湖南省常德市2022-2023高三数学上学期期末检测试卷（Word版附答案）

2022—2023学年度上学期常德市高三检测考试

数   学（试题卷）

本试卷满分150分，考试时间120分钟

注意事项：

1．所有试题的答案请在答题卡的指定区域内作答．

2．考试结束后，只交答题卡．

一、选择题：本题共8小题，每小题5分，共40分. 在每小题给出的四个选项中，只有一项是符合题目要求的.

1. 已知集合，，则

A.    B.  C.  D.

2. 已知复数，则复数在复平面内对应的点所在的象限为

A．第一象限   B.第二象限     C.第三象限    D.第四象限

3. 已知向量满足，且，则向量在向量上的投影向量为

A.   B.  C.   D.

4. 沙漏是我国古代的一种计时工具，是用两个完全相同的圆锥顶对顶叠放在一起组成的（如图）．在一个圆锥中装满沙子，放在上方，沙子就从顶点处漏到另一个圆锥中，假定沙子漏下来的速度是恒定的．已知一个沙漏中沙子全部从一个圆锥中漏到另一个圆锥中需用时80分钟．设经过t分钟沙漏上方圆锥中的沙子的高度与下方圆锥中的沙子的高度恰好相等（假定沙堆的底面是水平的），则t的值为

A.10      B.20       C.60      D.70

5. 在平面直角坐标系中，已知点为角终边上的点，则

A.    B.     C.   D.

6. 在平面直角坐标系中，已知直线与圆相交的弦长为，则

A.      B.      C.     D.

7. 已知，，，则

A.    B.  C.   D.

8. 已知双曲线的左右焦点分别为、，过的直线与曲线C的左右两支分别交于点M、N，且，则曲线C的离心率为

A.       B.   C.      D.

二、选择题：本题共4小题，每小题5分，共20分. 在每小题给出的选项中，有多项符合 题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分．

9. 已知抛物线，为坐标原点，点P为直线上一点，过点P作抛物线C的两条切线，切点分别为A，B，则

A.抛物线的焦点坐标为(0，1)     B.抛物线的准线方程为

C.直线AB一定过抛物线的焦点      D.

10. 已知定义在上的函数满足，且为奇函数，则下列说法一定正确的是

A.函数的周期为           B.函数的图像关于对称  

C.函数为偶函数            D.函数的图像关于对称

11．下列说法正确的是

A.数据6，5，3，4，2，7，8，9的上四分位数为7

B.若，且函数为偶函数，则

C.若随机事件A，B满足：，则A，B相互独立

D.已知采用分层抽样得到的样本数据由两部分组成，第一部分样本数据的平均数为，方差为；第二部分样本数据的平均数为，方差为，若总的样本方差为，则

12. 如图，已知正方体的棱长为2，分别是棱的中点，是侧面内(含边界)的动点，则下列说法正确的是

A.若直线与平面平行，则三棱锥P－AEF的

体积为

B.若直线与平面平行，则直线A1B1上存在唯一的点Q，使得DQ与A1P始终垂直

C.若，则EP的最小值为

D.若，则的最大值为

三、填空题：本题共4小题，每小题5分，共20分.

13. 已知函数，若曲线在点处的切线与直线垂直，则切线的方程为_____________．

14. 的展开式中的常数项为_____________．

15. 若函数在内存在唯一极值点，且在上单调递减，则的取值范围为_____________.

16. 已知数列满足首项，，则数列的前2n项的和为_____________．

四、解答题：本题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.

17．(本小题满分10分)

已知数列的首项，且满足．

(1)求证：数列是等比数列；

(2)若，求满足条件的最大整数的值．

18．(本小题满分12分)

如图，在梯形中，AD//BC，且，.

(1)若，，求梯形的面积；

(2)若，证明：为直角三角形．

19．(本小题满分12分)

如图所示的几何体是由等高的直三棱柱和半个圆柱组合而成，点G为的中点，DE为半个圆柱上底面的直径，且，．H为BC的中点．

(1)证明：平面DEH平面GCF；

(2)若Q是线段HE上一动点，求直线AQ与平面GCF所成角的正弦的最大值．

20．(本小题满分12分)

常益长高铁的试运营，标志着我省迈入“市市通高铁”的新时代.常益长高铁全线长157公里，共设有常德站、汉寿站、益阳南站、宁乡西站、长沙西站5个车站. 在试运营期间，铁路公司随机选取了乘坐常德开往长沙G6575次复兴号列车的名乘客，记录了他们的乘车情况，得到下表（单位：人）：

(用频率代替概率)

(1)从这200名乘客中任选一人，求该乘客仅乘坐一站的概率；

(2)在试营运期间，从常德上车的乘客中任选3人，设这3人到长沙下车的人数为X，求X的分布列，及其期望；

(3)已知德山经开区的居民到常德站乘车的概率为0.6，到汉寿站乘车的概率为0.4，若经过益阳南站后高铁上有一位来自德山经开区的乘客，求该乘客到长沙下车的概率.

