Question:
$\text{Given, } f(x+y) =xf(x) +2f(y)-x-y\text{. If } f(2019)=\frac{a}{b}, \text{then } a+b\text{ = ?} $
I started solving by replacing $y\to-x$, which yields $$f(0) = xf(x) +2f(-x)$$ then, replace $x\to-x$ $$f(0)=-xf(-x)+2f(x)$$ on solving, I get, $$(x+2)f(0)=(x^2+4)f(x)$$
But, I'm stuck here. How to move on?
If $x=0$ we get $f(y)=y$ for all $y$. Pluging this in to starting equation we get $$x+y=x^2+2y-x-y$$
which should be valid for all $x,y$. Since $x^2=2x$ ony for $x=2$ and $x=0$ this FE has no solution unlesss $f$ is defined on the set $\{0,2\}$.