福建省厦门市2022届高三毕业班第三次质量检测-数学试题【含答案】

高三诊断性测试数学参考答案及评分细则评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。4．只给整数分数。选择题和填空题不给中间分。一、选择题：本大题考查基础知识和基本运算。每小题5分，满分40分。1．B2．B3．D4．B5．D6．C7．A8．D二、选择题：本大题考查基础知识和基本运算。每小题5分，满分20分。全部选对的得5分，部分选对的得2分，有选错的得0分。9．ABD10．AC11．BD12．ACD三、填空题：本大题考查基础知识和基本运算。每小题5分，满分20分。1π311,x1,13．；14．,1；15．答案不唯一，如：fx1,fxx等；232xxx1,≤116．5；54π．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．本小题主要考查等差数列、等比数列、递推数列及数列求和等基础知识，考查运算求解能力、逻辑推理能力和创新能力等，考查化归与转化思想、分类与整合思想、函数与方程思想、特殊与一般思想等，考查逻辑推理、数学运算等核心素养，体现基础性和创新性．满分10分．解法一：（1）因为SSSn21,,nn成等差数列，所以SSSSnn21nn，···················1分所以an2an1an1，···········································································3分q即aann212，设an的公比为，则q2，··········································4分nn1所以an222．··································································6分2kk11**21（2）依题意，b2kk1kkNN,b2kk，·················7分22所以T10abab1133ab99abab2244ab1010································8分a12a35a9a22a45a10a12a35a92a12a35a9·······································9分a125a3a9122239523511所以4T10122252，数学参考答案及评分细则第1页（共16页）5214142两式相减得3T223252952115211211，101433142所以T211+3186．·································································10分1099解法二：（1）因为SSSn21,,nn成等差数列，所以SSSn12n2n，··········································································································1分设an的公比为q，①若q1，则ann2,S2n，Sn12Sn4n6,2Sn4n，所以SSSn12n2n，与矛盾，不合题意；··························································2分nnn+1+2aq11a1111qaq②若q1，则S，SS,，···············3分n1qnn1211qqn+1n+2na11qa11q2a11qn+1n+2n2所以，整理得，qq2q，即qq20，1q1q1q解得q1（舍去）或q2，··································································4分nn1所以an222．··································································6分2kk11**21（2）依题意，b2kk1kkNN,b2kk，·················7分22所以T10abab1122ab33ab44ab55ab66ab77ab88ab99ab1010····8分a1a22a3a43a5a6+4a7a85a9a10···························9分12222232432526427285292101222332542752921696512+25603186．·······················································································10分解法三：（1）因为SSSn21,,nn成等差数列，所以2SSSnn21n，·························1分当n1时，2SSS132，化简得aa322，···············································2分设an的公比为q，所以q2，·····························································4分n122n1222当时，S，因此2S，n3n3n122n322n242n2222SS++==，nn+2+13333满足，故符合题意．nn1所以an2(2)(2)．···································································6分数学参考答案及评分细则第2页（共16页）（2）依题意，b11，b21，b32，b42，b53，b63，b74，b84，b95，b105，··········································································································7分所以2345678910T102(2)2(2)2(2)3(2)3(2)4(2)4(2)5(2)5(2)··········································································································8分[2(2)2]2[(2)3(2)4]3[(2)5(2)6]4[(2)7(2)8]5[(2)9(2)]10...................................................................................................................................................9分2243254275292169651225603186．.................................................................................................................................10分18．本小题主要考查独立事件的概率、互斥事件的概率，二项分布、数学期望等基础知识；考查数学建模能力，运算求解能力，逻辑推理能力，创新能力以及阅读能力等；考查统计与概率思想、分类与整合思想等；考查数学抽象，数学建模和数学运算等核心素养；体现应用性和创新性．满分12分．解法一：（1）甲滑雪用时比乙多536180秒3分钟，因为前三次射击，甲、乙两人的被罚时间相同，所以在第四次射击中，甲至少要比乙多命中4发子弹．设“甲胜乙”为事件A，“在第四次射击中，甲有4发子弹命中目标，乙均未命中目标”为事件B，“在第四次射击中，甲有5发子弹命中目标，乙至多有1发子弹命中目标”为事件C，··········································································································1分依题意，事件B和事件C是互斥事件，A=B+C，·········································2分14415451514311，分PCPCB,C55························4554544469所以，PPPABC.1250069即甲胜乙的概率为.