Let f : R -{0,1} rightarrow R be a function s | vedclass

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Question

$f ( x )+ f \left(\frac{1}{1- x }\right)=1+ x$

$x =2 \Rightarrow f (2)+ f (-1)=3$

$x =-1 \Rightarrow f (-1)+ f \left(\frac{1}{2}\right)=0$

$x

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

$f ( x )+ f \left(\frac{1}{1- x }ight)=1+ x$

$x =2 \Rightarrow f (2)+ f (-1)=3$

$x =-1 \Rightarrow f (-1)+ f \left(\frac{1}{2}ight)=0$

$x =\frac{1}{2} \Rightarrow f \left(\frac{1}{2}ight)+ f (2)=\frac{3}{2}$

$(1)+(3)-(2) \Rightarrow 2 f (2)=\frac{9}{2}$

$	herefore f (2)=\frac{9}{4}$

Let $f : R -\{0,1\} \rightarrow R$ 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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