Question 1 |
Consider a system of linear equations:
x - 2 y + 3z = -1,
x - 3y + 4z = 1, and
-2x + 4 y - 6z = k.
The value of k for which the system has infinitely many solutions is ______.
x - 2 y + 3z = -1,
x - 3y + 4z = 1, and
-2x + 4 y - 6z = k.
The value of k for which the system has infinitely many solutions is ______.
1 | |
2 | |
3 | |
4 |
Question 1 Explanation:
\begin{aligned} x-2 y+3 z&=-1 \\ x-3 y+1 z&=1, \text { and } \\ -2 x+1 y-6 z&= k \\ \left[A:B \right]&=\left[\begin{array}{rrrrr} 1 & -2 & 3 :& -1 \\ 1 & -3 & 4 :& 1 \\ -2 & 4 & 6 :& k \end{array}\right]\\ R_{2} \rightarrow R_{2}-R_{1}, R_{3} &\rightarrow R_{1}+2 R_{1} \\ \left[\begin{array}{cccc} 1 & -2 & 3 :& -1 \\ 0 & -1 & 1 :& 2 \\ 0 & 0 & 0 :& k-2 \end{array}\right] \end{aligned}
For Infinlto may solulion
\begin{aligned} \rho(A: B)&=p(A)\\ &= r \lt \text{ number of variables}\\ \rho(A: B)&=2 \\ k-2&=0 \\ k&=2 \end{aligned}
For Infinlto may solulion
\begin{aligned} \rho(A: B)&=p(A)\\ &= r \lt \text{ number of variables}\\ \rho(A: B)&=2 \\ k-2&=0 \\ k&=2 \end{aligned}
Question 2 |
A function f(x) =1- x^{2}+x^{3} is defined in the closed interval [-1,1]. The value of x , in the open interval (-1,1) for which the mean value theorem is satisfied, is
-1/2 | |
-1/3 | |
1/3 | |
1/2 |
Question 2 Explanation:
Since f(1)\neq f(-1) Rolle's mean value theorem does not apply
By Lagrange moon voluo theorem
\begin{aligned} f^{\prime}(x) &=\frac{((1)-f(-1)}{1-(-1)}=\frac{2}{2}=1 \\ -2 x+3 x^{2} &=1 \\ x &=1-\frac{1}{3}\\ x \text{ lies in }(-1,1) \\ \Rightarrow \quad x&=-\frac{1}{3} \end{aligned}
By Lagrange moon voluo theorem
\begin{aligned} f^{\prime}(x) &=\frac{((1)-f(-1)}{1-(-1)}=\frac{2}{2}=1 \\ -2 x+3 x^{2} &=1 \\ x &=1-\frac{1}{3}\\ x \text{ lies in }(-1,1) \\ \Rightarrow \quad x&=-\frac{1}{3} \end{aligned}
Question 3 |
Suppose A and B are two independent events with probabilities P(A) \neq 0 and P(B) \neq 0 . Let \bar{A} and \bar{B} be their complements. Which one of the following statements is FALSE?
P(A\cap B)=P(A)P(B) | |
P(A|B)=P(A) | |
P(A\cup B)=P(A) + P(B) | |
P(\bar{A}\cup \bar{B})=P(\bar{A}) + P(\bar{B}) |
Question 3 Explanation:
P(A \cup B)=P(A)+P(B)-P(A \cap B)
since P(A \cap B)=P(A) \rho(B)
(not necessarily equal to zero).
So, P(A \cup B)=P(A)+P(B) is false.
since P(A \cap B)=P(A) \rho(B)
(not necessarily equal to zero).
So, P(A \cup B)=P(A)+P(B) is false.
Question 4 |
Let z = x + iy be a complex variable. Consider that contour integration is performed along the unit circle in anticlockwise direction. Which one of the following statements is NOT TRUE?
The residue of \frac{z}{z^{2}-1} at z=1 is 1/2 | |
\oint_{c}z^{2}dz=0 | |
\frac{1}{2\pi i} \oint_{c} \frac{1}{z} dz=1 | |
\bar{z} (complex conjugate of z ) is an analytical function |
Question 4 Explanation:
f(z)= \bar{z} =x-i y
\begin{array}{cc} u=x & v=-y \\ \Rightarrow u_{x}=1 & v_{x}=0 \\ u_{y}=0 & v_{y}=-1 \end{array}
u_{x} \neq v_{y} i.e. C-R not satisfied
\Rightarrow \bar{z} is not analytic function.
\begin{array}{cc} u=x & v=-y \\ \Rightarrow u_{x}=1 & v_{x}=0 \\ u_{y}=0 & v_{y}=-1 \end{array}
u_{x} \neq v_{y} i.e. C-R not satisfied
\Rightarrow \bar{z} is not analytic function.
