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【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，由小赞的店铺上传
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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．)1,216．31312222−+，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23
0m−−＜···················································6分362m−····················································
·····················7分又|z1|5m2＋925216m·················································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)∵()()()2,24,2
2,−−=−=−=mmOAOBAB,又()()()2,2,0−−=−=−=mnmnOBOCBC，·········································3分∵CBA、、三点共线∴()()022m2=−+−−mn
即022=+−mn······································6分（2）∵()()()2,0,2nmnmOCOB+=+=+，··················································8分∴()()422++=+=+nmO
COBOCOB()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−=aa·················································································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，kmEDBDB
E151292222=+=+=·································6分(2)在BAE中，32=BAD，15=BE.由余弦定理得BAEAEABAEABBE−+=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BQ→＝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→＝ACCI+=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＝53，∴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⊥，∴2220bbcca−+−=，222bcabc+−=.由余弦定理得2221c
os22bcaAbc+−==····1分∵(0,)A，∴3A=··············································································2分（Ⅰ）法一、∵D为边BC的中点
，∴12BDDCBC==，∵ABADDB=+，ACADDCADDB=+=−，·········································3分()()8ABACADDBADDB=+−=，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+−=，22214co
s122ADabADCADa+−=，∵ADBADC+=，∴2222221144112222ADacADabADaADa+−+−=−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=−，∵1A
E=，:2:BEECcb=，∴2ABEACESBEcSECb==，即11sin2211sin()23ccbb=−，即sin2sin()3=−，化简得2sin3cos=，即3tan
2=，····························································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2222ccAEABAC
ABcbcb−=−++即222cbAEACABcbcb=+++·················7分(2)2cbAEcACbAB+=+∴222(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
