题目内容
已知集合A={y|y=
},B={y|y=2x2+
},则A∩B=( )
| x2-x+1 |
| x2+1 |
| 1-x2 |
| A.[2,3] | B.[1,
| C.[2,
| D.[1,
|
∵函数y=
的定义域为R,∴方程y(x2+1)=x2-x+1有实数根,
即(1-y)x2-x+1-y=0,∴△=1-4(1-y)2≥0,
解得,
≤y≤
,∴A=[
,
],
设t=
,则0≤t≤1,x2=1-t2,代入函数y=2x2+
得
y=2(1-t2)+t=-2t2+t+2=-2(t-
)2+
,
∵0≤t≤1,∴1≤y≤
,∴B=[1,
],
∴则A∩B=[1,
].
故选B.
| 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.
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