If the function f:R−{1, −1} →A defined by f(x)=x21−x2, is surjective, then A is equal to

f(x)=x21−x2 is surjection then A= Range of flet x21−x2=y⇒ x2=y−x2y⇒ x2(1+y)=y⇒ x2=yy+1⇒ x=yy+1 is definediff yy+1 ≥0⇒ y(y+1)≥0 and y≠ −1∴ y≥0 and y<−1∴ A=R−[−1, 0)

If the function f:R−{1, −1} →A defined by f(x)=x21−x2, is surjective, then A is equal to

f(x)=x21−x2 is surjection then A= Range of flet x21−x2=y⇒ x2=y−x2y⇒ x2(1+y)=y⇒ x2=yy+1⇒ x=yy+1 is definediff yy+1 ≥0⇒ y(y+1)≥0 and y≠ −1∴ y≥0 and y<−1∴ A=R−[−1, 0)

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$I f t h e f u n c t i o n f : R - {1, - 1} \to A d e f i n e d b y f (x) = \frac{x^{2}}{1 - x^{2}}, i s s u r j e c t i v e, t h e n A i s e q u a l t o$

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$R - {- 1}$

$[0, \infty)$

$R - (- 1, 0)$

$R - [- 1, 0)$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{x^{2}}{1 - x^{2}} i s s u r j e c t i o n t h e n A = R a n g e o f f \\ l e t \frac{x^{2}}{1 - x^{2}} = y \\ \Rightarrow x^{2} = y - x^{2} y \\ \Rightarrow x^{2} (1 + y) = y \\ \Rightarrow x^{2} = \frac{y}{y + 1} \\ \Rightarrow x = \sqrt{\frac{y}{y + 1}} i s d e f i n e d \\ i f f \frac{y}{y + 1} \geq 0 \Rightarrow y (y + 1) \geq 0 a n d y \neq - 1 \\ ∴ y \geq 0 a n d y < - 1 \\ ∴ A = R - [- 1, 0) \end{array}$