【文档说明】湖南省五市十校教研教改共同体、三湘名校教育联盟、湖湘名校教育联合体2022-2023学年高二上学期期中考试数学试题 含答案.docx，共(19)页，1.009 MB，由envi的店铺上传

转载请保留链接：https://www.doc5u.com/view-1b9a8fa20f3971f643e8acfac452dd7f.html

以下为本文档部分文字说明：

绝密★启用前五市十校教研教改共同体三湘名校教育联盟湖湘名校教育联合体2022年下学期高二期中考试数学命题：双峰一中数学备课组审题：南县一中郭劲松永州一中数学备课组本试卷共4页。全卷满分150分，考试时间120分钟。注意事项：1.答题前，考生务必将自己的姓名、准考证号填写在本试卷和答题卡

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的答案标号涂黑，如有改动，用橡皮擦干净后，再选涂其他答案；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，将本试卷和答题卡一并交回。一、单项选择题：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.已知集合{||1|2}Axx=−，{1,2,3,4}B=，则AB=（）A.{1}B.{2}C.{1,2}D.{1,2,3}2.已知圆C的圆心坐标为(1,1)，且过坐标原点，则圆C的方程为

（）A.22(1)(1)2xy−+−=B.22(1)(1)2xy−+−=C.22(1)(1)2xy+++=D.222xy+=3.党的十八大报告指出，必须坚持在发展中保障和改善民生，不断实现人民对美好生活的向往，为响应中央号召，某社区决定在现有的

休闲广场内修建一个半径为4m的圆形水池来规划喷泉景观.设计如下：在水池中心竖直安装一根高出水面为2m的喷水管（水管半径忽略不计），它喷出的水柱呈抛物线型，要求水柱在与水池中心水平距离为3m2处达到最高，

且水柱刚好落在池内，则水柱的最大高度为（）A.8m3B.9m4C.25m8D.14m54.已知nS是等比数列na的前n项和，3S，9S，6S成等差数列，则下列结论正确的是（）A.2582aaa+=B.3693aaa+=C.2825aaa

=D.2936aaa=5.已知幂函数1yx−=的图象是等轴双曲线C，且它的焦点在直线yx=上，则下列曲线中，与曲线C的实轴长相等的双曲线是（）A.22122xy+=B.22122xy−=C.221xy−=D.22144xy−=

6.已知函数π()2sinsin3cos22fxxxx=++，下列说法正确的是（）A.函数()fx的最小正周期是2πB.函数()fx的最大值为7C.函数()fx的图象关于点π,012对称D.函数()fx在区间ππ,312−上单调递增7.如图水平放置的边长

为1的正方形ABCD沿x轴正向滚动，初始时顶点A在坐标原点，（沿x轴正向滚动指的是先以顶点B为中心顺时针旋转，再以顶点C为中心顺时针旋转，如此继续），设顶点(,)Axy的轨迹方程式()yfx=，则120222f=（）A.0B.1C.2D.728.已知三棱锥

PABC−中，1PAPB==，ABBC=，90APBABC==，若二面角PABC−−的大小为120°，则三棱锥PABC−的外接球的表面积为（）A.4πB.9π2C.14π3D.5π二、多项选择题：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.下列说法正确的是（）A.命题“(,0)x−，34xx…”的否定为“[0,)x+，34xx

”B.在ABC△中，若“AB”，则“sinsinAB”C.若0ab，则ab的充要条件是11abD.若直线30axy++=与2(1)10xaya+−++=平行，则1a=−或210.已知各项均为正数的等差数列

na单调递增，且52a=，则（）A.公差d的取值范围是10,2B.7922aa−=C.3746aaaaD.1911aa+的最小值为111.已知直线l与抛物线22ypx=（0p）交于A，B两点，ODAB⊥，OAOB⊥，则下列说法正确的是（）A.若点D的坐标为(2,1)，

则54p=B.直线AB过定点(2,0)pC.D点的轨迹方程为2220xypx+−=（原点除外）D.设AB与x轴交于点M，则ODM△的面积最大时，直线AB的斜率为112.在正方体1111ABCDABCD−中，1AB=，点M在正方体内部及表面上运动，下列说法正确的是（）A.若M为棱1CC的中

点，则直线1AC∥平面BDMB.若M在线段1BC上运动，则1CMMD+的最小值为22+C.当M与1D重合时，以M为球心，52为半径的球与侧面11BBCC的交线长为π4D.若M在线段1BD上运动，则M到直线1CC的最短距离为22三、填空题：本题共4小题，每小题5分，共20分.13.某中学高

