an IIT-JEE(indian institue of technology joint entrance exam) 2007 question

**grgrsanjay** · August 31st 2010, 06:49 AM

$\displaystyle\lim_{x\to \pi/4} \left(\frac{\int_{2}^{sec^2 x} \left(f(t) dt) \right}{x^2 - \frac{\pi^2}{16}}\right)$ equals

(A) $\frac{8}{\pi} f(2)$

(B) $\frac{2}{\pi} f(2)$

(C) $\frac{8}{\pi} f(\frac{1}{2})$

(D) $4f(2)$

**Defunkt** · August 31st 2010, 07:04 AM

Originally Posted by grgrsanjay

$\lim_{x\to \pi/4} \left(\frac{\int_{2}^{secx^2} \left(f(t) dt) \right}{x^2 - \frac{\pi^2}{16}}\right)$ equals

(A) $\frac{8}{\pi} f(2)$

(B) $\frac{2}{\pi} f(2)$

(C) $\frac{8}{\pi} f(/frac{1}{2})$

(D) $4f(2)$

Do you mean $sec^2(x)$ or $sec(x^2)$ ?

Edit: If the first, I think using L'hopital's rule along with the chain rule and the fundamental theorem should work.

**grgrsanjay** · August 31st 2010, 07:10 AM

its the first
can you show me the methods i did it but i failed to get it
i edited all my mistakes

**sa-ri-ga-ma** · August 31st 2010, 07:59 AM

When you use L'hopital's rule, you get

$\frac{f(sec^2(x))}{2x}$

Substitute the limit x = π/4 and find the result.

**grgrsanjay** · August 31st 2010, 08:16 AM

so is the answer coming out to be (A)??

**sa-ri-ga-ma** · August 31st 2010, 08:17 AM

I think it is (B)

**grgrsanjay** · August 31st 2010, 08:20 AM

but the answer is (A) given in all my books(4 books in total)

**Plato** · August 31st 2010, 08:26 AM

Originally Posted by sa-ri-ga-ma

When you use L'hopital's rule, you get

$\frac{f(sec^2(x))}{2x}$

Substitute the limit x = π/4 and find the result.

This reply is mistaken.

It should be $\displaystyle \frac{f(\sec^2(x))(2\sec^2(x)\tan(x))}{2x}$

So the answer is a.

**sa-ri-ga-ma** · August 31st 2010, 08:27 AM

OK. I will try again.

**grgrsanjay** · August 31st 2010, 08:28 AM

yea,yea got it thanks sa-ri-ga-ma and plato

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