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绝密★启用前保定市部分学校2022届高三上学期期中考试数学注意事项：1．答题前，考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。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，将本试卷和答题卡一并交回。4．本试卷主要考试内容：高考全部内容。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．若复数i(1i)z=+，则2z=（）A．2−B．2C．2i−D．2i2．已知集合{25}Mxx=−∣，{(1)(6)0}Nxxx=−−∣，则（）A．MNB．{26}MNxx

=−∣C．NMD．{56}MNxx=∣3．中国互联网络信息中心（CNNIC）发布了第46次《中国互联网络发展状况统计报告》，报告公布了截至2020年6月的中国互联网状况数据与对比数据，根据下图，下面结论不正确的是（）A．2020年6月我国网民规模接近

9.4亿，相比2020年3月新增网民3625万B．2020年6月我国互联网普及率达到67%，相比2020年3月增长2.5%C．2018年12月我国互联网普及率不到60%，经过半年后普及率超过60%D．2018年6月我国网民规模比2017年6月我国网民

规模增加的百分比大于7%4．圆2240xyx+−=上的点到直线3490xy−+=的距离的最小值为（）A．1B．2C．4D．55．已知120ABC=，2AB=，1BC=，则|2|ABBC−=（）A．23B．22C．2D．46．函数326(0)yxxx=−+…的最大值为（）A

．32B．27C．16D．407．中国古代数学的瑰宝《九章算术》中记载了一种称为“曲池”的几何体，该几何体为上、下底面均为扇环形的柱体（扇环是指圆环被扇形截得的部分）．现有一个如图所示的曲池，其高为3，1AA⊥底面

，底面扇环所对的圆心角为2，弧AD长度为弧BC长度的3倍，且2CD=，则该曲池的体积为（）A．92B．6C．112D．58．设函数()2sin()10,02fxx=+−剟的最小正周期为4，且()fx在[

0,5]内恰有3个零点，则的取值范围是（）A．50,312B．0,,432C．50,612D．0,,632二、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．已知tan4=，1tan4=−，则（）A．tan()tan1−=B．为锐角C．3tan45+=

D．tan2tan2=10．已知点()00,Pxy是抛物线2:4Cyx=上一动点，则（）A．C的焦点坐标为(2,0)B．C的准线方程为10x+=C．()2200011xxy+=−+D．02011xy++的最小

值为3411．已知()fx是定义域为(,0)(0,)−+的奇函数，函数1()()gxfxx=+，(1)1f=−，当210xx时，()()12111222xxfxxxxfxx−−恒成立，则（）A．()gx在(0,)+上单调递增B．()gx的图象与x轴有2个交点C．64(3)(

2)log2ff+−D．不等式()0gx的解集为(1,0)(0,1)−12．已知四面体ABCD的每个顶点都在球O（O为球心）的球面上，ABC△为等边三角形，M为AC的中点，2ABBD==，2AD=，且ACBD⊥，则（）A．BM⊥平面ACDB．O平面AB

CC．O到AC的距离为233D．二面角ACDO−−的正切值为63三、填空题：本题共4小题，每小题5分，共20分．把答案填在答题卡的相应位置．13．函数()lg162xfxx=+−的定义域为________．14．某公司要从7位男员工和6位女员工中选3人去外地学习，则至少选派2位男员工的不同选法种数

为________．15．在中国现代绘画史上，徐悲鸿的马独步画坛，无人能与之相颉颃．《八骏图》是徐悲鸿最著名的作品之一，画中刚劲矫健、剽悍的骏马，在人们心中是自由和力量的象征，鼓舞人们积极向上．现有8匹善于奔跑的马，它们奔跑的速度各

有差异．已知第(1,2,,7)ii=匹马的最长日行路程是第1i+匹马最长日行路程的1.1倍，且第8匹马的最长日行路程为400里，则这8匹马的最长日行路程之和为________里．（取81.12.14=）16．已知椭圆C的中心为坐标原点，焦点在

y轴上，1F，2F为C的两个焦点，C的短轴长为4，且C上存在一点P，使得126PFPF=，写出C的一个标准方程：________．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、证明过程或演算步骤．17．（10分）在等差数列na中，24a=，83

