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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)4h

Srhh==−·············································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2

ππ2T==．······························································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==−=−uuuruuuruuur

uuuruuuruuur．············1分所以2133ADABBDABAC=+=+uuuruuuruuuruuuruuur．··········2分所以111236AEADABAC==+uuuruuuruuuruuur．··············3分所以2136EBABAEABA

C=−=−uuuruuuruuuruuuruuur；·······························································4分（2）1536ECACAEABA

C=−=−+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=−uuuruuur

uuur(1,2)=−；·······························7分1536ECABAC=−+uuuruuuruuur15(0,3)(6,0)(5,1)36=−+=−．·······8分y

xEDBCAEDBAC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77130cos130526EBECEBECθ

⋅−===−×⋅uuuruuuruuuruuur．·····················································12分解法2：因为2136EBABAC=−uuuruuuru

uur，1536ECABAC=−+uuuruuuruuur．所以2115()()3636EBECABACABAC⋅=−⋅−+uuuruuuruuuruuuruuuruuur22251193618ABACABAC=−−+⋅uuuruuuruuuruuur

222511360793618=−×−×+×=−．···················································7分22221412||()369369EBABACABACABAC=−=+−⋅uuuruuuruuur

uuuruuuruuuruuur2241236059369=×+×−×=；···································9分222151255||()369369ECABACABACABAC=−+=+−⋅uuuruuu

ruuuruuuruuuruuuruuur221255360269369=×+×−×=．····························11分设ECuuur与EBuuur的夹角为θ，则77130cos130

526EBECEBECθ⋅−===−×⋅uuuruuuruuuruuur．·········12分20．（1）所有的样本点为：1234AAAA，1234AAAA，1234AAAA，1234AAAA；1234AAAA，1234AAAA，123

4AAAA，1234AAAA；1234AAAA，1234AAAA，1234AAAA，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()()()()()()P

BPAAAAPAAAAPAAAAPAAAAPAAAA=++++·································································································

······························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12G

FAB=．·················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P

AE，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P

HAE⊥，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，32AB，33AB，41AB，42AB，43AB，12BB，13

BB，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）[70

k∈，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；[90k∈，95)的频率为

0.2；[95k∈，100]的频率为0.1．·········································································9分11372.577.582.587.50.49

2.50.297.50.1201020k=×+×+×+×+×+×··································································································

···························10分87=．·············································································································12分