GATE EE 2012

-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}

\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}

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

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}

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