10真题
国考真题系列——10年真题及解析(打印版)
2010年中央、国家机关公务员录用考试《行政职业能力测验》试卷第一部分言语理解与表达（共40题，参考时限35分钟）请开始答题：1．“诗是不可译的，中国古典诗歌更是不可译的。
”爱好古典诗歌的中国人，包括不少作家、学者、翻译家常常如是说，语气中带着七分三分。
然而，话说回来，如果没有翻译。
中国古典诗歌如何走出国门，走向世界呢？依此填入划横线部分最恰当的一项是（）A．自豪遗憾B．无奈悲伤C．感伤埋怨D．骄傲惭愧2．茶艺与茶道精神是中国茶文化的核心，“艺”是指制茶、烹茶、品茶等艺茶之术，“道”是指艺茶过程中所的精神。
有道而无艺，那是的理论；有艺而无道，艺则无精、无神。
依此填入划横线部分最恰当的一项是（）A．传达虚浮B．包涵虚无C．贯穿空洞D．体现枯燥3．从20世纪90年代“人类基因工程”计划启动之日起，美国，日本，欧洲等展开了一场激烈的基因专利争夺战，因为谁拥有专利，就意味着谁能在国籍上获得基因产业的“王牌”，谁就能拥有今后基因开发的庞大市场，为此，美国等少数发达国家大量地将阶段性研究成果申请了专利。
依此填入划横线部分最恰当的一项是（）A．垄断抢先B．操作独立C．控制自发D．专营及时4．明代工艺品的名字大都先强调年号，然后再强调东西本身，但景泰蓝不是在景泰年间出现，而是在元代就出现了，到了景泰年间，皇家的重视使它，因此有了今天这样一个通俗易懂且带有文学色彩的名字──景泰蓝。
依此填入划横线部分最恰当的一项是（）A．如日中天B．名声大噪C．声名鹊起D．享誉中外5．五四运动后，许多人追求真理。
追求人们开始用新的眼光看中国、看世界，从对各种社会思潮，政治主张和政治力量的中认真思考，逐步看到西方的种种社会，开始怀疑资产阶级共和国的救国方案。
依此填入划横线部分最恰当的一项是（）A．识别通病B．甄别矛盾C．辨别现象D．鉴别弊端6．世界主要经济发展发达国家和地区目前已就发展低碳经济达成共识：以经济发展模式由“高碳”向“低碳”转型为，通过市场机制下的经济手段推动低碳经济的发展，以减缓人类活动对气候的破坏并逐渐达到一种互相的良性发展状态。
10年英语真题
2010 年全国硕士研究生入学统一考试英语试题Section I Use of EnglishDirections:Read the following text. Choose the best word(s) for each numbered blank and mark[A], [B], [C] or [D] on ANSWER SHEET 1. (10 points)In 1924 American’National Research Council sent to engine ers to supervise a series of industrial experiments at a large telephone-parts factory called the Hawthore Plant near Chicago.It hoped they would learn how stop-floor lignting__1__workers productivity. Instead, the studies ended __2___giving their name to the “Hawthorne effect”, the extremely influential idea that the very___3____to being experimentedupon changed subjects’ behavior.The idea arose because of the __4____behavior of the women in the Hawthorne plant.According to __5____of the experments, their hourly output rose when lighting was increased, but also when it was dimmed. It did not __6____what was done in the experiment; ___7_something was changed ,productivity rose. A(n)___8___that they were being experimented upon seemed to be ____9___to alte r workers’ behavior____10____itself.After several decades, the same data were _11__ to econometric the analysis. The Hawthorne experiments have another surprise in store: _12 __the descriptions on record, no systematic _13__ was found that levels of reproductivity were relatedto changes in lighting.It turns out that particular way of conducting the experiments may have led to__ 14__ interpretation of what happed.__ 15___ , lighting was always changed on a Sunday .When work started again on Monday, output __16___ rose compared with the previous Saturday and__ 17 _to rise for the next couple of days.__ 18__ a comparison with data for weeks when there was no experimentation showed that output always went up on Monday. Workers__ 19__ to be diligent for the first few days of the weeking week in any case , before __20 __a plateau and then slackening off. This suggests that the alleged “Hawthorne effect “ is hard to pin down.1. [A] affected [B] achieved [C] extracted [D] restored2. [A] at [B]up [C] with [D] off3. [A]truth [B]sight [C] act [D] proof4. [A] controversial [B] perplexing [C]mischievous [D] ambiguous5. [A]requirements [B]explanations [C] accounts [D] assessments6. [A] conclude [B] matter [C] indicate [D] work7. [A] as far as [B] for fear that [C] in case that [D] so long so8. [A] awareness [B] expectation [C] sentiment [D] illusion9. [A] suitable [B] excessive [C] enough [D] abundant10. [A] about [B] for [C] on [D] by11. [A] compared [B]shown [C] subjected [D] conveyed12. [A] contrary to [B] consistent with [C] parallel with [D] pealliar to13. [A] evidence [B]guidance [C]implication [D]source14. [A] disputable [B]enlightening [C]reliable [D]misleading15. [A] In contrast [B] For example [C] In consequence [D] As usual16. [A] duly [B]accidentally [C] unpredictably [D] suddenly17. [A]failed [B]ceased [C]started [D]continued20. [A]breaking [B]climbing [C]surpassing [D]hittingSection II Reading ComprehensionPart ADirections:Read the following four texts. Answer the questions below each text by choosing [A],[B], [C] or [D]. Mark your answers on ANSWER SHEET 1. (40 points)Text 1Text 2Over the past decade, thousands of patents have seen granted for what are called business methods. received one for its “one-click” online payment system. Merrill Lynch got legal protection for an asset allocation strategy. One inventor patented a technique for lifting a box.Now the nation’s top patent court appears completely rea dy to scale back on business-method patents, which have been controversial ever since they were first authorized 10 years ago. In a move that has intellectual-property lawyers abuzz the U.S. court of Appeals for the federal circuit said it would use a particular case to conduct a broad review of business-method patents. In re Bilski , as the case is known , is “a very big deal”, says Dennis’D. Crouch of the University of Missouri School of law. It “has the potential to eliminate an entire class ofpatents.”Curbs on business-method claims would be a dramatic about-face, because it was the federal circuit itself that introduced such patents with is 1998 decision in the so-called state Street Bank case, approving a patent on a way of pooling mutual-fund assets. That ruling produced an explosion in business-method patent filings, initially