2018届高三第一次模拟考试(一模)仿真卷(A卷)教师版

2018届高三第一次模拟考试仿真卷英 语 （A ）注意事项：1．答题前，先将自己的姓名、准考证号填写在试题卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。

2．选择题的作答：每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3．非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4．考试结束后，请将本试题卷和答题卡一并上交。

第Ⅰ卷第一部分 听力（共两节，满分 30 分）(略) 第二部分 阅读理解（共两节，满分40分） 第一节（共15小题：每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A 、B 、C 和D ）中选出最佳选项，并在答题卡上将该项涂黑。

AIn fairy tales, it's usually the princess that needs protecting. At Google in Silicon Valley, the princess is the one defending the castle. Parisa Tabri is a 31-year-old with perhaps the most unique job title in engineering- “Google Security Princess”. Her job is to hac into the most popular web browser （浏览器）on the planet, trying to find weanesses in the system before the “blac hats” do. To defeat Google's attacers, Tabri must firstly thin lie them.Tabri's role has evolved dramatically in the eight years since she first started woring at Google. Bac then, the young graduate from Illinois University was one of 50 security engineers---today there are over 500.Cybercrime （网络犯罪）has come a long way in the past decade - from the Nigerian Prince Scam to credit card theft. Tabri's biggest concern now is the people who find bugs in Google's software, and sell the information to governments or criminals. To fight against this, the company has set up a Vulnerability Rewards Program, paying anywhere from $100 to $ 20, 000 for reported mistaes.班级 姓名 准考证号 考场号 座位号It's a world away from Tabri's computer-free childhood home in Chicago. The daughter of an Iranian-American doctor father, and Polish-American nurse mother, Tabri had little contact with computers until she started studying engineering at college. Gae across a line-up of Google security staff today and you'll find women lie Tabri are few and far between（稀少的）--- though in the last few years she has hired more female tech geniuses. She admits there's an obvious gender disequilibrium in Silicon Valley.Funnily enough, during training sessions Tabri first ass new colleagues to hac into not a computer, but a vending machine. Tabri's job is as much about technological now-how（专门知识）as understanding the psychology of attacers.21. What can we learn about Tabri from the passage?A. She was the first female engineer at Google.B. She must thin differently so as to defeat the attacers.C. Her job relates to not only technology but also psychology.D. Her frequent contact with computers in childhood benefits her a lot.22. Why has Google set up a Vulnerability Rewards Program?A. To protect Google against cybercrime.B. To monitor the normal operation of Google.C. To help the government locate the cybercriminals.D. To raise people's awareness of personal information safety.23. What does the underlined word “disequilibrium” in Paragraph 4 refer to?A. Imbalance.B. Preference.C. Difference.D. Discrimination.24. Which of the following could be the best title of this passage?A. What leads to cybercrimeB. The “Security Princess” who guards GoogleC. Measures taen by Google to protect its usersD. How to become an ecellent security engineerBThe English have a difficult and, generally speaing, dysfunctional (怪异的) relationship with clothes. Their main problem is that they have a desperate need for rules, and are unable to get along without them.This helps to eplain why they have an international reputation for dressing in general very badly, but with specific areas of ecellence, such as high-class men’s suits, ceremonial costumes, and innovative (革新的) street fashion. In other words, we English dress best when we are “in uniform”.You may be surprised that I am including “innovative street fashion” in the category of the uniform. Surely the parrot-haired puns (朋克摇滚乐迷) or the Victorian vampire goths are being original, not following rules? It’s true that they all loo different and eccentric (古怪的) but in fact they all loo eccentric eactly in the same way. They are wearing a uniform. The only truly eccentric dresser in this country is the Queen, who pays no attention to fashion and continues to wear what she lies, a ind of 1950s fashion, with no regard for anyone else’s opinion. Howe ver, it is true that the styles invented by young English people are much more eccentric than any other nation’s street fashion. We may not be individually eccentric, apart from the Queen, but we have a sort of collective eccentricity, and \ye appreciate originality in dress even if we do not individually have it.Another “rule” of behavior I had discovered was that it is very important for the English not to tae themselves too seriously, to be able to laugh at themselves. However, it is well nown that most teenagers tend to tae themselves a bit too seriously.The goths, in their scary blac costumes, certainly loo as if they are taing themselves seriously. But when I got into conversation with them, I discovered that they too had a sense of humor. I was once chatting to a goth in the full vampire costume—with a white face, deep purple lipstic, and blac parrot-hair. I saw he was also wearing a T-shirt with “Goth”. “Why are you wearing that?” I ased. “In case you don’t realie I’m a goth.” he answered, pretending to be serious. We both burst out laughing.25. What can we now about the English people?A. They need rules to dress well.B. They are in need of uniforms.C. They are creative in general.D. They lead the world trend.26. Who is individually eccentric in dressing?A. A high-class man.B. A parrot-haired pun.C. The Queen.D. The fashion innovator.27. Which of the following can best describe the goths?A. They dress badly.B. They dress in an amusing way.C. They are unable to laugh at the way they dress.D. They are less fashionable than the other English people.28. What may be the best title for the tet?A. How the English DressB. How the English Admire FashionC. Why the English Lie UniformsD. Why the English Are Eccentric in DressCNot long ago, people thought babies were not able to learn things until they were five or si months old. Yet doctors in the United States say babies begin learning on their first day of life. Scientists note that babies are strongly influenced by their environment. They say a baby will smile if her mother does something the baby lies. A baby learns to get the best care possible by smiling to please her mother or other caregivers. This is how babies learn to connect and communicate with other human beings. One study shows that babies can learn before they are born. The researchers placed a tape recorder on the stomach of a pregnant woman. Then, they played a recording of a short story. On the day the baby was born, the researchers attempted to find if he new the sounds of the story repeated while in his mother. They did this by placing a device in the mouth of the newborn baby.The baby would hear the story if he moved his mouth one way. If the baby moved his mouth the other way, he would hear a different story. The researchers say the baby clearly lied the story he heard before he was born. They say the baby would move his mouth so he could hear the story again and again.Another study shows how mothers can strongly influence social development and language sills in their children. Researchers studied the children from the age of one month to three years. The researchers attempted to measure the sensitivity of the mothers. The women were considered sensitive if they supported their children’s activities and did not interfere(干预) unnecessarily. They tested the children for thining and language development when they were three years old. Also, the researchers observed the women for signs of depression. The children of depressed women did not do as well in tests as the children of women who did not suffer from depression. The children of depressed women did poorly in tests of language sills andunderstanding what they hear.These children also were less cooperative and had more problems dealing with other people. The researchers noted that the sensitivity of the mothers was important to the intelligence development of their children. Children did better when their mothers were caring, even when they suffered from depression. 29. According to the passage, which of the following is NOT the factor that influences intelligence development in babies?A. The environment.B. Their peersC. Mother’s sensitivityD. Education before birth30. What is the purpose of the eperiment in which newborn babies heard the stories?A. To prove that babies can learn on the first day they are bornB. To show mothers can strongly influence intelligence development in their babiesC. To indicate early education has a deep effect on the babies’ language sillsD. To prove that babies can learn before they are born31. Which group of children did the worst in tests of language sills?A. The children of women who did not suffer from depressionB. The children of depressed but caring mothersC. The children of depressed mothers who cared little for their childrenD. Children with high communication abilities32. What is the main idea of the passage?A. Scientific findings about how intelligence develops in babiesB. Scientific findings about how babies develop before birthC. Scientific findings about how time has an effect on babies’ intelligenceD. A study shows babies are not able to learn things until they are five or si months oldDOver seven months have passed since Panamanian officials launched an epansion of the world famous Pana ma Canal. Officials agreed to the epansion so that many of the world’s largest cargo ships (货船) could easily pass through the canal. Yet the $ 5.25-billion project has problems. It says ships still continue to rub against the canal’s walls and wear out its defenses designed to protect both shipping and the waterway.A dangerous systemThe canal lins two oceans-the Atlantic and the Pacific-through a system of locs (船闸). The locs are lie steps. They raise and lower ships from one part of the waterway to another on their trip from ocean to ocean.With the old locs, which are still in use, large ships would be tied to powerful engines on both sides. These engines help to eep the ships in the center of the canal. In the new locs, the ships are tied to tugboats (拖船). One tugboat is tied to the front of the ship, with the other tied to the bac. These boats then guide the ships through the canal.At first, pilots of the cargo ships and tugboat operators would sometimes try to rub the boats against the canal walls as a way to eep the ships straight. But this caused damage to rubber padding (垫料) lining the walls.Not enough trainingEven before the epanded canal opened in June 2016, tugboat operators had epressed concern about the new system. Many ased for more training. The fears and dangers remain, although the boats are going through.The Panama Canal Authority reports that, between June 2016 and January 2017, there were only 15 incidents that resulted in damage to locs or ships. That represents about 2 percent of the 700 times ships that have sailed through the epanded canal.Pilots have argued they should be replaced with a system of floating bumpers (减震) lie those used in some European locs. Officials say they plan to continue operating with the current system of defenses, but changes could happen in the future.33. What is the difference between the new locs and the old ones?A. The old locs don’t need rubber padding as defenses.B. The new locs need tugboats tied to both sides of the ships.C. The new locs are easier for the largest ships to pass through the canal.D. The old locs need powerful engines to drag the ships through the canal.34. What is the Panama Canal Authority’s attitude towards the epanded canal?A. Cautious.B. Critical.C. Positive.D. Doubtful.35. What can we learn about the current system of defenses?A. No ships shall rub against the canal walls to protect it.B. Nothing will be done at present to improve it.C. More training will be given to pilots for it.D. A new system will replace it.第二节（共 5 小题，每小题 2 分，满分 10 分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项，选项中有两项为多余选项。

