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数学参考答案第1页共12页2023年福州市普通高中毕业班质量检测数学试题参考答案与评分细则一、单项选择题：1-8DBCABDCD8【解析】因为()()32fxgx+−=，所以()()32++=fxgx，又()()12fxgx−+=，则有()()31+

=−fxfx，因为()1fx+是奇函数，所以()()11+=−−fxfx，可得()()31+=−+fxfx，即有()()2+=−fxfx与()()42+=−+fxfx，即()()4+=fxfx，所以()fx是周期为4的周期函数，故()gx也是周期为4的周期函数.因为()()2−−=

+fxfx，所以()()fxfx−=，所以()fx为偶函数.故A错误；由()1fx+是奇函数，则()10f=，所以()30f=，又()()()()24200+=+=ffff，所以()()()()()201512340kfkffff==++

+=，所以C选项错误；由()10f=得()02g=，所以B选项错误；因为()()()225212gff=−=−=，()()()()()(1)324264424ggffff+=−+−=−+=，所以()()()(

)01238gggg+++=，所以()()()()()2015012340kgkgggg==+++=，所以D选项正确.二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，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.

BCD10.BD11.BC12.AD12【解析一】由2213324xyy++=，令13cos,233sin,2xyy+==可得3cossin,2sin,xy=−=故223cos2sin2

sin23cos23,23xy+=−+=−，故A正确；B错误；()2232323cossin2sin54sin21,96xyxyxy+−=−=−−=−+，故C错误；D正确，故选AD.12【解析二

】令2xyt+=，则2ytx=−代入223xxyy++=，整理可得，223330xtxt−+−=，因为xR，所以()()22291233120ttt=−−=−−，解得2323t−，故A正确；B错数学参考答案第2页共12页误；由222

233xxyyxyxy+++==−，因为222xyxy+，所以32xyxy−，所以31xy−，又2232xxyyxy++=−，从而1329xy−，故C错误；D正确，故选AD.12【解析三】令xuv=+，y

uv=−，代入223xxyy++=，得2213vu+=，令cos,3sinuv==（0,2π）.s2π3cos3sin23in23,3323xyuv+=+−+=+=，故A正确，B错误；22222223scos9in18sin1,9xy

xyuv+−=++==+，故D正确，C错误；故选AD.三、填空题：本大题共4小题，每小题5分，共20分．13.247−14.0.6415.1116.40xy−+=15【解析一】：令32()362fxxxx=−++，则

()2366fxxx=−+，设(),1Pm，()(),Qnfn，依题意()()fmfn=，所以22366366mmnn−+=−+，则()222mnmn−=−，显然mn，则2mn+=，因为()()()31316fxx

x=−+−+，所以()fx的图象关于点()1,6中心对称，所以点P与点Q关于点()1,6对称，所以()162fn+=，则()11.fn=15【解析二】：令32()362fxxxx=−++，因为()()23130fxx=−+，故()fx在R上单调递增，

令323621xxx−++=，设其根为Px，得32361PPPxxx−+=−.由于()fx在点P处的切线与在点Q处的切线平行，得()fxk=存在两实根，其中一个为Px，设另一个为Qx.即2366xxk−+=两根为P

x，Qx，由韦达定理得2PQxx+=，则2QPxx=−，从而()3232362(2)3(2)6(2)2QQQQPPPfxxxxxxx=−++=−−−+−+322326128312121262(36)1011PPPPPPPPPxxxxxxxxx=−+−+−+−

+−+=−−++=.16【解析一】：由条件得B点为线段MN中点，设B点坐标为00(,)xy，得0(2,0)Mx、0(0,2)Ny，由||:||1:2MAAB=得A坐标为005(,)33xyA，将A、B坐标分别代入221126yx+=中，得22002

2001,126251,12969xyxy+=+=解得002,2,xy=−=则M、N坐标分别为(4,0)−、(0,4)，直线l的方程为40xy−+=.BANMOxy数学参考答案第3页共12页16【解析二】：设直线l方程为ykxm=+（,0k

m），可得,0mMk−，()0,Nm，根据||:||:||1:2:3MAABBN=得A,B坐标分别为5,66mmk−,,22mmk−.将A,B坐标代入C方程中，可得222222251,6361

