Question

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

• A. $(-\frac{\pi }{4},\frac{\pi }{4})$
• B. $(-\frac{\pi }{2},\frac{\pi }{2})$
• C. $(0,\frac{\pi }{2})$
• D. $(-\frac{\pi }{2},0)$

Answer 1

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

Let $x=\tan \left(a\right)$
Hence

$\Rightarrow$$f\left(x\right)={\tan }^{-1}\left(\frac{2\tan \left(a\right)}{1-{\tan }^2\left(a\right)}\right)$

$={\tan }^{-1}\left(\tan \right(2a\left)\right)$

$=2a$

$=2{\tan }^{-1}x$
Now, since f is one-one and onto,
$-1<x<1$

$\Rightarrow$$2{\tan }^{-1}(-1)<2{\tan }^{-1}x<2{\tan }^{-1}\left(1\right)$

$-\frac{\pi }{2}<f\left(x\right)<\frac{\pi }{2}$

So, the range of ${\tan }^{-1}\left(x\right)=(\frac{-\pi }{2},\frac{\pi }{2})$.

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