2021东莞高一下学期期末考试数学试题含答案

这是一份2021东莞高一下学期期末考试数学试题含答案，共14页。试卷主要包含了单项选择题，多项选择题，填空题，解答题等内容，欢迎下载使用。
东莞市学年度第二学期期末教学质量检查高一数学一、单项选择题：本大题共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.请把正确选项在答题卡中的相应位置涂黑.1.已知，是虚数单位，则复数z对应的点位于A.第一象限           B.第二象限         C.第三象限         D.第四象限2.已知向量，共线，则x的值为A.     B.0     C.1     D.23.图1是水平放置的的斜二测直观图，是的中点，则中长度最长的线段为
图1A.OA     B.OB     C.AD     D.AB4.已知一组数据为85，87，88，90，92，则这组数据的第60百分位数为A.87    B.     C.89     D.915.已知圆柱的底面半径为1，母线长为2，则该圆柱的外接球的体积为A.     B.     C.     D.6.“大美中国古建筑名塔”榴花塔以红石为基，用青砖灰沙砌筑建成.如图2，记榴花塔高为OT，测量小组选取与塔底O在同一水平面内的两个测量点A和B，现测得，，，在点B处测得塔顶T的仰角为，则塔高OT为 A.     B.      C.     D. 年3月，树人中学组织三个年级的学生进行“庆祝中国共产党成立100周年”党史知识竞赛.经统计，得到前200名学生分布的饼状图图和前200名中高一学生排名分布的频率条形图图前200名学生分布的饼状图

图1
前200名中高一学生排名分布的频率条形图
图2则下列命题错误的是（   ）A.成绩前200名的200人中，高一人数比高二人数多30人B.成绩第名的100人中，高一人数不超过一半C.成绩第名的50人中，高三最多有32人D.成绩第名的50人中，高二人数比高一的多8.在四边形ABCD中，，，，，则AD的长为（    ）A.     B.    C.     D.二、多项选择题：本大题共4小题，每小题5分，共20分.在每小题给出的四个选项中，有多项符合题目要求，全部选对的得5分，有选错的得0分，部分选对的得2分.请把正确选项在答题卡中的相应位置涂黑.9.袋子中有1个红球，1个黄球，1个蓝球，从中取两个球，每次取一个球，取球后不放回，设事件第一个球是红球，第二个球是黄球，则下列结论正确的是A.A与B互为对立事件          B.A与B互斥C.               D.10.已知复数，是z的共轭复数，则下列结论正确的是A.若，则           B.若，则C.若，则          D.若，则11.已知与均为单位向量，其夹角为，则下列结论正确的是       B.C.        D.12.如图，在正方体中，M，N，P分别为棱，，的中点，则下列结论正确的是
 A.平面B.点P与点D到平面的距离相等C.平面截正方体所得截面图形为等腰梯形D.平面将正方体分割成的上、下两部分的体积之比为 三、填空题：13.如图，由A，B两个元件组成一个串联电路，元件A，B正常工作的概率分别为和，则电路正常工作的概率为__________.

 14.广场上的玩具石凳是由正方体截去八个一样大的四面体得到的如图如果被截正方体的棱长为，那么玩具石凳的表面积为__________.

 15.已知点，，C为坐标平面内一点，且，则满足条件的点C的一个坐标为__________. 16.“牟和方盖”是我国古代数学家刘徽在研究球的体积的过程中构造的一个和谐优美的几何体，它是由两个相同的圆柱分别从纵横两个方向嵌入一个正方体时两圆柱公共部分形成的几何体如图如图2所示的“四脚帐篷”类似于“牟和方盖”的一部分，其中APC与BPD为相互垂直且全等的半圆面，它们的圆心为O，半径为用平行于底面ABCD的平面去截“四脚帐篷”所得的截面图形为__________;当平面经过OP的中点时，截面图形的面积为__________. 
图1                                         图2 四、解答题：本大题共6小题，第17题10分，18、19、20、21、22题各12分，共70分.解答应写出文字说明、证明过程或演算步骤.必须把解答过程写在答题卡相应题号指定的区域内，超出指定区域的答案无效.17.设复数，，记复数z与分别对应复平面内的点和(1)根据复数及其运算的几何意义，求Z和两点间的距离;(2)已知为正实数表示动点Z的集合是以点为圆心，r为半径的圆.那么满足条件的点Z的集合是什么图形?并求出该图形的面积.  18.在中，角A，B，C所对的边分别为a，b，现有下列四个条件：①，②，③，④(1)③④两个条件可以同时成立吗?请说明理由;(2)请在上述四个条件中选择使有解的三个条件，并求出的面积. 19.如图10，四棱锥中，ABCD为正方形，E为PC中点，平面平面ABCD，，
 (1)证明：平面(2)证明：(3)求三棱锥的体积. 20.如图，斜坐标系xOy中，，分别是与x轴、y轴正方向同向的单位向量，且，的夹角为，定义向量在斜坐标系xOy中的坐标为有序数对在斜坐标系xOy中完成下列问题：
 (1)若向量的坐标为，计算的大小;(2)若向量的坐标为，向量的坐标为，判断下列两个命题的真假，并说明理由.命题①若，则命题②若，则   月23日是世界读书日，树人中学为了解本校学生课外阅读情况，按性别进行分层，用分层随机抽样的方法从全校学生中抽出一个容量为100的样本，其中男生40名，女生60名.经调查统计，分别得到40名男生一周课外阅读时间单位：小时的频数分布表和60名女生一周课外阅读时间单位：小时的频率分布直方图：以各组的区间中点值代表该组的各个值女生一周自读时间颊率分布直方图(1)从一周课外阅读时间为的学生中按比例分配抽取6人，从这6人中任意抽取2人，求恰好一男一女的概率;(2)分别估计男生和女生一周课外阅读时间的平均数，(3)估计总样本的平均数和方差参考数据和公式：男生和女生一周课外阅读时间方差的估计值分别为和，和分别表示男生和女生一周阅读时间的样本，其中     22.如图1，“雪糕筒”为校园中常见的交通标识，其可以近似的看成一个圆锥.如图2，放置在水平地面上的某型号“雪糕筒”底面直径，母线，该“雪糕筒”绕点B被放倒后A、B、在同一条直线上.

