# GATE EC 2014 SET-4

 Question 1
The series $\sum_{^{n=0}}^{\infty }\frac{1}{n!}$ converges to
 A 2 ln 2 B $\sqrt{2}$ C 2 D e
Engineering Mathematics   Calculus
Question 1 Explanation:
Given,
\begin{aligned} \text { Let, } x(n) &=\sum_{n=0}^{\infty} \frac{1}{n !} \\ &=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots \\ &=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots \end{aligned}
Also we know that expression of $e^{x}$
\begin{aligned} e^{x}&=1+x+\frac{1}{2} x^{2}+\frac{1}{6} x^{3}+\frac{1}{24} x^{4}+\cdots\\ \text{Put }x&=1\text{ in above expression}\\ e^{1}&=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots \\ e&=\sum_{n=0}^{\infty} \frac{1}{n !} \end{aligned}
 Question 2
The magnitude of the gradient for the function $f(x,y,z)=x^{2}+3y^{2}+z^{3}$ at the point (1,1,1) is _____.
 A 5 B 6 C 7 D 8
Engineering Mathematics   Calculus
Question 2 Explanation:
\begin{aligned} f(v, y, z) &=x^{3}+3 y^{2}+z^{3} \\ \nabla f &=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{i}+\frac{\partial f}{\partial z} \hat{k} \\ &=2 x \hat{i}+6 y \hat{j}+3 z^{2} \hat{k} \\ (\Delta t)_{(1,1,1)} &=2 \hat{i}+6 \hat{j}+3 \hat{k} \\ |\Delta+|_{(1,1,1)} &=\sqrt{4+36+9}=\sqrt{49}=7 \end{aligned}
 Question 3
Let X be a zero mean unit variance Gaussian random variable. E[|X|] is equal to ______.
 A 0.8 B 1.8 C 0.2 D 1.2
Communication Systems   Random Processes
Question 3 Explanation:
For a Gaussian random variable,
\begin{aligned} f_{x}(x)&=\frac{1}{\sigma_{x} \sqrt{2 \pi}} e^{\left(\frac{\mu-x}{2 \sigma^{2}}\right)^{2}} \\ \text{For mean} (\mu)&=0 \text{ and Variance }\left(\sigma_{x}^{2}\right)=1 \\ f_{X}(x) &=\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} \\ E[|x|] &=\int_{-\infty}^{\infty}|x| f_{X}(x) d x \\ &=\int_{-\infty}^{\infty}|x| \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x \\ =& \int_{-\infty}^{0}-x \cdot \frac{1}{\sqrt{2 \pi}} e^{-x^{2}} d x+\int_{0}^{\infty} x \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x\\ &=2 \int_{0}^{\infty} x \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} x \cdot e^{\frac{-x^{2}}{2}} d x \\ &\text{Let }\quad -\frac{x^{2}}{2}=t \\ &\therefore \quad-\frac{2 x d x}{2}=d t \\ &\text{Hence }\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty}-e^{-t} d t=0.8 \\ &\therefore \quad E[|x|]=0.8 \\ \end{aligned}
 Question 4
If a and b are constants, the most general solution of the differential equation $\frac{d^{2}x}{dt^{2}}+2\frac{dx}{dt}+x=0$ is
 A $ae^{-t}$ B $ae^{-t}+bte^{-t}$ C $ae^{t}+bte^{-t}$ D $ae^{-2t}$
Engineering Mathematics   Differential Equations
Question 4 Explanation:
The differential equation is given as
\begin{aligned} \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+x&=0 \\ y&=C \cdot F+P \cdot I\\ \end{aligned}
since, Q=0, i.e. RHS term is zero, so there will
be no particular integral.
\begin{aligned} \therefore \quad y&=C \cdot F\\ \text{Let, }\quad \frac{\partial}{\partial x}&=D \\ \text{So. }\quad \left(D^{2}+2 D+1\right) x&=0 \\ \therefore \quad(D+1)^{2}&=0 \\ \therefore \quad y&=a e^{-t}+b t e^{t} \end{aligned}
 Question 5
The directional derivative of $f(x,y)=\frac{xy}{\sqrt{2}}(x+y)$ at (1,1) in the direction of the unit vector at an angle of $\frac{\pi }{4}$ with y-axis, is given by____.
