2016黄浦高三数学一模
黄浦区高考模拟考数学试卷
2016年黄浦区高考模拟考数 学 试 卷题号 一 二 三 总分 得分一、填空题（本大题共１２分，每小题４分，共４８分）15212i i+-＝ .2、已知：2,tg α=则(2)2tg πα+的值是 .3、若常数ｂ满足1,b >则21n 1lim n nb b b b -→∞+++=L .4、若x30.618=，且a [k k 1k Z ∈+∈，）（），则ｋ的值是 . 5、函数f (x)sin 2x sin 2x =+的最小正周期为 . 6、函数y sin x 3x =在区间0,2π⎡⎤⎢⎥⎣⎦上的最小值为 . 7、［理］（123x x （）展开式中，含ｘ正整数次项幂的项有 项. ［文］不等式x 12log 0-<的解集是 .１０、 某班有２１名男生，１５名女生.现从该班学生中任选两名作生活委员，则这两名生活委员性别相同的概率是 （结果用既约分数表示）.9、从集合{},111k k z k ∈≤≤中任选两个不同元素作为椭圆方程22221x y m n+=中的ｍ和ｎ，其中落在矩形Ｂ{}(,)11,9x y x y =<<内的椭圆有 个.１２、 已知双曲线2212y x +=的焦点为12,F F ，点Ｍ在双曲线上，且120,MF MF =u u u u r u u u u r g 则点Ｍ到ｘ轴的距离为 .11、已知四面体ABCD,沿棱AB 、AC 、AD 剪开，铺成平面图形， 得到123A A A △（如图），试写出四面体ABCD 应满足的一个性质： . １３、 已知集合A= ax b x>0cx d +⎧⎫⎨⎬+⎩⎭，这里a ，b,c,d 为实数，若{}012A ⊂，，，且{}2.52A=φI，-，则函数ax bcx d++可以是 （只有写出一个满足条件的函数）. 二、选择题（本大题共4题，每小题4分，共16分） 13、已知函数f(x)= 12log (0)ax a -≠满足(2)(2)f x f x -+=--，则实数a 的值为 （ ） A ． 1 B. 12-C. 14D. -1 14、“a=b ”直线2y x =+与圆22"()()2x a y b -+-=相切” 的 （ ） A. 充分不必要条件， B .必要不充分条件C. 充要条件D. 既不充分又不必要条件15、已知两线段2a =,b=22,若以a,b 为边作三角形，则a 边所对的角A 的取值范围为（ ） A.(,)63ππB .(0,]6π C. (0,)2π D. (0,]4π16、设b>0,二次函数221y ax bx a =++-的图像为下列之一，则a 的值为 （ ） A. 1 B. 1- C.15-- D. 15-+三、 解答题 （本大题共6题，第17、18题每题12分，第19、20题每题14分，第21题16分，第22题18分，共86分）17、已知向量{,12},{4,5},{,10},OA k OB OC k ===-u u u r u u u r u u u r且A 、B 、C 三点共线，求k 的值.18、已知数列{}n a 的通项公式为1133()[()1]()44n n n a n N --+=-∈.求 （1）求数列{}n a 中的最大项及其值； （2）求数列{}n a 中的最小项及其值.19、【理】在直棱柱111ABC A B C -中，已知01,,,90.AB a AC b AA c BAC ===∠= （1）求使11AB BC ⊥的充要条件（用,,a b c 表示）；（2）求证11B AC ∠为锐角；（3）若060,ABC ∠=则11B AC ∠是否可能为045？证明你的结论.【文】设a 为正数，直角坐标平面内的点集{(,)|,,}A x y x y a x y =--是三角形的三 （1）画出A 所表示的平面区域；（2）在平面直角坐标系中，规定,a Z y Z ∈∈且时，(,)x y 称为格点，当8a =时，A 内有几个格点（本小题只要直接写出结果即可）； （3）点集A连同它的边界构成的区域记为A ，若圆222{(,)|()()}(0)x y x p x q r A r -+-=⊆>，求r 的最大值.20、某厂2006年拟举行促销活动，经调查测算，该厂产品的年销售量（即该厂的年产量）x 万件与去年促销费m （万元）（0m ≥）满足231x m =-+.已知2006年生产的固定投入为8万元，每生产1万件该产品需要再投入16万元，厂家将每件产品的销售价格定为每件产品平均成本的1.5倍（产品成本包括固定投入和再投入两部分资金）. （1）将2006年该产品的利润y 万元表示为年促销费m （万元）的函数； （2）求2006年该产品利润的最大值，此时促销费为多少万元？21、已知抛物线2:2a p y x ax a =++-（a 为实常数）. （1）求所有抛物线a p 的公共点坐标；（2）当实数a 取遍一切实数时，求抛物线a p 的焦点方程.【理】（3）是否存在一条以y 轴为对称轴，且过点(1,1)--的开口向下的抛物线，使它与某个a p 只有一个公共点？若存在，求出所有这样的a ；若不存在，说明理由.【文】（3）是否存在直线y kx b =+（,k b 为实常数），使它与所有的抛物线a p 都有公共点？若存在，求出所有这样的直线；若不存在，说明理由.22、已知函数()y f x =的定义域为R +，对任意,x y R +∈，有恒等式()()()f xy f x f y =+；且当1x >时，()0f x <. （1）求(1)f 的值；（2）求证：当x R +∈时，恒有1()()f f x x=-； （3）求证：()(0,)f x +∞在上为减函数；【以下（4）小题选理科的学生做；选文科的学生不做】（4）由上一小题知：()(0,)f x +∞是上的减函数，因而()f x 的反函数1()f x -存在，试根据已知恒等式猜想1()f x -具有的性质，并给出证明.2006年黄浦区高考模拟考 数 学 试 卷参考答案题号 一 二 三 总分 得分一、填空题（本大题共１２分，每小题４分，共４８分）15212i i+-＝ i .2、已知：2,tg α=则(2)2tg πα+的值是4. 3、若常数ｂ满足1,b >则21n 1lim n n b b b b -→∞+++L 1b - . 4、若x30.618=，且a [k k 1k Z ∈+∈，）（），则ｋ的值是 1- . 5、函数f (x)sin 2x sin 2x =+的最小正周期为 π .6、函数y sin x 3x =在区间0,2π⎡⎤⎢⎥⎣⎦上的最小值为 1 . 7、［理］（123x x （）展开式中，含ｘ正整数次项幂的项有 3 项. ［文］不等式x 12log 0-<的解集是 ()()0,11,2⋃ .８、某班有２１名男生，１５名女生.现从该班学生中任选两名作生活委员，则这两名生活委员性别相同的概率是2（结果用既约分数表示）. 9、从集合{},111k k z k ∈≤≤中任选两个不同元素作为椭圆方程22221x y m n+=中的ｍ和ｎ，其中落在矩形Ｂ{}(,)11,9x y x y =<<内的椭圆有 72 个.１１、 已知双曲线2212y x +=的焦点为12,F F ，点Ｍ在双曲线上，且120,MF MF =u u u u r u u u u r g 则点Ｍ到ｘ轴的距离为3. 11、已知四面体ABCD,沿棱AB 、AC 、AD 剪开，铺成平面图形， 得到123A A A △（如图），试写出四面体ABCD 应满足的一个性质：四面体ABCD 的每组对棱相等（答案不唯一，可填“四面体ABCD 的四个面是四个全等三角形”；或填“四面体每个顶点为公共顶点的三个面角之和为π”） .１４、 已知集合A= ax b x>0cx d +⎧⎫⎨⎬+⎩⎭，这里a ，b,c,d 为实数，若{}012A ⊂，，，且{}2.52A=φI，-，则函数ax b cx d ++可以是 2.11xx -+ （只有写出一个满足条件的函数）.二、选择题（本大题共4题，每小题4分，共16分） 13、已知函数f(x)= 12log (0)ax a -≠满足(2)(2)f x f x -+=--，则实数a 的值为 （ B ） A ． 1 B. 12-C. 14D. -1 14、“a=b ”直线2y x =+与圆22"()()2x a y b -+-=相切” 的 （ A ） A. 充分不必要条件， B .必要不充分条件C. 充要条件D. 既不充分又不必要条件15、已知两线段2a =,b=22,若以a,b 为边作三角形，则a 边所对的角A 的取值范围为（D ）A.(,)63ππB .(0,]6π C. (0,)2π D. (0,]4π16、设b>0,二次函数221y ax bx a =++-的图像为下列之一，则a 的值为 （ B ） A. 1 B. 1- C.15-- D. 15-+三、 解答题 （本大题共6题，第17、18题每题12分，第19、20题每题14分，第21题16分，第22题18分，共86分）17、已知向量{,12},{4,5},{,10},OA k OB OC k ===-u u u r u u u r u u u r且A 、B 、C 三点共线，求k 的值.{}4,7AB OB OA k =-=--u u u r u u u r u u u r ，{}4,5BC OC OB k =-=--u u u r u u u r u u u r----------------------------4分,,A B C 三点共线，⇔存在实常数l ，使AB lBC =u u u r u u u r-----------------------------------------8分()4475k l k l -=--⎧⎪⇔⎨-=•⎪⎩2357k l ⎧=-⎪⎪⇔⎨⎪=-⎪⎩23k ∴=------------------------------------------------------------------------------------------------12分18、已知数列{}n a 的通项公式为1133()[()1]()44n n n a n N --+=-∈.求（1）求数列{}n a 中的最大项及其值； （2）求数列{}n a 中的最小项及其值.()1当2n ≥时，13310,44n -⎛⎫≤-< ⎪⎝⎭从而11331044n n n a --⎡⎤⎛⎫⎛⎫=-<⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦故10a =为数列{}n a 的最大项----------------------------------------------------------------------4分()211133311144424n n n n a ---⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫=-=--⎢⎥⎢⎥ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦134n -⎛⎫ ⎪⎝⎭Q 随n 的增大而减小，又32313424⎛⎫⎛⎫<< ⎪ ⎪⎝⎭⎝⎭--------------------------------------------8分134n -⎧⎫⎪⎪⎛⎫∴⎨⎬ ⎪⎝⎭⎪⎪⎩⎭中与12距离最近的数是234⎛⎫ ⎪⎝⎭故397631616256a ⎛⎫=•-=-⎪⎝⎭是数列{}n a 的最小项--------------------12分 19、【理】在直棱柱111ABC A B C -中，已知01,,,90.AB a AC b AA c BAC ===∠= （1）求使11AB BC ⊥的充要条件（用,,a b c 表示）； （2）求证11B AC ∠为锐角；（3）若060,ABC ∠=则11B AC ∠是否可能为045？证明你的结论.【文】设a 为正数，直角坐标平面内的点集{(,)|,,}A x y x y a x y =--是三角形的三 （1）画出A 所表示的平面区域；（2）在平面直角坐标系中，规定,a Z y Z ∈∈且时，(,)x y 称为格点，当8a =时，A 内有几个格点（本小题只要直接写出结果即可）；（3）点集A 连同它的边界构成的区域记为A ，若圆222{(,)|()()}(0)x y x p x q r A r -+-=⊆>，求r 的最大值.()()()1110,0,,,0,,0,,A c B a c C b c ------2分()1{}{}11,0,,0,,AB a c BC b c ==u u u r u u u u r11AB BC ⊥110AB BC ⇔•=u u u r u u u u r220a c a c ⇔-+=⇔=即11AB BC ⊥的充分条件是a c =---------------------------------------------6分()2{}10,,AC b c =u u u u r ，211111111cos 0AB AC c B AC AB AC AB AC •∠==>••u u u r u u u u rQ u u u r u u u ur u u u r u u u u r 11B AC ∴∠为锐角-------------------------------------------------------------------8分()301111222245cos 2B AC B AC a c b c ∠=⇔∠==+•+ 0060,tan 603,3bABC b a a∠=∴===Q2222232a c a c c ++=解得273a c -+=---------------------------------------------------------------11分 若060,ABC ∠=解当273a c -+=时，01145B AC ∠=--------------14分[文]()1,,x y a x y --是三角形三边长0,0,0()()x y a x y x y a x yx a x y yy a x y x >>-->⎧⎪+>--⎪⇔⎨+-->⎪⎪+-->⎩02022a x a y a x y ⎧<<⎪⎪⎪⇔<<⎨⎪⎪+>⎪⎩---------------------------------------8分 ∴点集A 构成的平面区域为等腰直角三角形ABC ，如上图阴影部分表示（不包括边界）。
2016年12月黄浦区高三一模试卷
