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数学参考答案及评分细则第1页(共20页)福建省2023届高中毕业班适应性练习卷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.D2.B3.C4.A5.D6.D7.A8.B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.BC10.ABD11.ACD12.BD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。13.2yx=−,2yx=−+(只需填其中的一个即可)14.215.1,2216.3a,221,233aa
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.本小题主要考查正弦定理、余弦定理、三角恒等变换、三角形面积及平面向量等基础知识,考查直观想象能力、逻辑推理能力、运算求解能力等,考查化归与转化思想、函数与方程思想、数形结合思想等,考查数学运算、
逻辑推理、直观想象等核心素养,体现基础性和综合性.满分10分.解法一:(1)因为π2sin6bcA=+,在ABC△中,由正弦定理得,sin2sinsin6BCA=+,...............
..................................................................................................1分又因为()()sinsinsi
nBACAC=−−=+,所以()sin2sinsin6ACCA+=+,..............................................................................
.................2分展开得31sincoscossin2sinsincos22ACACCAA+=+,...................................................
.....3分即sincos3sinsin0ACCA−=,因为sin0A,故cos3sinCC=,即3tan3C=.........................................................................4分又
因为()0,C,所以6C=...................................................................................................
........5分(2)设ABC△的外接圆的圆心为O,半径为R,数学参考答案及评分细则第2页(共20页)因为2BABDBA=,所以()0BABDBA−=,即0BAAD=,所以DABA⊥............
.........................................................................................................
..................6分故BD是O的直径,所以BCCD⊥.在ABC△中,1c=,122sinsin6cRBCA===,所以2BD=....................................................7分在ABD△中,223ADBDAB=−=
.设四边形ABCD的面积为S,BCx=,CDy=,则224xy+=,..............................................8分11312222ABDCBDSSSABADBCCDxy=+=
+=+△△.....................................................................9分2231312222xy++=+,当且仅当2xy==时,等号成立.所以四边形ABCD面积最大值为31
2+.........................................................................................10分解法二:(1)同解法一;......................
.............................................................................................5分(2)设ABC△的外接圆的圆心为O,半径为R,B
D在BA上的投影向量为BA,所以()2BABDBABABA==.又22BABDBABA==,所以1=,所以BD在BA上的投影向量为BA.所以DABA⊥....................
...................................................................................................................6分故BD是O的直径,所以BCCD⊥.在ABC△中,1c=,122
sinsin6cRBCA===,所以2BD=....................................................7分在ABD△中,223ADBDAB=−=.设四边形ABCD的面积为S,CBD=,π0,2,则2cosCB=,2sinCD
=,..................................................................................................
.......8分113sin2222ABDCBDSSSABADCBCD=+=+=+△△...................................................................9分当π22=时,S最大,所以四边形ABCD面积最大
值为312+.................................................10分解法三:(1)同解法一;..............................................................
.....................................................5分数学参考答案及评分细则第3页(共20页)(2)设ABC△的外接圆的圆心为O,半径为R,因为2BABDBA=,所以()0BABDBA−=,即0BAAD=,所以D
ABA⊥.....................................................................................................
..................................6分故BD是O的直径,所以BCCD⊥.在ABC△中,1c=,122sinsin6cRBCA===,所以2BD=.................................
...................7分在ABD△中,223ADBDAB=−=.设四边形ABCD的面积为S,点C到BD的距离为h,则113222ABDCBDSSSABADBDhh=+=+=+△△..........................
...............................................9分当1hR==时,S最大,所以四边形ABCD面积最大值为312+........................................
......10分解法四:(1)同解法一;...................................................................................................................5分(2
)设ABC△的外接圆的圆心为O,半径为R,在ABC△中,1c=,122sinsin6cRBCA===,................................................
.........................6分故ABC△外接圆O的半径R=1.即1OAOBAB===,所以π3AOB=.如图,以ABC△外接圆的圆心为原点,OB所在直线为x轴,建立平面直角坐标系xO
y,则13,22A,()1,0B.因为C,D为单位圆上的点,设()cos,sinC,()cos,sinD,其中()()0,2π0,2,.所以()13,cos1,sin22BABD=−=−,,...........................
........................................................7分代入2BABDBA=,即1BABD=,可得113cossin1222−++=,...................