2aa=．（1）求na的通项公式；（2）求数列()22111nana−+的前n项和nS．18．（12分）如图，在棱长为3的正方体1111ABCDABCD−中，E，F分别为棱AB，CD上一点，且1AECF==．（1

）证明：BF∥平面1CDE．（2）求1BC与平面1CDE所成角的正弦值．19．（12分）a，b，c分别为钝角ABC△内角A，B，C的对边．已知3coscoscosaAbCcB=+．（1）求cos4A+；（2）若2b=，cb，求c的取值范围．20．（12分）已知双曲线2

222:1(0,0)xyCabab−=的离心率为2，且过点(2,3)P．（1）求C的方程；（2）若斜率为55的直线l与C交于P，Q两点，且与x轴交于点M，若Q为PM的中点，求l的方程．21．（12分）新疆棉以绒长、品质好、产量高著称于世．现有两类以新疆长绒棉为主要原材料的均码

服装，A类服装为纯棉服饰，成本价为120元/件，总量中有30%将按照原价200元/件的价格销售给非会员顾客，有50%将按照8.5折的价格销售给会员顾客．B类服装为全棉服饰，成本价为160元/件，总量中有20%将按照原价300元/件的价格销售给非会员顾客，有40%将按照8.5折的价

格销售给会员顾客．这两类服装剩余部分将会在换季促销时按照原价6折的价格销售给顾客，并能全部售完．（1）通过计算比较这两类服装单件收益的期望（收益=售价-成本）；（2）某服装专卖店店庆当天，全场A，B两类服装均以会员价销售．假设每位来

店购买A，B两类服装的顾客只选其中一类购买，每位顾客限购1件，且购买了服装的顾客中购买A类服装的概率为13．已知该店店庆当天这两类服装共售出5件，设X为该店当天所售服装中B类服装的件数，Y为当天销售这两类服装带来的总收益．求当()0.5(N)PXnn„时，n可取的最

大值及Y的期望()EY．22．（12分）已知函数2()lnfxaxxb=+的图象在点(1,(1))f处的切线方程为22yx=−．（1）求()fx在(0,)ab+内的单调区间．（2）设函数24()e2elnxxgxxxx=−−，证明：()()1fxgx+．高三数学

考试参考答案1．C【解析】本题考查复数的四则运算，考查数学运算的核心素养．因为i(1i)1iz=+=−+，所以22iz=−．2．B【解析】本题考查一元二次不等式的解法与集合的交集、并集、子集，考查数学运算的核心素养．因为{25}Mxx=−∣，{16}Nxx=∣，所以{26}MNxx

=−∣．3．D【解析】本题考查统计图表的应用，考查数据分析的核心素养．2018年6月我国网民规模比2017年6月我国网民规模增加的比例为8016675116505050501010.0677%7511675116750001500−==．4．A【解析】本题

考查直线与圆，考查直观想象的核心素养．由2240xyx+−=，得22(2)4xy−+=，圆心为(2,0)，半径2r=，圆心到直线3490xy−+=的距离22|32409|334d−+==+，故圆22

40xyx+−=上的点到直线3490xy−+=的距离的最小值为1dr−=．5．C【解析】本题是一个考查平面向量数量积的易错题，易错点为AB与BC的夹角．因为2222221|2|44442421442A

BBCABABBCBCABBABCBC−=−+=++=+−+=，所以|2|2ABBC−=．6．A【解析】本题考查导数的应用，考查数学运算的核心素养．因为3(4)yxx=−−，所以当04x剟

时，0y…；当4x时，0y．因此，326(0)yxxx=−+…的最大值为3246432−+=．7．B【解析】本题考查数学文化与简单几何体的综合，考查空间想象能力．不妨设弧AD所在圆的半径为R，弧BC所在圆的半径为r，由弧

AD长度为弧BC长度的3倍可知3Rr=，22CDRrr=−==，即1r=．故该曲池的体积()22364VRr=−=．8．D【解析】本题考查三角函数的图象及其性质，考查推理论证能力与数形结合的数学思想．因为24T==，所以

12=．由()0fx=，得11sin22x+=．当[0,5]x时，15,22x++，又02剟，则55322+剟．因为1sin2yx=−在[0,3]上的零点为6，56，136

