MMP

1. The real matrix $A=\begin{pmatrix}a&-f&-g\\ f&a&h\\g&-h&a\end{pmatrix}$ is skew symmetric when
   1. $a=0$
   2. $f=0$
   3. $g=h$
   4. $f=g$ $
   Matrix $A$ is skew symmetric when $A_{ij}=-A_{ji}$. Hence, $a=0$
2. If $A$ and $B$ are matrices such that $AB=B$ and $BA=A$ then $A^2+B^2$ equals
1. 2AB
2. 2BA
3. A+B
4. AB
\begin{align*} A^2+B^2&=(BA)^2+(AB)^2\\ &=BABA+ABAB \end{align*} \begin{align*} A^2+B^2&=B\underline{AB}A+A\underline{BA}B\\ &=B\underline{BA}+A\underline{AB}\\ &=BA+AB=A+B \end{align*}
3. The average value of function $f(x)=4x^3$ in the interval 1 to 3 is
1. 15
2. 20
3. 30
4. 40
\begin{align*} <f(x)>&=\frac{\int_1^3f(x)dx}{\int_1^3dx}\\ &=\frac{\int_1^34^3dx}{\int_1^3dx}\\ &=\frac{[x^4]_1^3}{[x]_1^3}\\ &=\frac{81-1}{3-1}=40 \end{align*}
4. The matrix $\begin{pmatrix}8&x&0\\4&0&2\\12&6&0\end{pmatrix}$ will become singular if the value of $x$ is
   1. $4$
   2. $6$
   3. $8$
   4. $12$
   If the matrix is singular, its determinant is zero. $$det=-96+24x=0$$ Hence, $x=4$
5. The value of $$x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\dots} }}$$
1. $\sqrt{2}$
2. 1.6
3. $\sqrt{3}$
4. 0.8
$$x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\dots} }}=1+\frac{1}{x}$$ $$x^2-x-1=0$$ $$x=\frac{1\pm\sqrt{1+4}}{2}=1.6$$
6. The centre of the circle $\bar z z+(2+3i)\bar z+(2-3i)z+1=0$ is
   1. (2,3)
   2. (3,2)
   3. (-2,-3)
   4. (4,0)
   Let $z=x+iy$ \begin{align*} \therefore \bar z z+(2+3i)\bar z+(2-3i)z+1\\ =(x-iy)(x+iy)+(2+3i)(x-iy)\\+(2-3i)(x+iy)+1\\ =x^2+y^2+2x-2iy+3ix+3y\\+2x+2iy-3ix+3y+1\\ =x^2+y^2+4x+6y+1 \end{align*} $$\therefore\bar z z+(2+3i)\bar z+(2-3i)z+1=0$$ gives $$(x+2)^2+(y+3)^2-12=0$$ $(x+2)^2+(y+3)^2=12$ is equation of circle with centre $(-2,-3)$ and radius $\sqrt{12}$
7. Fourier transform of the function $f(x)=exp(-|x|)$ is
   1. $\frac{1}{\sqrt{2\pi}}\left[\frac{2}{1+k^2}\right]$
   2. $0$
   3. $\frac{1}{\sqrt{\pi}}\left[\frac{1}{k^2}\right]$
   4. $\frac{1}{\sqrt{2\pi}}\left[\frac{1}{1-k^2}\right]$
   $|x|=x$ if $x>0$ and $|x|=-x$ if $x<0$. Fourier transform of $f(x)$ is given by $$F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ikx}f(x)dx$$ \begin{align*} F(k)={\scriptstyle\frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{0}e^{-ikx}e^xdx+\int_{0}^{\infty}e^{-ikx}e^{-x}dx\right]} \end{align*} $$F(k)={\scriptstyle\frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{0}e^{x(1-ik)}dx+\int_{0}^{\infty}e^{-x(1+ik)}dx\right]}$$ $$F(k)=\frac{1}{\sqrt{2\pi}}\left[\frac{2}{1+k^2}\right]$$
8. The area of the triangle whose base is given by $\bar a=5\hat i-3\hat j+4\hat k$ and $\bar b=\hat j-\hat k$ is another side is :
   1. $\sqrt{50}/2$
   2. $\sqrt{61}/2$
   3. $\sqrt{14}/2$
   4. $\sqrt{51}/2$
   \begin{align*} &\text{Area of triangle }=\frac{1}{2}|\bar a\times\bar b|\\ &=\frac{1}{2}|-\hat i+5\hat j+5\hat k|=\sqrt{51}/2 \end{align*}
9. The position vector $\bar r=x\hat i+y\hat j+z\hat k$, $\bar\nabla.\left(r^2\bar r\right)$ is given by:
1. 0
2. $5r^2$
3. $r^2$
4. $3r^2$
$r^2=x^2+y^2+z^2$ $${\scriptscriptstyle\bar\nabla.\left(r^2\bar r\right)=\left(\hat i\frac{\partial}{\partial x}+\hat j\frac{\partial}{\partial y}+\hat k\frac{\partial}{\partial x}\right)\cdot\left(\left(x^2+y^2+z^2\right)\left(x\hat i+y\hat j+z\hat k\right)\right)}$$ $$\bar\nabla.\left(r^2\bar r\right)=5r^2$$
10. The eigenvalues of the matrix: $\begin{pmatrix}1&2&3\\0&4&7\\0&0&3\end{pmatrix}$ are:
    1. 1, 4, 3
    2. 3, 7, 3
    3. 7, 3, 2
    4. 1, 2, 3
    In upper or lower triangular matrix, eigenvalues are the diagonal elements, Hence (A)1, 4, 3
11. All solutions of the equation $e^z=-3$ are
1. $z=\ln \pi\ln 3,~n=\pm1,\pm2,\dots$
2. ${\scriptstyle z=\ln 3+i(2n+1)\pi,~n=0,\pm1,\pm2,\dots}$
3. ${\scriptstyle z=\ln 3+i~2n\pi,~n=0,\pm1,\pm2,\dots}$
4. $z=i3n\pi,~n=\pm1,\pm2,\dots$
Finding all solutions means to find all values of $z$ which will satify $e^z=-3$ \begin{align*} e^z&=-3\\ &=3(-1)\\ &={\scriptstyle3\left[\cos{(2n+1)\pi}+i\sin{(2n+1)\pi}\right]}\\ &=3e^{i(2n+1)\pi} \end{align*} Taking log on both sides we get $$z=\ln 3+i(2n+1)\pi~~~n=0,\pm1,\pm2,\dots$$
12. The solution of $\frac{dy}{dx}-y=e^{\lambda x}$ is :
1. $e^{-\lambda x}$
2. $\frac{1}{\lambda-1}e^{\lambda x}$
3. $e^{\lambda x}$
4. $\frac{1}{\lambda}e^{-\lambda x}$
Just by inspection or substituting $y=\frac{1}{\lambda-1}e^{\lambda x}$ in above equation one can verify that it is the solution. OR
The solution of $\frac{dy}{dx}+p(x)y=q(x)$ is given by $$y=\frac{\int u(x)q(x)dx+C}{u(x)}$$ where, $u(x)=exp\left(\int p(x)dx\right)$
Here $p(x)=-1$ and $q(x)=e^{\lambda x}$ $$u(x)=exp\left(-\int dx\right)=e^{-x}$$ \begin{align*} y&=\frac{\int e^{-x}e^{\lambda x}dx+C}{e^{-x}}\\ &=\frac{ e^{(\lambda-1) x}dx+C}{(\lambda-1)e^{-x}}\\ &=\frac{1}{\lambda-1}e^{\lambda x} \end{align*}
13. The function $f(z)=u(x,y)+iv(x,y)$ is analytic at $z=x+iy$. The value of $\nabla^2u$ at this point is:
1. 0
2. undefined
3. $\pi$
4. $e^{-\pi^2}$
The function $f(z)=u(x,y)+iv(x,y)$ is analytic at $z=x+iy$ means $u(x,y)$ and $v(x,y)$ satisfy Cauchy-Reiman equations namely $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$ $$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ $$\therefore\nabla^2u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$$ $$\nabla^2u=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}\right)+\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial y}\right)$$ \begin{align*} \nabla^2u&=\frac{\partial }{\partial x}\left(\frac{\partial v}{\partial y}\right)-\frac{\partial }{\partial y}\left(\frac{\partial v}{\partial x}\right)\\ &=\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial y\partial x}=0 \end{align*}
14. The value of the integral $\int\limits_Cdz\:z^2\:e^z$, where C is an open contour in the complex z-plane as shown in the figure below, is:
    