黄浦区2016学年度第一学期高三年级期终调研测试英语试卷（完卷时间： 120分钟满分： 150分）2016年12月9日上午第I卷（共100分）I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only 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. Six years ago. B. Seven years ago.C. Eight years ago.D. Nine years ago.2. A. See a film with the woman. B. Attend a charity show.C. Get ready for a charity show.D. Make a new movie.3. A. She is going to be the man’s neighbor. B. She has just moved into a new house.C. She is arranging a family trip.D. She arrived in Canada yesterday.4. A. How to pay the registration fee. B. Why to use a credit card.C. When to send a cheque.D. Where to pay cash.5. A. Film stars. B. Radio hosts.C. Pop singers.D. Composers.6. A. He drove her to the airport. B. He followed her to the airport.C. He bought her a map of the airport.D. He lined out the route to the airport on a map.7. A. The man should apply for a bank loan.B. The man should work in a bank to get money.C. The man should turn to someone available for help.D. The man should take other students’ approaches.8. A. Both the tennis courts and the table tennis tables are free.B. Neither of the tennis courts and table tennis tables are free.C. The table tennis tables are free, but the tennis courts are not.D. The tennis courts are free, but the table tennis tables are not.9. A. In a factory. B. In a school.C. In a gym.D. In a lab.10. A. A stationer’s. B. A paint shop.C. A bookstore.D. A drawing class.Section BDirections: In Section B, you will hear one short passage and two longer conversations. After each passage or conversation, you will be asked several questions. The passage and the conversations will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11.A. To arouse people’s interest in pop music. B. To provide more information about the music.C. To have it lined with the main building.D. To display a separate and different section.12.A. It once experienced serious damage. B. Its rebuilding was an easy job.C. It is owned by a rich family.D. It opens for 362 days every year.13.A. Museum visitors. B. Government workers.C. Music authors.D. Individual donators.Questions 14 through 16 are based on the following conversation.14. A. 4:00 p.m. in the classroom. B. 7:00 p.m. in the classroom.C. 4:00 p.m. in the garden.D. 7:00 p.m. in the garden.15. A. He has classes at that time. B. He is travelling abroad.C. He doesn’t like barbeque.D. He hasn’t prepared a gift.16. A. A watch. B. A card. C. A ball. D. A headphone.Questions 17 through 20 are based on the following conversation.17.A. The tickets have to be booked in advance. B. It will be performed at 6 p.m.C. There will be two performances.D. It will be on at the theatre.18.A. The Piazza. B. The Concert Hall.C. The theatre.D. The Town Hall.19.A. $8. B. $10. C. $15. D. $20.20.A. Art Exhibition. B. Ballet Performance. C. Mask Party. D. Living Theatre.II. Grammar and VocabularySection ADirections: After reading the passage below, fill in the blanks to make the passages 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.Infant Day Care, Good or Bad?The British psychoanalyst John Bowlby maintains that separation from the parents during the sensitive “attachment” period from birth to three may influence a child’s personality and lead to emotional problems in later life. Some people have drawn the conclusion from Bowlby’s work (21) _________ children should not be sent to day care before the age of three because of the parental separation (22) _________ involves, and many people do believe this. But there are also arguments (23) _________ such a strong conclusion.Firstly, experts point out that the isolated love affair between children and parents (24) _________ (find) in modern societies does not usually exist in traditional societies. For example, in some tribal societies, such as the Ngoni, the father and mother of a child did not raise their infant alone – far from it. Secondly, common sense tells us that day care would not be so widespread today (25) _________ parents and care-takers found children had problems with it. Statistical studies of this kind have not yet been carried out, and they have regularly reported that day care had a slightly positive effect on children’s development. But tests (26) ________ have been used to measure this development are not widely enough accepted to settle the issue.But Bowlby’s analysis raises the possibility that early day care has delayed effects. The possibility that such care might lead to, say, more mental illness or crime 15 or 20 years later can only be explored by the use of statistics. Whatever the long-term effects, parents sometimes find the immediate effects difficult (27) _________ (deal) with. Children under three are likely to protest at (28) _________ (leave) their parents and show unhappiness. At the age of three or three and a half almost all children find the change to nursery easy, and this is undoubtedly (29) _________ more and more parents make use of child care at this time. The matter, then, is far from clear-cut, though experience and available evidence (30) _________ (indicate) early care is reasonable for infants.Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.First Aid: Difference between Death and LifeFirst aid is emergency care for a victim of sudden illness or injury until more skillful medical treatment is available. It may save a life or improve certain ___31___ signs including pulse, temperature, and breathing. First aid must be ___32___ as quickly as possible. In the case of the critically injured, a few minutes can make the difference between complete recovery and loss of life.First-aid ___33___ depend upon a victim’s needs and the provider’s level of knowledge and skill. Knowing what not to do in an emergency is as important as knowing what to do. For example, ___34___ moving a person with a neck injury can lead to permanent health problems.Despite the variety of injuries possible, several ___35___ of first aid apply to all emergencies. The first step is to call for professional medical help. The victim, if conscious, should be reassured that medical aid has been requested, and asked for permission to provide any first aid. Next, ___36___ the scene, asking other people or the injured person’s family or friends about details of the injury or illness, any care that may have already been given, and ___37___ conditions such as heart trouble. Unless the accident scene becomes unsafe