}-y^{2}\\right) \\mathrm{d} y=0$ \n
such that $y(1)=1$. Then $\\left|(y(2))^{3}-12 y(2)\\right|$ is equal to :",
"options": [
{
"text": "64"
},
{
"text": "$16 \\sqrt{2}$"
},
{
"text": "32"
},
{
"text": "$32 \\sqrt{2}$"
}
],
"answer": "$32 \\sqrt{2}$",
"solution": "**Answer:** $32 \\sqrt{2}$\n\n$\\left(3 y^{2}-5 x^{2}\\right) y \\cdot d x+2 x\\left(x^{2}-y^{2}\\right) d y=0$\n\n
The area enclosed by the closed curve $$\\mathrm{C}$$ given by the differential equation
$$\\frac{d y}{d x}+\\frac{x+a}{y-2}=0, y(1)=0$$ is $$4 \\pi$$.
\n
Let $$P$$ and $$Q$$ be the points of intersection of the curve $$\\mathrm{C}$$ and the $$y$$-axis. If normals at $$P$$ and $$Q$$ on the curve $$\\mathrm{C}$$ intersect $$x$$-axis at points $$R$$ and $$S$$ respectively, then the length of the line segment $$R S$$ is :
Let $$y=f(x)$$ be the solution of the differential equation $$y(x+1)dx-x^2dy=0,y(1)=e$$. Then $$\\mathop {\\lim }\\limits_{x \\to {0^ + }} f(x)$$ is equal to
Let $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ be the solution curves of the differential equation $$\\frac{d y}{d x}=y+7$$ with initial conditions $$y_{1}(0)=0$$ and $$y_{2}(0)=1$$ respectively. Then the curves $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ intersect at
",
"options": [
{
"text": "no point"
},
{
"text": "two points"
},
{
"text": "infinite number of points"
},
{
"text": "one point"
}
],
"answer": "no point",
"solution": "**Answer:** no point\n\n
The given differential equation is
\n
$$\\frac{d y}{d x} = y + 7$$
\n
This is a first order linear differential equation and can be solved using an integrating factor.
\n
Rearrange the equation to the standard form of a linear differential equation :
\n
$$\\frac{d y}{d x} - y = 7$$
\n
The integrating factor is $e^{-\\int dx} = e^{-x}$.
\n
Multiplying each side of the equation by the integrating factor gives :
\n
$$e^{-x} \\frac{d y}{d x} - e^{-x} y = 7e^{-x}$$
\n
The left-hand side of the equation is the derivative of $(e^{-x}y)$ with respect to $x$. So we can write the equation as :
\n
$$\\frac{d}{d x}(e^{-x}y) = 7e^{-x}$$
\n
Integrate both sides with respect to $x$ :
\n
$$e^{-x}y = -7e^{-x} + C$$
\n
Multiply both sides by $e^{x}$ to isolate $y$ :
\n
$$y = -7 + Ce^{x}$$
\n
So, the general solution to the differential equation is $y = -7 + Ce^{x}$.
\n
Now, let's apply the initial conditions to find the particular solutions :
\n
For $y_{1}(0)=0$, we substitute into the general solution and solve for $C$ :
\n
$$0 = -7 + C$$
\n
So, $C = 7$, and the solution for $y_{1}$ is $y_{1}(x) = 7e^{x} - 7$.
\n
For $y_{2}(0)=1$, again substitute into the general solution:
\n
$$1 = -7 + C$$
\n
So, $C = 8$, and the solution for $y_{2}$ is $y_{2}(x) = 8e^{x} - 7$.
\n
The two curves intersect when $y_{1}(x) = y_{2}(x)$. Setting these equal and solving for $x$ gives :
\n
$$7e^{x} - 7 = 8e^{x} - 7$$
\n
$$e^{x} = 0$$
\n
But $e^{x} = 0$ has no solution, because the exponential function never equals zero.
\n
So, the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ do not intersect at any point.
Let $$y=y(x)$$ be the solution of the differential equation $$\\frac{d y}{d x}+\\frac{5}{x\\left(x^{5}+1\\right)} y=\\frac{\\left(x^{5}+1\\right)^{2}}{x^{7}}, x > 0$$. If $$y(1)=2$$, then $$y(2)$$ is equal to :
$$\n\\begin{aligned}\ny \\cdot \\frac{x^5}{1+x^5} & =\\int \\frac{x^5}{\\left(1+x^5\\right)} \\times \\frac{\\left(1+x^5\\right)^2}{x^7} d x \\\\\\\\\n& =\\int x^3 d x+\\int x^{-2} d x\n\\end{aligned}\n$$\n
Let $$y=y(x)$$ be a solution curve of the differential equation.
\n
$$\\left(1-x^{2} y^{2}\\right) d x=y d x+x d y$$.
\n
If the line $$x=1$$ intersects the curve $$y=y(x)$$ at $$y=2$$ and the line $$x=2$$ intersects the curve $$y=y(x)$$ at $$y=\\alpha$$, then a value of $$\\alpha$$ is :
Let the tangent at any point P on a curve passing through the points (1, 1) and $$\\left(\\frac{1}{10}, 100\\right)$$, intersect positive $$x$$-axis and $$y$$-axis at the points A and B respectively. If $$\\mathrm{PA}: \\mathrm{PB}=1: k$$ and $$y=y(x)$$ is the solution of the differential equation $$e^{\\frac{d y}{d x}}=k x+\\frac{k}{2}, y(0)=k$$, then $$4 y(1)-6 \\log _{\\mathrm{e}} 3$$ is equal to ____________.
",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nLet the equation of tangent to the curve at $(x, y)$. \n
Integrating factor $=\\mathrm{e}^{\\int-\\frac{1}{y} d y}=\\frac{1}{y}$\n
$\\begin{aligned} & x \\cdot \\frac{1}{y}=\\int \\frac{1-y^2}{y^2} d y \\\\\\\\ & \\frac{x}{y}=\\frac{-1}{y}-y+c \\\\\\\\ & x=-1-y^2+c y\\end{aligned}$\n
$\\begin{aligned} & x(1)=1 \\\\\\\\ & 1=-1-1+c \\Rightarrow c=3 \\\\\\\\ & x=-1-y^2+3 y \\\\\\\\ & 5 x(2)=5(-1-4+6) \\\\\\\\ & =5\\end{aligned}$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5038,
"subject": "General Science",
"question": "Let $y=y(x)$ be the solution of the differential equation \n
$\\begin{aligned} \\Rightarrow & \\frac{d t}{2 t^3-t}=x d x \\\\\\\\ & \\int \\frac{1}{2 t^3-t} d t=\\int x d x \\\\\\\\ \\Rightarrow & \\int \\frac{t}{2 t^4-t^2} d t=\\int x d x\\end{aligned}$\n
$\\begin{aligned} & t^2=z \\\\\\\\ & 2 t d t=d z \\\\\\\\ & \\frac{1}{2} \\int \\frac{d z}{2 z^2-z}=\\int x d x \\\\\\\\ & \\ln \\left|\\frac{z-\\frac{1}{2}}{z}\\right|=x^2+c\\end{aligned}$\n
If $$y=y(x)$$ is the solution curve of the differential equation $$\\left(x^2-4\\right) \\mathrm{d} y-\\left(y^2-3 y\\right) \\mathrm{d} x=0, x>2, y(4)=\\frac{3}{2}$$ and the slope of the curve is never zero, then the value of $$y(10)$$ equals :
If the solution curve, of the differential equation $$\\frac{\\mathrm{d} y}{\\mathrm{~d} x}=\\frac{x+y-2}{x-y}$$ passing through the point $$(2,1)$$ is $$\\tan ^{-1}\\left(\\frac{y-1}{x-1}\\right)-\\frac{1}{\\beta} \\log _{\\mathrm{e}}\\left(\\alpha+\\left(\\frac{y-1}{x-1}\\right)^2\\right)=\\log _{\\mathrm{e}}|x-1|$$, then $$5 \\beta+\\alpha$$ is equal to __________.
Let $$y=y(x)$$ be the solution of the differential equation $$\\frac{d y}{d x}=\\frac{(\\tan x)+y}{\\sin x(\\sec x-\\sin x \\tan x)}, x \\in\\left(0, \\frac{\\pi}{2}\\right)$$ satisfying the condition $$y\\left(\\frac{\\pi}{4}\\right)=2$$. Then, $$y\\left(\\frac{\\pi}{3}\\right)$$ is
The solution curve of the differential equation \n$$y \\frac{d x}{d y}=x\\left(\\log _e x-\\log _e y+1\\right), x>0, y>0$$ passing through the point $$(e, 1)$$ is
If the solution curve $$y=y(x)$$ of the differential equation $$\\left(1+y^2\\right)\\left(1+\\log _{\\mathrm{e}} x\\right) d x+x d y=0, x > 0$$ passes through the point $$(1,1)$$ and $$y(e)=\\frac{\\alpha-\\tan \\left(\\frac{3}{2}\\right)}{\\beta+\\tan \\left(\\frac{3}{2}\\right)}$$, then $$\\alpha+2 \\beta$$ is _________.
If $$\\sin \\left(\\frac{y}{x}\\right)=\\log _e|x|+\\frac{\\alpha}{2}$$ is the solution of the differential equation $$x \\cos \\left(\\frac{y}{x}\\right) \\frac{d y}{d x}=y \\cos \\left(\\frac{y}{x}\\right)+x$$ and $$y(1)=\\frac{\\pi}{3}$$, then $$\\alpha^2$$ is equal to
Let $$Y=Y(X)$$ be a curve lying in the first quadrant such that the area enclosed by the line $$Y-y=Y^{\\prime}(x)(X-x)$$ and the co-ordinate axes, where $$(x, y)$$ is any point on the curve, is always $$\\frac{-y^2}{2 Y^{\\prime}(x)}+1, Y^{\\prime}(x) \\neq 0$$. If $$Y(1)=1$$, then $$12 Y(2)$$ equals __________.
Let $$y=y(x)$$ be the solution of the differential equation $$\\sec x \\mathrm{~d} y+\\{2(1-x) \\tan x+x(2-x)\\} \\mathrm{d} x=0$$ such that $$y(0)=2$$. Then $$y(2)$$ is equal to:
If the solution $$y=y(x)$$ of the differential equation $$(x^4+2 x^3+3 x^2+2 x+2) \\mathrm{d} y-(2 x^2+2 x+3) \\mathrm{d} x=0$$ satisfies $$y(-1)=-\\frac{\\pi}{4}$$, then $$y(0)$$ is equal to :
Let $$y=y(x)$$ be the solution of the differential equation $$(x+y+2)^2 d x=d y, y(0)=-2$$. Let the maximum and minimum values of the function $$y=y(x)$$ in $$\\left[0, \\frac{\\pi}{3}\\right]$$ be $$\\alpha$$ and $$\\beta$$, respectively. If $$(3 \\alpha+\\pi)^2+\\beta^2=\\gamma+\\delta \\sqrt{3}, \\gamma, \\delta \\in \\mathbb{Z}$$, then $$\\gamma+\\delta$$ equals _________.
Let $$y=y(x)$$ be the solution curve of the differential equation $$\\sec y \\frac{\\mathrm{d} y}{\\mathrm{~d} x}+2 x \\sin y=x^3 \\cos y, y(1)=0$$. Then $$y(\\sqrt{3})$$ is equal to:
$$\\begin{aligned}\n& \\sec y \\frac{d y}{d x}+2 x \\sin y=x^3 \\cos y \\\\\n& \\Rightarrow \\sec ^2 y \\frac{d y}{d x}+2 x \\tan y=x^3\n\\end{aligned}$$
\n
Let $$z=\\tan y$$
\n
$$\\begin{aligned}\n& \\frac{d z}{d x}=\\sec ^2 y \\frac{d y}{d x} \\\\\n& \\Rightarrow \\frac{d z}{d x}+2 x z=x^3\n\\end{aligned}$$
\n
$$\\begin{aligned}\n& \\text { I.F. }=e^{x^2} \\\\\n& \\Rightarrow z . e^{x^2}=\\int e^{x^2} \\cdot x^3 d x+c\n\\end{aligned}$$
\n
$$\\Rightarrow \\tan y \\cdot e^{x^2}=\\frac{1}{2}\\left(x^2 e^{x^2}-e^{x^2}\\right)+c$$
Let $$\\alpha|x|=|y| \\mathrm{e}^{x y-\\beta}, \\alpha, \\beta \\in \\mathbf{N}$$ be the solution of the differential equation $$x \\mathrm{~d} y-y \\mathrm{~d} x+x y(x \\mathrm{~d} y+y \\mathrm{~d} x)=0,y(1)=2$$. Then $$\\alpha+\\beta$$ is equal to ________
$$\\begin{aligned}\n& \\alpha|x|=|y| e^{x y-\\beta} \\\\\n& \\frac{x d y-y d x}{y^2}+\\frac{x y(x d y+y d x)}{y^2}=0 \\\\\n& -d\\left(\\frac{x}{y}\\right)+\\frac{x}{y} d(x y)=0\n\\end{aligned}$$
Let $$y=y(x)$$ be the solution of the differential equation $$(1+y^2) e^{\\tan x} d x+\\cos ^2 x(1+e^{2 \\tan x}) d y=0, y(0)=1$$. Then $$y\\left(\\frac{\\pi}{4}\\right)$$ is equal to
Suppose the solution of the differential equation $$\\frac{d y}{d x}=\\frac{(2+\\alpha) x-\\beta y+2}{\\beta x-2 \\alpha y-(\\beta \\gamma-4 \\alpha)}$$ represents a circle passing through origin. Then the radius of this circle is :
$$\\begin{aligned}\n& \\frac{d y}{d x}=\\frac{(2+\\alpha) x-\\beta y+2}{\\beta x-2 \\alpha y-(\\beta \\gamma-4 \\alpha)} \\\\\n& \\beta x d y-2 \\alpha y d y-(\\beta \\gamma-4 \\alpha) d y \\\\\n& =2 x d x+\\alpha x d x-\\beta y d x+2 d x \\\\\n& \\beta \\int(x d y+y d y)-\\alpha y^2-(\\beta \\gamma-4 x) y=x^2+\\frac{\\alpha x^2}{2}+2 x \\\\\n& \\beta x y-\\alpha y^2-(\\beta \\gamma-4 \\alpha) y=x^2+\\frac{\\alpha x^2}{2}+2 x \\\\\n& \\left(1+\\frac{\\alpha}{2}\\right) x^2+\\alpha y^2-\\beta x y+2 x+(\\beta \\gamma-4 \\alpha) y=0 \\\\\n& \\because \\text { this represents circle passing through origin } \\\\\n& \\Rightarrow \\beta=0 \\text { and } 1+\\frac{\\alpha}{2}=\\alpha \\\\\n& \\Rightarrow \\alpha=2\n\\end{aligned}$$
If the solution $$y(x)$$ of the given differential equation $$\\left(e^y+1\\right) \\cos x \\mathrm{~d} x+\\mathrm{e}^y \\sin x \\mathrm{~d} y=0$$ passes through the point $$\\left(\\frac{\\pi}{2}, 0\\right)$$, then the value of $$e^{y\\left(\\frac{\\pi}{6}\\right)}$$ is equal to _________.
$$\\left(e^y + 1\\right) \\cos x \\, dx + e^y \\sin x \\, dy = 0,$$
\n\n
we aim to find the value of $ e^{y\\left(\\frac{\\pi}{6}\\right)} $ given that the solution $ y(x) $ passes through the point $\\left(\\frac{\\pi}{2}, 0\\right)$.
\n\n
First, we recognize that the differential equation can be rearranged as:
\n\n
$$d\\left(e^y \\sin x\\right) + \\cos x \\, dx = 0.$$
\n\n
Integrating this expression, we obtain:
\n\n
$$ e^y \\sin x + \\sin x = C,$$
\n\n
where $ C $ is a constant. Given that the solution passes through the point $\\left(\\frac{\\pi}{2}, 0\\right)$, we substitute these values into the equation to find $ C $:
\n\n
$$ e^0 \\sin\\left(\\frac{\\pi}{2}\\right) + \\sin\\left(\\frac{\\pi}{2}\\right) = C \\Rightarrow 1 + 1 = C \\Rightarrow C = 2.$$
\n\n
Thus, the equation simplifies to:
\n\n
$$e^y \\sin x + \\sin x = 2.$$
\n\n
We now need to determine the value of $ e^{y\\left(\\frac{\\pi}{6}\\right)} $. Substituting $ x = \\frac{\\pi}{6} $ into the equation, we get:
Let f : R $$\\to$$ R be defined as $$f(x) = {x^3} + x - 5$$. If g(x) is a function such that $$f(g(x)) = x,\\forall 'x' \\in R$$, then g'(63) is equal to ________________.
Let $$f^{1}(x)=\\frac{3 x+2}{2 x+3}, x \\in \\mathbf{R}-\\left\\{\\frac{-3}{2}\\right\\}$$ For $$\\mathrm{n} \\geq 2$$, define $$f^{\\mathrm{n}}(x)=f^{1} \\mathrm{o} f^{\\mathrm{n}-1}(x)$$. If $$f^{5}(x)=\\frac{\\mathrm{a} x+\\mathrm{b}}{\\mathrm{b} x+\\mathrm{a}}, \\operatorname{gcd}(\\mathrm{a}, \\mathrm{b})=1$$, then $$\\mathrm{a}+\\mathrm{b}$$ is equal to ____________.
Let $$f(x)=x^5+2 \\mathrm{e}^{x / 4}$$ for all $$x \\in \\mathbf{R}$$. Consider a function $$g(x)$$ such that $$(g \\circ f)(x)=x$$ for all $$x \\in \\mathbf{R}$$. Then the value of $$8 g^{\\prime}(2)$$ is :
Given that $$(g \\circ f)(x) = x$$ for all $$x \\in \\mathbf{R}$$. This means $$g(f(x)) = x$$ for all $$x \\in \\mathbf{R}$$. Differentiating both sides with respect to $$x$$, we get:\n\n
$$g'(f(x)) \\cdot f'(x) = 1$$
\n\n\n\n
Now, we want to find the value of $$8g'(2)$$. To do this, we need to find a value of $$x$$ such that $$f(x) = 2$$. Let's solve for $$x$$:\n\n
$$x^5 + 2e^{x/4} = 2$$
\n\n\n\n
By inspection, we see that $$x = 0$$ is a solution. Therefore, $$f(0) = 2$$. Now, we can substitute this into our differentiated equation:\n\n
Suppose for a differentiable function $$h, h(0)=0, h(1)=1$$ and $$h^{\\prime}(0)=h^{\\prime}(1)=2$$. If $$g(x)=h\\left(\\mathrm{e}^x\\right) \\mathrm{e}^{h(x)}$$, then $$g^{\\prime}(0)$$ is equal to:
To determine $$g^{\\prime}(0)$$, we start by applying the chain rule and product rule to find the derivative of the given function $$g(x) = h\\left(\\mathrm{e}^x\\right) \\mathrm{e}^{h(x)}$$.
\n\n
The product rule states that if we have two functions $$u(x)$$ and $$v(x)$$, then the derivative of their product is given by:
\n\n
$$ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) $$
\n\n
Let's denote $$u(x) = h(\\mathrm{e}^x)$$ and $$v(x) = \\mathrm{e}^{h(x)}$$.
\n\n
First, we need to find $$u'(x)$$ and $$v'(x)$$. Using the chain rule, we find:
Let $$y=f(x)=\\sin ^{3}\\left(\\frac{\\pi}{3}\\left(\\cos \\left(\\frac{\\pi}{3 \\sqrt{2}}\\left(-4 x^{3}+5 x^{2}+1\\right)^{\\frac{3}{2}}\\right)\\right)\\right)$$. Then, at x = 1,
Let $$f(x)=\\sum_\\limits{k=1}^{10} k x^{k}, x \\in \\mathbb{R}$$. If $$2 f(2)+f^{\\prime}(2)=119(2)^{\\mathrm{n}}+1$$ then $$\\mathrm{n}$$ is equal to ___________
Let $$f(x)=\\frac{\\sin x+\\cos x-\\sqrt{2}}{\\sin x-\\cos x}, x \\in[0, \\pi]-\\left\\{\\frac{\\pi}{4}\\right\\}$$. Then $$f\\left(\\frac{7 \\pi}{12}\\right) f^{\\prime \\prime}\\left(\\frac{7 \\pi}{12}\\right)$$ is equal to
Let $$f:(-\\infty, \\infty)-\\{0\\} \\rightarrow \\mathbb{R}$$ be a differentiable function such that $$f^{\\prime}(1)=\\lim _\\limits{a \\rightarrow \\infty} a^2 f\\left(\\frac{1}{a}\\right)$$. Then $$\\lim _\\limits{a \\rightarrow \\infty} \\frac{a(a+1)}{2} \\tan ^{-1}\\left(\\frac{1}{a}\\right)+a^2-2 \\log _e a$$ is equal to
y '$${\\left( {1 + \\ell n2x} \\right)^2} = \\left[ {{{x\\ell n2x - \\ell n2} \\over x}} \\right]$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5098,
"subject": "General Science",
"question": "If y = y(x) is an implicit function of x such that loge(x + y) = 4xy, then $${{{d^2}y} \\over {d{x^2}}}$$ at x = 0 is equal to ___________.",
"options": [],
"answer": "40",
"solution": "**Answer:** 40\n\nln(x + y) = 4xy (At x = 0, y = 1)
Let f : R $$\\to$$ R satisfy $$f(x + y) = {2^x}f(y) + {4^y}f(x)$$, $$\\forall$$x, y $$\\in$$ R. If f(2) = 3, then $$14.\\,{{f'(4)} \\over {f'(2)}}$$ is equal to ____________.
$$\\text { Let } y=\\log _e\\left(\\frac{1-x^2}{1+x^2}\\right),-1 < x<1 \\text {. Then at } x=\\frac{1}{2} \\text {, the value of } 225\\left(y^{\\prime}-y^{\\prime \\prime}\\right) \\text { is equal to }$$
This expression is simplified by recognizing that $\\sin 2 \\theta = 2\\sin \\theta \\cos \\theta$ and $\\cos 2 \\theta = 2\\cos^2 \\theta -1$, leading to the expression:\n\n
For simplification, let's denote the numerator as $$N = \\sin 2\\theta - \\sin \\theta$$ and the denominator as $$D = \\cos \\theta - \\cos 2\\theta$$.\n\n
We have to compute $$\\frac{d}{d\\theta}(\\frac{dy}{dx})$$ which is $$\\frac{d}{d\\theta}(\\frac{N}{D})$$.\n\n
We can use the quotient rule for differentiation, which states that if we have a function of the form $$\\frac{u}{v}$$, then its derivative is given by $$\\frac{vu' - uv'}{v^2}$$. \n\n
So here, $$N' = 2\\cos 2\\theta - \\cos \\theta$$ and $$D' = -\\sin \\theta + 2\\sin 2\\theta$$.\n\n
Applying the quotient rule :\n\n
$$\\frac{d}{d\\theta}(\\frac{dy}{dx}) = \\frac{D N' - N D'}{D^2}$$ \n\n
Substituting the expressions for $$N'$$, $$D'$$, $$N$$, and $$D$$ we get :\n\n
$$ = - {1 \\over {{{\\left( {{{dy} \\over {dx}}} \\right)}^3}}}{{{d^2}y} \\over {d{x^2}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5109,
"subject": "General Science",
"question": "Let ƒ and g be differentiable functions on R\nsuch that fog is the identity function. If for some\na, b $$ \\in $$ R, g'(a) = 5 and g(a) = b, then ƒ'(b) is\nequal to :",
"options": [
{
"text": "1"
},
{
"text": "5"
},
{
"text": "$${2 \\over 5}$$"
},
{
"text": "$${1 \\over 5}$$"
}
],
"answer": "$${1 \\over 5}$$",
"solution": "**Answer:** $${1 \\over 5}$$\n\nGiven the function composition f(g(x)) is the identity function, it means f(g(x)) = x for all x.\n\n
Let f and g be twice differentiable even functions on ($$-$$2, 2) such that $$f\\left( {{1 \\over 4}} \\right) = 0$$, $$f\\left( {{1 \\over 2}} \\right) = 0$$, $$f(1) = 1$$ and $$g\\left( {{3 \\over 4}} \\right) = 0$$, $$g(1) = 2$$. Then, the minimum number of solutions of $$f(x)g''(x) + f'(x)g'(x) = 0$$ in $$( - 2,2)$$ is equal to ________.
Let $$f:\\mathbb{R}\\to\\mathbb{R}$$ be a differentiable function that satisfies the relation $$f(x+y)=f(x)+f(y)-1,\\forall x,y\\in\\mathbb{R}$$. If $$f'(0)=2$$, then $$|f(-2)|$$ is equal to ___________.
For the differentiable function $$f: \\mathbb{R}-\\{0\\} \\rightarrow \\mathbb{R}$$, let $$3 f(x)+2 f\\left(\\frac{1}{x}\\right)=\\frac{1}{x}-10$$, then $$\\left|f(3)+f^{\\prime}\\left(\\frac{1}{4}\\right)\\right|$$ is equal to
Suppose $$f(x)=\\frac{\\left(2^x+2^{-x}\\right) \\tan x \\sqrt{\\tan ^{-1}\\left(x^2-x+1\\right)}}{\\left(7 x^2+3 x+1\\right)^3}$$. Then the value of $$f^{\\prime}(0)$$ is equal to
Let $$f: \\mathbb{R}-\\{0\\} \\rightarrow \\mathbb{R}$$ be a function satisfying $$f\\left(\\frac{x}{y}\\right)=\\frac{f(x)}{f(y)}$$ for all $$x, y, f(y) \\neq 0$$. If $$f^{\\prime}(1)=2024$$, then
Let $$g: \\mathbf{R} \\rightarrow \\mathbf{R}$$ be a non constant twice differentiable function such that $$\\mathrm{g}^{\\prime}\\left(\\frac{1}{2}\\right)=\\mathrm{g}^{\\prime}\\left(\\frac{3}{2}\\right)$$. If a real valued function $$f$$ is defined as $$f(x)=\\frac{1}{2}[g(x)+g(2-x)]$$, then
",
"options": [
{
"text": "$$f^{\\prime \\prime}(x)=0$$ for atleast two $$x$$ in $$(0,2)$$\n"
},
{
"text": "$$f^{\\prime}\\left(\\frac{3}{2}\\right)+f^{\\prime}\\left(\\frac{1}{2}\\right)=1$$\n"
},
{
"text": "$$f^{\\prime \\prime}(x)=0$$ for no $$x$$ in $$(0,1)$$\n"
},
{
"text": "$$f^{\\prime \\prime}(x)=0$$ for exactly one $$x$$ in $$(0,1)$$"
}
],
"answer": "$$f^{\\prime \\prime}(x)=0$$ for atleast two $$x$$ in $$(0,2)$$\n",
"solution": "**Answer:** $$f^{\\prime \\prime}(x)=0$$ for atleast two $$x$$ in $$(0,2)$$\n\n\n
Also $$\\mathrm{f}^{\\prime}\\left(\\frac{1}{2}\\right)=\\frac{\\mathrm{g}^{\\prime}\\left(\\frac{1}{2}\\right)-\\mathrm{g}^{\\prime}\\left(\\frac{3}{2}\\right)}{2}=0, \\mathrm{f}^{\\prime}\\left(\\frac{1}{2}\\right)=0$$
For the curve $$C:\\left(x^{2}+y^{2}-3\\right)+\\left(x^{2}-y^{2}-1\\right)^{5}=0$$, the value of $$3 y^{\\prime}-y^{3} y^{\\prime \\prime}$$, at the point $$(\\alpha, \\alpha)$$, $$\\alpha>0$$, on C, is equal to ____________.
If $$f(x)=x^{2}+g^{\\prime}(1) x+g^{\\prime \\prime}(2)$$ and $$g(x)=f(1) x^{2}+x f^{\\prime}(x)+f^{\\prime \\prime}(x)$$, then the value of $$f(4)-g(4)$$ is equal to ____________.
If $$f(x)=\\left|\\begin{array}{ccc}\n2 \\cos ^4 x & 2 \\sin ^4 x & 3+\\sin ^2 2 x \\\\\n3+2 \\cos ^4 x & 2 \\sin ^4 x & \\sin ^2 2 x \\\\\n2 \\cos ^4 x & 3+2 \\sin ^4 x & \\sin ^2 2 x\n\\end{array}\\right|,$$ then $$\\frac{1}{5} f^{\\prime}(0)=$$ is equal to :
Let $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$ be a thrice differentiable function such that $$f(0)=0, f(1)=1, f(2)=-1, f(3)=2$$ and $$f(4)=-2$$. Then, the minimum number of zeros of $$\\left(3 f^{\\prime} f^{\\prime \\prime}+f f^{\\prime \\prime \\prime}\\right)(x)$$ is __________.
If $$y(\\theta)=\\frac{2 \\cos \\theta+\\cos 2 \\theta}{\\cos 3 \\theta+4 \\cos 2 \\theta+5 \\cos \\theta+2}$$, then at $$\\theta=\\frac{\\pi}{2}, y^{\\prime \\prime}+y^{\\prime}+y$$ is equal to :
The line y = x + 1 meets the ellipse $${{{x^2}} \\over 4} + {{{y^2}} \\over 2} = 1$$ at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2 is equal to :
Consider ellipses $$\\mathrm{E}_{k}: k x^{2}+k^{2} y^{2}=1, k=1,2, \\ldots, 20$$. Let $$\\mathrm{C}_{k}$$ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse $$\\mathrm{E}_{k}$$. If $$r_{k}$$ is the radius of the circle $$\\mathrm{C}_{k}$$, then the value of $$\\sum_\\limits{k=1}^{20} \\frac{1}{r_{k}^{2}}$$ is :
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5138,
"subject": "General Science",
"question": "STATEMENT-1 : An equation of a common tangent to the parabola $${y^2} = 16\\sqrt 3 x$$ and the ellipse $$2{x^2} + {y^2} = 4$$ is $$y = 2x + 2\\sqrt 3 $$\n
STATEMENT-2 :If line $$y = mx + {{4\\sqrt 3 } \\over m},\\left( {m \\ne 0} \\right)$$ is a common tangent to the parabola $${y^2} = 16\\sqrt {3x} $$and the ellipse $$2{x^2} + {y^2} = 4$$, then $$m$$ satisfies $${m^4} + 2{m^2} = 24$$
",
"options": [
{
"text": "Statement-1 is false, Statement-2 is true."
},
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."
},
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."
},
{
"text": "Statement-1 is true, Statement-2 is false."
}
],
"answer": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.",
"solution": "**Answer:** Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.\n\nGiven equation of ellipse is $$2{x^2} + {y^2} = 4$$\n
$$ \\Rightarrow $$ $$\\alpha = 1 + \\sqrt 2 $$ and $$\\alpha = 1 - \\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5140,
"subject": "General Science",
"question": "Let L be a common tangent line to the curves
4x2 + 9y2 = 36 and (2x)2 + (2y)2 = 31. Then the
square of the slope of the line L is __________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nTangent to the curve $${{{x^2}} \\over 9} + {{{y^2}} \\over {14}} = 1$$ is
$$y = mx + \\sqrt {9{m^2} + 4} $$
and equation of tangent to the curve $${x^2} + {y^2} = {{31} \\over 4}$$ is
If m is the slope of a common tangent to the curves $${{{x^2}} \\over {16}} + {{{y^2}} \\over 9} = 1$$ and $${x^2} + {y^2} = 12$$, then $$12{m^2}$$ is equal to :
Let a circle of radius 4 be concentric to the ellipse $$15 x^{2}+19 y^{2}=285$$. Then the common tangents are inclined to the minor axis of the ellipse at the angle :
$$\n\\tan \\theta=\\frac{1}{\\sqrt{3}} \\Rightarrow \\theta=\\frac{\\pi}{6} \\text { with } X \\text {-axis or }\n$$$\\frac{\\pi}{3}$ with $Y$-axis\n
Hence, the common tangents are inclined to the minor axis of the ellipse at an angle of $\\frac{\\pi}{3}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5143,
"subject": "General Science",
"question": "The locus of the foot of perpendicular drawn from the centre of the ellipse $${x^2} + 3{y^2} = 6$$ on any tangent to it is :",
"options": [
{
"text": "$$\\left( {{x^2} + {y^2}} \\right) ^2 = 6{x^2} + 2{y^2}$$ "
},
{
"text": "$$\\left( {{x^2} + {y^2}} \\right) ^2 = 6{x^2} - 2{y^2}$$"
},
{
"text": "$$\\left( {{x^2} - {y^2}} \\right) ^2 = 6{x^2} + 2{y^2}$$ "
},
{
"text": "$$\\left( {{x^2} - {y^2}} \\right) ^2 = 6{x^2} - 2{y^2}$$ "
}
],
"answer": "$$\\left( {{x^2} + {y^2}} \\right) ^2 = 6{x^2} + 2{y^2}$$ ",
"solution": "**Answer:** $$\\left( {{x^2} + {y^2}} \\right) ^2 = 6{x^2} + 2{y^2}$$ \n\nGiven $$e{q^n}$$ of ellipse can be written as \n
The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse $${x^2} + 2{y^2} = 4$$ is an ellipse with eccentricity :
Let $$P(2\\cos \\theta ,\\,\\sqrt 2 \\sin \\theta )$$ be any point on ellipse $${{{x^2}} \\over 4} + {{{y^2}} \\over 2} = 1$$ and Q(4, 3) and let (h, k) be the mid point of PQ
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5146,
"subject": "General Science",
"question": "Let $\\mathrm{P}$ be a point on the ellipse $\\frac{x^2}{9}+\\frac{y^2}{4}=1$. Let the line passing through $\\mathrm{P}$ and parallel to $y$-axis meet the circle $x^2+y^2=9$ at point $\\mathrm{Q}$ such that $\\mathrm{P}$ and $\\mathrm{Q}$ are on the same side of the $x$-axis. Then, the eccentricity of the locus of the point $R$ on $P Q$ such that $P R: R Q=4: 3$ as $P$ moves on the ellipse, is :",
"options": [
{
"text": "$\\frac{13}{21}$"
},
{
"text": "$\\frac{\\sqrt{139}}{23}$"
},
{
"text": "$\\frac{\\sqrt{13}}{7}$"
},
{
"text": "$\\frac{11}{19}$"
}
],
"answer": "$\\frac{\\sqrt{13}}{7}$",
"solution": "**Answer:** $\\frac{\\sqrt{13}}{7}$\n\n\n\n $\\begin{aligned} & \\mathrm{h}=3 \\cos \\theta \\\\\\\\ & \\mathrm{k}=\\frac{18}{7} \\sin \\theta\\end{aligned}$\n
$\\begin{aligned} & \\therefore \\text { locus }=\\frac{\\mathrm{x}^2}{9}+\\frac{49 \\mathrm{y}^2}{324}=1 \\\\\\\\ & \\mathrm{e}=\\sqrt{1-\\frac{324}{49 \\times 9}}=\\frac{\\sqrt{117}}{21}=\\frac{\\sqrt{13}}{7}\\end{aligned}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5147,
"subject": "General Science",
"question": "The eccentricity of an ellipse whose centre is at the origin is $${1 \\over 2}$$. If one of its directrices is x = – 4, then the\nequation of the normal to it at $$\\left( {1,{3 \\over 2}} \\right)$$ is :",
"options": [
{
"text": "2y – x = 2"
},
{
"text": "4x – 2y = 1"
},
{
"text": "4x + 2y = 7"
},
{
"text": "x + 2y = 4"
}
],
"answer": "4x – 2y = 1",
"solution": "**Answer:** 4x – 2y = 1\n\nGiven e = $${1 \\over 2}$$ and $${a \\over e}$$ = 4\n
$$ \\therefore $$ $$a$$ = 2\n
We have b2 = $$a$$2 (1 – e2) = $$4\\left( {1 - {1 \\over 4}} \\right)$$ = 3\n
$$ \\therefore $$ Equation of ellipse is\n
$${{{x^2}} \\over 4} + {{{y^2}} \\over 3} = 1$$\n
Now, the equation of normal at $$\\left( {1,{3 \\over 2}} \\right)$$ is\n
\n$$ \\therefore $$ Area = $${1 \\over 2}(QR) \\times 2 = QR = {{68} \\over {15}}$$\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5149,
"subject": "General Science",
"question": "If the normal to the ellipse 3x2\n + 4y2\n = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent\nto the ellipse at P passes through Q(4,4) then PQ is equal to :",
"options": [
{
"text": "$${{\\sqrt {61} } \\over 2}$$"
},
{
"text": "$${{\\sqrt {221} } \\over 2}$$"
},
{
"text": "$${{\\sqrt {157} } \\over 2}$$"
},
{
"text": "$${{5\\sqrt 5 } \\over 2}$$"
}
],
"answer": "$${{5\\sqrt 5 } \\over 2}$$",
"solution": "**Answer:** $${{5\\sqrt 5 } \\over 2}$$\n\nEquation of ellipse is $${{{x^2}} \\over 4} + {{{y^2}} \\over 3} = 1$$
\nNormal at P(2 cos $$\\theta $$, $$\\sqrt 3 \\sin \\theta $$) is 2x sin$$\\theta $$ - $$\\sqrt 3 y\\,cos\\theta $$ = sin $$\\theta $$ cos $$\\theta $$ as the normal is parallel to 2x + y = 4
$$ \\Rightarrow $$ $$\\beta $$ = $${{\\sqrt 2 } \\over 3}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5151,
"subject": "General Science",
"question": "Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is $${1 \\over 2}$$. If P(1, $$\\beta $$), $$\\beta $$ > 0 is a point on this ellipse, then the equation of the normal to it at P is :",
"options": [
{
"text": "4x – 3y = 2\n"
},
{
"text": "8x – 2y = 5"
},
{
"text": "7x – 4y = 1 "
},
{
"text": "4x – 2y = 1"
}
],
"answer": "4x – 2y = 1",
"solution": "**Answer:** 4x – 2y = 1\n\n$$e = {1 \\over 2}$$
If the maximum distance of normal to the ellipse $$\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1, b < 2$$, from the origin is 1, then the eccentricity of the ellipse is :
Let the tangent and normal at the point $$(3 \\sqrt{3}, 1)$$ on the ellipse $$\\frac{x^{2}}{36}+\\frac{y^{2}}{4}=1$$ meet the $$y$$-axis at the points $$A$$ and $$B$$ respectively. Let the circle $$C$$ be drawn taking $$A B$$ as a diameter and the line $$x=2 \\sqrt{5}$$ intersect $$C$$ at the points $$P$$ and $$Q$$. If the tangents at the points $$P$$ and $$Q$$ on the circle intersect at the point $$(\\alpha, \\beta)$$, then $$\\alpha^{2}-\\beta^{2}$$ is equal to :
Let $$\\mathrm{P}\\left(\\frac{2 \\sqrt{3}}{\\sqrt{7}}, \\frac{6}{\\sqrt{7}}\\right), \\mathrm{Q}, \\mathrm{R}$$ and $$\\mathrm{S}$$ be four points on the ellipse $$9 x^{2}+4 y^{2}=36$$. Let $$\\mathrm{PQ}$$ and $$\\mathrm{RS}$$ be mutually perpendicular and pass through the origin. If $$\\frac{1}{(P Q)^{2}}+\\frac{1}{(R S)^{2}}=\\frac{p}{q}$$, where $$p$$ and $$q$$ are coprime, then $$p+q$$ is equal to :
",
"options": [
{
"text": "143"
},
{
"text": "147"
},
{
"text": "137"
},
{
"text": "157"
}
],
"answer": "157",
"solution": "**Answer:** 157\n\nGiven, points $P$ and $R$ are on the ellipse defined by $9x^2+4y^2=36$ which simplifies to $\\frac{x^2}{4} + \\frac{y^2}{9} = 1$. \nThis is the standard form of the equation of an ellipse centered at the origin, with semi-major axis $a=3$ along the $y$-axis and semi-minor axis $b=2$ along the $x$-axis. \n
OP is the distance from origin O to point P, which is given by :\n\n
1. Let's represent the given point $P\\left(\\frac{2 \\sqrt{3}}{\\sqrt{7}}, \\frac{6}{\\sqrt{7}}\\right)$ in polar coordinates. We can write $P$ as $(r_1 \\cos \\theta, r_1 \\sin \\theta)$. Since $P$ lies on the ellipse, it must satisfy the equation of the ellipse. Substituting $x=r_1 \\cos \\theta$ and $y=r_1 \\sin \\theta$ into the equation of the ellipse gives us :\n\n
2. Similarly, if we represent the point $R$ as $(-r_2 \\sin \\theta, r_2 \\cos \\theta)$, (the negative sign is due to the fact that line RS is perpendicular to line PQ. Since they are perpendicular, the angle between them is 90 degrees or $\\pi/2$ radians. In terms of sin and cos, $\\sin(\\theta + \\pi/2) = \\cos(\\theta)$ and $\\cos(\\theta + \\pi/2) = -\\sin(\\theta)$.) it too should satisfy the equation of the ellipse. We have :\n\n
7. So, the final answer $p+q = 13 + 144 = 157$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 5159,
"subject": "General Science",
"question": "
Let the line $$2 x+3 y-\\mathrm{k}=0, \\mathrm{k}>0$$, intersect the $$x$$-axis and $$y$$-axis at the points $$\\mathrm{A}$$ and $$\\mathrm{B}$$, respectively. If the equation of the circle having the line segment $$A B$$ as a diameter is $$x^2+y^2-3 x-2 y=0$$ and the length of the latus rectum of the ellipse $$x^2+9 y^2=k^2$$ is $$\\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime, then $$2 \\mathrm{~m}+\\mathrm{n}$$ is equal to
Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \\over 3},{b^2} = {{32} \\over 5}$$ \n
So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$ ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5166,
"subject": "General Science",
"question": "An ellipse is drawn by taking a diameter of thec circle $${\\left( {x - 1} \\right)^2} + {y^2} = 1$$ as its semi-minor axis and a diameter of the circle $${x^2} + {\\left( {y - 2} \\right)^2} = 4$$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :",
"options": [
{
"text": "$$4{x^2} + {y^2} = 4$$ "
},
{
"text": "$${x^2} + 4{y^2} = 8$$"
},
{
"text": "$$4{x^2} + {y^2} = 8$$"
},
{
"text": "$${x^2} + 4{y^2} = 16$$"
}
],
"answer": "$${x^2} + 4{y^2} = 16$$",
"solution": "**Answer:** $${x^2} + 4{y^2} = 16$$\n\nEquation of circle is $${\\left( {x - 1} \\right)^2} + {y^2} = 1$$\n
$$ \\Rightarrow $$ radius $$=1$$ and diameter $$=2$$ \n
$$\\therefore$$ Length of semi-minor axis is $$2.$$\n
Equation of circle is $${x^2} + {\\left( {y - 2} \\right)^2} = 4 = {\\left( 2 \\right)^2}$$\n
$$ \\Rightarrow $$ radius $$=2$$ and diameter $$=4$$ \n
$$\\therefore$$ Length of semi major axis is $$4$$ \n
$$ \\Rightarrow {x^2} + 4{y^2} = 16$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5167,
"subject": "General Science",
"question": "Consider an ellipse, whose center is at the origin and its major axis is along the x-axis. If its eccentricity is $${3 \\over 5}$$ and the distance between its foci is 6, then the area (in sq. units) of the quadrilatateral inscribed in the ellipse, with the vertices as the vertices of the ellipse, is : ",
"options": [
{
"text": "8"
},
{
"text": "32"
},
{
"text": "80"
},
{
"text": "40"
}
],
"answer": "40",
"solution": "**Answer:** 40\n\ne = 3/5 & 2ae = 6 $$ \\Rightarrow $$ a = 5\n
$$ \\because $$ b2 = a2 (1 $$-$$ e2)\n
$$ \\Rightarrow $$ b2 = 25(1 $$-$$ 9/25)\n
\n $$ \\Rightarrow $$ b = 4\n
$$ \\therefore $$ Area of required quadrilateral \n
= 4(1/2 ab) = 2ab = 40",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5168,
"subject": "General Science",
"question": "The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate\naxes and passing through the points (4, −1) and (−2, 2) is :\n",
"options": [
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${2 \\over {\\sqrt 5 }}$$ "
},
{
"text": "$${{\\sqrt 3 } \\over 2}$$"
},
{
"text": "$${{\\sqrt 3 } \\over 4}$$ "
}
],
"answer": "$${{\\sqrt 3 } \\over 2}$$",
"solution": "**Answer:** $${{\\sqrt 3 } \\over 2}$$\n\nCentre at (0, 0)\n
$$ \\Rightarrow $$ e = $${{\\sqrt 3 } \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5169,
"subject": "General Science",
"question": "If the length of the latus rectum of an ellipse is 4 units and the distance between a focus an its nearest vertex on the major axis is $${3 \\over 2}$$ units, then its eccentricity is : ",
"options": [
{
"text": "$${1 \\over 2}$$ "
},
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${2 \\over 3}$$"
},
{
"text": "$${1 \\over 9}$$"
}
],
"answer": "$${1 \\over 3}$$",
"solution": "**Answer:** $${1 \\over 3}$$\n\nIf the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively, \n
$$ = \\sqrt {{2 \\over {r + 1}}} = \\sqrt {{2 \\over {r + 1}}} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5171,
"subject": "General Science",
"question": "Let the length of the latus rectum of an ellipse with its major axis along x-axis and centre at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?",
"options": [
{
"text": "$$\\left( {4\\sqrt 2 ,2\\sqrt 3 } \\right)$$"
},
{
"text": "$$\\left( {4\\sqrt 3 ,2\\sqrt 3 } \\right)$$"
},
{
"text": "$$\\left( {4\\sqrt 3 ,2\\sqrt 2 } \\right)$$"
},
{
"text": "$$\\left( {4\\sqrt 2 ,2\\sqrt 2 } \\right)$$"
}
],
"answer": "$$\\left( {4\\sqrt 3 ,2\\sqrt 2 } \\right)$$",
"solution": "**Answer:** $$\\left( {4\\sqrt 3 ,2\\sqrt 2 } \\right)$$\n\n$${{2{b^2}} \\over a} = 8$$ and 2ae $$=$$ 2b\n
$$ \\Rightarrow $$ $${b \\over a}$$ = e and 1 $$-$$ e2 = e2 $$ \\Rightarrow $$ e $$=$$ $${1 \\over {\\sqrt 2 }}$$\n
$$ \\Rightarrow $$ b = 4$$\\sqrt 2 $$ and a $$=$$ 8\n
So equation of ellipse is $${{{x^2}} \\over {64}} + {{{y^2}} \\over {32}} = 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5172,
"subject": "General Science",
"question": "Let S and S' be the foci of an ellipse and B be any one of the extremities of its minor axis. If $$\\Delta $$S'BS is a right angled triangle with right angle at B and area ($$\\Delta $$S'BS) = 8 sq. units, then the length of a latus rectum of the ellipse is :\n",
"options": [
{
"text": "2"
},
{
"text": "4$$\\sqrt 2 $$"
},
{
"text": "4"
},
{
"text": "2$$\\sqrt 2 $$"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\n\n b2 = a2e2 . . . . . . (i)\n
$${1 \\over 2}$$ S'B.SB = 8\n
S'B.SB = 16\n
a2e2 + b2 = 16 . . . . .(ii)\n
b2 = a2 (1 $$-$$ e2) . . . . .(iii)\n
using (i), (ii), (iii) a = 4\n
b = $$2\\sqrt 2 $$\n
e = $${1 \\over {\\sqrt 2 }}$$\n
$$ \\therefore $$ $$\\ell $$ (L.R) $$=$$ $${{2{b^2}} \\over a} = 4$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5173,
"subject": "General Science",
"question": "In an ellipse, with centre at the origin, if the\ndifference of the lengths of major axis and minor\naxis is 10 and one of the foci is at (0,5$$\\sqrt 3$$), then\nthe length of its latus rectum is :",
"options": [
{
"text": "5"
},
{
"text": "8"
},
{
"text": "10"
},
{
"text": "6"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\nFocus (0, be) = (0, 5$$\\sqrt 3$$)\n
= $$3\\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5176,
"subject": "General Science",
"question": "Let $${{{x^2}} \\over {{a^2}}} + {{{y^2}} \\over {{b^2}}} = 1$$ (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, $$\\phi \\left( t \\right) = {5 \\over {12}} + t - {t^2}$$, then a2 + b2 is equal to :",
"options": [
{
"text": "145"
},
{
"text": "126"
},
{
"text": "135"
},
{
"text": "116"
}
],
"answer": "126",
"solution": "**Answer:** 126\n\nGiven ellipse $${{{x^2}} \\over {{a^2}}} + {{{y^2}} \\over {{b^2}}} = 1$$ (a > b)
$$ \\Rightarrow e = {{ - 1 + \\sqrt 5 } \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5178,
"subject": "General Science",
"question": "A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of an ellipse with eccentricity $${1 \\over 3}$$ and the distance of the nearer focus from this directrix is $${8 \\over {\\sqrt {53} }}$$, then the equation of the other directrix can be :",
"options": [
{
"text": "11x + 7y + 8 = 0 or 11x + 7y $$-$$ 15 = 0"
},
{
"text": "11x $$-$$ 7y $$-$$ 8 = 0 or 11x + 7y + 15 = 0"
},
{
"text": "2x $$-$$ 7y + 29 = 0 or 2x $$-$$ 7y $$-$$ 7 = 0"
},
{
"text": "2x $$-$$ 7y $$-$$ 39 = 0 or 2x $$-$$ 7y $$-$$ 7 = 0"
}
],
"answer": "2x $$-$$ 7y + 29 = 0 or 2x $$-$$ 7y $$-$$ 7 = 0",
"solution": "**Answer:** 2x $$-$$ 7y + 29 = 0 or 2x $$-$$ 7y $$-$$ 7 = 0\n\n
Let the eccentricity of an ellipse $${{{x^2}} \\over {{a^2}}} + {{{y^2}} \\over {{b^2}}} = 1$$, $$a > b$$, be $${1 \\over 4}$$. If this ellipse passes through the point $$\\left( { - 4\\sqrt {{2 \\over 5}} ,3} \\right)$$, then $${a^2} + {b^2}$$ is equal to :
Let the maximum area of the triangle that can be inscribed in the ellipse $${{{x^2}} \\over {{a^2}}} + {{{y^2}} \\over 4} = 1,\\,a > 2$$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $$6\\sqrt 3 $$. Then the eccentricity of the ellipse is :
Let the eccentricity of the ellipse $${x^2} + {a^2}{y^2} = 25{a^2}$$ be b times the eccentricity of the hyperbola $${x^2} - {a^2}{y^2} = 5$$, where a is the minimum distance between the curves y = ex and y = logex. Then $${a^2} + {1 \\over {{b^2}}}$$ is equal to :
If the ellipse $$\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$$ meets the line $$\\frac{x}{7}+\\frac{y}{2 \\sqrt{6}}=1$$ on the $$x$$-axis and the line $$\\frac{x}{7}-\\frac{y}{2 \\sqrt{6}}=1$$ on the $$y$$-axis, then the eccentricity of the ellipse is :
If the length of the latus rectum of the ellipse $$x^{2}+4 y^{2}+2 x+8 y-\\lambda=0$$ is 4 , and $$l$$ is the length of its major axis, then $$\\lambda+l$$ is equal to ____________.
Let a line L pass through the point of intersection of the lines $$b x+10 y-8=0$$ and $$2 x-3 y=0, \\mathrm{~b} \\in \\mathbf{R}-\\left\\{\\frac{4}{3}\\right\\}$$. If the line $$\\mathrm{L}$$ also passes through the point $$(1,1)$$ and touches the circle $$17\\left(x^{2}+y^{2}\\right)=16$$, then the eccentricity of the ellipse $$\\frac{x^{2}}{5}+\\frac{y^{2}}{\\mathrm{~b}^{2}}=1$$ is :
Let C be the largest circle centred at (2, 0) and inscribed in the ellipse $${{{x^2}} \\over {36}} + {{{y^2}} \\over {16}} = 1$$. If (1, $$\\alpha$$) lies on C, then 10 $$\\alpha^2$$ is equal to ____________
\nDistance between $(1, \\alpha)$ and $(2,0)$ should be $r$\n
\n$$\n\\begin{aligned}\n& 1+\\alpha^{2}=\\frac{128}{10} \\\\\\\\\n& \\alpha^{2}=\\frac{118}{10} \\\\\\\\\n&=118.00\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5187,
"subject": "General Science",
"question": "Let an ellipse with centre $(1,0)$ and latus rectum of length $\\frac{1}{2}$ have its major axis along $\\mathrm{x}$-axis. If its minor axis subtends an angle $60^{\\circ}$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to ____________.",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\n $$\n\\begin{aligned}\n& \\frac{2 b^2}{a}=\\frac{1}{2}, \\quad \\tan 30^{\\circ}=\\frac{b}{a e} \\\\\\\\\n& b^2=\\frac{a}{4}, \\frac{1}{3}=\\frac{b^2}{a^2-b^2} \\Rightarrow a^2-b^2=3 b^2 \\Rightarrow b^2=\\frac{a^2}{4} \\\\\\\\\n& \\Rightarrow \\quad a=1, b^2=\\frac{1}{4} \\Rightarrow b=\\frac{1}{2} \\\\\\\\\n& \\Rightarrow \\quad(2 a+2 b)^2=9\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5188,
"subject": "General Science",
"question": "
Let the ellipse $$E:{x^2} + 9{y^2} = 9$$ intersect the positive x and y-axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is $${m \\over n}$$, where m and n are coprime, then $$m - n$$ is equal to :
In a group of 100 persons 75 speak English and 40 speak Hindi. Each person speaks at least one of the two languages. If the number of persons, who speak only English is $$\\alpha$$ and the number of persons who speak only Hindi is $$\\beta$$, then the eccentricity of the ellipse $$25\\left(\\beta^{2} x^{2}+\\alpha^{2} y^{2}\\right)=\\alpha^{2} \\beta^{2}$$ is :
Let $$P$$ be a parabola with vertex $$(2,3)$$ and directrix $$2 x+y=6$$. Let an ellipse $$E: \\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1, a>b$$, of eccentricity $$\\frac{1}{\\sqrt{2}}$$ pass through the focus of the parabola $$P$$. Then, the square of the length of the latus rectum of $$E$$, is
Let $$A(\\alpha, 0)$$ and $$B(0, \\beta)$$ be the points on the line $$5 x+7 y=50$$. Let the point $$P$$ divide the line segment $$A B$$ internally in the ratio $$7:3$$. Let $$3 x-25=0$$ be a directrix of the ellipse $$E: \\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$$ and the corresponding focus be $$S$$. If from $$S$$, the perpendicular on the $$x$$-axis passes through $$P$$, then the length of the latus rectum of $$E$$ is equal to,
Let $$f(x)=x^2+9, g(x)=\\frac{x}{x-9}$$ and $$\\mathrm{a}=f \\circ g(10), \\mathrm{b}=g \\circ f(3)$$. If $$\\mathrm{e}$$ and $$l$$ denote the eccentricity and the length of the latus rectum of the ellipse $$\\frac{x^2}{\\mathrm{a}}+\\frac{y^2}{\\mathrm{~b}}=1$$, then $$8 \\mathrm{e}^2+l^2$$ is equal to.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5195,
"subject": "General Science",
"question": "The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $${{{x^2}} \\over 9} + {{{y^2}} \\over 5} = 1$$, is :",
"options": [
{
"text": "$${{27 \\over 2}}$$"
},
{
"text": "$$27$$ "
},
{
"text": "$${{27 \\over 4}}$$"
},
{
"text": "$$18$$"
}
],
"answer": "$$27$$ ",
"solution": "**Answer:** $$27$$ \n\nThe end point of latus rectum of ellipse\n
$${{{x^2}} \\over {{a^2}}} + {{{y^2}} \\over {{b^2}}} = 1$$ in first quadrant is \n
$$\\left( {ae,{{{b^2}} \\over a}} \\right)$$ and the tangent at this point \n
intersects $$x$$-axis at $$\\left( {{a \\over e},0} \\right)$$ and \n
$$y$$-axis at $$(0,a).$$\n
The given ellipse is $${{x{}^2} \\over 9} + {{{y^2}} \\over 5} = 1$$\n
$$\\therefore$$ end point of latus rectum in first quadrant is \n
$$L\\left( {2,\\,\\,5/3} \\right)$$ \n
Equation of tangent at $$L$$ is $${{2x} \\over 9} + {y \\over 3} - 1$$\n
It meets $$x$$-axis at $$A(9/2, 0)$$ \n
and $$y$$-axis at $$B(0,3)$$ \n
$$\\therefore$$ Area of $$\\Delta OAB = {1 \\over 2} \\times {9 \\over 2} \\times 3 = {{27} \\over 4}$$ \n
By symmetry area of quadrilateral \n
$$ = 4 \\times \\left( {Area\\,\\,\\Delta OAB} \\right) = 4 \\times {{27} \\over 4} = 27\\,\\,$$ sq. units.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5196,
"subject": "General Science",
"question": "If the tangent at a point on the ellipse $${{{x^2}} \\over {27}} + {{{y^2}} \\over 3} = 1$$ meets the coordinate axes at A and B, and O is the origin, then the\nminimum area (in sq. units) of the triangle OAB is :",
"options": [
{
"text": "$${9 \\over 2}$$ "
},
{
"text": "$$3\\sqrt 3 $$ "
},
{
"text": "$$9\\sqrt 3 $$"
},
{
"text": "9"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\nEquation of tangent to ellipse\n
also f'($$\\theta$$) changes sign $$-$$ to + hence minimum.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5205,
"subject": "General Science",
"question": "Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3, $$-$$4), one focus at (4, $$-$$4) and one vertex at (5, $$-$$4). If mx $$-$$ y = 4, m > 0 is a tangent to the ellipse E, then the value of 5m2 is equal to _____________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nGiven C(3, $$-$$4), S(4, $$-$$4)
So, 5m2 = 3.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5206,
"subject": "General Science",
"question": "If a tangent to the ellipse x2 + 4y2 = 4 meets the tangents at the extremities of it major axis at B and C, then the circle with BC as diameter passes through the point :",
"options": [
{
"text": "$$(\\sqrt 3 ,0)$$"
},
{
"text": "$$(\\sqrt 2 ,0)$$"
},
{
"text": "(1, 1)"
},
{
"text": "($$-$$1, 1)"
}
],
"answer": "$$(\\sqrt 3 ,0)$$",
"solution": "**Answer:** $$(\\sqrt 3 ,0)$$\n\n
$${{{x^2}} \\over 4} + {{{y^2}} \\over 1} = 1$$
Equation of tangent i (cos$$\\theta$$)x + 2sin$$\\theta$$y = 2
so, $$\\left( {\\sqrt 3 ,0} \\right)$$ satisfying option (1)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5207,
"subject": "General Science",
"question": "On the ellipse $${{{x^2}} \\over 8} + {{{y^2}} \\over 4} = 1$$ let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x + 2y = 0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5 $$-$$ e2). A is :",
"options": [
{
"text": "6"
},
{
"text": "12"
},
{
"text": "14"
},
{
"text": "24"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\n
Equation of tangent : y = 2x + 6 at P
$$\\therefore$$ P($$-$$8/3, 2/3)
$$e = {1 \\over {\\sqrt 2 }}$$
S & S' = ($$-$$2, 0) & (2, 0)
Area of $$\\Delta$$SPS' = $${1 \\over 2} \\times 4 \\times {2 \\over 3}$$
A = $${4 \\over 3}$$
$$\\therefore$$ (5 $$-$$ e2)A = $$\\left( {5 - {1 \\over 2}} \\right)$$$${4 \\over 3}$$ = 6",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5208,
"subject": "General Science",
"question": "If the minimum area of the triangle formed by a tangent to the ellipse $${{{x^2}} \\over {{b^2}}} + {{{y^2}} \\over {4{a^2}}} = 1$$ and the co-ordinate axis is kab, then k is equal to _______________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nTangent
If two tangents drawn from a point ($$\\alpha$$, $$\\beta$$) lying on the ellipse 25x2 + 4y2 = 1 to the parabola y2 = 4x are such that the slope of one tangent is four times the other, then the value of (10$$\\alpha$$ + 5)2 + (16$$\\beta$$2 + 50)2 equals ___________.
",
"options": [],
"answer": "2929",
"solution": "**Answer:** 2929\n\n$\\because(\\alpha, \\beta)$ lies on the given ellipse, $25 \\alpha^{2}+4 \\beta^{2}=1\\quad\\quad...(i)$\n
\nTangent to the parabola, $y=m x+\\frac{1}{m}$ passes through $(\\alpha, \\beta)$. So, $\\alpha m^{2}-\\beta m+1=0$ has roots $m_{1}$ and $4 m_{1}$,\n
\n$$\nm_{1}+4 m_{1}=\\frac{\\beta}{\\alpha} \\text { and } m_{1} \\cdot 4 m_{1}=\\frac{1}{\\alpha}\n$$\n
\nGives that $4 \\beta^{2}=25 \\alpha \\quad\\quad...(ii)$\n
Let the tangents at the points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ on the ellipse $$\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$$ meet at the point $$R(\\sqrt{2}, 2 \\sqrt{2}-2)$$. If $$\\mathrm{S}$$ is the focus of the ellipse on its negative major axis, then $$\\mathrm{SP}^{2}+\\mathrm{SQ}^{2}$$ is equal to ___________.
The line $$x=8$$ is the directrix of the ellipse $$\\mathrm{E}:\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$$ with the corresponding focus $$(2,0)$$. If the tangent to $$\\mathrm{E}$$ at the point $$\\mathrm{P}$$ in the first quadrant passes through the point $$(0,4\\sqrt3)$$ and intersects the $$x$$-axis at $$\\mathrm{Q}$$, then $$(3\\mathrm{PQ})^{2}$$ is equal to ____________.
Let a tangent to the curve $$9{x^2} + 16{y^2} = 144$$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is ________
This shows that it is an ellipse centered at (0,0) with semi-major axis a = 6 along the x-axis and semi-minor axis b = 3 along the y-axis.\n\n
The equation of a circle with center (2,0) and radius r is :\n\n
$$(x-2)^2 + y^2 = r^2$$\n\n
Substituting y^2 from the ellipse equation into the circle equation gives us :\n\n
$$x^2 - 4x + 4 + \\frac{36 - x^2}{4} = r^2$$\n\n
Solving this equation leads to :\n\n
$$3x^2 - 16x + 52 - 4r^2 = 0$$\n\n
For the roots of this quadratic equation to be real (which they must be, since they represent real intersection points), the discriminant (D) must be greater than or equal to zero :\n\n
Setting D = 0 gives the minimum value for r (the radius of the inscribed circle) :\n\n
$$256 - 624 + 48r^2 = 0$$\n
$$48r^2 = 368$$\n
$$r^2 = \\frac{368}{48} = \\frac{23}{3}$$\n\n
And we're asked for the value of 12r2, so :\n\n
$$12r^2 = 12 \\times \\frac{23}{3} = 92$$\n\n
So, the correct answer is Option B : 92.\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5218,
"subject": "General Science",
"question": "A function $$f$$ from the set of natural numbers to integers defined by\n$$$f\\left( n \\right) = \\left\\{ {\\matrix{\n {{{n - 1} \\over 2},\\,when\\,n\\,is\\,odd} \\cr \n { - {n \\over 2},\\,when\\,n\\,is\\,even} \\cr \n\n } } \\right.$$$\nis",
"options": [
{
"text": "neither one -one nor onto"
},
{
"text": "one-one but not onto"
},
{
"text": "onto but not one-one"
},
{
"text": "one-one and onto both"
}
],
"answer": "one-one and onto both",
"solution": "**Answer:** one-one and onto both\n\nWe have $$f:N \\to I$$\n
If $$x$$ and $$y$$ are two even natural numbers, \n
then $$f\\left( x \\right) = f\\left( y \\right) \\Rightarrow {{ - x} \\over 2} = {{ - y} \\over 2} \\Rightarrow x = y$$\n
Again if $$x$$ and $$y$$ are two odd natural numbers then \n
$$f\\left( x \\right) = f\\left( y \\right) \\Rightarrow {{x - 1} \\over 2} = {{y - 1} \\over 2} \\Rightarrow x = y$$\n
$$\\therefore$$ $$f$$ is onto. \n
Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number. \n
$$\\therefore$$ $$f$$ is onto. \n
Hence $$f$$ is one one and onto both.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5219,
"subject": "General Science",
"question": "If $$f:R \\to S$$, defined by\n $$f\\left( x \\right) = \\sin x - \\sqrt 3 \\cos x + 1$$,\n is onto, then the interval of $$S$$ is",
"options": [
{
"text": "[-1, 3]"
},
{
"text": "[-1, 1]"
},
{
"text": "[0, 1]"
},
{
"text": "[0, 3]"
}
],
"answer": "[-1, 3]",
"solution": "**Answer:** [-1, 3]\n\n$$f\\left( x \\right)$$ is onto \n
$$\\therefore$$ $$S=$$ range of $$f(x)$$\n
Now $$f\\left( x \\right) = \\sin \\,x - \\sqrt 3 \\,\\cos \\,x + 1$$\n
Clearly, range of $$f\\left( x \\right) = \\left( { - {\\pi \\over 2},{\\pi \\over 2}} \\right)$$\n
For $$f$$ to be onto, co-domain $$=$$ range\n
$$\\therefore$$ Co-domain of function $$ = B = \\left( { - {\\pi \\over 2},{\\pi \\over 2}} \\right).$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5221,
"subject": "General Science",
"question": "Let $$f:N \\to Y$$ be a function defined as f(x) = 4x + 3 where\n Y = { y $$ \\in $$ N, y = 4x + 3 for some x $$ \\in $$ N }.\n Show that f is invertible and its inverse is",
"options": [
{
"text": "$$g\\left( y \\right) = {{3y + 4} \\over 4}$$"
},
{
"text": "$$g\\left( y \\right) = 4 + {{y + 3} \\over 4}$$"
},
{
"text": "$$g\\left( y \\right) = {{y + 3} \\over 4}$$"
},
{
"text": "$$g\\left( y \\right) = {{y - 3} \\over 4}$$"
}
],
"answer": "$$g\\left( y \\right) = {{y - 3} \\over 4}$$",
"solution": "**Answer:** $$g\\left( y \\right) = {{y - 3} \\over 4}$$\n\nClearly $$f$$ is one one and onto, so invertible\n
Also $$f\\left( x \\right) = 4x + 3 = y \\Rightarrow x = {{y - 3} \\over 4}$$\n
$$\\therefore$$ The statement -$$1$$ is correct but statement- $$2$$ is false.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5223,
"subject": "General Science",
"question": "For real x, let f(x) = x3 + 5x + 1, then",
"options": [
{
"text": "f is one-one but not onto R"
},
{
"text": "f is onto R but not one-one"
},
{
"text": "f is one-one and onto R"
},
{
"text": "f is neither one-one nor onto R"
}
],
"answer": "f is one-one and onto R",
"solution": "**Answer:** f is one-one and onto R\n\nGiven that $$f\\left( x \\right) = {x^3} + 5x + 1$$\n
$$\\therefore$$ $$\\,\\,\\,\\,\\,$$ $$f'\\left( x \\right) = 3{x^2} + 5 > 0,\\,\\,\\,\\forall x \\in R$$\n
$$ \\Rightarrow f\\left( x \\right)\\,\\,$$ is strictly increasing on $$R$$\n
$$ \\Rightarrow f\\left( x \\right)$$ is one one \n
$$\\therefore$$ $$\\,\\,\\,\\,\\,\\,\\,$$ Being a polynomial $$f(x)$$ is cont. and inc.\n
on $$R$$ with $$\\mathop {\\lim }\\limits_{x \\to \\infty } \\,f\\left( x \\right) = - \\infty $$\n
and $$\\mathop {\\lim }\\limits_{x \\to \\infty } \\,f\\left( x \\right) = \\infty $$\n
$$\\therefore$$ $$\\,\\,\\,\\,\\,\\,\\,$$ Range of $$f = \\left( { - \\infty ,\\infty } \\right) = R$$\n
Hence $$f$$ is onto also, So, $$f$$ is one and onto $$R.$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5224,
"subject": "General Science",
"question": "The function $$f:R \\to \\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$ defined as\n
$$f\\left( x \\right) = {x \\over {1 + {x^2}}}$$, is",
"options": [
{
"text": "invertible"
},
{
"text": "injective but not surjective. "
},
{
"text": "surjective but not injective"
},
{
"text": "neither injective nor surjective."
}
],
"answer": "surjective but not injective",
"solution": "**Answer:** surjective but not injective\n\n$$f\\left( x \\right) = {x \\over {1 + {x^2}}}$$\n
$$ \\therefore $$ f(x) is surjective but not injective\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5225,
"subject": "General Science",
"question": "The function f : N $$ \\to $$ N defined by f (x) = x $$-$$ 5 $$\\left[ {{x \\over 5}} \\right],$$ Where N is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is : ",
"options": [
{
"text": "one-one and onto"
},
{
"text": "one-one but not onto."
},
{
"text": "onto but not one-one."
},
{
"text": "neither one-one nor onto."
}
],
"answer": "neither one-one nor onto.",
"solution": "**Answer:** neither one-one nor onto.\n\nf(1) = 1 - 5$$\\left[ {{1 \\over 5}} \\right]$$ = 1\n
f(10) = 10 - 5$$\\left[ {{10 \\over 5}} \\right]$$ = 0 which is not present in the set of natural numbers.\n
So this function is neither one-one nor onto.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5226,
"subject": "General Science",
"question": "Let A = {x $$ \\in $$ R : x is not a positive integer}.\n
Define a function $$f$$ : A $$ \\to $$ R as $$f(x)$$ = $${{2x} \\over {x - 1}}$$,\n
then $$f$$ is : ",
"options": [
{
"text": "not injective"
},
{
"text": "neither injective nor surjective"
},
{
"text": "surjective but not injective "
},
{
"text": "injective but not surjective"
}
],
"answer": "injective but not surjective",
"solution": "**Answer:** injective but not surjective\n\nf(x) = $${{2x} \\over {x - 1}}$$\n
So range $$ \\notin $$ R (in to function)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5227,
"subject": "General Science",
"question": "The number of functions f from {1, 2, 3, ...., 20} onto {1, 2, 3, ...., 20} such that f(k) is a multiple of 3,\nwhenever k is a multiple of 4, is :",
"options": [
{
"text": "65 $$ \\times $$ (15)!"
},
{
"text": "56 $$ \\times $$ 15"
},
{
"text": "(15)! $$ \\times $$ 6!"
},
{
"text": "5! $$ \\times $$ 6!"
}
],
"answer": "(15)! $$ \\times $$ 6!",
"solution": "**Answer:** (15)! $$ \\times $$ 6!\n\nGiven that $f(k)$ is a multiple of 3 whenever $k$ is a multiple of 4, we need to consider how to map elements from the domain {1, 2, 3, ..., 20} to the codomain {1, 2, 3, ..., 20} following this rule.\n\n
1. We first consider the subset of the domain that consists of multiples of 4: {4, 8, 12, 16, 20}. There are 5 elements in this subset.\n\n
2. We then consider the subset of the codomain that consists of multiples of 3: {3, 6, 9, 12, 15, 18}. There are 6 elements in this subset.\n\n
3. According to the given condition, each of the 5 multiples of 4 must be mapped to a multiple of 3. This can be done in ${ }^6C_5 \\cdot 5! = 6!$ ways, considering that there are 6 options for each of the 5 multiples of 4 (each choice constitutes a combination), and we then consider the permutations of these 5 choices.\n\n
4. The remaining 15 elements in the domain (20 original elements minus the 5 multiples of 4) can be mapped onto the remaining 15 elements in the codomain (20 original elements minus the 6 multiples of 3, plus one multiple of 3 that has been assigned to a multiple of 4). This can be done in $15!$ ways.\n\n
So, combining these two cases, the total number of onto functions $f$ is $6! \\times 15!$, which corresponds to option C.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5228,
"subject": "General Science",
"question": "Let a function f : (0, $$\\infty $$) $$ \\to $$ (0, $$\\infty $$) be defined by f(x) = $$\\left| {1 - {1 \\over x}} \\right|$$. Then f is :",
"options": [
{
"text": "not injective but it is surjective"
},
{
"text": "neiter injective nor surjective"
},
{
"text": "injective only"
},
{
"text": "both injective as well as surjective"
}
],
"answer": "neiter injective nor surjective",
"solution": "**Answer:** neiter injective nor surjective\n\n$$f\\left( x \\right) = \\left| {1 - {1 \\over x}} \\right| = {{\\left| {x - 1} \\right|} \\over x} = \\left\\{ {\\matrix{\n {{{1 - x} \\over x}} & {0 < x \\le 1} \\cr \n {{{x - 1} \\over x}} & {x \\ge 1} \\cr \n\n } } \\right.$$\n
\n $$ \\Rightarrow $$ f(x) is not injective\n
but range of function is $$\\left[ {0,\\infty } \\right)$$\n
Remarks : If co-domain is $$\\left[ {0,\\infty } \\right)$$, then f(x) will be surjective.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5229,
"subject": "General Science",
"question": "If the function ƒ : R – {1, –1} $$ \\to $$ A defined by \nƒ(x) = $${{{x^2}} \\over {1 - {x^2}}}$$ , is surjective, then A is equal to",
"options": [
{
"text": "R – (–1, 0)"
},
{
"text": "R – {–1}"
},
{
"text": "R – [–1, 0)"
},
{
"text": "[0, $$\\infty $$)"
}
],
"answer": "R – [–1, 0)",
"solution": "**Answer:** R – [–1, 0)\n\nLet ƒ(x) = $${{{x^2}} \\over {1 - {x^2}}}$$ = y\n
$$ \\therefore $$ A is R – [–1, 0).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5230,
"subject": "General Science",
"question": "Let A = {a, b, c} and B = {1, 2, 3, 4}. Then the\nnumber of elements in the set C = {f : A $$ \\to $$ B |\n2 $$ \\in $$ f(A) and f is not one-one} is ______.",
"options": [],
"answer": "19",
"solution": "**Answer:** 19\n\nThe desired functions will contain either one\nelement or two elements in its codomain of\nwhich '2' always belongs to f(A).\n
Case 1 : When 2 is the image of all element of set A.\n
Number of ways this is possible = 1\n
Case 2 : When one image is 2 and other one image is one of {1, 3, 4}.\n
Number of ways we can choose one of {1, 3, 4} is = 3C1.\n
Now divide 3 elements {a, b, c} of set A into two parts.\n We can do this $${{3!} \\over {2!1!}}$$ ways.\n
Now map one part of set A into the element 2 of set B and map other part of set A into one of {1, 3, 4} of set B.\n We can do that 2! ways.\n
So number of functions in this case \n = 3C1 $$ \\times $$ $${{3!} \\over {2!1!}}$$ $$ \\times $$ 2! = 18\n
$$ \\therefore $$ Total number of functions = 1 + 18 = 19",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5231,
"subject": "General Science",
"question": "Let f, g : N $$ \\to $$ N such that f(n + 1) = f(n) + f(1) $$\\forall $$ n$$\\in$$N and g be any arbitrary function. Which of the following statements is NOT true?",
"options": [
{
"text": "If g is onto, then fog is one-one"
},
{
"text": "f is one-one"
},
{
"text": "If f is onto, then f(n) = n $$\\forall $$n$$\\in$$N"
},
{
"text": "If fog is one-one, then g is one-one"
}
],
"answer": "If g is onto, then fog is one-one",
"solution": "**Answer:** If g is onto, then fog is one-one\n\n$$f(n + 1) = f(n) + 1$$
$$f(2) = 2f(1)$$
$$f(3) = 3f(1)$$
$$f(4) = 4f(1)$$
.....
$$f(n) = nf(1)$$
$$f(x)$$ is one-one",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5232,
"subject": "General Science",
"question": "Let x denote the total number of one-one functions from a set A with 3 elements to a set B with 5 elements and y denote the total number of one-one functions form the set A to the set A $$\\times$$ B. Then :",
"options": [
{
"text": "2y = 273x"
},
{
"text": "y = 91x"
},
{
"text": "2y = 91x"
},
{
"text": "y = 273x"
}
],
"answer": "2y = 91x",
"solution": "**Answer:** 2y = 91x\n\nNumber of elements in A = 3
Then the number of possible functions $$g:A \\to A$$ such that $$gof = f$$ is :",
"options": [
{
"text": "55"
},
{
"text": "105"
},
{
"text": "5!"
},
{
"text": "10C5"
}
],
"answer": "105",
"solution": "**Answer:** 105\n\n
f(1) = 2
f(2) = 2
f(3) = 4
f(4) = 4
f(5) = 6
f(6) = 6
f(7) = 8
f(8) = 8
f(9) = 10
f(10) = 10
$$ \\therefore $$ f(1) = f(2) = 2
f(3) = f(4) = 4
f(5) = f(6) = 6
f(7) = f(8) = 8
f(9) = f(10) = 10
Given, g(f(x)) = f(x)
when x = 1, g(f(1)) = f(1) $$ \\Rightarrow $$ g(2) = 2
Here, you can see for even terms mapping is fixed. But far odd terms 1, 3, 5, 7, 9 we can map to any one of the 10 elements.
$$ \\therefore $$ For 1, number of functions = 10
For 3, number of functions = 10
$$\\eqalign{\n & . \\cr \n & . \\cr \n & . \\cr \n \n & \\cr} $$ for 9, number of functions = 10
$$ \\therefore $$ Total number of functions = 10 $$\\times$$ 10 $$\\times$$ 10 $$\\times$$ 10 $$\\times$$ 10 = 105",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5234,
"subject": "General Science",
"question": "Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A $$\\to$$ A such that f(1) + f(2) = 3 $$-$$ f(3) is equal to ",
"options": [],
"answer": "720",
"solution": "**Answer:** 720\n\nf(1) + f(2) = 3 $$-$$ f(3)
$$\\Rightarrow$$ f(1) + f(2) = 3 + f(3) = 3
The only possibility is : 0 + 1 + 2 = 3
$$\\Rightarrow$$ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.
So number of bijective functions.
$$\\left| \\!{\\underline {\\,\n 3 \\,}} \\right. $$ $$\\times$$ $$\\left| \\!{\\underline {\\,\n 5 \\,}} \\right. $$ = 720",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5235,
"subject": "General Science",
"question": "Let g : N $$\\to$$ N be defined as
g(3n + 1) = 3n + 2,
g(3n + 2) = 3n + 3,
g(3n + 3) = 3n + 1, for all n $$\\ge$$ 0.
Then which of the following statements is true?",
"options": [
{
"text": "There exists an onto function f : N $$\\to$$ N such that fog = f"
},
{
"text": "There exists a one-one function f : N $$\\to$$ N such that fog = f"
},
{
"text": "gogog = g"
},
{
"text": "There exists a function : f : N $$\\to$$ N such that gof = f"
}
],
"answer": "There exists an onto function f : N $$\\to$$ N such that fog = f",
"solution": "**Answer:** There exists an onto function f : N $$\\to$$ N such that fog = f\n\ng : N $$\\to$$ N
Here f(x) is onto, also f(g(x)) = f(x) $$\\forall$$ x$$\\in$$N",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5236,
"subject": "General Science",
"question": "
Let S = {1, 2, 3, 4}. Then the number of elements in the set { f : S $$\\times$$ S $$\\to$$ S : f is onto and f (a, b) = f (b, a) $$\\ge$$ a $$\\forall$$ (a, b) $$\\in$$ S $$\\times$$ S } is ______________.
",
"options": [],
"answer": "37",
"solution": "**Answer:** 37\n\nThere are 16 ordered pairs in $S \\times S$. We write all these ordered pairs in 4 sets as follows.\n
",
"options": [
{
"text": "one-one but not onto"
},
{
"text": "onto but not one-one"
},
{
"text": "neither one-one nor onto"
},
{
"text": "one-one and onto"
}
],
"answer": "one-one and onto",
"solution": "**Answer:** one-one and onto\n\n
When n = 1, 5, 9, 13 then $${{n + 1} \\over 2}$$ will give all odd numbers.
\n
When n = 3, 7, 11, 15 .....
\n
n $$-$$ 1 will be even but not divisible by 4
\n
When n = 2, 4, 6, 8 .....
\n
Then 2n will give all multiples of 4
\n
So range will be N.
\n
And no two values of n give same y, so function is one-one and onto.
Now possible value of $$f(d) = 2,3,4,5,6,7,$$ and $$8$$.
\n
f(d) can't be 9 and 10 as if $$f(d) = 9$$ or 10 then $$f(b) = 2 + 9 = 11$$ or $$f(b) = 2 + 10 = 12$$, which is not possible as here any function's maximum value can be 10.
\n
$$\\therefore$$ Total possible functions when $$f(c) = 0$$ and $$f(a) = 1$$ are = 7
\n
(2) When $$f(c) = 0$$ and $$f(a) = 2$$ then
\n
$$3 \\times 0 + 2 \\times 2 + f(d) = f(b)$$
\n
$$ \\Rightarrow 4 + f(d) = f(b)$$
\n
$$\\therefore$$ possible value of $$f(d) = 1,3,4,5,6$$
\n
$$\\therefore$$ Total possible functions in this case = 5
\n
(3) When $$f(c) = 0$$ and $$f(a) = 3$$ then
\n
$$3 \\times 0 + 2 \\times 3 + f(d) = f(b)$$
\n
$$ \\Rightarrow 6 + f(d) = f(b)$$
\n
$$\\therefore$$ Possible value of $$f(d) = 1,2,4$$
\n
$$\\therefore$$ Total possible functions in this case = 3
\n
(4) When $$f(c) = 0$$ and $$f(a) = 4$$ then
\n
$$3 \\times 0 + 2 \\times 4 + f(d) = f(b)$$
\n
$$ \\Rightarrow 8 + f(d) = f(b)$$
\n
$$\\therefore$$ Possible value of $$f(d) = 1,2$$
\n
$$\\therefore$$ Total possible functions in this case = 2
\n
(5) When $$f(c) = 0$$ and $$f(a) = 5$$ then
\n
$$3 \\times 0 + 2 \\times 5 + f(d) = f(b)$$
\n
$$ \\Rightarrow 10 + f(d) = f(b)$$
\n
Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.
\n
$$\\therefore$$ No function is possible in this case.
\n
$$\\therefore$$ Total possible functions when $$f(c) = 0$$ and $$f(a) = 1,2,3$$ and $$4$$ are $$ = 7 + 5 + 3 + 2 = 17$$
\n
Case II:
\n
(1) When $$f(c) = 1$$ and $$f(a) = 0$$ then
\n
$$3 \\times 1 + 2 \\times 0 + f(d) = f(b)$$
\n
$$ \\Rightarrow 3 + f(d) = f(b)$$
\n
$$\\therefore$$ Possible value of $$f(d) = 2,3,4,5,6,7$$
\n
$$\\therefore$$ Total possible functions in this case = 6
\n
(2) When $$f(c) = 1$$ and $$f(a) = 2$$ then
\n
$$3 \\times 1 + 2 \\times 2 + f(d) = f(b)$$
\n
$$ \\Rightarrow 7 + f(d) = f(b)$$
\n
$$\\therefore$$ Possible value of $$f(d) = 0,3$$
\n
$$\\therefore$$ Total possible functions in this case = 2
\n
(3) When $$f(c) = 1$$ and $$f(a) = 3$$ then
\n
$$3 \\times 1 + 2 \\times 3 + f(d) = f(b)$$
\n
$$ \\Rightarrow 9 + f(d) = f(b)$$
\n
$$\\therefore$$ Possible value of $$f(d) = 0$$
\n
$$\\therefore$$ Total possible functions in this case = 1
\n
$$\\therefore$$ Total possible functions when $$f(c) = 1$$ and $$f(a) = 0,2$$ and $$3$$ are
\n
$$ = 6 + 2 + 1 = 9$$
\n
Case III:
\n
(1) When $$f(c) = 2$$ and $$f(a) = 0$$ then
\n
$$3 \\times 2 + 2 \\times 0 + f(d) = f(b)$$
\n
$$ \\Rightarrow 6 + f(d) = f(b)$$
\n
$$\\therefore$$ Possible values of $$f(d) = 1,3,4$$
\n
$$\\therefore$$ Total possible functions in this case = 3
\n
(2) When $$f(c) = 2$$ and $$f(a) = 1$$ then,
\n
$$3 \\times 2 + 2 \\times 1 + f(d) = f(b)$$
\n
$$ \\Rightarrow 8 + f(d) = f(b)$$
\n
$$\\therefore$$ Possible values of $$f(d) = 0$$
\n
$$\\therefore$$ Total possible function in this case = 1
\n
$$\\therefore$$ Total possible functions when $$f(c) = 2$$ and $$f(a) = 0,1$$ are
\n
$$ = 3 + 1 = 4$$
\n
Case IV:
\n
(1) When $$f(c) = 3$$ and $$f(a) = 0$$ then
\n
$$3 \\times 3 + 2 \\times 0 + f(d) = f(b)$$
\n
$$ \\Rightarrow 9 + f(d) = f(b)$$
\n
$$\\therefore$$ Possible values of $$f(d) = 1$$
\n
$$\\therefore$$ Total one-one functions from four cases
The number of bijective functions $$f:\\{1,3,5,7, \\ldots, 99\\} \\rightarrow\\{2,4,6,8, \\ldots .100\\}$$, such that $$f(3) \\geq f(9) \\geq f(15) \\geq f(21) \\geq \\ldots . . f(99)$$, is ____________.
The number of functions $$f$$, from the set $$\\mathrm{A}=\\left\\{x \\in \\mathbf{N}: x^{2}-10 x+9 \\leq 0\\right\\}$$ to the set $$\\mathrm{B}=\\left\\{\\mathrm{n}^{2}: \\mathrm{n} \\in \\mathbf{N}\\right\\}$$ such that $$f(x) \\leq(x-3)^{2}+1$$, for every $$x \\in \\mathrm{A}$$, is ___________.
Let $$S=\\{1,2,3,4,5,6\\}$$. Then the number of one-one functions $$f: \\mathrm{S} \\rightarrow \\mathrm{P}(\\mathrm{S})$$, where $$\\mathrm{P}(\\mathrm{S})$$ denote the power set of $$\\mathrm{S}$$, such that $$f(n) \\subset f(\\mathrm{~m})$$ where $$n < m$$ is ____________.
$$f(n)$$ corresponding a set having m elements which belongs to P(S), should be a subset of $$f(n+1)$$, so $$f(n+1)$$ should be a subset of P(S) having at least $$m+1$$ elements.
\n
Now, if f(1) has one element then f(2) has 3, f(3) has 3 and so on and f(6) has 6 elements. Total number of possible functions = 6! = 720 .... (1)\n
If f(1) has no elements (i.e. null set $$\\phi$$) then
\n
\n
Each index number represents the number of elements in respective rows
\n
Taking every series of arrow and counting number of such possible functions (sets)
Let $$f:R \\to R$$ be a function such that $$f(x) = {{{x^2} + 2x + 1} \\over {{x^2} + 1}}$$. Then
",
"options": [
{
"text": "$$f(x)$$ is many-one in $$( - \\infty , - 1)$$"
},
{
"text": "$$f(x)$$ is one-one in $$( - \\infty ,\\infty )$$"
},
{
"text": "$$f(x)$$ is one-one in $$[1,\\infty )$$ but not in $$( - \\infty ,\\infty )$$"
},
{
"text": "$$f(x)$$ is many-one in $$(1,\\infty )$$"
}
],
"answer": "$$f(x)$$ is one-one in $$[1,\\infty )$$ but not in $$( - \\infty ,\\infty )$$",
"solution": "**Answer:** $$f(x)$$ is one-one in $$[1,\\infty )$$ but not in $$( - \\infty ,\\infty )$$\n\n
$$ \\therefore $$ $$f'(x) = 0$$ at x = 1 and x = -1. So, x = 1 and x = -1 are point of maxima/minima.
\n
Option A : $$f(x)$$ is many-one in $$( - \\infty , - 1)$$.\n
As x = -1 is a point of maxima/minima. So it is a boundary point in the range $$( - \\infty , - 1)$$. So, $$f(x)$$ is one-one in $$( - \\infty , - 1)$$.\n
$$ \\therefore $$ Option A is incorrect.
\n
Option B : $$f(x)$$ is one-one in $$( - \\infty ,\\infty )$$.\n
As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $$( - \\infty ,\\infty )$$. So, $$f(x)$$ is many-one function in $$( - \\infty ,\\infty )$$.\n
$$ \\therefore $$ Option B is incorrect.
\n
Option C : $$f(x)$$ is one-one in $$[1,\\infty )$$ but not in $$( - \\infty ,\\infty )$$.\n
As x = 1 is a point of maxima/minima. So it is a boundary point in the range $$[1,\\infty )$$. So, $$f(x)$$ is one-one in $$[1,\\infty )$$.\n
As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $$( - \\infty ,\\infty )$$. So, $$f(x)$$ is many-one function in $$( - \\infty ,\\infty )$$ .\n
$$ \\therefore $$ Option C is correct.
\n
Option D : $$f(x)$$ is many-one in $$(1,\\infty )$$.\n
As x = 1 is a point of maxima/minima. So it is a boundary point in the range $$[1,\\infty )$$. So, $$f(x)$$ is one-one in $$[1,\\infty )$$.\n
$$ \\therefore $$ Option D is incorrect.
\n
Note :
\n
Methods to Check One-One Function :\n\n
(i) If a function is one-one, any line parallel to $x$-axis cuts the graph of the function maximum at one point.\n
(ii) Any continuous function which is entirely increasing or decreasing (no maxima/minima is present) in whole domain will always be one-one function.\n
\n$\\therefore$ Only two functions are possible.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5245,
"subject": "General Science",
"question": "
Let $$\\mathrm{A}=\\{1,2,3,4,5\\}$$ and $$\\mathrm{B}=\\{1,2,3,4,5,6\\}$$. Then the number of functions $$f: \\mathrm{A} \\rightarrow \\mathrm{B}$$ satisfying $$f(1)+f(2)=f(4)-1$$ is equal to __________.
",
"options": [],
"answer": "360",
"solution": "**Answer:** 360\n\nGiven that the function $$f : A \\rightarrow B$$ satisfies the condition $$f(1) + f(2) = f(4) - 1$$, where the set $$A = \\{1, 2, 3, 4, 5\\}$$ and the set $$B = \\{1, 2, 3, 4, 5, 6\\}$$. \n\n
We want to find out how many such functions exist.\n\n
First, observe that the condition $$f(1) + f(2) = f(4) - 1$$ can be rewritten as $$f(1) + f(2) + 1 = f(4)$$. So, the sum of $$f(1), f(2),$$ and 1 is equal to $$f(4)$$. Since $$f(4)$$ is a value in set B, it can take values from 1 to 6. \n\n
The maximum value of $$f(1) + f(2) + 1$$ can be $$6 + 6 + 1 = 13$$, but this is more than 6 (the maximum value of $$f(4)$$), so it's not possible. Thus, the maximum value of $$f(4)$$ in this case can be 6.\n\n
Let's now analyze the number of functions for each value of $$f(4)$$ from 3 to 6 (we start from 3 because $$f(1)$$ and $$f(2)$$ take values from set B and their minimum sum plus 1 is 3):\n\n
1. When $$f(4) = 6$$, then $$f(1) + f(2) = 5$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 4), (2, 3), (3, 2), (4, 1)$$. For each of these 4 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$4 \\times 6 \\times 6 = 144$$ functions.\n\n
2. When $$f(4) = 5$$, then $$f(1) + f(2) = 4$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 3), (2, 2), (3, 1)$$. For each of these 3 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$3 \\times 6 \\times 6 = 108$$ functions.\n\n
3. When $$f(4) = 4$$, then $$f(1) + f(2) = 3$$. The pairs $$(f(1), f(2))$$ that satisfy this equation are $$(1, 2), (2, 1)$$. For each of these 2 cases, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$2 \\times 6 \\times 6 = 72$$ functions.\n\n
4. When $$f(4) = 3$$, then $$f(1) + f(2) = 2$$. The only pair $$(f(1), f(2))$$ that satisfies this equation is $$(1, 1)$$. For this case, the functions $$f(3)$$ and $$f(5)$$ can each take 6 values from set B, resulting in $$1 \\times 6 \\times 6 = 36$$ functions.\n\n
Adding the numbers of functions from all these cases, we get a total of $$144 + 108 + 72 + 36 = 360$$ functions from $$A$$ to $$B$$ that satisfy the given condition.\n\n
Therefore, the number of functions $$f : A \\rightarrow B$$ satisfying the condition $$f(1) + f(2) = f(4) - 1$$ is 360.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5246,
"subject": "General Science",
"question": "
Let $$\\mathrm{R}=\\{\\mathrm{a}, \\mathrm{b}, \\mathrm{c}, \\mathrm{d}, \\mathrm{e}\\}$$ and $$\\mathrm{S}=\\{1,2,3,4\\}$$. Total number of onto functions $$f: \\mathrm{R} \\rightarrow \\mathrm{S}$$\nsuch that $$f(\\mathrm{a}) \\neq 1$$, is equal to ______________.
",
"options": [],
"answer": "180",
"solution": "**Answer:** 180\n\nTotal number of onto functions\n
$=240-60=180$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5247,
"subject": "General Science",
"question": "The function $f: \\mathbf{N}-\\{1\\} \\rightarrow \\mathbf{N}$; defined by $f(\\mathrm{n})=$ the highest prime factor of $\\mathrm{n}$, is :",
"options": [
{
"text": "one-one only"
},
{
"text": "neither one-one nor onto"
},
{
"text": "onto only"
},
{
"text": "both one-one and onto"
}
],
"answer": "neither one-one nor onto",
"solution": "**Answer:** neither one-one nor onto\n\n
$$\\begin{aligned}\n& \\mathrm{f}: \\mathrm{N}-\\{1\\} \\rightarrow \\mathrm{N} \\\\\n& \\mathrm{f}(\\mathrm{n})=\\text { The highest prime factor of } \\mathrm{n} . \\\\\n& \\mathrm{f}(2)=2 \\\\\n& \\mathrm{f}(4)=2 \\\\\n& \\Rightarrow \\text { many one } \\\\\n& 4 \\text { is not image of any element } \\\\\n& \\Rightarrow \\text { into }\n\\end{aligned}$$
Let $$\\mathrm{A}=\\{1,2,3, \\ldots, 7\\}$$ and let $$\\mathrm{P}(\\mathrm{A})$$ denote the power set of $$\\mathrm{A}$$. If the number of functions $$f: \\mathrm{A} \\rightarrow \\mathrm{P}(\\mathrm{A})$$ such that $$\\mathrm{a} \\in f(\\mathrm{a}), \\forall \\mathrm{a} \\in \\mathrm{A}$$ is $$\\mathrm{m}^{\\mathrm{n}}, \\mathrm{m}$$ and $$\\mathrm{n} \\in \\mathrm{N}$$ and $$\\mathrm{m}$$ is least, then $$\\mathrm{m}+\\mathrm{n}$$ is equal to _________.
Let $$A=\\{(x, y): 2 x+3 y=23, x, y \\in \\mathbb{N}\\}$$ and $$B=\\{x:(x, y) \\in A\\}$$. Then the number of one-one functions from $$A$$ to $$B$$ is equal to _________.
Let $$A=\\{1,3,7,9,11\\}$$ and $$B=\\{2,4,5,7,8,10,12\\}$$. Then the total number of one-one maps $$f: A \\rightarrow B$$, such that $$f(1)+f(3)=14$$, is :
= $${1 \\over {1 - x}} = {f_3}\\left( x \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5255,
"subject": "General Science",
"question": "Let N be the set of natural numbers and two functions f and g be defined as f, g : N $$ \\to $$ N such that\n
So, $$f\\left( { - {1 \\over 2}} \\right)$$ = $${{1 + {1 \\over 2}} \\over {1 - {1 \\over 2}}}$$ = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5261,
"subject": "General Science",
"question": "Let f : R → R be defined as f (x) = 2x – 1 and g : R - {1} → R be defined as g(x) =\n$${{x - {1 \\over 2}} \\over {x - 1}}$$.\nThen the composition function f(g(x)) is :",
"options": [
{
"text": "one-one but not onto"
},
{
"text": "onto but not one-one"
},
{
"text": "both one-one and onto"
},
{
"text": "neither one-one nor onto"
}
],
"answer": "one-one but not onto",
"solution": "**Answer:** one-one but not onto\n\nGiven, f(x) = 2x $$-$$ 1; f : R $$\\to$$ R
Now, draw the graph of $$1 + {1 \\over {x - 1'}}$$
$$\\because$$ Any horizontal line does not cut the graph at more than one points, so it is one-one and here, co-domain and range are not equal, so it is into.
Hence, the required function is one-one into.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5262,
"subject": "General Science",
"question": "If the functions are defined as $$f(x) = \\sqrt x $$ and $$g(x) = \\sqrt {1 - x} $$, then what is the common domain of the following functions :
f + g, f $$-$$ g, f/g, g/f, g $$-$$ f where $$(f \\pm g)(x) = f(x) \\pm g(x),(f/g)x = {{f(x)} \\over {g(x)}}$$",
"options": [
{
"text": "$$0 \\le x \\le 1$$"
},
{
"text": "$$0 \\le x < 1$$"
},
{
"text": "$$0 < x < 1$$"
},
{
"text": "$$0 < x \\le 1$$"
}
],
"answer": "$$0 < x < 1$$",
"solution": "**Answer:** $$0 < x < 1$$\n\n$$f + g = \\sqrt x + \\sqrt {1 - x} $$
$$ \\Rightarrow x \\ge 0$$ & $$1 - x \\ge 0 \\Rightarrow x \\in [0,1]$$
$$f - g = \\sqrt x - \\sqrt {1 - x} $$
$$ \\Rightarrow x \\ge 0$$ & $$1 - x \\ge 0 \\Rightarrow x \\in [0,1]$$
$$f/g = {{\\sqrt x } \\over {\\sqrt {1 - x} }}$$
$$ \\Rightarrow x \\ge 0$$ & $$1 - x > 0 \\Rightarrow x \\in [0,1)$$
$$g/f = {{\\sqrt {1 - x} } \\over {\\sqrt x }}$$
$$ \\Rightarrow 1 - x \\ge 0$$ & $$x > 0 \\Rightarrow x \\in (0,1]$$
$$g - f = \\sqrt {1 - x} - \\sqrt x $$
$$ \\Rightarrow 1 - x \\ge 0$$ & $$x \\ge 0 \\Rightarrow x \\in [0,1]$$
$$ \\Rightarrow $$ $$x \\in (0,1)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5263,
"subject": "General Science",
"question": "Let $$f:R - \\left\\{ {{\\alpha \\over 6}} \\right\\} \\to R$$ be defined by $$f(x) = {{5x + 3} \\over {6x - \\alpha }}$$. Then the value of $$\\alpha$$ for which (fof)(x) = x, for all $$x \\in R - \\left\\{ {{\\alpha \\over 6}} \\right\\}$$, is :",
"options": [
{
"text": "No such $$\\alpha$$ exists"
},
{
"text": "5"
},
{
"text": "8"
},
{
"text": "6"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\n$$f(x) = {{5x + 3} \\over {6x - \\alpha }} = y$$ ..... (i)
$$5x + 3 = 6xy - \\alpha y$$
$$x(6y - 5) = \\alpha y + 3$$
$$x = {{\\alpha y + 3} \\over {6y - 5}}$$
$${f^{ - 1}}(x) = {{\\alpha x + 3} \\over {6x - 5}}$$ ...... (ii)
fo $$f(x) = x$$
$$f(x) = {f^{ - 1}}(x)$$
From eqn (i) & (ii)
Clearly $$(\\alpha = 5)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5264,
"subject": "General Science",
"question": "Consider function f : A $$\\to$$ B and g : B $$\\to$$ C (A, B, C $$ \\subseteq $$ R) such that (gof)$$-$$1 exists, then :",
"options": [
{
"text": "f and g both are one-one"
},
{
"text": "f and g both are onto"
},
{
"text": "f is one-one and g is onto"
},
{
"text": "f is onto and g is one-one"
}
],
"answer": "f is one-one and g is onto",
"solution": "**Answer:** f is one-one and g is onto\n\n$$\\therefore$$ (gof)$$-$$1 exist $$\\Rightarrow$$ gof is bijective
$$\\Rightarrow$$ 'f' must be one-one and 'g' must be ONTO.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5265,
"subject": "General Science",
"question": "
Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If $$f(g(x)) = 8{x^2} - 2x$$ and $$g(f(x)) = 4{x^2} + 6x + 1$$, then the value of $$f(2) + g(2)$$ is _________.
Let g : S $$\\to$$ S be a function such that $$fog(n) = \\left\\{ {\\matrix{\n {n + 1} & , & {if\\,n\\,\\,is\\,odd} \\cr \n {n - 1} & , & {if\\,n\\,\\,is\\,even} \\cr \n\n } } \\right.$$.
\n
Then $$g(10)g(1) + g(2) + g(3) + g(4) + g(5))$$ is equal to _____________.
Let $$f(x) = {{x - 1} \\over {x + 1}},\\,x \\in R - \\{ 0, - 1,1\\} $$. If $${f^{n + 1}}(x) = f({f^n}(x))$$ for all n $$\\in$$ N, then $${f^6}(6) + {f^7}(7)$$ is equal to :
Let $$f:R \\to R$$ and $$g:R \\to R$$ be two functions defined by $$f(x) = {\\log _e}({x^2} + 1) - {e^{ - x}} + 1$$ and $$g(x) = {{1 - 2{e^{2x}}} \\over {{e^x}}}$$. Then, for which of the following range of $$\\alpha$$, the inequality $$f\\left( {g\\left( {{{{{(\\alpha - 1)}^2}} \\over 3}} \\right)} \\right) > f\\left( {g\\left( {\\alpha -{5 \\over 3}} \\right)} \\right)$$ holds ?
$$f(x) = {\\left( {2\\left( {1 - {{{x^{25}}} \\over 2}} \\right)(2 + {x^{25}})} \\right)^{{1 \\over {50}}}}$$. If the function $$g(x) = f(f(f(x))) + f(f(x))$$, then the greatest integer less than or equal to g(1) is ____________.
Let $$f(x)$$ be a quadratic polynomial with leading coefficient 1 such that $$f(0)=p, p \\neq 0$$, and $$f(1)=\\frac{1}{3}$$. If the equations $$f(x)=0$$ and $$f \\circ f \\circ f \\circ f(x)=0$$ have a common real root, then $$f(-3)$$ is equal to ________________.
Let $$f, g: \\mathbb{N}-\\{1\\} \\rightarrow \\mathbb{N}$$ be functions defined by $$f(a)=\\alpha$$, where $$\\alpha$$ is the maximum of the powers of those primes $$p$$ such that $$p^{\\alpha}$$ divides $$a$$, and $$g(a)=a+1$$, for all $$a \\in \\mathbb{N}-\\{1\\}$$. Then, the function $$f+g$$ is
",
"options": [
{
"text": "one-one but not onto"
},
{
"text": "onto but not one-one"
},
{
"text": "both one-one and onto"
},
{
"text": "neither one-one nor onto"
}
],
"answer": "neither one-one nor onto",
"solution": "**Answer:** neither one-one nor onto\n\n
$$f,g:N - \\{ 1\\} \\to N$$ defined as
\n
$$f(a) = \\alpha $$, where $$\\alpha$$ is the maximum power of those primes p such that p$$\\alpha$$ divides a.
Let $$\\alpha, \\beta$$ and $$\\gamma$$ be three positive real numbers. Let $$f(x)=\\alpha x^{5}+\\beta x^{3}+\\gamma x, x \\in \\mathbf{R}$$ and $$g: \\mathbf{R} \\rightarrow \\mathbf{R}$$ be such that $$g(f(x))=x$$ for all $$x \\in \\mathbf{R}$$. If $$\\mathrm{a}_{1}, \\mathrm{a}_{2}, \\mathrm{a}_{3}, \\ldots, \\mathrm{a}_{\\mathrm{n}}$$ be in arithmetic progression with mean zero, then the value of $$f\\left(g\\left(\\frac{1}{\\mathrm{n}} \\sum\\limits_{i=1}^{\\mathrm{n}} f\\left(\\mathrm{a}_{i}\\right)\\right)\\right)$$ is equal to :
For some a, b, c $$\\in\\mathbb{N}$$, let $$f(x) = ax - 3$$ and $$\\mathrm{g(x)=x^b+c,x\\in\\mathbb{R}}$$. If $${(fog)^{ - 1}}(x) = {\\left( {{{x - 7} \\over 2}} \\right)^{1/3}}$$, then $$(fog)(ac) + (gof)(b)$$ is equal to ____________.
For $$x \\in \\mathbb{R}$$, two real valued functions $$f(x)$$ and $$g(x)$$ are such that, $$g(x)=\\sqrt{x}+1$$ and $$f \\circ g(x)=x+3-\\sqrt{x}$$. Then $$f(0)$$ is equal to
But, if we consider the domain of the composite function $f \\circ g(x)$ then in that case $f(0)$ will be not defined as $\\mathrm{g}(\\mathrm{x})$ cannot be equal to zero.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5276,
"subject": "General Science",
"question": "Let $f: \\mathbf{R} \\rightarrow \\mathbf{R}$ and $g: \\mathbf{R} \\rightarrow \\mathbf{R}$ be defined as \n
$f(x)=\\left\\{\\begin{array}{ll}\\log _{\\mathrm{e}} x, & x>0 \\\\ \\mathrm{e}^{-x}, & x \\leq 0\\end{array}\\right.$ and \n
$g(x)=\\left\\{\\begin{array}{ll}x, & x \\geqslant 0 \\\\ \\mathrm{e}^x, & x<0\\end{array}\\right.$. Then, gof : $\\mathbf{R} \\rightarrow \\mathbf{R}$ is :",
"options": [
{
"text": "one-one but not onto"
},
{
"text": "neither one-one nor onto"
},
{
"text": "onto but not one-one"
},
{
"text": "both one-one and onto"
}
],
"answer": "neither one-one nor onto",
"solution": "**Answer:** neither one-one nor onto\n\nGiven, $f(x)=\\left\\{\\begin{array}{ll}\\log _{\\mathrm{e}} x, & x>0 \\\\ \\mathrm{e}^{-x}, & x \\leq 0\\end{array}\\right.$ and \n
$g(x)=\\left\\{\\begin{array}{ll}x, & x \\geqslant 0 \\\\ \\mathrm{e}^x, & x<0\\end{array}\\right.$\n
$\\mathrm{g}(\\mathrm{f}(\\mathrm{x})) \\Rightarrow$ Many one into",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5277,
"subject": "General Science",
"question": "
Let $$f: \\mathbf{R}-\\left\\{\\frac{-1}{2}\\right\\} \\rightarrow \\mathbf{R}$$ and $$g: \\mathbf{R}-\\left\\{\\frac{-5}{2}\\right\\} \\rightarrow \\mathbf{R}$$ be defined as $$f(x)=\\frac{2 x+3}{2 x+1}$$ and $$g(x)=\\frac{|x|+1}{2 x+5}$$. Then, the domain of the function fog is :
If $$f(x)=\\frac{4 x+3}{6 x-4}, x \\neq \\frac{2}{3}$$ and $$(f \\circ f)(x)=g(x)$$, where $$g: \\mathbb{R}-\\left\\{\\frac{2}{3}\\right\\} \\rightarrow \\mathbb{R}-\\left\\{\\frac{2}{3}\\right\\}$$, then $$(g ogog)(4)$$ is equal to
To find $$(g \\circ g \\circ g)(4),$$ we first need to understand the composition of $$f$$ with itself, i.e., $$(f \\circ f)(x) = f(f(x)) = g(x).$$ We can then repeatedly apply $$g$$ to get the given expression.
So, $$g(x) = x$$ for all $$x$$ in the domain of $$g$$, which is $$\\mathbb{R}-\\left\\{\\frac{2}{3}\\right\\}$$. It's important to note that the domain restriction is preserved through the composition because $$f(x)$$ has a vertical asymptote at $$x = \\frac{2}{3}$$ which doesn't intersect the graph.
\n\n
So, $$g(x)$$ is the identity function on its domain, which means that applying $$g$$ any number of times will result in the same input for $$x$$ in the given domain. Hence, we have:
\n\n
$$ (g \\circ g \\circ g \\circ g)(4) = g(g(g(g(4)))) = g(g(g(4))) = g(g(4)) = g(4) = 4 $$
If $$f(x)=\\left\\{\\begin{array}{cc}2+2 x, & -1 \\leq x < 0 \\\\ 1-\\frac{x}{3}, & 0 \\leq x \\leq 3\\end{array} ; g(x)=\\left\\{\\begin{array}{cc}-x, & -3 \\leq x \\leq 0 \\\\ x, & 0 < x \\leq 1\\end{array}\\right.\\right.$$, then range of $$(f o g)(x)$$ is
Consider the function $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$ defined by $$f(x)=\\frac{2 x}{\\sqrt{1+9 x^2}}$$. If the composition of $$f, \\underbrace{(f \\circ f \\circ f \\circ \\cdots \\circ f)}_{10 \\text { times }}(x)=\\frac{2^{10} x}{\\sqrt{1+9 \\alpha x^2}}$$, then the value of $$\\sqrt{3 \\alpha+1}$$ is equal to _______.
$$ \\Rightarrow \\left| x \\right| > x,\\,\\,\\, \\Rightarrow x < 0$$\n
Hence domain of $$f\\left( x \\right)$$ is $$\\left( { - \\infty ,0} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5287,
"subject": "General Science",
"question": "The domain of the definition of the function \n
x $$ \\in $$ (-1, 0) $$ \\cup $$ (1, 2) $$ \\cup $$ (2, $$\\infty $$)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5288,
"subject": "General Science",
"question": "The real valued function $$f(x) = {{\\cos e{c^{ - 1}}x} \\over {\\sqrt {x - [x]} }}$$, where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to :",
"options": [
{
"text": "all real except integers"
},
{
"text": "all non-integers except the interval [ $$-$$1, 1 ]"
},
{
"text": "all integers except 0, $$-$$1, 1"
},
{
"text": "all real except the interval [ $$-$$1, 1 ]"
}
],
"answer": "all non-integers except the interval [ $$-$$1, 1 ]",
"solution": "**Answer:** all non-integers except the interval [ $$-$$1, 1 ]\n\nDomain of $$\\cos e{c^{ - 1}}x$$ :
$$x \\in ( - \\infty , - 1] \\cup [1,\\infty )$$
and, $$x - [x] > 0$$
$$ \\Rightarrow \\{ x\\} > 0$$
$$ \\Rightarrow x \\ne I$$
$$ \\therefore $$ Required domain = $$( - \\infty , - 1] \\cup [1,\\infty ) - I$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5289,
"subject": "General Science",
"question": "Let [ x ] denote the greatest integer $$\\le$$ x, where x $$\\in$$ R. If the domain of the real valued function $$f(x) = \\sqrt {{{\\left| {[x]} \\right| - 2} \\over {\\left| {[x]} \\right| - 3}}} $$ is ($$-$$ $$\\infty$$, a) $$]\\cup$$ [b, c) $$\\cup$$ [4, $$\\infty$$), a < b < c, then the value of a + b + c is :",
"options": [
{
"text": "8"
},
{
"text": "1"
},
{
"text": "$$-$$2"
},
{
"text": "$$-$$3"
}
],
"answer": "$$-$$2",
"solution": "**Answer:** $$-$$2\n\nFor domain,
Let $$\\mathrm{D}$$ be the domain of the function $$f(x)=\\sin ^{-1}\\left(\\log _{3 x}\\left(\\frac{6+2 \\log _{3} x}{-5 x}\\right)\\right)$$. If the range of the function $$\\mathrm{g}: \\mathrm{D} \\rightarrow \\mathbb{R}$$ defined by $$\\mathrm{g}(x)=x-[x],([x]$$ is the greatest integer function), is $$(\\alpha, \\beta)$$, then $$\\alpha^{2}+\\frac{5}{\\beta}$$ is equal to
First, the function $f(x) = \\sin^{-1}(\\log_{3x}(\\frac{6 + 2 \\log_3{x}}{-5x}))$ has several restrictions :
\n\n
Since the arcsine function $\\sin^{-1}(x)$ is only defined for $-1\\leq x\\leq 1$, this means that $\\log_{3x}(\\frac{6 + 2 \\log _3 x}{-5 x})$ must be between -1 and 1.
\n
\n
For the logarithm to be defined, $3x > 0 \\Rightarrow x > 0$ .......(1)
because the base of a logarithm must be greater than 0 and not equal to 1. Also, $x \\neq \\frac{1}{3}$ .......(2)
as the base cannot be 1.
\n
\n
Moreover, the inner function of the logarithm $\\frac{6 + 2 \\log_3{x}}{-5x}$ must be greater than 0.
$$\n\\Rightarrow \\text { From (1), (2), (3), } 0 < x < \\frac{1}{27}\n$$
\n
\n\n
The next step is to solve the inequality for $-1 \\leq \\log_{3x}(\\frac{6+2 \\log _3 x}{-5 x}) \\leq 1$. To do this, we make the observation that for the logarithmic part to be within $[-1,1]$, it must be true that $3x \\leq \\frac{6+2 \\log_3 x}{-5x} \\leq \\frac{1}{3x}$.
\n
Solving the inequality $15x^2 + 6 + 2 \\log_3 x \\geq 0$, we find that $x \\in(0, \\frac{1}{27})$ ........(4)
\n
Likewise, solving the inequality $6+2 \\log_3 x + \\frac{5}{3} \\geq 0$, we find that $x \\geq 3^{-\\frac{23}{6}}$ ........(5)
\n
Combining Equations (3), (4), and (5), the intersection of all these intervals is $x \\in[3^{-\\frac{23}{6}}, \\frac{1}{27})$.
\n
Now, consider the function $g(x) = x - [x]$, where $[x]$ is the greatest integer function. For this function, the range is the fractional part of $x$. In this case, the range $(\\alpha, \\beta)$ is given by the minimum and maximum possible values of $x$ in its domain. Hence, $\\alpha = 3^{-\\frac{23}{6}}$ and $\\beta = \\frac{1}{27}$.
\n
Finally, substitute these values into the equation $\\alpha^{2}+\\frac{5}{\\beta}$:
The domain of the function $$f(x)=\\frac{1}{\\sqrt{[x]^{2}-3[x]-10}}$$ is : ( where $$[\\mathrm{x}]$$ denotes the greatest integer less than or equal to $$x$$ )
If domain of the function $$\\log _{e}\\left(\\frac{6 x^{2}+5 x+1}{2 x-1}\\right)+\\cos ^{-1}\\left(\\frac{2 x^{2}-3 x+4}{3 x-5}\\right)$$ is $$(\\alpha, \\beta) \\cup(\\gamma, \\delta]$$, then $$18\\left(\\alpha^{2}+\\beta^{2}+\\gamma^{2}+\\delta^{2}\\right)$$ is equal to ______________.
$$\n\\Rightarrow x \\in\\left[\\frac{-1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}\\right] \\cup\\left(\\frac{5}{3}, \\infty\\right)\n$$\n
So, common domain is $\\left(\\frac{-1}{2}, \\frac{-1}{3}\\right) \\cup\\left[\\frac{1}{2}, \\frac{1}{\\sqrt{2}}\\right]$\n
$$\n\\begin{aligned}\n& \\therefore 18\\left(\\alpha^2+\\beta^2+\\gamma^2+\\delta^2\\right)=18\\left(\\frac{1}{4}+\\frac{1}{9}+\\frac{1}{4}+\\frac{1}{2}\\right) \\\\\\\\\n& =18\\left(\\frac{9+4+9+18}{36}\\right)=\\frac{1}{2}(40)=20\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5294,
"subject": "General Science",
"question": "If the domain of the function \n
$f(x)=\\frac{\\sqrt{x^2-25}}{\\left(4-x^2\\right)}+\\log _{10}\\left(x^2+2 x-15\\right)$ is $(-\\infty, \\alpha) \\cup[\\beta, \\infty)$, then $\\alpha^2+\\beta^3$ is equal to :",
"options": [
{
"text": "140"
},
{
"text": "175"
},
{
"text": "125"
},
{
"text": "150"
}
],
"answer": "150",
"solution": "**Answer:** 150\n\n
To find the domain of the function \n$$f(x) = \\frac{\\sqrt{x^2-25}}{(4-x^2)}+\\log_{10}(x^2+2x-15),$$ \nwe need to consider the domain conditions for both the square root function and the logarithmic function.
\n
The square root function $\\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so\n$$x^2 - 25 \\geq 0.$$\nThis inequality is satisfied when\n$$x \\leq -5 \\quad \\text{or} \\quad x \\geq 5.$$
\n
The denominator of the rational part of $f(x)$, $(4-x^2)$, cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have\n$$4 - x^2 \\neq 0.$$\nThis inequality is violated when\n$$x = \\pm2.$$
\n
Combining these conditions gives us the domain for the rational part of the function:\n$$x \\in (-\\infty, -5] \\cup (5, \\infty) \\quad \\text{and} \\quad x \\neq 2,-2.$$
\n
Moving on to the logarithmic function, $\\log_{10}(x^2+2x-15)$, the argument must be positive:\n$$x^2 + 2x - 15 > 0.$$\nThis is a quadratic inequality, which we can factor to find the solution:\n$$(x+5)(x-3) > 0.$$\nFrom this, we see that the inequality is satisfied for\n$$x < -5 \\quad \\text{or} \\quad x > 3.$$
\n
The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us:\n$$x \\in (-\\infty, -5) \\cup (5, \\infty),$$
\n
Since the question states that the domain is of the form $(-\\infty, \\alpha) \\cup [\\beta, \\infty)$, we can infer that\n$$\\alpha = -5 \\quad \\text{and} \\quad \\beta = 5.$$
\n
We calculate $\\alpha^2 + \\beta^3$ as follows:\n$$\\alpha^2 + \\beta^3 = (-5)^2 + 5^3 = 25 + 125 = 150.$$
\n
So the correct answer, representing the sum of $\\alpha^2$ and $\\beta^3$, is:\nOption D $150$.
If the domain of the function $$f(x)=\\log _e\\left(\\frac{2 x+3}{4 x^2+x-3}\\right)+\\cos ^{-1}\\left(\\frac{2 x-1}{x+2}\\right)$$ is $$(\\alpha, \\beta]$$, then the value of $$5 \\beta-4 \\alpha$$ is equal to
If the domain of the function $$f(x)=\\cos ^{-1}\\left(\\frac{2-|x|}{4}\\right)+\\left\\{\\log _e(3-x)\\right\\}^{-1}$$ is $$[-\\alpha, \\beta)-\\{\\gamma\\}$$, then $$\\alpha+\\beta+\\gamma$$ is equal to :
If the domain of the function $$f(x)=\\sin ^{-1}\\left(\\frac{x-1}{2 x+3}\\right)$$ is $$\\mathbf{R}-(\\alpha, \\beta)$$, then $$12 \\alpha \\beta$$ is equal to :
Let $$f(x)=\\left\\{\\begin{array}{ccc}-\\mathrm{a} & \\text { if } & -\\mathrm{a} \\leq x \\leq 0 \\\\ x+\\mathrm{a} & \\text { if } & 0< x \\leq \\mathrm{a}\\end{array}\\right.$$ where $$\\mathrm{a}> 0$$ and $$\\mathrm{g}(x)=(f(|x|)-|f(x)|) / 2$$. Then the function $$g:[-a, a] \\rightarrow[-a, a]$$ is
",
"options": [
{
"text": "neither one-one nor onto.\n"
},
{
"text": "both one-one and onto.\n"
},
{
"text": "one-one.\n"
},
{
"text": "onto"
}
],
"answer": "neither one-one nor onto.\n",
"solution": "**Answer:** neither one-one nor onto.\n\n\n
$$\\begin{aligned}\n& f(x)=\\left\\{\\begin{array}{l}\n-a \\quad \\text { if }-a \\leq x \\leq 0 \\\\\nx+a \\quad \\text { if } 0< x \\leq a\n\\end{array}\\right. \\\\\n& f(|x|)=\\left\\{\\begin{array}{cc}\n-a & -a \\leq|x| \\leq 0 \\\\\n|x|+a & \\text { if } 0 < |x| \\leq a\n\\end{array}\\right.\n\\end{aligned}$$
\n
$$|x|<0$$ is not possible, so
\n
$$\\begin{aligned}\n& f(|x|)= \\begin{cases}x+a & -a \\leq x \\leq 0 \\\\\n-x+a & 0 < x \\leq a\\end{cases} \\\\\n& |f(x)|= \\begin{cases}a & -a \\leq x \\leq 0 \\\\\nx+a & 0 < x \\leq a\\end{cases} \\\\\n& h(x)= \\begin{cases}-\\frac{x}{2} & -a \\leq x \\leq 0 \\\\\n0 & 0 < x \\leq a\\end{cases}\n\\end{aligned}$$
On f(x + y) = f(x) f(y)\n * Put x = 1, y = 1\n f(2) = (f(1))2\n * Put x = 2, y = 1\n f(3) = f(2). f(1) = f((1))3\n * Put x = 2, y = 2\n f(4) = f((2))2 = f((1))4\n
$$ \\therefore $$ Sum = $$19 + {1 \\over 2} = {{39} \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5315,
"subject": "General Science",
"question": "If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x3) + x g(x3) is divisible by x2 + x + 1, then P(1) is equal to ___________.",
"options": [],
"answer": "0",
"solution": "**Answer:** 0\n\nGiven, p(x) = f(x3) + xg(x3)
We know, x2 + x + 1 = (x $$-$$ $$\\omega$$) (x $$-$$ $$\\omega$$2)
Given, p(x) is divisible by x2 + x + 1. So, roots of p(x) is $$\\omega$$ and $$\\omega$$2.
$$ = {{\\cot 1 - \\cot 21} \\over {\\sin 1}} = \\cos e{c^2}1\\cos ec(21).\\sin 20$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5317,
"subject": "General Science",
"question": "Let S = {1, 2, 3, 4, 5, 6, 7}. Then the number of possible functions f : S $$\\to$$ S such that f(m . n) = f(m) . f(n) for every m, n $$\\in$$ S and m . n $$\\in$$ S is equal to _____________.",
"options": [],
"answer": "490",
"solution": "**Answer:** 490\n\nF(mn) = f(m) . f(n)
Put m = 1 f(n) = f(1) . f(n) $$\\Rightarrow$$ f(1) = 1
Let c, k $$\\in$$ R. If $$f(x) = (c + 1){x^2} + (1 - {c^2})x + 2k$$ and $$f(x + y) = f(x) + f(y) - xy$$, for all x, y $$\\in$$ R, then the value of $$|2(f(1) + f(2) + f(3) + \\,\\,......\\,\\, + \\,\\,f(20))|$$ is equal to ____________.
Let f : R $$\\to$$ R be a function defined by $$f(x) = {{2{e^{2x}}} \\over {{e^{2x}} + e}}$$. Then $$f\\left( {{1 \\over {100}}} \\right) + f\\left( {{2 \\over {100}}} \\right) + f\\left( {{3 \\over {100}}} \\right) + \\,\\,\\,.....\\,\\,\\, + \\,\\,\\,f\\left( {{{99} \\over {100}}} \\right)$$ is equal to ______________.
Let f : N $$\\to$$ R be a function such that $$f(x + y) = 2f(x)f(y)$$ for natural numbers x and y. If f(1) = 2, then the value of $$\\alpha$$ for which
Let $$f(x)=2 x^{2}-x-1$$ and $$\\mathrm{S}=\\{n \\in \\mathbb{Z}:|f(n)| \\leq 800\\}$$. Then, the value of $$\\sum\\limits_{n \\in S} f(n)$$ is equal to ___________.
For $$\\mathrm{p}, \\mathrm{q} \\in \\mathbf{R}$$, consider the real valued function $$f(x)=(x-\\mathrm{p})^{2}-\\mathrm{q}, x \\in \\mathbf{R}$$ and $$\\mathrm{q}>0$$. Let $$\\mathrm{a}_{1}$$, $$\\mathrm{a}_{2^{\\prime}}$$ $$\\mathrm{a}_{3}$$ and $$\\mathrm{a}_{4}$$ be in an arithmetic progression with mean $$\\mathrm{p}$$ and positive common difference. If $$\\left|f\\left(\\mathrm{a}_{i}\\right)\\right|=500$$ for all $$i=1,2,3,4$$, then the absolute difference between the roots of $$f(x)=0$$ is ___________.
$$\n\\text { Let } f(x)=a x^{2}+b x+c \\text { be such that } f(1)=3, f(-2)=\\lambda \\text { and } $$ $$f(3)=4$$. If $$f(0)+f(1)+f(-2)+f(3)=14$$, then $$\\lambda$$ is equal to :
Consider a function $$f:\\mathbb{N}\\to\\mathbb{R}$$, satisfying $$f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\\ge2$$ with $$f(1)=1$$. Then $$\\frac{1}{f(2022)}+\\frac{1}{f(2028)}$$ is equal to
Suppose $$f$$ is a function satisfying $$f(x + y) = f(x) + f(y)$$ for all $$x,y \\in N$$ and $$f(1) = {1 \\over 5}$$. If $$\\sum\\limits_{n = 1}^m {{{f(n)} \\over {n(n + 1)(n + 2)}} = {1 \\over {12}}} $$, then $$m$$ is equal to __________.
Let $$f:\\mathbb{R}\\to\\mathbb{R}$$ be a function defined by $$f(x) = {\\log _{\\sqrt m }}\\{ \\sqrt 2 (\\sin x - \\cos x) + m - 2\\} $$, for some $$m$$, such that the range of $$f$$ is [0, 2]. Then the value of $$m$$ is _________
Let $$f(x) = 2{x^n} + \\lambda ,\\lambda \\in R,n \\in N$$, and $$f(4) = 133,f(5) = 255$$. Then the sum of all the positive integer divisors of $$(f(3) - f(2))$$ is
Let $$f(x)$$ be a function such that $$f(x+y)=f(x).f(y)$$ for all $$x,y\\in \\mathbb{N}$$. If $$f(1)=3$$ and $$\\sum\\limits_{k = 1}^n {f(k) = 3279} $$, then the value of n is
If $$f(x) = {{(\\tan 1^\\circ )x + {{\\log }_e}(123)} \\over {x{{\\log }_e}(1234) - (\\tan 1^\\circ )}},x > 0$$, then the least value of $$f(f(x)) + f\\left( {f\\left( {{4 \\over x}} \\right)} \\right)$$ is :
Hence, the least value is 4 .",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5336,
"subject": "General Science",
"question": "
If a function $$f$$ satisfies $$f(\\mathrm{~m}+\\mathrm{n})=f(\\mathrm{~m})+f(\\mathrm{n})$$ for all $$\\mathrm{m}, \\mathrm{n} \\in \\mathbf{N}$$ and $$f(1)=1$$, then the largest natural number $$\\lambda$$ such that $$\\sum_\\limits{\\mathrm{k}=1}^{2022} f(\\lambda+\\mathrm{k}) \\leq(2022)^2$$ is equal to _________.
If $$S=\\{a \\in \\mathbf{R}:|2 a-1|=3[a]+2\\{a \\}\\}$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$ and $$\\{t\\}$$ represents the fractional part of $$t$$, then $$72 \\sum_\\limits{a \\in S} a$$ is equal to _________.
$$ \\Rightarrow $$ x = $${{2y - 1} \\over {y - 1}}$$ = f -1(y)\n
As on the given domain the function is invertible and its inverse can be computed as shown above.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5339,
"subject": "General Science",
"question": "The inverse function of\n
$$ \\Rightarrow x = {y^{{1 \\over {\\log 5}}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5341,
"subject": "General Science",
"question": "Let f : R $$-$$ {3} $$ \\to $$ R $$-$$ {1} be defined by f(x) = $${{x - 2} \\over {x - 3}}$$.
Let g : R $$ \\to $$ R be given as g(x) = 2x $$-$$ 3. Then, the sum of all the values of x for which f$$-$$1(x) + g$$-$$1(x) = $${{13} \\over 2}$$ is equal to :",
"options": [
{
"text": "3"
},
{
"text": "5"
},
{
"text": "2"
},
{
"text": "7"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\nFinding inverse of f(x)
Note :\n (1) When $$n$$ is odd then the period of $${\\sin ^n}\\theta $$, $${\\cos ^n}\\theta $$, $${\\csc ^n}\\theta $$, $${\\sec ^n}\\theta $$ = $$2\\pi $$\n
(2) When $$n$$ is even then the period of $${\\sin ^n}\\theta $$, $${\\cos ^n}\\theta $$, $${\\csc ^n}\\theta $$, $${\\sec ^n}\\theta $$ = $$\\pi $$\n
(3) When $$n$$ is even/odd then the period of $${\\tan ^n}\\theta $$, $${\\cot ^n}\\theta $$ = $$\\pi $$\n
(3) When $$n$$ is even/odd then the period of $$\\left| {{{\\sin }^n}\\theta } \\right|$$, $$\\left| {{{\\cos }^n}\\theta } \\right|$$, $$\\left| {{{\\csc }^n}\\theta } \\right|$$, $$\\left| {{{\\sec }^n}\\theta } \\right|$$, $$\\left| {{{\\tan }^n}\\theta } \\right|$$, $$\\left| {{{\\cot }^n}\\theta } \\right|$$ = $$\\pi $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5344,
"subject": "General Science",
"question": "Which one is not periodic?",
"options": [
{
"text": "$$\\left| {\\sin 3x} \\right| + {\\sin ^2}x$$ "
},
{
"text": "$$\\cos \\sqrt x + {\\cos ^2}x$$ "
},
{
"text": "$$\\cos \\,4x + {\\tan ^2}x$$ "
},
{
"text": "$$cos\\,2x + \\sin x$$ "
}
],
"answer": "$$\\cos \\sqrt x + {\\cos ^2}x$$ ",
"solution": "**Answer:** $$\\cos \\sqrt x + {\\cos ^2}x$$ \n\n$$\\sqrt x $$ is non periodic function and $$\\cos \\left( {something} \\right)$$ is a periodic function so here in $$\\cos \\sqrt x $$ $$ \\to $$ inside periodic function there is non periodic function which always produce non periodic function.\n
$${{{\\cos }^2}x}$$ is a periodic function with period $$\\pi $$\n
Note :\n (1) When $$n$$ is odd then the period of $${\\sin ^n}\\theta $$, $${\\cos ^n}\\theta $$, $${\\csc ^n}\\theta $$, $${\\sec ^n}\\theta $$ = $$2\\pi $$\n
(2) When $$n$$ is even then the period of $${\\sin ^n}\\theta $$, $${\\cos ^n}\\theta $$, $${\\csc ^n}\\theta $$, $${\\sec ^n}\\theta $$ = $$\\pi $$\n
(3) When $$n$$ is even/odd then the period of $${\\tan ^n}\\theta $$, $${\\cot ^n}\\theta $$ = $$\\pi $$\n
(3) When $$n$$ is even/odd then the period of $$\\left| {{{\\sin }^n}\\theta } \\right|$$, $$\\left| {{{\\cos }^n}\\theta } \\right|$$, $$\\left| {{{\\csc }^n}\\theta } \\right|$$, $$\\left| {{{\\sec }^n}\\theta } \\right|$$, $$\\left| {{{\\tan }^n}\\theta } \\right|$$, $$\\left| {{{\\cot }^n}\\theta } \\right|$$ = $$\\pi $$\n
$$\\cos \\sqrt x + {\\cos ^2}x$$ = non periodic function + periodic function = non periodic function",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5345,
"subject": "General Science",
"question": "The range of the function f(x) = $${}^{7 - x}{P_{x - 3}}$$ is",
"options": [
{
"text": "{1, 2, 3, 4, 5}"
},
{
"text": "{1, 2, 3, 4, 5, 6}"
},
{
"text": "{1, 2, 3, 4}"
},
{
"text": "{1, 2, 3}"
}
],
"answer": "{1, 2, 3}",
"solution": "**Answer:** {1, 2, 3}\n\nThe range of the function $f(x) = {}^{7-x}P_{x-3}$ can be found by considering the possible values of $f(x)$ as $x$ varies over its domain.\n
\nThe domain of $f(x)$ is the set of all real numbers such that \n
(i) $x \\geq 3$ (since the permutation function is only defined for non-negative integers) \n
(ii) 7 - x > 0 $$ \\Rightarrow $$ x < 7 \n
(iii) x - 3 $$ \\le $$ 7 - x $$ \\Rightarrow $$ 2x $$ \\le $$ 10 $$ \\Rightarrow $$ x $$ \\le $$ 5.\n
\nTo find the range of $f(x)$, we need to consider what values the expression ${}^{7-x}P_{x-3}$ can take as $x$ varies over its domain.\n
\nFor $x=3$, we have ${}^{7-x}P_{x-3} = {}^{4}P_{0} = 1$.\n
\nFor $x=4$, we have ${}^{7-x}P_{x-3} = {}^{3}P_{1} = 3$.\n
\nFor $x=5$, we have ${}^{7-x}P_{x-3} = {}^{2}P_{2} = 2$.\n
\nTherefore, the range of $f(x)$ is the set of all non-negative integers less than or equal to 3, i.e., ${1, 2, 3}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5346,
"subject": "General Science",
"question": "Let f : R $$ \\to $$ R be defined by f(x) = $${x \\over {1 + {x^2}}},x \\in R$$. Then the range of f is : ",
"options": [
{
"text": "$$\\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$"
},
{
"text": "$$R - \\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$"
},
{
"text": "($$-$$ 1, 1) $$-$$ {0}"
},
{
"text": "R $$-$$ [$$-$$1, 1]"
}
],
"answer": "$$\\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$",
"solution": "**Answer:** $$\\left[ { - {1 \\over 2},{1 \\over 2}} \\right]$$\n\nf(0) = 0 & f(x) is odd\n
$$ \\therefore $$ Range is $$\\left( {{2 \\over 5},{1 \\over 2}} \\right) \\cup \\left( {{3 \\over 4},{4 \\over 5}} \\right]$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5348,
"subject": "General Science",
"question": "The range of a$$\\in$$R for which the
Let the sets A and B denote the domain and range respectively of the function $$f(x)=\\frac{1}{\\sqrt{\\lceil x\\rceil-x}}$$, where $$\\lceil x\\rceil$$ denotes the smallest integer greater than or equal to $$x$$. Then among the statements
\n
(S1) : $$A \\cap B=(1, \\infty)-\\mathbb{N}$$ and
\n
(S2) : $$A \\cup B=(1, \\infty)$$
",
"options": [
{
"text": "only $$(\\mathrm{S} 2)$$ is true"
},
{
"text": "only (S1) is true"
},
{
"text": "neither (S1) nor (S2) is true"
},
{
"text": "both (S1) and (S2) are true"
}
],
"answer": "only (S1) is true",
"solution": "**Answer:** only (S1) is true\n\n$$\nf(x)=\\frac{1}{\\sqrt{\\lceil x\\rceil-x}}\n$$\n
If $\\mathrm{x} \\in \\mathrm{I},\\lceil\\mathrm{x}\\rceil=[\\mathrm{x}]$ (greatest integer function)\n
$$\n\\begin{aligned}\n& B=(1, \\infty) \\\\\\\\\n& \\text { So, } A \\cap B=(1, \\infty)-N \\\\\\\\\n& A \\cup B \\neq(1, \\infty) \\\\\\\\\n& \\Rightarrow S 1 \\text { is only correct }\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5354,
"subject": "General Science",
"question": "
Let the range of the function $$f(x)=\\frac{1}{2+\\sin 3 x+\\cos 3 x}, x \\in \\mathbb{R}$$ be $$[a, b]$$. If $$\\alpha$$ and $$\\beta$$ ar respectively the A.M. and the G.M. of $$a$$ and $$b$$, then $$\\frac{\\alpha}{\\beta}$$ is equal to
If the range of $$f(\\theta)=\\frac{\\sin ^4 \\theta+3 \\cos ^2 \\theta}{\\sin ^4 \\theta+\\cos ^2 \\theta}, \\theta \\in \\mathbb{R}$$ is $$[\\alpha, \\beta]$$, then the sum of the infinite G.P., whose first term is 64 and the common ratio is $$\\frac{\\alpha}{\\beta}$$, is equal to __________.
