2016年上海普陀区高三一模作文立意解析课件
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2016年上海市普陀区高考语文一模试卷一、文言文(39分)1.(5分)默写(1)枯藤老树昏鸦,。
(《天净沙•秋思》)(2),匹马戍梁州。
(《诉衷情》)(3)月上柳梢头,。
(《生查子•元夕》)(4)政通人和,。
(《岳阳楼记》)(5),在乎山水之间也。
(《醉翁亭记》)2.(4分)阅读下面的小令,完成下列各题。
四块玉•别情【元】关汉卿自送别,心难舍,一点相思几时绝?凭阑袖拂杨花雪。
溪又斜,山又遮,人去也!(1)“绝”在文中的意思是。
(2)下列理解正确的一项是A、“自送别,心难舍”体现送别时心中难舍。
B、“一点相思”特别强调相思之情并不多。
C、“杨花雪”语出苏轼《少年游》,暗含离别之味。
D、这首小令似游子口吻描写离别相思之情。
3.(8分)阅读下文,完成下列各题。
余幼好书,家贫难致。
有张氏藏书甚富。
往借,不与,归而形诸梦。
其切如是。
故有所览辄省记。
通籍后,俸去书来,落落大满,素蟫灰丝时蒙卷轴。
然后叹借者之用心专,而少时之岁月为可惜也!今黄生贫类予,其借书亦类予;惟予之公书与张氏之吝书若不相类。
然则予固不幸而遇张乎,生固幸而遇予乎?知幸与不幸,则其读书也必专,而其归书也必速。
(1)《黄生借书说》的作者是朝的(人名)。
(2)用现代汉语翻译下面的句子。
往借,不与,归而形诸梦。
(3)下列理解不正确的一项是A、选文表述了“知幸与不幸,则其读书也必专,而其归书也必速。
”的观点。
B、作者感叹“可惜”是因为在那段少时岁月,借书读反而促使自己认真读书。
C、“若不相类”体现了“予之公书”与“张氏之吝书”的确是不一样的。
D、“固”解释为“本来”,表明“幸”与“不幸”在一定条件下可以转化。
4.(12分)阅读下文,完成下列各题不见长安晋明帝数岁,坐元帝膝上。
有人从长安来,元帝问洛下①消息,潸然流涕。
明帝问何以致泣,具以东渡意②告之。
因问明帝:“汝意谓长安何如日远?”答曰:“日远。
不闻人从日边来,居然可知。
”元帝异之。
明日,集群臣宴会,告以此意,便重问之。
2016届上海普陀高三一模考试时间:120分钟满分:150分英语试卷2016. 1考生注意:1. 考试时间120分钟,试卷满分150分。
2. 本考试设试卷和答题纸两部分。
试卷分为第I卷(第1-13页)和第II卷(第13-14页),全卷共14页。
所有答题必须涂(选择题)或写(非选择题)在答题纸上,做在试卷上一律不得分。
3. 答题前,务必在答题纸上填写准考证号和姓名。
第I卷(共103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. Teacher. B. Repairman. C. Shop assistant. D. Doctor.2. A. At home. B. At school. C. In the hospital. D. On the street.3. A. His new job is too difficult for him.B. He is used to his new job.C. He is still trying to get used to his new job.D. He doesn’t like his new job.4. A. 15 dollars. B. 14 dollars. C. 10 dollars. D. 12 dollars.5. A. She is surprised at her mum’s c oming back so soon.B. She thinks that she is too slow.C. She wants the man to be quick.D. She will go out herself.6. A. She used to be in poor health. B. She was popular among boys.C. She was somewhat overweight.D. She didn’t do well at high sc hool.7. A. At the airport. B. In a restaurant.C. In a booking office.D. At the hotel reception.8. A. Teaching her son by herself.B. Having confidence in her son.C. Asking the teacher for extra help.D. Telling her son not to worry.9. A. Have a short break. B. Take two weeks off.C. Continue her work outdoors.D. Go on vacation with the man.10. A. He is taking care of his twin brother.B. He has been feeling ill all week.C. He is worried about Rod.D. He has been in perfect condition.Section BDirections: In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages 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. Lawyer. B. Computer programmer.C. Blogger.D. Firefighter.12. A. 21% of all the employers. B. 1% of American teenagers.C. 79% of all the employers.D. 1% of American adults.13. A. With many bloggers, America is sure to win her reputation in the world.B. Washington is the city which has most bloggers in America.C. There are fewer employees of newspapers than a few years ago.D. The topics of blogging cover almost every area of people’s daily life.Questions 14 through 16 are based on the following news.14. A. In the Town Hall. B. In a community.C. In some body’s house.D. In a stadium.15. A. The equal job, the equal pay.B. The best way of cooking and cleaning.C. The women’s liberation movement.D. Women’s ability to be good leaders.16. A. Women’s responsibility of child raising.B. Women’s abili ty to do anything important.C. Not only concrete issues but also attitude and beliefs.D. How to take jobs and help others.Section CDirections:In Section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet.Blanks 17 through 20 are based on the following conversation.Complete the form. Write ONE WORD for each answer.SHOWPlace for next show: In New York.Time for next show: This __17__.Place for the woman’s work:At the __18__.The woman’s purpose to San Francisco:On __19__.Transportation: Driving in a big __20__.Blanks 21 through 24 are based on the following conversation.Complete the form. Write NO MORE THAN THREE WORDS for each answer.Coastal Redwood TreesCharacteristics: a.__21__bark against fires.b. a liking for a __22__ climate.Location of the tallest trees: Along the __23__California coast.Height of the tallest trees: More than 350 feet.Age of the oldest recorded tree: __24__.II. Grammar and Vocabulary (26分)Section ADirections:After reading the passages 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.(A)Different forms of hospitality (好客)I am a British woman social anthropologist (人类学家). I once spent a year in Moldova, in Eastern Europe,(25) ______ (study) everyday life in the country. I stayed with a Moldovan family to see from the inside how people managed their lives. I had a wonderful time and made many new friends. What I observed is of course based on my own experience at a particular place and time.I often found (26) _______ surprisingly difficult to see life there through the eyes of a Moldovan. This was(27) ______ the people I met were extremely hospitable and I was treated as an honoured guest at all times. As my hosts, they wanted me to enjoy myself, and not to get (28) ______ (involve) in shopping, cooking, or other domestic jobs. Most mornings I was encouraged to go out to explore the city, or carry out my research, and I returned later to find that my elderly landlady and her sister had travelled across the city on buses to the central market (29) ________ (bring) back heavy loads of potatoes, a whole lamb, or other large quantities of products.I was often invited to people‟s ho mes, and was always offered food on entering. Most of the adults I met enjoyed inviting friends, family, neighbours, colleagues and even strangers into their homes, (30) ______ they treated them to food, drink, and a lively hospitable atmosphere. Hosts hurried to serve guests as well and as quickly as possible. (31) ______ a household was expecting guest, large amounts of food were prepared in advance, usually by the women. Wine had already been made, generally by the men, (32) ______ were also responsible for pouring it. Unexpected visitors were still offered as much food and drink as the household (33) ______ provide in the circumstances.