江西省赣州市2023届高三数学（理）上学期1月期末考试试题（Word版附解析）

赣州市2022～2023学年度第一学期期末考试

 高三数学（理科）试卷 2023年1月

本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共150分，考试时间120分钟

第Ⅰ卷

一、选择题：本题共12小题，每小题5分，共60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.集合，，则（）

A. B. C. D.

2.函数则（）

A. B.0 C. D.2

3.若数列是等比数列，且，，则（）

A. B. C.62 D.64

4.为了研究某班学生的右手一拃长x（单位：厘米）和身高y（单位：厘米）的关系，从该班随机抽取了12名学生，根据测量数据的散点图可以看出y与x之间有线性相关关系，设其回归直线方程为，已知，，，若某学生的右手一拃长为22厘米，据此估计其身高为（）

A.175 B.179 C.183 D.187

5.若复数（a，，为其共轭复数），定义：.则对任意的复数，有下列命题：

：；：；：；：若，则为纯虚数.

其中正确的命题个数为（）

A.1 B.2 C.3 D.4

6.程大位（1533~1606），明朝人，珠算发明家.在其杰作《直指算法统宗》里，有这样一道题：荡秋千，平地秋千未起，踏板一尺离地，送行二步与人齐，五尺人高曾记。仕女佳人争蹴，终朝笑语欢嬉，良工高士素好奇，算出索长有几？将其译成现代汉语，其大意是，一架秋千当它静止不动时，踏板离地一尺，将它向前推两步（古人将一步算作五尺）即10尺，秋千的踏板就和人一样高，此人身高5尺，如果这时秋千的绳索拉得很直，请问绳索有多长？（）

A.14尺 B.14.5 尺C.15尺 D.15.5尺

7.已知过抛物线C：的焦点F的直线l被C截得的弦长为8，则坐标原点O到l的距离为（）

A. B. C. D.

8.若展开式的各项系数和为729，展开式中的系数为（）

A. B. C.30 D.90

9.直线与双曲线E：（，）交于M，N两点，若为直角三角形（其中O为坐标原点），则双曲线E的离心率为（）

A. B. C. D.

10.已知函数，的最小值为a，则实数a的值为（）

A. B. C. D.1

11.在三棱锥中，，，且，，，，则三棱锥的外接球的表面积为（）

A. B. C. D.

12.已知，，，则（）

A. B. C. D.

第Ⅱ卷

本卷包括必考题和选考题两部分，第13～21题为必考题，每个试题考生都必须作答，第22～23题为选考题，考生根据要求作答.

二、填空题：本题共4小题，每小题5分，共20分.

13.已知，均为单位向量且夹角为45°，则________.

14.已知tan，则__________.

15.如图，、、是同一平面内的三条平行直线，与间的距离是1，与间的距离是2，等腰直角三角形的三顶点分别在、l2、上，则的斜边长可以是__________（写出一个即可）.

16.斐波那契，意大利数学家，其中斐波那契数列是其代表作之一，即数列满足，且，则称数列为斐波那契数列.已知数列为斐波那契数列，数列满足，若数列的前12项和为86，则__________.

三、解答题：共70分.解答应写出文字说明、证明过程或演算步骤.第17～21题为必考题，每个试题考生都必须作答.第22、23题为选考题，考生根据要求作答.

17.（本小题满分12分）

在中，角A，B，C的对边分别为a，b，c，满足.

（1）求B的值；

（2）若与边上的高之比为3∶5，且，求的面积.

18.（本小题满分12分）

如图，在四棱锥中，底面，且，，.

（1）证明：平面平面；

（2）求平面与平面夹角的余弦值.

19.（本小题满分12分）

设有标号为1，2，3，…，n的n个小球（除标号不同外，其余均一样）和标号为1，2，3，…，n的n个盒子，将这n个小球任意地放入这n个盒子，每个盒子放一个小球，若i（，2，3，…，n）号球放入了i号盒子，则称该球放对了，否则称放错了.用表示放对了的球的个数.

（1）当时，求的概率；

（2）当时，求的分布列与数学期望.

20.（本小题满分12分）

已知椭圆C：（）过点且离心率为，过点作两条斜率之和为0的直线，，交C于A，B两点，交C于M，N两点.

（1）求椭圆C的方程；

（2）是否存在实数使得？若存在，请求出的值，若不存在，说明理由.

21.（本小题满分12分）

已知函数（其中e为自然对数的底数），且曲线在处的切线方程为.

