Question

Let$f\left(x\right)=min\{1,1+x\sin x\},0\le x\le 2\pi$. If m is the number of points, where f is not differentiable and n is the number of points, where f is not continuous, then the ordered pair $(m,n)$ is equal to

• A. $(1,0)$
• B. $(1,1)$
• C. $(2,1)$
• D. $(2,0)$

Answer 1

The correct option is A $(1,0)$

. $f\left(x\right)=min\{1,1+x\sin x\},0\le x\le 2\pi$
$f\left(x\right)=\{\begin{array}{cc}1& 0\le x<\pi \\ 1+x\sin x,& \pi \le x\le 2\pi \end{array}$
Now at $x=\pi ,\underset{x\to {\pi }^{-}}{lim}f\left(x\right)=1=\underset{x\to {\pi }^{+}}{lim}f\left(x\right)$
$\therefore f\left(x\right)$ is continuous in $[0,2\pi ]$ Now, at $x=\pi ,L.H.D=\underset{h\to 0}{lim}\frac{f(\pi -h)-f\left(\pi \right)}{-h}=0$
$R.H.D.=\underset{h\to 0}{lim}\frac{f(\pi +h)-f\left(\pi \right)}{h}=\frac{1-(\pi +h)\sin h-1}{h}$
$=-\pi$
$\therefore f\left(x\right)$ is not differentiable at $x=\pi$
$\therefore (m,n)=(1,0)$