(B)How English family life has evolved since the eighteenth century The majority of English families of the pre-industrial age, roughly until the mid-eighteen century, lived in a rural location. Many of them owned or had the use of a small piece of land, and actually all family members were busy with agricultural work in one form or another, usually (34) _______ (grow) food for their own consumption and sometimes also producing food or other goods for sale.The labour was controlled by the husband, (35) ______ _____ his wife and children, too, had an economi c value as their contributions to the family income were likely to make the difference between starvation andsurvival.Children worked from an early age, girls helping their mothers, and boys their fathers. School was an occasional factor in their lives. Instead, children learned by doing (36) _______ their parents show ed them. Knowledge of caring (37) ______ animals, sewing was handed down from parent to child.Also, most people engaged in handicraft production in the home, and the family (38) ______ (pay) to work with cloth, wood or leather. In general, this work could be put aside and taken up again when there was a break such as agricultural work.The process of industrialization in the second half of the eighteenth century and during the nineteenth transformed life for the majority of the population. It was the use of steam to power machinery (39) ______ required large buildings, and it resulted in the construction of numerous factories in many towns and cities. These in turn (40) _______(encourage)migration from the countryside in search of work. If electricity had preceded steam, domestic industry might have survived more fully.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.mentplexC.depressionD.expectE. equivalentF. holdG.mappedH.recommendedI.handleJ.notedK.severeBeing sociable looks like a good way to add years to your life. Relationships with family, friends, neighbours, even pets, will all do the trick, but the biggest longevity (长寿) boost seems to come from marriage or a(n) __41__ relationship. The effect was first __42__ in 1858by William Farr, who wrote that widows (寡妇)and widowers were at a much higher risk of dying than their married peers. Studies since then suggest that marriage could add as much as seven years to a man‟s life and two to a woman‟s. The effect s __43__ for all causes of death, whether illness, accident or self-harm.Even if the odds are stacked against you, marriage can more than compensate. Linda Waite of the University of Chicago has found that a married older man with heart disease can __44__ to live nearly four years longer than an unmarried man with a healthy heart. Likewise, a married man who smokes more than a pack a day is likely to live as long as a divorced man who doesn‟t smoke. There‟s a flip side, however, as partners are more likely to become ill or die in the couple of years following their spouse‟s death, and caring for a spouse with mental disorder can leave you with some of the same __45__ problems. Even so, the odds favour marriage. In a 30-year study of more than 10,000 people, Nicholas Christakis of Harvard Medical School describes how all kinds of social networks have similar effects.So how does it work? The effects are, __46__ affected by socio-economic factors, health-service provision, emotional support and other more physiological (生理的) mechanisms. For example, social contact can boost development of the brain and immune system, leading to better health and less chance of __47__ later in life. People in supportive relationships may __48__ stress better. Then there are the psychological benefits of a supportive partner.A life partner, children and good friends are all __49__ if you aim to live to 100. The ultimate social network is still being __50__ out, but Christakis says: “People are interconnected, so their health is interconnected.”III. Reading Comprehension(47分)Section ADirections: For each blank in the following passage 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.Who needs sleep?It‟s 2 a.m. The time when you should be in beds, sound asleep. But pull back the curtains and you might be surprised by the number of lights on in your street Night-time is __51__ just for sleeping. It has become the new daytime, offering us the chance to catch up on everything we didn‟t manage to finish during what used to be our __52__ hours. Now, __53__ sleeping, we can check our bank balances by phone, buy groceries, surf the net for cheap flights or go to the gym.Such flexibility, __54__, has a price. Our bodies are run by circadian rhythms (昼夜节律), a prehistori c internal clock that regulates when we feel sleepy or awake and affects our body temperature and level of alertness. It makes our brains and bodies __55__ during the day and allows them to recover through the night. So powerful is this clock that even two weeks on a nightshift without break will not __56__ its rhythm, and when scientists keep human volunteers in isolation, without any indication of what time it is in the day, they still show daily cycles of temperature changes, sleep and wakefulness, and hormone release. But, __57__ working against our body‟s natural rhythm is likely to cause ourselves both physical and psychological damage. Research also shows it may actually __58__ our risk of health problems such as stomach diseases.Consultant Tom Mackey believes that our normal circadian rhythms are increasingly being completely __59__. “More and more of us are being pressured into doing things at odd hours. This is going to have a(n) __60__ impact on quality and length of sleep. If people don‟t go to bed at a reasonable time, say around 11 p.m., and have between six and eight hours of sleep, they will be unable to concentrate. Y ou need sleep for rest and __61__. If you stuff your mind with information for too long, then everything gets disorganized ---you become __62__ to manage daytime activities.”The circadian rhythms that run the sleep/wake cycle are as old as __63__ itself. Our prehistoric ancestors would have needed their biological clock to get them out hunting during the day and probably in bed around nightfall to avoid intruders. Our night vision is not as fast as that of nocturnal (夜间活动的) animals ---our natural rhythm was to sleep as the sun went down. The invention of the electric light obviously __64__ that. Like most biological systems, circadian rhythms are not made to __65__. Our internal clock runs a bit longer than 24 hours, hence its Latin name, circadian, which means “about a day.”51. A. by all means B. on earth C. in no time D. to this day52. A. sleeping B. waking C. business D. rush53. A. in terms of B. regardless of C. as a result of D. instead of54. A. furthermore B. otherwise C. however D. somewhat55. A. active B. relaxing C. tiring D. conscious56. A. form B. destroy C. improve D. recover57. A. Efficiently B. Proudly C. Continually D. Independently58. A. minimize B. assess C. avoid D. increase59. A. broken B. enhanced C. emphasized D. misunderstood60. A. effective B. negative C. direct D. reliable61. A. reservation B. resetting C. repair