GATE EE 2010

P=\int_{0}^{1}xe^x\; dx
Integrating by parts:
\begin{aligned} \text{Let } \; u&= x,\\ dv&= e^x\; dx\\ du&= dx,\\ v&=\int e^x\; dx=e^x \\ \text{Now, } \int udv&= uv-\int v\; du\\ \therefore \;\;\int x e^x\; dx &=xe^x-\int e^x\; dx \\ &=xe^x-e^x+c \\ \int_{0}^{1} xe^x\; dx&= [xe^x-e^x]_0^1\\ &= (1 \times e^1 -e^1)-(0 \times e^0 -e^0)\\ &= 0-(-1)\\ &= 1 \end{aligned}

\begin{aligned} \vec{r} &= x\hat{i}+y\hat{j}+z\hat{k}\\ \text{div } \vec{r}&= \bigtriangledown \cdot \vec{r}\\ &=\left ( \hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z} \right )(x\hat{i}+y\hat{j}+z\hat{k})\\ &=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\\ &=1+1+1=3 \end{aligned}

If period = T, then x(t+T)=x(t)
Here, x(t)= 8 \sin(0.8 \pi t+\pi /4)
putting x(t+T)=x(t) we get
8 \sin(0.8\pi (t+T)+\pi /4)=8 \sin (0.8 \pi t+\pi /4)
\sin(0.8\pi (t+T)+\pi /4)= \sin (0.8 \pi t+ \pi /4)
\sin(\theta _1)=\sin(\theta _2)\Rightarrow \theta _1=\theta _2+2 \pi
\therefore \;\; 0.8\pi (t+T)+ \pi /4=0.8 \pi t+ \pi /4+2 \pi
0.8\pi T=2 \pi
T=2.5 s

Integrator is always a linear system.
\begin{aligned} \text{Since, }y(t)&=\int_{-\infty }^{5t}x(\tau )d\tau \;\;t \gt 0 \\ &\text{For t=1,} \\ y(t)&= \int_{-\infty }^{5}x(\tau )d\tau \end{aligned}
here value at t=1 depends on future values like at t=2,3,... of input x(t).
So, it is a noncausal systems.

As the switch has been closed for a long time, the circuit is in steady state. At steadystate, capacitor is open circuit,

Using KVL,
5-I-4I=0 I=1A
V_c(0^-)=4 \times 1=4V
As the voltage across capacitorcan not change abruptly,
So, V_c(0^+)=V_c(0^-)=4V
Circuit at t=0^+

Current through capacitor at t=0^+
I_c(0^+)=\frac{4}{4}=1A