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1、1 2015 年全国高中数学联合竞赛A 卷参考答案及评分标准一试说明：1.评阅试卷时，请依据本评分标冶填空题只设。分和香分两档；其他各题的评阅，请严格按照本评分标准的评分档次给分，不要增加其他中间档次.2.如果考生的解答方法和本解答不同，只要思路合理、步骤正确，在评卷时可参考本评分标准适当划分档次评分，解答题中第9 小题 4 分为一个档次，第 10、11 小题该分为一个档次，不要增加其他中间档次一、填空题：本大题共8 小题，每题8 分，总分值64 分1设ba,为不相等的实数，假设二次函数baxxxf2)(满足)()(bfaf，则)2(f答案：4.解：由己知条件及二次函数图像的轴对称性，可得22
2、aba，即20ab，所以(2)424fab2假设实数满足tancos，则4cossin1的值为答案：2.解：由条件知，sincos2，反复利用此结论，并注意到1sincos22，得)cos1)(sin1(sinsinsincoscossin1222242cossin223已知复数数列nz满足),2,1(1,111nnizzznn，其中 i 为虚数单位，nz表示nz的共轭复数，则2015z答案：2015+1007i解：由己知得，对一切正整数n，有211(1)11(1)2nnnnzznizninizi,于是201511007(2)20151007zzii4在矩形ABCD中，1,2 ADAB，边DC
3、上包含点D、C的动点P与CB延长线上包含点B的动点Q满足条件BQDP，则PQPA的最小值为答案34解：不妨设A(0,0),B(2,0),D(0,l)设P 的坐标为t,l)其中02t，则由|DPBQ得 Q 的坐标为 2，-t)，故(,1),(2,1)PAtPQtt,因此，22133()(2)(1)(1)1()244PA PQtttttt当12t时，min3()4PA PQ5在正方体中随机取三条棱，它们两两异面的概率为答案：255解：设正方体为ABCD-EFGH，它共有12 条棱，从中任意取出3 条棱的方法2 共有312C=220 种下面考虑使3 条棱两两异面的取法数由于正方体的棱共确定3 个互不
4、平行的方向 即AB、AD、AE 的方向，具有相同方向的4 条棱两两共面，因此取出的3 条棱必属于3个不同的方向可先取定AB 方向的棱，这有4 种取法不妨设取的棱就是AB，则 AD 方向只能取棱 EH 或棱 FG，共 2 种可能当AD 方向取棱是EH 或 FG 时，AE 方向取棱分别只能是 CG 或 DH 由上可知，3 条棱两两异面的取法数为42=8，故所求概率为82220556在平面直角坐标系xOy 中，点集0)63)(63(),(yxyxyx所对应的平面区域的面积为答案：24 解：设1(,)|3|60Kx yxy先 考 虑1K在 第 一 象 限 中 的 部 分，此 时 有36xy,故这些点对
5、应于图中的OCD 及其内部由对称性知，1K对应的区域是图中以原点O 为中心的菱形ABCD 及其内部同理，设2(,)|3|60Kx yxy，则2K对应的区域是图中以O 为中心的菱形EFGH及其内部由点集K的定义知，K所对应的平面区域是被1K、2K中恰好一个所覆盖的部分，因此此题所要求的即为图中阴影区域的面积S由于直线 CD 的方程为36xy，直线 GH 的方程为36xy，故它们的交点P 的坐标为3 3(,)2 2由对称性知，138842422CPGSS7设为正实数，假设存在实数)2(,baba，使得2sinsinba，则的取值范围为答 案：9 513,),)4 24w 解：2sinsinba知，
6、1sinsinba，而2,wwbasi，故题目条件等价于：存在整数,()k l kl，使得wlkw22222 当4w时，区间2,ww的长度不小于4，故必存在,k l满足式当04w时，注意到)8,0(2,ww，故仅需考虑如下几种情况：(i)ww2252，此时21w且45w无解；(ii)ww22925，此时2549w;(iii)ww221329，此时29413w,得4413w综合(i)、(ii)、(iii)，并注意到4w亦满足条件，可知9 513,),)4 24w8对四位数abcd(9d,0,91cba，)，假设,dccbba则称abcd为P类文档编码:CY4E6X7O4V1 HM4D6R7R8T
7、9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S
8、2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7
9、O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R
10、8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H1
11、0S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6
12、X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R
13、7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S23 数；假设dccbba,，则称abcd为Q类数，用 N(P)和 N(Q)分别表示P类数与Q类数的个数，则N(P)-N(Q)的值为答案：285解：分别记P 类数、Q 类数的全体为A、B，再将个位数为零的P 类数全体记为0A，个位数不等于零的尸类数全体记为1A对任一四位数1Aabcd，将其对应到四位数dcba，注意到1
14、,dccbba，故Bdcba反之，每个Bdcba唯一对应于从中的元素abcd这建立了1A与 B 之间的一一对应，因此有011()()|N PN QABAABA下面计算0|A对任一四位数00Aabc,b可取 0,1，9，对其中每个b，由9ab及9cb知，a和c分别有b9种取法，从而99220019 1019|(9)2856bkAbk因此，()()285N PN Q二、解答题：本大题共3 小题，总分值56 分，解答应写出文字说明、证明过程或演算步骤。9 此题总分值16 分假设实数cba,满足cbacba424,242，求c的最小值解：将2,2,2abc分别记为,x y z，则,0 x y z由条件
15、知，222,xyz xyz，故2222224()2zyxzyzy zy 8 分因此，结合平均值不等式可得，42233211111 13(2)3 222444yyzyyyyyy y 12 分当212yy，即312y时，z的最小值为3324此时相应的x值为3324，符合要求 由于2logcz，故c的最小值32235log(2)log 34316 分10 此题总分值20 分设4321,aaaa为四个有理数，使得：3,1,81,23,2,2441jiaaji，求4321aaaa的值解：由条件可知，(14)ija aij是 6 个互不相同的数，且其中没有两个为相反数，由 此 知，4321,aaaa的 绝
16、 对 值 互 不 相 等，不 妨 设|4321aaaa，则|(14)ijaaij中最小的与次小的两个数分别是12|aa及13|aa，最大与次大的两个数分别是34|aa及24|aa，从而必须有121324341,81,3,24,a aa aa aa a10 分于是2341112113,248aaaaaaa文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R
17、8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H1
18、0S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6
19、X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R
20、7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7
21、H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4
22、E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D
23、6R7R8T9 ZH5T2Y7H10S24 故2231412113,24 2,82a aa aaa，15 分结合1aQ，只可能114a由此易知，123411,4,642aaaa或者123411,4,642aaaa检验知这两组解均满足问题的条件故123494aaaa 20 分11 此题总分值20 分在平面坐标系xOy 中，21,FF分别为椭圆1222yx的左右焦点，设不经过焦点1F的直线l与椭圆交于两个不同的点BA,，焦点2F到直线l的距离为d，如果11,BFlAF的斜率依次成等差数列，求d的取值范围解：由条件知，点1F、2F的坐标分别为-1,0 和 l,0)设直线 l 的方程为ykxm，点 A
24、、B 的坐标分别为11(,)xy和22(,)xy，则12,x x满足方程22()12xkxm，即222(21)4(22)0kxkmxm由于点 A、B 不重合，且直线l 的斜率存在，故12,x x是方程的两个不同实根，因此有的判别式22222(4)4(21)(22)8(21)0kmkmkm，即2221km由直线11,BFlAF的斜率1212,11yykxx依次成等差数列知，1212211yykxx，又1122,ykxm ykxm，所以122112()(1)()(1)2(1)(1)kxmxkxmxk xx，化简并整理得，12()(2)0mkxx假设mk，则直线l 的方程为ykxk，即z 经过点1F
25、-1,0)，不符合条件因此必有1220 xx，故由方程及韦达定理知，1224()221kmxxk，即12mkk由、知，222121()2kmkk，化简得2214kk，这等价于2|2k反之，当,m k满足及2|2k时，l 必不经过点1F否则将导致mk，与矛盾,而此时,m k满足，故l 与椭圆有两个不同的交点A、B，同时也保证了1AF、1BF的斜率存在否则12,xx中的某一个为-l，结合1220 xx知121xx，与方程有两个不同的实根矛盾 10 分点2Fl,0到直线l:ykxm的距离为2222|1111|2|(2)221111kmdkkkkkk注意到2|2k，令211tk，则(1,3)t，上式可
26、改写为文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6
27、X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R
28、7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7
29、H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4
30、E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D
31、6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2