D. replacement62. A. bored B. willing C. likely D. unable63. A. evolution B. clock C. mystery D. hunting64. A. improved B. changed C. speeded D. followed65. A. measure B. reverse C. regulate D. discoverSection 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 just read.AWhen milk arrived on the doorstepWhen I was a boy growing up in New Jersey in the 1960s, we had a milkman delivering milk to our doorstep. His name was Mr. Basille. He wore a white cap and drove a white truck. As a 5-year-old boy, i couldn‟t take my eyes off the coin changer fixed to his belt. He noticed this one day during a delivery and gave me a quarter out of his coin changer.Of course, he delivered more than milk. There was cheese, eggs and so on. If we needed to change our order, my mother would pen a note - “Please add a bottle of buttermilk next delivery” - and place it in the box along with the empty bottles. And then, the buttermilk would magically appear.All of this was about more than convenience. There existed a close relationship between families and their milkmen. Mr. Basille even had a key to our house, for those times when it was so cold outside that we put the box indoors, so that the milk wouldn‟t freeze. And I remember Mr. Basille from time to time taking a break at our kitchen table, having a cup of tea and telling stories about his delivery.There is sadly no home milk delivery today. Big companies allowed the production of cheaper milk, thus making it difficult for milkmen to compete. Besides, milk is for sale everywhere, and it may just not have been practical to have a delivery service.Recently, an old milk box in the countryside I saw brought back my childhood memories. I took it home and planted it on the back porch (门廊). Every so often my son‟s friends will ask what it is. So I start telling stories of my boyhood, and of the milkman who brought us friendship along with his milk.66.Mr Basille gave the boy a quarter out of his coin changer _____.A. to satisfy his curiosity.B. to please his mother.C. to show his magical power.D. to pay for the delivery.67. What can be inferred from Para. 3?A. He preferred tea to coffee.B. He had a large sum of money.C. He was treated as a family member.D. He was a famous and popular person.68. Why does home milk delivery no longer exist?A. It is forbidden by law.B. Its service is getting poor.C. It has been driven out of the market.D. Nobody wants to be a milkman now.69.Why did the author bring back home an old milk box?A. He planted flowers in it.B. He missed the good old days.C. He needed it for his milk bottles.D. He was fond of telling interesting stories.BCWUThe communication unionHead of ResearchSalary: £55.271We are looking for a Head of Research to manage the CWU Research Department and Information Centre. Y ou would be required to exercise control of all research work of the department and manage a team of three researchers and four support staff.The person appointed would be expected to carry out research work of a strategic nature across the range of businesses in which the CWU has or seeks membership and to contribute to the strategic thinking and direction of the union as a whole.Y ou will need: proven line management skills, especially in managing and motivating a team; good research skills, holding a good degree in a related subject or other similar experience; a high level ofmathematical and calculating skills; the ability to produce high quality work under pressure; a commitment to and knowledge of the trade union movement and social democratic politics; and knowledge and/ or experience of the postal and/ or telecommunications industry.To apply, please request an application pack by ********************by telephoning HR (Human Resources ) on 020 8971 7482. When applying please state your source.Closing Date for Applications: 4th December 2015Anticipated interview date: 17th December 2015No agencies please70. In which column of a newspaper could we find this advertisement?A. Arts.B. Sales.C. Jobs.D. News.71. One of the duties of the person to be appointed is _____.A. taking charge of research work.B. seeking membership for the trade union.C. running a telecommunications company.D. managing a team of three or four members.72. If you want to apply for this position, you can do all EXCEPT _____.A. ask an agency for an application formB. dial 020 8971 7482 for more information*************************************D. send in your application before 4th December 201573. Which of the following applicants is most likely to be employed?A. A chemistry teacher with a master's degree.B. A clerk from a telecommunications company.C. A university graduate majoring in computer science.D. A director from a research centre with a master's degree.CA child's map often provides a much-needed rest for parents too. Time for an uninterrupted phone call, or a rest on the sofa. And naps have to be a good thing for preschools, surely, since they need to take a rest and get enough sleep for their brains to develop. Short naps have also been shown to be good for adults---improving alertness and reaction times.So it feels counterintuitive for a review of 26 studies to conclude that napping in children over two years of age may not be a good idea at all. The review says that after two years of age, napping is associated with going to sleep late at night, poorer quality sleep and waking earlier. So should we discourage naps in preschool children -even if they really seem to need one?Although the review talks about the effects of napping on two-year-olds, most of the evidence in review actually comes from studies on three-year-olds. Also, the authors of the view article are clear that the research on children‟s naps is of poor quality; some studies rely on parents‟ remembering how much their children slept, or are for very short periods.A study published in the journal of attention disorders in February suffers from some methodological shortcomings (for instance small numbers --only 28 children between the ages of three and four, and only for five days), but is at least a randomized (任意的) controlled trial, using actigraphs (活动记录仪) worn on the wrist to objectively measure sleep. It found that children who missed their naps slept better at night and scored higher in studies of memory and attention.Dr Mark Mahone, one of the authors, says that sleep at night may be of a better quality than duringdaytime napping. Having a sound sleep at night, he says, provides a greater proportion of the slow-wave, restorative sleep that promotes brain development and reinforces what has been