Question Number 191151 by TUN last updated on 19/Apr/23
$${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} \\ $$$$=>{f}\left({x}\right)=¿ \\ $$
Answered by Rasheed.Sindhi last updated on 19/Apr/23
$${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} …….\left({i}\right) \\ $$$${Replacing}\:{x}\:{by}\:\mathrm{1}−{x}: \\ $$$${f}\left(\mathrm{1}−{x}\right)+{f}\left(\:\mathrm{1}−\left(\mathrm{1}−{x}\right)\:\right)=\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}−{x}\right)+{f}\left(\:{x}\:\right)=\left(\mathrm{1}−{x}\right)^{\mathrm{2}} ……\left({ii}\right) \\ $$$$\left({i}\right)\:\:\:\&\:\:\:\left({ii}\right):\:{x}^{\mathrm{2}} =\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2}{x}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}\:{is}\:{constant} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{2}{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Commented by mr W last updated on 19/Apr/23
$${but}\:{in}\:{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} \:{x}\:{is}\:{not}\:{a} \\ $$$${constant}\:{but}\:{a}\:{variable}\:\left(\in{R}\right). \\ $$$${so}\:{i}\:{think}\:{the}\:{answer}\:{is}\:{that} \\ $$$${f}\left({x}\right)\:{satisfying}\:{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)={x}^{\mathrm{2}} \: \\ $$$${doesn}'{t}\:{exist}. \\ $$
Commented by Rasheed.Sindhi last updated on 19/Apr/23
$$\mathrm{Very}\:\mathrm{right}\:\mathrm{sir}!\:\boldsymbol{\mathrm{Thanks}}! \\ $$