Let \[f:( - 1,1) \to B\], be a function defined by
\[f\left( x \right) = {\tan ^{ - 1}}{{2x} \over {1 - {x^2}}}\],
then f is both one-one and onto when B is the interval
Solutions · 1
Official solution
Given \[\,\,f\left( x \right) = {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) = 2{\tan ^{ - 1}}x\]
for \[x \in \left( { - 1,1} \right)\]
If\[\,\,x \in \left( { - 1,1} \right) \Rightarrow {\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 4},{\pi \over 4}} \right)\]
\[ \Rightarrow 2{\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)\]
Clearly, range of \[f\left( x \right) = \left( { - {\pi \over 2},{\pi \over 2}} \right)\]
For f to be onto, co-domain = range
\[\therefore\] Co-domain of function \[ = B = \left( { - {\pi \over 2},{\pi \over 2}} \right).\]