    1. $\frac{5}{e}+e$
    2. $e-\frac{5}{e}$
    3. $\frac{5}{e}-e$
    4. $-\frac{5}{e}-e$
    \begin{align*} &\int_Cdz\:z^2\:e^z\\ &=\int_Cd(x+iy)\:(x+iy)^2\:e^{(x+iy)} \end{align*} Along path (1) $x=1$, $dx=0$ and $y$ changes 0 to 1 \begin{align*} I_1&=i\int_0^1 (1+iy)^2e^{(1+iy)}dy\\ &=i\int_0^1 (1-y^2+2iy)e^{(1+iy)}dy \end{align*} Along path (2) $y=1$, $dy=0$ and $x$ changes 1 to $-1$ \begin{align*} I_2&=\int_{-1}^1 (x+i)^2e^{(x+i)}dx\\ &=\int_{-1}^1 (x^2-1+2ix)e^{(x+i)}dx \end{align*} Along path (3) $x=-1$, $dx=0$ and $y$ changes 1 to 0 \begin{align*} I_3&=i\int_{-1}^0 (-1+iy)^2e^{(-1+iy)}dy\\ &=i\int_{-1}^0 (1-y^2-2iy)e^{(-1+iy)}dy \end{align*} $$I=I_1+I_2+I_3=\frac{5}{e}+e$$
15. Let $\vec{a}$ and $\vec{b}$ be two distinct three-dimensional vectors. Then the component of $\vec{b}$ that is perpendicular to $\vec{a}$ is given by
1. $\frac{\vec{a}\times(\vec{b}\times\vec{a})}{a^2}$
2. $\frac{\vec{b}\times(\vec{b}\times\vec{a})}{b^2}$
3. $\frac{(\vec{a}.\vec{b})\vec{b}}{b^2}$
4. $\frac{(\vec{b}.\vec{a})\vec{a}}{a^2}$
\begin{align*} \frac{\vec{a}\times(\vec{b}\times\vec{a})}{a^2}&=\frac{(\vec{a}\cdot\vec{a})\vec{b}-(\vec{a}\cdot\vec{b})\vec{a}}{a^2}\\ &=\frac{a^2\vec{b}-ab\cos\theta\vec{a}}{a^2} \end{align*} \begin{align*} & \vec{a}\cdot\frac{\vec{a}\times(\vec{b}\times\vec{a})}{a^2}\\ &=\frac{a^2\vec{a}\cdot\vec{b}-ab\cos\theta\vec{a}\cdot\vec{a}}{a^2}\\ &=\frac{a^3b\cos\theta-a^3b\cos\theta}{a^2}\\ &=0 \end{align*} Hence answer is (A)
16. Let $p_n(x)$ (where $n = 0,1,2,\dots$) be a polynomial of degree $n$ with real coefficients, defined in the interval $2\leq n\leq 4$. If $\int_2^4p_n(x)p_m(x)dx=\delta_{nm}$, then
1. $p_0(x)=\frac{1}{\sqrt{2}}$ and $p_1(x)=\sqrt{\frac{3}{2}}(-3-x)$
2. $p_0(x)=\frac{1}{\sqrt{2}}$ and $p_1(x)=\sqrt{3}(3+x)$
3. $p_0(x)=\frac{1}{2}$ and $p_1(x)=\sqrt{\frac{3}{2}}(3-x)$
4. $p_0(x)=\frac{1}{\sqrt{2}}$ and $p_1(x)=\sqrt{\frac{3}{2}}(3-x)$
For $p_0(x)=\frac{1}{\sqrt{2}}$, we have \begin{align*} \int_2^4p_0(x)p_0(x)dx&=\frac{1}{2}\int_2^4dx\\ &=\frac{1}{2}[x]_2^4=1 \end{align*} Hence, $p_0(x)=\frac{1}{\sqrt{2}}$ is correct option. For, $p_1(x)=\sqrt{\frac{3}{2}}(3-x)$, we have \begin{align*} &\int_2^4p_1(x)p_1(x)dx\\ &=\frac{3}{2}\int_2^4(3-x)^2dx\\ &=\frac{3}{2}\int_2^4(9-6x+x^2)dx \end{align*} \begin{align*} &\int_2^4p_1(x)p_1(x)dx\\ &=\frac{3}{2}\left[9x-3x^2+\frac{x^3}{3}\right]_2^4\\ &=1 \end{align*} \begin{align*} &\int_2^4p_0(x)p_1(x)dx\\ &=\sqrt{\frac{3}{4}}\int_2^4(3-x)dx\\ &=\sqrt{\frac{3}{4}}\left[3x-\frac{x^2}{2}\right]_2^4\\ &=0 \end{align*} Hence, orthonormality condition is satified for $p_0(x)=\frac{1}{\sqrt{2}}$ and $p_1(x)=\sqrt{\frac{3}{2}}(3-x)$. Hence, answer is (D)
17. Which of the following is an analytic function of the complex variable $z = x + iy$ in the domain $| z |< 2$?
1. $(3+x-iy)^7$
2. $(1+x+iy)^4(7-x-iy)^3$
3. $(1-2x-iy)^4(3-x-iy)^3$
4. $(x+iy-1)^{1/2}$
A function is analytic if it satisfies Cauchy–Riemann condition i.e. a function $f(z)$ is analytic if $\frac{\partial f(z)}{\partial\bar z}=0$ Above functions can be written in terms of $z$ using $z=x+iy$, $\bar z=x-iy$, $x=\frac{z+\bar z}{2}$ and $y=\frac{z-\bar z}{2i}$ \begin{align*} &(3+x-iy)^7=(3+\bar z)^7\\ &(1+x+iy)^4(7-x-iy)^4\\ &=(1+z)^4(7-z)^3\\ &(1-2x-iy)^4(3-x-iy)^3\\ &=\left(1-2\frac{z+\bar z}{2}-i\frac{z-\bar z}{2i}\right)^4(3-z)^3\\ &=\left(1-2z-\frac{\bar z}{2}\right)^4(3-z)^3\\ &(x+iy-1)^{1/2}=(z-1)^{1/2} \end{align*} For the functions which contain $\bar z$, $\frac{\partial f(z)}{\partial\bar z}\neq0$. Hence, the functions in the options (A) and (C) are not analytical.
For the functions which are functions of only $z$ have $\frac{\partial f(z)}{\partial\bar z}=0$. Hence, we have option either (B) or (C).
The option (B) i.e. $f(z)=(1+z)^4(7-z)^3$ is a polynomial in $z$ and polynomial in $z$ is analytic everywhere.
For the option (C) i.e. $f(z)=(z-1)^{1/2}$, $f'(z)=\frac{1}{2\sqrt{z-1}}$ and this function is analytic only for $|z|<1$. Our domain is $|z|<2$. Hence, it not analytical over the domain $|z|<2$.
Hence, the correct option is (B).
18. Consider the matrix $M=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$
1. The eigenvalues of $M$ are
1. 0, 1, 2
2. 0, 0, 3
3. 1, 1, 1
4. -1, 1, 3
2. The exponential of M simplifies to ($I$ is the $3\times3$ identity matrix)
1. $e^M=I+\left(\frac{e^3-1}{3}\right)M$
2. $e^M=I+M+\frac{M^2}{2!}$
