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参考答案第1页共9页2023～2024学年福州市高三年级4月份质量检测参考答案与评分细则一、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．C3．A4．C5．D6．B7．B8．A二、选择题：本大题考

查基础知识和基本运算．每小题6分，满分18分．全部选对的得6分，部分选对的得部分分，有选错的得0分．9．ABC10．BC11．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15分．12．2−13．814．6，22四、解答题：本大题共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.【考查意图】本小题主要考查递推数列与数列求和等基础知识，考查运算求解能力、推理论证能力等；考查分类与整合、化归与转化等思想方法；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13分.解：（1）因为12,2nnaann−=+�，所以12nnaan−−=，···································1分当2n�时，112211()()()nnnnnaaaaaaaa−−−=−

+−++−+L，所以22242nann=+−+++L，·························································3分所以(22),22nnnan+=�，所以2,2na

nnn=+�，··································4分又因为12a=，·················································

······························5分所以2*,nannn=+∈N.································································

······6分（2）由（1）可知2*(1),nannnnn=+=+∈N，·············································7分所以()111111nannnn=

=−++，····························································9分所以11111223(1)(1)nSnnnn=++++××−+L1111111122311nnnn=−+−++−+−−+L，··

···································11分所以111nSn=−+，·········································································1

2分又因为1n�，所以1nS<.·································································································13分参考答案第2页共9页16.【考查意图】本小题主要考查正态分布、全

概率公式、条件概率等基础知识，考查数学建模能力、逻辑思维能力和运算求解能力等，考查分类与整合思想、概率与统计思想等，考查数学建模、数据分析、数学运算等核心素养，体现基础性、综合性和应用性．满分15分.解：（1）依题意得，0,0.2µσ==，·······················

····································1分所以零件为合格品的概率为(0.60.6)(33)0.9973PXPXµσµσ−<<=−<<+=，···················································

················································2分零件为优等品的概率为(0.20.2)()0.6827PXPXµσµσ−<<=−<<+=，·····3分所以零件为合格品但非优等品的概率为0.99730.6827

0.3146P=−=，···········5分所以从该生产线上随机抽取100个零件，估计抽到合格品但非优等品的个数为1000.314631×≈.·························································

····················6分（2）设从这批零件中任取2个作检测，2个零件中有2个优等品为事件A，恰有1个优等品，1个为合格品但非优等品为事件B，从这批零件中任取1个检测是优等品为事件C，这批产品通过检测为事件D，··············

··············································8分则DABC=+，且A与BC互斥，·······················································9分所以()()()P

DPAPBC=+·································································10分()()(|)PAPBPCB=+··································

·······················11分221220.68270.68270.31460.6827CC=×+×××21.62920.6827=×，·······································

·····················12分所以这批零件通过检测时，检测了2个零件的概率为()(|)()PADPADPD=·········································

··································13分220.68271.62920.6827=×11.6292=0.61≈.···············································

······························15分答：这批零件通过检测时，检测了2个零件的概率约为0.61.17.【考查意图】本小题主要考查直线与平面平行的判定定理、直线与平面垂直的判定与性质定理、平面与平面的夹角、空间向量

、三角函数的概念等基础知识，考査直观想象能力、逻辑推理能力、运算求解能力等,考查数形结合思想、化归与转化思想等，考査直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性.满分15分.解法一：（1）在正方形ABEF中，连接AH并延长，交BE的延长线于点K，连

接PK.···································································································2分参考答案第3页共9页因为,GH分别为线段,APEF中点，所以HFHE

=，所以RtAFH△≌RtKEH△，所以AHKH=，·····························4分所以GHPK∥.································5分又因为,GHBCE

PKBCE⊄⊂面面，所以GHBCE∥面.···········································································7分（2）依题意得，

ABBCE⊥面，又因为BPBCE⊂面，所以ABBP⊥.又因为BPAE⊥,ABAEA=I,,ABAEABEF⊂面，所以BPABEF⊥面，································································

········8分又BEABEF⊂面，所以BPBE⊥，·····················································9分所以,,BPBEBA两两垂直.以B为原点，,,BPBEBA所在直线分别为,,xyz轴建立空间直角坐标系，如图所示.····

····························································10分不妨设1AB=,则31(1,0,0),(,,1)22PD−，()31=1,0,0,,,122BPBD��=−������uuuruuur，···········

··············································11分设平面BPD的法向量为(),,xyz=m，则0,0,BPBD�⋅=��⋅=��uuuruuurmm即0,310,22xxyz=���−+=��取2y=

