必修二第4单元测试

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必修二单元(4)

高中学生学科素质训练系列试题高一下学期化学单元测试（4）[新课标人教版] 命题范围第四章化学与可持续发展注意事项：1．本试卷包括第I卷（选择题）和第II卷（非选择题）两部分，满分100分。

考试时间90分钟。

2．答第I卷前，考生务必将自己的姓名、准考证号、考试科目用铅笔涂在答题卡上。

答第II卷时，须将密封线内的项目填写清楚。

3．考生请将第I卷选择题的答案填涂至答题卡上；第II卷的答案，用蓝、黑钢笔或圆珠笔直接答在各题的相应位置上。

考试结束后，只收答题卡和第II卷。

可能用到的相对原子质量：H 1 C 12 O 16 S 32 Cu 64 Fe 56第Ⅰ卷（选择题，共44分）一、选择题（本题包括10小题，每小题2分，共20分。

每小题只有一个选项符合题意）1．保护环境已经成为人类的共识。

人类应以可持续发展的方式使用资源，以合理的方式对废物进行处理并循环使用。

下列做法不利于．．．环境保护的是（）A．发电场的煤经脱硫处理B．将煤转化为水煤气作燃料C．回收并合理处理聚乙烯塑料废物D．电镀废液经中和后直接排放2．下列叙述合理的是（）A．金属材料都是导体，非金属材料都是绝缘体B．棉、麻、丝、毛及合成纤维完全燃烧都是只生成CO2和H2OC．水电站把机械能转化成电能，而核电站把化学能转化成电能D．我国规定自2008年6月1日起，商家不得无偿提供塑料袋，目的是减少“白色污染”3．植物及其废弃物可制成乙醇燃料，下列关于乙醇燃料的说法错误的是（）A．它是一种再生能源B．乙醇易燃烧，污染小C．乙醇只能在实验室内作燃料D．粮食作物是制乙醇的重要原料4．在冶金工业上，通常均不能用化学还原剂制得的金属组是（）A．Na、Ba、Mg、A1 B．Zn、Fe、Cu、AgC．Na、K、Zn、Fe D．Mg、Al、Zn、Fe5．下列说法正确的是（）A．石油的炼制过程都是化学变化过程B．石油分馏的目的是将含有碳原子数较多的烃先气化，经冷凝而分离出来C．石油经过常、减压分馏及裂化等工序炼制后即能得到纯净物D．石油分馏出来的各馏分仍是多种烃的混合物6．单质A能从盐溶液中置换出单质B，则下列说法中，不正确的是（）A．.当A、B都是金属时，A一定比B活动性强B．当A、B都是非金属时，A一定比B活动性强C．当A是金属时，B可能是金属，也可能是非金属D．当A是非金属时，B可能是金属，也可能是非金属7．下列属于高分子化合物且在自然界中原来并不存在的是（）A．淀粉B．纤维素C．聚乙烯D．蛋白质8．汽车排放的尾气中因含有氮的氧化物而污染大气。

2022-2023学年高中政治统编版选择性必修二第四单元社会争议解决测试卷

第四单元社会争议解决测试卷一、单选题1．民法典第1254条规定了高空抛物、坠物致害责任，对难以明确具体侵权人的，由可能的加害人（建筑物使用人）予以补偿，能够证明自己不是侵权人的除外。

某小区发生一起高空坠物事件，居民陈某被大风吹落的一盆花砸成重伤。

监控录像显示，花盆从五楼以上的住户家中坠落。

受害人将该楼所有住户告上法院，要求赔偿。

在此案中（）①原告和被告可以委托辩护人帮助自己进行诉讼②监控录像拍摄的相关视频属于视频资料，坠落的花盆属于物证③根据“谁主张，谁举证”的举证原则，陈某要证明花盆与住户之间的归属关系④法院如果立案，被告应承担自己没有过错的举证责任A．①②B．①③C．②④D．③④2．马某与杨某为了配合家里的拆迁于2020年12月登记离婚。

双方在婚姻登记机关签订的《自愿离婚协议书》约定:“婚生女萌萌由杨某抚养，马某支付相应的抚养费。

”一段时间后马某以双方系“假离婚”为由拒付抚养费。

萌萌于2022年1月诉至法院。

本案中（）①马某应按《自愿离婚协议书》的约定按时支付抚养费②《自愿离婚协议书》系双方真实的意思表示，合法有效③“假离婚”非双方真实离婚意愿，马某无须支付抚养费④判决生效后，当事人任何一方若不服，均可以提出上诉A．①②B．①③C．②④D．③④3．下列行为符合我国法律规定的是（）①消费者遭遇“双十一”先提价后打折的促销陷阱后可直接向仲裁机构申请仲裁②王某在送货途中因疫情引发的交通管制导致货物无法按期送达，应承担违约责任③某平台外卖配送员送外卖途中撞伤人且承担主要交通责任，用人单位应承担赔偿责任④儿子不赡养母亲戴某，遗赠扶养协议的扶养人可凭借与戴某订立的遗赠扶养协议继承其百万房产A．①②B．①④C．②③D．③④4．2022年4月，扬州市邗江区检察院依法对一起因口角矛盾引发的故意伤害案件犯罪嫌疑人作出不起诉决定。

面对没有积蓄的犯罪嫌疑人和急需手术的伤者，承办检察官及时启动司法救助程序，发放了3万元司法救助金用于伤者手术治疗。

2020年人教版必修二Unit 4 Wildlife protection单元测试

Unit 4 Wildlife protection话题诵读日积月累[导语]世界自然基金会(WWF)是在全球享有盛誉的、最大的独立的非政府的环境保护组织之一，自1961年成立以来，WWF一直致力于环保事业，拥有全世界将近520万的支持者和一个在100多个国家活跃着的网络。

北部白犀牛小嘴狐猴World Wildlife FundFor more than 45 years，World Wildlife Fund (WWF) has been protecting the future of nature.The largest multinational conservation organization in the world，WWF works in more than 100 countries and is supported by 1.2 million members in the United States and close to 5.2 million globally.WWF's unique way of working combines global reach with a foundation in science，involves action at every level，from local to global，and ensures the delivery of innovative (创新的) solutions that meet the needs of both people and nature.WWF's mission is the conservation of ing the best scientific knowledge available and advancing that knowledge，it works to preserve the diversity and abundance (丰富) of life on theearth and the health of ecological systems (生态系统)．WWF safeguards hundreds of species around the world，but it focuses special attention on the flagship species: giant pandas，tigers，polar bears，endangered whales (鲸) and dolphins，rhinos，elephants，marine turtles and great apes (大猩猩)．These species not only need special measures and extra protection in order to survive，they also serve as umbrella species: helping them helps lots of other species that live in the same habitats.In addition to the flagship animals，it works in several areas of species protection all around the world and works for the animals that live within the world's priority (重点) ecoregions.Its wildlife trade experts at TRAFFIC work to ensure that trade in wildlife products doesn't harm a species，while also fighting against illegal and unsustainable trade.By the year 2020 WWF will conserve (保存) 19 of the world's most important natural places and significantly change global markets to protect the future of nature.[词海拾贝]1．conservation n．保护2．global adj.地球的，全球的3．foundation. n．基金(会)；基础4．ensure v t.确保；使……获得5．mission n．代表团；使命[问题思考]1．What does WWF stand for?_______________________________________________________ 答案：WWF stands for World Wildlife Fund.2．What's WWF's task?_______________________________________________________ 答案：WWF's task is the conservation of nature.3．What should we do as a student?_______________________________________________________ 答案：略自主预习步步提高词汇识记Ⅰ.将单词与其正确释义配对1．decrease A．鉴赏；感激；意识到2．reserve B．安全的；可靠的3．hunt C．雇佣；利用4．respond D．减少；变小5．contain E．包含；容纳；容忍6．affect F．保护区7．appreciate G．成功；接替；继任8．succeed H．回答；响应；做出反应9．secure I．损害；危害10．income J．收入11．employ K．打猎；猎取；搜寻12．harm L．影响；感动；侵袭答案：1.D 2.F 3.K 4.H 5.E 6.L7.A8.G 9．B10.J11.C12.IⅡ.根据所给汉语提示写出单词1．________n．仁慈，宽恕；怜悯2．________v．咬；叮；刺痛3．________v t.检查；视察4．________n．事件；事变5．________adj.凶猛的；猛烈的答案：1.mercy 2.bite 3.inspect 4.incident 5.fierce Ⅲ.根据语境及汉语意思或首字母提示写出正确的单词1．He asked to be put under police________(保护)．2．As a child I learned to________(打猎) and fish. 3．Your opinion will not________(影响) my decision. 4．He had hoped to________(成功) as a violinist. 5．Hard work never did anyone any________(伤害)．6．Yogurt c________much less fat than cream. 7．Russia used to be a very p________country.8．He turned his a________back to the road again. 9．In time you'll a________the beauty of this language. 10．How many people does the company e________？答案：1.protection 2.hunt 3.affect 4.succeed 5.harm6．contains7.powerful8.attention9.appreciate10．employⅣ.选用短语的适当形式完成句子decrease to; in peace; in danger; burst into laughter；pay attention to; do harm to; protect... from; in relief；die out; at the mercy of1．Reading in bed for a long time will________________your eyes.2．Every student should________________what your teacher said in class.3．The two countries have been living together______．4．No one was injured in the accident; we all smiled_______．5．It's high time that we should take measures to protect these old traditions before they______________．6．The poor cat is________________the naughty boy.7．Hearing the funny story, all of the boys________________．8．Children's lives are____________________every time they cross the road.9．The government is doing the best to ________those rare animals________ being hunted.10．The number of the rare birds in this area has________aboutninety.答案：1.do harm to 2.pay attention to 3.in peace 4.in relief 5.die out 6.at the mercy of7.burst into laughter8．in danger9.protect from10.decreased to语篇理解课文表层理解Ⅰ.根据课文内容判断正(T)误(F)1．The antelope in Tibet becomes an endangered species because of its fur being used to make sweaters.()2．The living conditions of the elephants are worse than they were.()3．What the government of Zimbabwe did provide us with a good example of how to protect wildlife.()4．The millipede insect produces a drug naturally which can protect animals from mosquitoes.()5．Daisy returned home with the monkey and began to make her new drug.()答案：1.T 2.F 3.T 4.T 5.FⅡ.读课文，回答问题1．What does the word “it”refer to in Line 2 in the second paragraph?______________________________________________________________________________________________________________ 2．Fill in the blank with the words from the second paragraph.When Daisy heard that the elephant asked whether she came to take its photo，she laughed ________．3．Why did the monkey rub itself according to the third paragraph?_______________________________________________________ _______________________________________________________ 4．Was Daisy able to make her new drug，why?_______________________________________________________ _______________________________________________________ 答案：1.The elephant.2．in relief3．Because the monkey protected itself from mosquitoes.4．No，she wasn't.Because she was in her dream and everything had gone.Ⅲ.根据课文内容完成下表Character Means of transport AimDaisy(1)______________________ To see some (2)________Daisy's JourneyAnimals Places Situations (3)______ TibetThey are being killed for their(4)________An elephant (5)______ They used to be an endangered species because of farmers' too much (6)______ Their numbers are (7)________ because they get protection from the (8)________A monkeyA thickrainforest The monkey is rubbing a (n)(9)__________over its body to protect itself from mosquitoes，which shows plants，animals and human beings live together by(10)________each other答案：1.A flying carpet 2.endangered wildlife3．An antelope 4.soft fur 5.Zimbabwe 6.hunting7．lipede insect10．supporting课文深层理解Ⅰ.读课文，回答问题What's the main idea of the passage?It is about a strange dream that Daisy dreamed.In her dream，she took a ________ to travel and talked with many different________，which made her know the importance of ____________________．答案：flying carpet animals wildlife protectionⅡ.仔细阅读课文，选择最佳选项1．Daisy first learned about the antelope ________．A．from her sweater made of furB．once she was in TibetC．after she saw an antelope looking sadD．when she woke up from her dream2．There are not as many antelopes as it used to be in Tibet ________．A．because of pollutionB．due to the development of industryC．because they're being killed for furD．for lack of food3．Why did elephants use to be an endangered species in Zimbabwe?A．Tourists hunted too many elephants.B．Farmers hunted them without mercy.C．The government encouraged farmers to hunt them.D．Elephants were killed by the government.4．Farmers used to hunt and kill elephants because they________．A．didn't want to make more moneyB．wanted to protect their cropsC．wanted to get their fur and meatD．thought they were dangerous to them5．What does “No rainforest，no animal，no drugs”mean?A．There are no drugs in the rainforest.B．There will be no rainforest，no animals and no drugs in the future.C．Animals in the rainforest don't need drugs.D．The rainforest，animals and drugs are closely related to each other.答案：1.C 2.C 3.B 4.B 5.DⅢ.领会句子所用的句型并译成汉语1．This is what wildlife protection is all about._______________________________________________________ 2．We're being killed for the wool beneath our stomachs._______________________________________________________3. No rainforest, no animals，no drugs._______________________________________________________ 4．After a while she saw some zebra with black and white lines going beneath their stomachs._______________________________________________________ 答案：1.这正是野生动物保护的意义所在。

人教版高一数学必修二第四章圆与方程(单元测试,含答案).doc

与方程姓名：班级:一、选择题(共8小题；共40分)1Mx2 +尸一4x + 6y = 0的圆心坐标是()A (2,3)B (-2,3) C(-2,-3) D(2,-3)2OO的百径是3,百线1与OO相交，圆心0到百线1的距离是d,贝M应满足()Ad > 3 B 15 < d < 3 C 0 < d < 15 Dd < 0 3圆(x — 2)2 + (y- l)2 = 4与圆(x + l)2 + (y- 2)2 = 9的公切线有()条A1 B 2 C3 D4 4从原点向圆x2 + y2 一12y + 27 = 0作两条切线，则该圆夹在两条切线间的劣弧长为()A nB 2nC 4TTD 6TT5过点(1,1)的直线与圆(x - 2)2 + (y - 3)2 = 9相交于A, B两点，贝lj| AB |的最小值为() A2V3 B4 C2V5 D5 6已知圆C的半径为2, |员|心在x轴的正半轴上，直线3x + 4y + 4 = 0与圆C相切，贝I」圆C的方程为()Ax2 4-y2 - 2x - 3 = 0 B x2 4- y2 + 4x = 0Cx2 +y2 + 2x - 3 = 0 D x2 + y2 - 4x = 07耍在边长为16米的正方形草坪上安装喷水龙头，使整个草坪都能喷洒到水.假设每个喷水龙头的喷洒范閘都是半径为6米的圆面，则需安装这种喷水龙头的个数最少是()A6 B 5 C4 D38 已知圆:C1:(x-2)2 + (y-3)3 = 1,圆：C2:(x-3)2 + (y-4)2 = 9, M、N分别是圆C〔、C?上的动点，P为x轴上的动点,贝OIPMI + IPNI的最小值为()A5V2-4 B V17- 1 C6-2V2 D V17二、填空题(共7小题；共35分)9过点A(3,—4)与闘x2 +y2 = 25相切的直线方程是_______ .10如果单位圆X? +y2 = 1与圆C: (x — a)2 + (y - a)2 = 4相交，则实数a的取值范围为 ________ 11在空间直角坐标系,已知点A(l,0,2), B(l,-3,1),点M在y轴上,且M到A与到B的距离相等,则点M的坐标是 _____ ・12已知圆C: (x-2)2+y2 = l.若直线y二k(x+l)上存在点P,使得过P向圆C所作的州条切线所成的角为夕则实数k的取值范闌为 _______ .13如图，以棱长为a的止方体的三条棱所在的直线为坐标轴建立空间百角坐标系，若点P为对角线AB的点，点Q在棱CD上运动，则PQ的最小值为 .14在圆C:(x-2)2 + (y-2)2 = 8内,过点P(l,0)的最长的弦为AB,最短的弦为DE,贝9以边形ADBE的面积为____ •15据气象台预报：在A城正东方300km的海而B处有一台风心，正以每小时40km的速度向術北方向移动，在距台风心250km以内的地区将受其影响.从现在起经过约__________ h,台风将影响A城, 持续时间约为_______ h.(结果精确到Olh)三、解答题(共5小题；共65分)16若关于x, y的方程X? + y? - 4x + 4y + m = 0表示圆C.(1)求实数m的取值范围；(2)若圆C与圆M:x2 4-y2 = 2相离，求m的取值范囤.17已知圆C:x? + y? + 4x + 4y + m = 0,直线l:x + y 4- 2 = 0.(1)若I员IC与直线1相离，求m的取值范围；(2)若I员1D过点P(l,l), H.与恻C关丁•直线1对称，求I処D的方程.18如图，在平面直角坐标系xOy,点A(0,3),直线l:y = 2x-4.设圆C的半径为1,圆心在1上.(1)若圆心C也在直线y = x-l上，过点A作圆C的切线，求切线的方程;(2)若圆C上存在点M,使MA = 2M0,求圆心C的横坐标a的取值范|节|・19已知直线啲方程为2x+(l + m)y+2m = 0, m€R，点P的坐标为(-1,0).(1)求证：直线1恒过定点，并求出定点坐标；(2)求点P到直线1的距离的最大值；(3)设点P在直线1上的射影为点M, N的坐标为(2,1),求线段MN长的取值范闱.20 在平面直角坐标系xOy,已知圆Ci： (x + 3)2 + (y - I)2 = 4和圆C?： (x 一4)2 + (y — 5)2 = 4.(1)若直线1过点A(4,0), £L被圆C］截得的弦长为2孙，求直线啲方程；（2）设P为平面上的点，满足：存在过点P的无穷多对互相垂肖的肖线h和12,它们分别与圆C1 和圆C2相交，且直线h被圆C］截得的弦长与直线12被圆C2截得的弦长相等，试求所有满足条件的点p的坐标.答案第一部分I D 2 C 3 B 4 B 5 B 6 D 7 C 8 A第二部分9 3x-4y = 2510 -—< a < H J C —< a < —」 2 22 2 II (0,-1,0) 12 ［一普，晋］13 yal4 4V615 20； 66第三部分 16 (1) |w|C 化简为(x- 2)2 4-(y + 2)2 = 8-m,所以8 — m > 0,即m V 8.(2)圆C 的圆心为(2,-2),半径为V8^ (m<8),圆M 的圆心为(0,0),半径为返,由题意，得圆心距大于两圆的半径和，则“22 + 22 + 解得6<m<8.17 (1)圆Ux?+y2+4x + 4y + m = 0即(x 4- 2)2 + (y + 2)2 = 8 - m.圆心C(-2,—2)到直线啲距离d =三|旦=V2,若圆C 与直线1相离，则d > r,所以 * = 8 — m < 2即 m > 6乂严=8 - m > 0即m V 8.故m 的取值范围是(6,8).(2)设圆D 的圆心D 的坐标为(xo ，y ())，由于圆C 的圆心C(_2,_2), 依题意知点D 和点C 关于直线1对称，解牡:0 所以圆D 的方程为x 2+y 2 = r 2,而r=|DP |=V2,因此，圆D 的方程为x 2+y 2 = 2.18 (1)由题设，I 员I 心C 是直线y = 2x- 4和y = x- 1的交点, 解得点C(3,2),于是切线的斜率必存在.设过A(0,3)的圆C 的切线方稈为y = kx + 3由题意，得解得：k=0或—孑 4故所求切线方程为{Xo-2 Yo+2Xo+2 + 竽+2 = 0x (-1) = -1I 3k + 1 |Vk 2 + 1y = 3 或3x + 4y — 12 = 0(2)因为圆心在直线y = 2x —4上，所以圆C的方程为(x — a)2 3 + ［y — 2 (a — 2)］2 = 1 设点M(x,y),因为MA = 2M0,所以Jx2 + (y — 3)2 = 2jx2 +y2, 化简得x? + y2 + 2y — 3 = 0,即x2 + (y + l)2 = 4, 所以点M在以D(0,-l)为圆心，2为半径的圆上.由题意，点M(x,y)在圆C上，所以圆(：与圆D有公共点，贝I」12-11 < CD <2 + 1, 即l<Va2 + (2a-3)2<3 整理，得—8 S 5a2— 12a S 0由5a2-12a + 8>0,得a G R;S5a2 - 12a < 0,得12所以点C的横坐标a的取值范闌为［0,y .19(1)由2x + (l + m)y+2m = 0得2x + y + m(y + 2) = 0,所以直线1恒过直线2x + y= 0与直线y + 2 = 0交点Q.解方程组炸暮律得Q(l，-2)，所以直线1恒过定点，且定点为Q(l,-2).2 设点P在直线1上的射影为点M,贝IJIPMI < |PQ|,当且仅当直线1与PQ垂直时,等号成立, 所以点P到直线1的距离的最大值即为线段PQ的长度为2逅.3因为直线1绕着点Q(l,-2)旋转，所以点M在以线段PQ为直径的I员1上，其I员I心为点C(O.-l)，半径为说，因为N的坐标为(2,1),所以|CN| = 2V2,从而V2 < |MN| < 3V2.20(1)由于直线x = 4与圆C］不相交，所以直线1的斜率存在.设直线1的方程为y = k(x - 4),圆C］的I员I心到直线1的距离为d, 乂因为直线1被I员©截得的弦长为2箱，所以|l-k(-3-4)| d = ------- , ----Vl + k 2 y = 0 或 7x + 24y - 28 = 0 (2)设点P(a,b)满足条件，不妨设直线h 的方程为y — b = k(x — a), k H 0, 则直线】2的方程为山点到直线的距离公式得 d = J22 - (V3)2 = 1从而即所以直线1的方程k(24k + 7) = 0, 7 241因为圆Ci和C2的半径相等，及宜线I】被圆C］截得的弦长与直线-被【员丄2截得的弦长相等，所以I 员IC］的|员］心到直线1］的距离和圆C2的國心到直线】2的距离相等,即|1 一k(-3 - a) - b| |5 + £ (4 — a) — b|整理得|1 + 3k + ak — bl = |5k + 4 — a — bk|,从而1 + 3k + ak — b = 5k + 4 — a - bk,(a + b — 2)k — b — a + 3, 因为k的取值有无穷多个，所以(a + b — 2 = 0,戒(a — b + 8 = 0, (b - a + 3 = 0 严ia + b-5 = 0 解得这样点P只可能是点P］ (I，-扌)或点卩2 (-!，¥)• 经检验点P］和P2满足题口条件.。