3. $e^M=I+3^3M$
4. $e^M=(e-1)M$
(A) $\begin{vmatrix} 1-\lambda&1&1\\1&1-\lambda&1\\1&1&1-\lambda \end{vmatrix}=0$ gives \begin{align*} &(1-\lambda)\left\{(1-\lambda)^2-1\right\}\\ &-\left\{(1-\lambda)-1\right\}+\left\{1-(1-\lambda)\right\}=0 \end{align*} $$\text{i.e.}\quad \lambda^2(3-\lambda)=0$$ Hence, $\lambda=0, 0,3$
(B)$$e^M=I+M+\frac{M^2}{2!}+\frac{M^3}{3!}+\dots$$ But $M^2=3M$, $M^3=3^2M$, $M^4=3^2M$, $M^n=3^{n-1}M$ $$e^M=I+M+M\frac{3^1}{2!}+M\frac{3^2}{3!}+\dots$$ $$e^M=I+M\left(1+\frac{3^1}{2!}+\frac{3^2}{3!}+\dots\right)$$ $$e^M=I+\frac{M}{3}\left(3+\frac{3^2}{2!}+\frac{3^3}{3!}+\dots\right)$$ \begin{align*} e^M&=I\\ &+\frac{M}{3}\left(\left[1+3+\frac{3^2}{2!}+\frac{3^3}{3!}+\dots\right]-1\right)\\ &=I+\frac{M}{3}\left(e^3-1\right) \end{align*} Hence, answer is (i)
19. A $2\times2$ matrix $A$ has eigenvalues $e^{i\pi/5}$ and $e^{i\pi/6}$. The smallest value of $n$ such that $A^n = I$ is
1. 20
2. 30
3. 60
4. 120
As $e^{i\pi/5}$ and $e^{i\pi/6}$ are eigenvalues of matrix $A$, eigenvalues of matrix of matrix $A^n$ will be $e^{in\pi/5}$ and $e^{in\pi/6}$. However, $A^n$ is identity matrix whose eignevalues will be 1 and 1, the diagonal elements. Hence, $e^{in\pi/5}=1$ and $e^{ni\pi/6}=1$ $$e^{in\pi/5}=\cos{(n\pi/5)}+i\sin{(n\pi/5)}=1$$ $$\Rightarrow \cos{(n\pi/5)}=1 \text{ and } sin{(n\pi/5)}=0$$ Hence, $n\pi/5=2m\pi$, where $m=0,1,2,\dots$, hence, $n=10m$, where $m=0,1,2,\dots$ Hence, $n=0, 10,20, 30,40,50,60\dots$ $$e^{in\pi/6}=\cos{(n\pi/6)}+i\sin{(n\pi/6)}=1$$ $$\Rightarrow \cos{(n\pi/6)}=1 \text{ and } sin{(n\pi/6)}=0$$ Hence, $n\pi/6=2m\pi$, where $m=0,1,2,\dots$, hence, $n=12m$, where $m=0,1,2,\dots$ Hence, $n=0, 12,24, 36,48,60,72\dots$
The common minimum value of $n$ is 60. Hence, answer is (C)
20. The unit normal vector at the point $\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right)$ on the surface of the ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$
1. $\frac{bc\hat i+ca\hat j+ab\hat k}{\sqrt{a^2+b^2+c^2}}$
2. $\frac{a\hat i+b\hat j+c\hat k}{\sqrt{a^2+b^2+c^2}}$
3. $\frac{b\hat i+c\hat j+a\hat k}{\sqrt{a^2+b^2+c^2}}$
4. $\frac{\hat i+\hat j+\hat k}{\sqrt{3}}$
The unit normal is defined as $\hat n=\frac{\nabla f}{|\nabla f|}$ $$\nabla f=2\left(\frac{x}{a^2}\hat i+\frac{y}{b^2}\hat j+\frac{z}{c^2}\hat k\right)$$ \begin{align*} &\left(\nabla f\right)_{\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right)}\\ &=\frac{2}{\sqrt{3}}\left(\frac{\hat i}{a^2}+\frac{\hat j}{b^2}+\frac{\hat k}{c^2}\right) \end{align*} \begin{align*} &\left(|\nabla f|\right)_{\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right)}\\ &=\frac{2}{\sqrt{3}}\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}\\ &=\frac{2}{\sqrt{3}}\sqrt{\left(\frac{b^2c^2+a^2c^2+b^2a^2}{a^2b^2c^2}\right)} \end{align*} $$\frac{\nabla f}{|\nabla f|}=\frac{bc\:\hat i+ac\:\hat j+ab\:\hat k}{\sqrt{b^2c^2+a^2c^2+b^2a^2}}$$
21. The Taylor expansion of the function $\ln{(\cosh x)}$, where $x$ is real, about the point $x= 0$ starts with the following terms:
1. $-\frac{1}{2}x^2+\frac{1}{12}x^4+\cdots$
2. $\frac{1}{2}x^2-\frac{1}{12}x^4+\cdots$
3. $-\frac{1}{2}x^2+\frac{1}{6}x^4+\cdots$
4. $\frac{1}{2}x^2+\frac{1}{6}x^4+\cdots$
We will require following identities: $$\sinh{x}=\frac{e^x-e^{-x}}{2}$$ $$\cosh{x}=\frac{e^x+e^{-x}}{2}$$ $$\tanh{x}=\frac{\sinh{x}}{\cosh{x}}$$ $$\frac{d}{dx}\sinh{x}=\cosh{x}$$ $$\frac{d}{dx}\cosh{x}=\sinh{x}$$ $$\frac{d}{dx}\tanh{x}=\text{sech }^2{x}$$ $$\frac{d}{dx}\text{sech }{x}=-\text{sech }{x}\tanh{x}$$ Taylor series of $f(x)$ about $x=0$ is given by \begin{align*} f(x)&=f(x)+\frac{1}{1!}f'(0)x+\frac{1}{2!}f''(0)x^2\\ &+\frac{1}{3!}f'''(0)x^3+\frac{1}{4!}f''''(0)x^4+.. \end{align*} \begin{align*} &f(x)=\ln{(\cosh{x})}\\ &\Rightarrow f(0)=0\\ &f'(x)=\tanh{x} \\ &\Rightarrow f'(0)=0\\ &f''(x)=\text{sech }^2{x}=1-\tanh^2{x}\\ &\Rightarrow f''(0)=1\\ &f'''(x)=-2\text{sech }^2{x}\tanh{x}\\ &\Rightarrow f'''(0)=0\\ f''''(x)&=4\text{sech }^2{x}\tanh^2{x}-2\text{sech }^4{x} \\ &\Rightarrow f''''(0)=-2 \end{align*} $$f(x)=\frac{1}{2!}x^2-\frac{1}{12}x^4+\cdots$$ Hence, answer is (B)
22. The value of the integral $\int_C \frac{z^3dz}{z^2-5z+6}$, where $C$ is a closed contour defined by the equation $2|z|- 5 = 0$, traversed in the anti-clockwise direction, is
1. $-16\pi i$
2. $16\pi i$
3. $8\pi i$
4. $2\pi i$
