# Engineering Mathematics

 Question 1
The value of the integral
$\int \int _D 3(x^2+y^2)dxdy$
, where $D$ is the shaded triangular region shown in the diagram, is _____ (rounded off to the nearest integer).

 A 128 B 1024 C 512 D 64
GATE EC 2022      Calculus
Question 1 Explanation:
\begin{aligned} I&=\int_{0}^{4}\int_{-x}^{x}(3x^2+3y^2)dydx\\ &=\int_{0}^{4}\left [ 3x^y+\frac{3y^3}{3} \right ]_{-x}^xdx\\ &=\int_{0}^{4}[3x^2(2x)+2x^3]dx\\ &=\int_{0}^{4}8x^3 dx\\ &=2 \times 4^4\\ &=512 \end{aligned}
 Question 2
Consider the following series:
$\sum_{n=1}^{\infty }\frac{n^d}{c^n}$
For which of the following combinations of $c,d$ values does this series converge?
 A $c=1,d=-1$ B $c=2,d=1$ C $c=0.5,d=-10$ D $c=1,d=-2$
GATE EC 2022      Complex Variables
Question 2 Explanation:
\begin{aligned} \Sigma u_n&=\Sigma \frac{n}{2^n}\\ &\text{Ratio test; }\\ \lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{n+1}{2^{n+1}} \times \frac{2^n}{n}=\frac{1}{2}\\ \frac{1}{2}& \lt 2 \end{aligned}
$\therefore$ By ratio test, $\Sigma u_n$ is convergent.
(A) c=1, d=-1
$\Sigma u_n=\Sigma \frac{1}{n}$ is divergent by P-test
(B) c=0.5, d=-10
\begin{aligned} \Sigma u_n&=\Sigma \frac{n^{-10}}{(0.5)^n}\\ &\text{Ratio test; }\\ \lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{(n+1)^{-10}}{(0.5)^{n+1}} \times \frac{(0.5)^n}{n^{10}}\\ &=\frac{1}{0.5}=2\\ 2& \gt 1 \end{aligned}
$\therefore$ By ratio test, $\Sigma u_n$ is divergent.
(D) c=1,d=-2 $\Sigma u_n=\Sigma \frac{n^{-2}}{(1)^n}=\Sigma \frac{1}{n^2}$
$\Sigma u_n$ is convergent by P-test.
 Question 3
Let $\alpha ,\beta$ be two non-zero real numbers and $v_1,v_2$ be two non-zero real vectors of size 3 x 1. Suppose that $v_1$ and $v_2$ satisfy $v_1^Tv_2=0,v_1^Tv_1=1$, and $v_2^Tv_2=1$. Let $A$ be the 3x3 matrix given by:
$A=\alpha v_1v_1^T+ \beta v_2v_2^T$
The eigenvalues of $A$ are __________.
 A $0,\alpha ,\beta$ B $0,\alpha+\beta ,\alpha-\beta$ C $0,\frac{\alpha +\beta}{2},\sqrt{\alpha \beta }$ D $0,0,\sqrt{\alpha ^2+ \beta ^2}$
GATE EC 2022      Linear Algebra
Question 3 Explanation:
After multiply by $v_1$ on both sides,
\begin{aligned} Av_1&=\alpha v_1v_1^Tv_1+ \beta v_2v_2^Tv_1 Av_1&=\alpha v_1 \end{aligned}
$\alpha$ is an eigen value of A.

