Question

已知集合A={y|y^2-(a^2+a+1)y+a(a^2+1)＜0},B={y|y=1/2x^2-x+5/2,0≤x≤3},若A∩B=空集,求实数a的取值范围

需要过程谢谢

Answer 1

A={y|y^2-(a^2+a+1)y+a(a^2+1)＜0} = （a, a^2+1）
B={y|y=1/2x^2-x+5/2,0≤x≤3} = [2, 4]
A∩B=空集 => a>=4 或a^2+1<=2
a的取值范围: a>=4 或|a|<=1

Answer 2

A:y^2-(a^2+a+1)+a(a^2+1)>0
(y-a)[y-(a^2+1)]>0
B:y=1/2(x^2-2x+5)=1/2[(x-1)^2+4]
又 0<=x<=3
-1<=(x-1)<=2
故0<=(x-1)^2<=4
2<=y<=4
回到A:不管a是正数还是负数,都有a^2+1>a
所以有A=(y>a^2+1或者y<a)
又因为A交B是空集,所以(a^2+1>4,a<2),解得a<-√3

Answer 3

A:y^2-(a^2+a+1)+a(a^2+1)>0
(y-a)[y-(a^2+1)]>0
B:y=1/2(x^2-2x+5)=1/2[(x-1)^2+4]
又 0<=x<=3
-1<=(x-1)<=2
故0<=(x-1)^2<=4
2<=y<=4
回到A:不管a是正数还是负数,都有a^2+1>a
所以有A=(y>a^2+1或者y<a)
又因为A交B是空集,所以(a^2+1>4,a<2),解得a<-√3