第1页共11页三明市2024年普通高中高三毕业班质量检测数学参考答案及评分细则评分说明：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．选择题和填空题不给中间分．一、选择题：本大题考查基础知识和基本运算．每小题5分，满分40分．1．C2．C3．D4．A5．A6．B7．B8．C二、选择题：本大题考查基础知识和基本运算．每小题6分，满分18分．全部选对的得6分，部分选对的得部分分，有选错的得0分．9．BC10．ACD11．BCD三、填空题：本大题考查基础知识和基本运算．每小题5分，满分15分．12．613．1,3e14．6,7,8,9,21（第一空2分，第二空3分）四、解答题：本大题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15.解法一：（1）证明：取BD的中点M，连接PMMC、,·······················1分 BPD△和BCD△均为等边三角形，∴BDPM，BDCM.··································································2分又PMCMM，∴BD平面CPM，·········································································3分CP又平面CPM，∴BDCP.····················································································4分（2）以M为原点，,MBMC所在直线为,xy轴，过M作平面BCD的垂线所在直线为z轴，如图所示建立空间直角坐标系，···········································5分 平面ABD平面PBD，平面ABD平面PBDBD，PM平面PBD，PMBD∴PM平面ABD. PBD△和CBD△均为等边三角形，∴3PMMCPC，60PMC，{#{QQABDYaEoggoAJJAABhCQQUgCkAQkAEAAKoGwAAIMAAAiBFABCA=}#}第2页共11页∴330,,22P，0,3,0C，1,0,0B，··············································6分∴331,,22BP，1,3,0BC.330,,22MP设平面PBC的法向量为(,,)xyzm∴0,0BPBCmm即330,2230xyzxy取1z,则3,3,1m，···································································8分平面ABD的法向量330,,22MP，·················································10分设平面ABD与平面PBC的夹角为，∴coscos,MPMPMPnnn33913313··································12分∴平面ABD与平面PBC夹角的余弦值为3913.····································13分解法二：（1）同解法一······································································4分（2）如图，取MC的中点E为原点，连接PE，过点E作//EFMB,交BC于点F,由（1）知CMBD,EFMC,又由（1）知BD平面CPM，PE又平面CPM，∴BDPE， PBD△和CBD△均为等边三角形且棱长为2，∴3PMMCPC，PEMC,BDMCM...