Question

If $\left|{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right|<\frac{\pi }{3}$, then:

• A. $x\in [-\frac{1}{3},\frac{1}{\sqrt{3}}]$
• B. $x\in (-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$
• C. $x\in (0,\frac{1}{\sqrt{3}})$
  
• D. non of these

Answer 1

The correct option is B $x\in (-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

Que - 1

$|\frac{\pi }{2}-si{n}^{-1}\left(\frac{1-x^2}{1+x^2}\right)|<\frac{\pi }{3}$

$\Rightarrow \left|co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right|<\frac{\pi }{3}$

$\Rightarrow 0\le co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)<\frac{\pi }{3}$

$\Rightarrow 1\ge \frac{1-x^2}{1+x^2}>\frac{1}{2}$

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

$\Rightarrow \frac{1-x^2}{1+x^2}-1⩽0$

$\Rightarrow \frac{1-x^2-1-x^2}{1+x^2}⩽0$

$\Rightarrow \frac{-2x^2}{1+x^2}⩽0$

Always true

$x\in R$ ___ (1)

$\frac{2-2x^2}{1+x^2}-1>0$

$\frac{2-2x^2-1-x^2}{1+x^2}>0$

$\frac{1-3x^2}{1+x^2}>0$

$\Rightarrow \frac{3x^2-1}{1+x^2}<0$

$\Rightarrow \frac{(\sqrt{3}x-1)(\sqrt{3}x+1)}{1+x^2}<0$

$\Rightarrow x\in (-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ ___ (ii)

From (i) & (ii)

$x\in (-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$