To determine the range of the function $$f(\\theta)=\\frac{\\sin ^4 \\theta+3 \\cos ^2 \\theta}{\\sin ^4 \\theta+\\cos ^2 \\theta}$$, let's start by simplifying the expression. Let $$\\sin^2 \\theta = x$$, so $$\\cos^2 \\theta = 1 - x$$. The function then transforms into:
\n\n
$$ f(x) = \\frac{x^2 + 3(1-x)}{x^2 + (1-x)} $$
\n\n
Simplify the numerator and denominator separately:
Next, we need to find the range of this function. Let's analyze the function by testing specific values of $$x$$ in the interval $$[0, 1]$$ (since $$\\sin^2 \\theta$$ ranges from 0 to 1):
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5358,
"subject": "General Science",
"question": "A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is $${60^ \\circ }$$ and when he retires $$40$$ meters away from the tree the angle of elevation becomes $${30^ \\circ }$$. The breadth of the river is :",
"options": [
{
"text": "$$60\\,\\,m$$ "
},
{
"text": "$$30\\,\\,m$$"
},
{
"text": "$$40\\,\\,m$$"
},
{
"text": "$$20\\,\\,m$$"
}
],
"answer": "$$20\\,\\,m$$",
"solution": "**Answer:** $$20\\,\\,m$$\n\n\n
From the figure\n
$$\\tan {60^ \\circ } = {y \\over x}$$\n
$$ \\Rightarrow y = \\sqrt {3x} .......\\left( 1 \\right)$$\n
$$\\sqrt 3 x = {{x + 40} \\over {\\sqrt 3 }} \\Rightarrow x = 20m$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5359,
"subject": "General Science",
"question": "A tower stands at the centre of a circular park. $$A$$ and $$B$$ are two points on the boundary of the park such that $$AB(=a)$$ subtends an angle of $${60^ \\circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $$A$$ or $$B$$ is $${30^ \\circ }$$. The height of the tower is :",
"options": [
{
"text": "$$a/\\sqrt 3 $$ "
},
{
"text": "$$a\\sqrt 3 $$"
},
{
"text": "$$2a/\\sqrt 3 $$ "
},
{
"text": "$$2a\\sqrt 3 $$"
}
],
"answer": "$$a/\\sqrt 3 $$ ",
"solution": "**Answer:** $$a/\\sqrt 3 $$ \n\nIn the $$\\Delta AOB,\\,\\,\\angle AOB = {60^ \\circ },$$ and \n
$$\\angle OBA = \\angle OAB$$ \n
(since $$OA=OB=AB$$ radius of same circle). \n
$$\\therefore$$ $$\\Delta AOB$$ is a equilateral triangle. \n
Let the height of tower is $$h$$\n
\n
Given distance between two points $$A$$ & $$B$$ lie on boundary of \n
circular park, subtends an angle of $${60^ \\circ }$$ at the foot of the tower \n
is $$AB$$ i.e. $$AB$$$$=a.$$ A tower $$OC$$ stands at the center of a circular \n
park. Angle of elevation of the top of the tower from $$A$$ and $$B$$ is $${30^ \\circ }$$ .\n
In $$\\Delta OAC\\,\\,\\tan {30^ \\circ } = {h \\over a}$$\n
$$ \\Rightarrow {1 \\over {\\sqrt 3 }} = {h \\over a} \\Rightarrow h = {a \\over {\\sqrt 3 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5360,
"subject": "General Science",
"question": "$$AB$$ is a vertical pole with $$B$$ at the ground level and $$A$$ at the top. $$A$$ man finds that the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $${60^ \\circ }$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7$$ m. From $$D$$ the angle of elevation of the point $$A$$ is $${45^ \\circ }$$. Then the height of the pole is :",
"options": [
{
"text": "$${{7\\sqrt 3 } \\over 2} {1 \\over {\\sqrt {3 - 1} }}m$$ "
},
{
"text": "$${{7\\sqrt 3 } \\over 2}\\left( {\\sqrt {3 } + 1 } \\right)m$$ "
},
{
"text": "$${{7\\sqrt 3 } \\over 2}\\left( {\\sqrt {3 } - 1 } \\right)m$$"
},
{
"text": "$${{7\\sqrt 3 } \\over 2} {1 \\over {\\sqrt {3 + 1} }}m$$ "
}
],
"answer": "$${{7\\sqrt 3 } \\over 2}\\left( {\\sqrt {3 } + 1 } \\right)m$$ ",
"solution": "**Answer:** $${{7\\sqrt 3 } \\over 2}\\left( {\\sqrt {3 } + 1 } \\right)m$$ \n\n\n
(As $$\\cos \\alpha = {q \\over {\\sqrt {{p^2} + {q^2}} }}$$ and $$\\sin \\alpha = {p \\over {\\sqrt {{p^2} + {q^2}} }}$$ )",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5362,
"subject": "General Science",
"question": "A bird is sitting on the top of a vertical pole $$20$$ m high and its elevation from a point $$O$$ on the ground is $${45^ \\circ }$$. It files off horizontally straight away from the point $$O$$. After one second, the elevation of the bird from $$O$$ is reduced to $${30^ \\circ }$$. Then the speed (in m/s) of the bird is :",
"options": [
{
"text": "$$20\\sqrt 2 $$ "
},
{
"text": "$$20\\left( {\\sqrt 3 - 1} \\right)$$ "
},
{
"text": "$$40\\left( {\\sqrt 2 - 1} \\right)$$"
},
{
"text": "$$40\\left( {\\sqrt 3 - \\sqrt 2 } \\right)$$ "
}
],
"answer": "$$20\\left( {\\sqrt 3 - 1} \\right)$$ ",
"solution": "**Answer:** $$20\\left( {\\sqrt 3 - 1} \\right)$$ \n\nLet the speed be $$y$$ $$m/sec$$. \n
Let $$AC$$ be the vertical pole of height $$20$$ $$m.$$\n
Let $$O$$ be the point on the ground such that $$\\angle AOC = {45^ \\circ }$$\n
Let $$OC = x$$\n
\n
Time $$t=1$$ $$s$$ \n
From $$\\Delta AOC,\\,\\,\\tan {45^ \\circ } = {{20} \\over x}\\,\\,\\,\\,\\,\\,\\,.....\\left( i \\right)$$\n
and from $$\\Delta BOD,\\,\\,\\tan {30^ \\circ } = {{20} \\over {x + y}}...\\left( {ii} \\right)$$\n
So, $$y = 20\\left( {\\sqrt 3 - 1} \\right)\\,\\,i.e.,$$\n
speed $$ = 20\\left( {\\sqrt 3 - 1} \\right)m/s$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5363,
"subject": "General Science",
"question": "If the angles of elevation of the top of a tower from three collinear points $$A, B$$ and $$C,$$ on a line leading to the foot of the tower, are $${30^ \\circ }$$, $${45^ \\circ }$$ and $${60^ \\circ }$$ respectively, then the ratio, $$AB:BC,$$ is :",
"options": [
{
"text": "$$1:\\sqrt 3 $$ "
},
{
"text": "$$2:3$$"
},
{
"text": "$$\\sqrt 3 :1$$ "
},
{
"text": "$$\\sqrt 3 :\\sqrt 2 $$ "
}
],
"answer": "$$\\sqrt 3 :1$$ ",
"solution": "**Answer:** $$\\sqrt 3 :1$$ \n\n\n As $$PB$$ bisects $$\\angle APC,$$ therefore $$AB$$ $$:$$ $$BC$$ $$=PA:PC$$\n
Also in $$\\Delta APQ,\\sin {30^ \\circ } = {h \\over {PA}} \\Rightarrow PA = 2h$$ \n
and in $$\\Delta CPQ,$$ $$\\sin {60^ \\circ } = {h \\over {PC}} \\Rightarrow PC = {{2h} \\over {\\sqrt 3 }}$$\n
$$\\therefore$$ $$AB:BC = 2h:{{2h} \\over {\\sqrt 3 }} = \\sqrt 3 :1$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5364,
"subject": "General Science",
"question": "The angle of elevation of the top of a vertical tower from a point A, due east of it is 45o. The angle of elevation of the top of the same tower from a point B, due south of A is 30o. If the distance between A and B is $$54\\sqrt 2 \\,m,$$ then the height of the tower (in metres), is :",
"options": [
{
"text": "$$36\\sqrt 3 $$"
},
{
"text": "54"
},
{
"text": "$$54\\sqrt 3 $$ "
},
{
"text": "108"
}
],
"answer": "54",
"solution": "**Answer:** 54\n\n\n
$$ \\Rightarrow $$ h = 54 m",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5365,
"subject": "General Science",
"question": "A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30o. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60o. Then the time taken (in minutes) by him, from B to reach the pillar, is :",
"options": [
{
"text": "6"
},
{
"text": "10"
},
{
"text": "20"
},
{
"text": "5"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\nAccording to given information, we have the following figure
\n \nNow, from $\\triangle A C D$ and $\\triangle B C D$, we have
\n$\\therefore$ Distance $y$ will be cover in $5 \\mathrm{~min}$.
\n$\\because$ Distance $x$ covered in $10 \\mathrm{~min}$.
\n$\\therefore$ Distance $\\frac{x}{2}$ will be cover in $5 \\mathrm{~min}$.\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5366,
"subject": "General Science",
"question": "Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point\non the ground such that AP = 2AB. If $$\\angle $$BPC = $$\\beta $$, then tan$$\\beta $$ is equal to:",
"options": [
{
"text": "$${1 \\over 4}$$"
},
{
"text": "$${2 \\over 9}$$"
},
{
"text": "$${4 \\over 9}$$"
},
{
"text": "$${6 \\over 7}$$"
}
],
"answer": "$${2 \\over 9}$$",
"solution": "**Answer:** $${2 \\over 9}$$\n\n \n
Let the height of tower $$AB = x$$ and $$LCPA = \\propto $$\n
$$ \\Rightarrow \\tan \\beta = {2 \\over 9}$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5367,
"subject": "General Science",
"question": "PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles\nof elevation of the top of the tower at P, Q and R are respectively 45$$^\\circ $$, 30$$^\\circ $$ and 30$$^\\circ $$, then the height of the tower (in m) is :",
"options": [
{
"text": "$$50\\sqrt 2 $$"
},
{
"text": "100"
},
{
"text": "50"
},
{
"text": "$$100\\sqrt 3 $$"
}
],
"answer": "100",
"solution": "**Answer:** 100\n\n \n
$$\\therefore\\,\\,\\,$$ height of tower $$=100$$ $$m.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5368,
"subject": "General Science",
"question": "An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\\sqrt 3 $$ kmabove it, is obsered at an elevation of $${60^o}$$ from a point on the ground. If, after five seconds, its elevation from the same point, is $${30^o}$$, then the speed (in km / hr) of the aeroplane, is : ",
"options": [
{
"text": "1500"
},
{
"text": "1440"
},
{
"text": "750"
},
{
"text": "720"
}
],
"answer": "1440",
"solution": "**Answer:** 1440\n\nFor $$\\Delta $$OA, A, OA1 = $${{\\sqrt 3 } \\over {\\tan {{60}^o}}}$$ = 1 km\n
As, a distance of 3 $$-$$ 1 = 2 km is convered in 5 seconds.\n
Therefore the speed of the plane is \n
$${{2 \\times 3600} \\over 5}$$ = 1440 km/hr",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5369,
"subject": "General Science",
"question": "A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. Fromthe top of T1, if the angle of depression of the foot of T2 is twice the angle of elevation of the top of T2, then the width (in m) of the road between the feetof the towers T1 and T2 is : ",
"options": [
{
"text": "$$10\\sqrt 2 $$ "
},
{
"text": "$$10\\sqrt 3 $$"
},
{
"text": "$$20\\sqrt 3 $$"
},
{
"text": "$$20\\sqrt 2 $$"
}
],
"answer": "$$20\\sqrt 3 $$",
"solution": "**Answer:** $$20\\sqrt 3 $$\n\nLet the distance between T1 and T2 be x \n
\n
From the figure \n
EA = 60 m (T1) and DB = 80 m (T2)\n
$$\\angle DEC = \\theta $$ and $$\\angle BEC = 2\\theta $$\n
$$ \\Rightarrow $$ $${x^2} = 1200$$ $$ \\Rightarrow $$ $$x = 20\\sqrt 3 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5370,
"subject": "General Science",
"question": "A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30o to 45o ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is : ",
"options": [
{
"text": "$$9\\left( {1 + \\sqrt 3 } \\right)$$ "
},
{
"text": "$$18\\left( {1 + \\sqrt 3 } \\right)$$"
},
{
"text": "$$18\\left( {\\sqrt 3 - 1} \\right)$$"
},
{
"text": "$${9 \\over 2}\\left( {\\sqrt 3 - 1} \\right)$$"
}
],
"answer": "$$9\\left( {1 + \\sqrt 3 } \\right)$$ ",
"solution": "**Answer:** $$9\\left( {1 + \\sqrt 3 } \\right)$$ \n\n\n
= 9 $$\\left( {\\sqrt 3 + 1} \\right)$$ min.\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5371,
"subject": "General Science",
"question": "Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA = 6m. A vertical lamp-post at the mid point D of AC subtends an angle 30o at B. The height (in m) of the lamp-post is -",
"options": [
{
"text": "$$2\\sqrt {21} $$"
},
{
"text": "$${3 \\over 2}\\sqrt {21} $$"
},
{
"text": "$$7\\sqrt {3} $$"
},
{
"text": "$${2 \\over 3}\\sqrt {21} $$"
}
],
"answer": "$${2 \\over 3}\\sqrt {21} $$",
"solution": "**Answer:** $${2 \\over 3}\\sqrt {21} $$\n\n\n BD = hcot30o = h$$\\sqrt 3 $$\n
So, 72 + 52 = 2(h$$\\sqrt 3 $$)2 + 32)\n
$$ \\Rightarrow $$ 37 = 3h2 + 9\n
$$ \\Rightarrow $$ 3h2 = 28\n
$$ \\Rightarrow $$ h = $$\\sqrt {{{28} \\over 3}} = {2 \\over 3}\\sqrt {21} $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5372,
"subject": "General Science",
"question": "The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45o from\na point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the\ntop of the tower from B be 30o, then the distance (in m) of the foot of the tower from the point A is :",
"options": [
{
"text": "$$15\\left( {1 + \\sqrt 3 } \\right)$$"
},
{
"text": "$$15\\left( {3 - \\sqrt 3 } \\right)$$"
},
{
"text": "$$15\\left( {3 + \\sqrt 3 } \\right)$$"
},
{
"text": "$$15\\left( {5 - \\sqrt 3 } \\right)$$"
}
],
"answer": "$$15\\left( {3 + \\sqrt 3 } \\right)$$",
"solution": "**Answer:** $$15\\left( {3 + \\sqrt 3 } \\right)$$\n\n\n$${x \\over d} = \\tan {30^o} \\Rightarrow {x \\over d} = {1 \\over {\\sqrt 3 }} \\Rightarrow d = \\sqrt 3 x$$\n
\nand $${{x + 30} \\over d} = \\tan {45^o}$$ $$ \\Rightarrow $$ d = x + 30
\n$$ \\Rightarrow $$ d = 15 ( 3 + $$\\sqrt 3 $$)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5373,
"subject": "General Science",
"question": "ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If\nthe angles of elevation of the top of the tower at A and B are cot–1\n (3$$\\sqrt 2 $$ ) and cosec–1\n(2$$\\sqrt 2 $$ ) respectively,\nthen the height of the tower (in metres) is :",
"options": [
{
"text": "$${{100} \\over {3\\sqrt 3 }}$$"
},
{
"text": "25"
},
{
"text": "20"
},
{
"text": "10$$\\sqrt 5 $$"
}
],
"answer": "20",
"solution": "**Answer:** 20\n\n\n$$\\Delta $$APM
\n$${h \\over {AM}} = {1 \\over {3\\sqrt 2 }}$$
\n$$\\Delta $$BPM
\n$${h \\over {BM}} = {1 \\over {\\sqrt 7 }}$$
\n$$\\Delta $$ABM
\nAM2 + MB2 = (100)2
'\n$$ \\Rightarrow $$ 18h2 + 7h2 = 100 × 100
\n$$ \\Rightarrow $$ h2 = 4 × 100
\n$$ \\Rightarrow $$ h = 20",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5374,
"subject": "General Science",
"question": "Two poles standing on a horizontal ground are of\nheights 5m and 10 m respectively. The line joining\ntheir tops makes an angle of 15º with ground. Then\nthe distance (in m) between the poles, is :-",
"options": [
{
"text": "$$5\\left( {2 + \\sqrt 3 } \\right)$$"
},
{
"text": "$${5 \\over 2}\\left( {2 + \\sqrt 3 } \\right)$$"
},
{
"text": "$$10\\left( {\\sqrt3 - 1 } \\right)$$"
},
{
"text": "$$5\\left( {\\sqrt3 + 1 } \\right)$$"
}
],
"answer": "$$5\\left( {2 + \\sqrt 3 } \\right)$$",
"solution": "**Answer:** $$5\\left( {2 + \\sqrt 3 } \\right)$$\n\n\n \n$$\\tan {15^o} = {5 \\over d} \\Rightarrow d = {5 \\over {\\tan {{15}^o}}}$$
\n$$ = 5\\left( {2 + \\sqrt 3 } \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5375,
"subject": "General Science",
"question": "Two vertical poles of heights, 20 m and 80 m stand\na part on a horizontal plane. The height (in meters)\nof the point of intersection of the lines joining the\ntop of each pole to the foot of the other, from this\nhorizontal plane is :",
"options": [
{
"text": "12"
},
{
"text": "16"
},
{
"text": "15"
},
{
"text": "18"
}
],
"answer": "16",
"solution": "**Answer:** 16\n\n\n
$$ \\therefore $$ Height of intersection point is 16 m",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5376,
"subject": "General Science",
"question": "If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30o and the angle of depression of reflection of the cloud in the lake from P be 60o, then the height of the cloud (in meters) from\nthe surface of the lake is :",
"options": [
{
"text": "45"
},
{
"text": "42"
},
{
"text": "50"
},
{
"text": "60"
}
],
"answer": "50",
"solution": "**Answer:** 50\n\n\n tan 30o = $${x \\over y} \\Rightarrow y = \\sqrt 3 x\\,\\,\\,\\,....(i)$$\n
tan 60o = $${{25 + x + 25} \\over y}$$\n
$$ \\Rightarrow \\,\\,\\sqrt 3 y = 50 + x$$\n
$$ \\Rightarrow $$ $$3x = 50 + x$$\n
$$ \\Rightarrow $$ x = 25 m\n
$$ \\therefore $$ Height of cloud from surface \n
= 25 + 25 = 50m",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5377,
"subject": "General Science",
"question": "The angle of elevation of the summit of a\nmountain from a point on the ground is 45°.\nAfter climbing up one km towards the summit\nat an inclination of 30° from the ground, the\nangle of elevation of the summit is found to be\n60°. Then the height (in km) of the summit from\nthe ground is :",
"options": [
{
"text": "$${1 \\over {\\sqrt 3 - 1}}$$"
},
{
"text": "$${{\\sqrt 3 + 1} \\over {\\sqrt 3 - 1}}$$"
},
{
"text": "$${1 \\over {\\sqrt 3 + 1}}$$"
},
{
"text": "$${{\\sqrt 3 - 1} \\over {\\sqrt 3 + 1}}$$"
}
],
"answer": "$${1 \\over {\\sqrt 3 - 1}}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt 3 - 1}}$$\n\n\n
$$ \\Rightarrow $$ h = $${1 \\over {\\sqrt 3 - 1}}$$ km",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5378,
"subject": "General Science",
"question": "Let AD and BC be two vertical poles at A and B respectively on a horizontal ground. If\nAD = 8 m, BC = 11 m and AB = 10 m; then the distance (in meters) of a point M on AB from the point\nA such that MD2 + MC2 is minimum is ______.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n\n
It is minimum if h = 5.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5379,
"subject": "General Science",
"question": "The angle of elevation of the top of a hill from a point on the horizontal plane passing through the\nfoot of the hill is found to be 45o. After walking a distance of 80 meters towards the top, up a slope\ninclined at an angle of 30o to the horizontal plane, the angle of elevation of the top of the hill\nbecomes 75o. Then the height of the hill (in meters) is ____________.",
"options": [],
"answer": "80",
"solution": "**Answer:** 80\n\n\n
sin 30o = $${x \\over {80}}$$ $$ \\Rightarrow $$ x = 40\n
cos 30o = $${y \\over {80}}$$ $$ \\Rightarrow $$ y = $$40\\sqrt 3 $$\n
$$ \\Rightarrow $$ h = 80 m",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5380,
"subject": "General Science",
"question": "The angle of elevation of a cloud C from a point P, 200 m above a still lake is 30°. If the angle of depression of the image of C in the lake from the point P is 60°,then PC (in m) is equal to :\n",
"options": [
{
"text": "$$200\\sqrt 3 $$"
},
{
"text": "400"
},
{
"text": "100"
},
{
"text": "$$400\\sqrt 3 $$"
}
],
"answer": "400",
"solution": "**Answer:** 400\n\n\n Let PA = x
$$\\Rightarrow PC = 400$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5381,
"subject": "General Science",
"question": "Two vertical poles are 150 m apart and the height of one is three times that of the other. If\nfrom the middle point of the line joining their feet, an observer finds the angles of elevation of\ntheir tops to be complementary, then the height of the shorter pole (in meters) is :",
"options": [
{
"text": "30"
},
{
"text": "25"
},
{
"text": "20$$\\sqrt 3 $$"
},
{
"text": "25$$\\sqrt 3 $$"
}
],
"answer": "25$$\\sqrt 3 $$",
"solution": "**Answer:** 25$$\\sqrt 3 $$\n\n\n $$ \\therefore $$ $$\\tan (90 - \\theta ) = {x \\over {75}}$$
$$ \\Rightarrow x = 25\\sqrt 3 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5382,
"subject": "General Science",
"question": "The angle of elevation of a jet plane from a point A on the ground is 60$$^\\circ$$. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30$$^\\circ$$. If the jet plane is flying at a constant height, then its height is :",
"options": [
{
"text": "$$3600\\sqrt 3 $$ m"
},
{
"text": "$$1200\\sqrt 3 $$ m"
},
{
"text": "$$1800\\sqrt 3 $$ m"
},
{
"text": "$$2400\\sqrt 3 $$ m"
}
],
"answer": "$$1200\\sqrt 3 $$ m",
"solution": "**Answer:** $$1200\\sqrt 3 $$ m\n\n\n $$v = 432 \\times {{1000} \\over {60 \\times 60}}$$ m/sec = 120 m/sec
$$h = 1200\\sqrt 3 $$ meter\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5383,
"subject": "General Science",
"question": "A man is observing, from the top of a tower, a boat speeding towards the lower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30$$^\\circ$$ (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is 45$$^\\circ$$. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :",
"options": [
{
"text": "$$10(\\sqrt 3 + 1)$$"
},
{
"text": "$$10(\\sqrt 3 - 1)$$"
},
{
"text": "10"
},
{
"text": "$$10\\sqrt 3$$"
}
],
"answer": "$$10(\\sqrt 3 + 1)$$",
"solution": "**Answer:** $$10(\\sqrt 3 + 1)$$\n\n\n $${h \\over {x + y}} = \\tan 30^\\circ $$
$$ = 10\\left( {\\sqrt 3 + 1} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5384,
"subject": "General Science",
"question": "A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be $${\\pi \\over 3}$$. If the radius of the circumcircle of $$\\Delta$$ABC is 2, then the height of the pole is equal to :",
"options": [
{
"text": "$${{1 \\over {\\sqrt 3 }}}$$"
},
{
"text": "2$${\\sqrt 3 }$$"
},
{
"text": "$${\\sqrt 3 }$$"
},
{
"text": "$${{{2\\sqrt 3 } \\over 3}}$$"
}
],
"answer": "2$${\\sqrt 3 }$$",
"solution": "**Answer:** 2$${\\sqrt 3 }$$\n\n\n Let PD = h, R = 2
As angle of elevation of top of pole from A, B, C are equal. So D must be circumcentre of $$\\Delta$$ABC
$$h = R\\tan \\left( {{\\pi \\over 3}} \\right) = 2\\sqrt 3 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5385,
"subject": "General Science",
"question": "Let in a right angled triangle, the smallest angle be $$\\theta$$. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sin$$\\theta$$ is equal to :",
"options": [
{
"text": "$${{\\sqrt 5 + 1} \\over 4}$$"
},
{
"text": "$${{\\sqrt 5 - 1} \\over 2}$$"
},
{
"text": "$${{\\sqrt 2 - 1} \\over 2}$$"
},
{
"text": "$${{\\sqrt 5 - 1} \\over 4}$$"
}
],
"answer": "$${{\\sqrt 5 - 1} \\over 2}$$",
"solution": "**Answer:** $${{\\sqrt 5 - 1} \\over 2}$$\n\n Let a $$\\Delta$$ABC having C = 90$$^\\circ$$ and A = $$\\theta$$
$$ \\Rightarrow \\sin \\theta = {{\\sqrt 5 - 1} \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5386,
"subject": "General Science",
"question": "A spherical gas balloon of radius 16 meter subtends an angle 60$$^\\circ$$ at the eye of the observer A while the angle of elevation of its center from the eye of A is 75$$^\\circ$$. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :",
"options": [
{
"text": "$$8(2 + 2\\sqrt 3 + \\sqrt 2 )$$"
},
{
"text": "$$8(\\sqrt 6 + \\sqrt 2 + 2)$$"
},
{
"text": "$$8(\\sqrt 2 + 2 + \\sqrt 3 )$$"
},
{
"text": "$$8(\\sqrt 6 - \\sqrt 2 + 2)$$"
}
],
"answer": "$$8(\\sqrt 6 + \\sqrt 2 + 2)$$",
"solution": "**Answer:** $$8(\\sqrt 6 + \\sqrt 2 + 2)$$\n\n O $$\\to$$ centre of sphere
P, Q $$\\to$$ point of contact of tangents from A
Let T be top most point of balloon & R be foot of perpendicular from O to ground.
From triangle OAP, OA = 16cosec30$$^\\circ$$ = 32
From triangle ABO, OR = OA sin75$$^\\circ$$ = $$32{{\\left( {\\sqrt 3 + 1} \\right)} \\over {2\\sqrt 2 }}$$
So level of top most point = OR + OT
$$ = 8\\left( {\\sqrt 6 + \\sqrt 2 + 2} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5387,
"subject": "General Science",
"question": "Two poles, AB of length a metres and CD of length a + b (b $$\\ne$$ a) metres are erected at the same horizontal level with bases at B and D. If BD = x and tan$$\\angle$$ACB = $${1 \\over 2}$$, then :",
"options": [
{
"text": "x2 + 2(a + 2b)x $$-$$ b(a + b) = 0"
},
{
"text": "x2 + 2(a + 2b)x + a(a + b) = 0"
},
{
"text": "x2 $$-$$ 2ax + b(a + b) = 0"
},
{
"text": "x2 $$-$$ 2ax + a(a + b) = 0"
}
],
"answer": "x2 $$-$$ 2ax + b(a + b) = 0",
"solution": "**Answer:** x2 $$-$$ 2ax + b(a + b) = 0\n\n
$$ \\Rightarrow {x^2} - 2ax + ab + {b^2} = 0$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5388,
"subject": "General Science",
"question": "A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :",
"options": [
{
"text": "12$$\\sqrt {15} $$"
},
{
"text": "12$$\\sqrt {10} $$"
},
{
"text": "8$$\\sqrt {10} $$"
},
{
"text": "6$$\\sqrt {10} $$"
}
],
"answer": "12$$\\sqrt {10} $$",
"solution": "**Answer:** 12$$\\sqrt {10} $$\n\n
From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60$$^\\circ$$. The pole subtends an angle 30$$^\\circ$$ at the top of the tower. Then the height of the tower is :
Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let $${\\pi \\over 8}$$ and $$\\theta$$ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then tan2$$\\theta$$ is equal to
A tower PQ stands on a horizontal ground with base $$Q$$ on the ground. The point $$R$$ divides the tower in two parts such that $$Q R=15 \\mathrm{~m}$$. If from a point $$A$$ on the ground the angle of elevation of $$R$$ is $$60^{\\circ}$$ and the part $$P R$$ of the tower subtends an angle of $$15^{\\circ}$$ at $$A$$, then the height of the tower is :\n
Let a vertical tower $$A B$$ of height $$2 h$$ stands on a horizontal ground. Let from a point $$P%$$ on the ground a man can see upto height $$h$$ of the tower with an angle of elevation $$2 \\alpha$$. When from $$P$$, he moves a distance $$d$$ in the direction of $$\\overrightarrow{A P}$$, he can see the top $$B$$ of the tower with an angle of elevation $$\\alpha$$. If $$d=\\sqrt{7} h$$, then $$\\tan \\alpha$$ is equal to
The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is $$45^{\\circ}$$. Let R be a point on AQ and from a point B, vertically above $$\\mathrm{R}$$, the angle of elevation of $$\\mathrm{P}$$ is $$60^{\\circ}$$. If $$\\angle \\mathrm{BAQ}=30^{\\circ}, \\mathrm{AB}=\\mathrm{d}$$ and the area of the trapezium $$\\mathrm{PQRB}$$ is $$\\alpha$$, then the ordered pair $$(\\mathrm{d}, \\alpha)$$ is :
A horizontal park is in the shape of a triangle $$\\mathrm{OAB}$$ with $$\\mathrm{AB}=16$$. A vertical lamp post $$\\mathrm{OP}$$ is erected at the point $$\\mathrm{O}$$ such that $$\\angle \\mathrm{PAO}=\\angle \\mathrm{PBO}=15^{\\circ}$$ and $$\\angle \\mathrm{PCO}=45^{\\circ}$$, where $$\\mathrm{C}$$ is the midpoint of $$\\mathrm{AB}$$. Then $$(\\mathrm{OP})^{2}$$ is equal to :
The angle of elevation of the top of a tower from a point A due north of it is $$\\alpha$$ and from a point B at a distance of 9 units due west of A is $$\\cos ^{-1}\\left(\\frac{3}{\\sqrt{13}}\\right)$$. If the distance of the point B from the tower is 15 units, then $$\\cot \\alpha$$ is equal to :
The angle of elevation of the top $$\\mathrm{P}$$ of a tower from the feet of one person standing due South of the tower is $$45^{\\circ}$$ and from the feet of another person standing due west of the tower is $$30^{\\circ}$$. If the height of the tower is 5 meters, then the distance (in meters) between the two persons is equal to
",
"options": [
{
"text": "10"
},
{
"text": "$$\\frac{5}{2} \\sqrt{5}$$"
},
{
"text": "$$5 \\sqrt{5}$$"
},
{
"text": "5"
}
],
"answer": "10",
"solution": "**Answer:** 10\n\nLet's denote the person standing due south as S and the one standing due west as W. Also, let the tower be at point T.\n\n
From person S's perspective, we have a right triangle $\\triangle SPT$. The height of the tower PT is given as 5m, which is the opposite side for angle S. The angle at S is $45^\\circ$. From the tangent trigonometric ratio, we have :\n\n
Similarly, from person W's perspective, we have a right triangle $\\triangle WPT$. The angle at W is $30^\\circ$. From the tangent trigonometric ratio, we have:\n\n
Hence, the correct answer is 10 meters, which corresponds to Option A.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5397,
"subject": "General Science",
"question": "
From the top $$\\mathrm{A}$$ of a vertical wall $$\\mathrm{AB}$$ of height $$30 \\mathrm{~m}$$, the angles of depression of the top $$\\mathrm{P}$$ and bottom $$\\mathrm{Q}$$ of a vertical tower $$\\mathrm{PQ}$$ are $$15^{\\circ}$$ and $$60^{\\circ}$$ respectively, $$\\mathrm{B}$$ and $$\\mathrm{Q}$$ are on the same horizontal level. If $$\\mathrm{C}$$ is a point on $$\\mathrm{AB}$$ such that $$\\mathrm{CB}=\\mathrm{PQ}$$, then the area (in $$\\mathrm{m}^{2}$$ ) of the quadrilateral $$\\mathrm{BCPQ}$$ is equal to :
",
"options": [
{
"text": "$$200(3-\\sqrt{3})$$"
},
{
"text": "$$300(\\sqrt{3}-1)$$"
},
{
"text": "$$300(\\sqrt{3}+1)$$"
},
{
"text": "$$600(\\sqrt{3}-1)$$"
}
],
"answer": "$$600(\\sqrt{3}-1)$$",
"solution": "**Answer:** $$600(\\sqrt{3}-1)$$\n\nGiven, $A B$ be a vertical wall of height $30 \\mathrm{~m}$ and $P Q$ be a vertical tower.\n
Such that $\\angle B Q A=\\angle T A Q=60^{\\circ}$ (Alternate angles)\n
A common tangent $$\\mathrm{T}$$ to the curves $$\\mathrm{C}_{1}: \\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1$$ and $$C_{2}: \\frac{x^{2}}{42}-\\frac{y^{2}}{143}=1$$ does not pass through the fourth quadrant. If $$\\mathrm{T}$$ touches $$\\mathrm{C}_{1}$$ at $$\\left(x_{1}, y_{1}\\right)$$ and $$\\mathrm{C}_{2}$$ at $$\\left(x_{2}, y_{2}\\right)$$, then $$\\left|2 x_{1}+x_{2}\\right|$$ is equal to ______________.
For the hyperbola $$\\mathrm{H}: x^{2}-y^{2}=1$$ and the ellipse $$\\mathrm{E}: \\frac{x^{2}}{\\mathrm{a}^{2}}+\\frac{y^{2}}{\\mathrm{~b}^{2}}=1$$, a $$>\\mathrm{b}>0$$, let the
\n
(1) eccentricity of $$\\mathrm{E}$$ be reciprocal of the eccentricity of $$\\mathrm{H}$$, and
\n
(2) the line $$y=\\sqrt{\\frac{5}{2}} x+\\mathrm{K}$$ be a common tangent of $$\\mathrm{E}$$ and $$\\mathrm{H}$$.
\n
Then $$4\\left(\\mathrm{a}^{2}+\\mathrm{b}^{2}\\right)$$ is equal to _____________.
The equation of tangent to hyperbola $${x^2} - {y^2} = 1$$ within slope $$m$$ is equal to $$y = mx\\, \\pm \\,\\sqrt {{m^2} - 1} $$ ...... (i)
\n
And for same slope $$m$$, equation of tangent to ellipse $${{{x^2}} \\over {{a^2}}} + {{{y^2}} \\over {{b^2}}} = 1$$ is $$y = mx\\, \\pm \\,\\sqrt {{a^2}{m^2} + {b^2}} $$ ...... (ii)
From equation (i) and (ii) : $${a^2} = {1 \\over 2},\\,{b^2} = {1 \\over 4}$$
\n
$$\\therefore$$ $$4({a^2} + {b^2}) = 3$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5402,
"subject": "General Science",
"question": "The locus of the point of intersection of the straight lines, \n
tx $$-$$ 2y $$-$$ 3t = 0\n
x $$-$$ 2ty + 3 = 0 (t $$ \\in $$ R), is : ",
"options": [
{
"text": "an ellipse with eccentricity $${2 \\over {\\sqrt 5 }}$$"
},
{
"text": "an ellipse with the length of major axis 6 "
},
{
"text": "a hyperbola with eccentricity $$\\sqrt 5 $$ "
},
{
"text": "a hyperbola with the length of conjugate axis 3 "
}
],
"answer": "a hyperbola with the length of conjugate axis 3 ",
"solution": "**Answer:** a hyperbola with the length of conjugate axis 3 \n\nHere, tx $$-$$ 2y $$-$$ 3t = 0 & x $$-$$ 2ty + 3 = 0\n
x2 $$-$$ 4y2 $$-$$ 16x2 y2 = 0",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5404,
"subject": "General Science",
"question": "The locus of the point of intersection of the lines, $$\\sqrt 2 x - y + 4\\sqrt 2 k = 0$$ and $$\\sqrt 2 k\\,x + k\\,y - 4\\sqrt 2 = 0$$ (k is any non-zero real parameter), is : ",
"options": [
{
"text": "an ellipse whose eccentricity is $${1 \\over {\\sqrt 3 }}.$$"
},
{
"text": "an ellipse with length of its major axis $$8\\sqrt 2 .$$"
},
{
"text": "a hyperbola whose eccentricity is $$\\sqrt 3 .$$"
},
{
"text": "a hyperbola with length of its transverse axis $$8\\sqrt 2 .$$"
}
],
"answer": "a hyperbola with length of its transverse axis $$8\\sqrt 2 .$$",
"solution": "**Answer:** a hyperbola with length of its transverse axis $$8\\sqrt 2 .$$\n\nHere, lines are : \n
$$\\sqrt 2 x$$ $$-$$ y + 4$$\\sqrt 2 k$$ = 0\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ $$\\sqrt 2 x + 4\\sqrt 2 k = y\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,....$$(i)\n
and $$\\sqrt 2 kx + ky - 4\\sqrt 2 = 0\\,\\,\\,\\,\\,...\\left( {ii} \\right)$$\n
Hence, the locus is a hyperbola with length of its transverse axis equal to 8$${\\sqrt 2 }$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5405,
"subject": "General Science",
"question": "A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is :",
"options": [
{
"text": "an ellipse"
},
{
"text": "a parabola"
},
{
"text": "a hyperbola"
},
{
"text": "a straight line"
}
],
"answer": "a parabola",
"solution": "**Answer:** a parabola\n\nLet equation of circle is \n
x2 + y2 + 2fx + 2fy + e = 0, it passes through (0, 2b)\n
g2 $$-$$ c = 4a2 $$ \\Rightarrow $$ c = $$\\left( {{g^2} - 4{a^2}} \\right)$$\n
Putting in equation (1)\n
$$ \\Rightarrow $$ 4b2 + 4f + g2 $$-$$ 4a2 = 0\n
$$ \\Rightarrow $$ x2 + 4y + 4(b2 $$-$$ a2) = 0, it represent a hyperbola. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5406,
"subject": "General Science",
"question": "The locus of the point of intersection of the lines $$\\left( {\\sqrt 3 } \\right)kx + ky - 4\\sqrt 3 = 0$$ and $$\\sqrt 3 x - y - 4\\left( {\\sqrt 3 } \\right)k = 0$$ is a conic, whose eccentricity is _________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$\\sqrt 3 kx + ky = 4\\sqrt 3 $$ ........(1)
Let T and C respectively be the transverse and conjugate axes of the hyperbola $$16{x^2} - {y^2} + 64x + 4y + 44 = 0$$. Then the area of the region above the parabola $${x^2} = y + 4$$, below the transverse axis T and on the right of the conjugate axis C is :
Let R be a rectangle given by the lines $$x=0, x=2, y=0$$ and $$y=5$$. Let A$$(\\alpha,0)$$ and B$$(0,\\beta),\\alpha\\in[0,2]$$ and $$\\beta\\in[0,5]$$, be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the mid-point of AB lies on a :
",
"options": [
{
"text": "hyperbola"
},
{
"text": "straight line"
},
{
"text": "parabola"
},
{
"text": "circle"
}
],
"answer": "hyperbola",
"solution": "**Answer:** hyperbola\n\nWe have, $R$ be a rectangle formed by the lines $x=0$, $x=2, y=0$ and $y=5$\n
\n
Let $P Q R S$ be the rectangle such that\n
$$\nP(0,0), Q(2,0), R(2,5) \\text { and } S(0,5)\n$$\n
$P Q=S R=2$ units and $P S=Q R=5$ units\n
$\\therefore$ Area of rectangle $P Q R S=(2 \\times 5) \\mathrm{sq}$ units $=10 \\mathrm{sq}$ units\n
Area of $\\triangle P A B=\\frac{1}{5} \\times$ Area of rectangle\n
On substitute values of $\\alpha$ and $\\beta$ in Eq. (i), we get\n
$$\n\\begin{aligned}\n\\quad(2 h)(2 k) & =4 \\\\\\\\\n\\Rightarrow h k & =1 \\\\\\\\\n\\therefore \\text { Locus of } M \\text { is } x y & =1 \\text {, which is a hyperbola. }\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5412,
"subject": "General Science",
"question": "The normal to a curve at $$P(x,y)$$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscissa of $$P$$, then the curve is a :",
"options": [
{
"text": "circle "
},
{
"text": "hyperbola "
},
{
"text": "ellipse "
},
{
"text": "parabola"
}
],
"answer": "hyperbola ",
"solution": "**Answer:** hyperbola \n\nEquation of normal at $$P\\left( {x,y} \\right)$$ is $$Y - y = - {{dx} \\over {dy}}\\left( {x - x} \\right)$$ \n
Coordinate of $$G$$ at $$X$$ axis is $$\\left( {X,0} \\right)$$ (let)\n
$$\\therefore$$ the curve is a hyperbola ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5413,
"subject": "General Science",
"question": "A normal to the hyperbola, 4x2 $$-$$ 9y2 = 36 meets the co-ordinate axes $$x$$ and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the ocus of P is : ",
"options": [
{
"text": "4x2 + 9y2 = 121"
},
{
"text": "9x2 + 4y2 = 169"
},
{
"text": "4x2 $$-$$ 9y2 = 121"
},
{
"text": "9x2 $$-$$ 4y2 = 169"
}
],
"answer": "9x2 $$-$$ 4y2 = 169",
"solution": "**Answer:** 9x2 $$-$$ 4y2 = 169\n\nGiven, 4x2 $$-$$ 9y2 = 36\n
$$QR = 6\\sqrt 3 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5418,
"subject": "General Science",
"question": "Let A (sec$$\\theta$$, 2tan$$\\theta$$) and B (sec$$\\phi$$, 2tan$$\\phi$$), where $$\\theta$$ + $$\\phi$$ = $$\\pi$$/2, be two points on the hyperbola 2x2 $$-$$ y2 = 2. If ($$\\alpha$$, $$\\beta$$) is the point of the intersection of the normals to the hyperbola at A and B, then (2$$\\beta$$)2 is equal to ____________.",
"options": [],
"answer": "Bonus",
"solution": "**Answer:** Bonus\n\nSince, point A (sec$$\\theta$$, 2tan$$\\theta$$) lies on the hyperbola 2x2 $$-$$ y2 = 2
Let the eccentricity of the hyperbola $${{{x^2}} \\over {{a^2}}} - {{{y^2}} \\over {{b^2}}} = 1$$ be $${5 \\over 4}$$. If the equation of the normal at the point $$\\left( {{8 \\over {\\sqrt {5} }},{{12} \\over {5}}} \\right)$$ on the hyperbola is $$8\\sqrt 5 x + \\beta y = \\lambda $$, then $$\\lambda$$ $$-$$ $$\\beta$$ is equal to ___________.
Let the tangent drawn to the parabola $$y^{2}=24 x$$ at the point $$(\\alpha, \\beta)$$ is perpendicular to the line $$2 x+2 y=5$$. Then the normal to the hyperbola $$\\frac{x^{2}}{\\alpha^{2}}-\\frac{y^{2}}{\\beta^{2}}=1$$ at the point $$(\\alpha+4, \\beta+4)$$ does NOT pass through the point :
The vertices of a hyperbola H are ($$\\pm$$ 6, 0) and its eccentricity is $${{\\sqrt 5 } \\over 2}$$. Let N be the normal to H at a point in the first quadrant and parallel to the line $$\\sqrt 2 x + y = 2\\sqrt 2 $$. If d is the length of the line segment of N between H and the y-axis then d$$^2$$ is equal to _____________.
Let $$e_1$$ be the eccentricity of the hyperbola $$\\frac{x^2}{16}-\\frac{y^2}{9}=1$$ and $$e_2$$ be the eccentricity of the ellipse $$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1, \\mathrm{a} > \\mathrm{b}$$, which passes through the foci of the hyperbola. If $$\\mathrm{e}_1 \\mathrm{e}_2=1$$, then the length of the chord of the ellipse parallel to the $$x$$-axis and passing through $$(0,2)$$ is :
$$ \\Rightarrow 3{e^2} = 4 \\Rightarrow e = {2 \\over {\\sqrt 3 }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5427,
"subject": "General Science",
"question": "Let a and b respectively be the semitransverse and semi-conjugate axes of a\nhyperbola whose eccentricity satisfies the equation 9e2 − 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then a2 − b2 is equal to :\n",
"options": [
{
"text": "7"
},
{
"text": "$$-$$ 7"
},
{
"text": "5"
},
{
"text": "$$-$$ 5"
}
],
"answer": "$$-$$ 7",
"solution": "**Answer:** $$-$$ 7\n\nAs S(5, 0) is the focus.\n
$$ \\therefore $$ ae = 5 . . . (1)\n
As 5x = 9\n
$$ \\therefore $$ x = $${9 \\over 5}$$ is the directrix \n
The point (5, 2$$\\sqrt 3 $$) does not satisfy the above equation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5429,
"subject": "General Science",
"question": "Let $$0 < \\theta < {\\pi \\over 2}$$. If the eccentricity of the\n
Maximum value of LR = 2tan$${\\pi \\over 2}$$ sin$${\\pi \\over 2}$$\n
= 2$$\\left( \\infty \\right) \\times 1$$\n
= $$\\infty $$\n
$$ \\therefore $$ Interval of LR = (3, $$\\infty $$) \n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5430,
"subject": "General Science",
"question": "A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is : ",
"options": [
{
"text": "$${3 \\over 2}$$"
},
{
"text": "$$\\sqrt 3 $$"
},
{
"text": "2"
},
{
"text": "$${2 \\over {\\sqrt 3 }}$$"
}
],
"answer": "$${2 \\over {\\sqrt 3 }}$$",
"solution": "**Answer:** $${2 \\over {\\sqrt 3 }}$$\n\nLet the equation of hyperbola\n
Let $$H:{{{x^2}} \\over {{a^2}}} - {{{y^2}} \\over {{b^2}}} = 1$$, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $$4(2\\sqrt 2 + \\sqrt {14} )$$. If the eccentricity H is $${{\\sqrt {11} } \\over 2}$$, then the value of a2 + b2 is equal to __________.
Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola $${{{x^2}} \\over {{a^2}}} - {{{y^2}} \\over {{b^2}}} = 1$$. Let e' and l' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $${e^2} = {{11} \\over {14}}l$$ and $${\\left( {e'} \\right)^2} = {{11} \\over 8}l'$$, then the value of $$77a + 44b$$ is equal to :
Let the hyperbola $$H:{{{x^2}} \\over {{a^2}}} - {y^2} = 1$$ and the ellipse $$E:3{x^2} + 4{y^2} = 12$$ be such that the length of latus rectum of H is equal to the length of latus rectum of E. If $${e_H}$$ and $${e_E}$$ are the eccentricities of H and E respectively, then the value of $$12\\left( {e_H^2 + e_E^2} \\right)$$ is equal to ___________.
Let the foci of the ellipse $$\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$$ and the hyperbola $$\\frac{x^{2}}{144}-\\frac{y^{2}}{\\alpha}=\\frac{1}{25}$$ coincide. Then the length of the latus rectum of the hyperbola is :
An ellipse $$E: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$$ passes through the vertices of the hyperbola $$H: \\frac{x^{2}}{49}-\\frac{y^{2}}{64}=-1$$. Let the major and minor axes of the ellipse $$E$$ coincide with the transverse and conjugate axes of the hyperbola $$H$$, respectively. Let the product of the eccentricities of $$E$$ and $$H$$ be $$\\frac{1}{2}$$. If $$l$$ is the length of the latus rectum of the ellipse $$E$$, then the value of $$113 l$$ is equal to _____________.
Let the hyperbola $$H: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$$ pass through the point $$(2 \\sqrt{2},-2 \\sqrt{2})$$. A parabola is drawn whose focus is same as the focus of $$\\mathrm{H}$$ with positive abscissa and the directrix of the parabola passes through the other focus of $$\\mathrm{H}$$. If the length of the latus rectum of the parabola is e times the length of the latus rectum of $$\\mathrm{H}$$, where e is the eccentricity of H, then which of the following points lies on the parabola?
Let $$\\mathrm{H}_{\\mathrm{n}}: \\frac{x^{2}}{1+n}-\\frac{y^{2}}{3+n}=1, n \\in N$$. Let $$\\mathrm{k}$$ be the smallest even value of $$\\mathrm{n}$$ such that the eccentricity of $$\\mathrm{H}_{\\mathrm{k}}$$ is a rational number. If $$l$$ is the length of the latus rectum of $$\\mathrm{H}_{\\mathrm{k}}$$, then $$21 l$$ is equal to ____________.
Let the eccentricity of an ellipse $$\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$$ is reciprocal to that of the hyperbola $$2 x^{2}-2 y^{2}=1$$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is ___________.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nEquation of hyperbola is $2 x^2-2 y^2=1$\n
$\\therefore$ Length of the latus rectum $=\\frac{2 b^2}{a}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}$\n
$\\therefore$ Square of length of the latus rectum $=2$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5450,
"subject": "General Science",
"question": "For $0<\\theta<\\pi / 2$, if the eccentricity of the hyperbola \n
$x^2-y^2 \\operatorname{cosec}^2 \\theta=5$ is $\\sqrt{7}$ times eccentricity of the
ellipse $x^2 \\operatorname{cosec}^2 \\theta+y^2=5$, then the value of $\\theta$ is :",
"options": [
{
"text": "$\\frac{\\pi}{6}$"
},
{
"text": "$\\frac{5 \\pi}{12}$"
},
{
"text": "$\\frac{\\pi}{3}$"
},
{
"text": "$\\frac{\\pi}{4}$"
}
],
"answer": "$\\frac{\\pi}{3}$",
"solution": "**Answer:** $\\frac{\\pi}{3}$\n\n
To find the value of $\\theta$, we need to determine the relationship between the eccentricities of the given hyperbola and ellipse. Let's start by writing down the standard forms of ellipse and hyperbola and then relate them to the given equations.
\n\n
The standard form of an ellipse is:\n\n\n\n
$\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$
\n\n\n\n
The eccentricity $e$ of an ellipse is given by:
\n\n\n\n
$e = \\sqrt{1 - \\frac{b^2}{a^2}}\\ \\text{for}\\ a > b$
\n\n\n\n\n\n
For the given ellipse:
\n\n\n\n
$x^2 \\operatorname{cosec}^2 \\theta + y^2 = 5$
\n\n\n\n
We can compare it with the standard form by writing it as:
Looking at the options provided and knowing the sine values for common angles, we can see that $\\operatorname{sin} \\theta = \\frac{\\sqrt{3}}{2}$ corresponds to $\\theta = \\frac{\\pi}{3}$.\n\n\n\n
If the foci of a hyperbola are same as that of the ellipse $$\\frac{x^2}{9}+\\frac{y^2}{25}=1$$ and the eccentricity of the hyperbola is $$\\frac{15}{8}$$ times the eccentricity of the ellipse, then the smaller focal distance of the point $$\\left(\\sqrt{2}, \\frac{14}{3} \\sqrt{\\frac{2}{5}}\\right)$$ on the hyperbola, is equal to
Let the foci and length of the latus rectum of an ellipse $$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1, a>b b e( \\pm 5,0)$$ and $$\\sqrt{50}$$, respectively. Then, the square of the eccentricity of the hyperbola $$\\frac{x^2}{b^2}-\\frac{y^2}{a^2 b^2}=1$$ equals
Let $$P$$ be a point on the hyperbola $$H: \\frac{x^2}{9}-\\frac{y^2}{4}=1$$, in the first quadrant such that the area of triangle formed by $$P$$ and the two foci of $$H$$ is $$2 \\sqrt{13}$$. Then, the square of the distance of $$P$$ from the origin is
Let the latus rectum of the hyperbola $$\\frac{x^2}{9}-\\frac{y^2}{b^2}=1$$ subtend an angle of $$\\frac{\\pi}{3}$$ at the centre of the hyperbola. If $$\\mathrm{b}^2$$ is equal to $$\\frac{l}{\\mathrm{~m}}(1+\\sqrt{\\mathrm{n}})$$, where $$l$$ and $$\\mathrm{m}$$ are co-prime numbers, then $$\\mathrm{l}^2+\\mathrm{m}^2+\\mathrm{n}^2$$ is equal to ________.
Let the foci of a hyperbola $$H$$ coincide with the foci of the ellipse $$E: \\frac{(x-1)^2}{100}+\\frac{(y-1)^2}{75}=1$$ and the eccentricity of the hyperbola $$H$$ be the reciprocal of the eccentricity of the ellipse $$E$$. If the length of the transverse axis of $$H$$ is $$\\alpha$$ and the length of its conjugate axis is $$\\beta$$, then $$3 \\alpha^2+2 \\beta^2$$ is equal to
Let $$\\mathrm{S}$$ be the focus of the hyperbola $$\\frac{x^2}{3}-\\frac{y^2}{5}=1$$, on the positive $$x$$-axis. Let $$\\mathrm{C}$$ be the circle with its centre at $$\\mathrm{A}(\\sqrt{6}, \\sqrt{5})$$ and passing through the point $$\\mathrm{S}$$. If $$\\mathrm{O}$$ is the origin and $$\\mathrm{SAB}$$ is a diameter of $$\\mathrm{C}$$, then the square of the area of the triangle OSB is equal to __________.
Let $$H: \\frac{-x^2}{a^2}+\\frac{y^2}{b^2}=1$$ be the hyperbola, whose eccentricity is $$\\sqrt{3}$$ and the length of the latus rectum is $$4 \\sqrt{3}$$. Suppose the point $$(\\alpha, 6), \\alpha>0$$ lies on $$H$$. If $$\\beta$$ is the product of the focal distances of the point $$(\\alpha, 6)$$, then $$\\alpha^2+\\beta$$ is equal to
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5458,
"subject": "General Science",
"question": "The locus of a point $$P\\left( {\\alpha ,\\beta } \\right)$$ moving under the condition that the line $$y = \\alpha x + \\beta $$ is tangent to the hyperbola $${{{x^2}} \\over {{a^2}}} - {{{y^2}} \\over {{b^2}}} = 1$$ is :",
"options": [
{
"text": "an ellipse "
},
{
"text": "a circle "
},
{
"text": "a parabola "
},
{
"text": "a hyperbola "
}
],
"answer": "a hyperbola ",
"solution": "**Answer:** a hyperbola \n\nTangent to the hyperbola $${{{x^2}} \\over {{a^2}}} - {{{y^2}} \\over {{b^2}}} = 1$$ is \n
$$y = mx \\pm \\sqrt {{a^2}{m^2} - {b^2}} $$ \n
Given that $$y = \\alpha x + \\beta $$ is the tangent of hyperbola \n
$$ \\Rightarrow m = \\alpha $$ and $${a^2}{m^2} - {b^2} = {\\beta ^2}$$\n
Equation of tangent is $${{\\sqrt 2 x} \\over 1} - {{\\sqrt 3 y} \\over 3} = 1$$\n
It satisfy point $$\\left( {2\\sqrt 2 ,3\\sqrt 3 } \\right)$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5460,
"subject": "General Science",
"question": "Tangents are drawn to the hyperbola 4x2 - y2 = 36 at the points P and Q.\n
If these tangents intersect at the\npoint T(0, 3) then the area (in sq. units) of $$\\Delta $$PTQ is :",
"options": [
{
"text": "$$36\\sqrt 5 $$"
},
{
"text": "$$45\\sqrt 5 $$"
},
{
"text": "$$54\\sqrt 3 $$"
},
{
"text": "$$60\\sqrt 3 $$"
}
],
"answer": "$$45\\sqrt 5 $$",
"solution": "**Answer:** $$45\\sqrt 5 $$\n\n \n
TM is the height of the triangle Length of TM = 12 + 3 = 15\n
$$\\therefore\\,\\,\\,$$ Area of $$\\Delta PQT$$ = $${1 \\over 2} \\times 6\\sqrt 5 \\times 15$$\n
$$ = 45\\sqrt 5 $$ sq. units ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5461,
"subject": "General Science",
"question": "The equation of a tangent to the hyperbola 4x2 – 5y2 = 20 parallel to the line x – y = 2 is :",
"options": [
{
"text": "x $$-$$ y + 9 = 0"
},
{
"text": "x $$-$$ y $$-$$ 3 = 0"
},
{
"text": "x $$-$$ y + 1 = 0"
},
{
"text": "x $$-$$ y + 7 = 0"
}
],
"answer": "x $$-$$ y + 1 = 0",
"solution": "**Answer:** x $$-$$ y + 1 = 0\n\nHyperbola $${{{x^2}} \\over 5} - {{{y^2}} \\over 4} = 1$$\n
slope of tangent = 1 \n
equation of tangent y = x $$ \\pm $$ $$\\sqrt {5 - 4} $$\n
$$ \\Rightarrow $$ y = x $$ \\pm $$ 1\n
$$ \\Rightarrow $$ y = x + 1 \n
or\n
$$ \\Rightarrow $$ y = x $$-$$ 1",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5462,
"subject": "General Science",
"question": "If the eccentricity of the standard hyperbola\npassing through the point (4,6) is 2, then the\nequation of the tangent to the hyperbola at (4,6)\nis :",
"options": [
{
"text": "2x – y – 2 = 0"
},
{
"text": "3x – 2y = 0"
},
{
"text": "2x – 3y + 10 = 0"
},
{
"text": "x – 2y + 8 = 0"
}
],
"answer": "2x – y – 2 = 0",
"solution": "**Answer:** 2x – y – 2 = 0\n\nFormula for standard hyperbola :\n
= 6",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5464,
"subject": "General Science",
"question": "Consider a hyperbola H : x2 $$-$$ 2y2 = 4. Let the tangent at a point P(4, $${\\sqrt 6 }$$) meet the x-axis at Q and latus rectum at R(x1, y1), x1 > 0. If F is a focus of H which is nearer to the point P, then the area of $$\\Delta$$QFR is equal to :",
"options": [
{
"text": "$${\\sqrt 6 }$$ $$-$$ 1"
},
{
"text": "$${7 \\over {\\sqrt 6 }}$$ $$-$$ 2"
},
{
"text": "$${4\\sqrt 6 }$$ $$-$$ 1"
},
{
"text": "$${4\\sqrt 6 }$$"
}
],
"answer": "$${7 \\over {\\sqrt 6 }}$$ $$-$$ 2",
"solution": "**Answer:** $${7 \\over {\\sqrt 6 }}$$ $$-$$ 2\n\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5465,
"subject": "General Science",
"question": "Let a line L : 2x + y = k, k > 0 be a tangent to the hyperbola x2 $$-$$ y2 = 3. If L is also a tangent to the parabola y2 = $$\\alpha$$x, then $$\\alpha$$ is equal to :",
"options": [
{
"text": "12"
},
{
"text": "$$-$$12"
},
{
"text": "24"
},
{
"text": "$$-$$24"
}
],
"answer": "$$-$$24",
"solution": "**Answer:** $$-$$24\n\nTangent to hyperbola of
Let the eccentricity of the hyperbola $$H:{{{x^2}} \\over {{a^2}}} - {{{y^2}} \\over {{b^2}}} = 1$$ be $$\\sqrt {{5 \\over 2}} $$ and length of its latus rectum be $$6\\sqrt 2 $$. If $$y = 2x + c$$ is a tangent to the hyperbola H, then the value of c2 is equal to :
Let a line L1 be tangent to the hyperbola $${{{x^2}} \\over {16}} - {{{y^2}} \\over 4} = 1$$ and let L2 be the line passing through the origin and perpendicular to L1. If the locus of the point of intersection of L1 and L2 is $${({x^2} + {y^2})^2} = \\alpha {x^2} + \\beta {y^2}$$, then $$\\alpha$$ + $$\\beta$$ is equal to _____________.
Let the equation of two diameters of a circle $$x^{2}+y^{2}-2 x+2 f y+1=0$$ be $$2 p x-y=1$$ and $$2 x+p y=4 p$$. Then the slope m $$ \\in $$ $$(0, \\infty)$$ of the tangent to the hyperbola $$3 x^{2}-y^{2}=3$$ passing through the centre of the circle is equal to _______________.
Let $$\\mathrm{P}\\left(x_{0}, y_{0}\\right)$$ be the point on the hyperbola $$3 x^{2}-4 y^{2}=36$$, which is nearest to the line $$3 x+2 y=1$$. Then $$\\sqrt{2}\\left(y_{0}-x_{0}\\right)$$ is equal to :
",
"options": [
{
"text": "3"
},
{
"text": "$$-$$9"
},
{
"text": "$$-$$3"
},
{
"text": "9"
}
],
"answer": "$$-$$9",
"solution": "**Answer:** $$-$$9\n\nIf $\\left(x_0, y_0\\right)$ is point on hyperbola then tangent at $\\left(x_0, y_0\\right)$ is parallel to $3 x+2 y=1$\n
Equation of tangent $= \\frac{x x_0}{12}-\\frac{y y_0}{9}=2$\n
Slope of tangent $=\\frac{-3}{2}$\n
Equation of tangent in slope form\n
$y=\\frac{-3}{2} x \\pm \\sqrt{12 \\cdot \\frac{9}{4}-9}$\n
The foci of a hyperbola are $$( \\pm 2,0)$$ and its eccentricity is $$\\frac{3}{2}$$. A tangent, perpendicular to the line $$2 x+3 y=6$$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $$\\mathrm{x}$$ - and $$\\mathrm{y}$$-axes are $$\\mathrm{a}$$ and $$\\mathrm{b}$$ respectively, then $$|6 a|+|5 b|$$ is equal to __________
Given that the foci are at $(\\pm 2, 0)$, the distance between the foci is $2ae = 2\\times2 = 4$.
So, $a = \\frac{4}{3}$.
\n
The eccentricity of the hyperbola is given as $\\frac{3}{2}$. Thus, $e = \\frac{3}{2}$.
\n
For a hyperbola, $e^2 = 1 + \\frac{b^2}{a^2}$, which gives $b^2 = a^2(e^2 - 1)$.
Substituting the given values, we get $b^2 = \\frac{16}{9} - \\frac{16}{9} = \\frac{20}{9}$.
\n
The equation of the hyperbola is $\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$, or $\\frac{9x^2}{16} - \\frac{9y^2}{20} = 1$.
\n
The slope of the tangent line, which is perpendicular to the given line $2x + 3y = 6$, is $\\frac{3}{2}$.
\n
The equation of the tangent to a hyperbola is $y = mx \\pm \\sqrt{a^2 m^2 - b^2}$. Substituting the given values, we get $y = \\frac{3x}{2} \\pm \\sqrt{\\frac{16}{9}\\times\\frac{9}{4} - \\frac{20}{9}} = \\frac{3x}{2} \\pm \\frac{4}{3}$.
\n
The $x$-intercept occurs when $y = 0$, so $|a| = \\frac{8}{9}$, and the $y$-intercept occurs when $x = 0$, so $|b| = \\frac{4}{3}$.
Let $$m_{1}$$ and $$m_{2}$$ be the slopes of the tangents drawn from the point $$\\mathrm{P}(4,1)$$ to the hyperbola $$H: \\frac{y^{2}}{25}-\\frac{x^{2}}{16}=1$$. If $$\\mathrm{Q}$$ is the point from which the tangents drawn to $$\\mathrm{H}$$ have slopes $$\\left|m_{1}\\right|$$ and $$\\left|m_{2}\\right|$$ and they make positive intercepts $$\\alpha$$ and $$\\beta$$ on the $$x$$-axis, then $$\\frac{(P Q)^{2}}{\\alpha \\beta}$$ is equal to __________.
",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nEquation of tangent to the hyperbola $\\frac{y^2}{a^2}-\\frac{x^2}{b^2}=1$\n
$$\ny=m x \\pm \\sqrt{a^2-b^2 m^2}\n$$\n
Given the hyperbola $H: \\frac{y^2}{25} - \\frac{x^2}{16} = 1$, the equation of the tangent to this hyperbola can be written as :\n\n
$$y=mx \\pm \\sqrt{25 - 16m^2}$$\n\n
We know that the tangents pass through the point $P(4, 1)$, which gives us the equation :\n\n
$$1 = 4m \\pm \\sqrt{25 - 16m^2}$$\n\n
Squaring both sides to get rid of the square root, we obtain :\n\n
$$(4m - 1)^2 = 25 - 16m^2$$\n\n
which simplifies to the quadratic equation :\n\n
$$4m^2 - m - 3 = 0.$$\n\n
Solving this equation, we find the roots $m_1 = 1$ and $m_2 = -\\frac{3}{4}$, which are the slopes of the tangents. \n\n
Given that we are interested in the positive values of the slopes, we consider $|m_1| = 1$ and $|m_2| = \\frac{3}{4}$.\n\n
The equations of the tangents are then :\n\n
1) $y = x - 3$ and \n\n
2) $y = \\frac{3}{4}x - 4$.\n\n
The x-intercepts of these lines (when $y = 0$) are given by $x = \\alpha = 3$ for the first line and $x = \\beta = \\frac{16}{3}$ for the second line.\n\n
The intersection point $Q$ of these tangents is found by solving the system of equations $y = x - 3$ and $y = \\frac{3}{4}x - 4$, which gives $Q(-4, -7)$.\n\n
The square of the distance $PQ$ is then $(-4-4)^2 + (-7-1)^2 = 128$.\n\n
Consider a hyperbola $$\\mathrm{H}$$ having centre at the origin and foci on the $$\\mathrm{x}$$-axis. Let $$\\mathrm{C}_1$$ be the circle touching the hyperbola $$\\mathrm{H}$$ and having the centre at the origin. Let $$\\mathrm{C}_2$$ be the circle touching the hyperbola $$\\mathrm{H}$$ at its vertex and having the centre at one of its foci. If areas (in sq units) of $$C_1$$ and $$C_2$$ are $$36 \\pi$$ and $$4 \\pi$$, respectively, then the length (in units) of latus rectum of $$\\mathrm{H}$$ is
The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and $$x= \\pm \\frac{4}{\\sqrt{3}}$$, respectively. Let the line $$y-\\sqrt{3} x+\\sqrt{3}=0$$ touch this hyperbola at $$\\left(x_0, y_0\\right)$$. If $$\\mathrm{m}$$ is the product of the focal distances of the point $$\\left(x_0, y_0\\right)$$, then $$4 \\mathrm{e}^2+\\mathrm{m}$$ is equal to _________.
Let $$I(x)=\\int \\frac{x^{2}\\left(x \\sec ^{2} x+\\tan x\\right)}{(x \\tan x+1)^{2}} d x$$. If $$I(0)=0$$, then $$I\\left(\\frac{\\pi}{4}\\right)$$ is equal to :
$$\n\\begin{aligned}\nI(x)= & \\int \\frac{x^2\\left(x \\sec ^2 x+\\tan x\\right)}{(x \\tan x+1)^2} d x \\\\\\\\\n= & x^2 \\int \\frac{x \\sec ^2 x+\\tan x}{(x \\tan x+1)^2} d x \\\\\\\\\n& \\quad-\\int\\left\\{\\frac{d}{d x}\\left(x^2\\right) \\int \\frac{x \\sec ^2 x+\\tan x}{(x \\tan x+1)^2} d x\\right\\} d x \\text { (integration by parts) }\n\\end{aligned}\n$$\n
$$\n=x^2\\left(\\frac{-1}{x \\tan x+1}\\right)+\\int \\frac{2 x}{x \\tan x+1} d x\n$$\n
Now, let \n
$$\n\\begin{aligned}\nI_1 & =2 \\int \\frac{x}{x \\tan x+1} d x \\\\\\\\\n& =2 \\int \\frac{x \\cos x}{x \\sin x+\\cos x} d x\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\n& \\text { On putting } x \\sin x+\\cos x=t \\\\\\\\\n& \\Rightarrow (x \\cos x+\\sin x-\\sin x) d x=d t \\\\\\\\\n& \\Rightarrow x \\cos x d x=d t\n\\end{aligned}\n$$\n
If $$\\int \\operatorname{cosec}^5 x d x=\\alpha \\cot x \\operatorname{cosec} x\\left(\\operatorname{cosec}^2 x+\\frac{3}{2}\\right)+\\beta \\log _x\\left|\\tan \\frac{x}{2}\\right|+\\mathrm{C}$$\nwhere $$\\alpha, \\beta \\in \\mathbb{R}$$ and $$\\mathrm{C}$$ is the constant of integration, then the value of $$8(\\alpha+\\beta)$$ equals _________.
$$\\begin{aligned}\n& I=\\int(\\operatorname{cosec} x)^5 d x=\\int(\\operatorname{cosec} x)^3(\\operatorname{cosec} x)^2 d x \\\\\n& =(\\operatorname{cosec} x)^3 \\int \\operatorname{cosec}^2 x d x- \\\\\n& \\int\\left(\\frac{d}{d x}(\\operatorname{cosec} x)^3 \\int \\operatorname{cosec}^2 x d x\\right) d x\n\\end{aligned}$$
\n
$$\\begin{aligned}\n& =-\\cot x(\\operatorname{cosec} x)^3-\\int 3 \\operatorname{cosec}^2 x \\cdot(-\\operatorname{cosec} x \\cot x)(-\\cot x) d x \\\\\n& =-\\cot x(\\operatorname{cosec} x)^3-\\int 3 \\operatorname{cosec}^3 x \\cot ^2 x d x \\\\\n& =-\\cot x(\\operatorname{cosec} x)^3-3 \\int(\\operatorname{cosec} x)^3\\left(\\operatorname{cosec}^2 x-1\\right) d x \\\\\n& I=-\\cot x(\\operatorname{cosec} x)^3-3 I+3 \\int(\\operatorname{cosec} x)^3 d x \\\\\n& 4 I=-\\cot x(\\operatorname{cosec} x)^3+3 \\int(\\operatorname{cosec} x)^3 d x \\\\\n& =-\\cot x(\\operatorname{cosec} x)^3+3 I_1 \\\\\n& I_1=\\int \\operatorname{cosec} x \\cdot \\operatorname{cosec}^2 x d x=\\operatorname{cosec} x(-\\cot x)- \\\\\n& \\qquad \\int(-\\operatorname{cosec} x \\cot x)(-\\cot x) d x\n\\end{aligned}$$
\n
$$\\begin{aligned}\n& I_1=-\\operatorname{cosec} x \\cot x-\\int \\operatorname{cosec} x\\left(\\operatorname{cosec}^2 x-1\\right) d x \\\\\n& I_1=-\\operatorname{cosec} x \\cot x-I+\\int \\operatorname{cosec} x d x \\\\\n& 2 l_1=-\\operatorname{cosec} x \\cot x+\\ln \\left|\\tan \\frac{x}{2}\\right| \\\\\n& \\Rightarrow \\quad 4 l=-\\cot x(\\operatorname{cosec} x)^3-\\frac{3}{2} \\operatorname{cosec} x \\cot x +\\frac{3}{2} \\ln \\left|\\tan \\frac{x}{2}\\right|+C\n\\end{aligned}$$
Let $$I(x)=\\int \\sqrt{\\frac{x+7}{x}} \\mathrm{~d} x$$ and $$I(9)=12+7 \\log _{e} 7$$. If $$I(1)=\\alpha+7 \\log _{e}(1+2 \\sqrt{2})$$, then $$\\alpha^{4}$$ is equal to _________.
Let $$I(x)=\\int \\frac{(x+1)}{x\\left(1+x e^{x}\\right)^{2}} d x, x > 0$$. If $$\\lim_\\limits{x \\rightarrow \\infty} I(x)=0$$, then $$I(1)$$ is equal to :
",
"options": [
{
"text": "$$\\frac{e+1}{e+2}-\\log _{e}(e+1)$$"
},
{
"text": "$$\\frac{e+1}{e+2}+\\log _{e}(e+1)$$"
},
{
"text": "$$\\frac{e+2}{e+1}-\\log _{e}(e+1)$$"
},
{
"text": "$$\\frac{e+2}{e+1}+\\log _{e}(e+1)$$"
}
],
"answer": "$$\\frac{e+2}{e+1}-\\log _{e}(e+1)$$",
"solution": "**Answer:** $$\\frac{e+2}{e+1}-\\log _{e}(e+1)$$\n\n$$\n\\begin{aligned}\n& \\mathrm{I}=\\int \\frac{x+1}{x\\left(1+x e^x\\right)^2} d x \\\\\\\\\n& \\text { Put } 1+x e^x=t \\Rightarrow x e^x=t-1 \\\\\\\\\n& \\Rightarrow\\left(x e^x+e^x\\right) d x=d t \\\\\\\\\n& \\Rightarrow e^x(x+1) d x=d t\n\\end{aligned}\n$$\n
$$\n\\therefore I=\\int \\frac{d t}{e^x \\cdot x t^2}=\\int \\frac{d t}{(t-1) t^2}\n$$\n
Let $\\frac{1}{t^2(t-1)}=\\frac{\\mathrm{A}}{(t-1)}+\\frac{\\mathrm{B} t+\\mathrm{C}}{t^2}$\n
Comparing coefficients of $t^2, t$ and constant terms, we get\n
$$\nA+B=0, C-B=0,-C=1\n$$\n
On solving above equations, we get\n
$$\n\\begin{aligned}\n& \\mathrm{C}=-1,=\\mathrm{B}, \\mathrm{A}=1 \\\\\\\\\n& \\therefore \\mathrm{I}=\\int \\frac{1}{t-1} d t+\\int \\frac{-t-1}{t^2} d t \\\\\\\\\n& =\\int \\frac{1}{t-1} d t-\\int \\frac{1}{t} d t-\\int \\frac{1}{t^2} d t \\\\\\\\\n& =\\log |t-1|-\\log |t|+\\frac{1}{t}+\\mathrm{C} \\\\\\\\\n& \\Rightarrow \\mathrm{I}=\\log \\left|x e^x\\right|-\\log \\left|1+x e^x\\right|+\\frac{1}{1+x e^x}+c\n\\end{aligned}\n$$\n
$$\\text { The integral } \\int \\frac{\\left(x^8-x^2\\right) \\mathrm{d} x}{\\left(x^{12}+3 x^6+1\\right) \\tan ^{-1}\\left(x^3+\\frac{1}{x^3}\\right)} \\text { is equal to : }$$
For $$x \\in\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)$$, if $$y(x)=\\int \\frac{\\operatorname{cosec} x+\\sin x}{\\operatorname{cosec} x \\sec x+\\tan x \\sin ^2 x} d x$$, and $$\\lim _\\limits{x \\rightarrow\\left(\\frac{\\pi}{2}\\right)^{-}} y(x)=0$$ then $$y\\left(\\frac{\\pi}{4}\\right)$$ is equal to
If $$\\int \\frac{\\sin ^{\\frac{3}{2}} x+\\cos ^{\\frac{3}{2}} x}{\\sqrt{\\sin ^3 x \\cos ^3 x \\sin (x-\\theta)}} d x=A \\sqrt{\\cos \\theta \\tan x-\\sin \\theta}+B \\sqrt{\\cos \\theta-\\sin \\theta \\cot x}+C$$, where $$C$$ is the integration constant, then $$A B$$ is equal to
$$\\begin{aligned}\n& \\text {} \\int \\frac{\\sin ^{\\frac{3}{2}} x+\\cos ^{\\frac{3}{2}} x}{\\sqrt{\\sin ^3 x \\cos ^3 x \\sin (x-\\theta)}} d x \\\\\n& I=\\int \\frac{\\sin ^{\\frac{3}{2}} x+\\cos ^{\\frac{3}{2}} x}{\\sqrt{\\sin ^3 x \\cos ^3 x(\\sin x \\cos \\theta-\\cos x \\sin \\theta)}} d x \\\\\n& =\\int \\frac{\\sin ^{\\frac{3}{2}} x}{\\sin ^{\\frac{3}{2}} x \\cos ^2 x \\sqrt{\\tan x \\cos \\theta-\\sin \\theta}} d x+\\int \\frac{\\cos ^{\\frac{3}{2}} x}{\\sin ^2 x \\cos ^{\\frac{3}{2}} x \\sqrt{\\cos \\theta-\\cot x \\sin \\theta}} d x= \\\\\n& \\int \\frac{\\sec ^2 x}{\\sqrt{\\tan x \\cos \\theta-\\sin \\theta}} d x+\\int \\frac{\\operatorname{cosec}^2 x}{\\sqrt{\\cos \\theta-\\cot x \\sin \\theta}} d x \\\\\n&\n\\end{aligned}$$
Let $$\\int \\frac{2-\\tan x}{3+\\tan x} \\mathrm{~d} x=\\frac{1}{2}\\left(\\alpha x+\\log _e|\\beta \\sin x+\\gamma \\cos x|\\right)+C$$, where $$C$$ is the constant of integration. Then $$\\alpha+\\frac{\\gamma}{\\beta}$$ is equal to :
If $$\\int \\frac{1}{\\sqrt[5]{(x-1)^4(x+3)^6}} \\mathrm{~d} x=\\mathrm{A}\\left(\\frac{\\alpha x-1}{\\beta x+3}\\right)^B+\\mathrm{C}$$, where $$\\mathrm{C}$$ is the constant of integration, then the value of $$\\alpha+\\beta+20 \\mathrm{AB}$$ is _________.
$$\\begin{aligned}\n& I=\\int \\frac{1}{\\sqrt[5]{(x-1)^4(x+3)^6}} d x \\\\\n& \\frac{x+3}{x-1}=t \\Rightarrow d x=\\frac{-4}{(t-1)^2} d t \\Rightarrow x=\\left(\\frac{3+t}{t-1}\\right) \\\\\n& \\Rightarrow(x-1)^4(x+3)^6=(x-1)^5(x+3)^5\\left(\\frac{x+3}{x-1}\\right) \\\\\n& I=\\int \\frac{\\frac{-4}{(t-1)^2} d t}{t^{1 / 5}\\left(\\frac{3+t}{t-1}-1\\right)\\left(\\frac{3+t}{t-3}+3\\right)} \\\\\n& I=\\int \\frac{-4 d t}{t^{1 / 5}(16 t)}=\\frac{5}{4}\\left(\\frac{x-1}{x+3}\\right)^{1 / 5}+c\n\\end{aligned}$$
$$\\begin{aligned}\n& I(x)=\\int \\frac{6}{\\sin ^2 x(1-\\cot x)^2} d x \\\\\n& I(x)=\\int \\frac{6}{(\\sin x-\\cos x)^2} d x \\\\\n& =\\int \\frac{6 \\sec ^2 x}{(\\tan x-1)^2} d x\n\\end{aligned}$$
\n
$$\\begin{aligned}\n& \\text { Let } \\tan x=t \\Rightarrow \\sec ^2 x d x=d t \\\\\n& =\\int \\frac{6 d t}{(t-1)^2} \\\\\n& =-\\frac{6}{(t-1)}+c \\\\\n& =\\frac{-6}{(\\tan x-1)}+c \\\\\n& I(x)=\\frac{6}{1-\\tan x}+c \\\\\n& I(0)=3 \\\\\n& \\frac{6}{1-\\tan 0}+c=3 \\\\\n& c=-3\n\\end{aligned}$$
By comparing with the given equation in the question we get,\n
A = - 2 and B = - 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5538,
"subject": "General Science",
"question": "$$\\int {{e^{\\sec x}}}$$ $$(\\sec x\\tan xf(x) + \\sec x\\tan x + se{x^2}x)dx$$ \n= esecxf(x) + C then a possible choice of f(x) is :- ",
"options": [
{
"text": "x sec x + tan x + 1/2"
},
{
"text": "sec x + xtan x - 1/2"
},
{
"text": "sec x - tan x - 1/2"
},
{
"text": "sec x + tan x + 1/2"
}
],
"answer": "sec x + tan x + 1/2",
"solution": "**Answer:** sec x + tan x + 1/2\n\nGiven\n
$$\n\\text { The integral } \\int \\frac{\\left(1-\\frac{1}{\\sqrt{3}}\\right)(\\cos x-\\sin x)}{\\left(1+\\frac{2}{\\sqrt{3}} \\sin 2 x\\right)} d x \\text { is equal to }\n$$
For $$\\alpha, \\beta, \\gamma, \\delta \\in \\mathbb{N}$$, if $$\\int\\left(\\left(\\frac{x}{e}\\right)^{2 x}+\\left(\\frac{e}{x}\\right)^{2 x}\\right) \\log _{e} x d x=\\frac{1}{\\alpha}\\left(\\frac{x}{e}\\right)^{\\beta x}-\\frac{1}{\\gamma}\\left(\\frac{e}{x}\\right)^{\\delta x}+C$$\n\n, where $$e=\\sum_\\limits{n=0}^{\\infty} \\frac{1}{n !}$$ and $$\\mathrm{C}$$ is constant of integration, then $$\\alpha+2 \\beta+3 \\gamma-4 \\delta$$ is equal to :
$$\\int\\left(\\left(\\frac{x}{e}\\right)^{2 x}+\\left(\\frac{e}{x}\\right)^{2 x}\\right) \\log _{e} x d x=\\frac{1}{\\alpha}\\left(\\frac{x}{e}\\right)^{\\beta x}-\\frac{1}{\\gamma}\\left(\\frac{e}{x}\\right)^{\\delta x}+C$$\n
$$\n\\begin{aligned}\nI & =\\int\\left[\\left(\\frac{x}{e}\\right)^{2 x}+\\left(\\frac{e}{x}\\right)^{2 x}\\right] \\log _e x d x \\\\\\\\\n& =\\int\\left[e^{2\\left(x \\log _e x-x\\right)}+e^{-2\\left(x \\log _e x-x\\right)}\\right] \\log _e x d x ~~~~\\text{ [From Eqs. (i) and (ii)] }\n\\end{aligned}\n$$\n
The integral $$\n\\int\\left[\\left(\\frac{x}{2}\\right)^x+\\left(\\frac{2}{x}\\right)^x\\right] \\ln \\left(\\frac{e x}{2}\\right) d x\n$$ is equal to :
If $$\\int \\frac{1}{\\mathrm{a}^2 \\sin ^2 x+\\mathrm{b}^2 \\cos ^2 x} \\mathrm{~d} x=\\frac{1}{12} \\tan ^{-1}(3 \\tan x)+$$ constant, then the maximum value of $$\\mathrm{a} \\sin x+\\mathrm{b} \\cos x$$, is :
So, a = $${{{1 + \\sqrt {17} } \\over 2}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5550,
"subject": "General Science",
"question": "The number of solutions of the equation
$${\\sin ^{ - 1}}\\left[ {{x^2} + {1 \\over 3}} \\right] + {\\cos ^{ - 1}}\\left[ {{x^2} - {2 \\over 3}} \\right] = {x^2}$$, for x$$\\in$$[$$-$$1, 1], and [x] denotes the greatest integer less than or equal to x, is :",
"options": [
{
"text": "0"
},
{
"text": "Infinite"
},
{
"text": "2"
},
{
"text": "4"
}
],
"answer": "0",
"solution": "**Answer:** 0\n\nThere are three cases possible for $$x \\in [ - 1,1]$$
Case I : $$x \\in \\left[ { - 1, - \\sqrt {{2 \\over 3}} } \\right)$$
The domain of the function $$f(x)=\\sin ^{-1}\\left[2 x^{2}-3\\right]+\\log _{2}\\left(\\log _{\\frac{1}{2}}\\left(x^{2}-5 x+5\\right)\\right)$$, where [t] is the greatest integer function, is :
Thus, the domain of the function is $(\\alpha, \\beta] = \\left(-\\frac{3}{4}, -\\frac{1}{2}\\right]$. Now, we need to find the value of $36|\\alpha + \\beta|$:
If the domain of the function $$f(x)=\\sec ^{-1}\\left(\\frac{2 x}{5 x+3}\\right)$$ is $$[\\alpha, \\beta) \\mathrm{U}(\\gamma, \\delta]$$, then $$|3 \\alpha+10(\\beta+\\gamma)+21 \\delta|$$ is equal to _________.
$$\n\\begin{array}{rlrl} \n& x \\leq-\\frac{3}{7} \\text { or } x > -\\frac{3}{5} \\\\\\\\\n& \\Rightarrow -\\frac{3}{5} < x \\leq-\\frac{3}{7}\n\\end{array}\n$$\n
Case II : $7 x+3 \\geq 0$ and $5 x+3<0$\n
$$\nx \\geq-\\frac{3}{7} \\text { and } x<-\\frac{3}{5}\n$$\n
Which is not possible\n
Case III : $x+1 \\geq 0$ and $5 x+3<0$\n
$$\n\\begin{aligned}\n& x \\geq-1 \\text { and } x<-\\frac{3}{5} \\\\\\\\\n& \\Rightarrow -1 \\leq x<-\\frac{3}{5}\n\\end{aligned}\n$$\n
Case IV : $x+1 \\leq 0$ and $5 x+3 \\geq 0$\n
$$\nx \\leq-1 \\text { and } x \\geq-\\frac{3}{5}\n$$\n
Which is not possible\n
$\\therefore$ Domain is $\\left[-1,-\\frac{3}{5}\\right) \\cup\\left(-\\frac{3}{5},-\\frac{3}{7}\\right]$\n
If the domain of the function $$\\sin ^{-1}\\left(\\frac{3 x-22}{2 x-19}\\right)+\\log _{\\mathrm{e}}\\left(\\frac{3 x^2-8 x+5}{x^2-3 x-10}\\right)$$ is $$(\\alpha, \\beta]$$, then $$3 \\alpha+10 \\beta$$ is equal to:
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5562,
"subject": "General Science",
"question": "Considering only the principal values of inverse functions, the set\n A = { x $$ \\ge $$ 0: tan$$-$$1(2x) + tan$$-$$1(3x) = $${\\pi \\over 4}$$}",
"options": [
{
"text": "contains two elements "
},
{
"text": "contains more than two elements "
},
{
"text": "is an empty set "
},
{
"text": "is a singleton "
}
],
"answer": "is a singleton ",
"solution": "**Answer:** is a singleton \n\ntan$$-$$1(2x) + tan$$-$$1(3x) = $$\\pi $$/4\n
$$ \\because $$ x > 0",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 5563,
"subject": "General Science",
"question": "Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy
For $$k \\in \\mathbb{R}$$, let the solutions of the equation $$\\cos \\left(\\sin ^{-1}\\left(x \\cot \\left(\\tan ^{-1}\\left(\\cos \\left(\\sin ^{-1} x\\right)\\right)\\right)\\right)\\right)=k, 0<|x|<\\frac{1}{\\sqrt{2}}$$ be $$\\alpha$$ and $$\\beta$$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $$x^{2}-b x-5=0$$ are $$\\frac{1}{\\alpha^{2}}+\\frac{1}{\\beta^{2}}$$ and $$\\frac{\\alpha}{\\beta}$$, then $$\\frac{b}{k^{2}}$$ is equal to ____________.
Considering only the principal values of the inverse trigonometric functions, the domain of the function $$f(x)=\\cos ^{-1}\\left(\\frac{x^{2}-4 x+2}{x^{2}+3}\\right)$$ is :
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $$\\cos ^{-1}(x)-2 \\sin ^{-1}(x)=\\cos ^{-1}(2 x)$$ is equal to :
Considering only the principal values of inverse trigonometric functions, the number of positive real values of $$x$$ satisfying $$\\tan ^{-1}(x)+\\tan ^{-1}(2 x)=\\frac{\\pi}{4}$$ is :
Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $$2 \\sin ^{-1} x+3 \\cos ^{-1} x=\\frac{2 \\pi}{5}$$, is __________.
Given that the inverse trigonometric function assumes principal values only. Let $$x, y$$ be any two real numbers in $$[-1,1]$$ such that $$\\cos ^{-1} x-\\sin ^{-1} y=\\alpha, \\frac{-\\pi}{2} \\leq \\alpha \\leq \\pi$$.\nThen, the minimum value of $$x^2+y^2+2 x y \\sin \\alpha$$ is
$$ = {\\log _e}2$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5595,
"subject": "General Science",
"question": "The sum of possible values of x for
Let $$x * y = {x^2} + {y^3}$$ and $$(x * 1) * 1 = x * (1 * 1)$$.
\n
Then a value of $$2{\\sin ^{ - 1}}\\left( {{{{x^4} + {x^2} - 2} \\over {{x^4} + {x^2} + 2}}} \\right)$$ is :
",
"options": [
{
"text": "$${\\pi \\over 4}$$"
},
{
"text": "$${\\pi \\over 3}$$"
},
{
"text": "$${\\pi \\over 2}$$"
},
{
"text": "$${\\pi \\over 6}$$"
}
],
"answer": "$${\\pi \\over 3}$$",
"solution": "**Answer:** $${\\pi \\over 3}$$\n\nThe star \"*\" in this context represents a binary operation, similar to addition (+), subtraction (-), multiplication (×), and division (÷). It is a custom operation defined by the problem statement, and the specific rules of the operation are provided in the problem. \n
\nIn this case, the operation \"*\" is defined by the equation $x * y = x^2 + y^3$, which means if you have two numbers $x$ and $y$, then the result of applying the \"*\" operation to them is $x^2 + y^3$. \n
\nThe problem also specifies an additional rule for this operation: $(x * 1) * 1 = x * (1 * 1)$, which needs to be taken into account when solving the problem. This is a type of \"associativity\" condition.\n
Let m and M respectively be the minimum and the maximum values of $$f(x) = {\\sin ^{ - 1}}2x + \\sin 2x + {\\cos ^{ - 1}}2x + \\cos 2x,\\,x \\in \\left[ {0,{\\pi \\over 8}} \\right]$$. Then m + M is equal to :
If $$\\mathrm{n(S)}$$ denotes the number of elements in $$\\mathrm{S}$$ then :
",
"options": [
{
"text": "$$\\mathrm{n}(\\mathrm{S})=0$$"
},
{
"text": "$$\\mathrm{n}(\\mathrm{S})=1$$ and only one element in $$\\mathrm{S}$$ is less than $$\\frac{1}{2}$$."
},
{
"text": "$$\\mathrm{n}(\\mathrm{S})=1$$ and the elements in $$\\mathrm{S}$$ is more than $$\\frac{1}{2}$$."
},
{
"text": "$$\\mathrm{n}(\\mathrm{S})=1$$ and the element in $$\\mathrm{S}$$ is less than $$\\frac{1}{2}$$."
}
],
"answer": "$$\\mathrm{n}(\\mathrm{S})=1$$ and the element in $$\\mathrm{S}$$ is less than $$\\frac{1}{2}$$.",
"solution": "**Answer:** $$\\mathrm{n}(\\mathrm{S})=1$$ and the element in $$\\mathrm{S}$$ is less than $$\\frac{1}{2}$$.\n\n$$ {\\,2{{\\tan }^{ - 1}}\\left( {{{1 - x} \\over {1 + x}}} \\right) = {{\\cos }^{ - 1}}\\left( {{{1 - {x^2}} \\over {1 + {x^2}}}} \\right)}$$\n
Let $$S$$ be the set of all solutions of the equation $$\\cos ^{-1}(2 x)-2 \\cos ^{-1}\\left(\\sqrt{1-x^{2}}\\right)=\\pi, x \\in\\left[-\\frac{1}{2}, \\frac{1}{2}\\right]$$. Then $$\\sum_\\limits{x \\in S} 2 \\sin ^{-1}\\left(x^{2}-1\\right)$$ is equal to :
If the sum of all the solutions of $${\\tan ^{ - 1}}\\left( {{{2x} \\over {1 - {x^2}}}} \\right) + {\\cot ^{ - 1}}\\left( {{{1 - {x^2}} \\over {2x}}} \\right) = {\\pi \\over 3}, - 1 < x < 1,x \\ne 0$$, is $$\\alpha - {4 \\over {\\sqrt 3 }}$$, then $$\\alpha$$ is equal to _____________.
We're given the equation $\\sin^{-1}x = 2\\tan^{-1}x$ for $x$ in the interval $(-1, 1]$. We want to find the number of solutions.
\n
Step 1: Apply the sine and tangent functions to both sides :
\n
We can rewrite the equation by applying the sine function to both sides :
\n
$$\\sin(\\sin^{-1}x) = \\sin(2\\tan^{-1}x).$$
\n
This simplifies to:
\n
$$x = \\sin(2\\tan^{-1}x).$$
\n
Step 2: Use the double-angle identity for sine :
\n
Recall that $\\sin(2y) = 2\\sin(y)\\cos(y)$. Applying this identity to the right-hand side gives :
\n
$$x = 2\\sin(\\tan^{-1}x)\\cos(\\tan^{-1}x).$$
\n
Step 3: Use the identities for sine and cosine of an inverse tangent :
\n
Recall that $\\sin(\\tan^{-1}x) = \\frac{x}{\\sqrt{1 + x^2}}$ and $\\cos(\\tan^{-1}x) = \\frac{1}{\\sqrt{1 + x^2}}$. Substituting these into the equation gives :
If $$S=\\left\\{x \\in \\mathbb{R}: \\sin ^{-1}\\left(\\frac{x+1}{\\sqrt{x^{2}+2 x+2}}\\right)-\\sin ^{-1}\\left(\\frac{x}{\\sqrt{x^{2}+1}}\\right)=\\frac{\\pi}{4}\\right\\}$$, then $$\\sum_\\limits{x \\in s}\\left(\\sin \\left(\\left(x^{2}+x+5\\right) \\frac{\\pi}{2}\\right)-\\cos \\left(\\left(x^{2}+x+5\\right) \\pi\\right)\\right)$$ is equal to ____________.
Let $$x=\\frac{m}{n}$$ ($$m, n$$ are co-prime natural numbers) be a solution of the equation $$\\cos \\left(2 \\sin ^{-1} x\\right)=\\frac{1}{9}$$ and let $$\\alpha, \\beta(\\alpha >\\beta)$$ be the roots of the equation $$m x^2-n x-m+ n=0$$. Then the point $$(\\alpha, \\beta)$$ lies on the line
(Where tanA $$=$$ 20, tanB $$=$$ 1) $${{1\\left( {{1 \\over {20}}} \\right) + 1} \\over {1 - {1 \\over {20}}}} = {{21} \\over {19}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5620,
"subject": "General Science",
"question": "If S is the sum of the first 10 terms of the series
$$f\\left( x \\right) = x$$ if $$x$$ is rational\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$ = - x$$ if $$x$$ is irrational. Then",
"options": [
{
"text": "$$f(x)$$ is continuous at every x, except $$x = 0$$"
},
{
"text": "$$f(x)$$ is discontinuous at every $$x,$$ except $$x = 0$$"
},
{
"text": "$$f(x)$$ is continuous everywhere"
},
{
"text": "$$f(x)$$ is discontinuous everywhere"
}
],
"answer": "$$f(x)$$ is discontinuous at every $$x,$$ except $$x = 0$$",
"solution": "**Answer:** $$f(x)$$ is discontinuous at every $$x,$$ except $$x = 0$$\n\nLet a is a rational number other than $$0,$$ in \n
$$\\left[ { - 5,5} \\right],$$ then\n
$$f\\left( a \\right) = a$$ and $$\\mathop {\\lim }\\limits_{x \\to a} \\,\\,f\\left( x \\right) = - a$$\n
[ As in the immediate neighbourhood of a rational -\n
number, we find irrational numbers ]\n
$$\\therefore$$ $$f(x)$$ is not continuous at any rational number\n
If a is irrational number, then\n
$$\\,f\\left( a \\right) = - a$$ and $$\\mathop {\\lim }\\limits_{x \\to a} \\,f\\left( x \\right) = a$$ \n
$$\\therefore$$ $$f(x)$$ is not continuous at any irrational number clearly \n
and $$f\\left( 0 \\right) = q \\Rightarrow p = - {3 \\over 2},q = {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5627,
"subject": "General Science",
"question": "If $$f:R \\to R$$ is a function defined by\n
$$f\\left( x \\right) = \\left[ x \\right]\\cos \\left( {{{2x - 1} \\over 2}} \\right)\\pi $$,\n
where [x] denotes the greatest integer function, then $$f$$ is",
"options": [
{
"text": "continuous for every real $$x$$"
},
{
"text": "discontinuous only at $$x=0$$"
},
{
"text": "discontinuous only at non-zero integral values of $$x$$"
},
{
"text": "continuous only at $$x=0$$"
}
],
"answer": "continuous for every real $$x$$",
"solution": "**Answer:** continuous for every real $$x$$\n\nLet $$f\\left( x \\right) = \\left[ x \\right]\\cos \\left( {{{2x - 1} \\over 2}} \\right)$$\n
Now, value of the function at $$x = n$$ is $$f(n)=0$$\n
Since, $$L.H.L.$$ $$ = R.H.L. = f\\left( n \\right)$$\n
$$\\therefore$$ $$f\\left( x \\right) = \\left[ x \\right]\\cos \\left( {{{2x - 1} \\over 2}} \\right)$$ is continuous for every real $$x.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5628,
"subject": "General Science",
"question": "Let a, b $$ \\in $$ R, (a $$ \\ne $$ 0). If the function f defined as\n
$$ \\therefore $$ (a, b) = $$\\left( {\\sqrt 2 ,1 + \\sqrt 3 } \\right)$$ or $$\\left( {\\sqrt 2 ,1 - \\sqrt 3 } \\right)$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5629,
"subject": "General Science",
"question": "The value of k for which the function \n
Thevaue of k for which f s continuous at x = 2 is :",
"options": [
{
"text": "1"
},
{
"text": "e"
},
{
"text": "e-1"
},
{
"text": "e-2"
}
],
"answer": "e-1",
"solution": "**Answer:** e-1\n\nSince f(x) is continuous at x = 2.\n
$$ \\Rightarrow $$ k = e$$-$$1",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5631,
"subject": "General Science",
"question": "If the function f defined as \n
$$f\\left( x \\right) = {1 \\over x} - {{k - 1} \\over {{e^{2x}} - 1}},x \\ne 0,$$ is continuous at\n
x = 0, then the ordered pair (k, f(0)) is equal to : ",
"options": [
{
"text": "(3, 2)"
},
{
"text": "(3, 1)"
},
{
"text": "(2, 1)"
},
{
"text": "$$\\left( {{1 \\over 3},\\,2} \\right)$$"
}
],
"answer": "(3, 1)",
"solution": "**Answer:** (3, 1)\n\nIf the function is continuous at x = 0, then\n $$\\mathop {\\lim }\\limits_{x \\to 0} $$ f(x) will exist and f(0) = $$\\mathop {\\lim }\\limits_{x \\to 0} $$ f(x)\n
For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero\n
Hence, f(0) = 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5632,
"subject": "General Science",
"question": "Let f : R $$ \\to $$ R be a function defined as \n $$f(x) = \\left\\{ {\\matrix{\n 5 & ; & {x \\le 1} \\cr \n {a + bx} & ; & {1 < x < 3} \\cr \n {b + 5x} & ; & {3 \\le x < 5} \\cr \n {30} & ; & {x \\ge 5} \\cr \n\n } } \\right.$$ \n
Then, f is",
"options": [
{
"text": "continuous if a = 0 and b = 5"
},
{
"text": "continuous if a = –5 and b = 10"
},
{
"text": "continuous if a = 5 and b = 5 "
},
{
"text": "not continuous for any values of a and b"
}
],
"answer": "not continuous for any values of a and b",
"solution": "**Answer:** not continuous for any values of a and b\n\nChecking \n
if f(x) is continuous at x = 1 : \n
f(1$$-$$) = 5\n
f(1) = 5\n
f(1+) = a + b\n
if f(x) is continuous at x = 1, \n
then\n
f(1$$-$$) = f(1) = f(1+)\n
$$ \\Rightarrow $$ 5 = 5 = a + b\n
$$ \\therefore $$ a + b = 5 . . . . . . . . (1)\n
checking if f(x) is continuous at x = 3 :\n
f(3$$-$$) = a + 3b\n
f(3) = b + 15\n
f(3+) = b + 15\n
if f(x) = is continuous at x = 3\n
then, \n
f(3$$-$$) = f(3) = f(3+)\n
$$ \\Rightarrow $$ a + 3b = b + 15 = b + 15\n
$$ \\Rightarrow $$ a + 2b = 15 . . . . . (2)\n
checking if f(x) is continuous at x = 5 :\n
f(5$$-$$) = b + 25\n
f(5) = 30\n
f(5+) = 30\n
if f(x) is continuous at x = 5 then, \n
f(5$$-$$) = f(5) = f(5+)\n
$$ \\Rightarrow $$ b + 25 = 30 = 30\n
$$ \\Rightarrow $$ b = 5\n
By putting this value in equation (2), we get,\n
a + 2(5) = 15\n
$$ \\Rightarrow $$ a = 5\n
when a = 5 and b = 5 then equation (1)\n
a + b = 5 does not satisfy.\n
$$ \\therefore $$ f is not continuous for any value of a and b.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5633,
"subject": "General Science",
"question": "Let ƒ : [–1,3] $$ \\to $$ R be defined as
\n$$f(x) = \\left\\{ {\\matrix{\n {\\left| x \\right| + \\left[ x \\right]} & , & { - 1 \\le x < 1} \\cr \n {x + \\left| x \\right|} & , & {1 \\le x < 2} \\cr \n {x + \\left[ x \\right]} & , & {2 \\le x \\le 3} \\cr \n\n } } \\right.$$
\nwhere [t] denotes the greatest integer less than\nor equal to t. Then, ƒ is discontinuous at:",
"options": [
{
"text": "only three points"
},
{
"text": "four or more points"
},
{
"text": "only two points"
},
{
"text": "only one point"
}
],
"answer": "only three points",
"solution": "**Answer:** only three points\n\n$$f(x) = \\left\\{ {\\matrix{\n {\\left| x \\right| + \\left[ x \\right]} & , & { - 1 \\le x < 1} \\cr \n {x + \\left| x \\right|} & , & {1 \\le x < 2} \\cr \n {x + \\left[ x \\right]} & , & {2 \\le x \\le 3} \\cr \n\n } } \\right.$$\n
So f(0) = 5 = k",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5639,
"subject": "General Science",
"question": "Let $$f(x) = x.\\left[ {{x \\over 2}} \\right]$$, for -10< x < 10, where [t] denotes the greatest integer function. Then the number of points of discontinuity of f is equal to _____.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\n$$x \\in ( - 10,10)$$
$$ \\therefore $$ $$a$$ + 2b = -2 + 2 = 0",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5643,
"subject": "General Science",
"question": "If f : R $$ \\to $$ R is a function defined by f(x)= [x - 1] $$\\cos \\left( {{{2x - 1} \\over 2}} \\right)\\pi $$, where [.] denotes the greatest\ninteger function, then f is :",
"options": [
{
"text": "continuous for every real x"
},
{
"text": "discontinuous at all integral values of x except at x = 1"
},
{
"text": "discontinuous only at x = 1"
},
{
"text": "continuous only at x = 1"
}
],
"answer": "continuous for every real x",
"solution": "**Answer:** continuous for every real x\n\nGiven, $$f(x) = [x - 1]\\cos \\left( {{{2x - 1} \\over 2}} \\right)\\pi $$ where [ . ] is greatest integer function and f : R $$\\to$$ R
$$\\because$$ It is a greatest integer function then we need to check its continuity at x $$\\in$$ I except these it is continuous.
$$\\therefore$$ It is continuous at every integers.
Therefore, the given function is continuous for all real x.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5644,
"subject": "General Science",
"question": "Let f : R $$ \\to $$ R be defined as
As LHL $$ \\ne $$ RHL so f(x) is not continuous at x = 0",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5646,
"subject": "General Science",
"question": "Let f : R $$ \\to $$ R and g : R $$ \\to $$ R be defined as
where a, b are non-negative real numbers. If (gof) (x) is continuous for all x $$\\in$$ R, then a + b is equal to ____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$g[f(x)] = \\left[ {\\matrix{\n {f(x) + 1} & {f(x) < 0} \\cr \n {{{(f(x) - 1)}^2} + b} & {f(x) \\ge 0} \\cr \n\n } } \\right.$$
$$g[f(x)] = \\left[ {\\matrix{\n {x + a + 1} & {x + a < 0\\& x < 0} \\cr \n {|x - 1| + 1} & {|x - 1| < 0\\& x \\ge 0} \\cr \n {{{(x + a - 1)}^2} + b} & {x + a \\ge 0\\& x < 0} \\cr \n {{{(|x - 1| - 1)}^2} + b} & {|x - 1| \\ge 0\\& x \\ge 0} \\cr \n\n } } \\right.$$
\n$$\n\\begin{aligned}\n&\\frac{1}{2} =b=\\frac{a+1}{2}+1 \\Rightarrow b=\\frac{1}{2}, a=-2 \\\\\\\\\n&\\therefore \\quad a+b =\\frac{-3}{2}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5649,
"subject": "General Science",
"question": "Let a function f : R $$\\to$$ R be defined as $$f(x) = \\left\\{ {\\matrix{\n {\\sin x - {e^x}} & {if} & {x \\le 0} \\cr \n {a + [ - x]} & {if} & {0 < x < 1} \\cr \n {2x - b} & {if} & {x \\ge 1} \\cr \n\n } } \\right.$$
where [ x ] is the greatest integer less than or equal to x. If f is continuous on R, then (a + b) is equal to: ",
"options": [
{
"text": "4"
},
{
"text": "3"
},
{
"text": "2"
},
{
"text": "5"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\nContinuous x = 0
$$\\Rightarrow$$ b = 2 $$-$$ a + 1 $$\\Rightarrow$$ b = 3
$$\\therefore$$ a + b = 3",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5650,
"subject": "General Science",
"question": "Let f : R $$\\to$$ R be defined as $$f(x) = \\left\\{ {\\matrix{\n {{{{x^3}} \\over {{{(1 - \\cos 2x)}^2}}}{{\\log }_e}\\left( {{{1 + 2x{e^{ - 2x}}} \\over {{{(1 - x{e^{ - x}})}^2}}}} \\right),} & {x \\ne 0} \\cr \n {\\alpha ,} & {x = 0} \\cr \n\n } } \\right.$$
If f is continuous at x = 0, then $$\\alpha$$ is equal to :",
"options": [
{
"text": "1"
},
{
"text": "3"
},
{
"text": "0"
},
{
"text": "2"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nFor continuity
\n \n$$\n=\\left\\{\\begin{array}{cc}\n\\left|1+x-e^{x}\\right|, & x<0 \\\\\n1, & x=0 \\\\\n{\\left[(x-1)^{2}-1\\right],} & 0 < x < 2 \\\\\n\\left|2-(x-1)^{2}\\right|, & x \\geq 2\n\\end{array}\\right.\n$$\n
\nSo, $x=0,2$ are the two points where fog is discontinuous.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5657,
"subject": "General Science",
"question": "
where a, b, c $$\\in$$ R and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true?
",
"options": [
{
"text": "There exists a, b, c $$\\in$$ R such that f is continuous on R."
},
{
"text": "If f is discontinuous at exactly one point, then a + b + c = 1"
},
{
"text": "If f is discontinuous at exactly one point, then a + b + c $$\\ne$$ 1"
},
{
"text": "f is discontinuous at at least two points, for any values of a, b and c"
}
],
"answer": "If f is discontinuous at exactly one point, then a + b + c $$\\ne$$ 1",
"solution": "**Answer:** If f is discontinuous at exactly one point, then a + b + c $$\\ne$$ 1\n\n
Let $$f(x) = \\left[ {2{x^2} + 1} \\right]$$ and $$g(x) = \\left\\{ {\\matrix{\n {2x - 3,} & {x < 0} \\cr \n {2x + 3,} & {x \\ge 0} \\cr \n\n } } \\right.$$, where [t] is the greatest integer $$\\le$$ t. Then, in the open interval ($$-$$1, 1), the number of points where fog is discontinuous is equal to ______________.
If for $$\\mathrm{p} \\neq \\mathrm{q} \\neq 0$$, the function $$f(x)=\\frac{\\sqrt[7]{\\mathrm{p}(729+x)}-3}{\\sqrt[3]{729+\\mathrm{q} x}-9}$$ is continuous at $$x=0$$, then :
$$\n\\text { Let the function } f(x)=\\left\\{\\begin{array}{cl}\n\\frac{\\log _{e}(1+5 x)-\\log _{e}(1+\\alpha x)}{x} & ;\\text { if } x \\neq 0 \\\\\n10 & ; \\text { if } x=0\n\\end{array} \\text { be continuous at } x=0 .\\right.\n$$
Let $$[x]$$ be the greatest integer $$\\leq x$$. Then the number of points in the interval $$(-2,1)$$, where the function $$f(x)=|[x]|+\\sqrt{x-[x]}$$ is discontinuous, is ___________.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nThe function $f(x) = |[x]| + \\sqrt{x-[x]}$ is composed of two parts : the greatest integer function $[x]$, and the fractional part function $x-[x]$. \n\n
1. The greatest integer function $[x]$ is discontinuous at every integer, since it jumps from one integer to the next without taking any values in between. The absolute value does not affect where this function is continuous or discontinuous. So within the interval $(-2,1)$, $[x]$ is discontinuous at $-2, -1, 0$ and $1$. However, because the interval is open $(-2,1)$, the endpoints $-2$ and $1$ are not included.\n\n
2. The function $\\sqrt{x-[x]}$ is the square root of the fractional part of $x$. The fractional part function $x-[x]$ is continuous everywhere, as it always takes a value between $0$ and $1$ (inclusive of $0$, exclusive of $1$). However, the square root function $\\sqrt{x}$ is only defined for $x \\geq 0$, and so $\\sqrt{x-[x]}$ is discontinuous wherever $x-[x] < 0$. This happens exactly at the points where $x$ is a negative integer, as the fractional part of a negative integer is $1$ (considering that the \"fractional part\" is defined as the part of the number to the right of the decimal point, which for negative numbers works a bit differently). Within the interval $(-2,1)$, this is the case for $-2$ and $-1$. However, since the interval is open $(-2,1)$, the endpoint $-2$ is not included.\n\n
Now, let's analyze the discontinuities within the given interval $(-2,1)$:\n\n
At $x=-1$: $f(-1^{+})=1+0=1$ and $f(-1^{-})=2+1=3$. Since these two values are different, $f(x)$ is discontinuous at $x=-1$.\n\n
At $x=0$: $f(0^{+})=0+0=0$ and $f(0^{-})=1+1=2$. Again, these two values are different, so $f(x)$ is discontinuous at $x=0$.\n\n
At $x=1$: $f(1^{+})=1+0=1$ and $f(1^{-})=0+1=1$. These two values are the same, so $f(x)$ is continuous at $x=1$. However, this point is not within the open interval $(-2,1)$.\n\n
So within the interval $(-2,1)$, the function $f(x) = |[x]| + \\sqrt{x-[x]}$ is discontinuous at the points $-1$ and $0$ (2 points in total).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5668,
"subject": "General Science",
"question": "
Let $$f(x)=\\left[x^{2}-x\\right]+|-x+[x]|$$, where $$x \\in \\mathbb{R}$$ and $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then, $$f$$ is :
",
"options": [
{
"text": "continuous at $$x=0$$, but not continuous at $$x=1$$"
},
{
"text": "continuous at $$x=0$$ and $$x=1$$"
},
{
"text": "continuous at $$x=1$$, but not continuous at $$x=0$$"
},
{
"text": "not continuous at $$x=0$$ and $$x=1$$"
}
],
"answer": "continuous at $$x=1$$, but not continuous at $$x=0$$",
"solution": "**Answer:** continuous at $$x=1$$, but not continuous at $$x=0$$\n\nWe have,\n
Hence, $f(x)$ is continuous at $x=1$, but not continuous at $x=0$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5669,
"subject": "General Science",
"question": "Consider the function.\n
where $[x]$ denotes the greatest integer less than or equal to $x$. If $\\mathrm{S}$ denotes the set of all ordered pairs (a, b) such that $f(x)$ is continuous at $x=3$, then the number of elements in $\\mathrm{S}$ is : ",
"options": [
{
"text": "Infinitely many"
},
{
"text": "4"
},
{
"text": "2"
},
{
"text": "1"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\n
$$f\\left(3^{-}\\right)=\\frac{a}{b} \\frac{\\left(7 x-12-x^2\\right)}{\\left|x^2-7 x+12\\right|} \\quad$$ (for $$f(x)$$ to be cont.)
Let $$g(x)$$ be a linear function and $$f(x)=\\left\\{\\begin{array}{cl}g(x) & , x \\leq 0 \\\\ \\left(\\frac{1+x}{2+x}\\right)^{\\frac{1}{x}} & , x>0\\end{array}\\right.$$, is continuous at $$x=0$$. If $$f^{\\prime}(1)=f(-1)$$, then the value $$g(3)$$ is
Let $$f:(0, \\pi) \\rightarrow \\mathbf{R}$$ be a function given by $$f(x)=\\left\\{\\begin{array}{cc}\\left(\\frac{8}{7}\\right)^{\\frac{\\tan 8 x}{\\tan 7 x}}, & 0< x<\\frac{\\pi}{2} \\\\ \\mathrm{a}-8, & x=\\frac{\\pi}{2} \\\\ (1+\\mid \\cot x)^{\\frac{\\mathrm{b}}{\\mathrm{a}}|\\tan x|}, & \\frac{\\pi}{2} < x < \\pi\\end{array}\\right.$$
\n
where $$\\mathrm{a}, \\mathrm{b} \\in \\mathbf{Z}$$. If $$f$$ is continuous at $$x=\\frac{\\pi}{2}$$, then $$\\mathrm{a}^2+\\mathrm{b}^2$$ is equal to _________.
For $$\\mathrm{a}, \\mathrm{b}>0$$, let $$f(x)= \\begin{cases}\\frac{\\tan ((\\mathrm{a}+1) x)+\\mathrm{b} \\tan x}{x}, & x< 0 \\\\ 3, & x=0 \\\\ \\frac{\\sqrt{\\mathrm{a} x+\\mathrm{b}^2 x^2}-\\sqrt{\\mathrm{a} x}}{\\mathrm{~b} \\sqrt{\\mathrm{a}} x \\sqrt{x}}, & x> 0\\end{cases}$$\nbe a continuous function at $$x=0$$. Then $$\\frac{\\mathrm{b}}{\\mathrm{a}}$$ is equal to :
If the function $$f(x)=\\frac{\\sin 3 x+\\alpha \\sin x-\\beta \\cos 3 x}{x^3}, x \\in \\mathbf{R}$$, is continuous at $$x=0$$, then $$f(0)$$ is equal to :
Let ,$$f:[-1,2] \\rightarrow \\mathbf{R}$$ be given by $$f(x)=2 x^2+x+\\left[x^2\\right]-[x]$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. The number of points, where $$f$$ is not continuous, is :
Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Let $$f:[0, \\infty) \\rightarrow \\mathbf{R}$$ be a function defined by $$f(x)=\\left[\\frac{x}{2}+3\\right]-[\\sqrt{x}]$$. Let $$\\mathrm{S}$$ be the set of all points in the interval $$[0,8]$$ at which $$f$$ is not continuous. Then $$\\sum_\\limits{\\text {aes }} a$$ is equal to __________.
then the value of k is",
"options": [
{
"text": "0"
},
{
"text": "4"
},
{
"text": "2"
},
{
"text": "1"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\n$$\\mathop {\\lim }\\limits_{x \\to a} {{f\\left( a \\right)g'\\left( x \\right) - g\\left( a \\right)f'\\left( x \\right)} \\over {g'\\left( x \\right) - f'\\left( x \\right)}}$$ \n
then $$f(x)$$ is",
"options": [
{
"text": "discontinuous everywhere"
},
{
"text": "continuous as well as differentiable for all x"
},
{
"text": "continuous for all x but not differentiable at x = 0"
},
{
"text": "neither differentiable nor continuous at x = 0"
}
],
"answer": "continuous for all x but not differentiable at x = 0",
"solution": "**Answer:** continuous for all x but not differentiable at x = 0\n\n$$f\\left( 0 \\right) = 0;\\,\\,f\\left( x \\right) = x{e^{ - \\left( {{1 \\over {\\left| x \\right|}} + {1 \\over x}} \\right)}}$$\n
$$f(x)$$ is not differentiable at $$x=0.$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5682,
"subject": "General Science",
"question": "Suppose $$f(x)$$ is differentiable at x = 1 and\n
$$\\therefore$$ $$f'\\left( x \\right)$$ exist at everywhere.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5685,
"subject": "General Science",
"question": "Let $$f:R \\to R$$ be a function defined by\n
$$f(x) = \\min \\left\\{ {x + 1,\\left| x \\right| + 1} \\right\\}$$, then which of the following is true?",
"options": [
{
"text": "$$f(x)$$ is differentiale everywhere"
},
{
"text": "$$f(x)$$ is not differentiable at x = 0"
},
{
"text": "$$f(x) > 1$$ for all $$x \\in R$$"
},
{
"text": "$$f(x)$$ is not differentiable at x = 1"
}
],
"answer": "$$f(x)$$ is differentiale everywhere",
"solution": "**Answer:** $$f(x)$$ is differentiale everywhere\n\n$$f\\left( x \\right) = \\min \\left\\{ {x + 1,\\left| x \\right| + 1} \\right\\}$$\n
$$ \\Rightarrow f\\left( x \\right) = x + 1\\,\\forall \\,x \\in R$$\n
Then which one of the following is true?",
"options": [
{
"text": "$$f$$ is neither differentiable at x = 0 nor at x = 1"
},
{
"text": "$$f$$ is differentiable at x = 0 and at x = 1"
},
{
"text": "$$f$$ is differentiable at x = 0 but not at x = 1"
},
{
"text": "$$f$$ is differentiable at x = 1 but not at x = 0"
}
],
"answer": "$$f$$ is differentiable at x = 0 but not at x = 1",
"solution": "**Answer:** $$f$$ is differentiable at x = 0 but not at x = 1\n\nWe have $$f\\left( x \\right) = \\left\\{ {\\matrix{\n {\\left( {x - 1} \\right)\\sin \\left( {{1 \\over {x - 1}}} \\right),} & {if\\,\\,x \\ne 1} \\cr \n {0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,} & {if\\,\\,x = 1} \\cr \n\n } } \\right.$$\n
$$\\therefore$$ $$f$$ is differentiable at $$x=0$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5687,
"subject": "General Science",
"question": "Let $$f\\left( x \\right) = x\\left| x \\right|$$ and $$g\\left( x \\right) = \\sin x.$$\n Statement-1: gof is differentiable at $$x=0$$ and its derivative is continuous at that point. \n Statement-2: gof is twice differentiable at $$x=0$$. ",
"options": [
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."
},
{
"text": "Statement-1 is true, Statement-2 is false "
},
{
"text": "Statement-1 is false, Statement-2 is true "
},
{
"text": "Statement-1 is true, Statement-2 is true Statement-2 is a correct explanation for Statement-1"
}
],
"answer": "Statement-1 is true, Statement-2 is false ",
"solution": "**Answer:** Statement-1 is true, Statement-2 is false \n\nGiven that $$f\\left( x \\right) = x\\left| x \\right|\\,\\,$$ and $$\\,\\,g\\left( x \\right) = \\sin x$$\n
So that go \n
$$f\\left( x \\right) = g\\left( {f\\left( x \\right)} \\right)$$\n
$$ = g\\left( {x\\left| x \\right|} \\right) = \\sin x\\left| x \\right|$$\n
$$\\therefore$$ Statement - $$2$$ is also true but not a correct explanation for statement $$1.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5689,
"subject": "General Science",
"question": "If the function.\n
So, S $$ \\equiv $$ ($$\\lambda $$, $$\\mu $$) = {$$\\lambda $$ $$ \\in $$ R & $$\\mu $$ $$ \\in $$ [0, $$\\infty $$)}\n
Therefore set S is subset of R $$ \\times $$ [0, $$\\infty $$)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5694,
"subject": "General Science",
"question": "Let ƒ(x) = 15 – |x – 10|; x $$ \\in $$ R. Then the set\nof all values of x, at which the function,\ng(x) = ƒ(ƒ(x)) is not differentiable, is :",
"options": [
{
"text": "{10,15}"
},
{
"text": "{5,10,15,20}"
},
{
"text": "{10}"
},
{
"text": "{5,10,15}"
}
],
"answer": "{5,10,15}",
"solution": "**Answer:** {5,10,15}\n\nƒ(x) = 15 – |x – 10|\n
g(x) = ƒ(ƒ(x)) = 15 – |ƒ(x) – 10|\n
= 15 – |15 – |x – 10| – 10|\n
= 15 – |5 – |x – 10||\n
As this is a linear expression so it is non differentiable when value inside the modulus is zero.\n
So non differentiable when\n
x – 10 = 0 $$ \\Rightarrow $$ x = 10\n
and 5 – |x – 10| = 0\n
$$ \\Rightarrow $$ |x – 10| = 5\n
$$ \\Rightarrow $$ x - 10 = $$ \\pm $$ 5\n
$$ \\Rightarrow $$ x = 5, 15\n
$$ \\therefore $$ g(x) is not differentiable at x = 5, 10, 15. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5695,
"subject": "General Science",
"question": "Let f : R $$ \\to $$ R be differentiable at c $$ \\in $$ R and f(c) = 0. If g(x) = |f(x)| , then at x = c, g is :",
"options": [
{
"text": "differentiable if f '(c) = 0"
},
{
"text": "differentiable if f '(c) $$ \\ne $$ 0"
},
{
"text": "not differentiable"
},
{
"text": "not differentiable if f '(c) = 0"
}
],
"answer": "differentiable if f '(c) = 0",
"solution": "**Answer:** differentiable if f '(c) = 0\n\n$$g'(c) = \\mathop {\\lim }\\limits_{x \\to c} {{g(x) - g(c)} \\over {x - c}}$$
= 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5697,
"subject": "General Science",
"question": "Let f be a differentiable function such that f(1) = 2 and f '(x) = f(x) for all x $$ \\in $$ R R. If h(x) = f(f(x)), then h'(1) is equal to : ",
"options": [
{
"text": "4e"
},
{
"text": "2e2"
},
{
"text": "4e2"
},
{
"text": "2e"
}
],
"answer": "4e",
"solution": "**Answer:** 4e\n\n$${{f'(x)} \\over {f(x)}} = 1\\forall x \\in R$$\n
Intergrate & use f(1) = 2\n
f(x) = 2ex-1 $$ \\Rightarrow $$ f '(x) = 2ex$$-$$1\n
It is not differentiable at x = 1",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5701,
"subject": "General Science",
"question": "Let f : ($$-$$1, 1) $$ \\to $$ R be a function defined by f(x) = max $$\\left\\{ { - \\left| x \\right|, - \\sqrt {1 - {x^2}} } \\right\\}.$$ If K be the set of all points at which f is not differentiable, then K has exactly - ",
"options": [
{
"text": "one element"
},
{
"text": "three elements"
},
{
"text": "five elements"
},
{
"text": "two elements"
}
],
"answer": "three elements",
"solution": "**Answer:** three elements\n\nf : ($$-$$ 1, 1) $$ \\to $$ R\n
f(x) = max {$$-$$ $$\\left| x \\right|, - \\sqrt {1 - {x^2}} $$}\n
\n Non-derivable at 3 points in ($$-$$1, 1)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5702,
"subject": "General Science",
"question": "Let $$f\\left( x \\right) = \\left\\{ {\\matrix{\n {\\max \\left\\{ {\\left| x \\right|,{x^2}} \\right\\}} & {\\left| x \\right| \\le 2} \\cr \n {8 - 2\\left| x \\right|} & {2 < \\left| x \\right| \\le 4} \\cr \n\n } } \\right.$$\n
Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S",
"options": [
{
"text": "equals $$\\left\\{ { - 2, - 1,1,2} \\right\\}$$"
},
{
"text": "equals $$\\left\\{ { - 2, - 1,0,1,2} \\right\\}$$"
},
{
"text": "equals $$\\left\\{ { - 2,2} \\right\\}$$"
},
{
"text": "is an empty set"
}
],
"answer": "equals $$\\left\\{ { - 2, - 1,0,1,2} \\right\\}$$",
"solution": "**Answer:** equals $$\\left\\{ { - 2, - 1,0,1,2} \\right\\}$$\n\n$$f\\left( x \\right) = \\left\\{ {\\matrix{\n {8 + 2x,} & { - 4 \\le x \\le - 2} \\cr \n {{x^2},} & { - 2 \\le x \\le - 1} \\cr \n {\\left| x \\right|,} & { - 1 < x < 1} \\cr \n {{x^2},} & {1 \\le x \\le 2} \\cr \n {8 - 2x,} & {2 < x \\le 4} \\cr \n\n } } \\right.$$\n
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5703,
"subject": "General Science",
"question": "Let S be the set of points where the function, ƒ(x) = |2-|x-3||, x $$ \\in $$ R is not differentiable. Then $$\\sum\\limits_{x \\in S} {f(f(x))} $$ is equal to_____.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n\n
f(x) is non-differentiable at x = 1, 3, 5\n
$$ \\therefore $$ S is {1, 3, 5}\n
$$\\sum\\limits_{x \\in S} {f(f(x))} $$\n
= f(f(1)) + f(f(3)) + f(f (5))\n
= f(0) + f(2) + f(0)\n
= 1 + 1 + 1 = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5704,
"subject": "General Science",
"question": "Let S be the set of all functions ƒ : [0,1] $$ \\to $$ R,\nwhich are continuous on [0,1] and differentiable\non (0,1). Then for every ƒ in S, there exists a\nc $$ \\in $$ (0,1), depending on ƒ, such that",
"options": [
{
"text": "$$\\left| {f(c) - f(1)} \\right| < \\left| {f'(c)} \\right|$$"
},
{
"text": "$$\\left| {f(c) + f(1)} \\right| < \\left( {1 + c} \\right)\\left| {f'(c)} \\right|$$"
},
{
"text": "$$\\left| {f(c) - f(1)} \\right| < \\left( {1 - c} \\right)\\left| {f'(c)} \\right|$$"
},
{
"text": "None"
}
],
"answer": "None",
"solution": "**Answer:** None\n\n
If we consider the case where f(x) is a constant function, then its derivative f'(x) is equal to 0 for all x in the interval (0,1).
\n
Therefore, if we substitute this into the expressions provided in Options A, B and C, we would have :
\n
Option A : |f(c) - f(1)| < |f'(c)| would become |constant - constant| < |0|, which is 0 < 0. This is not true.
\n
Option B : |f(c) + f(1)| < (1 + c)|f'(c)| would become |constant + constant| < (1 + c)$$ \\times $$0, which is a positive number < 0. This is not true.
\n
Option C : |f(c) - f(1)| < (1 - c)|f'(c)| would become |constant - constant| < (1 - c)$$ \\times $$0, which is 0 < 0. This is not true.
\n
Hence, for the case where f(x) is a constant function, none of the options A, B and C are correct.
\n
So, the correct answer would be Option D : None.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5705,
"subject": "General Science",
"question": "Let ƒ be any function continuous on [a, b] and\ntwice differentiable on (a, b). If for all x $$ \\in $$ (a, b),\nƒ'(x) > 0 and ƒ''(x) < 0, then for any c $$ \\in $$ (a, b),\n$${{f(c) - f(a)} \\over {f(b) - f(c)}}$$ is greater than :",
"options": [
{
"text": "1"
},
{
"text": "$${{b - c} \\over {c - a}}$$"
},
{
"text": "$${{b + a} \\over {b - a}}$$"
},
{
"text": "$${{c - a} \\over {b - c}}$$"
}
],
"answer": "$${{c - a} \\over {b - c}}$$",
"solution": "**Answer:** $${{c - a} \\over {b - c}}$$\n\n\n
It is clear from graph that, slope of AC $$>$$ slope of CB\n
$$ \\Rightarrow $$ $${{f\\left( c \\right) - f\\left( a \\right)} \\over {c - a}}$$ $$>$$ $${{f\\left( b \\right) - f\\left( c \\right)} \\over {b - c}}$$\n
$$ \\Rightarrow $$ $${{f\\left( c \\right) - f\\left( a \\right)} \\over {f\\left( b \\right) - f\\left( c \\right)}}$$ $$ > {{c - a} \\over {b - c}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5706,
"subject": "General Science",
"question": "Suppose a differentiable function f(x) satisfies the identity f(x+y) = f(x) + f(y) + xy2 + x2y, for all real x and y. \n$$\\mathop {\\lim }\\limits_{x \\to 0} {{f\\left( x \\right)} \\over x} = 1$$, then f'(3) is equal to ______.\n",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\nGiven, f(x + y) = f(x) + f(y) + xy2 + x2y ...(1)
[x $$ \\ne $$ 0, f(x) $$ \\ne $$ 0 as given function $$f:\\left( {0,\\infty } \\right) \\to \\left( {0,\\infty } \\right)$$ only takes positive value as input and output]\n
The value of $$\\lambda $$ for which f ''(0) exists, is _______.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nIf g(x) = x5sin$$\\left( {{1 \\over x}} \\right)$$\n
and h(x) = x5cos$$\\left( {{1 \\over x}} \\right)$$\n
then g''(0) = 0 and h''(0) = 0\n
So, f''(0+\n) = g''(0+\n) + 10 = 10\n
and f''(0–) = h''(0–) + 2$$\\lambda $$ = f''(0+)\n
$$ \\Rightarrow $$ 2$$\\lambda $$ = 10\n
$$ \\Rightarrow $$ $$\\lambda $$ = 5",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5711,
"subject": "General Science",
"question": "Let f : R $$ \\to $$ R be a function defined by f(x) = max {x, x2}. Let S denote the set of all points in R, where f is not differentiable.\nThen :",
"options": [
{
"text": "{0, 1}"
},
{
"text": "{0}"
},
{
"text": "$$\\phi $$(an empty set)"
},
{
"text": "{1}"
}
],
"answer": "{0, 1}",
"solution": "**Answer:** {0, 1}\n\n\n From graph you can see, \n (1) when x < 0 then y = x2 is greater than y = x. That is why for f(x) that curved part is chosen.\n (2) when 0 $$ \\le $$ x < 1 then y = x is greater than y = x2. That is why for f(x) part of that straight line is chosen.\n (3) when x $$ \\ge $$ 1 then y = x2 is greater than y = x. That is why for f(x) that curved part is chosen.\n\n
Here on the graph of f(x) there is two sharp corner at x = 0 and x = 1. As we know no function is differentiable at the sharp corner. So f(x) is not differentiable at those two sharp corner.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5712,
"subject": "General Science",
"question": "For all twice differentiable functions f : R $$ \\to $$ R, \nwith f(0) = f(1) = f'(0) = 0",
"options": [
{
"text": "f''(x) $$ \\ne $$ 0, at every point x $$ \\in $$ (0, 1)\n"
},
{
"text": "f''(x) = 0, for some x $$ \\in $$ (0, 1)"
},
{
"text": "f''(0) = 0\n"
},
{
"text": "f''(x) = 0, at every point x $$ \\in $$ (0, 1)"
}
],
"answer": "f''(x) = 0, for some x $$ \\in $$ (0, 1)",
"solution": "**Answer:** f''(x) = 0, for some x $$ \\in $$ (0, 1)\n\nf : R $$ \\to $$ R, with f(0) = f(1) = 0\n and f'(0) = 0\n $$ \\because $$ f(x) is differentiable and continuous\n and f(0) = f(1) = 0\n
Applying Rolle’s theorem in [0, 1] for function f(x)\n f'(c) = 0, c $$ \\in $$ (0, 1)\n
Now again\n $$ \\because $$ f'(c) = 0, f'(0) = 0\n again applying Rolles theorem in [0, c] for function f'(x)\n f''(c1) = 0 for some c1 $$ \\in $$ (0, c) $$ \\in $$ (0, 1)\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5713,
"subject": "General Science",
"question": "The number of points, at which the function f(x) = | 2x + 1 | $$-$$ 3| x + 2 | + | x2 + x $$-$$ 2 |, x$$\\in$$R is not differentiable, is __________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$f(x) = |2x + 1| - 3|x + 2| + |{x^2} + x - 2|$$
Non. differentiable at x = 1 and $${{ - 1} \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5714,
"subject": "General Science",
"question": "A function f is defined on [$$-$$3, 3] as
$$f(x) = \\left\\{ {\\matrix{\n {\\min \\{ |x|,2 - {x^2}\\} ,} & { - 2 \\le x \\le 2} \\cr \n {[|x|],} & {2 < |x| \\le 3} \\cr \n\n } } \\right.$$ where [x] denotes the greatest integer $$ \\le $$ x. The number of points, where f is not differentiable in ($$-$$3, 3) is ___________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n\n Points of non-differentiability in ($$-$$3, 3) are at x = $$-$$2, $$-$$1, 0, 1, 2.
i.e. 5 points.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5715,
"subject": "General Science",
"question": "Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4.
$$f(a) - af'(a) = 4 - 2a$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5716,
"subject": "General Science",
"question": "Let the functions f : R $$ \\to $$ R and g : R $$ \\to $$ R be defined as :
Number of points of non-differentiability = 1",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5717,
"subject": "General Science",
"question": "Let f : S $$ \\to $$ S where S = (0, $$\\infty $$) be a twice differentiable function such that f(x + 1) = xf(x). If g : S $$ \\to $$ R be defined as g(x) = loge f(x), then the value of |g''(5) $$-$$ g''(1)| is equal to :",
"options": [
{
"text": "1"
},
{
"text": "$${{187} \\over {144}}$$"
},
{
"text": "$${{197} \\over {144}}$$"
},
{
"text": "$${{205} \\over {144}}$$"
}
],
"answer": "$${{205} \\over {144}}$$",
"solution": "**Answer:** $${{205} \\over {144}}$$\n\n$$f(x + 1) = xf(x)$$
From (1), $$b = {3 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5719,
"subject": "General Science",
"question": "Let f : R $$ \\to $$ R satisfy the equation f(x + y) = f(x) . f(y) for all x, y $$\\in$$R and f(x) $$\\ne$$ 0 for any x$$\\in$$R. If the function f is differentiable at x = 0 and f'(0) = 3, then
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5720,
"subject": "General Science",
"question": "Let a function g : [ 0, 4 ] $$\\to$$ R be defined as
$$g(x) = \\left\\{ {\\matrix{\n {\\mathop {\\max }\\limits_{0 \\le t \\le x} \\{ {t^3} - 6{t^2} + 9t - 3),} & {0 \\le x \\le 3} \\cr \n {4 - x,} & {3 < x \\le 4} \\cr \n\n } } \\right.$$, then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$f(x) = {x^3} - 6{x^2} + 9x - 3$$
g(x) is non-differentiable at x = 3",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5721,
"subject": "General Science",
"question": "Let f : R $$\\to$$ R be a function defined as $$f(x) = \\left\\{ {\\matrix{\n {3\\left( {1 - {{|x|} \\over 2}} \\right)} & {if} & {|x|\\, \\le 2} \\cr \n 0 & {if} & {|x|\\, > 2} \\cr \n\n } } \\right.$$
Let g : R $$\\to$$ R be given by $$g(x) = f(x + 2) - f(x - 2)$$. If n and m denote the number of points in R where g is not continuous and not differentiable, respectively, then n + m is equal to ______________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5722,
"subject": "General Science",
"question": "Let $$f:[0,\\infty ) \\to [0,3]$$ be a function defined by
Then which of the following is true?",
"options": [
{
"text": "f is continuous everywhere but not differentiable exactly at one point in (0, $$\\infty$$)"
},
{
"text": "f is differentiable everywhere in (0, $$\\infty$$)"
},
{
"text": "f is not continuous exactly at two points in (0, $$\\infty$$)"
},
{
"text": "f is continuous everywhere but not differentiable exactly at two points in (0, $$\\infty$$)"
}
],
"answer": "f is differentiable everywhere in (0, $$\\infty$$)",
"solution": "**Answer:** f is differentiable everywhere in (0, $$\\infty$$)\n\nGraph of $$\\max \\{ \\sin t:0 \\le t \\le x\\} $$ in $$x \\in [0,\\pi ]$$
f(x) is differentiable everywhere in (0, $$\\infty$$)",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 5723,
"subject": "General Science",
"question": "Let $$f:[0,3] \\to R$$ be defined by $$f(x) = \\min \\{ x - [x],1 + [x] - x\\} $$ where [x] is the greatest integer less than or equal to x. Let P denote the set containing all x $$\\in$$ [0, 3] where f i discontinuous, and Q denote the set containing all x $$\\in$$ (0, 3) where f is not differentiable. Then the sum of number of elements in P and Q is equal to ______________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n
1 $$-$$ {x} = 1 $$-$$ x; 0 $$\\le$$ x < 1
Non differentiable at
$$x = {1 \\over 2},1,{3 \\over 2},2,{5 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5724,
"subject": "General Science",
"question": "Let [t] denote the greatest integer less than or equal to t. Let f(x) = x $$-$$ [x], g(x) = 1 $$-$$ x + [x], and h(x) = min{f(x), g(x)}, x $$\\in$$ [$$-$$2, 2]. Then h is :",
"options": [
{
"text": "continuous in [$$-$$2, 2] but not differentiable at more than four points in ($$-$$2, 2)"
},
{
"text": "not continuous at exactly three points in [$$-$$2, 2]"
},
{
"text": "continuous in [$$-$$2, 2] but not differentiable at exactly three points in ($$-$$2, 2)"
},
{
"text": "not continuous at exactly four points in [$$-$$2, 2]"
}
],
"answer": "continuous in [$$-$$2, 2] but not differentiable at more than four points in ($$-$$2, 2)",
"solution": "**Answer:** continuous in [$$-$$2, 2] but not differentiable at more than four points in ($$-$$2, 2)\n\nmin{x $$-$$ [x], 1 $$-$$ x + [x]}
h(x) = min{x $$-$$ [x], 1 $$-$$ [x $$-$$ [x])}
$$\\Rightarrow$$ always continuous in [$$-$$2, 2] but not differentiable at 7 points.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5725,
"subject": "General Science",
"question": "The function
Clearly, non-differentiable at x = $$-$$1 & x = 3.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5726,
"subject": "General Science",
"question": "Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then",
"options": [
{
"text": "f''(x) = 0 for all x $$\\in$$ (0, 2)"
},
{
"text": "f''(x) = 0 for some x $$\\in$$ (0, 2)"
},
{
"text": "f'(x) = 0 for some x $$\\in$$ [0, 2]"
},
{
"text": "f''(x) > 0 for all x $$\\in$$ (0, 2)"
}
],
"answer": "f''(x) = 0 for some x $$\\in$$ (0, 2)",
"solution": "**Answer:** f''(x) = 0 for some x $$\\in$$ (0, 2)\n\n
f(0) = 0, f(1) = 1 and f(2) = 2
\n
Let h(x) = f(x) $$-$$ x
\n
Clearly h(x) is continuous and twice differentiable on (0, 2)
\n
Also, h(0) = h(1) = h(2) = 0
\n
$$\\therefore$$ h(x) satisfies all the condition of Rolle's theorem.
\n
$$\\therefore$$ there exist C1 $$\\in$$(0, 1) such that h'(c1) = 0
Let f, g : R $$\\to$$ R be two real valued functions defined as $$f(x) = \\left\\{ {\\matrix{\n { - |x + 3|} & , & {x < 0} \\cr \n {{e^x}} & , & {x \\ge 0} \\cr \n\n } } \\right.$$ and $$g(x) = \\left\\{ {\\matrix{\n {{x^2} + {k_1}x} & , & {x < 0} \\cr \n {4x + {k_2}} & , & {x \\ge 0} \\cr \n\n } } \\right.$$, where k1 and k2 are real constants. If (gof) is differentiable at x = 0, then (gof) ($$-$$ 4) + (gof) (4) is equal to :
Let f(x) = min {1, 1 + x sin x}, 0 $$\\le$$ x $$\\le$$ 2$$\\pi $$. If m is the number of points, where f is not differentiable and n is the number of points, where f is not continuous, then the ordered pair (m, n) is equal to
where [t] denotes greatest integer $$\\le$$ t. If m is the number of points where $$f$$ is not continuous and n is the number of points where $$f$$ is not differentiable, then the ordered pair (m, n) is :
Let $$f(x)=\\left\\{\\begin{array}{l}\\left|4 x^{2}-8 x+5\\right|, \\text { if } 8 x^{2}-6 x+1 \\geqslant 0 \\\\ {\\left[4 x^{2}-8 x+5\\right], \\text { if } 8 x^{2}-6 x+1<0,}\\end{array}\\right.$$ where $$[\\alpha]$$ denotes the greatest integer less than or equal to $$\\alpha$$. Then the number of points in $$\\mathbf{R}$$ where $$f$$ is not differentiable is ___________.
Let $$f:[0,1] \\rightarrow \\mathbf{R}$$ be a twice differentiable function in $$(0,1)$$ such that $$f(0)=3$$ and $$f(1)=5$$. If the line $$y=2 x+3$$ intersects the graph of $$f$$ at only two distinct points in $$(0,1)$$, then the least number of points $$x \\in(0,1)$$, at which $$f^{\\prime \\prime}(x)=0$$, is ____________.
If $$[t]$$ denotes the greatest integer $$\\leq t$$, then the number of points, at which the function $$f(x)=4|2 x+3|+9\\left[x+\\frac{1}{2}\\right]-12[x+20]$$ is not differentiable in the open interval $$(-20,20)$$, is __________.
$f(x)$ is non differentiable at $x=-\\frac{3}{2}$\n\n
and $f(x)$ is discontinuous at $\\{-19,-18, \\ldots ., 18,19\\}$
as well as $\\left\\{-\\frac{39}{2},-\\frac{37}{2}, \\ldots,-\\frac{3}{2},-\\frac{1}{2}, \\frac{1}{2}, \\ldots, \\frac{39}{2}\\right\\}$,
at same point they are also non differentiable\n\n
$$\n\\begin{aligned}\n\\therefore & \\text { Total number of points of non differentiability } \\\\\n&=39+40 \\\\\n&=79\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5734,
"subject": "General Science",
"question": "
Suppose $$f: \\mathbb{R} \\rightarrow(0, \\infty)$$ be a differentiable function such that $$5 f(x+y)=f(x) \\cdot f(y), \\forall x, y \\in \\mathbb{R}$$. If $$f(3)=320$$, then $$\\sum_\\limits{n=0}^{5} f(n)$$ is equal to :
",
"options": [
{
"text": "$$f$$ is continuous but $$f'$$ is not continuous"
},
{
"text": "$$f$$ and $$f'$$ both are continuous"
},
{
"text": "$$f$$ is continuous but not differentiable"
},
{
"text": "$$f'$$ is continuous but not differentiable"
}
],
"answer": "$$f$$ is continuous but $$f'$$ is not continuous",
"solution": "**Answer:** $$f$$ is continuous but $$f'$$ is not continuous\n\n
$$f(x)=\\left\\{\\begin{array}{cc}x+1, & x < 0 \\\\ |x-1|, & x \\geq 0\\end{array}\\right.$$ and $$\\mathrm{g}(x)=\\left\\{\\begin{array}{cc}x+1, & x < 0 \\\\ 1, & x \\geq 0\\end{array}\\right.$$
\n
Then $$(g \\circ f)(x)$$ is :
",
"options": [
{
"text": "continuous everywhere but not differentiable at $$x=1$$"
},
{
"text": "differentiable everywhere"
},
{
"text": "not continuous at $$x=-1$$"
},
{
"text": "continuous everywhere but not differentiable exactly at one point"
}
],
"answer": "continuous everywhere but not differentiable exactly at one point",
"solution": "**Answer:** continuous everywhere but not differentiable exactly at one point\n\n$$\n\\begin{aligned}\n& \\text { Sol. } f(x)=\\left\\{\\begin{array}{c}\nx+1, x<0 \\\\\\\n1-x, 0 \\leq x<1 \\\\\nx-1,1 \\leq x\n\\end{array}\\right. \\\\\\\\\n& g(x)=\\left\\{\\begin{array}{c}\nx+1, x<0 \\\\\n1, x \\geq 0\n\\end{array}\\right. \\\\\\\\\n& g(f(x))=\\left\\{\\begin{array}{c}\nx+2, x<-1 \\\\\n1, x \\geq-1\n\\end{array}\\right.\n\\end{aligned}\n$$\n
\n $\\therefore \\mathrm{g}(\\mathrm{f}(\\mathrm{x}))$ is continuous everywhere\n
$\\mathrm{g}(\\mathrm{f}(\\mathrm{x}))$ is not differentiable at $\\mathrm{x}=-1$\nDifferentiable everywhere else.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5738,
"subject": "General Science",
"question": "
Let $$f:( - 2,2) \\to R$$ be defined by $$f(x) = \\left\\{ {\\matrix{\n {x[x],} & { - 2 < x < 0} \\cr \n {(x - 1)[x],} & {0 \\le x \\le 2} \\cr \n\n } } \\right.$$ where $$[x]$$ denotes the greatest integer function. If m and n respectively are the number of points in $$( - 2,2)$$ at which $$y = |f(x)|$$ is not continuous and not differentiable, then $$m + n$$ is equal to ____________.