....................8分即1sin62−=.由()0,2可知ππ11π666−−,,所以解得ππ=66−或π5π=66−,即π3=或π=.数学参考答案及评分细则第4页(共20页)当π3=时,A,
D重合,舍去;当π=时,BD是O的直径.设四边形ABCD的面积为S,则1313sinsin2222ABDCBDSSSBDBD=+=+=+△△,...................................................
....9分由()0,2π知sin1,所以当3π2=时,即C的坐标为()0,1−时,S最大,所以四边形ABCD面积最大值为312+.....................................
....................................................10分18.本小题主要考查指数与对数基本运算、递推数列、等差数列、等比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、综合性和创新性.满分12分.解法一:(1)由212122lognnnaaa−++=,得212122nn
aana−++=,.............................................................2分则2123222nnaana++++=,从而2121212321212
32222222nnnnnnnaaaaaaannaa−+++−+++++++==,.........................................3分又21214222162nnaannaa+++==,..................
............................................................................................4分所以21212+32124nnnnaaaa−+++
+=,.....................................................................................................
.5分即212+3212nnnaaa−++=,所以21na−是等差数列.............................................................................6分(2)设等差数列21na−的公差为d.
当1n=时,1322logaaa+=,即321log8a+=,所以32a=,所以311daa=−=,..................................................
...................................................7分所以数列21na−是首项为1,公差为1的等差数列,所以21nan−=;..................................
.....................................................................................................8分又()21211212222nnnnaanna−+++++===;....
...................................................................................................9
分()()9123456789135792468Saaaaaaaaaaaaaaaaaa=++++++++=++++++++()()3579123452222156806952023=++++++++=+=,又1110910
695227432023SSa=+=+=;数学参考答案及评分细则第5页(共20页)又0na,则1nnSS+,且9102023SS,.............................................................................
.....11分所以n的最小值为10.........................................................................................................................12
分解法二:(1)由20na,且2122216nannaa++=,则()2122222loglog16nannaa++=,........................................................................
....................................2分得2222221loglog4nnnaaa+++=,...........................................................
...............................................4分因为212122lognnnaaa−++=,2123222lognnnaaa++++=,所以()()2121212321=4nnnnnaaaaa−+++++++,.....
...................................................................................5分即21232+12nnnaaa−++=,
所以21na−是等差数列...............................................................................6分(2)设等差数列
21na−的公差为d.当1n=时,1322logaaa+=,即321log8a+=,所以32a=,所以311daa=−=,......................................................................
...............................7分所以数列21na−是首项为1,公差为1的等差数列,所以21nan−=;.........................................................................
...........................................................8分又212122lognnnaaa−++=,所以()21211212222nnnnaanna−+++++===
;................................................................................................9分当kN时,21232kkSaaaa=++++()()135
212462kkaaaaaaaa−=+++++++++()()357211232222kk+=+++++++++()()841123kkk−+=+,()()()2121228411124822323kkkkkkkkkkSSa+−−++−=−=+−=+,数学参考答案及评分细则第6页(共20页
)所以5925156248695202323SS−−==+=,()51025841562743202323SS−==+=,又0na,则1nnSS+,且9102023SS,...................
...............................................................11分所以n的最小值为10.......................................
..................................................................................12分解法三:(1)同解法一;..................................
.................................................................................6分(2)设等差数列21na−的公差为d.当1n=时,1322logaaa+=,即321log8a+=,所以32a
=,所以311daa=−=,...................................................................................................
..7分所以数列21na−是首项为1,公差为1的等差数列,所以21nan−=;........................................................................................................
............................8分又()21211212222nnnnaanna−+++++===;..............................................................................
......................9分当kN时,2112321kkSaaaa−−=++++()()1352124622kkaaaaaaaa−−=+++++++++()()3572112322
22kk−=+++++++++()()()118411114821423kkkkkk−−−++−=+=+−,所以()4925184156695202323SS−−==+=,25110910=695227432
023SSa+=++=.又0na,则1nnSS+,且9102023SS,..........................................................................
........11分所以n的最小值为10.............................................................................................................
............12分19.本小题主要考查一元线性回归模型、条件概率与全概率公式等基础知识,考查数学建模能力、运算求解能力、逻辑推理能力、直观想象能力等,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:(1)由散点图判断ln(2012)xycd−+=适宜作为该机场飞往A地航班放行准点率y关于年份数x数学参考答案及评分细则第7页(共20页)的经验回归方程类型..........................
