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数学参考答案及评分细则第1页（共16页）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。13．π2；14．3,13；15

．答案不唯一，如：11,1,11,21,1xxfxfxxxx≤等；16．5；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21,,nnnSSS成等差数列，所以2

1nnnnSSSS，···················1分所以211nnnaaa，·································································

··········3分即212nnaa，设na的公比为q，则2q，··········································4分所以1222nnna．······································

····························6分（2）依题意，**21221121,22kkkkbkkbkkNN，·················7分所以1011339922441010Tabababa

babab································8分13924102525aaaaaa13913925225aaaaaa················

·······················9分13925aaa39122252所以3511104122252T，数学参考答案及评分细则第2页（共16页）两式相减得53591111111021414232222525221

433T，所以11101422+318699T．·······························································

··10分解法二：（1）因为21,,nnnSSS成等差数列，所以122nnnSSS，·····································································································

·····1分设na的公比为q，①若1q，则2,2nnaSn，1246,24nnnSSnSn，所以122nnnSSS，与122nnnSSS矛盾，不合题意；·································

·························2分②若1q，则111nnaqSq，+1+2111211,11nnnnaqaqSSqq，··············

·3分所以+1+21111121111nnnaqaqaqqqq，整理得，+1+22nnnqqq，即220qq，解得1q（舍去）或2q，·····················································

·············4分所以1222nnna．··································································6分（2）依题意，**21221121,22kkkkbkkbkk

NN，·················7分所以101122334455667788991010Tabababababababababab····8分12345678

91023+45aaaaaaaaaa···························9分2345612222232278910422522

3579122232425221696512+25603186．·······························································

························10分解法三：（1）因为21,,nnnSSS成等差数列，所以212nnnSSS，·························1分当1n时，1322SSS，化简得322a

a，···············································2分设na的公比为q，所以2q，·····················································

········4分当2q时，1223nnS，因此122223nnS，1322+2+1222222242++==3333nnnnnnSS，满足212nnnSSS

，故2q符合题意．所以12(2)(2)nnna．···································································6分数学参考答案及评分细则第3页（共16页）（2）依题意，11b，21b，

32b，42b，53b，63b，74b，84b，95b，105b，·····················································································

·····················7分所以2345678910102(2)2(2)2(2)3(2)3(2)4(2)4(2)5(2)5(2)T·

·········································································································8分2345678910[2(2)]2[(2)(2)]3[(2

)(2)]4[(2)(2)]5[(2)(2)].....................................................

..............................................................................................9分45792232425

2216965122560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536180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4555411551414113B,C5545

444PCPC，························4分所以，69ABC12500PPP.

即甲胜乙的概率为6912500.········································································5分（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的子弹数为Y，则1120,,20,54X

BYB，···········································7分所以甲被罚时间的期望为1112045EX（分钟），·················

···············8分乙被罚时间的期望为1112054EY（分钟），······································9分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4页（共16页）ONPDCBVAM（2）设甲在一次射击中命中目标的子弹数为，则45,5B

，所以4545E，所以甲在四次射击中命中目标的子弹数的期望为416E，·······························7分设乙在一次射击中命中目标的子弹数为，则35,4B，所以315544

E，所以乙在四次射击中命中目标的子弹数的期望为415E，·······························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PMVC，垂足为M，在△VBC内过M作MNVC交V

B于N，连结PN，则直线PN即为直线l.···························2分理由如下：因为PMVC，MNVC，PMMNM，所以VC平面PMN，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PMN与平面重合，因为平面PMN平面VABPN

，所以直线PN即为直线l.····························4分（2）因为VAB△和ABC△均为等边三角形，所以,VAVBACBC，又因为VCVC,所以VACVBC≌△△，所以PVMNVM，又V

MVM，所以RtRtVPMVNM△△≌，所以VPVN，所以23VNVB.·······5分如图，设AB的中点为D，连结,VDCD，因为VAB△和ABC△均为等边三角形，所以,VAVBACBC，所以,ABVDABCD

