设函数f(x)=lnx-px+1,其中p为常数.(Ⅰ)求函数f(x)的极值点;( 发布时间:2017-04-29 22:49 │ 来源:www.tikuol.com 题型:解答题 问题: 设函数f(x)=lnx-px+1,其中p为常数.(Ⅰ)求函数f(x)的极值点;(Ⅱ)当p>0时,若对任意的x>0,恒有在f(x)≤0,求p的取值范围;(Ⅲ)求证:ln2222+ln3232+…+lnn2n2<2n2-n-12(n+1)(n∈N,n≥2).
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