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Question Number 97648 by bobhans last updated on 09/Jun/20

Determine all function f : R / {0, 1} \to R

satisfying the functional relation

f (x) + f (\frac{1}{1 - x}) = \frac{2 (1 - 2 x)}{x (1 - x)}, x \neq 0, x \neq 1

Commented by bemath last updated on 09/Jun/20

nice question

Answered by bemath last updated on 09/Jun/20

putting y = \frac{1}{1 - x} \Rightarrow f (x) + f (y) = 2 (\frac{1}{x} - y) \dots (1)

s e t z = \frac{1}{1 - y} t h e n x = \frac{1}{1 - z}

\Rightarrow f (y) + f (z) = 2 (\frac{1}{y} - z) \dots (2)

a n d f (z) + f (x) = 2 (\frac{1}{z} - x) \dots (3)

(1) + (3) \Rightarrow 2 (f (x) + f (y) + f (z)) = 2 (\frac{1}{x} - x) - 2 y + \frac{2}{z}

u s i n g t h e s e c o n d r e l a t i o n, this

reduces to 2 f (x) = 2 (\frac{1}{x} - x) - 2 (\frac{1}{y} + y) + 2 (\frac{1}{z} + z)

n o w u s i n g y + \frac{1}{y} = 1 + \frac{1}{1 - x}

z + \frac{1}{z} = \frac{x - 1}{x} + \frac{x}{x - 1}, we get

f (x) = \frac{x + 1}{x - 1} .

Commented by bobhans last updated on 09/Jun/20

check . f (\frac{1}{1 - x}) = \frac{\frac{1}{1 - x} + 1}{\frac{1}{1 - x} - 1} = \frac{1 + 1 - x}{1 - 1 + x} = \frac{2 - x}{x}

\Leftrightarrow f (x) + f (\frac{1}{1 - x}) = \frac{x + 1}{x - 1} + \frac{2 - x}{x} = \frac{x^{2} + x + (x - 1) (2 - x)}{x (x - 1)}

= \frac{x^{2} + x - x^{2} + 3 x - 2}{x (x - 1)} = \frac{4 x - 2}{x (x - 1)} = \frac{2 (2 x - 1)}{x (x - 1)} (true) \dots

Commented by bemath last updated on 09/Jun/20

maybe sir w have a short method

Commented by john santu last updated on 09/Jun/20

very \dots . . coollll