",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nGiven function is $f(x)=\\left\\{\\begin{array}{cc}x[x], & -2 < x < 0 \\\\ (x-1)[x], & 0 \\leq x<2\\end{array}\\right.$ \n
When $[x]$ is denotes greatest integer function\n
\n
Clearly, $|f(x)|$ remains same.\n
Given that, $m$ and $n$ respectively are the number points in $(-2,2)$ at which $y=|f(x)|$ is not continuous and not differentiable\n
Let $$\\mathrm{k}$$ and $$\\mathrm{m}$$ be positive real numbers such that the function $$f(x)=\\left\\{\\begin{array}{cc}3 x^{2}+k \\sqrt{x+1}, & 0 < x < 1 \\\\ m x^{2}+k^{2}, & x \\geq 1\\end{array}\\right.$$ is differentiable for all $$x > 0$$. Then $$\\frac{8 f^{\\prime}(8)}{f^{\\prime}\\left(\\frac{1}{8}\\right)}$$ is equal to ____________.
",
"options": [],
"answer": "309",
"solution": "**Answer:** 309\n\nHere, $$f(x)=\\left\\{\\begin{array}{cc}3 x^{2}+k \\sqrt{x+1}, & 0 < x < 1 \\\\ m x^{2}+k^{2}, & x \\geq 1\\end{array}\\right.$$\n
$f(x)$ is differentiable at $x=a$, if $f^{\\prime}\\left(a^{-}\\right)=f'\\left(a^{+}\\right)$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 5740,
"subject": "General Science",
"question": "
Let $$a \\in \\mathbb{Z}$$ and $$[\\mathrm{t}]$$ be the greatest integer $$\\leq \\mathrm{t}$$. Then the number of points, where the function $$f(x)=[a+13 \\sin x], x \\in(0, \\pi)$$ is not differentiable, is __________.
Given that $ f(x) = [a + 13\\sin(x)] $, where $[t]$ is the greatest integer function and $ x \\in (0, \\pi) $.
\n
The function $[t]$ is not differentiable wherever $ t $ is an integer because at these points, the function has a jump discontinuity.
\n
For $ f(x) $ to have a point of non-differentiability, the value inside the greatest integer function, i.e., $ a + 13\\sin(x) $, should be an integer.
\n
So, we need to find the values of $ x $ in the interval $ (0, \\pi) $ for which $ a + 13\\sin(x) $ is an integer.
\n
Now, $ \\sin(x) $ varies from 0 to 1 in the interval $ (0, \\pi) $, so the maximum value of $ 13\\sin(x) $ in this interval is 13.
\n
For each integer value of $ 13\\sin(x) $ between 0 and 13, we'll have a corresponding value of $ x $. There will be two such values of $ x $ for each value of $ 13\\sin(x) $ (because of the periodic and symmetric nature of sine function over the interval $ (0, \\pi) $), except for the maximum value, 13, which will have only one corresponding value of $ x $ (namely $ x = \\frac{\\pi}{2} $).
\n
Thus, the total number of integer values of $ 13\\sin(x) $ between 0 and 13 is 13. Excluding the maximum value, we have 12 integer values, each giving rise to two values of $ x $. Including the maximum value, which gives one value of $ x $, we have :
\n
$ 12 \\times 2 + 1 = 25 $
\n
Thus, $ f(x) $ is not differentiable at 25 points in the interval $ (0, \\pi) $.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5741,
"subject": "General Science",
"question": "Let $f(x)=\\left|2 x^2+5\\right| x|-3|, x \\in \\mathbf{R}$. If $\\mathrm{m}$ and $\\mathrm{n}$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $\\mathrm{m}+\\mathrm{n}$ is equal to :",
"options": [
{
"text": "5"
},
{
"text": "3"
},
{
"text": "2"
},
{
"text": "0"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\n$\\begin{aligned} & f(x)=\\left|2 x^2+5\\right| x|-3| \\\\\\\\ & \\text { Graph of } y=\\left|2 x^2+5 x-3\\right|\\end{aligned}$\n
\n \n
Now $f(x)$ is continuous $\\forall x \\in R$ \n
but non- differentiable at $x=\\frac{-1}{2}, \\frac{1}{2}, 0$\n
$$\n\\begin{aligned}\n\\therefore m & =0 \\\\\\\\\nn & =3 \\\\\\\\\nm+n & =3\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5742,
"subject": "General Science",
"question": "Let $f: \\mathbf{R} \\rightarrow \\mathbf{R}$ be defined as :\n
If $f$ is continuous everywhere in $\\mathbf{R}$ and $m$ is the number of points where $f$ is NOT differential then $\\mathrm{m}+\\mathrm{a}+\\mathrm{b}+\\mathrm{c}$ equals :",
"options": [
{
"text": "1"
},
{
"text": "4"
},
{
"text": "3"
},
{
"text": "2"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nAt $\\mathrm{x}=1, \\mathrm{f}(\\mathrm{x})$ is continuous therefore,
Function is differentiable at every point in its domain\n
$$\n\\therefore \\mathrm{m}=0\n$$\n
m + a + b + c = 0 + 1 + 1 + 0 = 2 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5743,
"subject": "General Science",
"question": "
Consider the function $$f:(0,2) \\rightarrow \\mathbf{R}$$ defined by $$f(x)=\\frac{x}{2}+\\frac{2}{x}$$ and the function $$g(x)$$ defined by
\n
$$g(x)=\\left\\{\\begin{array}{ll}\n\\min \\lfloor f(t)\\}, & 0<\\mathrm{t} \\leq x \\text { and } 0 < x \\leq 1 \\\\\n\\frac{3}{2}+x, & 1 < x < 2\n\\end{array} .\\right. \\text { Then, }$$
",
"options": [
{
"text": "$$g$$ is continuous but not differentiable at $$x=1$$\n"
},
{
"text": "$$g$$ is continuous and differentiable for all $$x \\in(0,2)$$\n"
},
{
"text": "$$g$$ is not continuous for all $$x \\in(0,2)$$\n"
},
{
"text": "$$g$$ is neither continuous nor differentiable at $$x=1$$"
}
],
"answer": "$$g$$ is continuous but not differentiable at $$x=1$$\n",
"solution": "**Answer:** $$g$$ is continuous but not differentiable at $$x=1$$\n\n\n
$$\\begin{aligned}\n& f:(0,2) \\rightarrow R ; f(x)=\\frac{x}{2}+\\frac{2}{x} \\\\\n& f^{\\prime}(x)=\\frac{1}{2}-\\frac{2}{x^2}\n\\end{aligned}$$
\n
$$\\therefore \\mathrm{f}(\\mathrm{x})$$ is decreasing in domain.
\n
\n
$$g(x)= \\begin{cases}\\frac{x}{2}+\\frac{2}{x} & 0 < x \\leq 1 \\\\ \\frac{3}{2}+x & 1 < x < 2\\end{cases}$$
Consider the function $$f:(0, \\infty) \\rightarrow \\mathbb{R}$$ defined by $$f(x)=e^{-\\left|\\log _e x\\right|}$$. If $$m$$ and $$n$$ be respectively the number of points at which $$f$$ is not continuous and $$f$$ is not differentiable, then $$m+n$$ is
Let $$f(x)=\\sqrt{\\lim _\\limits{r \\rightarrow x}\\left\\{\\frac{2 r^2\\left[(f(r))^2-f(x) f(r)\\right]}{r^2-x^2}-r^3 e^{\\frac{f(r)}{r}}\\right\\}}$$ be differentiable in $$(-\\infty, 0) \\cup(0, \\infty)$$ and $$f(1)=1$$. Then the value of ea, such that $$f(a)=0$$, is equal to _________.
Let $$f$$ be a differentiable function in the interval $$(0, \\infty)$$ such that $$f(1)=1$$ and $$\\lim _\\limits{t \\rightarrow x} \\frac{t^2 f(x)-x^2 f(t)}{t-x}=1$$ for each $$x>0$$. Then $$2 f(2)+3 f(3)$$ is equal to _________.
= $${1 \\over {\\sqrt 2 }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5754,
"subject": "General Science",
"question": "For each t $$ \\in R$$, let [t] be the greatest integer less than or equal to t.\n
Then $$\\mathop {\\lim }\\limits_{x \\to {0^ + }} x\\left( {\\left[ {{1 \\over x}} \\right] + \\left[ {{2 \\over x}} \\right] + ..... + \\left[ {{{15} \\over x}} \\right]} \\right)$$",
"options": [
{
"text": "does not exist in R"
},
{
"text": "is equal to 0"
},
{
"text": "is equal to 15"
},
{
"text": "is equal to 120"
}
],
"answer": "is equal to 120",
"solution": "**Answer:** is equal to 120\n\nGiven, \n
If you put y = 0 at $${{\\sqrt {1 + \\sqrt {1 + {y^4}} } - \\sqrt 2 } \\over {{y^4}}}$$ it is in $${0 \\over 0}$$ form. So we can use L' Hospital's Rule. \n
$$\\mathop {\\lim }\\limits_{n \\to \\infty } {1 \\over n}\\left( {{{7n - 5} \\over 2}} \\right) = {7 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5767,
"subject": "General Science",
"question": "Let f : R $$\\to$$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $$\\mathop {\\lim }\\limits_{x \\to 2} {{{x^2}f(2) - 4f(x)} \\over {x - 2}}$$ is equal to :",
"options": [
{
"text": "4"
},
{
"text": "8"
},
{
"text": "16"
},
{
"text": "12"
}
],
"answer": "12",
"solution": "**Answer:** 12\n\nThis limit can be solved using L'Hopital's Rule, which states that for the limit of the form 0/0 or ±∞/±∞, the limit can be found by taking the derivative of the numerator and the derivative of the denominator separately.\n\n
$$\\mathop {\\lim }\\limits_{x \\to 2} {{{x^2}f(2) - 4f(x)} \\over {x - 2}}$$ is in the indeterminate form, and we are given that f(2) = 4 and f'(2) = 1, so we can apply L'Hopital's rule.\n\n
Taking the derivative of the numerator and the denominator, we get :\n\n
Numerator: derivative of $x^2 f(2) - 4f(x)$ is $2x f(2) - 4f'(x)$.\n
Let [t] denote the greatest integer $$\\le$$ t and {t} denote the fractional part of t. The integral value of $$\\alpha$$ for which the left hand limit of the function
\n
$$f(x) = [1 + x] + {{{\\alpha ^{2[x] + {\\{x\\}}}} + [x] - 1} \\over {2[x] + \\{ x\\} }}$$ at x = 0 is equal to $$\\alpha - {4 \\over 3}$$, is _____________.
Let a be an integer such that $$\\mathop {\\lim }\\limits_{x \\to 7} {{18 - [1 - x]} \\over {[x - 3a]}}$$ exists, where [t] is greatest integer $$\\le$$ t. Then a is equal to :
Let f(x) be a polynomial function such that $$f(x) + f'(x) + f''(x) = {x^5} + 64$$. Then, the value of $$\\mathop {\\lim }\\limits_{x \\to 1} {{f(x)} \\over {x - 1}}$$ is equal to:
Here power of \"n\" in the numerator is 2 and power of \"n\" in the denominator is 1.
\n
To get the value of limit equal to zero power of \"n\" should be equal in both numerator and denominator, otherwise value of limit will be infinite ($$\\infty $$).
\n
$$\\therefore$$ Coefficient of n2 should be 0 in this case.
\n
$$\\therefore$$ $$1 - {\\alpha ^2} = 0$$
\n
$$ \\Rightarrow \\alpha = \\, \\pm \\,1$$
\n
But $$\\alpha$$ should be < 0
\n
$$\\therefore$$ $$\\alpha = \\, + \\,1$$ not possible
The set of all values of $$a$$ for which $$\\mathop {\\lim }\\limits_{x \\to a} ([x - 5] - [2x + 2]) = 0$$, where [$$\\alpha$$] denotes the greatest integer less than or equal to $$\\alpha$$ is equal to
Let $$a_{1}, a_{2}, a_{3}, \\ldots, a_{\\mathrm{n}}$$ be $$\\mathrm{n}$$ positive consecutive terms of an arithmetic progression. If $$\\mathrm{d} > 0$$ is its common difference, then
\n
$$\\lim_\\limits{n \\rightarrow \\infty} \\sqrt{\\frac{d}{n}}\\left(\\frac{1}{\\sqrt{a_{1}}+\\sqrt{a_{2}}}+\\frac{1}{\\sqrt{a_{2}}+\\sqrt{a_{3}}}+\\ldots \\ldots \\ldots+\\frac{1}{\\sqrt{a_{n-1}}+\\sqrt{a_{n}}}\\right)$$ is
$\\begin{aligned} & =\\lim _{n \\rightarrow \\infty}\\left[\\frac{1}{\\sqrt{d}}\\left(\\frac{\\sqrt{a_1+(n-1) d}-\\sqrt{a_1}}{\\sqrt{n}}\\right)\\right] \\\\\\\\ & =\\lim _{n \\rightarrow \\infty}\\left[\\frac{1}{\\sqrt{d}}\\left(\\sqrt{\\frac{a_1}{n}+\\left(d-\\frac{d}{n}\\right)}-\\sqrt{\\frac{a_1}{n}}\\right)\\right] \\\\\\\\ & =\\frac{1}{\\sqrt{d}}(\\sqrt{0+d-0}-\\sqrt{0})=\\frac{\\sqrt{d}}{\\sqrt{d}}=1\\end{aligned}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5779,
"subject": "General Science",
"question": "Let $f(x)=\\left\\{\\begin{array}{l}x-1, x \\text { is even, } \\\\ 2 x, \\quad x \\text { is odd, }\\end{array} x \\in \\mathbf{N}\\right.$.\n
If for some $\\mathrm{a} \\in \\mathbf{N}, f(f(f(\\mathrm{a})))=21$, then $\\lim\\limits_{x \\rightarrow \\mathrm{a}^{-}}\\left\\{\\frac{|x|^3}{\\mathrm{a}}-\\left[\\frac{x}{\\mathrm{a}}\\right]\\right\\}$, where $[t]$ denotes the greatest integer less than or equal to $t$, is equal to :",
"options": [
{
"text": "169"
},
{
"text": "121"
},
{
"text": "225"
},
{
"text": "144"
}
],
"answer": "144",
"solution": "**Answer:** 144\n\n$f(x)=\\left\\{\\begin{array}{l}x-1, x \\text { is even, } \\\\ 2 x, \\quad x \\text { is odd, }\\end{array} x \\in \\mathbf{N}\\right.$\n
Let $$\\beta=\\mathop {\\lim }\\limits_{x \\to 0} \\frac{\\alpha x-\\left(e^{3 x}-1\\right)}{\\alpha x\\left(e^{3 x}-1\\right)}$$ for some $$\\alpha \\in \\mathbb{R}$$. Then the value of $$\\alpha+\\beta$$ is :
If $$\\lim _\\limits{x \\rightarrow 0} \\frac{a x^2 e^x-b \\log _e(1+x)+c x e^{-x}}{x^2 \\sin x}=1$$, then $$16\\left(a^2+b^2+c^2\\right)$$ is equal to ________.
If $$\\alpha=\\lim _\\limits{x \\rightarrow 0^{+}}\\left(\\frac{\\mathrm{e}^{\\sqrt{\\tan x}}-\\mathrm{e}^{\\sqrt{x}}}{\\sqrt{\\tan x}-\\sqrt{x}}\\right)$$ and $$\\beta=\\lim _\\limits{x \\rightarrow 0}(1+\\sin x)^{\\frac{1}{2} \\cot x}$$ are the roots of the quadratic equation $$\\mathrm{a} x^2+\\mathrm{b} x-\\sqrt{\\mathrm{e}}=0$$, then $$12 \\log _{\\mathrm{e}}(\\mathrm{a}+\\mathrm{b})$$ is equal to _________.
If the function $$f(x) = \\left\\{ {\\matrix{\n {{{{{\\log }_e}(1 - x + {x^2}) + {{\\log }_e}(1 + x + {x^2})} \\over {\\sec x - \\cos x}}} & , & {x \\in \\left( {{{ - \\pi } \\over 2},{\\pi \\over 2}} \\right) - \\{ 0\\} } \\cr \n k & , & {x = 0} \\cr \n\n } } \\right.$$ is continuous at x = 0, then k is equal to:
R.H.L. $$ \\ne $$ L.H.L.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5820,
"subject": "General Science",
"question": "For each t $$ \\in $$ R , let [t] be the greatest integer less than or equal to t\n
If $$\\mathop {\\lim }\\limits_{x \\to 1} {{\\sin (3{x^2} - 4x + 1) - {x^2} + 1} \\over {2{x^3} - 7{x^2} + ax + b}} = - 2$$, then the value of (a $$-$$ b) is equal to ___________.
If $$\\lim\\limits_{x \\rightarrow 0} \\frac{\\alpha \\mathrm{e}^{x}+\\beta \\mathrm{e}^{-x}+\\gamma \\sin x}{x \\sin ^{2} x}=\\frac{2}{3}$$, where $$\\alpha, \\beta, \\gamma \\in \\mathbf{R}$$, then which of the following is NOT correct?
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5843,
"subject": "General Science",
"question": "Let $f, g$ and $h$ be the real valued functions defined on $\\mathbb{R}$ as\n
$f(x)=\\left\\{\\begin{array}{cc}\\frac{x}{|x|}, & x \\neq 0 \\\\ 1, & x=0\\end{array}\\right.$\n
$g(x)=\\left\\{\\begin{array}{cc}\\frac{\\sin (x+1)}{(x+1)}, & x \\neq-1 \\\\ 1, & x=-1\\end{array}\\right.$\n\n
and $h(x)=2[x]-f(x)$, where $[x]$ is the greatest integer $\\leq x$.\nThen the
We know that $1 - \\cos 2x$ can be rewritten using the double-angle formula for cosine, which is $\\cos 2x = 1 - 2\\sin^2x$. So, $1 - \\cos 2x = 2\\sin^2x$.
Now, for the limit to exist, the coefficient of $x$ in the numerator must be zero, as the limit would be undefined otherwise. This gives us the equation:
\n
$$a-\\frac{c}{2} = 0 \\Rightarrow c=2a.$$
\n
Then, for the limit to equal $17$, the coefficient of $x^2$ in the numerator must equal $17 \\times 2 = 34$. This gives us the equation:
If $$\\alpha > \\beta > 0$$ are the roots of the equation $$a x^{2}+b x+1=0$$, and $$\\lim_\\limits{x \\rightarrow \\frac{1}{\\alpha}}\\left(\\frac{1-\\cos \\left(x^{2}+b x+a\\right)}{2(1-\\alpha x)^{2}}\\right)^{\\frac{1}{2}}=\\frac{1}{k}\\left(\\frac{1}{\\beta}-\\frac{1}{\\alpha}\\right), \\text { then } \\mathrm{k} \\text { is equal to }$$ :
The value of $$\\lim _\\limits{x \\rightarrow 0} 2\\left(\\frac{1-\\cos x \\sqrt{\\cos 2 x} \\sqrt[3]{\\cos 3 x} \\ldots \\ldots . \\sqrt[10]{\\cos 10 x}}{x^2}\\right)$$ is __________.
Let $$\\mathrm{a}>0$$ be a root of the equation $$2 x^2+x-2=0$$. If $$\\lim _\\limits{x \\rightarrow \\frac{1}{a}} \\frac{16\\left(1-\\cos \\left(2+x-2 x^2\\right)\\right)}{(1-a x)^2}=\\alpha+\\beta \\sqrt{17}$$, where $$\\alpha, \\beta \\in Z$$, then $$\\alpha+\\beta$$ is equal to _________.
But x $$ \\ne $$ 2 because it is not satisfying the domain of given equation i.e. log2(x $$-$$ 3) $$ \\to $$ its domain x > 3
finally x is 5
$$ \\therefore $$ No. of solutions = 1.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5856,
"subject": "General Science",
"question": "The number of solutions of the equation
No. of solution = 1",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5857,
"subject": "General Science",
"question": "
If the solution of the equation $$\\log _{\\cos x} \\cot x+4 \\log _{\\sin x} \\tan x=1, x \\in\\left(0, \\frac{\\pi}{2}\\right)$$, is $$\\sin ^{-1}\\left(\\frac{\\alpha+\\sqrt{\\beta}}{2}\\right)$$, where $$\\alpha$$, $$\\beta$$ are integers, then $$\\alpha+\\beta$$ is equal to :
Let $$S = \\left\\{ {\\alpha :{{\\log }_2}({9^{2\\alpha - 4}} + 13) - {{\\log }_2}\\left( {{5 \\over 2}.\\,{3^{2\\alpha - 4}} + 1} \\right) = 2} \\right\\}$$. Then the maximum value of $$\\beta$$ for which the equation $${x^2} - 2{\\left( {\\sum\\limits_{\\alpha \\in s} \\alpha } \\right)^2}x + \\sum\\limits_{\\alpha \\in s} {{{(\\alpha + 1)}^2}\\beta = 0} $$ has real roots, is ____________.
Let a, b, c be three distinct positive real numbers such that $${(2a)^{{{\\log }_e}a}} = {(bc)^{{{\\log }_e}b}}$$ and $${b^{{{\\log }_e}2}} = {a^{{{\\log }_e}c}}$$.
Then, 6a + 5bc is equal to ___________.
",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nGiven, $(2 a)^{\\ln a}=(b c)^{\\ln b}$, where $2 a>0, b c>0$\n
In this case, since x and y are natural logarithms of positive numbers a and b respectively, this implies that a = b. However, this cannot be true as a, b, and c are given to be distinct positive real numbers.
\n
Case 2 :\n
x = -y\n
In this case, $ \\ln a = -\\ln b $\n
$ \\Rightarrow a \\times b = 1$\n
$ \\Rightarrow b = \\frac{1}{a} $
\n
Also, y + z = 0 (from the equations above)\n
$ \\Rightarrow \\ln b + \\ln c = 0 $\n
$ \\Rightarrow \\ln (b \\times c) = 0 $\n
$ \\Rightarrow b \\times c = 1 $
\n
Given $ b = \\frac{1}{a} $ and $ b \\times c = 1 $\n$ \\Rightarrow c = a $
\n
Thus, in the case where x = -y, the possible values are :\n
$$\\therefore$$ Statement p $$\\to$$ (q $$\\to$$ p) is equivalent to p $$\\to$$ (p $$\\vee$$ q).
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5864,
"subject": "General Science",
"question": "Statement-1 : $$ \\sim \\left( {p \\leftrightarrow \\sim q} \\right)$$ is equivalent to $${p \\leftrightarrow q}$$.\n Statement-2 : $$ \\sim \\left( {p \\leftrightarrow \\sim q} \\right)$$ is a tautology.",
"options": [
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1"
},
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1"
},
{
"text": "Statement-1 is true, Statement-2 is false"
},
{
"text": "Statement-1 is false, Statement-2 is true"
}
],
"answer": "Statement-1 is true, Statement-2 is false",
"solution": "**Answer:** Statement-1 is true, Statement-2 is false\n\n\n
Therefore, S1 is true and S2 is false.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5865,
"subject": "General Science",
"question": "Consider :\n Statement − I : $$\\left( {p \\wedge \\sim q} \\right) \\wedge \\left( { \\sim p \\wedge q} \\right)$$ is a fallacy.\n Statement − II :$$\\left( {p \\to q} \\right) \\leftrightarrow \\left( { \\sim q \\to \\sim p} \\right)$$ is a tautology.",
"options": [
{
"text": "Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I"
},
{
"text": "Statement -I is True; Statement -II is False."
},
{
"text": "Statement -I is False; Statement -II is True"
},
{
"text": "Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I "
}
],
"answer": "Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I",
"solution": "**Answer:** Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I\n\n
Statement - I : $$(p \\wedge \\sim q) \\wedge ( \\sim p \\wedge q)$$
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n\n
\n
p
\n
q
\n
$$ \\sim p$$
\n
$$ \\sim q$$
\n
$$p \\wedge \\sim q$$
\n
$$ \\sim p \\wedge q$$
\n
$$(p \\wedge \\sim q) \\wedge ( \\sim p \\wedge q)$$
From the table you can see ~p and\n ~(p $$ \\vee $$ q) $$ \\vee $$ (~p $$ \\wedge $$ q) are equivalent.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5872,
"subject": "General Science",
"question": "If (p $$ \\wedge $$ $$ \\sim $$ q) $$ \\wedge $$ (p $$ \\wedge $$ r) $$ \\to $$ $$ \\sim $$ p $$ \\vee $$ q is false, then the truth values of $$p, q$$ and $$r$$ are, respectively :",
"options": [
{
"text": "F, T, F"
},
{
"text": "T, F, T"
},
{
"text": "T, T, T"
},
{
"text": "F, F, F"
}
],
"answer": "T, F, T",
"solution": "**Answer:** T, F, T\n\n \nFrom the truth table you can see (p $$ \\wedge $$ ~q) $$ \\wedge $$ (p $$ \\wedge $$ r) $$\\to$$ ~p $$\\vee$$ q is false only when values of (p, q, r) is (T, F, T).",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5873,
"subject": "General Science",
"question": "If p $$ \\to $$ ($$ \\sim $$ p$$ \\vee $$ $$ \\sim $$ q) is false, then the truth values of p and q are respectively : ",
"options": [
{
"text": "F, F"
},
{
"text": "T, F"
},
{
"text": "F, T"
},
{
"text": "T, T"
}
],
"answer": "T, T",
"solution": "**Answer:** T, T\n\n
\n
\n
p
\n
q
\n
~p
\n
~q
\n
~p $$ \\vee $$ ~q
\n
p $$ \\to $$ (~p $$\\vee$$ ~q)
\n
\n
\n
T
\n
T
\n
F
\n
F
\n
F
\n
F
\n
\n
\n
T
\n
F
\n
F
\n
T
\n
T
\n
T
\n
\n
\n
F
\n
T
\n
T
\n
F
\n
T
\n
T
\n
\n
\n
F
\n
F
\n
T
\n
T
\n
T
\n
T
\n
\n
\nFrom the truth table,
\np $$ \\to $$ (~p $$\\vee$$ ~q) is false only when p and q both are true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5874,
"subject": "General Science",
"question": "The logical statement\n
\n$$\\left( {{S_1}} \\right):\\left( {q \\vee p} \\right) \\to \\left( {p \\leftrightarrow \\sim q} \\right)$$ is a tautology
\n$$\\left( {{S_2}} \\right): \\,\\,\\sim q \\wedge \\left( { \\sim p \\leftrightarrow q} \\right)$$ is a fallacy. Then:",
"options": [
{
"text": "both (S1) and (S2) are not correct"
},
{
"text": "only (S1) is correct"
},
{
"text": "only (S2) is correct"
},
{
"text": "both (S1) and (S2) are correct"
}
],
"answer": "both (S1) and (S2) are not correct",
"solution": "**Answer:** both (S1) and (S2) are not correct\n\nTruth table for S1 :\n
$$ \\equiv $$ t\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5895,
"subject": "General Science",
"question": "Let A, B, C and D be four non-empty sets. The contrapositive statement of \"If A $$ \\subseteq $$ B and B $$ \\subseteq $$ D,\nthen A $$ \\subseteq $$ C\" is :",
"options": [
{
"text": "If A ⊈ C, then A ⊈ B or B ⊈ D"
},
{
"text": "If A ⊈ C, then A ⊈ B and B $$ \\subseteq $$ D"
},
{
"text": "If A $$ \\subseteq $$ C, then B $$ \\subset $$ A or D $$ \\subset $$ B"
},
{
"text": "If A ⊈ C, then A $$ \\subseteq $$ B and B $$ \\subseteq $$ D"
}
],
"answer": "If A ⊈ C, then A ⊈ B or B ⊈ D",
"solution": "**Answer:** If A ⊈ C, then A ⊈ B or B ⊈ D\n\nContrapositive of p $$ \\to $$ q is ~q $$ \\to $$ ~p\n
If A $$ \\subseteq $$ B and B $$ \\subseteq $$ D,\nthen A $$ \\subseteq $$ C means\n
(A $$ \\subseteq $$ B) $$ \\wedge $$ (B $$ \\subseteq $$ D) $$ \\to $$ (A $$ \\subseteq $$ C)\n
Then which of the following statements is correct?",
"options": [
{
"text": "(b) is a tautology but not (a)."
},
{
"text": "(a) and (b) both are not tautologies."
},
{
"text": "(a) and (b) both are tautologies."
},
{
"text": "(a) is a tautology but not (b)."
}
],
"answer": "(a) and (b) both are tautologies.",
"solution": "**Answer:** (a) and (b) both are tautologies.\n\n\n (b) is tautologies
$$ \\therefore $$ a & b are both tautologies.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5901,
"subject": "General Science",
"question": "The statement A $$ \\to $$ (B $$ \\to $$ A) is equivalent to :",
"options": [
{
"text": "A $$ \\to $$ (A $$\\mathrel{\\mathop{\\kern0pt\\longleftrightarrow}\n\\limits_{}} $$ B)"
},
{
"text": "A $$ \\to $$ (A $$ \\vee $$ B)"
},
{
"text": "A $$ \\to $$ (A $$ \\wedge $$ B)"
},
{
"text": "A $$ \\to $$ (A $$ \\to $$ B)"
}
],
"answer": "A $$ \\to $$ (A $$ \\vee $$ B)",
"solution": "**Answer:** A $$ \\to $$ (A $$ \\vee $$ B)\n\n$$A \\to (B \\to A)$$
$$ \\Rightarrow A \\to ( \\sim B \\vee A)$$
$$ \\Rightarrow \\, \\sim A \\vee ( \\sim B \\vee A)$$
$$ \\Rightarrow \\, \\sim B \\vee ( \\sim A \\vee A)$$
$$ \\Rightarrow \\, \\sim B \\vee t$$
= t (tantology)
From options :
(B) $$A \\to (A \\vee B)$$
$$ \\Rightarrow \\, \\sim A \\vee (A \\vee B)$$
$$ \\Rightarrow ( \\sim A \\vee A) \\vee B$$
$$ \\Rightarrow t \\vee B$$
$$ \\Rightarrow t$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5902,
"subject": "General Science",
"question": "Let F1(A, B, C) = (A $$ \\wedge $$ $$ \\sim $$ B) $$ \\vee $$ [$$\\sim$$C $$\\wedge$$ (A $$\\vee$$ B)] $$\\vee$$ $$\\sim$$ A and F2(A, B) = (A $$\\vee$$ B) $$\\vee$$ (B $$ \\to $$ $$\\sim$$A) be two logical expressions. Then :",
"options": [
{
"text": "Both F1 and F2 are not tautologies"
},
{
"text": "F1 and F2 both are tautologies"
},
{
"text": "F1 is not a tautology but F2 is a tautology"
},
{
"text": "F1 is a tautology but F2 is not a tautology"
}
],
"answer": "F1 is not a tautology but F2 is a tautology",
"solution": "**Answer:** F1 is not a tautology but F2 is a tautology\n\nTruth table for F1 :\n \n$$ \\therefore $$ F1 is not a tautology.\n
Truth table for F2 :\n \n $$ \\therefore $$ F2 is a tautology.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5903,
"subject": "General Science",
"question": "Which of the following Boolean expression is a tautology?",
"options": [
{
"text": "(p $$ \\wedge $$ q) $$ \\vee $$ (p $$ \\to $$ q)"
},
{
"text": "(p $$ \\wedge $$ q) $$ \\vee $$ (p $$\\vee$$ q)"
},
{
"text": "(p $$ \\wedge $$ q) $$ \\to $$ (p $$ \\to $$ q)"
},
{
"text": "(p $$ \\wedge $$ q) $$ \\wedge $$ (p $$ \\to $$ q)"
}
],
"answer": "(p $$ \\wedge $$ q) $$ \\to $$ (p $$ \\to $$ q)",
"solution": "**Answer:** (p $$ \\wedge $$ q) $$ \\to $$ (p $$ \\to $$ q)\n\n$$\\matrix{\n p & q & {p \\wedge q} & {p \\vee q} & {p \\to q} & {(p \\wedge q) \\to (p \\to q)} \\cr \n T & T & T & T & T & T \\cr \n F & T & F & T & T & T \\cr \n T & F & F & T & F & T \\cr \n F & F & F & F & T & T \\cr \n\n } $$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 5904,
"subject": "General Science",
"question": "If the Boolean expression (p $$ \\Rightarrow $$ q) $$ \\Leftrightarrow $$ (q * ($$ \\sim $$p) is a tautology, then the boolean expression (p * ($$ \\sim $$q)) is equivalent to :",
"options": [
{
"text": "q $$ \\Rightarrow $$ p"
},
{
"text": "p $$ \\Rightarrow $$ q"
},
{
"text": "p $$ \\Rightarrow $$ $$ \\sim $$ q"
},
{
"text": "$$ \\sim $$q $$ \\Rightarrow $$ p"
}
],
"answer": "q $$ \\Rightarrow $$ p",
"solution": "**Answer:** q $$ \\Rightarrow $$ p\n\n
The Boolean expression
\n
$(p \\Rightarrow q) \\Leftrightarrow\\left(q^*(\\sim p)\\right)$ is a tautology.
",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5911,
"subject": "General Science",
"question": "If the truth value of the Boolean expression $$\\left( {\\left( {p \\vee q} \\right) \\wedge \\left( {q \\to r} \\right) \\wedge \\left( { \\sim r} \\right)} \\right) \\to \\left( {p \\wedge q} \\right)$$ is false, then the truth values of the statements p, q, r respectively can be :",
"options": [
{
"text": "T F T"
},
{
"text": "F F T"
},
{
"text": "T F F "
},
{
"text": "F T F"
}
],
"answer": "T F F ",
"solution": "**Answer:** T F F \n\n
\n\n
\n
p
\n
q
\n
r
\n
$$\\underbrace {p \\vee q}_a$$
\n
$$\\underbrace {q \\to r}_b$$
\n
$${a \\wedge b}$$
\n
$${ \\sim r}$$
\n
$$\\underbrace {a \\wedge b \\wedge ( \\sim r)}_c$$
\n
$$\\underbrace {p \\wedge q}_d$$
\n
$$c \\to d$$
\n
\n\n\n
\n
T
\n
F
\n
T
\n
T
\n
T
\n
T
\n
F
\n
F
\n
F
\n
T
\n
\n
\n
F
\n
F
\n
T
\n
F
\n
T
\n
F
\n
F
\n
F
\n
F
\n
T
\n
\n
\n
T
\n
F
\n
F
\n
T
\n
T
\n
T
\n
T
\n
T
\n
F
\n
F
\n
\n
\n
F
\n
T
\n
F
\n
T
\n
F
\n
F
\n
T
\n
F
\n
F
\n
T
\n
\n\n
",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5912,
"subject": "General Science",
"question": "Consider the two statements :
(S1) : (p $$\\to$$ q) $$ \\vee $$ ($$ \\sim $$ q $$\\to$$ p) is a tautology .
(S2) : (p $$ \\wedge $$ $$ \\sim $$ q) $$ \\wedge $$ ($$\\sim$$ p $$\\wedge$$ q) is a fallacy.
Then :",
"options": [
{
"text": "only (S1) is true."
},
{
"text": "both (S1) and (S2) are false."
},
{
"text": "both (S1) and (S2) are true."
},
{
"text": "only (S2) is true."
}
],
"answer": "both (S1) and (S2) are true.",
"solution": "**Answer:** both (S1) and (S2) are true.\n\nS1 : ($$\\sim$$ p $$ \\vee $$ q) $$ \\vee $$ (q $$ \\vee $$ p) = (q $$ \\vee $$ $$\\sim$$ p) $$ \\vee $$ (q $$ \\vee $$ p)
S1 = q $$ \\vee $$ ($$\\sim$$ p $$ \\vee $$ p) = qvt = t = tautology
S2 : (p $$ \\wedge $$ $$\\sim$$ q) $$ \\wedge $$ ($$\\sim$$ p $$\\vee$$ q) = (p $$ \\wedge $$ $$\\sim$$ q) $$ \\wedge $$ $$\\sim$$ (p $$ \\wedge $$ $$\\sim$$ q) = C = fallacy",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5913,
"subject": "General Science",
"question": "The statement (p $$ \\wedge $$ (p $$\\to$$ q) $$\\wedge$$ (q $$\\to$$ r)) $$\\to$$ r is :",
"options": [
{
"text": "a tautology"
},
{
"text": "equivalent to p $$\\to$$ $$\\sim$$ r"
},
{
"text": "a fallacy"
},
{
"text": "equivalent to q $$\\to$$ $$\\sim$$ r"
}
],
"answer": "a tautology",
"solution": "**Answer:** a tautology\n\n(p $$ \\wedge $$ (p $$\\to$$ q) $$\\wedge$$ (q $$\\to$$ r)) $$\\to$$ r
$$\\equiv$$ (p $$\\wedge$$ ($$\\sim$$ p $$\\vee$$ q) $$\\vee$$ ($$\\sim$$ q $$\\vee$$ r)) $$\\to$$ r
$$\\equiv$$ ((p $$\\wedge$$ q) $$\\wedge$$ ($$\\sim$$ p $$\\vee$$ r)) $$\\to$$ r
$$\\equiv$$ (p $$\\wedge$$ q $$\\wedge$$ r) $$\\to$$ r
$$\\equiv$$ $$\\sim$$ (p $$\\wedge$$ q $$\\wedge$$ r) $$\\vee$$ r
Let $$\\Delta$$ $$\\in$$ {$$\\wedge$$, $$\\vee$$, $$\\Rightarrow$$, $$\\Leftrightarrow$$} be such that (p $$\\wedge$$ q) $$\\Delta$$ ((p $$\\vee$$ q) $$\\Rightarrow$$ q) is a tautology. Then $$\\Delta$$ is equal to :
\n$$\n\\begin{aligned}\n& \\sim(p \\vee q) \\vee q \\\\\\\\\n& =(\\sim p \\wedge \\sim q) \\vee q \\\\\\\\\n& =(\\sim p \\vee q) \\wedge(\\sim q \\vee q) \\\\\\\\\n& =(\\sim p \\vee q) \\wedge T \\\\\\\\\n& =\\sim p \\vee q\n\\end{aligned}\n$$\n
\nNow $(p \\wedge q) \\Delta(\\sim p \\vee q)$\n
\n$$\n\\begin{array}{cccccc}\np & q & \\sim p & p \\wedge q & \\sim p \\vee q & (p \\wedge q) \\Delta(\\sim p \\vee q) \\\\\nT & T & F & T & T & T \\\\\nT & F & F & F & F & T \\\\\nF & T & T & F & T & T \\\\\nF & F & T & F & T & T\n\\end{array}\n$$\n
The maximum number of compound propositions, out of p$$\\vee$$r$$\\vee$$s, p$$\\vee$$r$$\\vee$$$$\\sim$$s, p$$\\vee$$$$\\sim$$q$$\\vee$$s, $$\\sim$$p$$\\vee$$$$\\sim$$r$$\\vee$$s, $$\\sim$$p$$\\vee$$$$\\sim$$r$$\\vee$$$$\\sim$$s, $$\\sim$$p$$\\vee$$q$$\\vee$$$$\\sim$$s, q$$\\vee$$r$$\\vee$$$$\\sim$$s, q$$\\vee$$$$\\sim$$r$$\\vee$$$$\\sim$$s, $$\\sim$$p$$\\vee$$$$\\sim$$q$$\\vee$$$$\\sim$$s that can be made simultaneously true by an assignment of the truth values to p, q, r and s, is equal to __________.
",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\nThere are total 9 compound propositions, out of which 6 contain $\\sim s$. So if we assign $s$ as false, these 6 propositions will be true.\n
\nIn remaining 3 compound propositions, two contain $p$ and the third contains $\\sim r$. So if we assign $p$ and $r$ as true and false respectively, these 3 propositions will also be true.\n
\nHence maximum number of propositions that can be true are 9.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5921,
"subject": "General Science",
"question": "
Let p, q, r be three logical statements. Consider the compound statements
",
"options": [
{
"text": "If S2 is True, then S1 is True"
},
{
"text": "If S2 is False, then S1 is False"
},
{
"text": "If S2 is False, then S1 is True"
},
{
"text": "If S1 is False, then S2 is False"
}
],
"answer": "If S2 is False, then S1 is True",
"solution": "**Answer:** If S2 is False, then S1 is True\n\n
$${S_1}:( \\sim p \\vee q) \\vee ( \\sim p \\vee r)$$
Let $$\\Delta$$, $$\\nabla $$ $$\\in$$ {$$\\wedge$$, $$\\vee$$} be such that p $$\\nabla$$ q $$\\Rightarrow$$ ((p $$\\Delta$$ q) $$\\nabla$$ r) is a tautology. Then (p $$\\nabla$$ q) $$\\Delta$$ r is logically equivalent to :
If the proposition $$p \\to (( \\sim p) \\vee q)$$ is evaluated as FALSE, then :
",
"options": [
{
"text": "P1 is TRUE and P2 is FALSE"
},
{
"text": "P1 is FALSE and P2 is TRUE"
},
{
"text": "Both P1 and P2 are FALSE"
},
{
"text": "Both P1 and P2 are TRUE"
}
],
"answer": "Both P1 and P2 are FALSE",
"solution": "**Answer:** Both P1 and P2 are FALSE\n\nGiven $p \\rightarrow(\\sim p \\vee q)$ is false
$\\Rightarrow \\sim p \\vee q$ is false and $p$ is true
\nIf $p \\wedge q$ is $T$, then $(p \\wedge \\sim q) \\vee(\\sim p \\wedge q)$ is $F$\n
\nso $x$ cannot be a tautology.\n
\nCase-III\n
\nWhen $\\Delta$ is same as $\\Rightarrow$\n
\nThen $(p \\Rightarrow \\sim q) \\vee(\\sim p \\Rightarrow q)$ is same at $(\\sim p \\vee \\sim q) \\vee$ $(p \\vee q)$, which is always true, so $x$ becomes a tautology.\n
\n$p \\Leftrightarrow q$ is true when $p$ and $q$ have same truth values, then $p \\Leftrightarrow \\sim q$ and $\\sim p \\Leftrightarrow q$ both are false. Hence $x$ cannot be a tautology.\n
\nSo finally $x$ can be $\\vee$ or $\\Rightarrow$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5929,
"subject": "General Science",
"question": "
If the truth value of the statement $$(P \\wedge(\\sim R)) \\rightarrow((\\sim R) \\wedge Q)$$ is F, then the truth value of which of the following is $$\\mathrm{F}$$ ?
Let the operations $$*, \\odot \\in\\{\\wedge, \\vee\\}$$. If $$(\\mathrm{p} * \\mathrm{q}) \\odot(\\mathrm{p}\\, \\odot \\sim \\mathrm{q})$$ is a tautology, then the ordered pair $$(*, \\odot)$$ is :
\n",
"options": [
{
"text": "only (S1) is a tautology"
},
{
"text": "neither (S1) nor (S2) is a tautology"
},
{
"text": "both (S1) and (S2) are tautologies"
},
{
"text": "only (S2) is a tautology"
}
],
"answer": "neither (S1) nor (S2) is a tautology",
"solution": "**Answer:** neither (S1) nor (S2) is a tautology\n\n
If $$p,q$$ and $$r$$ are three propositions, then which of the following combination of truth values of $$p,q$$ and $$r$$ makes the logical expression $$\\left\\{ {(p \\vee q) \\wedge \\left( {( \\sim p) \\vee r} \\right)} \\right\\} \\to \\left( {( \\sim q) \\vee r} \\right)$$ false?
Now, we will use De Morgan's laws to simplify the expression. De Morgan's laws state:
\n\n
$\\sim (A \\wedge B) \\equiv \\sim A \\vee \\sim B$
\n
$\\sim (A \\vee B) \\equiv \\sim A \\wedge \\sim B$
\n\n
Applying the first De Morgan's law to the outer conjunction of the negated expression:
\n
$\\sim p \\vee \\sim (q \\wedge \\sim (p \\wedge q))$
\n
Now we need to address the inner conjunction $(q \\wedge \\sim (p \\wedge q))$. To simplify this part of the expression, we again apply the first De Morgan's law:\n
\n
$\\sim p \\vee (\\sim q \\vee (p \\wedge q))$\n
\n
Distribute the disjunction over the inner conjunction:
The statement $$(p \\wedge(\\sim q)) \\vee((\\sim p) \\wedge q) \\vee((\\sim p) \\wedge(\\sim q))$$ is equivalent to _________.
",
"options": [
{
"text": "$$(\\sim p) \\vee(\\sim q)$$"
},
{
"text": "$$p \\vee(\\sim q)$$"
},
{
"text": "$$\\mathrm{p} \\vee \\mathrm{q}$$"
},
{
"text": "$$(\\sim p) \\vee q$$"
}
],
"answer": "$$(\\sim p) \\vee(\\sim q)$$",
"solution": "**Answer:** $$(\\sim p) \\vee(\\sim q)$$\n\n$$\n\\begin{array}{|c|c|c|c|c|c|c|}\n\\hline \\mathbf{p} & \\mathbf{q} & \\sim \\mathbf{q} & \\sim \\mathbf{p} & \\mathbf{p} \\wedge \\sim \\mathbf{q} & \\mathbf{\\sim p} \\wedge \\mathbf{q} & \\mathbf{\\sim p} \\wedge \\mathbf{\\sim q} \\\\\n\\hline T & T & F & F & F & F & F \\\\\nT & F & T & F & T & F & F \\\\\nF & T & F & T & F & T & F \\\\\nF & F & T & T & F & F & T \\\\\n\\hline\n\\end{array}\n$$\n
Using the truth table, we can see that the given statement is equivalent to $$(\\sim p) \\vee(\\sim q)$$.\n
Alternate Method :\n
The given statement $$(p \\wedge(\\sim q)) \\vee((\\sim p) \\wedge q) \\vee((\\sim p) \\wedge(\\sim q))$$ can be simplified using the following steps:\n\n
Here, for every value of components compound statement (last column) are not true. Hence, $S_1$ is not a tautology. \n
$S_2:(q \\Rightarrow p) \\Rightarrow((\\sim p) \\wedge q)$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5959,
"subject": "General Science",
"question": "Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”,\nand r be the statement “x is a rational number iff y is a transcendental number”.\n
Statement –1: r is equivalent to either q or p.\n
Statement –2: r is equivalent to $$ \\sim \\left( {p \\leftrightarrow \\sim q} \\right)$$ ",
"options": [
{
"text": "Statement − 1 is false, Statement − 2 is false"
},
{
"text": "Statement −1 is false, Statement −2 is true"
},
{
"text": "Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1"
},
{
"text": "Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for\nStatement −1"
}
],
"answer": "Statement − 1 is false, Statement − 2 is false",
"solution": "**Answer:** Statement − 1 is false, Statement − 2 is false\n\n
p : x is an irrational number
\n
q : y is a transcendental number
\n
r : x is a rational number, if y is a transcendental number
\n
$$\\Rightarrow$$ r : $$\\sim$$ p $$\\leftrightarrow$$ q
\n
S1 : r $$\\equiv$$ q $$\\vee$$ p
\n
and S2 : r $$\\equiv$$ $$\\sim$$ (p $$\\leftrightarrow$$ $$\\sim$$ q)
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n\n
\n
p
\n
q
\n
$$ \\sim p$$
\n
$$ \\sim q$$
\n
r $$ \\sim p \\leftrightarrow q$$
\n
Statement I $$q \\vee p$$
\n
$$(p \\leftrightarrow \\sim q)$$
\n
Statement II $$ \\sim (p \\leftrightarrow \\sim q)$$
\n
\n\n\n
\n
T
\n
T
\n
F
\n
F
\n
F
\n
T
\n
F
\n
T
\n
\n
\n
T
\n
F
\n
F
\n
T
\n
T
\n
T
\n
T
\n
F
\n
\n
\n
F
\n
T
\n
T
\n
F
\n
T
\n
T
\n
T
\n
F
\n
\n
\n
F
\n
F
\n
T
\n
T
\n
F
\n
F
\n
F
\n
T
\n
\n\n
\n
It is clear from the table that r is not equivalent to either of the statements.
\n
Therefore, both statements are false.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5960,
"subject": "General Science",
"question": "Let S be a non-empty subset of R. Consider the following statement:\n P : There is a rational number x ∈ S such that x > 0.\n Which of the following statements is the negation of the statement P?",
"options": [
{
"text": "There is no rational number x ∈ S such that x ≤ 0"
},
{
"text": "Every rational number x ∈ S satisfies x ≤ 0"
},
{
"text": "x ∈ S and x ≤ 0 $$ \\Rightarrow $$ x is not rational"
},
{
"text": "There is a rational number x ∈ S such that x ≤ 0"
}
],
"answer": "Every rational number x ∈ S satisfies x ≤ 0",
"solution": "**Answer:** Every rational number x ∈ S satisfies x ≤ 0\n\n
Given that S is a non-empty subset of R.
\n
$$\\bullet$$ P : There is a rational number x $$\\in$$ S such that x > 0.
\n
Now, we need to find the negation of P. Clearly, P is equivalent to saying that \"There is a positive rational number in S.
\n
So, its negation ($$\\sim$$ P) is \"There is no positive rational number in S\".
\n
Thus, for $$\\sim$$ P : There exists no positive rational number in S.
\n
$$\\bullet$$ $$\\Leftrightarrow$$ $$\\sim$$ P : Every rational number x $$\\in$$ S satisfies x $$\\le$$ 0.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5961,
"subject": "General Science",
"question": "Consider the following statements\n P : Suman is brilliant\n Q : Suman is rich\n R : Suman is honest\n The negation of the statement,\n
“Suman is brilliant and dishonest if and only if Suman is rich” can be\nexpressed as :",
"options": [
{
"text": "$$ \\sim \\left[ {Q \\leftrightarrow \\left( {P \\wedge \\sim R} \\right)} \\right]$$"
},
{
"text": "$$ \\sim Q \\leftrightarrow P \\wedge R$$"
},
{
"text": "$$ \\sim \\left( {P \\wedge \\sim R} \\right) \\leftrightarrow Q$$"
},
{
"text": "$$ \\sim P \\wedge \\left( {Q \\leftrightarrow \\sim R} \\right)$$"
}
],
"answer": "$$ \\sim \\left[ {Q \\leftrightarrow \\left( {P \\wedge \\sim R} \\right)} \\right]$$",
"solution": "**Answer:** $$ \\sim \\left[ {Q \\leftrightarrow \\left( {P \\wedge \\sim R} \\right)} \\right]$$\n\n\"Suman is brilliant and dishonest\" an be expressed as : $${P \\wedge \\sim R}$$\n
So “Suman is brilliant and dishonest if and only if Suman is rich” can be expressed as :\n
Now negation of this = $$ \\sim \\left[ {\\left( {P \\wedge \\sim R} \\right) \\leftrightarrow Q} \\right]$$\n
You should know, $$p \\leftrightarrow q \\equiv q \\leftrightarrow p$$\n
So, $$ \\sim \\left[ {\\left( {P \\wedge \\sim R} \\right) \\leftrightarrow Q} \\right] \\equiv \\sim \\left[ {Q \\leftrightarrow \\left( {P \\wedge \\sim R} \\right)} \\right]$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5962,
"subject": "General Science",
"question": "The negation of the statement “If I become a teacher, then I will open a school” is :",
"options": [
{
"text": "I will become a teacher and I will not open a school"
},
{
"text": "Either I will not become a teacher or I will not open a school"
},
{
"text": "Neither I will become a teacher nor I will open a school"
},
{
"text": "I will not become a teacher or I will open a school"
}
],
"answer": "I will become a teacher and I will not open a school",
"solution": "**Answer:** I will become a teacher and I will not open a school\n\n
Let, p : I become a teacher
\n
q : I will open a school
\n
$$\\therefore$$ \"If I become a teacher, then I will open a school\", is p $$\\to$$ q
\n
So, negation of $$(p \\to q) = \\sim (p \\to q) = p \\wedge \\sim q$$ = I will become a teacher and I will not open a school
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5963,
"subject": "General Science",
"question": "Consider the following two statements : \n
P : If 7 is an odd number, then 7 is\ndivisible by 2.\n Q : If 7 is a prime number, then 7 is an\nodd number\n
If V1 is the truth value of the contrapositive of P and V2 is the truth value of contrapositive of Q, then the ordered pair (V1 , V2) equals :",
"options": [
{
"text": "(T, T)"
},
{
"text": "(T, F)"
},
{
"text": "(F, T)"
},
{
"text": "(F, F)"
}
],
"answer": "(F, T)",
"solution": "**Answer:** (F, T)\n\nContrapositive of P : If 7 is not divisible by 2, \nthen 7 is not an odd number.\n
This statement is false.\n
$$ \\therefore $$ V1 = False (F)\n
Contrapositive of Q : If 7 is not an odd number \n,then 7 is not a prime number.\n
This statement is true.\n
$$ \\therefore $$ V2 = True (T)\n
$$ \\therefore $$ (V1, V2) = (F, T).",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 5964,
"subject": "General Science",
"question": "The contrapositive of the following statement, \n
“If the side of a square doubles, then its area increases four times”, is :",
"options": [
{
"text": "If the side of a square is not doubled, then its area does not increase four times."
},
{
"text": "If the area of a square increases four times, then its side is doubled."
},
{
"text": "If the area of a square increases four times, then its side is not doubled."
},
{
"text": "If the area of a square does not increase four times, then its side is not doubled.\n"
}
],
"answer": "If the area of a square does not increase four times, then its side is not doubled.\n",
"solution": "**Answer:** If the area of a square does not increase four times, then its side is not doubled.\n\n\nContrapositive of p $$ \\to $$ q is $$ \\sim $$q $$ \\to $$ $$ \\sim $$p.\n
Here,\n
Let\n
p = Side of a square is doubles.\n
q = Area of square increases four times.\n
$$ \\therefore $$ $$ \\sim $$q $$ \\to $$ $$ \\sim $$p = If the area of a square does not increase four times, then its side is not doubled.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5965,
"subject": "General Science",
"question": "Contrapositive of the statement\n
‘If two numbers are not equal, then their squares are not equal’, is :",
"options": [
{
"text": "If the squares of two numbers are equal, then the numbers are equal."
},
{
"text": "If the squares of two numbers are equal, then the numbers are not equal."
},
{
"text": "If the squares of two numbers are not equal, then the numbers are not equal.\n"
},
{
"text": "If the squares of two numbers are not equal, then the numbers are equal."
}
],
"answer": "If the squares of two numbers are equal, then the numbers are equal.",
"solution": "**Answer:** If the squares of two numbers are equal, then the numbers are equal.\n\nLet,\n
p : two numbers are not equal\n
q : squares of two numbers are not equal\n
Contrapositive of p $$ \\to $$ q is $$ \\sim $$q $$ \\to $$ $$ \\sim $$p.\n
$$ \\therefore $$ $$ \\sim $$q $$ \\to $$ $$ \\sim $$p means \"If the squares of two numbers are equal, then the numbers are equal\".",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5966,
"subject": "General Science",
"question": "Consider the following two statements : \n
Statement p :\n The value of sin 120o can be derived by taking $$\\theta = {240^o}$$ in the equation \n 2sin$${\\theta \\over 2} = \\sqrt {1 + \\sin \\theta } - \\sqrt {1 - \\sin \\theta } $$\n
Statement q :\n The angles A, B, C and D of any quadrilateral ABCD satisfy the equation \n cos$$\\left( {{1 \\over 2}\\left( {A + C} \\right)} \\right) + \\cos \\left( {{1 \\over 2}\\left( {B + D} \\right)} \\right) = 0$$\n
Then the truth values of p and q are respectively : ",
"options": [
{
"text": "F, T"
},
{
"text": "T, F"
},
{
"text": "T, T"
},
{
"text": "F, F"
}
],
"answer": "F, T",
"solution": "**Answer:** F, T\n\nStatement p :\n sin 120o = cos 30o = $${{\\sqrt 3 } \\over 2}$$ $$ \\Rightarrow $$ 2 sin 120o = $$\\sqrt 3 $$\n
Therefore, statement p is false and statement q is true.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5967,
"subject": "General Science",
"question": "Consider the statement : \"P(n) : n2 – n + 41 is prime\". Then which one of the following is true ?",
"options": [
{
"text": "P(5) is false but P(3) is true"
},
{
"text": "Both P(3) and P(5) are true"
},
{
"text": "P(3) is false but P(5) is true"
},
{
"text": "Both P(3) and P(5) are false"
}
],
"answer": "Both P(3) and P(5) are true",
"solution": "**Answer:** Both P(3) and P(5) are true\n\nP(n) : n2 $$-$$ n + 41 is prime\n
P(5) = 61 which is prime\n
P(3) = 47 which is also prime",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 5968,
"subject": "General Science",
"question": "Consider the following three statements :\n
P : 5 is a prime number\n
Q : 7 is a factor of 192\n
R : L.C.M. of 5 and 7 is 35\n
Then the truth value of which one of the following statements is true ?",
"options": [
{
"text": "(P $$ \\wedge $$ Q) $$ \\vee $$ ($$ \\sim $$ R)"
},
{
"text": "P $$ \\vee $$ ($$ \\sim $$ Q $$ \\wedge $$ R)"
},
{
"text": "(~ P) $$ \\wedge $$ ($$ \\sim $$ Q $$ \\wedge $$ R)"
},
{
"text": "($$ \\sim $$ P) $$ \\vee $$ (Q $$ \\wedge $$ R)"
}
],
"answer": "P $$ \\vee $$ ($$ \\sim $$ Q $$ \\wedge $$ R)",
"solution": "**Answer:** P $$ \\vee $$ ($$ \\sim $$ Q $$ \\wedge $$ R)\n\nIt is obvious",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 5969,
"subject": "General Science",
"question": "Contrapositive of the statement \" If two numbers are not equal, then their squares are not equal.\" is :",
"options": [
{
"text": "If the squares of two numbers are equal, then the numbers are not equal"
},
{
"text": "If the squares of two numbers are equal, then the numbers are equal "
},
{
"text": "If the squares of two numbers are not equal, then the numbers are equal"
},
{
"text": "If the squares of two numbers are not equal, then the numbers are not equal"
}
],
"answer": "If the squares of two numbers are equal, then the numbers are equal ",
"solution": "**Answer:** If the squares of two numbers are equal, then the numbers are equal \n\nLet,\n
p : two numbers are not equal\n
q : squares of two numbers are not equal\n
Contrapositive of p $$ \\to $$ q is $$ \\sim $$q $$ \\to $$ $$ \\sim $$p.\n
$$ \\therefore $$ $$ \\sim $$q $$ \\to $$ $$ \\sim $$p means \"If the squares of two numbers are equal, then the numbers are equal\".",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5970,
"subject": "General Science",
"question": "The contrapositive of the statement \"If you are\nborn in India, then you are a citizen of India\", is :\n",
"options": [
{
"text": "If you are not a citizen of India, then you are\nnot born in India."
},
{
"text": "If you are born in India, then you are not a\ncitizen of India."
},
{
"text": "If you are a citizen of India, then you are born\nin India."
},
{
"text": "If you are not born in India, then you are not\na citizen of India"
}
],
"answer": "If you are not a citizen of India, then you are\nnot born in India.",
"solution": "**Answer:** If you are not a citizen of India, then you are\nnot born in India.\n\nLet p = you are born in India.\n
q = you are a citizen of India.\n
$$ \\therefore $$ $$ \\sim $$ p = you are not born in India.\n
$$ \\sim $$ q = you are not a citizen of India.\n
We know Contrapositive of p $$ \\to $$ q is ~q $$ \\to $$ ~p\n
So contrapositive of statement will be :\n
“If you are not a citizen of India, then you are\nnot born in India.”\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5971,
"subject": "General Science",
"question": "Negation of the statement :\n
$$\\sqrt 5 $$ is an integer or 5 is an irrational is :",
"options": [
{
"text": "$$\\sqrt 5 $$ is not an integer and 5 is not irrational."
},
{
"text": "$$\\sqrt 5 $$ is irrational or 5 is an integer."
},
{
"text": "$$\\sqrt 5 $$ is an integer and 5 is irrational."
},
{
"text": "$$\\sqrt 5 $$ is not an integer or 5 is not irrational."
}
],
"answer": "$$\\sqrt 5 $$ is not an integer and 5 is not irrational.",
"solution": "**Answer:** $$\\sqrt 5 $$ is not an integer and 5 is not irrational.\n\np = $$\\sqrt 5 $$ is an integer.\n
= $$\\sqrt 5 $$ is not an integer and 5 is not irrational.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5972,
"subject": "General Science",
"question": "Consider the statement : ‘‘For an integer n, if n3 – 1 is even, then n is odd.’’ The contrapositive statement of this statement is :\n",
"options": [
{
"text": "For an integer n, if n is even, then n3 – 1 is even."
},
{
"text": "For an integer n, if n3 – 1 is not even, then n is not odd."
},
{
"text": "For an integer n, if n is odd, then n3 – 1 is even."
},
{
"text": "For an integer n, if n is even, then n3 – 1 is odd."
}
],
"answer": "For an integer n, if n is even, then n3 – 1 is odd.",
"solution": "**Answer:** For an integer n, if n is even, then n3 – 1 is odd.\n\nLet,\np : n3–1 is even, \n q : n is odd\n
Contrapositive of p $$ \\to $$ q = $$ \\sim $$q $$ \\to $$ $$ \\sim $$p\n
$$ \\Rightarrow $$ If n is not odd then n3 – 1 is not even.\n
$$ \\Rightarrow $$ For an integer n, if n is even, then n3 – 1 is odd.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5973,
"subject": "General Science",
"question": "Contrapositive of the statement : \n‘If a function f is differentiable at a, then it is also continuous at a’, is:",
"options": [
{
"text": "If a function f is continuous at a, then it is not differentiable at a."
},
{
"text": "If a function f is not continuous at a, then it is differentiable at a."
},
{
"text": "If a function f is not continuous at a, then it is not differentiable at a.\n"
},
{
"text": "If a function f is continuous at a, then it is differentiable at a."
}
],
"answer": "If a function f is not continuous at a, then it is not differentiable at a.\n",
"solution": "**Answer:** If a function f is not continuous at a, then it is not differentiable at a.\n\n\np = function is differentiable at a\n
q = function is continuous at a\n
Contrapositive of statements p $$ \\to $$ q is\n
$$ \\sim $$q $$ \\to $$ $$ \\sim $$p\n
$$ \\therefore $$ Contrapositive statement is :\n
If a function f is not continuous at a, then it is not differentiable at a.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5974,
"subject": "General Science",
"question": "The contrapositive of the statement \"If I reach the\nstation in time, then I will catch the train\" is :",
"options": [
{
"text": "If I will catch the train, then I reach the station\nin time."
},
{
"text": "If I do not reach the station in time, then I will\nnot catch the train."
},
{
"text": "If I will not catch the train, then I do not reach\nthe station in time."
},
{
"text": "If I do not reach the station in time, then I will\ncatch the train."
}
],
"answer": "If I will not catch the train, then I do not reach\nthe station in time.",
"solution": "**Answer:** If I will not catch the train, then I do not reach\nthe station in time.\n\nLet p denotes statement\n
p : I reach the station in time.\n
q : I will catch the train.\n
Contrapositive of p $$ \\to $$ q\nis $$ \\sim $$q $$ \\to $$ $$ \\sim $$p\n
$$ \\sim $$q $$ \\to $$ $$ \\sim $$p : If I will not catch the train, then I do not reach the station in time.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5975,
"subject": "General Science",
"question": "The contrapositive of the statement \"If you will work, you will earn money\" is :",
"options": [
{
"text": "If you will not earn money, you will not work"
},
{
"text": "If you will earn money, you will work"
},
{
"text": "You will earn money, if you will not work"
},
{
"text": "To earn money, you need to work"
}
],
"answer": "If you will not earn money, you will not work",
"solution": "**Answer:** If you will not earn money, you will not work\n\nContrapositive of p $$ \\to $$ q is ~q $$ \\to $$ ~p\n
p : you will work\n
q : you will earn money\n
~q : you will not earn money\n
~p : you will not work\n
$$ \\therefore $$ ~q $$ \\to $$ ~p : If you will not earn money, you will not work",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5976,
"subject": "General Science",
"question": "Consider the following three statements :
(A) If 3 + 3 = 7 then 4 + 3 = 8
(B) If 5 + 3 = 8 then earth is flat.
(C) If both (A) and (B) are true then 5 + 6 = 17.
Then, which of the following statements is correct?",
"options": [
{
"text": "(A) is false, but (B) and (C) re true"
},
{
"text": "(A) and (C) are true while (B) is false"
},
{
"text": "(A) is true while (B) and (C) are false"
},
{
"text": "(A) and (B) are false while (C) is true"
}
],
"answer": "(A) and (C) are true while (B) is false",
"solution": "**Answer:** (A) and (C) are true while (B) is false\n\nTruth Table
\n
\n\n
\n
P
\n
q
\n
P$$ \\to $$q
\n
\n\n\n
\n
T
\n
T
\n
T
\n
\n
\n
T
\n
F
\n
F
\n
\n
\n
F
\n
T
\n
T
\n
\n
\n
F
\n
F
\n
T
\n
\n\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5977,
"subject": "General Science",
"question": "Which of the following is the negation of the statement \"for all M > 0, there exists x$$\\in$$S such that x $$\\ge$$ M\" ?",
"options": [
{
"text": "there exists M > 0, such that x < M for all x$$\\in$$S"
},
{
"text": "there exists M > 0, there exists x$$\\in$$S such that x $$\\ge$$ M"
},
{
"text": "there exists M > 0, there exists x$$\\in$$S such that x < M"
},
{
"text": "there exists M > 0, such that x $$\\ge$$ M for all x$$\\in$$S"
}
],
"answer": "there exists M > 0, such that x < M for all x$$\\in$$S",
"solution": "**Answer:** there exists M > 0, such that x < M for all x$$\\in$$S\n\nP : for all M > 0, there exists x$$\\in$$S such that x $$\\ge$$ M.
$$ \\sim $$ P : there exists M > 0, for all x$$\\in$$S
Such that x < M
Negation of 'there exists' is 'for all'.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5978,
"subject": "General Science",
"question": "Consider the statement \"The match will be played only if the weather is good and ground is not wet\". Select the correct negation from the following :",
"options": [
{
"text": "The match will not be played and weather is not good and ground is wet."
},
{
"text": "If the match will not be played, then either weather is not good or ground is wet."
},
{
"text": "The match will be played and weather is not good or ground is wet."
},
{
"text": "The match will not be played or weather is good and ground is not wet."
}
],
"answer": "The match will be played and weather is not good or ground is wet.",
"solution": "**Answer:** The match will be played and weather is not good or ground is wet.\n\np : weather is good
$$\\equiv$$ weather is not good or ground is wet",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5979,
"subject": "General Science",
"question": "
Consider the following statements:
\n
A : Rishi is a judge.
\n
B : Rishi is honest.
\n
C : Rishi is not arrogant.
\n
The negation of the statement \"if Rishi is a judge and he is not arrogant, then he is honest\" is
",
"options": [
{
"text": "B $$\\to$$ (A $$\\vee$$ C)"
},
{
"text": "($$\\sim$$B) $$\\wedge$$ (A $$\\wedge$$ C)"
},
{
"text": "B $$\\to$$ (($$\\sim$$A) $$\\vee$$ ($$\\sim$$C))"
},
{
"text": "B $$\\to$$ (A $$\\wedge$$ C)"
}
],
"answer": "($$\\sim$$B) $$\\wedge$$ (A $$\\wedge$$ C)",
"solution": "**Answer:** ($$\\sim$$B) $$\\wedge$$ (A $$\\wedge$$ C)\n\n
$$\\because$$ Given statement is
\n
(A $$\\wedge$$ C) $$\\to$$ B
\n
Then its negation is
\n
$$\\sim$$ {(A $$\\wedge$$ C) $$\\to$$ B}
\n
or $$\\sim$$ {$$\\sim$$ (A $$\\wedge$$ C) $$\\vee$$ B}
\n
$$\\therefore$$ (A $$\\wedge$$ C) $$\\wedge$$ ($$\\sim$$ B)
The number of ordered triplets of the truth values of $$p, q$$ and $$r$$ such that the truth value of the statement $$(p \\vee q) \\wedge(p \\vee r) \\Rightarrow(q \\vee r)$$ is True, is equal to ___________.
Let A be a $$3 \\times 3$$ matrix and $$\\operatorname{det}(A)=2$$. If $$n=\\operatorname{det}(\\underbrace{\\operatorname{adj}(\\operatorname{adj}(\\ldots . .(\\operatorname{adj} A))}_{2024-\\text { times }}))$$, then the remainder when $$n$$ is divided by 9 is equal to __________.
Let $$A$$ be a non-singular matrix of order 3. If $$\\operatorname{det}(3 \\operatorname{adj}(2 \\operatorname{adj}((\\operatorname{det} A) A)))=3^{-13} \\cdot 2^{-10}$$ and $$\\operatorname{det}(3\\operatorname{adj}(2 \\mathrm{A}))=2^{\\mathrm{m}} \\cdot 3^{\\mathrm{n}}$$, then $$|3 \\mathrm{~m}+2 \\mathrm{n}|$$ is equal to _________.
Let $$A=\\left[\\begin{array}{ll}1 & 2 \\\\ 0 & 1\\end{array}\\right]$$ and $$B=I+\\operatorname{adj}(A)+(\\operatorname{adj} A)^2+\\ldots+(\\operatorname{adj} A)^{10}$$.\nThen, the sum of all the elements of the matrix $$B$$ is:
Let $$\\alpha \\beta \\neq 0$$ and $$A=\\left[\\begin{array}{rrr}\\beta & \\alpha & 3 \\\\ \\alpha & \\alpha & \\beta \\\\ -\\beta & \\alpha & 2 \\alpha\\end{array}\\right]$$. If $$B=\\left[\\begin{array}{rrr}3 \\alpha & -9 & 3 \\alpha \\\\ -\\alpha & 7 & -2 \\alpha \\\\ -2 \\alpha & 5 & -2 \\beta\\end{array}\\right]$$ is the matrix of cofactors of the elements of $$A$$, then $$\\operatorname{det}(A B)$$ is equal to :
If $$A$$ is a square matrix of order 3 such that $$\\operatorname{det}(A)=3$$ and $$\\operatorname{det}\\left(\\operatorname{adj}\\left(-4 \\operatorname{adj}\\left(-3 \\operatorname{adj}\\left(3 \\operatorname{adj}\\left((2 \\mathrm{~A})^{-1}\\right)\\right)\\right)\\right)\\right)=2^{\\mathrm{m}} 3^{\\mathrm{n}}$$, then $$\\mathrm{m}+2 \\mathrm{n}$$ is equal to :
$$=$$ $$-$$ 25",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6006,
"subject": "General Science",
"question": "The number of distinct real roots of the equation, \n
$$ \\therefore $$ detB $$ \\in $$ (1, 2)\n",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6023,
"subject": "General Science",
"question": "Let m and M be respectively the minimum and maximum values of\n
Then the sum of the maximum values of $$\\alpha$$ and $$\\beta$$, such that $$\\mathrm{p}^{\\alpha}$$ and $$(\\mathrm{p}+2)^{\\beta}$$ divide $$\\Delta$$, is __________.
Let $$\\mathrm{A_1,A_2,A_3}$$ be the three A.P. with the same common difference d and having their first terms as $$\\mathrm{A,A+1,A+2}$$, respectively. Let a, b, c be the $$\\mathrm{7^{th},9^{th},17^{th}}$$ terms of $$\\mathrm{A_1,A_2,A_3}$$, respective such that $$\\left| {\\matrix{\n a & 7 & 1 \\cr \n {2b} & {17} & 1 \\cr \n c & {17} & 1 \\cr \n\n } } \\right| + 70 = 0$$.
\n
If $$a=29$$, then the sum of first 20 terms of an AP whose first term is $$c-a-b$$ and common difference is $$\\frac{d}{12}$$, is equal to ___________.
Let $$\\mathrm{D}_{\\mathrm{k}}=\\left|\\begin{array}{ccc}1 & 2 k & 2 k-1 \\\\\nn & n^{2}+n+2 & n^{2} \\\\\nn & n^{2}+n & n^{2}+n+2\\end{array}\\right|$$. If $$\\sum_\\limits{k=1}^{n} \\mathrm{D}_{\\mathrm{k}}=96$$, then $$n$$ is equal to _____________.
$$\\left|\\begin{array}{ccc}x+1 & x & x \\\\ x & x+\\lambda & x \\\\ x & x & x+\\lambda^{2}\\end{array}\\right|=\\frac{9}{8}(103 x+81)$$, then $$\\lambda, \\frac{\\lambda}{3}$$ are the roots of the equation :
",
"options": [
{
"text": "$$4 x^{2}+24 x-27=0$$"
},
{
"text": "$$4 x^{2}-24 x+27=0$$"
},
{
"text": "$$4 x^{2}-24 x-27=0$$"
},
{
"text": "$$4 x^{2}+24 x+27=0$$"
}
],
"answer": "$$4 x^{2}-24 x+27=0$$",
"solution": "**Answer:** $$4 x^{2}-24 x+27=0$$\n\n$$\\left|\\begin{array}{ccc}x+1 & x & x \\\\ x & x+\\lambda & x \\\\ x & x & x+\\lambda^{2}\\end{array}\\right|=\\frac{9}{8}(103 x+81)$$\n
The values of $$\\alpha$$, for which $$\\left|\\begin{array}{ccc}1 & \\frac{3}{2} & \\alpha+\\frac{3}{2} \\\\ 1 & \\frac{1}{3} & \\alpha+\\frac{1}{3} \\\\ 2 \\alpha+3 & 3 \\alpha+1 & 0\\end{array}\\right|=0$$, lie in the interval
$$\\text { Let } A=\\left[\\begin{array}{lll}\n1 & 0 & 0 \\\\\n0 & \\alpha & \\beta \\\\\n0 & \\beta & \\alpha\n\\end{array}\\right] \\text { and }|2 \\mathrm{~A}|^3=2^{21} \\text { where } \\alpha, \\beta \\in Z \\text {, Then a value of } \\alpha \\text { is }$$
For $$\\alpha, \\beta \\in \\mathbb{R}$$ and a natural number $$n$$, let $$A_r=\\left|\\begin{array}{ccc}r & 1 & \\frac{n^2}{2}+\\alpha \\\\ 2 r & 2 & n^2-\\beta \\\\ 3 r-2 & 3 & \\frac{n(3 n-1)}{2}\\end{array}\\right|$$. Then $$2 A_{10}-A_8$$ is
$$ = I.I = {I^2} = I$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6049,
"subject": "General Science",
"question": "Let A be a 3 $$ \\times $$ 3 matrix such that A2 $$-$$ 5A + 7I = 0\n
Statement - I : \n
A$$-$$1 = $${1 \\over 7}$$ (5I $$-$$ A).\n
Statement - II :\n
The polynomial A3 $$-$$ 2A2 $$-$$ 3A + I can be reduced to 5(A $$-$$ 4I).\n
Then : ",
"options": [
{
"text": "Statement-I is true, but Statement-II is false."
},
{
"text": "Statement-I is false, but Statement-II is true."
},
{
"text": "Both the statements are true."
},
{
"text": "Both the statements are false"
}
],
"answer": "Both the statements are true.",
"solution": "**Answer:** Both the statements are true.\n\nGiven,\n
So, statement 2 is also correct.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6050,
"subject": "General Science",
"question": "Suppose A is any 3$$ \\times $$ 3 non-singular matrix and ( A $$-$$ 3I) (A $$-$$ 5I) = O where I = I3 and O = O3. If $$\\alpha $$A + $$\\beta $$A-1 = 4I, then $$\\alpha $$ + $$\\beta $$ is equal to :",
"options": [
{
"text": "8"
},
{
"text": "7"
},
{
"text": "13"
},
{
"text": "12"
}
],
"answer": "8",
"solution": "**Answer:** 8\n\nGiven,\n
The number of matrices $$A=\\left(\\begin{array}{ll}a & b \\\\ c & d\\end{array}\\right)$$, where $$a, b, c, d \\in\\{-1,0,1,2,3, \\ldots \\ldots, 10\\}$$, such that $$A=A^{-1}$$, is ___________.
Let $$\\mathrm{A}$$ be a $$2 \\times 2$$ matrix with real entries such that $$\\mathrm{A}'=\\alpha \\mathrm{A}+\\mathrm{I}$$, where $$\\alpha \\in \\mathbb{R}-\\{-1,1\\}$$. If $$\\operatorname{det}\\left(A^{2}-A\\right)=4$$, then the sum of all possible values of $$\\alpha$$ is equal to :
If $2(1-\\alpha)^2=\\alpha$, then $2 \\alpha^2-5 \\alpha+2=0$\n
Sum of value of $\\alpha=\\frac{5}{2} \\quad\\left[\\because\\right.$ Sum of zero $\\left.=\\frac{\\text {-Coefficient of } x}{\\text { Coefficient of } x^2}\\right]$\n
If $2(1-\\alpha)^2=-\\alpha$, then $2 \\alpha^2-3 \\alpha+2=0$\n
$\\therefore$ No real value of $\\alpha$\n
Hence, sum of all values of $\\alpha=\\frac{5}{2}$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6060,
"subject": "General Science",
"question": "
If $$A=\\left[\\begin{array}{cc}1 & 5 \\\\ \\lambda & 10\\end{array}\\right], \\mathrm{A}^{-1}=\\alpha \\mathrm{A}+\\beta \\mathrm{I}$$ and $$\\alpha+\\beta=-2$$, then $$4 \\alpha^{2}+\\beta^{2}+\\lambda^{2}$$ is equal to :
$$\n\\begin{aligned}\n& \\text { So, } 5 \\lambda-10=5 \\Rightarrow \\lambda=3 \\\\\\\\\n& \\therefore 4 \\alpha^2+\\beta^2+\\lambda^2 \\\\\\\\\n& =4\\left(\\frac{1}{25}\\right)+\\left(\\frac{121}{25}\\right)+9 \\\\\\\\\n& =\\frac{125}{25}+9=14\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6061,
"subject": "General Science",
"question": "Consider the matrix $f(x)=\\left[\\begin{array}{ccc}\\cos x & -\\sin x & 0 \\\\ \\sin x & \\cos x & 0 \\\\ 0 & 0 & 1\\end{array}\\right]$.\n
Given below are two statements :\n
Statement I : $ f(-x)$ is the inverse of the matrix $f(x)$.\n
Statement II : $f(x) f(y)=f(x+y)$.\n
In the light of the above statements, choose the correct answer from the options given below :",
"options": [
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Statement I is true but Statement II is false"
}
],
"answer": "Both Statement I and Statement II are true",
"solution": "**Answer:** Both Statement I and Statement II are true\n\n
$$\\begin{aligned}\n& f(-x)=\\left[\\begin{array}{ccc}\n\\cos x & \\sin x & 0 \\\\\n-\\sin x & \\cos x & 0 \\\\\n0 & 0 & 1\n\\end{array}\\right] \\\\\n& f(x) \\cdot f(-x)=\\left[\\begin{array}{lll}\n1 & 0 & 0 \\\\\n0 & 1 & 0 \\\\\n0 & 0 & 1\n\\end{array}\\right]=I\n\\end{aligned}$$
Let $$B=\\left[\\begin{array}{ll}1 & 3 \\\\ 1 & 5\\end{array}\\right]$$ and $$A$$ be a $$2 \\times 2$$ matrix such that $$A B^{-1}=A^{-1}$$. If $$B C B^{-1}=A$$ and $$C^4+\\alpha C^2+\\beta I=O$$, then $$2 \\beta-\\alpha$$ is equal to
$$\\begin{aligned}\n\\Rightarrow C & =B^{-1} A B \\\\\n\\Rightarrow C^4 & =\\left(B^{-1} A B\\right)\\left(B^{-1} A B\\right)\\left(B^{-1} A B\\right)\\left(B^{-1} A B\\right) \\\\\n& =B^{-1} A^4 B \\\\\n& =B^{-1} B^2 B \\\\\n\\Rightarrow \\quad C^4 & =B^2\n\\end{aligned}$$
\n
Also, $$C^2=\\left(B^{-1} A B\\right)\\left(B^{-1} A B\\right)$$
\n
$$\\begin{aligned}\n& =B^{-1} A^2 B \\\\\n& =B^{-1} B B\n\\end{aligned}$$
Let $$A$$ be a $$2 \\times 2$$ symmetric matrix such that $$A\\left[\\begin{array}{l}1 \\\\ 1\\end{array}\\right]=\\left[\\begin{array}{l}3 \\\\ 7\\end{array}\\right]$$ and the determinant of $$A$$ be 1 . If $$A^{-1}=\\alpha A+\\beta I$$, where $$I$$ is an identity matrix of order $$2 \\times 2$$, then $$\\alpha+\\beta$$ equals _________.
$$ = 2021 - 1 = 2020$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6070,
"subject": "General Science",
"question": "Let $$A = \\left[ {\\matrix{\n 0 & 1 & 0 \\cr \n 1 & 0 & 0 \\cr \n 0 & 0 & 1 \\cr \n\n } } \\right]$$. Then the number of 3 $$\\times$$ 3 matrices B with entries from the set {1, 2, 3, 4, 5} and satisfying AB = BA is ____________.",
"options": [],
"answer": "3125",
"solution": "**Answer:** 3125\n\nLet matrix $$B = \\left[ {\\matrix{\n a & b & c \\cr \n d & e & f \\cr \n g & h & i \\cr \n\n } } \\right]$$
$$\\because$$ $$AB = BA$$
$$\\left[ {\\matrix{\n 0 & 1 & 0 \\cr \n 1 & 0 & 0 \\cr \n 0 & 0 & 1 \\cr \n\n } } \\right]\\left[ {\\matrix{\n a & b & c \\cr \n d & e & f \\cr \n g & h & i \\cr \n\n } } \\right] = \\left[ {\\matrix{\n a & b & c \\cr \n d & e & f \\cr \n g & h & i \\cr \n\n } } \\right]\\left[ {\\matrix{\n 0 & 1 & 0 \\cr \n 1 & 0 & 0 \\cr \n 0 & 0 & 1 \\cr \n\n } } \\right]$$
$$\\left[ {\\matrix{\n d & e & f \\cr \n a & b & c \\cr \n g & h & i \\cr \n\n } } \\right] = \\left[ {\\matrix{\n b & a & c \\cr \n e & d & f \\cr \n h & g & i \\cr \n\n } } \\right]$$
$$ \\Rightarrow d = b,e = a,f = c,g = h$$
$$\\therefore$$ Matrix $$B = \\left[ {\\matrix{\n a & b & c \\cr \n b & a & c \\cr \n g & g & i \\cr \n\n } } \\right]$$
Let $$A = \\left( {\\matrix{\n {1 + i} & 1 \\cr \n { - i} & 0 \\cr \n\n } } \\right)$$ where $$i = \\sqrt { - 1} $$. Then, the number of elements in the set { n $$\\in$$ {1, 2, ......, 100} : An = A } is ____________.
$$\n\\text { Let } A=\\left[\\begin{array}{l}\n1 \\\\\n1 \\\\\n1\n\\end{array}\\right] \\text { and } B=\\left[\\begin{array}{ccc}\n9^{2} & -10^{2} & 11^{2} \\\\\n12^{2} & 13^{2} & -14^{2} \\\\\n-15^{2} & 16^{2} & 17^{2}\n\\end{array}\\right] \\text {, then the value of } A^{\\prime} B A \\text { is: }\n$$
Let $$A=\\left(\\begin{array}{cc}1 & 2 \\\\ -2 & -5\\end{array}\\right)$$. Let $$\\alpha, \\beta \\in \\mathbb{R}$$ be such that $$\\alpha A^{2}+\\beta A=2 I$$. Then $$\\alpha+\\beta$$ is equal to
Let $$A=\\left[\\begin{array}{cc}1 & -1 \\\\ 2 & \\alpha\\end{array}\\right]$$ and $$B=\\left[\\begin{array}{cc}\\beta & 1 \\\\ 1 & 0\\end{array}\\right], \\alpha, \\beta \\in \\mathbf{R}$$. Let $$\\alpha_{1}$$ be the value of $$\\alpha$$ which satisfies $$(\\mathrm{A}+\\mathrm{B})^{2}=\\mathrm{A}^{2}+\\left[\\begin{array}{ll}2 & 2 \\\\ 2 & 2\\end{array}\\right]$$ and $$\\alpha_{2}$$ be the value of $$\\alpha$$ which satisfies $$(\\mathrm{A}+\\mathrm{B})^{2}=\\mathrm{B}^{2}$$. Then $$\\left|\\alpha_{1}-\\alpha_{2}\\right|$$ is equal to ___________.
Let $$A=\\left[\\begin{array}{cc}1 & \\frac{1}{51} \\\\ 0 & 1\\end{array}\\right]$$. If $$\\mathrm{B}=\\left[\\begin{array}{cc}1 & 2 \\\\ -1 & -1\\end{array}\\right] A\\left[\\begin{array}{cc}-1 & -2 \\\\ 1 & 1\\end{array}\\right]$$, then the sum of all the elements of the matrix $$\\sum_\\limits{n=1}^{50} B^{n}$$ is equal to
$$ \\therefore $$ Sum of the elements = 100",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6079,
"subject": "General Science",
"question": "Let $A=I_2-2 M M^T$, where $M$ is a real matrix of order $2 \\times 1$ such that the relation $M^T M=I_1$ holds. If $\\lambda$ is a real number such that the relation $A X=\\lambda X$ holds for some non-zero real matrix $X$ of order $2 \\times 1$, then the sum of squares of all possible values of $\\lambda$ is equal to __________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$\\begin{aligned} & A=I_2-2 M^T \\\\\\\\ & A^2=\\left(I_2-2 M M^T\\right)\\left(I_2-2 M^T\\right) \\\\\\\\ & =I_2-2 M^T-2 M M^T+4 M^T M^T \\\\\\\\ & =I_2-4 M M^T+4 M M^T \\\\\\\\ & =I_2\\end{aligned}$\n
If $\\alpha=|B|$ and $\\beta$ is the sum of all the diagonal elements of $B$, then $\\alpha^3+\\beta^3$ is equal to ____________.",
"options": [],
"answer": "28",
"solution": "**Answer:** 28\n\n
$$\\therefore$$ There can be infinitely many $$B's$$ \n
for which $$AB=BA$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6082,
"subject": "General Science",
"question": "If $$A$$ and $$B$$ are square matrices of size $$n\\, \\times \\,n$$ such that \n $${A^2} - {B^2} = \\left( {A - B} \\right)\\left( {A + B} \\right),$$ then which of the following will be always true?",
"options": [
{
"text": "$$A=B$$ "
},
{
"text": "$$AB=BA$$ "
},
{
"text": "either of $$A$$ or $$B$$ is a zero matrix"
},
{
"text": "either of $$A$$ or $$B$$ is identity matrix"
}
],
"answer": "$$AB=BA$$ ",
"solution": "**Answer:** $$AB=BA$$ \n\n$${A^2} - {B^2} = \\left( {A - B} \\right)\\left( {A + B} \\right)$$ \n
$${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$$ \n
$$ \\Rightarrow AB = BA$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6083,
"subject": "General Science",
"question": "Let P = $$\\left[ {\\matrix{\n 1 & 0 & 0 \\cr \n 3 & 1 & 0 \\cr \n 9 & 3 & 1 \\cr \n\n } } \\right]$$ and Q = [qij] be two 3 $$ \\times $$ 3 matrices such that Q – P5 = I3. \n
If $$a_1^2$$ + $$a_2^2$$ = $${2 \\over 3}$$(b$$_1^2$$ + b$$_2^2$$) and (k2 + 1) b$$_2^2$$ $$\\ne$$ $$-$$2b1b2, then the value of k is __________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$XB = A$$\n
$$ \\Rightarrow K = {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6094,
"subject": "General Science",
"question": "The number of elements in the set $$\\left\\{ {A = \\left( {\\matrix{\n a & b \\cr \n 0 & d \\cr \n\n } } \\right):a,b,d \\in \\{ - 1,0,1\\} \\,and\\,{{(I - A)}^3} = I - {A^3}} \\right\\}$$, where I is 2 $$\\times$$ 2 identity matrix, is :",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\n$${(I - A)^3} = {I^3} - {A^3} - 3A(I - A) = I - {A^3}$$
Let $$A = [{a_{ij}}]$$ be a square matrix of order 3 such that $${a_{ij}} = {2^{j - i}}$$, for all i, j = 1, 2, 3. Then, the matrix A2 + A3 + ...... + A10 is equal to :
Let $$M = \\left[ {\\matrix{\n 0 & { - \\alpha } \\cr \n \\alpha & 0 \\cr \n\n } } \\right]$$, where $$\\alpha$$ is a non-zero real number an $$N = \\sum\\limits_{k = 1}^{49} {{M^{2k}}} $$. If $$(I - {M^2})N = - 2I$$, then the positive integral value of $$\\alpha$$ is ____________.
Let $$A = \\left( {\\matrix{\n 2 & { - 2} \\cr \n 1 & { - 1} \\cr \n\n } } \\right)$$ and $$B = \\left( {\\matrix{\n { - 1} & 2 \\cr \n { - 1} & 2 \\cr \n\n } } \\right)$$. Then the number of elements in the set {(n, m) : n, m $$\\in$$ {1, 2, .........., 10} and nAn + mBm = I} is ____________.
Let $$S = \\left\\{ {\\left( {\\matrix{\n { - 1} & a \\cr \n 0 & b \\cr \n\n } } \\right);a,b \\in \\{ 1,2,3,....100\\} } \\right\\}$$ and let $${T_n} = \\{ A \\in S:{A^{n(n + 1)}} = I\\} $$. Then the number of elements in $$\\bigcap\\limits_{n = 1}^{100} {{T_n}} $$ is ___________.
\n$\\therefore$ In this case $\\mathrm{A}^2$ will become identity matrix and a can take any value from 1 to 100
\n$\\therefore$ Total number of common element will be 100 .",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6099,
"subject": "General Science",
"question": "
Let $$A=\\left(\\begin{array}{rrr}2 & -1 & -1 \\\\ 1 & 0 & -1 \\\\ 1 & -1 & 0\\end{array}\\right)$$ and $$B=A-I$$. If $$\\omega=\\frac{\\sqrt{3} i-1}{2}$$, then the number of elements in the $$\\operatorname{set}\\left\\{n \\in\\{1,2, \\ldots, 100\\}: A^{n}+(\\omega B)^{n}=A+B\\right\\}$$ is equal to ____________.
\n$$\n\\therefore A^{n}+(\\omega B)^{n}=A+(\\omega B)^{n} \\quad \\text { for } n \\in \\mathrm{N}\n$$\n
\nFor $\\omega^{n}$ to be unity $n$ shall be multiple of 3 and for $B^{n}$ to be $B . n$ shell be $3,5,7, \\ldots 99$\n
\n$\\therefore n=\\{3,9,15, \\ldots . .99\\}$\n
\nNumber of elements $=17$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6100,
"subject": "General Science",
"question": "
Which of the following matrices can NOT be obtained from the matrix $$\\left[\\begin{array}{cc}-1 & 2 \\\\ 1 & -1\\end{array}\\right]$$ by a single elementary row operation ?
Let $$A=\\left[\\begin{array}{lll}0 & 1 & 2 \\\\ a & 0 & 3 \\\\ 1 & c & 0\\end{array}\\right]$$, where $$a, c \\in \\mathbb{R}$$. If $$A^{3}=A$$ and the positive value of $$a$$ belongs to the interval $$(n-1, n]$$, where $$n \\in \\mathbb{N}$$, then $$n$$ is equal to ___________.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$\n\\text { We have, } A=\\left[\\begin{array}{lll}\n0 & 1 & 2 \\\\\na & 0 & 3 \\\\\n1 & c & 0\n\\end{array}\\right] \\text {, where } a, c \\in R\n$$\n
$$\n\\begin{aligned}\nA^2 & =\\left[\\begin{array}{lll}\n0 & 1 & 2 \\\\\na & 0 & 3 \\\\\n1 & c & 0\n\\end{array}\\right]\\left[\\begin{array}{lll}\n0 & 1 & 2 \\\\\na & 0 & 3 \\\\\n1 & c & 0\n\\end{array}\\right] \\\\\\\\\n& =\\left[\\begin{array}{ccc}\na+2 & 2 c & 3 \\\\\n3 & a+3 c & 2 a \\\\\na c & 1 & 2+3 c\n\\end{array}\\right]\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\nA^3 & =\\left[\\begin{array}{ccc}\na+2 & 2 c & 3 \\\\\n3 & a+3 c & 2 a \\\\\na c & 1 & 2+3 c\n\\end{array}\\right]\\left[\\begin{array}{ccc}\n0 & 1 & 2 \\\\\na & 0 & 3 \\\\\n1 & c & 0\n\\end{array}\\right] \\\\\\\\\n& =\\left[\\begin{array}{ccc}\n2 a c+3 & a+2+3 c & 2 a+4+6 c \\\\\na(a+3 c)+2 a & 3+2 a c & 6+3 a+9 c \\\\\na+2+3 c & a c+2 c+3 c^2 & 2 a c+3\n\\end{array}\\right]\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\n& A^3 =A [Given]\\\\\\\\\n& 2 a c+3= 0 \\text { and } a+2+3 c=1 \\\\\\\\\n& a^2+2 a+3 a c =a \\\\\\\\\n& \\Rightarrow a^2 +a+3\\left(-\\frac{3}{2}\\right)=0\\\\\\\\\n& \\Rightarrow 2 a^2+2 a-9=0\n\\end{aligned}\n$$\n
Let $$\\mathrm{A}=\\left[\\mathrm{a}_{\\mathrm{ij}}\\right]_{2 \\times 2}$$, where $$\\mathrm{a}_{\\mathrm{ij}} \\neq 0$$ for all $$\\mathrm{i}, \\mathrm{j}$$ and $$\\mathrm{A}^{2}=\\mathrm{I}$$. Let a be the sum of all diagonal elements of $$\\mathrm{A}$$ and $$\\mathrm{b}=|\\mathrm{A}|$$. Then $$3 a^{2}+4 b^{2}$$ is equal to :
Let $$P$$ be a square matrix such that $$P^{2}=I-P$$. For $$\\alpha, \\beta, \\gamma, \\delta \\in \\mathbb{N}$$, if $$P^{\\alpha}+P^{\\beta}=\\gamma I-29 P$$ and $$P^{\\alpha}-P^{\\beta}=\\delta I-13 P$$, then $$\\alpha+\\beta+\\gamma-\\delta$$ is equal to :
Let $$A=\\left[\\begin{array}{ccc}2 & 1 & 2 \\\\ 6 & 2 & 11 \\\\ 3 & 3 & 2\\end{array}\\right]$$ and $$P=\\left[\\begin{array}{lll}1 & 2 & 0 \\\\ 5 & 0 & 2 \\\\ 7 & 1 & 5\\end{array}\\right]$$. The sum of the prime factors of $$\\left|P^{-1} A P-2 I\\right|$$ is equal to
Let $$A=\\left[\\begin{array}{lll}2 & a & 0 \\\\ 1 & 3 & 1 \\\\ 0 & 5 & b\\end{array}\\right]$$. If $$A^3=4 A^2-A-21 I$$, where $$I$$ is the identity matrix of order $$3 \\times 3$$, then $$2 a+3 b$$ is equal to
$$ \\Rightarrow {\\left( {25\\alpha } \\right)^2} = 25 \\Rightarrow \\left| \\alpha \\right| = {1 \\over 5}$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6109,
"subject": "General Science",
"question": "Let $$A$$ be a square matrix all of whose entries are integers. \n Then which one of the following is true? ",
"options": [
{
"text": "If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are not necessarily integers"
},
{
"text": "If det $$A \\ne \\pm 1,$$ then $${A^{ - 1}}$$ exists and all its entries are non integers"
},
{
"text": "If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are integers"
},
{
"text": "If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ need not exists "
}
],
"answer": "If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are integers",
"solution": "**Answer:** If det $$A = \\pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are integers\n\nAs all entries of square matrix $$A$$ are integers, therefore all co-factors should also be integers. \n
If det $$A = \\pm 1\\,\\,$$ then $${A^{ - 1}}\\,\\,$$ exists. Also all entries of $${A^{ - 1}}$$ are integers. ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6110,
"subject": "General Science",
"question": "Let $$A$$ be a $$\\,2 \\times 2$$ matrix\n Statement - 1 : $$adj\\left( {adj\\,A} \\right) = A$$\n Statement - 2 :$$\\left| {adj\\,A} \\right| = \\left| A \\right|$$",
"options": [
{
"text": "statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1. "
},
{
"text": "statement - 1 is true, statement - 2 is false. "
},
{
"text": "statement - 1 is false, statement -2 is true "
},
{
"text": "statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1. "
}
],
"answer": "statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1. ",
"solution": "**Answer:** statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1. \n\nWe know that $$\\left| {adj\\left( {adj\\,\\,A} \\right)} \\right| = {\\left| {Adj\\,\\,A} \\right|^{2 - 1}}$$b \n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = {\\left| A \\right|^{2 - 1}} = \\left| A \\right|$$ \n
$$\\therefore$$ $$\\,\\,\\,\\,\\,\\,$$ Both the statements are true and statement $$-2$$ is a correct explanation for statement - $$1.$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6111,
"subject": "General Science",
"question": "Let $$P$$ and $$Q$$ be $$3 \\times 3$$ matrices $$P \\ne Q.$$ If $${P^3} = {Q^3}$$ and\n $${P^2}Q = {Q^2}P$$ then determinant of $$\\left( {{P^2} + {Q^2}} \\right)$$ is equal to :",
"options": [
{
"text": "$$-2$$ "
},
{
"text": "$$1$$ "
},
{
"text": "$$0$$ "
},
{
"text": "$$-1$$"
}
],
"answer": "$$0$$ ",
"solution": "**Answer:** $$0$$ \n\nGiven \n
Here correct option will be $$\\left( {3,\\,{1 \\over {81}}} \\right)$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6118,
"subject": "General Science",
"question": "Let $$A = \\left[ {\\matrix{\n x & y & z \\cr \n y & z & x \\cr \n z & x & y \\cr \n\n } } \\right]$$, where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If $${A^2} = {I_3}$$, then the value of $${x^3} + {y^3} + {z^3}$$ is ____________.",
"options": [],
"answer": "7",
"solution": "**Answer:** 7\n\n$$A = \\left[ {\\matrix{\n x & y & z \\cr \n y & z & x \\cr \n z & x & y \\cr \n\n } } \\right]$$ \n
$$\\omega = {{ - 1 + i\\sqrt 3 } \\over 2}$$, and I3 be the identity matrix of order 3. If the determinant of the matrix (P$$-$$1AP$$-$$I3)2 is $$\\alpha$$$$\\omega$$2, then the value of $$\\alpha$$ is equal to ______________.",
"options": [],
"answer": "36",
"solution": "**Answer:** 36\n\n$$|{P^{ - 1}}AP - I{|^2}$$
$$ = {3^3}{.2^2}|{A^{ - 1}}{|^2} = {3^3}{.2^2}.{1 \\over {|A{|^2}}} = {3^2}{.2^2}.{1 \\over {{4^2}}} = 108$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6123,
"subject": "General Science",
"question": "Let $$M = \\left\\{ {A = \\left( {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right):a,b,c,d \\in \\{ \\pm 3, \\pm 2, \\pm 1,0\\} } \\right\\}$$. Define f : M $$\\to$$ Z, as f(A) = det(A), for all A$$\\in$$M, where z is set of all integers. Then the number of A$$\\in$$M such that f(A) = 15 is equal to _____________.",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\n| A | = ad $$-$$ bc = 15
For ad possible pairs are (3, 3), ($$-$$3, $$-$$3)
For bc possible pairs are (3, $$-$$2), ($$-$$3, 2), ($$-$$2, 3), (2, $$-$$3)
So total matrix = 2 $$\\times$$ 4 = 8
Case II ad = 6 & bc = $$-$$9
Similarly total matrix = 2 $$\\times$$ 4 = 8
$$\\Rightarrow$$ Total such matrices are = 16",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6124,
"subject": "General Science",
"question": "Let A and B be two 3 $$\\times$$ 3 real matrices such that (A2 $$-$$ B2) is invertible matrix. If A5 = B5 and A3B2 = A2B3, then the value of the determinant of the matrix A3 + B3 is equal to :",
"options": [
{
"text": "2"
},
{
"text": "4"
},
{
"text": "1"
},
{
"text": "0"
}
],
"answer": "0",
"solution": "**Answer:** 0\n\nC = A2 $$-$$ B2; | C | $$\\ne$$ 0
Let $$A = \\left( {\\matrix{\n 2 & { - 1} \\cr \n 0 & 2 \\cr \n\n } } \\right)$$. If $$B = I - {}^5{C_1}(adj\\,A) + {}^5{C_2}{(adj\\,A)^2} - \\,\\,.....\\,\\, - {}^5{C_5}{(adj\\,A)^5}$$, then the sum of all elements of the matrix B is
Let $$f(x) = \\left| {\\matrix{\n a & { - 1} & 0 \\cr \n {ax} & a & { - 1} \\cr \n {a{x^2}} & {ax} & a \\cr \n\n } } \\right|,\\,a \\in R$$. Then the sum of the squares of all the values of a, for which $$2f'(10) - f'(5) + 100 = 0$$, is
Let A and B be two square matrices of order 2. If $$det\\,(A) = 2$$, $$det\\,(B) = 3$$ and $$\\det \\left( {(\\det \\,5(det\\,A)B){A^2}} \\right) = {2^a}{3^b}{5^c}$$ for some a, b, c, $$\\in$$ N, then a + b + c is equal to :
Consider a matrix $$A=\\left[\\begin{array}{ccc}\\alpha & \\beta & \\gamma \\\\ \\alpha^{2} & \\beta^{2} & \\gamma^{2} \\\\ \\beta+\\gamma & \\gamma+\\alpha & \\alpha+\\beta\\end{array}\\right]$$, where $$\\alpha, \\beta, \\gamma$$ are three distinct natural numbers.
\n
If $$\\frac{\\operatorname{det}(\\operatorname{adj}(\\operatorname{adj}(\\operatorname{adj}(\\operatorname{adj} A))))}{(\\alpha-\\beta)^{16}(\\beta-\\gamma)^{16}(\\gamma-\\alpha)^{16}}=2^{32} \\times 3^{16}$$, then the number of such 3 - tuples $$(\\alpha, \\beta, \\gamma)$$ is ____________.
Let the matrix $$A=\\left[\\begin{array}{lll}0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 1 & 0 & 0\\end{array}\\right]$$ and the matrix $$B_{0}=A^{49}+2 A^{98}$$. If $$B_{n}=A d j\\left(B_{n-1}\\right)$$ for all $$n \\geq 1$$, then $$\\operatorname{det}\\left(B_{4}\\right)$$ is equal to :
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6139,
"subject": "General Science",
"question": "Let A be a $n \\times n$ matrix such that $|\\mathrm{A}|=2$. If the determinant of the matrix\n\n$\\operatorname{Adj}\\left(2 \\cdot \\operatorname{Adj}\\left(2 \\mathrm{~A}^{-1}\\right)\\right) \\cdot$ is $2^{84}$, then $\\mathrm{n}$ is equal to :",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n$\\because\\left|\\operatorname{adj}\\left(2 \\cdot \\operatorname{adj}\\left(2 A^{-1}\\right)\\right)\\right|=2^{84}$\n\n
Let $$A=\\left(\\begin{array}{cc}\\mathrm{m} & \\mathrm{n} \\\\ \\mathrm{p} & \\mathrm{q}\\end{array}\\right), \\mathrm{d}=|\\mathrm{A}| \\neq 0$$ and $$\\mathrm{|A-d(A d j A)|=0}$$. Then
Let $$\\alpha$$ be a root of the equation $$(a - c){x^2} + (b - a)x + (c - b) = 0$$ where a, b, c are distinct real numbers such that the matrix $$\\left[ {\\matrix{\n {{\\alpha ^2}} & \\alpha & 1 \\cr \n 1 & 1 & 1 \\cr \n a & b & c \\cr \n\n } } \\right]$$ is singular. Then, the value of $${{{{(a - c)}^2}} \\over {(b - a)(c - b)}} + {{{{(b - a)}^2}} \\over {(a - c)(c - b)}} + {{{{(c - b)}^2}} \\over {(a - c)(b - a)}}$$ is
\n$$\n\\begin{aligned}\n& \\frac{(a-c)^2}{(b-a)(c-b)}+\\frac{(b-a)^2}{(a-c)(c-b)}+\\frac{(c-b)^2}{(a-c)(b-a)} \\\\\\\\\n& \\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} \\\\\\\\\n& =3 \\frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6146,
"subject": "General Science",
"question": "Let the determinant of a square matrix A of order $m$ be $m-n$, where $m$ and $n$
satisfy $4 m+n=22$ and $17 m+4 n=93$.
If $\\operatorname{det}(n \\operatorname{adj}(\\operatorname{adj}(m A)))=3^{a} 5^{b} 6^{c}$ then $a+b+c$ is equal to :",
"options": [
{
"text": "96"
},
{
"text": "84"
},
{
"text": "109"
},
{
"text": "101"
}
],
"answer": "96",
"solution": "**Answer:** 96\n\nGiven that $|A|=m-n$, and let's solve the system of linear equations to find the values of $m$ and $n$ :\n\n
$4m + n = 22$ ...... (1)\n
$17m + 4n = 93$ ....... (2)\n\n
We can multiply equation (1) by 4 to make the coefficients of $n$ in both equations equal:\n\n
$16m + 4n = 88$ ......... (3)\n\n
Now, subtract equation (3) from equation (2):\n\n
$(17m + 4n) - (16m + 4n) = 93 - 88$\n
$m = 5$\n\n
Now, we can find the value of $n$ by substituting $m$ into equation (1):\n\n
$4(5) + n = 22$\n
$20 + n = 22$\n
$n = 2$\n\n
Since the order of matrix A is $m$, the order is 5. Now, let's find the determinant of $n \\operatorname{adj}(\\operatorname{adj}(mA))$:\n\n
We know that $\\operatorname{det}(A) = m-n = 3$.\n\n
Let $$A$$ be a $$3 \\times 3$$ matrix of non-negative real elements such that $$A\\left[\\begin{array}{l}1 \\\\ 1 \\\\ 1\\end{array}\\right]=3\\left[\\begin{array}{l}1 \\\\ 1 \\\\ 1\\end{array}\\right]$$. Then the maximum value of $$\\operatorname{det}(\\mathrm{A})$$ is _________.
Let A and B be two square matrices of order 3 such that $$\\mathrm{|A|=3}$$ and $$\\mathrm{|B|=2}$$. Then $$|\\mathrm{A}^{\\mathrm{T}} \\mathrm{A}(\\operatorname{adj}(2 \\mathrm{~A}))^{-1}(\\operatorname{adj}(4 \\mathrm{~B}))(\\operatorname{adj}(\\mathrm{AB}))^{-1} \\mathrm{AA}^{\\mathrm{T}}|$$ is equal to :
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6153,
"subject": "General Science",
"question": "If the system of linear equations \n $$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\\,\\,x + 4cy + cz = 0;$$ \n has a non - zero solution, then $$a, b, c$$.",
"options": [
{
"text": "satisfy $$a+2b+3c=0$$"
},
{
"text": "are in A.P"
},
{
"text": "are in G.P"
},
{
"text": "are in H.P."
}
],
"answer": "are in H.P.",
"solution": "**Answer:** are in H.P.\n\nFor homogeneous system of equations to have non zero solution, $$\\Delta = 0$$\n
$$\\therefore$$ $$a,b,c$$ are in Harmonic Progression.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6154,
"subject": "General Science",
"question": "The system of equations \n
$$\\matrix{\n {\\alpha \\,x + y + z = \\alpha - 1} \\cr \n {x + \\alpha y + z = \\alpha - 1} \\cr \n {x + y + \\alpha \\,z = \\alpha - 1} \\cr \n\n } $$
But $$\\alpha \\ne 1.\\,\\,\\,$$ $$\\therefore$$ $$\\,\\,\\alpha = - 2$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6155,
"subject": "General Science",
"question": "Let $$a, b, c$$ be any real numbers. Suppose that there are real numbers $$x, y, z$$ not all zero such that $$x=cy+bz,$$ $$y=az+cx,$$ and $$z=bx+ay.$$ Then $${a^2} + {b^2} + {c^2} + 2abc$$ is equal to :",
"options": [
{
"text": "$$2$$ "
},
{
"text": "$$-1$$ "
},
{
"text": "$$0$$ "
},
{
"text": "$$1$$ "
}
],
"answer": "$$1$$ ",
"solution": "**Answer:** $$1$$ \n\nThe given equations are \n
$$\\matrix{\n { - x + cy + bz = 0} \\cr \n {cx - y + az = 0} \\cr \n {bx + ay - z = 0} \\cr \n\n } $$\n
As $$x,y,z$$ are not all zero\n
$$\\therefore$$ The above system should not have unique (zero) solution \n
$$ \\Rightarrow \\Delta = 0 \\Rightarrow \\left| {\\matrix{\n { - 1} & c & b \\cr \n c & { - 1} & a \\cr \n b & a & { - 1} \\cr \n\n } } \\right| = 0$$ \n
If $$k = 1$$ then $${8 \\over {1 + 3}} \\ne {{4.1} \\over 2}$$ which is false\n
And if $$k = 3$$\n
Then $${8 \\over 6} \\ne {{4.3} \\over {9 - 1}}$$ which is true, therefore $$k=3$$\n
Hence for only one value of $$k.$$ System has no solution.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6159,
"subject": "General Science",
"question": "The set of all values of $$\\lambda $$ for which the system of linear equations:
$$\\matrix{\n {x + \\lambda y - z = 0} \\cr \n {\\lambda x - y - z = 0} \\cr \n {x + y - \\lambda z = 0} \\cr \n\n } $$
\nhas a non-trivial solution for :",
"options": [
{
"text": "infinitely many values of $$\\lambda .$$ "
},
{
"text": "exactly one value of $$\\lambda .$$ "
},
{
"text": "exactly two values of $$\\lambda .$$ "
},
{
"text": "exactly three values of $$\\lambda .$$ "
}
],
"answer": "exactly three values of $$\\lambda .$$ ",
"solution": "**Answer:** exactly three values of $$\\lambda .$$ \n\n
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6161,
"subject": "General Science",
"question": "If S is the set of distinct values of 'b' for which the following system of linear equations \n
x + y + z = 1\n x + ay + z = 1\n ax + by + z = 0\n
has no solution, then S is :",
"options": [
{
"text": "an empty set"
},
{
"text": "an infinite set"
},
{
"text": "a finite set containing two or more elements"
},
{
"text": "a singleton "
}
],
"answer": "a singleton ",
"solution": "**Answer:** a singleton \n\n$$\\left| {\\matrix{\n 1 & 1 & 1 \\cr \n 1 & a & 1 \\cr \n a & b & 1 \\cr \n\n } } \\right| = 0$$\n
$$ \\Rightarrow $$ S is singleton set.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6162,
"subject": "General Science",
"question": "The number of real values of $$\\lambda $$ for which the system of linear equations\n
$$ = 10$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6164,
"subject": "General Science",
"question": "Let S be the set of all real values of k for which the systemof linear equations\n x + y + z = 2\n 2x + y $$-$$ z = 3\n 3x + 2y + kz = 4\n has a unique solution. Then S is : ",
"options": [
{
"text": "an empty set "
},
{
"text": "equal to {0}"
},
{
"text": "equal to R"
},
{
"text": "equal to R $$-$$ {0}"
}
],
"answer": "equal to R $$-$$ {0}",
"solution": "**Answer:** equal to R $$-$$ {0}\n\nAs system of linear equations have unique solutions so, determinant of coefficient $$ \\ne $$ 0
\n$$ \\therefore $$ k $$ \\in $$ R - {0}",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6165,
"subject": "General Science",
"question": "If the system of linear equations \n x + ay + z = 3\n x + 2y + 2z = 6\n x + 5y + 3z = b\n has no solution, then :",
"options": [
{
"text": "a = $$-$$ 1, b = 9"
},
{
"text": "a = $$-$$ 1, b $$ \\ne $$ 9"
},
{
"text": "a $$ \\ne $$ $$-$$ 1, b = 9"
},
{
"text": "a = 1, b $$ \\ne $$ 9"
}
],
"answer": "a = $$-$$ 1, b $$ \\ne $$ 9",
"solution": "**Answer:** a = $$-$$ 1, b $$ \\ne $$ 9\n\nAs the given system of equations has no solution then \n
$$\\Delta $$ = 0 and at least one of $$\\Delta $$1, $$\\Delta $$2 and $$\\Delta $$2 should not be zero.\n
$$ \\Rightarrow $$ b $$ \\ne $$ 0",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6166,
"subject": "General Science",
"question": "The number of values of k for which the system of linear equations, \n (k + 2)x + 10y = k\n kx + (k +3)y = k -1\n has no solution, is : ",
"options": [
{
"text": "1"
},
{
"text": "2"
},
{
"text": "3"
},
{
"text": "infinitely many"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nSystem of linear equation have no solution, \n
$$\\therefore\\,\\,\\,$$ determinant of coefficient = 0\n
$$ \\therefore $$ Greatest value of c is $${1 \\over 2}$$.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6172,
"subject": "General Science",
"question": "The set of all values of $$\\lambda $$ for which the system of linear equations\n x – 2y – 2z = $$\\lambda $$x\n x + 2y + z = $$\\lambda $$y\n – x – y = $$\\lambda $$z\n has a non-trivial solutions :",
"options": [
{
"text": "is an empty set"
},
{
"text": "contains more than two elements"
},
{
"text": "is a singleton"
},
{
"text": "contains exactly two elements\n"
}
],
"answer": "is a singleton",
"solution": "**Answer:** is a singleton\n\n$$\\left| {\\matrix{\n {\\lambda - 1} & 2 & 2 \\cr \n 1 & {2 - \\lambda } & 1 \\cr \n 1 & 1 & 1 \\cr \n\n } } \\right| = 0$$\n
$$ \\Rightarrow {\\left( {\\lambda - 1} \\right)^3} = 0 \\Rightarrow \\lambda = 1$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6173,
"subject": "General Science",
"question": "If the system of linear equations\n 2x + 2y + 3z = a\n 3x – y + 5z = b\n x – 3y + 2z = c\n where a, b, c are non zero real numbers, has more one solution, then : ",
"options": [
{
"text": "b – c – a = 0"
},
{
"text": "a + b + c = 0"
},
{
"text": "b – c + a = 0"
},
{
"text": "b + c – a = 0"
}
],
"answer": "b – c – a = 0",
"solution": "**Answer:** b – c – a = 0\n\nP1 : 2x + 2y + 3z = a\n
P2 : 3x $$-$$ y + 5z = b\n
P3 : x $$-$$ 3y + 2z = c\n
We find\n
P1 + P3 = P2 $$ \\Rightarrow $$ a + c = b\n",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6174,
"subject": "General Science",
"question": "The number of values of $$\\theta $$ $$ \\in $$ (0, $$\\pi $$) for which the system of linear equations\n
The system of equation has infinite non-trival solution.\n
Also adding equation (1) and 3$$ \\times $$(3), we get\n
10x = 20z\n
$$ \\Rightarrow $$ x = 2z",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6179,
"subject": "General Science",
"question": "The sum of distinct values of $$\\lambda $$ for which the\nsystem of equations
By checking all the options we find (C) is correct
$$2\\lambda + \\mu = 14$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6184,
"subject": "General Science",
"question": "Suppose the vectors x1, x2 and x3 are the solutions of the system of linear equations, Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. if
So, |A| = 2(1) = 2",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6185,
"subject": "General Science",
"question": "If the system of equations \nx - 2y + 3z = 9 \n2x + y + z = b \nx - 7y + az = 24, has infinitely many solutions, then a - b is equal to.........",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nD = 0
then the set A :",
"options": [
{
"text": "is an empty set.\n"
},
{
"text": "contains more than two elements."
},
{
"text": "contains exactly two elements."
},
{
"text": "is a singleton."
}
],
"answer": "contains exactly two elements.",
"solution": "**Answer:** contains exactly two elements.\n\nLet $$X = \\left[ {\\matrix{\n x \\cr \n y \\cr \n z \\cr \n\n } } \\right]$$
\n$$ \\Rightarrow x = \\pm {7 \\over {\\sqrt {171} }}$$
\n$$ \\Rightarrow z = \\pm {{11} \\over {\\sqrt {171} }}$$
\n$$ \\therefore $$ So, there are 2 solution set of (x, y, z)\n\n\n",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6187,
"subject": "General Science",
"question": "Let S be the set of all $$\\lambda $$ $$ \\in $$ R for which the system\nof linear equations\n
2x – y + 2z = 2\n x – 2y +\n$$\\lambda $$z = –4\n x +\n$$\\lambda $$y + z = 4\n
has no solution. Then the set S :",
"options": [
{
"text": "contains more than two elements."
},
{
"text": "contains exactly two elements."
},
{
"text": "is a singleton."
},
{
"text": "is an empty set."
}
],
"answer": "contains exactly two elements.",
"solution": "**Answer:** contains exactly two elements.\n\nFor no solution :\n
2x – y + 2z = 0 (contradiction) hence no solution.\n
$$ \\therefore $$ $$\\lambda $$ = 1 belongs to set S.\n
When $$\\lambda $$ = $$ - {1 \\over 2}$$\n
2x – y + 2z = 2 ...(1)\n
x – 2y $$ - {1 \\over 2}$$z\n = –4 ...(2)\n
x $$ - {1 \\over 2}$$y\n + z = 4 ...(3)\n
(1) and (3) contradict each other, hence no\nsolution.\n
$$ \\therefore $$ $$\\lambda $$ = $$ - {1 \\over 2}$$ belongs to set S.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6188,
"subject": "General Science",
"question": "If for some $$\\alpha $$ and $$\\beta $$ in R, the intersection of the\nfollowing three places \nx + 4y – 2z = 1 \nx + 7y – 5z = b \nx + 5y + $$\\alpha $$z = 5 \nis a line in R3, then $$\\alpha $$ + $$\\beta $$ is equal to :",
"options": [
{
"text": "-10"
},
{
"text": "0"
},
{
"text": "10"
},
{
"text": "2"
}
],
"answer": "10",
"solution": "**Answer:** 10\n\nFor planes to intersect on a line there should be infinite solution of the\ngiven system of equations.\n
Only ($$\\mu $$, $$\\delta $$) = (4, 3) does satisfy the equation (1).",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6191,
"subject": "General Science",
"question": "If the system of linear equations,\n x + y + z = 6\n x + 2y + 3z = 10\n 3x + 2y + $$\\lambda $$z = $$\\mu $$\n has more than two solutions, then $$\\mu $$ - $$\\lambda $$2\n is equal to ______.",
"options": [],
"answer": "13",
"solution": "**Answer:** 13\n\nGiven system of equation more than\n2 solutions.\nHence system of equation has infinite many\nsolution.\n
$$ \\therefore $$ $${1 \\over a},{1 \\over b},{1 \\over c}$$ are in A.P. ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6193,
"subject": "General Science",
"question": "Let S be the set of all integer solutions, (x, y, z),\nof the system of equations\n x – 2y + 5z = 0\n –2x + 4y + z = 0\n –7x + 14y + 9z = 0\n such that 15 $$ \\le $$ x2\n + y2\n + z2 $$ \\le $$ 150. Then, the\nnumber of elements in the set S is equal to\n______ .\n",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\n$$x - 2y + 5z = 0$$ ....(1)
(A) The system has unique solution if $$k \\ne 2,k \\ne - 2$$.
(B) The system has unique solution if k = $$-$$2
(C) The system has unique solution if k = 2
(D) The system has no solution if k = 2
(E) The system has infinite number of solutions if k $$ \\ne $$ $$-$$2.
Which of the following statements are correct?",
"options": [
{
"text": "(B) and (E) only"
},
{
"text": "(C) and (D) only"
},
{
"text": "(A) and (E) only"
},
{
"text": "(A) and (D) only"
}
],
"answer": "(A) and (D) only",
"solution": "**Answer:** (A) and (D) only\n\n$$x - 2y + 0.z = 1$$
$$ \\therefore $$ For K = 2; The system has no solution.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6196,
"subject": "General Science",
"question": "If the system of equations
kx + y + 2z = 1
3x $$-$$ y $$-$$ 2z = 2
$$-$$2x $$-$$2y $$-$$4z = 3
has infinitely many solutions, then k is equal to __________.",
"options": [],
"answer": "21",
"solution": "**Answer:** 21\n\nD = 0
$$ \\Rightarrow $$ k = 21",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6197,
"subject": "General Science",
"question": "The following system of linear equations
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x $$-$$ y + 4z = 8",
"options": [
{
"text": "does not have any solution"
},
{
"text": "has a solution ($$\\alpha$$, $$\\beta$$, $$\\gamma$$) satisfying $$\\alpha$$ + $$\\beta$$2 + $$\\gamma$$3 = 12"
},
{
"text": "has a unique solution"
},
{
"text": "has infinitely many solutions"
}
],
"answer": "has a unique solution",
"solution": "**Answer:** has a unique solution\n\n$$\\Delta = \\left| {\\matrix{\n 2 & 3 & 2 \\cr \n 3 & 2 & 2 \\cr \n 1 & { - 1} & 4 \\cr \n\n } } \\right| = - 20 \\ne 0$$ $$ \\therefore $$ unique solution
Unique solution : (0, 1, 2)",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6198,
"subject": "General Science",
"question": "Consider the following system of equations :
x + 2y $$-$$ 3z = a
2x + 6y $$-$$ 11z = b
x $$-$$ 2y + 7z = c,
where a, b and c are real constants. Then the system of equations :",
"options": [
{
"text": "has no solution for all a, b and c"
},
{
"text": "has a unique solution when 5a = 2b + c"
},
{
"text": "has infinite number of solutions when 5a = 2b + c"
},
{
"text": "has a unique solution for all a, b and c"
}
],
"answer": "has infinite number of solutions when 5a = 2b + c",
"solution": "**Answer:** has infinite number of solutions when 5a = 2b + c\n\n$$D = \\left| {\\matrix{\n 1 & 2 & { - 3} \\cr \n 2 & 6 & { - 11} \\cr \n 1 & { - 2} & 7 \\cr \n\n } } \\right|$$
for k = 1 equation identical so k = $$-$$2 for no solution.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6201,
"subject": "General Science",
"question": "Let $$\\alpha$$, $$\\beta$$, $$\\gamma$$ be the real roots of the equation, x3 + ax2 + bx + c = 0, (a, b, c $$\\in$$ R and a, b $$\\ne$$ 0). If the system of equations (in u, v, w) given by $$\\alpha$$u + $$\\beta$$v + $$\\gamma$$w = 0, $$\\beta$$u + $$\\gamma$$v + $$\\alpha$$w = 0; $$\\gamma$$u + $$\\alpha$$v + $$\\beta$$w = 0 has non-trivial solution, then the value of $${{{a^2}} \\over b}$$ is",
"options": [
{
"text": "5"
},
{
"text": "3"
},
{
"text": "1"
},
{
"text": "0"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\nx3 + ax2 + bx + c = 0
$$ \\Rightarrow \\lambda = - 2$$ and $$\\mu \\in R$$ or $$\\mu$$ = 6 and $$\\lambda \\in R$$.
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6203,
"subject": "General Science",
"question": "The value of k $$\\in$$R, for which the following system of linear equations
$$\\therefore$$ For no solution $$\\lambda$$ = 2 and $$\\mu$$ $$\\ne$$ 10.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6205,
"subject": "General Science",
"question": "The values of a and b, for which the system of equations
2x + 3y + 6z = 8
x + 2y + az = 5
3x + 5y + 9z = b
has no solution, are :",
"options": [
{
"text": "a = 3, b $$\\ne$$ 13"
},
{
"text": "a $$\\ne$$ 3, b $$\\ne$$ 13"
},
{
"text": "a $$\\ne$$ 3, b = 3"
},
{
"text": "a = 3, b = 13"
}
],
"answer": "a = 3, b $$\\ne$$ 13",
"solution": "**Answer:** a = 3, b $$\\ne$$ 13\n\n$$D = \\left| {\\matrix{\n 2 & 3 & 6 \\cr \n 1 & 2 & a \\cr \n 3 & 5 & 9 \\cr \n\n } } \\right| = 3 - a$$
If a = 3, b $$\\ne$$ 13, no solution.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6206,
"subject": "General Science",
"question": "For real numbers $$\\alpha$$ and $$\\beta$$, consider the following system of linear equations :
x + y $$-$$ z = 2, x + 2y + $$\\alpha$$z = 1, 2x $$-$$ y + z = $$\\beta$$. If the system has infinite solutions, then $$\\alpha$$ + $$\\beta$$ is equal to ______________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nFor infinite solutions
$$\\theta = {{7\\pi } \\over {18}}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6208,
"subject": "General Science",
"question": "If the system of linear equations
2x + y $$-$$ z = 3
x $$-$$ y $$-$$ z = $$\\alpha$$
3x + 3y + $$\\beta$$z = 3
has infinitely many solution, then $$\\alpha$$ + $$\\beta$$ $$-$$ $$\\alpha$$$$\\beta$$ is equal to _____________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n2 $$\\times$$ (i) $$-$$ (ii) $$-$$ (iii) gives :
Hence, $$\\alpha$$ + $$\\beta$$ $$-$$ $$\\alpha$$$$\\beta$$ = 5",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6209,
"subject": "General Science",
"question": "Let [$$\\lambda$$] be the greatest integer less than or equal to $$\\lambda$$. The set of all values of $$\\lambda$$ for which the system of linear equations x + y + z = 4, 3x + 2y + 5z = 3, 9x + 4y + (28 + [$$\\lambda$$])z = [$$\\lambda$$] has a solution is :",
"options": [
{
"text": "R"
},
{
"text": "($$-$$$$\\infty$$, $$-$$9) $$\\cup$$ ($$-$$9, $$\\infty$$)"
},
{
"text": "[$$-$$9, $$-$$8)"
},
{
"text": "($$-$$$$\\infty$$, $$-$$9) $$\\cup$$ [$$-$$8, $$\\infty$$)"
}
],
"answer": "R",
"solution": "**Answer:** R\n\n$$D = \\left| {\\matrix{\n 1 & 1 & 1 \\cr \n 3 & 2 & 5 \\cr \n 9 & 4 & {28 + [\\lambda ]} \\cr \n\n } } \\right| = - 24 - [\\lambda ] + 15 = - [\\lambda ] - 9$$
if $$[\\lambda ] + 9 \\ne 0$$ then unique solution
if $$[\\lambda ] + 9 = 0$$ then D1 = D2 = D3 = 0
so infinite solutions
Hence, $$\\lambda$$ can be any red number.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6210,
"subject": "General Science",
"question": "If the following system of linear equations
for $$a = {1 \\over 3},b \\ne {7 \\over 3}$$, system has no solutions.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6211,
"subject": "General Science",
"question": "If $$\\alpha$$ + $$\\beta$$ + $$\\gamma$$ = 2$$\\pi$$, then the system of equations
x + (cos $$\\gamma$$)y + (cos $$\\beta$$)z = 0
(cos $$\\gamma$$)x + y + (cos $$\\alpha$$)z = 0
(cos $$\\beta$$)x + (cos $$\\alpha$$)y + z = 0
has :",
"options": [
{
"text": "no solution"
},
{
"text": "infinitely many solution"
},
{
"text": "exactly two solutions"
},
{
"text": "a unique solution"
}
],
"answer": "infinitely many solution",
"solution": "**Answer:** infinitely many solution\n\n
Given $$\\alpha$$ + $$\\beta$$ + $$\\gamma$$ = 2$$\\pi$$
So, the system of equation has infinitely many solutions.
\n",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6212,
"subject": "General Science",
"question": "Consider the system of linear equations
$$-$$x + y + 2z = 0
3x $$-$$ ay + 5z = 1
2x $$-$$ 2y $$-$$ az = 7
Let S1 be the set of all a$$\\in$$R for which the system is inconsistent and S2 be the set of all a$$\\in$$R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then",
"options": [
{
"text": "n(S1) = 2, n(S2) = 2"
},
{
"text": "n(S1) = 1, n(S2) = 0"
},
{
"text": "n(S1) = 2, n(S2) = 0"
},
{
"text": "n(S1) = 0, n(S2) = 2"
}
],
"answer": "n(S1) = 2, n(S2) = 0",
"solution": "**Answer:** n(S1) = 2, n(S2) = 0\n\n$$\\Delta = \\left| {\\matrix{\n { - 1} & 1 & 2 \\cr \n 3 & { - a} & 5 \\cr \n 2 & { - 2} & { - a} \\cr \n\n } } \\right|$$
If the system of linear equations $$2x - 3y = \\gamma + 5$$, $$\\alpha x + 5y = \\beta + 1$$, where $$\\alpha$$, $$\\beta$$, $$\\gamma$$ $$\\in$$ R has infinitely many solutions then the value of | 9$$\\alpha$$ + 3$$\\beta$$ + 5$$\\gamma$$ | is equal to ____________.
Let the system of linear equations $$x + 2y + z = 2$$, $$\\alpha x + 3y - z = \\alpha $$, $$ - \\alpha x + y + 2z = - \\alpha $$ be inconsistent. Then $$\\alpha$$ is equal to :
If $$X = {({x_1},{x_2},{x_3})^T}$$ and I is an identity matrix of order 3, then the system $$(A - 2I)X = \\left( {\\matrix{\n 4 \\cr \n 1 \\cr \n 1 \\cr \n\n } } \\right)$$ has :
",
"options": [
{
"text": "no solution"
},
{
"text": "infinitely many solutions"
},
{
"text": "unique solution"
},
{
"text": "exactly two solutions"
}
],
"answer": "infinitely many solutions",
"solution": "**Answer:** infinitely many solutions\n\n
Let $$A = \\left[ {\\matrix{\n a & b & c \\cr \n d & e & f \\cr \n g & h & i \\cr \n\n } } \\right]$$
\n
$$A = \\left[ {\\matrix{\n 1 \\cr \n 1 \\cr \n 0 \\cr \n\n } } \\right] = \\left[ {\\matrix{\n 1 \\cr \n 1 \\cr \n 0 \\cr \n\n } } \\right] \\Rightarrow \\left[ {\\matrix{\n a & b & c \\cr \n d & e & f \\cr \n g & h & i \\cr \n\n } } \\right]\\left[ {\\matrix{\n 1 \\cr \n 1 \\cr \n 0 \\cr \n\n } } \\right] = \\left[ {\\matrix{\n 1 \\cr \n 1 \\cr \n 0 \\cr \n\n } } \\right] \\Rightarrow \\matrix{\n {a + b = 1} \\cr \n {d + e = 1} \\cr \n {g + h = 0} \\cr \n\n } $$
\n
$$A = \\left[ {\\matrix{\n 1 \\cr \n 0 \\cr \n 1 \\cr \n\n } } \\right] = \\left[ {\\matrix{\n { - 1} \\cr \n 0 \\cr \n 1 \\cr \n\n } } \\right] \\Rightarrow \\left[ {\\matrix{\n a & b & c \\cr \n d & e & f \\cr \n g & h & i \\cr \n\n } } \\right]\\left[ {\\matrix{\n 1 \\cr \n 0 \\cr \n 1 \\cr \n\n } } \\right] = \\left[ {\\matrix{\n { - 1} \\cr \n 0 \\cr \n 1 \\cr \n\n } } \\right] \\Rightarrow \\matrix{\n {a + c = - 1} \\cr \n {d + f = 1} \\cr \n {g + i = 0} \\cr \n\n } $$
have a unique solution (x$$^ * $$, y$$^ * $$, z$$^ * $$). If ($$\\alpha$$, x$$^ * $$), (y$$^ * $$, $$\\alpha$$) and (x$$^ * $$, $$-$$y$$^ * $$) are collinear points, then the sum of absolute values of all possible values of $$\\alpha$$ is
has infinitely many solutions, then the distance of the point $$\\left( {\\lambda ,\\mu , - {1 \\over 2}} \\right)$$ from the plane $$8x + y + 4z + 2 = 0$$ is :
Let A and B be two $$3 \\times 3$$ non-zero real matrices such that AB is a zero matrix. Then
",
"options": [
{
"text": "the system of linear equations $$A X=0$$ has a unique solution"
},
{
"text": "the system of linear equations $$A X=0$$ has infinitely many solutions"
},
{
"text": "B is an invertible matrix"
},
{
"text": "$$\\operatorname{adj}(\\mathrm{A})$$ is an invertible matrix"
}
],
"answer": "the system of linear equations $$A X=0$$ has infinitely many solutions",
"solution": "**Answer:** the system of linear equations $$A X=0$$ has infinitely many solutions\n\n
AB is zero matrix
\n
$$ \\Rightarrow |A| = |B| = 0$$
\n
So neither A nor B is invertible
\n
If $$|A| = 0$$
\n
$$ \\Rightarrow |\\mathrm{adj}\\,A| = 0$$ so $$\\mathrm{adj}\\,A$$
For the system of linear equations $$\\alpha x+y+z=1,x+\\alpha y+z=1,x+y+\\alpha z=\\beta$$, which one of the following statements is NOT correct?
",
"options": [
{
"text": "It has infinitely many solutions if $$\\alpha=1$$ and $$\\beta=1$$"
},
{
"text": "It has infinitely many solutions if $$\\alpha=2$$ and $$\\beta=-1$$"
},
{
"text": "$$x+y+z=\\frac{3}{4}$$ if $$\\alpha=2$$ and $$\\beta=1$$"
},
{
"text": "It has no solution if $$\\alpha=-2$$ and $$\\beta=1$$"
}
],
"answer": "It has infinitely many solutions if $$\\alpha=2$$ and $$\\beta=-1$$",
"solution": "**Answer:** It has infinitely many solutions if $$\\alpha=2$$ and $$\\beta=-1$$\n\nFor infinite solution $\\Delta=\\Delta_x=\\Delta_y=\\Delta_z=0$\n
",
"options": [
{
"text": "If $$\\alpha=\\beta=7$$, then the system has no solution"
},
{
"text": "For every point $$(\\alpha, \\beta) \\neq(7,7)$$ on the line $$x-2 y+7=0$$, the system has infinitely many solutions"
},
{
"text": "There is a unique point $$(\\alpha, \\beta)$$ on the line $$x+2 y+18=0$$ for which the system has infinitely many solutions"
},
{
"text": "If $$\\alpha=\\beta$$ and $$\\alpha \\neq 7$$, then the system has a unique solution"
}
],
"answer": "For every point $$(\\alpha, \\beta) \\neq(7,7)$$ on the line $$x-2 y+7=0$$, the system has infinitely many solutions",
"solution": "**Answer:** For every point $$(\\alpha, \\beta) \\neq(7,7)$$ on the line $$x-2 y+7=0$$, the system has infinitely many solutions\n\n$\\Delta=\\left|\\begin{array}{ccc}1 & 1 & 1 \\\\ \\alpha & \\beta & 7 \\\\ 1 & 2 & 3\\end{array}\\right|$\n\n
So, for $\\alpha=\\beta \\neq 7, \\Delta \\neq 0$ so unique solution.\n\n
$\\alpha=\\beta=7$, equation (i) and (ii) represent 2 parallel planes so no solution.\n\n
If $\\alpha-2 \\beta+7=0$, but $(\\alpha, \\beta) \\neq(7,7)$, then no solution.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6231,
"subject": "General Science",
"question": "For $\\alpha, \\beta \\in \\mathbb{R}$, suppose the system of linear equations\n
for some $$\\alpha,\\beta\\in \\mathbb{R}$$. Then which of the following is NOT correct.
",
"options": [
{
"text": "It has a solution for all $$\\alpha\\ne-1$$ and $$\\beta=2$$"
},
{
"text": "It has no solution if $$\\alpha=-1$$ and $$\\beta\\ne2$$"
},
{
"text": "It has no solution for $$\\alpha=-1$$ and for all $$\\beta \\in \\mathbb{R}$$"
},
{
"text": "It has no solution for $$\\alpha=3$$ and for all $$\\beta\\ne2$$"
}
],
"answer": "It has no solution for $$\\alpha=-1$$ and for all $$\\beta \\in \\mathbb{R}$$",
"solution": "**Answer:** It has no solution for $$\\alpha=-1$$ and for all $$\\beta \\in \\mathbb{R}$$\n\n$D=\\left|\\begin{array}{ccc}\\alpha & 2 & 1 \\\\ 2 \\alpha & 3 & 1 \\\\ 3 & \\alpha & 2\\end{array}\\right|=0 \\Rightarrow \\alpha=-1,3$\n
",
"options": [
{
"text": "It has infinitely many solutions if $$a=3, b=8$$"
},
{
"text": "It has infinitely many solutions if $$a=3, b=6$$"
},
{
"text": "It has unique solution if $$a=b=8$$"
},
{
"text": "It has unique solution if $$a=b=6$$"
}
],
"answer": "It has infinitely many solutions if $$a=3, b=6$$",
"solution": "**Answer:** It has infinitely many solutions if $$a=3, b=6$$\n\nThe given system of equations is :\n\n
1. For a=3 and b=8, we have Δ = 0 and Δ₃ = 0, which indicates an infinite number of solutions.\n\n
2. For a=3 and b=6, we have Δ = 0 and Δ₃ ≠ 0, which indicates no solution.\n\n
3. For a=8 and b=8, we have Δ ≠ 0, which indicates a unique solution.\n\n
4. For a=6 and b=6, we have Δ ≠ 0, which indicates a unique solution.\n\n
Therefore, the statement that is NOT correct is Option B: \"It has infinitely many solutions if a=3, b=6\", because in this case the system actually has no solution.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6238,
"subject": "General Science",
"question": "
Let $$\\mathrm{S}$$ be the set of values of $$\\lambda$$, for which the system of equations
$$6 \\lambda x-3 y+3 z=4 \\lambda^{2}$$,
$$2 x+6 \\lambda y+4 z=1$$,
$$3 x+2 y+3 \\lambda z=\\lambda$$ has no solution. Then $$12 \\sum_\\limits{i \\in S}|\\lambda|$$ is equal to ___________.
",
"options": [],
"answer": "24",
"solution": "**Answer:** 24\n\nGiven that $S$ be the set of values of $\\lambda$ for which given system of equations has no solution.\n
",
"options": [
{
"text": "The system has infinitely many solutions for $$\\alpha=-6$$ and $$\\beta=9$$"
},
{
"text": "The system has a unique solution for $$\\alpha$$ $$ \\ne $$ $$-5$$ and $$\\beta=8$$"
},
{
"text": "The system is inconsistent for $$\\alpha=-5$$ and $$\\beta=8$$"
},
{
"text": "The system has infinitely many solutions for $$\\alpha=-5$$ and $$\\beta=9$$"
}
],
"answer": "The system has infinitely many solutions for $$\\alpha=-6$$ and $$\\beta=9$$",
"solution": "**Answer:** The system has infinitely many solutions for $$\\alpha=-6$$ and $$\\beta=9$$\n\nGiven system of linear equation is\n
If $\\alpha=-5$ and $\\beta=8$, then $\\Delta$ equals zero but $\\Delta_3$ does not, which would imply the system is inconsistent for $\\alpha=-5$ and $\\beta=8$.\n\n
Therefore, the option \"The system is inconsistent for $\\alpha=-5$ and $\\beta=8$ \" is correct.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6241,
"subject": "General Science",
"question": "
Let S be the set of all values of $$\\theta \\in[-\\pi, \\pi]$$ for which the system of linear equations
\n
$$x+y+\\sqrt{3} z=0$$
\n
$$-x+(\\tan \\theta) y+\\sqrt{7} z=0$$
\n
$$x+y+(\\tan \\theta) z=0$$
\n
has non-trivial solution. Then $$\\frac{120}{\\pi} \\sum_\\limits{\\theta \\in \\mathrm{s}} \\theta$$ is equal to :
",
"options": [
{
"text": "40"
},
{
"text": "30"
},
{
"text": "10"
},
{
"text": "20"
}
],
"answer": "20",
"solution": "**Answer:** 20\n\nSince, the given system has a non trivial solution,\n
$$x+3 y+5 z=\\beta$$, which one of the following is NOT true?
