Question

设f(x)是定义在R上的函数,且满足f(0)=1,并且对任意实数x,y有f(x-y)=f(x)-y(2x-y+1),1.

Answer 1

令x=0有f(-y)=f(0)-y(-y+1)=1+y²-y,则有f(y)=y²+y+1即f(x)=x²+x+1

g(x)=x²+x+1-(4a+3)+a²=x²+x+a²-4a-2=(x+1/2)²+a²-4a-9/4,因为x∈[0,1]所以g(x)max=g(1)=a²-4a即h(a)=a²-4a=(a-2)²-2,于是值域[-2,+∞)