湖南省株洲市渌口区第三中学2022-2023学年高二上学期期末数学试题

2022―2023学年度高二年级上学期期末检测

数学试题

注意事项：

1.答卷前，考生务必将自己的姓名、考场号、座位号、准考证号填写在答题卡上。

2.回答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑。如需改动，用橡皮擦干净后，再选涂其他答案标号。回答非选择题时，将答案写在答题卡上。写在本试卷上无效。

3.考试结束后，将本试卷和答题卡一并交回。

考试时间为120分钟，满分150分

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.已知点是直线上一点，则直线的倾斜角为（    ）

A.30° B.60° C.120° D.150°

2.若点在圆的内部，则实数a的取值范围是（    ）

A. B. C. D.

3.已知数列：1，2，3，5，8，…则89是该数列的第（    ）项.

A.9 B.10 C.11 D.12

4.已知的三个顶点分别为，，，则边上的高等于（    ）

A. B. C. D.

5.空间四边形中，点M在上，且，N为的中点，则（    ）

A. B.

C. D.

6.在平面直角坐标系中，椭圆C的中心在原点，焦点，在x轴上，离心率为，过的直线l交椭圆C于A，B两点，且的周长为16，则椭圆C的方程为（    ）

A. B. C. D.

7.若直线与直线交于点M，则M到坐标原点距离的最小值为（    ）

A. B. C. D.

8.如图，在边长为的等边三角形中，圆与三条边相切，圆与圆相切且与、相切，…，圆与圆相切且与、相切，依次得到圆、、…、.当圆的半径小于时，n的最小值为（    ）

（参考数据：，）

A.4 B.5 C.6 D.7

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.已知直线l：，则下列结论正确的是（    ）

A.直线l在两坐标轴上的截距均为1

B.向量是直线l的一个法向量

C.过点与直线l平行的直线方程为

D.若直线m：，则

10.已知等差数列的公差，当且仅当时，的前n项和最小，则（    ）

A. B. C. D.

11.已知圆柱底面半径，高，下底面圆心为O，上底面圆心为，A是下底面圆上任意一点，B是上底面圆上任意一点，则下列命题中正确的是（    ）

A.向量与的夹角为定值 B.的最大值为

C.与夹角的最大值为 D.

12.已知到两定点，距离乘积为常数16的动点P的轨迹为C，则（    ）

A.C一定经过原点  B.C关于x轴、y轴对称

C.的面积的最大值为45 D.C在一个面积为64的矩形内

三、填空题：本题共4小题，每小题5分，共20分。

13.写出一个与x轴y轴都相切的圆的方程____________.

14.某学校一航模小组进行飞机模型飞行高度实验，飞机模型在第一分钟内上升了10米高度.若通过动力控制系统，可使飞机模型在以后的每一分钟内上升的高度都是它在前一分钟上升高度的80%.在此动力控制系统下，该飞机模型在第三分钟内上升的高度是____________米.

15.已知椭圆（），以原点为圆心，b为半径的圆经过椭圆的左右焦点，，M为椭圆短轴的一个端点，的延长线交椭圆于点N，____________.

16.已知棱长为1的正方体中，E为侧面中心，F在棱上运动，正方体表面上有一点P满足（，），则所有满足条件的P点构成图形的面积为____________.

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.（10分）已知圆Ｃ的方程为.

（1）若直线与圆C交于A，B两点，且，求m的值；

（2）当时，过点作圆Ｃ的切线l，求切线l的方程.

18.（12分）已知数列中，，，（，）.

（1）求数列的通项公式；

（2）为数列的前n项和，设，是数列的前n项和，求证：.

19.（12分）如图，在棱长都相等的平行六面体中，，，两两夹角均为60°.

（1）求的值；

（2）求证：平面.

20.（12分）如图，正方体的棱长为2，E、F分别为和的中点，P为棱上的动点.

（1）若点P与重合，求证：平面；

（2）当平面与平面所成锐二面角的正弦值最小时，求的值.

21.（12分）已知抛物线C：（）与圆O：交于A，B两点，且，直线l过C的焦点F，且与C交于M，N两点.

（1）抛物线C的方程；

（2）求的最小值.

22.（12分）已知曲线C为双曲线的右支，斜率为k的直线l过双曲线右焦点，且与曲线C相交于A，B两点.

（1）求斜率k的取值范围；

（2）在x轴上是否存在点M使得，如果存在，求出点M的坐标；如果不存在，请说明理由.

数学参考答案及评分意见

1.B  【解析】因为，故的倾斜角，故选B.

2.A  【解析】点在圆的内部，即，解得，故选A.

3.B  【解析】由题意可得数列从第3项起，每一项等于前两项的和，所以这个数列为：1，2，3，5，8，13，21，34，55，89，…所以89是该数列的第10项.故选B.

4.D  【解析】由题意，，，可得，，

，，所以边上的高.故选D.

5.B  【解析】，故选B.

6.D  【解析】的周长为，即，离心率为，故，，所以椭圆的方程为，故选D.

7.C  【解析】两直线满足，所以两直线垂直，

由得，过定点，

由得，过定点，

故交点在以为直径的圆上，其中，则线段的最小值为.故选C.