2361,64124mmkmmk+=+=即222222222521236,2124,mmkkmmkk+=+=两式相减得4mk=，将其代入22222124mmkk+=，得1k=，从而直线l的方程为40xy−+

=.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）【命题意图】本小题主要考查正弦定理、余弦定理、三角恒等变换、基本不等式等基础知识；考查运算求解能力；考查化归与转化思想、函数与方程思想等；导向对发

展逻辑推理、数学运算等核心素养的关注；体现基础性和综合性.满分10分.【解析一】（1）由余弦定理可得2222cos=+−bcaacB，代入2222bac−=，得到2222(2cos)2+−−=caacBac，·····

·································1分化简得22cos0+=cacB，即2cos0caB+=，·············································································

·······2分由正弦定理可得sin2sincos0CAB+=，即()sin2sincos0++=ABAB，····································································3分sincoscossin2

sincos0ABABAB++=，3sincoscossinABAB=−，·········································································4分可得tan3tanBA=−.··············

···········································································5分（2）由（1）可知tan3tanBA=−，所以ta

nA与tanB异号，又22220bac−=，故ba，则有BA，因为,(0,π)AB，故A为锐角，B为钝角，则C为锐角，···································6分所以()22tantantan3tan2tantantantantan13tan13

tan1ABAAACABABAA+−=−+===−−−+······················7分22313233tantanAA==+，·····································

····················8分等号当且仅当3tan3A=，即π6A=时成立，····················································9分所以C的最大值为π6.·················

·······························································10分数学参考答案第4页共12页【解析二】（1）由正弦定理可得：sinsinBbAa=，·············

·····································1分由余弦定理可得：222222cos2cos2bcaAbcacbBac+−=+−，·····················································

····2分则222222222222tansincos2tansincos2bcaBBAbbcabcacbAABaacbac+−+−===+−+−，···································

·····3分因为2222bac−=，所以2222tan23tan2BccAcc+==−−.···················································5分（2）222cos2+−=abcCab·············

································································6分222222−+−=baabab···········································

···························7分2234+=abab344=+abba321632=.··················································································

····8分所以π6C，等号当且仅当223=ba时，即3=ba时成立，·································9分所以C的最大值为π6·································

················································10分【解析三】（1）由2222bac−=，由正弦定理可得，222sinsin2sinBAC−=，··········1分因为22222222sinsinsinsinsinsinsinsinBABBA

BAA−=−+−22222222sin(1sin)sin(1sin)sincossincosBAABBAAB=−−−=−()()(sincoscossin)(sincoscossin)sinsinBABABAB

AABBA=+−=+−，···············3分又因为πABC++=，所以()sinsinCAB=+，所以()()()2sinsin2sin+−=+ABBAAB，因为0πAB+，所以()sin0AB+，所以()()sin2sinBAAB−=+

，化简得3sincossincosABBA−=，·································································4分可得tan3tanBA=−.········

·················································································5分（2）同解析一.································

························································10分数学参考答案第5页共12页FEBDCAPxGOFEBDCAPyz18.（本小题满分12分）【命题意图】本小题主要考查空间几何体点、线、面位置

关系、直线与平面所成角等基础知识；考查空间想象能力、逻辑推理能力、运算求解能力；考查化归与转化思想，数形结合思想等；导向对发展直观想象、逻辑推理和数学运算等核心素养的关注；体现基础性与综合性.满分12分.【解析】（1）取PD中点F，连接AF、EF.∵EFCD，12EFCD=，ABCD，12ABC

D=，∴EFAB，EFAB=，··································1分∴四边形ABEF为平行四边形，∴BEAF，·························································

··································3分又∵BE平面PAD，AF平面PAD，∴BE平面PAD.·················································································

····4分（2）取AD中点O，BC中点G，连接,POOG，可得POAD⊥，OGCD.∵平面PAD⊥平面ABCD，平面PAD平面ABCDAD=，POAD⊥，PO平面PAD，∴PO⊥平面ABCD.····················

······························································5分∵ADCD⊥，OGCD，∴OGOA⊥.·····················································6分以O为原点