········································································5分12500（2）依题意得，甲选手在比赛中未击中目标的子弹数为X，乙选手在比赛中未击中目标11的子弹数为Y，则XBYB20,,20,，···········································7分541所以甲被罚时间的期望为1EX1204（分钟），································8分51乙被罚时间的期望为1EY1205（分钟），······································9分4又在赛道上甲选手滑行时间慢3分钟，所以甲最终用时的期望比乙多2分钟.······················································11分因此，仅从最终用时考虑，乙选手水平更高.············································12分解法二：（1）同解法一．···············································································5分数学参考答案及评分细则第3页（共16页）44（2）设甲在一次射击中命中目标的子弹数为，则B5,，所以E54，所55以甲在四次射击中命中目标的子弹数的期望为4E16，·······························7分3315设乙在一次射击中命中目标的子弹数为，则B5,，所以E5，所444以乙在四次射击中命中目标的子弹数的期望为4E15，·······························9分所以在四次射击中，甲命中目标的子弹数的期望比乙多1，所以乙被罚时间的期望比甲多1分钟，又因为在赛道上甲的滑行时间比乙慢3分钟，所以甲最终用时的期望比乙多2分钟，······················································································11分因此，仅从最终用时考虑，乙选手水平更高.··············································12分19．本小题主要考查直线与直线、直线与平面、平面与平面的位置关系，直线与平面所成角、二面角等基础知识；考查空间想象能力，逻辑推理能力，运算求解能力等；考查化归与转化思想，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；体现基础性和综合性．满分12分．解法一：（1）如图，在△VAC内过P作PMVC，垂足为M，V在△VBC内过作MNVC交VB于N，M连结PN，则直线PN即为直线l.···························2分lP理由如下：因为，，PMMNM，N所以平面，AVCPMNC由于过空间一点与已知直线垂直的平面有且只有一个，所以平面与平面重合，B因为平面PMN平面VABPN，所以直线即为直线.····························4分（2）因为△VAB和△ABC均为等边三角形，所以VAVB,ACBC，又因为VCVC,所以△VAC≌△VBC，所以PVMNVM，2又VMVM，所以Rt△VPM≌Rt△VNM，所以VPVN，所以VNVB.·······5分3如图，设AB的中点为D，连结VD,CD，V因为和均为等边三角形，M所以，所以ABVD,ABCD，P又因为VDCDD，所以AB平面VCD，因为AB平面ABC，NA所以平面平面CABC.OD在△中，作VOCD，垂足为O，B因为平面ABC平面VCDCD，VO平面VCD，所以VO平面ABC，π所以VCD是直线VC与平面ABC所成的角，所以VCD.························7分3因为和均是边长为4的等边三角形，所以VDDC23,所以△VCD是等边三角形，所以VO3，DOOC3.以O为原点，分别以OC,OV的方向为y轴和z轴正方向建立如图所示的空间直角坐标数学参考答案及评分细则第4页（共16页）系Oxyz，则ABCV2,3,0,2,3,0,0,3,0,0,0,3，·······································8分所以CV0,3,3,CA2,23,0,AB4,0,0,1245328zCPCVCA,,1,PNAB,0,0.V333333过l及点C的平面为平面CPN，设平面CPN的法向量为n(,,)xyz，P453ANxyz0,yCPn0,C33O则即取n(0,3,5)，D8PNn0,Bx0.x3即平面的一个法向量为n(0,3,5).··················································10分易知，平面ABC的一个法向量为m(0,0,1)，···········································11分mn557所以cosmn,，mn271457所以过及点的平面与平面ABC所成的锐二面角的余弦值为．············12分14解法二：（1）如图，在△VAB内过P作PN∥AB，交VB于N，则直线PN即为直线l.·············································································2分理由如下：V取VC的中点Q，连结AQ,BQ，因为△VAB和△ABC均为等边三角形，QP所以VAAC,VBBC，所以VCAQ,VCBQ，lN又因为AQBQQ，所以VC平面ABQ，AC又因为平面，所以平面∥平面，又因为平面平面VABl，平面ABQ平面VABAB，B所以AB∥l，所以直线即为直线.························································4分22（2）由（1）知，PN∥AB，因为VPVA，所以VNVB.·····························5分33设AB的中点为D，连结VD，交PN于G，连结CG，V因为和均为等边三角形，所以，所以，VAVB,ACBCABVD,ABCDP又因为VDCDD，GN所以AB平面VCD，AB平面ABC，AOC所以平面ABC平面.DB在△中，作VOCD，垂足为O，因为平面ABC平面VCDCD，VO平面VCD，所以VO平面ABC，数学参考答案及评分细则第5页（共16页）π所以VCD是直线VC与平面ABC所成的角，所以VCD，······················7分3π因为△VAB和△ABC均是边长为4的等边三角形，所以VDDC23,VDC，3123因为AB∥PN，所以DGDV．由（1）知，过l及点C的平面为平面CPN，33因为AB平面，PN平面，所以AB∥平面，·····················8分设平面CPN平面ABCl'，因为AB平面ABC，所以AB∥l'，因为AB平面VCD，CG平面，CD平面，所以ABCG,ABCD，所以CGl',CDl'，又因为CG平面，CD平面，所以GCD为平面与平面ABC所成的锐二面角的平面角，····················10分221在△GCD中，由余弦定理得，CG2DG2DC22DGDCcosGDC，CG,3········································································································11分CG2DC2DG257所以cosGCD，2CGDC1457所以过及点的平面与平面所成的锐二面角的余弦值为．············12分1420．本小题主要考查正弦定理、余弦定理及三角恒等变换等基础知识，考查逻辑推理能力、运算求解能力等，考查化归与转化思想、函数与方程思想、数形结合思想等，考查数学运算、逻辑推理等核心素养，体现基础性和综合性．