，得0,1xz==，所以平面BPD的一个法向量是()0,2,1=m，·········································13分又平面BPA的一个法向量为()0,1,0=n.···········

·····································14分设平面BPD与平面BPA的夹角为θ，则225coscos,551θ⋅=<>===×mnmnmn.KFEPCDHAGBzyxBGAHDCPEF参考

答案第4页共9页所以平面DBP与平面BPA夹角的余弦值为255.····································15分解法二：（1）证明：取BP的中点Q，连接,GQEQ.·····1分因为,GH分别为线段

,APEF的中点，所以GQAB∥，12GQAB=，····························2分又因为,ABEFABEF=∥，所以,GQHEGQHE=∥，·································3分所以四边形GQEH是平行四边形，····

··················································4分所以GHQE∥，·············································································

·5分又因为,GHBCEQEBCE⊄⊂面面，所以GHBCE∥面.············································································7分（2）同解法一.··

··················································································15分解法三：（1）证明：取AB的中点I，连接,GIHI.········1分因为,GH分别为线段,APEF的中点，所以,GI

BPHIEB∥∥，又因为,GIBCEBPBCE⊄⊂面面，所以GIBCE∥面.············································3分因为,HIBCEBEBCE⊄⊂面面，所以HIBCE∥面.·······························

·············································5分又因为,,GIHIIGIGIHHIGIH=⊂⊂I面面，所以GIHBCE面∥面，·······································

······························6分又因为GHGIH⊂面，所以GHBCE∥面.············································································7分（2）同解法一.······

··············································································15分18.【考查意图】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，考査直观想象能力、逻辑推理能力、运算求解能力等,考查数形结合思想、化归与转化思想、分类与整合思想等，考査直观想象、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17分.解法一：（1）依题意，222bca+=．·························

·································1分IFEPCDHAGBQFEPCDHAGB参考答案第5页共9页COyxPF2F1IG设00(,)Pxy，则2200221xyab+=，0axa−<<，所以2

22222222020000022||()()(1)(1)2xbPFxcyxcbxcxbcaa=−+=−+−=−−++，所以22220002||2||ccPFxcxaxaaa=−+=−，······························

··············3分又ac>，所以0caxa>，20axc>，所以20||cPFaxa=−，20adxc=−所以0220||caxPFcaadaxc−==−，即2||PFd为定值，且这个定值为ca．··············

····4分（2）（ⅰ）依题意，00(,)33xyG，设直线IG与x轴交于点C，因为IG⊥x轴，所以0(,0)3xC，·······················5分所以001202||||()()333xxFCFCccx−=+−−=，·

·········································6分因为△PF1F2的内切圆与x轴切于点C，所以121202||||||||3PFPFFCFCx−=−=，······················

···························7分又因为12||||2PFPFa+=，解得02||3xPFa=−，··········································8分由（1）得20||cPFaxa=−，所以0

03xcaxaa−=−，所以椭圆E的离心率13cea==．·························10分（ⅱ）由26a=，得3a=，又13ca=，所以1c=，2228bac=−=，所以椭圆E的方程为22198xy+=．···············

·······································11分根据椭圆对称性，不妨设点P在第一象限或y轴正半轴上，即003x<�，0022y<�，又1(1,0)F−，2(1,0)F，所以直线PF1的方程为00(1)1yyxx=+

+，设直线IG与PF1交于点D，因为03Dxx=，所以000(3)3(1)Dyxyx+=+，△F1CD的面积1S与△PF1F2的面积S之比为000200100(3)1(1)233(1)(3)118(1)22xyxxxSSxy++×++==+××，···

·············································13分令2(3)()18(1)xfxx+=+（03x<�），则2(3)(1)(1))(18xxxfx−′++=，参考答案第6页共9页当[0,1)x∈，()0fx′<，当(1,3)x∈，()0fx′

>，所以函数()fx在[0,1)单调递减，在(1,3)单调递增.又因为1(0)2f=，4(1)9f=，1(3)2f=，所以()fx的值域是41[,]92，所以14192SS��，···································

·······································15分所以11415SSS−��，················································

·······················16分根据对称性，△PF1F2被直线IG分成两个部分的图形面积之比的取值范围是45[,]54．········17分解法二：（1）同解法一·································

··········································4分（2）（ⅰ）依题意，00(,)33xyG，设直线IG与x轴交于点C，因为IG⊥x轴，所以0(,0)3xC，·······

················5分所以001202||||()()333xxFCFCccx−=+−−=，··········································6分因为△PF1F2的内切圆与x轴切于点C，所以12120

