题目内容
在y=2x,y=log2x,y=x2这三个函数中,当0<x1<x2<1时,使f(
)>
恒成立的函数的个数是( )
| x1+x2 |
| 2 |
| f(x1+x2) |
| 2 |
| A.0个 | B.1个 | C.2个 | D.3个 |
对于y=2x有f(
)= 2
=
=2x1+x2-1
∵0<x1<x2<1,∴
>x1+x2-1
∴f(
)>
恒成立
对于y=log2x有f(
)= log2 (
),
=
=log2
∵0<x1<x2<1,
∴
<
,
∴f(
)<
故选B
| x1+x2 |
| 2 |
| x1+x2 |
| 2 |
| f(x1+x2) |
| 2 |
| 2x1+x2 |
| 2 |
∵0<x1<x2<1,∴
| x1+x2 |
| 2 |
∴f(
| x1+x2 |
| 2 |
| f(x1+x2) |
| 2 |
对于y=log2x有f(
| x1+x2 |
| 2 |
| x1+x2 |
| 2 |
| f(x1+x2) |
| 2 |
| log2( x1+x2) |
| 2 |
| x1+x2 |
∵0<x1<x2<1,
∴
| x1+x2 |
| 2 |
| x1+x2 |
∴f(
| x1+x2 |
| 2 |
| f(x1+x2) |
| 2 |
故选B
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