Question

⒈已知函数f(x)是定义在(0,∞)的增函数,且

Answer 1

1

1.令y=0,x=1,解得f(1)=0,f(x/y)=f(x)+f(1/y),令x=y,则f(1)=f(y)+f(1/y),所以-f(y)=f(1/y),所以原式得证

2.f(a)-f(a-1)>2,即f(a/(a-1))>2,2=f(3)+f(3)=f(9),所以a/(a-1)>9,所以1