【附20套高考模拟试题】2020届浙江省镇海中学高三校模拟考自选模块试卷含答案

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2020届浙江省宁波市镇海中学高三下学期高考适应性考试数学试卷及解析

2020届浙江省宁波市镇海中学高三下学期高考适应性考试数学试卷及解析

2020届浙江省宁波市镇海中学高三下学期高考适应性考试数学试卷★祝考试顺利★(含答案)第Ⅰ卷(选择题,共40分)一、选择题:本大题共10小题,每小题4分,共40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.已知集合{}1,2,3M =,{}2,3,4N =,则M N ⋃=( )A. {}1,2,3,4B. {}3,4C. {}1,4D. {}2,3 【答案】A【解析】根据并集定义计算.【详解】由题意{1,2,3,4}M N . 故选:A .2.已知复数z 满足()1210z i +-=,其中i 为虚数单位,则复数z 的虚部是( ) A. 12 B. 12- C. 12i D. 12i - 【答案】B【解析】由复数的综合运算求出z 后可得其虚部.【详解】由题意210i iz +-=,21(1)112222i i i z i i i --===--,其虚部为12-. 故选:B .3. 在△ABC 中“sinA>sinB”是“cosA<cosB”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 【答案】C【详解】试题解析:必要性在△ABC 中,“cosA>cosB”,由余弦函数在(0,π)是减函数,故有A <B ,若B 不是钝角,显然有“sinA<sinB”成立,若B 是钝角,因为A+B <π,故有A <π-B <2π,故有sinA <sin (π-B )=sinB 综上,“cosA>cosB”可以推出“sinA<sinB”:充分性:由“s inA <sinB”若B 是钝角,在△ABC 中,显然有0<A <B <π,可得,“cosA>cosB”若B 不是钝角,显然有0<A <B <2π,此时也有cosA >cosB 综上,“sinA<sinB”推出“cosA>cosB”成立故,“cosA>cosB”是“sinA<sinB”的充要条件4.若0a >,0b >,且11a b+=,则22a b +的最小值为( )A. 2B.C. 4D. 【答案】C【解析】已知等式应用基本不等式得到ab 的最小值,然后再在待求式应用基本不等式可得结论.【详解】∵0,0a b >>,∴11a b +=2ab ≥,当且仅当a b =,即a b ==等号成立,∴2224a b ab +≥≥,当且仅当a b =时等号成立,综上22a b +的最小值是4.故选:C .5.设m ,n 是两条异面直线,则下列命题中正确的是( )A. 过m 且与n 垂直的平面有且只有一个B. 过m 且与n 平行的平面有且只有一个C. 过空间一点P 与m ,n 都平行的平面有且只有一个D. 过空间一点P 与m ,n 都垂直的平面有且只有一个【答案】B【解析】。

2020届宁波市镇海中学高三英语模拟试卷及答案解析

2020届宁波市镇海中学高三英语模拟试卷及答案解析

2020届宁波市镇海中学高三英语模拟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A4 Best Drive--In Movie Theaters in the USColorado: Holiday Twin Drive--InAddress: 2206 S Overland Trail, Fort Collins, CO 80526, USAPhone: +1 970-221-1244The theater, open since 1968 and currently the most popular drive-in in the US, provides various special foods. The menu there even amazes meat-free customers. But please remember the outside food is forbidden here. Besides, the Rocky Mountains provide a pastoral (田园式的) backdrop to screenings, and sunsets usually don’t disappoint either. It also offers lots of unique events that go beyond the big screen.North Carolina: Hound’s Drive--InAddress: 114 Raven Cir, Kings Mountain, NC 28086, USAPhone: +1 704-739-4424Having only opened in 2016, it’s one of the newest theaters on the block. The drive-in features newer equipment and digital projection. People can bring their animal friends along.Florida: Fort Lauderdale Swap ShopAddress: 3291 W Sunrise Blvd, Fort Lauderdale, FL 33311, USAPhone: +1 954-791-7927The Florida favorite offers plenty ofways to have fun. With 14 screens, the self-proclaimed (自称的) world’s largest drive-in equals any indoor cinema in terms of capacity and programming. It also includes an 88-acre flea (跳蚤) market and a free Ferrari museum. It’s best to book tickets on the Internet ahead of time if you don’t want to wait in line.California: Mission Tiki Drive -InAddress: 10798 Ramona Ave, Montclair, CA 91763, USAPhone: +1 909-628-0511Let’s have fun in the old-school outdoor cinema in Montclair, California. Remember tickets are available at the ticket office only. It alternates (交替) up to eight new releases on four screens and hosts almost daily swapmeets where people can exchange things they no longer need. It also organizes classic car and lowrider meet-ups.1.What can people do in Holiday Twin Drive-In?A.Participate in somespecial activities.B.Enjoy the film with the latest equipment.C.Learn about the benefits of being meat free.D.Share home-made cookies while watching the film.2.Which of the following theaters is friendly to visitors with pets?A.Hound’s Drive- In.B.Mission Tiki Drive-In.C.Holiday Twin Drive-In.D.Fort Lauderdale Swap Shop.3.What do Fort Lauderdale Swap Shop and Mission Tiki Drive-In have in common?A.Both feature old-fashioned styles.B.Both allow booking tickets online.C.Both provide free museum exhibitions.D.Both offer chances to trade second-hand goods.BContrary to the long-held belief that plants in the natural world are always in competition, new research has found that in severe environments adult plants help smaller ones and grow well as a result.The research, led by Dr Rocio, studied adult and seedling (幼苗)plants in the ecological desert in the south-east of Spain. Dr Rocio said, “If you're a seedling in a poor land — the top of a mountain or a sand hill, for example-and you’re lucky enough to end up underneath a big plant, your chances of survival are certainly better than if you landed somewhere on your own. What we have found, which was surprising, is an established large plant, called a ‘nurse’, protects a seedling; it also produces more flowers than the same plants of similar large size growing on their own.”Other benefits of nurse-seedling partnerships include that more variety of plants growing together can have a positive effect on the environment. For example, vegetation areas with nurse plants with more flowers might be able to attract higher numbers of pollinators(传粉者)in an area, in turn supporting insect and soil life and even provide a greater range of different fruit types for birds and other animals.“The biggest winner for this system of nursing a plant is biodiversity(生物多样性),” Dr Rocio said. “The more biodiversean area, the greater number of species of plants, insect life, mammals and birds, and the better the chances of long-term healthy functioning of the environment and ecosystems. ” This system is win-win for adultand seedling plants in unfavorable environments.The research is of value to those who manage and protect plants in tough environments. Most home gardeners and farmers plan to ensure their soil and conditions are the best they can be for plant growth, but the findings might be of value to those who garden in bare places.4. What is a common understanding of plants?A. They can help each other.B. They can survive ill conditions.C. They compete with each other.D. They grow well on their own.5. What will happen to seedling plants if they grow under adult plants?A. They will produce more flowers.B. They will die owing to competition.C. They will make adult plants larger.D. They will get support from adult plants.6. What is the effect of the nurse-seedling partnership?A. It leads to unfavorable environments.B. It produces long-term healthy chances.C. It attracts higher and larger pollinators.D It provides a more variety of plant types.7. Who will benefit from the new research?A. People studying organic farming.B. People protecting plants on sand hills.C. People wanting to change biodiversity.D. People keeping more animals on the farm.CAt first glance, there is nothing unusual about BingoBox’s convenience store–shelves packed with snacks line the walls, attracting passers-by through the glass windows. But upon closer look, BingoBox is no ordinary store. The door unlocks only after customers scan (扫描) aQR code to enter, and there is no cashier — just a lone checkout counter (柜台) in a corner. The Shanghai-based company is one of many unmanned store operators (运营者) opening outlets all over China, hoping to improve slim profit by reducing staff costs.“Ifstaff costs rise quickly, that puts greater pressure on low-profit businesses like convenience stores and supermarkets,” said Andrew Song, an analyst at Guotai Junan Securities. “InChina, manpower costs have been rising ly quickly.”However, the future vision of shopping without a check-out person is still a work in progress. A Post reporterwho visited a BingoBox store inShanghaiwas briefly locked in when trying to exit without buying anything. Although a sign near the exit stated that empty-handed customers can leave by scanning a QR code, no QR code was to be found. Repeated calls to the customer service hotline went unanswered.The idea of unmanned stores first caught the world’s attention in December last year. Equipped with technology such as RFID tags, mobile payment systems and facial and movement recognition, such stores collect large amounts of data that give operators a better idea of consumer preferences and buying habits, which can then be used to optimize (使最优化) operations and make more efficient inventory decisions. For companies like BingoBox, lower operating costs also mean it can afford to expand its reach to areas with less foot traffic or fewer people, according to its founder and chief executive ChenZilin.8. What makes BingoBox store look like an ordinary convenience store?A. No cashier to check out.B. A lone checkout counter.C. Shelves packed with goods.D. Entering by scanning a QR code.9. Why are unmanned stores popular with operators?A. The customers prefer mobile payment systems.B. The unmanned stores help improve profit with lower labor costs.C. The employees focus on consumer preferences and buying habits.D. The operators care more about operations and inventory decisions.10. Why is the reporter’s case mentioned in the passage?A. To show his anger and dissatisfaction.B. To warn people not to go to a BingoBox store.C. To explain unmanned stores still have a long way to go.D. To complain that QR code service is not convenient at all..11. What can we infer from the chief executive Chen Zilin?A. Nowadays all stores should be equipped with advanced technology.B. The operators collect data about consumer preferences and buying habits.C. BingoBox made wiser decisions based on the data collected in those unmanned stores.D. The operators can open unmanned supermarkets in more distant places with low cost.DA team of engineers atHarvardUniversity in trying to create the first robotic fly. Designed to do what a fly does naturally, the tiny is the size of a fat housefly. Its mini wings allow it to stay in the air and perform controlled flight tasks."The added difficulty with a project like this is that actually none of its components is off the shelf and so we have to develop them all on our own’ said Robert Wood, a Harvard engineering professor.They engineered a series of systems to start and drive the robotic fly. “The seemingly simple system which just moves the wings hasa number of interdependencies (相互依赖)on the individual components, each of which individually has to perform well, but then has to be matched well to everything it d connected to,” said Wood.While this first robotic flyer is linked to a small, off-board power source, the goal is eventually to equip it with a built-in power source, so that it might someday perform data-gathering work at rescue sites,in farmers’ fields or on the battlefield. "Basically it should be able to take off, land and fly around,” he said.Wood says the design offers a new way to study flight mechanics and control at insect-scale. Yet, the power, sensing and computation technologies on board could have much broader applications.“You can start thinking about using them to answer open scientific questions, you know, to study biology in ways that would be difficult with the animal,but using these robots instead” he said. "So there are a lot of technologies and open interesting scientific questions that are really what drives us on a day-to-day basis.”12. What is the typical characteristic of the robotic fly?A. It's automatic.B.It's very small.C. It's controllable.D. It's quite powerful.13. We can infer from the passage that the robotic flyer can____ .A. act as a spy planeB. help do farm workC.fly at a very high speedD. answer many scientific questions14. What is Wood's idea about the robotic fly according to the last paragraph?A. It is highly questionable.B. It has wide practical applications.C. It gives scientists interest in flying machines.D. It points to a new direction in studying biology.15. What can be the best title for the passage?A. Harvand's Study in the Field of Insects.B. A Breakthrough in Engineering ScienceC. An Interesting Invention一Robotic FlyD. Robotic Fly一a Copy of Real Life Insect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。

浙江省宁波市镇海中学2020届高三6月考前模拟数学答案

浙江省宁波市镇海中学2020届高三6月考前模拟数学答案

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2020年浙江省宁波镇海中学高考物理选考模拟试卷(6月份) (含答案解析)

2020年浙江省宁波镇海中学高考物理选考模拟试卷(6月份) (含答案解析)

2020年浙江省宁波镇海中学高考物理选考模拟试卷(6月份)一、单选题(本大题共12小题,共36.0分)1.一个物体静止在斜面上时,正确表示斜面对物体作用力F方向的是图中的()A. B.C. D.2.物理学中研究问题有多种方法,有关研究问题的方法叙述错误的是()A. 在不需要考虑物体本身的大小和形状时,用质点代替物体的方法,采用了等效替代的思想B. 伽利略通过理想斜面实验指出力不是维持物体运动的原因C. 根据速度的定义式,当△t非常小时,就可以表示物体在t时刻的瞬时速度,该定义运用了极限思想法D. 探究加速度与力、质量三个物理量之间的定量关系,采用了控制变量的研究方法3.下列有关运动的描述中,参考系的选取符合描述正确的是()A. “小小竹排江中游”,是以竹排为参考系的B. “飞流直下三千尺”,是以飞流作为参考系的C. “轻舟已过万重山”,是以万重山为参考系的D. “钱塘江潮水扑面而来”,是以潮水为参考系的4.甲、乙两物体从同一点开始沿一直线运动,甲的x−t和乙的v−t图象如图所示,下列说法中错误的是()A. 甲在3s末回到出发点,甲运动过程中,距出发点的最大距离为4mB. 0〜6秒内,甲、乙两物体位移都为零C. 第3秒内甲、乙两物体速度方向相同D. 2〜4s内甲的位移大小为8m5.如图,轻绳一端固定,另一端系一小球。

小球在水平面内做匀速圆周运动,轻绳长度一定,与竖直方向夹角为θ,角速度为ω。

图中关于θ和ω的对应关系图像正确的是()A. B.C. D.6.2017年9月,我国控制“天舟一号”飞船离轨,使它进入大气层烧毁,残骸坠入南太平洋一处号称“航天器坟场”的远离大陆的深海区。

在受控坠落前,“天舟一号”在距离地面380km的圆轨道上飞行,则下列说法中正确的是()A. 在轨运行时,“天舟一号”的线速度大于第一宇宙速度B. 在轨运行时,“天舟一号”的角速度小于同步卫星的角速度C. 受控坠落时,应通过“反推”实现制动离轨D. “天舟一号”离轨后,在进入大气层前,运行速度不断减小7.手摇式发电机的线圈在匀强磁场中匀速转动时,其磁通量随时间按如图所示的正弦规律变化。

2020年浙江省宁波市镇海中学高考数学仿真试卷(有答案解析)

2020年浙江省宁波市镇海中学高考数学仿真试卷(有答案解析)

A. sinx
B. cosx
C. sin2x
D. cos2x
3. 满足线性约束条件
,的目标函数 z=x+y 的最大值是( )
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B.
C. 2
D. 3
4. 如图,网格纸上小正方形边长为 1,粗线画出的是某几何体的三视图,则该几何体的体积为( )
A.
B.
C. 4
D.
5. 某观察者站在点 O 观察练车场上匀速行驶的小车 P 的运动情况,小车从点 A 出发的运动轨迹如 图所示.设观察者从点 A 开始随动点 P 变化的视角为 θ=∠AOP,练车时间为 t,则函数 θ=f(t) 的图象大致为( )
出答案. 本题考查三视图求几何体的体积,由三视图正确复原几何体和补形是解题的关键,考查空间想象能 力.属于中档题.
5.答案:D
解析:【分析】 根据视角 θ=∠AOP 的值的变化趋势,可得函数图象的单调性特征,从而选出符合条件的选项. 本题主要考查利用函数的单调性判断函数的图象特征,属于基础题. 【解答】 解:根据小车从点 A 出发的运动轨迹可得,视角 θ=∠AOP 的值先是匀速增大,然后又减小,接着基 本保持不变,然后又减小,最后又快速增大, 故选:D.
且项数为偶数,设 n=2k,k∈N*,等差数列的公差设为 d,不妨设

