2015年江苏省高考数学试卷答案与解析.pdf

1 2015 年江苏省高考数学试卷参考答案与试题解析一、填空题本大题共14 小题，每题5分，共计70 分1 5 分 2015?江苏已知集合A=1，2，3，B=2，4，5，则集合A B 中元素的个数为5考点：并集及其运算专题：集合分析：求出 AB，再明确元素个数解答：解：集合A=1，2，3，B=2，4，5，则 AB=1，2，3，4，5；所以 AB 中元素的个数为5；故答案为：5 点评：题考查了集合的并集的运算，根据定义解答，注意元素不重复即可，属于基础题2 5 分 2015?江苏已知一组数据4，6，5，8，7，6，那么这组数据的平均数为6考点：众数、中位数、平均数专题：概率与统计分析：直接求解数据的平均数即可解答：解：数据4，6，5，8，7，6，那么这组数据的平均数为：=6故答案为：6点评：此题考查数据的均值的求法，基本知识的考查3 5 分 2015?江苏设复数z 满足 z2=3+4i i 是虚数单位，则 z 的模为考点：复数求模专题：数系的扩充和复数分析：直接利用复数的模的求解法则，化简求解即可2 解答：解：复数z 满足 z2=3+4i，可得|z|z|=|3+4i|=5，|z|=故答案为：点评：此题考查复数的模的求法，注意复数的模的运算法则的应用，考查计算能力4 5 分 2015?江苏根据如下图的伪代码，可知输出的结果S 为7考点：伪代码专题：图表型；算法和程序框图分析：模拟执行程序框图，依次写出每次循环得到的I，S 的值，当I=10 时不满足条件I8，退出循环，输出S的值为 7解答：解：模拟执行程序，可得S=1，I=1 满足条件I8，S=3，I=4 满足条件I8，S=5，I=7 满足条件I8，S=7，I=10 不满足条件I8，退出循环，输出S的值为 7故答案为：7点评：此题主要考查了循环结构的程序，正确判断退出循环的条件是解题的关键，属于基础题5 5 分 2015?江苏袋中有形状、大小都相同的4 只球，其中1 只白球、1 只红球、2只黄球，从中一次随机摸出2 只球，则这2 只球颜色不同的概率为考点：古典概型及其概率计算公式专题：概率与统计分析：根据题意，把4个小球分别编号，用列举法求出基本领件数，计算对应的概率即可解解：根据题意，记白球为A，红球为B，黄球为C1、C2，则文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P4文档编码:CW7D3E4C4F6 HR3O10Y2I5J10 ZQ7P4U1R4P43 答：一次取出2 只球，基本领件为AB、AC1、AC2、BC1、BC2、C1C2共 6 种，其中 2 只球的颜色不同的是AB、AC1、AC2、BC1、BC2共 5 种；所以所求的概率是P=故答案为：点评：此题考查了用列举法求古典概型的概率的应用问题，是基础题目6 5 分 2015?江苏已知向量=2，1，=1，2，假设 m+n=9，8 m，n R，则 m n 的值为3考点：平面向量的基本定理及其意义专题：平面向量及应用分析：直接利用向量的坐标运算，求解即可解答：解：向量=2，1，=1，2，假设m+n=9，8可得，解得 m=2，n=5，m n=3故答案为：3点评：此题考查向量的坐标运算，向量相等条件的应用，考查计算能力7 5 分 2015?江苏不等式24 的解集为 1，2考点：指、对数不等式的解法专题：函数的性质及应用；不等式的解法及应用分析：利用指数函数的单调性转化为x2x2，求解即可解答：解；24，x2x2，即 x2x2 0，解得：1x2 故答案为：1，2点此题考查了指数函数的性质，二次不等式的求解，属于简单的综合题目，难度不文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P14 评：大8 5 分 2015?江苏已知tan=2，tan+=，则 tan的值为3考点：两角和与差的正切函数专题：三角函数的求值分析：直接利用两角和的正切函数，求解即可解答：解：tan=2，tan+=，可知 tan +=，即=，解得 tan=3故答案为：3点评：此题考查两角和的正切函数，基本知识的考查9 5 分 2015?