# GATE EC 2016 SET-3

 Question 1
Consider a 2 x 2 square matrix

$A=\begin{bmatrix} \sigma & x\\ \omega & \sigma \end{bmatrix}$

where x is unknown. If the eigenvalues of the matrix A are $(\sigma +j\omega ) \; and \; (\sigma -j\omega )$, then $x$ is equal to
 A $+j\omega$ B $-j\omega$ C $+\omega$ D $-\omega$
Engineering Mathematics   Linear Algebra
Question 1 Explanation:
\begin{aligned} A &=\left[\begin{array}{ll} \sigma & x \\ \omega & \sigma \end{array}\right] \\ \text { Trace } &=\text { sum of eigen values } \\ 2 \sigma &=\sigma+j \omega+\sigma-j \omega \\ |A| &=\text { product of eigens } \\ \sigma^{2}-x \omega &=(\sigma+j \omega)(\sigma-j \omega)=\sigma^{2}+\omega^{2} \end{aligned}
which is possible only when, $x=-\omega$
 Question 2
For $f(z)=\frac{sin(z)}{z^{2}}$, the residue of the pole at z=0 is __________
 A 0.5 B 1 C 2 D 3
Engineering Mathematics   Complex Analysis
Question 2 Explanation:
Residue of $\frac{\sin z}{z^{2}}$
= coefficient of $\frac{1}{2}$ in $\left\{\frac{z-\frac{z^{3}}{3 !}+\frac{z^{5}}{5 !}---}{z^{2}}\right\}$
= coefficient of $\frac{1}{z}$ in $\left\{\frac{1}{z}-\frac{z}{3 !}+\frac{z^{3}}{5 !}---\right\}$
$=1$

 Question 3
The probability of getting a "head" in a single toss of a biased coin is 0.3. The coin is tossed repeatedly till a "head" is obtained. If the tosses are independent, then the probability of getting "head" for the first time in the fifth toss is __________
 A 0.01 B 0.04 C 0.07 D 0.1
Engineering Mathematics   Probability and Statistics
Question 3 Explanation:
\begin{aligned} P(H)&=0.3 \\ P(T)&=0.7 \end{aligned}
since all tosses are independent
so, probability of getting head for the first time in $5^{\text {th }}$ toss is
\begin{aligned} &=P(T) P(T) P(T) P(T) P(H) \\ &=0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.3=0.072 \end{aligned}
 Question 4
The integral $\int_{0}^{1}\frac{dx}{\sqrt{(1-x)}}$ is equal to __________
 A 1 B 2 C 3 D 4
Engineering Mathematics   Calculus
Question 4 Explanation:
\begin{aligned} \int_{0}^{1} \frac{1}{\sqrt{1-x}} d x &=-2 \int_{0}^{1} \frac{1}{2 \sqrt{1-x}} d x \\ &=-\left.2(\sqrt{1-x})\right|_{0} ^{1} \\ &=-2(0-1)=2 \end{aligned}
 Question 5
Consider the first order initial value problem

$y'=y+2x-x^{2},y(0)=1,(0\leq x\leq \infty )$

with exact solution $y(x)=x^{2}+e^{x}$. For x = 0.1, the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-size h = 0.1 is __________
 A 0.06 B 0.012 C 0.6 D 0.12
Engineering Mathematics   Numerical Methods
Question 5 Explanation:
\begin{aligned} \frac{d y}{d x} &=y+2 x-x^{2} \\ y(0) &=1 \quad \quad 0 \leq x \leq \infty \\ f(x, y) &=y+2 x-x^{2} \\ x_{0}=0 ; y_{0}=1 ; h &=0.1 \\ k_{1} &=h f\left(x_{0}, y_{0}\right) \\ &=0.1\left(1+2 \times 0-0^{2}\right)=0.1 \\ k_{3} &=h /\left(x_{0}+h_{1} y_{0}+k_{1}\right) \\ &=0.1\left(\left(y_{0}+k_{1}\right)+2\left(x_{0}+h\right)-\left(x_{0}+h\right)^{2}\right) \\ &\left.=0.1(11+0.1)+2(0.1)-(0.1)^{2}\right) \\ &=0.129\\ y_{1} &=y_{0}+\frac{1}{2}\left(k_{1}+k_{2}\right) \\ &=1+\frac{1}{2}(0.1+0.129) \\ &=1.1145 \\ \text { Exact solution } y(x) &=x^{2}+e^{x} \\ &=(0.1)^{2}+e^{0.1} \\ &=1.1152 \\ \text { Error } &=1.1152-1.1145=0.00069 \\ \% \text { error } &=0.06 \% \end{aligned}

There are 5 questions to complete.