福建省福州市2024-2025学年高三上学期8月第一次质量检测试题 数学 PDF版含答案

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{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}{#{QQABBQwEggA

AAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第1页(共14页)2024-2025学年福州市高三年级

第一次质量检测数学答案一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的。题目12345678答案DCCBADBC二、多项选择题:本题共3小题,每小题6分,共18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6分,部分选对的得部分分,有选错的得0分。题目91011答案ADABDBCD三、填空题:本大题共3小题,每小题5分,共15分。题目121314答案35()2425,四、解答题:本题共5小题

,共77分。解答应写出文字说明、证明过程或演算步骤。15.(13分)已知数列na满足12a=,132nnaa+=+.(1)证明:数列1na+是等比数列;(2)求na的前n项和nS.【解法一

】(1)证明:因为132nnaa+=+,且12a=,所以10na+,···················································································1分所以1132

111nnnnaaaa++++=++········································································3分3(1)31nnaa+==+,·········

···························································5分又113a+=,所以数列1na+是以3为首项,3为公比的等比数列.·······························6分(2)由

(1)得13nna+=,所以31nna=−,·············································8分所以()()()2313131nnS=−+−++−{#{QQABBQwEg

gAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第2页(共14页)()233333nn=++++−··································

··························10分13313nn+−=−−············································································12分133

.2nn+−=−············································································13分【解法二】(1)证明:因为132nnaa+=+,所以()113331nnnaaa++=+=+,··········

···················································2分因为12a=,所以1130a+=,所以10na+,···························

·············4分所以1131nnaa++=+,所以数列1na+是以3为首项,3为公比的等比数列.·······························6分(2)略,同解法一.16.(15分)已知A

BC△的内角,,ABC的对边分别为,,abc,且2cos3cos3cosaCbCcB=+.(1)求角C;(2)若4a=,3b=,D为AB中点,求CD的长.【解法一】(1)因为2cos3cos3cos

aCbCcB=+,由正弦定理,得2sincos3sincos3cossinACBCBC=+··············································2分()3sinBC=+·························

·········································4分()3sinπA=−3sinA=,······································································6分因为0πA,则sin0A

,所以3cos2C=,··········································7分由于0πC,则π6C=;······························

······································8分(2)因为D为AB中点,故()12CDCACB=+,······································10分

{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第3页(共14页)所以()2214CDCACB=+························································

··············11分22111πcos4426CACBCACB=++············································13分1113316344422=++314=,·············

····································································14分所以CD的长为312.·································

·····································15分【解法二】(1)因为2cos3cos3cosaCbCcB=+,由余弦定理,得2222222cos3322abcacbaCbcabac+−+−=

+···························2分=3a,····························································4分所以3cos2C=,···········

·····································································6分由于0πC,则π6C=;·····································

·······························8分(2)由(1)知,π6ACB=,在ABC△中,由余弦定理,得2222coscababACB=+−··············································

·····················10分2234(3)2432=+−7=,···········································································

················11分故7c=,·······················································································12分因为D为AB中点,所以coscos0ADCBD

C+=,故222222022ADCDACBDCDBCADCDBDCD+−+−+=,··········································13分所以()2222227734220772222CDCD

CDCD+−+−+=,{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第4页(共14页)解得2314=CD,

················································································14分故CD的长为312.·············

······························································15分【解法三】(1)略,同解法一或解法二;(2)由(1)知,π6ACB=,在ABC△中,由余弦定理,得2222coscab

abACB=+−···································································10分()223432432=+−7=,·········

··················································································11分故7c=,········································

···············································12分所以222cos2bcaAbc+−=()()222374237+−=37=−,····································

·········································13分在ACD△中,由余弦定理,得2222cosCDACADACADA=+−()22773323227=+−−314=,···········

············································································14分故CD的长为312.··············

·····························································15分17.(15分)如图,在四棱锥SABCD−中,BC⊥平面SAB,∥ADBC,1SABC==,2SB=,o45SBA=.(1)求

证:SA⊥平面ABCD;{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第5页(共14页)(2)若12AD=,求平面SCD与平面SAB的夹角的余弦值.【

