Answer:
Option B
Explanation:
Concept Involved To check differentiability at a point we use RHD and LHD at a point and if RHD = LHD, then f(x) is differentiable at the point.
Description of Situation
As, R
$\left\{f'(x)\right\}=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$
and $L\left\{f'(x)\right\}=\lim_{h \rightarrow 0}\frac{f(x-h)-f(x)}{-h}$
Here, students generaliy gets confused in defining modulus
Sol. To check differentiable at x=0
$R\left\{f'(0)\right\}=\lim_{h \rightarrow 0}\frac{f(0+h)-f(0)}{h}$
$=\lim_{h \rightarrow 0}\frac{h^{2}|\cos \frac{\pi}{h}|-0}{h}$
$=\lim_{h \rightarrow 0}h.|\cos \frac{\pi}{h}|=0$
$L\left\{f'(0)\right\}=\lim_{h \rightarrow 0}\frac{f(0-h)-f(0)}{-h}$
$=\lim_{h \rightarrow 0}\frac{h^{2}|\cos \left(-\frac{\pi}{h}\right)|-0}{-h}}$
=0
so, f(x) is differentiable at x=0
To check differentiability at x=2
$R\left\{f'(2)\right\}=\lim_{h \rightarrow 0}\frac{f(2+h)-f(2)}{h}$
$=\lim_{h \rightarrow 0}\frac{(2+h)^{2}| \cos \left(\frac{\pi}{2+h}\right)|-0}{h}$
$=\lim_{h \rightarrow 0}\frac{(2+h)^{2} \sin \left(\frac{\pi}{2}-\frac{\pi}{2+h}\right)}{h}$
$=\lim_{h \rightarrow 0}\frac{(2+h)^{2} \sin \left(\frac{\pi h}{2(2+h)}\right)}{h.\frac{\pi}{2(2+h)}}$
.$\frac{\pi}{2(2+h)}$
$\Rightarrow$ $R(f'(2))=\pi$
$L(f'(2))=\lim_{h \rightarrow 0}\frac{f(2-h)-f(2)}{-h}$
=$\lim_{h \rightarrow 0}\frac{(2-h)^{2}.|\cos \frac{\pi}{2-h}|-2^{2}.|\cos \frac{\pi}{2}|}{-h}$
=$\lim_{h \rightarrow 0}\frac{(2-h)^{2}-\left(-\cos \frac{\pi}{2-h}\right)-0}{-h}$
=$\lim_{h \rightarrow 0}\frac{-(2-h)^{2}.\sin\left( \frac{\pi}{2}-\frac{\pi}{2-h}\right)}{h}$
$\lim_{h \rightarrow 0}\frac{(2-h)^{2}.\sin\left( -\frac{\pi h}{2(2-h)}\right)}{h \times \frac{-\pi}{2(2-h)}}$
$\times \frac{-\pi}{2(2-h)}$
$\Rightarrow$ $L.F'(2))=-\pi$
Thus, f(x) is differentiable at x=0 but not at x=2
