Question

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

differentiable at $x=0$ but not differentiable at $x=2$

Differentiability at $x=0$:
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

$=\underset{h\to 0}{lim}\frac{h^2\mid \cos \frac{\pi }{h}\mid -0}{h}$

$=\underset{h\to 0}{lim}h\cos \left(\frac{\pi }{h}\right)=0$

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

$=\underset{h\to 0}{lim}\frac{h^2\mid \cos \frac{\pi }{-h}\mid -0}{-h}$

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

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

Differentiability at $x=2$

$R.H.D.=f^{\prime }\left(2^{+}\right)=\underset{h\to 0}{lim}\frac{f(2+h)-f\left(2\right)}{h}$

$=\underset{h\to 0}{lim}\frac{(2+h{)}^2\mid \cos \frac{\pi }{2+h}\mid -0}{h}$

$=\underset{h\to 0}{lim}\frac{(2+h{)}^2}{h}\sin [\frac{\pi }{2}-\frac{\pi }{(2+h)}]$

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

$=\pi$

$L.H.D.=f^{\prime }\left(2^{-}\right)=\underset{h\to 0}{lim}\frac{f(2-h)-f\left(2\right)}{-h}$

$=\underset{h\to 0}{lim}\frac{(2-h{)}^2\mid \cos (\frac{\pi }{2-h})\mid }{-h}$

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

$=\underset{h\to 0}{lim}\frac{(2-h{)}^2}{h}\sin [\frac{\pi }{2}-\frac{\pi }{2-h}]$

$=\underset{h\to 0}{lim}\frac{(2-h{)}^2}{h}\cdot \sin \left[\frac{-h\pi }{2(2-h)}\right]$

$=\underset{h\to 0}{lim}\frac{(2-h)}{h}\times \frac{-\pi h}{2(2-h)}\cdot \left[\frac{\sin \left[\frac{-\pi h}{2(2-h)}\right]}{\frac{-\pi h}{2(2-h)}}\right]$

$=-\pi$

Thus $f^{\prime }(2^{+}{)}^{=}\ne f^{\prime }\left(2^{-}\right)$.
Hence, the function is not differentiable at $x=2$.