Question

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

• A. $n=1,m=1$
• B. $n=1,m=-1$
• C. $n=2,m=2$
• D. $n>2,m=n$

Answer 1

Answer: (C)

$n=2,m=2$

${lim}_{x\to 1}g\left(x\right)={lim}_{x\to 1}\frac{{(x-1)}^n}{\log \left({\cos }^m(x-1)\right)}$

$={lim}_{x\to 1+h}\frac{{(1+h-1)}^n}{\log \left({\cos }^m(1+h-1)\right)}$

$={lim}_{x\to 1+h}\frac{h^n}{\log \left({\cos }^{m}h\right)}$

$={lim}_{h\to 0}\frac{n{h}^{n-1}}{\frac{1}{{\cos }^{m}h}\times m{\cos }^{m-1}h\times -\sin h}$ using L'Hospitals rule

$p={lim}_{x\to 1^{-}}\frac{|x-1|-(1-1)}{x-1-0}$

$p={lim}_{x\to 1^{-}}\frac{|x-1|-0}{x-1}$

$p={lim}_{x\to 1^{-}}\frac{|x-1|}{x-1}=-1$

Given:${lim}_{x\to 1^{+}}g\left(x\right)=p$

$\Rightarrow {lim}_{h\to 0}\frac{n{h}^{n-1}}{\frac{1}{{\cos }^{m}h}\times m{\cos }^{m-1}h\times -\sin h}=-1$

$\Rightarrow {lim}_{h\to 0}\frac{n{h}^{n-1}}{-m\sin h}=-1$ for $n-1=1$ then ${lim}_{h\to 0}\frac{\sin h}{h}=1$

$\Rightarrow {lim}_{h\to 0}-\frac{n}{m}\frac{h}{\sin h}=-1$

$\Rightarrow n=m$ and $n-1=1$ or $n=1+1=2$

$\therefore n=m=2$