则 a1<0,d>0,且 ak+1≤0,ak-1<0 即 ak≤-1, 由 ak+1-1≥0, 则-1+kd≥ak+kd≥1,即 kd≥2, 即有 d≥2, 则|a1|+|a2|+…+|an|=-a1-a2-…-ak+ak+1+…+a2k
解析:【分析】 本题主要考查了三角函数的周期性及其求法,考查了三角函数的奇偶性,为中档题. 分别把四个选项中的值代入 f(x)·sinx,逐一进行验证即可. 【解答】 解:若 f(x)=sinx,则 f(x)·sinx=sin2x 为偶函数,不符合题意;

2020年浙江省宁波市镇海中学高考数学模拟试卷(3月份)(带答案)

2020年浙江省宁波市镇海中学高考数学模拟试卷(3月份)(带答案)

2020 年浙江省宁波市镇海中学高考数学模拟试卷(3 月 份)题号 得分一二三总分一、选择题(本大题共 10 小题,共 40.0 分)1. 设集合 A={1,2,3,4},B={x∈N|-3≤x≤3},则 A∩B=( )A. {1,2,3,4}B. {-3,-2,-1,0,1,2,3,4}C. {1,2,3}D. {1,2}2. 双曲线的渐近线方程是( )A. 2x±y=0B. x±2y=03. 已知公差不为零的等差数列{an}满足值为( )C. 4x±y=0D. x±4y=0,Sn 为数列{an}的前 n 项和,则 的A.B.C.D.4. “a>0”是“”的( )A. 充分而不必要条件 C. 充要条件B. 必要而不充分条件 D. 既不充分也不必要条件5. 函数的图象可能是( )A.B.C.D.6. 某射手射击所得环数 ξ 的分布列如下:ξ 78910P x 0.1 0.3 y已知 ξ 的数学期望 E(ξ)=8.9,则 y 的值为( )A. 0.8B. 0.6C. 0.4D. 0.27. 已知正四棱柱 ABCD-A1B1C1D1 中,AB=2,CC1=2 ,E 为 CC1 的中点,则直线 AC1与平面 BED 的距离为( )A. 2B.C.D. 1第 1 页,共 14 页8. 对于定义域为 R 的函数 f(x),若存在非零实数 x0,使函数 f(x)在(-∞,x0)和(x0,+∞)上与 x 轴都有交点,则称 x0 为函数 f(x)的一个“界点”.则下列四个函数中,不存在“界点”的是( )A. f(x)=2x-x2B. f(x)=x2+bx-2(b∈R)C. f(x)=1-|x-2|D. f(x)=x-sinx9. 已知是平面内三个单位向量,若 ,则的最小值()A.B.C.D. 510. 已知数列{an}满足 2an≤an-1+an+1(n∈N*,n≥2),则( )A. a5≤4a2-3a1B. a2+a7≤a3+a6C. 3(a7-a6)≥a6-a3D. a2+a3≥a6+a7二、填空题(本大题共 7 小题,共 36.0 分)11. 设 i 为虚数单位,给定复数,则 z 的虚部为______,|z|=______.12. 某几何体的三视图如图所示,则该几何体的体积是______,表面积是______.13. 已知 x,y 满足条件则 2x+y 的最大值是______,原点到点 P(x,y)的距离的最小值是______ 14. 小明口袋中有 3 张 10 元,3 张 20 元(因纸币有编号认定每张纸币不同),现从中掏出纸币超过 45 元的方法有 种;若小明每次掏出纸币的概率是等可能的,不 放回地掏出 4 张,刚好是 50 元的概率为15. 在△ABC 中,∠BAC=120°,AD 为∠BAC 的平分线,AB=2AC,则 =______.16. 若函数在上有零点,则的最小值为17. 如图,椭圆的离心率为 e,F 是 Γ 的右焦点,点 P 是 Γ 上第一象限内任意一点,,,若 λ<e,则 e 的取值范围是______.第 2 页,共 14 页三、解答题(本大题共 5 小题,共 74.0 分)18. 已知函数.(Ⅰ)求函数 f(x)的单调递增区间;(Ⅱ)设△ABC 中的内角 A,B,C 所对的边分别为 a,b,c,若求 a2+c2 的取值范围.,且,19. 如图,四棱锥 P-ABCD 中,PC 垂直平面 ABCD,AB⊥AD,AB∥CD, PD=AB=2AD=2CD=2,E 为 PB 的中点. (Ⅰ)证明:平面 EAC⊥平面 PBC; (Ⅱ)求直线 PD 与平面 AEC 所成角的正弦值.20. 在数列{an}中,a1=1,a2=3,且对任意的 n∈N*,都有 an+2=3an+1-2an. (Ⅰ)证明数列{an+1-an}是等比数列,并求数列{an}的通项公式;(Ⅱ)设,记数列{bn}的前 n 项和为 Sn,若对任意的 n∈N*都有,求实数 m 的取值范围.第 3 页,共 14 页21. 已知椭圆的焦点坐标为 F1(-1,0),F2(1,0),过 F2 垂直于长轴的直线交椭圆 于 P、Q 两点,且|PQ|=3. (1)求椭圆的方程; (2)过 F2 的直线 l 与椭圆交于不同的两点 M、N,则△F1MN 的内切圆的面积是否 存在最大值?若存在,求出这个最大值及此时的直线方程;若不存在,请说明理由.22. 已知函数 f(x)=x2e3x (Ⅰ)若 x<0,求证:f(x)< (Ⅱ)若 x>0,恒有 f(x)≥(k+3)x+2lnx+1,求实数 k 的取值范围2020 年浙江省宁波市镇海中学高考数学模拟试卷(3 月 份)答案和解析【答案】1. C2. B3. A4. C5. D6. C7. D8. D9. A10. C11. 212. 144-12π 168+6π13. 614. 3215. 316. -17. (0, ]第 4 页,共 14 页18. 解:(Ⅰ)==所以,解得,k∈Z.所以函数 f(x)的单调递增区间为,k∈Z(Ⅱ)因为,所以.所以 .又因为,所以 3=a2+c2-ac,即 a2+c2=3+ac.而 a2+c2≥2ac,所以 ac≤3,即 a2+c2≤6.又因为 a2+c2=3+ac>3,所以 3<a2+c2≤6.19. (Ⅰ)证明:PC⊥平面 ABCD,故PC⊥AC.………………(2 分)又 AB=2,CD=1,AD⊥AB,所以 AC=BC= .故 AC2+BC2=AB2,即AC⊥BC.………………(4 分)所以 AC⊥平面 PBC,所以平面 ACE⊥平面PBC. …………………………(6 分)(Ⅱ)解:PC⊥平面 ABCD,故 PC⊥CD.又 PD=2,所以 PC= . …………(8 分)在平面 ACE 内,过点 P 作 PF 垂直 CE,垂足为 F.由(Ⅰ)知平面 ACE⊥平面 PBC,所以 PF 垂直平面 ACE.由面积法得:即.…………(10 分)又点 E 为 AB 的中点,.所以.……………………………………(12 分)又点 E 为 AB 的中点,所以点 P 到平面 ACE 的距离与点 B 到平面 ACE 的距离相等. 连结 BD 交 AC 于点 G,则 GB=2DG.所以点 D 到平面 ACE 的距离是点 B 到平面 ACE 的距离的一半,即 .所以直线 PD 与平面 AEC 所成角的正弦值为.……………………(15 分)另解:如图,取 AB 的中点 F,如图建立坐标系.因为 PD=2,所以.所以有:C(0,0,0),D(0,1,0),,A(1,1,0),B(1,-1,0),. …………(9 分).,.第 5 页,共 14 页设平面 ACE 的一个法量为 =(x,y,z),则取 x=1,得 y=-1,.即=.…………(13 分)设直线 PD 与平面 AEC 所成角为 θ,则 sinθ=|cos< , =. …………(15分)20. 解:(Ⅰ)由 an+2=3an+1-2an 可得 an+2-an+1=2(an+1-an). ………………(2 分)又 a1=1,a2=3,所以 a2-a1=2. 所以{an+1-an}是首项为 2,公比为 2 的等比数列.…………………(3 分)所以.…………………(4 分)所以 an=a1+(a2-a1)+…+(an-an-1)=1+2+22+…+2n=2n-1.…………(7 分)(Ⅱ)因为==.………(9 分)所以 Sn=b1+b2+…+bn= (12 分) 又因为对任意的 n∈N*都有,所以=.………恒成立,即,即当 n=1 时,.………(15 分)21. 解:(1)设椭圆方程为=1(a>b>0),由焦点坐标可得 c=1,由|PQ|=3,可得 =3,又 a2-b2=1,解得 a=2,b= ,故椭圆方程为=1,(2)设 M(x1,y1),N(x2,y2),不妨令 y1>0,y2<0,设△F1MN 的内切圆的半径 为 R,则△F1MN 的周长=4a=8,(|MN|+|F1M|+|F1N|)R=4R因此内切圆面积最大,R 就最大,此时也最大,由题知,直线 l 的斜率不为零,可设直线 l 的方程为 x=my+1,由得(3m2+4)y2+6my-9=0,得 则 令 t=, ,则 t≥1,,=,第 6 页,共 14 页则,令 f(t)=3t+ ,则 f′(t)=3- ,当 t≥1 时,f′(t) 0,f(t)在[1,+∞)上单调递增,有 f(t)≥f(1)=4,≤3,即当 t=1,m=0 时,≤3,=4R,∴Rmax= ,这时所求内切圆面积的最大值为 π.故直线 l 方程为:x=1,△F1MN 内切圆面积的最大值为 π.22. 证明:(Ⅰ)∵函数 f(x)=x2e3x,∴f′(x)=2xe3x+3x2e3x=x(3x+2)e3x.由 f′(x)>0,得 x<- 或 x>0;由 f′(x)<0,得-,∴f(x)在(-∞,- )内单调递增,在(- ,0)内单调递减,在(0,+∞)内单调递增,∴f(x)的极大值为 f(- )= ,∴当 x<0 时,f(x)≤f(- )= < = .解:(Ⅱ)∵x2e3x≥(k+3)x+2lnx+1,∴k≤,x>0,令 g(x)=,x>0,则 g′(x)=,令 h(x)=x2(1+3x)e3x+2lnx-1,则 h(x)在(0,+∞)上单调递增, 且 x→0+时,h(x)→-∞,h(1)=4e3+2lnx-1, ∴存在 x0∈(0,1),使得 h(x0)=0, ∴当 x∈(0,x0)时,g′(x)<0,g(x)单调递减, 当 x∈(x0,+∞)时,g′(x)>0,g(x)单调递增,∴g(x)在(0,+∞)上的最小值是 g(x0)=,∵h(x0)=+2lnx0-1,得=,令=t0,则 2lnx0+3x0=lnx0,且 φ(1)=0,∴t=1,∴g(x0)==,∴实数 k 的取值范围是(-∞,0]. 【解析】1. 【分析】可解出集合 B,然后进行交集的运算即可. 考查描述法、列举法的定义,以及交集的运算. 【解答】 解:B={0,1,2,3};第 7 页,共 14 页∴A∩B={1,2,3}. 故选:C.2. 【分析】本题考查了双曲线的渐进方程,把双曲线的标准方程中的“1”转化成“0”即可求出渐 进方程.属于基础题.渐近线方程是 -y2=0,整理后就得到双曲线的渐近线.【解答】解:双曲线其渐近线方程是 -y2=0整理得 x ±2y=0. 故选:B.3. 解:公差不为零的等差数列{an}满足,∴=a1(a1+3d),解得 a1=-4d, ∵Sn 为数列{an}的前 n 项和,∴==.故选:A.由公差不为零的等差数列{an}满足,利用等差数列的通项公式列方程求出a1=-4d,由此能求出 的值. 本题考查等差数列的前 3 项和公式和前 1 项和的比值的求法,考查等差数列的性质等基 础知识,考查运算求解能力,是基础题.4. 解:由 a>0,得 a+ ≥2 =2 ,是充分条件,由 a+ ≥2 ,得:a>0,故 a>0”是“”的充要条件,故选:C. 根据充分必要条件的定义结合不等式的性质判断即可. 本题考查了充分必要条件,考查不等式的性质,是一道基础题.5. 【分析】本题主要考查函数图象的识别和判断,利用函数的奇偶性和对称性的关系,根据函数零 点判断函数的正负是解决本题的关键. 判断函数 f(x)的奇偶性,结合图象的对称性以及函数在 x 轴右侧的函数零点判断函数 的正负进行判断即可 【解答】解:f(-x)=ln(-x+)cos(-2x)=lncos2x=-ln(x+)cos2x=-f(x),则函数 f(x)是奇函数,图象关于原点对称,排除 A,B,第 8 页,共 14 页令得或,所以 x 轴右侧的零点为,在上取,则,排除 C,故选:D.6. 解:由表格可知:x+0.1+0.3+y=1,7x+8×0.1+9×0.3+10×y=8.9 解得 y=0.4. 故选:C. 根据分布列的概率之和是 1,得到关于 x 和 y 之间的一个关系式,由变量的期望值,得 到另一个关于 x 和 y 的关系式,联立方程,解出要求的 y 的值. 本题是期望和分布列的简单应用,通过创设情境激发学生学习数学的情感,培养其严谨 治学的态度.在学生分析问题、解决问题的过程中培养其积极探索的精神,属于基础题.7. 解:如图:连接 AC,交 BD 于 O,在三角形 CC1A 中,易证 OE∥C1A,从而 C1A∥平面 BDE, ∴直线 AC1 与平面 BED 的距离即为点 A 到平面 BED 的距 离,设为 h,在三棱锥 E-ABD 中,VE-ABD= S△ABD×EC= × ×2×2× =在三棱锥 A-BDE 中,BD=2 ,BE= ,DE= ,∴S△EBD= ×2 ×=2∴VA-BDE= ×S△EBD×h= ×2 ×h=∴h=1 故选:D. 先利用线面平行的判定定理证明直线 C1A∥平面 BDE,再将线面距离转化为点面距离, 最后利用等体积法求点面距离即可 本题主要考查了线面平行的判定,线面距离与点面距离的转化,三棱锥的体积计算方法, 等体积法求点面距离的技巧,属基础题8. 解:满足“界点”的函数必须满足至少含有 2 个零点即可.A.f(x)=2x-x2 的两个零点为 2,4,当 x0=3,在(-∞,3)和(3,+∞)上与 x 轴都有 交点,满足条件. B.判别式△=b2+8>0 恒成立,即抛物线与 x 轴恒有两个交点,在两个零点之间的任何 一个 x0 都是“界点”. C.由 1-|x-2|=0 得|x-2|=1,得 x-2=1 或 x-2=-1,即 x=3 或 x=1,函数有两个零点 1,3,存 在“界点”. D.函数 f(x)的导数 f′(x)=1-cosx≥0,即函数 f′(x)在 R 上是增函数,不可能存 在两个零点,不存在“界点”. 故选:D. 满足“界点”的函数必须满足至少含有 2 个零点即可.结合条件判断函数的零点个数即 可. 本题主要考查函数零点个数的判断,结合满足“界点”的函数必须满足至少含有 2 个零 点是解决本题的关键.第 9 页,共 14 页9. 解:根据题意设 =(1,0), =(0,1), 对应的点 C 在单位圆上,( +2 )2-(2 + )2=3 2-3 2=0,所以| +2 |=|2 + |,|2 + |+|3 +2 - |表示 C 点到点(-2,0)和(3,2)的距离之和,过点(-2,0)和(3,2)的直线为 2x-5y+4=0,原点到直线 2x-5y+4=0 的距离为= <1,所以与单位圆相交,所以|2 + |+|3 +2 - |的最小值为点(-2,0)和(3,2)之间的距离,即 . 故选:A. ,所以可以把他们当成平面直角坐标系的基向量.| +2 |=2| + |,由阿波罗尼斯圆的性质,可以转化为| +2 |=|2 + |.本题考察平面向量的坐标运算,用到了平面几何中的阿波罗尼斯圆的结论、解析几何中 直线与圆的位置关系,综合性很强,属于中档题.10. 解:∵2an≤an-1+an+1(n∈N*,n≥2),∴an-an-1≤an+1-an, ∴a4-a3≤a5-a4≤a6-a5≤a7-a6, ∴a6-a3=a6-a5+a5-a4+a4-a3≤3(a7-a6), 即 3(a7-a6)≥a6-a3, 故选:C. 由已知可得 a4-a3≤a5-a4≤a6-a5≤a7-a6,则 a6-a3=a6-a5+a5-a4+a4-a3≤3(a7-a6),答案可求. 本题考查数列递推式,考查不等式的性质,是中档题.11. 解:=(1+i)(2 1+i)=2(i 1+i)=-2+2i,则 z 的虚部为 2,|z|==2 .故答案为:2,2 . 利用复数的运算法则、虚部的定义即可得出. 本题考查了复数的运算法则、虚部的定义,考查了推理能力与计算能力,属于基础题.12. 解:由已知中的三视图可得该几何体是一个正方体挖去一个圆锥所得的组合体,其表面积 S=2×6×6+4×4×4-9π+ ×6π×5=168+6π,几何体的体积为:=144-12π.故答案为:144-12π;168+6π. 由已知中的三视图可得该几何体是一个正方体挖去一个圆锥所得的组合体,利用公式求 解即可. 本题考查的知识点是由三视图求表面积,根据已 知中的三视图分析出几何体的形状,是解答的关 键.13. 解:作出 x,y 满足条件的可行域如图:第 10 页,共 14 页目标函数 z=2x+y 在的交点 A(2,2)处取最大值为 z=2×2+1×2=6.原点到点 P(x,y)的距离的最小值是:|OB|= . 故答案为:6; ; 画出约束条件表示的可行域,判断目标函数 z=2x+y 的位置,求出最大值.利用可行域 转化求解距离即可. 本题考查简单的线性规划的应用,正确画出可行域,判断目标函数经过的位置是解题的 关键.14. 【分析】本题考查概率的求法,考查分类讨论思想等基础知识,考查运算求解能力,是基础题. 现从中掏出纸币超过 45 元的方法有 8 种情况:①6 张全取;②1 张 10 元 3 张 20 元;③2 张 10 元 2 张 20 元;④3 张 10 元 1 张 20 元;⑤2 张 20 元 1 张 10 元;⑥3 张 20 元;⑦3 张 10 元 2 张 20 元;⑧2 张 10 元,3 张 20 元.由此能求出现从中掏出纸币超过 45 元的 方法总数;小明每次掏出纸币的概率是等可能的,不放回地掏出 4 张,基本事件总数N= =15,刚好是 50 元包含的基本事件个数 M= =3,由此能求出刚好是 50 元的概率. 【解答】 解:小明口袋中有 3 张 10 元,3 张 20 元(因纸币有编号认定每张纸币不同), 现从中掏出纸币超过 45 元的方法有 8 种情况: ①6 张全取;②1 张 10 元 3 张 20 元;③2 张 10 元 2 张 20 元; ④3 张 10 元 1 张 20 元;⑤2 张 20 元 1 张 10 元;⑥3 张 20 元;⑦3 张 10 元 2 张 20 元; ⑧2 张 10 元,3 张 20 元. ∴现从中掏出纸币超过 45 元的方法有n= + + + + + +=32.小明每次掏出纸币的概率是等可能的,不放回地掏出 4 张,基本事件总数 N= =15,刚好是 50 元包含的基本事件个数 M= =3,∴刚好是 50 元的概率 P= = = .故答案为:32; .15. 解:设 AC=x,则 AB=2x,在三角形 ABC 中由余弦定理得 BC2=x2+(2x)2-2•x•2x•cos120°=7x2,∴cosC== ,∴sinC= ,∴sin∠ADC=sin(60°+C)=sin60°cosC+cos60°sinC=.在△ADC 中由正弦定理得,∴,∴AD= x= × = ,故答案为:3. 设 AC=x 后,用余弦定理求出 BC,再求出 cosC,sinC,sin∠ADC,接着在△ADC 中用正第 11 页,共 14 页弦定理得 AD= AB,则 AB=3AD.本题考查了正弦定理、余弦定理,两角和的正弦,属中档题.16. 【分析】本题考查二次函数的零点问题解法,注意运用判别式大于等于 0,端点处的函数值的符 号,结合配方法,考查运算能力,属于中档题. 由题意可得△≥0,f(-1)≤0 或 f(1)≤0,化 a2-3b 为 a 的式子,由二次函数的最值求法, 可得最小值. 【解答】解:函数在[-1,1]上有零点,可得△≥0,即(a+ )2≥4b,且 f(-1)f(1)≤0,即( -a+b)( +a+b)≤0;或 f(-1)≥0,f(1)≥0,-1<- <1,即 a-b≤ ,a+b≥- ,-7<a<5.即有 a2-3b≥a2-= [(a-1)2- ]≥ ×(- )=- ,当且仅当 a=1 时,取得最小值- ,故答案为:- .17. 解:设直线 OP 的方程为 y=kx(k>0),代入椭圆方程可得 P(,),,可得 Q(,),由,可得 kFQ=- ,即为=- ,化为 λ=<e= ,可得 <2+k2,对 k>0 恒成立,由 2+k2>2,可得 a2≤2b2, 即为 a2≤2(a2-c2),可得 c≤ a,即 0<e≤ ,故答案为:(0, ].设直线 OP 的方程为 y=kx(k>0),代入椭圆方程求得 P,Q 的坐标,由向量数量积为 0 的等价条件可得 OP,FQ 的斜率之积为-1,整理,结合恒成立解法可得 a,b 的关系, 可得所求离心率的范围.第 12 页,共 14 页本题考查椭圆的方程和性质,主要是离心率的范围,考查直线方程和椭圆方程联立,化 简整理的运算能力,属于中档题.18. (Ⅰ)利用三角函数恒等变换的应用可求 f(x)=,利用正弦函数的单调性即可求解.(Ⅱ)由已知可求,求得 ,利用余弦定理,基本不等式可求 ac≤3,可得 a2+c2≤6,根据 a2+c2=3+ac>3,即可得解其取值范围. 本题主要考查了三角函数恒等变换的应用,正弦函数的单调性,余弦定理,基本不等式 等知识在解三角形中的应用,考查了数形结合思想,属于中档题.19. (Ⅰ)证明 PC⊥AC,AC⊥BC.推出 AC⊥平面 PBC,即可证明平面 ACE⊥平面 PBC.(Ⅱ)过点 P 作 PF 垂直 CE,垂足为 F.说明 PF 垂直平面 ACE.通过点 E 为 AB 的中 点,所以点 P 到平面 ACE 的距离与点 B 到平面 ACE 的距离相等.连结 BD 交 AC 于点 G, 则 GB=2DG.转化求解即可. 另解:建立坐标系.求出平面 ACE 的一个法量,利用空间向量的数量积求解直线 PD 与 平面 AEC 所成角即可. 本题考查直线与平面垂直的判断定理的应用,直线与平面所成角的求法,考查空间想象 能力以及计算能力.20. (Ⅰ)通过 an+2=3an+1-2an 可得 an+2-an+1=2(an+1-an).推出{an+1-an}是首项为 2,公比为 2 的等比数列然后求解通项公式.(Ⅱ)因为=,利用裂项消项法,求解数列的和,然后求解m 的范围. 本题考查数列的递推关系式的应用,数列求和,考查转化思想以及计算能力.21. 本题考查椭圆的标准方程,考查直线与椭圆的位置关系,考查三角形面积的计算,考查学生分析解决问题的能力,分析得出最大,R 就最大是关键.(1)设椭圆方程,由焦点坐标可得 c=1,由|PQ|=3,可得 =3,又 a2-b2=1,由此可求椭圆方程; (2)设 M(x1,y1),N(x2,y2),不妨 y1>0,y2<0,设△F1MN 的内切圆的径 R,则△F1MN 的周长=4a=8,(|MN|+|F1M|+|F1N|)R=4R,因此最大,R就最大.设直线 l 的方程为 x=my+1,与椭圆方程联立,从而可表示△F1MN 的面积,利 用换元法,借助于导数,即可求得结论.22. (Ⅰ)求出 f′(x)=2xe3x+3x2e3x=x(3x+2)e3x.从而 f(x)在(-∞,- )内单调递增,在(- ,0)内单调递减,在(0,+∞)内单调递增,进而 f(x)的极大值为 f(- )= ,由此能证明当 x<0 时,f(x)< .(Ⅱ)k≤,x>0,令 g(x)=,x>0,则 g′(x)=,令 h(x)=x2(1+3x)e3x+2lnx-1,则 h(x)在(0,+∞)上单调递增,推导出存在 x0∈(0,1),使得 h(x0)=0,g(x)在(0,+∞)上的最小值是 g(x0)=,由此能求出实数 k 的取值范围.第 13 页,共 14 页本题考查不等式的证明,考查实数的取值范围的求不地,考查导数性质、函数的单调性、 最值等基础知识,考运算求解能力,是中档题.第 14 页,共 14 页。

