Question

Find the value of $x$ which satisfy this inequality:
$|\frac{\pi }{2}-{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)|$ $\le$ $\frac{\pi }{3}$

• A. $-\frac{1}{\sqrt{2}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. both A and B
  
• D. None of these

Answer 1

The correct option is B $-\frac{1}{\sqrt{3}}$

We have $\frac{-\pi }{3}\le \frac{\pi }{2}-{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\le \frac{\pi }{3}$

$\frac{-5\pi }{6}\le -{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\le \frac{-\pi }{6}$

$\frac{5\pi }{6}\ge {\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\ge \frac{\pi }{6}$

${\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\frac{\pi }{6},\frac{5\pi }{6}$

$\frac{1-x^2}{1+x^2}=\sin \left(\frac{\pi }{6}\right)$

$\frac{1-x^2}{1+x^2}=\frac{1}{2}$

$2-2x^2=1+x^2$

$x^2=\frac{1}{3}$

$\therefore x=\frac{\pm 1}{\sqrt{3}}$