一年级有600人，高二年级有480人，高三年级有420人，因新冠疫情防控的需要，现用分层抽样从中抽取一个容量为300人的样本进行核酸检测，则高三年级被抽取的人数为___________.14.设双曲线C：22221xyab−=（0a，0b）的左、右焦点分别为1F、2F，P是渐近线上

一点，且满足212PFFF=，2120PFFF=，则双曲线C的离心率为___________.15.已知动点(,)Mxy在运动过程中总满足关系式2222(2)(2)8xyxy−++++=，记(4,0)A−，(2,3)B，则ABM△面积的最大值为___________.16.意大利数学

家斐波那契在研究兔子繁殖问题时发现了数列1，1，2，3，5，8，13，…，数列中的每一项被称为斐波那契数，用符号()Fn表示（*nN），已知(1)1F=，(2)1F=，()(1)(2)FnFnFn=−+−（3n…）.（1）若22(

5)(6)()FFFn+=，则n=___________（2分）；（2）若(2022)Fa=，则(1)(2)(2020)FFF+++=___________（3分）.四、解答题：本题共6小题，共70分.解答应写出必要的

文字说明、证明过程及演算步骤.17.（本小题满分10分）已知双曲线C：22221xyab−=（0a，0b）的左右焦点分别为1F，2F，点M在双曲线C的右支上，且122MFMF−=，离心率2e=.（1）求双曲线C的标准方程；（2）若1260FMF=，求12FMF△的面积.18.（

本小题满分12分）10月9日晚，2022年世界乒乓球团体锦标赛在中国成都落幕.中国队女团与男团分别完成了五连冠与十连冠的霸业.乒乓球运动在我国一直有着光荣历史，始终领先世界水平，被国人称为“国球”，在某次团体选拔赛中，甲乙两队进行比赛，

采取五局三胜制（即先胜三局的团队获得比赛的胜利），假设在一局比赛中，甲队获胜的概率为0.6，乙队获胜的概率为0.4，各局比赛结果相对独立.（1）求这场选拔赛三局结束的概率；（2）若第一局比赛乙队获胜，求比赛进入第五局的概率.19.（本小题满分12分）已知

锐角三角形ABC中，角A，B，C所对的边分别为a，b，c，向量(2sin,3)mA=−，(,)nba=，且mn⊥.（1）求角B的大小；（2）若3c=，求ABC△面积的取值范围.20.（本小题满分12分）已知数列na满足12a=，且11220nnnnaaaa+

++−=，数列nb是各项均为正数的等比数列，nS为nb的前n项和，满足14ba=，378S=.（1）求数列na的通项公式；（2）设nnnbCa=，记数列nC的前n项和为nT，求nT的取值范围.

21.（本小题满分12分）如图，在四棱锥PABCD−中，112PAPDDCBCAB=====，ABCD∥，90APDABC==，平面PAD⊥平面ABCD，E为PA中点.（1）求证：DE∥面PBC；（2）求证：PA⊥面PBD；（3）点Q在棱PB上，设PQPB=（01

），若二面角PADQ−−的余弦值为55，求.22.（本小题满分12分）已知椭圆C：22221xyab+=（0ab）过点31,2D，A为左顶点，且直线DA的斜率为12.（1）求椭圆C的标准方程；（2）设(,0)Mm在椭圆内部，(,0)T

t在椭圆外部，过Ｍ作斜率不为0的直线交椭圆C于P，Q两点，若PTMQTM=，求证：mt为定值，并求出这个定值.五市十校教研教改共同体三湘名校教育联盟湖湘名校教育联合体2022年下学期高二期中考试

数学参考答案、提示及评分细则一、选择题：（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.【答案】C【解析】∵(1,3)A=−，∴{1,2}AB=.2.【答案】B【解析】圆心(1,1)C，半径||2rOC==，故

圆C方程为22(1)(1)2xy−+−=.3.【答案】C【解析】取一截面建系如图，设抛物线方程为22xpy=−（0p），记最大高度为h，如图：(1.5,2)Ah−−，(2.5,)Bh−在抛物线上，故92(2)4252()4phph=−−=−−，两式相除有2

592hh=−，解得25m8h=.4.【答案】AB【解析】若公比1q=有313Sa=，616Sa=，919Sa=，此时9362SSS+，故公比1q，由题意9362SSS=+()()()9361112111111aqaqaqqqq−−−=+−−−，

化简有472qqq+=，故有2582aaa+=或3692aaa+=，选答案AB.5.【答案】B【解析】由双曲线几何性质知，双曲线的焦点在实轴上，实轴与双曲线的交点1(1,1)A−−，2(1,1)A是双曲线的顶

