已知a>0,函数f(x)=(1-ax)/x,x∈(0,+∞).设0<x1<2/a,记曲线y=f(x)在点M(x1,f(x1))处的切线为l
(1)求l的方程
(2)设l与x轴交点为(x2,0).求证:①0<x2≤1/a②若0<x1<1/a,则x1