After multiply by $v_2$ on both sides,
\begin{aligned} Av_2&=\alpha v_1v_1^Tv_2+ \beta v_2v_2^Tv_2 Av_2&=\beta v_2 \end{aligned}
$\beta$ is an eigen value of A.
$v_1v_1^T$ and $v_2v_2^T$ both are singular matrieces.
Therefore, A is also a singular matrix.
$|A|=0\Rightarrow \lambda _A=0$
Hence $\lambda _A= 0,\alpha ,\beta$
 Question 4
The function $f(x)=8 \log _e x-x^2+3$ attains its minimum over the interval $[1,e]$ at $x=$ _________.
(Here $\log _e x$ is the natural logarithm of $x=$.)
 A 2 B 1 C $e$ D $\frac{1+e}{2}$
GATE EC 2022      Calculus
Question 4 Explanation:
$f(x)=8 \log _e x-x^2+3$ when $x \in [1,e]$
Differentiating both side,
\begin{aligned} f(x)&=\frac{8}{x}-2x=0 \; where\; x \gt 0\\ f'(x)&=0\\ \frac{8}{x}-2x&=0\\ 8-2x^2&=0\\ x^2&=4\\ x&=\pm 2 \end{aligned}
\begin{aligned} f''(x)&=\frac{-8}{x^2}-2 \\ f''(2)&=-6 \lt 0\\ \end{aligned}
f(x) is maximum for x = 2
Minimum of f(x) will be in [1, e] = min [f(1), f(e)]
$f(e)=8 \ln e -e^2+3=3.61$
Hence, minimum value of f(x) occurs at x=1
 Question 5
A simple closed path $C$ in the complex plane is shown in the figure. If
$\oint C\frac{2^z}{z^2-1}dz=-i \pi A$
where $i=\sqrt{-1}$, then the value of $A$ is ______ (rounded off to two decimal places)

 A 0.2 B 0.4 C 0.5 D 0.6
GATE EC 2022      Complex Variables
Question 5 Explanation:
\begin{aligned} Roots\;\; (z-1)(z+1)&=0\\ z&=\pm 1\\ \because \; z&=-1 \text{ is in contour} \end{aligned}
\begin{aligned} \Rightarrow \oint _c \frac{2^z}{z^2-1}&=\left [ \lim_{z \to -1} \frac{(z+1)2^z}{(z+1)(z-1)} \right ] \times 2 \pi i\\ &=\frac{2^{-1}}{(-1-1)} \times 2 \pi i\\ &=-\frac{1}{2} \pi i\\ \Rightarrow \;\; A&=1/2=0.5 \end{aligned}
 Question 6
The bar graph shows the frequency of the number of wickets taken in a match by a bowler in her career. For example, in 17 of her matches, the bowler has taken 5 wickets each. The median number of wickets taken by the bowler in a match is __________ (rounded off to one decimal place)

 A 3.2 B 4 C 6.4 D 5
GATE EC 2022      Probability and Statistics
Question 6 Explanation:
Sum of frequency of No. of wickets = 5 +7 + 8 + 25 + 20 +17 + 8 + 4 + 3 + 2 + 1 = 100
Mean = Average of the 50th & 51th matches = 4
 Question 7
Consider the following partial differential equation (PDE)
$a\frac{\partial^2 f(x,y)}{\partial x^2} +b\frac{\partial^2 f(x,y)}{\partial y^2}=f(x,y)$
, where $a$and $b$ are distinct positive real numbers. Select the combination(s) of values of the real parameters $\xi$ and $\eta$ such that $f(x,y)=e^{(\xi x+\eta y)}$ is a solution of the given PDE.
 A $\xi =\frac{1}{\sqrt{2a}},\eta =\frac{1}{\sqrt{2b}}$ B $\xi =\frac{1}{\sqrt{a}},\eta =0$ C $\xi =0,\eta =0$ D $\xi =\frac{1}{\sqrt{a}},\eta =\frac{1}{\sqrt{b}}$
GATE EC 2022      Differential Equations
Question 7 Explanation:
Given
\begin{aligned} f(x,y)&=e^{(\xi x+\eta y)}\\ \frac{\partial f(x,y)}{\partial x}&=\xi \cdot e^{(\xi x+\eta y)}\\ \frac{\partial^2 f(x,y)}{\partial x^2}&=\xi ^2\cdot e^{(\xi x+\eta y)}\\ \end{aligned}
Similarly,
$\frac{\partial^2 f(x,y)}{\partial y^2}=\eta ^2\cdot e^{(\xi x+\eta y)}$
Now, as given
$a \frac{\partial^2 f(x,y)}{\partial x^2}+b\frac{\partial^2 f(x,y)}{\partial y^2}=f(x,y)$
or
$a \times \xi ^2\cdot e^{(\xi x+\eta y)}+b \times \eta ^2\cdot e^{(\xi x+\eta y)}= e^{(\xi x+\eta y)}$
$a\cdot \xi ^2+b\cdot \eta ^2=1$
Thus, $\xi =\frac{1}{\sqrt{2a}}, \eta = \frac{1}{\sqrt{2b}}$
$(\xi =\frac{1}{\sqrt{a}}, \eta = 0)$ and $(\xi =0, \eta = \frac{1}{\sqrt{b}})$ satisfy the above result.
 Question 8
Consider a system of linear equations $Ax=b$, where
$A=\begin{bmatrix} 1 & -\sqrt{2} & 3\\ -1& \sqrt{2}& -3 \end{bmatrix},b=\begin{bmatrix} 1\\ 3 \end{bmatrix}$
This system of equations admits ______.
 A a unique solution for x B infinitely many solutions for x C no solutions for x D exactly two solutions for x
GATE EC 2022      Linear Algebra
Question 8 Explanation:
Here equation will be
$x-\sqrt{2}y+3z=1$
$-x+\sqrt{2}y-3z=3$
therefore inconsistant solution i.e. there will not be any solution.
 Question 9
Consider the two-dimensional vector field $\vec{F}(x,y)=x\vec{i}+y\vec{j}$, where $\vec{i}$ and $\vec{j}$ denote the unit vectors along the x-axis and the y-axis, respectively. A contour C in the x-y plane, as shown in the figure, is composed of two horizontal lines connected at the two ends by two semicircular arcs of unit radius. The contour is traversed in the counter-clockwise sense. The value of the closed path integral
$\oint _c \vec{F}(x,y)\cdot (dx\vec{i}+dy\vec{j})$