2||||||||3PFPFFCFCx−=−=，·················································7分又因为12||||2PFPFa+=，得0102||,3||.3xPFaxPFa�=+����=−��··

·········································8分所以2200022000(),3(),3xxcyaxxcya�++=+����−+=−��两式平方后取差，得00443cxax=对任意0x成立，所以椭圆E的离心率13cea==

．························································10分（ⅱ）同解法一···········································································

········17分解法三：（1）同解法一···········································································4分（2）（ⅰ）依题意，00(,)33xy

G，因为IG⊥x轴，设点I坐标为0(,)3xt.··········5分可求直线1PF方程为00()yyxcxc=++，则点I到直线1PF的距离0002200()()3||()xyctxctyxc+−+=++，······················

···········6分即()222200000()()()3xyctxctyxc��+−+=++����，化简得22000002()()()033xxyttcxcyc+++−+=，①同理，由点I到直

线2PF的距离等于||t，可得参考答案第7页共9页22000002()()()033xxyttcxcyc+−−−−=，②············································7分将式①−式②，得

00084233tcxycx⋅=⋅，则04yt=.·····································8分将04yt=代入式①，得2200001()()()016233yxxcxcc+++−+=，化简得220022198xycc+=，得229ca=，所以椭

圆E的离心率13cea==．························································10分（ⅱ）同解法一···························································

························17分19.【考查意图】本小题主要考查集合、导数、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、数形结合思想等，考查数学抽象、逻辑推理、直观想象、数学运算等核心素养，体

现基础性、综合性与创新性.满分17分.解：（1）依题意，因为0(),()fxxlxL∈∈R，所以2,1xxxkx∀∈−++R�，且0xR，20001xxkx−+=+，····················1分令2()(1)1xxkxφ=−+−−，()214k∆=−−，则()0xφ�，且0

()0xφ=，所以0,0,∆��∆���所以0∆=，···································································3分即()2140k−−=，解得3k=或1−.········

······································································4分（2）（ⅰ）先证必要性.若直线()ylx=是曲线()ygx=的切线，设切点为()00,e1xx+，因为()exgx′=

，所以切线方程为()000e1e()xxyxx−+=−，即()000e(1)e1xxlxxx=+−+（*）.························································5分一方面，()()00g

xlx=，所以000,()()xgxlx∃∈=R，································6分参考答案第8页共9页另一方面，令()()()000ee(1)exxxGxgxlxxx=−=−−−，则()00Gx=

，因为()0eexxGx′=−，所以当0xx<时，()0Gx′<，()Gx在()0,x−∞单调递减，当0xx>时，()0Gx′>，()Gx在()0,x+∞单调递增，所以()()00GxGx=�，所以()()gxlx�.····························

··················7分即,()()xgxlx∀∈R�，所以()(),gxxlxT∈∈R，即()lx是函数()gx在R上的“最佳下界线”.···············8分再证充分性.若()lx是函数()gx在R上的“最佳下界线”，不妨设()lxkxb=+，由“最佳下界线”的定义，

()(),xgxlx∀∈R�，且()()000,xgxlx∃∈=R，····9分令()()()e1xHxgxlxkxb=−=+−−，则()0Hx�且()00Hx=，所以()min0Hx=.··················

·······················10分因为()exHxk′=−，①若0k�，则()0Hx′�，所以()Hx在R上单调递增，所以10xx∃<，使得()()100HxHx<=，故0k�不符合题意.·····················11分②若0k>，令()

0Hx′=，得lnxk=，当(),lnxk∈−∞时，()0Hx′<，得()Hx在(),lnk−∞单调递减，当()ln,xk∈+∞时，()0Hx>，得()Hx在()ln,k+∞单调递增，所以，当且仅当lnxk=时，()Hx取得最小值()lnHk.················

············12分又由()Hx在0x处取得最小值，()min0Hx=，所以0ln,(ln)0,xkHk=��=�·······················································

····················13分即000e,e10,xxkkxb�=��+−−=��解得0exk=，()001e1xbx=−+，····························14分所以()000e(1)e1xxlxxx=+−+，参考答案第9页共9

页由（*）式知直线()ylx=是曲线()ygx=在点()00,e1xx+处的切线.············15分综上所述，直线()ylx=是曲线()ygx=的一条切线的充要条件是直线()ylx=是函数()gx在R上的“最佳下界线”.（ⅱ）集合(),

(1,)(),hxxgxxLTRI∈+∞∈元素个数为2个.············································17分