【文档说明】福建省泉州市二检2022-2023学年高三数学试卷答案.pdf，共(21)页，1.552 MB，由管理员店铺上传

转载请保留链接：https://www.doc5u.com/view-897d1f4eab4e359ff634744ed88b446f.html

以下为本文档部分文字说明：

高三数学试题参考答案第1页(共23页)泉州市2023届高中毕业班质量监测（二）高三数学参考答案保密★使用前三、填空题：本题共4小题，每小题5分，共20分。13．已知复数z满足(1i)(0)zaa，||2z

，则a．【试题答案】2.14．已知圆222:(3)(4)Cxyr，(1,0)A，(1,0)B.若C上存在点P，使得90APB，则正数r可以是．（只要写出一个符合条件的r即可）【试题答案】填写4到6之间的任意一个数，均可得5分.15．已知函数()|1|lnfxxax的最小

值为0，则a的取值范围为．【试题答案】01a；{|01}aa；[0,1].16．在RtABC△中，90A，AB翻折至ACD△，使二面角ACDB为直二面角，且四面体ABCD的四个顶点都在3，AC1，D是

边AB上的一动点，沿CD将△ACD球O的球面上.当线段AB的长度最小时，球O的表面积为．【试题答案】163：153.每题给分板：根据正误，确定0分，5分两种判分，不设中间分。二、多选题：高三数学试题参考

答案第1页泉州市2023届高中毕业班质量监测（二）高三数学参考答案17．（10分）在梯形ABCD中，ADBC∥，ADCD，3BC，sin3cosACBCAABABC．（1）若ABC△的面积为33，求AC；（

2）若3CD，求tanBAC．【命题意图】本小题主要考查解三角形、三角恒等变换等基础知识；考查运算求解能力等；考查函数与方程思想，化归与转化思想等；体现基础性，导向对数学运算、逻辑推理、直观想象等核心素养的关注．【试题解析】解法一：解：（1）在ABC△中，由正弦定理，得sinsinACA

BABCACB，················1分【正弦定理】化简得sinsinACACBABABC，代入sin3cos=ACACBABABC，得到sin3cosABCABC，即tan3ABC．···························

···································2分【运算求解】因为0,ABC，所以3ABC．···················································

····3分由1sin3323ABCSABBC△，得到4AB．························4分【面积公式】在ABC△中，由余弦定理，得222222cos4343133ACABBCABBC，所以13AC．

···································································5分【余弦定理】（2）设BAC，因为ADBC∥，所以23CAD．··········

·························6分在ABC△中，由正弦定理，得sinsinACBCABC，所以332sinAC．·············7分高三数学试题参考答案第2页在RtADC△中，sinCDCADAC，所以32sin3AC．

···························8分所以33322sinsin3，即23sin2sin3，所以313cossin2sin22，·································9分【两角差

正弦公式】得到tan33，即tan33BAC．······················································10分解法二：解：（1）同解法一.·················

·············································································1分（2）过A作AEBC，垂足为E，在RtABE△中，3AECD，所以2sinAEABABC

，····························6分在ABC△中，由余弦定理，得2222212cos2322372ACABBCABBCABC，所以7AC，·································

····················································7分由余弦定理得2221cos227ABACBCBACABAC，·····································9分所以33sintan

3327BACBAC，.······················································10分高三数学试题参考答案第3页解法三：解：（1）同解法一．·······························

···························································5分（2）在RtBCD△中，3tan3CDCBDBC，所以6CBD，过A作AEBC

，垂足为E，在RtABE△中，3AECD，所以2sinAEABABC，··························6分在ABD△中，因为6ABDADB，所以2ADAB，······7分【平几知识】在RtADC△中，3tan2CD

CADAD，··················································8分所以3322tantan()3333132BACCAD.·····

···········································10分【两角差正切公式和运算求解，各1分】解法四：解：（1）同解法一．············································

··············································5分（2）过A作AEBC，垂足为E，在RtABE△中，3AECD，所以12,tan6AEBECEBCBEBAEAB

C，，································7分在RtAEC△中，223tan33CECAEAE，············································8分在ABC

