If the function is defined by $f(x+y) = kxy+f(x)+ 2y^2$ , where k is some constant for all real numbers and $g(x-g(y))=g(g(y))+xg(y)+g(x)-1$ for all $x,y$ belongs to real numbers then find range of $f(x) - g(\dfrac{1}{x})$ .....
My Approach : First of all to find $f(x)$ I substituted $x=0$ and got $f(y) = f(0)+2y^2$ and as value of $f(0)=0$ so i got $f(y)=2y^2$ but i am not getting how to simplify $g(x)$....
Please help..
Note that
\begin{align} g(g(y))&=g(0-(-g(y)))\\&=g(-g(y))+g(0)-1. \end{align}
Similarly, we have
\begin{align} g(0-g(y))=g(g(y))+g(0)-1. \end{align}
From these equations, we get $g(0)=1$. Next, put $y=0$ and use $g(0)=1$ to get
\begin{align} g(x-1)=g(1)+x+g(x)-1. \end{align}
Putting $x=1$, we get $g(1)=\frac{1}{2}$. Thus, we have
\begin{align} g(x-1)&=x+g(x)-\frac{1}{2}\\&\Rightarrow g(x)-g(x-1)=\frac{1}{2}-x. \end{align}
With this, we see that the form of $g(x)$ should be that of a quadratic term, followed by any periodic function of period $1$. Assume $g(x)=ax^{2}+bx+c+p(x)$, where $a,b,c\in \mathbb{R}$, and $p(x)$ is a periodic function of period $1$, i.e., $p(x-1)=p(x)$. Without loss of generality, assume $p(1)=0$ (if not, use $p(x)-p(1)$ in place of $p(x)$).
With $g(0)=1$, we see $c=1$. Also,
\begin{align} ax^{2}+bx+1-\left(a(x-1)^{2}+b(x-1)+1\right)=\frac{1}{2}-x. \end{align}
Equating the coefficients of $x^{2}$, $x$ and constant on both sides, we get $a=-\frac{1}{2}$ and $b=0$.
Thus, $g(x)=-\frac{1}{2}x^{2}+1+p(x)=\frac{2-x^{2}}{2}+p(x)$.