，又因为VDCDD，所以AB平面VCD，因为AB平面ABC，所以平面ABC平面VCD.在△VCD中，作VOCD，垂足为O，因为平面ABC平面VCDCD，VO平面VCD，所以VO平面ABC，所以VCD是直线VC与平面ABC所成的角，所以π3VCD.·

·······················7分因为VAB△和ABC△均是边长为4的等边三角形，所以23VDDC,所以VCD△是等边三角形，所以3VO，3DOOC.以O为原点，分别以,OCOV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lNPAVBCM数学参考

答案及评分细则第5页（共16页）lQNPAVBCGNPODAVBC系Oxyz，则2,3,0,2,3,0,0,3,0,0,0,3ABCV，·····················

··················8分所以0,3,3,2,23,0,4,0,0,CVCAAB12453,,13333CPCVCA,28,0,033PNAB.过l及点C的平面为平面C

PN，设平面CPN的法向量为(,,)xyzn，则0,0,CPPNnn即4530,3380.3xyzx取(0,3,5)n，即平面CPN的一个法向量为(0,3,5)n.·········

·········································10分易知，平面ABC的一个法向量为(0,0,1)m，···········································11分所以557cos,1427mnmnmn，所以过l

及点C的平面与平面ABC所成的锐二面角的余弦值为5714．············12分解法二：（1）如图，在△VAB内过P作PNAB∥，交VB于N，则直线PN即为直线l.·················

····························································2分理由如下：取VC的中点Q，连结,AQBQ，因为VAB△和ABC△均为等边三角形，所以,VAA

CVBBC，所以,VCAQVCBQ，又因为AQBQQ，所以VC平面ABQ，又因为VC平面，所以平面∥平面ABQ，又因为平面平面VABl，平面ABQ平面VABAB，所以ABl∥，所以直

线PN即为直线l.························································4分（2）由（1）知，PNAB∥，因为23VPVA，所以23VNVB.·····························5分设AB的中点为D，连

结VD，交PN于G，连结CG，因为VAB△和ABC△均为等边三角形，所以,VAVBACBC，所以,ABVDABCD，又因为VDCDD，所以AB平面VCD，AB平面ABC，所以平面ABC平面V

CD.在△VCD中，作VOCD，垂足为O，因为平面ABC平面VCDCD，VO平面VCD，所以VO平面ABC，yzNPODAVBCx数学参考答案及评分细则第6页（共16页）所以VCD是直线VC与平面ABC

所成的角，所以π3VCD，······················7分因为VAB△和ABC△均是边长为4的等边三角形，所以23VDDC,π3VDC，因为ABPN∥，所以12333DGDV．由（1）知，过l

及点C的平面为平面CPN，因为AB平面CPN，PN平面CPN，所以AB∥平面CPN，·····················8分设平面CPN平面'ABCl，因为AB平面ABC，所以'ABl∥，因为AB平面VCD，CG平面VCD，CD平面VCD，所以,ABCGABCD，所以','

CGlCDl，又因为CG平面CPN，CD平面ABC，所以GCD为平面CPN与平面ABC所成的锐二面角的平面角，····················10分在△GCD中，由余弦定理得，2222cosCGDGDCDGDCGDC，2213CG,········

································································································11分所以22257cos214CGDCDGGCDCGDC，所以过l及点C的平面与平面A

BC所成的锐二面角的余弦值为5714．············12分20．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．解法一：（1）在△ABC中，由余弦定理得222cos2acbBac，·····························1分又因为6a，12cos2bBc

，所以2221222acbbcac，···································································2分整理得2236bcbc.··········

·································································3分在△ABC中，由余弦定理得22362cosbcbcA，所以2cosbcbcA，即1cos2A.···

····························································4分又因为(0,)A，所以3A．·········································

·······················5分（2）选③．·········································································

······················6分因为M为△ABC的内心，所以1=26BADCADBAC，由ABCABDACDSSS，·······························

··················7分得111sinsinsin232626bccADbAD，因为=33AD，所以3133()22bcbc，即3bcbc..................