新外研版高中英语必修二Unit 4 Stage and screen单元测试(含答案)

新外研版高中英语必修二Unit 4 Stage and screen单元测试（考试时间：120分钟分值：150分）一、根据汉语提示写出正确的单词（每小题0.5分，共5分）1.You mustn’t miss that modern __________ (芭蕾舞剧), because it is extremely good.2.There was a __________（短暂的）pause in the conversation.3.The __________（纪录片）is about animal life in Africa.4.John Smith’s latest movie is a real __________ (失败之作).5.There are two __________（女性的）singers in the popular band.6.It’s __________（正常的）to feel tired after such a long trip.7.The extra money could be spent on improving public __________ (交通运输系统).8. A prisoner __________（逃离）from a jail in northern England last Monday.9. She felt she had looked at the problem from every __________ (方面).10.This musician has perfect __________（技巧，方法）but little expression.二、用所给单词的正确形式填空（每小题0.5分，共5分）1. A lot of workers and students took part in the peace __________ (move).2. Summer is a time when people feel __________ (energy).3. I myself enjoy novels, __________ (poet), eating out and long walks.4. My husband is __________ (extreme) long-sighted while I am very short-sighted.5. Why don’t you ever give me flowers? I wish you’d be more __________ (romance).6. There are special __________ (arrange) for people working overseas.7. The __________ (origin) castle was probably built about AD 860.8. There is __________ (absolute) no difference between the two pictures!9. Can you give me a __________ (definitely) answer by tomorrow?10. The men danced and the women __________ (clap).三、选择合适的短语，并用其正确形式，完成句子（每小题1分，共5分）1.John is a lawyer _______________.2.The weather has helped us _______________.3.Actually, when you are really excited or nervous about something, you can say that you are _______________.4.You should_______________ what your parents expect of you.5.What message are you trying to_______________ to the consumer?四、阅读理解（每小题2分，共30分）阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项。

2023-2024学年高中语文语文版必修二第四单元单元测试(含答案解析)

2023-2024学年语文版高中语文单元测试班级 __________ 姓名 __________ 考号 __________一、单选题（本大题共计3小题每题3分共计9分）1.下列各组句子中画线词的词类活用不相同的一项是（）A. 常以身翼蔽沛公良庖岁更刀B. 浴乎沂风乎舞雩然而不王者未之有也C. 项伯杀人臣活之抑王兴甲兵危士臣D. 老吾老以及人之老越国以鄙远【答案】D【解析】A项都是名词作状语B项都是名词作动词C项都是使动用法D项名词作动词/意动用法故选D2.下列对活用情况说明有误的一项是（）A. 吾得兄事之兄名词作状语像对待兄长一样常以身翼蔽沛公翼同上像翅膀一样B. 籍吏民封府库籍名词用作动词登记在册沛公军霸上军同上驻扎C. 先破秦入咸阳者王之王名词使动用法使……为王沛公欲王关中王同上D. 素善留侯张良善形容词用作动词交好拔剑撞而破之破形容词用作动词击碎、击破【答案】C【解析】C项“沛公欲王关中”句意为沛公想要在关中称王王名词用作动词称王3.下列各句中画线词的活用类型与例句相同的一项是（）例句既东封郑又欲肆其西封A. 且贰于楚也B. 范增数目项王C. 项伯乃夜驰之沛公军D. 越国以鄙远【答案】C【解析】东名词作状语在东面A项贰数量词作动词指从属二主B项目名词作动词用眼神示意C项夜名词作状语在晚上D项远形容词作名词远的地方故选C二、文言文阅读（本大题共计1小题每题15分共计15分）4.（1）下列画线的字解释正确的是（）4.（2）下列对《鸿门宴》一文中的文学常识的解说不正确的一项是（）4.（3）下列叙述不符合文意的一项是（）4.（4）把文中画横线的句子翻译成现代汉语①臣死且不避卮酒安足辞！夫秦王有虎狼之心杀人如不能举刑人如恐不胜天下皆叛之②故遣将守关者备他盗出入与非常也③大行不顾细谨大礼不辞小让如今人方为刀俎我为鱼肉何辞为？A. 交戟之卫士欲止不内同“纳”使进B. 沛公起如厕到往C. 道芷阳间行小径D. 窃为大王不取也私下【答案】C【解析】（1）C项句意取道芷阳抄小路秘密地走间秘密地【答案】B【解析】（2）B项山东崤山以东【答案】D【解析】（3）D项“显示了项羽‘为人不忍’的性格”错句中描述并不能显示项羽“为人不忍”的性格而是为了表现樊哙的勇武【答案】（4）①我死都不怕一杯酒有什么可推辞的？秦王有虎狼一样的心肠杀人惟恐不能杀尽惩罚人惟恐不能用尽酷刑所以天下人都背叛他②派遣将领把守关口是为了防备其他盗贼的进入和意外的变故③做大事不必注意细枝末节行大礼不必讲究小的谦让现在人家正好比是菜刀和砧板我们则好比是鱼和肉何必告辞呢？【解析】（4）①举、胜都有“尽”的意思②备为了防备非常意外的变故③让责备刀俎刀和砧板何辞为宾语前置句为语气助词用于句末表示反问三、古诗词鉴赏（本大题共计1小题每题15分共计15分）5.（1）下列对这首词的赏析不正确的一项是（）5.（2）请简要赏析“雄关漫道真如铁而今迈步从头越”两句（从修辞手法和思想感情两个角度考虑）A. 词的上片写拂晓时分红军急行军前往娄山关下片写黄昏时分战斗结束全篇没有从正面描写战争B. “西风烈长空雁叫霜晨月” 这两句极为精炼传神不但交代了红军进攻娄山关的时间、环境而且营造了一种壮烈的抒情氛围C. “马蹄声碎喇叭声咽” “碎”字表明了马蹄声急而低“咽”字表现了战士们内心的悲凉用字十分精妙D. 这是一首悲壮的革命战争的赞歌凝重与雄劲豪放相统一气氛悲壮但又蕴含着勃勃生机赞颂了红军的豪迈气概和英勇精神【答案】C【解析】（1）C项“表现了战士们内心的悲凉”错误“咽”指的是在昂首挺进的队伍中军号吹奏着沉郁低回的音调暗示了战斗的激烈【答案】（2）①“雄关漫道真如铁”使用了比喻的手法暗示出夺取娄山关的艰难②这两句词体现了红军夺取娄山关后的豪迈之情表现了革命队伍踏平艰难险阻的决心和勇气【解析】（2）这两句词的意思是“不要说娄山关坚硬如铁难以逾越而今让我们重振旗鼓大步向前” “雄关漫道真如铁”一句使用了比喻的修辞手法通过坚硬、沉重的“铁”这一物象艺术地把“艰辛”具体化、形象化暗示出夺取娄山关的艰难“而今迈步从头越”中的“迈步”指举大步“从头越”指重振旗鼓奋勇向前这一句形象地勾画出红军一往无前的战斗雄姿表达了红军再次越过娄山关的胜利豪情这两句词体现了红军夺取娄山关后的豪迈之情表现了红军跨越雄关、踏平险阻的坚强决心和无畏勇气四、现代文阅读（本大题共计2小题每题15分共计30分）6.（1）下列对小说相关内容和艺术特色的分析鉴赏不正确的一项是（）6.（2）小说中说赵一曼“身上弥漫着拔俗的文人气质和职业军人的冷峻” 请结合作品简要分析6.（3）小说中历史与现实交织穿插这种叙述方式有哪些好处？请结合作品简要分析A. 小说以“赵一曼女士”为题不同于以往烈士、同志、英雄等惯常用法称谓的陌生化既表达了对主人公的尊敬之意又引起了读者的注意B. “通过对此人的严厉审讯有可能澄清中共与苏联的关系”这既是大野泰治向上级提出的建议也暗示他已从赵一曼那里得到有价值的回答C. “他指着石碑说赵一曼？我说对赵一曼”两个陌生人之间有意无意的搭讪看似闲笔实则很有用心说明赵一曼仍活在人们的记忆里D. 医院是“我”与赵一曼的连接点小说由此切入主人公监禁期间鲜为人知的特殊生活经历在跨越时空的精神对话中再现了赵一曼的英雄本色【答案】B【解析】（1）B项“暗示他已从赵一曼那里得到有价值的回答”赏析有误根据文本第六段“滨江省警务厅的大野泰治对赵一曼女士进行了严刑拷问始终没有得到有价值的回答他觉得很没面子”可知大野泰治从赵一曼女士那里“始终没有得到有价值的回答”【答案】（2）①文人的气质喜欢丁香花情趣不俗时常深情、甜蜜地回忆战斗生活文雅浪漫用大义与真情感化青年智慧过人②军人的冷峻遭严刑拷打而不屈服意志坚定笑对即将到来的死亡从容淡定充满母爱又不忘大义理智沉稳【解析】（2）解答时在通读全文的基础上结合小说对人物外貌、语言、神态、动作等方面的描写分析人物形象的特点即可赵一曼“拔俗的文人气质”主要体现在以下几个方面根据第五段“白色的小柜上有一个玻璃花瓶里面插着丁香花赵一曼女士喜欢丁香花” 可知赵一曼虽然被监禁但仍然爱着丁香花可见其高雅的情怀根据第三段“冬夜里钟声会传得很远很远钟声里抗联的士兵在森林里烤火烤野味儿……这些都给躺在病床上的赵一曼女士留下清晰回忆” 可知赵一曼时常回忆往日的战斗生活富有浪漫精神根据第二十一段“对女护士韩勇义赵一曼女士采取的则是‘女人对女人’的攻心术”可知赵一曼能用大义和真情感化他人并采用了一些艺术化的方式具有知识分子的智慧赵一曼“职业军人的冷峻”主要体现在根据第六段“滨江省警务厅的大野泰治对赵一曼女士进行了严刑拷问始终没有得到有价值的回答他觉得很没面子” 可知赵一曼遭到严刑拷打但不屈服有钢铁般的意志根据“赵一曼女士淡淡地笑了” 可见赵一曼为国捐躯之际从容赴死有视死如归的气魄小说中赵一曼争取董警士和韩勇义时进行了长期的观察、思考并采取了相应的手段可见有军人般的洞察力和执行力小说最后的遗书则体现出赵一曼在充满母爱的同时也不忘国家大义以身作则【答案】（3）①既能表现当代人对赵一曼女士的尊敬之情又能表现赵一曼精神的当下意义使主题内蕴更深刻②可以拉开时间距离更加全面地认识英雄使人物形象更加立体③灵活使用文献档案与小说叙述相互印证使艺术描写更真实【解析】（3）解答时在通读全文的基础上从故事情节、人物形象、主题思想、艺术效果等角度分别分析小说历史与现实交织的作用即可文章以“现实”开头第一段“后来得知赵一曼女士曾在这里住过院我便翻阅了她的一些资料” 简单交代了现实情况之后紧接对历史中赵一曼的介绍描绘二者相互印证显得更加真实可信更有说服力在人物塑造方面文章既有直接描写历史中的赵一曼又通过对现实情况的描写层面衬托出赵一曼的形象如“我去的时候在那里遇到一位年迈的老人他指着石碑说赵一曼？我说对赵一曼” 通过现实中两人的搭讪侧面表现出赵一曼仍被人铭记使赵一曼的形象更加立体完整在故事情节方面历史与现实的交织使小说的情节更加紧凑如“《滨江省警务厅关于赵一曼的情况》扼要地介绍了赵一曼女士从市立医院逃走和被害的情况”就通过现实引出后文对赵一曼遇害的历史事实的描写推动了故事情节的发展在主题思想上历史与现实的交织使赵一曼的形象不仅仅停留在历史中更有了丰富的现实意义突出了对赵一曼英雄精神的赞美和敬佩之情使得文章内涵更深刻7.（1）下列关于原文内容的理解和分析正确的一项是（）7. （2）下列对原文论证的相关分析不正确的一项是（）7.（3）根据原文内容下列说法不正确的一项是（）A. 汉服以四千年前的礼仪文化为基础形成了独具华夏民族文化风貌的服装体系B. 汉服的领口分“交领”“直领”和“盘领” 但衣襟都是左侧压右侧都属“右衽”款式C. 汉服中作为礼服的“深衣”的袖子长度是有规制的其长度等于臂长的一倍半D. 汉服的上衣象征上升的清轻之气用鲜明的纯色下裳则用象征重浊之气的杂色【答案】C【解析】（1）A项“四千年前”理解错误原文是“从黄帝至明末近四千年中”B项“直领”的两扇衣襟是直接从胸前平行垂直下来的彼此不交叠所以不分左右衽D项文中只是说“上衣崇尚颜色端正纯一” 并没有说一定要“鲜明”故选C【答案】B【解析】（2）B项此处的引用只是让读者了解到了衣和裳的作用无法证明和传统文化之间的关系【答案】C【解析】（3）C项“襟三江而带五湖”中的“带”是指腰间的大带用以表明南昌处于五湖环绕的中心位置五、综合读写（本大题共计2小题每题15分共计30分）8.百年风雨兼程世纪沧桑巨变光明中学举行首届校园戏剧节诚挚邀请学子们寻找、挖掘并讲述契合时代的历史故事在收到的剧本中有不少句子措辞不当需精心校稿请找出并修改下面五个句子的错误①端午将至特为兄台奉上粽礼一份聊表心意还请惠存②就定于8月8日8时8分小店准时开张大吉欢迎各位届时光临③在下才疏学浅方才斗胆进言其间多有不妥之处难免大师见笑鄙人啊④老夫行动不便只能就此留步如蒙不弃欢迎随时光临寒舍⑤大人垂爱本当敬从但卑职疏懒已久不堪重用未敢俯就还请见谅【答案】①“惠存”改为“笑纳”②删去“大吉”③“见笑”改为“哂笑”“嘲笑”“取笑”等（或删去“鄙人”）④“留步”改为“恭送”⑤“俯就”改为“高就”【解析】①“惠存”是敬词意思是请保存多用于赠人照片、书籍等时所题的上款此处是送给对方“粽” 是送给对方礼物应使用“笑纳” “笑纳” 客套话用于请人收下自己的礼物②“大吉”意思是非常吉利表示祝福而句中要表述的是“小店开张”这一事实可以删除“大吉”③“见笑” 被人笑话（多用作谦辞）后面不能跟宾语可以改为“嘲笑” 也可删除“鄙人”这一词语④“留步” 客套话用于主人送客时客人请主人不要送出去不能说自己“留步” 应改为“恭送”⑤“俯就”是敬辞用于请对方同意担任职务不能用于自己可以改为“高就”9.（1）文中四个画线词语引号的用法与下面的句子中“金沙滩”一词引号的用法相同的一项是（）电视剧《山海情》根据上个世纪90年代“西海固”地区整村搬迁建设闽宁镇的感人故事改编闽宁镇能在短短20年变成“金沙滩” 是每一个扶贫干部、支教老师、吊庄农民用汗水和鲜血换来的9.（2）在上文横线处补写恰当的语句使整段文字语意完整连贯内容贴切逻辑严密（每句不超过15个字）A. “出逃”B. “得宝们”C. “不输在起跑线上”D. “起跑线”【答案】D【解析】（1）“金沙滩”中的引号表示特殊含义A项引号表示强调B项引号表示特定称谓C项引号表示引用D项引号表示特殊含义故选D【答案】（2）①剧中共安排了两次“出逃”的戏②但教育带给他们的精神启蒙/动力③注定不可能有后来的成功【解析】（2）①处根据“《山海情》的首尾很有意味”“一次是……一次是……这两次‘出逃’……”可知此处应填写与“剧中共安排了两次‘出逃’的戏”类似的句子②处根据“得宝、水旺等人都是白校长的学生上过学憧憬山外面的世界虽然他们没有受到完整的教育”“使得他们有勇气脚步朝外打破涌泉村陈旧的生存状态”可知是教育带给得宝、水旺等人精神启蒙让他们有勇气走出去此处应填写与“但教育带给他们的精神启蒙/动力”类似的句子③处联系实际并结合“他们像许许多多人一样落入‘不输在起跑线上’的谬误认知中倘若以‘起跑线’定输赢《山海情》中的‘得宝们’……”可知“得宝们”在这种“不输在起跑线上”的谬误认知中是不能获得成功的此处可填写与“注定不可能有后来的成功”类似的句子。

