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GATE [ME, 2017, 2 marks] Question : $P(0.3)$, $Q(0.5,4)$ and $R(1,5)$ are three points on the curve defined by $f(x)$. Numerical integration is carried out using both Trapezoidal rule and Simpson's rule within limits $x=0$ and $x=1$ for the curve. The difference between the two results will be (a) $0$ (b) $0.25$ (c) $0.5$ (d) $1$ Solution :   $x$     $0$     $0.5$     $1$     $y$     $3$     $4$     $5$   $h=\dfrac{1}{2}$ Using Trapezoidal rule $\displaystyle\int\limits_{a}^{b}F(x)dx=\dfrac{h}{2}[(y_0+y_{n})+2(y_1+y_2+\cdots+y_{n-1})]$ $\displaystyle\int\limits_{0}^{1}F(x)dx=\dfrac{1/2}{2}[(3+5)+2(4)]$ $\displaystyle\int\limits_{0}^{1}F(x)dx=4$ Using Simpson's rule $\displaystyle\int\limits_{a}^{b}F(x)dx=\dfrac{h}{3}[(y_0+y_n)+4(y_1+y_3+\cdots+y_{n-1})+2(y_2+y_4+\cdots+y_{n-2})]$ $\displaystyle\int\limits_{0}^{1}F(x)dx=\dfrac{1/2}{3}[(8+16)+4(4)]$ $\dis...

GATE [ME, 2016, 2 marks] Question : The error in numerically computing the integral $\displaystyle\int\limits_0^{\pi}(\sin x+\cos x)dx$ using the Trapezoidal rule with three intervals of equal lenght between $0$ and $\pi$ is____. Solution : Given $F(x)=\sin x+\cos x$ $h=\dfrac{b-a}{n}=\dfrac{\pi-0}{3}=\dfrac{\pi}{3}$  $x$   $0$   $\dfrac{\pi}{3}$   $\dfrac{2\pi}{3}$   $\pi$   $\sin x+\cos x$   $1$   $1.366$   $0.366$   $-1$  By Trapezoidal rule $\displaystyle\int\limits_{a}^{b}F(x)dx=\dfrac{h}{2}[(y_0+y_{n})+2(y_1+y_2+\cdots+y_{n-1})]$ $\displaystyle\int\limits_{0}^{\pi}\sin x+\cos xdx=\dfrac{\pi/3}{2}[(1-1)+2(1.366+0.366)]$ $\displaystyle\int\limits_{0}^{\pi}\sin x+\cos xdx=1.181$ By definite integration $\displaystyle\int\limits_{0}^{\pi}\sin x+\cos xdx=\left|-\cos x+\sin x\right|^{0}_{\pi}$ $\displaystyle\int\limits_{0}^{\pi}\sin...

GATE [EC, 2017, 1 mark] Question : Consider the $5\times 5$ matrix $$A=\begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 5 & 1 & 2 & 3 & 4\\ 4 & 5 & 1 & 2 & 3\\ 3 & 4 & 5 & 1 & 2\\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}$$ It is given that $A$ has only one real eigen value. Then the real eigen value of $A$ is (a) $-2.5$ (b) $0$ (c) $15$ (d) $25$ Solution : The characteristic equation is $|A-\lambda I|=0$ $\implies\begin{vmatrix} 1-\lambda & 2 & 3 & 4 & 5\\ 5 & 1-\lambda & 2 & 3 & 4\\ 4 & 5 & 1-\lambda & 2 & 3\\ 3 & 4 & 5 & 1-\lambda & 2\\ 2 & 3 & 4 & 5 & 1-\lambda \end{vmatrix}=0$ $R_1\to R_1+R_2+R_3+R_4+R_5$ $\implies\begin{vmatrix} 15-\lambda & 15-\lambda & 15-\lambda & 15-\lambda & 15-\lambda\\ 5 & 1-\lambda & 2 & 3 & 4\\ 4 & 5 & 1-\lambda & 2 & 3\\ 3 & 4 & 5 & 1-\lambda ...

