Engineering Mathematics

Given contour is a circle at centre (0,0) and radius 2 given function is \frac{1}{(z-i)(z+4i)} here z = i is a singular point lies now by caucus integral formula

\begin{aligned} &\int \frac{1}{(z-i)(z+4i)}dz=\int \frac{\left ( \frac{1}{z+4} \right )}{z-i}dz=2 \pi i \times f(i)\\ &f(z)=\frac{1}{z+4i}\\ &f(i)=\frac{1}{5i}\\ &2 \pi i \times f(i)=2 \pi i \times \frac{1}{5i}=\frac{2\pi}{5}\\ &\text{Now }\frac{1}{2 \pi}\int_{c}\frac{1}{(z-1)(z+4i)}dz=\frac{1}{2 \pi}+\frac{2 \pi}{5}=\frac{1}{5} \end{aligned}

From the property of eigen values,
Sum of eigen values = Trace of matrix
4=3+q
q=1

Product of eigen values = Determinant
\begin{aligned} -1&=\begin{vmatrix} 3 &p \\ p& q \end{vmatrix}\\ 3q-p^2&=-1\\ 3-p^2&=-1\\ p^2&=4\\ p&=\pm 2\\ \Rightarrow |p|&=2 \end{aligned}

Zero solution is always a solution of Ax = 0.
Option (b) is correct.
Given A is 3x5 real matrix and r(A) = 2 and Ax = 0 is a system of homogeneous linear equations since r(A) \lt number of unknown the system has infinite solution option (c) is also correct.

Given D.E. is \frac{du}{dx}=\frac{-xu^2}{2+x^2u}
\Rightarrow (2+x^2u)du+xu^2dx=0
Here M=xu^2 \;\;\;\;N=2+x^2u
\frac{\partial N}{\partial x}=2xu \;\;\;\;\;\frac{\partial M}{\partial x}=2xu
This is exact D.E
General solution is
\int_{\text{u constant}}Mdx+\int_{\text{term without 'x'}}Ndx=c
\Rightarrow \int (xu^2)dx+\int 2du=c
\therefore \; \frac{x^2u^2}{2}+2u=constant

Since \varphi (s) has a triple roots at s=-\sigma
Therefore, \varphi (s)=(s+\sigma )^3\psi (s)
It satisfies all the conditions in option (B) is correct.

\begin{aligned} &\text{ Let }(x+b)=t\\ &\Rightarrow \; dx=dt\\ &\text{When ,} x=-\infty ;t=-\infty \\ &\int_{-\infty }^{-\infty }e^{-n(x+b)^2}dx=\int_{-\infty }^{-\infty }e^{-at^2}dt\\ &\text{Let, }at^2=y^2\Rightarrow t=\frac{y}{\sqrt{a}}\\ &2at\;dt=3y\;dy\\ &dt=\frac{ydy}{at}=\frac{ydy}{a\frac{y}{\sqrt{a}}}=\frac{y}{\sqrt{a}}\\ &\int_{-\infty }^{-\infty }e^{-at^2}dt=\int_{-\infty }^{-\infty }e^{-y^2}\cdot \frac{dy}{\sqrt{a}}=\sqrt{\frac{\pi}{a}} \end{aligned}

Exact value
\begin{aligned} &=\int_{1}^{2} (4x^2+2x+6)dx\\ &=\frac{4x^3}{3}+\frac{2x^2}{2}+6x\\ &=\frac{4}{3}(\pi) \times (3)+6\\ &=\frac{28}{3}+9=\frac{55}{3} \end{aligned}
Approximate value
Here f(x) is a polynomial of degree 2, so Simpsons rule gives exact value with zero error
\begin{aligned} \therefore \;\; \text{Approx value}&=\frac{55}{3}\\ \frac{I_e-I_s}{I_e}&=0\\ \therefore \;\;e=\left (\frac{I_e-I_s}{I_e} \right )\times 100&=0 \end{aligned}

Given,
\begin{aligned} \bar{F} &=3x\bar{i}+5y\bar{j}+6z\bar{k} \\ \triangledown \cdot \bar{F}&= \frac{\partial }{\partial x}(3x)+\frac{\partial }{\partial x}(5y)+\frac{\partial }{\partial x}(6z)\\ &= 3+5+6=14 \end{aligned}
By gauss divers once Theorem
\begin{aligned} \int_{A}^{}\bar{F}\cdot dA &=\int \int \int (\triangledown \cdot F)dV =\int \int \int 14\; dv\\ &=14 \times \text{volume of a cube of side 1 unit } \\ &=14 \times (1)^3=14\end{aligned}

The constant term in the Fourier series expansion of F(t) is the average value of F(t) in one fundamental period i.e.,
\frac{\int_{0}^{1}4dt+\int_{1}^{2}0dt}{2}=\frac{4}{2}=2

Component of \vec{a} parallel to \vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{20}{\sqrt{46}}=2.95
Now,
\begin{aligned} |\vec{a}|^2&=2.95^2+(a\perp )^2\\ 78&=2.95^2+(a\perp )^2\\ a\perp&=8.32 \end{aligned}