高中地理必修二4、5章检测题

必修二第四、五章单元测试题本试题分第Ⅰ卷（选择题）和第Ⅱ卷（综合题）两部分。

满分100分，时间90分钟。

一、选择题（本大题30小题，每小题1.5分，共45分。

下列各题只有一个是最佳选项。

）下图中①②③表示同一类地理事象，a、b、c三轴分别表示不同地理要素对地理事象的影响程度，读图完成第1题。

1.若上图a、b、c分别表示原料、动力、市场要素对工业区位选择影响情况，则下列说法正确的是（）A.②类工业可以在沪、粤、晋、云、贵、川等地大力发展B.影响①②③工业区位的主导因素分别是市场、动力、原料C.能D.目前，①②③三类工业，在发展中国家所占比重较大，在发达国家中所占比重较小读下图，根据影响工业布局的因素对某工业部门的影响程度，分析判断2—3题。

2图中数字代表的工业布局情况，可能都正确的是（）A．①有色冶金②石油化工③电器装配④水产加工B．①印刷工业②木材加工③制糖厂④有色冶金C．①水产加工②印刷工业③精密仪表④服装加工D．①电器装配②水产加工③集成电路④有色冶金3．下列说法中，正确的是（）A．我国改革开放的初期，③型工业得到迅猛发展B．宝山钢铁厂、攀枝花钢铁厂等企业都可能属于②型C．对①、④类工业部门来说，形成的工业地域面积较小，发育程度较低D．随着交通运输业的发展，影响①、④型工业的传统因素逐渐增强4.读下图，关于连接两县城高速公路选线的方案，正确的是（）①方案1客货流量最大②方案2建设成本最高③方案3营运成本最大④宜选择方案3A.①②B.③④C.①③D.②④下图是我国公路、水运、铁路、民航分别在1985年、1996年货运周转量与客运周转量所占百分比示意图。

据此完成5～65.按W、X、Y、Z顺序分别对应的交通运输方式是（A. B.C. D.6.关于我国交通运输方式的叙述，正确的是（A.B.C.D.货运一直以铁路和公路为主，客运一直读下图，回答7－9题。

7．图中所示的四个城市中，工业兴起最早的是( )A．① B．② C．③ D．④8．二战后，利用当地廉价的劳动力，抓住经济发展机遇，形成中小企业集聚的独特工业小区的是( )A．① B．② C．③ D．④9．图中所示的四个城市，工业布局属于“资源型”的是( )A．①② B．③④ C．②④ D．①③下图是某地区土地利用现状示意图，读图过完成10～11题。

高中数学选择性必修二第四章数列单元测试(基础卷)(含答案)

第四章数列单元过关检测基础A 卷解析版学校:___________姓名：___________班级：___________考号：___________ 题型：8（单选）+4（多选）+4（填空）+6（解答），满分150分，时间：120分钟一、单选题1．已知数列{a n }的前4项为：l ，−12，13，−14，则数列{a n }的通项公式可能为（） A ．a n =1n B ．a n =−1nC ．a n =(−1)n nD ．a n =(−1)n−1n【答案】D 【解析】【分析】分母与项数一样，分子都是1，正负号相间出现，依此可得通项公式【详解】正负相间用(−1)n−1表示，∴a n =(−1)n−1n．故选D ．【点睛】本题考查数列的通项公式，属于基础题，关键是寻找规律，寻找与项数有关的规律． 2．记n S 为等差数列{}n a 的前n 项和，若33a =，621S =，则数列{}n a 的公差为（） A ．1 B ．-1C ．2D ．-2【答案】A【分析】利用等差数列{a n }的前n 项和与通项公式列方程组，求出首项和公差，由此能求出数列{a n }的公差．【详解】∴S n 为等差数列{a n }的前n 项和，a 3∴3∴S 6∴21∴∴316123656212a a d S a d =+=⎧⎪⎨⨯=+=⎪⎩∴ 解得a 1∴1∴d ∴1∴ ∴数列{a n }的公差为1．故选A ∴ 【点睛】本题考查数列的公差的求法，考查等差数列的前n 项和公式等基础知识，考查运算求解能力，是基础题．3．已知数列{}n a ，满足111n n a a +=-，若112a =，则2019a =（） A ．2 B ．12C ．1-D ．12-【答案】C 【分析】利用递推公式计算出数列{}n a 的前几项，找出数列{}n a 的周期，然后利用周期性求出2019a 的值. 【详解】111n n a a +=-，且112a =，211121112a a ∴===--，32111112a a ===---， 111a ===，所以，()a a n N *=∈，则数列{}n a 是以3为周期的周期数列，20193672331a a a ⨯+===-∴. 故选C. 【点睛】本题考查利用数列递推公式求数列中的项，推导出数列的周期是解本题的关键，考查分析问题和解决问题的能力，属于中等题.4．在等比数列{}n a 中，6124146,5a a a a ⋅=+=，则255a a =( ) A ．94或49B ．32C ．32或23 D ．32或94【答案】A 【分析】根据等比数列的性质得6124146a a a a ⋅=⋅=，又由4145a a +=，联立方程组，解得414,a a 的值，分类讨论求解，即可得到答案. 【详解】由题意，根据等比数列的性质，可得6124146a a a a ⋅=⋅=，又由4145a a +=，联立方程组，解得41423a a =⎧⎨=⎩或41432a a =⎧⎨=⎩，当41423a a =⎧⎨=⎩时，则1014432a q a ==，此时201022559()4a q q a ===；当41432a a =⎧⎨=⎩时，则1014423a q a ==，此时201022554()9a q q a ===，故选A. 【点睛】值是解答的关键，着重考查了运算与求解能力，属于基础题. 5．等比数列{}n a 中（） A ．若12a a <，则45a a <B ．若12a a <，则34a a <C ．若32S S >，则12a a <D ．若32S S >，则12a a >【答案】B 【分析】根据等比数列的通项公式及求和公式，等比数列的公比分析即可求出答案. 【详解】等比数列{}n a 中，20q >，∴当12a a <时，可得2212a q a q <，及34a a <，故B 正确；但341a a q =和352a a q =不能判断大小（3q 正负不确定），故A 错误；当32S S >时，则12312+++a a a a a >，可得30a >，即210a q >，可得10a >，由于q 不确定，不能确定12,a a 的大小，故CD 错误. 故选：B. 【点睛】本题考查等比数列通项公式和求和公式的应用，属于基础题.6．两等差数列{}n a 和{}n b ，前n 项和分别为n S ，n T ，且723n n S n T n +=+，则220715a ab b ++的值为（） A ．14924B ．7914C ．165D ．5110【分析】在{}n a 为等差数列中，当(m n p q m +=+，n ，p ，)q N +∈时，m n p q a a a a +=+．所以结合此性质可得：2202171521a a Sb b T +=+，再根据题意得到答案．【详解】解：在{}n a 为等差数列中，当(m n p q m +=+，n ，p ，)q N +∈时，m n p q a a a a +=+．所以1212202171521121121()2121()2a a a a Sb b T b b ⨯+⨯+==+⨯+⨯，又因为723n n S n T n +=+，所以22071514924a ab b +=+．故选：A ．【点睛】本题主要考查等差数列的下标和性质，属于中档题．7．函数()2cos 2f x x x =-的正数零点从小到大构成数列{}n a ，则3a =（）A ．1312π B ．54π C ．1712πD ．76π 【答案】B 【分析】先将函数化简为()2sin 26f x x π⎛⎫=- ⎪⎝⎭再解函数零点得4x k ππ=+或512x k ππ=+，k Z ∈，再求3a 即可. 【详解】解：∵()2cos 22sin 26f x x x x π⎛⎫=--=- ⎪⎝⎭∴ 令()0f x =得：2263x k πππ-=+或22263x k πππ-=+，k Z ∈， ∴4x k ππ=+或512x k ππ=+，k Z ∈，∴ 正数零点从小到大构成数列为：12355,,,4124a a a πππ===故选：B. 【点睛】本题考查三角函数的性质，数列的概念，考查数学运算求解能力，是中档题.8．已知函数3()13xxf x =+（x ∈R ），正项等比数列{}n a 满足501a =，则 1299(ln )(ln )(ln )f a f a f a +++=A ．99B ．101C ．992D ．1012【答案】C 【详解】因为函数31()()()11331x x xf x f x f x ---==∴+-=++（x ∈R ），正项等比数列{}n a 满足2501995011a a a a =∴==，9921ln ln ln ln ...0a a a a +=+=则1299(ln )(ln )(ln )f a f a f a +++=992，选C二、多选题A ．{}n a 可能为等差数列B ．{}n a 可能为等比数列C ．{}n a 中一定存在连续三项构成等差数列D ．{}n a 中一定存在连续三项构成等比数列【答案】AC 【分析】由2n S an bn c =++可求得n a 的表达式，利用定义判定得出答案．【详解】当1n =时，11a S a b c ==++．当2n ≥时，()()221112n n n a S S an bn c a n b n c an a b -=-=++-----=-+．当1n =时，上式＝+a b ．所以若{}n a 是等差数列，则0.a b a b c c +=++∴=所以当0c 时，{}n a 是等差数列，不可能是等比数列；当0c ≠时，{}n a 从第二项开始是等差数列．故选：AC 【点睛】本题只要考查等差数列前n 项和n S 与通项公式n a 的关系，利用n S 求通项公式，属于基础题． 10．已知数列{}n a 的首项为4，且满足()*12(1)0n n n a na n N++-=∈，则（）A ．n a n ⎧⎫⎨⎬⎩⎭为等差数列B ．{}n a 为递增数列C ．{}n a 的前n 项和1(1)24n n S n +=-⋅+D ．12n n a +⎧⎫⎨⎬⎩⎭的前n 项和22n n n T +=【答案】BD 【分析】由12(1)0n n n a na ++-=得121n n a a n n +=⨯+，所以可知数列n a n ⎧⎫⎨⎬⎩⎭是等比数列，从而可求出12n n a n +=⋅，可得数列{}n a 为递增数列，利用错位相减法可求得{}n a 的前n 项和，由于111222n nn n a n n +++⋅==，从而利用等差数列的求和公式可求出数列12n n a +⎧⎫⎨⎬⎩⎭的前n 项和.【详解】由12(1)0n n n a na ++-=得121n n a a n n +=⨯+，所以n a n ⎧⎫⎨⎬⎩⎭是以1141a a ==为首项，2为公比的等比数列，故A 错误；因为11422n n na n-+=⨯=，所以12n n a n +=⋅，显然递增，故B 正确；因为23112222n n S n +=⨯+⨯++⋅，342212222n n S n +=⨯+⨯++⋅，所以231212222n n n S n ++-=⨯+++-⋅()22212212nn n +-=-⋅-，故2(1)24n n S n +=-⨯+，故C 错误；因为111222n n n n a n n +++⋅==，所以12n n a +⎧⎫⎨⎬⎩⎭的前n 项和2(1)22n n n n n T ++==，故D 正确. 故选：BD本题考查等差数列、等比数列的综合应用，涉及到递推公式求通项，错位相减法求数列的和，等差数列前n 项和等，考查学生的数学运算能力，是一道中档题.11．已知无穷等差数列{}n a 的前n 项和为n S ，67S S <，且78S S >，则（） A ．在数列{}n a 中，1a 最大 B ．在数列{}n a 中，3a 或4a 最大 C ．310S S =D ．当8n ≥时，0n a <【答案】AD 【分析】由已知得到780,0a a ><,进而得到0d <,从而对ABD 作出判定.对于C,利用等差数列的和与项的关系可等价转化为160a d +=，可知不一定成立，从而判定C 错误. 【详解】由已知得：780,0a a ><,结合等差数列的性质可知，0d <,该等差数列是单调递减的数列， ∴A 正确，B 错误，D 正确，310S S =,等价于1030S S -=,即45100a a a ++⋯+=,等价于4100a a +=，即160a d +=,这在已知条件中是没有的，故C 错误. 故选:AD. 【点睛】本题考查等差数列的性质和前n 项和，属基础题,关键在于掌握和与项的关系.12．将2n 个数排成n 行n 列的一个数阵，如图：该数阵第一列的n 个数从上到下构成以m 为公差的等差数列，每一行的n 个数从左到右构成以m 为公比的等比数列（其中0m >）.已知112a =，1a a =+，记这2n 个数的和为S .下列结论正确的有（）1112131.n a a a a ⋯⋯ 2122232.n a a a a ⋯⋯ 3132333.n a a a a ⋯⋯……123.n n n nn a a a a ⋯⋯A ．3m =B ．767173a =⨯C ．()1313j ij a i -=-⨯ D ．()()131314n S n n =+- 【答案】ACD 【分析】根据等差数列和等比数列通项公式，结合13611a a =+可求得m ，同时确定67a 、ij a 的值、得到,,A B C 的正误；首先利用等比数列求和公式求得第i 行n 个数的和，再结合等差求和公式得到D 的正误. 【详解】对于A ，2213112a a m m =⋅=，6111525a a m m =+=+，2235m m ∴=+，又0m >，3m ∴=，A 正确；对于B ，612517a m =+=，666761173a a m ∴=⋅=⨯，B 错误；对于C ，()111131i a a i m i =+-=-，()111313j j ij i a a mi --∴=⋅=-⋅，C 正确；对于D ，第i 行n 个数的和()()()()()1131133131122n n n i a m i i S m-----'===--，()()()()()()3111131258313131312224n n nn n S n n n +∴=-⨯+++⋅⋅⋅+-=-⨯=+-⎡⎤⎣⎦，D 正确. 故选：ACD .本题考查数列中的新定义问题，解题关键是能够灵活应用等差和等比数列的通项公式和求和公式，将新定义的数阵转化为等差和等比数列的问题来进行求解.三、填空题13．已知{}n a 为等差数列,135246105,99a a a a a a ++=++=,{}n a 前n 项和n S 取得最大值时n 的值为___________. 【答案】20 【分析】先由条件求出1,a d ，算出n S ，然后利用二次函数的知识求出即可【详解】设{}n a 的公差为d ，由题意得135********d a a a a d a a ++++==++即1235a d +=，①2461113599a a a a d a d a d ++=+++++=即1333a d +=，②由①②联立得139,2a d ==-所以()()22139(2)40204002n S n n n n n n -=+⨯-=-+=--+故当20n =时，n S 取得最大值400 故答案为：20等差数列的n S 是关于n 的二次函数，但要注意n 只能取正整数.14．《九章算术》中有一个“两鼠穿墙”的问题：“今有垣厚五尺，两鼠对穿.大鼠日一尺，小鼠亦日一尺.大鼠日自倍，小鼠日自半.问几何日相逢？各穿几何？”其大意为：“今有一堵墙厚五尺，两只老鼠从墙的两边沿一条直线相对打洞穿墙，大老鼠第一天打洞1尺，以后每天是前一天的2倍；小老鼠第一天也打洞1尺，以后每天是前一天的12.问大、小老鼠几天后相遇？各自打洞几尺？”如果墙足够厚，S n 为前n 天两只老鼠打洞长度之和，则S n =_____尺.【答案】2n +1﹣21﹣n【分析】写出两只老鼠打洞的通项公式，利用分组求和即可得解. 【详解】根据题意大老鼠第n 天打洞12n na 尺，小老鼠第n 天打洞112n n b -⎛⎫= ⎪⎝⎭尺，所以11111242122n n n S --⎛⎫=+++⋅⋅⋅++++⋅⋅⋅+ ⎪⎝⎭111221112nn ⎛⎫- ⎪-⎝⎭=+--112122n n -⎛⎫=-+- ⎪⎝⎭1212n n -=+-故答案为：1212n n -+- 【点睛】此题考查等比数列的辨析，写出通项公式，根据求和公式求和，关键在于熟练掌握相关公式，涉及分组求和.15．我国古代，9是数字之极，代表尊贵之意，所以中国古代皇家建筑中包含许多与9相关的设计．例如，北京天坛圆丘的底面由扇环形的石板铺成（如图），最高一层是一块天心石，围绕它的第一圈有9块石板，从第二圈开始，每一圈比前一圈多9块，共有9圈，则前9圈的石板总数是__________．【答案】405 【分析】前9圈的石板数依次组成一个首项为9，公差为9的等差数列，9989994052S ⨯=⨯+⨯= 16．如图，互不相同的点12,,,n A A A 和12,,,,n B B B 分别在角O 的两条边上，所有n n A B 相互平行，且所有梯形11n n n n A B B A ++的面积均相等.设n n OA a =.若11a =，22a =，则数列{}n a 的通项公式是________.【答案】n a =【分析】根据三角形相似和所有梯形11n n n n A B B A ++的面积均相等，找到与n a 相关的递推公式，再由递推公式求得通项公式. 【详解】由于11//,n n n n A B A B ++ 所以11,n n n n OA B OA B ++梯形11n n n n A B B A ++ 的面积为11n n OA B ++∆的面积減去n n OA B △的面积，2222i i j jOA B i i OA B j jS OA a SOA a == 则可得 222211,n n n n a a a a +--=- 即递推公式为222112,n n n a a a +-=+故2{}n a 为等差数列，且公差d =2221a a -3=，故21(1)332n a n n =+-⨯=-，得n a =故答案为: n a 【点睛】本题主要考查数列在平面几何中的应用，根据几何关系寻找递推有关系是解决问题的关键，属于中档题.四、解答题17．设等差数列{}n a 的前n 项的和为n S ，且462S =-，675S =-，求：（1）求{}n a 的通项公式n a ；（2）求数列{}n a 的前14项和.【答案】（1）323n a n =-；（2）147. 【分析】（1）由已知条件列出关于1,a d 的方程组，求出1,a d 可得到n a ；（2）由通项公式n a 先判断数列{}n a 中项的正负，然后再化简数列{}n a 中的项，即可求出结果. 【详解】解：（1）设等差数列{}n a 的公差为d ，依题意得11434622656752a d a d ⨯⎧+=-⎪⎪⎨⨯⎪+=-⎪⎩，解得120,3a d =-=，∴()2013323n a n n =-+-⨯=-；（2）∵323n a n =-，∴由0n a <得8n <，22(20323)3433432222n n n n n S n n -+--===-∴123141278141472a a a a a a a a a S S ++++=----+++=-223433431414772222⎛⎫=⨯-⨯-⨯-⨯ ⎪⎝⎭()()7424372143147=---=.【点睛】此题考查等差数列的基本量计算，考查计算能力，属于基础题. 18．数列{}n a 满足11a =，22a =，2122n n n a a a ++=-+ （1）设1n n n b a a +=-，证明数列{}n b 是等差数列（2）求数列11n n b b +⎧⎫⎨⎬⎩⎭的前n 项和n S .【答案】（1）证明过程见详解；（2）21n nS n =+. 【分析】（1）先化简得到()()2112n n n n a a a a +++---=即12n n b b ，再求得1211b a a =-=，最后判断数列{}n b 是以1为首项，以2为公差的等差数列.（2）先求出数列{}n b 的通项公式21n b n =-，再运用“裂项相消法”求数列11n n b b +⎧⎫⎨⎬⎩⎭的前n 项和nS 即可. 【详解】解：（1）因为2122n n n a a a ++=-+，所以()()2112n n n n a a a a +++---= 因为1n n n b a a +=-，所以12nn b b ，且1211b a a =-=所以数列{}n b 是以1为首项，以2为公差的等差数列. （2）由（1）的()11221n b n n =+-⨯=-，所以()()111111212122121n n b b n n n n +⎛⎫==- ⎪-+-+⎝⎭所以12233411111n n n S b b b b b b b b +=++++11111111111121323525722121n n ⎛⎫⎛⎫⎛⎫⎛⎫=-+-+-++- ⎪ ⎪ ⎪ ⎪-+⎝⎭⎝⎭⎝⎭⎝⎭111.22121n n n ⎛⎫=-= ⎪++⎝⎭ 【点睛】本题考查利用定义求等差数列的通项公式、根据递推关系判断数列是等差数列、根据“裂项相消法”求和，还考查了转化的数学思维方式，是基础题.19．在①112n n a a +=-，②116n n a a +-=-，③18n n a a n +=+-这三个条件中任选一个，补充在下面的问题中，若问题中的n S 存在最大值，则求出最大值；若问题中的n S 不存在最大值，请说明理由.问题：设n S 是数列{}n a 的前n 项和，且14a =，__________，求{}n a 的通项公式，并判断n S 是否存在最大值.注：如果选择多个条件分别解答，按第一个解答计分. 【答案】答案见解析【分析】若选①，求出数列{}n a 是首项为4，公比为12-的等比数列，求出通项公式和前n 项和，通过讨论n 的奇偶性，求出其最大值即可；若选②，求出数列{}n a 是首项为4，公差为16-的等差数列，求出通项公式和前n 项和，求出其最大值即可；若选③，求出217242n n n a -+=，当16n ≥时，0n a >，故n S 不存在最大值.【详解】解：选①因为112n n a a +=-，14a =，所以{}n a 是首项为4.公比为12-的等比数列，所1211422n n n a --⎛⎫⎛⎫=⨯-=- ⎪⎪⎝⎭⎝⎭.当n 为奇数时，141281113212n n nS ⎡⎤⎛⎫--⎢⎥ ⎪⎝⎭⎢⎥⎛⎫⎣⎦==+ ⎪⎝⎭+，因为81132n ⎛⎫+ ⎪⎝⎭随着n 的增加而减少，所以此时n S 的最大值为14S =. 当n 为偶数时，81132n nS ⎛⎫=-⎪⎝⎭，且81814323n n S ⎛⎫=-<<⎪⎝⎭ 综上，n S 存在最大值，且最大值为4. 选②因为116n n a a +-=-，14a =.所以{}n a 是首项为4，公差为16-的等差数列，所以11254(1)666n a n n ⎛⎫=+--=-+ ⎪⎝⎭. 由125066n -+≥得25n ≤，所以n S 存在最大值.且最大值为25S （或24S ），因为25252412545026S ⨯⎛⎫=⨯+⨯-= ⎪⎝⎭，所以n S 的最大值为50. 选③因为18n n a a n +=+-，所以18n n a a n +-=-，所以217a a -=-，326a a -=-，…19n n a a n --=-，则2121321(79)(1)171622n n n n n n n a a a a a a a a --+---+=-+-+=-+-=，又14a =，所以217242n n n a -+=. 当16n ≥时，0n a >，故n S 不存在最大值. 【点睛】此题考查数列的通项公式和求和公式，考查等差数列和等比数列的性质，属于基础题 20．已知数列{}n a 的前n 项和为n S ，满足22n n S a =-．（1）求数列{}n a 的通项公式；（2）设()21n n b n a =-，求数列{}n b 的前n 项和n T ．【答案】（1）2nn a =；（2）()12326n n T n +=-⨯+【分析】（1）利用1(2)n n n a S S n -=-≥，11a S =，可得{}n a 为等比数列，利用等比数列的通项公式即可求得通项公式n a ；（2）利用错位相减法求和即可求n T ．【详解】（1）当1n =时，11122a S a ==-，解得12a =，当1n >时，由22n n S a =-可得1122n n S a --=-，1n >两式相减可得122n n n a a a -=-，即12nn a a -=，所以{}n a 是以2为首项，以2为公比的等比数列，所以1222n nn a -=⋅=（2）由（1）(21)2nn b n =-⋅，23123252(21)2n n T n =⨯+⨯+⨯++-⋅，则23412123252(23)2(21)2n n n T n n +=⨯+⨯+⨯++-⨯+-⨯，两式相减得2312222222(21)2n n n T n +-=+⨯+⨯++⨯--⨯()112118(12)2(21)226(21)2232612n n n n n n n n -++++-=+--⨯=---⨯=--⋅--，所以()12326n n T n +=-⨯+．【点睛】方法点睛：由数列前n 项和求通项公式时，一般根据11,2,1n n n S S n a S n --≥⎧=⎨=⎩求解，考查学生的计算能力.21．已知数列{}n a 的前n 项和为23122n S n n =-. （1）求数列{}n a 的通项公式；（2）数列[]lg n n b a =，[]x 表示不超过x 的最大整数，求{}n b 的前1000项和1000T . 【答案】（1）32n a n =-；（2）10002631T =. 【分析】（1）利用1n n n a S S -=-可求出；（2）根据数列特点采用分组求和法求解. 【详解】（1）当1n =时，111a S ==，当2n ≥时，()()221313111322222n n n a S S n n n n n -⎡⎤=-=-----=-⎢⎥⎣⎦，将1n =代入上式验证显然适合，所以32n a n =-. （2）因为410a =，34100a =，3341000a =，333410000a =，所以0,131,4332,343333,3341000n n n b n n ≤≤⎧⎪≤≤⎪=⎨≤≤⎪⎪≤≤⎩，所以100003130230036672631T =⨯+⨯+⨯+⨯=.【点睛】本题考查n a 和n S 的关系，考查分组求和法，属于基础题. 22．在①535S =，②13310a a +=，③113n a n a +=+这三个条件中任选一个，补充在下面问题中并作答.已知{}n a 是各项均为正数的等差数列，其前n 项和为n S ，________，且1a ，412a ，9a 成等比数列. （1）求数列{}n a 的通项公式；（2）设()1n n n b a =-，求1ni i b =∑.【答案】（1）32n a n =-；（2）13,213,2n i i n n b n n =⎧⎪⎪=⎨-⎪⎪⎩∑是偶数是奇数【分析】（1）利用1a ，412a ，9a 成等比数列∴可得221132690a a d d +-=，若选①：由535S =得：127a d +=，即可解出1a 和d 的值，即可求出{}n a 的通项公式；若选②：由13310a a +=可得152d a =-，即可解出1a 和d 的值，即可求出{}n a 的通项公式；若选③：由113n a n a +=+，可表示出419a a =+，9124a a =+，结合1a ，412a ，9a 成等比数列∴即可解出1a 和d 的值，即可求出{}n a 的通项公式；（2）由（1）可得()()132n n b n =--，分n 为奇数和偶数，利用并项求和即可求解.【详解】 {}n a 是各项均为正数的等差数列，1a ，412a ，9a 成等比数列. 所以241914a a a =⋅，即()()2111348a d a a d +=⋅+，整理可得221132690a a d d +-=，若选①：535S =，则1545352a d ⨯+=，即127a d +=，由127a d +=可得172a d =-代入221132690a a d d +-=可得：2230d d --=，解得3d =或1d =-（舍）所以11a =，所以()11332n a n n =+-⨯=-，若选②：13310a a +=，即152d a =-，代入221132690a a d d +-=得：2111762450a a -+=，即 ()()11117450a a --=解得：113a d =⎧⎨=⎩或145175017a d ⎧=⎪⎪⎨⎪=-<⎪⎩不符合题意；若选③：113n a n a +=+，则419a a =+，9124a a =+，代入241914a a a =⋅可得21126270a a +-= 解得：113a d =⎧⎨=⎩或1273a d =-⎧⎨=⎩不符合题意；综上所述：113a d =⎧⎨=⎩， 32n a n =-，（2）()()132n n b n =--， ()()()()()12311231111111n n n i n n i b a a a a a --==-+-+-+-+-∑ ()()()()114710135132n n n n -=-+-++--+-- 当n 为偶数时，13322n i i n n b ==⨯=∑，当n 为奇数时，()11131322n i i n n b =--=-+-⨯=∑，所以13,213,2n i i n n b n n =⎧⎪⎪=⎨-⎪⎪⎩∑是偶数是奇数. 【点睛】关键点点睛：本题得关键点是分别由条件①②③结合1a ，412a ，9a 成等比数列计算出1a 和d 的值，由{}n a 是各项均为正数的等差数列，所以10a >，0d >，第二问中()1n n n b a =-正负交错的数列求和，需要用奇偶并项求和，注意分n 为奇数和偶数讨论.。

