2024届四川省雅安市高三下学期第三次诊断性考试数学(文)答案

2024-05-09 · 4页 · 407.3 K

雅安市高2021级第三次诊断考试数学(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=}#}

VIP会员专享最低仅需0.2元/天

VIP会员免费下载,付费最高可省50%

开通VIP

导出为Word

图片预览模式

文字预览模式
版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报
预览说明:图片预览排版和原文档一致,但图片尺寸过小时会导致预览不清晰,文字预览已重新排版并隐藏图片
相关精选
查看更多
更多推荐