南师附中2020届高三年级第二学期期初检测试卷及答案(含附加题)
2020届南京师范大学附属中学高三英语二模试卷及答案解析
2020届南京师范大学附属中学高三英语二模试卷及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AThe COVID -19 pandemic has affected all aspects of life, including the way we travel. But for those who are looking to expand their horizons while still staying safe, the following three travel trends in 2021 may provide inspirations. Let’s take a look.StaycationWith many travel restrictions during the pandemic, people preferred traveling to nearby places in 2020. This trend continues in 2021. According to search data, 62 percent of people are interested in taking a vacation within driving distance of home. People who live in large cities want to get back in touch with nature. Travelers are looking for places different from their everyday accommodations, for example, farm stays, villas and cottages.Pod travelWhile 2020 saw a rise in solo travel and isolated adventures, 2021 shows that people want to be more connected. “Pod travel”, or gathering in isolated spaces with loved ones, is growing in popularity. 85 percent of survey respondents favor traveling with family or friends, and over half of the trips searched include three or more people. Pod travel is here to stay for those who want to safely be together while reducing risks associated with socializing with others.Remote working and travelingMany people worked and learned from home in 2020 because of the pandemic. Remote working blurs the line between working and traveling. There was a 128 percent increase in the mention of phrases such as “relocation”, “relocate”, “remote work” and “trying a new neighborhood”. People are actively booking longer stays (e. g. two plus week trips) in small to mid—size cities with access to immersive natural surroundings and wide—open spaces.1.What can we learn about Staycation?A.Travelling to the countryside.B.Taking an isolated adventure.CHaving holidays in nearby places. D.Staying indoors all by oneself.2.What’s special about Pod travel?A.Traveling alone.B.Traveling far away.C.Traveling while working.D.Traveling with loved ones.3.Where might we find the text in a magazine?A.Medicine.cation.C.TourismD.Career.BUnderstanding the link between a clean environment and human life is not a new concept. In fact, it was noticed as early as ancient Rome. Today we see how green living has infiluenced our everyday lives. There is a growing community of people who embrace a zero waste lifestyle and make changes to the way they live to reduce their carbon footprint.Living a zero waste lifestyle means doing one’s best to achieve the aim of not sending anything to a landfill. People who adopt this lifestyle ultimately cut down on their waste by reducing what they need and want. They reuse what they own, sending few things to be recycled.Many people who adopt the zero waste lifestyle claim to be frustrated by the many harmful chemical substances found in beauty and cleaning products. They also find the uses of disposable items and excessive packaging. For example, how many times have we had to peel away layers of plastic wrap and cardboard before finally taking out the item which we had bought? Instead of buying pre-packed food and goods, those who identify with the zero waste philosophy tend to shop in stores that allow them to make purchases and bring their own cloth bags and glass jars to store their purchases.Many people may have the misconception that it is easier to live a zero waste lifestyle in the West. Nevertheless, Malaysian environmental journalist, Ms. Aurora Tin, has proven that a zero waste lifestyle is possible even in the Asian context. Instead of going to the supermarket to buy pre-packaged foods, Ms. Tin now visits the wet market and brings her own bags for vegetables. She has even stopped using store-bought toothpaste and makes her own toothpaste from coconut oil and baking soda. This lifestyle may be too big a change for the average person, but we could follow her suit to make gradual changes to our own lives.4. Which of the following is a zero waste lifestyle?A. Bringing a resuable container to take away food.B. Choosing appliances that cost less