GATE [EE, 2007, 2 marks] Question : The linear operation $L(x)$ is defined by the cross product $L(x)=b\times X$, where $b=\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}^T$ and $X=\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}^T$ are three dimensional vectors. The $3\times 3$ matrix $M$ of this operation sstisfies $$L(x)=M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$$ Then the eigen values of $M$ are (a) $0,+1,-1$ (b) $1,-1,1$ (c) $i,-i,1$ (d) $i,-i,0$ Solution : As $L(x)=M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$ and $L(x)=b\times X$ $\therefore M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=b\times X \quad\quad\ldots(1)$ Since $M$ is $3\times 3$ matrix, let $M=\begin{bmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9 \end{bmatrix}$ and $b\times X=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 0\\ x_1 & x_2 & x_3 \end{vmatrix}$ $=x_3\hat{i}+0\hat{j}-x_1\hat{k}$ $\begin{bmatrix} x_3 & 0 ...

GATE [ME, 2017, 2 marks] Question : Consider the differential equation $3y''(x)+27y(x)=0$ with initial conditions $y(0)=0$ and $y'(0)=2000$. The value of $y$ at $x=1$ is____. Solution : Given $3y''(x)+27y(x)=0$ $\implies (3D^2+27)y=0$ The auxiliary equation is $3D^2+27=0$ $\implies D=\pm 3i$ General solution is $y=C_1\cos 3x+C_2\sin 3x\quad\quad\quad\ldots\text{(1)}$ differentiating equation $(1)$ $y'=-3C_1\sin 3x+3C_2\cos 3x$ $y(0)=0$ $\implies C_1=0$ $y'(0)=2000$ $3C_2=2000$ $\implies C_2=\dfrac{200}{3}$ $\therefore$ the particular solution is $y=\dfrac{2000\sin 3x}{3}$ At $x=1$, $y=\dfrac{2000\sin 3}{3}$ $y=94.08$ CORRECT ANSWER : 94.08

GATE [EE, 2010, 2 marks] Question : For the differential equation $\displaystyle\frac{d^2x}{dt^2}+6\frac{dx}{dt}+8x=0$ with initial conditions $x(0)=1$ and $\displaystyle\frac{dx}{dt}\Bigg|_{t=0}=0$, the solution is (a) $x(t)=2e^{-6t}-e^{-2t}$ (b) $x(t)=2e^{-2t}-e^{-4t}$ (c) $x(t)=-e^{-6t}+2e^{-4t}$ (d) $x(t)=e^{-2t}+2e^{-4t}$ Solution : Given $\dfrac{d^2x}{dt^2}+6\dfrac{dx}{dt}+8x=0$ Let $\dfrac{d}{dt}=D$ so the equation becomes $\implies (D^2+6D+8)x=0$ the auxiliary equation is $\implies D^2+6D+8=0$ $\implies (D+2)(D+4)=0$ $\implies D=-2,-4$ the solution is $\implies x=c_1e^{-2t}+c_2e^{-4t}\quad\quad\quad\quad\ldots(1)$ differentiting equation $(1)$ $\implies\dfrac{dx}{dt}=-2c_1e^{-2t}-4c_2e^{-4t}\quad\quad\quad\quad\ldots(2)$ since $x(0)=1$ and $\dfrac{dx}{dt}\Bigg|_{t=0}=0$ putting values in equations $(1)$ and $(2)$ respectively $1=c_1+c_2\quad\quad\quad\quad\ldots(3)$ $0=-2c_1-4c_2\quad\quad\quad\quad\ldots(4)$ solving $3$ and $(4)$ $\implies c_1=2, c_2...

GATE [EE, 2016, 1 mark] Question : The value of the integral $$\oint\limits_{C}\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}dz$$ over the contour $|z|=1$, taken in the anti-clockwise direction, would be (a) $\dfrac{24\pi i}{13}$ (b) $\dfrac{48\pi i}{13}$ (c) $\dfrac{24}{13}$ (d) $\dfrac{12}{13}$ Solution : Let $f(z)=\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}$ Poles of $f(z)$ are $\dfrac{1}{2},2+i$ and $2-i$ Only $z=\dfrac{1}{2}$ lies inside the circle $|z|=1$ $\displaystyle \text{Residue, }R=\lim_{z\to 1/2}\left[\left(z-\dfrac{1}{2}\right)f(z)\right]$ $\displaystyle R=\lim_{z\to 1/2}\left[\dfrac{2z+5}{z^2-4z+5}\right]$ $\displaystyle R=\dfrac{2\left(\dfrac{1}{2}\right)+5}{\left(\dfrac{1}{2}\right)^2-4\left(\dfrac{1}{2}\right)+5}$ $\displaystyle R=\dfrac{24}{13}$ $\therefore$ by Residue theorem, we get $\displaystyle\oint\limits_{C}\dfrac{2z+5}{\left(z-\dfrac{1}{2}\right)(z^2-4z+5)}dz=2\pi iR=\dfrac{48\pi i}{13}$ CORRECT ANSWER : B

