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2019年普通高等学校招生全国统一考试理科数学(北京卷)解析版.pdf

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1、绝密启用前2019 年普通高等学校招生全国统一考试（北京卷）理科数学本试卷共 5 页，150 分。考试时长 120 分钟。考生务必将答案答在答题卡上，在试卷上作答无效。考试结束后，将本试卷和答题卡一并交回。第一部分（选择题共 40 分）一、选择题共 8 小题，每小题 5 分，共 40 分。在每小题列出的四个选项中，选出符合题目要求的一项。1.已知复数z=2+i，则z zA.3B.5C.3 D.5【答案】D【解析】【分析】题先求得z，然后根据复数的乘法运算法则即得.【详解】z2i,z z(2i)(2i)5故选 D.【点睛】本容易题，注重了基础知识、基本计算能力的考查.2.执行如图所示的程序框图，

2、输出的s值为A.1 B.2 C.3 D.4【答案】B【解析】【分析】根据程序框图中的条件逐次运算即可.【详解】运行第一次，=1k，22 123 12s，运行第二次，2k，2222322s，运行第三次，3k，2222322s，结束循环，输出=2s，故选 B.【点睛】本题考查程序框图，属于容易题，注重基础知识、基本运算能力的考查.3.已知直线l 的参数方程为13,24xtyt（t 为参数），则点（1,0）到直线l 的距离是A.15B.25C.45D.65【答案】D【解析】【分析】首先将参数方程化为直角坐标方程，然后利用点到直线距离公式求解距离即可.【详解】直线l的普通方程为41320 xy,即 4

3、320 xy,点1,0到直线l的距离22|402|6543d,故选 D.【点睛】本题考查直线参数方程与普通方程转化,点到直线的距离,属于容易题,注重基础知识?基本运算能力的考查.4.已知椭圆22221xyab（a b0）的离心率为12，则文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W

4、1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文

5、档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W

6、1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文

7、档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W

8、1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文

9、档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1A.a2=2b2B.3a2=4b2C.a=2bD.3a=4b【答案】B【解析】【分析】由题意利用离心率的定义和,a b c

10、的关系可得满足题意的等式.【详解】椭圆的离心率2221,2cecaba,化简得2234ab,故选 B.【点睛】本题考查椭圆的标准方程与几何性质,属于容易题,注重基础知识?基本运算能力的考查.5.若 x，y 满足|1|xy，且 y-1，则 3x+y 的最大值为A.-7B.1 C.5 D.7【答案】C【解析】【分析】首先画出可行域，然后结合目标函数的几何意义确定其最值即可.【详解】由题意1,11yyxy作出可行域如图阴影部分所示.文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:

11、CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W

12、5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:

13、CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W

14、5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:

15、CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W

16、5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:

17、CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1设3,3zxy yzx,当直线0:3lyzx经过点2,1时,z 取最大值5.故选 C.【点睛】本题是简单线性规划问题的基本题型,根据“画?移?解”等步骤可得解.题目难度不大题,注重了基础知识?基本技能的考查.6.在天文学中，天体的明暗程度可以用星等或亮度来描述.两颗星的星等与亮度满足212152lgEmmE，其中星等为m1的星的亮度为E2（k=1,2）.已知太阳的星等是26.7，天狼星的星等是1.45，则太阳与天狼星的亮度的比值为A.1010.1B.10.1 C.lg10.1 D.1010.1【答案】D【解析】【分析】先求出

18、12lgEE，然后将对数式换为指数式求12EE，再求12EE.【详解】两颗星的星等与亮度满足12125lg2EmmE,令21.45m，126.7m，1212221g(1.4526.7)10.155EmmE，10.110.112211010EEEE，故选 D.【点睛】考查考生的数学应用意识、信息处理能力、阅读理解能力以及指数对数运算.7.设点 A，B，C 不共线，则“AB与AC的夹角为锐角”是“|ABACBC”的A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F

