Question 1 |
All the four entries of the 2x2 matrix P=\begin{bmatrix} P_{11} &P_{12} \\ P_{21} & P_{22} \end{bmatrix} are nonzero, and one of its eigenvalue is zero. Which of the following statements is true?
P_{11}P_{22}-P_{12}P_{21}=1 | |
P_{11}P_{22}-P_{12}P_{21}=-1 | |
P_{11}P_{22}-P_{12}P_{21}=0 | |
P_{11}P_{22}+P_{12}P_{21}=0 |
Question 1 Explanation:
Eigenvalues are the roots of the determinant formed by matrix [sI-P]
\begin{array}{l} [sI-P]=\left[\begin{array}{cc}s-P_{11} & -P_{12} \\-P_{21} & s-P_{22}\end{array}\right] \\\; [s I - P]=0 \\ \Rightarrow \quad \left(s-P_{11}\right)\left(s-P_{22}\right)-P_{12} P_{21}=0\\ \Rightarrow \quad s^{2}-\left(P_{11}+P_{22}\right) s+P_{11} P_{22}-P_{12} P_{21}=0 \end{array}
Since, one of the its eigenvalues is zero, therefore,
putting s=0
P_{11} P_{22}-P_{12} P_{21}=0
which is the desired condition.
\begin{array}{l} [sI-P]=\left[\begin{array}{cc}s-P_{11} & -P_{12} \\-P_{21} & s-P_{22}\end{array}\right] \\\; [s I - P]=0 \\ \Rightarrow \quad \left(s-P_{11}\right)\left(s-P_{22}\right)-P_{12} P_{21}=0\\ \Rightarrow \quad s^{2}-\left(P_{11}+P_{22}\right) s+P_{11} P_{22}-P_{12} P_{21}=0 \end{array}
Since, one of the its eigenvalues is zero, therefore,
putting s=0
P_{11} P_{22}-P_{12} P_{21}=0
which is the desired condition.
Question 2 |
The system of linear equations
4x + 2y = 7
2x + y = 6
has
4x + 2y = 7
2x + y = 6
has
a unique solution | |
no solution | |
an infinite number of solutions | |
exactly two distinct solutions |
Question 2 Explanation:
The system can be written in matrix from as
\left[\begin{array}{ll}4 & 2 \\ 2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}7 \\6 \end{array}\right]
The Augmentod matrix [AIB] is given by
\left[\begin{array}{ll|l}4 & 2 & 7 \\ 2 & 1 & 6\end{array}\right]
Performing Gauss elimination on this [A \mid B] as follows:
\left[\begin{array}{ll|l}4 & 2 & 7 \\2 & 1 & 6 \end{array}\right] \frac{R_{2}-\frac{2}{4} R_{1}}{=R_{2}-\frac{1}{2} R_{1}}\left[\begin{array}{ll|l}4 & 2 & 7 \\0 & 0 & 5 / 2 \end{array}\right]
Now Rank [A \mid B]=2
(The number of non-zero rows in [ A|B]
Rank [A]=1
(The number of non-zero rows in [A])
since, Rank [A \mid B] \neq \text{Rank}[A].
The system has no solution.
\left[\begin{array}{ll}4 & 2 \\ 2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}7 \\6 \end{array}\right]
The Augmentod matrix [AIB] is given by
\left[\begin{array}{ll|l}4 & 2 & 7 \\ 2 & 1 & 6\end{array}\right]
Performing Gauss elimination on this [A \mid B] as follows:
\left[\begin{array}{ll|l}4 & 2 & 7 \\2 & 1 & 6 \end{array}\right] \frac{R_{2}-\frac{2}{4} R_{1}}{=R_{2}-\frac{1}{2} R_{1}}\left[\begin{array}{ll|l}4 & 2 & 7 \\0 & 0 & 5 / 2 \end{array}\right]
Now Rank [A \mid B]=2
(The number of non-zero rows in [ A|B]
Rank [A]=1
(The number of non-zero rows in [A])
since, Rank [A \mid B] \neq \text{Rank}[A].
The system has no solution.
Question 3 |
The equation sin(z) = 10 has
no real or complex solution | |
exactly two distinct complex solutions | |
a unique solution | |
an infinite number of complex solutions |
Question 3 Explanation:
sinz can have value between -1 to +1. Thus no solution.