or the victim may suffer further injury, do not move the victim.First aid requires rapid assessment of victims to determine whether ___38___ conditions exist. One method for ___39___ a victim’s condition is known by the acronym ABC, which stands for:A – Airway: is it open and clear?B – Breathing: is the person breathing? Look, listen and feel for breathing.C – Circulation: is there a pulse? Is the person bleeding ___40___? Check skin color and temperature foradditional indications of circulation problems.III. Reading ComprehensionSection ADirections:For each blank in the following passages there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Animal RightsEvery conscious being has interests that should be respected. No being who is conscious of being alive should be devalued to thinghood, dominated, and used as a resource or ___41___. The key point of the idea known as animal rights is a movement to extend moral consideration to all ___42___ beings. Nobody should have to demonstrate a specific level of intelligence or be judged beautiful to be given moral consideration. No being should have to be useful to humanity or capable of accepting “duties” in order to be extende d moral consideration. ___43___, what other animals need from us is being free from duties to us.Animal rights is about letting animals live on their own terms. It can be written into our laws, but is not an actual list or bill of rights as we have for human society. It begins with our promises not to act like ___44___ of others. Animal rights is about justice ─ treating animals fairly.Why is animal rights ___45 ___? It is because we humans often act as though we are the only beings on the planet. Although we depend on other animals for our very survival, humans are the only animals that have upset the balance of nature. There are lots of ways that humans ___46___ animals. We domesticate them and use them for food, even though our nutritional needs can be completely supplied by a(n) ___47___ diet. Although other materials are available, we use animal’s skin and other body parts for clothing, furs, hats, boots, jewellery and even pet toys. Humans can talk about it but animals cannot. All animals wish to experience life in its fullness. Unlike many animals who have to kill to survive, humans do not. Why should humans cause ___48___ to other beings when it’s not necessary?As we do, animals protect their children; they feel fear; they warn each other of dangers; they play. We might differ from other animals in some ways, but that doesn’t give us the right to ___49___ them down, take their lands, pollute their waters, or use them for our conveniences. Animals also experience pain and it’s not difficult to observe ___50___ of pain in the way a conscious being reacts to it. We take advantage, cause distress, and act ___51___ when we use animals for amusement. Lots of pets are ___52___ on the streets when their owners no longer find it convenient or affordable to keep or care for them.Whether we admit it or not, it’s a prejudice to think we are ___53___ to animals and that it is our right to control them, which can only make people act mean, hateful or neglectful. However, each of us has within us the power to ___54___. We can adopt a different attitude, one that reshape our destiny. This will have wonderful effects on the planet’s other communities, for life is ___55___ avoiding suffering. It is interacting, singing, pursuing joy. We humans can learn to live responsibly, with respect, kindness and love.41. A. companies B. goods C. insects D. providers42. A. active B. conscious C. intelligent D. strange43. A. Indeed B. Moreover C. Nevertheless D. Otherwise44. A. followers B. friends C. masters D. tutors45. A. necessary B. neglected C. respected D. revolutionary46. A. distinguish B. eliminate C. exploit D. raise47. A. animal-free B. eco-friendly C. low-salt D. well-balanced48. A. conflict B. confusion C. isolation D. misery49. A. calm B. chase C. pull D. tear50. A. signs B. symbols C. symptoms D. performances51. A. differently B. enthusiastically C. gently D. unfairly52. A. abandoned B. chosen C. oppressed D. spoiled53. A. accessible B. appealing C. reasonable D. superior54. A. change B. dominate C. persist D. proceed55. A. contrary to B. more than C. owing to D. rather thanSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have read.(A)①Did English football finally find a new star? At the age of 19, Theo Walcott came onto the scene by scoring a hat-trick for England in a 4-1 victory over Croatia in 2010 World Cup.②Walcott’s lightning speed and accurate shooting turned the teenager into an overnight star. Many thought he was a new dawn for the England team. He was building his fame for his fast pace, with former Barcelona manager Pep declaring that “you would need a gun to stop him.” FIFA World Player of the Year winner Lionel Messi once also described Walcott as “one of the most dangerous players I have ever played against.” In addition to his speed, Walcott also possessed good balance, movement and technique.③It was symbolic that Walcott’s goals came from the right-wing. The position had been played by “golden boy” David Beckham for more than 10 years. No longer were the cheers for Beckham. The fans’hopes now rested on the shoulders of Walcott.④Walcott was born in London to a black British Jamaican father and a white English mother. He grew up as a Liverpool fan due to his father’s support of Liverpool. When Chelsea asked him to be a ball boy, he used the opportunity to meet his Liverpool idols.⑤The teenager’s rise to fame was not completely out of blue. He was part of England’s World Cup team in 2006, but he did not get to play a match. He also spent over two years at Arsenal, where he was fast becoming a key player.⑥But that year, few were expecting the wonderful performance between England and Croatia. The teenager was the first England player to score three goals in a game since Michael Owen did so seven years before.⑦Although England was full of superstars, they had a poor record in major tournaments. But things were beginning to change. The win against Croatia was sweet revenge. Croatia was the team which knocked England out of Euro 2008.