Engineering Mathematics

\begin{aligned} \vec{F}&=y\hat{a}_{x}-x\hat{a}_{y} \\ \vec{r}&=x\hat{i}+y\hat{j} \\ \vec{F}&=y\hat{i}-x\hat{j} \\ d\vec{r}&=dx\hat{i}+dy\hat{j} \\ &=\int _{c}F.d\vec{r}\\ &=\int_{c}F_{1}dx+F_{2}dy \\ &=\int _{c}ydx-xdy \\ &\text{Where C is,}\\ y&=x^{2} \\ dy&=2x dx \\ x &\text{ varies from -1 to 2,}\\ \int _{c}\vec{F}dl&=\int_{-1}^{2}x^{2}dx-x\cdot 2xdx \\ &=\int_{-1}^{2}(x^{2}-2x)dx \\ &=\int_{-1}^{2}-x^{2}dx=\left.\begin{matrix} -\frac{x^{3}}{3} \end{matrix}\right|_{-1}^{2} \\ &=-\frac{8}{3}-\frac{1}{3}=-\frac{9}{3}\\ &=-3 \end{aligned}

As per GATE official answer key MTA (Marks to ALL)
\begin{aligned} \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} I_{11} & 0 &0 \\ I_{21}&l_{22} & 0\\ l_{31} &l_{32} &l_{33} \end{bmatrix}\begin{bmatrix} u_{11} &u_{12} &u_{13} \\ 0 &u_{22} &u_{23} \\ 0 & 0 &u_{33} \end{bmatrix} \\ \text{consider, }u_{11}&=u_{22}=u_{33}=1 \\ \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} l_{11} & 0 &0 \\ l_{21}&l_{22} & 0\\ l_{31} &l_{32} &l_{33} \end{bmatrix}\begin{bmatrix} 1 &u_{12} &u_{13} \\ 0 &1 &u_{23} \\ 0 & 0 &1 \end{bmatrix} \\ \begin{bmatrix} 2 &3 &3 \\ 3 &2 &1 \\ 3 &1 &7 \end{bmatrix}&=\begin{bmatrix} l_{11} & l_{11}u_{12} &l_{11}u_{13} \\ l_{21}& l_{21}u_{12}+l_{22} & l_{21}u_{13}+ l_{22}u_{23}\\ l_{31} &l_{31}u_{12}+l_{32} & l_{31}u_{13}+ l_{32}u_{23}+l_{33}\end{bmatrix} \\ l_{11}&=2 \; \; l_{11}u_{12}=3 \; \; l_{11}u_{13}=3 \; \; \\ l_{21}&=3 \; \; 2u_{12}=3 \; \; 2u_{13}=3 \\ l_{31}&=3\; \; u_{12}=\frac{3}{2} \; \; u_{13}=\frac{3}{2} \\ l_{21}u_{12}+l_{22}&=2\; \; l_{21}u_{13}+l_{22}u_{23}=1 \\ (3)\left ( \frac{3}{2} \right )+l_{22}&=2 \; \; (3)\left ( \frac{3}{2} \right )+\left ( -\frac{5}{2} \right )u_{23}=1 \\ l_{22}&=-\frac{5}{2} \; \; u_{23}=-\frac{7}{5} \\ l_{31}u_{12}+l_{32}&=1 \; \; l_{31}u_{13}+l_{32}u_{23}+l_{33}=7\\ (3)(\frac{3}{2})+l_{32}&=1 \; \; (3)\left ( \frac{3}{2} \right )+\left ( -\frac{7}{2} \right )\left ( \frac{7}{5} \right )+l_{33}=7 \\ l_{32}&=-\frac{7}{2} \; \; l_{33}=-\frac{74}{10}\\ L&=\begin{bmatrix} 2 &0 &0 \\ 3 &-5/2 &0 \\ 3 &-7/2 & 74/10 \end{bmatrix} \end{aligned}
The number of purely real elements of lower triangular matrix are 9.

\begin{aligned} y&=3x^{2}+3x+1 \; \; in \; [-2,0] \\ \frac{\partial y}{\partial x}&=6x+3,\; \; \frac{\partial^2 y}{\partial x^2}=6 \\ \frac{\mathrm{d} y}{\mathrm{d} x}&=0\\ 6x+3&=0 \\ x&=\frac{-1}{2} \\ \frac{d^{2} y}{dx^{2}}&=6 \gt 0\text{ minimum} \end{aligned}