by emerging internet companies trying to stake out exclusive pinhts to specific types of online transactions. Later, move established companies raced to add such patents to their files, if only as a defensive move against rivals that might bent them to the punch. In 2005, IBM noted in a court filing that it had been issued more than 300 business-method patents despite the fact that it questioned the legal basis for granting them. Similarly, some Wall Street investment films armed themselves with patents for financial products, even as they took positions in courtcases opposing the practice.The Bilski case involves a claimed patent on a method for hedging risk in the energy market. The Federal circuit issued an unusual order stating that the case would be heard by all 12 of the court’s judges, rather than a typical panel of three, and that one issue it wants to evaluate is weather it should” reconsider” its statestreet Bank ruling.The Federal Circuit’s action comes in the wake of a series of recent decisions by the supreme Count that has nurrowed the scope of protections for patent holders. Last April, for example the justices signaled that too many patents were being upheld for “inventions” that are obvious. The judges on the Federal circuit are “reacting to the anti-patient trend at the supreme court” ,says Harole C.wegner, a partend attorney and professor at aeorge Washington University Law School.26. Business-method patents have recently aroused concern because of[A] their limited value to business[B] their connection with asset allocation[C] the possible restriction on their granting[D] the controversy over authorization27. Which of the following is true of the Bilski case?[A] Its rulling complies with the court decisions[B] It involves a very big business transaction[C] It has been dismissed by the Federal Circuit[D] It may change the legal practices in the U.S.28. The word “about-face” (Line 1, Paro 3) most probably means[A] loss of good will[B] increase of hostility[C] change of attitude[D] enhancement of disnity29. We learn from the last two paragraphs that business-method patents[A] are immune to legal challenges[B] are often unnecessarily issued[C] lower the esteem for patent holders[D] increase the incidence of risks30. Which of the following would be the subject of the text?[A] A looming threat to business-method patents[B] Protection for business-method patent holders[C] A legal case regarding business-method patents[D] A prevailing trend against business-method patentsText 3In his book The Tipping Point,Malcolm Aladuell argues that social epidemics are driven in large part by the acting of a tiny minority of special individuals,often called influentials,who are unusually informed,persuasive,or well-connected.The idea is intuitively compelling,but it doesn’t explain how ideas actually spread. The supposed importance of influentials derives from a plausible sounding but largely untested theory called the “two step flow of communication”: Information flows from the media to the influentials and from them to ereryone else.Marketers have embraced the two-step flow because it suggests that if they can just find and influence the influentials,those selected people will do most of the work for them. The theory also seems to explain the sudden and unexpected popularity of people was wearing, promoting or developing whaterver it is before anyone else paid attention. Anecdotal evidence of this kind fits nicely with the idea that only certain specialpeople can drive trends.In their recent work,however,some researchers have come up with the finding that influentials have far less impact on social epidemics than is generally supposed.In fact,they don’t seem to be required of all.The researchers’ argument stems from a simple obserrating about social influence,with the exception of a few celebrities like Oprah Winfrey —whose outsize presence is primarily a function of media,not interpersonal,influence — even the most influential members of a population simply don’t interact with that many others.Yet it is precisely these non-celebring influentials who,according to thetwo-step-flow theory,are supposed to drive social epidemics by influcencing their friends and colleagues directly.For a social epidemic to occur,however,each person so affected,must then influcence his or her own acquaintances,who must in turn influence theirs,and so on;and just how many others pay attention to each of these people has little to do with the initial influential.If people in the network just two degrees removed from the initial influential prove resistant,for example from the initial influential prove resistant,for example the casecade of change won’t propagate very far or affect many people.Building on the basic truth about interpersonal influence,the researchers studied the dynamics of populations manipulating a number of variables relating of populations,manipulating a number of variables relating to people’s ability to influence others and their tendence to be.31.By citing the book The Tipping Point,the author intends to[A]analyze the consequences of social epidemics[B]discuss influentials’ function in spreading ideas[C]exemplify people’s intuitive response to social epidemics[D]describe the essential characteristics of influentials.32.The author suggests that the “two-step-flow theory”[A]serves as a solution to marketing problems[B]has helped explain certain prevalent trends[C]has won support from influentials[D]requires solid evidence for its validity33.what the resarchers have observed recenty shows that[A] the power of influence goes with social interactions[B] interpersonal links can be enhanced through the media[C] influentials have more channels to reach the public[D] most celebrities enjoy wide media attention34.The underlined phrase “these people” in paragraph 4 refers to the ones who[A] stay outside the network of social influnce[B] have little contact with the source of influnence[C] are influenced and then influence others[D] are influenced by the initial influential35.what is the essential element in the dynamics of social influence?[A]The eagerness to be accepted[B]The impulse to influence others[C]The readiness to be influenced[D]The inclination to rely on othersText 4Bankers have been blaming themselves for their troubles in public. Behind the scenes, they have been taking aim at someone else: the accounting standard-setters. Their r ules, moan the banks, have forced them to report enormous losses, and it’s just not fair. These rules say they must value some assets at the price a third party would pay, not the price managers and regulators would like them to fetch.Unfortunately, bank s’ lobbying now seems to be working. The details may be unknowable, but the independence of standard-setters, essential to the proper functioning of capital markets, is being compromised. And, unless banks carry toxic assets at prices that attract buyers, reviving the banking system will be difficult. After a bruising encounter with Congress, America’s Financial Accounting Standards Board (FASB) rushed through rule changes. These gave banks more freedom to use models to value illiquid assets and more flexibility in recognizing losses on long-term assets in their income statement. Bob Herz, the FASB’s chairman, cried out against those who “question our motives.” Yet bank shares rose and the changes enhance what one lobby group politely calls “the use of judgment by management.” European ministers instantly demanded that the International Accounting Standards Board (IASB) do likewise. The IASB says it does not want to act without overall planning, but the pressure to fold when it completes it reconstruction of rules later this year is strong. Charlie McCreevy, a European commissioner, warned the IASB that it did “not live in a political vacuum” but “in the real word” and that Europecould yet develop different rules.It was banks that were on the wrong planet, with accounts that vastly overvalued assets. Today they argue that market prices overstate losses, because they largely reflect the temporary illiquidity of markets, not the likely extent of bad debts. The truth will not be known for years. But bank’s sh ares trade below their book value, suggesting that investors are skeptical. And dead markets partly reflect the paralysis of banks which will not sell assets for fear of booking losses, yet are reluctant to buy all those supposed bargains.To get the syst em working again, losses must be recognized and dealt with. America’s new plan to buy up toxic assets will not work unless banks mark assets to levels which buyers find attractive. Successful markets require independent and evencombative standard-setters. The FASB and IASB have been exactly that, cleaning up rules on stock options and pensions, for example, against hostility form special interests. But by giving in to critics now they are inviting pressure to make moreconcessions.36. Bankers complained that they were forced to[A] follow unfavorable asset evaluation rules[B]collect payments from third parties[C]cooperate with the price managers[D]reevaluate some of their assets.37.According to the author , the rule changes of the FASB may result in[A]the diminishing role of management[B]the revival of the banking system[C]the banks’ long-term asset losses[D]the weakening of its independence38.According to Paragraph 4, McCreevy objects to the IASB’s attempt to[A]keep away from political influences.[B]evade the pressure from their peers.[C]act on their own in rule-setting.[D]take gradual measures in reform.39.The author thinks the banks were “on the wrong planet ”in that they[A]misinterpreted market price indicators[B]exaggerated the real value of their assets[C]neglected the likely existence of bad debts.[D]denied booking losses in their sale of assets.40.The author’s attitude towards standard-setters is one of[A]satisfaction.[B]skepticism.[C]objectiveness[D]sympathyPart BDirections:For Questions 41-45, choose the most suitable paragraphs from the first A-G and fill them into the numbered boxes to from a coherent text. Paragraph E has been correctly placed. There is one paragraph which dose not fit in with the text. Mark your answerson ANSWER SHEET1. (10 points)[A]The first and more important is the consumer’s growing preference for eatingout;the consumption of food and drink in places other than homes has risen from about 32 percent of total consumption in 1995 to 35 percent in 2000 and is expected to approach 38 percent by 2005. This development is boosting wholesale demand from the food service segment by 4 to 5 percent a year across Europe,compared with growth in retail demand of 1 to 2 percent. Meanwhile,as the recession is looming large, people are getting anxious. They tend to keep a tighter hold on their purse and consider eating at home a realistic alternative.