A={x|(x-2)[x-(3a+1)]<0}
case 1: 3a+1<2 ie a< 1/3
A={x|(x-2)[x-(3a+1)]<0} = { x| 3a+1<x<2 }
a<1/3
(a-1)^2 >0
=> a^2+1 >2a
B={x| (x-2a)/[x-(a^2+1)]<0} = { x| 2a<x<a^2+1 }
A∩B=B
=> B is subset of A
2a≤3a+1 and a^2+1≥2
a≥-1 and "a≥1 or a≤-1"
a≥1 or a=-1
solution for case 1 : a=-1
case 2: a=1/3
A={x|(x-2)[x-(3a+1)]<0} ={ x| (x-2)^2<0 } =Φ
a=1/3
B={x| (x-2a)/[x-(a^2+1)]<0} = { x| (x- 2/3)/(x- 10/9) < 0 } = { x| 2/3<x<10/9 }
A∩B≠B
舍去： case 2:
case 3: a>1/3
A={x|(x-2)[x-(3a+1)]<0} = { x| 2<x<3a+1 }
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a=1
B={x| (x-2a)/[x-(a^2+1)]<0} = { x| (x-2)/(x-2) < 0 } = Φ
A∩B=B
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a> 1/3 and a≠1
(a-1)^2 >0
=> a^2+1 >2a
B={x| (x-2a)/[x-(a^2+1)]<0} = { x| 2a<x<a^2+1 }
A∩B=B
=> B is subset of A
2a≤2 and a^2+1≥3a+1
a≤1 and a^2-3a≥0
a≤1 and "a≥3 or a≤0"
a≤0
solution for case 3: a=1
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A∩B=B
=> case 1 or case 2 or case 3
=> a=1 or -1
