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2021年2020年广东省深圳市高考数学一模试卷答案解析.pdf

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1、2020 年广东省深圳市高考数学一模试卷答案解析1/19 2020年广东省深圳市高考数学一模试卷答案解析一、选择题（共12 题，共 60 分）1已知集合A0，1，2，3，Bx|x22x30，则 AB（）A（1，3）B（1，3C（0，3）D（0，3【解答】解：集合 A0，1，2，3，Bx|x22x 30（1，3），则 AB（1，3，故选：B2设 z，则 z的虚部为（）A 1B1C 2D2【解答】解：z，z 的虚部为1故选：B3某工厂生产的30 个零件编号为01，02，19，30，现利用如下随机数表从中抽取5个进行检测若从表中第1 行第 5 列的数字开始，从左往右依次读取数字，则抽取的第5个零件

2、编号为（）34 57 07 86 36 04 68 96 08 23 23 45 78 89 07 84 42 12 53 31 25 30 07 32 8632 21 18 34 29 78 64 54 07 32 52 42 06 44 38 12 23 43 56 77 35 78 90 56 42A25B23C12D07【解答】解：根据随机数的定义，1行的第 5 列数字开始由左向右依次选取两个数字，依次为 07，04，08，23，12，则抽取的第5 个零件编号为，12，|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 1 页，共 19 页2

3、020 年广东省深圳市高考数学一模试卷答案解析2/19 故选：C4记 Sn为等差数列 an 的前n项和，若a23，a59，则 S6为（）A36B32C28D24【解答】解：S63（3+9）36故选：A5若双曲线（a0，b0）的一条渐近线经过点（1，2），则该双曲线的离心率为（）ABCD2【解答】解：双曲线（a0，b0）的一条渐近线经过点（1，2），点（1，2）在直线上，则该双曲线的离心率为e故选：C6已知 tan 3，则（）ABCD【解答】解：因为 tan 3，则cos2|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 2 页，共 19 页文档编码

4、:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5

5、R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码

6、:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5

7、R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码

8、:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5

9、R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码

10、:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J42020 年广东省深圳市高考数学一模试卷答案解析3/19 故选：D7的展开式中x3的系数为（）A168B84C42D21【解答】解：由于的展开式的通项公式为Tr+1?（2）rx72r，则令 72r3，求得 r2，可得展开式中x3的系数为?4 84，故选：B8函数 f（x）ln|e2x1|x 的图象大致为（）ABCD【解答】解：，故排除CD；f（1）ln|e21|+1

11、ln（1e2）+lne，故排除B故选：A9如图，网格纸上小正方形的边长为1，粗线画出的是某四面体的三视图，则该四面体的外接球表面积为（）|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 3 页，共 19 页文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6

12、S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7

13、V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6

14、S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7

15、V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6

16、S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7

17、V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J42020 年广东省深圳市高考数学一模试卷答案解析4/19 AB32C36D48【解答】解：根据几何体的三视图

18、转换为几何体为三棱锥体ABCD：如图所示：设外接球的半径为r，则：（2r）242+42+42，解得 r2 12，所以：S4 1248 故选：D10已知动点M 在以 F1，F2为焦点的椭圆上，动点N 在以 M 为圆心，半径长为|MF1|的圆上，则|NF2|的最大值为（）A2B4C8D16【解答】解：由椭圆的方程可得焦点在y 轴上，a24，即 a2，由题意可得|NF2|F2M|+|MN|F2M|+|MF1|，当 N，M，F2三点共线时取得最大值|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 4 页，共 19 页文档编码:CL1F3S5N6S9 HD1

19、S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4

20、文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1

21、S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4

22、文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1

23、S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4

24、文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1

25、S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J42020 年广东省深圳市高考数学一模试卷答案解析5/19 而|F2M|+|MF1|2a4，所以|NF2|的最大值为4，故选：B11著名数学家欧拉提出了如下定理：三角形的外心、重心、垂心依次位于同一直线上，且重心到外心的距离是重心到垂心距离的一半此直线被称为三角形的欧拉线，该定理则被称为欧拉线定理设点 O，H 分别是 ABC 的外心、垂心，且 M 为 BC 中点，则（）ABCD【解答】解：如图所示的

