IIT-JEE 2008 · Paper 1 · Mathematics Q5

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MathematicsSingle correct+3 / -1Question 5 of 69

Let
$g(x)=\frac{(x-1)^n}{\log\cos^m(x-1)};\quad0<x<2,\ m\text{ and }n\text{ are integers},\ m\ne0,\ n>0,$
and let $p$ be the left hand derivative of $|x-1|$ at $x=1$ . If $\displaystyle\lim_{x\to1^+}g(x)=p$ , then

n=1,\ m=1

n=1,\ m=-1

n=2,\ m=2

n>2,\ m=n

Answer and worked solution

Official answer: C

Value of p. For $x<1$ , $|x-1|=1-x$ , whose derivative is $-1$ : $p=-1$ .

Limit. With $h=x-1\to0^+$ , $\log\cos^mh=m\log\cos h\approx-\frac{mh^2}2$ , so
$g\approx\frac{h^n}{-mh^2/2}=-\frac2mh^{n-2}.$
A finite non-zero limit needs $n=2$ , and then $-\frac2m=-1$ gives $m=2$ .

(C)

Topics:Limits Differentiability

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IIT-JEE 2008 · Paper 1, Mathematics question 5