2020年浙江省宁波市镇海中学高考数学模拟试卷(3月份)(带答案)

2020年浙江省宁波市镇海中学高考数学模拟试卷(3月份)(带答案)

2020 年浙江省宁波市镇海中学高考数学模拟试卷(3 月 份)题号 得分一二三总分一、选择题(本大题共 10 小题,共 40.0 分)1. 设集合 A={1,2,3,4},B={x∈N|-3≤x≤3},则 A∩B=( )A. {1,2,3,4}B. {-3,-2,-1,0,1,2,3,4}C. {1,2,3}D. {1,2}2. 双曲线的渐近线方程是( )A. 2x±y=0B. x±2y=03. 已知公差不为零的等差数列{an}满足值为( )C. 4x±y=0D. x±4y=0,Sn 为数列{an}的前 n 项和,则 的A.B.C.D.4. “a>0”是“”的( )A. 充分而不必要条件 C. 充要条件B. 必要而不充分条件 D. 既不充分也不必要条件5. 函数的图象可能是( )A.B.C.D.6. 某射手射击所得环数 ξ 的分布列如下:ξ 78910P x 0.1 0.3 y已知 ξ 的数学期望 E(ξ)=8.9,则 y 的值为( )A. 0.8B. 0.6C. 0.4D. 0.27. 已知正四棱柱 ABCD-A1B1C1D1 中,AB=2,CC1=2 ,E 为 CC1 的中点,则直线 AC1与平面 BED 的距离为( )A. 2B.C.D. 1第 1 页,共 14 页8. 对于定义域为 R 的函数 f(x),若存在非零实数 x0,使函数 f(x)在(-∞,x0)和(x0,+∞)上与 x 轴都有交点,则称 x0 为函数 f(x)的一个“界点”.则下列四个函数中,不存在“界点”的是( )A. f(x)=2x-x2B. f(x)=x2+bx-2(b∈R)C. f(x)=1-|x-2|D. f(x)=x-sinx9. 已知是平面内三个单位向量,若 ,则的最小值()A.B.C.D. 510. 已知数列{an}满足 2an≤an-1+an+1(n∈N*,n≥2),则( )A. a5≤4a2-3a1B. a2+a7≤a3+a6C. 3(a7-a6)≥a6-a3D. a2+a3≥a6+a7二、填空题(本大题共 7 小题,共 36.0 分)11. 设 i 为虚数单位,给定复数,则 z 的虚部为______,|z|=______.12. 某几何体的三视图如图所示,则该几何体的体积是______,表面积是______.13. 已知 x,y 满足条件则 2x+y 的最大值是______,原点到点 P(x,y)的距离的最小值是______ 14. 小明口袋中有 3 张 10 元,3 张 20 元(因纸币有编号认定每张纸币不同),现从中掏出纸币超过 45 元的方法有 种;若小明每次掏出纸币的概率是等可能的,不 放回地掏出 4 张,刚好是 50 元的概率为15. 在△ABC 中,∠BAC=120°,AD 为∠BAC 的平分线,AB=2AC,则 =______.16. 若函数在上有零点,则的最小值为17. 如图,椭圆的离心率为 e,F 是 Γ 的右焦点,点 P 是 Γ 上第一象限内任意一点,,,若 λ<e,则 e 的取值范围是______.第 2 页,共 14 页三、解答题(本大题共 5 小题,共 74.0 分)18. 已知函数.(Ⅰ)求函数 f(x)的单调递增区间;(Ⅱ)设△ABC 中的内角 A,B,C 所对的边分别为 a,b,c,若求 a2+c2 的取值范围.,且,19. 如图,四棱锥 P-ABCD 中,PC 垂直平面 ABCD,AB⊥AD,AB∥CD, PD=AB=2AD=2CD=2,E 为 PB 的中点. (Ⅰ)证明:平面 EAC⊥平面 PBC; (Ⅱ)求直线 PD 与平面 AEC 所成角的正弦值.20. 在数列{an}中,a1=1,a2=3,且对任意的 n∈N*,都有 an+2=3an+1-2an. (Ⅰ)证明数列{an+1-an}是等比数列,并求数列{an}的通项公式;(Ⅱ)设,记数列{bn}的前 n 项和为 Sn,若对任意的 n∈N*都有,求实数 m 的取值范围.第 3 页,共 14 页21. 已知椭圆的焦点坐标为 F1(-1,0),F2(1,0),过 F2 垂直于长轴的直线交椭圆 于 P、Q 两点,且|PQ|=3. (1)求椭圆的方程; (2)过 F2 的直线 l 与椭圆交于不同的两点 M、N,则△F1MN 的内切圆的面积是否 存在最大值?若存在,求出这个最大值及此时的直线方程;若不存在,请说明理由.22. 已知函数 f(x)=x2e3x (Ⅰ)若 x<0,求证:f(x)< (Ⅱ)若 x>0,恒有 f(x)≥(k+3)x+2lnx+1,求实数 k 的取值范围2020 年浙江省宁波市镇海中学高考数学模拟试卷(3 月 份)答案和解析【答案】1. C2. B3. A4. C5. D6. C7. D8. D9. A10. C11. 212. 144-12π 168+6π13. 614. 3215. 316. -17. (0, ]第 4 页,共 14 页18. 解:(Ⅰ)==所以,解得,k∈Z.所以函数 f(x)的单调递增区间为,k∈Z(Ⅱ)因为,所以.所以 .又因为,所以 3=a2+c2-ac,即 a2+c2=3+ac.而 a2+c2≥2ac,所以 ac≤3,即 a2+c2≤6.又因为 a2+c2=3+ac>3,所以 3<a2+c2≤6.19. (Ⅰ)证明:PC⊥平面 ABCD,故PC⊥AC.………………(2 分)又 AB=2,CD=1,AD⊥AB,所以 AC=BC= .故 AC2+BC2=AB2,即AC⊥BC.………………(4 分)所以 AC⊥平面 PBC,所以平面 ACE⊥平面PBC. …………………………(6 分)(Ⅱ)解:PC⊥平面 ABCD,故 PC⊥CD.又 PD=2,所以 PC= . …………(8 分)在平面 ACE 内,过点 P 作 PF 垂直 CE,垂足为 F.由(Ⅰ)知平面 ACE⊥平面 PBC,所以 PF 垂直平面 ACE.由面积法得:即.…………(10 分)又点 E 为 AB 的中点,.所以.……………………………………(12 分)又点 E 为 AB 的中点,所以点 P 到平面 ACE 的距离与点 B 到平面 ACE 的距离相等. 连结 BD 交 AC 于点 G,则 GB=2DG.所以点 D 到平面 ACE 的距离是点 B 到平面 ACE 的距离的一半,即 .所以直线 PD 与平面 AEC 所成角的正弦值为.……………………(15 分)另解:如图,取 AB 的中点 F,如图建立坐标系.因为 PD=2,所以.所以有:C(0,0,0),D(0,1,0),,A(1,1,0),B(1,-1,0),. …………(9 分).,.第 5 页,共 14 页设平面 ACE 的一个法量为 =(x,y,z),则取 x=1,得 y=-1,.即=.…………(13 分)设直线 PD 与平面 AEC 所成角为 θ,则 sinθ=|cos< , =. …………(15分)20. 解:(Ⅰ)由 an+2=3an+1-2an 可得 an+2-an+1=2(an+1-an). ………………(2 分)又 a1=1,a2=3,所以 a2-a1=2. 所以{an+1-an}是首项为 2,公比为 2 的等比数列.…………………(3 分)所以.…………………(4 分)所以 an=a1+(a2-a1)+…+(an-an-1)=1+2+22+…+2n=2n-1.…………(7 分)(Ⅱ)因为==.………(9 分)所以 Sn=b1+b2+…+bn= (12 分) 又因为对任意的 n∈N*都有,所以=.………恒成立,即,即当 n=1 时,.………(15 分)21. 解:(1)设椭圆方程为=1(a>b>0),由焦点坐标可得 c=1,由|PQ|=3,可得 =3,又 a2-b2=1,解得 a=2,b= ,故椭圆方程为=1,(2)设 M(x1,y1),N(x2,y2),不妨令 y1>0,y2<0,设△F1MN 的内切圆的半径 为 R,则△F1MN 的周长=4a=8,(|MN|+|F1M|+|F1N|)R=4R因此内切圆面积最大,R 就最大,此时也最大,由题知,直线 l 的斜率不为零,可设直线 l 的方程为 x=my+1,由得(3m2+4)y2+6my-9=0,得 则 令 t=, ,则 t≥1,,=,第 6 页,共 14 页则,令 f(t)=3t+ ,则 f′(t)=3- ,当 t≥1 时,f′(t) 0,f(t)在[1,+∞)上单调递增,有 f(t)≥f(1)=4,≤3,即当 t=1,m=0 时,≤3,=4R,∴Rmax= ,这时所求内切圆面积的最大值为 π.故直线 l 方程为:x=1,△F1MN 内切圆面积的最大值为 π.22. 证明:(Ⅰ)∵函数 f(x)=x2e3x,∴f′(x)=2xe3x+3x2e3x=x(3x+2)e3x.由 f′(x)>0,得 x<- 或 x>0;由 f′(x)<0,得-,∴f(x)在(-∞,- )内单调递增,在(- ,0)内单调递减,在(0,+∞)内单调递增,∴f(x)的极大值为 f(- )= ,∴当 x<0 时,f(x)≤f(- )= < = .解:(Ⅱ)∵x2e3x≥(k+3)x+2lnx+1,∴k≤,x>0,令 g(x)=,x>0,则 g′(x)=,令 h(x)=x2(1+3x)e3x+2lnx-1,则 h(x)在(0,+∞)上单调递增, 且 x→0+时,h(x)→-∞,h(1)=4e3+2lnx-1, ∴存在 x0∈(0,1),使得 h(x0)=0, ∴当 x∈(0,x0)时,g′(x)<0,g(x)单调递减, 当 x∈(x0,+∞)时,g′(x)>0,g(x)单调递增,∴g(x)在(0,+∞)上的最小值是 g(x0)=,∵h(x0)=+2lnx0-1,得=,令=t0,则 2lnx0+3x0=lnx0,且 φ(1)=0,∴t=1,∴g(x0)==,∴实数 k 的取值范围是(-∞,0]. 【解析】1. 【分析】可解出集合 B,然后进行交集的运算即可. 考查描述法、列举法的定义,以及交集的运算. 【解答】 解:B={0,1,2,3};第 7 页,共 14 页∴A∩B={1,2,3}. 故选:C.2. 【分析】本题考查了双曲线的渐进方程,把双曲线的标准方程中的“1”转化成“0”即可求出渐 进方程.属于基础题.渐近线方程是 -y2=0,整理后就得到双曲线的渐近线.【解答】解:双曲线其渐近线方程是 -y2=0整理得 x ±2y=0. 故选:B.3. 解:公差不为零的等差数列{an}满足,∴=a1(a1+3d),解得 a1=-4d, ∵Sn 为数列{an}的前 n 项和,∴==.故选:A.由公差不为零的等差数列{an}满足,利用等差数列的通项公式列方程求出a1=-4d,由此能求出 的值. 本题考查等差数列的前 3 项和公式和前 1 项和的比值的求法,考查等差数列的性质等基 础知识,考查运算求解能力,是基础题.4. 解:由 a>0,得 a+ ≥2 =2 ,是充分条件,由 a+ ≥2 ,得:a>0,故 a>0”是“”的充要条件,故选:C. 根据充分必要条件的定义结合不等式的性质判断即可. 本题考查了充分必要条件,考查不等式的性质,是一道基础题.5. 【分析】本题主要考查函数图象的识别和判断,利用函数的奇偶性和对称性的关系,根据函数零 点判断函数的正负是解决本题的关键. 判断函数 f(x)的奇偶性,结合图象的对称性以及函数在 x 轴右侧的函数零点判断函数 的正负进行判断即可 【解答】解:f(-x)=ln(-x+)cos(-2x)=lncos2x=-ln(x+)cos2x=-f(x),则函数 f(x)是奇函数,图象关于原点对称,排除 A,B,第 8 页,共 14 页令得或,所以 x 轴右侧的零点为,在上取,则,排除 C,故选:D.6. 解:由表格可知:x+0.1+0.3+y=1,7x+8×0.1+9×0.3+10×y=8.9 解得 y=0.4. 故选:C. 根据分布列的概率之和是 1,得到关于 x 和 y 之间的一个关系式,由变量的期望值,得 到另一个关于 x 和 y 的关系式,联立方程,解出要求的 y 的值. 本题是期望和分布列的简单应用,通过创设情境激发学生学习数学的情感,培养其严谨 治学的态度.在学生分析问题、解决问题的过程中培养其积极探索的精神,属于基础题.7. 解:如图:连接 AC,交 BD 于 O,在三角形 CC1A 中,易证 OE∥C1A,从而 C1A∥平面 BDE, ∴直线 AC1 与平面 BED 的距离即为点 A 到平面 BED 的距 离,设为 h,在三棱锥 E-ABD 中,VE-ABD= S△ABD×EC= × ×2×2× =在三棱锥 A-BDE 中,BD=2 ,BE= ,DE= ,∴S△EBD= ×2 ×=2∴VA-BDE= ×S△EBD×h= ×2 ×h=∴h=1 故选:D. 先利用线面平行的判定定理证明直线 C1A∥平面 BDE,再将线面距离转化为点面距离, 最后利用等体积法求点面距离即可 本题主要考查了线面平行的判定,线面距离与点面距离的转化,三棱锥的体积计算方法, 等体积法求点面距离的技巧,属基础题8. 解:满足“界点”的函数必须满足至少含有 2 个零点即可.A.f(x)=2x-x2 的两个零点为 2,4,当 x0=3,在(-∞,3)和(3,+∞)上与 x 轴都有 交点,满足条件. B.判别式△=b2+8>0 恒成立,即抛物线与 x 轴恒有两个交点,在两个零点之间的任何 一个 x0 都是“界点”. C.由 1-|x-2|=0 得|x-2|=1,得 x-2=1 或 x-2=-1,即 x=3 或 x=1,函数有两个零点 1,3,存 在“界点”. D.函数 f(x)的导数 f′(x)=1-cosx≥0,即函数 f′(x)在 R 上是增函数,不可能存 在两个零点,不存在“界点”. 故选:D. 满足“界点”的函数必须满足至少含有 2 个零点即可.结合条件判断函数的零点个数即 可. 本题主要考查函数零点个数的判断,结合满足“界点”的函数必须满足至少含有 2 个零 点是解决本题的关键.第 9 页,共 14 页9. 解:根据题意设 =(1,0), =(0,1), 对应的点 C 在单位圆上,( +2 )2-(2 + )2=3 2-3 2=0,所以| +2 |=|2 + |,|2 + |+|3 +2 - |表示 C 点到点(-2,0)和(3,2)的距离之和,过点(-2,0)和(3,2)的直线为 2x-5y+4=0,原点到直线 2x-5y+4=0 的距离为= <1,所以与单位圆相交,所以|2 + |+|3 +2 - |的最小值为点(-2,0)和(3,2)之间的距离,即 . 故选:A. ,所以可以把他们当成平面直角坐标系的基向量.| +2 |=2| + |,由阿波罗尼斯圆的性质,可以转化为| +2 |=|2 + |.本题考察平面向量的坐标运算,用到了平面几何中的阿波罗尼斯圆的结论、解析几何中 直线与圆的位置关系,综合性很强,属于中档题.10. 解:∵2an≤an-1+an+1(n∈N*,n≥2),∴an-an-1≤an+1-an, ∴a4-a3≤a5-a4≤a6-a5≤a7-a6, ∴a6-a3=a6-a5+a5-a4+a4-a3≤3(a7-a6), 即 3(a7-a6)≥a6-a3, 故选:C. 由已知可得 a4-a3≤a5-a4≤a6-a5≤a7-a6,则 a6-a3=a6-a5+a5-a4+a4-a3≤3(a7-a6),答案可求. 本题考查数列递推式,考查不等式的性质,是中档题.11. 解:=(1+i)(2 1+i)=2(i 1+i)=-2+2i,则 z 的虚部为 2,|z|==2 .故答案为:2,2 . 利用复数的运算法则、虚部的定义即可得出. 本题考查了复数的运算法则、虚部的定义,考查了推理能力与计算能力,属于基础题.12. 解:由已知中的三视图可得该几何体是一个正方体挖去一个圆锥所得的组合体,其表面积 S=2×6×6+4×4×4-9π+ ×6π×5=168+6π,几何体的体积为:=144-12π.故答案为:144-12π;168+6π. 由已知中的三视图可得该几何体是一个正方体挖去一个圆锥所得的组合体,利用公式求 解即可. 本题考查的知识点是由三视图求表面积,根据已 知中的三视图分析出几何体的形状,是解答的关 键.13. 解:作出 x,y 满足条件的可行域如图:第 10 页,共 14 页目标函数 z=2x+y 在的交点 A(2,2)处取最大值为 z=2×2+1×2=6.原点到点 P(x,y)的距离的最小值是:|OB|= . 故答案为:6; ; 画出约束条件表示的可行域,判断目标函数 z=2x+y 的位置,求出最大值.利用可行域 转化求解距离即可. 本题考查简单的线性规划的应用,正确画出可行域,判断目标函数经过的位置是解题的 关键.14. 【分析】本题考查概率的求法,考查分类讨论思想等基础知识,考查运算求解能力,是基础题. 现从中掏出纸币超过 45 元的方法有 8 种情况:①6 张全取;②1 张 10 元 3 张 20 元;③2 张 10 元 2 张 20 元;④3 张 10 元 1 张 20 元;⑤2 张 20 元 1 张 10 元;⑥3 张 20 元;⑦3 张 10 元 2 张 20 元;⑧2 张 10 元,3 张 20 元.由此能求出现从中掏出纸币超过 45 元的 方法总数;小明每次掏出纸币的概率是等可能的,不放回地掏出 4 张,基本事件总数N= =15,刚好是 50 元包含的基本事件个数 M= =3,由此能求出刚好是 50 元的概率. 【解答】 解:小明口袋中有 3 张 10 元,3 张 20 元(因纸币有编号认定每张纸币不同), 现从中掏出纸币超过 45 元的方法有 8 种情况: ①6 张全取;②1 张 10 元 3 张 20 元;③2 张 10 元 2 张 20 元; ④3 张 10 元 1 张 20 元;⑤2 张 20 元 1 张 10 元;⑥3 张 20 元;⑦3 张 10 元 2 张 20 元; ⑧2 张 10 元,3 张 20 元. ∴现从中掏出纸币超过 45 元的方法有n= + + + + + +=32.小明每次掏出纸币的概率是等可能的,不放回地掏出 4 张,基本事件总数 N= =15,刚好是 50 元包含的基本事件个数 M= =3,∴刚好是 50 元的概率 P= = = .故答案为:32; .15. 解:设 AC=x,则 AB=2x,在三角形 ABC 中由余弦定理得 BC2=x2+(2x)2-2•x•2x•cos120°=7x2,∴cosC== ,∴sinC= ,∴sin∠ADC=sin(60°+C)=sin60°cosC+cos60°sinC=.在△ADC 中由正弦定理得,∴,∴AD= x= × = ,故答案为:3. 设 AC=x 后,用余弦定理求出 BC,再求出 cosC,sinC,sin∠ADC,接着在△ADC 中用正第 11 页,共 14 页弦定理得 AD= AB,则 AB=3AD.本题考查了正弦定理、余弦定理,两角和的正弦,属中档题.16. 【分析】本题考查二次函数的零点问题解法,注意运用判别式大于等于 0,端点处的函数值的符 号,结合配方法,考查运算能力,属于中档题. 由题意可得△≥0,f(-1)≤0 或 f(1)≤0,化 a2-3b 为 a 的式子,由二次函数的最值求法, 可得最小值. 【解答】解:函数在[-1,1]上有零点,可得△≥0,即(a+ )2≥4b,且 f(-1)f(1)≤0,即( -a+b)( +a+b)≤0;或 f(-1)≥0,f(1)≥0,-1<- <1,即 a-b≤ ,a+b≥- ,-7<a<5.即有 a2-3b≥a2-= [(a-1)2- ]≥ ×(- )=- ,当且仅当 a=1 时,取得最小值- ,故答案为:- .17. 解:设直线 OP 的方程为 y=kx(k>0),代入椭圆方程可得 P(,),,可得 Q(,),由,可得 kFQ=- ,即为=- ,化为 λ=<e= ,可得 <2+k2,对 k>0 恒成立,由 2+k2>2,可得 a2≤2b2, 即为 a2≤2(a2-c2),可得 c≤ a,即 0<e≤ ,故答案为:(0, ].设直线 OP 的方程为 y=kx(k>0),代入椭圆方程求得 P,Q 的坐标,由向量数量积为 0 的等价条件可得 OP,FQ 的斜率之积为-1,整理,结合恒成立解法可得 a,b 的关系, 可得所求离心率的范围.第 12 页,共 14 页本题考查椭圆的方程和性质,主要是离心率的范围,考查直线方程和椭圆方程联立,化 简整理的运算能力,属于中档题.18. (Ⅰ)利用三角函数恒等变换的应用可求 f(x)=,利用正弦函数的单调性即可求解.(Ⅱ)由已知可求,求得 ,利用余弦定理,基本不等式可求 ac≤3,可得 a2+c2≤6,根据 a2+c2=3+ac>3,即可得解其取值范围. 本题主要考查了三角函数恒等变换的应用,正弦函数的单调性,余弦定理,基本不等式 等知识在解三角形中的应用,考查了数形结合思想,属于中档题.19. (Ⅰ)证明 PC⊥AC,AC⊥BC.推出 AC⊥平面 PBC,即可证明平面 ACE⊥平面 PBC.(Ⅱ)过点 P 作 PF 垂直 CE,垂足为 F.说明 PF 垂直平面 ACE.通过点 E 为 AB 的中 点,所以点 P 到平面 ACE 的距离与点 B 到平面 ACE 的距离相等.连结 BD 交 AC 于点 G, 则 GB=2DG.转化求解即可. 另解:建立坐标系.求出平面 ACE 的一个法量,利用空间向量的数量积求解直线 PD 与 平面 AEC 所成角即可. 本题考查直线与平面垂直的判断定理的应用,直线与平面所成角的求法,考查空间想象 能力以及计算能力.20. (Ⅰ)通过 an+2=3an+1-2an 可得 an+2-an+1=2(an+1-an).推出{an+1-an}是首项为 2,公比为 2 的等比数列然后求解通项公式.(Ⅱ)因为=,利用裂项消项法,求解数列的和,然后求解m 的范围. 本题考查数列的递推关系式的应用,数列求和,考查转化思想以及计算能力.21. 本题考查椭圆的标准方程,考查直线与椭圆的位置关系,考查三角形面积的计算,考查学生分析解决问题的能力,分析得出最大,R 就最大是关键.(1)设椭圆方程,由焦点坐标可得 c=1,由|PQ|=3,可得 =3,又 a2-b2=1,由此可求椭圆方程; (2)设 M(x1,y1),N(x2,y2),不妨 y1>0,y2<0,设△F1MN 的内切圆的径 R,则△F1MN 的周长=4a=8,(|MN|+|F1M|+|F1N|)R=4R,因此最大,R就最大.设直线 l 的方程为 x=my+1,与椭圆方程联立,从而可表示△F1MN 的面积,利 用换元法,借助于导数,即可求得结论.22. (Ⅰ)求出 f′(x)=2xe3x+3x2e3x=x(3x+2)e3x.从而 f(x)在(-∞,- )内单调递增,在(- ,0)内单调递减,在(0,+∞)内单调递增,进而 f(x)的极大值为 f(- )= ,由此能证明当 x<0 时,f(x)< .(Ⅱ)k≤,x>0,令 g(x)=,x>0,则 g′(x)=,令 h(x)=x2(1+3x)e3x+2lnx-1,则 h(x)在(0,+∞)上单调递增,推导出存在 x0∈(0,1),使得 h(x0)=0,g(x)在(0,+∞)上的最小值是 g(x0)=,由此能求出实数 k 的取值范围.第 13 页,共 14 页本题考查不等式的证明,考查实数的取值范围的求不地,考查导数性质、函数的单调性、 最值等基础知识,考运算求解能力,是中档题.第 14 页,共 14 页。