GATE [EC, 2015, 1 mark] Question : Let $z=x+iy$ be a complex variable. Consider that contour integartion is performed along the unit circle in anticlockwise direction. Which one of the following statements is NOT TRUE ? (a) The residue of $\dfrac{z}{z^2-1}$ at $z=1$ is $1/2$ (b) $\displaystyle\oint\limits_{C}z^2dz=0$ (c) $\displaystyle\dfrac{1}{2\pi i}\oint\limits_{C}\dfrac{1}{z}dz=1$ (d) $\bar{z}$ (complex conjugate of $z$) is analytical function Solution : $\rightarrow$ Residue of $\dfrac{z}{z^2-1}$ at $\displaystyle z=1=\lim_{z\to 1}(z-1)\dfrac{z}{z^2-1}$ $\displaystyle =\lim_{z\to 0}\dfrac{z}{z+1}$ $=\dfrac{1}{2}$ Option (A) is TRUE $\rightarrow$ $z^2$ is analytical at every point within and on circle $|z|=1$ by Cauchy's integral theorem, we get $\displaystyle\oint\limits_{C}z^2dz=0$ Option (B) is TRUE $\rightarrow\dfrac{1}{z}$ has a pole at $z=0$ $\therefore$ by Cauchy's integral theorem, we get $\displaystyle\oint\limits_{C}\dfrac{1}{z}dz=2\pi i\left[...

GATE [EC, EE, IN, 2012, 1 mark] Question : Given $\displaystyle f(z)=\dfrac{1}{z+1}-\dfrac{2}{z+3}$. If $C$ is a counter clockwise path in the $z$-plane such that $|z+1|=1$, the value of $\displaystyle\dfrac{1}{2\pi j}\oint\limits_{C}f(z)dz$ is (a) $-2$ (b) $-1$ (c) $1$ (d) $2$ Solution : Given $f(z)=\dfrac{1}{z+1}-\dfrac{2}{z+3}$ $\implies =\dfrac{-z+1}{(z+1)(z+3)}$ $\therefore f(z)$ has poles at $-1,-3$ $-3$ lie outside the circle $|z+1|=1$ $\therefore$ by Cauchy's Integral Formula and Cauchy's Integral Theorem, we have $\displaystyle\dfrac{1}{2\pi j}\oint\limits_{C}f(z)dz=\dfrac{1}{2\pi j}\left[\oint\limits_{C}\dfrac{1}{z+1}dz-0\right]$ $=\dfrac{2\pi j\left(\dfrac{-z+1}{z+3}\right)_{z=-1}}{2\pi j}$ $=\dfrac{-(-1)-1}{-1+3}$ $=\dfrac{2}{2}$ $=1$ CORRECT ANSWER : C

GATE [EE, 2013, 1 mark] Question : Square roots of $-i$, where $i=\sqrt{-1}$, are (a) $i, -i$ (b) $\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right),$ $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$ (c) $\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right),$ $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{\pi}{4}\right)$ (d) $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(-\dfrac{3\pi}{4}\right),$ $\cos\left(-\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$ Solution : As we know that $-i=i^{3}$ $=e^{i3\pi/2}$ $\therefore \sqrt{-i}=e^{i3\pi/2\times 1/2}$ $=\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$ Now $,-i=\dfrac{1}{i}=e^{-i\pi/2}$ $\implies (-i)^{1/2}=e^{-i\pi/2\times 1/2}$ $=e^{-i\pi/4}$ $=\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)$ CORRECT ANSWER : B