△中,32333tantan()tan()336323133BACBAECAECAE.················································10分【两角和正切公式和运算求解，各1分】高三数学试题参考答案第4页解法五：

解：（1）同解法一．··························································································5分（2）过

A作AEBC，垂足为E，在RtABE△中，3AECD，所以12tanAEBECEBCBEABC，，···············································6分以E为原点，BC所在直线为x

轴建立平面直角坐标系，如图所示：则(1,0),(2,0),(0,3)BCA，所以(1,3),(2,3)ABAC.························································7分在ABC△中,1cosco

s,27ABACBACABACABAC，····························9分所以33sintan3327BACBAC，.·············

·········································10分解法六：解：（1）同解法一．·························································

·································5分（2）过A作AEBC，垂足为E，在RtABE△中，3AECD，所以2sinAEABABC，···························6分高

三数学试题参考答案第5页设,BAaBCb，则12332abACba，，222()27ACbabbaa，··················

·························7分在ABC△中,()1coscos,2727ABACabaBACABACABAC，········9分所以33s

intan3327BACBAC，····················································10分【说明：本题在无其他解答得分的情况下，若能正确作出满足题干条件的直角梯形，可以给1分.】高三数学试题参

考答案第1页(共23页)泉州市2023届高中毕业班质量监测（二）高三数学参考答案18．（12分）已知数列na满足132a，11122nnnaa．（1）求2a，3a及na的通项公式；（2）求数列

||na的前n项和nS．【命题意图】本题主要考查数列的递推，数列前n项和等知识；考查运算求解能力，推理论证能力、抽象概括能力等；考查化归与转化思想，分类与整合思想等；体现考查逻辑推理，数学运算等核心素养的命题意图．【试题解析】解法一：

解：（1）132a，11122nnnaa，可得214a，··················································1分318a，··········

·····················································································2分由11122nnnaa可得11222nn

nnaa，··················································3分所以数列2nna是以123a为首项，公差为2的等差数列，·························4分于是231225nnann

，····························································5分所以252nnna*Nn．·······················

················································6分（2）252nnna，1,2n时0na，3n时，0na，于是132S，2317244S，···································

································7分当3n时，234131132725222222nnnnnS，高三数学试题参考答案第2页(共23页)234

511311327252222222nnnnnS，两式相减得：24121322250222222nnnnS，································9分331

11111112551255212211224224212nnnnnnnnnS，·······················10分所以52122nnnS（3n），············

····················································11分又274S也符合上式，综上：3,1,2521,2.22nnnSnn·················

·······12分（如果2n没有合并，不扣分）解法二：解：（1）因为132a，11122nnnaa，可得214a，······································

·····1分318a，······························································································2分猜

想：252nnna．···············································································3分以下用数学归纳法证明：①当1n时，132a符合题意；·······

······················································4分②假设当nk时，252kkka成立，则当1nk时，111215111251

232222222kkkkkkkkkkaa.符合猜想结论.············································································

5分综合①②得，数列na的通项公式为252nnna．·····································6分（2）同解法一．·······························

··························································12分解法三：高三数学试题参考答案第3页(共23页)解：（1）因为132a，11122nnnaa

，可得214a，···········································1分318a，········································································

······················2分由11122nnnaa可得：1111222nnnnnnaa···········································3分所以数列112nnna

是以100322a为首项，公比为12的等比数列，·············4分于是11131222nnnna，······························

·······································5分所以252nnna*Nn．···································································

····6分（2）同解法一．12分高三数学试题参考答案第1页(共23页)泉州市2023届高中毕业班质量监测（二）高三数学参考答案19．（12分）随着老年人消费需求从“生存型”向“发展型”转变，消费层次不断提升，“银发经济

”成为社会热门话题之一，被各企业持续关注.某企业为了解该地老年人消费能力情况，对该地年龄在[60,80)的老年人的年收入按年龄[60,70)，[70,80)分成两组进行分层抽样调查，已知抽取了年龄在[60,70)的老年人500人，年龄在[70,80)的老年人

