密云区2019-2020学年第二学期高三第二次阶段性测试

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【6月密云区高三二模语文】2020年北京市密云区2020届高三第二次阶段性测试语文试卷含答案(6.2)

【6月密云区高三二模语文】2020年北京市密云区2020届高三第二次阶段性测试语文试卷含答案(6.2)

密云区2019-2020学年第二学期第二次阶段性测试高三语文试卷2020.6考生须知:1.本试卷共8页,满分150分,考试时间150分钟。

2.考生务必将答案答在答题纸上,在试卷上作答无效。

3.考试结束后,将本试卷和答题纸一并交回。

一、本大题共6道小题,共18分。

阅读下面的材料,完成1-6题。

材料一故宫之所以“火”起来,是因为它勇于改变。

故宫在放飞自我的道路上也越走越远,陆续推出的《穿越故宫来看你》等爆款H5,可以让你立即体验故宫的魔性!用逗趣活泼的文案语言讲历史故事,让历史中的帝王生动起来,让内容更有可读性和传播性。

再结合H5这种比较具象的传播形式,故宫这一比较严肃的文化IP开始走出围墙,变得有趣、好玩,吸引了年轻人传播和互动的热情。

通过这种现代话术的解构,高冷、严肃的文化走进现代传播语境,故宫将枯燥的历史文化融入现代潮流文化,让文化有了情绪、态度,通过走近年轻群体,强化了大众对故宫的品牌感知。

除了传播内容的转变带来了故宫形象的转变之外,故宫IP年轻化的重要一步就是品牌形象的智慧化呈现,即借助现代多样化的传播技术手段打造故宫IP,强化大家对品牌的日常场景感知。

故宫和腾讯长期合作,除了用H5的形式活化故宫,还推出了有故宫IP元素的表情包、游戏等泛文娱作品,让故宫文物和各种元素重新“活”起来。

比如,在游戏《天天爱消除》里还原金水桥等故宫知名建筑景观等。

游戏《奇迹暖暖》分别以《清代皇后冬朝服》《十二美人图》以及养心殿文物为主题进行还原与再创作。

此外,腾讯地图和故宫博物院联合推出“玩转故宫”微信小程序1.0,以“轻应用”玩转“大故宫”,以“新方法”连接“新公众”。

腾讯地图基于位置的场景化服务与真实世界中的故宫连接起来,把真实的景点客观还原到手机地图上,用科技手段让游客体验不一样的故宫。

以深厚的文化底蕴为载体,借助新媒体传播手段,故宫将品牌感知融入人们生活的点点滴滴,不断更迭的互联网场景,就像一把钥匙,打开了历史的大门。

密云区2019-2020学年第二学期高三第二次阶段性测试20200602

密云区2019-2020学年第二学期高三第二次阶段性测试20200602

密云区2019-2020学年第二学期高三第二次阶段性测试数学试卷 2020.6一、选择题:本大题共10小题,每小题4分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项. 1.已知集合{|0}M x x =∈R ≥,N M ⊆,则在下列集合中符合条件的集合N 可能是 A. {0,1} B. 2{|1}x x = C. 2{|0}x x > D. R2.在下列函数中,定义域为实数集的偶函数为A.sin y x =B.cos y x =C.||y x x =D. ln ||y x = 3. 已知x y >,则下列各不等式中一定成立的是 A .22x y >B .11x y>C .11()()33x y >D .332x y -+>4.已知函数()y f x =满足(1)2()f x f x +=,且(5)3(3)4f f =+,则(4)f = A .16 B .8C .4D . 25.已知双曲线221(0)x y a a-=>的一条渐近线方程为20x y +=,则其离心率为C.D. 6.已知平面向量和a b ,则“||||=-b a b ”是“1()02-=g b a a ”的 A.充分而不必要条件 B.必要而不充分条件 C.充分必要条件 D.既不充分也不必要条件7.已知圆22:(1)2C x y +-=,若点P 在圆C 上,并且点P 到直线y x =的距离为2,则满足条件的点P 的个数为A .1B .2C .3D .48.设函数1()sin()2f x x ωϕ=+,x ∈R ,其中0ω>,||ϕ<π.若51()82f π=,()08f 11π=,且()f x 的最小正周期大于2π,则A .13ω=,24ϕ11π=-B .23ω=,12ϕπ= C .13ω=,24ϕ7π= D .23ω=,12ϕ11π=-9. 某三棱锥的三视图如图所示,则该三棱锥中最长的棱长为 AB .2C. D.10. 已知函数()f x 的定义域为 ,且满足下列三个条件:①对任意的 ,且 ,都有 ;② ;③是偶函数;若,,(2020)c f =,则 ,, 的大小关系正确的是 A .a b c << B .C .D .第9题图11主视图1俯视图2二、填空题:本大题共5小题,每小题5分,共25分.11.抛物线2()y mx m =为常数过点(1,1)-,则抛物线的焦点坐标为_______.12.在61()x x+的展开式中,常数项为_______.(用数字作答).13. 已知n S 是数列{n a }的前n 项和,且211(*)n S n n n =-∈N ,则1a =_________,n S 的最小值为_______. 14. 在ABC V 中,三边长分别为4a =,5b =,6c =,则ABC V 的最大内角的余弦值为_________,ABC V 的面积为_______.15. 已知集合22{,,A a a x y x y ==-∈∈Z Z}.给出如下四个结论: ①2A ∉,且3A ∈;②如果{|21,}B b b m m ==-∈N*,那么B A ⊆;③如果{|22,}C c c n n ==+∈N*,那么对于c C ∀∈,则有c A ∈;④如果1a A ∈,2a A ∈,那么12a a A ∈. 其中,正确结论的序号是__________.三、解答题: 本大题共6小题,共85分.解答应写出文字说明, 演算步骤或证明过程. 16.(本小题满分14分)如图,直三棱柱111ABC A B C -中,112AC BC AA ==,D 是棱1AA 的中点,1DC BD ⊥. (Ⅰ)证明:1DC BC ⊥;(Ⅱ)求二面角11A BD C --的大小.17.(本小题满分15分) 已知函数 .(Ⅰ)求函数的单调递增区间和最小正周期;(Ⅱ)若当π[0,]2x ∈时,关于x 的不等式()f x m ≥_______,求实数的取值范围.请选择①和②中的一个条件,补全问题(Ⅱ),并求解.其中,①有解;②恒成立. 注意:如果选择①和②两个条件解答,以解答过程中书写在前面的情况计分.18.(本小题满分14分)某健身机构统计了去年该机构所有消费者的消费金额(单位:元),如图所示:(Ⅰ)将去年的消费金额超过3200元的消费者称为“健身达人”,现从所有“健身达人”中随机抽取2人,求至少有1位消费者,其去年的消费金额超过4000元的概率;(Ⅱ)针对这些消费者,该健身机构今年欲实施入会制.规定:消费金额为2000元、2700元和3200元的消费者分别为普通会员、银卡会员和金卡会员.预计去年消费金额在(0,1600]、(1600,3200]、(3200,4800]内的消费者今年都将会分别申请办理普通会员、银卡会员和金卡会员.消费者在申请办理会员时,需一次性预先缴清相应等级的消费金额.该健身机构在今年年底将针对这些消费者举办消费返利活动,预设有如下两种方案:方案 按分层抽样从普通会员,银卡会员,金卡会员中总共抽取25位“幸运之星”给予奖励.其中,普通会员、银C 1 ABC A 1B 1第16题图D(800,1600] (1600,2400] (2400,3200] (4000,4800] (3200,4000] 消费金额/元 人数卡会员和金卡会员中的“幸运之星”每人分别奖励500元、600元和元.方案2 每位会员均可参加摸奖游戏,游戏规则如下:从一个装有3个白球、2个红球(球只有颜色不同)的箱子中,有放回地摸三次球,每次只能摸一个球.若摸到红球的总数为2,则可获得200元奖励金;若摸到红球的总数为3,则可获得300元奖励金;其他情况不给予奖励.如果每位普通会员均可参加1次摸奖游戏;每位银卡会员均可参加2次摸奖游戏;每位金卡会员均可参加3次摸奖游戏(每次摸奖的结果相互独立). 以方案的奖励金的数学期望为依据,请你预测哪一种方案投资较少?并说明理由.19.(本小题满分14分)已知椭圆:过点(1,2P ,设它的左、右焦点分别为,,左顶点为,上顶点为,.(Ⅰ)求椭圆C 的标准方程和离心率;(Ⅱ)过点6(,0)5Q -作不与轴垂直的直线交椭圆于,(异于点)两点,试判断的大小是否为定值,并说明理由.20.(本小题满分14分)已知函数()ln ,f x x a x a =-∈R .(Ⅰ)当1a =时,求曲线()f x 在1x =处的切线方程;(Ⅱ)设函数1()()ah x f x x+=+,试判断函数()h x 是否存在最小值,若存在,求出最小值,若不存在,请说明理由. (Ⅲ)当0x >时,写出ln x x 与2x x -的大小关系.21.(本小题满分14分)设n 为正整数,集合A =12{|(,,,),{0,1},1,2,,}n k t t t t k n αα=∈=L L .对于集合A 中的任意元素12(,,,)n x x x α=L 和12(,,,)n y y y β=L ,记111122221(,)[(||)(||)(||)]2n n n n M x y x y x y x y x y x y αβ=+-++-+++-+++L .(Ⅰ)当n =3时,若(0,1,1)α=,(0,0,1)β=,求(,)M αα和(,)M αβ的值; (Ⅱ)当4n =时,对于A 中的任意两个不同的元素,αβ,证明:(,)(,)(,)M M M αβααββ+≤.(Ⅲ)给定不小于2的正整数n ,设B 是A 的子集,且满足:对于B 中的任意两个不同元素α,β,(,)(,)(,)M M M αβααββ=+.写出一个集合B ,使其元素个数最多,并说明理由.。

2024-2025学年人教A版(2019)高三地理下册阶段测试试卷8

2024-2025学年人教A版(2019)高三地理下册阶段测试试卷8

2024-2025学年人教A版(2019)高三地理下册阶段测试试卷8考试试卷考试范围:全部知识点;考试时间:120分钟学校:______ 姓名:______ 班级:______ 考号:______总分栏题号一二三四五六总分得分评卷人得分一、选择题(共6题,共12分)1、读两区域示意图.甲、乙两地附近的气候状况是A. 甲地受信风、山脉的影响,形成热带雨林气候B. 甲地深受洋流、山脉的影响,气候带呈南北狭长分布C. 乙地主要受海陆热力性质差异的影响,夏季降水丰沛D. 乙地受东北信风、山脉的影响,形成热带沙漠气候2、【题文】读图,判断下列说法正确的是①由图甲可知,目前我国进口的石油主要来源于中东地区和俄罗斯②图乙我国构建的东北、西北、西南、东南四大油气进口通道中,目前石油运输量最大的是东南通道③我国石油储备基地——大连、黄岛、舟山、镇海,共同的区位优势是海运便利、靠近市场④我国西北油气资源进口通道的源地是俄罗斯A. ①②B. ②③C. ③④D. ①③3、某跨国公司专为黑种人配方而生产的化妆品,下列国家最有可能有销售市场的是()A. 沙特阿拉伯B. 印度C. 日本D. 俄罗斯4、修建水库主要影响的水循环环节是A. 植物蒸腾B. 大气降水C. 地表径流D. 水汽输送5、据观测,喜马拉雅山目前仍在以一定的速度抬升,其原因是A. 亚欧板块与印度洋板块相互碰撞B. 亚欧板块与非洲板块相互碰撞C. 亚欧板块与太平洋板块相互碰撞D. 亚欧板块与美洲板块相互碰撞6、一般情况下,海水中的浮游植物数量与营养盐、光照、水温呈正相关,但在不同的季节、海域,影响浮游植物生长繁殖的主导因素不同。

如图示意长江口附近海域某年8月浮游植物密度的水平分布,据此完成9~11题。

夏季图示海域浮游植物密度自西向东()A. 递减B. 先减后增C. 先增后减D. 递增评卷人得分二、填空题(共7题,共14分)7、在可持续发展复合系统中,生态持续发展是基础,经济持续发展是目的..8、(2013秋•红花岗区校级期中)读图,完成:(1)写出 C点的地理坐标C .(2)判断方向:B在D点的方向,C在A点的方向(2)①~④点,属于中纬度的是.(3)E~H四点中,属于西半球的是.(4)A点的对跖点的地理坐标是.9、为防止全球进一步变暖,应完全禁止燃烧化石燃料,并大面积植树造林..10、图为“亚洲冬夏季风示意图”,读图回答下列各题(1)表示亚洲夏季的是(甲、乙)图,判断理由是:(2)读图完成下表:图名比较项目甲乙气压中心名称A地风向A地降水(多、少)(3)东亚季风形成的主要原因是什么?11、读2012年4月12日8时和13日8时某区域海平面等压线图(单位百帕),完成下列各题.(1)图中 a、b、c、d四地,风向变化最大的是.(2)简析该时段内甲地天气变化情况.12、(2012秋•临渭区校级期中)读北半球某地区的海平面等压线图,回答下列问题.(1)影响甲处的天气系统是.在该系统的控制下,甲处的天气特点是.(2)图中①处的风向为.①处与⑤处相比,风力较大的是处.(3)过若干小时,②将受(天气系统)影响,若此天气系统出现在我国的冬季,受其影响的地区可能出现等天气现象.A.天气晴朗B.大风C.气温下降D.气温升高(4)图中属于暖气团控制的地点是①、②、③、④四处中的处.(5)图中②、③、④处有可能出现连续性降水天气的是处.(6)当台风中心位于厦门的东部时,厦门吹风,台风中心的天气为.13、阅读下面某地区地层剖面图,回答下列问题:(5分)(1)说出该地区前后共经历过哪些地质过程。

江苏省扬州市2019—2020学年度第二学期高三历史阶段性检测(二)(含答案)

江苏省扬州市2019—2020学年度第二学期高三历史阶段性检测(二)(含答案)

2019—2020学年度第二学期阶段性检测(二)高三历史注意事项考生在答题前请认真阅读本注意事项及各题答题要求1.本试卷包含选择题(第1题-第20题,共20题)、非选择题(第21题-第25题,共5题)两部分。

本次考试满分为120分,考试时间为100分钟。

考试结束后,请将答题纸交回。

2.答题前,请务必将自己的姓名、考试号等用书写黑色字迹的0.5毫米签字笔填写在答题纸上。

3.请认真核对答题纸表头规定填写或填涂的项目是否准确。

4.作答非选择题必须用书写黑色字迹的0.5毫米签字笔写在答题纸上的指定位置,在其它位置作答一律无效。

第Ⅰ卷(选择题共60分)一、选择题:本大题共20小题,每小题3分,共计60分。

在每小题列出的四个选项中,只有一项是最符合题目要求的。

1.《诗经·大雅》祭祀乐歌有谓:“文王在上,於昭于天。

周虽旧邦,其命维新。

……文王陟(升)降,在帝左右。

……文王孙子,本支百世。

凡周之士,不显亦世。

”对此理解有误的是A.早期政治中神权与王权结合B.按血缘关系来分配政治权力C.“家天下”的局面开始出现D.体现了现实主义的艺术风格2.右侧邮票是汉代画像石中的农事图。

此图可以用来说明当时A.土地公有制下的集体劳作B.精耕细作农业的不断发展C.男耕女织的生产劳作状态D.曲辕犁已经普及全国各地3.“夫其为物,厥美可珍。

廉方有则,体洁性贞。

含章蕴藻,实好斯文。

取彼之弊,以为此新。

”关于材料所赞叹的技术成果,说法正确的是A.有利于信息的记录与传播B.唐代经海路外传至阿拉伯C.直接推动了西欧思想解放D.两汉时使用已经十分普遍4.《续资治通鉴·宋宁宗嘉定十一年》记载:“(金)户部尚书瓜勒佳必喇为翰林学士承旨、权参知政事,行省于辽东。

”陆游在《纵笔》一诗中也写道:“行省当年驻陇头,腐儒随牒亦西游。

”有人据此认为行省制度开始于宋代。

下列说法最为合理的是A.年代久远导致行省制度确立时间莫衷一是B.只通过历史文献就可以准确还原历史真相C.以诗证史的方法可证明宋代确立行省制度D.元代的行省制存在对前代地方官制的继承5.下图是《货郎图》(局部),描绘了一位老货郎挑满玩具百货杂物的挑担抵达村头,众多妇女儿童争购围观的热闹场面,表现了南宋时钱塘一带的风土人情,具有浓厚的生活情趣。