......................8分数学参考答案及评分细则第7页（共16页）由（1）可得2236bcbc，即2()336bcbc，·······················

··············9分所以2()33609bcbc，即(9)(4)09bcbc，········································10分又因为0bc，

所以36bc，·································································11分所以113sin36932322ABCSbc．·························

·······················12分解法二：（1）因为6a，12cos2bBc，所以2cos2baBc，···········································································1分

在△ABC中，由正弦定理得sin2sincos2sinBABC，即sin2sincos2sin()BABAB，·························································2分即sin2sincos2sincos2cossinBABAB

AB，即sin2cossinBAB，··········································································3分因为

0,B，所以sin0B，故1cos2A.·····································································4分又因为(0,)A，所以3A．·····················

···········································5分（2）选②．·····························································

··································6分因为M为△ABC的垂心，所以222BMDMBDACBACB，又3MD，·················7分所以在△MBD中，tan3tanBDMDBM

DACB，同理可得3tanCDABC，·····································8分又因为6BDCD，所以3tan+3tan6ABCACB，即tan+tan23ABCACB，··········

····························9分又因为在△ABC中，tantan3ABCACBBAC，所以tantan31tantanABCACBABCACB，因此tantan

=3ABCACB．··································································10分故tantanABCACB，为方程22330xx两根，即tan=

tan=3ABCACB，因为0,ABCACB，，所以==3ABCACB，所以△ABC为等边三角形，···································11分所以21369322ABCS．········

······················································12分解法三：（1）同解法一.····················································

·····························5分（2）选②．·······························································································

6分因为M为△ABC的垂心，数学参考答案及评分细则第8页（共16页）所以AMBACB，26ABMBAC，·····································8分所以在△

ABM中，由正弦定理得sinsinAMABABMAMB，即sinsin6AMABACB．··········································9分又因为在△ABC中，由正弦定理得

sinsin3ABBCACB，··························10分所以sinsin63AMBC，因为6a，所以23AM.········································11分又因

为3MD，所以116(233)9322ABCSaAD.···················12分解法四：（1）同解法一.········································

·········································5分（2）选③．··················································································

·············6分因为M为△ABC的内心，所以1=26BADCADBAC.在△ABD中，由正弦定理得sinsin6BDADB，因为33AD，所以332sinBDB，同理可得33332sinsin()3CDCB.········

·····················································7分又因为6BDCD，所以333312sinsin()3BB，即4sinsin()3sinsin()33BBBB，即134sinsin()3(s

insincos)322BBBBB，·······································8分即4sinsin()3sin()36BBB，即4sin[()]sin[()]3s

in()66666BBB，·········································9分即31314sin()cos()sin()cos()3sin()262626266BBBBB

，数学参考答案及评分细则第9页（共16页）即22314sin()cos()3sin()46466BBB，即24sin()3sin()1066BB，即sin()14sin()1066BB

，····10分因为3A,所以203B，所以5(,)666B，所以sin()06B，故sin()16B，即62B，即3B，所以△ABC为等边三角形，··································11分所

以21369322ABCS.·································································12分解法五：（1）同解法一.··································

···············································5分（2）选③．·············································

··················································6分因为M为△ABC的内心，所以1=26BADCADBAC.又因为33AD，在△ABD中，由余

弦定理得2223=272339272BDcccc，同理可得22927CDbb．······························7分又因为1πsin26=1πsin26ABDACDABADSBDABCDSACACA

D，所以2222927927cccbbb，即()[3()]0bcbcbc，故bc或3bcbc.························································

·······················8分（i）当bc时，△ABC为等边三角形，所以21369322ABCS.（ii）当3bcbc时，由（1）知2236bcbc，即2()336bcbc，··········9分所以2()33609bcbc

，即(9)(4)09bcbc，········································10分因为0bc，所以36bc.···························

···········································11分又因为3A，所以113sin36932322ABCSbc．综上所述，93ABCS．········

······························································12分说明：设△ABC的外接圆半径为R，则在△ABC中，由正弦定理得数学参考答案及评分细则第10页（共16页）6243sinsi

n3BCRA，即23R，因为M为外心，所以23AM，与4AM矛盾，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，依题意得2222,2661,33,cababc解得

2,1,1,abc··········3分所以C的方程为2212xy．···································································4分（2）因为直

