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Let f:\left ( -1,1 \right )\rightarrow B, be a function defined by f\left ( x \right )= \tan ^{-1}\left ( \frac{2x}{1-x^{2}} \right )

then f is both one­-one and onto when B is the interval

  • Option 1)

  • Option 2)

    \left ( 0,\frac{ \pi }{2} \right )

  • Option 3)

    \left ( -\frac{ \pi }{2},\frac{ \pi }{2} \right )

  • Option 4)

    \left [- \frac{ \pi }{2},\frac{ \pi }{2} \right ]

 

Answers (1)

best_answer

As we learnt in 

Domains and Ranges of Inverse Trigonometric Functions -

For \tan ^{-1}x

Domain \epsilon R

Range \epsilon\left [ -\frac{\pi }{2},\frac{\pi }{2} \\ \right ]

-

 

  f(x)=\frac{tan^{-1}2x}{1-x^{2}}= \tan ^{-1}\frac{x+x}{1-x^{2}}=2\tan ^{-1}x

Thus, range of f(x) is same as range of 2 tan-1x in the given domain. As domain is (-1,1), range of 2tan-1x will lie between \left \{ 2\tan ^{-1}\left ( -1 \right ), \: 2\tan ^{-1}1 \right \} \: \: i.e. \: \left ( \frac{-\pi }{2}, \frac{\pi }{2} \right )


Option 1)

This option is incorrect

Option 2)

\left ( 0,\frac{ \pi }{2} \right )

This option is incorrect

Option 3)

\left ( -\frac{ \pi }{2},\frac{ \pi }{2} \right )

This option is correct

Option 4)

\left [- \frac{ \pi }{2},\frac{ \pi }{2} \right ]

This option is incorrect

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