Question

Let $f:[1,1]\to R$ be a function defined by
$f\left(x\right)=\{\begin{array}{cc}{x}^2\left|\cos \frac{\pi }{x}\right|& \text{for}x\ne 0,\\ 0& \text{for}x=0.\end{array}$
The set of points where $f$ is $not$ differentiable is

• A. $\{x\in [-1,1]:x\ne 0\}$
• B. $\{x\in [-1,1]:x=0orx=\frac{2}{2n+1},n\in Z\}$
• C. $\{x\in [-1,1]:x=\frac{2}{2n+1},n\in Z\}$
• D. $[-1,1]$

Answer 1

The correct option is C $\{x\in [-1,1]:x=\frac{2}{2n+1},n\in Z\}$

Given :
$f\left(x\right)=\{\begin{array}{cc}{x}^2\left|\cos \frac{\pi }{x}\right|& \text{for}x\ne 0,\\ 0& \text{for}x=0.\end{array}$
Checking at $x=0\text{and}\frac{2}{2n+1}$
For $x=0$
RHD, $f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h-0}=\underset{h\to 0}{lim}\frac{h^2\cos \frac{\pi }{h}-0}{h-0}=0$
LHD, $f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{0-h}=\underset{h\to 0}{lim}\frac{h^2\cos \frac{\pi }{h}-0}{-h}=0$
So the function is differentiable at $x=0$,
Modulus will change its value at $x=\frac{2}{2n+1}$
Checking for $n=0\Rightarrow x=2$
RHD, $f^{\prime }\left(2^{+}\right)=\underset{h\to 0}{lim}\frac{(2+h{)}^2(\cos \frac{\pi }{2+h})-2^2\cos \frac{\pi }{2}}{h-0}=\underset{h\to 0}{lim}\frac{(2+h{)}^2(\cos \frac{\pi }{2+h})}{h}=\underset{h\to 0}{lim}4\frac{\cos \frac{\pi }{2+h}}{h}$
Using L Hospital rule,
$=\underset{h\to 0}{lim}4\pi \frac{\sin \frac{\pi }{2+h}(\frac{1}{2+h}{)}^2}{1}=\pi$

LHD, $f^{\prime }\left(2^{-}\right)=\underset{h\to 0}{lim}\frac{(2-h{)}^2(-\cos \frac{\pi }{2-h})+2^2\cos \frac{\pi }{2}}{0-h}=\underset{h\to 0}{lim}\frac{(2-h{)}^2(-\cos \frac{\pi }{2-h})}{-h}=\underset{h\to 0}{lim}4\frac{\cos \frac{\pi }{2-h}}{h}$
Using L Hospital rule,
$=\underset{h\to 0}{lim}-4\pi \frac{\sin \frac{\pi }{2-h}(\frac{1}{2-h}{)}^2}{1}=-\pi$

So the function will be non differentiable at $x=\frac{2}{2n+1}$ and differentiable at all other points.