Question Number 80861 by jagoll last updated on 07/Feb/20
$${for}\:{x},{y}\:\in\mathbb{R} \\ $$$${given}\:{f}\left({x}\right)+{f}\left(\mathrm{2}{x}+{y}\right)+\mathrm{5}{xy}= \\ $$$${f}\left(\mathrm{3}{x}−{y}\right)+{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${find}\:{f}\left(\mathrm{10}\right) \\ $$
Commented by jagoll last updated on 07/Feb/20
$${i}\:{work}\:{by}\: \\ $$$${put}\:{y}\:=\:\mathrm{0} \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{2}{x}\right)={f}\left(\mathrm{3}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${let}\:\mathrm{3}{x}\:=\:{t}\: \\ $$$${f}\left(\frac{{t}}{\mathrm{3}}\right)+{f}\left(\frac{{t}}{\mathrm{2}}\right)={f}\left({t}\right)+\frac{{t}^{\mathrm{2}} }{\mathrm{9}}+\mathrm{1} \\ $$
Commented by jagoll last updated on 07/Feb/20
$${my}\:{way}\:{is}\:{right}? \\ $$
Commented by ~blr237~ last updated on 07/Feb/20
$${yes} \\ $$
Commented by mr W last updated on 07/Feb/20
$${f}\left({x}\right)+{f}\left(\mathrm{2}{x}+{y}\right)+\mathrm{5}{xy}={f}\left(\mathrm{3}{x}−{y}\right)+{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${y}=\mathrm{0}: \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{2}{x}\right)={f}\left(\mathrm{3}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1}\:\:\:…\left({i}\right) \\ $$$${y}={x}: \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{3}{x}\right)+\mathrm{5}{x}^{\mathrm{2}} ={f}\left(\mathrm{2}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}{f}\left({x}\right)+\mathrm{5}{x}^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{1}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{10}\right)=\mathrm{1}−\frac{\mathrm{3}×\mathrm{10}^{\mathrm{2}} }{\mathrm{2}}=−\mathrm{149} \\ $$
Commented by jagoll last updated on 07/Feb/20
$${thank}\:{you}\:{mr}\:{w}\:{and}\:{mr}\:{blr} \\ $$
Answered by ~blr237~ last updated on 07/Feb/20
$${taking}\:{y}=\mathrm{0}\:\:\:{we}\:{have}\:{f}\left({x}\right)+{f}\left(\mathrm{2}{x}\right)={f}\left(\mathrm{3}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1}\:\:\:\left(\mathrm{1}\right) \\ $$$${taking}\:{x}={y}\:{we}\:{have}\:{f}\left({x}\right)+{f}\left(\mathrm{3}{x}\right)+\mathrm{5}{x}^{\mathrm{2}} ={f}\left(\mathrm{2}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow\:\mathrm{2}{f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}−\mathrm{5}{x}^{\mathrm{2}} \: \\ $$$${f}\left({x}\right)=−\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\mathrm{10}\right)=−\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{100}+\mathrm{1}=−\mathrm{149} \\ $$