绝密★启用前|2018年11月浙江省普通高中学业水平考试英语仿真模拟试题A 卷（考试时间：90分钟 试卷满分：120分）选择题部分第一部分 听力（略）第一部分 阅读理解（共两节，满分35分） 第一节（共10小题；每小题2.5分，满分25分）阅读下列短文，从每题所给的A 、B 、C 和D 四个选项中，选出最佳选项，并在答题纸上将该项涂黑。

AThanks to the vast territory and huge population of China, different living habits with geographical features have been an ever-lasting topic among Chinese. Netizens (网民) went wild again as they recently discovered another striking difference in the ways the country’s northerners and southerners shop for food.According to netizens from southern China, they usually buy what they need for just one or two meals. “It is quite normal for us to buy one or half of a Chinese cabbage (大白菜) at a time,” said a netizen. However, one may be looked down upon if he does this in the north. “We l ove to store up on food. Buying 25 kilograms of Chinese cabbage is normal,” said a netizen from the north.Besides, in the south, it is said that winter melon (冬瓜) is always sold in slices, “One slice is enough for two meals!” one said. While in the north, it is simply not possible. “Of course we buy a whole one, otherwise nobody is willing to buy the other half,” said a northern Chinese.Experts say that the difference is due to the varied climate. Food can be stored for a relatively long time in the north as it is usually cold and dry there. However, the comparatively hot and humid weather in the south forces southern Chinese to buy just enough for one meal or one day, otherwise it might go bad.Besides obvious ways of food shopping, other north-south differences in China include the flavor of tofu (豆腐) (northerners prefer a salty taste while southerners like it sweet), and the way to give directions (northerners prefer say “go east/west,” but southerners usually say “go left/right”). 21. In what case is a buyer looked down upon by a northerner?A. When he buys a whole winter melon.B. When he buys too small an amount of food.C. When he buys a bag of a Chinese cabbage at a time.D. When he buys food just enough for one or two persons.22. Why do experts think there is much difference in how much food to buy between northerners and southerners?A. Due to different habits.B. Due to different customs.C. Due to different climates.D. Due to different characters.23. How do southerners prefer tofu? A. They prefer it to taste sweet. B. They prefer to have it taste salty.C. They prefer it much more than northerners.D. They prefer to have it more often to keep healthy.BFifteen-year-old David had spent six months planning a hike to the bottom of the Grand Canyon (大峡谷) with his classmates. No one could have known it would end in tragedy.When the group arrived at the starting point for the hike, they found out there was a problem with their permit. To solve it, the five boys and three leaders walk back up to the ranger station (护林站) — only to find out it was closed.When the boys got moving, they’d already drunk most of their water. In the burning sun, temperatures reached 115°F. Even the leaders were having trouble walking on. But the group finally made it down to the bottom of the Grand Canyon. Heat evil was occurring. ＂My son made it about 100 yards from the river and then dropped,＂ says his father, Robin. Two boys of them raced to the river to get him water.When the boys arrived, the last rafting (筏运) trip of the day was passing by. As luck would have it, among the waiting riders were a few doctors. They rushed to give David first aid before going for help. But it was too late. David passed away from the heat evil.It’s been 21 years since the boy’s death, and hiking still holds a strong memory of David for his family. The summer before the tragedy, his father had a long time to devote to his family. David and his dad spent the summer hiking all over the High Uintas, where snow-topped peaks reflect tower over glacier (冰川)-formed lakes.The photo Robin took of Christmas Meadows there won the Reader’s Digest ＂From Sea to Shining Sea ＂ photo contest and appeared on the July 2017 magazine cover. To Robin, it isn’t just a fascinating landscape — it’s a lso a powerful reminder of his late son.此卷只装订不密封班级 姓名 准考证号 考场号 座位号24. How many people does David’s group consist of?A. 11B. 10C. 9D. 825. What does the sentence ＂Even the leaders were having trouble walking on.＂in Paragraph 3 suggest?A. The leaders were taller and older.B. The leaders covered shorter distance.C. The leaders were healthier and stronger.D. The trouble the group faced was so great as to be beyond expectation.26. Where did Robin take the photo which won the Reader’s Digest＂From Sea to Shining Sea＂photo contest?A. On the High Uintas.B. Where David had been to.C. In the Grand Canyon.D. In the Christmas Meadows.27. What does the passage intend to tell us mainly?A. Take enough water with you when hiking.B. Don’t hike to the Grand Canyon in too hot weather.C. Robin still misses his late son, David.D. Doctors can’t necessarily succeed in saving every patient.CIn 1975, George Carlin appeared on a popular TV show, Saturday Night Live, with his famous words about blue food. ＂Why is there no blue food? I can’t find blue food—I can’t find the flavor of blue! I mean yellow is lemon; orange is orange and red is cherry. Where is the blue food? ＂Well, Carlin almost has it right—there’s not no blue food, but there’s certainly not a lot of it. Fresh-picked blueberries are blue, though they become purple when they are turned into jam. The blue in blueberries—like the purple in grapes and the red in tomatoes—is found in nature. But it isn’t a hot color for food. People don’t seem to pr efer blue food. Some diet programs even suggest that those determined to lose weight should make their food blue.Eating, in part, begins with our eyes. Charles Spence—an experimental psychologist from Oxford University—points out that color can change our taste experience. Commonly, we consider red-colored foods up to 20% sweeter than they actually are and green foods as being more sour. Spence suggests: human expectations may be influenced by our long history of watching—green fruits can become sweet as they grow up and turn red.But what about blue? Except blueberries, much of the blue food we see these days is dyed blue artificially. Food producers argue that artificial color doesn’t do much harm to health. A lot of research shows that some physical problems of kids are related to food dyes—while other studies show no effects at all. Blue birthday cake or even blue-dyed chicken can be served on the dinner table.28. What’s the purpose of the first paragraph?A. To introduce the topic.B. To introduce a TV show.C. To state the author’s opinion.D. To explain why people dislike blue.29. What can we learn about blueberries from Paragraph 2?A. The blue in blueberries is found natural.B. Blueberries are still blue when processed.C. Most people don’t eat blueberries owing to the color.D. People often feel cold when they eat blueberries.30. What does the author mean by saying eating sometimes begins with our eyes?A. People prefer red-colored food to green food.B. We should examine food carefully before eating.C. Our taste experience can be influenced by color.D. Red-colored foods look sweeter than they actually are.第二节（共5小题，每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

