Question 1 |
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 1 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 2 |
Solution of \triangledown^2T=0 in a square domain (0 \lt x \lt 1
and 0 \lt y \lt 1) with boundary conditions:
T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y is
T(x, 0) = x; T(0, y) = y; T(x, 1) = 1 + x; T(1, y) = 1 + y is
T(x,y)=x-xy+y | |
T(x,y)=x+y | |
T(x,y)=-x+y | |
T(x,y)=x+xy+y |
Question 2 Explanation:
T(x, 0) = x \Rightarrow
option (c) is not correct.
T(0, y) = y \Rightarrow all options satisfied.
T(x, 1) = 1 + x; \Rightarrow only option (b) is satisfied.
T(1, y) = 1 + y is \Rightarrow only option (b) is satisfied.
T(0, y) = y \Rightarrow all options satisfied.
T(x, 1) = 1 + x; \Rightarrow only option (b) is satisfied.
T(1, y) = 1 + y is \Rightarrow only option (b) is satisfied.
Question 3 |
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 3 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 4 |
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 4 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 5 |
If A=\begin{bmatrix}
10 &2k+5 \\
3k-3 & k+5
\end{bmatrix} is a symmetric matrix, the
value of k is ___________.
8 | |
5 | |
-0.4 | |
\frac{1+\sqrt{1561}}{12} |
Question 5 Explanation:
A=\begin{bmatrix}
10 & 2k+5\\
3k-3 & k+5
\end{bmatrix}
(2k + 5) = (3k - 3)
k=8
(2k + 5) = (3k - 3)
k=8
There are 5 questions to complete.