2021届高三文科数学《大题精练》 (10)

这是一份2021届高三文科数学《大题精练》 (10)，共11页。试卷主要包含了选修等内容，欢迎下载使用。
2021届高三数学（文）“大题精练”10  1.（本小题满分12分）已知数列满足，，设．（1）求数列的通项公式；（2）若，求数列的前项和．  2.（本小题满分12分）如图，四棱柱的底面为菱形，．（1）证明：平面；（2）设，若平面，
求三棱锥的体积．   3.（本小题满分12分）世界互联网大会是由中国倡导并每年在浙江省嘉兴市桐乡乌镇举办的世界性互联网盛会，大会旨在搭建中国与世界互联互通的国际平台和国际互联网共享共治的中国平台，让各国在争议中求共识、在共识中谋合作、在合作中创共赢．2019年10月20日至22日，第六届世界互联网大会如期举行，为了大会顺利召开，组委会特招募了1 000名志愿者．某部门为了了解志愿者的基本情况，调查了其中100名志愿者的年龄，得到了他们年龄的中位数为34岁，年龄在岁内的人数为15，并根据调查结果画出如图所示的频率分布直方图：（1）求，的值并估算出志愿者的平均年龄（同一组的数据用该组区间的中点值代表）；（2）这次大会志愿者主要通过现场报名和登录大会官网报名，即现场和网络两种方式报名调查．这100位志愿者的报名方式部分数据如下表所示，完善下面的表格，通过计算说明能否在犯错误的概率不超过0.001的前提下，认为“选择哪种报名方式与性别有关系”？ 男性女性总计现场报名  50网络报名31  总计 50 参考公式及数据：，其中.0.050.010.0050.0013.8416.6357.87910.828    4.（本小题满分12分）已知．（1）当时，求曲线在处的切线方程；（2）若存在，使得成立，求的取值范围． 5.（本小题满分12分）已知椭圆()的离心率为，以的短轴为直径的圆与直线相切．（1）求的方程；（2）直线交椭圆于，两点，且．已知上存在点，使得是以为顶角的等腰直角三角形．若在直线右下方，求的值．   （二）选考题：共10分．请考生在第22，23两题中任选一题作答．如果多做，则按所做第一个题目计分，作答时请用2B铅笔在答题卡上将所选题号后的方框涂黑．6.（本小题满分10分）选修：坐标系与参数方程已知直角坐标系中，曲线的参数方程为（为参数）．以为极点，轴的正半轴为极轴，建立极坐标系，曲线的极坐标方程为．（1）写出的普通方程和的直角坐标方程；（2）设点为上的任意一点，求到距离的取值范围． 7.（本小题满分10分）选修：不等式选讲已知，且．（1）求的取值范围；（2）求证：．2021届高三数学（文）“大题精练”10（答案解析）  1.（本小题满分12分）已知数列满足，，设．（1）求数列的通项公式；（2）若，求数列的前项和．【解析】（1）因为，所以，·······································1分又因为，所以，即，····················································3分所以为等差数列，················································4分其首项为，公差．···············································5分所以．·······················································7分（2）由（1）及题设得，，········································8分所以数列的前项和······························································9分··························································11分．························································12分2.（本小题满分12分）如图，四棱柱的底面为菱形，．（1）证明：平面；（2）设，若平面，
求三棱锥的体积． 【解析】（1）证明：依题意，，且，∴，·························································1分∴四边形是平行四边形，·········································2分∴，·························································3分∵平面，平面，∴平面．······················································5分（2）依题意，，在中，，······················································6分所以三棱锥的体积．···························································8分由（1）知平面，∴··························································10分·····················································11分．···················································12分3.（本小题满分12分）世界互联网大会是由中国倡导并每年在浙江省嘉兴市桐乡乌镇举办的世界性互联网盛会，大会旨在搭建中国与世界互联互通的国际平台和国际互联网共享共治的中国平台，让各国在争议中求共识、在共识中谋合作、在合作中创共赢．2019年10月20日至22日，第六届世界互联网大会如期举行，为了大会顺利召开，组委会特招募了1 000名志愿者．某部门为了了解志愿者的基本情况，调查了其中100名志愿者的年龄，得到了他们年龄的中位数为34岁，年龄在岁内的人数为15，并根据调查结果画出如图所示的频率分布直方图：（1）求，的值并估算出志愿者的平均年龄（同一组的数据用该组区间的中点值代表）；（2）这次大会志愿者主要通过现场报名和登录大会官网报名，即现场和网络两种方式报名调查．这100位志愿者的报名方式部分数据如下表所示，完善下面的表格，通过计算说明能否在犯错误的概率不超过0.001的前提下，认为“选择哪种报名方式与性别有关系”？ 