Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f^{\prime }\left(0^{+}\right)={\lim }_{h\to 0}\frac{f(0+h)-f(0)}{h}$ $\begin{array}{rl}=\underset{h\to 0}{\lim }\frac{h^2\left|\cos \frac{\pi }{h}\right|}{h}& =\underset{h\to 0}{\lim }h\left|\cos \frac{\pi }{h}\right|\\ & =0\times \text{ some finite value }=0\end{array}$ $\begin{array}{c}=0\times \text{ some finite value }=0\\ \text{ and }{f}^{\prime }\left(0^{-}\right)={\lim }_{h\to 0}\frac{f(0-h)-f(0)}{-h}={\lim }_{h\to 0}\frac{h^2|\cos \frac{\pi }{-h}|}{-h}\\ ={\lim }_{h\to 0}-h\left|\cos \frac{\pi }{h}\right|=0\times \text{ some finite value }=0\end{array}$ $\begin{array}{l}\because f^{\prime }\left(0^{+}\right)=f^{\prime }\left(0^{-}\right)\therefore f\text{ is differentiable at }x=0\\ \text{ Now }{f}^{\prime }\left(2^{+}\right)={\lim }_{h\to 0}\frac{f(2+h)-f(2)}{h}\\ ={\lim }_{h\to 0}\frac{(2+h{)}^2|\cos \frac{\pi }{2+h}|-4|\cos \frac{\pi }{2}|}{h}\end{array}$ $={\lim }_{h\to 0}\frac{(2+h{)}^2\left(\cos \frac{\pi }{2+h}\right)}{h}$ $={\lim }_{h\to 0}\frac{(2+h{)}^2}{h}\sin \left(\frac{\pi }{2}-\frac{\pi }{2+h}\right)$ $={\lim }_{h\to 0}\frac{(2+h{)}^2}{h}\sin \left(\frac{\pi h}{2(2+h)}\right)$ $={\lim }_{h\to 0}\frac{(2+h{)}^2}{h}\times \frac{\sin \left(\frac{\pi h}{2(2+h)}\right)}{\left(\frac{\pi h}{2(2+h)}\right)}\times \frac{\pi h}{2(2+h)}=\pi$ and $f^{\prime }\left(2^{-}\right)={\lim }_{h\to 0}\frac{f(2-h)-f(2)}{-h}$ $={\lim }_{h\to 0}\frac{(2-h{)}^2|\cos \left(\frac{\pi }{2-h}\right)|-0}{-h}$ $={\lim }_{h\to 0}\frac{-(2-h{)}^2\cos \left(\frac{\pi }{2-h}\right)}{-h}$ $={\lim }_{h\to 0}\frac{(2-h{)}^2\sin \left(\frac{\pi }{2}-\frac{\pi }{2-h}\right)}{h}$ $={\lim }_{h\to 0}\frac{(2-h{)}^2}{h}\times \frac{\sin \left(\frac{-\pi h}{2(2-h)}\right)}{\left(\frac{-\pi h}{2(2-h)}\right)}\times \left(\frac{-\pi h}{2(2-h)}\right)=-\pi$ $\because f^{\prime }\left(2^{+}\right)\ne f^{\prime }\left(2^{-}\right),\therefore f$ is not differentiable at $x=2.$