Question

The greatest and the least values of ${\left[{\sin }^{-1}\left(x\right)\right]}^3+{\left[{\cos }^{-1}\left(x\right)\right]}^3$ are

• A. $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$
• B. $\left(-\frac{{\mathrm{\pi }}^3}{8}\right)$
• C. $\left(\frac{7{\mathrm{\pi }}^3}{8},\frac{{\mathrm{\pi }}^3}{32}\right)$
• D. None of these

Answer 1

The correct option is C

$\left(\frac{7{\mathrm{\pi }}^3}{8},\frac{{\mathrm{\pi }}^3}{32}\right)$

Explanation for the correct option

Step 1: Simplify the given expression

Given expression is ${\left[{\sin }^{-1}\left(x\right)\right]}^3+{\left[{\cos }^{-1}\left(x\right)\right]}^3$

We have,
$\begin{array}{cc}{\left[{\sin }^{-1}\left(x\right)\right]}^3+{\left[{\cos }^{-1}\left(x\right)\right]}^3=\left[{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)\right]\left\{{\left[{\sin }^{-1}\left(x\right)\right]}^2+{\left[{\cos }^{-1}\left(x\right)\right]}^2-{\sin }^{-1}\left(x\right){\cos }^{-1}\left(x\right)\right\}& \left[\because a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\right]\\ =\left(\frac{\mathrm{\pi }}{2}\right)\left\{{\left[{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)\right]}^2-2{\sin }^{-1}\left(x\right){\cos }^{-1}\left(x\right)-{\sin }^{-1}\left(x\right){\cos }^{-1}\left(x\right)\right\}& \left[\because a^2+b^2={\left(a+b\right)}^2-2ab,{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)=\frac{\mathrm{\pi }}{2}\right]\\ =\left(\frac{\mathrm{\pi }}{2}\right)\left\{{\left(\frac{\mathrm{\pi }}{2}\right)}^2-3{\sin }^{-1}\left(x\right)\left[\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\left(x\right)\right]\right\}& \\ =\left(\frac{\mathrm{\pi }}{2}\right)\left[\frac{{\mathrm{\pi }}^2}{4}-3a\left(\frac{\mathrm{\pi }}{2}-a\right)\right]& \left[a={\sin }^{-1}\left(x\right)\right]\\ =\left(\frac{\mathrm{\pi }}{2}\right)\left[\frac{{\mathrm{\pi }}^2}{4}-\frac{12a\left({\displaystyle \frac{\mathrm{\pi }}{2}}-a\right)}{4}\right]& \\ =\left(\frac{12\mathrm{\pi }}{8}\right)\left[\frac{{\mathrm{\pi }}^2}{12}-\frac{\mathrm{a\pi }}{2}+{\mathrm{a}}^2\right]& \\ =\left(\frac{3\mathrm{\pi }}{2}\right)\left[a^2-\frac{a\mathrm{\pi }}{2}+\frac{{\mathrm{\pi }}^2}{12}+\frac{{\mathrm{\pi }}^2}{48}-\frac{{\mathrm{\pi }}^2}{48}\right]& \\ =\left(\frac{3\mathrm{\pi }}{2}\right)\left[{\left(a-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{48}\right]& \\ =\left(\frac{3\mathrm{\pi }}{2}\right)\left[{\left({\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{48}\right]& \left[\because a={\sin }^{-1}\left(x\right)\right]\end{array}$

Step 2: Solve for the required greatest and least values

The above expression will be greatest when ${\left[{\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right]}^2$ will be greatest and least when ${\left[{\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right]}^2$ will be least

We know that,
$-\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}\left(x\right)\le \frac{\mathrm{\pi }}{2}$

So, ${\left[{\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right]}^2$ will be greatest when ${\sin }^{-1}\left(x\right)=-\frac{\mathrm{\pi }}{2}$. Then,

$\begin{array}{rcl}\left(\frac{3\mathrm{\pi }}{2}\right)\left[{\left({\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{48}\right]& =& \left(\frac{3\mathrm{\pi }}{2}\right)\left[{\left(-\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{48}\right]\\ & =& \left(\frac{3\mathrm{\pi }}{2}\right)\left[{\left(-\frac{3\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{48}\right]\\ & =& \left(\frac{3\mathrm{\pi }}{2}\right)\left[\frac{9{\mathrm{\pi }}^2}{16}+\frac{{\mathrm{\pi }}^2}{48}\right]\\ & =& \left(\frac{3\mathrm{\pi }}{2}\right)\left(\frac{28{\mathrm{\pi }}^2}{48}\right)\\ & =& \frac{7{\mathrm{\pi }}^3}{8}\end{array}$

${\left[{\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right]}^2$ will be least when ${\sin }^{-1}\left(x\right)=\frac{\mathrm{\pi }}{4}$. Then,
$\begin{array}{rcl}\left(\frac{3\mathrm{\pi }}{2}\right)\left[{\left({\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{48}\right]& =& \left(\frac{3\mathrm{\pi }}{2}\right)\left[{\left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{48}\right]\\ & =& \left(\frac{3\mathrm{\pi }}{2}\right)\left[{\left(0\right)}^2+\frac{{\mathrm{\pi }}^2}{48}\right]\\ & =& \frac{{\mathrm{\pi }}^3}{32}\end{array}$

Thus, the greatest and least values of ${\left[{\sin }^{-1}\left(x\right)\right]}^3+{\left[{\cos }^{-1}\left(x\right)\right]}^3$ are $\frac{7{\mathrm{\pi }}^3}{8}$ and $\frac{{\mathrm{\pi }}^3}{32}$ respectively.

Hence, option(C) i.e. $\left(\frac{7{\mathrm{\pi }}^3}{8},\frac{{\mathrm{\pi }}^3}{32}\right)$ is correct.