Question

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

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