定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x3)=12f(x),且当0≤x1<x2≤1时,有f(x1)≤f(x2),则f(12010)的值为()A.1256B.1128

题目简介

定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x3)=12f(x),且当0≤x1<x2≤1时,有f(x1)≤f(x2),则f(12010)的值为()A.1256B.1128

题目详情

定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(
x
3
)=
1
2
f(x)
,且当0≤x1<x2≤1时,有f(x1)≤f(x2),则f(
1
2010
)
的值为(  )
A.
1
256
B.
1
128
C.
1
64
D.
1
32
题型:单选题难度:中档来源:钟祥市模拟

答案

∵定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(class="stub"x
3
)=class="stub"1
2
f(x)

∴f(1)+f(0)=1,∴f(1)=1
f(class="stub"1
2
)+f(1-class="stub"1
2
)=1,∴f(class="stub"1
2
)=class="stub"1
2

f(class="stub"1
3
)=class="stub"1
2
f(1),∴f(class="stub"1
2
)=class="stub"1
2

f(class="stub"1
3
)=class="stub"1
2

class="stub"1
1458
>class="stub"1
2010
>class="stub"1
2187
,且当0≤x1<x2≤1时,有f(x1)≤f(x2),
f(class="stub"1
1458
)<f(class="stub"1
2010
)<f(class="stub"1
2187
)

又∵f(class="stub"1
1458
)=class="stub"1
2
f(class="stub"1
486
)=class="stub"1
22
f(
.
162
)=…=class="stub"1
26
f(class="stub"1
2
)=class="stub"1
27

f(class="stub"1
37
)=class="stub"1
2
f(class="stub"1
36
)=class="stub"1
22
f(class="stub"1
35
)=…=class="stub"1
27
f(1)=class="stub"1
27

f(class="stub"1
2010
)
=class="stub"1
27
=class="stub"1
128

故选B

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