2019-2020学年河南省实验中学高三生物第二次联考试题及答案解析

2019-2020学年河南省实验中学高三生物第二次联考试题及答案解析

2019-2020学年河南省实验中学高三生物第二次联考试题及答案解析一、选择题:本题共15小题,每小题2分,共30分。

每小题只有一个选项符合题目要求。

1.溶酶体是细胞的“消化车间”,内部含有多种水解酶。

下列叙述错误的是A. 溶酶体执行功能时会伴随其膜组分的更新B. 溶酶体通过胞吞作用吞噬侵入细胞的病毒或病菌C. 衰老或损伤的细胞器可被溶酶体消化分解D. 溶酶体属于生物膜系统,能合成水解酶执行其功能2.大豆子叶的颜色受一对等位基因控制,基因型为AA的个体呈深绿色,基因型为Aa的个体呈浅绿色,基因型为aa的个体呈黄色,黄色个体在幼苗阶段死亡。

下列说法错误的是()A. 浅绿色植株连续自交n代,成熟后代中深绿色个体的概率为(2n-1)/(2n+1)B. 浅绿色植株自由交配n代,成熟后代中深绿色个体的概率为n/(n+2)C. 浅绿色植株与深绿色植株杂交,其成熟后代有深绿色和浅绿色,且比例为l:1D. 浅绿色植株自花传粉,不会产生黄色子代3.下列选项不能说明神经系统分级调节的是()A. 指尖采血时,针刺指尖不能引起缩手反射B. 运动员听到枪声时迅速起跑C. 司机看见路人过斑马线时停车等候D. 婴儿膀胱充盈时,引起膀胱排尿4.研究人员找到一种抗体,可让急性髓性白血病细胞成熟为树突状细胞。

树突状细胞如长时间暴露于该抗体下,再加上特定的培养条件,还能进一步分化为与自然杀伤(NK)细胞(该细胞识别靶细胞是非特异性的)高度相似的细胞——诱导NK细胞。

下列相关叙述中,不正确的是()A. 树突状细胞在特异性免疫中的作用是摄取和呈递抗原B. 在二次免疫中,抗体可由浆细胞与记忆B细胞合成并分泌C. 细胞毒性T细胞识别抗原具有特异性,而诱导NK细胞则无特异性D. 该诱导过程可能涉及到细胞基因的选择性表达5.“人类基因组计划”对于人类深入了解自身的基因结构和功能有重要意义,其目的是测定人类基因组的()A.mRNA的碱基序列B.DNA的碱基序列C.tRNA的碱基序列D.rRNA的碱基序列6.下列是对“一对相对性状的杂交实验”中性状分离现象的各项假设性解释,其中错误的是()A.生物的性状是由细胞中的遗传因子决定的B.体细胞中的遗传因子成对存在,互不融合C.在配子中只含有每对遗传因子中的一个D.生物的雌雄配子数量相等,且随机结合7.如图表示生长素浓度与植物生长的关系,P点和Q点可分别表示()A.植株横放时根弯曲部位的近地侧和远地侧B.植株横放时茎弯曲部位的远地侧和近地侧C.具顶端优势植株的顶芽部位和侧芽部位D.胚芽鞘向光弯曲生长时尖端下部的向光侧和背光侧8.图甲、图乙表示基因型为AaBb的某雄性二倍体生物一次减数分裂不同时期的图像,图丙为染色体7的放大图,下列相关叙述正确的是()A.该减数分裂只发生了交叉互换导致基因重组,而没有发生自由组合B.图甲为次级精母细胞,基因型为aaBb,图乙表示精细胞,基因型为ABC.该减数分裂基因B、b一定是在进行减数第一次分裂时发生分离D.甲中有2个染色体组,图乙中有2个染色体组9.某科研小组探究pH对月鳢胰蛋白酶活性的影响,结果如图所示,下列有关叙述正确的是()A.pH过低和过高时胰蛋白酶活性降低的原因相同B.pH过高或过低会促使胰蛋白酶水解成氨基酸,从而使其失活C.该实验可利用双缩脲试剂检测因pH过高导致的胰蛋白酶被破坏的程度D.pH由过低升高到过高时,酶的活性先升高后降低10.下丘脑是调节内脏活动和内分泌活动的较高级神经中枢。

密云区2019-2020学年第二学期高三第二次阶段性测试答案20200602

密云区2019-2020学年第二学期高三第二次阶段性测试答案20200602

密云区2019-2020学年第二学期高三第二次阶段性测试数学试卷 2020.6一、选择题:本大题共10小题,每小题4分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项.1.已知集合{|0}M x x =∈R ≥,N M ⊆,则在下列集合中符合条件的集合N 可能是 A. {0,1} B. 2{|1}x x = C. 2{|0}x x > D. R2.在下列函数中,定义域为实数集的偶函数为A.sin y x =B.cos y x =C.||y x x =D. ln ||y x =3. 已知x y >,则下列各不等式中一定成立的是A .22x y >B .11x y> C .11()()33x y >D .332x y -+>4.已知函数()y f x =满足(1)2()f x f x +=,且(5)3(3)4f f =+,则(4)f = A .16 B .8 C .4 D . 25.已知双曲线221(0)x y a a-=>的一条渐近线方程为20x y +=,则其离心率为 A.52 B.174 C. 32 D. 1546.已知平面向量和a b ,则“||||=-b a b ”是“1()02-=g b a a ”的 A.充分而不必要条件 B.必要而不充分条件 C.充分必要条件 D.既不充分也不必要条件7.已知圆22:(1)2C x y +-=,若点P 在圆C 上,并且点P 到直线y x =的距离为22,则满足条件的点P 的个数为A .1B .2C .3D .48.设函数1()sin()2f x x ωϕ=+,x ∈R ,其中0ω>,||ϕ<π.若51()82f π=,()08f 11π=,且()f x 的最小正周期大于2π,则 A .13ω=,24ϕ11π=-B .23ω=,12ϕπ= C .13ω=,24ϕ7π= D .23ω=,12ϕ11π=-9. 某三棱锥的三视图如图所示,则该三棱锥中最长的棱长为1A .2B .2C .22D .2310. 已知函数()f x 的定义域为 ,且满足下列三个条件:①对任意的 ,且,都有;② ;③ 是偶函数;若,,(2020)c f =,则 ,, 的大小关系正确的是 A .a b c << B .C .D .二、填空题:本大题共5小题,每小题5分,共25分.11.抛物线2()y mx m =为常数过点(1,1)-,则抛物线的焦点坐标为_______.12.在61()x x+的展开式中,常数项为_______.(用数字作答).13. 已知n S 是数列{n a }的前n 项和,且211(*)n S n n n =-∈N ,则1a =_________,n S 的最小值为_______.14. 在ABC V 中,三边长分别为4a =,5b =,6c =,则ABC V 的最大内角的余弦值为_________,ABC V 的面积为_______.15. 已知集合22{,,A a a x y x y ==-∈∈Z Z}.给出如下四个结论: ①2A ∉,且3A ∈;②如果{|21,}B b b m m ==-∈N*,那么B A ⊆;③如果{|22,}C c c n n ==+∈N*,那么对于c C ∀∈,则有c A ∈; ④如果1a A ∈,2a A ∈,那么12a a A ∈. 其中,正确结论的序号是__________.三、解答题: 本大题共6小题,共85分.解答应写出文字说明, 演算步骤或证明过程.16.(本小题满分14分)C 1A 1B 1如图,直三棱柱111ABC A B C -中,112AC BC AA ==,D 是棱1AA 的中点,1DC BD ⊥. (Ⅰ)证明:1DC BC ⊥;(Ⅱ)求二面角11A BD C --的大小.17.(本小题满分15分)已知函数 .(Ⅰ)求函数的单调递增区间和最小正周期;(Ⅱ)若当π[0,]2x ∈时,关于x 的不等式()f x m ≥_______,求实数的取值范围.请选择①和②中的一个条件,补全问题(Ⅱ),并求解.其中,①有解;②恒成立. 注意:如果选择①和②两个条件解答,以解答过程中书写在前面的情况计分.18.(本小题满分14分)某健身机构统计了去年该机构所有消费者的消费金额(单位:元),如图所示:(Ⅰ)将去年的消费金额超过3200元的消费者称为“健身达人”,现从所有“健身达人”中随机抽取2人,求至少有1位消费者,其去年的消费金额超过4000元的概率;(Ⅱ)针对这些消费者,该健身机构今年欲实施入会制.规定:消费金额为2000元、2700元和3200元的消费者分别为普通会员、银卡会员和金卡会员.预计去年消费金额在(0,1600]、(1600,3200]、(3200,4800]内的消费者今年都将会分别申请办理普通会员、银卡会员和金卡会员.消费者在申请办理会员时,需一次性预先缴清相应等级的消费金额.该健身机构在今年年底将针对这些消费者举办消费返利活动,预设有如下两种方案:方案 按分层抽样从普通会员,银卡会员,金卡会员中总共抽取25位“幸运之星”给予奖励.其中,普通会员、银卡会员和金卡会员中的“幸运之星”每人分别奖励500元、600元和元.方案2 每位会员均可参加摸奖游戏,游戏规则如下:从一个装有3个白球、2个红球(球只有颜色不同)的箱子中,有放回地摸三次球,每次只能摸一个球.若摸到红球的总数为2,则可获得200元奖励金;若摸到红球的总数为3,则可获得300元奖励金;其他情况不给予奖励.如果每位普通会员均可参加1次摸奖游戏;每位银卡会员均可参加2次摸奖游戏;每位金卡会员均可参加3次摸奖游戏(每次摸奖的结果相互独立).以方案的奖励金的数学期望为依据,请你预测哪一种方案投资较少?并说明理由.19.(本小题满分14分) 已知椭圆:过点3(1,)2P ,设它的左、右焦点分别为,,左顶点为,(800,1600] 40 30 20 10 0[0,800](1600,2400] (2400,3200] (4000,4800](3200,4000] 820253584消费金额/元人数上顶点为,且满足.(Ⅰ)求椭圆C 的标准方程和离心率;(Ⅱ)过点6(,0)5Q -作不与轴垂直的直线交椭圆于,(异于点)两点,试判断 的大小是否为定值,并说明理由.20.(本小题满分14分)已知函数()ln ,f x x a x a =-∈R .(Ⅰ)当1a =时,求曲线()f x 在1x =处的切线方程; (Ⅱ)设函数1()()ah x f x x+=+,试判断函数()h x 是否存在最小值,若存在,求出最小值,若不存在,请说明理由.(Ⅲ)当0x >时,写出ln x x 与2x x -的大小关系.21.(本小题满分14分)设n 为正整数,集合A =12{|(,,,),{0,1},1,2,,}n k t t t t k n αα=∈=L L .对于集合A 中的任意元素12(,,,)n x x x α=L 和12(,,,)n y y y β=L ,记111122221(,)[(||)(||)(||)]2n n n n M x y x y x y x y x y x y αβ=+-++-+++-+++L .(Ⅰ)当n =3时,若(0,1,1)α=,(0,0,1)β=,求(,)M αα和(,)M αβ的值; (Ⅱ)当4n =时,对于A 中的任意两个不同的元素,αβ,证明:(,)(,)(,)M M M αβααββ+≤.(Ⅲ)给定不小于2的正整数n ,设B 是A 的子集,且满足:对于B 中的任意两个不同元素α,β,(,)(,)(,)M M M αβααββ=+.写出一个集合B ,使其元素个数最多,并说明理由.(考生务必将答案答在答题卡上,在试卷上作答无效)密云区2019-2020学年第二学期高三第二次阶段性测试数学试卷参考答案 2020.6一、选择题:共10小题,每小题4分,共40分.题号 1 2 3 4 5 6 7 8 9 10 答案ABDBACCBDD二、填空题:共5小题,每小题5分,共25分.11.1(,0)4- 12.20 13.10-;30- 14.18;157415. ①②④. 备注:(1)若小题有两问,第一问3分,第二问2分;(2)第15题答案为①②④之一,3分;为①②④之二,4分;为①②④,5分;其它答案0分.三、解答题:共6小题,共85分.解答应写出文字说明,演算步骤或证明过程. 16.(本小题满分14分)(Ⅰ)证明:在直三棱柱111ABC A B C -中,侧面11ACC A 为矩形.因为112AC BC AA ==,D 是棱1AA 的中点,所以ADC ∆和11A DC ∆均为等腰直角三角形.所以o1145ADC A DC ∠=∠=. 因此o190C DC ∠=,即1C D DC ⊥. 因为1DC BD ⊥,BD DC D =I , 所以1DC ⊥平面BCD . 因为BC ⊂平面BCD ,所以1DC BC ⊥.(Ⅱ)解:因为1CC ⊥平面ABC ,AC ⊂平面ABC ,BC ⊂平面ABC ,所以1CC AC ⊥,1CC BC ⊥. 又因为1DC BC ⊥,111CC DC C =I , 所以BC ⊥平面11ACC A .因为AC ⊂平面11ACC A ,所以BC AC ⊥ 以C 为原点建立空间直角坐标系,如图所示. 不妨设1AC =,则(0,0,0)C ,(1,0,0)A ,(010)B ,,,(101)D ,,,1(102)A ,,,1(0,0,2)C , 所以1(0,0,1)A D =-u u u u r ,1(1,1,2)A B =--u u u r ,1(1,0,1)C D =-u u u u r ,1(0,1,2)C B =-u u u r. 设平面1A BD 的法向量()x y z =,,m ,由1100.A D AB ⎧⋅=⎪⎨⋅=⎪⎩u u u u r u u u r,m m 得020.z x y z -=⎧⎨-+-=⎩, 令1x =,则(1,1,0)=m .设平面1C BD 的法向量()x y z =,,n ,由1100.C D C B ⎧⋅=⎪⎨⋅=⎪⎩u u u u ru u u r ,n n 得020.x z y z -=⎧⎨-=⎩,令1x =,则(1,2,1)=n .则有1112013cos ,.||||226⋅⨯+⨯+⨯<>===⋅⨯m n m n m n因为二面角1A BD C --为锐角,C 1ABC A 1 B 1第16题图DDC 1 AB C A 1 B 1第16题图zxy所以二面角1A BD C --的大小为π6. 17. (本小题满分15分)(Ⅰ)解:因为22()=23sin cos cos sin f x x x x x +-=3sin 2cos 2x x + =π2sin(2)6x +.所以函数()f x 的最小正周期πT =. 因为函数sin y x =的的单调增区间为ππ[2π,2π],22k k k -++∈Z , 所以πππ2π22π,262k x k k -+++∈Z ≤≤, 解得ππππ,36k x k k -++∈Z ≤≤.所以函数数()f x 的的单调增区间为ππ[π,π],36k k k -++∈Z ,(Ⅱ)解:若选择①由题意可知,不等式()f x m ≥有解,即max ()m f x ≤.因为π[0,]2x ∈,所以ππ7π2666x +≤≤. 故当ππ262x +=,即π6x =时,()f x 取得最大值,且最大值为π()26f =.所以2m ≤.若选择②由题意可知,不等式()f x m ≥恒成立,即min ()m f x ≤.因为π[0,]2x ∈,所以ππ7π2666x +≤≤. 故当π7π266x +=,即π2x =时,()f x 取得最小值,且最小值为π()12f =-.所以1m -≤.18.(本小题满分14分)(Ⅰ)解:记“在抽取的2人中至少有1位消费者在去年的消费超过4000元”为事件A.由图可知,去年消费金额在(3200,4000]内的有8人,在(4000,4800]内的有4人, 消费金额超过3200元的“健身达人”共有 8+4=12(人),从这12人中抽取2人,共有212C 种不同方法,其中抽取的2人中至少含有1位消费者在去年的消费超过4000元,共有112844C C C +种不同方法.所以,()P A =11284421219=33C C C C +. (Ⅱ)解:方案1 按分层抽样从普通会员,银卡会员,金卡会员中总共抽取25位“幸运之星”,则“幸运之星”中的普通会员、银卡会员、金卡会员的人数分别为820257100+⨯=,25352515100+⨯=,12253100⨯=, 按照方案1奖励的总金额为1750015600380014900ξ=⨯+⨯+⨯=(元).方案2 设η表示参加一次摸奖游戏所获得的奖励金,则η的可能取值为0,200,300.由题意,每摸球1次,摸到红球的概率为121525C P C ==,所以03012133323281(0)()()()()5555125P C C η==+=, 21233236(200)()()55125P C η===, 3033328(300)()()55125P C η===. 所以η的分布列为:数学期望为81368020030076.8125125125E η=⨯+⨯+⨯=(元), 按照方案2奖励的总金额为2(28602123)76.814131.2ξ=+⨯+⨯⨯=(元),因为由12ξξ>,所以施行方案2投资较少.19.(本小题满分14分)(Ⅰ)解:根据题意得2222222131,4152,6.a b a b c a b c ⎧+=⎪⎪⎪+=⨯⎨⎪⎪=+⎪⎩解得2,1,3.a b c ⎧=⎪=⎨⎪=⎩所以椭圆C 的方程为2214x y +=,离心率3е2=.(Ⅱ)解:方法一因为直线不与轴垂直,所以直线的斜率不为. 设直线的方程为:65x ty =-, 联立方程226,51.4x ty x y ⎧=-⎪⎪⎨⎪+=⎪⎩化简得221264(4)0525t y ty +--=.显然点6(,0)5Q -在椭圆C 的内部,所以0∆>.设11(,)M x y ,22(,)N x y ,则122125(4)t y y t +=+,1226425(4)y y t =-+. 又因为(2,0)A -,所以11(2,)AM x y =+u u u u r ,22(2,)AN x y =+u u u r.所以1212(2)(2)AM AN x x y y =+++u u u u r u u u rg12122121222266(2)(2)55416(1)()5256441216(1)()25(4)55(4)25ty tx y y t y y t y y t t t t t =-+-++=++++=+⨯-+⨯+++=0 所以AM AN ⊥u u u u r u u u r ,即o90MAN ∠=是定值.方法二(1)当直线垂直于x 轴时 解得M 与N 的坐标为64(,)55-±.由点(2,0)A -,易证o90MAN ∠=. (2)当直线斜率存在时设直线的方程为:6(),0.5y k x k =+≠,联立方程226(),51.4y k x x y ⎧=+⎪⎪⎨⎪+=⎪⎩化简得2222484(3625)(14)0525k k x k x -+++=. B AM N Qxy显然点6(,0)5Q -在椭圆C 的内部,所以0∆>.设11(,)M x y ,22(,)N x y ,则2122485(14)k x x k +=-+,21224(3625)25(14)k x x k -=+.又因为(2,0)A -,所以11(2,)AM x y =+u u u u r ,22(2,)AN x y =+u u u r.所以1212(2)(2)AM AN x x y y =+++u u u u r u u u rg12122221212222222266(2)(2)()()55636(1)(2)()45254(3625)64836(1)(2)425(14)55(14)25x x k x k x k k x x k x x k k k k k k k =+++++=++++++--=+⨯++⨯++++=0所以AM AN ⊥u u u u r u u u r ,即o90MAN ∠=是定值.20.(本小题满分14分)(Ⅰ)解:当1a =时,()ln ,0f x x x x =->,所以1'()1,0f x x x=->,因此'(1)0k f ==. 又因为(1)1f =,所以切点为(1,1).所以切线方程为1y =.(Ⅱ)解:1()ln 0ah x x a x x a x+=-+>∈R ,,. 所以221(1)(1)'()10a a x x a h x x x x x ++--=-->=,. 因为0x >,所以10x +>. (1)当10a +≤,即a ≤-1时因为0x >,所以(1)0x a -+>,故'()0h x >.此时函数()h x 在(0,)+∞上单调递增.所以函数()h x 不存在最小值. (2)当10a +>,即a >-1时令'()0h x =,因为0x >,所以1x a =+.()h x 与'()h x 在(0,)+∞上的变化情况如下:x(0,1)a +1a +(1,)a ++∞'()h x − 0 + ()h x↘极小值↗所以当1x a =+时,()h x 有极小值,也是最小值,并且min ()(1)2ln(1)h x h a a a a =+=+-+. 综上所述,当a ≤-1时,函数()h x 不存在最小值;当1a >-时,函数()h x 有最小值2ln(1)a a a +-+.(Ⅲ)解:当0x >时,2ln x x x x -≤.21.(本小题满分14分)(Ⅰ)解:因为(0,1,1)α=,(0,0,1)β=,所以1(,)[(00|00|)(11|11|)(11|11|)]22M αα=++-+++-+++-=,1(,)[(00|00|)(10|10|)(11|11|)]22M αβ=++-+++-+++-=.(Ⅱ)证明:当4n =时,对于A 中的任意两个不同的元素,αβ,设12341234(,,,)(,,,)x x x x y y y y αβ==,,有12341234(,)(,)M x x x x M y y y y ααββ=+++=+++,.对于任意的,i i x y ,1,2,3,4i =,当i i x y ≥时,有11(||)[()]22i i i i i i i i i x y x y x y x y x ++-=++-=, 当i i x y ≤时,有11(||)[()]22i i i i i i i i i x y x y x y x y y ++-=+--=. 即1(||)max{,}2i i i i i i x y x y x y ++-=. 所以,有11223344(,)max{,}max{,}max{,}max{,}M x y x y x y x y αβ=+++. 又因为,{0,1}i i x y ∈,所以max{,}i i i i x y x y ≤+,1,2,3,4i =,当且仅当0i i x y =时等号成立. 所以,11223344max{,}max{,}max{,}max{,}x y x y x y x y +++11223344()()()()x y x y x y x y ≤+++++++ 12341234()()x x x x y y y y =+++++++,即(,)(,)(,)M M M αβααββ≤+,当且仅当0i i x y =(1,2,3,4i =)时等号成立.(Ⅲ)解:由(Ⅱ)问,可证,对于任意的123123(,,,,)(,,,,)n n x x x x y y y y αβ==L L ,,若(,)(,)(,)M M M αβααββ=+,则0i i x y =,1,2,3,,i n =L 成立. 所以,考虑设012312{(,,,,)|,0}n n A x x x x x x x =====L L , 11231{(,,,,)|1,{0,1},2,3,,}n i A x x x x x x i n ==∈=L L ,对于任意的2,3,,k n =L ,123123121{(,,,,)|(,,,,),0,1}k n n k k A x x x x x x x x A x x x x -=∈=====L L L .所以01n A A A A =U UL U .高三数学试题参考答案 第11页共11页 假设满足条件的集合B 中元素个数不少于2n +, 则至少存在两个元素在某个集合k A (1,2,,1k n =-L )中,不妨设为123123(,,,,)(,,,,)n n x x x x y y y y αβ==L L ,,则1k k x y ==. 与假设矛盾,所以满足条件的集合B 中元素个数不多于1n +. 取0(0,0,0)e =L ;对于1,2,,1k n =-L ,取123(,,,,)k n k e x x x x A =∈L ,且10k n x x +===L ;n n e A ∈. 令01{,,,}n B e e e =L ,则集合B 满足条件,且元素个数为1n +.故B 是一个满足条件且元素个数最多的集合.。

