已知f(x)=ax-lnx(x∈(0,e]),其中e是自然常数,a∈R
(Ⅰ)当a=1时,求f(x)的单调区间和极值;
(Ⅱ)是否存在实数a,使f(x)的最小值是3,若存在,求出a的值;若不存在,说明理由.
考点分析:
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