Question

Let $f:R\to R$ be a continuous function. Then $\underset{x\to \frac{\pi }{4}}{lim}\frac{\frac{\pi }{4}\underset{2}{\overset{{\sec }^{2}x}{\int }}f\left(x\right)dx}{x^2-\frac{{\pi }^2}{16}}$ is equal to

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

Answer 1

The correct option is C $2f\left(2\right)$

$\underset{x\to \frac{\pi }{4}}{lim}\frac{\frac{\pi }{4}\underset{2}{\overset{{\sec }^{2}x}{\int }}f\left(x\right)dx}{x^2-\frac{{\pi }^2}{16}}\left[\frac{0}{0}\text{form}\right]$
Using L-Hospital's rule, we get
$=\frac{\pi }{4}\underset{x\to \frac{\pi }{4}}{lim}\frac{2{\sec }^{2}x\cdot \tan x\cdot f\left({\sec }^{2}x\right)}{2x}$
$=\frac{\pi }{4}\cdot \frac{2\cdot 1\cdot f\left(2\right)}{\frac{\pi }{4}}$
$=2f\left(2\right)$