Question

已知集合 A={x|x^2+ax+1<=0}, 且B= {x|x^2-3x+2<=0},A包含于B,求实数a的取值范围

Answer 1

B= {x|x^2-3x+2<=0}
={ x| (x-1)(x-2)<=0 }
= { x| 1<=x<=2}
A={x|x^2+ax+1<=0}
case1:
if a^2-4 <0 ( no real root for x^2+ax+1 =0 }
ie -2<a< 2
=>A = 空集
=> A is subset of B

case 2:
if a<=-2 or a>=2
A = { x| (-a-√[a^2-4])/2 <= x <= (-a+√[a^2-4])/2 }
A is subset of B
=> (-a-√[a^2-4])/2 > 1 and (-a+√[a^2-4])/2 <2
=> √[a^2-4] < -a-2 and √[a^2-4] < 4+a
solving
(√[a^2-4] < -a-2) and (a<=-2 or a>=2)
=>[ (√[a^2-4] < -a-2) and a>=2] or [ (√[a^2-4] < -a-2) and a<=-2]
=> False or [ (√[a^2-4] < -a-2) and a<=-2]
=> a^2-4 < (-a-2)^2
=> 4a > -8
=> a > -2

solving
√[a^2-4] < 4+a and [a<=-2 or a>=2]
=>[ √[a^2-4] < 4+a and a<=-2 ] or [ √[a^2-4] < 4+a and a>=2 ]
=>[ a^2-4 < (4+a)^2 and -4<a<=-2 ] or a^2-4< (4+a)^2
=> [8a > -20 and -4<a<=2] or 8a>-20
=> a> -5/2
solution of case 2:
a > -2 and a> -5/2
ie a > -2

case 1 or case 2
-2<a< 2 or a> -2
ie -1 <a< 2

Answer 2

我读高中的时候，这类题都做烂了！先把a当成已知数，将两个不等式分别求解，然后再划数轴，根据包含关系结合数轴得出答案！……希望能够帮助你！

追问

谢谢，很有帮助呢！！！

Answer 3

a<=-5/2

追问

waincharles ，恭喜你，做错了！

追答

开始理解错了。
由B解得1≤x≤2
当x取1时，A=1^2+a*1+1=a+2<=0 ,则a<=-2
当x取2时，A=2^2+a*2+1=2a+5<=0 ,则a<=-5/2
在数轴上可知，-2≤a≤-5/2 时，包含于B

追问

不好意思，又错了，-5/2不对

追答

标准答案是？