# GATE EE 2014 SET 1

 Question 1
Given a system of equations
x + 2y + 2z = $b_1$
5x + y + 3z = $b_2$
What of the following is true regarding its solutions
 A The system has a unique solution for any given $b_1 \; and \; b_2$ B The system will have infinitely many solutions for any given $b_1 \; and \; b_2$ C Whether or not a solution exists depends on the given $b_1 \; and \; b_2$ D The systems would have no solution for any values of $b_1 \; and \; b_2$
Engineering Mathematics   Linear Algebra
 Question 2
Let $f(x)=xe^{-x}$. The maximum value of the function in the interval $(0,\infty )$ is
 A $e^{-1}$ B e C $1-e^{-1}$ D $1+e^{-1}$
Engineering Mathematics   Calculus
Question 2 Explanation:
\begin{aligned} f(x)&=xe^{-x}\\ f'(x)&=e^{-x}-xe^{-x}=0\\ e^{-x}(1-x)&=0\\ x&=1\\ f''(x)&=-e^{-x}-e^{-x}+xe^{-x}\\ &=e^{-x}(x-2)\\ f''(1)&=e^{-1}(-1)=-e^{-1} \lt 0 \end{aligned}
Hence f(x) has maximum value at x=1
$f(1)=1\cdot e^{-1}=e^{-1}$
 Question 3
The solution for the differential equation
$\frac{d^{2}x}{dt^{2}}=-9x$
with initial conditions x(0)=1 and $\frac{dx}{dt}|_{t=0}=1$, is
 A $t^{2}+t+1$ B $sin 3t+ \frac{1}{3} cos3t+\frac{2}{3}$ C $\frac{1}{3} sin 3t+cos3t$ D $cos3t+t$
Engineering Mathematics   Differential Equations
Question 3 Explanation:
\begin{aligned} \frac{d^2x}{dt^2}&=-9x\\ \frac{d^2x}{dt^2}+9x&=0\\ (D^2+9)x &=0 \end{aligned}
Auxiliary equation is $m^2+9=0$
\begin{aligned} m&= \pm 3i\\ x&= C_1 \cos 3t+C_2 \sin 3t\;\;...(i)\\ x(0) &=1\;\;i.e.\;x\rightarrow 1\; when t\rightarrow 0 \\ 1&=C_1 \\ \frac{dx}{dt}&=-3C_1 \sin 3t+3C_2 \cos 3t\;\;...(ii) \\ x'(0)&=1\;\; i.e.\;x'\rightarrow 1\; when \; t\rightarrow 0 \\ 1&=3C_2 \\ C_2&=\frac{1}{3}\\ \therefore \;x&=\cos 3t+\frac{1}{3}\sin 3t \end{aligned}
 Question 4
Let $X(s)=\frac{3s+5}{s^{2}+10s+21}$ be the Laplace Transform of a signal x(t). Then, x($0^+$) is
 A 0 B 3 C 5 D 21
Signals and Systems   Laplace Transform
Question 4 Explanation:
Given, $X(s)=\left [ \frac{3s+5}{s^2+10s+21} \right ]$
Using initial value theorem,
\begin{aligned} x(0^+) &= \lim_{s \to \infty }[sX(s)]\\ x(0^+) &= \lim_{s \to \infty } \left [ \frac{s(3s+5)}{s^2+10+21} \right ] \\ &= \lim_{s \to \infty }\left [ \frac{3+\frac{5}{s}}{1+\frac{10}{s}+\frac{21}{s^2}} \right ]=3 \end{aligned}
 Question 5
Let S be the set of points in the complex plane corresponding to the unit circle. (That is, S={z:|z|=1}). Consider the function f(z)=zz* where z* denotes the complex conjugate of z. The f(z) maps S to which one of the following in the complex plane
 A unit circle B horizontal axis line segment from origin to (1, 0) C the point (1, 0) D the entire horizontal axis
Engineering Mathematics   Complex Variables
Question 5 Explanation:

\begin{aligned} z&=x+iy\\ z^*&=x-iy\\ zz^*&=(x+iy)(x-iy)\\ &=x^2+y^2 \end{aligned}
which is equal to (1) always as given
\begin{aligned} |z|&=1\\ zz^*&=x^2+y^2 \end{aligned}

 Question 6
The three circuit elements shown in the figure are part of an electric circuit. The total power absorbed by the three circuit elements in watts is _____.
 A 110 B 250 C 330 D 360
Electric Circuits   Basics
Question 6 Explanation:
Given electrical circuit is shown below:

Applying KCL at node, current through 15V voltage source =2 A
Power absorbed by 100 V voltage source = 10x100=1000 Watt.
Power absorbed by 80 V voltage source =-(80x8)=-640 Watts and power absorbed by 15 V voltage source = -(15x2)=-30 Watt.
Therefore, total power absorbed by the three circuit element=(100-640-30_ watts =330 Watts
 Question 7
$C_0$ is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If, half of the entire gap as shown in figure (b) is filled with a dielectric of permittivity $\epsilon _r$ , the expression for the modified capacitance is
 A $\frac{C_{0}}{2}(1+\epsilon _{r})$ B $(C_{0}+\epsilon _{r})$ C $\frac{C_{0}}{2}(\epsilon _{r})$ D $C_{0}(1+\epsilon _{r})$
Electromagnetic Fields   Electrostatic Fields
Question 7 Explanation:
Let A be the area of the parallel plate capacitor and d be the distance between the plates.

