Question

Let $f\left(x\right)=\{\begin{array}{c}{x}^2\left|\cos \frac{\pi }{x}\right|,x\ne 0\\ 0,x=0\end{array}$$,x\in R.$ Then $f$ is

Answer 1

For differentiability at $x=0$

$f^{\prime }(0+h)=\underset{h\to 0}{lim}\frac{h^2\left|\cos \frac{\pi }{h}\right|-0}{h-0}=0$
$f^{\prime }(0-h)=\underset{h\to 0}{lim}\frac{h^2\left|\cos \frac{\pi }{h}\right|-0}{-h}=0$

$\because f^{\prime }\left(0^{+}\right)=f^{\prime }\left(0^{-}\right)=0$ (finite)

So $f\left(x\right)$ is differentiable at $x=0$

Now,For differentiability at $x=2$

$f^{\prime }(2+h)=\underset{h\to 0}{lim}\frac{(2+h{)}^2\left|\cos \frac{\pi }{2+h}\right|-0}{h-0}$
$f^{\prime }(2+h)=\underset{h\to 0}{lim}\frac{(2+h{)}^2\cdot \sin (\frac{\pi }{2}-\frac{\pi }{2+h})}{h}$

$f^{\prime }(2+h)=\underset{h\to 0}{lim}\frac{(2+h{)}^2\cdot \sin \left(\frac{\pi h}{2(2+h)}\right)}{\left(\frac{\pi }{2(2+h)}\right)h}\times \frac{\pi }{2(2+h)}$
$f^{\prime }(2+h)=(2{)}^2\cdot \frac{\pi }{2\cdot 2}=\pi$

$f^{\prime }(2-h)=\underset{h\to 0}{lim}\frac{(2-h{)}^2\left|\cos \frac{\pi }{2-h}\right|-0}{-h}=$

$f^{\prime }(2-h)=\underset{h\to 0}{lim}\frac{(2-h{)}^2(-\cos \left(\frac{\pi }{2-h}\right))}{-h}$

$f^{\prime }(2-h)=\underset{h\to 0}{lim}\frac{(2-h{)}^2\cdot \sin (\frac{\pi }{2}-\frac{\pi }{2-h})}{h}=-\pi$


$\therefore f^{\prime }\left(2^{+}\right)\ne f^{\prime }\left(2^{-}\right)$ .

So $f\left(x\right)$ is not differentiable at $x=2.$