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=}#}{#{QQABBQwEggAAAIAAARgCAw

mKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}{#{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132nn

aa+=+,且12a=,所以10na+,···················································································1分所以1132111nnnnaaaa++++=++············

····························································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=−+−++−{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFA

BAA=}#}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ABC△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2cos3co

s3cosaCbCcB=+,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1113

316344422=++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2222222cos3322abcac

baCbcabac+−+−=+···························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coscos0ADCBDC+=,故222222022ADCDACBDCD

BCADCDBDCD+−+−+=,··········································13分所以()2222227734220772222CDCDCDCD+−+−

+=,{#{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2222coscababACB=+−················

···················································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=+−()22773323

227=+−−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0180SA

B,所以90SAB=,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)解:由(1)知SA⊥平面ABCD,又,ABAD平面ABCD,所以

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

以,,SAADAB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(,,)xyz=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AB

x=,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2222cosSASBABSBSABBA=+−,······················································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110,10,2SCxyzCDxy=−+−==−=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SBM△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,········································································1

1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{#{Q

QABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}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22143xy+=;·

·······························································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222122

2134yyxx−=−−,···············································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km212121212221222134−+==−=−−−+yyyy

xxxxyyxx,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1203Px

xx++=,······························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()22223849343

43434PDPDmmyymxxmmmmm=−−=−+=−++++,·······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【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−+=+=xmyxy,得()2234690mymy+−−=,·······································6分所以()()22236+3

63+414410mmm==+,且122634myym+=+,·················7分所以122634myym+=+,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()()2

12121222163348344yyyykxxxxmmmm−−+===−−+=−−++,······································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,12263

4myym+=+,·······························································12分因为PAB△的重心G在y轴上,所以1203Pxxx++=,······························13分所以()()()12

1212228112234634Pxxxmymymyymmmm=−+=−−+−=−++=−+=++,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3330xy+=.·······17分19.(17分)阅读以下材料:①设()fx为函数()fx的导函数.若()fx在区间D上单调递增,则称()fx为区间D{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFAB

AA=}#}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()()4323213fxa

xxaxx=+−+−+.(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()()4323213fxaxxaxx=+−+−+,所以()()32436211fxaxxax=+−+−,··················································1分令()()32436211hxaxxax=+−+−,所以()()()()2

12662162211hxaxxaaxax=+−+=++−.·····························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,{#{QQABBQwEggAAAIAAARgCAwmKCkIQ

kBGCCQgOgFAIMAAAQRFABAA=}#}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104a−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,()3233fxxxx=−−+,()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()()23236133vttutttt=−−−+−−+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{#{QQABBQwEggAA

AIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}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()3232101,44333344xxxyxyxxxxxxyx−−−+−−+−

或;······································································································14分(

2)点Q的集合为()2242204,44eee0eexxxxxxxyxxxyyxxy−−−−++−−或或.···························································

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

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