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高一数学答案第1页共6页2019～2020学年度第二学期质量检测高一数学答案及评分标准2020．7一、单项选择题：本大题共8小题，每小题5分，共40分．1~4ADCB5~8DCBD二、多项选择题：本大题共4小题，每小题5分，共20分．全部选对的得5分，部分选对的得3分，有选错的得0分．

9．ACD10．ACD11．BC12．BCD三、填空题：本题共4小题，每小题5分，共20分．13．{|(21)π,}xxkk≠+∈Z14．①②③15．0或316．0.65四、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．（本题满分10分）解：（1）圆锥SO的母线长22345l=+=．······························································1分圆锥SO的侧面积π3515πS=××=侧．·

·······················································3分圆锥SO的体积21(π3)412π3V=×××=．···············································

·····5分（2）设圆锥SO的内接圆柱OO′的底面半径为r，则hRrHR−=，即343hr−=．解得334hr=−．········································7分内接圆柱OO′的侧面积32π2π

(3)4hSrhh==−·············································8分侧面积2233(4)[4(2)]22Shhhππ=−+=−−．当2h=时，S有最大值6π，即圆柱侧面积的最大值为6π．···············

···10分18．（1）2ππ()3sincos(sincos)(sincoscossin)244fxxxxxxx=++−················1分31sin2cos222xx=−·······················

·····················································3分πsin(2)6x=−················································

·········································4分所以()fx的最小正周期2ππ2T==．··············································

················5分BOASO'C高一数学答案第2页共6页令πππ2π22π()262kxkk−−+∈Z��，解得ππππ()63kxkk−+∈Z��．所以()fx的单调递增区间是ππ[π,π]()63kkk−+∈Z．··

·······························6分（2）由（1）及题设条件知π1()sin()263AfA=−=．因为π02A<<，所以πππ663A−<−<．····················

·········································7分从而22ππ122cos()1sin()1()6633AA−=−−=−=．·······················

··············8分所以ππcoscos[()]66AA=−+··············································································10分ππππcos()cossin()sin6666AA=−−−·

·················································11分2231126132326−=×−×=．···············································

··········12分19．（本小题满分12分）解：（1）1111()3333BDBCACABACAB==−=−uuuruuuruuuruuuruuuruuur．············1分所以2133ADABBDABAC=+=

+uuuruuuruuuruuuruuur．··········2分所以111236AEADABAC==+uuuruuuruuuruuur．··············3分所以2136EBABAEABAC=−=−uu

uruuuruuuruuuruuur；·······························································4分（2）1536ECACAEABAC=−=−+uuuruuuruuuruuuruuur．·····················

··················································5分解法1：如图，以ACuuur，ABuuur所在方向分别为x轴，y轴的正方向，建立平面直角坐标系xAy，则(0,3)B，(6,0)C．·······························

·························6分则2136EBABAC=−uuuruuuruuur(1,2)=−；·······························7分1536ECABAC=−+uuuruuuruuur15

(0,3)(6,0)(5,1)36=−+=−．·······8分yxEDBCAEDBAC高一数学答案第3页共6页所以·(-1)52(-1)-7EBEC=×+×=uuuruuur，····················································

················9分22(1)25EB=−+=uuur，225(1)26EC=+−=uuur．······································11分设ECuuur与EBuuur的夹角为θ，则77130cos130526EB

ECEBECθ⋅−===−×⋅uuuruuuruuuruuur．·····················································12分解法2：因为2136EBABAC=−uuuruuuruuur，1536ECABAC=−+uu

uruuuruuur．所以2115()()3636EBECABACABAC⋅=−⋅−+uuuruuuruuuruuuruuuruuur22251193618ABACABAC=−−+⋅uuuruuuruuuruuur222511360793618

=−×−×+×=−．···················································7分22221412||()369369EBABACABACABAC=−=+−⋅uuuruuuruuuruu

uruuuruuuruuur2241236059369=×+×−×=；···································9分222151255||()369369ECABACABACABAC=−+=+−

⋅uuuruuuruuuruuuruuuruuuruuur221255360269369=×+×−×=．····························11分设ECuuur与EBuuur的夹角为θ，则77130cos130526EBECEBECθ⋅−

===−×⋅uuuruuuruuuruuur．·········12分20．（1）所有的样本点为：1234AAAA，1234AAAA，1234AAAA，1234AAAA；1234AAAA，1234AAAA，1234AAAA，1234AAAA；1234AAAA，1234AAAA，1234A