\begin{align*} &\int_C\frac{z^3dz}{z^2-5z+6} \quad C:|z|=\frac{5}{2}=2.5\\ &= \int_C\frac{z^3dz}{(z-2)(z-3)}\\ &= \int_C\frac{\frac{z^3}{(z-3)}}{(z-2)}dz \end{align*} The function has a pole of order 1 at $z=2$ which lies inside $C$. According to residue theorem \begin{align*} &\oint\limits_Cf(z)dz\\ &=2\pi i\times&\text{sum of residues at}\\ &&\text{ poles inside }C \end{align*} In general the residue at a pole of order $m$ at $z=z_0$ is \begin{eqnarray} \begin{array}{c}R=\frac{1}{(m-1)!}\lim\limits_{z\to z_0}\left\{\frac{d^{m-1}\left((z-z_0)^mf(z)\right)}{dz^{m-1}}\right\}_{z=z_0}\end{array} \end{eqnarray} For simple pole of order 1 $$R=\lim\limits_{z\to z_0}\left\{(z-z_0)f(z_0)\right\}$$ $$R=\lim\limits_{z\to 2}\left\{(z-2)\frac{\frac{z^3}{(z-3)}}{(z-2)}\right\}=-8$$ $$\oint_Cf(z)dz=2\pi i\times(-8)=-16\pi i$$ Hence, answer is (A)
23. The function $f(x)$ obeys the differential equation $\frac{d^2f}{dx^2}-(3 - 2i)f = 0$ and satisfies the conditions $f(0) = 1$ and $f(x)\rightarrow \infty$ as $x\rightarrow 0$. The value of $f(\pi)$ is
1. $e^{2\pi}$
2. $e^{-2\pi}$
3. $-e^{-2\pi}$
4. $-e^{2\pi i}$
24. If $y=\frac{1}{\tanh x}$, then $x$ is
1. $\ln{\left(\frac{y+1}{y-1}\right)}$
2. $\ln{\left(\frac{y-1}{y+1}\right)}$
3. $\ln{\sqrt{\frac{y-1}{y+1}}}$
4. $\ln{\sqrt{\frac{y+1}{y-1}}}$
$$\sinh{x}=\frac{e^x-e^{-x}}{2}$$ $$\cosh{x}=\frac{e^x+e^{-x}}{2}$$ $$\tanh{x}=\frac{\sinh{x}}{\cosh{x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ $$y=\frac{1}{\tanh{x}}=\frac{e^x+e^{-x}}{e^x-e^{-x}}$$ $$\frac{y+1}{y-1}=\frac{\frac{e^x+e^{-x}}{e^x-e^{-x}}+1}{\frac{e^x+e^{-x}}{e^x-e^{-x}}-1}=e^{2x}$$ $$2x=\ln{\left(\frac{y+1}{y-1}\right)}$$ $$x=\ln{\sqrt{\frac{y+1}{y-1}}}$$ Hence, answer is (D)
25. The function $\frac{z}{\sin{\pi z^2}}$ of a complex variable $z$ has
1. a simple pole at $0$ and poles of order $2$ at $\pm\sqrt{n}$ for $n=1,2,3,\dots$
2. a simple pole at $0$ and poles of order $2$ at $\pm\sqrt{n}$ and $\pm i\sqrt{n}$ for $n=1,2,3,\dots$
3. poles of order $2$ at $\pm\sqrt{n}$ for $n=0,1,2,3,\dots$
4. poles of order $2$ at $\pm n$ for $n=0,1,2,3,\dots$
$$\sin{\pm n\pi}=0,\quad \text{for } n=0,1,2,\dots$$ $$z^2=\pm n$$ $$z=\pm\sqrt{n} \quad\text{or } z=\pm i\sqrt{n}$$ Hence, answer is (B)
26. The Fourier transform of $f(x)$ is $\tilde{f}(k)=\int_{-\infty}^{\infty}dx\:e^{ikx}f(x)$. If $f(x)=\alpha\delta(x)+\beta\delta'(x)+\gamma\delta''(x)$, where $\delta(x)$ is the Dirac delta-function (and prime denotes derivative), what is $\tilde{f}(k)$?
1. $\alpha+i\beta k+i\gamma k^2$
2. $\alpha+\beta k-\gamma k^2$
3. $\alpha-i\beta k-\gamma k^2$
4. $i\alpha+\beta k-i\gamma k^2$
For Dirac delta function, we have, $$\int\limits f(x)\delta(x-y)\:dx=f(y)$$ $$\int\limits f(x)\delta'(x-y)\:dx=-f'(y)$$ $$\int\limits f(x)\delta''(x-y)\:dx=(-1)^2f''(y)$$ $$\int\limits f(x)\delta^n(x-y)\:dx=(-1)^nf^n(y)$$ \begin{align*} \tilde{f}(k)&={\scriptstyle\int\limits_{-\infty}^{\infty}dx\:e^{ikx}\left(\alpha\delta(x)+\beta\delta'(x)+\gamma\delta''(x)\right)}\\ &=\alpha f(0)+\beta(-f'(0))+\gamma f''(0)\\ &=\alpha-i\beta k-\gamma k^2 \end{align*} Hence, answer is (C)
27. The solution of the differential equation $\frac{dx}{dt}=2\sqrt{1-x^2}$, with initial condition $x=0$ at $t=0$ is
1. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ \sinh{2t},\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
2. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{2}\\ 1,\quad t\geq\frac{\pi}{2} \end{cases} \end{align*}
3. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ 1,\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
4. $x=1-\cos{2t},\quad t\geq 0$
$$\int\frac{dx}{\sqrt{1-x^2}}=2\int dt$$ Let us make substitution $x=\sin\theta$, $dx=\cos\theta$ $$\int d\theta=2t+c_1$$ $$\theta=2t+c_1$$ $$\sin^{-1}x=c_1+2t$$ $$x=c+\sin{2t}$$ At $t=0$, $x=0$, hence, $c=0$ $$x=\sin{2t}$$ However, maximum value of $\sin(2t)$ is 1 $$\sin{2t}=1\quad \text{when } 2t=\frac{\pi}{2}\quad\text{ or }t=\frac{\pi}{4}$$ Hence, \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{2}\\ 1,\quad t\geq\frac{\pi}{2} \end{cases} \end{align*} Hence, answer is (B).
28. Linearly independent solution of the differential equation $$\frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=0$$ are:
1. $e^{-x}$, $e^{-2x}$
2. $e^{-x}$, $e^{2x}$
3. $e^{-2x}$, $e^{x}$
4. $e^{2x}$, $e^{x}$
Auxiliary equation is $$D^2+3D+2=0$$ $$(D+1)(D+2)=0$$ Hence, roots are $-1$ and $-2$

