山东省青岛胶州市2020-2021学年高一下学期期中考试：数学试题及答案

这是一份山东省青岛胶州市2020-2021学年高一下学期期中考试：数学试题及答案，共11页。试卷主要包含了考试结束后，请将答题卡上交，设则大小关系正确的是，已知平面向量，且，则等内容，欢迎下载使用。
2020—2021学年度第二学期期中学业水平检测高一数学试题本试卷共6页，22题．全卷满分150分．考试用时120分钟．注意事项：1．答卷前，考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置上，并将条形码粘贴在答题卡指定位置上。2．回答选择题时，选出每小题答案后，用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，请将答题卡上交。 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． 1．的值为（    ） A． B． C． D． 2．四边形为矩形，对角线长为，若，则（    ） A． B． C． D．  3．已知为虚数单位，下列与相等的是（    ） A．  B． C．  D．  4．已知，为坐标原点，则下列说法正确的是（    ） A．  B．三点共线 C．三点共线  D． 5．已知角是的内角，向量 且与共线，则可以判断的形状为（    ） A．等腰三角形 B．等腰直角三角形 C．直角三角形 D．等边三角形  6．已知复数，，其中为虚数单位，，若为纯虚数，则下列说法正确的是（    ） A． B．复数在复平面内对应的点在第一象限 C． D．  7．如图所示，为测量山高，选择和另一座山的山顶为测量观测点，从点测得点的仰角，点的仰角以及，从点测得，若山高米，则山高等于（    ）  A．米 B．米 C．米 D．米  8．设则大小关系正确的是（    ）  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，选对但不全的得3分，有选错的得0分． 9．关于一组样本数据的平均数、中位数、频率分布直方图和方差，下列说法正确的是（    ） A．改变其中一个数据，平均数和中位数都会发生改变 B．频率分布直方图中，中位数左边和右边的直方图的面积应该相等 C．若数据的频率分布直方图为单峰不对称，且在左边“拖尾”，则平均数小于中位数D．样本数据的方差越小，说明样本数据的离散程度越小  10．已知平面向量，且，则（    ） A．  B．或 C．与夹角的大小为 D． 11．在中，角的对边分别是，，．则下列说法正确的是（    ） A．为锐角三角形 B．面积为或 C．长度为  D．外接圆的面积为 12．下列说法正确的是（    ） A．在中，是的充要条件 B．将函数的图象向右平移个单位长度得到函数的图象 C．存在实数，使得等式成立D．在中，若，则是钝角三角形    三、填空题：本大题共4小题，每小题5分，共20分．13．某人任意统计5次上班步行到单位所花的时间（单位：分钟）分别为．则这组数据的标准差为        ． 14．函数在区间上的值域为        ．15．在中, 已知，则        ．16．右图为某校名高一学生的体育测试成绩的频率分布直方图，如果要按照分层抽样方式抽取名学生进行分析，则要抽取的之间的学生人数是        ；估计这名学生的体育测试平均成绩为        ． （本小题第一空2分，第二空3分）   四、解答题：本大题共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）某蔬菜基地准备对现有地铺管道设施加装智能控制水泵系统，对大棚蔬菜进行自动控制根部滴灌，可以大大节省人力和资金，地铺管道在达到满水最大压状态后，水泵自动停机，管道可以自动连续进行滴注工作至最小工作压，然后水泵会重新开启，不同性能的管道系统根据最小工作压需要配置与之压力性能相对应的水泵系统，不同品种的蔬菜由于需要的滴注速度和强度不同，可以选择配备不同性能的管道系统和水泵系统．为充分了解基地内既有管道系统的总体情况，现随机抽取不同蔬菜棚内的若干条管道进行满水测试，对这些管道的最小工作压数据（单位：千帕）分组为，将其按从左到右的顺序分别编号为第一组，第二组，，第五组．下图是根据实验数据制成的频率分布直方图，已知，． （1）求的值；（2）已知最小工作压在以上的管道系统都需要分别配备台大功率水泵，若第一组与第二组共有条管道，求该基地需要配备的大功率水泵的台数．             18．（本题满分12分）已知向量，，在复平面坐标系中，为虚数单位，复数对应的点为．（1）求；（2）为曲线(为的共轭复数)上的动点，求与之间的最小距离；（3）若，求在上的投影向量．  19．（本题满分12分）已知函数．（1）将函数化为形式，求的最小正周期和单调递增区间；（2）若为的内角，恰为的最大值，求；（3）若，求． 20．（本题满分12分）在中，分别为角的对边，，．（1）求；（2）若，为延长线上一点，连接，，求的面积．  21．（本题满分12分）在中，为所在平面内的两点，，，．（1）以和作为一组基底表示，并求；（2）为直线上一点，设，若直线经过的垂心，求．  22．（本题满分12分）在平面直角坐标系中，已知，．（1）若，为轴上的一动点，点．（ⅰ）当三点共线时，求点的坐标；    （ⅱ）求的最小值；（2）若，，且与的夹角，求的取值范围．
2020—2021学年度第二学期期中学业水平检测高一数学 答案及评分标准一、单项选择题：本大题共8小题．每小题5分，共40分．1--8：B C D B     A C A D  二、多项选择题：本大题共4小题．每小题5分，共20分．9．BCD；     10．AC；     11．BD；     12．ABD．三、填空题：本大题共4小题，每小题5分，共20分．13．；     14．；     15．；     16．（1），（2）．四、解答题：本大题共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．（本小题满分10分）解：（1）根据分布列的特点，有·········································3分 因为，所以，························································5分（2）第一组和第二组的频率为···········································6分所以，管道总条数为条·················································7分所以最小工作压在以上的管道系统有条·····································9分所以该基地需要配备的大功率水泵的台数··································10分 