Question 1 |
The maximum value attained by the function f(x) = x(x- 1)(x - 2) in the interval [1, 2] is _____.
0 | |
1 | |
2 | |
4 |
Question 1 Explanation:
\begin{aligned} f(x) &=x^3-3x^2+2x \\ f'(x)&=3x^2-6x+2\\ f'(x)&=0 \;\text{for stationary point} \end{aligned}
stationary points are 1+\frac{1}{\sqrt{3}}
only 1+\frac{1}{\sqrt{3}} lies in [1,2]
\begin{aligned} f(1) &=0 \\ f(2)&=0 \\ f\left ( 1+\frac{1}{\sqrt{3}} \right )&=-\frac{2}{3\sqrt{3}} \end{aligned}
Maximum value is 0.
stationary points are 1+\frac{1}{\sqrt{3}}
only 1+\frac{1}{\sqrt{3}} lies in [1,2]
\begin{aligned} f(1) &=0 \\ f(2)&=0 \\ f\left ( 1+\frac{1}{\sqrt{3}} \right )&=-\frac{2}{3\sqrt{3}} \end{aligned}
Maximum value is 0.
Question 2 |
Consider a 3 x 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.
1 | |
2 | |
3 | |
4 |
Question 2 Explanation:
A=\begin{bmatrix} 1 & 1 &1 \\ 1 & 1 &1 \\ 1& 1 & 1 \end{bmatrix} Eigen value are 0,0,3
Question 3 |
The Laplace Transform of f(t)=e^{2t}sin(5t)u(t) is
\frac{5}{s^{2}-4s+29} | |
\frac{5}{s^{2}+5} | |
\frac{s-2}{s^{2}-4s+29} | |
\frac{5}{s+5} |
Question 3 Explanation:
Laplace transform of \sin 5t u(t)\rightarrow \frac{5}{s^2+25}
e^{2t}\sin 5t u(t)\rightarrow \frac{5}{(s-2)^2+25}=\frac{5}{s^2-4s+29}
e^{2t}\sin 5t u(t)\rightarrow \frac{5}{(s-2)^2+25}=\frac{5}{s^2-4s+29}
Question 4 |
A function y(t), such that y(0)=1 and y(1)=3e^{-1}, is a solution of the differential equation \frac{d^{2}y}{dt^{2}}+2\frac{dy}{dt}+y=0. Then y(2) is
5e^{-1} | |
5e^{-2} | |
7e^{-1} | |
7e^{-2} |
Question 4 Explanation:
Auxiliary equation,
\begin{aligned} m^2+2m+1&=0 \\ m&= -1,-1\\ y&=(c_1+c_2t)e^{-t} \\ y(0) &=1 \\ c_1&=1 \\ y&= (1+c_2t)e^{-t}\\ y(1)&=3e^{-1} \\ \Rightarrow \; (1+c_2)e^{-3} &=3e^{-3} \\ c_2&=2 \\ y&=(1+2t)e^{-t} \\ y(2)&=5e^{-2} \end{aligned}
\begin{aligned} m^2+2m+1&=0 \\ m&= -1,-1\\ y&=(c_1+c_2t)e^{-t} \\ y(0) &=1 \\ c_1&=1 \\ y&= (1+c_2t)e^{-t}\\ y(1)&=3e^{-1} \\ \Rightarrow \; (1+c_2)e^{-3} &=3e^{-3} \\ c_2&=2 \\ y&=(1+2t)e^{-t} \\ y(2)&=5e^{-2} \end{aligned}
Question 5 |
The value of the integral \oint_{c}\frac{2z+5}{(z-\frac{1}{2})(z^{2}-4z+5)}dz over the contour |z|=1, taken in the anti-clockwise direction, would be
\frac{24 \pi i}{13} | |
\frac{48 \pi i}{13} | |
\frac{24}{13} | |
\frac{12}{13} |
Question 5 Explanation:
Singlarilies, Z=\frac{1}{2}, 2\pm i
Only, Z=\frac{1}{2} lies inside C
By residue theorem,
\oint _C=2 \pi i(R)=\frac{48 \pi i}{13}
Residue at \frac{1}{2}
=R_{1/2}=\lim_{Z \to 1/2}\left [ \left ( Z-\frac{1}{2} \right )\cdot \frac{2Z+5}{\left ( Z-\frac{1}{2} \right )(Z^2+4Z+5)} \right ]=\frac{24}{13}
Only, Z=\frac{1}{2} lies inside C
By residue theorem,
\oint _C=2 \pi i(R)=\frac{48 \pi i}{13}
Residue at \frac{1}{2}
=R_{1/2}=\lim_{Z \to 1/2}\left [ \left ( Z-\frac{1}{2} \right )\cdot \frac{2Z+5}{\left ( Z-\frac{1}{2} \right )(Z^2+4Z+5)} \right ]=\frac{24}{13}
There are 5 questions to complete.