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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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Verified Answer
The correct answer is:
$R-[-1,0)$
For $f(x)$ to be a surjective function $A=\operatorname{range}$ of $f(x)$
now, $y=\frac{x^2}{1-x^2}$
$$
\begin{aligned}
& \Rightarrow x^2=\frac{y}{1+y} \\
& \Rightarrow x=\sqrt{\frac{y}{1+y}}
\end{aligned}
$$
for $x$ to be real $\frac{y}{1+y} \geq 0$
$$
\begin{aligned}
& \Rightarrow y \in(-\infty,-1) \cup[0, \infty) \\
& \Rightarrow \text { Range of } f(x) \text { is } R \sim[-1,0) \\
& \Rightarrow A \text { is } R \sim[-1,0)
\end{aligned}
$$
now, $y=\frac{x^2}{1-x^2}$
$$
\begin{aligned}
& \Rightarrow x^2=\frac{y}{1+y} \\
& \Rightarrow x=\sqrt{\frac{y}{1+y}}
\end{aligned}
$$
for $x$ to be real $\frac{y}{1+y} \geq 0$
$$
\begin{aligned}
& \Rightarrow y \in(-\infty,-1) \cup[0, \infty) \\
& \Rightarrow \text { Range of } f(x) \text { is } R \sim[-1,0) \\
& \Rightarrow A \text { is } R \sim[-1,0)
\end{aligned}
$$
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