Mathematics - Continuity - Differentiability Question with Solution | TestHub

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