Question Number 194896 by cortano12 last updated on 19/Jul/23
$$\:\:\:\:\:\:\:\cancel{ } \\ $$
Answered by AST last updated on 19/Jul/23
$$\left({x}^{\mathrm{2}} −{f}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} {f}\left({x}\right)=\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} {f}\left({x}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{f}\left({x}\right)=\underset{−} {+}\mathrm{1}\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}} \overset{−} {+}\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow{f}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$