Explanation

(d)
$f^{'}\left( 1^{-} \right) = \lim_{h \rightarrow 0}\ \ \frac{f(1 - h) - f(1)}{- h}$
$= \lim_{h \rightarrow 0}\ \ \frac{(1 - h - 1).\sin\left( \frac{1}{1 - h - 1} \right) - 0}{- h}$
$= - \lim_{h \rightarrow 0}\ \sin\frac{1}{h}$
And $f^{'}\left( 1^{+} \right) = \lim_{h \rightarrow 0}\ \ \frac{f(1 + h) - f(1)}{h}$
$= \lim_{h \rightarrow 0}\ \ \frac{(1 + h - 1)\sin{\left( \frac{1}{1 + h - 1} \right) - 0}}{h}$
$= \lim_{h \rightarrow 0}\ \sin\frac{1}{h}$
$\therefore\ \ \ f^{'}\left( 1^{-} \right) \neq f^{'}\left( 1^{+} \right)$
$f\$is not differentiable at$\ x = 1$
Again, now
$f^{'}\left( 0^{+} \right) = \lim_{h \rightarrow 0}\ \frac{(0 + h - 1)\sin{\left( \frac{1}{0 + h - 1} \right) - \sin 1}}{h}$
$= \lim_{h \rightarrow 0}\ \ \frac{\left\lbrack - \left\{ (h - 1)\cos{\left( \frac{1}{h - 1} \right) \times \left( \frac{1}{(h - 1)^{2}} \right)} \right\} + \sin\left( \frac{1}{h - 1} \right) \right\rbrack}{1}$
[using L ‘Hospital’s rule]
$= \cos{1 - \sin 1}$
And $f^{'}\left( 0^{-} \right) = \lim_{h \rightarrow 0}\ \frac{(0 - h - 1)\sin{\left( \frac{1}{0 - h - 1} \right) - \sin 1}}{- h}$
$= \lim_{h \rightarrow 0}\ \ \frac{( - h - 1)\cos{\left( \frac{1}{- h - 1} \right)\left( \frac{1}{( - h - 1)^{2}} \right)} - \sin\left( \frac{1}{- h - 1} \right)}{- 1}$
[using L ‘Hospital’s rule]
$= \cos{1 - \sin 1}$
$\Rightarrow \ \ \ f^{'}\left( 0^{-} \right) = f^{'}\left( 0^{+} \right)$
$\therefore\ \ f$ is differentiable at $x = 0$