【文档说明】福建省莆田市2020-2021学年高一下学期期末质量监测数学试题答案.pdf，共(7)页，293.806 KB，由小赞的店铺上传

转载请保留链接：https://www.doc5u.com/view-7cc0efd8532be082f9f9c6aa58a333ae.html

以下为本文档部分文字说明：

高一数学试题参考答案第1页（共7页）莆田市2020-2021学年下学期期末质量监测高一数学试题参考解答及评分标准评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则．2．对计算题，当考生的解

答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，

表示考生正确做到这一步应得的累加分数．4．只给整数分数．单项选择题和单空填空题不给中间分．一、选择题：本题考查基础知识和基本运算．每小题5分，满分40分．1．A2．C3．B4．B5．A6．D7．D8．

B二、选择题：本题考查基础知识和基本运算．每小题5分，满分20分．（本题为多项选择题，每小题中，全部选对的得5分，部分选对的得2分，有选错的得0分）9．AC10．AB11．ABD12．ACD三、填空题：本题考查基础知识和基本运算．每小题5

分，满分20分．13．7014．2315．211416．5652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．本小题主要考查平面向量的数量积、平面向量夹角与模的运算等基础知识；考查推理论证能力、运算求解能力等；考

查化归与转化思想、函数与方程思想等；考查数学运算、逻辑推理等核心素养，体现基础性、综合性．满分10分．解：（1）由221abab，得222321aabb．··································2分又1a，2b，所以

2341ab，则1ab．···························3分所以12cos22abab．··························································

·5分因为0,π，所以π4．······························································6分（2）因为222244abaabb·····························

····························8分1485，····························································9分所以25ab．········

··································································10分18．本小题主要考查直线与平面平行和垂直的判定与性质、平面与平面平行和垂直的判定与性

质等基础知识；考查空间想象能力、推理论证能力等；考查化归与转化思想等；考查直观想象、逻辑推理等核心素养，体现基础性、综合性．满分12分．解：证法一：高一数学试题参考答案第2页（共7页）（1）在BC上取点P，使2BPPC，连接DP，AP．········

························1分因为2=BDDE，所以∥DPCE，23=DPCE．·······································2分又因为∥AFCE，23=AFCE，所以∥AFDP，AFDP．·····················

··········································3分所以四边形AFDP为平行四边形．所以∥DFAP．····································································

·············4分又AP平面ABC，DF平面ABC，所以∥DF平面ABC．································································

·······6分（2）因为平面ACEF平面ABC,AFAC,平面ACEF平面ABC=AC,所以AF平面ABC．·······································································8分又

AP平面ABC，所以⊥AFAP．··················································10分由（1）知∥DFAP，∥CEAF．·······················································11分所以

DFCE．··············································································12分证法二：（1）在CE上取点Q，使2CQQE，连接DQ，FQ，则AFCQ．····················

·······································1分又∥AFCQ，所以四边形AFQC为平行四边形．所以∥FQAC．····································

·············································2分因为FQ平面ABC，AC平面ABC，所以∥FQ平面ABC．·····························································

··········3分因为2BDDE，2CQQE，所以∥DQBC.同理，∥DQ平面ABC．····································································

4分又FQ，DQ平面DFQ，FQDQQ，所以平面∥DFQ平面ABC．·······························································5分因为DF平面DFQ，所以∥DF

平面ABC．······································6分（2）因为平面ACEF平面ABC,AFAC,且平面ACEF平面ABC=AC,高一数学试题参考答案第3页（共7页）所以AF平面ABC．··········

·····························································8分因为AC，BC平面ABC，所以AFAC，AFBC．·····················9分由（1）知∥AFCE，∥FQAC，

∥DQBC，所以⊥CEFQ，⊥CEDQ．······························································10分又FQDQQ，FQ，DQ平面DFQ，所以⊥CE平面DFQ．·······

·······························································11分因为DF平面DFQ，所以⊥CEDF．························

······················12分19．本小题主要考查正弦定理、余弦定理、三角形面积公式及三角恒等变换等基础知识；考查推理论证能力、运算求解能力等；考查化归与转化思想、函数与方程思想等；

考查数学运算、逻辑推理等核心素养，体现基础性、综合性．满分12分．解：（1）解法一：选①mcoscos，AB，n2，bca，且mn，由mn得2coscos0bcAaB．····································

···········1分由正弦定理得sin2sincossincos0BCAAB，····························2分即sincoscossin2cossin0ABABAC．所以sin2cossin0ABAC．······

···············································3分又πABC，所以sinsinABC．所以sin2cossin0CAC．显然sin0C，所以1cos2A．···

············5分因为0π＜＜A，所以π3A．·······························································6分解法二：选②coscos23cossinaAaBCbAC，因为πABC，所以coscos

ABC．····································1分则coscos23cossinaBCaBCbAC．由正弦定理得sincossincos23cossi

nsinABCABCABC．·································································································

······2分即sincoscossinsincoscossinsinABCBCBCBC23cossinsinABC，高一数学试题参考答案第4页（共7页）所以2sinsinsin23cossinsinABCABC.································

········4分显然sinsin0BC，所以sin3cosAA，则tan3A.·····················5分因为0π＜＜A，所以π3A．·····················································

··········6分解法三：选③22(sinsin)sinsinsinBCBAC，由已知得222sin2sinsinsinsinsinsinBBCCABC，即222sinsinsinsinsinBCABC．·························

······················2分由正弦定理得222bcabc．···························································3分由余弦

定理得2221cos222bcabcAbcbc．·······································5分因为0π＜＜A，所以π3A．······················