男性女性总计现场报名  50网络报名31  总计 50 参考公式及数据：，其中.0.050.010.0050.0013.8416.6357.87910.828 【解析】（1）因为志愿者年龄在内的人数为，所以志愿者年龄在内的频率为：；··································1分由频率分布直方图得：，即，①·······················································3分由中位数为可得，即，②·······················································4分由①②解得，.·················································5分志愿者的平均年龄为（岁）．······················································7分（2）根据题意得到列联表： 男性女性总计现场报名网络报名总计····························································9分所以的观测值，··························································11分所以不能在犯错误的概率不超过的前提下，认为选择哪种报名方式与性别有关系．12分说明：第（1）小题中，方程①②列对一个给2分，两个都列对给3分．4.（本小题满分12分）已知．（1）当时，求曲线在处的切线方程；（2）若存在，使得成立，求的取值范围．【解析】．······················································1分（1）当时，，所以，·························································3分所以曲线在处的切线方程为，即．····································5分（2）存在，使得成立，等价于不等式在有解．···········································6分设，则，······················································7分当时，，为增函数；当时，，为减函数．·····························8分又，，故·····················································10分所以当时，，·················································11分所以，即的取值范围为．········································12分5.（本小题满分12分）已知椭圆()的离心率为，以的短轴为直径的圆与直线相切．（1）求的方程；（2）直线交椭圆于，两点，且．已知上存在点，使得是以为顶角的等腰直角三角形．若在直线右下方，求的值．【解析】（1）依题意，，········································2分因为离心率，所以，解得，··················································4分所以椭圆的标准方程为．·········································5分（2）因为直线的倾斜角为，且是以为顶角的等腰直角三角形，在直线右下方，所以轴． 6分过作的垂线，垂足为，则为线段的中点，所以，故，····················7分所以，即，整理得．①···································8分由得.所以，解得，····················································9分所以，②，③························································10分由①-②得，，④将④代入②得，⑤··············································11分将④⑤代入③得，解得．综上，的值为．···············································12分（二）选考题：共10分．请考生在第22，23两题中任选一题作答．如果多做，则按所做第一个题目计分，作答时请用2B铅笔在答题卡上将所选题号后的方框涂黑．6.（本小题满分10分）选修：坐标系与参数方程已知直角坐标系中，曲线的参数方程为（为参数）．以为极点，轴的正半轴为极轴，建立极坐标系，曲线的极坐标方程为．（1）写出的普通方程和的直角坐标方程；（2）设点为上的任意一点，求到距离的取值范围．【解析】（1）的普通方程为，即．···································2分曲线的直角坐标方程为，即．········································5分（2）由（1）知，是以为圆心，半径的圆，·····························6分圆心到的距离，··················································7分所以直线与圆相离，到曲线距离的最小值为；最大值，····················9分所以到曲线距离的取值范围为．·····································10分7.（本小题满分10分）选修：不等式选讲已知，且．（1）求的取值范围；（2）求证：．【解析】（1）依题意，，故．·······································1分所以，·························································3分所以，即的取值范围为．···········································5分（2）因为，所以···························································7分···········································8分．·········································9分又因为，所以．························································10分