JEE Main 2022 — Mathematics Calculus

Given,

$$f(x)=\mathrm{min}{1,1+x\mathrm{sin}x}x\in [0,2\pi ]$$f(x)=min1,1+xsinxx∈[0,2π]

f(x)={\begin{matrix}1,0\leq x\leq \pi \\ 1+x\mathrm{sin}x,\pi \leq x\leq 2\pi \end{matrix}

Now, at $x=\pi ,\underset{x\rightarrow {\pi }^{-}}{\mathrm{lim}}f(x)=1=\underset{x\rightarrow {\pi }^{+}}{\mathrm{lim}}f(x)$x=π,x→π−lim​f(x)=1=x→π+lim​f(x)

$\therefore f(x)$∴f(x) is continuous in $[0,2\pi ]$[0,2π]

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

Now, at $x=\pi$x=π

L.H.D $=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(\pi -h)-f(\pi )}{-h}=0$=h→0lim​−hf(π−h)−f(π)​=0

R.H.D $=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(\pi +h)-f(\pi )}{h}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{1-(\pi +h)\mathrm{sin}h-1}{h}$=h→0lim​hf(π+h)−f(π)​=h→0lim​h1−(π+h)sinh−1​

$$=-\pi$$=−π

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

$$\therefore (m,n)=(1,0)$$∴(m,n)=(1,0)