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阿拉善盟第一中学2022~2023学年度第一学期高三年级期末考试数学（理科）考生注意：1.本试卷分选择题和非选择题两部分。满分150分，考试时间120分钟。2.答题前，考生务必用直径0.5毫米黑色墨水签字笔将密封线内项目填写清楚。3.考生作答

时，请将答案答在答题卡上。选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答，超出答题区域书写的答案无效．．．．．．．．．

．．．．，在试题卷．．．．、草稿纸上作答无效．．．．．．．．。4.本卷命题范围：高考范围。一、选择题：本题共12小题，每小题5分，共60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.记集合{|||2}Mxx=，()2|ln3N

xyxx==−，则MN=（）A.{|23}xx„B.{|3xx或2}x−C.{|02}xx„D.{|23}xx−„2.已知复数1iz=+（i是虚数单位），则izzz=+（）A.31i55+B.11i55+C.31i55−+D.11i55−+3.命题“2a…，()2

fxxax=−x是奇函数”的否定是（）A.2a…，()2fxxax=−是偶函数B.2a…，()2fxxax=−不是奇函数C.2a，()2fxxax=−是偶函数D.2a，()2fxxax=−不是奇函数4.若()4sin5+=−

，则()cos2−=（）A.35B.35−C.725D.725−5.若双曲线2221xym−=（0m）的渐近线与圆22610xyy+−+=相切，则m=（）A.24B.2C.322D.226.端午节为每年农历五月初五，又称端阳节

、午日节、五月节等.端午节是中国汉族人民纪念屈原的传统节日，以围绕才华横溢、遗世独立的楚国大夫屈原而展开，传播至华夏各地，民俗文化共享，屈原之名人尽皆知，追怀华夏民族的高洁情怀.小华的妈妈为小华煮了8个粽子，其中5个甜茶粽和3个艾香粽，小华随机取出两个，事件A“取到的两个为同一种馅”，事件B“

取到的两个都是艾香粽”，则()|PBA=（）A.35B.313C.58D.13287.正方体1111ABCDABCD−中，E为1CC的中点，则异面直线1BE与1CD所成角的余弦值为（）A.1010B.1010−C.104D.104−8.某地锰矿石原有储量为a万吨，计划每年的开采量

为本年年初储量的m（01m，且m为常数）倍，第n（*nN）年开采后剩余储量为(1)nam−，按该计划使用10年时间开采到剩余储量为原有储量的一半.若开采到剩余储量为原有储量的70%，则需开采约（参考数据：1027）（）A.3年B.4年C.5年D.6年9.在平行四边形ABCD

中，4AB=，3AD=，13AEEB=，2DFFC=，且6BFCE=−，则平行四边形ABCD的面积为（）A.2465B.1265C.245D.12510.更相减损术是出自中国古代数学专著《九章算术》的一种算法，其内容如下：“可半者半之，不可半者，副置分母、子之数，以

少减多，更相减损，求其等也，以等数约之”，如图是该算法的程序框图，如果输入99a=，231b=，则输出的a是（）A.23B.33C.37D.4211.已知函数()()sinfxAx=+（0A，0，π0−）的部分图象

如图所示，下列说法中错误的是（）A.函数()fx的图象关于点2π,03−对称B.函数()fx的图象关于直线11π12x=−对称C.函数()fx在ππ,42上单调递增D.函数()fx的图象向右平移π3个单位可得函数2sin2yx=−的图象12.若e是自然

对数的底数，()elnxxm+，则整数m的最大值为（）A.0B.1C.2D.3二、填空题：本题共4小题，每小题5分，共20分。13.我国关于人工智能领域的研究十分密集，发文量激增，在视觉、语音、自然语言处理等基础智能任务实现全球领先，并且拥有一批追求算法技术极致优化的人工智能企业，如

图是过去十年人工智能领域高水平论文发表量前十国家及发表的论文数.现有如下说法：①这十个国家的论文发表数量平均值为0.87；②这十个国家的论文发表数量的中位数为0.4；③这十个国家的论文发表数量的众数为0

.4；④德国发表论文数量约占美国的32%.其中正确的是___________.（填序号）14.如图，中华中学某班级课外学习兴趣小组为了测量某座山峰的高气度，先在山脚A处测得山顶C处的仰角为60°，又利用无人机在离地面高400m的M处（即