learned the day before. The study also found that the children who went without naps did not sleep more at weekends.I could never get my children to nap, but for parents who can, there is no reason to stop. Mahone says that more research is needed before anyone starts making recommendation, and children‟s sleep requirements are known to be variable.74.The word ‘’counterintuitive‟‟ in para. 2 most probably means _____.A. difficultB. unreasonableC. revolutionaryD. meaningless75.What does the author think of the study published in Attention Disoders?A. Its objects are too young to take the testB. Its findings are reliable due to the actigraphsC. It has enough objects to produce the resultsD. It provides various ways to help adults to take naps76.According to Dr Mark Mahone, a better-quality night sleep will result in ______.A. frequent rests on the sofaB. higher efficiency of learningC. more serious attention disorderD. fewer daytime naps at weekends77.The author talks mainly about his or her_______.A. opinions on whether children needs napsB. various systemic studies on children‟s napsC. comments on some studies on children‟s napsD. understanding of the functions of children‟s napsSection CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.Are bees happier in cities?We often think of them as living happily in wildflower meadows (草场) and rolling fields. But new research suggests Britain‟s bees are happier near towns and cities.A new study of wildlife sites across four English counties has found that most are home to fewer species of bee today than they were in the past. It found that the expansion of farmland has actually been more damaging to Britain‟s bee population than the concreting over (铺设混凝土) of the countryside for housing. For instance, meadows near Milton Keynes now boast more species of bee than sites in more rural areas.Reading University researcher Dr Deepa Senapathi believes intensive agriculture is to blame. Climate change could be destroying the relationship between bees and plants. That's according to a study that said warmer springs can change the life cycles of bees, which can throw them out of the plants they rely on. The research is the first clear example of the potential for climate change to destroy such critical relationships between species.While the gardens, parks and churchyards of towns and cities provide bees with a variety of plants to forage on and an extended flowering season, popular crops such as oilseed rape only bloom for a few weeks.Dr senapathi said: …While concreting over the countryside may appear to be bad news for nature, we‟ve found that progressive urbanization may be much less damaging than intensive agriculture.Urban areas may benefit bees more than farmland by providing a wide variety of flowering plants and anextended flowering season, according to the researcher.“Over the past century rural landscapes in Britain have become increasingly dominated by large expanses of monoculture - the growing of a single type of plant, which has helped boost crop production. But without a mixture of habitat and food sources, rural areas can sometimes be little better than green deserts for biodiversity (生物多样性).”Scientists around the country are trying to work out why populations of bees and other insects are falling. Pesticides, climate change and disease may, like intensive farming, be playing a role.(Note: Answer the questions or complete the statements in NO MORE THAN TEN WORDS)78. According to the new study of wildlife sites across four English countries, the number of bee species is_____79. According to Dr Senapathi, the development of cities causes __________________to bees than farmlands in suburbs.80. What are the two advantages of urban areas over farmland when it comes to housing for bees?81. Green deserts in the passage refers to the place where there is ______________________第II卷(共47分)I.TranslationDirections: Translate the following sentences into English, using the word given in the brackets.1.学生应该在课堂上举手回答问题。
真题还原:写作70分26.根据以下材料,自选角度,自拟题目,写一篇不少于800字的文章(不要写成诗歌)。
人的心中总有一些坚硬的东西,也有一些柔软的东西。
如何对待它们,将关系到能否造就和谐的自我。
命题分析趋势及点评:从2015年上海高考的命题来分析来看,今年的命题依旧延续了之前五年来的命题思路和风格。
从材料特点来看,将比喻型材料和关系型材料合二为一,并且是在设置好的范围中谈关系。
例如2014年上海高考作文是在“沙漠”这个范围中谈“自由”和“不自由”,而今年的作文是在“造就和谐的自我”的范围中谈“坚硬”和“柔软”。
从形式上来看,仍旧是“一行半”的形式。
整个材料阅读起来难度不大,但是观点倾向十分清晰明了,让同学们在落笔之前无限拓展思考的角度。
材料变短了,审题的难度降低了,但是题目内容越来越科学,依然留给考生较大的发挥空间。
从命题思维上来看,上海高考作文仍旧体现了对人性和个体生命的关注,更强调个体的主观能动性对自身的塑造,同时体现了哲学的思辨性。
那是不是上海今年的高考并没有体现出变化呢?周老师认为并非如此,综观近几年的上海高考作文题,可以看出命题者的别具匠心。
首先是将常规的各种类型作文题考查点综合起来,这一综合,就拓展了考生的思路和写作的范围。
比如今年的材料中,对于“坚硬”、“柔软”的分析可以从不同层面引发出去,也可以和社会现象、当今人们的心理结合起来,这样就可以把这个材料写“活”,体现作文的思想性。
其次是观点倾向更清晰,本则材料明确提出了我们要“对待”自己心中的“坚硬”和“柔软”,以达到“造就和谐的自我”的目的,所以过程和方法是多样的,而最后的结果导向是明确的。
这样看来,今年的作文题目相比于前几年就更加科学了。
材料详细解读:1.材料本身的重点词句及关键信息:考生都比较容易找到材料中的关键词——“坚硬”、“柔软”、“和谐的自我”。
这里,周老师要提醒同学们的是:在材料中还有一句不可忽视,造就和谐的自我并不是把心中原有的东西随意组合就可以了,关键是“如何对待它们”。
答案:25. from 26. They 27.who 28. cleaner 29. that/which 30. had removed 31. Seeing 32. to clean分析:25. 本题考查的搭配问题remove.... from …, 做对本题的关键是要了解这个词语的搭配问题。
26. 本题考的是代词,要注意前后的语境及指代。
27. 本题考查的是非限制性定语从句的关系代词问题,比较容易。
28. 本题是属于给出限定词的题目,要看到前面的a lot, 这样就很容易填出clean的比较级。
29. 本题考查的是定语从句,也需要填关系代词,指代前面的先行词garbage, 用that和which都可以。
30.本题考查的时态问题,通过前面的by the end of…可知本题应填过去完成时,本题较容易。
31. enough 后往往要接to do 形式,所以本题答案为to clean ,难度不算太大。
答案:33. driving 34. an 35. when 36. were touched 37. what 38. Because of/ Owing to/ Dueto 39.checking 40. Could分析:33. 本题考查的是非谓语作定语,job跟drive是主动关系,所以填driving,本题很多学生会填to drive ,我们可以认为to do这种形式往往表示动作尚未发生,而这里表示的是从事的工作丢掉了,这不可能未发生,故填to drive是不合适的。
34. 本题填的是an, 要提醒学生看到单数可数名词时要考虑填冠词。
35. 本题考查的是状语从句的用法,根据句意可以看出应该是表示时间的,所以填when合适。
36. 本题填的是谓语动词,主语后面要填谓语,根据分析,本空要填被动语态。
37.38. 结合本句的句意,表达的应该是原因,又因为后面加的是名词,所以填以上几个表示原因的词组都可以。
2016年高考上海卷真题呈现随着现代社会的发展,人们的生活更容易进入大众视野,评价他人的生活变得越来越常见,这些评价对个人和社会的影响越来越大。
人们对“评价他人的生活”这种现象的看法不尽相同,请写一篇文章,谈谈你对这种现象的思考。
要求:(1)自拟题目;(2)不少于800字。
命题分析这是一则材料作文,材料的关键词是“评价”。
评价,有面对面的或公开的评价,如对面交谈,工作互评,网络评论等;也有私下里的不公开评价,“谁人背后无人说,哪个背后不说人”,自古皆然。
评价有好有坏,有中肯的有违心的,评价内容涉及广泛,可以说应有尽有,有的有用,有的没用,有的让人如沐春风,有的让人痛不欲生。
有时候说者无意听者有心,有时候说过即烟消云散。
参考立意:①评价无处不在,评价优劣都有,要正确看待,说的对的我们照着做,说的不对的我们引以为戒,这样才能不断进步。
②他人的中肯评价能够帮助我们更好地适应环境,掌握社会规则,赢得社会的认可。
我们虽然很累,但依然会把公交车上的位置让给老人,因为我们期望在别人眼里是有礼貌的,有教养的;到一个新班级上学,我们会注意身边同学的一言一行,模仿他们待人接物的态度和方式,以免懵然无知中触犯众怒。
③过度在意他人评价总是会让人患得患失,是不自信的表现。
特别在意别人的负面评价,会让一个人无所适从,情绪低落。
人,不管何时都要做到自信而不自负。
④可以几种观点兼而有之,稍有侧重即可。
也可以从如何评价他人等方面来考虑。
写作指导材料作文审题立意的几种方法:1.从抓关键句入手。
有的材料为突出中心,有时会在材料中设置关键句(开头句、结尾句、反复出现的句子),抓住这些关键句,就能把握材料主旨,准确理解材料,正确立意。
比如本则材料就可以抓住“评价”来做文章。
2.从分析原因入手。
任何事物的产生、变化和发展,都有其内在或外在的原因。
因此,阅读分析材料的因果联系,从原因切入立意,是行之有效的方法。
3.从材料情感倾向入手。
有的材料在叙述、说明或评论某个事物时,明显地流露出作者的情感倾向,这样我们可从材料的情感倾向入手来审题立意。