solutions are $e^{-x}$, $e^{-2x}$

Hence, answer is (A)

29. The Hermite polynomial $H_n(x)$ satisfies the differential equation $$\frac{d^2H_n}{dx^2}-2x\frac{dH_n}{dx}+2nH_n(x)=0$$ . The corresponding generating function $G(x,t)=\sum\limits_{n=0}^{\infty}\frac{1}{n!}H_n(x)t^n$ satisfies the
1. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2t\frac{\partial G}{\partial t}=0$
2. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}-2t^2\frac{\partial G}{\partial t}=0$
3. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2\frac{\partial G}{\partial t}=0$
4. $\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2\frac{\partial^2 G}{\partial x\partial t}=0$
Let us consider option (A)
$\frac{\partial^2G}{\partial x^2}=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\frac{d^2H_n}{d x^2}t^n$, $\frac{\partial G}{\partial x}=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\frac{dH_n}{d x}t^n$, $2t\frac{\partial G}{\partial t}=2n\sum\limits_{n=0}^{\infty}\frac{1}{n!}H_n( x)t^n$ \begin{align*} &\frac{\partial^2G}{\partial x^2}-2x\frac{\partial G}{\partial x}+2t\frac{\partial G}{\partial t}\\ &{\scriptstyle =\sum\limits_{n=0}^{\infty}\frac{1}{n!}\left(\frac{d^2H_n}{d x^2}-2x\frac{1}{n!}\frac{dH_n}{d x}+2n\frac{1}{n!}H_n( x)\right)t^n}\\ &=0 \end{align*} Hence, answer is (A)

30. The value of the integral $\int_0^8\frac{1}{x^2+5}dx$, evaluated using Simpson’s $\frac{1}{3}$ rule with $h=2$, is
1. 0.565
2. 0.620
3. 0.698
4. 0.736
According to Simpson's 1/3 rule \begin{align*} I&=\frac{h}{3}\left[y_0+y_n\right.\\ &+4\left(y_1+y_3+\dots+y_{n-1}\right)\\ &\left.+2\left(y_2+y_4\dots+y_{n-2}\right)\right] \end{align*} $$n=\frac{x_n-x_0}{h}=\frac{8-0}{2}=4$$ $$I=\frac{h}{3}\left[y_0+y_4+4\left(y_1+y_3\right)+2\left(y_2\right)\right]$$ $$y_0=f(x_0)=f(0)=\frac{1}{5}$$ $$y_1=f(x_1)=f(2)=\frac{1}{9}$$ $$y_2=f(x_2)=f(4)=\frac{1}{21}$$ $$y_3=f(x_3)=f(6)=\frac{1}{41}$$ $$y_4=f(x_4)=f(8)=\frac{1}{69}$$ \begin{align*} I&={\scriptstyle \frac{2}{3}\left[\frac{1}{5}+\frac{1}{69}+4\left(\frac{1}{9}+\frac{1}{41}\right)+2\left(\frac{1}{21}\right)\right]}\\ &=0.568
31. The trace of $3\times3$ matrix is 2. Two of its eigenvalues are 1 and 2. The third eigenvalue is
1. -1
2. 0
3. 1
4. 2

The trace of a matrix is the sum of the eigenvalues. Hence, $1+2+x=2$, hence $x=-1$

Hence, answer is (A)

32. A linear transformation $T$, defined as $T\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}x_1+x_2\\x_2-x_3\end{pmatrix}$, transforms a vector $\vec x$ from a three-dimensional real space to a two-dimensional real space. The transformation matrix $T$ is
1. $\begin{pmatrix}1&1&0\\0&1&-1\end{pmatrix}$
2. $\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}$
3. $\begin{pmatrix}1&1&1\\-1&1&1\end{pmatrix}$
4. $\begin{pmatrix}1&0&0\\0&0&1\end{pmatrix}$

\begin{align*} T\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}&=\begin{pmatrix}1&1&0\\0&1&-1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\\ &=\begin{pmatrix}x_1+x_2\\x_2-x_3\end{pmatrix} \end{align*}

Hence, answer is (A)

33. The value of $\oint\limits_C\frac{e^{2z}}{(z+1)^4}dz$, where $C$ is circle defined by $|z|=3$, is
1. $\frac{8\pi i}{3}e^{-2}$
2. $\frac{8\pi i}{3}e^{-1}$
3. $\frac{8\pi i}{3}e$
4. $\frac{8\pi i}{3}e^{2}$

