Question

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

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

Answer 1

The correct option is C $f$ is neither differentiable at $x=0$ nor at $x=1$

$f^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}$

$\Rightarrow f^{\prime }\left(1\right)=\underset{h\to 0}{lim}\frac{(1+h-1)\sin \left(\frac{1}{1+h-1}\right)-0}{h}=\underset{h\to 0}{lim}\frac{h}{h}\sin \left(\frac{1}{h}\right)$

$\Rightarrow f^{\prime }\left(1\right)=\underset{h\to 0}{lim}\sin \left(\frac{1}{h}\right)$

$\therefore f$ is not differentiable at $x=1$

Similarly, $f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{(h-1)\sin \left(\frac{1}{h-1}\right)-\sin 1}{h}$

$\Rightarrow f$ is also not differentiable at $x=0$