2016高三语文一模作文(共10篇)篇一:2016上海各区高三语文一模作文立意解析2016各区一模作文立意解析长宁区根据以下材料,自选角度,自拟题目,写一篇不少于800字的文章(不要写成诗歌)。
有人说:“痛苦是财富,这话是扯淡;痛苦就是痛苦,对痛苦的思考才是财富。
”①正视痛苦,思考出路如材料所述,痛苦就是痛苦,当我们经历痛苦时,最明智的做法是思考痛苦产生的原因,思考降低痛苦的方法,这样便能从痛苦中收获一笔宝贵的财富。
②体验痛苦,收获财富事实也不尽如材料所言,痛苦本身也可以是财富,它是一种真真切切的体验,代表着仍然鲜活的、拒绝麻木的心灵。
文明的历史就是一部痛苦的历史。
③规避痛苦,乐享人生乐而不淫,哀而不伤,学习规避痛苦,或者不沉湎在痛苦之中,是一种值得提倡的健康积极的生活态度。
④直面痛苦,承担人生生活中的痛苦往往是不可规避的,因为它们是伴随责任而来的,这就要求我们在最开始就能预见痛苦,并且敢于直面它,承担起生活的分量。
普陀区根据以下材料,自选角度,自拟题目,写一篇不少于800字的文章(不要写成诗歌)。
有人说坚持做同一件事,坚持10年就能够成功,坚持20年就能够创造出奇迹。
而事实上,绝大部分人同一件事情做了一辈子,依旧平凡无奇。
①做好规划,审慎坚持坚持了多年却没有成功,可能是一开始坚持的方向就有了问题,或者是没有一定的物质基础,所谓“无物以相之,亦不能至也”,也有可能是坚持的方式有问题,因为坚持并非是简单的重复,而是要有变化、有创新。
所以在坚持之前应该做好规划,解决掉一些问题:坚持什么,凭借什么坚持,怎么坚持。
②谋事在人,成事在天有了正确的坚持不代表理所应当就会成功,成功都是有一定的偶然性因素的,需要把握时机。
③勇于坚持,拒绝盲从坚持本身就是一种成功,成功不一定是名利上的,也可以是精神修养上的,这种情况下,坚持住就是一切。
崇明区古希腊哲学家伊壁鸠鲁说:“认识错误就是拯救自己的第一步。
”然而当代哲学家卡尔波普尔又说:“如果我们过于爽快的承认错误,就可能使自己发觉不了我们非常接近于正确。
2016年上海市普陀区高考数学一模试卷一、填空题(本大题56分)本大题共有14小题,要求直接将结果填写在答题纸对应的空格中,每小空格填对得4分,填错或不正确的位置一律得零分.1.若全集U=R,集合M={x|x(x﹣2)≤0},N={1,2,3,4},则N∩∁U M= .2.若函数,,则f(x)+g(x)= .3.在(2x﹣1)7的二项展开式中,第四项的系数为.4.在,则函数y=tanx的值域为.5.若数列{a n}中,a1=1,a n+1=2a n+1(n∈N*),则数列的各项和为.6.若函数f(x)=(x≥0)的反函数是f﹣1(x),则不等式f﹣1(x)>f(x)的解集为.7.设O为坐标原点,若直线与曲线相交于A、B点,则扇形AOB的面积为.8.若正六棱柱的底面边长为10,侧面积为180,则这个棱柱的体积为.9.若在北纬45°的纬度圈上有A、B两地,经度差为90°,则A、B两地的球面距离与地球半径的比值为.10.方程的解x= .11.设P是双曲线上的动点,若P到两条渐近线的距离分别为d1,d2,则d1•d2= .12.如图,已知正方体ABCD﹣A1B1C1D,若在其12条棱中随机地取3条,则这三条棱两两是异面直线的概率是(结果用最简分数表示)13.若F是抛物线y2=4x的焦点,点P i(i=1,2,3,…,10)在抛物线上,且,则= .14.若函数最大值记为g(t),则函数g(t)的最小值为.二、选择题(本大题20分)本大题共有4小题,每小题有且只有一个结论是正确的,必须把正确结论的代号写在答题纸相应的空格中,每题选对得5分,不选、选错或选出的代号超过一个(不论是否都写在空格内),或者没有填写在题号对应的空格内,一律得零分. 15.下列命题中的假命题是()A.若a<b<0,则B.若,则0<a<1C.若a>b>0,则a4>b4D.若a<1,则16.若集合,则“x∈A”是“x∈B”成立的()A.充分非必要条件B.必要非充分条件C.充要条件 D.既不充分也不必要条件17.如图,在四面体ABCD中,AB⊥BD,CD⊥DB,若AB与CD所成的角的大小为60°,则二面角C﹣BD﹣A的大小为()A.60°或90°B.60° C.60°或120°D.30°或150°18.若函数,关于x的方程f2(x)﹣(a+1)f(x)+a=0,给出下列结论:①存在这样的实数a,使得方程由3个不同的实根;②不存在这样的实数a,使得方程由4个不同的实根;③存在这样的实数a,使得方程由5个不同的实数根;④不存在这样的实数a,使得方程由6个不同的实数根.其中正确的个数是()A.1个B.2个C.3个D.4个三、解答题(本大题74分)本大题共有5小题,解答下列各题必须在答题纸规定的方框内写出必要的步骤.19.如图,椭圆+=1的左、右两个焦点分别为F1、F2,A为椭圆的右顶点,点P在椭圆上且∠PF1F2=arccos(1)计算|PF1|的值x(2)求△PF1A的面积.20.某种“笼具”由内,外两层组成,无下底面,内层和外层分别是一个圆锥和圆柱,其中圆柱与圆锥的底面周长相等,圆柱有上底面,制作时需要将圆锥的顶端剪去,剪去部分和接头忽略不计,已知圆柱的底面周长为24πcm,高为30cm,圆锥的母线长为20cm.(1)求这种“笼具”的体积(结果精确到0.1cm3);(2)现要使用一种纱网材料制作50个“笼具”,该材料的造价为每平方米8元,共需多少元?21.已知函数f(x)=2sin2x+sin2x﹣1.(1)求函数f(x)的单调递增区间;(2)设,其中0<x0<π,求tanx0的值.22.已知n∈N*,数列{a n}的前n项和为S n,且2a n﹣S n=1.(1)求证:数列{a n}是等比数列,并求出通项公式;(2)对于任意a i、a j∈{a1,a2,…,a n}(其中1≤i≤n,1≤j≤n,i、j均为正整数),若a i和a j的所有乘积a i•a j的和记为T n,试求的值;(3)设,若数列{c n}的前n项和为C n,是否存在这样的实数t,使得对于所有的n都有成立,若存在,求出t的取值范围;若不存在,请说明理由.23.已知集合M是满足下列性制的函数f(x)的全体,存在实数a、k(k≠0),对于定义域内的任意x均有f(a+x)=kf(a﹣x)成立,称数对(a,k)为函数f(x)的“伴随数对”.(1)判断f(x)=x2是否属于集合M,并说明理由;(2)若函数f(x)=sinx∈M,求满足条件的函数f(x)的所有“伴随数对”;(3)若(1,1),(2,﹣1)都是函数f(x)的“伴随数对”,当1≤x<2时,f(x)=cos(x);当x=2时,f(x)=0,求当2014≤x≤2016时,函数y=f(x)的解析式和零点.2016年上海市普陀区高考数学一模试卷参考答案与试题解析一、填空题(本大题56分)本大题共有14小题,要求直接将结果填写在答题纸对应的空格中,每小空格填对得4分,填错或不正确的位置一律得零分.1.若全集U=R,集合M={x|x(x﹣2)≤0},N={1,2,3,4},则N∩∁U M= {3,4} .【考点】交、并、补集的混合运算.【分析】求解一元二次不等式化简M,求出其补集,再由交集运算得答案.【解答】解:∵M={x|x(x﹣2)≤0}={x|0≤x≤2},∴∁U M={x|x<0或x>2},又N={1,2,3,4},∴N∩∁U M={3,4}.故答案为:{3,4}.2.若函数,,则f(x)+g(x)= 1(0≤x≤1).【考点】函数解析式的求解及常用方法.【分析】容易求出f(x),g(x)的定义域,求交集便可得出f(x)+g(x)的定义域,并可求得f(x)+g(x)=.【解答】解:;解得,0≤x≤1;∴(0≤x≤1).故答案为:.3.在(2x﹣1)7的二项展开式中,第四项的系数为﹣560 .【考点】二项式系数的性质.【分析】直接利用二项式定理写出结果即可即可.【解答】解:在(2x﹣1)7的二项展开式中,第四项的系数为: =﹣560.故答案为:﹣560.4.在,则函数y=tanx的值域为[﹣1,1] .【考点】正切函数的图象.【分析】根据正切函数的图象与性质,求出x∈[﹣,]时函数y=tanx的值域即可.【解答】解:∵,∴﹣1≤tanx≤1,∴函数y=tanx的值域为[﹣1,1].故答案为:[﹣1,1].5.若数列{a n}中,a1=1,a n+1=2a n+1(n∈N*),则数列的各项和为1﹣.【考点】数列的求和.【分析】a1=1,a n+1=2a n+1(n∈N*),变形为a n+1+1=2(a n+1),利用等比数列的通项公式可得:1+a n,再利用等比数列的前n项和公式可得的前n项和.【解答】解:a1=1,a n+1=2a n+1(n∈N*),∴a n+1+1=2(a n+1),∴数列{a n+1}是等比数列,首项为2,公比为2.∴1+a n=2n,∴=,∴数列的首项为,公比为.∴数列的各项和为: =1﹣.故答案为:1﹣.6.若函数f(x)=(x≥0)的反函数是f﹣1(x),则不等式f﹣1(x)>f(x)的解集为{x|x>1} .【考点】反函数.【分析】由y=f(x)=(x≥0),求出f﹣1(x)=x3,x≥0,由此能求出不等式f﹣1(x)>f(x)的解集.【解答】解:设y=f(x)=(x≥0),则x=y3,x,y互换,得f﹣1(x)=x3,x≥0,∵f﹣1(x)>f(x),∴,∴x9>x,∴x8>1,解得x>1.∴不等式f﹣1(x)>f(x)的解集为{x|x>1}.故答案为:{x|x>1}.7.设O为坐标原点,若直线与曲线相交于A、B点,则扇形AOB的面积为.【考点】直线与圆的位置关系;扇形面积公式.【分析】通过曲线方程确定曲线表示单位圆在x轴上方的部分(含于x轴的交点),y=时,∠AOB=π,即可求出扇形AOB的面积.【解答】解:由曲线,得x2+y2=1(y≥0)∴曲线表示単位圆在x轴上方的部分(含于x轴的交点)y=时,∠AOB=π,扇形AOB的面积为=.故答案为:.8.若正六棱柱的底面边长为10,侧面积为180,则这个棱柱的体积为450.【考点】棱柱、棱锥、棱台的体积.【分析】根据侧面积公式求出棱柱的高,根据底面边长求出底面积,代入体积公式得出体积.【解答】解:设棱柱的底面边长为a,高为h,则S侧=6ah=60h=180,解得h=3.S底==150.∴正六棱柱的体积V=S底h=450.故答案为:450.9.若在北纬45°的纬度圈上有A、B两地,经度差为90°,则A、B两地的球面距离与地球半径的比值为.【考点】球面距离及相关计算.【分析】求出球心角,然后A、B两点的距离,求出两点间的球面距离,即可求出A、B两地的球面距离与地球半径的比值.【解答】解:地球的半径为R,在北纬45°,而AB=R,所以A、B的球心角为:,所以两点间的球面距离是:,所以A、B两地的球面距离与地球半径的比值为;故答案为:.10.方程的解x= log23 .【考点】对数的运算性质.【分析】化简可得4x﹣5=4(2x﹣2),从而可得(2x)2﹣4•2x+3=0,从而解得.【解答】解:∵,∴4x﹣5=4(2x﹣2),即(2x)2﹣4•2x+3=0,∴2x=1(舍去)或2x=3;∴x=log23,故答案为:log23.11.设P是双曲线上的动点,若P到两条渐近线的距离分别为d1,d2,则d1•d2=.【考点】双曲线的简单性质.【分析】先确定两条渐近线方程,设双曲线C上的点P(x,y),求出点P到两条渐近线的距离,结合P在双曲线C上,即可求d1•d2的值.【解答】解:由条件可知:两条渐近线分别为x±y=0设双曲线C上的点P(x,y),则点P到两条渐近线的距离分别为d1=,d2=所以d1•d2=•==.故答案为:.12.如图,已知正方体ABCD﹣A1B1C1D,若在其12条棱中随机地取3条,则这三条棱两两是异面直线的概率是(结果用最简分数表示)【考点】列举法计算基本事件数及事件发生的概率;空间中直线与直线之间的位置关系.【分析】正方体ABCD﹣A1B1C1D,在其12条棱中随机地取3条,先求出基本事件总数,再求出这三条棱两两是异面直线包含的基本事件个数,由此能求出这三条棱两两是异面直线的概率.【解答】解:正方体ABCD﹣A1B1C1D,在其12条棱中随机地取3条,基本事件总数n==220,这三条棱两两是异面直线包含的基本事件个数m=8,∴这三条棱两两是异面直线的概率是p===.故答案为:.13.若F是抛物线y2=4x的焦点,点P i(i=1,2,3,…,10)在抛物线上,且,则= 200 .【考点】抛物线的简单性质.【分析】根据抛物线的定义得抛物线上的点到焦点的距离等于该点到准线的距离,因此求出抛物线的准线方程,结合题中数据加以计算,即可得到本题答案.【解答】解:∵抛物线y2=4x的焦点为F(1,0),准线为x=﹣1,∴根据抛物线的定义,P i(i=1,2,3,…,2015)到焦点的距离等于P i到准线的距离,即|P i F|=x i+1,,可得1﹣x1+1﹣x2+…+1﹣x100=0,∴x1+x2+…+x100=100∴|P1F|+|P2F|+…|P100F|=(x1+1)+(x2+1)+…+(x100+1)=(x1+x2+…+x100)+100=100+100=200.故答案为:200.14.若函数最大值记为g(t),则函数g(t)的最小值为.【考点】函数的最值及其几何意义.【分析】化简sinx+=sinx+3+﹣3,从而可得0≤sinx+3+﹣3≤,从而求得g(t)=f max(x)=,从而求值.【解答】解:∵sinx+=sinx+3+﹣3,∵﹣1≤sinx≤1,∴2≤sinx+3≤4,∴3≤sinx+3+≤,∴0≤sinx+3+﹣3≤,∴g(t)=f max(x)=,∴当t=时,函数g (t )有最小值为;故答案为;.二、选择题(本大题20分)本大题共有4小题,每小题有且只有一个结论是正确的,必须把正确结论的代号写在答题纸相应的空格中,每题选对得5分,不选、选错或选出的代号超过一个(不论是否都写在空格内),或者没有填写在题号对应的空格内,一律得零分. 15.下列命题中的假命题是( )A .若a <b <0,则B .若,则0<a <1C .若a >b >0,则a 4>b 4D .若a <1,则【考点】命题的真假判断与应用.【分析】正确选项进行证明,不正确选项,举出反例即可.【解答】解:对于A ,a <b <0,则•a<•b,∴,正确对于B ,,则>0,∴0<a <1,正确对于C ,a >b >0,a 4>b 4,正确;对于D ,a=, =2>1,不正确, 故选:D .16.若集合,则“x∈A”是“x∈B”成立的( )A .充分非必要条件B .必要非充分条件C .充要条件D .既不充分也不必要条件【考点】必要条件、充分条件与充要条件的判断.【分析】先分别求出集合A ,B ,然后根据必要条件、充分条件和充要条件的定义进行判断.【解答】解:∵≥0,∴0≤x <3, ∴A=(0,3],∵lg|2x ﹣3|<0=lg1,∴|2x ﹣3|<1,且2x ﹣3≠0,∴1<x <2,且x ≠∴B=(1,)∪(,2),∴“x∈A”是“x∈B”成立的必要非充分条件, 故选:B .17.如图,在四面体ABCD中,AB⊥BD,CD⊥DB,若AB与CD所成的角的大小为60°,则二面角C﹣BD﹣A的大小为()A.60°或90°B.60° C.60°或120°D.30°或150°【考点】二面角的平面角及求法.【分析】过D在平面ABD内作AB的平行线DE,则∠CDE或∠CDE的补角为二面角C﹣BD﹣A 的平面角,由此能求出二面角C﹣BD﹣A的大小.