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If $R$ is the set of all real numbers and $f: R-\{2\} \rightarrow R$ is defined by $f(x)=\frac{2+x}{2-x}$ for $x \in R-\{2\}$
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Verified Answer
The correct answer is:
$R-\{-1\}$
Let $y=f(x)=\frac{2+x}{2-x}$
$$
\begin{aligned}
& \Rightarrow \quad 2 y-x y=2+x \\
& \Rightarrow \quad x(1+y)=2 y-2 \\
& \Rightarrow \quad x=\frac{2(y-1)}{1+y}, y \neq-1
\end{aligned}
$$
$\therefore$ Domain of $y$ is $R-\{-1\}$.
Hence, range of $f(x)$ is $R-\{-1\}$.
$$
\begin{aligned}
& \Rightarrow \quad 2 y-x y=2+x \\
& \Rightarrow \quad x(1+y)=2 y-2 \\
& \Rightarrow \quad x=\frac{2(y-1)}{1+y}, y \neq-1
\end{aligned}
$$
$\therefore$ Domain of $y$ is $R-\{-1\}$.
Hence, range of $f(x)$ is $R-\{-1\}$.
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