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高二数学参考答案第1页（共8页）2022—2023学年下期第一次联考高二数学参考答案一、选择题（本大题共12小题，每小题5分，共60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）1．D【解析】因为()cos33fππ=为常数，所以()03fπ′=.故选D

.2．C【解析】2418≠×，A选项错误；2214≠−×，B选项错误.因为642963==，所以9，6，4依次成等比数列，C选项正确.2648≠×，D选项错误.故选C.3．A【解析】直线130kxyk−+−=可化为()13ykx−

=−，表示过点()3,1，斜率为k的直线，所以所有直线都通过定点()3,1.故选A.4．B【解析】若方程22151xymm+=−−表示焦点在y轴上的椭圆，则150mm−>−>，解得：35m<<．所以p成立的充要条件是：35m<<．结合四个选项可知：p成立的充分不必要

条件是45m<<，故选B.5．C【解析】由题意得()()()00000limlim()xxfxxfxfxabxax∆→∆→+∆−′==+∆=∆，故选C.6．D【解析】∵a，b，c成等差数列，∴2b＝a＋c,又a＋1，b，c与a，b，c＋2都分别成等比数列，∴b2＝(a

＋1)·c，b2＝a·(c＋2)．联立222,(1),(2),bacbacbac=+=+=+解得b＝12.故选D.7．A【解析】()1112CMAMACABAD=−=+�����������������������-()1ABBCCC++�������������＝1122−−

−���abc，111122MCCMabc=−+=+�������������.故选A．8．D高二数学参考答案第2页（共8页）【解析】根据题意：()0,4M−，()23,8A−−，故6412116m−=，解

得4m=，即221164yx−=，当水面宽度为46米时，即26x=−时，47y=−，拱顶M到水面的距离为()474−.故选D.9．A【解析】设等差数列{}na的公差为d，()()()110105661055111202aaSaaa+==

+=+=，所以613a=，则652daa=−=，所以()52521naann=+−=+，所以1111122123nnnbaann+==−⋅++，所以()11111111112355721232323323

nnTnnnn=−+−++−=−=++++⋯，因为17kT=，所以()13237kk=+，解得9k=.故选A.10．C【解析】因为'()fa、'(1)fa+分别是函数()fx在xa=、1xa=+处的切线斜率，由图可知'(1)'()0fafa+<<，又

0(1)()(1)()'()(1)fafafafafxaa+−+−==+−，0(,1)xaa∈+，所以()()()()11fafafafa′′++−＜＜，故选C.11．C【解析】依题意，()3,0,3S，()3,0,0A

，()0,1,0B，()0,1,0C−，313,,222E；若13SDSC=，则231,,233D−,则333,,222CE=����,1333,,222CE=����∵，故A正确；232,,233CD=

����，()3,1,0AC=−−����，313,,222AE=−����，故D点到直线CE的距离228721CDCEdCDCE⋅=−=����������������，故B正确；设(

),,nxyz=�为平面ACE的法向量，则00ACnAEn⋅=⋅=����������，即303130222xyxyz−−=−++=，令2z=−，则高二数学参考答案第3页（共8页）()3,

3,2n=−−�为平面ACE的一个法向量，故C错误；而232,,233CD=����，故点D到平面ACE的距离11CDndn⋅==������，故D正确.故选C.12．A【解析】如图所示，设点Q所在椭圆的另一焦点为F，则44445QPQOQPQFPFPO=≤++−+=−+=.故选

A.二、填空题（本大题共4小题，每小题5分，共20分）13．514．20xye−+=15．224ayxb=16．1024013．【解析】(2PA=−����，0，1)−，点P到平面α的距离为|||401|5||5nPAn⋅−+−==��������．14．【解析】由题意，(

)()1ln()1lnfxxxxx−+=′−=+−−故()()e12,eekflnef=−=+=−=−′，则曲线()yfx=在ex=−处的切线方程为：()2y+e=x+e,即：20x-y+e=15.【解析】设抛物线方程为()220ypxp=>，依题意,2aAb

，代入()220ypxp=>得222,244aapbpb==，所以抛物线标准方程为224ayxb=.16.【解析】由题意得2142Sa=+，所以12142aaa+=+，解得28a=，又因为221144nnn

nnaSSaa++++=−=−，于是()211222nnnnaaaa+++−=−，高二数学参考答案第4页（共8页）因此数列{}12nnaa+−是以2124aa−=为首项、2为公比的等比数列，故1112422nnnnaa−++−=×=，于是11122nnnnaa++−=，因此数列2nna

是以1为首项、1为公差的等差数列，故1(1)2nnann=+−=，故2nnan=⋅,所以101010241020a==⋅.三、解答题（本大题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤）17．【解析】（1）由020xyxy−=+−=可得11xy==，·

·························································2分所以1l，2l的交点为()1,1，································································

····3分故过()1,1且斜率为2的直线的方程为()121yx−=−即210xy−−=.················5分（2）根据正弦定理由2sinsinBA=可得222ba=，·········

·························7分设顶点C的坐标为(),xy，则()()2222222122xyxyxy++=++⇒+=，·············································8分

因为构成三角形，故0y≠，··································································9分故轨迹方程为()2202xyy+≠=.·········

····················································10分18．【解析】（1）设等差数列{}na的公差为d，由19a=得到436624Sd=+=，解得2d=−，······································

···············································2分得112nan=−，21()102nnaanSnn+==−+.·······································

·········6分（2）当5n≤时，2121210nnnnTaaaaaaSnn=+++=+++==−+⋯⋯．····························7分当6n≥时，1212567nnnTaaaaaaaaa=+++=+++−−−−⋯⋯⋯2521050nSSnn

