A function $f$ defined on $\mathbb R^2$ satisfying $f(x+t,y) = f(x,y) + ty$ and $f(x,t+y) = f(x,y) + tx$

Question

Suppose that a function $f$ defined on $\mathbb R^2$ satisfies the following conditions:

$ f(x+t,y) = f(x,y) + ty \ ; \ f(x,t+y) = f(x,y) + tx \ ; \ f(0,0) = K$.

Then $\forall \ x,y \in \mathbb R, f(x,y) = $

$ \ (a) \ K (x+y) $
$ \ (b) \ K - xy $
$ \ (c) \ K + xy $
$ \ (d) $ None of these

Attempt: I tried a number of ways to separate $f(x,y)$ from the two equations given firstly, but couldn't reach a conclusive step.

Could someone please give me a hint on how to move forward with this problem.

Thank you for your help.

Answer 1

As said above,

$$K = f(x+(-x),\ y+(-y)) = f(x,y + (-y)) - x(y-y)$$ $$=f(x,y)-xy-0$$ $$f(x,y) = K+xy$$

So, option C is correct.

Answer 2

Hint: What is $f(x-x,y-y)$? You can use the first two statements to simplify it one way and the last statement to simplify it another way.

Answer 3

$$\begin{align}f(x,y)&=f(\color{red}0+\color{green}x,\color{blue}y)=f(\color{red}0,\color{blue}y)+\color{green}x\color{blue}y&\text{i.e., $x=\color{red}0,t=\color{green}x,y=\color{blue}y$}\\&=f(\color{blue}0,\color{red}0+\color{green}y)+xy=f(\color{blue}0,\color{red}0)+\color{green}y\color{blue}0+xy&\text{i.e., $y=\color{red}0,t=\color{green}y, x=\color{blue}0$}\\&=K+xy \end{align}$$