记f(x)=(ax+1)/(x-2)
设x1＜x2＜2
f(x1)-f(x2)
=(ax1+1)/(x1-2)-(ax2+1)/(x2-2)
=[(ax1+1)(x2-2)]/[(x1-2)(x2-2)]-[(ax2+1)(x1-2)]/[(x1-2)(x2-2)]
=[(x2-x1)+2a(x2-x1)]/[(x1-2)(x2-2)]
=[(1+2a)(x2-x1)]/[(x1-2)(x2-2)]
因为x1＜x2＜2,且f(x)在（负无穷,2）为增函数
所以f(x1)-f(x2)＜0
即[(1+2a)(x2-x1)]/[(x1-2)(x2-2)]＜0
而(x1-2)(x2-2)＞0,x2-x1＞0
所以1+2a＜0
解得a＜-1/2
所以a的取值范围为（负无穷,-1/2）