GATE EE 2015 SET 1


Question 1
A random variable X has probability density function f(x) as given below:

f(x)=\left\{\begin{matrix} a+bx & for \; 0 \lt x \lt 1\\ 0& otherwise \end{matrix}\right.

If the expected value E[X]=2/3, then Pr[X \lt 0.5] is _____________.
A
0.25
B
0.5
C
0.75
D
1
Engineering Mathematics   Probability and Statistics
Question 1 Explanation: 
\begin{aligned} f(x)&=\left\{\begin{matrix} a+bx & \text{for }0 \lt x \lt 1\\ 0 & \text{otherwise} \end{matrix}\right.&\\ \text{Now given }E(X)&=2/3\\ \int_{0}^{1}xf(x)dx&=\frac{2}{3}\\ \int_{0}^{1}x(a+bx)dx&=\frac{2}{3}\\ a\left ( \frac{x^2}{2} \right )_0^1+b\left ( \frac{x^3}{3} \right )_0^1&=\frac{2}{3}\\ a\left ( \frac{1}{2} \right )+b\left ( \frac{1}{3} \right )&=\frac{2}{3}\\ 3a+2b&=4\;\;...(i)\\ \text{Now, }\int_{0}^{1}f(x)dx &=1\\ (\text{total probability} & \text{ is always equal to 1})\\ \int_{0}^{1}(a+bx)dx&=\left ( ax+\frac{bx^2}{2} \right )_0^1=1\\ a+\frac{b}{2}&=1\\ 2a+b&=2\;\;...(ii)\\ \text{Now solving } & \text{ (i) and (ii), we get}\\ a&=0,b=2\\ So,\; \;f(x)&=\left\{\begin{matrix} 2x & \text{for }0 \lt x lt 1\\ 0 & \text{otherwise} \end{matrix}\right.&\\ \text{Now we need}&\\ P(x \lt 0.5)&=\int_{0}^{0.5}2xdx=2\left ( \frac{x^2}{2} \right )_0^{0.5}\\ &=0.5^2-0^2=0.25 \end{aligned}
Question 2
If a continuous function f(x) does not have a root in the interval [a,b], then which one of the following statements is TRUE?
A
f(a)\cdot f(b)=0
B
f(a)\cdot f(b) \lt 0
C
f(a)\cdot f(b) \gt 0
D
f(a)/ f(b)\leq 0
Engineering Mathematics   Calculus
Question 2 Explanation: 
Intermediate value theorem states that if a function is continious and f(a) \cdot f(b) \lt 0, then surely there is a root in (a,b). The contrapositive of this theorem is that if a function is continious and has no root in (a,b) then surely f(a) \cdot f(b) \geq 0. But since it is given that there is no root in the closed interval [a,b] it means f(a) \cdot f(b) \neq 0.
So surely f(a) \cdot f(b) \gt 0 which is choise(C).


Question 3
If the sum of the diagonal elements of a 2x2 matrix is -6, then the maximum possible value of determinant of the matrix is ________.
A
6
B
8
C
9
D
12
Engineering Mathematics   Linear Algebra
Question 3 Explanation: 
Consider a symmetric matrix A =\begin{bmatrix} a & b\\ b & d \end{bmatrix}
Given a+d=-6
|A|=ad-b^2
Now since b^2 is always non-negative, maximum determinant will come when b^2=0.
So we need to maximize
\begin{aligned} |A|&=ad-0=ad=a \times -(6+a) \\ &= -a^2-6a\\ \frac{d|A|}{da} &=-2a-6=0 \\ \Rightarrow \;\; a &=-3 \text{ is the only stationary point} \end{aligned}
Since, \left [\frac{d^2|A|}{da} \right ]_{a=-3}=-2 \lt 0, we have a maximum at a=-3.
Since a+d=-6, corresponding value of d=-3.
|A|=ad=-3 \times -3=9
Question 4
Consider a function \vec{f}=\frac{1}{r^{2}}\hat{r}, where r is the distance from the origin and \hat{r} is the unit vector in the radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is
A
0
B
2\pi
C
4\pi
D
R\pi
Electromagnetic Fields   Coordinate Systems and Vector Calculus
Question 4 Explanation: 
\bar{f}=\frac{1}{r^2}\hat{r}
From divergence theorem as we know,
\begin{aligned} \int _{vol.} (\triangledown \cdot \bar{f})dV &=\oint \oint _S \bar{f}\cdot d\bar{S}\\ \oint \oint _S \bar{f}\cdot d\bar{S} &= \oint \oint _S \left ( \frac{1}{r^2} \cdot \hat{r}\right ) \times r^2 \sin \theta \cdot d\theta \cdot d\phi \cdot \hat{r} \\ &= \oint \oint _S \sin \theta \cdot d\theta \cdot d\phi=4\pi \end{aligned}
Question 5
When the Wheatstone bridge shown in the figure is used to find the value of resistor R_X, the galvanometer G indicates zero current when R_1=50\Omega ,R_2=65\Omega and R_3=100\Omega. If R_3 is known with \pm 5% tolerance on its nominal value of 100 \Omega , what is the range of R_X in Ohms?
A
[123.50, 136.50]
B
[125.89, 134.12]
C
[117.00, 143.00]
D
[120.25, 139.75]
Electrical and Electronic Measurements   Characteristics of Instruments and Measurement Systems
Question 5 Explanation: 
R_1=50\Omega
R_2=60\Omega
R_3=100\pm 5%\Omega
The value of R_3 \text{with} \pm 5% of tollerance,
\begin{aligned} R_3&= 100\pm 5\% \\ &= 100+100 \times \frac{5}{100}=105 \Omega \\ &= 100-100 \times \frac{5}{100}=95 \Omega \end{aligned}
In both condition, the bridge is balance, so under balance condition,
\begin{aligned} \frac{R_1}{R_3}&=\frac{R_2}{R_x} \\ R_x&=\frac{R_2R_3}{R_1} \\ (i) \; \text{when}, \; R_3&=105\Omega \\ \therefore \;\;R_x &= \frac{105 \times 65}{50}=136.50\Omega \\ (ii)\; \text{when}, \; R_3&=95 \Omega \\ \therefore \;\;R_x &= \frac{95 \times 65}{50}=123.50\Omega \end{aligned}




There are 5 questions to complete.