=−=−+．··································································9分高二数学参考答案第5页（共8页）故2210,51050,6nnnnnT

nnnn∗∗−+≤∈=−+≥∈NN且且；·····················································10分所以603600600503050T=−+=.（不求Tn，直接求T60

也得分）···················12分19．【解析】（1）如图过点O作AB的平行线OD交劣弧�BC于点D，连接1OO，1OD，因为1OO∥1AA，1AA⊂平面1AAB，1OO⊄平面1AAB，则1OO∥平面1AAB同理可证

OD∥平面1AAB，1OOODO=∩，且1OO⊂平面1OOD，OD⊂平面1OOD所以平面1AAB∥平面1OOD，又因为1OD⊂平面1OOD，所以1OD∥平面1AAB故存在点D满足题意.因为ABC△为底面O⊙

的内接正三角形，所以3BACπ∠=，即6ABOBODπ∠=∠=，又因为3AB=，所以O⊙的半径为332sin3π=，所以劣弧�BD的长度为362326ππππ××=.·································

··············6分（2）如图取BC的中点为M，连接MA，以MB为x轴，MA为y轴，过M作1OO平行线为z轴，建立空间直角坐标系，又因为13AAAB==，设AB中点为N.故()0,0,0M，3,0,02B，330,,02

A，3,0,02C−，30,,02O，130,,32O，333,,044N，易知平面1AAB的法向量33,,044ON=����设平面1CBO的法向量为(),,nxyz=�

，又因为130,,32MO=�����，3,0,02MB=����故1·0·0nMOnMB==�����������即3302302yzx+==，令23y=得()0,23,1n=−�高二数学参考答案第6页（共8页）易知平面1CB

O和平面1BAA夹角为锐角，所以平面1CBO和平面1BAA夹角的余弦值为33921323134nONnON⋅==⋅×����������··········12分20.【解析】（1）依题意动点M到定点()1,0F的距离等于动点M到直线1x

=−的距离，··············2分由抛物线的定义可知，动点M的轨迹是以点F为焦点，直线1x=−为准线的抛物线，所以曲线C的方程为24yx=.·····························

··········································5分（2）联立方程2214yxyx=−=，整理得24810xx−+=.············································6分设()()1122,,,AxyBxy，则

有122xx+=，1214xx=.于是212115ABkxx=+−=.··································································8分设(),0,Pa

P到直线AB的距离为d，因为:210ABlxy−−=，由点到直线的距离公式得215ad−=.··························································9分又12ABPSABd=⋅△，所以2ABPSdAB=△，于是932212515a×−

=.·························10分解得5a=或4a=−，············································································11分故点P的坐标为()5,0或(

)4,0−．·····························································12分21．【解析】（1）2nnaSn+=∵①1122,2nnaSnn−−∴+=−≥②①-②得12nnnaaa−+=−，即122nnaa−=+，··········

··································2分变形可得11222nnaa−=−−，······································································

·3分又112aS+=，得11a=故数列{}2na−是以-1为首项，12为公比的等比数列，·····························4分由等比数列的通项公式可得1122nna−−=−，············································5分高

二数学参考答案第7页（共8页）*1122nnanN−∴=−∈，.·······································································6分（2）令()()()232nfnna=−−，则()

1232nnfn−−=·············································7分()()12123521222nnnnnnfnfn−−−−∴+−=−=·······

··········································8分当1n=或2n=时，()()10fnfn+−>，当3,nnN≥∈时，()()10fnfn+−<又()334f=，()max34fn∴

=，····························································10分因为不等式()()22232nnaλλ−>−−对任意的正整数n恒成立，2324λλ∴−>，························

······················································11分解得λ13<<22.·················································

··························12分22．【解析】（1）设直线:lykxm=+，显然0k≠，令0y=，得mxk=−，则(,0)mNk−，因为M是线段PN的中点，所以Pmxk=，2Pym=，所

以(,2)mPmk，(,2)mQmk−，又200QQymmmkmxk−−−′==−−3k=−，························································4分所以3kk′=−为定值.············

····························································5分（2）联立22142xyykxm+==+，消去y并整理得222()124240kxmkx

m++−=+，········6分则2222164(12)(24)0mkkm∆=−+−>，则2242mk<+，根据韦达定理可得222412APmxxk−⋅=+，所以22(24)(12)Akmxmk−=

+，高二数学参考答案第8页（共8页）所以222(24)(12)AAkmykxmmmk−=+=++2222244(12)kmkmmk−+=+，所以2222222(24)44,(12)(12)kmkmkmAmkmk−−+++

，··············································7分由（1）知，3kk′=−，所以直线QM的方程为3ykxm=−+，联立221423xyykxm+==−+，消去y并整理得2

22(118)12240kxkmxm+−+−=，则22221444(118)(24)0mkkm∆=−+−>，则22362mk<+，根据韦达定理可得2224118BQmxxk−=+，所以22(24)(118)Bkmxmk−=+，·····················8分所以2223

(24)3(118)BBkmykxmmmk−=−+=−++222221212(118)kmkmmk++=+，所以2222222(24)1212,(118)(118)kmkmkmBmkmk−++++，···········································

9分所以ABABAByykxx−=−22222222222222441212(12)(118)(24)(24)(12)(118)kmkmkmkmmkmkkmkmmkmk−+++−++=−−−++2222261224(2)kmmkkm+−−=−222

26(2)24(2)kmmkm−+−=−2614kk+=1342kk=+，··········································································

·····11分因为0k>，所以13136242422kkkk+≥×=，当且仅当66k=时取等号，所以62ABk≥；即tanθ的最小值为62.····································

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