线AB与以OA为直径的圆的一个交点在圆O上，所以直线AB与圆O相切．······································································5分（i）当直线AB垂直于x轴时，不妨设6666,,,3333AB

，此时0OAOB，所以OAOB，故以AB为直径的圆过点O．····················6分（ii）当直线AB不垂直于x轴时，设直线AB的方程为ykxm，1122,,,AxyBxy．因为AB与圆O相切，所以O到直线AB的距离26

31mk，即223220mk．·············································································7分由22,1,2xykxmy

得222214220kxkmxm，·····································8分所以2121222422,2121kmmxxxxkk，·····································

····················9分1212OAOBxxyy1212xxkxmkxm2212121kxxkmxxm2222222412121mkmkkmmkk

xyABO数学参考答案及评分细则第11页（共16页）2222212242121kmkmkmmkk22232221mkk0，所以OAOB，故以AB为直径的圆过点O．综上，以AB为直径的圆过点O．···

························································12分解法二：（1）同解法一．·······················································

······················4分（2）因为直线AB与以OA为直径的圆的一个交点在圆O上，所以直线AB与圆O相切．······································

································5分设直线AB与圆O相切于点00,Mxy．（i）当00y时，直线AB垂直于x轴，不妨设6666,,,3333AB，此时0OAOB，所以OAOB，故以AB为直径的

圆过点O．····················6分（ii）当00y时，直线AB的方程为0000xyyxxy，因为220023xy，所以直线AB的方程为00023xyxyy．·····················

·······························7分设1122,,,AxyBxy，由000222,312xyxyyxy得22220000189248180xyxxxy，··························

······························8分所以20012122222000024818,189189xyxxxxxyxy，·················································9分因为220023

xy，所以2001212220024184,6969xxxxxxxx，222OAOBAB22222OMAMOMBMAMBM222OMAMBM423AMBM220020010042113xxxxxxyy

22012120004213xxxxxxxy数学参考答案及评分细则第12页（共16页）22220000220001842442136969xxxxyxx

22220000222000184244212369693xxxxxxx40220049422323694433xxx0.所以222OAOBA

B，即OAOB，故以AB为直径的圆过点O．综上，以AB为直径的圆过点O．···························································12分解法三：（1）同解法一．················

·······························································4分（2）因为直线AB与以OA为直径的圆的一个交点在圆O上，所以直线AB与圆O相切．·······················

···············································5分（i）当直线AB不垂直于x轴时，设直线AB的方程为ykxm，1122,,,AxyBxy．因为AB与圆O相切，所以O到直线AB的距离2631mk，即223220mk．······

·······································································6分由22,1,2xykxmy得222214220kxk

mxm，·····································7分所以2121222422,2121kmmxxxxkk，·······························

··························8分121222221myykxxmk，2222121212122221mkyykxmkxmkxxmkxxmk，2222212122222223220212121mmkmkxxyykkk

．设,Pxy是以AB为直径的圆N上的任意一点，由0PAPB，得12120xxxxyyyy，····························10分化简得22

121212120xyxxxyyyxxyy，故圆N的方程为22224202121kmmxyxykk，它过定点O．···················11分数学参考答案及评分细则第13页（共16页）（ii）当直线AB垂直于x轴时，不妨设6666,,,33

33AB，此时0OAOB，所以OAOB，故以AB为直径的圆过点O．综上，以AB为直径的圆过点O．···························································1

2分解法四：（1）同解法一．···············································································4分（2）因为直线AB与以OA为直径的圆的一个交点在圆O上

，所以直线AB与圆O相切．······································································5分（i）当直线AB不垂直于x轴时，设直线AB的方程为ykxm

，1122,,,AxyBxy．因为AB与圆O相切，所以O到直线AB的距离2631mk，即223220mk．············································

·································6分由22,1,2xykxmy得222214220kxkmxm，·····································7分所以2

121222422,2121kmmxxxxkk，·························································8分121222221myykxxmk