2019-2020学年人教A版四川省蓉城名校联盟高三第二学期第二次联考(理科)数学试卷 含解析

2019-2020学年人教A版四川省蓉城名校联盟高三第二学期第二次联考(理科)数学试卷 含解析

2019-2020学年高三第二学期第二次联考数学试卷(理科)一、选择题1.已知集合A={﹣1,1,3,4},集合B={x|x2﹣4x+3>0},则A∩B=()A.{﹣1,4}B.{﹣1,1,4}C.{﹣1,3,4}D.(﹣∞,1)∪(3,+∞)2.已知复数z=,则|z|=()A.1B.C.2D.33.已知实数0<a<b,则下列说法正确的是()A.>B.ac2<bc2C.lna<lnb D.()a<()b4.已知命题p:x<2m+1,q:x2﹣5x+6<0,且p是q的必要不充分条件,则实数m的取值范围为()A.m>B.m≥C.m>1D.m≥15.若数列{a n}为等差数列,且满足3+a5=a3+a8,S n为数列{a n}的前n项和,则S11=()A.27B.33C.39D.446.已知α,β是空间中两个不同的平面,m,n是空间中两条不同的直线,则下列说法正确的是()A.若m⊂α,n⊂β,且α⊥β,则m⊥nB.若m⊂α,n⊂α,且m∥β,n∥β,则α∥βC.若m⊥α,n∥β,且α⊥β,则m⊥nD.若m⊥α,n∥β,且α∥β,则m⊥n7.已知抛物线y2=20x的焦点与双曲线﹣=1(a>0,b>0)的一个焦点重合,且抛物线的准线被双曲线截得的线段长为,那么该双曲线的离心率为()A.B.C.D.8.如图,在△ABC中,=,P是BN上的一点,若m=﹣,则实数m 的值为()A.B.C.1D.29.已知实数a>0,b>1满足a+b=5,则+的最小值为()A.B.C.D.10.已知集合A={1,2,3,4,5,6}的所有三个元素的子集记为B1,B2,B3…,B n,n∈N*.记b i为集合B i中的最大元素,则b1+b2+b3+…+b n=()A.45B.105C.150D.21011.关于圆周率π,数学发展史上出现过许多很有创意的求法,如著名的浦丰实验和查理斯实验.受其启发,我们也可以通过设计下面的实验来估计π的值:先请全校m名同学每人随机写下一个都小于1的正实数对(x,y);再统计两数能与1构成钝角三形三边的数对(x,y)的个数a;最后再根据统计数a估计π的值,那么可以估计π的值约为()A.B.C.D.12.已知=(2sin,cos),=(cos,2cos),函数f(x)=•在区间[0,]上恰有3个极值点,则正实数ω的取值范围为()A.[,)B.(,]C.[,)D.(,2]二、填空题13.实数x,y满足,则z=2x+y的最大值为.14.成都市某次高三统考,成绩X经统计分析,近似服从正态分布X~N(100,σ2),且P(86<X≤100)=0.15,若该市有8000人参考,则估计成都市该次统考中成绩X大于114分的人数为.15.已知函数f(x)=﹣x3+x+a,x∈[,e]与g(x)=3lnx﹣x﹣1的图象上存在关于x轴对称的点,则a的取值范围为.16.在四面体ABCD中,AB=CD=,AC=BD=,AD=BC=5,E,F分别是AD,BC的中点.则下述结论:①四面体ABCD的体积为20;②异面直线AC,BD所成角的正弦值为;③四面体ABCD外接球的表面积为50π;④若用一个与直线EF垂直,且与四面体的每个面都相交的平面α去截该四面体,由此得到一个多边形截面,则该多边形截面面积最大值为6.其中正确的有.(填写所有正确结论的编号)三、解答题:共70分。

