题目内容
已知A = {x|3≤2x + 3≤11},B ={y|y = x2 1,1≤x≤2},求解析:由3≤2x + 3≤11,得0≤x≤4,∴A = [0,4]
由y = x2 1,1≤x≤2得x = 0时ymax = 1;x = 2时,ymin = 5,
∴5≤y≤1,即B = [5,1] ∴A∩B =
, ∴
= R.
练习册系列答案
相关题目
已知A={x|3-x≥
},B={
-(a+1)x+a≤0},当A
B时,a的取值范围是
[ ]
|
A.a≥1 B.1<a<2 |
C.a>2 D.a≥2 |