400MD=），观测到山顶C处的仰角为15°，山脚A处的俯角为45°，则山高BC=___________m.15.在三棱锥PABC−中，PBC△是等边三角形，平面PBC⊥平面ABC，ABAC⊥，ABAC=，且三棱锥PAB

C−的所有顶点都在半径为4的球O的球面上，则三棱锥PABC−的体积为___________.16.已知椭圆C：22221xyab+=（0ab）的左、右焦点分别为1F，2F，C的下顶点为A，离心率为12，过2F且垂直于1

AF的直线与C交于D，E两点，8DE=，则ADE△的周长为___________.三、解答题：共70分。解答应写出文字说明、证明过程或演算步骤。第17~21题为必考题，每个试题考生都必须作答。第22、23题为

选考题，考生根据要求作答。（一）必考题：共60分。17.（本小题满分12分）已知数列na的前n项和21nSn=+.（1）求na；（2）令241nnba=−，若对于任意*nN，数列nb的前n项和nTm恒成立，求实数m的取值范围.18.（本小题

满分12分）盲盒里面通常装的是动漫、影视作品的周边，或者设计师单独设计出来的玩偶.由于盒子上没有标注，购买者只有打开后才会知道自己买到了什么，因此这种惊喜吸引了众多年轻人，形成了“盲盒经济”.某款盲盒内装有正版海贼王手办，且每个盲盒只装一个.

某销售网点为调查该款盲盒的受欢迎程度，随机抽取了400人进行问卷调查，并全部收回.经统计，有30%的人购买了该款盲盒，在这些购买者当中，男生占一；而在未购买者当中，男生、女生各占50%.（1）完成下面的22

列联表，并判断是否有99.5%的把握认为是否购买该款盲盒与性别有关？女生男生总计购买未购买总计（2）从购买该款盲盒的人中按性别用分层抽样的方法随机抽取6人，再从这6人中随机抽取3人发放优惠券，记X为抽到的3人中女生

的人数，求X的分布列和数学期望.参考公式：()()()()22()nadbcKabcdacbd−=++++，其中nabcd=+++.参考数据：()20PKk…0.100.050.0250.0100.0050.0010k2.7063.8415.0246.6357.87910.828

19.（本小题满分12分）如图，在直四棱柱1111ABCDABCD−中，四边形ABCD是菱形，2AB=，13AA=，60BAD=，点E是棱1BB上的一点（与B，1B不重合）.（1）求证：1ACDE⊥；（2）若二面角AECB−−的余弦值为64，求直线1BC与平面AEC所成角的正弦值.2

0.（本小题满分12分）已知抛物线C：22ypx=（0p）的焦点为F，过点()0,4P−的直线l与C相交于A，B两点.当直线l经过点Ｆ时，点A恰好为线段PF的中点.（1）求C的方程；（2）是否存在定点T，使得TATB为常数？若存在，求出点T的坐标及该常数；若不存在，

说明理由.21.（本小题满分12分）设向量1ln,2max=，()21,nx=，()()1fxmnax=−+，（aR）.（1）讨论函数()fx的单调性；（2）设函数()()2822gxfxxaxx=−++，

若()gx存在两个极值点1x，2x（12xx），证明：()1gx−()()()21222gxaxx−−.（二）选考题：共10分。请考生在第22、23两题中任选一题作答。如果多做，则按所做的第一题计分。22.（本小题满分10分）选修4-4：坐标系与参数方程在直角坐标系xOy中，曲线C的参数

方程为2cos2,3sinxtyt==（t为参数），以O为极点，x轴的正半轴为极轴建立极坐标系，直线l的极坐标方程为πsin06m++=.（1）求l的直角坐标方程；（2）若l与C有公共点，求m的取值范围.23.

（本小题满分10分）选修4-5：不等式选讲已知0a，0b，0c，且3332224abc++=.证明：（1）169abc„；（2）2abcbcacababc+++++„.阿拉善盟第一中学2022~2023学年度第一学期高三年级期末考试·数学

（理科）参考答案、提示及评分细则1.B集合{|2Mxx=或2}x−，2|30{|0Nxxxxx=−=或3}x，所以{|3MNxx=或2}x−.故选B.2.A由题意知()()()()()()1i2i1i31ii1i1ii2i2i55zzz+−+===+++−++−，故选A.3.