点，故双曲线C的实轴长1222AA==，选答案B.6.【答案】D【解析】由π()2sincos3cos2sin23cos22sin23fxxxxxxx=+=+=+知A，B错误.由π212f=，所以C错误.当ππ,312x−时，πππππ2,,33222x

+−−，所以D正确.7.【答案】D【解析】A点运动轨迹最终构成图象如图：由图可知4T=.故1572022222ff==，在B→D段时，A点的轨迹方程为22(2)2xy−+=（1

2y剟），∴5722f=.8.【答案】C【解析】由题意，取AC中点1O，AB中点2O，连接2PO，12OO则1O，2O分别是ABC△与PAB△的外心，且21120POO=，分别过1O，2O

作1l⊥面ABC，2l⊥面PAB，记12llO=，则O为外接球球心，在12RtOOO△中，1126tan306OOOO==，∴2221176ROOOA=+=，故7144ππ63S==球，选C.二、多选题（本题共4小题，每小题5分，在每小题给出的选项中，有多项符合要求，

全部选对的得5分，有选错的得0分，部分选对的得2分）9.【答案】BC【解析】对A：否定为：0(,0)x−，0034xx，所以A错误；对D，当2a=时，两直线重合，所以D错误.10.【答案】AB【解

析】由题意得0d，1102402add−，∴10,2d，故A正确；由()79599522aaaaaa−=+−==，故B正确；由23746(22)(22)(2)(2)30aaaaddddd−

=−+−−+=−，知3746aaaa故C错误；由19524aaa+==有()91191919191111112144aaaaaaaaaa+=++=++…，当且仅当19aa=时取到等号，但19aa，故不能取“=”，所以D错.11.【答案】ABC【解析

】(2,1)D，由ABOD⊥知AB方程为25yx=−+，联立22ypx=，消去x有250ypyp+−=，记()11,Axy，()22,Bxy，则125yyp=−，由212121241OAOByypkkxxyy===−，∴54p=，故A正确；对选项BCD，可设AB：x

myt=+，代入22ypx=有2220ypmypt−−=，则122yypt=−，由212412OAOBpkktpyy==−=，故直线AB为2xmyp=+，过定点(2,0)p，即(2,0)Mp，故B

正确；由ODDM⊥，得D在以OM为直径的圆：2220xypx+−=上运动（原点除外），故C正确；当(,)Dpp时，RtODM△面积最大，此时，有1ABk=，故D错误.12.【答案】ACD【解析】易知A，D正确；对选项B：展开11BCD△与1BCC△到同一平面上如图.知2

2111121cos13522CMMDCD+=+−=+…，故B错误；对选项C：M与1D重合时，在侧面11BBCC上的射影为1C，故交线是以1C为圆心的一段圆弧（14个圆），且圆半径2251122r=−=，故圆弧长1π2π44r==，所以C正确.三、填空题（

本题共4小题，每小题5分，共20分）13.【答案】84【解析】由分层抽样易得.14.【答案】5【解析】不妨设P在第一象限，则,bcPca，依题意：22bcbcaa==，∴离线率2215cb

eaa==+=.15.【答案】18【解析】易得M在椭圆2211612xy+=上运动，且B在椭圆上，A为左顶点，由AB方程：240xy−+=，设直线l：20xym−+=与椭圆相切于点M.联立222011612xymxy−+=+=，消去x得2216123480ymym

−+−=，由08m==，依题意，8m=−时，ABM△面积最大，此时直线l与AB距离为1255d=，又35AB=，∴1125351825ABMS==△.16.【答案】（1）11（2分）（2）1a−（

3分）【解析】（1）222211(5)(6)5889FFa+=+==，∴11n=；（2）(1)(2)(2020)[(3)(2)][(4)(3)][(2022)(2021)](2022)FFFFFFFFFF+++=−+−++−=(2)1Fa−=−

.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.【答案】（1）2213yx−=（2）33【解析】（1）由题意122MFMFa−=，············································

·····································1分∴221aa==，·······························································································

···············2分又22aecc===，···································································································

······3分∴2223bca=−=，·············································································································4分故双曲线C的

方程为2213yx−=；··························································································5分（2）1MFm=，2MFn=，则由双曲

线定义可得2mn−=①，由三角形余弦定理得2222cos60416mnmnc+−==②，······················································7分2−①②有12mn=，·······················

···················································································9分∴12FMF△的面积1sin60332Smn==.·······························

··············································10分18.【答案】（1）0.28（2）0.432【解析】设“第i局甲胜”为事件iA，“第j局乙胜”为事件jB（i，1j=，2，3，4，5），（1）记A=

“三局结束比赛”，则123123AAAABBB=+，····························································2分∴()()()()()()()()123123123123()PAPAAAP