⑧Walcott’s wonderful performance lighted England fans’ hope for World Cup victory in South Africa in 2010, since England had not lifted the cup since 1966.⑨But before England fans got too carried away, our reflection on the past history told us that placing a country’s hopes on one teenager was dangerous.56. Which of the following CANNOT account for Walcott’s increasing fame?A. Fast speed.B. Masterly skills.C. Positional sense.D. Unusual family.57. Why did the author mention David Beckham in the 3rd paragraph?A. To show that England football once had a glorious history.B. To illustrate that Walcot t could be entitled “golden boy”.C. To indicate that England fans were difficult to please.D. To imply that people had high expectation on Walcott.58. In the 5th paragraph, the underlined phrase “out of blue” most probably means “________”.A. impoliteB. unexpectedC. impossibleD. unintentional59. What is the author most likely to agree with?A. Walcott might not live up to fans’ expectation.B. Walcott might transfer from Arsenal to Liverpool.C. Croatia might change the history of the World Cup.D. England might be defeated by the opponent in the next round.(B)✓OverviewExplore Stewart Island and the surrounding bays in our modern mini-buses. Our guides enjoy sharing their local knowledge of the history and environment of Stewart Island. Highlights include Lee Bay, the gateway to Rakiura National Park, beautiful Horseshoe Bay and amazing views of Paterson Inlet from Observation Rock.✧More information♦Departure location: Oban Visitor Centre.♦What to bring: Comfortable walking shoes or boots, waterproof jacket, warm sweater or fleece jacket, sunscreen or sunglasses, insect repellent and camera.♦Car parking: Vehicle parking is available at Oban (extra cost—reservations recommended).♦Wheelchair access: Available.♦Children ticket: Children under ten go free for travel as long as they are accompanied by an adult.✧Reviews♦“There was so much to see and learn that it was hard to take everything in. The bays we stopped at were beautiful with golden sandy beaches, the forests were overpowering and we expecteddinosaurs to appear at any time, the views from lookout point were splendid and the anchor pointwith Bluff brought a smile. Thank you to Chris and the experienced team for such an informativetour.”Ron P♦“Any visitor to Stewart Island could do no better than take one of the guided tours from the Oban Visitor Centre—especially if you only have limited time available. We had the delightful andextremely informative Kylie conduct a small number on one of the village tours. This is a beautifulplace—a few fascinating shops and restaurants, wonderful walks and warm and friendly people.”Michael Mason ♦“I love finding out about places and the guide was full of information and stories as we visited every interesting place and view in Oban (it didn’t take too long...). A great way to start a visit as it helpsyou know where everything is.”Kiwieric60. If a traveler plans to leave a car at Oban, he had better ________.A. refer to the guides firstB. use wheelchair accessC. make a reservationD. walk to the center in advance61. Herry, a six-year-old boy, wanted to have a sightseeing of the Stewart Island with his parents. How muchshould they pay for the mini-bus tour?A. $135.B. $90.C. $ 45.D. Free.62. If a traveler takes the guided tour, he can experience all the following EXCEPT ________.A. breath-taking sceneryB. charming walksC. dinosaur samplesD. detailed tour guide(C)①What does it say about the future of meat when the country’s largest processor of chicken, pork, and beef buys a stake(股份) in a start-up that aims to “perfectly replace animal protein with plant protein”?②Tyson Foods announced this week that it purchased a 5 percent stake in Beyond Meat, the Southern California-based food-tech start-up that made headlines earlier this year with its veggie burger that reportedly cooks and tastes like real beef.③To be sure, Beyond Meat’s meatless creations have yet to take the country by storm. Although the 100 percent plant-based burgers have achieved plenty of positive press since they appeared for the first time in May, so far they’re only available at Whole Foods stores in seven states. Even though the company’s “chicken” strips, “beef” pies, and meatless frozen dinners are available nationwide, Beyond Meat is hardly a household name.④That may be what makes the news of Tyson’s investment all the more noteworthy. While the two companies declined to give details about the deal, it’s doubtful that Tyson’s 5 percent stake made much of dent(凹陷) in the meat giant’s coffers(金库). The company posted $41.4 billion in sales last year; prior to the deal with Tyson, Beyond Meat had reportedly raised $64 million in project capital funding—about what Tyson earns before lunch on any given day.⑤Tyson is doing pretty great. The company reported record third-quarter earnings per share in August and says that it expects overall meat production to increase 2 to 3 percent during the next financial year. But like a big oil company shelling out cash to invest in wind power, Tyson’s toe-in-the-water move to team up with a start-up devoted to bringing more plant-based protein to American dinner tables seems to suggest the meat industry is starting to see which way the winds are blowing.⑥Sales of plant-based protein, which totaled an estimated $5 billion last year, continue to pale compared with the market for meat in America—but vegetarian alternatives to meat are booming, with sales growing at more than double the rate for food products overall. The steady drumbeat of news about the negative health impacts, environmental problems, and animal welfare concerns associated with meat consumption appears to be sinking in. According to a survey released in April, more than half of Americans surveyed said they plan to eat more plant-based foods in the coming year.63. Beyond Meat’s veggie burger made headlines probably because __________.A. it makes perfect use of animal proteinB. it uses high tech in the making processC. it tastes as good as a genuine beef burgerD. it represents the diet trend in South California64. Which of the following statements is TRUE regarding the state of Beyond Meat?A. It is the creator of the country’s first 100 percent plant-based burgers.B. It has been well received as