试题解析：（1）依题意，f?x?的定义域为?0,???，
11?3x2?2x32当a??3，b?1时，f?x??lnx?x?2x，f??x???3x?2?，
xx2由 f??x??0，得3x2?2x?1?0，解得?1?x?解得x?1；由 f??x??0，得3x2?2x?1?0，31?1??1?或x??1.?f?x?在?0,?单调递增，在?,???单调递减； 所以f?x?的极大3?3??3?5，此即为最大值；无最小值； 6值为f????ln3??1??3?（2）F?x??lnx?x?a1a?1??1?，x??,3?，则有F?(x0)?02?在x0??,3?上有解， x4?2??2?x0121712?1?a?(?x0?x0)min，x0??,3?所以当x0?时，?x0?x0取得最小值，所以
22162?2?a?7 161?ln2 221．（I）?（II）当x?0时，不等式f(x)?k即k?(x?1)?1?ln(x?1)?对于x?0恒成立
x(x?1)?1?ln(x?1)?x?1?ln(x?1)设h(x)?，则h?(x)?
xx21xg?(x)?1???0，g(x)?x?1?ln(x?1)在区间?0,???上是增函数，
x?1x?1且g(x)?0存在唯一实数根a,满足a?(2,3)，即a?1?ln(a?1) 由x?a时，g(x)?0,h?(x)?0；0?x?a时，g(x)?0,h?(x)?0 知h(x)(x?0)的最小值为h(a)?故正整数k的最大值为3。 22. (1)
g?(x)?0恒成立 x(a?1)?1?ln(a?1)??a?1?(3,4)
a?f?x??2014g?x?in?1i?1n?n????f?xi????2014g?xn?1??min ?i?1?max?f?x?,g?x?均为增函数
1?3?n?????f?xi???n?2???n-
2?2??i?1?max??2014g?xn?1???min?2014?2?4028?32n?4028,?n?2685?13 ?n的最大值为2685. (2) H?x??2ln?x?1??x2?a
H'?x??2x?1?2x H?x1??H?x2??2??ln?x1?1??ln?x2?1?????x1?x2??x1?x2?xx 1?x21?x2?2xxlnx1?1??x1?x2?
1?2x2?1H'??x1?x2??2???4x??x1?1?x2?1x2?
原式?1x1?1???x2?xlnx1?1?2?lnx1?1??1?1?x2?x2?1x1?x2?2x?22?1?x1?1???x2?1?令
x1?1x?t?t??1,???? 2?1①式lnt?2?t?1t?1?2???1?2?t?1?? 令u?t??lnt?4t?1?2 u'?t??1t?4?t?1?2??t?1?2t?t?1?2?0?u?t?在?1,???上是增函数 ?u?t??u?1??0?u?t?无零点,故A、B两点不存在
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吉林省延边第二中学2024学年高二数学下学期期中试题 理
