Calculus

 Question 1
Given $\int_{-\infty }^{\infty }e^{-x^2}dx=\sqrt{\pi}$
If a and b are positive integers, the value of $\int_{-\infty }^{\infty }e^{-a(x+b)^2}dx$ is ___.
 A $\sqrt{\pi a}$ B $\sqrt{\frac{\pi}{a}}$ C $b \sqrt{\pi a}$ D $b \sqrt{\frac{\pi}{a}}$
GATE ME 2022 SET-2   Engineering Mathematics
Question 1 Explanation:
\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}
 Question 2
Consider a cube of unit edge length and sides parallel to co-ordinate axes, with its centroid at the point (1, 2, 3). The surface integral $\int_{A}^{}\vec{F}.d\vec{A}$ of a vector field $\vec{F}=3x\hat{i}+5y\hat{j}+6z\hat{k}$ over the entire surface $A$ of the cube is ______.
 A 14 B 27 C 28 D 31
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Question 2 Explanation:
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}
 Question 3
$F(t)$ is a periodic square wave function as shown. It takes only two values, 4 and 0, and stays at each of these values for 1 second before changing. What is the constant term in the Fourier series expansion of $F(t)$?

 A 1 B 2 C 3 D 4
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Question 3 Explanation:
The constant term in the Fourier series expansion of $F(t)$ is the average value of $F(t)$ in one fundamental period i.e.,
$\frac{\int_{0}^{1}4dt+\int_{1}^{2}0dt}{2}=\frac{4}{2}=2$
 Question 4
Consider two vectors:
$\vec{a}=5i+7j+2k$
$\vec{b}=3i-j+6k$
Magnitude of the component of $\vec{a}$ orthogonal to $\vec{b}$ in the plane containing the vectors $\vec{a}$ and $\vec{b}$ is __________ (round off to 2 decimal places).
 A 2.95 B 8.32 C 12.65 D 5.23
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Question 4 Explanation:

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}
 Question 5
The Fourier series expansion of $x^3$ in the interval $-1\leq x\leq 1$ with periodic continuation has
 A only sine terms B only cosine terms C both sine and cosine terms D only sine terms and a non-zero constant
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Question 5 Explanation:
$f(x)=x^3, \;\; -1 \leq x \leq 1$
It is an odd function
Fourier series contains only sine terms.
 Question 6
Given a function $\varphi =\frac{1}{2}(x^2+y^2+z^2)$ in threedimensional Cartesian space, the value of the surface integral
$\oiint_{S}{\hat{n}.\triangledown \varphi dS}$
where S is the surface of a sphere of unit radius and $\hat{n}$ is the outward unit normal vector on S, is
 A $4 \pi$ B $3 \pi$ C $4 \pi/3$ D $0$
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Question 6 Explanation:
\begin{aligned} \varphi &=\frac{1}{2}(x^2+y^2+z^2)\\ \triangledown \varphi &=(x\hat{i}+y\hat{j}+z\hat{k})=\bar{F}\\ \oiint_{S}(\triangledown \varphi\cdot \bar{n})dS&=\int \int _v\int Div\; \bar{F} dv\\ &=\int \int \int 3dv\\ &=3v\\ &=3\left ( \frac{4}{3} \pi \right )=4\pi \end{aligned}
 Question 7
The limit
$p=\lim_{x \to \pi}\left ( \frac{x^2+\alpha x+2\pi ^2}{x-\pi+2 \sin x } \right )$
has a finite value for a real $\alpha$. The value of $\alpha$ and the corresponding limit $p$ are
 A $\alpha =-3\pi, \text{ and }p= \pi$ B $\alpha =-2\pi, \text{ and }p= 2\pi$ C $\alpha =\pi, \text{ and }p= \pi$ D $\alpha =2\pi, \text{ and }p= 3\pi$
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Question 7 Explanation:
\begin{aligned} p&=\lim_{x \to \pi}\left ( \frac{x^2+\alpha x+2 \pi ^2}{x-\pi+2 \sin x} \right ) \\ p&=\left ( \frac{\pi ^2+\alpha \pi +2 \pi ^2}{\pi-\pi+2 \sin \pi } \right ) \\ &= \frac{2 \pi ^2+\alpha \pi}{0}\\ \therefore \;\; \alpha &= -3 \pi\\ p&=\lim_{x \to \pi}\left ( \frac{x^2- 3 \pi x+2 \pi ^2}{x-\pi+2 \sin x} \right ) \\ &=\lim_{x \to \pi}\left ( \frac{2x- 3 \pi }{1+2 \cos \pi} \right ) \\ &= \frac{2 \pi-3 \pi}{1-2}=\frac{-\pi}{-1}=\pi\\ \therefore \; \alpha &=-3 \pi \text{ and }p= \pi \end{aligned}
 Question 8
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 8 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 9
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 $r$
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Question 9 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 10
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 10 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}

There are 10 questions to complete.

13 thoughts on “Calculus”

1. The question posted have minor mistakes in it, in answers it show up something else other than what mentioned in question. Please look into it ,otherwise its a good platform to practise

• Thank you for your suggestions.
Can you please share the questions numbers for the above suggestions.

• question number 10

• question no 9 mein b vector, wrong type ho gaya hai

2. in que 10 x->1 not zero

• Thank You Jashwanth Reddy Earla,
We have updated the question.

3. que 35 f(x)= 2x^3-3x^2

• Thank You Jashwanth Reddy Earla,
We have updated the question.