·········································6分（2）解法一：由余弦定理得2222cosabcbcA．····························

·······7分所以22π42cos3bcbc，即224bcbc，则243bcbc．····8分显然0＞b，0＞c，因为22≤bcbc，·······································

··········9分所以222243324≥bcbcbcbcbc．·······················10分则216≤bc，4≤bc，当且仅当2bc时等号成立．··

·················11分所以6≤abc，即△ABC周长的最大值为6．···································12分解法二：由正弦定理得sinsinsinabcABC．································

·······7分因为2a，π3A，所以243πsinsin3sin3bcBC．则43sin3bB，43sin3cC．······················································8分又2πsinsi

n3CB，高一数学试题参考答案第5页（共7页）所以△ABC周长43432πsinsin2333abcBB··················9分434331sincossin223sin2cos

23322BBBBB31π4sincos24sin2226BBB．·································10分因为2π03＜＜B，所以ππ5π666＜＜B，所以πsin16

≤B，当且仅当ππ62=B，即π3=B时等号成立．·············11分则6≤abc，即△ABC周长的最大值为6．······································12分20．本小题主要考查独立事件、互斥事件的概率、古典

概型等基础知识；考查运算求解能力、推理论证能力等；考查统计与概率思想、分类与整合思想、化归与转化思想等；考查数学运算、逻辑推理、数学建模、数学抽象等核心素养，体现基础性、应用性．满分12分．解：（1）记“选择方

案一，甲获胜”为事件A，“第i局甲获胜”为事件iA（1,2,3i），则1A，2A，3A两两相互独立，且12123123AAAAAAAAA．················2分因为12AA，123AAA，123AAA为互斥事件，所以12123123A

AAAPAPAPAPAA··································3分12123123PAPAAPAAAPPPPPAA····4分222121222033333332

7．所以选择方案一，甲获胜的概率为2027．··················································6分（2）记硬币正面朝上为1，反面朝上为0．掷3枚硬币，样本空间为123,,|0,1;1

,2,3ixxxxi，包含8个等可能的样本点．···································································8分记“掷3枚硬币，恰有

2枚正面朝上”为事件B．则1,1,0,1,0,1,0,1,1B，························································10分所以3182PB．··········

·······························································11分高一数学试题参考答案第6页（共7页）因此方案二被选择的可能性更大．················································

········12分21．本小题主要考查直线与平面、平面与平面垂直的判定与性质、异面直线所成的角、三棱锥的体积等基础知识；考查空间想象能力、推理论证能力、运算求解能力等；考查函数与方程思想、化归与转化思想等；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性．满分1

2分．解：（1）证明：Rt△ABC中，ADDB，所以PDDB．又F为线段PB的中点，所以DFPB．··············································1分在Rt△ABC中，D，E分别是AB，AC的中点，所以∥DEBC．因

为90B，所以DEAB．则DEPD，DEDB．又PD，DB平面PDB，PDDBD，所以DE平面PDB．····························································

···········2分因为DF平面PDB，所以DEDF．··············································3分因为∥DEBC，所以DFBC．··························

·······························4分因为PB，BC平面PBC，PBBCB，所以DF平面PBC．·······································································5分（2）因

为PBDEEPBDVV，·········································································6分又DE平面PDB，112DEBC，又111sin332△EPBD

PDBVSDEPDDBPDBDE11331sinsin62PDBPDB，····································7分所以11sin22PDB，即sin1PDB．所以90PDB，即B

DPD．·······················································8分又BDDE，PD，DE平面PDE，PDDED，所以BD平面PDE．延长DE到G，使得DGBC，因为∥DGBC，所以四边形BDGC为平行四

边形．所以∥BDCG．················································································9分则PCG为直线BD与PC所成角或其补角．···································

·····10分因为BD平面PDE，又PG平面PDE，所以BDPG，则CGPG．·························································

11分在Rt△PDG中，3PD，2DGBC，所以7PG．高一数学试题参考答案第7页（共7页）在Rt△PGC中，3CGBD，所以721tan33PGPCGCG．即直线BD与PC所成角的正切值为213．·······································

····12分22．本小题主要考查频率分布直方图、平均数等基础知识；考查运算求解能力、推理论证能力、应用意识、创新意识等；考查统计与概率思想、分类与整合思想等；考查数据分析、数学运算等核心素养，体现综合性、应用性、创新性．满分12分．解：（1）由频率分布直方图得20(

0.010.0130.0070.002)1a，即0.018a.···················································································2分所以100.2300.26500.36700.14

900.0427.8189.83.641.2．故估计该市年龄100岁及以下居民的平均年龄为41.2岁．·······················4分（2）依题意可得，该市各

年龄分组区间居民知晓防骗知识的人数如表所示．年龄分组区间知晓人数（万人）[0,20)3000.20.3420.4[20,40)3000.260.4535.1[40,60)3000.360.5458.32[60,80)3000.140.6527.3[80

,100]3000.040.748.88····································································································9分（各年龄分组区间居

民知晓防骗知识的人数每计算正确1个得1分，共5分.）所以估计该市年龄100岁及以下居民对防骗知识的知晓率为20.435.158.3227.38.88150100%100%50%300300．··········10分（

3）本题结论开放，只要考生能从统计学的角度做出合理的分析即可．如：①一次调查未必能客观反映总体；②样本容量过小也可能影响估计的准确性；③调查的样本不一定具代表性，也会对估计产生影响；④对于敏感性问题，被调查者不一定会

提供真实信息．等等.············12分