Explanation

(c)
Given, $g(x) = \frac{(x - 1)^{n}}{\log{\cos^{m}{(x - 1)}}};\ \ 0 < x < 2,\ \ \ \ m \neq 0,\ \ \ n$ are integers and $|x - 1| = \left\{ \begin{array}{r}
x - 1;\ \ x \geq 1 \\
1 - x;\ \ \ x < 1
\end{array} \right.\$
The left hand derivative of $|x - 1|$ at $x = 1$ is $p = - 1$
Also, $\lim_{x \rightarrow 1^{+}}\ \ g(x) = p = - 1$
$\Rightarrow \ \ \ \ \lim_{h \rightarrow 0}\ \ \frac{(1 + h - 1)^{n}}{\log{\cos^{m}{(1 + h - 1)}}} = - 1$
$\Rightarrow \ \lim_{h \rightarrow 0}\ \frac{h^{n}}{m\log{\cos h}} = - 1$
$\Rightarrow \ \lim_{h \rightarrow 0}\frac{n.h^{n - 1}}{m\frac{1}{\cos h}( - \sin h)} = - 1$
[using L ‘Hospital’s rule]
$\Rightarrow \ \ \ \ \left( \frac{n}{m} \right)\lim_{h \rightarrow 0}\frac{h^{n - 2}}{\left( \frac{\tan h}{h} \right)} = 1$
$\Rightarrow \ \ n = 2$ and $\frac{n}{m} = 1$
$\Rightarrow \ \ m = n = 2$