高中历史选择性必修二第四单元综合复习测试题(含答案)

高中历史选择性必修二第四单元综合复习测试题（含答案）总分：100分时间：45分钟一、单项选择题(每小题4分，共60分)1．二里头文化的聚落分布，由王都（大邑）、区域性中心聚落（大族邑）、次级中心聚落（小族邑）及众多村落（属邑）组成，形成金字塔式的聚落结构和众星捧月式的聚落空间分布格局。

这种格局本质上反映了（）A．早期王权国家的形成B．家国一体观念的出现C．中华文明的多元一体D．国野分立的城乡之别2．秦都咸阳宫殿建筑区有较为完善的陶水管道、排水池以及散水等遗物遗迹；汉代长安城内的宫殿、官署、邸第、里居等有众多井台，普遍饮用井水；秦汉时期的厕所有“厕”、“囿”“涵”“清”“轩”“更衣之室”等众多名称。

这反映秦汉时期（）A．国家组织能力增强B．城市的生活服务设施丰富C．居民卫生意识提高D．政府重视民众的日常生活3．隋文帝建立新都大兴城.皇宫、政府机构位于高坡之上，皇城“不设杂居”，使得“公私有辨，风俗齐整”；各区域有明确界线，使得“坊有墉（墙），墉有门，逮亡奸伪无所容足。

”全城划分为109坊和东、西两市。

大兴城的布局设计反映出（）A．当时城市治安问题严重B．坊市界限已经被完全打破C．实用与安全并重的理念D．城市建设中突出民本思想4．下图为1800年—2015年世界城镇化趋势图，反映了城镇人口占世界总人口比例的变化情况。

据此可知，下列有关世界城镇化进程的解释正确的是（）注：上图根据联合国《世界城镇化展望(2018年版》等资料编制，反映了从1800年到2015年，城镇人口占世界总人口比例的变化情况。

A．1800―1850年增长缓慢的原因是各国政府不重视城镇化建设B．1850—1925年迅速增长的原因是欧美国家推动亚非拉城镇化C．1925—1950年增长较快的原因是持久和平与稳定的世界环境D．1950—2015年持续增长的主要原因是发展中国家城镇化加快5．下图为元大都街道胡同复原图,据此可知,当时（）A．城市人口大量增加B．胡同地摊经济潜滋暗长C．城市经济沿袭前代D．城市建设体现蒙古风格6．福建土楼是出于族群安全而采取的一种自卫式的居住样式，往往将供奉祖先牌位的上堂安置于建筑的最里边；徽州古村落聚族而居，以宗祠为中心来营建民宅、园林、牌坊等。

人教版高中英语必修二Unit4单元测试题2套（有答案,无听力）

⼈教版⾼中英语必修⼆Unit4单元测试题2套（有答案,⽆听⼒）Unit 4 单元测试题第⼆部分：阅读理解（共两节，满分40）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和D四个选项中，选出最佳选项。