300人.现作出年龄在[60,70)的老年人年收入的频率分布直方图（如下图所示）.（1）根据频率分布直方图，估计该地年龄在[60,70)的老年人年收入的平均数及第95百分位数；（2）已知年龄在[60,70)的老

年人年收入的方差为3，年龄在[70,80)的老年人年收入的平均数和方差分别为3.75和1.4，试估计年龄在[60,80)的老年人年收入的方差.【命题意图】本题考查频率分布直方图、平均数、百分位数、方差等基础知识；考查数据处理、运算求解等基本能力；导向教学对数据处理、数学运算等素养

的关注；体高三数学试题参考答案第2页(共23页)现基础性、应用性．【试题解析】解：（1）频率分布直方图中，该地年龄在[60,70)的老年人年收入的平均数约为：0.042+0.083+0.184+0.265+0.206+0.157+0.058+0.049=5.35

，······································································2分【列式与计算结果各1分】由频率分布直方图，年收入在8.5万元以下的老年人所占比例为10.0410.96，年收入在7.5万元以下

的老年人所占比例为1(0.0510.041)0.91，···········3分因此，第95百分位数一定位于[7.5,8.5)内，·······························

··················4分由0.950.917.518.30.05，···························5分【列式或体现计算方法，1分】可以估计该地年龄在[60,70)的老年人年收入的第95百分位

数为8.3.················6分（2）把年龄在[60,70)的老年人样本的平均数记为x，方差记为2xs；年龄在[70,80)的老年人样本的平均数记为y，方差记为2ys；年龄在[60,80

)的老年人样本的平均数记为z，方差记为2s.·················································································7分由（1）得，5.35x，由题意得，23xs，3.75y，21.4ys

，则5003004.75500300500300zxy，························9分【列式与计算各1分】由222221{500[()]300[()]}800xyssxzsyz，···········

···························10分可得2221{500[3(5.354.75)]300[1.4(3.754.75)]}3800s，【列式与计算结果各1分】····································

·································12分即估计该地年龄在[60,80)的老年人的年收入方差为3.··································12分高三数学试

题参考答案第1页(共23页)泉州市2023届高中毕业班质量监测（二）高三数学参考答案20．（12分）如图，四棱锥PABCD中，BC平面PAB，ADBC∥，2PAABBC，1AD，E为AB的中点，且PEEC.（1）求证：平面PBD平面P

EC；（2）求二面角DPCE的余弦值.【命题意图】本题考查空间点、直线与平面间的位置关系等知识；考查推理论证、运算求解等能力；考查数形结合思想、化归与转化思想等；体现应用性、创新性、综合性，导向对直观想象、数学运算等核心素养的关注.【试题解析】解法一：解：（1）

BC平面PAB，PE平面PAB，BCPE.························································································1分又PEEC，ECBCC，PE

平面BCD.···············································································2分BD平面BCD，PEBD.································

························································3分又1tantan2ABDBCE，ABDBCE，90ABDCEB，即BDCE.··

····························4分PECEE，BD平面PEC.···············································································5分高三数学试

题参考答案第2页(共23页)又BD平面PBD，平面PBD平面PEC.······································································

6分（2）由（1）得PEAB，且E为AB的中点，2PBPAAB.·············································································7分以E为坐标原点，,EPEA所在的直线分别为x轴，y轴建立如

图所示的空间直角坐标系Exyz，··················································································8分则(3,0,0)P，(0,1,0)A，(0,1,0)B，(0

,1,1)D，(0,1,2)C，(3,1,2)PC，(3,1,1)PD，(3,0,0)PE.···································9分设平面PCD的一个法向量为(,)xy,zn.由0PCn=，0

PDn=得32030xyzxyz，，令1y，则2z，3x，即(3,12),n.···················10分设平面PCE的一个法向量为()a,b,cm.由

0PCm=，0PEm=得32030abca，，令1c,可得(0,2,1)m.·············································11

分410cos585mnm,nmn.二面角DPCE的余弦值为105.····················································12分解法二：解：（1）依题意得AD平面PAB，以A为坐标原点，AB

