# GATE EC 2015 SET-3

 Question 1
For $A=\begin{bmatrix} 1 & tan x\\ -tan x &1 \end{bmatrix}$, the Determinant of $A^{T}A^{-1}$ is
 A $sec^{2}x$ B cos4x C 1 D 0
Engineering Mathematics   Linear Algebra
Question 1 Explanation:
Long Method:
\begin{aligned} A &=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \\ A^{T} &=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ \operatorname{ad} A &=\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right] \\ A^{-1} &=\frac{1}{|A|}[a d j(A)]^{T} \\ &=\frac{1}{1+\tan ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ &=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \end{aligned}
Here,
\begin{aligned} A^{\top} A^{-1} &=\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right] \\ &=\frac{1}{\sec ^{2} x}\left[\begin{array}{cc} 1-\tan ^{2} x & -2 \tan x \\ 2 \tan x & 1-\tan ^{2} x \end{array}\right] \\ A^{\top} A^{-1} &=\left[\begin{array}{cc} \frac{1-\tan ^{2} x}{\sec ^{2} x} & \frac{-2 \tan x}{\sec ^{2} x} \\ \frac{2 \tan x}{\sec ^{2} x} & \frac{1-\tan ^{2} x}{\sec ^{2} x} \end{array}\right] \\ \left|A^{T} A^{-1}\right| &=\left(\frac{1-\tan ^{2} x}{\sec ^{2} x}\right)^{2}+\left(\frac{2 \tan x}{\sec ^{2} x}\right)^{2} \\ &=\frac{1+\tan ^{4} x-2 \tan ^{2} x+4 \tan ^{2} x}{\sec ^{4} x}\\ &=1 \end{aligned}
Or
Short Method:
\begin{aligned} \text{since, }|A B|&=|A||B| \\ \left|A^{T} A^{-1}\right| &=\left|A^{T}\right|\left|A^{-1}\right| \\ &=|A| \times \frac{1}{|A|}=1 \\ &\left(\text { Note }:\left|A^{T}\right|=|A| \text { and }\left|A^{-1}\right|=\frac{1}{|A|} \right) \end{aligned}
 Question 2
The contour on the x-y plane, where the partial derivative of $x^{2}+y^{2}$ with respect to y is equal to the partial derivative of 6y + 4x with respect to x, is
 A y=2 B x=2 C x+y=4 D x-y=0
Engineering Mathematics   Calculus
Question 2 Explanation:
Partial derivative w.r.t.
$\frac{\partial}{\partial y}\left(x^{2}+y^{2}\right)=2 y$
Partial derivative w.r.t. x
$\frac{\partial}{\partial x}(6 y+4 x)=4$
From given condition
\begin{aligned} 2 y&=4 \\ \Rightarrow y\quad &=2 \end{aligned}

 Question 3
If C is a circle of radius r with centre $z_{0}$, in the complex z-plane and if n is a non-zero integer, then $\oint_{C}\frac{dz}{(z-z_{0})^{n+1}}$ equals
 A $2\pi nj$ B 0 C $\frac{nj}{2\pi }$ D $2\pi n$
Engineering Mathematics   Complex Analysis
Question 3 Explanation:
By Cauchy integral formula
$\begin{array}{l} \oint \frac{f(z)}{\left(z-z_{0}\right)^{n+1} d z}=\frac{2 \pi i f^{*}\left(z_{0}\right)}{n !} \\ \oint \frac{d z}{\left(z-z_{0}\right)^{n+1}}=\frac{2 \pi i}{n !} \cdot 0=0 \end{array}$
 Question 4
Consider the function $g(t)=e^{-t}sin(2\pi t)u(t)$ where u(t) is the unit step function. The area under g(t) is _______.
 A 0.15 B 0.34 C 0.5 D 1
Signals and Systems   Fourier Transforms, Frequency Response and Correlation
Question 4 Explanation:
\begin{aligned} g(t) &=e^{-t} \sin (2 \pi t) u(t) \\ \text { Area } &=\int_{-\infty}^{\infty} g(t) d t \\ &=\int_{-\infty}^{\infty} \sin (2 \pi t) u(t) e^{-t} d t=\left.G(s)\right|_{s=0} \\ G(s) &=\frac{2 \pi}{(s+1)^{2}+4 \pi^{2}} \\ \text { Area } &=\left.\frac{2 \pi}{(s+1)^{2}+4 \pi^{2}}\right|_{S=0}=0.155 \end{aligned}
 Question 5
The value $\sum_{n=0}^{\infty }n(\frac{1}{2})^{n}$ is ______.
 A 1 B 2 C 3 D 4
Signals and Systems   DTFS, DTFT and DFT
Question 5 Explanation:
\begin{aligned} \sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n} &=\sum_{n=-\infty}^{\infty} n\left(\frac{1}{2}\right)^{n} u(n) \\ &=Z \cdot T\left[n\left(\frac{1}{2}\right)^{n} u(n)\right]_{z=1} \\ =& \frac{\frac{1}{2} Z^{-1}}{\left(1-\frac{1}{2} z^{-1}\right)^{2}}=\left.\frac{\frac{Z}{2}}{\left(z-\frac{1}{2}\right)^{2}}\right|_{z=1} \\ =& \frac{1 / 2}{1 / 4}=2 \end{aligned}

There are 5 questions to complete.