.................................................................................................1分令ln(2012)tx=−,先建立y关于t的线性回归方程.由于1011022211
01226.8101.580.4ˆ427.7101.510iiiiitytyctt==−−===−−,......................................................................
..2分ˆˆ80.441.574.4dyct=−=−=,.......................................................................................................3分该机场飞往A
地航班放行准点率y关于t的线性回归方程为7.444ˆyt+=,因此y关于年份数x的回归方程为ln(2012)7ˆ444.xy−+=............................................................4分所以当2023x=时,该机场飞往A地航班放
行准点率y的预报值为ln(20232012)74.44ln1174.4ˆ42.4074.4844y−+=++==.所以2023年该机场飞往A地航班放行准点率y的预报值为84%................
.................................5分(2)设1A=“该航班飞往A地”,2A=“该航班飞往B地”,3A=“该航班飞往其他地区”,C=“该航班准点放行”,.....................
................................................................................................6分则()10.2PA=,()20.2PA=,()30.6P
A=,()10.84PCA=,()20.8PCA=,()30.75PCA=..........................................................................7分(i
)由全概率公式得,()()()()()()()112233PCPAPCAPAPCAPAPCA=++.......................................................
.........8分0.840.20.80.20.750.60.778=++=,所以该航班准点放行的概率为0.778.............................................................
...................................9分(ii)()()()()()()11110.20.840.778PAPCAPACPACPCPC===,()()()()()()22220.20.80.778PAPC
APACPACPCPC===,()()()()()()33330.60.750.778PAPCAPACPACPCPC===,..................................................
............11分因为0.60.750.20.840.20.8,所以可判断该航班飞往其他地区的可能性最大......................................................................12分20.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,空间几何体的体积、平面与平面的夹角等基础知识;考查直观想象能力,逻辑推理能力,运算求解能力等;考查化归与转化思想,数数学参考答案及评分细则第8页(共20页)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1,取AB中点O,连接PO,CO.因为2PAPB==,2AB=,所以POAB⊥,1PO=,1BO=.又因为
ABCD是菱形,60ABC=,所以COAB⊥,3CO=.因为2PC=,所以222PCPOCO=+,所以POCO⊥.又因为AB平面ABCD,CO平面ABCD,ABCOO=,所以PO⊥平面ABCD.........
.........................................................................................................
......2分因为ADBC∥,BCPBC平面,ADPBC平面,所以ADPBC∥平面,所以1133143343DPBCAPBCPABCABCVVVPOS−−−=====△................3分因为3162MPBCDPBCVV−−==,.....
........................................................................................................4分所以点
M到平面PBC的距离是点D到平面PBC的距离的12,所以PMMD=...............................................................................................
......................................5分(图1)(图2)(2)由(1)知,BOCO⊥,POBO⊥,POCO⊥,如图2,以O为坐标原点,OC,OB,OP的方向分别为x轴,y轴,z轴正方向建立空间直角坐标系,.....................
..................................................................................................................................6分则()0,1,0A−,()0
,1,0B,()3,0,0C,()3,2,0D−,()0,0,1P,所以31,1,22M−.则()3,1,0AC=,()3,1,0BC=−,()3,3,0BD=−,()0,1,1AP=,31,1,22CM=−−.因为QAP,设()0,,AQAP
==,则()3,1,CQAQAC=−=−−,数学参考答案及评分细则第9页(共20页)因为BD∥,Q,C,M,故存在实数,ab,使得CQaCMbBD=+,...............7分所以333,231,,2ababa−+=−−−=−=解得4
,31,32.3ab==−=所以123,,33CQ=−−............................................................................
