Question

The solution set of $({\cos }^{-1}x{)}^4-({\sin }^{-1}x{)}^4>0$ is

• A. $[0,\frac{1}{\sqrt{2}})$
• B. $[-1,\sqrt{2})$
• C. $[-1,\frac{1}{\sqrt{2}})$
• D. $[-1,\frac{1}{\sqrt{2}}]$

Answer 1

The correct option is C $[-1,\frac{1}{\sqrt{2}})$

Given:
$({\cos }^{-1}x{)}^4-({\sin }^{-1}x{)}^4>0$
$\Rightarrow (({\cos }^{-1}x{)}^2-({\sin }^{-1}x{)}^2)(({\cos }^{-1}x{)}^2+({\sin }^{-1}x{)}^2)>0$
$(\because ({\cos }^{-1}x{)}^2+({\sin }^{-1}x{)}^2)>0)$
$\Rightarrow ({\cos }^{-1}x{)}^2-({\sin }^{-1}x{)}^2>0$
$\Rightarrow ({\cos }^{-1}x-{\sin }^{-1}x)({\cos }^{-1}x+{\sin }^{-1}x)>0$
$\Rightarrow ({\cos }^{-1}x-{\sin }^{-1}x)\cdot \frac{\pi }{2}>0$
$\Rightarrow ({\cos }^{-1}x-{\sin }^{-1}x)>0$
$\Rightarrow \frac{\pi }{2}-2{\sin }^{-1}x>0$
$\Rightarrow {\sin }^{-1}x<\frac{\pi }{4}$
$\Rightarrow x<\frac{1}{\sqrt{2}}$
Since, ${\cos }^{-1}x$ and ${\sin }^{-1}x$ is defined if and only if $-1\le x\le 1$
$\therefore x\in [-1,\frac{1}{\sqrt{2}})$