Question

Let $f:(-1,1)\to B$, be a function defined by
$f\left(x\right)=ta{n}^{-1}\frac{2x}{1-x^2}$, then $f$ is both one-one and onto when $B$ is in the interval

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

Answer 1

The correct option is A $(-\frac{\pi }{2},\frac{\pi }{2})$

Let $x=\tan \theta$
$f\left(x\right)={\tan }^{-1}\left(\frac{2tan\theta }{1-{\tan }^2\theta }\right)$
$={\tan }^{-1}\left(\tan 2\theta \right)$
$=2\theta$
$=2{\tan }^{-1}x$
Now $f:(-1,1)\to B$
$-1=\tan \theta$
$\theta =\frac{-\pi }{4}$
Also $1=\tan \theta$
$\theta =\frac{\pi }{4}$
Therefore $\theta \in (\frac{-\pi }{4},\frac{\pi }{4})$
$2\theta \in (\frac{-\pi }{2},\frac{\pi }{2})$
$2{\tan }^{-1}\left(x\right)\in (\frac{-\pi }{2},\frac{\pi }{2})$
$f\left(x\right)\in (\frac{-\pi }{2},\frac{\pi }{2})$
Therefore $f:(-1,1)\to (\frac{-\pi }{2},\frac{\pi }{2})$
Hence $B=(\frac{-\pi }{2},\frac{\pi }{2})$