已知定义在R上的函数f(x)满足f(1)=1,f(1-x)=1-f(x),2f(x)=f(4x),且当0≤x1<x2≤1时,f(x1)≤f(x2),则f(133)等于()A.14B.18C.116D.

题目简介

已知定义在R上的函数f(x)满足f(1)=1,f(1-x)=1-f(x),2f(x)=f(4x),且当0≤x1<x2≤1时,f(x1)≤f(x2),则f(133)等于()A.14B.18C.116D.

题目详情

已知定义在R上的函数f(x)满足f(1)=1,f(1-x)=1-f(x),2f(x)=f(4x),且当0≤x1<x2≤1时,f(x1)≤f(x2),则f(
1
33
)等于(  )
A.
1
4
B.
1
8
C.
1
16
D.
1
32
题型:单选题难度:偏易来源:绵阳一模

答案

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

∵2f(x)=f(4x)
∴f(x)=class="stub"1
2
f(4x)
在f(x)=class="stub"1
2
f(4x)中,令x=class="stub"1
4
可得f(class="stub"1
4
)=class="stub"1
2
f(1)
=class="stub"1
2

在f(1-x)+f(x)=1中,令x=class="stub"1
4
可得f(class="stub"1
4
)+f(class="stub"3
4
)=1即f(class="stub"3
4
)=class="stub"1
2

同理可求f(class="stub"1
8
)=class="stub"1
2
f(class="stub"1
2
)=class="stub"1
4
,f(class="stub"7
8
)=1-f(class="stub"1
8
)=class="stub"3
4

f(class="stub"1
16
)=class="stub"1
2
f(class="stub"1
4
)
=class="stub"1
4
,f(class="stub"15
16
)=1-f(class="stub"1
16
)=class="stub"3
4

f(class="stub"1
32
)=class="stub"1
2
f(class="stub"1
8
)
=class="stub"1
8
,f(class="stub"31
32
)=1-f(class="stub"1
32
)=class="stub"7
8

f(class="stub"1
64
)
=class="stub"1
2
f(class="stub"1
16
)
=class="stub"1
8
,f(class="stub"63
64
)=1-class="stub"1
8
=class="stub"7
8

∵当0≤x1≤x2≤1时,f(x1)≤f(x2),
class="stub"1
8
=f(class="stub"1
64
)≤f(class="stub"1
33
)≤
class="stub"1
32
=class="stub"1
8

∴f(class="stub"1
33
)
=class="stub"1
8

故选B

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