AAA，1234AAAA；1234AAAA，1234AAAA，1234AAAA，1234AAAA．·······································4分注：少一个样本点不得分．EDBAC高一数学答案第4页共6页（2）用B表示事件“李明通过面

试”，则李明恰好答对3道题或答对4道题．所以12341234123412341234()()()()()BAAAAAAAAAAAAAAAAAAAA=UUUU．·················································

··············································································6分由互斥事件的概率加法公式与互为独立事件的概率乘法公式，得12341234123412341234()()()()()()PBPA

AAAPAAAAPAAAAPAAAAPAAAA=++++··········································································································

·····················8分1114113112111111111123452345234523452345=×××+×××+×××+×××+×××·······························································

····················································10分111111(43211)2345120=××××++++=．················································12分注

：（2）严格按步骤评分．21．（本小题满分12分）证明：（1）如图，设G为PA的中点，连接GE，GF．又F为PB的中点，所以GF为PAB△的中位线．所以GFABP，且12GFAB=．·················1分矩形ABCD中，点E是CD的中点

，故ECABP，且12ECAB=．··········································································2分所以GFECP，且GFEC

=．所以四边形GFCE为平行四边形．···································································3分所以CFEGP．····························

·····································································4分又CF⊄平面PAE，EG⊂平面PAE，所以CFP平面PAE．··········

···········6分GFPECAB高一数学答案第5页共6页（2）设H为AE的中点，连接PH，BH．在矩形ABCD中，24ABAD==，点E是CD的中点，故2PAPE==，90APE∠=°，故PHAE⊥，且2PH=．····················8分在ABH△中，2AH

=，4AB=，45HAB∠=°，由余弦定理，得2222cos45BHABAHABAH=+−⋅⋅°2162242102=+−×××=．······························································9分在

PHB△中，23PB=，2PH=，10BH=，故222PBPHBH=+，故PHBH⊥．··························································10分又PHAE⊥，BHAEH=I，BH⊂平面ABCE，AE⊂平面ABCE，故PH⊥

平面ABCE．······················································································11分又PH⊂平面PAE，故平面PAE⊥平面ABCE．·····

···································12分22．（本小题满分12分）解：（1）根据题意，[70k∈，100)，按组距为5可分成6个小区间，分别是[70，75)，[75，80)，[80，85)，[85，90)，[90，95)，[95，100)

．因为70100k<�，由55(1)nkn<+�，*n∈N，·················································1分所以14n=，15，16，17，18，19．····································

·································2分每个小区间对应的频率值分别是20339,14,15,16,56052,17,18,19.nnnYan−−�=�=��⋅=�所以3(14151639)5(8+4+2)160a×++

−+=．························································3分解得1100a=．·········································································

·····························4分FPHECBA高一数学答案第6页共6页（2）由（1）中的数据，得[85k∈，90)的频率为20171250.4100−××=；[90k∈，95)的频率为20181250.2100−××=；[95k∈，100]的频

率为20191250.1100−××=．利用按比例分配分层随机抽样抽取的7件产品中，[85k∈，90)的有4件，分别记作1A，2A，3A，4A；[90k∈，100)的有3件，分别记作1B，2B，3B．

················6分从抽取的7件产品中任取2件产品，则样本空间{Ω=12AA，13AA，14AA，11AB，12AB，13AB，23AA，24AA，21AB，22AB，23AB，34AA，31AB，3

2AB，33AB，41AB，42AB，43AB，12BB，13BB，23}BB，所以()21nΩ=．·························································7分事件A

=“随机抽取的2件产品中至少有一件A级品”，则{A=11AB，12AB，13AB，21AB，22AB，23AB，31AB，32AB，33AB，41AB，42AB，43AB，12BB，13BB，23}BB，所以()15nA=．由古典概型公式，得()155()()217nPAnAΩ===．······

··························8分（3）[70k∈，75)的频率为3143916020×−=；[75k∈，80)的频率为3153916010×−=；[80k∈，85)的频率为3163936020×−=；[85k∈，90)的频率为0.4；[90

k∈，95)的频率为0.2；[95k∈，100]的频率为0.1．·········································································9分11372.577.582.587.50.492.50.

297.50.1201020k=×+×+×+×+×+×·························································································

····································10分87=．·································································································

············12分