定义在R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(12008)等于?-数学

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(x5)=12f(x),且当0≤x1<x2≤1时,f(x1)≤f(x2).则f(12008)等于?-数学

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
5
)=
1
2
f(x)
,且当0≤x1<x2≤1时,f(x1)≤f(x2).则f(
1
2008
)
等于?
题型:解答题难度: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
2008
<class="stub"1
1250
<1
所以有f(class="stub"1
2008
)≥f(class="stub"1
3125
)=class="stub"1
32

       f(class="stub"1
2008
)≤f(class="stub"1
1250
)=class="stub"1
32

f(class="stub"1
2008
)
=class="stub"1
32

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