2018年普通高等学校招生全国统一考试广东省第一次模拟语文试卷注意事项：1．答题前，考生务必用0.5毫米黑色字迹签字笔将自己所在的县（市、区）、学校以及自己的姓名、考生号、考场号和座位号填写在答题卡和试卷的指定位置，并用2B铅笔在答题卡的“考生号”处填涂考生号。

2．回答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号。

回答非选择题时，用0.5毫米黑色字迹签字笔将答案写在答题卡上。

写在本试卷上无效。

3．考试结束后，将本试卷和答题卡一并交回。

一、现代文阅读（35分）（一）论述类文本阅读（本题共3小题，9分）阅读下面的文字，完成1～3题。

审美活动的动力机制有情感、想象、超越三种。

与之相应，审美活动所产生的美有三种存在的形态：情象、意象、境界。

审美活动最基本的动力是情感。

从本质上来说，美是情感的对象化。

这里说的情感是审美主体的情感，而不是审美对象的情感。

这里说的对象，有两种形态：一是原生形态。

即原本有一个对象，但不是审美对象，是审美主体将其情感赋予给它，使对象成为主体情感的载体，从而成为审美对象。

二是自创形态。

即原本无对象，因为有情，需要外化，于是就创造一个形象，让其成为情感的载体。

这种情况在艺术创作中居多。

以上两种情况都可以说是情感造形。

情感造形的产物就是情象。

情象是美的基础形态。

凡审美都有情感造形存在，只是审美中的情感造形，除艺术创作外，都是不自觉的，而是直觉的。

刘勰说诗人创作时“登山则情满于山，观海则意溢于海”。

这种情况不只是艺术创作时有，只要是进入审美形态的人都有。

情象是审美主体的创造，只是这创造主要表现在情感的赋予上，而当想象参与后，则就有很大的不同。

想象的最重要的功能是创造新事物，这新事物是现实中不存在的。

想象的创造，不只有情感在起作用，还有人的意识、文化修养在起作用。

于是，这新创造的形象就具有丰富的意蕴，这具有丰富意蕴的形象我们叫它“意象”。

宁夏银川一中2018届高三第一次模拟考试语文试题及答案解析注意事项：1．答题前，考生务必先将自己的姓名、准考证号填写在答题卡上，认真核对条形码上的姓名、准考证号，并将条形码粘贴在答题卡的指定位置上。

