Let f: (−1, 1)→B, be a function defined by f(x)=tan−1(2x1−x2), then f is both one−one and onto when B is the interval

f(x)=Tan−1(2x1−x2)put x =tanθ=tan−1(2tanθ1−tan2θ)=tan−1(tan 2θ)=2θ=2 tan−1xNow −1<x<1⇒ tan−1(−1)<tan−1x<tan−1(1)⇒ −π4<tan−1x<π4⇒ −π2<2tan−1x<π2Range f(x)=(−π2, π2)Given f is onto∴ B=Range of f(x)B=(−π2, π2)

Let f: (−1, 1)→B, be a function defined by f(x)=tan−1(2x1−x2), then f is both one−one and onto when B is the interval

f(x)=Tan−1(2x1−x2)put x =tanθ=tan−1(2tanθ1−tan2θ)=tan−1(tan 2θ)=2θ=2 tan−1xNow −1<x<1⇒ tan−1(−1)<tan−1x<tan−1(1)⇒ −π4<tan−1x<π4⇒ −π2<2tan−1x<π2Range f(x)=(−π2, π2)Given f is onto∴ B=Range of f(x)B=(−π2, π2)

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$\begin{array}{l} L e t f : (- 1, 1) \to B, b e a f u n c t i o n d e f i n e d b y f (x) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}), t h e n f i s b o t h o n e - o n e a n d \\ o n t o w h e n B i s t h e int e r v a l \end{array}$

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$[0, \frac{π}{2})$

$(- \frac{π}{2}, \frac{π}{2})$

$[- \frac{π}{2}, \frac{π}{2}]$

$(0, \frac{π}{2})$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = {Tan}^{- 1} (\frac{2 x}{1 - x^{2}}) \\ p u t x = \tan θ \\ = \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) \\ = \tan^{- 1} (\tan 2 θ) \\ = 2 θ \\ = 2 \tan^{- 1} x \\ N o w - 1 < x < 1 \\ \Rightarrow \tan^{- 1} (- 1) < \tan^{- 1} x < \tan^{- 1} (1) \\ \Rightarrow \frac{- π}{4} < \tan^{- 1} x < \frac{π}{4} \\ \Rightarrow \frac{- π}{2} < 2 \tan^{- 1} x < \frac{π}{2} \\ R a n g e f (x) = (\frac{- π}{2}, \frac{π}{2}) \\ G i v e n f i s o n t o \\ ∴ B = R a n g e o f f (x) \\ B = (\frac{- π}{2}, \frac{π}{2}) \end{array}$