发布时间 : 星期六 文章福建省厦门双十中学2015届高三上学期期中考试试题更新完毕开始阅读
令f'(x)?0，所以1?lnx?0，解得0?x?e；令f(')x0所以1?lnx?0，解得x?e， ?，
所以f(x)的增区间为：(0,e)，减区间为：(e,??). ···················································································· 5分 (2)
解法一：
由(1)知，函数f(x)在(e,??)上单调递减，所以f(2014)?f(2015)，即
ln2014ln2015?20142015?2015ln2014?2014ln2015 ?ln20142015?ln20152014?20142015?20152014 ··············································································································································································· 9分 解法二：
201420152014?2015????20142015?2014??2015????2014?2014?1，因为 201420141????1???2014?121312014232014?1?1?C2014()?C2014()??C2014()201420142014111?2????2!3!2014!111 ?2????1?22?32013?201411111?2?(1?)?(?)??(?)223201320141?3?2014?3201520143??1，所以20152014?20142015. ·所以················································································· 9分 201520142014lnx2lnx2(3)若kx?f(x)?2对任意x?0恒成立，则k?2?，记g(x)?2?，只需
xxxxk?g(x)max.
1?2lnx21?2x?2lnx?2?又g'(x)?， ································································································· 10分 x3xx32记h(x)?1?2x?2lnx，则h'(x)??2??0，所以h(x)在(0,??)上单调递减.
x2232又h(1)??1?0，h()?1?2?2ln?1?2?ln2?1??ln2?ln?0，
222e2,1)，使得h(x0)?0，即1?2x0?2lnx0?0， ·所以存在唯一x0?(················································· 11分 2当x?0时，h(x),g'(x),g(x)的变化情况如下：
x (0,x0) x0 (x0,??) ↗ 极大值 ↘ ············································································································································································· 12分
h(x) g'(x) g(x) ＋ ＋ 0 0 － － 2x0?lnx0，又因为1?2x0?2lnx0?0，所以2x0?2lnx0?1， 2x0(2x0?2lnx0)?2x01?2x01121所以g(x0)????()?， 222x02x02x0x0所以g(x)max?g(x0)?因为x0?(321························································· 13分 ,1)，所以?(1,2)，所以?g(x0)?1?2， ·
22x0又g(x)max?g(1)?2，所以2?g(x0)?1?2，
因为k?g(x)max，即k?g(x0)，且k∈Z，故k的最小整数值为3.
所以存在最小整数k?3，使得kx?f(x)?2对任意x?0恒成立. ··························································· 14分