设函数f(x)=x-1x-alnx(a∈R).(1)当a=2时,求f(x)的单调区间;(2)若f...
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发布时间：2024-10-23 15:27
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时间：2024-10-24 19:30
（1）f（x）定义域为（0，+∞），
f′（x）=x2?2x+1x2≥0，故f（x）在（0，+∞）上单调递增；
（2）由f（x）有两个极值点x1和x2，知，a＞2．
∵f（x1）-f（x2）=（x1-x2）+x1?x2x1x2-a（lnx1-lnx2），
∴k=f(x1)?f(x2)x1?x2=1+1x1x2-a?lnx1?lnx2x1?x2，
又x1x2=1．于是k=2-a?lnx1?lnx2x1?x2，
若存在a，使得k=2-a，则lnx1?lnx2x1?x2=1，即lnx1-lnx2=x1-x2，
亦即x2?1x2?2lnx2=0（*）
再由（1）知，函数h（t）=t-1t-2lnt在（0，+∞）上单调递增，
而x2＞1，
∴x2?1x2?2lnx2＞1-1-2ln1=0，这与（*）式矛盾，
故不存在a，使得k=2-a．
（3）∵nk?2lnk?1k+1=ln2n(n+1)，
∴nk?2lnk?1k+已赞过已踩过你对这个回答的评价是？评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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