2020年镇海中学浙江省高考语文模拟卷及答案

2020年镇海中学浙江省高考语文模拟卷及答案

2020年镇海中学浙江省高考语文模拟卷语文(含答案)考生须知:1.本试题卷共8页,四部分,24题。

满分150分,考试时间150分钟。

2.考生答题前,务必将自己的姓名、准考证号用黑色字迹的签字笔或钢笔填写在答题卷上。

3.选择题的答案须用2B铅笔将答题卷上对应题目的答案标号涂黑,如要改动,须将原填涂处用橡皮擦净。

4.非选择题的答案须用黑色字迹的签宇笔或钢笔写在答题卷的相应区域内,写在试卷上无效。

一、语言文字运用(共20分)1.下列各句中,没有错别字且加点字的注音全都正确的一项是(3分)A.细盐似的雪粒,窸窸窣. (sū) 窣地打在身上,在乱石中跳跃;紧接着,雪珠变成了雪霰.(xiàn) :雪霰又变成了纷纷洋洋的飞絮,天空转而变得阴暗,沉黑。

B.风火墙是我国传统建筑中的一种墙垣.(yuán) ,它是人字形坡顶房屋两端的山墙,通常要比屋面高出三至六尺,有防止火势蔓.(màn) 延的功能,形式多种多样。

C.秋冬季节的芦苇.(wěi)荡,南来北往的白鹭在此嬉戏,修长而饱满的苇穗,像一支支画笔,饱蘸.(zàn) 天地间的风霜雨雪,在河洲上涂鸦出一幅绚丽的画卷。

D.先生对马有着特珠的感情,观察细微,骨骼.(gé)神态,皆了然于心,加之先生对造型和水墨的独特理解,故展纸运笔,逶迤.(yǐ)顿挫,出神入化,一气呵成。

阅读下面的文字,完成2~3题。

【甲】“水立方”如何变成“冰立方”?这可不是在平地上浇筑冰层那么简单!【乙】中建一局建设发展公司副总经理侯本才介绍,转换过程需经过5道工序。

即把水抽干,搭设钢架和支撑板,铺上保温层和防水层、安装可拆装制冰系统。

输送载冷剂将水面冰化,完成水冰转换。

改造最大的难度在于控制精度。

冰壶比赛对平整度要求非常高,每平方米承受150公斤重量的情况下支撑结构变形不能超过1毫米。

实现复杂的温度和湿度分层控制,也是..“水变冰”的难点。

水上项目要求环境高温高湿,冰上项目要求低温低湿,两者需求天壤之别..,赛场的温度必....。

【20套精选试卷合集】浙江省镇海中学2019-2020学年高考英语模拟试卷含答案

【20套精选试卷合集】浙江省镇海中学2019-2020学年高考英语模拟试卷含答案

高考模拟英语试卷本试卷分为第I卷(选择题)和第II卷(非选择题)两部分。

第I卷(选择题共100分)第一部分听力(共两节,共20小题;每小题1.5分,满分30分)做题时,先将答案标在试卷上。

录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

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, 并标在试卷的相应位置。

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.What do we learn from the conversation?A. The man wants to go to Shanghai.B. The man wants to go to Nanjing.C. There are no trains to Shanghai for the rest of the day.2.Why can’t the woman go to the party?A. Because she has a cooking class on Saturday night.B. Because she doesn’t like football.C. Because she has to work.3.What does the woman mean?A. She doesn’t need the job.B. She hasn’t got a job.C. She has got a good job.4.When did the woman last see the man’s sister?A. Yesterday.B. Three days ago.C. Two days ago.5.What happened to the man?A. He had to work overtime.B. He was stuck in traffic.C. He met a traffic accident.第二节(共15小题;每小题1.5分,满分22.5分)听下面五段对话或独白,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,并标在试卷的相应位置。

2020年浙江省宁波镇海中学高考物理选考模拟试卷(6月份)(含解析)

2020年浙江省宁波镇海中学高考物理选考模拟试卷(6月份)(含解析)

2020年浙江省宁波镇海中学高考物理选考模拟试卷(6月份)一、单选题(本大题共12小题,共36.0分)1.一个物体静止在斜面上时,正确表示斜面对物体作用力F方向的是图中的()A. B.C. D.2.下列有关研究物理问题的思想或方法的描述错误的是()A. 在探究加速度与力、质量的关系时,利用了控制变量的思想方法B. 物理学中引入“质点”的模型时,采用了理想化方法C. 物理学中建立“加速度”的概念时,采用了等效法D. 伽利略研究力与运动的关系时,采用了理想实验法3.关于参考系的选取,下列说法正确的是()A. 参考系必须选取静止不动的物体B. 参考系必须是和地面联系在一起的C. 在空中运动的物体不能作为参考系D. 任何物体都可以作为参考系4.如图所示四个运动图象中,表示物体在做匀速直线运动的是()A. B.C. D.5.如图两根长度不同的细线下面分别悬挂两个小球A和B,细线上端固定在同一点,若两个小球绕竖直轴做匀速圆周运动时恰好在同一高度的水平面内,则下列说法中正确的是()A. 线速度v A=v BB. 角速度ωA>ωBC. 加速度a A=a BD. 周期T A=T B6.我国发射的“神州六号”载人飞船,与“神州五号”飞船相比,它在更高的轨道上绕地球做匀速圆周运动,如图所示,下列说法中正确的是()A. “神州六号”的速度较小B. “神州六号”的速度与“神州五号”的相同C. “神州六号”的周期更短D. “神州六号”的周期与“神州五号”的相同7.如图甲所示,矩形线圈abcd在匀强磁场中逆时针匀速转动时,线圈中产生的交流电如图乙所示,设沿abcda方向为电流正方向,则()A. 乙图中Oa时间段对应甲图中A至B图的过程B. 乙图中c时刻对应甲图中的C图C. 若乙图中d等于0.02s,则1 s内电流的方向改变了50次D. 若乙图中b等于0.02s,则交流电的频率为50Hz8.下列与a粒子相关的说法中正确的是()A. 天然放射性现象中产生的a射线速度与光速相当,贯穿能力很强B. 丹麦物理学家玻尔进行了a粒子散射实验并首先提出了原子的核式结构模型C. 铀238)核放出一个a粒子后就变为钍234)D. 高速a粒子轰击氮核可从氮核中打出中子,核反应方程为 24He+714N→816O+01n9.如图所示,一固定杆与水平方向夹角为θ,将一质量为m1的滑块套在杆上,通过轻绳悬挂一个质量为m2的小球,杆与滑块之间的动摩擦因数为μ,若滑块与小球保持相对静止以相同的加速度a一起运动,此时绳子与竖直方向夹角为β,且θ<β,则滑块的运动情况是()A. 沿着杆加速下滑B. 沿着杆减速上滑C. 沿着杆减速下滑D. 沿着杆加速上滑10.如图所示,水平桌面上的A点处有一个质量为m的物体以初速度v0被抛出,不计空气阻力,当它到达B点时,以地面为零势能面,其机械能的表达式正确的是()A. 12mv02+mgH B. 12mv02+mgℎC. mgH−mgℎD. 12mv02+mg(H−ℎ)11.如图所示,导线框中电流为I,导线框垂直于磁场放置,匀强磁场的磁感应强度为B,AB与CD相距为d,则MN所受安培力大小为()A. F=BIdB. F=BIdsinθC. F=BIdsinθD. F=BIdcosθ12.为监测某化工厂的污水排放量,技术人员在该厂的排污管末端安装了如图所示的长方体流量计。

【附20套高考模拟试题】浙江省学军、镇海等名校协作体2020届高三3月联考英语试题含答案

【附20套高考模拟试题】浙江省学军、镇海等名校协作体2020届高三3月联考英语试题含答案

【附20套高考模拟试题】浙江省学军、镇海等名校协作体2020届高三3月联考英语试题含答案浙江省学军、镇海等名校协作体2020届高三3月联考英语试题第一部分(共20小题每,小题1.5分,满分30分)1.Your donation greatly appreciated and the money will be used to help the students from poor families.A.has been B.isC.was D.had been2.Take the medicine right away! ______ it yesterday, you would be quite all right now.A.Had you taken B.Would you takeC.Should you take D.Were you to take3.We arranged to meet at the cinema at 7:30, but Jack failed to ______.A.break up B.set upC.turn up D.give up4.My teacher often says that success in making money is not always a good ______ of success in life. A.belief B.element C.criterion D.instance5.We’ll go early.,we may not get a seat.A.Otherwise B.Meanwhile C.However D.Besides6.--- What caused the party to be put off? --- ______ the invitations.A.T om delayed sending B.T om’s delaying sendingC.T om delaying to send D.Tom delayed to send7.--- Did you watch the final match of China Open yesterday?---Sure. I it so attentively that I forgot to cook supper.A.watched B.had watchedC.was watching D.was to watch8.Membership of this club is open to those who are its aim.A.in salute to B.in honour ofC.in line with D.in sympathy with9.Jane went to her teacher just now. She ________ about the solution to the problem.A.wondered B.was wondering C.had wondered D.would wonder10.There are various things on sale, so you can choose ______ interests you.A.whoever B.no matter whoC.whatever D.no matter what11.﹣Mom,I'll stay in to accompany my grandpa this evening.﹣________!A.With pleasure B.Never mindC.Suit yourself D.It depends12.Some pre-school children go to a day care center, __________ they learn simple games and songs. A.then B.there C.while D.where13.Our dream is to _______ a World Cup that makes you, your grandchildren and everyone in football really proud.A.stage B.chairC.found D.watch14.Even a small personal computer store vast amounts of information.A.might B.canC.ought to D.has to15.A heavy sandstorm is going to envelop our city. It is unwise to have your car .A.wash B.washedC.washing D.to wash16.Look over there! There is a long, winding path ________ up to the house.A.lead B.leadingC.led D.to lead17.I wish I ______ at my sister’s wedding last Tuesday, but I was on a business trip in New York then. A.will be B.would be C.have been D.had been18.This book is said to be a special one, as it ____ many events not found in other history books. A.writes B.prints C.covers D.reads19.In the "moon garden" onboard the Chang'e 4, the shoots of cotton marked the first live matter ever_____ on the moon.A.having grown B.to be grownC.being grown D.grown20.— BoB.could I use your computer this evening?—Sorry. I a report on it then.A.will be writing B.have been writingC.have written D.will have written第二部分阅读理解(满分40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。

2020年宁波市镇海中学高三语文模拟试题及参考答案

2020年宁波市镇海中学高三语文模拟试题及参考答案

2020年宁波市镇海中学高三语文模拟试题及参考答案一、现代文阅读(36分)(一)现代文阅读I(9分)阅读下面文字,完成下列小题。

回家马兰莲火车像一条长龙,穿行在连绵不断的山脉中,雪花零零散散地从天空飘落下来,像飞,像蝉翼,清澈洁净,晶莹剔透。

肖锦云坐在靠窗的座位,怀里抱着一个大大的包裹。

她表情淡漠,甚至有些木然,她的眼睛里流出一种奇妙的神色,说不上是喜悦还是忧伤。

她一动不动,静静地坐着,像一尊雕塑。

她的对面坐着一对年轻夫妻,好像刚结婚不久,女人依偎在男人的怀里,甜甜地睡着了。

从她嘴角露出的微笑,足以证明,她正做着美梦。

男人一手楼着女人,一手拿着手机,拇指不停地在手机屏幕上滑动着,女人动了一下,盖在身上的衣服掉了下去。

男人放下手机,捡起衣服盖在女人的身上。

肖锦云看着他们,脸上呈现出一种复杂的表情,说不出是羡慕还是嫉妒,男人抬起头,刚好与她的目光相撞,便问道:“大姐,你去哪?”肖锦云显然没想到男人会跟她搭讪,怔了一下,淡淡地说:“回家。

”男人还想说话,肖锦云却把目光移向了窗外。

火车转过一个弯儿,“呜呜”地鸣叫着,钻进一个隧道里,闹哄哄的车厢里安静了许多。

肖锦云的脑子里开始翻江倒海了……六年前一个飘雪的日子,肖锦云接到男朋友打来的电话,筹备了大半年的婚事又要推后。

这已经是第三次了。

肖锦云知道男朋友的工作性质特殊,她决定自己去南方把婚事办了。

下了半个月的雪丝毫没有停下来的意思,肖锦云被两家的父母送上了南下的火车……“呜——”火车发出一声长鸣,咣当咣当地跑出隧道,外面的雪似乎有些大了,山坡上、树枝上落满了白茸茸的雪花。

女人醒了。

她揉了揉眼睛,惊喜地叫道:“老公,雪下得好大,雪花真美。

”男人轻描淡写地说:“这算什么,老家的雪比这漂亮多了。

”女人撒起娇来,搂着男人的脖子说:“老公,我想吃鸭脖。

”“你就是个馋猫。

”男人从背包里拿出一个袋子,又拿出两瓶冰红茶。

女人像饿狼扑食一样自顾自地吃起来。

男人抬头看了看肖锦云,试探着问:“大姐,你喝点儿水吧。

浙江省镇海中学2020届高三校内模拟考试物理试卷含解析【含高考模拟卷15套】

浙江省镇海中学2020届高三校内模拟考试物理试卷含解析【含高考模拟卷15套】

浙江省镇海中学2020届高三校内模拟考试物理试卷一、单项选择题:本题共6小题,每小题4分,共24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、一个质量为m 的小球,以大小为v 0的初速度被竖直向上抛出,从抛出到落地的过程中,重力对小球做功为mv 02。

