JEE Main 2022 — Mathematics Calculus

Given,

f(x)={\begin{matrix}\frac{\mathrm{sin}(x-[x])}{x-[x]}, & x\in (-2,-1) \\ \mathrm{max}(2x,3[|x|]), & |x|<1 \\ 1, & \mathrm{otherwise}\end{matrix}

On simplifying we get,

f(x)={\begin{matrix}\frac{\mathrm{sin}(x+2)}{x+2}, & x\in (-2,-1) \\ \mathrm{max}{2x,0} & ,x\in (-1,1) \\ 1 & \mathrm{otherwise}\end{matrix}

$$f(-{2}^{+})=\underset{h\rightarrow 0}{\mathrm{lim}}f(-2+h)=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}(-2+h+2)}{-2+h+2}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sinh}}{h}=1$$f(−2+)=h→0lim​f(−2+h)=h→0lim​−2+h+2sin(−2+h+2)​=h→0lim​hsinh​=1

$f$f is continuous at $x=-2$x=−2

$$f(-{1}^{-})=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{sin}(-1-h+2)}{(-1-h+2)}=\mathrm{sin}1$$f(−1−)=h→0lim​(−1−h+2)sin(−1−h+2)​=sin1

$$f(-1)=f(-{1}^{*})=0$$f(−1)=f(−1∗)=0

f({1}^{+})=1&f({1}^{-})=0\Rightarrow f is not continuous at $x=1$x=1

$f$f is continuous but not diff. at $x=0$x=0

\begin{matrix}\Rightarrow f\mathrm{is}\mathrm{discontinuous}\mathrm{at}x=-1&1 \\ &f\mathrm{is}\mathrm{not}\mathrm{diff}.\mathrm{at}x=-1,0&1\end{matrix}}\Rightarrow \begin{matrix}m=2 \\ n=3\end{matrix}