# Calculus

 Question 1
For two n-dimensional real vectors $P$ and $Q$, the operation $s(P,Q)$ is defined as follows:

$s(P,Q) = \displaystyle \sum_{i=1}^n (P[i] \cdot Q[i])$

Let $\mathcal{L}$ be a set of 10-dimensional non-zero real vectors such that for every pair of distinct vectors $P,Q \in \mathcal{L}, s(p,Q)=0$. What is the maximum cardinality possible for the set $\mathcal{L}$?
 A 9 B 10 C 11 D 100
GATE CSE 2021 SET-2   Engineering Mathematics
Question 1 Explanation:
 Question 2
A function is defined in Cartesian coordinate system as $f(x, y)=x e^{y}$. The value of the directional derivative of the function (in integer) at the point (2,0) along the direction of the straight line segment from point (2,0) to point $\left(\frac{1}{2}, 2\right)$ is _______
 A 0.5 B 1 C 1.5 D 2.2
GATE CE 2021 SET-2   Engineering Mathematics
Question 2 Explanation:
\begin{aligned} f(x, y)&=x e^{y}\\ \mathrm{P}(2,0) \text{ and } Q\left(\frac{1}{2}, 2\right)\\ \operatorname{grad} f &=\hat{i}\left(e^{y}\right)+\hat{j}\left(x e^{y}\right)+\hat{k}(0) \\ \Rightarrow \qquad \qquad \qquad (\operatorname{grad} f)_{p} &=\hat{i}+2 \hat{j} ]\\ \overline{P Q}&=\left(\frac{1}{2}-2\right) \hat{i}+(2-0) \hat{j}=-\frac{3}{2} \hat{i}+2 \hat{j}\\ \text{Required directional derivative} &=(\text { grad } f)_{P} \widehat{P Q}\\ &=(\hat{i}+2 \hat{j}) \times \frac{\left(-\frac{3}{2} \hat{i}+2 \hat{j}\right)}{\sqrt{\frac{9}{4}+4}}=\frac{\frac{-3}{2}+4}{\sqrt{\frac{25}{4}}} \\ &=\frac{\frac{-3+8}{2}}{\left(\frac{5}{2}\right)}=1 \end{aligned}
 Question 3
Find the positive real root of $x^3-x-3=0$ using Newton-Raphson method. If the starting guess $(x_0)$ is 2, the numerical value of the root after two iterations $(x_2)$ is _______ (round off to two decimal places).
 A 1.67 B 1.12 C 2.44 D 3.25
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Question 3 Explanation:
\begin{aligned} \text { Given, }\qquad\quad f(x) &=x^{3}-x-3, \quad x_{0}=2 \\ f^{\prime}(x)&=3 x^{2}-1\\ \text { Iteration 1: } \quad x_{1}&=x_{0}-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)}=2-\frac{(8-2-3)}{3(4)-1}=1.72\\ \text { Iteration 2: } \quad x_{2}&=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}=1.72-\frac{\left(1.72^{3}-1.72-3\right)}{3(1.72)^{2}-1}=1.67 \end{aligned}
 Question 4
Let the superscript T represent the transpose operation. Consider the function $f(x)=\frac{1}{2}x^T Qx=r^Tx, \; \text{ where } x \text{ and }r \text{ are }n \times 1$ vectors and $Q$ is a symmetric $n \times n$ matrix. The stationary point of $f(x)$ is
 A $Q^Tr$ B $Q^{-1}r$ C $\frac{r}{r^Tr}$ D $r$
GATE ME 2021 SET-2   Engineering Mathematics
Question 4 Explanation:
\begin{aligned} \text{Let}\qquad Q=\left[\begin{array}{ll}a & c \\c & b\end{array}\right], x&=\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right], R=\left[\begin{array}{l}r_{1} \\r_{2}\end{array}\right] \\ F(x)&=\frac{1}{2}\left(x_{1}, x_{2}\right)\left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]-\left[r_{1} r_{2}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right] \\ &=\frac{1}{2}\left[a x_{1}^{2}+b x_{2}^{2}+2 c x_{1} x_{2}\right]-\left[r_{1} x_{1}+r_{2} x_{2}\right]\\ \text{i.e.}\qquad \qquad U\left(x_{1}, x_{2}\right)&=\frac{1}{2} a x_{1}^{2}+\frac{1}{2} b x_{2}^{2}+c_{1} x_{1} x_{2}-r_{1} x_{1}-r_{2} x_{2} \end{aligned}
Now, for critical point, $\frac{\partial u}{\partial x_{1}}=0$ and $\frac{\partial u}{\partial x_{2}}=0$
$\Rightarrow \quad a_{1} x_{1}+c x_{2}-r_{1}=0 \quad \text { and } c x_{2}+c x_{1}-r_{2}=0$
In matrix form we can write it as
\begin{aligned} \left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l} x_{1} \\m_{2}\end{array}\right] &=\left[\begin{array}{l}r_{1} \\ r_{2}\end{array}\right] \\\Rightarrow \qquad Q x &=r \end{aligned}
By multiplying both side by $Q^{-1}$
$x=Q^{-1} r$
 Question 5
The value of $\int_{0}^{\pi /2}\int_{0}^{\cos \theta }r \sin \theta dr d\theta$ is
 A 0 B $\frac{1}{6}$ C $\frac{4}{3}$ D $\pi$
GATE ME 2021 SET-2   Engineering Mathematics
Question 5 Explanation:
\begin{aligned} I &=\int_{\theta=0}^{\theta=\frac{\pi}{2}} \int_{r=0}^{r=\cos \theta} r \sin \theta d r d \theta \\ &=\int_{\theta=0}^{\frac{\pi}{2}}\left[\frac{r^{2}}{2}\right]_{0}^{\infty \cos \theta} \times \sin \theta d \theta \\ &=\frac{1}{2} \int_{\theta}^{\frac{\pi}{2}} \sin \theta \cdot \cos ^{2} \theta d \theta \\ \text{Let}, \qquad\cos \theta &=t\\ -\sin \theta d \theta &=d t \\ \cos \theta &=t \\ \text{at},\qquad\theta &=\frac{\pi}{2} ; t=0 \\ \theta &=0, t=1 \\ &=\int_{1}^{0} \frac{-t^{2}}{2} d t \\ &=\frac{-1}{2}\left[\frac{t^{3}}{3}\right]_{1}^{0}=\frac{-1}{2} \times\left(\frac{-1}{3}\right) \\ &=\frac{-1}{2}\left[\frac{-1}{3}\right]=\frac{1}{6} \end{aligned}
 Question 6
The value (round off to one decimal place) of $\int_{-1}^{1} x e^{|x|} d x$ is ________
 A 1.2 B 3.1 C 2.4 D 0
GATE CE 2021 SET-2   Engineering Mathematics
Question 6 Explanation:
\begin{aligned} \int_{-1}^{1} x e^{-|x|} d x&=0\\ \text { As } \qquad \qquad \qquad f(-x)&=x e^{x} \text { is an odd function. } \end{aligned}
 Question 7
The unit normal vector to the surface $X^{2}+Y^{2}+Z^{2}-48=0$ at the point (4,4,4) is
 A $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$ B $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ C $\frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}, \frac{2}{\sqrt{2}}$ D $\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}$
GATE CE 2021 SET-2   Engineering Mathematics
Question 7 Explanation:
\begin{aligned} \phi &=x^{2}+y^{2}+z^{2}-48, P(4,4,4) \\ \operatorname{grad} \phi &=\vec{\nabla} \phi=\hat{i} \frac{\partial \phi}{\partial x}+\hat{j} \frac{\partial \phi}{\partial y}+\hat{k} \frac{\partial \phi}{\partial z} \\ &=(2 x) \hat{i}+(2 y) \hat{j}+(2 z) \hat{k} \\ \vec{n} &=(\operatorname{grad} \phi)_{P}=8 \hat{i}+8 \hat{j}+8 \hat{k} \\ \hat{n} &=\frac{\vec{n}}{|\vec{n}|}=\frac{8 \hat{i}+8 \hat{j}+8 \hat{k}}{\sqrt{64+64+64}}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \\ & \simeq\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},\right) \end{aligned}
 Question 8
The value of $\lim _{x \rightarrow \infty} \frac{x \ln (x)}{1+x^{2}}$ is
 A 0 B 1 C 0.5 D $\infty$
GATE CE 2021 SET-2   Engineering Mathematics
Question 8 Explanation:
\begin{aligned} &\lim _{x \rightarrow \infty}\left(\frac{x \ln x}{x^{2}+1}\right) \qquad \qquad \qquad \qquad \qquad \left(\frac{\infty}{\infty} \text { form }\right)\\ &=\lim _{x \rightarrow \infty}\left(\frac{x\left(\frac{1}{x}\right)+\ln x}{2 x}\right) \qquad \qquad \qquad \left(\frac{\infty}{\infty} \text { form }\right)\\ \lim _{x \rightarrow \infty}\left(\frac{0+\frac{1}{x}}{2}\right)&=\lim _{x \rightarrow \infty}\left(\frac{1}{2 x}\right)=\frac{1}{2 \times \infty}=0 \end{aligned}
 Question 9
Let $f(x)=x^2-2x+2$ be a continuous function defined on $x \in [1,3]$. The point $x$ at which the tangent of $f(x)$ becomes parallel to the straight line joining $f(1)$ and $f(3)$ is
 A 0 B 1 C 2 D 3
GATE ME 2021 SET-1   Engineering Mathematics
Question 9 Explanation:
By Lagrangian mean value theorem,
\begin{aligned} \frac{f(3)-f(1)}{3-1} &=f^{\prime}(c) \\ \Rightarrow\quad \frac{5-1}{3-1} &=2 x-2\\ \Rightarrow\quad 2 x-2&=4\\ \Rightarrow\quad x&=2 \in(1,3) \end{aligned}
 Question 10
Consider the integral
$\oint _{c}\frac{sin\left ( x \right )}{x^{2}\left ( x^{2}+4 \right )}dx$
where C is a counter-clockwise oriented circle defined as $\left | x-i \right |=2$. The value of the integral is
 A $-\frac{\pi }{8}\sin\left ( 2i \right )$ B $\frac{\pi }{8}\sin\left ( 2i \right )$ C $-\frac{\pi }{4}\sin\left ( 2i \right )$ D $\frac{\pi }{4}\sin\left ( 2i \right )$
GATE EC 2021   Engineering Mathematics
Question 10 Explanation:
MARKS TO ALL AS PER IIT ANSWER KEY
$\oint_{c} \frac{\sin x}{x^{2}\left(x^{2}+4\right)} d x, c:|x-i|=2$
Poles are given by $x^{2}=0$ and $x^{2}+4=0$
$\Rightarrow\qquad x=0$ is a pole of order '2'
$x=2$ i are simple nodes
$x=0$ lies inside 'c'
$x=2$ i lies inside 'c'
$x=-2$ i lies outside 'c'
\begin{aligned} \text{Res}_{0} &=\frac{1}{(2-1)} \lim _{x \rightarrow 0} \frac{d}{d z}\left[(x-0)^{2} \frac{\sin x}{x^{2}\left(x^{2}+4\right)}\right] \\ &=\lim _{x \rightarrow 0} \frac{\left(x^{2}+4\right) \cos x-\sin x(2 x)}{\left(x^{2}+4\right)^{2}}=\frac{1}{4} \\ \text{Res}_{2 i} &=\lim _{x \rightarrow 2 i}(x-2 i) \frac{\sin x}{x^{2}(x-2 i)(x+2 i)}=\frac{\sin (2 i)}{(-4)(4 i)} \\ \text{By CRT}\quad \oint_{c} f d x &=2 \pi i\left[\text{Res}_{0}+\text{Res}_{2 i}\right]=2 \pi i\left[\frac{1}{4}+\frac{\sin (2 i)}{-16}\right] \end{aligned}

There are 10 questions to complete.

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