$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\mathop \smallint \nolimits_2^{{{\sec }^2}x} f\left( t \right)dt}}{{{x^2} - \frac{{{\pi ^2}}}{{16}}}}$ equals

$\frac{2}{\pi }f\left( 2 \right)$

$\frac{2}{\pi }f\left( {\frac{1}{2}} \right)$

JEE MATHEMATICS MCQ: LIMITS

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Application +4 | -1

\lim\limits_{x \to \frac{\pi }{4}} \frac{{\int \nolimits_2^{{{\sec }^2}x} f\left( t \right)dt}}{{{x^2} - \frac{{{\pi ^2}}}{{16}}}}

equals

View answer and explanation

Correct option: A

\frac{8}{\pi }f\left( 2 \right)

Option A (

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

) is correct because it results from applying L'Hopital's rule and evaluating the derivative of

\sec^2 x

x = \frac{\pi}{4}

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