Question

If $f\left(x\right)=({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2$, then

• A. $f\left(x\right)$ has the least value of $\frac{{\pi }^2}{8}$
• B. $f\left(x\right)$ has the greatest value of $\frac{5{\pi }^2}{8}$
• C. $f\left(x\right)$ has the least value of $\frac{{\pi }^2}{16}$
• D. $f\left(x\right)$ has the greatest value of $\frac{5{\pi }^2}{4}$

Answer 1

Answer: (A), (D)

The correct options are
A $f\left(x\right)$ has the least value of $\frac{{\pi }^2}{8}$
D $f\left(x\right)$ has the greatest value of $\frac{5{\pi }^2}{4}$

$f\left(x\right)=({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2$
$=({\sin }^{-1}x+{\cos }^{-1}x{)}^2-2{\sin }^{-1}x{\cos }^{-1}x$
$={\left(\frac{\pi }{2}\right)}^2-2{\cos }^{-1}x(\frac{\pi }{2}-{\cos }^{-1}x)$
$=2\left[\right({\cos }^{-1}x{)}^2-\frac{\pi }{2}{\cos }^{-1}x+\frac{{\pi }^2}{8}]$
$=2\left[\right({\cos }^{-1}x{)}^2-2\frac{\pi }{4}{\cos }^{-1}x+\frac{{\pi }^2}{16}-\frac{{\pi }^2}{16}+\frac{{\pi }^2}{8}]$
$f\left(x\right)=2[{({\cos }^{-1}x-\frac{\pi }{4})}^2+\frac{{\pi }^2}{16}]$.....(i)
we know that $0\le {\cos }^{-1}x\le \pi$
$\Rightarrow -\frac{\pi }{4}\le {\cos }^{-1}x-\frac{\pi }{4}\le \frac{3\pi }{4}$.......(ii)
Max $f\left(x\right)=2[{\left(\frac{3\pi }{4}\right)}^2+\frac{{\pi }^2}{16}]$
$=\frac{5{\pi }^2}{4}$
From eq (ii)
$\Rightarrow -\frac{\pi }{4}\le {\cos }^{-1}x-\frac{\pi }{4}\le \frac{3\pi }{4}$
$\Rightarrow 0\le {({\cos }^{-1}x-\frac{\pi }{4})}^2\le {\left(\frac{3\pi }{4}\right)}^2$.......(iii)
From (i) and (iii)
Min $f\left(x\right)=2(0+\frac{{\pi }^2}{16})$
$=\frac{{\pi }^2}{8}$