不计空气阻力,则此过程重力对小球的冲量大小为A .0(21)mv -B .0(21)mv +C .0(31)mv -D .0(31)mv +2、下列说法正确的是( )A .在光电效应中,增加入射光的强度,饱和光电流不变B .β衰变现象说明电子是原子的组成部分C .两个氘核的聚变反应方程式为22311120H H He n +→+D .处于基态的氢原子吸收光子发生跃迁后动能增加3、如图所示,一个质量为=9.1×10-31kg 、电荷量为e=1.6×10-19C 的电子,以4×106m/s 的速度从M 点垂直电场线方向飞入匀强电场,电子只在电场力的作用下运动,在N 点离开电场时,其速度方向与电场线成150°角,则M 与N 两点间的电势差约为( )A .-1.0×102VB .-1.4×102VC .1.8×102VD .2.2×102V4、如图所示,在光滑的水平桌面上有一弹簧振子,弹簧劲度系数为k ,开始时,振子被拉到平衡位置O 的右侧A 处,此时拉力大小为F ,然后释放振子从静止开始向左运动,经过时间t 后第一次到达平衡位置O 处,此时振子的速度为v ,在这个过程中振子的平均速度为A .等于B .大于C .小于D .05、下列各力中按照力的效果命名的是( )A .支持力B .电场力C .分子力D .摩擦力6、如图所示,在铁芯上、下分别绕有匝数n 1=800和n 2=200的两个线圈,上线圈两端u=51sin314tV 的交流电源相连,将下线圈两端接交流电压表,则交流电压表的读数可能是A.2.0V B.9.0VC.12.7V D.144.0V二、多项选择题:本题共4小题,每小题5分,共20分。

2020年浙江省宁波市镇海中学高考数学模拟试卷(5月份) (含答案解析)

2020年浙江省宁波市镇海中学高考数学模拟试卷(5月份) (含答案解析)

2020年浙江省宁波市镇海中学高考数学模拟试卷(5月份)一、选择题(本大题共10小题,共40.0分)1. 已知集合A ={3,2,1,0},B ={−1,0,1},则A ∩B =( )A. {1,0}B. {2,1,0}C. {3,2,1}D. {2,1}2. 已知函数f(x)=axsinx +xcosx(a ∈R)为奇函数,则f(−π3)=( )A. −π6B. −√3π6C. π6D. √3π63. 已知x ,y 满足{x ≥1x +y ≤4ax +by +c ≤0且目标函数z =2x +y 的最大值为7,最小值为1,则a+b+ca = ( )A. 2B. 1C. −1D. −24. 如图,网格纸上每个小正方形的边长均为1,粗线画出的是某棱锥的三视图,则该棱锥的体积为( )A. 32B. 3C. 23D. 435. 函数f(x)=xx 2+1的图象大致是( ).A.B.C.D.6. 将函数f (x )=cos (4x −π3)的图像上各点的横坐标伸长到原来的2倍,纵坐标不变,得到函数y =g(x)的图像,则g(x)的最小正周期是( )A. π2 B. π C. 2π D. 4π7. 在△ABC 中,已知|AB⃗⃗⃗⃗⃗ |=|BC ⃗⃗⃗⃗⃗ |=|CA ⃗⃗⃗⃗⃗ |=2,则向量AB ⃗⃗⃗⃗⃗ ⋅BC ⃗⃗⃗⃗⃗ =( ) A. 2B. −2C. 2√3D. −2√38. 已知双曲线C :x 2a 2−y 2b 2=1(a >0,b >0)的右焦点为F ,P 是第一象限C 上的点,Q 为第二象限C 上的点,O 是坐标原点,若OF ⃗⃗⃗⃗⃗ +OQ ⃗⃗⃗⃗⃗⃗ =OP ⃗⃗⃗⃗⃗ ,则双曲线C 的离心率e 的取值范围是( )A. (1,+∞)B. (2,+∞)C. [2,2√3)D. (√3,2)9. 函数f(x)=e x sin x 在区间[0,π2]上的值域为( )A. [0,e π2]B. (0,e π2) C. [0,e π2) D. (0,e π2] 10. 设数列{a n }的通项公式为a n =2n −7(n ∈N ∗)则|a 1|+|a 2|+⋯+|a 7|=( )A. 7B. 0C. 18D. 25二、填空题(本大题共7小题,共21.0分)11. 已知复数z 满足(1+2i )z =3−4i ,i 为虚数单位,则z 的虚部是________,|z |=________. 12. 已知随机变量X 的分布列如表:若EX =2,则a =_____.13. 已知ab >0 , a +b =5,则2a+1+1b+1的最小值为__________.14. 若(2x +1x )n 的二项展开式中的所有二项式系数之和等于256,则该展开式中常数项的值为______.15. 已知椭圆C :x 216+y 2b 2=1(4>b >0)的左右焦点为F 1,F 2,离心率为√32,若P 为椭圆上一点,且∠F 1PF 2=90°,则△F 1PF 2面积为______16. 2019年国际篮联篮球世界杯于8月31日到9月15日在8个城市的场馆举行,甲、乙、丙、丁四位同事拟购票去看比赛,该比赛的某购票点为他们提供四种结账方式:现金、支付宝、微信、银联卡.若甲没有银联卡,乙只带了现金,丙、丁用哪种方式结账都可以,甲、乙、丙、丁购票后,恰好用了其中的三种结账方式,那么他们结账方式的可能情况有________种.17.在四面体P−ABC中,若PA=3,PB=4,PC=5,底面△ABC是边长为2√3的正三角形,O为△ABC的中心,则∠PAO的余弦值为______.三、解答题(本大题共5小题,共74.0分)18.在△ABC中,C−A=π2,sinB=13.(1)求sin A的值;(2)设AC=√6,求△ABC的面积.19.如图,平面ABCD⊥平面CDEF,且四边形ABCD是梯形,四边形CDEF是矩形,∠BAD=∠CDA=90∘,AB=AD=DE=12CD,M是线段DE上的点,满足DM=2ME.(1)证明:BE//平面MAC;(2)求直线BF与平面MAC所成角的正弦值.20.已知数列{a n}为等差数列,a2=5,a6=13,{b n}为等比数列,b2=a4,b n+1=3b n.(1)求通项公式a n,b n;(2)求{a n⋅b n}前n项和S n.21.在平面直角坐标系xOy中,P(x0,y0)(y0≠0)是椭圆C:x22λ2+y2λ2=1(λ>0)上的点,过点P的直线l的方程为x0x2λ2+y0yλ2=1.(Ⅰ)求椭圆C的离心率;(Ⅱ)当λ=1时,设直线l与x轴、y轴分别相交于A,B两点,求△OAB面积的最小值;(Ⅲ)设椭圆C的左、右焦点分别为F1,F2,点Q与点F1关于直线l对称,求证:点Q,P,F2三点共线.22.已知函数f(x)=(ax+1)lnx−x2+1.(1)令g(x)=f′(x),判断g(x)的单调性;(2)当x>1时,f(x)<0,求a的取值范围.-------- 答案与解析 --------1.答案:A解析:【分析】本题考查了集合的交集运算,根据集合A,B,得到其交集,属于基础题.【解答】解:由题意可得:A∩B={0,1}.故选A.2.答案:A解析:【分析】本题考查了正弦、余弦函数,函数的奇偶性,属于基础题.利用函数的奇偶性可求出a的值,进而可得答案.【解答】解:因为f(x)=axsinx+xcosx(a∈R)为奇函数,所以,即,所以a=0,所以,所以.故选A.3.答案:D解析:【分析】先根据约束条件画出可行域,再利用几何意义求最值,z=2x+y表示直线在y轴上的截距,只需求出可行域直线在y轴上的截距最大最小值时所在的顶点即可.本题主要考查了简单的线性规划,以及利用几何意义求最值的方法,属于基础题.【解答】解:由题意得:目标函数z=2x+y在点B取得最大值为7,在点A处取得最小值为1,∴A(1,−1),B(3,1),∴直线AB的方程是:x−y−2=0,∴则a+b+ca=−2.故选D.4.答案:A解析:【分析】本题考查了空间几何体的三视图以及三棱锥的体积公式,属于基础题.如图所示:三棱锥N−B1MB即为所求三棱锥,根据三棱锥的体积公式即可求得其值.【解答】解:如图所示:正方体ABCD−A1B1C1D1的边长为3,M,N分别为AB,DD1的三等分点,且BM=D1N=1.三棱锥N−B1MB即为所求三棱锥,V=13×(12×1×3)×3=32,故选A.5.答案:A解析:【分析】本题考查由解析式选择函数的图象,解题关键是研究函数的性质,如单调性、奇偶性等,研究图象的特殊点,函数的值正负及变化趋势.【解答】解:由f(x)=xx2+1,当x >0时,f(x)>0,x <0时,f(x)<0,只有A 符合. 故选A .6.答案:B解析:【分析】本题考查三角函数图像的伸缩变换. 【解答】解:由题意得g (x )=cos (12×4x −π3)=cos (2x −π3),∴T =2π2=π.故选B .7.答案:B解析:解:AB ⃗⃗⃗⃗⃗⋅BC ⃗⃗⃗⃗⃗ =|AB ⃗⃗⃗⃗⃗ |⋅|BC ⃗⃗⃗⃗⃗ |cos(π−π3)=2×2×(−12)=−2 故选B直接利用向量的数量积的定义即可求解本题主要考查了向量的数量积的定义的简单应用,属于基础试题8.答案:B解析: 【分析】本题考查向量加法的平行四边形法则,以及双曲线的性质. 【解答】解:由已知F (c,0),P (x 1,y 1),因为OF ⃗⃗⃗⃗⃗ +OQ ⃗⃗⃗⃗⃗⃗ =OP ⃗⃗⃗⃗⃗ ,由向量加法的平行四边形法则,QP ⃗⃗⃗⃗⃗ =0F ⃗⃗⃗⃗⃗ ,所以Q (−x 1,y 1) 所以(2x 1,0)=(c,0),2x 1=c,x 1=c2,因为P 是第一象限C 上的点,所以x 1>a, 即c2>a,所以e =ca >2. 故选B .9.答案:A解析:【分析】利用导数判断函数f(x)在[0,π2]上是增函数,由此能求出函数f(x)=e x sinx在区间[0,π2]上的值域.【解答】解:∵f(x)=e x sinx,∴f′(x)=e x(sinx+cosx),∵x∈[0,π2],∴f′(x)>0,∴f(x)在[0,π2]上是增函数,∴f(x)min=f(0)=0,f(x)max=f(π2)=eπ2.∴函数f(x)=e x sinx在区间[0,π2]上的值域为[0,eπ2].故选A.10.答案:D解析:解:∵数列{a n}的通项公式为a n=2n−7(n∈N∗),∴由a n=2n−7≥0,得n≥72,∴|a1|+|a2|+⋯+|a7|=−a1−a2−a3+a4+a5+a6+a7=−(2×1−7)−(2×2−7)−(2×3−7)+2×4−7+2×5−7+2×6−7+2×7−7=25.故选:D.|a1|+|a2|+⋯+|a7|=−a1−a2−a3+a4+a5+a6+a7,由此能求出结果.本题考查数列的前7项的绝对值的求法,是基础题,解题时要认真审题,注意数列的通项公式的合理运用.11.答案:−2;√5解析:【分析】本题考查复数代数形式的乘除运算,考查复数的基本概念及复数模的求法,是基础题.把已知等式变形,利用复数代数形式的乘除运算化简求得z的虚部,再由复数模的公式求|z|.【解答】解:由(1+2i)z=3−4i,得z=3−4i1+2i =(3−4i)(1−2i)(1+2i)(1−2i)=−1−2i,∴z的虚部是−2,|z|=√5.故答案为−2;√5.12.答案:0解析:【分析】本题主要考查了离散型随机变量的分布列、数学期望等知识,属于基础题,先根据概率和=1求出b,然后根据EX=2,可求出a.【解答】解:根据题意可知13+b+16+14=1,解得b=14,所以EX=13a+14×2+16×3+14×4=2,解得a=0,故答案为0.13.答案:3+2√27解析:【分析】本题考查利用基本不等式求最值,属于一般题.由已知得a+1+b+1=7,然后利用基本不等式求解即可.【解答】解:因为ab>0 , a+b=5,所以a+1+b+1=7,a>0,b>0所以2a+1+1b+1=17(a+1+b+1)(2a+1+1b+1)=1(3+2(b+1)+a+1)≥17(3+2√2(b+1)a+1×a+1b+1)=3+2√27,当且仅当a+1=√2(b+1)时取等号,所以2a+1+1b+1的最小值为3+2√27.故答案为3+2√27.14.答案:1120 解析:【分析】本题考查二项式系数的性质,熟练掌握二项展开式的通项是关键,是基础题.由已知求得n值,写出二项展开式的通项,由x的指数为0求得r值,则答案可求.【解答】解:由题意可知,2n=256,解得n=8.∴(2x+1x )n=(2x+1x)8,其展开式的通项T r+1=C8r⋅(2x)8−r⋅(1x)r=28−r⋅C8r⋅x8−2r,令8−2r=0,得r=4.∴该展开式中常数项的值为T5=24⋅C84=1120.故答案为1120.15.答案:4解析:【分析】本题考查了椭圆的定义、勾股定理、三角形的面积计算公式,属于中档题.先根据离心率求出b,c,设|PF1|=m,|PF2|=n.在Rt△PF1F2中,由勾股定理可得m2+n2=(2c)2,利用椭圆的定义可得m+n=2a,联立解得mn即可.【解答】解:椭圆C:x216+y2b2=1(4>b>0)的左右焦点为F1,F2,离心率为√32,∴e2=c2a2=1−b2a2=1−b216=(√32)2,∴b2=4,∴c=2√3,∴|F1F2|=2c=4√3,设|PF1|=m,|PF2|=n.在Rt△PF1F2中,由勾股定理可得m2+n2=(2c)2=48,又|PF1|+|PF2|=2a,∴m+n=8.则mn=(m+n)2−(m2+n2)2=8.∴△F1PF2的面积S=12mn=4.