money.C. Turning off a device to stop using power.D. Classifying the garbage before throwing it away.5. What may disappoint a person who adopts a zero waste lifestyle?A. Recycable carboard.B. Excessive packaging.C. Glass jars to store purchases.D. Natural substances in cleaning products.6. What is the main idea of the last paragraph?A. How do people live a zero waste lifestyle.B. Why Ms. Tin chooses to live a zero waste lifestyle..C. We can also practice a zero waste lifestyle in Asia.D. It is easy to live a zero waste lifestyle in the West.7. What is the best title of the passage?A. Living a zero waste lifestyle.B. Going green ismore than a fashion.C. A zero waste lifetyle is easy to achieve.D. Making environmentally-conscious decisions.CThree years ago, a group of researchers at Cornell University Food and Brand Lab had a guess. They knew that many apples being served to kids as part of the National School Lunch Program were ending up in the trash, almost untouched. But unlike others, they wondered if the reason was more complicated than simply that the kids didn’t want the fruit.They thought the fact that the apples were being whole, rather than sliced (切片的), was doing the fruit no favor. And they wanted to prove this idea.They carried out a study at eight schools and found that fruit consumption (消费) jumped by more than 60 percent when apples were served sliced. They then made a follow-up study at six other schools, which not only showed the same result, but further supported. The whole apple consumption at schools that served sliced apples reduced by about 50 percent than before.Based on these results, it seems that even the simplest forms of inconvenience can influence us. Sliced apples just make more sense for kids. The hardest part is getting kids to start eating fruit, and that’s exactly why sliced apples are the way to go. A child holding a whole apple has to break the skin, eat around the core (果核), and deal with the inconvenience of holding a large fruit. These problems might seem silly or unimportant, but they’re important when you’re missing teeth or have braces (牙箍), as so many kids do.Sometimes what seems like a really small inconvenience actually makes a huge difference.8. What can we learn from Paragraph 1?A. Lots of apples served to kids were wasted.B. Researcher supported the lunch program.C. Apples are not the only acceptable fruit for kids.D. The lunch program was well worth spreading.9. What did researchers want to prove?A. Kid didn’t want the fruits.B. Kids favored the whole apples.C. Kids preferred to the sliced apples.D. Kid’s eating habits influenced fruit consumption.10. According to the study, we can know________.A. 14 schools served sliced fruitsB. kids have problems eating the whole applesC. eating fruits must be healthy for childrenD. few kids are missing teeth or have braces11. What’s the main idea of the passage?A. Form matters more than taste.B. Apples are beneficial to each kid.C. The way fruit is eaten links kids’ growth.D. Inconvenience probably makes a difference.DHave 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 the world. 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 67million 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.12. 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.13. 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.14. Where does the word “pourriel” come from?A. French.B. An email.C. A holiday.D. Canada.15. 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.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。
江苏省南京师范附属中学2020届高三年级第二学期期初试卷英语参考答案