BBBPAPAPAPBPBPB=+=+0.60.60.60.40.40.4=+0.28=；·······································································

·····················································6分（2）记B=“决胜局进入第五局比赛”，则234234234BAABABABAA=++，··············

····················8分∴()()()234234234()PBPAAAPABAPBAA=++0.60.60.40.60.40.60.40.60.6=++0.432=.··································

·························································································12分19.【答案】（1

）π3（2）9393,82【解析】（1）由302sin30sin2mnmnbAabAa⊥=−==，··································2分由正弦定理得sin3sin2bABa==，······················

·································································4分又π0,2B，∴π3B=，·······················································

·············································6分（2）解法一：在锐角ABC△中，由（1）知，π3B=，有2π3AC+=，令π3A=+，则π3C=−，ππ66−，由正弦定理

得sinsincAaC=，ABC△的面积π3sin11π3sin3sinπ223sin3SacB+==−··················8分933193cossin(3tan)422271934

44313tan3tancossin22++===−−−−，·································10分由ππ66−得33tan33−，23433tan33−，则133,423tan

−，于是得939382S，所以ABC△面积的取值范围是9393,82.···········································································12分解法二

：由（1）可知，π3B=，故2π3AC+=，又因为3sinsinaAC=，所以ππ3sinπ3sin3sin333133sinsinsin22tanCCAaCCCC−−+====+，··································

·····8分又因为π02C，ππ0π32C−−，所以ππ62C，故3tan3C，即有103tanC，则362a，·····································

·················································10分又由13333224Saa==，即939382S，所以ABC△面积的取值范围是9393,82.···························

················································12分20【答案】（1）2nan=（2）1,14【解析】（1）由2112111222nnnnnnnaaaaaaa+++===++，·········

·····················································1分∴11112nnaa+−=（常数），···········································

··························································2分故数列1na是以12为公差的等差数列，且首项为1112a=，············

····································································································3分∴111(1)222nnna=+−=，··

··································································································4分故2nan=；········································

·················································································5分（2）设nb公比为q（0q），由题意：1412ba==，∴223111744302228Sqqqq=++=+−=

，解得12q=或32−（舍），∴1112nnnbbq−==，∴12nnnnanCb+==，·······································································

·······································7分∴23411232222nnnT+=++++，有3412112122222nnnnnT++−=++++，两式相减得2312111122222nnnnT++=+++−2211122

1212nnn+−=−−21222nn++=−，····································································································

··················9分∴1212nnnT++=−，···············································································································

10分由12102nnnnTT+++−=，知nT在*nN上单调递增，·································································11分∴1,14nT.···················

·································································································12分21.【答案】（1）略（2）略（3）12【

解析】（1）证明：取PB中点F，连接EF，CF，则//12EFAB，又//12DCAB，∴//EFDC，∴四边形EFCD是平行四边形，∴DECF∥，又DE面PBC，CF面PBC，∴DE∥面PBC；···

···········································································································4分（2）证明：由题意：1BCC

D==，90BCD=，∴2BD=，同理2AD=，又2AB=，∴222BDADAB+=，∴BDAD⊥，·······································································

·············································6分又面PAD⊥面ABCD，∴BD⊥面PAD，∴BDDA⊥.又PAPB⊥且BDPBB=，∴PA⊥面PBD；··················

·····························································································8分（3）以D为原点，建立如

图所示的空间直角坐标系，则(0,0,0)D，(2,0,0)A，(0,2,0)B，22,0,22P，∴(2,0,0)DA=，22,0,22DP=，22,2,22PB=−−，由PQPB=，有22(1),2,(1)22DQDPPQDP

PB=+=+=−−，························10分令(,,)nxyz=是面ADQ的法向量，则20,0,2(1)2(1)02022xnDAnDQxyz==−−=++

=，令1y=，有20,1,1n=−，······························································································1

1分取面PAD的法向量(0,1,0)m=，由51|cos,|52nm==.···········································································

·····················12分22.【答案】（1）22143xy+=（2）mt为定值4，证明略【解析】（1）由题意：(,0)Aa−，∴22191,42331212ababa+====+

，故椭圆C的标准方程为22143xy+=；·······················································································4分（2）设PQ：xnym=+，联立22143xy+=消去x，有()222346

3120nymnym+++−=，记()11,Pxy，()22,Qxy，则0且122634mnyyn−+=+，212231234myyn−=+，······································································7分若PTM

QTM=，则0TPTQkk+=····················································································9分()()1212211200yyynymty

nymtxtxt+=+−++−=−−()12122()0nyymtyy+−+=()22223126()03434nmmnmtnn−−−+=++2460nmnt−+=（0n），∴4mt=（定值），综上：mt为定

值4.············································································································

...12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com