Question 1 |
If y(x) satisfies the differential equation
(\sin x)\frac{dy}{dx}+y \cos x =1
subject to the condition y(\pi /2)=\pi /2, then y(\pi /6) is
(\sin x)\frac{dy}{dx}+y \cos x =1
subject to the condition y(\pi /2)=\pi /2, then y(\pi /6) is
0 | |
\frac{\pi}{6} | |
\frac{\pi}{3} | |
\frac{\pi}{2} |
Question 1 Explanation:
\begin{aligned} \frac{d y}{d x}+y \cot x&=\text{cosec} x\\ 1.F. \qquad&=e^{\int \cot x d x}=e^{\log \sin x}=\sin x\\ \Rightarrow \quad y(\sin x)&=\int \text{cosec} x \sin x d x+c\\ \Rightarrow \qquad y \sin x&=x+c\\ \Rightarrow \qquad \frac{\pi}{2} \sin \frac{\pi}{2} & =\frac{\pi}{2}+c \\ \Rightarrow \qquad \frac{\pi}{2} & =\frac{\pi}{2}+c \quad \Rightarrow c=0 \\ \Rightarrow \qquad y \sin x & =x\\ \Rightarrow \qquad y \sin \frac{\pi}{6}&=\frac{\pi}{6}\\ \Rightarrow \qquad y\left(\frac{1}{2}\right) &=\frac{\pi}{6} \\ \Rightarrow y &=\frac{\pi}{3} \end{aligned}
Question 2 |
The value of \lim_{x \to 0}\left ( \frac{1- \cos x}{x^2} \right ) is
\frac{1}{4} | |
\frac{1}{3} | |
\frac{1}{2} | |
1 |
Question 2 Explanation:
\begin{aligned} \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)&=? \;\;\;\;\;\;\left(\frac{0}{0} \text { form }\right) \\ \text { Applying } L \cdot H \text { rule } & =\lim _{x \rightarrow 0} \frac{\sin x}{2 x}\left(\frac{0}{0}\right)=\lim _{x \rightarrow 0} \frac{\cos x}{2}=\frac{1}{2} \end{aligned}
Question 3 |
The Dirac-delta function (\delta (t-t_0)) \text{ for }t,t_0 \in \mathbb{R}, has the following property
\int_{a}^{b}\varphi (t)\delta (t-t_0)dt=\left\{\begin{matrix} \varphi (t_0) & a \lt t_0 \lt b\\ 0 &\text{otherwise} \end{matrix}\right.
The Laplace transform of the Dirac-delta function \delta (t-a) for a \gt 0;
\mathcal{L} (\delta (t-a))=F(s) is
\int_{a}^{b}\varphi (t)\delta (t-t_0)dt=\left\{\begin{matrix} \varphi (t_0) & a \lt t_0 \lt b\\ 0 &\text{otherwise} \end{matrix}\right.
The Laplace transform of the Dirac-delta function \delta (t-a) for a \gt 0;
\mathcal{L} (\delta (t-a))=F(s) is
0 | |
\infty | |
e^{sa} | |
e^{-sa} |
Question 3 Explanation:
\begin{aligned} \because \qquad \int_{0}^{-} f(t) \delta(t-a) d t&=f(a) \\ \therefore \qquad L\{\delta(t-a)\}&=\int_{0}^{-} e^{-s t} \delta(t-a) d t=e^{-a s} \end{aligned}
Question 4 |
The ordinary differential equation \frac{dy}{dt}=-\pi y subject to an initial condition y(0)=1 is solved numerically using the following scheme:
\frac{y(t_{n+1})-y(t_n)}{h}=-\pi y(t_n)
where h is the time step, t_n=nh, and n=0,1,2,.... This numerical scheme is stable for all values of h in the interval.
\frac{y(t_{n+1})-y(t_n)}{h}=-\pi y(t_n)
where h is the time step, t_n=nh, and n=0,1,2,.... This numerical scheme is stable for all values of h in the interval.
0 \lt h \lt \frac{2}{\pi} | |
0 \lt h \lt 1 | |
0 \lt h \lt \frac{\pi}{2} | |
for all h \gt 0 |
Question 4 Explanation:
\begin{aligned} \frac{y\left(t_{n+1}\right)-y\left(t_{n}\right)}{h} &=-\pi y\left(t_{n}\right) \\ y_{n+1} &=-\pi / y_{n}+y_{n}=(-\pi h+1) y_{n} \end{aligned}
It is recursion relation between y_{n+1} and y_{n}
So solution will be stable if
\begin{aligned} |-\pi h+1| & \lt 1 \\ -1 \lt -\pi h+1 & \lt 1 \\ -2 \lt -\pi h & \lt 0 \\ 0 & \lt \pi h \lt 2 \\ 0 & \lt h \lt \frac{2}{\pi} \end{aligned}
Therefore option (A) is correct.
It is recursion relation between y_{n+1} and y_{n}
So solution will be stable if
\begin{aligned} |-\pi h+1| & \lt 1 \\ -1 \lt -\pi h+1 & \lt 1 \\ -2 \lt -\pi h & \lt 0 \\ 0 & \lt \pi h \lt 2 \\ 0 & \lt h \lt \frac{2}{\pi} \end{aligned}
Therefore option (A) is correct.
Question 5 |
Consider a binomial random variable X. If X_1,X_2,..., X_n are independent and identically distributed samples from the distribution of X with sum Y=\sum_{i=1}^{n}X_i, then the distribution of Y as n\rightarrow \infty can be approximated as
Exponential | |
Bernoulli | |
Binomial | |
Normal |
There are 5 questions to complete.
Great job. Can you add the marks for each question so we can differentiate between 2- and 1-mark questions?