Q95 / 225 · JEE Main 2005 · Mathematics

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

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

Solution

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 $=B=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Hence, the correct answer is option: 'D'.

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