its products are available nationwide.C. It is far from being a match to real food processing giants like Tyson.D. It provides high-quality dining experience in selected Whole Foods stores.65. What can we infer from paragraph 4?A. The purchase of the stake barely costs a thing for Tyson.B. The 5 percent stake in Beyond Meat means a lot to Tyson.C. Tyson’s investment hasn’t caught the attention of the media as expected.D. Tyson is relying on this investment to raise more project capital funding.66. What does the passage mainly talk about?A. Meat will still take over the market in spite of other alternatives.B. A major American meat company is betting on plant-based protein.C. Tyson and Beyond Meat work together to build a global meat giant.D. Plants have been found to contain protein that does more good to human beings.Section CDirections:Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.Would You B ully(欺负) a Driverless Car or Show It Respect?Say you’re driving down a two-way street and there’s a truck unloading a delivery in the opposite lane. The oncoming traffic needs to pull out into your lane to overtake.What do you do?___67___ Eventually one of us feels charitable and slows down to allow the oncoming car to overtake and give permission with a quick flash of headlights or a wave of the hand.But what if the car waiting patiently behind the parked truck is a driverless or autonomous vehicle (A V)? Will this robot car be able to understand what you mean when you flash your lights or wave your hands?Its sensors could decide that it’s only safe to overtake when there’s no oncoming traffic at all. On a busy road at school home time, this may be never, leading to increasingly angry drivers queuing behind. ___68___ This is one of the conclusions to be drawn from research carried out by Dr Chris Tennant of the psychological and behavioural science department at the London School of Economics.His Europe-wide survey finds that nearly two-thirds of drivers think machines won’t have enough common sense to interact with human drivers, and more than two-fifths think a robot car would remain stuck behind our assumed parked truck for a long time.Driving isn’t just about technology and engineering, it’s about human interactions and psychology. The road is a social space. ___69___ “If you view the road as a social space, you will consciously negotiate your journey with other drivers. People who like that negotiation process appear to feel less comfortable engaging with AVs than with human drivers,” says Mr Tennant in his report.___70___ A statistic often trotted out(动不动就搬出) is that human error is responsible for more than 90% of accidents, with our tendency to road anger, tiredness and lack of concentration.IV. Summary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.Super Size MeFast food, otherwise known as junk food, is a huge passion for a large number of people across the Western world. But what would happen if you ate lots of junk food, every day? Would it seriously damage your health? These were the questions which led Morgan Spurlock, an independent film-maker, to do an experiment, which he made into a documentary film entitled Super Size Me.The main basis of his experiment was that Spurlock promised to eat three McDonald’s meals a day, every day, for a month. He could only eat food from McDonald’s and every time an employee asked if he would like to “super size” the meal, he had to agree. “Super sizing” refers to the fact that with this type of meal you get a considerable larger portion of everything.Spurlock knew that by eating three McDonald’s meals a day, he would consume a lot of fat and a great deal of salt and sugar in each meal—much more than he needed. Although Spurlock knew he would put on a bit of weight, and that this diet was unhealthy, he wasn’t quite prepared for just how unhealthy it turned out to be. The changes in his body were horrifying in the first week, he put on 4.5 kilos and by the end of the thirty days he had gained nearly 14 kilos, bringing his total weight to a massive 98kg.Spurlock says “I’d love people to walk out of the movie and say, ’Next time I’m not going to “super size”. Maybe I’m not going have any junk food at all. I’m going to sit down and eat dinner with my kids, with the TV off, so that we can eat healthy food, talk about what we’re eating and have a relationship with each other.’” Food for thought indeed.第II卷（共40分）V. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1.这款手表不防水。
上海高三数学一模汇总(杨浦、青浦)
杨浦区2016学年度第一学期期末高三年级质量调研数学学科试卷考生注意： 1．答卷前，考生务必在答题纸写上姓名、考号, 并将核对后的条形码贴在指定位置上．2．本试卷共有21道题，满分150分，考试时间120分钟．一．填空题（本大题满分54分）本大题共有12题，1-6每题4分，7-12每题5分。
考生应在答题纸相应编号的空格内直接填写结果．1、 若“a b >”，则“33a b >”是________命题．（填：真、假）2、 已知(0]A =-∞,，()B a =+∞,，若A B =R ，则a 的取值范围是________．3、 294z z i +=+（i 为虚数单位），则||z =________．4、 若ABC △中，4a b +=，30C ∠=︒，则ABC △面积的最大值是_________．5、 若函数2()log 1x af x x -=+的反函数的图像过点(2,3)-，则a =________． 6、 过半径为2的球O 表面上一点A 作球O 的截面，若OA 与该截面所成的角是60︒，则该截面的面积是__________．7、 抛掷一枚均匀的骰子（刻有1、2、3、4、5、6）三次，得到的数字依次记作a 、b 、c ，则a bi +（i 为虚数单位）是方程220x x c -+=的根的概率是___________．8、 设常数0a >，9(x展开式中6x 的系数为4，则2lim()n n a a a →∞++⋅⋅⋅+=_______．9、 已知直线l 经过点(0)且方向向量为(21)-,，则原点O 到直线l 的距离为__________．10、 若双曲线的一条渐近线为20x y +=，且双曲线与抛物线2y x =的准线仅有一个公共点，则此双曲线的标准方程为_________．11、 平面直角坐标系中，给出点(1,0)A ，(4,0)B ，若直线10x my +-=上存在点P ，使得||2||PA PB =，则实数m 的取值范围是___________．12、 函数()y f x =是最小正周期为4的偶函数，且在[]2,0x ∈-时，()21f x x =+，若存在12,,,n x x x 满足120n x x x ≤<<<，且()()()()1223f x f x f x f x -+-+()()12016n n f x f x -+-=，则n n x +最小值为__________．二、选择题（本大题满分20分）本大题共有4题，每题有且只有一个正确答案，考生应在答题纸的相应编号上，填上正确的答案，选对得5分，否则一律得零分. 13、若a 与b c -都是非零向量，则“a b a c ⋅=⋅”是“()a b c ⊥-”的()(A) 充分但非必要条件 (B) 必要但非充分条件 (C) 充要条件(D) 既非充分也非必要条件14、行列式147258369中，元素7的代数余子式的值为()(A) 15-(B) 3-(C) 3(D) 1215、一个公司有8名员工，其中6位员工的月工资分别为5200，5300，5500，6100，6500，6600，另两位员工数据不清楚。
09-16年上海黄浦区数学一模考点汇编及试卷
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二、填空题（共 12 小题，每小题 4 分，满分 48 分）
7．（4 分）若 ，则
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8．（4 分）已知
，则
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9．（4 分）若两个相似三角形的面积之比为 4：9，则这两个相似三角形的周长之比为 ． 10．（4 分）如图，D、E 是△ABC 边 AB、AC 上的两点，且 DE∥BC，AD：DB=1：2，则
填空题
5 平面向量 6 解直角三角形的应用-方向角问题 7 分式的基本性质 8 三角形面积的求法 9 平行线分线段成比例 10 相似三角形的判定与性质 11 相似三角形的判定与性质 12 三角函数的应用 13 特殊角的三角函数值 14 解直角三角形的应用-仰角俯角问题
