题目
题型:单选题难度:一般来源:不详
x2-x+1 |
x2+1 |
1-x2 |
A.[2,3] | B.[1,
| C.[2,
| D.[1,
|
答案
x2-x+1 |
x2+1 |
即(1-y)x2-x+1-y=0,∴△=1-4(1-y)2≥0,
解得,
1 |
2 |
3 |
2 |
1 |
2 |
3 |
2 |
设t=
1-x2 |
1-x2 |
y=2(1-t2)+t=-2t2+t+2=-2(t-
1 |
4 |
9 |
4 |
∵0≤t≤1,∴1≤y≤
9 |
4 |
9 |
4 |
∴则A∩B=[1,
3 |
2 |
故选B.
核心考点
试题【已知集合A={y|y=x2-x+1x2+1},B={y|y=2x2+1-x2},则A∩B=( )A.[2,3]B.[1,32]C.[2,178]D.[1,17】;主要考察你对集合运算等知识点的理解。[详细]
举一反三
A.{3,5} | B.{1,2,3,4,5,7} |
C.{6,8} | D.{1,2,4,6,7,8} |
4-x |
x-2 |
(1)若A=B,求a的值.
(2)若B⊆A,,且a>0,求a的取值范围.
A.{3} | B.{4,5} | C.Q | D.{1,4,5} |