，176，且()fx在[0,5]内恰有3个零点，所以0,613517626+剟„或,62175,62+„„解得0,,632．9．ACD【解析】本题考查三角恒等变换，考查数学运算的核心素养．因为tan

4=，1tan4=−，所以tan()tantantan1−=−=，未必是锐角（比如可以是第三象限角），1tan3tan41tan5++==−，8tan2tan215

==−．10．BCD【解析】本题考查抛物线的定义与性质以及基本不等式的应用，考查逻辑推理与数学运算的核心素养．C的焦点坐标为(1,0)，C的准线方程为10x+=，根据抛物线的定义可得P到焦点的距离等于P到准线的距离，即()22000

11xxy+=−+．因为2004yx=，所以22000222000111111132141414444yyxyyy++=+=+−−=+++…，当且仅当20201141yy+=+，即201y=时，等号成立，故02011

xy++的最小值为34．11．BC【解析】本题考查函数性质的综合，考查推理论证能力与抽象概括能力．()()12111222xxfxxxxfxx−−，两边同时除以12xx得()()122111fxfxxx−−，即()()121

211fxfxxx++，得()()12gxgx，则()gx在(0,)+上单调递减，A错误．因为()fx是定义域为(,0)(0,)−+的奇函数，且(1)0g=，所以()gx在(,0)−上单调递减，且(1)(1)0gg=−=，B正确．由(3

)(2)gg得11(3)(2)32ff++，即641(3)(2)log26ff−=，即64(3)(2)log2ff+−，C正确．不等式()0gx的解集为(,1)(0,1)−−，D错误．12．AD【解析】本题考查四面体的外接球与二

面角，考查空间想象能力与推理论证能力．设ABC△的中心为G，过点G作直线l⊥平面ABC，则球心O在l上．由M为AC的中点，得BMAC⊥．因为ACBD⊥，所以AC⊥平面BDM，则ACDM⊥，所以2ADD

C==，所以222ADDCAC+=，所以90ADC=，112DMAC==，所以222DMBMBD+=，所以BMDM⊥，可得BM⊥平面ACD，所以球心O在直线MB上，因此O与G重合．过M作MHCD⊥于H，连接OH

，则OHCD⊥，从而OHM为二面角ACDO−−的平面角．因为333BMOM==，222ADHM==，所以O到AC的距离为33，且6tan3OMOHMHM==．13．(0,4]【解析】本题考查函数的定义域与基本初等函数，考查数学运算的核心素养．由1620x−

…，得4x„，因为0x，所以04x„．14．161【解析】本题考查排列组合的应用，考查分类讨论的数学思想．若派2位男员工去学习，则有2176CC126=6种选法；若派3位男员工去学习，则有37C35=种选法．故至少选派2位男员工的选法种数

为12635161+=．15．4560【解析】本题考查等比数列的应用，考查抽象概括能力．依题意可得，第8匹马、第7匹马、……、第1匹马的最长日行路程里数依次成等比数列，且首项为400，公比为1.1，故这8匹马的最长日行路程之和为()840011.14000

(2.141)456011.1−=−=−里．16．22149xy+=（答案不唯一，只要形如22214xya+=，且2496a…即可）【解析】本题考查椭圆的方程与性质，考查逻辑推理与数学运算的核心素养．因为126PFPF=，所以12272

PFPFPFa+==，则227aPF=．又因为2acPFac−+剟，所以27aac−…，即57ca…．根据题意可设C的方程为22221(0)xyabba+=，则24b=，2b=，由57ca…，得22245117baa−=−…，解得2496a

…．17．解：（1）设na的公差为d，因为24a=，832aa=，所以()1114,722,adadad+=+=+2分解得13,1,ad==·····································

··························································4分所以1(1)2naandn=+−=+．·································

·····································5分（2）由（1）知，()222211111(2)(3)nannna−=−+++，·····································7分故2222222221

111111163445(2)(3)9(3)9(3)nnnSnnnn+=−+−++−=−=++++．·····10分18．（1）证明：在正方形ABCD中，因为AECF=，所以BEDF=且BEDF∥，1分所以四边形BEDF为平行四边形，···

·······························································2分从而BFDE∥，·······································

··················································3分又BF平面1CDE，DE平面1CDE，··················································

······4分所以BF∥平面1CDE．···············································································5分（2）解：以D