Maximum value of y in [-2, 0] is maximum {f(-2), f(0)}
max{7, 1} = 7

Minimum value of y in [-2, 0]
min\begin{Bmatrix} f(-2) & f(0) &f(-\frac{1}{2}) \\ \downarrow, &\downarrow, &\downarrow \\ 7& 1 & \frac{1}{4} \end{Bmatrix}+=\frac{1}{4}
Maximum value 7, minimum value \frac{1}{4}

\begin{aligned} \frac{\mathrm{d} y}{\mathrm{d} x}&=2x-y \\ y(0)&=1, \; y \text{ at } x=\ln 2\\ \frac{\mathrm{d} y}{\mathrm{d} x}+y&=2x \\ P&=1, \;\; Q=2x \\ I.F.&=e^{\int Pdx}=e^{\int 1dx}=e^{x}\\ \text{Solution, } Y(I.F)&=\int Q(I.F.)dx\\ye^{x}&=\int 2x\cdot e^{x}dx \\&=2(xe^{x}-e^{x})+C \\ y&=2x-2+ce^{-x} \\ y(0)&=1 \\ 1&=0-2+C \\ C&=3 \\ \therefore \; \; y&=2x-2+3e^{-x} \\ \text{At }x&= \ln 2 \\ y&=2(\ln 2)-2+3e^{-\ln 2} \\ &=1.386-2+\frac{3}{2}\\ &=0.886\end{aligned}

\begin{aligned} I&=\int _C \frac{z^2+1}{z^2-2z}dz\;\;\;|z|=1 \\ \text{Using } & \text{Cauchy's integral theorem}\\ \int _C\frac{F(z)}{z-a}dz&=2 \pi i (Re_{(z=a)})\;\;\;...(i)\\ I&=\int _C \frac{z^2+1}{z(z-2)}dz \end{aligned}
Poles are at z=0 and 2 but only z=0 lies inside the unit circle.
Residue at (z=0)=\lim_{z \to 0}\frac{z^2+1}{z(z-2)}
Re_{(z=0)}=-\frac{1}{2}
Using equation (i)
\int _C \frac{z^2+1}{z^2-2z}dz=2 \pi i \times \left ( \frac{-1}{2} \right )=-\pi i

\begin{aligned} \because \; p& \neq q\\ &\frac{1}{2\pi}\int_{-\pi}^{\pi} \sin p\theta \cos q\theta d\theta \\ &=\frac{1}{2\pi}\cdot \frac{1}{2}\int_{-\pi}^{\pi} [\sin (p+q)\theta + \sin (p-q)\theta] d\theta \\ &=\frac{1}{4\pi}\left [ \frac{-1}{(p+q)}\cos (p+q)\theta -\frac{1}{(p-q)}\cos (p-q)\theta \right ]_{-\pi}^{\pi}\\ &=\frac{-1}{4\pi} \left \{ \frac{1}{(p+q)}(\cos (p+q) \pi -\cos (p+q)(-\pi)) \right.\\ &+\left. \frac{1}{(p-q)}(\cos (p-q) \pi -\cos (p-q)(-\pi)) \right \}\\ &=0 \end{aligned}

Given Polynomial ax^{3}+bx^{2}+cx+d=0;\; \; \; a\neq 0

Option (A):
If d=0, then the polynomial equation becomes
\begin{aligned} ax^3+bx^2+cx&=0\\ x(ax^2+bx+c)&=0 \\ x=0 \text{ or } ax^2+bx+c&=0 \end{aligned}
d can be choosen to ensure x=0 is a root of given polynomial.
Hence, Option (A) is correct.

Option B:
A third degree polynomial equation with all root equal is given by
(x+\alpha )^3=0
Thus, by selecting suitable values of a, b, c and d we can have all roots identical.
Hence, option (B) is incorrect.

Option (C): Complex roots always occurs in pairs,
So, the given polynomial will have maximum of 2 complex roots and 1 real root.
Hence, option (C) is incorrect.

Option (D): Nature or roots depends on other coefficients also apart from coefficient 'c'.
Hence, option (D) is correct.
Hence, the correct options are (A) and (D).

\begin{aligned} p&=0.02\\ n&=50\\ \lambda &=np=50(0.02)=1\\ p(x\geq 2)&=1-p(x \lt 2)\\ &=1-(p(x=0)+p(x=1))\\ &=1-\left ( \frac{e^{-\lambda }\lambda ^0}{0!}+\frac{e^{-\lambda }\lambda ^1}{1!} \right )\\ &=1-e^{-\lambda }(1+\lambda )\\ &=1-e^{-1}(1+1)=0.26 \end{aligned}

\begin{aligned} \bigtriangledown \cdot (uA)&=u(\bigtriangledown \cdot A)+(\bigtriangledown A)F\\ &=(x^2+y^2+z^2)[2+3+4]\\ &+(2x\hat{i}+2y\hat{j}+2z\hat{k})(2x\hat{i}+3y\hat{j}+4z\hat{k})\\ &=9(x^2+y^2+z^2)+(4x^2+6y^2+8z^2)\\ At\; (1,1,1)&=9(3)+[4+6+8]\\ &=27+18=45 \end{aligned}

\begin{aligned} \oint _{|z|=5}\frac{z^3+z^2+8}{2+2}dz&=2 \pi j \text{ (sum of residues)} \\ &=2 \pi j \times \left [ \lim_{z \to -2} \frac{(z+2)(z^3+z^2+8)}{(z+2)}\right ] \\ &= 2 \pi j \times \left [ \frac{-8+4+8}{1} \right ]=8 \pi j \end{aligned}