Question

Solution set of the inequality ${\sin }^{-1}x\le {\cos }^{-1}x$ is

• A. $(0,\frac{1}{\sqrt{2}}]$
• B. $[-1,\frac{1}{\sqrt{3}}]$
• C. $[-1,\frac{1}{\sqrt{2}}]$
• D. $(-1,\frac{1}{\sqrt{3}})$

Answer 1

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

${\cos }^{-1}x\ge {\sin }^{-1}x\Rightarrow \frac{\pi }{2}\ge 2{\sin }^{-1}x(\because {\cos }^{-1}x+{\sin }^{-1}x=\frac{\pi }{2})\Rightarrow {\sin }^{-1}x\le \frac{\pi }{4}\Rightarrow -1\le x\le \sin \left(\frac{\pi }{4}\right)\therefore x\in [-1,\frac{1}{\sqrt{2}}]$