Let $f: R \rightarrow R$ be a continuous function. The $\lim _{x \rightarrow \pi / 4} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(x) d x}{x^{2}-\frac{\pi^{2}}{16}}$ is equal to:

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Limits

Let

f : R \to R

be a continuous function. The

lim_{x \to π /4} \frac{\frac{π}{4} \int _{2}^{s e c^{2} x} f ( x ) d x}{x ^{2} - \frac{π ^{2}}{16}}

is equal to:

Answer: (A)

This is a

\frac{0}{0}

form, so apply L'Hôpital's rule.

Let

I (x) = \int_{2}^{s e c^{2} x} f (t) d t

. By Leibniz rule:

\frac{d I}{d x} = f (sec^{2} x) \cdot 2 sec^{2} x tan x

The numerator's derivative is:

\frac{π}{4} \cdot f (sec^{2} x) \cdot 2 sec^{2} x tan x

The denominator's derivative is

2 x

.

Applying L'Hôpital:

lim_{x \to π /4} \frac{\frac{π}{4} \cdot f ( s e c ^{2} x ) \cdot 2 s e c ^{2} x t a n x}{2 x}

x = π /4

sec^{2} (π /4) = 2

tan (π /4) = 1

= \frac{\frac{π}{4} \cdot f ( 2 ) \cdot 2 \cdot 2 \cdot 1}{2 \cdot \frac{π}{4}} = \frac{π f ( 2 )}{\frac{π}{2}} = 2 f (2)

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option