Question:
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a continuous function. Then $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(x) d x}{x^{2}-\frac{\pi^{2}}{16}}$ is equal to :
Correct Option: , 2
Solution:
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int_{2}^{\sec ^{2} x} f(x) d x}{x^{2}-\frac{\pi^{2}}{16}}$
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\pi}{4} \cdot \frac{\left[f\left(\sec ^{2} x\right) \cdot 2 \sec x \cdot \sec x \tan x\right]}{2 x}$
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\pi}{4} f\left(\sec ^{2} x\right) \cdot \sec ^{3} x \cdot \frac{\sin x}{x}$
$\frac{\pi}{4} \mathrm{f}(2) \cdot(\sqrt{2})^{3} \cdot \frac{1}{\sqrt{2}} \times \frac{4}{\pi}$
$\Rightarrow 2 \mathrm{f}(2)$