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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