Question

函数f(x)=ax ²+bx+1(a＞0)

Answer 1

a-b+1=0,(4a-b^2)/(4a)>=0,

b=a+1带入上式有4a-(a+1)^2>=0

解得a=1,b=2

g(x)=x^2+(2-k)x+1,对x∈[-1,1],

(2-k)/(-2)>=1或(2-k)/(-2)=4或k