[B]Retail sales of food and drink in Europe’s largest markets are at a standstill, leaving European grocery retailers hungry for opportunities to grow. Most leading retailers have already tried e-commerce, with limited success, and expansion abroad. But almost all have ignored the big, profitable opportunity in their own backyard: the wholesale food and drink trade, which appears to be just the kind of marketretailers need.[C]Will such variations bring about a change in the overall structure of the food and drink market? Definitely not. The functioning of the market is based on flexible trends dominated by potential buyers.In other words,it is up to the buyer,tather than the seller,to decide what to buy .At any rate,this change will ultimately be acclaimed by an ever-growing number of both domestic and international consumers,regardless of how long the current consummer pattern will take hold.[D]All in all, this clearly seems to be a market in which big retailers that master the intricacies of wholesaling in Europe may well expect to rake in substantial profits there by. At least, that is how it looks as a whole. Closer inspection reveals import differences among the biggest national markets, especially in their customer segments and wholesale structures, as well as the competitive dynamics of individual food and drink categories. Big retailers must understand these differences before they can identify the segments of European whloesaling in which particular abilities might unseat smaller but entrenched competitors. New skills and unfamiliar businessmodels are needed too.[E]Despite variations in detail, wholesale markets in the countries that have been closely examined—France, Germany, Italy, and Spain—are made out of same building blocks. Demand comes mainly from two sources: independent morn-and-pop grocery stores which, unlike large retail chains, are two small to buy straight from producers, and food service operators range from snack machines to large institutional catering ventures, but most of these businesses are known in the trade as “horeca”: hotels, restaurants, and cafes. Overall, Europe’s retail wholesale market, but the figures, when added together, mask two opposing trends.[F]For example, wholesale food and drink sales come to ＄ 268 billion in France, Germany, Italy, Spain, and the United Kingdom in 2000- more than 40 percent of retail sales. Moreover, average overall margins are higher in wholesale than in retail; wholesale demand from the food service sector is growing quickly as more Europeans eat out more often; and changes in the competitive dynamics of this fragmented industry are at last making it feasible for wholesalers to consolidate. [G]However, none of these requirements should deter large retails and even some large good producers and existing wholesalers from trying their hand, for those that master the intricacies of wholesaling in Europe stand to reap considerable gains.41 → 42 → 43 → 44 → E → 45Part CDirections:Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written carefully on ANSWER SHEET 2. (10 points) One basic weakness in a comservation system based wholly one economic motives is that most members of the munity have no economic value.Yet these ereatures are members of the biotic community and ,if its stability depends on itsinteyrity,they are entitled to continuance.When one of these noneconomic categories is threatened and,if we happen to love it .We invert excuses to give it economic importance.At the beginning of century songbiras were supposed to be disappearing.(46) Scinentists jumped to the rescue with some distinctly shaky evidence to the effect that insects would eat us up if birds failed to control them,the evideuce had to be comic in order to be valid.It is pamful to read these round about accounts today .We have no land ethic yet ,(47) but we have at least drawn near the point of admitting that birds should continue survival as a matter of intrinsic right,regardless of the presence or absence ofeconomic advantage to us.A panallel situation exists in respect of predatory mamals and fish-eating birds .(48) Time was when biologists somewhat over worded the evidence that these creatures preserve the health of game by killing the physically weak,or that they prey onlyon “worthless species”.Some species of tree have been read out of the party by economics-minded foresters because they grow too slowly .or have too low a sale vale to pay as imeber crops (49) In Europe ,where forestry is ecologically more advanced ,the Non-commercial tree species are recognized as members of native forest community ,to be preservedas such ,within reason.To sum up:a system of conservation based solely on economic self-interest is hopelessly lopsided.