北京市密云区2019-2020学年第二学期高三第二次阶段性测试数学试卷及评分标准20200602

北京市密云区2019-2020学年第二学期高三第二次阶段性测试数学试卷及评分标准20200602

密云区2019-2020学年第二学期高三第二次阶段性测试数学试卷 2020.6一、选择题:本大题共10小题,每小题4分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项.1.已知集合{|0}M x x =∈R ≥,N M ⊆,则在下列集合中符合条件的集合N 可能是 A.{0,1} B.2{|1}x x = C. 2{|0}x x > D. R2.在下列函数中,定义域为实数集的偶函数为A.sin y x =B.cos y x =C.||y x x =D.ln ||y x =3.已知x y >,则下列各不等式中一定成立的是A .22x y >B .11x y> C .11()()33x y > D .332x y -+>4.已知函数()y f x =满足(1)2()f x f x +=,且(5)3(3)4f f =+,则(4)f = A .16 B .8 C .4 D . 25.已知双曲线221(0)x y a a-=>的一条渐近线方程为20x y +=,则其离心率为6.已知平面向量和a b ,则“||||=-b a b ”是“1()02-=g b a a ”的 A.充分而不必要条件B.必要而不充分条件 C.充分必要条件D.既不充分也不必要条件7.已知圆22:(1)2C x y +-=,若点P 在圆C 上,并且点P 到直线y x =的距离为2,则满足条件的点P 的个数为 A .1B .2C .3D .48.设函数1()sin()2f x x ωϕ=+,x ∈R ,其中0ω>,||ϕ<π.若51()82f π=,()08f 11π=,且()f x 的最小正周期大于2π,则 A .13ω=,24ϕ11π=-B .23ω=,12ϕπ= C .13ω=,24ϕ7π= D .23ω=,12ϕ11π=-9.某三棱锥的三视图如图所示,则该三棱锥中最长的棱长为 AB .2 C. D.10.已知函数()f x 的定义域为,且满足下列三个条件: ①对任意的,且,都有;②;③是偶函数;若,,(2020)c f =,则,,的大小关系正确的是 A .a b c <<B .C .D .二、填空题:本大题共5小题,每小题5分,共25分.11.抛物线2()y mx m =为常数过点(1,1)-,则抛物线的焦点坐标为_______.12.在61()x x+的展开式中,常数项为_______.(用数字作答).13.已知n S是数列{n a }的前n 项和,且211(*)n S n n n =-∈N ,则1a =_________,n S 的最小值为_______.14. 在ABC V 中,三边长分别为4a =,5b =,6c =,则ABC V 的最大内角的余弦值为_________,ABC V 的面积为_______.15. 已知集合22{,,A a a x y x y ==-∈∈Z Z}.给出如下四个结论: ①2A ∉,且3A ∈;②如果{|21,}B b b m m ==-∈N*,那么B A ⊆;③如果{|22,}C c c n n ==+∈N*,那么对于c C ∀∈,则有c A ∈; ④如果1a A ∈,2a A ∈,那么12a a A ∈. 其中,正确结论的序号是__________.第9题图11主视图1俯视图2三、解答题: 本大题共6小题,共85分.解答应写出文字说明, 演算步骤或证明过程. 16.(本小题满分14分)如图,直三棱柱111ABC A B C -中,112AC BC AA ==,D 是棱1AA 的中点,1DC BD ⊥.(Ⅰ)证明:1DC BC ⊥;(Ⅱ)求二面角11A BD C --的大小.17.(本小题满分15分)已知函数.(Ⅰ)求函数的单调递增区间和最小正周期;(Ⅱ)若当π[0,]2x ∈时,关于x 的不等式()f x m ≥_______,求实数的取值范围. 请选择①和②中的一个条件,补全问题(Ⅱ),并求解.其中,①有解;②恒成立. 注意:如果选择①和②两个条件解答,以解答过程中书写在前面的情况计分.18.(本小题满分14分)某健身机构统计了去年该机构所有消费者的消费金额(单位:元),如图所示:(Ⅰ)将去年的消费金额超过3200元的消费者称为“健身达人”,现从所有“健身达人”中随机抽取2人,求至少有1位消费者,其去年的消费金额超过4000元的概率; (Ⅱ)针对这些消费者,该健身机构今年欲实施入会制.规定:消费金额为2000元、2700元和3200元的消费者分别为普通会员、银卡会员和金卡会员.预计去年消费金额在(0,1600]、(1600,3200]、(3200,4800]内的消费者今年都将会分别申请办理普通会员、银卡会员和金卡会员.消费者在申请办理会员时,需一次性预先缴清相应等级的消费金额.该健身机构在今年年底将针对这些消费者举办消费返利活动,预设有如下两种方案: 方案按分层抽样从普通会员,银卡会员,金卡会员中总共抽取25位“幸运之星”给予奖励.其中,普通会员、银卡会员和金卡会员中的“幸运之星”每人分别奖励500元、600元和元.方案2每位会员均可参加摸奖游戏,游戏规则如下:从一个装有3个白球、2个红球(球只有颜色不同)的箱子中,有放回地摸三次球,每次只能摸一个球.若摸到红球的总数为2,则可获得200元奖励金;若摸到红球的总数为3,则可获得300元奖励金;其他情况不给予奖励.如果每位普通会员均可参加1次摸奖游戏;每位银卡会员C 1 A BC A 1B 1第16题图D(800,1600] (1600,2400] (2400,3200] (4000,4800](3200,4000] 消费金额/元人数均可参加2次摸奖游戏;每位金卡会员均可参加3次摸奖游戏(每次摸奖的结果相互独立).以方案的奖励金的数学期望为依据,请你预测哪一种方案投资较少?并说明理由.19.(本小题满分14分)已知椭圆:过点P ,设它的左、右焦点分别为,,左顶点为,上顶点为.(Ⅰ)求椭圆C 的标准方程和离心率;(Ⅱ)过点6(,0)5Q -作不与轴垂直的直线交椭圆于,(异于点)两点,试判断的大小是否为定值,并说明理由.20.(本小题满分14分)已知函数()ln ,f x x a x a =-∈R .(Ⅰ)当1a =时,求曲线()f x 在1x =处的切线方程; (Ⅱ)设函数1()()ah x f x x+=+,试判断函数()h x 是否存在最小值,若存在,求出最小值,若不存在,请说明理由.(Ⅲ)当0x >时,写出ln x x 与2x x -的大小关系.21.(本小题满分14分)设n 为正整数,集合A =12{|(,,,),{0,1},1,2,,}n k t t t t k n αα=∈=L L .对于集合A 中的任意元素12(,,,)n x x x α=L 和12(,,,)n y y y β=L ,记111122221(,)[(||)(||)(||)]2n n n n M x y x y x y x y x y x y αβ=+-++-+++-+++L .(Ⅰ)当n =3时,若(0,1,1)α=,(0,0,1)β=,求(,)M αα和(,)M αβ的值; (Ⅱ)当4n =时,对于A 中的任意两个不同的元素,αβ,证明:(,)(,)(,)M M M αβααββ+≤.(Ⅲ)给定不小于2的正整数n ,设B 是A 的子集,且满足:对于B 中的任意两个不同元素α,β,(,)(,)(,)M M M αβααββ=+.写出一个集合B ,使其元素个数最多,并说明理由.(考生务必将答案答在答题卡上,在试卷上作答无效)密云区2019-2020学年第二学期高三第二次阶段性测试数学试卷参考答案 2020.6一、选择题:共10小题,每小题4分,共40分.二、填空题:共5小题,每小题5分,共25分.11.1(,0)4- 12.20 13.10-;30- 14.18 15. ①②④. 备注:(1)若小题有两问,第一问3分,第二问2分;(2)第15题答案为①②④之一,3分;为①②④之二,4分;为①②④,5分;其它答案0分.三、解答题:共6小题,共85分.解答应写出文字说明,演算步骤或证明过程. 16.(本小题满分14分)(Ⅰ)证明:在直三棱柱111ABC A B C -中,侧面11ACC A 为矩形.因为112AC BC AA ==,D 是棱1AA 的中点,所以ADC ∆和11A DC ∆均为等腰直角三角形.所以o1145ADC A DC ∠=∠=.因此o190C DC ∠=,即1C D DC ⊥.因为1DC BD ⊥,BD DC D =I , 所以1DC ⊥平面BCD . 因为BC ⊂平面BCD ,所以1DC BC ⊥.(Ⅱ)解:因为1CC ⊥平面ABC ,AC ⊂平面ABC ,BC ⊂平面ABC ,所以1CC AC ⊥,1CC BC ⊥. 又因为1DC BC ⊥,111CC DC C =I , 所以BC ⊥平面11ACC A .因为AC ⊂平面11ACC A ,所以BC AC ⊥ 以C 为原点建立空间直角坐标系,如图所示. 不妨设1AC =,则(0,0,0)C ,(1,0,0)A ,(010)B ,,,(101)D ,,,1(102)A ,,,1(0,0,2)C , C 1ABC A 1 B 1所以1(0,0,1)A D =-u u u u r ,1(1,1,2)A B =--u u u r ,1(1,0,1)C D =-u u u u r ,1(0,1,2)C B =-u u u r . 设平面1A BD 的法向量()x y z =,,m ,由1100.A D AB ⎧⋅=⎪⎨⋅=⎪⎩u u u u r u u u r,m m 得020.z x y z -=⎧⎨-+-=⎩, 令1x =,则(1,1,0)=m .设平面1C BD 的法向量()x y z =,,n ,由1100.C D C B ⎧⋅=⎪⎨⋅=⎪⎩u u u u r u u u r ,n n 得020.x z y z -=⎧⎨-=⎩,令1x =,则(1,2,1)=n .则有cos ,||||⋅<>===⋅m n m n m n因为二面角1A BD C --为锐角, 所以二面角1A BD C --的大小为π6. 17. (本小题满分15分)(Ⅰ)解:因为22(cos cos sin f x x x x x +-2cos 2x x + =π2sin(2)6x +.所以函数()f x 的最小正周期πT =. 因为函数sin y x =的的单调增区间为ππ[2π,2π],22k k k -++∈Z , 所以πππ2π22π,262k x k k -+++∈Z ≤≤, 解得ππππ,36k x k k -++∈Z ≤≤.所以函数数()f x 的的单调增区间为ππ[π,π],36k k k -++∈Z ,(Ⅱ)解:若选择①由题意可知,不等式()f x m ≥有解,即max ()m f x ≤.因为π[0,]2x ∈,所以ππ7π2666x +≤≤. 故当ππ262x +=,即π6x =时,()f x 取得最大值,且最大值为π()26f =.所以2m ≤.若选择②由题意可知,不等式()f x m ≥恒成立,即min ()m f x ≤.因为π[0,]2x ∈,所以ππ7π2666x +≤≤. 故当π7π266x +=,即π2x =时,()f x 取得最小值,且最小值为π()12f =-.所以1m -≤.18.(本小题满分14分)(Ⅰ)解:记“在抽取的2人中至少有1位消费者在去年的消费超过4000元”为事件A.由图可知,去年消费金额在(3200,4000]内的有8人,在(4000,4800]内的有4人,消费金额超过3200元的“健身达人”共有 8+4=12(人),从这12人中抽取2人,共有212C 种不同方法,其中抽取的2人中至少含有1位消费者在去年的消费超过4000元,共有112844C C C +种不同方法.所以,()P A =11284421219=33C C C C +. (Ⅱ)解:方案1 按分层抽样从普通会员,银卡会员,金卡会员中总共抽取25位“幸运之星”,则“幸运之星”中的普通会员、银卡会员、金卡会员的人数分别为820257100+⨯=,25352515100+⨯=,12253100⨯=, 按照方案1奖励的总金额为1750015600380014900ξ=⨯+⨯+⨯=(元).方案2 设η表示参加一次摸奖游戏所获得的奖励金,则η的可能取值为0,200,300.由题意,每摸球1次,摸到红球的概率为121525C P C ==,所以03012133323281(0)()()()()5555125P C C η==+=, 21233236(200)()()55125P C η===, 3033328(300)()()55125P C η===. 所以η的分布列为:数学期望为81368020030076.8125125125E η=⨯+⨯+⨯=(元), 按照方案2奖励的总金额为2(28602123)76.814131.2ξ=+⨯+⨯⨯=(元),因为由12ξξ>,所以施行方案2投资较少.19.(本小题满分14分)(Ⅰ)解:根据题意得22222131,42,6.a b c a b c ⎧+=⎪⎪=⨯⎪=+⎪⎩解得2,1,a b c ⎧=⎪=⎨⎪=⎩所以椭圆C 的方程为2214x y +=,离心率е=(Ⅱ)解:方法一因为直线不与轴垂直,所以直线设直线的方程为:65x ty =-, 联立方程226,51.4x ty x y ⎧=-⎪⎪⎨⎪+=⎪⎩化简得2212(4)0525t y ty +--=.显然点6(,0)5Q -在椭圆C 的内部,所以0∆>.设11(,)M x y ,22(,)N x y ,则122125(4)t y y t +=+,1226425(4)y y t =-+. 又因为(2,0)A -,所以11(2,)AM x y =+u u u u r ,22(2,)AN x y =+u u u r.所以1212(2)(2)AM AN x x y y =+++u u u u r u u u rg12122121222266(2)(2)55416(1)()5256441216(1)()25(4)55(4)25ty tx y y t y y t y y t t t t t =-+-++=++++=+⨯-+⨯+++=0所以AM AN ⊥u u u u r u u u r,即o 90MAN ∠=是定值.方法二(1)当直线垂直于x 轴时 解得M 与N 的坐标为64(,)55-±.由点(2,0)A -,易证o 90MAN ∠=. (2)当直线斜率存在时设直线的方程为:6(),0.5y k x k =+≠,联立方程226(),51.4y k x x y ⎧=+⎪⎪⎨⎪+=⎪⎩化简得2222484(3625)(14)0525k k x k x -+++=. 显然点6(,0)5Q -在椭圆C 的内部,所以0∆>.设11(,)M x y ,22(,)N x y ,则2122485(14)k x x k +=-+,21224(3625)25(14)k x x k -=+.又因为(2,0)A -,所以11(2,)AM x y =+u u u u r ,22(2,)AN x y =+u u u r.所以1212(2)(2)AM AN x x y y =+++u u u u r u u u rg12122221212222222266(2)(2)()()55636(1)(2)()45254(3625)64836(1)(2)425(14)55(14)25x x k x k x k k x x k x x k k k k k k k =+++++=++++++--=+⨯++⨯++++=0所以AM AN ⊥u u u u r u u u r ,即o90MAN ∠=是定值.20.(本小题满分14分)(Ⅰ)解:当1a =时,()ln ,0f x x x x =->,所以1'()1,0f x x x=->,因此'(1)0k f ==. 又因为(1)1f =,所以切点为(1,1).所以切线方程为1y =.(Ⅱ)解:1()ln 0ah x x a x x a x+=-+>∈R ,,. 所以221(1)(1)'()10a a x x a h x x x x x++--=-->=,. 因为0x >,所以10x +>. (1)当10a +≤,即a ≤-1时因为0x >,所以(1)0x a -+>,故'()0h x >.此时函数()h x 在(0,)+∞上单调递增.所以函数()h x 不存在最小值. (2)当10a +>,即a >-1时令'()0h x =,因为0x >,所以1x a =+.()h x 与'()h x 在(0,)+∞上的变化情况如下:所以当1x a =+时,()h x 有极小值,也是最小值,并且min ()(1)2ln(1)h x h a a a a =+=+-+. 综上所述,当a ≤-1时,函数()h x 不存在最小值;当1a >-时,函数()h x 有最小值2ln(1)a a a +-+.(Ⅲ)解:当0x >时,2ln x x x x -≤.21.(本小题满分14分)(Ⅰ)解:因为(0,1,1)α=,(0,0,1)β=,所以1(,)[(00|00|)(11|11|)(11|11|)]22M αα=++-+++-+++-=,1(,)[(00|00|)(10|10|)(11|11|)]22M αβ=++-+++-+++-=.(Ⅱ)证明:当4n =时,对于A 中的任意两个不同的元素,αβ,设12341234(,,,)(,,,)x x x x y y y y αβ==,,有12341234(,)(,)M x x x x M y y y y ααββ=+++=+++,.对于任意的,i i x y ,1,2,3,4i =,高三数学第二次阶段性测试试题第11页共11页当i i x y ≥时,有11(||)[()]22i i i i i i i i i x y x y x y x y x ++-=++-=, 当i i x y ≤时,有11(||)[()]22i i i i i i i i i x y x y x y x y y ++-=+--=. 即1(||)max{,}2i i i i i i x y x y x y ++-=. 所以,有11223344(,)max{,}max{,}max{,}max{,}M x y x y x y x y αβ=+++. 又因为,{0,1}i i x y ∈,所以max{,}i i i i x y x y ≤+,1,2,3,4i =,当且仅当0i i x y =时等号成立. 所以,11223344max{,}max{,}max{,}max{,}x y x y x y x y +++11223344()()()()x y x y x y x y ≤+++++++ 12341234()()x x x x y y y y =+++++++,即(,)(,)(,)M M M αβααββ≤+,当且仅当0i i x y =(1,2,3,4i =)时等号成立.(Ⅲ)解:由(Ⅱ)问,可证,对于任意的123123(,,,,)(,,,,)n n x x x x y y y y αβ==L L ,,若(,)(,)(,)M M M αβααββ=+,则0i i x y =,1,2,3,,i n =L 成立. 所以,考虑设012312{(,,,,)|,0}n n A x x x x x x x =====L L , 11231{(,,,,)|1,{0,1},2,3,,}n i A x x x x x x i n ==∈=L L ,对于任意的2,3,,k n =L ,123123121{(,,,,)|(,,,,),0,1}k n n k k A x x x x x x x x A x x x x -=∈=====L L L .所以01n A A A A =U UL U .假设满足条件的集合B 中元素个数不少于2n +, 则至少存在两个元素在某个集合k A (1,2,,1k n =-L )中, 不妨设为123123(,,,,)(,,,,)n n x x x x y y y y αβ==L L ,,则1k k x y ==. 与假设矛盾,所以满足条件的集合B 中元素个数不多于1n +. 取0(0,0,0)e =L ;对于1,2,,1k n =-L ,取123(,,,,)k n k e x x x x A =∈L ,且10k n x x +===L ;n n e A ∈. 令01{,,,}n B e e e =L ,则集合B 满足条件,且元素个数为1n +.故B 是一个满足条件且元素个数最多的集合.。

2019-2020学年高三英语第二次联考试题及参考答案

2019-2020学年高三英语第二次联考试题及参考答案

2019-2020学年高三英语第二次联考试题及参考答案第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项AThailand is a country with a long and rich history. It is also one of those countries which have many traditions which modern times fortunately have not affected. Thailand is famous for its unique culture. It is well worth noting that Thai culture hasbeen handed down from one generation to the next.Thai Classical DanceThe inspiring culture includes local music and wonderful Thai dances. The dances of course have something to do with its deep-rooted Buddhist religion, fighting arts and beautiful clothing. Thai classical dance performances are generally performed by gracious (高雅的) Thai ladies wearing beautiful Thai local costumes. Most resort (旅游胜地) areas and many hotels frequently offer these Thai culture dance shows for foreign visitors.Thai GreetingThe unique Thai gesture of greeting another person, the wai, is especiallyone of the great aspects of Thai culture. The wai is when a person joins both hands to either head or chest level while bending their head slightly towards his hands. This way of greeting is especially done when a younger person greets an older person and it indicates a sign of respect to their elders. Employees would also wai their managers even if the manager would be younger than themselves.BangkokBangkok is the culture center of Thailand and has been the Thai capital since the end of the eighteenth century. Observing Thai culture in Bangkok can be great experience as the combination of modern times and traditions have created a kind of unique atmosphere. Bangkok offers a package of Thai culture which is shown by numerous beautiful Buddhist temples and many examples of modern Thai architecture.Bangkok National MuseumAnother location in Bangkok where one can enjoy and see Thai culture is at the famed Bangkok national museum, which offers tourists an opportunity to view national treasures and unique Thai art pieces with its culture feature dating back as early as the late sixteenth century.1.Thai classical dance is related to ________.A.its living level and educationB.its history and architectural styleC.its customsD.its religious belief2.In Thailand a worker uses the gesture, the wai , to greet________.A.his close friend.B.a young stranger.C.his younger colleagues.D.his young boss.3.What make Thai culture in Bangkok so unique?A.The long history and fine weather of Bangkok.B.The mixture of the modern culture and traditions.C.A number of beautiful Thai Buddhist templesD.Many examples of modern Thai architecture.BRock and pop hitmaker Jim Steinman, who wrote and composed music for Meat Loaf, Bonnie Tyler, Celine Dion, and more, died Monday in Danbury, Connecticut. He was 73. Steinman's brother Billconfirmed that the cause of death was kidney (肾) failure.A statement posted on Steinman's Facebook page read, “It's with a heavy heart that I can confirm Jim's passing. There will be much more to say in the coming hours and days as we prepare to honor this giant of a human being and his glorious legacy.”Steinman's wholly unique career found him working as a composer, lyricist, and producer for many artists in a variety of styles. According to a biography on his website, the records he's worked on have sold more than 190 million copies worldwide. He was nominated (提名) for four Grammys, and won Album of the Year for his work on Dion's 1996 smash,Falling Into You.Steinman began his career in a musical theater while in college, writing and starring in a rock musical calledThe Dream Engine, which gained the attentionof New York theatrical producer Joe Papp. After graduating, Steinman worked at the Public Theater (which was established by Papp). In 1973, Yvonne Elliman recorded Steinman's song “Happy Ending”, which became Steinman's first commercially released tune. That same year, the Public Theater staged his musicalMore Than You Deserve.One of the actors who auditioned forMore Than You Deservewas Meat Loaf, and he and Steinman soon struck up a close personal and professional relationship. The two began working on Meat Loaf's solo album,Bat Out of Hell, in the early 70s, but it wouldn't be released until 1977. It wasn't until about one year later — after Meat Loaf performed onSaturday Night Live— that the album became a hit.“There is no other songwriter ever like him,” Meat Loaf said. “I can never repay him. He has been such aninfluence, in fact, the biggest influence on my life, and I learned so much from him that there would be no way I could ever repay Mr. Jim Steinman.”4. What caused Jim Steinman's death?A. A hit on the head.B. A kidney problem.C. A failed operation.D. A lack of blood supply.5. What is the correct order of the following events?a. Meat Loaf performed onSaturday Night Live.b. Jim released his first commercial tuneHappy Ending.c. Jim and Meat Loaf released the albumBat Out of Hell.d. Jim starred in a rock musical calledThe Dream Engine.e. Jim's musicalMore Than You Deservewas put on show.f. Jim won Album of the Year for his work onFalling Into You.A. dbecafB. dbcfaeC. fdcaebD. fbceda6. Why was Jim Steinman important to Meat Loaf?A. Because Jim was an extraordinary songwriter.B. Because Jim allowed him not to repay his debt.C. Because Jim was influential in the music industry.D. Because Jim gave him friendship and career support.7. What do we learn about Jim Steinman?A. He mainly focused his career on acting in musicals.B. He is a well-known and widely respected musician.C. His closest friends were Bill, Papp and Meat Loaf.D. He won four Grammys in the course of his career.CRichard Campbell is a secondary school student. He is15 years old. He lives in a small town in the north ofEngland. Every morning, he gets up at eight o’clock, puts on his uniform and walks to school.One hour later, the lessons start. The students usually study maths, English, history and geography in the morning. They usually study music and drawing and they play sports after lunch. They have a ten-minute break between classes. They also spend a long time in the school library, reading books and doing their homework.Richard likes his school very much. His favourite subjects are English and geography, but he doesn’t like maths because he is not good at it.Richard and all his friends spend the whole day at school. Lunch is at one o’clock. He doesn’t like the food that the school serves. This is why he often brings a packed lunch from home. He always has his lunch in the dining hall, but some of his friends sometimes eat in the courtyard or outside the school gate.At weekends, he always goes out with his friends because he doesn’t go to school. On Saturday, he always goes to the cinema or to the sports centre. On Sunday, he just goes for a walk with his dog.8. What time do Richard’s lessons begin in the morning?A. At 8:00.B. At 8:30.C. At 9:00.D. At 9:10.9. What does Richard like best?A. Maths and English.B. English and geography.C. History and maths.D. Geography and history.10. Where does Richard have lunch on weekdays?A. In the dining hall.B. At home.C. In the courtyard.D. Outside his school gate.11. What does Richard always do on Saturday?A. He reads and does homework.B. He goes to see films or does sports.C. He walks his dog or does exercise.D. He spends a long time in the library.DThe Great Barrier Reef's outlook remains “very poor” despite coral (珊瑚) recovery over the past year, Australian government scientistssaid Monday, just days before a UNESCO ruling on the site's world heritage (遗产) status.The United Nations cultural agency recommended last month that the world's largest reef (珊瑚礁) system be placed on its endangered list because of damage to the corals largely caused by climate change.The Australian Institute of Marine Science (AIMS) said the corals were now in a “recovery window” after a decade of harmful heat stress and cyclones (旋风). But such opportunities were becoming rarer due to the influence ofclimate change, the government agency, which has monitored the reef for 35 years, said in its annualreport released today. “The increasing emergence of climate-related extreme weather events and starfish outbreaks is causing more severe and frequent pressures, giving the reef fewer opportunities like this to recover,”CEO Paul Hardisty said. The scientists surveyed 127 reef sites in 2021 and found hard coral cover had increased at 69 of the 81 locations surveyed in the past two years.Separate scientific research released last October found the 2, 300-kilometre (1, 400 miles) system had lost half its corals since 1995, with a series of ocean heatwaves causing mass coral death.Britta Schaffelke, research program director at AIMS, said the latest findings provided a slight hope that the reef still has the power of recovering. But she added that its future is still very poor because of the dangers of climate change and other factors that are affecting the reef.UNESCO has urged Australia to take urgent climate action but the government has long resisted calls to commit to net zero emissions (排放) by 2050. The government has said it hopes to meet the target “as soon as possible” without harming its economy, insisting dealing with climate change requires a global effort. The reef was worth about US $4. 8 billion a year in tourism for the Australian economy and there are fears that an “in danger” listing could weaken its tourist appeal.12. What is the major cause of the damage to the corals?A. The climate change.B. Lack of money.C. Over development.D. Too many tourists.13. What is mainly talked about in Paragraph 3?A. The result of the survey.B. The efforts AIMS has made.C. The slight chance of the recovery.D. The terrible situation of the climate.14. What is Britta Schaffelke's attitude towards the future of the reef?A. Unclear.B. Positive.C Intolerant. D. Anxious.15. What can we infer from the last paragraph?A. Australia wants to put the reef on the endangered list.B. The Australian government has ignored UNESCO's demand.C. Australia hopes to keep a balance between emission target and its economy.D. The Australian government refuses to take its share of responsibility of climate change.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