【解答】解:过D在平面ABD内作AB的平行线DE,∵在四面体ABCD中,AB⊥BD,CD⊥DB,∴DE⊥BD,∴∠CDE或∠CDE的补角为二面角C﹣BD﹣A的平面角,∵AB与CD所成的角的大小为60°,∴∠CDE=60°或∠CDE=120°,∴二面角C﹣BD﹣A的大小为60°或120°.故选:C.18.若函数,关于x的方程f2(x)﹣(a+1)f(x)+a=0,给出下列结论:①存在这样的实数a,使得方程由3个不同的实根;②不存在这样的实数a,使得方程由4个不同的实根;③存在这样的实数a,使得方程由5个不同的实数根;④不存在这样的实数a,使得方程由6个不同的实数根.其中正确的个数是()A.1个B.2个C.3个D.4个【考点】根的存在性及根的个数判断.【分析】由f2(x)﹣(a+1)f(x)+a=0可解得f(x)=1或f(x)=a,作函数的图象,从而讨论求解.【解答】解:∵f2(x)﹣(a+1)f(x)+a=0,∴f(x)=1或f(x)=a,作函数的图象如下,,当a=1时,方程有3个不同的实根,故①正确;当a>1或a≤﹣1时,方程有6个不同的实根,故④不正确;当﹣1<a<1时,方程有5个不同的实根,故③正确;综上可知,不存在这样的实数a,使得方程由4个不同的实根;故②正确;故选:C.三、解答题(本大题74分)本大题共有5小题,解答下列各题必须在答题纸规定的方框内写出必要的步骤.19.如图,椭圆+=1的左、右两个焦点分别为F1、F2,A为椭圆的右顶点,点P在椭圆上且∠PF1F2=arccos(1)计算|PF1|的值x(2)求△PF1A的面积.【考点】椭圆的简单性质.【分析】(1)根据椭圆的性质,可得|PF1|=x,则|PF2|=10﹣x,|F1F2|=2=8,结合已知可余弦定理构造方程,解得x值;(2)由出sin∠PF1F2,进而计算△PF1F2的面积,可得P到x轴的距离d,结合△PF1A的底边|F1A|=a+c=9,可得三角形面积.【解答】解:(1)∵椭圆+=1的左、右两个焦点分别为F1、F2,P为椭圆上一点,|PF1|=x,则|PF2|=10﹣x,|F1F2|=2=8,∵∠PF1F2=arccos,故cos∠PF1F2==,解得:x=6,(2)由∠PF1F2=arccos,可得:sin∠PF1F2==,故△PF1F2的面积S=(5+)•(5﹣)•=,故P到x轴的距离d==,由|F1A|=a+c=9,可得△PF1A的面积为:×=20.某种“笼具”由内,外两层组成,无下底面,内层和外层分别是一个圆锥和圆柱,其中圆柱与圆锥的底面周长相等,圆柱有上底面,制作时需要将圆锥的顶端剪去,剪去部分和接头忽略不计,已知圆柱的底面周长为24πcm,高为30cm,圆锥的母线长为20cm.(1)求这种“笼具”的体积(结果精确到0.1cm3);(2)现要使用一种纱网材料制作50个“笼具”,该材料的造价为每平方米8元,共需多少元?【考点】棱柱、棱锥、棱台的体积.【分析】(1)笼具的体积等于圆柱的体积减去圆锥的体积;(2)求出笼具的表面积即可,笼具的表面积包括圆柱的侧面,上底面和圆锥的侧面.【解答】解:(1)设圆柱的底面半径为r,高为h,圆锥的母线长为l,高为h1,则2πr=24π,解得r=12cm.h1=cm.∴笼具的体积V=πr2h﹣=π×=3552π≈11158.9cm3.(2)圆柱的侧面积S1=2πrh=720cm2,圆柱的底面积S2=πr2=144πcm2,圆锥的侧面积为πrl=240πcm2.故笼具的表面积S=S1+S2+S3=1104πcm2.故制造50个这样的笼具总造价为:元.答:这种笼具的体积约为11158.9cm3,生产50个笼具需要元.21.已知函数f(x)=2sin2x+sin2x﹣1.(1)求函数f(x)的单调递增区间;(2)设,其中0<x0<π,求tanx0的值.【考点】三角函数中的恒等变换应用;三角函数的最值.【分析】(1)利用三角函数的关系结合辅助角公式进行化简,即可求函数f(x)的单调递增区间;(2)化简条件,利用同角的三角函数的关系式建立方程关系进行求解即可.【解答】解:(1)f(x)=2sin2x+sin2x﹣1=sin2x﹣cos2x=sin(2x﹣).由2kπ﹣≤2x﹣≤2kπ+,k∈Z,得2kπ≤x≤2kπ,k∈Z,得kπ﹣≤x≤kπ+,k∈Z,即函数f(x)的单调递增区间是[kπ﹣,kπ+],k∈Z;(2)cos(+α)cos(﹣α)+sin2α=(cos cosα)2﹣(sin sinα)2+sin2α=cos2α﹣sin2α+sin2α=,即f()=sin(2×﹣)=sin(x0﹣)=,即sinx0﹣cosx0=,①平方得2sinx0cosx0=,∵0<x0<π,∴cosx0>0,则sinx0+cosx0==②,由①②得sinx0=,cosx0=,则tanx0==.22.已知n∈N*,数列{a n}的前n项和为S n,且2a n﹣S n=1.(1)求证:数列{a n}是等比数列,并求出通项公式;(2)对于任意a i、a j∈{a1,a2,…,a n}(其中1≤i≤n,1≤j≤n,i、j均为正整数),若a i和a j的所有乘积a i•a j的和记为T n,试求的值;(3)设,若数列{c n}的前n项和为C n,是否存在这样的实数t,使得对于所有的n都有成立,若存在,求出t的取值范围;若不存在,请说明理由.【考点】数列与不等式的综合;等比数列的通项公式;数列的极限.【分析】(1)当n≥2时通过2a n﹣S n=1与2a n﹣1﹣S n﹣1=1作差,进而计算可得结论;(2)通过(1)可得T n的表达式,进而计算即得结论;(3)通过(1)可知数列{c n}的通项公式,利用并项相加、分n为奇数、偶数两种情况讨论即可.【解答】(1)证明:∵2a n﹣S n=1,∴当n≥2时,2a n﹣1﹣S n﹣1=1,两式相减,整理得:a n=2a n﹣1(n≥2),又∵2a1﹣S1=1,即a1=1,∴数列{a n}是首项为1、公比为2的等比数列,∴a n=2n﹣1;(2)解:∵T n=(1+2+22+…+2n﹣1)(1+2+22+…+2n﹣1)=•=4n﹣2•2n+1,∴==1;(3)结论:存在这样的实数t,使得对于所有的n都有成立.理由如下:由(1)可知,1+b n=3log2a n=3n﹣3,即b n=3n﹣4,b n+1=3n﹣1,故c n=(﹣1)n+1b n•b n+1=(﹣1)n+1(3n﹣4)(3n﹣1),c n+1=(﹣1)n+2(3n﹣1)(3n+2),特别地,当n为奇数时,有n+1为偶数,此时c n+c n+1=(3n﹣4)(3n﹣1)﹣(3n﹣1)(3n+2)=﹣6(3n﹣1),①若n为偶数,则C n=(c1+c2)+(c3+c4)+…+(c n﹣1+c n)=﹣6×[2+8+…+(3n﹣4)]=﹣n(3n﹣2),由可知t≤﹣(3﹣)对所有正偶数n都成立,故t≤﹣;②若n为奇数,则C n=C n﹣1+c n(n≥2),由①可知C n=﹣(n﹣1)(3n﹣5)+(3n﹣4)(3n﹣1)=n2﹣3n﹣,其中C1=﹣2满足上式;由①②可得实数t的取值范围是:t≤﹣,所以存在这样的实数t,使得对于所有的n都有成立.23.已知集合M是满足下列性制的函数f(x)的全体,存在实数a、k(k≠0),对于定义域内的任意x均有f(a+x)=kf(a﹣x)成立,称数对(a,k)为函数f(x)的“伴随数对”.(1)判断f(x)=x2是否属于集合M,并说明理由;(2)若函数f(x)=sinx∈M,求满足条件的函数f(x)的所有“伴随数对”;(3)若(1,1),(2,﹣1)都是函数f(x)的“伴随数对”,当1≤x<2时,f(x)=cos(x);当x=2时,f(x)=0,求当2014≤x≤2016时,函数y=f(x)的解析式和零点.【考点】函数的值.【分析】(1)f(x)=x2的定义域为R.假设存在实数a、k(k≠0),对于定义域内的任意x 均有f(a+x)=kf(a﹣x)成立,则(a+x)2=k(a﹣x)2,化为:(k﹣1)x2﹣2a(k+1)x+a2(k﹣1)=0,由于上式对于任意实数x都成立,可得,解得k,a.即可得出.(2)函数f(x)=sinx∈M,可得:sin(a+x)=ksin(a﹣x),展开化为:sin(x+φ)=0,由于∀x∈R都成立,可得k2+2kcos2a+1=0,变形cos2a=,利用基本不等式的性质与三角函数的单调性即可得出.(3)由于(1,1),(2,﹣1)都是函数f(x)的“伴随数对”,可得f(1+x)=f(1﹣x),f(2+x)=﹣f(2﹣x),因此f(x+4)=f(x),T=4.对x分类讨论可得:即可得出解析式,进而得出零点.【解答】解:(1)f(x)=x2的定义域为R.假设存在实数a、k(k≠0),对于定义域内的任意x均有f(a+x)=kf(a﹣x)成立,则(a+x)2=k(a﹣x)2,化为:(k﹣1)x2﹣2a(k+1)x+a2(k﹣1)=0,由于上式对于任意实数x都成立,∴,解得k=1,a=0.∴(0,1)是函数f(x)的“伴随数对”,f(x)∈M.(2)∵函数f(x)=sinx∈M,∴sin(a+x)=ksin(a﹣x),∴(1+k)cosasinx+(1﹣k)sinacosx=0,∴sin(x+φ)=0,∵∀x∈R都成立,∴k2+2kcos2a+1=0,∴cos2a=,≥2,∴|cos2a|≥1,又|cos2a|≤1,故|cos2a|=1.当k=1时,cos2a=﹣1,a=nπ+,n∈Z.当k=﹣1时,cos2a=1,a=nπ,n∈Z.∴f(x)的“伴随数对”为(nπ+,1),(nπ,﹣1),n∈Z.(3)∵(1,1),(2,﹣1)都是函数f(x)的“伴随数对”,∴f(1+x)=f(1﹣x),f(2+x)=﹣f(2﹣x),∴f(x+4)=f(x),T=4.当0<x<1时,则1<2﹣x<2,此时f(x)=f(2﹣x)=﹣cos;当2<x<3时,则1<4﹣x<2,此时f(x)=﹣f(4﹣x)=﹣cos;当3<x<4时,则0<4﹣x<1,此时f(x)=﹣f(4﹣x)=cos.∴f(x)=.∴f(x)=.∴当2014≤x≤2016时,函数y=f(x)的零点为2014,2015,2016.。
2016年高考上海卷真题呈现随着现代社会的发展,人们的生活更容易进入大众视野,评价他人的生活变得越来越常见,这些评价对个人和社会的影响越来越大。
人们对“评价他人的生活”这种现象的看法不尽相同,请写一篇文章,谈谈你对这种现象的思考。
要求:(1)自拟题目;(2)不少于800字。
命题分析这是一则材料作文,材料的关键词是“评价”。
评价,有面对面的或公开的评价,如对面交谈,工作互评,网络评论等;也有私下里的不公开评价,“谁人背后无人说,哪个背后不说人”,自古皆然。
评价有好有坏,有中肯的有违心的,评价内容涉及广泛,可以说应有尽有,有的有用,有的没用,有的让人如沐春风,有的让人痛不欲生。
有时候说者无意听者有心,有时候说过即烟消云散。
参考立意:①评价无处不在,评价优劣都有,要正确看待,说的对的我们照着做,说的不对的我们引以为戒,这样才能不断进步。
②他人的中肯评价能够帮助我们更好地适应环境,掌握社会规则,赢得社会的认可。
我们虽然很累,但依然会把公交车上的位置让给老人,因为我们期望在别人眼里是有礼貌的,有教养的;到一个新班级上学,我们会注意身边同学的一言一行,模仿他们待人接物的态度和方式,以免懵然无知中触犯众怒。
③过度在意他人评价总是会让人患得患失,是不自信的表现。
特别在意别人的负面评价,会让一个人无所适从,情绪低落。
人,不管何时都要做到自信而不自负。
④可以几种观点兼而有之,稍有侧重即可。
也可以从如何评价他人等方面来考虑。
写作指导材料作文审题立意的几种方法:1.从抓关键句入手。
有的材料为突出中心,有时会在材料中设置关键句(开头句、结尾句、反复出现的句子),抓住这些关键句,就能把握材料主旨,准确理解材料,正确立意。
比如本则材料就可以抓住“评价”来做文章。
2.从分析原因入手。
任何事物的产生、变化和发展,都有其内在或外在的原因。
因此,阅读分析材料的因果联系,从原因切入立意,是行之有效的方法。
3.从材料情感倾向入手。
有的材料在叙述、说明或评论某个事物时,明显地流露出作者的情感倾向,这样我们可从材料的情感倾向入手来审题立意。
2016年上海高考语文作文题:随着现代社会的发展,人们的生活更容易进入大众视野,评价他人生活变得越来越常见,这些评价对个人和社会的影响也越来越大。
人们对“评价他人的生活”这种现象的看法不尽相同,请写一篇文章,谈谈你对这种现象的思考。
嚎叫后的自省金斯伯格在《嚎叫》一书中曾写道:“……鼻子神圣,皮肤神圣,世界神圣,一切神圣……”而在他连用十几个“神圣”对当时社会制度与风尚进行疯狂批驳的嚎叫声后,却流露着他对自然平淡的坚守与对精神明净的追求。
随着现代社会的发展,人们的生活更容易进入大众视野,评价他人的生活变得越来越常见。
明星私生活在众人嘴里却变得污浊不堪;总统的公益讲演在旁人眼中却只是为挣取更多的收入;艺术人为创新灵感而奔放的举动却在外人评论中成为颓落腐败的标志……种种不良过分的评价不仅显露了你自己不纯之心,更伤害了当事人,影响了社会的稳定发展。
评价他人的生活,这是人之常情,评价是为了表达自己的看法,抒发心中的情感,使他人得以收到你的心意,或守之,或改之,以达到促进互利的作用。
评价,首先需要的是你的客观理性的思考,而不带有一丝物欲的追求。
我曾看过一篇对音乐人李健生活的评价:“李健,是一名有天赋有主见的音乐人,不演商业性的音乐会,毅然退出风流一世的水木组合,可见其对音乐纯净的坚守,但在现代社会,他还缺少了一份包容其它形式音乐作品的宽心。
”这样的评价我认为十分合理,不仅客观而且带有爱护之心,这才是合理的评价。
然而,评价并非只是给他人建议而已,更是一种自省!苏格拉底曾说过:“不经省察过的人生是不值得过的。
”这里的省察不仅是对自己的反思改进,也包含有对他人生活的评价与警示。
大文豪托尔斯泰曾有一群农奴,从前的他每日观察着这些农奴的生活,并作有笔录,“我看到了他们的辛劳而却看不到对自由的渴望,我希望他们与我一样去享受自由,靠自己的双手来养活自己。
”不断地观察、评价着底层人民的生活,托翁并非只是如常人般给予可怜的眼光,而是省察了自己的行为,于是他给了每个农奴一笔钱,释放他们回乡,并开始了对精神之道的追求,周游各方,评价他人,省察自己,得到精神的成长。
2016上海各区高三语文一模题型比较与分析D在几代考古工作者50多年的不懈努力下,二里头遗址的全貌愈发清晰:这里发现了最早的城市干道网络,发现了最早的宫城,发现了最早的中轴线布局的宫殿建筑群……作为中国最早的王朝都城,二里头遗址开城市布局和规划的先河,。
A.被中国城市建设史留下浓墨重彩的一笔B.被中国城市建设史留下淋漓尽致的一笔C.在中国城市建设史留下浓墨重彩的一笔D.在中国城市建设史留下淋漓尽致的一笔【2016学年徐汇区一模】积累应用(10分)1、填空题。
(5分)(1)宋词有借助意象写秋的,如柳永《雨霖铃》首句“”,李清照《声声慢》下阙首句“”;有直接写秋的,如辛弃疾《水龙吟·登建康赏心亭》开篇句“”。
(3分)(2)不少成语出自经典作品,如“豁然开朗”出自《》,“”源自《秋水》中河伯自省被有学问的人耻笑。
(2分)2、选择题。
(5分)(1)你乘游轮“夜游浦江”时想用诗句描述所见美景分享到微信“朋友圈”,下列最贴切的一项是()。
(1分)A.灯火万家城四畔,星河一道水中央。
B.旌旗环水次,舟楫泛中流。
C.江流宛转绕芳甸,月照花林皆似霰。
D.星垂平野阔,月涌大江流。
(2)电影《刮痧》讲述了一个发生在美国的故事。
中国人许大同的儿子丹尼斯发烧,家人用中医刮痧疗法给丹尼斯治病,。
