2023年福州市普通高中毕业班第二次质量检测数学答案与评分细则---定稿

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fx()+gx(−32)=，所以fx(+32)+gx()=，又f(12−x)+gx()=，则有fx(+31)=f(−x)，因为fx(+1)是奇函数，所以fx(+11)=−f(−x)，可得fx(+31)=−fx(+)，即有fx(+2)=−fx()与fx(+42)=−fx(+)，即fx(+=4)fx()，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−f(−x)=fx(+2)，所以f(−=x)fx()，所以fx()为偶函数.故A错误；由fx(+1)是奇函数，则f(10)=，所以f(30)=，又f(2)+f(4)=f(2)+f(0)=0，20所以fk()=5f(1)+f(2)+f(3)+f(4)=0，所以C选项错误；k=1由f(10)=得g(02)=，所以B选项错误；因为g(2)=−2f(5)=−2f(1)=2，g(1)+g(3)=2−f(4)+2−f(6)=−4f(4)+f(2)=4，20所以g(0)+g(1)+g(2)+g(3)=8，所以gk()=5g(0)+g(1)+g(2)+g(3)=40，k=1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.AD12xy+=3cos,1322x=−3cossin,12【解析一】由x+y+y=3，令可得243y=2sin,y=3sin,2故2xy+=23cos−2sin+2sin=23cos−23,23，故A正确；B错误；22x+y−xy=−32xy=−32(3cos−sin)2sin=−54sin2+1,9，故C错误；D6正确，故选AD.222212【解析二】令2x+=yt，则y=−t2x代入x+xy+y=3，整理可得，3x−3txt+−=30，222因为xR，所以=9t−12(t−3)=−3(t−12)0，解得−23t23，故A正确；B错数学参考答案第1页共12页 误；222222由x+xy+y=33x+y=−xy，因为x+y2xy，所以32−xyxy，所以22−31xy，又x+xy+y=−32xy，从而132−xy9，故C错误；D正确，故选AD.2222v12【解析三】令x=+uv，y=−uv，代入x+xy+y=3，得u+=1，3令uv==cos,3sin（0,2π）.π2x+y=3u+=v3cos+3sin=23ins+−23,23，故A正确，B错误；32222222x+y−xy=u+3v=cos+9ins=18sin+1,9，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．2413.−14.0.6415.1116.xy−+=40732215【解析一】：令fx()=x−3x+6x+2，则f(x)=3x−6x+6，设Pm(,1)，Qnfn(,())，2222依题意fm()=fn()，所以3m−6m+=63n−6n+6，则m−n=2(m−n)，显然mn，3则mn+=2，因为fx()=(x−1)+3(x−1)+6，所以fx()的图象关于点(1,6)中心对称，所fn()+1以点P与点Q关于点(1,6)对称，所以=6，则fn()=11.232215【解析二】：令fx()=x−3x+6x+2，因为f(x)=3(x−1)+30，故fx()在R上单调3232递增，令x−3x+6x+=21，设其根为x，得x−3x+6x=−1.由于fx()在点P处的切线PPPP与在点Q处的切线平行，得f(x)=k存在两实根，其中一个为x，设另一个为x.即PQ23x−6x+=6k两根为xP，xQ，由韦达定理得xxPQ+=2，则xxQP=−2，从而3232fx(Q)=xQ−3xQ+6xQ+=2(2−xP)−3(2−xP)+6(2−xP)2+32232=−x+6x−12x+−83x+12x−12126+−x+=−2(x−3x+6x)1011+=.PPPPPPPPP16【解析一】：由条件得B点为线段MN中点，设B点坐标为y(,xy)，得Mx(2,0)、Ny(0,2)，由|MA|:|AB|1:2=得A坐标0000N225xy00xy为A(,)，将A、B坐标分别代入+=1中，得B33126A22MxxyO00+=1,126x0=−2,解得则M、N坐标分别为(4,0)−、22y=2,25xy000+=1,12969(0,4)，直线l的方程为xy−+=40.数学参考答案第2页共12页 m16【解析二】：设直线l方程为y=+kxm（km,0），可得M−,0，Nm(0,)，根据k5mmmm|MA|:|AB|:|BN|1:2:3=得A,B坐标分别为−,,−,.将A,B坐标代入C方66k22k程中，可得2225mm2+=1,25m2+2mk22=1236,k21236k636即两式相减得mk=4，222222mm+=1,m+2mk=124,k124264k2222将其代入m+2mk=124k，得k=1，从而直线l的方程为xy−+=40.四、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.222【解析一】（1）由余弦定理可得b=c+a−2accosB，2222222代入b−=a2c，得到(c+a−2accos)B−a=2c，······································1分2化简得c+=2accosB0，即c+=2cosaB0，····················································································2分由正弦定理可得sinC+=2sincosAB0，即sin(AB+)+2sincosAB=0，····································································3分sincosAB+cossinAB+2sincosAB=0，3sincosAB=−cossinAB，·········································································4分tanB可得=−3.·························································································5分tanA（2）由（1）可知tanBA=−3tan，所以tanA与tanB异号，222又b−a=20c，故ba，则有BA，因为AB,(0,π)，故A为锐角，B为钝角，则C为锐角，···································6分tanA+−tanBtanA3tanA2tanA所以tanC=−tan(AB+)===······················7分22tantanAB−1−3tanA−13tanA+1223==，·························································8分12333tanA+tanA3π等号当且仅当tanA=，即A=时成立，····················································9分36π所以C的最大值为.················································································10分6数学参考答案第3页共12页 sinBb【解析二】（1）由正弦定理可得：=，··················································1分sinAa222b+−cacosA2bc由余弦定理可得：=，·························································2分222cosBa+−cb2ac222b+−ca222tanBsinBcosAb2bcb+−ca则===，········································3分222222tanAsinAcosBaa+−cba+−cb2ac22222tanBc+2c因为b−=a2c，所以==−3.