Question Number 226134 by fantastic2 last updated on 20/Nov/25
$${Q}\mathrm{226015} \\ $$
Answered by fantastic2 last updated on 20/Nov/25
$${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}\:…{i} \\ $$$${put}\:{x}=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\left.{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}…\:{ii} \\ $$$${put}\:{x}=\frac{{x}−\mathrm{1}}{{x}}\:{in}\:{i} \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+{f}\left({x}\right)=\frac{{x}−\mathrm{1}}{{x}}\:…{iii} \\ $$$$\left.{ii}\:−{iii}\right)\:{then}\:{iv}+{i} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{{x}^{\mathrm{3}} −{x}+\mathrm{1}}{\mathrm{2}{x}\left(\mathrm{1}−{x}\right)} \\ $$