2012 年上海市黄浦区一模数学试卷 考察知识点 特殊角的三角函数值 相似三角形的性质 二次函数图象的性质与系数的关系 二次函数的图象与系数的关系及二次函数的 性质 平面向量 平行线分线段成比例 平面向量 特殊角的三角函数值 二次函数图象与几何变换 解直角三角形的应用-仰角俯角问题 二次函数的性质 比例线段 解直角三角形的应用-坡度坡角问题 锐角三角函数的定义；坐标与图形性质
DE：BC= ． 11．（4 分）已知线段 AB 的长为 2，P 是线段 AB 的一个黄金分割点，且 PA＜PB，则 PA 的 长为 ． 12．（4 分）若二次函数 y=x2﹣3x+k 的图象与 x 轴有公共点，则实数 k 的取值范围是 ．
第 1页（共 18页）
13．（4 分）将二次函数 y=x2+2 的图象沿 y 轴方向向下平移 3 个单位，则所得图象的函数解 析式是 ． 14．（4 分）已知抛物线 y=ax2+bx+c 的开口向下，对称轴为直线 x=1，若点 A（﹣1，y1）与 B（﹣2，y2）是此抛物线上的两点，则 y1 y2．
2016届黄浦区高三一模数学卷
上海市黄浦区2016届高三一模数学试卷2016.01一. 填空题（本大题共14题，每题4分，共56分）1. 不等式|1|1x -<的解集用区间表示为 ；2. 函数22cos sin y x x =-的最小正周期是 ；3. 直线312=yx 的一个方向向量可以是 ；4. 若将两个半径为1的铁球熔化后铸成一个球，则该球的半径为 ；5. 若无穷数列中的任意一项均等于其之后所有项的和，则其公比为 ；6. 若函数x a y sin +=在区间[,2]ππ上有且只有一个零点，则a = ；7. 若函数22()1f x x a x =-+-为偶函数且非奇函数，则实数a 的取值范围为 ；8. 若0a >且1a ≠，函数2()x f x a+=的反函数的图像过点P ，则点P 的坐标是 ； 9.（理）在()n a b +的二项式展开式中，若二项式系数的和为128，则二项式系数的最大值为 ；（结果用数字作答）（文）在()n a b +的二项式展开式中，若二项式系数的和为256，则二项式系数的最大值为 ；（结果用数字作答）10. 在ABC △中，若cos(2)sin()2A C B B C A +-++-=，且2=AB ，则=BC ；11. 为强化安全意识，某学校拟在未来的连续5天中，随机抽取2天进行紧急疏散演练，那么选择的2天恰好为连续两天的概率是 ；（结果用最简分数表示）12. 已知*∈N k ，若曲线222k y x =+与曲线xy k =无交点，则k = ； 13. 已知点(,0)M m (0)m >和抛物线2:4C y x =，过C 的焦点F 的直线与C 交于两点A 、B 两点，若2AF FB =，且||||MF MA =，则m = ；14. 若非零向量,,a b c 满足230a b c ++=，且a b b c c a ⋅=⋅=⋅，则b 与c 夹角为 ；二. 选择题（本大题共4题，每题5分，共20分）15. 已知复数z ，“0z z +=”是“z 为纯虚数”的（ ）A. 充分非必要条件B. 必要非充分条件C. 充要条件D. 既非充分也非必要条件16. 已知R x ∈，下列不等式中正确的是（ ）A. y x 3121>B. 111122++>+-x x x xC. 211122+>+x x D. 2112||1x x >+ 17. 已知P 为直线b kx y +=上一动点，若点P 与原点均在直线02=+-y x 的同侧，则k 、b 满足的条件分别为（ ）A. 1k =，2b <B. 1k =，2b >C. 1k ≠，2b <D. 1k ≠，2b >18. 已知1234,,,a a a a 是各项均为正数的等差数列，其公差d 大于零，若直线1234,,,l l l l 的长 分别为1234,,,a a a a ，则（ ）A. 对任意的d ，均存在以321,,l l l 为三边的三角形B. 对任意的d ，均不存在以321,,l l l 为三边的三角形C. 对任意的d ，均存在以432,,l l l 为三边的三角形D. 对任意的d ，均不存在以432,,l l l 为三边的三角形三. 解答题（本大题共5题，共12+12+14+18+18=74分）19. 已知三棱柱111ABC A B C -的底面为直角三角形，两条直角边AC 和BC 的长分别为4 和3，侧棱1AA 的长为10；（1）若侧棱1AA 垂直于底面，求该三棱柱的表面积；（2）若侧棱1AA 与底面所成的角为60︒，求该三棱柱的体积；20. 如图，已知点A 是单位圆上一点，且位于第一象限，以x 轴的正半轴为始边，OA 为终 边的角设为α，将OA 绕坐标原点逆时针旋转2π至OB ； （1）用α表示A 、B 两点的坐标；（2）M 为x 轴上异于O 的点，若MA MB ⊥，求点M 的横坐标的取值范围；21. 如图，某地要在矩形区域OABC 内建造三角形池塘OEF ，E 、F 分别在AB 、BC 边 上，5OA =米，4OC =米，4EOF π∠=，设CF x =，AE y =；（1）试用解析式将y 表示成x 的函数；（2）求三角形池塘OEF 面积S 的最小值及此时x 的值；22.（理）已知椭圆2222:1x y a bΓ+=(0)a b >>，过原点的两条直线1l 和2l 分别与Γ交于A 、 B 和C 、D ，得到平行四边形ACBD ；（1）当ACBD 为正方形，求该正方形的面积S ；（2）若直线2l 和1l 关于y 轴对称，Γ上任意一点P 到1l 和2l 的距离分别为1d 和2d ，当 2212d d +为定值时，求此时直线1l 和2l 的斜率及该定值；（3）当ACBD 为菱形，且圆221x y +=内切于菱形ACBD 时，求a 、b 满足的关系式；（文）已知12,,,n a a a 是由n ()n N *∈个整数1,2,,n 按任意次序排列而成的数列，数列{}n b 满足1k k b n a =+-(1,2,,)k n =；（1）当3n =时，写出数列{}n a 和{}n b ，使得223a b =；（2）证明：当n 为正偶数时，不存在满足k k a b =(1,2,,)k n =的数列{}n a ； （3）若12,,,n c c c 是1,2,,n 从大到小的顺序排列而成的数列，写出k c (1,2,,)k n =， 并用含n 的式子表示122n c c nc +++； 参考公式：222112(1)(21)6n n n n +++=++23.（理）已知12,,,n a a a 是由n ()n N *∈个整数1,2,,n 按任意次序排列而成的数列，数列{}n b 满足1k k b n a =+-(1,2,,)k n =，12,,,n c c c 是1,2,,n 从大到小的顺序排列而成的数列，记122n n S c c nc =+++； （1）证明：当n 为正偶数时，不存在满足k k a b =(1,2,,)k n =的数列{}n a ； （2）写出k c (1,2,,)k n =，并用含n 的式子表示n S ；（3）利用22212(1)(2)()0n b b n b -+-++-≥， 证明：1212(1)(21)6n b b nb n n n +++≤++及122n n a a na S +++≥；参考公式：222112(1)(21)6n n n n +++=++（文）已知椭圆2222:1x y a bΓ+=(0)a b >>，过原点的两条直线1l 和2l 分别与Γ交于A 、B 和C 、D ，得到平行四边形ACBD ；（1）若4a =，3b =，且ACBD 为正方形，求该正方形的面积S ；（2）若直线1l 的方程为0bx ay -=，2l 和1l 关于y 轴对称，Γ上任意一点P 到1l 和2l 的距 离分别为1d 和2d ，证明：222212222a b d d a b +=+； （3）当ACBD 为菱形，且圆221x y +=内切于菱形ACBD 时，求a 、b 满足的关系式；。
2016年上海市黄浦区高考一模数学试卷(理科)【解析版】
2016年上海市黄浦区高考数学一模试卷（理科）一、填空题（共14小题，每小题4分，满分56分）1．（4分）不等式|x﹣1|＜1的解集用区间表示为．2．（4分）函数y＝cos2x﹣sin2x的最小正周期T＝．3．（4分）直线＝3的一个方向向量可以是．4．（4分）两个半径为1的铁球，熔化后铸成一个大球，这个大球的半径为．5．（4分）若无穷等比数列中任意一项均等于其之后所有项的和，则其公比为．6．（4分）若函数y＝a+sin x在区间[π，2π]上有且只有一个零点，则a＝．7．（4分）若函数f（x）＝+为偶函数且非奇函数，则实数a的取值范围为．8．（4分）若对任意不等于1的正数a，函数f（x）＝a x+2的反函数的图象都经过点P，则点P的坐标是．9．（4分）在（a+b）n的二项展开式中，若奇数项的二项式系数的和为128，则二项式系数的最大值为（结果用数字作答）．10．（4分）在△ABC中，若cos（A+2C﹣B）+sin（B+C﹣A）＝2，且AB＝2，则BC＝．11．（4分）为强化安全意识，某学校拟在未来的连续5天中随机抽取2天进行紧急疏散演练，那么选择的2天恰好为连续2天的概率是（结果用最简分数表示）．12．（4分）已知k∈Z，若曲线x2+y2＝k2与曲线xy＝k无交点，则k＝．13．（4分）已知点M（m，0），m＞0和抛物线C：y2＝4x．过C的焦点F的直线与C交于A，B两点，若＝2，且||＝||，则m＝．14．（4分）若非零向量，，满足+2+3＝，且•＝•＝•，则与的夹角为．二、选择题（共4小题，每小题5分，满分20分）15．（5分）已知复数z，“z+＝0”是“z为纯虚数”的（）A．充分非必要条件B．必要非充分条件C．充要条件D．既非充分也不必要条件16．（5分）已知x∈R，下列不等式中正确的是（）A．＞B．＞C．＞D．＞17．（5分）已知P为直线y＝kx+b上一动点，若点P与原点均在直线x﹣y+2＝0的同侧，则k，b满足的条件分别为（）A．k＝1，b＜2B．k＝1，b＞2C．k≠1，b＜2D．k≠1，b＞2 18．（5分）已知a1，a2，a3，a4是各项均为正数的等差数列，其公差d大于零，若线段l1，l2，l3，l4的长分别为a1，a2，a3，a4，则（）A．对任意的d，均存在以l1，l2，l3为三边的三角形B．对任意的d，均不存在以为l1，l2，l3三边的三角形C．对任意的d，均存在以l2，l3，l4为三边的三角形D．对任意的d，均不存在以l2，l3，l4为三边的三角形三、解答题（共5小题，满分74分）19．（12分）已知三棱柱ABC﹣A′B′C′的底面为直角三角形，两条直角边AC和BC的长分别为4和3，侧棱AA′的长为10．（1）若侧棱AA′垂直于底面，求该三棱柱的表面积；（2）若侧棱AA′与底面所成的角为60°，求该三棱柱的体积．20．（12分）如图，已知点A是单位圆上一点，且位于第一象限，以x轴的正半轴为始边，OA为终边的角设为α，将OA绕坐标原点逆时针旋转至OB．（1）用α表示A，B两点的坐标；（2）M为x轴上异于O的点，若MA⊥MB，求点M横坐标的取值范围．21．（14分）如图，某地要在矩形区域OABC内建造三角形池塘OEF，E，F分别在AB，BC边上，OA＝5米，OC＝4米，∠EOF＝，设CF＝x，AE＝y．（1）试用解析式将y表示成x的函数；（2）求三角形池塘OEF面积S的最小值及此时x的值．22．（18分）已知椭圆Γ：+＝1（a＞b＞0），过原点的两条直线l1和l2分别与Γ交于点A、B和C、D，得到平行四边形ACBD．（1）当ACBD为正方形时，求该正方形的面积S；（2）若直线l1和l2关于y轴对称，Γ上任意一点P到l1和l2的距离分别为d1和d2，当d12+d22为定值时，求此时直线l1和l2的斜率及该定值．（3）当ACBD为菱形，且圆x2+y2＝1内切于菱形ACBD时，求a，b满足的关系式．23．（18分）已知a1，a2，…，a n是由n（n∈N*）个整数1，2，…，n按任意次序排列而成的数列．数列{b n}满足b k＝n+1﹣a k（k＝1，2，…，n），c1，c2，…，c n是1，2，…，n按从大到小的顺序排列而成的数列，记S n＝c1+2c2+…+n c n．（1）证明：当n为正偶数时，不存在满足a k＝b k（k＝1，2，…，n）的数列{a n}；（2）写出c k（k＝1，2，…，n），并用含n的式子表示S n；（3）利用（1﹣b1）2+（2﹣b2）2+…+（n﹣b n）2≥0，证明：b1+2b2+…+nb n≤n（n+1）（2n+1）及a1+2a2+…+na n≥S n．（参考：12+22+…+n2＝n（n+1）（2n+1））2016年上海市黄浦区高考数学一模试卷（理科）参考答案与试题解析一、填空题（共14小题，每小题4分，满分56分）1．（4分）不等式|x﹣1|＜1的解集用区间表示为（0，2）．【解答】解：不等式|x﹣1|＜1等价为：﹣1＜x﹣1＜1，解得，0＜x＜2，即原不等式的解集为{x|0＜x＜2}，用区间表示为：（0，2），故答案为：（0，2）．2．（4分）函数y＝cos2x﹣sin2x的最小正周期T＝π．【解答】解：y＝cos2x﹣sin2x＝cos2x，∴函数y＝cos2x﹣sin2x的最小正周期T＝＝π．故答案为：π．3．（4分）直线＝3的一个方向向量可以是（﹣2，﹣1）．．【解答】解：∵直线＝3，∴x﹣2y﹣3＝0．∴直线＝3的一个方向向量可以是（﹣2，﹣1）．故答案为：（﹣2，﹣1）．4．（4分）两个半径为1的铁球，熔化后铸成一个大球，这个大球的半径为．【解答】解：设大球的半径为r，则根据体积相同，可知，即．故答案为：．5．（4分）若无穷等比数列中任意一项均等于其之后所有项的和，则其公比为．【解答】解：设数列中的任意一项为a，由无穷等比数列中的每一项都等于它后面所有各项的和，得a＝，即1﹣q＝q∴q＝．故答案为：．6．（4分）若函数y＝a+sin x在区间[π，2π]上有且只有一个零点，则a＝1．【解答】解：作函数y＝sin x在区间[π，2π]上的图象如下，，结合图象可知，若函数y＝a+sin x在区间[π，2π]上有且只有一个零点，则a﹣1＝0，故a＝1；故答案为：1．7．（4分）若函数f（x）＝+为偶函数且非奇函数，则实数a的取值范围为a＞1．【解答】解：∵函数f（x）＝+为偶函数且非奇函数，∴f（﹣x）＝f（x），且f（﹣x）≠﹣f（x），又，∴a≥1．a＝1，函数f（x）＝+为偶函数且奇函数，故答案为：a＞1．8．（4分）若对任意不等于1的正数a，函数f（x）＝a x+2的反函数的图象都经过点P，则点P的坐标是（1，﹣2）．【解答】解：∵当x+2＝0，即x＝﹣2时，总有a0＝1，∴函数f（x）＝a x+2的图象都经过点（﹣2，1），∴其反函数的图象必经过点P（1，﹣2）故答案为：（1，﹣2）9．