GATE [EC, 2012, 2 mark] Question : If $f(z)=C_0+C_1z^{-1}$, then $\displaystyle\oint\limits_{\text{unit circle}}\dfrac{1+f(z)}{z}dz$ is given by (a) $2\pi C_1$ (b) $2\pi(1+C_0)$ (c) $2\pi jC_1$ (d) $2\pi j(1+C_0)$ Solution : Let $\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{1+f(z)}{z}dz$ Given $f(z)=C_0+C_1z^{-1}$ $\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{1+C_0+C_1/z}{z}dz$ $\displaystyle I=\oint\limits_{\text{unit circle}}\dfrac{z(1+C_0)+C_1}{z^2}dz$ Singularities for $I$ is $z=0$ and $z=0$ lies inside the unit circle. $\therefore$ by Cauchy's integral theorem we have, $I=2\pi j\times(\text{residue of }\dfrac{z(1+C_0)+C_1}{z^2}\text{ at } z=0)$ $=2\pi j\left\{\dfrac{1}{1!}\dfrac{d}{dz}z^2\dfrac{z(1+C_0)+C_1}{z^2}\right\}_{z=0}$ $=2\pi j\{1+C_0\}_{z=0}$ $\therefore$ Answer $=2\pi j(1+C_0)$ CORRECT ANSWER : D

GATE [EC, 2010, 2 marks] Question : The residues of a comlex function $\displaystyle X(z)=\dfrac{1-2z}{z(z-1)(z-2)}$ at its poles are (a) $\displaystyle\dfrac{1}{2},-\dfrac{1}{2}$ and $1$ (b) $\displaystyle\dfrac{1}{2},\dfrac{1}{2}$ and $-1$ (c) $\displaystyle\dfrac{1}{2},1$ and $\displaystyle -\dfrac{3}{2}$ (d) $\displaystyle\dfrac{1}{2},-1$ and $\displaystyle\dfrac{3}{2}$ Solution : $X(z)=\dfrac{1-2z}{z(z-1)(z-2)}$ The poles are $z=0,1,2$ Residue at $z=0$ $\displaystyle =\lim_{z\to 0}(z-0)\frac{1-2z}{z(z-1)(z-2)}$ $=\dfrac{1-2\times 0}{(0-1)(0-2)}$ $=\dfrac{1}{2}$ Residue at $z=1$ $\displaystyle =\lim_{z\to 1}(z-1)\frac{1-2z}{z(z-1)(z-2)}$ $\dfrac{1-2\times 1}{1(1-2)}$ $=1$ Residue at $z=2$ $\displaystyle =\lim_{z\to 2}(z-2)\frac{1-2z}{z(z-1)(z-2)}$ $\dfrac{1-2\times 2}{2(2-1)}$ $=-\dfrac{3}{2}$ CORRECT ANSWER : C

GATE [CE, 2009, 2 marks] Question : The value of the integral $\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz$ (where $C$ is a closed curve given by $|z|=1$ is (a) $-\pi i$ (b) $\displaystyle\dfrac{\pi i}{5}$ (c) $\displaystyle\dfrac{2\pi i}{5}$ (d) $\pi i$ Solution : $\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz$ putting $(2z-1)(z-3)=0$ so the singularities are $z=\dfrac{1}{2}$ and $z=3$ but only $z=\dfrac{1}{2}$ lies inside the circle $|z|=1$ (the closed curve) hence by Cauchy's integral theorem $\displaystyle\dfrac{1}{2}\int\limits_{C}\dfrac{\left[\dfrac{\cos (2\pi z)}{(z-3)}\right]}{\left(z-\dfrac{1}{2}\right)}dz=\dfrac{1}{2}2\pi if(\dfrac{1}{2})$ where $f(z)=\dfrac{\cos (2\pi z)}{(z-3)}$ $\therefore\displaystyle\int\limits_{C}\dfrac{\cos (2\pi z)}{(2z-1)(z-3)}dz=\dfrac{2\pi i}{5}$ CORRECT ANSWER : C

