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

$$forx,y\in \mathbb{R}$$$$givenf\left(x\right)+f(2x+y)+5xy=$$$$f(3x-y)+x^2+1$$$$findf\left(10\right)$$

Commented by jagoll last updated on 07/Feb/20

$$iworkby$$$$puty=0$$$$f\left(x\right)+f\left(2x\right)=f\left(3x\right)+x^2+1$$$$let3x=t$$$$f\left(\frac{t}{3}\right)+f\left(\frac{t}{2}\right)=f\left(t\right)+\frac{t^2}{9}+1$$

Commented by jagoll last updated on 07/Feb/20

$$mywayisright?$$

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

$$yes$$

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

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

Commented by jagoll last updated on 07/Feb/20

$$thankyoumrwandmrblr$$

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

$$takingy=0wehavef\left(x\right)+f\left(2x\right)=f\left(3x\right)+x^2+1\left(1\right)$$$$takingx=ywehavef\left(x\right)+f\left(3x\right)+5x^2=f\left(2x\right)+x^2+1\left(2\right)$$$$\left(1\right)+\left(2\right)\Rightarrow 2f\left(x\right)=2x^2+2-5x^2$$$$f\left(x\right)=-\frac{3}{2}{x}^2+1$$$$f\left(10\right)=-\frac{3}{2}\times 100+1=-149$$

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