（4分）在（a+b）n的二项展开式中，若奇数项的二项式系数的和为128，则二项式系数的最大值为70（结果用数字作答）．【解答】解：在（a+b）n的展开式中，奇数项的二项式系数的和等于偶数项的二项式系数的和，∴2n＝256，解得n＝8，展开式共n+1＝8+1＝9项，据中间项的二项式系数最大，故展开式中系数最大的项是第5项，最大值为＝70．故答案为：70．10．（4分）在△ABC中，若cos（A+2C﹣B）+sin（B+C﹣A）＝2，且AB＝2，则BC＝2．【解答】解：∵cos（A+2C﹣B）+sin（B+C﹣A）＝2，cos（A+2C﹣B）≤1，sin （B+C﹣A）≤1，∴cos（A+2C﹣B）＝1，sin（B+C﹣A）＝1，∵A，B，C∈（0，π），∴A+2C﹣B∈（﹣π，3π），B+C﹣A∈（﹣π，2π），∴由正弦函数，余弦函数的图象和性质可得：A+2C﹣B＝0或2π，B+C﹣A＝，∴结合三角形内角和定理可得：①，或②，由①可得：A＝，B＝，C＝，由②可得：A＝，B＝﹣，C＝，（舍去），∴由AB＝2，利用正弦定理可得：，解得：BC＝2．故答案为：2．11．（4分）为强化安全意识，某学校拟在未来的连续5天中随机抽取2天进行紧急疏散演练，那么选择的2天恰好为连续2天的概率是（结果用最简分数表示）．【解答】解：某学校拟在未来的连续5天中随机抽取2天进行紧急疏散演练，基本事件总数为n＝＝10，选择的2天恰好为连续2天包含的基本事件个数m＝4，∴选择的2天恰好为连续2天的概率p＝．故答案为：．12．（4分）已知k∈Z，若曲线x2+y2＝k2与曲线xy＝k无交点，则k＝±1．【解答】解：曲线x2+y2＝k2，令x＝k cosθ，y＝k sinθ，代入曲线xy＝k，曲线x2+y2＝k2与曲线xy＝k无交点，可得k2sinθcosθ＝k，不成立．即sin2θ＝不成立，1，k∈Z，可得k＝±1．故答案为：±1．13．（4分）已知点M（m，0），m＞0和抛物线C：y2＝4x．过C的焦点F的直线与C交于A，B两点，若＝2，且||＝||，则m＝．【解答】解：由题意可知：F（1，0），由抛物线定义可知A（x1，y1），可知B（x2，y2），∵＝2，可得：2（x2﹣1，y2）＝（1﹣x1，﹣y1），可得y2＝﹣，x2＝，，解得x1＝2，y1＝±2．||＝||，可得|m﹣1|＝，解得m＝．故答案为：．14．（4分）若非零向量，，满足+2+3＝，且•＝•＝•，则与的夹角为．【解答】解：由+2+3＝，得，代入•＝•，得，即．再代入•＝•，得，即．∴cos＝＝＝﹣．∴与的夹角为．故答案为：．二、选择题（共4小题，每小题5分，满分20分）15．（5分）已知复数z，“z+＝0”是“z为纯虚数”的（）A．充分非必要条件B．必要非充分条件C．充要条件D．既非充分也不必要条件【解答】解：对于复数z，若z+＝0，z不一定为纯虚数，可以为0，反之，若z 为纯虚数，则z+＝0．∴“z+＝0”是“z为纯虚数”的必要非充分条件．故选：B．16．（5分）已知x∈R，下列不等式中正确的是（）A．＞B．＞C．＞D．＞【解答】解：取x＝0可得＝1＝，故A错误；取x＝0可得＝1＝，故B错误；取x＝1可得＝＝，故D错误；选项C，∵x2+2＞x2+1＞0，∴＞，故正确．故选：C．17．（5分）已知P为直线y＝kx+b上一动点，若点P与原点均在直线x﹣y+2＝0的同侧，则k，b满足的条件分别为（）A．k＝1，b＜2B．k＝1，b＞2C．k≠1，b＜2D．k≠1，b＞2【解答】解：∵P为直线y＝kx+b上一动点，∴设P（x，kx+b），∵点P与原点均在直线x﹣y+2＝0的同侧，∴（x﹣kx﹣b+2）（0﹣0+2）＞0，即2[（1﹣k）x+2﹣b]＞0恒成立，即（1﹣k）x+2﹣b＞0恒成立，则1﹣k＝0，此时2﹣b＞0，得k＝1且b＜2，故选：A．18．（5分）已知a1，a2，a3，a4是各项均为正数的等差数列，其公差d大于零，若线段l1，l2，l3，l4的长分别为a1，a2，a3，a4，则（）A．对任意的d，均存在以l1，l2，l3为三边的三角形B．对任意的d，均不存在以为l1，l2，l3三边的三角形C．对任意的d，均存在以l2，l3，l4为三边的三角形D．对任意的d，均不存在以l2，l3，l4为三边的三角形【解答】解：A：对任意的d，假设均存在以l1，l2，l3为三边的三角形，∵a1，a2，a3，a4是各项均为正数的等差数列，其公差d大于零，∴a2+a3＞a1，a3+a1＝2a2＞a2，而a1+a2﹣a3＝a1﹣d不一定大于0，因此不一定存在以为l1，l2，l3三边的三角形，故不正确；B：由A可知：当a1﹣d＞0时，存在以为l1，l2，l3三边的三角形，因此不正确；C：对任意的d，由于a3+a4，＞a2，a2+a4＝2a1+4d＝a1+2d+a3＞0，a2+a3﹣a4＝a1＞0，因此均存在以l2，l3，l4为三边的三角形，正确；D．由C可知不正确．故选：C．三、解答题（共5小题，满分74分）19．（12分）已知三棱柱ABC﹣A′B′C′的底面为直角三角形，两条直角边AC和BC的长分别为4和3，侧棱AA′的长为10．（1）若侧棱AA′垂直于底面，求该三棱柱的表面积；（2）若侧棱AA′与底面所成的角为60°，求该三棱柱的体积．【解答】解：（1）因为侧棱AA′⊥底面ABC，所以三棱柱的高h等于侧棱AA′的长，而底面三角形ABC的面积S＝AC•BC＝6，周长c＝4+3+5＝12，于是三棱柱的表面积S全＝ch+2S△ABC＝132．（2）如图，过A作平面ABC的垂线，垂足为H，A′H为三棱柱的高．因为侧棱AA′与底面ABC所长的角为60°，所以∠A′AH＝60°，又底面三角形ABC的面积S＝6，故三棱柱的体积V＝S•A′H＝6×＝30．20．（12分）如图，已知点A是单位圆上一点，且位于第一象限，以x轴的正半轴为始边，OA为终边的角设为α，将OA绕坐标原点逆时针旋转至OB．（1）用α表示A，B两点的坐标；（2）M为x轴上异于O的点，若MA⊥MB，求点M横坐标的取值范围．【解答】解：（1）点A是单位圆上一点，且位于第一象限，以x轴的正半轴为始边，OA为终边的角设为α，α∈（0，）可得A（cosα，sinα），将OA绕坐标原点逆时针旋转至OB．可得B（cos（），sin（）），即B（﹣sinα，cosα）．（2）设M（x，0），x≠0，＝（cosα﹣x，sinα），＝（﹣sinα﹣x，cosα）．MA⊥MB，可得（cosα﹣x）（﹣sinα﹣x）+sinαcosα＝0．x sinα﹣x cosα+x2＝0，可得﹣x＝sinα﹣cosα＝sin（）∈（﹣1，1）．综上x∈（﹣1，0）∪（0，1）．点M横坐标的取值范围：（﹣1，0）∪（0，1）．21．（14分）如图，某地要在矩形区域OABC内建造三角形池塘OEF，E，F分别在AB，BC边上，OA＝5米，OC＝4米，∠EOF＝，设CF＝x，AE＝y．（1）试用解析式将y表示成x的函数；（2）求三角形池塘OEF面积S的最小值及此时x的值．【解答】解：（1）由∠EOF＝，可得∠COF+∠AOE＝，即有tan∠COF＝，tan∠AOE＝，则tan（∠COF+∠AOE）＝＝1，即有y＝，由y≤4，解得x≥，则函数的解析式为y＝，（≤x≤4）；（2）三角形池塘OEF面积S＝S矩形OABC ﹣S△AOE﹣S△COF﹣S△BEF＝4×5﹣×5y﹣×4x﹣×（4﹣y）（5﹣x）＝20﹣•﹣2x﹣（5﹣x）•＝20+（≤x≤4），令t＝x+4（≤t≤8），即有S＝20+（5t+﹣80）≥20+（2﹣80）＝20﹣20．当且仅当5t＝即t＝4，此时x＝4﹣4，△OEF的面积取得最小值，且为20﹣20．22．（18分）已知椭圆Γ：+＝1（a＞b＞0），过原点的两条直线l1和l2分别与Γ交于点A、B和C、D，得到平行四边形ACBD．（1）当ACBD为正方形时，求该正方形的面积S；（2）若直线l1和l2关于y轴对称，Γ上任意一点P到l1和l2的距离分别为d1和d2，当d12+d22为定值时，求此时直线l1和l2的斜率及该定值．（3）当ACBD为菱形，且圆x2+y2＝1内切于菱形ACBD时，求a，b满足的关系式．【解答】解：（1）∵ACBD为正方形，∴直线l1和l2的方程为y＝x和y＝﹣x，设点A、B的坐标为（x1，y1）、（x2，y2），解方程组，得＝＝，由对称性可知，S＝4＝；（2）由题意，不妨设直线l1的方程为y＝kx，则直线l2的方程为y＝﹣kx，设P（x0，y0），则+＝1，又∵d1＝，d2＝，∴+＝+＝，将＝b2（1﹣）代入上式，得+＝，∵d12+d22为定值，∴k2﹣＝0，即k＝±，于是直线l1和l2的斜率分别为和﹣，此时+＝；（3）设AC与圆x2+y2＝1相切的切点坐标为（x0，y0），则切线AC的方程为：x0x+y0y＝1，点A、C的坐标为（x1，y1）、（x2，y2）为方程组的实数解．①当x0＝0或y0＝0时，ACBD均为正方形，椭圆均过点（1，1），于是有+＝1；②当x0≠0或y0≠0时，将y＝（1﹣x0x）代入+＝1，整理得：（a2+b2）x2﹣2a2x0x﹣a2（1+b2）＝0，由韦达定理可知x1x2＝，同理可知y1y2＝，∵ACBD为菱形，∴AO⊥CO，即x1x2+y1y2＝0，∴+＝0，整理得：a2+b2＝a2b2（+），又∵+＝1，∴a2+b2＝a2b2，即+＝1；综上所述，a，b满足的关系式为+＝1．23．（18分）已知a1，a2，…，a n是由n（n∈N*）个整数1，2，…，n按任意次序排列而成的数列．数列{b n}满足b k＝n+1﹣a k（k＝1，2，…，n），c1，c2，…，c n是1，2，…，n按从大到小的顺序排列而成的数列，记S n＝c1+2c2+…+n c n．（1）证明：当n为正偶数时，不存在满足a k＝b k（k＝1，2，…，n）的数列{a n}；（2）写出c k（k＝1，2，…，n），并用含n的式子表示S n；（3）利用（1﹣b1）2+（2﹣b2）2+…+（n﹣b n）2≥0，证明：b1+2b2+…+nb n≤n（n+1）（2n+1）及a1+2a2+…+na n≥S n．（参考：12+22+…+n2＝n（n+1）（2n+1））【解答】解：（1）证明：当n为正偶数时，存在满足a k＝b k（k＝1，2，…，n）的数列{a n}，由b k＝n+1﹣a k（k＝1，2，…，n），可得a k＝，由n为正偶数，可得n+1为奇数，不为整数，a k为整数，故不成立，则当n为正偶数时，不存在满足a k＝b k（k＝1，2，…，n）的数列{a n}；（2）{c k}：n，n﹣1，n﹣2， (1)＝3+，n 由S1＝1，S2﹣S1＝3，S3﹣S2＝6，S4﹣S3＝10，…，S n﹣S n﹣1＞1．累加可得，S n＝1+3+6+10+…+[3+]＝（12+22+…+n2）+（1+2+…+n）]＝×n（n+1）（2n+1）+n（n+1）＝n（n+1）（n+2）；（3）证明：由（1﹣b1）2+（2﹣b2）2+…+（n﹣b n）2≥0，可得12+22+…+n2﹣2（b1+2b2+…+nb n）+（b12+b22+…+b n2）≥0，即有b1+2b2+…+nb n≤[（12+22+…+n2）+（b12+b22+…+b n2）]＝12+22+…+n2＝n（n+1）（2n+1）；由排序定理可得，乱序之和不小于倒序之和，由a1+2a2+…+na n为乱序之和，S n＝c1+2c2+…+n c n为倒序之和．即可得到a1+2a2+…+na n≥S n．。
上海市浦东新区2016年第一学期高三数学一模试卷
浦东新区2016年第一学期期末质量测试高三数学试卷 （含答案） 2016.1注意：1. 答卷前，考生务必在答题纸上指定位置将学校、姓名、考号填写清楚. 2. 本试卷共有32道试题，满分150分，考试时间130分钟.一、填空题（本大题共有12题，满分36分）只要求直接填写结果，每个空格填对得3分，否则一律得零分.注：填写其他等价形式则得分1．已知集合{}{}=3,2A x x B x x ≤=<，则R AC B = []2,32．已知向量()2,1,(1,)a b m =-=平行，则m = 12- 3．关于,x y 的一元二次方程组23122x y x y +=⎧⎨-=⎩的系数矩阵2312⎛⎫⎪-⎝⎭4．计算：1132lim 32n nnn n ++→∞-+ 3 5．若复数z 满足1012ii z=-（i 为虚数单位），则z6．()1021x +的二项展开式中的第八项为 3960x7．某船在海平面A 处测得灯塔B 在北偏东30︒方向，与A 相距6.0海里.船由A 向正北方向航行8.1海里达到C 处，这时灯塔B 与船相距_____4.2______海里（精确到0.1海里） 8．已知3cos(),,252ππααπ⎛⎫-=∈ ⎪⎝⎭，则sin 3πα⎛⎫+= ⎪⎝⎭310- 9．如图，已知正方体1111D C B A ABCD -，21=AA ，E 为棱1CC 的中点，则AE与平面11BCC B 所成的角为552arctan．（2arcsin 3，arccos 3）（结果用反三角表示）10．已知函数()f x 的图像与()2xg x =的图像关于直线y x =对称，令()(1)h x f x =-，则关于函数()h x 有下列命题：①()h x 的图像关于原点对称； ②()h x 的图像关于y 轴对称； ③()h x 的最大值为0； ④()h x 在区间(1,1)-上单调递增。
其中正确命题的序号为____②③_____（写出所有正确命题的序号）。
2016年高三第一次联合模拟考试 数理答案
2016年高三第一次联合模拟考试理科数学答案ABDACB BBACDC （注：11题4,e >∴D 选项也不对，此题无答案。