B命题“2a…，()2fxxax=−是奇函数”的否定是：2a…，()2fxxax=−不是奇函数.故选B.4.C由()4sinπsin5+=−=−，得4sin5=，所以()224cosπ2cos22sin1215−=−=−=−725=.故选C.

5.D双曲线2221xym−=（0m）的渐近线为xym=，即0xmy=.不妨取0xmy+=，圆22610xyy+−+=，即22(3)8xy+−=，所以圆心为()0,3，半径22r=，依题意圆心()0,3到渐近线0

xmy+=的距离23221mdm==+，解得22m=或22m=−（舍去）.故选D.6.B由题意，()225328CC13C28PA+==，()2328C3C28PAB==，所以()()()3328|1313

28PABPBAPA===.故选B.7.A平移1CD到1BA，再连接AE，则1ABE或其补角为异面直线1BE与1CD所成的角，设正方体的棱长为2，易得1122CDBA==，3AE=，15BE=，由余弦定理得22211111cos2ABBEAEABEABBE+−==8591010

410+−=.故选A.8.C设第n年开采后剩余储量为y，则(1)nyam=−，当10n=时，12ya=，所以101(1)2aam=−，0a，故1101011(1)122mm=−−=，进而1012nya

=，设第n年时，70%ya=，故1071710210naa=10122217101logloglog22101072nn====，故5n.故选C.9.A因为13AEEB=，2DFFC=，所以()()BFCECFCBCBBE

=−+=1334CDCBCBCD−+22221515||3434cos55cos6412412CBCDCBCDBCDBCD=−+−=−+−=−−=−，解得1cos5BCD=，则26sin5BCD=，所以26246sin4355AB

CDSCBCDBCD===.故选A.10.B根据程序框图，输入的99a=，231b=，因为ab，且ab，所以23199132b=−=；第二次循环，1329933b=−=；第三次循环，993366

a=−=；第四次循环，663333a=−=，此时33ab==，输出33a=.故选B.11.D易得2A=，π1243TT+=即πT=，所以2π2T==，将π,03代入得2π3=−，所以()fx=2π2sin23x−

.对于A选项，令2π2π3xk−=，kZ，解得ππ32kx=+，kZ，对称中心为ππ,032k+，kZ，当2k=−时，对称中心为2π,03−，故A正确；对于B选项，根据2ππ2π32xk−=+，kZ，解得7π

π122kx=+，kZ，当3k=−时，11π12x=−，故B正确；对于C选项，由π2ππ2π22π232kxk−+−+剟，得()fx的单调递增区间为π7ππ,π1212kk++，kZ，又πππ7π,π,π421212kk++

，kZ，故C正确；对于D选项，函数()fx图象上所有的点向右平移π3个单位，得到函数()π2π2sin233gxx=−−π2sin23x=−−，故D错误.故选D.12.C令e0xt=，则()elnxxm+

等价于()lntxm+，即etxm+，从而eelnttmxt−=−.设()elntgtt=−，则()1etgtt=−.易知()gt在()0,+上单调递增，且1e202g=−，()1e10g=−.所以存在唯一的01,12t，使得()

00gt=，即001ett=，则00lntt=−.当()00,tt时，()()00gtgt=，()gt在()00,t上单调递减；当()0,tt+时，()()00gtgt=，()gt在()0,t+上单调递增.从而()0min00001()elntgtgtttt==−=+

，而001tt+在1,12上是减函数，所以00152,2tt+.因此()gt的最小值()052,2gt，从而整数m的最大值是2.故选C.13.①②结合图表易得该组数据的众数为0.3，德国发表论文数量约占美国的18%，故只有①②正确.14.600

由题意知45AMD=，则24002AMMD==，又由60CAB=，所以18060MAC=−4575−=，180756045ACM=−−=，在MAC△中，由正弦定理得sinsinACMAAMCACM=，即4002sin6

