f is differentiable function such that$f(x + y) - f(x) = f(y) + 2xy$$x,y \in R$$f(x) = ?$

We have, $f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{2xh + f(h)}}{h}$

$ = 2x + \mathop {\lim }\limits_{h \to 0} \frac{{f(h)}}{h}$

Putting y=0 in the original equation yields f(0) = 0. So, the limit is of $\frac{0}{0}$ type.

$ = 2x + \mathop {\lim }\limits_{h \to 0} f'(h)$ [Using L.H. Rule]

$ = 2x + f'(0)$

Now, $f(x) = \int {2x + f'(0)dx} $

$f(x) = {x^2} + f'(0)x + C$

$\because f(0) = 0,C = 0$

$\therefore f(x) = {x^2} + f'(0)x = {x^2} + Bx$ where B is some constant.