Question

If $\alpha$ and $\beta$ are the least and the greatest values of $f\left(x\right) =\left(\sin^{-1}x\right)^{2} + \left(\cos^{-1}x\right)^{2} $ for all $x \in [R$ respectively, then $8 (\alpha + \beta) = $

• A. $\pi^2$
• B. $ 11 \pi^2$
• C. $ 9 \pi^2$
• D. $ 25 \pi^2$

Answer 1

Answer: (B)

$f(x)=\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}$
Let $\sin ^{-1} x=a$ and $\cos ^{-1} x=b$
Then, $f(x)=a^{2}+b^{2}$
$=(a+b)^{2}-2 a b$
Put the value of $a$ and $b$
$f(x)=\left(\sin ^{-1} x+\cos ^{-1} x\right)^{2}-2 \sin ^{-1} x \cos ^{-1} x$
$=\frac{\pi^{2}}{4}-2 \sin ^{-1} x \cos ^{-1} x\left[\because \sin ^{-1} x+\cos ^{-1} x=\pi / 2\right]$
$=\frac{\pi^{2}}{4}-2 \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right)$
$=\frac{\pi^{2}}{4}-\pi \sin ^{-1} x+2\left(\sin ^{-1} x\right)^{2}$
For minimum and maximum value,
$f'(x)=0-\pi \cdot \frac{1}{\sqrt{1-x^{2}}}+4 \sin ^{-1} x \frac{1}{\sqrt{1-x^{2}}}=0 $
$=\frac{1}{\sqrt{1-x^{2}}}\left[4 \sin ^{-1} x-\pi\right]=0=\sin ^{-1} x=\pi / 4 $
$x=\sin \pi / 4=\frac{1}{\sqrt{2}}$
Therefore, $ f'' \left(\frac{1}{\sqrt{2}}\right)=+$ ve
$f(x)_{\min } =$ when $ x=1 / \sqrt{2} $
$\therefore (x)_{\min } =\frac{\pi^{2}}{4}-2 \sin ^{-1} x \cos ^{-1} x $
$ \therefore f(x)_{\min } =\frac{\pi^{2}}{4}-2 \cdot \frac{\pi}{4} \cdot \frac{\pi}{4} $
$\alpha =\frac{\pi^{2}}{8} $
$\therefore (x)_{\max }=\frac{\pi^{2}}{4} -2 \sin ^{-1} x \cos ^{-1} x$
We can see that, $f(x)$ is maximum, when $x=-1$
$ f(x)_{\max }= \frac{\pi^{2}}{4}-2\left(-\frac{\pi}{2}\right)(\pi) $
$ \beta= \frac{\pi^{2}}{4}+\pi^{2}=\frac{5 \pi^{2}}{4} $
$\Rightarrow \beta=\frac{5 \pi^{2}}{4} $
Hence, $ 8(\alpha+\beta) =8\left[\frac{\pi^{2}}{8}+\frac{5 \pi^{2}}{4}\right]$
$=8\left[\frac{\pi^{2}+10 \pi^{2}}{8}\right] $
$=11 \pi^{2}$