GATE [CS, 2015, 2 marks] Question : If for non-zero $x$, $af(x)+bf\left(\dfrac{1}{x}\right)=\dfrac{1}{x}-25$ where $a\ne b$ then $\displaystyle\int\limits_1^2f(x)dx$ is (a) $\displaystyle\frac{1}{a^2-b^2}\left[a(\ln 2-25)\frac{47b}{2}\right]$ (b) $\displaystyle\frac{1}{a^2-b^2}\left[a(2\ln 2-25)-\frac{47b}{2}\right]$ (c) $\displaystyle\frac{1}{a^2-b^2}\left[a(2\ln 2-25)+\frac{47b}{2}\right]$ (d) $\displaystyle\frac{1}{a^2-b^2}\left[a(\ln 2-25)-\frac{47b}{2}\right]$ Solution : Given $af(x)+bf\left(\dfrac{1}{x}\right)=\dfrac{1}{x}-25\quad\quad\quad\ldots(1)$ replacing $\dfrac{1}{x}$ by $x$ in $(1)$ $\implies af\left(\dfrac{1}{x}\right)+bf(x)=x-25\quad\quad\quad\ldots(2)$ Solving $(1)$ and $(2)$ $\implies f(x)=\dfrac{1}{a^2-b^2}\left[a\left(\dfrac{1}{x}-25\right)-b(x-25)\right]$ $\displaystyle\int\limits_1^2f(x)dx$ $=\dfrac{1}{a^2-b^2}\left[a(\ln x-25x)-\left(\dfrac{x^2}{2}-25x\right)\right]_1^2$ $=\displaystyle\frac{1}{a^2-b^2}\left[a(\ln 2-25)\frac{47b}{2}\right]$ CORR...

GATE [EE, 2014, 2 marks] Question : To evaluate the double integral $\displaystyle\int\limits_0^8\left(\int\limits_{y/2}^{y/2+1}\left(\frac{2x-y}{2}\right)dx\right)dy$, we make the subsitution $\displaystyle u=\left(\frac{2x-y}{2}\right)$ and $\displaystyle v=\frac{y}{2}$. The integral will reduce to (a) $\displaystyle\int\limits_0^4\left(\int\limits_0^2 2u du\right)dv$ (b) $\displaystyle\int\limits_0^4\left(\int\limits_0^1 2u du\right)dv$ (c) $\displaystyle\int\limits_0^4\left(\int\limits_0^1 u du\right)dv$ (d) $\displaystyle\int\limits_0^4\left(\int\limits_0^2 u du\right)dv$ Solution : Given $\displaystyle\int\limits_0^8\left(\int\limits_{y/2}^{y/2+1}\left(\frac{2x-y}{2}\right)dx\right)dy$ $u=\left(\dfrac{2x-y}{2}\right)$ $\implies du=dx$ $x=y/2 \implies u=0$ $x=y/2+1 \implies u=1$ $v=\dfrac{y}{2}$ $\implies dv=\dfrac{dy}{2}$ $x=0\implies v=0$ $x=0\implies v=4$ $\therefore$ the integral reduces to $\displaystyle\int\limits_0^4\left(\int\limits_0^1 2u du\right)d...

GATE [EC, 2017, 2 marks] Question : The values of the integrals $\displaystyle\int\limits_0^1\left(\int\limits_0^1\frac{x-y}{(x+y)^3}dy\right)dx$ and $\displaystyle\int\limits_0^1\left(\int\limits_0^1\frac{x-y}{(x+y)^3}dx\right)dy$ are (a) same and equal to $0.5$ (b) same and equal to $-0.5$ (c) $0.5$ and $-0.5$ respectively (d) $-0.5$ and $0.5$ respectively Solution : $\displaystyle\int\limits_0^1\left(\int\limits_0^1\frac{x-y}{(x+y)^3}dy\right)dx$ $\displaystyle=\int\limits_0^1\left[\int\limits_0^1\left(\dfrac{2x}{(x+y)^3}-\dfrac{1}{(x+y)^2}\right)dy\right]dx$ $\displaystyle=\int\limits_0^1\left[\dfrac{-x}{(x+y)^2}+\dfrac{1}{x+y}\right]_0^1dx$ $\displaystyle=\int\limits_0^1\dfrac{1}{(x+1)^2}dx$ $=\left[\dfrac{-1}{x+1}\right]_0^1$ $=-\dfrac{1}{2}+1$ $=\dfrac{1}{2}=0.5$ Now $\displaystyle\int\limits_0^1\left(\int\limits_0^1\frac{x-y}{(x+y)^3}dx\right)dy$ $\displaystyle=\int\limits_0^1\left[\int\limits_0^1\left(\dfrac{1}{(x+y)^2}-\dfrac{2y}{(x+y)^3}\right)dx\right]...

