Question

Let $f(x) = \begin{cases} x^2|cos\frac{\pi}{x}|, & x \ne 0 \\ , & x \in \\ 0, & x = 0 \end{cases} $ then $f$ is

• A. differentiable both at $x=0$ and at $x=2$
• B. differentiable at $x=0$ but not differentiable at $x=2$
• C. not differentiable at $x=0$ but differentiable at $x=2$
• D. differentiable neither at $x=0$ nor at $x=2$

Answer 1

Answer: (B)

$f'(0)=\displaystyle\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$
$\displaystyle\lim _{h \rightarrow 0} \frac{h^{2}\left|\cos \frac{\pi}{h}\right|-0}{h}$
$=\displaystyle\lim _{h \rightarrow 0} h \cos \left(\frac{\pi}{h}\right)=0$
so, $f(x)$ is differentiable at $x=0$
$f'\left(2^{+}\right)=\displaystyle\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$
$\displaystyle\lim _{h \rightarrow 0} \frac{(2+h)^{2}\left|\cos \frac{\pi}{2+h}\right|-0}{h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{(2+h)^{2} \cos \left(\frac{\pi}{2+h}\right)}{h}$
$f'\left(2^{+}\right)=\displaystyle\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left(\frac{\pi}{2}-\frac{\pi}{2+h}\right)$
$=\displaystyle\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left[\frac{\pi \cdot h}{2(2+h)}\right]$
$=\displaystyle\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{\frac{\pi h}{2(2+h)}} \sin \frac{\pi h}{2(2+h)} \times \frac{\pi}{2(2+h)}=\pi$
Again, $f'\left(2^{-}\right)=\displaystyle\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{(2-h)^{2}\left|\cos \left(\frac{\pi}{2-h}\right)\right|}{-h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{-(2-h)^{2} \cos \left(\frac{\pi}{2-h}\right)}{-h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{(2-h)^{2} \sin \left[\frac{\pi}{2}-\frac{\pi}{2-h}\right]}{h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{(2-h)^{2}}{h} \cdot \sin \left[\frac{-\pi h}{2(2-h)}\right]$
$=\displaystyle\lim _{h \rightarrow 0} \frac{(2-h)^{2}}{\frac{\pi h}{2(2-h)}} \cdot \sin \frac{\pi h}{2(2-h)} \times \frac{\pi}{2(2-h)}=-\pi$