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 ___.
If a and b are positive integers, the value of \int_{-\infty }^{\infty }e^{-a(x+b)^2}dx is ___.
\sqrt{\pi a} | |
\sqrt{\frac{\pi}{a}} | |
b \sqrt{\pi a} | |
b \sqrt{\frac{\pi}{a}} |
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 ______.
14 | |
27 | |
28 | |
31 |
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}
\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)?
1 | |
2 | |
3 | |
4 |
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
\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).
\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).
2.95 | |
8.32 | |
12.65 | |
5.23 |
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
only sine terms | |
only cosine terms | |
both sine and cosine terms | |
only sine terms and a non-zero constant |
Question 5 Explanation:
f(x)=x^3, \;\; -1 \leq x \leq 1
It is an odd function
Fourier series contains only sine terms.
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
\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
4 \pi | |
3 \pi | |
4 \pi/3 | |
0 |
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
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
\alpha =-3\pi, \text{ and }p= \pi | |
\alpha =-2\pi, \text{ and }p= 2\pi | |
\alpha =\pi, \text{ and }p= \pi | |
\alpha =2\pi, \text{ and }p= 3\pi |
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).
1.67 | |
1.12 | |
2.44 | |
3.25 |
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
Q^Tr | |
Q^{-1}r | |
\frac{r}{r^Tr} | |
r |
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
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
0 | |
\frac{1}{6} | |
\frac{4}{3} | |
\pi |
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.
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
in que 10 x->1 not zero
Thank You Jashwanth Reddy Earla,
We have updated the question.
que 35 f(x)= 2x^3-3x^2
Thank You Jashwanth Reddy Earla,
We have updated the question.
Question 101- Answer is maximum(x,-x)
Thank You Ajay,
We have updated the answer.
Error in statement of Q.40
error in que 40
Please correct question number 6 , Function have extra Y