Question

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

• A. $\frac{{\mathrm{\pi }}^2}{4}$ and $0$
• B. $\frac{{\mathrm{\pi }}^2}{4}$ and $-\frac{\mathrm{\pi }}{2}$
• C. $\frac{5{\mathrm{\pi }}^2}{4}$ and $\frac{{\mathrm{\pi }}^2}{8}$
• D. $\frac{{\mathrm{\pi }}^2}{4}$ and $-\frac{{\mathrm{\pi }}^2}{4}$

Answer 1

The correct option is C

$\frac{5{\mathrm{\pi }}^2}{4}$ and $\frac{{\mathrm{\pi }}^2}{8}$

Explanation of the correct answer.

Step $1$: Simplify the equation.

Given : ${\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2$

Add and subtract $2{\sin }^{-1}x{\cos }^{-1}x$,

$\Rightarrow$${\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2+2{\sin }^{-1}x{\cos }^{-1}x-2{\sin }^{-1}x{\cos }^{-1}x$

$\Rightarrow {\left({\sin }^{-1}x+{\cos }^{-1}x\right)}^2-2{\sin }^{-1}x{\cos }^{-1}x$ $\left[\because {(a+b)}^2=a^2+b^2+2ab\right]$

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

$\Rightarrow \left(\frac{{\mathrm{\pi }}^2}{4}\right)-\mathrm{\pi }{\sin }^{-1}x+2{\left({\sin }^{-1}x\right)}^2$

$\Rightarrow 2\left[{\left({\sin }^{-1}x\right)}^2-\frac{\mathrm{\pi }}{2}{\sin }^{-1}x+\left(\frac{{\mathrm{\pi }}^2}{8}\right)\right]$

$\Rightarrow 2\left[{\left({\sin }^{-1}x\right)}^2-\frac{\mathrm{\pi }}{2}{\sin }^{-1}x+\left(\frac{{\mathrm{\pi }}^2}{8}\right)\right]$

Add and subtract ${\left(\frac{\mathrm{\pi }}{4}\right)}^2$,

$\Rightarrow 2\left[{\left({\sin }^{-1}x\right)}^2-\frac{\mathrm{\pi }}{2}{\sin }^{-1}x+{\left(\frac{\mathrm{\pi }}{4}\right)}^2-{\left(\frac{\mathrm{\pi }}{4}\right)}^2+\left(\frac{{\mathrm{\pi }}^2}{8}\right)\right]$

$\Rightarrow 2\left[{\left({\sin }^{-1}x-\frac{\mathrm{\pi }}{4}\right)}^2+\left(\frac{{\mathrm{\pi }}^2}{16}\right)\right]$

Step $2$: Compute the least value.

For least value put ${\sin }^{-1}x=\frac{\mathrm{\pi }}{4}$,

$\Rightarrow 2\left[0+\left(\frac{{\mathrm{\pi }}^2}{16}\right)\right]$

$\Rightarrow \frac{{\mathrm{\pi }}^2}{8}$

Step $3$: Compute the greatest value.

For greatest value put ${\sin }^{-1}x=-\frac{\mathrm{\pi }}{2}$,

$$\begin{gathered} \Rightarrow 2\left[{\left(-\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}\right)}^2+\left(\frac{{\mathrm{\pi }}^2}{16}\right)\right] \\ \Rightarrow 2\left[\frac{9{\mathrm{\pi }}^2}{16}+\frac{{\mathrm{\pi }}^2}{16}\right] \\ \Rightarrow \frac{20{\mathrm{\pi }}^2}{16} \\ \Rightarrow \frac{5{\mathrm{\pi }}^2}{4} \end{gathered}$$

Therefore, the greatest and least value of ${\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2$are respectively $\frac{5{\mathrm{\pi }}^2}{4}$ and $\frac{{\mathrm{\pi }}^2}{8}$.

Hence, option $\left(\mathrm{C}\right)$ is the correct option.