(50) It tends to ignore, and thus eventually to eliminate, many elements in the land community that lack commercial value, but that are essential to its healthy functioning.Without the uneconomic pats.Section Ⅲ WritingPart A51. Directions:You are supposed to write for the postgraduate association a notice to recruit volunteers for an international conference on globalization, you should conclude the basic qualification of applicant and the other information you think relative. You should write about 100 words. Do not sign your own name at the end of the letter. Use “postgraduate association” instead.Part B52. Directions:Write an essay of 160-200 words based on the following drawing. In your essay, you should1) describe the drawing briefly,2) explain its intended meaning, and then3) give your comments.You should write neatly on ANSHWER SHEET 2. (20 points)。
2022年10月自考真题
2022年10月自考真题2022年10月自考真题一、选择题（每小题3分，共30分）1.以下哪个语言不属于面向对象编程语言？A.C++B.C#C.JavaD.Fortran2.某Web应用程序在Microsoft Internet Information Server （IIS）中的运行模式是什么？A.独立模式B.网络模式C.安全模式D.安装模式3.要让所有台式机都安装某应用程序，可以采用哪种方法？A.分发B.数据库C.网络D.客户端4.配置IIS有何作用？A.用于定义虚拟站点B.用于管理工作流C.用于跟踪Web服务器的流量D.用于分配IP地址5.以下哪项内容涵盖了UML中的构造视图？A.类图B.计量图C.状态图D.流程图二、简答题（每小题10分，共20分）6.请说明TCP/IP (Transmission Control Protocol /Internet Protocol) 的功能。
TCP/IP是传输控制协议/因特网协议的缩写，是应用于互联网的主要协议，是应用程序如何彼此发送信息的标准。
它指定了数据如何在网络中发送和接收，包括IP协议概念、地址格式，TCP协议概念、连接管理、流量控制和错误检测等等。
它的功能主要有：（1）定义了在网络中传输数据的格式；（2）提供了将信息在网络中发送的流程；（3）实现了数据传输的安全性，稳定性与可靠性；（4）支持点对点通信。
7.请简述面向对象编程（Object-Oriented Programming，OOP）的三大特征。
面向对象编程（Object-Oriented Programming，OOP）的三大特征是：（1）封装：将对象的属性和行为封装在一起，使其外部只能通过调用其提供的方法来访问，隐藏状态数据；（2）继承：使用继承技术来实现子类继承父类的属性和行为，实现代码复用，并可以根据实际需要进行扩展和修改；（3）多态：指对象具有多种形态，不同对象响应不同的消息，实现不同的行为，这样，程序的可维护性和可扩展性会更强。
10年真题英语一答案解析
10年真题英语一答案解析一、阅读理解部分1. A对于这一题，我们需要识别关键词：“international education”、“share global perspectives”、“study abroad”、“broaden their horizons”。
通过分析这些关键词，我们可以得出理解这段文字的关键信息为：国际教育的重要性，通过留学可以拓宽视野，促进自身发展。
答案是A。
2. B这一题的关键信息在第二段的最后一句：“language exposure”。
根据这个信息，我们可以得出答案为B。
3. D此题的答案位于第三段的最后一句：“the majority of immigrants end up....where many feel they do not belong”。
根据这一信息我们可以得出答案为D。
4. C这一题的答案可以从第四段的最后一句中得到：“show impressive declinations in violence”。
根据这一信息我们可以得出答案为C。
5. A这一题的关键信息在第五段的第一句：“the U.N. has proclaimed that access to sanitary facilities is a human right”。
根据这个信息，我们可以得到答案是A。
二、完型填空部分6. B根据题干我们可以得知这一题目为一个对比题。
它与前面的“figures of speech”相呼应，我们需要选出与“figures of speech”相对应的词语。
选项B相对应于“statistics”。
答案是B。
7. D题干中提到了“realize the value of the individual”，我们需要识别关键词找出与之相呼应的人物。
选项D中的“leaders”与题干意思相呼应。
答案是D。
8. C题干中的关键信息为“administrators....classes”，选项C中的“are in charge of”与之意思相对应。
10年高考真题-平面向量的数量积及其应用
5.2 平面向量的数量积及其应用考点一 平面向量的数量积1.(2020课标Ⅲ理,6,5分)已知向量a,b 满足|a|=5,|b|=6,a ·b=-6,则cos<a,a+b>=( ) A.-3135 B.-1935 C.1735 D.1935 答案 D 由题意得cos<a,a+b>=a ·(a+b)|a|·|a+b|=2|a|·√a 2+b +2a ·b=5×√25+36-12=1935.故选D.2.(2020课标Ⅱ,5,5分)已知单位向量a,b 的夹角为60°,则在下列向量中,与b 垂直的是( ) A.a+2b B.2a+b C.a-2b D.2a-b答案 D 解法一:要判断A 、B 、C 、D 四个选项中的向量哪个与b 垂直,只需判断这四个向量哪个与b 的数量积为零即可.A.(a+2b)·b=a ·b+2b 2=|a||b|cos 60°+2|b|2=1×1×cos 60°+2×12=52≠0. B.(2a+b)·b=2a ·b+b 2=2|a||b|cos 60°+|b|2=2×1×1×cos 60°+12=2≠0. C.(a-2b)·b=a ·b-2b 2=|a||b|cos 60°-2|b|2=1×1×cos 60°-2×12=-32≠0. D.(2a-b)·b=2a ·b-b 2=2|a||b|cos 60°-|b|2=2×1×1×cos 60°-12=0.故选D. 解法二:由于a 与b 均为单位向量,且夹角为60°,所以可设a=(1,0),b=(12,√32). 对于选项A,a+2b=(2,√3),则b ·(a+2b)=12×2+√32×√3=52≠0,所以b 与a+2b 不垂直; 对于选项B,2a+b=(52,√32),则b ·(2a+b)=12×52+√32×√32=2≠0,所以b 与2a+b 不垂直; 对于选项C,a-2b=(0,-√3),则b ·(a-2b)=12×0-√32×√3=-32≠0,所以b 与a-2b 不垂直; 对于选项D,2a-b=(32,-√32),则b ·(2a-b)=12×32-√32×√32=0,所以b 与2a-b 垂直. 故选D.3.(2019课标Ⅱ,3,5分)已知向量a=(2,3),b=(3,2),则|a-b|=( ) A.√2 B.2 C.5√2 D.50答案 A 本题主要考查平面向量的坐标运算以及向量模的计算;考查数学运算的核心素养. ∵a=(2,3),b=(3,2),∴|a|2=13,|b|2=13,a ·b=12,则|a-b|=√a 2-2a ·b +b 2=√13-2×12+13=√2.故选A. 一题多解 ∵a=(2,3),b=(3,2),∴a-b=(-1,1),∴|a-b|=√(-1)2+12=√2,故选A.4.(2016课标Ⅲ,3,5分)已知向量BA ⃗⃗⃗⃗⃗ =(12,√32),BC ⃗⃗⃗⃗⃗ =(√32,12),则∠ABC=( ) A.30° B.45° C.60° D.120° 答案 A 由已知得cos ∠ABC=BA ⃗⃗⃗⃗⃗⃗ ·BC⃗⃗⃗⃗⃗ |BA ⃗⃗⃗⃗⃗⃗|·|BC ⃗⃗⃗⃗⃗|=√32,所以∠ABC=30°,故选A.5.(2016天津文,7,5分)已知△ABC 是边长为1的等边三角形,点D,E 分别是边AB,BC 的中点,连接DE 并延长到点F,使得DE=2EF,则AF ⃗⃗⃗⃗⃗ ·BC ⃗⃗⃗⃗⃗ 的值为( ) A.-58 B.18 C.14 D.118 答案 B 建立如图所示的平面直角坐标系.则B (-12,0),C (12,0),A (0,√32),所以BC⃗⃗⃗⃗⃗ =(1,0). 易知DE=12AC,∠FEC=∠ACE=60°,则EF=14AC=14, 所以点F 的坐标为(18,-√38),所以AF⃗⃗⃗⃗⃗ =(18,-5√38), 所以AF ⃗⃗⃗⃗⃗ ·BC ⃗⃗⃗⃗⃗ =(18,-5√38)·(1,0)=18.故选B. 疑难突破 利用公式a ·b=|a||b|cos<a,b>求解十分困难,可以考虑建立适当的平面直角坐标系,利用坐标运算求解.确定点F 的坐标是解题的关键.评析 本题考查了向量的坐标运算和向量的数量积,考查运算求解能力和数形结合思想. 6.(2016课标Ⅱ理,3,5分)已知向量a=(1,m),b=(3,-2),且(a+b)⊥b,则m=( ) A.-8 B.-6 C.6 D.8答案 D 由题可得a+b=(4,m-2),又(a+b)⊥b,∴4×3-2×(m-2)=0,∴m=8.故选D. 7.(2015山东理,4,5分)已知菱形ABCD 的边长为a,∠ABC=60°,则BD ⃗⃗⃗⃗⃗⃗ ·CD ⃗⃗⃗⃗⃗ =( ) A.-32a 2 B.-34a 2 C.34a 2 D.32a 2答案 D BD ⃗⃗⃗⃗⃗⃗ ·CD ⃗⃗⃗⃗⃗ =(BC ⃗⃗⃗⃗⃗ +CD ⃗⃗⃗⃗⃗ )·CD ⃗⃗⃗⃗⃗ =BC ⃗⃗⃗⃗⃗ ·CD ⃗⃗⃗⃗⃗ +CD ⃗⃗⃗⃗⃗ 2=12a 2+a 2=32a 2.8.(2015课标Ⅱ文,4,5分)向量a=(1,-1),b=(-1,2),则(2a+b)·a=( ) A.-1 B.0 C.1 D.2答案 C 因为2a+b=2(1,-1)+(-1,2)=(2,-2)+(-1,2)=(1,0),所以(2a+b)·a=(1,0)·(1,-1)=1×1+0×(-1)=1.故选C. 9.(2015四川理,7,5分)设四边形ABCD 为平行四边形,|AB ⃗⃗⃗⃗⃗ |=6,|AD ⃗⃗⃗⃗⃗ |=4.若点M,N 满足BM ⃗⃗⃗⃗⃗⃗ =3MC ⃗⃗⃗⃗⃗⃗ ,DN ⃗⃗⃗⃗⃗⃗ =2NC ⃗⃗⃗⃗⃗ ,则AM ⃗⃗⃗⃗⃗⃗ ·NM⃗⃗⃗⃗⃗⃗⃗ =( ) A.20 B.15 C.9 D.6答案 C 依题意有AM ⃗⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +BM ⃗⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +34BC ⃗⃗⃗⃗⃗ ,NM ⃗⃗⃗⃗⃗⃗⃗ =NC ⃗⃗⃗⃗⃗ +CM ⃗⃗⃗⃗⃗⃗ =13DC ⃗⃗⃗⃗⃗ -14BC ⃗⃗⃗⃗⃗ =13AB ⃗⃗⃗⃗⃗ -14BC ⃗⃗⃗⃗⃗ ,所以AM ⃗⃗⃗⃗⃗⃗ ·NM ⃗⃗⃗⃗⃗⃗⃗ =(AB ⃗⃗⃗⃗⃗ +34BC ⃗⃗⃗⃗⃗ )·(13AB ⃗⃗⃗⃗⃗ -14BC ⃗⃗⃗⃗⃗ )=13AB ⃗⃗⃗⃗⃗ 2-316BC ⃗⃗⃗⃗⃗ 2=9.