···················································5分22tanAc−2c222a+−bc（2）cosC=·············································································6分2ab2222ba−ab+−=2······································································7分2ab223ab+3ab==+4ab44ba32163=.······················································································8分2π22所以C，等号当且仅当ba=3时，即ba=3时成立，·································9分6π所以C的最大值为·················································································10分6222222【解析三】（1）由b−=a2c，由正弦定理可得，sinB−=sinA2sinC，··········1分22222222因为sinB−sinA=sinB−sinBsinA+sinBsinA−sinA22222222=sinB(1sin−A)sin−A(1sin−B)=sinBcosA−sinAcosB=(sincosBA+cossin)(sincosBABA−cossin)BA=sin(AB+)sin(B−A)，···············3分又因为A+BC+=π，所以sinC=+sin(AB)，2所以sin(A+B)sin(B−A)=2sin(A+B)，因为0+ABπ，所以sin(AB+)0，所以sin(B−A)=2sin(AB+)，化简得−=3sincosABsincosBA，·································································4分tanB可得=−3.·························································································5分tanA（2）同解析一.························································································10分数学参考答案第4页共12页 18.（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.P11∵EFCD，EF=CD，ABCD，AB=CD，FE22∴EFAB，EF=AB，··································1分DC∴四边形ABEF为平行四边形，AB∴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PO⊥AD，OGCD.∵平面PAD⊥平面ABCD，平面PAD平面ABCD=AD，PO⊥AD，PO平面PAD，∴PO⊥平面ABCD.··················································································5分z∵AD⊥CD，OGCD，P∴OG⊥OA.·····················································6分FE以O为原点，以ADOGOP,,所在直线为x轴、y轴、z轴，DC建立如图所示空间直角坐标系.······························7分yOGA因为AB=2,CD=4,AD=23，△PAD是等边三角形，Bx122所以PD==AD23，OA==AD3，PO=PD−OD=3，2所以A(3,0,0)，B(3,2,0)，C(−3,4,0)，P(0,0,3).则AB=(0,2,0)，BC=−(23,2,0)，BP=−(3,2,3−).········································9分设平面PBC的法向量为n=(xyz,,)，由BC=n0，BP=n0，−23xy+2=0,可得令x=1，可得yz==3,3，−3x−2y+3z=0,从而n=(1,3,3)是平面PBC的一个法向量.··················································10分ABn2321则cosAB,n===，·····················································11分|AB|||n27721所以直线AB与平面PBC所成角的正弦值为.·············································12分7数学参考答案第5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，22则(3)=3−=36；·················································································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，33则(3)=3−=918.··················································································4分n−1n（2）因为3kk−2,3（k=1,2,,3）中与3互质的正整数只有32k−与31k−两个，nnn−1所以1,3中与3互质的正整数个数为23，nn−111nn−−11n所以(3)=23，a=(3)=23=3，········································7分n22logan−13n所以=.····················································································8分n−1a3nloga3n设数列的前n项和为Tn.an12n−1112n−1T=+0+++，T=+0+++，···································9分n12n−1n23n3333333111−1111nn−−133n−1311−T=+++−=−······································10分3n31323n−13n13n1−311n−1=−−nn−1223312n+1=−，··············································································11分n223312nn++1321则T=−=−.·····································································12分nnn−12223443n说明：不超过3，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11nnn−−11n排除掉是3的倍数的数即可，可得a=(3)=(3−3)=3，学生若这样写，同样可n22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6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.2【解析一】（1）把男性样本记为xx,,,x，其平均数记为x，方差记为s；把女性样本12120x222记为yy,,,y，其平均数记为y，方差记为s.