...........................................8分设平面BCQ的法向量为1(,,)xyz=n,则110,0,CQBC==nn即230,3330yzxxy−−+=−=.取1x=,得到平面
BCQ的一个法向量()11,3,23=n.............................................................10分设平面BCQ与平面ABCD夹角是,又因为()20,0,1=n是
平面ABCD的一个法向量,.........................................................................11分则1212123coscos,2===nnnnnn.所以平面BC
Q与平面ABCD夹角的余弦值是32......................................................................12分解法二:(1)如图3,取AB中点O,连接PO,CO,因为2PAPB==,2AB=,所
以POAB⊥,1PO=,1BO=,又因为ABCD是菱形,60ABC=,所以COAB⊥,3CO=.因为2PC=,所以222PCPOCO=+,所以POCO⊥.因为AB平面PAB,PO平面PAB,ABPOO=,所以CO
⊥平面PAB...........................................................................................................................2分1113322
3323APBCCABPABPVVCOS−−====△................................................................3分过M作
MNAD∥交AP于点N,ADBC∥,所以MNBC∥,又BCPBC平面,MNPBC平面,(图3)数学参考答案及评分细则第10页(共20页)所以MNPBC∥平面,所以1336MPBCNPBCCNBPNBPVVVCOS−−−====△,
因为13CABPABPVCOS−=△,13CNBPNBPVCOS−=△所以2ABPNBPSS=△△,............................................................................................
..............................4分所以N是PA的中点,所以M是PD的中点,所以PMMD=............................................
..........5分(2)在平面ABCD内,过C作EFBD∥交AD延长线于点E,交AB延长线于点F,因为ABCD是菱形,所以ADDE=.如图4,在平面PAD内,作PPAE∥交EM的延长线于点P,设EP交AP于点Q.所以,四边形EDPP是平行四边形,,PPDEPPDE=∥,所以QPP
QAE△∽△,所以12PQPPAQAE==,所以点Q是线段PA上靠近P的三等分点.........................................................................................7分如图5,在平面PAB内,作QTP
O∥,交AB于T,因为PO⊥平面ABCD,所以QT⊥平面ABCD,所以QT⊥BC,因为1PO=,2233QTPO==,....................................................................................
.....................8分在平面ABCD内,作TNBC⊥,交BC于点N,连接QN,过A作AKTN∥交BC于K,在ABK△中,2AB=,60ABK=,所以332AKAB==,(图5)所以22333TNAK==,............................
........................................................................................9分因为QT⊥BC,TN
BC⊥,QTTNT=,所以BC⊥平面QTN,因为QN平面QTN,所以BCQN⊥.所以QNT是二面角ABCQ−−的平面角..................................................
................................11分数学参考答案及评分细则第11页(共20页)在RtQTN△中,3tan3QTQNTNT==,所以3cos2QNT=.所以平面BCQ与平面ABCD夹角的余弦值是32.................
.....................................................12分解法三:(1)同解法一;.....................................
............................................................................................5分(2)由(1)知,BOCO⊥,POBO⊥,POCO⊥,如图2,以O为坐标原点,OC,OB,OP的方
向分别为x轴,y轴,z轴正方向建立空间直角坐标系,.............................................................................
..........................................................................6分则(0,1,0)A−,(0,1,0)B,(3,0,0)C,(3,2,0)D−,(0,0,
1)P,所以31(,1,)22M−.则(3,1,0)AC=,(3,1,0)BC=−,(3,3,0)BD=−,(0,1,1)AP=,31(,1,)22CM=−−.设平面的法向量为(,,)xyz=n,则0,0BDCM==,nn即330,31022x
yxyz−=−−+=.取1y=,得到平面的一个法向量()3,1,5=n..........................................................................7分因为
QAP,设()0,,AQAP==,则()3,1,CQAQAC=−=−−,因为3150CQ=−+−+=n,所以23=,所以123,,33CQ=−−...............................
........8分设平面BCQ的法向量为1111(,,)xyz=n,则110,0CQBC==,nn即11111230,3330yzxxy−−+=−=.取11x=,得到平面BCQ的一个法向量()
11,3,23=n............................................................10分设平面BCQ与平面ABCD夹角是,又因为()20,0,1=n是平面ABCD的一个法向量,....................
.....................................................11分则1212123coscos,2===nnnnnn.所以平面BCQ与平面ABCD夹角的余弦值是32.................................
.....................................12分21.本小题主要考查圆、椭圆的标准方程及简单几何性质,直线与椭圆的位置关系等基础知识;考查运算数学参考答案及评分细则第12页(共20页)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)由题意得,()11,0A−,()21,0A.因为D为BC中点,所以1ADBC⊥,即12ADAC⊥,.
