题目内容
【题目】设f(x)是[0,1]上的不减函数,即对于0≤x1≤x2≤1有f(x1)≤f(x2),且满足(1)f(0)=0;(2)f( )=
f(x);(3)f(1﹣x)=1﹣f(x),则f(
)=( )
A.
B.
C.
D.
【答案】C
【解析】解:∵(1)f(0)=0;(2)f( )=
f(x);(3)f(1﹣x)=1﹣f(x),
∴f(1)=1﹣f(0)=1,
f( )=
f(1)=
,f(1﹣
)=1﹣f(
).即f(
)=1﹣
=
,
f( )=
f(
)=
×
=
,f(
)=
f(
)=
×
=
f( )=
f(
)=
×
=
,f(
)=
f(
)=
×
=
,
f( )=
f(
)=
×
=
,f(
)=
f(
)=
×
=
,
f( )=
f(
)=
×
=
,f(
)=
f(
)=
×
=
,
f( )=
f(
)=
×
=
,f(
)=
f(
)=
×
=
,
f( )=
f(
)=
×
=
,f(
)=
f(
)=
×
=
,
∵对于0≤x1≤x2≤1有f(x1)≤f(x2),
∴当 ≤x≤
时,f(x)=
,
∵ ∈[
,
]时,∴f(
)=
,
故选:C.
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