2019-2020学年北京第二十中学高三英语第二次联考试题及参考答案

2019-2020学年北京第二十中学高三英语第二次联考试题及参考答案

2019-2020学年北京第二十中学高三英语第二次联考试题及参考答案第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项AIf you are planning to visit the historic capital city of Scotland, Edinburgh, a travel destination that people crowd to from around the world, and want to attend one Festival while you are there, keep on reading to discover more information.AKA. Imaginate Festival When: 22 May – 2 June 2021Where: Traverse Theater, Assembly RoxyA festival where kids take overEdinburgh. With a whole range of free pop-up performances, take your kids to see some of the most inspiring theatre and dance from a whole range of talented performers.EdinburghInternational Film Festival When: 19 June – 29 June 2021Where: Film House, Festival TheaterOriginally the very best in international film, it was established in 1947. The dynamic programme features everything from documentaries to shorts, along with a range of experimental cinema, in an attractive setting with a spray of red carpet charm.EdinburghArt Festival When: 25 July – 25 August 2021Where: City ArtCenter, The Scottish GalleryWith over 40 exhibitions to attend, the Edinburgh Art Festival is theUK’s largest visual arts event where you can see everything from historical works to contemporary masterpieces.The RoyalEdinburghMilitary Tattoo When: 2 – 24 August 2021Where:EdinburghCastleWith a different theme every year, over 200,000 visitors crowd toEdinburghto see the military bands and the symbolic piper set against the backdrop ofEdinburghCastle.1. Who is the AKA. Imaginate Festival intended for?A. Children.B. Talented performers.C. Parents.D. Dancers.2. What’s special about Edinburgh Art Festival?A. It includes all forms of arts.B. It is about great works in history.C. It is the largest festival in the world.D. It lasts for the longest time.3. Which Festival offers performances by soldiers?A. Edinburgh Art FestivalB. AKA. Imaginate FestivalC. The RoyalEdinburghMilitary TattooD.EdinburghInternational Film FestivalBCraig Blackburn, a father and car fan, built a Batmobile for his son’s hope for using the vehicle to brighten the lives of sick children. And now he hopes to use it for more than just his sons hope after seeing the childrens reaction to the Batmobile.Based on the number of failures he had seen in car groups, he estimated that only about one in 50 attempted constructions was actually finished and he realized what an incredible opportunity he had.Mr. Blackburn started the project at the beginning of 2018 after hearing a friend in the US was doing the same thing. It started with importing an outer shell overseas, before picking brains of a friend who had a background as a worker in a car factory to gain knowledge of how to build the car. With the help of his friend, Mr. Blackburn built the Batmobile in 18 months with the cost reaching six figures.Mr.Blackburn hoped to add a flamethrower(喷火器)onto the back of the vehicle and said he had thought about building the more recent Batman Tumbler from the series film Dark Knight. Though Mr. Blackburn encountered plenty of difficulties to get over during the construction, in September 2019, the carmade its first show at the Carnival of Flowers in Toowoomba, before being used by Blackburn’s son for his hope.“It was great. It was so good to see the kids’ and adults’ excitement at seeing the Batmobile.” Mr. Blackburn said. As a result, the car lovers hope to make the car work on the roads as soon as possible, so he can visit sick children and take them out with his son.4. What is Craig Blackburn’s initial purpose of making the Batmobile?A. To realize his son’s dream.B. To donate it to sick children.C. To pay his respects to the film Dark Knight.D. To show off at the Carnival of Flowers in Toowoomba.5. How did Blackburn feel about the car-making at first?A. Hopeful.B. Confused.C. Impossible.D. Unsure.6. How did Blackburn’s friend help him?A. By making an outer shell for him.B. By offering him financial support.C. By sharing the knowledge of building cars.D. By telling him the background of the car factory.7. What is the car lovers’ expectation of the Batmobile?A. It will be driven soon on the roads.B. It can be displayed around the world.C. It can change the lives of sick children.D. It will appear in the next film about Batman.CIn the Hollywood industry, most of the studios are using AI to make movies for various reasons such as getting the actors out of danger, replacing the actors indifferent scenes, forming appealing atmosphere to enhance(增强)the views, etc.Directors and producers are using VFX (Visual effects), animations and AI to makea model that looks like the actor and replaces the actor's place. In particular, AI is used in scenes where actor shave to produce certain stunts (特技)to develop setups in the studio artificially. VFX is used to later change the internal studio backgroundsto a different place in the movie.There are many movies with two characters of the same actor. When AI was not a part of the film, editors used different methods to show the two aspects of the same actor in one scene, but now AI is being used to form the second character of the same actor and is being performed to the viewers.By a perfect combination of animation, VFX and AI, realistic models are being created. And the most fun part is the fictitious character can hold the face of the actor but the age, hairstyles, and clothing can be changed to create more enhanced looks according to the movie scene and story. With the help of AI, the directors recreating appealing scenes to enhance the thrill and excitement. InJurassic Park,no dinosaurs were running but with the help of AI and visual effects, we could enjoy the scenes and the atmospheres.The directors and the producers direct to form a green screen including the obstacles and those green screens get replaced with the views that are made from AI and VFX, and the actors make the scenes alive anddeliver the most suitable action-packed movie scenes. This method also enhances the viewers' experience, which makes the movie a blockbuster(大片).8. What does the author intend to tell us in paragraph 1?A. How AI helps actors.B. What AI brings to movies.C. Why AI is applied to movies.D. Where AI is made full use of.9. What can we infer from paragraph 3?A. AI can be used as an editor.B. Editors consider AI irreplaceable.C. Editors used many methods to replace AI.D. AI makes what used to be complex scenes easier.10. What's the author's attitude to using VFX, animations and AI in movies?A. Supportive.B. Sceptical.C. Unconcerned.D. Enthusiastic.11. Which of the following could be the best title for thetext?A. What AI Brings Out Hollywood MoviesB. Why AI Is Applied to Hollywood MoviesC. Which Hollywood Movies Make Much Use of AID. How AI Is Being Applied to Hollywood MoviesD36-year-old Juan Dual likes to joke that he’s empty inside. Juan’s story began when he was only 13. It was then that he was diagnosed with a terrible disease, which left him with a 99.8% chance of developing cancer of the digestive system. At age 19, right after finishing high-school, Juan underwent a tough operation to take away his colon and rectum. Sadly, it was only the beginning. By age 28, Juan’s disease had affected his stomach and gallbladder so he had to go under the knife again.Having just recovered from several serious surgeries, Juan Dual decided to accept the invitation of some friends of his parents and travel to Japan. It was there that things started to change for the better. He didn’t speak a word of Japanese, so he spent most of his time walking his dog. One day, the dog pulled harder, and Juan realized that he was still able to jog, and he started to do just that.Months later, he found himself working in a small, peaceful town in England. There was little in terms of entertainment, but the town was surrounded by hills, so he devoted even more of his time to running. He befriended some like-minded folks and told them what he’d been through, and they seemed amazed at the fact that he was still alive, let alone that he was pushing himself to exercise. That’s when the idea of focusing on motivating others took root in his mind.With the help of Pepa, a nutritionist, Juan Dual slowly relearned how to eat to keep his energy level high enough to sustain him during physical activity. Eight months after his last operation, he finished the Barcelona half marathon in two hours. He then started training for mountain running and ultra-marathons.12. Why does Juan Dual say he is empty inside?A. Because he has no desire for anything.B. Because he doesn’t have much knowledge.C. Because he always suffers from great hunger.D. Because many of his organs have been removed.13. What made Juan Dual aware that he could still run?A. His parents’ support.B. A walk with his dog.C. The idea of challenging himself.D. His quick recovery from surgeries.14. When did Juan Dual decide to inspire others with his story?A. After finishing the Barcelona half marathon.B After being introduced to a nutritionist named Pepa.C. After sharing it with his friends in an English town.D. After making friends with people with similar sufferings.15. Which of the following words can best describe Juan Dual?A. Ambitious and intelligent.B. Inspiring and responsible.C. Unfortunate but determined.D. Confident but stubborn.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

2020北京密云高三二模试题(地理)

2020北京密云高三二模试题(地理)

密云区2019-2020学年度第二学期二模考试高三年级地理试卷第Ⅰ卷 选择题(每题3分,共45分)气象学上入春,入夏,入秋,入冬皆有实际数值上的标准,若连续5日平均温度稳定在10℃,即为入春,高于22℃为入夏,10℃-22℃之间为春秋,低于10℃为入冬。

图1为2020年3月17日全国入春进程图,读图完成1题。

1.以下叙述正确的是A .处于春季的地区均位于我国地势的第三级阶梯B .纬度是影响东部地区入春进程差异的主要因素C .处于冬季的地区均位于我国的非季风区D .影响武汉和拉萨入春差异的因素是降水图2为“我国东北地区局部沿46°N 地形剖面与1月平均气温分布图”。

据此完成2、3题。

图 2图12.图中气温最低点的海拔约为A.600 m B.850 m C.1000 m D.1350 m3.图中1月份平均气温曲线最大峰值出现的主要原因是,该处A.臭氧聚集,吸收紫外线增温 B.为暖锋过境以后,气温上升快C.人类活动密集,热岛效应强D.冬季风背风坡,气流下沉增温塞内加尔河是西非一条国际性河流,发源于几内亚富塔贾隆高原,流经几内亚、马里、塞内加尔等国家,据悉该河海水可向上游倒灌 200km 以上。

1972 年,塞内加尔、马里等国联合成立了塞内加尔河流域治理开发委员会,规划建设了两座水利枢纽,分别是位于塞内加尔境内的迪亚马坝和位于马里境内的马南塔里坝。

读图3完成4、5题。

4.图示区域A.位于西半球、低纬度B.年降水量自南向北逐渐减少C.河流汛期长,有凌汛D. 热带沙漠气候为主,地势低5. 与马南塔里坝相比,迪亚马坝独特的功能是A. 旅游观光B.防止海水倒灌C. 航运灌溉D.防治洪涝灾害丹霞地貌为“有陡崖的陆相红层地貌”,最突出的特点是“赤壁丹崖”,其颜色丹红,奇峰秀美,高不可攀。

图4为河北省承德市的双塔山丹霞地貌景观。

据此完成6、7题。

6.图示双塔山地貌景观A.属于旅游景区的吸引物B.便于登高俯视周围美景C.体现自然环境的生产功能和平衡功能D.因历史文化价值突出被列入世界遗产7.下列丹霞地貌的形成过程,正确的是图4 图3A.甲乙丙丁B.乙甲丁丙C.丙丁乙甲D.丁丙甲乙2020年3月28日,搭载支援欧洲医用无纺布、医用桌布等防疫物资的中欧班列从武汉(30.52°N,114.31°E)中心站始发,预计15天后抵达德国杜伊斯堡(51.44°N,6.76°E)。