Question 5 |
The value of p such that the vector \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} is an eigenvector of the matrix \begin{bmatrix} 4 & 1& 2\\ p& 2& 1\\ 14&-4 &10 \end{bmatrix} is _______.
12 | |
17 | |
20 | |
22 |
Question 5 Explanation:
\begin{array}{r} A x=\lambda N \\ {\left[\begin{array}{lll} 4 & 1 & 2 \\ p & 2 & 1 \\ 14 & -4 & 10 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]=\lambda\left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right]} \\ {\left[\begin{array}{c} 12 \\ p+7 \\ 36 \end{array}\right]=\lambda\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]} \\ \frac{p+7}{12}=2 \Rightarrow p=17 \end{array}
Question 6 |
In the circuit shown, at resonance, the amplitude of the sinusoidal voltage (in Volts) across the capacitor is ________.
20 | |
23 | |
25 | |
28 |
Question 6 Explanation:
\begin{array}{l} \text { At resonance, } I=\frac{10}{4} \\ \qquad \begin{aligned} \omega&=\frac{1}{\sqrt{0.1 \times 10^{-3} \times 10^{-6}}}=10^{5} \text { rad/sec } \\ X_{C}&=\frac{1}{\omega C}=\frac{1}{10^{5} \times 1 \times 10^{-6}}=10 \Omega \\ V_{C}&=I X_{C}=10 \times \frac{10}{4}=25 \mathrm{V} \end{aligned} \end{array}
Question 7 |
In the network shown in the figure, all resistors are identical with R = 300 \Omega. The resistance R_{ab} (in \Omega) of the network is ______.
95 | |
100 | |
110 | |
120 |
Question 7 Explanation:
Modifying the given circuit
\begin{aligned} R_{a b} &=\left(\frac{1}{2 R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{2 R}\right)^{-1} \\ &=\frac{R}{3}=\frac{300}{3}=100 \Omega \end{aligned}
\begin{aligned} R_{a b} &=\left(\frac{1}{2 R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{2 R}\right)^{-1} \\ &=\frac{R}{3}=\frac{300}{3}=100 \Omega \end{aligned}
Question 8 |
In the given circuit, the values of V_{1} \; and \; V_{2} respectively are
5 V, 25 V | |
10 V,30 V | |
15 V,30 V | |
0 V , 20V |
Question 8 Explanation:
Current flowing through both the parallel 4 \Omega will
be I.
\text { So, } V_{2}=4(I+I+2 I)+4 I \quad \text { by KVL }
I+I+2 I=5 \quad \text{by KCL}
\begin{aligned} I &=\frac{5}{4} A \\ V_{2} &=4 \times 5+\frac{4 \times 5}{4}=25 \mathrm{V} \\ V_{1} &=4 I=\frac{4 \times 5}{4}=5 \mathrm{V} \end{aligned}
Question 9 |
A region of negative differential resistance is observed in the current voltage characteristics of a silicon PN junction if
both the P-region and the N-region are heavily doped | |
the N-region is heavily doped compared to the P-region | |
the P-region is heavily doped compared to the N-region | |
an intrinsic silicon region is inserted between the P-region and the N-region |
Question 10 |
A silicon sample is uniformly doped with donor type impurities with a concentration of 10^{16} /cm^{3}. The electron and hole mobilities in the sample are 1200 cm^{2}/V-s and 400 cm^{2}/V-s respectively. Assume complete ionization of impurities. The charge of an electron is 1.6 \times 10^{-19} C. The resistivity of the sample (in \Omega-cm) is ____________.
0.2 | |
0.5 | |
0.7 | |
1 |
Question 10 Explanation:
\begin{aligned} &=\frac{1}{\sigma_{N}}=\frac{1}{N_{D} q \mu_{n}} \\ &=\frac{1}{10^{16} \times 1.6 \times 10^{-19} \times 1200}=0.52 \Omega-\mathrm{cm} \end{aligned}
There are 10 questions to complete.
