【文档说明】黑龙江省齐齐哈尔市2020-2021学年高一下学期期末考试数学答案.pdf，共(4)页，1.379 MB，由管理员店铺上传

转载请保留链接：https://www.doc5u.com/view-a65ec7cc5a4cbfdfadfa16a61ec87961.html

以下为本文档部分文字说明：

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BBAABcossinsinsinsin2······················································2分0sins

in0sin,sin0BABABA，，21cosB·······················································3分3B······

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

················································5分选②BbcasincsinCsinA)(，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.（1

2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321254

1)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,∴∠DAB=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EPBEDOS

DOVVEBPEBPDDBPE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009

0602，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86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),(a2，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GBO

GOGB二面角BPAC的余弦值为33··················································································································6分（2）过点N作B

MNQ//交CP于点NQANBMANQ，3CMCPCM在MCB中，33360cos22222MCBCMCBCBM)1(32BM设yPBPNMBQ

NyBPBN1，)1(3)1(,32yQNyBN···········································8分在NBA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yyyNQ

ANAQ②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）