Thank you for reporting, we will resolve it shortly
Q.
Let $f : (-1 ,1) \to B$ , be a function defined by $f(x) = \tan^{-1} \frac{2x}{1 - x^2}$ , then f is both one - one and onto when B is the interval
Relations and Functions - Part 2
Solution:
Given $f\left(x\right) = \tan^{-1} \left(\frac{2x}{1-x^{2}}\right) = 2 \tan^{-1}x $ for $x \in\left(-1,1\right) $
If $x \in\left(-1,1\right) \Rightarrow \tan^{-1} x \in\left(\frac{- \pi}{4}, \frac{\pi}{4}\right) $
$\Rightarrow 2 \tan^{-1} x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
Clearly, range of $ f\left(x\right) = \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) $
For $f$ to be onto, codomain = range
$\therefore $ Co-domain of function $= B = \left(- \frac{\pi }{2} , \frac{\pi }{2}\right) $