2019-2020学年上师大附中高三下英语阶段测试

2019-2020学年上师大附中高三下英语阶段测试

II. Grammar and VocabularySection ADirections: After reading the passage below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.V oice for the PlanetIt's the voice you notice first. In person, David Attenborough speaks in the same awestruck hush he has used in dozens of nature documentaries, a crisp half whisper (21) ________ is often imitated but seldom matched. Sitting in his home in the Richmond neighbourhood of west London for one in a series of conversations, I feel obliged to drink (22) ________ second cup of tea when he offers. It somehow seems wrong to say no.In his native U. K. Attenborough is held in the kind of respect usually (23) ________ (reserve) for royalty. Over decades - first as a television executive, then as a wildlife filmmaker and recently as a kind of elder statesman for the planet -- he has achieved near saintly(圣人的)status. He was knighted(封为骑士)by the Queen in 1985 and (24) ________ (usually refer) to as Sir David.Attenborough pioneered a style of wildlife film-making that brought viewers to remote landscapes and gave them a close perspective on the wonders of nature. In the autumn of his life, Attenborough has largely moved away from (25) ________ these films are made but lends his storytelling abilities to wildlife documentaries in collaboration with filmmakers his storytelling abilities to wildlife documentaries in collaboration with filmmakers he has mentored. His most famous work, the 2006 BBC series Planet Earth, set a benchmark in the use of high-definition cameras and had a budget equal to (26) ________ of a Hollywood movie. Among its highlights (27) ________ (be) the first footage of a snow leopard, the impossibly rare Asian wildcat that hunts high in the Himalayas. More than a decade (28) ________ it was first released, Planet Earth remains among the all-time best-selling nonfiction DVDs.Now Attenborough is putting his voice and the weight of authority he has accumulated to greater moral purpose. In recent months he has stood (29) ________ ________ ________ powerful audiences at the 2018 U. N. climate talks in Katowice, Poland, and the 2019 World Economic Forum at Davos, in Switzerland, to urge them into action on climate change. These kinds of event are not his chosen habitat, Attenborough tells TIME. "I would much prefer not to be a sign- (30) ________ (carry) conservationist. My life is the natural world. But I can't not carry a placard(标语牌)if I see what's happening."Section BDirections: Fill in each blank with a proper word chosen from the box. Each word can only be used once. Note that there is one word more than you need.Speeding Into the Future With BiometricsImagine checking in for a flight by simply walking onto the plane. Or buying groceries with a nod of your head, or withdrawing funds from a ATM with a single glance. Thanks to biometric technology - which uses voice, face, fingerprint, or other physical or behavioral characteristic recognition to __31__ verify an individual's identity -- those scenarios are not far from becoming reality.Rapid user identification and more intuitive(简便的)shopping __32__ are just a couple of the emerging capabilities of biometrics, which is ready to blossom into a $ 59.31 billion global industry by 2025, according to Grand View Research. Whether it's providing better ways to fight fraud or delivering smarter customer service, the biometrics industry is expanding the __33__ applications of the technology and transforming human interactions across business and government.Tools that enable voice recognition, iris recognition, and other methods of identification offer faster and more __34__ ways to access electronic devices or physical locations by creating unique identifiers such as "voice-prints" and "eye-prints" that can't be faked. They also allow users to maintain consistent digital identities wherever they go.Millions of smart-phone and tablet users already rely on finger-print scanners and facial recognition systems to unlock or manipulate their devices, but more sophisticated solutions are being __35__ or are in development across a variety of industries."The future of business lies in highly intelligent and automated transactions and smart, faultless customer interactions," says Brett Beranek, general manager of security and biometrics at Nuance Communications. "Among today's fastest-growing trends, the __36__ toward more nautral and personalized exchanges is quickly makingbiometrics a go-to technology for firms all around the world, including Fortune 500 leaders."Nuance has been developing its __37__ biometrics technologies over the past two decades. Its tools go beyond identifying simple characteristics, such as the sound of an individual's voice, to learning speaking, typing, and behavioral patterns and memorizing __38__ vocabulary. Such capabilities allow the technologies to recognize when someone is trying to impersonate a customer on the phone or during digital transactions, stopping fraud in its tracks.Nuance's technology is not only capable of identifying who is __39__ with it, but also what the user is likely looking for, enabling personalized interactions from the moments they begin. This friction-less experience can be a competitive advantage - driving __40__ and improving customer satisfaction.III. Reading ComprehensionsSection ADirections: For each blank in the following passage, there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Friendly LaughterMost people can share a laugh with a total stranger. But there are subtle - and __41__ -- differences in our laughs with friends.Greg Bryant, a cognitive scientist at the University of California, Los Angeles, and his colleagues previously found that adults from 24 societies around the world can distinguish simultaneous "co-laughter" between friends from that between strangers. The findings suggested that his ability may be __42__ used to help read social interactions. So the researchers wondered: Can babies distinguish such laughter, too?Bryant and his fellow researcher Athena V ouloumanos, a developmental psychologist at New York University, played recording of co-laughter between __43__ of either friends or strangers to 24 five-month-old infants in New York City. The babies listened __44__ to the laughs shared between buddies - suggesting they could tell the two types apart, according to a study published in March in Scientific Reports.The researchers then showed the babies short videos of two people acting either like friends or strangers and paired those with the __45__ recordings. The babies stared for longer at clips paired with a mismatched recording -for example, if they saw friends __46__ but heard strangers laughing."There's something about co-laughter that is giving __47__ to even a five-month-old about the social relationship between the individuals," Bryant says. Exactly what components of laughter the infants are detecting remains to be seen, but prior work by Bryant's team provides __48__. Laughs between friends tend to include greater variations in pitch and __49__, for example.Such characteristics also distinguish __50__ laughs from fake ones. Many scientists think heartfelt laughter most likely __51__ from play vocalizations, which are also produced by nonhuman primates, rodens and other mammals. Fake laughter probably emerged later in humans, __52__ that ability to produce a wide range of speech sounds. The researchers suggest that we may be __53__ to spontaneous(自发的)laughter during development because of its long evolutionary history.It's really cool to see how early infants are distinguishing between different forms of laughter," says Adrienne Wood, a psychologist at the University of Virginia, who was not involved in the study. "Almost every __54__ moment is a social interaction for babies. Therefore it __55__ that they are becoming very much accustomed to their social worlds."41. A. distinct B. invisible C. detectable D. conscious42. A. universally B. apparently C. fairly D. precisely43. A. groups B. pairs C. rivals D. partners44. A. shorter B. longer C. less patiently D. more diligently45. A. friendly B. strange C. visual D. audio46. A. interacting B. reflecting C. clubbing D. interpreting47. A. value B. meaning C. information D. friendship48. A. accounts B. implications C. routes D. hints49. A. engagement B. frequency C. intensity D. length50. A. obliged B. involuntary C. encouraged D. internal51. A. evolved B. heaped C. sprang D. originated52. A. apart from B. along with C. as against D. ahead of53. A. available B. crucial C. sensitive D. neutral54. A. screaming B. kicking C. shifting D. waking55. A. turns out B. comes true C. rings hollow D. makes senseSection BDirections: Read the following two passage. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)These days, nobody needs to cook. Families graze on high-cholesterol(胆固醇)take-aways and microwaved ready-meals. Cooking is an occasional hobby and a vehicle for celebrity chefs, which makes it odd that the kitchen has become the heart of the modern house. What the great hall was to the medieval castle, the kitchen is to the 21st - century home.The money spent on kitchens has risen with their status. In America the kitchen market is now worth $ 170 billion, five times the country's film industry. In the year to August 2007, the Swedish furniture chain IKEA sold over one million kitchens worldwide. The average budget for a "major" kitchen overhaul in 2006, calculates Remodeling magazine, was a staggering $ 54,000, even a "minor" improvement cost on average $ 18,000.Exclusivity, more familiar in the world of high fashion, has reached the kitchen: Robinson & Cornish, a British manufacturer of custom-made-kitchens, offers a Georgian-style one, which would cost 145,000 to 155,000 pounds -- excluding building, plumbing and electrical work. Its big selling point is that nobody else will have it: "You won't see this kitchen anywhere else in the word."The elevation of the room that once belonged only to the servants for the modern family tells the story of a century of social change. Right into the early 20th century, kitchens were smoky, noisy places, generally located underground, or to the back of the house, as far from living space as possible. That was as it should be: kitchens were for servants, and the aspiring middle classes wanted nothing to do with them.But as the working classes prospered and the servant shortage set in, housekeeping became a matter of interest to the educated classes. One of the pioneers of a radical new way of thinking about the kitchen was Catharine Esther Beecher, sister of Harriet Beecher Stowe. In American Human's Home, published in 1869, the Beecher sisters recommended a scientific approach to household management, designed to enhance the efficiency of a woman's work and promote order. Many contemporary ideas about kitchen design can be traced back to another American, Chris Frederick, who set about enhancing the efficiency of the housewife. Her 1919 work, House-Engineering: Scientific Management in the Home, was based on detailed observation of a wife's daily routine. She borrowed the principle of efficiency on the factory floor and applied mystic tasks on the kitchen floor.Frederick's central idea, that "stove, sink and kitchen table must be placed in such a relation that useless steps are avoided entirely," inspired the first fully fitted kitchen, designed in the 1920s by Mangarete Schutter Libotsky. It was a modernist triumph, and many elements remain central features of today's kitchen.56. What does the author say about the kitchen of today?A. It is where housewives display their cooking skills.B. It is where the family entertains important guests.C. It has become something odd a modern house.D. It is regarded as the center of a modern home.57. Why does the Georgian-style kitchen sell at a very high price?A. It is believed to have tremendous artistic value.B. There will be no kitchen exactly the same anywhere.C. It is manufactured by a famous British company.D. No other manufacturer can produce anything like it.58. What was the Beecher sisters' idea of a kitchen?A. A place where women could work more efficiently.B. A place where high technology could be applied.C. A place of interest to the educated people.D. A place to experiment with new ideas.59. What do we learn about today's kitchen?A. It represents the rapid technological advance in people's daily life.B. Many of its central features are no different from those of the 1920s.C. It has been transformed beyond recognition.D. Many of its functions have changed greatly.(B)Occupational Licenses with the Biggest Bang for BuckSome 1.8 million American were laid off or discharged from their jobs each month on average in 2019, according to data from the U. S. Bureau of Labor statistics. People who lose their jobs often confront a difficult choice: should they take a new job that pays less, or should they make a costly investment in gaining new skills so that they can compete for another similar job or an even better one?If they do decide on retraining ,which programs and occupational licenses are worth their while? In general, the highest-paying jobs tend to have the most difficult education/ training and experience requirements. But that is not always the case. The following are five occupational licenses with the biggest bang for your buck.Drone Pilots: If you want to become a drone pilot, all you need to do is be above 16 years old, pass the Federal Aviation Administration's Remote Pilot Certificate exam (which requires about 15 to 20 hours of studying), and pay a $ 150 licensing fee. Pay for drone pilots averages $ 56,426 per year, and jobs are growing rapidly across a range of industries. For example, companies like UPS are making substantial investments in drone delivery and will need to hire thousands of drone pilots in the coming years.Home Inspectors: If you need a job that makes about $ 60K per year, you might want to consider becoming a Home Inspector. Both Home Inspectors and HV AC Contractors earn about $ 61K per year, on average, but getting a state HV AC Contractor license typically requires about 4,000 hours of training and experience (those systems arebecoming even more complex), whereas a Home Inspector license only requires 360 hours of training and experience, and much of the training can be gained free of charge on the job.Massage Therapists: On average, Manicurists/Pedicurists are required to complete more hours of training than Massage Therapists (700 hours versus 500 hours), but Massage Therapists earn almost twice as much, on average ($54,639 versus $ 32,509).Radiologic Technologists: Licensing requirements for cosmetologists(美容师)have become so onerous that candidates now need 2,700 hours of training and experience on average. That's not much less than the requirement for becoming a Radiologic or MRI Technologist (3,300 hours), a job which is growing considerably faster than average, is more recession - proof, and pays twice as much ($ 56,162 versus $ 28,608).Dental Hygienists: Among jobs that require a two-year associate's degree granted by a college or university, some pay substantially more than others. The average state licensing fee for becoming a Dental Hygienist is a hefty $ 1,600, but the pay bump you'll receive will likely make up for it ten times over in the first year.60. The underlined expression "the biggest bang for your buck" in Paragraph 2 probably means ________.A. the job loss for your hesitation to investB. a good income resulting from your skillsC. a good return for the money you have spentD. the great efforts you'd make to change your life.61. Which of the following statements is true according to the passage?A. Among the drone operators, those who work for delivery services can earn the most.B. Compared to a home inspector, being an HV AC Contractor is more cost-effective.C. As an MRI technologist, you'd be less likely to be jobless during an economic crisis.D. Higher education isn't a compulsory requirement if you want to be a dental hygienist.62. Which of the following matching for the chart is correct according to the passage?A. ① Radiologic Technologist; ① Cosmetologist; ①General ContractorB. ① Drone Pilot; ① General Contractor; ① Dental HygienistsC. ① Message Therapist; ① Radiologic Technologist; ① CosmetologistD. ① Drone Piolt; ① Cosmetologist; ① Radiologic Technologist(C)I have been wondering lately why I should teach my newborn son English. Everyone I know speaks English, but would Peter be better off learning a more sensible, mellifluous(流畅的)language, like maybe Italian? It is, I admit, a stupid question. But stupid questions can contain the seeds of great insights. This particular stupid question leads to the frontier of economic theory, as well as to the intellectual foundation for the government's antitrust(反垄断)case against Microsoft.Why are American children taught English? The answer is that everyone learns English because everyone else learns it. In this respect, language is a perfect example of what economic theorists call a network. In a network, the benefit one person gets from using some good - in this case, English -- depends on the number of other people using it.Networks fascinate economic theorists because they don't fit nearby into the standard model of how markets work. In most cases, economists are defenders of free markets. People left to their own devices, we argue, will typically achieve an outcome that is good society as a whole - the vaunted(被大肆吹捧的)invisible hand.In the case of networks, however, this logic doesn't seem to work. It is easy to imagine that people might getstuck with a network that, once established, is hard to replace. Parents deciding what language to teach their children, for instance, don't really have much choice. How else can we explain why the Chinese keep speaking Chinese when less complicated languages are available?For a while supporters of the new economics of networks pointed to what seemed to be a compelling example of the problem - the QWERTY keyboard. As the story goes, this arrangement of letters was originally designed to prevent typists from jamming the keys on early typewriters. Despite the availability of superior designs and the fact that jamming keys is no longer an issue, QWERTY remains the standard. This, theorists argued, was a network-driven market failure: People still type on this inefficient keyboard just because that's what everybody else does.This debate over networks, keyboards, and market failure might seem like arcana(奥秘)only economists can love, but it is having a profound influence on public policy. Many academics who have written about the theory of networks have worked for the Justice Department and other federal agencies. A frequent claim is that computer operating system are like languages: Once a standard becomes dominant, it is practically impossible for anyone to consider an alternative, even a better one. The only difference between English and Windows, the argument goes, is that English is free.63. Which of the following examples best illustrates the idea "network" mentioned in the passage?A. Microsoft limits reasonable competition through its aggressive pricing mechanism.B. Some scholars speak out against the fundamental economic theory in a journal.C. Peter chooses to learn Italian for the purpose of an early promotion in his company.D. Families sit together to watch the Spring Festival Gala on New Year's Eve.64. The underlined phrase "this logic" in Paragraph 4 refers to the idea that ________.A. networks don't fit into the standard model of how markets workB. less governmental intervention is good for societyC. what language we learn depends on the environment we live inD. the market wouldn't operate properly without the "invisible hand"65. What can we infer from the passage?A. Free market contributes most to a prosperous economy.B. The QWERTY keyboard reflects a network failure in business.C. Convenience gave way to efficiency in the design of the keyboard.D. Personal preferences may well be determined by how others act.66. Which of the following can be the best title for this passage?A. The Dominance of Microsoft Is to BlameB. Networks Make Substitutes ImpossibleC. The Language We Use Depends on NetworksD. Policy-making Is Subject to Public OpinionsSection CDirections: Read the passage carefully. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.Einstein's Opinions on Creative Thinking"The greatest scientists are artists as well," said Albert Einstein, one of the greatest physicists and an amateur pianist and violinist.For Einstein, insight did not come from logic or mathematics. ___67___ As he told one friend, "When Iexamine myself and my methods of thought. I find that the gift of imagination has meant more to me than any talent for absorbing absolute knowledge. All great achievements of science must start from intuitive knowledge. Imagination is more important than knowledge.But how did art differ from science for Einstein? Surprisingly, it wasn't the content of an idea, or its subject, that determined whether something was art or science, but how the idea was expressed. If what is seen and experienced is described in the language of logic, then it is science. If it is communicated and recognized intuitively, then it is art. ___68___ That's why he said that great scientists were also artists. Einstein first described his intuitive thought processes at a physics conference in Kyoto in 1922 when he indicated that he used images and feelings to solve his problems and found words, logical symbols or mathematical equations later.____69____ "If I were not a physicist," he once said, "I would probably be a musician. I often think in music and I see my life in terms of music. I get most joy in life out of music. Whenever I feel that I have come to the end of the road or into a difficult situation in my work. I would bury myself in music, and that would usually solve all my difficulties."Music provided Einstein with a connection between time and space which both combine spatial and structural aspects. "The theory of relativity occurred to me my intuition and music is the driving force behind this intuition", said Einstein. "My parents had me study the violin from the time I was six. ____70____. "IV. Summary WritingDirections: Read the following three passages. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.Should You Disinfect Your Online Shopping?People who are self-isolating are increasingly relying on grocery deliveries. This raises a new worry: whether delivered goods carry the new coronavirus. Research suggests it can spread via particles in the air, but also via surfaces. How long can the virus survive and how can we protect ourselves?COVID-19 is a respiratory(呼吸道的)illness and is largely transmitted via drops in the air from coughing or sneezing, says John Lednicky at the University of Florida. The new virus has also been found to persist on surfaces.A team led by Vincent Munster at the US National Institute of persist on surfaces. A team led by Vincent Munster at the US National Institute of Allergy and Infectious Diseases in Montana found it may survive on plastic and stainless steel for up to 72 hours.But other research suggests the coronaviruses that cause SARS and MERS can persist on metal, glass and plastic for up to nine days. Research by the US Centers for Disease Control and Prevention suggested that traces of the new coronavirus could be on surfaces for even longer; its RNA was detected in cabins of people who had vacated the Diamond Princess cruise ship 17 days earlier, including those without symptoms.This doesn't necessarily mean these virus particles could still infect other people, says Lednicky. How long virus particles remain active depends on various factors. Those coughed or sneezed out may be covered in a layer of mucus (黏液)that helps them survive longer. Surface survival may also be affected by UV light, which can destroy the ability of some viruses to reinfect us. Heat and higher humidity can also inactivate viruses.Is it worth trying to disinfect your shopping? Lednicky doesn't think so. Most household cleaning products won't kill coronaviruses, he says. Even if you use one that does, you're unlikely to be able to clean every nook and cranny of, for example, a bunch of grapes. It is more practical to practise social distancing and good personal hygiene, he says.V. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.72. 这事他越想越不对劲。