故选C.10.(2015福建文,7,5分)设a=(1,2),b=(1,1),c=a+kb.若b ⊥c,则实数k 的值等于( ) A.-32B.-53C.53D.32答案 A c=a+kb=(1+k,2+k).由b ⊥c,得b ·c=0,即1+k+2+k=0,解得k=-32.故选A.11.(2015广东文,9,5分)在平面直角坐标系xOy 中,已知四边形ABCD 是平行四边形,AB ⃗⃗⃗⃗⃗ =(1,-2),AD ⃗⃗⃗⃗⃗ =(2,1),则AD ⃗⃗⃗⃗⃗ ·AC⃗⃗⃗⃗⃗ =( ) A.5 B.4 C.3 D.2答案 A ∵四边形ABCD 是平行四边形,∴AC ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ =(3,-1),∴AD ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =2×3+1×(-1)=5.选A. 12.(2015重庆理,6,5分)若非零向量a,b 满足|a|=2√23|b|,且(a-b)⊥(3a+2b),则a 与b 的夹角为( )A.π4 B.π2 C.3π4 D.π答案 A ∵(a-b)⊥(3a+2b),∴(a-b)·(3a+2b)=0⇒3|a|2-a ·b-2|b|2=0⇒3|a|2-|a|·|b|·cos <a,b>-2|b|2=0. 又∵|a|=2√23|b|,∴83|b|2-2√23|b|2·cos <a,b>-2|b|2=0.∴cos <a,b>=√22.∵<a,b>∈[0,π], ∴<a,b>=π4.选A.13.(2015重庆文,7,5分)已知非零向量a,b 满足|b|=4|a|,且a ⊥(2a+b),则a 与b 的夹角为( ) A.π3 B.π2 C.2π3 D.5π6 答案 C 因为a ⊥(2a+b),所以a ·(2a+b)=0, 得到a ·b=-2|a|2,设a 与b 的夹角为θ,则cos θ=a ·b |a||b|=-2|a|24|a|2=-12,又0≤θ≤π,所以θ=2π3,故选C.14.(2014课标Ⅱ,理3,文4,5分)设向量a,b 满足|a+b|=√10,|a-b|=√6,则a ·b=( ) A.1 B.2 C.3 D.5 答案 A ∵|a+b|=√10,∴a 2+2a ·b+b 2=10.① 又|a-b|=√6,∴a 2-2a ·b+b 2=6.② ①-②,得4a ·b=4,即a ·b=1,故选A.15.(2014大纲全国文,6,5分)已知a 、b 为单位向量,其夹角为60°,则(2a-b)·b=( ) A.-1 B.0 C.1 D.2答案 B (2a-b)·b=2a ·b-|b|2=2×1×1×cos 60°-12=0,故选B.16.(2014大纲全国理,4,5分)若向量a 、b 满足:|a|=1,(a+b)⊥a,(2a+b)⊥b,则|b|=( ) A.2 B.√2 C.1 D.√22答案 B 由题意得{(a +b)·a =a 2+a ·b =0,(2a +b)·b =2a ·b +b 2=0⇒-2a 2+b 2=0,即-2|a|2+|b|2=0,又|a|=1,∴|b|=√2.故选B. 17.(2020课标Ⅱ理,13,5分)已知单位向量a,b 的夹角为45°,ka-b 与a 垂直,则k= . 答案√22解析 因为(ka-b)·a=ka 2-a ·b=0,且单位向量a,b 的夹角为45°,所以k-√22=0,即k=√22.18.(2020课标Ⅰ理,14,5分)设a,b 为单位向量,且|a+b|=1,则|a-b|= . 答案 √3解析 由|a+b|=1,得|a+b|2=1,即a 2+b 2+2a ·b=1,而|a|=|b|=1,故a ·b=-12,|a-b|=√|a -b|2=√a 2+b 2-2a ·b =√1+1+1=√3.19.(2020浙江,17,4分)已知平面单位向量e 1,e 2,满足|2e 1-e 2|≤√2.设a=e 1+e 2,b=3e 1+e 2,向量a,b 的夹角为θ,则cos 2θ的最小值是 . 答案2829解析 由题可知{a =e 1+e 2,b =3e 1+e 2⇔{e 1=b -a2,e 2=3a -b 2,从而{ |b -a2|=1,|3a -b2|=1,|3b -5a 2|≤√2⇔{|b -a|=2,|3a -b|=2,|3b -5a|≤2√2 ⇔{a 2-2a ·b +b 2=4①,9a 2-6a ·b +b 2=4②,25a 2-30a ·b +9b 2≤8③,由①②可得{a ·b =2a 2④,b 2=4+3a 2⑤,代入③可得a 2≥72, 从而cos θ=a ·b |a||b|=2a 2|a||b|=2|a||b|=2√|a|24+3|a|2=2√14|a|2+3≥2√729,所以cos 2θ≥2829,故cos 2θ的最小值为2829.20.(2018上海,8,5分)在平面直角坐标系中,已知点A(-1,0)、B(2,0),E 、F 是y 轴上的两个动点,且|EF ⃗⃗⃗⃗⃗ |=2,则AE ⃗⃗⃗⃗⃗ ·BF ⃗⃗⃗⃗⃗ 的最小值为 . 答案 -3解析 本题主要考查数量积的运算以及二次函数的最值问题.设E(0,m),F(0,n),又A(-1,0),B(2,0), ∴AE ⃗⃗⃗⃗⃗ =(1,m),BF⃗⃗⃗⃗⃗ =(-2,n). ∴AE ⃗⃗⃗⃗⃗ ·BF ⃗⃗⃗⃗⃗ =-2+mn,又知|EF ⃗⃗⃗⃗⃗ |=2, ∴|m-n|=2.①当m=n+2时,AE ⃗⃗⃗⃗⃗ ·BF ⃗⃗⃗⃗⃗ =mn-2=(n+2)n-2=n 2+2n-2=(n+1)2-3. ∴当n=-1,即E(0,1),F(0,-1)时,AE ⃗⃗⃗⃗⃗ ·BF ⃗⃗⃗⃗⃗ 取得最小值-3. ②当m=n-2时,AE ⃗⃗⃗⃗⃗ ·BF ⃗⃗⃗⃗⃗ =mn-2=(n-2)n-2=n 2-2n-2=(n-1)2-3. ∴当n=1,即E(0,-1),F(0,1)时,AE ⃗⃗⃗⃗⃗ ·BF ⃗⃗⃗⃗⃗ 取得最小值-3. 综上可知,AE ⃗⃗⃗⃗⃗ ·BF ⃗⃗⃗⃗⃗ 的最小值为-3.21.(2017山东理,12,5分)已知e 1,e 2是互相垂直的单位向量.若√3e 1-e 2与e 1+λe 2的夹角为60°,则实数λ的值是 . 答案√33解析 本题考查向量的坐标运算和向量的夹角公式.由题意不妨设e 1=(1,0),e 2=(0,1),则√3e 1-e 2=(√3,-1),e 1+λe 2=(1,λ).根据向量的夹角公式得cos =√3,2√1+λ=√3-2√1+λ=12,所以√3-λ=√1+λ2,解得λ=√33.疑难突破 根据“e 1,e 2是互相垂直的单位向量”将原问题转化为向量的坐标运算是解决本题的突破口. 易错警示 对向量的夹角公式掌握不牢而致错.22.(2017课标Ⅰ理,13,5分)已知向量a,b 的夹角为60°,|a|=2,|b|=1,则|a+2b|= . 答案 2√3解析 本题考查向量数量积的计算.由题意知a ·b=|a|·|b|cos 60°=2×1×12=1,则|a+2b|2=(a+2b)2=|a|2+4|b|2+4a ·b=4+4+4=12. 所以|a+2b|=2√3.23.(2016江苏,13,5分)如图,在△ABC 中,D 是BC 的中点,E,F 是AD 上的两个三等分点,BA ⃗⃗⃗⃗⃗ ·CA ⃗⃗⃗⃗⃗ =4,BF ⃗⃗⃗⃗⃗ ·CF ⃗⃗⃗⃗⃗ =-1,则BE ⃗⃗⃗⃗⃗ ·CE⃗⃗⃗⃗⃗ 的值是 .答案78解析 解法一:(坐标法) 建立直角坐标系,设D(0,0),A(3b,3c),B(-a,0),C(a,0),E(2b,2c),F(b,c),则AB⃗⃗⃗⃗⃗ =(-a-3b,-3c),AC ⃗⃗⃗⃗⃗ =(a-3b,-3c),BA ⃗⃗⃗⃗⃗ ·CA ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =9b 2+9c 2-a 2=4,FB ⃗⃗⃗⃗⃗ =(-a-b,-c),FC ⃗⃗⃗⃗⃗ =(a-b,-c),BF ⃗⃗⃗⃗⃗ ·CF ⃗⃗⃗⃗⃗ =FB ⃗⃗⃗⃗⃗ ·FC ⃗⃗⃗⃗⃗ =b 2+c 2-a 2=-1,EB ⃗⃗⃗⃗⃗ =(-a-2b,-2c),EC ⃗⃗⃗⃗⃗ =(a-2b,-2c),BE ⃗⃗⃗⃗⃗ ·CE ⃗⃗⃗⃗⃗ =EB ⃗⃗⃗⃗⃗ ·EC ⃗⃗⃗⃗⃗ =4b 2+4c 2-a 2, 由{9b 2+9c 2-a 2=4,b 2+c 2-a 2=-1,得b 2+c 2=58,a 2=138, 所以BE ⃗⃗⃗⃗⃗ ·CE ⃗⃗⃗⃗⃗ =4×58-138=78. 解法二:先证明一个三角形中与中点有关的向量公式.BA ⃗⃗⃗⃗⃗ ·CA ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =(AD ⃗⃗⃗⃗⃗ +DB ⃗⃗⃗⃗⃗⃗ )·(AD ⃗⃗⃗⃗⃗ +DC ⃗⃗⃗⃗⃗ )=(AD ⃗⃗⃗⃗⃗ +DB ⃗⃗⃗⃗⃗⃗ )·(AD ⃗⃗⃗⃗⃗ -DB ⃗⃗⃗⃗⃗⃗ )=|AD ⃗⃗⃗⃗⃗ |2-|DB ⃗⃗⃗⃗⃗⃗ |2, BF ⃗⃗⃗⃗⃗ ·CF ⃗⃗⃗⃗⃗ =FB ⃗⃗⃗⃗⃗ ·FC ⃗⃗⃗⃗⃗ =(FD ⃗⃗⃗⃗⃗ +DB ⃗⃗⃗⃗⃗⃗ )·(FD ⃗⃗⃗⃗⃗ +DC ⃗⃗⃗⃗⃗ )=(FD ⃗⃗⃗⃗⃗ +DB ⃗⃗⃗⃗⃗⃗ )·(FD ⃗⃗⃗⃗⃗ -DB ⃗⃗⃗⃗⃗⃗ )=|FD ⃗⃗⃗⃗⃗ |2-|DB ⃗⃗⃗⃗⃗⃗ |2=19|AD ⃗⃗⃗⃗⃗ |2-|DB ⃗⃗⃗⃗⃗⃗ |2,BE ⃗⃗⃗⃗⃗ ·CE ⃗⃗⃗⃗⃗ =EB ⃗⃗⃗⃗⃗ ·EC ⃗⃗⃗⃗⃗ =(ED ⃗⃗⃗⃗⃗ +DB ⃗⃗⃗⃗⃗⃗ )·(ED ⃗⃗⃗⃗⃗ +DC ⃗⃗⃗⃗⃗ )=(ED ⃗⃗⃗⃗⃗ +DB ⃗⃗⃗⃗⃗⃗ )·(ED ⃗⃗⃗⃗⃗ -DB ⃗⃗⃗⃗⃗⃗ )=|ED ⃗⃗⃗⃗⃗ |2-|DB ⃗⃗⃗⃗⃗⃗ |2=49|AD ⃗⃗⃗⃗⃗ |2-|DB ⃗⃗⃗⃗⃗⃗ |2,设|AD ⃗⃗⃗⃗⃗ |=m,|BD ⃗⃗⃗⃗⃗⃗ |=n,由题意可得{m 2-n 2=4,m 29-n 2=-1, 解得m 2=458,n 2=138,所以BE ⃗⃗⃗⃗⃗ ·CE ⃗⃗⃗⃗⃗ =49m 2-n 2=49×458-138=78.24.(2016课标Ⅰ,理13,5分)设向量a=(m,1),b=(1,2),且|a+b|2=|a|2+|b|2,则m= . 答案 -2解析 由|a+b|2=|a|2+|b|2,知a ⊥b,∴a ·b=m+2=0,∴m=-2. 评析 本题考查向量数量积及向量的模,难度不大.25.(2016课标Ⅰ文,13,5分)设向量a=(x,x+1),b=(1,2),且a ⊥b,则x= . 答案 -23解析 因为a ⊥b,所以x+2(x+1)=0,解得x=-23.易错警示 混淆两向量平行与垂直的条件是造成失分的主要原因.26.(2016山东文,13,5分)已知向量a=(1,-1),b=(6,-4).若a ⊥(ta+b),则实数t 的值为 . 答案 -5解析 因为a ⊥(ta+b),所以a ·(ta+b)=0,即ta 2+a ·b=0,又因为a=(1,-1),b=(6,-4),所以|a|=√2,a ·b=1×6+(-1)×(-4)=10,因此可得2t+10=0,解得t=-5.评析 本题主要考查向量的数量积运算,向量的模以及两向量垂直的充要条件等基础知识,考查学生的运算求解能力以及方程思想的应用.27.(2016北京文,9,5分)已知向量a=(1,√3),b=(√3,1),则a 与b 夹角的大小为 . 答案π6解析 ∵cos<a,b>=a ·b|a|·|b|=1×√3+√3×12×2=√32, ∴a 与b 夹角的大小为π6.28.(2015浙江,13,4分)已知e 1,e 2是平面单位向量,且e 1·e 2=12.若平面向量b 满足b ·e 1=b ·e 2=1,则|b|= . 答案23√3 解析 令e 1与e 2的夹角为θ,∴e 1·e 2=|e 1|·|e 2|cos θ=cos θ=12,又0°≤θ≤180°,∴θ=60°.