Question

高中数学，如图第二小题

Answer 1

f(x)=x^2-ax+b

（1）a/2>=1,a>=2

  (2) 分三种情况

1. a/2>=b函数在［0 b］单调递减

   f(0)<=6,b<=6

   f(b)>=2,b^2-ab+b>=2,b(b-a+1)>=2,因为a>=2b,b^2-b+2<=0,无解

2. a/2<=0函数在［0 b］单调递增

   f(0)>=2,b>=2

   f(b)<=6,b^2-ab+b<=6,b(b-a+1)<=6,a<0,b(b+1)<=6,b^2+b-6<=6,(b+3)(b-2)<=0,b在[-3,2]

   综上b最大为2，a为0

3. b>a/2>0函数在a/2处有极小值

   f(b)<=6，b(b-a+1)<=6,a>=b-6/b+1

   f(0)<=6，b<=6

   f(a/2)>=2,a^2/4-a^2/2+b>=2,b-a^2/4>=2, a<=2sqrt(b-2)

    通过画图(a>=b-6/b+1和a<=2sqrt(b-2))发现b＝3，a＝2

综上所述b最大值为3，此时的a为2

Answer 2

发图过来啊