为坐标原点，DA的方向为x轴的正方向建立空间直角坐标系Dxyz−，如图所示，则(0,0,0)D，(3,1,0)E，(3,3,0)B，1(0,3,3)C，································6分(3,1,0)DE=，

1(0,3,3)DC=，1(3,0,3)BC=−．············································7分设平面1CDE的法向量为(,,)nxyz=，则10nDEnDC==，即30,330,xyyz+=+=····················

························································8分令1x=，得(1,3,3)n=−．···················································

·························9分因此1113938cos,191932||nBCnBCnBC−+===，··········································11分故1BC与平面1CDE所成角的正弦值为3819．·········

·········································12分19．解：（1）因为3coscoscosaAbCcB=+，所以3sincossincossincosAABCCB=+，···········

·······································1分即3sincossin()sinAABCA=+=，·····························································2分又sin0A，

所以c1cos3A=，·····································································3分且22sin3A=，············································

·············································4分故224cos(cossin)426AAA−+=−=．···························

·····················6分（2）因为1cos03A=，所以A为锐角，························································7分又cb，所以CB，因为ABC△为钝角三角形，所以C为钝角．··········

···········································8分因为222242cos43abcbcAcc=+−=−+，··················································

·10分所以2224803abcc+−=−，·····································································11分解得6c．························

·······································································12分20．解：（1）因为212cbeaa==+=，所以3b

a=，即3ba=．················2分将点P的坐标代入222213xyaa−=，得224913aa−=，···········································3分解得21a=

，故C的方程为2213yx−=．·························································4分（2）设()11,Pxy，()22,Qxy，(,0)Mm，因为Q为PM的中点，所以122yy=．················

············································5分因为直线l的斜率为55，所以可设l的方程为5xyt=+，·······························6分联立221,35,yxxyt−==+得()22146

5310ytyt++−=，···········································7分()()222(65)4143112140ttt=−−=+，············································8分由韦达定

理可得12357tyy+=−，()2123114tyy−=．········································9分因为122yy=，所以1223537tyyy+==−，解得257ty=−

，·························10分()22212231522714ttyyy−==−=，解得221t=，······································11分即21t=，故l的方程为5210xy−=．······

········································12分21．解：（1）设A类服装、B类服装的单件收益分别为1X元，2X元，则()10.32000.52000.850.22000.612049EX=++−=，········

···············2分()20.23000.43000.850.43000.616074EX=++−=，·······················4分()()12EXEX，故B类服装单件收益的期望更高．··········

·······························5分（2）由题意可知，2~5,3XB，·································································6分511(0)3

243PX===，14152110(1)C33243PX===，23252140(2)C33243PX===，32352180(3)C33243PX===，424521

80(4)C33243PX===．·································································7分因为1104017(3)0.524381PX++==，1104080131(

4)0.5243243PX+++==，8分所以当()0.5()PXnnN„时，n可取的最大值为3．······································9分(2000.85120)(5)(3000.85160)25045YXXX=−−+−=+（元

），·············10分因为210()533EX==，·············································································11分所以()(25045

)25045()400EYEXEX=+=+=（元）．································12分22．（1）解：因为()(2ln1)fxaxx=+，·····················································

···1分所以(1)2fa==．·····················································································2分又(1)0fb=

=，所以2ab+=．···································································3分当e0ex时，()0fx；当e2ex时，()0fx．·························

····4分所以()fx在(0,)ab+内的单调递减区间为e0,e，单调递增区间为e,2e．····5分（2）证明：由（1）知()()()22222()()2lne2elne2lnxxxfxgxxxxxxxxx+=+−−=−

−．·················································································································6分设函数2()2l

nxxx=−，2()2xxx=−．当01x时，()0x；当1x时，()0x．所以min()()(1)1xx==…，·7分所以2()()exfxgxx+−…．····················································

·······················8分设函数2()e(0)xhxxx=−，则()e2(0)xhxxx=−，设()e2(0)xpxxx=−，则()e2(0)xpxx=−，令()0p

x=，得ln2x=，·········································9分则min()(ln2)2(1ln2)0pxp==−，····································

·······················10分所以()0hx，从而()hx为增函数，则()(0)1hxh=，··································11分因此2()()e1xfxgxx+−…，故(

)()1fxgx+．··········································12分