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

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$\min i m u m v a l u e o f y = \frac{π^{3}}{8}$

$\min i n i m u m v a l u e o f y = \frac{π^{3}}{32}$

$\max i m u m v a l u e o f y = \frac{π^{3}}{8}$

$ma x i m u m v a l u e o f y = \frac{7 π^{3}}{32}$

answer is B.

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Detailed Solution

$y = {(\sin^{- 1} x)}^{3} + {(\cos^{- 1} x)}^{3}$
$\begin{array}{l} f (x) = {(\sin^{- 1} x)}^{3} + {(\cos^{- 1} x)}^{3} \\ = {(\sin^{- 1} x + \cos^{- 1} x)}^{3} - 3 (\sin^{- 1} x + \cos^{- 1} x) \sin^{- 1} x \cos^{- 1} x \end{array}$
$\begin{array}{l} = {(\frac{π}{2})}^{3} - 3 \frac{π}{2} \sin^{- 1} x \cos^{- 1} x \\ = \frac{π^{3}}{8} - 3 \frac{π}{2} \sin^{- 1} x (\frac{π}{2} - \sin^{- 1} x) \\ = \frac{π}{2} \{\frac{π^{2}}{4} - 3 a (\frac{π}{2} - a)\} where a = \sin^{- 1} x \\ = \frac{π}{2} \{3 a^{2} - \frac{3 a π}{2} + \frac{π^{2}}{4}\} \\ = \frac{π}{8} \{12 α^{2} - 6 π a + π^{2}\} \\ = \frac{12 π}{8} \{α^{2} - \frac{1}{2} π a + \frac{π^{2}}{12}\} \\ = \frac{12 π}{8} \{{(a - \frac{π}{4})}^{2} + \frac{π^{2}}{48}\} \end{array}$
$\frac{π^{3}}{32} \leq (\sin^{- 1} x) + (\cos^{- 1} x) \leq \frac{7 π^{3}}{8}$