The locus of the centres of the circles,
Question: The locus of the centres of the circles, which touch the circle, $x^{2}+y^{2}=1$ externally, also touch the $\mathrm{y}$-axis and lie in the first quadrant, is :$y=\sqrt{1+4 x}, x \geq 0$$x=\sqrt{1+4 y}, y \geq 0$$x=\sqrt{1+2 y}, y \geq 0$$\mathrm{y}=\sqrt{1+2 \mathrm{x}}, \mathrm{x} \geq 0$Correct Option: , 4 Solution: $\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}=|\mathrm{h}|+1$ $\Rightarrow x^{2}+y^{2}=x^{2}+1+2 x$ $\Rightarrow y^{2}=1+2 x$ $\Rightarrow y=\sqrt{1+2 x} ; \quad x \geq 0$...
Read More →Solve the following
Question: If $\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+c$, where $c$ is $a$ constant of integration, then $g(-1)$ is equal to :$-\frac{5}{2}$1$-\frac{1}{2}$$-1$Correct Option: 1 Solution: Let $x^{2}=t$ $2 x d x=d t$ $\Rightarrow \frac{1}{2} \int \mathrm{t}^{2} \cdot \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\frac{1}{2}\left[-\mathrm{t}^{2} \cdot \mathrm{e}^{-\mathrm{t}}+\int 2 \mathrm{t} \cdot \mathrm{e}^{-\mathrm{t}} \cdot \mathrm{dt}\right]$ $=\frac{1}{2}\left(-t^{2} \cdot e^{-t}\right)+\left(-t \...
Read More →The value of
Question: If $\cos ^{-1} \mathrm{x}-\cos ^{-1} \frac{\mathrm{y}}{2}=\alpha$ where $-1 \leq \mathrm{x} \leq 1,-2 \leq \mathrm{y} \leq 2, \mathrm{x} \leq \frac{\mathrm{y}}{2}$ then for all $x, y, 4 x^{2}-4 x y \cos \alpha+y^{2}$ is equal to$4 \sin ^{2} \alpha-2 x^{2} y^{2}$$4 \cos ^{2} \alpha+2 \mathrm{x}^{2} \mathrm{y}^{2}$$4 \sin ^{2} \alpha$$2 \sin ^{2} \alpha$Correct Option: , 3 Solution: $\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha$ $\cos \left(\cos ^{-1} x-\cos ^{-1} \frac{y}{2}\right)=\cos \...
Read More →If the line a x+y=c, touches both the curves
Question: If the line $a x+y=c$, touches both the curves $x^{2}+y^{2}=1$ and $y^{2}=4 \sqrt{2} x$, then $|c|$ is equal to :$1 / 2$2$\sqrt{2}$$\frac{1}{\sqrt{2}}$Correct Option: , 3 Solution: Tangent to $\mathrm{y}^{2}=4 \sqrt{2} \mathrm{x}$ is $\mathrm{y}=\mathrm{mx}+\frac{\sqrt{2}}{\mathrm{~m}}$ it is also tangent to $x^{2}+y^{2}=1$ $\Rightarrow\left|\frac{\sqrt{2} / \mathrm{m}}{\sqrt{1+\mathrm{m}^{2}}}\right|=1 \Rightarrow \mathrm{m}=\pm 1$ $\Rightarrow$ Tagent will be $y=x+\sqrt{2}$ or $y=-x-...
Read More →Prove the following
Question: The sum $1+\frac{1^{3}+2^{3}}{1+2}+\frac{1^{3}+2^{3}+3^{3}}{1+2+3}+\ldots$ $\left.+\frac{1^{3}+2^{3}+3^{3}+\ldots .+15^{3}}{1+2+3+\ldots .+15}-\frac{1}{2}(1+2+3+\ldots .+15)\right]$12401860660620Correct Option: , 4 Solution: $\operatorname{Sum}=\sum_{n=1}^{15} \frac{1^{3}+2^{3}+\ldots n^{3}}{1+2+\ldots .+n}-\frac{1}{2} \cdot \frac{15.16}{2}$ $=\sum_{n=1}^{15} \frac{n(n+1)}{2}-60$ $=\sum_{n=1}^{15} \frac{n(n+1)(n+2-(n-1))}{6}-60$ $=\frac{15.16 .17}{6}-60=620$...