满分12分．a2c2b2解法一：（1）在△ABC中，由余弦定理得cosB，·····························1分2ac又因为a6，b12cosB2c，a2c2b2所以bc122，···································································2分2ac整理得b22c36bc.···········································································3分在△中，由余弦定理得b22c362bccosA，1所以bc2bccosA，即cosA.·······························································4分2又因为A(0,)，所以A．································································5分3（2）选③．·······························································································6分1因为M为△的内心，所以BADCAD=BAC，26由SSSABCABDACD，·················································7分111得bcsincADsinbADsin，23262631bc因为AD=33，所以bc33(bc)，即bc........................................8分223数学参考答案及评分细则第6页（共16页）由（1）可得b22c36bc，即(bc)23bc36，·····································9分()bc2bc所以3bc360，即(bc9)(4)0，········································10分99又因为bc0，所以bc36，·································································11分113所以Sbcsin3693．················································12分ABC2322解法二：（1）因为a6，b12cosB2c，所以b2acosB2c，···········································································1分在△ABC中，由正弦定理得sinBABC2sincos2sin，即sinBABAB2sincos2sin()，·························································2分即sinBABABAB2sincos2sincos2cossin，即sinBAB2cossin，··········································································3分因为B0,，1所以sinB0，故cosA.·····································································4分2又因为A(0,)，所以A．································································5分3（2）选②．·······························································································6分因为M为△的垂心，所以BMDMBDACBACB，又MD3，·················7分222所以在△MBD中，BDMDtanBMD3tanACB，同理可得CD3tanABC，·····································8分又因为BDCD6，所以3tanABC+3tanACB6，即tanABC+tanACB23，······································9分又因为在△中，tanABCACBtanBAC3，tanABCtanACB所以3，1tanABCtanACB因此tanABCtanACB=3．··································································10分故tanABC，tanACB为方程xx22330两根，即tanABC=tanACB=3，因为ABC，ACB0,，所以ABC==ACB，所以△ABC为等边三角形，···································11分313所以S6293．······························································12分ABC22解法三：（1）同解法一.·················································································5分（2）选②．·······························································································6分因为为△的垂心，数学参考答案及评分细则第7页（共16页）所以AMBACB，ABMBAC，·····································8分26AMAB所以在△ABM中，由正弦定理得，sinABMsinAMBAMAB即．··········································9分sinsinACB6又因为在△ABC中，ABBC由正弦定理得，··························10分sinACBsin3AMBC所以，因为a6，所以AM23.········································11分sinsin6311又因为MD3，所以SaAD6(233)93.···················12分ABC22解法四：（1）同解法一.·················································································5分（2）选③．·······························································································6分1因为M为△ABC的内心，所以BADCAD=BAC.26BDAD33在△ABD中，由正弦定理得，因为AD33，所以2BD，sinsinBsinB63333同理可得2CD.·····························································7分sinCsin(B)33333又因为BDCD6，所以12，sinBsin(B)3即4sinBBBBsin()3sinsin()，3313即4sinBBBBBsin()3(sinsincos)，·······································8分322即4sinBBBsin()3sin()，36即4sin[(BBB)]sin[()]3sin()，·········································9分666663131即4sin(BBBBB)cos()sin()cos()3sin()，262626266数学参考答案及评分细则第8页（共16页）3122即4sin(BBB)cos()3sin()，464662即4sin(BB)3sin()10，即sin(BB)14sin()10，····10分666625因为A,所以0B，所以B(,)，33666所以sin(B)0，故sin(B)1，66即B，即B，所以△ABC为等边三角形，··································11分62313所以S6293.