Let the line L be the projection of the line
Question: Let the line $\mathrm{L}$ be the projection of the line $\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$ in the plane $x-2 y-z=3$. If $d$ is the distance of the point $(0,0,6)$ from $\mathrm{L}$, then $\mathrm{d}^{2}$ is equal to__________ Solution: $L_{1}: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$ for foot of $\perp \mathrm{r}$ of $(1,3,4)$ on $\mathrm{x}-2 \mathrm{y}-\mathrm{z}-3=0$ $(1+t)-2(3-2 t)-(4-t)-3=0$ $\Rightarrow \mathrm{t}=2$ So foot of $\perp r \triangleq(3,-1,2)$ point of int...
Read More →The sum of the roots of the equation
Question: The sum of the roots of the equation $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$, is :$\log _{2} 14$$\log _{2} 11$$\log _{2} 12$$\log _{2} 13$Correct Option: , 2 Solution: $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$ $\log _{2}\left(2^{x+1}\right)-\log _{2}\left(3+2^{x}\right)^{2}+\log _{2}\left(10-2^{-x}\right)=0$ $\log _{2}\left(\frac{2^{x+1} \cdot\left(10-2^{-x}\right)}{\left(3+2^{x}\right)^{2}}\right)=0$ $\frac{2\left(10.2^...
Read More →Prove the following
Question: If $y \frac{d y}{d x}=x\left[\frac{y^{2}}{x^{2}}+\frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi^{\prime}\left(\frac{y^{2}}{x^{2}}\right)}\right], x0, \phi0$, and $y(1)=-1$, then $\phi\left(\frac{y^{2}}{4}\right)$ is equal to:$4 \phi(2)$$4 \phi(1)$$2 \phi$ (1)$\phi$ (1)Correct Option: , 2 Solution: Let, $\mathrm{y}=\mathrm{tx}$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{t}+\mathrm{x} \frac{\mathrm{dt}}{\mathrm{dx}}$ $\therefore \operatorname{tx}\left(\mathrm{t}+\mathrm{x} \frac{\mathrm{...
Read More →The sum of all integral values
Question: The sum of all integral values of $\mathrm{k}(\mathrm{k} \neq 0)$ for which the equation $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$ in $x$ has no real roots, is____________ Solution: $\frac{2}{x-1}-\frac{1}{x-2}=\frac{2}{k}$ $x \in R-\{1,2\}$ $\Rightarrow \mathrm{k}(2 \mathrm{x}-4-\mathrm{x}+1)=2\left(\mathrm{x}^{2}-3 \mathrm{x}+2\right)$ $\Rightarrow \mathrm{k}(\mathrm{x}-3)=2\left(\mathrm{x}^{2}-3 \mathrm{x}+2\right)$ for $x \neq 3, \quad k=2\left(x-3+\frac{2}{x-3}+3\right)$ $x-3+\fra...
Read More →An angle of intersection of the curves,
Question: An angle of intersection of the curves, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and $x^{2}+y^{2}=a b, ab$, is :$\tan ^{-1}\left(\frac{a+b}{\sqrt{a b}}\right)$$\tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)$$\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$$\tan ^{-1}(2 \sqrt{\mathrm{ab}})$Correct Option: , 3 Solution: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, x^{2}+y^{2}=a b$ $\frac{2 x_{1}}{a^{2}}+\frac{2 y_{1} y^{\prime}}{b^{2}}=0$ $\Rightarrow \mathrm{y}_{1}{ }^{\prime}=\frac{-\...
Read More →Prove the following
Question: If $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{\mathrm{x}} \mathrm{y}+2^{\mathrm{y}} \cdot 2^{\mathrm{x}}}{2^{\mathrm{x}}+2^{\mathrm{x}+\mathrm{y}} \log _{e} 2}, \mathrm{y}(0)=0$, then for $\mathrm{y}=1$$(1,2)$$\left(\frac{1}{2}, 1\right)$$(2,3)$$\left(0, \frac{1}{2}\right]$Correct Option: 1 Solution: $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{\mathrm{x}}\left(\mathrm{y}+2^{\mathrm{y}}\right)}{2^{\mathrm{x}}\left(1+2^{\mathrm{y}} \ell \mathrm{n} 2\right)}$ $\Rightarrow \int \frac{\left(1+...
