Question Number 7781 by 314159 last updated on 15/Sep/16
Commented by prakash jain last updated on 15/Sep/16
$${f}\left(\mathrm{1}\right)=\mathrm{2} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{8} \\ $$$${f}\left({x}+{y}\right)−{kxy}={f}\left({x}\right)+\mathrm{2}{y}^{\mathrm{2}} \\ $$$${x}=\mathrm{0} \\ $$$${f}\left({y}\right)={f}\left(\mathrm{0}\right)+\mathrm{2}{y}^{\mathrm{2}} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow\mathrm{2}={f}\left(\mathrm{0}\right)+\mathrm{2}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({y}\right)=\mathrm{2}{y}^{\mathrm{2}} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} \\ $$$${f}\left({x}+{y}\right)=\mathrm{2}\left({x}+{y}\right)^{\mathrm{2}} \\ $$$${f}\left({x}+{y}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}{xy} \\ $$$${f}\left({x}+{y}\right)−\mathrm{4}{xy}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} \\ $$$${f}\left({x}+{y}\right)−\mathrm{4}{xy}={f}\left({x}\right)+\mathrm{2}{y}^{\mathrm{2}} \\ $$$${k}=\mathrm{4} \\ $$$${f}\left({x}+{y}\right){f}\left(\frac{\mathrm{1}}{{x}+{y}}\right)=\mathrm{2}\left({x}+{y}\right)^{\mathrm{2}} ×\mathrm{2}\frac{\mathrm{1}}{\left({x}+{y}\right)^{\mathrm{2}} }=\mathrm{4}={k} \\ $$