 A 1 B 2 C 3 D 4
Engineering Mathematics   Calculus
Question 5 Explanation:
\begin{aligned} f(x, y) &=\frac{x y}{\sqrt{2}}(x+y)=\frac{x^{2} y+x y^{2}}{\sqrt{2}} \\ \nabla f &=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j} \\ &=\frac{2 x y+y^{2}}{\sqrt{2}}\hat{i}+\frac{(x^{2}+2xy)}{\sqrt{2}}\hat{j} \end{aligned}
The direction is $\hat{n}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$
Directional derivative in direction of $\hat{n}$ is
$\begin{array}{l} =\left(\frac{2 x y+y^{2}}{2}+\frac{x^{2}+2 x y}{2}\right)_{(1,1)} \\ =\frac{3}{2}+\frac{3}{2}=3 \end{array}$
 Question 6
The circuit shown in the figure represents a
 A voltage controlled voltage source B voltage controlled current source C current controlled current source D current controlled voltage source
Network Theory   Basics of Network Analysis
 Question 7
The magnitude of current (in mA) through the resistor $R_{2}$ in the figure shown is______.
 A 1.2 B 2.1 C 2.8 D 3.6
Network Theory   Basics of Network Analysis
Question 7 Explanation:
Using source transformation, we get,

Applying KVL in above circuit, we get,
$20-2 I-I-4 I+8-3 I=0$
or $28=10 I$
or $\quad I=2.8 \mathrm{mA}$
 Question 8
At T = 300 K, the band gap and the intrinsic carrier concentration of GaAs are 1.42 eV and $10^{6}cm^{-3}$, respectively. In order to generate electron hole pairs in GaAs, which one of the wavelength $(\lambda _{C})$ ranges of incident radiation, is most suitable ? (Given that : Plank's constant is $6.62 \times 10^{-34}$J-s, velocity of light is $3 \times 10^{10} cm/s$ and charge of electron is $1.6*10^{-19}$C)
 A $0.42 \mu m \lt \lambda _{c} \lt 0.87\mu m$ B $0.87 \mu m \lt \lambda _{c} \lt 1.42\mu m$ C $1.42 \mu m \lt \lambda _{c} \lt 1.62\mu m$ D $1.62 \mu m \lt \lambda _{c} \lt 6.62\mu m$
Electronic Devices   Basic Semiconductor Physics
Question 8 Explanation:
$\begin{array}{l} \lambda_{c} \leq \frac{1.24}{E_{g}(e V)} \mu \mathrm{m} \\ \lambda_{c} \leq \frac{1.24}{1.42} \\ \lambda_{c} \leq 0.87 \mathrm{m} \\ \end{array}$
 Question 9
In the figure, $ln(\rho _{i}$) is plotted as a function of 1/T , where $\rho _{i}$ is the intrinsic resistivity of silicon, T is the temperature, and the plot is almost linear. The slope of the line can be used to estimate
 A band gap energy of silicon $(E_{g})$ B sum of electron and hole mobility in silicon $(\mu _{n}+\mu _{p})$ C reciprocal of the sum of electron and hole mobility in silicon $(\mu _{n}+\mu _{p})^{-1}$ D intrinsic carrier concentration of silicon $(n_{i})$
Electronic Devices   Basic Semiconductor Physics
 Question 10
The cut-off wavelength (in $\mu$ m) of light that can be used for intrinsic excitation of a semiconductor material of bandgap $E_{g}$ = 1.1 eV is _____.
 A 0.12 B 1.12 C 2.12 D 3
Electronic Devices   Basic Semiconductor Physics
Question 10 Explanation:
\begin{aligned} \quad \lambda_{c}&=\frac{1.24}{E_{g}(\mathrm{eV})} \mu \mathrm{m}=\frac{1.24}{1.1} \mu \mathrm{m} \\ \lambda_{c}&=1.12 \mu \mathrm{m} \end{aligned}
There are 10 questions to complete.