Read More →Solve this
Question: Let $\mathrm{z}=\frac{1-i \sqrt{3}}{2}, i=\sqrt{-1}$. Then the value of $21+\left(z+\frac{1}{z}\right)^{3}+\left(z^{2}+\frac{1}{z^{2}}\right)^{3}+\left(z^{3}+\frac{1}{z^{3}}\right)^{3}+\ldots+\left(z^{21}+\frac{1}{z^{21}}\right)^{3}$ is____________ Solution: $Z=\frac{1-\sqrt{3} i}{2}=e^{-i \frac{\pi}{3}}$ $z^{r}+\frac{1}{z^{r}}=2 \cos \left(-\frac{\pi}{3}\right) r=2 \cos \frac{r \pi}{3}$ $\Rightarrow 21+\sum_{\mathrm{r}=1}^{21}\left(\mathrm{z}^{\mathrm{r}}+\frac{1}{\mathrm{z}^{\mathrm{...
Read More →The Boolean expression
Question: The Boolean expression $(p \wedge \sim q) \Rightarrow(q \vee \sim p)$ is equivalent to:$\mathrm{q} \Rightarrow \mathrm{p}$$\mathrm{p} \Rightarrow \mathrm{q}$$\sim \mathrm{q} \Rightarrow \mathrm{p}$$\mathrm{p} \Rightarrow \sim \mathrm{q}$Correct Option: , 2 Solution: $\therefore(\mathrm{p} \wedge \sim \mathrm{q}) \Rightarrow(\mathrm{q} \vee \sim \mathrm{p})$ $\equiv p \Rightarrow q$ So, option (2) is correct....
Read More →If the truth value of the Boolean expression
Question: If the truth value of the Boolean expression $((\mathrm{p} \vee \mathrm{q}) \wedge(\mathrm{q} \rightarrow \mathrm{r}) \wedge(\sim \mathrm{r})) \rightarrow(\mathrm{p} \wedge \mathrm{q}) \quad$ is false then the truth values of the statements $p, q, r$ respectively can be:T F TF F TT F FF T FCorrect Option: 3 Solution:...
Read More →Let ABC be a triangle with A (-3,1)
Question: Let $\mathrm{ABC}$ be a triangle with $\mathrm{A}(-3,1)$ and $\angle \mathrm{ACB}=\theta, 0\theta\frac{\pi}{2} .$ If the equation of the median through $\mathrm{B}$ is $2 \mathrm{x}+\mathrm{y}-3=0$ and the equation of angle bisector of $C$ is $7 x-4 y-1=0$, then $\tan \theta$ is equal to:$\frac{1}{2}$$\frac{3}{4}$$\frac{4}{3}$2Correct Option: , 3 Solution: $\therefore \quad M\left(\frac{a-3}{2}, \frac{b+1}{2}\right)$ lies on $2 x+y-3=0$ $\Rightarrow 2 \mathrm{a}+\mathrm{b}=11$ .....(i)...
Read More →Solve this
Question: The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{t=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ is:$\frac{1}{2} \tan ^{-1}(2)$$\frac{1}{2} \tan ^{-1}(4)$$\tan ^{-1}(4)$$\frac{1}{4} \tan ^{-1}$Correct Option: , 2 Solution: $L=\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \sum_{r=0}^{2 n-1} \frac{1}{1+4\left(\frac{r}{n}\right)^{2}}$ $\Rightarrow \mathrm{L}=\int_{0}^{2} \frac{1}{1+4 \mathrm{x}^{2}} \mathrm{dx}$ $\Rightarrow \mathrm{L}=\left.\frac{1}{2} \tan ^{-1}(2 \mathrm{x})\rig...
Read More →If the sum of an infinite GP
Question: If the sum of an infinite GP a, ar, ar $^{2}$, ar $^{3}, \ldots$ is 15 and the sum of the squares of its each term is 150 , then the sum of $a r^{2}, a r^{4}, a r^{6}, \ldots$ is :$\frac{5}{2}$$\frac{1}{2}$$\frac{25}{2}$$\frac{9}{2}$Correct Option: , 2 Solution: Sum of infinite terms : $\frac{a}{1-r}=15$ .....(i) Series formed by square of terms: $a^{2}, a^{2} r^{2}, a^{2} r^{4}, a^{2} r^{6} \ldots . .$ Sum $=\frac{\mathrm{a}^{2}}{1-\mathrm{r}^{2}}=150$ $\Rightarrow \frac{\mathrm{a}}{1...
