Question

An extremum value of the function $f\left(x\right)={\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3,$ $(-1<x<1)$ is

• A. $\frac{7{\pi }^3}{8}$
• B. $\frac{{\pi }^3}{8}$
• C. $\frac{{\pi }^3}{32}$
• D. $\frac{{\pi }^3}{16}$

Answer 1

Answer: (C)

$\frac{{\pi }^3}{32}$

$f^{\prime }\left(x\right)=\frac{3[{\left({\sin }^{-1}x\right)}^2-{\left({\cos }^{-1}x\right)}^2]}{\left(\sqrt{1-x^2}\right)}$
For extremum value, we have $f^{\prime }\left(x\right)=0$
So ${\left({\sin }^{-1}x\right)}^2-{\left({\cos }^{-1}x\right)}^2=0$

i.e ${\sin }^{-1}x-{\cos }^{-1}x=0$ or ${\sin }^{-1}x+{\cos }^{-1}x=0$
But since ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$
Hence, ${\sin }^{-1}x={\cos }^{-1}x$
Hence $x=\sqrt{2}$ thus $f\left(\pi /4\right)={\left(\pi /4\right)}^3+{\left(\pi /4\right)}^3={\pi }^3/32$