Question

The greatest and the least value of ${\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3$ are

• A. $-\frac{\pi }{2},\frac{\pi }{2}$
• B. $-\frac{{\pi }^3}{8},\frac{{\pi }^3}{8}$
• C. $\frac{7{\pi }^3}{8},\frac{{\pi }^3}{32}$
• D. $\frac{7{\pi }^3}{8},\frac{{\pi }^3}{8}$

Answer 1

The correct option is C $\frac{7{\pi }^3}{8},\frac{{\pi }^3}{32}$

We have ${\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3={({\sin }^{-1}x+{\cos }^{-1}x)}^3-3{\sin }^{-1}x{\cos }^{-1}x({\sin }^{-1}x+{\cos }^{-1}x)=\frac{{\pi }^3}{8}-3\left({\sin }^{-1}x{\cos }^{-1}x\right)\frac{\pi }{2}=\frac{{\pi }^3}{8}-\frac{3\pi }{2}{\sin }^{-1}x(\frac{\pi }{2}-{\sin }^{-1}x)=\frac{{\pi }^3}{8}+\frac{3\pi }{2}[{\left({\sin }^{-1}x\right)}^2-\frac{\pi }{2}{\sin }^{-1}x]=\frac{{\pi }^3}{8}+\frac{3\pi }{2}\left[{({\sin }^{-1}x-\frac{\pi }{4})}^2\right]-\frac{3{\pi }^3}{32}=\frac{{\pi }^3}{32}+\frac{3\pi }{2}{({\sin }^{-1}x-\frac{\pi }{4})}^2$
The lease value is $\frac{{\pi }^3}{32}$

We know that
${({\sin }^{-1}x-\frac{\pi }{4})}^2\le {\left(\frac{3\pi }{4}\right)}^2$

The greatest value is $\frac{{\pi }^3}{32}+\frac{3\pi }{2}\times \frac{9{\pi }^2}{16}=\frac{7{\pi }^3}{8}$