的方向为x轴正方向，建立如图所示的空间直角坐标系Exyz，····················································1分设((0,))PAB，则(2,0,0)B，(2cos,2sin,0)P，(1,0,0)E，(2,

0,2)C，(0,0,1)D，···················3分(12cos,2sin,0)PE，(1,0,2)CE.················································4分高三数学试题参考

答案第3页(共23页)PECE，2cos10PECE，1cos2，所以3.·········································6分(13

0)P，，，(1,3,2)PC，(0,3,0)PE，(1,3,1)PD.···················7分设平面PCE的一个法向量为()a,b,cm.由0PCm=，0PEm=得32

030abcb，，令1c，则(2,0,1)m.··········································································8分由(2,0,1)BD

m,BD平面PEC.·····················································9分又BD平面PBD，平面PBD平面PEC.··················································

······················10分（2）设平面PCD的一个法向量为(,)xy,zn.由0PCn=，0PDn=得32030xyzxyz，，令1x，则2z，3y，即(1,3,2)

n.···········································11分410cos585mnm,nmn.二面角DPCE的余弦值为105.·····································

··················12分高三数学试题参考答案第1页(共23页)泉州市2023届高中毕业班质量监测（二）高三数学参考答案21．（12分）在上海举办的第五届中国国际进口博览会中，硬币大小的无导线

心脏起搏器引起广大参会者的关注.这种起搏器体积只有传统起搏器的110，其无线充电器的使用更是避免了传统起搏器囊袋及导线引发的相关并发症.在起搏器研发后期，某企业快速启动无线充电器主控芯片试生产，试产期同步进行产品检测，检测包括智能检测与人工抽检.智能检测在生产线上自动完成，包含安全检测、电池检测、

性能检测等三项指标，人工抽检仅对智能检测三项指标均达标的产品进行抽样检测，且仅设置一个综合指标，四项指标均达标的产品才能视为合格品.已知试产期的产品，智能检测三项指标的达标率约为99100，9899，9798，设人工抽检的综合指标不．达标率为(01)pp.（1）

求每个芯片智能检测不达标的概率；（2）人工抽检30个芯片，记恰有1个不达标的概率为()p，求()p的极大值点0p；（3）若芯片的合格率不超过96%，则需对生产工序进行改良.以（2）中确定的0p作为p的值，判断该企业是否需对生产工序进行改良.【命题意图】本题考查对立事件、相互独

立事件、条件概率等知识；考查抽象概括、运算求解、推理论证等能力；考查化归与转化、函数与方程思想、必然与或然思想等；体现综合性、应用性、创新性，体现检测数学建模、数学抽象、数学运算等素养的考查意图.高三数学试题参考答案第2页(共23页)【试题解析】解

：（1）每个芯片智能检测中安全检测、电池检测、性能检测三项指标达标的概率分别记为123PPP，，，并记芯片智能检测不达标为事件A.视指标的达标率为任取一件新产品，该项指标达标的概率，则有199100P，29899P，39798P.················1

分【前述内容均未交代，此分必扣】根据对立事件的性质及事件独立性的定义得，···················2分【公式成立的条件】1239998973()111009998100PAPPP，··············4分【公式与计算，各占1分】所以，每个芯片智能检测不达标的

概率为3100.·············································4分（2）人工抽检30个芯片恰有1个不合格品的概率为112930()(1)pCpp，···············6分因此129281283030()[(1)29(1)]

(1)(130)'pCpppCpp，·························7分令()0'p，得130p.当1(0,)30p时，()0'p；当1(,1)30p时，()0'p.··········

······················8分所以()p有唯一的极大值点0130p.·························································9分（3）设芯片人工抽检达标为事件B，则工人在流水线进行人工抽检时，抽检

一个芯片恰为合格品为事件|BA，由（2）得，29(|)130PBAp，···························································10分由（1）得9

7()100PA，2997()()(|)93.89630100PABPAPBA％％，···11分【计算得93.8％，即得分】因此，该企业需对生产工序进行改良.·····················································

··12分高三数学试题参考答案第1页(共23页)泉州市2023届高中毕业班质量监测（二）高三数学参考答案22．（12分）已知点M为圆O：221xy上的动点，点1(20)F，，2(20)F，，延长