江苏现有橡皮泥制作的底面半径为5，高为 4 的圆锥和底面半径为2，高为 8 的圆柱各一个，假设将它们重新制作成总体积与高均保持不变，但底面半径相同的新的圆锥和圆柱各一个，则新的底面半径为考点：棱柱、棱锥、棱台的体积专题：计算题；空间位置关系与距离分析：由题意求出原来圆柱和圆锥的体积，设出新的圆柱和圆锥的底面半径r，求出体积，由前后体积相等列式求得r解答：解：由题意可知，原来圆锥和圆柱的体积和为：设新圆锥和圆柱的底面半径为r，则新圆锥和圆柱的体积和为：，解得：故答案为：点评：此题考查了圆柱与圆锥的体积公式，是基础的计算题文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P15 10 5 分 2015?江苏在平面直角坐标系xOy 中，以点 1，0为圆心且与直线mxy2m1=0m R相切的所有圆中，半径最大的圆的标准方程为x12+y2=2考点：圆的标准方程；圆的切线方程专题：计算题；直线与圆分析：求出圆心到直线的距离d 的最大值，即可求出所求圆的标准方程解答：解：圆心到直线的距离d=，m=1 时，圆的半径最大为，所求圆的标准方程为x12+y2=2故答案为：x12+y2=2点评：此题考查所圆的标准方程，考查点到直线的距离公式，考查学生的计算能力，比较基础11 5 分 2015?江苏设数列 an满足 a1=1，且 an+1 an=n+1n N*，则数列 的前10 项的和为考点：数列的求和；数列递推式专题：等差数列与等比数列分析：数列 an 满足 a1=1，且 an+1an=n+1n N*，利用“累加求和”可得an=再利用“裂项求和”即可得出解答：解：数列 an满足 a1=1，且 an+1an=n+1n N*，当 n 2 时，an=anan1+a2a1+a1=+n+2+1=当 n=1 时，上式也成立，an=2 数列 的前 n 项的和 Sn=文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P16 数列 的前 10 项的和为故答案为：点评：此题考查了数列的“累加求和”方法、“裂项求和”方法、等差数列的前n 项和公式，考查了推理能力与计算能力，属于中档题12 5 分 2015?江苏在平面直角坐标系xOy 中，P 为双曲线x2y2=1 右支上的一个动点，假设点P 到直线 xy+1=0 的距离大于c 恒成立，则实数c 的最大值为考点：双曲线的简单性质专题：计算题；圆锥曲线的定义、性质与方程分析：双曲线 x2y2=1 的渐近线方程为x y=0，c 的最大值为直线xy+1=0 与直线 xy=0的距离解答：解：由题意，双曲线x2y2=1 的渐近线方程为x y=0，因为点 P 到直线 x y+1=0 的距离大于c 恒成立，所以 c 的最大值为直线xy+1=0 与直线 x y=0 的距离，即故答案为：点评：此题考查双曲线的性质，考查学生的计算能力，比较基础13 5 分 2015?江苏已知函数f x=|lnx|，gx=，则方程|fx+gx|=1 实根的个数为4考点：根的存在性及根的个数判断专题：综合题；函数的性质及应用分析：由|f x+g x|=1 可得 gx=fx 1，分别作出函数的图象，即可得出结论解答：解：由|fx+gx|=1 可得 gx=f x 1gx与 hx=fx+1 的图象如下图，图象有两个交点；文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P17 gx与 x=fx 1 的图象如下图，图象有两个交点；所以方程|fx+gx|=1 实根的个数为4故答案为：4点评：此题考查求方程|fx+gx|=1 实根的个数，考查数形结合的数学思想，考查学生分析解决问题的能力，属于中档题14 5 分 2015?江苏设向量=cos，sin+cos k=0，1，2，12，则ak?ak+1的值为考点：数列的求和专题：等差数列与等比数列；平面向量及应用分析：利用向量数量积运算性质、两角和差的正弦公式、积化和差公式、三角函数的周期性即可得出解解：文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P1文档编码:CP5E10J4P1M1 HH6G4E3O1H6 ZE1T8A4Z8P18 答：=+=+=+