【文档说明】福建省福州市2024届高三下学期4月末三模试题 数学 --简略答案4月9日.pdf,共(9)页,196.692 KB,由envi的店铺上传
转载请保留链接:https://www.doc5u.com/view-fd5e0fbc1149ab1606fe84a36cbaea7d.html
以下为本文档部分文字说明:
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2,2nannn
=+�,··································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=−+,················
·························································12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0.99730.68270.3146P=−=
,···········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()()()PDPAPBC=+·································································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,GHBCEPKBCE⊄⊂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BP
AE⊥,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,GIBPHIEB∥∥,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,222b
ca+=.··························································1分IFEPCDHAGBQFEPCDHAGB参考答案第5页共9页COyxPF2F1IG设00(,)Pxy,则2200221xyab+=,
0axa−<<,所以222222222020000022||()()(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003xcaxaa−=−,所以椭圆E的离心率13cea==.·························10分(ⅱ)由26a=,得3a=,又13ca=,所以1c=,2228ba
c=−=,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121202||||||||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(,)33xyG,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1
PF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()()()016233yxx
cxcc+++−+=,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()0
00e1e()xxyxx−+=−,即()000e(1)e1xxlxxx=+−+(*).························································5分一方面,()()00gxlx=,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