Question

Let $f\left(x\right)=\{\begin{array}{ccc}(x-1)\sin \left(\frac{1}{x-1}\right)& \text{if}& x\ne 1\\ 0,& \text{if}& x\ne 1\end{array}$.
Then which one of the following is true?

• A. $f$ is differentiable neither at $x=0$ nor at $x=1$
• B. $f$ is differentiable at $x=0$ and $x=1$
• C. $f$ is differentiable at $x=0$ but not at $x=1$
• D. $f$ is differentiable at $x=1$ but not at $x=0$

Answer 1

The correct option is C $f$ is differentiable at $x=0$ but not at $x=1$

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

$f^{\prime }\left(0\right)=\sin (-1)+(-1)\cos (-1)(-1)$

$=\cos 1-\sin 1$

As $x\to 1,f^{\prime }\left(1\right)\to \infty$ $(\because \frac{1}{x-1}\to \infty )$

$\therefore f\left(x\right)$ is differentiable at $x=0$, but not at $x=1$