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Question Number 80861 by jagoll last updated on 07/Feb/20

for x,y ∈R given f(x)+f(2x+y)+5xy= f(3x−y)+x^2 +1 find f(10)

$${for}\:{x},{y}\:\in\mathbb{R} \\ $$$${given}\:{f}\left({x}\right)+{f}\left(\mathrm{2}{x}+{y}\right)+\mathrm{5}{xy}= \\ $$$${f}\left(\mathrm{3}{x}−{y}\right)+{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${find}\:{f}\left(\mathrm{10}\right) \\ $$

Commented by jagoll last updated on 07/Feb/20

i work by put y = 0 f(x)+f(2x)=f(3x)+x^2 +1 let 3x = t f((t/3))+f((t/2))=f(t)+(t^2 /9)+1

$${i}\:{work}\:{by}\: \\ $$$${put}\:{y}\:=\:\mathrm{0} \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{2}{x}\right)={f}\left(\mathrm{3}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${let}\:\mathrm{3}{x}\:=\:{t}\: \\ $$$${f}\left(\frac{{t}}{\mathrm{3}}\right)+{f}\left(\frac{{t}}{\mathrm{2}}\right)={f}\left({t}\right)+\frac{{t}^{\mathrm{2}} }{\mathrm{9}}+\mathrm{1} \\ $$

Commented by jagoll last updated on 07/Feb/20

my way is right?

$${my}\:{way}\:{is}\:{right}? \\ $$

Commented by ~blr237~ last updated on 07/Feb/20

yes

$${yes} \\ $$

Commented by mr W last updated on 07/Feb/20

f(x)+f(2x+y)+5xy=f(3x−y)+x^2 +1 y=0: f(x)+f(2x)=f(3x)+x^2 +1 ...(i) y=x: f(x)+f(3x)+5x^2 =f(2x)+x^2 +1 ...(ii) (i)+(ii): 2f(x)+5x^2 =2(x^2 +1) ⇒f(x)=1−((3x^2 )/2) ⇒f(10)=1−((3×10^2 )/2)=−149

$${f}\left({x}\right)+{f}\left(\mathrm{2}{x}+{y}\right)+\mathrm{5}{xy}={f}\left(\mathrm{3}{x}−{y}\right)+{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${y}=\mathrm{0}: \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{2}{x}\right)={f}\left(\mathrm{3}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1}\:\:\:...\left({i}\right) \\ $$$${y}={x}: \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{3}{x}\right)+\mathrm{5}{x}^{\mathrm{2}} ={f}\left(\mathrm{2}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{2}{f}\left({x}\right)+\mathrm{5}{x}^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{1}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{10}\right)=\mathrm{1}−\frac{\mathrm{3}×\mathrm{10}^{\mathrm{2}} }{\mathrm{2}}=−\mathrm{149} \\ $$

Commented by jagoll last updated on 07/Feb/20

thank you mr w and mr blr

$${thank}\:{you}\:{mr}\:{w}\:{and}\:{mr}\:{blr} \\ $$

Answered by ~blr237~ last updated on 07/Feb/20

taking y=0 we have f(x)+f(2x)=f(3x)+x^2 +1 (1) taking x=y we have f(x)+f(3x)+5x^2 =f(2x)+x^2 +1 (2) (1)+(2)⇒ 2f(x)=2x^2 +2−5x^2 f(x)=−(3/2)x^2 +1 f(10)=−(3/2)×100+1=−149

$${taking}\:{y}=\mathrm{0}\:\:\:{we}\:{have}\:{f}\left({x}\right)+{f}\left(\mathrm{2}{x}\right)={f}\left(\mathrm{3}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1}\:\:\:\left(\mathrm{1}\right) \\ $$$${taking}\:{x}={y}\:{we}\:{have}\:{f}\left({x}\right)+{f}\left(\mathrm{3}{x}\right)+\mathrm{5}{x}^{\mathrm{2}} ={f}\left(\mathrm{2}{x}\right)+{x}^{\mathrm{2}} +\mathrm{1}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow\:\mathrm{2}{f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}−\mathrm{5}{x}^{\mathrm{2}} \: \\ $$$${f}\left({x}\right)=−\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\mathrm{10}\right)=−\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{100}+\mathrm{1}=−\mathrm{149} \\ $$

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