Let $C_1$ be the capacitance of half portion with air as dielectic medium and $C_2$ be capacitance with a dielectic of permitivity, $\varepsilon _r$.
Then, $C_1=\frac{\varepsilon _0 \left ( \frac{A}{2} \right )}{d}$ (As area become half)
and $C_2=\frac{\varepsilon _0 \varepsilon _r \left ( \frac{A}{2} \right )}{d}$ (As area becomes half)
Now, these two capacitance will be in parallel if a voltage is applied between the plates as same potential difference will be there between both the capacitance.

Equivalent capacitance is
\begin{aligned} C_{eqv} &= C_1+C_2\\ &=\frac{\varepsilon _0 A}{2d}+ \frac{\varepsilon _0 \varepsilon _r A}{2d}\\ &= \frac{\varepsilon _0 A}{2d}(1+ \varepsilon _r)=\frac{C_0}{2}(1+\varepsilon _r) \end{aligned}
Therefore, modified capacitance,
$C_{eqv}=\frac{C_0}{2}(1+\varepsilon _r)$
 Question 8
A combination of 1 $\mu$F capacitor with an initial voltage $v_c(0)=-2V$ in series with a 100 $\Omega$ resistor is connected to a 20 mA ideal dc current source by operating both switches at t=0s as shown. Which of the following graphs shown in the options approximates the voltage $v_s$ across the current source over the next few seconds ?

 A A B B C C D D
Electric Circuits   Transients and Steady State Response
Question 8 Explanation:
Given $C=1\mu F, V_c(0)=-2V$, $R=100\Omega , I=20mA$. Circuit fot the given condition at time $t \gt 0$ is shown below:

Applying KVL, we have,$V_s(s)=\left ( \frac{-2}{s} \right )+\frac{I}{s} \left ( R+\frac{I}{Cs} \right )$
$\;\;=\frac{1}{s}\left [ -2+IR+\frac{I}{Cs} \right ]$
$\;\;=\frac{1}{s}\left [ (IR-2) +\frac{I}{Cs}\right ]$
Putting values of R, C and I, we get,
$V_s(s)=\frac{1}{s}\left [ (20 \times 10^{-3} \times 200-2)+\left ( \frac{20 \times 10^{-3}}{10^{-6}} \right ) \times \frac{1}{s} \right ]$
$\;\;=\frac{1}{s}\left [ (2-2)+20 \times 10^3 \times \frac{1}{s} \right ]$
$\;\;=\frac{20 \times 10^3}{s^2}$
$\therefore \;\; V_s(s)=\frac{20 \times 10^3}{s^2}$
$V_s(t)=20000 t u(t)$
$\therefore \;\; V_s(s)=(20000)tu(t)$
Which is equation of a straight line passing through origin. Hence option (C) is correct.
 Question 9
x(t) is nonzero only for $T_x \lt t \lt T'_x$, and similarly, y(t) is nonzero only for $T_y \lt t \lt T'_y$. Let z(t) be convolution of x(t) and y(t). Which one of the following statements is TRUE ?
 A z(t) can be nonzero over an unbounded interval. B z(t) is nonzero for $t \lt T_{x}+T_{y}$ C z(t) is zero outside of $T_{x}+T_{y} \lt t \lt T'_{x}+T'_{y}$ D z(t) is nonzero for $t\gt T'_{x}+T'_{y}$
Signals and Systems   Linear Time Invariant Systems
Question 9 Explanation:
$x(t)$ is non-zero for $T_x \lt t \lt T'_x$
and $y(t)$ is non-zero for $T_y \lt t \lt T'_y$
$\because \;\; z(t)=x(t)\otimes y(t)$
then the limits of the resultant signal is
$T_x+T_y \lt t \lt T'_x+T'_y$
i.e. $z(t)$ is non-zero for $T_x+T_y \lt t \lt T'_x+T'_y$
 Question 10
For a periodic square wave, which one of the following statements is TRUE ?
 A The Fourier series coefficients do not exist B The Fourier series coefficients exist but the reconstruction converges at no point C The Fourier series coefficients exist but the reconstruction converges at most point D The Fourier series coefficients exist and the reconstruction converges at every point.
Signals and Systems   Fourier Series
Question 10 Explanation:

Reconstruction of signal by its Fourier series coefficient is not possible at those points where signal is discontinuous.
In the above figure, at integer multiples of 'T/2', signal recovery is not possible by using its coefficient.
Therefore, reconstruction of x(t) by using its coefficient is possible at most of the points except those instants where x(t) is discontinous.
There are 10 questions to complete.