0sin45AC=，解得4003AC=，则sin60600BCAC==.15.24因为ABAC⊥，所以BC为ABC△所在截面圆1O的直径，又平面PBC⊥平面ABC，PBC△为等边三角形，所以O在1PO上，如图所示.设PB

x=（0x），则112BOx=，132POx=，所以132POx==21141642OOx+=−+，解得43x=，所以134362PO==，43BC=，又ABAC⊥，ABAC=，所以1126

261222ABCSABAC===△，所以1111262433PABCABCVSPO−===△.16.523因为C的离心率12cea==，所以2ac=，22223bacc=−=，所以C的方程为2222143xyc

c+=，即22234120xyc+−=.在12AFF△中，122AFAFac===，122FFc=，所以12AFF△为正三角形.过2F且垂直于1AF的直线与C交于D，E两点，所以DE为线段1AF的垂直平分线，直线DE的斜率为33，所以直线D

E的方程为()33yxc=−.设()11,Dxy，()22,Exy，由()22234120,33xycyxc+−==−得22138320xcxc−−=，所以12813cxx+=，2123213cxx=−，所以2221218324811483131313ccDEkxxc

=+−=+−−==，解得136c=，所以1323ac==.因为DE为线段1AF的垂直平分线，所以1ADDF=，1AEEF=，所以ADE△的周长为1212522243ADAEDEDFDFEFEFaaa++=+++=+==.17.解：（1）当1n

=时，211112aS==+=；··············································································2分当2n…时，2211(1)121nnnaSSn

nn−=−=−−−−=−.∵12a=不满足上式，∴2,1,21,2.nnann==−…··········································································

······5分（2）由（1）可得1211413ba==−，··············································································

············.6分当2n…时，()24111(21)111nbnnnnn===−−−−−，··································································8分∴41

111111131223211nTnnnn=+−+−++−+−−−−41131n=+−71733n=−，·······································································

·············································10分又nTm恒成立，∴73m…，即实数m的取值范围为7,3+.········································

·····················································12分18.解：（1）女生男生总计购买8040120未购买140140280总计220180400·······································

····································································································2分根据列联表中的数据，可得22400(8014

040140)28009.428220180120280297K−==，································4分因为9.4287.879，所以有99.5%的把握认为是否购买该款盲盒与性别有关.··········

···························6分（2）抽取6人中，女生有：80648040=+（人），男生有：40628040=+（人）.X的所有可能取值为1，2，3，··························

········································································7分()212436CC11C5PX===，()122436CC32C5PX===，()3436C13C5PX===，所以

X的分布列为：X123P153515··········································································································

·······························10分所以1311232555EX=++=.·······································································

···················12分19.（1）证明：连接BD，11BD，如图所示.因为四边形ABCD是菱形，所以ACBD⊥，···················1分因为直四棱柱1111ABCDABCD−，所以1BB⊥平面ABCD，又AC平面ABCD，所以1BBAC

⊥，··2分又1BDBBB=，BD，1BB平面11DBBD，所以AC⊥平面11DBBD，······································4分又1DE平面11DBBD，所以1ACDE⊥.······

·············································································5分（2）解：记AC与BD的交点为O，连接11AC交11BD于点1O.以O为坐标原点，分别以OA，OB，1OO所在的直线为x轴，y轴，x轴

建立空间直角坐标系，如图所示，所以()3,0,0A，()3,0,0C−，()13,0,3C−，()0,1,0B，设BEm=（03m），故()0,1,Em，所以()13,1,3BC=−−，()3,1,CEm=，()23,0,0C

A=，()0,0,BEm=.设平面ACE的一个法向量(),,nxyz=，所以30,230,nCExymznCAx=++===令ym=，解得0x=，1z=−，所以平面ACE的一个法向量()0,,1nm=−.··························

··········6分设平面BCE的一个法向量(),,mabc=，所以30,0,mCEabmcmBEmc=++===令1a=，解得3b=−，0c=，所以平面BCE的一个法向量()1,3,0m=−，······························7分

所以236cos,4113nmmnmnmm===++，·······································································9

分解得1m=，································································································

·······················10分所以平面ACE的一个法向量为()0,1,1n=−.设直线1BC与平面AEC所成的角为，所以114226sin1311319nBCnBC===+++，所以直线1BC与平面AEC所成角的正弦值为22613.·················

·················································12分20.解：（1）因为,02pF，()0,4P−，且点A恰好为线段PF中点，所以,24pA−，·····················1分又

