Question

The number of points where $f\left(x\right)=\{\begin{array}{cc}{(x-1)}^2\cos \left(\frac{1}{x-1}\right)-|x|,& x\ne 1\\ -1,& x=1\end{array}$ is not differentiable is

Answer 1

$f\left(x\right)={(x-1)}^2\cos \left(\frac{1}{x-1}\right)-|x|$
Evaluating limit at $x=1,$
Clearly $-1<\cos \left(\frac{1}{(x-1)}\right)<1$ for any value of $x$ ,
$\underset{x\to 1}{lim}f\left(x\right)={(x-1)}^2\cos \left(\frac{1}{x-1}\right)-|x|=-1=f\left(1\right)$
$\Rightarrow f\left(x\right)$ is continuous function at $x=1$
Now,
$R.H.D.=f^{\prime }\left(1^{+}\right)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}$
$=\underset{h\to 0}{lim}\frac{{\left(h\right)}^2\cos \left(\frac{1}{h}\right)-|1+h|+1}{h}=\underset{h\to 0}{lim}(h\cos \frac{1}{h}-\frac{h}{h})=-1$
$L.H.D.=f^{\prime }\left(1^{-}\right)=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{{\left(h\right)}^2\cos \left(\frac{1}{h}\right)-|1-h|+1}{-h}=\underset{h\to 0}{lim}(-h\cos \frac{1}{h}-\frac{h}{h})=-1$

Clearly, $L.H.D.=R.H.D.$
Hence, $f\left(x\right)$ is differentiable at $x=1.$
We know that if $g\left(x\right),h\left(x\right)$ are differntiable and non-differentiable at $x=a$ respectively , then $g\left(x\right)\pm h\left(x\right)$ will also be non-differentiable at $x=a.$

Here $|x|$ is not differentiable at $x=0$
So, $f\left(x\right)$ is non differentiable at $x=0$.
Hence, $f\left(x\right)$ is non differentiable at exactly one point.