【文档说明】福建省永泰县第一中学2020-2021学年高一下学期期中考试 数学答案.doc，共(8)页，539.500 KB，由管理员店铺上传

转载请保留链接：https://www.doc5u.com/view-0c87cf0f8dd29554cdddd0dae05478dc.html

以下为本文档部分文字说明：

参考答案一、单项选择题(本题共8小题，每小题5分，共40分)题号12345678答案CBCBABDD二、多项选择题(本题共4小题，每小题5分，共20分)题号9101112答案ABCABCDBC三、填空题(本大题共4小题，每小题5分，共20分)13．221+14．1615．)1,216．3

1312222−+，四、解答题(本大题共6小题，共70分)17.（本题满分10分）解：12zz=()()()()3123121212miimiiii−−−=++−···········································

····1分()()2262312mim−+−−=+·····················································3分()()62355mmi−−−=+···············

·····································4分12zz是实数，所以60m−=，6m=········································5分(2)

12zz在复平面内对应的点在第三象限，可得60m−＜，且230m−−＜···················································6分362m−·································

········································7分又|z1|5m2＋925216m···············································

··8分44mm−或·····································································9分综上，实数a的取值范围为)46，．·····································

·····10分18．（本题满分12分）解：(1)∵()()()2,24,22,−−=−=−=mmOAOBAB,又()()()2,2,0−−=−=−=mnmnOBOCBC，·········································3分∵CB

A、、三点共线∴()()022m2=−+−−mn即022=+−mn······································6分（2）∵()()()2,0,2nmnmOCOB+=+=+，······

············································8分∴()()422++=+=+nmOCOBOCOB()423942222+−=+−+=mmm·····················

·······················································································11分∴OCOB+取最小值为2·······················

··················································12分19.（本题满分12分）【解析】（1）由()2coscoscbAaB−=结合正弦定理可得()2sinsincossincosCBAAB−=·····················

·········································2分所以()2sincossincoscossinsinsinCAABABABC=+=+=······················3分因为sin0C，所以1cos2A=．·············

····················································4分因为()A0,，所以A=3．·····································································5分[选择条件

①的答案]因为acb3=+.又因为ABC的面积为32，所以32sin21==AbcSABC，·································································6分所以323sinbc21=

···········································································7分可得bc=8.···················································

································8分所以由余弦定理可得bccbbcbccbAbccba3)(3cos22)(cos222222−+=−−+=−+=················10分即24)3(22−=a

a·················································································11分解得32=a·················································

·····························12分[选择条件②的答案]又因为ABC的面积为32所以32sin21==AbcSABC··································································

···6分所以323sinbc21=···········································································7分可得bc=8.···············

················································································8分所以由余弦定理可得bccbbcbccbAbccba+

−=−+−=−+=22222)(3cos22)(cos2·················10分即8)33(22+=aa·········································································

·········11分解得32=a·························································································12分[选择条件③的答案]由sin3cos2CC+=得2sin23C

+=所以sin13C+=··················································································8分因为()

0,C，所以32C+=．所以6C=··············································9分．所以π2B=，即ABC是直角三角形···················································

·····10分又因为ABC的面积为32，所以21123223ABCSaca===△，·····························································11分所以32=a·············

············································································12分20.（本题满分12分）解：（1）连接BD在BCD中，

由余弦定理得BCDCDBCCDBCBD−+=cos2222············1分∴()()32cos333323333222−+=BD···2分即9km=BD·····················

······························3分∵在BCD中，CDBC=，32=BCD，∴6==CDBCBD················4分∴在BDE中，2632=−=−=CDBCDEBDE······················

··········5分在BDERt中，kmEDBDBE151292222=+=+=·································6分(2)在BAE中，32=BAD，15=BE.由余弦定理得BAEAEA

BAEABBE−+=cos2222····································7分即AEABAEAB++=22225····································································8分

故()222225+=−+AEABAEABAEAB·············································10分从而()225432+AEAB，即310+AEAB·····························

···············11分当且仅当AEAB=时，等号成立，即设计为AEAB=时步行道BAE最长·························································12分21.（本题满分12分）解法一：(1)设BIBQCICR==，，由B

