# 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.