Question

定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x/5)=1/2f(x),且当0≤x1

Answer 1

f(0)=0, f(x)+f(1-x)=1 ==> f(1/2)=1/2, f(1)=1

f(1)=1, f(x/5)=1/2f(x), f(x)+f(1-x)=1 ==> f(1/5)=1/2, f(4/5)=1/2

f(1/5)=1/2, f(1/2)=1/2, f(4/5)=1/2, f(x1)≤f(x2){0≤x1 [1/5,4/5]区间f(x)=1/2

f(x/5)=1/2f(x) ==> [1/5^n, 4/5^n]区间上f(x)=1/2^n

1/2020 属于 [1/5^5, 4/5^5] ==> f(1/2020) = 1/2^5 =1/32