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

Determine all function f:R/{0,1}→R  satisfying the functional relation  f(x) + f((1/(1−x))) = ((2(1−2x))/(x(1−x))) , x≠0, x≠1

$$\mathrm{Determine}\:\mathrm{all}\:\mathrm{function}\:\mathrm{f}:\mathrm{R}/\left\{\mathrm{0},\mathrm{1}\right\}\rightarrow\mathrm{R} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{functional}\:\mathrm{relation} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2x}\right)}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}\right)}\:,\:\mathrm{x}\neq\mathrm{0},\:\mathrm{x}\neq\mathrm{1} \\ $$

Commented by bemath last updated on 09/Jun/20

nice question

$$\mathrm{nice}\:\mathrm{question} \\ $$

Answered by bemath last updated on 09/Jun/20

putting y=(1/(1−x)) ⇒f(x)+f(y)=2((1/x)−y)...(1)  set z = (1/(1−y)) then x = (1/(1−z))  ⇒f(y)+f(z)=2((1/y)−z)...(2)  and f(z)+f(x)=2((1/z)−x)...(3)  (1)+(3)⇒ 2(f(x)+f(y)+f(z))= 2((1/x)−x)−2y+(2/z)  using the second relation, this  reduces to 2f(x)=2((1/x)−x)−2((1/y)+y)+2((1/z)+z)  now using y+(1/y)=1+(1/(1−x))  z+(1/z)=((x−1)/x)+(x/(x−1)) , we get  f(x) = ((x+1)/(x−1)) .

$$\mathrm{putting}\:\mathrm{y}=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{f}\left({x}\right)+{f}\left({y}\right)=\mathrm{2}\left(\frac{\mathrm{1}}{{x}}−{y}\right)...\left(\mathrm{1}\right) \\ $$$${set}\:{z}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{y}}\:{then}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{1}−{z}} \\ $$$$\Rightarrow{f}\left({y}\right)+{f}\left({z}\right)=\mathrm{2}\left(\frac{\mathrm{1}}{{y}}−{z}\right)...\left(\mathrm{2}\right) \\ $$$${and}\:{f}\left({z}\right)+{f}\left({x}\right)=\mathrm{2}\left(\frac{\mathrm{1}}{{z}}−{x}\right)...\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{3}\right)\Rightarrow\:\mathrm{2}\left({f}\left({x}\right)+{f}\left({y}\right)+{f}\left({z}\right)\right)=\:\mathrm{2}\left(\frac{\mathrm{1}}{{x}}−{x}\right)−\mathrm{2}{y}+\frac{\mathrm{2}}{{z}} \\ $$$${using}\:{the}\:{second}\:{relation},\:\mathrm{this} \\ $$$$\mathrm{reduces}\:\mathrm{to}\:\mathrm{2}{f}\left({x}\right)=\mathrm{2}\left(\frac{\mathrm{1}}{{x}}−{x}\right)−\mathrm{2}\left(\frac{\mathrm{1}}{{y}}+{y}\right)+\mathrm{2}\left(\frac{\mathrm{1}}{{z}}+{z}\right) \\ $$$${now}\:{using}\:{y}+\frac{\mathrm{1}}{{y}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${z}+\frac{\mathrm{1}}{{z}}=\frac{{x}−\mathrm{1}}{{x}}+\frac{{x}}{{x}−\mathrm{1}}\:,\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{f}\left({x}\right)\:=\:\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\:. \\ $$

Commented by bobhans last updated on 09/Jun/20

check. f((1/(1−x)))=(((1/(1−x))+1)/((1/(1−x))−1)) = ((1+1−x)/(1−1+x)) =((2−x)/x)  ⇔ f(x)+f((1/(1−x))) = ((x+1)/(x−1))+((2−x)/x) = ((x^2 +x+(x−1)(2−x))/(x(x−1)))  = ((x^2 +x−x^2 +3x−2)/(x(x−1))) = ((4x−2)/(x(x−1)))=((2(2x−1))/(x(x−1)))  (true)...

$$\mathrm{check}.\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)=\frac{\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}−\mathrm{1}}\:=\:\frac{\mathrm{1}+\mathrm{1}−\mathrm{x}}{\mathrm{1}−\mathrm{1}+\mathrm{x}}\:=\frac{\mathrm{2}−\mathrm{x}}{\mathrm{x}} \\ $$$$\Leftrightarrow\:\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}+\frac{\mathrm{2}−\mathrm{x}}{\mathrm{x}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{2}−\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{2}}{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)}\:=\:\frac{\mathrm{4x}−\mathrm{2}}{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)}=\frac{\mathrm{2}\left(\mathrm{2x}−\mathrm{1}\right)}{\mathrm{x}\left(\mathrm{x}−\mathrm{1}\right)}\:\:\left(\mathrm{true}\right)... \\ $$

Commented by bemath last updated on 09/Jun/20

maybe sir w have a short method

$$\mathrm{maybe}\:\mathrm{sir}\:\mathrm{w}\:\mathrm{have}\:\mathrm{a}\:\mathrm{short}\:\mathrm{method} \\ $$

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

very.....coollll

$$\mathrm{very}.....\mathrm{coollll} \\ $$

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