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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．ππsincos1212+的值为（）A．32B．62C．64D．342．四边形ABCD为矩形，对角线长为4

，若,,ABaADbBDc===，则||abc−−=（）A．0B．6C．8D．103．已知i为虚数单位，下列与i相等的是（）A．1iB．(1i)(1i)−+C．1i1i−+D．2342021iiiii+++++4．已知3π

5π(2,1),(1,4),(sin,cos)23ABC−，O为坐标原点，则下列说法正确的是（）A．(1,5)AB=−B．,,AOC三点共线C．,,ABC三点共线D．3OAOBOC+=5．已知角,,ABC是ABC的内角，向量(sin,sin),(cos,cos)mABnAB==且m与n

共线，则可以判断ABC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221(1)(23)izmmm=−++−，23izm=+，其中i为虚数单位，Rm，若1z为纯虚数，则下列说法正确的是（）A．1m=

B．复数2z在复平面内对应的点在第一象限C．2||2z=D．2212||||zz=7．如图所示，为测量山高MN，选择A和另一座山的山顶C为测量观测点，从A点测得M点的仰角60MAN=，C点的仰角30CAB=以及75MAC=，从

C点测得60MCA=，若山高1002BC=米，则山高MN等于（）A．300米B．360米C．240米D．320米8．设2132tan121sin40cos4sin4,,,221tan122abc−=−

==+则,,abc大小关系正确的是（）A．cbaB．abcC．acbD．bca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(3,),(1,3)amb=−=−，且|2|||abab+=−，则（）A．3m=B．33

m=或3−C．a与b夹角的大小为5π6D．||||||abab+=+11．在ABC中，角,,ABC的对边分别是,,abc，222bcbca=++，8,213ba==．则下列说法正确的是（）A．ABC为锐角三角形B．ABC面积为43或123C．AB长度为6D．ABC外接圆

的面积为52π312．下列说法正确的是（）A．在ABC中，BAsinsin是BCAC的充要条件B．将函数xy2sin=的图象向右平移π6个单位长度得到函数πsin(2)3yx=−的图象C．存在实数x，使得等式3sincos2xx−=成立D．在ABC

中，若222sinsinsinABC+，则ABC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8,12,10,11,9．则这组数据的标准差为．1

4．函数sin3cosyxx=−在区间[0,π]上的值域为．15．在ABC中,已知3,2,10ABACBC===，则ABBC=．16．右图为某校1000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20

0名学生进行分析，则要抽取的[80,90)之间的学生人数是；估计这1000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）0.02成绩1009080700.0050.0460500.03频率/组距四、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[4,6),[6,8),[8,10),[10,12),[12,14)，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23ba=，,Rab．（1

）求,ab的值；（2）已知最小工作压在10以上的管道系统都需要分别配备2台大功率水泵，若第一组与第二组共有200条管道，求该基地需要配备的大功率水泵的台数n．0.08最小工作压（千帕）14121080.04b64a频率/组距18．（本题满分12分）已知向量2(cos2,2),(1,sin

)ab=−=−，2mab=+，在复平面坐标系中，i为虚数单位，复数1i1imz+=−对应的点为1Z．（1）求1||z；（2）Z为曲线1|2|1zz−=(1z为1z的共轭复数)上的动点，求Z与1Z之间的最小距离；（3）若π6=，求a在b上的投影向量n．19．（本题满分12分）已知函数2π

5π()2sin()3sin(2)(R)126fxxxx=−−+．（1）将函数()fx化为sin()Axk++形式，求()fx的最小正周期T和单调递增区间；（2）若为ABC的内角，()f恰为()

fx的最大值，求；（3）若πtan()26−=，求()f．20．（本题满分12分）在ABC中，,,abc分别为角,,ABC的对边，2cosacB=，2sin3B=．（1）求cosA；（2）若25

a=，P为CA延长线上一点，连接PB，BPBC=，求PBC的面积．21．（本题满分12分）在ABC中，,MN为ABC所在平面内的两点，π3,22,4ABACBAC===，13MCBC=，0NANC+=．（1）以AB和AC作为一组基底表示NM，并求||N

M；（2）D为直线MN上一点，设(,R)CDxAByACxy=+，若直线CD经过ABC的垂心，求,xy．22．（本题满分12分）在平面直角坐标系中，已知23(,),(8,8),(7,0)2AtBmmCmt−−−，,R,0tmt．（1）若1

,4tm==，P为x轴上的一动点，点(1,2)A−．（ⅰ）当,,APB三点共线时，求点P的坐标；（ⅱ）求||||PAPB+的最小值；（2）若sint=，(0,π)，且CA与CB的夹角π[0,)2，求m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：BCDBACAD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．13．2；14．[3,2]−；15．152−；16．（1）40，（2）73．四、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(0

.080.080.04)21ab++++=·······················3分因为23ba=，所以0.18b=，0.12a=··························································5分（2）第一组和第二组

的频率为(0.120.08)20.4+=········································6分所以，管道总条数为2005000.4=条····················································

··············7分所以最小工作压在10以上的管道系统有500(0.080.04)2120+=条·················9分所以该基地需要配备的大功率水泵的台数1202240n==···········

·····················10分18.（本小题满分12分）解：（1）22222(cos2,2)(1,sin)cos22sincossin2sinab=−−=+=−+22cossin1=+=····························