2．选择题答案使用2B铅笔填涂,如需改动，用橡皮擦干净后，再选涂其他答案的标号；非选择题答案使用0．5毫米的黑色中性（签字）笔或碳素笔书写，字体工整、笔迹清楚。

3．考生必须按照题号在答题卡各题号相对应的答题区域(黑色线框)内作答，写出草稿纸上、超出答题区域或非题号对应的答题区域的答案一律无效。

4．保持卡面清洁，不折叠，不破损。

5．做选考题时，考生按照题目要求作答，并用2B铅笔在答题卡上把所选题目对应的题目涂黑。

第Ⅰ卷阅读题一、现代文阅读(35分)(一)论述文阅读阅读下面的文字，完成1～3题。

（9分，每小题3分）在努力实现中华民族伟大复兴中国梦的今天,为了更好地传承发展中华优秀传统文化，需要对传统文化进行创造性转化和创新性发展。

如何进行创造性转化和创新性发展?深入挖掘中华优秀传统文化的价值追求是一条重要途径，有助于克服在传承发展传统文化中容易出现的形式化倾向。

历史悠久的中华传统文化，在殷周之际出现了重要的人文转向，即由神本向人本转向以及人道主义思潮出现。

这为儒、墨、道、法等诸子百家提供了最直接的思想来源，也规定了此后中华传统文化的基本精神和主要走向。

在博大精深的中华传统文化中，儒、道两家最具代表性。

儒家致力于以仁为核心、仁礼相辅互动的理论构建。

与儒家讲仁爱形成对照的是，道家更强调道法自然，认为文明的发展要注意克服虚伪性和工具化的倾向。

道家身国同构、经国理身的理念,表现出其根本价值取向是向往真正符合人性的和谐社会与美好人生。

由此，儒、道两家虽然对社会和人生的理想形成了不同的致思路向，但在拥有共同价值追求的基础上可以相融互补。

在魏晋玄学那里，通过名教与自然之辨，二者实现了辩证的统一，为容纳和吸收外来的佛教准备了思想文化条件。

齐齐哈尔市2018届高三第一次模拟考试语文试卷2018．3 考生注意：1．本试卷满分150分，考试时间150分钟。

2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

回答非选择题时，将答案写在答题卡上，写在本试卷上无效。

考试结束后，将本试卷和答题卡一并交回。

3．本卷命题范围：高考范围。

一、现代文阅读（35分）（一）论述类文本阅读（本题共3小题，9分）阅读下面的文字，完成1—3题。

诗意，是所有优秀艺术的共同特征。

凡是好的文学作品，无论小说、散文、戏剧等，其本质都是诗，或者说，都是充满诗意的。

读完莫美的小小说集——《印象》，我的这种感觉就更加鲜明。

我相信，莫美在创作每一篇小小说的时候，一定是作为一首首诗在写的；否则，是难以如此诗意盎然的。

这种诗意，最直接的感觉是来自他的语言。

莫美小小说的语言，以质朴简洁、真实生动、自然流畅见长，不雕饰、不做作、不隐晦，娓娓道来，在舒缓散淡的叙述中，使作品体现出一种诗化的特色，呈现出一种特别的意境，展現出一种自然天成之美。

如《牛不知道自己的力气有多大》中的“夕阳西沉，红霞满天”“山脚下，小河旁，水田里，一头牛，一个人，似乎未动，其实在动”“压耙带泥”“抽了老黄牛一鞭”等语句反复出现，如同《诗经》的“一叹三唱”，韵味十足。