..................................................................1分又1PEAD∥,所以2PEAC⊥,又E为2AC的中点,所以2PAPC=,
所以1211124PAPAPAPCACAA+=+==,所以点P的轨迹是以12,AA为焦点的椭圆(左、右顶点除外)................................................2分设:22221xyab+=(xa),其中0ab,222abc−
=.则24a=,2a=,1c=,223bac=−=..............................................................................3分故:22143xy+=(2x)...........
...............................................................................................4分(2)结论③正确.下证:12QCC△的面积是定值........
...............................................................5分由题意得,()12,0B−,()22,0B,()10,1C−,()20,1C,且直线2l的斜率不为0,可设直线2l:1xm
y=−,()()1122,,,MxyNxy,且12x,22x.由221,431,xyxmy+==−得()2234690mymy+−−=,........................
.........................................................6分所以12122269,3434myyyymm−+==++,...................
.........................................................................7分所以()121223myyyy=−+.直线1BM的方程为:()1122yyxx=++,直
线2BN的方程为:()2222yyxx=−−,....................8分由()()11222,22,2yyxxyyxx=++=−−数学参考答案及评分细则第13页(共20页)得()()21122222yxxxyx++=−−.............
.......................................................................................................
........9分()()211213ymyymy+=−1221213myyymyyy+=−()()12212132332yyyyyy−++=−+−121231229322yyyy−−=−−13=,解得x4=−...................................
...................................................................................................11分故点Q在直线4x=−,所以Q到12CC的距离4d=,因此12QCC△的面积是定值,为121
124422CCd==.........................................................12分解法二:(1)同解法一...........................
.........................................................................................4分(2)结论③正确.下证:12QCC△的面积是定值....
...................................................................5分由题意得,()12,0B−,()22,0B,()10,1C−,()20,1C,且直线2l的斜率不为0,可设直线2l:1xmy=−,()()1
122,,,MxyNxy,且12x,22x.由221,431,xyxmy+==−得()2234690mymy+−−=,...............................................
..................................6分所以12122269,3434myyyymm−+==++,.............................................
...............................................7分所以()121223myyyy=−+.直线1BM的方程为:()1122yyxx=++,直线2BN的方程为:()2222yyxx=−−,....................
8分由()()11222,22,2yyxxyyxx=++=−−得()()()()2112211222222yxyxxyxyx++−=+−−.........................................................
...............................................9分()()()()2112211213213ymyymyymyymy++−=+−−12212123
23myyyyyy+−=+数学参考答案及评分细则第14页(共20页)()()121221212323243myyyyyyyy++−+==−+..............................................................
..............11分故点Q在直线4x=−,所以Q到12CC的距离4d=,因此12QCC△的面积是定值,为121124422CCd==..............................
...........................12分解法三:(1)同解法一..............................................................................
......................................4分(2)结论③正确.下证:12QCC△的面积是定值......................................................................
.5分由题意得,()12,0B−,()22,0B,()10,1C−,()20,1C,直线2l的斜率不为0.(i)当直线2l垂直于x轴时,2l:1x=−,由221,431xyx+==−得1,32xy=−=−或1,3.2xy=−=不妨设331,,1,
22MN−−−,则直线1BM的方程为:()322yx=+,直线2BN的方程为:()122yx=−,由()()32,2122yxyx=+=−得4,3,xy=−=−所以(4,3
)Q−−,故Q到12CC的距离4d=,此时△12QCC的面积是121124422CCd==.............................6分(ii)当直线2l不垂直于x轴时,设直线l:()1ykx=+,
()()1122,,,MxyNxy,且12x,22x.由()221,431,xyykx+==+得()()22224384120kxkxk+++−=,...................
..............................................7分所以221212228412,4343kkxxxxkk−−+==++...........................................
..................................................8分直线1MB的方程为:()1122yyxx=++,直线2MB的方程为:()2222yyxx=−−,.............
........9分由()()11222,22,2yyxxyyxx=++=−−数学参考答案及评分细则第15页(共20页)得()()()()2112211222222yxyxxyxyx++
−=+−−.............................................................................................
.........10分()()()()()()()()21122112121221212kxxkxxkxxkxx++++−=++−+−12121242634xxxxxx−+=++.下证:121212426434xxxxxx−+=−++.即证()1212124264
34xxxxxx−+=−++,即证()121241016xxxx=−+−,即证22224128410164343kkkk−−=−−++,即证()()()22244121081643kkk−=−−−+,上式显然成立,....