建议：任意选项均可给分）13. 2; 14.14; 15.8; 16. []1,3 17.解：（Ⅰ）证明：113133()222+-=-=-n n n a a a …….3分12111=-=a b 31=∴+n n b b ， 所以数列{}n b 是以1为首项，以3为公比的等比数列；….6分（Ⅱ）解：由（1）知，13-=n n b ，由111n n b m b ++≤-得13131n n m -+≤-，即()143331nm +≤-，…9分 设()143331=+-n nc ，所以数列{}n c 为减数列，()1max 1==n c c ， 1∴≥m …….12分18解：（Ⅰ）平均数为500.051500.12500.153500.34500.155500.26500.05370⨯+⨯+⨯+⨯+⨯+⨯+⨯=………….4分 （Ⅱ）X 的所有取值为0,1,2,3,4． ……….5分由题意，购买一个灯管，且这个灯管是优等品的概率为0.200.050.25+=，且1~4,4X B ⎛⎫ ⎪⎝⎭4413()(0,1,2,3,4)44-⎛⎫⎛⎫==⋅= ⎪⎪⎝⎭⎝⎭kkk P X k C k所以044181(0)C (1)4256P X ==⨯-=， 1341110827(1)C (1)4425664P X ==⨯⨯-==， 2224115427(2)C ()(1)44256128P X ==⨯-==， 331411123(3)C ()(1)4425664P X ==⨯-==， 4404111(4)C ()(1)44256P X ==⨯-=． 以随机变量X 的分布列为：X0 1 2 34 P81256 2764 27128 3641256……………………….10分所以X 的数学期望1()414E X =⨯=．…….12分 19.（Ⅰ）证明：四边形ABCD 是菱形，BD AC ∴⊥.⊥AE 平面ABCD ,BD ⊂平面ABCDBD AE ∴⊥.⋂=AC AE A ,BD ∴⊥平面ACFE .………….4分 （Ⅱ）解：如图以O 为原点，,OA OB 为,x y 轴正向，z 轴过O 且平行于CF ，建立空间直角坐标系.则(0,3,0),(0,3,0),(1,0,2),(1,0,)(0)B D E F a a -->，(1,0,)=-OF a .…………6分设平面EDB 的法向量为(,,)=n x y z ， 则有00⎧⋅=⎪⎨⋅=⎪⎩n OB n OE ，即3020y x z ⎧=⎪⎨+=⎪⎩令1z =，(2,0,1)=-n .…………8分由题意o2||2sin 45|cos ,|2||||15⋅=<>===+OF n OF n OF n a 解得3a =或13-. 由0>a ，得3=a . …….12分20. 解：（Ⅰ）由题意得22222,3122 1.a b c ca a b⎧⎪⎪=+⎪⎪=⎨⎪⎪⎪+=⎪⎩解得 2.1,3.a b c ⎧=⎪=⎨⎪=⎩所以C 的方程为2214x y +=. …….4分（Ⅱ）存在0x .当04x =时符合题意. 当直线l 斜率不存在时，0x 可以为任意值.设直线l 的方程为(1)y k x =-,点A ,B 满足：22(1),1.4y k x x y =-⎧⎪⎨+=⎪⎩所以A x ,B x 满足2224(1)4x k x +-=,即2222(41)8440k x k x k +-+-=. 所以22222222(8)4(41)(44)0,8,4144.41A B A B k k k k x x k k x x k ⎧⎪∆=-++>⎪⎪+=⎨+⎪⎪-=⎪+⎩………8分 不妨设1A x >>B x ，因为||||A B d PB d PA ⋅-⋅=2001[|||1||||1|]A B B A k x x x x x x +-⋅---⋅-2001[2(1)()2]0A B A B k x x x x x x =+-+++=从而2200228(1)8(1)204141x k k x k k +--+=++.整理得0280x -=，即04x =. 综上，04=x 时符合题意.…….12分21.解：（Ⅰ）'()2xf x e ax =-，由题设得，'(1)2f e a b =-=，(1)1f e a b =-=+， 解得，1,2a b e ==-． …….4分（Ⅱ）法1：由（Ⅰ）知，[]2(),'()21210,0,1xxf x e x f x e x x x x x =-∴=-≥+-=-≥∈，故()f x 在[]0,1上单调递增，所以，max ()(1)1f x f e ==-．法2：由（Ⅰ）知，2(),'()2,''()2xxxf x e x f x e x f x e =-∴=-=-，'()f x ∴在()0,ln 2上单调递减，在()ln 2,+∞上单调递增，所以，'()'(ln 2)22ln 20f x f ≥=->，所以，()f x 在[]0,1上单调递增，所以，max ()(1)1f x f e ==-． …….7分（Ⅲ）因为(0)1f =，又由（Ⅱ）知，()f x 过点(1,1)e -，且()y f x =在1x =处的切线方程为(2)1y e x =-+，故可猜测：当0,1x x >≠时，()f x 的图象恒在切线(2)1y e x =-+的上方．下证：当0x >时，()(2)1f x e x ≥-+．设()()(2)1,0g x f x e x x =--->，则'()2(2),''()2x xg x e x e g x e =---=-， 由（Ⅱ）知，'()g x 在()0,ln 2上单调递减，在()ln 2,+∞上单调递增， 又'(0)30,'(1)0,0ln 21,'(ln 2)0g e g g =->=<<∴<， 所以，存在()00,1x ∈，使得'()0g x =， 所以，当()()00,1,x x ∈+∞时，'()0g x >；当0(,1)x x ∈，'()0g x <，故()g x 在()00,x 上单调递增，在()0,1x 上单调递减，在()1,+∞上单调递增． 又2(0)(1)0,()(2)10xg g g x e x e x ==∴=----≥，当且仅当1x =时取等号．故(2)1,0x e e x x x x+--≥>． 由（Ⅱ）知，1xe x ≥+，故ln(1),1ln x x x x ≥+∴-≥，当且仅当1x =时取等号．所以，(2)1ln 1x e e x x x x+--≥≥+． 即(2)1ln 1x e e x x x+--≥+．所以，(2)1ln x e e x x x x +--≥+， 即(1)ln 10x e e x x x +---≥成立，当1x =时等号成立． …….12分 22. 解：（Ⅰ）作'AA EF ⊥交EF 于点'A ，作'BB EF ⊥交EF 于点'B .因为''A M OA OM =-，''B M OB OM =+， 所以2222''2'2A M B M OA OM +=+.从而222222''''AM BM AA A M BB B M +=+++2222('')AA OA OM =++.故22222()AM BM r m +=+ ……5分（Ⅱ）因为EM r m =-，FM r m =+，所以22AM CM BM DM EM FM r m ⋅=⋅=⋅=-.因为2222AM BM AM BM AM BM CM DM AM CM BM DM EM FM ++=+=⋅⋅⋅ 所以22222()AM BM r m CM DM r m++=-. 又因为3=r m ，所以52+=AM BM CM DM ． …………….10分 23.解：（Ⅰ）直线l 的极坐标方程分别是8sin =θρ. 圆C 的普通方程分别是22(2)4x y +-=，所以圆C 的极坐标方程分别是θρsin 4=. …….5分（Ⅱ）依题意得，点M P ,的极坐标分别为⎩⎨⎧==,,sin 4αθαρ和⎩⎨⎧==.,8sin αθαρ 所以αsin 4||=OP ，αsin 8||=OM ， 从而2||4sin sin 8||2sin OP OM ααα==.同理，2sin ()||2||2OQ ON πα+=. 所以||||||||OP OQ OM ON ⋅222sin ()sin sin (2)22216πααα+=⋅=， 故当4πα=时，||||||||OP OQ OM ON ⋅的值最大，该最大值是161. …10分 24.解 ：（Ⅰ）由已知得32x m -<-，得51m x m -<<+，即3m = …… 5分（Ⅱ）()x a f x -≥得33x x a -+-≥恒成立33()3x x a x x a a -+-≥---=-（当且仅当(3)()0--≤x x a 时取到等号）33∴-≥a 解得6a ≥或0a ≤故a 的取值范围为 0a ≤或6a ≥ …… 10分。
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上海市黄浦区2016届高三一模数学试卷
2016.01
一. 填空题（本大题共14题，每题4分，共56分）
1. 不等式|1|1x -<的解集用区间表示为 ；
2. 函数22cos sin y x x =-的最小正周期是 ；
3. 直线312=y
x 的一个方向向量可以是 ；
4. 若将两个半径为1的铁球熔化后铸成一个球，则该球的半径为 ；
5. 若无穷数列中的任意一项均等于其之后所有项的和，则其公比为 ；
6. 若函数x a y sin +=在区间[,2]ππ上有且只有一个零点，则a = ；
7. 若函数()f x =则实数a 的取值范围为 ；
8. 若0a >且1a ≠，函数2()x f x a +=的反函数的图像过点P ，则点P 的坐标是 ；
9.（理）在()n a b +的二项式展开式中，若二项式系数的和为128，则二项式系数的最大值为 ；（结果用数字作答）
（文）在()n a b +的二项式展开式中，若二项式系数的和为256，则二项式系数的最大值为 ；（结果用数字作答）
10. 在ABC △中，若cos(2)sin()2A C B B C A +-++-=，且2=AB ，则=BC ；
11. 为强化安全意识，某学校拟在未来的连续5天中，随机抽取2天进行紧急疏散演练，那么选择的2天恰好为连续两天的概率是 ；（结果用最简分数表示）
12. 已知*∈N k ，若曲线222k y x =+与曲线xy k =无交点，则k = ；
13. 已知点(,0)M m (0)m >和抛物线2
:4C y x =，过C 的焦点F 的直线与C 交于两点A 、B 两点，若2AF FB = ，且||||MF MA = ，则m = ；
14. 若非零向量,,a b c 满足230a b c ++= ，且a b b c c a ⋅=⋅=⋅ ，则b 与c 夹角为 ；
二. 选择题（本大题共4题，每题5分，共20分）
15. 已知复数z ，“0z z +=”是“z 为纯虚数”的（ ）
A. 充分非必要条件
B. 必要非充分条件
C. 充要条件
D. 既非充分也非必要条件
16. 已知R x ∈，下列不等式中正确的是（ ）
A. y x 3121>
B. 111122++>+-x x x x
C. 211122+>+x x
D. 2112||1
x x >+
17. 已知P 为直线b kx y +=上一动点，若点P 与原点均在直线02=+-y x 的同侧，则k 、 b 满足的条件分别为（ ）
A. 1k =，2b <
B. 1k =，2b >
C. 1k ≠，2b <
D. 1k ≠，2b >
18. 已知1234,,,a a a a 是各项均为正数的等差数列，其公差d 大于零，若直线1234,,,l l l l 的长 分别为1234,,,a a a a ，则（ ）
A. 对任意的d ，均存在以321,,l l l 为三边的三角形
B. 对任意的d ，均不存在以321,,l l l 为三边的三角形
C. 对任意的d ，均存在以432,,l l l 为三边的三角形
D. 对任意的d ，均不存在以432,,l l l 为三边的三角形
三. 解答题（本大题共5题，共12+12+14+18+18=74分）
19. 已知三棱柱111ABC A B C -的底面为直角三角形，两条直角边AC 和BC 的长分别为4 和3，侧棱1AA 的长为10；
（1）若侧棱1AA 垂直于底面，求该三棱柱的表面积；
（2）若侧棱1AA 与底面所成的角为60︒，求该三棱柱的体积；
20. 如图，已知点A 是单位圆上一点，且位于第一象限，以x 轴的正半轴为始边，OA 为终 边的角设为α，将OA 绕坐标原点逆时针旋转
2
π至OB ； （1）用α表示A 、B 两点的坐标；
（2）M 为x 轴上异于O 的点，若MA MB ⊥，求点M 的横坐标的取值范围；
21. 如图，某地要在矩形区域OABC 内建造三角形池塘OEF ，E 、F 分别在AB 、BC 边 上，5OA =米，4OC =米，4EOF π∠=
，设CF x =，AE y =；
（1）试用解析式将y 表示成x 的函数；
（2）求三角形池塘OEF 面积S 的最小值及此时x 的值；
22.（理）已知椭圆22
22:1x y a b
Γ+=(0)a b >>，过原点的两条直线1l 和2l 分别与Γ交于A 、 B 和C 、D ，得到平行四边形ACBD ；
（1）当ACBD 为正方形，求该正方形的面积S ；
（2）若直线2l 和1l 关于y 轴对称，Γ上任意一点P 到1l 和2l 的距离分别为1d 和2d ，当 2212d d +为定值时，求此时直线1l 和2l 的斜率及该定值；
（3）当ACBD 为菱形，且圆221x y +=内切于菱形ACBD 时，求a 、b 满足的关系式；
（文）已知12,,,n a a a 是由n ()n N *
∈个整数1,2,,n 按任意次序排列而成的数列，数列{}n b 满足1k k b n a =+-(1,2,,)k n = ；
（1）当3n =时，写出数列{}n a 和{}n b ，使得223a b =；
（2）证明：当n 为正偶数时，不存在满足k k a b =(1,2,,)k n = 的数列{}n a ；
（3）若12,,,n c c c 是1,2,,n 从大到小的顺序排列而成的数列，写出k c (1,2,,)k n = ， 并用含n 的式子表示122n c c nc +++ ； 参考公式：222112(1)(21)6
n n n n +++=
++
23.（理）已知12,,,n a a a 是由n ()n N *∈个整数1,2,,n 按任意次序排列而成的数列，数列{}n b 满足1k k b n a =+-(1,2,,)k n = ，12,,,n c c c 是1,2,,n 从大到小的顺序排列而成的数列，记122n n S c c nc =+++ ；
（1）证明：当n 为正偶数时，不存在满足k k a b =(1,2,,)k n = 的数列{}n a ；
（2）写出k c (1,2,,)k n = ，并用含n 的式子表示n S ；
（3）利用22212(1)(2)()0n b b n b -+-++-≥ ， 证明：1212(1)(21)6
n b b nb n n n +++≤
++ 及122n n a a na S +++≥ ； 参考公式：222112(1)(21)6n n n n +++=++
（文）已知椭圆22
22:1x y a b
Γ+=(0)a b >>，过原点的两条直线1l 和2l 分别与Γ交于A 、B 和C 、D ，得到平行四边形ACBD ；
（1）若4a =，3b =，且ACBD 为正方形，求该正方形的面积S ；
（2）若直线1l 的方程为0bx ay -=，2l 和1l 关于y 轴对称，Γ上任意一点P 到1l 和2l 的距
离分别为1d 和2d ，证明：22
2
2
12222a b d d a b +=+； （3）当ACBD 为菱形，且圆22
1x y +=内切于菱形ACBD 时，求a 、b 满足的关系式；。