Menu Close

for-x-y-R-given-f-x-f-2x-y-5xy-f-3x-y-x-2-1-find-f-10-




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

Leave a Reply

Your email address will not be published. Required fields are marked *