Let f:\mathbb{R}-{0,1} o \mathbb{R} be a function such that f(x)+f\left(\frac{1}{1-x} ight)=1+x. Then f(2) is equal to

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Let f:\mathbb{R}-{0,1}	o \mathbb{R} be a function such that f(x)+f\left(\frac{1}{1-x}ight)=1+x. Then f(2) is equal to

Accepted Answer

Answer: A. \frac{9}{4} \begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{1-\mathrm{x}}ight)=1+\mathrm{x} \\ & \mathrm{x}=2 \Rightarrow \mathrm{f}(2)+\mathrm{f}(-1)=3 ........(1) \\ & \mathrm{x}=-1 \Rightarrow \mathrm{f}(-1)+\mathrm{f}\left(\frac{1}{2}ight)=0 .........(2) \\ & \mathrm{x}=\frac{1}{2} \Rightarrow \mathrm{f}\left(\frac{1}{2}ight)+\mathrm{f}(2)=\frac{3}{2} ........(3) \\ & (1)+(3)-(2) \Rightarrow 2 \mathrm{f}(2)=\frac{9}{2} \\ & 	herefore \mathrm{f}(2)=\frac{9}{4}\end{aligned}

Let $$f:\mathbb{R}-{0,1}\to \mathbb{R}$$ be a function such that $$f(x)+f\left(\frac{1}{1-x}\right)=1+x$$. Then $$f(2)$$ is equal to

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