Inside circle $|z|=3$, $z=-1$ is a pole of order 4. \begin{align*} &\oint\limits_Cf(z)dz\\ &=2\pi i\times&\text{sum of residues at}\\ &&\text{ poles inside }C \end{align*} In general the residue at a pole of order $m$ at $z=z_0$ is \begin{eqnarray} \begin{array}{c}R=\frac{1}{(m-1)!}\lim\limits_{z\to z_0}\left\{\frac{d^{m-1}\left((z-z_0)^mf(z)\right)}{dz^{m-1}}\right\}_{\!z=z_0}\end{array} \end{eqnarray} $$R=\frac{1}{3!}\lim\limits_{z\to -1}\left\{\frac{d^3\left((z+1)^4e^{2z}\right)}{dz^3}\right\}$$ $$R=\frac{8}{6}e^{-2}$$ $$\oint_Cf(z)dz=2\pi i\times\frac{8}{6}e^{-2}=\frac{8\pi i}{3}e^{-2} $$

Hence, answer is (A)

34. Consider the differential equation $\frac{d^2x}{dt^2}-3\frac{dx}{dt}+2x=0$. If $x=0$ at $t=0$ and $x=1$ at $t=1$, the value of $x$ at $t=2$ is
1. $e^2+1$
2. $e^2+e$
3. $e+2$
4. $2e$

Characteristic equation is $$m^2-3m+2=0$$ $$\Rightarrow (m-1)(m-2)=0$$ $$\Rightarrow m=1,2$$ Hence solution is $$x=c_1e^{t}+c_2e^{2t}$$ $x=0$ at $t=0$ $\Rightarrow c_1=-c_2$

and $x=1$ at $t=1$ $\Rightarrow c_1e+c_2e^{2}=1$ $$\Rightarrow c_1=\frac{1}{e-e^2}$$ $$ c_2=-\frac{1}{e-e^2}$$ At $t=2$ \begin{align*} x&=c_1e^{2}+c_2e^{4}\\ &=\frac{e^{2}}{e-e^2}-\frac{e^{4}}{e-e^2}\\ &=e+e^2 \end{align*}

Hence, answer is (B)

35. The rank-2 tensor $x_ix_j$, where $x_i$ are the Cartesian coordinates of the position vector in three dimensions, has 6 independent elements. Under rotation, these 6 elements decompose into irreducible sets (that is, the elements of each set transform only into linear combinations of elements in that set) containing
1. 4 and 2 elements
2. 5 and 1 elements
3. 3, 2 and 1 elements
4. 4, 1 and 1 elements

The rank-2 tensor $T_{ij}=x_ix_j$ can be decomposed into into irreducible representations in the following way $$T_{ij}=T_{ij}^{(0)}+T_{ij}^{(1)}+T_{ij}^{(2)}$$ where,

$T_{ij}^{(0)}$ is a tensor of rank 0 and has 1 independent component only.

$T_{ij}^{(1)}$ is a tensor of rank 1 and has 3 independent components.

$T_{ij}^{(2)}$ is a tensor of rank 2 and has 5 independent components.

Hence, answer is (B)

36. Consider the differential equation $\frac{dy}{dx}=x^2-y$ with initial condition $y=2$ at $x=0$. Let $y_{(1)}$ and $y_{(1/2)}$ be the solutions at $x=1$ obtained using Euler's forward algorithm with step size 1 and $\frac{1}{2}$ respectively. The value of $\left(y_{(1)}-y_{(1/2)}\right)/y_{(1/2)}$ is
1. $-1/2$
2. $-1$
3. $1/2$
4. 1

Euler's forward formula is $$y_{n+1}=y_n+hf(x_n,y_n)$$ where, $f(x,y)=x^2-y$. \begin{align*} y_{1}&=y_0+f(x_0,y_0)\\ &=2+f(0,2)\\ &=0 \end{align*} \begin{align*} y_{1/2}&=y_0+\frac{1}{2}f(x_0,y_0)\\ &=2+\frac{1}{2}f(0,2)\\ &=1 \end{align*} $$\left(y_{(1)}-y_{(1/2)}\right)/y_{(1/2)}=-1$$

Hence, answer is (B)

37. A rod of length $L$ carries a total charge $Q$ distributed uniformly. If this is observed in a frame moving with a speed $v$ along the rod, the change per unit length (as measured by the moving observer) is
1. $\frac{Q}{L}\left(1-\frac{v^2}{c^2}\right)$
2. $\frac{Q}{L}\sqrt{1-\frac{v^2}{c^2}}$
3. $\frac{Q}{L\sqrt{1-\frac{v^2}{c^2}}}$
4. $\frac{Q}{L\left(1-\frac{v^2}{c^2}\right)}$

Length contraction is given by $$L'=L \sqrt{1-\frac{v^2}{c^2}}$$ Hence, charge per unit length is given by $$\frac{Q}{L\sqrt{1-\frac{v^2}{c^2}}}$$

Hence, answer is (C)

38. One of the eigen values of the matrix $\begin{pmatrix}2&3&0\\3&2&0\\0&0&1\end{pmatrix}$ is 5
1. The other two eigenvalues are
1. 0 and 0
2. 1 and 1
3. 1 and -1
4. -1 and -1
2. The normalized eigenvector corresponding to the eigenvalue 5 is
1. $\frac{1}{\sqrt{2}} \begin{pmatrix}0\\-1\\1\end{pmatrix}$
2. $\frac{1}{\sqrt{2}} \begin{pmatrix}-1\\1\\0\end{pmatrix}$
3. $\frac{1}{\sqrt{2}} \begin{pmatrix}1\\0\\-1\end{pmatrix}$
4. $\frac{1}{\sqrt{2}} \begin{pmatrix}1\\1\\0\end{pmatrix}$

1. Determinant of matrix = product of eigenvalues.

   Let $\lambda_1$ and $\lambda_2$ be other eigenvalues $$Det=5\times\lambda_1\times\lambda_2=-5$$ $$\lambda_1\times\lambda_2=-1\quad---(1)$$ Trace of matrix=sum of eigenvalues $$5+\lambda_1+\lambda_2=5$$ $$\lambda_1+\lambda_2=0\quad---(2)$$ Solving equation (1) and (2) we get $$\lambda_1=1\quad\lambda_2=-1$$

   Hence, answer is (C)

2. Eigenvalue equation for $\lambda=5$ is $A-\lambda=0$ $$\begin{pmatrix}-3&3&0\\3&-3&0\\0&0&-4\end{pmatrix}\!\begin{pmatrix}x\\y\\z\end{pmatrix}\!=\!\begin{pmatrix}0\\0\\0\end{pmatrix}$$ $$-3x+3y=0$$ $$3x-3y=0$$ $$-4z=0$$ $$\Rightarrow z=0, x=y=c$$ Applying normalization condition $$|c|^2+|c|^2+0=1$$ $$\Rightarrow c=\frac{1}{\sqrt{2}}$$ $$\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\\0\end{pmatrix}$$

   Hence, answer is (d)

