Question

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

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

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

Explanation for the correct option

Step 1: Simplify the given expression

Given, ${\left[{\sin }^{-1}\left(x\right)\right]}^2+{\left[{\cos }^{-1}\left(x\right)\right]}^2$
$\begin{array}{cc}{\left[{\sin }^{-1}\left(x\right)\right]}^2+{\left[{\cos }^{-1}\left(x\right)\right]}^2={\left[{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)\right]}^2-2{\sin }^{-1}\left(x\right){\cos }^{-1}\left(x\right)& \left[\because a^2+b^2={\left(a+b\right)}^2-2ab\right]\\ ={\left(\frac{\mathrm{\pi }}{2}\right)}^2-2{\sin }^{-1}\left(x\right)\left[\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\left(x\right)\right]& \left[\because {\cos }^{-1}\left(x\right)=\frac{\mathrm{\pi }}{2}-{\sin }^{-1}\left(x\right)\right]\\ =\frac{{\mathrm{\pi }}^2}{4}-{\mathrm{\pi sin}}^{-1}\left(\mathrm{x}\right)+2{\left[{\sin }^{-1}\left(\mathrm{x}\right)\right]}^2& \\ =2\left\{\frac{{\mathrm{\pi }}^2}{8}-\frac{\mathrm{\pi }}{2}{\sin }^{-1}\left(x\right)+{\left[{\sin }^{-1}\left(x\right)\right]}^2\right\}& \\ =2\left\{{\left[{\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right]}^2+\frac{{\mathrm{\pi }}^2}{16}\right\}& \end{array}$

Step 2: Solve for the least value

The only term in the expression that contains a variable is the term ${\left[{\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right]}^2$.
The function has the least value when above expression is $0$, i.e.,
$\begin{array}{cr}& {\left[{\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}\right]}^2=0\\ \Rightarrow & {\sin }^{-1}\left(x\right)-\frac{\mathrm{\pi }}{4}=0\\ \Rightarrow & {\sin }^{-1}\left(x\right)=\frac{\mathrm{\pi }}{4}\end{array}$

So the least value is,
$2\left\{{\left[\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right]}^2+\frac{{\mathrm{\pi }}^2}{16}\right\}=\frac{{\mathrm{\pi }}^2}{8}$

Step 3: Solve for the greatest value

When ${\sin }^{-1}\left(x\right)=-\frac{\mathrm{\pi }}{2}$, the function has the greatest value i.e.,
$\begin{array}{rcl}2\left[{\left(-\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{16}\right]& =& 2\left[{\left(-\frac{3\mathrm{\pi }}{4}\right)}^2+\frac{{\mathrm{\pi }}^2}{16}\right]\\ & =& 2\left(\frac{9{\mathrm{\pi }}^2+{\mathrm{\pi }}^2}{16}\right)\\ & =& \frac{5{\mathrm{\pi }}^2}{4}\end{array}$

Therefore, the greatest and least values of the given functions are $\frac{5{\mathrm{\pi }}^2}{4}$ and $\frac{{\mathrm{\pi }}^2}{8}$.

Hence, option(C) i.e. $\frac{5{\mathrm{\pi }}^4}{4}$ and $\frac{{\mathrm{\pi }}^2}{8}$ is correct.