··········································1分所以2123mab=+=+=··········································································2分所以13i(3i

)(3i)(1i)33ii124i12i1i(1i)(1i)222z++++++−+======+−−+·············3分所以1|||12i|5z=+=······························

···················································4分（2）112iz=−··························································

································5分曲线|2|1zz−=，即|(24i)|1z−−=，因此曲线是复平面内以0(2,4)Z−圆心，半径为1的圆·········································6分故0Z与1Z

之间的距离为22(21)(2(4))37−+−−=······································7分所以Z与1Z之间的最小距离为371−·························································

··8分（3）因为π6=，所以11(,2),(1,)24ab=−=−················································9分此时1ab=，a与b的夹角余弦为8cos17||||abab==··········

··························10分与b方向相同的单位向量为41(1,|7)|41beb==−···········································

11分所以a在b上的投影向量(||cos)164(,)1717nae==−······································12分19．（本小题满分12分）解：（1）2π5π()2sin()3sin(2)(R)126fxxxx=−

−+π1cos(2)πππ623sin(2)3sin(2)cos(2)12666xxxx−−=+−=−−−+3π1π2[sin(2)cos(2)]12626xx=−−−+πππ2sin(2)12sin(2)1663xx=−−+=−+············

··········································3分所以()fx的最小正周期2ππ2T==·······························································4分由πππ

2π22π232kxk−+−+，得π5πππ1212kxk−++所以()fx的单调递增区间为π5π[π,π]()1212kkkZ−++····································5

分（2）π()2sin(2)13f=−+因为0π，所以ππ52π333−−·······················································6分所以当ππ232−=······················

································································7分即当5π12=时，()f恰为()fx的最大值························

······························8分（3）222πππ4sin()cos()4tan()π666()2sin(2)111πππ3sin()cos()tan()1666f−−−=−+=+=+−+−−+························

······················································································10分因为πtan()26−=，所以2π4tan()8176()11π55tan()16f

−=+=+=−+······················12分20．（本小题满分12分）解：（1）由题意2cosacB=，根据正弦定理，可得：sin2sincosACB=···········1分所以sin()sincoscossin2sincosBCBCBCCB+=+=即

sincossincosBCCB=，即sincoscossin0BCBC−=也即sin()0BC−=······················································

·······························2分因为BC−−，所以0BC−=，即BC=············································

·3分所以241coscos()cos2(12sin)2199ABCBB=−+=−=−−=−=−·················5分（2）BC=，所以bc=由余弦定理知，222221202cos22()9abcbcAbb=

=+−=−−···························7分解得3bc==····························································································8

分因为1coscos9PABBAC=−=································································9分在PAB中，因为BPBC=，所以BPCBCP=所以ππ2π2()2BACPBCBCABAC−=−

=−=··································10分又因为45sin9BAC=············································································11分所以

1145sin2525229PBCSPBCBBAC===4059····················12分21．（本小题满分12分）解：（1）由13MCBC=，所以M为线段BC上靠近C的三等分点·······················1分

由0NANC+=，所以N为线段AC的中点····················································2分1111()2323NMNCCMACCBACABAC=+=+=+−1136ABAC=+·······

······4分因为2||||cos32262ABACABACBAC===··································5分所以2221111117||()3693693NMABACABACABAC=+=++=

···············6分（2）D为直线MN上一点，设NDkNM=则1111()2236CDCNNDACkNMACkABAC=+=−+=−++111()362kABkAC=+−·················

·····························································7分2111111(())()362362CDABkABkACABkABkACAB=+−=+−111π9()322sin3624kk=+−··

····························································8分因为CD直线经过ABC的垂心，所以CDAB⊥，即0CDAB=·····················9分所以111π()9()322sin03624CDABk

k=+−=解得34k=·······························································································10分所以11113()36248CDkA

BkACABAC=+−=−因为CDxAByAC=+，所以13,48xy==−·················································12分22.（本小题满分12分）解：（1）（ⅰ）设(,0)Px，1,

4tm==，所以(1,2),(4,2)AB······························1分因为(1,2),(3,4)APxAB=−=，AP与AB共线··········································2分所

以4(1)6x−=，解得52x=所以,,APB三点共线时，点P的坐标为5(,0)2················································3分（ⅱ）因为(1,2)A关于x轴的对称点为(1,2)A

−···············································4分所以||||||||PAPBPAPB+=+······························································

······5分所以当,,APB三点共线时，||||PAPB+取得最小值········································6分最小值即22||345AB=+=··················································

·····················7分所以||||PAPB+取得最小值5（2）sint=，所以2(sin,)sinA，23(sin7,),(1,8)sin2CAmCBm=+−=−因为CA与CB的夹角π[0,)2，所以0

CACB恒成立··································8分所以32sin7(8)02sinCACBmm=+−+−，又因为(0,π),sin0，可得2sin7sinsin1630mm

−++−即2(3sin)sin7sin16m−−+恒成立，又因为3sin0−，可得22sin7sin16(3sin)(3sin)43sin3sinm−+−+−+=−−恒成立·········

··············9分令3sink−=，(0,π),sin(0,1]，23k所以2441kkmkkk++=++·····································································10分因为412415k

k+++=，其中等号当且仅当2k=成立································11分所以2k=时，41kk++有最小值5，所以m的取值范围是：5m···················································

····················12分