Question 1 |
The rank of the matrix \begin{bmatrix} 1& 1& 1\\ 1& -1&0 \\ 1& 1& 1 \end{bmatrix} is
0 | |
1 | |
2 | |
3 |
Question 1 Explanation:
\begin{array}{l} R_{3} \rightarrow R_{1}-R_{3} \\ {\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right]}\\ \therefore \text{Rank}=2 \end{array}
Question 2 |
\bigtriangledown \times \bigtriangledown \times P, where P is a vector, is equal to
P \times \bigtriangledown \times P-\bigtriangledown ^{2}P | |
\bigtriangledown^{2} P+\bigtriangledown (\bigtriangledown \cdot P) | |
\bigtriangledown^{2} P+\bigtriangledown \times P | |
\bigtriangledown (\bigtriangledown \cdot P)-\bigtriangledown ^{2}P |
Question 2 Explanation:
From vector triple product.
\begin{aligned} A \times(B \times C) &=B(A . C)-C(A . B) \\ A &=\nabla . B=\nabla, C=P \\ \nabla \times \nabla \times P &=(\nabla P)-P(\nabla \nabla)=\nabla(\nabla P)-\nabla^{2} P \end{aligned}
\begin{aligned} A \times(B \times C) &=B(A . C)-C(A . B) \\ A &=\nabla . B=\nabla, C=P \\ \nabla \times \nabla \times P &=(\nabla P)-P(\nabla \nabla)=\nabla(\nabla P)-\nabla^{2} P \end{aligned}
Question 3 |
\int \int (\bigtriangledown \times P)\cdot ds, where P is a vector, is equal to
\oint P\cdot dl | |
\oint \bigtriangledown \times \bigtriangledown \times P\cdot dl | |
\oint \bigtriangledown \times P\cdot dl | |
\int \int \int \bigtriangledown \cdot Pdv |
Question 3 Explanation:
\iint(\Delta x P) d s=\oint P .d l (strokes Theorem)
Question 4 |
A probability density function is of the form
p(x)=Ke^{-\alpha |x|},x\in (-\infty ,\infty )
The value of K is
p(x)=Ke^{-\alpha |x|},x\in (-\infty ,\infty )
The value of K is
0.5 | |
1 | |
0.5\alpha | |
\alpha |
Question 4 Explanation:
\begin{aligned} \int_{-\infty}^{\infty} p(x) d x&=1 \\ \int_{-\infty}^{\infty} K e^{-\alpha|x|} d x&=1 \\ \int_{-\infty}^{0} K e^{\mu x} d x+\int_{0}^{\infty} K e^{-u x}&=1\\ \Rightarrow \quad\frac{K}{\alpha}\left[e^{\alpha x}\right]_{-\infty}^{0}+\frac{K}{-\alpha}\left[e^{-\alpha x}\right]_{0}^{\infty}&=1 \\ \Rightarrow \quad \frac{K}{\alpha}+\frac{K}{\alpha}&=1\\ 2 K&=\alpha \\ \Rightarrow \quad K&=0.5 \alpha \end{aligned}
Question 5 |
A solution for the differential equation
\dot{x}(t)+2x(t)=\delta (t)
with initial condition x(0^-)=0 is
\dot{x}(t)+2x(t)=\delta (t)
with initial condition x(0^-)=0 is
e^{-2t}u(t) | |
e^{2t}u(t) | |
e^{-t}u(t) | |
e^{t}u(t) |
Question 5 Explanation:
\dot{x}(t)+2 x(t)=\delta(t)
Taking L.T. on both sides
\begin{aligned} s X(s)-x(0)+2 X(s)&=1 \\ X(s)[s+2]&=1 \\ X(s)&=\frac{1}{s+2} \\ x(t) &=e^{-2 t} u(t) \end{aligned}
Taking L.T. on both sides
\begin{aligned} s X(s)-x(0)+2 X(s)&=1 \\ X(s)[s+2]&=1 \\ X(s)&=\frac{1}{s+2} \\ x(t) &=e^{-2 t} u(t) \end{aligned}
There are 5 questions to complete.