GATE EE 2015 SET 1

\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}

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).

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

\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}

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}

I_{fs}=50A,
V_m=0.1V,
R_m=\frac{0.1}{50}=2 \times 10^{-3}\Omega
\because \; m=10
R_{sh}=\frac{R_m}{(m-1)}
\;\;\;=\frac{2 \times 10^{-3}}{9}=0.22\Omega

Step down chopper
\begin{aligned} \Rightarrow \; V_0&=\alpha V_s\\ V_0&=I_0R+E\\ I_0&=\frac{V_0-E}{R}\\ I_0&=\frac{\alpha V_s-E}{R}\\ &=\frac{(0.4)20-5}{3}=1A \end{aligned}

System input: \sin \omega _0 t
\begin{aligned} f_0&=\frac{1}{2T}Hz\\ \text{Therefore,}\\ \omega _0&=2 \pi f_0=\frac{2 \pi}{2T}=\frac{\pi}{T}\; rad/sec\\ y(t)&=\frac{1}{T}\int_{t-T}^{t}\sin \omega _0\tau \; d\tau\\ &=\frac{1}{T}[-\cos\omega _0 t ]_{t-T}^t\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\omega _0 (t-T) ]\\ &\left [ \because \; \omega _0=\frac{t}{T} \right ]\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\left ( \frac{\pi t}{T}- \pi \right ) ]\\ \\ &=-\frac{1}{T}[\cos\omega _0 t +\cos\omega _0 t ]\\ &=-\frac{2}{T} \cos\omega _0 t \\ &=\frac{2}{T}\sin (\omega _0 t -90^{\circ}) \\ &=\text{Output is lagging by }90^{\circ} \text{ w.r.t. input.} \end{aligned}

\begin{aligned} g(t)&=u(t)-u(t-1)\\ G(s)&=\frac{1}{s}-\frac{e^{-s}}{s}\\ G(s) \times G(s)&=g(t)* g(t) \end{aligned}

Maximum value =1