19、3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X

20、7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F

21、3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X

22、7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F

23、3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X

24、7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F

26、21|xyx y就是其中之一（如图）.给出下列三个结论：曲线 C 恰好经过6个整点（即横、纵坐标均为整数的点）；曲线 C 上任意一点到原点的距离都不超过2；曲线 C 所围成的“心形”区域的面积小于3.其中，所有正确结论的序号是A.B.C.D.【答案】C【解析】【分析】将所给方程进行等价变形确定x 的范围可得整点坐标和个数，结合均值不等式可得曲线上的文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:

27、CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W

28、5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:

29、CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W

30、5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:

31、CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W

32、5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1点到坐标原

33、点距离的最值和范围，利用图形的对称性和整点的坐标可确定图形面积的范围.【详解】由221xyx y得,221yx yx,2222|3341,10,2443xxxyx厔,所以x可为的整数有0,-1,1,从而曲线22:1C xyx y恰好经过(0,1),(0,-1),(1,0),(1,1),(-1,0),(-1,1)六个整点,结论正确.由221xyx y得,222212xyxy,解得222xy,所以曲线C上任意一点到原点的距离都不超过2.结论正确.如图所示,易知0,1,1,0,1,1,0,1ABCD,四边形ABCD的面积131 1 1 122ABCDS,很明显“心形”区域的面积大于2ABCDS,即“

34、心形”区域的面积大于3,说法错误.故选 C.【点睛】本题考查曲线与方程?曲线的几何性质，基本不等式及其应用,属于难题,注重基础知识?基本运算能力及分析问题解决问题的能力考查,渗透“美育思想”.第二部分（非选择题共 110 分）二、填空题共 6 小题，每小题 5 分，共 30 分。9.函数 f(x)=sin22x 的最小正周期是_文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O

35、8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I

36、8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O

37、8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I

38、8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O

39、8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I

40、8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1【答案】2.【解析】【分析】将

41、所给的函数利用降幂公式进行恒等变形，然后求解其最小正周期即可.【详解】函数2sin 2fxx142cos x,周期为2【点睛】本题主要考查二倍角的三角函数公式?三角函数的最小正周期公式，属于基础题.10.设等差数列 an的前 n 项和为 Sn，若 a2=-3，S5=-10，则 a5=_，Sn的最小值为_【答案】(1).0.(2).-10.【解析】【分析】首先确定公差,然后由通项公式可得5a的值，进一步研究数列中正项?负项的变化规律,得到和的最小值.【详解】等差数列na中,53510Sa,得322,3aa,公差321daa,5320aad,由等差数列na的性质得5n时,0na,6n时,na大

42、于 0,所以nS的最小值为4S或5S,即为10.【点睛】本题考查等差数列的通项公式?求和公式?等差数列的性质,难度不大,注重重要知识?基础知识?基本运算能力的考查.11.某几何体是由一个正方体去掉一个四棱柱所得，其三视图如图所示如果网格纸上小正方形的边长为1，那么该几何体的体积为_文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z

43、1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB

44、6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z

45、1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB

46、6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z

47、1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB

48、6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1【答案】40.【解析】【分析】画出三视图对应的几何体，应用割补法求几何体的体积

49、.【详解】在正方体中还原该几何体，如图所示几何体的体积V=43-12（2+4）2 4=40【点睛】易错点有二，一是不能正确还原几何体；二是计算体积有误.为避免出错，应注重多观察、细心算.12.已知 l，m是平面外的两条不同直线给出下列三个论断：lm；m；l以其中的两个论断作为条件，余下的一个论断作为结论，写出一个正确的命题：_【答案】如果 l，m，则 lm.文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1

50、文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6W1C2W5V9 ZO5X7I8Q8Z1文档编码:CS2A1A5F3O8 HB6

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本文标题：2019年普通高等学校招生全国统一考试理科数学(北京卷)解析版.pdf
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