1.A={x|x²+4x=0}B={x|x²+2(a+1)x+a²-1=0,a∈R,x∈R},若A∪B=A,求a的取值范围2.设a＞-2,A={x|-2≤x≤a},B={y|y=2x+3,x∈A},C={z|z=x²,x∈A},且C⊆B,求实数a取值范围

Question

1.A={x|x²+4x=0} B={x|x²+2(a+1)x+a²-1=0,a∈R,x∈R},若A∪B=A,求a的取值范围
2.设a＞-2,A={x|-2≤x≤a},B={y|y=2x+3,x∈A},C={z|z=x²,x∈A},且C⊆B,求实数a取值范围

Answer 1

1.由题意可得,A ={0,-4}
那么,B={0}或者{-4}或者{0,-4}或者空集
当B={0}时,a=-1
当B={-4}时,a =7
当B={0,-4}时,a=1
当B为空集时,a综上,a的范围为：a<=-1或者a=1或者a =7
2.由题意可知,B={y|-1<=y<=2a+3},C包含B
当-2=4,得a>=1/2,所以1/2<=a<2
当a>=2时,C={z|4<=z<=a ^2}有2a+3>=a^2,得-1<=a<=3,所以2<=a<=3
综上,1/2<=a<=3