Let f(x)={(x−1)sin(1x−1),x≠10,x=1, then

f'(1+)=limh→0f(1+h)−f(1)h=limh→0hsin(1h)h=limh→0sin(1h)= non-existent∴f is not differentiable at x=1.Forx=0, since (x−1) and sin(1x−1) both are differentiable at x=0∴(x−1)sin(1x−1) is differentiable at x=0

Let f(x)={(x−1)sin(1x−1),x≠10,x=1, then

f'(1+)=limh→0f(1+h)−f(1)h=limh→0hsin(1h)h=limh→0sin(1h)= non-existent∴f is not differentiable at x=1.Forx=0, since (x−1) and sin(1x−1) both are differentiable at x=0∴(x−1)sin(1x−1) is differentiable at x=0

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Let $f (x) = {\begin{matrix} (x - 1) \sin (\frac{1}{x - 1}), & x \neq 1 \\ 0, & x = 1 \end{matrix},$ then

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$f$ is neither differentiable at $x = 0$ nor at $x = 1$

$f$ is differentiable at $x = 0$ and at $x = 1$

$f$ is differentiable at $x = 0$ but non-differentiable at $x = 1$

$f$ is differentiable at $x = 1$ but non-differentiable at $x = 0$

answer is C.

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Detailed Solution

$f' (1^{+}) = \lim_{h \to 0} \frac{f (1 + h) - f (1)}{h} = \lim_{h \to 0} \frac{h \sin (\frac{1}{h})}{h} = \lim_{h \to 0} \sin (\frac{1}{h})$
= non-existent
$∴ f$ is not differentiable at $x = 1.$
For $x = 0,$ since $(x - 1)$ and $\sin (\frac{1}{x - 1})$ both are differentiable at $x = 0$
$∴ (x - 1) \sin (\frac{1}{x - 1})$ is differentiable at $x = 0$