Q68 / 125 · BITSAT 2014 · Mathematics
Let $f(x)=\left\{\begin{array}{c}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \quad \text { if } x=1\end{array}\right.$
Then which one of the following is true?
- A$f$ is differentiable at $x=0$ and $x=1$
- Correct answerB$f$ is differentiable at $x=0$ but not at $x=1$
- C$f$ is differentiable at $x=1$ but not at $x=0$
- D$f$ is neither differentiable at $x=0$ nor at $x=1$
Solution
We have; $f(x)=\left\{\begin{array}{ll}(x-1) \sin \left(\frac{1}{x-1}\right) & \text { if } x \neq 1 \\ 0 & \text { if } x=1\end{array}\right.$
$R f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$
$=\lim _{h \rightarrow 0} \frac{h \sin \frac{1}{h}-0}{h}=\lim _{h \rightarrow 0} \sin \frac{1}{h}$
which does not exist.
$\therefore f$ is not differentiable at $x=1$ Also $\left.f^{\prime}(0)=\sin \frac{1}{(x-1)}-\frac{x-1}{(x-1)^{2}} \cos \left(\frac{1}{x-1}\right)\right]_{x=0}$
$=-\sin 1+\cos 1$
$\therefore f$ is differentiable at $x=0$
Hence, the correct answer is option: 'B'.