故答案为:4.16.答案:26解析:【分析】本题主要考查分类计数原理,考查排列与组合的应用,属于中档题.根据题意结账方式可分为三3类:第一类,当甲、丙、丁都不选微信时,则甲有2种选择,当甲选择现金,其余2人有A22=2(种)结账方式,当甲选择支付宝时,丙、丁可以银联卡,或者其中一人选择银联卡,另一人只能选支付宝或现金,故有1+C21C21=5(种)结账方式,即2+5=7(种)结账方式;第二类,当甲、丙、丁都不选支付宝时,则甲有2种选择,当甲选择现时,其余2人有A22=2(种)结账方式,当甲选择微信时,丙、丁可以是银联卡,或者其中一人选择银联卡,另一人只能选微信或现金,故有1+C21C21=5(种)结账方式,即2+5=7(种)结账方式;第三类,当甲、丙、丁都不选银联卡时,若有人使用现金,则有C31A22′=6(种)结账方式;若没有人使用现金,则有C32A22=6(种)结账方式,故有6+6=12(种)结账方式,再根据分类计数原理相加即可得结果.【解答】解:甲没有银联卡,乙只带了现金,丙、丁用哪种方式结账都可以,可分为三3类,第一类,当甲、丙、丁都不选微信时,则甲有2种选择: ①当甲选择现金,其余2人有A22=2(种)结账方式; ②当甲选择支付宝时,丙、丁可以银联卡,或者其中一人选择银联卡,另一人只能选支付宝或现金,故有1+C21C21=5(种)结账方式.综上,有2+5=7(种)结账方式,第二类,当甲、丙、丁都不选支付宝时,则甲有2种选择: ①当甲选择现时,其余2人有A22=2(种)结账方式; ②当甲选择微信时,丙、丁可以是银联卡,或者其中一人选择银联卡,另一人只能选微信或现金,故有1+C21C21=5(种)结账方式.综上,有2+5=7(种)结账方式.第三类,当甲、丙、丁都不选银联卡时,若有人使用现金,则有C31A22′=6(种)结账方式;若没有人使用现金,则有C32A22=6(种)结账方式,故有6+6=12(种)结账方式,根据分类计数原理可得共有7+7+12=26(种)结账方式.17.答案:136解析:【分析】本题考查了空间线线角的计算,重点考查了余弦定理的应用,属于中档题.【解答】解:如图:在△ABC中,连接AO并延长交BC于D,∵O为△ABC的中心,∴AD为BC边上的中线,又AB=BC=AC=2√3,∴AD=3.在△PBC中,∵PB=4,PC=5,BC=2√3,由余弦定理,在△PDC中,由余弦定理=52+(√3)2−2×5×√3×2120√3=352,在△PAD中,由余弦定理,故答案为136.18.答案:解:(1)因为C−A=π2且C+A=π−B,所以A=π4−B2,所以,即,又sinA>0,所以sinA=√33;(2)由题意可知A为锐角,故,又,∴A>B,则B为锐角,,由正弦定理得ACsinB =BCsinA,所以BC=AC·sinAsinB=3√2,又因为sinC=sin(A+B)=sinAcosB+cosAsinB=√33×2√23+√63×13=√63,所以.解析: 【分析】本题考查了正弦定理、三角形面积公式和两角和与差的三角函数公式,是中档题.(1)要求sin A 的值,应该用题目中的已知条件,将A 表示出来,可以得到A =π4−B2,进一步可以求出sin A ;(2)已知AC 的长度,可以根据正弦定理求出BC 的长度,再根据三角形面积公式,即可求得答案.19.答案:解:(1)连接BD ,交AC 于N ,连接MN ,由于AB =12CD ,所以DNNB =2,所以MN//BE ,由于MN ⊂平面MAC ,BE ⊄平面MAC , 所以BE//平面MAC.(2)因为平面ABCD ⊥平面CDEF ,DE ⊥CD ,所以DE ⊥平面ABCD ,可知AD,CD,DE 两两垂直,分别以DA ⃗⃗⃗⃗⃗ ,DC ⃗⃗⃗⃗⃗ ,CE⃗⃗⃗⃗⃗ 的方向为x,y,z 轴,建立空间直角坐标系D −xyz . 设AB =1则C (0,2,0),M (0,0,23),F (0,2,1),B (1,1,0),A (1,0,0),MA ⃗⃗⃗⃗⃗⃗ =(1,0,−23),AC ⃗⃗⃗⃗⃗ =(−1,2,0).设平面MAC 的法向量n ⃗ =(x,y,z ),则{n ⃗ ·MA ⃗⃗⃗⃗⃗⃗ =x −23z =0n ⃗ ·AC ⃗⃗⃗⃗⃗ =−x +2y =0,令z =3,得平面MAC 的一个法向量n⃗ =(2,1,3),而BF ⃗⃗⃗⃗⃗ =(−1,1,1),设所求角为θ,则sinθ=|cos⟨n ⃗ ,BF ⃗⃗⃗⃗⃗ ⟩|=√4221, 故直线BF 与平面MAC 所成的角的正弦值为√4221.解析:本题考查线面平行的证明,考查线面角的正弦值的求法,考查空间中线线、线面、面面间的位置关系等基础知识,考查空间想象能力与思维能力,考查运算求解能力,是中档题. (1)连结BD ,交AC 于N ,连结MN ,推导出MN//BE ,由此能证明BE//平面MAC ;(2)推导出DE ⊥平面ABCD ,从而AD ,CD ,DE 两两垂直,以D 为原点建立空间直角坐标系D −xyz ,利用向量法能求出直线BF 与平面MAC 所成角的正弦值.20.答案:解:(1)∵数列{a n }为等差数列,a 2=5,a 6=13,设公差为d ,∴{a 1+d =5a 1+5d =13, 解得a 1=3,d =2,∴a n =3+(n −1)×2=2n +1. ∵{b n }为等比数列,b 2=a 4,b n+1=3b n . ∴b 2=2×4+1=9,q =b n+1b n=3,∴b 1=3,∴b n =3n . (2)a n ⋅b n =(2n +1)·3n ,S n =3·3+5·32+7·33+⋯+3n ·(2n +1)①3S n =3⋅32+5⋅33+7⋅34+⋯+(2n +1)⋅3n+1,② ①−②,得:−2S n =9+2(32+33+⋯+3n )−(2n +1)·3n+1=9+2×9×(1−3n−1)1−3−(2n +1)·3n+1=3n+1−(2n +1)·3n+1, ∴S n =n ·3n+1.解析:(1)由已知条件利用等差数列的通项公式,求出a 1=3,d =2,从而a n =2n +1.由{b n }为等比数列,结合已知条件求得b n =3n .(2)由a n ⋅b n =(2n +1)⋅3n ,利用错位相减法能求出{a n ⋅b n }前n 项和S n .本题考查数列的通项公式的求法,考查数列的前n 项和的求法,解题时要认真审题,注意错位相减法的合理运用.21.答案:(本小题满分14分)解:(Ⅰ)依题a =√2λ,c =√2λ2−λ2=λ, 所以椭圆C 离心率为e =√2λ=√22.…(3分) (Ⅱ)依题意x 0≠0,令y =0,由x 0x 2+y 0y =1,得x =2x 0,则A(2x 0,0).令x =0,由x 0x 2+y 0y =1,得y =1y 0,则B(0,1y 0).则△OAB 的面积S △OAB =12|OA||OB|=12|2x 0y 0|=1|x0y 0|.因为P(x 0,y 0)在椭圆C :x 22+y 2=1上,所以x 022+y 02=1. 所以1=x 022+y 02≥00√2,即|x 0y 0|≤√22,则1|x 0y 0|≥√2.所以S △OAB =12|OA||OB|=1|x0y 0|≥√2.当且仅当x 022=y 02,即x 0=±1,y 0=±√22时,△OAB 面积的最小值为√2. …(8分)(Ⅲ)由y 02λ2=1−x 022λ2>0,解得−√2λ<x 0<√2λ. ①当x 0=0时,P(0,λ),Q(−λ,2λ),此时k F 2P =−1,k F 2Q =−1. 因为k F 2Q =k F 2P ,所以三点Q ,P ,F 2共线. 当P(0,−λ)时,也满足.②当x 0≠0时,设Q(m,n),m ≠−λ,F 1Q 的中点为M ,则M(m−λ2,n 2),代入直线l 的方程,得:x 0m +2y 0n −x 0λ−4λ2=0.设直线F 1Q 的斜率为k ,则k =nm+λ=2y 0x 0,所以2y 0m −x 0n +2y 0λ=0.由{x 0m +2y 0n −x 0λ−4λ2=02y 0m −x 0n +2y 0λ=0,解得m =2x 02λ+4x 0λ24y 02+x 02−λ,n =4x 0y 0λ+8y 0λ24y 02+x 02.所以Q(2x 02λ+4x 0λ24y 02+x 02−λ,4x 0y 0λ+8y 0λ24y 02+x 02).当点P 的横坐标与点F 2的横坐标相等时,把x 0=λ,y 02=λ22代入m =2x 02λ+4x 0λ24y 02+x 02−λ,得m =λ,则P ,Q ,F 2三点共线.当点P 的横坐标与点F 2的横坐标不相等时,直线F 2P 的斜率为k F 2P =yx 0−λ.由−√2λ≤x 0≤√2λ,x 0≠−2λ. 所以直线F 2Q 的斜率为k F 2Q =4x 0y 0λ+8y 0λ24y 02+x 022x 02λ+4x 0λ24y 02+x 02−2λ=4x 0y 0λ+8y 0λ22x 02λ+4x 0λ2−8y 02λ−2x 02λ=4x 0y 0λ+8y 0λ24x 0λ2−8y 02λ=x 0y 0+2y 0λx 0λ−2y 02=y 0(x 0+2λ)x 02+λx 0−2λ2=y 0(x 0+2λ)(x 0−λ)(x 0+2λ)=y 0x 0−λ.因为k F 2Q =k F 2P ,所以Q ,P ,F 2三点共线. 综上所述Q ,P ,F 2三点共线.…(14分)解析:(Ⅰ)利用椭圆方程,求出a ,c ,即可求椭圆C 的离心率; (Ⅱ)由x 0x 2+y 0y =1,求出A 的坐标,然后求解B 的坐标,表示三角形的面积,通过P(x 0,y 0)在椭圆C 上,利用基本不等式求解三角形OAB 面积的最小值. (Ⅲ)由x 22λ2+y 2λ2=1,求出−√2λ<x 0<√2λ.①当x 0=0时,求出P(0,λ),Q(−λ,2λ),证明三点Q ,P ,F 2共线.②当x 0≠0时,设Q(m,n),m ≠−λ,F 1Q 的中点为M ,则M(m−λ2,n 2),代入直线l 的方程,求出Q 坐标,通过点P 的横坐标与点F 2的横坐标相等时,说明P ,Q ,F 2三点共线.点P 的横坐标与点F2的横坐标不相等时,证明k F2Q =k F2P,说明Q,P,F2三点共线.本题考查直线与椭圆的综合应用,椭圆的简单性质的应用,考查转化思想以及分类讨论思想的应用,考查计算能力.22.答案:解:(1)由f(x)=(ax+1)lnx−x2+1,则g(x)=f′(x)=alnx+1x−2x+a,所以g′(x)=−2x2+ax−1x(x>0).①当a≤0时,g′(x)<0,g(x)为(0,+∞)上的减函数;②当a>0时,若a2−8≤0,即0<a≤2√2时,g′(x)≤0,g(x)为(0,+∞)上的减函数;若a2−8>0,即a>2√2时,由g′(x)=0有两根,得x1=a−√a2−84>0,x2=a+√a2−84>0,∴在x∈(0,x1)上,g′(x)<0,g(x)为减函数;在x∈(x1,x2)上g′(x)>0,g(x)为增函数;在x∈(x2,+∞)上,g′(x)<0,g(x)为减函数.综上:当a≤2√2时,g(x)为(0,+∞)上的减函数;当a>2√2时,g(x)在(0,x1)和(x2,+∞)为减函数,在(x1,x2)上为增函数;(2)由(1)知,对a讨论如下,①当a≤0时,g′(x)<0,则f′(x)为(1,+∞)上的减函数,则f′(x)<f′(1)=−1+a<0,故f(x)为(1,+∞)上的减函数,由于f(1)=0,所以f(x)<f(1)=0,即a≤0时满足题意.②当a>0时,由于f′(1)=−1+a,对其讨论如下:(A)若f′(1)=−1+a≤0,即a≤1,则由(1)知,f′(x)为(1,+∞)上的减函数,则f′(x)<f′(1)=−1+a<0,所以f(x)为(1,+∞)的减函数,由于f(1)=0,所以f(x)<f(1)=0,即0<a≤1时满足题意.(B)若f′(1)=−1+a>0,即a>1,则由(1)知,当1<a≤2√2时,f′(x)为(1,+∞)上的减函数,<0,又f′(e a)=−2e a+a+a2+1e a所以存在x0∈(1,e a),使得在x∈(1,x0)时,f′(x)>0,于是f(x)为(1,x0)上的增函数,因为f(1)=(a+1)ln1−12+1=0,所以f(x)>f(1)=0,即1<a≤2√2时不满足题意.当a>2√2时,由于x1<1,所以对x2与1的大小关系讨论如下,1)如果x2≤1,即2√2<a≤3时,由(1)知,f′(x)为(1,+∞)上的减函数,<0,又f′(e a)=−2e a+a+a2+1e则存在x0∈(1,e a),使得在x∈(1,x0)时,f′(x)>0,于是f(x)为(1,x0)上的增函数,又f(1)=0,则f(x)>f(1)=0,即2√2<a≤3时不满足题意.2)如果x2>1,即a>3,那么由(1)知,f′(x)为(1,x2)上的增函数,则当x∈(1,x2)时,f′(x)>0,于是f(x)为(1,x2)上的增函数,又f(1)=0,则f(x)>f(1)=0,即a>3时不满足题意.综上所述,a的取值范围为(−∞,1].解析:本题考查了利用导数研究函数的单调性,考查导数中的函数不等式问题,考查导数的应用以及分类讨论思想,属于难题.(1)求出函数的导数,通过讨论a的范围,求出函数的单调区间即可;(2)通过讨论a的范围,结合函数的单调性确定a的范围即可.。