因为A在C上，所以2(2)24pp−=，即28p=，····································································3分解得22p=，所以C的方程为242yx=.·······························

···············································4分（2）设(),Tmn，由题意可知直线l斜率存在，设直线l的方程为4ykx=−，()11,Axy，()22,Bxy，由24

2,4yxykx==−得2421620kyy−−=，所以1242yyk+=，12162yyk=−，·························6分所以()()()()1212TATBxmxmynyn=−−+−−()()()(

)22222222121212121212221288328ymymynynyymyymyynyyn=−−+−−=−+++−++222216232322162428nmmnkkkkk=−++−−+2221642816242mmnm

nkk−++=−++.·············································································9分令8162420,16420,mnm++=

−=····································································································10分解得22,8mn==−即()22,8T−，··

··························································································11分此时2272TATBmn=+=.··················

·················································································12分21.（1）解：根据已知得()()21ln12fxaxxax=+−+，0x，则()()()()()2111xax

axxaafxxaxxx−++−−=+−+==，························································1分若0a„，当01x时，()0fx；当1x时，(

)0fx，所以()fx在()0,1上单调递减，在()1,+上单调递增.································································

·.2分若01a，由()0fx，得0xa或1x；由()0fx，得1ax，所以()fx在()0,a，()1,+上单调递增，在(),1a上单调递减.··················

·····································3分若1a=，()0fx…恒成立，所以()fx在()0,+上单调递增.··························

······························4分若1a，由()0fx，得01x或xa；由()0fx，得1xa，所以()fx在()0,1，(),a+上单调递增，在()1,a上单

调递减..······················································5分（2）证明：由已知得()82ln2gxaxxx=−+，从而()()22224282xaxagxxxx−−+=−−=，0x.····

6分当44a−剟时，()0gx„恒成立，函数()gx不可能有两个极值点；当4a−时，()0gx=有两个根1x，2x，因为124xxa+=−，与1x，2x都是正数相矛盾，不合题意；·········································

··································································································7分当4a时，()0gx=有两个根1x，2x，因为120xxa

+=，且124xx=，所以两根1x，2x均为正数，故()gx有两个极值点，·····························································································

················8分因为12xx，由124xx=知102x，22x，因为()()1212212121222lnlnln42ln822244gxgxxxxaaxxxxxxxx−−−=−−=−−−−，所以()()()()121222

gxgxaxx−−−等价于222ln42ln224xaaxx−−，即22242ln2ln2xxx−+.········································································

······························10分令()42lnhxxxx=−+（2x），()22242(1)310xhxxxx−−−=−−+=，所以()hx在()2,+上单调递减，又()22ln2h=，所以当2x

时，()2ln2hx.故()()()()121222gxgxaxx−−−成立.················································································12分

22.解：（1）因为l：πsin06m++=，所以31sincos022m++=，又因为siny=，cosx=，得31022yxm++=，即l的直角坐标方程为320xym++=

.····················································································3分（2）将2cos2,3si

nxtyt==代入320xym++=中，得2cos23sin20ttm++=，所以()2212sin3sin20ttm−++=，即24sin3sin220ttm−−−=，要使l与C有公共点，则24sin3sin220ttm−−−=有解，即224si

n3sin2mtt=−−有解.···················5分令sinta=，则1,1a−，令()2432faaa=−−，11a−剟，所以()fa在31,8−上单调递减，在3,18上单调递增，所以min39941()2816816faf

==−−=−，又()14321f=−−=−，()14325f−=+−=，所以412516m−剟，·······························································

················································8分解得415322m−剟，即m的取值范围是415,322−.··········································

····························10分23.证明，（1）因为0a，0b，0c，则320a，320b，320c，所以33333322232223abcabc++…，即124()3abc„，所以

169abc„，当且仅当333222abc==，即3169abc===时取等号.·····························································

······5分（2）因为0a，0b，0c，所以2bcbc+…，2acac+…，2abab+…，所以3222aaabcbcabc=+„，3222bbbacacabc=+„，3222cccabababc=+„，所以33333322222222222abcabcabcb

cacababcabcabcabcabc++++++==+++„，当且仅当abc==，即3169abc===时取等号.··········································

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