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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