南师附中2020届高三年级第二学期期初检测试卷英语试题参考答案及评分标准听力：CBBCB ACBAB ABCAB CBCAB单项选择：CDBCA DACBB DABAC完形填空：BDCAC DDBCC DCBAB ABBDC阅读理解：BCDBA CABCD CABBD任务型阅读：71.Harm72. unsatisfied/dissatisfied73. popularity74. removing/handling75. Reconsider/Rethinkparisons77.expect78. objective79. opinions/ideas/views/viewpoints80. exist作文：Possible version one:Young Internet users are always coming up with new slang to differentiate themselves. Butthe rapidly developing Internet lingo, some people are concerned that it's ruining the Chinese language.In my opinion, it's unnecessary for older generations to fear the use of slang is weakeningthe people's capability to write and remember Chinese characters. For one thing, new terms add variety to a language as they give a group of people a sense of belonging. For another, youth lingo, which is a source of identity for young people, has existed long before the Internet. As a matter of fact, youth lingo has not made the Chinese language disappear but has impacted the language.To conclude, while language development is unavoidable, Chinese Internet lingo is unlikelyto replace the existing use of Mandarin Chinese, but will rather be enriching it by adding another layer.Possible Version two:Nowadays, young people often create some new slangs to differentiate themselves. However, with the rapid development of Internet lingo, many critics, especially the seniors, believe that the emergence of such lingo will damage the foundation of Chinese language.In my opinion, it is not necessary to worry about the new slang that will likely weaken the new generation’s capability of writing or remembering Chinese characters. There are two reasons. Firstly, the purpose of language is to convey ideas and share opinions. Thus, the prevailing youth lingo,as a more concise and efficient way of expression, is in fact an enrichment of modern Chinese language. Secondly, since every era has its own young generation,the youth lingo, can bea new identity for young people,cannot pollute or extinct Chinese language but by contrast, it may become an emblem or living fossil of this specific era.To conclude, while language development is inevitable, Chinese internet lingo is unlikely to replace the existing use of Mandarin Chinese but will be enriching it by adding a new layer or dimension to it.听力试题原文Text 1W: Simon, could you return the tools I lent you for building the bookshelf last month?M: Oh, well, I hate to tell you this, but I can’t seem to find them.Text 2W: Well, you said you didn’t care very much for you job. Have you every thought of giving it up?M: Not really. You see I have a lot of friends in the office. I’m part of the family there.Text 3M: Hi, Miss Green. Sorry, I failed your subject again but I really tried my best.W: No one is sure to succeed every time. Cheer up, young man. You will have more chances.M: Thank you, Miss Green. I’ll try harder next time.Text 4W: I don’t know how you can eat so much yet never put on any weight, son. Your father’s got the same luck. I can’t take a bite without calculating how many calories I am taking.M: But remember Aunt Louise, Mom? She ate a lot and never gained a pound.Text 5M: Don’t you like the coat you just tried on?W: Well, I like the color and fabric.M: And it is really nice and reasonably priced.W: Yes. I would have bought it right away if they had had it in my size.Text 6M: Can you describe the room to me, please?W: Certainly. Let’s see, first there’s a big double bed, and of course there’s a telephone by thebed, and you have the radio alarm next to that. Then there’s a TV…M: Oh, well, are the bed sheets changed every day?W: Yes. They’re changed every day. And in fact the pillows are filled with a special material instead of cotton, which can help you fall asleep.M: I’m allergic to flowers.W: I’m sure they’re not dried flowers. But I’m sorry I can’t remember the name.M: Well, that seems to be just fine.Text 7M: Tell me about yourself and your past experience.W: I have worked as an executive secretary for 5 years, first for a trading company, then a trust company.M: How fast can you type?W: I can type 100 Chinese words per minute and 150 English words a minute.M: Can you operate computers skillfully?W: Yes, I can. I have received some special training in computers. Besides I am good at operating common office machines like fax.M: Sometimes we are very busy and need to work overtime. How do you feel about that?W: Could you tell me how often and how many hours I should work overtime?M: It just depends.If we have important visiting delegations, you have to stay with us. It’s not unusual.W: That’s all right.Text 8M: Hi, Asha. Is your mother feeling any better?W: No, I’m afraid not. She’s very homesick. She misses her family and friends back in India.M: Do you think she will ever move back to India?W: I’m not sure. It’s very difficult for her, because my father and my sisters all live here in California, so she doesn’t want to leave us.M: Does she ever go visit her family in India?W: She goes back once a year to visit her family and friends. But when she’s in India, she missesher family in the United States.M: It must be very hard for her, to always be torn between two places.W: Yes, it is. And she’s always telling us not to forg et our Indian culture. So, next year, my sisters and I are planning on traveling to India with her for a few months, so that we can all experience the culture together.M: I think that’s a wonderful idea. She’ll be so happy!Text 9M: Hello, I’m a new stud ent on Economics. Can I know some information about the courses?W: Well, I lecture on that program. What do you need to know?M: I’d like to know how many lectures a week I have to attend?W: Two lectures a week, on Tuesday and Thursday. Besides,there’s o ne group discussion every Friday.M: So what time?W: Let me see …You should have known this information yesterday, at the meeting for the new students.M: Oh, was there a meeting yesterday? I didn’t know about that …no one mentioned…W: Never mind. Now lectures are at four in the afternoon.M: Four’s bit late. I’ve got a part time job that starts at four thirty.W: Well, but attendance at lectures is necessary. We expect at least 80% attendance at this university, you know.M: 80%! That’s high.W: But I’m afraid you have to meet that requirement, otherwise, you can’t get enough credits to graduate. And what’s more, if you want to earn the scholarship, 90% attendance rate or above is a must.M: Ok, I see. Thank you for your information.W: You’re welcome.Text 10I grew up in a small town until I was 18 and then moved to a big city, so I have experienced the good and bad sides of both. I never thought that I would like living in a big city, but I was wrong. After ten years of living in one, I can’t imag ine ever living in a small town again. Surely small towns and big cities both have some problems in terms of transport. In a small town, youhave to own a car to make life comfortable. You can’t get around without one because there isn’t any kind of public transport. Big cities generally have heavy traffic and expensive parking, but there you have a choice of taking public transport, which is cheaper than driving. So, if you don’t have a car, you’d better live in the city. I also love the exciting life in b ig cities. I can always enjoy a lot of films, concerts, and other wonderful shows. However, these things are not common in small towns. The final thing I like about large cities is that you can meet different kinds of people. I think that living in an area where everyone was just like me would quickly become dull. Of course, safety should be considered, and that’s one area where small towns are better than big cities. Still, I would rather be a bit more careful and live in a large city than to feel safe but dull.。
江苏省南京师范大学附属中学2020届高三第二学期第一次模拟考试数学试题含附加题(word版含答案)
江苏省南师附中2020届高三年级第一次模拟考试数学I一、填空题：本大题共14小题，每小题5分，共70分，请把答案直接填写在答题卡相应，位置上。
1.集合{0,},x A e B =={-1,0,1},若A ∪B=B,则x=__.2.复数12i z i+=(i 是虚数单位)的虚部是___. 243.log 4log 2+=___.4.执行如图所示的程序框图,输出的S 值为____.5.在△ABC 中，a=4, b=5, c=6,则sin 2sin A C=____. 6.已知函数()sin()3),f x x x ϕϕ=+++0≤φ≤π.若f(x)是奇函数,则()6f π的值为___. 7.已知3()|log |,f x x =若a ，b 满足f(a-1)= f(2b-1),且a≠2b, 则a +b 的最小值为____.8.将黑白2个小球随机放入编号为1, 2, 3的三个盒子中,则黑白两球均不在1号盒子的概率为___.9.若抛物线24x y =的焦点到双曲线C:22221(x y a a b -=> 0, b>0)的渐近线距离等于1,3则双曲线C 的离心率为___.10. 设m,n 为空间两条不同的直线，,αβ为空间两个不同的平面，给出下列命题:①若m ∥α,m ∥β,则α∥β;②若m ⊥α,m ∥β,则α⊥β;③若m ∥α,m ∥n,则n ∥α;④若m ⊥α,α∥β,则m ⊥β.其中的正确命题序号是____.11.设x>0,y>0,向量a = (1-x,4), b = (x,-y),若a ||b , 则x+y 的最小值为___.12.在△ABC 中,点P 是边AB 的中点,己知2||3,||4,,3CP CA ACB π==∠=则CP CA ⋅=u u u r u u u r ___. 13.已知正数a, b, c 满足22(),b a c b ac b ++-=b a c+的最大值为___. 14.若210(0)1m x m mx -<≠+对一切x≥4恒成立,则实数m 的取值范围是___. 二、解答题:本大题共6小题，共90分.请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤.15.如图,在四棱锥P- ABCD 中,已知底面ABCD 为矩形,且2,AB =BC=1,E,F 分别是AB,PC 的中点，PA ⊥DE.