",
"options": [
{
"text": "System has a unique solution for $$\\alpha=3,\\beta\\ne14$$."
},
{
"text": "System has infinitely many solutions for $$\\alpha=3, \\beta=14$$."
},
{
"text": "System has no solution for $$\\alpha=3, \\beta=24$$."
},
{
"text": "System has a unique solution for $$\\alpha=-3, \\beta=14$$."
}
],
"answer": "System has a unique solution for $$\\alpha=3,\\beta\\ne14$$.",
"solution": "**Answer:** System has a unique solution for $$\\alpha=3,\\beta\\ne14$$.\n\nGiven system of equations,\n
$$\n\\begin{aligned}\nx+y+z & =6 ........(i)\\\\\\\\\nx+2 y+\\alpha z & =10 ........(ii)\\\\\\\\\nx+3 y+5 z & =\\beta ........(iii)\n\\end{aligned}\n$$\n
For $\\lambda=-13, \\mu=15$ system of equation has infinite solution hence $\\lambda+2 \\mu=17$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6245,
"subject": "General Science",
"question": "If the system of equations\n
Then, the system $$(A-3 I)\\left(\\begin{array}{l}x \\\\ y \\\\ z\\end{array}\\right)=\\left(\\begin{array}{l}1 \\\\ 2 \\\\ 3\\end{array}\\right)$$ has :
Let for any three distinct consecutive terms $$a, b, c$$ of an A.P, the lines $$a x+b y+c=0$$ be concurrent at the point $$P$$ and $$Q(\\alpha, \\beta)$$ be a point such that the system of equations
Consider the system of linear equations $$x+y+z=5, x+2 y+\\lambda^2 z=9, x+3 y+\\lambda z=\\mu$$, where $$\\lambda, \\mu \\in \\mathbb{R}$$. Then, which of the following statement is NOT correct?
",
"options": [
{
"text": "System is consistent if $$\\lambda \\neq 1$$ and $$\\mu=13$$\n"
},
{
"text": "System is inconsistent if $$\\lambda=1$$ and $$\\mu \\neq 13$$\n"
},
{
"text": "System has unique solution if $$\\lambda \\neq 1$$ and $$\\mu \\neq 13$$\n"
},
{
"text": "System has infinite number of solutions if $$\\lambda=1$$ and $$\\mu=13$$"
}
],
"answer": "System has unique solution if $$\\lambda \\neq 1$$ and $$\\mu \\neq 13$$\n",
"solution": "**Answer:** System has unique solution if $$\\lambda \\neq 1$$ and $$\\mu \\neq 13$$\n\n\n
Consider the system of linear equations $$x+y+z=4 \\mu, x+2 y+2 \\lambda z=10 \\mu, x+3 y+4 \\lambda^2 z=\\mu^2+15$$ where $$\\lambda, \\mu \\in \\mathbf{R}$$. Which one of the following statements is NOT correct ?
",
"options": [
{
"text": "The system has unique solution if $$\\lambda \\neq \\frac{1}{2}$$ and $$\\mu \\neq 1,15$$"
},
{
"text": "The system has infinite number of solutions if $$\\lambda=\\frac{1}{2}$$ and $$\\mu=15$$\n"
},
{
"text": "The system is consistent if $$\\lambda \\neq \\frac{1}{2}$$\n"
},
{
"text": "The system is inconsistent if $$\\lambda=\\frac{1}{2}$$ and $$\\mu \\neq 1$$"
}
],
"answer": "The system is inconsistent if $$\\lambda=\\frac{1}{2}$$ and $$\\mu \\neq 1$$",
"solution": "**Answer:** The system is inconsistent if $$\\lambda=\\frac{1}{2}$$ and $$\\mu \\neq 1$$\n\n
Consider the matrices : $$A=\\left[\\begin{array}{cc}2 & -5 \\\\ 3 & m\\end{array}\\right], B=\\left[\\begin{array}{l}20 \\\\ m\\end{array}\\right]$$ and $$X=\\left[\\begin{array}{l}x \\\\ y\\end{array}\\right]$$. Let the set of all $$m$$, for which the system of equations $$A X=B$$ has a negative solution (i.e., $$x<0$$ and $$y<0$$), be the interval $$(a, b)$$. Then $$8 \\int_\\limits a^b|A| d m$$ is equal to _________.
If the system of equations $$x+4 y-z=\\lambda, 7 x+9 y+\\mu z=-3,5 x+y+2 z=-1$$ has infinitely many solutions, then $$(2 \\mu+3 \\lambda)$$ is equal to :
Let $$\\alpha \\beta \\gamma=45 ; \\alpha, \\beta, \\gamma \\in \\mathbb{R}$$. If $$x(\\alpha, 1,2)+y(1, \\beta, 2)+z(2,3, \\gamma)=(0,0,0)$$ for some $$x, y, z \\in \\mathbb{R}, x y z \\neq 0$$, then $$6 \\alpha+4 \\beta+\\gamma$$ is equal to _________.
Given that $\\alpha \\beta \\gamma = 45$ and $\\alpha, \\beta, \\gamma \\in \\mathbb{R}$, consider the equation $x(\\alpha, 1, 2) + y(1, \\beta, 2) + z(2, 3, \\gamma) = (0, 0, 0)$ for some $x, y, z \\in \\mathbb{R}$ where $x y z \\neq 0$. To find the value of $6 \\alpha + 4 \\beta + \\gamma$, follow these steps:
\n\n\n
Express the given equation in matrix form:
\n\n
$ \\begin{aligned} & \\alpha x + y + 2 z = 0 \\\\ & x + \\beta y + 3 z = 0 \\\\ & 2 x + 2 y + \\gamma z = 0 \\end{aligned} $
\n\n\n
Since $x, y, z \\neq 0$, the determinant of the coefficients matrix must be zero:
Given $\\alpha \\beta \\gamma = 45$, substitute this value into the equation:
\n\n
$ 45 - 6 \\alpha - 4 \\beta - \\gamma + 10 = 0 $
\n\n\n
Simplify the equation:
\n\n
$ 45 + 10 = 6 \\alpha + 4 \\beta + \\gamma $
\n\n\n
Thus,
\n\n
$ 6 \\alpha + 4 \\beta + \\gamma = 55 $
\n\n
So, the value of $6 \\alpha + 4 \\beta + \\gamma$ is 55.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6259,
"subject": "General Science",
"question": "Let $$A$$ and $$B$$ be two symmetric matrices of order $$3$$. \n
Statement - 1 : $$A(BA)$$ and $$(AB)$$$$A$$ are symmetric matrices. \n
Statement - 2 : $$AB$$ is symmetric matrix if matrix multiplication of $$A$$ with $$B$$ is commutative.",
"options": [
{
"text": "statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1. "
},
{
"text": "statement - 1 is true, statement - 2 is false. "
},
{
"text": "statement - 1 is false, statement -2 is true "
},
{
"text": "statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1. "
}
],
"answer": "statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1. ",
"solution": "**Answer:** statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1. \n\n$$\\therefore$$ $$A' = A,B' = B$$ \n
\nSo $$AB = \\left( {\\left[ {\\matrix{\n 2 & 4 \\cr \n 4 & { - 1} \\cr \n\n } } \\right]\\left[ {\\matrix{\n 0 & { - 1} \\cr \n 1 & 0 \\cr \n\n } } \\right]} \\right) = \\left[ {\\matrix{\n 4 & { - 2} \\cr \n { - 1} & { - 4} \\cr \n\n } } \\right]$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6261,
"subject": "General Science",
"question": "Let A and B be 3 $$\\times$$ 3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A2B2 $$-$$ B2A2) X = O, where X is a 3 $$\\times$$ 1 column matrix of unknown variables and O is a 3 $$\\times$$ 1 null matrix, has :",
"options": [
{
"text": "no solution"
},
{
"text": "exactly two solutions"
},
{
"text": "infinitely many solutions"
},
{
"text": "a unique solution"
}
],
"answer": "infinitely many solutions",
"solution": "**Answer:** infinitely many solutions\n\nAT = A, BT = $$-$$B
Let A2B2 $$-$$ B2A2 = P
PT = (A2B2 $$-$$ B2A2)T = (A2B2)T $$-$$ (B2A2)T
= (B2)T (A2)T $$-$$ (A2)T (B2)T
= B2A2 $$-$$ A2B2
$$ \\Rightarrow $$ P is skew-symmetric matrix
$$\\left[ {\\matrix{\n 0 & a & b \\cr \n { - a} & 0 & c \\cr \n { - b} & { - c} & 0 \\cr \n\n } } \\right]\\left[ {\\matrix{\n x \\cr \n y \\cr \n z \\cr \n\n } } \\right] = \\left[ {\\matrix{\n 0 \\cr \n 0 \\cr \n 0 \\cr \n\n } } \\right]$$
$$ \\therefore $$ equation have infinite number of solution",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6262,
"subject": "General Science",
"question": "Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A2 is 1, then the possible number of such matrices is :",
"options": [
{
"text": "6"
},
{
"text": "4"
},
{
"text": "1"
},
{
"text": "12"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\nLet $$A = \\left[ {\\matrix{\n a & b \\cr \n b & c \\cr \n\n } } \\right]$$
$${A^2} = \\left[ {\\matrix{\n a & b \\cr \n b & c \\cr \n\n } } \\right]\\left[ {\\matrix{\n a & b \\cr \n b & c \\cr \n\n } } \\right] = \\left[ {\\matrix{\n {{a^2} + {b^2}} & {ab + bc} \\cr \n {ab + bc} & {{c^2} + {b^2}} \\cr \n\n } } \\right]$$
$$ = {a^2} + 2{b^2} + {c^2} = 1$$
$$a = 1,b = 0,c = 0$$
$$a = 0,b = 0,c = 1$$
$$a = - 1,b = 0,c = 0$$
$$c = - 1,b = 0,a = 0$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6263,
"subject": "General Science",
"question": "Let $$A = \\left[ {\\matrix{\n 2 & 3 \\cr \n a & 0 \\cr \n\n } } \\right]$$, a$$\\in$$R be written as P + Q where P is a symmetric matrix and Q is skew symmetric matrix. If det(Q) = 9, then the modulus of the sum of all possible values of determinant of P is equal to :",
"options": [
{
"text": "36"
},
{
"text": "24"
},
{
"text": "45"
},
{
"text": "18"
}
],
"answer": "36",
"solution": "**Answer:** 36\n\n$$A = \\left[ {\\matrix{\n 2 & 3 \\cr \n a & 0 \\cr \n\n } } \\right]$$, $${A^T} = \\left[ {\\matrix{\n 2 & a \\cr \n 3 & 0 \\cr \n\n } } \\right]$$
Let $$A = \\left[ {\\matrix{\n 0 & { - 2} \\cr \n 2 & 0 \\cr \n\n } } \\right]$$. If M and N are two matrices given by $$M = \\sum\\limits_{k = 1}^{10} {{A^{2k}}} $$ and $$N = \\sum\\limits_{k = 1}^{10} {{A^{2k - 1}}} $$ then MN2 is :
",
"options": [
{
"text": "a non-identity symmetric matrix"
},
{
"text": "a skew-symmetric matrix"
},
{
"text": "neither symmetric nor skew-symmetric matrix"
},
{
"text": "an identity matrix"
}
],
"answer": "a non-identity symmetric matrix",
"solution": "**Answer:** a non-identity symmetric matrix\n\n
Let $$\\mathrm{A}$$ and $$\\mathrm{B}$$ be any two $$3 \\times 3$$ symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?
",
"options": [
{
"text": "$$\\mathrm{A}^{4}-\\mathrm{B}^{4}$$ is a smmetric matrix"
},
{
"text": "$$\\mathrm{AB}-\\mathrm{BA}$$ is a symmetric matrix"
},
{
"text": "$$\\mathrm{B}^{5}-\\mathrm{A}^{5}$$ is a skew-symmetric matrix"
},
{
"text": "$$\\mathrm{AB}+\\mathrm{BA}$$ is a skew-symmetric matrix"
}
],
"answer": "$$\\mathrm{B}^{5}-\\mathrm{A}^{5}$$ is a skew-symmetric matrix",
"solution": "**Answer:** $$\\mathrm{B}^{5}-\\mathrm{A}^{5}$$ is a skew-symmetric matrix\n\n
Let A be a symmetric matrix such that $$\\mathrm{|A|=2}$$ and $$\\left[ {\\matrix{\n 2 & 1 \\cr \n 3 & {{3 \\over 2}} \\cr \n\n } } \\right]A = \\left[ {\\matrix{\n 1 & 2 \\cr \n \\alpha & \\beta \\cr \n\n } } \\right]$$. If the sum of the diagonal elements of A is $$s$$, then $$\\frac{\\beta s}{\\alpha^2}$$ is equal to __________.
Sure! A symmetric matrix is a square matrix that is equal to its transpose. For a matrix to be symmetric, the element at row i and column j must be equal to the element at row j and column i. In other words, $$A_{ij} = A_{ji}$$.
\n
For a 3 $$ \\times $$ 3 symmetric matrix, it looks like this:
\n
$$\n\\begin{pmatrix}\na & b & c \\\\\nb & d & e \\\\\nc & e & f \\\\\n\\end{pmatrix}\n$$\n
\n
Notice that there are only 6 unique elements we need to fill because of the symmetry:
\n\n
$$a$$ in the (1,1) position
\n
$$b$$ in the (1,2) and (2,1) positions
\n
$$c$$ in the (1,3) and (3,1) positions
\n
$$d$$ in the (2,2) position
\n
$$e$$ in the (2,3) and (3,2) positions
\n
$$f$$ in the (3,3) position
\n\n
Each of these unique elements can take a value from the set $${0,1,2,3,4,5,6,7,8,9}$$, which has 10 elements.
\n
We have 10 choices for each of the 6 unique elements, so the total number of symmetric matrices can be calculated as:
Thus, the total number of symmetric matrices of order 3 with entries from this set is $$10^{6}$$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6269,
"subject": "General Science",
"question": "Let $$A$$ be $$a\\,2 \\times 2$$ matrix with real entries. Let $$I$$ be the $$2 \\times 2$$ identity matrix. Denote by tr$$(A)$$, the sum of diagonal entries of $$a$$. Assume that $${a^2} = I.$$ \n Statement-1 : If $$A \\ne I$$ and $$A \\ne - I$$, then det$$(A)=-1$$\n Statement- 2 : If $$A \\ne I$$ and $$A \\ne - I$$, then tr $$(A)$$ $$ \\ne 0$$.",
"options": [
{
"text": "statement - 1 is false, statement -2 is true "
},
{
"text": "statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1."
},
{
"text": "statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1."
},
{
"text": "statement - 1 is true, statement - 2 is false."
}
],
"answer": "statement - 1 is true, statement - 2 is false.",
"solution": "**Answer:** statement - 1 is true, statement - 2 is false.\n\nLet $$A = \\left[ {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right]$$ $$\\,\\,\\,$$ then $${A^2} = 1$$\n
$$ \\Rightarrow a + d = 0 \\Rightarrow Tr\\left( A \\right) = 0$$\n
$$\\left| A \\right| = ad - bc = - {a^2} - bc = - 1$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6271,
"subject": "General Science",
"question": "The number of all 3 × 3 matrices A, with\nenteries from the set {–1, 0, 1} such that the sum\nof the diagonal elements of AAT is 3, is",
"options": [],
"answer": "672",
"solution": "**Answer:** 672\n\nLet A = $$\\left[ {\\matrix{\n {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\cr \n {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\cr \n {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \\cr \n\n } } \\right]$$\n
This is only possible when three enteries must be either 1 or – 1 and all other six enteries are 0.\n
$$ \\therefore $$ Number of matrices = 9C3 $$ \\times $$ 2 $$ \\times $$ 2 $$ \\times $$ 2\n
= 672",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6272,
"subject": "General Science",
"question": "Let A be a 2 $$ \\times $$ 2 real matrix with entries from\n{0, 1} and |A|\n$$ \\ne $$ 0. Consider the following two\nstatements :\n
(P) If A $$ \\ne $$ I2\n, then |A| = –1\n (Q) If |A| = 1, then tr(A) = 2,\n
where I2\n denotes 2 $$ \\times $$ 2 identity matrix and tr(A)\ndenotes the sum of the diagonal entries of A. Then :\n",
"options": [
{
"text": "(P) is true and (Q) is false"
},
{
"text": "Both (P) and (Q) are false"
},
{
"text": "Both (P) and (Q) are true"
},
{
"text": "(P) is false and (Q) is true"
}
],
"answer": "(P) is false and (Q) is true",
"solution": "**Answer:** (P) is false and (Q) is true\n\nLet A = $$\\left[ {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right]$$, where a, b, c, d $$ \\in $$ {0, 1}\n
$$ \\Rightarrow $$ |A| = ad – bc\n
$$ \\therefore $$ ad = 0 or 1 and bc = 0 or 1\n
So possible values of |A| are 1, 0 or –1\n
(P) If A $$ \\ne $$ I2\n then |A| is either 0 or –1\n
(Q) If |A| = 1 then ad = 1 and bc = 0\n
$$ \\Rightarrow $$ a = d = 1 $$ \\Rightarrow $$ Tr(A) = 2",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6273,
"subject": "General Science",
"question": "The total number of 3 $$\\times$$ 3 matrices A having entries from the set {0, 1, 2, 3} such that the sum of all the diagonal entries of AAT is 9, is equal to _____________.",
"options": [],
"answer": "766",
"solution": "**Answer:** 766\n\n$$A{A^T} = \\left[ {\\matrix{\n x & y & z \\cr \n a & b & c \\cr \n d & e & f \\cr \n\n } } \\right]\\left[ {\\matrix{\n x & a & d \\cr \n y & b & e \\cr \n z & c & f \\cr \n\n } } \\right]$$
Let $$A = \\left( {\\matrix{\n 1 & 0 & 0 \\cr \n 0 & 4 & { - 1} \\cr \n 0 & {12} & { - 3} \\cr \n\n } } \\right)$$. Then the sum of the diagonal elements of the matrix $${(A + I)^{11}}$$ is equal to :
Let $$A$$ be a $$2 \\times 2$$ real matrix and $$I$$ be the identity matrix of order 2. If the roots of the equation $$|\\mathrm{A}-x \\mathrm{I}|=0$$ be $$-1$$ and 3, then the sum of the diagonal elements of the matrix $$\\mathrm{A}^2$$ is
Let $$A=\\left[\\begin{array}{ll}a & b \\\\ c & d\\end{array}\\right]$$
\n
We have $$\\mathrm{a}+\\mathrm{d}=2$$
\n
$$\\mathrm{ad}-\\mathrm{bc}=-3$$
\n
$$A^2=\\left[\\begin{array}{ll}a & b \\\\\nc & d\n\\end{array}\\right] \\times\\left[\\begin{array}{ll}\na & b \\\\\nc & d\n\\end{array}\\right]=\\left[\\begin{array}{ll}\na^2+b c & a b+b d \\\\\na c+c d & b c+d^2\n\\end{array}\\right]$$
\n
We need $$a^2+b c+b c+d^2$$
\n
$$\\begin{aligned}\n& =a^2+2 b c+d^2 \\\\\n& =(a+d)^2-2 a d+2 b c \\\\\n& =4-2(a d-b c) \\\\\n& =4-2(-3) \\\\\n& =4+6 \\\\\n& =10\n\\end{aligned}$$
Let $$R=\\left(\\begin{array}{ccc}x & 0 & 0 \\\\ 0 & y & 0 \\\\ 0 & 0 & z\\end{array}\\right)$$ be a non-zero $$3 \\times 3$$ matrix, where $$x \\sin \\theta=y \\sin \\left(\\theta+\\frac{2 \\pi}{3}\\right)=z \\sin \\left(\\theta+\\frac{4 \\pi}{3}\\right) \\neq 0, \\theta \\in(0,2 \\pi)$$. For a square matrix $$M$$, let trace $$(M)$$ denote the sum of all the diagonal entries of $$M$$. Then, among the statements:
\n
(I) Trace $$(R)=0$$
\n
(II) If trace $$(\\operatorname{adj}(\\operatorname{adj}(R))=0$$, then $$R$$ has exactly one non-zero entry.
",
"options": [
{
"text": "Only (I) is true\n"
},
{
"text": "Only (II) is true\n"
},
{
"text": "Both (I) and (II) are true\n"
},
{
"text": "Neither (I) nor (II) is true"
}
],
"answer": "Neither (I) nor (II) is true",
"solution": "**Answer:** Neither (I) nor (II) is true\n\n
$$\\begin{aligned}\n& x \\sin \\theta=y \\sin \\left(\\theta+\\frac{2 \\pi}{3}\\right)=z \\sin \\left(\\theta+\\frac{4 \\pi}{3}\\right) \\neq 0 \\\\\n& \\Rightarrow x, y, z \\neq 0\n\\end{aligned}$$
Let $$A$$ be a square matrix of order 2 such that $$|A|=2$$ and the sum of its diagonal elements is $$-$$3 . If the points $$(x, y)$$ satisfying $$\\mathrm{A}^2+x \\mathrm{~A}+y \\mathrm{I}=\\mathrm{O}$$ lie on a hyperbola, whose transverse axis is parallel to the $$x$$-axis, eccentricity is $$\\mathrm{e}$$ and the length of the latus rectum is $$l$$, then $$\\mathrm{e}^4+l^4$$ is equal to ________.
Let $$A=\\left[\\begin{array}{cc}2 & -1 \\\\ 1 & 1\\end{array}\\right]$$. If the sum of the diagonal elements of $$A^{13}$$ is $$3^n$$, then $$n$$ is equal to ________.
Total possible combination of x and y = 2 $$ \\times $$ 2 = 4",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6285,
"subject": "General Science",
"question": "Let a, b, c $$ \\in $$ R be all non-zero and satisfy\n a3 + b3 + c3 = 2. If the matrix\n
A = $$\\left( {\\matrix{\n a & b & c \\cr \n b & c & a \\cr \n c & a & b \\cr \n\n } } \\right)$$\n
satisfies ATA = I, then a value of abc can be :",
"options": [
{
"text": "3"
},
{
"text": "$${1 \\over 3}$$"
},
{
"text": "-$${1 \\over 3}$$"
},
{
"text": "$${2 \\over 3}$$"
}
],
"answer": "$${1 \\over 3}$$",
"solution": "**Answer:** $${1 \\over 3}$$\n\nGiven, \n$${a^3} + {b^3} + {c^3} = 2$$
\n$${A^T}A = I$$
\n$$ \\Rightarrow \\left[ {\\matrix{\n a & b & c \\cr \n b & c & a \\cr \n c & a & b \\cr \n\n } } \\right]\\left[ {\\matrix{\n a & b & c \\cr \n b & c & a \\cr \n c & a & b \\cr \n\n } } \\right] = \\left[ {\\matrix{\n 1 & 0 & 0 \\cr \n 0 & 1 & 0 \\cr \n 0 & 0 & 1 \\cr \n\n } } \\right]$$
\n$$ \\Rightarrow {a^2} + {b^2} + {c^2} = 1$$
\nand ab + bc + ca = 0
\nNow (a + b + c)2 = 1
\n$$ \\Rightarrow (a + b + c) = \\pm 1$$
\nSo, $${a^3} + {b^3} + {c^3} - 3abc$$
\n$$ \\Rightarrow (a + b + c)\\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \\right)$$
\n$$ \\Rightarrow \\pm 1(1 - 0) = \\pm 1$$
$$ \\therefore 2 - 3abc = \\pm 1$$
\n$$ \\Rightarrow 3abc = 2 \\pm 1 = 3,1$$
\n$$ \\Rightarrow abc = 1,{1 \\over 3}$$\n",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6286,
"subject": "General Science",
"question": "Let M be any 3 $$ \\times $$ 3 matrix with entries from the set {0, 1, 2}. The maximum number of such matrices, for which the sum of diagonal elements of MTM is seven, is ________.",
"options": [],
"answer": "540",
"solution": "**Answer:** 540\n\n$$\\left[ {\\matrix{\n a & b & c \\cr \n d & e & f \\cr \n g & h & i \\cr \n\n } } \\right]\\left[ {\\matrix{\n a & d & g \\cr \n b & e & h \\cr \n c & f & i \\cr \n\n } } \\right]$$
Let $$S$$ be the set containing all $$3 \\times 3$$ matrices with entries from $$\\{-1,0,1\\}$$. The total number of matrices $$A \\in S$$ such that the sum of all the diagonal elements of $$A^{\\mathrm{T}} A$$ is 6 is ____________.
Let $$P=\\left[\\begin{array}{cc}\\frac{\\sqrt{3}}{2} & \\frac{1}{2} \\\\ -\\frac{1}{2} & \\frac{\\sqrt{3}}{2}\\end{array}\\right], A=\\left[\\begin{array}{ll}1 & 1 \\\\ 0 & 1\\end{array}\\right]$$ and $$Q=P A P^{T}$$. If $$P^{T} Q^{2007} P=\\left[\\begin{array}{ll}a & b \\\\ c & d\\end{array}\\right]$$, then $$2 a+b-3 c-4 d$$ equal to :
The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant. Given $\\mathrm{A}$, $\\mathrm{B}$, and that $\\mathrm{C} = \\mathrm{ABA}^{\\mathrm{T}}$, and $\\mathrm{X} = \\mathrm{A}^{\\mathrm{T}} \\mathrm{C}^2 \\mathrm{A}$, we find $\\operatorname{det} \\mathrm{X}$ as follows:
\n
First, we express $|\\mathrm{C}|$, the determinant of $\\mathrm{C}$, in terms of the determinants of $\\mathrm{A}$ and $\\mathrm{B}$, using the property that $|\\mathrm{ABC}| = |\\mathrm{A}| \\cdot |\\mathrm{B}| \\cdot |\\mathrm{C}|$:
Next, we find $|\\mathrm{X}|$, using the property $|\\mathrm{ABC}| = |\\mathrm{A}| \\cdot |\\mathrm{B}| \\cdot |\\mathrm{C}|$ and substituting $|\\mathrm{C}|$:
Let $$\\mathrm{A}$$ be a square matrix such that $$\\mathrm{AA}^{\\mathrm{T}}=\\mathrm{I}$$. Then $$\\frac{1}{2} A\\left[\\left(A+A^T\\right)^2+\\left(A-A^T\\right)^2\\right]$$ is equal to
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 6292,
"subject": "General Science",
"question": "Tangents drawn from the point ($$-$$8, 0) to the parabola y2 = 8x touch the parabola at $$P$$ and $$Q.$$ If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :",
"options": [
{
"text": "24"
},
{
"text": "32"
},
{
"text": "48"
},
{
"text": "64"
}
],
"answer": "48",
"solution": "**Answer:** 48\n\nEquation of the chord of contact PQ is given by : T=0 \n
or T $$ \\equiv $$ yy1 $$-$$ 4(x + x1), where (x1, y1) $$ \\equiv $$ ($$-$$8, 0)\n
$$\\therefore\\,\\,\\,$$Equation becomes : x = 8\n
& chord of contact is x = 8\n
$$\\therefore\\,\\,\\,$$ Coordinates of point P and Q are (8, 8) and (8, $$-$$ 8) \n
$$ = \\sqrt {108} = 6\\sqrt 3 $$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6295,
"subject": "General Science",
"question": "If one end of a focal chord of the parabola,\ny2 = 16x is at (1, 4), then the length of this focal\nchord is :",
"options": [
{
"text": "24"
},
{
"text": "20"
},
{
"text": "25"
},
{
"text": "22"
}
],
"answer": "25",
"solution": "**Answer:** 25\n\n\n For this parabola y2 = 16x,\n
a = 4\n
Here PQ is focal cord.\n
Let P(at12, 2at1) and Q(at22, 2at2).\n
Given P(1, 4),\n
$$ \\therefore $$ at12 = 1\n
$$ \\Rightarrow $$ 4t12 = 1\n
$$ \\Rightarrow $$ t12 = $${1 \\over 4}$$\n
$$ \\Rightarrow $$ t1 = $${1 \\over 2}$$\n
In parabola if the parameter of one end point of the focal cord is t1 then parameter of the other end point t2 = $$ - {1 \\over {{t_1}}}$$\n
Here parameter for point Q t2 = - 2\n
$$ \\therefore $$ Length of focal cord \n
|PQ| = a$${\\left( {{t_1} - {t_2}} \\right)^2}$$ = 4$${\\left( {{1 \\over 2} + 2} \\right)^2}$$ = 25",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6296,
"subject": "General Science",
"question": "Let the latus ractum of the parabola y2\n = 4x be\nthe common chord to the circles C1\n and C2\neach of them having radius 2$$\\sqrt 5 $$. Then, the\ndistance between the centres of the circles C1\nand C2\n is :",
"options": [
{
"text": "8"
},
{
"text": "12"
},
{
"text": "$$8\\sqrt 5 $$"
},
{
"text": "$$4\\sqrt 5 $$"
}
],
"answer": "8",
"solution": "**Answer:** 8\n\n\n For parabola y2 = 4x, \n
Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of $${\\pi \\over 2}$$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse $$E:{{{x^2}} \\over {{a^2}}} + {{{y^2}} \\over {{b^2}}} = 1$$, $${a^2} > {b^2}$$. If e is the eccentricity of the ellipse E, then the value of $${1 \\over {{e^2}}}$$ is equal to :
Let PQ be a focal chord of length 6.25 units of the parabola y2 = 4x. If O is the vertex of the parabola, then 10 times the area (in sq. units) of $$\\Delta$$POQ is equal to ___________.
Let the focal chord of the parabola $$\\mathrm{P}: y^{2}=4 x$$ along the line $$\\mathrm{L}: y=\\mathrm{m} x+\\mathrm{c}, \\mathrm{m}>0$$ meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola $$\\mathrm{H}: x^{2}-y^{2}=4$$. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is :
Let $$\\mathrm{PQ}$$ be a focal chord of the parabola $$y^{2}=36 x$$ of length 100 , making an acute angle with the positive $$x$$-axis. Let the ordinate of $$\\mathrm{P}$$ be positive and $$\\mathrm{M}$$ be the point on the line segment PQ such that PM : MQ = 3 : 1. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line $$\\mathrm{PQ}$$?
",
"options": [
{
"text": "$$(6,29)$$"
},
{
"text": "$$(-3,43)$$"
},
{
"text": "$$(3,33)$$"
},
{
"text": "$$(-6,45)$$"
}
],
"answer": "$$(-3,43)$$",
"solution": "**Answer:** $$(-3,43)$$\n\nThe given parabola is of the form $$y^2 = 4ax$$. Here, 4a = 36, which means a = 9.\n
Length of focal chord at $(t)=a\\left(t+\\frac{1}{t}\\right)^2=100$\n
$$\n\\therefore t=3 \\text { or } \\frac{1}{3}\n$$\n \n
For each value of t, we can find the coordinates of the points P and Q on the parabola using the parametric form of the parabola $y^2=4ax$, where a is 9. The parametric form is given by $x=at^2$ and $y=2at$.
\n
Let's consider $t=3$ first.
\n\n
For $t=3$, the coordinates of P are given by $P(at^2, 2at) = P(81, 54)$.
\n
\n
Since $t_1t_2=-1$ for a focal chord, the parameter value for Q is $t_2=-1/t_1=-1/3$. So the coordinates of Q are given by $Q(at_2^2, 2at_2) = Q(1, -6)$.
\n
\n\n
Now, let's find the coordinates of the point M that divides the line segment PQ in the ratio 3:1.
\n\n
The x-coordinate of M is given by $\\frac{3Q_x+P_x}{4} = \\frac{3*1+81}{4} = 21$.
\n
\n
The y-coordinate of M is given by $\\frac{3Q_y+P_y}{4} = \\frac{3*(-6)+54}{4} = 9$.
\n
\n\n
So, the coordinates of M are M$(21, 9)$.
\n
We can repeat these steps for $t=1/3$ to find the other set of points P, Q, and M. However, since the ordinate of P is positive and PQ makes an acute angle with the positive x-axis, $t=3$ is the appropriate choice in this context.
\n
The slope of the line PQ is $m_{PQ} = \\frac{-6 - 54}{1 - 81} = \\frac{-60}{-80} = \\frac{3}{4}$. The slope of the line perpendicular to PQ is $m_{\\perp PQ} = -1/m_{PQ} = -\\frac{4}{3}$. Thus, the equation of the line through M and perpendicular to PQ is $(y - 9) = -\\frac{4}{3}(x - 21)$, or $4x + 3y = 111$.
\n
Now, we check which of the given points do NOT lie on this line.
\n
Option A : $(6, 29)$\nSubstitute these values into the equation :
\n
$4\\times6 + 3\\times29 = 111?$
\n
$24 + 87 = 111?$
\n
$111 = 111$
\n
So, option A does lie on the line.
\n
Option B : $(-3, 43)$\nSubstitute these values into the equation :
\n
$4\\times(-3) + 3\\times43 = 111?$
\n
$-12 + 129 = 111?$
\n
$117 = 111$
\n
So, option B does NOT lie on the line.
\n
Option C : $(3, 33)$\nSubstitute these values into the equation :
\n
$4\\times3 + 3\\times33 = 111?$
\n
$12 + 99 = 111?$
\n
$111 = 111$
\n
So, option C does lie on the line.
\n
Option D : $(-6, 45)$\nSubstitute these values into the equation :
\n
$4\\times(-6) + 3\\times45 = 111?$
\n
$-24 + 135 = 111?$
\n
$111 = 111$
\n
So, option D does lie on the line.
\n
Therefore, $(-3, 43)$ which does not lie on the line passing through M and perpendicular to the line PQ.
Let $$P(\\alpha, \\beta)$$ be a point on the parabola $$y^2=4 x$$. If $$P$$ also lies on the chord of the parabola $$x^2=8 y$$ whose mid point is $$\\left(1, \\frac{5}{4}\\right)$$, then $$(\\alpha-28)(\\beta-8)$$ is equal to _________.
Let the length of the focal chord PQ of the parabola $$y^2=12 x$$ be 15 units. If the distance of $$\\mathrm{PQ}$$ from the origin is $$\\mathrm{p}$$, then $$10 \\mathrm{p}^2$$ is equal to __________.
Let $$P Q$$ be a chord of the parabola $$y^2=12 x$$ and the midpoint of $$P Q$$ be at $$(4,1)$$. Then, which of the following point lies on the line passing through the points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ ?
Suppose $$\\mathrm{AB}$$ is a focal chord of the parabola $$y^2=12 x$$ of length $$l$$ and slope $$\\mathrm{m}<\\sqrt{3}$$. If the distance of the chord $$\\mathrm{AB}$$ from the origin is $$\\mathrm{d}$$, then $$l \\mathrm{~d}^2$$ is equal to _________.
So from $$(i),$$ $$y = \\pm \\left( {x + 2a} \\right).$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6308,
"subject": "General Science",
"question": "Given : A circle, $$2{x^2} + 2{y^2} = 5$$ and a parabola, $${y^2} = 4\\sqrt 5 x$$. \n Statement-1 : An equation of a common tangent to these curves is $$y = x + \\sqrt 5 $$.\n
Statement-2 : If the line, $$y = mx + {{\\sqrt 5 } \\over m}\\left( {m \\ne 0} \\right)$$ is their common tangent, then $$m$$ satiesfies $${m^4} - 3{m^2} + 2 = 0$$.
",
"options": [
{
"text": "Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1."
},
{
"text": "Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."
},
{
"text": "Statement-1 is true; Statement-2 is false."
},
{
"text": "Statement-1 is false Statement-2 is true."
}
],
"answer": "Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.",
"solution": "**Answer:** Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.\n\nLet common tangent be \n
$$y = mx + {{\\sqrt 5 } \\over m}$$\n
Since, perpendicular distance from center of the circle to \n
the common tangent is equal to radius of the circle, \n
$$ \\Rightarrow $$ m2 = 2 + 2$$\\sqrt 2 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6310,
"subject": "General Science",
"question": "Two parabolas with a common vertex and with axes along x-axis and $$y$$-axis, respectively intersect each other in the first quadrant. If the length of the latus rectum of each parabola is $$3$$, then the equation of the common tangent to the two parabolas is :",
"options": [
{
"text": "4(x + y) + 3 = 0"
},
{
"text": "3(x + y) + 4 = 0"
},
{
"text": "8(2x + y) + 3 = 0"
},
{
"text": "x + 2y + 3 = 0"
}
],
"answer": "4(x + y) + 3 = 0",
"solution": "**Answer:** 4(x + y) + 3 = 0\n\nAs origin is the only common point to x-axis and y-axis, so origin is the common vertex \n
Let the equation of two of parabolas be y2 = 4ax and x2 = 4by\n
Now latus rectum of both parabolas = 3\n
$$\\therefore\\,\\,\\,$$ 4a = 4b = 3\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ a = b = $${3 \\over 4}$$\n
$$\\therefore\\,\\,\\,$$ Two parabolas are y2 = 3x and x2 = 3y\n
\n$$ \\therefore $$ tangent are y = x + $$\\sqrt 2$$ & y = – x – $$\\sqrt 2$$
\nCompare with y = – ax + c
\n$$ \\Rightarrow a = \\pm 1 $$ & $$ c = \\pm \\sqrt 2 $$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6313,
"subject": "General Science",
"question": "Let P be the point of intersection of the common tangents to the parabola y2\n = 12x and the hyperbola\n8x2\n– y2\n = 8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS'\nin a ratio : \n",
"options": [
{
"text": "14 : 13"
},
{
"text": "13 : 11"
},
{
"text": "5 : 4"
},
{
"text": "2 : 1"
}
],
"answer": "5 : 4",
"solution": "**Answer:** 5 : 4\n\nLet equation of common tangent is y = mx + $${3 \\over m}$$
\ny = x + 4 or x – y + 4 = 0",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6315,
"subject": "General Science",
"question": "If the common tangent to the parabolas, y2 = 4x and x2 = 4y also touches the circle, x2 + y2 = c2, then c is equal to :",
"options": [
{
"text": "$${1 \\over {\\sqrt 2 }}$$"
},
{
"text": "$${1 \\over {2\\sqrt 2 }}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${1 \\over 4}$$"
}
],
"answer": "$${1 \\over {\\sqrt 2 }}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt 2 }}$$\n\n$$y = mx + {1 \\over m}$$ (tangent at y2\n = 4x)\n
y = mx – m2\n (tangent at x2\n = 4y)\n
$${1 \\over m} = - {m^2}$$ (for common tangent)\n
m3\n = – 1\n
$$ \\Rightarrow $$ m = - 1\n
$$ \\therefore $$ Equation of tangent\n
y = –x –1\n
x + y + 1 = 0\n
This line touches circle whose center at (0, 0),\n
$$ \\therefore $$ apply p( Distance from center of the circle of the line ) = r ( Radius of the circle )\n
c = $$\\left| {{{0 + 0 + 1} \\over {\\sqrt 2 }}} \\right| = {1 \\over {\\sqrt 2 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6316,
"subject": "General Science",
"question": "A line is a common tangent to the circle (x $$-$$ 3)2 + y2 = 9 and the parabola y2 = 4x. If the two points of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a + c) is equal to _________.",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\nCircle : (x $$-$$ 3)2 + y2 = 9
$$ = 2\\left( {{3 \\over 2} + {6 \\over 2}} \\right) = 9$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6317,
"subject": "General Science",
"question": "Let L be a tangent line to the parabola y2 = 4x $$-$$ 20 at (6, 2). If L is also a tangent to the ellipse $${{{x^2}} \\over 2} + {{{y^2}} \\over b} = 1$$, then the value of b is equal to :",
"options": [
{
"text": "20"
},
{
"text": "14"
},
{
"text": "16"
},
{
"text": "11"
}
],
"answer": "14",
"solution": "**Answer:** 14\n\nParabola y2 = 4x $$-$$ 20
This is also tangent to ellipse $${{{x^2}} \\over 2} + {{{y^2}} \\over b} = 1$$
Apply c2 = a2m2 + b2
($$-$$4)2 = (2)(1) + b
b = 14",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6318,
"subject": "General Science",
"question": "A tangent line L is drawn at the point (2, $$-$$4) on the parabola y2 = 8x. If the line L is also tangent to the circle x2 + y2 = a, then 'a' is equal to ___________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\ntangent of y2 = 8x is y = mx + $${2 \\over m}$$
Let the common tangents to the curves $$4({x^2} + {y^2}) = 9$$ and $${y^2} = 4x$$ intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and l respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $${l \\over {{e^2}}}$$ is equal to ______________.
If $$y = {m_1}x + {c_1}$$ and $$y = {m_2}x + {c_2}$$, $${m_1} \\ne {m_2}$$ are two common tangents of circle $${x^2} + {y^2} = 2$$ and parabola y2 = x, then the value of $$8|{m_1}{m_2}|$$ is equal to :
Two tangent lines $$l_{1}$$ and $$l_{2}$$ are drawn from the point $$(2,0)$$ to the parabola $$2 \\mathrm{y}^{2}=-x$$. If the lines $$l_{1}$$ and $$l_{2}$$ are also tangent to the circle $$(x-5)^{2}+y^{2}=r$$, then 17r is equal to ___________.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6324,
"subject": "General Science",
"question": "Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16 x$. If one of these tangents touches the two curves at $Q$ and $R$, then $(Q R)^2$ is equal to :",
"options": [
{
"text": "76"
},
{
"text": "81"
},
{
"text": "64"
},
{
"text": "72"
}
],
"answer": "72",
"solution": "**Answer:** 72\n\n
Let a tangent on $${y^2} = 16x$$ be $$y = mx + {4 \\over m}$$
\nDistance of point $(6,-2 \\sqrt{2})$ from common tangent $=5$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6326,
"subject": "General Science",
"question": "
Let a common tangent to the curves $${y^2} = 4x$$ and $${(x - 4)^2} + {y^2} = 16$$ touch the curves at the points P and Q. Then $${(PQ)^2}$$ is equal to __________.
",
"options": [],
"answer": "32",
"solution": "**Answer:** 32\n\nTangent of slope $m$ to the parabola \n
$y^2=4 x$ is given by $y=m x+\\frac{1}{m}$ and
Tangent of slope $m$ to the circle $(x-4)^2+y^2=16$ is given by\n
Then, the point of contact on parabola is $(8,4 \\sqrt{2})$ \n
Length of tangent $P Q$ from $(8,4 \\sqrt{2})$ on the circle is\n
$$\n\\begin{array}{ll}\n&\\Rightarrow P Q=\\sqrt{(8-4)^2+(4 \\sqrt{2})^2-16} \\\\\\\\\n&\\Rightarrow P Q=\\sqrt{16+32-16} \\\\\\\\\n&\\Rightarrow P Q=\\sqrt{32} \\Rightarrow(P Q)^2=32\n\\end{array}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6327,
"subject": "General Science",
"question": "Let the line $\\mathrm{L}: \\sqrt{2} x+y=\\alpha$ pass through the point of the intersection $\\mathrm{P}$ (in the first quadrant) of the circle $x^2+y^2=3$ and the parabola $x^2=2 y$. Let the line $\\mathrm{L}$ touch two circles $\\mathrm{C}_1$ and $\\mathrm{C}_2$ of equal radius $2 \\sqrt{3}$. If the centres $Q_1$ and $Q_2$ of the circles $C_1$ and $C_2$ lie on the $y$-axis, then the square of the area of the triangle $\\mathrm{PQ}_1 \\mathrm{Q}_2$ is equal to ___________.",
"options": [],
"answer": "72",
"solution": "**Answer:** 72\n\n
$x^2+y^2=3$ and $x^2=2 y$
\n\n$y^2+2 y-3=0 $\n\n
$\\Rightarrow(y+3)(y-1)=0$\n\n
$y=-3$ (Rejected) or $y=1$\n\n
For $\\mathrm{y}=1, \\mathrm{x}=\\sqrt{2} \\Rightarrow P(\\sqrt{2}, 1)$\n\n
Let $$C$$ be the circle of minimum area touching the parabola $$y=6-x^2$$ and the lines $$y=\\sqrt{3}|x|$$. Then, which one of the following points lies on the circle $$C$$ ?
$$ \\Rightarrow {y^2} - 4x + 2 = 0.$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6330,
"subject": "General Science",
"question": "If two tangents drawn from a point $$P$$ to the parabola $${y^2} = 4x$$ are at right angles, then the locus of $$P$$ is ",
"options": [
{
"text": "$$2x+1=0$$"
},
{
"text": "$$x=-1$$ "
},
{
"text": "$$2x-1=0$$ "
},
{
"text": "$$x=1$$ "
}
],
"answer": "$$x=-1$$ ",
"solution": "**Answer:** $$x=-1$$ \n\nThe locus of perpendicular tangents is directrix \n
i.e., $$x=-1$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6331,
"subject": "General Science",
"question": "Let $$O$$ be the vertex and $$Q$$ be any point on the parabola, $${{x^2} = 8y}$$. If the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, then locus of $$P$$ is :",
"options": [
{
"text": "$${y^2} = 2x$$ "
},
{
"text": "$${{x^2} = 2y}$$ "
},
{
"text": "$${{x^2} = y}$$"
},
{
"text": "$${y^2} = x$$ "
}
],
"answer": "$${{x^2} = 2y}$$ ",
"solution": "**Answer:** $${{x^2} = 2y}$$ \n\n
Let the coordinates of Q and P be (x1, y1) and (h, k) respectively.
\n
$$\\because$$ Q lies on x2 = 8y,
\n
$$\\therefore$$ x$$_1^2$$ = 8y ....... (1)
\n
Again, P divides OQ internally in the ratio 1 : 3.
\n
$$\\therefore$$ $$h = {{{x_1} + 0} \\over 4} = {{{x_1}} \\over 4}$$ or x1 = 4h and
Replace (a, b) as (x, y) $$\\Rightarrow$$ (3y $$-$$ x) = 2(3x $$-$$ y)2 + 2
or $$2{(3x - y)^2} + (x - 3y) + 2 = 0$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6335,
"subject": "General Science",
"question": "If two tangents drawn from a point P to the parabola y2 = 16(x $$-$$ 3) are at right angles, then the locus of point P is :",
"options": [
{
"text": "x + 3 = 0"
},
{
"text": "x + 1 = 0"
},
{
"text": "x + 2 = 0"
},
{
"text": "x + 4 = 0"
}
],
"answer": "x + 1 = 0",
"solution": "**Answer:** x + 1 = 0\n\nLocus is directrix of parabola
x $$-$$ 3 + 4 = 0 $$\\Rightarrow$$ x + 1 = 0.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6336,
"subject": "General Science",
"question": "The parabolas : $a x^2+2 b x+c y=0$ and $d x^2+2 e x+f y=0$ intersect on the line $y=1$. If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in G.P., then :",
"options": [
{
"text": "$\\frac{d}{a}, \\frac{e}{b}, \\frac{f}{c}$ are in A.P."
},
{
"text": "$\\frac{d}{a}, \\frac{e}{b}, \\frac{f}{c}$ are in G.P."
},
{
"text": "$d, e, f$ are in A.P."
},
{
"text": "$d, e, f$ are in G.P."
}
],
"answer": "$\\frac{d}{a}, \\frac{e}{b}, \\frac{f}{c}$ are in A.P.",
"solution": "**Answer:** $\\frac{d}{a}, \\frac{e}{b}, \\frac{f}{c}$ are in A.P.\n\n
The equations of two sides of a variable triangle are $$x=0$$ and $$y=3$$, and its third side is a tangent to the parabola $$y^2=6x$$. The locus of its circumcentre is :
$$ \\Rightarrow {x^2} + y{}^2 - 4x + 8y + 12 = 0$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6340,
"subject": "General Science",
"question": "P and Q are two distinct points on the parabola, y2 = 4x, with parameters t and t1 respectively. If the normal at P passes through Q, then the minimum value of $$t_1^2$$ is :",
"options": [
{
"text": "2"
},
{
"text": "4"
},
{
"text": "6"
},
{
"text": "8"
}
],
"answer": "8",
"solution": "**Answer:** 8\n\nt1 = $$-$$ t $$-$$ $${2 \\over t}$$\n
$$\\left| c \\right|$$ = 10$$\\sqrt 3 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6342,
"subject": "General Science",
"question": "Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the\nparabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and $$\\angle $$CPB =\n$$\\theta $$, then a value of tan$$\\theta $$ is :",
"options": [
{
"text": "$${4 \\over 3}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "2"
},
{
"text": "3"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n\n
As equation of tangent PA at (x1, y1) on the parabola y2 = 4ax, \n
yy1 = 2a (x + x1)\n
here (x1, y1) = ( 16, 16)\n
y . 16 = 2.4 (x + 16)\n
$$ \\Rightarrow $$ 2y = x + 16 .....(1)\n
At pont A value of y = 0\n
putting y = 0 in equation (1) we get, \n
0 = x + 16\n
$$ \\Rightarrow $$ x = $$-$$ 16\n
$$\\therefore\\,\\,\\,$$ Coordinate of point A = ($$-$$ 16, 0)\n
Slope of line P A :\n
As$$\\,\\,\\,\\,$$ 2y = x + 16\n
$$\\therefore\\,\\,\\,$$ y = $${1 \\over 2}\\,$$ x + 8\n
$$ \\Rightarrow $$ tan $$\\theta $$ = 2 \n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6343,
"subject": "General Science",
"question": "If the parabolas y2 = 4b(x – c) and y2 = 8ax have a common normal, then which on of the following is a valid choice for the ordered triad (a, b, c)?",
"options": [
{
"text": "(1, 1, 3)"
},
{
"text": "(1, 1, 0)"
},
{
"text": "$$\\left( {{1 \\over 2},2,0} \\right)$$"
},
{
"text": "$$\\left( {{1 \\over 2},2,3} \\right)$$"
}
],
"answer": "(1, 1, 3)",
"solution": "**Answer:** (1, 1, 3)\n\nNormal to the two given curves are\n
By checking all options we found (A) is right option.\n
Note : We get that all the options are correct for m = 0\ni.e., when common normal is x-axis. But may be in question they want common\nnormal other than x – axis, hence answer is (A).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6344,
"subject": "General Science",
"question": "The area (in sq. units) of the smaller of the two\ncircles that touch the parabola, y2 = 4x at the point\n(1, 2) and the x-axis is :-",
"options": [
{
"text": "$$4\\pi \\left( {3 +\\sqrt 2 } \\right)$$"
},
{
"text": "$$8\\pi \\left( {2 - \\sqrt 2 } \\right)$$"
},
{
"text": "$$8\\pi \\left( {3 - 2\\sqrt 2 } \\right)$$"
},
{
"text": "$$4\\pi \\left( {2 - \\sqrt 2 } \\right)$$"
}
],
"answer": "$$8\\pi \\left( {3 - 2\\sqrt 2 } \\right)$$",
"solution": "**Answer:** $$8\\pi \\left( {3 - 2\\sqrt 2 } \\right)$$\n\n\n
Equation of tangent to the\nparabola y2 = 4x at P(1, 2),\n
T = 0\n
2y = $$4\\left( {{{x + 1} \\over 2}} \\right)$$\n
$$ \\Rightarrow $$ y = x + 1\n
Equation of normal of the tangent at point P(1, 2)\n
y - 2 = (-1)(x - 1)\n
$$ \\Rightarrow $$ y - 2 = - x + 1\n
$$ \\Rightarrow $$ x + y - 3 = 0\n
This normal also passes through the center (h, r) of the circle.\n
$$ \\therefore $$ r = 4 + $${2\\sqrt 2 }$$ and 4 - $${2\\sqrt 2 }$$\n
If r = 4 + $${2\\sqrt 2 }$$ then center of the circle is \n
(-1 - $${2\\sqrt 2 }$$, 4 + $${2\\sqrt 2 }$$). From the diagram you can see both the x coordinate and y coordinate of the circle should be positive but here x coordinate is negative.\n
So possible value of radius r = 4 - $${2\\sqrt 2 }$$\n
Then area of the circle\n
= $$\\pi $$r2\n
= $$\\pi $$(4 - $${2\\sqrt 2 }$$)2\n
= $$\\pi $$(16 + 8 - $${16\\sqrt 2 }$$)\n
= $$8\\pi \\left( {3 - 2\\sqrt 2 } \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6345,
"subject": "General Science",
"question": "If the tangent to the curve, y = ex\n at a point\n(c, ec) and the normal to the parabola, y2 = 4x\nat the point (1, 2) intersect at the same point on\nthe x-axis, then the value of c is ________ .\n",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nFor $$y = {e^x}$$
$$ \\Rightarrow $$ c = 4",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6346,
"subject": "General Science",
"question": "If the three normals drawn to the parabola, y2 = 2x pass through the point (a, 0) a $$\\ne$$ 0, then 'a' must be greater than :",
"options": [
{
"text": "$${1 \\over 2}$$"
},
{
"text": "1"
},
{
"text": "$$-$$1"
},
{
"text": "$$-$$$${1 \\over 2}$$"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nLet the equation of the normal is
y = mx $$-$$ 2am $$-$$ am3
here 4a = 2 $$ \\Rightarrow $$ a = $${1 \\over 2}$$
y = mx $$-$$ m $$-$$ $${1 \\over 2}$$m3
It passing through A(a, 0) then
0 = am $$-$$ m $$-$$ $${1 \\over 2}$$m3
m = 0, a $$-$$ 1 $$-$$ $${1 \\over 2}$$m2 = 0
m2 = 2(a $$-$$ 1) > 0
$$ \\Rightarrow $$ a > 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6347,
"subject": "General Science",
"question": "Let the tangent to the parabola S : y2 = 2x at the point P(2, 2) meet the x-axis at Q and normal at it meet the parabola S at the point R. Then the area (in sq. units) of the triangle PQR is equal to :",
"options": [
{
"text": "$${{25} \\over 2}$$"
},
{
"text": "$${{35} \\over 2}$$"
},
{
"text": "$${{15} \\over 2}$$"
},
{
"text": "25"
}
],
"answer": "$${{25} \\over 2}$$",
"solution": "**Answer:** $${{25} \\over 2}$$\n\n
Tangent at P : y(2) = 2(1/2) (x + 2)
$$\\Rightarrow$$ 2y = x + 2
$$\\therefore$$ Q = ($$-$$2, 0)
Normal at P : y $$-$$ 2 = $$-$$ $${{(2)} \\over {2.{1 \\over 2}}}(x - 2)$$
$$ = {{25} \\over 2}$$ sq. units.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6348,
"subject": "General Science",
"question": "A tangent and a normal are drawn at the point P(2, $$-$$4) on the parabola y2 = 8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is equal to :",
"options": [
{
"text": "$$-$$16"
},
{
"text": "$$-$$18"
},
{
"text": "$$-$$12"
},
{
"text": "$$-$$20"
}
],
"answer": "$$-$$16",
"solution": "**Answer:** $$-$$16\n\n
Given, parabola
\n
$${y^2} = 8x$$ ...... (i)
\n
Equation of tangent at $$P(2, - 4)$$ is
\n
$$ - 4y = 4(x + 2)$$
\n
or, $$x + y + 2 = 0$$ ..... (ii)
\n
and Equation of normal to the parabola is
\n
$$x - y + C = 0$$
\n
$$\\therefore$$ Normal passes through $$(2, - 4)$$
\n
$$\\therefore$$ $$C = - 6$$
\n
Normal : $$x - y = 6$$ ..... (iii)
\n
Equation of directrix of parabola
\n
$$x = - 2$$ ..... (iv)
\n
Point of intersection of tangent and normal with directrix are $$x = - 2$$ at $$A( - 2,0)$$ and $$B( - 2, - 8)$$ respectively.
\n
$$Q(a,b)$$ and $$P(2, - 4)$$ are given and AQBP is a square.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6349,
"subject": "General Science",
"question": "Consider the parabola with vertex $$\\left( {{1 \\over 2},{3 \\over 4}} \\right)$$ and the directrix $$y = {1 \\over 2}$$. Let P be the point where the parabola meets the line $$x = - {1 \\over 2}$$. If the normal to the parabola at P intersects the parabola again at the point Q, then (PQ)2 is equal to :",
"options": [
{
"text": "$${{75} \\over 8}$$"
},
{
"text": "$${{125} \\over {16}}$$"
},
{
"text": "$${{25} \\over 2}$$"
},
{
"text": "$${{15} \\over 2}$$"
}
],
"answer": "$${{125} \\over {16}}$$",
"solution": "**Answer:** $${{125} \\over {16}}$$\n\n $$\\left( {y - {3 \\over 4}} \\right) = {\\left( {x - {1 \\over 2}} \\right)^2}$$ .... (1)
Let the normal at the point on the parabola y2 = 6x pass through the point (5, $$-$$8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :
If $$\\mathrm{P}(\\mathrm{h}, \\mathrm{k})$$ be a point on the parabola $$x=4 y^{2}$$, which is nearest to the point $$\\mathrm{Q}(0,33)$$, then the distance of $$\\mathrm{P}$$ from the directrix of the parabola $$\\quad y^{2}=4(x+y)$$ is equal to :
Let the tangent to the parabola $$\\mathrm{y}^{2}=12 \\mathrm{x}$$ at the point $$(3, \\alpha)$$ be perpendicular to the line $$2 x+2 y=3$$. Then the square of distance of the point $$(6,-4)$$ from the normal to the hyperbola $$\\alpha^{2} x^{2}-9 y^{2}=9 \\alpha^{2}$$ at its point $$(\\alpha-1, \\alpha+2)$$ is equal to _________.
The ordinates of the points P and $$\\mathrm{Q}$$ on the parabola with focus $$(3,0)$$ and directrix $$x=-3$$ are in the ratio $$3: 1$$. If $$\\mathrm{R}(\\alpha, \\beta)$$ is the point of intersection of the tangents to the parabola at $$\\mathrm{P}$$ and $$\\mathrm{Q}$$, then $$\\frac{\\beta^{2}}{\\alpha}$$ is equal to _______________.
",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\n$$\n\\begin{aligned}\n& \\text { Give parabola is : } y^2=12 x \\quad(\\because a=3) \\\\\\\\\n& \\text { So, } \\mathrm{P} \\equiv\\left(a t_1^2, 2 a t_1\\right) \\\\\\\\\n& \\mathrm{Q} \\equiv\\left(a t_2^2, 2 a t_2\\right) \\\\\\\\\n& \\text { So, point } \\mathrm{R}(\\alpha, \\beta) \\equiv\\left(a t_1 t_2, a\\left(t_1+t_2\\right)\\right) \\\\\\\\\n& \\equiv((3 t)(3 t), 3(t+3 t))=\\left(9 t^2, 12 t\\right) \\\\\\\\\n& \\therefore \\frac{\\beta^2}{\\alpha}=\\frac{144 t^2}{9 t^2}=16\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6355,
"subject": "General Science",
"question": "Let A(4, $$-$$ 4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of $$\\Delta $$ACB is maximum. Then, the area (in sq. units) of $$\\Delta $$ACB, is : ",
"options": [
{
"text": "$$31{1 \\over 4}$$"
},
{
"text": "$$30{1 \\over 2}$$"
},
{
"text": "32"
},
{
"text": "$$31{3 \\over 4}$$"
}
],
"answer": "$$31{1 \\over 4}$$",
"solution": "**Answer:** $$31{1 \\over 4}$$\n\n\n
$$ = {{125} \\over 2} = 31{1 \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6356,
"subject": "General Science",
"question": "If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :",
"options": [
{
"text": "$$5\\sqrt 5 $$"
},
{
"text": "$${\\left( {10} \\right)^{2/3}}$$"
},
{
"text": "$$5\\left( {{2^{1/3}}} \\right)$$"
},
{
"text": "5"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\nVertex is (a2, 0)\n
y2 $$=$$ $$-$$(x $$-$$ a2) and x $$=$$ 0 $$ \\Rightarrow $$ (0, $$ \\pm $$ 2a)\n
Area of triangle is $$ = {1 \\over 2}.$$4a.(a2) = 250\n
$$ \\Rightarrow $$ a3 = 125 or a = 5",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 6357,
"subject": "General Science",
"question": "The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x2 such that the rectangle lies inside the parabola, is : \n",
"options": [
{
"text": "36"
},
{
"text": "20$$\\sqrt 2 $$"
},
{
"text": "18$$\\sqrt 3 $$"
},
{
"text": "32"
}
],
"answer": "32",
"solution": "**Answer:** 32\n\n\n f (a) = 2a(12 $$-$$ a)2\n
f '(a) = 2(12 $$-$$ 3a2)\n
Maximum at a = 2\n
maximum area = f(2) = 32",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6358,
"subject": "General Science",
"question": "Let P(4, –4) and Q(9, 6) be two points on the parabola, y2 = 4x and let x be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of $$\\Delta $$PXQ is maximum. Then this maximum area (in sq. units) is : ",
"options": [
{
"text": "$${{625} \\over 4}$$"
},
{
"text": "$${{125} \\over 4}$$"
},
{
"text": "$${{75} \\over 2}$$"
},
{
"text": "$${{125} \\over 2}$$"
}
],
"answer": "$${{125} \\over 4}$$",
"solution": "**Answer:** $${{125} \\over 4}$$\n\n\n y2 = 4x\n
Area = $${1 \\over 2}\\left| {\\matrix{\n {{1 \\over 4}} & 1 & 1 \\cr \n 9 & 6 & 1 \\cr \n 4 & { - 4} & 1 \\cr \n\n } } \\right| = {{225} \\over 4}$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 6359,
"subject": "General Science",
"question": "The area (in sq. units) of an equilateral triangle\ninscribed in the parabola y2 = 8x, with one of\nits vertices on the vertex of this parabola, is :",
"options": [
{
"text": "$$256\\sqrt 3 $$"
},
{
"text": "$$64\\sqrt 3 $$"
},
{
"text": "$$128\\sqrt 3 $$"
},
{
"text": "$$192\\sqrt 3 $$"
}
],
"answer": "$$192\\sqrt 3 $$",
"solution": "**Answer:** $$192\\sqrt 3 $$\n\nLet A = (2t2\n, 4t)\n
B = (2t2\n, -4t) (by symmetry as equilateral\ntriangle)\n\n \n
tan 30o = $$\\frac{4t}{2t^{2}} $$ = $$\\frac{2}{t} $$\n
$$ \\Rightarrow $$ t = 2$$\\sqrt{3} $$\n
AB = 8t = 16$$\\sqrt{3} $$\n
Area of equilateral = $$\\frac{1}{2} \\times 16\\sqrt{3} \\times 24$$\n
= $$192\\sqrt 3 $$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 6360,
"subject": "General Science",
"question": "Let P be a point on the parabola, y2\n = 12x and\nN be the foot of the perpendicular drawn from\nP on the axis of the parabola. A line is now\ndrawn through the mid-point M of PN, parallel\nto its axis which meets the parabola at Q. If the\ny-intercept of the line NQ is $${4 \\over 3}$$,\n then :",
"options": [
{
"text": "MQ = $${1 \\over 3}$$"
},
{
"text": "PN = 4"
},
{
"text": "PN = 3"
},
{
"text": "MQ = $${1 \\over 4}$$"
}
],
"answer": "MQ = $${1 \\over 4}$$",
"solution": "**Answer:** MQ = $${1 \\over 4}$$\n\n\n
Let P = (3t2\n , 6t) ; N = (3t2\n, 0)\n
M = (3t2\n, 3t)\n
Let point Q = (h, 3t) which lies on the parabola y2 = 12x\n
PN = 6t = 2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6361,
"subject": "General Science",
"question": "If P is a point on the parabola y = x2 + 4 which is closest to the straight line y = 4x $$-$$ 1, then the co-ordinates of P are :",
"options": [
{
"text": "($$-$$2, 8)"
},
{
"text": "(2, 8)"
},
{
"text": "(1, 5)"
},
{
"text": "(3, 13)"
}
],
"answer": "(2, 8)",
"solution": "**Answer:** (2, 8)\n\nGiven, curve y = x2 + 4
Consider the circle $$C: x^2+y^2=4$$ and the parabola $$P: y^2=8 x$$. If the set of all values of $$\\alpha$$, for which three chords of the circle $$C$$ on three distinct lines passing through the point $$(\\alpha, 0)$$ are bisected by the parabola $$P$$ is the interval $$(p, q)$$, then $$(2 q-p)^2$$ is equal to __________.
$$ \\Rightarrow xy = {{105} \\over {64}}$$ which is the required locus.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6366,
"subject": "General Science",
"question": "A parabola has the origin as its focus and the line $$x=2$$ as the directrix. Then the vertex of the parabola is at :",
"options": [
{
"text": "$$(0,2)$$ "
},
{
"text": "$$(1,0)$$ "
},
{
"text": "$$(0,1)$$ "
},
{
"text": "$$(2,0)$$ "
}
],
"answer": "$$(1,0)$$ ",
"solution": "**Answer:** $$(1,0)$$ \n\n\n
Vertex of a parabola is the mid point of focus and the point \n
where directrix meets the axis of the parabola. \n
Here focus is $$O\\left( {0,0} \\right)$$ and directrix meets the axis at $$B\\left( {2,0} \\right)$$\n
$$\\therefore$$ Vertex of the parabola is $$(1,0)$$ ",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 6367,
"subject": "General Science",
"question": "Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the\norigin, on the positive x-axis then which of the following points does not lie on it?",
"options": [
{
"text": "(5, 2$$\\sqrt 6$$) "
},
{
"text": "(6, 4$$\\sqrt 2$$) "
},
{
"text": "(8, 6)"
},
{
"text": "(4, -4)"
}
],
"answer": "(8, 6)",
"solution": "**Answer:** (8, 6)\n\n\n
By checking each options you can see. point (8, 6) does not lie on the parabola. ",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 6368,
"subject": "General Science",
"question": "If $$\\theta $$ denotes the acute angle between the curves, y = 10 – x2 and y = 2 + x2 at a point of their intersection, the |tan $$\\theta $$| is equal to :",
"options": [
{
"text": "$$8 \\over 15$$"
},
{
"text": "$$4 \\over 9$$"
},
{
"text": "$$7 \\over 17$$"
},
{
"text": "$$8 \\over 17$$"
}
],
"answer": "$$8 \\over 15$$",
"solution": "**Answer:** $$8 \\over 15$$\n\n\n
Angle between the curves is the acute angle between the tangents at the point of intersection. \n
y = 10 $$-$$ x2 (for curve 1)\n
and y = 2 + x2 (for curve 2)\n
$$ \\therefore $$ 10 $$-$$ x2 = 2 + x2\n
$$ \\Rightarrow $$ 2x2 = 8\n
$$ \\Rightarrow $$ x2 = 4\n
$$ \\Rightarrow $$ x = 2, $$-$$ 2\n
$$ \\therefore $$ points of intersection (2, 6) and ($$-$$ 2, 6)\n
$${{dy} \\over {dx}}$$ for curve 1 = $$-$$ 2x\n
$$ \\therefore $$ Slope(m1) of curve 1 is = $$-$$ 2(2) = $$-$$ 4\n
= $${8 \\over {15}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6369,
"subject": "General Science",
"question": "The length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S (> R) respectively from the origin, is :",
"options": [
{
"text": "4(S + R)"
},
{
"text": "2(S $$-$$ R)"
},
{
"text": "4(S $$-$$ R)"
},
{
"text": "2(S + R)"
}
],
"answer": "4(S $$-$$ R)",
"solution": "**Answer:** 4(S $$-$$ R)\n\n
If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $$3x + y - 29 = 0$$, is $${x^2} + a{y^2} + bxy + cx + dy + k = 0$$, then $$a + b + c + d + k$$ is equal to :
Let $$x = 2t$$, $$y = {{{t^2}} \\over 3}$$ be a conic. Let S be the focus and B be the point on the axis of the conic such that $$SA \\bot BA$$, where A is any point on the conic. If k is the ordinate of the centroid of the $$\\Delta$$SAB, then $$\\mathop {\\lim }\\limits_{t \\to 1} k$$ is equal to :
A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with :
Let P1 be a parabola with vertex (3, 2) and focus (4, 4) and P2 be its mirror image with respect to the line x + 2y = 6. Then the directrix of P2 is x + 2y = ____________.
Let $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ be any points on the curves $$(x-1)^{2}+(y+1)^{2}=1$$ and $$y=x^{2}$$, respectively. The distance between $$P$$ and $$Q$$ is minimum for some value of the abscissa of $$P$$ in the interval :
Let $$A, B$$ and $$C$$ be three points on the parabola $$y^2=6 x$$ and let the line segment $$A B$$ meet the line $$L$$ through $$C$$ parallel to the $$x$$-axis at the point $$D$$. Let $$M$$ and $$N$$ respectively be the feet of the perpendiculars from $$A$$ and $$B$$ on $$L$$. Then $$\\left(\\frac{A M \\cdot B N}{C D}\\right)^2$$ is equal to __________.
$$\\begin{aligned}\n& \\text { For } D, y=2 a t_3 \\\\\n& \\Rightarrow x=a\\left(t_1 t_3+t_2 t_3-t_1 t_2\\right) \\\\\n& C D=\\left|a\\left(t_1 t_3+t_2 t_3-t_1 t_3\\right)-a t_3^2\\right| \\\\\n& A M=\\left|2 a t_1-2 a t_3\\right| \\\\\n& B N=\\left|2 a t_3-2 a t_2\\right| \\\\\n& \\left(\\frac{A M \\cdot B N}{C D}\\right)^2=\\left(\\frac{4 a^2\\left(t_1-t_3\\right)\\left(t_3-t_2\\right)}{a\\left(t_1 t_3+t_2 t_3-t_1 t_3-t_3^2\\right)}\\right)^2 \\\\\n& =\\left(\\frac{4 a^2\\left(t_1-t_3\\right)\\left(t_3-t_2\\right)}{a\\left(t_1-t_3\\right)\\left(t_2-t_3\\right)}\\right)^2 \\\\\n& =16 a^2=16 \\cdot\\left(\\frac{3}{2}\\right)^2=36\n\\end{aligned}$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6377,
"subject": "General Science",
"question": "The equation of a tangent to the parabola $${y^2} = 8x$$ is $$y=x+2$$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is :",
"options": [
{
"text": "$$(2,4)$$"
},
{
"text": "$$(-2,0)$$"
},
{
"text": "$$(-1,1)$$"
},
{
"text": "$$(0,2)$$"
}
],
"answer": "$$(-2,0)$$",
"solution": "**Answer:** $$(-2,0)$$\n\nParabola $${y^2} = 8x$$\n
\n
We know that the locus of point of intersection of two perpendicular tangents to a \n
parabola is its directrix. Point must be on the directrix of parabola\n
as equation of directrix $$x + 2 = 0 \\Rightarrow x = - 2$$\n
Hence the point is $$\\left( { - 2,0} \\right)$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6378,
"subject": "General Science",
"question": "The slope of the line touching both the parabolas $${y^2} = 4x$$ and $${x^2} = - 32y$$ is ",
"options": [
{
"text": "$${{1 \\over 8}}$$"
},
{
"text": "$${{2 \\over 3}}$$"
},
{
"text": "$${{1 \\over 2}}$$"
},
{
"text": "$${{3 \\over 2}}$$"
}
],
"answer": "$${{1 \\over 2}}$$",
"solution": "**Answer:** $${{1 \\over 2}}$$\n\nLet tangent to $${y^2} = 4x$$ be $$y = mx + {1 \\over m}$$\n
$$ \\Rightarrow {m^3} = {4 \\over {32}} \\Rightarrow m = {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6379,
"subject": "General Science",
"question": "Let P be a point on the parabola, x2 = 4y. If the distance of P from the center of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is : ",
"options": [
{
"text": "x + 4y $$-$$ 2 = 0"
},
{
"text": "x $$-$$ y + 3 = 0"
},
{
"text": "x + y +1 = 0"
},
{
"text": "x + 2y = 0"
}
],
"answer": "x + y +1 = 0",
"solution": "**Answer:** x + y +1 = 0\n\nLet P(2t, t2) be any point on the parabola.\n
Center of the given circle C = ($$-$$ g, $$-$$f) = ($$-$$3, 0)\n
For PC to be minimum, it must be the normal to the parabola at P.\n
Slope of line PC = $${{{y_2} - {y_1}} \\over {{x_2} - {x_1}}}$$ = $${{{t^2} - 0} \\over {2t + 3}}$$\n
Also, slope of tangent to parabola at P = $${{dy} \\over {dx}}$$ = $${x \\over 2}$$ = t\n
$$ \\therefore $$ Slope of normal = $${{ - 1} \\over t}$$\n
$$\\therefore\\,\\,\\,$$ Real roots of above equation is\n
t = $$-$$ 1\n
Coordinate of P = (2t, t2) = ($$-$$2, 1)\n
Slope of tangent to parabola at P = t = $$-$$ 1\n
Therefore, equation of tangent is : \n
(y $$-$$ 1) = ($$-$$ 1) (x + 2)\n
$$ \\Rightarrow $$ x + y + 1 = 0",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6380,
"subject": "General Science",
"question": "The equation of a tangent to the parabola, x2\n = 8y, which makes an angle $$\\theta $$ with the positive directions of x-axis, is :",
"options": [
{
"text": "x = y cot $$\\theta $$ – 2 tan $$\\theta $$"
},
{
"text": "y = x tan $$\\theta $$ + 2 cot $$\\theta $$"
},
{
"text": "x = y cot $$\\theta $$ + 2 tan $$\\theta $$"
},
{
"text": "y = x tan $$\\theta $$ – 2 cot $$\\theta $$\n"
}
],
"answer": "x = y cot $$\\theta $$ + 2 tan $$\\theta $$",
"solution": "**Answer:** x = y cot $$\\theta $$ + 2 tan $$\\theta $$\n\nx2 = 8y\n
$$ \\Rightarrow $$ x = y cot $$\\theta $$ + 2 tan $$\\theta $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6381,
"subject": "General Science",
"question": "The shortest distance between the line y = x and\nthe curve y2 = x – 2 is :",
"options": [
{
"text": "$$7\\over 4 \\sqrt2$$"
},
{
"text": "$$7\\over8$$"
},
{
"text": "$$11\\over 4 \\sqrt2$$"
},
{
"text": "2"
}
],
"answer": "$$7\\over 4 \\sqrt2$$",
"solution": "**Answer:** $$7\\over 4 \\sqrt2$$\n\n\nSlope of the line y = x is 1.\n
$$ \\therefore $$ The shortest distance between line y = x and\nparabola y2 = x – 2 is the distance between line y = x and\ntangent of parabola having slope 1.\n
Slope of tangent at point P on the parabola y2 = x – 2 is,\n
\n$$ \\therefore $$ C = ($$5 \\over 2$$, -1)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6384,
"subject": "General Science",
"question": "If y = mx + 4 is a tangent to both the parabolas, y2 = 4x and x2 = 2by, then b is equal to :",
"options": [
{
"text": "-128"
},
{
"text": "128"
},
{
"text": "-64"
},
{
"text": "-32"
}
],
"answer": "-128",
"solution": "**Answer:** -128\n\nGiven y = mx + 4 is tangent to both the parabolas.\n
$$ \\therefore $$ Applying condition of tangent\n for y2\n = 4x, we get\n
$${1 \\over m}$$ = 4\n
$$ \\Rightarrow $$ m = $${1 \\over 4}$$\n
For x2\n = 2by line y = $${x \\over 4}$$ + 4 is tangent\n
$$ \\Rightarrow $$ b = -128",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6385,
"subject": "General Science",
"question": "Let a line y = mx (m > 0) intersect the parabola,\ny2 = x at a point P, other than the origin. Let\nthe tangent to it at P meet the x-axis at the point\nQ. If area ($$\\Delta $$OPQ) = 4 sq. units, then m is equal\nto __________.",
"options": [],
"answer": "0.5",
"solution": "**Answer:** 0.5\n\n\n
$$ \\therefore $$ Centre is $$\\left( {{{ - 16} \\over 5},{{53} \\over {10}}} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6389,
"subject": "General Science",
"question": "A tangent is drawn to the parabola y2 = 6x which is perpendicular to the line 2x + y = 1. Which of the following points does NOT lie on it?",
"options": [
{
"text": "(0, 3)"
},
{
"text": "($$-$$6, 0)"
},
{
"text": "(4, 5)"
},
{
"text": "(5, 4)"
}
],
"answer": "(5, 4)",
"solution": "**Answer:** (5, 4)\n\nEquation of tangent : $$y = mx + {3 \\over {2m}}$$
$${m_T} = {1 \\over 2}$$ ($$\\because$$ perpendicular to line $$2x + y = 1$$)
$$ \\Rightarrow x - 2y + 6 = 0$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6390,
"subject": "General Science",
"question": "Let C be the locus of the mirror image of a point on the parabola y2 = 4x with respect to the line y = x. Then the equation of tangent to C at P(2, 1) is :",
"options": [
{
"text": "x $$-$$ y = 1"
},
{
"text": "2x + y = 5"
},
{
"text": "x + 3y = 5"
},
{
"text": "x + 2y = 4"
}
],
"answer": "x $$-$$ y = 1",
"solution": "**Answer:** x $$-$$ y = 1\n\n\n Image of y2 = 4x w.r.t. y = x is x2 = 4y
tangent from (2, 1)
xx1 = 2(y + y1)
2x = 2(y + 1)
x = y + 1",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6391,
"subject": "General Science",
"question": "Let y = mx + c, m > 0 be the focal chord of y2 = $$-$$ 64x, which is tangent to (x + 10)2 + y2 = 4. Then, the value of 4$$\\sqrt 2 $$ (m + c) is equal to _____________.",
"options": [],
"answer": "34",
"solution": "**Answer:** 34\n\ny2 = $$-$$64x
$$\\Rightarrow$$ m = $${1 \\over {2\\sqrt 2 }}$$ & c = $${8 \\over {\\sqrt 2 }}$$
$$4\\sqrt 2 (m + c) = 4\\sqrt 2 \\left( {{{17} \\over {2\\sqrt 2 }}} \\right)$$ = 34",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6392,
"subject": "General Science",
"question": "Let a parabola b be such that its vertex and focus lie on the positive x-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from O(0, 0) to the parabola P which meet P at S and R, then the area (in sq. units) of $$\\Delta$$SOR is equal to :",
"options": [
{
"text": "$$16\\sqrt 2 $$"
},
{
"text": "16"
},
{
"text": "32"
},
{
"text": "$$8\\sqrt 2 $$"
}
],
"answer": "16",
"solution": "**Answer:** 16\n\n Clearly RS is latus-rectum
Let P : y2 = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of $${\\pi \\over 4}$$ with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is :
",
"options": [
{
"text": "8 only"
},
{
"text": "2 only"
},
{
"text": "$${1 \\over 4}$$ only"
},
{
"text": "any a > 0"
}
],
"answer": "any a > 0",
"solution": "**Answer:** any a > 0\n\n
$$ \\Rightarrow m - 3 = 1 + 3m$$ and $$m - 3 = - 1 - 3m$$
\n
$$ \\Rightarrow 2m = - 4$$ and $$4m = 2$$
\n
$$m = - 2$$ and $$m = {1 \\over 2}$$
\n
We know, Equation of tangent to the parabola $${y^2} = 4m$$ is $$y = mx + {a \\over m}$$ and point of contact is $$\\left( {{a \\over {{m^2}}},\\,{{2a} \\over m}} \\right)$$
A circle of radius 2 unit passes through the vertex and the focus of the parabola y2 = 2x and touches the parabola $$y = {\\left( {x - {1 \\over 4}} \\right)^2} + \\alpha $$, where $$\\alpha$$ > 0. Then (4$$\\alpha$$ $$-$$ 8)2 is equal to ______________.
If the line $$y = 4 + kx,\\,k > 0$$, is the tangent to the parabola $$y = x - {x^2}$$ at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :
The sum of diameters of the circles that touch (i) the parabola $$75 x^{2}=64(5 y-3)$$ at the point $$\\left(\\frac{8}{5}, \\frac{6}{5}\\right)$$ and (ii) the $$y$$-axis, is equal to ______________.
",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\n \nEquation of tangent to the parabola at $P\\left(\\frac{8}{5}, \\frac{6}{5}\\right)$\n
\n$$\n\\begin{aligned}\n&75 x \\cdot \\frac{8}{5}=160\\left(y+\\frac{6}{5}\\right)-192 \\\\\\\\\n&\\Rightarrow 120 x=160 y \\\\\\\\\n&\\Rightarrow 3 x=4 y\n\\end{aligned}\n$$
\nEquation of circle touching the given parabola at $\\mathrm{P}$ can be taken as
The tangents at the points $$A(1,3)$$ and $$B(1,-1)$$ on the parabola $$y^{2}-2 x-2 y=1$$ meet at the point $$P$$. Then the area (in unit $${ }^{2}$$ ) of the triangle $$P A B$$ is :
Let $$P(a, b)$$ be a point on the parabola $$y^{2}=8 x$$ such that the tangent at $$P$$ passes through the centre of the circle $$x^{2}+y^{2}-10 x-14 y+65=0$$. Let $$A$$ be the product of all possible values of $$a$$ and $$B$$ be the product of all possible values of $$b$$. Then the value of $$A+B$$ is equal to :
If the length of the latus rectum of a parabola, whose focus is $$(a, a)$$ and the tangent at its vertex is $$x+y=a$$, is 16, then $$|a|$$ is equal to :
If the tangents drawn at the points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ on the parabola $$y^{2}=2 x-3$$ intersect at the point $$R(0,1)$$, then the orthocentre of the triangle $$P Q R$$ is :
Intersection of PQ with parabola $$P:(6,3)\\,Q:(2, - 1)$$
\n
Slope of $$RQ = - 1$$ & Slope of $$PQ = 1$$
\n
Therefore $$\\angle PQR = 90^\\circ \\Rightarrow $$ Orthocentre is at $$Q:(2, - 1)$$
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6401,
"subject": "General Science",
"question": "Let $\\mathrm{S}$ be the set of all $\\mathrm{a} \\in \\mathrm{N}$ such that the area of the triangle formed by the tangent at the point $\\mathrm{P}(\\mathrm{b}$, c), b, c $\\in \\mathbb{N}$, on the parabola $y^{2}=2 \\mathrm{a} x$ and the lines $x=\\mathrm{b}, y=0$ is $16 $ unit2, then $\\sum\\limits_{\\mathrm{a} \\in \\mathrm{S}} \\mathrm{a}$ is equal to :",
"options": [],
"answer": "146",
"solution": "**Answer:** 146\n\n\n As $\\mathrm{P}(\\mathrm{b}, \\mathrm{c})$ lies on parabola so $\\mathrm{c}^{2}=2 \\mathrm{ab}---(1)$\n\n
Now equation of tangent to parabola $\\mathrm{y}^{2}=2 \\mathrm{ax}$ in point form is
For point $\\mathrm{B}$, put $\\mathrm{y}=0$, now $\\mathrm{x}=-\\mathrm{b}$\n\n
So, area of $\\triangle \\mathrm{PBA}, \\frac{1}{2} \\times \\mathrm{AB} \\times \\mathrm{AP}=16$\n
$$\n\\begin{aligned}\n& \\Rightarrow \\frac{1}{2} \\times 2 b \\times c=16 \\\\\\\\\n& \\Rightarrow b c=16\n\\end{aligned}\n$$\n
As $\\mathrm{b}$ and $\\mathrm{c}$ are natural number so possible values of $(b, c)$ are $(1,16),(2,8),(4,4),(8,2)$ and $(16,1)$\n
Now from equation (1) $\\mathrm{a}=\\frac{\\mathrm{c}^2}{2 \\mathrm{~b}}$ and $\\mathrm{a} \\in \\mathrm{N}$, so values of $(b, c)$ are $(1,16),(2,8)$ and $(4,4)$ now values of are 128,16 and 2 .\n
Hence sum of values of $a$ is 146 .",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6402,
"subject": "General Science",
"question": "
Let $$\\mathrm{y}=f(x)$$ represent a parabola with focus $$\\left(-\\frac{1}{2}, 0\\right)$$ and directrix $$y=-\\frac{1}{2}$$. Then \n
If the tangent at a point P on the parabola $$y^2=3x$$ is parallel to the line $$x+2y=1$$ and the tangents at the points Q and R on the ellipse $$\\frac{x^2}{4}+\\frac{y^2}{1}=1$$ are perpendicular to the line $$x-y=2$$, then the area of the triangle PQR is :
A triangle is formed by the tangents at the point (2, 2) on the curves $$y^2=2x$$ and $$x^2+y^2=4x$$, and the line $$x+y+2=0$$. If $$r$$ is the radius of its circumcircle, then $$r^2$$ is equal to ___________.
The equations of the sides AB and AC of a triangle ABC are $$(\\lambda+1)x+\\lambda y=4$$ and $$\\lambda x+(1-\\lambda)y+\\lambda=0$$ respectively. Its vertex A is on the y-axis and its orthocentre is (1, 2). The length of the tangent from the point C to the part of the parabola $$y^2=6x$$ in the first quadrant is :
Let a tangent to the curve $$\\mathrm{y^2=24x}$$ meet the curve $$xy = 2$$ at the points A and B. Then the mid points of such line segments AB lie on a parabola with the :
Let $$\\mathrm{A}(0,1), \\mathrm{B}(1,1)$$ and $$\\mathrm{C}(1,0)$$ be the mid-points of the sides of a triangle with incentre at the point $$\\mathrm{D}$$. If the focus of the parabola $$y^{2}=4 \\mathrm{ax}$$ passing through $$\\mathrm{D}$$ is $$(\\alpha+\\beta \\sqrt{2}, 0)$$, where $$\\alpha$$ and $$\\beta$$ are rational numbers, then $$\\frac{\\alpha}{\\beta^{2}}$$ is equal to :
Let the tangent to the curve $$x^{2}+2 x-4 y+9=0$$ at the point $$\\mathrm{P}(1,3)$$ on it meet the $$y$$-axis at $$\\mathrm{A}$$. Let the line passing through $$\\mathrm{P}$$ and parallel to the line $$x-3 y=6$$ meet the parabola $$y^{2}=4 x$$ at $$\\mathrm{B}$$. If $$\\mathrm{B}$$ lies on the line $$2 x-3 y=8$$, then $$(\\mathrm{AB})^{2}$$ is equal to ___________.
Let a line perpendicular to the line $$2 x-y=10$$ touch the parabola $$y^2=4(x-9)$$ at the point P. The distance of the point P from the centre of the circle $$x^2+y^2-14 x-8 y+56=0$$ is __________.
Let a conic $$C$$ pass through the point $$(4,-2)$$ and $$P(x, y), x \\geq 3$$, be any point on $$C$$. Let the slope of the line touching the conic $$C$$ only at a single point $$P$$ be half the slope of the line joining the points $$P$$ and $$(3,-5)$$. If the focal distance of the point $$(7,1)$$ on $$C$$ is $$d$$, then $$12 d$$ equals ________.
Let $$L_1, L_2$$ be the lines passing through the point $$P(0,1)$$ and touching the parabola $$9 x^2+12 x+18 y-14=0$$. Let $$Q$$ and $$R$$ be the points on the lines $$L_1$$ and $$L_2$$ such that the $$\\triangle P Q R$$ is an isosceles triangle with base $$Q R$$. If the slopes of the lines $$Q R$$ are $$m_1$$ and $$m_2$$, then $$16\\left(m_1^2+m_2^2\\right)$$ is equal to __________.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6412,
"subject": "General Science",
"question": "Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is : ",
"options": [
{
"text": "9"
},
{
"text": "18"
},
{
"text": "36"
},
{
"text": "32"
}
],
"answer": "36",
"solution": "**Answer:** 36\n\n\n
all are possible so that total no. of positive case \n
9 + 9 + 9 + 9 = 36",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6413,
"subject": "General Science",
"question": "Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the\ntop of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total\nnumber of beams is :",
"options": [
{
"text": "180"
},
{
"text": "210"
},
{
"text": "170"
},
{
"text": "190"
}
],
"answer": "170",
"solution": "**Answer:** 170\n\n\nAny two non-adjacent pillers are joined by beams
\n$$ \\therefore $$ number of beams = number of diagonals = 20C2 - 20 = 170",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6414,
"subject": "General Science",
"question": "Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let $$\\alpha$$ be the number of triangles having these points from different sides as vertices and $$\\beta$$ be the number of quadrilaterals having these points from different sides as vertices. Then ($$\\beta$$ $$-$$ $$\\alpha$$) is equal to :",
"options": [
{
"text": "717"
},
{
"text": "795"
},
{
"text": "1890"
},
{
"text": "1173"
}
],
"answer": "717",
"solution": "**Answer:** 717\n\n$$\\alpha = {}^6{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^7{C_1} $$\n
$$ \\therefore $$ $$ \\beta - \\alpha = 1890 - 1173 = 717$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6415,
"subject": "General Science",
"question": "If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :",
"options": [
{
"text": "240"
},
{
"text": "360"
},
{
"text": "333"
},
{
"text": "364"
}
],
"answer": "333",
"solution": "**Answer:** 333\n\n\n Total number of triangles
A triangle is formed by X-axis, Y-axis and the line $$3x+4y=60$$. Then the number of points P(a, b) which lie strictly inside the triangle, where a is an integer and b is a multiple of a, is ____________.
",
"options": [],
"answer": "31",
"solution": "**Answer:** 31\n\nIf x = 1, y = $57 \\over 4 $ = 14.25
Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total number of persons, who participated in the tournament, is ___________.
",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\nLet, $n$ be the total number of couples who participated in the tournament.\n
According to the question, $2 \\times{ }^n C_2 \\times{ }^{n-2} C_2=840$\n
Hence, total number of players who participated in the tournament $=2 \\times 8=16$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6419,
"subject": "General Science",
"question": "
The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is
To solve this problem, we need to determine the number of triangles formed by the vertices of a regular octagon such that none of the sides of the triangle is also a side of the octagon.
\n\n
Let's start by counting the total number of triangles that can be formed using the 8 vertices of the octagon. The number of ways to choose 3 vertices out of 8 is given by the combination formula:
Next, we need to exclude those triangles that have at least one side coinciding with a side of the octagon. Let's analyze how many such invalid triangles there can be.
\n\n
Consider each side of the octagon. For any given side, there are exactly 5 other vertices remaining (since we must exclude the two vertices that form the current side). Out of these 5 vertices, we can choose any 1 to form a triangle that has one side common with the octagon. Hence, for each side of the octagon, there are 5 such triangles.
\n\n
Since the octagon has 8 sides, the total number of triangles that have at least one side as a side of the octagon is:
\n\n
\n\n
$$ 8 \\times 5 = 40 $$
\n\n\n\n
Therefore, the number of triangles whose sides do not coincide with any sides of the octagon is:
\n\n
\n\n
$$ 56 - 40 = 16 $$
\n\n\n\n
So, the correct answer is:
\n\n
Option B: 16
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6420,
"subject": "General Science",
"question": "The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by",
"options": [
{
"text": "$$7!\\, \\times 5!\\,\\,$$ "
},
{
"text": "$$6!\\, \\times 5!$$ "
},
{
"text": "$$30!$$ "
},
{
"text": "$$5!\\, \\times 4!$$ "
}
],
"answer": "$$6!\\, \\times 5!$$ ",
"solution": "**Answer:** $$6!\\, \\times 5!$$ \n\n\n
6 men can sit at the round table = $$\\left( {6 - 1} \\right)! = 5!$$ ways\n
Now at the round table among 6 men there are 6 empty places and 5 women can sit at those 6 empty positions.
\n
So total no of ways 6 men and 5 women can dine at the round table\n
= $$5!\\, \\times {}^6{C_5} \\times 5!$$\n
= $$5!\\, \\times 6 \\times 5!$$\n
= $$5!\\, \\times 6!$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6421,
"subject": "General Science",
"question": "The number of ways in which 5 boys and 3 girls can be seated on a round table if a\nparticular boy B1 and a particular girl G1 never sit adjacent to each other, is :",
"options": [
{
"text": "5 $$ \\times $$ 6!"
},
{
"text": "6 $$ \\times $$ 6!"
},
{
"text": "7!"
},
{
"text": "5 $$ \\times $$ 7!"
}
],
"answer": "5 $$ \\times $$ 6!",
"solution": "**Answer:** 5 $$ \\times $$ 6!\n\nNumber of ways = Total - when B1 and G1 sit together \n
Total ways to seat 8 people on round table = (8 - 1)! = 7!\n
When B1 and G1 sit together then assume B1 and G1 are one people, so total 7 people are there and among B1 and G1 they can sit 2! ways.\n
So total no of ways when B1 and G1 sit together\n = (7 - 1)! $$ \\times $$ 2! = 6! $$ \\times $$ 2!\n
Number of ways = 7! - 6! $$ \\times $$ 2! = 6!$$ \\times $$(7 - 2) = 5 $$ \\times $$ 6!",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6422,
"subject": "General Science",
"question": "Let n > 2 be an integer. Suppose that there are\nn Metro stations in a city located along a\ncircular path. Each pair of stations is connected\nby a straight track only. Further, each pair of\nnearest stations is connected by blue line,\nwhereas all remaining pairs of stations are\nconnected by red line. If the number of red lines\nis 99 times the number of blue lines, then the\nvalue of n is :",
"options": [
{
"text": "201"
},
{
"text": "199"
},
{
"text": "101"
},
{
"text": "200"
}
],
"answer": "201",
"solution": "**Answer:** 201\n\n \nNumber of blue lines = Number of sides = n
\nNumber of red lines = number of diagonals = nC2 – n
around the table $=6$ ! and 5 girls can be arranged in 7 gaps in ${ }^7 \\mathrm{P}_5$ ways.\n
$\\therefore$ Required no. of ways $=6 ! \\times{ }^7 \\mathrm{P}_5$ $=126 \\times(5 !)^2$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6424,
"subject": "General Science",
"question": "A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is",
"options": [
{
"text": "346"
},
{
"text": "140"
},
{
"text": "196"
},
{
"text": "280"
}
],
"answer": "196",
"solution": "**Answer:** 196\n\n\n
Case 1 :\n
No of ways student can answer 10 questions = $${}^5{C_4} \\times {}^8{C_6}$$ = 140\n
Case 2 :\n
No of ways student can answer 10 questions = $${}^5{C_5} \\times {}^8{C_5}$$ = 56\n
$$\\therefore$$ Total ways = 140 + 56 = 196",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6425,
"subject": "General Science",
"question": "Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can\nbe formed from this class, if there are two specific boys A and B, who refuse to be the members of the same\nteam, is :\n",
"options": [
{
"text": "500"
},
{
"text": "350"
},
{
"text": "200"
},
{
"text": "300"
}
],
"answer": "300",
"solution": "**Answer:** 300\n\nFrom 5 girls 2 girls can be selected \n
= 5C2 ways\n
From 7 boys 3 boys can be selected \n
= 7C3 way\n
$$ \\therefore $$ Total number of ways we can select 2 girls and 3 boys \n
= 5C2 $$ \\times $$ 7C3 ways \n
When two boys A and B are chosen in a team then one more boy will be chosen from remaining 5 boys.\n
So, no of ways 3 boys can be chosen when A and B should must be chosen = 5C1 ways\n
$$ \\therefore $$ Total number of ways a team of 2 girl and 3 boys can be made where boy A and B must be in the team = 5C1 $$ \\times $$ 5C2 ways\n
$$ \\therefore $$ Required number of ways \n = Total number of ways $$-$$ when A and B are always included. \n
= 300",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6426,
"subject": "General Science",
"question": "A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can\nrandomly be selected from this group such that there is at least one boy and at least one girl in each team, is\n1750, then n is equal to : ",
"options": [
{
"text": "24"
},
{
"text": "25"
},
{
"text": "27"
},
{
"text": "28"
}
],
"answer": "25",
"solution": "**Answer:** 25\n\nGiven that 5 Boy, n girls.
\n$$ \\Rightarrow n = 25, -28$$\n",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6427,
"subject": "General Science",
"question": "A committee of 11 members is to be formed from\n8 males and 5 females. If m is the number of ways\nthe committee is formed with at least 6 males and\nn is the number of ways the committee is formed\nwith at least 3 females, then :",
"options": [
{
"text": "n = m – 8"
},
{
"text": "m = n = 78"
},
{
"text": "m + n = 68"
},
{
"text": "m = n = 68"
}
],
"answer": "m = n = 78",
"solution": "**Answer:** m = n = 78\n\nAt least 6 males means in the committee there can be 6 males or 7 males or 8 males.\n
So, m = n = 78",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6428,
"subject": "General Science",
"question": "Four fair dice are thrown independently 27 times. Then the expected number of times, at\nleast two dice show up a three or a five, is _________.",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\n4 dice are independently thrown. Each die has probability to show 3 or 5 is
= 11",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6429,
"subject": "General Science",
"question": "There are 3 sections in a question paper and\neach section contains 5 questions. A candidate\nhas to answer a total of 5 questions, choosing\nat least one question from each section. Then\nthe number of ways, in which the candidate\ncan choose the questions, is :",
"options": [
{
"text": "2250"
},
{
"text": "2255"
},
{
"text": "3000"
},
{
"text": "1500"
}
],
"answer": "2250",
"solution": "**Answer:** 2250\n\n\n
$$ \\therefore $$ Total number of selection of 5 questions\n
= 2250",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6430,
"subject": "General Science",
"question": "A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at\nleast 2 Indians and double the number of foreigners as Indians. Then the number of ways,\nthe committee can be formed, is :",
"options": [
{
"text": "1050"
},
{
"text": "575"
},
{
"text": "560"
},
{
"text": "1625"
}
],
"answer": "1625",
"solution": "**Answer:** 1625\n\nGiven,
Number of Indians = 6
Number of foreigners = 8
Committee of at least 2 Indians and double number of foreigners is to be formed. Hence, the required cases are
= 1050 + 560 + 15 = 1625",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6431,
"subject": "General Science",
"question": "The students S1, S2, ....., S10 are to be divided into 3 groups A, B and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is ___________.",
"options": [],
"answer": "31650",
"solution": "**Answer:** 31650\n\nIf group C has one student then number of\ngroups\n
= 10C1\n[29\n – 2] = 5100\n
If group C has two students then number of\ngroups\n
= 10C2\n[28\n – 2] = 11430\n
If group C has three students then number of\ngroups\n
= 10C3\n × [27\n – 2] = 15120\n
So total groups = 31650",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6432,
"subject": "General Science",
"question": "The total number of positive integral solutions (x, y, z) such that xyz = 24 is :",
"options": [
{
"text": "36"
},
{
"text": "24"
},
{
"text": "45"
},
{
"text": "30"
}
],
"answer": "30",
"solution": "**Answer:** 30\n\n$$x.y.z = 24$$
$$x.y.z = {2^3}.\\,{3^1}$$
Three 2 has to be distributed among x, y and z
Each may receive none, one or two
$$\\therefore$$ Number of ways = $${}^{3 + 3 - 1}{C_{3 - 1}}$$ = $$^5{C_2}$$ ways
Similarly one 3 has to be distributed among x, y and z
$$ \\therefore $$ Number of ways = $${}^{1 + 3 - 1}{C_{3 - 1}}$$ = $$^3{C_2}$$ ways
Total ways = $$^5{C_2}\\,.{\\,^3}{C_2}$$ = 30",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6433,
"subject": "General Science",
"question": "The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is :",
"options": [
{
"text": "35"
},
{
"text": "42"
},
{
"text": "82"
},
{
"text": "77"
}
],
"answer": "77",
"solution": "**Answer:** 77\n\n(I) First possibility is 1, 1, 1, 1, 1, 2, 3
Total = 42 + 35 = 77",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6434,
"subject": "General Science",
"question": "There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsman and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsman and 1 wicketkeeper, is ______________.",
"options": [],
"answer": "777",
"solution": "**Answer:** 777\n\n15 : Players
$$ = 777$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6435,
"subject": "General Science",
"question": "There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100 k, then k is equal to _____________.",
"options": [],
"answer": "238",
"solution": "**Answer:** 238\n\nClass $$\\matrix{\n {{{10}^{th}}} & {{{11}^{th}}} & {{{12}^{th}}} \\cr \n\n } $$
$$\\Rightarrow$$ K = 238",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6436,
"subject": "General Science",
"question": "Let S = {1, 2, 3, 4, 5, 6, 9}. Then the number of elements in the set T = {A $$ \\subseteq $$ S : A $$\\ne$$ $$\\phi$$ and the sum of all the elements of A is not a multiple of 3} is _______________.",
"options": [],
"answer": "80",
"solution": "**Answer:** 80\n\n3n type $$\\to$$ 3, 6, 9 = P
3n $$-$$ 1 type $$\\to$$ 2, 5 = Q
3n $$-$$ 2 type $$\\to$$ 1, 4 = R
number of subset of S containing one element which are not divisible by 3 = $${}^2$$C1 + $${}^2$$C1 = 4
number of subset of S containing two numbers whose some is not divisible by 3
number of subsets of S containing 6 elements whose sum is not divisible by 3 = 4
$$\\Rightarrow$$ Total subsets of Set A whose sum of digits is not divisible by 3 = 4 + 14 + 22 + 22 + 14 + 4 = 80.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6437,
"subject": "General Science",
"question": "
The number of ways to distribute 30 identical candies among four children C1, C2, C3 and C4 so that C2 receives at least 4 and at most 7 candies, C3 receives at least 2 and at most 6 candies, is equal to :
Let A be a matrix of order 2 $$\\times$$ 2, whose entries are from the set {0, 1, 2, 3, 4, 5}. If the sum of all the entries of A is a prime number p, 2 < p < 8, then the number of such matrices A is ___________.
The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is _____________.
",
"options": [],
"answer": "56",
"solution": "**Answer:** 56\n\nFirst we arrange 5 red cubes in a row and assume\nx1, x2, x3, x4, x5 and x6 number of blue cubes\nbetween them
\n
\nHere, $x_1+x_2+x_3+x_4+x_5+x_6=11$ and $x_2, x_3, x_4, x_5 \\geq 2$
There are ten boys B1, B2, ......., B10 and five girls G1, G2, ........, G5 in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B1 and B2 together should not be the members of a group, is ___________.
Let A be a 3 $$\\times$$ 3 matrix having entries from the set {$$-$$1, 0, 1}. The number of all such matrices A having sum of all the entries equal to 5, is ___________.
$\\begin{aligned} 1 & \\rightarrow 7 \\text { times } \\\\\\\\ \\text { and }-1 & \\rightarrow 2 \\text { times } \\end{aligned}$
\nnumber of possible marrix $=\\frac{9 !}{7 ! 2 !}=36$
\nCase-II:
$1 \\rightarrow 6$ times,
\n$-1 \\rightarrow 1$ times
\nand $0 \\rightarrow 2$ times
\nnumber of possible marrix $=\\frac{9 !}{6 ! 2 !}=252$
\nCase-III:
$ 1 \\rightarrow 5$ times,
\nand $0 \\rightarrow 4$ times
\nnumber of possible marrix $=\\frac{9 !}{5 ! 4 !}=126$
\nHence total number of all such matrix $A$ $=414$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6442,
"subject": "General Science",
"question": "
In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, $$-$$2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is ____________.
",
"options": [],
"answer": "40",
"solution": "**Answer:** 40\n\nLet student marks $x$ correct answers and $y$ incorrect. So\n
\n$3 x-2 y=5$ and $x+y \\leq 5$ where $x, y \\in \\mathrm{W}$\n
\nOnly possible solution is $(x, y)=(3,2)$\n
\nStudents can mark correct answers by only one choice but for an incorrect answer, there are two choices. So total number of ways of scoring 5 marks $={ }^{5} C_{3}(1)^{3} \\cdot(2)^{2}=40$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6443,
"subject": "General Science",
"question": "
Number of integral solutions to the equation $$x+y+z=21$$, where $$x \\ge 1,y\\ge3,z\\ge4$$, is equal to ____________.
",
"options": [],
"answer": "105",
"solution": "**Answer:** 105\n\n$\\begin{aligned} & x+y+z=21 \\\\\\\\ & \\because \\quad x \\geq 1, y \\geq 3, y \\geq 4 \\\\\\\\ & \\therefore \\quad x_1+y_1+z_1=13 \\\\\\\\ & \\text { Number of solutions }={ }^{13+3-1} C_{3-1} \\\\\\\\ & ={ }^{15} C_2=\\frac{15 \\times 14}{2}=7 \\times 15 \\\\\\\\ & =105\\end{aligned}$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6444,
"subject": "General Science",
"question": "The number of ways of selecting two numbers $a$ and $b, a \\in\\{2,4,6, \\ldots ., 100\\}$ and $b \\in\\{1,3,5, \\ldots . ., 99\\}$ such that 2 is the remainder when $a+b$ is divided by 23 is :",
"options": [
{
"text": "186"
},
{
"text": "54"
},
{
"text": "108"
},
{
"text": "268"
}
],
"answer": "108",
"solution": "**Answer:** 108\n\n
$$a+b=23\\lambda+2$$
\n
$$\\lambda=0,1,2,$$ ...., but $$\\lambda$$ cannot be even as $$a+b$$ is odd
Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is ____________
",
"options": [],
"answer": "6860 OR 3",
"solution": "**Answer:** 6860 OR 3\n\nTotal 8 oranges, 5 white apple and 7 red apple. 5 fruits needs to be selected.\n
\nCase I: 3 orange $+1$ red apple $+1$ white apple\n
The number of square matrices of order 5 with entries from the set {0, 1}, such that the sum of all the elements in each row is 1 and the sum of all the elements in each column is also 1, is :
A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is __________
The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is ______________.
The number of seven digit positive integers formed using the digits $$1,2,3$$ and $$4$$ only and sum of the digits equal to $$12$$ is ___________.
",
"options": [],
"answer": "413",
"solution": "**Answer:** 413\n\n$$\nx_1+x_2+x_3+\\ldots x_7=12\n$$. This equation represents the number of ways to distribute 12 identical items (the sum of the digits) into 7 distinct boxes (the seven digits of the number), where each box can contain one of the numbers 1, 2, 3, or 4.\n
Number of solutions\n
$$\n\\begin{aligned}\n& =\\text { Coefficient of } x^{12} \\text { in }\\left(x^1+x^2+x^3+x^4\\right)^7 \\\\\\\\\n& =\\text { Coefficient of } x^5 \\text { in }\\left(1+x+x^2+x^3\\right)^7 \\\\\\\\\n& =\\text { Coefficient of } x^5 \\text { in }\\left(1-x^4\\right)^7(1-x)^{-7} \\\\\\\\\n& =\\text { Coefficient of } x^5 \\text { in }\\left(1-7 x^4\\right)(1-x)^{-7} \\\\\\\\\n& =\\text { Coefficient of } x^5 \\text { in }\\left(1-7 x^4\\right) \\sum_{r=0}^{\\infty}{ }^{7+r-1} C_r \\cdot x^r \\\\\\\\\n& ={ }^{11} C_5-7 \\times{ }^7 C_1 \\\\\\\\\n& =462-49=413\n\\end{aligned}\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6451,
"subject": "General Science",
"question": "
The number of triplets $$(x, \\mathrm{y}, \\mathrm{z})$$, where $$x, \\mathrm{y}, \\mathrm{z}$$ are distinct non negative integers satisfying $$x+y+z=15$$, is :
$$\n\\begin{aligned}\n\\text { Total number of solution } & ={ }^{15+3-1} C_{3-1} \\\\\\\\\n& ={ }^{17} C_2=\\frac{17 \\times 16}{1 \\times 2}=136\n\\end{aligned}\n$$\n
Now, we need to exclude the solutions where two of $(x, y, z)$ are the same.\n\n
1) For the case $x = y \\neq z$ :\n
$ 2x + z = 15 $\n
The solutions are :\n
$ x = 0, z = 15 $\n
$ x = 1, z = 13 $\n
$ x = 2, z = 11 $\n
$ x = 3, z = 9 $\n
$ x = 4, z = 7 $\n
$ x = 5, z = 5 $ (Not valid as all are the same)\n
$ x = 6, z = 3 $\n
$ x = 7, z = 1 $\n
Out of these, 7 are valid.\n\n
Similarly, for the cases $y = z \\neq x$ and $z = x \\neq y$, there will be 7 valid solutions for each, so a total of $ 7 \\times 3 = 21 $ solutions where two of the variables are equal.\n\n
Thus, the number of triplets where all are distinct is :\n
$ 136 - 21 = 115 $\n
There is one solution in which $\\mathrm{x}=\\mathrm{y}=\\mathrm{z}$\n
There are $3^{20}$ ways to distribute the oranges to the 3 children.
\n
\n
Number of ways one child receives no orange :\n
Choose 1 child out of the 3 to not receive any orange in ${ }^3 C_1 = 3$ ways. Distribute 20 oranges to the remaining 2 children in $2^{20}$ ways. However, we've included the scenarios where the 2 children each get all the oranges. So, we subtract the 2 ways where one of the two remaining children gets all the oranges.\n$ { }^3 C_1(2^{20} - 2) $
\n
\n
Number of ways two children receive no orange :\n
Choose 2 children out of the 3 to not receive any oranges in ${ }^3 C_2 = 3$ ways. The third child will receive all 20 oranges in $1^{20} = 1$ way.\n$ { }^3 C_2 \\times 1^{20} = 3 $
\n
\n\n Number of ways
$=$ Total $-($ One child receive no orange + two child receive no orange)\n
To solve this problem, we can use a classic combinatorics method known as \"stars and bars\" (or \"balls and bins\"), which is a way to solve problems involving distributing identical items into distinct groups with certain restrictions.
\n\n
First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That accounts for 6 apples (2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to distribute freely among the three children.
\n\n
The \"stars and bars\" technique involves representing the apples as stars (*) and the divisions between children as bars (|). For example, if we had 5 apples to distribute among three children, one possible distribution could be represented as **|*|**. This means the first child gets 2 apples, the second child gets 1 apple, and the third child gets 2 apples.
\n\n
In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars corresponds to a distribution of the apples.
\n\n
The total number of objects we're arranging is 15 apples + 2 bars = 17 objects. We need to choose 2 positions out of these 17 to place the bars. The remaining positions will be occupied by the stars (apples).
\n\n
The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient:
Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child gets at least 2 apples. The correct answer is Option B: 136.
The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to:
$$\\begin{aligned}\n& 2 M \\\\\n& 2 A \\\\\n& 2 T \\\\\n& H, E, I, C, S\n\\end{aligned}$$
\n
Case-I
\n
2 Alike 2 Alike 1 Diff
\n
$${ }^3 C_2 \\times{ }^6 C_1=18$$
\n
Case-II
\n
2 Alike + 3 Diff
\n
$${ }^3 C_1 \\times{ }^7 C_3=105$$
\n
Case-III
\n
All different
\n
$${ }^8 C_5=56$$
\n
Total ways $$=179$$
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6456,
"subject": "General Science",
"question": "Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is :",
"options": [
{
"text": "125"
},
{
"text": "105"
},
{
"text": "374"
},
{
"text": "625"
}
],
"answer": "374",
"solution": "**Answer:** 374\n\nThere are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed\n
Case 1 : First digit is 1 = 1 _ _ _\n
Possible numbers starting with 1 = 1$$ \\times $$5$$ \\times $$5$$ \\times $$5 = 125\n
But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.\n
Case 2 : First digit is 2 = 2 _ _ _\n
Possible numbers starting with 2 = 1$$ \\times $$5$$ \\times $$5$$ \\times $$5 = 125\n
Case 3 : First digit is 3 = 3 _ _ _\n
Possible numbers starting with 3 = 1$$ \\times $$5$$ \\times $$5$$ \\times $$5 = 125\n
Total possible numbers = 124 + 125 + 125 = 374",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6457,
"subject": "General Science",
"question": "Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are :",
"options": [
{
"text": "216"
},
{
"text": "375"
},
{
"text": "400"
},
{
"text": "720"
}
],
"answer": "720",
"solution": "**Answer:** 720\n\n\n $$\\therefore$$ Total no of ways = 5$$ \\times $$6$$ \\times $$6$$ \\times $$$${}^4{C_1}$$ = 720",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6458,
"subject": "General Science",
"question": "Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are :",
"options": [
{
"text": "312"
},
{
"text": "3125"
},
{
"text": "120"
},
{
"text": "216"
}
],
"answer": "216",
"solution": "**Answer:** 216\n\nNote : For a number to be divisible by 3, the sum of digits should be divisible by 3.\n
Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.\n
Set 1 : Set is = (1, 2, 3, 4, 5). Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3)\n
\n
So total no of arrangement = 1$$ \\times $$2$$ \\times $$3$$ \\times $$4$$ \\times $$5 = 5!\n
Set 2 : Set is = (0, 1, 2, 4, 5). Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3)\n\n
So total no of arrangement = 4$$ \\times $$4$$ \\times $$3$$ \\times $$2$$ \\times $$1 = 4.4!\n
$$\\therefore$$ Total arrangement = 5! + 4.4! = 216",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6459,
"subject": "General Science",
"question": "How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order ",
"options": [
{
"text": "480"
},
{
"text": "240"
},
{
"text": "360"
},
{
"text": "120"
}
],
"answer": "360",
"solution": "**Answer:** 360\n\nIn the word ''GARDEN'', there are two vowels A and E present, and A should come always before E.\n
$$\\therefore\\,\\,\\,$$ Total no of ways = $${{6!} \\over {2!}}$$ = 360\n
Here A and E has fixed order that is why we divide by 2!. ",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6460,
"subject": "General Science",
"question": "How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?",
"options": [
{
"text": "$$8.{}^6{C_4}.{}^7{C_4}$$ "
},
{
"text": "$$6.7.{}^8{C_4}$$ "
},
{
"text": "$$6.8.{}^7{C_4}$$. "
},
{
"text": "$$7.{}^6{C_4}.{}^8{C_4}$$ "
}
],
"answer": "$$7.{}^6{C_4}.{}^8{C_4}$$ ",
"solution": "**Answer:** $$7.{}^6{C_4}.{}^8{C_4}$$ \n\n
This problem is solved using gap method. As here no 'S' is adjacent to each other so we have to put them in the gap. So first write all the letters other than 'S' such a way that there is a gap between two letters.
\n
Given word is MISSISSIPPI.
\n
Here, I = 4 times, S = 4 times, P = 2 times, M = 1 time
\n
_M_I_I_I_I_P_P_
\n
Those seven letters M, I, I, I, I, P, P can be arranged in $${{7!} \\over {4!2!}}$$ ways
\n
Those seven letters creates 8 gaps and we have to choose 4 gaps from those 8 gaps to put those four 'S' letters.
\n
This can be done $${}^8{C_4}$$ ways.
\n
After placing those four 'S' letters we can arrange them in $${{4!} \\over {4!}}$$ ways.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6461,
"subject": "General Science",
"question": "If the four letter words (need not be meaningful ) are to be formed using the\nletters from the word “MEDITERRANEAN” such that the first letter is R and the fourth letter is E, then the total number of all such words is :\n",
"options": [
{
"text": "$${{11!} \\over {{{\\left( {2!} \\right)}^3}}}$$"
},
{
"text": "110"
},
{
"text": "56"
},
{
"text": "59"
}
],
"answer": "59",
"solution": "**Answer:** 59\n\nHere total no of different letters present are,\n
(1) One M\n
(2) Three E (E E E)\n
(3) One D\n
(4) One I\n
(5) One T\n
(6) Two R (R R)\n
(7) Two A (A A)\n
(8) Two N (N N)\n
In the four letter word first letter is R and last letter is E.\n
$$ \\therefore $$ Word is = R _ _ E\n
Now remaining letters are, \n
M, EE, D, I, T, R, AA, NN\n
Those 2 empty places can be filled with identical letters [EE, AA, NN] in 3 ways.\n
Or two empty places can be filled with distinct letters [M, E, D, I, T, R, A, N] in $${}^8{C_2} \\times 2!$$ ways.\n
$$ \\therefore $$ Total no of words = 3 + $${}^8{C_2} \\times 2!$$ = 59",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6462,
"subject": "General Science",
"question": "The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repitition of digits allowed) is equal to : ",
"options": [
{
"text": "374"
},
{
"text": "372"
},
{
"text": "375"
},
{
"text": "250"
}
],
"answer": "374",
"solution": "**Answer:** 374\n\nTotal no 1 digit numbers possible = 4 (allowed digits 1, 3, 7, 9)\n
Total no 2 digit numbers possible = 4$$ \\times $$5 = 20\n
Total no 3 digit numbers possible = 4$$ \\times $$5$$ \\times $$5 = 100\n
Total no 4 digit numbers possible = 2$$ \\times $$5$$ \\times $$5$$ \\times $$5 = 250\n
So the number of natural numbers less than 7,000 possible are\n = 4 + 20 + 100 + 250 = 374",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6463,
"subject": "General Science",
"question": "All possible numbers are formed using the digits\n1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number\nof such numbers in which the odd digits occupy\neven places is :",
"options": [
{
"text": "175"
},
{
"text": "162"
},
{
"text": "160"
},
{
"text": "180"
}
],
"answer": "180",
"solution": "**Answer:** 180\n\n\n
For those three odd digit numbers 1, 1, 3 we can choose any three positions out of the four even positions.\n
$$ \\therefore $$ No of ways we can choose 3 positions out of the 4 positions = 4C3\n
After choosing those three positions, number of ways we can arrange three odd digit numbers = 4C3 $$ \\times $$ $${{3!} \\over {2!}}$$\n
Then the remaining 6 digits can be arrange = $${{6!} \\over {2!4!}}$$ ways\n
$$ \\therefore $$ Total number of 9 digit numbers = 4C3 $$ \\times $$ $${{3!} \\over {2!}}$$ $$ \\times $$ $${{6!} \\over {2!4!}}$$ = 180\n\n\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6464,
"subject": "General Science",
"question": "The number of four-digit numbers strictly greater\nthan 4321 that can be formed using the digits\n0,1,2,3,4,5 (repetition of digits is allowed) is :",
"options": [
{
"text": "306"
},
{
"text": "288"
},
{
"text": "310"
},
{
"text": "360"
}
],
"answer": "310",
"solution": "**Answer:** 310\n\n\n\n\n\n So total 4 digit numbers possibe = 216 + 72 + 18 + 4 = 310",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6465,
"subject": "General Science",
"question": "The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by\n11 and no digit is repeated is :",
"options": [
{
"text": "36"
},
{
"text": "60"
},
{
"text": "72"
},
{
"text": "48"
}
],
"answer": "60",
"solution": "**Answer:** 60\n\n \n digit 0, 1, 2, 5, 7, 9
\n= 60",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6466,
"subject": "General Science",
"question": "The number of words (with or without meaning)\nthat can be formed from all the letters of the\nword “LETTER” in which vowels never come\ntogether is ________ .",
"options": [],
"answer": "120",
"solution": "**Answer:** 120\n\nConsonants $$ \\to $$ LTTR\n Vowels $$ \\to $$ EE\n
Total No of words = $${{6!} \\over {2!2!}}$$ = 180\n
Total no of words if vowels are together\n = $${{5!} \\over {2!}}$$ = 60\n
$$ \\therefore $$ Total no of words where vowels never come together = 180 – 60 = 120.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6467,
"subject": "General Science",
"question": "Two families with three members each and one family with four members are to be seated in a row.\nIn how many ways can they be seated so that the same family members are not separated?",
"options": [
{
"text": "2! 3! 4!"
},
{
"text": "(3!)3.(4!) "
},
{
"text": "3! (4!)3"
},
{
"text": "(3!)2.(4!)"
}
],
"answer": "(3!)3.(4!) ",
"solution": "**Answer:** (3!)3.(4!) \n\nF1 $$ \\to $$ 3 members\n F2 $$ \\to $$ 3 members\n F3 $$ \\to $$ 4 members\n
Total arrangements of three families = 3!\n
Arrangement between members of F1 family = 3!\n
Arrangement between members of F2 family = 3!\n
Arrangement between members of F3 family = 4!\n
$$ \\therefore $$ Total numbers of ways can they be seated so that the same family members are not separated\n
= (3!)3.(4!) ",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6468,
"subject": "General Science",
"question": "The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word ’SYLLABUS’ such that two letters are distinct and two letters are alike, is :\n",
"options": [],
"answer": "240",
"solution": "**Answer:** 240\n\nIn 'SYLLABUS' word\n
1. Two S letters\n
2. Two L letters\n
3. One Y letter\n
4. One A letter\n
5. One B letter\n
6. One U letter\n
Number of ways we can select two alike\nletters = 2C1\n
Then number of ways we can select two distinct\nletters = 5C2\n
Then total arrangement of\nselected letters = $${{4!} \\over {2!}}$$\n
So total number of words, with or without meaning, that can be formed\n
= 2C1 $$ \\times $$ 5C2 $$ \\times $$ $${{4!} \\over {2!}}$$ = 240",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6469,
"subject": "General Science",
"question": "If the number of five digit numbers with distinct\ndigits and 2 at the 10th place is 336 k, then k\nis equal to :",
"options": [
{
"text": "6"
},
{
"text": "8"
},
{
"text": "4"
},
{
"text": "7"
}
],
"answer": "8",
"solution": "**Answer:** 8\n\n\n
Except 1 and 2 there are eight options on first place and only one option at fourth place possible which is digit '2'.\n
Remaining three places can be filled by any three digits out of 1, 3, 4, 5, 6, 7, 8 and 9.\n
$$ \\Rightarrow $$ k = 8",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6470,
"subject": "General Science",
"question": "The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is _____________.",
"options": [],
"answer": "32",
"solution": "**Answer:** 32\n\nThe numbers are lying between 100 and 1000\nthen each number is of three digits.\n
The possible combination of 3 digits numbers\nare\n
The possible combination of numbers which are divisible by 3 are 1, 2,\n3; 3, 4, 5; 1, 3, 5 and 2, 3, 4.\n (If sum of digits of a number is divisible by 3 then the number is divisible by 3)\n
$$ \\therefore $$ Total number of numbers = 4 × 3! = 24\n
The possible combination of numbers divisible by 5 are 1, 2, 5; 2, 3, 5; 3, 4, 5; 1, 3, 5;\n1, 4, 5 and 2, 4, 5.\n (If the last digit of a number is 0 or 5 then the number is divisible by 5)\n
$$ \\therefore $$ Total number of numbers = 6 × 2! = 12\n
The possible combination of number divisible by both 3 and 5 are 1, 3, 5 and 3, 4, 5.\n\n
$$ \\therefore $$ Total number of numbers = 2 $$ \\times $$ 2! = 4\n
$$ \\therefore $$ Total required number = 24 + 12 - 4 = 32",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6471,
"subject": "General Science",
"question": "The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is :",
"options": [
{
"text": "26664"
},
{
"text": "122664"
},
{
"text": "122234"
},
{
"text": "22264"
}
],
"answer": "26664",
"solution": "**Answer:** 26664\n\nTotal possible numbers using 1, 2, 2 and 3 is\n
= $${{4!} \\over {2!}}$$ = 12\n
When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are\n
= $${{3!} \\over {2!}}$$ = 3\n
When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are\n
= 3! = 6\n
When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are\n
= $${{3!} \\over {2!}}$$ = 3\n
$$ \\therefore $$ Sum of unit places of all (3 + 6 + 3) 12 numbers is \n
= (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 $$ \\times $$ 24 = 26664",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6472,
"subject": "General Science",
"question": "The number of times the digit 3 will be written when listing the integers from 1 to 1000 is :",
"options": [],
"answer": "300",
"solution": "**Answer:** 300\n\nIn single digit numbers = 1\n
In double digit numbers = 10 + 9 = 19\n
In triple digit numbers = 100 + 90 + 90 = 280\n
Total = 300 times",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6473,
"subject": "General Science",
"question": "If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number of all numbers greater than 10,000 is equal to _____________.",
"options": [],
"answer": "96",
"solution": "**Answer:** 96\n\n = 4 $$\\times$$ 4 $$\\times$$ 3 $$\\times$$ 2 = 96",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6474,
"subject": "General Science",
"question": "The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, is ______________.",
"options": [],
"answer": "52",
"solution": "**Answer:** 52\n\n(i) When '0' is at unit place
Number of numbers = 20
(ii) When 4 or 6 are at unit place
Number of numbers = 32
Total three digit even number = 20 + 32 = 52",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6475,
"subject": "General Science",
"question": "A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is ____________.",
"options": [],
"answer": "100",
"solution": "**Answer:** 100\n\n
\n\n
\n
5
\n
a
\n
b
\n
b
\n
a
\n
5
\n
\n\n
\n
For divisible by 55 it shall be divisible by 11 and 5\nboth, for divisibility by 5 unit digit shall be 0 or 5 but\nas the number is six digit palindrome unit digit is 5.\n\n
A number is divisible by 11 if the difference between sum of the digits in the odd places and the sum of the digits in the even places is a multiple of 11 or zero.\n
Sum of the digits in the even place = a + b + 5\n
Sum of the digits in the odd places = a + b + 5\n
Difference between the two sums = (a + b + 5 ) - (a + b + 5) = 0\n
0 is divisible by 11.\n
Hence, 5abba5 is divisible by 11.\n
So, required number = 10 $$\\times$$ 10 = 100",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6476,
"subject": "General Science",
"question": "The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is ___________.",
"options": [],
"answer": "576",
"solution": "**Answer:** 576\n\nTotal possible words = 6! = 720\n
When 4 consonants are together (V, W, L, S)\n such cases = 3! ⋅ 4! = 144\n
All consonants should not be together
= Total $$-$$ All consonants together,
= 6! $$-$$ 3! 4! = 576",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6477,
"subject": "General Science",
"question": "All the arrangements, with or without meaning, of the word FARMER are written excluding any word that has two R appearing together. The arrangements are listed serially in the alphabetic order as in the English dictionary. Then the serial number of the word FARMER in this list is ___________.",
"options": [],
"answer": "77",
"solution": "**Answer:** 77\n\nFirst find all possible words and then subtract words\nfrom each case that have both R together.\n
Let b1b2b3b4 be a 4-element permutation with bi $$\\in$$ {1, 2, 3, ........, 100} for 1 $$\\le$$ i $$\\le$$ 4 and bi $$\\ne$$ bj for i $$\\ne$$ j, such that either b1, b2, b3 are consecutive integers or b2, b3, b4 are consecutive integers. Then the number of such permutations b1b2b3b4 is equal to ____________.
There are 98 sets of three consecutive integer and 97 sets of four consecutive integers.
\n
Using the principle of inclusion and exclusion,
\n
Number of permutations of $b_{1} b_{2} b_{3} b_{4}=$ Number of permutations when $b_{1} b_{2} b_{3}$ are consecutive + Number of permutations when $b_{2} b_{3} b_{4}$ are consecutive - Number of permutations when $b_{1} b_{2}$ $b_{3} b_{4}$ are consecutive
The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is :
",
"options": [
{
"text": "36"
},
{
"text": "48"
},
{
"text": "60"
},
{
"text": "72"
}
],
"answer": "72",
"solution": "**Answer:** 72\n\nTo make a no. divisible by 3 we can use the digits\n1,2,5,6,7 or 1,2,3,5,7.
\n Using 1,2,5,6,7, number of even numbers is\n = 4 × 3 × 2 × 1 × 2 = 48
\n Using 1,2,3,5,7, number of even numbers is\n = 4 × 3 × 2 × 1 × 1 = 24
Numbers are to be formed between 1000 and 3000 , which are divisible by 4 , using the digits $$1,2,3,4,5$$ and 6 without repetition of digits. Then the total number of such numbers is ____________.
",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\nHere 1st digit is 1 or 2 only
\nCase-I
\nIf first digit is 1
\nThen last two digits can be 24, 32, 36, 52, 56, 64
\n \nCase – II
\nIf first digit is 2 then last two digit can be 16, 36,\n56, 64
The number of natural numbers lying between 1012 and 23421 that can be formed using the digits $$2,3,4,5,6$$ (repetition of digits is not allowed) and divisible by 55 is _________.
When number is 4-digit number $(\\overline{a b c d})$ here $d$ is fixed as 5\n
\nSo, $(a, b, c)$ can be $(6,4,3),(3,4,6),(2,3,6)$, $(6,3,2),(3,2,4)$ or $(4,2,3)$\n
\n$\\Rightarrow 6$ numbers\n
\nCase-II
No number possible",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6488,
"subject": "General Science",
"question": "
The total number of six digit numbers, formed using the digits 4, 5, 9 only and divisible by 6, is ____________.
",
"options": [],
"answer": "81",
"solution": "**Answer:** 81\n\nA number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 and sum of all digits of the number is divisible by 3 .\n
Units, place must be occupied by 4 and hence, at \nleast one 4 must be there.\n
Possible combination of 4, 5, 9 are as follows :\n
The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ___________.
Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to ____________.
",
"options": [],
"answer": "710",
"solution": "**Answer:** 710\n\nNumbers which are divisible by 3 (4 digit) and less than or equal to 2800\n\n
$=\\frac{2799-1002}{3}+1=600$\n\n
Numbers which are divisible by 11 (4 digit) and less than or equal to 2800\n\n
$=\\frac{2794-1001}{11}+1=164$\n\n
Numbers which are divisible by 33 (4 digit) and less than or equal to 2800\n\n
$=\\frac{2772-1023}{33}+1=54$\n\n
$\\therefore$ Total numbers = $600+164-54=710$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6491,
"subject": "General Science",
"question": "The number of seven digits odd numbers, that can\nbe formed using all the
Number of 4-digit numbers (the repetition of digits is allowed) which are made using the digits 1, 2, 3 and 5, and are divisible by 15, is equal to ___________.
If all the six digit numbers $$x_1\\,x_2\\,x_3\\,x_4\\,x_5\\,x_6$$ with $$0< x_1 < x_2 < x_3 < x_4 < x_5 < x_6$$ are arranged in the increasing order, then the sum of the digits in the $$\\mathrm{72^{th}}$$ number is _____________.
\n \nTotal Numbers = 3 × 4 × 3 × 2 = 72",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6495,
"subject": "General Science",
"question": "The total number of three-digit numbers, divisible by 3, which can be formed using the digits $1,3,5,8$, if repetition of digits is allowed, is :",
"options": [
{
"text": "21"
},
{
"text": "22"
},
{
"text": "18"
},
{
"text": "20"
}
],
"answer": "22",
"solution": "**Answer:** 22\n\nThe number of three-digit numbers divisible by 3 by considering the possible sums of digits that are divisible by 3. Your approach is as follows:\n\n
1. Sum of digits is 3: $(1, 1, 1)$ - 1 possible number\n
2. Sum of digits is 9: $(1, 3, 5)$ and $(3, 3, 3)$ -\n
Let's consider the cases separately :\n\n
a. Sum of digits is 9: $(1, 3, 5)$\n For this case, we can arrange the digits in $3!$ ways :\n 135, 153, 315, 351, 513, and 531.\n\n
b. Sum of digits is 9: $(3, 3, 3)$\n For this case, since all the digits are the same, there is only 1 possible number :\n 333.\n\n
Now, the total number of possible numbers when the sum of digits is 9 is :\n\n
$$\n3! + 1 = 6 + 1 = 7\n$$\n\n
3. Sum of digits is 12: $(1, 3, 8)$ - $3!$ possible numbers\n
4. Sum of digits is 15: $(5, 5, 5)$ - 1 possible number\n
5. Sum of digits is 18: $(5, 5, 8)$ - $\\frac{3!}{2!}$ possible numbers (since 5 is repeated)\n
6. Sum of digits is 21: $(5, 8, 8)$ - $\\frac{3!}{2!}$ possible numbers (since 8 is repeated)\n
7. Sum of digits is 24: $(8, 8, 8)$ - 1 possible number\n\n
Adding up the possible numbers for each case, we get:\n\n
So, there are a total of 22 three-digit numbers divisible by 3 that can be formed using the digits $1, 3, 5, 8$ with repetition allowed.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6496,
"subject": "General Science",
"question": "A person forgets his 4-digit ATM pin code. But he remembers that in the code all the digits are different, the greatest digit is 7 and the sum of the first two digits is equal to the sum of the last two digits. Then the maximum number of trials necessary to obtain the correct code is ___________.",
"options": [],
"answer": "72",
"solution": "**Answer:** 72\n\nLet the 4-digit ATM pin code be represented by the digits $$abxy$$, where all digits are different, the greatest digit is 7, and the sum of the first two digits is equal to the sum of the last two digits: $$a + b = x + y$$.\n\n
Since the greatest digit is 7, the possible digits for the pin code are $$0, 1, 2, 3, 4, 5, 6,$$ and $$7$$.\n\n
1. Let's denote the sum of the first two digits and the sum of the last two digits as $$\\lambda$$. Since the largest digit is 7, we can analyze different possible values for $$\\lambda$$.\n\n
We will analyze different possible values for $$\\lambda$$ and the corresponding possible pairs of digits:\n\n
- $$\\lambda = 7$$: The pairs of digits that have a sum of 7 are $(0,7),(7,0),(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. Since 7 must be present, we consider the pairs with 7: $(0,7),(7,0)$. \n\n
When the pair $(0,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $$07xy$$ and $$07yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.\n\n
When the pair $(7,0)$ is chosen, the other two digits can also be any one of the remaining pairs: $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$. For each of these pairs, there are 2 possible pin codes: $$70xy$$ and $$70yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 12 possible pin codes for this case.\n\n
So, there are 24 possible pin codes for $$\\lambda = 7$$. \n \n\n
- $$\\lambda = 8$$: The pairs of digits that have a sum of 8 are $(1,7),(7,1),(2,6),(6,2),(3,5),(5,3)$. Since 7 must be present, we consider the pairs with 7: $(1,7),(7,1)$.\n\n
When the pair $(1,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $$17xy$$ and $$17yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.\n\n
When the pair $(7,1)$ is chosen, the other two digits can also be any one of the remaining pairs: $(2,6),(6,2),(3,5),(5,3)$. For each of these pairs, there are 2 possible pin codes: $$71xy$$ and $$71yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.\n\n
So, there are 16 possible pin codes for $$\\lambda = 8$$.\n
\n\n
- $$\\lambda = 9$$: The pairs of digits that have a sum of 9 are $(2,7),(7,2),(3,6),(6,3),(4,5),(5,4)$. Since 7 must be present, we consider the pairs with 7: $(2,7),(7,2)$.\n\n
When the pair $(2,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $$27xy$$ and $$27yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.\n\n
When the pair $(7,2)$ is chosen, the other two digits can also be any one of the remaining pairs: $(3,6),(6,3),(4,5),(5,4)$. For each of these pairs, there are 2 possible pin codes: $$72xy$$ and $$72yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 8 possible pin codes for this case.\n\n
So, there are 16 possible pin codes for $$\\lambda = 9$$.\n\n
\n\n
- $$\\lambda = 10$$: The pairs of digits that have a sum of 10 are $(3,7),(7,3),(4,6),(6,4)$. Since 7 must be present, we consider the pairs with 7: $(3,7),(7,3)$.\n\n
When the pair $(3,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $$37xy$$ and $$37yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.\n\n
When the pair $(7,3)$ is chosen, the other two digits can also be any one of the remaining pairs: $(4,6),(6,4)$. For each of these pairs, there are 2 possible pin codes: $$73xy$$ and $$73yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.\n\n
So, there are 8 possible pin codes for $$\\lambda = 10$$.\n
\n\n
- $$\\lambda = 11$$: The pairs of digits that have a sum of 11 are $(4,7),(7,4),(5,6),(6,5)$. Since 7 must be present, we consider the pairs with 7: $(4,7),(7,4)$.\n\n
When the pair $(4,7)$ is chosen, the other two digits can be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $$47xy$$ and $$47yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.\n\n
When the pair $(7,4)$ is chosen, the other two digits can also be any one of the remaining pairs: $(5,6),(6,5)$. For each of these pairs, there are 2 possible pin codes: $$74xy$$ and $$74yx$$, where $$x$$ and $$y$$ are the digits from the chosen pair. In total, there are 4 possible pin codes for this case.\n\n
So, there are 8 possible pin codes for $$\\lambda = 11$$.\n\n
\n\n
- $$\\lambda = 12$$: The only pair of digits that have a sum of 12 is $(5,7)$. Since 7 must be present, this case is not possible.\n\n
- $$\\lambda = 13$$: The only pair of digits that have a sum of 13 is $(6,7)$. Since 7 must be present, this case is not possible.\n\n
- $$\\lambda = 14$$: The only pair of digits that have a sum of 14 is $(7,7)$. Since all the digits must be different, this case is not possible.\n\n
Adding up all the possible pin codes for each value of $$\\lambda$$, we get:\n\n
$$16 + 24 + 24 + 16 + 8 = 72$$\n\n
Therefore, the maximum number of trials necessary to obtain the correct code is 72.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6497,
"subject": "General Science",
"question": "
Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits $$1,2,3,4,5$$ with repetition, is _________.
",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\nA number is divisible by 6 if it is divisible by both 2 and 3. A number is divisible by 2 if its last digit is even, which means it must be either 2 or 4 from the given digits. A number is divisible by 3 if the sum of its digits is divisible by 3.\n
The number of five digit numbers, greater than 40000 and divisible by 5 , which can be formed using the digits $$0,1,3,5,7$$ and 9 without repetition, is equal to :
",
"options": [
{
"text": "132"
},
{
"text": "72"
},
{
"text": "120"
},
{
"text": "96"
}
],
"answer": "120",
"solution": "**Answer:** 120\n\nSince the five-digit number must be greater than 40000, the only options for the first digit are 5, 7, or 9. That leaves 3 remaining choices for the first digit. \n\n
Since the number has to be divisible by 5, the last digit must be 0 or 5. If the first digit is 5, the last digit can only be 0, since digits cannot be repeated. If the first digit is 7 or 9, the last digit can be 0 or 5, so there are 2 choices. \n\n
Five possible configurations for the first and last digits :\n\n
- 5xxx0\n
- 7xxx0\n
- 7xxx5\n
- 9xxx0\n
- 9xxx5\n\n
For each of these configurations, there are three places in the middle that can be filled with the remaining 4 unused digits. \n\n
Since repetition isn't allowed, there are 4 options for the second digit, 3 for the third, and 2 for the fourth, which can be represented as 4P3, or permutations of 4 items taken 3 at a time. This is equal to 4 $$ \\times $$ 3 $$ \\times $$ 2 = 24.\n\n
So, for each of the 5 configurations, there are 24 ways to arrange the middle three digits. Therefore, the total number of five-digit numbers that meet the criteria is 5 $$ \\times $$ 24 = 120.\n",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6499,
"subject": "General Science",
"question": "
Let the digits a, b, c be in A. P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?
The problem involves forming nine-digit numbers from three digits a, b, c which are in Arithmetic Progression (AP), used three times each, such that at least once, three consecutive digits are in AP.
\n
We have the two possible sequences for the AP :
\n\n
a, b, c
\n
c, b, a
\n\n
This shows the flexibility in ordering the three digits that are in AP in our nine-digit number.
\n
The next step is to choose the location of this sequence of three numbers within our nine-digit number.
\n
Since there are nine places in our number and our sequence takes up three places, we have seven different starting points for our sequence : it can start at the first place, the second place, and so on, up to the seventh place.
\n
Therefore, the number of ways to select 3 consecutive places out of the 9 places for the AP sequence is 7.
\n
However, we also have to account for the fact that our sequence can be in one of two orders (a, b, c or c, b, a). So, we multiply the number of starting points by 2 to get $^7C_1 \\times 2 = 14$ ways to arrange the sequence within our nine-digit number.
\n
The remaining 6 digits (two 'a', two 'b', two 'c\") can be arranged in $\\frac{6!}{2!2!2!}$ ways.
\n
Therefore, the total number of such nine-digit numbers is $^7C_1 \\times 2 \\times \\frac{6!}{2!2!2!} = 1260$.
The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ___________.
",
"options": [],
"answer": "4898",
"solution": "**Answer:** 4898\n\nGiven that digits are $1,2,3,4,5,6,7$\n
If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which $$\\mathrm{C}$$ and $$\\mathrm{S}$$ do not come together, is $$(6 !) \\mathrm{k}$$, then $$\\mathrm{k}$$ is equal to :
",
"options": [
{
"text": "5670"
},
{
"text": "1890"
},
{
"text": "2835"
},
{
"text": "945"
}
],
"answer": "5670",
"solution": "**Answer:** 5670\n\n$$\n\\text { Total number of words }=\\frac{11 !}{2 ! 2 ! 2 !}\n$$\n
Number of words in which $\\mathrm{C}$ and $\\mathrm{S}$ are together\n
Out of $n$ objects, if $r$ things are same, so number of ways $=\\frac{n !}{r !}$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6503,
"subject": "General Science",
"question": "
The number of 4-letter words, with or without meaning, each consisting of 2 vowels and 2 consonants, which can be formed from the letters of the word UNIVERSE without repetition is __________.