·································································12分ABC22解法五：（1）同解法一.·················································································5分（2）选③．·······························································································6分1因为M为△ABC的内心，所以BADCAD=BAC.又因为AD33，263在△ABD中，由余弦定理得BD2=c227233cc29c27，2同理可得CD22b9b27．······························7分1πABADsinBDSAB又因为ABD=26，CDS1πACACDACADsin26c229c27c所以，即(bc)[3(bc)bc]0，b229b27bbc故bc或bc.···············································································8分313（i）当bc时，△为等边三角形，所以S6293.ABC22（ii）当时，由（1）知b22c36bc，即(bc)23bc36，··········9分()bc2bc所以3bc360，即(bc9)(4)0，········································10分99因为bc0，所以bc36.······································································11分113又因为A，所以Sbcsin3693．3ABC2322综上所述，SABC93．······································································12分说明：设△的外接圆半径为R，则在△中，由正弦定理得数学参考答案及评分细则第9页（共16页）BC62R43，即R23，因为M为外心，所以AM23，与AM4矛sinAsin3盾，故不能选①．21．本小题主要考查椭圆的标准方程及简单几何性质，直线与圆、椭圆的位置关系，平面向量等基础知识；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；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．满分12分．解法一：（1）以O为坐标原点，椭圆C的长轴、短轴所在直线分别为x轴、y轴，建立平面直角坐标系，如图.··············································································1分设椭圆的长半轴为a,短半轴为b,半焦距为c，c2y,Aa2a2,66x依题意得b1,解得b1,··········3分O33c1,Ba2b2c2,x2所以的方程为y21．···································································4分2（2）因为直线AB与以OA为直径的圆的一个交点在圆上，所以直线AB与圆相切．······································································5分6666x（i）当直线垂直于轴时，不妨设AB,,,，3333此时OAOB0，所以OAOB，故以为直径的圆过点O．····················6分（ii）当直线不垂直于轴时，设直线的方程为ykxm，Ax1,,,y1Bx2y2．m6因为与圆相切，所以到直线的距离，k213即3mk22220．·············································································7分ykxm,22k21x24kmx2m220由x2得，·····································8分y1,24km2m22所以xx,xx，·························································9分122kk221122122OAOBxxyyxxkxmkxm1kx1x2kmx1x2m121212122222m24km1k22kmm2kk121数学参考答案及评分细则第10页（共16页）1k22m22km4kmm22k2121k23mk22220，21k2所以OAOB，故以AB为直径的圆过点O．综上，以为直径的圆过点．···························································12分解法二：（1）同解法一．·············································································4分（2）因为直线与以OA为直径的圆的一个交点在圆O上，所以直线AB与圆相切．······································································5分设直线与圆相切于点Mx00,y．6666（）当时，直线垂直于x轴，不妨设AB,,,，iy003333此时OAOB0，所以，故以为直径的圆过点．····················6分x0222（ii）当y00时，直线的方程为yy00xx，因为xy00，y03x2所以直线的方程为yx0．····················································7分yy003x02yx,yy003设Ax1,,,y1Bx2y2，由得x2y212222218x09y0x24x0x818y00，························································8分224xy00818所以x1x222,x1x222，·················································9分18x09y018x09y0224xx00184因为，所以x1x222,x1x2，69xx006922222OA2OB2AB2OMAMOMBMAMBM422OM2AMBM2AMBM3224xx0021x2x01x0x13yy0024x0221x1x2x1x2x0x03y0数学参考答案及评分细则第11页（共16页）2224x018x0424x0221x03y69x2269x0002224x018x0424x022122x032269xx69x003044249x0222323xx006944330.所以OA2OB2AB2，即OAOB，故以AB为直径的圆过点O．综上，以为直径的圆过点．···························································12分解法三：（1）同解法一．···············································································4分（2）因为直线与以OA为直径的圆的一个交点在圆O上，所以直线AB与圆相切．······································································5分（i）当直线不垂直于x轴时，设直线的方程为ykxm，Ax1,,,y1Bx2y2．m6因为与圆相切，所以到直线的距离，k213即3mk22220．