Read More →If 5 x+9=0 is the directrix of the hyperbola
Question: If $5 x+9=0$ is the directrix of the hyperbola $16 x^{2}-9 y^{2}=144$, then its corresponding focus is :$\left(-\frac{5}{3}, 0\right)$$(5,0)$$(-5,0)$$\left(\frac{5}{3}, 0\right)$Correct Option: , 3 Solution: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ $\mathrm{a}=3, \mathrm{~b}=4 \ \mathrm{e}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$ corresponding focus will be $(-$ ae, 0$)$ i.e., $(-5,0)$....
Read More →A spherical iron ball of radius 10 cm is coated with a layer of
Question: A spherical iron ball of radius $10 \mathrm{~cm}$ is coated with a layer of ice of uniform thickness that melts at a rate of $50 \mathrm{~cm}^{3} / \mathrm{min}$. When the thickness of the ice is $5 \mathrm{~cm}$, then the rate at which the thickness (in $\mathrm{cm} / \mathrm{min}$ ) of the ice decreases, is :$\frac{1}{9 \pi}$$\frac{5}{6 \pi}$$\frac{1}{18 \pi}$$\frac{1}{36 \pi}$Correct Option: , 3 Solution: $\mathrm{V}=\frac{4}{3} \pi\left((10+\mathrm{h})^{3}-10^{3}\right)$ $\frac{\ma...
Read More →The smallest natural number n,
Question: The smallest natural number $\mathrm{n}$, such that the coefficient of $x$ in the expansion of $\left(x^{2}+\frac{1}{x^{3}}\right)^{n}$ is ${ }^{\mathrm{n}} \mathrm{C}_{23}$, is :35382358Correct Option: , 2 Solution: $\mathrm{T}_{\mathrm{r}}=\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-2 \mathrm{r}} \cdot \mathrm{x}^{-3 \mathrm{r}}$ $2 n-5 r=1 \Rightarrow 2 n=5 r+1$ for $\mathrm{r}=15 . \mathrm{n}=38$ smallest value of $\mathrm{n}$ i...
Read More →Let λ be a real number for which the system of linear equations
Question: Let $\lambda$ be a real number for which the system of linear equations $x+y+z=6$ $4 x+\lambda y-\lambda z=\lambda-2$ $3 x+2 y-4 z=-5$ has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation.$\lambda^{2}-3 \lambda-4=0$$\lambda^{2}-\lambda-6=0$$\lambda^{2}+3 \lambda-4=0$$\lambda^{2}+\lambda-6=0$Correct Option: , 2 Solution: $\mathrm{D}=0$ $\left|\begin{array}{ccc}1 1 1 \\ 4 \lambda -\lambda \\ 3 2 -4\end{array}\right|=0 \Rightarrow \lambda=3$...
Read More →The tangent and normal to the ellipse
Question: The tangent and normal to the ellipse $3 x^{2}+5 y^{2}=32$ at the point $P(2,2)$ meet the $x$-axis at $Q$ and $R$, respectively. Then the area (in sq. units) of the triangle $P Q R$ is :$\frac{14}{3}$$\frac{16}{3}$$\frac{68}{15}$$\frac{34}{15}$Correct Option: , 3 Solution: $3 x^{2}+5 y^{2}=32$ $\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(2,2)}=-\frac{3}{5}$ Tangent : $\mathrm{y}-2=-\frac{3}{5}(\mathrm{x}-2) \Rightarrow \mathrm{Q}\left(\frac{16}{3}, 0\right)$ Normal : $\mathrm{y}-2=\...
