Let $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0<x<2$, m and n are integers, m ≠ 0, n >0, and let p the left h and derivative of |x - 1| at x = 1. If $\lim\limits_{x \rightarrow 1^{+}} g(x)=p$, then
Answer & explanation
Correct answer: option 3
We have,
$f(x)=|x-1|= \begin{cases}x-1, & x \geq 1 \\ 1-x, & x<1\end{cases}$
∴ p = Left hand derivative of f(x) at x = 1
$\Rightarrow p=\lim\limits_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim\limits_{x \rightarrow 1^{-}} \frac{1-x-0}{x-1}=-1$
Now,
$\lim\limits_{x \rightarrow 1^{+}} g(x)=p$
$\Rightarrow \lim\limits_{h \rightarrow 0} g(1+h)=-1$
$\Rightarrow \lim\limits_{h \rightarrow 0} \frac{h^n}{m \log \cos h}=-1$
$\Rightarrow \frac{1}{m} \lim\limits_{h \rightarrow 0} \frac{n h^{n-1}}{-\tan h}=-1$ [ Using L'Hospital's rule on LHS]
$\Rightarrow \frac{n}{m} \lim\limits_{h \rightarrow 0} \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=1$
⇒ n = 2 and in that case $\frac{n}{m}=1$
⇒ m = n = 2