if $f(x)=\frac{x-1}{x+1}$ , x∈R-{-1}, then f^{-1}(x) is equal to
$a. \frac{x-1}{x+1} b.\frac{x+1}{x-1} c.\frac{2}{1+x}$ d.Does Not exist
f(1) = 0
f(x)
$\begin{align*} &\Rightarrow\frac{y}{1} =\frac{x-1}{x+1} \\ &\Rightarrow \frac{y-1}{y+1} =\frac{(x-1)-(x+1)}{(x-1)+(x+1)} \\ &\Rightarrow\frac{y-1}{y+1} = \frac{-2}{2x} \\ &\text{x and y are dummy only} \\ &\Rightarrow f^{-1}(x) = \frac{1+x}{1-x} \\ \end{align*}$
f^{-1}(x) does not exist for x=1 so option D is correct..
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