Question

设函数y=f(x)是定义在R*上的减函数,f(xy)=f(x)+f(y),f(1/3)=1,如果f(x)+f(2-x)小

Answer 1

f(xy)=f(x)+f(y),

f(x)=f(x)+f(1)

f(1)=0

f(3*1/3)=f(1)=f(3)+f(1/3)=0

f(3)=-1

f(1/9*3)=f(1/3)=f(1/9)+f(3)=1

f(1/9)=2

f(x)+f(2-x)=f(x*(2-x))1/9

(3-2√2)/3