Question

The maximum value of $({\sec }^{-1}x{)}^2+(cose{c}^{-1}x{)}^2$ is equal to

• A. $\frac{{\pi }^2}{4}$
• B. $\frac{11{\pi }^2}{8}$
• C. ${\pi }^2$
• D. $\frac{{\pi }^2}{2}$

Answer 1

The correct option is B $\frac{11{\pi }^2}{8}$

Let $I=({\sec }^{-1}x{)}^2+(cose{c}^{-1}x{)}^2$
$=({\sec }^{-1}x+cose{c}^{-1}x{)}^2-2{\sec }^{-1}xcose{c}^{-1}x$
$=\frac{{\pi }^2}{4}-2{\sec }^{-1}x(\frac{\pi }{2}-{\sec }^{-1}x)$
$=\frac{{\pi }^2}{4}+2\left(\right({\sec }^{-1}x{)}^2-\frac{\pi }{2}\left({\sec }^{-1}x\right)+\frac{{\pi }^2}{16}-\frac{{\pi }^2}{16})$
$=\frac{{\pi }^2}{4}+2{({\sec }^{-1}x-\frac{\pi }{4})}^2$
$\Rightarrow I_{max}=\frac{{\pi }^2}{4}+2\times \frac{9{\pi }^2}{16}$
$={\frac{11\pi }{8}}^2$