..........................................................................................................................
...11分故点Q在直线4x=−,所以Q到12CC的距离4d=,此时12QCC△的面积是定值,为121124422CCd==.由(i)(ii)可知,12QCC△的面积为定值..............
....................................................................12分解法四:(1)同解法一.........................
...........................................................................................4分(2)结论③正确.下证:12QCC△的面积是定值.................
......................................................5分由题意得,()12,0B−,()22,0B,()10,1C−,()20,1C,且直线2l的斜率不为0,可
设直线2l:1xmy=−,()()1122,,,MxyNxy,且12x,22x.由221,431,xyxmy+==−得()2234690mymy+−−=,.....................................................
............................6分所以12122269,3434myyyymm−+==++.............................................................................
................7分直线1BM的方程为:()1122yyxx=++,直线2BN的方程为:()2222yyxx=−−,....................8分数学参考答案及评分细则第16页(共20页)因为222214
3xy+=,所以222yx−22234xy+=−,故直线2BN的方程为:()222324xyxy+=−−.由()()11222,2232,4yyxxxyxy=+++=−−得()()1212422322yyxxxx−=−+++........
.......................................................................................................9分()()121
24311yymxmy=−++()1221212431yymyymyy=−+++()2224939634mmm−=−−+++3=,解得x4=−................................................
......................................................................................11分故点Q在直线4x=−,所以Q到12CC的距离4d=,因此12QCC△的面积
是定值,为121124422CCd==.........................................................12分22.本小题主要考查导数及其应用、函数的单调性、不等式等基础知识
,考查逻辑推理能力、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)()(
)1exfxxa=++,.................................................................................................1分故1xa−−时,()0fx;1xa−−时,()0fx..
..................................................................2分当10a−−,即1a−时,()fx在()0,1a−−单调递减,在
()1,a−−+单调递增;当10a−−,即1a−≥时,()fx在()0,+单调递增.综上,当1a−时,()fx在()0,1a−−单调递减,在()1,a−−+单调递增;当1a−≥时,()fx在()0,+单调递增..
...........................................................................................4分(2)不存在01
,,axx,且01xx,使得曲线()yfx=在0xx=和1xx=处有相同的切线.............5分证明如下:假设存在满足条件的01,,axx,因为()fx在()()00,xfx处的切线方程为()()()000yfxfxxx−=−数学参考答案及评分细则第1
7页(共20页)即()()0020001eexxyxaxaxax=+++−−,..............................................................
.........................6分同理()fx在()()11,xfx处的切线方程为()()1121111eexxyxaxaxax=+++−−,且它们重合,所以()()()()0101012200111e1e,ee,xxxxxaxaaxaxaxax++=++−−=−−
...................................................................7分整理得()()()()2201110011xaaxaxxaaxax++−−=++−−,即()()201
01120xxaxxaa+++++=,()()()20101111xxaxxa+++++=,所以()()01111xaxa++++=,...................................
.......................................................................8分由0101(1)e(1)exxxaxa++=++两边同乘以1ea+,得011101(1)e(1)exaxaxaxa++++++=++,.........
.....................................................................................9分令001txa=++,111txa=++,则
010101ee,1,tttttt==且01tt,由011tt=得011tt=,代入0101eetttt=得11121eettt=,两边取对数得11112lnttt=+...........................10分令1()2lngtttt=+−,当0t时,
1()2lngtttt=+−,()222121()10tgtttt+=++=≥,所以()gt在(0,)+上单调递增,又()10g=,所以11t=,从而01t=,与01tt矛盾;.........11分当0t时,()1()2lngtttt=−+−,()222121()10tgtttt+
=++=≥,所以()gt在(,0)−上单调递增,又()10g−=,所以11t=−,从而01t=−,与01tt矛盾;综上,不存在01,tt,使得010101ee,1,tttttt==且01tt.故不存在01,,axx且01xx,使得曲线()yfx=在0xx=和
1xx=处有相同的切线.....................12分解法二:(1)同解法一;....................................................................................
