Question

设集合A={x|x^2+4x<0,x∈R},B={x|x^2+2(a+1)x+a^2-1<0,x∈R}.若B包含于A,求实数a的取值范围.

Answer 1

A={x|x^2+4x<0,x∈R}
= {x|-4<x<0}
B={x|x^2+2(a+1)x+a^2-1<0,x∈R}
={x| -(a+1)-√(2a+2) < x < -(a+1)+√(2a+2) }

B is subset of A

case 1: B=Φ , ie a≤ -1

case 2: B≠Φ , ie a> -1
=>
-(a+1)-√(2a+2)≥-4 and -(a+1)+√(2a+2)≤0
-a+3≥√(2a+2) (1)and
√(2a+2)≤ a+1 (2)

from (1)
-a+3≥√(2a+2) ( a<3 )
a^2-8a+7≥0
a ≤ 1

from (2)
√(2a+2)≤ a+1
2a+2 ≤ a^2+2a+1
a^2-1 ≥ 0
a ≥ 1 or a≤-1
solution for case 2:

"a>-1" and "a ≤ 1" and "a ≥ 1 or a≤-1"
ie a=1

B包含于A,求实数a的取值范围
case 1 or case 2
ie a≤-1 or a=1