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{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}{#{QQ

ABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第1页（共14页）2024-2

025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.（1

3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=−+−++−{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}数学试题第2页（共14页）()23333

3nn=++++−····························································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2cos3cos3cosaCbCcB=+，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()12C

DCACB=+，······································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2cos3cos3c

osaCbCcB=+，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)2

432=+−7=，···························································································11分故7c=，······························

·························································12分因为D为AB中点，所以coscos0ADCBDC+=，故222222022ADCDACBDCDBCADCDBDCD+−+−+=，··················

························13分所以()2222227734220772222CDCDCDCD+−+−+=，{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCC

QgOgFAIMAAAQRFABAA=}#}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=+−()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；{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAI

MAAAQRFABAA=}#}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sins

inSASBSBASAB=，·········································································1分所以12sin45sinSAB=，···························

···········································2分所以sin1SAB=，因为0180SAB，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ABx=，在△SAB中，因为1SA=，o45SBA=，2SB=，由余弦定理，得2222cosSASB

ABSBSABBA=+−，······················································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S

M，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，········································································11分又SC平面SBC，所以SMS

C⊥，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{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAI

MAAAQRFABAA=}#}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2221222134yyxx−=−−，····································

···········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−+==−=−−−+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()2

234690mymy+−−=，················11分所以()()22236+363+414410mmm==+，且122634myym+=+，···············12分{#{QQ

ABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}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()()()12121222811

2234634Pxxxmymymyymmmm=−+=−−+−=−++=−+=++，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2289,3

434mPmm−++，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(,)Cx

y−−，由2211043−+=+=xmyxy，得()2234690mymy+−−=，·······································6分所以()()22236+363+414410mmm==+，且122634myym+=+，

·················7分所以122634myym+=+，所以()121212228112236344xxmymymyymmmm+=−+−=+−=−=−++，·······8分{#{QQABBQwEggAAAIAAARg

CAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}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{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQ

gOgFAIMAAAQRFABAA=}#}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()()4323213fxaxxaxx=+−+−+，所以()()32436211fxaxxax=+−+−，·················································

·1分令()()32436211hxaxxax=+−+−，所以()()()()212662162211hxaxxaaxax=+−+=++−．·····························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，{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}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{#{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()3232101,44333344xxxyxyxxxxxxyx−−−+−−+−

或；·······························································································

·······14分（2）点Q的集合为()2242204,44eee0eexxxxxxxyxxxyyxxy−−−−++−−或或．·············································

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