Question

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

• A. $f(2)$
• B. $2 f(2)$
• C. $2 f(\sqrt{2})$
• D. $4 f(2)$

Answer 1

Answer: (B)

$\displaystyle\lim _{x \rightarrow \frac{\pi}{4}} \frac{\frac{\pi}{4} \int\limits_{2}^{\sec ^{2} x} f(x) d x}{x^{2}-\frac{\pi^{2}}{16}}$
$\displaystyle\lim _{x \rightarrow \frac{\pi}{4}} \frac{\pi}{4} \cdot \frac{\left. f \left(\sec ^{2} x \right) \cdot 2 \sec x \cdot \sec x \tan x \right]}{2 x }$
$\displaystyle\lim _{ x \rightarrow \frac{\pi}{4}} \frac{\pi}{4} f \left(\sec ^{2} x \right) \cdot \sec ^{3} x \cdot \frac{\sin x }{ x }$
$\frac{\pi}{4} f (2) \cdot(\sqrt{2})^{3} \cdot \frac{1}{\sqrt{2}} \times \frac{4}{\pi}$
$\Rightarrow 2 f (2)$