",
"options": [],
"answer": "432",
"solution": "**Answer:** 432\n\nGiven, word is UNIVERSE\n\n
Here, vowels are E, I, U and consonants are N, R, S, V \n
$\\therefore$ Required number of 4-letters words, with or without meaning, \n
In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is _________.
",
"options": [],
"answer": "44",
"solution": "**Answer:** 44\n\nThis problem can be solved using the concept of derangements, which is a permutation of objects where no object appears in its original position. In this case, we have 5 students who should not sit in their allotted seats. \n\n
The formula for calculating the number of derangements (also known as subfactorials) is :\n\n
So, there are 44 ways in which none of the students sits on the allotted seat.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6505,
"subject": "General Science",
"question": "The sum of integers from 1 to 100 that are divisible by 2 or 5 is :",
"options": [
{
"text": "3000"
},
{
"text": "3050"
},
{
"text": "3600"
},
{
"text": "3250"
}
],
"answer": "3050",
"solution": "**Answer:** 3050\n\nAccording to this question, any number between 1 to 100 should be divisible by 2 or 5 but not by 2$$ \\times $$5 = 10.\n
Possible numbers between 1 to 100 divisible by 2 are 2, 4, 6, .... , 100\n
This is an A.P where first term = 2, last term = 100 and total terms = 50.\n
$$ \\therefore $$ Sum of the numbers divisible by 2\n
= $${{50} \\over 2}\\left[ {2 + 100} \\right]$$\n
= 25$$ \\times $$102\n
= 2550\n
Possible numbers between 1 to 100 divisible by 5 are 5, 10, 15, .... , 100\n
$$ \\therefore $$ Sum of the numbers divisible by 5\n
= $${{20} \\over 2}\\left[ {5 + 100} \\right]$$\n
= 10$$ \\times $$105\n
= 1050\n
And possible numbers between 1 to 100 divisible by 10 are 10, 20, 30, .... , 100\n
$$ \\therefore $$ Sum of the numbers divisible by 10\n
$$ \\therefore $$ Required sum = 2550 + 1050 - 550 = 3050",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6506,
"subject": "General Science",
"question": "The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is : ",
"options": [
{
"text": "24"
},
{
"text": "30"
},
{
"text": "36"
},
{
"text": "48"
}
],
"answer": "30",
"solution": "**Answer:** 30\n\nHere number should be divisible by 3, that means sum of numbers should be divisible by 3.\n
Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are \n
(1)$$\\,\\,\\,\\,$$ (0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3)\n
(2) $$\\,\\,\\,\\,$$ (0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3 = 6 (divisible by 3)\n
Case 1 : \n
When 4 digits are (0, 2, 3, 4) then\n
$$\\therefore\\,\\,\\,\\,$$ Total possible numbers = $$^3{C_1}$$ $$ \\times $$ $$^3{C_1}$$ $$ \\times $$ $$^2{C_1}$$ $$ \\times $$ $$^1{C_1}$$ \n
$$\\therefore\\,\\,\\,\\,$$ Total possible number in this case = $$^2{C_1}$$ $$ \\times $$ $$^3{C_1}$$ $$ \\times $$ $$^2{C_1}$$ $$ \\times $$ $$^1{C_1}$$\n
$$\\therefore\\,\\,\\,\\,$$ Total possible numbers will be = 18 + 12 = 30 ",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6507,
"subject": "General Science",
"question": "A natural number has prime factorization given by n = 2x3y5z, where y and z are such that y + z = 5 and y$$-$$1 + z$$-$$1 = $${5 \\over 6}$$, y > z. Then the number of odd divisions of n, including 1, is :",
"options": [
{
"text": "11"
},
{
"text": "6"
},
{
"text": "12"
},
{
"text": "6x"
}
],
"answer": "12",
"solution": "**Answer:** 12\n\ny + z = 5 ....... (1)
$$ \\Rightarrow $$ Total possible cases = (3050 + 3150 + 3250 + 3350 + .... + 3352)
$$ \\therefore $$ Total odd divisors must be (3 + 1) ( 2 + 1) = 12",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6508,
"subject": "General Science",
"question": "Let n be a non-negative integer. Then the number of divisors of the form \"4n + 1\" of the number (10)10 . (11)11 . (13)13 is equal to __________.",
"options": [],
"answer": "924",
"solution": "**Answer:** 924\n\nN = 210 $$\\times$$ 510 $$\\times$$ 1111 $$\\times$$ 1313
Then $${x \\over 2}$$ $$\\in$$ [50, 499], $${x \\over 2}$$ $$\\in$$ N
\n
Number whose G.C.D. with 18 is 1 in this range have the required condition. There are 6 such number from 18 $$\\times$$ 3 to 18 $$\\times$$ 4. Similarly from 18 $$\\times$$ 4 to 18 $$\\times$$ 5 ......., 26 $$\\times$$ 18 to 27 $$\\times$$ 18
If for some $$m, n ;{ }^6 C_m+2\\left({ }^6 C_{m+1}\\right)+{ }^6 C_{m+2}>{ }^8 C_3$$ and $${ }^{n-1} P_3:{ }^n P_4=1: 8$$, then $${ }^n P_{m+1}+{ }^{\\mathrm{n}+1} C_m$$ is equal to
$$ = {}^{n + 2}{C_{r + 1}}$$ ",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6515,
"subject": "General Science",
"question": "The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is ",
"options": [
{
"text": "$${}^8{C_3}$$ "
},
{
"text": "21"
},
{
"text": "$${3^8}$$ "
},
{
"text": "5 "
}
],
"answer": "21",
"solution": "**Answer:** 21\n\nTo distribute n objects among p people where everyone should get atleast one object, then number of ways to distribute those n objects\n
= $${}^{n - 1}{C_{p - 1}}$$\n
For this question, n = 8 and p = 3\n
$$ \\therefore $$ Number of ways = $${}^{8 - 1}{C_{3 - 1}}$$ = $${}^7{C_2}$$ = 21",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6516,
"subject": "General Science",
"question": "At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is ",
"options": [
{
"text": "5040"
},
{
"text": "6210"
},
{
"text": "385"
},
{
"text": "1110"
}
],
"answer": "385",
"solution": "**Answer:** 385\n\nA voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.\n
Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$\n
Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$\n
Case 3 : When he give vote to 3 candidates then no ways = $${}^{10}{C_3}$$\n
Case 4 : When he give vote to 4 candidates then no ways = $${}^{10}{C_4}$$\n So, total no of ways he can give votes \n = $${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$$\n = 385\n
Note : Here we use addition rule as he can vote any one of those four rules. Whenever there is \"or\" choices, we use addition rule.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6517,
"subject": "General Science",
"question": "The set S = {1, 2, 3, ........., 12} is to be partitioned into three sets A, B, C of equal size. Thus $$A \\cup B \\cup C = S,\\,A \\cap B = B \\cap C = A \\cap C = \\phi $$. The number of ways to partition S is ",
"options": [
{
"text": "$${{12!} \\over {{{(4!)}^3}}}\\,\\,$$ "
},
{
"text": "$${{12!} \\over {{{(4!)}^4}}}\\,\\,$$ "
},
{
"text": "$${{12!} \\over {3!\\,\\,{{(4!)}^3}}}$$ "
},
{
"text": "$${{12!} \\over {3!\\,\\,{{(4!)}^4}}}$$ "
}
],
"answer": "$${{12!} \\over {{{(4!)}^3}}}\\,\\,$$ ",
"solution": "**Answer:** $${{12!} \\over {{{(4!)}^3}}}\\,\\,$$ \n\n
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6518,
"subject": "General Science",
"question": "In a shop there are five types of ice-cream available. A child buys six ice-cream.\n Statement - 1: The number of different ways the child can buy the six ice-cream is $${}^{10}{C_5}$$.\n Statement - 2: The number of different ways the child can buy the six ice-cream is equal to the number of different ways of arranging 6 A and 4 B's in a row.",
"options": [
{
"text": "Statement - 1 is false, Statement - 2 is true "
},
{
"text": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1"
},
{
"text": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1"
},
{
"text": "Statement - 1 is true, Statement - 2 is false"
}
],
"answer": "Statement - 1 is false, Statement - 2 is true ",
"solution": "**Answer:** Statement - 1 is false, Statement - 2 is true \n\nNote : n items can be distribute among p persons are $${}^{n + p - 1}{C_{p - 1}}$$ ways.\n
Here n = 6 ice-cream\n
p = 5 types of ice-cream\n
Each ice-cream belongs to one of the 5 ice-cream type. So chosen 6 ice-crean can be divide into 5 types of ice-cream.\n
$$ \\therefore $$ The number of different ways the child can buy the six ice-cream is = $${}^{6 + 5 - 1}{C_{5 - 1}}$$ = $${}^{10}{C_4}$$\n
$$ \\therefore $$ Statement - 1 is false.\n
Number of different ways of arranging 6 A and 4 B's in a row\n
= $${{10!} \\over {6!4!}} = {}^{10}{C_4}$$\n
$$ \\therefore $$ Statement - 2 is true.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6519,
"subject": "General Science",
"question": "There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is",
"options": [
{
"text": "36"
},
{
"text": "66"
},
{
"text": "108"
},
{
"text": "3"
}
],
"answer": "108",
"solution": "**Answer:** 108\n\n
Thus number of ways $$ = ({}^3{C_2}) \\times ({}^9{C_2}) = 3 \\times {{9 \\times 8} \\over 2} = 108$$
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6520,
"subject": "General Science",
"question": "These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then:",
"options": [
{
"text": "$$N \\le 100$$ "
},
{
"text": "$$100 < N \\le 140$$ "
},
{
"text": "$$140 < N \\le 190\\,$$ "
},
{
"text": "$$N > 190$$ "
}
],
"answer": "$$N \\le 100$$ ",
"solution": "**Answer:** $$N \\le 100$$ \n\n
We need 3 points to create a triangle. With 10 points number of triangle possible $${}^{10}{C_3}$$
\n
Here 6 points are on the same line so we can't make any triangle with those 6 points.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6521,
"subject": "General Science",
"question": " Statement - 1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is emply is $${}^9{C_3}$$.\n Statement - 2: The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$.",
"options": [
{
"text": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1."
},
{
"text": "Statement - 1 is true, Statement - 2 is false."
},
{
"text": "Statement - 1 is false, Statement - 2 is true."
},
{
"text": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1."
}
],
"answer": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1.",
"solution": "**Answer:** Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1.\n\n
Let XA, XB, XC and XD represent number of balls present in box A, B, C and D respectively.
\n
As no box can be empty so,
\n
XA $$\\ge$$ 1, XB $$\\ge$$ 1, XC $$\\ge$$ 1 and XD $$\\ge$$ 1
Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6
\n
From formula we know, n things can be distributed among r people in $${}^{n + r - 1}{C_{r - 1}}$$ ways where each people can have either 0 or more things.
\n
$$\\therefore$$ 6 balls can be distributed among 4 boxes in $${}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}$$ ways where each box can have either 0 or more balls.
\n
Therefore, Statement 1 is correct. The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$ ways. But Statement - 2 is not the correct explanation of Statement - 1.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6522,
"subject": "General Science",
"question": "Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is:",
"options": [
{
"text": "880"
},
{
"text": "629"
},
{
"text": "630"
},
{
"text": "879"
}
],
"answer": "879",
"solution": "**Answer:** 879\n\n
For alike n objects, number of ways we can select zero or more objects = n + 1 and number of ways we can select at least one object = n
\n
Given 10 identical white balls, 9 identical green balls and 7 identical black balls.
\n
To find number of ways for selecting atleast one ball.
\n
Number of ways to choose zero or more white balls = (10 + 1) [since, all white balls are mutually identical]
\n
Number of ways to choose zero or more green balls = (9 + 1) [since, all green balls are mutually identical]
\n
Number of ways to choose zero or more black balls = (7 + 1) [since, all black balls are mutually identical]
\n
Hence, number of ways to choose zero or more balls of any colour = (10 + 1) (9 + 1) (7 + 1)
\n
Also, number of ways to choose a total of zero balls = 1
\n
Hence, the number, if ways to choose at least one ball (irrespective of any colour) = (10 + 1) (9 + 1) (7 + 1) $$-$$ 1 = 879
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6523,
"subject": "General Science",
"question": "Let A and B be two sets containing 2 elements and\n4 elements respectively. The number of subsets of\nA $$ \\times $$ B having 3 or more elements is :",
"options": [
{
"text": "219"
},
{
"text": "211"
},
{
"text": "256"
},
{
"text": "220"
}
],
"answer": "219",
"solution": "**Answer:** 219\n\nA $$ \\times $$ B will have 2 $$ \\times $$ 4 = 8 elements.\n
The number of subsets having atleast 3 elements\n
= 8C3 + 8C4 + 8C5 + 8C6 + 8C7 + 8C8 \n
= 28 – (8C0 + 8C1 + 8C2) = 256 – 1 – 8 – 28 = 219 ",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6524,
"subject": "General Science",
"question": "Let $${T_n}$$ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If $${T_{n + 1}} - {T_n}$$ = 10, then the value of n is :",
"options": [
{
"text": "7 "
},
{
"text": "5 "
},
{
"text": "10 "
},
{
"text": "8"
}
],
"answer": "5 ",
"solution": "**Answer:** 5 \n\nNumber of possible triangle using n vertices = nC3\n
n2 + 3n $$-$$ 108 = 0",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6527,
"subject": "General Science",
"question": "A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are\nladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X\nand Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in\nthis party, is:",
"options": [
{
"text": "468"
},
{
"text": "469"
},
{
"text": "484"
},
{
"text": "485"
}
],
"answer": "485",
"solution": "**Answer:** 485\n\n
\n
\n
\n
X(7 Friends)
\n
Y(7 Friends)
\n
\n
\n
\n
4 Ladies
\n
3 Men
\n
3 Ladies
\n
4 Men
\n
\n
\n
Case 1
\n
3
\n
0
\n
0
\n
3
\n
\n
\n
Case 2
\n
0
\n
3
\n
3
\n
0
\n
\n
\n
Case 3
\n
2
\n
1
\n
1
\n
2
\n
\n
\n
Case 4
\n
1
\n
2
\n
2
\n
1
\n
\n
\n
In Case 1, Case 2, Case 3 and Case 4, total 6 friends are present and 3 from X and 3 from Y and among those 6 friend 3 are ladies and 3 are men in every case.\n
$$\\therefore$$ No of ways 6 friends can be invited =\n
$$ \\therefore $$ $$K = {2^{25}}$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6529,
"subject": "General Science",
"question": "Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is : ",
"options": [
{
"text": "164"
},
{
"text": "240"
},
{
"text": "82"
},
{
"text": "120"
}
],
"answer": "120",
"solution": "**Answer:** 120\n\nNumber of ways = 10C3 = 120",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6530,
"subject": "General Science",
"question": "There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is :",
"options": [
{
"text": "12"
},
{
"text": "9"
},
{
"text": "7"
},
{
"text": "11"
}
],
"answer": "12",
"solution": "**Answer:** 12\n\nLet m-men, 2-women\n
$$ \\Rightarrow $$ $${a \\over {11}} = {b \\over {22}} = {c \\over {42}}$$\n",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6534,
"subject": "General Science",
"question": "An urn contains 5 red marbles, 4 black marbles\nand 3 white marbles. Then the number of ways\nin which 4 marbles can be drawn so that at the\nmost three of them are red is ___________.",
"options": [],
"answer": "490",
"solution": "**Answer:** 490\n\nHere 5 red marbels and 7 non red marbels presents.\n
No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.\n
= 490",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6535,
"subject": "General Science",
"question": "A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is __________.",
"options": [],
"answer": "135",
"solution": "**Answer:** 135\n\nSelect any 4 questions in 6C4\n ways which are\ncorrect.\n
Answering right option for each question is possible in 1 way.\n
So ways of choosing right option for 4 questions = 1.1.1.1 = (1)4\n
Number of ways of choosing wrong option for each question = 3\n
So ways of choosing wrong option for 2 questions = (3)2\n
$$ \\therefore $$ Required number of ways = 6C4.(1)4.(3)2 = 135",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6536,
"subject": "General Science",
"question": "Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to :",
"options": [
{
"text": "5"
},
{
"text": "2"
},
{
"text": "4"
},
{
"text": "6"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\nTotal matches between boys of both team = $${}^7{C_1} \\times {}^4{C_1} = 28$$
Total matches between girls of both team = $${}^n{C_1}\\,{}^6{C_1} = 6n$$
Now, 28 + 6n = 52
$$ \\Rightarrow $$ n = 4",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6537,
"subject": "General Science",
"question": "If $${}^n{P_r} = {}^n{P_{r + 1}}$$ and $${}^n{C_r} = {}^n{C_{r - 1}}$$, then the value of r is equal to :",
"options": [
{
"text": "1"
},
{
"text": "4"
},
{
"text": "2"
},
{
"text": "3"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n$${}^n{P_r} = {}^n{P_{r + 1}} \\Rightarrow {{n!} \\over {(n - r)!}} = {{n!} \\over {(n - r - 1)!}}$$
From (1) and (2), $$ 2r - 1 - r = 1 \\Rightarrow r = 2$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6538,
"subject": "General Science",
"question": "
A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then $$\\mathrm{b}+3 \\mathrm{~g}$$ is equal to ____________.
The number of matrices A such that the sum of all entries is a prime number $\\mathrm{p} \\in(2,13)$ is __________.",
"options": [],
"answer": "204",
"solution": "**Answer:** 204\n\n$A=\\left[\\begin{array}{ll}a_{11} & a_{12} \\\\ a_{13} & a_{14}\\end{array}\\right]$\n\n
Such that $\\Sigma a_{i i}=3,5,7$ or 11\n\n
Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.\n\n
Then total number of possible matrices\n\n$$\n=4+12+4\n$$\n\n$=20$\n\n\n
For sum 5 the possible entries are $(0,0,1,4)$, $(0,0,2,3),(0,1,2,2),(0,1,1,3)$ and $(1,1,1,2)$.\n\n
$\\therefore $ Total possible matrices $=12+12+12+12+4=52$\n\n
For sum 7 the possible entries are $(0,0,3,4)$, $(0,2,2,3),(0,1,2,4),(0,1,3,3),(1,2,2,2)$, $(1,1,2,3)$ and $(1,1,1,4)$.\n\n
$\\therefore $ Total possible matrices $=80$\n\n
For sum 11 the possible entries are $(0,3,4,4)$, $(1,2,4,4),(2,3,3,3),(2,2,3,4)$.\n\n
Let $$x$$ and $$y$$ be distinct integers where $$1 \\le x \\le 25$$ and $$1 \\le y \\le 25$$. Then, the number of ways of choosing $$x$$ and $$y$$, such that $$x+y$$ is divisible by 5, is ____________.
Note : In first case total number of ways = 20 as in the question given that the chosen $$x$$ and $$y$$ are distinct integers each time. So when you choose x as 5 then you can't choose y as 5. Possible values of y are (10, 15, 20, 25). So, here four possible pairs of (x, y) possible { (5, 10), (5, 15), (5, 20), (5, 25)}.\n
Similarly, four possible pairs of (x, y) possible each time when x = 10, 15, 20 and 25.\n
$$ \\therefore $$ Total number of ways in the first case = 5 $$ \\times $$ 4 = 20.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6543,
"subject": "General Science",
"question": "
Eight persons are to be transported from city A to city B in three cars of different makes. If each car can accommodate at most three persons, then the number of ways, in which they can be transported, is :
",
"options": [
{
"text": "560"
},
{
"text": "1680"
},
{
"text": "3360"
},
{
"text": "1120"
}
],
"answer": "1680",
"solution": "**Answer:** 1680\n\nLet $C_1, C_2$ and $C_3$ be the three cars in which 8 person are to be transported from city $A$ to city $B$.\n
\n Let the number of persons transported in cars $C_1, C_2$ and $C_3$ are 3,3 and 2 respectively.\n
There are total $\\frac{8 !}{3 ! 3 ! 2 ! 2 !}$ group\n
Let the number of elements in sets $$A$$ and $$B$$ be five and two respectively. Then the number of subsets of $$A \\times B$$ each having at least 3 and at most 6 elements is :
First, let's determine the number of elements in the Cartesian product $$A \\times B$$. If set $$A$$ has 5 elements and set $$B$$ has 2 elements, then the number of elements in $$A \\times B$$ is:
We need to find the number of subsets of $$A \\times B$$ with at least 3 elements and at most 6 elements. The total number of elements in $$A \\times B$$ is 10, so we must consider the subsets containing 3, 4, 5, and 6 elements.
\n\n
The number of ways to choose $$r$$ elements from a set of 10 elements is given by the binomial coefficient:
\n\n
\n\n
$$\\binom{10}{r} = \\frac{10!}{r!(10-r)!}$$
\n\n\n\n
We will find the sum of the binomial coefficients for $$r = 3, 4, 5, 6$$:
Therefore, the number of subsets of $$A \\times B$$ each having at least 3 and at most 6 elements is:
\n\n
Option D - 792
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6545,
"subject": "General Science",
"question": "If $\\mathrm{n}$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $\\mathrm{n}$ is equal to :",
"options": [
{
"text": "47"
},
{
"text": "53"
},
{
"text": "51"
},
{
"text": "43"
}
],
"answer": "51",
"solution": "**Answer:** 51\n\n
To determine the number of ways in which five different employees can be seated in four indistinguishable offices, we will use the concept of partitioning of an integer.
\n\n
The problem is equivalent to partitioning the number 5 (representing the 5 different employees) into at most 4 parts (representing the 4 indistinguishable offices) where each part represents the number of employees in each office.
\n\n
The partitions of 5 into at most 4 parts are as follows :
\n\n\n
$(5, 0, 0, 0)$ – One office has 5 employees, and the other three have none.
\n
$(4, 1, 0, 0)$ – One office has 4 employees, another one has 1, and the other two have none.
\n
$(3, 2, 0, 0)$ – One office has 3 employees, another one has 2, and the other two have none.
\n
$(3, 1, 1, 0)$ – One office has 3 employees, two have 1 each, and one has none.
\n
$(2, 2, 1, 0)$ – Two offices have 2 employees each, one has 1, and one has none.
\n
$(2, 1, 1, 1)$ – One office has 2 employees, and the other three have 1 each.
\n\n
However, since the offices are indistinguishable, we must not count partitions that differ only by the order of the parts. This means that partitions like $(5, 0, 0, 0)$, $(0, 5, 0, 0)$, $(0, 0, 5, 0)$, and $(0, 0, 0, 5)$ all represent the same scenario and thus are counted as one.
\n\n
So we end up with 6 distinct partition scenarios.
\n\n
Now, for each partition, we need to find the number of ways to assign the employees according to each partition :
\n\n\n
$(5, 0, 0, 0)$ – All employees are in one office. There is 1 way to do this because the offices are indistinguishable.
\n
$(4, 1, 0, 0)$ - For the group with 4 employees, there are $${}^5{C_4}$$ ways to choose which 4 out of the 5 will be together. Then the remaining employee is automatically assigned to the second office. So, there are $${}^5{C_4}$$ ways for this partition.
\n
$(3, 2, 0, 0)$ - For the group with 3 employees, there are ${}^5{C_3}$ ways to choose them, and for the group with 2 employees, there are $${}^2{C_2}$$ ways to choose them (which is essentially 1 way since the last 2 employees are automatically grouped). But the 2 offices these groups can occupy are indistinguishable, so we do not multiply by 2. So, there are $${}^5{C_3}$$ ways for this partition.
\n
$(3, 1, 1, 0)$ - For the group with 3 employees, there are $${}^5{C_3}$$ ways. Then we have two indistinguishable offices, each with 1 employee. The order does not matter as offices are indistinguishable. So, there are simply $${}^5{C_3}$$ ways.
\n
$(2, 2, 1, 0)$ – For the first group of 2 employees, there are $${}^5{C_2}$$ ways to choose them. For the next group of 2 employees, there are $${}^3{C_2}$$ ways from the remaining 3. Then, the last employee is alone, and since the offices are indistinguishable, there are no further combinations to consider. However, we have double-counted since the two groups of two are indistinguishable. To adjust for this, we divide by 2. This gives us $\\frac{{}^5{C_2} \\cdot {}^3{C_2}}{2}$ ways.
\n
$(2, 1, 1, 1)$ - For the group with 2 employees, there are ${}^5{C_2}$ ways to choose them. The other three employees are each in their own office, with no further combinations since the offices are indistinguishable, so there are ${}^5{C_2}$ ways for this partition.
\n\n
Let's compute these cases :
\n\n
\n
$(5, 0, 0, 0)$ corresponds to $1$ way.
\n
$(4, 1, 0, 0)$ corresponds to ${}^5{C_4} = 5$ ways.
\n
$(3, 2, 0, 0)$ corresponds to ${}^5{C_3} = 10$ ways.
\n
$(3, 1, 1, 0)$ corresponds to ${}^5{C_3} = 10$ ways.
$(2, 1, 1, 1)$ corresponds to ${}^5{C_2} = 10$ ways.
\n
\n
Adding up all the ways we get :
\n\n
$$ 1 + 5 + 10 + 10 + 15 + 10 = 51 $$
\n\n
Therefore, the number of ways five different employees can sit into four indistinguishable offices ($\\mathrm{n}$) is 51, which corresponds to Option C.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6546,
"subject": "General Science",
"question": "The number of elements in the set $\\mathrm{S}=\\{(x, y, z): x, y, z \\in \\mathbf{Z}, x+2 y+3 z=42, x, y, z \\geqslant 0\\}$ equals __________.",
"options": [],
"answer": "169",
"solution": "**Answer:** 169\n\n$$\nx+2 y+3 z=42\n$$\n
In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections : $$A, B$$ and $$C$$. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section $$A$$ has 8 questions, section $$B$$ has 6 questions and section $$C$$ has 6 questions, then the total number of ways a student can select 15 questions is __________.
The problem involves choosing 15 questions out of a total of 20 available questions, with the constraint that at least 4 questions must be chosen from each of the three sections A, B, and C. To evaluate the total number of ways a student can select these questions, we need to consider every possible combination of questions from sections A, B, and C that sum up to 15 questions while respecting the constraints.
Section A has 8 questions, Section B and C each have 6 questions. The student must choose at least 4 questions from each section, which satisfies the minimum requirement. However, since the student is to attempt a total of 15 questions, there are several combinations to consider, as outlined below:
Choosing 4 questions from A, 5 from B, and 6 from C
Choosing 4 questions from A, 6 from B, and 5 from C
Choosing 7 questions from A, 4 from B, and 4 from C
Choosing 6 questions from A, 5 from B, and 4 from C
Choosing 6 questions from A, 4 from B, and 5 from C
Choosing 5 questions from A, 5 from B, and 5 from C
Choosing 5 questions from A, 6 from B, and 4 from C
Choosing 5 questions from A, 4 from B, and 6 from C
Total ways of select $=11376$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6549,
"subject": "General Science",
"question": "
There are 5 points $$P_1, P_2, P_3, P_4, P_5$$ on the side $$A B$$, excluding $$A$$ and $$B$$, of a triangle $$A B C$$. Similarly there are 6 points $$\\mathrm{P}_6, \\mathrm{P}_7, \\ldots, \\mathrm{P}_{11}$$ on the side $$\\mathrm{BC}$$ and 7 points $$\\mathrm{P}_{12}, \\mathrm{P}_{13}, \\ldots, \\mathrm{P}_{18}$$ on the side $$\\mathrm{CA}$$ of the triangle. The number of triangles, that can be formed using the points $$\\mathrm{P}_1, \\mathrm{P}_2, \\ldots, \\mathrm{P}_{18}$$ as vertices, is:
There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is ________.
Let the set $$S=\\{2,4,8,16, \\ldots, 512\\}$$ be partitioned into 3 sets $$A, B, C$$ with equal number of elements such that $$\\mathrm{A} \\cup \\mathrm{B} \\cup \\mathrm{C}=\\mathrm{S}$$ and $$\\mathrm{A} \\cap \\mathrm{B}=\\mathrm{B} \\cap \\mathrm{C}=\\mathrm{A} \\cap \\mathrm{C}=\\phi$$. The maximum number of such possible partitions of $$S$$ is equal to:
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6554,
"subject": "General Science",
"question": "If the letter of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number",
"options": [
{
"text": "601"
},
{
"text": "600"
},
{
"text": "603"
},
{
"text": "602"
}
],
"answer": "601",
"solution": "**Answer:** 601\n\n\n\n\n\n\n\n",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6555,
"subject": "General Science",
"question": "From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is :",
"options": [
{
"text": "at least 500 but less than 750"
},
{
"text": "at least 750 but less than 1000"
},
{
"text": "at least 1000"
},
{
"text": "less than 500"
}
],
"answer": "at least 1000",
"solution": "**Answer:** at least 1000\n\nFrom 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways\n
And from 4 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways\n
$$\\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \\times {}^3{C_1}$$ ways\n
Let 4 novels are N1, N2, N3, N4 and 1 dictionary is D1.\n
Dictionary should be in the middle. So the arrangement will be like this\n
_ _ D1 _ _\n
On those 4 blank places 4 novels N1, N2, N3, N4 can be placed. And 4 novels can be arrange $$4!$$ ways.\n
$$\\therefore$$ Total no of ways = $${}^6{C_4} \\times {}^3{C_1}$$$$ \\times 4!$$ = 1080",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6556,
"subject": "General Science",
"question": "The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is:",
"options": [
{
"text": "120"
},
{
"text": "72"
},
{
"text": "216"
},
{
"text": "192"
}
],
"answer": "192",
"solution": "**Answer:** 192\n\n
For a four digit number the first place can be filled in 3 ways with 6 or 7 or 8 and the remaining four places in 4! ways i.e., 3 $$\\times$$ 4! = 72.
\n
For a five digit number it can be arranged in 5! ways,
\n
$$\\therefore$$ total number of integers = (72 + 120) = 192.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6557,
"subject": "General Science",
"question": "If all the words (with or without meaning) having five letters,formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is :",
"options": [
{
"text": "$${46^{th}}$$ "
},
{
"text": "$${59^{th}}$$"
},
{
"text": "$${52^{nd}}$$"
},
{
"text": "$${58^{th}}$$"
}
],
"answer": "$${58^{th}}$$",
"solution": "**Answer:** $${58^{th}}$$\n\n
Clearly, number of words start with $$A = {{4!} \\over {2!}} = 12$$
\n
Number of words start with $$L = 4! = 24$$
\n
Number of words start with $$M = {{4!} \\over {2!}} = 12$$
\n
Number of words start with $$SA = {{3!} \\over {2!}} = 3$$
\n
Number of words start with $$SL = 3! = 6$$
\n
Note that, next word will be \"SMALL\"
\n
Hence, the position of word \"SMALL\" is 58th.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6558,
"subject": "General Science",
"question": "If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is : ",
"options": [
{
"text": "44th "
},
{
"text": "45th "
},
{
"text": "46th "
},
{
"text": "47th "
}
],
"answer": "46th ",
"solution": "**Answer:** 46th \n\n
To find the position of the word QUEEN:
\n
$$\\bullet$$ The number of words starting with E is 4! = 24.
\n
$$\\bullet$$ The number of words starting with N is $${{4!} \\over 2} = 12$$.
\n
$$\\bullet$$ The number of words starting with QE is 3! = 6.
\n
$$\\bullet$$ Number of words starting with QN is $${{3!} \\over 2} = 3$$.
\n
Therefore, the position of the word QUEEN is next to the sum, 24 + 12 + 6 + 3 = 45.
\n
That is, the word 'QUEEN' will be on 46th position.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6559,
"subject": "General Science",
"question": "From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and\narranged in a row on a shelf so that the dictionary is always in the middle. The number of such\narrangements is :",
"options": [
{
"text": "at least 750 but less than 1000"
},
{
"text": "at least 1000"
},
{
"text": "less than 500"
},
{
"text": "at least 500 but less than 750"
}
],
"answer": "at least 1000",
"solution": "**Answer:** at least 1000\n\nFrom 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways\n
And from 3 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways\n
$$\\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \\times {}^3{C_1}$$ ways\n
Let 4 novels are N1, N2, N3, N4 and 1 dictionary is D1.\n
Dictionary should be in the middle. So the arrangement will be like this\n
_ _ D1 _ _\n
On those 4 blank places 4 novels N1, N2, N3, N4 can be placed. And 4 novels can be arrange $$4!$$ ways.\n
$$\\therefore$$ Total no of ways = $${}^6{C_4} \\times {}^3{C_1}$$$$ \\times 4!$$ = 1080",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6560,
"subject": "General Science",
"question": "n$$-$$digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is : ",
"options": [
{
"text": "6"
},
{
"text": "7"
},
{
"text": "8"
},
{
"text": "9"
}
],
"answer": "7",
"solution": "**Answer:** 7\n\nIn n digit number first place can be filled with any one of 2, 5, 7. So no of ways first digit can be filled = 3
\nSimilarly, \nno of ways 2nd digit can be filled = 3 ways \n. \n. \n. \n. \n- - - - - - nth - - - - - - - = 3 ways \n \n$$ \\therefore $$ Total numbers = 3 $$ \\times $$ 3 $$ \\times $$ 3 .... n times \n= 3n \n$$ \\therefore $$ According to question, for smallest value of n,
\n3n > 900
\n36 = 729 < 900
\n37 = 2187 > 900
\n$$ \\therefore $$ n = 7",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6561,
"subject": "General Science",
"question": "The number of four letter words that can be formed using the letters of the word BARRACK is : ",
"options": [
{
"text": "120"
},
{
"text": "144"
},
{
"text": "264"
},
{
"text": "270"
}
],
"answer": "270",
"solution": "**Answer:** 270\n\nCase 1 :\n
When all the four letters different then no of words\n = 5C4 $$ \\times $$4!\n
Case 2 :\n
When out of four letters two letters are R and other two different letters are chosen from B, A, C, K then the no of words = 4C2 $$ \\times $$ $${{4!} \\over {2!}}$$ = 72\n
Case 3 :\n
When out of four letters two letters are A and other two different letters are chosen from B, R, C, K then the no of words = 4C2 $$ \\times $$ $${{4!} \\over {2!}}$$ = 72\n
Case 4 :\n
When word is formed using two R and two A then number of words = $${{4!} \\over {2!2!}}$$ = 6\n
So, total number of 4 letters words possible = 120 + 72 + 72 + 6 = 270",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6562,
"subject": "General Science",
"question": "Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :",
"options": [
{
"text": "$${5 \\over 2}\\left( {6!} \\right)$$"
},
{
"text": "$${6!}$$"
},
{
"text": "56"
},
{
"text": "$${1 \\over 2}\\left( {6!} \\right)$$"
}
],
"answer": "$${5 \\over 2}\\left( {6!} \\right)$$",
"solution": "**Answer:** $${5 \\over 2}\\left( {6!} \\right)$$\n\nHere none number repeats more than once.\n
We can choose the number which repeats more than once among 1, 3, 5, 7, 9 in 5C1 ways.\n
Let number 3 repeats more than once. So six digits are 1, 3, 3, 5, 7, 9.\n
We can arrange those six digits in $${{6!} \\over {2!}}$$ ways.\n
$$ \\therefore $$ Total six digit numbers = 5C1 $$ \\times $$ $${{6!} \\over {2!}}$$ = $${5 \\over 2}\\left( {6!} \\right)$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6563,
"subject": "General Science",
"question": "The number of 4 letter words (with or without\nmeaning) that can be formed from the eleven\nletters of the word 'EXAMINATION' is\n_______.",
"options": [],
"answer": "2454",
"solution": "**Answer:** 2454\n\n2A, 2I, 2N, E, X, M, T, O\n
The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is _____________.
Give a number to each alphabet starting from 0 in alphabetical order.
\n
\n
Step 3 :
\n
In word \"MANKIND\" first alphabet is \"M\" which is present in the 4th position in the Alphabet Box. And after \"M\", in the word \"MANKIND\" there are 6 alphabets \"ANKIND\" which can be arrange in $${{6!} \\over {2!}}$$ ways. Here $$2!$$ present because two N presents.
\n
So for alphabet \"M\" we write $$4.{{6!} \\over {2!}}$$
\n
Step 4 :
\n
As in word \"MANKIND\" we calculated \"M\", so delete \"M\" from alphabet box and check remaining alphabets got right numbers or not in alphabet box.
\n
As you can see after deleting \"M\", 4th position in the alphabet box is missing so we create a new alphabet box.
\n
\n
Step 5 :
\n
Now next alphabet after \"M\" in word \"MANKIND\" is \"A\" which is present in the 0th position in the new alphabet box. And after \"A\" in the word \"MANKIND\" there are 5 alphabets \"NKIND\" which can be arrange in $${{5!} \\over {2!}}$$ ways.
\n
So, for alphabet \"A\" we write $$0.{{5!} \\over {2!}}$$
\n
Step 6 :
\n
Now in word \"MANKIND\" we calculated \"A\", so delete \"A\" from previous new alphabet box and check remaining alphabets got right numbers or not in alphabet box.
\n
As you can see after deleting \"A\", 0th positioin in the alphabet box is missing, so we have to create a new alphabet box.
\n
\n
Step 7 :
\n
Now next alphabet after \"A\" in word \"MANKIND\" is \"N\" which is present in the 3rd and 4th position in the alphabet box. We have to choose minimum position for \"N\" in the alphabet box. So, we choose 3rd position \"N\". And after \"N\" in the word \"MANKIND\" there are 4 alphabets \"KIND\" which can be arrange $${{4!} \\over {2!}}$$ ways. Here $$2!$$ used because of 2 N's (one \"N\" is that \"N\" which we are considering and other \"N\" which is present in the word \"KIND\")
\n
So for alphabet \"N\" we write $$3\\, \\times \\,{{4!} \\over {2!}}$$
\n
Step 8 :
\n
Now in word \"MANKIND\" we calculated \"N\", so delete 3rd \"N\" from alphabet box and check remaining alphabets got right number or not in alphabet box. As you can see after deleting 3rd \"N\", 3rd position is missing in the alphabet box so we have to create a new alphabet box.
\n
\n
Step 9 :
\n
Now next alphabet after \"N\" in word \"MANKIND\" is \"K\" which is present at the 2nd position in the new alphabet box. And after \"K\" in the word \"MANKIND\" there are 3 alphabets \"IND\" which we can arrange $$3!$$ ways.
\n
So for alphabet \"K\" we write $$2\\times3!$$
\n
Step 10 :
\n
Now, in word \"MANKIND\" we calculated \"K\", so delete \"K\" from alphabet box and check remaining alphabets got right number or not in alphabet box. As you can see after deleting \"K\", 2nd position in the alphabet box is missing so we have to create a new alphabet box.
\n
\n
Step 11 :
\n
Now next alphabet after \"K\" in word \"MANKIND\" is \"I\" which is present at 1st position in the new alphabet box. And after \"I\" in the word \"MANKIND\" there are 2 alphabets \"ND\" which we can arrange in $$2!$$ ways.
\n
So for alphabet \"I\" we write $$1\\times2!$$
\n
Step 12 :
\n
Now in word \"MANKIND\" we calculated \"I\", so delete \"I\" from alphabet box and check remaining alphabets got right numbers or not in alphabet box. As you can see after deleting \"I\", 1st position in the alphabet box is missing so we have to create a new alphabet box.
\n
\n
Step 13 :
\n
Now next alphabet after \"I\" in word \"MANKIND\" is \"N\" which is present at 1st position in the new alphabet box. And after \"N\" in the word \"MANKIND\" there is 1 alphabet \"D\" which we can arrange in $$1!$$ ways.
\n
So for alphabet \"N\" we write $$1\\times1!$$.
\n
Step 14 :
\n
Now in word \"MANKIND\" we calculated \"N\", so delete \"N\" from alphabet box and check remaining alphabets got right numbering or not in alphabet box. As you can see after deleting \"N\", remaining alphabet \"D\" got correct numbering so new alphabet box will be same.
\n
\n
Step 15 :
\n
Now next alphabet after \"N\" in word \"MANKIND\" is \"D\" which is present at the 0th position in the new alphabet box. And after \"D\" in the word \"MANKIND\" there is no alphabet so it is the last alphabet.
\n
So for last alphabet we always write $$0!$$
\n
$$\\therefore$$ Position of the word \"MANKIND\" in the dictionary
Let $$S$$ be the set of all passwords which are six to eight characters long, where each character is either an alphabet from $$\\{A, B, C, D, E\\}$$ or a number from $$\\{1,2,3,4,5\\}$$ with the repetition of characters allowed. If the number of passwords in $$S$$ whose at least one character is a number from $$\\{1,2,3,4,5\\}$$ is $$\\alpha \\times 5^{6}$$, then $$\\alpha$$ is equal to ___________.
Let 5 digit numbers be constructed using the digits $$0,2,3,4,7,9$$ with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is __________.
The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is :
T O U G H $\\rightarrow 1$ !\n ____________________________\n Total = 89",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6576,
"subject": "General Science",
"question": "
Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is ____________.
\nTotal number greater than 7000 $=120+48=168$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6578,
"subject": "General Science",
"question": "
All words, with or without meaning, are made using all the letters of the word MONDAY. These words are written as in a dictionary with serial numbers. The serial number of the word MONDAY is :
The letters of "MONDAY" arranged in alphabetical order are: A, D, M, N, O, Y.
\n\n
Fix A as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways.
\n
Fix D as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways.
\n\n
So far, we have 120 (for A) + 120 (for D) = 240 words.
\n
Now, let's proceed with words starting with M:
\n\n
Fix MA as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.
\n
Fix MD as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.
\n
Fix MN as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.
\n\n
Now, we have 240 (for A and D) + 24 (for MA) + 24 (for MD) + 24 (for MN) = 312 words.
\n
After MN, we consider words that start with MO:
\n\n
Fix MOA as the first three letters, we can arrange the remaining 3 letters in 3! ways = 6 ways.
\n
Fix MOD as the first three letters, we can arrange the remaining 3 letters in 3! ways = 6 ways.
\n\n
Adding these to the total, we have 312 (previous total) + 6 (for MOA) + 6 (for MOD) = 324 words.
\n\n
Next, we consider words that start with 'MON'. The word 'MONDAY' comes after 'MONAD' in dictionary order, so:
\n\n
Fix 'MONA' as the first four letters. The remaining 2 letters ('D' and 'Y') can be arranged in 2! ways, which gives us 2 more words: 'MONADY' and 'MONAYD'.
\n\n
So, we have 324 (previous total) + 2 (for 'MONADY' and 'MONAYD') = 326 words.
\n
Finally, we have the word 'MONDAY' itself, which is the 327th word.
If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial numbers, then the serial number of the word THAMS is :
",
"options": [
{
"text": "103"
},
{
"text": "104"
},
{
"text": "102"
},
{
"text": "101"
}
],
"answer": "103",
"solution": "**Answer:** 103\n\nTo solve this problem, we start by finding the number of permutations before we reach a word that begins with T.\n\n
We have 5 letters in the word MATHS. If we fix the first letter, there are 4! (=4$$ \\times $$3$$ \\times $$2$$ \\times $$1=24) ways to arrange the remaining letters. \n\n
The letters before T in alphabetical order are A, H, M, and S. For each of these 4 letters, there are 4! ways to arrange the rest of the word, giving 4$$ \\times $$4! = 96 words that come before any words starting with T.\n\n
Then, in the section of words that start with 'T', we start with 'TA'. There are 3 remaining letters after 'TA', hence the permutations for words that start with 'TA' are (5-2)!=3!=6.\n\n
So, the total permutations for all words that come before 'TH' is 96 + 6 = 102.\n\n\n
Now, we need to count the words that come before 'THAMS'. After 'TH', the letters left are 'AMS'. Arranged in dictionary order, they would be 'AMS', 'ASM', 'MAS', 'MSA', 'SAM', 'SMA'. So, 'THAMS' is the first word in the 'TH' category, so it is the 103rd word overall when counting from the beginning.\n\n
So, the correct answer is 103, which corresponds to Option A.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6580,
"subject": "General Science",
"question": "
The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to __________.
",
"options": [],
"answer": "26664",
"solution": "**Answer:** 26664\n\nWe have, four digits are $2,1,2,3$.\n
Total numbers when 1 is at unit digit $=\\frac{3 !}{2 !}=3$\n
Total number when 2 is at unit digit $=3 !=6$\n
Total numbers when 3 is at unit digit $=\\frac{3 !}{2 !}=3$\n
Sum of digits at unit place $=3 \\times 1+6 \\times 2+3 \\times 3=24$\n
All the letters of the word PUBLIC are written in all possible orders and these words are written as in a dictionary with serial numbers. Then the serial number of the word PUBLIC is :
The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to __________.
All the letters of the word \"GTWENTY\" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word \"GTWENTY\" is _________.
Let $$[t]$$ be the greatest integer less than or equal to $$t$$. Let $$A$$ be the set of all prime factors of 2310 and $$f: A \\rightarrow \\mathbb{Z}$$ be the function $$f(x)=\\left[\\log _2\\left(x^2+\\left[\\frac{x^3}{5}\\right]\\right)\\right]$$. The number of one-to-one functions from $$A$$ to the range of $$f$$ is
The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3 , is equal to ________.
To solve this problem, we need to find the number of 3-digit numbers formed using the digits 2, 3, 4, 5, and 7, with no repetition of digits allowed, and these numbers should not be divisible by 3. Let's break down the solution step-by-step:
\n\n
1. Calculating the total number of 3-digit numbers without repetition:
\n\n
The number of ways to form a 3-digit number from 5 unique digits (2, 3, 4, 5, 7) without repetition can be calculated using permutations:
\n\n
The total number of permutations for choosing 3 digits out of 5 and arranging them is given by:
So, there are 60 possible 3-digit numbers that can be formed from the digits 2, 3, 4, 5, and 7 without repetition.
\n\n
2. Finding the 3-digit numbers divisible by 3:
\n\n
A number is divisible by 3 if the sum of its digits is divisible by 3. Let's consider the sums of every combination of these three digits to find out which sums are divisible by 3:
\n\n
Possible sums of combinations:
\n\n
\n\n
2 + 3 + 4 = 9 (divisible by 3)
\n\n
2 + 4 + 7 = 13
\n\n
2 + 5 + 7 = 14
\n\n
2 + 3 + 5 = 10
\n\n
2 + 3 + 7 = 12 (divisible by 3)
\n\n
3 + 4 + 5 = 12 (divisible by 3)
\n\n
3 + 4 + 7 = 14
\n\n
3 + 5 + 7 = 15 (divisible by 3)
\n\n
4 + 5 + 7 = 16
\n\n
\n\n
The combinations whose sums are divisible by 3 are:
\n\n
\n\n
2, 3, 4
\n\n
2, 3, 7
\n\n
3, 4, 5
\n\n
3, 5, 7
\n\n
\n\n
Since the sum of the digits is divisible by 3 for these combinations, any permutation of these sets will yield a number divisible by 3:
\n\n
The number of 3-digit numbers that can be formed from each set of 3 digits is:
\n\n
$$3! = 6$$
\n\n
So, the total number of 3-digit numbers divisible by 3 is:
3. Calculating the 3-digit numbers not divisible by 3:
\n\n
To find the 3-digit numbers not divisible by 3, we subtract the number of those divisible by 3 from the total number of 3-digit numbers:
\n\n
$$60 - 24 = 36$$
\n\n
Therefore, the number of 3-digit numbers that can be formed using the digits 2, 3, 4, 5, and 7 without repetition, and which are not divisible by 3, is equal to 36.
60 words can be made using all the letters of the word $$\\mathrm{BHBJO}$$, with or without meaning. If these words are written as in a dictionary, then the $$50^{\\text {th }}$$ word is:
To find the $$50^{\\text{th}}$$ word formed by the letters of \"BHBJO\" as if listed in a dictionary, let's analyze the arrangement methodically. Given the letters are B, H, B, J, O, there are some repetitions with the letter B appearing twice.
\n\n
First, calculate the total number of permutations of these letters:
\n\n
$$ \\frac{5!}{2!} = 60 $$
\n\n
Let's arrange the letters in alphabetical order first: B, B, H, J, O.
\n\n
We need to systematically count the words while following dictionary order:
\n\n
1. Words starting with B:
\n\n
\n\n
Next position letters: B, H, J, O
\n\n
Number of permutations: $$\\frac{4!}{1!} = 24$$ words
\n\n
\n\n
Since 24 words starting with 'B' exist and are less than 50, Move to next starting letter alphabetically.
\n\n
2. Words starting with H:
\n\n
\n\n
Next position letters: B, B, J, O
\n\n
Number of permutations: $$\\frac{4!}{2!} = 12$$ words
\n\n
\n\n
After 'B', tally becomes 24 (B-words) + 12 (H-words) = 36, needs more.
\n\n
3. Words starting with J:
\n\n
\n\n
Next position letters: B, B, H, O
\n\n
Number of permutations: $$\\frac{4!}{2!} = 12$$ words
\n\n
\n\n
Now tally is 36 + 12 = 48 words.
\n\n
Still 2 more to reach 50.
\n\n
4. Words starting with O:
\n\n
\n\n
Next position letters: B, B, H, J
\n\n
Number of permutations: $$\\frac{4!}{2!} = 12$$ words
\n\n
Our answer must be here since 48 + 2 more = 50 total.
If all the words with or without meaning made using all the letters of the word \"NAGPUR\" are arranged as in a dictionary, then the word at $$315^{\\text {th }}$$ position in this arrangement is :
$$313^{\\text {th }} \\text { word }=\\text { NRAGPU }$$
\n
$$314^{\\text {th }}$$ word $$=$$ $$\\mathrm{NRAGUP}$$
\n
$$315^{\\text {th }} \\text { word }=\\text { NRAPGU }$$
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6588,
"subject": "General Science",
"question": "A box 'A' contains $$2$$ white, $$3$$ red and $$2$$ black balls. Another box 'B' contains $$4$$ white, $$2$$ red and $$3$$ black balls. If two balls are drawn at random, without eplacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is : ",
"options": [
{
"text": "$${9 \\over {16}}$$ "
},
{
"text": "$${7 \\over {16}}$$"
},
{
"text": "$${9 \\over {32}}$$"
},
{
"text": "$${7 \\over {8}}$$"
}
],
"answer": "$${7 \\over {16}}$$",
"solution": "**Answer:** $${7 \\over {16}}$$\n\nProbability of drawing a white ball and then a red ball \n
from bag B is given by \n$${{{}^4{C_1} \\times {}^2{C_1}} \\over {{}^9{C_2}}}$$ = $${2 \\over 9}$$\n
Probability of drawing a white ball and then a red ball \n
from bag A is given by $${{{}^2{C_1} \\times {}^3{C_1}} \\over {{}^7{C_2}}}$$ = $${2 \\over 7}$$\n
Hence, the probability of drawing a white ball and then \n
a red ball from bag B = $${{{2 \\over 9}} \\over {{2 \\over 7} + {2 \\over 9}}}$$ = $${{2 \\times 7} \\over {18 + 14}}$$ = $${7 \\over {16}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6589,
"subject": "General Science",
"question": "Box I contains 30 cards numbered 1 to 30 and\nBox II contains 20 cards numbered 31 to 50. A\nbox is selected at random and a card is drawn\nfrom it. The number on the card is found to be\na non-prime number. The probability that the\ncard was drawn from Box I is :",
"options": [
{
"text": "$${8 \\over {17}}$$"
},
{
"text": "$${2 \\over 3}$$"
},
{
"text": "$${2 \\over 5}$$"
},
{
"text": "$${4 \\over {17}}$$"
}
],
"answer": "$${8 \\over {17}}$$",
"solution": "**Answer:** $${8 \\over {17}}$$\n\nLet B1 be the event where Box-I is selected.\n
And B2 be the event where Box-II is selected.\n
P(B1) = P(B2) = $${1 \\over 2}$$\n
Let E be the event where selected card is non prime.\n
For B1 : Prime numbers: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}\n
= $${8 \\over {17}}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6590,
"subject": "General Science",
"question": "In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is :",
"options": [
{
"text": "$${{14} \\over {45}}$$"
},
{
"text": "$${{8} \\over {45}}$$"
},
{
"text": "$${{7} \\over {45}}$$"
},
{
"text": "$${{28} \\over {45}}$$"
}
],
"answer": "$${{28} \\over {45}}$$",
"solution": "**Answer:** $${{28} \\over {45}}$$\n\nConsider following events
Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $${6 \\over {11}}$$, then n is equal to __________.
\n\n$$\n\\begin{aligned}\n&\\Rightarrow \\frac{11}{10}=\\frac{6}{10}+\\frac{36}{(n+5)(n+4)} \\\\\\\\\n&\\Rightarrow \\frac{5}{10 \\times 36}=\\frac{1}{(n+5)(n+4)} \\\\\\\\\n&\\Rightarrow n^{2}+9 n-52=0 \\\\\\\\\n&\\Rightarrow n=4 \\text { is only possible value }\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6592,
"subject": "General Science",
"question": "
Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :
",
"options": [
{
"text": "$$\\frac{4}{9}$$"
},
{
"text": "$$\\frac{5}{18}$$"
},
{
"text": "$$\\frac{1}{6}$$"
},
{
"text": "$$\\frac{3}{10}$$"
}
],
"answer": "$$\\frac{5}{18}$$",
"solution": "**Answer:** $$\\frac{5}{18}$$\n\nLet $E \\rightarrow$ Ball drawn from Bag II is black.\n
\n$E_{R} \\rightarrow$ Bag I to Bag II red ball transferred.\n
\n$E_{B} \\rightarrow$ Bag I to Bag II black ball transferred.\n
\n$E_{w} \\rightarrow$ Bag I to Bag II white ball transferred.\n
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is :
",
"options": [
{
"text": "$$\\frac{3}{7}$$"
},
{
"text": "$$\\frac{5}{6}$$"
},
{
"text": "$$\\frac{5}{7}$$"
},
{
"text": "$$\\frac{2}{7}$$"
}
],
"answer": "$$\\frac{5}{7}$$",
"solution": "**Answer:** $$\\frac{5}{7}$$\n\nLet $E_{i} \\rightarrow$ Bag have at least $i$ black balls \n
$E \\rightarrow 2$ balls are drawn & both black\n\n
Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and $$\\lambda$$ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $$y^2=\\lambda x$$ with one vertex at the vertex of the parabola, is :
",
"options": [],
"answer": "432",
"solution": "**Answer:** 432\n\n$E_{1}$ : Ball is drawn from urn $A(4 R+6 B)$\n
\n$E_{2}:$ Ball is drawn from urn $B(5 R+5 B)$\n
\n$E_{3}$ : Ball is drawn from urn $C(\\lambda R+4 B)$\n
\n$A \\rightarrow$ Ball drawn is red.\n
\nRequired probability $=P\\left(\\frac{E_{3}}{A}\\right)$\n
Bag A contains 3 white, 7 red balls and Bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is
Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn $$\\mathrm{A}$$ is :
There are three bags $$X, Y$$ and $$Z$$. Bag $$X$$ contains 5 one-rupee coins and 4 five-rupee coins; Bag $$Y$$ contains 4 one-rupee coins and 5 five-rupee coins and Bag $$Z$$ contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag $$\\mathrm{Y}$$, is :
To solve this problem, we use Bayes' theorem. Let's define the events:
\n\n
\n\n
$$X$$: Selecting bag $$X$$
\n\n
$$Y$$: Selecting bag $$Y$$
\n\n
$$Z$$: Selecting bag $$Z$$
\n\n
$$A$$: Drawing a one-rupee coin
\n\n
\n\n
We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:
\n\n
\n\n
$$P(X) = P(Y) = P(Z) = \\frac{1}{3}$$
\n\n\n\n
Next, we need the probability of drawing a one-rupee coin from each bag:
\n\n
\n\n
From bag $$X$$: $$P(A|X) = \\frac{5}{5+4} = \\frac{5}{9}$$
\n\n
From bag $$Y$$: $$P(A|Y) = \\frac{4}{4+5} = \\frac{4}{9}$$
\n\n
From bag $$Z$$: $$P(A|Z) = \\frac{3}{3+6} = \\frac{3}{9} = \\frac{1}{3}$$
\n\n
\n\n
We need to find the probability that the coin came from bag $$Y$$ given that a one-rupee coin was drawn, i.e., we need $$P(Y|A)$$. Using Bayes' theorem:
\n\n
\n\n
$$P(Y|A) = \\frac{P(A|Y) \\cdot P(Y)}{P(A)}$$
\n\n\n\n
To find $$P(A)$$, the total probability of drawing a one-rupee coin can be calculated as follows:
A company has two plants $$A$$ and $$B$$ to manufacture motorcycles. $$60 \\%$$ motorcycles are manufactured at plant $$A$$ and the remaining are manufactured at plant $$B .80 \\%$$ of the motorcycles manufactured at plant $$A$$ are rated of the standard quality, while $$90 \\%$$ of the motorcycles manufactured at plant $$B$$ are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If $$p$$ is the probability that it was manufactured at plant $$B$$, then $$126 p$$ is
$$\\begin{aligned}\n& P(\\text { standard automobile from } A)=\\frac{6}{10} \\times \\frac{8}{10}=\\frac{12}{25} \\\\\n& P(\\text { standard automobile from } B)=\\frac{4}{10} \\times \\frac{9}{10}=\\frac{9}{25}\n\\end{aligned}$$
\n
$$\\begin{aligned}\n& \\text { Required Probability } \\frac{\\frac{9}{25}}{\\frac{12}{25}+\\frac{9}{25}} \\\\\n& P=\\frac{9}{21}=\\frac{3}{7}\n\\end{aligned}$$
\n
So, $$126 P=126 \\times \\frac{3}{7}=54$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6599,
"subject": "General Science",
"question": "A dice is tossed $$5$$ times. Getting an odd number is considered a success. Then the variance of distribution of success is :",
"options": [
{
"text": "$$8/3$$"
},
{
"text": "$$3/8$$"
},
{
"text": "$$4/5$$ "
},
{
"text": "$$5/4$$"
}
],
"answer": "$$5/4$$",
"solution": "**Answer:** $$5/4$$\n\nTotal no of trials = 5\n $$\\therefore$$ n = 5\n Odd no possibe are = {1, 3, 5} =3\n Sample space = {1, 2, 3, 4, 5, 6} = 6\n
Probablity of getting odd number = $${3 \\over 6}$$ = $${1 \\over 2}$$\n $$\\therefore$$ p = $${1 \\over 2}$$\n
Formula for variance = npq\n where q = 1 - p = 1 - $${1 \\over 2}$$ = $${1 \\over 2}$$\n $$\\therefore$$ Variance = $$5 \\times {1 \\over 2} \\times {1 \\over 2}$$ = $${5 \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6600,
"subject": "General Science",
"question": "The mean and variance of a random variable $$X$$ having binomial distribution are $$4$$ and $$2$$ respectively, then $$P(X=1)$$ is :",
"options": [
{
"text": "$${1 \\over 4}$$ "
},
{
"text": "$${1 \\over 32}$$"
},
{
"text": "$${1 \\over 16}$$"
},
{
"text": "$${1 \\over 8}$$"
}
],
"answer": "$${1 \\over 32}$$",
"solution": "**Answer:** $${1 \\over 32}$$\n\nMean = $$np$$ = 4\n Variance = $$npq$$ = 2\n
Note: $$p$$ can not be less than 0 as probability is always $$\\ge$$ 0 and $$\\le$$ 1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6605,
"subject": "General Science",
"question": "A multiple choice examination has $$5$$ questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get $$4$$ or more correct answers just by guessing is :",
"options": [
{
"text": "$${{17} \\over {{3^5}}}$$ "
},
{
"text": "$${{13} \\over {{3^5}}}$$"
},
{
"text": "$${{11} \\over {{3^5}}}$$"
},
{
"text": "$${{10} \\over {{3^5}}}$$"
}
],
"answer": "$${{11} \\over {{3^5}}}$$",
"solution": "**Answer:** $${{11} \\over {{3^5}}}$$\n\nEach question has 3 alternative and exactly one is correct.\n
$$\\therefore$$ Probability of giving correct answer P(correct) = $${1 \\over 3}$$\n
$$\\therefore$$ Probability of giving wrong answer P(wrong) = $${2 \\over 3}$$\n
Here student give 4 or more correct answer.\n
$$\\therefore$$ Student give either 4 correct answer or 5 correct answer.\n
Using Binomial Distribution,\n
Required probability = $${}^5{C_4}{\\left( {{1 \\over 3}} \\right)^4}{\\left( {{2 \\over 3}} \\right)^1}$$ + $${}^5{C_5}{\\left( {{1 \\over 3}} \\right)^5}{\\left( {{2 \\over 3}} \\right)^0}$$ = $${{11} \\over {{3^5}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6606,
"subject": "General Science",
"question": "An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is :",
"options": [
{
"text": "$${{240} \\over {729}}$$ "
},
{
"text": "$${{192} \\over {729}}$$"
},
{
"text": "$${{256} \\over {729}}$$"
},
{
"text": "$${{496} \\over {729}}$$"
}
],
"answer": "$${{256} \\over {729}}$$",
"solution": "**Answer:** $${{256} \\over {729}}$$\n\nProbability of fail, P(F) = p\n
$$ \\therefore $$ Probability of success, P(S) = 2p\n
We know, \n
p + 2Pp = 1\n
$$ \\Rightarrow $$ p = $${1 \\over 3}$$\n
$$ \\therefore $$ Now, probability of at least 5 success in 6 trials,\n
= $$10 \\times {3 \\over 5} \\times {2 \\over 5}$$ = $${{12} \\over 5}$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6608,
"subject": "General Science",
"question": "If the probability of hitting a target by a shooter, in any shot, is $${1 \\over 3}$$, then the minimum number of independent\nshots at the target required by him so that the probability of hitting the target atleast once is greater than $${5 \\over 6}$$ is :\n",
"options": [
{
"text": "4"
},
{
"text": "6"
},
{
"text": "5"
},
{
"text": "3"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\n$$1 - {}^n{C_0}{\\left( {{1 \\over 3}} \\right)^0}{\\left( {{2 \\over 3}} \\right)^n} > {5 \\over 6}$$\n
$${n_{\\min }} = 5$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6609,
"subject": "General Science",
"question": "For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability\nthat the candidate solve any problem is $${4 \\over 5}$$\n, then the probability that he is unable to solve less than two\nproblems is : ",
"options": [
{
"text": "$${{164} \\over {25}}{\\left( {{1 \\over 5}} \\right)^{48}}$$"
},
{
"text": "$${{316} \\over {25}}{\\left( {{4 \\over 5}} \\right)^{48}}$$"
},
{
"text": "$${{201} \\over 5}{\\left( {{1 \\over 5}} \\right)^{49}}$$"
},
{
"text": "$${{54} \\over 5}{\\left( {{4 \\over 5}} \\right)^{49}}$$"
}
],
"answer": "$${{54} \\over 5}{\\left( {{4 \\over 5}} \\right)^{49}}$$",
"solution": "**Answer:** $${{54} \\over 5}{\\left( {{4 \\over 5}} \\right)^{49}}$$\n\nThere are 50 questions in an exam.
\nSo Probability of each question to be correct is p = $${4 \\over 5}$$
\nand also probability of each question to be incorrect is q = $${1 \\over 5}$$
\nLet Y is the number of correct question in 50 questions, hence required probability is
\n= $${{54} \\over 5}{\\left( {{4 \\over 5}} \\right)^{49}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6610,
"subject": "General Science",
"question": "Let a random variable X have a binomial distribution with mean 8 and variance 4. If $$P\\left( {X \\le 2} \\right) = {k \\over {{2^{16}}}}$$, then k\nis equal to :",
"options": [
{
"text": "17"
},
{
"text": "1"
},
{
"text": "137"
},
{
"text": "121"
}
],
"answer": "137",
"solution": "**Answer:** 137\n\nLet number of trials be n and probability of success = p, probability of failure = q
\nGiven np = 8, npq = 4
\n$$ \\Rightarrow $$ q = $${1 \\over 2}$$, p = $${1 \\over 2}$$, n = 16 (as p + q = 1)
\np$$\\left( {x \\le 2} \\right)$$ = $${{{}^{16}{C_0} + {}^{16}{C_1} + {}^{16}{C_2}} \\over {{2^{16}}}} = {{1 + 16 + 120} \\over {{2^{16}}}} = {{137} \\over {{2^{16}}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6611,
"subject": "General Science",
"question": "A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then $$\\left( {{{mean\\,\\,of\\,X} \\over {s\\tan dard\\,\\,deviation\\,\\,of\\,X}}} \\right)$$ is equal to : ",
"options": [
{
"text": "4"
},
{
"text": "$$3\\sqrt 2 $$"
},
{
"text": "$${{4\\sqrt 3 } \\over 3}$$"
},
{
"text": "$$4\\sqrt 3 $$"
}
],
"answer": "$$4\\sqrt 3 $$",
"solution": "**Answer:** $$4\\sqrt 3 $$\n\np (probability of getting white ball) = $${{30} \\over {40}}$$\n
\n$$ \\because $$ $$mean \\over standard\\,deviation$$ = $$12\\over{\\sqrt3}$$ = $$4{\\sqrt3}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6612,
"subject": "General Science",
"question": "Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P (X = 2) equals :",
"options": [
{
"text": "$$25 \\over 169$$"
},
{
"text": "$$49\\over 169$$"
},
{
"text": "$$24 \\over 169$$"
},
{
"text": "$$52 \\over 169$$"
}
],
"answer": "$$25 \\over 169$$",
"solution": "**Answer:** $$25 \\over 169$$\n\nP (X = 1) means out of two drawn cards one card is ace. \n
and P(X = 2) means both the drawn cards are ace.\n
$$ \\therefore $$ P(X = 1) = first card is ace or 2nd card is ace.\n
= $${{25} \\over {169}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6613,
"subject": "General Science",
"question": "In a workshop, there are five machines and the probability of any one of them to be out of service on a day is $${{1 \\over 4}}$$\n. If the probability that at most two machines will be out of service on the same day is $${\\left( {{3 \\over 4}} \\right)^3}k$$, then k is equal to :",
"options": [
{
"text": "$${{{17} \\over 4}}$$"
},
{
"text": "$${{{17} \\over 2}}$$"
},
{
"text": "$${{{17} \\over 8}}$$"
},
{
"text": "4"
}
],
"answer": "$${{{17} \\over 8}}$$",
"solution": "**Answer:** $${{{17} \\over 8}}$$\n\nProbablity of at most two machines will be out of service = $${\\left( {{3 \\over 4}} \\right)^3}k$$\n
$$ \\Rightarrow $$ k = $${{17} \\over 8}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6614,
"subject": "General Science",
"question": "In a bombing attack, there is 50% chance that\na bomb will hit the target. Atleast two\nindependent hits are required to destroy the\ntarget completely. Then the minimum number\nof bombs, that must be dropped to ensure that\nthere is at least 99% chance of completely\ndestroying the target, is __________.",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\nLet n is total no. of bombs being dropped\n
n = 11",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6615,
"subject": "General Science",
"question": "An ordinary dice is rolled for a certain number of times. If the probability of getting an odd\nnumber 2 times is equal to the probability of getting an even number 3 times, then the\nprobability of getting an odd number for odd number of times is :",
"options": [
{
"text": "$${5 \\over {36}}$$"
},
{
"text": "$${3 \\over {16}}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${1 \\over {32}}$$"
}
],
"answer": "$${1 \\over 2}$$",
"solution": "**Answer:** $${1 \\over 2}$$\n\nP(odd no. twice) = P(even no. thrice)
$$ = {{16} \\over {{2^5}}} = {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6616,
"subject": "General Science",
"question": "A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :",
"options": [
{
"text": "$${{15} \\over {{2^8}}}$$"
},
{
"text": "$${{15} \\over {{2^{12}}}}$$"
},
{
"text": "$${{15} \\over {{2^{13}}}}$$"
},
{
"text": "$${{15} \\over {{2^{14}}}}$$"
}
],
"answer": "$${{15} \\over {{2^{13}}}}$$",
"solution": "**Answer:** $${{15} \\over {{2^{13}}}}$$\n\nLet the coin be tossed n-times
$$ = 10 \\times {1 \\over {125}} \\times {{16} \\over {25}} = {{32} \\over {625}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6618,
"subject": "General Science",
"question": "A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. \nthe smallest value of n, so that the probability of guessing at least 'n' correct answers is less than $${1 \\over 2}$$, is :",
"options": [
{
"text": "5"
},
{
"text": "6"
},
{
"text": "3"
},
{
"text": "4"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\n$$P(E) < {1 \\over 2}$$
$$ \\Rightarrow n \\ge 5$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6619,
"subject": "General Science",
"question": "Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is :",
"options": [
{
"text": "$${1 \\over 8}$$"
},
{
"text": "$${5 \\over 8}$$"
},
{
"text": "$${5 \\over 16}$$"
},
{
"text": "1"
}
],
"answer": "$${5 \\over 16}$$",
"solution": "**Answer:** $${5 \\over 16}$$\n\n
Let x be the number of heads obtained by A, and y be the number of heads obtained by B.
\n
Note that x and y are binomial variable with parameters n = 3 and p = $${1 \\over 2}$$
\n
$$\\therefore$$ Probability that both A and B obtained the same number of heads is
Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is :
In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability $${3 \\over 4}$$ and the remaining 6 questions correctly with probability $${1 \\over 4}$$. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is $${{{{27}k}} \\over {{4^{10}}}}$$, then k is equal to ___________.
If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :\n
",
"options": [
{
"text": "$$\n\\frac{33}{2^{32}}\n$$"
},
{
"text": "$$\\frac{33}{2^{29}}$$"
},
{
"text": "$$\\frac{33}{2^{28}}$$"
},
{
"text": "$$\\frac{33}{2^{27}}$$"
}
],
"answer": "$$\\frac{33}{2^{28}}$$",
"solution": "**Answer:** $$\\frac{33}{2^{28}}$$\n\n If $n$ is number of trails, $p$ is probability of success and $q$ is probability of unsuccess then,\n
\n$$\n\\begin{aligned}\n& \\text { Mean }=n p \\text { and variance }=n p q \\text {. } \\\\\\\\\n& \\text { Here }\\\\\\\\\n& n p+n p q=24 \\quad \\dots(i)\\\\\\\\\n& n p . n p q=128 \\quad \\dots(ii)\\\\\\\\\n&\\text { and } q=1-p \\quad \\dots(iii)\n\\end{aligned}\n$$\n
\nfrom eq. (i), (ii) and (iii) : $p=q=\\frac{1}{2}$ and $n=32$.\n
\n$\\therefore$ Required probability $=p(X=1)+p(X=2)$\n
The mean and variance of a binomial distribution are $$\\alpha$$ and $$\\frac{\\alpha}{3}$$ respectively. If $$\\mathrm{P}(X=1)=\\frac{4}{243}$$, then $$\\mathrm{P}(X=4$$ or 5$$)$$ is equal to :
Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If $$P(X>n-3)=\\frac{k}{2^{n}}$$, then k is equal to :
The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is ____________.
",
"options": [],
"answer": "96",
"solution": "**Answer:** 96\n\nLet two roots of a quadratic equation are mean = np and variance = npq.\n
The random variable $$\\mathrm{X}$$ follows binomial distribution $$\\mathrm{B}(\\mathrm{n}, \\mathrm{p})$$, for which the difference of the mean and the variance is 1 . If $$2 \\mathrm{P}(\\mathrm{X}=2)=3 \\mathrm{P}(\\mathrm{X}=1)$$, then $$n^{2} \\mathrm{P}(\\mathrm{X}>1)$$ is equal to :
",
"options": [
{
"text": "15"
},
{
"text": "12"
},
{
"text": "11"
},
{
"text": "16"
}
],
"answer": "11",
"solution": "**Answer:** 11\n\n$$\n\\begin{aligned}\n& n p-n p q=1 \\\\\\\\\n\\Rightarrow & n p(1-q)=1 \\\\\\\\\n\\Rightarrow & n p^2=1\n\\end{aligned}\n$$\n
Let a die be rolled $$n$$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $$\\frac{k}{2^{15}}$$, then $$\\mathrm{k}$$ is equal to :
A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $$\\frac{k}{3^{11}}$$, then $$k$$ is equal to :
",
"options": [
{
"text": "82"
},
{
"text": "164"
},
{
"text": "123"
},
{
"text": "75"
}
],
"answer": "123",
"solution": "**Answer:** 123\n\nGiven, a pair of dice is thrown once, then number of outcomes $=6 \\times 6=36$\n
A total of 5 are $\\{(1,4),(2,3),(4,1),(3,2)\\}$\n
$$\n\\begin{aligned}\n& \\therefore p =p(\\text { success })=\\frac{4}{36}=\\frac{1}{9} \\\\\\\\\n& q =1-\\frac{1}{9}=\\frac{8}{9}\n\\end{aligned}\n$$\n
In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $$\\frac{1}{3}$$ and $$\\frac{2}{3}$$ respectively. Let $$x$$ be the number of matches that the team wins, and $y$ be the number of matches that team loses. If the probability $$\\mathrm{P}(|x-y| \\leq 2)$$ is $$p$$, then $$3^9 p$$ equals _________.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6635,
"subject": "General Science",
"question": "Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is :",
"options": [
{
"text": "$${{2 \\over 5}}$$"
},
{
"text": "$${{4 \\over 5}}$$"
},
{
"text": "$${{3 \\over 5}}$$"
},
{
"text": "$${{1 \\over 5}}$$"
}
],
"answer": "$${{2 \\over 5}}$$",
"solution": "**Answer:** $${{2 \\over 5}}$$\n\n
X : Mr. A selected wining horse
\n
$$\\overline X $$ : Mr. A did not select wining horse
\n
Mr. A selected two horses, now probability of not wining the first horse which Mr. A choses = $${4 \\over 5}$$
\n
And probability of not wining the second horse also which Mr. A choses = $${3 \\over 4}$$ (Here Mr. A out of remaining 4 horses choses one horse among 3 horses which did not win)
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6636,
"subject": "General Science",
"question": "The probability that $$A$$ speaks truth is $${4 \\over 5},$$ while the probability for $$B$$ is $${3 \\over 4}.$$ The probability that they contradict each other when asked to speak on a fact is :",
"options": [
{
"text": "$${4 \\over 5}$$ "
},
{
"text": "$${1 \\over 5}$$"
},
{
"text": "$${7 \\over 20}$$"
},
{
"text": "$${3 \\over 20}$$"
}
],
"answer": "$${7 \\over 20}$$",
"solution": "**Answer:** $${7 \\over 20}$$\n\n
The probability of speaking truth by A, P(A) = $${4 \\over 5}$$. The probability of not speaking truth by A, P($$\\overline A $$) = 1 $$-$$ $${4 \\over 5} = {1 \\over 5}$$.
\n
The probability of speaking truth by B, P(B) = $${3 \\over 4}$$. The probability of not speaking truth of B, P($$\\overline B $$) $$ = {1 \\over 4}$$.
\n
The probability that they contradict each other
\n
$$ = P(A) \\times P(\\overline B ) + P(\\overline A ) \\times P(B)$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6637,
"subject": "General Science",
"question": "Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :",
"options": [
{
"text": "$${2 \\over 9}$$"
},
{
"text": "$${1 \\over 9}$$"
},
{
"text": "$${8 \\over 9}$$"
},
{
"text": "$${7 \\over 9}$$"
}
],
"answer": "$${1 \\over 9}$$",
"solution": "**Answer:** $${1 \\over 9}$$\n\n
Person 1st has three options to apply.
\n
Similarly, person 2nd has three options t apply
\n
and person 3rd has three options to apply.
\n
Total cases = 33
\n
Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3)
\n
So, probability $$ = {3 \\over {{3^3}}} = {1 \\over 9}$$.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6638,
"subject": "General Science",
"question": "A die is thrown. Let $$A$$ be the event that the number obtained is greater than $$3.$$ Let $$B$$ be the event that the number obtained is less than $$5.$$ Then $$P\\left( {A \\cup B} \\right)$$ is :",
"options": [
{
"text": "$${3 \\over 5}$$ "
},
{
"text": "$$0$$"
},
{
"text": "$$1$$"
},
{
"text": "$${2 \\over 5}$$"
}
],
"answer": "$$1$$",
"solution": "**Answer:** $$1$$\n\nNo of outcome for a die = { 1, 2, 3, 4, 5, 6 }\n According to the question,\n
A = { 4, 5, 6 }\n
$$\\therefore$$ P(A) = $${3 \\over 6}$$\n
B = { 1, 2, 3, 4 }\n
$$\\therefore$$ P(A) = $${4 \\over 6}$$\n
A $$ \\cap $$ B = { 4 }\n
So P(A $$ \\cap $$ B) = $${1 \\over 6}$$\n
We know, $$P\\left( {A \\cup B} \\right)$$ = $$P\\left( A \\right)$$ + $$P\\left( B \\right)$$ - $$P\\left( {A \\cap B} \\right)$$\n
$$\\therefore$$ Option (C) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6639,
"subject": "General Science",
"question": "If two different numbers are taken from the set {0, 1, 2, 3, ........, 10}; then the probability that their sum as\nwell as absolute difference are both multiple of 4, is :",
"options": [
{
"text": "$${{12} \\over {55}}$$"
},
{
"text": "$${{14} \\over {45}}$$"
},
{
"text": "$${{7} \\over {55}}$$"
},
{
"text": "$${{6} \\over {55}}$$"
}
],
"answer": "$${{6} \\over {55}}$$",
"solution": "**Answer:** $${{6} \\over {55}}$$\n\nLet A = {0, 1, 2, 3, 4, ......., 10}\n
Total number of ways of selecting 2 different numbers from A is\n
n (S) = 11C2 = 55, where 'S' denotes sample space\n
$$ \\therefore $$ P(E) = $${{n\\left( E \\right)} \\over {n\\left( S \\right)}}$$ = $${6 \\over {55}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6640,
"subject": "General Science",
"question": "Three persons P, Q and R independently try to hit a target. I the probabilities of their hitting the target are $${3 \\over 4},{1 \\over 2}$$ and $${5 \\over 8}$$ respectively, then the probability that the target is hit by P or Q but not by R is : ",
"options": [
{
"text": "$${{21} \\over {64}}$$ "
},
{
"text": "$${{9} \\over {64}}$$"
},
{
"text": "$${{15} \\over {64}}$$"
},
{
"text": "$${{39} \\over {64}}$$"
}
],
"answer": "$${{21} \\over {64}}$$ ",
"solution": "**Answer:** $${{21} \\over {64}}$$ \n\n
We have the following probabilities:
\n
$$\\bullet$$ The probability that the target is hit by the person P is $${3 \\over 4}$$.
\n
$$\\bullet$$ The probability that the target is not hit by the person P is $$1 - {3 \\over 4} = {1 \\over 4}$$.
\n
$$\\bullet$$ The probability that the target is hit by the person Q is $${1 \\over 2}$$.
\n
$$\\bullet$$ The probability that the target is not hit by the person Q is $$1 - {1 \\over 2} = {1 \\over 2}$$.
\n
$$\\bullet$$ The probability that the target is hit by the person R is $${5 \\over 8}$$.
\n
$$\\bullet$$ The probability that the target is not hit by the person R is $$1 - {5 \\over 8} = {3 \\over 8}$$.
\n
Here, we have used the fact that if the probability of occurrence of an event is p, then the probability of non-occurrence of an event is $$q = 1 - p$$.
\n
Therefore, the probability that the target is hit by P or Q and not by R is
\n
(Probability that the target is hit by P and not by Q and R) + (Probability that the target is hit by Q and not by P and R) + (Probability that the target is hit by both P and Q and not by R)
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6641,
"subject": "General Science",
"question": "An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is : ",
"options": [
{
"text": "$${{255} \\over {256}}$$"
},
{
"text": "$${{127} \\over {128}}$$"
},
{
"text": "$${{63} \\over {64}}$$"
},
{
"text": "$${{1} \\over {2}}$$"
}
],
"answer": "$${{127} \\over {128}}$$",
"solution": "**Answer:** $${{127} \\over {128}}$$\n\nAn unbiased coin is tossed 8 times which is same as 8 coins tossed 1 times. \n
$$\\therefore\\,\\,\\,$$ Possible no. of out come = 28\n
$$\\therefore\\,\\,\\,$$ Sample space = 28\n
Here in this condition, all head or all tail out come is not acceptable.\n
= $${{127} \\over {128}}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6642,
"subject": "General Science",
"question": "In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to : ",
"options": [
{
"text": "$${{200} \\over {{6^5}}}$$"
},
{
"text": "$${{225} \\over {{6^5}}}$$"
},
{
"text": "$${{150} \\over {{6^5}}}$$"
},
{
"text": "$${{175} \\over {{6^5}}}$$"
}
],
"answer": "$${{175} \\over {{6^5}}}$$",
"solution": "**Answer:** $${{175} \\over {{6^5}}}$$\n\n$$\\underline {} \\,\\,\\,\\underline {} \\,\\,\\,\\underline {} \\,\\,\\underline 4 \\,\\,\\underline 4 $$\n
$${1 \\over {{6^2}}}\\left( {{{{5^3}} \\over {{6^3}}} + {{2{C_1}{{.5}^2}} \\over {{6^3}}}} \\right) = {{175} \\over {{6^5}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6643,
"subject": "General Science",
"question": "In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :\n",
"options": [
{
"text": "$${{400} \\over 3}$$ loss"
},
{
"text": "0"
},
{
"text": "$${{400} \\over 9}$$ loss"
},
{
"text": "$${{400} \\over 3}$$ gain"
}
],
"answer": "0",
"solution": "**Answer:** 0\n\nExpected Gain/Loss = \n
here w denotes probability that outcome 5 or 6 (w = $${2 \\over 6} = {1 \\over 3}$$)\n
here L denotes probability that outcome\n
1,2,3,4 (L = $${4 \\over 6}$$ = $${2 \\over 3}$$)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6644,
"subject": "General Science",
"question": "The minimum number of times one has to toss a\nfair coin so that the probability of observing at least\none head is at least 90% is :",
"options": [
{
"text": "2"
},
{
"text": "3"
},
{
"text": "4"
},
{
"text": "5"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\nProbablity of getting head P(H) = $${1 \\over 2}$$\n
$$ \\therefore $$ Minimum value of n = 4\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6645,
"subject": "General Science",
"question": "Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is\nmore than 99% is :",
"options": [
{
"text": "6"
},
{
"text": "5"
},
{
"text": "8"
},
{
"text": "7"
}
],
"answer": "7",
"solution": "**Answer:** 7\n\n$$1 - {\\left( {{1 \\over 2}} \\right)^n} > {{99} \\over {100}}$$
\n$$ \\Rightarrow $$ n = 7",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6646,
"subject": "General Science",
"question": "The probability of a man hitting a target is $${1 \\over {10}}$$. The least number of shots required, so that the\nprobability of his hitting the target at least once is greater than $${1 \\over {4}}$$, is ____________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nWe have, $$1 - $$(probability of all shots results in failure out of n trials) > $${1 \\over 4}$$
$$ \\Rightarrow {3 \\over 4} > {\\left( {{9 \\over {10}}} \\right)^n} \\Rightarrow n \\ge 3$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6647,
"subject": "General Science",
"question": "In a box, there are 20 cards, out of which 10\nare lebelled as A and the remaining 10 are\nlabelled as B. Cards are drawn at random, one\nafter the other and with replacement, till a\nsecond A-card is obtained. The probability that\nthe second A-card appears before the third\nB-card is :",
"options": [
{
"text": "$${{13} \\over {16}}$$"
},
{
"text": "$${{11} \\over {16}}$$"
},
{
"text": "$${{15} \\over {16}}$$"
},
{
"text": "$${{9} \\over {16}}$$"
}
],
"answer": "$${{11} \\over {16}}$$",
"solution": "**Answer:** $${{11} \\over {16}}$$\n\nPossibilities that the second A card appears before the third B card are \n =AA + ABA + BAA + ABBA + BBAA + BABA\n
= $${{11} \\over {16}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6648,
"subject": "General Science",
"question": "Let Bi (i = 1, 2, 3) be three independent events in a sample space. The probability that only B1 occur is $$\\alpha $$, only B2 occurs is $$\\beta $$ and only B3 occurs is $$\\gamma $$. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations $$\\left( {\\alpha - 2\\beta } \\right)p = \\alpha \\beta $$ and $$\\left( {\\beta - 3\\gamma } \\right)p = 2\\beta \\gamma $$ (All the probabilities are assumed to lie in the interval (0, 1)). Then $${{P\\left( {{B_1}} \\right)} \\over {P\\left( {{B_3}} \\right)}}$$ is equal to ________.",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nLet x, y, z be probability of B1, B2, B3 respectively\n
$${x \\over z} = 6$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6649,
"subject": "General Science",
"question": "The coefficients a, b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :",
"options": [
{
"text": "$${1 \\over {72}}$$"
},
{
"text": "$${5 \\over {216}}$$"
},
{
"text": "$${1 \\over {36}}$$"
},
{
"text": "$${1 \\over {54}}$$"
}
],
"answer": "$${5 \\over {216}}$$",
"solution": "**Answer:** $${5 \\over {216}}$$\n\nax2 + bx + c = 0\n
a, b, c $$ \\in $$ {1,2,3,4,5,6}\n
n(s) = 6 × 6 × 6 = 216\n
For equal roots, D = 0 $$ \\Rightarrow $$ b2 = 4ac\n
$$ \\Rightarrow $$ ac = $${{{b^2}} \\over 4}$$\n
Favourable case :\n
If b = 2, ac = 1 $$ \\Rightarrow $$ a = 1, c = 1\n
If b = 4, ac = 4 :\n a = 1, c = 4\n a = 4, c = 1\n a = 2, c = 2\n
If b = 6, ac = 9 $$ \\Rightarrow $$ a = 3, c = 3\n
$$ \\therefore $$ Favorable cases = 5\n
$$ \\therefore $$ Required probability = $${5 \\over {216}}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6650,
"subject": "General Science",
"question": "Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to :",
"options": [
{
"text": "$${4 \\over {9}}$$"
},
{
"text": "$${9 \\over {56}}$$"
},
{
"text": "$${11 \\over {27}}$$"
},
{
"text": "$${3 \\over {7}}$$"
}
],
"answer": "$${4 \\over {9}}$$",
"solution": "**Answer:** $${4 \\over {9}}$$\n\nTotal cases :
Number divisible by 3 $$ \\equiv $$ Sum of digits must be divisible by 3
Case - I
1, 2, 3, 4, 5, 6
Number of ways = 6!
Case - II
0, 1, 2, 4, 5, 6
Number of ways = 5 . 5!
Case - III
0, 1, 2, 3, 4, 5
Number of ways = 5 . 5!
n(favourable) = 6! + 2 . 5 . 5!
$$P = {{6! + 2.\\,5.\\,5!} \\over {6\\,.\\,6!}} = {4 \\over 9}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6651,
"subject": "General Science",
"question": "Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is :",
"options": [
{
"text": "$${4 \\over 9}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${5 \\over {12}}$$"
},
{
"text": "$${17 \\over {36}}$$"
}
],
"answer": "$${17 \\over {36}}$$",
"solution": "**Answer:** $${17 \\over {36}}$$\n\nn(S) = 36
Probability = $${{17} \\over {36}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6652,
"subject": "General Science",
"question": "Let there be three independent events E1, E2 and E3. The probability that only E1 occurs is $$\\alpha$$, only E2 occurs is $$\\beta$$ and only E3 occurs is $$\\gamma$$. Let 'p' denote the probability of none of events occurs that satisfies the equations ($$\\alpha$$ $$-$$ 2$$\\beta$$)p = $$\\alpha$$$$\\beta$$ and ($$\\beta$$ $$-$$ 3$$\\gamma$$)p = 2$$\\beta$$$$\\gamma$$. All the given probabilities are assumed to lie in the interval (0, 1).
Then, $$\\frac{Probability\\ of\\ occurrence\\ of\\ E_{1}}{Probability\\ of\\ occurrence\\ of\\ E_{3}} $$ is equal to _____________.",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nLet P(E1) = x, P(E2) = y and P(E3) = z\n
$$\\alpha $$ = P$$\\left( {{E_1} \\cap {{\\overline E }_2} \\cap {{\\overline E }_3}} \\right)$$ = $$P\\left( {{E_1}} \\right).P\\left( {{{\\overline E }_2}} \\right).P\\left( {{{\\overline E }_3}} \\right)$$\n
$${{P\\left( {{E_1}} \\right)} \\over {P\\left( {{E_3}} \\right)}}$$ = 6",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6653,
"subject": "General Science",
"question": "Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be $${1 \\over 2}$$ and probability of occurrence of 0 at the odd place be $${1 \\over 3}$$. Then the probability that '10' is followed by '01' is equal to :",
"options": [
{
"text": "$${1 \\over 18}$$"
},
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${1 \\over 9}$$"
},
{
"text": "$${1 \\over 6}$$"
}
],
"answer": "$${1 \\over 9}$$",
"solution": "**Answer:** $${1 \\over 9}$$\n\nP(0 at even place) = $${1 \\over 2}$$\n
$$\\therefore$$ Possible a : {$$-$$5, $$-$$4, ............., 3}
$$\\therefore$$ Required probability = $${8 \\over {36}}$$ = $${2 \\over {9}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6655,
"subject": "General Science",
"question": "A fair coin is tossed n-times such that the probability of getting at least one head is at least 0.9. Then the minimum value of n is ______________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nP(Head) = $${1 \\over 2}$$
$$({2^n} - 2)$$ is multiply of 3 only when n is odd
Required Probability $$ = {{45} \\over {90}} = {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6657,
"subject": "General Science",
"question": "Two fair dice are thrown. The numbers on them are taken as $$\\lambda$$ and $$\\mu$$, and a system of linear equations
x + y + z = 5
x + 2y + 3z = $$\\mu$$
x + 3y + $$\\lambda$$z = 1
is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then :",
"options": [
{
"text": "$$p = {1 \\over 6}$$ and $$q = {1 \\over 36}$$"
},
{
"text": "$$p = {5 \\over 6}$$ and $$q = {5 \\over 36}$$"
},
{
"text": "$$p = {5 \\over 6}$$ and $$q = {1 \\over 36}$$"
},
{
"text": "$$p = {1 \\over 6}$$ and $$q = {5 \\over 36}$$"
}
],
"answer": "$$p = {5 \\over 6}$$ and $$q = {5 \\over 36}$$",
"solution": "**Answer:** $$p = {5 \\over 6}$$ and $$q = {5 \\over 36}$$\n\n$$D \\ne 0 \\Rightarrow \\left| {\\matrix{\n 1 & 1 & 1 \\cr \n 1 & 2 & 3 \\cr \n 1 & 3 & \\lambda \\cr \n\n } } \\right| \\ne 0 \\Rightarrow \\lambda \\ne 5$$
For no solution D = 0 $$\\Rightarrow$$ $$\\lambda$$ = 5
Let S = {E1, E2, ........., E8} be a sample space of a random experiment such that $$P({E_n}) = {n \\over {36}}$$ for every n = 1, 2, ........, 8. Then the number of elements in the set $$\\left\\{ {A \\subseteq S:P(A) \\ge {4 \\over 5}} \\right\\}$$ is ___________.
If the numbers appeared on the two throws of a fair six faced die are $$\\alpha$$ and $$\\beta$$, then the probability that $$x^{2}+\\alpha x+\\beta>0$$, for all $$x \\in \\mathbf{R}$$, is :\n
",
"options": [
{
"text": "$$\\frac{17}{36}$$"
},
{
"text": "$$\n\\frac{4}{9}\n$$\n"
},
{
"text": "$$\\frac{1}{2}$$"
},
{
"text": "$$\\frac{19}{36}$$"
}
],
"answer": "$$\\frac{17}{36}$$",
"solution": "**Answer:** $$\\frac{17}{36}$$\n\nFor $x^{2}+\\alpha x+\\beta>0 \\forall x \\in R$ to hold, we should have $\\alpha^{2}-4 \\beta<0$\n
\nIf $\\alpha=1, \\beta$ can be 1, 2, 3, 4, 5, 6 i.e., 6 choices\n
\nIf $\\alpha=2, \\beta$ can be 2, 3, 4, 5, 6 i.e., 5 choices\n
\nIf $\\alpha=3, \\beta$ can be $3,4,5,6$ i.e., 4 choices\n
\nIf $\\alpha=4, \\beta$ can be 5 or 6 i.e., 2 choices\n
\nIf $\\alpha=6$, No possible value for $\\beta$ i.e., 0 choices
Let $$S$$ be the sample space of all five digit numbers. It $$p$$ is the probability that a randomly selected number from $$S$$, is a multiple of 7 but not divisible by 5 , then $$9 p$$ is equal to :
First number divisible by 7 is 10003 and last is 99995.
\n
$$\\Rightarrow$$ Number of numbers divisible by 7.
\n
$$ = {{99995 - 10003} \\over 7} + 1$$
\n
$$ = 12857$$
\n
First number divisible by 35 is 10010 and last is 99995.
\n
$$\\Rightarrow$$ Number of numbers divisible by 35
\n
$$ = {{99995 - 10010} \\over {35}} + 1$$
\n
$$ = 2572$$
\n
Hence number of number divisible by 7 but not by 5
\n
$$ = 12857 - 2572$$
\n
$$ = 10285$$
\n
$$9P. = {{10285} \\over {90000}} \\times 9$$
\n
$$ = 1.0285$$
",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6663,
"subject": "General Science",
"question": "Let A be the event that the absolute difference between two randomly choosen real numbers in the sample space $[0,60]$ is less than or equal to a . If $\\mathrm{P}(\\mathrm{A})=\\frac{11}{36}$, then $\\mathrm{a}$ is equal to _______.",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\n$$\n\\begin{aligned}\n& |\\mathrm{x}-\\mathrm{y}|<\\mathrm{a} \\Rightarrow-\\mathrm{a}<\\mathrm{x}-\\mathrm{y}<\\mathrm{a} \\\\\\\\\n& \\Rightarrow \\mathrm{x}-\\mathrm{y}<\\mathrm{a} \\text { and } \\mathrm{x}-\\mathrm{y}>-\\mathrm{a}\n\\end{aligned}\n$$\n
\n
$$\n\\begin{aligned}\n& \\mathrm{P}(\\mathrm{A})=\\frac{\\operatorname{ar}(\\mathrm{OACDEG})}{(\\mathrm{OBDF})} \\\\\\\\\n& =\\frac{\\operatorname{ar}(\\mathrm{OBDF})-\\operatorname{ar}(\\mathrm{ABC})-\\operatorname{ar}(\\mathrm{EFG})}{\\operatorname{ar}(\\mathrm{OBDF})} \\\\\\\\\n& \\Rightarrow \\frac{11}{36}=\\frac{(60)^2-\\frac{1}{2}(60-\\mathrm{a})^2-\\frac{1}{2}(60-\\mathrm{a})^2}{3600} \\\\\\\\\n& \\Rightarrow 1100=3600-(60-\\mathrm{a})^2 \\\\\\\\\n& \\Rightarrow (60-\\mathrm{a})^2=2500 \\Rightarrow 60-\\mathrm{a}=50 \\\\\\\\\n& \\Rightarrow \\mathrm{a}=10\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6664,
"subject": "General Science",
"question": "A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is p. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colour is $q$. If $p: q=m: n$, where $m$ and $n$ are coprime, then $m+n$ is equal to :",
"options": [],
"answer": "14",
"solution": "**Answer:** 14\n\n
If an unbiased die, marked with $$-2,-1,0,1,2,3$$ on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is :
Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that $$N-2,\\sqrt{3N},N+2$$ are in geometric progression be $$\\frac{k}{48}$$. Then the value of k is :
Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $$S = \\left\\{ {x \\in \\mathbb{Z}:x(66 - x) \\ge {5 \\over 9}M} \\right\\}$$ and the event $$\\mathrm{A = \\{ x \\in S:x\\,is\\,a\\,multiple\\,of\\,3\\}}$$. Then P(A) is equal to :
\n$\\therefore $ Probability of not getting 2 or 3 in a throw of dice $=\\frac{2}{3}$\n
\nAs given $\\frac{2}{3}=\\frac{k}{6} \\Rightarrow k=4$\n
\n$\\therefore$ Required value $=1+4+5+6+4=20$\n",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6669,
"subject": "General Science",
"question": "
Let $$\\Omega$$ be the sample space and $$\\mathrm{A \\subseteq \\Omega}$$ be an event.
\n
Given below are two statements :
\n
(S1) : If P(A) = 0, then A = $$\\phi$$
\n
(S2) : If P(A) = 1, then A = $$\\Omega$$
\n
Then :
",
"options": [
{
"text": "both (S1) and (S2) are true"
},
{
"text": "both (S1) and (S2) are false"
},
{
"text": "only (S2) is true"
},
{
"text": "only (S1) is true"
}
],
"answer": "both (S1) and (S2) are false",
"solution": "**Answer:** both (S1) and (S2) are false\n\n$\\Omega=$ sample space
\n$\\mathrm{A}=$ be an event
\n$ \\Omega$ = A wire of length 1 which starts at point 0 and ends at point 1 on the coordinate axis = $[0,1]$
\n$\\mathrm{A}=\\left\\{\\frac{1}{2}\\right\\}$ = Selecting a point on the wire which is at $\\left\\{\\frac{1} {2}\\right\\}$ or 0.5\n
As wire is an 1-D object so from geometrical probability\n
$P(A)=\\frac{\\text { Favourable Length }}{\\text { Total Length }}$\n
Here total length of wire = 1 unit\n
and point has zero length so point A at $\\left\\{\\frac{1} {2}\\right\\}$ or 0.5 has length = 0.\n
Let $$S=\\left\\{M=\\left[a_{i j}\\right], a_{i j} \\in\\{0,1,2\\}, 1 \\leq i, j \\leq 2\\right\\}$$ be a sample space and $$A=\\{M \\in S: M$$ is invertible $$\\}$$ be an event. Then $$P(A)$$ is equal to :
Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that $${2^N} < N!$$ is $${m \\over n}$$, where m and n are coprime, then $$4m-3n$$ is equal to :
",
"options": [
{
"text": "12"
},
{
"text": "6"
},
{
"text": "8"
},
{
"text": "10"
}
],
"answer": "8",
"solution": "**Answer:** 8\n\n$N$ denote the sum of the numbers obtained when two dice are rolled.\n
Such that $2^N < N$!\n
$$\n\\text { i.e., } 4 \\leq N \\leq 12 \\text { i.e., } N \\in\\{4,5,6, \\ldots 12\\}\n$$\n
Three dice are rolled. If the probability of getting different numbers on the three dice is $$\\frac{p}{q}$$, where $$p$$ and $$q$$ are co-prime, then $$q-p$$ is equal to :
An integer is chosen at random from the integers $$1,2,3, \\ldots, 50$$. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is
If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $$i^{\\text {th }}$$ roll than the number obtained in the $$(i-1)^{\\text {th }}$$ roll, $$i=2,3$$, is equal to
Let's denote the outcomes of the three rolls as $$X_1$$, $$X_2$$, and $$X_3$$, where $$X_i$$ represents the number obtained in the $$i^{\\text{th}}$$ roll. We are looking for the probability that:
\n\n
$$X_2 > X_1 \\text{ and } X_3 > X_2$$
\n\n
The total number of outcomes when rolling a dice three times is:
\n\n
$$6^3 = 216$$
\n\n
Let's count the favorable outcomes. For each satisfying outcome, we must ensure that both inequalities are adhered to. Consider the possible sequences where every subsequent roll's number is higher than the previous roll's number. These sequences are:
\n\n
\n\n
(1, 2, 3)
\n\n
(1, 2, 4)
\n\n
(1, 2, 5)
\n\n
(1, 2, 6)
\n\n
(1, 3, 4)
\n\n
(1, 3, 5)
\n\n
(1, 3, 6)
\n\n
(1, 4, 5)
\n\n
(1, 4, 6)
\n\n
(1, 5, 6)
\n\n
(2, 3, 4)
\n\n
(2, 3, 5)
\n\n
(2, 3, 6)
\n\n
(2, 4, 5)
\n\n
(2, 4, 6)
\n\n
(2, 5, 6)
\n\n
(3, 4, 5)
\n\n
(3, 4, 6)
\n\n
(3, 5, 6)
\n\n
(4, 5, 6)
\n\n
\n\n
Clearly, there are 20 such favorable outcomes. Hence, the probability is given by:
\n\n
$$ \\frac{\\text{Number of favorable outcomes}}{\\text{Total number of outcomes}} = \\frac{20}{216} = \\frac{5}{54} $$
Let $$\\mathrm{a}, \\mathrm{b}$$ and $$\\mathrm{c}$$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked $$1,2,3,4$$. If the probability that $$a x^2+b x+c=0$$ has all real roots is $$\\frac{m}{n}, \\operatorname{gcd}(\\mathrm{m}, \\mathrm{n})=1$$, then $$\\mathrm{m}+\\mathrm{n}$$ is equal to _________.
A quadratic equation $$ax^2 + bx + c = 0$$ has real roots if and only if its discriminant is non-negative. The discriminant $$\\Delta$$ of the quadratic equation is given by:
\n\n
$$\\Delta = b^2 - 4ac$$
\n\n
For the quadratic equation to have all real roots, the discriminant must be non-negative:
\n\n
$$\\Delta \\geq 0$$
\n\n
That means:
\n\n
$$b^2 - 4ac \\geq 0$$
\n\n
Given that $$a, b, c$$ are the outcomes of rolling a fair tetrahedral die, they can each be one of the numbers 1, 2, 3, or 4. Our task is to determine the probability that this condition holds.
\n\n
We need to analyze the cases where $$b^2 \\geq 4ac$$.
\n\n
Let’s consider all possible values for $$a$$, $$b$$, and $$c$$, and count how many of them satisfy the condition. Since there are 4 choices for each of the variables, there are a total of $$4 \\times 4 \\times 4 = 64$$ possible combinations.
\n\n
Now, we count the valid combinations where $$b^2 \\geq 4ac$$:
\n\n
\n
For $$a = 1$$: $$b^2 \\geq 4c$$
\n
\n
$$b = 1: 1 \\geq 4c \\rightarrow \\text{(Not possible since } c \\ \\text{must be } \\geq 1 \\text{ and not zero)}$$
\n
$$b = 2: 4 \\geq 4c \\rightarrow c \\leq 1 \\rightarrow c = 1$$ (1 case)
\n
$$b = 3: 9 \\geq 4c \\rightarrow c \\leq 2 \\rightarrow c = 1 \\text{ or } 2$$ (2 cases)
\n
$$b = 4: 16 \\geq 4c \\rightarrow c \\leq 4 \\rightarrow c = 1, 2, 3, 4$$ (4 cases)
Let the sum of two positive integers be 24 . If the probability, that their product is not less than $$\\frac{3}{4}$$ times their greatest possible product, is $$\\frac{m}{n}$$, where $$\\operatorname{gcd}(m, n)=1$$, then $$n$$-$$m$$ equals
The coefficients $$a, b, c$$ in the quadratic equation $$a x^2+b x+c=0$$ are chosen from the set $$\\{1,2,3,4,5,6,7,8\\}$$. The probability of this equation having repeated roots is :
If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is :
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6680,
"subject": "General Science",
"question": "Two aeroplanes $${\\rm I}$$ and $${\\rm I}$$$${\\rm I}$$ bomb a target in succession. The probabilities of $${\\rm I}$$ and $${\\rm I}$$$${\\rm I}$$ scoring a hit correctly are $$0.3$$ and $$0.2,$$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is :",
"options": [
{
"text": "$$0.2$$ "
},
{
"text": "$$0.7$$ "
},
{
"text": "$$0.06$$ "
},
{
"text": "0.32"
}
],
"answer": "0.32",
"solution": "**Answer:** 0.32\n\n
$$\\therefore$$ Option (B) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6682,
"subject": "General Science",
"question": "One ticket is selected at random from $$50$$ tickets numbered $$00, 01, 02, ...., 49.$$ Then the probability that the sum of the digits on the selected ticket is $$8$$, given that the product of these digits is zer, equals :",
"options": [
{
"text": "$${1 \\over 7}$$ "
},
{
"text": "$${5 \\over 14}$$"
},
{
"text": "$${1 \\over 50}$$"
},
{
"text": "$${1 \\over 14}$$"
}
],
"answer": "$${1 \\over 14}$$",
"solution": "**Answer:** $${1 \\over 14}$$\n\nSample space = {00, 01, 02, 03, ..........49} = 50 tickets\n
$$\\therefore$$ Option (D) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6683,
"subject": "General Science",
"question": "If $$C$$ and $$D$$ are two events such that $$C \\subset D$$ and $$P\\left( D \\right) \\ne 0,$$ then the correct statement among the following is :",
"options": [
{
"text": "$$P\\left( {{C \\over D}} \\right)$$$$ \\ge P\\left( C \\right)$$ "
},
{
"text": "$$P\\left( {{C \\over D}} \\right)$$$$ < P\\left( C \\right)$$ "
},
{
"text": "$$P\\left( {{C \\over D}} \\right)$$$$ = {{P\\left( D \\right)} \\over {P\\left( C \\right)}}$$ "
},
{
"text": "$$P\\left( {{C \\over D}} \\right)$$$$ = P\\left( C \\right)$$ "
}
],
"answer": "$$P\\left( {{C \\over D}} \\right)$$$$ \\ge P\\left( C \\right)$$ ",
"solution": "**Answer:** $$P\\left( {{C \\over D}} \\right)$$$$ \\ge P\\left( C \\right)$$ \n\nGiven that $$C \\subset D$$ means $$C$$ is present entirely inside $$D$$. Which is shown below.\n\n
$$P\\left( {{C \\over D}} \\right)$$ = $${{P\\left( {C \\cap D} \\right)} \\over {P\\left( D \\right)}}$$ = $${{P\\left( C \\right)} \\over {P\\left( D \\right)}}$$\n
As $$C \\cap D$$ means common part of events C and D which is equal to C.\n\n
$$0 \\le P\\left( D \\right) \\le 1$$\n
$$\\therefore$$ $${{P\\left( C \\right)} \\over {P\\left( D \\right)}} \\ge P\\left( C \\right)$$\n
Note: Here we are dividing with $${P\\left( D \\right)}$$ which is $$ \\le 1$$ and $$ \\ge 0$$, as we know on dividing with a number n in the range $$0 \\le n \\le 1$$ we get always more than or equal to the original number. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6684,
"subject": "General Science",
"question": "Three numbers are chosen at random without replacement from $$\\left\\{ {1,2,3,..8} \\right\\}.$$ The probability that their minimum is $$3,$$ given that their maximum is $$6,$$ is :",
"options": [
{
"text": "$${3 \\over 8}$$ "
},
{
"text": "$${1 \\over 5}$$"
},
{
"text": "$${1 \\over 4}$$"
},
{
"text": "$${2 \\over 5}$$"
}
],
"answer": "$${1 \\over 5}$$",
"solution": "**Answer:** $${1 \\over 5}$$\n\nGiven set S = $$\\left\\{ {1,2,3,..8} \\right\\}$$\n
Choosing 3 numbers from 8 numbers can be done $${{}^8{C_3}}$$ ways.\n
Choosing 3 numbers from 8 numbers while minimum no is 3 can be done $$1 \\times {}^5{C_2}$$ ways.\n
= $${{1 \\over 5}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6685,
"subject": "General Science",
"question": "Let $$A$$ and $$B$$ be two events such that $$P\\left( {\\overline {A \\cup B} } \\right) = {1 \\over 6},\\,P\\left( { {A \\cap B} } \\right) = {1 \\over 4}$$ and $$P\\left( {\\overline A } \\right) = {1 \\over 4},$$ where $$\\overline A $$ stands for the complement of the event $$A$$. Then the events $$A$$ and $$B$$ are :",
"options": [
{
"text": "independent but not equally likely. "
},
{
"text": "independent and equally likely. "
},
{
"text": "mutually exclusive and independent."
},
{
"text": "equally likely but not independent."
}
],
"answer": "independent but not equally likely. ",
"solution": "**Answer:** independent but not equally likely. \n\n
$$P(\\overline {A \\cup B} ) = {1 \\over 6}$$
\n
or, $$1 - P(A \\cup B) = {1 \\over 6}$$
\n
$$\\therefore$$ $$P(A \\cup B) = 1 - {1 \\over 6} = {5 \\over 6};$$ $$P(A \\cap B) = {1 \\over 4}$$; and $$P(\\overline A ) = {1 \\over 4};$$
Since the probability of A and B are different, so they are not equally likely events.
\n
Therefore, (A) is the correct option.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6686,
"subject": "General Science",
"question": "Let two fair six-faced dice $$A$$ and $$B$$ be thrown simultaneously. If $${E_1}$$ is the event that die $$A$$ shows up four, $${E_2}$$ is the event that die $$B$$ shows up two and $${E_3}$$ is the event that the sum of numbers on both dice is odd, then which of the following statements is $$NOT$$ true?",
"options": [
{
"text": "$${E_1}$$ and $${E_2}$$ are independent."
},
{
"text": "$${E_2}$$ and $${E_3}$$ are independent."
},
{
"text": "$${E_1}$$ and $${E_3}$$ are independent."
},
{
"text": "$${E_1},$$ $${E_2}$$ and $${E_3}$$ are independent."
}
],
"answer": "$${E_1},$$ $${E_2}$$ and $${E_3}$$ are independent.",
"solution": "**Answer:** $${E_1},$$ $${E_2}$$ and $${E_3}$$ are independent.\n\nTotal possible outcome with two six faced dice = 62 = 36\n
When dice A shows up 4, the possible cases are\n E1 = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases\n $$\\therefore$$ $$P\\left( {{E_1}} \\right) = {6 \\over {36}} = {1 \\over 6}$$\n
When dice B shows up 2, the possible cases are\n E2 = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases\n
$$\\therefore$$ $${E_1},$$ $${E_2}$$ and $${E_3}$$ are not independent.\n
$$\\therefore$$ Option (D) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6687,
"subject": "General Science",
"question": "If A and B are any two events such that P(A) = $${2 \\over 5}$$ and P (A $$ \\cap $$ B) = $${3 \\over {20}}$$, hen the conditional probability, P(A $$\\left| {} \\right.$$(A' $$ \\cup $$ B')), where A' denotes the complement of A, is equal to :",
"options": [
{
"text": "$${1 \\over 4}$$ "
},
{
"text": "$${5 \\over 17}$$"
},
{
"text": "$${8 \\over 17}$$"
},
{
"text": "$${11 \\over 20}$$"
}
],
"answer": "$${5 \\over 17}$$",
"solution": "**Answer:** $${5 \\over 17}$$\n\n$$P\\left( {{A \\over {A' \\cup B'}}} \\right)$$\n
$$ = {5 \\over {17}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6688,
"subject": "General Science",
"question": "Let E and F be two independent events. The probability that both E and F happen is $${1 \\over {12}}$$ and the probability that neither E nor F happens is $${1 \\over {2}}$$, then a value of $${{P\\left( E \\right)} \\over {P\\left( F \\right)}}$$ is : ",
"options": [
{
"text": "$${4 \\over 3}$$"
},
{
"text": "$${3 \\over 2}$$"
},
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${5 \\over 12}$$"
}
],
"answer": "$${4 \\over 3}$$",
"solution": "**Answer:** $${4 \\over 3}$$\n\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6689,
"subject": "General Science",
"question": "A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p' is : ",
"options": [
{
"text": "$${1 \\over 5}$$"
},
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${2 \\over 5}$$"
},
{
"text": "$${1 \\over 4}$$"
}
],
"answer": "$${1 \\over 3}$$",
"solution": "**Answer:** $${1 \\over 3}$$\n\nP(X getting head) = p\n
$$ \\Rightarrow $$ p = $${1 \\over 3}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6690,
"subject": "General Science",
"question": "Let A, B and C be three events, which are pair-wise independent and $$\\overrightarrow E $$ denotes the completement of an event E. If $$P\\left( {A \\cap B \\cap C} \\right) = 0$$ and $$P\\left( C \\right) > 0,$$ then $$P\\left[ {\\left( {\\overline A \\cap \\overline B } \\right)\\left| C \\right.} \\right]$$ is equal to :",
"options": [
{
"text": "$$P\\left( {\\overline A } \\right) - P\\left( B \\right)$$"
},
{
"text": "$$P\\left( A \\right) + P\\left( {\\overline B } \\right)$$"
},
{
"text": "$$P\\left( {\\overline A } \\right) - P\\left( {\\overline B } \\right)$$"
},
{
"text": "$$P\\left( {\\overline A } \\right) + P\\left( {\\overline B } \\right)$$"
}
],
"answer": "$$P\\left( {\\overline A } \\right) - P\\left( B \\right)$$",
"solution": "**Answer:** $$P\\left( {\\overline A } \\right) - P\\left( B \\right)$$\n\nHere, $$P\\left( {\\overline A \\cap \\overline B \\left| C \\right.} \\right) = {{P\\left( {\\overline A \\cap \\overline B \\cap C} \\right)} \\over {P\\left( C \\right)}}$$\n
= $${{P\\left( C \\right) - P\\left( A \\right).P(C) - P\\left( B \\right).P(C)} \\over {P\\left( C \\right)}}$$\n
[$$ \\because $$ A, B and C are independent events]\n
= 1 - P(A) - P(B)\n
= $$P\\left( {\\overline A } \\right)$$ - P(B) or $$P\\left( {\\overline B } \\right)$$ - P(A)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6691,
"subject": "General Science",
"question": "An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is : ",
"options": [
{
"text": "$${{21} \\over {49}}$$"
},
{
"text": "$${{27} \\over {49}}$$"
},
{
"text": "$${{26} \\over {49}}$$"
},
{
"text": "$${{32} \\over {49}}$$"
}
],
"answer": "$${{32} \\over {49}}$$",
"solution": "**Answer:** $${{32} \\over {49}}$$\n\n5 Red and 2 green balls\n
P (2nd red ball) = $${5 \\over 7}$$ $$ \\times {4 \\over 7} + {2 \\over 7} \\times {6 \\over 7}$$ = $${{32} \\over {49}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6692,
"subject": "General Science",
"question": "An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, ……, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is :",
"options": [
{
"text": "$${{19} \\over {36}}$$"
},
{
"text": "$${{15} \\over {72}}$$"
},
{
"text": "$${{13} \\over {36}}$$"
},
{
"text": "$${{19} \\over {72}}$$"
}
],
"answer": "$${{19} \\over {72}}$$",
"solution": "**Answer:** $${{19} \\over {72}}$$\n\n\n $$P\\left( A \\right) = {1 \\over 2} \\times {{11} \\over {36}} + {1 \\over 2} \\times {2 \\over 9} = {{19} \\over {72}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6693,
"subject": "General Science",
"question": "Two integers are selected at random from the set {1, 2, ...., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :",
"options": [
{
"text": "$${2 \\over 5}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${7 \\over 10}$$"
},
{
"text": "$${3 \\over 5}$$"
}
],
"answer": "$${2 \\over 5}$$",
"solution": "**Answer:** $${2 \\over 5}$$\n\nSince sum of two numbers is even so either both are odd or both are even. Hence number of elements in reduced samples space = 5C2 + 6C2\n
So, required probability = $${{{}^5{C_2}} \\over {{}^5{C_2} + {}^6{C_2}}}$$ = $${2 \\over 5}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6694,
"subject": "General Science",
"question": "Let A and B be two non-null events such that\nA $$ \\subset $$ B . Then, which of the following statements\nis always correct?",
"options": [
{
"text": "P(A|B) = 1"
},
{
"text": "P(A|B) = P(B) – P(A)"
},
{
"text": "P(A|B) $$ \\le $$ P(A)\n"
},
{
"text": "P(A|B) $$ \\ge $$ P(A)\n"
}
],
"answer": "P(A|B) $$ \\ge $$ P(A)\n",
"solution": "**Answer:** P(A|B) $$ \\ge $$ P(A)\n\n\n$$P\\left( {{A \\over B}} \\right) = {{P\\left( {A \\cap B} \\right)} \\over {P\\left( B \\right)}}$$\n
As A $$ \\subset $$ B,\n
then P(A$$ \\cap $$B) = P(A)\n
$$ \\therefore $$ $$P\\left( {{A \\over B}} \\right) = {{P\\left( A \\right)} \\over {P\\left( B \\right)}}$$\n
As P(B) $$ \\le $$ 1\n
$$ \\therefore $$ $${{P\\left( A \\right)} \\over {P\\left( B \\right)}}$$ $$ \\ge $$ P(A)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6695,
"subject": "General Science",
"question": "Four persons can hit a target correctly with\nprobabilities\n$${1 \\over 2}$$, $${1 \\over 3}$$, $${1 \\over 4}$$ and\n$${1 \\over 8}$$ respectively. if all hit\nat the target independently, then the probability that\nthe target would be hit, is :",
"options": [
{
"text": "$${{25} \\over {32}}$$"
},
{
"text": "$${{25} \\over {192}}$$"
},
{
"text": "$${{1} \\over {192}}$$"
},
{
"text": "$${{7} \\over {32}}$$"
}
],
"answer": "$${{25} \\over {32}}$$",
"solution": "**Answer:** $${{25} \\over {32}}$$\n\nLet four persons are A, B, C and D.\n
Probablity of hitting a target by them,\n
P(A) = $${1 \\over 2}$$\n
P(B) = $${1 \\over 3}$$\n
P(C) = $${1 \\over 4}$$\n
P(D) = $${1 \\over 8}$$\n
Probablity of hitting target atleast once = 1 - Probablity of not hitting by anybody\n
P(Hit) = 1 - $$P\\left( {\\overline A \\cap \\overline B \\cap \\overline C \\cap \\overline D } \\right)$$\n
= 1 - $$P\\left( {\\overline A } \\right).P\\left( {\\overline B } \\right).P\\left( {\\overline C } \\right).P\\left( {\\overline D } \\right)$$\n
= $${{25} \\over {32}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6696,
"subject": "General Science",
"question": "Assume that each born child is equally likely to be a boy or a girl. If two families have two children each,\nthen the conditional probability that all children are girls given that at least two are girls is :",
"options": [
{
"text": "$${1 \\over {10}}$$"
},
{
"text": "$${1 \\over {17}}$$"
},
{
"text": "$${1 \\over {11}}$$"
},
{
"text": "$${1 \\over {12}}$$"
}
],
"answer": "$${1 \\over {11}}$$",
"solution": "**Answer:** $${1 \\over {11}}$$\n\nA = At least two girls
\n$$ \\Rightarrow {1 \\over {16 - 1 - 4}} = {1 \\over {11}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6697,
"subject": "General Science",
"question": "In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws total a of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is :\n",
"options": [
{
"text": "$${5 \\over {6}}$$"
},
{
"text": "$${5 \\over {31}}$$"
},
{
"text": "$${31 \\over {61}}$$"
},
{
"text": "$${30 \\over {61}}$$"
}
],
"answer": "$${30 \\over {61}}$$",
"solution": "**Answer:** $${30 \\over {61}}$$\n\nSum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}\n
Game ends and A wins if A throws 6 in 1st\nthrow or A don’t throw 6 in 1st throw, B don’t\nthrow 7 in 1st throw and then A throw 6 in his\n2\nnd chance and so on.\n\n
P(A) = A + $$\\overline A \\overline B A$$ + $$\\overline A \\overline B \\overline A \\overline B A$$\n
= $${{P\\left( A \\right) - P\\left( {A \\cap B} \\right)} \\over {1 - P\\left( B \\right)}}$$\n
= $${{{1 \\over 3} - {1 \\over {18}}} \\over {1 - {1 \\over 6}}}$$ = $${{1 \\over 3}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6700,
"subject": "General Science",
"question": "A dice is thrown two times and the sum of the\nscores appearing on the die is observed to be\na multiple of 4. Then the conditional probability\nthat the score 4 has appeared atleast once is :",
"options": [
{
"text": "$${1 \\over 8}$$"
},
{
"text": "$${1 \\over 9}$$"
},
{
"text": "$${1 \\over 4}$$"
},
{
"text": "$${1 \\over 3}$$"
}
],
"answer": "$${1 \\over 9}$$",
"solution": "**Answer:** $${1 \\over 9}$$\n\nLet A is the event for getting score a multiple\nof 4.\n
So, Required probability = $${1 \\over 9}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 6701,
"subject": "General Science",
"question": "When a missile is fired from a ship, the probability that it is intercepted is $${1 \\over 3}$$ and the probability that the missile hits the target, given that it is not intercepted, is $${3 \\over 4}$$. If three missiles are fired independently from the ship, then the probability that all three hit the target, is :",
"options": [
{
"text": "$${3 \\over 4}$$"
},
{
"text": "$${3 \\over 8}$$"
},
{
"text": "$${1 \\over 27}$$"
},
{
"text": "$${1 \\over 8}$$"
}
],
"answer": "$${1 \\over 8}$$",
"solution": "**Answer:** $${1 \\over 8}$$\n\nProbability of not getting intercepted = $${2 \\over 3}$$
When it is not intercepted, probability of missile hitting target = $${3 \\over 4}$$
$$\\therefore$$ So when such 3 missiles launched\nthen P (all 3 hitting the target) \n
$$= {1 \\over 8}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6702,
"subject": "General Science",
"question": "A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :",
"options": [
{
"text": "$${{39} \\over {50}}$$"
},
{
"text": "$${{3} \\over {4}}$$"
},
{
"text": "$${{22} \\over {425}}$$"
},
{
"text": "$${{52} \\over {867}}$$"
}
],
"answer": "$${{39} \\over {50}}$$",
"solution": "**Answer:** $${{39} \\over {50}}$$\n\nConsider the events,\n
= $${{39} \\over {50}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6703,
"subject": "General Science",
"question": "Let A and B be independent events such that P(A) = p, P(B) = 2p. The largest value of p, for which P (exactly one of A, B occurs) = $${5 \\over 9}$$, is :",
"options": [
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${2 \\over 9}$$"
},
{
"text": "$${4 \\over 9}$$"
},
{
"text": "$${5 \\over 12}$$"
}
],
"answer": "$${5 \\over 12}$$",
"solution": "**Answer:** $${5 \\over 12}$$\n\nP (Exactly one of A or B)
$$ = P\\left( {A \\cap \\overline B } \\right) + \\left( {\\overline A \\cap B} \\right) = {5 \\over 9}$$
$$ = P(A)P(\\overline B ) + P(\\overline A )P(B) = {5 \\over 9}$$
=$${{{{{{\\left( {{5 \\over 6}} \\right)}^4}.{1 \\over 5}} \\over {1 - {5 \\over 6}}}} \\over {{{{{\\left( {{5 \\over 6}} \\right)}^2}.{1 \\over 6}} \\over {1 - {5 \\over 6}}}}} = {\\left( {{5 \\over 6}} \\right)^2} = {{25} \\over {36}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6705,
"subject": "General Science",
"question": "When a certain biased die is rolled, a particular face occurs with probability $${1 \\over 6} - x$$ and its opposite face occurs with probability $${1 \\over 6} + x$$. All other faces occur with probability $${1 \\over 6}$$. Note that opposite faces sum to 7 in any die. If 0 < x < $${1 \\over 6}$$, and the probability of obtaining total sum = 7, when such a die is rolled twice, is $${13 \\over 96}$$, then the value of x is :",
"options": [
{
"text": "$${1 \\over 16}$$"
},
{
"text": "$${1 \\over 8}$$"
},
{
"text": "$${1 \\over 9}$$"
},
{
"text": "$${1 \\over 12}$$"
}
],
"answer": "$${1 \\over 8}$$",
"solution": "**Answer:** $${1 \\over 8}$$\n\nProbability of obtaining total sum 7 = probability of getting opposite faces.
$$ \\Rightarrow $$ $$x = {1 \\over 8}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6706,
"subject": "General Science",
"question": "An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is p, then 98 p is equal to _____________.",
"options": [],
"answer": "28",
"solution": "**Answer:** 28\n\nI1 = first unit is functioning
Let E1 and E2 be two events such that the conditional probabilities $$P({E_1}|{E_2}) = {1 \\over 2}$$, $$P({E_2}|{E_1}) = {3 \\over 4}$$ and $$P({E_1} \\cap {E_2}) = {1 \\over 8}$$. Then :
If $$A$$ and $$B$$ are two events such that $$P(A)=\\frac{1}{3}, P(B)=\\frac{1}{5}$$ and $$P(A \\cup B)=\\frac{1}{2}$$, then $$P\\left(A \\mid B^{\\prime}\\right)+P\\left(B \\mid A^{\\prime}\\right)$$ is equal to :
Out of $$60 \\%$$ female and $$40 \\%$$ male candidates appearing in an exam, $$60 \\%$$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :
Let $$\\mathrm{A}$$ and $$\\mathrm{B}$$ be two events such that $$P(B \\mid A)=\\frac{2}{5}, P(A \\mid B)=\\frac{1}{7}$$ and $$P(A \\cap B)=\\frac{1}{9} \\cdot$$ Consider
",
"options": [
{
"text": "Both (S1) and (S2) are true"
},
{
"text": "Both (S1) and (S2) are false"
},
{
"text": "Only (S1) is true"
},
{
"text": "Only (S2) is true"
}
],
"answer": "Both (S1) and (S2) are true",
"solution": "**Answer:** Both (S1) and (S2) are true\n\n
25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $$\\frac{k}{10}%$$. Then the value of k is __________.
",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\nProbability of a person being smoker $=\\frac{1}{4}$\n
\nProbability of a person being non-smoker $=\\frac{3}{4}$\n
\n$P\\left(\\frac{\\text { Person is smoker }}{\\text { Person diagonsed with cancer }}\\right)=\\frac{\\frac{1}{4} \\cdot 27 P}{\\frac{1}{4} \\cdot 27 P+\\frac{3 P}{4}}$\n
In a bolt factory, machines $$A, B$$ and $$C$$ manufacture respectively $$20 \\%, 30 \\%$$ and $$50 \\%$$ of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine $$C$$ is :
$$\n\\begin{aligned}\n& =\\frac{P\\left(\\frac{E}{C}\\right) \\times P(C)}{P\\left(\\frac{E}{A}\\right) \\times P(A)+P\\left(\\frac{E}{B}\\right) \\times P(B)+P\\left(\\frac{E}{C}\\right) \\times P(C)} \\\\\\\\\n& =\\frac{\\frac{5}{10} \\times \\frac{2}{100}}{\\frac{3}{100} \\times \\frac{2}{10}+\\frac{4}{100} \\times \\frac{3}{10}+\\frac{2}{100} \\times \\frac{5}{10}} \\\\\\\\\n& =\\frac{10}{6+12+10}=\\frac{10}{28}=\\frac{5}{14}\n\\end{aligned}\n$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6713,
"subject": "General Science",
"question": "Let Ajay will not appear in JEE exam with probability $\\mathrm{p}=\\frac{2}{7}$, while both Ajay and Vijay will appear in the exam with probability $\\mathrm{q}=\\frac{1}{5}$. Then the probability, that Ajay will appear in the exam and Vijay will not appear is :",
"options": [
{
"text": "$\\frac{9}{35}$"
},
{
"text": "$\\frac{3}{35}$"
},
{
"text": "$\\frac{24}{35}$"
},
{
"text": "$\\frac{18}{35}$"
}
],
"answer": "$\\frac{18}{35}$",
"solution": "**Answer:** $\\frac{18}{35}$\n\n
We are given that the probability of Ajay not appearing in the JEE exam is $\\mathrm{p}=\\frac{2}{7}$, and the probability that both Ajay and Vijay will appear in the exam is $\\mathrm{q}=\\frac{1}{5}$.
\n\n
We are asked to find the probability that Ajay will appear in the exam and Vijay will not. Let's denote this probability as $\\mathrm{r}$.
\n\n
To find $\\mathrm{r}$, we need to use the concept of complementary events. The probability that Ajay will appear in the exam is the complement of the probability that he will not appear. So,
The event that both Ajay and Vijay appear in the exam is independent of the event that only Ajay appears (and Vijay does not). Therefore, we can express the probability that only Ajay will appear (and Vijay will not) as the difference of Ajay appearing minus both Ajay and Vijay appearing, because the probability of both appearing ($\\mathrm{q}$) is included in the probability of Ajay appearing:
Therefore, the probability that Ajay will appear in the exam and Vijay will not appear is $\\mathrm{r} = \\frac{18}{35}$, which corresponds to Option D.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6714,
"subject": "General Science",
"question": "A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let
$a=P(X=3), b=P(X \\geqslant 3)$ and $c=P(X \\geqslant 6 \\mid X>3)$. Then $\\frac{b+c}{a}$ is equal to __________.",
"options": [],
"answer": "12",
"solution": "**Answer:** 12\n\n
To solve this problem, we need to compute the probabilities $a$, $b$, and $c$, and then plug those values into the expression $\\frac{b+c}{a}$.
\n\n
Let's begin by defining each of the variables:
\n\n
\n\n
$a = P(X=3)$: This is the probability that the first six appears on the third toss.
\n\n
$b = P(X \\geqslant 3)$: This is the probability that the first six appears on the third toss or later.
\n\n
$c = P(X \\geqslant 6 \\mid X>3)$: This is the probability that the first six appears on the sixth toss or later, given that it has not appeared in the first three tosses.
\n\n
\n\n
Since we're dealing with a fair die, each side has an equal probability of $\\frac{1}{6}$ of landing face up. Let's find the probabilities step by step:
\n\n
Calculating $a$:
\n\n
\n\n
The probability of rolling anything other than a six is $\\frac{5}{6}$. So for the first six to show up exactly on the third roll, the sequence of rolls must be NN6, where N is anything but a six (i.e., the results of the first two rolls). Thus,
For the first six to appear on the third roll or later, we can think of two cases: when the first six appears on the third roll (which we've already calculated, $a$), and when it appears after the third roll. To combine these probabilities, we can use the fact that $P(X \\geqslant 3) = 1 - P(X < 3)$, where $P(X < 3)$ is the probability that the first six appears on either the first or the second roll. So we calculate the latter first:
This is the probability that the first six appears on or after the sixth roll, given that it hasn't appeared in the first three rolls. Since $X>3$, the first three outcomes must not be a six, which occurs with probability $\\left(\\frac{5}{6}\\right)^3$. The subsequent outcomes until (and including) the fifth roll also must not be a six. So,
Notice here, we did not include the probability of rolling a six, because we are looking for the probability that we have not yet rolled a six after the fifth roll.
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is
To solve this problem, we need to calculate the probability of two independent events occurring in succession: the first marble drawn is red, and the second marble drawn is white. Since the drawing is with replacement, the number of marbles of each color remains the same for both draws.
\n\n
The total number of marbles in the box is the sum of red, white, blue, and orange marbles:
The probability of drawing a red marble in the first draw is the number of red marbles divided by the total number of marbles:
\n$$\nP(\\text{First is red}) = \\frac{10}{75}.\n$$\n\n
Since the marble is replaced, the probability of drawing a white marble in the second draw remains as the number of white marbles divided by the total number of marbles:
\n$$\nP(\\text{Second is white}) = \\frac{30}{75}.\n$$\n\n
The probability of both independent events occurring in succession (drawing a red marble first and then a white marble) is the product of their individual probabilities:
\n$$\nP(\\text{First is red and second is white}) = P(\\text{First is red}) \\times P(\\text{Second is white}) = \\frac{10}{75} \\times \\frac{30}{75}.\n$$\n\n
Now, let's calculate this probability:
\n
$$\nP(\\text{First is red and second is white}) = \\frac{10 \\times 30}{75 \\times 75} = \\frac{300}{5625}= \\frac{4}{75}.\n$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6717,
"subject": "General Science",
"question": "An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is :",
"options": [
{
"text": "$${2 \\over 7}$$"
},
{
"text": "$${1 \\over 21}$$"
},
{
"text": "$${1 \\over 23}$$"
},
{
"text": "$${1 \\over 3}$$"
}
],
"answer": "$${2 \\over 7}$$",
"solution": "**Answer:** $${2 \\over 7}$$\n\nOut of nine balls three balls can be chosen = $${}^9{C_3}$$ ways\n
According to the question, all three ball should be different. So out of 3 red balls 1 is chosen and out of 4 blue 1 is chosen and out of 2 green 1 is chosen.\n
$$\\therefore$$ Probability = $${{24} \\over {84}}$$ = $${{2} \\over {7}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6718,
"subject": "General Science",
"question": "Four numbers are chosen at random (without replacement) from the set $$\\left\\{ {1,2,3,....20} \\right\\}.$$ \n
Statement - 1: The probability that the chosen numbers when arranged in some order will form an AP is $${1 \\over {85}}.$$\n
Statement - 2: If the four chosen numbers form an AP, then the set of all possible values of common difference is $$\\left( { \\pm 1, \\pm 2, \\pm 3, \\pm 4, \\pm 5} \\right).$$
",
"options": [
{
"text": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1."
},
{
"text": "Statement - 1 is true, Statement - 2 is false."
},
{
"text": "Statement - 1 is false, Statement -2 is true."
},
{
"text": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1."
}
],
"answer": "Statement - 1 is true, Statement - 2 is false.",
"solution": "**Answer:** Statement - 1 is true, Statement - 2 is false.\n\nFour numbers can be chosen $${}^{20}{C_4}$$ ways.\n
When common difference d = 1 then the possible sets are (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) = 17 sets\n So when d = 1 then 17 different AP's are possible with 4 numbers.\n
Now let's create a table of all possible sets -\n\n
$$\\therefore$$ Statement - 2: is false as common difference can also be $$ \\pm 6$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6719,
"subject": "General Science",
"question": "If $$12$$ different balls are to be placed in $$3$$ identical boxes, then the probability that one of the boxes contains exactly $$3$$ balls is :",
"options": [
{
"text": "$$220{\\left( {{1 \\over 3}} \\right)^{12}}$$ "
},
{
"text": "$$22{\\left( {{1 \\over 3}} \\right)^{11}}$$"
},
{
"text": "$${{55} \\over 3}{\\left( {{2 \\over 3}} \\right)^{11}}$$ "
},
{
"text": "$$55{\\left( {{2 \\over 3}} \\right)^{10}}$$"
}
],
"answer": "$${{55} \\over 3}{\\left( {{2 \\over 3}} \\right)^{11}}$$ ",
"solution": "**Answer:** $${{55} \\over 3}{\\left( {{2 \\over 3}} \\right)^{11}}$$ \n\n1st ball can go any of the 3 boxes. So total choices for 1st ball = 3\n
2nd ball can also go any of the 3 boxes. So total choices for 2nd ball = 3\n .\n .\n .\n .\n 12th ball can go any of the 3 boxes. So total choices for 12th ball = 3\n
Total choices for all 12 balls = $$3 \\times $$$$3 \\times $$$$3 \\times $$.................12 times = 312.\n
Now question says choose 3 balls from 12 balls. So no of ways = $${}^{12}{C_3}$$ ways.\n And then put it in a box. No of ways we can put = $${}^{12}{C_3} \\times 1$$ ways.\n
Now we have 9 balls left and we have to put those 9 balls in the remaining 2 boxes.\n
Each ball can go to any of the 2 boxes, so for each ball there is 2 choices.\n
$$\\therefore$$ Total ways for 9 balls = 29\n
$$\\therefore$$ Total ways we can put those 12 balls in the boxes = $${}^{12}{C_3} \\times 1 \\times {2^9}$$\n
So option (C) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6720,
"subject": "General Science",
"question": "From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :\n",
"options": [
{
"text": "$${{21} \\over {220}}$$ "
},
{
"text": "$${{3} \\over {11}}$$ "
},
{
"text": "$${{1} \\over {11}}$$ "
},
{
"text": "$${{2} \\over {23}}$$ "
}
],
"answer": "$${{1} \\over {11}}$$ ",
"solution": "**Answer:** $${{1} \\over {11}}$$ \n\n
The number of ways to form a committee having at least one woman is
Therefore, the probability for the committees to have more women than men is
\n
$${{105} \\over {1155}} = {1 \\over {11}}$$
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 6721,
"subject": "General Science",
"question": "Two different families A and B are blessed with equal numbe of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is $${1 \\over {12}},$$ then the number of children in each family is : ",
"options": [
{
"text": "3"
},
{
"text": "4"
},
{
"text": "5"
},
{
"text": "6"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\nLet the number of children in each family be x. \n
Thus the total number of children in both the families are 2x \n
Now, it is given that 3 tickets are distributed amongst the children of these two families. \n
Thus, the probability that all the three tickets go to the children in family B\n
Thus, the number of children in each family is 5.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6722,
"subject": "General Science",
"question": "Let S = {1, 2, . . . . . ., 20}. A subset B of S is said to be \"nice\", if the sum of the elements of B is 203. Then the probability that a randonly chosen subset of S is \"nice\" is :\n",
"options": [
{
"text": "$${5 \\over {{2^{20}}}}$$"
},
{
"text": "$${7 \\over {{2^{20}}}}$$"
},
{
"text": "$${4 \\over {{2^{20}}}}$$"
},
{
"text": "$${6 \\over {{2^{20}}}}$$"
}
],
"answer": "$${5 \\over {{2^{20}}}}$$",
"solution": "**Answer:** $${5 \\over {{2^{20}}}}$$\n\n
We can solve this problem by counting the number of \"nice\" subsets in the set S = {1, 2, $\\ldots$, 20 }, and then dividing that number by the total number of possible subsets of S.
Since a \"nice\" subset must sum to 203, the elements not in the subset must sum to 210 - 203 = 7.
\n
Now we need to find the ways to make the sum of 7 using the elements of S. The combinations are :
\n\n
1. 7
\n
2. 1 + 6
\n
3. 2 + 5
\n
4. 3 + 4
\n
5. 1 + 2 + 4
\n
6. 1 + 3 + 3(This doesn't work since 3 is repeated)
\n
7. 2 + 2 + 3(This doesn't work since 2 is repeated)
\n\n
So, there are 5 \"nice\" subsets.
\n
Since the set S has 20 elements, there are $2^{20}$ possible subsets (including the empty set and the set itself). The probability of randomly choosing a \"nice\" subset is therefore :\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6723,
"subject": "General Science",
"question": "If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle\nformed with these chosen vertices is equilateral is :",
"options": [
{
"text": "$${1 \\over {10}}$$"
},
{
"text": "$${3 \\over {10}}$$"
},
{
"text": "$${3 \\over {20}}$$"
},
{
"text": "$${1 \\over {5}}$$"
}
],
"answer": "$${1 \\over {10}}$$",
"solution": "**Answer:** $${1 \\over {10}}$$\n\n \nChoosing vertices of a regular hexagon alternate, here A1, A3, A5 or A2, A4, A6 will result in an equilateral triangle.