·············································································6分ykxm,22k21x24kmx2m220由x2得，·····································7分y1,24km2m22所以xx,xx，·························································8分122kk22112212myykxx2m，121221k2mk222yykxmkxmkxx22mkxxm，1212121221k22m22m22k23m22k22xxyy0．12122k212k212k21设Px,y是以为直径的圆N上的任意一点，由PAPB0，得xx1xx2yy1yy20，····························10分22化简得xyxxxyyyxx121212yy120，42kmm故圆的方程为x22yxy0，它过定点．···················11分2kk22121数学参考答案及评分细则第12页（共16页）6666x（ii）当直线AB垂直于轴时，不妨设AB,,,，3333此时OAOB0，所以OAOB，故以AB为直径的圆过点O．综上，以为直径的圆过点．···························································12分解法四：（1）同解法一．···············································································4分（2）因为直线与以OA为直径的圆的一个交点在圆O上，所以直线与圆相切．······································································5分（i）当直线不垂直于轴时，设直线的方程为ykxm，Ax1,,,y1Bx2y2．m6因为与圆相切，所以到直线的距离，k213即3mk22220．·············································································6分ykxm,22k21x24kmx2m220由x2得，·····································7分y1,24km2m22所以xx,xx，·························································8分122kk22112212myykxx2m，121221k2xxyy2kmm1212,以为直径的圆N的圆心为,，即22．222kk121AB1半径r1k2xx222122221221k16km8m814kx1x2x1x2222221k22k11k216k28m281k24k22m22，22kk22121以为直径的圆的方程为2222kmm1k24k22m22xy，2222k12k12k142kmm整理得x22yxy0，2kk22121故以为直径的圆过定点．······························································11分（ii）当直线垂直于轴时，不妨设，此时，所以，故以为直径的圆过点．综上，以为直径的圆过点．···························································12分22．本小题主要考查导数，函数的单调性、零点、不等式等基础知识；考查逻辑推理能力，数学参考答案及评分细则第13页（共16页）直观想象能力，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，分类与整合思想等；考查逻辑推理，直观想象，数学运算等核心素养；体现基础性、综合性和创新性．满分12分．解法一：（1）依题意，fx的定义域为0,，a11a1xa1由fxlnxaR，得f'x，······················1分xxx22x①当a1时，f'x0恒成立，所以在0,单调递增；··········································································································2分②当a1时，令f'x0，得xa1，当xa0,1时，f'x0，所以在0,a1单调递减；当xa1,时，f'x0，所以在a1,单调递增；综上，当时，在单调递增；当a1时，在单调递减，在单调递增．···········4分xa1ax3（2）设hfxxgx，则hf''xxax3e1x，x21ex··········································································································5分①当x3时，h'x0恒成立，所以hx在3,单调递增，5a1aaa又因为0a，所以h3ln31ln310，33e2e2e2所以hx0，hx在不存在零点；············································6分②当03x时，设xxex1，则'xex11，当01x时，'x0，所以x在0,1单调递减；当13x时，'x0，所以x在1,3单调递增；11所以x10，即ex1x，因为x0，所以，···················7分ex1x5ax33ax又因为0a且，所以ax3<0，所以，3ex1xxa1ax3ax213axa1所以h'x，·························8分x22xx131a当0a时，函数xax213axa1的对称轴为x0，32a所以x在0,3单调递增，所以xa010，所以h'x0，所以hx在单调递增；···········································9分数学参考答案及评分细则第14页（共16页）15216当a时，13a4aa15a210a10，339所以x0，所以h'x0，所以hx在0,3单调递增；··················10分5综上可知，当0a时，均有在单调递增，3又因为h1aa110，所以hx在恰有一个零点1,···············11分5故当0a时，在0+，恰有一个零点，3因此不存在xx12,，且xx12，使得fxiigxi1,2．··························12分解法二：（1）同解法一；···············································································4分a1（2）记Fxfxgx，则Fxlnxax2e1x1，x11axa1exaex3x2则Fxax3e1x，························5分xx22xexx22xx记haxa1eaxxe3eexxax31e，··················7分设xxeex，则'xexe，当01x时，'x0，所以x在0,1单调递减；当x1时，'x0，所以x在1,+单调递增；所以x10，即eexx，··························································8分58x5285212所以，hxee3xxxxxxxxxee3e5128，333333225因为12458160，所以5xx1280，所以h0，·······9分3又hx01ex05所以当0ax,0时，ha0，·····················································10分3所以Fx0，Fx在0,上单调递增，又因为F10，所以Fx在上恰有1个零点1，··················································11分因此不存在，且，使得．··························12分解法三：（1）同解法一；xa1ax3（2）设hfxxgx，则hf''xxax3e1x，x21ex······························