Question

The set of values of x satisfying $\left|{\sin }^{-1}x\right|<\left|{\cos }^{-1}x\right|$, is

• A. $[-1,1/\sqrt{2}]$
• B. $[-1,-1/\sqrt{2}]\cup [1/\sqrt{2},1]$
• C. $(-1,1/\sqrt{2})$
• D. None of these

Answer 1

Answer: (A)

We know that $0\le {\cos }^{-1}\le \pi$.
$\therefore \left|{\cos }^{-1}x\right|={\cos }^{-1}x$

It is evident from the graphs of $y=\left|{\sin }^{-1}x\right|$
and $y=\left|{\cos }^{-1}x\right|$ that $\left|{\sin }^{-1}x\right|<\left|{\cos }^{-1}x\right|$
for all $x\in [-1,1/\sqrt{2}]$