JEE Advanced 2008 Paper 1Question 1
Let $g(x)=\dfrac{(x-1)^n}{\log\cos^m(x-1)}$; $0<x<2$, $m$ and $n$ are integers, $m\neq0$, $n>0$, and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\lim_{x\to1^+}g(x)=p$, then
Limits & Continuity
Solution
Concept. Find $p$, then the behaviour of $g$ as $h=x-1\to0^+$. Formulas used. (1) $|x-1|=1-x$ for $x<1$, so its left derivative at $1$ is $-1$(2) $\lim_{h\to0}\frac{\ln\cos h}{h^2}=-\frac12$ (since $\ln(1+u)\sim u$, $\cos h-1\sim-\frac{h^2}{2}$)Step 1 - $p$ (1). $p=-1$. Step 2 - Rewrite. $\ln\cos^mh=m\ln\cos h$, so $$g=\frac{h^n}{m\ln\cos h}=\frac{h^{n-2}}{m}\cdot\frac{h^2}{\ln\cos h}$$ Step 3 - Use (2). $\frac{h^2}{\ln\cos h}\to-2$, so $g\sim-\frac{2}{m}h^{n-2}$. Step 4 - Match $-1$. $n=1$: unbounded; $n>2$: $\to0$. So $n=2$ and $-\frac2m=-1\Rightarrow m=2$. Answer: C