21．(本小题满分12分)

已知点为椭圆上的一点，椭圆C的离心率为.

(1)求椭圆C的方程；

(2)如图，过点P作直线l1、l2，分别交椭圆于另一点M、R，直线l1，l2交直线l：x=3于N，S，设直线l1，l2的斜率分别为k1，k2，且k1+k2=0，若面积是面积的2倍，求直线l1的方程.

22．(本小题满分12分)

已知函数.

(1)讨论函数的零点个数；

(2)证明：当，.

2022-2023学年度上学期常德市高三检测考试

数  学（参考答案）

一、选择题：本题共8小题，每小题5分，共40分．

二、选择题：本题共4小题，每小题5分，共20分．全部选对的得5分，部分选对的得2分，有选错的得0分.

三、填空题：本题共4小题，每小题5分，共20分．

13.       14.       15.        16.

三、解答题：本大题共70分，解答应写出文字说明、证明过程或演算步骤．

17. (本小题满分10分)

解：(1),·······························································2分

，

是以为首项，为公比的等比数列········································5分

(2)由(1)可得：，

即：·······························································.8分

令，当n时，，单调递增；

又，，

满足不等式的最大整数·················································10分

18．（本小题满分12分）

解：(1)在中，由余弦定理得················································1分

在中，由余弦定理得················································2分

由有，，解得······················································3分

，又，···························································4分

梯形的面积

································································6分

法二：(1)取BC的中点E，连AE，则BE=CE=2································1分

AD=EC，AD//EC，四边形AECD为平行四边形·······························2分

AE=DC=2··························································3分

在中，

又，····························································4分

梯形的面积·······················································6分

(2)设，，则，，

在中，由正弦定理得，即①

在中，由正弦定理得，即②··············································8分

由①②得：··························································9分

化简得，

又

所以································································11分

又

所以，，为直角三角形··················································12分

法二：取BC的中点E，连AE，则BE=CE=2

AD=EC，AD//EC，四边形AECD为平行四边形··································8分

···································································10分

，为直角三角形·······················································12分

19．（本小题满分12分）

(1)证明：取DE的中点M，连MG、MH·········································1分

···································································2分

，又平面CGF

···································································3分

，又平面CGF

···································································4分

···································································5分

(2)如图，以C为原点，CB为x轴，CF为y轴，CD为z轴建立空间直角坐标系，

则A(2,0,2)，C(0,0,0)，F(0,2,0)，G(―1,1,2)，H(1,0,0)，E(0,2,2)······················6分

则,设面CGF的法向量

令得,

即·······························8分

································9分

设所求线面角为，

则·································································11分

所以当时，取得最大值为················································12分

20．（本小题满分12分）

解：(1)仅乘坐一站的乘客有10+10+10+30=60人该乘客仅乘坐一站的概率···············2分

(2)从常德上车的乘客到长沙下车的概率·······································3分

故这3人到长沙下车的人数，·············································5分

·····································································7分

···································································8分

(3) 记事件A：该乘客在过益阳南站后到长沙站下车，记事件B1:该乘客在常德站上车，记事件B2:该乘客在汉寿站上车.

，·································································10分

···································································12分

(阅卷说明：①直接得结果的不扣分；

②的本问给1分)

21．（本小题满分12分）

解：(1)由题可知·························································2分

解得：，

椭圆C的方程为·····················································4分

(2)记，设，

则直线l1：；直线l2：

联立消y得：

则，即······························································6分

又

···································································8分

同理································································9分

；

；即································································10分

，解得：

直线l1的方程为：即····················································12分

22．（本小题满分12分）

解：(1)由题意的零点

即为方程的实数解，即：·············································1分

令，

则······························································2分

令，；

当时，，单调递增；

当时，，单调递减.

································································3分

，在单调递减，

又，；···························································4分

所以，当，与函数有一个交点， 有一个零点；

当，与函数没有交点，无零点····································5分

(2)令，

···································································6分

在单调递减，

···································································8分

···································································9分

即·································································12分