 A 0 B 1 C $8+2 \pi$ D -1
GATE EC 2022      Calculus
Question 9 Explanation:
$\oint \vec{F} (x,y)\cdot [dx\vec{i}+dy\vec{j}]$
Given $\vec{F} (x,y)=x\vec{i}+y\vec{j}$
$\therefore \int_{c}xdx+ydy=0$
Because here vector is conservative.
If the integral function is the total derivative over the closed contoure then it will be zero
 Question 10
In a high school having equal number of boy students and girl students, $75\%$ of the students study Science and the remaining $25\%$ students study Commerce. Commerce students are two times more likely to be a boy than are Science students. The amount of information gained in knowing that a randomly selected girl student studies Commerce (rounded off to three decimal places) is ______ bits.
 A 1.25 B 3.32 C 5.45 D 6.55
GATE EC 2021      Probability and Statistics
Question 10 Explanation:
\begin{aligned} \text{Given,}\qquad P(G)&=\frac{1}{2} \text { and } P(B)=\frac{1}{2} \\ P(C)&=\frac{1}{4} \text { and } P(S)=\frac{3}{4} \end{aligned}
Let probability of selected science student is a boy is $P(B/S) = x$
Given that Commerce students are two time more likely to be a boy than are Science students.
Then, probability of selected Commerce student is a boy
$P\left(\frac{B}{C}\right)=2 x$
We have to find probability of randomly selected girl studies Commerce i.e.
\begin{aligned} P\left(\frac{C}{G}\right)&=\frac{P(C \cap G)}{P(G)}=\frac{P(C) P(G / C)}{P(G)} \\ P\left(\frac{C}{G}\right)&=\frac{\frac{1}{4} \times P\left(\frac{G}{C}\right)}{\frac{1}{2}} &\ldots(i) \end{aligned}
To find $P(G / C)$, first we have to find $P(B/C).$
The probability of selected student is a boy
\begin{aligned} P(B) &=P(S) P(B / S)+P(C) \times P(B / C) \\ \frac{1}{2} &=\frac{3}{4} \times x+\frac{1}{4} \times 2 x \\ \Rightarrow\qquad\qquad x &=\frac{2}{5}\\ \text{Then},\qquad \qquad P(B / C)&=2 x=\frac{4}{5}\\ \text{We know that},\qquad \\ P\left(\frac{B}{C}\right)+P\left(\frac{G}{C}\right)&=1 \\ P\left(\frac{G}{C}\right) =1- \frac{4}{5}&=\frac{1}{5} \\ \text{From equation (i),} \\ P\left(\frac{G}{C}\right) = \frac{1}{4} \times \frac{1}{\frac{5}{1/2}} &= \frac{1}{10} \end{aligned}
The amount of information gained in knowing that a randomly selected girl student studies Commerce
$I\left ( \frac{C}{G} \right )=\log _2 \frac{1}{P\left ( \frac{C}{G} \right )}= \log _2 10=3.32$

There are 10 questions to complete.