发布时间 : 星期五 文章【数学】福州市2020届高三毕业班适应性练习卷 文科数学更新完毕开始阅读6deca7c1df80d4d8d15abe23482fb4daa48d1d2e
37. 在三棱锥P?ABC中，PA?底面ABC，AB?AC,AB?6,AC?8，D是线段AC上一点，
且AD?3DC．三棱锥P?ABC的各个顶点都在球O表面上，过点D作球O的截面，则所得截面圆的面积的最小值为 ．
【命题意图】本题主要考查直线与直线、直线与平面的位置关系、球体与截面等基础知识，意在考查直观想象、逻辑推理与数学运算等数学核心素养． 【答案】12?．
【解答】将三棱锥P?ABC补成直三棱柱，则三棱锥和该直三棱柱的外接球都是球O，记三角形ABC的外心为O1，设球的半径为R，PA?2x，则球心O到平面ABC的距离为x，即OO1?x，连接O1A，则O1A?1BC?5，2ABEOP所以R2?x2?25．在△ABC中，取AC的中点为E，连接O1D,O1E，则O1E?DC11AB?3，DE?AC?2， 24O1所以O1D?13．在Rt△OO1D中，OD?x2?13，由题意得到当截面与直线OD垂直时，截面面积最小，设此时截面圆的半径为，则r2?R2?OD2?x2?25??x2?13??12， 所以最小截面圆的面积为12?．
三、解答题：本大题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．第
17～21题为必考题，每个试题考生都必须作答．第22、23题为选考题，考生根据要求作答．
（一）必考题：共60分． 38. （本小题满分12分）
已知数列?an?满足a1?1，nan?1??n?1?an?n?n?1?，设bn?（1）求数列?bn?的通项公式；
（2）若cn?2bn?n，求数列?cn?的前n项和．
【命题意图】本题主要考查等差数列、等比数列等基础知识，意在考查逻辑推理、数学运算等数学核心素养．满分12分．
【解答】（1）因为bn?an． nan，所以an?nbn， ··············································· 1分 n又因为nan?1??n?1?an?n?n?1?，
所以n?n?1?bn?1??n?1?nbn?n?n?1?，即bn?1?bn?1， ······························· 3分 所以?bn?为等差数列， ·········································································· 4分 其首项为b1?a1?1，公差d?1． ···························································· 5分 所以bn?1??n?1??n． ········································································ 7分 （2）由（1）及题设得，cn?2n?n， ······················································ 8分 所以数列?cn?的前n项和
············································· 9分 Sn??2?22?23?L?2n???1?2?3?L?n? ·
2?2n?2n?1?n? ······································································ 11分 ??1?22?2n?1n2?n······································································· 12分 ??2． ·
239. （本小题满分12分）
如图，四棱柱ABCD?A1B1C1D1的底面为菱形，ACIBD?O． （1）证明：B1C∥平面A1BD； （2）设AB?AA1?2,?BAD?求三棱锥B1?A1BD的体积．
A1D1B1C1?，若A1O?平面ABCD， 3ADOBC【命题意图】本题主要考查直线与直线、直线与平面的位置关系、多面体的体积等基础知识，意在考查直观想象、逻辑推理与数学运算等数学核心素养．满分12分．
【解析】（1）证明：依题意，A1B1//AB，且AB//CD，
···················································································· 1分 ∴A1B1//CD， ·
··························································· 2分 ∴四边形A1B1CD是平行四边形， ·
···················································································· 3分 ∴B1C∥A1D，
∵B1C?平面A1BD，A1D?平面A1BD，
·········································································· 5分 ∴B1C∥平面A1BD． ·
（2）依题意，AA1?2,AO?3，
················································ 6分 在Rt△AAO中，AO?AA12?AO2?1， ·11所以三棱锥A1?BCD的体积
?1?312VA1?BCD?S△BCD?AO???2?1?3． ·??········································ 8分 1??3?433?由（1）知B1C∥平面A1BD，
············································································· 10分 ∴VB1?A1BD?VC?A1BD ·
?VA1?BCD ·············································································· 11分
?3·············································································· 12分 ． ·340. （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[40,45)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：
频率/组距 2n2m0.0200.010O20253035404550年龄/岁
（1）求m，n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”？
现场报名 网络报名 总计 男性 31 女性 50 总计 50 n(ad?bc)2参考公式及数据：K?，其中n?a?b?c?d.
(a?b)(c?d)(a?c)(b?d)2P?K2…k0? 0.05 3.841 0.01 6.635 0.005 7.879 0.001 10.828 k0
【命题意图】本题主要考查概率与统计等基础知识，考查学生的创新意识和应用意识，意在考查数学建模、数学抽象、数学运算、数据分析等数学核心素养．满分12分．
【解答】（1）因为志愿者年龄在[40,45)内的人数为15， 所以志愿者年龄在[40,45)内的频率为：
15·································· 1分 ?0.15； ·
100由频率分布直方图得：(0.020?2m?4n?0.010)?5?0.15?1，
即m?2n?0.07，① ············································································· 3分 由中位数为34可得0.020?5?2m?5?2n?(34?30)?0.5，
即5m?4n?0.2，② ············································································· 4分 由①②解得m?0.020，n?0.025. ···························································· 5分 志愿者的平均年龄为
(22.5?0.020?27.5?0.040?32.5?0.050?37.5?0.050?42.5?0.030?47.5?0.010)?5?34（岁）． ········································································ 7分
（2）根据题意得到列联表：
现场报名 网络报名 总计 男性 19 31 女性 31 19 总计 50 50 100 50 50 ········································································································ 9分 所以K2的观测值
100?(19?19?31?31)22????19?31???19?31????5.76＜10.828， ·?······· 11分 k?50?50?5050?50?50?502所以不能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． ······································································································ 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．