则xs==14,6；ys==21,17.记总样本1290yxy2数据的平均数为z，方差为s.由x=14，y=21，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12090关系，可得总样本平均数为z=+xy·········································2分12090++12090120149021+=210=17，······························································3分根据方差的定义，总样本方差为112022902s=(xii−z)+(y−z)·································································4分210ii==1111202290=(xii−+−xxz)+(y−+yy−z)，210ii==11120120120120由(xii−x)=x−120x=0，可得2(xii−xx)(−z)=2(x−z)(x−x)=0，ii==11ii==119090同理，2(yii−y)(y−z)=2(y−z)(y−y)=0，············································5分ii==11因此，1120212029029022s=(xii−x)+(x−z)+(y−y)+(y−z)210i=1i=1i=1i=112222=120sxy+(x−z)+90s+(y−z)，···············································6分2102122所以s=1206+(1417−)+9017+(2117−)23，210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（2）由（1）知=23，所以XN(17,23)，又因为234.8，所以P(12.2X21.8)=P(174.8−X174.8+)0.6827，······························8分1PX(12.2)(10.6827−)=0.15865，························································9分2数学参考答案第7页共12页 因为XB(3,0.15865)，···········································································10分33所以PX(=3)=C0.158650.004.····························································11分3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2【解析二】（1）把男性样本记为xx,,,x，其平均数记为x，方差记为s；把女性样本12120x222记为yy,,,y，其平均数记为y，方差记为s.则xs==14,6；ys==21,17.记总样本1290yxy2数据的平均数为z，方差为s.由x=14，y=21，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，12090可得总样本平均数为z=+xy··················································2分12090++12090120149021+=210=17，········································································3分根据方差的定义，总样本方差为112022902s=(xii−z)+(y−z)·································································4分210ii==111120120221209022=xi−2zxi+120z+yi−2zyi+90z210i=1i=1i=1i=1120901222=xii+y−2120z(x+90y)+210z210ii==11120901222=xii+y−2z210z+210z210ii==11120901222=xii+y−210z210ii==11120901222=xii+y−z，①····································································5分210ii==11111202120212022222由sx=(xi−x)=xi−x，可得xix=+120(sx)，②120ii==11120i=190222同理yiy=+90(sy)，③i=1将②③代入①得2122222s=120s+x+90s+y−z··························································6分(xy)()210=1+2++2−2120(614)90(1721)1723，210所以总样本的均值为17，方差为23，数学参考答案第8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.22【解析一】（1）当l的斜率为时，得l方程为yx=+(2)，·····························1分11332y=2,px2由2消元得y−3py+4p=0，······················································2分yx=+(2),3322由弦长公式得|AB|=1()+(3)p−16p=13，···········································3分222即9pp−=162，解得p=2或p=−（舍去），92从而E的标准方程为yx=4.·······································································4分22yyyy−41221（2）设Ay(,)，By(,)，得k==，12AB2244yyyy+2112−4424y1直线AB方程为y=()x−+y，即4x−(y+yy)+yy=0，····················5分11212yy+412又直线AB过点(2,0)−，将该点坐标代入直线方程，得yy=8.·····························6分1222yy34设Cy(,)，Dy(,)，同理可得yy=8，·················································7分343444直线AD方程为4x−(y+yy)+yy=0，·······················································8分1414直线BC方程为4x−(y+yy)+yy=0，·······················································9分2323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.yB4x−(y+yy)+yy=0,1414由消去y，A4x−(y2+yy3)+yy23=0,GxCD因为直线AD与BC相交，所以y+yy+y，2314yyy(+y)−yyy(+y)23141423解得x=，···························································10分4[(y+y)(−y+y)]2314数学参考答案第9页共12页 yyy+yyy−yyy−yyy123234124134=4[(y+y)(−y+y)]23148y+8y−8y−8y3241=4[(y+y)(−y+y)]2314=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x=2上.···················12分【解析二】（1）同解析一.