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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)12345678B

ACBBDDA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()1112312232262342ABDAADAABABDSSSSS=+++=+=+△△△△

.····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1

8．(本小题满分12分)（1）π3A=；（2）3c=或5c=【解析】（1）由余弦定理222cos2cabBac+−=，222cos2abcCab+−=，得2222222coscoscos22abccabaAbCcBbc

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

········4分∵0πA，·······················································································5分∴π3A=.········

······················································································6分（方法二）由正弦定理得2sinaRA=，2sinbRB=，2s

incRC=，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28150

cc−+=，················································································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19．(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）213cos13CPD=−【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，90CEP=，2CE=，223PE=，∴263CP=，···············

·······································································10分∴222133cos13263CPE==，··················

··············································11分∴()213coscos180cos13CPDCPECPE=−=−=−.························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222acbac+−=，由余弦定理得2221cos222acbac

Bacac+−===，···········································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()sinsinsincoscossinABCBCBC=+=+，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311sincoscos222BBB+=+，··············2分即311sincos222BB−=，所以π1sin6

2B−=，······································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8

ac+，·····································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