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Let $f:(-1,1) \rightarrow B$, be a function defined by $f(x)=\tan ^{-1} \frac{2 x}{1-x^{2}}$, then $f$ is both one-one and onto when $B$ is the interval.

1) $\left(0, \frac{\pi}{2}\right)$
2) $\left[0, \frac{\pi}{2}\right)$
3) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
4) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
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(4).

Explanation

Given $f(x)=\tan ^{1}\left(\frac{2 x}{1-x^{2}}\right)$ for $x \in(-1,1)$
clearly range of $f(x)=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\therefore$ co-domain of function $=B=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
put $x=a, y=x-a$

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