Read More →The locus of mid-points of the line segments
Question: The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is :$9 x^{2}+4 y^{2}+18 x+8 y+145=0$$36 \mathrm{x}^{2}+16 \mathrm{y}^{2}+90 \mathrm{x}+56 \mathrm{y}+145=0$$36 x^{2}+16 y^{2}+108 x+80 y+145=0$$36 x^{2}+16 y^{2}+72 x+32 y+145=0$Correct Option: , 3 Solution: General point on $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is $A(2 \cos \theta, 3 \sin \theta)$ given $\mathrm{B}(-3,-5)$ midpoint $C\left(\frac{2 \cos \theta-...
Read More →Solve the Following Questions
Question: If $\int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} d x=\frac{1}{14}\left(u x+v \log _{e}\left(4 e^{x}+7 e^{-x}\right)\right)+C$ where $C$ is a constant of integration, then $u+v$ is equal to Solution: $\int \frac{2 e^{x}}{4 e^{x}+7 e^{-x}} d x+3 \int \frac{e^{-x}}{4 e^{x}+7 e^{-x}} d x$ $=\int \frac{2 \mathrm{e}^{2 \mathrm{x}}}{4 \mathrm{e}^{2 \mathrm{x}}+7} \mathrm{~d} \mathrm{x}+3 \int \frac{\mathrm{e}^{-2 \mathrm{x}}}{4+7 \mathrm{e}^{-2 \mathrm{x}}} \mathrm{dx}$ Let $\quad 4 \mathrm...
Read More →Solve this
Question: If $\mathrm{A}=\left(\begin{array}{ll}\frac{1}{\sqrt{5}} \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} \frac{1}{\sqrt{5}}\end{array}\right), \mathrm{B}=\left(\begin{array}{ll}1 0 \\ i 1\end{array}\right), i=\sqrt{-1}$, and $Q=A^{\mathrm{T}} B A$, then the inverse of the matrix $\mathrm{A} \mathrm{Q}^{2021} \mathrm{~A}^{\mathrm{T}}$ is equal to :$\left(\begin{array}{cc}\frac{1}{\sqrt{5}} -2021 \\ 2021 \frac{1}{\sqrt{5}}\end{array}\right)$$\left(\begin{array}{cc}1 0 \\ -2021 i 1\end{array}\r...
Read More →Prove the following
Question: If $\alpha=\lim _{x \rightarrow \pi / 4} \frac{\tan ^{3} x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}$ and $\beta=\lim _{x \rightarrow 0}(\cos x)^{\cot x}$ are the roots of the equation, $a x^{2}+b x-4=0$, then the ordered pair $(a, b)$ is :$(1,-3)$$(-1,3)$$(-1,-3)$$(1,3)$Correct Option: , 4 Solution: $\alpha=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan ^{3} x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)} ; \frac{0}{0}$ form Using L Hopital rule $\alpha=\lim _{x \rightarrow \frac{\pi}{4...
Read More →Negation of the statement
Question: Negation of the statement $(p \vee r) \Rightarrow(q \vee r)$ is: $\mathrm{p} \wedge \sim \mathrm{q} \wedge \sim \mathrm{r}$$\sim \mathrm{p} \wedge \mathrm{q} \wedge \sim \mathrm{r}$$\sim \mathrm{p} \wedge \mathrm{q} \wedge \mathrm{r}$$\mathrm{p} \wedge \mathrm{q} \wedge \mathrm{r}$Correct Option: 1 Solution: $\because \sim(\mathrm{A} \Rightarrow \mathrm{B})=\mathrm{A} \wedge \sim \mathrm{B}$ $\therefore \sim((p \vee r) \Rightarrow(q \vee r))$ $=(p \vee r) \wedge(\sim q \wedge \sim r)$ ...
Read More →An online exam is attempted by 50 candidates
Question: An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2 . The variance of marks obtained by 30 girls is also 2 . The average marks of all 50 candidates is 15 . If $\mu$ is the average marks of girls and $\sigma^{2}$ is the variance of marks of 50 candidates, then $\mu+\sigma^{2}$ is equal to Solution: $\sigma_{b}^{2}=2$ (variance of boys) $n_{1}=$ no. of boys $\bar{x}_{b}=12$ $\mathrm{n}_{2}=$ no. of girls $\sigm...