北京市密云区 2019-2020 学年高三第二学期第二次阶段性测试英语试卷及答案

北京市密云区 2019-2020 学年高三第二学期第二次阶段性测试英语试卷及答案

密云区2019-2020 学年第二学期第二次阶段性测试第一部分:知识运用(共两节,45 分)第一节语法填空(共10 小题;每小题 1.5 分,共15 分)阅读下列短文,根据短文内容填空。

在未给提示词的空白处仅填写 1 个适当的单词,在给出提示词的空白处用括号内所给词的正确形式填空。

ALast week, our class was on duty for student self-management. On the first day, I was shocked to see so much leftover food thrown away by students. What a waste! Being concerned about it, my classmates and I had a heated discussion on how to solve the problem. Finally, we all 1 (agree) that the wall newspaper would be the best choice. The next day, we put our idea into reality. Towards lunch time, we put 2 a wall newspaper outside the school cafeteria, calling on students not to waste food. Many students gathered around to read and expressed their support. To my great delight, there were changes soon. In the cafeteria, I found the trays returned after lunch all empty without any leftover. Food 3 (save) and the dining hall was cleaner.BAs we know, the global water shortage is becoming increasingly severe mainly due to global warming, environmental pollution and the ever-increasing population. Therefore, it’s high time we did something about it. Firstly, an 4 (effect) way, I think, is to reserve water in a scientific way for future use. Secondly, new methods need to be developed to use the existing water resources, for example, 5 (turn) sea water into fresh water. Thirdly, we must stop water pollution by law. Last but not least, it’s everyone’s responsibility 6 (make) good use of water, such as recycling and saving water in our daily life. In conclusion, people around the world should be aware of the real situation of water shortage, protect the present water resources and explore potential ones scientifically.CThe Palace Museum is working to take cultural relics into people’s daily life and bring their cultural value into full play by selling cultural and creative products, on the theme of “Bring the Palace Museum culture home”. The creative products mostly are creative daily necessities, like stationaries, bags, decorations and so on. 7 (Base) on the treasures in the museum, the Palace Museum has developed products such as Qianli Jiangshan series and Qingming Shanghe Tu series, Palace Dolls, folding fans, 8 are very popular with young people. The Palace Museum now 9 (change) the traditional way of communication, learns to use a variety of ways to publicize excellent traditional culture, and lets the Palace Museum cultural heritage resources live. The culture creative products are definitely brilliant choices for 10 (gift) that bear unique royal features.第二节完形填空(共20 小题;每小题 1.5 分,共30 分)阅读下面短文,掌握其大意,从每题所给的A、B、C、D 四个选项中,选出最佳选项,并在答题卡上将该项涂黑。

2019-2020学年密云区第二中学高三语文期中试卷及答案解析

2019-2020学年密云区第二中学高三语文期中试卷及答案解析

2019-2020学年密云区第二中学高三语文期中试卷及答案解析一、现代文阅读(36分)(一)现代文阅读I(9分)阅读下面的文字,完成小题。

材料一:数据统计显示,全世界垃圾年均增速为8.42%,而中国垃圾增速超10%。

中国城市生活垃圾累积堆存量已达70亿吨。

目前,全国已有2/3的大中城市陷入垃圾包围中,且有1/4城市已无合适场所堆放垃圾。

随着城市化进程和经济社会的高速发展,垃圾问题已成为近年热议的话题。

对于生活垃圾、农业垃圾、建筑垃圾等,如何实施无害化处理,变废为宝,成为每个城市实现可持续发展、建设科学生态系统的重要工作。

国内外广泛采用的城市生活垃圾处理方式主要有卫生填埋、焚烧发电等。

其中,继传统的卫生填埋之后,考虑到垃圾增量、土地资源紧张、循环利用等因素,不少国家开始加大焚烧发电的规划。

从20世纪70年代起,一些发达国家便着手通过焚烧垃圾来发电。

据统计,目前日本、丹麦、瑞士等国家的生活垃圾焚烧率达到70%~80%。

不过,焚烧发电也并非是直接“变废为宝”。

焚烧是一种能够处理混合垃圾的典型技术,垃圾分类是焚烧的充分条件,它可以起到减少垃圾处理量、减少污染排放量、改善燃烧工况、提高发电效率等作用。

受技术和工艺制约,发电时燃烧产生的有毒废气如果得不到有效处理,将严重威胁居民生命健康,这也是居民担忧并导致焚烧厂建设受阻的原因。

另外,垃圾发电原理是将纸张、塑料、菜叶等生活垃圾经过分拣、干燥等工序处理后,进行高温焚烧,将焚烧中产生的热能转化为高温蒸汽,推动汽轮发电机发电,发电所需助燃物量大,因此垃圾发电成本很高,投资惊人。

目前垃圾分拣存在很大难度,世界上采用垃圾焚烧的城市中约有一半城市没有做到垃圾完全分类。

给垃圾分类是解决问题的有效手段,是世界一些发达国家通行做法。

我国垃圾分类仍然困难很大。

一方面,巨型垃圾场内建筑与生活垃圾混倒,无必要的分类,使垃圾处理难度加大;一方面,民间自发拾荒大军,在一定程度上变废品为资源,但大多缺乏规范和检验,使垃圾在捡拾、收集、运输、加工过程中造成严重的二次污染。

密云区2019-2020学年第二学期高三第一次阶段性测试答案20200403 - 副本(1)

密云区2019-2020学年第二学期高三第一次阶段性测试答案20200403 - 副本(1)

密云区2019-2020学年第二学期高三第一次阶段性测试数学试卷参考答案及评分标准一、选择题:共10小题,每小题4分,共40分.题号 1 2 3 4 5 6 7 8 9 10 答案CCBADBDDCC二、填空题:共5小题,每小题5分,共25分.11.10- 12.02±(,);y x =± 13.16;2114.π;π[+π,π],2k k k -∈Z 15.(,3)-∞. 备注:若小题有两问,第一问3分,第二问2分.三、解答题:共6小题,共85分.解答应写出文字说明,演算步骤或证明过程. 16.(本小题满分14分)(Ⅰ)解:由余弦定理得2221cos 22b c a A bc +-==, 在ABC ∆中,0πA <<,所以π3A =. 若选择①和②方法一 将7a =,2b =代入222b c a bc +-=化简得2230c c --=.所以1c =-(舍),或3c =. 因此11333sin 232222ABC S bc A ∆==⨯⨯⨯=. 方法二 由正弦定理得sin sin a bA B=, 所以72sin 32B =,因此3sin 7B =. 在ABC ∆中,因为a b >,所以A B >. 因此B 为锐角,所以2cos 7B =. 所以33sin sin()sin cos cos sin 27C A B A B A B =+=+=.因此133sin 22ABC S ab C ∆==. 若选择①和③由sin 2sin C B =得2sin 22sin R C R B =⨯(R 为ABC ∆外接圆的半径), 所以2c b =.将7a =,2c b =代入222b c a bc +-=解得73b =. 所以273c =. 所以11727373sin 222633ABC S bc A ∆==⨯⨯⨯=. 若选择②和③由sin 2sin C B =得2sin 22sin R C R B =⨯(R 为ABC ∆外接圆的半径), 所以2c b =.因为2b =,所以4c =.所以113sin 2423222ABC S bc A ∆==⨯⨯⨯=. (Ⅱ)解:因为π3A =,所以2π3B C +=.所以2πcos cos cos cos()3B C B B +=+-2π2πcos cos cos sin sin 33B B B =++ 31πsin cos sin()226B B B =+=+. 因为2π03B <<,所以π5π66B <<. 所以当π3B =时,cos cos B C +有最大值1.17. (本小题满分14分)(Ⅰ)解:记“选取的这份试卷的调查结果是膳食合理状况类中习惯良好者”为事件A.有效问卷共有 380+550+330+410+400+430=2500(份), 受访者中膳食合理习惯良好的人数是4000.65260⨯=人,所以,()P A =260=0.1042500. (Ⅱ)解:记事件A 为“该区卫生习惯良好者”,事件B 为“该区体育锻炼状况习惯良好者”,事件C 为“该区膳食合理习惯良好者”, 由题意,估计可知()=0.6()=0.8()=0.65P A P B P C ,,,设事件E 为“该居民在“卫生习惯状况类、体育锻炼状况类、膳食合理状况类”三类习惯中,至少具备2个良好习惯”. 由题意知,()()()()E ABC ABC ABC ABC =U U U所以事件E 的概率()()()()()P E P ABC P ABC P ABC P ABC =+++=()()()()()()()()()()()()P A P B P C P A P B P C P A P B P C P A P B P C +++=0.60.80.35+0.60.20.65+0.40.80.65+0.60.80.65⨯⨯⨯⨯⨯⨯⨯⨯ =0.168+0.078+0.208+0.312=0.766所以该居民在“卫生习惯状况类、体育锻炼状况类、膳食合理状况类”三类习惯中,至少具备2 个良好习惯的概率为0.766. (Ⅲ)解:615432>D D D D D D ξξξξξξ=>>>.18.(本小题满分15分)(Ⅰ)解:取AD 中点为O ,连接OP ,OC 和AC .因为PAD ∆为等边三角形, 所以PO OD ⊥.因为平面P AD ⊥平面ABCD ,PO ⊂平面P AD ,所以PO ⊥平面ABCD . 因为OC ⊂平面ABCD ,所以PO OC ⊥. 在菱形ABCD 中,AD CD =,60ADC ∠=o, 所以ADC ∆为正三角形,因此OC AD ⊥.以O 为原点建立空间直角坐标系,如图所示.则(0,0,0)O ,(100)A ,,,(230)B ,,,(030)C ,,,(1,0,0)D -, (0,0,3)P ,13(,0,)22M -,(1,3,0)N . 所以13(,3,)22CM =--u u u u r ,(1,3,0)AB =u u u r ,(1,0,3)AP =-u u u r . 设平面PAB 的法向量()x y z =,,m ,由00.AB AP ⎧⋅=⎪⎨⋅=⎪⎩u u u ru u u r ,m m 得3030.x y x z ⎧+=⎪⎨-+=⎪⎩,NPABC DMxyzO令3x =,则(3,1,1)=-m .设直线CM 与平面PAB 所成角为θ,则有||315sin |cos ,|.10||||25CM CM CM θ⋅=<>===⋅⨯u u u u ru u u u r u u u ur m m m 所以直线CM 与平面PAB 所成角的正弦值为1510. (Ⅱ)解:因为,OC AD OC PO ⊥⊥,所以OC ⊥平面P AD .所以(0,3,0)OC =u u u r是平面P AD 的法向量,则有35cos ,5||||53OC OC OC ⋅-<>===-⋅⋅u u u r u u u ru u u r m m m ,因为二面角B AP D --的平面角为钝角, 所以二面角B AP D --的余弦值为55-. (Ⅲ)解:结论MN //平面PAB .因为33(,3,)22MN =-u u u u r , 所以3333(1)()1022MN =⨯+⨯-+-⨯=u u u u r g m . 因此MN ⊥u u u u rm .又因为直线MN ⊄平面PAB , 所以MN //平面PAB .19.(本小题满分14分) (Ⅰ)解:因为()()e1xf x ax =+,x ∈R , 所以()'()e1xf x ax a x =++∈R ,.'(0)1k f a ==+,又因为(0)1f =,所以切线方程为=(+1)1y a x +.(Ⅱ)解:因为()'()e 1xf x ax a x a =++∈∈R R ,,,(1)当0a =时因为'()e 0,xf x x =>∈R ,所以()f x 的单调增区间是(),-∞+∞,无单调减区间. (2)当0a ≠时令'()0f x =,则11x a=--. ① 当0a >时,()f x 与'()f x 在R 上的变化情况如下:x 1()a -∞,-1- 11a -- 1(1,)a--+∞'f x () — 0 + f x ()↘↗所以()f x 的单调减区间是1()a -∞,-1-,单调增区间是1(1,)a--+∞. ②当0a <时,()f x 与'()f x 在R 上的变化情况如下:x1()a-∞,-1-11a --1(1,)a--+∞'f x () + 0 — f x ()↗↘所以()f x 的单调增区间是1()a-∞,-1-,单调减区间是1(1,)a--+∞. 综上所述,当0a =时,()f x 的单调增区间是(),-∞+∞,无单调减区间;当0a >时,()f x的单调减区间是1()a -∞,-1-,单调增区间是1(1,)a--+∞;当0a <时,()f x 的单调增区间是1()a -∞,-1-,单调减区间是1(1,)a--+∞.(Ⅲ)解:方法一因为()()e1,xf x ax x =+∈R ,所以令()0f x =,得10ax +=. (1)当0a =时,方程无解,此时函数()f x 无零点; (2)当0a ≠时,解得1x a=-, 此时函数()f x 有唯一的一个零点.综上所述,当0a =时,函数()f x 无零点;当0a ≠时,函数()f x 有一个零点. 方法二(1)当0a =时因为()e 0xf x =>,所以函数()f x 无零点;(2)当0a >时因为10a <-1-,(0)10f =>,()f x 在区间1(1,)a--+∞单调递增,所以()f x 在区间1(1,)a--+∞内有且仅有唯一的零点; 若1(,1)x a ∈-∞--,则11(1)10ax a a a+<--+=-<,又因为e 0x >,所以()()e 10xf x ax =+<.即函数()f x 在区间1()a-∞,-1-内没有零点.故当0a >时,()f x 有且仅有唯一的零点. (3)当0a <时因为111(1)е()0a f a a ----=->,111(1)е0a f a a--=<,并且()f x 在区间1(1,)a --+∞单调递减,所以()f x 在区间1(1,)a--+∞内有且仅有唯一的零点;若1(,1)x a ∈-∞--,则11(1)10ax a a a+>--+=->,又因为e 0x >,所以()()e 10xf x ax =+>.即函数()f x 在区间1()a-∞,-1-内没有零点.故当0a <时,()f x 有且仅有唯一的零点.综上所述:当0a =时,函数()f x 无零点;当0a ≠时,函数()f x 有一个零点.21.(本小题满分14分)(Ⅰ)解:根据题意得2221,3,2.b ca abc =⎧⎪⎪=⎨⎪⎪=+⎩解得2,1,3.a b c ⎧=⎪=⎨⎪=⎩ 所以椭圆M 的方程为2214x y +=. (Ⅱ)解:方法一 点M 在以OD 为直径的圆上.设点00(,)P x y ,则00x ≠,01y ≠± ,并且220014x y +=, 0(0,)Q y ,00(,)2x M y . 因此000012(1)2AM y y k x x --==. 所以直线AM 的方程为002(1)1y y x x -=+.令1y =-,解得01x x y =-. 所以00(,1)1x N y --,00(,1)2(1)x D y --. 所以00000000(,1)(,1)2(1)22(1)x x x yMD y y y y =---=----u u u u r.因为00(,)2xMO y =--u u u u r ,所以000000()(1)2(1)2x y xMD MO y y y =⨯-++-u u u u r u u u u r g 200000(1)4(1)x y y y y =-⨯++-.因为220014x y +=,所以220014x y =-. 所以200000(1)(1)01y MD MO y y y y =--⨯++=-u u u u r u u u u r g . 因此MD MO ⊥u u u u r u u u u r.所以点M 在以OD 为直径的圆上. 方法二 点M 在以OD 为直径的圆上. 设点00(,)P x y ,则220014x y +=,并且0(0,)Q y ,00(,)2x M y . 因此000012(1)2AM y y k x x --==. 所以直线AM 的方程为002(1)1y y x x -=+. 令1y =-,解得01x x y =-. 所以00(,1)1x N y --,00(,1)2(1)x D y --.设E 为线段OD 的中点,则001(,)4(1)2x E y --.所以2ME =2200001()()4(1)22x x y y -++-=22200020(21)1()16(1)2x y y y -++-. 设以OD 为直径的圆的半径为r ,则222020116(1)4x r OE y ==+- . 所以22222200002200(21)11()16(1)16(1)42x x y r ME y y y --=-+-+-- 222000020()11()4(1)42x y y y y -=⨯+-+- 因为220014x y +=,所以220014x y =-. 所以22222000020()11(1)()0(1)42y y r ME y y y --=-⨯+-+=-. 因此||r ME =.所以点M 在以OD 为直径的圆上.22.(本小题满分14分)(Ⅰ)解:由题意,数列{}n a 的通项公式为55=-n a n ,数列{}n b 的通项公式为99=-n b n . 得,,(55)(99)5914=-+-=+-i j c i j i j ,则2,650=c ,396,62020=c . 得,,(55)[9(1)9]595=--+-=--i j d i j i j ,则2,649=-d .(Ⅱ)证明:已知6a =,7b =,得数列{}n a 的通项公式为66=-n a n ,数列{}n b 的通项公式为77=-n b n .所以,,6(1)7(1)6713=-+-=+-i j c i j i j ,,∈∈**N N i j . 所以,,(66)[7(1)7]676i j d i j i j =--+-=--,17,,∈∈**N N i i j ≤≤. 所以,若∈t M ,则存在,∈∈N N u v ,使67=+t u v . 若*t M ∈,则存在,6,∈∈*N N u u v ≤,使67=-t u v . 因此,对于整数t ,考虑集合0{|6,,6}==-∈N M x x t u u u ≤, 即{t ,6t -,12t -,18t -,24t -,30t -,36}-t . 下面证明:集合0M 中至少有一元素是7的倍数. 反证法:假设集合0M 中任何一个元素,都不是7的倍数, 则集合0M 中每一元素关于7的余数可以为1,2,3,4,5,6.又因为集合0M 中共有7个元素,所以集合0M 中至少存在两个元素关于7的余数相同, 不妨设为126,6--t u t u ,其中1212,,6∈<N u u u u ≤.则这两个元素的差为7的倍数,即21126(6)6()---=-t u t u u u . 所以120-=u u ,与12u u <矛盾.所以假设不成立,即原命题成立. 即集合0M 中至少有一元素是7的倍数,不妨设该元素为0006,,6-∈N t u u u ≤. 则存在∈Z s ,使00067,,6t u s u u -=∈N ≤,即00067,,6,=+∈∈N Z t u s u u s ≤. 由已证可知,若∈t M ,则存在,∈∈N N u v ,使67=+t u v .而t M ∉,所以s 为负整数,设v s =-,则*N v ∈,且00067,,6,=-∈∈*N N t u v u u v ≤. 所以,当6a =,7b =时,对于整数t ,若t M ∉,则*t M ∈成立.(Ⅲ)解:下面用反证法证明:若对于整数t ,*t M ∈,则t M ∉.假设命题不成立,即*t M ∈,且t M ∈.则对于整数t ,存在,∈∈N N n m ,,6,∈∈*N N u u v ≤,使6767=-=+t u v n m 成立. 整理,得6()7()u n m v -=+.又因为∈N m ,∈*N v ,所以7()06-=+>u n m v 且u n -是7的倍数.因为,6∈N u u ≤,所以6-u n ≤,所以矛盾,即假设不成立. 所以,对于整数t ,若*t M ∈,则t M ∉. 又由第二问,对于整数t ,t M ∉,则*t M ∈. 所以t 的最大值,就是集合*M 中元素的最大值. 又因为67,,6*N,N t u v u v u =-∈∈≤,所以*max max ()667129t M ==⨯-⨯=.。