法庭上,面对控方律师嘲讽式的提问及对中国传统医学的“误读”,许大同感到百口莫辩。
①填入上面语段空缺处最恰当的一项是()。
(2分)A.虽然孩子病好了但许大同被指控B.因为此事许大同被指控虐待孩子C.这成了许大同虐待孩子的证据被指控D.许大同因为虐待孩子被指控②下列为控方律师设计了提问,最贴合情景的一项是()。
(2分)A.许先生,刮痧疗法怎么没能治好中国那些东亚病夫?B.许先生,难道丹尼斯背上的印痕是你爱孩子的吻印?C.许先生,你说“经脉穴位”能治病,能解释一下吗?D.许先生,你就是靠这种野蛮治病方法活到今天的吗?【题型变化解读】学生和家长一直关心的高考改革终于在一模考试中初露端倪。
2016 各区一模作文立意分析xx依据以下资料,自选角度,自拟题目,写一篇许多于 800 字的文章(不要写成诗歌)。
有人说:“难过是财产,这话是扯淡;难过就是难过,对难过的思虑才是财产。
”①正视难过,思虑出路如资料所述,难过就是难过,当我们经历难过时,最理智的做法是思虑难过产生的原由,思虑降低难过的方法,这样便能从难过中收获一笔可贵的财产。
② 体验难过,收获财产事实也不尽如资料所言,难过自己也能够是财产,它是一种真真实切的体验,代表着仍旧鲜活的、拒绝麻痹的心灵。
文明的历史就是一部难过的历史。
③躲避难过, xx 人生乐而不淫,哀而不伤,学习躲避难过,或许不沉沦在难过之中,是一种值得倡导的健康踊跃的生活态度。
④ 直面难过,肩负人生生活中的难过常常是不行躲避的,因为它们是陪伴责任而来的,这就要求我们在最开始就能预示难过,而且敢于直面它,肩负起生活的重量。
xx依据以下资料,自选角度,自拟题目,写一篇许多于 800 字的文章(不要写成诗歌)。
有人说坚持做同一件事,坚持 10 年便可以成功,坚持 20 年便可以创建出奇观。
而事实上,绝大多数人同一件事情做了一辈子,仍旧平庸无奇。
① 做好规划,谨慎坚持坚持了多年却没有成功,可能是一开始坚持的方向就有了问题,或许是没有必定的物质基础,所谓“无物以相之,亦不可以至也”,也有可能是坚持的方式有问题,因为坚持并不是是简单的重复,而是要有变化、有创新。
因此在坚持以前应当做好规划,解决掉一些问题:坚持什么,依赖什么坚持,怎么坚持。
② 谋事在人,成事在天有了正确的坚持不代表理所应当就会成功,成功都是有必定的有时性因素的,需要掌握机遇。
③ 勇于坚持,拒绝盲从坚持自己就是一种成功,成功不必定是名利上的,也能够是精神涵养上的,这类状况下,坚持住就是全部。
崇明区古希腊哲学家伊壁鸠鲁说:“认识错误就是挽救自己的第一步。
”但是今世哲学家卡尔波普尔又说:“假如我们过于痛快的认可错误,便可能使自己觉察不了我们特别靠近于正确。
2016年上海市普陀区高考数学一模试卷一、填空题(本大题56分)本大题共有14小题,要求直接将结果填写在答题纸对应的空格中,每小空格填对得4分,填错或不正确的位置一律得零分. 1.(4分)若全集U=R,集合M={x|x(x﹣2)≤0},N={1,2,3,4},则N∩∁U M=.2.(4分)若函数,,则f(x)+g(x)=.3.(4分)在(2x﹣1)7的二项展开式中,第四项的系数为.4.(4分)在,则函数y=tanx的值域为.5.(4分)若数列{a n}中,a1=1,a n+1=2a n+1(n∈N*),则数列的各项和为.6.(4分)若函数f(x)=(x≥0)的反函数是f﹣1(x),则不等式f﹣1(x)>f(x)的解集为.7.(4分)设O为坐标原点,若直线与曲线相交于A、B点,则扇形AOB的面积为.8.(4分)若正六棱柱的底面边长为10,侧面积为180,则这个棱柱的体积为.9.(4分)若在北纬45°的纬度圈上有A、B两地,经度差为90°,则A、B两地的球面距离与地球半径的比值为.10.(4分)方程的解x=.11.(4分)设P是双曲线上的动点,若P到两条渐近线的距离分别为d1,d2,则d1•d2=.12.(4分)如图,已知正方体ABCD﹣A1B1C1D,若在其12条棱中随机地取3条,则这三条棱两两是异面直线的概率是(结果用最简分数表示)13.(4分)若F是抛物线y2=4x的焦点,点P i(i=1,2,3,…,10)在抛物线上,且,则=.14.(4分)若函数最大值记为g(t),则函数g(t)的最小值为.二、选择题(本大题20分)本大题共有4小题,每小题有且只有一个结论是正确的,必须把正确结论的代号写在答题纸相应的空格中,每题选对得5分,不选、选错或选出的代号超过一个(不论是否都写在空格内),或者没有填写在题号对应的空格内,一律得零分.15.(5分)下列命题中的假命题是()A.若a<b<0,则B.若,则0<a<1C.若a>b>0,则a4>b4D.若a<1,则16.(5分)若集合,则“x∈A”是“x∈B”成立的()A.充分非必要条件B.必要非充分条件C.充要条件D.既不充分也不必要条件17.(5分)如图,在四面体ABCD中,AB⊥BD,CD⊥DB,若AB与CD所成的角的大小为60°,则二面角C﹣BD﹣A的大小为()A.60°或90°B.60°C.60°或120°D.30°或150°18.(5分)若函数,关于x的方程f2(x)﹣(a+1)f(x)+a=0,给出下列结论:①存在这样的实数a,使得方程由3个不同的实根;②不存在这样的实数a,使得方程由4个不同的实根;③存在这样的实数a,使得方程由5个不同的实数根;④不存在这样的实数a,使得方程由6个不同的实数根.其中正确的个数是()A.1个B.2个C.3个D.4个三、解答题(本大题74分)本大题共有5小题,解答下列各题必须在答题纸规定的方框内写出必要的步骤.19.(12分)如图,椭圆+=1的左、右两个焦点分别为F1、F2,A为椭圆的右顶点,点P在椭圆上且∠PF1F2=arccos(1)计算|PF1|的值x(2)求△PF1A的面积.20.(14分)某种“笼具”由内,外两层组成,无下底面,内层和外层分别是一个圆锥和圆柱,其中圆柱与圆锥的底面周长相等,圆柱有上底面,制作时需要将圆锥的顶端剪去,剪去部分和接头忽略不计,已知圆柱的底面周长为24πcm,高为30cm,圆锥的母线长为20cm.(1)求这种“笼具”的体积(结果精确到0.1cm3);(2)现要使用一种纱网材料制作50个“笼具”,该材料的造价为每平方米8元,共需多少元?21.(14分)已知函数f(x)=2sin2x+sin2x﹣1.(1)求函数f(x)的单调递增区间;(2)设,其中0<x0<π,求tanx0的值.22.(16分)已知n∈N*,数列{a n}的前n项和为S n,且2a n﹣S n=1.(1)求证:数列{a n}是等比数列,并求出通项公式;(2)对于任意a i、a j∈{a1,a2,…,a n}(其中1≤i≤n,1≤j≤n,i、j均为正整数),若a i和a j的所有乘积a i•a j的和记为T n,试求的值;(3)设,若数列{c n}的前n项和为C n,是否存在这样的实数t,使得对于所有的n都有成立,若存在,求出t的取值范围;若不存在,请说明理由.23.(18分)已知集合M是满足下列性制的函数f(x)的全体,存在实数a、k (k≠0),对于定义域内的任意x均有f(a+x)=kf(a﹣x)成立,称数对(a,k)为函数f(x)的“伴随数对”.(1)判断f(x)=x2是否属于集合M,并说明理由;(2)若函数f(x)=sinx∈M,求满足条件的函数f(x)的所有“伴随数对”;(3)若(1,1),(2,﹣1)都是函数f(x)的“伴随数对”,当1≤x<2时,f(x)=cos(x);当x=2时,f(x)=0,求当2014≤x≤2016时,函数y=f(x)的解析式和零点.2016年上海市普陀区高考数学一模试卷参考答案与试题解析一、填空题(本大题56分)本大题共有14小题,要求直接将结果填写在答题纸对应的空格中,每小空格填对得4分,填错或不正确的位置一律得零分. 1.(4分)若全集U=R,集合M={x|x(x﹣2)≤0},N={1,2,3,4},则N∩∁U M={3,4} .【考点】1H:交、并、补集的混合运算.【专题】11:计算题;37:集合思想;4A:数学模型法;5J:集合.【分析】求解一元二次不等式化简M,求出其补集,再由交集运算得答案.【解答】解:∵M={x|x(x﹣2)≤0}={x|0≤x≤2},∴∁U M={x|x<0或x>2},又N={1,2,3,4},∴N∩∁U M={3,4}.故答案为:{3,4}.【点评】本题考查一元二次不等式的解法,考查了交、并、补集的混合运算,是基础的计算题.2.(4分)若函数,,则f(x)+g(x)=1(0≤x≤1).【考点】36:函数解析式的求解及常用方法.【专题】33:函数思想;49:综合法;51:函数的性质及应用.【分析】容易求出f(x),g(x)的定义域,求交集便可得出f(x)+g(x)的定义域,并可求得f(x)+g(x)=.【解答】解:;解得,0≤x≤1;∴(0≤x≤1).故答案为:.【点评】考查函数定义域的概念,清楚f(x)+g(x)的定义域为f(x)和g(x)定义域的交集.3.(4分)在(2x﹣1)7的二项展开式中,第四项的系数为﹣560.【考点】DA:二项式定理.【专题】11:计算题;33:函数思想;34:方程思想;5P:二项式定理.【分析】直接利用二项式定理写出结果即可即可.【解答】解:在(2x﹣1)7的二项展开式中,第四项的系数为:=﹣560.故答案为:﹣560.【点评】本题考查二项式定理的应用,系数的求法,注意二项式系数与项的系数的区别.4.(4分)在,则函数y=tanx的值域为[﹣1,1] .【考点】HC:正切函数的图象.【专题】38:对应思想;4O:定义法;57:三角函数的图像与性质.【分析】根据正切函数的图象与性质,求出x∈[﹣,]时函数y=tanx的值域即可.【解答】解:∵,∴﹣1≤tanx≤1,∴函数y=tanx的值域为[﹣1,1].故答案为:[﹣1,1].【点评】本题考查了正切函数的图象与性质的应用问题,是基础题目.5.(4分)若数列{a n}中,a1=1,a n+1=2a n+1(n∈N*),则数列的各项和为1.【考点】8E:数列的求和.【专题】35:转化思想;49:综合法;54:等差数列与等比数列.【分析】a1=1,a n+1=2a n+1(n∈N*),变形为a n+1+1=2(a n+1),利用等比数列的通项公式可得:1+a n,再利用等比数列的前n项和公式可得的前n项和.【解答】解:a1=1,a n+1=2a n+1(n∈N*),∴a n+1=2(a n+1),+1∴数列{a n+1}是等比数列,首项为2,公比为2.∴1+a n=2n,∴=,∴数列的首项为,公比为.∴数列的各项和为:=1.故答案为:1.【点评】本题考查了等比数列的通项公式及其前n项和公式、递推关系,考查了推理能力与计算能力,属于中档题.6.(4分)若函数f(x)=(x≥0)的反函数是f﹣1(x),则不等式f﹣1(x)>f(x)的解集为{x|x>1} .【考点】4R:反函数.【专题】11:计算题;35:转化思想;49:综合法;51:函数的性质及应用;59:不等式的解法及应用.【分析】由y=f(x)=(x≥0),求出f﹣1(x)=x3,x≥0,由此能求出不等式f﹣1(x)>f(x)的解集.【解答】解:设y=f(x)=(x≥0),则x=y3,x,y互换,得f﹣1(x)=x3,x≥0,∵f﹣1(x)>f(x),∴,∴x9>x,∴x8>1,解得x>1.∴不等式f﹣1(x)>f(x)的解集为{x|x>1}.故答案为:{x|x>1}.【点评】本题考查不等式的解集的求法,是基础题,解题时要认真审题,注意不等式性质、反函数性质的合理运用.7.(4分)设O为坐标原点,若直线与曲线相交于A、B点,则扇形AOB的面积为.【考点】G8:扇形面积公式;J9:直线与圆的位置关系.【专题】11:计算题;34:方程思想;49:综合法;5B:直线与圆.【分析】通过曲线方程确定曲线表示单位圆在x轴上方的部分(含于x轴的交点),y=时,∠AOB=π,即可求出扇形AOB的面积.【解答】解:由曲线,得x2+y2=1(y≥0)∴曲线表示単位圆在x轴上方的部分(含于x轴的交点)y=时,∠AOB=π,扇形AOB的面积为=.故答案为:.【点评】本题考查直线与圆的位置关系,利用数形结合,属于中档题.8.(4分)若正六棱柱的底面边长为10,侧面积为180,则这个棱柱的体积为450.【考点】LF:棱柱、棱锥、棱台的体积.【专题】31:数形结合;44:数形结合法;5Q:立体几何.【分析】根据侧面积公式求出棱柱的高,根据底面边长求出底面积,代入体积公式得出体积.=6ah=60h=180,【解答】解:设棱柱的底面边长为a,高为h,则S侧解得h=3.S底==150.h=450.∴正六棱柱的体积V=S底故答案为:450.【点评】本题考查了正棱柱的结构特征和体积计算,属于基础题.9.(4分)若在北纬45°的纬度圈上有A、B两地,经度差为90°,则A、B两地的球面距离与地球半径的比值为.【考点】L*:球面距离及相关计算.【专题】15:综合题;34:方程思想;49:综合法;5F:空间位置关系与距离.【分析】求出球心角,然后A、B两点的距离,求出两点间的球面距离,即可求出A、B两地的球面距离与地球半径的比值.【解答】解:地球的半径为R,在北纬45°,而AB=R,所以A、B的球心角为:,所以两点间的球面距离是:R,所以A、B两地的球面距离与地球半径的比值为;故答案为:.【点评】本小题主要考查球面距离及相关计算、经纬度等基础知识,考查运算求解能力,考查空间想象能力,属于基础题.10.(4分)方程的解x=log23.【考点】4H:对数的运算性质.【专题】11:计算题;51:函数的性质及应用.【分析】化简可得4x﹣5=4(2x﹣2),从而可得(2x)2﹣4•2x+3=0,从而解得.【解答】解:∵,∴4x﹣5=4(2x﹣2),即(2x)2﹣4•2x+3=0,∴2x=1(舍去)或2x=3;∴x=log23,故答案为:log23.【点评】本题考查了对数运算及幂运算的应用,同时考查了指数式与对数式的互化.11.(4分)设P是双曲线上的动点,若P到两条渐近线的距离分别为d1,d2,则d1•d2=.【考点】KC:双曲线的性质.【专题】11:计算题;34:方程思想;49:综合法;5D:圆锥曲线的定义、性质与方程.【分析】先确定两条渐近线方程,设双曲线C上的点P(x,y),求出点P到两条渐近线的距离,结合P在双曲线C上,即可求d1•d2的值.【解答】解:由条件可知:两条渐近线分别为x±y=0设双曲线C上的点P(x,y),则点P到两条渐近线的距离分别为d1=,d2=所以d1•d2=•==.故答案为:.【点评】本题考查双曲线的标准方程,考查双曲线的几何性质,求出点P到两条渐近线的距离是关键,属于中档题.12.(4分)如图,已知正方体ABCD﹣A1B1C1D,若在其12条棱中随机地取3条,则这三条棱两两是异面直线的概率是(结果用最简分数表示)【考点】CC:列举法计算基本事件数及事件发生的概率;LO:空间中直线与直线之间的位置关系.【专题】11:计算题;35:转化思想;49:综合法;5F:空间位置关系与距离.【分析】正方体ABCD﹣A1B1C1D,在其12条棱中随机地取3条,先求出基本事件总数,再求出这三条棱两两是异面直线包含的基本事件个数,由此能求出这三条棱两两是异面直线的概率.【解答】解:正方体ABCD﹣A1B1C1D,在其12条棱中随机地取3条,基本事件总数n==220,这三条棱两两是异面直线包含的基本事件个数m=8,∴这三条棱两两是异面直线的概率是p===.故答案为:.【点评】本题考查概率的求法,是基础题,解题时要认真审题,注意正方体的结构特征、等可能事件概率计算公式的合理运用.13.