Question

Let $f:(-1,1)\to B$ is defined as $f(x)={\tan }^{-1}\frac{2x}{1-x^2}.$ Function $f$ is one-one and onto, then the interval B is

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

Answer 1

Answer: (D)

$x\in (-1,1)\Rightarrow {\tan }^{-1}x\in \left(-\frac{\pi }{4},\frac{\pi }{4}\right)$
$\Rightarrow$ $2{\tan }^{-1}x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
and $f(x)={\tan }^{-1}\frac{2x}{1-x^2}=2{\tan }^{-1}x,(x^2<1)$
$\therefore$ $f(x)\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
Function is one-one onto. $x\in (-1,1)\Rightarrow {\tan }^{-1}x\in \left(-\frac{\pi }{4},\frac{\pi }{4}\right)$
$\Rightarrow$ $2{\tan }^{-1}x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
and $f(x)={\tan }^{-1}\frac{2x}{1-x^2}=2{\tan }^{-1}x,(x^2<1)$
$\therefore$ $f(x)\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
Function is one-one onto.