有的本身就是一首诗，如《温海垂钓》中对温海迷人景色的描写：“湖周青山环绕，森林茂密，空气清新，一尘不染。

湖水自然澄澈，颜色多变，春湛蓝，夏碧绿，秋斑驳，冬苍翠……”就是一首山水诗。

这种诗意，骨子里来自于作者对历史、社会、生活的深刻思考。

中国作协原副主席谭谈说：“莫美先生不以人云亦云、芸芸众生眼中的‘美’为审美标准，这种‘莫美’的文学价值追求，值得称道。

”像经典名篇《牛不知道自己的力气有多大》就饱含着作者对“三农”问题、人的自我认识和觉醒等主题的现实追问和哲学思量，不但深刻，而且显示出“多义多解的长篇气象”。

2018—2018年高三第一次模拟考试 数学试题参考解答及评分标准一、选择题：本大题主要考查基本知识和基本运算．共10小题，每小题5分，满分50分．二、填空题：本大题主要考查基本知识和基本运算．共4小题，每小题5分，满分20分．第（14）小题的第一个空2分、第二个空3分． （11）0.82 （12）3（13）320 （14）030 三、解答题（15）本小题主要考查等比数列的概念、通项公式及前n 项和公式等基本知识，考查运算求解能力．满分12分．解：由已知，得51113162,(13)242,13n a a -⎧⋅=⎪⎨-=⎪-⎩①②由①得 181162a =，解得 12a =． …9分将12a =代入②得()21324213n=-－，即 3243n=，解得 n ＝5．∴ 数列{}n a 的首项12a =，项数n ＝5． …12分 （16）本小题主要考查三角函数的图象和性质等基本知识以及利用三角公式进行恒等变换的技能，考查运算求解能力．满分14分．解：（Ⅰ）()f x ＝22sin 2cos sin x x x +-＝sin 2cos 2x x +…3分sin 2cos 222x x ⎫+⎪⎪⎭sin 2cos cos 2sin44x x ππ⎫+⎪⎭)4x π+． …7分（Ⅱ）列表如下：…3分…6分…14分（17）本小题主要考查空间线面关系，考查空间想像能力和推理论证能力．满分14分．证明：（Ⅰ）∵底面ABCD是正方形，∴BD⊥AC．…2分∵C1C⊥底面ABCD，BD⊂底面ABCD，∴BD⊥C1C．∵AC ⊂平面A1ACC1，C1C⊂平面A1ACC1，且AC∩C1C＝C，∴BD⊥平面A1ACC1．…5分∵BD⊂平面A1BD，∴平面1A BD⊥平面11A ACC．…7分（Ⅱ）连B1C．…9分在△1A BD中，∵O是BD的中点，M是BA1的中点，∴MO∥A1D．…10分∵A1 B1∥DC，且A1 B1＝DC，∴四边形A1 DC B1为平行四边形．∴ A 1D ∥B 1C ． …12分 ∴ MO ∥B 1C ， 且B 1C ⊂平面11B BCC ，MO ⊄平面11B BCC ，∴ MO //平面11B BCC ． …14分说明：直线在平面内，既可用符号“”表示，也可用符号“⊂”表示，而且应特别让学生知道后一种表示． （18）本小题主要考查运用数学知识解决实际问题的能力．满分12分．解：（1）⎪⎪⎪⎩⎪⎪⎪⎨⎧≤<-≤<-≤<=时当时当时当50002000,1252032000500,251015000,201)(x x x x x x x f ……………6 分（2）∵ 17516525<<，∴ 老李2018年12月份的应纳税金额在500~2000元之间由16525101=-x ，得1900=x ， ……………………9分∴ 老李12月份的工资总收入为3500元，∴ 老李2018年1月份的工资总收入为4200%)201(3500=+⋅（元），应纳税金额为260016004200=-=x （元）， …………11分 ∴ 2651252600203)2600(=-⋅=f （元），即老李2018年1月份应缴纳个人所得税265元（12分） （19）本小题主要考查直线的方程、圆的方程、直线与圆的位置关系等基本知识，考查综合运用数学知识分析和 解决问题的能力．满分14分．解：（Ⅰ）依题意，可设圆C 的方程为()()222x a y b r -+-=，且a 、b 满足方程组()3330,2231 1.3a b b a --⎧++=⎪⎪⎨+⎪⨯-=-⎪+⎩由此解得 0a b ==． …5分 又因为点P (1,1)在圆C 上，所以()()()()222221110102r a b =-+-=+++=．故圆C 的方程为222x y +=． …7分 （Ⅱ）由题意可知，直线PA 和直线PB 的斜率存在且互为相反数，故可设PA 所在的直线方程为1(1)y k x -=-，PB 所在的直线方程为1(1)y k x -=--．由221(1),2y k x x y -=-⎧⎨+=⎩ 消去y ，并整理得222(1)2(1)(1)20k x k k x k ++-+--=． ① …10分 设()11,A x y ，又已知P (1,1)，则1x 、1为方程①的两相异实数根，由根与系数的关系得()2121211k x k --=+，即212211k k x k --=+．同理，若设点B 22(,)x y ，则可得222211k k x k +-=+． …12分于是 12121212(1)(1)AB y y k x k x k x x x x --+-==--＝1212()2k x x k x x +--=1． 而直线OP 的斜率也是1，且两直线不重合，因此，直线OP 与AB 平行． …14分 （20）本小题主要考查函数、方程、不等式等基本知识，考查综合运用数学知识分析和解决问题的能力．满分14 分．解：（Ⅰ）当0a =时，1()2f x x c =-+．由(1)0f =得：102c -+=，即12c =，∴ 11()22f x x =-+．显然x ＞1时，()f x ＜0，这与条件②相矛盾，不合题意． ∴ 0a ≠，函数21()2f x ax x c =-+是二次函数． …2分 由于对一切x ∈R ，都有()0f x ≥，于是由二次函数的性质可得20140.2a ac >⎧⎪⎨⎛⎫≤ ⎪⎪⎝⎭⎩，－－ 即010.(*)16a ac >⎧⎪⎨≥>⎪⎩，…4分 由(1)0f =得 12a c +=，即12c a =-，代入（*）得 11216a a ⎛⎫-≥ ⎪⎝⎭． 整理得 2110216a a -+≤，即2104a ⎛⎫-≤ ⎪⎝⎭． 而2104a ⎛⎫-≥ ⎪⎝⎭，∴ 14a =．将14a =代入（*）得，14c =， ∴ 14a c ==． …7分另解：（Ⅰ）当0a =时，1()2f x x c =-+． 由(1)0f =得 102c -+=，即12c =， ∴ 11()22f x x =-+．显然x ＞1时，()f x ＜0，这与条件②相矛盾，∴ 0a ≠，因而函数21()2f x ax x c =-+是二次函数． …2分 由于对一切x ∈R ，都有()0f x ≥，于是由二次函数的性质可得20140.2a ac >⎧⎪⎨⎛⎫≤ ⎪⎪⎝⎭⎩，－－ 即010.16a ac >⎧⎪⎨≥>⎪⎩， …4分由此可知 a c >>0，0，∴ 22a c ac +⎛⎫≤ ⎪⎝⎭．由(1)0f =，得 12a c +=，代入上式得 116ac ≤． 但前面已推得 116ac ≥， ∴ 116ac =． 由 1,161,2ac a c ⎧=⎪⎪⎨⎪+=⎪⎩ 解得 14a c ==． …7分（Ⅱ）∵ 14a c ==， ∴ 2111()424f x x x =-+． ∴ 2111()()424g x f x m x x m x ⎛⎫=-=-++ ⎪⎝⎭． 该函数图象开口向上，且对称轴为21x m =+． …8分 假设存在实数m 使函数2111()()424g x f x mx x m x ⎛⎫=-=-++ ⎪⎝⎭在区间[],2m m +上有最小值－5． ① 当m ＜－1时，21m +＜m ，函数()g x 在区间[],2m m +上是递增的， ∴ ()g m ＝－5，即21115424m m m ⎛⎫-++=- ⎪⎝⎭， 解得 m ＝－3或m ＝73． ∵ 73＞－1， ∴ m ＝73舍去． …10分② 当－1≤m ＜1时，m ≤21m +＜m ＋1，函数()g x 在区间[],21m m +上是递减的，而在区间[]21,2m m ++上是递增的，∴ ()21g m +＝－5，即()()211121215424m m m ⎛⎫+-+++=- ⎪⎝⎭．解得 m ＝12--m ＝12-+均应舍去． …12分 ③当m ≥1时，21m +≥m ＋2，函数()g x 在区间[],2m m +上是递减的， ∴ ()2g m +＝－5，即()()2111225424m m m ⎛⎫+-+++=- ⎪⎝⎭．解得 m ＝1--m ＝1-+m ＝1--应舍去．综上可得，当m ＝－3或m ＝1-+()()g x f x mx =-在区间[],2m m +上有最小值－5． …14分。

