2015年重庆市高考数学试卷(理科)答案与解析.pdf

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1、1 2015 年重庆市高考数学试卷理科参考答案与试题解析一、选择题：本大题共10 小题，每题5分，共 50 分，在每题给出的四个选项中，只有一项是符合题目要求的.1 5 分 2015?重庆已知集合A=1，2，3，B=2，3，则AA=B BA B=?CAB DBA 考点：子 集与真子集专题：集 合分析：直 接利用集合的运算法则求解即可解答：解：集合 A=1，2，3，B=2，3，可得 A B，AB=2，3，BA，所以 D 正确故选：D点评：此 题考查集合的基本运算，基本知识的考查2 5 分 2015?重庆在等差数列an中，假设a2=4，a4=2，则 a6=A1 B0C1D6考点：等 差数列的性质专

2、题：等 差数列与等比数列分析：直 接利用等差中项求解即可解答：解：在等差数列an 中，假设 a2=4，a4=2，则 a4=a2+a6=2，解得 a6=0故选：B点评：此 题考查等差数列的性质，等差中项个数的应用，考查计算能力3 5 分 2015?重庆重庆市2013 年各月的平均气温数据的茎叶图如，则这组数据的中位数是A19 B20 CD23 考点：茎 叶图专题：概 率与统计分析：根 据中位数的定义进行求解即可2 解答：解：样本数据有12 个，位于中间的两个数为20，20，则中位数为，故选：B 点评：此 题主要考查茎叶图的应用，根据中位数的定义是解决此题的关键比较基础4 5 分 2015?重庆“

3、x1”是“x+2 0”的A充要条件B充分而不必要条件C必要而不充分条件D既不充分也不必要条件考点：充 要条件专题：简 易逻辑分析：解“x+2 0”，求出其充要条件，再和x1 比较，从而求出答案解答：解：由“x+2 0”得：x+21，解得：x 1，故“x1”是“x+2 0”的充分不必要条件，故选：B点评：此 题考察了充分必要条件，考察对数函数的性质，是一道基础题5 5 分 2015?重庆某几何体的三视图如下图，则该几何体的体积为ABCD考点：由 三视图求面积、体积专题：空 间位置关系与距离分析：判 断三视图对应的几何体的形状，利用三视图的数据，求解几何体的体积即可解答：解：由三视图可知，几何体是

4、组合体，左侧是三棱锥，底面是等腰三角形，腰长为，高为 1，一个侧面与底面垂直，并且垂直底面三角形的斜边，右侧是半圆柱，底面半径为 1，高为 2，所求几何体的体积为：=故选：A点评：此 题考查三视图与直观图的关系，组合体的体积的求法，判断几何体的形状是解题的关键文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8

5、Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P

6、8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6

7、P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R

8、6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6

9、R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O

10、6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9

11、O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y93 6 5 分 2015?重庆假设非零向量，满足|=|，且3+2，则与的夹角为ABCD考点：数 量积表示两个向量的夹角专题：平 面向量及应用分析：根 据向量垂直的等价条件以及向量数量积的应用进行求解即可解答：解：3+2，?3+2=0，即 3222?=0，即?=3222=2，cos，=，即，=，故选：A 点评：此 题主要考查向量夹角的求解，利用向量数量积的应用以及向量垂直的等价条件是解决此题的关键7 5 分 2

12、015?重庆执行如下图的程序框图，假设输出k 的值为 8，则判断框图可填入的条件是文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文

13、档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9

14、文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y

15、9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8

16、Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P

17、8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6

18、P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R

19、6P8Y94 AsBsCsDs考点：循 环结构专题：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k，S的值，当S时，退出循环，输出 k 的值为 8，故判断框图可填入的条件是S解答：解：模拟执行程序框图，k 的值依次为0，2，4，6，8，因此 S=此时 k=6，因此可填：S故选：C点评：此 题考查了当型循环结构的程序框图，根据框图的流程判断程序运行的S 值是解题的关键8 5 分 2015?重庆已知直线l：x+ay1=0a R是圆 C：x2+y24x2y+1=0 的对称轴过点A 4，a作圆 C 的一条切线，切点为B，则|AB|=A2BC6D考点：直 线与圆的位置关系专题：

20、直 线与圆分析：求 出圆的标准方程可得圆心和半径，由直线l：x+ay1=0 经过圆 C 的圆心 2，1，求得 a 的值，可得点A 的坐标，再利用直线和圆相切的性质求得|AB|的值解答：解：圆 C：x2+y24x2y+1=0，即 x22+y 12=4，表示以C2，1为圆心、半径等于2的圆由题意可得，直线l：x+ay 1=0 经过圆 C 的圆心 2，1，故有 2+a1=0，a=1，点 A 4，1 由于 AC=2，CB=R=2，切线的长|AB|=6，故选：C点评：此 题主要考查圆的标准方程，直线和圆相切的性质，属于基础题9 5 分 2015?重庆假设tan=2tan，则=A1B2C3D4考点：三角函

