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1、2012 杨浦区高考数学质量抽查试卷数学（理科）20XX 年 3 月一、填空题（每小题 4分，满分 56 分）1若线性方程组的增广矩阵为135246，则其对应的线性方程组是25(1)x+的展开式中2x的系数是（结果用数字作答）.3若双曲线2221(0)9xyaa的一条渐近线方程为023yx，则a=_4计算：2111lim(1)333nnL .5若直线l过点(2,0)，且与圆221xy相切，则直线l的斜率是 .6函数2)cos(sin)(xxxf的最小正周期为 .7一支田径队有男运动员48人，女运动员36人，若用分层抽样的方法从该队的全体运动员中抽取一个容量为21的样本，则抽取男运动员的人数为_
2、.8若行列式093114212xx，则x9如图，测量河对岸的塔高AB时，可以选与塔底B在同一水平面内的两个测点C与D.测得75BCD，60BDC，30CD米，并在点C测得塔顶A的仰角为60，则塔高AB_米.10.在不考虑空气阻力的条件下，火箭的最大速度v（米/秒）和燃料的质量M（千克）、火箭（除燃料外）的质量m（千克）的关系式是)1ln(2000mMv.当燃料质量与火箭（除燃料外）的质量之比为时，火箭的最大速度可达12（千米/秒）11.圆柱形容器内部盛有高度为8cm的水，若放入三个相同的球（球的半径与圆柱的底面半径相同）后，水恰好淹没最上面的球（如图所示)，则球的半径是cm12.设 幂 函 数
3、3)(xxf，若 数 列na满 足：20121a，且)(1nnafa，)(Nn则 数 列 的 通 项na13.对任意一个非零复数z，定义集合NnzAnz,，设是方程012x的一个根，若在A中任取两个不同的数，则其和为零的概率为P=(结果用分数表示)14函数11yx的图像与函数2sinyx)42(x的图像所有交点的横坐标之和等于_.二、选择题（每小题5 分，满分20 分）(11 题图)(9 题图)1AABECD1B1C1D15下列函数中既是奇函数，又在区间1,1上是增函数的为()AyxBsinyxCxxyeeD3yx16执行如图所示的程序框图，输出的S值为（）A1.B1.C2.D0.17“3ta
4、n3x”是“56x”()A充分非必要条件.B必要非充分条件.C充要条件.D既非充分也非必要条件.18已知点(1,1)A若曲线G上存在两点,B C，使ABC为正三角形，则称G为型曲线给定下列三条曲线：3(03)yxx；22(20)yxx；1(0)yxx其中，型曲线的个数是()A.0B.1C.2D.3三解答题（本大题满分74 分）本大题共5 题，解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.19（本题满分12 分）本题共有2 个小题，第1 小题满分 4 分，第 2 小题满分8 分 已知关于x的不等式022mxx解集为2,1.（1）求实数m的值；（2）若复数sincos,221izim
5、z，且21zz为纯虚数，求2tan的值.20（本题满分14 分）本题共有2 个小题，第1 小题满分 7 分，第 2 小题满分7 分 如图所示,直四棱柱1111ABCDA B C D的侧棱1AA长为a,底面ABCD是边长2ABa,BCa的矩形,E为11C D的中点,(1)求证:DE平面EBC；(2)求点C到平面EBD的距离.21（本题满分14 分）本题共有2 个小题，第1 小题满分 6 分，第 2 小题满分8 分设Ra,122)(2xxaaxf为奇函数.(16 题图)文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6
6、P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3
7、G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J
8、9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7
9、N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10
10、Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP
11、2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9
12、M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8文档编码:CP2J9D1J2X4 HB9M7N6P8X4 ZY5X10Y3G9X8（1）求函数11242)()(xxxfxF的零点；（2）设)1(log2)(2kxxg,若不等式1()()fxg x在区间1 2,2 3上恒成立,求实数k的取值范围.22（本题满分16 分）本题共有3 个小题，第1 小题满分 4 分，第 2 小题满分6 分，第 3 小题满分6 分.已知数列12:,nnAa aaL.如果数列12:,nnBb bbL满足1nba，11kkkkbaab，其中2,3
13、,knL，则称nB为nA的“生成数列”.（1）若数列41234:,Aa aaa的“生成数列”是4:5,2,7,2B，求4A；（2）若n为偶数，且nA的“生成数列”是nB，证明：nB的“生成数列”是nA；（3）若n为奇数，且nA的“生成数列”是nB，nB的“生成数列”是nC，.依次将数列nA，nB，nC，的第(1,2,)i inL项取出，构成数列:,iiiia b c L.探究：数列i是否为等差数列,并说明理由.23（本题满分18 分）本题共有2 个小题，第1 小题满分 4 分，第 2 小题的满分6 分;满分 8 分.如图，椭圆14:221yxC，x轴被曲线22:Cyxb截得的线段长等于1C的长
14、半轴长.（1）求实数b的值；（2）设2C与y轴的交点为M，过坐标原点O的直线l与2C相交于点BA、，直线MBMA、分别与1C相交与、DE.证明：0MEMD记BMA，MDE的面积分别是12,S S.若21SS=，求的取值范围.、(23 题图)文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6
15、R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 H
16、A6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6
17、W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10
18、ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B
19、6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8
20、文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:C
21、J5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H820XX 年杨浦区高三年级二模数学试卷(理科)参考答案和评分标准说明：1、本解答仅列出试题的一种解法，如果考生的解法与所列解答不同，可参考解答中的评分精神进行评分2、评阅试卷，应坚持每题评阅到底，不要因为考生的解答中出现错误而中断对该题的评阅，当考生的解答在某一步出现错误，影响了后继部分，但该步以后的解答未改变这一题的内容和难度时，可视影响程度决定后面部分的给分，这时原则上不应超过后面部分应给分数之半，如果有较严重的概念性错误，就不给分一填空题（本大题满分56 分）1.64253yxyx；2.5；3.2；4.23；5.33
