# GATE EE 2016 SET 1

 Question 1
The maximum value attained by the function $f(x) = x(x- 1)(x - 2)$ in the interval [1, 2] is _____.
 A 0 B 1 C 2 D 4
Engineering Mathematics   Calculus
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.
 Question 2
Consider a 3 x 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.
 A 1 B 2 C 3 D 4
Engineering Mathematics   Linear Algebra
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
 A $\frac{5}{s^{2}-4s+29}$ B $\frac{5}{s^{2}+5}$ C $\frac{s-2}{s^{2}-4s+29}$ D $\frac{5}{s+5}$
Signals and Systems   Laplace Transform
Question 3 Explanation:
Laplace transform of $\sin 5t u(t)\rightarrow \frac{5}{s^2+25}$
$e^{2t}\sin 5t u(t)\rightarrow \frac{5}{(s-2)^2+25}=\frac{5}{s^2-4s+29}$
 Question 4
A function y(t), such that y(0)=1 and y(1)=3$e^{-1}$, is a solution of the differential equation $\frac{d^{2}y}{dt^{2}}+2\frac{dy}{dt}+y=0$. Then y(2) is
 A $5e^{-1}$ B $5e^{-2}$ C $7e^{-1}$ D $7e^{-2}$
Engineering Mathematics   Calculus
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}
 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
 A $\frac{24 \pi i}{13}$ B $\frac{48 \pi i}{13}$ C $\frac{24}{13}$ D $\frac{12}{13}$
Engineering Mathematics   Calculus
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 $\frac{1}{2}$
$=R_{1/2}=\lim_{Z \to 1/2}\left [ \left ( Z-\frac{1}{2} \right )\cdot \frac{2Z+5}{\left ( Z-\frac{1}{2} \right )(Z^2+4Z+5)} \right ]=\frac{24}{13}$

There are 5 questions to complete.