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.