NIMCET 2017 Previous Year Questions (PYQs) – Page 8 of 12

$f'(0)=\lim_{x\to0}\dfrac{f(x)-f(0)}{x}$ 

$=\lim_{x\to0}\dfrac{x^2\sin(1/x)}{x}$ 

$=\lim_{x\to0}x\sin(\frac{1}{x})=0$

For $x\ne0$,

$f'(x)=2x\sin\left(\dfrac{1}{x}\right)-\cos\left(\dfrac{1}{x}\right)$

Answer: $\boxed{f'(0)=0}$ ✅

Let $I=\displaystyle\int_{0}^{\pi}x^{3}\sin x\,dx$.

Using integration by parts: let $u=x^{3},\; dv=\sin x\,dx$ $\Rightarrow du=3x^{2}dx,\; v=-\cos x$

$I = [-x^{3}\cos x]_0^{\pi} + \displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx$

Now, let $J=\displaystyle\int_{0}^{\pi}x^{2}\cos x\,dx$

Again, by parts: $u=x^{2},\; dv=\cos x\,dx \Rightarrow du=2x\,dx,\; v=\sin x$

$J = [x^{2}\sin x]_0^{\pi} - \displaystyle\int_{0}^{\pi}2x\sin x\,dx$

Let $K=\displaystyle\int_{0}^{\pi}x\sin x\,dx$ By parts: $u=x,\; dv=\sin x\,dx \Rightarrow du=dx,\; v=-\cos x$

$K=[-x\cos x]_0^{\pi}+\displaystyle\int_{0}^{\pi}\cos x\,dx = \pi$

So $J=0-2K=-2\pi$

Then $\displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx = 3J = -6\pi$

$I = [-x^{3}\cos x]_0^{\pi} + 3J = (-\pi^{3}\cos\pi - 0) - 6\pi = \pi^{3} - 6\pi$

Answer: $\boxed{\pi^{3} - 6\pi}$ ✅

$4\sin^2x + 3\cos^2x = 3(\sin^2x + \cos^2x) + \sin^2x = 3 + \sin^2x \le 4$

$\sin\frac{x}{2} + \cos\frac{x}{2} \le \sqrt{2}$ with equality when $\dfrac{x}{2}=\dfrac{\pi}{4}+2n\pi$

When $\sin^2x=1$ and $\sin\frac{x}{2}+\cos\frac{x}{2}=\sqrt{2}$, both maxima occur simultaneously.

Therefore, maximum value $= 4 + \sqrt{2}$

Answer: $\boxed{4+\sqrt{2}}$ ✅

Question: Solve $(e^x+1)\,y\,dy=(y+1)\,e^x\,dx$.

Solution:

Separate: $\dfrac{y}{y+1}\,dy=\dfrac{e^x}{e^x+1}\,dx$

Integrate: $y-\ln(y+1)=\ln(e^x+1)+C$

Answer (implicit): $\boxed{\,y-\ln(1+y)=\ln(1+e^x)+C\,}$

Equivalently: $\dfrac{e^{y}}{y+1}=K(1+e^x)$.

Answer : $e^y=k(y+1)(1+e^x)$

Compute partial derivatives: 

$f_x=2x-2,\; f_y=4y+4$.

Set $f_x=f_y=0 \Rightarrow 2x-2=0,\; 4y+4=0 \Rightarrow (x,y)=(1,-1)$.

Hessian: $H=\begin{pmatrix}2 & 0\\ 0 & 4\end{pmatrix}$ (positive definite since $2>0$ and $\det=8>0$).

Therefore, $(1,-1)$ is a strict (global) minimum. Also, by completing squares: $f(x,y)=(x-1)^2+2(y+1)^2-5 \Rightarrow f_{\min}=-5$ at $(1,-1)$.

Answer: Critical point $\boxed{(1,-1)}$; nature: $\boxed{\text{minimum}}$; minimum value $\boxed{-5}$.