26、RtABC，其中角B 为直角，则垂心H 与 B 重合，O 为 ABC 的外心，OAOC，即 O 为斜边 AC 的中点，又 M 为 BC 中点，M 为 BC 中点，故选：D12已知定义在0，上的函数f（x）sin（x）（0）的最大值为，则正实数的取值个数最多为（）A4B3C2D1【解答】解：定义在 0，上的函数f（x）sin（x）（0）的最大值为，|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 5 页，共 19 页文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A

27、6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL

28、1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A

29、6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL

30、1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A

31、6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL

32、1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A

33、6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J42020 年广东省深圳市高考数学一模试卷答案解析6/19 0 1，解得 0 3，x 0时，则 sin（），令 g（）sin（），ysin（）在（0，上单调递增，g（0）0，g（）10，因此存在唯一实数，使得 sin（）3，sin（x）1，必须 3，x综上可得：正实数的取值个数最多为2 个故选：C二、填空题（共4 题，共 20 分）13若 x，y满足约束条件，则 zx2y 的最小值为3【解答】解：画出 x，y 满足约束条件，表示的平面区域，如图所示；结合图象知目标函数zx2y 过 A 时，

34、z取得最小值，由，解得 A（1，2），所以 z 的最小值为z122 3故答案为：3|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 6 页，共 19 页文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文

35、档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S

36、9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文

37、档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S

38、9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文

39、档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S

40、9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J42020 年广东省深圳市高考数学一模试卷答案解析7/19 14设数列 an的前n项和为 Sn，若 Sn2an n，则 a663【解答】解：数列 an的前 n 项和

41、为 Sn，由于 Sn2ann，所以当 n2 时，Sn1 2an1（n1），得：an2an1+1，整理得（an+1）2（an1+1），所以（常数），所以数列 an+1 是以 2为首项，2 为公比的等比数列所以，整理得所以故答案为：6315很多网站利用验证码来防止恶意登录，以提升网络安全某马拉松赛事报名网站的登录验证码由0，1，2，9 中的四个数字随机组成，将从左往右数字依次增大的验证码称为“递增型验证码”（如 0123），已知某人收到了一个“递增型验证码”，则该验证码的首位数字是1 的概率为【解答】解：基本事件的总数为，其中该验证码的首位数字是1 的包括的事件个数为|精.|品.|可.|编.|辑.

42、|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 7 页，共 19 页文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9

43、HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G

44、3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9

45、HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G

46、3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9

47、HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G

48、3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J42020 年广东省深圳市高考数学一模试卷答案解析8/19 该验证码的首位数字是1 的概率故答案为：16已知点M（m，m）和点 N（n，n）（mn），若线段MN 上的任意一点P 都满足：经过点P 的所有直线中恰好有两条直线与曲线C：y+x（1x 3）相切，则|mn|的最大值为

49、【解答】解：由点 M（m，m）和点 N（n，n），可得 M，N 在直线 yx上，联立曲线C：y+x（1x3），可得x2，无实数解，由 y+x 的导数为y x+1，可得曲线C 在 x 1 处的切线的斜率为0，可得切线的方程为y，即有与直线y x的交点 E（0，），同样可得曲线C 在 x3 处切线的斜率为4，切线的方程为y4x，联立直线y x，可得交点F（，），此时可设M（0，），N（，），则由图象可得|mn|的最大值为 0，|精.|品.|可.|编.|辑.|学.|习.|资.|料.*|*|*|*|欢.|迎.|下.|载.第 8 页，共 19 页文档编码:CL1F3S5N6S9 HD1S9M5R7A6

50、ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F3S5N6S9 HD1S9M5R7A6 ZL5N7V6G3J4文档编码:CL1F

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本文标题：2021年2020年广东省深圳市高考数学一模试卷答案解析.pdf
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