···········································································4分（2）设直线AB的方程为y=+kx(2)，1y=+kx(2),12由消去x得ky11−4y+8k=0，···················································5分2yx=4,22yy12设Ay(,)，By(,)，则yy=8.·····························································6分12124422yy34设直线CD的方程为y=+kx(2)，Cy(,)，Dy(,)，23444同理可得yy=8.······················································································7分342y−yy4yy41114直线AD方程为y−y=()x−，即yx=+，122yy4y++yyy41−414144化简得4x−(y+yy)+yy=0，··································································8分1414同理，直线BC方程为4x−(y+yy)+yy=0，··············································9分2323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.4x−(y+yy)+yy=0,1414由消去y，4x−(y+yy)+yy=0,2323因为直线AD与BC相交，所以y+yy+y，2314yyy(+y)−yyy(+y)23141423解得x=，···························································10分4[(y+y)(−y+y)]2314yyy+yyy−yyy−yyy123234124134=4[(y+y)(−y+y)]23148y+8y−8y−8y3241=4[(y+y)(−y+y)]2314=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x=2上.···················12分【解析三】（1）同解析一.···········································································4分数学参考答案第10页共12页 x=−my2,2（2）设直线AB方程为x=−my2，由消去x得y−4my+=80，········5分2yx=4,22yy12设Ay(,)，By(,)，则yy=8.·····························································6分12124422yy34设直线CD的方程为x=−ny2，Cy(,)，Dy(,)，3444同理可得yy=8.······················································································7分342y−yy4yy41114直线AD方程为y−y=()x−，即yx=+，122yy4y++yyy41−414144化简得4x−(y+yy)+yy=0，··································································8分1414同理，直线BC方程为4x−(y+yy)+yy=0，··············································9分2323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.4x−(y+yy)+yy=0,1414由，消去y，4x−(y+yy)+yy=02323因为直线AD与BC相交，所以y+yy+y，2314yyy(+y)−yyy(+y)23141423解得x=，···························································10分4[(y+y)(−y+y)]2314yyy+yyy−yyy−yyy123234124134=4[(y+y)(−y+y)]23148y+8y−8y−8y3241=4[(y+y)(−y+y)]2314=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x=2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.1【解析】（1）f(x)=lnx+−1，·································································1分x111x−1记gx()=lnx+−1，则gx()=−=，22xxxx数学参考答案第11页共12页 所以x(0,1)，gx()0，所以gx()单调递减；x(1,+)，gx()0，所以gx()单调递增；··················································2分所以gx()==g(10)，所以gx()0，即fx()0，且仅有f(10)=，min所以fx()为(0,+)上的增函数.····································································3分1x−1（2）（ⅰ）由（1）得f(x)=lnx++−1a，gx()=，2xx因为x(1,+)，所以gx()0，所以fx()单调递增，所以f(x)f(12)=−a，··········································································4分①当a2时，fx()0，所以fx()为递增函数，所以fx()=f(10)，满足题意；·································································5分a1②当a2时，f(x)=−20a，f(e)1=+0，ae又由fx()单调递增，所以fx()有唯一零点x，··············································6分0则xx(0,)时，fx()0，fx()单调递减，0所以fx()=f(10)，不合题意，舍去.0综上，a−(,2.·····················································································7分737（ⅱ）经计算：a=0.5，a=(0.5,0.6)，a=(0.6,0.7).···························8分123126011111因为aa−=+−=−0，所以数列a单调递增，nn+1n2n+12n+2n+12n+12n+2所以，当n=1或2时，0.5a0.6；当n3时，aa0.6.···························9分nn321(x−)当a=2时，由（ⅰ）可知，此时fx()0，即lnxx(1)，x+121(x−)121k+12k+1令=，则x=，则有ln，···········································10分xk+121k−kk21−令k=+n1,n+2,,2n，则有1112n+32n+54n+14n+1+++ln+ln++ln=ln，····················11分n+1n+22n2n+12n+34n−12n+14n+11因为ln=ln2−ln20.7，所以当n3时，0.6a0.7.n2nn++121所以，当n=1或2时，10a=5；当n3时，10a=6.··································12分nn数学参考答案第12页共12页