Question

已知集合A=﹛x||x-1|<2﹜,集合B=﹛x|x²-bx+a<0,其中a,b∈R﹜. ﹙1﹚若A=B,求a,b的值；

﹙2﹚若b=a+1,且A∪B=A，求a的取值范围
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Answer 1

A={x||x-1|<2}
={ x| -1<x<3 }
B={x|x²-bx+a<0 a,b∈R }
={ x| [b-√(b^2-4a)]/2 <x < [b+√(b^2-4a)]/2 }
(1)
A=B
=> -1 = [b-√(b^2-4a)]/2 and 3=[b+√(b^2-4a)]/2
=> b-√(b^2-4a) = -2 (1) and
b +√(b^2-4a) = 6 (2)
(2) +(1)
2b=4
b =2
a=-3
(2)
b=a+1

B={x|x²-(a+1)x+a<0 a,b∈R }

A∪B=A

=> B is subset of A

case 1: if a>1
B={ x| 1 <x < a }
solution for case 1: 1<a ≤3

case 2: if a<1
B={ x| a <x < 1 }
solution for case 2: -3≤a <1

case 3: if a=1, B=Φ
=> B is subset of A

a的取值范围:

case 1 or case 2 or case 3
1<a ≤3 or -3≤a <1 or a=1
ie -3≤a≤3