Q→＝BA→＋AQ→＝－AB→＋12AC→=12ab−+·························································1分可得AI→＝ABBI+=AB→＋λBQ→＝1()(1)22aabab+−+=−+················

··········2分由CR→＝CA→＋AR→＝－AC→＋13AB→=13ab−可得·····················································3分AI→＝A

CCI+=AC→＋μCR→=1()(1)33babab+−=+−··································4分(1)(1)23abab−+=+−又a与b不共线，∴1－λ＝13μ，12λ＝

1－μ，解得λ＝45，μ＝35.·······························································6分所以AI→＝15a＋25b·························

·····························································7分(2)由(1)知AI→＝15a＋25b∴BP→＝AP→－AB→＝nAI→－AB→＝n

(15a＋25b)－a＝n5－1a＋2n5·b···························8分又∵BP→＝mBC→＝mAC→－mAB→=mamb−+·······················································

··9分∵a与b不共线，∴1525nmnm−=−=································································10分解得m＝23，n＝5

3，∴1nm−=··········································································12分解法二：（1）同解法一．············

································································7分（2）因为ABPBPA=−mBCnAI=−+···························

································8分()12125555mnmnmn=−−++=++−+baabab()mnm=−a，···9分又AB=a，所以115205mnmn+=−+=，，··

································································10分解得m＝23，n＝53，∴1nm−=··················

························································12分22.（本题满分12分）解：222()()()mnbbcaccabbcca=−++−=−+−，∵mn⊥，∴22

20bbcca−+−=，222bcabc+−=.由余弦定理得2221cos22bcaAbc+−==····1分∵(0,)A，∴3A=······························································

················2分（Ⅰ）法一、∵D为边BC的中点，∴12BDDCBC==，∵ABADDB=+，ACADDCADDB=+=−，·········································3分()()8ABAC

ADDBADDB=+−=，即228ADDB−=······························4分22184ADBC=+，∵8BCa==，∴21648244AD=+=，∴26AD=··

·························································································5分∴中线AD的长为26························

·······················································6分法二、∵8ABAC=，∴cos8ABACbcA==，∵3A=，∴1cos82bcAbc==，16bc=···································

································································3分由余弦定理2222cosabcbcA=+−，得22226416bcbcbc=+−=+

−，2280bc+=，··········································4分由余弦定理，22214cos122ADacADBADa+−=，22214cos122ADabADCADa+−=，∵ADBADC+=，∴2

222221144112222ADacADabADaADa+−+−=−即22221616ADcADb+−=−−+，222232ADbc=+−，·····························4分∴21(8032)242AD=−=，26AD=···················

·····································5分∴中线AD的长为26································································

···············6分（Ⅱ）法一、设(0)3BAE=，则3CAE=−，∵1AE=，:2:BEECcb=，∴2ABEACESBEcSECb==，即11sin2211sin()23ccbb=−，即sin2sin()3

=−，化简得2sin3cos=，即3tan2=，····························································7分故321sin77==，1sin()sin32−=.∵ABCABEACESSS=+，

∴111sinsinsin()23223bccb=+−，·····················································8分即13()sin22cbbc+=，即217bc+=，······························

····················9分.∴2111222(2)()(41)77bcbcbcbccb+=++=+++···································10分122197(52)

(54)777bccb+=+=（当且仅当bc=时取等号）·············11分.∴2bc+的最小值977·····················································

·······················12分法二、∵,,BEC三点共线，:2:BEECcb=，∴22cBEBCcb=+，即2222ccAEABACABcbcb−=−++即222cbAEACABcbcb=+++·················7分(2)2cbAEcACbAB+=+∴22

2(2)(2)cbAEbABcAC+=+，∴22222(2)44coscbbccbbcbcA+=++，即222222222(2)427cbbcbcbcbc+=++=，···················

······························8分∵0,0bc，∴27cbbc+=，即217bc+=，······························································9分∴2111222(2)()

(41)77bcbcbcbccb+=++=+++···································10分122197(52)(54)777bccb+=+=（当且仅当bc=时取等号）·············11分.∴2bc+的最

小值977············································································12分