IIT-JEE 2007 · Paper 1 · Mathematics Q49

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MathematicsSingle correct+3 / -1Question 49 of 66

$\lim_{x\to\frac\pi4}\frac{\displaystyle\int_2^{\sec^2x}f(t)\,dt}{x^2-\frac{\pi^2}{16}}$
equals

\dfrac8\pi f(2)

\dfrac2\pi f(2)

\dfrac2\pi f\left(\frac12\right)

4f(2)

Answer and worked solution

Official answer: A

Form 0/0 ( $\sec^2\frac\pi4=2$ ). By L'Hôpital's rule and the Leibniz rule,
$\lim_{x\to\pi/4}\frac{f(\sec^2x)\cdot2\sec^2x\tan x}{2x}=\frac{f(2)\cdot2\cdot2\cdot1}{\pi/2}=\frac8\pi f(2).$

(A)

Topics:L'Hospital's rule Leibniz rule

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IIT-JEE 2007 · Paper 1, Mathematics question 49