设f(x)=ax−ln(x)−1f(x)=ax-ln(x)-1f(x)=ax−ln(x)−1,若f(x)>=0f(x)>=0f(x)>=0对x>0x>0x>0恒成立,求 a 的最小值。A.1/eB.1C.eD.0A. 1/eB. 1C. eD. 0A.1/eB.1C.eD.0