(4分)若F是抛物线y2=4x的焦点,点P i(i=1,2,3,…,10)在抛物线上,且,则=200.【考点】K8:抛物线的性质.【专题】15:综合题;34:方程思想;49:综合法;5D:圆锥曲线的定义、性质与方程.【分析】根据抛物线的定义得抛物线上的点到焦点的距离等于该点到准线的距离,因此求出抛物线的准线方程,结合题中数据加以计算,即可得到本题答案.【解答】解:∵抛物线y2=4x的焦点为F(1,0),准线为x=﹣1,∴根据抛物线的定义,P i(i=1,2,3,…,2015)到焦点的距离等于P i到准线的距离,即|P i F|=x i+1,,可得1﹣x1+1﹣x2+…+1﹣x100=0,∴x1+x2+…+x100=100∴|P1F|+|P2F|+...|P100F|=(x1+1)+(x2+1)+...+(x100+1)=(x1+x2+ (x100)+100=100+100=200.故答案为:200.【点评】本题考查了抛物线的定义、标准方程和简单几何性质,考查向量等知识,属于中档题.14.(4分)若函数最大值记为g(t),则函数g(t)的最小值为.【考点】3H:函数的最值及其几何意义.【专题】11:计算题;32:分类讨论;36:整体思想;51:函数的性质及应用.【分析】化简sinx+=sinx+3+﹣3,从而可得0≤sinx+3+﹣3≤,区间[0,]的中点值为,故讨论t与的大小,从而求得g(t)=f max(x)=,从而求值.【解答】解:∵sinx+=sinx+3+﹣3,∵﹣1≤sinx≤1,∴2≤sinx+3≤4,∴3≤sinx+3+≤,∴0≤sinx+3+﹣3≤,∴g(t)=f max(x)=,∴当t=时,函数g(t)有最小值为;故答案为;.【点评】本题考查了对勾函数的应用及分段函数的应用,同时考查了正弦函数的性质及整体思想与分类讨论的思想.二、选择题(本大题20分)本大题共有4小题,每小题有且只有一个结论是正确的,必须把正确结论的代号写在答题纸相应的空格中,每题选对得5分,不选、选错或选出的代号超过一个(不论是否都写在空格内),或者没有填写在题号对应的空格内,一律得零分.15.(5分)下列命题中的假命题是()A.若a<b<0,则B.若,则0<a<1C.若a>b>0,则a4>b4D.若a<1,则【考点】2K:命题的真假判断与应用.【专题】15:综合题;49:综合法;5:高考数学专题;5L:简易逻辑.【分析】正确选项进行证明,不正确选项,举出反例即可.【解答】解:对于A,a<b<0,则•a<•b,∴,正确对于B,,则>0,∴0<a<1,正确对于C,a>b>0,a4>b4,正确;对于D,a=,=2>1,不正确,故选:D.【点评】本题考查不等式的性质,考查命题的真假判断,属于中档题.16.(5分)若集合,则“x∈A”是“x∈B”成立的()A.充分非必要条件B.必要非充分条件C.充要条件D.既不充分也不必要条件【考点】29:充分条件、必要条件、充要条件.【专题】11:计算题;34:方程思想;4O:定义法;5L:简易逻辑.【分析】先分别求出集合A,B,然后根据必要条件、充分条件和充要条件的定义进行判断.【解答】解:∵≥0,∴0≤x<3,∴A=[0,3),∵lg|2x﹣3|<0=lg1,∴|2x﹣3|<1,且2x﹣3≠0,∴1<x<2,且x≠∴B=(1,)∪(,2),∴“x∈A”是“x∈B”成立的必要非充分条件,故选:B.【点评】此题主要考查必要条件、充分条件和充要条件的定义,是一道基础题.17.(5分)如图,在四面体ABCD中,AB⊥BD,CD⊥DB,若AB与CD所成的角的大小为60°,则二面角C﹣BD﹣A的大小为()A.60°或90°B.60°C.60°或120°D.30°或150°【考点】MJ:二面角的平面角及求法.【专题】5G:空间角.【分析】过D在平面ABD内作AB的平行线DE,则∠CDE或∠CDE的补角为二面角C﹣BD﹣A的平面角,由此能求出二面角C﹣BD﹣A的大小.【解答】解:过D在平面ABD内作AB的平行线DE,∵在四面体ABCD中,AB⊥BD,CD⊥DB,∴DE⊥BD,∴∠CDE或∠CDE的补角为二面角C﹣BD﹣A的平面角,∵AB与CD所成的角的大小为60°,∴∠CDE=60°或∠CDE=120°,∴二面角C﹣BD﹣A的大小为60°或120°.故选:C.【点评】本题考查二面角的大小的求法,是基础题,解题时要认真审题,注意空间思维能力的培养.18.(5分)若函数,关于x的方程f2(x)﹣(a+1)f(x)+a=0,给出下列结论:①存在这样的实数a,使得方程由3个不同的实根;②不存在这样的实数a,使得方程由4个不同的实根;③存在这样的实数a,使得方程由5个不同的实数根;④不存在这样的实数a,使得方程由6个不同的实数根.其中正确的个数是()A.1个B.2个C.3个D.4个【考点】53:函数的零点与方程根的关系.【专题】13:作图题;32:分类讨论;44:数形结合法;51:函数的性质及应用.【分析】由f2(x)﹣(a+1)f(x)+a=0可解得f(x)=1或f(x)=a,作函数的图象,从而讨论求解.【解答】解:∵f2(x)﹣(a+1)f(x)+a=0,∴f(x)=1或f(x)=a,作函数的图象如下,,当a=1时,方程有3个不同的实根,故①正确;当﹣1<a<1时,方程有6个不同的实根,故④不正确;当a>1或a≤﹣1时,方程有5个不同的实根,故③正确;综上可知,不存在这样的实数a,使得方程由4个不同的实根;故②正确;故选:C.【点评】本题考查了复合函数的应用及数形结合的思想应用及分段函数.三、解答题(本大题74分)本大题共有5小题,解答下列各题必须在答题纸规定的方框内写出必要的步骤.19.(12分)如图,椭圆+=1的左、右两个焦点分别为F1、F2,A为椭圆的右顶点,点P在椭圆上且∠PF1F2=arccos(1)计算|PF1|的值x(2)求△PF1A的面积.【考点】K4:椭圆的性质.【专题】11:计算题;35:转化思想;49:综合法;58:解三角形;5D:圆锥曲线的定义、性质与方程.【分析】(1)根据椭圆的性质,可得|PF1|=x,则|PF2|=10﹣x,|F1F2|=2=8,结合已知可余弦定理构造方程,解得x值;(2)由出sin∠PF1F2,进而计算△PF1F2的面积,可得P到x轴的距离d,结合△PF1A的底边|F1A|=a+c=9,可得三角形面积.【解答】解:(1)∵椭圆+=1的左、右两个焦点分别为F1、F2,P为椭圆上一点,|PF1|=x,则|PF2|=10﹣x,|F1F2|=2=8,∵∠PF1F2=arccos,故cos∠PF1F2==,解得:x=6,(2)由∠PF1F2=arccos,可得:sin∠PF1F2==,故△PF1F2的面积S=(5+)•(5﹣)•=,故P到x轴的距离d==,由|F1A|=a+c=9,可得△PF1A的面积为:×=【点评】本题考查的知识点是椭圆的简单性质,余弦定理,三角形面积公式,难度中档.20.(14分)某种“笼具”由内,外两层组成,无下底面,内层和外层分别是一个圆锥和圆柱,其中圆柱与圆锥的底面周长相等,圆柱有上底面,制作时需要将圆锥的顶端剪去,剪去部分和接头忽略不计,已知圆柱的底面周长为24πcm,高为30cm,圆锥的母线长为20cm.(1)求这种“笼具”的体积(结果精确到0.1cm3);(2)现要使用一种纱网材料制作50个“笼具”,该材料的造价为每平方米8元,共需多少元?【考点】LF:棱柱、棱锥、棱台的体积.【专题】31:数形结合;44:数形结合法;5Q:立体几何.【分析】(1)笼具的体积等于圆柱的体积减去圆锥的体积;(2)求出笼具的表面积即可,笼具的表面积包括圆柱的侧面,上底面和圆锥的侧面.【解答】解:(1)设圆柱的底面半径为r,高为h,圆锥的母线长为l,高为h1,则2πr=24π,解得r=12cm.h1=cm.∴笼具的体积V=πr2h﹣=π×(122×30﹣×122×16)=3552π≈11158.9cm3.(2)圆柱的侧面积S1=2πrh=720cm2,圆柱的底面积S2=πr2=144πcm2,圆锥的侧面积为πrl=240πcm2.故笼具的表面积S=S1+S2+S3=1104πcm2.故制造50个这样的笼具总造价为:元.答:这种笼具的体积约为11158.9cm3,生产50个笼具需要元.【点评】本题考查了圆柱,圆锥的表面积和体积计算,属于基础题.21.(14分)已知函数f(x)=2sin2x+sin2x﹣1.(1)求函数f(x)的单调递增区间;(2)设,其中0<x0<π,求tanx0的值.【考点】GL:三角函数中的恒等变换应用;HW:三角函数的最值.【专题】34:方程思想;4R:转化法;56:三角函数的求值.【分析】(1)利用三角函数的关系结合辅助角公式进行化简,即可求函数f(x)的单调递增区间;(2)化简条件,利用同角的三角函数的关系式建立方程关系进行求解即可.【解答】解:(1)f(x)=2sin2x+sin2x﹣1=sin2x﹣cos2x=sin(2x﹣).由2kπ﹣≤2x﹣≤2kπ+,k∈Z,得2kπ≤x≤2kπ,k∈Z,得kπ﹣≤x≤kπ+,k∈Z,即函数f(x)的单调递增区间是[kπ﹣,kπ+],k∈Z;(2)cos(+α)cos(﹣α)+sin2α=(cos cosα)2﹣(sin sinα)2+sin2α=cos2α﹣sin2α+sin2α=,即f()=sin(2×﹣)=sin(x0﹣)=,即sinx0﹣cosx0=,①平方得2sinx0cosx0=,∵0<x0<π,∴cosx0>0,则sinx0+cosx0==②,由①②得sinx0=,cosx0=,则tanx0==.【点评】本题主要考查三角函数的化简和三角函数的图象和性质,利用辅助角公式进行化简以及利用三角函数的同角的基本关系式是解决本题的关键.22.(16分)已知n∈N*,数列{a n}的前n项和为S n,且2a n﹣S n=1.(1)求证:数列{a n}是等比数列,并求出通项公式;(2)对于任意a i、a j∈{a1,a2,…,a n}(其中1≤i≤n,1≤j≤n,i、j均为正整数),若a i和a j的所有乘积a i•a j的和记为T n,试求的值;(3)设,若数列{c n}的前n项和为C n,是否存在这样的实数t,使得对于所有的n都有成立,若存在,求出t的取值范围;若不存在,请说明理由.【考点】88:等比数列的通项公式;8J:数列的极限;8K:数列与不等式的综合.【专题】11:计算题;35:转化思想;49:综合法;55:点列、递归数列与数学归纳法.【分析】(1)当n≥2时通过2a n﹣S n=1与2a n﹣1﹣S n﹣1=1作差,进而计算可得结论;(2)通过(1)可得T n的表达式,进而计算即得结论;(3)通过(1)可知数列{c n}的通项公式,利用并项相加、分n为奇数、偶数两种情况讨论即可.【解答】(1)证明:∵2a n﹣S n=1,﹣S n﹣1=1,∴当n≥2时,2a n﹣1两式相减,整理得:a n=2a n﹣1(n≥2),又∵2a1﹣S1=1,即a1=1,∴数列{a n}是首项为1、公比为2的等比数列,∴a n=2n﹣1;(2)解:∵T n=(1+2+22+…+2n﹣1)(1+2+22+…+2n﹣1)=•=4n﹣2•2n+1,∴==1;(3)结论:存在这样的实数t,使得对于所有的n都有成立.理由如下:由(1)可知,1+b n=3log2a n=3n﹣3,即b n=3n﹣4,b n+1=3n﹣1,故c n=(﹣1)n+1b n•b n+1=(﹣1)n+1(3n﹣4)(3n﹣1),c n+1=(﹣1)n+2(3n﹣1)(3n+2),特别地,当n为奇数时,有n+1为偶数,此时c n+c n+1=(3n﹣4)(3n﹣1)﹣(3n﹣1)(3n+2)=﹣6(3n﹣1),①若n为偶数,则C n=(c1+c2)+(c3+c4)+…+(c n﹣1+c n)=﹣6×[2+8+…+(3n﹣4)]=﹣n(3n﹣2),由可知t≤﹣(3﹣)对所有正偶数n都成立,故t≤﹣;②若n为奇数,则C n=C n﹣1+c n(n≥2),由①可知C n=﹣(n﹣1)(3n﹣5)+(3n﹣4)(3n﹣1)=n2﹣3n﹣,其中C1=﹣2满足上式;由①②可得实数t的取值范围是:t≤﹣,所以存在这样的实数t,使得对于所有的n都有成立.【点评】本题是一道关于数列与不等式的综合题,考查分类讨论的思想,考查运算求解能力,注意解题方法的积累,属于中档题.23.(18分)已知集合M是满足下列性制的函数f(x)的全体,存在实数a、k (k≠0),对于定义域内的任意x均有f(a+x)=kf(a﹣x)成立,称数对(a,k)为函数f(x)的“伴随数对”.(1)判断f(x)=x2是否属于集合M,并说明理由;(2)若函数f(x)=sinx∈M,求满足条件的函数f(x)的所有“伴随数对”;(3)若(1,1),(2,﹣1)都是函数f(x)的“伴随数对”,当1≤x<2时,f(x)=cos(x);当x=2时,f(x)=0,求当2014≤x≤2016时,函数y=f(x)的解析式和零点.【考点】3T:函数的值.【专题】31:数形结合;32:分类讨论;35:转化思想;51:函数的性质及应用.【分析】(1)f(x)=x2的定义域为R.假设存在实数a、k(k≠0),对于定义域内的任意x均有f(a+x)=kf(a﹣x)成立,则(a+x)2=k(a﹣x)2,化为:(k ﹣1)x2﹣2a(k+1)x+a2(k﹣1)=0,由于上式对于任意实数x都成立,可得,解得k,a.即可得出.(2)函数f(x)=sinx∈M,可得:sin(a+x)=ksin(a﹣x),展开化为:sin (x+φ)=0,由于∀x∈R都成立,可得k2+2kcos2a+1=0,变形cos2a=,利用基本不等式的性质与三角函数的单调性即可得出.(3)由于(1,1),(2,﹣1)都是函数f(x)的“伴随数对”,可得f(1+x)=f (1﹣x),f(2+x)=﹣f(2﹣x),因此f(x+4)=f(x),T=4.对x分类讨论可得:即可得出解析式,进而得出零点.【解答】解:(1)f(x)=x2的定义域为R.假设存在实数a、k(k≠0),对于定义域内的任意x均有f(a+x)=kf(a﹣x)成立,则(a+x)2=k(a﹣x)2,化为:(k﹣1)x2﹣2a(k+1)x+a2(k﹣1)=0,由于上式对于任意实数x都成立,∴,解得k=1,a=0.∴(0,1)是函数f(x)的“伴随数对”,f(x)∈M.(2)∵函数f(x)=sinx∈M,∴sin(a+x)=ksin(a﹣x),∴(1+k)cosasinx+(1﹣k)sinacosx=0,∴sin(x+φ)=0,∵∀x∈R都成立,∴k2+2kcos2a+1=0,∴cos2a=,≥2,∴|cos2a|≥1,又|cos2a|≤1,故|cos2a|=1.当k=1时,cos2a=﹣1,a=nπ+,n∈Z.当k=﹣1时,cos2a=1,a=nπ,n∈Z.∴f(x)的“伴随数对”为(nπ+,1),(nπ,﹣1),n∈Z.(3)∵(1,1),(2,﹣1)都是函数f(x)的“伴随数对”,∴f(1+x)=f(1﹣x),f(2+x)=﹣f(2﹣x),∴f(x+4)=f(x),T=4.当0<x<1时,则1<2﹣x<2,此时f(x)=f(2﹣x)=﹣cos;当2<x<3时,则1<4﹣x<2,此时f(x)=﹣f(4﹣x)=﹣cos;当3<x<4时,则0<4﹣x<1,此时f(x)=﹣f(4﹣x)=cos.∴f(x)=.∴f(x)=.∴当2014≤x≤2016时,函数y=f(x)的零点为2014,2015,2016.【点评】本题考查了新定义“伴随数对”、三角函数的单调性、基本不等式的性质,考查了分类讨论思想方法、推理能力与计算能力,属于难题.。