GATE [CS, 2017, 1 mark] Question : If $\displaystyle f(x)=R\sin\left(\frac{\pi x}{2}\right)+S, f'\left(\frac{1}{2}\right)=\sqrt{2}$ and $\displaystyle\int\limits_0^1 f(x)dx=\frac{2R}{\pi}$, then the constants $R$ and $S$ are, respectively (a) $\displaystyle\frac{2}{\pi}$ and $\displaystyle\frac{16}{\pi}$ (b) $\displaystyle\frac{2}{\pi}$ and $0$ (c) $\displaystyle\frac{4}{\pi}$ and $0$ (d) $\displaystyle\frac{4}{\pi}$ and $\displaystyle\frac{16}{\pi}$ Solution : Given $f(x)=R\sin\left(\dfrac{\pi x}{2}\right)+S$ $\implies f'(x)=R\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{\pi x}{2}\right)$ Now $f'\left(\dfrac{1}{2}\right)=\sqrt{2}$ $\implies R\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{\pi}{4}\right)=\sqrt{2}$ $\implies R\left(\dfrac{\pi}{2}\right)\left(\dfrac{1}{\sqrt{2}}\right)=\sqrt{2}$ $\implies R=\dfrac{4}{\pi}$ We have $\displaystyle\int\limits_0^1 f(x)dx=\frac{2R}{\pi}$ $\implies\displaystyle\int\limits_0^1 R\sin\left(\dfrac{\pi x}{2}\right)+S dx=\fra...

Question : If $\displaystyle\int\limits_{0}^{1}\frac{dx}{1+x^{3/2}}$, then (a) $\ln 2 < I < \dfrac{\pi}{4}$ (b) $\log_{10} 2 < I < \ln 2$ (c) $\dfrac{\pi}{4} <I, \ln 2 < I$ (d) $\dfrac{\pi}{4} >I, \ln 2> I$ Solution : Given $I=\displaystyle\int\dfrac{x^3\sin^{-1}x^2}{\sqrt{1-x^4}}dx$ Let $\sin^{-1}x^2=t\implies x^2=\sin t$ $\implies d\sin^{-1}x^2=dt$ $\implies\dfrac{1}{\sqrt{1-x^4}}\cdot 2xdx=dt$ $\therefore I=\displaystyle\int\frac{t\sin t}{2}dt$ $\implies I=\dfrac{\sin t-t\cos t}{2}+C$ $\implies I=\dfrac{x^2-\left(\sin^{-1}x^2\right)\cdot\cos\left(\sin^{-1}x^2\right)}{2}+C$ $\implies I=\dfrac{x^2-\left(\sin^{-1}x^2\right)\cdot\cos\left(\cos^{-1}\sqrt{1-x^4}\right)}{2}+C$ $\implies I=\dfrac{x^2-\left(\sin^{-1}x^2\right)\sqrt{1-x^4}}{2}+C$

Question : An electron gun is placed inside a long solenoid of radius $R$ on its axis. The solenoid has $n \text{ turns/length}$ and carries a current $I$. The electron gun shoots an electron along the radius of the solenoid with speed $v$. If the electron does not hit the surface of the solenoid, maximum possible value of $v$ is (all symbols have their standard meaning) : (a) $\dfrac{2e\mu_0nIR}{m}$ (b) $\dfrac{e\mu_0nIR}{4m}$ (c) $\dfrac{e\mu_0nIR}{2m}$ (d) $\dfrac{e\mu_0nIR}{m}$ Solution : Top view of solenoid As we know maximum possible radius of electron $=\dfrac{R}{2}$ $\implies\dfrac{R}{2}=\dfrac{mv}{qB}=\dfrac{mv_{max}}{e(\mu_0nI)}$ $\therefore v_{max}=\dfrac{R}{2}\dfrac{e\mu_0nI}{m}$ $\fbox{$\therefore v_{max}=\dfrac{e\mu_0nIR}{2m}$}$

