Question 1 |
Which one of the following equations is a correct identity for arbitrary 3x3 real matrices P, Q and R?
P(Q+R)=PQ+RP | |
(P-Q)^{2}=P^{2}-2PQ+Q^{2} | |
det(P+Q)=det \; P+det \; Q | |
(P+Q)^{2}=P^{2}+PQ+QP+Q^{2} |
Question 1 Explanation:
\begin{aligned} (P+Q)^{2} &=P^{2}+P Q+Q P+Q^{2} \\ &=P \cdot P+P \cdot Q+Q \cdot P+Q \cdot Q \\ &=P^{2}+P Q+Q P+Q^{2} \end{aligned}
Question 2 |
The value of the integral \int_{0}^{2}\frac{(x-1)^{2}\sin(x-1)}{(x-1)^{2}+\cos(x-1)}dx is
3 | |
0 | |
-1 | |
-2 |
Question 2 Explanation:
\begin{aligned} I &= \int_{0}^{2}\frac{(x-1)^2 \sin (x-1)}{(x-1)^2 + \cos (x-1)}dx\\ \text{Taking}\; x-1=z&\Rightarrow dx=dz \\ \text{for}\; x=0,&z\rightarrow -1\; \text{and}\; x=2, z \rightarrow 1 \\ \therefore \; I &=\int_{-1}^{1} \frac{z^2 \sin z}{z^2+\cos z}dz\\ \text{let}\;\; f(z)&=\frac{z^2 \sin z}{z^2+ \cos z} dz\\ f(-z) &= \frac{z^2 \sin z}{z^2+ \cos z}\\ f(z)&= -f(z) \; \text{function is ODD}\\ \therefore \;\; I&=0 \end{aligned}
Question 3 |
The solution of the initial value problem \frac{\mathrm{d} y}{\mathrm{d} x}= -2xy; y(0)=2 is
1+e^{-x^{2}} | |
2e^{-x^{2}} | |
1+e^{x^{2}} | |
2e^{x^{2}} |
Question 3 Explanation:
\frac{d y}{d x}=2 x y=0 \qquad ...(1)
I F.=e^{\int 2 x d x}=e^{x^{2}}
Multiplying I.F. to both side of equation (1)
e^{x^{2}}\left[\frac{d y}{d x}+2 x y\right]=0
\Rightarrow \frac{d}{d x}\left(e^{x^{2}} y\right)=0
e^{x^{2}} y=c
from the given boundary condition, C=2
\therefore e^{x^{2}} y=2
y=2 e^{-x^{2}}
I F.=e^{\int 2 x d x}=e^{x^{2}}
Multiplying I.F. to both side of equation (1)
e^{x^{2}}\left[\frac{d y}{d x}+2 x y\right]=0
\Rightarrow \frac{d}{d x}\left(e^{x^{2}} y\right)=0
e^{x^{2}} y=c
from the given boundary condition, C=2
\therefore e^{x^{2}} y=2
y=2 e^{-x^{2}}
Question 4 |
A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs. 500 is _______
12% | |
65% | |
98% | |
50% |
Question 4 Explanation:
49 to 51
Z=\frac{x-\mu}{\sigma}
\text { Given, } x=500, \mu=500, \sigma=50
\therefore \quad z=\frac{500-500}{50}=0
\therefore P(0)=50 \%
Z=\frac{x-\mu}{\sigma}
\text { Given, } x=500, \mu=500, \sigma=50
\therefore \quad z=\frac{500-500}{50}=0
\therefore P(0)=50 \%
Question 5 |
Laplace transform of cos(\omega t) is \frac{s}{s^{2}+\omega ^{2}}. The Laplace transform of e^{-2t}cos(4t) is
\frac{s-2}{(s-2)^{2}+16} | |
\frac{s+2}{(s-2)^{2}+16} | |
\frac{s-2}{(s+2)^{2}+16} | |
\frac{s+2}{(s+2)^{2}+16} |
Question 5 Explanation:
Given L \cos (\omega t)=\frac{s}{s^{2}+\omega^{2}}
\Rightarrow L\left(e^{-2 t} \cos 4 t\right)=?
By the given formula
\Rightarrow \quad L(\cos 4 t)=\frac{s}{s^{2}+16}
\Rightarrow L\left(e^{2 t} \cos 4 t\right)=\frac{s+2}{(s+2)^{2}+16}
( \because By using first shifting property of Laplace)
Hence option should be (D).
\Rightarrow L\left(e^{-2 t} \cos 4 t\right)=?
By the given formula
\Rightarrow \quad L(\cos 4 t)=\frac{s}{s^{2}+16}
\Rightarrow L\left(e^{2 t} \cos 4 t\right)=\frac{s+2}{(s+2)^{2}+16}
( \because By using first shifting property of Laplace)
Hence option should be (D).
There are 5 questions to complete.