Question 1 |
Consider a 3x3 real symmetric matrix S such that two of its eigenvalues are a\neq 0,\; b\neq 0 with respective eigenvectors \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix},\begin{bmatrix} y_{1}\\ y_{2}\\ y_{3} \end{bmatrix}. If a\neq b then x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3} equals
a | |
b | |
ab | |
0 |
Question 1 Explanation:
3\times 3 real symmetric matrix such that two of its
eigen value are a \neq 0 b \neq 0 with respective eigen
vectors \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]\left[\begin{array}{l}y_{1} \\ y_{2} \\ y_{3}\end{array}\right] \text{ if } a \neq b then
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0 because they are orthogonal.
\therefore \qquad x^{T} y =0 \quad(\text { since } a \neq b)
\left[x_{1} x_{2} x_{3}\right]\left[\begin{array}{c}y_{1} \\y_{2} \\y_{3}\end{array}\right]=0
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0
eigen value are a \neq 0 b \neq 0 with respective eigen
vectors \left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right]\left[\begin{array}{l}y_{1} \\ y_{2} \\ y_{3}\end{array}\right] \text{ if } a \neq b then
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0 because they are orthogonal.
\therefore \qquad x^{T} y =0 \quad(\text { since } a \neq b)
\left[x_{1} x_{2} x_{3}\right]\left[\begin{array}{c}y_{1} \\y_{2} \\y_{3}\end{array}\right]=0
x_{1} y_{1}+x_{2} y_{2}+x_{3} y_{3}=0
Question 2 |
If a function is continuous at a point,
the limit of the function may not exist at the point | |
the function must be derivable at the point | |
the limit of the function at the point tends to infinity | |
the limit must exist at the point and the value of limit should be same as the value of the function at that point |
Question 2 Explanation:
f(x) is continuous at any point
\text{if} \quad \lim_{x \rightarrow a^{-}} f(x)=\lim_{x \rightarrow a^{+}} f(x)=f(a)
\text{if} \quad \lim_{x \rightarrow a^{-}} f(x)=\lim_{x \rightarrow a^{+}} f(x)=f(a)
Question 3 |
Divergence of the vector field x^{2}z\hat{i}+xy\hat{j}-yz^{2}\hat{k} at (1,-1,1) is
0 | |
1 | |
5 | |
6 |
Question 3 Explanation:
\begin{aligned} \vec{F} &=x^{2} z \hat{i}+x y \hat{j}-y z^{2} \hat{k} \\ \nabla \cdot \vec{F} &=\frac{\partial}{\partial x}\left(x^{2} z\right)+\frac{\partial}{\partial y}(x y)-\frac{\partial}{\partial z}\left(y z^{2}\right) \\ \nabla \cdot \vec{F} &=2 x z+x-2 y z \\ \left.\therefore \nabla \cdot \vec{F}\right|_{1,-1,1} &=2 \times 1 \times 1+1-2 \times-1 \times 1 \\ &=2+1+2=5 \end{aligned}
Question 4 |
A group consists of equal number of men and women. Of this group 20% of the men and 50% of the women are unemployed. If a person is selected at random from this group, the probability of the selected person being employed is _______
0.25 | |
1 | |
4.5 | |
0.65 |
Question 4 Explanation:
Total \% of employed person
\begin{aligned} &=100-\frac{20+50}{2} \% \\ &=100-35 \%=65 \% \\ \text { Required prob } &=\frac{65}{100}=0.65 \end{aligned}
\begin{aligned} &=100-\frac{20+50}{2} \% \\ &=100-35 \%=65 \% \\ \text { Required prob } &=\frac{65}{100}=0.65 \end{aligned}
Question 5 |
The definite integral \int_{1}^{3}\frac{ 1}{x}\mathrm{d}x is evaluated using Trapezoidal rule with a step size of 1. The correct answer is _______
1.165 | |
2.365 | |
8.254 | |
9.548 |
Question 5 Explanation:
\begin{aligned} I &=\int_{1}^{3} \frac{1}{x} \mathrm{d} x=\frac{h}{2}\left[\left(y_{0}+y_{2}\right)+2 y_{1}\right] \\ &=\frac{1}{2}[1+0.33+2 \times 0.5]=\frac{2.33}{2}=1.165 \end{aligned}
There are 5 questions to complete.