Let \(g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 < x < 2, m\) and \(n\) are integers, \(m \neq 0, n>0\) and let \(p\) be the left hand derivative of \(|x-1|\) at \(x=1\). If \(\lim _{x \rightarrow 1^{+}} g(x)=p\), then

Given, \(g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 < x < 2, m \neq 0, n\) are integers and \(|x-1|=\left\{\begin{array}{l}x-1 ; x \geq 1 \\ 1-x ; x < 1\end{array}\right.\)
The left hand derivative of \(|x-1|\) at \(x=1\) is \(p=-1\).
Also,
\[
\begin{aligned}
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{(1+h-1)^n}{\log \cos ^m(1+h-1)}=-1 \Rightarrow \lim _{h \rightarrow 0} \frac{h^n}{\log \cos ^m h}=-1 \\
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{h^n}{m \log \cos h}=-1
\end{aligned}
\]
[Using L' Hospital rule]
\[
\begin{array}{ll}
\Rightarrow & \quad \lim _{h \rightarrow 0} \frac{n \cdot h^{n-1}}{m \frac{1}{\cos h}(-\sin h)}=-1 \\
\Rightarrow \quad & \quad \lim _{h \rightarrow 0}\left(-\frac{n}{m}\right) \cdot \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=-1 \\
\Rightarrow \quad & \quad\left(\frac{n}{m}\right) \lim _{h \rightarrow 0} \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=1 \\
\Rightarrow & n=2 \text { and } \frac{n}{m}=1 \\
\therefore & \quad m=n=2
\end{array}
\]