Question

Let $g\left(x\right)=\frac{{(x-1)}^n}{\log {\cos }^m(x-1)}$; 0< x< 2 and 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 ${lim}_{x\to 1}g\left(x\right)=p$ then

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

Answer 1

The correct option is C n=2, m=2

Clearly $p=-1$ and
$\underset{x\to 1}{lim}g\left(x\right)=\underset{h\to 0}{lim}g(1+h)=\underset{h\to 0}{lim}\frac{h^n}{m\log \cos h}$
$=-\frac{n}{m}\underset{h\to 0}{lim}\frac{h^{n-1}}{\tan h}$, ($\frac{0}{0}$ form)
$=-\frac{n}{m}\underset{h\to 0}{lim}\frac{h}{\tan h}\times h^{n-2}$
The last limit is nonzero only if $n=2$ and so $n/m=1$ $\Rightarrow m=2.$