\nHence the required probability = $${2 \\over {{}^6{C_3}}}$$ = $${1 \\over {10}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6724,
"subject": "General Science",
"question": "If 10 different balls are to be placed in 4 distinct\nboxes at random, then the probability that two\nof these boxes contain exactly 2 and 3 balls is :",
"options": [
{
"text": "$${{965} \\over {{2^{11}}}}$$"
},
{
"text": "$${{965} \\over {{2^{10}}}}$$"
},
{
"text": "$${{945} \\over {{2^{11}}}}$$"
},
{
"text": "$${{945} \\over {{2^{10}}}}$$"
}
],
"answer": "$${{945} \\over {{2^{10}}}}$$",
"solution": "**Answer:** $${{945} \\over {{2^{10}}}}$$\n\nTotal ways of distribution = 410 = 220\n
Number of ways selecting two boxes out of four = 4C2\n
Then number of ways selecting 5 balls out of 10 = 10C5\n
Then no of ways of distributing 5 balls into two groups of 2 balls and 3 balls = 5C3.2!\n
Then number of ways to distributing remaining balls into two boxes = 25\n
Number of ways placing exactly 2 and 3 balls in two of these boxes \n
$$ = {{15 + 120} \\over {{{10}^4}}} = {{135} \\over {{{10}^4}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6726,
"subject": "General Science",
"question": "Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :",
"options": [
{
"text": "$${{10} \\over {99}}$$"
},
{
"text": "$${{5} \\over {33}}$$"
},
{
"text": "$${{15} \\over {101}}$$"
},
{
"text": "$${{5} \\over {101}}$$"
}
],
"answer": "$${{5} \\over {33}}$$",
"solution": "**Answer:** $${{5} \\over {33}}$$\n\nOut of 11 consecutive natural numbers either\n6 even and 5 odd numbers or 5 even and 6\nodd numbers.\n
Let, E = Even\n O = Odd\n
Case-1 :\n
E, O, E, O, E, O, E, O, E, O, E\n
2b = a + c $$ \\Rightarrow $$ Even\n
$$ \\Rightarrow $$ Both a and c should be either even or odd.\n
Total probability = $${1 \\over 2} \\times {5 \\over {33}}$$ + $${1 \\over 2} \\times {5 \\over {33}}$$ = $${5 \\over {33}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6727,
"subject": "General Science",
"question": "The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is :",
"options": [
{
"text": "$${{135} \\over {{2^9}}}$$"
},
{
"text": "$${{65} \\over {{2^8}}}$$"
},
{
"text": "$${{65} \\over {{2^7}}}$$"
},
{
"text": "$${{35} \\over {{2^7}}}$$"
}
],
"answer": "$${{135} \\over {{2^9}}}$$",
"solution": "**Answer:** $${{135} \\over {{2^9}}}$$\n\nGiven, set P = {1, 2, 3, 4, 5}
Let the two subsets be A and B
Then, n (A $$\\cap$$ B) = 2 (as given in question)\n
We can choose two elements from set P in 5C2 ways.\n
After choosing two common elements for set A and B, each of remaining three elements from set P have three choice (1) It can go to set A (2) It can go to set B (3) It don't go to any sets it stays at set P.\n
$$ \\therefore $$ Total ways for the three elements = 3 $$ \\times $$ 3 $$ \\times $$ 3 = 33\n
$$\\therefore$$ Required probability = $${{{}^5{C_2} \\times {3^3}} \\over {\\left( {{2^5}} \\right)\\left( {{2^5}} \\right)}}$$ = $${{{}^5{C_2} \\times {3^3}} \\over {{4^5}}} = {{10 \\times 27} \\over {{2^{10}}}} = {{135} \\over {{2^9}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6728,
"subject": "General Science",
"question": "Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is :",
"options": [
{
"text": "$${2 \\over 9}$$"
},
{
"text": "$${1 \\over 5}$$"
},
{
"text": "$${122 \\over 297}$$"
},
{
"text": "$${97 \\over 297}$$"
}
],
"answer": "$${97 \\over 297}$$",
"solution": "**Answer:** $${97 \\over 297}$$\n\nn(s) = n(when 7 appears on thousands place)
$$\\therefore$$ $$P(E) = {{8 \\times 9 \\times 9 + 9 \\times 25} \\over {33 \\times 9 \\times 9}} = {{97} \\over {297}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6729,
"subject": "General Science",
"question": "A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is :",
"options": [
{
"text": "$${1 \\over 7}$$"
},
{
"text": "$${4 \\over 7}$$"
},
{
"text": "$${6 \\over 7}$$"
},
{
"text": "$${3 \\over 7}$$"
}
],
"answer": "$${3 \\over 7}$$",
"solution": "**Answer:** $${3 \\over 7}$$\n\nDigits = 3, 3, 4, 4, 4, 5, 5
Total 7 digit numbers = $${{7!} \\over {2!2!3!}}$$
Number of 7 digit number divisible by 2 $$ \\Rightarrow $$ last digit = 4
Now 7 digit numbers which are divisible by 2 = $${{6!} \\over {2!2!2!}}$$
Required probability = $${{{{6!} \\over {2!2!2!}}} \\over {{{7!} \\over {3!2!2!}}}} = {3 \\over 7}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6730,
"subject": "General Science",
"question": "Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is :",
"options": [
{
"text": "$${1 \\over {66}}$$"
},
{
"text": "$${1 \\over {11}}$$"
},
{
"text": "$${1 \\over {9}}$$"
},
{
"text": "$${2 \\over {11}}$$"
}
],
"answer": "$${1 \\over {11}}$$",
"solution": "**Answer:** $${1 \\over {11}}$$\n\nAAEIIMNNOTX
----------------M----------------
Total words with M at fourth Place = $${{10!} \\over {2!2!2!}}$$
Total words = $${{11!} \\over {2!2!2!}}$$
Required probability = $${{10!} \\over {11!}}$$ = $${{1} \\over {11}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6731,
"subject": "General Science",
"question": "Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2 $$\\times$$ 2 matrices. The probability that such formed matrix have all different entries and are non-singular, is :",
"options": [
{
"text": "$${{45} \\over {162}}$$"
},
{
"text": "$${{21} \\over {81}}$$"
},
{
"text": "$${{22} \\over {81}}$$"
},
{
"text": "$${{43} \\over {162}}$$"
}
],
"answer": "$${{43} \\over {162}}$$",
"solution": "**Answer:** $${{43} \\over {162}}$$\n\n$$A = \\left| {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right|$$
| A | = ad $$-$$ bc
Total case = 64
For non-singular matrix | A | $$\\ne$$ 0 $$\\Rightarrow$$ ad $$-$$ bc $$\\ne$$ 0
$$\\Rightarrow$$ ad $$\\ne$$ bc
And a, b, c, d are all different numbers in the set {1, 2, 3, 4, 5, 6}
$$A = \\left[ {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right]$$
\n
Given that matrix A is singular.
\n
$$\\therefore$$ | A | = 0
\n
$$ \\Rightarrow \\left| {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right| = 0$$
\n
$$ \\Rightarrow ad = bc$$
\n
Case I :
\n
ad = bc condition satisfy when a = b = c = d.
\n
For ex when a = 2, b = 2, c = 2, d = 2, then ad = bc satisfy.
\n
Now there are 10 prime numbers.
\n
We can choose any one of the 10 prime number in $${}^{10}C_{1}$$ = 10 ways and put them in the four positions of the matrix and matrix will be singular.
\n
$$\\therefore$$ In this case, total favorable case = 10
\n
Case 2 :
\n
ad = bc condition satisfies when
\n
(1)
\n
a = 2, d - 3 then
\n
(a) b = 2, c = 3
\n
(b) b = 3, c = 2
\n
or
\n
a = 3, d = 2 then
\n
(a) b = 2, c = 3
\n
(b) b = 3, c = 2
\n
So you can see for two different prime number for a and d there are 4 possible value of b and c which satisfy ad = bc condition.
\n
Two different values of a and d can be chosen from 10 prime numbers = $${}^{10}C_{2}$$ ways
\n
And for each combination of a and d there are 4 possible values of b and c.
\n
$$\\therefore$$ Total possible values = $${}^{10}C_{2}$$ $$\\times$$ 4
\n
From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition
\n
= 10 + $${}^{10}C_{2}$$ $$\\times$$ 4
\n
For sample space,
\n
Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number = $${}^{10}C_{1}$$ ways
The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies f(a) + 2f(b) $$-$$ f(c) = f(d) is :
",
"options": [
{
"text": "$${1 \\over {24}}$$"
},
{
"text": "$${1 \\over {40}}$$"
},
{
"text": "$${1 \\over {30}}$$"
},
{
"text": "$${1 \\over {20}}$$"
}
],
"answer": "$${1 \\over {20}}$$",
"solution": "**Answer:** $${1 \\over {20}}$$\n\nNumber of one-one function from $\\{a, b, c, d\\}$ to set $\\{1,2,3,4,5\\}$ is ${ }^{5} P_{4}=120 n(s)$.\n
\nThe required possible set of value (f(a), $f(b), f(c), f(d))$ such that $f(a)+2 f(b)-f(c)=f(d)$ are $(5,3,2,1),(5,1,2,3),(4,1,3,5),(3,1,4,5)$, $(5,4,3,2)$ and $(3,4,5,2)$\n
Five numbers $${x_1},{x_2},{x_3},{x_4},{x_5}$$ are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order $$({x_1} < {x_2} < {x_3} < {x_4} < {x_5})$$. The probability that $${x_2} = 7$$ and $${x_4} = 11$$ is :
If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to _______________.
Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is :
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :
A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is
To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).
\n
Since a head is twice as likely to occur as a tail, we can denote the probability of getting a tail as $$ P(T) = p $$ and the probability of getting a head as $$ P(H) = 2p $$.
\n
These probabilities must sum to 1 because those are the only two possible outcomes for each coin toss :\n
$$ P(H) + P(T) = 1 $$\n
$$ 2p + p = 1 $$\n
$$ 3p = 1 $$\n
$$ p = \\frac{1}{3} $$
\n
Therefore, the probability of getting a tail (T) is $$ P(T) = \\frac{1}{3} $$ and the probability of getting a head (H) is $$ P(H) = 2 \\times \\frac{1}{3} = \\frac{2}{3} $$.
\n
Now to find the probability of getting two tails and one head, we need to consider the different sequences in which this can occur. There are three unique sequences: TTH, THT, and HTT.
\n
The probability of each sequence is found by multiplying the probabilities of each individual event since each coin toss is independent:\n
The coefficients $$\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$$ in the quadratic equation $$\\mathrm{a} x^2+\\mathrm{bx}+\\mathrm{c}=0$$ are from the set $$\\{1,2,3,4,5,6\\}$$. If the probability of this equation having one real root bigger than the other is p, then 216p equals :
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6743,
"subject": "General Science",
"question": "At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of $$5$$ phone calls during $$10$$ minute time intervals. The probability that there is at the most one phone call during a $$10$$-minute time period is :",
"options": [
{
"text": "$${6 \\over {{5^e}}}$$ "
},
{
"text": "$${5 \\over 6}$$ "
},
{
"text": "$${6 \\over 55}$$"
},
{
"text": "$${6 \\over {{e^5}}}$$ "
}
],
"answer": "$${6 \\over {{e^5}}}$$ ",
"solution": "**Answer:** $${6 \\over {{e^5}}}$$ \n\n
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 6744,
"subject": "General Science",
"question": "A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins\nRs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the\nexpected gain/loss (in Rs.) of the person is :",
"options": [
{
"text": "$${1 \\over 4}$$ loss"
},
{
"text": "$${1 \\over 2}$$ gain"
},
{
"text": "$${1 \\over 2}$$ loss"
},
{
"text": "2 gain"
}
],
"answer": "$${1 \\over 2}$$ loss",
"solution": "**Answer:** $${1 \\over 2}$$ loss\n\nWhen two dice are thrown then sample space will {(1, 1), (2, 2) ....... (6, 6)} contain total 36 elements number of cases.
\nThen the expectation will be $${6 \\over {36}} \\times 15 \\times {4 \\over {36}} \\times 12 - {{26} \\over {36}} \\times 6$$
\n$${{90 + 48 - 156} \\over {36}} = - {1 \\over 2}$$ = $$ {1 \\over 2}$$ loss\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6745,
"subject": "General Science",
"question": "An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value of k when k\nconsecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value -1. Then the expected\nvalue of X, is :\n",
"options": [
{
"text": "$$ - {3 \\over {16}}$$"
},
{
"text": "$$ - {1 \\over 8}$$"
},
{
"text": "$${1 \\over 8}$$"
},
{
"text": "$${3 \\over {16}}$$"
}
],
"answer": "$${1 \\over 8}$$",
"solution": "**Answer:** $${1 \\over 8}$$\n\nNumber of ways 3 consecutive heads can appers\n
(1) HHHT_\n
(2) _THHH\n
(3) THHHT\n
$$ \\therefore $$ Probablity of getting 3 consecutive heads\n
= $${1 \\over 8}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6746,
"subject": "General Science",
"question": "A random variable X has the following\nprobability distribution :
= $${{23} \\over {36}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 6747,
"subject": "General Science",
"question": "Let X be a random variable such that the probability function of a distribution is given by $$P(X = 0) = {1 \\over 2},P(X = j) = {1 \\over {{3^j}}}(j = 1,2,3,...,\\infty )$$. Then the mean of the distribution and P(X is positive and even) respectively are :",
"options": [
{
"text": "$${3 \\over 8}$$ and $${1 \\over 8}$$"
},
{
"text": "$${3 \\over 4}$$ and $${1 \\over 8}$$"
},
{
"text": "$${3 \\over 4}$$ and $${1 \\over 9}$$"
},
{
"text": "$${3 \\over 4}$$ and $${1 \\over 16}$$"
}
],
"answer": "$${3 \\over 4}$$ and $${1 \\over 8}$$",
"solution": "**Answer:** $${3 \\over 4}$$ and $${1 \\over 8}$$\n\nMean = $$\\sum {{X_i}{P_i}} = \\sum\\limits_{r = 0}^\\infty {r.{1 \\over {{3^r}}} = {3 \\over 4}} $$
$$ = {{{1 \\over 9}} \\over {1 - {1 \\over 9}}} = {{1/9} \\over {8/9}} = {1 \\over 8}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6748,
"subject": "General Science",
"question": "The probability distribution of random variable X is given by :
\n
\n\n
\n
X
\n
1
\n
2
\n
3
\n
4
\n
5
\n
\n\n\n
\n
P(X)
\n
K
\n
2K
\n
2K
\n
3K
\n
K
\n
\n\n
Let p = P(1 < X < 4 | X < 3). If 5p = $$\\lambda$$K, then $$\\lambda$$ equal to ___________.",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\n$$\\sum {P(X) = 1 \\Rightarrow k + 2k + 3} k + k = 1$$
$$ \\Rightarrow \\lambda = 30$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 6749,
"subject": "General Science",
"question": "Let X be a random variable with distribution.
\n
\n\n
\n
x
\n
$$ - $$2
\n
$$ - $$1
\n
3
\n
4
\n
6
\n
\n\n\n
\n
P(X = x)
\n
$${1 \\over 5}$$
\n
a
\n
$${1 \\over 3}$$
\n
$${1 \\over 5}$$
\n
b
\n
\n\n
If the mean of X is 2.3 and variance of X is $$\\sigma$$2, then 100 $$\\sigma$$2 is equal to :",
"options": [],
"answer": "781",
"solution": "**Answer:** 781\n\n
A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is $${1 \\over n}$$. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :
Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $$\\mathrm{X}$$ be the number of white balls, among the drawn balls. If $$\\sigma^{2}$$ is the variance of $$\\mathrm{X}$$, then $$100 \\sigma^{2}$$ is equal to ________.
Let $$\\mathrm{S} = \\{ {w_1},{w_2},......\\} $$ be the sample space associated to a random experiment. Let $$P({w_n}) = {{P({w_{n - 1}})} \\over 2},n \\ge 2$$. Let $$A = \\{ 2k + 3l:k,l \\in N\\} $$ and $$B = \\{ {w_n}:n \\in A\\} $$. Then P(B) is equal to :
A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $$\\mathrm{X}$$ denotes the number of tosses of the coin, then the mean of $$\\mathrm{X}$$ is :
",
"options": [
{
"text": "$$\\frac{81}{64}$$"
},
{
"text": "$$\\frac{37}{16}$$"
},
{
"text": "$$\\frac{21}{16}$$"
},
{
"text": "$$\\frac{15}{16}$$"
}
],
"answer": "$$\\frac{21}{16}$$",
"solution": "**Answer:** $$\\frac{21}{16}$$\n\nThe given probabilities for getting a head (H) and a tail (T) are as follows:\n\n
The random variable X can take the values 1, 2, or 3. These correspond to the following events:\n\n
- X = 1 : A head is obtained on the first toss. This happens with probability $P(H) = \\frac{3}{4}$.\n
- X = 2 : A tail is obtained on the first toss and a head on the second. This happens with probability $P(T)P(H) = \\frac{1}{4} \\times \\frac{3}{4}$.\n
- X = 3 : Either two tails and then a head are obtained, or three tails are obtained.\n This happens with probability $P(T)P(T)P(H) + P(T)P(T)P(T) = \\left(\\frac{1}{4}\\right)^2 \\times \\frac{3}{4} + \\left(\\frac{1}{4}\\right)^3$.\n\n
Now, we calculate the mean (expected value) of X:\n\n
Two dice A and B are rolled. Let the numbers obtained on A and B be $$\\alpha$$ and $$\\beta$$ respectively. If the variance of $$\\alpha-\\beta$$ is $$\\frac{p}{q}$$, where $$p$$ and $$q$$ are co-prime, then the sum of the positive divisors of $$p$$ is equal to :
Next, we calculate the expected value of the differences. The expected value is calculated as the sum of the products of each outcome and its corresponding probability. Given that the table is symmetric around 0, the expected value is 0.\n\n
Now, we can calculate the variance, which is the expected value of the squared differences minus the square of the expected value of the differences :\n\n
Here, $$p = 35$$ and $$q = 6$$, and they are co-prime.\n\n
The positive divisors of 35 are 1, 5, 7, and 35. The sum of these divisors is $$1 + 5 + 7 + 35 = 48$$. ",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6758,
"subject": "General Science",
"question": "
A fair $$n(n > 1)$$ faces die is rolled repeatedly until a number less than $$n$$ appears. If the mean of the number of tosses required is $$\\frac{n}{9}$$, then $$n$$ is equal to ____________.
",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\nIn this case, the success is defined as getting a number less than $n$ when rolling an $n$-sided die. The probability of success, $p$, in each roll is then $(n-1)/n$, and the probability of failure, $q = 1 - p$, is $1/n$.\n
$$\n\\text { Mean }=\\sum\\limits_{i=1}^{\\infty} p_i x_i=1 \\cdot \\frac{n-1}{n}+\\frac{2}{n} \\cdot\\left(\\frac{n-1}{n}\\right)+\\frac{3}{n^2}\\left(\\frac{n-1}{n}\\right)+\\ldots\n$$\n
Let the probability of getting head for a biased coin be $$\\frac{1}{4}$$. It is tossed repeatedly until a head appears. Let $$\\mathrm{N}$$ be the number of tosses required. If the probability that the equation $$64 \\mathrm{x}^{2}+5 \\mathrm{Nx}+1=0$$ has no real root is $$\\frac{\\mathrm{p}}{\\mathrm{q}}$$, where $$\\mathrm{p}$$ and $$\\mathrm{q}$$ are coprime, then $$q-p$$ is equal to ________.
",
"options": [],
"answer": "27",
"solution": "**Answer:** 27\n\nWe have the quadratic equation $64x^2 + 5Nx + 1 = 0$. For it to have no real roots, the discriminant ($b^2 - 4ac$) should be less than 0. Here, $a = 64$, $b = 5N$, and $c = 1$. \n\n
This gives us :\n\n
$(5N)^2 - 4\\times64\\times1 < 0$\n\n
$\\Rightarrow 25N^2 < 256$\n\n
$\\Rightarrow N^2 < \\frac{256}{25}$\n\n
$\\Rightarrow N < \\sqrt{\\frac{256}{25}} = \\frac{16}{5}$\n\n
Since $N$ must be an integer (as it represents the number of tosses), the possible values of $N$ are 1, 2, or 3. \n\n
The probability of getting the first head on the $n$-th toss (given the probability of getting a head is $1/4$) is given by the geometric distribution formula, $(1 - p)^{n-1}\\times p$.\n\n
So, the probability for our specific values of $N$ is:\n\n
If the probability that the random variable $$\\mathrm{X}$$ takes values $$x$$ is given by $$\\mathrm{P}(\\mathrm{X}=x)=\\mathrm{k}(x+1) 3^{-x}, x=0,1,2,3, \\ldots$$, where $$\\mathrm{k}$$ is a constant, then $$\\mathrm{P}(\\mathrm{X} \\geq 2)$$ is equal to :
",
"options": [
{
"text": "$$\\frac{7}{18}$$"
},
{
"text": "$$\\frac{20}{27}$$"
},
{
"text": "$$\\frac{7}{27}$$"
},
{
"text": "$$\\frac{11}{18}$$"
}
],
"answer": "$$\\frac{7}{27}$$",
"solution": "**Answer:** $$\\frac{7}{27}$$\n\nAs, we know that sum of all the probabilities $=1$\n
Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable $$x$$ to be the number of rotten apples in a draw of two apples, the variance of $$x$$ is
Three balls are drawn at random from a bag containing 5 blue and 4 yellow balls. Let the random variables $$X$$ and $$Y$$ respectively denote the number of blue and yellow balls. If $$\\bar{X}$$ and $$\\bar{Y}$$ are the means of $$X$$ and $$Y$$ respectively, then $$7 \\bar{X}+4 \\bar{Y}$$ is equal to ___________.
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $$X$$ denote the number of defective items in the sample. If the variance of $$X$$ is $$\\sigma^2$$, then $$96 \\sigma^2$$ is equal to __________.
From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable $$X$$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $$X$$ is $$\\frac{m}{n}$$, where $$\\operatorname{gcd}(m, n)=1$$, then $$n-m$$ is equal to _________.
",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 6766,
"subject": "General Science",
"question": "A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and\nthis ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at\nrandom from the bag, then the probability that this drawn ball is red, is :",
"options": [
{
"text": "$${3 \\over 4}$$"
},
{
"text": "$${3 \\over 10}$$"
},
{
"text": "$${2 \\over 5}$$"
},
{
"text": "$${1 \\over 5}$$"
}
],
"answer": "$${2 \\over 5}$$",
"solution": "**Answer:** $${2 \\over 5}$$\n\n \n
If we follow path 1, then probability of getting 1st ball black $$ = {6 \\over {10}}$$ and probability of getting 2nd ball red when there is 4 R and 8 B balls = $${4 \\over {12}}$$.\n
So, the probability of getting 1st ball black and 2nd ball red = $${6 \\over {10}} \\times {4 \\over {12}}$$.\n
If we follow path 2, then the probability of getting 1st ball red $$ = {4 \\over {10}}$$ and probability of getting 2nd ball red when in the bag there is 6 red and 6 black balls = $${6 \\over {12}}$$\n
$$\\therefore\\,\\,\\,$$ Probability of getting 2nd ball as red \n
$$ = {2 \\over 5}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6767,
"subject": "General Science",
"question": "A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is :",
"options": [
{
"text": "$\\frac{1}{4}$"
},
{
"text": "$\\frac{9}{50}$"
},
{
"text": "$\\frac{1}{5}$"
},
{
"text": "$\\frac{11}{50}$"
}
],
"answer": "$\\frac{1}{5}$",
"solution": "**Answer:** $\\frac{1}{5}$\n\nLet $X$ be the number rolled on the die, and let $W$ be the event that all balls drawn are white. We want to find the probability $P(W)$, which can be calculated using the law of total probability as follows :\n\n
The probability of rolling any number from 1 to 6 on the die is equal, so $P(X=x) = \\frac{1}{6}$ for all $x \\in \\{1, 2, 3, 4, 5, 6\\}$. \n\n
Now let's calculate the conditional probabilities $P(W|X=x)$ for each possible value of $x$ :\n\n
1. $P(W|X=1) = {{{}^6{C_1}} \\over {{}^{10}{C_1}}}= \\frac{6}{10} = \\frac{3}{5}$, since there are 6 white balls out of a total of 10 balls.\n
2. $P(W|X=2) = {{{}^6{C_2}} \\over {{}^{10}{C_2}}} = \\frac{15}{45} = \\frac{1}{3}$, since there are 15 ways to choose 2 white balls out of 6, and 45 ways to choose 2 balls out of 10.\n
3. $P(W|X=3) = {{{}^6{C_3}} \\over {{}^{10}{C_3}}} = \\frac{20}{120} = \\frac{1}{6}$, since there are 20 ways to choose 3 white balls out of 6, and 120 ways to choose 3 balls out of 10.\n
4. $P(W|X=4) = {{{}^6{C_4}} \\over {{}^{10}{C_4}}} = \\frac{15}{210} = \\frac{1}{14}$, since there are 15 ways to choose 4 white balls out of 6, and 210 ways to choose 4 balls out of 10.\n
5. $P(W|X=5) = {{{}^6{C_5}} \\over {{}^{10}{C_5}}} = \\frac{6}{252} = \\frac{1}{42}$, since there are 6 ways to choose 5 white balls out of 6, and 252 ways to choose 5 balls out of 10.\n
6. $P(W|X=6) = {{{}^6{C_6}} \\over {{}^{10}{C_6}}} = \\frac{1}{210}$, since there are 1 ways to choose 6 white balls out of 6, and 210 ways to choose 6 balls out of 10.\n\n
$$P(W) = \\frac{1}{6}\\left(\\frac{126+70+35+15+5+1}{210}\\right)=\\frac{42}{210}=\\frac{1}{5}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6768,
"subject": "General Science",
"question": "A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is :",
"options": [
{
"text": "$\\frac{2}{5}$"
},
{
"text": "$\\frac{2}{7}$"
},
{
"text": "$\\frac{1}{7}$"
},
{
"text": "$\\frac{1}{5}$"
}
],
"answer": "$\\frac{2}{7}$",
"solution": "**Answer:** $\\frac{2}{7}$\n\n$\\begin{aligned} & \\mathrm{P}(4 \\mathrm{~W} 4 \\mathrm{~B} / 2 \\mathrm{~W} 2 \\mathrm{~B})= \\\\\\\\ & \\frac{P(4 W 4 B) \\times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \\times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \\times P(2 W 2 B / 3 W 5 B)} \\\\ & +\\ldots \\ldots \\ldots \\ldots+P(6 W 2 B) \\times P(2 W 2 B / 6 W 2 B)\\end{aligned}$\n
$\\begin{aligned} & =\\frac{\\frac{1}{5} \\times \\frac{{ }^4 \\mathrm{C}_2 \\times{ }^4 \\mathrm{C}_2}{{ }^8 \\mathrm{C}_4}}{\\frac{1}{5} \\times \\frac{{ }^2 \\mathrm{C}_2 \\times{ }^6 \\mathrm{C}_2}{{ }^8 \\mathrm{C}_4}+\\frac{1}{5} \\times \\frac{{ }^3 \\mathrm{C}_2 \\times{ }^5 \\mathrm{C}_2}{{ }^8 \\mathrm{C}_4}+\\ldots+\\frac{1}{5} \\times \\frac{{ }^6 \\mathrm{C}_2 \\times{ }^2 \\mathrm{C}_2}{{ }^8 \\mathrm{C}_4}} \\\\\\\\ & =\\frac{2}{7}\\end{aligned}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6769,
"subject": "General Science",
"question": "A problem in mathematics is given to three students $$A,B,C$$ and their respective probability of solving the problem is $${1 \\over 2},{1 \\over 3}$$ and $${1 \\over 4}.$$ Probability that the problem is solved is :",
"options": [
{
"text": "$${3 \\over 4}$$"
},
{
"text": "$${1 \\over 2}$$ "
},
{
"text": "$${2 \\over 3}$$"
},
{
"text": "$${1 \\over 3}$$"
}
],
"answer": "$${3 \\over 4}$$",
"solution": "**Answer:** $${3 \\over 4}$$\n\nGiven $$P\\left( A \\right) = {1 \\over 2}$$, $$P\\left( B \\right) = {1 \\over 3}$$, $$P\\left( C \\right) = {1 \\over 4}$$\n
So, $$P\\left( {\\overline A } \\right) = {1 \\over 2}$$ (Probablity that the problem can't be solve by A)\n $$P\\left( {\\overline B } \\right) = {2 \\over 3}$$ (Probablity that the problem can't be solve by B)\n and $$P\\left( {\\overline C } \\right) = {3 \\over 4}$$ (Probablity that the problem can't be solve by C)\n
Now the probablity that the problem is solved by any one student of A, B and C = 1 - the probablity that the problem is solved by none of the students of A, B and C\n
$$P\\left( {A \\cup B \\cup C} \\right)$$ = 1 - $$P\\left( {\\overline A } \\right)P\\left( {\\overline B } \\right)P\\left( {\\overline C } \\right)$$\n
In the question given that A, B and C are mutually exclusive\n So $$P\\left( {A \\cup B \\cup C} \\right)$$ = $$P\\left( {A} \\right)$$ + $$P\\left( {B} \\right)$$ + $$P\\left( {C} \\right)$$\n
so, the events A and B are independent events but not equally likely.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6773,
"subject": "General Science",
"question": "For three events A, B and C,
P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P\n(Exactly one of C or A occurs) = $${1 \\over 4}$$\n and P(All the three events occur simultaneously) = $${1 \\over {16}}$$.\n
Then the\nprobability that at least one of the events occurs, is :",
"options": [
{
"text": "$${7 \\over {16}}$$"
},
{
"text": "$${7 \\over {64}}$$"
},
{
"text": "$${3 \\over {16}}$$"
},
{
"text": "$${7 \\over {32}}$$"
}
],
"answer": "$${7 \\over {16}}$$",
"solution": "**Answer:** $${7 \\over {16}}$$\n\nGiven, P (A $$ \\cap $$ B $$ \\cap $$ C) = $${1 \\over {16}}$$\n
P (exactly one of A or B occurs)\n
= P(A) + P (B) – 2P (A $$ \\cap $$ B) = $${1 \\over 4}$$ .....(1)\n
P (Exactly one of B or C occurs)\n
= P(B) + P (C) – 2P (B $$ \\cap $$ C) = $${1 \\over 4}$$ .....(2)\n
- P (B $$ \\cap $$ C) - P (C $$ \\cap $$ A)] = $${3 \\over 4}$$\n
$$ \\Rightarrow $$ P(A) + P(B) + P (C) - P (A $$ \\cap $$ B)\n
- P (B $$ \\cap $$ C) - P (C $$ \\cap $$ A) = $${3 \\over 8}$$\n
$$ \\therefore $$ P(atleast one event occurs)\n
= P (A $$ \\cup $$ B $$ \\cup $$ C)\n
= P(A) + P(B) + P (C) - P (A $$ \\cap $$ B)\n
- P (B $$ \\cap $$ C) - P (C $$ \\cap $$ A) + P (A $$ \\cap $$ B $$ \\cap $$ C)\n
= $${3 \\over 8} + {1 \\over {16}}$$ = $${7 \\over {16}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6774,
"subject": "General Science",
"question": "In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the students selected has opted neither for NCC\nnor for NSS is :",
"options": [
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${1 \\over 6}$$"
},
{
"text": "$${2 \\over 3}$$"
},
{
"text": "$${5 \\over 6}$$"
}
],
"answer": "$${1 \\over 6}$$",
"solution": "**Answer:** $${1 \\over 6}$$\n\n\n A $$ \\to $$ opted NCC\n
B $$ \\to $$ opted NSS\n
$$ \\therefore $$ P (nither A nor B) $$=$$ $${{10} \\over {60}} = z{1 \\over 6}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6775,
"subject": "General Science",
"question": "Let A and B be two events such that the\nprobability that exactly one of them occurs is $${2 \\over 5}$$ and the probability that A or B occurs is $${1 \\over 2}$$ ,\nthen the probability of both of them occur\ntogether is :",
"options": [
{
"text": "0.20"
},
{
"text": "0.02"
},
{
"text": "0.01"
},
{
"text": "0.10"
}
],
"answer": "0.10",
"solution": "**Answer:** 0.10\n\nProbability that exactly one of them occurs\n
P(A) + P(B) – 2P (A $$ \\cap $$ B) = $${2 \\over 5}$$ .....(1)\n
$$ \\Rightarrow $$ 0.25 $$ \\le \\beta \\le $$ 0.35",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6777,
"subject": "General Science",
"question": "Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 $$-$$ k), the probability that exactly one of B and C occurs is (1 $$-$$ 2k), the probability that exactly one of C and A occurs is (1 $$-$$ k) and the probability of all A, B and C occur simultaneously is k2, where 0 < k < 1. Then the probability that at least one of A, B and C occur is :",
"options": [
{
"text": "greater than $${1 \\over 8}$$ but less than $${1 \\over 4}$$"
},
{
"text": "greater than $${1 \\over 2}$$"
},
{
"text": "greater than $${1 \\over 4}$$ but less than $${1 \\over 2}$$"
},
{
"text": "exactly equal to $${1 \\over 2}$$"
}
],
"answer": "greater than $${1 \\over 2}$$",
"solution": "**Answer:** greater than $${1 \\over 2}$$\n\n$$P(\\overline A \\cap B) + P(A \\cap \\overline B ) = 1 - k$$
$$P(\\overline A \\cap C) + P(A \\cap \\overline C ) = 1 - 2k$$
$$P(\\overline B \\cap C) + P(B \\cap \\overline C ) = 1 - k$$
$$P(A \\cap B \\cap C) = {k^2}$$
$$P(A) + P(B) - 2P(A \\cap B) = 1 - k$$ .....(i)
$$P(B) + P(C) - 2P(B \\cap C) = 1 - k$$ ..... (ii)
Let $$\\mathrm{E}_{1}, \\mathrm{E}_{2}, \\mathrm{E}_{3}$$ be three mutually exclusive events such that $$\\mathrm{P}\\left(\\mathrm{E}_{1}\\right)=\\frac{2+3 \\mathrm{p}}{6}, \\mathrm{P}\\left(\\mathrm{E}_{2}\\right)=\\frac{2-\\mathrm{p}}{8}$$ and $$\\mathrm{P}\\left(\\mathrm{E}_{3}\\right)=\\frac{1-\\mathrm{p}}{2}$$. If the maximum and minimum values of $$\\mathrm{p}$$ are $$\\mathrm{p}_{1}$$ and $$\\mathrm{p}_{2}$$, then $$\\left(\\mathrm{p}_{1}+\\mathrm{p}_{2}\\right)$$ is equal to :
Let $$S=\\{1,2,3, \\ldots, 2022\\}$$. Then the probability, that a randomly chosen number n from the set S such that $$\\mathrm{HCF}\\,(\\mathrm{n}, 2022)=1$$, is :
Two dice are thrown independently. Let $$\\mathrm{A}$$ be the event that the number appeared on the $$1^{\\text {st }}$$ die is less than the number appeared on the $$2^{\\text {nd }}$$ die, $$\\mathrm{B}$$ be the event that the number appeared on the $$1^{\\text {st }}$$ die is even and that on the second die is odd, and $$\\mathrm{C}$$ be the event that the number appeared on the $$1^{\\text {st }}$$ die is odd and that on the $$2^{\\text {nd }}$$ is even. Then :
",
"options": [
{
"text": "A and B are mutually exclusive"
},
{
"text": "the number of favourable cases of the events A, B and C are 15, 6 and 6 respectively"
},
{
"text": "B and C are independent"
},
{
"text": "the number of favourable cases of the event $$(\\mathrm{A\\cup B)\\cap C}$$ is 6"
}
],
"answer": "the number of favourable cases of the event $$(\\mathrm{A\\cup B)\\cap C}$$ is 6",
"solution": "**Answer:** the number of favourable cases of the event $$(\\mathrm{A\\cup B)\\cap C}$$ is 6\n\n$\\begin{aligned} & A=\\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\\} \\\\\\\\ & n(A)=15 \\\\\\\\ & B=\\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\\} \\\\\\\\ & n(B)=9 \\\\\\\\ & C=\\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\\} \\\\\\\\ & n(C)=9\\end{aligned}$\n
$$\n(4,5) \\in A \\text { and }(4,5) \\in B\n$$\n
$\\therefore A$ and $B$ are not exclusive events\n
$$\n\\begin{aligned}\n& n((A \\cup B) \\cap C)=n(A \\cap C)+n(B \\cap C)-n(A \\cap B \\cap C) \\\\\\\\\n= & 3+3-0 \\\\\\\\\n= & 6\n\\end{aligned}\n$$\n
Option (D) is correct.\n
$$\n\\begin{aligned}\n& n(B)=\\frac{9}{36}, n(C)=\\frac{9}{36}, n(B \\cap C)=0 \\\\\\\\\n& \\Rightarrow n(B) \\cdot n(C) \\neq n(B \\cap C) \\\\\\\\\n& \\therefore B \\text { and } C \\text { are not independent. }\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6782,
"subject": "General Science",
"question": "
A group of 40 students appeared in an examination of 3 subjects - Mathematics, Physics and Chemistry. It was found that all students passed in atleast one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, atmost 11 students passed in both Mathematics and Physics, atmost 15 students passed in both Physics and Chemistry, atmost 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is _________.
$$ = 1 + 2\\sqrt 5 $$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 6786,
"subject": "General Science",
"question": "The triangle of maximum area that can be inscribed in a given circle of radius 'r' is :",
"options": [
{
"text": "An equilateral triangle having each of its side of length $$\\sqrt 3 $$r."
},
{
"text": "An equilateral triangle of height $${{2r} \\over 3}$$."
},
{
"text": "A right angle triangle having two of its sides of length 2r and r."
},
{
"text": "An isosceles triangle with base equal to 2r."
}
],
"answer": "An equilateral triangle having each of its side of length $$\\sqrt 3 $$r.",
"solution": "**Answer:** An equilateral triangle having each of its side of length $$\\sqrt 3 $$r.\n\nArea of triangle ABC
Sign change of $${{dA} \\over {dx}}$$ at $$x = {r \\over 2}$$
$$ \\Rightarrow $$ A has maximum at $$x = {r \\over 2}$$\n
$$BC = 2\\sqrt {{r^2} - {x^2}} = \\sqrt 3 r$$,
$$AM = r + {1 \\over 2}r$$ = $${3 \\over 2}r$$
$$ \\Rightarrow AB = AC = \\sqrt 3 r$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6787,
"subject": "General Science",
"question": "Let ABCD be a square of side of unit length. Let a circle C1 centered at A with unit radius is drawn. Another circle C2 which touches C1 and the lines AD and AB are tangent to it, is also drawn. Let a tangent line from the point C to the circle C2 meet the side AB at E. If the length of EB is $$\\alpha$$ + $${\\sqrt 3 }$$ $$\\beta$$, where $$\\alpha$$, $$\\beta$$ are integers, then $$\\alpha$$ + $$\\beta$$ is equal to ____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n
$$ \\therefore $$ $$\\alpha$$ = 2, $$\\beta$$ = $$-$$1 $$ \\Rightarrow $$ $$\\alpha$$ + $$\\beta$$ = 1",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6788,
"subject": "General Science",
"question": "In $$\\Delta$$ABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $$\\Delta$$ABC is 30 cm2 and R and r are respectively the radii of circumcircle and incircle of $$\\Delta$$ABC, then the value of 2R + r (in cm) is equal to ___________.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\n
The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 $$-$$ 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to :
A straight line cuts off the intercepts $$\\mathrm{OA}=\\mathrm{a}$$ and $$\\mathrm{OB}=\\mathrm{b}$$ on the positive directions of $$x$$-axis and $$y$$ axis respectively. If the perpendicular from origin $$O$$ to this line makes an angle of $$\\frac{\\pi}{6}$$ with positive direction of $$y$$-axis and the area of $$\\triangle \\mathrm{OAB}$$ is $$\\frac{98}{3} \\sqrt{3}$$, then $$\\mathrm{a}^{2}-\\mathrm{b}^{2}$$ is equal to :
Let $$\\left(5, \\frac{a}{4}\\right)$$ be the circumcenter of a triangle with vertices $$\\mathrm{A}(a,-2), \\mathrm{B}(a, 6)$$ and $$C\\left(\\frac{a}{4},-2\\right)$$. Let $$\\alpha$$ denote the circumradius, $$\\beta$$ denote the area and $$\\gamma$$ denote the perimeter of the triangle. Then $$\\alpha+\\beta+\\gamma$$ is
Two vertices of a triangle $$\\mathrm{ABC}$$ are $$\\mathrm{A}(3,-1)$$ and $$\\mathrm{B}(-2,3)$$, and its orthocentre is $$\\mathrm{P}(1,1)$$. If the coordinates of the point $$\\mathrm{C}$$ are $$(\\alpha, \\beta)$$ and the centre of the of the circle circumscribing the triangle $$\\mathrm{PAB}$$ is $$(\\mathrm{h}, \\mathrm{k})$$, then the value of $$(\\alpha+\\beta)+2(\\mathrm{~h}+\\mathrm{k})$$ equals
Consider a triangle $$\\mathrm{ABC}$$ having the vertices $$\\mathrm{A}(1,2), \\mathrm{B}(\\alpha, \\beta)$$ and $$\\mathrm{C}(\\gamma, \\delta)$$ and angles $$\\angle A B C=\\frac{\\pi}{6}$$ and $$\\angle B A C=\\frac{2 \\pi}{3}$$. If the points $$\\mathrm{B}$$ and $$\\mathrm{C}$$ lie on the line $$y=x+4$$, then $$\\alpha^2+\\gamma^2$$ is equal to _______.
In a triangle $$\\mathrm{ABC}, \\mathrm{BC}=7, \\mathrm{AC}=8, \\mathrm{AB}=\\alpha \\in \\mathrm{N}$$ and $$\\cos \\mathrm{A}=\\frac{2}{3}$$. If $$49 \\cos (3 \\mathrm{C})+42=\\frac{\\mathrm{m}}{\\mathrm{n}}$$, where $$\\operatorname{gcd}(m, n)=1$$, then $$m+n$$ is equal to _________.
or $$a,b,c$$ are in $$A.P.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6798,
"subject": "General Science",
"question": "For a regular polygon, let $$r$$ and $$R$$ be the radii of the inscribed and the circumscribed circles. A $$false$$ statement among the following is :",
"options": [
{
"text": "There is a regular polygon with $${r \\over R} = {1 \\over {\\sqrt 2 }}$$ "
},
{
"text": "There is a regular polygon with $${r \\over R} = {2 \\over 3}$$ "
},
{
"text": "There is a regular polygon with $${r \\over R} = {{\\sqrt 3 } \\over 2}$$ "
},
{
"text": "There is a regular polygon with $${r \\over R} = {1 \\over 2}$$ "
}
],
"answer": "There is a regular polygon with $${r \\over R} = {2 \\over 3}$$ ",
"solution": "**Answer:** There is a regular polygon with $${r \\over R} = {2 \\over 3}$$ \n\n\n
$$= {{121} \\over {256}} - {3 \\over 4} = - {{71} \\over {256}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6803,
"subject": "General Science",
"question": "In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x2 – c2 = y, where c is the length of the third side of the triangle, then the circumradius of the triangle is :",
"options": [
{
"text": "$${y \\over {\\sqrt 3 }}$$"
},
{
"text": "$${c \\over 3}$$"
},
{
"text": "$${c \\over {\\sqrt 3 }}$$"
},
{
"text": "$${3 \\over 2}$$y"
}
],
"answer": "$${c \\over {\\sqrt 3 }}$$",
"solution": "**Answer:** $${c \\over {\\sqrt 3 }}$$\n\nGiven a + b = x and ab = y\n
If x2 $$-$$ c2 = y $$ \\Rightarrow $$ (a + b)2 $$-$$ c2 = ab\n
$$ \\Rightarrow \\angle C = {{2\\pi } \\over 3}$$\n
$$R = {c \\over {2\\sin C}} = {c \\over {\\sqrt 3 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6804,
"subject": "General Science",
"question": "The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 : $$\\sqrt 3 $$. If c = 4 cm, then the area (in sq. cm)\nof this triangle is :",
"options": [
{
"text": "2$$\\sqrt 3 $$"
},
{
"text": "4$$\\sqrt 3 $$"
},
{
"text": "$${4 \\over {\\sqrt 3 }}$$"
},
{
"text": "$${2 \\over {\\sqrt 3 }}$$"
}
],
"answer": "2$$\\sqrt 3 $$",
"solution": "**Answer:** 2$$\\sqrt 3 $$\n\nFrom the question 2B = A + C & A + B + C = $$\\pi $$
\n$$ \\Rightarrow $$ 3B = $$\\pi $$
\n$$ \\Rightarrow $$ B = $${\\pi \\over 3}$$
\n$$ \\therefore A + C = {{2\\pi } \\over 3}\\sigma $$
$$\\Rightarrow$$ Option (3) is correct.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6806,
"subject": "General Science",
"question": "Let $${{\\sin A} \\over {\\sin B}} = {{\\sin (A - C)} \\over {\\sin (C - B)}}$$, where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :",
"options": [
{
"text": "b2 $$-$$ a2 = a2 + c2"
},
{
"text": "b2, c2, a2 are in A.P."
},
{
"text": "c2, a2, b2 are in A.P."
},
{
"text": "a2, b2, c2 are in A.P."
}
],
"answer": "b2, c2, a2 are in A.P.",
"solution": "**Answer:** b2, c2, a2 are in A.P.\n\n$${{\\sin A} \\over {\\sin B}} = {{\\sin (A - C)} \\over {\\sin (C - B)}}$$
As A, B, C are angles of triangle.
A + B + C = $$\\pi$$
A = $$\\pi$$ $$-$$ (B + C) ...... (1)
Similarly sinB = sin(A + C) ..... (2)
From (1) and (2)
$${{\\sin (B + C)} \\over {\\sin (A + C)}} = {{\\sin (A - C)} \\over {\\sin (C - B)}}$$
$$\\sin (C + B).\\sin (C - B) = \\sin (A - C)\\sin (A + C)$$
$$ \\Rightarrow $$ (ex – 1)2 = 0 $$ \\Rightarrow $$ x = 0.\n
$$ \\therefore $$ Number of real roots = 1",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6808,
"subject": "General Science",
"question": "The sum of 162th power of the roots of the equation x3 $$-$$ 2x2 + 2x $$-$$ 1 = 0 is ________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nx3 $$-$$ 2x2 + 2x $$-$$ 1 = 0
$$\\Rightarrow$$ No. of real roots = 2",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6810,
"subject": "General Science",
"question": "The number of real roots of the equation e4x $$-$$ e3x $$-$$ 4e2x $$-$$ ex + 1 = 0 is equal to ______________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nt4 $$-$$ t3 $$-$$ 4t2 $$-$$ t + 1 = 0, ex = t > 0
$$\\Rightarrow$$ The number of real roots = 2",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6811,
"subject": "General Science",
"question": "Let $$\\alpha$$, $$\\beta$$ be two roots of the
$$ \\Rightarrow $$ $$\\alpha$$8 + $$\\beta$$8 = 50",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6812,
"subject": "General Science",
"question": "The sum of the roots of the equation
$$\\Rightarrow x_1+x_2+x_3 =\\log _e 45=\\log _e p $$\n
$$\\Rightarrow p = 45$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6816,
"subject": "General Science",
"question": "
If $$\\alpha, \\beta, \\gamma, \\delta$$ are the roots of the equation $$x^{4}+x^{3}+x^{2}+x+1=0$$, then $$\\alpha^{2021}+\\beta^{2021}+\\gamma^{2021}+\\delta^{2021}$$ is equal to :
Let the set $$C=\\left\\{(x, y) \\mid x^2-2^y=2023, x, y \\in \\mathbb{N}\\right\\}$$. Then $$\\sum_\\limits{(x, y) \\in C}(x+y)$$ is equal to _________.
First, let's consider the equation $$x^2 - 2^y = 2023$$ where $$x$$ and $$y$$ are natural numbers. Our goal is to find all the pairs $$(x, y)$$ that satisfy this equation and then sum the values of $$x+y$$ for each pair in set $$C$$.\n\n
Since $$2023$$ is an odd number, and $$x^2$$, the square of any natural number, is even when $$x$$ is even and odd when $$x$$ is odd, we can determine that for the left-hand side of the equation to be odd (thus equal to $$2023$$), $$x$$ must be odd since the right-hand side of the equation ($$2^y$$) is always even as it represents a power of two.
\n\n
Also, $$2023$$ can be factored into prime factors to further analyze the possible solutions:
\n\n
$$2023 = 7 \\times 17 \\times 17$$
\n\n
Thus, allowing us to rewrite the equation as:
\n\n
$$x^2 - 2^y = 7 \\times 17^2$$
\n\n
The next step is to check for potential values of $$x$$ that would fit the equation, keeping in mind that $$x$$ must be odd. We can try to express $$x^2$$ as $$7 \\times 17^2$$ plus a power of $$2$$, recognizing that we are looking for the decomposition of the form:
\n\n
$$x^2 = 7 \\times 17^2 + 2^y$$
\n\n
By examining the powers of $$2$$ and keeping in mind that they grow very quickly, we can reason that $$y$$ cannot be very large because $$x^2$$ must not exceed $$2023$$ by a large margin.
\n\n
Let's start by trying the lowest values for $$y$$ since that would make $$2^y$$ small and $$x$$ has a better chance of being a natural number:
\n\n\n
For $$y=1$$:
\n\n
$$x^2 = 2023 + 2^1 = 2023 + 2 = 2025$$
\n\n
Surprisingly, we find a perfect square since $$45^2 = 2025$$. Therefore, $$(x, y) = (45, 1)$$ is one solution.
\n\n\n
For $$y=2$$ or higher:
\n\n
$$2^y$$ becomes at least $$4$$ and increases exponentially, so $$x^2$$ must be at least $$2027$$ or higher in such cases. There's no natural number between $$45$$ and $$46$$, and $$46^2$$ far exceeds the target (2116), making it impossible for $$x^2$$ to be less than $$2116$$ for any larger $$y$$.
\n\n
Hence, it appears there is only one possible solution: $$(x, y) = (45, 1)$$.
\n\n
Therefore, the sum $$\\sum_\\limits{(x, y) \\in C}(x+y)$$ for this set will consist of only this one pair:
Let $$x_1, x_2, x_3, x_4$$ be the solution of the equation $$4 x^4+8 x^3-17 x^2-12 x+9=0$$ and $$\\left(4+x_1^2\\right)\\left(4+x_2^2\\right)\\left(4+x_3^2\\right)\\left(4+x_4^2\\right)=\\frac{125}{16} m$$. Then the value of $$m$$ is _________.
$$\\begin{aligned}\n& (4+1)(4+9)\\left(4+\\frac{1}{4}\\right)\\left(4+\\frac{9}{4}\\right)=\\frac{125}{16} m \\\\\n& 5 \\times 13 \\times\\left(\\frac{17}{4}\\right) \\times\\left(\\frac{25}{4}\\right)=\\frac{125}{16} m \\\\\n& \\frac{125}{16} \\times[13 \\times 17]=\\frac{125}{16} m \\\\\n& m=13 \\times 17 \\\\\n& m=221\n\\end{aligned}$$
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6820,
"subject": "General Science",
"question": "If one root of the equation $${x^2} + px + 12 = 0$$ is 4, while the equation $${x^2} + px + q = 0$$ has equal roots, \n then the value of $$'q'$$ is ",
"options": [
{
"text": "4 "
},
{
"text": "12 "
},
{
"text": "3 "
},
{
"text": "$${{49} \\over 4}$$ "
}
],
"answer": "$${{49} \\over 4}$$ ",
"solution": "**Answer:** $${{49} \\over 4}$$ \n\n$$4$$ is a root of $${x^2} + px + 12 = 0$$\n
$$ \\Rightarrow 16 + 4p + 12 = 0$$\n
$$ \\Rightarrow p = - 7$$\n
Now, the equation $${x^2} + px + q = 0$$\n
has equal roots. \n
$$\\therefore$$ $${p^2} - 4q = 0$$ $$ \\Rightarrow q = {{{p^2}} \\over 4} = {{49} \\over 4}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6821,
"subject": "General Science",
"question": "The quadratic equations $${x^2} - 6x + a = 0$$ and $${x^2} - cx + 6 = 0$$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is ",
"options": [
{
"text": "1 "
},
{
"text": "4 "
},
{
"text": "3 "
},
{
"text": "2"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nLet the roots of equation $${x^2} - 6x + a = 0$$ be $$\\alpha $$ \n
and $$4$$ $$\\beta $$ and that of the equation \n
$${x^2} - cx + 6 = 0$$ be $$\\alpha $$ and $$3\\beta .$$ Then\n
$$ \\therefore $$ $${\\alpha ^2} + {\\beta ^2}$$ = 20 + 5 = 25",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6826,
"subject": "General Science",
"question": "Let $$\\lambda$$ $$\\ne$$ 0 be in R. If $$\\alpha$$ and $$\\beta$$ are the roots of the equation x2 $$-$$ x + 2$$\\lambda$$ = 0, and $$\\alpha$$ and $$\\gamma$$ are the roots of equation 3x2 $$-$$ 10x + 27$$\\lambda$$ = 0, then $${{\\beta \\gamma } \\over \\lambda }$$ is equal to ____________.",
"options": [],
"answer": "18",
"solution": "**Answer:** 18\n\n3$$\\alpha$$2 $$-$$ 10$$\\alpha$$ + 27$$\\lambda$$ = 0 ..... (1)
Let a, b $$\\in$$ R be such that the equation $$a{x^2} - 2bx + 15 = 0$$ has a repeated root $$\\alpha$$. If $$\\alpha$$ and $$\\beta$$ are the roots of the equation $${x^2} - 2bx + 21 = 0$$, then $${\\alpha ^2} + {\\beta ^2}$$ is equal to :
If for some $$\\mathrm{p}, \\mathrm{q}, \\mathrm{r} \\in \\mathbf{R}$$, not all have same sign, one of the roots of the equation $$\\left(\\mathrm{p}^{2}+\\mathrm{q}^{2}\\right) x^{2}-2 \\mathrm{q}(\\mathrm{p}+\\mathrm{r}) x+\\mathrm{q}^{2}+\\mathrm{r}^{2}=0$$ is also a root of the equation $$x^{2}+2 x-8=0$$, then $$\\frac{\\mathrm{q}^{2}+\\mathrm{r}^{2}}{\\mathrm{p}^{2}}$$ is equal to ____________,
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6829,
"subject": "General Science",
"question": "If the value of real number $a>0$ for which $x^2-5 a x+1=0$ and $x^2-a x-5=0$\n
have a common real root is $\\frac{3}{\\sqrt{2 \\beta}}$ then $\\beta$ is equal to ___________.",
"options": [],
"answer": "13",
"solution": "**Answer:** 13\n\n
$${x^2} - 5\\alpha x + 1 = 0$$ ..... (1)
\n
$${x^2} - \\alpha x - 5 = 0$$ ...... (2)
\n
have a common root.
\n
Subtracting (1) with (2) we'll get $$x = {6 \\over {4\\alpha }}$$
$$\\because$$$$ + \\sqrt {4 + 12{a^2}} $$ is always positive which is greater than any negative no. So can ignore the inequality $$ + \\sqrt {4 + 12{a^2}} > - 4$$\n
If $$a$$ = 0 then $$ - 3{\\left\\{ x \\right\\}^2} + 2\\left\\{ x \\right\\} = 0$$ so $$\\left\\{ x \\right\\}$$ becomes 0 but question says $$\\left\\{ x \\right\\}$$ $$ \\ne $$ 0.\n So $$a$$ can't be 0.\n
So, the range of $$a$$ is = $$\\left( { - 1,0} \\right) \\cup \\left( {0,1} \\right)$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6832,
"subject": "General Science",
"question": "Let [t] denote the greatest integer $$ \\le $$ t. Then the equation in x,\n [x]2 + 2[x+2] - 7 = 0 has :",
"options": [
{
"text": "no integral solution."
},
{
"text": "exactly two solutions."
},
{
"text": "exactly four integral solutions."
},
{
"text": "infinitely many solutions.\n"
}
],
"answer": "infinitely many solutions.\n",
"solution": "**Answer:** infinitely many solutions.\n\n\n$${[x]^2} + 2[x + 2] - 7 = 0$$\n
or $$[{e^x}] = 1$$ $$\\therefore$$ $$1 \\le {e^x} < 2$$
$$ \\Rightarrow \\ln (1) \\le x < \\ln (2)$$
$$ \\Rightarrow 0 \\le x < \\ln (2)$$
$$ \\Rightarrow x \\in [0,\\ln 2)$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6834,
"subject": "General Science",
"question": "If [x] be the greatest integer less than or equal to x,
The equation $${x^2} - 4x + [x] + 3 = x[x]$$, where $$[x]$$ denotes the greatest integer function, has :
",
"options": [
{
"text": "exactly two solutions in ($$-\\infty,\\infty$$)"
},
{
"text": "no solution"
},
{
"text": "a unique solution in ($$-\\infty,\\infty$$)"
},
{
"text": "a unique solution in ($$-\\infty,1$$)"
}
],
"answer": "a unique solution in ($$-\\infty,\\infty$$)",
"solution": "**Answer:** a unique solution in ($$-\\infty,\\infty$$)\n\n
Let $$[\\alpha]$$ denote the greatest integer $$\\leq \\alpha$$. Then $$[\\sqrt{1}]+[\\sqrt{2}]+[\\sqrt{3}]+\\ldots+[\\sqrt{120}]$$ is equal to __________
Hence both the statements are correct and statement $$2$$ is a correct explanation of statement $$-1.$$ ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6840,
"subject": "General Science",
"question": "All the pairs (x, y) that satisfy the inequality\n $${2^{\\sqrt {{{\\sin }^2}x - 2\\sin x + 5} }}.{1 \\over {{4^{{{\\sin }^2}y}}}} \\le 1$$\n also satisfy the equation",
"options": [
{
"text": "sin x = |sin y|"
},
{
"text": "sin x = 2sin y"
},
{
"text": "2 sin x = sin y"
},
{
"text": "2 |sin x | = 3 sin y"
}
],
"answer": "sin x = |sin y|",
"solution": "**Answer:** sin x = |sin y|\n\n$${2^{\\sqrt {{{\\sin }^2}x - 2\\sin x + 5} }} \\le {2^{2{{\\sin }^2}y}}$$
then k = 3",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6843,
"subject": "General Science",
"question": "If a + b + c = 1, ab + bc + ca = 2 and abc = 3, then the value of a4 + b4 + c4 is equal to ______________.",
"options": [],
"answer": "13",
"solution": "**Answer:** 13\n\n(a + b + c)2 = 1\n
Let $$\\mathrm{S}$$ be the set of positive integral values of $$a$$ for which $$\\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32} < 0, \\forall x \\in \\mathbb{R}$$. Then, the number of elements in $$\\mathrm{S}$$ is :
",
"options": [
{
"text": "0"
},
{
"text": "$$\\infty$$"
},
{
"text": "3"
},
{
"text": "1"
}
],
"answer": "0",
"solution": "**Answer:** 0\n\n$x^2-8 x+32>0 \\forall x \\in R$ as discriminant of this quadratic is $64-4 \\times 32<0$\n
$$\n\\Rightarrow a x^2+2(a+1) x+9 a+4<0 \\forall x \\in R\n$$\n
$\\Rightarrow$ Only possible when $a<0$ and $D<0$\n
$\\Rightarrow$ Since $S$ is set of positive\nvalues of $a \\Rightarrow S$ is a null set\n
$$\n\\Rightarrow n(S)=0\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6846,
"subject": "General Science",
"question": "All the values of $$m$$ for which both roots of the equation $${x^2} - 2mx + {m^2} - 1 = 0$$ are greater than $$ - 2$$ but less then 4, lie in the interval ",
"options": [
{
"text": "$$ - 2 < m < 0$$ "
},
{
"text": "$$m > 3$$ "
},
{
"text": "$$ - 1 < m < 3$$ "
},
{
"text": "$$1 < m < 4$$ "
}
],
"answer": "$$ - 1 < m < 3$$ ",
"solution": "**Answer:** $$ - 1 < m < 3$$ \n\nEquation $${x^2} - 2mx + {m^2} - 1 = 0$$\n
$${\\left( {x - m} \\right)^2} - 1 = 0$$\n
or $$\\left( {x - m + 1} \\right)\\left( {x - m - 1} \\right) = 0$$\n
$$x = m - 1,m + 1$$\n
$$m - 1 > - 2$$ and $$m + 1 < 4$$\n
$$ \\Rightarrow m > - 1$$ and $$m<3$$ \n
or $$\\,\\,\\, - 1 < m < 3$$\n",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6847,
"subject": "General Science",
"question": "If both the roots of the quadratic equation x2 $$-$$ mx + 4 = 0 are real and distinct and they lie in the interval [1, 5], then m lies in the interval : ",
"options": [
{
"text": "($$-$$5, $$-$$4)"
},
{
"text": "(4, 5)"
},
{
"text": "(5, 6)"
},
{
"text": "(3, 4)"
}
],
"answer": "(4, 5)",
"solution": "**Answer:** (4, 5)\n\nx2 $$-$$mx + 4 = 0\n
So given option is (4, 5)\n",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6848,
"subject": "General Science",
"question": "Consider the quadratic equation (c – 5)x2 – 2cx + (c – 4) = 0, c $$ \\ne $$ 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is -",
"options": [
{
"text": "12"
},
{
"text": "18"
},
{
"text": "10"
},
{
"text": "11"
}
],
"answer": "11",
"solution": "**Answer:** 11\n\n\n
Number of elements in set S = 11",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6849,
"subject": "General Science",
"question": "The set of all real values of $$\\lambda $$ for which the\nquadratic equations, ($$\\lambda $$2\n + 1)x2\n – 4$$\\lambda $$x + 2 = 0\nalways have exactly one root in the interval\n(0, 1) is :",
"options": [
{
"text": "(–3, –1)"
},
{
"text": "(2, 4]"
},
{
"text": "(0, 2)"
},
{
"text": "(1, 3]"
}
],
"answer": "(1, 3]",
"solution": "**Answer:** (1, 3]\n\nGiven quadratic equation,
The number of integral values of k, for which one root of the equation $$2x^2-8x+k=0$$ lies in the interval (1, 2) and its other root lies in the interval (2, 3), is :
Only 1 integral value of $k$ is 7.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6851,
"subject": "General Science",
"question": "Product of real roots of equation $${t^2}{x^2} + \\left| x \\right| + 9 = 0$$ ",
"options": [
{
"text": "is always positive "
},
{
"text": "is always negative "
},
{
"text": "does not exist"
},
{
"text": "none of these "
}
],
"answer": "is always positive ",
"solution": "**Answer:** is always positive \n\nProduct of real roots $$ = {9 \\over {{t^2}}} > 0,\\forall \\,t\\, \\in R$$\n
$$\\therefore$$ Product of real roots is always positive. ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6852,
"subject": "General Science",
"question": "The number of real solutions of the equation $${x^2} - 3\\left| x \\right| + 2 = 0$$ is ",
"options": [
{
"text": "3 "
},
{
"text": "2 "
},
{
"text": "4 "
},
{
"text": "1 "
}
],
"answer": "4 ",
"solution": "**Answer:** 4 \n\n$${x^2} - 3\\left| x \\right| + 2 = 0$$\n
$$ \\Rightarrow {\\left| x \\right|^2} - 3\\left| x \\right| + 2 = 0$$\n
$$\\left( {\\left| x \\right| - 2} \\right)\\left( {\\left| x \\right| - 1} \\right) = 0$$\n
$$\\left| x \\right| = 1,2$$ or $$x = \\pm 1, \\pm 2$$\n
$$\\therefore$$ No. of solution $$=4$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6853,
"subject": "General Science",
"question": "Let S = { $$x$$ $$ \\in $$ R : $$x$$ $$ \\ge $$ 0 and\n
$$2\\left| {\\sqrt x - 3} \\right| + \\sqrt x \\left( {\\sqrt x - 6} \\right) + 6 = 0$$}. Then S",
"options": [
{
"text": "contains exactly four elements"
},
{
"text": "is an empty set"
},
{
"text": "contains exactly one element"
},
{
"text": "contains exactly two elements"
}
],
"answer": "contains exactly two elements",
"solution": "**Answer:** contains exactly two elements\n\nGiven, \n
$$2\\left| {\\sqrt x - 3} \\right| + \\sqrt x \\left( {\\sqrt x - 6} \\right) + 6 = 0$$\n
Case 1 : \n
When $$\\sqrt x - 3 \\ge 0,$$ then equation becomes \n
$$2\\left( {\\sqrt x - 3} \\right) + \\sqrt x \\left( {\\sqrt x - 6} \\right) + 6 = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,2\\sqrt x - 6 + x - 6\\sqrt x + 6 = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,x\\, - 4\\sqrt x = 0$$ \n
$$ \\Rightarrow \\,\\,\\,\\,\\sqrt x \\left( {\\sqrt x - 4} \\right) = 0$$\n
$$\\therefore\\,\\,\\,$$ $$\\sqrt x = 0,4$$\n
but as $$\\sqrt x - 3 \\ge 0$$ or $$\\sqrt x \\ge 3$$ then $$\\sqrt x \\ne 0$$ \n
$$\\therefore\\,\\,\\,$$ $$\\sqrt x = 4$$ value is possible. \n
Case 2 : \n
When $$\\sqrt x - 3 < 0$$ or $$\\sqrt x < 3.$$ There equation becomes\n
$$ - 2\\left( {\\sqrt x - 3} \\right) + \\sqrt x \\left( {\\sqrt x - 6} \\right) + 6 = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,\\, - 2\\sqrt x + 6 + x - 6\\sqrt x + 6 = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,x - 8\\sqrt x + 12 = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,x - 6\\sqrt x - 2\\sqrt x + 12 = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,\\sqrt x \\left( {\\sqrt x - 6} \\right) - 2\\left( {\\sqrt x - 6} \\right) = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,\\left( {\\sqrt x - 2} \\right)\\left( {\\sqrt x - 6} \\right) = 0$$\n
$$\\therefore\\,\\,\\,$$ $$\\sqrt x = 2,6$$ \n
as $$\\sqrt x < 3$$ so $$\\sqrt x \\ne 6$$ \n
$$\\therefore\\,\\,\\,$$ $$\\sqrt x = 2$$ is possible. \n
So, total possible value of $$\\sqrt x = 2,4$$\n
or for x possible values are 4, 16.\n
$$\\therefore\\,\\,\\,$$ Set S contains exactly two elements 4 and 16.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6854,
"subject": "General Science",
"question": "The sum of the solutions of the equation \n$$\\left| {\\sqrt x - 2} \\right| + \\sqrt x \\left( {\\sqrt x - 4} \\right) + 2 = 0$$ \n(x > 0) is equal to:",
"options": [
{
"text": "9"
},
{
"text": "12"
},
{
"text": "4"
},
{
"text": "10"
}
],
"answer": "10",
"solution": "**Answer:** 10\n\nCase 1 : When $$\\sqrt x \\ge 2$$\n
then $$\\left| {\\sqrt x - 2} \\right| = \\sqrt x - 2$$\n
$$ \\therefore $$ The given equation becomes,\n
$$\\left( {\\sqrt x - 2} \\right)$$ + $$\\sqrt x \\left( {\\sqrt x - 4} \\right) + 2$$ = 0\n
$$ \\Rightarrow $$ $$\\left( {\\sqrt x - 2} \\right)$$ + $$x - 4\\sqrt x $$ + 2 = 0\n
$$ \\Rightarrow $$ $$x - 3\\sqrt x $$ = 0\n
$$ \\Rightarrow $$ $$\\sqrt x \\left( {\\sqrt x - 3} \\right)$$ = 0\n
$$ \\therefore $$ $$\\sqrt x $$ = 0 or 3\n
$$\\sqrt x $$ = 0 is not possible as $$\\sqrt x \\ge 2$$.\n
So, $$\\sqrt x $$ = 3\n
or $$x$$ = 9\n
Case 2 : When $$\\sqrt x < 2$$\n
then $$\\left| {\\sqrt x - 2} \\right| = $$$$ - \\left( {\\sqrt x - 2} \\right)$$ = $$2 - \\sqrt x $$\n
$$ \\therefore $$ The given equation becomes,\n
$$\\left( {2 - \\sqrt x } \\right)$$ + $$\\sqrt x \\left( {\\sqrt x - 4} \\right) + 2$$ = 0\n
$$ \\Rightarrow $$ $${2 - \\sqrt x }$$ + $$x - 4\\sqrt x $$ + 2 = 0\n
$$ \\Rightarrow $$ $$x - 5\\sqrt x + 4$$ = 0\n
$$ \\Rightarrow $$ $$x - 4\\sqrt x - \\sqrt x + 4$$ = 0\n
$$ \\Rightarrow $$ $$\\sqrt x \\left( {\\sqrt x - 4} \\right)$$$$-\\left( {\\sqrt x - 4} \\right)$$ = 0\n
$$ \\Rightarrow $$ $$\\left( {\\sqrt x - 4} \\right)$$$$\\left( {\\sqrt x - 1} \\right)$$ = 0\n
$$ \\therefore $$ $$\\sqrt x $$ = 4 or 1\n
$$\\sqrt x $$ = 4 is not possible as $$\\sqrt x < 2$$.\n
$$ \\therefore $$ $$\\sqrt x $$ = 1\n
or $$x$$ = 1\n
So, Sum of all solutions = 9 + 1 = 10",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6855,
"subject": "General Science",
"question": "The number of real roots of the equation\n 5 + |2x\n– 1| = 2x\n(2x\n– 2) is",
"options": [
{
"text": "2"
},
{
"text": "1"
},
{
"text": "3"
},
{
"text": "4"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nWhen 2x $$ \\ge $$ 1
\n5 + 2x –1 = 2x (2x – 2)
\nLet 2x = t
\n$$ \\Rightarrow $$5 + t – 1 = t (t – 2)
\n$$ \\Rightarrow $$ t = 4, – 1(rejected)
\n$$ \\Rightarrow $$ 2x = 4
\n$$ \\Rightarrow $$ x = 2
\nNow when 2x < 1
\n5 + 1 – 2x = 2x (2x – 2)
\nLet 2x = t
\n$$ \\Rightarrow $$ 5 + 1 – t = t (t – 2)
\n$$ \\Rightarrow $$ 0 = t2 – t – 6
\n$$ \\Rightarrow $$ 0 = (t – 3) (t – 2)
\n$$ \\Rightarrow $$ t = 3, – 2
\n2x = 3, 2x = – 2 (rejected)
\n$$ \\therefore $$ Only one real roots.\n",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6856,
"subject": "General Science",
"question": "Let S be the set of all real roots of the equation, \n3x(3x – 1) + 2 = |3x – 1| + |3x – 2|. Then S :",
"options": [
{
"text": "contains exactly two elements."
},
{
"text": "is an empty set."
},
{
"text": "is a singleton."
},
{
"text": "contains at least four elements."
}
],
"answer": "is a singleton.",
"solution": "**Answer:** is a singleton.\n\nLet 3x = t ; t $$>$$ 0\n
Let $$\\lambda \\in \\mathbb{R}$$ and let the equation E be $$|x{|^2} - 2|x| + |\\lambda - 3| = 0$$. Then the largest element in the set S = {$$x+\\lambda:x$$ is an integer solution of E} is ______
The given equation is $$x(x^2+3|x|+5|x-1|+6|x-2|)=0$$, which can be solved by analyzing it in parts. It can be broken down into: $$x=0$$ and $$x^2+3|x|+5|x-1|+6|x-2|=0$$.
For $$x=0$$, it's clear that it is a solution to the equation since it makes the entire expression equal to zero.
$\\because$ All roots are positive $\\Rightarrow$ no solution\n
Case (II)\n
$$\n\\begin{aligned}\n& 0 < x < 1 \\\\\\\\\n& x^2+3 x-5(x-1)-6(x-2)=0 \\\\\\\\\n& x^2-8 x+17=0 \\\\\\\\\n& \\because D < 0 \\Rightarrow \\text { no solution }\n\\end{aligned}\n$$\n\n
Case (III)\n
$$\n1 < x < 2\n$$\n
$$\nx^2+3 x+5(x-1)-6(x-2)=0\n$$\n
$$\nx^2+2 x+7=0\n$$\n
$\\Rightarrow$ no solution\n
Case (IV)\n
$$\n\\begin{aligned}\n& x > 2 \\\\\\\\\n& x^2+3 x+5(x-1)+6(x-2)=0 \\\\\\\\\n& x^2+14 x-19=0\n\\end{aligned}\n$$\n\n
All roots less than 2\n
$\\Rightarrow$ no solution\n
Here $x=0$ is only solution.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6866,
"subject": "General Science",
"question": "
The number of distinct real roots of the equation $$|x+1||x+3|-4|x+2|+5=0$$, is _______
$$ = 1.$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6871,
"subject": "General Science",
"question": "If the roots of the equation $$b{x^2} + cx + a = 0$$ imaginary, then for all real values of $$x$$, the expression $$3{b^2}{x^2} + 6bcx + 2{c^2}$$ is :",
"options": [
{
"text": "less than $$4ab$$ "
},
{
"text": "greater than $$-4ab$$"
},
{
"text": "less than $$-4ab$$"
},
{
"text": "greater than $$4ab$$"
}
],
"answer": "greater than $$-4ab$$",
"solution": "**Answer:** greater than $$-4ab$$\n\nGiven that roots of the equation \n
$$b{x^2} + cx + a = 0$$ are imaginary\n
$$\\therefore$$ $${c^2} - 4ab < 0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( i \\right)$$\n
and $$\\,\\,\\,\\,\\,\\,\\sin x = \\ln \\left( {2 + \\sqrt 5 } \\right) > 1$$ So, rejected\n
Hence given equation has no solution.\n
$$\\therefore$$ The equation has no real roots.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6873,
"subject": "General Science",
"question": "The sum of all the real values of x satisfying the equation \n 2(x$$-$$1)(x2 + 5x $$-$$ 50) = 1 is : ",
"options": [
{
"text": "16"
},
{
"text": "14"
},
{
"text": "$$-$$4"
},
{
"text": "$$-$$ 5"
}
],
"answer": "$$-$$4",
"solution": "**Answer:** $$-$$4\n\nWe know, 2x = 1 only when x = 0.\n
Sum of real values of x = 1 + (-10) + 5 = -4",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6874,
"subject": "General Science",
"question": "The number of all possible positive integral values of $$\\alpha $$ for which the roots of the quadratic equation, 6x2 $$-$$ 11x + $$\\alpha $$ = 0 are rational numbers is : ",
"options": [
{
"text": "3"
},
{
"text": "2"
},
{
"text": "4"
},
{
"text": "5"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\nFor rational D must be perfect square\n
D = 121 $$-$$ 24$$\\alpha $$\n
for 121 $$-$$ 24$$\\alpha $$ to be perfect square a must be 3, 4, 5 \n
So, ans $$\\alpha $$ = 3",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6875,
"subject": "General Science",
"question": "The number of integral values of m for which the\nequation\n
$$ \\therefore $$ m > 0 and m $$ \\ne $$ $${{1 \\over 2}}$$\n
So we can say number of integral values of m are infinitely\nmany.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6876,
"subject": "General Science",
"question": "If m is chosen in the quadratic equation \n
(m2 + 1)\nx2 – 3x + (m2 + 1)2 = 0 \n
such that the sum of its\nroots is greatest, then the absolute difference of\nthe cubes of its roots is :-",
"options": [
{
"text": "$$4\\sqrt 3 $$"
},
{
"text": "$$8\\sqrt 3 $$"
},
{
"text": "$$8\\sqrt 5 $$"
},
{
"text": "$$10\\sqrt 5 $$"
}
],
"answer": "$$8\\sqrt 5 $$",
"solution": "**Answer:** $$8\\sqrt 5 $$\n\nGiven quadratic equation \n
(m2 + 1)\nx2 – 3x + (m2 + 1)2 = 0\n
Let roots of the equation $$\\alpha $$ and $$\\beta $$.\n\n
$$ \\therefore $$ Sum of roots = $$\\alpha $$ + $$\\beta $$ = $${3 \\over {{m^2} + 1}}$$\n
$$= \\sqrt 5 (9 - 1) = 8\\sqrt 5 $$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6877,
"subject": "General Science",
"question": "The least positive value of 'a' for which the\nequation
2x2 + (a – 10)x + $${{33} \\over 2}$$\n= 2a has real\nroots is",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nFor real roots Discriminate $$ \\ge $$ 0.\n
$$ \\Rightarrow $$ a $$ \\le $$ -4 $$ \\cup $$ a $$ \\ge $$ 8\n
$$ \\therefore $$ least positive a = 8",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6878,
"subject": "General Science",
"question": "Let f(x) be a quadratic polynomial such that\n f(–1) + f(2) = 0. If one of the roots of f(x) = 0\n is 3, then its other root lies in :",
"options": [
{
"text": "(–3, –1)"
},
{
"text": "(1, 3)"
},
{
"text": "(–1, 0)"
},
{
"text": "(0, 1)"
}
],
"answer": "(–1, 0)",
"solution": "**Answer:** (–1, 0)\n\nLet the other root is $$\\alpha $$.\n
$$ \\therefore $$ $$\\alpha $$ $$ \\in $$ (–1, 0)",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6879,
"subject": "General Science",
"question": "The sum of all integral values of k (k $$\\ne$$ 0) for which the equation $${2 \\over {x - 1}} - {1 \\over {x - 2}} = {2 \\over k}$$ in x has no real roots, is ____________.",
"options": [],
"answer": "66",
"solution": "**Answer:** 66\n\n$${2 \\over {x - 1}} - {1 \\over {x - 2}} = {2 \\over k}$$
$$\\therefore$$ $$k \\in \\left( {1,{5 \\over 2}} \\right]$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6881,
"subject": "General Science",
"question": "The numbers of pairs (a, b) of real numbers, such that whenever $$\\alpha$$ is a root of the equation x2 + ax + b = 0, $$\\alpha$$2 $$-$$ 2 is also a root of this equation, is :",
"options": [
{
"text": "6"
},
{
"text": "2"
},
{
"text": "4"
},
{
"text": "8"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\nConsider the equation x2 + ax + b = 0
If has two roots (not necessarily real $$\\alpha$$ & $$\\beta$$)
Either $$\\alpha$$ = $$\\beta$$ or $$\\alpha$$ $$\\ne$$ $$\\beta$$
Case (1) If $$\\alpha$$ = $$\\beta$$, then it is repeated root. Given that $$\\alpha$$2 $$-$$ 2 is also a root
Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then $${\\left( {{1 \\over p} + {1 \\over q}} \\right)^{ - 2}}$$ is equal to _________.
Let S be the set of all integral values of $$\\alpha$$ for which the sum of squares of two real roots of the quadratic equation $$3{x^2} + (\\alpha - 6)x + (\\alpha + 3) = 0$$ is minimum. Then S :
",
"options": [
{
"text": "is an empty set"
},
{
"text": "is a singleton"
},
{
"text": "contains exactly two elements"
},
{
"text": "contains more than two elements"
}
],
"answer": "is an empty set",
"solution": "**Answer:** is an empty set\n\n
$$\\therefore$$ $$x = \\, \\pm \\,1$$ are roots of above equation and $${x^5} + 3x - 1$$ is a monotonic term hence vanishs at exactly one value of x other than 1 or $$-$$1.
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6890,
"subject": "General Science",
"question": "The equation $\\mathrm{e}^{4 x}+8 \\mathrm{e}^{3 x}+13 \\mathrm{e}^{2 x}-8 \\mathrm{e}^{x}+1=0, x \\in \\mathbb{R}$ has :",
"options": [
{
"text": "two solutions and both are negative"
},
{
"text": "two solutions and only one of them is negative"
},
{
"text": "four solutions two of which are negative"
},
{
"text": "no solution"
}
],
"answer": "two solutions and both are negative",
"solution": "**Answer:** two solutions and both are negative\n\n$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0$\n\n
So, $\\mathrm{t}-\\frac{1}{\\mathrm{t}}=-3$ or $\\mathrm{t}-\\frac{1}{\\mathrm{t}}=-5$\n\n
$t^{2}+3 t-1=0$ or $t^{2}+5 t-1=0$\n\n
$\\mathrm{t}=\\frac{-3 \\pm \\sqrt{13}}{2}$ or $\\mathrm{t}=\\frac{-5 \\pm \\sqrt{29}}{2}$\n\n
as $t=e^{x}$ so $t$ must be positive,\n\n
$$\nt=\\frac{\\sqrt{13}-3}{2} \\text { or } \\frac{\\sqrt{29}-5}{2}\n$$\n\n
So, $x=\\ln \\left(\\frac{\\sqrt{13}-3}{2}\\right)$ or $x=\\ln \\left(\\frac{\\sqrt{29}-5}{2}\\right)$\n\n
Hence two solutions and both are negative.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6891,
"subject": "General Science",
"question": "
The number of real roots of the equation $$\\sqrt{x^{2}-4 x+3}+\\sqrt{x^{2}-9}=\\sqrt{4 x^{2}-14 x+6}$$, is :
\nNo real value of $t \\Rightarrow$ no real value of $x$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6893,
"subject": "General Science",
"question": "
Let m and $$\\mathrm{n}$$ be the numbers of real roots of the quadratic equations $$x^{2}-12 x+[x]+31=0$$ and $$x^{2}-5|x+2|-4=0$$ respectively, where $$[x]$$ denotes the greatest integer $$\\leq x$$. Then $$\\mathrm{m}^{2}+\\mathrm{mn}+\\mathrm{n}^{2}$$ is equal to __________.
$$\\therefore$$ Max value of $$y$$ is $$41$$ ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6895,
"subject": "General Science",
"question": "The sum of all real values of $$x$$ satisfying the equation $${\\left( {{x^2} - 5x + 5} \\right)^{{x^2} + 4x - 60}}\\, = 1$$ is : ",
"options": [
{
"text": "$$6$$"
},
{
"text": "$$5$$ "
},
{
"text": "$$3$$"
},
{
"text": "$$-4$$ "
}
],
"answer": "$$3$$",
"solution": "**Answer:** $$3$$\n\nGiven equation,\n$${\\left( {{x^2} - 5x + 5} \\right)^{{x^2} + 4x - 60}} = 1$$\n
Case 1 : When x2 - 5x + 5 = 1 and x2 + 4x - 60 is any real no then this equation satisfy.\n Note : When we put any real number as a power of 1 the value stays always 1 (1 any real no = 1).\n x2 - 5x + 5 = 1\n (x - 1)(x - 4) = 0\n $$\\therefore$$ x = 1, 4\n
Case 2 : When x2 - 5x + 5 is a real no and x2 + 4x - 60 = 0 then the given equation satisfy. As we know if power of any real no is zero then it will become 1((any real number)0 = 1).\n For, x2 + 4x - 60 = 0\n (x - 6)(x + 10) = 0\n $$\\therefore$$ x = 6, -10\n
Case 3 : When x2 - 5x + 5 = -1 and x2 + 4x - 60 is even this equation satisfy. As we know (-1)even = 1.\n For, x2 - 5x + 5 = -1\n (x - 2)(x - 3) = 0\n $$\\therefore$$ x = 2, 3\n But x can't be 3 because when x = 3 the value of x2 + 4x - 60 becomes 32 + 4.3 - 60 = - 39 which is an odd number, then (-1)-39 = -1. So for x = 3 equation does not satisfy.\n
$$\\therefore$$ The sum of all the real values = 1 + 4 + 6 + (-10) + 2 = 3
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6896,
"subject": "General Science",
"question": "The number of integral values of m for which the quadratic expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), x $$ \\in $$ R, is always positive, is :",
"options": [
{
"text": "7"
},
{
"text": "8"
},
{
"text": "3"
},
{
"text": "6"
}
],
"answer": "7",
"solution": "**Answer:** 7\n\nExpression is always positive it\n
$$ \\therefore $$ Intgral value of m {0, 1, 2, 3, 4, 5, 6}",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6897,
"subject": "General Science",
"question": "
Let $$a, b, c$$ be the lengths of three sides of a triangle satistying the condition $$\\left(a^2+b^2\\right) x^2-2 b(a+c) x+\\left(b^2+c^2\\right)=0$$. If the set of all possible values of $$x$$ is the interval $$(\\alpha, \\beta)$$, then $$12\\left(\\alpha^2+\\beta^2\\right)$$ is equal to __________.
$$\\begin{aligned}\n& \\Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0 \\\\\n& \\Rightarrow(a x-b)^2+(b x-c)^2=0 \\\\\n& \\Rightarrow a x-b=0, \\quad b x-c=0 \\\\\n& \\Rightarrow a+b>c \\quad b+c>a \\quad c+a>b\n\\end{aligned}$$
\n
$$\\begin{array}{l|l|l}\na+a x>b x & a x+b x>a & a x^2+a>a x \\\\\na+a x>a x^2 & a x+a x^2>a & x^2-x+1>0 \\\\\nx^2-x-1<0 & x^2+x-1>0 & \\text { always true }\n\\end{array}$$
\n
$$\\begin{aligned}\n& \\frac{1-\\sqrt{5}}{2}< x<\\frac{1+\\sqrt{5}}{2} \\\\\n& x< \\frac{-1-\\sqrt{5}}{2}, \\text { or } x >\\frac{-1+\\sqrt{5}}{2}\n\\end{aligned}$$
$$ \\Rightarrow - b = a - c$$ or $$c = a + b.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6906,
"subject": "General Science",
"question": "The value of $$a$$ for which the sum of the squares of the roots of the equation $${x^2} - \\left( {a - 2} \\right)x - a - 1 = 0$$ \n assume the least value is :",
"options": [
{
"text": "$$1$$ "
},
{
"text": "$$0$$ "
},
{
"text": "$$3$$ "
},
{
"text": "$$2$$ "
}
],
"answer": "$$1$$ ",
"solution": "**Answer:** $$1$$ \n\n$${x^2} - \\left( {a - 2} \\right)x - a - 1 = 0$$\n
For min. value of $${\\alpha ^2} + {\\beta ^2}$$ where $$\\alpha $$ is an integer\n
$$ \\Rightarrow \\,\\,a = 1.$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6907,
"subject": "General Science",
"question": "The value of $$a$$ for which the sum of the squares of the roots of the equation $${x^2} - \\left( {a - 2} \\right)x - a - 1 = 0$$ assume the least value is ",
"options": [
{
"text": "$$1$$ "
},
{
"text": "$$0$$ "
},
{
"text": "$$3$$ "
},
{
"text": "$$2$$ "
}
],
"answer": "$$1$$ ",
"solution": "**Answer:** $$1$$ \n\nGiven quadratic equation,\n
$$\\therefore\\,\\,\\,$$ tan A = $$-$$ $$\\sqrt {{{\\sec }^2}A - 1} = - {4 \\over 3}.$$\n
Hence, roots of the equation are sec A and tan A.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6918,
"subject": "General Science",
"question": "Let p, q and r be real numbers (p $$ \\ne $$ q, r $$ \\ne $$ 0), such that the roots of the equation $${1 \\over {x + p}} + {1 \\over {x + q}} = {1 \\over r}$$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to : ",
"options": [
{
"text": "$${{{p^2} + {q^2}} \\over 2}$$"
},
{
"text": "p2 + q2 "
},
{
"text": "2(p2 + q2)"
},
{
"text": "p2 + q2 + r2"
}
],
"answer": "p2 + q2 ",
"solution": "**Answer:** p2 + q2 \n\nGiven, \n
= p2 + q2 ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6919,
"subject": "General Science",
"question": "If f(x) is a quadratic expression such that f (1) + f (2) = 0, and $$-$$ 1 is a root of f (x) = 0, then the other root of f(x) = 0 is : ",
"options": [
{
"text": "$$-$$ $${5 \\over 8}$$"
},
{
"text": "$$-$$ $${8 \\over 5}$$"
},
{
"text": "$${5 \\over 8}$$"
},
{
"text": "$${8 \\over 5}$$"
}
],
"answer": "$${8 \\over 5}$$",
"solution": "**Answer:** $${8 \\over 5}$$\n\nLet $$\\alpha $$ and $$\\beta $$ = - 1 are the roots of the polynomial, then we get\n
$$ \\Rightarrow $$ $$\\alpha $$ = $${8 \\over 5}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6920,
"subject": "General Science",
"question": "If $$\\lambda $$ $$ \\in $$ R is such that the sum of the cubes of the roots of the equation, \n x2 + (2 $$-$$ $$\\lambda $$) x + (10 $$-$$ $$\\lambda $$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is : ",
"options": [
{
"text": "$$4\\sqrt 2 $$"
},
{
"text": "$$2\\sqrt 5 $$"
},
{
"text": "$$2\\sqrt 7 $$"
},
{
"text": "20"
}
],
"answer": "$$2\\sqrt 5 $$",
"solution": "**Answer:** $$2\\sqrt 5 $$\n\nLet $$\\alpha $$, $$\\beta $$ are the roots of the equation, \n
= $${{27} \\over {16}}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6930,
"subject": "General Science",
"question": "Let $$\\lambda \\ne 0$$ be in R. If $$\\alpha $$ and $$\\beta $$ are the roots of the equation, x2 - x + 2$$\\lambda $$ = 0 and $$\\alpha $$ and $$\\gamma $$ are the roots of the equation, $$3{x^2} - 10x + 27\\lambda = 0$$, then $${{\\beta \\gamma } \\over \\lambda }$$ is equal to:\n",
"options": [
{
"text": "36"
},
{
"text": "9"
},
{
"text": "27"
},
{
"text": "18"
}
],
"answer": "18",
"solution": "**Answer:** 18\n\n$$\\alpha $$ and $$\\beta $$ are the roots of the equation x2 - x + 2$$\\lambda $$ = 0 .....(1)\n
So option (D) is wrong.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6937,
"subject": "General Science",
"question": "Let $$\\alpha$$ and $$\\beta$$ be the roots of x2 $$-$$ 6x $$-$$ 2 = 0. If an = $$\\alpha$$n $$-$$ $$\\beta$$n for n $$ \\ge $$ 1, then the value of $${{{a_{10}} - 2{a_8}} \\over {3{a_9}}}$$ is :",
"options": [
{
"text": "3"
},
{
"text": "2"
},
{
"text": "4"
},
{
"text": "1"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven, $$\\alpha$$ and $$\\beta$$ be the roots of $${x^2} - 6x - 2 = 0$$
Let f(x) be a quadratic polynomial such that f($$-$$2) + f(3) = 0. If one of the roots of f(x) = 0 is $$-$$1, then the sum of the roots of f(x) = 0 is equal to :
Let $$\\alpha$$, $$\\beta$$ be the roots of the equation $${x^2} - 4\\lambda x + 5 = 0$$ and $$\\alpha$$, $$\\gamma$$ be the roots of the equation $${x^2} - \\left( {3\\sqrt 2 + 2\\sqrt 3 } \\right)x + 7 + 3\\lambda \\sqrt 3 = 0$$, $$\\lambda$$ > 0. If $$\\beta + \\gamma = 3\\sqrt 2 $$, then $${(\\alpha + 2\\beta + \\gamma )^2}$$ is equal to __________.
Let $$\\alpha$$, $$\\beta$$ be the roots of the equation $$x^{2}-\\sqrt{2} x+\\sqrt{6}=0$$ and $$\\frac{1}{\\alpha^{2}}+1, \\frac{1}{\\beta^{2}}+1$$ be the roots of the equation $$x^{2}+a x+b=0$$. Then the roots of the equation $$x^{2}-(a+b-2) x+(a+b+2)=0$$ are :
",
"options": [
{
"text": "non-real complex numbers"
},
{
"text": "real and both negative"
},
{
"text": "real and both positive"
},
{
"text": "real and exactly one of them is positive"
}
],
"answer": "real and both negative",
"solution": "**Answer:** real and both negative\n\n
If $$\\frac{1}{(20-a)(40-a)}+\\frac{1}{(40-a)(60-a)}+\\ldots+\\frac{1}{(180-a)(200-a)}=\\frac{1}{256}$$, then the maximum value of $$\\mathrm{a}$$ is :
Let $$\\alpha, \\beta(\\alpha>\\beta)$$ be the roots of the quadratic equation $$x^{2}-x-4=0 .$$ If $$P_{n}=\\alpha^{n}-\\beta^{n}$$, $$n \\in \\mathrm{N}$$, then $$\\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^{2}+P_{14} P_{15}}{P_{13} P_{14}}$$ is equal to __________.
Let $$\\alpha_1,\\alpha_2,....,\\alpha_7$$ be the roots of the equation $${x^7} + 3{x^5} - 13{x^3} - 15x = 0$$ and $$|{\\alpha _1}| \\ge |{\\alpha _2}| \\ge \\,...\\, \\ge \\,|{\\alpha _7}|$$. Then $$\\alpha_1\\alpha_2-\\alpha_3\\alpha_4+\\alpha_5\\alpha_6$$ is equal to _________.
Let $$\\lambda \\ne 0$$ be a real number. Let $$\\alpha,\\beta$$ be the roots of the equation $$14{x^2} - 31x + 3\\lambda = 0$$ and $$\\alpha,\\gamma$$ be the roots of the equation $$35{x^2} - 53x + 4\\lambda = 0$$. Then $${{3\\alpha } \\over \\beta }$$ and $${{4\\alpha } \\over \\gamma }$$ are the roots of the equation
Let $$\\alpha \\in\\mathbb{R}$$ and let $$\\alpha,\\beta$$ be the roots of the equation $${x^2} + {60^{{1 \\over 4}}}x + a = 0$$. If $${\\alpha ^4} + {\\beta ^4} = - 30$$, then the product of all possible values of $$a$$ is ____________.
\nProduct of value of $a=45$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6958,
"subject": "General Science",
"question": "
Let $$\\alpha, \\beta$$ be the roots of the equation $$x^{2}-\\sqrt{2} x+2=0$$. Then $$\\alpha^{14}+\\beta^{14}$$ is equal to
The quadratic formula is $$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$. For the quadratic equation $$x^2 - \\sqrt{2}x + 2 = 0$$, we have $$a = 1, b = -\\sqrt{2}, c = 2$$. Plugging these into the quadratic formula gives:\n
We can express complex numbers in the form $$re^{i\\theta}$$. For $\\alpha$ and $\\beta$, we find the magnitude $$r = \\sqrt{2}$$ and the arguments $$\\theta = \\frac{\\pi}{3}, -\\frac{\\pi}{3}$$ respectively. So, we have:\n
$$\\alpha = \\sqrt{2}e^{i\\frac{\\pi}{3}},$$\n
$$\\beta = \\sqrt{2}e^{-i\\frac{\\pi}{3}}.$$
\n
\n
Calculate the 14th power of the roots:
\n
To find $$\\alpha^{14}$$ and $$\\beta^{14}$$, we use the property of exponents which says that $$(a^m)^n = a^{mn}$$. So, we have:\n
Let $$\\alpha, \\beta$$ be the roots of the quadratic equation $$x^{2}+\\sqrt{6} x+3=0$$. Then $$\\frac{\\alpha^{23}+\\beta^{23}+\\alpha^{14}+\\beta^{14}}{\\alpha^{15}+\\beta^{15}+\\alpha^{10}+\\beta^{10}}$$ is equal to :
Here, the exponential power of $$\\sqrt{3}$$ in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by $$(\\sqrt{3})^{8} = 81$$ to simplify:\n\n
The number of points, where the curve $$f(x)=\\mathrm{e}^{8 x}-\\mathrm{e}^{6 x}-3 \\mathrm{e}^{4 x}-\\mathrm{e}^{2 x}+1, x \\in \\mathbb{R}$$ cuts $$x$$-axis, is equal to _________.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nFirstly, we know that the given function
$f(x)=e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1$ intersects the x-axis
where $f(x) = 0$. Setting $f(x)$ equal to zero gives us :\n\n
$e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0.$\n\n
Let $t = e^{2x}$. The equation now becomes :\n\n
$t^4 - t^3 - 3t^2 - t + 1 = 0.$\n\n
Dividing by $t^2$ and rearranging the equation gives :\n\n
This quadratic formula for $t$ yields two possible values (t = 2.37 and t = 0.42). Both of them are positive, thus both are acceptable values for $t = e^{2x}$.\n\n
Finally, to get $x$, take the natural log of both sides and divide by 2 :\n\n
$x = \\frac{1}{2}\\ln(t).$\n\n
This gives us two solutions for $x$, corresponding to the two positive solutions for $t$. So, the number of points where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$ intersects the x-axis is 2.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6961,
"subject": "General Science",
"question": "
If $$a$$ and $$b$$ are the roots of the equation $$x^{2}-7 x-1=0$$, then the value of $$\\frac{a^{21}+b^{21}+a^{17}+b^{17}}{a^{19}+b^{19}}$$ is equal to _____________.
",
"options": [],
"answer": "51",
"solution": "**Answer:** 51\n\nWe have, $a$ and $b$ are the roots of the equation\n
$$\n\\begin{aligned}\n& =\\frac{a^{17}\\left(51 a^2\\right)+b^{17}\\left(51 b^2\\right)}{a^{19}+b^{19}} \\quad[\\because \\text { From Eq. (i) and (ii) }] \\\\\\\\\n& =\\frac{51\\left[a^{19}+b^{19}\\right]}{a^{19}+b^{19}}=51\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6962,
"subject": "General Science",
"question": "
Let $$\\alpha, \\beta, \\gamma$$ be the three roots of the equation $$x^{3}+b x+c=0$$. If $$\\beta \\gamma=1=-\\alpha$$, then $$b^{3}+2 c^{3}-3 \\alpha^{3}-6 \\beta^{3}-8 \\gamma^{3}$$ is equal to :
$$\n\\therefore \\text { Sum of roots }=(5+\\sqrt{3})+4=9+\\sqrt{3}\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6964,
"subject": "General Science",
"question": "Let $\\alpha$ and $\\beta$ be the roots of the equation $p x^2+q x-r=0$, where $p \\neq 0$. If $p, q$ and $r$ be the consecutive terms of a non constant G.P. and $\\frac{1}{\\alpha}+\\frac{1}{\\beta}=\\frac{3}{4}$, then the value of $(\\alpha-\\beta)^2$ is :",
"options": [
{
"text": "8"
},
{
"text": "9"
},
{
"text": "$\\frac{20}{3}$"
},
{
"text": "$\\frac{80}{9}$"
}
],
"answer": "$\\frac{80}{9}$",
"solution": "**Answer:** $\\frac{80}{9}$\n\nGiven : $p x^2+q x-r=0$\n
Since $a = (\\sqrt{3} + \\sqrt{2})^x $, we have two cases :
\n\n\n
$(\\sqrt{3} + \\sqrt{2})^x = 5 + 2\\sqrt{6}$ = $(\\sqrt{3} + \\sqrt{2})^2$. There is one real solution for x in this case which is x = 2.
\n
$(\\sqrt{3} + \\sqrt{2})^x = 5 - 2\\sqrt{6}$ = $(\\sqrt{3} - \\sqrt{2})^2= \\frac{1}{(\\sqrt{3} + \\sqrt{2})^2} $ = $(\\sqrt{3} + \\sqrt{2})^{-2}$. There is one real solution for x in this case which is x = - 2.
\n\n Conclusion\n\n
There are two real solutions for *x*. Therefore, the number of elements in the set S is 2. This corresponds with option (C).
Let $$\\alpha, \\beta \\in \\mathbf{N}$$ be roots of the equation $$x^2-70 x+\\lambda=0$$, where $$\\frac{\\lambda}{2}, \\frac{\\lambda}{3} \\notin \\mathbf{N}$$. If $$\\lambda$$ assumes the minimum possible value, then $$\\frac{(\\sqrt{\\alpha-1}+\\sqrt{\\beta-1})(\\lambda+35)}{|\\alpha-\\beta|}$$ is equal to :
Let $$\\alpha, \\beta ; \\alpha>\\beta$$, be the roots of the equation $$x^2-\\sqrt{2} x-\\sqrt{3}=0$$. Let $$\\mathrm{P}_n=\\alpha^n-\\beta^n, n \\in \\mathrm{N}$$. Then $$(11 \\sqrt{3}-10 \\sqrt{2}) \\mathrm{P}_{10}+(11 \\sqrt{2}+10) \\mathrm{P}_{11}-11 \\mathrm{P}_{12}$$ is equal to
Let $$\\alpha, \\beta$$ be the roots of the equation $$x^2+2 \\sqrt{2} x-1=0$$. The quadratic equation, whose roots are $$\\alpha^4+\\beta^4$$ and $$\\frac{1}{10}(\\alpha^6+\\beta^6)$$, is:
If 2 and 6 are the roots of the equation $$a x^2+b x+1=0$$, then the quadratic equation, whose roots are $$\\frac{1}{2 a+b}$$ and $$\\frac{1}{6 a+b}$$, is :
Given that the roots of the quadratic equation are $2$ and $6$, we can use Vieta's formulas which relate the coefficients of the polynomial to sums and products of its roots.
First, let's start by substituting $$y = (8)^x$$ in the given equation. By substituting, the equation $$8^{2x} - 16 \\cdot 8^x + 48 = 0$$ will be transformed into
\n\n
$$ y^2 - 16y + 48 = 0 $$
\n\n
Now, we have a quadratic equation in $$y$$. To find the roots of this quadratic equation, we can use the quadratic formula:
\n\n
$$ y = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} $$
\n\n
In this case, $$a = 1$$, $$b = -16$$, and $$c = 48$$. Substituting these values into the formula, we get:
\n\n
$$ y = \\frac{16 \\pm \\sqrt{256 - 192}}{2} $$
\n\n
$$ y = \\frac{16 \\pm \\sqrt{64}}{2} $$
\n\n
$$ y = \\frac{16 \\pm 8}{2} $$
\n\n
Solving for the two possible values of $$y$$, we have:
\n\n
$$ y = \\frac{16 + 8}{2} = 12 $$
\n\n
$$ y = \\frac{16 - 8}{2} = 4 $$
\n\n
Now, recall that we substituted $$y = (8)^x$$. So, we need to solve for $$x$$ when $$y = 12$$ and $$y = 4$$:
\n\n
$$ 8^x = 12 $$
\n\n
$$ x = \\log_8(12) $$
\n\n
$$ 8^x = 4 $$
\n\n
$$ x = \\log_8(4) $$
\n\n
Therefore, the solutions for $$x$$ are $$\\log_8(12)$$ and $$\\log_8(4)$$. The sum of these solutions is:
\n\n
$$ \\log_8(12) + \\log_8(4) $$
\n\n
Using the logarithmic property that $$\\log_b(m) + \\log_b(n) = \\log_b(m \\cdot n)$$, we get:
Let $$\\alpha, \\beta$$ be roots of $$x^2+\\sqrt{2} x-8=0$$. If $$\\mathrm{U}_{\\mathrm{n}}=\\alpha^{\\mathrm{n}}+\\beta^{\\mathrm{n}}$$, then $$\\frac{\\mathrm{U}_{10}+\\sqrt{2} \\mathrm{U}_9}{2 \\mathrm{U}_8}$$ is equal to ________.
Let $$\\alpha, \\beta$$ be the distinct roots of the equation $$x^2-\\left(t^2-5 t+6\\right) x+1=0, t \\in \\mathbb{R}$$ and $$a_n=\\alpha^n+\\beta^n$$. Then the minimum value of $$\\frac{a_{2023}+a_{2025}}{a_{2024}}$$ is
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 6973,
"subject": "General Science",
"question": "If m is the A.M. of two distinct real numbers l and n $$(l,n > 1)$$ and $${G_1},{G_2}$$ and $${G_3}$$ are three geometric means between $$l$$ and n, then $$G_1^4\\, + 2G_2^4\\, + G_3^4$$ equals:",
"options": [
{
"text": "$$4\\,lm{n^2}$$ "
},
{
"text": "$$4\\,{l^2}{m^2}{n^2}$$ "
},
{
"text": "$$4\\,{l^2}m\\,n$$ "
},
{
"text": "$$4\\,l\\,{m^2}n$$ "
}
],
"answer": "$$4\\,l\\,{m^2}n$$ ",
"solution": "**Answer:** $$4\\,l\\,{m^2}n$$ \n\n$$m = {{l + n} \\over 2}$$ and common ratio of \n
= $$ - \\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6979,
"subject": "General Science",
"question": "If m arithmetic means (A.Ms) and three\ngeometric means (G.Ms) are inserted between\n3 and 243 such that 4th A.M. is equal to 2nd\nG.M., then m is equal to _________ .",
"options": [],
"answer": "39",
"solution": "**Answer:** 39\n\nGiven m arithmetic means (A.Ms) present between 3 and 243
$$ - \\sqrt 2 $$ $$ \\le $$ sin x + cos x $$ \\le $$ $$\\sqrt 2 $$.\n
So Minimum value of sin x + cos x = $$ - \\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6981,
"subject": "General Science",
"question": "The minimum value of $$f(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}$$, where a, $$x \\in R$$ and a > 0, is equal to :",
"options": [
{
"text": "$$a + {1 \\over a}$$"
},
{
"text": "2a"
},
{
"text": "a + 1"
},
{
"text": "$$2\\sqrt a $$"
}
],
"answer": "$$2\\sqrt a $$",
"solution": "**Answer:** $$2\\sqrt a $$\n\nWe know, $$AM \\ge GM$$
If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is :
Consider two G.Ps. 2, 22, 23, ..... and 4, 42, 43, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is $${(2)^{{{225} \\over 8}}}$$, then $$\\sum\\limits_{k = 1}^n {k(n - k)} $$ is equal to :
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6986,
"subject": "General Science",
"question": "Let $A_{1}$ and $A_{2}$ be two arithmetic means and $G_{1}, G_{2}, G_{3}$ be three geometric
means of two distinct positive numbers. Then $G_{1}^{4}+G_{2}^{4}+G_{3}^{4}+G_{1}^{2} G_{3}^{2}$ is equal to :",
"options": [
{
"text": "$\\left(A_{1}+A_{2}\\right)^{2} G_{1} G_{3}$"
},
{
"text": "$\\left(A_{1}+A_{2}\\right) G_{1}^{2} G_{3}^{2}$"
},
{
"text": "$2\\left(A_{1}+A_{2}\\right) G_{1}^{2} G_{3}^{2}$"
},
{
"text": "$2\\left(A_{1}+A_{2}\\right) G_{1} G_{3}$"
}
],
"answer": "$\\left(A_{1}+A_{2}\\right)^{2} G_{1} G_{3}$",
"solution": "**Answer:** $\\left(A_{1}+A_{2}\\right)^{2} G_{1} G_{3}$\n\n
Now, we have the following relations :
\n
Arithmetic progression :
\n
Since $A_1$ and $A_2$ are arithmetic means between $a$ and $b$, we can say that $a$, $A_1$, $A_2$, and $b$ are in an arithmetic progression. This means there are three equal intervals between $a$ and $b$, which are represented by the common difference $d$.
\n\n
To find the value of $d$, we can use the following equation :
\n\n
$$\nb - a = 3d\n$$
\n\n
From this equation, we can find the value of $d$ :\n
Let $$a, b, c$$ and $$d$$ be positive real numbers such that $$a+b+c+d=11$$. If the maximum value of $$a^{5} b^{3} c^{2} d$$ is $$3750 \\beta$$, then the value of $$\\beta$$ is
",
"options": [
{
"text": "110"
},
{
"text": "108"
},
{
"text": "90"
},
{
"text": "55"
}
],
"answer": "90",
"solution": "**Answer:** 90\n\nGiven that $$a+b+c+d=11$$ and the maximum value of $$a^5 b^3 c^2 d$$ is $$3750\\beta$$, you assumed the numbers to be $$\\frac{a}{5}, \\frac{a}{5}, \\frac{a}{5}, \\frac{a}{5}, \\frac{a}{5}, \\frac{b}{3}, \\frac{b}{3}, \\frac{b}{3}, \\frac{c}{2}, \\frac{c}{2}, d$$.\n\n
So, the value of $$\\beta$$ is 90.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6988,
"subject": "General Science",
"question": "Let $3, a, b, c$ be in A.P. and $3, a-1, b+1, c+9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :",
"options": [
{
"text": "-4"
},
{
"text": "-1"
},
{
"text": "13"
},
{
"text": "11"
}
],
"answer": "11",
"solution": "**Answer:** 11\n\n
Since $3, a, b, c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:
\n\n
$$ d = a - 3 $$
\n\n
The nth term of an A.P. is given by the formula:
\n\n
$$ T_n = a + (n-1)d $$
\n\n
So, using this formula, we can express $b$ and $c$ in terms of $a$ and $d$:
\n\n
$$ b = a + d $$
\n\n
$$ c = a + 2d $$
\n\n
Substituting $d = a - 3$ into these expressions:
\n\n
$$ b = a + (a - 3) $$
\n\n
$$ c = a + 2(a - 3) $$
\n\n
Therefore:
\n\n
$$ b = 2a - 3 $$
\n\n
$$ c = 3a - 6 $$
\n\n
Now, let's consider that $3, a-1, b+1, c+9$ are in geometric progression (G.P.). For terms in a G.P., the ratio (common ratio, r) between consecutive terms is constant. So:
Now, we will establish the relation between the terms using the property of G.P.:
\n\n
$$ \\frac{a - 1}{3} = \\frac{b + 1}{a - 1} $$
\n\n
$$ (a - 1)^2 = 3(b + 1) $$
\n\n
$$ a^2 - 2a + 1 = 3b + 3 $$
\n\n
Substituting $b = 2a - 3$, we get:
\n\n
$$ a^2 - 2a + 1 = 3(2a - 3) + 3 $$
\n\n
$$ a^2 - 2a + 1 = 6a - 9 + 3 $$
\n\n
$$ a^2 - 8a + 7 = 0 $$
\n\n
Solving this quadratic equation:
\n\n
$$ (a - 7)(a - 1) = 0 $$
\n\n
Hence, $a = 7$ or $a = 1$. However, if $a = 1$, the terms $3, a-1, b+1, c+9$ cannot form a G.P. as it would involve division by zero. Therefore, $a = 7$. We use this value to find $b$ and $c$:
\n\n
$$ b = 2a - 3 = 2(7) - 3 = 14 - 3 = 11 $$
\n\n
$$ c = 3a - 6 = 3(7) - 6 = 21 - 6 = 15 $$
\n\n
Now we can find the arithmetic mean ($A$) of $a$, $b$, and $c$:
\n\n
$$ A = \\frac{a + b + c}{3} $$
\n\n
$$ A = \\frac{7 + 11 + 15}{3} $$
\n\n
$$ A = \\frac{33}{3} $$
\n\n
$$ A = 11 $$
\n\n
Hence, the arithmetic mean of $a$, $b$, and $c$ is $11$, which corresponds to Option D.
For $$0 < c < b < a$$, let $$(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0$$ and $$\\alpha \\neq 1$$ be one of its root. Then, among the two statements
\n
(I) If $$\\alpha \\in(-1,0)$$, then $$b$$ cannot be the geometric mean of $a$ and $$c$$
\n
(II) If $$\\alpha \\in(0,1)$$, then $$b$$ may be the geometric mean of $$a$$ and $$c$$
",
"options": [
{
"text": "only (II) is true\n"
},
{
"text": "Both (I) and (II) are true\n"
},
{
"text": "only (I) is true\n"
},
{
"text": "Neither (I) nor (II) is true"
}
],
"answer": "Both (I) and (II) are true\n",
"solution": "**Answer:** Both (I) and (II) are true\n\n\n
$$\\begin{aligned}\n& f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\\\\n& f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\\\\n& f(1)=0 \\\\\n& \\therefore \\alpha \\cdot 1=\\frac{c+a-2 b}{a+b-2 c} \\\\\n& \\alpha=\\frac{c+a-2 b}{a+b-2 c} \\\\\n& \\text { If, }-1<\\alpha<0 \\\\\n& -1<\\frac{c+a-2 b}{a+b-2 c}<0 \\\\\n& b+c<2 a \\text { and } b>\\frac{a+c}{2}\n\\end{aligned}$$
Let three real numbers $$a, b, c$$ be in arithmetic progression and $$a+1, b, c+3$$ be in geometric progression. If $$a>10$$ and the arithmetic mean of $$a, b$$ and $$c$$ is 8, then the cube of the geometric mean of $$a, b$$ and $$c$$ is
$$ \\Rightarrow x = {\\log _3}\\left( {{3 \\over 4}} \\right)$$\n
or $$x = {\\log _3}3 - {\\log _3}4$$\n
$$ \\Rightarrow x = 1 - {\\log _3}4$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6992,
"subject": "General Science",
"question": "Let $${{T_r}}$$ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, $$m \\ne n,\\,\\,{T_m} = {1 \\over n}\\,\\,and\\,{T_n} = {1 \\over m},\\,$$ then a - d equals",
"options": [
{
"text": "$${1 \\over m} + {1 \\over n}$$ "
},
{
"text": "1 "
},
{
"text": "$${1 \\over {m\\,n}}$$ "
},
{
"text": "0 "
}
],
"answer": "0 ",
"solution": "**Answer:** 0 \n\n$${T_m} = a + \\left( {m - 1} \\right)d = {1 \\over n}...........\\left( 1 \\right)$$\n
For $${{{a_6}} \\over {a{}_{21}}},\\,\\,p = 11,\\,q = 41$$\n
$$ \\Rightarrow {{{a_6}} \\over {a{}_{21}}} = {{11} \\over {41}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6994,
"subject": "General Science",
"question": "A person is to count 4500 currency notes. Let $${a_n}$$ denote the number of notes he counts in the $${n^{th}}$$ minute. If $${a_1}$$ = $${a_2}$$ = ....= $${a_{10}}$$= 150 and $${a_{10}}$$, $${a_{11}}$$,.... are in an AP with common difference - 2, then the time taken by him to count all notes is ",
"options": [
{
"text": "34 minutes"
},
{
"text": "125 minutes"
},
{
"text": "135 minutes "
},
{
"text": "24 minutes "
}
],
"answer": "34 minutes",
"solution": "**Answer:** 34 minutes\n\nTill $$10$$th minute number of counted notes $$ = 1500$$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$ = n\\left[ {148 - n + 1} \\right]$$\n
$$ \\Rightarrow $$$${n^2} - 149n + 3000 = 0$$\n
$$ \\Rightarrow n = 125,24$$\n
But $$n=125$$ is not possible\n
$$\\therefore$$ total time $$ = 24 + 10 = 34$$ minutes. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6995,
"subject": "General Science",
"question": "A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after",
"options": [
{
"text": "19 months"
},
{
"text": "20 months"
},
{
"text": "21 months "
},
{
"text": "18 months "
}
],
"answer": "21 months ",
"solution": "**Answer:** 21 months \n\nLet required number of months $$=n$$\n
then the sum of its first 17 terms is equal to :",
"options": [
{
"text": "306"
},
{
"text": "153"
},
{
"text": "612"
},
{
"text": "204"
}
],
"answer": "306",
"solution": "**Answer:** 306\n\nAs a1 a2 . . . . . an . . . . . are in A.P.\n
= 306",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6997,
"subject": "General Science",
"question": "If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :",
"options": [
{
"text": "2"
},
{
"text": "4$${^{{1 \\over 3}}}$$"
},
{
"text": "4$${^{{2 \\over 3}}}$$"
},
{
"text": "4"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\na, b and c are in AP.\n
\n$$\\, \\Rightarrow d = {8 \\over 5}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7005,
"subject": "General Science",
"question": "If the sum and product of the first three term in\nan A.P. are 33 and 1155, respectively, then a value\nof its 11th term is :-",
"options": [
{
"text": "–25"
},
{
"text": "–36"
},
{
"text": "25"
},
{
"text": "–35"
}
],
"answer": "–25",
"solution": "**Answer:** –25\n\nLet the three terms are a - d, a, a + d\n
= 52",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7011,
"subject": "General Science",
"question": "Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also three consecutive terms of a G.P., then $${a \\over c}$$ equal to : ",
"options": [
{
"text": "2"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${7 \\over 13}$$"
},
{
"text": "4"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\nT7 = A + 6d = a; T11 = A + 10d = b; T13 = A + 12d = c\n
Now a, b, c are in G.P. \n
$$ \\therefore $$ b2 = ac\n
$$ \\Rightarrow $$ (A + 10d)2 = (A + 6d) (A + 12d)\n
$$ \\therefore $$ n = 14",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7014,
"subject": "General Science",
"question": "The common difference of the A.P. b1, b2, … , bm\n is 2 more than the common difference of A.P. a1, a2, …, an. If a40 = –159, a100 = –399 and\nb100 = a70, then b1\n is equal to :",
"options": [
{
"text": "127"
},
{
"text": "81"
},
{
"text": "–127"
},
{
"text": "-81"
}
],
"answer": "-81",
"solution": "**Answer:** -81\n\nLet common difference of series\n a1\n, a2\n, a3\n,..., an\n be d.\n
$$ \\because $$ a40 = a1 + 39d == –159 ...(i)\n
and a100 = a1 + 99d = –399 ...(ii)\n
From eqn. (ii) and (i)\n d = –4 and a1\n = –3.\n
The common difference of the A.P. b1, b2, … , bm\n is 2 more than the common difference of A.P. a1, a2, …, an.\n
$$ \\therefore $$ Common difference of b1\n, b2\n, b3\n, ..., be (–2).\n
$$ \\because $$ b100 = a70\n
$$ \\therefore $$ b1\n + 99(–2) = (–3) + 69(–4)\n
$$ \\therefore $$ b1\n = 198 – 279\n
$$ \\therefore $$ b1\n = – 81\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7015,
"subject": "General Science",
"question": "If $${3^{2\\sin 2\\alpha - 1}}$$, 14 and $${3^{4 - 2\\sin 2\\alpha }}$$ are the first three terms of an A.P. for some $$\\alpha $$, then the sixth\nterms of this A.P. is:",
"options": [
{
"text": "66"
},
{
"text": "81"
},
{
"text": "65"
},
{
"text": "78"
}
],
"answer": "66",
"solution": "**Answer:** 66\n\nGiven that
$${S_{n - 4}} = {{20} \\over 2}\\{ 1 + 248\\} = 2490$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7017,
"subject": "General Science",
"question": "If the first term of an A.P. is 3 and the sum of\nits first 25 terms is equal to the sum of its next\n15 terms, then the common difference of this\nA.P. is :",
"options": [
{
"text": "$${1 \\over 4}$$"
},
{
"text": "$${1 \\over 5}$$"
},
{
"text": "$${1 \\over 7}$$"
},
{
"text": "$${1 \\over 6}$$"
}
],
"answer": "$${1 \\over 6}$$",
"solution": "**Answer:** $${1 \\over 6}$$\n\nFirst 25 terms = a, a + d, .......,a + 24d\n
Next 15 terms = a + 25d, a + 26d, ......, a + 39d\n
$$ \\Rightarrow $$ d = $${1 \\over 6}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7018,
"subject": "General Science",
"question": "If the sum of first 11 terms of an A.P.,\n a1, a2, a3, ....\nis 0 (a $$ \\ne $$ 0), then the sum of the A.P.,\n a1\n, a3\n, a5\n,....., a23 is ka1\n, where k is equal to :\n",
"options": [
{
"text": "$${{121} \\over {10}}$$"
},
{
"text": "-$${{121} \\over {10}}$$"
},
{
"text": "$${{72} \\over 5}$$"
},
{
"text": "-$${{72} \\over 5}$$"
}
],
"answer": "-$${{72} \\over 5}$$",
"solution": "**Answer:** -$${{72} \\over 5}$$\n\nLet common difference be d.\n
= $${{201} \\over 2}$$ = $$100{1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7020,
"subject": "General Science",
"question": "Let ƒ : R $$ \\to $$ R be such that for all \nx $$ \\in $$ R (21+x + 21–x), ƒ(x) and (3x + 3–x) are in\nA.P., then the minimum value of ƒ(x) is",
"options": [
{
"text": "2"
},
{
"text": "0"
},
{
"text": "3"
},
{
"text": "4"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\nf(x) = $${{2\\left( {{2^x} + {2^{ - x}}} \\right) + \\left( {{3^x} + {3^{ - x}}} \\right)} \\over 2} \\ge 3$$\n
As we know, A.M > G.M",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7021,
"subject": "General Science",
"question": "Five numbers are in A.P. whose sum is 25 and product is 2520. If one of these five numbers is -$${1 \\over 2}$$ , then the greatest number amongst them is:\n",
"options": [
{
"text": "$${{21} \\over 2}$$"
},
{
"text": "27"
},
{
"text": "7"
},
{
"text": "16"
}
],
"answer": "16",
"solution": "**Answer:** 16\n\nLet the A.P is\n a - 2d, a - d, a, a + d, a + 2d\n $$ \\because $$ sum = 25 $$ \\Rightarrow $$ 5a = 25 $$ \\Rightarrow $$ a = 5\n
$$ \\Rightarrow $$ d = $$ \\pm $$1, d = $$ \\pm {{11} \\over 2}$$\n
When d = $$ \\pm $$1 we can't get any fraction term like -$${1 \\over 2}$$.\n
$$ \\therefore $$ d = $$ \\pm {{11} \\over 2}$$\n
And when d = $${{11} \\over 2}$$\n
we get largest term = 5 + 2d = 5 + 11 = 16",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7022,
"subject": "General Science",
"question": "The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is _________.",
"options": [],
"answer": "1000",
"solution": "**Answer:** 1000\n\nLet N be the four digit number
gcd(N, 18) = 3
Hence N is an odd integer which is divisible by 3 but not by 9.
4 digit odd multiples of 3
1005, 1011, ..........., 9999 $$ \\to $$ 1500
4 digit odd multiples of 9
1017, 1035, ..........., 9999 $$ \\to $$ 500
Hence number of such N = 1000",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7023,
"subject": "General Science",
"question": "Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2 $$-$$ S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to :",
"options": [
{
"text": "7000"
},
{
"text": "1000"
},
{
"text": "3000"
},
{
"text": "5000"
}
],
"answer": "3000",
"solution": "**Answer:** 3000\n\nS1 = $${{2n} \\over 2}$$[2a + (2n $$-$$ 1)d]
S6 = $${{6n} \\over 2}$$[2a + (6n $$-$$ 1)d] = 3(S2 $$-$$ S1) = 3000",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7024,
"subject": "General Science",
"question": "Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 $$-$$ S6 is equal to:",
"options": [
{
"text": "1862"
},
{
"text": "1842"
},
{
"text": "1852"
},
{
"text": "1872"
}
],
"answer": "1862",
"solution": "**Answer:** 1862\n\nLet first term of A.P. be a and common difference is d.
$$ = 1862$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7025,
"subject": "General Science",
"question": "The sum of all the elements in the set {n$$\\in$$ {1, 2, ....., 100} | H.C.F. of n and 2040 is 1} is equal to _____________.",
"options": [],
"answer": "1251",
"solution": "**Answer:** 1251\n\n2040 = 23 $$\\times$$ 3 $$\\times$$ 5 $$\\times$$ 17
= 1251",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7026,
"subject": "General Science",
"question": "Let Sn be the sum of the first n terms of an arithmetic progression. If S3n = 3S2n, then the value of $${{{S_{4n}}} \\over {{S_{2n}}}}$$ is :",
"options": [
{
"text": "6"
},
{
"text": "4"
},
{
"text": "2"
},
{
"text": "8"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\nLet a be first term and d be common diff. of this A.P.
Or x = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7028,
"subject": "General Science",
"question": "The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit \"1\" and they all are multiple of 11, is _____________.",
"options": [],
"answer": "7744",
"solution": "**Answer:** 7744\n\n209, 220, 231, ..........., 495
Let 3, 6, 9, 12, ....... upto 78 terms and 5, 9, 13, 17, ...... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ________.
Let A = {1, a1, a2 ....... a18, 77} be a set of integers with 1 < a1 < a2 < ....... < a18 < 77.
Let the set A + A = {x + y : x, y $$\\in$$ A} contain exactly 39 elements. Then, the value of a1 + a2 + ...... + a18 is equal to _____________.
",
"options": [],
"answer": "702",
"solution": "**Answer:** 702\n\nIf we write the elements of $A+A$, we can certainly find 39 distinct elements as $1+1,1+a_{1}, 1+a_{2}, \\ldots .1$ $+a_{18}, 1+77, a_{1}+77, a_{2}+77, \\ldots \\ldots a_{18}+77,77+77$.
It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P.\n
If $$\\{ {a_i}\\} _{i = 1}^n$$, where n is an even integer, is an arithmetic progression with common difference 1, and $$\\sum\\limits_{i = 1}^n {{a_i} = 192} ,\\,\\sum\\limits_{i = 1}^{n/2} {{a_{2i}} = 120} $$, then n is equal to :
Let $$a, b$$ be two non-zero real numbers. If $$p$$ and $$r$$ are the roots of the equation $$x^{2}-8 \\mathrm{a} x+2 \\mathrm{a}=0$$ and $$\\mathrm{q}$$ and s are the roots of the equation $$x^{2}+12 \\mathrm{~b} x+6 \\mathrm{~b}=0$$, such that $$\\frac{1}{\\mathrm{p}}, \\frac{1}{\\mathrm{q}}, \\frac{1}{\\mathrm{r}}, \\frac{1}{\\mathrm{~s}}$$ are in A.P., then $$\\mathrm{a}^{-1}-\\mathrm{b}^{-1}$$ is equal to _____________.
",
"options": [],
"answer": "38",
"solution": "**Answer:** 38\n\n$\\because$ Roots of $2 a x^{2}-8 a x+1=0$ are $\\frac{1}{p}$ and $\\frac{1}{r}$ and roots of $6 b x^{2}+12 b x+1=0$ are $\\frac{1}{q}$ and $\\frac{1}{s}$.\n
Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ___________.
Suppose $$a_{1}, a_{2}, \\ldots, a_{n}$$, .. be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is $$5: 17$$ and , $$110 < {a_{15}} < 120$$, then the sum of the first ten terms of the progression is equal to
Sum $=321$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7041,
"subject": "General Science",
"question": "Let $a_1, a_2, a_3, \\ldots$ be an A.P. If $a_7=3$, the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero, then $n !-4 a_{n(n+2)}$ is equal to :",
"options": [
{
"text": "24"
},
{
"text": "$\\frac{381}{4}$"
},
{
"text": "9"
},
{
"text": "$\\frac{33}{4}$"
}
],
"answer": "24",
"solution": "**Answer:** 24\n\n$a_{7}=3 \\Rightarrow a+6 d=3 \\Rightarrow a=3-6 d$\n\n
Let $$a_{1}=8, a_{2}, a_{3}, \\ldots, a_{n}$$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is ___________.
Let $$a_{1}, a_{2}, \\ldots, a_{n}$$ be in A.P. If $$a_{5}=2 a_{7}$$ and $$a_{11}=18$$, then \n
$$12\\left(\\frac{1}{\\sqrt{a_{10}}+\\sqrt{a_{11}}}+\\frac{1}{\\sqrt{a_{11}}+\\sqrt{a_{12}}}+\\ldots+\\frac{1}{\\sqrt{a_{17}}+\\sqrt{a_{18}}}\\right)$$ is equal to ____________.
For three positive integers p, q, r, $${x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}$$ and r = pq + 1 such that 3, 3 log$$_yx$$, 3 log$$_zy$$, 7 log$$_xz$$ are in A.P. with common difference $$\\frac{1}{2}$$. Then r-p-q is equal to
Let $$s_{1}, s_{2}, s_{3}, \\ldots, s_{10}$$ respectively be the sum to 12 terms of 10 A.P. s whose first terms are $$1,2,3, \\ldots .10$$ and the common differences are $$1,3,5, \\ldots \\ldots, 19$$ respectively. Then $$\\sum_\\limits{i=1}^{10} s_{i}$$ is equal to :
",
"options": [
{
"text": "7360"
},
{
"text": "7220"
},
{
"text": "7260"
},
{
"text": "7380"
}
],
"answer": "7260",
"solution": "**Answer:** 7260\n\nWe have 10 arithmetic progressions (A.P.s) with the first terms $$a_i$$ and the common differences $$d_i$$, where $$i = 1, 2, \\ldots, 10$$.\n\n
The first terms are $$a_i = i$$ and the common differences are $$d_i = 2i - 1$$.\n\n
Now, we need to find the sum of the first 12 terms for each A.P. The formula for the sum of the first n terms of an A.P. is:\n\n
Thus, the sum $$\\sum\\limits_{i=1}^{10} s_i$$ is equal to 7260.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7046,
"subject": "General Science",
"question": "
Let $$x_{1}, x_{2}, \\ldots, x_{100}$$ be in an arithmetic progression, with $$x_{1}=2$$ and their mean equal to 200 . If $$y_{i}=i\\left(x_{i}-i\\right), 1 \\leq i \\leq 100$$, then the mean of $$y_{1}, y_{2}, \\ldots, y_{100}$$ is :
Let $$S_{K}=\\frac{1+2+\\ldots+K}{K}$$ and $$\\sum_\\limits{j=1}^{n} S_{j}^{2}=\\frac{n}{A}\\left(B n^{2}+C n+D\\right)$$, where $$A, B, C, D \\in \\mathbb{N}$$ and $$A$$ has least value. Then
",
"options": [
{
"text": "$$A+B+C+D$$ is divisible by 5"
},
{
"text": "$$A+C+D$$ is not divisible by $$B$$"
},
{
"text": "$$A+B=5(D-C)$$"
},
{
"text": "$$A+B$$ is divisible by $$\\mathrm{D}$$"
}
],
"answer": "$$A+B$$ is divisible by $$\\mathrm{D}$$",
"solution": "**Answer:** $$A+B$$ is divisible by $$\\mathrm{D}$$\n\n$$\n\\begin{aligned}\n& \\because S_k=\\frac{1+2+\\ldots+k}{k} \\\\\\\\\n& =\\frac{k(k+1)}{2 k}=\\frac{k+1}{2}\n\\end{aligned}\n$$\n
(D) $A+B=24+2=26$, divisible by 13.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7048,
"subject": "General Science",
"question": "Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10}=390$ and the ratio of the tenth and the fifth terms is $15: 7$, then $\\mathrm{S}_{15}-\\mathrm{S}_5$ is equal to :",
"options": [
{
"text": "800"
},
{
"text": "890"
},
{
"text": "790"
},
{
"text": "690"
}
],
"answer": "790",
"solution": "**Answer:** 790\n\n
To solve this problem, we will start by using the properties of an arithmetic progression (AP).
\n\n
The sum of the first $n$ terms of an AP can be calculated using the formula:\n$$ S_n = \\frac{n}{2} (2a + (n-1)d) $$\nwhere $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference between the terms.
\n\n
Given the information:\n$$ S_{10} = 390 $$
\n\n
We can plug $n=10$ into the sum formula to get:
\n\n
$$ S_{10} = \\frac{10}{2} (2a + (10-1)d) $$
\n\n
$$ 390 = 5(2a + 9d) $$
\n\n
$$ 390 = 10a + 45d $$
\n\n
$$ 78 = 2a + 9d \\quad .........\\text{(1)} $$
\n\n
Next, we're given the ratio of the tenth term ($T_{10}$) to the fifth term ($T_5$):\n$$ \\frac{T_{10}}{T_5} = \\frac{15}{7} $$
\n\n
The $n$th term of an AP is given by:\n
$$ T_n = a + (n-1)d $$
\n\n
So, for the tenth term:\n$$ T_{10} = a + (10-1)d = a + 9d $$
\n\n
And for the fifth term:\n
$$ T_5 = a + (5-1)d = a + 4d $$
\n\n
Now we can write the ratio as:
\n\n
$$ \\frac{a + 9d}{a + 4d} = \\frac{15}{7} $$
\n\n
$$ 7(a + 9d) = 15(a + 4d) $$
\n\n
$$ 7a + 63d = 15a + 60d $$
\n\n
$$ 63d - 60d = 15a - 7a $$
\n\n
$$ 3d = 8a \\quad .........\\text{(2)} $$
\n\n
Now we have two equations (1) and (2):
\n\n
$$ 78 = 2a + 9d \\quad \\text{(1)} $$
\n\n
$$ 3d = 8a \\quad \\text{(2)} $$
\n\n
We can solve these equations simultaneously.
\n\n
From equation (2):
\n\n
$$ d = \\frac{8}{3}a $$
\n\n
Plugging this back into (1):
\n\n
$$ 78 = 2a + 9\\left(\\frac{8}{3}a\\right) $$
\n\n
$$ 78 = 2a + 24a $$
\n\n
$$ 78 = 26a $$
\n\n
$$ a = 3 $$
\n\n
Now we can find $d$:
\n\n
$$ d = \\frac{8}{3}a $$
\n\n
$$ d = \\frac{8}{3} \\times 3 $$
\n\n
$$ d = 8 $$
\n\n
Now we can find $S_{15}$ and $S_5$ using the formula for the sum of an AP.
Therefore, the correct answer is Option C, which is 790.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7049,
"subject": "General Science",
"question": "Let $3,7,11,15, \\ldots, 403$ and $2,5,8,11, \\ldots, 404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to ___________.",
"options": [],
"answer": "6699",
"solution": "**Answer:** 6699\n\n
To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).\n\n
The first AP is:
\n\n
$$3, 7, 11, 15, \\ldots, 403$$
\n\n
The common difference ($d_1$) for the first AP can be calculated by subtracting the first term from the second term:
\n\n
$$d_1 = 7 - 3 = 4$$
\n\n
The second AP is:
\n\n
$$2, 5, 8, 11, \\ldots, 404$$
\n\n
The common difference ($d_2$) for the second AP is:
\n\n
$$d_2 = 5 - 2 = 3$$
\n\n
To find the terms common to both sequences, we need to find a term that appears in both sequences. Any common term must be of the form $3 + 4k$ and $2 + 3l$ for some integers $k$ and $l$. We want to find when these two forms will give us the same number, so we set them equal to each other:
\n\n
$$3 + 4k = 2 + 3l$$
\n\n
Rearranging the terms gives us:
\n\n
$$4k - 3l = 2 - 3$$
\n\n
This simplifies to:
\n\n
$$4k - 3l = -1 ....... (1)$$
\n\n
The solutions to equation $(1)$ will give us the common terms. Notice this is a Diophantine equation (A Diophantine equation is a polynomial equation, usually with two or more variables,) and has an infinite number of solutions. Let's find one such solution. We can see that:
Since $11$ is a common term, we can assert that every common term in both APs will be of the form $11 + m(4 \\times 3)$, where $m$ is a non-negative integer, and $4 \\times 3 = 12$ is the LCM of the common differences of the two APs. Thus, the general form for the common terms would be:
\n\n
$$11 + 12m$$
\n\n
Now we are to find all terms that are common up to $403$ in the first sequence and up to $404$ in the second sequence. Because the first sequence doesn't exceed $403$, we'll use this as our limit:
\n\n
$$11 + 12m \\leq 403$$
\n\n
To find the largest possible integer value for $m$, we solve the inequality:
\n\n
$$12m \\leq 403 - 11$$
\n\n
$$12m \\leq 392$$
\n\n
$$m \\leq 32\\frac{2}{3}$$
\n\n
Since $m$ has to be an integer, the largest possible value for $m$ is $32$. Therefore, the common terms are generated by $m = 0, 1, 2, \\ldots, 32$. There are $32 + 1 = 33$ terms in total.\n\n
We will now sum these up. The sum of an AP is given by the formula:
\n\n
$$S = \\frac{n}{2}(a_1 + a_n)$$
\n\n
Where $S$ is the sum, $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term. Using the formula:
\n\n
$$S = \\frac{33}{2}(11 + (11 + 12 \\times 32))$$
\n\n
$$S = \\frac{33}{2}(11 + 11 + 384)$$
\n\n
$$S = \\frac{33}{2}(11 + 11 + 384)$$
\n\n
$$S = \\frac{33}{2}(406)$$
\n\n
$$S = 33 \\times 203$$
\n\n
$$S = 6699$$
\n\n
Therefore, the sum of the common terms in the two arithmetic progressions is 6699.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7050,
"subject": "General Science",
"question": "The number of common terms in the progressions
$4,9,14,19, \\ldots \\ldots$, up to $25^{\\text {th }}$ term and
$3,6,9,12, \\ldots \\ldots$, up to $37^{\\text {th }}$ term is :",
"options": [
{
"text": "9"
},
{
"text": "8"
},
{
"text": "5"
},
{
"text": "7"
}
],
"answer": "7",
"solution": "**Answer:** 7\n\n
$$4,9,14,19, \\ldots$$, up to $$25^{\\text {th }}$$ term
\n
$$\\mathrm{T}_{25}=4+(25-1) 5=4+120=124$$
\n
$$3,6,9,12, \\ldots$$, up to $$37^{\\text {th }}$$ term
\n
$$\\mathrm{T}_{37}=3+(37-1) 3=3+108=111$$
\n
Common difference of $$\\mathrm{I}^{\\text {st }}$$ series $$\\mathrm{d}_1=5$$
\n
Common difference of $$\\mathrm{II}^{\\text {nd }}$$ series $$\\mathrm{d}_2=3$$
\n
First common term $$=9$$, and their common difference $$=15\\left(\\operatorname{LCM}\\right.$$ of $$\\mathrm{d}_1$$ and $$\\left.\\mathrm{d}_2\\right)$$ then common terms are $$9,24,39,54,69,84,99$$
$$\\text { The } 20^{\\text {th }} \\text { term from the end of the progression } 20,19 \\frac{1}{4}, 18 \\frac{1}{2}, 17 \\frac{3}{4}, \\ldots,-129 \\frac{1}{4} \\text { is : }$$
If $$\\log _e \\mathrm{a}, \\log _e \\mathrm{~b}, \\log _e \\mathrm{c}$$ are in an A.P. and $$\\log _e \\mathrm{a}-\\log _e 2 \\mathrm{~b}, \\log _e 2 \\mathrm{~b}-\\log _e 3 \\mathrm{c}, \\log _e 3 \\mathrm{c} -\\log _e$$ a are also in an A.P, then $$a: b: c$$ is equal to
Let $$S_n$$ denote the sum of first $$n$$ terms of an arithmetic progression. If $$S_{20}=790$$ and $$S_{10}=145$$, then $$\\mathrm{S}_{15}-\\mathrm{S}_5$$ is :
Let $$a_1, a_2, a_3, \\ldots$$ be in an arithmetic progression of positive terms.
\n
Let $$A_k=a_1^2-a_2^2+a_3^2-a_4^2+\\ldots+a_{2 k-1}^2-a_{2 k}^2$$.
\n
If $$\\mathrm{A}_3=-153, \\mathrm{~A}_5=-435$$ and $$\\mathrm{a}_1^2+\\mathrm{a}_2^2+\\mathrm{a}_3^2=66$$, then $$\\mathrm{a}_{17}-\\mathrm{A}_7$$ is equal to ________.
For $$x \\geqslant 0$$, the least value of $$\\mathrm{K}$$, for which $$4^{1+x}+4^{1-x}, \\frac{\\mathrm{K}}{2}, 16^x+16^{-x}$$ are three consecutive terms of an A.P., is equal to :
To determine the least value of $$\\mathrm{K}$$ for which the terms $$4^{1+x} + 4^{1-x}, \\frac{\\mathrm{K}}{2}, 16^x + 16^{-x}$$ form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.
\n\n
For three numbers to be in an arithmetic progression, the middle term must be the average of the other two terms. Therefore, we can write:
A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $$\\mathrm{m}$$ is equal to:
To determine the value of $$\\mathrm{m}$$, we need to formulate the problem using some basic concepts of arithmetic progression and work. Let's first understand the nature of the problem:
\n\n
Initially, there are $$\\mathrm{m}$$ computers, and it is estimated that with these $$\\mathrm{m}$$ computers, the assignment can be completed in 17 days.
\n\n
However, due to the crash of 4 computers every day starting from the second day onward, the total time taken extends by 8 days, making it 25 days in total.
\n\n
To begin with, let's define the total work (W) in terms of the number of computers and days:
\n\n
The total work (W) is given by:
\n\n
The amount of work completed each day with $$\\mathrm{m}$$ computers for 17 days:
\n\n
$$W = 17m$$
\n\n
When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are $$\\mathrm{m}$$ computers. On the second day, there are $$\\mathrm{m} - 4$$ computers, on the third day, there are $$\\mathrm{m} - 8$$ computers, and so on. We need to sum this series until 25 days are completed.
\n\n
This can be formulated as:
\n\n
Total work done over 25 days with decrement in the number of computers:
where $$n$$ is the number of days. Here, $$n = 25$$.
\n\n
Notice that we form an arithmetic series where the first term (a) is $$\\mathrm{m}$$ and the common difference (d) is -4. The sum of the first n terms of an arithmetic series is:
Let for $$f(x) = {a_0}{x^2} + {a_1}x + {a_2},\\,f'(0) = 1$$ and $$f'(1) = 0$$. If a0, a1, a2 are in an arithmatico-geometric progression, whose corresponding A.P. has common difference 1 and corresponding G.P. has common ratio 2, then f(4) is equal to _____________.
Suppose $$a_{1}, a_{2}, 2, a_{3}, a_{4}$$ be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $$\\frac{49}{2}$$, then $$a_{4}$$ is equal to __________.
",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\nSince, common ratio of A.G.P. is 2 therefore A.G.P. can be taken as\n
Let $$0 < z < y < x$$ be three real numbers such that $$\\frac{1}{x}, \\frac{1}{y}, \\frac{1}{z}$$ are in an arithmetic progression and $$x, \\sqrt{2} y, z$$ are in a geometric progression. If $$x y+y z+z x=\\frac{3}{\\sqrt{2}} x y z$$ , then $$3(x+y+z)^{2}$$ is equal to ____________.
