Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given,

$f\left(x\right)=\left\{\begin{array}{ll}\frac{\sin \left(x-\left[x\right]\right)}{x-\left[x\right]},& x\in \left(-2,-1\right)\\ \max \left(2x,3\left[\left|x\right|\right]\right),& \left|x\right|<1\\ 1,& \mathrm{otherwise}\end{array}\right.$

On simplifying we get,

$f\left(x\right)=\left\{\begin{array}{cl}\frac{\sin \left(x+2\right)}{x+2},& x\in \left(-2,-1\right)\\ \max \left\{2x,0\right\}& , x\in \left(-1,1\right)\\ 1& \mathrm{otherwise}\end{array}\right.$

$f\left(-2^{+}\right)=\underset{h\to 0}{\lim }f\left(-2+h\right)=\underset{h\to 0}{\lim }\frac{\sin \left(-2+h+2\right)}{-2+h+2}=\underset{h\to 0}{\lim }\frac{\sinh }{h}=1$

$f$ is continuous at $x=-2$

$f\left(-1^{-}\right)=\underset{h\to 0}{\lim }\frac{\sin \left(-1-h+2\right)}{\left(-1-h+2\right)}=\sin 1$

$f\left(-1\right)=f\left(-1^{*}\right)=0$

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

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

$\left.\begin{array}{r}\Rightarrow \mathrm{f} \mathrm{is} \mathrm{discontinuous} \mathrm{at} \mathrm{x}=-1 \& 1\\ \& f \mathrm{is} \mathrm{not} \mathrm{diff}. \mathrm{at} x=-1,0 \& 1\end{array}\right\}\Rightarrow \begin{array}{c}m=2\\ n=3\end{array}$