# GATE EC 2014 SET-3

 Question 1
The maximum value of the function $f(x)=ln(1+x)-x(\; where \; x>-1)$ occurs at x = ____.
 A 0 B 1 C 0.5 D -1
Engineering Mathematics   Calculus
Question 1 Explanation:
\begin{aligned} f^{\prime}(x) &=\frac{1}{1+x}-1=0 \\ \frac{1-1-x}{1+x} &=0 \\ \frac{x}{1+x} &=0 \\ x &=0\\ f^{\prime \prime}(x)&=\frac{-1}{(1+x)^{2}} \\ f^{\prime}(0)&=-1 \lt 0 \end{aligned}
f(x) have maximum value at x=0
\begin{aligned} f(0)&=\ln (1+0)-0=0 \\ t_{\max }&=0 \end{aligned}
 Question 2
Which ONE of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively ?
 A $\frac{dy}{dx}+xy=e^{-x}$ B $\frac{dy}{dx}+xy=0$ C $\frac{dy}{dx}+xy=e^{-y}$ D $\frac{dy}{dx}+e^{-y}=0$
Engineering Mathematics   Differential Equations
Question 2 Explanation:
General form of linear differential equation
$\frac{d y}{d x}+p y=\theta$ when P and $\theta$ can be function of x
Only option (A) is in this form.
 Question 3
Match the application to appropriate numerical method.
 A P1-M3, P2-M2, P3-M4, P4-M1 B P1-M3, P2-M1, P3-M4, P4-M2 C P1-M4, P2-M1, P3-M3, P4-M2 D P1-M2, P2-M1, P3-M3, P4-M4
Engineering Mathematics   Numerical Methods
 Question 4
An unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss is
 A 0.067 B 0.073 C 0.082 D 0.091
Engineering Mathematics   Probability and Statistics
Question 4 Explanation:
It means 3-head appears in $1^{\text {st }}$ 9 trials.
Probability of getting exactly 3 head in 1^{\text {st }} 9 trials
\begin{aligned} &=\text{ Coefficient } p^{3}\text{ in }(4+p) 9 \\ \text{When, }&[\overline{p+q=1}] \\ &={ }^{9} C_{3} q^{6} p^{3}\\ \text{when, }\quad \rho&= \text{ probability of occure of head}\\ q &=\text { probability of occure of tail } \\ &={ }^{9} C_{3} \times\left(\frac{1}{2}\right)^{6}\left(\frac{1}{2}\right)^{3} \\ &={ }^{9} C_{3} \times\left(\frac{1}{2}\right)^{9} \end{aligned}
and in $10^{\text {th }}$ trial head must appears.
So required probability
\begin{aligned} &={ }^{9} C_{3}\left(\frac{1}{2}\right)^{9} \times \frac{1}{2} \\ &=\frac{9 \times 8 \times 7}{3 !} \times\left(\frac{1}{2}\right)^{10}=\frac{84}{1024}=0.082 \end{aligned}
 Question 5
If z = xy ln(xy), then
 A $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$ B $y\frac{\partial z}{\partial x}=x\frac{\partial z}{\partial y}$ C $x\frac{\partial z}{\partial x}=y\frac{\partial z}{\partial y}$ D $y\frac{\partial z}{\partial x}+x\frac{\partial z}{\partial y}=0$
Engineering Mathematics   Calculus
Question 5 Explanation:
\begin{aligned} \frac{\partial z}{\partial x}&=yln(x y)+\frac{x y}{x y} y\\ \frac{\partial z}{\partial x}&=y[\ln (x y)+1] &\ldots(i)\\ \frac{\partial z}{\partial y}&=x \ln (x y)+\frac{x y}{x y} \times x\\ \frac{\partial z}{\partial y}&=x[\ln (x y)+1]\\ \text{Here}\quad x\frac{\partial z}{\partial x}&=y \frac{\partial z}{\partial y} \end{aligned}
 Question 6
A series RC circuit is connected to a DC voltage source at time t = 0. The relation between the source voltage $V_{S}$, the resistance R, the capacitance C, and the current $i(t)$ is given below :
$V_{S}=Ri(t)+\frac{1}{C}\int_{0}^{t}i(u)du$
Which one of the following represents the current $i(t)$ ?
 A A B B C C D D
Network Theory   Sinusoidal Steady State Analysis
Question 6 Explanation:
Given that:
$V_{s}=R i(t)+\frac{1}{C} \int_{0}^{t} i(t) d t\qquad\ldots(i)$
Using Laplace transform,
\begin{aligned} \text { WS) }&=R I(s)+\frac{1}{C s} I(s) &\ldots(ii)\\ \text { or } I(s)&=\frac{V(s)}{\left(R+\frac{1}{C S}\right)} &\ldots(iii)\\ \text { For, } V(s)&=\frac{1}{s}&\ldots(iv) \end{aligned}
From equation (iii) and (iv),
$I(s)=\frac{C}{(R C s+1)}=\frac{1}{R\left(s+\frac{1}{R C}\right)}\qquad\ldots(v)$
Using inverse Laplace transform in equation (v), we get,
$i(t)=\frac{1}{R} e^{-t / R C}$
Thus, option (A) is correct.
 Question 7
In the figure shown, the value of the current I (in Amperes) is_____.
 A 0 B 0.25 C 0.5 D 1
Network Theory   Network Theorems
Question 7 Explanation:

Using super position theorem, when 5 V source acting alone, we get

$I_{1}=\frac{V}{R_{\mathrm{eq}}}=\frac{5}{10+5+5}=\frac{1}{4} \mathrm{A}\quad \ldots(i)$
When 1 A source acting alone, we get

$I_{2}=\frac{1 \times 5}{5+10+5}=\frac{5}{20}=\frac{1}{4} \mathrm{A}\quad \ldots(ii)$
Therefore,
$I=I_{1}+I_{2}=\frac{1}{2} A=0.5 \mathrm{A}$
 Question 8
In MOSFET fabrication, the channel length is defined during the process of
 A isolation oxide growth B channel stop implantation C poly-silicon gate patterning D lithography step leading to the contact pads
Electronic Devices   IC Fabrication
Question 8 Explanation:
Channel length is defined during the poly-silicon gate pattering.
 Question 9
A thin P-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to
 A the minority carrier mobility B the minority carrier recombination lifetime C the majority carrier concentration D the excess minority carrier concentration
Electronic Devices   Basic Semiconductor Physics
 Question 10
At T = 300 K, the hole mobility of a semiconductor $\mu _{P}=500 cm^{2}/V-s$ and $\frac{KT}{q}=26mV$. The hole diffusion constant $D_{P} \; in \; cm^{2}/s$ is _____.
 A 11 B 12 C 13 D 14
Electronic Devices   Basic Semiconductor Physics
Question 10 Explanation:
$\begin{array}{l} \frac{D_{p}}{\mu_{p}}=v_{T} \\ D_{p}=\mu_{p} V_{T}=500 \times 26 \times 10^{-3} \\ D_{p}=13 \mathrm{cm}^{2} / \mathrm{s} \end{array}$
There are 10 questions to complete.