",
"options": [],
"answer": "150",
"solution": "**Answer:** 150\n\n$\\because \\frac{1}{x}, \\frac{1}{y}, \\frac{1}{z}$ are in A.P.\n
If $$1+\\frac{\\sqrt{3}-\\sqrt{2}}{2 \\sqrt{3}}+\\frac{5-2 \\sqrt{6}}{18}+\\frac{9 \\sqrt{3}-11 \\sqrt{2}}{36 \\sqrt{3}}+\\frac{49-20 \\sqrt{6}}{180}+\\ldots$$ upto $$\\infty=2+\\left(\\sqrt{\\frac{b}{a}}+1\\right) \\log _e\\left(\\frac{a}{b}\\right)$$, where a and b are integers with $$\\operatorname{gcd}(a, b)=1$$, then $$\\mathrm{11 a+18 b}$$ is equal to __________.
Let the first term of a series be $$T_1=6$$ and its $$r^{\\text {th }}$$ term $$T_r=3 T_{r-1}+6^r, r=2,3$$,\n............ $$n$$. If the sum of the first $$n$$ terms of this series is $$\\frac{1}{5}\\left(n^2-12 n+39\\right)\\left(4 \\cdot 6^n-5 \\cdot 3^n+1\\right)$$, then $$n$$ is equal to ___________.
$$ = \\left| {\\matrix{\n 0 & p & 1 \\cr \n 0 & q & 1 \\cr \n 0 & r & 1 \\cr \n\n } } \\right| = 0$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7066,
"subject": "General Science",
"question": "Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is",
"options": [
{
"text": "5 "
},
{
"text": "3/5 "
},
{
"text": "8/5 "
},
{
"text": "1/5 "
}
],
"answer": "3/5 ",
"solution": "**Answer:** 3/5 \n\nLet $$a=$$ first team of $$G.P.$$ and $$r=$$ common ratio of $$G.P.;$$\n
$$\\therefore$$ $$r$$ should be positive] ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7068,
"subject": "General Science",
"question": "The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is ",
"options": [
{
"text": "- 4"
},
{
"text": "- 12"
},
{
"text": "12"
},
{
"text": "4"
}
],
"answer": "- 12",
"solution": "**Answer:** - 12\n\nAs per question, \n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,a + ar = 12\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( 1 \\right)$$\n
$$ \\Rightarrow {r^2} = 4, \\Rightarrow r = - 2$$\n
(As terms are $$=+ve$$ and $$-ve$$ alternately)\n
$$ \\Rightarrow a = - 12$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7069,
"subject": "General Science",
"question": "Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is :",
"options": [
{
"text": "$$2 - \\sqrt 3 $$ "
},
{
"text": "$$2 + \\sqrt 3 $$ "
},
{
"text": "$$\\sqrt 2 + \\sqrt 3 $$ "
},
{
"text": "$$3 + \\sqrt 2 $$ "
}
],
"answer": "$$2 + \\sqrt 3 $$ ",
"solution": "**Answer:** $$2 + \\sqrt 3 $$ \n\nLet $$a,ar,a{r^2}$$ are in $$G.P.$$ \n
$$ \\therefore $$$$\\,\\,\\,$$ a = $${1 \\over 4} - {1 \\over {2\\sqrt 2 }}$$ ($$ \\because $$ a < b)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7073,
"subject": "General Science",
"question": "If three distinct numbers a, b, c are in G.P. and the\nequations ax2\n + 2bx + c = 0 and\ndx2\n + 2ex + ƒ = 0 have a common root, then\nwhich one of the following statements is\ncorrect?",
"options": [
{
"text": "$$d \\over a$$, $$e \\over b$$, $$f \\over c$$ are in G.P."
},
{
"text": "d, e, ƒ are in A.P"
},
{
"text": "d, e, ƒ are in G.P"
},
{
"text": "$$d \\over a$$, $$e \\over b$$, $$f \\over c$$ are in A.P."
}
],
"answer": "$$d \\over a$$, $$e \\over b$$, $$f \\over c$$ are in A.P.",
"solution": "**Answer:** $$d \\over a$$, $$e \\over b$$, $$f \\over c$$ are in A.P.\n\nGiven, a, b, c are in G.P.\n
$$ \\therefore $$ b2 = ac\n
In this equation ax2\n + 2bx + c = 0,\n
Discrimant, D = 4b2 - 4ac\n
= 4ac - 4ac\n
= 0\n
Discrimant = 0 meand roots of the equation are equal.\n
Let both the roots of the equation = $$\\alpha $$\n
$$ \\therefore $$ \n$$d \\over a$$, $$e \\over b$$, $$f \\over c$$ are in A.P.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7074,
"subject": "General Science",
"question": "Let $$a$$, b and c be in G.P. with common ratio r, where $$a$$ $$ \\ne $$ 0 and 0 < r $$ \\le $$ $${1 \\over 2}$$\n. If 3$$a$$, 7b and 15c are the first three\nterms of an A.P., then the 4th term of this A.P. is : ",
"options": [
{
"text": "$$a$$"
},
{
"text": "$${7 \\over 3}a$$"
},
{
"text": "5$$a$$"
},
{
"text": "$${2 \\over 3}a$$"
}
],
"answer": "$$a$$",
"solution": "**Answer:** $$a$$\n\na = a, b = ar and c = ar2
\n3a, 7b, 15c $$ \\to $$ A.P.
\n14b = 3a + 15c
\n14(ar) = 3a + 15(ar2)
\n15r2 – 14r + 3 = 0
\n$$ \\Rightarrow r = {1 \\over 3},{3 \\over 5}(rejected)$$
\n4th term is $$ \\Rightarrow $$ $$15c - {2 \\over 3}a = {{15} \\over 9}a - {2 \\over 3}a = a$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7075,
"subject": "General Science",
"question": "The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is : ",
"options": [
{
"text": "36"
},
{
"text": "28"
},
{
"text": "32"
},
{
"text": "24"
}
],
"answer": "28",
"solution": "**Answer:** 28\n\nLet terms are $${a \\over r},a,ar \\to G.P$$\n
$${{{a_9}} \\over {{a_5}}} = {{{a_1}{r^8}} \\over {{a_1}{r^4}}} = {r^4} = {5^4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7077,
"subject": "General Science",
"question": "The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $${{27} \\over {19}}$$.Then the common ratio of this series is :",
"options": [
{
"text": "$${4 \\over 9}$$"
},
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${2 \\over 3}$$"
},
{
"text": "$${2 \\over 9}$$"
}
],
"answer": "$${2 \\over 3}$$",
"solution": "**Answer:** $${2 \\over 3}$$\n\n$${a \\over {1 - r}} = 3\\,\\,\\,\\,\\,\\,\\,.....(1)$$\n
as $$\\left| r \\right| < 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7078,
"subject": "General Science",
"question": "Let a1, a2, a3, ..... a10 be in G.P. with ai > 0 for i = 1, 2, ….., 10 and S be the set of pairs (r, k), r, k $$ \\in $$ N (the set of natural numbers) for which\n
Then the number of elements in S, is -",
"options": [
{
"text": "10"
},
{
"text": "4"
},
{
"text": "2"
},
{
"text": "infinitely many"
}
],
"answer": "infinitely many",
"solution": "**Answer:** infinitely many\n\nApply\n
C3 $$ \\to $$ C3 $$-$$ C2\n
C2 $$ \\to $$ C2 $$-$$ C1\n
We get D = 0",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7079,
"subject": "General Science",
"question": "If a, b, c be three distinct real numbers in G.P. and a + b + c = xb , then x cannot be",
"options": [
{
"text": "2"
},
{
"text": "-3"
},
{
"text": "4"
},
{
"text": "-2"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\na, b, c are in G.P.\n
$$ \\Rightarrow $$ $$\\lambda $$ = -171",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7081,
"subject": "General Science",
"question": "Let an be the nth term of a G.P. of positive terms.
= $${1 \\over {26}}\\left( {{3^{50}} - 1} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7085,
"subject": "General Science",
"question": "Let a , b, c , d and p be any non zero distinct real numbers such that\n (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :",
"options": [
{
"text": "a, c, p are in G.P."
},
{
"text": "a, b, c, d are in G.P."
},
{
"text": "a, b, c, d are in A.P. "
},
{
"text": "a, c, p are in A.P."
}
],
"answer": "a, b, c, d are in G.P.",
"solution": "**Answer:** a, b, c, d are in G.P.\n\n(a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0\n
$$ \\therefore $$ a, b, c, d are in G.P.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7086,
"subject": "General Science",
"question": "The sum of first four terms of a geometric progression (G. P.) is $${{65} \\over {12}}$$ and the sum of their respective reciprocals is $${{65} \\over {18}}$$. If the product of first three terms of the G.P. is 1, and the third term is $$\\alpha$$, then 2$$\\alpha$$ is _________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nLet the terms are $$a,ar,a{r^2},a{r^3}$$
$$2\\alpha = 3$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7087,
"subject": "General Science",
"question": "Let A1, A2, A3, ....... be squares such that for each n $$ \\ge $$ 1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one, is __________.",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\n\n
Smallest n = 9",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7088,
"subject": "General Science",
"question": "In an increasing geometric series, the sum of the second and the sixth term is $${{25} \\over 2}$$ and the product of the third and fifth term is 25. Then, the sum of 4th, 6th and 8th terms is equal to :",
"options": [
{
"text": "30"
},
{
"text": "32"
},
{
"text": "26"
},
{
"text": "35"
}
],
"answer": "35",
"solution": "**Answer:** 35\n\na, ar, ar2, .....
$$ = 5(1 + 2 + 4) = 35$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7089,
"subject": "General Science",
"question": "Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to ___________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nA.P. from the set will be 11, 16, 21, 26 .....\n
G.P. from the set will be 4, 8, 16, 32, 64, 128, 256,\n512, 1024, 2048, 4096, 8192 .....\n
So common terms are 16, 256, 4096.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7090,
"subject": "General Science",
"question": "Let $${1 \\over {16}}$$, a and b be in G.P. and $${1 \\over a}$$, $${1 \\over b}$$, 6 be in A.P., where a, b > 0. Then 72(a + b) is equal to ___________.",
"options": [],
"answer": "14",
"solution": "**Answer:** 14\n\n$${a^2} = {b \\over {16}}$$ and $${2 \\over b} = {1 \\over a} + 6$$
Solving, we get $$a = {1 \\over {12}}$$ or $$a = - {1 \\over 4}$$ [rejected]
if $$a = {1 \\over {12}} \\Rightarrow b = {1 \\over 9}$$
$$ \\therefore $$ $$72(a + b) = 72\\left( {{1 \\over {12}} + {1 \\over 9}} \\right) = 14$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7091,
"subject": "General Science",
"question": "If the sum of an infinite GP a, ar, ar2, ar3, ....... is 15 and the sum of the squares of its each term is 150, then the sum of ar2, ar4, ar6, ....... is :",
"options": [
{
"text": "$${5 \\over 2}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${25 \\over 2}$$"
},
{
"text": "$${9 \\over 2}$$"
}
],
"answer": "$${1 \\over 2}$$",
"solution": "**Answer:** $${1 \\over 2}$$\n\nSum of infinite terms :
Sum = $${{a{r^2}} \\over {1 - {r^2}}} = {{12.\\left( {{1 \\over {25}}} \\right)} \\over {1 - {1 \\over {25}}}} = {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7092,
"subject": "General Science",
"question": "Let a1, a2, ......., a10 be an AP with common difference $$-$$ 3 and b1, b2, ........., b10 be a GP with common ratio 2. Let ck = ak + bk, k = 1, 2, ......, 10. If c2 = 12 and c3 = 13, then $$\\sum\\limits_{k = 1}^{10} {{c_k}} $$ is equal to _________.",
"options": [],
"answer": "2021",
"solution": "**Answer:** 2021\n\n$$a_{1}, a_{2}, a_{3}, \\ldots, a_{10}$$ are in AP common difference $$=-3$$
$$b_{1}, b_{2}, b_{3}, \\ldots, b_{10}$$ are in GP common ratio $$=2$$
\n$$\n\\begin{aligned}\n&=\\left(\\frac{10}{2}\\right)[22+9(-3)]+2\\left(\\frac{2^{10}-1}{2-1}\\right) \\\\\n&=5(22-27)+2(1023)=2046-25 \\\\\n&=2021\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7093,
"subject": "General Science",
"question": "Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r2, then r2 $$-$$ d is equal to :",
"options": [
{
"text": "7 $$-$$ 7$$\\sqrt 3 $$"
},
{
"text": "7 + $$\\sqrt 3 $$"
},
{
"text": "7 $$-$$ $$\\sqrt 3 $$"
},
{
"text": "7 + 3$$\\sqrt 3 $$"
}
],
"answer": "7 + $$\\sqrt 3 $$",
"solution": "**Answer:** 7 + $$\\sqrt 3 $$\n\nLet numbers be $${a \\over r}$$, a, ar $$\\to$$ G.P.
$${a \\over r}$$, 2a, ar $$\\to$$ A.P. $$\\Rightarrow$$ 4a = $${a \\over r}$$ + ar $$\\Rightarrow$$ r + $${1 \\over r}$$ = 4
r = 2 $$\\pm$$ $$\\sqrt 3 $$
4th form of G.P. = 3r2 $$\\Rightarrow$$ ar2 = 3r2 $$\\Rightarrow$$ a = 3
r = 2 + $$\\sqrt 3 $$, a = 3, d = 2a $$-$$ $${a \\over r}$$ = 3$$\\sqrt 3 $$
Let for n = 1, 2, ......, 50, Sn be the sum of the infinite geometric progression whose first term is n2 and whose common ratio is $${1 \\over {{{(n + 1)}^2}}}$$. Then the value of
$${1 \\over {26}} + \\sum\\limits_{n = 1}^{50} {\\left( {{S_n} + {2 \\over {n + 1}} - n - 1} \\right)} $$ is equal to ___________.
Let A1, A2, A3, ....... be an increasing geometric progression of positive real numbers. If A1A3A5A7 = $${1 \\over {1296}}$$ and A2 + A4 = $${7 \\over {36}}$$, then the value of A6 + A8 + A10 is equal to
Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be $$\\frac{98}{25}$$. Then the sum of the first 21 terms of an AP, whose first term is $$10\\mathrm{a r}, \\mathrm{n}^{\\text {th }}$$ term is $$\\mathrm{a}_{\\mathrm{n}}$$ and the common difference is $$10 \\mathrm{ar}^{2}$$, is equal to :
$\\Rightarrow r_{1}+r_{2}=7$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7099,
"subject": "General Science",
"question": "Let $a, b, c>1, a^3, b^3$ and $c^3$ be in A.P., and $\\log _a b, \\log _c a$ and $\\log _b c$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $\\frac{a+4 b+c}{3}$ and the common difference is $\\frac{a-8 b+c}{10}$ is $-444$, then $a b c$ is equal to :",
"options": [
{
"text": "343"
},
{
"text": "216"
},
{
"text": "$\\frac{343}{8}$"
},
{
"text": "$\\frac{125}{8}$"
}
],
"answer": "216",
"solution": "**Answer:** 216\n\n
Let $$\\{ {a_k}\\} $$ and $$\\{ {b_k}\\} ,k \\in N$$, be two G.P.s with common ratios $${r_1}$$ and $${r_2}$$ respectively such that $${a_1} = {b_1} = 4$$ and $${r_1} < {r_2}$$. Let $${c_k} = {a_k} + {b_k},k \\in N$$. If $${c_2} = 5$$ and $${c_3} = {{13} \\over 4}$$ then $$\\sum\\limits_{k = 1}^\\infty {{c_k} - (12{a_6} + 8{b_4})} $$ is equal to __________.
Let $$a_1,a_2,a_3,...$$ be a $$GP$$ of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $$a_1a_9+a_2a_4a_9+a_5+a_7$$ is equal to __________.
",
"options": [],
"answer": "60",
"solution": "**Answer:** 60\n\nLet $r$ be the common ratio of the G.P\n
For the two positive numbers $$a,b,$$ if $$a,b$$ and $$\\frac{1}{18}$$ are in a geometric progression, while $$\\frac{1}{a},10$$ and $$\\frac{1}{b}$$ are in an arithmetic progression, then $$16a+12b$$ is equal to _________.
The 4$$^\\mathrm{th}$$ term of GP is 500 and its common ratio is $$\\frac{1}{m},m\\in\\mathbb{N}$$. Let $$\\mathrm{S_n}$$ denote the sum of the first n terms of this GP. If $$\\mathrm{S_6 > S_5 + 1}$$ and $$\\mathrm{S_7 < S_6 + \\frac{1}{2}}$$, then the number of possible values of m is ___________
Let a$$_1$$, a$$_2$$, a$$_3$$, .... be a G.P. of increasing positive numbers. Let the sum of its 6th and 8th terms be 2 and the product of its 3rd and 5th terms be $$\\frac{1}{9}$$. Then $$6(a_2+a_4)(a_4+a_6)$$ is equal to
Let the first term $$\\alpha$$ and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to
",
"options": [
{
"text": "241"
},
{
"text": "231"
},
{
"text": "220"
},
{
"text": "210"
}
],
"answer": "231",
"solution": "**Answer:** 231\n\nGiven that the first term $a$ and common ratio $r$ of a geometric progression be positive integer. So, their 1st three terms are $a, a r, a r^2$\n
According to the question, $a^2+a^2 r^2+a^2 r^4=33033$\n
So, sum of these first three terms is $a+a r+a r^2$ $=11+44+176=231$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7106,
"subject": "General Science",
"question": "
On subtracting Equation (ii) from Equation (i), we get\n
$$\n\\begin{aligned}\n& k\\left(\\frac{-1}{20}\\right)=1+\\frac{21}{20}+\\left(\\frac{21}{20}\\right)^2+\\ldots \\ldots+\\left(\\frac{21}{20}\\right)^{19}-20\\left(\\frac{21}{20}\\right)^{20} \\\\\\\\\n& \\Rightarrow k\\left(\\frac{-1}{20}\\right)=\\frac{1\\left(\\left(\\frac{21}{20}\\right)^{20}-1\\right)}{\\frac{21}{20}-1}-20\\left(\\frac{21}{20}\\right)^{20} \\\\\\\\\n& \\Rightarrow k\\left(\\frac{-1}{20}\\right)=20\\left(\\left(\\frac{21}{20}\\right)^{20}-1\\right)-20\\left(\\frac{21}{20}\\right)^{20} \\\\\\\\\n& =20\\left(\\frac{21}{20}\\right)^{20}-20-20\\left(\\frac{21}{20}\\right)^{20} \\\\\\\\\n& \\Rightarrow k\\left(\\frac{-1}{20}\\right)=-20 \\\\\\\\\n& \\Rightarrow k=400\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7107,
"subject": "General Science",
"question": "If three successive terms of a G.P. with common ratio $\\mathrm{r}(\\mathrm{r}>1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to _____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n
To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$, $ar$, and $ar^2$, where $a$ is the first term and $r > 1$. These three terms represent the lengths of the sides of a triangle.
\n
According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, for the three terms to form a triangle, the following inequalities must hold:
\n
$$\n1) \\ a + ar > ar^2 \\\\\\\\\n$$\n
$$2) \\ a + ar^2 > ar $$\n
$$\n3) \\ ar + ar^2 > a\n$$
\n
Given that $r > 1$, inequalities 2 and 3 will always hold because:
\n
$$\nar < ar^2 \\ \\text{and} \\ a < ar,\n$$
\n
indicating that both $a + ar^2$ and $ar + ar^2$ will be greater than $ar$ and $a$ respectively. Therefore, we only need to check the first inequality to ensure that the three terms can form a triangle:
\n
$$\na + ar > ar^2\n$$
\n
Simplifying this, we get:
\n
$$\na(1 + r) > a r^2\n$$
\n
Since $a$ is positive (as it represents the length of a side of a triangle), we can divide both sides of the inequality by $a$ without changing the direction of the inequality:
\n
$$\n1 + r > r^2\n$$
\n
We can then subtract $r$ from both sides:
\n
$$\n1 > r^2 - r\n$$
\n
Simplifying the right side by factoring $r$:
\n
$$\n1 > r(r - 1)\n$$
\n
Given that $r > 1$, the quantity $(r - 1)$ is positive; hence, $r(r - 1)$ is also positive. This means the actual value for $r$ to satisfy the inequality is within the interval $(1, \\sqrt{2})$ because $r(r - 1)$ increases with increasing $r$, and it would be 1 when $r = \\sqrt{2}$. It should be greater than 1, and less than $\\sqrt{2}$ such that $r^2 - r$ stays below 1.
\n
Now let’s consider the expressions $[r]$ and $[-r]$. The symbol $[x]$ denotes the greatest integer less than or equal to $x$ (also known as the floor function).
\n
Since $1 < r < \\sqrt{2}$, $[r] = 1$, because 1 is the greatest integer less than $r$ within that interval.
\n
For $[-r]$, we need the greatest integer less than or equal to $-r$. Since $-r$ is negative and less than $-1$ (because $r > 1$), $[-r] = -2$, as this is the greatest integer that does not exceed the negative value of $r$ (which lies between $-\\sqrt{2}$ and $-1$).
\n
Now we can substitute these values into the expression:
Let $$2^{\\text {nd }}, 8^{\\text {th }}$$ and $$44^{\\text {th }}$$ terms of a non-constant A. P. be respectively the $$1^{\\text {st }}, 2^{\\text {nd }}$$ and $$3^{\\text {rd }}$$ terms of a G. P. If the first term of the A. P. is 1, then the sum of its first 20 terms is equal to -
If each term of a geometric progression $$a_1, a_2, a_3, \\ldots$$ with $$a_1=\\frac{1}{8}$$ and $$a_2 \\neq a_1$$, is the arithmetic mean of the next two terms and $$S_n=a_1+a_2+\\ldots . .+a_n$$, then $$S_{20}-S_{18}$$ is equal to
Let $$a$$ and $$b$$ be be two distinct positive real numbers. Let $$11^{\\text {th }}$$ term of a GP, whose first term is $$a$$ and third term is $$b$$, is equal to $$p^{\\text {th }}$$ term of another GP, whose first term is $$a$$ and fifth term is $$b$$. Then $$p$$ is equal to
The problem involves finding a relation between terms of two different geometric progressions (GPs) which share common first terms but have different terms equated to the same value. We solve this by setting up equations based on the given conditions for each GP and comparing the terms specified to be equal.
For the first GP, with first term $$a$$ and third term $$b$$:
The third term is given by $$t_3 = ar^2 = b$$, leading to $$r^2 = \\frac{b}{a}$$.
The eleventh term is $$t_{11} = ar^{10} = a\\left(\\frac{b}{a}\\right)^5$$.
For the second GP, with first term $$a$$ and fifth term $$b$$:
The fifth term is $$T_5 = ar^4 = b$$, yielding $$r^4 = \\frac{b}{a}$$ and $$r = \\left(\\frac{b}{a}\\right)^{1/4}$$.
The $$p^{\\text{th}}$$ term is thus $$T_p = ar^{p-1} = a\\left(\\frac{b}{a}\\right)^{\\frac{p-1}{4}}$$.
Equating the eleventh term of the first GP to the $$p^{\\text{th}}$$ term of the second GP gives: $$a\\left(\\frac{b}{a}\\right)^5 = a\\left(\\frac{b}{a}\\right)^{\\frac{p-1}{4}}$$.
Simplifying, we find that $$5 = \\frac{p-1}{4}$$, leading to $$p = 21$$, which is the solution.
Let $$a, a r, a r^2$$, ............ be an infinite G.P. If $$\\sum_\\limits{n=0}^{\\infty} a r^n=57$$ and $$\\sum_\\limits{n=0}^{\\infty} a^3 r^{3 n}=9747$$, then $$a+18 r$$ is equal to
Let the first three terms 2, p and q, with $$q \\neq 2$$, of a G.P. be respectively the $$7^{\\text {th }}, 8^{\\text {th }}$$ and $$13^{\\text {th }}$$ terms of an A.P. If the $$5^{\\text {th }}$$ term of the G.P. is the $$n^{\\text {th }}$$ term of the A.P., then $n$ is equal to:
In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $$\\frac{70}{3}$$ and the product of the third and fifth terms is 49. Then the sum of the $$4^{\\text {th }}, 6^{\\text {th }}$$ and $$8^{\\text {th }}$$ terms is equal to:
Let's denote the first term of the geometric progression by $$a$$ and the common ratio by $$r$$. The terms of the geometric progression can be written as follows:
\n\n
First term: $$a$$
\n\n
Second term: $$ar$$
\n\n
Third term: $$(ar^2)$$
\n\n
Fourth term: $$(ar^3)$$
\n\n
Fifth term: $$(ar^4)$$
\n\n
Sixth term: $$(ar^5)$$
\n\n
Eighth term: $$(ar^7)$$
\n\n
We are given two key pieces of information:
\n\n
1. The sum of the second and sixth terms is $$\\frac{70}{3}$$:
\n\n
$$ar + ar^5 = \\frac{70}{3}$$
\n\n
2. The product of the third and fifth terms is 49:
\n\n
$$(ar^2) \\cdot (ar^4) = 49$$
\n\n
$$a^2 r^6 = 49$$
\n\n
$$a^2 = \\frac{49}{r^6}$$
\n\n
$$a = \\frac{7}{r^3}$$
\n\n
Substituting $$a = \\frac{7}{r^3}$$ into the first equation:
\n\n
$$\\frac{7}{r^3} \\cdot r + \\frac{7}{r^3} \\cdot r^5 = \\frac{70}{3}$$
Solve this quadratic equation using the quadratic formula:
\n\n
$$x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$
\n\n
Where $$a = 3$$, $$b = -10$$, and $$c = 3$$. Thus:
\n\n
$$x = \\frac{10 \\pm \\sqrt{100 - 36}}{6}$$
\n\n
$$x = \\frac{10 \\pm \\sqrt{64}}{6}$$
\n\n
$$x = \\frac{10 \\pm 8}{6}$$
\n\n
$$x = 3$$ or $$x = \\frac{1}{3}$$
\n\n
Since $$x = r^2$$ and $$r$$ is positive, we get $$r = \\sqrt{3}$$ or $$r = \\frac{1}{\\sqrt{3}}$$. We need to choose the value that results in positive, increasing terms:
Let $$A B C$$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $$A B C$$ and the same process is repeated infinitely many times. If $$\\mathrm{P}$$ is the sum of perimeters and $$Q$$ is be the sum of areas of all the triangles formed in this process, then :
Statement-1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) +.....+ (361 + 380 + 400) is 8000. \n
Statement-2: $$\\sum\\limits_{k = 1}^n {\\left( {{k^3} - {{(k - 1)}^3}} \\right)} = {n^3}$$, for any natural number n.
\n",
"options": [
{
"text": "Statement-1 is false, Statement-2 is true."
},
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."
},
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."
},
{
"text": "Statement-1 is true, Statement-2 is false."
}
],
"answer": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.",
"solution": "**Answer:** Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.\n\n$$n$$th term of the given series\n
Given $$ \\Rightarrow k = 100$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7130,
"subject": "General Science",
"question": "The sum of first 9 terms of the series. \n
$$ \\Rightarrow m = 101.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7132,
"subject": "General Science",
"question": "Let z = 1 + ai be a complex number, a > 0, such that z3 is a real number. \n
$$ \\Rightarrow $$ $${{{2^{12}} - 1} \\over {\\sqrt 3 i}} = {{4095} \\over {\\sqrt 3 i}} = - {{4095} \\over 3}\\sqrt 3 i = - 1365\\sqrt 3 i$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7133,
"subject": "General Science",
"question": "For any three positive real numbers a, b and c,\n
9(25$${a^2}$$ + b2) + 25(c2 - 3$$a$$c) = 15b(3$$a$$ + c).\n Then",
"options": [
{
"text": "b, c and $$a$$ are in G.P."
},
{
"text": "b, c and $$a$$ are in A.P."
},
{
"text": "$$a$$, b and c are in A.P."
},
{
"text": "$$a$$, b and c are in G.P."
}
],
"answer": "b, c and $$a$$ are in A.P.",
"solution": "**Answer:** b, c and $$a$$ are in A.P.\n\n9(25$${a^2}$$ + b2) + 25(c2 - 3$$a$$c) = 15b(3$$a$$ + c)\n
$$\\therefore\\,\\,\\,$$ n = $${{1 + 59} \\over 4}$$ = 15 or $${{1 - 59} \\over 4}$$ = $$-$$ 14.5\n
$$\\therefore\\,\\,\\,$$ n = 15 (as n can't be $$-$$ve)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7136,
"subject": "General Science",
"question": "Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series\n 12 + 2.22 + 32 + 2.42 + 52 + 2.62 ...........\n If B - 2A = 100$$\\lambda $$, then $$\\lambda $$ is equal to",
"options": [
{
"text": "496"
},
{
"text": "232"
},
{
"text": "248"
},
{
"text": "464"
}
],
"answer": "248",
"solution": "**Answer:** 248\n\nNote : \n
$$ \\Rightarrow \\,\\,\\,\\lambda $$ = 248",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7137,
"subject": "General Science",
"question": "The sum of the first 20 terms of the series \n
505 $$ = {5 \\over 3}$$ A, A = 303",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7146,
"subject": "General Science",
"question": "The sum of the following series\n
Sum upto 15 terms $$ \\Rightarrow $$ then put $$n$$ = 15\n
$$ = {1 \\over 2}\\left( {{{{{\\left( {15 \\times 16} \\right)}^2}} \\over 4} + {{15 \\times 16 \\times 31} \\over 6}} \\right) = 7820$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7147,
"subject": "General Science",
"question": "Some identical balls are arranged in rows to form\nan equilateral triangle. The first row consists of one\nball, the second row consists of two balls and so\non. If 99 more identical balls are addded to the total\nnumber of balls used in forming the equilaterial\ntriangle, then all these balls can be arranged in a\nsquare whose each side contains exactly 2 balls\nless than the number of balls each side of the\ntriangle contains. Then the number of balls used to\nform the equilateral triangle is :-",
"options": [
{
"text": "262"
},
{
"text": "190"
},
{
"text": "157"
},
{
"text": "225"
}
],
"answer": "190",
"solution": "**Answer:** 190\n\n\n
As triangle is equilateral so each side of triangle has n-balls.\n
At the top of the triangle there is 1 ball then next line has 2 balls. Similarly last line has n balls.\n
$$ \\therefore $$ Total no of balls used to create this triangle\n
= 1 + 2 + 3 + ..............+ n\n
= $${{n\\left( {n + 1} \\right)} \\over 2}$$\n\n Here Number of balls in each side of square is = (n – 2)\n
$$ \\therefore $$ Total no of balls used to create this square\n
= 1540",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7156,
"subject": "General Science",
"question": "If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ..... is (102)m, then m is equal to :",
"options": [
{
"text": "20"
},
{
"text": "5"
},
{
"text": "10"
},
{
"text": "25"
}
],
"answer": "20",
"solution": "**Answer:** 20\n\n3 + 4 + 8 + 9 + 13 + 14 +…….upto 40 terms\n
$$ \\Rightarrow $$ 7 + 17 + 27 +…….20 terms\n
S = $${{20} \\over 2}$$ [2 × 7 + 19 × 10]\n
= 102 × 20 = 102 m\n
$$ \\therefore $$ m = 20",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7157,
"subject": "General Science",
"question": "If |x| < 1, |y| < 1 and x $$ \\ne $$ y, then the sum to infinity\nof the following series\n
$$ \\Rightarrow $$ S = $${13 \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7160,
"subject": "General Science",
"question": "The sum of the series
$$\\Rightarrow$$ l2 = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7168,
"subject": "General Science",
"question": "The sum of the series
Let $$\\{ {a_n}\\} _{n = 0}^\\infty $$ be a sequence such that $${a_0} = {a_1} = 0$$ and $${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$$ for all n $$\\ge$$ 0. Then, $$\\sum\\limits_{n = 2}^\\infty {{{{a_n}} \\over {{7^n}}}} $$ is equal to:
For a natural number n, let $${\\alpha _n} = {19^n} - {12^n}$$. Then, the value of $${{31{\\alpha _9} - {\\alpha _{10}}} \\over {57{\\alpha _8}}}$$ is ___________.
The greatest integer less than or equal to the sum of first 100 terms of the sequence $${1 \\over 3},{5 \\over 9},{{19} \\over {27}},{{65} \\over {81}},$$ ...... is equal to ___________.
The series of positive multiples of 3 is divided into sets : $$\\{3\\},\\{6,9,12\\},\\{15,18,21,24,27\\}, \\ldots$$ Then the sum of the elements in the $$11^{\\text {th }}$$ set is equal to ____________.
Consider the sequence $$a_{1}, a_{2}, a_{3}, \\ldots$$ such that $$a_{1}=1, a_{2}=2$$ and $$a_{n+2}=\\frac{2}{a_{n+1}}+a_{n}$$ for $$\\mathrm{n}=1,2,3, \\ldots .$$ If $$\\left(\\frac{\\mathrm{a}_{1}+\\frac{1}{\\mathrm{a}_{2}}}{\\mathrm{a}_{3}}\\right) \\cdot\\left(\\frac{\\mathrm{a}_{2}+\\frac{1}{\\mathrm{a}_{3}}}{\\mathrm{a}_{4}}\\right) \\cdot\\left(\\frac{\\mathrm{a}_{3}+\\frac{1}{\\mathrm{a}_{4}}}{\\mathrm{a}_{5}}\\right) \\ldots\\left(\\frac{\\mathrm{a}_{30}+\\frac{1}{\\mathrm{a}_{31}}}{\\mathrm{a}_{32}}\\right)=2^{\\alpha}\\left({ }^{61} \\mathrm{C}_{31}\\right)$$, then $$\\alpha$$ is equal to :
If $$\\frac{1}{2 \\times 3 \\times 4}+\\frac{1}{3 \\times 4 \\times 5}+\\frac{1}{4 \\times 5 \\times 6}+\\ldots+\\frac{1}{100 \\times 101 \\times 102}=\\frac{\\mathrm{k}}{101}$$, then 34 k is equal to _________.
$$\n\\begin{aligned}\n&\\text { Let }\\left\\{a_{n}\\right\\}_{n=0}^{\\infty} \\text { be a sequence such that } a_{0}=a_{1}=0 \\text { and } \\\\\\\\\n&a_{n+2}=3 a_{n+1}-2 a_{n}+1, \\forall n \\geq 0 .\n\\end{aligned}\n$$
\n
Then $$a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$$ is equal to
Let $$\\sum_\\limits{n=0}^{\\infty} \\frac{\\mathrm{n}^{3}((2 \\mathrm{n}) !)+(2 \\mathrm{n}-1)(\\mathrm{n} !)}{(\\mathrm{n} !)((2 \\mathrm{n}) !)}=\\mathrm{ae}+\\frac{\\mathrm{b}}{\\mathrm{e}}+\\mathrm{c}$$, where $$\\mathrm{a}, \\mathrm{b}, \\mathrm{c} \\in \\mathbb{Z}$$ and $$e=\\sum_\\limits{\\mathrm{n}=0}^{\\infty} \\frac{1}{\\mathrm{n} !}$$ Then $$\\mathrm{a}^{2}-\\mathrm{b}+\\mathrm{c}$$ is equal to ____________.
If $${{{1^3} + {2^3} + {3^3}\\, + \\,...\\,up\\,to\\,n\\,terms} \\over {1\\,.\\,3 + 2\\,.\\,5 + 3\\,.\\,7\\, + \\,...\\,up\\,to\\,n\\,terms}} = {9 \\over 5}$$, then the value of $$n$$ is
",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nGiven $\\frac{1^{3}+2^{3}+3^{3}+\\ldots \\text { up to } n \\text { terms }}{1.3+2.5+3.7+\\ldots \\text { up to } n \\text { terms }}=\\frac{9}{5}$\n
is $\\frac{\\alpha}{\\beta}$, where $\\alpha$ and $\\beta$ are co-prime, then $\\alpha+3 \\beta$ is equal to __________.",
"options": [],
"answer": "7",
"solution": "**Answer:** 7\n\nWe can rewrite the given series as follows :\n\n
The series inside the parentheses is an infinite geometric series with the first term $a = - \\frac{1}{3}$ and the common ratio $r = -\\frac{1}{3}$:\n\n
Let $$< a_{\\mathrm{n}} > $$ be a sequence such that $$a_{1}+a_{2}+\\ldots+a_{n}=\\frac{n^{2}+3 n}{(n+1)(n+2)}$$. If $$28 \\sum_\\limits{k=1}^{10} \\frac{1}{a_{k}}=p_{1} p_{2} p_{3} \\ldots p_{m}$$, where $$\\mathrm{p}_{1}, \\mathrm{p}_{2}, \\ldots ., \\mathrm{p}_{\\mathrm{m}}$$ are the first $$\\mathrm{m}$$ prime numbers, then $$\\mathrm{m}$$ is equal to
",
"options": [
{
"text": "5"
},
{
"text": "7"
},
{
"text": "6"
},
{
"text": "8"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\nGiven the sum of the first n terms, $S_n = \\frac{n^2+3n}{(n+1)(n+2)}$, we can find the nth term $a_n$ as the difference between the sum of the first n terms and the sum of the first n-1 terms :\n\n
The prime factorization of 30030 is $2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13$, which consists of 6 primes.\n\n
Therefore, m is equal to 6.\n",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 7204,
"subject": "General Science",
"question": "
For $$k \\in \\mathbb{N}$$, if the sum of the series $$1+\\frac{4}{k}+\\frac{8}{k^{2}}+\\frac{13}{k^{3}}+\\frac{19}{k^{4}}+\\ldots$$ is 10 , then the value of $$k$$ is _________.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nFrom the given series :\n
In this equation, you've treated the infinite series on the right-hand side as a geometric series with a ratio of $1/k$, so the sum of this series can be expressed as $\\frac{1/k^3}{1 - 1/k}$.\n\n
So, the solutions are $k = 2$ or $k = 0.8$. However, since the question mentions $k \\in \\mathbb{N}$, i.e., $k$ is a natural number, the only valid solution is $k = 2$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7205,
"subject": "General Science",
"question": "
Let $$S=109+\\frac{108}{5}+\\frac{107}{5^{2}}+\\ldots .+\\frac{2}{5^{107}}+\\frac{1}{5^{108}}$$. Then the value of $$\\left(16 S-(25)^{-54}\\right)$$ is equal to ___________.
The sum of all those terms, of the arithmetic progression 3, 8, 13, ...., 373, which are not divisible by 3, is equal to ____________.
",
"options": [],
"answer": "9525",
"solution": "**Answer:** 9525\n\nThe sum of all those terms, of the arithmetic progression $3,8,13 \\ldots, 373$.\n
Let $$\\mathrm{a}_{\\mathrm{n}}$$ be the $$\\mathrm{n}^{\\text {th }}$$ term of the series $$5+8+14+23+35+50+\\ldots$$ and $$\\mathrm{S}_{\\mathrm{n}}=\\sum_\\limits{k=1}^{n} a_{k}$$. Then $$\\mathrm{S}_{30}-a_{40}$$ is equal to :
If $$\\operatorname{gcd}~(\\mathrm{m}, \\mathrm{n})=1$$ and $$1^{2}-2^{2}+3^{2}-4^{2}+\\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n$$ then $$m^{2}-n^{2}$$ is equal to :
Let $$S_n$$ be the sum to $$n$$-terms of an arithmetic progression $$3,7,11$$,\nIf $$40<\\left(\\frac{6}{n(n+1)} \\sum_\\limits{k=1}^n S_k\\right)<42$$, then $$n$$ equals ________.
Let $$\\alpha=1^2+4^2+8^2+13^2+19^2+26^2+\\ldots$$ upto 10 terms and $$\\beta=\\sum_\\limits{n=1}^{10} n^4$$. If $$4 \\alpha-\\beta=55 k+40$$, then $$\\mathrm{k}$$ is equal to __________.
If $$\\left(\\frac{1}{\\alpha+1}+\\frac{1}{\\alpha+2}+\\ldots . .+\\frac{1}{\\alpha+1012}\\right)-\\left(\\frac{1}{2 \\cdot 1}+\\frac{1}{4 \\cdot 3}+\\frac{1}{6 \\cdot 5}+\\ldots \\ldots+\\frac{1}{2024 \\cdot 2023}\\right)=\\frac{1}{2024}$$, then $$\\alpha$$ is equal to ___________.
If the sum of the series $$\\frac{1}{1 \\cdot(1+\\mathrm{d})}+\\frac{1}{(1+\\mathrm{d})(1+2 \\mathrm{~d})}+\\ldots+\\frac{1}{(1+9 \\mathrm{~d})(1+10 \\mathrm{~d})}$$ is equal to 5, then $$50 \\mathrm{~d}$$ is equal to :
Let $$\\alpha=\\sum_\\limits{r=0}^n\\left(4 r^2+2 r+1\\right){ }^n C_r$$ and $$\\beta=\\left(\\sum_\\limits{r=0}^n \\frac{{ }^n C_r}{r+1}\\right)+\\frac{1}{n+1}$$. If $$140<\\frac{2 \\alpha}{\\beta}<281$$, then the value of $$n$$ is _________.
If $$\\frac{1}{\\sqrt{1}+\\sqrt{2}}+\\frac{1}{\\sqrt{2}+\\sqrt{3}}+\\ldots+\\frac{1}{\\sqrt{99}+\\sqrt{100}}=m$$ and $$\\frac{1}{1 \\cdot 2}+\\frac{1}{2 \\cdot 3}+\\ldots+\\frac{1}{99 \\cdot 100}=\\mathrm{n}$$, then the point $$(\\mathrm{m}, \\mathrm{n})$$ lies on the line
If $$\\mathrm{S}(x)=(1+x)+2(1+x)^2+3(1+x)^3+\\cdots+60(1+x)^{60}, x \\neq 0$$, and $$(60)^2 \\mathrm{~S}(60)=\\mathrm{a}(\\mathrm{b})^{\\mathrm{b}}+\\mathrm{b}$$, where $$a, b \\in N$$, then $$(a+b)$$ equal to _________.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7220,
"subject": "General Science",
"question": "Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can be formed such that Y $$ \\subseteq $$ X, Z $$ \\subseteq $$ X and Y $$ \\cap $$ Z is empty, is :",
"options": [
{
"text": "35"
},
{
"text": "25"
},
{
"text": "53"
},
{
"text": "52"
}
],
"answer": "35",
"solution": "**Answer:** 35\n\nFor any element xi present in X, 4 cases arises while making subsets Y and Z.\n
Case- 1 : xi $$ \\in $$ Y, xi $$ \\in $$ Z $$ \\Rightarrow $$ Y $$ \\cap $$ Z $$ \\ne $$ $$\\phi $$\n
Case- 2 : xi $$ \\in $$ Y, xi $$ \\notin $$ Z $$ \\Rightarrow $$ Y $$ \\cap $$ Z = $$\\phi $$\n
Case- 3 : xi $$ \\notin $$ Y, xi $$ \\in $$ Z $$ \\Rightarrow $$ Y $$ \\cap $$ Z = $$\\phi $$\n
Case- 4 : xi $$ \\notin $$ Y, xi $$ \\notin $$ Z $$ \\Rightarrow $$ Y $$ \\cap $$ Z = $$\\phi $$\n
$$ \\therefore $$ For every element, number of ways = 3 for which Y $$ \\cap $$ Z = $$\\phi $$\n
$$ \\Rightarrow $$ Total ways = 3 × 3 × 3 × 3 × 3 [$$ \\because $$ no. of elements in set X = 5]\n
= 35",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7221,
"subject": "General Science",
"question": "Let A and B be two sets containing four and\ntwo elements respectively. Then, the number\nof subsets of the set A $\\times$ B , each having atleast\nthree elements are",
"options": [
{
"text": "219"
},
{
"text": "256"
},
{
"text": "275"
},
{
"text": "510"
}
],
"answer": "219",
"solution": "**Answer:** 219\n\nGiven,
\n$$\n\\begin{aligned}\n&n(A)=4, n(B) =2 \\\\\\\\\n&\\Rightarrow n(A \\times B) =8\n\\end{aligned}\n$$
\nTotal number of subsets of set $(A \\times B)=2^8$
\nNumber of subsets of set $A \\times B$ having no element (i.e. $\\phi)=1$
\nNumber of subsets of set $A \\times B$ having one element $={ }^8 C_1$
\nNumber of subsets of set $A \\times B$ having two elements $={ }^8 C_2$
\n$\\therefore$ Number of subsets having atleast three elements
So, P = Q",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7223,
"subject": "General Science",
"question": "Two sets A and B are as under :\n
A = {($$a$$, b) $$ \\in $$ R $$ \\times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};\n
B = {($$a$$, b) $$ \\in $$ R $$ \\times $$ R : 4($$a$$ - 6)2 + 9(b - 5)2 $$ \\le $$ 36 };\n
Then",
"options": [
{
"text": "neither A $$ \\subset $$ B nor B $$ \\subset $$ A"
},
{
"text": "B $$ \\subset $$ A"
},
{
"text": "A $$ \\subset $$ B"
},
{
"text": "A $$ \\cap $$ B = $$\\phi $$ ( an empty set )"
}
],
"answer": "A $$ \\subset $$ B",
"solution": "**Answer:** A $$ \\subset $$ B\n\nGiven, \n
$$\\therefore\\,\\,\\,$$ $$\\left| y \\right| < 1$$\n
$$ \\Rightarrow \\,\\,\\,\\, - 1 < y < 1$$\n
This will look like this, \n
\n
By combining both those graphs it will look like this,\n
\n
To check entire set A is inside of B or not put any paint of set A on the equation $${{{x^2}} \\over 9} + {{{y^2}} \\over 4} \\le 1$$ and if this inequality satisfies, then it is inside B. \n
By looking at the graph you can surely say (0, 1) or (0, $$-$$1) inside the graph. But to check point ($$-2$$, 1) or ($$-$$2, $$-$$1) is inside of the ellipse or not, put on the $${{{x^2}} \\over 9} + {{{y^2}} \\over 4} \\le 1.$$ \n
B = {x $$\\varepsilon $$ Z : $$-$$ < 2x $$-$$ 1 < 9}\n
B = {0, 1, 2, 3, 4}\n
A $$ \\times $$ B has is 15 elements so number of subsets of A $$ \\times $$ B is 215.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7226,
"subject": "General Science",
"question": "Let X = {n $$ \\in $$ N : 1 $$ \\le $$ n $$ \\le $$ 50}. If\n A = {n $$ \\in $$ X: n is a multiple of 2} and\n B = {n $$ \\in $$ X: n is a multiple of 7}, then the number of elements in the smallest subset of X\ncontaining both A and B is ________.",
"options": [],
"answer": "29",
"solution": "**Answer:** 29\n\nX = {1, 2, 3, 4, …, 50}\n
Option A : A ∩ B = {–3}\nThe intersection of sets A and B would be the set of elements common to both sets. In this case, the only common element is -3. So, option A is true.
\n
Option B : B – A = (–3, 5)\nThe subtraction (or difference) of sets A from B is the set of elements that are in B but not in A. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting A from B would leave an open interval (-3, 5), not including -3 and 5. So, option B is also true.
\n
Option C : A ∪ B = R\nThe union of sets A and B is the set of elements that are in A, or B, or both. Here, A U B would cover all real numbers. So, option C is true.
\n
Option D : A - B = (-∞, -3) ∪ (5, ∞)\nThe subtraction (or difference) of set B from A is the set of elements that are in A but not in B. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting B from A would leave (-∞, -3) U [5, ∞), not including -3 and 5. But according to the convention for writing intervals, it should be (-∞, -3) U (5, ∞). So, option D is not true.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7229,
"subject": "General Science",
"question": "Let $$\\mathop \\cup \\limits_{i = 1}^{50} {X_i} = \\mathop \\cup \\limits_{i = 1}^n {Y_i} = T$$ where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi’s and exactly 6 of sets Yi’s, then n is equal to :\n",
"options": [
{
"text": "30"
},
{
"text": "50"
},
{
"text": "15"
},
{
"text": "45"
}
],
"answer": "30",
"solution": "**Answer:** 30\n\n$$\\mathop \\cup \\limits_{i = 1}^{50} {X_i} = $$ X1, X2,....., X50 = 50 sets. Given each sets having 10 elements.\n
So total elements = 50 $$ \\times $$ 10\n
$$\\mathop \\cup \\limits_{i = 1}^n {Y_i} =$$ $$ Y1, Y2,....., Yn = n sets. Given each sets having 5 elements.\n
So total elements = 5 $$ \\times $$ n\n
Now each element of set T contains exactly 20 of sets Xi.\n
So number of effective elements in set T = $${{50 \\times 10} \\over {20}}$$\n
Also each element of set T contains exactly 6 of sets Yi.\n
So number of effective elements in set T = $${{50 \\times 10} \\over {20}}$$\n
$$ \\Rightarrow $$ n = 30",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7230,
"subject": "General Science",
"question": "Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more\nthan the total number of subsets of B, then the value of m.n is ______.",
"options": [],
"answer": "28",
"solution": "**Answer:** 28\n\nNumber of subsets of A = 2m\n
Number of subsets of B = 2n\n
Given = 2m – 2n\n = 112\n
$$ \\therefore $$ m = 7, n = 4 (27 – 24\n = 112)\n
$$ \\therefore $$ m $$ \\times $$ n = 7 $$ \\times $$ 4 = 28",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7231,
"subject": "General Science",
"question": "Let A = {n $$ \\in $$ N: n is a 3-digit number}
\n B = {9k + 2: k $$ \\in $$ N}\n
and C = {9k + $$l$$: k $$ \\in $$ N} for some $$l ( 0 < l < 9)$$
\nIf the sum of all the elements of the set A $$ \\cap $$ (B $$ \\cup $$ C) is 274 $$ \\times $$ 400, then $$l$$ is equal to ________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nIn this problem, we're dealing with 3-digit numbers in set $A$, and subsets $B$ and $C$ which represent numbers of specific forms.\n\n
1. First, we consider the numbers of the form $9k + 2$ (Set $B$) within the 3-digit range, which starts at 101 and ends at 992.\n\n
2. We calculate the sum of these numbers, denoted as $s_1$. To calculate $s_1$, you use the formula for the sum of an arithmetic series : \n\n
Here, $n$ is the total count of such numbers. These are 3-digit numbers of the form $9k + 2$, and we can find the total count by subtracting the smallest such number (101) from the largest (992), dividing the result by 9 (because we're considering numbers with a difference of 9), and then adding 1.\n\n
The sum $s_1$ is calculated as follows :\n\n
$$(100/2) \\times (101 + 992) = 54650$$\n\n
3. According to the problem, the sum of all elements of the set $A \\cap (B \\cup C)$ is $274 \\times 400 = 109600$. \n\n
Since the set $A \\cap (B \\cup C)$ is the union of two disjoint sets (the set of all three-digit numbers of form $9k + 2$ and the set of all three-digit numbers of form $9k + l$), we can write this sum as :\n\n
$$s_1 $$(for numbers of the form 9k + 2) + $$s_2$$ (for numbers of the form 9k + l) = 109600\n\n
4. Solving this equation for $s_2$ (the sum of numbers of the form $9k + l$), we get :\n\n
5. The sum $s_2$ can be expressed as $(n/2) \\times (\\text{{first term}} + \\text{{last term}})$, where $n$ is the count of numbers of the form $9k + l$. The first term here is the smallest 3-digit number of this form, which is $99 + l$, and the last term is the largest such number, which is $990 + l$.\n\n
We equate this to $s_2$ to solve for $l$ :\n\n
$$54950 = (100/2)[(99 + l) + (990 + l)]$$\n\n
6. Simplifying this equation, we get :\n\n
$$2l + 1089 = 1099$$\n\n
Solving for $l$, we find :\n\n
$$l = 5$$\n\n
So, the correct answer is 5.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7232,
"subject": "General Science",
"question": "The number of elements in the set {x $$\\in$$ R : (|x| $$-$$ 3) |x + 4| = 6} is equal to :",
"options": [
{
"text": "4"
},
{
"text": "2"
},
{
"text": "3"
},
{
"text": "1"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nCase 1 :
$$ \\therefore $$ x = $${{\\sqrt {73} - 1} \\over 2}$$ only",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7233,
"subject": "General Science",
"question": "Let A = {2, 3, 4, 5, ....., 30} and '$$ \\simeq $$' be an equivalence relation on A $$\\times$$ A, defined by (a, b) $$ \\simeq $$ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :",
"options": [
{
"text": "5"
},
{
"text": "6"
},
{
"text": "8"
},
{
"text": "7"
}
],
"answer": "7",
"solution": "**Answer:** 7\n\nad = bc
$$ \\Rightarrow $$ 7 ordered pair",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7234,
"subject": "General Science",
"question": "Let A = {n $$\\in$$ N | n2 $$\\le$$ n + 10,000}, B = {3k + 1 | k$$\\in$$ N} an dC = {2k | k$$\\in$$N}, then the sum of all the elements of the set A $$\\cap$$(B $$-$$ C) is equal to _____________.",
"options": [],
"answer": "832",
"solution": "**Answer:** 832\n\nB $$-$$ C $$ \\equiv $$ {7, 13, 19, ......, 97, .......}
So, A$$\\cap$$(B $$-$$ C) = {7, 13, 19, ......., 97}
Hence, sum = $${{16} \\over 2}(7 + 97) = 832$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7235,
"subject": "General Science",
"question": "If A = {x $$\\in$$ R : |x $$-$$ 2| > 1}, B = {x $$\\in$$ R : $$\\sqrt {{x^2} - 3} $$ > 1}, C = {x $$\\in$$ R : |x $$-$$ 4| $$\\ge$$ 2} and Z is the set of all integers, then the number of subsets of the set (A $$\\cap$$ B $$\\cap$$ C)c $$\\cap$$ Z is ________________.",
"options": [],
"answer": "256",
"solution": "**Answer:** 256\n\nA = ($$-$$$$\\infty$$, 1) $$\\cup$$ (3, $$\\infty$$)
B = ($$-$$$$\\infty$$, $$-$$2) $$\\cup$$ (2, $$\\infty$$)
C = ($$-$$$$\\infty$$, 2] $$\\cup$$ [6, $$\\infty$$)
So, A $$\\cap$$ B $$\\cap$$ C = ($$-$$$$\\infty$$, $$-$$2) $$\\cup$$ [6, $$\\infty$$)
z $$\\cap$$ (A $$\\cap$$ B $$\\cap$$ C)' = {$$-$$2, $$-$$1, 0, $$-$$1, 2, 3, 4, 5}
Let R1 and R2 be relations on the set {1, 2, ......., 50} such that
\n
R1 = {(p, pn) : p is a prime and n $$\\ge$$ 0 is an integer} and
\n
R2 = {(p, pn) : p is a prime and n = 0 or 1}.
\n
Then, the number of elements in R1 $$-$$ R2 is _______________.
",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nGiven, ${R}_1=\\left\\{\\left(p, p^n\\right): p\\right.$ is a Prime and $n \\geq 0$ is an integer $\\}$\n
and, set $A=\\{1,2,3 \\ldots \\ldots .50\\}$\n
$p$ is a Prime number which can take 15 values $2,3,5,7,11,13,17,19,23,29,31,37,41,43$ and 47\n
$\\therefore$ We can calculate no. of elements in $\\mathrm{R}_1$\n
$$\n\\mathrm{R}_1=\\left(2,2^0\\right),\\left(2,2^1\\right),\\left(2,2^2\\right)\\left(2,2^3\\right) \\ldots \\ldots\\left(2,2^5\\right)=6$$ number of ordered pairs\n
$\\left(3,3^0\\right),\\left(3,3^1\\right),\\left(3,3^2\\right) \\ldots \\ldots . .\\left(3,3^3\\right)=4$ number of order paris\n
$\\left(5,5^0\\right),\\left(5,5^1\\right),\\left(5,5^2\\right) \\ldots \\ldots \\ldots . .=3$ number of order paris\n
$\\left(7,7^0\\right) \\ldots \\ldots .\\left(7,7^2\\right) \\ldots \\ldots \\ldots=3$ number of order paris\n
$\\left(11,11^0\\right)$ and $\\left(11,11^1\\right)=2$ number of order paris\n
$\\left(13,13^0\\right)$ and $\\left(13,13^1\\right)=2$ number of order paris\n
$$ \\therefore $$ For the 11 prime numbers ($11,13,17,19,23,29,31,37,41,43$ and 47), $n$ can only be 0, 1 (two pairs each).\n
Let $$A=\\{1,2,3,4,5,6,7\\}$$. Define $$B=\\{T \\subseteq A$$ : either $$1 \\notin T$$ or $$2 \\in T\\}$$ and $$C=\\{T \\subseteq A: T$$ the sum of all the elements of $$T$$ is a prime number $$\\}$$. Then the number of elements in the set $$B \\cup C$$ is ________________.
Let $$A=\\{1,2,3,4,5,6,7\\}$$ and $$B=\\{3,6,7,9\\}$$. Then the number of elements in the set $$\\{C \\subseteq A: C \\cap B \\neq \\phi\\}$$ is ___________.
Let R be a relation from the set $$\\{1,2,3, \\ldots, 60\\}$$ to itself such that $$R=\\{(a, b): b=p q$$, where $$p, q \\geqslant 3$$ are prime numbers}. Then, the number of elements in R is :
We have a set S = {1, 2, 3, ..., 60}, and a relation R defined on the set S. An element (a, b) belongs to the relation R if and only if b can be expressed as the product of two prime numbers p and q, where both p and q are greater than or equal to 3.
\n
In terms of number theory, prime numbers are integers greater than 1 that have no divisors other than 1 and themselves. We are interested in prime numbers that are greater than or equal to 3, because p and q must both be greater than or equal to 3.
\n
The primes greater than or equal to 3 and less than or equal to 60 are {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59}.
\n
We need to find all possible values of b = p $$ \\times $$ q such that b belongs to the set S. We start by choosing the smallest prime number (which is 3) and keep multiplying it with all the prime numbers until the product exceeds 60 :
\n
\n
1. If we start with p = 3, we can choose q to be 3, 5, 7, 11, 13, 17, or 19. This gives us 7 valid products that are less than or equal to 60.
\n
2. If we start with p = 5, we can choose q to be 5, 7, or 11. This gives us 3 valid products that are less than or equal to 60.
\n
3. If we start with p = 7, we can choose q to be 7. This gives us 1 valid product that is less than or equal to 60.
\n
\n
So, we have a total of 7 + 3 + 1 = 11 possible values for b = p $$ \\times $$ q that satisfy the conditions.
\n
Since a can be any number in the set S, there are 60 possible values for a for each of the 11 values of b. Therefore, the total number of elements in the relation R is 60 $$ \\times $$ 11 = 660.
Let $$S=\\{4,6,9\\}$$ and $$T=\\{9,10,11, \\ldots, 1000\\}$$. If $$A=\\left\\{a_{1}+a_{2}+\\ldots+a_{k}: k \\in \\mathbf{N}, a_{1}, a_{2}, a_{3}, \\ldots, a_{k}\\right.$$ $$\\epsilon S\\}$$, then the sum of all the elements in the set $$T-A$$ is equal to __________.
And $$T = \\{ 9,10,11,\\,\\,......,\\,\\,1000\\} $$.
\n
We have to find all numbers in the form of $$4x + 6y + 9z$$, where $$x,y,z \\in \\{ 0,1,2,\\,......\\} $$.
\n
If a and b are coprime number then the least number from which all the number more than or equal to it can be express as $$ax + by$$ where $$x,y \\in \\{ 0,1,2,\\,......\\} $$ is $$(a - 1)\\,.\\,(b - 1)$$.
\n
Then for $$6y + 9z = 3(2y + 3z)$$
\n
All the number from $$(2 - 1)\\,.\\,(3 - 1) = 2$$ and above can be express as $$2x + 3z$$ (say t).
\n
Now $$4x + 6y + 9z = 4x + 3(t + 2)$$
\n
$$ = 4x + 3t + 6$$
\n
again by same rule $$4x + 3t$$, all the number from $$(4 - 1)\\,(3 - 1) = 6$$ and above can be express from $$4x + 3t$$.
Then $$4x + 6y + 9z$$ express all the numbers from 12 and above.
\n
again 9 and 10 can be express in form $$4x + 6y + 9z$$.
\n
Then set $$A = \\{ 9,10,12,13,\\,....,\\,1000\\} .$$
\nSubsets containing six elements $\\mathrm{S}_6=1$
\nSubsets containing seven elements $\\mathrm{S}_7=1$
\n$$\n\\Rightarrow \\text { sum }=43\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7245,
"subject": "General Science",
"question": "The number of elements in the set
$\\left\\{n \\in \\mathbb{N}: 10 \\leq n \\leq 100\\right.$ and $3^{n}-3$ is a multiple of 7$\\}$ is ___________.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\nTo determine the number of elements in the given set, we need to find how many natural numbers $n$ between $10$ and $100$ (inclusive) satisfy the condition that $3^n - 3$ is a multiple of $7$.\n\n
Recall that for any integers $a$ and $b$, $a$ is a multiple of $b$ if there exists an integer $k$ such that $a = bk$. So in our case, we need to find how many $n$ satisfy the equation $3^n - 3 = 7k$ for some integer $k$.\n\n
Notice that $3^n - 3 = 3(3^{n-1} - 1)$. We want this expression to be a multiple of 7. Let's explore a few powers of 3 modulo 7:\n\n
$3^1 \\equiv 3 \\pmod{7}$\n
$3^2 \\equiv 9 \\equiv 2 \\pmod{7}$\n
$3^3 \\equiv 27 \\equiv 6 \\pmod{7}$\n
$3^4 \\equiv 81 \\equiv 4 \\pmod{7}$\n
$3^5 \\equiv 243 \\equiv 5 \\pmod{7}$\n
$3^6 \\equiv 729 \\equiv 1 \\pmod{7}$\n
We observe that $3^n \\pmod{7}$ follows a cycle of length 6. So, $3^{n-1} \\pmod{7}$ also follows the same cycle, but shifted:\n\n
$3^0 \\equiv 1 \\pmod{7}$\n
$3^1 \\equiv 3 \\pmod{7}$\n
$3^2 \\equiv 2 \\pmod{7}$\n
$3^3 \\equiv 6 \\pmod{7}$\n
$3^4 \\equiv 4 \\pmod{7}$\n
$3^5 \\equiv 5 \\pmod{7}$\n
We want $3(3^{n-1} - 1) \\equiv 0 \\pmod{7}$, which means that $3^{n-1} - 1 \\equiv 0 \\pmod{7}$. From the cycle above, we see that this is true when $n-1$ is a multiple of 6, or equivalently, when $n$ is one more than a multiple of 6.\n\n
Now let's find the multiples of 6 between 10 and 100:\n\n
There are 15 elements in this set. Therefore, the number of elements in the given set is $\\boxed{15}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7246,
"subject": "General Science",
"question": "Let $A=\\{1,2,3,4\\}$ and $\\mathrm{R}$ be a relation on the set $A \\times A$ defined by
$R=\\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\\}$. Then the number of elements in $\\mathrm{R}$ is ____________.",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\n$$\n2a + 3b = 4c + 5d\n$$\n
\nGiven A = {1, 2, 3, 4}, the maximum value of $2a + 3b$ is 20, when (a, b) = (4, 4), and the minimum value of $4c + 5d$ is 9, when (c, d) = (1, 1). Therefore, the possible values for $2a + 3b = 4c + 5d$ are 9, 13, 14, 17, 18, and 19.\n
\nNow, let's find the combinations of (a, b), (c, d) that satisfy the given equation:\n
\nThere are a total of 6 elements in the relation R for the given equation with the specified values of a, b, c, and d.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7247,
"subject": "General Science",
"question": "
Let $$\\mathrm{A}=\\{1,3,4,6,9\\}$$ and $$\\mathrm{B}=\\{2,4,5,8,10\\}$$. Let $$\\mathrm{R}$$ be a relation defined on $$\\mathrm{A} \\times \\mathrm{B}$$ such that $$\\mathrm{R}=\\left\\{\\left(\\left(a_{1}, b_{1}\\right),\\left(a_{2}, b_{2}\\right)\\right): a_{1} \\leq b_{2}\\right.$$ and $$\\left.b_{1} \\leq a_{2}\\right\\}$$. Then the number of elements in the set R is :
",
"options": [
{
"text": "180"
},
{
"text": "26"
},
{
"text": "52"
},
{
"text": "160"
}
],
"answer": "160",
"solution": "**Answer:** 160\n\nGiven that the sets are $A = \\{1, 3, 4, 6, 9\\}$ and $B = \\{2, 4, 5, 8, 10\\}$, for the relation $\\mathrm{R}$ on the set $A \\times B$, we need to find the combinations of pairs that satisfy the conditions $a_1 \\leq b_2$ and $b_1 \\leq a_2$. \n\n
We find the number of combinations by considering the possible values for $b_2$ for each $a_1$ and the possible values for $a_2$ for each $b_1$ :\n\n
For each $a_1$ in $A = \\{1, 3, 4, 6, 9\\}$, the number of valid $b_2$ values in $B = \\{2, 4, 5, 8, 10\\}$ are :\n
- For $a_1 = 1$, there are 5 choices for $b_2$.\n
- For $a_1 = 3$, there are 4 choices for $b_2$.\n
- For $a_1 = 4$, there are 4 choices for $b_2$.\n
- For $a_1 = 6$, there are 2 choices for $b_2$.\n
- For $a_1 = 9$, there is 1 choice for $b_2$.\n\n
This results in a total of $5+4+4+2+1 = 16$ possible pairs $(a_1, b_2)$.\n\n
Similarly, for each $b_1$ in $B$, the number of valid $a_2$ values in $A$ are :\n
- For $b_1 = 2$, there are 4 choices for $a_2$.\n
- For $b_1 = 4$, there are 3 choices for $a_2$.\n
- For $b_1 = 5$, there are 2 choices for $a_2$.\n
- For $b_1 = 8$, there is 1 choice for $a_2$.\n
- For $b_1 = 10$, there are no choices for $a_2$.\n\n
This results in a total of $4+3+2+1+0 = 10$ possible pairs $(b_1, a_2)$.\n\n
Therefore, the total number of elements in the relation $\\mathrm{R}$, which satisfies the given conditions, is $16 \\times 10 = 160$.\n\n
So, the correct answer is 160.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7248,
"subject": "General Science",
"question": "
Let $$\\mathrm{A}=\\{2,3,4\\}$$ and $$\\mathrm{B}=\\{8,9,12\\}$$. Then the number of elements in the relation\n\n$$\\mathrm{R}=\\left\\{\\left(\\left(a_{1}, \\mathrm{~b}_{1}\\right),\\left(a_{2}, \\mathrm{~b}_{2}\\right)\\right) \\in(A \\times B, A \\times B): a_{1}\\right.$$ divides $$\\mathrm{b}_{2}$$ and $$\\mathrm{a}_{2}$$ divides $$\\left.\\mathrm{b}_{1}\\right\\}$$ is :
We want to find the number of elements of the form $( (a_1, b_1), (a_2, b_2) )$ such that :
\n\n
$ a_1 $ divides $ b_2 $
\n
$ a_2 $ divides $ b_1 $
\n\n
For the first condition :\n $ a_1 $ divides $ b_2 $\n Given $ a_1 \\in A $ and $ b_2 \\in B $, we can list the pairs:\n $ (a_1, b_2) \\in {(2,8),(2,12),(3,9),(3,12),(4,8),(4,12)} $\n This gives 6 pairs.
\n
For the second condition, the pairs are the same, because it's just the reversed relation. So :\n $ a_2 $ divides $ b_1 $\n Again has 6 valid pairs.
\n
Now, for every pair from the first condition, we can have any pair from the second condition. This leads to :\n $ 6 \\times 6 = 36 $\nrelations.
$$\n\\begin{array}{lr}\n& \\therefore n \\in[5-2 \\sqrt{3}, 5+2 \\sqrt{3}] \\\\\\\\\n& \\therefore n \\in[13,8.3] \\\\\\\\\n& \\therefore n=2,3,4,5,6,7,8\n\\end{array}\n$$\n
Thus, number of element in the set is ' 6 '",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7250,
"subject": "General Science",
"question": "
Let $$\\mathrm{A}=\\{1,2,3,4, \\ldots ., 10\\}$$ and $$\\mathrm{B}=\\{0,1,2,3,4\\}$$. The number of elements in the relation $$R=\\left\\{(a, b) \\in A \\times A: 2(a-b)^{2}+3(a-b) \\in B\\right\\}$$ is ___________.
For any other $ a-b $ value, the quadratic will grow larger than the maximum value in B, so we don't need to consider them.
\n
In total, we have $ 10 + 8 = 18 $ pairs in the relation $ R $.
\n
Therefore, the number of elements in the relation $ R $ is 18.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7251,
"subject": "General Science",
"question": "Let $A=\\{1,2,3, \\ldots, 20\\}$. Let $R_1$ and $R_2$ two relation on $A$ such that \n
$R_1=\\{(a, b): b$ is divisible by $a\\}$\n
$R_2=\\{(a, b): a$ is an integral multiple of $b\\}$.\n
Then, number of elements in $R_1-R_2$ is equal to _____________.",
"options": [],
"answer": "46",
"solution": "**Answer:** 46\n\n
To determine the number of elements in $R_1 - R_2$, let's first articulate the meaning of both relations on set $A = \\{1, 2, 3, \\ldots, 20\\}$:
$R_1$ includes pairs $(a, b)$ where $b$ is divisible by $a$. This includes pairs like $(1,1), (1,2), \\ldots, (1,20)$ for $1$; similar series for $2$ up to $(2,20)$ (excluding odd numbers); for $3$ up to $(3,18)$; and so on, reflecting the divisibility condition.
$R_2$ consists of pairs $(a, b)$ where $a$ is an integral multiple of $b$. Essentially, this relationship is the reverse of $R_1$. However, for the essence of $R_1 - R_2$, the key overlap comes with pairs where $a = b$, since those are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in $A$ from 1 to 20, resulting in the common elements between $R_1$ and $R_2$ (the intersection $R_1 \\cap R_2$) being 20 pairs.
The difference $R_1 - R_2$ seeks elements present in $R_1$ but not in $R_2$. Given that $R_1$ and $R_2$ share 20 elements that are identical, to find $R_1 - R_2$, we subtract these 20 common elements from the total in $R_1$, resulting in $66 - 20 = 46$ pairs.
Let $$A$$ and $$B$$ be two finite sets with $$m$$ and $$n$$ elements respectively. The total number of subsets of the set $$A$$ is 56 more than the total number of subsets of $$B$$. Then the distance of the point $$P(m, n)$$ from the point $$Q(-2,-3)$$ is :
Let $$A=\\{1,2,3, \\ldots \\ldots \\ldots \\ldots, 100\\}$$. Let $$R$$ be a relation on $$\\mathrm{A}$$ defined by $$(x, y) \\in R$$ if and only if $$2 x=3 y$$. Let $$R_1$$ be a symmetric relation on $$A$$ such that $$R \\subset R_1$$ and the number of elements in $$R_1$$ is $$\\mathrm{n}$$. Then, the minimum value of $$\\mathrm{n}$$ is _________.
Let $$A=\\{1,2,3,4\\}$$ and $$R=\\{(1,2),(2,3),(1,4)\\}$$ be a relation on $$\\mathrm{A}$$. Let $$\\mathrm{S}$$ be the equivalence relation on $$\\mathrm{A}$$ such that $$R \\subset S$$ and the number of elements in $$\\mathrm{S}$$ is $$\\mathrm{n}$$. Then, the minimum value of $$n$$ is __________.
If R is the smallest equivalence relation on the set $$\\{1,2,3,4\\}$$ such that $$\\{(1,2),(1,3)\\} \\subset \\mathrm{R}$$, then the number of elements in $$\\mathrm{R}$$ is __________.
To find the number of symmetric relations on the set $$\\{1,2,3,4\\}$$ that are not reflexive, we first calculate the total number of symmetric relations and then subtract the count of those that are both symmetric and reflexive.
A symmetric relation involves pairs where if a pair (x, y) is in the relation, then (y, x) is also in the relation. For a set with $$n$$ elements, there are $$\\frac{n(n+1)}{2}$$ slots in the relation matrix that can independently be occupied or not, corresponding to a total of $$2^{\\frac{n(n+1)}{2}}$$ possible symmetric relations.
A relation is reflexive if every element is related to itself, requiring all diagonal slots of the relation matrix (n of them) to be filled. The remaining $$\\frac{n(n-1)}{2}$$ slots can be filled in any manner, leading to $$2^{\\frac{n(n-1)}{2}}$$ reflexive (and possibly symmetric) relations.
For the set $$\\{1,2,3,4\\}$$ ($$n=4$$):
Total symmetric relations: $$2^{\\frac{4(4+1)}{2}} = 2^{10} = 1024$$
Symmetric and reflexive relations: $$2^{\\frac{4(4-1)}{2}} = 2^{6} = 64$$
Therefore, the number of symmetric relations that are not reflexive: $$1024 - 64 = 960$$.
Let $$A=\\{2,3,6,7\\}$$ and $$B=\\{4,5,6,8\\}$$. Let $$R$$ be a relation defined on $$A \\times B$$ by $$(a_1, b_1) R(a_2, b_2)$$ if and only if $$a_1+a_2=b_1+b_2$$. Then the number of elements in $$R$$ is __________.
To find the number of elements in the relation $$R$$ defined on $$A \\times B$$, we need to determine all pairs $$((a_1, b_1), (a_2, b_2))$$ such that $$a_1 + a_2 = b_1 + b_2$$, where $$a_1, a_2 \\in A$$ and $$b_1, b_2 \\in B$$.
\n\n
First, consider all possible sums of pairs from set $$A$$ and set $$B$$.
\n\n
Possible sums from set $$A = \\{2, 3, 6, 7\\}$$:
\n\n
\n\n
$$2 + 2 = 4$$
\n\n
$$2 + 3 = 5$$
\n\n
$$2 + 6 = 8$$
\n\n
$$2 + 7 = 9$$
\n\n
$$3 + 2 = 5$$
\n\n
$$3 + 3 = 6$$
\n\n
$$3 + 6 = 9$$
\n\n
$$3 + 7 = 10$$
\n\n
$$6 + 2 = 8$$
\n\n
$$6 + 3 = 9$$
\n\n
$$6 + 6 = 12$$
\n\n
$$6 + 7 = 13$$
\n\n
$$7 + 2 = 9$$
\n\n
$$7 + 3 = 10$$
\n\n
$$7 + 6 = 13$$
\n\n
$$7 + 7 = 14$$
\n\n
\n\n
Possible sums from set $$B = \\{4, 5, 6, 8\\}$$:
\n\n
\n\n
$$4 + 4 = 8$$
\n\n
$$4 + 5 = 9$$
\n\n
$$4 + 6 = 10$$
\n\n
$$4 + 8 = 12$$
\n\n
$$5 + 4 = 9$$
\n\n
$$5 + 5 = 10$$
\n\n
$$5 + 6 = 11$$
\n\n
$$5 + 8 = 13$$
\n\n
$$6 + 4 = 10$$
\n\n
$$6 + 5 = 11$$
\n\n
$$6 + 6 = 12$$
\n\n
$$6 + 8 = 14$$
\n\n
$$8 + 4 = 12$$
\n\n
$$8 + 5 = 13$$
\n\n
$$8 + 6 = 14$$
\n\n
$$8 + 8 = 16$$
\n\n
\n\n
Now, identify the common sums from both sets:
\n\n
Common sums: $$8, 9, 10, 12, 13, 14$$
\n\n
For each common sum, count the pairs from set $$A$$ and set $$B$$ that produce these sums:
\n\n
\n\n
Sum = 8: From $$A$$: {(2,6), (6,2)} - 2 pairs; From $$B$$: {(4,4)} - 1 pair; Hence, 2 * 1 = 2 pairs
\n\n
Sum = 9: From $$A$$: {(2,7), (3,6), (6,3), (7,2)} - 4 pairs; From $$B$$: {(4,5), (5,4)} - 2 pairs; Hence, 4 * 2 = 8 pairs
\n\n
Sum = 10: From $$A$$: {(3,7), (7,3)} - 2 pairs; From $$B$$: {(4,6), (5,5), (6,4)} - 3 pairs; Hence, 2 * 3 = 6 pairs
\n\n
Sum = 12: From $$A$$: {(6,6)} - 1 pair; From $$B$$: {(4,8), (6,6), (8,4)} - 3 pairs; Hence, 1 * 3 = 3 pairs
\n\n
Sum = 13: From $$A$$: {(6,7), (7,6)} - 2 pairs; From $$B$$: {(5,8), (8,5)} - 2 pairs; Hence, 2 * 2 = 4 pairs
\n\n
Sum = 14: From $$A$$: {(7,7)} - 1 pair; From $$B$$: {(6,8), (8,6)} - 2 pairs; Hence, 1 * 2 = 2 pairs
\n\n
\n\n
Adding all these, we get the number of elements in the relation $$R$$:
\n\n
2 + 8 + 6 + 3 + 4 + 2 = 25
\n\n
Thus, the number of elements in $$R$$ is 25.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7258,
"subject": "General Science",
"question": "Let $R=\\{(1,3),(4,2),(2,4),(2,3),(3,1)\\}$ be a relation on the set $A=\\{1,2,3,4\\}$. The relation $R$ is :",
"options": [
{
"text": "a function"
},
{
"text": "transitive"
},
{
"text": "not symmetric"
},
{
"text": "reflexive"
}
],
"answer": "not symmetric",
"solution": "**Answer:** not symmetric\n\n
Let's evaluate each of the properties for the relation $R$.
\n
Relation R : $R=\\{(1,3),(4,2),(2,4),(2,3),(3,1)\\}$
\n
\n
A relation is a function if each element in the domain is related to exactly one element in the codomain. In this case, for example, 2 is related to both 4 and 3, so $R$ is not a function.
\n
\n
A relation is transitive if for every pair of elements $(x,y)$ and $(y,z)$ in the relation, $(x,z)$ is also in the relation. In this case, for example, $(1,3)$ and $(3,1)$ are in the relation but $(1,1)$ is not, so $R$ is not transitive.
\n
\n
A relation is symmetric if for every pair $(x,y)$ in the relation, $(y,x)$ is also in the relation. In this case, for example, $(1,3)$ is in the relation but $(3,1)$ is not, so $R$ is not symmetric.
\n
\n
A relation is reflexive if every element is related to itself, i.e., if all pairs of the form $(x,x)$ are in the relation for all $x$ in the set. In this case, none of the pairs are of the form $(x,x)$, so $R$ is not reflexive.
\n
\n
\n
Therefore, the correct answer is Option C : $R$ is not symmetric.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7259,
"subject": "General Science",
"question": "Let $R=\\{(3,3),(6,6),(9,9),(12,12),(6,12)$, $(3,9),(3,12),(3,6)\\}$ be a relation on the set $A=\\{3,6,9,12\\}$. The relation is :",
"options": [
{
"text": "reflexive and symmetric only"
},
{
"text": "an equivalence relation"
},
{
"text": "reflexive only"
},
{
"text": "reflexive and transitive only"
}
],
"answer": "reflexive and transitive only",
"solution": "**Answer:** reflexive and transitive only\n\n
We have to examine whether the relation $R$ satisfies the properties of reflexivity, symmetry, and transitivity.
\n
Relation R : $R=\\{(3,3),(6,6),(9,9),(12,12),(6,12)$, $(3,9),(3,12),(3,6)\\}$ on set $A=\\{3,6,9,12\\}$.
\n
We will evaluate each of the three properties :
\n
\n
Reflexivity : A relation is reflexive if all elements in the set are related to themselves. Here, (3,3), (6,6), (9,9), and (12,12) are all present in $R$, so $R$ is reflexive.
\n
\n
Symmetry : A relation is symmetric if for every pair (a,b) in the relation, the pair (b,a) is also in the relation. In $R$, we have pairs such as (6,12) and (3,6), but their corresponding reverse pairs (12,6) and (6,3) are not in $R$, so $R$ is not symmetric.
\n
\n
Transitivity : A relation is transitive if for every pair of pairs ((a,b), (b,c)) in the relation, the pair (a,c) is also in the relation. For the pairs (3,6) and (6,12) in $R$, we have (3,12) in $R$, and for the pairs (3,6) and (6,6) in $R$, we also have (3,6) in $R$. Therefore, $R$ is transitive.
\n
\n
\n
Therefore, the relation $R$ is reflexive and transitive, but not symmetric.
\n
So, the correct answer is Option D : $R$ is reflexive and transitive only.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7260,
"subject": "General Science",
"question": "Let $W$ denote the words in the English dictionary. Define the relation $R$ by \n
$R=\\{(x, y) \\in W \\times W \\mid$ the words $x$ and $y$ have at least one letter in common}. Then, $R$ is",
"options": [
{
"text": "reflexive, symmetric and not transitive"
},
{
"text": "reflexive, symmetric and transitive"
},
{
"text": "reflexive, not symmetric and transitive"
},
{
"text": "not reflexive, symmetric and transitive"
}
],
"answer": "reflexive, symmetric and not transitive",
"solution": "**Answer:** reflexive, symmetric and not transitive\n\n
Let's evaluate the relation $R$ for the properties of reflexivity, symmetry, and transitivity.
\n
Relation R : $R={(x, y) \\in W \\times W \\mid}$ the words $x$ and $y$ have at least one letter in common}.
\n
\n
Reflexivity : Each word in English obviously has at least one letter in common with itself, so the relation is reflexive.
\n
Symmetry : If a word $x$ has at least one letter in common with a word $y$, then $y$ necessarily has that same letter in common with $x$. So, the relation is symmetric.
\n
Transitivity : This property is not always satisfied. For instance, consider the three words 'cat', 'bat', and 'bee'. 'Cat' and 'bat' share a letter (the 'a'), and 'bat' and 'bee' share a letter (the 'b'), but 'cat' and 'bee' do not share any letters. Therefore, even though 'cat' is related to 'bat' and 'bat' is related to 'bee', 'cat' is not related to 'bee', so the relation is not transitive.
\n
\n
In conclusion, the correct answer is Option A : $R$ is reflexive, symmetric and not transitive.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7261,
"subject": "General Science",
"question": "Let R be the real line. Consider the following subsets of the plane $$R \\times R$$ :\n $$S = \\left\\{ {(x,y):y = x + 1\\,\\,and\\,\\,0 < x < 2} \\right\\}$$\n $$T = \\left\\{ {(x,y): x - y\\,\\,\\,is\\,\\,an\\,\\,{\\mathop{\\rm int}} eger\\,} \\right\\}$$,\n
Which one of the following is true ?
",
"options": [
{
"text": "Neither S nor T is an equivalence relation on R"
},
{
"text": "Both S and T are equivalence relation on R"
},
{
"text": "S is an equivalence relation on R but T is not"
},
{
"text": "T is an equivalence relation on R but S is not"
}
],
"answer": "T is an equivalence relation on R but S is not",
"solution": "**Answer:** T is an equivalence relation on R but S is not\n\nGiven $$S = \\left\\{ {\\left( {x,y} \\right):y = x + 1\\,\\,} \\right.\\,$$ \n
and $$\\,\\,\\,\\left. {0 < x < 2} \\right\\}$$\n
As $$\\,\\,\\,\\,x \\ne x + 1\\,\\,\\,$$ \n
for any $$\\,\\,\\,x \\in \\left( {0,2} \\right) \\Rightarrow \\left( {x,x} \\right) \\notin S$$ \n
$$\\therefore$$ $$S$$ is not reflexive. \n
Hence $$S$$ in not an equivalence relation. \n
Also $$\\,\\,\\,T = \\left\\{ {x,\\left. y \\right)} \\right.:x - y$$ is an integer $$\\left. {} \\right\\}$$ \n
as $$x - x = 0$$ is an integer $$\\forall x \\in R$$ \n
$$\\therefore$$ $$T$$ is reflexive.\n
If $$x-y$$ is an integer then $$y-x$$ is also an integer \n
$$\\therefore$$ $$T$$ is symmetric \n
If $$x-y$$ is an integer and $$y - z$$ is an integer then \n
$$(x-y)+(y-z)=x-z$$ is also an integer. \n
$$\\therefore$$ $$T$$ is transitive \n
Hence $$T$$ is an equivalence relation ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7262,
"subject": "General Science",
"question": "Consider the following relations \n
$R=\\{(x, y) \\mid x, y$ are real numbers and $x=w y$ for some rational number $w\\}$; \n
$S=\\left\\{\\left(\\frac{m}{n}, \\frac{p}{q}\\right) \\mid m, n, p\\right.$ and $q$ are integers such that $n, q \\neq 0$ and $q m=p m\\}$. Then",
"options": [
{
"text": "$R$ is an equivalence relation but $S$ is not an equivalence relation\n"
},
{
"text": "Neither $R$ nor $S$ is an equivalence relation"
},
{
"text": "$S$ is an equivalence relation but $R$ is not an equivalence relation"
},
{
"text": "$R$ and $S$ both are equivalence relations"
}
],
"answer": "$S$ is an equivalence relation but $R$ is not an equivalence relation",
"solution": "**Answer:** $S$ is an equivalence relation but $R$ is not an equivalence relation\n\n
Let's evaluate each relation for the properties of an equivalence relation: reflexivity, symmetry, and transitivity.
\n
Relation R : $R=(x, y) \\mid x, y$ are real numbers and $x=w y$ for some rational number $w$.
\n
\n
Reflexivity : For all $x$ in $R$, $x = 1x$. Since 1 is a rational number, every element is related to itself.
\n
Symmetry : For all $x, y$ in $R$, if $x = w y$ for some rational $w$, then $y = \\frac{1}{w}x$. However, if $w = 0$, then $\\frac{1}{w}$ is undefined, and therefore, $R$ doesn't satisfy symmetry.
\n
Transitivity : If $x = w y$ and $y = v z$ for some rational numbers $w$ and $v$, then $x = (w v)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.
\n
\n
Therefore, $R$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.
\n
Relation S : $S=\\left\\{\\left(\\frac{m}{n}, \\frac{p}{q}\\right) \\mid m, n, p\\right.$ and $q$ are integers such that $n, q \\neq 0$ and $q m=p m\\}$
\n
\n
Reflexivity : For all $\\frac{m}{n}$, $\\frac{m}{n} = \\frac{m}{n}$. Since $n \\neq 0$ and $m = m$, every element is related to itself.
\n
Symmetry : For all $\\frac{m}{n}$, $\\frac{p}{q}$, if $q m = p n$, then $n p = m n$. So if $\\frac{m}{n}$ is related to $\\frac{p}{q}$, then $\\frac{p}{q}$ is related to $\\frac{m}{n}$.
\n
Transitivity : \n
$$\n\\begin{array}{rlr}\n \\frac{m}{n} R \\frac{p}{q} \\text { and } \\frac{p}{q} R \\frac{r}{s} \\\\\\\\\n\\Rightarrow m q=n p \\text { and } p s=r q \\\\\\\\\n\\Rightarrow m q \\cdot p s=n p \\cdot r q \\\\\\\\\n\\Rightarrow \\quad m s=n r \\\\\\\\\n\\Rightarrow \\frac{m}{n}=\\frac{r}{s} \\Rightarrow \\frac{m}{n} R \\frac{r}{s}\n\\end{array}\n$$\n\n
So if $\\frac{m}{n}$ is related to $\\frac{p}{q}$ and $\\frac{p}{q}$ is related to $\\frac{r}{s}$, then $\\frac{m}{n}$ is related to $\\frac{r}{s}$. The relation $S$ is transitive.
\n
\n
Therefore, $S$ is an equivalence relation on the set of all fractions where denominator is not zero.
\n
In conclusion, the correct answer is \n
Option C : $S$ is an equivalence relation but $R$ is not an equivalence relation.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7263,
"subject": "General Science",
"question": "Let $R$ be the set of real numbers.\n
Statement I : $A=\\{(x, y) \\in R \\times R: y-x$ is an integer $\\}$ is an equivalence relation on $R$.\n
Statement II : $ B=\\{(x, y) \\in R \\times R: x=\\alpha y$ for some rational number $\\alpha\\}$ is an equivalence relation on $R$.",
"options": [
{
"text": "Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I."
},
{
"text": "Statement I is true, Statement II is false."
},
{
"text": "Statement I is false, Statement II is true."
},
{
"text": "Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I."
}
],
"answer": "Statement I is true, Statement II is false.",
"solution": "**Answer:** Statement I is true, Statement II is false.\n\n
An equivalence relation on a set must satisfy three properties: reflexivity (every element is related to itself), symmetry (if an element is related to a second, the second is related to the first), and transitivity (if a first element is related to a second, and the second is related to a third, then the first is related to the third).
\n
Statement I : $A={(x, y) \\in R \\times R: }$ y-x is an integer .
\n
\n
Reflexivity : For all $x$ in $R$, $x - x = 0$ which is an integer. So, every element is related to itself.
\n
Symmetry : For all $x, y$ in $R$, if $y - x$ is an integer, then $x - y = - (y - x)$ is also an integer. So, if $x$ is related to $y$, then $y$ is related to $x$.
\n
Transitivity : For all $x, y, z$ in $R$, if $y - x$ and $z - y$ are integers, then $(z - y) + (y - x) = z - x$ is also an integer. So, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.
\n
\n
Therefore, $A$ is an equivalence relation on $R$.
\n
Statement II : $ B={(x, y) \\in R \\times R: x=\\alpha y}$ for some rational number $\\alpha$.
\n
\n
Reflexivity : For all $x$ in $R$, $x = 1x$. Since 1 is a rational number, every element is related to itself.
\n
Symmetry : For all $x, y$ in $R$, if $x = \\alpha y$ for some rational $\\alpha$, then $y = \\frac{1}{\\alpha}x$. However, if $\\alpha = 0$, then $\\frac{1}{\\alpha}$ is undefined, and therefore, $B$ doesn't satisfy symmetry.
\n
Transitivity : If $x = \\alpha y$ and $y = \\beta z$ for some rational numbers $\\alpha$ and $\\beta$, then $x = (\\alpha \\beta)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.
\n
\n
Therefore, $B$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.
\n
In conclusion, the correct answer is
Option B : Statement I is true, Statement II is false.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7264,
"subject": "General Science",
"question": "Consider the following two binary relations on the set A = {a, b, c} : \n R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and \n R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}. \n Then :\n",
"options": [
{
"text": "both R1 and R2 are not symmetric."
},
{
"text": "R1 is not symmetric but it is transitive."
},
{
"text": "R2 is symmetric but it is not transitive. "
},
{
"text": "both R1 and R2 are transitive."
}
],
"answer": "R2 is symmetric but it is not transitive. ",
"solution": "**Answer:** R2 is symmetric but it is not transitive. \n\nHere both R1 and R2 are symmetric as for any (x, y) $$ \\in $$ R1, we have (y, x) $$ \\in $$ R1 and similarly for any (x, y) $$ \\in $$ R2, we have (y, x) $$ \\in $$ R2
\nIn R1, (b, c) $$ \\in $$ R1, (c, a) $$ \\in $$ R1 but (b,a) $$ \\notin $$ R1
\nSimilarly in R2, (b, a) $$ \\in $$ R2, (a, c) $$ \\in $$ R2 but (b, c) $$ \\notin $$ R2
\n$$ \\therefore $$ R1 and R2 are not transitive.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7265,
"subject": "General Science",
"question": "Let N denote the set of all natural numbers. Define two binary relations on N as R = {(x, y) $$ \\in $$ N $$ \\times $$ N : 2x + y = 10} and R2 = {(x, y) $$ \\in $$ N $$ \\times $$ N : x + 2y = 10}. Then : ",
"options": [
{
"text": "Range of R1 is {2, 4, 8)."
},
{
"text": "Range of R2 is {1, 2, 3, 4}."
},
{
"text": "Both R1 and R2 are symmetric relations."
},
{
"text": "Both R1 and R2 are transitive relations."
}
],
"answer": "Range of R2 is {1, 2, 3, 4}.",
"solution": "**Answer:** Range of R2 is {1, 2, 3, 4}.\n\nFor R1; 2x + y = 10 and x, y $$ \\in $$ N possible values for x and y are : \n
$$\\therefore\\,\\,\\,$$ Range of R2 = $$\\left\\{ {1,2,3,4} \\right\\}$$\n
R2 is not symmetric and transitive",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7266,
"subject": "General Science",
"question": "Let R1\n and R2\n be two relation defined as\nfollows :\n R1\n = {(a, b) $$ \\in $$ R2\n : a2\n + b2 $$ \\in $$ Q} and\n R2\n = {(a, b) $$ \\in $$ R2\n : a2\n + b2 $$ \\notin $$ Q}, \n where Q is the\nset of all rational numbers. Then :",
"options": [
{
"text": "Neither R1\n nor R2\n is transitive."
},
{
"text": "R2\n is transitive but R1\n is not transitive."
},
{
"text": "R1\n and R2\n are both transitive."
},
{
"text": "R1\n is transitive but R2\n is not transitive."
}
],
"answer": "Neither R1\n nor R2\n is transitive.",
"solution": "**Answer:** Neither R1\n nor R2\n is transitive.\n\nFor R1 :
Let a = 1 + $$\\sqrt 2 $$, b = 1 $$-$$ $$\\sqrt 2 $$, c = $${8^{{1 \\over 4}}}$$
$$ \\therefore $$ R2 is not transitive.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7267,
"subject": "General Science",
"question": "Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1, $$-$$1) is the set :",
"options": [
{
"text": "$$S = \\{ (x,y)|{x^2} + {y^2} = \\sqrt 2 \\} $$"
},
{
"text": "$$S = \\{ (x,y)|{x^2} + {y^2} = 2\\} $$"
},
{
"text": "$$S = \\{ (x,y)|{x^2} + {y^2} = 1\\} $$"
},
{
"text": "$$S = \\{ (x,y)|{x^2} + {y^2} = 4\\} $$"
}
],
"answer": "$$S = \\{ (x,y)|{x^2} + {y^2} = 2\\} $$",
"solution": "**Answer:** $$S = \\{ (x,y)|{x^2} + {y^2} = 2\\} $$\n\nGiven R = {(P, Q) | P and Q are at the same distance from the origin}.
Then equivalence class of (1, $$-$$1) will contain al such points which lies on circumference of the circle of centre at origin and passing through point (1, $$-$$1).
i.e., radius of circle = $$\\sqrt {{1^2} + {1^2}} = \\sqrt 2 $$
$$ \\therefore $$ Required equivalence class of (S)
$$ = \\{ (x,y)|{x^2} + {y^2} = 2\\} $$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7268,
"subject": "General Science",
"question": "Define a relation R over a class of n $$\\times$$ n real matrices A and B as
\"ARB iff there exists a non-singular matrix P such that PAP$$-$$1 = B\".
Then which of the following is true?",
"options": [
{
"text": "R is reflexive, transitive but not symmetric"
},
{
"text": "R is symmetric, transitive but not reflexive."
},
{
"text": "R is reflexive, symmetric but not transitive"
},
{
"text": "R is an equivalence relation"
}
],
"answer": "R is an equivalence relation",
"solution": "**Answer:** R is an equivalence relation\n\nFor reflexive relation,
$\\forall(A, A) \\in R$ for matrix $P$.
\n$\\Rightarrow A=P A P^{-1}$ is true for $P=1$
\nSo, $R$ is reflexive relation.
\nFor symmetric relation,
\nLet $(A, B) \\in R$ for matrix $P$.
\n$$\n\\Rightarrow \\quad A=P B P^{-1}\n$$
\nAfter pre-multiply by $P^{-1}$ and post-multiply by $P$, we get
\n$$\nP^{-1} A P=B\n$$
\nSo, $(B, A) \\in R$ for matrix $P^{-1}$.
\nSo, $R$ is a symmetric relation.
\nFor transitive relation,
\nLet $A R B$ and $B R C$
\nSo, $A=P B P^{-1}$ and $B=P C P^{-1}$
\nNow, $A=P\\left(P C P^{-1}\\right) P^{-1}$
\n$\\Rightarrow A=(P)^2 C\\left(P^{-1}\\right)^2 \\Rightarrow A=(P)^2 \\cdot C \\cdot\\left(P^2\\right)^{-1}$
\n$\\therefore(A, C) \\in R$ for matrix $P^2$.
\n$\\therefore R$ is transitive relation.
\nHence, $R$ is an equivalence relation.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7269,
"subject": "General Science",
"question": "Let N be the set of natural numbers and a relation R on N be defined by $$R = \\{ (x,y) \\in N \\times N:{x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0\\} $$. Then the relation R is :",
"options": [
{
"text": "symmetric but neither reflexive nor transitive"
},
{
"text": "reflexive but neither symmetric nor transitive"
},
{
"text": "reflexive and symmetric, but not transitive"
},
{
"text": "an equivalence relation"
}
],
"answer": "reflexive but neither symmetric nor transitive",
"solution": "**Answer:** reflexive but neither symmetric nor transitive\n\n$${x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0$$
Now, x = y $$\\forall$$(x, y) $$\\in$$N $$\\times$$ N so reflexive but not symmetric & transitive.
See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) & (1, $$-$$1) satisfies but (3, $$-$$1) does not.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7270,
"subject": "General Science",
"question": "Which of the following is not correct for relation R on the set of real numbers ?",
"options": [
{
"text": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ 0 < |x| $$-$$ |y| $$\\le$$ 1 is neither transitive nor symmetric."
},
{
"text": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ 0 < |x $$-$$ y| $$\\le$$ 1 is symmetric and transitive."
},
{
"text": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ |x| $$-$$ |y| $$\\le$$ 1 is reflexive but not symmetric."
},
{
"text": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ |x $$-$$ y| $$\\le$$ 1 is reflexive nd symmetric."
}
],
"answer": "(x, y) $$\\in$$ R $$ \\Leftrightarrow $$ 0 < |x $$-$$ y| $$\\le$$ 1 is symmetric and transitive.",
"solution": "**Answer:** (x, y) $$\\in$$ R $$ \\Leftrightarrow $$ 0 < |x $$-$$ y| $$\\le$$ 1 is symmetric and transitive.\n\nNote that (a, b) and (b, c) satisfy 0 < |x $$-$$ y| $$\\le$$ 1 but (a, c) does not satisfy it so 0 $$\\le$$ |x $$-$$ y| $$\\le$$ 1 is symmetric but not transitive. \n
For example,\n
x = 0.2, y = 0.9, z = 1.5\n
0 ≤ |x – y| = 0.7 ≤ 1\n
0 ≤ |y – z| = 0.6 ≤ 1\n
But |x – z| = 1.3 > 1\n
So, (b) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7271,
"subject": "General Science",
"question": "
Let a set A = A1 $$\\cup$$ A2 $$\\cup$$ ..... $$\\cup$$ Ak, where Ai $$\\cap$$ Aj = $$\\phi$$ for i $$\\ne$$ j, 1 $$\\le$$ j, j $$\\le$$ k. Define the relation R from A to A by R = {(x, y) : y $$\\in$$ Ai if and only if x $$\\in$$ Ai, 1 $$\\le$$ i $$\\le$$ k}. Then, R is :
",
"options": [
{
"text": "reflexive, symmetric but not transitive."
},
{
"text": "reflexive, transitive but not symmetric."
},
{
"text": "reflexive but not symmetric and transitive."
},
{
"text": "an equivalence relation."
}
],
"answer": "an equivalence relation.",
"solution": "**Answer:** an equivalence relation.\n\n
Let R1 = {(a, b) $$\\in$$ N $$\\times$$ N : |a $$-$$ b| $$\\le$$ 13} and
\n
R2 = {(a, b) $$\\in$$ N $$\\times$$ N : |a $$-$$ b| $$\\ne$$ 13}. Then on N :
",
"options": [
{
"text": "Both R1 and R2 are equivalence relations"
},
{
"text": "Neither R1 nor R2 is an equivalence relation"
},
{
"text": "R1 is an equivalence relation but R2 is not"
},
{
"text": "R2 is an equivalence relation but R1 is not"
}
],
"answer": "Neither R1 nor R2 is an equivalence relation",
"solution": "**Answer:** Neither R1 nor R2 is an equivalence relation\n\n$R_{1}=\\{(a, b) \\in N \\times N:|a-b| \\leq 13\\}$ and\n
\n$R_{2}=\\{(a, b) \\in N \\times N:|a-b| \\neq 13\\}$\n
\nIn $R_{1}: \\because|2-11|=9 \\leq 13$\n
\n$\\therefore \\quad(2,11) \\in R_{1}$ and $(11,19) \\in R_{1}$ but $(2,19) \\notin R_{1}$\n
\n$\\therefore \\quad R_{1}$ is not transitive\n
\nHence $R_{1}$ is not equivalence\n\n
In $R_{2}:(13,3) \\in R_{2}$ and $(3,26) \\in R_{2}$ but $(13,26) \\notin R_{2}$\n \n$(\\because|13-26|=13)$\n
\n$\\therefore R_{2}$ is not transitive\n
\nHence $R_{2}$ is not equivalence.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7273,
"subject": "General Science",
"question": "
Let $$R_{1}$$ and $$R_{2}$$ be two relations defined on $$\\mathbb{R}$$ by
\n
$$a \\,R_{1} \\,b \\Leftrightarrow a b \\geq 0$$ and $$a \\,R_{2} \\,b \\Leftrightarrow a \\geq b$$
\n
Then,
",
"options": [
{
"text": "$$R_{1}$$ is an equivalence relation but not $$R_{2}$$"
},
{
"text": "$$R_{2}$$ is an equivalence relation but not $$R_{1}$$"
},
{
"text": "both $$R_{1}$$ and $$R_{2}$$ are equivalence relations"
},
{
"text": "neither $$R_{1}$$ nor $$R_{2}$$ is an equivalence relation"
}
],
"answer": "neither $$R_{1}$$ nor $$R_{2}$$ is an equivalence relation",
"solution": "**Answer:** neither $$R_{1}$$ nor $$R_{2}$$ is an equivalence relation\n\n
$$a\\,{R_1}\\,b \\Leftrightarrow ab \\ge 0$$
\n
So, definitely $$(a,a) \\in {R_1}$$ as $${a^2} \\ge 0$$
\n
If $$(a,b) \\in {R_1} \\Rightarrow (b,a) \\in {R_1}$$
\n
But if $$(a,b) \\in {R_1},(b,c) \\in {R_1}$$
\n
$$\\Rightarrow$$ Then $$(a,c)$$ may or may not belong to R1
\n
{Consider $$a = - 5,b = 0,c = 5$$ so $$(a,b)$$ and $$(b,c) \\in {R_1}$$ but $$ac < 0$$}
\n
So, R1 is not equivalence relation
\n
$$a\\,{R_2}\\,b \\Leftrightarrow a \\ge b$$
\n
$$(a,a) \\in {R_2} \\Rightarrow $$ so reflexive relation
\n
If $$(a,b) \\in {R_2}$$ then $$(b,a)$$ may or may not belong to R2
For $$\\alpha \\in \\mathbf{N}$$, consider a relation $$\\mathrm{R}$$ on $$\\mathbf{N}$$ given by $$\\mathrm{R}=\\{(x, y): 3 x+\\alpha y$$ is a multiple of 7$$\\}$$. The relation $$R$$ is an equivalence relation if and only if :
",
"options": [
{
"text": "$$\\alpha=14$$"
},
{
"text": "$$\\alpha$$ is a multiple of 4"
},
{
"text": "4 is the remainder when $$\\alpha$$ is divided by 10"
},
{
"text": "4 is the remainder when $$\\alpha$$ is divided by 7"
}
],
"answer": "4 is the remainder when $$\\alpha$$ is divided by 7",
"solution": "**Answer:** 4 is the remainder when $$\\alpha$$ is divided by 7\n\n
$$R = \\{ (x,y):3x + \\alpha y$$ is multiple of 7$$\\} $$, now R to be an equivalence relation
\n
(1) R should be reflexive : $$(a,a) \\in R\\,\\forall \\,a \\in N$$
Let $$P(S)$$ denote the power set of $$S=\\{1,2,3, \\ldots ., 10\\}$$. Define the relations $$R_{1}$$ and $$R_{2}$$ on $$P(S)$$ as $$\\mathrm{AR}_{1} \\mathrm{~B}$$ if $$\\left(\\mathrm{A} \\cap \\mathrm{B}^{\\mathrm{c}}\\right) \\cup\\left(\\mathrm{B} \\cap \\mathrm{A}^{\\mathrm{c}}\\right)=\\emptyset$$ and $$\\mathrm{AR}_{2} \\mathrm{~B}$$ if $$\\mathrm{A} \\cup \\mathrm{B}^{\\mathrm{c}}=\\mathrm{B} \\cup \\mathrm{A}^{\\mathrm{c}}, \\forall \\mathrm{A}, \\mathrm{B} \\in \\mathrm{P}(\\mathrm{S})$$. Then :
",
"options": [
{
"text": "only $$R_{2}$$ is an equivalence relation"
},
{
"text": "both $$R_{1}$$ and $$R_{2}$$ are not equivalence relations"
},
{
"text": "both $$R_{1}$$ and $$R_{2}$$ are equivalence relations"
},
{
"text": "only $$R_{1}$$ is an equivalence relation"
}
],
"answer": "both $$R_{1}$$ and $$R_{2}$$ are equivalence relations",
"solution": "**Answer:** both $$R_{1}$$ and $$R_{2}$$ are equivalence relations\n\n$\\begin{aligned} & \\mathrm{S}=\\{1,2,3, \\ldots \\ldots 10\\} \\\\\\\\ & \\mathrm{P}(\\mathrm{S})=\\text { power set of } \\mathrm{S} \\\\\\\\ & \\mathrm{AR}_1 \\mathrm{B} \\Rightarrow(\\mathrm{A} \\cap \\overline{\\mathrm{B}}) \\cup(\\overline{\\mathrm{A}} \\cap \\mathrm{B})=\\phi \\\\\\\\ & \\mathrm{R}_1 \\text { is reflexive, symmetric } \\\\\\\\ & \\text { For transitive } \\\\\\\\ & (\\mathrm{A} \\cap \\overline{\\mathrm{B}}) \\cup(\\overline{\\mathrm{A}} \\cap \\mathrm{B})=\\phi ;\\{\\mathrm{a}\\}=\\phi=\\{\\mathrm{b}\\} \\therefore \\mathrm{A}=\\mathrm{B} \\\\\\\\ & (\\mathrm{B} \\cap \\overline{\\mathrm{C}}) \\cup(\\overline{\\mathrm{B}} \\cap \\mathrm{C})=\\phi \\therefore \\mathrm{B}=\\mathrm{C} \\\\\\\\ & \\therefore \\mathrm{A}=\\mathrm{C} \\text { equivalence. }\\end{aligned}$\n\n
and $\\mathrm{T}=\\left\\{(\\mathrm{a}, \\mathrm{b}): \\mathrm{a}, \\mathrm{b} \\in \\mathbb{R}, \\mathrm{a}^{2}-\\mathrm{b}^{2} \\in \\mathbb{Z}\\right\\}$,",
"options": [
{
"text": "$\\mathrm{S}$ is transitive but $\\mathrm{T}$ is not\n"
},
{
"text": "both $\\mathrm{S}$ and $\\mathrm{T}$ are symmetric"
},
{
"text": "neither $S$ nor $T$ is transitive"
},
{
"text": "$T$ is symmetric but $S$ is not"
}
],
"answer": "$T$ is symmetric but $S$ is not",
"solution": "**Answer:** $T$ is symmetric but $S$ is not\n\nFor relation $\\mathrm{T}=\\mathrm{a}^{2}-\\mathrm{b}^{2}=-\\mathrm{I}$\n\n
Then, $(\\mathrm{b}, \\mathrm{a})$ on relation $\\mathrm{R}$\n\n
$2+\\frac{\\mathrm{b}}{\\mathrm{a}}$ not necessarily positive\n\n
$\\therefore \\mathrm{S}$ is not symmetric",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7277,
"subject": "General Science",
"question": "
Let $$R$$ be a relation on $$\\mathbb{R}$$, given by $$R=\\{(a, b): 3 a-3 b+\\sqrt{7}$$ is an irrational number $$\\}$$. Then $$R$$ is
",
"options": [
{
"text": "an equivalence relation"
},
{
"text": "reflexive and symmetric but not transitive"
},
{
"text": "reflexive and transitive but not symmetric"
},
{
"text": "reflexive but neither symmetric nor transitive"
}
],
"answer": "reflexive but neither symmetric nor transitive",
"solution": "**Answer:** reflexive but neither symmetric nor transitive\n\nFor reflexive :\n\n
$3 a-3 a+\\sqrt{7}$ is an irrational number $\\forall a \\in R R$ is reflexive\n\n
For symmetric :\n\n
Let $3 a-3 b+\\sqrt{7}$ is an irrational number\n\n
$\\Rightarrow 3 b-3 a+\\sqrt{7}$ is an irrational number\n\n
For example, Let $3 a-3 b=\\sqrt{7}$\n\n
$\\sqrt{7}+\\sqrt{7}$ is irrational but $-\\sqrt{7}+\\sqrt{7}$ is not.\n\n
$\\therefore R$ is not symmetric\n\n
For transitive :\n\n
Let $3 a-3 b+\\sqrt{7}$ is irrational and $3 b-3 c+\\sqrt{7}$ is irrational.\n\n
$\\Rightarrow 3 a-3 c+\\sqrt{7}$ is irrational.\n\n
For example, take $a=0, b=-\\sqrt{7}, c=\\frac{\\sqrt{7}}{3}$\n\n
$R$ is not transitive.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7278,
"subject": "General Science",
"question": "
Let $$\\mathrm{R}$$ be a relation on $$\\mathrm{N} \\times \\mathbb{N}$$ defined by $$(a, b) ~\\mathrm{R}~(c, d)$$ if and only if $$a d(b-c)=b c(a-d)$$. Then $$\\mathrm{R}$$ is
",
"options": [
{
"text": "symmetric and transitive but not reflexive"
},
{
"text": "reflexive and symmetric but not transitive"
},
{
"text": "transitive but neither reflexive nor symmetric"
},
{
"text": "symmetric but neither reflexive nor transitive"
}
],
"answer": "symmetric but neither reflexive nor transitive",
"solution": "**Answer:** symmetric but neither reflexive nor transitive\n\nGiven, $(a, b) R(c, d) \\Rightarrow a d(b-c)=b c(a-d)$\n\n
Symmetric :\n\n
(c, d) $R(a, b) \\Rightarrow \\operatorname{cb}(\\mathrm{d}-\\mathrm{a})=\\mathrm{da}(\\mathrm{c}-\\mathrm{b}) $\n\n
$\\Rightarrow$ Symmetric.\n\n
Reflexive :\n\n
(a, b) R (a, b) $\\Rightarrow a b(b-a) \\neq b a(a-b) $\n\n
$\\Rightarrow$ Not reflexive.\n\n
Transitive : \n
$(2,3) \\mathrm{R}(3,2)$ and $(3,2) \\mathrm{R}(5,30)$ but\n\n
The minimum number of elements that must be added to the relation $$ \\mathrm{R}=\\{(\\mathrm{a}, \\mathrm{b}),(\\mathrm{b}, \\mathrm{c})\\}$$ on the set $$\\{a, b, c\\}$$ so that it becomes symmetric and transitive is :
The minimum number of elements that must be added to the relation R = {(a, b), (b, c), (b, d)} on the set {a, b, c, d} so that it is an equivalence relation, is __________.
\nAdding $(a, a),(b, b),(c, c),(d, d)$ make reflexive.\n
\nAdding $(b, a),(c, b),(d, b)$ make Symmetric\n
\nAnd adding $(a, d),(a, c)$ to make transitive\n
\nFurther $(d, a) \\&(c, a)$ to be added to make Symmetricity.\n
\nFurther $(c, d) \\&(d, c)$ also be added.\n
\nSo total 13 elements to be added to make equivalence.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7282,
"subject": "General Science",
"question": "
The relation $$\\mathrm{R = \\{ (a,b):\\gcd (a,b) = 1,2a \\ne b,a,b \\in \\mathbb{Z}\\}}$$ is :
",
"options": [
{
"text": "reflexive but not symmetric"
},
{
"text": "transitive but not reflexive"
},
{
"text": "symmetric but not transitive"
},
{
"text": "neither symmetric nor transitive"
}
],
"answer": "neither symmetric nor transitive",
"solution": "**Answer:** neither symmetric nor transitive\n\n
Given,
\n
(a, b) belongs to relation R if $$\\gcd (a,b) = 1, 2a \\ne b$$.
\n
Here $$\\gcd $$ means greatest common divisor. $$\\gcd $$ of two numbers is the largest number that divides both of them.
Let $$\\mathrm{A}=\\{-4,-3,-2,0,1,3,4\\}$$ and $$\\mathrm{R}=\\left\\{(a, b) \\in \\mathrm{A} \\times \\mathrm{A}: b=|a|\\right.$$ or $$\\left.b^{2}=a+1\\right\\}$$ be a relation on $$\\mathrm{A}$$. Then the minimum number of elements, that must be added to the relation $$\\mathrm{R}$$ so that it becomes reflexive and symmetric, is __________
Relation to be reflexive $(a, a) \\in R \\forall a \\in A$\n
$\\Rightarrow (-4,-4),(-3,-3),(-2,-2)$ also should be added in $R$.\n\n
Relation to be symmetric if $(a, b) \\in R$, then $(b, a) \\in R \\forall a, b \\in A$\n
$\\Rightarrow (4,-4),(3,-3),(1,0),(-2,3)$ also should be added in $R$\n
$\\Rightarrow$ Minimum number of elements to be added to $R=3+4=7$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7284,
"subject": "General Science",
"question": "
The number of relations, on the set $$\\{1,2,3\\}$$ containing $$(1,2)$$ and $$(2,3)$$, which are reflexive and transitive but not symmetric, is __________.
To find the number of such relations, let's first understand what it means for a relation to be reflexive, transitive, and not symmetric.
\n\n
A relation $$R$$ on a set $$S$$ is reflexive if every element is related to itself. That is, $$(a, a) \\in R$$ for all $$a \\in S$$.
\n\n
A relation is transitive if whenever $$(a, b) \\in R$$ and $$(b, c) \\in R$$, then it must also be the case that $$(a, c) \\in R$$.
\n\n
A relation is symmetric if whenever $$(a, b) \\in R$$, then $$(b, a) \\in R$$ also.
\n\n
Since we are looking for relations that are reflexive and transitive but not symmetric, we will need to include certain elements and exclude others.
\n\n
First, for the relation to be reflexive on the set $$\\{1,2,3\\}$$, it must contain $$(1,1), (2,2),$$ and $$(3,3)$$.
\n\n
Given that the relation must contain $$(1,2)$$ and $$(2,3)$$ and be transitive, it must also contain $$(1,3)$$ because if $$(1,2)$$ and $$(2,3)$$ are included, then to maintain transitivity, $$(1,3)$$ must be included as well.
\n\n
So far, the must-have elements of the desired relation are:
\n\n
\n
For reflexivity: $$(1,1), (2,2), (3,3)$$
\n
Given in the problem: $$(1,2), (2,3)$$
\n
For transitivity (induced by given elements): $$(1,3)$$
\n
\n\n
\"To maintain a relation that is reflexive, transitive, and not symmetric, we must carefully select additional pairs beyond the reflexive minimum $(1,1), (2,2), (3,3)$, and the given $(1,2), (2,3)$, which also necessitates $(1,3)$ due to transitivity. While including pairs such as $(2,1)$, $(3,1)$, or $(3,2)$ could potentially introduce symmetry, we can include some of these pairs as long as the resulting relation does not fulfill the condition for symmetry for all elements. This means we can include one or more of these pairs if doing so does not result in every pair being mirrored (i.e., for every $(a,b) \\in R$, $(b,a) \\in R$ is not required), thus keeping the relation not symmetric. The key is ensuring that the inclusion of any such pair does not lead to a situation where for every $(a,b)$ in the relation, the reverse $(b,a)$ also exists, which would make the relation symmetric, contradicting our requirement.\"
Here, only (3,2) is in R3, and neither (2,1) nor (3,1) are in R3.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7285,
"subject": "General Science",
"question": "
Let $$\\mathrm{A}=\\{1,2,3,4,5,6,7\\}$$. Then the relation $$\\mathrm{R}=\\{(x, y) \\in \\mathrm{A} \\times \\mathrm{A}: x+y=7\\}$$ is :
",
"options": [
{
"text": "reflexive but neither symmetric nor transitive"
},
{
"text": "transitive but neither symmetric nor reflexive"
},
{
"text": "symmetric but neither reflexive nor transitive"
},
{
"text": "an equivalence relation"
}
],
"answer": "symmetric but neither reflexive nor transitive",
"solution": "**Answer:** symmetric but neither reflexive nor transitive\n\nHere, $A=\\{1,2,3,4,5,6,7\\}$\n
Since, $x+y=7 \\Rightarrow y=7-x$\n
So, $\\mathrm{R}=\\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\\}$\n
$\\because(a, b) \\in \\mathrm{R} \\Rightarrow(b, a) \\in \\mathrm{R}$\n
Let $$A=\\{0,3,4,6,7,8,9,10\\}$$ and $$R$$ be the relation defined on $$A$$ such that $$R=\\{(x, y) \\in A \\times A: x-y$$ is odd positive integer or $$x-y=2\\}$$. The minimum number of elements that must be added to the relation $$R$$, so that it is a symmetric relation, is equal to ____________.
Case I : $x-y$ is odd, if one is odd and one is even and $x>y$. \n
$\\therefore$ Possibilites are $\\{(3,0),(4,3),(6,3),(7,6),(7,4)$, $(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10$, $7),(10,3)\\}$ \n
No. of cases $=15$ \n
Case II : $x-y=2$ \n
$\\therefore$ Possibilities are $\\{(6,4),(8,6),(9,7),(10,8)\\}$ \n
$\\therefore$ No. of cases $=4$ \n
So, minimum ordered pair to be added $=15+4=19$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7287,
"subject": "General Science",
"question": "Consider the relations $R_1$ and $R_2$ defined as $a R_1 b \\Leftrightarrow a^2+b^2=1$ for all $a, b \\in \\mathbf{R}$ and $(a, b) R_2(c, d) \\Leftrightarrow$ $a+d=b+c$ for all $(a, b),(c, d) \\in \\mathbf{N} \\times \\mathbf{N}$. Then :",
"options": [
{
"text": "$R_1$ and $R_2$ both are equivalence relations"
},
{
"text": "Only $R_1$ is an equivalence relation"
},
{
"text": "Only $R_2$ is an equivalence relation"
},
{
"text": "Neither $R_1$ nor $R_2$ is an equivalence relation"
}
],
"answer": "Only $R_2$ is an equivalence relation",
"solution": "**Answer:** Only $R_2$ is an equivalence relation\n\n
To determine if the given relations $R_1$ and $R_2$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity.
\n\n
Let's start by analysing $R_1$:
\n\n
Reflexivity: A relation $R$ on a set $S$ is reflexive if every element is related to itself, that is, for every $a \\in S$, the pair $(a,a)$ is in $R$. In the case of $R_1$, for an arbitrary $a \\in \\mathbf{R}$, we need to check if $a R_1 a$ holds, which translates to checking if $a^2 + a^2 = 1$. This would mean $2a^2 = 1$ or $a^2 = \\frac{1}{2}$. Since this is not true for every real number $a$, $R_1$ is not reflexive.
\n\n
Symmetry: A relation $R$ is symmetric if whenever $a R b$, then $b R a$. For $R_1$, if $a^2 + b^2 = 1$, then it is also true that $b^2 + a^2 = 1$. Thus, $R_1$ is symmetric.
\n\n
Transitivity: A relation $R$ is transitive if whenever $a R b$ and $b R c$, then $a R c$. For $R_1$, suppose $a R_1 b$ and $b R_1 c$, this means $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$. However, if we add these two, we get $a^2 + 2b^2 + c^2 = 2$. There is no guarantee that $a^2 + c^2 = 1$. Therefore, $R_1$ is not transitive.
\n\n
Conclusion: Relation $R_1$ is not reflexive or transitive, and hence it is not an equivalence relation.
\n\n
Now let's consider $R_2$:
\n\n
Reflexivity: For any $(a, b) \\in \\mathbf{N} \\times \\mathbf{N}$, it's clear that $a + b = b + a$, which is just a reiteration of the commutative property of addition. Therefore, $(a, b) R_2 (a, b)$, and $R_2$ is reflexive.
\n\n
Symmetry: If $(a, b) R_2 (c, d)$, meaning $a + d = b + c$, then by reordering the terms we can similarly have $c + b = d + a$, which means $(c, d) R_2 (a, b)$, so $R_2$ is symmetric.
\n\n
Transitivity: Suppose $(a, b) R_2 (c, d)$ and $(c, d) R_2 (e, f)$, i.e., $a + d = b + c$ and $c + f = d + e$, we want to show that $(a, b) R_2 (e, f)$. Adding the two equations, we get $a + d + c + f = b + c + d + e$. By rearranging and simplifying, we get $a + f = b + e$, thus, $(a, b) R_2 (e, f)$.
\n\n
So $R_2$ is reflexive, symmetric, and transitive, and therefore it is an equivalence relation.
\n\n
Conclusion: According to the above examination, only $R_2$ is an equivalence relation. Thus, the correct answer is:
\n\n
Option C: Only $R_2$ is an equivalence relation.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7288,
"subject": "General Science",
"question": "Let $S=\\{1,2,3, \\ldots, 10\\}$. Suppose $M$ is the set of all the subsets of $S$, then the relation
Let $$R$$ be a relation on $$Z \\times Z$$ defined by $$(a, b) R(c, d)$$ if and only if $$a d-b c$$ is divisible by 5. Then $$R$$ is
",
"options": [
{
"text": "Reflexive and transitive but not symmetric\n"
},
{
"text": "Reflexive and symmetric but not transitive\n"
},
{
"text": "Reflexive but neither symmetric nor transitive\n"
},
{
"text": "Reflexive, symmetric and transitive"
}
],
"answer": "Reflexive and symmetric but not transitive\n",
"solution": "**Answer:** Reflexive and symmetric but not transitive\n\n\n
$$(a, b) R(a, b)$$ as $$a b-a b=0$$
\n
Therefore reflexive
\n
Let $$(a, b) R(c, d) \\Rightarrow a d-b c$$ is divisible by 5
\n
$$\\Rightarrow \\mathrm{bc}-\\mathrm{ad}$$ is divisible by $$5 \\Rightarrow(\\mathrm{c}, \\mathrm{d}) \\mathrm{R}(\\mathrm{a}, \\mathrm{b})$$
\n
Therefore symmetric
\n
Relation not transitive as $$(3,1) \\mathrm{R}(10,5)$$ and $$(10,5) \\mathrm{R}(1,1)$$ but $$(3,1)$$ is not related to $$(1,1)$$
Let a relation $$\\mathrm{R}$$ on $$\\mathrm{N} \\times \\mathbb{N}$$ be defined as: $$\\left(x_1, y_1\\right) \\mathrm{R}\\left(x_2, y_2\\right)$$ if and only if $$x_1 \\leq x_2$$ or $$y_1 \\leq y_2$$.\nConsider the two statements:
\n
(I) $$\\mathrm{R}$$ is reflexive but not symmetric.
\n
(II) $$\\mathrm{R}$$ is transitive
\n
Then which one of the following is true?
",
"options": [
{
"text": "Only (II) is correct.\n"
},
{
"text": "Both (I) and (II) are correct.\n"
},
{
"text": "Neither (I) nor (II) is correct.\n"
},
{
"text": "Only (I) is correct."
}
],
"answer": "Only (I) is correct.",
"solution": "**Answer:** Only (I) is correct.\n\n
$$\\begin{aligned}\n& \\left(x_1, y_1\\right) R\\left(x_2, y_2\\right) \\\\\n& \\text { If } x_1 \\leq x_2 \\text { or } y_1 \\leq y_2\n\\end{aligned}$$
Let $$A=\\{2,3,6,8,9,11\\}$$ and $$B=\\{1,4,5,10,15\\}$$. Let $$R$$ be a relation on $$A \\times B$$ defined by\n$$(a, b) R(c, d)$$ if and only if $$3 a d-7 b c$$ is an even integer. Then the relation $$R$$ is
",
"options": [
{
"text": "reflexive but not symmetric.\n"
},
{
"text": "an equivalence relation.\n"
},
{
"text": "reflexive and symmetric but not transitive.\n"
},
{
"text": "transitive but not symmetric."
}
],
"answer": "reflexive and symmetric but not transitive.\n",
"solution": "**Answer:** reflexive and symmetric but not transitive.\n\n\n
$$(a, b) R(c, d) \\Rightarrow 3 a d-7 b c \\in$$ even
\n
For reflexive
\n
$$(a, b) R(a, b) \\Rightarrow 3 a b-7 b a=-4 a b \\in$$ even
\n
For symmetric
\n
$$(a, b) R(c, d)$$
\n
\n
then
\n
$$(c, d) R(a, b)=3 b c-7 a d$$
\n
$$\\in$$ even
\n
Now check for transitive
\n
$$\\begin{aligned}\n& \\begin{array}{ll}\n(a, b) R(c, d) \\text { and }(c, d) R(e, f) \\text { then }(a, b) R(e, f) \\\\\n3 a d-7 b c=2 m \\quad \\Rightarrow (2,5) R(6,8) \\text { and } \\\\\n3 c f-7 e d=2 n \\quad (6,8) R(9,4) \\\\\n\\text { then } 3 a f-7 e b \\neq \\text { even } \\notin(2,5) R(9,4) \\\\\n\\Rightarrow \\text { Not transitive option }(3)\n\\end{array}\n\\end{aligned}$$
Let $$\\mathrm{A}=\\{1,2,3,4,5\\}$$. Let $$\\mathrm{R}$$ be a relation on $$\\mathrm{A}$$ defined by $$x \\mathrm{R} y$$ if and only if $$4 x \\leq 5 \\mathrm{y}$$. Let $$\\mathrm{m}$$ be the number of elements in $$\\mathrm{R}$$ and $$\\mathrm{n}$$ be the minimum number of elements from $$\\mathrm{A} \\times \\mathrm{A}$$ that are required to be added to R to make it a symmetric relation. Then m + n is equal to :
$$\\begin{aligned}\n& A=\\{1,2,3,4,5\\} \\\\\n& x R y \\Leftrightarrow 4 x \\leq 5 y \\\\\n& 4 x \\leq 5 y \\quad \\Rightarrow \\quad \\frac{x}{y} \\leq \\frac{5}{4} \\quad \\Rightarrow \\frac{x}{y} \\leq 1.25\n\\end{aligned}$$
Let the relations $$R_1$$ and $$R_2$$ on the set $$X=\\{1,2,3, \\ldots, 20\\}$$ be given by $$R_1=\\{(x, y): 2 x-3 y=2\\}$$ and $$R_2=\\{(x, y):-5 x+4 y=0\\}$$. If $$M$$ and $$N$$ be the minimum number of elements required to be added in $$R_1$$ and $$R_2$$, respectively, in order to make the relations symmetric, then $$M+N$$ equals
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7294,
"subject": "General Science",
"question": "If $A, B$ and $C$ are three sets such that $A \\cap B=A \\cap C$ and $A \\cup B=A \\cup C$, then :",
"options": [
{
"text": "$A=C$"
},
{
"text": "$B=C$"
},
{
"text": "$A \\cap B=\\phi$"
},
{
"text": "$A=B$"
}
],
"answer": "$B=C$",
"solution": "**Answer:** $B=C$\n\n
From the given conditions, we have :
\n\n
A ∩ B = A ∩ C : The intersection of set A with set B is the same as the intersection of set A with set C. This indicates that all elements common to A and B are also common to A and C, and vice versa.
\n
\n
A ∪ B = A ∪ C : The union of set A with set B is the same as the union of set A with set C. This indicates that all elements in A, B, and C are the same.
\n
\n\n
From these two conditions, we can infer that set B is equal to set C because every element of B is also an element of C and vice versa. Hence, Option B : B = C is the correct answer.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7295,
"subject": "General Science",
"question": "In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is ",
"options": [
{
"text": "42"
},
{
"text": "102"
},
{
"text": "1"
},
{
"text": "38"
}
],
"answer": "38",
"solution": "**Answer:** 38\n\nWe're given that there are 140 students numbered from 1 to 140. \n\n
1. Define the set $A$ to be the set of even numbered students. The cardinality of $A$ (the number of elements in $A$), denoted as $n(A)$, can be computed as the greatest integer less than or equal to $140/2$. Hence, $n(A) = \\left[\\frac{140}{2}\\right] = 70$. ([.] denotes greatest integer function)\n\n
2. Similarly, let $B$ be the set of students whose number is divisible by 3. Hence, $n(B) = \\left[\\frac{140}{3}\\right] = 46$. ([.] denotes greatest integer function)\n\n
3. Let $C$ be the set of students whose number is divisible by 5. Hence, $n(C) = \\left[\\frac{140}{5}\\right] = 28$.\n\n
So far, we've found the number of students who opted for Mathematics ($n(A)$), Physics ($n(B)$), and Chemistry ($n(C)$).\n\n
We also need to consider the students who have opted for multiple subjects :\n\n
1. $n(A \\cap B)$ represents the count of numbers that are divisible by both 2 and 3 (i.e., divisible by 6). \n
So, $n(A \\cap B) = \\left[\\frac{140}{6}\\right] = 23$.\n\n
2. $n(B \\cap C)$ represents the count of numbers that are divisible by both 3 and 5 (i.e., divisible by 15). \n
So, $n(B \\cap C) = \\left[\\frac{140}{15}\\right] = 9$.\n\n
3. $n(C \\cap A)$ represents the count of numbers that are divisible by both 2 and 5 (i.e., divisible by 10). \n
So, $n(C \\cap A) = \\left[\\frac{140}{10}\\right] = 14$.\n\n
Finally, $n(A \\cap B \\cap C)$ represents the count of numbers that are divisible by 2, 3, and 5 (i.e., divisible by 30). \n
So, $n(A \\cap B \\cap C) = \\left[\\frac{140}{30}\\right] = 4$.\n\n
Now we use the principle of inclusion and exclusion to compute the number of students who have opted for at least one subject. The principle states :\n\n
$$n(A \\cup B \\cup C) = n(A) + n(B) + n(C) - n(A \\cap B) - n(B \\cap C) - n(C \\cap A) + n(A \\cap B \\cap C)$$\n\n
Substituting the values we calculated above :\n\n
$$n(A \\cup B \\cup C) = (70+46+28) - (23+9+14) + 4 = 102$$\n\n
Hence, the number of students who opted for at least one subject is 102. Therefore, the number of students who did not opt for any of the subjects is
Total $$-$$ n(A $$ \\cup $$ B $$ \\cup $$ C)\n
= 140 $$-$$ 102 = 38",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 7296,
"subject": "General Science",
"question": "Two newspapers A and B are published in a city.\nIt is known that 25% of the city populations reads\nA and 20% reads B while 8% reads both A and\nB. Further, 30% of those who read A but not B\nlook into advertisements and 40% of those who\nread B but not A also look into advertisements,\nwhile 50% of those who read both A and B look\ninto advertisements. Then the percentage of the\npopulation who look into advertisement is :-",
"options": [
{
"text": "13.5"
},
{
"text": "13"
},
{
"text": "12.8"
},
{
"text": "13.9"
}
],
"answer": "13.9",
"solution": "**Answer:** 13.9\n\n\n
The total population to be 100 (for simplicity's sake) and the percentages can be treated as actual numbers of people in this context.
\n
The percentage of people who read newspaper A is given as 25. However, among these, there are people who read both newspapers A and B, given as 8. To find the number of people who read only newspaper A, we subtract the number of people who read both from the total number of people who read A. That is,
\n
n(A only) = 25 – 8 = 17
\n
Similarly, the number of people who read only newspaper B is calculated as :
\n
n(B only) = 20 – 8 = 12
\n
Now, we are given the percentage of each of these groups that look into the advertisements:
\n
\n
30% of those who read A but not B,
\n
40% of those who read B but not A,
\n
50% of those who read both A and B.
\n
\n
To find the total percentage of the population that looks into advertisements, we add up the contributions from each of these groups. We calculate each group's contribution by multiplying the size of the group by the percentage of that group that looks at advertisements :
\n
= $${{30} \\over {100}} \\times 17$$ (from A only) + $${{40} \\over {100}} \\times 12 $$(from B only) + $${{50} \\over {100}} \\times 8 $$(from both A and B)
\n
= 5.1 (from A only) + 4.8 (from B only) + 4 (from both A and B)
\n
Adding these up, we get
\n
= 13.9
\n
This means that 13.9% of the total population looks into the advertisements.
\n
So, the correct answer is :
\n
Option D : 13.9.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7297,
"subject": "General Science",
"question": "Let A, B and C be sets such that $$\\phi $$ $$ \\ne $$ A $$ \\cap $$ B $$ \\subseteq $$ C. Then which of the following statements is not true ?",
"options": [
{
"text": "If (A – B) $$ \\subseteq $$ C, then A $$ \\subseteq $$ C"
},
{
"text": "B $$ \\cap $$ C $$ \\ne $$ $$\\phi $$"
},
{
"text": "(C $$ \\cup $$ A) $$ \\cap $$ (C $$ \\cup $$ B) = C"
},
{
"text": "If (A – C) $$ \\subseteq $$ B, then A $$ \\subseteq $$ B "
}
],
"answer": "If (A – C) $$ \\subseteq $$ B, then A $$ \\subseteq $$ B ",
"solution": "**Answer:** If (A – C) $$ \\subseteq $$ B, then A $$ \\subseteq $$ B \n\nAccording to the question, we have the following Venn diagram.\n
Here, $A \\cap B \\subseteq C$ and $A \\cap B \\neq \\phi$\n
\n Now, from the Venn diagram, it is clear that $B \\cap C \\neq \\phi$, is true\n
Also, $(C \\cup A) \\cap(C \\cup B)=C \\cup(A \\cap B)=C$ is true.\n
If $(A-B) \\subseteq C$, for this statement the Venn diagram is\n
\n From the Venn diagram, it is clear that if $A-B \\subseteq C$, then $A \\subseteq C$. \n
Now, if $(A-C) \\subseteq B$, for this statement the Venn diagram.\n
\n From the Venn diagram, it is clear that \n
$A \\cap B \\neq \\phi, A \\cap B \\subseteq C$ and $A-C=\\phi \\subseteq B$ but $A \\subseteq B$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7298,
"subject": "General Science",
"question": "If A = {x $$ \\in $$ R : |x| < 2} and B = {x $$ \\in $$ R : |x – 2| $$ \\ge $$ 3};\nthen :",
"options": [
{
"text": "A – B = [–1, 2)"
},
{
"text": "A $$ \\cup $$ B = R – (2, 5)"
},
{
"text": "A $$ \\cap $$ B = (–2, –1)"
},
{
"text": "B – A = R – (–2, 5)"
}
],
"answer": "B – A = R – (–2, 5)",
"solution": "**Answer:** B – A = R – (–2, 5)\n\nA : x $$ \\in $$ (–2, 2); \n
B : x $$ \\in $$ (–$$\\infty $$, –1] $$ \\cup $$ [5, $$\\infty $$)\n
$$ \\Rightarrow $$ B – A = R – (–2, 5)\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7299,
"subject": "General Science",
"question": "A survey shows that 63% of the people in a city read newspaper A whereas 76% read\nnewspaper B. If x% of the people read both the newspapers, then a possible value of x can be:",
"options": [
{
"text": "37"
},
{
"text": "65"
},
{
"text": "29"
},
{
"text": "55"
}
],
"answer": "55",
"solution": "**Answer:** 55\n\n\n
A $$ \\cup $$ B = 63 - x + x + 76 - x = 139 - x\n
As 139 - x $$ \\le $$ 100\n
$$ \\Rightarrow $$ x $$ \\ge $$ 39\n
From venn diagram, you can see x should be less than 76 and 63 otherwise only A newspaper or only B newspaper reader will be negative number.\n
Intersection of x $$ \\le $$ 63 and x $$ \\le $$76 is = x $$ \\le $$ 63.\n
$$ \\therefore $$ 39 $$ \\le $$ x $$ \\le $$ 63\n
From options possible value of x = 55.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7300,
"subject": "General Science",
"question": "A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be :",
"options": [
{
"text": "63"
},
{
"text": "36"
},
{
"text": "54"
},
{
"text": "38"
}
],
"answer": "36",
"solution": "**Answer:** 36\n\nC $$ \\to $$ person like coffee\n
73 – x $$ \\ge $$ 0 $$ \\Rightarrow $$ x $$ \\le $$ 73\n
65 – x $$ \\ge $$ 0 $$ \\Rightarrow $$ x $$ \\le $$ 65\n
$$ \\therefore $$ 38 $$ \\le $$ x $$ \\le $$ 65",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7301,
"subject": "General Science",
"question": "Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If K% of them are suffering from both ailments, then K can not belong to the set :",
"options": [
{
"text": "{80, 83, 86, 89}"
},
{
"text": "{84, 86, 88, 90}"
},
{
"text": "{79, 81, 83, 85}"
},
{
"text": "{84, 87, 90, 93}"
}
],
"answer": "{79, 81, 83, 85}",
"solution": "**Answer:** {79, 81, 83, 85}\n\n
This solution begins by applying the principle of inclusion and exclusion, which in the context of this problem, is represented by the formula :
\n
n(A ∪ B) ≥ n(A) + n(B) - n(A ∩ B)
\n
Here, n(A ∪ B) represents the total number of patients in the hospital, which is 100%. n(A) represents the proportion of patients with a heart ailment (89%), and n(B) represents the proportion of patients with a lung infection (98%).
\n
By rearranging this formula, the solution establishes an inequality for n(A ∩ B), the proportion of patients suffering from both ailments :
\n
100% ≥ 89% + 98% - n(A ∩ B)
\n
Therefore,
\n
n(A ∩ B) ≥ 87%
\n
Next, the solution notes that n(A ∩ B) cannot be greater than the smaller of n(A) and n(B), since it cannot be larger than the smallest group. Thus, we have another inequality :
\n
n(A ∩ B) ≤ 89%
\n
Combining these two inequalities gives :
\n
87% ≤ n(A ∩ B) ≤ 89%
\n
Hence, the proportion of patients suffering from both ailments must be a value between 87% and 89% inclusive. So, the set of values {79,81,83,85} which are all less than 87% are values that n(A ∩ B) cannot belong to. Therefore, option C is the correct answer.
An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?
We are given the number of medals for events A, B, and C which are 48, 25, and 18 respectively. We are also given that the total number of unique medal recipients across all events is 60 and that 5 people received a medal in all three events.
\n
\n
Using the Principle of Inclusion and Exclusion (PIE), we know that the total number of unique medal recipients can be calculated by adding the number of medal recipients in each event, subtracting the number of people who received a medal in any two events (to correct for double counting), and then adding back the number of people who received a medal in all three events (since we subtracted these people too much).
However, we want to find the total number of people who received a medal in any two events (which is represented by |A ∩ B| + |B ∩ C| + |C ∩ A| in the equation).
\n \n
\n
To find this, we rearrange the PIE formula to solve for |A ∩ B| + |B ∩ C| + |C ∩ A| :
Substituting the given values, we find that the total number of people who received a medal in any two events is 48 + 25 + 18 + 5 - 60 = 36.
\n
\n
However, this includes people who received a medal in all three events, and we want to find the number of people who received a medal in exactly two events. Therefore, we need to subtract the people who received a medal in all three events from our calculated value.
\n
Since each person who received a medal in all three events is counted three times in |A ∩ B| + |B ∩ C| + |C ∩ A| (once for each pair of events), we subtract three times the number of people who received a medal in all three events from our calculated value:
\n
Number of people who received a medal in exactly two events = |A ∩ B| + |B ∩ C| + |C ∩ A| - 3 $$ \\times $$ |A ∩ B ∩ C|
\n
Substituting the values we know, we find that the number of people who received a medal in exactly two events is 36 - 3 $$ \\times $$ 5 = 21.
\n
\n\n
Therefore, 21 people received a medal in exactly two of the three events.
