Question 1 |
Let p and q be real numbers such that p^{2}+q^{2}=1. The eigenvalues of the matrix \begin{bmatrix} p & q\\ q& -p \end{bmatrix}
are
1 and 1 | |
1 and -1 | |
j and -j | |
pq and -pq |
Question 1 Explanation:
Characteristic equation of A
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
Question 2 |
Let p\left ( z\right )=z^{3}+\left ( 1+j \right )z^{2}+\left ( 2+j \right )z+3, where z is a complex number.
Which one of the following is true?
Which one of the following is true?
\text{conjugate}\:\left \{ p\left ( z \right ) \right \}=p\left ( \text{conjugate} \left \{ z \right \} \right ) for all z | |
The sum of the roots of p\left ( z \right )=0 is a real number | |
The complex roots of the equation p\left ( z \right )=0
come in conjugate pairs | |
All the roots cannot be real |
Question 2 Explanation:
Since sum of the roots is a complex number
\Rightarrow absent one root is complex
So all the roots cannot be real.
\Rightarrow absent one root is complex
So all the roots cannot be real.
Question 3 |
Let f\left ( x \right )
be a real-valued function such that {f}'\left ( x_{0} \right )=0
for some x _{0} \in\left ( 0,1 \right ), and {f}''\left ( x \right )> 0
for all x \in \left ( 0,1 \right ). Then f\left ( x \right )
has
no local minimum in (0,1) | |
one local maximum in (0,1) | |
exactly one local minimum in (0,1) | |
two distinct local minima in (0,1) |
Question 3 Explanation:
x_{0} \in(0,1), where f(x)=0 is stationary point
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
Question 4 |
For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is
\text{10 V} in series with 12\:\Omega
| |
\text{65 V} in series with 15\:\Omega
| |
\text{50 V} in series with 2\:\Omega
| |
\text{35 V} in series with 2\:\Omega
|
Question 4 Explanation:
Given circuit can be resolved as shown below,
V_{T H}=15+50=65 \mathrm{~V}
\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}
V_{T H}=15+50=65 \mathrm{~V}
\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}
Question 5 |
Which one of the following vector functions represents a magnetic field \overrightarrow{B}?
(\hat{X}, \hat{Y} and \hat{Z} are unit vectors along x-axis, y-axis, and z-axis, respectively)
(\hat{X}, \hat{Y} and \hat{Z} are unit vectors along x-axis, y-axis, and z-axis, respectively)
10x\hat{X}+20y\hat{Y}-30z\hat{Z} | |
10y\hat{X}+20x\hat{Y}-10z\hat{Z} | |
10z\hat{X}+20y\hat{Y}-30x\hat{Z} | |
10x\hat{X}-30z\hat{Y}+20y\hat{Z} |
Question 5 Explanation:
If \vec{B} is magnetic flux density then \vec{\nabla} \cdot \vec{B}=0
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
There are 5 questions to complete.