定义在R上函数f(x)满足f(0)=0,f(x)+f(1-x)=1,且f(x5)=12f(x)当0≤x1<x2≤1时,f(x1)≤f(x2),则f(12011)=()A.12B.116C.132D.1

题目简介

定义在R上函数f(x)满足f(0)=0,f(x)+f(1-x)=1,且f(x5)=12f(x)当0≤x1<x2≤1时,f(x1)≤f(x2),则f(12011)=()A.12B.116C.132D.1

题目详情

定义在R上函数f(x)满足f(0)=0,f(x)+f(1-x)=1,且f(
x
5
)=
1
2
f(x)
当0≤x1<x2≤1时,f(x1)≤f(x2),则f(
1
2011
)
=(  )
A.
1
2
B.
1
16
C.
1
32
D.
1
64
题型:单选题难度:偏易来源:不详

答案

∵f(x)+f(1-x)=1
令x=1得的f(1)=1,x=class="stub"1
2
得f(class="stub"1
2
)=class="stub"1
2

∵f(class="stub"x
5
)=class="stub"1
2
f(x)得,
f(class="stub"1
5
)=class="stub"1
2
f(1)=class="stub"1
2
,f(class="stub"1
25
)=class="stub"1
2
f(class="stub"1
5
)=class="stub"1
4
,f(class="stub"1
125
)=class="stub"1
2
f(class="stub"1
25
)=class="stub"1
8
,f(class="stub"1
625
)=class="stub"1
2
f(class="stub"1
125
)=class="stub"1
16
,f(class="stub"1
3125
)=class="stub"1
2
f(class="stub"1
625
)=class="stub"1
32

由f(class="stub"x
5
)=class="stub"1
2
f(x)得
f(class="stub"1
10
)=class="stub"1
2
f(class="stub"1
2
)=class="stub"1
4
,f(class="stub"1
50
)=class="stub"1
2
f(class="stub"1
10
)=class="stub"1
8
,f(class="stub"1
250
)=class="stub"1
2
f(class="stub"1
50
)=class="stub"1
16
,f(class="stub"1
1250
)=class="stub"1
2
f(class="stub"1
250
)=class="stub"1
32

∵0≤x1<x2≤1时,f(x1)≤f(x2),
由f(class="stub"1
3125
)≤f(class="stub"1
2011
)≤f(class="stub"1
1250
)及f(class="stub"1
3125
)=f(class="stub"1
1250
)=class="stub"1
32
得f(class="stub"1
2011
)=class="stub"1
32

故选C.

更多内容推荐