Question 1 |
Two matrices A and B are given below:
A=\begin{bmatrix} p &q \\ r& s \end{bmatrix}; B=\begin{bmatrix} p^2+q^2 & pr +qs \\ pr+qs & r^2+s^2 \end{bmatrix}
If the rank of matrix A is N, then the rank of matrix B is
A=\begin{bmatrix} p &q \\ r& s \end{bmatrix}; B=\begin{bmatrix} p^2+q^2 & pr +qs \\ pr+qs & r^2+s^2 \end{bmatrix}
If the rank of matrix A is N, then the rank of matrix B is
N/2 | |
N-1 | |
N | |
2N |
Question 1 Explanation:
\begin{aligned} A &=\begin{bmatrix} p & q\\ r & s \end{bmatrix} \\ A \times A=A^2&=\begin{bmatrix} p^2+q^2 & pr+qs\\ pr+qs & r^2+s^2 \end{bmatrix} =B \\ A^2 &=B \end{aligned}
Rank of amtrix does not change when we squaring the matrix, hence rank of B = rank of A=N.
Rank of amtrix does not change when we squaring the matrix, hence rank of B = rank of A=N.
Question 2 |
A particle, starting from origin at t=0s, is traveling along x-axis with velocity
v=\frac{\pi}{2}\cos (\frac{\pi}{2}t)m/s
At t=3s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is_____
v=\frac{\pi}{2}\cos (\frac{\pi}{2}t)m/s
At t=3s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is_____
1 | |
2 | |
3 | |
4 |
Question 3 |
Let \triangledown \cdot (f v)=x^2y+y^2z+z^2x, where f and v are scalar and vector fields respectively. If v=yi+zj+xk , then v\cdot \triangledown f is
x^2y+y^2z+z^2x | |
2xy+2yz+2zx | |
x+y+z | |
0 |
Question 3 Explanation:
\begin{aligned} \vec{V}&=y\hat{i}+z\hat{j}+x\hat{k}\\ \hat{i}\frac{\partial (fV)}{\partial x}+\hat{j}\frac{\partial (fV)}{\partial y}+\hat{k}\frac{\partial (fV)}{\partial z}&=x^2y+y^2z+z^2x\\ y\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}&=x^2y+y^2z+z^2x\;\;...(i)\\ \vec{V}\cdot \Delta f&=y\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}\;\;...(ii)\\ \text{From equations (i) and (ii)}\\ \vec{V}\cdot \Delta f&=x^2y+y^2z+z^2x \end{aligned}
Question 4 |
Lifetime of an electric bulb is a random variable with density f(x)=kx^2, where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then the value of k is _____
0.85 | |
0.42 | |
0.25 | |
0.75 |
Question 4 Explanation:
Life time of an electric bulb with density
f(x)=Kx^2
If minimum and maximum lifetimes of bulb are 1 and 2 years respectively then
\begin{aligned} \int_{1}^{2}Kx^2dx &=1\\ \left.\begin{matrix} K\frac{x^3}{3} \end{matrix}\right|_1^2&=1\\ K\left ( \frac{8}{3}-\frac{1}{3} \right )&=1\\ \frac{7K}{3}&=1\\ K&=\frac{3}{7}=0.42 \end{aligned}
f(x)=Kx^2
If minimum and maximum lifetimes of bulb are 1 and 2 years respectively then
\begin{aligned} \int_{1}^{2}Kx^2dx &=1\\ \left.\begin{matrix} K\frac{x^3}{3} \end{matrix}\right|_1^2&=1\\ K\left ( \frac{8}{3}-\frac{1}{3} \right )&=1\\ \frac{7K}{3}&=1\\ K&=\frac{3}{7}=0.42 \end{aligned}
Question 5 |
A function f(t) is shown in the figure.
The Fourier transform F(\omega) of f(t) is
The Fourier transform F(\omega) of f(t) is
real and even function of w | |
real and odd function of w | |
imaginary and odd function of w | |
imaginary and even function of w |
Question 5 Explanation:
Fiven signal f(t) is an odd signal. Hence, F(\omega ) is imaginary and odd function of \omega .
There are 5 questions to complete.