对于0≤x1≤x2≤1有f(x1)≤f(x2),且f(0)=0,f(x/3)=0.5f(x),f(1-x)=1-f(x)

答案：1/128
f(1-x)=1-f(x)得到f(1)=1-f(0)=1；f(x/3)=f(x)/2得到f(1/3)=f(1)/2=1/2,f(2/3)=1-f(1/3)=1/2;而0≤x1≤x2≤1有f(x1)≤f(x2),所以当1/3