The greatest and least values of ${(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2}$ are respectively

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$\frac{π^{2}}{4}$

$\frac{π^{2}}{8}$

$\frac{π}{2}$

$\frac{5 π^{2}}{4}$

answer is A, C.

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Detailed Solution

$\begin{array}{l} f (x) = {(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2} \\ = {(\sin^{- 1} x + \cos^{- 1} x)}^{2} - 2 \sin^{- 1} x \cos^{- 1} x \\ = \frac{π^{2}}{4} - 2 \sin^{- 1} x [\frac{π}{2} - \sin^{- 1} x] \\ = \frac{π^{2}}{4} - π \sin^{- 1} x + 2 {(\sin^{- 1} x)}^{2} \\ = 2 {(\sin^{- 1} x - \frac{π}{4})}^{2} + 2 [\frac{π^{2}}{16}] \end{array}$

$since - \frac{π}{2} \leq \sin^{- 1} x \leq \frac{π}{2}$

$\begin{array}{l} \Rightarrow - \frac{3 π}{4} \leq \sin^{- 1} x - \frac{π}{4} \leq \frac{π}{4} \\ \Rightarrow 0 \leq {(\sin^{- 1} x - \frac{π}{4})}^{2} \leq \frac{9 π^{2}}{16} \\ \Rightarrow 0 \leq 2 {(\sin^{- 1} x - \frac{π}{4})}^{2} \leq \frac{9 π^{2}}{8} \\ \Rightarrow \frac{π^{2}}{8} \leq 2 {(\sin^{- 1} x - \frac{π}{4})}^{2} + \frac{π^{2}}{8} \leq \frac{5 π^{2}}{4} \end{array}$