If f:R→R be a differentiable function, such that f(x+2y)=f(x)+f(2y)+4xy for all x,y∈R then

f(x+2y)=f(x)+f(2y)+4xy for x,y∈R putting x=y=0, we  get f(0)=0 now, f(x+2y)=f(x)+f(2y)+4xy⇒f(x+2y)−f(x)2y=2x+f(2y)2y⇒limy→0 f(x+2y)−f(x)2y=limy→0 2x+f(2y)−f(0)2y                                 =limy→0 2x+f(2y)−02y⇒f′(x)=2x+f′(0) for all X⇒f′(1)=2+f′(0)