     图1 图2(1)求“雪糕筒”被放倒后最高点离水平地面的距离;(2)求直线PB与圆面所成的角的余弦值;(3)若放倒后的“雪糕简”绕点沿水平地面旋转一周，请说明旋转一周形成的曲面所围成的旋转体的特征不用说明理由                东莞市2020－2021学年度第二学期期末教学质量检查高一数学 参考答案一、单项选择题题号12345678答案DADC BADB 二、多项选择题（全部选对的得5分，选对但不全的得2分，有选错的得0分）题号9101112答案CDABACBCD 三、填空题（16题第一空2分，第二空3分）13. 0.72    14.    15. （形如的坐标均正确）  16. 正方形，  四、解答题17. 解：（1）解法1：因为点对应的复数分别为，·································1分所以点和之间的距离······················································3分. ·····································································5分解法2：复数，分别对应向量，···············································1分所以 ··································································3分 . ·····································································5分说明：直接写出结果，只给2分.（2）由题知方程表示的动点的集合是以点为圆心、1为半径的圆，····················6分方程表示的动点的集合是以点为圆心、3为半径的圆，·····························7分故不等式表示的动点的集合是以点为圆心、半径分别为1和3的两个圆所形成的圆环形图形（含边界） ， ······9分所以该圆环形图形的面积为. ···············································10分18.解：（1）由条件③及正弦定理得，··············································1分由，得，·······························································2分展开得，即，···································································3分因为，所以，即，························································4分因为，所以.·····························································5分由条件④得， ，··························································7分因为，所以  ，···························································8分在中，不符合题意，所以不能同时满足条件③④．································9分  （2）因为同时满足上述四个条件中的三个，不能同时满足，则满足三角形有解的所有组合为或．若选择组合：由，得，解得 ，·························································10分  因为所以，即为直角三角形，解得，·········································11分  所以的面积.····························································12分  若选择组合：由，即，解得，·························································10分  由，且得，····························································11分  所以的面积．···························································12分19.证明：（1）连接交于点，连接,············································1分因为四边形为正方形，所以点为的中点，又为的中点，所以,························································2分又因为，, ······························································3分所以 .·································································4分（2）因为四边形为正方形，所以，···········································5分又因为平面平面，平面平面,平面，············································6分所以平面，·····························································7分因为平面，所以.··································································8分解：（3）取中点，连接，因为，所以，又因为平面平面，平面平面，平面，所以平面，·····························································9分由，，得，，································································10分记点到平面的距离为，因为为的中点，所以，··································11分所以.··································································12分说明：若平面在第2问已经证明，则前两问共9分.20.解：（1）由题知，·····················································1分故····································································3分.······································································4分（2）由题知， 命题①是真命题.a. 当时，即，显然;························································5分b. 当时，即至少一个不为，不妨设,若，则存在，使得，故,····················································6分即，因为不共线，所以，······················································7分由，带入得，即.综上所述，命题“若，则”是真命题. ·········································8分命题②是假命题.解法1：若，则，···································································10分当时，结论不成立，······················································11分所以，命题“若，则”是假命题. ············································12分解法2：令的坐标为，的坐标为，则，·········································9分因为，所以，···························································10分此时，································································11分所以，命题“若，则”是假命题. ············································12分21. 解：（1）一周课外阅读时间为的学生中男生有3人，女生有人， 若从中按比例分配抽取6人，则男生有1人，记为,女生有5人，记为，，，，，·················1分则样本空间， ···························································2分记事件“恰好一男一女”，则，···············································3分所以，所以从这6人中任意抽取2人恰好一男一女的概率为.································4分（2）估计男生一周课外阅读时间平均数；······································6分估计女生一周课外阅读时间的平均数. ·········································8分（3）估计总样本的平均数，················································9分，，，，··································································10分，，··································································11分， 所以估计总样本的平均数和方差分别是3.6和3. ·································12分22. 解：（1）由图可知，“雪糕筒”绕点被放倒后、、在同一条直线上，所以轴截面与地面垂直，故点为放倒后的最高点， ······1分在中，，所以，··························································2分所以， ································································3分解得，即最高点离水平地面的距离为.··········································4分（2）由题知，轴截面圆面，且轴截面水平地面，轴截面圆面，且轴截面水平地面，因为、、在同一条直线上，所以五点共面，·····································5分故平面圆面，即直线在圆面内的射影为直线，所以为直线与圆面所成的角，···············································6分解法1：在直角梯形中，，··················································7分可得，·································································8分在中，， 所以直线与圆面所成的角的余弦值为.··········································9分解法2：设，则,··························································7分在中，，·······························································8分所以， 所以直线与圆面所成的角的余弦值为.··········································9分（3）如图所示，放倒后的“雪糕筒”绕点沿水平地面旋转一周形成的曲面所围成的旋转体为半球（第一个得分点）被平行于底面的平面截去平面上方的部分，留下截面与底面之间的部分（第二个得分点），再挖去一个以截面圆为底面、球心为顶点的圆锥（第三个的分点）所剩下的部分. ······12分说明：每个得分要点1分，共3分.