已知f(x)满足f(ax-1)=lg(x+2)/(x-3),其中a为实数，且a≠0,若f(x)为非奇非偶函数，则a的范围？

令t=ax-1
x=(t+1)/a
f(t)=lg(t+1+2a)/(t+1-3a)
设f(t)=f(-t)
f(-t)=lg(1+2a-t)/(1-3a-t)=f(t)=lg(t+1+2a)/(t+1-3a)
(t-1-2a)/(t+3a-1)=(t+1+2a)/(t+1-3a)
t^2+5at+(1+2a)(3a-1)=t^2-5at+(3a-1)(2a+1)
10at=0
因为a≠0,无解，所以f(t)≠f(-t)
设f(t)=-f(-t)
f(-t)=lg(1+2a-t)/(1-3a-t)=-f(t)=lg(t+1-3a)/(t+1+2a)
(t-1-2a)/(t+3a-1)=(t+1-3a)/(t+1+2a)
t^2-(3a-1)^2=t^2-(2a+1)^2
(3a-1)^2=(2a+1)^2
3a-1=2a+1
a=2
或3a-1=-(2a+1)
a=0
所以当a≠2时f(t)≠-f(-t)
若f(x)为非奇非偶函数，则a≠0,a≠2