18. （本小题满分12分）解：（1）···························································1分所以·······························································2分所以·······························································3分所以·······························································4分（2）······························································5分曲线，即，因此曲线是复平面内以圆心，半径为的圆···································6分故与之间的距离为·····················································7分所以与之间的最小距离为···············································8分（3）因为，所以······················································9分此时，与的夹角余弦为················································10分与方向相同的单位向量为 ··············································11分所以在上的投影向量··················································12分 19．（本小题满分12分）解：（1） ··································································3分所以的最小正周期·····················································4分由，得 所以的单调递增区间为·················································5分（2）因为，所以··························································6分所以当·····························································7分即当时，恰为的最大值·················································8分（3）··································································10分因为，所以·························································12分 20．（本小题满分12分）解：（1）由题意，根据正弦定理，可得：····································1分 所以即，即也即·······························································2分因为，所以，即······················································3分所以·······························································5分（2），所以由余弦定理知，······················································7分解得·······························································8分因为·······························································9分在中，因为，所以所以······························································10分又因为····························································11分所以=·····························································12分 21．（本小题满分12分）解：（1）由，所以为线段上靠近的三等分点·································1分由，所以为线段的中点·················································2分···································································4分因为·······························································5分所以·······························································6分（2）为直线上一点，设则···································································7分···································································8分因为直线经过的垂心，所以， 即··········································9分所以解得······························································10分所以因为，所以·························································12分 22. （本小题满分12分）解：（1）（ⅰ）设，，所以·············································1分因为，与共线························································2分所以，解得所以三点共线时，点的坐标为············································3分 （ⅱ）因为关于轴的对称点为············································4分所以·······························································5分 所以当三点共线时，取得最小值··········································6分最小值即····························································7分所以取得最小值（2），所以， 因为与的夹角，所以恒成立··············································8分所以，又因为，可得 即恒成立，又因为，可得恒成立··························································9分令，， 所以······························································10分因为，其中等号当且仅当成立···········································11分所以时，有最小值，所以的取值范围是：··················································12分