因为b ·(e 1-e 2)=0,所以b 与e 1、e 2的夹角均为30°,从而|b|=1cos30°=23√3.29.(2014重庆文,12,5分)已知向量a 与b 的夹角为60°,且a=(-2,-6),|b|=√10,则a ·b= . 答案 10解析 由a=(-2,-6),得|a|=√(-2)2+(-6)2=2√10, ∴a ·b=|a||b|cos<a,b>=2√10×√10×cos 60°=10. 30.(2014课标Ⅰ理,15,5分)已知A,B,C 为圆O 上的三点,若AO ⃗⃗⃗⃗⃗ =12(AB ⃗⃗⃗⃗⃗ +AC ⃗⃗⃗⃗⃗ ),则AB ⃗⃗⃗⃗⃗ 与AC ⃗⃗⃗⃗⃗ 的夹角为 . 答案 90°解析 由AO ⃗⃗⃗⃗⃗ =12(AB ⃗⃗⃗⃗⃗ +AC⃗⃗⃗⃗⃗ )可知O 为BC 的中点,即BC 为圆O 的直径,又因为直径所对的圆周角为直角,所以∠BAC=90°,所以AB ⃗⃗⃗⃗⃗ 与AC ⃗⃗⃗⃗⃗ 的夹角为90°.31.(2014湖北文,12,5分)若向量OA ⃗⃗⃗⃗⃗ =(1,-3),|OA ⃗⃗⃗⃗⃗ |=|OB ⃗⃗⃗⃗⃗ |,OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ =0,则|AB ⃗⃗⃗⃗⃗ |= . 答案 2√5解析 |AB ⃗⃗⃗⃗⃗ |=|OB ⃗⃗⃗⃗⃗ -OA ⃗⃗⃗⃗⃗ |=√OA ⃗⃗⃗⃗⃗ 2+OB ⃗⃗⃗⃗⃗ 2-2OB ⃗⃗⃗⃗⃗ ·OA ⃗⃗⃗⃗⃗ , ∵|OA ⃗⃗⃗⃗⃗ |=|OB ⃗⃗⃗⃗⃗ |=√12+(-3)2=√10,OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ =0, ∴|AB ⃗⃗⃗⃗⃗ |=√20=2√5,故答案为2√5.32.(2014湖北理,11,5分)设向量a=(3,3),b=(1,-1).若(a +λb)⊥(a-λb),则实数λ= . 答案 ±3解析 |a|=3√2,|b|=√2,a ·b=3×1+3×(-1)=0.因为(a +λb)⊥(a-λb),所以(a +λb)·(a-λb)=|a|2-λ2|b|2=18-2λ2=0.故λ=±3.33.(2013课标Ⅱ,理13,文14,5分)已知正方形ABCD 的边长为2,E 为CD 的中点,则AE ⃗⃗⃗⃗⃗ ·BD ⃗⃗⃗⃗⃗⃗ = . 答案 2解析 解法一:AE ⃗⃗⃗⃗⃗ ·BD ⃗⃗⃗⃗⃗⃗ =(AD ⃗⃗⃗⃗⃗ +12AB ⃗⃗⃗⃗⃗ )·(AD ⃗⃗⃗⃗⃗ -AB ⃗⃗⃗⃗⃗ )=AD ⃗⃗⃗⃗⃗ 2-12AB ⃗⃗⃗⃗⃗ 2=22-12×22=2.解法二:以A 为原点建立平面直角坐标系(如图),可得A(0,0),E(1,2),B(2,0),C(2,2),D(0,2),AE ⃗⃗⃗⃗⃗ =(1,2),BD ⃗⃗⃗⃗⃗⃗ =(-2,2),则AE ⃗⃗⃗⃗⃗ ·BD ⃗⃗⃗⃗⃗⃗ =1×(-2)+2×2=2.34.(2013天津,理12,5分)在平行四边形ABCD 中,AD=1,∠BAD=60°,E 为CD 的中点.若AC ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ =1,则AB 的长为 . 答案12解析 易知AC ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ ,BE ⃗⃗⃗⃗⃗ =AE ⃗⃗⃗⃗⃗ -AB ⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗ +DE ⃗⃗⃗⃗⃗ -AB ⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗ +12AB ⃗⃗⃗⃗⃗ -AB ⃗⃗⃗⃗⃗ =-12AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ ,于是AC ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ =(AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ )·(-12AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ )=-12|AB ⃗⃗⃗⃗⃗ |2+12AB ⃗⃗⃗⃗⃗ ·AD ⃗⃗⃗⃗⃗ +|AD ⃗⃗⃗⃗⃗ |2=-12|AB ⃗⃗⃗⃗⃗ |2+12·|AB ⃗⃗⃗⃗⃗ |·1·cos 60°+12=-12|AB ⃗⃗⃗⃗⃗ |2+14|AB ⃗⃗⃗⃗⃗ |+1,由已知AC ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ =1,可得-12|AB ⃗⃗⃗⃗⃗ |2+14|AB ⃗⃗⃗⃗⃗ |+1=1,解得|AB ⃗⃗⃗⃗⃗ |=12.35.(2013课标Ⅰ,理13,文3,5分)已知两个单位向量a,b 的夹角为60°,c=ta+(1-t)b.若b ·c=0,则t= . 答案 2解析 解法一:∵b ·c=0,∴b ·[ta+(1-t)b]=0,ta ·b+(1-t)·b 2=0, 又∵|a|=|b|=1,<a,b>=60°, ∴12t+1-t=0,t=2.解法二:由t+(1-t)=1知向量a 、b 、c 的终点A 、B 、C 共线,在平面直角坐标系中设a=(1,0),b=(12,√32), 则c=(32,-√32).把a 、b 、c 的坐标代入c=ta+(1-t)b,得t=2.评析 本题考查了向量的运算,利用三点共线的条件得到c 的坐标是解题关键.36.(2012课标,理13,文13,5分)已知向量a,b 夹角为45°,且|a|=1,|2a-b|=√10,则|b|= . 答案 3√2解析 |2a-b|=√10两边平方得 4|a|2-4|a|·|b|cos 45°+|b|2=10. ∵|a|=1,∴|b|2-2√2|b|-6=0.∴|b|=3√2或|b|=-√2(舍去).评析 本题考查了向量的基本运算,考查了方程的思想.通过“平方”把向量转化为向量的数量积是求解的关键.37.(2012安徽文,11,5分)设向量a=(1,2m),b=(m+1,1),c=(2,m),若(a+c)⊥b,则|a|= . 答案 √2 解析 a+c=(3,3m), ∵(a+c)⊥b, ∴(a+c)·b=0, ∴3m+3+3m=0, ∴m=-12, ∴a=(1,-1),∴|a|=√12+(-1)2=√2.评析 本题主要考查向量的基本运算,考查了向量垂直的充要条件.38.(2011课标,文13,5分)已知a 与b 为两个不共线的单位向量,k 为实数,若向量a+b 与向量ka-b 垂直,则k= . 答案 1解析 由题意知|a|=1,|b|=1,<a,b>≠0且<a,b>≠π. 由a+b 与向量ka-b 垂直,得(a+b)·(ka-b)=0, 即k|a|2+(k-1)|a||b|·cos<a,b>-|b|2=0, (k-1)(1+cos<a,b>)=0.又1+cos<a,b>≠0, ∴k-1=0,k=1.评析 本题考查向量的模、向量的数量积等相关知识,考查学生的运算求解能力,属中等难度试题.考点二 平面向量数量积的应用1.(2018天津理,8,5分)如图,在平面四边形ABCD 中,AB ⊥BC,AD ⊥CD,∠BAD=120°,AB=AD=1.若点E 为边CD 上的动点,则AE ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ 的最小值为( )A.2116 B.32 C.2516 D.3 答案 A 本题主要考查数量积的综合应用.解法一:如图,以D 为原点,DA 所在直线为x 轴,DC 所在直线为y 轴,建立平面直角坐标系,则A(1,0),B (32,√32),C(0,√3),令E(0,t),t ∈[0,√3],∴AE ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ =(-1,t)·(-32,t -√32)=t 2-√32t+32,∵t ∈[0,√3],∴当t=--√322×1=√34时,AE ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ 取得最小值,(AE ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ )min =316-√32×√34+32=2116.故选A.解法二:令DE ⃗⃗⃗⃗⃗ =λDC ⃗⃗⃗⃗⃗ (0≤λ≤1),由已知可得DC=√3, ∵AE ⃗⃗⃗⃗⃗ =AD ⃗⃗⃗⃗⃗ +λDC ⃗⃗⃗⃗⃗ ,∴BE ⃗⃗⃗⃗⃗ =BA ⃗⃗⃗⃗⃗ +AE ⃗⃗⃗⃗⃗ =BA ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ +λDC ⃗⃗⃗⃗⃗ , ∴AE ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ =(AD ⃗⃗⃗⃗⃗ +λDC ⃗⃗⃗⃗⃗ )·(BA ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ +λDC ⃗⃗⃗⃗⃗ ) =AD ⃗⃗⃗⃗⃗ ·BA ⃗⃗⃗⃗⃗ +|AD ⃗⃗⃗⃗⃗ |2+λDC ⃗⃗⃗⃗⃗ ·BA ⃗⃗⃗⃗⃗ +λ2|DC ⃗⃗⃗⃗⃗ |2 =3λ2-32λ+32.当λ=--322×3=14时,AE ⃗⃗⃗⃗⃗ ·BE ⃗⃗⃗⃗⃗ 取得最小值2116.故选A.方法总结 向量的最值问题常用数形结合的方法和函数的思想方法求解,建立函数关系时,可用平面向量基本定理,也可利用向量的坐标运算.2.(2017课标Ⅱ理,12,5分)已知△ABC 是边长为2的等边三角形,P 为平面ABC 内一点,则PA ⃗⃗⃗⃗⃗ ·(PB ⃗⃗⃗⃗⃗ +PC ⃗⃗⃗⃗⃗ )的最小值是( )A.-2B.-32 C.-43 D.-1答案 B 设BC 的中点为D,AD 的中点为E,则有PB ⃗⃗⃗⃗⃗ +PC ⃗⃗⃗⃗⃗ =2PD ⃗⃗⃗⃗⃗ , 则PA⃗⃗⃗⃗⃗ ·(PB ⃗⃗⃗⃗⃗ +PC ⃗⃗⃗⃗⃗ )=2PA ⃗⃗⃗⃗⃗ ·PD ⃗⃗⃗⃗⃗ =2(PE ⃗⃗⃗⃗⃗ +EA ⃗⃗⃗⃗⃗ )·(PE ⃗⃗⃗⃗⃗ -EA ⃗⃗⃗⃗⃗ )=2(PE ⃗⃗⃗⃗⃗ 2-EA ⃗⃗⃗⃗⃗ 2). 而AE ⃗⃗⃗⃗⃗ 2=(√32)2=34,当P 与E 重合时,PE ⃗⃗⃗⃗⃗ 2有最小值0,故此时PA ⃗⃗⃗⃗⃗ ·(PB ⃗⃗⃗⃗⃗ +PC⃗⃗⃗⃗⃗ )取最小值, 最小值为-2EA ⃗⃗⃗⃗⃗ 2=-2×34=-32.方法总结 在求向量数量积的最值时,常用取中点的方法,如本题中利用PA ⃗⃗⃗⃗⃗ ·PD ⃗⃗⃗⃗⃗ =PE ⃗⃗⃗⃗⃗ 2-EA ⃗⃗⃗⃗⃗ 2可快速求出最值. 一题多解 以AB 所在直线为x 轴,AB 的中点为原点建立平面直角坐标系,如图,则A(-1,0),B(1,0),C(0,√3),设P(x,y),取BC 的中点D,则D (12,√32).PA ⃗⃗⃗⃗⃗ ·(PB ⃗⃗⃗⃗⃗ +PC ⃗⃗⃗⃗⃗ )=2PA ⃗⃗⃗⃗⃗ ·PD ⃗⃗⃗⃗⃗ =2(-1-x,-y)·(12-x,√32-y)=2[(x+1)·(x -12)+y ·(y -√32)]=2[(x+14)2+(y -√34)2-34]. 因此,当x=-14,y=√34时,PA ⃗⃗⃗⃗⃗ ·(PB ⃗⃗⃗⃗⃗ +PC ⃗⃗⃗⃗⃗ )取得最小值,为2×(-34)=-32,故选B.3.(2017浙江,10,4分)如图,已知平面四边形ABCD,AB ⊥BC,AB=BC=AD=2,CD=3,AC 与BD 交于点O.记I 1=OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ ,I 2=OB ⃗⃗⃗⃗⃗ ·OC ⃗⃗⃗⃗⃗ ,I 3=OC ⃗⃗⃗⃗⃗ ·OD⃗⃗⃗⃗⃗⃗ ,则( )A.I 1<I 2<I 3B.I 1<I 3<I 2C.I 3<I 1<I 2D.I 2<I 1<I 3 答案 C 如图,建立直角坐标系,则B(0,0),A(0,2),C(2,0).