C= 1 from f(0)= 1。
And f(x+ 1)-f(x)=2x.
2ax+a+b=2x
∴a= 1 b=- 1
∴f(x)=x^2-x+ 1
(2)∫ inequality f (x) > 2x+m holds for any x belonging to-1.
The inequality f (x)-2x > m holds for any x that belongs to-1.
∴m<; [f(x)-2x] minimum value
∵[f(x)-2x]min=[x^2-3x+ 1]min=- 1
∴m<; - 1
2.( 1) solution: let x=y=0, then there is f(0+0)=f(0)+f(0).
∴f(0)=0
(2) Prove that if x=-y, then f(x-x)=f(x)+f(-x).
∴ f(-x)=-f(x)
(3) Solution: ∫f( 1)= 1
∴f(2)=f( 1+ 1)=f( 1)+f( 1)=2
∴f(a- 1)+2=f(a- 1)+f(2)=f(a+ 1)
∫ f (2a) > f (a-1)+2 = f (a+1), and the function f(x) is a decreasing function on R.
∴2a>; a+ 1
∴a>; 1