Let g(x)=∫x0f(t)dt and f(x) satisfies the equation f(x+y)=f(x)+f(y)+2xy−1 for all x, yϵR and f′(0)=2 then
g(x) increases on (0,∞)
f′(x)=limh→0f(x+h)−f(x)h
=limh→0f(x)+f(h)+2xh−1−f(x)h
=limh→0f(h)−1h+2x
[putting x=y=0 in the given condition, we obtain f(0)=1]
∴f′(x)=limh→0f(0+h)−f(0)h+2x
∴f′(x)=f′(0)+2x=2(x+1)
⇒f(x)=(x+1)2+C, Putting x=0 we obtain C=0
Thus, f(x)=(x+1)2. But g′(x)=f(x)=(x+1)2>0
So, g increase on (−∞,∞) in particular on (0,∞)