Question
Let $$f:\left( { - 1,1} \right) \to B,$$ be a function 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.
$$\left( {0,\frac{\pi }{2}} \right)$$
B.
$$\left[ {0,\left. {\frac{\pi }{2}} \right)} \right.$$
C.
$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$
D.
$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
Answer :
$$\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
Solution :
$$\eqalign{
& {\text{Given }}f\left( x \right) = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) = 2{\tan ^{ - 1}}x{\text{ for }}x \in \left( { - 1,1} \right) \cr
& {\text{If }}x \in \left( { - 1,1} \right) \Rightarrow {\tan ^{ - 1}}x \in \left( {\frac{{ - \pi }}{4},\frac{\pi }{4}} \right) \cr
& \Rightarrow 2{\tan ^{ - 1}}x \in \left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right) \cr} $$
Clearly, range of $$f\left( x \right) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$
For $$f$$ to be onto, co-domain = range
Co-domain of function $$ = B = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$$