Question

Let $f\left(x\right)=\{\begin{array}{c}{(x-1)}^2\cos \left(\frac{1}{x-1}\right)-\left|x\right|,x\ne 1\\ -1,x=1\end{array}$

The set of points where $f\left(x\right)$ is not differentiable is

• A. $1$
• B. $0,1$
• C. $0$
• D. none of these

Answer 1

The correct option is A $1$

$f\left(x\right)={(x-1)}^2\cos \left(\frac{1}{x-1}\right)-\left|x\right|$

Evaluating limit at $x=1$,

Clearly $-1<\cos \left(1/\right(x-1)<1$ for any value of $x$, and hence

$\underset{x\to 1}{lim}f\left(x\right)={(x-1)}^2\cos \left(\frac{1}{x-1}\right)-\left|x\right|=-1$

Hence limit exists

Now differentiating

$f^{\prime }\left(x\right)=\sin \left(\frac{1}{x-1}\right)-1$

$\underset{x\to 1}{lim}{f}^{\prime }\left(x\right)$ does not exist