39. The Gauss hypergeometric function $F(a,b,c;z)$, defined by the Taylor series expansion around $z=0$ as $${\textstyle F(a,b,c;z)=\sum\limits_{n=0}^\infty\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)n!}z^n}$$ satisfies the recursion relation
1. $\frac{d}{dz}F(a,b,c;z)=\frac{c}{ab}F(a-1,b-1,c-1;z)$
2. $\frac{d}{dz}F(a,b,c;z)=\frac{c}{ab}F(a+1,b+1,c+1;z)$
3. $\frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}F(a-1,b-1,c-1;z)$
4. $\frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}F(a+1,b+1,c+1;z)$

$${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=1}^\infty n\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)n!}z^{n-1}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=1}^\infty\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{c(c+1)\cdots(c+n-1)(n-1)!}z^{n-1}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\sum\limits_{n=0}^\infty\frac{a(a+1)\cdots(a+n)b(b+1)\cdots(b+n)}{c(c+1)\cdots(c+n)(n)!}z^{n}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}\sum\limits_{n=0}^\infty\frac{(a+1)\cdots(a+n)(b+1)\cdots(b+n)}{(c+1)\cdots(c+n)(n)!}z^{n}}$$ $${\scriptstyle \frac{d}{dz}F(a,b,c;z)=\frac{ab}{c}\frac{d}{dz}F(a+1,b+1,c+1;z)}$$

Hence, answer is (D)

40. Let $(x,y)$ and $(x',y')$ be the coordinate systems used by the observers $O$ and $O'$, respectively. Observer moves with a velocity $v= \beta c$ along their common positive $x$-axis. If $x_+=x+ct$ and $x_-=x-ct$ are the linear combinations of the coordinates, the Lorentz transformation relating $O$ and $O'$ takes the form
1. $x_+'=\frac{x_--\beta x_+}{\sqrt{1-\beta^2}}$ and $x_-'=\frac{x_+-\beta x_-}{\sqrt{1-\beta^2}}$
2. $x_+'=\sqrt{\frac{1+\beta}{1-\beta}}x_+$ and $x_-'=\sqrt{\frac{1-\beta}{1+\beta}}x_-$
3. $x_+'=\frac{x_+-\beta x_-}{\sqrt{1-\beta^2}}$ and $x_-'=\frac{x_--\beta x_+}{\sqrt{1-\beta^2}}$
4. $x_+'=\sqrt{\frac{1-\beta}{1+\beta}}x_+$ and $x_-'=\sqrt{\frac{1+\beta}{1-\beta}}x_-$

According to Lorentz transformation $$x'=\gamma(x-vt)$$ $$y'=y$$ $$z'=z$$ $$t'=\gamma(t-xv/c^2)$$ where, $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$. Hence, for $v= \beta c$, $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ \begin{align*} x_+'&=x'+ct'\\ &=\gamma(x-\beta ct)+c\gamma(t-x\beta/c)\\ &=\gamma\left\{x-\beta ct+ct-x\beta\right\}\\ &=\gamma\left\{x_+-\beta (x+ct)\right\}\\ &=\gamma\left\{1-\beta \right\}x_+\\ &=\sqrt{\frac{1-\beta}{1+\beta}}x_+ \end{align*} \begin{align*} x_-'&=x'-ct'\\ &=\gamma(x-\beta ct)-c\gamma(t-x\beta/c)\\ &=\gamma\left\{x-\beta ct-ct+x\beta\right\}\\ &=\gamma\left\{x_-+\beta (x-ct)\right\}\\ &=\gamma\left\{1+\beta \right\}x_-\\ &=\sqrt{\frac{1+\beta}{1-\beta}}x_- \end{align*}

Hence, answer is (D)

41. In finding the roots of the polynomial $f(x)=3x^3-4x-5$ using the iterative Newton-Raphson method, the initial guess is taken to be $x=2$. In the next iteration its value is nearest to
1. 1.671
2. 1.656
3. 1.559
4. 1.551

Newton-Raphson formula is \begin{align*} x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\\ &=2-\frac{3\times2^3-4\times2-5}{9\times2^2-4}\\ &=1.65625 \end{align*}

Hence, answer is (B)

42. For a particle of energy $E$ and $P$ momentum (in a frame $F$), the rapidity $y$ is defined as $y=\frac{1}{2}\ln{\left(\frac{E+p_3c}{E-p_3c}\right)}$. In a frame $F'$ moving with velocity $v=(0,0,\beta c)$ with respect to $F$, the rapidity $y'$ will be
1. $y'=y+\frac{1}{2}\ln{\left(1-\beta^2\right)}$
2. $y'=y-\frac{1}{2}\ln{\left(\frac{1+\beta}{1-\beta}\right)}$
3. $y'=y+\ln{\left(\frac{1+\beta}{1-\beta}\right)}$
4. $y'=y+2\ln{\left(\frac{1+\beta}{1-\beta}\right)}$

$$y=\frac{1}{2}\ln{\left(\frac{E+p_3c}{E-p_3c}\right)}$$ $$y'=\frac{1}{2}\ln{\left(\frac{E'+p_3'c}{E'-p_3'c}\right)}$$ According to Lorentz transformation (boost) $$p_3'=\gamma\left(p_3-\frac{vE}{c^2}\right)$$ $$E'=\gamma\left(E-vp_3\right)$$ Using $v=\beta c$ $$p_3'=\gamma\left(p_3-\frac{\beta E}{c}\right)$$ $$E'=\gamma\left(E-\beta cp_3\right)$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\gamma\left(E-\beta cp_3\right)+\gamma\left(p_3-\frac{\beta E}{c}\right)c}{\gamma\left(E-\beta cp_3\right)-\gamma\left(p_3-\frac{\beta E}{c}\right)c}\right)}}$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\left(E-\beta cp_3\right)+\left(cp_3-\beta E\right)}{\left(E-\beta cp_3\right)-\left(cp_3-\beta E\right)}\right)}}$$ $$y'=\frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)-\beta\left(E+cp_3\right)}{\left(E- cp_3\right)+\beta\left(E- cp_3\right)}\right)}$$ $$y'=\frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)}{\left(E- cp_3\right)}\frac{1-\beta}{1+\beta}\right)}$$ $$y'={\textstyle \frac{1}{2}\ln{\left(\frac{\left(E+cp_3\right)}{\left(E- cp_3\right)}\right)} + \frac{1}{2}\ln{\left(\frac{1-\beta}{1+\beta}\right)}}$$ $$y'=y - \frac{1}{2}\ln{\left(\frac{1+\beta}{1-\beta}\right)}$$

Hence, answer is (B)

