Question 1 |
Find the positive real root of x^3-x-3=0 using Newton-Raphson method. If the starting guess (x_0) is 2, the numerical value of the root after two iterations (x_2) is _______ (round off to two decimal places).
1.67 | |
1.12 | |
2.44 | |
3.25 |
Question 1 Explanation:
\begin{aligned} \text { Given, }\qquad\quad f(x) &=x^{3}-x-3, \quad x_{0}=2 \\ f^{\prime}(x)&=3 x^{2}-1\\ \text { Iteration 1: } \quad x_{1}&=x_{0}-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)}=2-\frac{(8-2-3)}{3(4)-1}=1.72\\ \text { Iteration 2: } \quad x_{2}&=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}=1.72-\frac{\left(1.72^{3}-1.72-3\right)}{3(1.72)^{2}-1}=1.67 \end{aligned}
Question 2 |
Consider the following differential equation
(1+y)\frac{dy}{dx}=y
The solution of the equation that satisfies the condition is y(1)=1 is
(1+y)\frac{dy}{dx}=y
The solution of the equation that satisfies the condition is y(1)=1 is
2ye^y=e^x+e | |
y^2e^y=e^x | |
ye^y=e^x | |
(1+y)e^y=2e^x |
Question 2 Explanation:
\begin{aligned} (1+y) \frac{d y}{d x} &=y \\ \Rightarrow\qquad \left(\frac{1}{y}+1\right) d y &=d x \\ \Rightarrow\qquad \log y+y &=x+c \\ \text { Using, } \qquad y(1) &=1 \\ \quad \log 1+1 &=1+c \quad \Rightarrow c=0 \\ \text { Hence, } \quad \log y+y &=x \\ \Rightarrow\qquad \log y+y \operatorname{loge} &=x \\ \log _{\mathrm{e}}\left(y \cdot e^{y}\right) &=x \\ \Rightarrow\qquad y e^{y} &=e^{x} \end{aligned}
Question 3 |
Let the superscript T represent the transpose operation. Consider the function f(x)=\frac{1}{2}x^T Qx=r^Tx, \; \text{ where } x \text{ and }r \text{ are }n \times 1 vectors and Q is a symmetric n \times n matrix. The stationary point of f(x) is
Q^Tr | |
Q^{-1}r | |
\frac{r}{r^Tr} | |
r |
Question 3 Explanation:
\begin{aligned} \text{Let}\qquad Q=\left[\begin{array}{ll}a & c \\c & b\end{array}\right], x&=\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right], R=\left[\begin{array}{l}r_{1} \\r_{2}\end{array}\right] \\ F(x)&=\frac{1}{2}\left(x_{1}, x_{2}\right)\left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]-\left[r_{1} r_{2}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right] \\ &=\frac{1}{2}\left[a x_{1}^{2}+b x_{2}^{2}+2 c x_{1} x_{2}\right]-\left[r_{1} x_{1}+r_{2} x_{2}\right]\\ \text{i.e.}\qquad \qquad U\left(x_{1}, x_{2}\right)&=\frac{1}{2} a x_{1}^{2}+\frac{1}{2} b x_{2}^{2}+c_{1} x_{1} x_{2}-r_{1} x_{1}-r_{2} x_{2} \end{aligned}
Now, for critical point, \frac{\partial u}{\partial x_{1}}=0 and \frac{\partial u}{\partial x_{2}}=0
\Rightarrow \quad a_{1} x_{1}+c x_{2}-r_{1}=0 \quad \text { and } c x_{2}+c x_{1}-r_{2}=0
In matrix form we can write it as
\begin{aligned} \left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l} x_{1} \\m_{2}\end{array}\right] &=\left[\begin{array}{l}r_{1} \\ r_{2}\end{array}\right] \\\Rightarrow \qquad Q x &=r \end{aligned}
By multiplying both side by Q^{-1}
x=Q^{-1} r
Now, for critical point, \frac{\partial u}{\partial x_{1}}=0 and \frac{\partial u}{\partial x_{2}}=0
\Rightarrow \quad a_{1} x_{1}+c x_{2}-r_{1}=0 \quad \text { and } c x_{2}+c x_{1}-r_{2}=0
In matrix form we can write it as
