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)413.414.1515.316.29三、解答题(本大题共6小题，共70分)117.(1)∵bacosCc，21由正弦定理得sinBsinAcosCsinCsin(AC)sinAcosCcosAsinC，21∴sinCcosAsinC.∵sinC0，21∴cosA，∴A.··················································6分23(2)由已知得ABACcbcosA3，故bc6.由余弦定理得a2b2c22bccosAb2c2bc2bcbcbc6，当且仅当bc6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200.02810300.03010400.02210500.0201033.4，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0.02810a250.0300.5，解得a32.3.··························6分（2）补全的22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4004008002800145295105255所以K29.3096.635．400400550250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，数学（文科）答案1/4{#{QQABaYSAggiAQJJAARgCQQkACAMQkAGAAKoOQBAMMAAAiRFABCA=}#}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，311在正PAC中，PM，又因为DMBC，222510在RtPDM中，PD2PM2DM2，PD，223PM2310sinPDM.PD10102310直线PD与平面ABCD所成角的正弦值为.························12分1020.（1）由已知得，.········································1分kMFy2c2c设，则203，．··················3分k2cy2cMF10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，，，，，，数学（文科）答案2/4{#{QQABaYSAggiAQJJAARgCQQkACAMQkAGAAKoOQBAMMAAAiRFABCA=}#}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，．·························································4分（2）依题意，由，得，即，················································6分设，，则设，则，当时，，单调递增；所以在上，，且，当a0，即a时，h'(x)0，h(x)在0,上单调递减，h(x)h(0)，不符合22题意，舍去，······························································8分②当，即时，（I）若，即，数学（文科）答案3/4{#{QQABaYSAggiAQJJAARgCQQkACAMQkAGAAKoOQBAMMAAAiRFABCA=}#}，使得，当时，，h(x)在内单调递减，，不符合题意，舍去，········································10分（II）若，即，恒成立，在上单调递增，则，符合题意．综上，实数的取值范围为．···········································12分22.（1）曲线是以为圆心的半圆，所以半圆的极坐标方程为，曲线以为圆心的圆，转换为极坐标方程为.故半圆，圆的极坐标方程分别为：，5分（2）由（1）得：.点到直线的距离.所以其中，当时，的面积的最大值为4.···········10分23.（1）当时，不等式为，当时，可以化为，解得；当时，可以化为，得；当时，可以化为，解得，不等式不成立；综上,可得不等式的解集为·································5分（2）当时，当时等号成立，由可得（舍）或，故,由柯西不等式可得，即得当且仅当时，即时取等号．·······························10分数学（文科）答案4/4{#{QQABaYSAggiAQJJAARgCQQkACAMQkAGAAKoOQBAMMAAAiRFABCA=}#}