Read More →A perpendicular is drawn from a point on the line
Question: A perpendicular is drawn from a point on the line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}$ to the plane $x+y+z=3$ such that the foot of the perpendicular $Q$ also lies on the plane $x-y+z=3$. Then the co-ordinates of Q are:$(2,0,1)$$(4,0,-1)$$(-1,0,4)$$(1,0,2)$Correct Option: 1 Solution: Let point $P$ on the line is $(2 \lambda+1,-\lambda-1, \lambda)$ foot of perpendicular $Q$ is given by $\frac{\mathrm{x}-2 \lambda-1}{1}=\frac{\mathrm{y}+\lambda+1}{1}=\frac{\mathrm{z}-\lambda}{1}=\f...
Read More →If both the mean and the standard deviation
Question: If both the mean and the standard deviation of 50 observations $\mathrm{x}_{1}, \mathrm{x}_{2} \ldots, \mathrm{x}_{50}$ are equal to 16 , then the mean of $\left(x_{1}-4\right)^{2},\left(x_{2}-4\right)^{2}, \ldots . .\left(x_{50}-4\right)^{2}$ is :525380480400Correct Option: , 4 Solution: $\operatorname{Mean}(\mu)=\frac{\sum \mathrm{x}_{\mathrm{i}}}{50}=16$ standard deviation $(\sigma)=\sqrt{\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{50}-(\mu)^{2}}=16$ $\Rightarrow(256) \times 2=\frac{\su...
Read More →The distance of the point having position vector
Question: The distance of the point having position vector $-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ from the straight line passing through the point $(2,3,-4)$ and parallel to the vector, $6 \hat{i}+3 \hat{j}-4 \hat{k}$ is :7$4 \sqrt{3}$$2 \sqrt{13}$6Correct Option: 1 Solution: $\mathrm{AD}=\left|\frac{\overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}\right|=\sqrt{61}$ $\Rightarrow \quad P D=\sqrt{A P^{2}-A D^{2}}=\sqrt{110-61}=7$...
Read More →If a < 0 and z = (1 + i)2 / a - i,
Question: If $a0$ and $z=\frac{(1+i)^{2}}{a-i}$, has magnitude $\sqrt{\frac{2}{5}}$, then $\bar{z}$ is equal to :$-\frac{3}{5}-\frac{1}{5} i$$-\frac{1}{5}+\frac{3}{5} \mathrm{i}$$-\frac{1}{5}-\frac{3}{5} \mathrm{i}$$\frac{1}{5}-\frac{3}{5} \mathrm{i}$Correct Option: , 3 Solution: Given $a0$ $z=\frac{(1+i)^{2}}{a-i}=\frac{2 i(a+i)}{a^{2}+1}$ Also $|z|=\sqrt{\frac{2}{5}} \Rightarrow \frac{2}{\sqrt{a^{2}+1}}=\sqrt{\frac{2}{5}} \Rightarrow a=3$ So $\bar{z}=\frac{-2 i(3-i)}{10}=\frac{-1-3 i}{5}$...
Read More →If the coefficients of
Question: If the coefficients of $x^{2}$ and $x^{3}$ are both zero, in the expansion of the expression $\left(1+a x+b x^{2}\right)(1-3 x)^{15}$ in powers of $x$, then the ordered pair $(a, b)$ is equal to :$(28,315)$$(-54,315)$$(-21,714)$$(24,861)$Correct Option: 1 Solution: Coefiicient of $\mathrm{x}^{2}={ }^{15} \mathrm{C}_{2} \times 9-3 \mathrm{a}\left({ }^{15} \mathrm{C}_{1}\right)+\mathrm{b}=0$ $\Rightarrow-45 \mathrm{a}+\mathrm{b}+{ }^{15} \mathrm{C}_{2} \times 9=0$ ..............(i) Also,...