32、Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S25 21313()()222tdttt考虑到函数13()()2f ttt在1,3上上单调递减，故由得，(3)(1)fdf，即(3,2)d 20 分加试1 此题总分值40 分设)2(,21naaan是实数，证明：可以选取1,1,21n，使得)(1()()(12
33、2121niiiniiniianaa证法一：我们证明：22221112()(1)()nnnniijiniiijaaana，即对1,2,2ni，取1i，对1,2nin，取1i符合要求这里，x表示实数x的整数部分 10 分事实上，的左边为2222222111 1 1 122222nnnnnnijijijnnniiijjjaaaaaa2221 122222nnijnijnnana柯西不等式30 分2221 1212222nnijnijnnaa利用122nnn2221 12(1)nnijnijnana利用 xx21(1)()niina所以 得证，从而此题得证证法二：首先，由于问题中12,na aa的对
34、称性，可设12naaa此外，假设将12,na aa中的负数均改变符号，则问题中的不等式左边的21)(niia不减，而右边的21niia不 变，并 且 这 一 手 续 不 影 响1i的 选 取，因 此 我 们 可 进 一 步 设120naaa10 分文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7
35、O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R
36、8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H1
37、0S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6
38、X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R
39、7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7
40、H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S26 引理：设12
41、0naaa，则1110(1)niiiaa事实上，由于1(1,2,1)iiaain，故当n是偶数时，1123411(1)()()()0niinniaaaaaaa，11232111(1)()()niinnniaaaaaaaa当n是奇数时，11234211(1)()()()0niinnniaaaaaaaa，1123111(1)()()niinniaaaaaaa引理得证30 分回到原题，由柯西不等式及上面引理可知22122211111(1)(1)nnnniiiiiiiiiaanaana，这就证明了结论40 分证 法 三：加 强 命 题：设12,na aa2n 是 实 数，证 明：可 以 选 取12,1
42、,1 n，使得2221111()()()()nnniiiiiiiaanan.证明不妨设22212naaa，以下分n为奇数和n为偶数两种情况证明.当n为奇数时，取12121n，13221nnn，于是有12221112()()()nnniijniijaaa12221122()+()nnijnijaa1222112112()+2()()22nnijnijnnana应用柯西不等式.1222112(1)()+(1)()nnijnijnana另外，由于22212naaa，易证有122211211(1)(1)nnijnijaann，因此，由式即得到1222112(1)()+(1)()nnijnijnana2
43、11()()niinan，故n为奇数时，原命题成立，而且由证明过程可知，当且仅当12121n，13221nnn，且12naaa时取等号.文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T
44、9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S
45、2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7
46、O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R
47、8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H1
48、0S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6
49、X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S2文档编码:CY4E6X7O4V1 HM4D6R7R8T9 ZH5T2Y7H10S27 当n为偶数时，取1221n，24221nnn，于是有2222112()()()nnniijniijaaa22222122()
50、+()nnijnijaa2222122()+2()()22nnijnijnnana应用柯西不等式.222212()+()nnijnijnaa22111()()()nniiiinanan，故n为偶数时，原命题也成立，而且由证明过程可知，当且仅当120naaa时取等号，假设12,na aa不全为零，则取不到等号.综上，联赛加试题一的加强命题获证.2 此题总分值40 分设,21nAAAS其中nAAA,21是n个互不相同的有限集合)2(n，满足对任意的SAAji,，均有SAAji，假设2min1iniAk，证明：存在iniAx1，使得x属于nAAA,21中的至少kn个集合证明：不妨设1|Ak设在12,