Question

已知函数f(x)=ax^2-2ax 2 b（a不等于0）在区间【2，3】

Answer 1

(1)f(x)=ax^2-2ax+2+b(a>0)
=a(x-1)^2-a+2+b
f(2)=2,f(3)=5
4a-4a+2+b=2,
9a-6a+2+b=5
a=1,b=0
(2)g(x)=x^2-2x+2-(m+1)x=x^2-(m+3)x+2
对称轴x=(m+3)/2.
[2，4]单调
(m+3)/2>=4或(m+3)/2<=2
m>=5或m<=1.

Answer 2

4a-4a+2+b=5,9a-6a+2+b=2,a=-1,b=3或4a-4a+2+b=2,9a-6a+2+b=5,a=1,b=0可知a=1,b=0,∴f(x)=x²-2x+2,g（x）=x²-（2+2^m)x+2若单调,则对称轴应该在x=2的左边或x=4的右边所以1+2^(m-1)＜2或＞4即m＜1或m＞（log23）+1