Question

$\underset{x\to \frac{\pi }{4}}{lim}\frac{{\int }_2^{{\sec }^{2}x}f\left(t\right)dt}{x^2-\frac{{\pi }^2}{16}}$

• A. $\frac{8}{\pi }f\left(2\right)$
• B. $\frac{2}{\pi }f\left(2\right)$
• C. $\frac{2}{\pi }f\left(\frac{1}{2}\right)$
• D. $4f\left(2\right)$

Answer 1

The correct option is A $\frac{8}{\pi }f\left(2\right)$

Let $L=\underset{x\to \frac{\pi }{4}}{lim}\frac{{\int }_2^{{\sec }^{2}x}f\left(t\right)dt}{x^2-\frac{{\pi }^2}{16}}$ $(\frac{0}{0}$ form $)$
Thus applying L-hospital's rule,
$L=\underset{x\to \frac{\pi }{4}}{lim}\frac{f\left({\sec }^{2}x\right)2\sec x}{2x}\sec x\tan x$
.$\therefore L=\frac{2f\left(2\right)}{\pi /4}=\frac{8f\left(2\right)}{\pi }$