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, $f(x)=a^2+b^2$
$=(a+b{)}^2-2ab$
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,
$f^{\prime }(x)=0-\pi \cdot \frac{1}{\sqrt{1-x^2}}+4{\sin }^{-1}x\frac{1}{\sqrt{1-x^2}}=0$
$=\frac{1}{\sqrt{1-x^2}}\left[4{\sin }^{-1}x-\pi \right]=0={\sin }^{-1}x=\pi /4$
$x=\sin \pi /4=\frac{1}{\sqrt{2}}$
Therefore, $f^{\prime \prime }\left(\frac{1}{\sqrt{2}}\right)=+$ 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, $8(\alpha +\beta )=8\left[\frac{{\pi }^2}{8}+\frac{5{\pi }^2}{4}\right]$
$=8\left[\frac{{\pi }^2+10{\pi }^2}{8}\right]$
$=11{\pi }^2$