(审核版)浙江省镇海中学2020届高三校模拟考语文试题(含答案解析).doc

(审核版)浙江省镇海中学2020届高三校模拟考语文试题(含答案解析).doc

一、语言文字运用(共24分,其中选择题每小题3分)1.下列词语中加点的字,注音没有错误的一项是()A.辐辏.(zòu)骠.骑(piào)拧.螺丝(nǐng)枵.腹从公(xiāo)B.划.拨(huà)股.肱(gōu)癫.疯病(diān)相机行事(xiāng)C.困扎.(zā)应.允(yīng)黑黢.黢(qū)生杀予.夺(yǔ)D.侪.辈(chài)逶.迤(wēi)炸.鸡块(zhà)滂.沱大雨(pāng)2.下列各句中,没有错别字的一项是()A.柳宗元过世的时候,韩愈为他写了非常感人的墓志铭,然而在韩愈的立场上,柳宗元是一个典型的知识分子,是非常值得歌诵的。

B.如果关汉卿真是一个“蒸不烂、煮不透、捶不扁”的人,我相信他不会这么软弱,他的戏一定泼辣、野性,会真正构成民间感动的力量。

C.纪念何克希同志诞辰110周年座谈会暨央视文献纪录片《不朽的番号——新四军浙东游击纵队》开机仪式在余姚梁弄举行。

D.明朝后期,“公安派”和“竟陵派”以“性灵”为主张,认为写文章应该直接抒发自己的心灵、情感,应探利得珠,反对虚假的道德文章。

3.下列个句中,加点的词语运用不正确的一项是()A.《白鹿原》是一个整体性的世界,自足的世界,饱满丰富的世界,它正是这样的凝重,浑厚的风范跻身于我国当代杰出的长篇小说的行列。

