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

Question

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

Dr. Abubacker · Accepted Answer

(a)
Given, $f(x) = 	an^{- 1}{\frac{2x}{1 - x^{2}} = 2	an^{- 1}{x\left( x^{2} < 1 ight)}}$
Since, $\ \ \ \ \ \ \ x \in ( - 1,\ 1)$.
$\Rightarrow \ \ \ 	an^{- 1}{x \in \left( - \frac{\pi}{4},\ \frac{\pi}{4} ight)}$
$\Rightarrow \ 2	an^{- 1}{x \in \left( - \frac{\pi}{2},\ \frac{\pi}{2} ight)}$
So, $f(x) \in \left( - \frac{\pi}{2},\ \frac{\pi}{2} ight)$

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

Explanation

Reviewed by Dr. Abubacker

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