Question : An evacuated vessel weighs $50g$ when empty, $144g$ when filled with liquid of density $0.47 g/ml$ and $50.5 g$ when filled with an ideal gas at $760\text{ mm Hg }$ at $300K$. The molar mass of the ideal gas is (Given $R=0.0821\text{ L atm }K^{-1}mol^{-1}$) (a) $61.575$ (b) $130.98$ (c) $123.75$ (d) $47.87$ Solution : Weight of fluid $=144-50=94 g$ Density of liquid $=0.47 g/ml$ So, Volume of liquid $=\dfrac{94}{0.47}=200 ml=0.2 L$ Weight of glass $=50.5-50=0.5 g$ Now $PV=nRT$ $\implies 1\times0.2=\dfrac{0.5}{M}\times 0.0821\times 300$ $\implies M=61.575$

UPSC [CSE 2019 P1] Question 1(d) : If $$A=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix} \text{and } B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$$ then show that $AB=6I_3$. Use this result to solve the following system of equations : $$\begin{array}{r} 2x+y+z=5\\ x-y=0\\ 2x+y-z=1\\ \end{array}$$ Solution : Given $A=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix} \text{and } B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$ $\implies AB=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix}\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$ $\implies AB=\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 6 \end{bmatrix}=6\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}=6I...

UPSC [CSE 2019 P1] Question 1(a) : Let $f:\left[0,\dfrac{\pi}{2}\right]\to\textbf{R}$ be a continuous function such that $$f(x)=\frac{\cos ^2 x}{4x^2-\pi^2}, 0\le x<\frac{\pi}{2}$$ Find the value of $f\left(\dfrac{\pi}{2}\right).$  Solution : Given that $f:\left[0,\dfrac{\pi}{2}\right]\to\textbf{R}$ is a continuous function such that $f(x)=\dfrac{\cos ^2 x}{4x^2-\pi^2}, 0\le x<\dfrac{\pi}{2}$. Now, since $f$ is continuous on $\left[0,\dfrac{\pi}{2}\right]$, We have, $\displaystyle\lim_{x\to\pi/2^{-}}f(x)=f\left(\frac{\pi}{2}\right)$ $\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{\cos ^2 x}{4x^2-\pi^2}\quad\quad\ldots\left(\frac{0}{0}\right)$ form $\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{2\cos x(-\sin x)}{8x}$ $\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{-2\sin 2x}{8x}$ $\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\fr...

Question : For $m > 0, n > 0$, let $\displaystyle I_{m,n}=\int\limits_{0}^{1}x^m(\log x)^ndx$, then $I_{5,5}$ is given by (a) $-\dfrac{5!}{6^5}$ (b) $-\dfrac{5!}{5^5}$ (c) $-\dfrac{5!}{6^6}$ (d) $\dfrac{5!}{6^6}$ Solution : Given $\displaystyle I_{m,n}=\int\limits_{0}^{1}x^m(\log x)^ndx$ put $\log x =-t$ We get $\displaystyle I_{m,n}=(-1)^n\int\limits_{0}^{\infty}t^ne^{-(m+1)t}dt$ now put $(m+1)t=u$ $\implies\displaystyle I_{m,n}=(-1)^n\int\limits_{0}^{\infty}\left(\dfrac{u}{m+1}\right)^ne^{-u}\dfrac{du}{m+1}$ $\implies\displaystyle I_{m,n}=\frac{(-1)^n}{(m+1)^{n+1}}\int\limits_{0}^{\infty}u^ne^{-u}du$ As we know Gamma Function is $\displaystyle\left[\int\limits_{0}^{\infty}x^{(z-1)}e^{-x}dx=\Gamma(z)=(z-1)!\right]$ $\implies I_{m,n}=\dfrac{(-1)^n\Gamma(n+1)}{(m+1)^{n+1}}$ $\implies I_{5,5}=\dfrac{(-1)^5\Gamma(5+1)}{(5+1)^{5+1}}$ $\implies I_{5,5}=-\dfrac{5!}{6^6}$