Read More →If Q(0,-1,-3) is the image of the point
Question: If $\mathrm{Q}(0,-1,-3)$ is the image of the point $\mathrm{P}$ in the plane $3 x-y+4 z=2$ and $R$ is the point $(3,-1,-2)$, then the area (in sq. units) of $\triangle \mathrm{PQR}$$\frac{\sqrt{65}}{2}$$\frac{\sqrt{91}}{4}$$2 \sqrt{13}$$\frac{\sqrt{91}}{2}$Correct Option: , 4 Solution: $\mathrm{R}$ lies on the plane. $\mathrm{DQ}=\frac{|1-12-2|}{\sqrt{9+1+16}}=\frac{13}{\sqrt{26}}=\sqrt{\frac{13}{2}}$ $\Rightarrow P Q=\sqrt{26}$ Now, $\mathrm{RQ}=\sqrt{9+1}=\sqrt{10}$ $\Rightarrow \mat...
Read More →Prove the following
Question: $\lim _{n \rightarrow \infty}\left(\frac{(n+1)^{1 / 3}}{n^{4 / 3}}+\frac{(n+2)^{1 / 3}}{n^{4 / 3}}+\ldots \ldots+\frac{(2 n)^{1 / 3}}{n^{4 / 3}}\right)$ is equal to :$\frac{4}{3}(2)^{4 / 3}$$\frac{3}{4}(2)^{4 / 3}-\frac{4}{3}$$\frac{3}{4}(2)^{4 / 3}-\frac{3}{4}$$\frac{4}{3}(2)^{3 / 4}$Correct Option: , 3 Solution: $\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n}\left(\frac{n+r}{n}\right)^{1 / 3}$ $=\int_{0}^{1}(1+x)^{1 / 3} \mathrm{dx}=\frac{3}{4}\left(2^{4 / 3}-1\right)$...
Read More →If a1, a2, a3, …., an are in A.P. and
Question: If $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots \ldots . . \mathrm{a}_{\mathrm{n}}$ are in A.P. and $a_{1}+a_{4}+a_{7}+\ldots \ldots \ldots+a_{16}=114$, then $\mathrm{a}_{1}+\mathrm{a}_{6}+\mathrm{a}_{11}+\mathrm{a}_{16}$ is equal to :38987664Correct Option: , 3 Solution: $a_{1}+a_{4}+a_{7}+a_{10}+a_{13}+a_{16}=114$ $\Rightarrow \frac{6}{2}\left(a_{1}+a_{16}\right)=114$ $\Rightarrow a_{1}+a_{16}=38$ So, $\mathrm{a}_{1}+\mathrm{a}_{6}+\mathrm{a}_{11}+\mathrm{a}_{16}=\frac{4}{...
Read More →ABC is a triangular park with AB = AC = 100 metres.
Question: $\mathrm{ABC}$ is a triangular park with $\mathrm{AB}=\mathrm{AC}=100$ metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at $\mathrm{A}$ and $\mathrm{B}$ are $\cot ^{-1}(3 \sqrt{2})$ and $\operatorname{cosec}^{-1}(2 \sqrt{2})$ respectively, then the height of the tower (in metres) is :$10 \sqrt{5}$$\frac{100}{3 \sqrt{3}}$2025Correct Option: , 3 Solution: $\cot \alpha=3 \sqrt{2}$ $\ \operatorname{cosec} \beta=2 \sqrt{2}$ So, ...
Read More →Prove the following
Question: If $\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}$ $=\mathrm{A}\left(\tan ^{-1}\left(\frac{\mathrm{x}-1}{3}\right)+\frac{f(\mathrm{x})}{\mathrm{x}^{2}-2 \mathrm{x}+10}\right)+\mathrm{C}$ where $\mathrm{C}$ is a constant of integration, then :$\mathrm{A}=\frac{1}{27}$ and $f(\mathrm{x})=9(\mathrm{x}-1)$$\mathrm{A}=\frac{1}{81}$ and $f(\mathrm{x})=3(\mathrm{x}-1)$$\mathrm{A}=\frac{1}{54}$ and $f(\mathrm{x})=9(\mathrm{x}-1)^{2}$$\mathrm{A}=\frac{1}{54}$ and $f(\mathrm{x})=3(\mathrm{x}-1)...
