Question

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

(2) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

(3) $\left[0, \frac{\pi}{2}\right)$

(4) $\left(0, \frac{\pi}{2}\right)$.

Answer 1

SOLUTION :  $f(x)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=2 \tan ^{-1} x$ for $x \in(-1,1)$

If $x \in(-1,1) \Rightarrow \tan ^{-1} x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \Rightarrow 2 \tan ^{-1} x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Clearly, range of $f(x)=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for $f$ to be onto, co-domain $=$ range

$\therefore$ Codomain of function $=\mathrm{B}=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. 

Above correct Option is Option(1).