In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let $$m$$ and $$n$$ respectively be the least and the most number of students who studied all the three subjects. Then $$\\mathrm{m}+\\mathrm{n}$$ is equal to ___________.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7305,
"subject": "General Science",
"question": "In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average marks of the girls?",
"options": [
{
"text": "73"
},
{
"text": "65"
},
{
"text": "68"
},
{
"text": "74"
}
],
"answer": "65",
"solution": "**Answer:** 65\n\nGiven that, total students = 100\n
and number of boys = 70\n
$$\\therefore$$ No. of girls = 100 - 70 = 30\n
Average of 100 students = 72 \n
$$\\therefore$$ Total marks of 100 students = 100 $$ \\times $$ 72 = 7200 \n
Average of 70 boys = 75 \n
$$\\therefore$$ Total marks of 70 boys = 70 $$ \\times $$ 75 = 5250\n
$$\\therefore$$ Total marks of 30 girls = 7200 $$ - $$ 5250 = 1950\n
$$\\therefore$$ Average marks of 30 girls = $${{1950} \\over {30}}$$ = 65 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7306,
"subject": "General Science",
"question": "The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of\nthe set is increased by 2, then the median of the new set :",
"options": [
{
"text": "is increased by 2"
},
{
"text": "is decreased by 2"
},
{
"text": "is two times the original median"
},
{
"text": "remains the same as that of the original set "
}
],
"answer": "remains the same as that of the original set ",
"solution": "**Answer:** remains the same as that of the original set \n\nHere total no of observation is 9 which is a odd number. As we know for odd number 9 the median will be the 5th term.\n
Now question says, you increase largest 4 number by 2 which does not affect the 5th term so the new median will be the same.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7307,
"subject": "General Science",
"question": "If in a frequency distribution, the mean and median are 21 and 22 respectively, then\nits mode is approximately :",
"options": [
{
"text": "20.5"
},
{
"text": "22.0"
},
{
"text": "24.0"
},
{
"text": "25.5"
}
],
"answer": "24.0",
"solution": "**Answer:** 24.0\n\nGiven that, \n
= 24 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7308,
"subject": "General Science",
"question": "The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys\nand girls combined is 50. The percentage of boys in the class is",
"options": [
{
"text": "80"
},
{
"text": "60"
},
{
"text": "40"
},
{
"text": "20"
}
],
"answer": "80",
"solution": "**Answer:** 80\n\nLet x and y are number of boys and girls in a class respectively. \n
$$\\therefore$$ 52x + 42y = 50 (x + y)\n
$$ \\Rightarrow$$ 52x + 42y = 50x + 50y\n
$$ \\Rightarrow$$ 2x = 8y\n
$$ \\Rightarrow$$ x = 4y\n
$$\\therefore$$ Total no. of students = x + y = 4y + y = 5y\n
$$\\therefore$$ Percentage of boys = $${{4y} \\over {5y}} \\times 100\\,\\% $$ = 80%",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7309,
"subject": "General Science",
"question": "The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted\nand three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is :",
"options": [
{
"text": "15.8"
},
{
"text": "14.0"
},
{
"text": "16.8"
},
{
"text": "16.0"
}
],
"answer": "14.0",
"solution": "**Answer:** 14.0\n\nInitially we have $$16$$ observations and among them one is $$16.$$ \n
So, we have $$15$$ unknowns. Let those are $${a_1},a{}_2,{a_3}.....{a_{15}}$$ \n
$$\\therefore\\,\\,\\,$$ Mean of $$16$$ datal set \n
$$=14$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7310,
"subject": "General Science",
"question": "The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is : ",
"options": [
{
"text": "25"
},
{
"text": "30"
},
{
"text": "35"
},
{
"text": "40"
}
],
"answer": "35",
"solution": "**Answer:** 35\n\nMean $$\\left( {\\overline x } \\right)$$ = $${{{x_1} + {x_2}..... + {x_n}} \\over n}$$ = $${{\\sum x } \\over n}$$\n
Here, Mean = 40 of 25 teachers\n
$$\\therefore$$ 40 = $${{\\sum x } \\over {25}}$$\n
After retireing of a 60 year old teacher, total age of 24 teachers, \n
x1 + x2 + . . . . . .x24 = 1000 $$-$$ 60 = 940\n
Now a new teacher of age A year is appointed. \n
$$\\therefore$$ Now total age of this 25 teachers \n
x1 + x2 + x3 + . . . . . + x25 = 940 + A\n
$$\\therefore$$ Mean age = $${{940 + A} \\over {25}}$$\n
According to question,\n
$${{940 + A} \\over {25}}$$ = 39\n
$$ \\Rightarrow $$ A = 35",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7311,
"subject": "General Science",
"question": "The mean of set of 30 observations is 75. If each observation is multiplied by a non-zero number $$\\lambda $$ and then each of them is decreased by 25, their mean remains the same. Then $$\\lambda $$ is equal to :",
"options": [
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${2 \\over 3}$$"
},
{
"text": "$${4 \\over 3}$$"
},
{
"text": "$${10 \\over 3}$$"
}
],
"answer": "$${4 \\over 3}$$",
"solution": "**Answer:** $${4 \\over 3}$$\n\nAs mean is a linear operation, so if each observation is multiplied by $$\\lambda $$ and decreased by 25 then the mean becomes 75$$\\lambda $$$$-$$25.\n
According to the question,\n
75$$\\lambda $$ $$-$$ 25 = 75 $$ \\Rightarrow $$ $$\\lambda $$ = $${4 \\over 3}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7312,
"subject": "General Science",
"question": "The mean and the median of the following ten\nnumbers in increasing order 10, 22, 26, 29, 34, x,\n42, 67, 70, y are 42 and 35 respectively, then $${y \\over x}$$ is equal to",
"options": [
{
"text": "$${7 \\over 2}$$"
},
{
"text": "$${8 \\over 3}$$"
},
{
"text": "$${9 \\over 4}$$"
},
{
"text": "$${7 \\over 3}$$"
}
],
"answer": "$${7 \\over 3}$$",
"solution": "**Answer:** $${7 \\over 3}$$\n\nGiven ten numbers are 10, 22, 26, 29, 34, x,\n42, 67, 70, y.\n
$$ \\therefore $$ $${y \\over x} = {{84} \\over {36}}$$ = $${7 \\over 3}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7313,
"subject": "General Science",
"question": "If for some x $$ \\in $$ R, the frequency distribution of the marks obtained by 20 students in a test is :
\n\n
\n
\n
Marks
\n
2
\n
3
\n
5
\n
7
\n
\n
\n
Frequency
\n
(x + 1)2
\n
2x - 5
\n
x2 - 3x
\n
x
\n
\n
\n then the mean of the marks is",
"options": [
{
"text": "3.0"
},
{
"text": "2.8"
},
{
"text": "2.5"
},
{
"text": "3.2"
}
],
"answer": "2.8",
"solution": "**Answer:** 2.8\n\nNumber of students
\nAverage marks = $${{32 + 3 + 21} \\over {20}} = {{56} \\over {20}} = 2.8$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 7314,
"subject": "General Science",
"question": "Consider the data on x taking the values 0, 2, 4,\n8,....., 2n with frequencies nC0\n,\nnC1\n,\nnC2\n,....,\nnCn\nrespectively. If the mean of this data is $${{728} \\over {{2^n}}}$$, then n is equal to _________ .\n",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nMean = $${{\\sum {{x_1}.{f_1}} } \\over {\\sum {{f_1}} }}$$\n
$$ \\Rightarrow $$ n = 6",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7315,
"subject": "General Science",
"question": "The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If the mean age of the teachers in this school now is 39 years, then the age (in years) of the newly appointed teacher is _________.",
"options": [],
"answer": "35",
"solution": "**Answer:** 35\n\nMean $$\\left( {\\overline x } \\right)$$ = $${{{x_1} + {x_2}..... + {x_n}} \\over n}$$ = $${{\\sum x } \\over n}$$\n
Here, Mean = 40 of 25 teachers\n
$$\\therefore$$ 40 = $${{\\sum x } \\over {25}}$$\n
After retireing of a 60 year old teacher, total age of 24 teachers, \n
x1 + x2 + . . . . . .x24 = 1000 $$-$$ 60 = 940\n
Now a new teacher of age A year is appointed. \n
$$\\therefore$$ Now total age of this 25 teachers \n
x1 + x2 + x3 + . . . . . + x25 = 940 + A\n
$$\\therefore$$ Mean age = $${{940 + A} \\over {25}}$$\n
According to question,\n
$${{940 + A} \\over {25}}$$ = 39\n
$$ \\Rightarrow $$ A = 35",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7316,
"subject": "General Science",
"question": "Consider the following frequency distribution :
\n
\n\n
\n
Class :
\n
0-6
\n
6-12
\n
12-18
\n
18-24
\n
24-30
\n
\n\n\n
\n
Frequency :
\n
$$a $$
\n
$$b$$
\n
12
\n
9
\n
5
\n
\n\n
If mean = $${{309} \\over {22}}$$ and median = 14, then the value (a $$-$$ b)2 is equal to _____________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n
$$\\therefore$$ $${(a - b)^2} = {(8 - 10)^2}$ = 4",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 7317,
"subject": "General Science",
"question": "Consider the following frequency distribution :
\n
\n\n
\n
Class :
\n
10-20
\n
20-30
\n
30-40
\n
40-50
\n
50-60
\n
\n\n\n
\n
Frequency :
\n
$$\\alpha $$
\n
110
\n
54
\n
30
\n
$$\\beta $$
\n
\n\n
If the sum of all frequencies is 584 and median is 45, then | $$\\alpha$$ $$-$$ $$\\beta$$ | is equal to _______________.",
"options": [],
"answer": "164",
"solution": "**Answer:** 164\n\n$$\\because$$ Sum of frequencies = 584
Let $$x_{1}, x_{2}, x_{3}, \\ldots, x_{20}$$ be in geometric progression with $$x_{1}=3$$ and the common ratio $$\\frac{1}{2}$$. A new data is constructed replacing each $$x_{i}$$ by $$\\left(x_{i}-i\\right)^{2}$$. If $$\\bar{x}$$ is the mean of new data, then the greatest integer less than or equal to $$\\bar{x}$$ is ____________.
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 7321,
"subject": "General Science",
"question": "In an experiment with 15 observations on $$x$$, then following results were available:\n $$\\sum {{x^2}} = 2830$$, $$\\sum x = 170$$\n One observation that was 20 was found to be wrong and was replaced by the correct value\n30. Then the corrected variance is :",
"options": [
{
"text": "188.66"
},
{
"text": "177.33"
},
{
"text": "8.33"
},
{
"text": "78.00"
}
],
"answer": "78.00",
"solution": "**Answer:** 78.00\n\nGiven that, \n
N = 15, $$\\sum {x{}^2} = 2830,\\,\\sum x = 170$$ \n
$$ = 78$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7322,
"subject": "General Science",
"question": "Consider the following statements:\n (a) Mode can be computed from histogram\n (b) Median is not independent of change of scale\n (c) Variance is independent of change of origin and scale.\n Which of these is/are correct?",
"options": [
{
"text": "only (a)"
},
{
"text": "only (b)"
},
{
"text": "only (a) and (b)"
},
{
"text": "(a), (b) and (c) "
}
],
"answer": "only (a) and (b)",
"solution": "**Answer:** only (a) and (b)\n\n
The statements are analyzed as follows :
\n
(a) Mode can be computed from histogram : This is correct. The mode is the value that appears most frequently in a data set. A histogram provides a graphical representation of the frequency of each data value. The data value corresponding to the highest bar in a histogram is the mode.
\n
(b) Median is not independent of change of scale : This is correct. The median is the middle value in a data set when the values are arranged in ascending or descending order. When you change the scale (e.g., by multiplying all data points by a constant), the median changes as well.
\n
(c) Variance is independent of change of origin and scale : This is not correct. Variance, which measures the dispersion of a set of data points, is affected by changes in scale (it is not independent of scale). If you multiply all the values in a dataset by a constant, the variance gets multiplied by the square of that constant. However, it is true that variance is independent of the change of origin (it remains the same if you add or subtract a constant from all data points in the set).
\n
Therefore, the correct answer is Option C : only (a) and (b) are correct.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7323,
"subject": "General Science",
"question": "In a series of 2n observations, half of them equal $$a$$ and remaining half equal $$–a$$. If the\nstandard deviation of the observations is 2, then $$|a|$$ equals",
"options": [
{
"text": "2"
},
{
"text": "$$\\sqrt 2 $$"
},
{
"text": "$${1 \\over n}$$"
},
{
"text": "$${{\\sqrt 2 } \\over n}$$"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nMean $$\\left( A \\right) = {{a - a} \\over {2n}} = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\left| a \\right| = 2$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7324,
"subject": "General Science",
"question": "Let x1, x2,...........,xn be n observations such that\n
$$\\sum {x_i^2} = 400$$ and $$\\sum {{x_i}} = 80$$. Then a\npossible value of n among the following is",
"options": [
{
"text": "18"
},
{
"text": "15"
},
{
"text": "12"
},
{
"text": "9"
}
],
"answer": "18",
"solution": "**Answer:** 18\n\nAs we know,\n
$$\\therefore\\,\\,\\,$$ Possible value of n according to the option is = 18 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7325,
"subject": "General Science",
"question": "Suppose a population A has 100 observations 101, 102,........, 200, and another\npopulation B has 100 observations 151, 152,......., 250. If VA and VB represent the\nvariances of the two populations, respectively, then $${{{V_A}} \\over {{V_B}}}$$ is",
"options": [
{
"text": "1"
},
{
"text": "$${9 \\over 4}$$"
},
{
"text": "$${4 \\over 9}$$"
},
{
"text": "$${2 \\over 3}$$"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nSeries A = 101, 102 ............ 200\n
Series B = 151, 152 ............ 250\n
Here series B can be obtained if we change the origin of A by 50 units. \n
And we know the variance does not change by changing the origin. \n
So, $$\\,\\,\\,\\,$$ $${V_A} = {V_B}$$ \n
$$ \\Rightarrow \\,\\,\\,\\,\\,{{{V_A}} \\over {{V_B}}} = 1$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7326,
"subject": "General Science",
"question": "The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following\ngives possible values of a and b?",
"options": [
{
"text": "a = 0, b = 7"
},
{
"text": "a = 5, b = 2"
},
{
"text": "a = 1, b = 6"
},
{
"text": "a = 3, b = 4"
}
],
"answer": "a = 3, b = 4",
"solution": "**Answer:** a = 3, b = 4\n\nGiven that, \n
then b = 4, 3. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7327,
"subject": "General Science",
"question": "Statement - 1 : The variance of first n even natural numbers is $${{{n^2} - 1} \\over 4}$$\n
Statement - 2 : The sum of first n natural numbers is $${{n\\left( {n + 1} \\right)} \\over 2}$$ and the sum of squares of first n natural numbers is $${{n\\left( {n + 1} \\right)\\left( {2n + 1} \\right)} \\over 6}$$",
"options": [
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1"
},
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1"
},
{
"text": "Statement-1 is true, Statement-2 is false"
},
{
"text": "Statement-1 is false, Statement-2 is true"
}
],
"answer": "Statement-1 is false, Statement-2 is true",
"solution": "**Answer:** Statement-1 is false, Statement-2 is true\n\nLet first n even natural numbers = 2,4, 6, 8 ...... 2n\n
$$\\therefore$$ Sum of those num = 2 + 4 + 6 + ..... 2n\n
= 2 (1 + 2 + ..... n)\n
= $$2.{{n\\left( {n + 1} \\right)} \\over 2}$$ \n
= n (n + 1)\n
$$\\therefore\\,\\,\\,$$ Mean $$\\left( {\\overline x } \\right) = {{n\\left( {n + 1} \\right)} \\over n}$$ $$ = n + 1$$ \n
Statement 2 is true as those are standard formula. \n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7328,
"subject": "General Science",
"question": "If the mean deviation of number 1, 1 + d, 1 + 2d,........, 1 + 100d from their mean is 255, then the d is\nequal to",
"options": [
{
"text": "20.0"
},
{
"text": "10.1"
},
{
"text": "20.2"
},
{
"text": "10.0"
}
],
"answer": "10.1",
"solution": "**Answer:** 10.1\n\nMean $$\\,\\,\\,\\,\\left( {\\overline x } \\right) = {{Sum\\,\\,of\\,\\,numbers} \\over n}$$ \n
$$ = 10.1$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7329,
"subject": "General Science",
"question": "For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding\nmeans are given to be 2 and 4, respectively. The variance of the combined data set is",
"options": [
{
"text": "$${5 \\over 2}$$"
},
{
"text": "$${11 \\over 2}$$"
},
{
"text": "6"
},
{
"text": "$${13 \\over 2}$$"
}
],
"answer": "$${11 \\over 2}$$",
"solution": "**Answer:** $${11 \\over 2}$$\n\nGiven that,\n
$${\\sigma _1}^2 = 4$$ \n
and $${\\sigma _2}^2 = 5$$\n
And also given, \n
$$\\overline x = 2\\,\\,$$ and $$\\overline y = 4\\,\\,$$\n
So, $$\\,\\,\\,$$ $${{\\sum {{x_i}} } \\over 5} = 2$$ \n
$$ \\Rightarrow \\left| a \\right| = {{2500} \\over {25 \\times 25}} = 4$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7331,
"subject": "General Science",
"question": "Let x1, x2,........., xn be n observations, and let $$\\overline x $$ be their arithematic mean and $${\\sigma ^2}$$ be their variance.\n
Statement 1 : Variance of 2x1, 2x2,......., 2xn is 4$${\\sigma ^2}$$.\n Statement 2 : : Arithmetic mean of 2x1, 2x2,......, 2xn is 4$$\\overline x $$.",
"options": [
{
"text": "Statement 1 is false, statement 2 is true"
},
{
"text": "Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1"
},
{
"text": "Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1"
},
{
"text": "Statement 1 is true, statement 2 is false"
}
],
"answer": "Statement 1 is true, statement 2 is false",
"solution": "**Answer:** Statement 1 is true, statement 2 is false\n\nGiven that, \n
for $${x_1},{x_2},....{x_n},$$ $$A.M = \\overline x $$ \n
So, statement $${\\rm I}$$ is correct. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7332,
"subject": "General Science",
"question": "All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10\nto each of the students. Which of the following statistical measures will not change even after the grace\nmarks were given? ",
"options": [
{
"text": "median"
},
{
"text": "mode"
},
{
"text": "variance"
},
{
"text": "mean "
}
],
"answer": "variance",
"solution": "**Answer:** variance\n\nAs we know variance does not change with the change of origin. So, here even after adding grace marks $$10$$, the variance will be same. \n
$$ \\Rightarrow 3{a^2} - 32a + 84 = 0$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7335,
"subject": "General Science",
"question": "The mean of 5 observations is 5 and their variance is 124. If three of the observations\nare 1, 2 and 6 ; then the mean deviation from the mean of the data is : ",
"options": [
{
"text": "2.4"
},
{
"text": "2.8"
},
{
"text": "2.5"
},
{
"text": "2.6"
}
],
"answer": "2.8",
"solution": "**Answer:** 2.8\n\nLet 5 observations are x1, x2, x3, x4, x5\n
= 2.8",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7336,
"subject": "General Science",
"question": "If the mean deviation of the numbers 1, 1 + d, ..., 1 +100d from their mean is 255, then a value of d is :",
"options": [
{
"text": "10.1"
},
{
"text": "20.2"
},
{
"text": "10"
},
{
"text": "5.05"
}
],
"answer": "10.1",
"solution": "**Answer:** 10.1\n\nGiven numbers are, \n
$${{2550} \\over {101}}\\left| d \\right|$$ = 255\n
$$ \\Rightarrow $$ $$\\left| d \\right|$$ = 10.1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7337,
"subject": "General Science",
"question": "The sum of 100 observations and the sum of their squares are 400 and 2475,\nrespectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If\nthe incorrect observations are omitted, then the variance of the remaining observations\nis :",
"options": [
{
"text": "8.25 "
},
{
"text": "8.50"
},
{
"text": "8.00"
},
{
"text": "9.00"
}
],
"answer": "9.00",
"solution": "**Answer:** 9.00\n\n
Hence, the variance is 1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7340,
"subject": "General Science",
"question": "The mean and the standard deviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is : ",
"options": [
{
"text": "0"
},
{
"text": "1"
},
{
"text": "2"
},
{
"text": "4"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nHere mean = $$\\overline x $$ = 9\n
= $$\\sqrt {{{4{{\\left( {9 - 10} \\right)}^2} + {{\\left( {14 - 10} \\right)}^2}} \\over 5}} $$ = 2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7341,
"subject": "General Science",
"question": "A student scores the following marks in five tests\n: \n
45, 54, 41, 57, 43. \n
His score is not known for the\nsixth test. If the mean score is 48 in the six tests,\nthen the standard deviation of the marks in six tests\nis",
"options": [
{
"text": "$$100 \\over {\\sqrt 3}$$"
},
{
"text": "$$10 \\over {\\sqrt 3}$$"
},
{
"text": "$$10 \\over3$$"
},
{
"text": "$$100 \\over3$$"
}
],
"answer": "$$10 \\over {\\sqrt 3}$$",
"solution": "**Answer:** $$10 \\over {\\sqrt 3}$$\n\nLet the score in the sixth test = x\n
= $${{10} \\over {\\sqrt 3 }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7342,
"subject": "General Science",
"question": "If the data x1, x2,......., x10 is such that the mean of first four of these is 11, the mean of the remaining six is\n16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is : ",
"options": [
{
"text": "$$\\sqrt 2 $$"
},
{
"text": "2"
},
{
"text": "2$$\\sqrt 2 $$"
},
{
"text": "4"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n$${\\sigma ^2} = {{\\sum {x_i^2} } \\over {10}} - {\\left( {{{\\sum {{x_i}} } \\over {10}}} \\right)^2} \\to (i)$$
$$ \\Rightarrow $$ xy = 48\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7346,
"subject": "General Science",
"question": "The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are\n3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :",
"options": [
{
"text": "1"
},
{
"text": "7"
},
{
"text": "3"
},
{
"text": "5"
}
],
"answer": "7",
"solution": "**Answer:** 7\n\nmean $$\\overline x $$ = 4, $$\\sigma $$2 = 5.2, n = 5, . x1 = 3 x2 = 4 = x3\n
$$\\left| {{x_4} - {x_5}} \\right| = 7$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7347,
"subject": "General Science",
"question": "If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is : ",
"options": [
{
"text": "31"
},
{
"text": "50"
},
{
"text": "51"
},
{
"text": "30"
}
],
"answer": "31",
"solution": "**Answer:** 31\n\n$$\\sum\\limits_{i = 1}^{50} {\\left( {{x_i} - 30} \\right) = 50} $$\n
$$\\sum {{x_i}} = 50 \\times 30 = 50$$\n
$$\\sum {{x_i}} = 50 + 50 + 30$$\n
Mean $$ = \\overline x = {{\\sum {{x_i}} } \\over n} = {{50 \\times 30 + 50} \\over {50}}$$\n
$$ = 30 + 1 = 31$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7348,
"subject": "General Science",
"question": "The outcome of each of 30 items was observed; 10 items gave an outcome $${1 \\over 2}$$ – d each, 10 items gave outcome $${1 \\over 2}$$ each and the remaining 10 items gave outcome $${1 \\over 2}$$+ d each. If the variance of this outcome data is $${4 \\over 3}$$ then |d| equals :\n",
"options": [
{
"text": "$${2 \\over 3}$$"
},
{
"text": "$${{\\sqrt 5 } \\over 2}$$"
},
{
"text": "$${\\sqrt 2 }$$"
},
{
"text": "2"
}
],
"answer": "$${\\sqrt 2 }$$",
"solution": "**Answer:** $${\\sqrt 2 }$$\n\nVariance is independent of region. So we shift the given data by $${1 \\over 2}$$.\n
$$ = \\,\\,507.5$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7350,
"subject": "General Science",
"question": "The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -",
"options": [
{
"text": "6 : 7"
},
{
"text": "10 : 3"
},
{
"text": "4 : 9"
},
{
"text": "5 : 8"
}
],
"answer": "4 : 9",
"solution": "**Answer:** 4 : 9\n\nLet two observations are x1 & x2\n
$$\\sigma $$ $$ = \\sqrt 5 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7352,
"subject": "General Science",
"question": "5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :",
"options": [
{
"text": "16"
},
{
"text": "22"
},
{
"text": "20"
},
{
"text": "18"
}
],
"answer": "20",
"solution": "**Answer:** 20\n\nAverage height of 5 students, \n
$$ = 20$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7353,
"subject": "General Science",
"question": "Let X = {x\n$$ \\in $$ N : 1\n$$ \\le $$ x\n$$ \\le $$ 17} and\n Y = {ax + b: x\n$$ \\in $$ X and a, b $$ \\in $$ R, a > 0}. If mean\n and variance of elements of Y are 17 and 216\n respectively then a + b is equal to :",
"options": [
{
"text": "7"
},
{
"text": "9"
},
{
"text": "-7"
},
{
"text": "-27"
}
],
"answer": "-7",
"solution": "**Answer:** -7\n\nMean of X = $${{\\sum\\limits_{x = 1}^{17} x } \\over {17}}$$ = $${{17 \\times 18} \\over {17 \\times 2}}$$ = 9\n
Mean of Y = $${{\\sum\\limits_{x = 1}^{17} {\\left( {ax + b} \\right)} } \\over {17}}$$ = 17\n
$$ \\Rightarrow $$ x = 4",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7358,
"subject": "General Science",
"question": "The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations\nare 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :",
"options": [
{
"text": "5"
},
{
"text": "3"
},
{
"text": "7"
},
{
"text": "9"
}
],
"answer": "7",
"solution": "**Answer:** 7\n\nLet the two remaining observations be x and y.
$$ = {9 \\over {10}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7360,
"subject": "General Science",
"question": "If the variance of the terms in an increasing A.P., b1\n, b2\n, b3\n,....,b11 is 90, then the common\ndifference of this A.P. is_______.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nLet the common difference = d
\nand $${b_1} = a$$ \n$${b_2} = a + d$$ \n$${b_3} = a + 2d$$ \n... $${b_{11}} = a + 10d$$
= 12 - 9 = 3",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7362,
"subject": "General Science",
"question": "The mean and variance of 20 observations are\nfound to be 10 and 4, respectively. On\nrechecking, it was found that an observation 9\nwas incorrect and the correct observation was\n11. Then the correct variance is",
"options": [
{
"text": "3.98"
},
{
"text": "3.99"
},
{
"text": "4.01"
},
{
"text": "4.02"
}
],
"answer": "3.99",
"solution": "**Answer:** 3.99\n\nLet 20 observation be x1\n , x2\n ,....., x20\n
= 3.99\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7363,
"subject": "General Science",
"question": "The mean and the standard deviation (s.d.) of\n10 observations are 20 and 2 resepectively.\nEach of these 10 observations is multiplied by\np and then reduced by q, where p $$ \\ne $$ 0 and\nq $$ \\ne $$ 0. If the new mean and new s.d. become\nhalf of their original values, then q is equal to",
"options": [
{
"text": "10"
},
{
"text": "-20"
},
{
"text": "-10"
},
{
"text": "-5"
}
],
"answer": "-20",
"solution": "**Answer:** -20\n\nLet observations are\nx1, x2, ...., x10\n
Here mean = 20 and standard deviation(S.D) = 2\n
When each of these 10\nobservations is multiplied by p then new observations are\npx1, px2, ....., px10\n and new mean = 20p and new standard deviation(S.D) = 2|p|\n
Now when Reduced by q then new observations are\n px1 - q, px2 - q, ....., px10 - q\n
and new mean = 20p - q and new standard deviation(S.D) = 2|p|\n
Given 20p - q = $${{20} \\over 2}$$ = 10\n and 2|p| = $${2 \\over 2}$$ = 1\n
$$ \\Rightarrow $$ x.y = 54 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7365,
"subject": "General Science",
"question": "If the variance of the first n natural numbers is 10 and the variance of the first m even natural\nnumbers is 16, then m + n is equal to_____.\n",
"options": [],
"answer": "18",
"solution": "**Answer:** 18\n\nVariance $${\\sigma ^2} = {{\\sum {x_i^2} } \\over N} - {\\mu ^2}$$\n
$$ \\therefore $$ $$\\sigma $$ $$ \\ne $$ 6",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7367,
"subject": "General Science",
"question": "If the variance of 10 natural numbers 1, 1, 1, ....., 1, k is less than 10, then the maximum possible value of k is ________.",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\n$${\\sigma ^2} = {{\\sum {{x^2}} } \\over n} - {\\left( {{{\\sum x } \\over n}} \\right)^2}$$
Maximum integral value of k = 11.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7368,
"subject": "General Science",
"question": "Let X1, X2, ......., X18 be eighteen observations such that $$\\sum\\limits_{i = 1}^{18} {({X_i} - } \\alpha ) = 36$$ and $$\\sum\\limits_{i = 1}^{18} {({X_i} - } \\beta {)^2} = 90$$, where $$\\alpha$$ and $$\\beta$$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of | $$\\alpha$$ $$-$$ $$\\beta$$ | is ____________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nGiven, $$\\sum\\limits_{i = 1}^{18} {({x_1} - \\alpha ) = 36} $$
$${b^2} = 3{a^2} + 3{c^2} - 9{d^2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7370,
"subject": "General Science",
"question": "Consider the statistics of two sets of observations as follows :
\n\n
\n
\n
Size
\n
Mean
\n
Variance
\n
\n\n\n
\n
Observation I
\n
10
\n
2
\n
2
\n
\n
\n
Observation II
\n
n
\n
3
\n
1
\n
\n\n
If the variance of the combined set of these two observations is $${{17} \\over 9}$$, then the value of n is equal to ___________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nFor group - 1 : $${{\\sum {{x_i}} } \\over {10}} = 2 \\Rightarrow \\sum {{x_i}} = 20$$
$$ \\Rightarrow $$ 2n2 $$-$$ 5n $$-$$ 25 = 0 $$ \\Rightarrow $$ n = 5",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7371,
"subject": "General Science",
"question": "Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to __________.",
"options": [],
"answer": "68",
"solution": "**Answer:** 68\n\nLet first 2n observations are x1, x2 ...................., x2n
and last n observations are y1, y2 ....................., yn
Therefore, the correct answer is 68.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7372,
"subject": "General Science",
"question": "Let in a series of 2n observations, half of them are equal to a and remaining half are equal to $$-$$a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a2 + b2 is equal to :",
"options": [
{
"text": "425"
},
{
"text": "250"
},
{
"text": "925"
},
{
"text": "650"
}
],
"answer": "425",
"solution": "**Answer:** 425\n\nGiven series
(a, a, a, ........ n times), ($$-$$a, $$-$$a, $$-$$a, ...... n times)
Now $$\\overline x $$ = $${{\\sum {{x_i}} } \\over {2n}} = 0$$
as, xi $$ \\to $$ xi + b
then $$\\overline x $$ $$ \\to $$ $$\\overline x $$ + b
So, $$\\overline x $$ + b = 5 $$ \\Rightarrow $$ b = 5
No change in S.D. due to change in origin
Standard deviation ($$\\sigma$$) = $$\\sqrt {{{\\sum\\limits_{i = 1}^{2n} {{{({x_i} - \\overline x )}^2}} } \\over {2n}}} $$
$$ \\therefore $$ a2 + b2 = 425",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7373,
"subject": "General Science",
"question": "The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are :",
"options": [
{
"text": "10, 11"
},
{
"text": "3, 18"
},
{
"text": "8, 13"
},
{
"text": "1, 20"
}
],
"answer": "10, 11",
"solution": "**Answer:** 10, 11\n\nLet other two numbers be a, (21 $$-$$ a)
$$\\therefore$$ a = 10 and (21 $$-$$ a) = 21 $$-$$ 10 = 11
So, remaining two observations are 10, 11.
$$\\Rightarrow$$ Option (1) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7374,
"subject": "General Science",
"question": "If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and $${{20} \\over 3}$$, respectively, then the value of | a $$-$$ b | is equal to :",
"options": [
{
"text": "9"
},
{
"text": "11"
},
{
"text": "7"
},
{
"text": "1"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\n$$10 = {{7 + 10 + 11 + 15 + a + b} \\over 6}$$
| a $$-$$ b | = 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7375,
"subject": "General Science",
"question": "Let the mean and variance of the frequency distribution
be 6 and 6.8 respectively. If x3 is changed from 8 to 7, then the mean for the new data will be :",
"options": [
{
"text": "4"
},
{
"text": "5"
},
{
"text": "$${{17} \\over 3}$$"
},
{
"text": "$${{16} \\over 3}$$"
}
],
"answer": "$${{17} \\over 3}$$",
"solution": "**Answer:** $${{17} \\over 3}$$\n\nGiven 32 + 8$$\\alpha$$ + 9$$\\beta$$ = (8 + $$\\alpha$$ + $$\\beta$$) $$\\times$$ 6
$$\\Rightarrow$$ 14$$\\alpha$$ $$-$$ 11$$\\beta$$ = 48 ..... (ii)
from (i) & (ii)
$$\\alpha$$ = 5 & $$\\beta$$ = 2
So, new mean = $${{32 + 35 + 18} \\over {15}} = {{85} \\over {15}} = {{17} \\over 3}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7376,
"subject": "General Science",
"question": "The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation $$\\sqrt {13.44} $$, then the standard deviation of the second sample is :",
"options": [
{
"text": "8"
},
{
"text": "6"
},
{
"text": "4"
},
{
"text": "5"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\nn1 = 100
$$\\Rightarrow$$ $${\\sigma _2}^2$$ = 16 $$\\Rightarrow$$ $$\\sigma$$2 = 4",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 7377,
"subject": "General Science",
"question": "If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and $${{37} \\over 4}$$
respectively, then (a $$-$$ b)2 is equal to :",
"options": [
{
"text": "24"
},
{
"text": "12"
},
{
"text": "32"
},
{
"text": "16"
}
],
"answer": "16",
"solution": "**Answer:** 16\n\nMean = $${{6 + 10 + 7 + 13 + a + 12 + b + 12} \\over 8} = 9$$
$$ \\therefore $$ $${(a - b)^2} = {a^2} + {b^2} - 2ab = 80 - 64 = 16$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7378,
"subject": "General Science",
"question": "The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. if $$\\alpha$$ and $$\\sqrt \\beta $$ are the mean and standard deviation respectively for correct data, then ($$\\alpha$$, $$\\beta$$) is :",
"options": [
{
"text": "(11, 26)"
},
{
"text": "(10.5, 25)"
},
{
"text": "(11, 25)"
},
{
"text": "(10.5, 26)"
}
],
"answer": "(10.5, 26)",
"solution": "**Answer:** (10.5, 26)\n\nGiven :
Mean $$(\\overline x ) = {{\\sum {{x_i}} } \\over {20}} = 10$$
$$\\therefore$$ $$\\alpha $$ = 10.5, $$\\beta$$ = 26",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7379,
"subject": "General Science",
"question": "Let the mean and variance of four numbers 3, 7, x and y(x > y) be 5 and 10 respectively. Then the mean of four numbers 3 + 2x, 7 + 2y, x + y and x $$-$$ y is ______________.",
"options": [],
"answer": "12",
"solution": "**Answer:** 12\n\n$$5 = {{3 + 7 + x + y} \\over 4} \\Rightarrow x + y = 10$$
Mean = $${{48} \\over 4}$$ = 12",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7380,
"subject": "General Science",
"question": "Let n be an odd natural number such that the variance of 1, 2, 3, 4, ......, n is 14. Then n is equal to _____________.",
"options": [],
"answer": "13",
"solution": "**Answer:** 13\n\n$${{{n^2} - 1} \\over {12}} = 14 \\Rightarrow n = 13$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7381,
"subject": "General Science",
"question": "An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If $$\\mu$$ is the average marks of girls and $$\\sigma$$2 is the variance of marks of 50 candidates, then $$\\mu$$ + $$\\sigma$$2 is equal to ________________.",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\n$$\\sigma _b^2$$ = 2 (variance of boys)
Let the mean and the variance of 5 observations x1, x2, x3, x4, x5 be $${24 \\over 5}$$ and $${194 \\over 25}$$ respectively. If the mean and variance of the first 4 observation are $${7 \\over 2}$$ and a respectively, then (4a + x5) is equal to:
Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is _________.
The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to _____________.
The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to :
The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :
If the mean deviation about the mean of the numbers 1, 2, 3, .........., n, where n is odd, is $${{5(n + 1)} \\over n}$$, then n is equal to ______________.
The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $$\\sigma$$ is the standard deviation of the data after omitting the two wrong observations from the data, then $$38 \\sigma^{2}$$ is equal to ___________.
The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is _____________.
Let the mean and the variance of 20 observations $$x_{1}, x_{2}, \\ldots, x_{20}$$ be 15 and 9 , respectively. For $$\\alpha \\in \\mathbf{R}$$, if the mean of $$\\left(x_{1}+\\alpha\\right)^{2},\\left(x_{2}+\\alpha\\right)^{2}, \\ldots,\\left(x_{20}+\\alpha\\right)^{2}$$ is 178 , then the square of the maximum value of $$\\alpha$$ is equal to ________.
Let $$9=x_{1} < x_{2} < \\ldots < x_{7}$$ be in an A.P. with common difference d. If the standard deviation of $$x_{1}, x_{2}..., x_{7}$$ is 4 and the mean is $$\\bar{x}$$, then $$\\bar{x}+x_{6}$$ is equal to :
",
"options": [
{
"text": "$$2\\left(9+\\frac{8}{\\sqrt{7}}\\right)$$"
},
{
"text": "25"
},
{
"text": "$$18\\left(1+\\frac{1}{\\sqrt{3}}\\right)$$"
},
{
"text": "34"
}
],
"answer": "34",
"solution": "**Answer:** 34\n\n$\\begin{aligned} & \\text { Mean } \\Rightarrow \\bar{x}=\\frac{\\sum\\limits_{i=1}^7 x_i}{7}=\\frac{\\frac{7}{2}[2 a+6 d]}{7}=a+3 d=x_4 \\\\\\\\ & \\text { Variance }=\\frac{\\sum\\limits_{i=1}^7\\left(x_i-\\bar{x}\\right)^2}{7}=(4)^2 \\Rightarrow \\frac{\\sum\\limits_{i=1}^7\\left(x_i-x_4\\right)^2}{7}=16 \\\\\\\\ & \\Rightarrow \\frac{(3 d)^2+(2 d)^2+d^2+0+d^2+(2 d)^2+(3 d)^2}{7}=16 \\\\\\\\ & =4 d^2=16 \\Rightarrow d=2 \\\\\\\\ & \\Rightarrow \\bar{x}=9+3(2)=15 \\\\\\\\ & x_6=a+5 d=9+5(2)=19 \\Rightarrow \\bar{x}+x_6=34\\end{aligned}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7396,
"subject": "General Science",
"question": "Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and $\\alpha(>$\n\n0 ), and the mean and standard deviation of marks of class $B$ of $n$ students be respectively 55 and 30\n\n$-\\alpha$. If the mean and variance of the marks of the combined class of $100+\\mathrm{n}$ studants are\n\nrespectively 50 and 350 , then the sum of variances of classes $A$ and $B$ is :",
"options": [
{
"text": "450"
},
{
"text": "900"
},
{
"text": "650"
},
{
"text": "500"
}
],
"answer": "500",
"solution": "**Answer:** 500\n\n$$\n\\begin{array}{cll}\n\\quad \\mathbf{A} & \\mathbf{B} & \\mathbf{A}+\\mathbf{B} \\\\\n\\overline{\\mathrm{x}}_{1}=40 & \\overline{\\mathrm{x}}_{2}=55 & \\overline{\\mathrm{x}}=50 \\\\\n\\sigma_{1}=\\alpha & \\sigma_{2}=30-\\alpha & \\sigma^{2}=350 \\\\\n\\mathrm{n}_{1}=100 & \\mathrm{n}_{2}=\\mathrm{n} & 100+\\mathrm{n} \\\\\\\\\n\\overline{\\mathrm{x}}=\\frac{100 \\times 40+55 \\mathrm{n}}{100+\\mathrm{n}} &\n\\end{array}\n$$\n\n
The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :
$\\Rightarrow 3 \\alpha=15 \\Rightarrow \\alpha=5$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 7399,
"subject": "General Science",
"question": "Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of 100 consecutive positive integers $a_1, a_2, a_3, \\ldots ., a_{100}$ is 25 . Then $S$ is :",
"options": [
{
"text": "$\\{9\\}$"
},
{
"text": "$\\phi$"
},
{
"text": "$\\{99\\}$"
},
{
"text": "N"
}
],
"answer": "N",
"solution": "**Answer:** N\n\n
Let $${a_1} = a \\Rightarrow {a_2} = a + 1,\\,.....\\,{a_{100}} = a + 90$$
The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observation, then $$\\mathrm{a+3 b-5}$$ is equal to ___________.
Let $$X=\\{11,12,13,....,40,41\\}$$ and $$Y=\\{61,62,63,....,90,91\\}$$ be the two sets of observations. If $$\\overline x $$ and $$\\overline y $$ are their respective means and $$\\sigma^2$$ is the variance of all the observations in $$\\mathrm{X\\cup Y}$$, then $$\\left| {\\overline x + \\overline y - {\\sigma ^2}} \\right|$$ is equal to ____________.
Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If $$\\mu$$ and $$\\sigma^2$$ represent mean and variance of X, respectively, then $$10(\\mu^2+\\sigma^2)$$ is equal to :
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :
Let the six numbers $$\\mathrm{a_1,a_2,a_3,a_4,a_5,a_6}$$, be in A.P. and $$\\mathrm{a_1+a_3=10}$$. If the mean of these six numbers is $$\\frac{19}{2}$$ and their variance is $$\\sigma^2$$, then 8$$\\sigma^2$$ is equal to :
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7405,
"subject": "General Science",
"question": "The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is :",
"options": [
{
"text": "11"
},
{
"text": "12"
},
{
"text": "13"
},
{
"text": "14"
}
],
"answer": "13",
"solution": "**Answer:** 13\n\n1. Calculate the sum of the original observations:\n
$$\\frac{x_1+x_2+\\ldots+x_9+50}{10}=20$$\n
$$x_1+x_2+\\ldots+x_9=150$$\n\n
2. Calculate the sum of the squares of the original observations using the original variance:\n
The correct variance after correcting the error is 13 (Option C).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7406,
"subject": "General Science",
"question": "
The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _________
We later realize that two marks were incorrectly read as 45 and 50, when they should have been 20 and 25. Therefore, the corrected sum of the marks is:
The corrected sum of the squares of the marks is calculated by subtracting the squares of the incorrect marks and adding the squares of the correct marks:
be 5. If $$m$$ and $$\\sigma^{2}$$ are respectively the mean deviation about the mean and the variance of the data, then $$\\frac{3 \\alpha}{m+\\sigma^{2}}$$ is equal to __________
Let the positive numbers $$a_{1}, a_{2}, a_{3}, a_{4}$$ and $$a_{5}$$ be in a G.P. Let their mean and variance be $$\\frac{31}{10}$$ and $$\\frac{m}{n}$$ respectively, where $$m$$ and $$n$$ are co-prime. If the mean of their reciprocals is $$\\frac{31}{40}$$ and $$a_{3}+a_{4}+a_{5}=14$$, then $$m+n$$ is equal to ___________.
",
"options": [],
"answer": "211",
"solution": "**Answer:** 211\n\nSince $a_1, a_2, a_3, a_4, a_5$ are in geometric progression, we can write :\n\n
$a_2 = r a$, \n\n
$a_3 = r^2 a$,\n\n
$a_4 = r^3 a$,\n\n
$a_5 = r^4 a$.\n\n
where $r$ is the common ratio and $a_1$ = $$a$$ is the first term.\n\n
Given that the mean of the series is $\\frac{31}{10}$, we get\n\n
Let the mean of 6 observations $$1,2,4,5, \\mathrm{x}$$ and $$\\mathrm{y}$$ be 5 and their variance be 10 .\n\nThen their mean deviation about the mean is equal to :
By solving the system of equations $(1)$ and $(2)$, we get $x = 8$ and $y = 10$.
\n
Now, the mean deviation about the mean ($\\bar{x} = 5$) is calculated by taking the average of the absolute differences of each observation from the mean:
Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of $$\\mathrm{A}$$ and adding 2 to each element of $$\\mathrm{B}$$. Then the sum of the mean and variance of the elements of $$\\mathrm{C}$$ is ___________.
To solve this problem, let's break it down step by step :
\n
Step 1 : Determine the mean of set C
\n
The mean of set A = $5$\n
The mean of set B = $8$
\n
After subtracting 3 from each element in set A, the new mean becomes $5 - 3 = 2$.\nAfter adding 2 to each element in set B, the new mean becomes $8 + 2 = 10$.
\n
When we combine both modified sets to form set C, the mean of set C is the weighted average of the means of these modified sets:
\n
Mean of C = $\\frac{5 \\times 2 + 5 \\times 10}{10} = \\frac{10 + 50}{10} = 6$
\n
Step 2 : Determine the variance of set C
\n
Variance is defined as the expectation of the squared deviation of a random variable from its mean. One property of variance is that, if you add (or subtract) a constant from each data point in a set, the variance of the set does not change.
\n
Thus, the variance of the elements in set A remains $12$ even after subtracting 3 from each element, and the variance of the elements in set B remains $20$ even after adding 2 to each element.
\n
Now, when combining variances from two datasets into one :\n
Variance of C \n
= $\\frac{n_1 \\times \\text{Variance of A} + n_1 \\times (\\text{Mean of modified A} - \\text{Mean of C})^2 + n_2 \\times \\text{Variance of B} + n_2 \\times (\\text{Mean of modified B} - \\text{Mean of C})^2}{n_1 + n_2}$
\n
Given :\n
$n_1 = n_2 = 5$\n
Variance of A = $12$, Variance of B = $20$\n
Mean of modified A = $2$, Mean of modified B = $10$, Mean of C = $6$
Let the mean and variance of 12 observations be $$\\frac{9}{2}$$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $$\\frac{m}{n}$$, where $$\\mathrm{m}$$ and $$\\mathrm{n}$$ are coprime, then $$\\mathrm{m}+\\mathrm{n}$$ is equal to :
Let the mean and variance of 8 numbers $$x, y, 10,12,6,12,4,8$$ be $$9$$ and $$9.25$$ respectively. If $$x > y$$, then $$3 x-2 y$$ is equal to _____________.
The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $$\\sigma^{2}$$ respectively. If the variance of all the 30 numbers in the two sets is 13 , then $$\\sigma^{2}$$ is equal to :
",
"options": [
{
"text": "12"
},
{
"text": "11"
},
{
"text": "10"
},
{
"text": "9"
}
],
"answer": "10",
"solution": "**Answer:** 10\n\nWe know that if $n_1, n_2$ are the sizes, $\\bar{X}_1, \\bar{X}_2$ are the means and $\\sigma_1, \\sigma_2$ are the standard deviation of the series, then the combine variance of the series.\n
$$\n\\frac{\\beta}{\\alpha}=\\frac{2}{1}=2\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7418,
"subject": "General Science",
"question": "Let the median and the mean deviation about the median of 7 observation $170,125,230,190,210$, a, b be 170 and $\\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is :",
"options": [
{
"text": "31"
},
{
"text": "28"
},
{
"text": "30"
},
{
"text": "32"
}
],
"answer": "30",
"solution": "**Answer:** 30\n\n$$\n\\text { Median }=170 \\Rightarrow 125, \\mathrm{a}, \\mathrm{b}, 170,190,210,230\n$$\n\n
Mean deviation about\nMedian $=$\n
$$\n\\begin{aligned}\n& \\frac{0+45+60+20+40+170-a+170-b}{7}=\\frac{205}{7} \\\\\\\\\n& \\Rightarrow \\mathrm{a}+\\mathrm{b}=300 \\\\\\\\\n& \\text { Mean }=\\frac{170+125+230+190+210+a+b}{7}=175\n\\end{aligned}\n$$\n\n
Mean deviation\nAbout mean $=$\n
$$\n\\frac{50+175-a+175-b+5+15+35+55}{7}=30\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7419,
"subject": "General Science",
"question": "Let $\\mathrm{a}_1, \\mathrm{a}_2, \\ldots \\mathrm{a}_{10}$ be 10 observations such that $\\sum\\limits_{\\mathrm{k}=1}^{10} \\mathrm{a}_{\\mathrm{k}}=50$ and $\\sum\\limits_{\\forall \\mathrm{k} < \\mathrm{j}} \\mathrm{a}_{\\mathrm{k}} \\cdot \\mathrm{a}_{\\mathrm{j}}=1100$. Then the standard deviation of $\\mathrm{a}_1, \\mathrm{a}_2, \\ldots, \\mathrm{a}_{10}$ is equal to :",
"options": [
{
"text": "5"
},
{
"text": "$\\sqrt{115}$"
},
{
"text": "10"
},
{
"text": "$\\sqrt{5}$"
}
],
"answer": "$\\sqrt{5}$",
"solution": "**Answer:** $\\sqrt{5}$\n\n
The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If $$\\mu$$ and $$\\sigma^2$$ denote the mean and variance of the correct observations respectively, then $$15\\left(\\mu+\\mu^2+\\sigma^2\\right)$$ is equal to __________.
If the mean and variance of the data $$65,68,58,44,48,45,60, \\alpha, \\beta, 60$$ where $$\\alpha> \\beta$$, are 56 and 66.2 respectively, then $$\\alpha^2+\\beta^2$$ is equal to _________.
If the mean and variance of five observations are $$\\frac{24}{5}$$ and $$\\frac{194}{25}$$ respectively and the mean of the first four observations is $$\\frac{7}{2}$$, then the variance of the first four observations in equal to
Let $$\\alpha, \\beta \\in \\mathbf{R}$$. Let the mean and the variance of 6 observations $$-3,4,7,-6, \\alpha, \\beta$$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :
Let $$\\mathrm{a}, \\mathrm{b}, \\mathrm{c} \\in \\mathbf{N}$$ and $$\\mathrm{a}< \\mathrm{b}< \\mathrm{c}$$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $$9,25, a, b, c$$ be 18, 4 and $$\\frac{136}{5}$$, respectively. Then $$2 a+b-c$$ is equal to ________
$$\\begin{aligned}\n& a, b, c \\in N \\\\\n& a< b < c \\\\\n& \\text { Mean }=18 \\\\\n& \\frac{9+25+a+b+c}{5}=18 \\\\\n& 34+a+b+c=90 \\\\\n& a+b+c=56\n\\end{aligned}$$
The mean and standard deviation of 20 observations are found to be 10 and 2 , respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is
To find the correct standard deviation, we first need to adjust the mean and sum of squares of the observations based on the correction of the erroneous entry. The original erroneous observation was 8, and the correct observation is 12. The mean and standard deviation of the 20 observations before correction were 10 and 2, respectively.
\n\n
The sum of all 20 original observations ($S_{\\text{original}}$) can be calculated from the mean formula:\n\n
$ \\text{Mean} = \\frac{\\text{Sum of all observations}}{\\text{Number of observations}} $
\n\n
$ 10 = \\frac{S_{\\text{original}}}{20} $
\n\n
$ S_{\\text{original}} = 10 \\times 20 = 200 $
\n\n
The corrected sum of observations ($S_{\\text{correct}}$) will replace the incorrect observation (8) with the correct one (12):\n\n
To find the corrected standard deviation, we need the sum of squares of the deviations from the mean for both the original and corrected data. The original sum of squares ($SS_{\\text{original}}$) is calculated from the original standard deviation formula, where $\\sigma = 2$:\n\n
$ \\sigma^2 = \\frac{SS}{n} $
\n\n
$ 4 = \\frac{SS_{\\text{original}}}{20} $
\n\n
$ SS_{\\text{original}} = 4 \\times 20 = 80 $
\n\n
To calculate the corrected sum of squares ($SS_{\\text{correct}}$), we need to adjust $SS_{\\text{original}}$ by removing the square of the deviation of the incorrect observation and adding the square of the deviation of the correct observation:\n\n
The corrected standard deviation is closest to the value given in Option B, which is\n\n
$ \\sqrt{3.96} $
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7431,
"subject": "General Science",
"question": "The lines $$p\\left( {{p^2} + 1} \\right)x - y + q = 0$$ and $$\\left( {{p^2} + 1} \\right){}^2x + \\left( {{p^2} + 1} \\right)y + 2q$$ $$=0$$ are perpendicular to a common line for : ",
"options": [
{
"text": "exactly one values of $$p$$"
},
{
"text": "exactly two values of $$p$$ "
},
{
"text": "more than two values of $$p$$ "
},
{
"text": "no value of $$p$$ "
}
],
"answer": "exactly one values of $$p$$",
"solution": "**Answer:** exactly one values of $$p$$\n\nIf the lines $$p\\left( {{p^2} + 1} \\right)x - y + q = 0$$\n
$$y = {x \\over {\\sqrt 3 }} - 1$$ or $$y\\sqrt 3 = x - \\sqrt 3 $$
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7433,
"subject": "General Science",
"question": "A ray of light is incident along a line which meets another line, 7x − y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is :",
"options": [
{
"text": "41x − 38y + 38 = 0"
},
{
"text": "41x + 25y − 25 = 0"
},
{
"text": "41x + 38y − 38 = 0"
},
{
"text": "41x − 25y + 25 = 0"
}
],
"answer": "41x − 38y + 38 = 0",
"solution": "**Answer:** 41x − 38y + 38 = 0\n\nLet slope of incident ray be m.\n
$$ \\therefore $$ angle of incidence = angle of reflection \n
$$ \\Rightarrow 4\\sqrt 2 x + 5y - (15 + 4\\sqrt 2 ) = 0$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7438,
"subject": "General Science",
"question": "Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \\angle A=\\frac{2 \\pi}{3}, A B=A C$ and $B$ is on the positve $x$-axis. If $\\mathrm{BC}=4 \\sqrt{3}$ and the line $\\mathrm{BC}$ intersects the line $y=x+3$ at $(\\alpha, \\beta)$, then $\\frac{\\beta^4}{\\alpha^2}$ is __________.",
"options": [],
"answer": "36",
"solution": "**Answer:** 36\n\n\n
$\\frac{\\mathrm{c}}{\\sin 30^{\\circ}}=\\frac{4 \\sqrt{3}}{\\sin 120^{\\circ}}[$ By sine rule $]$\n
$\\frac{\\beta^4}{\\alpha^2}=36$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7439,
"subject": "General Science",
"question": "The portion of the line $4 x+5 y=20$ in the first quadrant is trisected by the lines $\\mathrm{L}_1$ and $\\mathrm{L}_2$ passing through the origin. The tangent of an angle between the lines $\\mathrm{L}_1$ and $\\mathrm{L}_2$ is :",
"options": [
{
"text": "$\\frac{30}{41}$"
},
{
"text": "$\\frac{8}{5}$"
},
{
"text": "$\\frac{2}{5}$"
},
{
"text": "$\\frac{25}{41}$"
}
],
"answer": "$\\frac{30}{41}$",
"solution": "**Answer:** $\\frac{30}{41}$\n\n
Co-ordinates of $$\\mathrm{A}=\\left(\\frac{5}{3}, \\frac{8}{3}\\right)$$
\n
Co-ordinates of $$\\mathrm{B}=\\left(\\frac{10}{3}, \\frac{4}{3}\\right)$$
\n
Slope of $$\\mathrm{OA}=\\mathrm{m}_1=\\frac{8}{5}$$
\n
Slope of $$\\mathrm{OB}=\\mathrm{m}_2=\\frac{2}{5}$$
If the line segment joining the points $$(5,2)$$ and $$(2, a)$$ subtends an angle $$\\frac{\\pi}{4}$$ at the origin, then the absolute value of the product of all possible values of $a$ is :
Let a ray of light passing through the point $$(3,10)$$ reflects on the line $$2 x+y=6$$ and the reflected ray passes through the point $$(7,2)$$. If the equation of the incident ray is $$a x+b y+1=0$$, then $$a^2+b^2+3 a b$$ is equal to _________.
Let two straight lines drawn from the origin $$\\mathrm{O}$$ intersect the line $$3 x+4 y=12$$ at the points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ such that $$\\triangle \\mathrm{OPQ}$$ is an isosceles triangle and $$\\angle \\mathrm{POQ}=90^{\\circ}$$. If $$l=\\mathrm{OP}^2+\\mathrm{PQ}^2+\\mathrm{QO}^2$$, then the greatest integer less than or equal to $$l$$ is :
$$\\therefore$$ Slope of the line $$QM=tan$$ $${{2\\pi } \\over 3} = - \\sqrt 3 $$\n
$$\\therefore$$ Equation of line $$QM$$ is $$\\left( {y - 0} \\right) = - \\sqrt 3 \\left( {x - 0} \\right)$$\n
$$ \\Rightarrow y = - \\sqrt 3 \\,x \\Rightarrow \\sqrt 3 x + y = 0$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7444,
"subject": "General Science",
"question": "The lines $${L_1}:y - x = 0$$ and $${L_2}:2x + y = 0$$ intersect the line $${L_3}:y + 2 = 0$$ at $$P$$ and $$Q$$ respectively. The bisector of the acute angle between $${L_1}$$ and $${L_2}$$ intersects $${L_3}$$ at $$R$$. \n
Statement-1: The ratio $$PR$$ : $$RQ$$ equals $$2\\sqrt 2 :\\sqrt 5 $$\n Statement-2: In any triangle, bisector of an angle divide the triangle into two similar triangles.
",
"options": [
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."
},
{
"text": "Statement-1 is true, Statement-2 is false."
},
{
"text": "Statement-1 is false, Statement-2 is true."
},
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."
}
],
"answer": "Statement-1 is true, Statement-2 is false.",
"solution": "**Answer:** Statement-1 is true, Statement-2 is false.\n\n\n
$${L_1}:y - x = 0$$\n
$${L_2}:2x + y = 0$$\n
$${L_3}:y + 2 = 0$$\n
On solving the equation of line $${L_1}$$ and $${L_2}$$ we get their point of \n
intersection $$(0, 0)$$ i.e., origin $$O.$$ \n
On solving the equation of line $${L_1}$$ and $${L_3},$$\n
we get $$P=(-2, -2).$$\n
Similarly, we get $$Q = \\left( { - 1, - 2} \\right)$$\n
We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [ Angle Bisector Theorem of a Triangle ]\n
If the line $$l_{1}: 3 y-2 x=3$$ is the angular bisector of the lines $$l_{2}: x-y+1=0$$ and $$l_{3}: \\alpha x+\\beta y+17=0$$, then $$\\alpha^{2}+\\beta^{2}-\\alpha-\\beta$$ is equal to _________.
In a $$\\triangle A B C$$, suppose $$y=x$$ is the equation of the bisector of the angle $$B$$ and the equation of the side $$A C$$ is $$2 x-y=2$$. If $$2 A B=B C$$ and the points $$A$$ and $$B$$ are respectively $$(4,6)$$ and $$(\\alpha, \\beta)$$, then $$\\alpha+2 \\beta$$ is equal to
$$\\begin{aligned}\n& A D: D C=1: 2 \\\\\n& \\frac{4-\\alpha}{6-\\alpha}=\\frac{10}{8} \\\\\n& \\alpha=\\beta \\\\\n& \\alpha=14 \\text { and } \\beta=14\n\\end{aligned}$$
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7449,
"subject": "General Science",
"question": "Let A $$\\left( {h,k} \\right)$$, B$$\\left( {1,1} \\right)$$ and C $$(2, 1)$$ be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is $$1$$ square unit, then the set of values which $$'k'$$ can take is given by :",
"options": [
{
"text": "$$\\left\\{ { - 1,3} \\right\\}$$ "
},
{
"text": "$$\\left\\{ { - 3, - 2} \\right\\}$$ "
},
{
"text": "$$\\left\\{ { 1,3} \\right\\}$$"
},
{
"text": "$$\\left\\{ {0,2} \\right\\}$$ "
}
],
"answer": "$$\\left\\{ { - 1,3} \\right\\}$$ ",
"solution": "**Answer:** $$\\left\\{ { - 1,3} \\right\\}$$ \n\nGiven : The vertices of a right angled triangle $$A\\left( {1,k} \\right),$$\n
$$B\\left( {1,1} \\right)$$ and $$C\\left( {2,1} \\right)$$ and area of $$\\Delta ABC = 1$$ square unit\n
$$ \\Rightarrow \\pm \\left( {k - 1} \\right) = 2 \\Rightarrow k = - 1,3$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7450,
"subject": "General Science",
"question": "The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $$(0, 0)$$ $$(0, 41)$$ and $$(41, 0)$$ is : ",
"options": [
{
"text": "820 "
},
{
"text": "780 "
},
{
"text": "901 "
},
{
"text": "861"
}
],
"answer": "780 ",
"solution": "**Answer:** 780 \n\n
\n
The number of integral points lie inside the triangle are
\n
1. If x = 1, then y may be 1, 2, 3, ....., 39
\n
2. If x = 2, then y may be 1, 2, 3, ....., 38
\n
3. If x = 3, then y may be 1, 2, 3, ....., 37
\n
$$ \\vdots $$
\n
39. If x = 39, then the value of y is 1.
\n
Hence, the number of interior points are
\n$$1 + 2 + 3 + .... + 39 = {{39 \\times 40} \\over 2} = 780$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7451,
"subject": "General Science",
"question": "In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x + y = 5 and x = 4. Then area of $$\\Delta $$ ABC (in sq. units) is :",
"options": [
{
"text": "12"
},
{
"text": "4"
},
{
"text": "5"
},
{
"text": "9"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\nMedian through C is x = 4\n
So the coordinate of C is 4. Let C = (4, y), then the midpoint of A(1, 2) and \n
C(4, y) is D which lies on the median through B.\n
The centroid of the triangle is the intersection of the mesians, Here the medians x = 4 and x + 4 and x + y = 5 intersect at G (4, 1). \n
The area of triangle $$\\Delta $$ABC = 3 $$ \\times $$ $$\\Delta $$AGC\n
= 3 $$ \\times $$ $${1 \\over 2}$$ [1(1 $$-$$ 3) + 4(3 $$-$$ 2) + 4(2 $$-$$ 1)] = 9.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7452,
"subject": "General Science",
"question": "If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to :",
"options": [
{
"text": "$${7 \\over 4}$$"
},
{
"text": "$${5 \\over 4}$$"
},
{
"text": "$${3 \\over 4}$$"
},
{
"text": "$${3 \\over 2}$$"
}
],
"answer": "$${7 \\over 4}$$",
"solution": "**Answer:** $${7 \\over 4}$$\n\n
For three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This is known as the triangle inequality theorem.
\n
Let's apply this to our lengths: $5$, $5r$, and $5r^2$. We have three inequalities to check:
\n\n
$5 + 5r > 5r^2$
\n
$5 + 5r^2 > 5r$
\n
$5r + 5r^2 > 5$
\n\n
Solving these inequalities will give us a range of values for $r$. Any value outside of this range will mean that $r$ cannot form a triangle with the given sides.
The roots of this quadratic equation are $\\frac{1-\\sqrt{5}}{2}$ and $\\frac{1+\\sqrt{5}}{2}$, so the inequality holds for $r$ in the interval $\\left(\\frac{1-\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}\\right)$.
\n\n
2. $1 + r^2 > r$
\n\n
We can rewrite this as:\n$r^2 - r + 1 > 0$
\n\n$$\n\\begin{array}{lc}\n\n\\Rightarrow r^2-2 \\cdot \\frac{1}{2} r+\\left(\\frac{1}{2}\\right)^2+1-\\left(\\frac{1}{2}\\right)^2>0 \\\\\\\\\n\\Rightarrow \\left(r-\\frac{1}{2}\\right)^2+\\frac{3}{4}>0 \\\\\\\\\n\\Rightarrow r \\in R \n\\end{array}\n$$\n\n
This inequality is true for all real numbers since the left-hand side is always positive, which means the inequality holds for all real $R$.
Taking the intersection of the intervals from the three inequalities, we get $\\left(\\frac{1-\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}\\right)$ as the valid range for $r$.
\n\n
Given the solution of the inequalities $r \\in \\left(\\frac{1-\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}\\right)$, we need to determine which of the options are outside this interval.
\n\n
Option A: $\\frac{7}{4}$
\n\n
The decimal representation of $\\frac{7}{4}$ is 1.75, which is outside the interval $\\left(\\frac{1-\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}\\right)$ or approximately (-0.618, 1.618). Hence, r cannot be equal to $\\frac{7}{4}$.
\n\n
Option B: $\\frac{5}{4}$
\n\n
The decimal representation of $\\frac{5}{4}$ is 1.25, which is inside the interval $\\left(\\frac{1-\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}\\right)$. Hence, r can be equal to $\\frac{5}{4}$.
\n\n
Option C: $\\frac{3}{4}$
\n\n
The decimal representation of $\\frac{3}{4}$ is 0.75, which is also inside the interval $\\left(\\frac{1-\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}\\right)$. Hence, r can be equal to $\\frac{3}{4}$.
\n\n
Option D: $\\frac{3}{2}$
\n\n
The decimal representation of $\\frac{3}{2}$ is 1.5, which is also inside the interval $\\left(\\frac{1-\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}\\right)$. Hence, r can be equal to $\\frac{3}{2}$.
\n
Therefore, the only value that r cannot be equal to, among the given options, is $\\frac{7}{4}$. So, the correct answer is:
\n
Option A: $\\frac{7}{4}$
\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7453,
"subject": "General Science",
"question": "Let two points be A(1, –1) and B(0, 2). If a point\nP(x', y') be such that the area of $$\\Delta $$PAB = 5 sq.\nunits and it lies on the line, 3x + y – 4$$\\lambda $$ = 0,\nthen a value of $$\\lambda $$ is :",
"options": [
{
"text": "4"
},
{
"text": "1"
},
{
"text": "-3"
},
{
"text": "3"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\nPoint P(x', y') lies on the line 3x + y – 4$$\\lambda $$ = 0\n
$$ \\Rightarrow $$ $$\\lambda $$ = -2 or $$\\lambda $$ = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7454,
"subject": "General Science",
"question": "Let A($$-$$1, 1), B(3, 4) and C(2, 0) be given three points. A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the areas of $$\\Delta$$ABC and $$\\Delta$$PQC respectively, such that A1 = 3A2, then the value of m is equal to :",
"options": [
{
"text": "1"
},
{
"text": "3"
},
{
"text": "2"
},
{
"text": "$${4 \\over {15}}$$"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\n
9m2 + 11m + 4 = 0 (Rejected $$ \\because $$ m is imaginary)
taking $$-$$ve sign
15m2 $$-$$ 11m $$-$$ 4 = 0
$$m = 1, - {4 \\over {15}}$$\n
$$ \\therefore $$ m = 1 (As given m > 0)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7455,
"subject": "General Science",
"question": "Let A be the set of all points ($$\\alpha$$, $$\\beta$$) such that the area of triangle formed by the points (5, 6), (3, 2) and ($$\\alpha$$, $$\\beta$$) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is :",
"options": [
{
"text": "$${4 \\over {\\sqrt 5 }}$$"
},
{
"text": "$${16 \\over {\\sqrt 5 }}$$"
},
{
"text": "$${8 \\over {\\sqrt 5 }}$$"
},
{
"text": "$${12 \\over {\\sqrt 5 }}$$"
}
],
"answer": "$${8 \\over {\\sqrt 5 }}$$",
"solution": "**Answer:** $${8 \\over {\\sqrt 5 }}$$\n\n
$$\\left| {{{0 - 0 + 8} \\over {\\sqrt 5 }}} \\right| = {8 \\over {\\sqrt 5 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7456,
"subject": "General Science",
"question": "Let the points of intersections of the lines x $$-$$ y + 1 = 0, x $$-$$ 2y + 3 = 0 and 2x $$-$$ 5y + 11 = 0 are the mid points of the sides of a triangle $$\\Delta $$ABC. Then, the area of the $$\\Delta $$ABC is _____________.",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nIntersection point of given lines are (1, 2), (7, 5), (2, 3)
Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 : $$-$$4x + 3y = 12 and the line L3, which passes through the point P(2, 3), intersects L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :
Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $$\\Delta$$PQR is :
Let the area of the triangle with vertices A(1, $$\\alpha$$), B($$\\alpha$$, 0) and C(0, $$\\alpha$$) be 4 sq. units. If the points ($$\\alpha$$, $$-$$$$\\alpha$$), ($$-$$$$\\alpha$$, $$\\alpha$$) and ($$\\alpha$$2, $$\\beta$$) are collinear, then $$\\beta$$ is equal to :
Let $$A\\left( {{3 \\over {\\sqrt a }},\\sqrt a } \\right),\\,a > 0$$, be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If $$D(3\\cos \\theta ,a\\sin \\theta )$$ is a point in the fourth quadrant such that the maximum area of $$\\Delta$$ACD is 12 square units, then a is equal to ____________.
",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nClearly $B$ is $\\left(-\\frac{3}{\\sqrt{a}},+\\sqrt{a}\\right)$ and $C$ is $\\left(-\\frac{3}{\\sqrt{a}},-\\sqrt{a}\\right)$\n
Let $$\\alpha$$1, $$\\alpha$$2 ($$\\alpha$$1 < $$\\alpha$$2) be the values of $$\\alpha$$ fo the points ($$\\alpha$$, $$-$$3), (2, 0) and (1, $$\\alpha$$) to be collinear. Then the equation of the line, passing through ($$\\alpha$$1, $$\\alpha$$2) and making an angle of $${\\pi \\over 3}$$ with the positive direction of the x-axis, is :
",
"options": [
{
"text": "$$x - \\sqrt 3 y - 3\\sqrt 3 + 1 = 0$$"
},
{
"text": "$$\\sqrt 3 x - y + \\sqrt 3 + 3 = 0$$"
},
{
"text": "$$x - \\sqrt 3 y + 3\\sqrt 3 + 1 = 0$$"
},
{
"text": "$$\\sqrt 3 x - y + \\sqrt 3 - 3 = 0$$"
}
],
"answer": "$$\\sqrt 3 x - y + \\sqrt 3 + 3 = 0$$",
"solution": "**Answer:** $$\\sqrt 3 x - y + \\sqrt 3 + 3 = 0$$\n\n
Points A($$\\alpha$$, $$-$$3), B(2, 0) and C(1, $$\\alpha$$) are collinear.
Let $$A(1,1), B(-4,3), C(-2,-5)$$ be vertices of a triangle $$A B C, P$$ be a point on side $$B C$$, and $$\\Delta_{1}$$ and $$\\Delta_{2}$$ be the areas of triangles $$A P B$$ and $$A B C$$, respectively. If $$\\Delta_{1}: \\Delta_{2}=4: 7$$, then the area enclosed by the lines $$A P, A C$$ and the $$x$$-axis is :
Let $$B$$ and $$C$$ be the two points on the line $$y+x=0$$ such that $$B$$ and $$C$$ are symmetric with respect to the origin. Suppose $$A$$ is a point on $$y-2 x=2$$ such that $$\\triangle A B C$$ is an equilateral triangle. Then, the area of the $$\\triangle A B C$$ is :
A variable line $$\\mathrm{L}$$ passes through the point $$(3,5)$$ and intersects the positive coordinate axes at the points $$\\mathrm{A}$$ and $$\\mathrm{B}$$. The minimum area of the triangle $$\\mathrm{OAB}$$, where $$\\mathrm{O}$$ is the origin, is :
$$ \\Rightarrow $$ 26x $$-$$ 122y $$-$$ 1675 = 0",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7469,
"subject": "General Science",
"question": "A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R (3, – 2) are fixed points, then the locus of the centroid of $$\\Delta $$PQR is a line :",
"options": [
{
"text": "parallel to y-axis"
},
{
"text": "with slope $${2 \\over 3}$$"
},
{
"text": "parallel to x-axis"
},
{
"text": "with slope $${3 \\over 2}$$"
}
],
"answer": "with slope $${2 \\over 3}$$",
"solution": "**Answer:** with slope $${2 \\over 3}$$\n\nLet the centroid of $$\\Delta $$PQR is (h, k) & P is ($$\\alpha $$, $$\\beta $$), then\n
$$ \\Rightarrow $$ locus is 6x $$-$$ 9y + 2 = 0",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7470,
"subject": "General Science",
"question": "If the line 3x + 4y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is :",
"options": [
{
"text": "(3, 4)"
},
{
"text": "(2, 2)"
},
{
"text": "(4, 4)"
},
{
"text": "(4, 3)"
}
],
"answer": "(2, 2)",
"solution": "**Answer:** (2, 2)\n\n\n
then incentre is $$(2,2)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7471,
"subject": "General Science",
"question": "Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant :",
"options": [
{
"text": "third"
},
{
"text": "fourth"
},
{
"text": "second"
},
{
"text": "first"
}
],
"answer": "second",
"solution": "**Answer:** second\n\nmBD $$ \\times $$ mAD = $$-$$ 1\n
$$ \\therefore $$ IInd quadrant.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 7472,
"subject": "General Science",
"question": "Let A(1, 0), B(6, 2) and C $$\\left( {{3 \\over 2},6} \\right)$$ be the vertices of a triangle ABC. If P is a Point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point $$\\left( { - {7 \\over 6}, - {1 \\over 3}} \\right)$$, is ________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n\n P is centroid of the triangle ABC.\n
Othocentre = (-3, 3)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7475,
"subject": "General Science",
"question": "In a triangle PQR, the co-ordinates of the points P and Q are ($$-$$2, 4) and (4, $$-$$2) respectively. If the equation of the perpendicular bisector of PR is 2x $$-$$ y + 2 = 0, then the centre of the circumcircle of the $$\\Delta$$PQR is :",
"options": [
{
"text": "($$-$$1, 0)"
},
{
"text": "(1, 4)"
},
{
"text": "(0, 2)"
},
{
"text": "($$-$$2, $$-$$2)"
}
],
"answer": "($$-$$2, $$-$$2)",
"solution": "**Answer:** ($$-$$2, $$-$$2)\n\nMid point of $$PQ \\equiv \\left( {{{ - 2 + 4} \\over 2},{{4 - 2} \\over 2}} \\right) \\equiv (1,1)$$
Circumcentre is ($$-$$2, $$-$$2)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7476,
"subject": "General Science",
"question": "Let tan$$\\alpha$$, tan$$\\beta$$ and tan$$\\gamma$$; $$\\alpha$$, $$\\beta$$, $$\\gamma$$ $$\\ne$$ $${{(2n - 1)\\pi } \\over 2}$$, n$$\\in$$N be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of $$\\Delta$$ABC coincides with origin and its orthocentre lies on y-axis, then the value of $${\\left( {{{\\cos 3\\alpha + \\cos 3\\beta + \\cos 3\\gamma } \\over {\\cos \\alpha \\cos \\beta \\cos \\gamma }}} \\right)^2}$$ is equal to ____________.",
"options": [],
"answer": "144",
"solution": "**Answer:** 144\n\nSince orthocentre and circumcentre both lies on y-axis.
then, $${\\left( {{{\\cos 3\\alpha + \\cos 3\\beta + \\cos 3\\gamma } \\over {\\cos \\alpha \\cos \\beta \\cos \\gamma }}} \\right)^2} = 144$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7477,
"subject": "General Science",
"question": "Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of $$\\Delta$$ABC, then (R + r) is equal to :",
"options": [
{
"text": "$$7\\sqrt 2 $$"
},
{
"text": "$${9 \\over {\\sqrt 2 }}$$"
},
{
"text": "$$2\\sqrt 2 $$"
},
{
"text": "$$3\\sqrt 2 $$"
}
],
"answer": "$${9 \\over {\\sqrt 2 }}$$",
"solution": "**Answer:** $${9 \\over {\\sqrt 2 }}$$\n\n\n
So, $$r + R = 3r = 3 \\times \\left( {{3 \\over {\\sqrt 2 }}} \\right) = {9 \\over {\\sqrt 2 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7478,
"subject": "General Science",
"question": "Consider a triangle having vertices A($$-$$2, 3), B(1, 9) and C(3, 8). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point $$\\left( {0,{\\alpha \\over 2}} \\right)$$, then the value of real number $$\\alpha$$ is ________________.",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\n $${\\left( {\\sqrt {50} } \\right)^2} = {\\left( {\\sqrt {45} } \\right)^2} + {\\left( {\\sqrt 5 } \\right)^2}$$
The distance of the origin from the centroid of the triangle whose two sides have the equations $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$ and whose orthocenter is $$\\left( {{7 \\over 3},{7 \\over 3}} \\right)$$ is :
In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If ($$\\alpha$$, $$\\beta$$) is the centroid of $$\\Delta$$ABC, then 15($$\\alpha$$ + $$\\beta$$) is equal to :
The equations of the sides $$\\mathrm{AB}, \\mathrm{BC}$$ and $$\\mathrm{CA}$$ of a triangle $$\\mathrm{ABC}$$ are $$2 x+y=0, x+\\mathrm{p} y=15 \\mathrm{a}$$ and $$x-y=3$$ respectively. If its orthocentre is $$(2, a),-\\frac{1}{2}<\\mathrm{a}<2$$, then $$\\mathrm{p}$$ is equal to ______________.
The equations of the sides $$\\mathrm{AB}, \\mathrm{BC}$$ and CA of a triangle ABC are $$2 x+y=0, x+\\mathrm{p} y=39$$ and $$x-y=3$$ respectively and $$\\mathrm{P}(2,3)$$ is its circumcentre. Then which of the following is NOT true?
Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q$$\\left(k_{1}, k_{2}\\right)$$, then $$k_{1}+k_{2}$$ is equal to :
Let $$\\mathrm{A}(\\alpha,-2), \\mathrm{B}(\\alpha, 6)$$ and $$\\mathrm{C}\\left(\\frac{\\alpha}{4},-2\\right)$$ be vertices of a $$\\triangle \\mathrm{ABC}$$. If $$\\left(5, \\frac{\\alpha}{4}\\right)$$ is the circumcentre of $$\\triangle \\mathrm{ABC}$$, then which of the following is NOT correct about $$\\triangle \\mathrm{ABC}$$?
",
"options": [
{
"text": "area is 24"
},
{
"text": "perimeter is 25"
},
{
"text": "circumradius is 5"
},
{
"text": "inradius is 2"
}
],
"answer": "perimeter is 25",
"solution": "**Answer:** perimeter is 25\n\n\n
If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $$(\\alpha,\\beta)$$, then the quadratic equation whose roots are $$\\alpha+4\\beta$$ and $$4\\alpha+\\beta$$, is :
The equations of the sides AB, BC and CA of a triangle ABC are : $$2x+y=0,x+py=21a,(a\\pm0)$$ and $$x-y=3$$ respectively. Let P(2, a) be the centroid of $$\\Delta$$ABC. Then (BC)$$^2$$ is equal to ___________.
Let $$(\\alpha, \\beta)$$ be the centroid of the triangle formed by the lines $$15 x-y=82,6 x-5 y=-4$$ and $$9 x+4 y=17$$. Then $$\\alpha+2 \\beta$$ and $$2 \\alpha-\\beta$$ are the roots of the equation :
Solve the equations $15x - y = 82$ and $6x - 5y = -4$
\n\n
Multiply the first equation by 5 and the second by 1 and then subtract the second from the first:
\n
$75x - 5y = 410$
\n
$6x - 5y = -4$
\n
Subtracting these gives $69x = 414$ which leads to $x = 6$
\n
Substitute $x = 6$ into the first equation to get $y = 8$
\n
So, the first vertex is $(6,8)$.
\n\n
Solve the equations $15x - y = 82$ and $9x + 4y = 17$
\n\n
Multiply the first equation by 4 and the second by 1 and then add:
\n
$60x - 4y = 328$
\n
$9x + 4y = 17$
\n
Adding these gives $69x = 345$ which leads to $x = 5$
\n
Substitute $x = 5$ into the second equation to get $y = -7$
\n
So, the second vertex is $(5,-7)$.
\n\n
Solve the equations $6x - 5y = -4$ and $9x + 4y = 17$
\n\n
Multiply the first equation by 4 and the second by 5 and then add:
\n
$24x - 20y = -16$
\n
$45x + 20y = 85$
\n
Adding these gives $69x = 69$ which leads to $x = 1$
\n
Substitute $x = 1$ into the first equation to get $y = 2$
\n
So, the third vertex is $(1,2)$.
\n
So, the vertices of the triangle are $(6,8)$, $(5,-7)$, and $(1,2)$.
\n\n
Now that we have the vertices of the triangle, we can find the centroid. The centroid $(\\alpha, \\beta)$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by :
\n
$$\n\\alpha=\\frac{x_1+x_2+x_3}{3}$$,\n
$$\n\\beta=\\frac{y_1+y_2+y_3}{3}\n$$
\n
Substituting the coordinates of the vertices into these equations, we get:
\n
$$\n\\alpha=\\frac{6+5+1}{3}=4 $$,\n
$$\n\\beta=\\frac{8-7+2}{3}=1\n$$
\n
Then, we find $\\alpha+2\\beta$ and $2\\alpha-\\beta$:
If the orthocentre of the triangle formed by the lines $$2 x+3 y-1=0, x+2 y-1=0$$ and $$a x+b y-1=0$$, is the centroid of another triangle, whose circumcentre and orthocentre respectively are $$(3,4)$$ and $$(-6,-8)$$, then the value of $$|a-b|$$ is _________.
So, the given triangle is right angled and also isosceles.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7492,
"subject": "General Science",
"question": "A man starts walking from the point P($$-$$3, 4), touches the x-axis at R, and then turns to reach at the point Q(0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $$50\\left( {{{(PR)}^2} + {{(RQ)}^2}} \\right)$$ is equal to ____________.",
"options": [],
"answer": "1250",
"solution": "**Answer:** 1250\n\n\n For minimum $(P R+R Q)$\n
$R$ lies on $P Q^{\\prime}$ (where $Q^{\\prime}$ is image of $Q$ in $X$-axis)\n
The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle $${\\pi \\over 4}$$ at the origin, is equal to :
$$ \\Rightarrow b = - 20 \\Rightarrow c = - {3 \\over 4}$$\n
Equation of $$K:$$ $$y - 4x = 3$$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$ \\Rightarrow 4x - y + 3 = 0$$\n
Distance between $$L$$ and $$K$$ \n
$$ = {{\\left| {52 - 32 + 3} \\right|} \\over {\\sqrt {17} }} = {{23} \\over {\\sqrt {17} }}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7496,
"subject": "General Science",
"question": "The foot of the perpendicular drawn from the origin, on the line, 3x + y = $$\\lambda $$ ($$\\lambda $$ $$ \\ne $$ 0) is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is : ",
"options": [
{
"text": "1 : 3"
},
{
"text": "3 : 1"
},
{
"text": "1 : 9"
},
{
"text": "9 : 1"
}
],
"answer": "9 : 1",
"solution": "**Answer:** 9 : 1\n\nEquation of the line, which is perpendicular to the line, \n
3x + y = $$\\lambda $$($$\\lambda $$ $$ \\ne $$0) and passing through origin , \n
is given by $${{x - 0} \\over 3} = {{y - 0} \\over 1} = r$$\n
\nBy Putting $$\\left( { - {1 \\over 4},{2 \\over 3}} \\right)$$ on both lines it satisfy.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7498,
"subject": "General Science",
"question": "If the line, 2x - y + 3 = 0 is at a distance $${1 \\over {\\sqrt 5 }}$$\nand $${2 \\over {\\sqrt 5 }}$$ from the lines 4x - 2y + $$\\alpha $$ = 0 and 6x - 3y + $$\\beta $$ = 0, respectively, then the sum of all possible values of $$\\alpha $$ and $$\\beta $$ is :\n",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\nApply distance between parallel line formula
$$ \\therefore $$ Sum of all values of $$\\alpha$$ and $$\\beta$$ is = 30",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7499,
"subject": "General Science",
"question": "If p and q are the lengths of the perpendiculars from the origin on the lines,
x cosec $$\\alpha$$ $$-$$ y sec $$\\alpha$$ = k cot 2$$\\alpha$$ and
x sin$$\\alpha$$ + y cos$$\\alpha$$ = k sin2$$\\alpha$$
Let the point $$P(\\alpha, \\beta)$$ be at a unit distance from each of the two lines $$L_{1}: 3 x-4 y+12=0$$, and $$L_{2}: 8 x+6 y+11=0$$. If $$P$$ lies below $$L_{1}$$ and above $${ }{L_{2}}$$, then $$100(\\alpha+\\beta)$$ is equal to :
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxkuuf/8206ce56-06e1-4b71-9f96-f1cdb4f19baf/cd6cd370-ccf1-11ee-a330-494dca5e9a63/file-6y3zli1lsoxkuug.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265056/exam_images/fendqfkdavmkbahhpwrf.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkjbalj/b11eb6ae-312f-49ae-887d-c92b77fa1799/cc33a060-6789-11ee-b7ba-e3cd7ea3cf86/file-6y3zli1lnkjbalk.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267182/exam_images/gheykswaxc1dfck1kir0.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln5ufvyu/8420caac-c5f0-403f-a175-e82f7eb8d93d/8ea00c50-5f75-11ee-8999-67742721d03c/file-6y3zli1ln5ufvyv.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqdjrqo/772aa650-3e8c-46a8-adce-0608922ed01d/09ef1300-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdjrqp.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqdl51c/d5a5c6d5-b0af-43b9-b7e2-7efa69d9eeff/30039c00-cdbd-11ee-a926-9fabe9a328d8/file-6y3zli1lsqdl51d.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263921/exam_images/s76qynlxlel1kkck45r2.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267198/exam_images/lmh3h5mktsgvuq0vvk3v.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwgi30yz/efd8a8c4-b19c-464d-9aab-50da9ed84968/ccf9e7b0-177f-11ef-97dc-2d80937d5077/file-1lwgi30z0.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263637/exam_images/evoxpwc1lpvv4diyvenq.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264328/exam_images/fxzt3slso48o3sotippm.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265178/exam_images/ymnjbdlge7ihvtgsvwzr.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265271/exam_images/qlwwjazgwqnl9oqbsmlw.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263675/exam_images/bt2q58jwtsfdf5ghjlxc.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267797/exam_images/t9q8kuirdplfx1cvehdq.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263588/exam_images/ffu3bdx7j47o0huvdfzw.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5v5uvjp/f35361f2-1b79-408c-b8ee-7c34d92e23d8/e7463d60-0905-11ed-a790-b11fa70c8a36/file-1l5v5uvjq.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l60jx84m/7dfa7ac3-4b51-48a5-b102-580ffe2e0760/d3aa7060-0bfc-11ed-8671-c161c1aef925/file-1l60jx84n.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgrmkz0q/2a7d38c3-ba55-42ad-a453-25c4fef11776/8da18aa0-e0da-11ed-9ecd-e999028462e7/file-1lgrmkz0r.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbcvqbx/b63739e8-c768-420d-a989-3fb42c2b49f4/96ca2bd0-627d-11ee-b4f9-75aabed442fa/file-6y3zli1lnbcvqby.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo75i5g8/e4ea036e-437d-4176-8624-2ce86ca94e2d/d2049270-73f9-11ee-b2c1-dbe2f99c6f1d/file-6y3zli1lo75i5g9.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwesuo/97ad35c7-e930-402f-8ae3-bf6649f70495/50a78900-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwesup.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxo287/d8da4921-3aa3-467d-a404-6687dd0b31e6/268e3570-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxo288.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91nk9em/51951291-ca13-4e57-8386-abbd403322ab/7e6fcbe0-47fc-11ed-8757-0f869593f41f/file-1l91nk9en.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264499/exam_images/jkfizacjmqkgdkpzypg0.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265485/exam_images/w9uwtpcdiyqyzwuecyz5.webp\")
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![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266918/exam_images/r8q6vmkbouo5l2gusylp.webp\")
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\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264770/exam_images/jzoaua9vt8av712nmhev.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266205/exam_images/qm3llf0rmyqukfzhahci.webp\")
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![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266416/exam_images/zvq4sh4mjcxsga2twdtf.webp\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ltao3w7v/14f352dd-3dd6-4237-8545-8edf0028c3df/01db21c0-d8e6-11ee-a868-79761dafd483/file-6y3zli1ltao3w7w.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2y3n7n/57ad7e01-e1c9-4f79-b41c-30dff98ab61e/d0fe9930-d4a6-11ee-bdd1-01c80c3e2d9a/file-1lt2y3n7o.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwelk3uw/46e00973-b9cb-4a02-b3d4-b1b1a19ec9fa/d20cd080-1673-11ef-97a1-518a89cfb45d/file-1lwelk3ux.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwemmte7/0b8f7144-df01-42b2-8694-08cedef5dd1e/06860df0-1678-11ef-922a-09b27e16ac69/file-1lwemmte8.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264992/exam_images/p877upzzzrjlz3axmszg.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266956/exam_images/cwlfbycr4bty1astkwr9.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxumdb/8342ad24-c8e3-4299-8fcb-c60fcb809b21/dcf834f0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxumdc.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3c14xb/42504263-4110-4311-9e4d-11eb95a83473/f131dbf0-1041-11ef-9f6c-75804a813f04/file-jaoe38c1lw3c14xc.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3c32ar/ca8579c0-7ba5-4526-8059-fbf044a6256c/26c8bb30-1042-11ef-9f6c-75804a813f04/file-jaoe38c1lw3c32as.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263707/exam_images/uiwdp9jmte0l8nkstssm.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4k23pz/525fa5bf-d94c-4b57-ab0f-dd8a7e2507fb/1dc1b670-10ee-11ef-8836-ed8ff380bbe9/file-1lw4k23q0.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265502/exam_images/egbnkczftvlxditjtcbz.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267731/exam_images/mtwwbj0355rmby4b8ntt.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266701/exam_images/izrw5r6ggvq1o2nql6cu.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266634/exam_images/yvyztb8wfjakqr2ttuvt.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267056/exam_images/ahksb1duj8ufepfas9sz.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91njbgq/c332d856-caa2-46a4-8d1d-eda9ad96fe55/6437c2a0-47fc-11ed-8757-0f869593f41f/file-1l91njbgr.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267746/exam_images/sfbzoig20kahtq3la2hf.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l87czgsa/9146332a-d177-4510-8b69-14ebf4d88eae/9d5eea90-3753-11ed-9417-1312e45a73c7/file-1l87czgsb.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263839/exam_images/lz3a9qs9vzwnjl9bmeiu.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266852/exam_images/nlqfkpsnhv5bqrg6xjyo.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266827/exam_images/o576rgrvrso9cx8btgxu.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264796/exam_images/a99rugdbqpo21gvuvpmm.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267455/exam_images/gwuen82slkerzet1ztko.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267190/exam_images/txlbrnxh7uu1cyfihe9v.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267678/exam_images/ssjeqbfwbhvdrbkypihv.webp\")
\n$${x \\over d} = \\tan {30^o} \\Rightarrow {x \\over d} = {1 \\over {\\sqrt 3 }} \\Rightarrow d = \\sqrt 3 x$$\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264081/exam_images/seb5qh2vead30zbnfys2.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264351/exam_images/b08oyn8k1lljfispvted.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264407/exam_images/wgqikoy7kceictq8u4qg.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266845/exam_images/jc6rgs0t7o340wghq0oy.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265755/exam_images/vceiuoriishdtliybgle.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266427/exam_images/aq1jmc4l3u0dm4zx0orb.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264592/exam_images/wpspq1xvs3grsccoimwk.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266261/exam_images/mad9v6jss642a5iyxeks.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267031/exam_images/aszqjhyd9ytrutgsg8cr.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266806/exam_images/yei81ivy9h4mpf0cbjzv.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266282/exam_images/cyid3ux4etov3mikiruk.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264042/exam_images/c7wzo3oqwrimegfuybce.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwld6t6g/11cc28b4-5cf0-4c13-9cdb-f2eaaa5c2649/ca276b80-5174-11ec-b1ae-c1b029d25cc4/file-1kwld6t6h.jpeg\")
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5386,
"subject": "General Science",
"question": "A spherical gas balloon of radius 16 meter subtends an angle 60$$^\\circ$$ at the eye of the observer A while the angle of elevation of its center from the eye of A is 75$$^\\circ$$. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :",
"options": [
{
"text": "$$8(2 + 2\\sqrt 3 + \\sqrt 2 )$$"
},
{
"text": "$$8(\\sqrt 6 + \\sqrt 2 + 2)$$"
},
{
"text": "$$8(\\sqrt 2 + 2 + \\sqrt 3 )$$"
},
{
"text": "$$8(\\sqrt 6 - \\sqrt 2 + 2)$$"
}
],
"answer": "$$8(\\sqrt 6 + \\sqrt 2 + 2)$$",
"solution": "**Answer:** $$8(\\sqrt 6 + \\sqrt 2 + 2)$$\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwld2tjm/54d13fc9-0216-4da9-b4ee-1d4f67320097/5b332f20-5174-11ec-b1ae-c1b029d25cc4/file-1kwld2tjn.png\")
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 5387,
"subject": "General Science",
"question": "Two poles, AB of length a metres and CD of length a + b (b $$\\ne$$ a) metres are erected at the same horizontal level with bases at B and D. If BD = x and tan$$\\angle$$ACB = $${1 \\over 2}$$, then :",
"options": [
{
"text": "x2 + 2(a + 2b)x $$-$$ b(a + b) = 0"
},
{
"text": "x2 + 2(a + 2b)x + a(a + b) = 0"
},
{
"text": "x2 $$-$$ 2ax + b(a + b) = 0"
},
{
"text": "x2 $$-$$ 2ax + a(a + b) = 0"
}
],
"answer": "x2 $$-$$ 2ax + b(a + b) = 0",
"solution": "**Answer:** x2 $$-$$ 2ax + b(a + b) = 0\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264596/exam_images/ojvw2ffxtio4f1jffdyo.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266417/exam_images/igmotqewqjtulhutbk8z.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5flljg6/71a3df8b-f1e7-4fa5-9624-d4e70f01a8df/eb523e50-0076-11ed-bd71-b57b399a7926/file-1l5flljg7.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5oc0rd9/9c4c842e-b702-4002-a405-e2ab09551342/852123d0-0544-11ed-987f-3938cfc0f7f1/file-1l5oc0rda.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79fj3e6/3edc2d44-ec0d-4c06-831f-a466d782744d/eb6330e0-24aa-11ed-8d2e-5f0df5271c2d/file-1l79fj3e7.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7psamcq/756ef252-1665-43d7-bc8f-4dd4bf8b8c96/60d220a0-2da9-11ed-8542-f96181a425b5/file-1l7psamcr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7qafj8i/a7a027fb-787f-4a20-838e-deb7a4fd2b6b/4dc88020-2df0-11ed-a744-1fb8f3709cfa/file-1l7qafj8j.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7rri7rq/1c703de6-5ecf-4544-826f-2ec23ff8478d/dc32c760-2ebf-11ed-b92e-01f1dabc9173/file-1l7rri7rr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssh0oh/2868d3c3-c2c7-439f-b463-f42683b289b7/6cb0f920-2f50-11ed-88f3-17ddb055f60b/file-1l7ssh0oi.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnoi86a2/300ca189-bd8f-4def-ab5a-3cd9c710ee8f/aee81b90-69b8-11ee-a65f-213955fd9d72/file-6y3zli1lnoi86a3.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267464/exam_images/h22mhjtns4c9oikrrw1r.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lef56ked/6c51cd9e-287f-4e86-804d-923bd55624d6/ab431f50-b264-11ed-a126-5dfa1a9d5fb8/file-1lef56kee.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln228t36/fd6163aa-9150-4285-8e02-935227613d64/f02c4220-5d60-11ee-8969-dbcde4c067b7/file-6y3zli1ln228t37.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265061/exam_images/w1lkvrlkm2a0iclvpvya.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lebzl8r1/919fa76e-b803-49d8-86c9-4e9df9241622/72ada3d0-b0a8-11ed-98af-57b0199700ba/file-1lebzl8r2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1vjyru/5d14555c-4b38-417d-b959-37b3238917a8/12d5a2a0-d410-11ee-b9d5-0585032231f0/file-1lt1vjyrv.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264616/exam_images/y5muchtddz9gnxydcpim.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266288/exam_images/sqc6djvwoxgp7yny26un.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266897/exam_images/bkhe3kwsqinvpy72llmd.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxrj67/57cc4f4f-ed29-40fc-903d-a6afb2548230/871275f0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxrj68.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnnba9/95b74993-a660-473c-9d3c-c4fef0db6ac2/87c00510-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnnbaa.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4m7nur/89a66068-e09d-4eb2-bb5b-6e5807946115/8a9cdd30-10f6-11ef-8553-fdfc6347789d/file-1lw4m7nus.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265692/exam_images/dqhksg6rhe7itnkgtxn3.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0ez3v/5fa06fb6-2e73-4b95-aa16-dacb000be7c4/83af90b0-d658-11ec-9a06-bd4ec5b93eb4/file-1l3b0ez3w.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwhfif9k/1a3e3d85-d2e7-4099-8558-58b863f1c6ee/866ac480-1802-11ef-b156-f754785ad3ce/file-1lwhfif9l.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwaskftu/d526b7cc-575f-4620-8047-7c036e1ade54/189a8020-145c-11ef-a76a-45e5478b32e1/file-1lwaskftv.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265699/exam_images/wxftevmnlm0p8wyfumnh.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ps10fx/454d4b24-e609-4a62-af81-65229ca491c4/559cd2d0-2da8-11ed-8542-f96181a425b5/file-1l7ps10fy.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0v2ql/1fa8d111-320f-4de3-8dab-6d23cbe7b0e2/436fe4d0-d65a-11ec-9a06-bd4ec5b93eb4/file-1l3b0v2qm.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0vpg8/606245c5-bf1b-4633-a0a9-2c4dec6f8121/54fb5180-d65a-11ec-9a06-bd4ec5b93eb4/file-1l3b0vpg9.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265142/exam_images/vmbo5oe8fbvyyyiih8ml.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5d349j2/93a986e3-81fb-491a-9045-ba91b8ee8ccd/15480400-ff15-11ec-8a3d-a39bde0e21e9/file-1l5d349j5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l99t3fun/ee0d4d39-3197-4ca1-984e-cfb744e32a27/66a131e0-4c78-11ed-b94d-45a8040c2a81/file-1l99t3fuo.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266784/exam_images/jxar1vgdorciobrxdspe.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266217/exam_images/fghim8fqhfaqmfos46vi.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264785/exam_images/rhqvelsmxghhl3cztv4v.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5699,
"subject": "General Science",
"question": "Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – $$\\pi $$) cos |x| is not differentiable. Then the set K is equal to :",
"options": [
{
"text": "{0, $$\\pi $$}"
},
{
"text": "$$\\phi $$ (an empty set)"
},
{
"text": "{ r }"
},
{
"text": "{0}"
}
],
"answer": "$$\\phi $$ (an empty set)",
"solution": "**Answer:** $$\\phi $$ (an empty set)\n\nf(x) = sin$$\\left| x \\right| - \\left| x \\right|$$ + 2(x $$-$$ $$\\pi $$) cosx\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264960/exam_images/j9hfstro1mtiidz6xe3y.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266646/exam_images/sowb6jz8ffftgdwa0zm0.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267767/exam_images/g4hsbmpaxkynaavxvqy0.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264640/exam_images/thggkxyle8cgcwavqqvd.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266844/exam_images/nsxa64efsvtbdkfgqlui.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263292/exam_images/hlto1ikhz1kx3bmvaaoo.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267003/exam_images/njxdcddt27d4labpnhcm.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266177/exam_images/mlkrr07qnsgtcedtgrri.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267341/exam_images/xx4dogq2jwbvxilj1i8d.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267638/exam_images/uj4pdcgnkf2sk9ywcaok.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266706/exam_images/ojrovdzbjtospfgtjkfj.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263692/exam_images/yav6eilmb28mctcjaeyf.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97st8z9/f16176cf-a04a-4df8-9ae6-30f66a17f773/b9ee7e50-4b5d-11ed-bfde-e1cb3fafe700/file-1l97st8za.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb5odf/8441c3d2-e56e-4faa-a526-4c7b6253ae3c/ed702f30-2e7f-11ed-8702-156c00ced081/file-1l7rb5odg.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lifi9pn2/0d3d2e66-f8f7-4969-8143-a77c378e96f7/b6eee2e0-01c8-11ee-91de-e7d2360ea0c8/file-1lifi9pn3.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnd9enbh/f19f578f-55e0-486b-a43d-c6ee1891f168/919e63d0-6389-11ee-b38e-7becbda30131/file-6y3zli1lnd9enbi.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsq9xozp/ba71121b-98fa-43fb-8b6d-a25522270b09/e89ef840-cdae-11ee-8e42-2f10126c4c2c/file-6y3zli1lsq9xozq.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqa8zwn/f12f5f83-13c0-4aa6-8f79-7a22074474aa/22f45480-cdb0-11ee-8e42-2f10126c4c2c/file-6y3zli1lsqa8zwo.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1qvsrr/82daf6e5-d235-46d9-8017-461f6f164b34/ce430770-d3fd-11ee-acd6-03dea0d2c848/file-1lt1qvsrs.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1qwxt1/0909b125-954a-4e40-9ed4-1383ae43484b/edf5f550-d3fd-11ee-acd6-03dea0d2c848/file-1lt1qwxt2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwnla3/94c1149f-a7b9-41cc-b180-4bf0cc496ba7/451674b0-ca2e-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwnla4.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267071/exam_images/hbbibexjhoou7gayvkwp.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5816,
"subject": "General Science",
"question": "$$\\mathop {\\lim }\\limits_{x \\to 0} {{{{\\sin }^2}x} \\over {\\sqrt 2 - \\sqrt {1 + \\cos x} }}$$ equals:",
"options": [
{
"text": "$$ \\sqrt 2$$"
},
{
"text": "$$2 \\sqrt 2$$"
},
{
"text": "4"
},
{
"text": "$$4 \\sqrt 2$$"
}
],
"answer": "$$4 \\sqrt 2$$",
"solution": "**Answer:** $$4 \\sqrt 2$$\n\n$$\\mathop {\\lim }\\limits_{x \\to 0} {{{{\\sin }^2}x} \\over {\\sqrt 2 - \\sqrt {1 + \\cos x} }}$$\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267636/exam_images/ndn0paa8angdbmk4hb5t.webp\")
\n![\"AIEEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3fdb05p/c9d4c926-25a7-47f2-b2d6-27e5c7d3dd72/0e89a3d0-d8be-11ec-bed7-8f98fee06619/file-1l3fdb05q.png?format=png\")
\n![\"JEE](\"https://gateclass.static.cdn.examgoal.net/2OAgJSCgoisOIuHQi/0RUPnVokAM5r5pw4wQPIququqMxsm/6RBqA2leE55pxx7XCRommU/uploadfile.jpg\")
\n![\"JEE](\"https://gateclass.cdn.examgoal.net/W3iUUKAK091ok1iYR/8HSPWtDptuwd7CcOkhxrHgDo4hy7f/t0aUb5WYW03FsXNGzU2q58/uploadfile.jpg\")
![\"JEE](\"https://gateclass.static.cdn.examgoal.net/5zTbH8zwu0wJZaFke/tyBqLcGYS9Pq5EOlU5qNL491PuOUh/gEmh8rTirgAroJSmV6GPRz/uploadfile.jpg\")
",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5871,
"subject": "General Science",
"question": "The Boolean expression\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266818/exam_images/u0czhffenck7fqfdrxfd.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264718/exam_images/klwq1gy6wqlgukcr0vja.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263729/exam_images/sjp2t7hj2rwdsx9ukbot.webp\")
",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 5875,
"subject": "General Science",
"question": "The Boolean expression ~(p $$ \\Rightarrow $$ (~q)) is equivalent to :",
"options": [
{
"text": "p $$ \\wedge $$ q"
},
{
"text": "q $$ \\Rightarrow $$ ~p"
},
{
"text": "p $$ \\vee $$ q"
},
{
"text": "(~p) $$ \\Rightarrow $$ q"
}
],
"answer": "p $$ \\wedge $$ q",
"solution": "**Answer:** p $$ \\wedge $$ q\n\n$$ \\sim \\left( {p \\to \\sim q} \\right) \\Rightarrow \\sim \\left( { \\sim p \\vee \\sim q} \\right) \\Rightarrow p \\wedge q$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 5876,
"subject": "General Science",
"question": "If the truth value of the statement p $$ \\to $$ (~q $$ \\vee $$ r) is false (F), then the truth values of the statements p, q, r are\nrespectively : ",
"options": [
{
"text": "T, F, T"
},
{
"text": "F, T, T"
},
{
"text": "T, T, F"
},
{
"text": "T, F, F"
}
],
"answer": "T, T, F",
"solution": "**Answer:** T, T, F\n\np $$ \\to $$ ( ~ q $$ \\vee $$ r) $$ \\equiv $$ ~ p $$ \\wedge $$ ( ~ q $$ \\vee $$ r) $$ \\equiv $$ ( ~ p $$ \\vee $$ ~ q) $$ \\vee $$ r $$ \\equiv $$ ~ ( p $$ \\wedge $$ q ) $$ \\vee $$ r![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265973/exam_images/zxx1e4fhowtletn8zjkm.webp\")
",
"topic": "Discrete Mathematics",
"subtopic": "Logic"
},
{
"id": 5883,
"subject": "General Science",
"question": "The Boolean expression ((p $$ \\wedge $$ q) $$ \\vee $$ (p $$ \\vee $$ $$ \\sim $$ q)) $$ \\wedge $$ ($$ \\sim $$ p $$ \\wedge $$ $$ \\sim $$ q) is equivalent to :",
"options": [
{
"text": "p $$ \\wedge $$ q"
},
{
"text": "p $$ \\wedge $$ ($$ \\sim $$ q)"
},
{
"text": "p $$ \\vee $$ ($$ \\sim $$ q)"
},
{
"text": "($$ \\sim $$ p) $$ \\wedge $$ ($$ \\sim $$ q)"
}
],
"answer": "($$ \\sim $$ p) $$ \\wedge $$ ($$ \\sim $$ q)",
"solution": "**Answer:** ($$ \\sim $$ p) $$ \\wedge $$ ($$ \\sim $$ q)\n\n((p $$ \\wedge $$ q) $$ \\vee $$ (p $$ \\vee $$ $$ \\sim $$ q)) $$ \\wedge $$ ($$ \\sim $$ p $$ \\wedge $$ $$ \\sim $$ q)\n![\"JEE](\"https://gateclass.cdn.examgoal.net/qTCfGCsgnLJmurfJO/mTRBMelO4xBQhIbdNCj3SfmvgSIpr/yQ9Y64EkKKeaEzp8Qw7Tue/uploadfile.jpg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267009/exam_images/ug8ifnvkt1ne5r5zuiz4.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265134/exam_images/emajs7ltkurmpe8hmtbv.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264386/exam_images/pkgg8abdxwnuxm3mksuq.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266697/exam_images/hyq0krwnkwrcah7t3agk.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267518/exam_images/h1zyjcoqy8xi3yytdif8.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265647/exam_images/ofzxonxbghnfr4htgcmr.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0mtai/207863b7-e6a3-4fcf-9b41-05c186319670/5daac9b0-d659-11ec-9a06-bd4ec5b93eb4/file-1l3b0mtaj.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266046/exam_images/d18zz8tkcr3s2bkasyjb.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97slo37/df9ec0ac-a06c-4e60-b3b5-9f73d35cb598/e7215830-4b5c-11ed-bfde-e1cb3fafe700/file-1l97slo38.png?format=png\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 5931,
"subject": "General Science",
"question": "![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lek3mdx9/896912c9-3981-4e34-b46a-5223559fea7c/7c31fad0-b51e-11ed-accc-792fddd82133/file-1lek3mdxa.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lek3nk24/09e41b34-c9d0-4e6b-a3aa-b005fb90770b/9cbe58c0-b51e-11ed-8c61-834212db4998/file-1lek3nk25.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264644/exam_images/sd5znxchfmvbwrpympou.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3bip55r/c48ed0bf-90ff-469a-a2d8-184046732c50/02c6fef0-d6a0-11ec-a354-cb09522b689a/file-1l3bip55s.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267738/exam_images/dj4d0ou4immj3t18yq1g.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266175/exam_images/abwa7naonbnbqkrjrncp.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267183/exam_images/gu3kbdwmjr2pxmljyejz.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266972/exam_images/bzjcfypz4zfwsyd69i2a.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5niod7a/d422eadb-ccc2-408b-8388-e3a333afd463/c3bb6e70-04d1-11ed-93b8-936002ac8631/file-1l5niod7b.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6dmy0zj/4987e864-b411-4a25-aaa9-467edd1daa02/d5c9d0f0-132e-11ed-941a-4dd6502f33e3/file-1l6dmy0zk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssn4pj/b0858943-62f1-4335-ac1e-b62256a92364/16a87a70-2f51-11ed-85dd-19dc023e9ad1/file-1l7ssn4pk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lh8cf7fa/cc7db376-968a-4280-8ff8-3d1538f92ccf/25044f60-ea0c-11ed-9dfd-37e68ac0819e/file-1lh8cf7fb.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwk9u5jo/5c19a59f-8f6e-4ba6-8e8c-779764eeedda/afcaae30-1992-11ef-bba0-ebdcce35fbed/file-1lwk9u5jp.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267100/exam_images/dqtfzq9ngy0k8tl3a08o.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264098/exam_images/o0bfsfl0vlazf4qy2pvy.webp\")
\n$$ \\therefore $$ $${{PS} \\over {PS}} = {{3 + {1 \\over 3}} \\over { - {1 \\over 3} + 3}} = {5 \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 6314,
"subject": "General Science",
"question": "The equation of common tangent to the curves y2\n = 16x and xy = –4, is :\n",
"options": [
{
"text": "x – y + 4 = 0"
},
{
"text": "x + y + 4 = 0"
},
{
"text": "x – 2y + 16 = 0"
},
{
"text": "2x – y + 2 = 0"
}
],
"answer": "x – y + 4 = 0",
"solution": "**Answer:** x – y + 4 = 0\n\nLet the equation of tangent to parabola![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264738/exam_images/lvif1mocwwwtckxww1qh.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwd1ocxr/d26061ca-43e1-4f4a-a0fc-77af60f91e91/491160f0-1599-11ef-acc0-2dadf5616f59/file-1lwd1ocxs.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxw9mg20/83fce52e-2630-4574-a56c-ffe9c399a8da/aef7f480-6b3f-11ec-8608-9b519146e1b7/file-1kxw9mg21.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266223/exam_images/w7aqioqqabs5dzmfr7a7.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxof1w7l/840d5c1f-ce31-4701-93d3-208583e5d683/a5091c10-66ee-11ec-9866-0df874a38238/file-1kxof1w7m.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l09usogd/0e8d490d-75bf-4d50-a3ef-e9477504eeb8/1b3947d0-9a51-11ec-b1c5-39b9b722e9af/file-1l09usoge.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lef4vkb3/455513c9-54ce-4923-8412-8892bb374d49/794a91f0-b263-11ed-a6d1-894f1caef997/file-1lef4vkb4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lef4wwy4/94c63b59-5b17-404e-a886-f1440050aed4/9edd19b0-b263-11ed-a126-5dfa1a9d5fb8/file-1lef4wwy5.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264958/exam_images/exuu3huqxlmuv2m9xipb.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263693/exam_images/be9bgy7e2oqgaj3q4lr8.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263531/exam_images/y0zkxse1bq2gklpforj2.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264518/exam_images/csbm6jz6ovhaimejwhpu.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266392/exam_images/kfqvzltdhtj2yuoqxjth.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwopy91l/b12d9de1-4cc2-4e41-a53c-978e58e1dcd9/0cc11390-534d-11ec-9cbb-695a838b20fb/file-1kwopy91m.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264184/exam_images/dtyf9rtine6siiyeobjo.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267052/exam_images/mnslvvuc5ia6lxqcjf12.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266547/exam_images/ql9pnjqathvtux4uc0ar.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263641/exam_images/fiy4gcor4i4z43cmqerx.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265153/exam_images/uwoptt2srzgkzl7wy0ci.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxoae8y6/3b0cfa8b-90f4-4e91-ba41-db601a79f96b/6ef08ee0-66dc-11ec-8eea-2b236925a651/file-1kxoae8y7.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264020/exam_images/ipwtluvs8xeh3hc4yv0v.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw1nqf2u/3f5eb39b-71fd-4a3c-a3b6-2db1d6eda7bc/23637760-0f56-11ef-91cd-f19f7dc20f18/file-1lw1nqf2v.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266932/exam_images/otkuicrgj5mighgdx4s8.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263284/exam_images/idjgwu7hbabliruceroz.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267822/exam_images/urqnnuedt8c4vkagbrve.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264985/exam_images/rrktibvdvq5qvsqqrcjk.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw1p45wf/b8641c5f-23f4-4335-a99c-bb1d21171afc/8ac176f0-0f5b-11ef-8792-3d19bf2e18a4/file-1lw1p45wg.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263639/exam_images/o1prumkf5cogacvzecia.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263645/exam_images/qbyz275lq6l05oar7ina.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263897/exam_images/dzexdmaraqpgczbig0ch.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264947/exam_images/axoyq3cd2jeffcuzpst3.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264154/exam_images/jj7xl6x9z7rdt8utfb0x.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5scup2c/50969c63-d4f1-466d-8464-4fc5bb572b2d/e838e240-077a-11ed-94f6-83604aa63acb/file-1l5scup2d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5qa7f14/20481fc2-2f60-4c10-b833-b651bfbff5ae/fd4dc580-0656-11ed-903e-c9687588b3f3/file-1l5qa7f15.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97t0fy1/82aaf9b5-d36f-45fe-b435-ffafbaba2be0/81fa0a90-4b5e-11ed-bfde-e1cb3fafe700/file-1l97t0fy2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7bummn7/86c95a3b-4fff-4748-b770-5b447b8c82d7/87ebf730-25ff-11ed-9c74-c5a04899a045/file-1l7bummn8.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7raxquq/8f209afe-272a-4b1d-9894-cf78a8a0ee0c/10e1a120-2e7f-11ed-8702-156c00ced081/file-1l7raxqur.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lefxfag8/8b4831e9-648d-41a1-9ba1-f0f0ff15128a/1d79d280-b2d3-11ed-8169-e1635469e777/file-1lefxfag9.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1let847b1/44924ea8-65a6-4741-9b68-fa3cc8917eba/1e026ad0-ba23-11ed-b1c7-2f0d4a78b053/file-1let847b2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5hgrn5/44884087-ba3d-4667-bbb2-3f2a5b14abd3/de014610-ad14-11ed-8a8c-4d67f5492755/file-1le5hgrn6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5hhow2/b6a28f6c-8c4b-40a8-8d7e-2bb4e657d95c/f7b02220-ad14-11ed-8a8c-4d67f5492755/file-1le5hhow3.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2n7sz9/699005b2-a3c2-4567-bfdf-626597fbe0f4/0194ea50-ab85-11ed-bcb9-87e2bc2e0c49/file-1le2n7sza.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmyvdpob/0009489a-90f2-422f-89e7-7c971c77552a/bebae0b0-5b9f-11ee-b31c-37f6bf9b942e/file-6y3zli1lmyvdpoc.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwd36y63/122dc7d3-2b8c-4a18-982f-a395f4f07780/372994b0-159f-11ef-aabb-5de744d2ad81/file-1lwd36y64.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266952/exam_images/idxhu7dj3nmxsx1ackba.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264617/exam_images/eimc4dj1il40k9qseeao.webp\")
\nAny two non-adjacent pillers are joined by beams![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266804/exam_images/l90mpotax9t4vio9u3us.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwoq1bj3/a1cd0f3a-86a8-4d72-98ac-6d132a8be6f4/621a22f0-534d-11ec-9cbb-695a838b20fb/file-1kwoq1bj4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lef6kjms/0eda98c6-e556-46f8-a286-86afedde79c5/1920a240-b26a-11ed-a7d3-67cb923c1f9d/file-1lef6kjmt.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266135/exam_images/ihwshbrvypvoyv0j1fui.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266027/exam_images/pu5xcxrx6roxmo36hk63.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267655/exam_images/sqrjvd7hnsr1xwffmqw7.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmltts5a/3a0ee79a-7b48-44a1-98c7-740fb4585474/7ae256d0-5473-11ee-9283-c929f40dddd4/file-6y3zli1lmltts5b.png?format=png\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264044/exam_images/uld4peekswqjropomsze.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265719/exam_images/cbjd8rqt5qx4eis3hsmv.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc3zecxi/95ec386f-1c04-46f9-b1e8-eca213646d74/ff4a6070-84a8-11ed-9e2c-bf48676ab5eb/file-1lc3zecxj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5h0vkb/6ab5456d-64a1-4237-89f3-e400627057ab/241652a0-ad13-11ed-8a8c-4d67f5492755/file-1le5h0vkc.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2pxjq6/e260bc6f-c1c6-485e-abbc-aa68e85a10dd/9fbb03e0-ab8f-11ed-a599-53c07234da0d/file-1le2pxjq7.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263746/exam_images/x99fiwum87l1tzmfgxkv.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267485/exam_images/cosyxpmaifzqeuqzagre.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264157/exam_images/nmuqaapkyzrj5ufunbg2.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267399/exam_images/xgnodv9vbpjc1qwfcggz.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265433/exam_images/cfevw9axkaw11ccg7uvl.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266444/exam_images/lzgxhaamyintp77qz02x.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265599/exam_images/a06mbwqkrmeayldp4uew.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263667/exam_images/au1aj9g5bmhlq9ar6avi.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263587/exam_images/gylahnddae97tru1yytr.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265718/exam_images/p2ps2ybooqoc7gyxtqdc.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264389/exam_images/aiakh8f4ehgt1n8uagk2.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264026/exam_images/psga0gi4ukcamvr6fhaw.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267736/exam_images/adpwow8qovowjur8hlhv.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwoqbke8/bc58d634-a369-4a42-b528-57c54b3ada02/7f0aa000-534e-11ec-9cbb-695a838b20fb/file-1kwoqbke9.png?format=png\")
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6478,
"subject": "General Science",
"question": "![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l65uqr2z/3aa8732e-2bc7-4c8a-a45b-f53f925f4291/0f4fb6b0-0ee7-11ed-a7de-eff776fdb55c/file-1l65uqr30.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97twac2/f9689a1f-abde-44d3-bf0b-1a1289bdd7bc/f788ec10-4b61-11ed-80b9-4154b7faa509/file-1l97twac3.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97tyhd1/d57389e9-902a-407f-843c-e06ca2c6e0b2/3494f950-4b62-11ed-80b9-4154b7faa509/file-1l97tyhd2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lf61sqp4/9a9d5f7b-a4f5-43a8-8dd7-0303f78b9d4c/98f24a80-c130-11ed-8b61-5f07d5ca97fb/file-1lf61sqp5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lef5alut/f8dc7152-4573-493d-a2cc-0b09243a1255/1b9fc550-b265-11ed-a126-5dfa1a9d5fb8/file-1lef5aluu.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgv2wsca/cbc5754b-29f9-4539-a370-acd512a8c4f5/c2331800-e2c0-11ed-920b-3fff2b1cb8d0/file-1lgv2wscg.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgv2zu9s/f697b5a7-ed4b-4299-82b8-a7a0d3445b64/171d6000-e2c1-11ed-920b-3fff2b1cb8d0/file-1lgv2zu9t.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgv31bpl/c0980147-5172-44a2-a58d-c6cb14a3f1ab/4066ae30-e2c1-11ed-99f0-2347181d377e/file-1lgv31bpy.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgv32tqq/b0d7959c-9018-4ef6-a138-3892ec03995b/6a221f20-e2c1-11ed-99f0-2347181d377e/file-1lgv32tqr.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgv34b44/8ad108c1-684d-453a-8d6e-de913d462fc1/935ca040-e2c1-11ed-99f0-2347181d377e/file-1lgv34b45.png?format=png\")
\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265112/exam_images/crszvfmrucm793mhf0u4.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263381/exam_images/wbv3ybxvwrwbq6hh5suh.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnj292pt/52c17b6b-83cf-405a-ba54-57beea816528/4aa6a710-66ba-11ee-bc75-8500166e483f/file-6y3zli1lnj292pu.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwk6ko9r/1d69134a-fee4-473a-84cb-efe47fa9ec3c/ecc716f0-1985-11ef-a7bd-376696e028ce/file-1lwk6ko9s.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265634/exam_images/trl438xafzvaoxkw1zrx.webp\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266793/exam_images/m8iisj7kowslkqzbmaox.webp\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264419/exam_images/tgwhspuv73d1hl6mxgsx.webp\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267734/exam_images/vnt57cckvux7ausi0knw.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267595/exam_images/iqfugye1vg9tyuyciht6.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265517/exam_images/k0f0rfyuhhvanqmcnfpu.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265490/exam_images/ustv4otfdmz6sfaiqxez.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263614/exam_images/ey8ucpg1jd1dowuj5uf6.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79egidm/a6690954-7aeb-421e-841b-1c4509718aab/ba67f6a0-24a6-11ed-8d2e-5f0df5271c2d/file-1l79egidn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79eif4t/9c168fac-5b9a-40c7-962f-9e1a2098a816/ef849cd0-24a6-11ed-8d2e-5f0df5271c2d/file-1l79eif4u.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79em0gg/97d3fb7a-f64e-4c46-bfbe-808b4622d7eb/536acd00-24a7-11ed-8d2e-5f0df5271c2d/file-1l79em0gh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79ep4sk/1aa80241-a0dc-436c-8af5-3fbdc96cf0a2/aa31c440-24a7-11ed-8d2e-5f0df5271c2d/file-1l79ep4sl.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79ewvv3/7e732731-8eaf-4f1e-91a2-d2f8288f1660/81c4fdf0-24a8-11ed-8d2e-5f0df5271c2d/file-1l79ewvv4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79f1t81/275df426-ebcd-42da-bccc-f9c7f09fddb4/0ac6b120-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f1t82.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79f4997/9a6ce487-3cbd-4e4f-822e-42b6ff66ac73/4ec74ab0-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f4998.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79f79sr/bdabc9fc-26b6-4f3a-9eaa-4b6346d15d96/a2a022b0-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f79ss.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le9xbqbh/6a03e2bb-e7c5-4885-abb8-b77d1c53361f/0699bad0-af86-11ed-8a80-c79017866e24/file-1le9xbqbi.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lelo42n9/53ce0458-9680-4b44-8466-08f8e9ec8d51/6747a240-b5fb-11ed-9b52-9b6c0357d10e/file-1lelo42na.png?format=png\")
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 6575,
"subject": "General Science",
"question": "![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2g13sg/30377b8c-aded-4085-8900-acce8af1dc3b/e75d6700-ab68-11ed-b566-111c81fc645a/file-1le2g13sh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2g1tke/43c45f16-3db2-4d80-a59f-b27d342fc2b4/fb46c0e0-ab68-11ed-b88f-0b6f40cc9ae6/file-1le2g1tkf.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2g2ax4/3d119479-b1ac-4838-9b4e-8989ddd3f6de/08ae7480-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g2ax5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2g2w2h/254b12e3-a7f2-4f3b-a4c0-bba7e60d20fc/1904bc90-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g2w2i.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2g3fcm/c842699b-7c36-45e1-856a-a43d02492a13/27e9c160-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g3fcn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2g48bm/5e3286ff-1f06-4186-a954-c8121f8cf64f/3e4b4730-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g48bn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2g4qkb/9fafb3cc-ba1e-4c03-a575-be2b1b45a714/4c6274b0-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g4qkc.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw8oycbm/cc620a6d-aa22-48e0-ad06-9f4595901b72/65685c20-1334-11ef-9f8d-838c388c326d/file-1lw8oycbn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l8la6rik/8b350f4d-948a-4095-a004-5c8116776062/a60740c0-3efb-11ed-8e30-11b9f1cb84fb/file-1l8la6ril.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lfq9svjl/bc19241f-0f31-4930-9f1d-f97a67d8717c/92b63210-cc4f-11ed-8d1a-d560c3385a80/file-1lfq9svjm.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxmoxg/bddbd042-adc7-47f2-8492-6edfb8c4f19c/00793740-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxmoxh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc8eefiq/e8de7882-e991-47ff-999d-2ea9ce2e3c3f/ccd33f20-8716-11ed-92bb-e57e64e1a06d/file-1lc8eefir.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7g5yy/9381a58a-1f21-4b26-abca-dc94a3e72b11/6c8ce9a0-05bf-11ed-8617-d71e6444d1a0/file-1l5p7g5yz.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leeeen4s/b4b3dede-ba28-4c8d-af23-57eb35c25759/f55c2dc0-b1fb-11ed-afd2-4f73ef263f1c/file-1leeeen4t.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw8pero8/601e15c3-bb4a-487e-b3ff-b7b116b4f031/2e363180-1336-11ef-9f8d-838c388c326d/file-1lw8pero9.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/Fyl9Vf0KWiyB0u0XJ/tctGFeitNT0FkdGQUNU4VngNUHo8n/XoAkkZjSps5tTrqaC7VVGN/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/3cwPhLtSMMbYWfmCP/EDmxTKI6c4GYJb2RPfMLqbjGkhNpF/mgqePz0BAb9zmlk9xrkpWi/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265335/exam_images/oszgg4ddvwq2jkpn9psx.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7buiqub/55e764a3-4414-4eff-841a-7ce0120a8c5f/1bed5420-25ff-11ed-9c74-c5a04899a045/file-1l7buiquc.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263858/exam_images/lnhcuunfwlwjtih9updz.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263722/exam_images/n4gew2oz06txcsbjoxw5.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266109/exam_images/qg9czmccpl9q9jom4wds.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6dn1sg2/70ac6607-e35b-496d-816f-33268be95f9a/3e6d8930-132f-11ed-941a-4dd6502f33e3/file-1l6dn1sg3.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264051/exam_images/ijwqqvgif3jlthy0b2tu.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264827/exam_images/atidu8b68l8usxxqwsp4.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnvub5/791e93e6-65e8-4ec4-acf1-6dac1a0eab27/74eb5a10-cde5-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnvub6.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264717/exam_images/ycmesahtltfopnsuq3ta.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266363/exam_images/t4j7msbaucclnkye1pro.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266007/exam_images/gqh1muo7zwcf1cjiz3fa.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264566/exam_images/c0o3n3zizv5saixvrv1t.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264122/exam_images/hwpbih4rxxssh7b0owih.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9umae/dcd93d6b-ea7f-4626-9b1a-432648434cd3/99642f60-0655-11ed-903e-c9687588b3f3/file-1l5q9umaf.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leq0ppnt/4e6ea50d-9a35-4fc8-8434-5621ca292f68/d132e190-b85f-11ed-8195-4f3c56fa1eb5/file-1leq0ppnu.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw1m11ba/909a9c7c-64f8-4f41-9021-c0d2d0a9d656/7865b860-0f4f-11ef-ad52-af909612c772/file-1lw1m11bb.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwhhd8gk/a86d447c-ca28-4c88-a2d0-2170f95b6e56/c85f5d40-1809-11ef-b156-f754785ad3ce/file-1lwhhd8gl.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwhhhojv/60e45a27-3bf3-4540-bde6-9ba4a53c1429/440a25b0-180a-11ef-b156-f754785ad3ce/file-1lwhhhojw.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwap0izi/fd6cee3f-e437-4a65-830c-d11ed7b1c7a4/3378e3e0-144e-11ef-860c-d121cbcdd1fc/file-1lwap0izj.png?format=png\")
![\"AIEEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwj5c728/3dbfb32f-aec8-426a-86ca-495696731c70/859754f0-503c-11ec-87d6-07ff3bfe6155/file-1kwj5c729.png\")
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 6797,
"subject": "General Science",
"question": "If in a $$\\Delta ABC$$ $$a\\,{\\cos ^2}\\left( {{C \\over 2}} \\right) + c\\,{\\cos ^2}\\left( {{A \\over 2}} \\right) = {{3b} \\over 2},$$ then the sides $$a, b$$ and $$c$$ :",
"options": [
{
"text": "satisfy $$a+b=c$$"
},
{
"text": "are in A.P"
},
{
"text": "are in G.P "
},
{
"text": "are in H.P"
}
],
"answer": "are in A.P",
"solution": "**Answer:** are in A.P\n\nIf $$a\\,{\\cos ^2}\\left( {{C \\over 2}} \\right) + c\\,{\\cos ^2}\\left( {{A \\over 2}} \\right) = {{3b} \\over 2}$$\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265911/exam_images/oijmrfmejq2eg8qdfmzd.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264957/exam_images/mahs6riio0a6znddsa2r.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265818/exam_images/u883z0iudbpn9rsvcpfy.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l14vgj3y/f007ae64-29a9-4ba7-a06b-961129209c7b/fea546d0-ab5f-11ec-ba67-db8ddd9d738c/file-1l14vgj3z.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263594/exam_images/bip7efus8gpxzmiurj3w.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwd5jqj1/376c6606-3ddf-4373-89b3-5a3bdffa64e6/6d099bd0-15a8-11ef-99cc-63deb36dcc4a/file-1lwd5jqj2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbuiql/3cbb58ae-c17d-4fc8-9ae2-606226e918d5/7db780d0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbuiqm.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267782/exam_images/zwhawncrjdcscooeyd20.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5xuvzep/d1d9533f-57ee-41d4-a9fe-c531cfba2e5b/5b97b000-0a81-11ed-808f-7334e319a570/file-1l5xuvzeq.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5xuwndt/2f07a224-3889-4add-a96c-03b381708852/6e1d0d10-0a81-11ed-808f-7334e319a570/file-1l5xuwndu.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264994/exam_images/bkaah5pafcac92p7jsty.webp\")
\nCase-I :\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267172/exam_images/rziw7accnofwe87kl8ml.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266183/exam_images/zoc6r8adwxvl4esfzzpb.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5hv330/3bf12c29-312e-4d43-a5a3-c05d3e6a864c/6c2ba9c0-ad16-11ed-8a8c-4d67f5492755/file-1le5hv331.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4nlbkg/60bbafa8-f162-4063-8d33-6f852bb91330/ef943d00-10fb-11ef-aaa0-17ca36a32505/file-1lw4nlbkh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lwgkotns/48dc3e18-87b2-4fd8-a3b9-f30ede35fcf3/fd635a80-1789-11ef-b185-b7bdfa18e39b/file-jaoe38c1lwgkotnt.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwexk63n/0315e6c4-33b6-4400-9e04-b71afe90ed53/c1552330-16a2-11ef-afc3-d53e649859cd/file-1lwexk63o.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265538/exam_images/ms3gmbxipmmuyc7ugdvd.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9sp3c/20ce78f3-fabd-4839-b609-84bc56bb070f/63f06880-0655-11ed-903e-c9687588b3f3/file-1l5q9sp3d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5v5wwhx/6e6ad712-3c5b-4e48-ba35-59fbeae28bd0/1fa05b50-0906-11ed-a790-b11fa70c8a36/file-1l5v5wwhy.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5ykeybm/9c9e29ec-41a0-441a-a6fa-621f2f2c715d/2f5b3b10-0ae5-11ed-a51c-73986e88f75f/file-1l5ykeybn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7nny14a/71f56c77-eb10-4874-b074-072a3d7a554d/cce478a0-2c7e-11ed-a18d-5933e4fde865/file-1l7nny14b.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7q9vxzm/482edcf8-ee96-45ef-83c7-507c2a3fb4dc/2cff2b20-2dee-11ed-a744-1fb8f3709cfa/file-1l7q9vxzn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkn5295/e47cf7ce-efd4-4752-9bbb-34637d3af9f9/c36b1990-6798-11ee-97fe-41fa1903ca4f/file-6y3zli1lnkn5296.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwapcytl/545aa1a6-edf5-4e15-9137-baf3d0f62d19/8d6a2390-144f-11ef-860c-d121cbcdd1fc/file-1lwapcytm.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2xo2gn/47c8e21d-8001-480c-b87e-4bf1dfc70a37/1fd43f80-d4a5-11ee-bdd1-01c80c3e2d9a/file-1lt2xo2go.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266517/exam_images/rfwj0y3zp4ddisjyhq5w.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwac04ho/c0187de2-d2dd-45ec-8dc9-5238b7084f24/51988fc0-141b-11ef-aad1-15919f5484e3/file-1lwac04hp.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264552/exam_images/ooaanpllsv7wqqhp4zns.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264168/exam_images/piff2idkgob8rasda4s6.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbx3j2/7964fa43-5af9-4fc1-96c1-0a4102aef97d/c56543e0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbx3j3.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwawri93/7135ddd1-f250-4342-8c29-b0037fcb56d9/81a82a80-146c-11ef-a0c7-b1f23fa7cdc4/file-1lwawri94.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264141/exam_images/ldgk0sluqe0qfofg0aga.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264060/exam_images/ievtu0uoxd4s4ljgzkmw.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264557/exam_images/vr7gusba4nyfa18dc5vf.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5faj1z/fe5f5c16-ed0d-474b-8146-8a10ec18f85d/5e420070-ad0c-11ed-a86d-8dfe0389db88/file-1le5faj20.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4jkysz/b1310f77-4268-428b-aeb5-13316c958a9a/413519a0-10ec-11ef-8836-ed8ff380bbe9/file-1lw4jkyt6.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264435/exam_images/to8fimmktaygs2umiznl.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lj7pt113/8d56bc6b-c920-4fcb-9e3c-2bea678573a5/22404c90-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pt13t.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lj7pw2t5/50f84c21-de84-4885-acda-aeb2d0014e55/76fea290-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pw2t6.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lj7pvyrh/71535ada-ad4c-4284-8e05-1212a5834ebf/73de76d0-114c-11ee-b657-a1fe60b76246/file-6y3zli1lj7pvyri.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267710/exam_images/ndmd3v807ggdl7pdczek.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7299,
"subject": "General Science",
"question": "A survey shows that 63% of the people in a city read newspaper A whereas 76% read\nnewspaper B. If x% of the people read both the newspapers, then a possible value of x can be:",
"options": [
{
"text": "37"
},
{
"text": "65"
},
{
"text": "29"
},
{
"text": "55"
}
],
"answer": "55",
"solution": "**Answer:** 55\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266258/exam_images/s2o6kxbybhgir34w3g6x.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljaginws/88d6b048-49b9-4f92-be73-c0b995c4c9c3/29c085b0-12ce-11ee-9fec-35dc3a4c73b2/file-6y3zli1ljaginwt.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lx36wai5/74d61a18-6cda-4479-99f2-da165f8c36c1/14310dd0-23fa-11ef-9cd5-572f1863d2cb/file-6y3zli1lx36wai6.png?format=png\")
\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264798/exam_images/cehho8pjccdpteeui8ch.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lek5dmcv/3fd2ee94-309d-415f-8593-3aa8daeaf873/5aaf9ff0-b525-11ed-ba8b-d3643e26f082/file-1lek5dmcw.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxw9ix7m/ee1f9082-9d8b-46f7-b6e1-9aa6d676fa80/4cfc2f30-6b3f-11ec-8608-9b519146e1b7/file-1kxw9ix7n.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266113/exam_images/yoallvxjvujp2wtehbma.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoggtre/0457091e-f79d-43cd-92f6-312124caeaea/e227d0a0-ccae-11ee-aa98-13f456b8f7af/file-6y3zli1lsoggtrf.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1e6sji/400a79ad-eefc-44fe-9456-362749aed957/294e71e0-d3cc-11ee-a50b-bb659a2e1d74/file-1lt1e6sjj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4q8kfi/88bdaca2-4aca-49af-a105-f44c4a414e43/484b38e0-1106-11ef-9d25-fd88fb6fcd30/file-1lw4q8kfj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwgf0viw/7dffe883-cd31-494e-b982-28a57ce172bd/d5c35180-1773-11ef-bded-616e4abb66b9/file-1lwgf0vix.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266475/exam_images/xlctrwy5tbqvjebu61yj.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265867/exam_images/lvhph7vjigkfkhimhf1y.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265739/exam_images/vsdoxcyq1slggciks1wh.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2yegk8/aaaae12e-3775-4328-9765-0804dbc280e4/fdc1fd80-d4a7-11ee-bdd1-01c80c3e2d9a/file-1lt2yegk9.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264479/exam_images/nm6k8ccujpr0qdcb1imw.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxw9pbpa/316e0bc4-d33f-4ef5-89c3-8879b661f4ca/ff0879e0-6b3f-11ec-8608-9b519146e1b7/file-1kxw9pbpb.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264030/exam_images/c4whvpojuhs4abrlpy4l.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265203/exam_images/f7qd8qi2cxp1fjerej1f.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265668/exam_images/sxvhou3saadfcy0c0w3l.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwjjr9ui/956a8127-b9f9-4bb1-93a5-f77f19c4361a/e8b04fa0-5074-11ec-8f7b-45f8e5a35226/file-1kwjjr9uj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l99tn43d/ebd74024-c280-415b-bc94-0859668a0f7c/89bb0690-4c7a-11ed-b94d-45a8040c2a81/file-1l99tn43e.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l59j3d6d/8a4b3248-26d5-486b-9475-fbc8a7d508b1/6bc27690-fd20-11ec-8035-4341fba75e09/file-1l59j3d6h.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5yktc03/7dba7f27-1a1c-4810-bc97-87e79ff18cdf/bf409530-0ae6-11ed-a51c-73986e88f75f/file-1l5yktc04.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7psetoj/75213353-b6a6-48ee-b558-8da28250e29e/d5b7d130-2da9-11ed-8542-f96181a425b5/file-1l7psetok.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw3avn6c/d5ed79f8-059d-4be3-a5b7-c41490773adc/6f4ef630-103d-11ef-94ca-5b39b659c816/file-1lw3avn6d.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263971/exam_images/d94fpagqlokdwur1hypq.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264731/exam_images/mfh7nksrzqack2i9ykdk.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265540/exam_images/uwujwuudzsbrhbcutpgm.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264905/exam_images/ubp63holkjbubzmm73xt.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265477/exam_images/migtvndabwdpqcwl9h0o.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267693/exam_images/loayka4cghxne7uqkofa.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265627/exam_images/uw3nb7jjk16aarugxdjz.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267104/exam_images/idr7pfamynl59pilcm5t.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264797/exam_images/ajzrnczcpsrafxwpsdbk.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwjjv06e/aee1807a-0488-47e6-945f-12a768cd6dab/5074a460-5075-11ec-8f7b-45f8e5a35226/file-1kwjjv06f.jpeg\")
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7479,
"subject": "General Science",
"question": "![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5gmnar6/e2444175-9dd2-4655-a827-d624ec162bae/ce00a120-0107-11ed-a5c5-a7461ac77783/file-1l5gmnar7.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9xuo0/da6e885c-cdde-4480-b9d1-c22df8a701e5/f34a6800-0655-11ed-903e-c9687588b3f3/file-1l5q9xuo1.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbzt6e/130dc0b4-c7c5-40cb-813c-5bded9b079b4/10d39160-2c50-11ed-9dc0-a1792fcc650d/file-1l7nbzt6f.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7qa6xtq/139436a6-31d3-4607-a05c-66e2b0bc9d4d/5ec55de0-2def-11ed-a744-1fb8f3709cfa/file-1l7qa6xtr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssjo3c/b55f47a7-9a89-4d60-875b-d115e046f61f/b6651380-2f50-11ed-85dd-19dc023e9ad1/file-1l7ssjo3d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7xrag3x/1171f510-f750-4a4d-8bf7-c934d96c9030/b8102bd0-320b-11ed-b25b-a342e8d3e498/file-1l7xrag3y.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lf41l718/e00587c4-d3cd-4c76-ac3c-9d17149d6abd/35cb41c0-c016-11ed-89c7-f9c9b186ec3c/file-1lf41l719.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5hxc6e/38f5f2ee-c758-49c2-a1e2-99e79ae87394/aad05860-ad16-11ed-8a8c-4d67f5492755/file-1le5hxc6f.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgxuayzx/c68e4c1e-f2ed-4189-96d3-77fa5b78bc40/7473b5d0-e445-11ed-97cc-4f9b4bf32610/file-1lgxuayzy.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljg2jcjv/40b71be8-c597-4834-92a0-31af71c5e4d0/30e031b0-15e4-11ee-a5f8-af44f69476e0/file-6y3zli1ljg2jcjw.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw8vj8jo/f80a9888-77a1-4003-a6a8-8ff4c790fc36/21421350-134e-11ef-8bbe-1b4949638519/file-1lw8vj8jp.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw8vkatq/d9a6291d-037a-470f-95f4-e32e235ad67c/3ed3cee0-134e-11ef-8bbe-1b4949638519/file-1lw8vkatr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwjjmzcw/b86a6d68-7b5e-434f-a89c-eef8431d96b0/715950f0-5074-11ec-8a49-997c456a58f9/file-1kwjjmzcx.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7bukhhd/39a595d1-0f44-4de9-a640-89709cea02f6/4c50df10-25ff-11ed-9c74-c5a04899a045/file-1l7bukhhe.png?format=png\")