[A]The World Around UsChannel (频道): C8A study of the ancient Egyptian pyramids (⾦字塔) and the area around the River Nile in Egypt. The filming of this programme is a work of art as it is so thoughtfully done. As well as the obvious camels, there are also many other desert animals and plant life.Summer HolidaysChannel: M6Some of the best summer holidays are on offer this year. Tonight’s programme shows a weekend in Disneyland in Paris, cheap sailing holidays in the Mediterranean and a shopping and sightseeing trip to New York.The Creative MindChannel: C4One of the most popular programmes on TV at the moment, The Creative Mind has different artistic themes (主题) from exhibition reviews, information about major and smaller museums, and interviews with artists, writers, actors and musicians. Prepared for Special MomentsChannel: TF1The fun cookery programme that offers lots of exciting ideas from children’s birthday parties to that candlelit dinner for the boss and his wife. Easy to follow step by step instructions that show you how to make your dinner party a little bit special. Born to RunChannel: W9An interesting story of a young man with learning difficulties who overcame (克服) the problems in his life, through his great talent (天赋) for athletics. This is a true story of how one person made the most of his life and also helped many other people with similar problems.21. Which channel should you turn to if you are very interested in wildlife?A. C4.B. M6.C. C8.D. W9.22. Which programme is suitable for those who like watching TV interviews?A. The World Around Us.B. Summer Holidays.C. The Creative Mind.D. Prepared for Special Moments.23. Which of the following words can best describe the hero of Born to Run?A. Creative.B. Dishonest.C. Self-centered.D. Strong-minded.[B]Madeleine Hamilton’s school in Franklin, Tennessee, recently made a big decision: Recess (课间休息) would come back. Madeleine and her sixth-grade classmates would go outside for 20 minutes every day. They could catch up with friends and kick a soccer ball around. “Recess is the one break in the day to relax and have fun,” she says.Most middle school students don’t get that chance. That’s be cause most middle schools don’t have recess anymore. Experts say that about 40 percent of U.S. school districts have cut recess time.Now, more and more experts agree: Schools should put recess back on the schedule.Students have been heading outside for recess since at least the 1800s. But in the 1980s, some U.S. schools started to cut recess. Why? American students weren’t doing as well in school as students in other countries. To catch up, many schools in the U.S. cut recess. This made more time in the schedule for important subjects like science and math.But research shows that taking a break can actually help students learn. Your brain can get overwhelmed (应接不暇的) when you move from subject to subject without a break. That makes it harder for you to remember new information. Scientists say that resting your brain helps it work better.Even adults (成年⼈) have started taking breaks to rest their brains — and play. At Google, employees can stop working to use the office’s LEGO station. And workers at Faceboo k can play video games in the company’s arcade (游戏机厅).Recess doesn’t just help students learn. Moving around can also help keep kids healthy. People who exercise daily can avoid extreme sadness. They are also less likely to become overweight.Of course, kids can get hurt playing sports like dodgeball. Some schools have cut recess to avoid broken arms and legs. Schools also need adults to keep students safe on the playground during recess. Some schools can’t afford to pay anyone to do this job.And 73 percent of school-age kids in America play sports or do other athletic activities after school. Do they really need to take time out of busy school days for recess?24. How does Madeleine feel about her school’s big decision?A. Worried.B. Excited.C. Moved.D. Surprised.25. Why did many U.S. schools cut recess in the 1980s?A. To ensure students’ safety.B. To shorten students’ school days.C. To save time for students to study.D. To let students play sports after school.26. What’s the real purpose of the L EGO station at Google?A. To help workers relax.B. To keep workers strong.C. To provide new information for workers.D. To help workers learn how LEGO works.27. What does the underlined part “this job” refer to?A. Teaching students how to play sports like dodgeball.B. Protecting students from getting hurt during a break.C. Designing after-school athletic activities for students.D. Giving students medical care if they are hurt at school.[C]What do you usually do when a flying insect lands on your arm? Most likely you blow it away. But what if that bug were a ladybug (瓢⾍)? Then chances are you would let it stay. What is it about ladybugs that we like?Farmers once thought ladybugs were a good luck sign. A ladybug in the field meant that the crops (庄稼) would be successful and the weather would be good. The farmers may have been exaggerating (夸⼤) about what a very small bug can do, but in fact ladybugs do help out with the crops. They eat harmful insects.Actually, it is not the adult ladybug that eats t he other insects. It’s the ladybug larvae (幼⾍). Ladybugs lay their eggs on leaves that are covered with aphids or other insects. When the larvae come out of the eggs, they are very hungry and will eat mites, aphids, mealybugs, green flies, and other crop-destroyers.A single ladybug larva can eat over 1,000 aphids in one day. In the 1800s, ladybugs were brought all the way from Australia to California and set free among orange trees that were dying from being eaten by insects. The little red and black “eating machines” completely saved all the trees.Ladybugs go about their daily business without worrying much about anything. Very few birds or insects will eat a ladybug because it gives off something that isdangerous to eat. Ladybugs also protect themselves by playing dead. They fall off leaves and remain very still until their enemy goes away.So next time you see a ladybug, think about what it may be up to. Maybe it is off to save an orange tree. Maybe its next move will be to fall off a leaf and play dead or stand up to a bird about 100 times its size. Maybe it’s letting you know that the weather will be good tomorrow. Who knows? Ladybugs are special that way.28. What do we know about mites and aphids?A. They do harm to the crops.B. They eat more than they should.C. They are food for adult ladybugs.D. They are the enemy of ladybug larvae.29. Why were ladybugs brought from Australia to California in the 1800s?A. To report the weather for farmers.B. To increase their number.C. To do research on them.D. To save orange trees.30. What does the underlined part “stand up to” in the last paragraph probably mean?A. Play with.B. Worry about.C. Get along with.D. Fight back against.31. What does the author think of the ladybug?A. It is a crop-destroyer.B. It is a good luck sign.C. It is a little flying helper.D. It is a big eating machine.[D]Each year, people are becoming more and more worried about the future of our planet. People are now starting to agree —global warming will certainly make a dif ference to the world’s weather. While some parts of the world may have morefloods (洪⽔), some other parts might have long periods of dry weather. All of this will have a serious influence on the world’s agriculture (农业). And that could make it difficult for us to get enough food to survive.This explains the idea behind the Svalbard Global Seed Vault (种⼦库). The Seed Vault is built into a mountain on an island near the North Pole. The purpose is to protect the world’s agriculture from future disasters. This may not just be a question of global warming. Perhaps there will be a major nuclear disaster or war one day. These things won’t be a problem for Svalbard.Some experts think that the population of the earth will probably go up by 50% in the next 40 years. Because of climate change, it probably won’t be easy for some plants to survive. Having many different types of plants will be necessary for agriculture to adapt to (适应) changes in the environment. In the Svalbard Global Seed Vault, there is space for 4.5 million types of seed. At the moment there are over 770,000 different types there.Any country in the world can leave seeds in Svalbard without paying. The seeds of some plants, such as peas, only survive for 20 to 30 years. But others will last for hundreds of years.In 2028, the Svalbard Vault will celebrate its 20th birthday. 200 years in the future, it’s possible that the ice caps will disappear. Even if this happens, Svalbard will be safe. That’s because it is on a mountain and will be above the sea level. And because it is so close to the North Pole, even if their electricity stops working, the ice will keep a cool temperature of -3.5℃ to keep the seeds in good condition. But the best protection is the fact that it is in such a faraway place. After all, maybe one day this will be the most important building in the whole world.32. What does the first paragraph mainly introduce?A. The role of agriculture.B. The reason for global warming.C. The future problems we’ll face.D. The planet where we’ll liv e in the future.33. What is the Seed Vault mainly used for?A. Storing plant seeds.B. Growing more plants.C. Discovering new plants.D. Providing humans with food.34. What do we know about the Seed Vault?A. It needs more space.B. It provides services for free.C. It is built above a mountain.D. It will close in 200 years.35. If there is an electrical problem at the Seed Vault, what will happen to the seeds?A. They will be sent away.B. They will still be safe.C. They will be of no use.D. They will be planted at once.第⼆节（共5⼩题；每⼩题2分，满分10分）根据短⽂内容，从短⽂后的选项中选出能填⼊空⽩处的最佳选项。

高中数学选择性必修二第四章《数列》单元测试卷

高中数学选择性必修二第四章《数列》单元测试卷注意事项：1．答题前，先将自己的姓名、准考证号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。

2．选择题的作答：每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3．非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4．考试结束后，请将本试题卷和答题卡一并上交。

一、选择题（本大题共12小题，每小题5分，共60分，在每小题给出的四个选项中，只有一项是符合题目要求的）1．数列{}n a 的通项公式为()()132nn a n =--，则{}n a 的第5项是（）A ．13B ．13-C ．15-D ．152．记n S 为数列{}n a 的前n 项和．“任意正整数n ，均有0n a >”是“{}n S 为递增数列” 的（） A ．充分不必要条件 B ．必要不充分条件 C ．充要条件D ．既不充分也不必要条件3．如图，将一个边长为1的正三角形的每条边三等分，以中间一段为边向外作正三角形，并擦去中间一段，得图（2），如此继续下去，得图（3）…，设第n 个图形的边长为n a ，则数列{}n a 的通项公式为（）A ．13nB ．131n - C ．13nD ．113n - 4．若数列{}n a 满足12a =，111nn na a a ++=-，则2018a 的值为（）A ．2B ．3-C ．12-D ．135．数列{}n a 满足()11nn n a a n ++=-⋅，则数列{}n a 的前20项的和为（） A ．100-B ．100C ．110-D ．1106．已知数列{}n a 的前n 项和1233n n S a =+，则{}n a 的通项公式n a =（）A ．12nn a ⎛⎫=- ⎪⎝⎭B ．112n n a -⎛⎫=- ⎪⎝⎭C ．112n n a -⎛⎫= ⎪⎝⎭D ．112n n a +⎛⎫=- ⎪⎝⎭7．在数列{}n a 中，11a =-，20a =，21n n n a a a ++=+，则5a 等于（） A ．0B ．1-C ．2-D ．3-8．程大位《算法统宗》里有诗云“九百九十六斤棉，赠分八子做盘缠．次第每人多十七，要将第八数来言．务要分明依次弟，孝和休惹外人传．”意为：996斤棉花，分别赠送给个子女做旅费，从第一个开始，以后每人依次多17斤，直到第八个孩子为止．分配时一定要等级分明，使孝顺子女的美德外传，则第八个孩子分得斤数为（） A ．65B ．184C ．183D ．1769．已知数列{}n a 的各项均为整数，82a =-，134a =，前12项依次成等差数列，从第11项起依次成等比数列，则15a =（） A ．8B ．16C ．64D ．12810．设数列{}n a 的前n 项和为n S ，若2n S n n =--，则数列()21n n a ⎧⎫⎪⎪⎨⎬+⎪⎪⎩⎭的前40项的和为（）A ．3940B ．3940-C ．4041D ．4041-11．已知等差数列{}n a 的前n 项和为n T ，34a =，627T =，数列{}n b 满足1123n n b b b b b +=+++⋅⋅⋅+，121b b ==，设n n n c a b =+，则数列{}n c 的前11项和为（）A ．1062B ．2124C ．1101D ．110012．已知数列{}n a 满足11a =，()12n n a a n +-≥∈*N ，则（） A ．21n a n ≥+B ．2n S n ≥C ．12n n a -≥D ．12n n S -≥二、填空题（本大题有4小题，每小题5分，共20分．请把答案填在题中横线上） 13．记n S 为数列{}n a 的前n 项和，若21n n S a =+，则6S =_____________．14．已知数列{}n a 的首项12a =，且()11122n n a a n +=+∈*N ，则数列11n a ⎧⎫⎨⎬-⎩⎭的前10项的和为_____．15．已知数列{}n a 前n 项和为n S ，若22n n n S a =-，则=n S _________．16．已知n S 为数列{}n a 的前n 项和，10a =，若()()1112n nn n a a +⎡⎤=+-+-⎣⎦，则100S =_____．三、解答题（本大题有6小题，共70分．解答应写出文字说明、证明过程或演算步骤） 17．（10分）已知数列{}n a 的前n 项和为n S ，且1，n a ，n S 成等差数列．（1）求数列{}n a 的通项公式；（2）若数列{}n b 满足12n n n a b na ⋅=+，求数列{}n b 的前n 项和n T ．18．（12分）设等差数列{}n a 的前n 项和为n S ，且3S ，52S ，4S 成等差数列，521322a a a =+-．（1）求数列{}n a 的通项公式；（2）设12n n b -=，求数列n n a b ⎧⎫⎨⎬⎩⎭的前n 项和n T ．19．（12分）已知公差不为0的等差数列{}n a 的首项11a =，且1a ，2a ，6a 成等比数列．（1）求数列{}n a 的通项公式；（2）记11n n n b a a +=，求数列{}n b 的前n 项和n S ．20．（12分）设正项数列{}n a 的前n 项和n S满足1n a +，()n ∈*N ．（1）求数列{}n a 的通项公式；（2）设11n n n b a a +=⋅，数列{}n b 的前n 项和为n T ，求n T 的取值范围．21．（12分）已知正项等比数列{}n a 的前n 项和为n S ，且()21n n S a n =-∈*N ．（1）求数列{}n a 的通项公式；（2）若lg n n b a =，求数列{}n n a b +的前n 项和n T ．22．（12分）若数列{}n a 的前n 项和n S 满足2n n S a λ=-（0n λ>∈*N ，）．（1）证明：数列{}n a 为等比数列，并求n a ；（2）若24,log n n n a n b a n λ⎧⎪==⎨⎪⎩，是奇，是偶，（n ∈*N ），求数列{}n b 的前n 项和n T ．高中数学选择性必修二第四章《数列》单元测试卷一、选择题（本大题共12小题，每小题5分，共60分，在每小题给出的四个选项中，只有一项是符合题目要求的） 1．【答案】B【解析】求数列{}n a 的某一项，只要把n 的值代入数列的通项即得该项． 2．【答案】A【解析】∵“0n a >”⇒“数列{}n S 是递增数列”，所以“0n a >”是“数列{}n S 是递增数列”的充分条件．如数列{}n a 为1-，0，1，2，3，4，…，显然数列{}n S 是递增数列，但是n a 不一定大于零，还有可能小于等于零，所以“数列{}n S 是递增数列”不能推出“0n a >”， ∴“0n a >”是“数列{}n S 是递增数列”的不必要条件．∴“0n a >”是“数列{}n S 是递增数列”的充分不必要条件．故答案为A ． 3．【答案】D【解析】本题主要考查了等比数列的判定和等比数列的通项的求法，属于基础题． 4．【答案】B【解析】12a =由题，111n n n a a a ++=-，所以121131a a a +==--，2321112a a a +==--，3431113a a a +==-，454121a a a +==-，故数列{}n a 是以4为周期的周期数列，故20185044223a a a ⨯+===-．故选B ． 5．【答案】A【解析】由()11nn n a a n ++=-，得211a a +=-，343a a +=-，565a a +=-，…192019a a +=-， ∴n a 的前20A ． 6．【答案】B【解析】令1n =，11a =，代入选项，排除A ，D 选项．令2n =，解得212a =-，排除C 选项．故选B ． 7．【答案】C【解析】因为21n n n a a a ++=+，所以3121a a a =+=-，4321a a a =+=-，5432a a a =+=-．故选C ． 8．【答案】B【解析】由题意可得，8个孩子所得的棉花构成公差为17的等差数列，且前8项和为996，设首项为1a ，结合等差数列前n 项和公式：811878828179962S a d a ⨯=+=+⨯=，解得：165a =，则81765717184a a d =+=+⨯=．即第八个孩子分得斤数为184．本题选择B 选项． 9．【答案】B【解析】设由前12项构成的等差数列的公差为d ，从第11项起构成的等比数列的公比为q ，由()2212131124d 423d a a a -+===-+，解得1d =或34d =，又数列{}n a 的各项均为整数，故1d =，所以13122a q a ==，所以111012213n n n n a n --≤⎧=⎨≥⎩，，，故415216a ==，故选B ．10．【答案】D【解析】根据2n S n n =--，可知当2n ≥时，()()221112n n n a S S n n n n n -⎡⎤=-=-------=-⎣⎦，当1n =时，112a S ==-，上式成立，所以2n a n =-，所以()221112(+11nn a n n n n ⎛⎫=-=-- ⎪++⎝⎭），所以其前n 项和11111111234+111n n T n n n n ⎛⎫⎛⎫=--+-++-=--=-⎪ ⎪++⎝⎭⎝⎭，所以其前40项和为404041T =-．故选D ． 11．【答案】C【解析】设数列{}n a 的公差为d ，则112461527a d a d +=+=⎧⎨⎩，解得121a d ==⎧⎨⎩，数列{}n a 的通项公式为1n a n =+，当2n ≥时，1n n n b b b +-=，∴12n n b b +=，即{}n b 从第二项起为等比数列，∴()222n n b n -=≥，数列{}n b 的通项公式为：21,1 2,2n n n b n -=⎧⎪=⎨≥⎪⎩，分组求和可得数列{}n c 的前11项和为()()29101123412112227721101S =+++++++++=+=+．本题选择C 选项．12．【答案】B【解析】由题得212a a -≥，322a a -≥，432a a -≥，432a a -≥， ∴()213243121n n a a a a a a a a n --+-+-++-≥-，∴()121n a a n -≥-，21n a n ∴≥-，∵123135a a a ≥≥≥，，，，21n a n ≥-，∴12313521n a a a a n ++++≥++++-，∴()21212n nS n n ≥+-=．故选B ．二、填空题（本大题有4小题，每小题5分，共20分．请把答案填在题中横线上） 13．【答案】63-【解析】根据21n n S a =+，可得1121n n S a ++=+，两式相减得1122n n n a a a ++=-，即12n n a a +=，当1n =时，11121S a a ==+，解得11a =-，所以数列{}n a 是以1-为首项，以2为公比的等比数列，所以()66126312S --==--．故答案是63-．14．【答案】1023【解析】由11122n n a a +=+，得()11112n n a a +-=-，∴{}1n a -为等比数列，()111111122n n n a a --⎛⎫⎛⎫-=-⨯= ⎪⎪⎝⎭⎝⎭，1121n n a -=-，101012102312S -==-，故答案为1023． 15．【答案】2n n ⋅【解析】∵1n n n a S S -=-，故()122n n n n S S S -=--，整理得到122n n n S S -=+，也即是11122n n n n S S --=+，故2n n S ⎧⎫⎨⎬⎩⎭为等差数列．又12a =，∴()11122n n S a n n =+-⨯=即·2n n S n =． 16．【答案】101223-【解析】由()()()1112n nn n a a n +⎡⎤=+-+-∈⎣⎦*N ，当n 为奇数时，有()12nn a +=-；当n 为偶数时，有122n n n a a +=+， ∴数列{}n a 的所有偶数项构成以2-为首项，以4为公比的等比数列，()()10013599246100S a a a a a a a a =+++⋯+++++⋯+()()()246982469824610022222a a a a a a a a =+++⋯+++++⋯+++++⋯+()()24698246100100322222a a a a a =+++⋯+-++++⋯+()()()5049101992144142232214143----=⨯-⨯-+=--．故答案是101223-．三、解答题（本大题有6小题，共70分．解答应写出文字说明、证明过程或演算步骤） 17．【答案】（1）12n n a -=；（2）21122n n T n n -++-=．【解析】（1）由已知1，n a n S 成等差数列得21n n a S =+，① 当1n =时，111211a S a =+=+，∴11a =，当 2n ≥时，1121n n a S --=+，② ①-②得122n n n a a a --=，∴12nn a a -=， ∴数列{}n a 是以1为首项，2为公比的等比数列，∴1111122n n n n a a q ---==⨯=．（2）由12n n n a b na ⋅=+得12n nb n a =+， ∴1212111242n n nT b b b n a a a =+++=++++++()12111242nn a a a ⎛⎫=+++++++ ⎪⎝⎭()21112212212212n n n n n n --+=+=++--． 18．【答案】（1）21n a n =-，()n ∈*N ；（2）12362n n n T -+=-．【解析】设等差数列{}n a 的首项为1a ，公差为d ，由3S ，52S ，4S 成等差数列，可知345S S S +=，由521322a a a =+-得：120a d -=，1420a d --= 解得:11a =，2d =，因此21n a n =-，()n ∈*N ．（2）令()11212n n n n a c n b -⎛⎫==- ⎪⎝⎭．则12n n T c c c =++⋯+，∴()21111113521222n n T n -⎛⎫⎛⎫=⋅+⋅+⋅++-⋅ ⎪ ⎪⎝⎭⎝⎭，①()23111111352122222nn T n ⎛⎫⎛⎫⎛⎫=⋅+⋅+⋅++-⋅ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，②①-②，得()2111111122122222n nn T n -⎡⎤⎛⎫⎛⎫⎛⎫=+⋅+++--⋅⎢⎥ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦()1111212122n nn -⎡⎤⎛⎫⎛⎫=+---⋅⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦2332n n +=-∴12362n n n T -+=-． 19．【答案】（1）32n a n =-；（2）31n nS n =+．【解析】（1）设等差数列{}n a 的公差为()0d d ≠，∵1a ，2a ，6a 成等比数列，∴2216a a a =⋅∴()()21115a d a a d +=⋅+， ∵11a =，∴23d d =，∵0d ≠，∴3d =，∴32n a n =-．（2）由（1）知()()1111323133231bn n n n n ⎛⎫==- ⎪-+-+⎝⎭，∴1211111111113447323133131n n n S b b b n n n n ⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫=++⋅⋅⋅+=-+-+⋅⋅⋅+-=-=⎪ ⎪ ⎪ ⎪⎢⎥-+++⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦． 20．【答案】（1）21n a n =-，()n ∈*N ；（2）1132n T ⎡⎫∈⎪⎢⎣⎭，．【解析】（1）①1n =时，由11a +，得11a =，②2n ≥时，由已知，得()241n n S a =+，∴()21141n n S a --=+，两式作差，得()()1120n n n n a a a a --+--=，又∵{}n a 是正项数列，∴12n n a a --=，∴数列{}n a 是以1为首项，2为公差的等差数列．∴21n a n =-，()n ∈*N ．（2）∵()()111111212122121n n n b a a n n n n +⎛⎫===- ⎪⋅-+-+⎝⎭，∴12111111111111123235221212212n n T b b b n n n ⎛⎫⎛⎫⎛⎫⎛⎫=+++=-+-++-=-< ⎪ ⎪ ⎪ ⎪-++⎝⎭⎝⎭⎝⎭⎝⎭．又因为数列{}n T 是递增数列，当1n =时n T 最小，113T =，∴1132n T ⎡⎫∈⎪⎢⎣⎭，．21．【答案】（1）12n n a -=；（2）()1lg2212n n n n T -=+-．【解析】（1）由()21n n S a n =-∈*N ，可得1121S a =-，∴1121a a =-，∴11a =．又2221S a =-，∴12221a a a +=-，∴22a =． ∵数列{}n a 是等比数列，∴公比212a q a ==，∴数列{}n a 的通项公式为12n n a -=．（2）由（1）可知，()lg 1lg2n n b a n ==-，∴数列{}n n b a +的前n 项和()()()1122n n n T b a b a b a =++++++()()()-101lg221lg22n n ⎡⎤=+++++-+⎣⎦()()1lg22lg21lg2122n n -=+++-++++⎡⎤⎣⎦()1lg2212n n n -=+-． 22．【答案】（1）12n n a λ-=⨯；（2）()()()()()14214344211334n n n n n n T n n n +⎧-+⎪+⎪=⎨--+⎪+⎪⎩，是偶，是奇．【解析】（1）由题意可知112S a λ=-，即1a λ=；当2n ≥时，()()1112222n n n n n n n a S S a a a a λλ---=-=---=-，即12n n a a -=； ∴数列{}n a 是首项为λ，公比为2的等比数列，∴12n n a λ-=⨯．（2）由（1）可知当4λ=时12n n a +=，从而121n n n b n n +⎧⎪=⎨+⎪⎩，是奇，是偶， n 为偶数时，()2414312142n n n n T ⎛⎫- ⎪++⎝⎭=+-； n 为奇数时，()()1211141431122142n n n n n n T T b n +++⎛⎫+- ⎪+++⎝⎭=-=+-+- ()()()142115234n n n n +-++=+-- ()()()14211334n n n +--+=+，综上，()()()()()14214344211334n n n n n n T n n n +⎧-+⎪+⎪=⎨--+⎪+⎪⎩，是偶，是奇．。