22、；6.；7.12；8.2或3；9.245；10.16e；11.4；12.132012n；13.31；14.8；二、选择题（本大题满分20 分）本大题共有4 题 15.B；16.D；17.B；18.C；三、解答题（本大题满分74 分）本大题共5 题19.解：(1)4 2m20，解得 m=1 (2)21zz=(cos 2sin)(sin 2cos)i为纯虚数所以，cos 2sin 0，tan=12,所以，2tan=4320.(1)证明:由2ECEDa,2CDaECED,2 分BC平面11CC D DBCDE,4 分即 DE垂直于平面EBC中两条相交直线,因此 DE平面 EBC,7 分(2)解 1:
23、结合第(1)问得，由aDEaDB2,5，8 分aBE3，BEDE，所以，2263221aaaSBED 10 分又由BCDEBEDCVV得32312631aah 12 分文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:C
24、J5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6
25、R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 H
26、A6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6
27、W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10
28、ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B
29、6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8
30、1AABEC()D O1B1C1Dxyz故 C到平面 BDE的距离为ah36 14 分解 2:如图建立直角坐标系,则(0,)Ea a,(0,)OEa auuu r,(,2,0)B aa,(,2,0)OBaauuu r,9 分因此平面EBD的一个法向量可取为(2,1,1)nr,由(0,2,0)C,得(1,0,0)BCuuu r,11 分因此 C到平面 BDE的距离为|63|n BCdanr uuu rr.（其他解法,可根据【解1】的评分标准给分）21.解：由 f(x)是奇函数，可得 a=1，所以，f（x）2121xx（1）F（x）2121xx42121xx2(2)2621xxx由2(2)26xx
31、0，可得2x 2，所以，x=1，即 F（x）的零点为x1。（2）f1（x）21log1xx，在区间1 2,2 3上，由1()()fxg x恒成立，即21log1xx212log()xk恒成立，即2111xxxk恒成立即221 21,2 3kxx，259k，所以，5533k22.（1）解：由题意得：541ab；52122aab；57133aab；52144aab4:2,1,4,5A.（2）证法一：证明：由已知，111()nbaaa，212121()nbaabaaa.因此，猜想1(1)()iiinbaaa.当1i时，111()nbaaa，猜想成立；假设*()ik kN时，1(1)()kkknbaa
32、a.当1ik时，11kkkkbaab11(1)()kkkknaaaaa文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R
33、8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA
34、6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W
35、2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 Z
36、A8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6
37、M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文
38、档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H811(1)()kkkknaaaaa111(1)()kknaaa故当1ik时猜想也成立.由、可知，对
39、于任意正整数i，有1(1)()iiinbaaa.设数列nB的“生成数列”为nC，则由以上结论可知111(1)()(1)()(1)()iiiiininncbbbaaabb，其中1,2,3,inL.由于n为偶数，所以11(1)()nnnnbaaaa，所以11(1)()(1)()iiiinnicaaaaaa，其中1,2,3,inL.因此，数列nC即是数列nA.证法二：因为1nba，1212bbaa，2323bbaa，11nnnnbbaa，由于n为偶数，将上述n个等式中的第2,4,6,nL这2n个式子都乘以1，相加得11223112231()()()()()()nnnnnbbbbbbbaaaaaaaL
40、L即1nba，1nba.由于1nab，11(2,3,)iiiiabbainL，根据“生成数列”的定义知，数列nA是nB的“生成数列”.（3）证法一：证明：设数列nX,nY,nZ中后者是前者的“生成数列”.欲证i成等差数列，只需证明,iiix y z成等差数列，即只要证明2(1,2,3,)iiiyxzinL即可.由（2）中结论可知1(1)()iiinyxxx，1(1)()iiinzyyy11(1)()(1)()iiinnxxxyy11(1)()(1)(1)()iininnnnxxxxxxx11(1)()(1)()iiinnxxxxx文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA
41、8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M
42、10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档
43、编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5
44、N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8
45、P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6
46、K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2
47、B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H812(1)()iinxxx，所以，122(1)()2iiiinixzxxxy，即,iiix y z成等差数列，所以i是等差数列.证法二：因为11(2,3,4,)iiiibaabinL，所以11()(2,3,4,)iiiibabain
48、L.所以欲证i成等差数列，只需证明1成等差数列即可.对于数列nA及其“生成数列”nB，因为1nba，1212bbaa，2323bbaa，11nnnnbbaa，由于n为奇数，将上述n个等式中的第2,4,6,1nL这12n个式子都乘以1，相加得11223112231()()()()()()nnnnnbbbbbbbaaaaaaaLL即112nnnnbaaaaa.设数列nB的“生成数列”为nC，因为1nba，112nncbaa，所以1112bac，即111,a b c成等差数列.同理可证，111111,;,b c d c d e L也成等差数列.即1是等差数列.所以i成等差数列.23（1）由题意知：半
49、长轴为2，则有22 b1b（2）由题意知，直线l的斜率存在，设为k，则直线l的方程为ykx.由21ykxyx得210 xkx，设1122(,),(,)A xyB xy，则12,x x是上述方程的两个实根，于是1212,1xxk x x。又点M的坐标为(0,1)，所以文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H
50、8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码:CJ5N3O6R8P4 HA6K6Y6W2B10 ZA8K4B6M10H8文档编码: