2019年上海南模中学高三三模(2019.05)

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2019-2020学年上海市南洋模范中学高三生物三模试卷及参考答案

2019-2020学年上海市南洋模范中学高三生物三模试卷及参考答案一、选择题：本题共15小题，每小题2分，共30分。

每小题只有一个选项符合题目要求。

1. 下列关于生长素发现过程的实验，说法不正确的是（）A.达尔文实验证明，胚芽鞘的感光部位是尖端B.鲍森·詹森的实验证明，胚芽鞘尖端产生的影响可以透过琼脂片传递给下部C.拜尔的实验证明，胚芽鞘的弯曲生长是由于生长素生长分布不均匀造成的D.温特实验证明，胚芽鞘尖端能产生促进生长的化学物质2. 2018年诺贝尔生理学或医学奖授予美国免疫学家詹姆斯·艾利森和日本免疫学家本庶佑，以表彰他们在癌症免疫治疗方面所作出的贡献。

下列叙述错误的是A. 癌变的细胞内，负责调节细胞周期的基因功能异常B. 细胞癌变后，细胞膜的成分发生改变，细胞间黏着性降低C. 人体免疫系统能够消灭癌细胞体现了免疫系统的防卫功能D. 人体免疫系统识别癌细胞的过程受到干扰会导致癌细胞“逃逸”3. 下列不一定发生在人体内环境中过程有A.正常机体内抗体与抗原的结合B.某种激素与激素受体结合C.肝脏细胞代谢产生的尿素运输到肾小球D.各种神经递质从突触前膜到达突触后膜4. 生物大分子通常都有一定的分子结构规律，即由一定的基本组成单位，按一定的排列顺序和连接方式形成多聚体，下列表述正确的是（）A.若该图为一段肽链的结构模式图，则1表示肽键，2表示中心碳原子，3的种类约有21种B.若该图为一段RNA的结构模式图，则1表示核糖，2表示磷酸基团，3的种类有4种C.若该图为一段单链DNA的结构模式图，则1表示磷酸基团，2表示脱氧核糖，3的种类有4种D.若该图表示多糖结构模式图，淀粉、纤维素和糖原的空间结构是相同的5. 如图表示人体细胞间信息传递的一种方式。

下列描述错误的是（）A. 该调节方式的信息传递速度缓慢、作用范围较广泛B. 若靶细胞属于垂体，则分泌细胞一定是下丘脑细胞C. 若分泌细胞是垂体细胞，则靶器官可能是甲状腺、性腺、肾上腺D. 体温调节和血糖平衡调节可能涉及这种调节方式6. 下列关于脂质功能的叙述错误的是（）A. 胆固醇在人体内参与血液中脂质的运输B. 磷脂是构成细胞膜和细胞器膜的重要成分C. 维生素D能有效地促进人和动物肠道对钙和磷的吸收D. 脂肪中含H比糖类多，是主要的能源物质7. 图1表示芽的生长与生长素浓度的关系，图2表示幼苗横放时，根的生长与生长素浓度的关系。

上海市2019届高三高考数学系列模拟卷(3)及答案解析

上海市2019届高考数学模拟试卷3考生注意：1．每位考生应同时领到试卷与答题纸两份材料，所有解答必须写在答题纸上规定位置，写在试卷上或答题纸上非规定位置一律无效；2．答卷前，考生务必将学校、姓名、学号等相关信息在答题纸上填写清楚； 3．本试卷共23道试题，满分150分，考试时间120分钟。

一、填空题（本大题满分56分，共14小题，每小题满分4分）集合{}2M x x =<，{}lg (1)N x y x ==-，则M N = 1.若知i 是虚数单位，使(1)ni +为实数的最小正整数n 为2.已3.若对于任意实数x ，不等式a x x >--+|1||2|恒成立，则实数a 的取值范围是4.在ABC ∆中，若120=∠A ，5=AB ，7=BC ，则三角形ABC 的面积=S5.若直线),(042R n m ny mx ∈=-+始终平分圆042422=---+y x y x 的周长，则mn 的取值范围是_________________6.设f n k ()=（其中n N ∈*），k 是2的小数点后第n 位数字，…74142135623.12=，则{}f f f f …个[()]88的值等于____________7.已知矩阵sin cos 0sin cos 1ααββ+⎛⎫⎪+⎝⎭为单位向量，且,,2παβπ⎡⎫∈⎪⎢⎣⎭，()sin αβ-的值 8.一个圆锥的侧面展开图是圆心角为43π，半径为18 cm 的扇形，则圆锥母线与底面所成角的余弦值为________9.在矩形ABCD 中，AB =1BC =，E 是CD 上一点，且1AE AB ⋅=，则AE AC ⋅的值为10.设函数f x ()是定义在R 上的奇函数，若f x ()的最小正周期为3，且f ()11>，f m m ()2231=-+，则m 的取值范围是_________________11.若π220≤≤x ，则使x x 2cos 2sin 12=-成立的x 的取值范围是 12.已知集合{1,2,,}U n =，n *∈N ．设集合A 同时满足下列三个条件：①A U ⊆；②若x A ∈，则2x A ∉；③若U x C A ∈，则2U x C A ∉．当7n =时，满足条件的集合A 的个数为______13.对任意的120x x <<，若函数()f x a x =的大致图像为如图所示的一条折线（两侧的射线均平行于x 轴），试写出a 、b 条件是14.已知数列{}n a 满足11,2,n n n a n a a n ++⎧=⎨-⎩为奇数，为偶数，设222n n n b a a +=-，则数列{}n b 的通项公式为________________二、选择题（本大题满分20分，共4小题，每小题满分5分）15.“21a >” 是“方程2221x y a+=表示椭圆”的A. 充分而不必要条件B. 必要而不充分条件C. 充分必要条件D. 既不充分也不必要条件16.设3144322314)1()1(+++=++++x b x a x a x a x a x 22)1(++x b 413)1(b x b +++定义),,,(),,,(43214321b b b b a a a a f =，则),1,2,3,4(f 等于A ．)4,3,2,1(B ．)0,4,3,0(C ．)2,2,0,1(--D ．)1,4,3,0(--17.互不相等的三个正数321,,x x x 成等比数列，且P 1(1log a x ,1log b y )，P 2(2log a x ，2log b y )，)log ,(log 333y x P b a 三点共线(其中0a >，1a ≠，0b >，1b ≠)，则1y ，2y ，3yA. 等差数列，但不等比数列；B. 等比数列而非等差数列C. 等比数列，也可能成等差数列D. 既不是等比数列，又不是等差数列 18.设函数)sin()sin()sin()(2211n n x a x a x a x f ααα+⋅+++⋅++⋅= ，其中)2,,,,2,1(*≥∈=n N n n i i α为已知实常数，R x ∈，则下列A ．若0)2()0(==πf f ，则0)(=x f 对任意实数x 恒成立；B ．若0)0(=f ，则函数)(x f 为奇函数；C ．若0)2(=πf ，则函数)(x f 为偶函数；D ．当0)2()0(22≠+πf f 时，若0)()(21==x f x f ，则)(221Z k k x x ∈=-π．74分，共5小题）19.(本题满分12分）第（1）小题6分，第（2）小题6分.在直三棱柱111C B A ABC -中，1==AC AB ，090=∠BAC ，且异面直线B A 1与11C B 所成的角等于060，设a AA =1.（1）求a 的值；（2）设D 是11C B 上的任意一点，求D 到平面BC A 1的距离.20.(本题满分14分）第（1）小题7分，第（2）小题7分.已知函数.3cos 33cos 3sin )(2xx x x f +=（1）将)(x f 写成)sin(φω+x A +B 的形式，并求其图象对称中心的横坐标；（2）如果△ABC 的三边a 、b 、c 满足ac b =2，且边b 所对的角为x ，试求x 的范围及此时函数)(x f 的值域。

2019-2020学年上海市南洋模范中学高三英语第三次联考试题及答案解析

2019-2020学年上海市南洋模范中学高三英语第三次联考试题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 atthe ticket office only. It alternates (交替) up to eight new releases on four screens and hosts almost daily swap meets 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.BOnce small farmers in Masii, a remote village in Kenya, have picked their crops, all they can do is wait until a buyer trucks through. The system works fairly well for beans and corn, but mangoes-the area’s other main crop-spoil (腐烂) more quickly. If the trader is late, they rot.However, a simple coating could change that. A company, SmartTech, has created a product that doubles the shelf life of fresh produce, enabling farmers to access far-off, larger markets. More time forfresh produce on grocers’ shelves also means less food waste-a $2.6 trillion problem, according to the United Nations’ Food and Agriculture Organization (FAO).James Rogers, CEO of SmartTech, wanted to solve the problem for food much in the same way that oxide barriers preventing rust (锈) have achieved for steel. Fortunately, researchers have found when plants made the jump from water to land, they developed cutin(蜡质), a barrier which is made of fatty acids that link together to form a seal around the plant, helping keep water in.The cutin was such a grand strategy that today you’ll still find it across the plant kingdom. SmartTech discovered through researches that an orange can last longer than a strawberry not so much because of the thickness of its skin, but because of the difference in the arrangement of those cutin molecules (分子)on thesurface. After extensive trials, Rogers and his team developed a natural and tasteless protective coating from plant material-stems, leaves and skins. The product extends the sweet spot between ripening and rot. And best of all, the treated produce doesn’t require refrigeration.“SmartTech has huge potential to turn poor farmers in Africa into commercial farmers,” says Rogers. “That means more money in pockets, and more food in stomachs.” But whether the company can cost-effectively reach small farmers in far-off areas still remains a challenge.4. The author mentions the small farmers in Kenya to ________.A. stress their need for preserving produceB. show their challenge in harvesting cropsC. express their wish to reach larger marketsD. evaluate their loss caused by slow transport5. What can we learn about SmartTech’s product?A. It is financially supported by FAO.B. It is intended to replace refrigeration.C. It is designed to thicken produce’s skin.D. It is based on plants’own defence system.6. What will James Rogers probably focus on next?A. How to expand farms.B. How to earn more money.C. How to produce more tasty food.D. How to profit farmers in remote area.7. The main purpose of the passage is to ________.A. promote a productB. present a technologyC. advertise SmartTechD. introduce James RogersCThe COVID-19 vaccination（接种疫苗）rate in the US has fallen to newlows in recent weeks, threatening President Joe Biden’s goal of having 70 percent of American adults with at least one shot by July 4.With just less than one month from July 4, the current vaccination rate will put the US at somewhere between 67 percent and 68 percent of the adult population with at least one dose（剂量）by Independence Day. To reach 70 percent by July 4, around 1.6 percent of the population needs to get their first dose per week from nowuntil July 4.The Centers for Disease Control and Prevention（CDC）reported last week that 63 percent of adults hadreceived their first dose of a COVID-19 vaccine. That was up slightly from 62 percent from the report a week before. The additional 1 percent of adults completing their first dose is the lowest since the CDC started tracking the vaccination rate in mid-February.On average, fewer than 1 million shots are given out per day, a decline of more than two-thirds from the peak of 3.4 million in April, The Washington Post reported. In South Carolina, about 71,000 residents got a shot in the week leading up to June 3, compared to a high of nearly 300,000 in one week in early April, according to data from the CDC.The slowdown is moreprominentacross the South and Midwest. Twelve states have seen vaccinations fall to 15 daily shots per 10,000 residents. Less than a quarter of black Americans had received their first COVID-19 shot as of June 7.James Hildreth, CEO of Meharry Medical College, told Politico, “We need to make a stronger effort to bring the vaccine to the communities, rather than relying on the communities to come to vaccination centers.”The sharp decline in vaccination began in mid-April when federal officials temporarily stopped the use of the Johnson＆Johnson vaccine while they investigated rare blood-clotting（凝血）reactions.The “low-hanging fruit—thosepeople who ly want to get vaccinated without you telling them anything” have already been vaccinated, which has led to the slowdown, Dr Anthony Fauci, director of the National Institute of Allergy and Infectious Diseases, said on a White House-organized call with community leaders last week, according to the Post.8. What can we learn from the text?A. Dr Anthony Fauci is in charge of the Centers for Disease Control and Prevention.B. In South Carolina, about 300,000 residents got a shot in the week leading up to June 3C. In mid-April federal officials temporarily suspended the Johnson＆Johnson vaccine.D. Less than one fourth of Americans had received their first COVID-19 shot by June 7.9. Which of the following can replace the underlined word “prominent” in Para. 5?A. Meaningful.B. Obvious.C. Inspiring.D. Complex.10. How can America increase the COVID-19 vaccination rate according to James Hildreth?A. By giving the vaccine shot at the communities.B. By offering the vaccine to the public for free.C. By frequently informing the public of the vaccine.D. By urging the communities to come to vaccination centers.11. What can be the best title for the text?A. Biden wishes to have 70% of adults with one shot by July 4B. CDC has been trackingthe vaccination rate since mid-FebruaryC. Some Americans need to get vaccinated without telling them toD. Biden’s July 4th vaccine goal may be missedDIf you believe in thermometers (温度计), you should believe in global warming. And if you believe in eating French fries with your hamburger, you should worry more about carbonemission.It won't happen overnight, but as the planet’s climate changes, the growth cycles of main crops will all be heavily affected. Here are three of the crops that might not beat the heat.The potato actually needs a great deal of beauty rest to develop properly. “They need a cool nighttime temperature in order to start growing the tuber, the part that we eat,” Nelson says.Temperature and altitude are two primary concerns in coffee growth, and farmers are stuck between a rock and a hard place. “We know that coffee grows in a certain temperature range,” Nelson says. “They've been moving up the mountains, but at some point you run out of mountains to move up.” Higher-grade strains of Arabica coffee in Central American regions are at particular risk due to the need for lower growing temperature, especially since diseases and pests are also becoming more of a factor as temperatures rise.Even if you're looking at the direct effects of temperature alone, rice is in trouble. “As nighttime temperatures go up, the rice is going to have a problem flowering and won't make as many seeds,” Nelson says. In addition to this direct heat consequence, rising sea levels will flood many rice paddies and destroy water salinity levels, while droughts will lower production, raise prices, and further shame westerners who are clumsy with chopsticks to be able to clear their plates once a new grain is adopted.12. What is the closest meaning to the underlined word “emission” in Paragraph 1?A. Mixture.B. Pollution.C. Giving off.D. Going down.13. What does the potatoes' development need according to Nelson?A. Cool temperature at night.B. Sudden rise of the temperature.C. Enough water and salt.D. Much sunshine and fresh air.14. The rice is difficult to flower,probably because ________.A. sea levels go up suddenlyB. many seeds can't be producedC. the temperature rises at nightD. both temperature and altitude rise15. What is the best title for the text?A. How Does Global Warming Affect WesternersB. How Does the Cool Temperature Affect RiceC. What Is the Real Reason For Crops’ GrowthD. Three Crops That Won't Survive Climate Changes第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019届上海市高考模拟卷(三)数学试题(解析版)

2019届上海市高考模拟卷（三）数学试题一、单选题1．设x ∈R ，则“|x －2|<1”是“x 2＋x －2>0”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】A【解析】根据不等式的性质，结合充分条件和必要条件的定义进行判断即可．【详解】解：由“|x ﹣2|＜1”得1＜x ＜3，由x 2+x ﹣2＞0得x ＞1或x ＜﹣2，即“|x ﹣2|＜1”是“x 2+x ﹣2＞0”的充分不必要条件，故选：A ．【考点】必要条件、充分条件与充要条件的判断．2．已知集合{(,)|||||1}P x y x y =+…，{}22(,)|1Q x y x y =+…，则有（）A ．P Q =B ．PQ C ．P Q P = D ．P Q Q ⋂=【答案】B【解析】根据两个集合分别表示的平面区域分析可得答案. 【详解】因为{(,)|||||1}P x y x y =+…表示四个顶点分别为(1,0),(0,1),(1,0),(0,1)--的正方形围成的区域(包括边界),而{}22(,)|1Q x y x y =+…表示的圆心为原点,半径为1的圆围成的区域(包括边界),所以P Q .故选:B 【点睛】本题考查了集合之间的真子集关系,属于基础题.3．将向量1a =(1x ，1y )，2a =(2x ，2y )，…n a =(n x ,n y )组成的系列称为向量列{n a }，并定义向量列{n a }的前n 项和12n n S a a a =++⋅⋅⋅+．如果一个向量列从第二项起,每一项与前一项的差都等于同一个向量，那么称这样的向量列为等差向量列。

若向量列{n a }是等差向量列，那么下述四个向量中，与21S 一定平行的向量是 ( ) A ．10a B ．11aC ．20aD ．21a【答案】B【解析】依题意，当{}n a 为等差向量列时，设每一项与前一项的差都等于d ，则可求出通项公式1(1)n a a n d =+- ，所以{}n a 前21项和211221111111()(20)2121021S a a a a a d a d a d a =+++=+++++=+= ，故与21S 平行的向量是11a ，选B.点睛: 本题主要考查新定义: 等差向量列的理解和应用, 属于中档题. 解题思路：设每一项与前一项的差都等于d ，运用类似等差数列的通项和求和公式，计算可得211121S a =，由向量共线定理，可得出结论. 考查类比的数学思想方法和向量共线定理的运用.4．设集合A ＝[0，12)，B ＝[12，1]，函数()()1,221,x x Af x x x B⎧+∈⎪=⎨⎪-∈⎩，若x 0∈A ，且f[f(x 0)]∈A ，则x 0的取值范围是( ) A ．(0，14] B ．(14，12) C ．(14，12] D ．[0，38]【答案】B 【解析】【详解】 ∵x 0∈A ，∴f(x 0)＝x 0＋12∈B. ∴f[f(x 0)]＝f(x 0＋12)＝2(1－x 0－12)＝1－2x 0. 又因为f[f(x 0)]∈A ，∴0≤1－2x 0<12，解得14<x 0≤12，又0≤x 0<12.∴14<x 0<12，故选B.二、填空题5．函数sin cos cos sin 44y x x x x ππ⎛⎫⎛⎫=+++ ⎪ ⎪⎝⎭⎝⎭的最小正周期T =___________.【答案】π【解析】利用两角和的正弦公式化简函数表达式，由此求得函数的最小正周期. 【详解】依题意ππsin sin 244y x x x ⎛⎫⎛⎫=++=+ ⎪ ⎪⎝⎭⎝⎭，故函数的周期2ππ2T ==. 故填：π. 【点睛】本小题主要考查两角和的正弦公式，考查三角函数最小正周期的求法，属于基础题.6．若函数21()12x f x =，(0,)x ∈+∞，则其反函数1()f x -=_________.【答案】2log (1)1x +-,(1,)x ∈+∞【解析】计算二阶行列式化简()f x ,再根据求反函数的步骤可求得反函数. 【详解】因为21()12x f x =1221121x x +=⨯-⨯=-,因为x ∈(0,)+∞,所以()(1,)f x ∈+∞, 所以由121x y +=-得21log (1)x y +=+,所以2log (1)1x y =+-,交换,x y 可得2log (1)1y x =+-, 所以12()log (1)1fx x -=+-,(1,)x ∈+∞,故答案为:2log (1)1x +-, (1,)x ∈+∞. 【点睛】本题考查了二阶行列式的计算,反函数的求法,属于基础题.7．在614x x ⎛⎫- ⎪⎝⎭的展开式中，2x 的系数为 .【答案】1516【解析】614x x ⎛⎫- ⎪⎝⎭展开式的通项为6621661144rrr r r r r T C x C x x --+⎛⎫⎛⎫=-=- ⎪ ⎪⎝⎭⎝⎭，由622r -=得2r =，所以222236115416T C x x ⎛⎫=-= ⎪⎝⎭，所以该项系数为1516.【考点】二项式定理及二项展开式的通项.8．过原点且与圆22420x y x y ++-=相切的直线方程为_______. 【答案】20x y -=【解析】切线的斜率显然存在,设出切线方程,利用圆心到直线的距离等于半径,列方程可解得答案. 【详解】由22420x y x y ++-=得22(2)(1)5++-=x y ,所以圆心为(2,1)-,因为圆心到y 轴的距离为2,所以所求切线的斜率一定存在, 所以设所求切线方程为y kx =,即0kx y -=,=解得2k =,所以所求切线方程为20x y -=. 故答案为:20x y -=. 【点睛】本题考查了求圆的切线方程,属于基础题.9．我国古代数学名著《九章算术》有“米谷粒分”题：粮仓开仓放粮，有人送来米1534石，验得米内夹谷，抽样取米一把，数得254粒内夹谷28粒，则这批米内夹谷约为__________石；（结果四舍五入，精确到各位）．【答案】169【解析】根据古典概型概率公式可得这批米内夹谷的概率约为28254，所以这批米内夹谷约为281534169254⨯≈石，故答案为169. 10．抛物线x 2=2py （p ＞0）的焦点为F ，其准线与双曲线-=1相交于A ，B 两点，若△ABF 为等边三角形，则p=___________. 【答案】6【解析】因为抛物线x 2=2py 的准线2py =-和双曲线-=1相交交点横坐标为=, 6.2x p p =∴=由等边三角形得解得【考点】本题主要考查抛物线的概念、标准方程、几何性质，考查分析问题解决问题的能力.11．若复数z x yi =+（x ，y ∈R ，i 为虚数单位）满足|||22|z z i =--，则33x y +的最小值为_______. 【答案】6【解析】根据复数模的计算公式将|||22|z z i =--化为2y x =-,将其代入到33x y +后,利用基本不等式可求得答案. 【详解】由|||22|z z i =--=化简得2x y +=,即2y x =-, 所以33x y +233x x -=+932363x x =+≥=⨯=,当且仅当 1.1x y ==时等号成立. 故答案为:6 【点睛】本题考查了复数的模的公式,基本不等式求最小值,属于基础题. 12．一个等差数列{}n a 中，2nna a 是一个与n 无关的常数，则此常数的集合为．【答案】11,2⎧⎫⎨⎬⎩⎭【解析】试题分析：设数列的首项为1a ，公差为d ，()()1211,21n n a a n d a a n d ∴=+-=+-1212n n a a d nd a a d nd-+∴=-+ 2n n a a 是一个与n 无关的常数10a d ∴-=或0d =，所以比值常数为11,2⎧⎫⎨⎬⎩⎭【考点】等差数列通项公式13．已知直三棱柱111ABC A B C -的6个顶点都在球O 的球面上，若3AB =，4AC =，AB AC ⊥，112AA =，则球的表面积为______.【答案】169π【解析】把直三棱柱111ABC A B C -的补成一个长方体，则直三棱柱111ABC A B C -的外接球和长方体的外接球是同一个球，由长方体的对角线长等于球的直径，求得球的半径，再利用球的表面积公式，即可求解．【详解】由题意，直三棱柱111ABC A B C -的底面ABC ∆为直角三角形，可把直三棱柱111ABC A B C -的补成一个长方体，则直三棱柱111ABC A B C -的外接球和长方体的外接球是同一个球，又由长方体的对角线长等于球的直径，且13,4,12AB AC AA ===，即213R ===，即132R =，所以球的表面积为221344()1692S R πππ==⨯=．故答案为：169π 【点睛】本题主要考查了直三棱柱与球的组合体问题，以及球的表面积的计算，其中解答中根据组合体的结构特征，求得球的半径是解答的关键，着重考查了运算与求解能力，属于基础题．14．新一季“中国好声音”开唱，开场节目是四位导师各选一首自己的代表作供其他导师演唱，每人恰好都是唱别人的歌.假设四首歌已选定，则有______种不同演唱方式. 【答案】9【解析】将问题转化为四个元素填四个空的全错位排列后,再按照元素1的位置分3类讨论计算结果相加即可得到. 【详解】将四位导师抽象为四个元素,设为1,2,3,4,四首歌抽象为四个空位,设为1,2,3,4,依题意转化为四个元素填四个空的全错位排列,第一类:元素1填在2号空位,则元素2有3种填法,元素3,4填法唯一,此时共有3种填法; 第二类,元素1填在3号空位,则元素3有3种填法,元素2,4填法唯一,此时共有3种填法;第三类,元素1填在4号空位,则元素4有3种填法,元素2,3填法唯一,此时共有3种填法; 根据分类计算原理可得共有3+3+3=9种填法. 综上所述,共有9种不同的演唱方式. 故答案为:9 【点睛】本题考查了有限制条件的排列问题,属于中档题.15．若函数()2(1)y x x ax b =+++的图象关于点()20，成中心对称，则a b +=______. 【答案】3【解析】在函数()2(1)y x x ax b =+++的图象上取两点(1,0)-,(0,)b ,求出它们关于点(2,0)对称的点(5,0),(4,)b -后,代入()2(1)y x x ax b =+++,解方程组可得答案.【详解】在函数()2(1)y x x ax b =+++的图象上取两点(1,0)-,(0,)b ,则它们关于点(2,0)对称的点(5,0),(4,)b -也在函数()2(1)y x x ax b =+++的图象上, 即(51)(255)0(41)(164)a b a b b +++=⎧⎨+++=-⎩,即52510340a b a b +=-⎧⎨+=-⎩,解得7,10a b =-=,所以3a b +=. 故答案为:3 【点睛】本题考查了函数图象的对称中心的性质,属于基础题.16．在平面直角坐标系中，O 是坐标原点，两定点,A B 满足·2OA OB OAOB===，由点集{|,1,,}P OP OA OB R λμλμλμ=++≤∈所表示的区域的面积是__________．【答案】【解析】【详解】由|OA |＝|OB |＝OA ·OB ＝2，知cos ∠AOB ＝12，又0≤∠AOB ≤π，则∠AOB ＝3π，又A ，B 是两定点，可设A 1)，B (0,2)，P (x ，y )，由OP ＝λOA ＋μOB，可得{2x y λμ，＝＋⇒{26x y x λμ==-.因为|λ|＋|μ|≤1x＋2y x -≤1，等价于由可行域可得S 0＝12×P 所表示的区域面积S ＝4S 0＝三、解答题17．已知(sin ,1)a α=，(cos ,2)b α=，0,2πα⎛⎫∈ ⎪⎝⎭. （1）若//a b ，求sin 2α的值；（2）在（1）的条件下，若5cos()13αβ+=，0,2πβ⎛⎫∈ ⎪⎝⎭，求sinβ的值. 【答案】(1)45,(2)65【解析】(1)由//a b 可得1tan 2α=,再由万能公式可得sin 2α的值, (2)利用sin sin()βαβα=+-sin()cos cos()sin αβααβα=+-+可得答案. 【详解】(1)因为 //a b ,所以2sin cos 0αα-=,即1tan 2α=, 所以2222sin cos 2tan sin 22sin cos sin cos tan 1ααααααααα===++2124215()12⨯==+. (2)由(1)知,cos 2sin αα= ,且(0,)2πα∈,所以22sin (2sin )1αα+=,所以21sin 5α=,所以sin α,cos α=, 又(0,)2πβ∈,所以(0,)αβπ+∈,所以12sin()13αβ+===, 所以sin sin()sin()cos cos()sin βαβααβααβα=+-=+-+1251313=-=【点睛】本题考查了向量平行的坐标表示,二倍角的正弦公式,同角公式,两角差的正弦公式,属于基础题.18．如图，正四棱锥P ABCD -内接于圆锥，圆锥的轴截面是边长为10cm 的正三角形.（1）求异面直线PA 与BC 所成角的大小；（2）若正四棱锥由圆锥削去一部分得到，则需要削去部分的体积为多少？（精确到30.1cm ）【答案】(1)arccos4,(2)382.3cm .【解析】(1)根据//AD BC 可知, PAD ∠就是异面直线P A 与BC 所成的角,在三角形PAD 中由余弦定理可求得,(2)用圆锥的体积减去正四棱锥的体积即可得到答案. 【详解】(1)在正四棱锥P ABCD -中,//AD BC ,所以PAD ∠就是异面直线P A 与BC 所成的角,在正方形ABCD 中,10AC =,所以AD =, 在三角形PAD 中,10PA PD ==,所以222cos2PA AD PD PAD PA AD +-∠=⨯⨯2224==,所以PAD ∠=,所以异面直线P A 与BC 所成角的大小为.(2)在直角三角形PAO 中,PO ===所以圆锥的体积211133V PO AO π=⋅⋅⋅=⨯25⨯=,正四棱锥P ABCD -的体积221133V PO AD =⋅⋅=⨯23=,所以需要削去部分的体积为12(2)333V V π-=-=-82.3≈. 所以需要削去部分的体积约为82.33cm . 【点睛】本题考查了正四棱锥的结构特征,异面直线所成角,椎体的体积公式,属于中档题. 19．首项为12的无穷等比数列{}n a 所有项的和为1，n S 为{}n a 的前n 项和，又()25log 1n n b S t +-=，常数*t N ∈，数列{}n c 满足n n n c a b =⋅.（1）求数列{}n a 的通项公式；（2）若{}n c 是递减数列，求t 的最小值. 【答案】(1)12n na =,(2)1【解析】(1)根据无穷等比数列{}n a 所有项的和为1,求出公比12q =,再根据等比数列的通项公式可得;(2)求出n S 后代入可得5n b n t =+,1(5)2n n c n t =+⋅,然后根据数列递减可得10n n c c +-<恒成立,由不等式恒成立可得答案.【详解】(1)设无穷等比数列{}n a 的公比为q ,则111a q =-,所以1211q=-,解得12q =,所以111111()222n n n n a a q--==⨯=, (2)因为11(1)22112n n S -=-112n =-,所以215log (11)2n n b t +-+=, 所以5n b n t =+,所以1(5)2n n n n c a b n t ==+⋅,因为{}n c 是递减数列, 所以1111(55)(5)22n n n n c c n t n t ++-=++⋅-+⋅11(55102)2n n t n t +=++--⋅ +11(55)2n n t =--⋅0< 恒成立,所以550n t --<恒成立,所以55t n >-+恒成立,因为()55f n n =-+为递减函数,所以1n =时,()f n 取得最大值(1)550f =-+=, 所以0t >,又因为*t N ∈,所以t 的最小值为1. 【点睛】本题考查了无穷等比数列的和,等比数列的通项公式和前n 项和,数列的单调性,属于中档题.20．设S 、T 是R 的两个非空子集，如果函数()y f x =满足：①{()|}T f x x S =∈；②对任意1x ，2x S ∈，当12x x <时，恒有()()12f x f x <，那么称函数()y f x =为集合S 到集合T 的“保序同构函数”.（1）试写出集合{|01}A x x =<<到集合R 的一个“保序同构函数”；（2）求证：不存在从集合Z 到集合Q 的“保序同构函数”；（3）已知2()1xf x x =+是集合[]0,s 到集合[]0,t 的“保序同构函数”，求s 和t 的最大值.【答案】(1) ()tan()2f x x ππ=-(01)x <<,(2)证明见解析,(3)s 的最大值为1,t 的最大值为12【解析】(1)直接由题意写出()tan()2f x x ππ=-(01)x <<即可;(2)用反证法证明即可;(3)用定义证明()f x 在[0,1]上递增,在[1,)+∞上递减后,可得1s ≤,(1)t f ≤. 【详解】(1)取()tan()2f x x ππ=-(01)x <<,该函数是集合{|01}A x x =<<到集合R 的一个“保序同构函数”; 证明:任取1201x x <<<, 则122222x x ππππππ-<-<-<,因为tan y x =在(,)22ππ-上为增函数,所以12tan()tan()22x x ππππ-<-, 即12()()f x f x <,由定义可知, 函数()tan()2f x x ππ=-是集合{|01}A x x =<<到集合R 的一个“保序同构函数”.(2)证明:假设存在一个从集合Z 到集合Q 的“保序同构函数”,由“保序同构函数”的定义可知,集合Z 和集合Q 中的元素必须是一一对应的,不妨设整数0和1在Q 中的像分别为a 和b ,根据保序性,因为0<1,所以a b <,又2a b +也是有理数,但是2a b+没有确定的原像,因为0和1之间没有另外的整数了,故假设不成立,故不存在从集合Z 到集合Q 的“保序同构函数”.(3)设120x x <<,则12122212()()11x x f x f x x x -=-++21122212()(1)(1)(1)x x x x x x --++, 所以当1201x x <<≤时,21120,10x x x x ->-<,所以12())0(f x f x -<,即12()()f x f x <,所以()f x 在[0,1]上递增,当211x x >≥时, 21120,10x x x x ->->,所以12())0(f x f x ->,即12()()f x f x >, 所以()f x 在[1,)+∞上递减, 因为2()1xf x x =+是集合[]0,s 到集合[]0,t 的“保序同构函数”,所以()f x 在[0,]s 上递增,所以1s ≤,所以s 的最大值为1,t 的最大值为11(1)112f ==+. 【点睛】本题考查了正切函数的单调性,函数单调性的定义,利用单调性求函数的最值,属于难题.。

2019-2020学年上海南洋模范高级中学高三生物第三次联考试题及参考答案

2019-2020学年上海南洋模范高级中学高三生物第三次联考试题及参考答案一、选择题：本题共15小题，每小题2分，共30分。

每小题只有一个选项符合题目要求。

1. 对下列图示的生物学实验的叙述正确的有几项（）a．若图①表示将显微镜镜头由a转换成b，则视野中观察到的细胞数目减少b．若图①是显微镜下洋葱根尖某视野的图像，则向右移装片能观察清楚c细胞的特点c．若在显微镜下观察图①细胞质流动时，发现细胞质的流动方向是顺时针，则细胞质的实际流动方向是逆时针d．若图①是在显微镜下目镜为10×，物镜为10×，视野中被相连的64个分生组织细胞所充满的图像。

目镜不变，物镜换成40×时，则在视野中可检测到的分生组织细胞数为4个e．标本染色较深，观察时应选用凹面反光镜和大光圈f．若转换高倍镜，需要先升镜筒，以免镜头破坏玻片标本A.一项B.两项C.三项D.四项2. 灵菌红素具有抗肿瘤和抑制T细胞介导的免疫反应的功能。

科研人员在橡胶树内生菌ITBBB5-1中发现一个新细胞器——马阔囊胞，该囊胞位于杆状细菌末端。

电镜观察发现，灵菌红素位于马阔囊胞和其分泌的胞外小囊泡中。

下列说法错误的是（）A.橡胶树内生菌ITBBB5-1的核膜包含不连续的四层磷脂分子B.新型细胞器马阔囊胞可能具有储存和分泌灵菌红素的功能C.马阔囊胞分泌灵菌红素过程体现了生物膜的结构特点D.灵菌红素在癌症治疗和器官移植中具有潜在的应用价值3. 下列实例中，能说明生命活动离不开细胞的是（）①流感病人打喷嚏时，会有大量流感病毒随飞沫散布于空气中①手触碰到盛有沸水的电水壶会迅速缩回①体操运动员完成单杠动作离不开肌肉的收缩和舒张①人的胚胎发育过程中，细胞不断地进行分裂A.①①①B.①①①C.①①①D.①①①①4. 甲、乙两图为生物组织的呼吸方式实验装置图，图中植物呼吸作用释放的CO2被NaOH溶液吸收，使容器内气体压强减小，刻度管内的红墨水滴会发生移动。

2019-2020学年上海市南洋模范中学高三英语第三次联考试卷及答案

2019-2020学年上海市南洋模范中学高三英语第三次联考试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AIn his 402nd anniversary year, Shakespeare is still rightly celebrated as a great language master and writer. But he was not the only great master of play writing to die in 1616, and he is certainly not the only writer to have left a lasting influence on theater.While less known worldwide, Tang Xianzu is considered one of Chinas greatest playwrights and is highly spoken of in that country of ancient literary and dramatic traditions.Tang was born in 1550 inLinchuan,Jiangxiprovince. Unlike Shakespeare's large body of plays,poems and sonnets (十四行诗), Tang wrote only four major plays: The Purple Hairpin, Peony Pavilion (《牡丹亭》), A Dream under the Southern bough, and Dream of Handan. The latter three were constructed around a dream narrative, a way through which Tang unlocked the emotional dimension of human desires and ambitions and explored human nature beyond the social and political limits of that time.Similar to Shakespeare, Tang's success rode the wave of a renaissance (复兴) in theater as an artistic practice. As in Shakespeare'sEngland, Tang's works became hugely popular inChinatoo. During Tang'sChina, his plays were enjoyed performed, and changed. Kunqu Opera, a form of musical drama, spread from southernChinato the whole nation and became a symbol of Chinese culture. Combining northern tune and southern music, kunqu Opera was known for its poetic language, music, dance movements and gestures. Tang's works benefited greatly from the popularity of kunqu Opera, and his plays are considered classics of kunqu Opera.While Tang and Shakespeare lived in a world away from each other, there are many things they share in common, such e humanity of their drama, their heroic figures, their love for poetic language, a lasting popularity and the anniversary during which we still celebrate them.1. Why is Shakespeare mentioned in the first paragraph?A. To describe Shakespeare's anniversary.B. To introduce the existence of Tang Xianzu.C. To explain the importance of Shakespeare.D. To suggest the less popularity of Tang Xianzu.2. What's possibly one of the main theme of Tang's works?A. Social reality.B. Female dreams.C. Human emotions.D. Political environment.3. What does the author mainly tell us in Paragraph 4?A. The influence of Kunqu Opera on Tang's works.B. Tang's success in copying Shakespeare's styles.C. The way Kunqu Opera became a symbol of Chinese culture.D. Tang's popularity for his poetic language and music.BIf you struggle to fall asleep quickly, you’re not alone! Fortunately, thereare plenty of solutions you can try. With a few changes, you can fall asleep fast every night!Keep your room dark. Turn off all the lights above your head when you go to bed. Any bright light can make you believe it is too early in the day for sleep. If you want to read or write before bed, try using a small book light. Now that blue lights can keep you awake, red ones are a great choice.If you can, keep noise in and around your room the lowest at night. If you have an old clock that ticks loudly and keeps you awake, replace it with a silent one. If you share your home with anyone else, request that they keep noises like talking, music, or TV shows at the lowest while you are trying to sleep. It is difficult to fall asleep if you live near a busy road or hear other boring sounds after bedtime. You could get a white noise machine or play recordings of nature sounds, like waves or whales’ singing. You could also listen to soft, relaxing music.Read a book in bed if you have difficulty in falling asleep. Staying in bed doing nothing when you’re having trouble falling asleep may keep you wide awake. While reading in bed may be slightly harmful to your eyes, it can distract (分散) you from your thoughts and help you feel sleepy. But remember to read from a print book rather than something with a screen. The light from electronic screens can keep you awake.Lowering your body temperature helps you sleep, so set the room temperature between 15.5℃-21℃could do the trick.4. What color1 book light should you choose toread before bed?A. Red.B. Blue.C. White.D. Orange.5. What is the author’s attitude towards reading in bed?A. Doubtful.B. Worried.C. Favorable.D. Uncaring.6. What can we inferred from the text?A. Reading on cellphones sometimes helps you fall asleep.B. The lower your temperature while sleeping is, the better.C. Bright lights are better for your reading before going to bed.D. Playing recordings like birds’ singing can improve your sleep.7. How does the author organize the text?A. By givingexamples.B. By asking questions.C. By offering suggestions.D. By listing research results.CBack about 20 months ago I started college and just struggled with everything, such as classes and friends. I quickly became depressed and angry at myself for not being about to do better in school, in addition to lack of friends due to poor social and communication skills.This went on for months until my 19th birthday. My parents sent me a cake, which was a great cake. But I remember having this large cake and ly no one to share it with. I ended throwing out the cake after having one piece, with about 90 % of it leftover. That night I was depressed that I decided to go outside to the freezing temperature of the winter and run. I put my earphones in, went outside and ran about 2 miles at 11 p.m. on my birthday.When I got back inside I was content. I was proud of what I was able to do. The next night I did the same. I wasn’t quick or fit but you know that I went outside and did something. This continued for about 2 months until I finally worked up the courage to go to the gym, where I started swimming again as I used to in high school. A month went by and I started lifting weights and continually running.Looking back I can see that exercise helped cure my depression but it didn’t only do that. At the gym I met new friends and back at my dorm I grew confidence to go to the end of the hall seeing people playing Super Smash Brothers and ask if I could join.So go forward to present now. I exercise every day and look forward to that hour and a half I get daily to do what I love with people who love it as well. I hope this helps someone who may be or have been in a similar situation.8. What made the author decide to run at night?A. His l9th birthday.B. His parents’ cake.C. His loneliness.D. His friends.9. Which of the following best describes the author?A. Traditional.B. Determined.C. Humorous.D. Generous.10. What is the biggest benefit of the author’s running?A. Regaining his confidence.B. Losing his weight.C. Playing with his brothers.D. Joining other activities.11. Why do you think the author wrote this passage?A. To recall his life in college.B. To show his gratitude to his parents.C. To emphasize the importance of friendship.D. To share his experience of dealing with hardship.DMark Twain,the famous American writer,was once traveling in France.He went by trainto Dijon.He was very tired and wanted to sleep.He therefore asked the conductor to wake him up when the train came to Dijon.But first he explained he was a very heavy sleeper,“I may possibly protest(抗议)loudly when you try to wake me up,” he said to the conductor.“But don’t take any notice of what I say.Just put me off the train anyway.”Then Mark Twain went to ter,when he woke up it was night time and the train had reached Paris already.He realized at once that the conductor had forgotten to wake him up at Dijon.He was so angry that he ran to the conductor and began to shout at him.“I have never been so angry in my life,” Mark Twain said.The conductor looked at him calmly(平静地).“You are not half so angry as the American whom I put off the train atDijon,” he said.12. Mark Twain knew that he was a heavy sleeper,so ________.A. he protested loudly to the conductorB. he did not sleep before he arrived inDijonC. he told the conductor to wake him up no matter how loudly he might protestD. he slept lightly that time13. The conductor didn’t wake up Mark Twain atDijonbecause ________.A. he didn’t take Mark Twain’s words seriouslyB. he forgot Mark Twain’s words when the train came toDijonC. he did not want to bear his protestD. he mistook another American traveler for Mark Twain14. The American whom the conductor put off the train ________.A. did not want to get off atDijonB. wanted to get off atParisC. wanted to get off atDijonD. did not want to get off atParis15. Which of the following is TRUE?A. The conductor didn’t take Mark Twain’s words seriously.B. The conductor did take Mark Twain’s words seriously.C. The conductor was a heavy sleeper.D. Mark Twain must get off atParis.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019-2020学年上海南洋模范高级中学高三英语第三次联考试题及答案

2019-2020学年上海南洋模范高级中学高三英语第三次联考试题及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ANo one knows when the first printing press was invented or who invented it. but the oldest known printed text originated in China during the first millennium (千年) AD.The Diamond Sutra (《金刚经》), a Buddhist book from Dunhuang, China during the Tang Dynasty, is said to be the oldest known printed book.The Diamond Sutrawas created with a method known as block printing (雕版印刷), which used boards of hand-carved wood blocks in reverse.It was said that the moveable type was developed by Bi Sheng. He was fromYingshan,Hubei,China, living from 970 to 1051 AD. His method replaced panels of printing blocks with moveable individual Chinese characters that could be reused. The first moveable Chinese Characters were carved into clay and baked into hard blocks that were then arranged onto an iron frame that was pressed against an iron plate.The earliest mention of Bi Sheng’s printing press is in the bookDream Pool Essays, written in 1086 by Shen Kuo, who noted that his nephews came into possession of Bi Sheng’s typefaces (字体) after his death. Shen Kuo explained that Bi Sheng did not use wood because the texture is inconsistent (不一致的) and absorbs wetness too easily.By the time of the Southern Song Dynasty, which ruled from 1127 to 1279 AD, books had become popular in society and helped create a scholarly class of citizens who had the capabilities to become civil servants. Large printed book collections also became a status symbol for the wealthy class.1. When was Bi Sheng’s printing press first introduced in history?A. After Bi Sheng died and his nephews owned his typefaces.B. When books became popular in the Southern Song Dynasty.C. After the block printing was replaced by the moveable type printing.D. WhenThe Diamond Sutrawas printed into a book.2. What can we infer from the passage?A. Shen Kuo made great contributions to printing.B. The moveable type printing was invented earlier than block printing.C. Printed books were hard to get in the Song Dynasty.D. By the Southern Song Dynasty, books had helped people get to higher social positions.3. Why does the author write this passage?A. To show that Buddhism was popular in the Tang Dynasty.B. To introduce the early history of printing.C. To memorize Bi Sheng, developing the moveable type printing.D. To indicate the advantages of moveable type printing.BIt was the first day of spring here but it didn't feel like it. The COVID-19 epidemic (流行病) was sweeping the globe and everywhere there were feelings of fear and loneliness. “Social Distancing” had become the new norm. Here the schools had been closed, the restaurant dining rooms had been shut, and people had been told to work from home whenever they could. Even the sheltered workshop where my oldest son worked had been closed until further notice. People had made a run on the stores and large areas of the shelves were bare. On the news the numbers of the sick and dead continued to rise. It felt like there was a weight on the souls of everyone in the world.My family were staying at home as much as possible and as I looked out of my window I wondered how long this crisis would last. It was then, however, that I saw something that lifted that weight off of my soul, made me smile, and made my heart feel happy again. On the street below my house there was an old friend of mine from high school who was a teacher there. With him was my younger son's former aide from the high school as well. They were delivering the school lunches door to door to the hungry children who were stuck at home. Watching them made me think of something everyone's childhood television neighbor, Mr. Roger's once said: "In the bad times, always look for the helpers."I have no doubt that this crisis will pass as all the crisis before. But it is our choice on whether it brings out the best in us or the worst in us. Let it bring out the best in you. Use it to strengthen your faith. Use it to free yourself from fear. Use it to grow kinder, more giving, and more loving. Become a helper to all those in need and you will be a happy person today and all the days to come.4. Why didn't the author like the first day of spring?A. He had to work at home instead of at office.B. There wasn't social distance between each other.C. There was an epidemic everywhere in the world.D. His own company had been closed for a long time.5. How did people feel in the face of the epidemic?A. Disappointed.B. TenseC. Confident.D. Cheerful.6. What made the author happy?A. Seeing someone delivering lunches from door to door.B. Talking with his old friend in the street.C. Watching a childhood television at home.D. Staying together with his family.7. What do Mr. Roger's words actually mean?A. It's not necessary for us to find helpers in bad times.B. In good times we needn't do much to help others.C. We should ask more people to help us in bad times.D. We are supposed to help each other in bad times.CPeople saved a 20-foot orca (虎鲸) that was stuck between rocks on an Alaskan shore by continuously pouring water over it and protecting it from birds who circled above the defenseless whale.The whale was ultimately saved after a six-hour, labor-intensive life-saving operation. Someone spotted the large whale on the Prince of Wales Island near the coast of British Columbia on the morning of July 29th. The Coast Guard was called around 9 a.m. local time. Chance Strickland, the captain of a private yacht in Alaska, and his crew anchored and began life-saving action that were videoed by Aroon Melane and posted on the Internet.Strickland could hear the orca calling out to killer whales swimming in the area. People on other boats stopped with water and buckets to pour water over the animal. “There were tears coming out of its eyes,” Mr. Strickland told the local newspaper. “It was pretty sad.”The group of people formed a chain that passed buckets of seawater back and forth and poured the water on the orca, which seemed to liven it up. It made a noise and raised its tail when it got water.The National Oceanic and Atmospheric Administration (NOAA) was called in, which can be seen on the video using a machine to spray amist of seawater on the orca, which doubled as a way to keep the whale cool and scare the large group of birds that were hoping to feast on the beast.Melane said in her video that the orca was stranded (搁浅) for about six hours until the tide came in andswept it back into the ocean. The group efforts of Strickland’s crew and the NOAA saved the 13-year-old killerwhale.8. Why did birds circle above the orca?A. They were eager to eat it.B. They wished to protect it.C. They were attracted by the people.D. They wanted to find a place to rest.9. What did Strickland do immediately after finding the whale?A. Posted pictures online.B. Called friends for help.C. Took action to save it.D. Videoed the trapped animal.10. Why did the whale make a noise and raise its tail?A. To express its eager for water.B. To extend its thanks to people.C. To call out to its fellow whales.D. To show its power and sadness.11. Which can be the best title for the text?A. Killer Whale Got SavedB. The Orca Inspired KindnessC. Combined Efforts WantedD. Animals and Humans UnitedDMichele Gentile, an Italian bookseller, is offering free books to children in exchange for plastic bottles to recycle.Michele owns Ex Libris Cafe in Polla，a small town in southern Italy. He said he thought of the recycling program, because he wanted to inspire children in the small town to read and pay attention to the environment.''My goal is to spread the passion and love for books among those people in Italy who do not usually read while at the time helping the environment," Gentile explained.The idea for the initiative (倡议) came after Gentile collaborated with a nearby middle school on an aluminum recycling project. Working together, the schoolchildren and Gentile collected enough cans to purchase books for an entire classroom. His new program too off from there and has already spread into northern Italy: Gentile's hopes his work will continue to make headlines and become a worldwide initiative.The free books come from customers in Gentile's shop who have donated money to purchase a ''suspended" book. The idea comes from a World War II practice in which customers would buy two Coffees: one for themselves and another for the next person in line. Gentile has been using the extra books as part of his recycling initiative, While Gentile's program is a great way to recycle and get kids to read, it also brings awareness to the growing problem of plastic waste. Single-use plastics make up around 26 percent of all the plastics in the world, only 14 percent of which are recycled. Plastics that end up in landfills take around 500 years to decompose (分解),is amajor concern for environmentalists.Cutting down on plastic waste is important if we want to better the environment for future generations, and recycling programs like Gentile's book giveaway are a great way to meet that goal.12. What is the purpose of Gentile's book giveaway?A. To sell more coffee and books.B. To attract more customers.C. To collect money for a new project.D. To inspire reading and recycling.13. How did Gentile come up with the idea?A. Working with a school to recycle cans.B. Seeing school kids don't like reading.C. Donating books for a local school.D. Buying a “suspended” book for a child.14. Who pay for the books of the project?A. The local government.B. Gentile himself.C. Some publishers.D. His customers.15. Whichof the following is the best title for the text?A. An Italian's Reading InitiativeB. Recycling Can Get Kids Free BooksC. Michele's Way of Doing BusinessD. A New Way to Deal with Plastic Waste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。

上海市南洋模范中学2019届高三下学期3月月考数学试题(解析版)

2018学年南模中学高三年级三月份月考卷数学2019.3.6一、填空题。

1.已知全集，若集合，则_________.【答案】【解析】【分析】求出集合A，即可求解∁U A【详解】全集U＝R，集合A＝{x|x＞1或x<0}则＝故答案为【点睛】本题考查集合的基本运算，补集的求法，分式不等式解法，准确计算是关键，是基础题．2.双曲线的焦距为__________.【答案】6【解析】【分析】将双曲线的方程化为标准方程，求得a，b，c，可得焦距2c的值．【详解】双曲线2x2﹣y2＝6即为1，可得a，b，c3，即焦距为2c＝6．故答案为：6．【点睛】本题考查双曲线的简单几何性质，焦距的求法，注意将双曲线的方程化为标准方程，运用双曲线的基本量的关系，考查运算能力，属于基础题．3.已知二项展开式中的第五项系数为，则正实数_____.【答案】【解析】【分析】由二项式定理的通项公式可得：，解出即可得出．【详解】T5x﹣2，∴，a＞0．解得a．故答案为：．【点睛】本题考查了二项式定理的应用，考查了推理能力与计算能力，准确计算是关键，属于基础题．4.已知函数的图像与它的反函数的图像重合，则实数的值为___.【答案】-3【解析】【分析】先求反函数：y，利用函数f（x）（a）图象与它的反函数图象重合，即为同一个函数即可得出．【详解】由y（a），解得x（y≠3），把x与y互换可得：y，∵函数f（x）（a）图象与它的反函数图象重合，∴﹣a＝3，解得a＝﹣3．故答案为：﹣3．【点睛】本题考查了反函数的求法及其性质，考查了推理能力与计算能力，属于中档题．5.设，满足约束条件，则目标函数的最大值为_____.【答案】14【解析】【分析】画出可行域，通过向上平移基准直线到可行域边界的位置，由此求得目标函数的最大值.【详解】画出可行域如下图所示，由图可知，目标函数在点处取得最大值，且最大值为.【点睛】本小题主要考查利用线性规划求线性目标函数的最大值.这种类型题目的主要思路是：首先根据题目所给的约束条件，画出可行域；其次是求得线性目标函数的基准函数；接着画出基准函数对应的基准直线；然后通过平移基准直线到可行域边界的位置；最后求出所求的最值.属于基础题.6.从集合中随机选取一个数记为，从集合中随机选取一个数记为，则直线不经过第三象限的概率为_____.【答案】【解析】【分析】将试验发生包含的事件（k，b）的所有可能的结果列举，满足条件的事件直线不经过第三象限，符合条件的（k，b）有2种结果，根据古典概型概率公式得到结果．【详解】试验发生包含的事件（k，b）的取值所有可能的结果有：（﹣1，﹣2）；（﹣1，1）；（﹣1，2）；（1，﹣2）；（1，1）；（1，2）；（2，﹣2）；（2，1）；（2，2）共9种结果．而当时，直线不经过第三象限，符合条件的（k，b）有2种结果，∴直线不过第三象限的概率P，故答案为．【点睛】本题考查古典概型，古典概型要求能够列举出所有事件和发生事件的个数，属于基础题．7.设，是双曲线的两个焦点，是双曲线上的一点，且，则的周长为___. 【答案】24【解析】【分析】先由双曲线的方程求出|F1F2|＝10，再由3|PF1|＝4|PF2|，运用双曲线的定义，求出|PF1|＝8，|PF2|＝6，由此能求出△PF1F2的周长．【详解】双曲线x21的a＝1，c5，两个焦点F1（﹣5，0），F2（5，0），即|F1F2|＝10，由3|PF1|＝4|PF2|，设|PF2|＝x，则|PF1|x，由双曲线的定义知，x﹣x＝2，解得x＝6．∴|PF1|＝8，|PF2|＝6，|F1F2|＝10，则△PF1F2的周长为|PF1|+|PF2|+|F1F2|＝8+6+10＝24．故答案为：24．【点睛】本题考查双曲线的定义和性质的应用，考查三角形周长的计算，熟练运用定义是关键，属于基础题．8.已知四面体中，，，分别为，的中点，且异面直线与所成的角为，则____. 【答案】1或【解析】【分析】取BD中点O，连结EO、FO，推导出EO＝FO＝1，，或，由此能求出EF．【详解】取BD中点O，连结EO、FO，∵四面体ABCD中，AB＝CD＝2，E、F分别为BC、AD的中点，且异面直线AB与CD所成的角为，∴EO∥CD，且EO，FO∥AB，且FO1，∴∠EOF是异面直线AB与CD所成的角或其补角，∴，或，当∠EOF时，△EOF是等边三角形，∴EF＝1．当时，EF．故答案为：1或．【点睛】本题考查异面直线所成角的应用，注意做平行线找到角是关键，解题时要认真审题，注意空间思维能力的培养，是易错题9.已知函数是定义在上的奇函数，当时，，则时，不等式的解集为____. 【答案】【解析】【分析】由奇函数的性质可得x＞0时的解析式，再解不等式即可．【详解】∵函数f（x）是定义在R上的奇函数，∴当x＞0时，﹣x＜0，∴f（﹣x）＝x2﹣6，由奇函数可得f（x）＝﹣x2+6，∴不等式f（x）＜x可化为，解得x＞2∴x＞0时，不等式f（x）＜x的解集为：（2，+∞）故答案为：（2，+∞）【点睛】本题考查函数的奇偶性，涉及不等式的解法，熟记奇函数得定义是关键，属基础题．10.关于的方程在上的解的个数是____.【答案】7【解析】【分析】化简y=从而作函数的图像，从而可解【详解】化简y=,作函数在上的图像如下：结合图像可知，两个图像共有7 个交点故答案为7【点睛】本题考查函数与方程，函数的性质，三角函数，准确作图是关键，是中档题11.任意实数，，定义，设函数，数列是公比大于0的等比数列，且，，则____.【答案】4【解析】【分析】f（x）＝，及其数列{a n}是公比大于0的等比数列，且＝1，对公比q分类讨论，再利用对数的运算性质即可得出．【详解】由题，∵数列{a n}是公比大于0的等比数列，且，①1＜q时，，，…，∈（0，1），，，∈（1，+∞），1．∴，分别为：，，…，，1，q，…，q4．∵∴0++…+＝，∴q4q q2．∴2．左边小于0，右边大于0，不成立，舍去．②0＜q＜1时，1，∴，分别为：，，…，，1，q，…，q4，，，…，∈（1，+∞），，，∈（0，1），∵∴log2q2．∴2．∴4，∴a1＝4．③q＝1时，＝…＝＝…＝＝1，不满足舍去．综上可得：＝4．故答案为：4．【点睛】本题考查了等比数列的通项公式及其性质、对数的运算性质，考查了分类讨论方法、推理能力与计算能力，属于难题．12.以正方形的四个顶点分别作为椭圆的两个焦点和短轴的两个端点，，，是椭圆上的任意三点（异于椭圆顶点），若存在锐角，使，（0为坐标原点）则直线，的斜率乘积为___. 【答案】或-2【解析】设椭圆方程为，A（，），B（，），从而得到的坐标表示，然后，再根据M点在该椭圆上，建立关系式，结合A、B点在也该椭圆上，得到，，从而得到相应的结果，同理当椭圆方程为可得答案【详解】由题意可设椭圆方程为，又设A（，），B（，），因为M点在该椭圆上，∴，则又因为A、B点在也该椭圆上，∴，∴，即直线OA、OB的斜率乘积为，同理当椭圆方程为时直线OA、OB的斜率乘积为﹣2．故答案为：或﹣2．【点睛】本题重点考查椭圆综合，平面向量的坐标运算，注意审题仔细，要注意分类讨论椭圆的焦点位置，属于中档题．二、选择题。

2019届上海市南洋模范中学三模数学试题(解析版)

∵tan∠A1CC1＝＝＝，
∴ ＝．
∴异面直线A1C与DD1所成角的大小为；
【点睛】
本题考查三棱锥的体积的求法，考查异面直线所成角的求法，是中档题，解题时要认真审题，注空间思维能力的培养．
18．已知函数．
（1）若不等式的解集为，求a的值；
（2）在（1）的条件下，若存在，使，求t的取值范围．
可得
，代入曲线方程，可得
2x02﹣y02＝1，即为2（2x）2﹣（2y）2＝1，
即为8x2﹣4y2＝1．
故答案为：8x2﹣4y2＝1．
【点睛】
本题考查中点的轨迹方程的求法，注意运用代入法和中点坐标公式，考查参数方程和普通方程的互化，注意运用同角的平方关系，考查运算能力，属于中档题．
12．在等差数列中，首项，公差，若某学生对其中连续10项进行求和，在遗漏掉一项的情况下，求得余下9项的和为185，则此连续10项的和为．
14．已知数列的通项公式为，则这个数列的前n项和 _____．
【答案】
【解析】分n为奇数、偶数两种情况讨论，利用分组求和法计算即得结论．
【详解】
当n为偶数时，Sn＝[（﹣1+2）+（﹣3+4）+…+（﹣n+1+n）]+（2+22+…+2n）
＝
＝2n+1+ ﹣2；
当n为奇数时，Sn＝[（﹣1+2）+（﹣3+4）+…+（﹣n+2+n﹣1）﹣n]+（2+22+…+2n）
不等式f（x）＜6的解集为（﹣1，3），
∴|2x﹣a|＜6﹣a的解集为（﹣1，3），

2018-2019年上海市南洋模范中学高三下三模教师版

南模中学高三三模数学试卷2019.05一. 填空题1. 若集合{|310}A x x =+>，{||1|2}B x x =-<，则A B =I 1(,3)3-2. 若复数z 满足1ii z -=-，其中i 为虚数单位，则z = 1i - 3. 若函数1()1f x x =+（0x >）的反函数为1()f x -，则不等式1()2f x ->的解集为 3(1,)24. 试写出71()x x -展开式中系数最大的项 35x5.若函数4y =a ，最大值为b ，则2lim 34n n n n n a b a b →∞-=- 126. 已知平面上三点A 、B 、C满足||AB =u u u r，||BC =u u u r||CA =u u u r，则AB BC BC CA CA AB ⋅+⋅+⋅u u u r u u u r u u u r u u u r u u u r u u u r的值等于 8-由|4B ||Bi ||CA u u u r|＝222AB BC CA +=u u u r u u u r u u u r即有△ABC 为直角三角形，由0AB BC CA ++=u u u r u u u r u u u r r两边平方可得，222ABBC BCCA CAAB 2()0AB BC CA +++++=u u u r u u u u u u r u u u r u u u r r u u u r u u u r u u u r u u u r即有AB BC BC CA CA AB ⋅+⋅+⋅u u u r u u u r u u u r u u u r u u u r u u u r 222||+||+||1=-2AB BC CA u u u r u u u r u u u r （）＝﹣12×（3+5+8）＝﹣8．7. 设P是曲线tan x y θθ⎧=⎪⎨⎪=⎩（θ为参数）上的一动点，O 为坐标原点，M 为线段OP 的中点，则点M 的轨迹的普通方程为 22841x y -=8. 在等差数列{}n a 中，首项13a =，公差2d =，若某学生对其中连续10项进行求和，在遗漏掉一项的情况下，求得余下9项的和为185，则此连续10项的和为 200等差数列{}n a 中的连续10项为*+129,,,,,()x x x x a a a a x N ++⋯∈，遗漏的项为*+,x n a n N ∈且19,n ≤≤则9()10(18)10(2)22x x x x x n x a a a a a a n +++⨯++⨯-=-+，化简得4494352x n ≤=+≤，所以5x =，511a =，则连续10项的和为(1111+18)10=2002+⨯.9. 从集合{1,2,3,4,5,6,7,8,9,10}A =中任取两个数，欲使取到的一个数大k 于，另一个数小于k （k A ∈）的概率为25，则k = 4或7 从集合{}1,2,3,4,5,6,7,8,9,10中任取两个数的基本事件有21045C =种，取到的一个数大于k ，另一个数小于k ，比k 小的数有1k -个，比k 大的数有10k -个，故一共有11110(1)(10)k k C C k k --⋅=--个基本事件，由题意可得(1)(10)2455k k --=，即(1)(10)18k k --=，整理得211280k k -+=，解得4k =或7k =，10. 已知数列{}n a 的通项公式为(1)2n n n a n =-⋅+（*n ∈N ），则这个数列的前n 项和为n S = 111222122222n n n n n k S n n k ++--⎧+-=-⎪⎪=⎨⎪+-=⎪⎩，*k ∈N当n 为偶数时，S n ＝[（﹣1+2）+（﹣3+4）+…+（﹣n+1+n ）]+（2+22+…+2n ）＝()212212n n --＝2n+1+2n﹣2；当n 为奇数时，S n ＝[（﹣1+2）+（﹣3+4）+…+（﹣n+2+n ﹣1）﹣n]+（2+22+…+2n ）＝12n -﹣n+()21212n --＝2n+1﹣2n ﹣52；11. 已知函数1()f x x x=-，数列{}n a 是公比大于0的等比数列，且61a =， 1239101()()()()()f a f a f a f a f a a +++⋅⋅⋅++=-，则1a =2设数列{}n a 的公比为0q >，则1n a ⎧⎫⎨⎬⎩⎭是首项为11a ，公比为1q 的等比数列，由()()()()()1239101f a f a f a f a f a a +++⋅⋅⋅++=-得121011210111a a a a a a a ⎛⎫+++-+++=- ⎪⎝⎭L L ，即()10101111111111a q a q a q q⎛⎫-⎪-⎝⎭-=---△，由61a =，得511a q =△，联立△△解得1a =. 12. 定义在R 上的奇函数()f x ，当0x ≥时，12log (1)[0,1)()1|3|[1,)x x f x x x +∈⎧⎪=⎨⎪--∈+∞⎩，则关于x 的函数()()F x f x a =-（01a <<）的所有零点之和为（结果用a 表示）12n -△当x≥0时，f （x ）＝12log (1),[0,1)13,[1,)x x x x +∈⎧⎪⎨⎪--∈+∞⎩即x△[0，1）时，f （x ）＝12log （x+1）△（﹣1，0]；x△[1，3]时，f （x ）＝x ﹣2△[﹣1，1]；x△（3，+∞）时，f （x ）＝4﹣x△（﹣∞，﹣1）；画出x≥0时f （x ）的图象，再利用奇函数的对称性，画出x ＜0时f （x ）的图象，如图所示；则直线y ＝a ，与y ＝f （x ）的图象有5个交点，则方程f （x ）﹣a ＝0共有五个实根，最左边两根之和为﹣6，最右边两根之和为6， △x△（﹣1，0）时，﹣x△（0，1），△f （﹣x ）＝12log （﹣x+1），又f （﹣x ）＝﹣f （x ），△f （x ）＝﹣12log （﹣x+1）＝12log （1﹣x ）﹣1＝log 2（1﹣x ），△中间的一个根满足log 2（1﹣x ）＝a ，即1﹣x ＝2a ，解得x ＝1﹣2a ， △所有根的和为1﹣2a ．二. 选择题13. 已知非零向量a r 、b r ，“函数2()()f x ax b =+r r 为偶函数”是“a b ⊥r r ”的（ C ）条件A. 充分非必要B. 必要非充分C. 充要D. 既不充分也不必要因为()2222()2f x ax b a x a bx b⋅r r r rr r ＝＋＝＋＋，所以若()2()f x ax b +r r ＝为偶数，则0a b ⋅rr ＝，即a b ⊥r r .若a b ⊥r r ，则有0a b ⋅rr ＝，所以()2222222()2f x ax b a x a bx b a x b⋅=r r r r rr r r ＝＋＝＋＋＋，为偶函数．14. 若a 、b 表示两条直线，α表示平面，下列命题中的真命题为（ C ） A. 若a α⊥，a b ⊥，则b ∥α B. 若a ∥α，a b ⊥，则b α⊥ C. 若a α⊥，b α⊆，则a b ⊥ D. 若a ∥α，b ∥α，则a ∥b 15. 抛物线24y x =的焦点为F ，点(,)P x y 为抛物线上的动点，又点(1,0)A -，则||||PF PA 的最小值是（ B ） A.12B. 22C. 3D. 33由题意可知，抛物线的准线方程为x=﹣1，A （﹣1，0），过P 作PN 垂直直线x=﹣1于N ，由抛物线的定义可知PF=PN ，连结PA ，当PA 是抛物线的切线时，PF PA有最小值，则△APN 最大，即△PAF 最大，就是直线PA 的斜率最大，设在PA 的方程为：y=k （x+1），所以214y k x y x（）=+⎧⎨=⎩，解得：k 2x 2+（2k 2﹣4）x+k 2=0，所以△=（2k 2﹣4）2﹣4k 4=0，解得k=±1，所以△NPA=45°，PF PA=cos△NPA=2． 16. 已知1x 、2x 是关于x 的方程220x mx m m ++-=的两个不相等的实数根，则经过两点211(,)A x x 、222(,)B x x 的直线与圆22(1)(1)1x y -++=的位置关系是（ C ） A. 相离 B. 相切 C. 相交 D. 随m 的变化而变化22212121,ABx x k x x x x -==+∴-Q 直线AB 的方程为21121()()y x x x x x -=+-. 即1212()y x x x x x =+-，所以直线AB 的方程为2222242,1111m m m y mx m m d m m m m -+-=-+-===+++, 因为2240,4()0,03m m m m ∆>∴-->∴<<,所以221999225,(),(,),()()161616256t g t t t t g t g m =>∴=+∈+∞>=令, 所以1615()225256d g t =<=,所以直线AB 与圆可能相交，也可能相切，也可能相离. 三. 解答题17. 如图，长方体1111ABCD A B C D -中，2AB BC ==，13AA =. （1）求四棱锥1A ABCD -的体积；（2）求异面直线1A C 与1DD 所成角的大小.【解】（1）△长方体ABCD ﹣A 1B 1C 1D 1中，AB ＝BC ＝2，AA 1＝3，△四棱锥A1﹣ABCD 的体积：1A ABCD V -11=3ABCD S AA ⨯矩形＝113AB AD AA ⨯⨯⨯＝12233⨯⨯⨯＝4．（2）△DD 1△CC 1，△△A 1CC 1是异面直线A 1C 与DD 1所成角（或所成角补角），△tan△A 1CC 1＝111AC CC＝22223+＝22，△11A CC ∠＝22arctan3． △异面直线A 1C 与DD 1所成角的大小为22arctan3； 18. 已知函数()|2|f x x a a =-+.（1）若不等式()6f x <的解集为(1,3)-，求a 的值；（2）在（1）的条件下，若存在0x ∈R ，使00()()f x t f x ≤--，求t 的取值范围. 【解】（1）△函数f （x ）＝|2x ﹣a|+a ，不等式f （x ）＜6的解集为（﹣1，3），△|2x ﹣a|＜6﹣a 的解集为（﹣1，3），由|2x ﹣a|＜6﹣a ，可得a ﹣6＜2x+a ＜6﹣a ，求得a ﹣3≤x≤3，故有a ﹣3＝﹣1，a ＝2．（2）在（1）的条件下，f （x ）＝|2x ﹣2|+2，令g （x ）＝f （x ）+f （﹣x ）＝|2x ﹣2|+|2x+2|+4＝44,18,1144,1x x x x x --⎧⎪-<<⎨⎪+⎩„… 故g （x ）的最小值为8，故使f （x ）≤t ﹣f （﹣x ）有解的实数t 的范围为[8，+∞）．19. 某景区欲建两条圆形观景步道1M 、2M （宽度忽略不计），如图所示，已知AB AC ⊥，60AB AC AD ===（单位：米），要求圆1M 与AB 、AD 分别相切于点B 、D ，圆2M与AC 、AD 分别相切于点C 、D . （1）若3BAD π∠=，求圆1M 、2M 的半径（结果精确到0.1米）；（2）若观景步道1M 、2M 的造价分别为每米0.8千元与每米0.9千米，则当BAD ∠多大时，总造价最低？最低总造价是多少？（结果分别精确到0.1°和0.1千元）【解】（1）连结M 1M 2，AM 1，AM 2，△圆M 1与AB ，AD 相切于B ，D ，圆M 2与AC ，AD 分别相切于点C ，D ， △M 1，M 2△AD ，△M 1AD ＝12△BAD ＝6π，△M 2AD ＝12π，△M1B ＝ABtan△M1AB ＝（米）， △tan6π＝22tan121tan12ππ-△tan 12π＝2，同理可得：M 2D ＝60×tan 12π＝60（2）≈16.1（米）．（2）设△BAD ＝2α（0＜α＜4π），由（1）可知圆M 1的半径为60tanα，圆M 2的半径为60tan （45°﹣α），设观景步道总造价为y 千元，则y ＝0.8•2π•60tanα+0.9•2π•60tan （45°﹣α）＝96πtanα+108π•1tan 1tan αα-+，设1+tanα＝x ，则tanα＝x ﹣1，且1＜x ＜2． △y ＝96π（x ﹣1）+108π（21x -）＝12π•（8x+18x﹣17）≥84π≈263.8，当且仅当8x ＝18x 即x ＝32时取等号，当x ＝32时，tanα＝12，△α≈26.6°，2α≈53.2°．△当△BAD 为53.2°时，观景步道造价最低，最低造价为263.8千元．20. 已知椭圆2222:1x y C a b +=（0a b >>）的右焦点为(1,0)F ，且点3(1,)2P 在椭圆C 上.（1）求椭圆C 的标准方程；（2）过椭圆22122:153x y C a b +=-上异于其顶点的任意一点Q 作圆224:3O x y +=的两条切线，切点分别为M 、N （M 、N 不在坐标轴上），若直线MN 在x 轴、y 轴上的截距分别为m 、n ，证明：22113m n +为定值；（3）若1P 、2P 是椭圆222223:1x y C a b+=上不同的两点，12PP x ⊥轴，圆E 过1P 、2P ，且椭圆2C 上任意一点都不在圆E 内，则称圆E 为该椭圆的一个内圆，试问：椭圆2C 是否存在过左焦点1F 的内切圆？若存在，求出圆心E 的坐标，若不存在，请说明理由. 【解】（1）△椭圆C 的右焦点为F （1，0），且点P （1,32）在椭圆C 上； △2222211914c a b a b c=⎧⎪⎪+=⎨⎪=+⎪⎩，解得a ＝2，b△椭圆C 的标准方程为22143x y +=．（2）由题意：C 1：223144x y +=，设点P （x 1，y 1），M （x 2，y 2），N （x 3，y 3）， △M ，N 不在坐标轴上，△k PM ＝﹣OM 1k ＝﹣22x y ，△直线PM 的方程为y ﹣y 2＝﹣22x y （x ﹣x 2），化简得：x 2x+y 2y ＝43，△，同理可得直线PN 的方程为x 3x+y 3y ＝43，△，把P 点的坐标代入△、△得212131314343x x y y x x y y ⎧+=⎪⎪⎨⎪+=⎪⎩，△直线MN 的方程为x 1x+y 1y ＝43，令y ＝0，得m ＝143x ，令x ＝0得n ＝143y ，△x 1＝43m ，y 1＝43n，又点P 在椭圆C 1上，△（43m ）2+3（43n）2＝4，则2211+3m n ＝34为定值．（3）由椭圆的对称性，可以设P 1（m ，n ），P 2（m ，﹣n ），点E 在x 轴上，设点E （t ，0），则圆E 的方程为：（x ﹣t ）2+y 2＝（m ﹣t ）2+n 2，由内切圆定义知道，椭圆上的点到点E 距离的最小值是|P 1E|，设点M （x ，y ）是椭圆C 上任意一点，则|ME|2＝（x ﹣t ）2+y 2＝223214x tx t -++，当x ＝m 时，|ME|2最小，△m ＝﹣2433t t-=，△，又圆E 过点F ，△t ）2＝（m ﹣t ）2+n 2，△点P 1在椭圆上，△2214m n =-，△由△△△，解得：tt，又tm＝﹣3＜﹣2，不合题意，综上：椭圆C 存在符合条件的内切圆，点E的坐标是（﹣，0）．21. 若{}n c 是递增数列，数列{}n a 满足：对任意*n ∈N ，存在*m ∈N ，使得10m nm n a c a c +-≤-，则称{}n a 是{}n c 的“分隔数列”.（1）设2n c n =，1n a n =+，证明：数列{}n a 是{}n c 的分隔数列；（2）设4n c n =-，n S 是{}n c 的前n 项和，32n n d c -=，判断数列{}n S 是否是数列{}n d 的分隔数列，并说明理由；（3）设1n n c aq -=，n T 是{}n c 的前n 项和，若数列{}n T 是{}n c 的分隔数列，求实数a 、q 的取值范围.【解】（1）△{c n }是递增数列，数列{a n }满足：对任意n△N *，存在m△N *，使得10m nm n a c a c +--„， △c n ≤a m ＜c n+1，△c n ＝2n ，a m ＝m+1，△2n≤m+1＜2n+2，△2n ﹣1＜m≤2n+1，△m ＝2n ，△对任意n△N *，存在m ＝2n△N *，使得10m nm n a c a c +--„，则称{a n }是{c n }的分隔数列；（2）c n ＝n ﹣4，S n 是{c n }的前n 项和，d n ＝c 3n ﹣2，△d n ＝（3n ﹣2）﹣4＝3n ﹣6，△d 1＝﹣3， △S n ＝(34)2n n -+-＝12n （n ﹣7），若数列{S n }是数列{d n }的分隔数列，△3n ﹣6≤12m （m ﹣7）＜3n ﹣3，即6（n ﹣2）≤m （m ﹣7）＜6（n ﹣1），由于n ＝4时，12≤m （m ﹣7）＜18，不存在自然数m ，使得不等式成立， △数列{S n }不是数列{d n }的分隔数列；（3）设1n n c aq -=，T n 是{c n }的前n 项和，△数列{T n }是{c n }的分隔数列，则{c n }为递增，当a ＞0时，q ＞1，△aq n ﹣1≤()11ma q q--＜aq n，即有q m ﹣1＜q n （q ﹣1），且q m ﹣1≥q n ﹣1（q ﹣1），当1＜q ＜2时，数列最小项可以得到m 不存在； q ＞2时，由m ＝n ，q m ﹣1≥q n ﹣1（q ﹣1）成立；q n ﹣1＜q n （q ﹣1）成立，可得n ＝2时，q 2﹣1＜q 2（q ﹣1），解得q ＞2，对n ＞3也成立；当a＜0时，0＜q＜1时，aq n﹣1≤()11ma qq--＜aq n，即有1﹣q m＞q n（1﹣q），且1﹣q m≤q n﹣1（1﹣q），取m＝n+1，可得1﹣q m＞q n（1﹣q）成立，1﹣q n+1≤q n﹣1（1﹣q）成立，可得q＝0恒成立，则a＜0，0＜q＜1不成立，综上可得，a＞0，q＞2．。

2019年上海市高三高考仿真模拟考试英语试卷(三)及解析

2019 年上海市高三高考仿真模拟考试英语试卷（三）★祝考试顺利★I. Listening ComprehensionSection A (10%)Directions : In Section A, you will hear ten short conversations betweentwo speakers. At the end of each conversation，a question will be asked about -what was said. The conversations and the questions will be spokenonly once. After you hear a conversation and the question about it，read the four possible answers on your paper，and decide which one is the best answer to the question you have heard.1.A. A waitress.B. An actress.C. A policewoman.D. A driver.【答案】 C2.A. In a furniture store.B. In a restaurant.C. At a bank.D. In a factory.【答案】 B3.A. The good weather tomorrow.B. The vacation plan.C. Getting up early in the morning.D. The possible heavy traffic.【答案】 B4.A. 100 dollars.B. 200 dollars.C. 300 dollars.D. 400 dollars.【答案】 A5.A. She works in a very nice restaurant.B. Restaurant food is dirty and has made her ill.C. Her job involves a lot of business dinners.D. She feels tired when preparing meals at home.【答案】 C6.A. She is at home now.C. She lost her hotel card.【答案】 B7.A. He is asking for leave.C. He is quitting his job. money.【答案】 C8.B. She is a tourist there.D. She doesn ’t know the man.B. He is asking for reasons. D. He is investing time andA. She would rather stay at home listening to CDs.B. She likes to listen to new music CDs.C. She wants the man to buy CDs for her.D. She can get new music CDs for free.【答案】 B9.A. Next Tuesday.B. Next Wednesday.C. Next Thursday.D. NextFriday.【答案】 A10.A. China.B. France.C. Italy.D. Japan.【答案】 ASection B (15%)Directions: In Section B, you will hear several longer conversation (s)and short passage (s), and you will be asked several questions on eachof the conversation (s) and the passage (s). The conversation(s) and thepassage (s) will be read twice ， but the questions will be spoken only once. Whenyou hear a question ，read the four possible answers on your paper anddecide which one is the best answer to the question you have heard.Questions are based on the following passage.11..A. A rental apartment usually has a kitchen and a bathroom.B. Rental apartments have no more than two bedrooms.C. One-room apartments are too small to have a bathroom.D. A living room is essential for all the rental apartments12.A. According to the number of bedrooms.B. The prices of all houses or apartments are fixed.C. Based on the location and the quality.D. The owners decide the prices themselves.13.A. Security deposit is equal to two months’ rent.B. It costs you a month’s rent to repair the damage.C. You are required by law to pay rent for a full year.D. You can renew the agreement by reducing the rent.【答案】 11. A 12. C13. AQuestions are based on the following passage.14.A. By staying alert to see something happening in nature.B. By possessing two essential characteristics of great physicists.C. By totally understanding the tricky make-up of the world.D. By succeeding in transferring the skills into the wider social area.15.A. His contributions have advanced progress in the society.B. He suggested and created many modern power systems.C. He laid out important principles in his scientific works.D. His contributed to helping people evolve and survive.16.A. Develop the observation of natural phenomena.B. Help humankind to survive in the world.C. Drive important changes to the society.D. Inspire many other great physicists.【答案】 14. D 15. B16. BQuestions are based on the following passage.17. A. Draw. B. Fold paper. C. Sing. D. Tell stories.18.A. At the age of four.B. At the age of five.C. At the age of six.D. At the age of seven.【答案】 17. B18. BQuestions are based on the following passage.19.A. His college life is too boring.B. Some principles of science are too difficult for him to understand.C. Students can’t quite follow his teaching.D. He can ’t concentrate when talking about some basics.20.A. Freshmen are usually highly motivated.B. Jane White teaches chemistry in college.C. Susan is popular with her students.D. The man will try something new.【答案】 19. C20. DII.Grammar andVocabulary Section A (10%)Directions : After reading the passage below, fill in the blanks to makethe passage coherent and grammatically correct.For the blanks with, a given word， fill in each blank with the proper form of the given word ； for the other blanks ，use one word that best fits each blank.What is an insect? Insects of all kinds have existed on earth formillions of years. Sometimes we think of all insects as pests.Humanlife, however,would have a difficult time continuing without insects,___21___ they pollinate plants and are a food source for many animals. Besides,insects, especially the social ___22___，are fascinating.One very interesting and useful social insect is the honey bee. Bees___23___ (consider) social insects because they live and work together. Thebee lives in a beehive with other worker bees, and with the queen, who canlay thousands of eggs a day. The worker bees make wax used ___24___(create) new cells inside the hive. Humans use this valuable beeswax___25___ candles ，cosmetics,expensive furniture polish,and other products. They also produce honey, a valuable food for humans (as well as for bees).Another fascinating social insect is ___26___ant. Many types of ants live in colonies underground. There thousands of them build and maintaintheir“city ”. They also cooperate to provide food,sometimes in a manner that looks like human. For example,some ants actually raise their own food. One type of ant“farmer ” keeps and cares for tiny insects that suck sweet sap ( 汁液 )from plants.Ant farmers milk them from the sap,just like actual farmers milk cows. The sap is then given as food to young ants and to thequeen, ___27___main duty is to lay eggs. Another type of ant that produces its own food does so by gardening. These farmer ants take pieces of leaves into the colony and chew them up. The soft ， ___28___ (chew)up leaves then sprout (涌出 )a fungus that the worker ants use to feed the queen and young ants.There are many other ways that ants cooperate to maintain and protect their colonies.There are also many other kinds of social insects. ___29___makes them all so fascinating is their organization. ___30___each insect has certain tasks to perform,the entire hive or colony appears to function as one living organism.【答案】 21. because22. ones23. are considered24.to create25.in 26. the27. whose28. chewed29. What30. Although【解析】【分析】这是一篇说明文。

2019-2020学年上海市上南中学高三英语三模试题及答案

2019-2020学年上海市上南中学高三英语三模试题及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AAs the MOOC craze continues to explode, anyone interested in taking an online course faces a tricky question: Which course to take? Here are five aspects that you should consider before you start.(1)What is your learning style?Many MOOCs are video-based. Other courses use presentation formats. Some also require participation in group work. If you want to stay motivated during your course, think about how you enjoy learning.Are you a visual learner, preferring to use images to understand a topic? If so, a video-based course will work well for you. If you are a verbal learner who gains new information by speaking and writing, try a text-based course with lots of note-taking. Social learners, meanwhile, will thrive in forum discussions and project-based assessment.(2)Are you ready to become a full-time student?Be realistic about the time that you can commit to your online studies. Participating in an online course can take as much time and commitment as a class-based program. Check the course requirements and make a plan around your current schedule.(3) Does the course really meet your needs?Whether you are interested in a professional qualification or want to take a personal development course like yoga, there is a MOOC for you. It's easy to get caught up in the excitement of taking lots of free courses in everything that you ever wanted to learn. Before you start a course, think about the end goal. Is the course aimed at beginners or advanced learners? Why do you need this qualification?(4) Do you need a support group?Some people learn best from seeing how others approach the problem. If you are such as learner, you will need to supplement your online lessons with an in-person support group.(5) What kind of certification will you get?Take time to find out what kind of certification is available upon completion of the course, and how you can prove your learning to others - for example, certificates, transcripts or digital badges.1. What kind of MOOC's does the author recommend to verbal learners?A. A video-based course.B. A text-based course.C. A forum-based discussion.D. A project-based assessment.2. What kind of learners need an in-person support group?A. Learners who prefer individual work.B. Learners who are in great need of a certificate.C. Learners who are too busy to become a full-time student.D. Learners who learn best from observing how others address the problem.3. What is the passage mainly about?A. Picking the right MOOC's for you.B. Deciding your learning style.C. Taking the right course you need.D. Choosing a suitable support group.BIn the U. S., speaking more than one language fluently is not very common except in Los Angeles, California.The city has one of the largest population in the U. S. of young people between the ages of 18 and 34. This generation is often called millennials (新千年一代). More than half of millennials in Los Angeles are bilingual (双语的), which means they speak more than one language.Maria Elena Burgos is cooking a Mexican breakfast. She says making Mexican food is just one of the many traditions in her home. Another is speaking Spanish to her children.“We want them to be bilingual. We want to keep the Spanish somewhere in their learning too, not only at home.”When Ms Burgos first came to the United States from Mexico, she learned English. She knew her children would learn English quickly. So she wanted them to speak Spanish at home and study the language at school.She says being bilingual will give them more opportunities in the future. Knowing Spanish also means the children can talk with their s in Mexico.“When we had our children, one of the decisions we as parents made was to name them with a name that was easily pronounced in English and Spanish.”Elizabeth wants to know her family’s culture.” “The culture-to go back to our roots because that’s part of who we are.”Monica wants to pass on the culture to her children“It’s nice to know our culture and then to be able to pass it onto our children and grand-children and everyone to let them know where we come from.”And, Monica says she does not speak only English and Spanish. She has even learned some Korean in school.4. Which of the following is a tradition of Maria?A. Cooking breakfast for her s.B. Talking with her children in Spanish.C. Talking with her s in Spanish.D. Forcing her children to speak Spanish.5. What can be inferred from Monica’s words?A. She doesn’t care where she comes from.B. She looks forward to learning Spanish.C. Everyone should know his own culture.D. She doesn’t like American culture.6. Which of the following is mentioned about children speaking more languages?A. It can give children more chances in the future.B. It can make children feel proud before their friends.C. It can give the children a chance to go to a good college.D. It can let children go to the places where they want to go.7. What’s the main purpose of this passage?A. To entertain the readers with a funny story.B. To inform the readers of a shocking experiment result.C. To encourage more persons to learn a second language.D. To tell a truth that many people in Los Angeles are bilingual.CHave you ever thrown a camel? Camels are large and heavy animals so it would be hard to throw. But in the French-speaking Democratic Republic ofthe Congo, “to throw a camel” is a way of saying “to make a spelling mistake”.In the past, a phrase like that was not accepted by the French government as an official French term. But recently, the French Ministry of Culture worked on a new kind of dictionary that accepts the idea that many people outside of France speak the language. The language has changed over time and is different in places like Ivory Coast in West Africa or Quebec in Cana-da, compared to how it is in ParisA new online dictionary, called the French speakers’ dictionary, includes new French words from around theworld. It was released on March 18—just in time for International French Speaker’s Day on March 20.Supporters say the new Internet dictionary is more democratic than earlier French dictionaries that only showed the way highly educated French people spoke. The new dictionary includes unofficial words like “pourriel”, which means an unwanted email if you are in Canada. It is a word based on “courriel”, which just means an email.French President Emmanuel Macron proposed the idea of the dictionary in 2018. It now contains about 600.000 terms, Roselyne Bachelot is the French Culture Minister. She said the dictionary is not just for France's 67 million citizens, but for the 300 million French speakers worldwide.The aim, supporters say, is to recognize the way language changes. Words and expressions included in the dictionary come from over 50 countries—even from the United States Some people in the southern U. S. state of Louisiana speak French.People can see the dictionary on a website or with an app. Users can also send in new words they think should be included. Official dictionaries produced by the French Academy in Paris were first published hundreds of years ago and are regularly updated.8. What do we mean if we say a man “throws a camel” in French?A. He talks big.B. He is very strong.C. He makes a joke.D. He misspells a word.9. What is the purpose of the publication date of the new online dictionary?A. To celebrate an event.B. To honor foreign users.C. To draw more attention.D. To meet an urgent need.10. Where does the word “pourriel” come from?A. French.B. An email.C. A holiday.D. Canada.11. What can people do with the new dictionary?A. Update it regularly.B. Recommend new words to it.C. Change the meaning of its words.D. Regard it as an official dictionary.DA new study has discovered that meditation (冥想) and oxygen sport together reduce depression. The Rutgers University study found that this mind and body combination, done twice a week for only two months,reduced the symptoms for a group of students by 40 percent.“We are excited by the findings because we saw such a meaningful improvement in both clinically depressed and non-depressed students,” said lead author Dr. Brandon Alderman. “It is the first time that both of these twobehavioral ways have been looked at together for dealing with depression.”Researchers believe the two activities have an interactive effect on combatingdepression. Alderman and Dr. Tracey Shors discovered that a combination of mental and physical training (MAP) enabled students with major depressive disorder not to let problems or negative thoughts defeat them.Rutgers researchers say those who participated in the study began with 30 minutes of focused attention meditation followed by 30 minutes of oxygen sport. They were told that if their thoughts drifted to the past or the future they should refocus on their breathing, enabling those with depression to accept moment-to-moment changes in attention.Shors, who studies the productionof new brain cells in the hippocampus—part of the brain involved in memory and learning—says scientists have shown in animal models that oxygen sport exercise keeps a large number of certain cells alive.The idea for the human intervention (干预) came fromher laboratory studies, she says, with the main goal of helping individuals acquire new skills so that they can learn to recover from stressful life events.By learning to focus their attention and exercise, people who are fighting depression can acquire new learning skills that can help them process information and reduce the overwhelming recollection of memories from the past, Shors says.“We know these treatments can be practiced over a lifetime and that they will be effective in improving mental health.” said Alderman. “The good news is that this intervention can be practiced by anyone at any time and at no cost.”12. What made the research so different?A. Adopting a way of meaningful talk.B. Combining the two behavioral ways to treat depression.C. Treating depression with special medicine.D. Comparing the depressed with the non-depressed.13. The underlined word “combating” in Paragraph 3 can be replaced by ______.A. fightingB. identifyingC. distinguishingD. examining14. What did the participants do in the research?A. They did oxygen sport half an hour before thinking.B. They thought quietly and then took exercise.C. They took exercise longer than they thought.D. They took exercise while thinking quietly.15. What is Shors’ main purpose of her studies?A. To find out certain brain cells of humans.B. To study the production of new brain cells.C. To offer people a new method to treat stress.D. To decide the links between stress and exercise.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。

上海市南洋模范中学2018-2019学年高三上学期第三次月考试卷数学含答案

上海市南洋模范中学2018-2019学年高三上学期第三次月考试卷数学含答案班级__________ 座号_____ 姓名__________ 分数__________一、选择题（本大题共12小题，每小题5分，共60分.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e 为自然对数的底数，若对任意的1[,1]x e∈，总存在唯一的[1,1]y ∈-，使得2ln 1y x x a y e -++= 成立，则实数a 的取值范围是（）A.1[,]e eB.2(,]e eC.2(,)e +∞D.21(,)e e e+【命题意图】本题考查导数与函数的单调性，函数的最值的关系，函数与方程的关系等基础知识，意在考查运用转化与化归思想、综合分析问题与解决问题的能力．2．在数列{}n a 中，115a =，*1332()n n a a n N +=-∈，则该数列中相邻两项的乘积为负数的项是（）A ．21a 和22aB ．22a 和23aC ．23a 和24aD ．24a 和25a 3．沿一个正方体三个面的对角线截得几何体如图所示，则该几何体的侧视图为（）A ．B ．C ．D ．4．函数f （x ）＝sin （ωx ＋φ）（ω＞0，－π2≤φ≤π2）的部分图象如图所示，则φω的值为（）A.18 B ．14C.12D ．15．已知圆M 过定点)1,0(且圆心M 在抛物线y x 22=上运动，若x 轴截圆M 所得的弦为||PQ ，则弦长||PQ 等于（）A ．2B ．3C ．4D ．与点位置有关的值【命题意图】本题考查了抛物线的标准方程、圆的几何性质，对数形结合能力与逻辑推理运算能力要求较高，难度较大.6．已知抛物线C ：24y x =的焦点为F ，定点(0,2)A ，若射线FA 与抛物线C 交于点M ，与抛物线C 的准线交于点N ，则||:||MN FN 的值是（）A ．B ．C ．1:D (1 7．如图是某几何体的三视图，正视图是等腰梯形，俯视图中的曲线是两个同心的半圆组成的半圆环，侧视图是直角梯形．则该几何体表面积等于（）A ．12+B ．12+23πC ．12+24πD ．12+π8．设函数()''y f x =是()'y f x =的导数.某同学经过探究发现，任意一个三次函数()()320f x ax bx cx d a =+++≠都有对称中心()()00,x f x ，其中0x 满足()0''0f x =.已知函数()3211533212f x x x x =-+-，则1232016...2017201720172017f f f f ⎛⎫⎛⎫⎛⎫⎛⎫++++= ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭（）A ．2013B ．2014 C ．2015 D ．20161111] 9．集合{}{}2|ln 0,|9A x x B x x =≥=<,则AB =（）A ．()1,3B ．[)1,3C ．[]1,+∞D ．[],3e 10．“24x ππ-<≤”是“tan 1x ≤”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【命题意图】本题主要考查充分必要条件的概念与判定方法，正切函数的性质和图象，重点是单调性. 11．棱长为2的正方体被一个平面截去一部分后所得的几何体的三视图如图所示，则该几何体的表面积为（）A ．B ．18C ．D ．12．由直线与曲线所围成的封闭图形的面积为（）A B1C D二、填空题（本大题共4小题，每小题5分，共20分.把答案填写在横线上）13．函数()2log f x x =在点()1,2A 处切线的斜率为 ▲ ．14．1F ，2F 分别为双曲线22221x y a b-=（a ，0b >）的左、右焦点，点P 在双曲线上，满足120PF PF ⋅=，若12PF F ∆______________.【命题意图】本题考查双曲线的几何性质，直角三角形内切圆半径与外接圆半径的计算等基础知识，意在考查基本运算能力及推理能力．15．圆上的点（2，1）关于直线x+y=0的对称点仍在圆上，且圆与直线x ﹣y+1=0相交所得的弦长为，则圆的方程为．16．在（1+2x ）10的展开式中，x 2项的系数为（结果用数值表示）．三、解答题（本大共6小题，共70分。

上海市南洋模范中学2019年高三下学期数学三模试卷(简答)

南模中学高三三模数学试卷2019.05一. 填空题1. 若集合，，则{|310}A x x =+>{||1|2}B x x =-<A B = 2. 若复数满足，其中为虚数单位，则 z 1ii z -=-i z =3. 若函数（）的反函数为，则不等式的解集为1()1f x x =+0x >1()f x -1()2f x ->4. 试写出展开式中系数最大的项71(x x-5. 若函数，最大值为，则4y =a b 2lim 34n n n nn a b a b →∞-=-6.已知平面上三点、、满足，，则A BC ||AB =||BC = ||CA =的值等于AB BC BC CA CA AB ⋅+⋅+⋅7. 设是曲线（为参数）上的一动点，为坐标原点，为线段的P tan x y θθ⎧=⎪⎨⎪=⎩θO M OP 中点，则点的轨迹的普通方程为M 8. 在等差数列中，首项，公差，若某学生对其中连续10项进行求和，在{}n a 13a =2d =遗漏掉一项的情况下，求得余下9项的和为185，则此连续10项的和为9. 从集合中任取两个数，欲使取到的一个数大于，另一个数{1,2,3,4,5,6,7,8,9,10}A =k 小于（）的概率为，则 k k A ∈25k =10. 已知数列的通项公式为（），则这个数列的前项和为{}n a (1)2n n n a n =-⋅+*n ∈N nn S =11. 已知函数，数列是公比大于0的等比数列，且，1()f x x x=-{}n a 61a =，则1239101()()()()()f a f a f a f a f a a +++⋅⋅⋅++=-1a =12. 定义在上的奇函数，当时，，则关于R ()f x 0x ≥12log (1)[0,1)()1|3|[1,)x x f x x x +∈⎧⎪=⎨⎪--∈+∞⎩的x 函数（）的所有零点之和为（结果用表示）()()F x f x a =-01a <<a 二. 选择题13. 已知非零向量、，“函数为偶函数”是“”的（）条a b 2()()f x ax b =+ a b ⊥ 件A. 充分非必要B. 必要非充分C. 充要D. 既不充分也不必要14. 若、表示两条直线，表示平面，下列命题中的真命题为（）a b αA. 若，，则∥ B. 若∥，，则a α⊥a b ⊥b αa αa b ⊥b α⊥C. 若，，则D. 若∥，∥，则∥a α⊥b α⊆a b ⊥a αb αa b15. 抛物线的焦点为，点为抛物线上的动点，又点，则24y x =F (,)P x y (1,0)A -的||||PF PA 最小值是（）A.1216. 已知、是关于的方程的两个不相等的实数根，则经过两点1x 2x x 220x mx m m ++-=、的直线与圆的位置关系是（）211(,)A x x 222(,)B x x 22(1)(1)1x y -++=A. 相离 B. 相切 C. 相交 D. 随的变化而变化m 三. 解答题17. 如图，长方体中，，.1111ABCD A B C D -2AB BC ==13AA =（1）求四棱锥的体积；1A ABCD -（2）求异面直线与所成角的大小.1A C 1DD 18. 已知函数.()|2|f x x a a =-+（1）若不等式的解集为，求的值；()6f x <(1,3)-a （2）在（1）的条件下，若存在，使，求的取值范围.0x ∈R 00()()f x t f x ≤--t19. 某景区欲建两条圆形观景步道、（宽度忽略不计），如图所示，已知1M 2M ，AB AC ⊥（单位：米），要求圆与、分别相切于点、，圆60AB AC AD ===1M AB AD B D 2M 与、分别相切于点、.AC AD C D （1）若，求圆、的半径（结果精确到0.1米）；3BAD π∠=1M 2M （2）若观景步道、的造价分别为每米0.8千元与每米0.9千米，则当多大1M 2M BAD ∠时，总造价最低？最低总造价是多少？（结果分别精确到0.1°和0.1千元）20. 已知椭圆（）的右焦点为，且点在椭圆上.2222:1x y C a b+=0a b >>(1,0)F 3(1,2P C （1）求椭圆的标准方程；C （2）过椭圆上异于其顶点的任意一点作圆的两条切22122:153x y C a b +=-Q 224:3O x y +=线，切点分别为、（、不在坐标轴上），若直线在轴、轴上的截距分M N M N MN x y 别为、，证明：为定值；m n 22113m n +（3）若、是椭圆上不同的两点，轴，圆过、，且1P 2P 222223:1x y C a b+=12PP x ⊥E 1P 2P 椭圆上任意一点都不在圆内，则称圆为该椭圆的一个内圆，试问：椭圆是否存在2C E E 2C 过左焦点的内切圆？若存在，求出圆心的坐标，若不存在，请说明理由.1F E21. 若是递增数列，数列满足：对任意，存在，使得，{}n c {}n a *n ∈N *m ∈N 10m nm n a c a c +-≤-则称是的“分隔数列”.{}n a {}n c （1）设，，证明：数列是的分隔数列；2n c n =1n a n =+{}n a {}n c （2）设，是的前项和，，判断数列是否是数列的4n c n =-n S {}n c n 32n n d c -={}n S {}n d 分隔数列，并说明理由；（3）设，是的前项和，若数列是的分隔数列，求实数、1n n c aq -=n T {}n c n {}n T {}n c a 的取值范围.q参考答案一. 填空题1. 2. 3.4.5.1(,3)3-1i -3(1,235x126.7.8. 200 9. 4或78-22841x y -=10. ，12. 111222122222n n n n n k S n n k ++--⎧+-=-⎪⎪=⎨⎪+-=⎪⎩*k ∈N 12n-二. 选择题13. C14. C15. B16. C三. 解答题17.（1）4；（2）.18.（1）；（2）.2a =8t ≥19.（1），；（2）54.1°，867.1.1r =260(2r =20.（1）；（2）；（3）.22143x y +=34(E 21.（1）证明略；（2）不是，反例：时，无解；（3）.4n =m 02a q >⎧⎨≥⎩。

上海市南洋模范中学2019届高三下学期3月月考数学试题

2021学年南模中学高三年级三月份月考卷 2021。

3。

6一、填空题。

八 — .... x _ …1。

全集U R ，假设集合A x —— 0 ，那么C U A x 12 22。

双曲线2x y 6的焦距为 6 1 ___ 一^■一， 3。

ax 1 二项展开式中的第五项系数为 x 15 ，、一 15，那么正实数a 24。

函数f x 3x 1 1 ... ............... 一一一，一* ……3x 」a 1的图像与它的反函数的图像重合，那么实数 x a 3 a 的值为 5。

设x ， y 满足约束条件3x y 6 0 x y 2 0 ，那么目标函数z 2x y 的最大值为 x 0，y 0 6。

从集合A 1，1，2中随机选取一个数记为 k ，从集合B 2，1，2中随机选取一个数记为 b ，那么直线y 不经过第三象限的概率为 2 7。

设F I ， F 2是双曲线x 2 -y- 1的两个焦点， 24P 是双曲线上的一点，且 3 PFI 4PF 2，那么 PF 1F 2的周长 8。

四面体ABCD 中，AB CD 那么 EF 。

2， E ， F 分别为BC ， AD 的中点，且异面直线AB 与CD 所成的角为一，39。

函数f x 是定义在 R 上的奇函数，x 0时，f x 2x 6，那么x 0时，不等式f x x 的解集为1 1 ................... ..... ......... 10。

关于x 的方程——，一 sin- x 在 4，4上的解的个数是 ||x 1| 1 211。

任意实数a ， b ，定义a b ab ，ab 0a ，设函数f x log 2x x ，数列a n 是公比大于0的等比数歹U ，，ab 0 b 且 a 6 1， f a 1 f a 2 f a 3 L f a 9 f a 10 2a 1，贝U a 1 12。

以正方形的四个顶点分别作为椭圆的两个焦点和短轴的两个端点， A ， B ， M 是椭圆上的任意三点〔异于椭圆顶点〕，假设存在锐角uuur uur ，使 OM cos OA sinuur OB ，那么直线OA ， OB 的斜率乘积为14。

2019-2020学年上南辅读学校高三英语三模试题及答案解析

2019-2020学年上南辅读学校高三英语三模试题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ACome and enjoy Vivaldi's TheFour Seasonsperformed by live musicians!Tickets△Zone A Sating (Excellent Visibility, $75)△Zone B Seating (Great Visibility, $60)△Zone C Seating (Good Visibility, $45)△Zone D Seating (Restricted Visibility, 30)Zone A and Zone B audiences will get the chance to take pictures with the performers on the stage after the show.Highlights* A beautiful venue bathed in candlelight.*Classical music performance by the Angel Strings quartet*A safe and socially-distanced event, ensuring you are comfortable and at ease.General Info*Dates and times: Various dates, at 6:30 pm and 8:30 pm (select during purchase).*How long: 65 minutes. Doors open 45 minutes before the start time. We recommend you arrive at least 30 minutes before the start of the event, as late entry is not permitted.*Where: Events on Oxlade*Age requirement: Must be 8 years old or older to attend. Anyone under the age of 16 must be accompanied by an adult.*Please note: The 6:30 pm seating will take place during daylight hours outdoors, and the space will not be that dark. In the case of rain, the event will be moved to the indoor area of the venue.DescriptionWhether you're looking for a beautifully unique classical music performance or a romantic candlelit experience, this performance is for you. You don't need to know all things about Vivaldi to enjoy the evening; simply sit back and admire the wonderful atmosphere and the pieces you'll hear.Join our musicians for an evening under the stars, and prepare to be taken into the clouds with Vivaldi' s most treasured masterpieces!1.What can someone with a $45 ticket do?A.Perform on the stage.B.Enjoy good visibility.C.Select a seat in Zone B.D.Take photos with the musicians.2.What should potential audiences keep in mind?A.Arrive at the venue on time.B.Learn about Vivaldi in advance.C.The performance lasts 45 minutes.D.The event will be canceled if it rains.3.What do we know about the 8:30 p.m. performance?A.It welcomes children under the age of 8.B.Its performers differ on different dates.C.Its stage will be decorated with candles.D.It will be shown in the indoor area of the venue.BNostalgia (怀旧) has become increasingly common in our current climate of accelerated, unexpected change. More and more Americans are turning back with longing towhat feels like simpler, sweeter times. They collect cassette tapes, manual typewriters even decades-old video games.Is it a mistake to get too obsessed with the past? Some psychologists warn that too much devotion to the so-called good old days is an escape from reality; it can indicate loneliness or that a person is having a difficult time coping in the present. Psychologist Stephanie Coontz argues that nostalgia distracts us from addressing the problems of modern life and contribute to anxiety, depression , insomnia etc.But new studies suggest that a modest dose of nostalgia is not only harmless, but actually beneficial. They suggest it helps strengthen our sense of identity and makes us feel more optimistic and inspired. It is also a tool for self — discovery and memories are a psychological immune response that is triggered when you want to take a break from negativity. Interestingly, those happy memories can be particularly beneficial both to kids in their teens and to society's elders. Recalling our childhood reminds us of “the times when we were accepted and loved unconditionally," says Krystine Batcho, a psychologist. "That is such a powerfully comforting phenomenon, knowing that there was a time in life when we didn't have to earn our love." Nostalgia can transform even the most ordinary past into legends which warms the heart and the body. Let's not forget that nostalgia has been a source of inspiration to innumerable American writers. Mark Twain recalled his boyhood, writing, "after all theseyears, I can picture that old time to myself now, just as it was then：The white town drowsing in the sunshine of a summer's morning."So go ahead, daydream a little about your best childhood friend, your first car, a long - gone family pct. As Dr. Sedikidessays，"Nostalgia is ly central to human experience. "But at the same time, keep these words of wisdom from the great inventor Charles Kettering in mind as well："You can't have a better tomorrow if you are thinking about yesterday all the time. "34. What did some psychologists in paragraph 2 probably agree?A. Nostalgia will cause some mental problems.B. Nostalgia makes us devoted to the good old days.C. Nostalgia shows you are trying to get rid of loneliness.D. Nostalgia helps us cope with the difficult time we are going through.5. There are many benefits of nostalgia except ________A. It can enable us to know ourselves better.B. It can bring us some comfort when we recall.C. We are likely to gain attention if we recall the happy childhood.D. We can sometimes break away from negativity with happy memories.6. What will be talked about in the following paragraph?A. The bad influence of too much devotion to nostalgia.B. The reasons why we should avoid nostalgia.C. The bad memories that always stick around you.D. The great changes nostalgia will bring to you.7. What's the best title of the passage?A. We all have a soft spot for nostalgia.B. Nostalgia is actually good for you.C. Don't be carried away by nostalgia.D. There are many times when we like to recall.CCuckoos don’t bother building their own nests—they just lay eggs that perfectly imitate those of other birds and take over their nests. But other birds are wishing up, evolving some seriously impressive tricks to spot the cuckoo eggs.Cuckoos are often know asparasites, meaning that they hide their eggs in the nest of other species. To avoid detection, the cuckoos have evolved so that eggs seem reproduction of those of their preferred targets. If the host bird doesn’t notice the strange egg in its nest, the little cuckoo will actually take the entire nest for itself after it comes out, taking the other eggs on its back and dropping them out of the nest.To avoid this unpleasant fate for their young, the other birds have evolved a few smart ways to spot the fakes, which we’re only now beginning to fully understand. One of the most amazing finds is that birds have an extra colour-sensitive cell in their eyes, which makes them far more sensitive to ultraviolet wavelengths and allows them to see a far greater range of colours than humans can. This allows cautious birds to detect a fake egg which might be exactly the same to our eyes.Fascinatingly, we’re actually able to observe different bird species at very different points in their evolutionary war with the cuckoos. For instance, some cuckoos lay their eggs in the nests of the redstarts. The blue eggs these cuckoos lay are practically alike to those of the redstarts, and yet they are still sometimes rejected. Compare that with cuckoos who target dunnocks. While those birds lay perfectly blue eggs, their cuckoo invaders just lay white eggs with brown irregular shaped spots. And yet dunnocks barely ever seem to notice the obvious trick.Biologists suspect these more easily fooled species like the dunnocks are on the same evolutionary path as the redstarts, but they have a long way to go until they evolve the same levels of suspicion. What’s remarkable is that the dunnocks fakes are so bad and the redstart ones so good, and yet cuckoos are still more successful with the former than the latter.It speaks to just how thoroughly a species’ behavior can be changed by the pressures of natural selection, or it might just be a bit of strategic cooperation on the part of the dunnocks. Biologists have suggested that these birds are willing to tolerate a parasite every so often because they don’t want to risk accidentally getting rid of one of their own eggs.8. This passage can be most likely found in a ________.A. science surveyB. nature magazineC. zoo advertisementD. travel journal9. What does the underlined word “parasite” in paragraph 2 most probably refer to?A. Animals that work together to raise young.B Small harmful animals such as worms or mice.C. Animals that can adapt to changing environments.D. Animals which live on or inside other host animals.10. Which of the following is TRUE about the dunnock according to the passage?A. It is colour-blind and therefore cannot identify foreign eggs in the nest.B. It can easily remove cuckoo eggs from the nest because fakes are so bad.C. It is a host bird that is more likely to raise a cuckoo chick than the redstart.D. It is unable to evolve and hence accepts cuckoo eggs that appear in the nest.11. Which of the following can be inferred from the passage?A. Dunnocks may eventually learn to recognise foreign eggs.B. Redstarts seem to be less suspicious compared to dunnocks.C. Cuckoo birds are good at taking responsibility for their own young.D. It is very easy for cuckoos to imitate the colouring of the dunnock’s egg.DThink ofJapanin the spring and the image that comes to mind is likely the country’s famous cherry blossoms, also known as “Sakura” — white and pink flowers, blooming across cities and mountains.The flowers, which experience a “peak bloom” that only lasts a few days, have been loved inJapanfor more than a thousand years. Crowds celebrate with viewing parties,flockingto the most popular locations to take photos and have picnics underneath the branches.But this year, cherry blossom season has come and gone in the blink of an eye, in one of the earliest blooms on record. Scientists warn it’s a symptom of the larger climate crisis threatening ecosystems everywhere.Yasuyuki Aono, a researcher atOsakaPrefectureUniversity, has gathered records fromKyotoback to 812 AD from historical documents and diaries. In the central city ofKyoto, cherry blossoms peaked on March 26, the earliest in more than 1,200 years, Aono said. And in the capitalTokyo, cherry blossoms reached full bloom on March 22, the second-earliest date on record.The peak bloom dates shift every year, depending on numerous factors including weather and rainfall, but have shown a general trend of moving earlier and earlier. InKyoto, the peak date stayed around mid-April for centuries, but began moving into early April during the 1800s. The date has only dipped into late March a handful of times in recorded history.“Sakura blooms are very temperature sensitive,” said Aono. “Flowering and full bloom could be earlier or later depending on the temperature alone,” he said. “The temperature was low in the 1820s, but it has risen by about 3.5 degrees Celsius to this day.”This year’s seasons in particular influenced the blossom dates, he added. The winter was very cold, but the spring came fast and unusually warm.12. What is the best title of the passage?A. Cherry blossom celebrations.B. Warning of a climate crisis.C. A strong love for cherry blossom.D. Cherry blossom season coming earlier.13. What does the underlined word “flocking” mean?A. Blocking.B. Flooding.C. Running.D. Following.14. What can we infer from paragraph 5?A. The peak blossom dates fall on a fixed date.B. The cherry blossom rarely peaks in March.C. The peak bloom dates mainly depend on weather and rainfall.D. Cherry blossom peaks around mid-April inTokyo.15. What is the author’s purpose in writing the passage?A. To inform people the date of cherry blossom.B. To show a study on cherry blossom dates.C. To present a Japanese tradition of cherry blossom celebration.D. To make people aware of the influence of climate change on cherry blossom.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019届上海市南洋模范中学高三下学期3月月考数学试题(解析版)

2019届上海市南洋模范中学高三下学期3月月考数学试题一、单选题1．“”是“不等式成立”的（）A．充分条件B．必要条件C．充分必要条件D．既非充分也不必要条件【答案】A【解析】解不等式|x﹣1|＜1，再由充分必要条件即可判断出结论．【详解】不等式|x﹣1|＜1成立，化为﹣1＜x﹣1＜1，解得0＜x＜2，∴“”是“不等式|x﹣1|＜1成立”的充分条件．故选：A．【点睛】本题考查了充分必要条件，绝对值不等式的解法、简易逻辑的判定方法，考查了推理能力与计算能力，属于基础题．2．给出下列命题，其中正确的命题为（）A．若直线和共面，直线和共面，则和共面；B．直线与平面不垂直，则与平面内的所有直线都不垂直；C．直线与平面不平行，则与平面内的所有直线都不平行；D．异面直线，不垂直，则过的任何平面与都不垂直.【答案】D【解析】试题分析：A：直线共面不具有传递性，故A错误；B：根据线面垂直的判定可知B错误；C：若直线，满足直线与平面不平行，故C错误；D：假设存在过的平面与垂直，则可知，∴假设不成立，故D正确，故选D．【考点】空间中点、线、面的位置关系及其判定．3．已知数列的通项公式为，其前项和，则双曲线的渐近线方程为（）A．B．C．D．【答案】C【解析】试题分析：根据数列的通项公式为,其前项和，那么可知，可知n=9,那么根据可知a=,b=3,故可知双曲线的渐近线方程为，选C.【考点】数列的求和，双曲线的性质点评：主要是考查了数列的通项公式和双曲线的性质的运用，属于基础题。

4．已知平面直角坐标系中两个定点，，如果对于常数，在函数，的图像上有且只有6个不同的点，使得成立，那么的取值范围是（）A．B．C．D．【答案】C【解析】画出函数y＝|x+2|+|x﹣2|﹣4在[﹣4，4]的图象，讨论若P在AB上，设P（x，﹣2x﹣4）；若P在BC上，设P（x，0）；若P在CD上，设P（x，2x﹣4）．求得向量PE，PF的坐标，求得数量积，由二次函数的最值的求法，求得取值范围，讨论交点个数，即可得到所求范围．【详解】函数y＝|x+2|+|x﹣2|﹣4，（1）若P在AB上，设P（x，﹣2x﹣4），﹣4≤x≤﹣2．∴（3﹣x，6+2x），（﹣3﹣x，6+2x）．∴x2﹣9+（6+2x）2＝5x2+24x+27=，∵x∈[﹣4，﹣2]，∴λ≤11．∴当λ或时有一解，当λ≤9时有两解；（2）若P在BC上，设P（x，0），﹣2＜x≤2．∴（3﹣x，2），（﹣3﹣x，2）．∴x2﹣9+4＝x2﹣5，∵﹣2＜x≤2，∴﹣5≤λ≤﹣1．∴当λ＝﹣5或﹣1时有一解，当﹣5＜λ＜﹣1时有两解；（3）若P在CD上，设P（x，2x﹣4），2＜x≤4．（3﹣x，6﹣2x），（﹣3﹣x，6﹣2x），∴x2﹣9+（6﹣2x）2＝5x2﹣24x+27，∵2＜x≤4，∴λ≤11．∴当λ或时有一解，当λ<9时有两解；综上，可得有且只有6个不同的点P的情况是λ＜﹣1．故选：C．【点睛】本题考查平面向量的数量积的坐标运算，二次函数的根的个数判断，注意运用分类讨论的思想方法，属于中档题．二、填空题5．已知全集，若集合，则_________.【答案】【解析】求出集合A，即可求解∁U A【详解】全集U＝R，集合A＝{x|x＞1或x<0}则＝故答案为【点睛】本题考查集合的基本运算，补集的求法，分式不等式解法，准确计算是关键，是基础题．6．双曲线的焦距为__________.【答案】6【解析】将双曲线的方程化为标准方程，求得a，b，c，可得焦距2c的值．【详解】双曲线2x2﹣y2＝6即为1，可得a，b，c3，即焦距为2c＝6．故答案为：6．【点睛】本题考查双曲线的简单几何性质，焦距的求法，注意将双曲线的方程化为标准方程，运用双曲线的基本量的关系，考查运算能力，属于基础题．7．已知二项展开式中的第五项系数为，则正实数_____.【答案】【解析】由二项式定理的通项公式可得：，解出即可得出．【详解】T5x﹣2，∴，a＞0．解得a．故答案为：．【点睛】本题考查了二项式定理的应用，考查了推理能力与计算能力，准确计算是关键，属于基础题．8．已知函数的图像与它的反函数的图像重合，则实数的值为___.【答案】-3【解析】先求反函数：y，利用函数f（x）（a）图象与它的反函数图象重合，即为同一个函数即可得出．【详解】由y（a），解得x（y≠3），把x与y互换可得：y，∵函数f（x）（a）图象与它的反函数图象重合，∴﹣a＝3，解得a＝﹣3．故答案为：﹣3．【点睛】本题考查了反函数的求法及其性质，考查了推理能力与计算能力，属于中档题．9．设，满足约束条件，则目标函数的最大值为_____.【答案】14【解析】画出可行域，通过向上平移基准直线到可行域边界的位置，由此求得目标函数的最大值.【详解】画出可行域如下图所示，由图可知，目标函数在点处取得最大值，且最大值为.【点睛】本小题主要考查利用线性规划求线性目标函数的最大值.这种类型题目的主要思路是：首先根据题目所给的约束条件，画出可行域；其次是求得线性目标函数的基准函数；接着画出基准函数对应的基准直线；然后通过平移基准直线到可行域边界的位置；最后求出所求的最值.属于基础题.10．从集合中随机选取一个数记为，从集合中随机选取一个数记为，则直线不经过第三象限的概率为_____.【答案】【解析】将试验发生包含的事件（k，b）的所有可能的结果列举，满足条件的事件直线不经过第三象限，符合条件的（k，b）有2种结果，根据古典概型概率公式得到结果．【详解】试验发生包含的事件（k，b）的取值所有可能的结果有：（﹣1，﹣2）；（﹣1，1）；（﹣1，2）；（1，﹣2）；（1，1）；（1，2）；（2，﹣2）；（2，1）；（2，2）共9种结果．而当时，直线不经过第三象限，符合条件的（k，b）有2种结果，∴直线不过第三象限的概率P，故答案为．【点睛】本题考查古典概型，古典概型要求能够列举出所有事件和发生事件的个数，属于基础题．11．设，是双曲线的两个焦点，是双曲线上的一点，且，则的周长为___.【答案】24【解析】先由双曲线的方程求出|F1F2|＝10，再由3|PF1|＝4|PF2|，运用双曲线的定义，求出|PF1|＝8，|PF2|＝6，由此能求出△PF1F2的周长．【详解】双曲线x21的a＝1，c5，两个焦点F1（﹣5，0），F2（5，0），即|F1F2|＝10，由3|PF1|＝4|PF2|，设|PF2|＝x，则|PF1|x，由双曲线的定义知，x﹣x＝2，解得x＝6．∴|PF1|＝8，|PF2|＝6，|F1F2|＝10，则△PF1F2的周长为|PF1|+|PF2|+|F1F2|＝8+6+10＝24．故答案为：24．【点睛】本题考查双曲线的定义和性质的应用，考查三角形周长的计算，熟练运用定义是关键，属于基础题．12．已知四面体中，，，分别为，的中点，且异面直线与所成的角为，则____.【答案】1或【解析】取BD中点O，连结EO、FO，推导出EO＝FO＝1，，或，由此能求出EF．【详解】取BD中点O，连结EO、FO，∵四面体ABCD中，AB＝CD＝2，E、F分别为BC、AD的中点，且异面直线AB与CD所成的角为，∴EO∥CD，且EO，FO∥AB，且FO1，∴∠EOF是异面直线AB与CD所成的角或其补角，∴，或，当∠EOF时，△EOF是等边三角形，∴EF＝1．当时，EF．故答案为：1或．【点睛】本题考查异面直线所成角的应用，注意做平行线找到角是关键，解题时要认真审题，注意空间思维能力的培养，是易错题13．已知函数是定义在上的奇函数，当时，，则时，不等式的解集为____.【答案】【解析】由奇函数的性质可得x＞0时的解析式，再解不等式即可．【详解】∵函数f（x）是定义在R上的奇函数，∴当x＞0时，﹣x＜0，∴f（﹣x）＝x2﹣6，由奇函数可得f（x）＝﹣x2+6，∴不等式f（x）＜x可化为，解得x＞2∴x＞0时，不等式f（x）＜x的解集为：（2，+∞）故答案为：（2，+∞）【点睛】本题考查函数的奇偶性，涉及不等式的解法，熟记奇函数得定义是关键，属基础题．14．关于的方程在上的解的个数是____.【答案】7【解析】化简y=从而作函数的图像，从而可解【详解】化简y=,作函数在上的图像如下：结合图像可知，两个图像共有7 个交点故答案为7【点睛】本题考查函数与方程，函数的性质，三角函数，准确作图是关键，是中档题15．任意实数，，定义，设函数，数列是公比大于0的等比数列，且，，则____.【答案】4【解析】f（x）＝，及其数列{a n}是公比大于0的等比数列，且＝1，对公比q分类讨论，再利用对数的运算性质即可得出．【详解】由题，∵数列{a n}是公比大于0的等比数列，且，①1＜q时，，，…，∈（0，1），，，∈（1，+∞），1．∴，分别为：，，…，，1，q，…，q4．∵∴0++…+＝，∴q4q q2．∴2．左边小于0，右边大于0，不成立，舍去．②0＜q＜1时，1，∴，分别为：，，…，，1，q，…，q4，，，…，∈（1，+∞），，，∈（0，1），∵∴log2q2．∴2．∴4，∴a1＝4．③q＝1时，＝…＝＝…＝＝1，不满足舍去．综上可得：＝4．故答案为：4．【点睛】本题考查了等比数列的通项公式及其性质、对数的运算性质，考查了分类讨论方法、推理能力与计算能力，属于难题．16．以正方形的四个顶点分别作为椭圆的两个焦点和短轴的两个端点，，，是椭圆上的任意三点（异于椭圆顶点），若存在锐角，使，（0为坐标原点）则直线，的斜率乘积为___.【答案】或-2【解析】设椭圆方程为，A（，），B（，），从而得到的坐标表示，然后，再根据M点在该椭圆上，建立关系式，结合A、B点在也该椭圆上，得到，，从而得到相应的结果，同理当椭圆方程为可得答案【详解】由题意可设椭圆方程为，又设A（，），B（，），因为M点在该椭圆上，∴，则又因为A、B点在也该椭圆上，∴，∴，即直线OA、OB的斜率乘积为，同理当椭圆方程为时直线OA、OB的斜率乘积为﹣2．故答案为：或﹣2．【点睛】本题重点考查椭圆综合，平面向量的坐标运算，注意审题仔细，要注意分类讨论椭圆的焦点位置，属于中档题．三、解答题17．如图，在圆锥中，为底面圆的直径，点为弧AB的中点，.（1）证明：平面；（2）若点为母线的中点，求与平面所成的角.（结果用反三角函数表示）【答案】（1）见证明；（2）【解析】（1）由圆的性质得出AB⊥OC，由SO⊥平面ABC得出SO⊥AB，故而AB⊥平面SOC；（2）连结OD，由AB⊥平面SOC可知∠ADO为所求角，设圆锥底面半径为a，求出OD，得出tan∠ADO,得解【详解】（1）证明：在圆锥中，∵点为弧AB的中点，∴∴由平面（2）联结，∵平面∴为与平面所成的角设，则，∴∴在中，∴【点睛】本题考查了线面垂直的判定，线面角的计算，熟记判断定理，准确找到所成角是关键，属于中档题．18．已知函数.（1）求的最小正周期及判断函数的奇偶性；（2）在中，，，，若任意实数恒有，求面积的最大值.【答案】（1），函数是非奇非偶函数;（2）【解析】（1）由化简得则周期可求，计算；，可判奇偶性；（2）由题得将平方，得t的二次不等式，利用，得，进而得由求得最大值【详解】（1）所以，的最小正周期为；，所以，函数是非奇非偶函数.（2）由得因为是的内角，所以0<A<，由，得两边平方，整理得，对任意实数恒成立所以得则有且所以（当且仅当等号成立）所以，当时，面积的最大值为【点睛】本题考查三角恒等变换，向量数量积，三角形面积，熟记三角公式，灵活运用二次不等式转化是关键，是中档题19．数列满足：，，且，，成等差数列，其中.（1）求实数的值及数列的通项公式；（2）若不等式成立的自然数恰有4个，求正整数的值.【答案】（1），（2）【解析】（1）由题意和等差中项的性质列出方程求出λ，再利用累加法求出数列{a n}的通项公式；（2）结合条件对n进行分类讨论，当n≥3时利用分离常数法化简得p，利用取特值和做商法判断出的单调性，再判断出的单调性，根据条件即可求出正整数p的值．【详解】（1）由题意：，∵，，成等差数列，∴，解得：∵，，∴，检验n=1 成立故（2）解：∵，∴∵，∴显然成立当时，，设∴当时，；当时，；又，，，，若还需有2解，则，即，解得，所以正整数【点睛】本题考查了等差中项的性质，累加法求数列的通项公式，以及数列单调性的判断与应用，考查方程思想与分类讨论思想的应用，是难题20．教材曾有介绍：圆上的点处的切线方程为。

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南模中学高三三模数学试卷
2019.05
一. 填空题
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AB BC BC CA CA AB 的值等于
2
19. 某景区欲建两条圆形观景步道 M1 、M 2（宽度忽略不计），如图所示，已知 AB AC ，
AB AC AD 60 （单位：米），要求圆 M1 与 AB 、 AD 分别相切于点 B 、 D ，圆 M 2
与 AC 、 AD 分别相切于点 C 、 D .
（1）若
BAD
3
，求圆
M1
、
M2
A. 相离
B. 相切
C. 相交
D. 随 m 的变化而变化
三. 解答题 17. 如图，长方体 ABCD A1B1C1D1 中， AB BC 2 ， AA1 3 . （1）求四棱锥 A1 ABCD 的体积；（2）求异面直线 A1C 与 DD1 所成角的大小.
18. 已知函数 f (x) | 2x a | a . （1）若不等式 f (x) 6 的解集为 (1,3) ，求 a 的值；（2）在（1）的条件下，若存在 x0 R ，使 f (x0 ) t f (x0 ) ，求 t 的取值范围.
的半径（结果精确到
0.1
米）；
（2）若观景步道 M1 、M 2 的造价分别为每米 0.8 千元与每米 0.9 千米，则当 BAD 多大时，
总造价最低？最低总造价是多少？（结果分别精确到 0.1°和 0.1 千元）
20.
已知椭圆
C
:
x2 a2
y2 b2
1 （ a b 0 ）的右焦点为 F (1,0) ，且点 P(1, 3) 在椭圆 C 上. 2
7.
设
P
是曲线
x
2 2
sec
（
为参数）上的一动点，O
为坐标原点， M
为线段
OP
的中
y tan
点，则点 M 的轨迹的普通方程为
8. 在等差数列{an} 中，首项 a1 3 ，公差 d 2 ，若某学生对其中连续 10 项进行求和，在
遗漏掉一项的情况下，求得余下 9 项的和为 185，则此连续 10 项的和为
分隔数列，并说明理由；
（3）设 cn aqn1 ，Tn 是{cn} 的前 n 项和，若数列{Tn} 是{cn} 的分隔数列，求实数 a 、q 的
取值范围.
4
参考答案
一. 填空题
1. ( 1 ,3) 3
6. 8
1 | x 3 | x [1, )
函数 F (x) f (x) a （ 0 a 1 ）的所有零点之和为
（结果用 a 表示）
二. 选择题
13. 已知非零向量 a 、 b ，“函数 f (x) (ax b)2 为偶函数”是“ a b ”的（
）条件
A. 充分非必要
B. 必要非充分
C. 充要
{an}
满足：对任意
n
N*
，存在
m
N*
，使得
am cn am cn1
0，
则称{an} 是{cn} 的“分隔数列”.
（1）设 cn 2n ， an n 1，证明：数列{an} 是{cn} 的分隔数列；
（2）设 cn n 4 ， Sn 是{cn} 的前 n 项和， dn c3n2 ，判断数列{Sn} 是否是数列{dn} 的