广东省东莞市东华高级中学2022-2023学年高一下学期2月月考数学试题

这是一份广东省东莞市东华高级中学2022-2023学年高一下学期2月月考数学试题，共9页。试卷主要包含了单选题，多选题，解答题等内容，欢迎下载使用。
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. 2022B. 2018 C. 4036D. 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. （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123456789101112DACDBABDBDABCBCBCD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：0010-1003010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