（1）求实数m，n的值；

（2）证明：对任意的，有.

请考生在第22、23题中任选一题作答.如果多做，则按所做的第一题计分.

22.[选修4-4：坐标系与参数方程]

在平面直角坐标系中，已知直线l的参数方程为（t为参数），以坐标原点O为极点，x轴的正半轴为极轴，取相同的单位长度建立极坐标系，曲线C的极坐标方程为.

（1）求直线l的普通方程和曲线C的直角坐标方程；

（2）设点，直线l与曲线C的交点为A，B，求的值.

23.[选修4-5：不等式选讲]

已知函数的最小值为m.

（1）求m的值；

（2）设a，b，c为正数，且，求证：.

赣州市2022～2023学年度第一学期期末考试

高三理科数学参考答案

一、选择题

12.解：由，得，由得，即，

当时，令，，

则，则在上单调递增，

则，即，即，则.

或由对数平均不等式得，即，

即.

二、填空题

13.1；14.；15.或或（三个任选一个）；16.8.

16.解析：斐波那契数列：1，1，2，3，5，8，13，21，34，55，89，144，…….

（特征：每三项中前两项为奇数后一项为偶数）

由得：，，，

则，

同理：，，，，，

得：，，，，

则，，

则，则.

三、解答题

17.解：（1）由，

由内角和定理得：····································································1分

进而得：···········································································2分

解得：，（舍去）····································································4分

从而得.·············································································6分

（2）由题设知，不妨设，······························································7分

由余弦定理得：······································································8分

联立得：···········································································9分

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

故，从而的面积·····································································12分

18.解：（1）连接，由题设，，得.························································1分

又，故，由余弦定理可得·······························································2分

从而有·············································································3分

又面且面，得········································································4分

，故面·············································································5分

又面，所以面面······································································6分

（2）法一：由（1）同理可得面，························································6分

以为坐标原点，,分别为x，y轴的正方向建立如图所示的空间直角坐标系，

则，，，，，········································································7分

由（1）知面，故面的一个法向量为·······················································8分

设面的法向量为，

由，，

结合，令，得········································································9分

故的一个法向量为···································································10分

记平面与平面夹角为，

则有··············································································12分

（注：取的中点，连接，易证面，则为平面的一个法向量）

法二：以为坐标原点，，分别为，轴的正方向建系，

可得面的一个法向量为，面的一个法向量为

19.证：（1）由题设知：································································4分

（2）的所有可能取值为0，1，2，3，5·····················································5分

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

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

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

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

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

故的分布列为

故的数学期望为·····································································12分

20.解：（1）解得·····································································3分

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

（2）∵直线与的斜率之和为0，令直线的斜率为，则直线的斜率为，

令，，则的方程为，

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

则，···············································································7分

则直线的斜率为，令，，

同理：，···········································································8分

，

，

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

同理：

·················································································11分

，所以存在实数·····································································12分

方法二（2）∵直线与的斜率之和为0，令直线的倾斜角为，则直线的斜率为倾斜角为··················5分

令直线的参数方程为（t为参数）·························································6分

代入得：

即·················································································7分

令点，对应的参数分别为，，则，

则·················································································9分

同理：令直线的参数方程为，（为参数）

代入得：

即

令点，对应的参数分别为，，则·························································10分

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

则，所以存在实数···································································12分

21.解：（1）由，得···································································1分

由题意知：··········································································3分

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

（2）记············································································5分

则

记

记·················································································6分

显然在上单调递增，

且当时，；当时，；

所以在上递减，在上递增，

故·················································································7分

（注：这里的符号也可以通过来说明.）

又，

故存在唯一，，使得···································································8分

故当时，；当时，；

即在与上单调递增，在上单调递减························································9分

且，，故存在唯一，使································································10分

故当时，；当时，.

所以在与单调递减，在与上递增·························································11分

故是与中的较小值，又且，

故恒成立.证毕·······································································12分

22.解：（1）直线l的普通方程为.·························································2分

由得，则

曲线C的直角坐标方程为································································5分

（2）直线l的普通方程为（t为参数）······················································6分

设点，对应的参数分别为，，

将代入得···········································································7分

则，，则，··········································································8分

则················································································10分

23.解：（1）由题意得·································································3分

从而函数在递减，在上递增·····························································4分

故，即.·············································································5分

（2）由（1）知：，又a，b，c为正数，

由，，·············································································6分

法一：·············································································7分

而·················································································8分

所以··············································································10分

法二：由，，，

，，···············································································8分

得················································································10