Question Number 198276 by essaad last updated on 16/Oct/23
Answered by witcher3 last updated on 16/Oct/23
$$\left(\mathrm{x}=\mathrm{y}\right)\Rightarrow\mathrm{f}\left(\mathrm{2x}\right)+\mathrm{2f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{f}\left(\mathrm{2x}\right)=\mathrm{1}−\mathrm{2f}\left(\mathrm{0}\right) \\ $$$$\mathrm{x}\rightarrow\mathrm{2x}\:\mathrm{surjective}\Leftrightarrow\forall\mathrm{t}\in\mathbb{R}\:\mathrm{f}\left(\mathrm{t}\right)=\mathrm{1}−\mathrm{2f}\left(\mathrm{0}\right)\:\mathrm{constant} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 16/Oct/23
$${f}\left({x}+{y}\right)+\mathrm{2}{f}\left({x}−{y}\right)=\frac{{f}\left({x}\right)}{{f}\left({y}\right)}…..\left({i}\right) \\ $$$${x}={y}=\mathrm{0}: \\ $$$$\left({i}\right)\Rightarrow\mathrm{3}{f}\left(\mathrm{0}\right)=\frac{{f}\left(\mathrm{0}\right)}{{f}\left(\mathrm{0}\right)}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow{f}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${y}={x}: \\ $$$$\left({i}\right)\Rightarrow{f}\left(\mathrm{2}{x}\right)+\mathrm{2}{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{2}{x}\right)=\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$