B.你看,有一个愿意我活几天的。

那力量就这么大,然而现在是没有了,连这一个也没有了。

同时我自己也觉得不配活下去。

别人呢?C.独念东汉党人,千古盛世,然郑康成教猱升本....,模楷儒冠,而名字不在党籍。

余抱杜门,论治不缘政党,谈艺不入文社。

D.吴三桂镇守海关,此地一夫当关万夫莫开........,然而吴三桂并不是可靠的人由他把守海关只会让明政府深受其害。

4.下列各句中,没有语病的一项是()A.西行姑娘楼佳悦将心爱的老马正与玄奘,让老马代替他伴随取经人迢迢万里行是电影《大唐玄奘》中最令人心碎的场景。

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2020届浙江省镇海中学高三校模拟考自选模块试卷第一部分(共20小题每,小题1.5分,满分30分)1.I'm very grateful to my high school teachers,without _____________help I wouldn't be so excellent. A.their B.whomC.whose D.which2.—You know quite a lot about the fashion show.—Well, Cathy ________ it to me during lunch.A.introduces B.introducedC.had introduced D.will introduce3.The warmth of ____ coat will mostly be determined by ____ soft of cloth used.A.the; a B.a; the C./; the D.a; a4.Why does she always drive to work ____ she could easily take the train?A.unless B.untilC.before D.when5.— Don't put the waste on the ground, young man.— Oh, I'm sorry. I ______ the dustbin there.A.hadn't seen B.haven’t seenC.didn't see D.wasn't seeing6.Join us and you will discover an environment ______ you can make the most of your skills and talents. A.that B.whereC.how D.what7.Mr. White, who ________ in Shanghai for seven years, is a manager of a company in Beijing. A.was working B.workedC.had worked D.has worked8.____ which university to attend, the girl asked her teacher for advice.A.Not knowing B.Knowing not C.Not known D.Known not9.Not until _________________ the better qualities in ourselves ____________ expect to find them in others.A.have we developed; can weB.we have developed; that can weC.we have developed; can weD.can we develop; that we will10.-Did Tom go back late last night?—No. It was just nine o'clock he arrived home.A.when B.after C.until D.that11.The house ______ I live in is very small.A.that B.whom C.when D.what12.— I got that job I wanted at the public library.—___________! That’s good news.A.Go ahead B.Cheers C.Congratulations D.Come on13.— ________ your purchases in time, make sure the express company knows your address exactly. A.To receive B.ReceivedC.Receiving D.Having received14.Although Jane agrees with me on most points, there was one on which she was unwilling to ______. A.give out B.give inC.give away D.give off15.Last year I applied to Princeton University.I ____ they would say yes—but they did, and now here I am.A.never think B.am never thinkingC.have never thought D.never thought16.The recently released film Kong:Skull Island successfully ________ the audience to the adventure with Dolby 3-D technology.A.transports B.adjustsC.transforms D.relates17.Many writers are drawn to building a world, _____ readers are somewhat familiar with but also feel distant from our normal lives.A.it B.one C.that D.the one18.His children were his pride, and being a devoted father became a top ______ in his life.A.capacity B.anxiety C.priority D.opportunity19._____ is often the case, we have worked out the production plan.A.Which B.When C.What D.As20.While his approach was a complete ________ from established practices, the result was satisfactory. A.separation B.prohibition C.departure D.judgment第二部分阅读理解(满分40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。

21.(6分)Researchers in China and the United States have developed a new cataract(白内障)treatment with cells that has restored vision in babies in a trial and may eventually be used in adults.The treatment-by doctors and staff members at the University of California, San Diego School of Medicine and Sichuan and Sun Yat-sen universities in China-was published in March 9 edition of the scientific journal Nature.A cataract is a clouding of the normally clear lens(晶体)of an eye. Typical cataract operation involves the removal of the cloudy lens and the insertion of an artificial one. The new operation has been tested in animals and during a small, human trial. It resulted in fewer complications(并发症)than the currentharmful operation, and in regrown lenses with superior visual function in all 12 of the baby cataract patients who received the procedure.A congenital cataract-lens clouding that occurs at birth or shortly after- is important cause of blindness in children. In the new research, K and Zhang, head of ophthalmic genetics at US San Die go’s Shiley Eye Institute, and his colleagues relied on the regrown potential of endogenous(同源的)stem cells.According to Zhang, endogenous stem cells are different from other stem cells that are typically grown in a laboratory, transplanted into a patient, and can have risks of immune(免疫的)rejection, infection or cancers. Zhang told CBS News, “We invented a new operation to make a very small opening at the side of a cataractous lens bag, remove the cataract inside, allow the opening to heal, and promote potential lens stem cells to regrow an entirely new lens with vision.”The human trial involved 12 babies under the age of 2 who were treated with the new method, while 25 babies received the standard operation care. The latter group experienced a higher incidence ofpost-operation danger, early- onset eye high blood pressure and increased lens clouding. The scientists reported fewer complications and faster healing among the 12 babies who has the new procedure.1、What is the text mainly about?A.A new cataract treatment with stem cells.B.The concept of the cataract.C.Bad effects of post-operation in the cataract.D.The reasons why the cataract comes into being.2、Which of the following best describe the new cataract treatment according to the passage? A.Convenient. B.Comfortable.C.Safe. D.Cheap.3、What can we learn about the new cataract treatment?A.It may be used widely. B.It has more risks.C.It has been put into practice widely. D.It can only restore vision in babies.4、What does the underlined word “congenital” in the fourth paragraph mean?A.Strange. B.Born.C.Serious. D.Dangerous.22.(8分)Electric devices can seem like a “third party” in some relationships because some partners spent more time on them than with each other.When Amanda Gao, a 26-year-old white collar worker in Beijing, went to a hotpot restaurant with her boyfriend on Friday night several weeks ago, she expected that they would have a good time together. To her disappointment, however, it did not turn out that later. As soon as they were led to their seats and shebegan to order dishes, he buried himself in his mobile phone.“It seemed that his phone was making its way between us. A date that should have belonged to usturn ed into one where my boyfriend dated a third party and I felt left out.” Gao said. Some people, like her, have found electronics have been sabotaging(破坏) their romantic relationships.A study, published in the journal Psychology of Popular Media Culture, in April, 2017, questioned nearly 200 college aged adults who were in committed(真诚的) relationships to report on their and their partner’s smartphone dependency. The results showed people who were more dependent on their phones were less sure about their relationships, and people considered their partners excessively(过度地) dependent on their devices were less satisfied in their relationship.Lin Yuan, a relationship advisor in Beijing, noted that as more and more electronics come out and spice up people’s live s, they are at the same time becoming a third party in relationships, especially for young people.Lin said she knew of some people who suggest that electronics should be kept out of bedrooms, which she considered challenging and hard to be put into practice for most couples. She recommended that if people are feeling neglected in their relationship, they need to respectfully let their partners know their feeling. “Communication is always the best and the most efficient way.” she said.1、What is Gao’s feeling when entering the restaurant with her boyfriend?A.Expectant. B.Disappointed.C.Annoyed. D.Uneasy.2、Which of the following may Lin Yuan agree with?A.Gao’s boyfriend must be addicted to playing games.B.Most couples can practice keeping electronics out of bedrooms.C.Partners should communicate more to understand each other better.D.Couples should restrict the use of electronics to avoid possible problems.3、Why was the case of Amanda Gao mentioned?A.To explain who the “third party” is.B.To stress the importance of electronic devices.C.To make advisors know more about the matter.D.To introduce peoples’ dependence on electronics.4、Which of the following can be the best title for the text?A.A third party—electronic devicesB.Are electronic devices killing romance?C.Do smart phones ruin partnership?D.Couples and smart phones23.(8分)It’s 8 a.m., Tuesday, Nov. 18, 2028, and you are headed for a business appointment 300 miles away. Y ou step into your circle, two- passenger air-cushion car, press a series of buttons and the national traffic computer notes your destination, figures out the current traffic situation and signals your car to slide out of the garage. Hands free, you sit back and begin to read the morning paper — which is flashed on a flat TV screen over the car’s dashboard. Tapping a button changes the page.The car speeds up to 150 mph in the city’s countryside, and then hits 250 mph in less built­up areas, driving over the smooth plastic road. You fly past a string of cities, many of them covered by the new domes (圆屋顶) that keep them evenly climatized all year round. Traffic is heavy, typically, but there’s no need to worry. The traffic computer, which sends and receives signals to and from all cars on the road between cities, keeps vehicles at least 50 yds apart. There hasn’t been an accident since the system began.Suddenly your TV phone buzzes. A business partner wants a sketch of a new kind of impeller your firm is putting out for sports boats. You reach for your case and draw the diagram with a pencil-thin infrared flashlight (红外线闪光灯) on what looks like a TV screen lining the back of the case. The diagram is sent to a similar screen in your partner’s office, 200 miles away. He presses a button and a fixed copy of the sketch rolls out of the machine. He wishes you good luck at the coming meeting and signs off.Ninety minutes after leaving your home, you slide beneath the dome of your destination city. Your car slows down and heads for an outer-core office building where you’ll meet your colleagues. After you get out, the vehicle parks itself in a garage to await your return. Private cars aren’t allowed inside most city cores. Moving sidewalks and electrams (电车) carry the public from one location to another.1、The traffic computer in your car can ________.A.keep your car at a safe distance from other carsB.keep your car at the same speed in different situationsC.keep your car receiving signals of TV programsD.keep your car driving avoiding heavy traffic2、Why are the cities covered by the new domes?A.To prevent people from being wet in the rain.B.To stop the climate of the cities changing violently all year.C.To protect the travelers against the strong sunshine.D.To make the city have the same weather all year.3、What will the city be like in the future?A.No accidents will happen because of heavy traffic.B.The sidewalk can move itself up and down.C.The road is built with the plastic material.D.The car parks itself on a dome to wait for your return.4、The third paragraph mainly tells ________.A.you are lucky to sell products of your companyB.you receive best wishes from your business partnerC.you can do business with a newly invented pencilD.you can do business even on the road in the future24.(8分)Sharpshooters refer to those who are skilled at firing a gun and accurately hitting what they are aiming at. However, some fish are also called amazing sharpshooters. People wonder how they hit their targets without using tools? By spitting! The fish shoot so accurately that they rarely miss their target.These amazing fish are archerfish (印度射水鱼).There are several different types of archerfish living in Southeast Asia. They spit at their dinner to catch it. To catch dinner, a fish first sets itself in a certain place, for example just below the surface of the water. It puts the tip of its mouth so it barely breaks through the water. Then, the fish waits for an unsuspecting insect to land on a leaf or branch hanging over the water.When an unsuspecting, or unknowing insect lands, the fish attacks. The fish squeezes its gill (腮)covers very rapidly, which forces water into the fish's mouth. In the roof of its mouth the fish has a groove (槽)• The fish forms the small groove and its tongue into a narrow tube by pressing its tongue up against its groove. The water forced into the fish's mouth is forcefully pushed out of the fish's moulh through the tube. Archerfish have been known to send a stream up to 5 meters (16 feet). However, archerfish can only shoot insects up to 1-2 meters (3-6 feet) away due to their limited accuracy.When the unsuspecting insect is hit with a sudden stream of water, it is knocked into or falls into the water. Once in the water, the insect becomes dinner. Archerfish learn how to spit when they are young. They get more accurate with practice. Adult archerfish rarely miss.Within a meter, archerfish will often spring out of the water and grab an insect in their mouths.1、Paragraph 1 flinctions as a(n) .A.background. B.comment.C.explanation. D.introduction.2、Which of the following is the right steps of archerfish’s hunting for an insect?a. It has its tongue pressed against the roof of its mouth.b. It gathers water into its mouth by squeezing its gill covers quickly.c. It shoots a stream of water at the insect above water.d. It stays underwater for an insect to land nearby.A.adcb B.dbacC.dbca D.adbc3、If an insect is 3 feet away, an archerfish will often to catch it.A.jump out of the water B.spit a kind of oily liquidC.shoot a stream of water at it D.push its tongue out of its mouth4、Archerfish can’t hit insects 5 meters away for the reason that .A.their eyesight is limitedB.they can’t hit faraway insects pr eciselyC.their mouths are under the surface of waterD.the longest distance they can hit is at most 5 meters25.(10分)Humans and many other mammals have unusually efficient internal temperature regulating systems that automatically maintain stable core body temperatures in cold winters and warm summers. In addition, people have developed cultural patterns and technologies that help them adjust to extremes of temperature and humidity (湿度).In very cold climates, there is a constant danger of developing hypothermia, which is a life-threatening drop in core body temperature to below normal levels. The normal temperature for humans is about37.0°C.However, differences in persons and even the time of day can cause it to be as much as 6°C higher or lower in healthy individuals. It is also normal for core body temperature to be lower in elderly people. Hypothermia begins to occur when the core body temperature drops to 34.4°C.Below 29.4°C, the body cools more rapidly because its natural temperature regulating system usually fails. The rapid decline in core body temperature is likely to result in death. However, there have been rare cases in which people have been saved after their temperatures had dropped to 13.9-15.6°C.This happened in 1999 to a Swedish woman who was trapped under an ice sheet in freezing water for 80 minutes. She was found unconscious, not breathing, and her heart had stopped beating, yet she was eventually saved despite the fact that her temperature had dropped to 13.7°C.In extremely hot climates or as a result of uncontrollable infections, core body temperatures can rise to equally dangerous levels. This is hyperthermia. Life-threatening hyperthermia typically starts in humans when their temperatures rise to 40.6-41.7°C.Only a few days at this extraordinarily high temperature level is likely to result in the worsening of internal organs and death.1、What keeps our body temperature stable?A.Culture and technologies. B.The stable earth temperature.C.Our strong determination. D.Some kind of in-body system.2、What is a Swedish woman mentioned for in the text?A.Proving the strength of life.B.Arguing against some conclusion.C.Showing the limit on humans’ body temperature.D.Introducing an exceptional case about our body temperature.3、Which of the following may cause hyperthermia?A.Extreme climates. B.Very cold climates.C.Controllable infections. D.Temperatures below 29.4°C.4、What can be a suitable title for the text?A.Humans’ Temperature Regulating System B.Changes of Body TemperaturesC.Huma ns’ Temperature D.A Ice Trap Survivor第三部分语言知识运用(共两节)第一节(每小题1.5分,满分30分)阅读下面短文,从短文后各题所给的A、B、C和D四个选项中,选出可以填入空白处的最佳选项.26.(30分)Communication is an important part of any relationship. Many of us are 1 to share our experiences or emotions with our friends. But when it's our turn to lend a(n) 2 ,we soon become bored or are short of idea on how to 3 and offer advice.That's because of what researchers call "listener burnout(倦怠)”.A friend might talk to us 4 often complaining about the same 5 problems. When we offer quick advice to 6 the situation, we may be unconsciously trying to 7 urselves from bumout. However, good listeners 8 their natural tendency to solve the other's problems hurriedly and to keep the conversation brief.To be a good 9 ,you need to use "active listening". It starts with the real 10 to help others and think through their feelings. Don't 11 things. Y ou can start by putting your phone 12 and sitting close to your friend. Let your facial expressions 13 一what he or she is saying 14 you are able to fully understand,acknowledge the other person's 15 by reflecting them back:"That must be really hard for you.”Use16 words or even sounds such as“yes”,"right”,and"hmm"”to17 the other person to continue.0f course,a 18 can be extremely hard if the other person is too critical.But don't get defensive. Effective listeners don't 19 negative criticism.Instead,they listen and understand what the person ls trying to convey 20 responding.1、A.afraid B.hesitant C.shy D.eager2、A.shoulder B.hand C.ear D.eye3、A.respond B.explain C.argue D.quit4、A.aimlessly B.endlessly C.deliberately D.cautiously5、A.difficult B.old C.acute D.sensitive6、A.fix B.discuss C.create D.describe7、A.forgive B.protect C.discourage D.prevent8、A.follow B.display C.form D.overcome9、A.reader B.partner C.listener D.speaker10、A.demand B.habit C.desire D.ability11、A.skip B.rush C.overlook D.postpone12、A.away B.off C.out D.up13、A.record B.restrict C.reflect D.replace14、A.Whether B.Since C.While D.If15、A.suggestions B.purposes C.responses D.feelings16、A.big B.tough C.strong D.shone17、A.force B.remind C.encourage D.convince18、A.conversation B.suggestion C.problem D.lecture19、A.give up B.make up C.leave out D.block out20、A.after B.before C.while D.once第二节(每小题1.5分,满分15分)阅读下面材料,在空白处填入1个适当的单词或括号内单词的正确形式。

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