................................4分数学参考答案及评分细则第18页(共20页)(2)不存在01,,axx,且01xx,使得曲线()yfx=在0xx=和1xx=处有相同的切线.............5分证明如下:假设存在满足条件的01,,axx,因为
()fx在()()00,xfx处的切线方程为()()()000yfxfxxx−=−即()()()00000001ee1exxxyxaxxaxxa=++++−++,..............................
........................................6分同理()fx在()()11,xfx处的切线方程为()()()11111111ee1exxxyxaxxaxxa=++++−++,且它们重合,所以()
()()()()()010011010001111e1e,e1ee1e,xxxxxxxaxaxaxxaxaxxa++=+++−++=+−++..........................7分整理得()()
011110001(1)1(1)xaxaxxaxaxaxxa+++−++=+++−++,令001txa=++,111txa=++,可得011tt=.......................................
.............................................8分由0101(1)e(1)exxxaxa++=++两边同乘以1ea+,得011101(1)e(1)exaxaxaxa++++++=++,则010101ee,1
,tttttt==且01tt,....................................................9分令()ethtt=,则()()01htht=,且01tt.由
(1)知,当1t−时,()ht单调递增,当1t−时,()ht单调递减,又当0t时,()0ht,当0t时,()0ht,所以若01,tt存在,不妨设1010tt−,设10tmt=,1m,又011tt=,所以
201tm=,则01tm=−,由0110eetttt=,得0000eemttmtt=即00eemttm=,则00lnmmtt+=,所以0ln1mtm=−,所以1ln1mmm−=−,即1ln0mmm+−=,.....................
...........................................................11分令1()2lngxxxx=−+,1x≥,则22221(1)()10xgxxxx−=−−=−,所以()gx在(1,)+上单调递减,所以当1x时,()(1)0gxg
=,数学参考答案及评分细则第19页(共20页)即12lnxxx−,取xm=,即1ln0mmm+−,所以1ln0mmm+−=在1m时无解,综上,不存在01,tt,使得010101ee,1,tttt
tt==且01tt.故不存在01,,axx且01xx,使得曲线()yfx=在0xx=和1xx=处有相同的切线.....................12分解法三:(1)同解法一;..........................
..........................................................................................4分(2)不存在01,,axx,且01xx,
使得曲线()yfx=在0xx=和1xx=处有相同的切线.............5分证明如下:假设存在满足条件的01,,axx,因为()fx在()()00,xfx处的切线方程为()()()000yfxfxxx−=−
即()()0020001eexxyxaxaxax=+++−−,.......................................................................................6分同理()fx在()()11,xfx处的切线方程为(
)()1121111eexxyxaxaxax=+++−−,且它们重合,所以()()()()0101012200111e1e,ee,xxxxxaxaaxaxaxax++=++−−=−−........
...........................................................7分整理得()()()()2201110011xaaxaxxaaxax++−−=++−−,即()()20101
120xxaxxaa+++++=,()()()20101111xxaxxa+++++=,所以()()01111xaxa++++=,....................................................
......................................................8分由0101(1)e(1)exxxaxa++=++两边同乘以1ea+,得011101(1)e(1)exaxaxaxa
++++++=++,..............................................................................................9分令001txa=++,111txa=++,则010101ee
,1,tttttt==且01tt,令()ethtt=,则()()01htht=,且01tt.由(1)知,当1t−时,()ht单调递增,当1t−时,()ht单调递减,又当0t时,()0ht,当0
t时,()0ht,数学参考答案及评分细则第20页(共20页)所以若01,tt存在,不妨设1010tt−,则0101eetttt−=−,()()0011lnlntttt−+=−+,所以()()()()01011lnlntttt−−−=
−−−....................................................................................................................11分以下证明
()()()()()()010101lnlntttttt−−−−−−−−.令1()2lngxxxx=−+,1x≥,则22221(1)()10xgxxxx−=−−=−,所以()gx在(1,)+上单调递减,所以当1x时,()(1)0gxg=,因为101tt−−,所以100tg
t−−,011001ln0tttttt−−−−+−−−,整理得()()()()()()010101lnlntttttt−−−−−−−−.因为()()()()01011lnlntttt−−−=−−−,所以()()011tt−−
,与011tt=矛盾;所以不存在01,tt,使得010101ee,1,tttttt==且01tt.故不存在01,,axx且01xx,使得曲线()yfx=在0xx=和1xx=处有相同的切线...............
......12分