数学（理科）答案1/6雅安市高2021级第三次诊断性考试数学（理科）参考答案一、选择题（本大题共12小题，每小题5分，共60分）AACDAABCDDBB二、填空题(本大题共4小题，每小题5分，共20分)13.1514.-115.29416.14三、解答题(本大题共6小题，共70分)17.（1）2211nnaSa，当12221124,2naaaaa时，，，122,nnnSa当时，两式相减得12(2)nnaan，，，数列是以2为首项，2为公比的等比数列，.··································································6分（2）由（1）可知nabnn1log12，记nnnnnbac21，∴nnnT2)1(242322321，14322)1(2423222nnnT，两式相减得111213222)1(2121242)1(2224nnnnnnnnnT.∴12nnnT.····························································12分18．（1）补全的列联表如下：年龄段Ⅰ年龄段Ⅱ合计使用频率高150110260使用频率低250290540合计400400800···································································3分所以，{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}数学（理科）答案2/6所以有99%的把握认为跑腿服务的使用频率高低与年龄有关.······················6分（2）由数表知，利用分层抽样的方法抽取的8人中，年龄在，内的人数分别为5，3，依题意，的所有可能取值分别为为-2,0,2，所以，，，所以的分布列为：-202P所以的数学期望为.·················12分19.（1）证明：因为，为线段的中点，所以，·························································1分在等腰梯形中，作于，则由得，所以，所以，因为，所以所以∽，所以，所以，所以，······································3分因为，，所以平面，···················································4分因为在平面内，所以，··································5分因为，在平面内，所以平面.·········6分（2）解：因为，，所以，，{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}数学（理科）答案3/6取的中点，连接，则，因为平面，所以，又所以平面，··································7分所以如图，以为原点，以所在的直线为轴，以所在的直线为轴，建立空间直角坐标系，·························································8分则，C(0,0,0)，令平面PCD法向量xzxyzxmCPyxmCD3330232302123，取133，，m，由（1）知平面，则平面的法向量，设二面角DPCA所成平面角为，则13133cosnmnm，所以二面角DPCA的余弦值为13133.································12分20.（1）由已知得，.··········································1分设，则322220012ccyccykkMFMF，．····················3分所以椭圆的方程为．·············································4分（2）①当直线的斜率为0时，的方程：，不妨设，，，{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}数学（理科）答案4/6，，，所以；······························...