数学（文科）答案1/4雅安市高2021级第三次诊断性考试数学（文科）参考答案一、选择题（本大题共12小题，每小题5分，共60分）BDDDACCBCABD二、填空题(本大题共4小题，每小题5分，共20分)13.414.1515.316.294三、解答题(本大题共6小题，共70分)17.(1) cCab21cos，由正弦定理得CACACACCABsincoscossin)sin(sin21cossinsin，∴CACsincossin21. 0sinC，∴21cosA，∴3A.··················································6分(2)由已知得3cosAcbACAB，故6bc.由余弦定理得62cos222222bcbcbcbccbAbccba，当且仅当6cb时，a有最小值6.····································12分18.（1）每月使用跑腿服务不低于5次的消费者年龄的平均数为200.02810300.03010400.02210500.0201033.4，设每月使用同城配送服务不低于5的消费者年龄的中位数为a，则0.02810250.0300.5a，解得32.3a.··························6分（2）补全的22列联表如下：年龄段I年龄段II合计使用同城配送服务频率高145105250使用同城配送服务频率低255295550合计400400800所以228001452951052559.3096.635400400550250K．所以，有99%的把握认为同城配送服务的使用频率高低与年龄段有关．·············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，{#{QQABaYSAggiAQJJAARgCQQkACAMQkAGAAKoOQBAMMAAAiRFABCA=}#}数学（文科）答案2/4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）因为，，所以，，取的中点，连接，则，因为平面，所以，又所以平面，·································7分PDM为直线PD与平面ABCD所成的角，在正PAC中，32PM，又因为1122DMBC，在RtPDM中，22251022PDPMDMPD，，33102sin10102PMPDMPD.直线PD与平面ABCD所成角的正弦值为31010.························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，，，，，，{#{QQABaYSAggiAQJJAARgCQQkACAMQkAGAAKoOQBAMMAAAiRFABCA=}#}数学（文科）答案3/4所以；·························································5分②当直线的斜率不为0时，如图，设的方程：，，．由，得．则，.········································7分,·····························10分又，所以．···············································11分综上，．所以，，成等差数列.···························12分21.(1)依题意，，令，得，因为，所以当时，，在上单调递减；当时，，故在上单调递增；当时，有解，综上，．·················...