，以ADOGOP,,所在直线为x轴、y轴、z轴，建立如图所示空间直角坐标系.······························7分因为2,4,23ABCDAD===，PAD△是等边三角形，所以23PDAD==，132OAAD==，223POP

DOD=−=，所以()3,0,0A，()3,2,0B，()3,4,0C−，()0,0,3P.则()0,2,0AB=，()23,2,0BC=−，()3,2,3BP=−−.········································9分设平面PB

C的法向量为(),,xyz=n，由0BC=n，0BP=n，可得2320,3230,xyxyz−+=−−+=令1x=，可得3,3yz==，从而()1,3,3=n是平面PBC的一个法向量.··························

························10分则2321cos,727||||ABABAB===nnn，·············································

········11分所以直线AB与平面PBC所成角的正弦值为217.·············································12分数学参考答案第6页共12页19.（本小题满分12分）【命题意图】本小题主要考查数列通项，数列求和等基础知

识；考查欧拉函数概念的理解和应用；考查逻辑推理能力、运算求解能力等；考查化归与转化思想、特殊与一般思想等；导向对发展数学抽象、逻辑推理、数学运算等核心素养的关注；体现基础性、应用性与创新性.满分12分.【解析】（1）不超过9，且与其互质的数即为1,9中排

除掉3,6,9剩下的正整数，则()223336=−=；·················································································2分不超过27，且与其互质的数即

为1,27中排除掉3,6,9,12,15,18,21,24,27剩下的正整数，则()3333918=−=.············································

······································4分（2）因为32,3kk−（11,2,,3nk−=）中与3n互质的正整数只有32k−与31k−两个，所以1,3n中与3n互质的正整数个数为123

n−，所以()1323nn−=，()1111323322nnnna−−===，········································7分所以31log13nnnana−−=.·················

···································································8分设数列3lognnaa的前n项和为nT.1211210333

nnnT−−=++++，23112103333nnnT−=++++，···································9分11211111111113331133333313nnnnnnnT−−−−−−=+++−=−−················

······················10分11112233nnn−−=−−121223nn+=−，·································································

·············11分则131213212223443nnnnnT−++=−=−.·····································································12分说明：不超

过3n，且与其互质的正整数即为排除掉3的倍数的数，在每相邻的三个正整数中排除掉是3的倍数的数即可，可得()11113(33)322nnnnna−−==−=，学生若这样写，同样可以给3分.20.（本

小题满分12分）【命题意图】本小题主要考查分层抽样得到样本的均值与方差、正态分布和二项分布等基础数学参考答案第7页共12页知识；考查数据处理能力、应用意识和创新意识等，考查统计与概率思想；导向对发展逻辑推理、数学运算、数学建模、数据分析等核心素养的关注；体现

综合性、应用性与创新性.满分12分.【解析一】（1）把男性样本记为12120,,,xxx，其平均数记为x，方差记为2xs；把女性样本记为1290,,,yyy，其平均数记为y，方差记为2ys.则214,6xxs==；221,17yys==.记总样本数据的平均数为z，方差为2s.由14x=

，21y=，根据按比例分配的分层随机抽样总样本平均数与各层样本平均数的关系，可得总样本平均数为120901209012090=+++zxy······································

···2分120149021210+=17=，······························································3分根据方差的定义，总样本方差为()()12090222111210iiiisxzyz==

=−+−·································································4分()()1209022111210iiiixxxzyyyz===−+−+−+−，由()120120111200iiiixx

xx==−=−=，可得()()()()12012011220iiiixxxzxzxx==−−=−−=，同理，()()()()909011220iiiiyyyzyzyy==−−=−−=，······································

······5分因此，()()()()12012090902222211111210iiiiiisxxxzyyyz=====−+−+−+−()()2222112090210xysxzsyz

=+−++−，···············································6分所以()()2221120614179017211723210s=+−++−，所以总

样本的均值为17，方差为23，并据此估计该项健身活动全体参与者的脂肪含量的总体均值为17，方差为23.··········7分（2）由（1）知223=，所以()17,23XN，又因为234.8，所以()()12.221.8174.8174.80.6827PXPX=−+，

······························8分()()112.210.68270.158652PX−=，····················································

