Question

If $f:(-1,1)\to B$ , is a function defined by $f\left(x\right)={\tan }^{-1}\frac{2x}{1-x^2}$, then find $B$ when $f\left(x\right)$ is both one-one and onto function.

• A. $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$
• B. $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
• C. $\left(0,\frac{\pi }{2}\right)$
• D. $\left[0,\frac{\pi }{2}\right)$

Answer 1

The correct option is B $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

For $xϵ(-1,1)$, we have

$f\left(x\right)={\tan }^{-1}\left[\frac{2x}{1-x^2}\right]$

Substituting $x=\tan \theta$ in above equation.

Therefore, $f\left(\tan \theta \right)={\tan }^{-1}\left[\frac{2\tan \theta }{1-{\tan }^2\theta }\right]$

$={\tan }^{-1}\tan \left(2\theta \right)=2\theta$

$=2{\tan }^{-1}x$

Thus $-\frac{\pi }{2}<{\tan }^{-1}\left[\frac{2x}{1-x^2}\right]<\frac{\pi }{2}$

Thus option B is correct.