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高一数学试卷参考答案及评分标准第1页（共4页）2020-2021学年度下学期期末质量监测高一数学试卷参考答案及评分标准一.单选题：本大题共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.题号12345678答案ADCCBCAD

二.多选题：本大题共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，部分选对得2分.题号9101112答案ACDCDBCAC三.填空题：本大题共4小题，每小题5分，共20分.把正确答案写在答题卡相应题的横

线上.13、114、22815、9445（注：第一空2分；第二空3分）16、25四.解答题：共70分，解答应写出文字说明、解答过程或演算步骤.17.（10分）解：选①BabtansinA2，由正弦定理可得BBAABcossinsin

sinsin2······················································2分0sinsin0sin,sin0BABABA，，21cos

B·······················································3分3B·····································································

··········································································································5分选②Bbc

asincsinCsinA)(，由正弦定理可得acbca222·················································2分212cos222acbcaB····································

·······································································································3分30BB·

·····················································································································································

·5分（2）acBacSABC43sin21·················································································································

··················6分acacacaccabB221cos222····························································

······································8分243bS2max43bSca时当且仅当······················································

···········································9分23432bb·························································

·························································································10分高一数学试卷参考答案及评分标准第2页（共4页）18.（12分）解：（1）设事件A为甲

第一轮活动中猜对1个谜语，事件B为乙第一轮活动中猜对1个谜语事件C为“星队”在第一轮活动中猜对1个谜语······································································

······························1分则PBPAP)(43)(，PBPAP1)(41)(，···········································································

·············2分则BABAC，因为：BABA与互斥，且A与B，A与B分别相互独立·······································3分所以3212541)1(43)()()()()()()

(PPPBPAPBPAPBAPBAPCPP的值为32··········································································

···························································································6分（2）设事件21AA，分别为甲两轮猜对1个，2个谜语，事件21BB，分别为乙两轮猜对1

个，2个谜语.事件D为两轮活动“星队”猜对3个谜语······························································································

··················7分则，，169)43()(8341432)(221APAP.94)32()(9431322)(221BPBP，························

···················································································9分则，1221BABAD且21BA与12BA互斥，1A与2B，2A与1B分别相互独立，········

·····························10分所以)()()()()()()(12211221BPAPBPAPBAPBAPDP.125941699483······················11

分因此，“星队”在两轮活动中猜对3个谜语的概率是.125·················································································12分19.（12分）解：（1）证明:取BP中点F，

连结FE，FA，∵E是CP中点CBEF21//····················································································1分∵AD//BC,∠ABC=900,∴∠DA

B=900∵AD=AB=1,∴BD=2,∠ADB=450∴∠DBC=450又∵CD=2,∴∠CDB=90°∴CB=��2+��2=2CBAD21//·············································

······························································································································3分ADEF/

/∴四边形ADEF为平行四边形∴DE//AF·································································································5分又∵DE平面ABP，AF⊂平

面ABP∴DE//平面ABP·····························································································6分（2）取BC中点O，

连DO∴DO⊥CB∵平面PBC⊥平面ABCD，平面PBC∩平面ABCD=CB，∴DO⊥平面PBC∴DO为三棱锥D-EBP的高·································································

···········································9分又∵BC=BP=2∴BE⊥CP∴BE=12CP=2··········································

··········································································10分∴31)2221(131)21(3131EPBEDOSDOVVEBPEB

PDDBPE三棱锥三棱锥三棱锥E-DBP的体积为31··········································································

·································································12分高一数学试卷参考答案及评分标准第3页（共4页）20.（12分）解：（1）连EC

，由余弦定理可得EC2=ED2＋DC2－2ED·DC·cos120°=3，∴EC=3·····························2分由DC=ED,∠CDE=120°∴∠ECD=30°∵∠DCF=90°∴∠ECF=60°····

····························································3分在ECF中,CFECEFCFECECF2cos222∴0632

CFCF∴32),(3或舍CF∴道路CF的长度为km32·········································································

·································································6分（2）设∠FCA=α，在△CFA中，由正弦定理可得42332)60sin(sin0ACFACCF)60sin(40AC

·····································································································································

··············8分cos)60sin(4cos,//9000ACABCABCFABABCDCAB3)602sin(232sincos23cossin2cos3202

AB·······························10分0000024060260900∴当0090602，即015时，AB取最大值为32故AB两地的最大距离为km)32

(························································································································12分21.（12分）解：(1)估计该组数据的众数为

8628884········································································································

···········2分该组数据的频率统计为因为0.08+0.12+0.15=0.35<0.5，0.08+0.12+0.15+0.3=0.65>0.5,所以中位数落在区间[84,88)内，设中位数为，x由，5.0)84(075.035.0x得到8

6x,因此中位数为86·················································································································

························6分(2)因为[72,76)与[76,80)的党员人数的比值为2∶3,采用分层随机抽样法抽取5人,则在[72,76)中抽2人,[76,80)中抽3人.···········································

·················7分设[72,76)抽取的2人编号为a1，a2;[76,80)抽取的3人编号b1，b2，b3;则5人中任选2人进行问卷调查对应的样本空间Ω={(a1，a2),(a1，b1),(a1，b2),(a1，b3),(a2，b1),(a

2，b2),(a2，b3),(b1，b2),(b1，b3),(b2，b3)}，共有10个样本点································································································

··································································9分至少有1人成绩低于76分的有：(a1，a2),(a1，b1),(a1，b2)

,(a1，b3),(a2，b1),(a2，b2),(a2，b3)共有7个样本点所以这2人中至少有1人成绩低于76分的概率为7.0107····················································

······························12分区间[72,76)[76,80)[80,84)[84,88)[88,92)[92,96)[96,100]频率0.080.120.150.30.160.10.09高一数学试卷参考答案及评分

标准第4页（共4页）22.（12分）解：（1）取PA中点G，AC中点O,连接GO,GB,OB在PACRt中，APOGCPOGAPCP//，，在等边PAB中，APBGOGB为二面角BPAC的平面角·····

·····································································································3分AB32321CPOG23BG，26OB，

90222GOBGBOBOG33cosGBOGOGB二面角BPAC的余弦值为33······················································

····························································6分（2）过点N作BMNQ//交CP于点NQANBMANQ，3CMCPCM在MCB

中，33360cos22222MCBCMCBCBM)1(32BM设yPBPNMBQNyBPBN1，)1(3)1(,32yQNyBN···········································8分在N

BA中，yyBNABBNABAN33360cos22222)1(32yyAN)1)(1(3)1(1yPMyPQyPMPQ······························································

·····················10分在PAQRt中，22222)1()1(33yPQAPAQ①在ANQRt中，)1(

)1(3)1(3222222yyyNQANAQ②由①=②，得yy1]5241[]3231[，，yBPBN的取值范围为]5241[，·····················································

················································································12分（其他解法，请酌情给分）