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高一年期中数学参考答案【简解版】第1页共5页福清市高中联合体2020-2021学年第二学期高一年期中考试数学参考答案【简解版】一、单选题(本题共8小题,每小题5分,共40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)12345678BACBBDDA二、多选题(本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,有选错的得0分,部分选对的得3分)9101112ABADBCABC三、填空题(本题共4小题,每小题5分,共20分)（15题第

一空2分，第二空3分）13．2014．2715．20，162323−16．3四、解答题(本题共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．(本小题满分10分)（1）623+；（2）203【解析

】（1）由正方体的特点可知三棱锥1AABD−中，1ABD△是边长为22的等边三角形，1AAD△、1AAB△、ABD△都是直角边为2的等腰直角三角形，所以截去的三棱锥1AABD−的表面积()1112312232262342ABDAADAABABDSS

SSS=+++=+=+△△△△.····5分（2）正方体的体积为328=，······························································

····6分三棱锥1AABD−的体积为111142223323ABDSAA==△，··················8分所以剩余的几何体1111ABCDDBC−的体积为420833−=.····

······························10分高一年期中数学参考答案【简解版】第2页共5页18．(本小题满分12分)（1）π3A=；（2）3c=或5c=【解析】（1）由余弦定理222cos2cabBac+−=，222cos2abcCab+−=，得2222222coscosc

os22abccabaAbCcBbcaabac+−+−=+=+=，·············3分∴1cos2A=，··················································

······································4分∵0πA，···································································

····················5分∴π3A=.········································································

······················6分（方法二）由正弦定理得2sinaRA=，2sinbRB=，2sincRC=，结合2coscoscosaAbCcB=+得4sincos2(sincossincos)RAARBCCB=+()()2sin2siπnRBCRA=+=−2sinRA

=，··································································3分由于sin0A，所以1cos2A=，······················································

·········4分∵0πA，···················································································

····5分∴π3A=.·····························································································

·6分（2）由余弦定理得2222cosabcbcA=+−，则222178282cc=+−，······9分即28150cc−+=，····················································

····························10分解得3c=或5c=.··················································································12分高一年期中数学参考答案【简解版】第3页共5页1

9．(本小题满分12分)（1）1iz=−+；（2）【解析】(1)设i(,)zxyxy=+R，由2z=，2zz+=−，且z在复平面内对应的点在第二象限，得222,22,0,0xyxxy+==−

，·······································································3分解得1,1,xy=−=··················

·······································································5分所以1iz=−+.······················································

·································6分(2)由(1)知1iz=−+，所以12ii(2i1)ii12i12izz−−−−===+−+−+，·····························8分所

以1i1izz−==+，所以复数满足11−.············································9分由复数的几何意义，得在复平面内对应的点的集合构成的图形是以(1,

0)为圆心，1为半径的圆面，····························································································11分所以其面积为2π1π=.······

······································································12分20．(本小题满分12分)（1）12DCADAB=−+；（2）213cos1

3CPD=−【解析】（1）因为2ADBC=，所以1122DCDAABBCADABADADAB=++=−++=−+.··························4分（2）由已知2ADBC=，2AEEB=，1EBBC==得2AD=，2AE=，在

ADE△中，90A=，2AEAD==，∴45AEDADE==，22DE=.·····················································6分在BCE△中，90B

=，1BEBC==，∴45BCEBEC==，2CE=，∴90CEP=，··································································

·················8分又∵2DPPE=，∴223PE=，····························································9分高一年期中数学参考答案【简解版】第4页共5页在CEP△中，9

0CEP=，2CE=，223PE=，∴263CP=，······················································································10分

∴222133cos13263CPE==，································································11分∴()213coscos180cos13C

PDCPECPE=−=−=−.························12分21．(本小题满分12分)（1）19BD=；（2）2sin77CAD=【解析】（1）在RtABC△中，tan2ACABABC==.··························

··············1分在RtACD△中，tan3CDCADAD==，所以60CAD=，····················3分所以cos1ADACCAD==.··················

···············································4分在ABD△中，由余弦定理得2222cos19BDABADABADBAD=+−=，所以19BD=.······················

································································6分（2）设CAD=，则60ABD=−，2cosAD=，···········

···············8分在ABD△中，由正弦定理得()2cos23sin60sin30=−，化简得3cossin2=，····················································

·······················································10分代入22sincos1+=，得24sin7=，····················································11分又

为锐角，所以2sin77=，即2sin77CAD=.·································12分22．(本小题满分12分)条件选择见解析；（1）π3B=；（2）12【解析】（1）选①∵(),mabca=+−，

(),nabc=−，且mn⊥，∴()()()0ababcca+−+−=，······························································2分高一年期中数学参考答案【简解版】第5页共5页化简得222

acbac+−=，由余弦定理得2221cos222acbacBacac+−===，···········································4分又因为0πB，·······························

··················································5分∴π3B=．···············································

·············································6分选②根据正弦定理，由22cosacbC−=得2sinsin2sincosACBC−=，·······2分又因为()sinsinsincosc

ossinABCBCBC=+=+，所以2cossinsinBCC=，又因为sin0C，所以1cos2B=，···························································4分又因

为()0,πB，·················································································5分所以π3B=．········

·················································································6分选③由π1sincos62BB+=+，得311sincoscos222BB

B+=+，··············2分即311sincos222BB−=，所以π1sin62B−=，······································4分又因为()0,πB，所以ππ66B−=，因此π3B=．······

·································6分（2）由余弦定理2222cosbacacB=+−，得()2163acac=+−．················7分又∵2aacc+，∴

()24acac+，························································8分当且仅当ac=时等号成立，·························

············································9分∴()()2233164acacac+=+−，解得8ac+，·························

············11分当且仅当4ac==时，等号成立．∴8412abc+++=，ABC△的周长的最大值为12．·······························12分