21、数的积化和差公式；三角函数的化简求值文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B

22、10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6

23、B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F

24、6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1

25、F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA

26、1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 H

27、A1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y95 专题：三角函数的求值分析：直接利

28、用两角和与差的三角函数化简所求表达式，利用同角三角函数的基本关系式结合已知条件以及积化和差个数化简求解即可解答：解：tan=2tan，则=3故答案为：3点评：此题考查两角和与差的三角函数，积化和差以及诱导公式的应用，考查计算能力10 5 分 2015?重庆设双曲线=1a0，b0的右焦点为F，右顶点为A，过 F 作 AF 的垂线与双曲线交于B，C 两点，过B，C 分别作 AC，AB 的垂线，两垂线交于点 D假设 D 到直线 BC 的距离小于a+，则该双曲线的渐近线斜率的取值范围是A 1，00，1B，1 1，+C，00，D，+考点：双 曲线的简单性质专题：计 算题；创新题型；圆锥曲线的定义、性质与

29、方程文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P

30、8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6

31、P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R

32、6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6

33、R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O

34、6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9

35、O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y96 分析：由双曲线的对称性知D 在 x 轴上，设 D x，0，则由

36、BD AC 得，求出 cx，利用 D 到直线 BC 的距离小于a+，即可得出结论解答：解：由题意，Aa，0，Bc，Cc，由双曲线的对称性知D 在 x轴上，设 Dx，0，则由 BDAC 得，cx=，D 到直线 BC 的距离小于a+，cx=a+，c2a2=b2，01，双曲线的渐近线斜率的取值范围是1，0 0，1 故选：A点评：此 题考查双曲线的性质，考查学生的计算能力，确定D 到直线 BC 的距离是关键二、填空题：本大题共3 小题，考生作答5 小题，每题5 分，共 25 分.把答案填写在答题卡相应位置上.11 5 分 2015?重庆设复数a+bia，b R的模为，则 a+bi abi=3考点：复

37、数代数形式的乘除运算；复数求模专题：数 系的扩充和复数分析：将 所求利用平方差公式展开得到a2+b2，恰好为已知复数的模的平方解答：解：因为复数a+bia，b R的模为，所以 a2+b2=3，则 a+bi abi=a2+b2=3；故答案为：3点评：此 题考查了复数的模以及复数的乘法运算；属于基础题12 5 分 2015?重庆的展开式中x8的系数是用数字作答 文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV

38、9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 Z

39、V9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10

40、ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10

41、 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B1

42、0 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B

43、10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1

44、B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y97 考点：二 项式定理专题：二 项式定理分析：先 求出二项式展开式的通项公式，再令x 的幂指数等于8，求得 r 的值，即可求得展开式中的x8的系数解答：解：由于的展开式的通项公式为Tr+1=?，令 15=8，求得 r=2，故开式中x8的系数是?=，故答案为：点评：此 题主要考查二项式定理的应用，二项式展开式的通

45、项公式，属于基础题13 5 分 2015?重庆在 ABC 中，B=120，AB=，A 的角平分线AD=，则 AC=考点：余 弦定理的应用专题：解 三角形分析：利 用已知条件求出A，C，然后利用正弦定理求出AC 即可解答：解：由题意以及正弦定理可知：，即，ADB=45 ，A=180 120 45，可得 A=30 ，则 C=30，三角形ABC 是等腰三角形，AC=2=故答案为：点评：此 题考查正弦定理以及余弦定理的应用，三角形的解法，考查计算能力三、考生注意：14、15、16三题为选做题，请从中任选两题作答，假设三题全做，则按前两题给分14 5 分 2015?重庆如题图，圆O 的弦 AB，CD 相

46、交于点 E，过点 A 作圆 O 的切线与 DC 的延长线交于点P，假设 PA=6，AE=9，PC=3，CE：ED=2：1，则 BE=2考点：与 圆有关的比例线段专题：选 作题；推理和证明分析：利 用切割线定理计算CE，利用相交弦定理求出BE 即可解答：解：设 CE=2x，ED=x，则 过点 A 作圆 O 的切线与DC 的延长线交于点P，由切割线定理可得PA2=PC?PD，即 36=3 3+3x，文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5

47、U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B

48、5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4

49、B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B

50、4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10B4B5U1 HA1F6B10R1B10 ZV9O6R6P8Y9文档编码:CW4V10