# GATE EC 2014 SET-2

 Question 1
The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of matrix AB is ______.
 A 200 B 100 C 50 D 45
Engineering Mathematics   Linear Algebra
Question 1 Explanation:
Determinant of A=5
Determinant of B=40
Determinant of AB=|A||B|
$\begin{array}{l} \quad=5 \times 40 \\ \quad=200 \end{array}$
 Question 2
Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E [X], is _____.
 A 100 B 50 C 25 D 10
Engineering Mathematics   Probability and Statistics
Question 2 Explanation:
$E[X]=\frac{1+2+3+\cdots 99}{50}=\frac{2500}{50}=50$
 Question 3
For $0\leq t \leq \infty$ , the maximum value of the function $f(t) =e^{-t}-2e^{-2t}$ occurs at
 A $t=log_{e}4$ B $t=log_{e}2$ C t=0 D $t=log_{e}8$
Engineering Mathematics   Calculus
Question 3 Explanation:
\begin{aligned} f(t)&=e^{-t}-2 e^{-2 t} \\ f^{\prime}(t)&=-e^{-t}+4 e^{-2 t} \\ \text{For maximum value} &P(t)=0 \\ f^{\prime}(t)&=0=-e^{-t}+4 e^{-2 t} \\ \Rightarrow \quad 4 e^{-2 t}&=e^{t} \\ 4 e^{t}&=1 \\ \therefore \quad t&=\log _{e} 4 \end{aligned}
 Question 4
The value of
$\lim_{x\rightarrow \infty }(1+\frac{1}{x})^{x}$
is
 A ln 2 B 1 C e D $\infty$
Engineering Mathematics   Calculus
Question 4 Explanation:
$\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e^{\lim _{x \rightarrow \infty} \frac{1}{x} \cdot x}=e^{1}=e$
 Question 5
If the characteristic equation of the differential equation
$\frac{d^{2}y}{dx^{2}}+2\alpha \frac{dy}{dx}+y=0$
has two equal roots, then the values of a are
 A $\pm 1$ B 0,0 C $\pm j$ D $\pm 1/2$
Engineering Mathematics   Differential Equations
Question 5 Explanation:
$\frac{d^{2} y}{d x^{2}}+2 \alpha\frac{d y}{d x}+y=0$
The characteristic equation is given as
\begin{aligned} \left(m^{2}+2(x)+1\right) &=0 \\ m_{1}, m_{2} &=\frac{-2 x_{1} \pm \sqrt{4 x^{2}-4}}{2} \end{aligned}\\ \text{since both roots are equal i.e.} \\ \begin{aligned} m_{1}=& m_{2} \\ \frac{-2 \alpha+\sqrt{4 \alpha^{2}-4}}{2} &=\frac{-2\alpha \cdot-\sqrt{4 a^{2}-4}}{2} \\ \sqrt{4\left(1^{2}-4\right.} &=-\sqrt{4 c^{2}-4} \\ 2 \sqrt{4 c^{2}-4} &=0 \\ 4 \alpha^{2}-4 &=0 \\ \alpha^{2} &=1 \\ \alpha &=\pm 1 \end{aligned}
 Question 6
Norton's theorem states that a complex network connected to a load can be replaced with an equivalent impedance
 A in series with a current source B in parallel with a voltage source C in series with a voltage source D in parallel with a voltage source
Network Theory   Network Theorems
Question 6 Explanation:
Norton's theorem states that a complex network connected to a load can be replaced with an equivalent impedance in parallel with a current source

 Question 7
In the figure shown, the ideal switch has been open for a long time. If it is closed at t = 0, then the magnitude of the current (in mA) through the 4 k$\Omega$ resistance at t = $0^{+}$ is _____.

 A 0.75 B 1.25 C 0.5 D 1.5
Network Theory   Transient Analysis
Question 7 Explanation:
At steady state $t = 0_{-}$

\begin{aligned} \therefore V_{c}(0) &=V_{c}\left(0^{+}\right)=5 \mathrm{V} \\ I_{L}\left(0^{-}\right) &=I_{L}\left(0^{+}\right)=1 \mathrm{mA} \end{aligned}
At, t=0, switch get closed

Thus, the current through $4\Omega$ resistance is
$I=\frac{5}{4 \times 10^{3}}=1.25 \mathrm{mA}$
 Question 8
A silicon bar is doped with donor impurities $N_{D}=2.25\times10^{15} atom/cm^{3}$. Given the intrinsic carrier concentration of silicon at T = 300 K is $n_{i}=1.5\times10^{10}cm^{-3}$. Assuming complete impurity ionization, the equilibrium election and hole concentrations are
 A $n_{0}=1.5 \times 10^{16}cm^{-3},$ $p_{0}=1.5 \times 10^{5}cm^{-3}$ B $n_{0}=1.5 \times 10^{10}cm^{-3},$ $p_{0}=1.5 \times 10^{15}cm^{-3}$ C $n_{0}=2.25 \times 10^{15}cm^{-3},$ $p_{0}=1.5 \times 10^{10}cm^{-3}$ D $n_{0}=2.25 \times 10^{15}cm^{-3},$ $p_{0}=1 \times 10^{5}cm^{-3}$
Electronic Devices   Basic Semiconductor Physics
Question 8 Explanation:
since $N_{D} \gt \gt n_{i}$
therefore equilibrium electron concentration is
$n \simeq N_{D}=2.25 \times 10^{15} \mathrm{cm}^{-3}$
And equilibrium hole concentration is given by mass action law
$p=\frac{n_{i}^{2}}{N_{D}}=\frac{\left(1.5 \times 10^{10}\right)^{2}}{2.25 \times 10^{15}}=1 \times 10^{5} \mathrm{cm}^{-3}$
 Question 9
An increase in the base recombination of a BJT will increase
 A the common emitter dc current gain $\beta$ B the breakdown voltage $BV_{CEO}$ C the unity-gain cut-off frequency $f_{T}$ D the transconductance $g_{m}$
Analog Circuits   BJT Analysis
 Question 10
In CMOS technology, shallow P-well or N -well regions can be formed using
 A low pressure chemical vapour deposition B low energy sputtering C low temperature dry oxidation D low energy ion-implantation
Electronic Devices   IC Fabrication
Question 10 Explanation:
Ion implanation/diffusion is used for well implantation.
There are 10 questions to complete.