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法则、用导数研究函数图像的意识、考查数形结合思想，考查考生的计算推理能力及分析问题、解决问题的能力．
解：（1）?f?x?是二次函数, 且关于x的不等式f?x??0的解集为
?x?1?x?3,x?R?,
·························· 4分 ?f?x??a(x?1)(x?3)?ax2?2ax?3a, 且a?0. ·
2?a?0,f?x??a??(x?1)?4????4,且f?1???4a,
······························································ 6分 ?f(x)min??4a??4,a?1.
故函数f?x?的解析式为f?x??x?2x?3.
2x2?2x?33?4lnx?x??4lnx?2(x?0), (2) ?g(x)?xx?g?(x)?1?34(x?1)(x?3)??. ·························································· 8分 x2xx2?(x),g(x) x,g的取值变化情况如下：
x
g?(x) g(x)
(0,1)
1
(1,3) ?
单调减少
3
(3,??)
?
单调增加
0
极大值
0
极小值
?
单调增加
·················································································································· 11分 当0?x?3时, g?x??g?1???4?0； ···················································· 12分 又ge???e55?3?20?2?25?1?22?9?0. ······································· 13分 5e故函数g(x)只有1个零点,且零点x0?(3,e5).············································ 14分
20．（本小题满分14分）
如图，M,N是抛物线C1:x?4y上的两动点（M,N异于原点O），且?OMN的角平分线垂直于y轴，直线MN与x轴，y轴分别相交于A,B.
2?????????????(1) 求实数?,?的值，使得OB??OM??ON；
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(2）若中心在原点，焦点在x轴上的椭圆C2经过A,M. 求椭圆C2焦距的最大值及此时
C2的方程.
【说明】本题主要考查直线的斜率、抛物线的切线、
两直线平行的位置关系，椭圆的基本性质， 考查学生运算能力、推理论证以及分析问 题、解决问题的能力，考查数形结合思想、 化归与转化思想．
C1yNBMAOC2x第20题图 x12x22),N(x2,),x1?x2?0,x1?x2. 解: (1) 设M(x1,44 由?OMN的角平分线垂直于y轴知，直线OM与直线MN的倾斜角互补，
x12x22x12?444?0,化简得x??2x. ·?从而斜率之和等于0，即············ 3分 21·x1x2?x1x12x12x2),N(?2x1,x1)知直线MN的方程为y???1(x?x1). 由点M(x1,444x12). ·分别在其中令y?0及x?0得A(2x1,0),B(0,···································· 5分 2?0??x1??(?2x1)??????????????2将B,M,N的坐标代入OB??OM??ON中得?x2, x211???x1?????24???2?即?, ······························································································· 7分
??4??2?所以??21,??. ·························································································· 8分 33x2y2 (2) 设椭圆C2的方程为2?2?1(a?b?0),
abx124x12x12x14?1, ·)代入,得2?1,2?将A(2x1,0),M(x1,····························· 9分 2aa16b4第12页 / 共14页
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x1422解得a?4x,b?, 由a?b得0?x12?48. ································ 10分
122212椭圆C2的焦距
33x12?(48?x12)222c?2a?b?x1(48?x1)???83
33222（或333······· 12分 x12(48?x12)??(x12?24)2?242??24?83） ·
333当且仅当x12?48?x12,x12?24?48时,上式取等号, 故(2c)max?83, ··· 13分
x2y2??1. ·此时椭圆C2的方程为····························································· 14分 9648 21．（本小题满分14分）
定义数列?an?： a1?1,a2?2，且对任意正整数n，有
nn?1?an?2??2?(?1)a?(?1)?1.记数列?an?前n项和为Sn. n??(1) 求数列?an?的通项公式与前n项和Sn；
（2）问是否存在正整数m,n，使得S2n?mS2n?1？若存在，则求出所有的正整数对
(m,n)；若不存在，则加以证明.
【说明】考查了等差、等比数列的通项公式、求和公式，数列的分组求和等知识，考查
了学生变形的能力，推理能力，探究问题的能力，分类讨论的数学思想、化归与转化的思想以及创新意识．
2k?12k?2?(?1)a?(?1)?1?a2k?1?2， 解：（1）对任意正整数k， a2k?1??2k?1??2k2k?1?a2k?2??2?(?1)a?(?1)?1?3a2k. ········································· 1分 2k?? 所以数列?a2k?1?是首项a1?1,公差为2等差数列；数列?a2k?是首项
·································································· 2分 a2?2,公比为3的等比数列. ·
对任意正整数k,a2k?1?2k?1,a2k?2?3k?1. ········································· 3分
n?2k?1??2k?1,?,k?N. 所以数列?an?的通项公式an??k?1??2?3,n?2k第13页 / 共14页
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n?2k?1?n,??,k?N. ·或an??······················································· 4分 n?12??2?3,n?2k对任意正整数k,S2k?(a1?a3???a2k?1)?(a2?a4???a2k)
k(1?2k?1)2(1?3k)???3k?k2?1. ··············································· 5分
21?3·························· 6分 S2k?1?S2k?a2k?k2?3k?1?2?3k?1?3k?1?k2?1 ·
k?12??3?k?1,n?2k?1?,k?N所以数列?an?的前n项和为Sn??k. 2??3?k?1,n?2k?1?n2n2?2n?33?,n?2k?1??4?或 Sn?? ········································ 7分 ,k?Nn2?32?n?1,n?2k??4 (2) S2n?mS2n?1?3n?n2?1?m(3n?1?n2?1)
?3n?1(3?m)?(m?1)(n2?1),
从而m?3,由m?N知m?1,2,3.···························································· 8分 ①当m?1时, 3n?1?·········· 9分 (3?m)?0?(m?1)(n2?1),即S2n?mS2n?1； ·
2②当m?3时, 2(n?1)?0,n?1,即S2?3S1； ···································· 10分 ③当m?2时, 3kn?1?n2?1?(n?1)(n?1),则存在k1,k2?N,k1?k2,
k使得n?1?31,n?1?32,k1?k2?n?1, 从而32?31?31(32kkkk?k1?1)?2,得3k1?1,3k2?k1?1?2,
········································· 13分 k1?0,k2?k1?1，得n?2,即S4?2S3. ·
综上可知,符合条件的正整数对(m,n)只有两对：(2,2)与(3,1). ………………14分
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