\begin{aligned} \left[\begin{array}{ll}a & c \\c & b\end{array}\right]\left[\begin{array}{l} x_{1} \\m_{2}\end{array}\right] &=\left[\begin{array}{l}r_{1} \\ r_{2}\end{array}\right] \\\Rightarrow \qquad Q x &=r \end{aligned}
By multiplying both side by Q^{-1}
x=Q^{-1} r
Question 4 |
The value of \int_{0}^{\pi /2}\int_{0}^{\cos \theta }r \sin \theta dr d\theta is
0 | |
\frac{1}{6} | |
\frac{4}{3} | |
\pi |
Question 4 Explanation:
\begin{aligned} I &=\int_{\theta=0}^{\theta=\frac{\pi}{2}} \int_{r=0}^{r=\cos \theta} r \sin \theta d r d \theta \\ &=\int_{\theta=0}^{\frac{\pi}{2}}\left[\frac{r^{2}}{2}\right]_{0}^{\infty \cos \theta} \times \sin \theta d \theta \\ &=\frac{1}{2} \int_{\theta}^{\frac{\pi}{2}} \sin \theta \cdot \cos ^{2} \theta d \theta \\ \text{Let}, \qquad\cos \theta &=t\\ -\sin \theta d \theta &=d t \\ \cos \theta &=t \\ \text{at},\qquad\theta &=\frac{\pi}{2} ; t=0 \\ \theta &=0, t=1 \\ &=\int_{1}^{0} \frac{-t^{2}}{2} d t \\ &=\frac{-1}{2}\left[\frac{t^{3}}{3}\right]_{1}^{0}=\frac{-1}{2} \times\left(\frac{-1}{3}\right) \\ &=\frac{-1}{2}\left[\frac{-1}{3}\right]=\frac{1}{6} \end{aligned}
Question 5 |
Value of (1+i)^8, where i=\sqrt{-1}, is equal to
4 | |
16 | |
4i | |
16i |
Question 5 Explanation:
\begin{aligned} (1+i)^{8} & \\ \mathrm{z} &=1+i \mathrm{r}=|z|=\sqrt{2} \\ \theta &=\frac{\pi}{4} \\ (1+i)^{8} &=\left(\sqrt{2} e^{i \frac{\pi}{4}}\right)^{8} \\ &=16 \times e^{i \times 2 \pi} \\ &=16(\cos 2 \pi+i \sin 2 \pi) \\ &=16 \times 1=16 \end{aligned}
Question 6 |
Value of \int_{4}^{5.2}\ln x \; dx using Simpson's one-third rule with interval size 0.3 is
1.83 | |
1.6 | |
1.51 | |
1.06 |
Question 6 Explanation:
\begin{aligned} \text { Here } \qquad f(x)&=\log x\\ a=4, \quad b=5.2, h=0.3\\ \text { So, } \qquad n&=\frac{b-a}{h}=\frac{5.2-4}{0.3}=4\\ \end{aligned}
\begin{array}{cccccc} \hline x & 4 & 4.3 & 4.6 & 4.9 & 5.2 \\ \hline y & \log 4 & \log 4.3 & \log 4.6 & \log 4.9 & \log 5.2 \\ \hline & y_{0} & y_{1} & y_{2} & y_{3} & y_{4} \end{array}
As per Simpson's 1 / 3^{\text {rd }} rule
\begin{aligned} \int_{4}^{5.2} \log x d x &=\frac{h}{3}\left[y_{0}+y_{4}+4\left(y_{1}+y_{3}\right)+2\left(y_{2}\right)\right] \\ &=1.8272 \simeq 1.83 \end{aligned}
\begin{array}{cccccc} \hline x & 4 & 4.3 & 4.6 & 4.9 & 5.2 \\ \hline y & \log 4 & \log 4.3 & \log 4.6 & \log 4.9 & \log 5.2 \\ \hline & y_{0} & y_{1} & y_{2} & y_{3} & y_{4} \end{array}
As per Simpson's 1 / 3^{\text {rd }} rule
\begin{aligned} \int_{4}^{5.2} \log x d x &=\frac{h}{3}\left[y_{0}+y_{4}+4\left(y_{1}+y_{3}\right)+2\left(y_{2}\right)\right] \\ &=1.8272 \simeq 1.83 \end{aligned}
Question 7 |
The mean and variance, respectively, of a binomial distribution for n
independent trials with the probability of success as p, are
\sqrt{np},np(1-2p) | |
\sqrt{np}, \sqrt{np(1-p)} | |
np,np | |
np,np(1-p) |
Question 7 Explanation:
Mean= np
Variance = npq = np(1 - p)
Variance = npq = np(1 - p)
Question 8 |
If the Laplace transform of a function f(t)
is given by \frac{s+3}{(s+1)(s+2)} , then f(0) is
0 | |
\frac{1}{2} | |
1 | |
\frac{3}{2} |
Question 8 Explanation:
By using partial fraction concept.
