设集合A={x|x^2-3x+2=0},B={x|x^2+2(a+1)x+(a^2-5)-0}1.若A∩B={2},求实数a的值；2.若A∪B=A,求实数a的取值范围

Question

设集合A={x|x^2-3x+2=0},B={x|x^2+2(a+1)x+(a^2-5)-0}
1.若A∩B={2},求实数a的值；
2.若A∪B=A,求实数a的取值范围

Answer 1

1
x²-3x+2=0
(x-1)(x-2)=0
解得x=1或x=2
所以A={1,2}
因为A∩B={2},
所以2∈A,2∈B
故2²+2(a+1)×2+(a²-5)=0
即a²+4a+3=0
(a+1)(a+3)=0
解得a=-1或a=-3
①当a=-1时,B={2,-2}此时A∩B={2},满足
②当=-3时,B={2},此时A∩B={2,},满足
综上：a=-1或-3
2、
由1知,当a=-3时,B={2},满足A∪B=A
若A∪B=A
①B=∅,那么△=[2(a+1)]²-4(a²-5)＜0,解得a＜-3
②B≠∅,那么2(a+1)=-3,且a²-5=2,解得a无解.
综上：a≤-3
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