Question

If minimum value of ${\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2$is$\frac{{\pi }^2}{k}$, then value of k is

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is C $8$

Let $f\left(x\right)={\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2$

$f\left(x\right)=={({\sin }^{-1}x+{\cos }^{-1}x)}^2-2{\sin }^{-1}x{\cos }^{-1}x=\frac{{\pi }^2}{4}-2{\sin }^{-1}x{\cos }^{-1}x$

At $x=\frac{1}{\sqrt{2}}$ $f\left(x\right)$ is minimum because ${\sin }^{-1}x$ and ${\cos }^{-1}x$ will give maximum value product

$f\left(x\right)=\frac{{\pi }^2}{4}-2\left(\frac{{\pi }^2}{16}\right)=\frac{{\pi }^2}{4}-\frac{{\pi }^2}{8}=\frac{{\pi }^2}{8}$

$\frac{{\pi }^2}{8}=\frac{{\pi }^2}{k}\Rightarrow k=8$