2020年(人教版)英语单元测试【必修2】Unit 4(I)(含答案)

2020年精编人教版英语资料必修二Unit 4 单元测试Wildlife protectionI.阅读理解AIreland has had a very difficult history. The problems started in the 16th century when English rulers tried to conquer (征服) Ireland. For hundreds of years, the Irish people fought against the English. Finally, in 1921, the British government was forced to give independence to the south of Ireland. The result is that today there are two “Irelands”. Northern Ireland, in the north, i s part of the United Kingdom. The Republic of Ireland, in the south, is an independent country.In the 1840s the main crop, potatoes, was affected by disease and about 750,000 people died of hunger. This, and a shortage (短缺) of work, forced many people to leave Ireland and live in the USA, the UK, Australia and Canada. As a result of these problems, the population fell from 8.2 million in 1841 to 6.6 million in 1851.For many years, the majority of Irish people earned their living as farmers. Today, many people still work on the land but more and more people are moving to the cities to work in factories and offices. Life in the cities is very different from life in the countryside, where things move at a quieter and slower pace.The Irish are famous for being warm-hearted and friendly. Oscar Wilde, a famous Irish writer, once said that the Irish were “the greatest talkers since the Greeks”. Since independence, Ireland has revived (复兴) its own culture of music, language, literature and singing. Different areas have different styles of old Irish songs which are sung without instruments. Other kinds of Irish music use many different instruments such as the violin, whistles, etc..1. What does the author tell us in Paragraph 1?A. How the Irish fought against the English.B. How Ireland gained independence.C. How English rulers tried to conquer Ireland.D. How two “Irelands” came into being.2. We learn from the text that in Ireland ________.A. food shortages in the 1840s led to a decline in populationB. people are moving to the cities for lack of work in the countrysideC. it is harder to make a living as a farmer than as a factory workerD. different kinds of old Irish songs are all sung with instruments3. The last paragraph is mainly about ________.A. the Irish characterB. Irish cultureC. Irish musical instrumentsD. a famous Irish writer4. What can be the best title for the text?A. Life in IrelandB. A Very Difficult HistoryC. Ireland, Past and PresentD. The Independence of IrelandBMrs. Amatuli was my teacher in the fourth grade. One day at lunch time, I was getting ready to eat my same old tuna fish (金枪鱼)sandwich and suddenly Mrs. Amatuli asked me if she could buy my sandwich from me. She explained that I could use the money to buy a hot lunch from the cafeteria.I was thrilled. I never bought my lunch at the cafeteria. It was too expensive for my family, and I always carried my lunch and brought the bag back home to use again the next day. My sandwiches were either bologna (大香肠)or tuna fish. It rarely varied beyond that.You can understand my delight when I had the opportunity to buy a hot lunch.When we finished lunch that day, Mrs. Amatuli took me aside and said she wanted to explain why she had bought my sandwich. I really didn't care why, but it gave me a few minutes of her precious attention so I was very quiet as she explained. You see, she was Catholic and she told methat Catholics didn't eat red meat on Fridays—they ate fish on Fridays.Oh, I couldn't wait to get home and tell my mama that from now on I wanted tuna fish on Fridays. After my mama understood why, she gladly fixed tuna fish for me on Fridays. She even fixed it on brown bread because she know Mrs. Amatuli liked brown bread. From then on, every Friday I could get in line with the rest of the kids for a hot lunch. I didn't care how many of the kids complained about cafeteria food—it tasted divine to me!I realize now that Mrs. Amatuli could have fixed herself a tuna sandwich of Friday. But she bought my sandwich because she saw a little girl who was thrilled over the simple act of having a hot lunch.I will never forget her for her compassion and generosity and what I should do is to follow her example.5．Mrs. Amatuli bought the writer's sandwich because ______.A．she was tired of cafeteria foodB．she hated getting in line with kidsC．she didn't eat red meat on FridayD．she wanted to show care to the writer6．What can we learn about the cafeteria food?A．It was terrific. B．It was terrible.C．The writer enjoyed it. D．Most kids were fond of it.7．The underlined word“divine”in Paragraph 5 means ________.A．perfect B．sweetC．unpleasant D．special8．Which of the following can best show the theme of this text?A．It is hard to please all.B．Better to give than to receive.C．Love makes the world go around.D．The more you offer, the more you are rewarded.CDear Dr. Cory：I have allergies (过敏反应). It feels like something is in my throat. What is it and what should I do to help it?Christa Dear Christa：You might be feeling mucous in your throat. The membranes there make mucus to keep nasal passages (鼻腔) moist and clean and to warm the air that we breathe. But sometimes our bodies make too much mucus, and it can collect at the back of the throat. Drinking lots of liquids, especially warm ones, may help.Dear Dr. Cory：Sometimes when I don't wear any socks with my shoes I get blisters (水疱). What are blisters? How can I avoid them?Jacy Dear Jacy：Blisters occur when something (like the back of your shoe) rubs against your skin and causes tiny blood vessels to leak fluid (液体). The blister acts like a pillow to protect damaged skin while it repairs itself.The best way to prevent blisters is to avoid“rubbing”spots. If the problem is your shoes, wear socks and shoes that fit to help prevent blisters.Dear Dr. Cory：I like reading your column. It tells me a lot about the body and how it works. I have a question for you：I have not been airing my feet out lately and the bottoms of them have been turning crisp (起皱的)and the skin is peeling. Why?Aaron Dear Aaron：People whose feet get wet or sweaty while wearing shoes may develop athlete's foot, caused bya fungus that likes warm, damp skin. The fungus is easily passed from one person to another, so public places such as locker rooms or showers are real hot spots. It's most common during summer in people who rarely remove their shoes, and in people who sweat a lot. It can cause bumps, redness, itching, and peeling of the skin, often between the toes.Anti-fungus cream, now available without a prescription (处方), cures the infection. Your doctor or health care provider can help find the cause of the problem.9．What should Christa do with her trouble?A．Drink plenty of warm water.B．Let it be for it can repair itself.C．Consult a doctor immediately.D．Remove the mucus immediately.10．What can we infer from the passage?A．Jacy doesn't like wearing socks.B．Dr. Cory often writes some books about health.C．Athletes are more likely to have foot problems.D．You can cure infection without the help of doctors.11．The following statements are true about blisters EXCEPT that ________.A．the blisters are helpful to skin's recoveryB．wearing suitable socks and shoes can prevent blistersC．blood vessels will leak fluid if rubbedD．the best way to cure blisters is to avoid rubbing spots12．Doctor Cory thinks the most direct reason of Aaron's trouble is ________.A．sweating B．not airing his feet outC．fungus D．rarely removing his shoes13．The letters are about ________.A．ways to keep fit B．health problemsC．suggestions about diet D．knowledge about medicineDShopping in the United States changes a lot. About ninety years ago most people shopped in small stores that were owned by one person or a family. Women went from the bakery to the butcher's to the grocer and on to the fruit and vegetable seller in order to get their food for the week.Then about sixty years ago, supermarkets were born. In a supermarket, people could get all the different kinds of food they needed without going to different stores.The next big change in shopping in the United States was the shopping mall. A shopping mall is a group of stores under one roof. Because malls allowed people to shop without worrying about the weather, they soon became very popular. The mall became a place for people to socialize in addition to shopping. If you walk through a mall, you will see older people sitting, chatting and drinking coffee. Malls are places for teenagers to hang out. Many teens will often just“go to the mall”and spend time with their friends.The recent change in American shopping was the superstore. Large chain stores such as WalMart, Office Depot and Toys“R”Us have been built all across the United States. Because t hey are so large, they can buy goods at a great discount and sell them much cheaper than smaller stores.Sometimes, when they are built near small towns, many of the small town stores have to close. They just cannot compete with their giant neighbors.And now, online shopping is becoming more and more popular all over the States. People are too busy to go to the physical stores, so they go shopping over the Internet. Online shopping has lots of advantages. For instance, online stores are usually available 24 hours a day. Searching or browsing online shops can be faster than browsing the physical stores. While, online shopping also has its disadvantages. People are at higher risk of being cheated on the part of the merchant than in a physical store. And privacy of personal information may be let out.14．Which of the following shows the right order of shopping development in the United States?①small stores②superstores③shopping malls④shopping online⑤supermarketsA．①②③④⑤B．①⑤②③④C．①⑤③②④D．②③④①⑤15．Which is the place for people to spend time with others according to the passage?A．Shopping malls. B．Small stores.C．Supermarkets. D．Superstores.16．Why can the superstores sell products at much lower prices?A．Because they are built near small towns.B．Because they are across the United States.C．Because they sell all kinds of products people need.D．Because they can buy goods at a reduction in the prices.17．What's the disadvantage of online shopping according to the passage?A．Wasting time.B．Leaking personal information.C．Fixed prices.D．No chance to do physical activities.II.七选五Millions of people all over the world use the word“OK”. In fact, some people say the word“OK”means all right or acceptable; it expresses agreement or approval.___18___ Some people say it came from the Native American Indian tribe known as the Choctaw. The Choctaw word “oken”means the same as the American word“okay”. Experts say early explorers in the American West spoke the Choctaw language in the 19th century.But many people doubt this. Language expert Allen Walker Read wrote about the word“OK”in reports published in the 1960s. He said the word began being used in the 1830s. ___19___ Some foreignborn people wrote“all correct”as“o-l-l k-o-r-r-e-c-t”，and used the letters OK. Other people say a railroad worker named Obadiah Kelly invented the word long ago. They said he put the first letters of his names—O and K—on each object people gave him to send on the train._20___ The organization supported Martin Van Buren for president in 1840. They called their group the OK Club. The letters were taken from the name of the town where Martin Van Buren was born－Old Kinderhook, New York.Then there is the expression A-OK. It is a space-age expression. It was used in 1961 during the flight of astronaut Alan Shepard. He was the first American to be launched into space. His flight ended when his spacecraft landed in the ocean, as planned. Shepard reported，“Everything is A-OK.”__21___ One story says it was first used during the early days of the telephone to tell an operator that a message had been received.There are also funny ways to say okay. __22__ These expressions were first used in the 1930s. Today, a character on the American television series, “The Simpsons”，says it another way. He says okely-dokely.A．Some people say okey-dokey or okey-doke.B．Still others say a political organization invented the word.C．Therefore, it has become popular in that area from then on.D．But many experts don't agree on what the expression means.E．Language experts do not agree about where the word came from.F．It was a short way of writing a different spelling of the words“all correct”．G．However, some experts say the expression did not begin with the space age.Ⅲ.完形填空After graduation from college, my son began to seek a job. Due to the financial crisis, the competition of obtaining employment was very __23__. My son ___24___ a company located in a coastal city, but it was _25__ to employ only one person. I said to my son，“Don't lay too much hope to __26__，or you would feel terribly disappointed when you fail.” He said to me, “I am not afraid of __27__. As long as I try, there will always be hope！”After the first and second __28_，at last there were only 3 left including my son, who could enter the final ___29_ in a week.On the day of the __30__，unexpectedly, the interview was unbelievably _31__，and theinterview strategies (策略) everyone prepared were not used at all. The interviewer told them to go home and ___32_ the result in 3 days. On the morning of the third day, my son received a note from the company that he was not employed. My son felt dreadfully disappointed, but he still sent a note in reply politely.In nightfall that day, my son suddenly phoned me __33__ that he had been employed! Greatly surprised, I could not wait to know the __34_：Formally the first note sent to my son was also part of the test in the interview. Three men received the ___35_ note and gave the different answers. One did not reply. The other sai d “goodbye” and my son said “thank you”. And only my son's reply was ___36_ to the company, so he was employed.Oh! What a(n) ___37__ test!___38，it is a piece of cake to say “thank you”. But whether you are willing to say it in a peaceful and grateful manner does _39__. Saying “thank you” shows ___40___ for others' work and demonstrates your grand generosity. Therefore, ___41_ under the same condition, you will get the upper hand ___42__ opportunities compared with others.23．A.unfair B．reasonableC．painful D．fierce24．A.inspected B．targetedC．interviewed D．entered25．A.intended B．believedC．assumed D．allowed26．A.accomplish B．obtainC．succeed D．survive27．A.struggle B．disappointmentC．objection D．pressure28．A.presentation B．communicationC．introduction D．interview29．A.round B．performanceC．campaign D．process30．A.match B．contestC．final D．competition31．A.simple B．difficultC．tense D．casual32．A.take in B．get throughC．respond to D．wait for33．A.seriously B．nervouslyC．excitedly D．sincerely34．A.mystery B．truthC．procedure D．outcome35．A.different B．uniqueC．original D．same36．A.satisfying B．challengingC．moving D．inspiring37．A.abnormal B．skillfulC．special D．informal38．A.Actually B．LogicallyC．Fortunately D．Technically39．A.function B．meanC．mind D．count40．A.faith B．respectC．reward D．preference41．A.when B．unlessC．since D．until42．A.in need of B．by means ofC．in terms of D．in spite ofIV.语法填空：Michelle has finally realized one of her lifelong __43__ (dream). She always wanted to attend the __44__ (cook) school that her adviser, Mr. Brown, attended. The school, __45__ is in Paris, accepts only 80 students every year. The students who __46__ (accept) generally have __47__ least three years of cooking experience. Michelle has five years of paid experience.The project __48__ attracted Mr. Brown’s attention, however, was a cake Michelle made for a wedding. Mr. Brown had never tasted a cake that was __49__ high and creamy as __50__ one Michelle made. He gave Michelle his card and invited her to bake cakes at his restaurants __51__ (begin) the next month. It was his invitation __52_ began her career baking for a large restaurant. Now she is waiting for the plane to Paris. There she will begin her three-month course that Mr. Brown once attended.V.短文改错Last Sunday morning, when I was having a walk in the park near my home, I came across a crew make a new film with one of my favourite actor. I didn't have my camera with me at that time，but I rushed back home to get. Unfortunately, by the time I got back, they have finished the scene and the actor couldn't be seen everywhere. I was really disappointing and about to leave when he walked out a building. He was right there in the front of me! I couldn't believe my luck — not only did I had my photo taken with him，but he signed his name on my shirt!VI.书面表达假如你是李华,你刚得知你们班的英国朋友Bob出了事故,对此你和你的同学们都感到很难过,现在同学们决定让你代表他们给Bob写一封慰问信。