2019-2020学年密云区第二中学高三语文第三次联考试卷及答案

2019-2020学年密云区第二中学高三语文第三次联考试卷及答案

2019-2020学年密云区第二中学高三语文第三次联考试卷及答案一、现代文阅读(36分)(一)现代文阅读I(9分)阅读下面的文字,完成下面小题。

新媒体是继报刊、广播、电视等传统媒体以后发展起来的新的媒体形态,是利用数字、网络技术、移动技术,通过互联网、无线通信网、有线网络等渠道以及电脑、手机、数字电视机等终端,向用户提供信息和娱乐的传播形态和媒体形态。

目前,新媒体主要包括电脑网络媒体、手机媒体、网络电视、电子出版物等。

相对于传统媒体,新媒体具有传播与更新速度快、信息量大、全球性和跨文化性、检索便捷、多煤体、超文本、互动性、成本低的优势,给传统纸质媒体带来巨大冲击。

媒体的命运取决于它在社会上扮演的角色和所起到的作用。

如果纸质媒体的功能完全被其他媒体替代,它将无法生存,如果纸质媒体能够具有其他媒体所不具有的功能,就有存在的价值。

阅读是人类获取信息的主要途径,纸质媒体和新媒体可以看成完全不同的两种阅读方式。

稳定的空间感,逻辑严密、线索清晰、便于记忆和思考,是纸质媒体阅读的优势;新媒体的阅读者更多的是一种信息的快速获取,记忆和思考退居次要地位。

从这个意义上看纸媒体与电子媒体的关系,就不是一个取代另一个的单向演变,而是一种阅读方式与另一种阅读方式的区别。

两种阅读方式的不同也给纸媒体和新媒体带来了长期竞争共存的空间。

借助于现代计算机技术,新媒体实现了高密度的信息储存,单位信息存储成本低,可以方便地实现大量信息的自动化处理,使得部分编辑工作为机器所替代,极大地降低了制作成本;电子信号的易复制性和快速传播性,使其发行成本低廉,使大范围信息传播免费或廉价提供成为可能。

但分析新媒体传播内容可见,新媒体传播的信息具有传统媒体的一般特征,和传统媒体没有质的区别。

大部分内容源于传统媒体,最突出之处,也仅是在传统媒体的基础上做了一些计算机技术的加工,实现了声音、画面、文字的同时传播。

可见新媒体在传播中所具有的纸质媒体难以实现的优势,主要源于其对先进的传播技术的应用。

2019-2020学年密云区第二中学高三语文第三次联考试卷及答案

2019-2020学年密云区第二中学高三语文第三次联考试卷及答案

2019-2020学年密云区第二中学高三语文第三次联考试卷及答案一、现代文阅读(36分)(一)现代文阅读I(9分)从古至今,丝绸之路上的交流和互动,都有艺术相伴。

艺术是温润和滋养丝绸之路的精神源泉和情感溪流,也是贯通丝绸之路的文化血脉。

丝绸织锦、陶瓷器、乐器、歌舞、建筑、绘画等,以其物质与艺术属性的和谐统一、实用与审美功能的相得益彰满足了丝绸之路沿线人们的各种需求。

同时,艺术的交流和相互影响,沟通了民族情感,化解了文化冲突,丝绸之路由此千年不绝,绵延至今。

人类的交流从物质交换开始,通过各种生活用品和器物的交换,沟通着不同族群人与人的关系,而物质被赋予艺术审美特性,则改变和丰富着人们的生活方式和美学风尚,也化解了人与人之间的隔阂。

中国丝绸、瓷器被西方人推崇,印度、波斯、中亚音乐舞蹈对中国乐舞的影响等等,是在艺术差异性中建立的新型审美关系,并发挥了长久而特殊的功能。

丝绸之路艺术从物质实用性与精神审美性两方面满足了东方与西方社会的需求,也穿越了国家民族地域界限。

丝绸之路艺术史与以往的国别艺术史、区域艺术史、世界艺术史不同之处之一,就在于它不是“纯艺术”的历史,而是与物质结合在一起的交流史。

从学术的角度说,在人类艺术史乃至人类史的视域中研究丝绸之路艺术,避免了在狭义的艺术视域下建构艺术史的发展逻辑,同时,提出一个艺术理论问题——“物的艺术表达”。

“物的艺术表达”的概念,意指物质交流负载艺术元素,同时,艺术创作及其传播对于物质载体、材料、质地的要求、利用和催生,使得“物”蕴含丰富的艺术性和审美性,也具有了艺术表达功能和“文本”叙事功能。

“物的艺术表达”的研究将使丝绸之路艺术的意蕴获得新的理解,也使得丝绸之路物质与艺术之关系获得新的阐释。

瓷器和其他器物,丝绸和织物,乐舞和乐器,建筑风格和雕塑手法等等,不仅以物的实用性而且以其艺术性沟通人类审美情感,为人类命运共同体的构建提供精神滋养。

可以说,蕴蓄艺术审美性的丝绸之路物质交流,或者说具有艺术品格的物质交流,就是心灵对话,就是审美意识的表达,它在一定程度上弥合了经济利益和文化冲突造成的裂痕,这是人类艺术史上特殊的艺术现象,是丝绸之路艺术的独特意义之所在。

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密云区2019-2020学年第二学期高三第二次阶段性测试
数学试卷
一、选择题:本大题共10小题,每小题4分,共40分.在每小题列出的四个选项中,选出符合题目要求的一项.
1.已知集合{|0}M x x =∈R ≥,N M ⊆,则在下列集合中符合条件的集合N 可能是( ) A. {0,1}
B. 2
{|1}x x =
C. 2
{|0}x x >
D. R
2.在下列函数中,定义域为实数集的偶函数为( ) A. sin y x =
B. cos y x =
C. ||y x x =
D.
ln ||y x =
3.已知x y >,则下列各不等式中一定成立的是( ) A. 2
2
x y >
B.
11
x y
> C. 11()()33
x y
>
D.
332x y -+>
4.已知函数()y f x =满足(1)2()f x f x +=,且(5)3(3)4f f =+,则(4)f =( ) A. 16
B. 8
C. 4
D. 2
5.已知双曲线2
21(0)x y a a
-=>的一条渐近线方程为20x y +=,则其离心率为( )
A.
B.
C.
D.
6.已知平面向量a r 和b r ,则“||||b a b =-r r
r ”是“1()02
b a a -⋅=r r r ”的( )
A. 充分而不必要条件
B. 必要而不充分条件
C. 充分必要条件
D. 既不充分也不必要条件
7.已知圆2
2
:(1)2C x y +-=,若点P 在圆C 上,并且点P 到直线y x =的距离为2
,则满足条件的点P 的个数为( )
A. 1
B. 2
C. 3
D. 4
8.设函数1
()sin()2
f x x ωϕ=+,x ∈R ,
其中0>ω,||ϕπ<.若5182
f π⎛⎫= ⎪⎝⎭,08f 11π⎛⎫= ⎪⎝⎭
,且()f x 的最小正周期大于2π,则( )
A. 13ω=,24ϕ11π
=-
B. 23ω=
,12π
ϕ= C. 13ω=,724
πϕ=
D. 23
ω=,12ϕ11π
=-
9.某三棱锥的三视图如图所示,则该三棱锥中最长的棱长为( )
B. 2
C.
D. 10.已知函数()f x 的定义域为 R ,且满足下列三个条件: ①对任意的[]12,4,8x x ∈ ,且 12x x ≠,都有()
1212
()0f x f x x x ->- ;
②(8)()f x f x += ; ③(4)y f x =+ 是偶函数;
若(7),(11)a f b f =-=,(2020)c f =,则,,a b c 的大小关系正确的是( ) A. a b c <<
B. b a c <<
C. b c a <<
D.
c b a <<
二、填空题:本大题共5小题,每小题5分,共25分.
11.抛物线2
(y mx m =为常数)过点(1,1)-,则抛物线的焦点坐标为_______.
12.在6
1()
x x
+展开式中,常数项为________.(用数字作答)
13.已知n S 是数列{}n a 的前n 项和,且(
)2
11n S n n n *
=-∈N ,则1
a
=_________,n S 的
最小值为_______.
14.在ABC V 中,三边长分别为4a =,5b =,6c =,则ABC V 的最大内角的余弦值为_________,ABC V 的面积为_______.
15.已知集合{
}
22
,,A a a x y x Z y Z ==-∈∈.给出如下四个结论: ①2A ∉,且3A ∈;
②如果{|21,}B b b m m ==-∈N*,那么B A ⊆;
③如果{|22,}C c c n n ==+∈N*,那么对于c C ∀∈,则有c A Î; ④如果1a A ∈,2a A ∈,那么12a a A ∈. 其中,正确结论的序号是__________.
三、解答题: 本大题共6小题,共85分.解答应写出文字说明, 演算步骤或证明过程.
16.如图,
直三棱柱111ABC A B C -中,11
2
AC BC AA ==,D 是棱1AA 的中点,1DC BD ⊥.
(1)证明:1DC BC ⊥; (2)求二面角11A BD C --的大小.
17.已知函数2()cos cos )sin f x x x x x =+- . (Ⅰ)求函数()f x 的单调递增区间和最小正周期;
(Ⅱ)若当[0,]2
x π
∈时,关于x 的不等式()f x m ≥,求实数M 的取值范围. 18.某健身机构统计了去年该机构所有消费者消费金额(单位:元)
,如下图所示:
(1)将去年的消费金额超过3200 元的消费者称为“健身达人”,现从所有“健身达人”中随机抽取2 人,求至少有1 位消费者,其去年的消费金额超过4000 元的概率;
(2)针对这些消费者,该健身机构今年欲实施入会制,详情如下表:
0,1600内的消费者今年都将会申请办理普通会员,消费金额在
预计去年消费金额在(]
(]
3200,4800内的消费者1600,3200内的消费者都将会申请办理银卡会员,消费金额在(]
都将会申请办理金卡会员. 消费者在申请办理会员时,需-次性缴清相应等级的消费金额.该健身机构在今年底将针对这些消费者举办消费返利活动,现有如下两种预设方案:
方案1:按分层抽样从普通会员,银卡会员,金卡会员中总共抽取25 位“幸运之星”给予奖励: 普通会员中的“幸运之星”每人奖励500 元;银卡会员中的“幸运之星”每人奖励600 元;金卡会员中的“幸运之星”每人奖励800 元.
方案2:每位会员均可参加摸奖游戏,游戏规则如下:从-个装有3 个白球、2 个红球(球只有颜色不同)的箱子中,有放回地摸三次球,每次只能摸-个球.若摸到红球的总数消费金额/元为2,则可获得200 元奖励金;若摸到红球的总数为3,则可获得300 元奖励金;其他情况不给予奖励. 规定每位普通会员均可参加1 次摸奖游戏;每位银卡会员均可参加2 次摸奖游戏;每位金卡会员均可参加3 次摸奖游戏(每次摸奖的结果相互独立) .
以方案 2 的奖励金的数学期望为依据,请你预测哪-种方案投资较少?并说明理由.
19.已知椭圆()22
22:10x y C a b a b +=>>
过点1,2P ⎛ ⎝⎭
,设它的左、右焦点分别为1F 、2F ,左顶点为A ,上顶点为B
,且满足12AB F =. (Ⅰ)求椭圆C 标准方程和离心率;
(Ⅰ)过点6,05Q ⎛⎫
-
⎪⎝⎭
作不与y 轴垂直直线交椭圆C 于M 、N (异于点A )两点,试判断MAN ∠的大小是否为定值,并说明理由. 20.已知函数()ln f x x a x =-,a R ∈.
(Ⅰ)当1a =时,求曲线()f x 在1x =处的切线方程; (Ⅱ)设函数1()()a
h x f x x
+=+
,试判断函数()h x 是否存在最小值,若存在,求出最小值,若不存在,请说明理由.
(Ⅲ)当0x >时,写出ln x x 与2x x -的大小关系.
21.设n 为正整数,集合A =12{|(,,,)n t t t αα=L ,{0,1}k t ∈,1k =,2,L ,}n .对于集合A 中任意元素12(,,,)n x x x α=L 和12(,,,)n y y y β=L ,记
111122221
(,)[(||)(||)(||)]2
n n n n M x y x y x y x y x y x y αβ=+-++-+++-+++L .
(Ⅰ)当n =3时,若(0,1,1)α=,(0,0,1)β=,求(,)M αα和(,)M αβ的值; (Ⅱ)当4n =时,对于A 中的任意两个不同的元素α,β,证明:(,)(,)(,)M M M αβααββ+≤.
(Ⅲ)给定不小于2的正整数n ,设B 是A 的子集,且满足:对于B 中的任意两个不同元素
α,β,(,)(,)(,)M M M αβααββ=+.写出一个集合B ,使其元素个数最多,并说明由.
的的的。

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