\begin{aligned} f(t) &=L^{-1}\left[\frac{s+3}{(s+1)(s+2)}\right] \\ &=L^{-1}\left[\frac{2}{s+1}-\frac{1}{s+2}\right] \\ \Rightarrow \qquad f(t) &=2 e^{-t}-e^{-2 t} \\ \text { So, } \qquad f(c)&=2 e^{0}-e^{0}=2-1=1 \end{aligned}
\begin{aligned} f(t) &=L^{-1}\left[\frac{s+3}{(s+1)(s+2)}\right] \\ &=L^{-1}\left[\frac{2}{s+1}-\frac{1}{s+2}\right] \\ \Rightarrow \qquad f(t) &=2 e^{-t}-e^{-2 t} \\ \text { So, } \qquad f(c)&=2 e^{0}-e^{0}=2-1=1 \end{aligned}
Question 9 |
Consider an n x n matrix A and a non-zero n x 1 vector p. Their product Ap=\alpha ^2p, where \alpha \in \mathbb{R} and \alpha \notin \{-1,0,1\}. Based on the given information, the eigen value of A^2 is:
\alpha | |
\alpha ^2 | |
\sqrt{\alpha } | |
\alpha ^4 |
Question 9 Explanation:
Given, A P=\alpha^{2} P
By comparison with A X=\lambda X \Rightarrow
\Rightarrow \quad \lambda=\alpha^{2}
Hence, eigen value of A is \alpha^{2}, so eigen value of A^{2} is \alpha^{4}.
By comparison with A X=\lambda X \Rightarrow
\Rightarrow \quad \lambda=\alpha^{2}
Hence, eigen value of A is \alpha^{2}, so eigen value of A^{2} is \alpha^{4}.
Question 10 |
Consider a single machine workstation to which jobs arrive according to a Poisson distribution with a mean arrival rate of 12 jobs/hour. The process time of the workstation is exponentially distributed with a mean of 4 minutes. The expected number of jobs at the workstation at any given point of time is ________ (round off to the nearest integer).
3 | |
4 | |
6 | |
8 |
Question 10 Explanation:
\lambda=12 per hour, \frac{1}{\mu}=4 \min / \mathrm{Jobs}, \mu=15 \mathrm{~Jobs} / \mathrm{hr}
\begin{aligned} \rho &=\frac{12}{15}=\frac{4}{5} \\ L_{s} &=\frac{\rho}{1-\rho}=\frac{\frac{4}{5}}{1-\frac{4}{5}}=\frac{\frac{4}{5}}{\frac{1}{5}} \\ \frac{4}{5} \times \frac{5}{1} &=4 \text { Jobs } \end{aligned}
\begin{aligned} \rho &=\frac{12}{15}=\frac{4}{5} \\ L_{s} &=\frac{\rho}{1-\rho}=\frac{\frac{4}{5}}{1-\frac{4}{5}}=\frac{\frac{4}{5}}{\frac{1}{5}} \\ \frac{4}{5} \times \frac{5}{1} &=4 \text { Jobs } \end{aligned}
There are 10 questions to complete.
