因为f(x)是奇函数,从而 f(-2)=-f(2),
又f(x)的周期为3,所以 f(-2)=f(1)
所以 f(2)=-f(1)≤-1
即 (2a-3)/(a+1) ≤-1
[(2a-3)+(a+1)]/(a+1)≤0
(3a-2)/(a+1)≤0
等价于(3a-2)(a+1)≤0且a+1≠0
解得 -1