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

• Aug 31st 2010, 07:49 AM
grgrsanjay
an IIT-JEE(indian institue of technology joint entrance exam) 2007 question
$\displaystyle\lim_{x\to \pi/4} \left(\frac{\int_{2}^{sec^2 x} \left(f(t) dt) \right}{x^2 - \frac{\pi^2}{16}}\right)$ equals

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

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

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

(D) $4f(2)$
• Aug 31st 2010, 08:04 AM
Defunkt
Quote:

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.
• Aug 31st 2010, 08:10 AM
grgrsanjay
its the first
can you show me the methods i did it but i failed to get it
i edited all my mistakes
• Aug 31st 2010, 08:59 AM
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.
• Aug 31st 2010, 09:16 AM
grgrsanjay
so is the answer coming out to be (A)??
• Aug 31st 2010, 09:17 AM
sa-ri-ga-ma
I think it is (B)
• Aug 31st 2010, 09:20 AM
grgrsanjay
but the answer is (A) given in all my books(4 books in total)
• Aug 31st 2010, 09:26 AM
Plato
Quote:

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.

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