Question 1 |
The series \sum_{^{n=0}}^{\infty }\frac{1}{n!} converges to
2 ln 2 | |
\sqrt{2} | |
2 | |
e |
Question 1 Explanation:
Given,
\begin{aligned} \text { Let, } x(n) &=\sum_{n=0}^{\infty} \frac{1}{n !} \\ &=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots \\ &=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots \end{aligned}
Also we know that expression of e^{x}
\begin{aligned} e^{x}&=1+x+\frac{1}{2} x^{2}+\frac{1}{6} x^{3}+\frac{1}{24} x^{4}+\cdots\\ \text{Put }x&=1\text{ in above expression}\\ e^{1}&=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots \\ e&=\sum_{n=0}^{\infty} \frac{1}{n !} \end{aligned}
\begin{aligned} \text { Let, } x(n) &=\sum_{n=0}^{\infty} \frac{1}{n !} \\ &=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots \\ &=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots \end{aligned}
Also we know that expression of e^{x}
\begin{aligned} e^{x}&=1+x+\frac{1}{2} x^{2}+\frac{1}{6} x^{3}+\frac{1}{24} x^{4}+\cdots\\ \text{Put }x&=1\text{ in above expression}\\ e^{1}&=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\cdots \\ e&=\sum_{n=0}^{\infty} \frac{1}{n !} \end{aligned}
Question 2 |
The magnitude of the gradient for the function f(x,y,z)=x^{2}+3y^{2}+z^{3} at the point (1,1,1) is _____.
5 | |
6 | |
7 | |
8 |
Question 2 Explanation:
\begin{aligned} f(v, y, z) &=x^{3}+3 y^{2}+z^{3} \\ \nabla f &=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{i}+\frac{\partial f}{\partial z} \hat{k} \\ &=2 x \hat{i}+6 y \hat{j}+3 z^{2} \hat{k} \\ (\Delta t)_{(1,1,1)} &=2 \hat{i}+6 \hat{j}+3 \hat{k} \\ |\Delta+|_{(1,1,1)} &=\sqrt{4+36+9}=\sqrt{49}=7 \end{aligned}
Question 3 |
Let X be a zero mean unit variance Gaussian random variable. E[|X|] is equal to ______.
0.8 | |
1.8 | |
0.2 | |
1.2 |
Question 3 Explanation:
For a Gaussian random variable,
\begin{aligned} f_{x}(x)&=\frac{1}{\sigma_{x} \sqrt{2 \pi}} e^{\left(\frac{\mu-x}{2 \sigma^{2}}\right)^{2}} \\ \text{For mean} (\mu)&=0 \text{ and Variance }\left(\sigma_{x}^{2}\right)=1 \\ f_{X}(x) &=\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} \\ E[|x|] &=\int_{-\infty}^{\infty}|x| f_{X}(x) d x \\ &=\int_{-\infty}^{\infty}|x| \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x \\ =& \int_{-\infty}^{0}-x \cdot \frac{1}{\sqrt{2 \pi}} e^{-x^{2}} d x+\int_{0}^{\infty} x \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x\\ &=2 \int_{0}^{\infty} x \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} x \cdot e^{\frac{-x^{2}}{2}} d x \\ &\text{Let }\quad -\frac{x^{2}}{2}=t \\ &\therefore \quad-\frac{2 x d x}{2}=d t \\ &\text{Hence }\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty}-e^{-t} d t=0.8 \\ &\therefore \quad E[|x|]=0.8 \\ \end{aligned}
\begin{aligned} f_{x}(x)&=\frac{1}{\sigma_{x} \sqrt{2 \pi}} e^{\left(\frac{\mu-x}{2 \sigma^{2}}\right)^{2}} \\ \text{For mean} (\mu)&=0 \text{ and Variance }\left(\sigma_{x}^{2}\right)=1 \\ f_{X}(x) &=\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} \\ E[|x|] &=\int_{-\infty}^{\infty}|x| f_{X}(x) d x \\ &=\int_{-\infty}^{\infty}|x| \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x \\ =& \int_{-\infty}^{0}-x \cdot \frac{1}{\sqrt{2 \pi}} e^{-x^{2}} d x+\int_{0}^{\infty} x \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x\\ &=2 \int_{0}^{\infty} x \cdot \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^{2}}{2}} d x=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} x \cdot e^{\frac{-x^{2}}{2}} d x \\ &\text{Let }\quad -\frac{x^{2}}{2}=t \\ &\therefore \quad-\frac{2 x d x}{2}=d t \\ &\text{Hence }\frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty}-e^{-t} d t=0.8 \\ &\therefore \quad E[|x|]=0.8 \\ \end{aligned}
Question 4 |
If a and b are constants, the most general solution of the differential equation \frac{d^{2}x}{dt^{2}}+2\frac{dx}{dt}+x=0 is
ae^{-t} | |
ae^{-t}+bte^{-t} | |
ae^{t}+bte^{-t} | |
ae^{-2t} |
Question 4 Explanation:
The differential equation is given as
\begin{aligned} \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+x&=0 \\ y&=C \cdot F+P \cdot I\\ \end{aligned}
since, Q=0, i.e. RHS term is zero, so there will
be no particular integral.
\begin{aligned} \therefore \quad y&=C \cdot F\\ \text{Let, }\quad \frac{\partial}{\partial x}&=D \\ \text{So. }\quad \left(D^{2}+2 D+1\right) x&=0 \\ \therefore \quad(D+1)^{2}&=0 \\ \therefore \quad y&=a e^{-t}+b t e^{t} \end{aligned}
\begin{aligned} \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+x&=0 \\ y&=C \cdot F+P \cdot I\\ \end{aligned}
since, Q=0, i.e. RHS term is zero, so there will
be no particular integral.
\begin{aligned} \therefore \quad y&=C \cdot F\\ \text{Let, }\quad \frac{\partial}{\partial x}&=D \\ \text{So. }\quad \left(D^{2}+2 D+1\right) x&=0 \\ \therefore \quad(D+1)^{2}&=0 \\ \therefore \quad y&=a e^{-t}+b t e^{t} \end{aligned}
Question 5 |
The directional derivative of f(x,y)=\frac{xy}{\sqrt{2}}(x+y) at (1,1) in the direction of the unit vector at an angle of \frac{\pi }{4} with y-axis, is given by____.
1 | |
2 | |
3 | |
4 |
Question 5 Explanation:
\begin{aligned} f(x, y) &=\frac{x y}{\sqrt{2}}(x+y)=\frac{x^{2} y+x y^{2}}{\sqrt{2}} \\ \nabla f &=\frac{\partial f}{\partial x} \hat{i}+\frac{\partial f}{\partial y} \hat{j} \\ &=\frac{2 x y+y^{2}}{\sqrt{2}}\hat{i}+\frac{(x^{2}+2xy)}{\sqrt{2}}\hat{j} \end{aligned}
The direction is \hat{n}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}
Directional derivative in direction of \hat{n} is
\begin{array}{l} =\left(\frac{2 x y+y^{2}}{2}+\frac{x^{2}+2 x y}{2}\right)_{(1,1)} \\ =\frac{3}{2}+\frac{3}{2}=3 \end{array}
The direction is \hat{n}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}
Directional derivative in direction of \hat{n} is
\begin{array}{l} =\left(\frac{2 x y+y^{2}}{2}+\frac{x^{2}+2 x y}{2}\right)_{(1,1)} \\ =\frac{3}{2}+\frac{3}{2}=3 \end{array}
There are 5 questions to complete.