1FM至N，使得1||||MNFM，线段1FN的垂直平分线交直线2FN于点P，记P的轨迹为．（1）求的方程；（2）直线l与交于,AB两点，且OAOB，求OAB△的面积的最小值．【命题意图】本小题主要考查直线的方程，双曲线的定义及标准方程，直线与双曲线

的位置关系等知识；考查运算求解能力，推理论证能力，直观想象能力和创新能力等；考查数形结合思想，函数与方程思想等；考查直观想象，逻辑推理，数学运算等核心素养；体现基础性，综合性与创新性．【试题解析】解法一：

（1）连结MO，1PF．因为线段1FN的垂直平分线交直线2FN于点P，所以1||||PFPN．所以2122||||||||||PFPFPFPNNF．【若后续未得分，至此可回补1分】在12NFF△中，1||||FMMN，12||||FOOF

，所以2||2||2NFOM，即2112||||2||PFPFFF.·········································2分【指出距离差绝对值为定值2，给2分；未加绝对值扣1分.】所以点P的轨迹是以1F，2F为焦点，实轴长为2的双曲线

.···3分【指出双曲线1分】由已知12,0F，22,0F，故的方程为2213yx.·····················································4分【方程正确1分】高三数学试题参考答案第2页(共23页)（2）当直线l的

斜率不存在时，由双曲线的对称性，不妨设点A坐标为(,)tt，则2213tt，232t.所以232OABSt△.·············································

········5分当直线l的斜率存在时，设直线l的方程为ykxb，11(,)Axy，22(,)Bxy.由22,1,3yykxbx消去y，整理得222(3)2(3)0kxkbxb，·······

····················6分当判别式222244(3)(3)0kbkb，即223bk时，由韦达定理，得12223kbxxk，212233bxxk.··············································7分

因为OAOB，所以12120xxyy，即1212()()0xxkxbkxb，因此221212(1)()0kxxkbxxb，所以2222232(1)033bkbkkbbkk，故22332kb.·····························

·····8分因为2222212112||(1)()(1)[()4]ABkxxkxxxx222222244(3)(1)[](3)3kbbkkk·················································

······9分222224(339)(1)(3)bkkk22226(9)(1)(3)kkk高三数学试题参考答案第3页(共23页)2229666(3)kk,当且仅当0k时，等号成立.······················

······10分点O到直线l的距离为2621bdk，·····················································11分所以1163||62222OABSABd△.综上，OAB△的面积的最小值为32（面积取得最小值时，直线l

的方程为6622xy或）.·························12分【取得最值的条件，此次未写不扣分】解法二：（1）连结MO，1PF．因为线段1FN的垂直平分线交直线2FN于点P，所以1|

|||PFPN．所以2122||||||||||PFPFPFPNNF．····1分【若后续未得分，至此可回补1分】在12NFF△中，1||||FMMN，12||||FOOF，所以2||2||2NFOM，即21||||

2PFPF.·················································2分【指出距离差绝对值为定值2，给2分；未加绝对值扣1分.】设(,)Pxy．由已知点1(20)F，，2(20)F，，所以2222222xyxy

（）（）.·································3分【距离公式1分】化简，得2213yx，故的方程为2213yx.·············································

·4分（2）依题意，设直线OA的方程为1ykx，直线OB的方程为2ykx，11(,)Axy，22(,)Bxy.因为OAOB，所以121kk.·····························································

········5分高三数学试题参考答案第4页(共23页)由11221111,3,yykxx解得212133xk，··································································6分所以22222111

11213(1)||(1)3kOAxykxk，···············································7分同理22223(1)||3kOBk，······

·········································································8分所以1||||2OABSOAOB△221222123(1)3(1)1233kkkk·

··························9分【代入1分】221231612309()3kk········································10分【恒

等变形1分】32，···············································11分【不等式运算得结果值，1分】当且仅当121,1kk或121,1kk时，等号成立.···········

·················································11分【未写取得最值的条件，此次不扣分】故OAB△的面积的最小值为32.························

··············································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