Question

The greatest and least values of $(si{n}^{-1}x{)}^3+(co{s}^{-1}x{)}^3$ are

• A. $\frac{-\pi }{2},\frac{\pi }{2}$
• B. $\frac{-{\pi }^3}{2},\frac{{\pi }^3}{2}$
• C. $\frac{{\pi }^3}{32},\frac{7{\pi }^3}{8}$
• D. None

Answer 1

The correct option is C

$\frac{{\pi }^3}{32},\frac{7{\pi }^3}{8}$

$=(si{n}^{-1}x+co{s}^{-1}x{)}^3-3si{n}^{-1}xco{s}^{-1}x(si{n}^{-1}x+co{s}^{-1}x)=\frac{{\pi }^3}{8}-3(si{n}^{-1}xco{s}^{-1}x)\frac{\pi }{2}=\frac{{\pi }^3}{8}-\frac{3\pi }{2}si{n}^{-1}x(\frac{\pi }{2}-si{n}^{-1}x)\Rightarrow \frac{{\pi }^3}{8}+\frac{3\pi }{2}\left[\right(si{n}^{-1}x{)}^2-\frac{\pi }{2}si{n}^{-1}x]\frac{{\pi }^3}{8}+\frac{3\pi }{2}\left[{(si{n}^{-1}x-\frac{\pi }{4})}^2\right]-\frac{3{\pi }^3}{32}=\frac{{\pi }^3}{32}+\frac{3\pi }{2}{(si{n}^{-1}x-\frac{\pi }{4})}^2$
So, the least value is $\frac{{\pi }^3}{32}$.

${(si{n}^{-1}x-\frac{\pi }{4})}^2\le {\left(\frac{3\pi }{4}\right)}^2$
Because when $x=-1$ $si{n}^{-1}x-\frac{\pi }{4}=-\frac{\pi }{2}-\frac{\pi }{4}=-\frac{3\pi }{4}$
Greatest value is $\frac{{\pi }^3}{32}+\frac{9{\pi }^2}{16}\times \frac{3\pi }{2}=\frac{7{\pi }^3}{8}$