····9分数学参考答案第8页共12页因为()3,0.15865XB，···········································································10分所以()3333C0.158650.004PX==

.····························································11分所以3位参与者的脂肪含量均小于12.2%的概率为0.004.······················

··············12分【解析二】（1）把男性样本记为12120,,,xxx，其平均数记为x，方差记为2xs；把女性样本记为1290,,,yyy，其平均数记为y，方差记为2ys.则214,6xxs=

=；221,17yys==.记总样本数据的平均数为z，方差为2s.由14x=，21y=，根据按比例分配的分层随机抽样总样本平均数与各层样本平均数的关系，可得总样本平均数为120901209012090=+++zxy··························

························2分120149021210+=17=，········································································3分根据方差的定义，总样本方差为

()()12090222111210iiiisxzyz===−+−·································································4分221201201209022111112120290210iiiiiii

ixzxzyzyz=====−++−+()12090222111212090210210iiiixyzxyz===+−++120902221112210210210iiiixyzzz===+−+1209022211121021

0iiiixyz===+−12090222111210iiiixyz===+−，①···························································

·········5分由()12012022221111120120xiiiisxxxx===−=−，可得()1202221120ixixsx==+，②同理()90222190iyiysy==+，③将②③代入①得()()222222112

090210xyssxsyz=+++−··························································6分()()22211206149017211723210=+++−

，所以总样本的均值为17，方差为23，数学参考答案第9页共12页并据此估计该项健身活动全体参与者的脂肪含量的总体均值为17，方差为23.··········7分（2）同解析一.·································

·······················································12分21.（本小题满分12分）【命题意图】本小题主要考查抛物线的标准方程及简单几何性质，直线与抛物线的位置关系等基础知识；考查运算求解能力，逻辑推理能力，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，化归与转化思想等；导向对发展直观想象，逻辑推理，数学运算等核心素养的关注；体现综合性与创新性.满分12分.【解析一】（1）当1l的斜率为23时，得1l方程为2(2)3yx=+，···

··························1分由22,2(2),3ypxyx==+消元得2340ypyp−+=，······················································2分由弦长公式得223||1()(3)16132

ABpp=+−=，···········································3分即29162pp−=，解得2p=或29p=−（舍去），从而E的标准方程为24yx=.····································

···································4分（2）设211(,)4yAy，222(,)4yBy，得21222112444AByykyyyy−==+−，直线AB方程为211124()4yyxyyy=−++，即12124()0xyyyyy−++=，·····

···············5分又直线AB过点(2,0)−，将该点坐标代入直线方程，得128yy=.·····························6分设233(,)4yCy，244(,)4yDy，同理可得348yy=，·····················

····························7分直线AD方程为14144()0xyyyyy−++=，·······················································8分直线BC方程为2

3234()0xyyyyy−++=，·······················································9分因为(2,0)−在抛物线的对称轴上，由对称性可知，交

点G必在垂直于x轴的直线上，所以只需证G的横坐标为定值即可.由141423234()0,4()0,xyyyyyxyyyyy−++=−++=消去y，因为直线AD与BC相交，所以2314yyyy++，解得231414232314()()4[()()

]yyyyyyyyxyyyy+−+=+−+，···························································10分xyGCABD数学参考答案第10页共12页123234

12413423144[()()]yyyyyyyyyyyyyyyy+−−=+−+3241231488884[()()]yyyyyyyy+−−=+−+2=，·······························

····························································11分所以点G的横坐标为2，即直线AD与BC的交点G在定直线2x=上.···················12分【解析二】（1）同解析一.········

···································································4分（2）设直线AB的方程为1(2)ykx=+，由12(2),4,ykxyx=+=消去x得211480kyyk−+=，···

················································5分设211(,)4yAy，222(,)4yBy，则128yy=.·························

····································6分设直线CD的方程为2(2)ykx=+，233(,)4yCy，244(,)4yDy，同理可得348yy=.·······································

···············································7分直线AD方程为241112241()444yyyyyxyy−−=−−，即1441414yyyxyyyy=+++

，化简得14144()0xyyyyy−++=，··································································8分同理，直线BC方程为23234()0xyyyyy−++=，··············