解法一】(1)在△SAB中,因为1SA=,o45SBA=,2SB=,由正弦定理,得sinsinSASBSBASAB=,····························································

·············1分所以12sin45sinSAB=,······································································2分所以sin1SAB=,因为018

0SAB,所以90SAB=,所以SAAB⊥.···················································································4分因为BC⊥平面SAB,SA平面SAB,所以BCS

A⊥,···················································································5分又BCABB=,所以SA⊥平面ABCD;··············

···························································6分(2)解:由(1)知SA⊥平面ABCD,又,ABAD平面ABCD,所以SAAB⊥,SAA

D⊥,因为BC⊥平面SAB,···········································································7分AB平面SAB,所以BCAB⊥,因为∥ADBC,所以ADAB⊥,所以,,SAAD

AB两两垂直.···································································8分以点A为原点,分别以AD,AB,AS所在直线为x轴,y轴,z轴建立如图所示的空间直角坐标系,·······················

·························································9分则1(1,1,0),,0,0,2(0,0,1),DSC所以()1,1,1SC=−,1,0,12SD=−,设平面SCD的法向量为1(,,)xy

z=n,则11,,SCSD⊥⊥nn即110,10,2SCxyzSDxz=+−==−=nn取2x=,则()12,1,1=−n,··············································

····················11分{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第6页(共14页)显然平面SAB的一个法向量()21,0,0=n,·········

·······································12分所以cos=121212nnn,nnn·········································

····························13分()2222211=+−+63=,·········································································14分所以平面SCD与平面SAB

的夹角的余弦值为63.···································15分【解法二】(1)证明:设ABx=,在△SAB中,因为1SA=,o45SBA=,2SB=,由余弦定理,得2222cosSASBABSBSABBA=+−,···················

···································1分所以21222co5s4xx=+−,································································2分所以2222212xx+−

=,所以2210xx−+=,解得1x=.···················································································

·····3分所以2222SAABSB+==,所以SAAB⊥.················································4分因为BC⊥平面SAB,SA平面SAB,所以BCSA⊥,····································

···············································5分又BCABB=,所以SA⊥平面ABCD;···································

······································6分(2)略,同解法一.【解法三】(1)设ABx=,在△SAB中,因为1SA=,o45SBA=,2SB=,由余弦定理,得222

2cosSASBABSBSABBA=+−,······················································1分所以21222co5s4xx=+−,············································

····················2分{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第7页(共14页)所以2222212xx+−=,

所以2210xx−+=,解得1x=.························································································3分

所以2222SAABSB+==,所以SAAB⊥.················································4分因为BC⊥平面SAB,BC平面ABCD,所以平面ABCD⊥平面SAB;·································

································5分又平面ABCD平面SABAB=,SAAB⊥,SA平面SAB,所以SA⊥平面ABCD;····················································

·····················6分(2)由(1)知SA⊥平面ABCD,过B作BMSA,则BM⊥平面ABCD,又,ABBC平面ABCD,所以BMAB⊥,BMBC⊥,因为BC⊥平面SAB,······

·····································································7分又AB平面SAB,所以BCAB⊥,所以,,BMBABC两两垂直.·········································

·························8分以点B为原点,分别以BA,BC,BM所在直线为x轴,y轴,z轴建立如图所示的空间直角坐标系,······································································

··········9分则1(0,1,0),1,(,0,21,0,1),CSD所以()1,1,1SC=−−,11,,02CD=−,设平面SCD的法向量为1(,,)xyz=n,则11,,SCCD⊥⊥nn即110,1

0,2SCxyzCDxy=−+−==−=nn取2y=,则()11,2,1=n,····································································11分显然平面SAB的一个法向量()20,1,0=n,

···············································12分所以cos=121212nnn,nnn······························································

·······13分2222121=++{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第8页(共14页)63=,····················································

·····················14分所以平面SCD与平面SAB的夹角的余弦值为63.···································15分【解法四】(1)略,同解法一或解法二或

解法三;(2)延长CD、BA交于点M,连接SM,则平面SCD平面SABSM=,·······························································7分在SB

M△中,2SB=,45SBA=,2BM=,由余弦定理,得2222cosSMSBMBSBSMBBM=+−,所以()22222222222SM=+−=,······································

