Question

Let $y=({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3$ then

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

Answer 1

Answer: (B), (C)

The correct options are
B $maxy=\frac{7{\pi }^3}{8}$
C $miny=\frac{{\pi }^3}{32}$

$y=({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3$

$=({\sin }^{-1}x+{\cos }^{-1}x{)}^3-3{\sin }^{-1}x{\cos }^{-1}x({\sin }^{-1}x+{\cos }^{-1}x)$

$=\frac{{\pi }^3}{2^3}-3{\sin }^{-1}x{\cos }^{-1}x\left(\frac{\pi }{2}\right)$ $(\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2})$

$=\frac{{\pi }^3}{8}-3{\sin }^{-1}x{\cos }^{-1}x\left(\frac{\pi }{2}\right)$

$=\frac{{\pi }^3}{8}-\frac{3\pi }{2}{\sin }^{-1}x(\frac{\pi }{2}-{\sin }^{-1}x)$

$=\frac{3\pi }{2}{\left({\sin }^{-1}x\right)}^2-\frac{3{\pi }^2}{4}{\sin }^{-1}x+\frac{{\pi }^3}{8}$

$=\frac{3\pi }{2}({\left({\sin }^{-1}x\right)}^2-\frac{\pi }{2}{\sin }^{-1}x+\frac{{\pi }^2}{12})$

$=\frac{3\pi }{2}({({\sin }^{-1}x-\frac{\pi }{4})}^2+\frac{{\pi }^2}{48})$

For minimum value, put ${\sin }^{-1}x=\frac{\pi }{4}$ (to make the square term zero)

We get $y_{min.}=\frac{{\pi }^3}{32}$

Similarly for maximum value we put ${\sin }^{-1}x=-\frac{\pi }{2}$ (to make the square term maximum)

We get $y_{max.}=\frac{7{\pi }^3}{8}$