Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given,

$f\left(x\right)=\min \left\{1,1+x\sin x\right\}x\in \left[0,2\pi \right]$

$f\left(x\right)=\left\{\begin{array}{c}1, 0\le x\le \pi \\ 1+x\sin x,\pi \le x\le 2\pi \end{array}\right.$

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 $\left[0,2\pi \right]$

So, number of points of discontinuity is zero or $n=0$

Now, at $x=\pi$

L.H.D $=\underset{h\to 0}{\lim }\frac{f\left(\pi -h\right)-f\left(\pi \right)}{-h}=0$

R.H.D $=\underset{h\to 0}{\lim }\frac{f\left(\pi +h\right)-f\left(\pi \right)}{h}=\underset{h\to 0}{\lim }\frac{1-\left(\pi +h\right)\sin h-1}{h}$

$=-\pi$

$\therefore f\left(x\right)$ is not differentiable at $x=\pi$ so, $m=1$

$\therefore \left(m,n\right)=\left(1,0\right)$