43. A function $f(x)$ satisfies the differential equation $\frac{d^2f}{dx^2}-\omega^2f=-\delta(x-a)$, where $\omega$ is positive. The Fourier transform $\tilde{f}(k)=\int_{-\infty}^{\infty}dx\:e^{ikx}f(x)$ of $f$, and the solution of the equation are, respectively,
1. $\frac{e^{ika}}{k^2+\omega^2}$ and $\frac{1}{2\omega}\left(e^{-\omega|x-a|}+e^{\omega|x-a|}\right)$
2. $\frac{e^{ika}}{k^2+\omega^2}$ and $\frac{1}{2\omega}e^{-\omega|x-a|}$
3. $\frac{e^{ika}}{k^2-\omega^2}$ and $\frac{1}{2\omega}\left(e^{-i\omega|x-a|}+e^{i\omega|x-a|}\right)$
4. $\frac{e^{ika}}{k^2-\omega^2}$ and $\frac{1}{2i\omega}\left(e^{-i\omega|x-a|}-e^{i\omega|x-a|}\right)$
$$\frac{d^2f(x)}{dx^2}-\omega^2f(x)=-\delta(x-a)$$ Taking Fourier transform $$F\!\!\left[\frac{d^2f(x)}{dx^2}\right]\!\!-\!\!F\!\left[\omega^2f(x)\right]\!\!=\!\!-F\!\left[\!\delta(\!x\!-\!a\!)\!\right]$$ $$F\left[\frac{d^2f(x)}{dx^2}\right]=-k^2f(k)$$ $$F\left[\delta(x-a)\right]=e^{ika}$$ $$-k^2f(k)-\omega^2f(k)=-e^{ika}$$ Hence, Fourier transform of $f(x)$ is $$f(k)=\frac{e^{ika}}{k^2+\omega^2}$$ Taking Inverse Fourier transform $$f(x)=F^{-1}\left[\frac{e^{ika}}{k^2+\omega^2}\right]$$ $$f(x)=\frac{1}{2\omega}F^{-1}\left[e^{ika}\frac{2\omega}{k^2+\omega^2}\right]$$ $$f(x)=\!\!\int F^{-1}\left[\frac{e^{ika}}{2\omega}\right]F^{-1}\left[\frac{2\omega}{k^2+\omega^2}\right]$$ $$F^{-1}\left[e^{ika}\right]=\delta(x-a)$$ $$F^{-1}\left[\frac{2\omega}{k^2+\omega^2}\right]=e^{-\omega |x|}$$ Using Convolution theorem, $$f(x)=\frac{1}{2\omega}\int \delta(k-a)e^{-\omega |x-k|}dk$$ $$f(x)=\frac{1}{2\omega}e^{-\omega |x-a|}$$

Hence, answer is (B)

44. Eigenvalues of the matrix $\begin{bmatrix}1&-1\\1&1\end{bmatrix}$
1. 1, -1
2. -1, $-i$
3. $i$, $-i$
4. $1+i$, $1-i$

Determinant of matrix is equal to product of eigenvalues. $$\begin{vmatrix}1&-1\\1&1\end{vmatrix}=2$$ Clearly $$(1+i)(1-i)=2$$

Hence, answer is (D)

45. Particular integral of first order linear differential $\frac{dy}{dx}=x+y$ is given by:
    1. $y(x)=-x-1$
    2. $y(x)=x+1$
    3. $y(x)=x-1$
    4. $y(x)=-x+1$
    Linear differential equation is given by $$\frac{dy}{dx}+P(x)y=Q(x)$$ Multiply both sides by the Integrating Factor $e^{\int P(x)dx}$ $${\scriptstyle \frac{dy}{dx}e^{\int P(x)dx}+P(x)ye^{\int P(x)dx}=Q(x)e^{\int P(x)dx}}$$ $$\frac{d\left(ye^{\int P(x)dx}\right)}{dx}=Q(x)e^{\int P(x)dx}$$ On integrating we get $$ye^{\int P(x)dx}=\int Q(x)e^{\int P(x)dx}dx+c$$ $$\frac{dy}{dx}-y=x$$ $P(x)=-1$, $Q(x)=x$ $$ye^{-\int dx}=\int xe^{-\int dx}dx+c$$ $$ye^{-x}=\int xe^{-x}dx+c$$ Integrating by parts $$ye^{-x}=-xe^{-x}+\int e^{-x}dx+c$$ $$ye^{-x}=-xe^{-x}-e^{-x}+c$$ $$y=-x-1$$

    Hence, answer is (A)

46. The Fourier transform of a Gaussian function is of the form:
1. Exponential
2. Lorentzian
3. Gaussian
4. Screened coulomb

Hence, answer is (C)

47. The real part of $\log{(3+4i)}$ is :
1. $\log2$
2. $\log3$
3. $\log4$
4. $\log5$

Let $z=x+iy$, $r=\sqrt{x^2+y^2}$, $\theta=\tan^{-1}{\frac{y}{x}}$ and $z=re^{i\theta}$

For $z=3+4i$, $r=\sqrt{3^2+16^2}=5$, $\theta=\tan^{-1}{\frac{4}{3}}=0.92729$ and $z=5e^{i0.92729}$ \begin{align*} \log{(3+4i)}&=\log{5e^{i0.92729}}\\ &=\log5+i0.92729 \end{align*}

Hence, answer is (D)

48. Consider three vectors $\vec a=\hat i+\hat j+\hat k$, $\vec b=\hat i-\hat j+\hat k$ and $\vec c=\hat i-\hat j-\hat k$. Which of the following statement is true?
1. $\vec a$, $\vec b$, $\vec c$ are linearly independent
2. $\vec a$, $\vec b$ are linearly independent
3. $\vec b$ and $\vec c$ are right angle to each other
4. $\vec a$ and $\vec c$ are parallel

The vectors in a subset $S = \{ \vec v_1 ,\vec v_2 ,\cdots,\vec v_n \}$ of a vector space $V$ are said to be linearly dependent, if there exist a finite number of distinct vectors $ \vec {v}_{1},\vec {v}_{2},\dots ,\vec {v}_{k}$ in $S$ and scalars $a_{1},a_{2},\dots ,a_{k}$, not all zero, such that $$a_{1}\vec {v}_{1}+a_{2}\vec {v}_{2}+\cdots +a_{k}\vec {v}_{k}=\vec {0}$$ where zero denotes the zero vector. Let $$a_1\vec a+a_2\vec b+a_3\vec c=0$$ $$\begin{bmatrix}1&1&1\\1&-1&-1\\1&1&-1\end{bmatrix}\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$ If $\vec a$, $\vec b$ and $\vec c$ are linearly independent, then the only solution to this system of equations is the trivial solution, $a_1=a_2=a_3=0$. For homogeneous systems this happens precisely when the determinant is non-zero. If determinant is zero then vectors are linearly dependent. $$\begin{vmatrix}1&1&1\\1&-1&-1\\1&1&-1\end{vmatrix}=4$$ Hence, $\vec a$, $\vec b$ and $\vec c$ are linearly independent.

Hence, answer is (A)