2023东莞东华高级中学高一下学期2月月考数学试题含答案

东华高级中学  东华松山湖高级中学

2022—2023学年第二学期高一2月考数学试卷

一、单选题：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. 命题“，”的否定是（    ）

A. ， B. ，

C. ， D. ，

2. 设，则“”是“”的（　　）.

A. 充分不必要条件 B. 必要不充分条件 C. 充分必要条件       D. 既不充分也不必要条件

3. 函数的零点所在的区间为（    ）

A.               B.               C.             D.

4．若，，向量与向量的夹角为150°，则向量在向量上的投影向量为（    ）

A．    B．        C．          D．

5. 设，，则（    ）

A. 且  B. 且

C. 且  D. 且

6. 要得到函数的图象，只需将函数的图象进行如下变换得到（    ）

A. 向左平移个单位    B. 向右平移个单位    C. 向右平移个单位   D. 向左平移个单位

7．已知，是方程的两根，且，，则的值为（    ）

A．     B．        C．或          D．或

8. 若定义上的函数满足：对任意有若的最大值和最小值分别为，则的值为（    ）

A. 2022                 B. 2018  C. 4036              D. 4044

二、多选题：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．在中，为中点，且，则（    ）

A．  B.    C．    D．

10．已知函数，则（    ）

A．的最大值为                           B．直线是图象的一条对称轴

C．在区间上单调递减               D．的图象关于点对称

11. 若，则下列关系式中一定成立的是（    ）

A.  B. （）

C. （是第一象限角） D.

12. 已知函数，若方程有四个不同的根，且，则下列结论正确的是（    ）

A.  B.

C.  D.

三、填空题：本题共4小题，每小题5分，共20分．

13．已知向量，满足，，，则______．

14. 请写出一个函数，使它同时满足下列条件：（1）的最小正周期是4；（2）的最大值为2．____________．

15. 若是定义在R上的奇函数，当时，(为常数)，则当时，_________.

16. 木雕是我国古建筑雕刻中很重要一种艺术形式，传统木雕精致细腻､气韵生动､极富书卷气．如图是一扇环形木雕，可视为扇形OCD截去同心扇形OAB所得部分．已知，，，则该扇环形木雕的面积为________．

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.（本题满分10分）

已知集合

(1)    求集合     (2)  若,求实数的取值范围.

18. （本题满分12分）

在平面直角坐标系中，是坐标原点，角的终边与单位圆的交点坐标为，射线绕点按逆时针方向旋转弧度后交单位圆于点，点的纵坐标关于的函数为.（1）求函数的解析式，并求的值；

（2）若，，求的值.

19.（本题满分12分）

函数

（1）请用五点作图法画出函数在上的图象（先列表，再画图）

（2）设，，当时，试研究函数的零点的情况.

20.（本题满分12分）

2020年我国面对前所未知，突如其来，来势汹汹的新冠肺炎疫情，中央出台了一系列助力复工复产好政策．城市快递行业运输能力迅速得到恢复，市民的网络购物也越来越便利．根据大数据统计，某条快递线路运行时，发车时间间隔t（单位：分钟）满足：，，平均每趟快递车辆的载件个数(单位：个)与发车时间间隔t近似地满足，其中．

（1）若平均每趟快递车辆的载件个数不超过1600个，试求发车时间间隔t的值；

（2）若平均每趟快递车辆每分钟的净收益(单位：元)，问当发车时间间隔t为多少时，平均每趟快递车辆每分钟的净收益最大？并求出最大净收益(结果取整数)．

21 .（本题满分12分）

已知函数是定义域上的奇函数，且满足

（1）          判断函数在区间上的单调性，并用定义证明

（2）          已知，且，若，证明：

22. （本题满分12分）

若函数对定义域内的每一个值，在其定义域内都存在唯一的，使成立，则称函数具有性质．

（1）判断函数是否具有性质，并说明理由；

（2）若函数的定义域为且且具有性质，求的值；

（3）已知，函数的定义域为且具有性质，若存在实数，使得对任意的，不等式都成立，求实数的取值范围．

东华高级中学  东华松山湖高级中学

2022—2023学年第二学期高一2月考数学答案

一、选择题

二、填空题     13.；     14. （答案不唯一）   15. ；    16.

三、解答题

17.解：（1），·································································4分

（2）由题意，若，则，···························································5分

①时，，解得； ································································ 6分

②时，，…………………… 8分       解得；…………………………………………………9分

综上，的取值范围为.····························································10分

18.解：（1）因为，且，所以，···················································2分

由此得········································································4分

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

（2）由知，即··································································7分

由于，得，与此同时，所以

由平方关系解得：，····························································9分

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

19、（1）·····································································2分

按五个关键点列表：

描点并将它们用光滑的曲线连接起来如图1：

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

（2）因为，

所以的零点个数等价于与图象交点的个数，············································8分

设，，则······································································9分

当，即时，有2个零点；

当，即时，有1个零点；

当，即时，有0个零点.      ······················································12分

20、解：（1）当时，，不满足题意，舍去． ···········································1分

当时，，即．·································································3分

解得（舍）或．·······························································4分

∵且，∴．···································································5分

所以发车时间间隔为5分钟．······················································6分

（2）由题意可得．····························································8分

当时，（元），·······························································9分

当且仅当，即时，等号成立，····················································10分

当时，单调递减，时，（元）····················································11分

所以发车时间间隔为6分钟时，净收益最大为140（元）．·······························12分

21.解：（1）由为奇函数，可得；·····················································1分

又，得；······································································2分

所以.

在上单调递增，理由如下：························································3分

，且，则······································································4分

因为，所以，，，

所以，，在上单调递增 ···························································6分

（2）证法一：由题意，，则有·····················································8分

因为，所以，即，······························································10分

所以，得证.··································································12 分          

证法二：由（1）知，在上单调递增，同理可证在上单调递减.

因为，，

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

要证，即证，

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

代入解析式得，即证

化简整理得，即证，····························································10分

因为，显然成立，······························································11分

所以原不等式得证，所以.························································ 12 分

22、解：（1）对于函数的定义域内任意的，

取，则，····································································1分

结合的图象可知对内任意的，是唯一存在的，········································2分

所以函数具有性质.

（2）因为，且，所以在上是增函数，···············································3分

又函数具有性质，所以，即，·····················································4分

因为，所以且，又，

所以，解得，所以．····························································5分

（3）因为，所以，且在定义域上单调递增，

又因为，在上单调递增，

所以在上单调递增，····························································6分

又因为具有性质，

从而，即，所以，

解得或（舍去），·····························································7分

因为存在实数，使得对任意的，不等式都成立，

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

因为在上单调递增，所以

即对任意的恒成立．····························································9分

所以或，···································································11分

解得或，       综上可得实数的取值范围是………………12分