8.B  【解析】由题，正三角形内切圆半径与边长满足：①，

，由正三角形性质易得，边上的高过圆心，且圆的切线（），其中，则组成正三角形，即，

则边长与高满足：，即②，

由①②得，，

由，∴数列为首项为2，公比为的等比数列，故.

由，得，解得，

因为且，故的最小值为5.故选B.

9.BCD  【解析】对于A，令，则，令，则，故A错误；

对于B，因为直线的方向向量为或，则，所以向量是直线的一个法向量；故B正确.

对于C，设与直线平行的直线方程为（），因为直线过，所以，所以过点与直线平行的直线方程为，C正确.

对于D，直线：的斜率为1，直线的斜率为，斜率，所以两直线垂直，故D正确.故选BCD.

10.ABD  【解析】当且仅当时，最小，∴当时，；当时，，

∴解得，

∴；

∴；；；ABD正确；

∵，则当时，；C错误.故选ABD.

11.ACD  【解析】以下底面互相垂直的两条直径和上下底面圆心连线为坐标轴建立空间直角坐标系；

设，，，

，因为，所以，故正确；

，所以的最大值为1，故B错误；

，

所以，所以与夹角的最大值为，故C正确；

，，，故，故D正确.故选ACD.

12.BCD  【解析】设点的坐标为，由题意可得.

对于A，将原点坐标代入方程得，所以A错误；

对于B，点关于轴、轴的对称点分别为、.

∵，

∵，

则点、都在曲线上，，所以曲线关于轴、轴对称，B正确；

对于C，设，，，则，

由余弦定理得，

当且仅当时等号成立，则，所以，

则的面积为，C正确；

对于D，，

可得，得，解得，

由C知，，得，

曲线在一个面积为的矩形内，D正确.故选BCD.

13.（答案不唯一）

14.  【解析】由题意，飞机模型每分钟上升的高度构成，公比的等比数列，则米.即飞机模型在第三分钟内上升的高度是米.

15.  【解析】由题意，且三角形为等腰直角三角形，

因为，，.

设，则，根据，得，解得.

故.故答案为.

16.  【解析】∵（，），，，，四点共面，

设，，，四点确定的平面，则与平面的交线与平行，

①当与重合时，取的中点，连接，，则，

则此时的轨迹为折线；

②当与重合时，，

此时的轨迹为折线，

∴当在棱上运动时，符合条件的点在正方体表面围成的图形为，直角梯形，.

∴.故答案为.

17.（1）  （2）或

【解析】（1）由题意知，圆的标准方程为（），·············································1分

所以圆心，半径，····································································2分

所以圆心到直线的距离.································································3分

由题意知，，所以.解得.·································································4分

（2）当时，圆半径为2.································································5分

当切线的斜率不存在时，切线的方程为.····················································6分

当切线的斜率存在时，设切线的方程为，

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

由题意知，.解得.······································································8分

此时切线的方程为.····································································9分

综上，切线的方程为或.································································10分

18.（1）  （2）证明见解析

【解析】（1）∵（，），

∴，.··············································································1分

故数列是首项为1，公比为2的等比数列，，·················································2分

∴时，·············································································4分

时也成立···········································································5分

∴.················································································6分

（2）∵，

∴数列是首项为1，公比为2的等比数列.····················································7分

∴数列的前项和为.····································································8分

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

∴数列的前项和，

∴.···············································································12分

19.（1）0  （2）见解析

【解析】设平行六面体的棱长为1.

（1）令，，，

则，.··············································································1分

则有，·············································································2分

故.················································································3分

故，

.··················································································5分

（2），············································································6分

.··················································································7分

故，

.··················································································9分

故，即.············································································10分

又由（1）知，，····································································11分

所以平面.··········································································12分

20.（1）见解析  （2）

【解析】（1）在正方体中，建立如图所示的空间直角坐标系，

因为、分别为和的中点，与重合，

则，，，，，

，，，·············································································2分

由，即，···········································································3分

由，即，···········································································4分

而，且，平面，因此，平面，····························································5分

（2）在（1）的空间直角坐标系中，设，，，，··············································6分

令平面的法向量为，

则令，得，··········································································8分

取平面的法向量，····································································9分

设平面与平面所成的锐二面角为，

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

当且仅当时取等号，

因此，当，即时，

，，当且仅当时取等号，······························································11分

所以当，即平面与平面所成锐二面角的正弦值最小时，此时.····································12分

21.（1）  （2）

【解析】（1）设，根据抛物线和圆的对称性得，由，解得，····································1分

故点在抛物线：上，···································································2分

所以，解得，········································································3分

所以：.·············································································4分

（2）由抛物线：，得，

设直线：，与联立

消去得，···········································································6分

设，，所以，，······································································7分

，

，·················································································9分

则，

当且仅当，时等号成立，······························································11分

故的最小值为.·······································································12分

22.（1）  （2）存在，定点

【解析】（1）由题意得双曲线的右焦点为，

设直线1的方程为：，联立方程···························································1分

整理得，···········································································2分

因为直线与曲线有两个交点，设，，

所以···············································································4分

解得或，

故斜率的取值范围为.··································································5分

（2）由（1）可得，，又，，

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

假设在轴上存在点满足，

则，，，

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

展开可得即.·········································································8分

因为斜率的取值范围为，

所以，即，··········································································9分

整理可得，·········································································10分

即，解得，·········································································11分

所以轴上存在点使得，且.······························································12分