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From the Indian National Mathematics Olympiad 1992:

Determine all functions $f: \mathbb R -[0,1] \rightarrow \mathbb R$, satisfying the functional relation:

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

where $x$ is a real number and not equal to $0$ or $1$

buzaku
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    including your attempt increases the chances of more replies – Vikram Jan 31 '15 at 11:12
  • Well, yes, but I have spent some 3 days on this, and have nothing concrete to take home except f(2) + f(1/2) = 0. Also, I am just circling around functions which are not defined at x = 0 and x = 1, and looking at the right hand side these functions might have x(1 - x) in the denominator. Basically, I am quite comprehensively clueless about what to do! – buzaku Feb 01 '15 at 18:30

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First plug in $$x=\frac{1}{1-x}$$ Then plug in $$x=\frac{x-1}{x}$$ You'll get a system of 3 equations which you can add and substract to get $f(x)$,since on the olympiad you have much time I'll leave the replacing of the x's part to you.

kingW3
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  • Yes, we can indeed solve for f(x) using the above substitutions. Hmm... I was completely along a wrong road! – buzaku Feb 02 '15 at 03:27