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

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

Commented by bemath last updated on 09/Jun/20

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

Answered by bemath last updated on 09/Jun/20

$$\mathrm{putting}\mathrm{y}=\frac{1}{1-x}\Rightarrow f\left(x\right)+f\left(y\right)=2(\frac{1}{x}-y)...\left(1\right)$$$$setz=\frac{1}{1-y}thenx=\frac{1}{1-z}$$$$\Rightarrow f\left(y\right)+f\left(z\right)=2(\frac{1}{y}-z)...\left(2\right)$$$$andf\left(z\right)+f\left(x\right)=2(\frac{1}{z}-x)...\left(3\right)$$$$\left(1\right)+\left(3\right)\Rightarrow 2(f\left(x\right)+f\left(y\right)+f\left(z\right))=2(\frac{1}{x}-x)-2y+\frac{2}{z}$$$$usingthesecondrelation,\mathrm{this}$$$$\mathrm{reduces}\mathrm{to}2f\left(x\right)=2(\frac{1}{x}-x)-2(\frac{1}{y}+y)+2(\frac{1}{z}+z)$$$$nowusingy+\frac{1}{y}=1+\frac{1}{1-x}$$$$z+\frac{1}{z}=\frac{x-1}{x}+\frac{x}{x-1},\mathrm{we}\mathrm{get}$$$$\mathrm{f}\left(x\right)=\frac{x+1}{x-1}.$$

Commented by bobhans last updated on 09/Jun/20

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

Commented by bemath last updated on 09/Jun/20

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

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

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

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