Read More →A plane P contains the line
Question: A plane P contains the line $x+2 y+3 z+1=0=x-y-z-6$ and is perpendicular to the plane $-2 x+y+z+8=0$. Then which of the following points lies on P?$(-1,1,2)$$(0,1,1)$$(1,0,1)$$(2,-1,1)$Correct Option: , 2 Solution: Equation of plane P can be assumed as $P: x+2 y+3 z+1+\lambda(x-y-z-6)=0$ $\Rightarrow \mathrm{P}:(1+\lambda) \mathrm{x}+(2-\lambda) \mathrm{y}+(3-\lambda) \mathrm{z}+1-6 \lambda=0$ $\Rightarrow \overrightarrow{\mathrm{n}}_{1}=(1+\lambda) \hat{\mathrm{i}}+(2-\lambda) \hat{\m...
Read More →The distance of the point (-1,2,-2)
Question: The distance of the point $(-1,2,-2)$ from the line of intersection of the planes $2 x+3 y+2 z=0$ and $x-2 y+z=0$ is :$\frac{1}{\sqrt{2}}$$\frac{5}{2}$$\frac{\sqrt{42}}{2}$$\frac{\sqrt{34}}{2}$Correct Option: , 4 Solution: $P_{1}: 2 x+3 y+2 z=0$ $\Rightarrow \overrightarrow{\mathrm{n}}_{1}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ $P_{2}: x-2 y+z=0$ $\Rightarrow \overrightarrow{\mathrm{n}}_{2}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\mathrm{k}$ Direction vector of line $L$ ...
Read More →Solve the Following Questions
Question: $3 \times 7^{22}+2 \times 10^{22}-44$ when divided by 18 leaves the remainder Solution: $3(1+6)^{22}+2 \cdot(1+9)^{22}-44=(3+2-44)=18 . \mathrm{I}$ $=-39+18 . \mathrm{I}$ $=(54-39)+18(\mathrm{I}-3)$ $=15+18 \mathrm{I}_{1}$ $\Rightarrow$ Remainder $=15$...
Read More →Two circles each of radius 5
Question: Two circles each of radius 5 units touch each other at the point $(1,2)$. If the equation of their common tangent is $4 \mathrm{x}+3 \mathrm{y}=10$, and $\mathrm{C}_{1}(\alpha, \beta)$ and $\mathrm{C}_{2}(\gamma, \delta)$, $\mathrm{C}_{1} \neq \mathrm{C}_{2}$ are their centres, then $|(\alpha+\beta)(\gamma+\delta)|$ is equal to Solution: Slope of line joining centres of circles $=\frac{4}{3}=\tan \theta$ $\Rightarrow \cos \theta=\frac{3}{5}, \sin \theta=\frac{4}{5}$ Now using parametri...
Read More →Let f : N → N be a function such that
Question: Let $\mathrm{f}: \mathbf{N} \rightarrow \mathbf{N}$ be a function such that $\mathrm{f}(\mathrm{m}+\mathrm{n})=\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})$ for every $\mathrm{m}, \mathrm{n} \in \mathbf{N} .$ If $\mathrm{f}(6)=18$ then $f(2) \cdot f(3)$ is equal to :6541836Correct Option: , 2 Solution: $f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$ Put $\mathrm{m}=1, \mathrm{n}=1$ $f(2)=2 f(1)$ Put $\mathrm{m}=2, \mathrm{n}=1$ $f(3)=f(2)+f(1)=3 f(1)$ $f(6)=2 f(3) \Rightarrow...
Read More →The value
Question: The value of $\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left(\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right)^{1 / 2} d x$ is:$\log _{4} 4$$\log _{e} 16$$2 \log _{\varepsilon} 16$$4 \log _{\mathrm{e}}(3+2 \sqrt{2})$Correct Option: , 2 Solution: $I=\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^{2}\right)^{1 / 2} d x$ $I=\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left|\frac{4 x}{x^{2}-1}\right| d x \Rightarrow I=2.4 \int_{0}^{1 / \sqrt...
Read More →Let A
Question: Let $A(\sec \theta, 2 \tan \theta)$ and $B(\sec \phi, 2 \tan \phi)$, where $\theta+\phi=\pi / 2$, be two points on the hyperbola $2 x^{2}-y^{2}=2$. 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 Solution: Since, point $A(\sec \theta, 2 \tan \theta)$ lies on the hyperbola $2 x^{2}-y^{2}=2$ Therefore, $2 \sec ^{2} \theta-4 \tan ^{2} \theta=2$ $\Rightarrow 2+2 \tan ^{2} \theta-4 \tan ^{2} \theta=2$ $\R...
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