题目内容
定义在R上的函数满足f(0)=0 ,f(x)+f(1-x)=1 , f(
)=
f(x),且当0≤x1<x2≤1时,f(x1)≤f(x2),则f(
)=
.
x |
5 |
1 |
2 |
1 |
2012 |
1 |
32 |
1 |
32 |
分析:依题意,可求得f(1)=1,f(
)=
,再分别利用f(
)=
f(x),可求得f(
)=f(
)=
,结合已知,即可求得答案.
1 |
2 |
1 |
2 |
x |
5 |
1 |
2 |
1 |
3125 |
1 |
1250 |
1 |
32 |
解答:解:依题意知,f(1)=1,由f(
)+(1-
)=1得:f(
)=
,
又f(
)=
f(x),
∴f(
)=f(
)=
,
f(
)=f(
)=
,
f(
)=f(
)=
,
f(
)=f(
)=
,
f(
)=f(
)=
,
∵
<
<
,
当0≤x1<x2≤1时,f(x1)≤f(x2),
∴f(
)=
.
故答案为:
.
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
又f(
x |
5 |
1 |
2 |
∴f(
1 |
5 |
1 |
2 |
1 |
2 |
f(
1 |
25 |
1 |
10 |
1 |
4 |
f(
1 |
125 |
1 |
50 |
1 |
8 |
f(
1 |
625 |
1 |
250 |
1 |
16 |
f(
1 |
3125 |
1 |
1250 |
1 |
32 |
∵
1 |
3125 |
1 |
2012 |
1 |
1250 |
当0≤x1<x2≤1时,f(x1)≤f(x2),
∴f(
1 |
2012 |
1 |
32 |
故答案为:
1 |
32 |
点评:本题考查抽象函数及其应用,着重考查赋值法求值,考查递推关系式的灵活应用,属于中档题.
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