If the function f: R - { 1, - 1} to A defined | vedclass

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(A) Let $y = \frac{x^2}{1 - x^2}$.We want to find the range of $f(x)$.$y(1 - x^2) = x^2$$y - yx^2 = x^2$$y = x^2(1 + y)$$x^2 = \frac{y}{1 + y}$.Since

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$R - [-1, 0)$

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(A) Let $y = \frac{x^2}{1 - x^2}$.We want to find the range of $f(x)$.$y(1 - x^2) = x^2$$y - yx^2 = x^2$$y = x^2(1 + y)$$x^2 = \frac{y}{1 + y}$.Since $x^2 \ge 0$,we must have $\frac{y}{1 + y} \ge 0$.Using the sign scheme for the inequality,we get $y \in (-\infty, -1) \cup [0, \infty)$.Also,$x^2 = \frac{y}{1 + y} 
eq 1$ (since $x 
eq \pm 1$),so $\frac{y}{1 + y} 
eq 1 \implies y 
eq y + 1$,which is always true.Thus,the range is $R - [-1, 0)$.For the function to be surjective,the codomain $A$ must be equal to the range.Therefore,$A = R - [-1, 0)$.

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

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