定义在[0,1]上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x5)=12f(x),且当0≤x1<x2≤1时f(x1)≤f(x2),则f(12010)等于()A.12B.116C.

题目简介

定义在[0,1]上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x5)=12f(x),且当0≤x1<x2≤1时f(x1)≤f(x2),则f(12010)等于()A.12B.116C.

题目详情

定义在[0,1]上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(
x
5
)=
1
2
f(x),且当0≤x1<x2≤1时f(x1)≤f(x2),则f(
1
2010
)等于(  )
A.
1
2
B.
1
16
C.
1
32
D.
1
64
题型:单选题难度:中档来源:不详

答案

∵f(0)=0,f(x)+f(1-x)=1,令x=1得:f(1)=1,
又f(class="stub"x
5
)=class="stub"1
2
f(x),
∴当x=1时,f(class="stub"1
5
)=class="stub"1
2
f(1)=class="stub"1
2

令x=class="stub"1
5
,由f(class="stub"x
5
)=class="stub"1
2
f(x)得:
f(class="stub"1
25
)=class="stub"1
2
f(class="stub"1
5
)=class="stub"1
4

同理可求: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

再令x=class="stub"1
2
,由f(x)+f(1-x)=1,可求得f(class="stub"1
2
)=class="stub"1
2

∴f(class="stub"1
2
)+f(1-class="stub"1
2
)=1,解得f(class="stub"1
2
)=class="stub"1
2

令x=class="stub"1
2
,同理反复利用f(class="stub"x
5
)=class="stub"1
2
f(x),
可得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
1250
)=class="stub"1
2
f(class="stub"1
250
)=class="stub"1
32

由①②可得:,有f(class="stub"1
1250
)=f(class="stub"1
3125
)=class="stub"1
32

∵0≤x1<x2≤1时f(x1)≤f(x2),而0<class="stub"1
3125
<class="stub"1
2010
<class="stub"1
1250
<1
所以有f(class="stub"1
2010
)≥f(class="stub"1
3125
)=class="stub"1
32

       f(class="stub"1
2010
)≤f(class="stub"1
1250
)=class="stub"1
32

故f(class="stub"1
2010
)=class="stub"1
32

故选C.

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