# GATE EE 2012

 Question 1
Two independent random variables X and Y are uniformly distributed in the interval [-1,1]. The probability that max [X,Y] is less than 1/2 is
 A $3/4$ B $9/16$ C $1/4$ D $2/3$
Engineering Mathematics   Probability and Statistics
Question 1 Explanation:
$-1 \leq x\leq 1 \text{ and } -1 \leq y\leq 1$ is the entire rectangle.The region in which maximum of {x,y} is less than 1/2 is shown below as shaded regioninside this rectangle.

\begin{aligned} P\left ( max{x,y} \lt \frac{1}{2} \right )&=\frac{\text{Area of shaded region}}{\text{Area of entire rectangle}} \\ &= \frac{\frac{3}{2} \times \frac{3}{2}}{2 \times 2}=\frac{9}{16} \end{aligned}
 Question 2
If $x = \sqrt{-1}$, then the value of $x^{x}$ is
 A $e^{-\pi/2}$ B $e^{\pi/2}$ C $x$ D 1
Engineering Mathematics   Complex Variables
Question 2 Explanation:
\begin{aligned} x&=i, \text{ then in polar coordinates,} \\ x &=\cos \frac{\pi}{2} +i \sin \frac{\pi}{2}=e^{\frac{\pi}{2}i}\\ \text{Now, } x^x&=i^i =(e^{\frac{\pi}{2}i})^i=e^{i^2 \frac{\pi}{2}}=e^{-\frac{\pi}{2}} \end{aligned}

 Question 3
Given $f(z)=\frac{1}{z+1}-\frac{2}{z+3}$. If C is a counter clockwise path in the z-plane such that |z+1|=1, the value of $\frac{1}{2 \pi j}\oint_{C}f(z)dz$ is
 A -2 B -1 C 1 D 2
Engineering Mathematics   Complex Variables
Question 3 Explanation:
Given:
\begin{aligned} f(z)&=\frac{1}{z+1}-\frac{2}{z+3}\\ &=\frac{(Z+3)-2(Z+1)}{(Z+1)(Z+3)}\\ &=\frac{-Z+1}{(Z+1)(Z+3)} \end{aligned}
Poles are at -1 and -3 i.e. (-1,0) and (-3,0).
From figure below of |Z+1|=1,
we see that (-1,0) is inside the circle and (-3,0) is outside the circle.

Residue theorem says,
$\frac{1}{2 \pi j}\oint _Cf(z)dz=$Residue of those poles which are inside C.
So the required integral $\frac{1}{2 \pi j}\oint _Cf(z)dz$ is given by the residue of function at poles (-1,0) ( which is inside the circle).
This residue is $=\frac{-(-1)+1}{(-1+3)}=\frac{2}{2}=1$
 Question 4
In the circuit shown below, the current through the inductor is
 A $\frac{2}{1+j}A$ B $\frac{-1}{1+j}A$ C $\frac{1}{1+j}A$ D 0A
Electric Circuits   Basics
Question 4 Explanation:

Apply KCL node at 'A'
so, current flowing through $1\Omega$ is $(1-I_2)$
Applying KVL in ABCD loop,
$1\angle 0-1\angle 0+1(1-I_2)-jI_2=0$
$I_2=\frac{1}{1+j}$
 Question 5
The impedance looking into nodes 1 and 2 in the given circuit is
 A 50 $\Omega$ B 100 $\Omega$ C 5 k$\Omega$ D 10.1 k$\Omega$
Electric Circuits   Network Theorems
Question 5 Explanation:

To find thevenin impedance across node 1 and node 2. Connect a 1 V source and find the current through voltage source.
Then, $Z_{Th}=\frac{1}{I_{Th}}$
By applying KCL at node B and A
$i_{AB}+99i_{b}=I_{Th}$
$i_b==i_A+i_{AB}$
$\Rightarrow \;\; i_b-i_A+99i_b=I_{Th}$
$\Rightarrow \;\; 100i_b-i_A=I_{Th}\;...(i)$
By applying KVL in outer loop
$10 \times 10^3i_b=1$
$i_b=10^{-4}A$
and $10 \times 10^{3}i_b=-100i_A$
$i_A=-100i_b$
$\therefore$ From equation (i),
$100i_b+100i_b=I_{Th}$
$\Rightarrow \; I_{Th}=200i_b$
$\;\;=200 \times 10^{-4}=0.02$
$\therefore \;\;Z_{Th}=\frac{1}{I_{Th}}=\frac{1}{0.02}=50\Omega$

There are 5 questions to complete.