Let \(\int(x)=\left\{\begin{array}{r}x^{2}\left|\cos \frac{\pi}{x}\right|, \quad x \neq 0 \\ 0, \quad x=0\end{array}, x \in R\right.\) then \(\int\) is

\(f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\)

\(\begin{aligned}

=\lim _{h \rightarrow 0} \frac{h^{2}\left|\cos \frac{\pi}{h}\right|}{h} &=\lim _{h \rightarrow 0} h\left|\cos \frac{\pi}{h}\right| \\

&=0 \times \text { some finite value }=0

\end{aligned}\)

\(\begin{array}{c}

=0 \times \text { some finite value }=0 \\

\text { and } f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{h^{2}\left|\cos \frac{\pi}{-h}\right|}{-h} \\

=\lim _{h \rightarrow 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 \rightarrow 0} \frac{f(2+h)-f(2)}{h} \\

\quad=\lim _{h \rightarrow 0} \frac{(2+h)^{2}\left|\cos \frac{\pi}{2+h}\right|-4\left|\cos \frac{\pi}{2}\right|}{h}

\end{array}\)

\(=\lim _{h \rightarrow 0} \frac{(2+h)^{2}\left(\cos \frac{\pi}{2+h}\right)}{h}\)

\(=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left(\frac{\pi}{2}-\frac{\pi}{2+h}\right)\)

\(=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left(\frac{\pi h}{2(2+h)}\right)\)

\(=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \times \frac{\sin \left(\frac{\pi h}{2(2+h)}\right)}{\left(\frac{\pi h}{2(2+h)}\right)} \times \frac{\pi h}{2(2+h)}=\pi\)

and \(f^{\prime}\left(2^{-}\right)=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}\)

\(=\lim _{h \rightarrow 0} \frac{(2-h)^{2}\left|\cos \left(\frac{\pi}{2-h}\right)\right|-0}{-h}\)

\(=\lim _{h \rightarrow 0} \frac{-(2-h)^{2} \cos \left(\frac{\pi}{2-h}\right)}{-h}\)

\(=\lim _{h \rightarrow 0} \frac{(2-h)^{2} \sin \left(\frac{\pi}{2}-\frac{\pi}{2-h}\right)}{h}\)

\(=\lim _{h \rightarrow 0} \frac{(2-h)^{2}}{h} \times \frac{\sin \left(\frac{-\pi h}{2(2-h)}\right)}{\left(\frac{-\pi h}{2(2-h)}\right)} \times\left(\frac{-\pi h}{2(2-h)}\right)=-\pi\)

\(\because f^{\prime}\left(2^{+}\right) \neq f^{\prime}\left(2^{-}\right), \therefore f\) is not differentiable at \(x=2 .\)