································9分因为(2,0)−在抛物线的对称轴上，由对称性可知，交点G必在垂直于x轴的直线上，所以只需证G的横坐标为定值即可.由141423234()0,4(

)0,xyyyyyxyyyyy−++=−++=消去y，因为直线AD与BC相交，所以2314yyyy++，解得231414232314()()4[()()]yyyyyyyyxyyyy+−+=+−+，·······························

····························10分12323412413423144[()()]yyyyyyyyyyyyyyyy+−−=+−+3241231488884[()()]yyyyyyyy+−−=+−+2=，················

···········································································11分所以点G的横坐标为2，即直线AD与BC的交点G在定直线2x=上.···················12分【解析三】（1

）同解析一.···········································································4分数学参考答案第11页共12页（2）设

直线AB方程为2xmy=−，由22,4,xmyyx=−=消去x得2480ymy−+=，········5分设211(,)4yAy，222(,)4yBy，则128yy=.················

·············································6分设直线CD的方程为2xny=−，233(,)4yCy，244(,)4yDy，同理可得348yy=.··········

············································································7分直线AD方程为241112241()444yyyyyxyy−−=−

−，即1441414yyyxyyyy=+++，化简得14144()0xyyyyy−++=，··································································8分同理，直线BC方程

为23234()0xyyyyy−++=，··············································9分因为(2,0)−在抛物线的对称轴上，由对称性可知，交点G必在垂直于x轴的直线上，所以只需证G的横坐标为定值即可.由141423234()0,4

()0xyyyyyxyyyyy−++=−++=，消去y，因为直线AD与BC相交，所以2314yyyy++，解得231414232314()()4[()()]yyyyyyyyxyyyy+−+=+−+，···························

································10分12323412413423144[()()]yyyyyyyyyyyyyyyy+−−=+−+3241231488884[()()]

yyyyyyyy+−−=+−+2=，···························································································11分所以点G的横坐标为2，

即直线AD与BC的交点G在定直线2x=上.···················12分22.（本小题满分12分）【命题意图】本小题主要考查函数的单调性、最值，不等式等基础知识；考查逻辑推理能力，运算求解能力和创新能力等；考查函数与方程思想，化归与转化思想，数形结合思想，分类与整合思想，

特殊与一般思想等；导向对数学抽象，数学建模，数学运算核心素养的关注；体现综合性和创新性.满分12分.【解析】（1）()1ln1fxxx=+−，·································································1分记1

()ln1gxxx=+−，则()22111xgxxxx−=−=，数学参考答案第12页共12页所以()0,1x，()0gx，所以()gx单调递减；()1,x+，()0gx，所以()gx单调递增；················

··································2分所以()()min10gxg==，所以()0gx，即()0fx，且仅有()10f=，所以()fx为()0,+上的增函数.····················

················································3分（2）（ⅰ）由（1）得()1ln1fxxax=++−，()21xgxx−=，因为()1,x+，所以()0gx，所以()fx单调递增，所以()(

)12fxfa=−，··········································································4分①当2a时，()0fx，所以()fx为递增函数，所以()()10fxf=，满足题意；·

································································5分②当2a时，()20fxa=−，1(e)10eaaf=+，又由()fx单调递增，所以()fx有唯一零点

0x，··············································6分则()00,xx时，()0fx，()fx单调递减，所以()()010fxf=，不合题意，舍去.综上，(,2a−.····························

·························································7分（ⅱ）经计算：10.5a=，()270.5,0.612a=，()3370.6,0.760a=.····················

·······8分因为1111110212212122nnaannnnn+−=+−=−+++++，所以数列na单调递增，所以，当1n=或2时，0.50.6na；当3n时，30.6naa.·························

··9分当2a=时，由（ⅰ）可知，此时()0fx，即()()21ln11xxxx−+，令()2111xxk−=+，则2121kxk+=−，则有121ln21kkk+−，····························

···············10分令1,2,,2knnn=++，则有11123254141lnlnlnln12221234121nnnnnnnnnnn++++++++++=++++−+，·················

···11分因为411lnln2ln20.72121nnn+=−++，所以当3n时，0.60.7na.所以，当1n=或2时，105na=；当3n时，106.na=··································12分