密云区2019-2020学年第二学期高三第二次阶段性测试20200602
【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
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. 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. 16B. 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( )A.B.C.D.6.已知平面向量a r 和b r ,则“||||b a b =-r rr ”是“1()02b 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0>ω,||ϕπ<.若5182f π⎛⎫= ⎪⎝⎭,08f 11π⎛⎫= ⎪⎝⎭,且()f x 的最小正周期大于2π,则( )A. 13ω=,24ϕ11π=-B. 23ω=,12πϕ= C. 13ω=,724πϕ=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( )B. 2C.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61()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()211n 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 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,112AC 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,]2x 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()2222:10x 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()()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.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(,)[(||)(||)(||)]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。
【精准解析】北京市密云区2020届高三二模英语试题+Word版含解析
密云区2019-2020学年第二学期第二次阶段性测试高三英语试卷第一部分: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)A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。
Last 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.【答案】1.agreed2.up3.was saved【解析】本文是记叙文。
2019-2020学年北京市密云区东邵渠中学高三语文二模试题及答案
2019-2020学年北京市密云区东邵渠中学高三语文二模试题及答案一、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。
夏敬观以为苏东坡有一类词,是天风海涛之曲,而中多幽咽怨断之音的,那是他最好的词。
至于豪放激荡的词,乃其第二乘也。
苏东坡的词摆脱绸缪婉转之态,举首高歌,写了浩气逸怀,这对于词是很大的开拓。
可是,在当时很多人不承认他这种风格,说他好像是教坊雷大使之舞,虽然跳得很好,极天下之工,要非本色。
因为词自五代《花间集》以来,都是写闺房儿女的,而苏东坡所写的是“大江东去”之类的词,因此被认为不是本色。
他的词是词的发展史上把词诗化的一个高峰。
可是,词毕竟是词,不管他写了多少豪杰的壮志,他最好的词,都应该有一种曲折幽微的美,要把浩气逸怀结合了词的曲折幽微的特点,这才是他第一等的作品。
《念奴娇·赤壁怀古》豪放的地方比较多,《满庭芳》也使大家感动了,但都不是他最好的作品。
苏东坡在新党当政时曾被迁贬,下过乌台狱,几乎被处死,被迁谪到黄州。
后来,新党失败了,旧党上台,苏东坡被召回朝廷,他与旧党司马光虽是很好的朋友,可是,在论政之间,他不苟且随声附和。
一个人一定应该如此,该放过去的放过去,该持守住的持守住。
他既然与旧党的人论政不合,于是出官到杭州。
后来又被召回汴京,《八声甘州》就是离杭回汴京时写的。
你看他这首词:“有情风万里卷潮来,无情送潮归。
”写得真是很好,有超越的一面,也有悲慨的一面。
那多情的风卷起钱塘江潮涌来,又无情地送潮归去,宇宙万物都是如此的。
“问钱塘江上,西兴浦口,几度斜晖?”钱塘江上,西兴浦口,有多少次的潮去潮回,有多少次的日升日落。
“不用思量今古,俯仰昔人非。
”我们不用说今古的变化,就是宋朝党争之中,有多少人起来,又有多少人倒下去了。
“谁似东坡老,白首忘机。
”现在我年岁已经老大了,把一切都置之度外了。
“忘机”则是说把得失荣辱的机智巧诈之心都忘记了。
后边你看他的转折。
2020届北京市密云区高三二模英语试题(学生版)
密云区 2019-2020 学年第二学期第二次阶段性测试高三英语试卷第一部分: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)A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。
Last 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.B阅读下列短文,根据短文内容填空。
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.。
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)。
北京市密云区2020届高三二模英语试题 (含答案)
北京市密云区2019-2020学年第二学期第二次阶段性测试高三英语试卷2020.06 考生须知1、本试卷满分120分。
2、本试卷考试时间100分钟。
3、在本试卷答题纸上认真填写考生信息。
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 seeso much leftover food thrown away by students. What a waste! Being concerned about it, my classmates andI 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 put2 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. Food3 (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 aboutit. 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 waterinto 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 resourcesand 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 suchas 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。
密云区2020届高三二模数学试题及答案
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C. D.6.已知平面向量和a b ,则“||||=-b a b ”是“1()02-=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,则满足条件的点P 的个数为 A .1 B .2 C .3 D .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 A .2 B .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 中,三边长分别为4a =,5b =,6c =,则ABC 的最大内角的余弦值为_________,ABC 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3111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] 40 30 20 10 0(1600,2400] (2400,3200] (4000,4800](3200,4000] 820253584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=∈=.对于集合A 中的任意元素12(,,,)n x x x α=和12(,,,)n y y y β=,记111122221(,)[(||)(||)(||)]2n n n n M x y x y x y x y x y x y αβ=+-++-+++-+++.(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 ⊥,BDDC D =,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 =,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第16题图DDC 1AB C A 1 B 1第16题图zx y所以1(0,0,1)A D =-,1(1,1,2)A B =--,1(1,0,1)C D =-,1(0,1,2)C B =-. 设平面1A BD 的法向量()x y z =,,m ,由1100.A D AB ⎧⋅=⎪⎨⋅=⎪⎩,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 ⎧⋅=⎪⎨⋅=⎪⎩,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 =+,22(2,)AN x y =+. 所以1212(2)(2)AM AN x x y y =+++B AM N Qxy12122121222266(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 ⊥,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 -+++=. 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 =+,22(2,)AN x y =+. 所以1212(2)(2)AM AN x x y y =+++12122221212222222266(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 ⊥,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==,,若(,)(,)(,)M M M αβααββ=+,则0i i x y =,1,2,3,,i n =成立. 所以,考虑设012312{(,,,,)|,0}n n A x x x x x x x =====,11231{(,,,,)|1,{0,1},2,3,,}n i A x x x x x x i n ==∈=,对于任意的2,3,,k n =,123123121{(,,,,)|(,,,,),0,1}k n n k k A x x x x x x x x A x x x x -=∈=====.所以01n A A A A =.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 =-)中, 不妨设为123123(,,,,)(,,,,)n n x x x x y y y y αβ==,,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 =;对于1,2,,1k n =-,取123(,,,,)k n k e x x x x A =∈,且10k n x x +===;n n e A ∈.令01{,,,}n B e e e =,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.。
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.。
2020年6月北京市密云区普通高中2020届高三下学期第二次阶段性测试(二模)英语试题及答案
绝密★启用前北京市密云区普通高中2020届高三毕业班下学期第二次阶段性测试(二模)英语试题第一部分: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 resp onsibility 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。
2020届北京市密云区高三二模英语试题(解析版)
密云区2019-2020 学年第二学期第二次阶段性测试高三英语试卷第一部分: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)A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。
Last 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.【答案】1. agreed2. up3. was saved【解析】本文是记叙文。
2019-2020学年北京密云二中高三英语二模试卷及答案
2019-2020学年北京密云二中高三英语二模试卷及答案第一部分阅读(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AIn his 402nd anniversary year, Shakespeare is still rightly celebrated as a great language master and writer. But he was not the only great master of play writing to die in 1616, and he is certainly not the only writer to have left a lasting influence on theater.While less known worldwide, Tang Xianzu is considered one of Chinas greatest playwrights and is highly spoken of in that country of ancient literary and dramatic traditions.Tang was born in 1550 inLinchuan,Jiangxiprovince. Unlike Shakespeare's large body of plays,poems and sonnets (十四行诗), Tang wrote only four major plays: The Purple Hairpin, Peony Pavilion (《牡丹亭》), A Dream under the Southern bough, and Dream of Handan. The latter three were constructed around a dream narrative, a way through which Tang unlocked the emotional dimension of human desires and ambitions and explored human nature beyond the social and political limits of that time.Similar to Shakespeare, Tang's success rode the wave of a renaissance (复兴) in theater as an artistic practice. As in Shakespeare'sEngland, Tang's works became hugely popular inChinatoo. During Tang'sChina, his plays were enjoyed performed, and changed. Kunqu Opera, a form of musical drama, spread from southernChinato the whole nation and became a symbol of Chinese culture. Combining northern tune and southern music, kunqu Opera was known for its poetic language, music, dance movements and gestures. Tang's works benefited greatly from the popularity of kunqu Opera, and his plays are considered classics of kunqu Opera.While Tang and Shakespeare lived in a world away from each other, there are many things they share in common, such e humanity of their drama, their heroic figures, their love for poetic language, a lasting popularity and