Question

A continuous function $f\left(x\right)$ on $R\to R$ satisfies the relation $f\left(x\right)+f(2x+y)+5xy=f(3x-y)+2x^2+1$ for $\forall x,y\in R$. Then find $f\left(x\right)$.

Answer 1

$f\left(x\right)+f(2x+y)+5xy=f(3x-y)+2x^2+1$ $yϵR$

$x=y=0\Rightarrow f\left(0\right)+f\left(0\right)+0=f\left(0\right)+0+1$

$f\left(0\right)=1$................................. (1)

$x=0,y=y\Rightarrow f\left(0\right)+f\left(y\right)+0=f(-y)+0+1$

$\Rightarrow 1+f\left(y\right)=f(-y)+1$

$\Rightarrow f\left(y\right)=f(-y)$ $\forall yϵR$

$\Rightarrow$ f is even .......................(2)

$x=2y\Rightarrow f\left(2y\right)+f\left(5y\right)+10y^2=f\left(5y\right)+8y^2+1$

$\Rightarrow f\left(2y\right)=-2y+1$

$f\left(x\right)=-2(\frac{x}{2}{)}^2+1=\frac{-x^2}{2}+1$