Read More →If the length of the perpendicular from
Question: If the length of the perpendicular from the point $(\beta, 0, \beta)(\beta \neq 0)$ to the line, $\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$ is $\sqrt{\frac{3}{2}}$, then $\beta$ is equal to :$-1$2$-2$1Correct Option: 1 Solution: One of the point on line is $\mathrm{P}(0,1,-1)$ and given point is $Q(\beta, 0, \beta)$. So, $\overrightarrow{\mathrm{PQ}}=\beta \hat{\mathrm{i}}-\hat{\mathrm{j}}+(\beta+1) \hat{\mathrm{k}}$ Hence, $\beta^{2}+1+(\beta+1)^{2}-\frac{(\beta-\beta-1)^{2}}{2}=\frac...
Read More →If α and β are the roots
Question: If $\alpha$ and $\beta$ are the roots of the quadratic equation, $x^{2}+x \sin \theta-2 \sin \theta=0, \theta \in\left(0, \frac{\pi}{2}\right)$, then $\frac{\alpha^{12}+\beta^{12}}{\left(\alpha^{-12}+\beta^{-12}\right)(\alpha-\beta)^{24}}$ is equal to :$\frac{2^{6}}{(\sin \theta+8)^{12}}$$\frac{2^{12}}{(\sin \theta-8)^{6}}$$\frac{2^{12}}{(\sin \theta-4)^{12}}$$\frac{2^{12}}{(\sin \theta+8)^{12}}$Correct Option: , 4 Solution: $\frac{\alpha^{12}+\beta^{12}}{\left(\frac{1}{\alpha^{12}}+\f...
Read More →Let f : R → R be differentiable at
Question: Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be differentiable at $\mathrm{c} \in \mathrm{R}$ and $f(c)=0$. If $\mathrm{g}(\mathrm{x})=|f(\mathrm{x})|$, then at $\mathrm{x}=\mathrm{c}, \mathrm{g}$ is :differentiable if $f^{\prime}(\mathrm{c})=0$not differentiabledifferentiable if $f^{\prime}$ (c) $\neq 0$not differentiable if $f^{\prime}(\mathrm{c})=0$Correct Option: 1 Solution: $g^{\prime}(c)=\lim _{h \rightarrow 0} \frac{|f(c+h)|-|f(c)|}{h}$ $=\lim _{h \rightarrow 0} \frac{|f(c+h)|}{h}...
Read More →Let A (3, 0, -1), B (2, 10, 6) and C (1, 2, 1) be the vertices
Question: Let $\mathrm{A}(3,0,-1), \mathrm{B}(2,10,6)$ and $\mathrm{C}(1,2,1)$ be the vertices of a triangle and $\mathrm{M}$ be the midpoint of $\mathrm{AC}$. If $\mathrm{G}$ divides $\mathrm{BM}$ in the ratio, $2: 1$, then $\cos (\angle \mathrm{GOA})$ (O being the origin) is equal to :$\frac{1}{\sqrt{30}}$$\frac{1}{6 \sqrt{10}}$$\frac{1}{\sqrt{15}}$$\frac{1}{2 \sqrt{15}}$Correct Option: , 3 Solution: $G$ is the centroid of $\triangle A B C$ $G \equiv(2,4,2)$ $\overrightarrow{\mathrm{OG}}=2 \ha...
Read More →The line x=y touches a circle at the point (1,1).
Question: The line $x=y$ touches a circle at the point $(1,1)$. If the circle also passes through the point $(1,-3)$, then its radius is :$3 \sqrt{2}$3$2 \sqrt{2}$2Correct Option: , 3 Solution: Equation of circle can be written as $(\mathrm{x}-1)^{2}+(\mathrm{y}-1)^{2}+\lambda(\mathrm{x}-\mathrm{y})=0$ It passes through $(1,-3)$ $16+\lambda(4)=0 \Rightarrow \lambda=-4$ So $(x-1)^{2}+(y-1)^{2}-4(x-y)=0$ $\Rightarrow x^{2}+y^{2}-6 x+2 y+2=0$ $\Rightarrow r=2 \sqrt{2}$...
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