the anniversary during which we still celebrate them.1. Why is Shakespeare mentioned in the first paragraph?A. To describe Shakespeare's anniversary.B. To introduce the existence of Tang Xianzu.C. To explain the importance of Shakespeare.D. To suggest the less popularity of Tang Xianzu.2. What's possibly one of the main theme of Tang's works?A. Social reality.B. Female dreams.C. Human emotions.D. Political environment.3. What does the author mainly tell us in Paragraph 4?A. The influence of Kunqu Opera on Tang's works.B. Tang's success in copying Shakespeare's styles.C. The way Kunqu Opera became a symbol of Chinese culture.D. Tang's popularity for his poetic language and music.BPaper is an important part of modern life. People use it in school, at work, to make artwork and books, to wrap presents and much more. Trees are the most common material for paper these days.So how do people make paper out of trees today? People first cut trees, load them onto trucks and bring them to a factory. Machines cut open the outer coverings of the trees, and cut the trees into pieces. Those pieces are boiled into a soup. After that, it is hit flat, dried and cut up into sheets of paper.The entire process, from planting a small tree to buying your school notebook, takes a very long time. Just growing the trees takes 10 to 20 years.Making tons of paper from trees can harm the planet. Humans cut down 80, 000 to 160,000 trees around the world every day, and use many of them to make paper. Some of those trees come from tree farms. But people also cut down forests for paper, which means that animals and birds lose their homes.Cutting forests down also contributes to climate change, and paper factories pollute the air. After you throw paper, it often takes the paper six to nine years to break down. That's why recycling is important. It saves a lot of trees, slows climate change and helps protect endangered animals, birds and all creatures that rely on forests for their homes and food.So if paper isn't good for the environment, why don't people write on something else?The answer: They do. With computers, tablets and cellphones, people use much less paper than in the past. Maybe a day will come when we won't use paper at all — or will save it for very special books and artworks.4. What can we know about making paper out of trees?A. It costs much money.B. It takes a lot of time.C. It is very easy and fast.D. It is dangerous and difficult.5. What is the impact of paper production?A. It promotes the recycling.B. It does harm to the environment.C. It slows down the climate change.D. It protects the animals from losing homes.6. How will we use paper someday in the future according to the text?A. Use it for books only.B. Use the recycled paper.C. Treasure it occasionally.D. Use it for artworks.7. What idea does the author want to express from the text?A. The influence of making paper on environment.B. The wonderful experience of making paper.C. The necessary process of making paper.D. The good reasons for making paper.CLight pollution is a significant but overlooked driver of the rapid decline of insect populations, according to the most comprehensive review of the scientific evidence to date.Artificial light at night can affect every aspect of insects' lives, the researchers said. "We strongly believe artificial light at night — in combination with habitat loss, chemical pollution.invasive (入侵的) species, and climate change — is driving insect declines, " the scientists concluded after assessing more than 150 studies.Insect population collapses have been reported around the world, and the first global scientific review published in February,said widespread declines threatened to cause a "catastrophic collapse of nature's ecosystems".There are thought to be millions of insect species, most still unknown to science, and about half are active at night. Those active in the day may also be disturbed by light at night when they are at rest.The most familiar impact of light pollution is moths (飞蛾) flapping around a bulb, mistaking it for the moon. Some insects use the polarisation of light to find the water they need to breed, as light waves line up after reflecting from a smooth surface. But artificial light can scupper (使泡汤) this. Insects areimportant prey (猎物) for many species, but light pollution can tip the balance in favour of the predator if it traps insects around lights. Such increases in predation risk were likely to cause the rapid extinction of affected species, the researchers said.The researchers said most human-caused threats to insects have analogues in nature, such as climate change and invasive species. But light pollution is particularly hard for insects to deal with.However, unlike other drivers of decline, light pollution is ly easy to prevent. Simply turning off lights that arenot needed is the most obvious action, he said, while making lights motion-activated also cuts light pollution. Shading lights so only the area needed is lit up is important. It is the same with avoiding blue-white lights, which interfere with daily rhythms. LED lights also offer hope as they can be easily tuned to avoid harmful colours and flicker rates.8. What is discussed in the passage?A. Causes of declining insect populations.B. Consequences of insect population collapses.C. Light pollution: the key bringer of insect declines.D. Insect declines: the driver of the collapsed ecosystem.9. What is the 5th paragraph mainly about?A. How light travels in space.B. How light helps insects find food.C. How the food chain is interrelated.D. How light pollution affects insects.10. What does the underlined word"analogues"in Paragraph 6probably mean?A. Selective things.B. Similar things.C. Variations.D. Limitations.11. What is the purpose of the last paragraph?A. To offer solutions.B. To give examples.C. To make comparisons.D. To present arguments.DWith graduation days being celebrated all over the country, a student who has to use a wheelchair honored his mother on his graduation day in a special way. Easley High School graduate, Alex Mays surprised people present when he got up and walked across the stage at Clemson's Littlejohn Coliseum.“I was really happy—it made me feel good,” Alex said.Alex was not given a chance to live right from his birth. He was born at 25 weeks and weighed just 1 pound, 10 ounces at birth. When he was very young, he had a disease and lost the ability to walk. After his mother's death in 2013, Alex had several other difficult life changes until he came to live with his grandparents, Dousay and her husband, Dewayne. Dousay said that when Alex came to live with them, they decided to bring him up in thebest possible way they could.Last fall, Alex said that he would walk across the stage to get his diploma to honor his late mother. He practiced hard and worked with a physical therapist for 9 months to complete his plan.The only help Alex got was from his mom's best friend, Tonya Johnson, who pushed his wheelchair to the stage wearing one of his mother's favorite shirts. “I had support from my family. I couldn't have done it without them,” Alex said.“Alex made everyone in the building feel encouraged that day” Pickens County School District public information specialist John Eby said. “The school teachers knew he was going to get up to get his diploma, but the distance he walked was a surprise, even to them,” Eby said.“Some of life's most important tests aren’t given in a classroom; Alex tested himself and passed with flying color1 s,” Eby added.12. In what way did Alex honor his late mother on his graduation day?A. By dressing like her.B. By saying sorry to her.C. By inviting her best friend.D. By walking to get his diploma.13. What can we learn from Paragraph 3?A. Alex was born healthy.B. Alex went through a lot.C. Alex had a purpose in life as a child.D. Alex has lived with his grandparents all the time.14. What did Alex also express on his graduation day?A. His big regret in life.B. His feelings for hisschool.C. His thanks for his family.D. His will to complete his study.15. Which of the following words can best describe Alex?A. Strong-minded.B. Warm-hearted.C. Cool-headed.D. Easy-going.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学年密云区第二中学高三生物下学期期末考试试卷及答案
2019-2020学年密云区第二中学高三生物下学期期末考试试卷及答案一、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. 关于组成生物体元素和化合物的叙述,错误的是()A.不同细胞的元素组成基本相同,但含量可能会有差异B.淀粉、糖原、纤维素彻底水解后,得到的单体可能不同C.在显微镜下,可观察到生物组织中脂肪被苏丹Ⅲ染液染成橘黄色D.用双缩脲试剂检测蛋白质时,先加入双缩脲试剂A液,再滴入B液2. 电影《我不是药神》中涉及的慢性粒细胞白血病,是一种白细胞异常增多的恶性肿瘤。
其病因是9号染色体上的原癌基因(ABL)插入到22号染色体上的BCR基因内部,形成了BCR﹣ABL融合基因。
该融合基因的表达使酪氨酸激酶活化,导致细胞癌变。
患者服用靶向药物“格列卫”能有效控制病情。
下列叙述错误的是()A.原癌基因主要负责调节细胞周期,控制细胞生长和分裂的进程B.患者骨髓中的造血干细胞增殖分化发生了异常C.造血干细胞分化为白细胞导致遗传信息的执行情况不同D.“格列卫”可能含酪氨酸激酶的抗体而起到靶向治疗的作用3. 如图为叶绿体结构与功能示意图,A、B、C、D表示叶绿体的结构,ⅢⅢⅢⅢⅢ表示有关物质。
下列叙述中不正确的是A.参与光合作用的色素位于图中的A部分B. CO2进入叶绿体后,通过固定形成Ⅲ物质C.若突然停止光照,C3的含量会升高D.用14C标记CO2进行光合作用,可在ⅢⅢ中测到较高放射性4. 下列有关探索DNA是遗传物质实验的叙述,错误的是A.格里菲思的实验指出了R型肺炎双球菌转化为S型是转化因子作用的结果B.艾弗里实验证明了DNA是肺炎双球菌的遗传物质C.赫尔希和蔡斯实验中标记T2噬菌体的D N A时利用了含32P的大肠杆菌D.搅拌不充分会使35S标记的噬菌体这一组的沉淀物放射性偏低5. 下列有关生物膜系统的描述错误的是()A.原核细胞不具有生物膜系统B.生物膜系统由具膜的细胞器构成C.各种生物膜的化学组成和结构相似D.丰富的生物膜为酶的附着提供了大量位点6. 下列动植物特征中,最可能属于荒漠生物群落的是()A. 耐寒的多年生草本植物B. 擅长挖洞的小型动物C. 营树栖生活的动物D. 根系发达的耐旱植物7. 下列关于酶的叙述,不正确的是()A.细胞内的酶都是在核糖体上合成的B.酶能降低反应的活化能C.所有的酶都具有专一性D.酶可以在细胞外起催化作用8. 1982年我国科学家在世界上第一次用人工方法合成具有生物活性的酵母丙氨酸转运核糖核酸(用tRNAyAla表示)。
2019-2020学年密云区第二中学高三英语二模试卷及答案
2019-2020学年密云区第二中学高三英语二模试卷及答案第一部分阅读(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AThese wonderful NYC attractions offer pay — what — you — wish days, free entry hours/days and other great stuff for local families.Staten IslandZooThere are plenty of creatures who call NYC home—the Staten Island Zoo is one of them. Once you’ve finished learning about the wildlife in the animal nursery, reptile (爬行动物) side rooms, horse barn and other areas of the attraction, make sure to mark your schedule for fun seasonal happenings, such as the Easter Egg Games and the scary, crazy Halloween Shows.Entry on Wednesdays is by suggested donation; children aged two and under free.Children’s Museum of the ArtsThe Children’s Museum of the Arts welcomes 135,000 little visitors each year through its doors. Once inside, the whole family can enjoy interactive programs, exhibitions (展览) and events that celebrate the changed power of the arts on youngsters and grown-ups alike.Pay-as-you-wish Thursdays, 3-6 p. m.Wave HillEveryone needs a few hours of calm now and then-kids included-and you’d be hard- pressed to find a more peaceful spot within city limits than Wave Hill the broad grounds located above the river, covering 28 acres of public gardens, plus woodlands and grasses to wander. Jump in on nature walks, story times and family art projects often led by local artists and free with general admission.Pay — as — you — wish Tuesdays and Saturdays,9 a. m — noon.New York Hall of ScienceNaturally, kids love it when the New York Hall of Science pleases them with neat exhibits and fun hands-on activities. The museum’s playground is themost attractivetochildren A tube slide (管道滑梯) will give little ones the knowledge on science topics, while the climbing area mirrors a giant spider web. There are also wind pipes, metal drums, sand- boxes and much more. What better way to make the mostout of science?Free entry Sep-Jun on Fridays, 2 — 5 p. m,and Sundays, 10 —11 a. m.1. What can children do in Staten Island Zoo?A. Feed injured animals.B. Join in seasonal activities.C. Build a home for creatures.D. Deal with the donations to the zoo.2. What do Children’s Museum of the Arts and Wave Hill have in common?A. They both have peaceful spots.B. They both are located by a river.C. They both have public gardens.D. They both have activities about arts.3. Which place can be free of charge for all?A. Wave Hill.B.Staten IslandZoo.C. New York Hall of Science.D. Children’s Museum of the Arts.BNostalgia (怀旧) has become increasingly common in our current climate of accelerated, unexpected change. More and more Americans are turning back with longing towhat feels like simpler, sweeter times. They collect cassette tapes, manual typewriters even decades-old video games.Is it a mistake to get too obsessed with the past? Some psychologists warn that too much devotion to the so-called good old days is an escape from reality; it can indicate loneliness or that a person is having a difficult time coping in the present. Psychologist Stephanie Coontz argues that nostalgia distracts us from addressing the problems of modern life and contribute to anxiety, depression , insomnia etc.But new studies suggest that a modest dose of nostalgia is not only harmless, but actually beneficial. They suggest it helps strengthen our sense of identity and makes us feel more optimistic and inspired. It is also a tool for self — discovery and memories are a psychological immune response that is triggered when you want to take a break from negativity. Interestingly, those happy memories can be particularly beneficial both to kids in their teens and to society's elders. Recalling our childhood reminds us of “the times when we were accepted and loved unconditionally," says Krystine Batcho, a psychologist. "That is such a powerfully comforting phenomenon, knowing that there was a time in life when we didn't have to earn our love." Nostalgia can transform even themost ordinary past into legends which warms the heart and the body. Let's not forget that nostalgia has been a source of inspiration to innumerable American writers. Mark Twain recalled his boyhood, writing, "after all these years, I can picture that old time to myself now, just as it was then:The white town drowsing in the sunshine of a summer's morning."So go ahead, daydream a little about your best childhood friend, your first car, a long - gone family pct. As Dr. Sedikidessays,"Nostalgia is ly central to human experience. "But at the same time, keep these words of wisdom from the great inventor Charles Kettering in mind as well:"You can't have a better tomorrow if you are thinking about yesterday all the time. "34. What did some psychologists in paragraph 2 probably agree?A. Nostalgia will cause some mental problems.B. Nostalgia makes us devoted to the good old days.C. Nostalgia shows you are trying to get rid of loneliness.D. Nostalgia helps us cope with the difficult time we are going through.5. There are many benefits of nostalgia except ________A. It can enable us to know ourselves better.B. It can bring us some comfort when we recall.C. We are likely to gain attention if we recall the happy childhood.D. We can sometimes break away from negativity with happy memories.6. What will be talked about in the following paragraph?A. The bad influence of too much devotion to nostalgia.B. The reasons why we should avoid nostalgia.C. The bad memories that always stick around you.D. The great changes nostalgia will bring to you.7. What's the best title of the passage?A. We all have a soft spot for nostalgia.B. Nostalgia is actually good for you.C. Don't be carried away by nostalgia.D. There are many times when we like to recall.CThis past year, I've found myself returning again and again to lines of poetry by Emily Dickinson. Like manypeople, I've needed the curing effects of reading more than ever. As scientists and psychologists will tell you, books are good for the brain and their benefits are particularly vital now.Books expand our world, providing an escape and offering novelty, surprise and excitement. They broaden our view and help us connect with others. Books can also distract us and help reduce ourmental chatter.When we hit the “flow state" of reading where we're fully lost in a book, our brain's mode network calms down. It's a network of brain that is active and gets absorbed in thinking and worrying endlessly when we are not doing anything else.There is so much noise in the world right now and the very act of reading is kind of meditation. You disconnect from the chaos around you.You reconnect with yourself when you are reading. And there's no more noise.In 2020, the NPD Group recorded the best year of book sales since 2004. Yet even as people are buying more books,many are reporting they're having a harder time getting through them. It's difficult for your brain to focus on a book when it's constantly scanning for threats to keep you alive.Our fight-or-flight response has been consistently activated.Sometimes I picture my brain as a cartoon brain with little arms and legs, fighting with a book I am holding and screaming: “Can't you see I'm busy!” Anxiety causes our brain to produce a flood of stress,which consumes our energy and makes it harder to concentrate.Then one day in December sitting on my couch, I remembered how much I like to read"The House of Mirth" every few years around the holidays. The memory inspired me to pick up the familiar book, opened it up and started reading.I just kept going.The comfort and distraction and brain-opening experience gave me peace.So return to something familiar.8. What does the underlined part “mental chatter” in Paragraph 2 mean?A. Getting lost in a book.B. Non-stop inner anxiety.C. Chatting with the author.D. Powerful network of brain.9. What do we know about reading according to the text?A. It can treat our headache.B. It can calm down the noisy people.C. It forces us to concentrate.on thinking.D. It makes us communicate with ourselves.10. Why was it difficult for people to finish reading books in 2020?A. People bought too many books.B. The books were too difficult to understand.C. People just wanted to escape from the threat.D. The life threat disturbed people's focus on books.11. Why is the author's experience mentioned in the last paragraph?A. To rid people of concern for safety.B. To present an effective reading way.C. To wake up memories of an old book.D. To recommend the book he/she reads.DI once had my Chinese MBA students brainstorming on “two-hour business plans.” I separated them into six groups and gave them an example: a restaurant chain. The more original their ideas, the better, I said. Finally, five of the six groups presented plans for restaurant chains. The sixth proposed a catering service. Though I admitted the time limit had been difficult, I expressed my disappointment.My students were middle managers, financial analysts and financiers from state-owned enterprises and global companies. They were without talent or opinions, but they had been shaped by an educational system that rarely stressed or rewarded critical thinking or inventiveness. The scene I just described came in different forms during my two years teaching at the school. Papers were often copied from the Web and the Harvard Business Review. Case study debates were written up and just memorized. Students frequently said that copying is a superior business strategy, better than inventing and creating.InChina, every product you can imagine has been made and sold. But so few well-developed marketing and management minds have been raised that it will be a long time before most people in the world can name a Chinese brand.With this problem in mind, partnerships with institutions like Yale and MIT have been established. And then there's the “thousand talent scheme.” this new government program is intended to improve technological modernization by attracting top foreign-trained scientists to the mainland with big money. But there are worries aboutChina's research environment. It's hardly known for producing independent thinking and openness, and even big salary offers may not be attractive enough to overcome this.At last, forChina, becoming a major world creator is not just about setting up partnership with top western universities. Nor is it about gathering a group of well-educated people and telling them to think creatively. It's about establishing a rich learning environment for young minds. It's not that simple.12. Why does the writer feel disappointed at his students?A. Because there is one group presenting a catering service.B. Because the six groups did not cooperate well in the brainstorm.C. Because all the students copied a case for the difficult topic.D. Because the students' ideas were lacking in creativeness.13. Which of the following scenes is NOT considered as lack of creation?A. Papers were often downloaded from the Internet.B. Students often said that copying is a preferable business strategy.C. Students combine knowledge and critical thoughts to solve a problem.D. Case study debates were written up as well as recited.14. We can infer form the passage that ________.A. China can make and sell any product all over the world from its own creation.B. high pay may not solve the problem ofChina's research environment.C. cooperation with institutions has been set up to make a Chinese brand.D. the new government program is aimed at encouraging imagination.15. Which is the best title for the passage?A Look for a new way of learning B. Reward creative thinkingC. How to become a creatorD. Establish a technical environment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学年密云区第二中学高三生物下学期期末试卷及参考答案
2019-2020学年密云区第二中学高三生物下学期期末试卷及参考答案一、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. 棉花的长绒和短绒是一对相对性状,下列实验能确定显隐性关系的是()A. 长绒×长绒→长绒B. 长绒×长线→152株长绒+49株短绒C. 短绒×短绒→短绒D. 长绒×短绒→135株长绒+132株短绒2. 绿色植物在光合作用时,CO2的固定发生在()A.无光条件下B.有光条件下C.暗反应过程中D.上述都对3. 下列对动物体内酶、激素和ATP这三类有机物的相关叙述,错误的是()A.这三类有机物一定都含有C、H、OB.酶和A TP均可以在细胞内外发挥作用,而激素只能在细胞内发挥作用C.能合成酶的细胞一定都能合成A TPD.成年男性体内,精原细胞形成精子的过程中有酶、激素和ATP的参与4. 生物大分子是构成生命的基础物质,下列有关生命活动主要承担者——蛋白质的说法错误的是()A.蛋白质的营养价值主要取决于其含有的非必需氨基酸的种类B.蛋白质与某种RNA结合,可成为“生产蛋白质的机器”C.蛋白质与糖类结合,可成为细胞间互相联络的“语言”D.某些蛋白质具有的螺旋结构,决定了其特定的功能5. 下列关于“观察洋葱根尖细胞有丝分裂”实验的叙述,正确的是A.通过观察细胞内染色体的行为判断细胞所处的时期B.装片的制作流程是:解离→染色→漂洗→制片C.显微镜下观察到处于有丝分裂中期的细胞最多D.显微镜下观察到的分生区细胞呈正方形,细胞继续分裂6. 下列各项表示细胞结构与其主要组成成分的对应关系,错误的是()A. 中心体——蛋白质B. 核糖体——蛋白质和RNAC. 细胞骨架——多糖D. 细胞壁——纤维素7. 新交规于2013年1月1日正式实施,新交规中一次性扣12分的11种情形中就包含饮酒后驾驶机动车,其中明确规定不得酒后驾车、醉酒驾车。
2019-2020学年北京密云二中高三英语月考试卷及答案
2019-2020学年北京密云二中高三英语月考试卷及答案第一部分阅读(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AExperts say that if food were a country, it would rank second behind theUSas one of the biggest greenhouse gas polluters. The reason is the rising demand for meat. Animal farming is responsible for 14.5 percent of global methane emissions. While cowsare the worst contributors, pigs, sheep, donkeys and other animals play a part as well.Animal agriculture also causes land to become damaged, water to be polluted and forests to get destroyed. With the world population forecast to rise to 9.8 billion by 2050, things are only looking worse for our already decreasing natural resources. While going vegetarian would help, meat consumption is too deep-rooted in most Western diets to allow for such a sharp, permanent change. That is why experts are advocating substituting some of the beef, chicken, or pork with ordinary insects!Insects, which grow into adults within a matter of months, if not weeks, are ready for consumption much faster than domestic animals. They also require much less room, use less water and food, and produce far less greenhouse gas than animals.Of the 1.1 million insect species worldwide, scientists have identified 1,700 as eatable. Among them are ants, grasshoppers, grubs, and earthworms. Just like animals, each insect has a different taste. Tree worms taste just like pork, and grubs are similar to smoked meat.While eating insects might be a new concept for Western people, over 2 billion people worldwide consume insects as a regular part of their diet. Besides being delicious, insects are high in protein, have very few calories, and are free of the saturated fat found in animal meat. Insects can be prepared in many ways. Creative cooks can use them to cook protein-rich soup, make baked treats, and even fry a few with vegetables. So eat insects--- both your body and Mother Earth will thank you for it!1. Which of the following animals contribute the most to global methane emissions?A. Sheep.B. Donkeys.C. Cows.D. Pigs.2. How is the third paragraph developed?A. By making comparisons.B. By providing examples.C. By listing data.D. By asking questions.3. What can be inferred from the last paragraph?A.Few people eat insects regularly.B. Ordinary insects are high in fat.C. Insects contain various vitamins.D. Saturated fat is harmful to health.BWolves have a certain undeserved reputation: fierce, dangerous, good forhunting down deer and farmers’ livestock. However, wolves have a softer, more social side, one that has been embraced by a heart-warming new initiative.In a bid to save some of Europe’s last wolves, scientists have explored the willingness of these supposedly fierce creatures to help others of their kind. Female wolves, the scientists have discovered, make excellent foster parents to wolf cubs that are not their own. The study, published in Zoo Biology, suggests that captive-bred wolfcubs(幼兽)could be placed with wild wolf families, boosting the wild population.The gray wolf was once the world’s most widely distributed mammal, but it became extinct as a result of widespread habitat destruction and the deliberate killing of wolves suspectedof preying on livestock. Fear and hatred of the wolf have since become culturally rooted, fuelled by myths, fables and stories.In Scandinavia, the gray wolf is endangered, the remaining population found by just five animals. As a result, European wolves are severely inbred and have little geneticvariability(变异性), making them vulnerable to threats, such as outbreaks of disease that they can’t adapt to quickly. So Inger Scharis and Mats Amundin of Linkoping University, in Sweden, started Europe’s first gray wolf-fostering program. They worked with wolves keptat seven zoos across Scandinavia. Eight wolf cubs between four and six days old were removed from their natural parents and placed with other wolf packs in other zoos. The foster mothers accepted the new cubs placed in their midst.The welfare of the foster cubs and the wolves’ natural behavior were monitored using a system of surveillance cameras. The foster cubs had a similar growth rate as their step siblings in the recipient litter, as well as their biological siblings in the source litter. The foster cubs had a better overall survival rate, with 73% surviving until 33 weeks, than their biological siblings left behind, of which 63% survived. That rate of survival is similar to that seen in wild wolf cubs. Scientists believe that wolves can recognize their young, but this study suggests they can only do so once cubs are somewhere between three to seven weeks of age.If captive-bred cubs can be placed with wild-living families, which already have cubs of a similar age, not onlywill they have a good chance of survival, but they could help dramatically increase the diversity of the wild population, say the researchers. Just like the wild wolves they would join, these foster cubs would need protection from hunting. Their arrival could help preserve the future of one of nature’s most iconic and polarizing animals.4. What’s the theme of the passage?A. Giving wolf cubs a new lifeB. Foster wolf parents and foster cubsC. The fate of wild wolvesD. Changing diversity of wild wolves5. Which of the following flow chart best demonstrates the relationship between the wolves?A. B.C. D.6. Which of the following statements is true?A. Female wolves are willing to raise wolf cubs of 3 to 7 weeks old.B. Foster cubs are accepted by foster parents and are well bred.C. Man’s hostile attitude towards wolves roots in myths, fables and stories.D. Foster cubs and their biological siblings have similar growth rate and survival rate.7. What’s the purpose of the research?A. To help wolves survive various threatsB. To improve wolves’ habitat and stop deliberate killingC. To save endangered wolves by increasing their populationD. To raise man’s awareness of protecting wolvesCOn March 18, 34-year-old Lance Karlson was walking on the beach and looking for somewhere toswim in Geographe Bay when he saw what he thought was a stingray (黄貂鱼) leap from the water.Realizing the creature was, in fact, an octopus (章鱼), he started filming it — just in time for the angryoctopus to launch itself at him. He immediately felt a sharp pain across his left arm, followed by a second strike across his neck and upper back. His goggles (护目镜) fogged and the water around him turned dark with what he thought might have been octopus ink as he struggled back toward the shore."I was confused — it was more of a shock than a fright," said Karlson, "I might have hit on its home." Within a minute, a perfect imprint of an octopus tentacle (触手) appeared on Karlson's neck and back.A former volunteer lifeguard, Karlson rushed back to his hotelroom to find something acidic to put on the wound. All his family could grab was Coca-Cola, which his wife poured over his back and the pain disappeared."The pain went away and more than anything since then, it's been more the physical hit that was painful.... The imprint on my neck is more from the physical hit, and I guess it makes complete sense when you look at the video I took," he said.Karlson said he'd never seen an octopus that close before and watched Netflix documentary "My Octopus Teacher" after the incident to learn more about the species."They are beautiful creatures and I really hope this promotes more interest in octopuses as opposed to fear of them. I think this is a fascinating creature with clearly some very strong emotions just like we do as humans," he added.8. When did the octopus attack Karlson?A. When he was swimming in the bay.B. When he was shooting the octopus.C. When he was looking for some fish.D. When he was fighting against the octopus.9. What plays an essential role in reducing Karlson's pain?A. The lifeguard's timely help.B. Karson's wife's quick action.C. The family members' efforts.D. Karlson’s knowledge of first aid.10. What does Karlson learn from his experience?A. The octopus is dangerous.B. People should get away from the octopus.C. People need know more about the octopus.D. The physical hit from the octopus is painful.11. Which of the following might be the best title?A. Pain from Strong AttackB. First Aid for Octopus' StrikeC. Face to Face with Angry OctopusD. Under Sea with Dangerous CreatureDScientists at the Massachusetts Institute of Technology (麻省理工学院) have turned spider webs into music——creating an strange soundtrack that could help them better understand how the spiders output their complex creations and even how they communicate.The MIT team worked with Berlin-based artist Tomas Saraceno to take 2D (two-dimensional) laser (激光) scans of a spider web, which were linked together and made into a mathematical model that could recreate the web in 3Din VR (virtual reality). They also worked with MIT’s music department to create the virtual instrument.“Even though the web looks really random (随机),there actually are a lot of inside structures and you can visualize (可视化) them and you can look at them, but it’s really hard to grasp for the human imagination or human brain to understand all these structural details,” said MIT engineering professor Markus Buehler, who presented the work on Monday at a virtual meeting of the American Chemical Society.Listening to the music while moving through the VR spider web lets you see and hear these structural changes and gives a better idea of how spiders see the world, he told CNN. “Spiders use vibrations (振动) as a way to locate themselves, to communicate with other spiders and so the idea of thinking really like a spider would experience the world was something that was very important to us as spider material scientists,” Buehler said.Spiders are able to build their webs without shelves or supports, so having a better idea of how they work could lead to the development of advanced new 3D printing techniques. “The reason why I did that is I wanted to be able to get information really from the spider world, which is very weird and mysterious,” Buehler explained. In addition to the scientific value, Buehler said the webs are musically interesting and that you can hear the sounds the spider creates during construction. “It’s unusual and eerie and scary, but finally beautiful.” he described.12. What have MIT scientists done according to the passage?A. They have translated spider webs into sounds.B. They have made a mathematical model to produce webs.C. They have created a soundtrack to catch spiders.D. They have known how spiders communicate.13. What can we know about spider webs from paragraph 3?A. Their structures are beautiful and clear.B. Professor Markus Buehler knows them well.C. The American Chemical Society presents the result.D. They are complex for people to figure it out.14. In which field will the study be helpful?A. virtual realityB. printingC. paintingD. film-making15. What is the main idea of the passage?A. It tells us that the music created by spiders is scary.B. It shows how the researchers carry out the experiment.C. It presents a new and creative way to study spiders.D. It explains why scientists did the experiment.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学年北京密云二中高三英语第二次联考试卷及答案解析
2019-2020学年北京密云二中高三英语第二次联考试卷及答案解析第一部分阅读(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ACourtyard Stay with Your DogsAs autumn approaches, we are inviting you to join us on our August dog event—Courtyard Stay with Yoga & Hike from August 21st to August 22nd nearQinglongLakein Fangshan district. This event will continue to raise fund for our Pre-treatment Guidance Project (PGP).Check out our full itinerary below:Day 1: Saturday, August 21stAt 9 am in the morning, you will be picked up by car or bus and head to Fangshan district. We have arranged a private courtyard house for a comfortable stay for the Saturday night. We will enjoy some local dishes for lunch, while enjoying the peaceful views of theQinglongLake. After lunch, we will go together for a light hike to the nearby mountain and water tracks with our dogs. For dinner, enjoy the coziest, home-style BBQ with both vegetable and meat options.Day 2: Sunday, August 22ndAfter breakfast, we will join a yoga teacher for a deeply relaxing yoga session, with the company of the morning sun. You can choose to hike a bit more afterwards in theforest park. We will be checking out around 4 pm in the afternoon, returning to our homes.Reservation: Please add our event manager, Diana, on Wechat to reserve a spot for yourself and your dog, a full payment will be required upon signing up. If you don’t have a dog but would still like to hang out with dogs, you are more than welcome to join too!Pricing:Early Bird Fee: 1400 RMB per human for the first four participants only before 7th August.Children Price: 800 RMB (4—13 years old).Full Price: 1600 RMB per human.All surplus funds and donations will go to our Pre-treatment Guidance Program.Please bring your ID, your dog’s ID and vaccine proof for any potential security check.We can’t wait to see you soon!1. Who would be most interested in the weekend activity?A. Taxi drivers.B. Yoga coaches.C. Pet dog owners.D. The PGP members.2. According to the itinerary, on August 22nd, you can ________.A. join a teacher to do some yogaB. have some local dishes for lunchC. enjoy the coziest, home-style BBQD. go together for a hike to the water tracks3. The purpose of the countryside stay activity is to________.A. promote theQinglongLakeB. raise fund for a public projectC. teach skills of BBQ and keeping dogsD. provide access to dogs for people without petsBMost teenagers are still trying to find their passion and purposes in life. However, not Gitanjali Rao. The 15-year-old girl has been coming up with innovative solutions to worldwide problems since she was ten. It is, therefore, not surprising that the teenager has won the honor of “America's Top Young Scientist”.In the third grade, Rao was inspired to build a device after witnessing the shocking story unfold in Flint, Michigan, where cost-cutting measures led to the use of a polluted river as the city's primary water supply and incredibly high levels of lead made their way into people's drinking water.After two months' research, Rao designed a small and portable device that used sensors to instantly detect lead in water. Called Tethys, after the Greek Goddess(女神) of freshwater, it attaches to a cellphone and informs the residents via an app if their drinking water contains lead. The design earned her the 3M Young Scientist Challenge in 2017. She is currently working with scientists and medical professionals to test Tethys' potential and hopes the device will be ready for commercial use by 2022.Later, Rao took on another social issue-drug addiction. Her app, called Epione, which won the Health Pillar Prize at the TCS Ignite Innovation Student Challenge in May 2019, is designed to catch drug addiction in young adults before it's too late.More recently, the teenager has developed an app named Kindly, which usesartificial intelligence technology to detect possible signs of cyberbullying(网上欺凌). When users type in a word or phrase, Kindly is able to pick it up if it's bullying, and then it gives the option to edit it or send it the way it is. It gives them the chance to rethink what they are saying so that they know what to do next time.All kinds of awesome, Gitanjali Rao has been selected from 5,000 equally impressive nominees(被提名人) for TIME Magazine's first-ever “Kid of the Year”.4. What gave Rao the idea of inventing the device Tethys?A. The incident of lead pollution.B. The issue with drug addiction.C. The shortage of water supplies.D. The high cost of purifying water.5. What is Rao expecting of Tethys?A. It'll remove metal from water.B. It'll make it to market soon.C. It'll win her a higher prize.D. It'll be fitted to cellphones.6. What will Kindly allow users to do?A. Receive pre-warning signals of threat.B. Input words into a computer automatically.C. Choose from secure social networking sites.D. Weigh their words before posting them online.7. Which of the following can best describe Gitanjali Rao as a young scientist?A. Ambitious and humble.B. Optimistic and adventurous.C. Talkative and outstanding.D. Creative and productive.CCuckoos don’t bother building their own nests—they just lay eggs that perfectly imitate those of other birds and take over their nests. But other birds are wishing up, evolving some seriously impressive tricks to spot the cuckoo eggs.Cuckoos are often know asparasites, meaning that they hide their eggs in the nest of other species. To avoid detection, the cuckoos have evolved so that eggs seem reproduction of those of their preferred targets. If the host bird doesn’t notice the strange egg in its nest, the little cuckoo will actually take the entire nest for itself after it comes out, taking the other eggs on its back and dropping them out of the nest.To avoid this unpleasant fate for their young, the other birds have evolved a few smart ways to spot the fakes, which we’re only now beginning to fully understand. One of the most amazing finds is that birds have an extra colour-sensitive cell in their eyes, which makes them far more sensitive to ultraviolet wavelengths and allows them to see a far greater range of colours than humans can. This allows cautious birds to detect a fake egg whichmight be exactly the same to our eyes.Fascinatingly, we’re actually able to observe different bird species at very different points in their evolutionary war with the cuckoos. For instance, some cuckoos lay their eggs in the nests of the redstarts. The blue eggs these cuckoos lay are practically alike to those of the redstarts, and yet they are still sometimes rejected. Compare that with cuckoos who target dunnocks. While those birds lay perfectly blue eggs, their cuckoo invaders just lay white eggs with brown irregular shaped spots. And yet dunnocks barely ever seem to notice the obvious trick.Biologists suspect these more easily fooled species like the dunnocks are on the same evolutionary path as the redstarts, but they have a long way to go until they evolve the same levels of suspicion. What’s remarkable is that the dunnocks fakes are so bad and the redstart ones so good, and yet cuckoos are still more successful with the former than the latter.It speaks to just how thoroughly a species’ behavior can be changed by the pressures of natural selection, or it might just be a bit of strategic cooperation on the part of the dunnocks. Biologists have suggested that these birds are willing to tolerate a parasite every so often because they don’t want to risk accidentally getting rid of one of their own eggs.8. This passage can be most likely found in a ________.A. science surveyB. nature magazineC. zoo advertisementD. travel journal9. What does the underlined word “parasite” in paragraph 2 most probably refer to?A. Animals that work together to raise young.B. Small harmful animals such as worms or mice.C. Animals that can adapt to changing environments.D. Animals which live on or inside other host animals.10. Which of the following is TRUE about the dunnock according to the passage?A. It is colour-blind and therefore cannot identify foreign eggs in the nest.B. It can easily remove cuckoo eggs from the nest because fakes are so bad.C. It is a host bird that is more likely to raise a cuckoo chick than the redstart.D. It is unable to evolve and hence accepts cuckoo eggs that appear in the nest.11. Which of the following can be inferred from the passage?A. Dunnocks may eventually learn to recognise foreign eggs.B. Redstarts seem to be less suspicious compared to dunnocks.C. Cuckoo birds are good at taking responsibility for their own young.D. It is very easy for cuckoos to imitate the colouring of the dunnock’s egg.DA year ago I received a full scholarship to attend the University of San Francisco. All of my hard work paid off. My mom had spent a lot on my attending a private high school, so I made sure to push myself: I volunteered, took part in various clubs, and graduated with honors. I was so excited to start a new part of my life.Soon enough, the big day came, but it wasn't like what I had thought. The first two weeks were the most difficult days of my entire life. Every night I would cry myself to sleep. I was missing my family, my home and everything in my hometown so much and I didn’t know how to deal with my broken heart.To distract myself, I threw myself into my studies. I also found a ton of jobs. In any free time, I started forcing myself to go to the gym. I wanted to keep every part of my day busy so I wouldn’t think about how lonely I felt. Soon after, I began to control my eating, considering it another solution to my homesickness. But soon there was something wrong with me.Finally, I went to see a doctor. When the doctor told me I had no choice but to take time away from school, I started to fear. How could I stop? School was what I was best at. “I’m not so bad,” I thought in my head. But the result was that I was taken to hospital again a month later and my mother camewoefully. I had to take a semester off from school, and go to the treatment center near my home.If there are girls who are suffering similarly, I hope you know that there is hope and that you should have a positive attitude towards life. Though you may feel alone, there are so many people who can understand your struggle. That’s why I want to share my story.12. Why did the author push herself during high school?A. She wanted to attend the University of San Francisco.B. It cost too much to study in a private school.C. Her parents controlled much of her life.D. Her family put her under pressure,13. What can we know about the author in the first two weeks?A. She couldn't fall asleep because of pressure.B. She couldn't pay attention to her study.C. She couldn't deal with her homesickness.D. She couldn't catch up with others.14. What does the underlined word “woefully” in paragraph 4 mean?A. Sadly.B. Surprisingly.C. Curiously.D. Happily.15. What is the author's purpose in writing this text?A. To look back on her past life.B. To increase her own confidence.C. To express appreciation to her mother.D. To encourage other girls like her to be positive.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学年北京密云二中高三英语月考试卷及参考答案
2019-2020学年北京密云二中高三英语月考试卷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AHow to Look at ShapeTake a seat at the virtual(虚拟的) table. At our new, monthly membership program, you'll join MoMA staff and fellow members for lively discussions about our collection and exhibitions. Ask questions, share your thoughts, and broaden the mind. A live Q&A, with Michelle Kuo and Anny Aviram, about shape and its role in MoMA's collection is also arranged.Draw, Write, and Connect with OthersExperiment with drawing and writing exercises as ways to connect with others, even when physically distant, in this 45-minute online workshop. This workshop is part of the Creativity Lab at Home plan. This session is led by Francis Estrada, Assistant Educator, and Hannah Fagin, Coordinator. Ifs open to anyone, but registration is limited and space is available on a first-come, first-served basis.Storytelling Through ArtThere are many ways to tell a story — through words through theater and dance, or through visual art, for example. Discover how artist Jacob Lawrence shared the history of an important event by combining words and art in a series of paintings calledThe Migration Series. For kids ages seven to fourteen. Parent participation is encouraged in this online event. Don' miss the opportunity to spend meaningful time with them.The Human ShelterIn 2016, MoMA opened Insecurities: Tracing Displacement and Shelter, an exhibition that examined how contemporary architecture arid design addressed ideas of shelter in light of global refugee(难民) emergencies. Danish Boris Benjamin Bertram documented the exhibition, and the result is a movie by him asking what makes a home, and, perhaps more importantly, when shelter becomes home. This online event is part of Member events.1.What is special about How to Look at Shape?A.It provides an interactive part.B.It is accessible to everyone.C.It is organized by Michelle Kuo.D.It focuses on MoMA's new collection.2.Which event is family-friendly?A.The Human Shelter.B.How to Look at Shape.C.Storytelling Through Art.D.Draw, Write, and Connect with Others.3.In which aspect might Bertram do well?A.Shelter design.B.Storytelling.C.Art education.D.Film-making.BOne rainy afternoon, I was on a crosstown bus when ayoung woman jumped on. She had a child with her who must have been about 3 or 4 years old.The bus was full, bumpy, and it soon got noisy as her kid began crying because he couldn’t sit next to his mother. There were a couple of open seats, but they weren’t together. She wasflusteredand looked embarrassed.Then another woman, a little older, stood up and moved so that the mother and child could sit together. The mom smiled as a thank-you. And then three words came out of the older woman’s mouth that elevated the entire energy of that bus ride: “I’ve been there”.Simple, undramatic and honest. In that moment, it seemed to unite people. Why? Because almost all experiences are shared human experiences. We forget that, as we forge (前进) through life, focused onour own troubles and needs—which are actually less unique than we think. How can these three words create more connection in your life? Ask yourself: “Where am I holding back?One thing I know for sure is this: Healing others helps heal yourself. I noticed this recently with my friend, Tracy, who took a new friend who had suffered a miscarriage under her wing. Tracy had three of them before having her daughter two years ago. Our intellect needs a doctor to explain the medical side of things, yes. But our souls need human connection to help us along. No one can do that better than someone who has been exactly where you are.Can the essence of these three words help you make a small difference right now? It can be as simple as volunteering your seat, sharing some helpful advice or even lightening the mood with a joke when you notice that someone’s uncomfortable—because we’re all in this together.4. The underlined word “flustered” in the second paragraph is closest in meaning to _______.A. angryB. anxiousC.scaredD. upset5. What does the woman mean by saying “ I’ve been there”in the third paragraph?A. The woman was on the bus and saw what had happened to the boy.B. The woman got to her destination and was ready to get off the bus.C. The woman once had the similar experience with that mother.D. The woman took the exact seat that the boy was on just now.6. Which of the following statements is TRUE according to the passage?A. Everyone has his or her own unique problem that is difficult to solve.B. Doctors can help us get through when we have mental or physical problems.C. The author’s friend Tracy felt better after she was comforted by her new friend.D. One can indeed make a difference to those in need of help by doing simple things.7. The passage isintended to _______.A. show a harmonious world by telling some touching storiesB. praise those who are willing to help others in emergenciesC. appeal to readers to give timely help to those in needD. illustrate some ways of helping others in detailCIn a recent survey of 2000 Americans, housecleaning was shown to have some mood-boosting effects — but that doesn't mean everybody is willing to do it.The majority of respondents (受访者) said cleaning gave them a sense of accomplishment (65%) and helped them clear their mind (63%). Half of these adults said they are most often motivated to clean when they're happy. In fact, 63% of those surveyed find the experience of cleaning to be relaxing - even more so than getting fresh air (61%).But that's not the only reason people clean. A big 70% admitted that tidying their home was a way of putting off having to do other things, with the average procrastinator (拖延者) using that trick four times a week. The survey showed that 86% of respondents do feel on top of their housework, but the last deep clean of their kitchen happened over a week and a half ago. That's no surprise because the kitchen is most terrible of all.Conducted by OnePoll on behalf of DishFish, the survey investigated people's attitudes toward dirty dishes and how they get through tricky task. More than two-thirds of people (69%) let their dishes pile up between washings with 20% saying “always” letting them be placed in the sink, which left them feeling stressed. More than any other room, the kitchen was rated as “very difficult” to cope with. And most people enjoy cleaning their toilet or taking out the garbage more than washing dishes by hand.How do they get through it? 66% listen to music while they clean. 72% have a best-loved song that they sounded while tidying up their home, with “Uptown Funk,”“Read All About It” and “Work” being the three favorite tunes on America's cleaning playlist.8. What is the result of the survey?A. Housecleaning may contribute to a good mood to some extent.B. Housecleaning may strengthen people's willingness to volunteer.C. Housecleaning may cause anxiety and concern for some people.D. Housecleaning may improve people's motivation to other housework.9. What is the top reason why people undertake housecleaning?A. They can entertain themselves.B. They can take in fresh air.C. They get a sense of achievement.D. They can delay other things.10. What are respondents' attitudes to dirty dishes?A. Many would rather wash dishes than throw out the rubbish.B. Half are under pressure with dirty dishes lying in the sink.C. A quarter will let dirty dishes pile up after their meals.D. Most prefer cleaning their toiletto washing dishes by hand.11. What column does the text belong to?A. Feature Story.B. Family Life.C. Scientific Hotspot.D. Finance Focus.DWatching what you eat can be easier said than done, but a recent study shows it might not just be about what's on your plate — it could be about how quickly it disappears.Japanese researchers followed 1,083 adultsfor five years, splitting them into three categories based on how quickly they ate: slow, normal, and fast. They also answered a questionnaire at the beginning of the study, sharing their diet, physical activity, and medical history. In the beginning, none of the volunteers had metabolic syndrome (新陈代谢综合征) - meaning at least three risk factors — which can lead to health problems like heart conditions and diabetes.When the participants reported back five years later 84 had been diagnosed (诊断) with metabolic syndrome — and their eating speed was a major predictor, according to the results in the journal Circulation. The fast eaterswere 89 percent more likely to have metabolic syndrome than slow and normal eaters. Just 2.3 percent of slow eaters received the diagnosis, compared to 11.6 percent of fast eaters. But that's not all. Fast eaters also saw more weight gain, larger waistlines, and higher blood sugar levels than slow eaters.The researchers saygobblingmakes it easier not to take notice of fullness before your body has a chance to signal you to stop. “So when people eat fast they are more likely to overeat,” said Takayuki Yamaji, MD, study author and cardiologist at Hiroshima University in Japan in a statement.Previous research backs up the weight benefits of slow eating, too. One study of New Zealand women found fast eaters have higher body-mass indexes (指数), and a Chinese study found that both healthy and fat men ate less when told to chew 40 times instead of 15 times before swallowing. Initial research even suggests chewing your food longer could bum more calories - up to about 1,000 extra every month.12. What are the participants divided by?A. Medical history.B. Health condition.C. Physical activity.D. Eating speed.13. Which may be the result of the study?A. Fast eaters are 4 times more likely to have metabolic syndrome.B. Normal and slow eaters don’t have metabolic illness.C. 89% of fast eaters have higher blood pressure.D. Slow caters are healthier than fast eaters.14. What does the underlined word “gobbling” in Paragraph 4 best mean?A. Tasting slowly.B. Digesting quickly.C. Eating greedily.D. Cooking carefully.15. What does the last paragraph tell us?A. The importance of eating speed.B. The advantage of eating slowly.C. The result of a Chinese study.D. Fast eating and overeating.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。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
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. 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 .22x y >
B .11x 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
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0ω>,||ϕ<π.若51()82f π=,(
)08f 11π
=,且()f x 的最小正周期大于2π,则
A .13
ω=,24ϕ11π
=-
B .23ω=
,12
ϕπ
= C .13
ω=,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 A
B .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 .
第9题图
1
1
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6
1
()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2
11(*)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2
2
{,,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,11
2
AC 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,]2
x ∈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 A
B
C A 1
B 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,
2
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)
5
Q -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()()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.(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
(,)[(||)(||)(||)]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,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.。