········9分所以222SMSBBM+=,所以SMSB⊥,················································································10分因

为BC⊥平面SAB,SM平面SAB,所以SMBC⊥,又SMSB⊥,SBBCB=,所以SM⊥平面SBC,········································································11分又SC平面SBC,

所以SMSC⊥,所以BSC为平面SCD与平面SAB的夹角,············································12分因为BC⊥平面SAB,SB平面SAB,所以BCSB⊥,因为2,1SBBC==,得3SC=,···

····················································13分所以26cos33SBBSCSC===,所以平面SCD与平面SAB的夹角的余弦值为63.····

·······························15分18.(17分)已知椭圆W:()222210xyabab+=的离心率为12,且过点()2,0.(1)求W的方程;(2)直线()100−+=xmym交W于A,B两点.ABCDMS{#{QQABBQwEggA

AAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第9页(共14页)(i)点A关于原点的对称点为C,直线BC的斜率为k,证明:km为定值;(ii)若W上存在点P使得AP,PB在AB上的投影向量相等,且△PAB的重

心在y轴上,求直线AB的方程.【解法一】(1)依题意,得222122caabac===−,·················································

··3分解得23ab==,····················································································4分所以W的方程为221

43xy+=;································································5分(2)依题意可设点11(,)Axy,22(,)Bxy,且12xx,(ⅰ)证明:因为点A关于原点的

对称点为C,所以11(,)Cxy−−,因为点A,B在W上,所以22112222143143xyxy+=+=,················································6分所以2222212143xxyy−−=−,即222

1222134yyxx−=−−,···············································8分因为直线AB:()100−+=xmym的斜率为1m,直线BC的斜率为k,·········9分所以km21212

1212221222134−+==−=−−−+yyyyxxxxyyxx,即km为定值34−;··························································

···················11分(ⅱ)设弦AB的中点D的坐标为(,)DDxy,点P的坐标为(,)PPxy,PAB△的重心G的坐标为(,)GGxy,由2211043−+=+=xmyxy,得()2234690mymy+−−=,··········

······11分所以()()22236+363+414410mmm==+,且122634myym+=+,···············12分{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试

题第10页(共14页)因为PAB△的重心G在y轴上,所以1203Pxxx++=,······························13分所以()()()121212228112234634Pxxxmymymyymmmm=−

+=−−+−=−++=−+=++,所以1224234Dxxxm+==−+,1223234Dyymym+==+,······························14分因为AP,PB在AB上的投影向

量相等,所以PAPB=,且PDAB⊥,所以直线PD的方程为()DDyymxx−=−−,所以()2222384934343434PDPDmmyymxxmmmmm=−−=−+=−++++,·······15分所以点2289,3434mPmm−++,又点P在W

上,所以2222894334143++−+=mmm,································16分即()22310mm−=,又因为0m,所以33m=,所以直线AB的方程为3330

xy+=.·······17分【解法二】(1)略,同解法一;(2)依题意可设点11(,)Axy,22(,)Bxy,且12xx,(ⅰ)证明:因为点A关于原点的对称点为C,所以11(,)Cxy−−,由2211043−+=+=xmy

xy,得()2234690mymy+−−=,·······································6分所以()()22236+363+414410mmm==+,且122634myym+=+,·················7分所以122634m

yym+=+,所以()121212228112236344xxmymymyymmmm+=−+−=+−=−=−++,·······8分{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题

第11页(共14页)因为直线BC的斜率为k,所以()()212121222163348344yyyykxxxxmmmm−−+===−−+=−−++,······································10分所以km为定值34−;···············

···························································11分(ⅱ)设弦AB的中点D的坐标为(,)DDxy,点P的坐标为(,)PPxy,PAB△的重心G的坐标为(

,)GGxy,由(ⅰ)知,122634myym+=+,·······························································12分因为PAB△的重心G在y轴上,所以1203

Pxxx++=,······························13分所以()()()121212228112234634Pxxxmymymyymmmm=−+=−−+−=−++=−+=++,所以1224234Dxxxm+