人教版试题试卷高二生物(人教版)第四章基因的表达单元检测

高二生物必修二第四章——基因的表达单元测试题一．选择( 共72分)1．下列哪一项不属于遗传信息转录所必需的条件（）A．解旋酶B．ATP C．模板链D．氨基酸2．关于遗传密码或遗传密码子的叙述正确的是（）A．遗传密码是DNA上能够决定一个氨基酸的3个相邻的碱基B．绝大多数生物通用一套遗传密码子C．因为有20种氨基酸，所以遗传密码子也一定有20种D．遗传密码的形成主要是在核糖体上3．DNA分子基本骨架的组成成分是（）A．核糖、磷酸B．脱氧核糖、磷酸、碱基C．核糖、磷酸、碱基D．脱氧核糖、磷酸4．下面是几个同学对有关蛋白质和核酸之间关系的总结，你认为其中错误的是（）A.同一生物体内，不同的体细胞核中DNA分子是相同的，但蛋白质和RNA是不同的B.基因中的遗传信息通过mRNA传递到蛋白质，遗传信息通过蛋白质中的氨基酸的排列顺序得到表达C．在蛋白质合成旺盛的细胞中，DNA分子多，转录成的mRNA分子也多，从而翻译成的蛋白质就多，如胰腺细胞与口腔上皮细胞相比较就是如此D．真核细胞中，DNA的复制和转录主要在细胞核中完成，而蛋白质的合成均在细胞质内完成5.信使RNA上决定氨基酸的某个密码子的一个碱基发生替换，则识别该密码子的tRNA及转运的氨基酸发生的变化是（）A.tRNA一定改变，氨基酸一定改变B．tRNA不一定改变，氨基酸不一定改变C．tRNA一定改变，氨基酸不一定改变D．tRNA不一定改变，氨基酸一定改变6．DNA分子的解旋发生在下列那些过程中（）A．复制B．转录C．翻译D．复制和转录7．胰岛B细胞内某个DNA分子，已知在该DNA分子的一条链上G+C占60%，A占24%，则另一条链上A占整个DNA分子的碱基比例为（）A．60% B．24% C．8% D．16%8．用a表示DNA，b表示基因，c表示脱氧核苷酸，d表示碱基，下图中四者关系中正确的是A．a b c d B．b c d a C． c d a b D．d c b a9．分析一个DNA分子时，发现30%的脱氧核苷酸含有腺嘌呤，由此可知该分子中，一条链上的鸟嘌呤含量最大值可占此链碱基总数的（）A．20% B．30% C．40% D．70%10．组成人体蛋白质的20种氨基酸所对应的密码子共有（）A．4种B．20种C．61种D．64种11．一条多肽链中有氨基酸1000个，则作为合成该多肽的模板信使RNA和用来转录信使RNA的DNA分子分别至少要有碱基多少个（）A．3000个和3000个B．1000个和2000个C．3000个和6000个D．2000个和4000个12．一个基因中碱基组A+T占70%，它转录形成的mRNA的U占25%，则mRNA上的碱基A占（）A ．15%B ．25%C ．30%D ．45%13．下列对tRNA 的描述，正确的是（）A ．每种tRNA 能识别并转运多种氨基酸B ．每种氨基酸只有一种tRNA 能转运它C ．tRNA 能识别mRNA 上的密码子D ．tRNA 转运氨基酸到细胞核内14．甲、乙、丙、丁为大肠杆菌的四个失去合成色氨酸能力的突变体。

高中数学选择性必修二第四章数列单元测试(提升卷)(无答案)

第四章数列单元过关检测能力提升B 卷学校:___________姓名：___________班级：___________考号：___________ 题型：8（单选）+4（多选）+4（填空）+6（解答），满分150分，时间：120分钟一、单选题1．已知等差数列{}n a 的公差和首项都不为零，且2a ，4a ，8a 成等比数列，则1324a a a a （）A ．13B ．23C ．53D ．22．设正项等比数列{}n a 的前n 项和为n S ，10103020102(21)0S S S -++=，则公比q 等于（）A ．12B ．13C ．14D ．23．两个等差数列{}n a 和{}n b ，其前n 项和分别为n S 、n T ，且723n n S n T n +=+，则220715a ab b +=+（）A ．49B ．378C ．7914 D ．149244．已知数列{}n a 为等差数列，135102a a a ++=-，24699a a a ++=-，以n S 表示{}n a 的前n 项和，则使得n S 达到最小值的n 是（） A ．37和38B ．38C ．37D ．36和375．十九世纪下半叶集合论的创立，奠定了现代数学的基础．著名的“康托三分集”是数学理性思维的构造产物，具有典型的分形特征，其操作过程如下：将闭区间[0,1]均分为三段，去掉中间的区间段12(,)33，记为第一次操作；再将剩下的两个区间1[0,]3，2[,1]3分别均分为三段，并各自去掉中间的区间段，记为第二次操作；…，如此这样，每次在上一次操作的基础上，将剩下的各个区间分别均分为三段，同样各自去掉中间的区间段.操作过程不断地进行下去，以至无穷，剩下的区间集合即是“康托三分集”．若使去掉的各区间长度之和不小于910，则需要操作的次数n 的最小值为（）（参考数据：lg 20.3010=，lg30.4771=） A ．4B ．5C ．6D ．76．已知数列1、1、2、1、2、4、1、2、4、8、1、2、4、8、16、…，其中第一项是02，接下来的两项是02、12，再接下来的三项是02、12、22，以此类推，若100N >且该数列的前N 项和为2的整数幂，则N 的最小值为（） A ．440B ．330C ．220D ．1107．等差数列12,,,n a a a ⋅⋅⋅()*3,n n ≥∈N ，满足121|||||||1|n a a a a ++⋅⋅⋅+=+2|1|a ++|1|n a +⋅⋅⋅++12|2||2||2|2019n a a a =-+-+⋅⋅⋅+-=，则（）A ．n 的最大值为50B ．n 的最小值为50C ．n 的最大值为51D ．n 的最小值为518．已知数列{}n a 满足()2*1232n n a a a a n N =∈，且对任意的*n N ∈都有12111nt a a a +++<，则实数t 的取值范围是（）A ．1+3⎛⎫∞ ⎪⎝⎭，B ．1+3⎡⎫∞⎪⎢⎣⎭，C ．2+3⎛⎫∞ ⎪⎝⎭，D ．2+3⎡⎫∞⎪⎢⎣⎭，二、多选题9．(多选)已知单调递增的等差数列{}n a 满足123101+a +a +0a a ⋯+=，则下列各式一定成立的有（） A ．11010a a +>B ．21000a a +=C ．31000a a +≤D ．510a =10．设等比数列{}n a 的公比为q ，其前n 项和为n S ，前n 项积为n T ，并且满足条件11a >，99101011,01a a a a -><-则下列结论正确的是（） A ．01q <<B ．10111a a >C ．n S 的最大值为10SD ．n T 的最大值为9T11．黄金螺旋线又名等角螺线，是自然界最美的鬼斧神工．在一个黄金矩形（宽长比约等于0.618）里先以宽为边长做正方形，然后在剩下小的矩形里以其宽为边长做正方形，如此循环下去，再在每个正方形里画出一段四分之一圆弧，最后顺次连接，就可得到一条“黄金螺旋线”．达·芬奇的《蒙娜丽莎》，希腊雅典卫城的帕特农神庙等都符合这个曲线．现将每一段黄金螺旋线与其所在的正方形所围成的扇形半径设为a n (n ∈N *)，数列{a n }满足a 1＝a 2＝1，a n ＝a n －1＋a n －2 (n ≥3)．再将扇形面积设为b n (n ∈N *)，则（）A ．4(b 2020－b 2019)＝πa 2018·a 2021B ．a 1＋a 2＋a 3＋…＋a 2019＝a 2021－1C ．a 12＋a 22＋a 32…＋(a 2020)2＝2a 2019·a 2021D ．a 2019·a 2021－(a 2020)2＋a 2018·a 2020－(a 2019)2＝012．如图，已知点E 是ABCD 的边AB 的中点，()*n F n ∈N 为边BC 上的一列点，连接nAF 交BD于n G ，点()*n G n ∈N满足()1223nn n n n G D aG A a G E +=⋅-+⋅，其中数列{}n a 是首项为1的正项数列，n S 是数列{}n a 的前n 项和，则下列结论正确的是（）A ．313a =B ．数列{}3n a +是等比数列C ．43n a n =-D ．122n n S n +=--三、填空题13．已知数列{}n a 满足142n na a +=-且14a =，n S 为数列{}n a 的前项和，则2020S =__________. 14．设数列{}n a 的前n 项和为n S ，若()*11111n n n n N S S a +⎛⎫-=∈ ⎪⎝⎭，且112a =-，则20191S =_______. 15．已知函数()331xx f x =+，()x R ∈，正项等比数列{}n a 满足501a =，则()()()1299f lna f lna f lna ++⋯+等于______．16．已知等比数列{}n a 中11a =，48a =，在n a 与1n a +两项之间依次插入12n -个正整数，得到数列{}n b ，即12345,1,,2,3,,4,5,6,7,,8,9,10,11,12,13,14,15,,a a a a a ⋅⋅⋅．则数列{}n b 的前2013项之和2013S =_______(用数字作答)．四、解答题17．在①对任意1n >，满足()1121n n n S S S +-+=+，②12n n n S S a +-=+，③()11n n S na n n +=-+这三个条件中任选一个，补充在下面问题中．问题：已知数列{}n a 的前n 项和为n S ，24a =，______，若数列{}n a 是等差数列，求数列{}n a 的通项公式；若数列{}n a 不一定是等差数列，说明理由．18．根据预测，疫情期间，某医院第()N n n *∈天口罩供应量和消耗量分别为n a 和n b （单位：个），其中4515,1310470,4n n n a n n ⎧+≤≤=⎨-+≥⎩，5n b n =+，第n 天末的口罩保有量是前n 天的累计供应量与消耗量的差．（1）求该医院第4天末的口罩保有量；（2）已知该医院口罩仓库在第n 天末的口罩容纳量()24468800n S n =--+（单位：个）．设在某天末，口罩保有量达到最大，问该保有量是否超出了此时仓库的口罩容纳量？19．已知正项数列{}n a 的前n 项和为2*111,1,,n n n n S a S S a n N ++=+=∈.（1）求{}n a 的通项公式；（2）若数列{}n b 满足：1122222...22n n n n a b a b a b a b +++++=-，求数列221log n n a b +⎧⎫⎪⎪⎨⎬⋅⎪⎪⎩⎭的前n 项和n T .20．已知数列{}n a 的前n 项和为n S ，11a =，且1a 为2a 与2S 的等差中项，当2n ≥时，总有11230n n n S S S +--+=.（1）求数列{}n a 的通项公式；（2）记m b 为1n a ⎧⎫⎨⎬⎩⎭在区间(()1*0,4m m N -⎤∈⎦内的个数，记数列(){}21m m b -⋅的前m 项和为m W ，求20W .21．已知项数为*,(2)m m N m ∈≥的数列{}n a 为递增数列，且满足*n a N ∈，若()12 (1)m n na a a ab m +++-=-，且*nb N ∈，则称{}n b 为{}n a 的“伴随数列”.（1）数列1,5,9,13,17是否存在“伴随数列”，若存在，写出其“伴随数列”若不存在，请说明理由；（2）若{}n b 为{}n a 的“伴随数列"，证明: 12m b b b >>⋯>；（3）已知数列{}n a 存在“伴随数列{}n b ，且11,2049m a a ==，求m 的最大值. 22．设数列{}n a 的前n 项和为n S ，已知()1*1221N n n n S a n ++=-+∈，且25a =．（1）证明12n n a ⎧⎫+⎨⎬⎩⎭为等比数列，并求数列{}n a 的通项公式；（2）设()3log 2nn n b a =+，且22212111n nT b b b =++⋅⋅⋅+，证明2n T <；（3）在（2）的条件下，若对于任意的*N n ∈不等式()()1260n n b n n b λ+-+-<恒成立，求实数的取值范围．。

人教版新课标高中英语 2019-2020学年必修二 Unit 4 Wildlife protection 单元测试含答案详解

Unit 4Wildlife protection(时间：100分钟分值：120分)第一部分阅读理解(共两节，满分35分)第一节(共10小题；每小题2.5分，满分25分)AIn Canada you can find dogs, cats, horses, etc.in almost all families.These are their pets.People love these pets and have them as their good friends.Before they keep pets in their houses, they take them to animal hospitals to give them injections(注射) so that they won't carry disease.They have special animal food stores, though they can get animal food in almost every kind of stores.Some people spend around two hundred Canadian dollars a month on animal food.When you visit people's houses, they would be very glad to show you their pets and they are very proud of them.You will also find that almost every family has a bird feeder in the garden.All kinds of birds are welcomed to come and have a good meal.They are free to come and go and nobody is allowed to kill any animal in Canada.They have a law against killing wild animals.If you killed an animal, you would be punished.If an animal happened to get run over by a car, people would be very sad about it.People in Canada have many reasons to like animals.One of them might be: their family ties are not as close as ours.When children grow up, they leave their parents and start their own life.Then the old will feel lonely.But pets can solve this problem.They can be good friends and never leave them alone.1．The passage mainly talks about ________．A．how to keep pets away from diseaseB．pets in CanadaC．how to take good care of petsD．life of the old in Canada2．Canadians give their pets injections before keeping them in their houses because ________．A．the pets are sickB．the pets are wildC．they want to stop them from carrying diseaseD．they want them to sleep on the way home3．In Canada, children leave their parents when they grow up because ________．A．their parents' houses are too smallB．they don't love their parents any moreC．they can only find jobs far from their parentsD．they wouldn't depend on their parents any moreBThe United States is full of cars.There are still many families without cars, but some families have two or more.However, cars are used for more than pleasure.They are a necessary part of life.Cars are used for business.They are driven to offices and factories by workers who have no other way to get to their jobs.When salesmen are sent to different parts of the city, they have to drive in order to carry their products.Farmers have to drive into the city in order to get supplies.Sometimes small children must be driven to school.In some cities school buses are used only when children live more than a mile away from the school.When the children are too young to walk too far, their mothers take turns in driving them to school.One mother drives on Mondays, taking her own children and the neighbours' as well.Another mother drives on Tuesdays, another on Wednesdays, and so on.This is called forming a car pool.Men also form car pools, with three or four men taking turns in driving to the place where they all work.More car pools should be formed in order to put fewer cars on the road and use less gas.Too many cars are being driven.Something will have to be done about the use of cars.4．The United States is filled with cars, ________．A．but not every family has a carB．but few families have two cars or even moreC．and every American has a carD．and every family has a car5．Mothers form car pools in order that ________．A．they can drive to schoolB．they can take turns in driving their children to schoolC．they reach school quicklyD．they can drive their children to school in time6．The traffic in and around cities is a great problem because ________．A．too many cars are being drivenB．there are too many car poolsC．people put fewer cars on the roadD．there is less gasCIn choosing a friend, one should be very careful.A good friend can help you study.You can have fun together and make each other happy.Sometimes you will meet fair-weather friends.They will be with you as long as you have money or luck, but when you are down, they will run away.How do I know when I have found a good friend? I look for certain qualities of character, especially understanding, honesty and reliability(可靠性)．Above all else, I look for understanding in a friend.A good friend tries to understand how another person is feeling.He is not quick to judge.Instead, he tries to learn from others.He puts himself in the other person's place, and he tries to think of ways to be helpful.He is also a good listener.At the same time, however, a good friend is honest.He does not look for faults in others.He notices their good points.In short, a friend will try to understand me and accept me.Another quality of a friend is reliability.I can always depend on a good friend.If he tells me he will meet me somewhere at a certain time, I can be sure that he will be there.If I need a favour, he will do his best to help me.If I am in trouble, he will not run away from me.There is a fourth quality that makes a friend special.A special friend is someone with whom we can have fun.We should enjoy our lives, and we would enjoy our friendship.That is why I especially like friends who are fun to be with.A good friend likes the same things I like.We share experience and learn from each other.A good friend has a good sense of humour, too.He likes to laugh with me.That is how we share the joy of being friends.And I know that he is looking for the same quality in me.When I meet someone who is reliable, honest, and understanding, I know I've found a friend!7．Which of the following qualities the writer thinks is the most important in choosing a friend?A．Understanding. B．Honesty.C．Reliability. D．A sense of humour.8．If you have fair-weather friends, ________．A．they will give you all that they have when you need helpB．you will be refused when you get into troubleC．you will become rich if you stay with themD．you can be sure that you get real friends9．Good friends need to ________．A．always point out each other's mistakesB．be helped with moneyC．understand each other's feelingsD．have money or luck10．This passage mainly discusses ________．A．the qualities of a friendB．where to choose friendsC．how to get along with friendsD．the importance of having a friend第二节(共5小题；每小题2分，满分10分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

