Let f : R→R be a continuous function. Then limx→π/4(π4∫2sec^2xf(x)dx)/(x^2-π^216) is equal to:

Correct answer: (C) 2f(2). limx→(π)/4(π4∫2sec^2xfxdx)/(x^2-π^216) Using L-Hospital's rule, we get =limx→(π)/4(π4fsec^2x·2·secx·secx·tanx-0)/2x =(π4·2(√2)^2·(1)·f(2))/(2·π4) =2f2

Q35 / 90 · JEE Main 2021 (01 Sep Shift 2) · Mathematics

Let $f : R \to R$ be a continuous function. Then $\lim_{x \to π / 4} \frac{\frac{π}{4} \int_{2}^{\sec^{2} x} f (x) d x}{x^{2} - \frac{π^{2}}{16}}$ is equal to:

A
$f (2)$
B
$2 f (\sqrt{2})$
Correct answer
C
$2 f (2)$
D
$4 f (2)$

Solution

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

Using L-Hospital's rule, we get

$= \lim_{x \to \frac{π}{4}} \frac{\frac{π}{4} f (\sec^{2} x) \cdot 2 \cdot \sec x \cdot \sec x \cdot \tan x - 0}{2 x}$

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

$= 2 f (2)$

Hence, the correct answer is option: 'C'.

Let f : R→R be a continuous function. Then limx→π/4π4∫2sec2xf(x)dxx2-π216 is equal to:

Solution

Relevant Questions

Let $f : R \to R$ be a continuous function. Then $\lim_{x \to π / 4} \frac{\frac{π}{4} \int_{2}^{\sec^{2} x} f (x) d x}{x^{2} - \frac{π^{2}}{16}}$ is equal to: