Question

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

From graph, $p=-1$
$\Rightarrow \underset{s\to 1^{+}}{\lim }g(x)=-1$
$\Rightarrow \underset{h\to 0}{\lim }g(1+h)=-1$
$\Rightarrow \underset{h\to 0}{\lim }\left(\frac{h^n}{\log {\cos }^{m}h}\right)=-1$
$\Rightarrow \underset{h\to 0}{\lim }\frac{n\cdot h^{n-1}}{m\cdot (-\tanh )}=-\left(\frac{n}{m}\right)\underset{h\to 0}{\lim }\left(\frac{h^{n-1}}{\tanh }\right)=-1$,
which holds if $n=m=2.$