高中数学必修二第四章直线与方程单元测试卷

第四章单元测试卷第Ⅰ卷(选择题，共60分)一、选择题(本大题共12小题，每题5分，共60分)1．已知圆的方程是(x －2)2＋(y －3)2＝4，则点P (3,2)( )A ．是圆心B ．在圆上C ．在圆内D ．在圆外2．圆x 2＋y 2＋Dx ＋Ey ＋F ＝0(D 2＋E 2－4F >0)关于直线x ＋y ＝0对称，则下列等式中成立的是( )A ．D ＋E ＋F ＝0B ．D ＋F ＝0C ．D ＋E ＝0 D ．E ＋F ＝03．由直线y ＝x ＋1上的一点向圆(x －3)2＋y 2＝1引切线，则切线的最小值为( )A ．1B ．22 C.7 D ．34．已知圆C ：(x －a )2＋(y －2)2＝4(a >0)及直线l ：x －y ＋3＝0，当直线l 被圆C 截得的弦长为23时，a 的值为( ) A. 2 B ．2－2 C.2－1 D.2＋15．若点P (3，－1)为圆(x －2)2＋y 2＝25的弦AB 的中点，则直线AB 的方程是( )A. x ＋y －2＝0 B ．2x －y －7＝0 C ．2x ＋y －5＝0 D ．x －y －4＝06．从点P (4，－1)向圆x 2＋y 2－4y －5＝0作切线PT (T 为切点)，则|PT |等于( )A ．5B ．4C ．3 D.107．圆x 2＋y 2－2x －8＝0和圆x 2＋y 2＋2x －4y －4＝0的公共弦所在的直线方程为( )A ．x －y －3＝0B ．x －y ＋1＝0C ．x ＋y －3＝0D ．x ＋y ＋1＝08．(2010·湖北)若直线y ＝x ＋b 与曲线y ＝3－4x －x 2有公共点，则b 的取值范围是( )A ．[1－22，1＋22]B ．[1－22，3)C ．[－1,1＋22]D ．[1－22，3]9．以P (2,3)为圆心，并与直线x ＋y －3＝0相切的圆的方程为( )A ．(x ＋2)2＋(y ＋3)2＝2B ．(x －2)2＋(y －3)2＝2C ．(x ＋2)2＋(y ＋3)2＝ 2D ．(x －2)2＋(y －3)2＝ 210．过直线x ＝－72上一点P 分别作圆C 1：x 2＋y 2＝1和圆C 2：(x －1)2＋y 2＝9的切线，切点分别为M 、N ，则|PM |与|PN |的大小关系是( )A ．|PM |＞|PN |B ．|PM |＜|PN |C ．|PM |＝|PN |D ．不能确定11．(2010·江西)直线y ＝kx ＋3与圆(x －2)2＋(y －3)2＝4相交于M ，N 两点，若|MN |≥23，则k 的取值范围是( )A ．[－34 ,0]B ．[－33，33]C ．[－3，3]D ．[－23，0] 12．已知点A (－2,0)，B (0,2)，C 是圆x 2＋y 2－2x ＝0上任一点，则△ABC 面积的最大值是( )A ．3＋ 2B ．3－2C ．6D ．4第Ⅱ卷(非选择题，共90分)二、填空题(本大题共4小题，每小题5分，共20分)13．圆心在x 轴上，半径为5，且过点A (2，－3)的圆的标准方程为________．14．圆x 2＋y 2＋4y －1＝0关于原点(0,0)对称的圆的方程为__________．15．已知圆C ：(x ＋5)2＋y 2＝r 2(r >0)和直线l ：3x ＋y ＋5＝0.若圆C 与直线l 没有公共点，则r 的取值范围是__________．16．已知圆(x －2)2＋(y －3)2＝13和圆(x －3)2＋y 2＝9交于A 、B 两点，则弦AB 的垂直平分线的方程是__________．三、解答题(本大题共6小题，共70分)17．(10分)求圆心在3x ＋y ＝0上，过原点且被y 轴截得的弦长为6的圆的方程．18．(12分)过点P (－1,2)作圆x 2＋y 2－2x ＋4y －15＝0的切线，求切线方程．19．(12分)一束光线从A (0,0,2)发出后经过xOy 平面反射，然后照到点B (1,1,2)处，则光线由A 反射后到B 点所经过的路程为多少？20．(12分)已知圆C ：(x －2)2＋(y －3)2＝4与直线l ：(m ＋2)x ＋(2m ＋1)y ＝7m ＋8，当直线l 被圆C 截得的弦长最短时，求m 的值．21．(12分)求以圆C 1：x 2＋y 2－12x －2y －13＝0和圆C 2：x 2＋y 2＋12x ＋16y －25＝0的公共弦为直径的圆的方程．22．(12分)已知点P (－2,2)和圆C ：x 2＋y 2＋2x ＝0.(1)求过P 点的圆C 的切线方程； (2)若(x ，y )是圆C 上一动点，由(1)所得写出y －2x ＋2的取值范围．。

【精品高中历史】第四单元中国特色社会主义道路的建设单元测试2 新人教版必修2+答案

必修二第4单元检测卷一、选择题1．新中国成立前夕，国民经济全面崩溃。

但仅过三年时间，工农业生产就超过历史最高水平。

这一历史现象出现的根本前提是（）A ．新中国的成立B ．过渡时期总路线的提出C ．“一五”计划的执行D ．“社会主义三大改造的完成2．右图所反映的历史现象的出现，主要得益于（）A ．国民经济的恢复与发展B ．社会主义工业化的总方针C ．对资本主义工商业的社会主义改造我国制造的第一批喷气式飞机D ．社会主义经济体系的建立3.右图是一本连环画的封面，该书讲述了孙家庄农民孙志刚的故事，图中的“社”指的是（）A.农业生产的合作组织B.公私合营的生产组织C.人民公社的生产组织D.包产到组的生产组织4．请判断出现下表中局面的原因是（）A ．社会主义建设总路线 B. 人民公社化运动C. 调整国民经济的“八字方针”D. 文化大革命5．造成右图漫画中所反映的历史运动的最主要因素（）A ．新中国经济基础薄弱华东人民美术出版社1953年4月出版B．缺乏经济建设的经验C．三大改造过急过粗D．指导思想的“左”倾错误6.中共八大是我国建设社会主义道路的一次成功探索，是因为（）①正确分析了中国社会的主要矛盾②提出了社会主义改造的任务③提出尽快使中国从落后的农业国变为先进工业国的任务④提出多快好省建设社会主义的总路线A .①③ B. ②④ C. ①② D. ②③7.“鼓足干劲、力争上游、多快好省地建设社会主义”的总路线（）①反映了广大人民要求改变我国经济文化落后状况的愿望②具有“左”倾冒进的思想倾向③忽视了经济发展的客观规律④直接导致了“大跃进”运动A. ①②③B. ②③④C. ①②④D. ①②③④8.中共发动“大跃进”和人民公社化运动的主要原因有是（）①缺乏社会主义建设经验②对国情认识不足③遵循了客观经济规律④急于求成A .①②③ B. ②③④ C. ①②④ D. ①③④9.1960年，中共中央开始纠正大跃进和人民公社化运动中出现的一些错误。

北师大版英语选择性必修二Unit 4 单元测试卷(含答案)

Unit 4 Humor测试卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.When does the restaurant close?A.At 9:00 pm.B.At 9:30 pm.C.At 11:00 pm.2.Where are the speakers standing?A.In Tom’s room.B.In Jack’s room.C.In John’s room.3.How old is the woman?A.18.B.20.C.22.4.What colour is John’s hair?A.Gray.B.Black.C.Brown.5.What does the woman plan to do tonight?A.Give a performance.B.Watch a movie.C.Practice the piano.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料,回答第6、7题。

6.What day is it today?A.Wednesday.B.Thursday.C.Friday.7.What will the man do this afternoon?A.Clean the blackboard.B.Clean the windows.C.Sweep the floor.听第7段材料,回答第8、9题。

8.What do people in the woman’s country love eating?A.Noodles.B.Rice.C.Bread.9.Where did Alisa grow up?A.In China.B.In Thailand.C.In France.听第8段材料,回答第10至12题。

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地理在线测试（四）一、选择题（共20题每小题只有一个正确答案，每题2分，共40分）1、东亚、东南亚和南亚水稻种植业地区人多地少、生产规模小，下列水稻种植业的特点形成与此无关的选项是：（）A.小农经营B.单位面积产量高，但商品率低C.机械化水平低D.水利工程量大2、下图是A、B、C、D四地生产产品的成本分析资料，某企业集团计划发展软件开发、移动通信器材。

最适合的地点是：（）3、有关邮电通信的叙述正确的是：（）A.邮电通信可以分为邮电和通信两大部分B.电信通信的主要任务是传递信函和包裹C.邮政通信受地形障碍影响大D.地形障碍可能影响电信通信的通信质量4、读“欧洲货物四种运输方式运费与运距相关曲线示意图”，下列叙述错误的是：（）A.空运价格始终最高，水运价格始终最低B.仅从运费考虑，中长距离应尽量选择水运和和铁路C.80千米 < 运距 < 550千米时，最廉价的运输方式是铁路D.运距 < 80千米时，水运价格高于公路和铁路5、有关京九铁路的叙述正确的是：（）A.京九铁路位于京沪线和京广线的东侧B.京九铁路经过大别山、井冈山、延安等革命老区C.京九铁路的建设起决定性作用的是自然因素，而非社会经济因素D.建设京九铁路能激活全国铁路网，带动沿线地区经济发展6、关于以莱茵河为主干的西欧航道网的说法正确的有：（）A.莱茵河流经瑞士、俄罗斯、德国、荷兰B.该航道网是由莱茵河、易北河、多瑙河等天然河流组成的C.莱茵河流经鲁尔区的河段是该航道网运输最繁忙的河段D.莱茵河通过中部运河等与多瑙河相连7、虽然我国自然资源种类多、总量大、类型齐全，但是还存在着资源的相对短缺。

我国资源的相对短缺不表现在：（）A.庞大的人口压力B.经济发展的需求C.生产技术和工艺水平的落后D.深刻的环境危机8、以下选项中，属于在铁路枢纽上兴起的城市，且其相应的铁路线都正确的一组是：（）A.开封、洛阳——陇海线B.石家庄、郑州——京广线C.包头、大同——京包线D.蚌埠——京九线要把握地理事物、地理现象的发展规律以及彼此之间的因果联系，就应掌握按地理事物的发展顺序，确定正确的排序。

据此回答9～12小题9、某国经济处在持续的发展过程中，以下是该国70年代、80年代和90年代三大产业构成变化表，下列判断正确的是：（）A. a为70年代B. b为70年代C. c为90年代D. a为90年代10、钢铁工业区的区位指向发生变化的顺序是：（）11、下面4幅图中，黑点表示乡镇、圆表示城市，按从左向右的顺序能正确反映城市发展一般规律的是：（）12、按下列事物的发展顺序，排列正确的是：（）A.信息网络不通畅、国际互联网、多媒体技术、信息高速公路B.多媒体技术、信息高速公路、信息网络不通畅、国际互联网C.多媒体技术、信息网络不通畅、信息高速公路、国际互联网D.国际互联网、信息网络不通畅、信息高速公路、多媒体技术发展绿色食品，避免“白色污染”，增强环保意识，是保护环境，提高人类生存质量的重要措施。

请回答13～15小题13、绿色食品是指：（）A.绿颜色的营养食品B.有叶绿素的营养食品C.经济附加值高的营养食品D.安全无公害的营养食品14、通常所说的“白色污染”是指：（）A.冶炼厂的白色烟尘B.石灰窑的白色粉尘C.聚乙烯等白色塑料垃圾D.白色建筑材料15、倡导“免赠贺卡”、“免用一次性木筷”的出发点是：（）A.减少个人经济支出B.节约木材，保护森林C.减少固体垃圾D.移风易俗16、下列关于环境问题和可持续发展的叙述正确的是：（）A.环境问题就其表现来看就是环境污染B.总体上讲，环境问题的本质就是发展问题C.实现可持续发展的过程就是改造自然、征服自然的过程D.经济可持续发展是可持续发展的目的17、如右图所示，某海港城市要建一港口，从影响港口建设的陆域条件和水域条件综合分析，图中A、B、C、D四个地点中，最佳位置是：（）18、制约环境人口容量的首要因素是：（）A.资源状况B.科技发展水平C.人口的文化 D.生活消费水平19、右图表示各类土地利用付租能力随距离递减示意图，从下面选出正确的选项：（）A.a为商业、b为工业、c为住宅B.a为商业、b为住宅、c为工业C.a为住宅、b为工业、c为商业D.a为住宅、b为商业、c为工业20、7世纪中期，日本推行“大化改新”，一些日本贵族和学者向从大唐归来的“遣唐使”学习中国文化的过程，属于文化扩散中的：（）A.迁移扩散B.刺激扩散C.等级扩散D.传染扩散二、不定项选择题（每小题4分，共计40分。

每个题目有一个或二个正确答案，多选、少选均不给分）21、昆明附近地区近几年将烟草业逐渐转向发展花卉业（“彩色农业”），该产业转化的意义有：A.减少运输成本B.使烟草的产量增加C.既有益生态，又美化环境D.消除不可持续发展的隐患22、将农业的输入要素按自然与人文两类归纳，同属一类的是：A.化肥、技术、机械、农药、劳动B.降水、化肥、机械、劳动、土地C.光照、降水、土地、灌溉、化肥D.种子、降水、劳动、土地、资本《晏子春秋》中说：“橘生淮南则为橘，生于淮北则为枳”。

据此回答23～24题：23、这句话说明农业生产具有：A.季节性B.周期性C.地域性D.不可预测性24、造成这一差异的主要因素是：A.地形B.气候C.土壤D.种植方法25、下图中各地段较合理的土地利用模式是：A. ① 村落② 林地③ 耕地B. ① 林地② 果园③ 耕地C. ① 耕地② 林地③ 村落D. ① 林地② 村落③ 耕地我国铁路运输能力不足，制约了国民经济的发展。

在每年的“春运高峰”和“五一”、“十一”的“假日消费黄金周”，出行难总是人们议论的焦点话题。

因此，我们需要建设更多的铁路，包括发展高速铁路。

据此回答26～27题26、贯穿我国南北的铁路大动脉中，将修建高速铁路的是：A.宝成线B.京广线C.京九线D.京沪线27、在西部大开发中，铁路等基础建设必须先行。

这主要是因为铁路建设：A.投资规模大，建设周期较长B.提供就业机会，提高人均收入C.加快商品流通，促进共同发展D.投资回报较高，实现可持续发展28、第五次人口普查数据显示：上海市中心区近几年人口减少比较明显，而原来的郊区人口却增长迅速，主要原因是：A.中心区和原郊区人口再生产类型不同B.中心区环境恶化，人口大量外迁C.原郊区外来人口过多，导致高自然增长率D.原郊区经济发展快，吸引大量外来人口29、如下图，公路一侧建有公交车站，现欲在公路对面设一公交车站，有`A、B、C、D四个可选地址，其中最合理的是：30、形成商业中心的主要条件是：A.交通便利，方便商品的集散B.在社会发展中具有一定的政治、历史地位C.环境优美，名胜古迹多D.经济发达，人口稠密，有稳定的货源和销售区海宁市宏达高级中学04~05学年第二学期期中试卷高一地理05/04/18第Ⅰ卷（选择题共50分）一、单项选择题（本大题共40小题，1~30题每小题1分，31~40小题每小题2分，共50分。

各小题均有四个选项，其中只有一项是符合题意要求的。

）1．不准行人和非机动车进入的公路线是A．国道B．省道C．县道D．高速公路2．热带雨林迁移农业属于（）A．畜牧业B．种植业C．密集农业D．商品农业3．在我国东部丘陵地区修筑梯田，改造的自然因素主要是（）A．光照B．地形C．湿度D．水分4．潘帕斯草原上的大牧场牧牛业属于（）A．游牧业B．粗放农业C．商品农业D．自给农业5．“南南合作”是指（）A．南半球国家之间的合作B．发达国家与发展中国家之间的合作C．发达国家之间的合作D．发展中国家之间的合作6．下列国家中，其出口产品主要为农矿产品的国家是（）A．美国B．印度尼西亚C．日本D．德国7．墨累—达令盆地发展农牧业的限制性条件是（）A．降雨稀少B．光照不足C．地形崎岖D．交通不便8．城市交通的主要特点有（）①交通点、线、面结合紧密②交通运输方式多样、极具综合性③客流量在早晚上下班时间较大④道路建设不会占用大量的城市用地A．①②B．①③C．②③D．②④9．新技术革命出现的时间是（）A．工业革命初期B．19世纪70年代C．20世纪50年代D．20世纪70年代10．某学生想得到名师的指点，来提高自己学习质量，其采用的最理想的传输手段是（）A．长途电话B．电报C．国际互联网D．寄信欧洲某企业在我国投资建立某个新厂，其产品主要出口欧洲，这一决策使得其生产零部件的一些生产厂陆续迁至该厂附近。

该企业设想：早晨接到订单，立即组织生产，下午发货，次日产品运到欧洲。

回答11－12题。

11．该厂产品可能是（）A．生物药品B．电子产品C．化工产品D．建筑陶瓷12．该工厂布局首先须考虑靠近（）A．公路枢纽B．铁路枢纽C．海港D．航空港13．电子邮件所用的地址与（）A．具体地点有关B．具体地点无关C．一般电话场所的通信方式一样D．一般传真场所一致14．以下城市为世界著名的国际金融中心的是（）A．纽约、华盛顿B．香港、上海C．伦敦、苏黎世D．加尔各答、巴黎15．形成鲁尔区的优越区位条件有（）A．温和湿润的气候B．丰富的劳动力资源C．便捷的水陆交通D．丰富的铁矿资源16．鲁尔区经过综合整治后，新建或迁入的企业多为（）A．大型企业B．钢铁工业C．中小型企业D．煤炭工业17．美国“硅谷”位于（）A．旧金山市东南部B．休斯敦市东南部C．美国西北部D．美国东北部18．现在，我国出口的主要商品是（）A．纺织品、服装B．小麦、稻米C．煤炭、铁矿石D．汽车、茶叶19．当商品交换由偶然性发展为经常性时，便会出现（）A．乡村B．城市C．集市D．聚落20．下列城市中，位于河口处的是（）A．上海B．宜宾C．重庆D．昆明21．下列属于世界城市稠密地区的是（）A．东南亚的日本、朝鲜半岛B．我国的西部地区C．美国的东北部地区D．欧洲的东部地区22．下列城市中，因科技而兴起的科技城是（）A．美国旧金山B．英国阿伯丁C．德国杜伊斯堡D．日本筑波23．下列城市中，位于京沪线和陇海线交汇处的是（）A．济南B．郑州C．徐州D．兰州24．最早出现逆城市化现象的国家是（）A．日本B．德国C．英国D．加拿大25．目前，发展中国家的城市人口占世界城市总人口的比重为（）A．60％B．50％C．40％D．70％26．在城市附近修建机场，其位置对自然条件要求较严格，下列说法正确的是（）①地形平坦开阔，坡度适当，以保证排水②地质条件好，保证地基稳定牢固③对气候条件要求不严，因为风向、降水不影响飞机飞行④与城市距离、交通联系、环境状况关系不大A．①②B．③④C．①③D．②④27．中国古代烽火台的主要功能是（）A．通信B．运输C．邮政D．贸易28．对交通运输线起决定作用的区位因素有（）A．地貌、水文B．社会经济C．科学技术D．气候河流29．世界上一些主要河流的两岸布满大大小小的城市，主要是因为（）A．河流两岸风景优美B．河流两岸气候湿润C．河流两岸土壤肥沃D．河流提供便利的水运30．运具和线路合二为一的新型运输方式是（）A．航空B．管道C．铁路D．水路图2是某地区1950－1980年人口增长图，读图1回答31－32题。