河北省石家庄市2018届高考一模考试数学理科试题（A ）含答案石家庄市2018届高中毕业班模拟考试（一）理科数学（A 卷）一、选择题：本大题共12个小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.已知集合{1,2,3,4,5,6,7}U =，{|3,}A x x x N =≥∈，则U C A =（ ） A ．{1,2} B ．{3,4,5,6,7} C ．{1,3,4,7} D ．{1,4,7}2.已知i 为虚数单位，(1)2i x yi +=+，其中,x y R ∈，则x yi +=（ ）A ．B C ．2 D ．43.函数()2(0)xf x x =<，其值域为D ，在区间(1,2)-上随机取一个数x ，则x D ∈的概率是（ ） A ．12 B ．13 C ．14 D ．234.点B 是以线段AC 为直径的圆上的一点，其中2AB =，则AC AB ⋅=（ ） A ．1 B ．2 C ．3 D ．45. x ，y 满足约束条件：11y xx y y ≤⎧⎪+≤⎨⎪≥-⎩，则2z x y =+的最大值为（ ）A ．-3B ．32C ．3D ．4 6.程序框图如图所示，该程序运行的结果为25s =，则判断框中可填写的关于i 的条件是（ ）A ．4?i ≤B ．4?i ≥C ．5?i ≤ D．5?i ≥7.南宋数学家秦九韶早在《数书九章》中就独立创造了已知三角形三边求其面积的公式：“以小斜幂并大斜幂，减中斜幂，余半之，自乘于上，以小斜幂乘大斜幂减之，以四约之，为实，一为从隅，开方得积.”（即：S =a b c >>），并举例“问沙田一段，有三斜（边），其小斜一十三里，中斜一十四里，大斜一十五里，欲知为田几何？”则该三角形田面积为（ ）A ．82平方里B ．83平方里C ．84平方里D ．85平方里8.如图，网格纸上小正方形的边长为1，粗线画出的是某几何体的三视图，则该几何体的表面积为（ ）A ．83π+B ．84π+C ．85π+D ．86π+9.已知()f x 是定义在[2,1]b b -+上的偶函数，且在[2,0]b -上为增函数，则(1)(2)f x f x -≤的解集为（ ） A ．2[1,]3- B ．1[1,]3- C ．[1,1]- D ．1[,1]310.在ABC ∆中，2AB =，6C π=，则3AC BC +的最大值为（ ）A B ．．．11.过抛物线214y x =焦点F 的直线交抛物线于A ，B 两点，点C 在直线1y =-上，若ABC ∆为正三角形，则其边长为（ ）A ．11B ．12C ．13D ．1412.设xOy ，''x Oy 为两个平面直角坐标系，它们具有相同的原点，Ox 正方向到'Ox 正方向的角度为θ，那么对于任意的点M ，在xOy 下的坐标为(,)x y ，那么它在''x Oy 坐标系下的坐标(',')x y 可以表示为：'cos sin x x y θθ=+，'cos sin y y x θθ=-.根据以上知识求得椭圆223'''5'10x y y -+-=的离心率为（ ）A ．3．4．3 D ．4二、填空题：本大题共4小题，每题5分，共20分.13.命题p ：01x ∃≥，200230x x --<的否定为 ．14.甲、乙、丙三位同学，其中一位是班长，一位是体育委员，一位是学习委员，已知丙的年龄比学委的大，甲与体委的年龄不同，体委比乙年龄小.据此推断班长是 ．15.一个直角三角形的三个顶点分别在底面棱长为2的正三棱柱的侧棱上，则该直角三角形斜边的最小值为 ．16.已知函数31()1x x f x x -+=-，ln ()xg x x =，若函数(())y f g x a =+有三个不同的零点1x ，2x ，3x （其中123x x x <<），则1232()()()g x g x g x ++的取值范围为 ．三、解答题：共70分.解答应写出文字说明、证明过程或演算步骤.第17～21题为必考题，每个试题考生都必须作答.第22、23题为选考题，考生根据要求作答. （一）必考题：共60分17.已知等比数列{}n a 的前n 项和为n S ，且满足122()n n S m m R +=+∈.（Ⅰ）求数列{}n a 的通项公式； （Ⅱ）若数列{}n b 满足211(21)log ()n n n b n a a +=+⋅，求数列{}n b 的前n 项和n T .18.四棱锥S ABCD -的底面ABCD 为直角梯形，//AB CD ，AB BC ⊥，222AB BC CD ===，SAD ∆为正三角形.（Ⅰ）点M 为棱AB 上一点，若//BC 平面SDM ，AM AB λ=，求实数λ的值； （Ⅱ）若BC SD ⊥，求二面角A SB C --的余弦值.19.小明在石家庄市某物流派送公司找到了一份派送员的工作，该公司给出了两种日薪薪酬方案.甲方案：底薪100元，每派送一单奖励1元；乙方案：底薪140元，每日前55单没有奖励，超过55单的部分每单奖励12元. （Ⅰ）请分别求出甲、乙两种薪酬方案中日薪y （单位：元）与送货单数n 的函数关系式；（Ⅱ）根据该公司所有派送员100天的派送记录，发现派送员的日平均派送单数满足以下条件：在这100天中的派送量指标满足如图所示的直方图，其中当某天的派送量指标在2(1)2(,]1010n n-(1,2,3,4,5)n =时，日平均派送量为502n +单.若将频率视为概率，回答下列问题：①根据以上数据，设每名派送员的日薪为X （单位：元），试分别求出甲、乙两种方案的日薪X 的分布列，数学期望及方差；②结合①中的数据，根据统计学的思想，帮助小明分析，他选择哪种薪酬方案比较合适，并说明你的理由. （参考数据：20.60.36=，21.4 1.96=，22.6 6.76=，23.411.56=，23.612.96=，24.621.16=，215.6243.36=，220.4416.16=，244.41971.36=）20.已知椭圆C ：22221(0)x y a b a b+=>>的左、右焦点分别为1F ，2F ，M 为椭圆上任意一点，当1290F MF ∠=时，12F MF ∆的面积为1. （Ⅰ）求椭圆C 的方程；（Ⅱ）已知点A 是椭圆C 上异于椭圆顶点的一点，延长直线1AF ，2AF 分别与椭圆交于点B ，D ，设直线BD 的斜率为1k ，直线OA 的斜率为2k ，求证：12k k ⋅为定值.21.已知函数()()()xf x x b e a =+-，(0)b >，在(1,(1))f --处的切线方程为(1)10e x ey e -++-=. （Ⅰ）求a ，b ；（Ⅱ）若方程()f x m =有两个实数根1x ，2x ，且12x x <，证明：21(12)11m e x x e--≤+-.（二）选考题：共10分，请考生在22、23题中任选一题作答，并用2B 铅笔将答题卡上所选题目对应的题号右侧方框涂黑，按所涂题号进行评分；多涂、多答，按所涂的首题进行评分；不涂，按本选考题的首题进行评分. 22.[选修4-4：坐标系与参数方程]在平面直角坐标系xOy 中，曲线C 的参数方程为cos 1sin x r y r ϕϕ⎧=⎪⎨=+⎪⎩（0r >，ϕ为参数），以坐标原点O 为极点，x 轴正半轴为极轴建立极坐标系，直线l 的极坐标方程为sin()13πρθ-=，若直线l 与曲线C 相切；（Ⅰ）求曲线C 的极坐标方程；（Ⅱ）在曲线C 上取两点M ，N 与原点O 构成MON ∆，且满足6MON π∠=，求面积MON ∆的最大值.23.[选修4-5：不等式选讲]已知函数()f x =R ；（Ⅰ）求实数m 的取值范围；（Ⅱ）设实数t 为m 的最大值，若实数a ，b ，c 满足2222a b c t ++=，求222111123a b c +++++的最小值.石家庄市2017-2018学年高中毕业班第一次模拟考试试题理科数学答案一、选择题1-5: AABDC 6-10: CCDBD 11、12：BA 二、填空题13. 2:1,230p x x x ⌝∀≥--≥ 14. 乙 15. 22,0e e ⎛⎫-⎪-⎝⎭三、解答题 17解：(1) 法一：由122()n n S m m R +=+∈得122()nn S m m R -=+∈，当当2n ≥时，12222n n n n a S S -=-=，即12(2)n n a n -=≥， 又1122ma S ==+，当2m =-时符合上式，所以通项公式为12n n a -=. 法二：由122()n n S m m R +=+∈得1232;4;8()S m S m S m m R =+⎧⎪=+⎨⎪=+∈⎩，从而有2213322,4a S S a S S =-==-=， 所以等比数列公比322a q a ==，首项11a =，因此通项公式为12n n a -=. （2）由（1）可得1212log ()log (22)21n n n n a a n -+⋅=⋅=-，1111()(21)(21)22121n b n n n n ∴==-+--+，12111111(1)2335212121n n nT b b b n n n ∴=+++=-+-++-=-++. 18.（1）因为//BC 平面SDM ，BC ⊂平面ABCD ，平面SDM 平面ABCD=DM ， 所以DM BC //，因为DC AB //,所以四边形BCDM 为平行四边形， 又CD AB 2=,所以M 为AB 的中点. 