For $\mathrm{x} \in \mathrm{R}-\{0,1\}$, let $\mathrm{f}_{1}(\mathrm{x})=\frac{1}{\mathrm{x}}$
$\mathrm{f}_{2}(\mathrm{x})=1-\mathrm{x}$ and $\mathrm{f}_{3}(\mathrm{x})=\frac{1}{1-\mathrm{x}}$ be three
given functions. If a function, $\mathrm{J}(\mathrm{x})$ satisfies $\left(f_{2}{ }^{\circ}{ }^{\circ} \mathrm{f}_{1}\right)(\mathrm{x})=\mathrm{f}_{3}(\mathrm{x})$ then $\mathrm{J}(\mathrm{x})$ is equal to :-
Correct Option: 1
Given $\mathrm{f}_{1}(\mathrm{x})=\frac{1}{\mathrm{x}}, \mathrm{f}_{2}(\mathrm{x})=1-\mathrm{x}$ and $\mathrm{f}_{3}(\mathrm{x})=\frac{1}{1-\mathrm{x}}$
$\left(\mathrm{f}_{2} \circ \mathrm{Jof} \mathrm{f}_{1}\right)(\mathrm{x})=\mathrm{f}_{3}(\mathrm{x})$
$\mathrm{f}_{2} \circ\left(\mathrm{J}\left(\mathrm{f}_{1}(\mathrm{x})\right)\right)=\mathrm{f}_{3}(\mathrm{x})$
$f_{2} \circ\left(J\left(\frac{1}{x}\right)\right)=\frac{1}{1-x}$
$1-J\left(\frac{1}{x}\right)=\frac{1}{1-x}$
$J\left(\frac{1}{x}\right)=1-\frac{1}{1-x}=\frac{-x}{1-x}=\frac{x}{x-1}$
Now $x \rightarrow \frac{1}{x}$
$J(x)=\frac{\frac{1}{x}}{\frac{1}{x}-1}=\frac{1}{1-x}=f_{3}(x)$
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