Question 4 |
For real values of x, the minimum value of the function f(x)=exp(x)+exp(-x) is
2 | |
1 | |
0.5 | |
0 |
Question 4 Explanation:
f(x)=e^{x}+e^{-x}=e^{x}+\frac{1}{e^{x}}
Arithmetic mean of e^{x} and \frac{1}{e^{x}} is
\mathrm{A.M},=\frac{e^{x}+\frac{1}{e^{x}}}{2}
Geometric mean of e^{x} and \frac{1}{e^{x}} is
G.M. =e^{x} \cdot \frac{1}{e^{x}}=1
It is known that A.M. \geq G.M
\frac{\left(e^{x}+\frac{1}{e^{x}}\right)}{2} \geq 1
e^{x}+\frac{1}{e^{x}} \geq 2
Therefore, \left(e^{x}+e^{-x}\right)_{\min }=2
Arithmetic mean of e^{x} and \frac{1}{e^{x}} is
\mathrm{A.M},=\frac{e^{x}+\frac{1}{e^{x}}}{2}
Geometric mean of e^{x} and \frac{1}{e^{x}} is
G.M. =e^{x} \cdot \frac{1}{e^{x}}=1
It is known that A.M. \geq G.M
\frac{\left(e^{x}+\frac{1}{e^{x}}\right)}{2} \geq 1
e^{x}+\frac{1}{e^{x}} \geq 2
Therefore, \left(e^{x}+e^{-x}\right)_{\min }=2
Question 5 |
Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x = 0 ?
sin(x^{3}) | |
sin(x^{2}) | |
cos(x^{3}) | |
cos(x^{2}) |
Question 5 Explanation:
\begin{array}{l} \sin x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots . \\ \cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{4}-\frac{x^{6}}{6}+\ldots . \end{array}
Question 6 |
Which of the following is a solution to the differential equation \frac{dx(t)}{dt}+3x(t)=0 ?
x(t)=3e^{-t} | |
x(t)=2e^{-3t} | |
x(t)=-\frac{3}{2}t^{2} | |
x(t)=3t^{2} |
Question 6 Explanation:
\begin{array}{rr} & (D+3) x(t)=0 \\ \Rightarrow \quad & D=-3 \\ \text { So, } & x(t)=C e^{-3 t} \end{array}
Question 7 |
In the following graph, the number of trees (P) and the number of cut-set (Q)
are
P = 2,Q = 2 | |
P = 2,Q = 6 | |
P = 4,Q = 6 | |
P = 4,Q = 10 |
Question 7 Explanation:
Different trees (P) are shown below
Different cut-sets (Q) are shown below:
Different cut-sets (Q) are shown below:
Question 8 |
In the following circuit, the switch S is closed at t = 0. The rate of change of
current \frac{di}{dt}(0^{+}) is given by
0 | |
\frac{R_{s}I_{s}}{L} | |
\frac{(R+R_{s})I_{s}}{L} | |
\infty |
Question 8 Explanation:
At t = o, the inductor behaves as an open circuit
\begin{aligned} \text { So, } \quad V_{L}&=I_{S} R_{S} \\ \qquad V_{L}&=L \frac{d i}{d t}\left(0^{+}\right) \\ \Rightarrow \quad \frac{d i}{d t}\left(0^{+}\right)&=\frac{V_{L}}{L}=\frac{I_{s} R_{S}}{L} \end{aligned}
\begin{aligned} \text { So, } \quad V_{L}&=I_{S} R_{S} \\ \qquad V_{L}&=L \frac{d i}{d t}\left(0^{+}\right) \\ \Rightarrow \quad \frac{d i}{d t}\left(0^{+}\right)&=\frac{V_{L}}{L}=\frac{I_{s} R_{S}}{L} \end{aligned}
Question 9 |
The input and output of a continuous time system are respectively denoted
by x(t) and y(t). Which of the following descriptions corresponds to a causal
system ?
y(t) = x(t - 2) + x(t + 4) | |
y(t) = (t - 4)x(t + 1) | |
y(t) = (t + 4)x(t - 1) | |
y(t) = (t + 5)x(t + 5) |
Question 9 Explanation:
A system is casual if the output at any time
depends only on values of the input at the present
time and in the past.
Question 10 |
The impulse response h(t) of a linear time invariant continuous time system is described by h(t) = exp(\alphat)u(t) + exp(\betat)u(-t), where u(-t) denotes the unit step function, and \alpha and \beta are real constants. This system is stable if
\alpha is positive and \beta is positive | |
\alpha is negative and \beta is negative | |
\alpha is positive and \beta is negative | |
\alpha is negatine and \beta is positive |
Question 10 Explanation:
h(t)=e^{\alpha t} u(t)+e^{\beta t} u(-t)
For the system to be stable, \int_{-\infty}^{\infty} h(t) d t \lt \infty
For the above condition, h(t) should be as shown below.
For the system to be stable, \int_{-\infty}^{\infty} h(t) d t \lt \infty
For the above condition, h(t) should be as shown below.
There are 10 questions to complete.