因为λ=，12λ∴=.（2）因为BC ⊥SD ， BC ⊥CD ，所以BC ⊥平面SCD ， 又因为BC ⊂平面ABCD ， 所以平面SCD ⊥平面ABCD ， 平面SCD平面ABCD CD =，在平面SCD 内过点S 作SE ⊥直线CD 于点E ， 则SE ⊥平面ABCD ， 在Rt SEA 和Rt SED 中， 因为SA SD =，所以AE DE ===，又由题知45EDA ∠=， 所以AE ED ⊥所以1AE ED SE ===， 以下建系求解.以点E 为坐标原点，EA 方向为X 轴，EC 方向为Y 轴，ES 方向为Z 轴建立如图所示空间坐标系，则(0,0,0)E ，(0,0,1)S ，(1,0,0)A ，(1,2,0)B ，(0,2,0)C ，(1,0,1)SA =-，(0,2,0)AB =，(0,2,1)SC =-，(1,0,0)CB =，设平面SAB 的法向量1(,,)n x y z =，则110n SA n AB ⎧⋅=⎪⎨⋅=⎪⎩，所以020x z y -=⎧⎨=⎩，令1x =得1(1,0,1)n =为平面SAB 的一个法向量，同理得2(0,1,2)n =为平面SBC 的一个法向量，12121210cos ,5||||n n n n n n ⋅<>==⋅，因为二面角A SB C --为钝角， 所以二面角A SB C --余弦值为5-.19.解：（1）甲方案中派送员日薪y （单位：元）与送单数n 的函数关系式为： N ,100∈+=n n y ， 乙方案中派送员日薪y （单位：元）与送单数n 的函数关系式为：⎩⎨⎧∈>-∈≤=N),55(,52012N),55(,140n n n n n y ， （2）①由已知，在这100天中，该公司派送员日平均派送单数满足如下表格：所以X 甲的分布列为：所以()=1520.21540.31560.21580.21600.1155.4E X ⨯+⨯+⨯+⨯+⨯=甲，()()()()()222222=0.2152155.4+0.3154155.4+0.2156155.4+0.2158155.4+0.1160155.4=6.44S ⨯-⨯-⨯-⨯-⨯-甲，所以X 乙的分布列为：所以()=1400.51520.21760.22000.1=155.6E X ⨯+⨯+⨯+⨯乙，()()()()22222=0.5140155.6+0.2152155.6+0.2176155.6+0.1200155.6=404.64S ⨯-⨯-⨯-⨯-乙，②答案一：由以上的计算可知，虽然()()E X E X <乙甲，但两者相差不大，且2S 甲远小于2S 乙，即甲方案日工资收入波动相对较小，所以小明应选择甲方案.答案二：由以上的计算结果可以看出，()()E X E X <乙甲，即甲方案日工资期望小于乙方案日工资期望，所以小明应选择乙方案. 20解：（1）设,,2211r MF r MF ==由题122221212224112c e a r r ar r c r r ⎧==⎪⎪+=⎪⎨+=⎪⎪⋅=⎪⎩，解得1a c ==，则21b =，∴椭圆C 的方程为2212x y +=.（2）设0000(,)(0)A x y x y ⋅≠，1122(,),(,)B x y C x y ， 当直线1AF 的斜率不存在时，设2(1,)2A -，则2(1,)2B --， 直线2AF的方程为(1)4y x =--代入2212x y +=，可得25270x x --= 275x ∴=，210y =-，则7(,510D -∴直线BD的斜率为1(10276(1)5k -==--，直线OA的斜率为22k =-121(626k k ∴⋅=-=-， 当直线2AF 的斜率不存在时，同理可得1216k k ⋅=-. 当直线1AF 、2AF 的斜率存在时，10±≠x设直线1AF 的方程为00(1)1y y x x =++，则由0022(1)112y y x x x y ⎧=+⎪+⎪⎨⎪+=⎪⎩消去x 可得：22222200000[(1)2]422(1)0x y x y x y x ++++-+=， 又220012x y +=，则220022y x =-，代入上述方程可得2220000(32)2(2)340x x x x x x ++---=，2000101003434,3232x x x x x x x x ----∴⋅=∴=++，则000100034(1)13232y x y y x x x --=+=-+++ 000034(,)2323x y B x x +∴--++，设直线2AF 的方程为00(1)1y y x x =--，同理可得000034(,)2323x y D x x ---， ∴直线BD 的斜率为000000001220000002323434341224362323y y x x x y x y k x x x x x x +-+===-+--+-+， 直线OA 的斜率为020y k x =， ∴20200001222200001123636366x x y y y k k x x x x -⋅=⋅===----. 所以，直线BD 与OA 的斜率之积为定值16-，即1216k k ⋅=-. 21．解：（Ⅰ）由题意()10f -=，所以()1(1)10f b a e ⎛⎫-=-+-=⎪⎝⎭， 又()()1x f x x b e a '=++-，所以1(1)1b f a e e'-=-=-+， 若1a e=，则20b e =-<，与0b >矛盾，故1a =，1b =.（Ⅱ）由（Ⅰ）可知()()()11x f x x e =+-， (0)0,(1)0f f =-=， 设)(x f 在(-1，0)处的切线方程为)(x h ， 易得，()1()11h x x e ⎛⎫=-+ ⎪⎝⎭，令()()()F x f x h x =- 即()()()1()1111x F x x e x e ⎛⎫=+---+ ⎪⎝⎭，()1()2x F x x e e '=+-， 当2x ≤-时，()11()20x F x x e e e '=+-<-< 当2x >-时，设()1()()2x G x F x x e e'==+-， ()()30x G x x e '=+>， 故函数()F x '在()2,-+∞上单调递增，又(1)0F '-=，所以当(),1x ∈-∞-时，()0F x '<，当()1,x ∈-+∞时，()0F x '>， 所以函数()F x 在区间(),1-∞-上单调递减，在区间()1,-+∞上单调递增， 故0)1()(=-≥F x F ，11()()f x h x ≥，设()h x m =的根为1x '，则111me x e'=-+-， 又函数()h x 单调递减，故111()()()h x f x h x '=≥，故11x x '≤，设()y f x =在(0,0)处的切线方程为()y t x =，易得()t x x =，令()()()()()11x T x f x t x x e x =-=+--，()()22x T x x e '=+-， 当2x ≤-时，()()2220x T x x e '=+-<-<，当2x >-时，故函数()T x '在()2,-+∞上单调递增，又(0)0T '=，所以当(),0x ∈-∞时，()0T x '<，当()0,x ∈+∞时，()0T x '>，所以函数()T x 在区间(),0-∞上单调递减，在区间()0,+∞上单调递增， 0)0()(=≥T x T ，22()()f x t x ≥ ，设()t x m =的根为2x '，则2x m '=，又函数()t x 单调递增，故222()()()t x f x t x '=≥，故22x x '≥， 又11x x '≤，2121(12)1111me m e x x x x m e e -⎛⎫''-≤-=--+=+ ⎪--⎝⎭. 选作题22（1）由题意可知直线l 的直角坐标方程为2y =+，曲线C 是圆心为，半径为r 的圆，直线l 与曲线C 相切，可得：2r ==；可知曲线C 的方程为22((1)4x y +-=，所以曲线C 的极坐标方程为2cos 2sin 0ρθρθ--=， 即4sin()3ρθπ=+. (2)由（1）不妨设M （1,ρθ），)6,(2πθρ+N ，（120,0ρρ>>）6πS MON =∆. 当12πθ=时, 32+≤∆MON S ，所以△MON 面积的最大值为2.23. 【解析】（1）由题意可知32x x m --≥恒成立，令3()2x g x x -=-， 去绝对值可得：36,(3)()263,(03)6,(0)x x x g x x x x x x --≥⎧⎪=-=-<<⎨⎪-≤⎩，画图可知()g x 的最小值为-3，所以实数m 的取值范围为3m ≤-；（2）由（1）可知2229a b c ++=，所以22212315a b c +++++=，222222222111()(123)11112312315a b c a b c a b c ++⋅++++++++++=+++ 22222222222221313239312132315155b a c a c b a b a c b c ++++++++++++++++++=≥=， 当且仅当2221235a b c +=+=+=，即2224,3,2a b c ===等号成立， 所以222111123a b c +++++的最小值为35.。

