f:(−1, 1)→B defined by f(x)=tan−1(2x1−x2) is bijection then B=

f(x)=Tan−1(2x1−x2)put x=Tanθ⇒ θ=Tan−1x=Tan−1(2Tanθ1−Tan2θ)=Tan−1(Tan2θ)= 2θ=2Tan−1x∴ f is bijection then B= Range∴ −1<x<1⇒ Tan−1(−1)<Tan−1x<Tan−1(1)⇒ −π4<Tan−1x<π2⇒ −π2<2Tan−1x<π2∴ B=Range = (−π2, π2)

f:(−1, 1)→B defined by f(x)=tan−1(2x1−x2) is bijection then B=

f(x)=Tan−1(2x1−x2)put x=Tanθ⇒ θ=Tan−1x=Tan−1(2Tanθ1−Tan2θ)=Tan−1(Tan2θ)= 2θ=2Tan−1x∴ f is bijection then B= Range∴ −1<x<1⇒ Tan−1(−1)<Tan−1x<Tan−1(1)⇒ −π4<Tan−1x<π2⇒ −π2<2Tan−1x<π2∴ B=Range = (−π2, π2)

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$f : (- 1, 1) \to B d e f i n e d b y f (x) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) i s b i j e c t i o n t h e n B =$

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$(\frac{- π}{2}, \frac{π}{2})$

$[- \frac{π}{2}, \frac{π}{2}]$

$[0, \frac{π}{2})$

$(0, \frac{π}{2})$

answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = {Tan}^{- 1} (\frac{2 x}{1 - x^{2}}) \\ p u t x = Tan θ \Rightarrow θ = {Tan}^{- 1} x \\ = {Tan}^{- 1} (\frac{2 Tan θ}{1 - {Tan}^{2} θ}) \\ = {Tan}^{- 1} (Tan 2 θ) \\ = 2 θ \\ = 2 {Tan}^{- 1} x \end{array}$

$\begin{array}{l} ∴ f i s b i j e c t i o n t h e n B = R a n g e \\ ∴ - 1 < x < 1 \\ \Rightarrow {Tan}^{- 1} (- 1) < {Tan}^{- 1} x < {Tan}^{- 1} (1) \\ \Rightarrow \frac{- π}{4} < {Tan}^{- 1} x < \frac{π}{2} \\ \Rightarrow \frac{- π}{2} < 2 {Tan}^{- 1} x < \frac{π}{2} \\ ∴ B = R a n g e = (\frac{- π}{2}, \frac{π}{2}) \end{array}$