==−+,1223234Dyymym+==+,······························14分因为AP,PB在AB上的投影向量相等,所以PAPB=,且PDAB⊥,所以直线PD的方程为()DDyymxx−=−−,所以()2222384934343434PDPDmmyymxx

mmmmm=−−=−+=−++++,·······15分所以点2289,3434mPmm−++,又点P在W上,所以2222894334143++−+=mmm,······················

··········16分即()22310mm−=,又因为0m,所以33m=,所以直线AB的方程为3330xy+=.·······17分19.(17分)阅读以下材料:①设()fx为函数()fx的导函数.若()f

x在区间D上单调递增,则称()fx为区间D{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第12页(共14页)上的凹函数;若()fx在区间D上单调递减,则称()

fx为区间D上的凸函数.②平面直角坐标系中的点P称为函数()fx的“k切点”,当且仅当过点P恰好能作曲线()yfx=的k条切线,其中kN.(1)已知函数()()4323213fxaxxaxx=+−+−+.(i)当0a时,讨论()fx的凹凸性;

(ii)当0a=时,点P在y轴右侧且为()fx的“3切点”,求点P的集合;(2)已知函数()exgxx=,点Q在y轴左侧且为()gx的“3切点”,写出点Q的集合(不需要写出求解过程).【解析】(1)因为()()4323213

fxaxxaxx=+−+−+,所以()()32436211fxaxxax=+−+−,··················································1分令()()32436211hxaxxax=+−+−,所以()()()()212662162211hxaxx

aaxax=+−+=++−.·····························2分(i)当0a=时,()()61hxx=−,令()0hx,解得1x;令()0hx,解得1x;故()fx为区间)1,+上的凹函

数,为区间(,1−上的凸函数;·····················3分当104a−时,令()0hx,解得2112+−axa,令()0hx,解得1x或212+−axa,故()fx为区间

211,2aa+−上的凹函数,为区间(,1−和21,2aa+−+上的凸函数;····························································································

··················4分当14a=−时,()()2310hxx=−−,故()fx为区间(),−+上的凸函数;······5分当14a−时,令()0hx,解得2112+−axa,{#{QQABBQwEggAAAI

AAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第13页(共14页)令()0hx,解得1x或212+−axa,故()fx为区间21,12aa+−上的凹函数,为区间

21,2aa+−−和)1,+上的凸函数;综上所述,当14a−时,()fx为区间21,12aa+−上的凹函数,为区间21,2aa+−−和)1,+上的凸函数;当14a=−时,()fx为区间(),−+上的凸函数;当10

4a−时,()fx为区间211,2aa+−上的凹函数,为区间(,1−和21,2aa+−+上的凸函数;当0a=时,()fx为区间)1,+上的凹函数,为区间(,1−上的凸函数;········6分(ii)当0a=时,()3233fxxx

x=−−+,()2361fxxx=−−,故在点(),()tft处的切线方程为()()23236133yttxtttt=−−−+−−+.············7分设()(),0Puvu为()fx的“3切点”,则关于t的方程()()2323613

3vttutttt=−−−+−−+有三个不同的解,即关于t的方程()3223363vtututu=−++−+−有三个不同的解,令()()3223363Fttututu=−++−+−,所以直线yv=与曲线()yFt=恰有三个不同的交点.····················

··············8分()()()()2661661Fttututtu=−++−=−−−.············································9分当1u时,()(

),FtFt随t变化情况如下:t(),1−1()1,uu(),u+()Ft−0+0−()Ft减极小值44u−增极大值3233uuu−−+减故324433uvuuu−−−+;···················································

···········11分{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第14页(共14页)当1u=时,()()2610Ftt=−−,()Ft单调递减,不符合题意;·················12

分当01u时,()(),FtFt随t变化情况如下:t(),u−u(),1u1()1,+()Ft−0+0−()Ft减极小值3233uuu−−+增极大值44u−减故323344uuuvu−−+−;综上所述,点P的集合为()323210

1,44333344xxxyxyxxxxxxyx−−−+−−+−或;··················································

····················································14分(2)点Q的集合为()2242204,44eee0eexxxxxxxyxxxyyxxy−−−−

++−−或或.················································································

······················17分{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}

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