安徽师大附中2018届高三第一次模拟考试数 学 试 卷(文)本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.满分150分.考试时间120分钟.第Ⅰ卷(选择题 共60分)一、选择题:本大题共12小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符号题目要求的.1．集合{}0,2,A a =,{}21,B a =,若{}0,1,2,4,16AB =,则a 的值为 （ ）A. 0B. 1C. 2D. 42. 若1a→=,b→=a b a →→→⎛⎫-⊥ ⎪⎝⎭,则a →与b →的夹角为 ( ) A. 030 B. 045 C. 060 D. 075 3. 已知函数2,0,()2,0,x x f x x x +⎧=⎨-+>⎩≤则不等式2()f x x ≥的解集为 （ ）A. []1,1-B. []2,2-C. []2,1-D. []1,2-4. 已知等比数列}{n a 的公比为正数，且3a ·9a =225a ，2a =1，则1a = （ ） A.12 B.C.D. 25. 函数sin sin y x x =+的值域是 （ ） A. B. C. D.6. 已知n S 为等比数列{}n a 的前n 项和，21=a ，若数列{}n a +1也是等比数列，则n S 等于 （ ）A. 2nB. 3nC. 122n +-D. 31n-7．函数f (x )＝⎩⎪⎨⎪⎧－x ＋3a ， x <0a x， x ≥0(a >0且a ≠1)是R 上的减函数，则a 的取值范围是( )A ． (0,1)B ． [13，1)C ． (0，13]D ． (0，23]8．设a ∈R ，函数()xxf x e a e-=+⋅的导函数是()f x '，且()f x '是奇函数 ． 若曲线()y f x =的一条切线的斜率是32，则切点的横坐标为 （ ） A ． ln 22- B ． ln 2- C ． ln 22D ． ln 29. ABC ∆为钝角三角形的充分不必要条件是 （ ）()1AB AC CA CB →→→→⎛⎫⎛⎫⋅⋅ ⎪⎪⎝⎭⎝⎭<0 ()2AB AC BA BC →→→→⎛⎫⎛⎫⋅⋅ ⎪⎪⎝⎭⎝⎭<0 ()3BA BC CA CB →→→→⎛⎫⎛⎫⋅⋅ ⎪⎪⎝⎭⎝⎭<0 ()4AB AC BA BC CA CB →→→→→→⎛⎫⎛⎫⎛⎫⋅⋅⋅ ⎪⎪⎪⎝⎭⎝⎭⎝⎭<0 A. （1）（4） B. （2）（4） C. （3）（4） D. （1）（2）（3）10. 设0,0.a b >>1133aba b+与的等比中项，则的最小值为 ( ) A . 8 B . 4 C . 1 D . 1411. 已知点p 在曲线14+=x e y 上，α为曲线在点p 处的切线的倾斜角，则α的取值范围是 （ ) A. 0,4π⎡⎫⎪⎢⎣⎭ B. ,42ππ⎡⎫⎪⎢⎣⎭ C. 3,24ππ⎛⎤ ⎥⎝⎦ D. 3,4ππ⎡⎫⎪⎢⎣⎭12. 已知ω为正实数，函数()2sin f x x ω=在区间,34ππ⎡⎤-⎢⎥⎣⎦上递增，那么 （ ）A. 0<ω≤247B. 0<ω≤2C. 0<ω≤32D. ω≥32安徽师大附中2018届高三第一次模拟考试数学答题卷第Ⅰ卷(选择题 共60分)一、选择题:本大题共12小题，每小题5分，共60分.在每小题给出的四个选项中，只有一第Ⅱ卷(非选择题 共90分)二、填空题:本大题共4小题，每小题5分，共20分.把正确答案填写在题中的横线上. 13. 已知f(x)=sinx ﹣3cosx,则f(x)的最大值为 ．14. 实数,x y 满足不等式组2,24,0,x y x y x y +≥⎧⎪-≤⎨⎪-≥⎩则23x y +的最小值是 ．15. 设f (x )是连续的偶函数，且当x >0时f (x )是单调函数，则满足f (x )＝f (x ＋3x ＋4)的所有x 之和为________．16. 已知数列{}n a 满足1133,2,n n a a a n +=-=则na n的最小值为__________. 三、解答题:本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤. 17．（本小题满分10分）在△ABC 中，角A 、B 、C 的对边分别为a 、b 、c .已知5a b +=，c =7，且.272cos 2sin 42=-+C B A (1) 求角C 的大小； （2）求△ABC 的面积.18．(本小题满分10分)已知函数32()f x x ax b =++的图像在点(1,(1))P f 处的切线为330x y +-=。