If the function f: R-{-1, 1} rightarrow A def | vedclass

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(A) For the function $f(x)$ to be surjective,the codomain $A$ must be equal to the range of $f(x)$.Let $y = \frac{x^2}{1-x^2}$.Rearranging for $x^2$,w

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$R-[-1, 0)$

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(A) For the function $f(x)$ to be surjective,the codomain $A$ must be equal to the range of $f(x)$.Let $y = \frac{x^2}{1-x^2}$.Rearranging for $x^2$,we get $y(1-x^2) = x^2$,which implies $y - yx^2 = x^2$,so $y = x^2(1+y)$.Thus,$x^2 = \frac{y}{1+y}$.Since $x^2 \geq 0$ and $x 
eq \pm 1$,we must have $\frac{y}{1+y} \geq 0$ and $x^2 
eq 1$ (which implies $\frac{y}{1+y} 
eq 1$,so $y 
eq 1+y$,which is always true for any finite $y$).The inequality $\frac{y}{1+y} \geq 0$ holds when $y \in (-\infty, -1) \cup [0, \infty)$.Therefore,the range of $f(x)$ is $(-\infty, -1) \cup [0, \infty)$,which can be written as $R - [-1, 0)$.Hence,$A = R - [-1, 0)$.

If the function $f: R-\{-1, 1\} \rightarrow A$ defined by $f(x) = \frac{x^2}{1-x^2}$ is surjective,then $A$ is equal to

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