Question

已知函数f(x)=loga(1-mx/x-1),(a>0,且a≠1)的图像关于原点对称.

Answer 1

f(-x)=-f(x)

loga[(1+mx)/(-x-1)]=-loga[(1-mx)/(x-1)]

(1+mx)/(-x-1)=(x-1)/(1-mx)

m²x²-x²=0

m=±1

当x=1时,(1-mx)/(x-1)=-1,不成立

∴m=-1

f(x)=loga[(x+1)/(x-1)]

(a-2+1)/(a-2-1)=a

a²-4a+1=0

a=2±√3

∵a>1

∴a=2+√3

r-1=0

∴r=1