设D(m,n),由AD=2和CD=3,得{m 2+(n -2)2=4,(m -2)2+n 2=9, 从而有n-m=54>0,∴n>m . 从而∠DBC>45°,又∠BCO=45°, ∴∠BOC 为锐角.从而∠AOB 为钝角.故I 1<0,I 3<0,I 2>0. 又OA<OC,OB<OD,故可设OD ⃗⃗⃗⃗⃗⃗ =-λ1OB ⃗⃗⃗⃗⃗ (λ1>1),OC ⃗⃗⃗⃗⃗ =-λ2OA ⃗⃗⃗⃗⃗ (λ2>1), 从而I 3=OC ⃗⃗⃗⃗⃗ ·OD ⃗⃗⃗⃗⃗⃗ =λ1λ2OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ =λ1λ2I 1, 又λ1λ2>1,I 1<0,I 3<0,∴I 3<I 1,∴I 3<I 1<I 2.故选C.4.(2016四川文,9,5分)已知正三角形ABC 的边长为2√3,平面ABC 内的动点P,M 满足|AP ⃗⃗⃗⃗⃗ |=1,PM ⃗⃗⃗⃗⃗⃗ =MC ⃗⃗⃗⃗⃗⃗ ,则|BM ⃗⃗⃗⃗⃗⃗ |2的最大值是( ) A.434 B.494 C.37+6√34D.37+2√334答案 B 以A 为坐标原点,建立如图所示的平面直角坐标系,则A(0,0),C(2√3,0),B(√3,3). 设P(x,y),∵|AP ⃗⃗⃗⃗⃗ |=1, ∴x 2+y 2=1, ∵PM ⃗⃗⃗⃗⃗⃗ =MC ⃗⃗⃗⃗⃗⃗ , ∴M 为PC 的中点, ∴M (x+2√32,y2), ∴|BM ⃗⃗⃗⃗⃗⃗ |2=(x+2√32-√3)2+(y2-3)2=x 24+y 24-3y+9 =14-3y+9=374-3y, 又∵-1≤y ≤1,∴当y=-1时,|BM ⃗⃗⃗⃗⃗⃗ |2取得最大值,且最大值为494. 思路分析 由△ABC 为正三角形,|AP⃗⃗⃗⃗⃗ |=1,考虑到用建立平面直角坐标系的方法来解决向量问题. 评析 本题考查了向量的坐标运算,运用了转化与化归思想.5.(2015福建理,9,5分)已知AB ⃗⃗⃗⃗⃗ ⊥AC ⃗⃗⃗⃗⃗ ,|AB ⃗⃗⃗⃗⃗ |=1t ,|AC ⃗⃗⃗⃗⃗ |=t.若点P 是△ABC 所在平面内的一点,且AP ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗⃗|AB ⃗⃗⃗⃗⃗⃗ |+4AC ⃗⃗⃗⃗⃗|AC ⃗⃗⃗⃗⃗|,则PB ⃗⃗⃗⃗⃗ ·PC⃗⃗⃗⃗⃗ 的最大值等于( ) A.13 B.15 C.19 D.21答案 A 以A 为原点,AB 所在直线为x 轴,AC 所在直线为y 轴建立平面直角坐标系,则B (1t,0)(t>0),C(0,t),P(1,4),PB ⃗⃗⃗⃗⃗ ·PC ⃗⃗⃗⃗⃗ =(1t-1,-4)·(-1,t-4)=17-(4t +1t)≤17-2×2=13(当且仅当t =12时,取“=”),故PB ⃗⃗⃗⃗⃗ ·PC⃗⃗⃗⃗⃗ 的最大值为13,故选A. 6.(2019浙江,17,6分)已知正方形ABCD 的边长为1.当每个λi (i=1,2,3,4,5,6)取遍±1时,|λ1AB ⃗⃗⃗⃗⃗ +λ2BC ⃗⃗⃗⃗⃗ +λ3CD ⃗⃗⃗⃗⃗ +λ4DA ⃗⃗⃗⃗⃗ +λ5AC ⃗⃗⃗⃗⃗ +λ6BD ⃗⃗⃗⃗⃗⃗ |的最小值是 ,最大值是 . 答案 0;2√5解析 本题考查平面向量的坐标表示及坐标运算,在向量的坐标运算中涉及多个未知数据以此来考查学生的数据处理能力,数学运算及数据分析的核心素养. 如图,建立平面直角坐标系,则A(0,0),B(1,0),C(1,1),D(0,1),∴AB ⃗⃗⃗⃗⃗ =(1,0),BC⃗⃗⃗⃗⃗ =(0,1),CD ⃗⃗⃗⃗⃗ =(-1,0),DA ⃗⃗⃗⃗⃗ =(0,-1),AC ⃗⃗⃗⃗⃗ =(1,1),BD ⃗⃗⃗⃗⃗⃗ =(-1,1), 故|λ1AB ⃗⃗⃗⃗⃗ +λ2BC ⃗⃗⃗⃗⃗ +λ3CD ⃗⃗⃗⃗⃗ +λ4DA ⃗⃗⃗⃗⃗ +λ5AC ⃗⃗⃗⃗⃗ +λ6BD ⃗⃗⃗⃗⃗⃗ | =|(λ1-λ3+λ5-λ6,λ2-λ4+λ5+λ6)|=√(λ1-λ3+λ5-λ6)2+(λ2-λ4+λ5+λ6)2.(*)显然(*)式中第一个括号中的λ1,λ3与第二个括号中的λ2,λ4的取值互不影响,∴只需讨论λ5与λ6的取值情况即可,当λ5与λ6同号时,不妨取λ5=1,λ6=1, 则(*)式即为√(λ1-λ3)2+(λ2-λ4+2)2,∵λ1,λ2,λ3,λ4∈{-1,1},∴λ1=λ3,λ2-λ4=-2(λ2=-1,λ4=1)时,(*)式取最小值0,当|λ1-λ3|=2(如λ1=1,λ3=-1),λ2-λ4=2(λ2=1,λ4=-1)时,(*)式取最大值2√5,当λ5与λ6异号时,不妨取λ5=1,λ6=-1,则(*)式即为√(λ1-λ3+2)2+(λ2-λ4)2. 同理可得最小值仍为0,最大值仍为2√5, 综上,最小值为0,最大值为2√5.解题关键 本题未知量比较多,所以给学生的第一感觉是难,而实际上注意到图形为规则的正方形,λi (i=1,2,3,4,5,6)的取值只有两种可能(1或-1),这就给建系及讨论λi 的值创造了条件,也是求解本题的突破口.7.(2013北京文,14,5分)已知点A(1,-1),B(3,0),C(2,1).若平面区域D 由所有满足AP ⃗⃗⃗⃗⃗ =λAB ⃗⃗⃗⃗⃗ +μAC ⃗⃗⃗⃗⃗ (1≤λ≤2,0≤μ≤1)的点P 组成,则D 的面积为 . 答案 3解析 AB ⃗⃗⃗⃗⃗ =(2,1),AC ⃗⃗⃗⃗⃗ =(1,2). 设P(x,y),由AP⃗⃗⃗⃗⃗ =λAB ⃗⃗⃗⃗⃗ +μAC ⃗⃗⃗⃗⃗ , 得{x -1=2λ+μ,y +1=λ+2μ,故有{λ=2x -y -33,μ=-x+2y+33.又λ∈[1,2],μ∈[0,1],故有 {1≤2x -y -33≤2,0≤2y -x+33≤1,即{3≤2x -y -3≤6,0≤2y -x +3≤3.则平面区域D 如图中阴影部分所示.易知其面积为3.评析 本题考查了平面向量的坐标运算、线性规划等知识;同时又考查了转化及数形结合思想,综合能力要求较高.。
保险考试真题10
单选题第1题某日天下大雪，一行人被A、B两车相撞致死。
后经交通警察查实，该事故是因A车驾驶员酒后驾车所致。
那么，在该起事故中行人死亡的近因是（）。
A、大雪天气B、酒后驾车C、A车撞击D、B车撞击正确答案：B 您的答案：单选题第2题在财产保险的承保中，如果保险标的低于承保标准，保险人通常采用减少保险金额，或者使用较高的免陪额或较高的保险费率的方式承保。
这一承保决策属于（）。
A、正常承保B、优惠承保C、有条件承保D、拒保正确答案：C 您的答案：单选题第3题根据利益的量的限定，对于确定财产保险的保险金额得有关规定是（）。
A、保险金额须以保险人的意愿为限，超过部分无效B、保险金额须以财产的预期价值为限，超过财产的预期价值部分无效C、保险金额须以财产的实际价值为限，超过财产的实际价值部分无效D、保险金额须以财产的预期价值为限，超过财产的实际价值部分无效正确答案：C 您的答案：单选题第4题精神病人甲在妻子的陪同下散步时发病，其妻子虽然尽力阻止，但甲还是打伤了路人乙，根据《民法通则》的规定，对此损害应该承担民事责任的是（）。
A、甲B、乙C、甲的妻子和乙D、甲的妻子，但是可以适当减轻其责任正确答案：D 您的答案：单选题第5题保险代理人与保险经纪人的区别之一是委托人不同。
其中，保险经纪人的委托人是（）。
A、保险人B、投保人C、受益人D、被保险人正确答案：B 您的答案：单选题第6题根据我国消费者权益保护法的规定，消费者协会履行的职能之一是（）。
A、向消费者提供消费信息和咨询服务B、从事营利性服务C、从事商品经营D、以牟利为目的向社会推荐商品和服务正确答案：A 您的答案：单选题第7题同一公司不同险种的死亡率经验（）。
A、有所不同B、基本相同C、完全一致D、差异很大正确答案：A 您的答案：单选题第8题虽然社会保险与商业保险有许多共同之处，但就实施方式而言，各国法律一般规定社会保险采取（）。
A、自愿方式B、互助方式C、强制方式D、协商方式正确答案：C 您的答案：单选题第9题风险管理中，风险识别是指（）。
10年高考真题-大一统国家的建立与巩固——秦汉
第二单元大一统国家的建立与巩固——秦汉考点一中央集权制度的建立与巩固1.(2020浙江7月选考,3,2分)史载:“秦兼天下,建皇帝之号,立百官之职。
汉因循而不革,明简易,随时宜也。
其后颇有所改。
”其中“颇有所改”的是( )①设丞相②设内朝③设御史大夫④设司隶校尉A.①③B.①④C.②③D.②④答案 D 题干材料反映了汉承秦制,也有改动。
秦朝和汉朝都设有丞相,故排除①;汉武帝时期开始设立内朝,秦朝没有设立内朝,故②符合题意;秦朝和汉朝都设有御史大夫,故排除③;汉武帝时期设置司隶校尉,秦朝未设置,故④符合题意。
本题正确答案为D项。
2.(2019浙江4月选考,27,2分)史载,汉武帝“元封五年,初置部刺史,掌奉诏条察州”,有学者认为此部刺史是由秦代的监(御史)嬗变而成。
秦时除置一守一尉外,又置一监。
汉兴,省监不置。
惠帝三年,“相国奏御史监三辅”。
武帝置刺史,“郡守不得面奏事,而刺史得面奏事”。
这说明( )①秦与汉在地方均设有专门负责监察的官员②秦与汉监察方向有别,一在中央,一在地方③秦汉监察体制的实际效能有限④秦汉监察机构的设置受到最高执政集团的重视A.①④B.②③C.①③④D.②③④答案 A 根据材料“元封五年,初置部刺史,掌奉诏条察州”“秦代的监(御史)”,可见秦与汉在地方均设有专门负责监察的官员,①正确;材料“郡守不得面奏事,而刺史得面奏事”体现了君主对监察的重视,④正确;“秦时除置一守一尉外,又置一监”说明秦代地方也有监察,②错误;材料无法体现秦汉监察体制的实际效能有限,③错误。
故A项正确。
3.(2015课标Ⅰ,25,4分)两汉时期,皇帝的舅舅、外祖父按例封侯;若皇帝幼小,执政大臣也主要从他们之中选择。
这被当时人视为“安宗庙,重社稷”的“汉家之制”。
汉代出现外戚干政的背景是( )A.皇帝依靠外戚抑制相权B.“家天下”观念根深蒂固C.母族亲属关系受到重视D.刘氏同姓诸侯王势力强大答案 C 两汉时期,皇帝的舅舅、外祖父按例封侯,如皇帝幼小,这些外戚往往成为执政大臣,体现了汉代对母族亲属关系的重视,故选C项。
2010.10考研管理类联考初数真题(有答案)
2010年10月真题一、问题求解：第1～15小题，每小题3分，共45分。
下列每题给出的,,,,A B C D E 五个选项中，只有一项是符合试题要求的.请在答题卡上将所选项的字母涂黑.1. 若13x x+=，则2421x x x =++( ） （A ）18- （B ）16 (C ）14（D ）1-4 (E ）18 2。
若实数,,a b c 满足：2229a b c ++=，则代数式222()()()a b b c c a -+-+-的最大值是（ ）(A)21 （B ）27 （C ）29 （D)32 （E)393. 某地震灾区现居民住房的总面积为a 平方米,当地政府计划每年以10%的住房增长率建设新房,并决定每年拆除固定数量的危旧房，如果10年后该地的住房总面积正好比现有住房面积增加一倍，那么，每年应该拆除危旧房的面积是（ ）平方米?（注：910111.1 2.4,1.1 2.6,1.1 2.9≈≈≈精确到小数点后一位)（A)180a （B ）140a （C ）380a （D)120a (E ）以上结论都不正确4. 某学生在军训时进行打靶测试，共射击10次。
他的第6、7、8、9次射击分别射中9.0环、8.4环、8.1环、9.3环，他的前9次射击的平均环数高于前5次的平均环数。
若要使10次射击的平均环数超过8.8环，则他第10次射击至少应该射中（ ）环？(打靶成绩精确到0。
1环）（A ）9。
0 (B)9。
2 （C)9。
4 （D ）9。
5 （E)9.95. 某种同样的商品装成一箱，每个商品的重量都超过1kg ，并且是1kg 的整数倍，去掉箱子重量后净重210kg,拿出若干个商品后，净重183kg ，则每个商品的重量为( ）kg 。
(A ）1 （B)2 （C ）3 (D)4 （E ）56. 在一条与铁路平行的公路上有一行人与一骑车人同向行进，行人速度为3。
6km/h ，骑车人速度为10.8km/h 。