，以AB为直径的圆N的圆心为N1212,22xxyy，即222,2121kmmkk．半径2ABr221112kxx222222121222211168814222121kkmmkx

xxxkk2222222211688142222121kkmkkmkk，以AB为直径的圆的方程为22222222221422212121kmmkkmxykkk

，整理得22224202121kmmxyxykk，故以AB为直径的圆过定点O．······························································11分（ii）当直线AB垂直于x轴时，不妨设6666

,,,3333AB，此时0OAOB，所以OAOB，故以AB为直径的圆过点O．综上，以AB为直径的圆过点O．·····················································

······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14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，fx的定义域为0,，由1lnafxxaxR，得22'111faxaxxxx，······

················1分①当1a时，'0fx恒成立，所以fx在0,单调递增；·······················································································

···················2分②当1a时，令'0fx，得1xa，当0,1xa时，'0fx，所以fx在0,1a单调递减；当1,xa时，'0fx，所以fx在1,a单调递增；综上，当

1a时，fx在0,单调递增；当1a时，fx在0,1a单调递减，在1,a单调递增．···········4分（2）设hfxxgx，则

121''313eexxhfaxxaxxaxx，··········································································································5分①当

3x时，'0hx恒成立，所以hx在3,单调递增，又因为503a，所以22213ln31ln3103eeehaaaa，所以0hx，hx在3,不存在零点；·······················

·····················6分②当03x时，设1exxx，则1'e1xx，当01x时，'0x，所以x在0,1单调递减；当13x时，'0x，所以x在1,3单调递增；所以1

0x，即1exx，因为0x，所以111exx，···················7分又因为503a且03x，所以3<0ax，所以133exaxaxx，所以222'31311haxaxaxaxaxxxx

，·························8分当103a时，函数2131xaxaxa的对称轴为3102axa，所以x在0,3单调递增，所以010xa，所以'0hx，所以hx在0,3单调递增；·

··········································9分数学参考答案及评分细则第15页（共16页）当1533a时，22161341510109aaaaa，所以0x，所以'0hx

，所以hx在0,3单调递增；··················10分综上可知，当503a时，均有hx在0,3单调递增，又因为1110haa，所以hx在0,3恰有一个零点1,····

···········11分故当503a时，hx在0+，恰有一个零点1，因此不存在12,xx，且12xx，使得1,2iifxgxi．··························12分解法二：（1）同解法一；·····················

··························································4分（2）记Fxfxgx，则11ln2e1xaFxxaxx，则2

1221ee3113eexxxxaaxxaFxaxxxx，························5分记221ee3ee31exxxhaxaaxxxxax

，··················7分设eexxx，则'eexx，当01x时，'0x，所以x在0,1单调递减；当1x时，'0x，所以x在1,+单调递增；所以10x

，即eexx，··························································8分所以，222585851ee3ee3e5128333333xhxxxxxxxxxx

，因为212458160，所以251280xx，所以503h，·······9分又01e0xhx所以当50,03ax时，0ha，··········································

···········10分所以0Fx，Fx在0,上单调递增，又因为10F，所以Fx在0,上恰有1个零点1，··················································11分因

此不存在12,xx，且12xx，使得1,2iifxgxi．··························12分解法三：（1）同解法一；（2）设hfxxgx，则121''313eexxhfax

xaxxaxx，················································································································5分数学参考答案及评分细则第16页

（共16页）①当3x时，'0hx恒成立，所以hx在3,单调递增，又因为503a，所以22213ln31ln3103eeehaaaa，所以0hx，hx在3,不存在零点；··················

··························6分②当03x时，设1exxx，则1'e1xx，当01x时，'0x，所以x在0,1单调递减；当13x

时，'0x，所以x在1,3单调递增；所以10x，即1exx，因为0x，所以111exx，·················7分又因为503a且03x，所以

3<0ax，所以133exaxaxx，所以222'31131haxxxaxxaxxxx，····································8分设2311aaxxx，则010x

，·······································9分2226455551285531103333xxxxxx，所以0a，所以'0hx，综上可知，当503a时，均有hx在0,3单调

递增，····························10分又因为1110haa，所以hx在0,3恰有一个零点1，················11分故当503a时，hx在0+

，恰有一个零点1，因此不存在12,xx，且12xx，使得1,2iifxgxi．··························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