（1）求证：EF//平面 PAD:（2）求证：平面PAC ⊥平面PDE.16.在三角形ABC 中，已知110tan ,cos 210B C ==- (1)求角A 的值;(2)若△ABC 的面积为3,10求边BC 的长.17.建造一个容积为8m³、深为2m 的无盖长方体形的水池，已知池底和池壁的造价分别为120元/m²和80元/m².(1)求总造价y (单位:元)关于底边一边长x (单位: m)的函数解析式，并指出函数的定义域;(2)如果要求总造价不超过2080元，求x 的取值范围;(3)求总造价y 的最小值.18.在直角坐标系xOy 中，已知椭圆C ：221,63x y +=若圆O 222:(0)x y R R +=>的一条切线与椭圆C 有两个交点A,B,且0.OA OB ⋅=u u u r u u u r(1)求圆O 的方程;(2)已知椭圆C 的上顶点为M ，点N 在圆O 上，直线MN 与椭圆C 相交于另一点Q ，且2,MN NQ =u u u u r u u u r 求直线MN 的方程.19.已知函数f(22()(2)ln 1()2a x ax x x x a R =+++∈ (1)若曲线y= f(x)在x=1处的切线的斜率为2,求函数f (x)的单调区间;(2)若函数f(x)在区间(1,e)上有零点,求实数a 的取值范围. (e 是自然对数的底数, e≈2.71828..)20.已知数列{}{}{},n n n a b c 、、对于给定的正整数k,记*,)(n n n k n n n k b a a c a a n N ++=-=+∈.若对任意的正整数n 满足:1,n n b b +≤且{}n c 是等差数列,则称数列{}n a 为“()H k ”数列.(1)若数列{}n a 的前n 项和为2,n S n =证明:{}n a 为H(k)数列;(2)若数列{}n a 为H(1)数列,且1121,1,5,a b c ==-=求数列{}n a 的通项公式;(3)若数列{}n a 为H(2)数列,证明:{}n a 是等差数列.数学II(附加题)21.[选做题]本题包括A 、B 、 C 三小题,请选定其中两题,并在相应的答题区域内作答.若多做，则按作答的前两题评分.解答时应写出文字说明、证明过程或演算步骤.A. [选修4-2:矩阵与变换] (本小题满分10 分)已知矩阵102,0201a A B ⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦，且AB= BA (1)求实数a;(2)求矩阵B 的特征值.B. [选修4- 4: 坐标系与参数方程] (本小题满分10分)在平面直角坐标系xOy 中,己知直线35:45x t l y t ⎧=⎪⎪⎨⎪=⎪⎩，(t 为参数).现以坐标原点O 为极点,以x 轴非负半轴为极轴建立极坐标系，设圆C 的极坐标方程为ρ=2cosθ,直线l 与圆C 交于A,B 两点，求弦AB 的长.C.[选修4- -5:不等式选讲]已知123,,(0,)x x x ∈+∞且满足1231233x x x x x x ++=，证明:1223313x x x x x x ++≥.[必做题]第22题、第23题，每题10分，共计20分.请在答卷卡指定区域内作答.解答应写出文字说明、证明过程或演算步骤.22. (本小题满分10分)如图,在四棱锥P- ABCD 中,已知棱AB, AD, AP 两两垂直,长度分别为1, 2, 2.若(,DC AB R λλ=∈u u u r 且向量PC uuu r 与BD u u u r 夹角的余弦值为15.15(1)求λ的值;(2)求直线PB 与平面PCD 所成角的正弦值.23.已知(2122101221(1),n n n x a a x a x a x ++++=++++L n ∈N *.记0(21).k n n k nT k a =-=+∑(1)求2T 的值;(2)化简n T 的表达式,并证明:对任意的*,n N ∈n T 都能被4n+2整除.。
江苏省南京师范大学附属中学2020届高三下学期期初检测试题英语试题及答案
南京师范大学附中2020届高三年级第二学期期初检测试卷英语试题第二部分英语知识运用（共两节，满分35分）第一节单项填空（共15小题，每小题1分，满分15分）请认真阅读下面各题，从题中所给的A、B、C、D四个选项中，选出最佳选项，并在答题卡上将该项涂黑。
21. My teacher often says that success in making money is not always a good ______ of success in life.A. beliefB. elementC. criterionD. instance22. Tom’s comments on this issue are confusing because they appear to ______ the remarks on the same issue made earlier by him.A. violateB. induceC. clarifyD. contradict23. —Was the proposal passed at yesterday’s meetin g?—Yes, but some members of the committee expressed ______ .A. associationB. reservationC. authorityD. corporation24. The natural environment should be protected because of the important role ______ plays in human survival and development.A. whichB. the oneC. itD. that25. It may be necessary to stop ______ in the learning process and go back to the difficult points in the lessons.A. at intervalsB. at randomC. at easeD. at length26. This car is important to our family. We would repair it at our expense ______ it break down within the first year.A. couldB. wouldC. mightD. should27. ______ both sides accept the agreement ______ a lasting peace be established in this region.A. Only if; willB. If only; wouldC. Should; willD. Unless; would28. But for the belief that things ______ better soon, many people couldn’t have survived the rough time.A. had gotB. gotC. would getD. will get29. Most spending that results in debt is like a drug: a quick hit of pleasure that ______ , only to drag you down for years to come.A. takes offB. wears offC. sets offD. shows off30. “If," Johnson added,” ______ enough time, we would certainly have done the job much better.A. to giveB. givenC. I was givenD. to be given31. The explosion in the chemical plant ______ avoided had the county authorities, aiming too much at short－term interests, performed their duties from the beginning.A. should beB. might beC. must have beenD. could have been32. —What's the recent progress in Xiong'an New Area, first announced two years ago?—It ______ increasing attention from foreign companies given its huge development potential.A. has been attractingB. had attractedC. would be attractingD. was attracting33. —Shall we go to the seven o'clock performance or the eight?— ______ suits you fine.A. WhateverB. WhicheverC. WheneverD. Wherever34. Modesty is a kind of good quality, which is ______ it takes to be your personal best.A. whatB. whereC. whichD. when35. —Which classical Chinese poem do you like best?—Don't laugh if we lay drunken on the battleground; how many soldiers ever came back ______ ?A. old and youngB.up and downC. safe and soundD. right and wrong第二节完形填空（共20小题；每小题1分，满分20分）请认真阅读下面短文，从短文后各题所给的A、B、C、D四个选项中，选出最佳选项，并在答题卡上将该项涂黑。
南师附中2020届高三年级第二学期期初检测试卷 数学(含附加题)数学附加题
南师附中2020届高三年级第二学期期初检测试卷
数学试题
第Ⅱ卷（选做题，40分）
21．【选做题】在A 、B 、C 三小题中只能选做2题，每小题10分，共计20分．请在答卷卡指定区域内作答．解答应写出文字说明、证明过程或演算步骤． A ．选修4—2：矩阵与变换
已知矩阵M ＝⎣⎡⎦⎤
2 1 1 2．
（1）求M 2；
（2）求矩阵M 的特征值和特征向量．
B ．选修4—4：坐标系与参数方程
在极坐标系() (02π)ρθθ<≤, 中，求曲线2sin ρθ=与cos 1ρθ=的交点Q 的极坐标．
【必做题】第22题、第23题，每题10分，共计20分．请在答卷卡指定区域内作答．解答应写出文字说明、证明过程或演算步骤． 22．（本小题满分10分）
在平面直角坐标系xOy 中，已知抛物线y 2＝2px (p ＞0)及点M (2，0)，动直线l 过点M 交抛物线于A ，B 两点，当l 垂直于x 轴时，AB ＝4．
（1）求p 的值；
（2）若l 与x 轴不垂直，设线段AB 中点为C ，直线l 1经过点C 且垂直于y 轴，直线l 2经过点M 且垂直于直线l ，记l 1，l 2相交于点P ，求证：点P 在定直线上．
O
y
B
x
M A C P
l l 1
l 2
23．（本小题满分10分）
对于给定正整数n ，设n
n
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n k
k S a ==∑．
（1）计算1234S S S S ，，，的值；
（2）求n S ．。
【附15套高考模拟卷】江苏省南师附中等五校2020届高三下学期期初教学质量调研物理试卷含解析
江苏省南师附中等五校2020届高三下学期期初教学质量调研物理试卷一、单项选择题：本题共6小题，每小题4分，共24分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1、如图所示，劲度系数为k 的轻弹簧的一端固定在墙上，另一端与置于水平面上质量为m 的物体P 接触，但未与物体P 连接，弹簧水平且无形变。
现对物体P 施加一个水平向右的瞬间冲量，大小为I 0，测得物体P 向右运动的最大距离为x 0，之后物体P 被弹簧弹回最终停在距离初始位置左侧2x 0处。
已知弹簧始终在弹簧弹性限度内，物体P 与水平面间的动摩擦因数为μ，重力加速度为g ，下列说法中正确的是 （ ）A ．物体P 与弹簧作用的过程中，系统的最大弹性势能20032P I E mgx mμ=- B ．弹簧被压缩成最短之后的过程，P 先做加速度减小的加速运动，再做加速度减小的减速运动，最后做匀减速运动C ．最初对物体P 施加的瞬时冲量0022I m gx μ=D ．物体P 整个运动过程，摩擦力的冲量与弹簧弹力的冲量大小相等、方向相反2、如图所示，D 是一只理想二极管（正向电阻为零，反向电阻无穷大），电流只能从a 流向b ，A 、B 为间距很小且正对的平行金属板，现有一带电粒子（不计重力），从B 板的边缘沿平行B 板的方向射入极板中，刚好落到A 板正中央，以E 表示两极板间的电场强度，U 表示两极板间的电压，p E ∆表示粒子电势能的减少量，若保持极板B 不动，粒子射入板间的初速度0v 不变，仅将极板A 稍向上平移，则下列说法中正确的是A ．E 变小B ．U 变大C ．p E ∆不变D ．若极板间距加倍，粒子刚好落到A 板边缘3、观看科幻电影《流浪地球》后，某同学设想地球仅在木星引力作用下沿椭圆轨道通过木星的情景，如图所示，轨道上P 点距木星最近（距木星表面的高度可忽略）。
则A ．地球靠近木星的过程中运行速度减小B ．地球远离木星的过程中加速度增大C ．地球远离木星的过程中角速度增大D ．地球在P 点的运行速度大于木星第一宇宙速度4、如图所示，一条质量分布均匀的柔软细绳平放在水平地面上，捏住绳的一端用恒力F 竖直向上提起，直到全部离开地面时，绳的速度为v ，重力势能为E p (重力势能均取地面为参考平面)。
南师附中2020届高三年级第二学期期初检测试卷数学(含附加题)数学参考答案及评分标准
A ．选修 4—２：矩阵与变换
解：（ 1） M 2＝ 2 1 12
21
54
＝
．………… 4 分
12
45
λ－ 2 － 1 （2）矩阵 M 的特征多项式为 f(λ)＝ － 1 λ－ 2 ＝ (λ－1)( λ－ 3)．
令 f(λ)＝ 0，解得 M 的特征值为 λ1＝1， λ2＝ 3．………… 6 分
21 Ⅰ当 λ＝ 1 时，
高三数学参考答案 第 1 页 共 11 页
又因为 A， C 是ⅠABC 的内角，所以 A＝ C．………… 6 分
（2）由（ 1）知，因为
A ＝ C ，所以
a＝ c，所以
cosB
＝
a2＋ c2－ 2ac
b2 ＝
a
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2
．…………
8分
因为 →BA ·→BC＝ 1，所以 a2cosB＝ a2－2＝ 1，所以 a2＝3．………… 10 分
代入切线方程，得 b＝－ 2，
所以 a＝－ 1， b＝－ 2．………… 2 分
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江苏省南京师范大学附属中学2020届高三第二学期第一次模拟考试数学试题含附加题(解析版)
2020年高考模拟高考数学一模试卷一、填空题1．集合A＝{0，e x}，B＝{﹣1，0，1}，若A∪B＝B，则x＝．2．已知复数z＝（i是虚数单位）则z的虚部是．3．log24+log42＝．4．执行如图所示的程序框图，输出的s值为．5．在△ABC中，a＝4，b＝5，c＝6，则＝．6．已知函数，0≤φ≤π．若f（x）是奇函数，则的值为．7．已知f（x）＝|log3x|，若a，b满足f（a﹣1）＝f（2b﹣1），且a≠2b，则a+b的最小值为．8．将黑白2个小球随机放入编号为1，2，3的三个盒子中，则黑白两球均不在1号盒子的概率为9．若抛物线x2＝4y的焦点到双曲线C：（a＞0，b＞0）的渐近线距离等于，则双曲线C的离心率为．10．设m，n为空间两条不同的直线，α，β为空间两个不同的平面，给出下列命题：①若m∥α，m∥β，则α∥β；②若m⊥α，m∥β，则α⊥β；③若m∥α，m∥n，则n∥α；④若m⊥α，α∥β，则m⊥β．其中的正确命题序号是．11．设x＞0，y＞0，向量＝（1﹣x，4），＝（x，﹣y），若∥，则x+y的最小值为．12．在△ABC中，点P是边AB的中点，已知||＝，||＝4，∠ACB＝，则•＝．13．已知正数a，b，c满足b2+2（a+c）b﹣ac＝0，则的最大值为．14．若（m≠0）对一切x≥4恒成立，则实数m的取值范围是．二、解答题：共6小题，共90分．请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤．15．如图，四棱锥P﹣ABCD的底面为矩形，AB＝，BC＝1，E，F分别是AB，PC的中点，DE⊥PA．（Ⅰ）求证：EF∥平面PAD；（Ⅱ）求证：平面PAC⊥平面PDE．16．在三角形ABC中，已知，．（1）求角A的值；（2）若△ABC的面积为，求边BC的长．17．建造一个容积为8m3、深为2m的无盖长方体形的水池，已知池底和池壁的造价分别为120元/m2和80元/m2．（1）求总造价y（单位：元）关于底边一边长x（单位：m）的函数解析式，并指出函数的定义域；（2）如果要求总造价不超过2080元，求x的取值范围；（3）求总造价y的最小值．18．在直角坐标系xOy中，已知椭圆＝1，若圆O：x2+y2＝R2（R＞O）的一条切线与椭圆C有两个交点A，B，且•＝0．（1）求圆O的方程；（2）已知椭圆C的上顶点为M，点N在圆O上，直线MN与椭圆C相交于另一点Q，且＝2，求直线MN的方程．19．已知函数．（1）若曲线y＝f（x）在x＝1处的切线的斜率为2，求函数f（x）的单调区间；（2）若函数f（x）在区间（1，e）上有零点，求实数a的取值范围．20．已知数列{a n}、{b n}、{c n}，对于给定的正整数k，记b n＝a n﹣a n+k，c n＝a n+a n+k（n∈N*）．若对任意的正整数n满足：b n≤b n+1，且{c n}是等差数列，则称数列{a n}为“H（k）”数列．（1）若数列{a n}的前n项和为S n＝n2，证明：{a n}为H（k）数列；（2）若数列{a n}为H（1）数列，且a1＝1，b1＝﹣1，c2＝5，求数列{a n}的通项公式；（3）若数列{a n}为H（2）数列，证明：{a n}是等差数列．【选做题】本题包括A、B、C三小题，请选定其中两题，并在相应的答题区域内作答．若多做，则按作答的前两题评分．解答时应写出文字说明、证明过程或演算步骤．[选修4-2：矩阵与变换]21．已知矩阵A＝，B＝，且AB＝BA．（1）求实数a；（2）求矩阵B的特征值．[选修4-4：坐标系与参数方程]22．在平面直角坐标系xOy中，已知直线为参数）．现以坐标原点O为极点，以x轴非负半轴为极轴建立极坐标系，设圆C的极坐标方程为ρ＝2cosθ，直线l与圆C交于A，B两点，求弦AB的长．[选修4-5：不等式选讲]23．已知x1，x2，x3∈（0，+∞），且满足x1+x2+x3＝3x1x2x3，证明：x1x2+x2x3+x3x1≥3．【必做题】第22题、第23题，每题10分，共计20分．请在答卷卡指定区域内作答．解答应写出文字说明、证明过程或演算步骤．24．如图，在四棱锥P﹣ABCD中，已知棱AB，AD，AP两两垂直，长度分别为1，2，2．若＝λ，且向量与夹角的余弦值为．（1）求实数λ的值；（2）求直线PB与平面PCD所成角的正弦值．25．已知（1+x）2n+1＝a0+a1x+a2x2+…+a2n+1x2n+1，n∈N*．记T n＝（2k+1）a n﹣k．（1）求T2的值；（2）化简T n的表达式，并证明：对任意的n∈N*，T n都能被4n+2整除．参考答案一、填空题：共14小题，每小题5分，共70分．请把答案直接填写在答题卡相应位置上．1．集合A＝{0，e x}，B＝{﹣1，0，1}，若A∪B＝B，则x＝0．【分析】推导出A⊆B，e x＞0，从而e x＝1，由此能求出结果．解：因为集合A＝{0，e x}，B＝{﹣1，0，1}，A∪B＝B，所以A⊆B，又e x＞0，所以e x＝1，所以x＝0．故答案为：0．2．已知复数z＝（i是虚数单位）则z的虚部是﹣1．【分析】直接利用复数代数形式的乘除运算化简得答案．解：∵z＝＝，∴复数z＝的虚部是﹣1．故答案为：﹣1．3．log24+log42＝．【分析】利用对数运算性质即可得出．解：原式＝2+＝2+＝．故答案为：．4．执行如图所示的程序框图，输出的s值为．【分析】由已知中的程序语句可知：该程序的功能是利用循环结构计算并输出变量s的值，模拟程序的运行过程，分析循环中各变量值的变化情况，可得答案．解：模拟程序的运行过程，可得：第一次运行：k＝1时，，第二次运行：k＝2时，，第三次运行：此时k＝3满足k≥3，退出循环，输出，故答案为：．5．在△ABC中，a＝4，b＝5，c＝6，则＝1．【分析】利用余弦定理求出cos C，cos A，即可得出结论．解：∵△ABC中，a＝4，b＝5，c＝6，∴cos C＝＝，cos A＝＝∴sin C＝，sin A＝，∴＝＝1．故答案为：1．6．已知函数，0≤φ≤π．若f（x）是奇函数，则的值为﹣1．【分析】利用两角和的正弦公式化简f（x）的解析式，再根据三角函数的奇偶性，求出φ的值，可得函数的解析式，从而求得的值．解：∵函数＝2sin（x+φ+），0≤φ≤π，若f（x）是奇函数，则φ＝，∴f（x）＝2sin（x+π）＝﹣2sin x，则＝﹣2sin＝﹣1，故答案为：﹣1．7．已知f（x）＝|log3x|，若a，b满足f（a﹣1）＝f（2b﹣1），且a≠2b，则a+b的最小值为．【分析】若a，b满足f（a﹣1）＝f（2b﹣1），且a≠2b，则（a﹣1）（2b﹣1）＝1，则b＝且a＞1，即a+b＝，构造函数，利用导数法，可得函数的最小值．解：∵f（x）＝|log3x|，若a，b满足f（a﹣1）＝f（2b﹣1），且a≠2b，则（a﹣1）（2b﹣1）＝1，则b＝且a﹣1＞0，即a＞1即a+b＝a+＝，由令g（a）＝，则g′（a）＝，令g′（a）＝0，则a＝1±，当a∈（1，1+）时，g′（a）＜0，当a∈（1+，+∞）时，g′（a）＞0，故当a＝1+时，g（a）取最小值，故答案为：．8．将黑白2个小球随机放入编号为1，2，3的三个盒子中，则黑白两球均不在1号盒子的概率为【分析】基本事件总数n＝3×3＝9，黑白两球均不在1号盒子包含的基本事件总数m＝2×2＝4，由此能求出黑白两球均不在1号盒子的概率．解：将黑白2个小球随机放入编号为1，2，3的三个盒子中，基本事件总数n＝3×3＝9，黑白两球均不在1号盒子包含的基本事件总数m＝2×2＝4，∴黑白两球均不在1号盒子的概率为p＝＝．故答案为：．9．若抛物线x2＝4y的焦点到双曲线C：（a＞0，b＞0）的渐近线距离等于，则双曲线C的离心率为3．【分析】先求出抛物线x2＝4y的焦点坐标为（0，1），和双曲线的一条渐近线方程为y ＝x，根据点到直线的距离公式和离心率公式即可求出．解：抛物线x2＝4y的焦点坐标为（0，1），双曲线C：（a＞0，b＞0）的一条渐近线方程为y＝x，∴＝＝，∴e＝＝3，故答案为：3．10．设m，n为空间两条不同的直线，α，β为空间两个不同的平面，给出下列命题：①若m∥α，m∥β，则α∥β；②若m⊥α，m∥β，则α⊥β；③若m∥α，m∥n，则n∥α；④若m⊥α，α∥β，则m⊥β．其中的正确命题序号是②④．【分析】在①中，α与β相交或平行；在②中，由面面垂直的判断定理得α⊥β；在③中，n∥α或n⊂α；在④中，由线面垂直的判定定理得m⊥β．解：由m，n为空间两条不同的直线，α，β为空间两个不同的平面，知：在①中，若m∥α，m∥β，则α与β相交或平行，故①错误；在②中，若m⊥α，m∥β，则由面面垂直的判断定理得α⊥β，故②正确；在③中，若m∥α，m∥n，则n∥α或n⊂α，故③错误；在④中，若m⊥α，α∥β，则由线面垂直的判定定理得m⊥β，故④正确．故答案为：②④．11．设x＞0，y＞0，向量＝（1﹣x，4），＝（x，﹣y），若∥，则x+y的最小值为9．【分析】先根据向量平行得到+＝1，再利用基本不等式即可求出最值．解：因为∥，所以4x+（1﹣x）y＝0，又x＞0，y＞0，所以+＝1，故x+y＝（+）（x+y）＝5++≥9．当＝，+＝1同时成立，即x＝3，y＝6时，等号成立．（x+y）min＝9．故答案为：9．12．在△ABC中，点P是边AB的中点，已知||＝，||＝4，∠ACB＝，则•＝6．【分析】用表示出，根据CP＝计算CB，再计算•的值．解：∵点P是边AB的中点，∴＝+，∴＝++，∴3＝4+×cos+||2，∴||＝2，∴＝4×2×cos＝﹣4，∴•＝（+）＝+＝6．故答案为：6．13．已知正数a，b，c满足b2+2（a+c）b﹣ac＝0，则的最大值为．【分析】由b2+2（a+c）b﹣ac＝0得（b+a+c）2＝ac+（a+c）2≤+（a+c）2＝（a+c）2再解关于b的不等式即可．解：由b2+2（a+c）b﹣ac＝0得（b+a+c）2＝ac+（a+c）2≤+（a+c）2＝（a+c）2，∴b+a+c≤（a+c），∴b≤（a+c），∴≤，当且仅当a＝c时取等．故答案为14．若（m≠0）对一切x≥4恒成立，则实数m的取值范围是（﹣∞，﹣）．【分析】等价于（m2x﹣1）（mx+1）＜0，m分﹣1＜m＜0，及m＝﹣1两类讨论，利用函数的单调性即可求得答案．解：等价于（m2x﹣1）（mx+1）＜0，x1＝，x2＝﹣，若（m≠0）对一切x≥4恒成立，则m＜0，当﹣1≤m＜0时，≥﹣，则＜4，解得﹣1≤m＜﹣，当m＜﹣1时，＜﹣，则﹣＜4，解得m＜﹣1．故答案为：（﹣∞，﹣）．二、解答题：共6小题，共90分．请在答题卡指定区域内作答.解答时应写出文字说明、证明过程或演算步骤．15．如图，四棱锥P﹣ABCD的底面为矩形，AB＝，BC＝1，E，F分别是AB，PC的中点，DE⊥PA．（Ⅰ）求证：EF∥平面PAD；（Ⅱ）求证：平面PAC⊥平面PDE．【分析】（Ⅰ）连接EC，并延长与DA的延长线交于N，则E是AC的中点，可得EF ∥PA，即可证明EF∥平面PAD；（Ⅱ）证明DE⊥平面PAC，再证明平面PAC⊥平面PDE．【解答】证明：（Ⅰ）连接EC，并延长与DA的延长线交于N，则E是AB的中点，因为F是PC的中点，…所以EF∥PN，又EF⊄平面PAD，PN⊂平面PAD，故EF∥平面PAD．…（Ⅱ）设AC∩DE＝G，由△AEG∽△CDG及E为AB中点得＝，又因为AB＝，BC＝1，所以AC＝，AG＝AC＝．所以，又∠BAC为公共角，所以△GAE∽△BAC．所以∠AGE＝∠ABC＝90°，即DE⊥AC．…又DE⊥PA，PA∩AC＝A，所以DE⊥平面PAC．…又DE⊂平面PDE，所以平面PAC⊥面PDE．…16．在三角形ABC中，已知，．（1）求角A的值；（2）若△ABC的面积为，求边BC的长．【分析】（1）先根据已知条件求出tan C，再由tan A＝﹣tan（B+C）求出tan A，从而求出角A；（2）设BC＝a，利用正弦定理得求出AB，再利用tan B＝求出sin B，所以△ABC的面积为：S＝＝＝，所以a＝1，即BC＝1．解：（1）在△ABC中，tan B＝，cos C＝﹣，C∈（，π），∴sin C＝，故tan C＝﹣3，所以，∵0＜A＜π，所以A＝；（2）由（1）知A＝450，设BC＝a，利用正弦定理：得：AB＝，又，解得sin B＝，所以△ABC的面积为：S＝＝＝＝，所以a＝1，即BC＝1．17．建造一个容积为8m3、深为2m的无盖长方体形的水池，已知池底和池壁的造价分别为120元/m2和80元/m2．（1）求总造价y（单位：元）关于底边一边长x（单位：m）的函数解析式，并指出函数的定义域；（2）如果要求总造价不超过2080元，求x的取值范围；（3）求总造价y的最小值．【分析】（1）底边一边长x，则另一边长为，由题意可知y＝320（x+）+480 （x ＞0）；（2）令y≤2080即可求出x的取值范围；（3）利用基本不等式求得x+，当且仅当x＝，即x＝2时，等号成立，从而求出总造价y的最小值．解：（1）底边一边长x，则另一边长为，∴y＝2（x+）×＝320（x+）+480，∴总造价y关于底边一边长x的函数解析式为：y＝320（x+）+480 （x＞0）；（2）由（1）可知：y＝320（x+）+480，∴令y≤2080得，320（x+）+480≤2080，解得：1≤x≤4，∴当x∈[1，4]时，总造价不超过2080元；（3）∵x＞0，∴x+，当且仅当x＝，即x＝2时，等号成立，∴y＝320（x+）+480≥320×4+480＝1760，∴当x＝2时，总造价y的值最小，最小值为1760元．18．在直角坐标系xOy中，已知椭圆＝1，若圆O：x2+y2＝R2（R＞O）的一条切线与椭圆C有两个交点A，B，且•＝0．（1）求圆O的方程；（2）已知椭圆C的上顶点为M，点N在圆O上，直线MN与椭圆C相交于另一点Q，且＝2，求直线MN的方程．【分析】（1）假设圆的切线，与椭圆联立，得出两根之和及两根之积，由数量积为零得圆的半径，即求出圆的方程；（2）设Q，N的坐标，在曲线上，写出坐标之间的关系，写出向量的坐标，利用它们的关系求出坐标，进而求出直线方程．解：（1）假设圆的切线的斜率存在时，设切线方程y＝kx+b，设A（x，y），B（x'，y'）．联立与椭圆的方程整理：（1+2k2）x2+4kbx+2b2﹣6＝0，x+x'＝，xx'＝，∴yy'＝k2xx'+kb（x+x'）+b2＝﹣+＝，因为：＝0，所以：xx'+yy'＝0，∴可得2b2﹣6+b2﹣6k2＝0，∴b2＝2+2k2；①又与圆相切，所以＝R，∴b2＝R2（1+k2）②，由①②得，2+2k2＝2k2R2+R2，∴R2＝2，所以圆的方程x2+y2＝2；（2）由题意得M（0，），设Q（m，n），N（a，b），＝（a，b﹣），＝（m﹣a，n﹣b），由题意得：，∴a＝，b＝；而又由题意：，解得：4n2﹣4﹣9＝0，∴n＝（舍），n＝﹣，m＝±，∴a＝±，b＝0，即N（±，0），所以直线MN的方程±＝1，即直线MN的方程+﹣＝0，﹣y+＝0．19．已知函数．（1）若曲线y＝f（x）在x＝1处的切线的斜率为2，求函数f（x）的单调区间；（2）若函数f（x）在区间（1，e）上有零点，求实数a的取值范围．【分析】（1）求导，由导数的结合意义可求得a＝0，进而得到函数解析式，再解关于导函数的不等式即可得到单调区间；（2）分类讨论，利用零点的存在性定理建立不等式即可求解．解：（1）函数f（x）的定义域为（0，+∞），，则f′（1）＝2（a+1）＝2，解得a＝0，∴f（x）＝2xlnx+1（x＞0），f′（x）＝2（lnx+1），令f′（x）＞0，解得；令f′（x）＜0，解得；∴函数f（x）的单调递减区间为，单调递增区间为；（2）函数在区间（1，e）上是一条不间断的曲线，由（1）知，f′（x）＝2（ax+1）（lnx+1），①当a≥0时，对任意x∈（1，e），ax+1＞0，lnx+1＞0，则f′（x）＞0，故函数f（x）在（1，e）上单调递增，此时对任意的x∈（1，e），都有成立，从而函数f（x）在区间（1，e）上无零点；②当a＜0时，令f′（x）＝0，解得或，其中，（i）若，即a≤﹣1，则对任意x∈（1，e），f′（x）＜0，故函数f（x）在区间（1，e）上单调递减，由题意可得，解得，其中，即，故a的取值范围为﹣2＜a≤﹣1；②若，即，则对任意x∈（1，e），f′（x）＞0，所以函数f（x）在区间（1，e）上单调递增，此时对任意x∈（1，e），都有成立，从而函数f（x）在区间（1，e）上无零点；③若，即，则对任意，所以函数在区间上单调递增，对任意，函数f（x）在区间上单调递减，由题意可得，解得，其中，即，所以a的取值范围为，综上所述，实数a的取值范围为．20．已知数列{a n}、{b n}、{c n}，对于给定的正整数k，记b n＝a n﹣a n+k，c n＝a n+a n+k（n∈N*）．若对任意的正整数n满足：b n≤b n+1，且{c n}是等差数列，则称数列{a n}为“H（k）”数列．（1）若数列{a n}的前n项和为S n＝n2，证明：{a n}为H（k）数列；（2）若数列{a n}为H（1）数列，且a1＝1，b1＝﹣1，c2＝5，求数列{a n}的通项公式；（3）若数列{a n}为H（2）数列，证明：{a n}是等差数列．【分析】（1）直接利用定义法证明数列为H（k）数列．（2）利用赋值法和定义法进行证明，进一步求出数列的通项公式．（3）直接利用代换法和定义法证明数列为等差数列．【解答】证明：（1）当n≥2时，＝2n﹣1．当n＝1时，a1＝S1＝1符合上式，则：a n＝2n﹣1所以：b n＝a n﹣a n+k，整理得：b n＝﹣2k，c n＝a n+a n+k＝4n﹣2k﹣2．则b n≤b n+1，c n+1﹣c n＝4．对任意的正整数n满足b n≤b n+1，且数列{c n}，是公差为4的等差数列，所以：数列{a n}为H（k）数列；（2）由于a1＝1，b1＝﹣1，c2＝5，由数列{a n}为H（1）数列，则数列{c n}是等差数列，且c1＝3，c2＝5，所以：c n＝2n+1．即a n+a n+1＝2n+1所以：a n+1﹣（n+1）＝a n﹣n，则{a n﹣n}是常数列所以：a1﹣1＝0，则：a n＝n．验证：b n＝a n﹣a n﹣1＝﹣1，所以：b n≤b n+1对任意正整数n都成立所以：a n＝n．附：a n+a n+1＝2n+1①，a n+1+a n+2＝2n+3②，②﹣①得：a n+2﹣a n＝2所以：a2k﹣1＝a1+2（k﹣1）＝2k﹣1．a2k＝a2+2（k﹣1）＝2k，所以：a n＝n．证明：（3）由数列{a n}为H（2）数列可知：{c n}是等差数列，记公差为d c n+2﹣c n＝（a n+2+a n+4）﹣（a n+a n+2）＝﹣b n﹣b n+2＝2d，所以：﹣b n+1﹣b n+3＝2d．则：（b n﹣b n+1）+（b n+2﹣b n+3）＝2d﹣2d＝0又b n≤b n+1，所以：b n＝b n+1，所以：数列{b n}为常数列，则b n＝a n﹣a n+2＝b1所以：c n＝a n+a n+2＝2a n﹣b1．由c n+1﹣c n＝2（a n+1﹣a n）＝d，所以：．所以：{a n}是等差数列．【选做题】本题包括A、B、C三小题，请选定其中两题，并在相应的答题区域内作答．若多做，则按作答的前两题评分．解答时应写出文字说明、证明过程或演算步骤．[选修4-2：矩阵与变换]21．已知矩阵A＝，B＝，且AB＝BA．（1）求实数a；（2）求矩阵B的特征值．【分析】（1）AB＝，BA＝，进而求解；（2）矩阵B的特征多项式为f（λ）＝（λ﹣2）（λ﹣1），令f（λ）＝0，进而求解．解：（1）因为AB＝＝，BA＝＝，且AB＝BA，所以a＝0；（2）因为B＝，矩阵B的特征多项式为f（λ）＝＝（λ﹣2）（λ﹣1），令f（λ）＝0，解得λ＝2，λ＝1．[选修4-4：坐标系与参数方程]22．在平面直角坐标系xOy中，已知直线为参数）．现以坐标原点O为极点，以x轴非负半轴为极轴建立极坐标系，设圆C的极坐标方程为ρ＝2cosθ，直线l与圆C交于A，B两点，求弦AB的长．【分析】直线为参数）化为普通方程，圆C的极坐标方程ρ＝2cosθ化为直角坐标方程，求出圆C的圆心到直线l的距离，即可求弦AB的长．解：直线为参数）化为普通方程为4x﹣3y＝0，…圆C的极坐标方程ρ＝2cosθ化为直角坐标方程为（x﹣1）2+y2＝1，…则圆C的圆心到直线l的距离为，…所以．…[选修4-5：不等式选讲]23．已知x1，x2，x3∈（0，+∞），且满足x1+x2+x3＝3x1x2x3，证明：x1x2+x2x3+x3x1≥3．【分析】依题意，，再利用柯西不等式即可得证．【解答】证明：∵x1+x2+x3＝3x1x2x3，∴，∴，当且仅当“x1＝x2＝x3＝1”时取等号，故x1x2+x2x3+x3x1≥3，即得证．【必做题】第22题、第23题，每题10分，共计20分．请在答卷卡指定区域内作答．解答应写出文字说明、证明过程或演算步骤．24．如图，在四棱锥P﹣ABCD中，已知棱AB，AD，AP两两垂直，长度分别为1，2，2．若＝λ，且向量与夹角的余弦值为．（1）求实数λ的值；（2）求直线PB与平面PCD所成角的正弦值．【分析】（1）根据已知条件即可建立坐标系：以A为坐标原点，分别以边AB，AD，AP所在直线为x，y，z轴建立空间直角坐标系，然后即可根据已知条件求出点P，A，B，C，D点的坐标，利用向量与夹角的余弦值为求出λ的值．（2）求出平面PCD的法向量，利用向量夹角的余弦公式求解直线PB与平面PCD所成角的正弦值．解：以A为坐标原点，分别以AB，AD，AP为x，y，z轴建立如图所示空间直角坐标系；则：A（0，0，0），B（1，0，0），D（0，2，0），P（0，0，2）；＝λ，可得C （λ，2，0）．（1）＝（λ，2，﹣2），＝（﹣1，2，0），向量与夹角的余弦值为．可得＝，解得λ＝10（舍去）或λ＝2．实数λ的值为2．；（2）＝（2，2，﹣2），＝（0，2，﹣2），平面PCD的法向量＝（x，y，z）．则且，即：x+y﹣z＝0，y﹣z＝0，∴x＝0，不妨去y＝z＝1，平面PCD的法向量＝（0，1，1）．又＝（1，0，2）．故cos＝＝．直线PB与平面PCD所成角的正弦值为：．25．已知（1+x）2n+1＝a0+a1x+a2x2+…+a2n+1x2n+1，n∈N*．记T n＝（2k+1）a n﹣k．（1）求T2的值；（2）化简T n的表达式，并证明：对任意的n∈N*，T n都能被4n+2整除．【分析】（1）由二项式定理得a i＝，利用公式计算T2的值；（2）由组合数公式化简T n，把T n化为（4n+2）的整数倍即可．解：由二项式定理，得a i＝（i＝0，1，2，…，2n+1）；（1）T2＝a2+3a1+5a0＝+3+5＝30；……（2）因为（n+1+k）＝（n+1+k）•＝＝（2n+1），……所以T n＝（2k+1）a n﹣k＝（2k+1）＝（2k+1）＝[2（n+1+k）﹣（2n+1）]＝2（n+1+k）﹣（2n+1）＝2（2n+1）﹣（2n+1）＝2（2n+1）••（22n+）﹣（2n+1）••22n+1＝（2n+1）；……T n＝（2n+1）＝（2n+1）（+）＝2（2n+1）；因为∈N*，所以T n能被4n+2整除；……注意：只要得出T n＝（2n+1），就给，不必要看过程．。