【2014潍坊二模】山东省潍坊市2014届高三4月模拟考试 理科4份(语数英理综)Word版含答案

2014年潍坊市高三第二次模拟考试2014.4第I卷（共105分）第一部分英语知识运用（共两节，满分55分）第一节单项填空（共10小题；每小题1.5分，满分15分）1. If you want to speak English without ______ foreigner’s accent, you have to learn ______ correct pronunciation of those words.A. the; aB. a; theC. a; 不填D. the; 不填2. The MH370 from Malaysia crashed in the southern Indian Ocean, ____ global concern about flight safety. A. to cause B. caused C. causing D. having caused3. A1l the candidates are required to be at the Conference Hall at 2:00 pm ______ otherwise informed.A. unlessB. if .C. whileD. though4. Although faced with stress from constant exams, she lost________ of her enthusiasm for study.A. someB. neitherC. noneD. all5. —________you go to the movies with me tonight? —I’d love to.A. WillB. MustC. MightD. Shall6. It is not quite clear_______ our local government will do with air pollution.A. whetherB. howC. whatD. that7. —How long have you been here , Alice? —Only a few minutes. Susan_______ me here.A. walksB. walkedC. has walkedD. had walked8. Left_______ in the reading room alone quite by accident, the boy felt nervous.A. lockingB. having lockedC. to lockD. locked9. I'd like to place things__________ it is convenient to get them.A. in whichB. whatC. whereD. that10.—I guess you find the speech boring. —_______. I like it.A. Not in the leastB. Don't mention itC. No doubtD. I couldn't agree more第二节完形填空(共30小题;11-20题每小题1分, 21-40小题 1.5分, 满分40分)AA few weeks ago, I was taking a quiet walk in my backyard 11 I saw a baby bird lying in the grass. It must have fallen off the nest high in the branches. I was sure it was dead. But as I tried to pick it up, 12 it moved.To put it back, I placed a ladder 13 the tree and tried to get it near the 14 with the nest. It was then that I saw a neighborhood cat in the leaves. Immediately it came to me that he must have 15 the bird out of the nest when climbing up along the branches and 16 himself up in the tree as well. Climbing up through the branches, I seized him, gave him a 17 “ Hiss!” and dropped him softly tothe ground. He ran off as fast as he could.I gently 18 the baby bird, climbed up the ladder and, stretching my arm out as far as I could,19 it in the nest. Then I said a little prayer for it.In the weeks that followed I looked up at the nest several times ,but only saw the mother bird. I may never know whether my efforts have made a 20 or not. But trying was enough. Sometimes love has to be its own reward.11. A. while B. as C. after D. when12. A. suddenly B. finally C. nearly D. actually13. A. off B. against C. on D. beside14. A. tree B. leaves C. wall D. branch15. A. led B. shaken C. tricked D. blown16. A. trapped B. wounded C. seized D. stored17. A. worrying B. puzzling C. scolding D. surprising18. A. tried on B. picked up C. took out D. turned over19. A. placed B. caught C. fed D. threw20. A. choice B. decision C. difference D. conclusionBA 58-year-old man was sharing his memories of his father. He said his father had been a workaholic who was always busy with his 21 . He 22 came to his son's sports games or activities. 23 , the son said his dad almost never spoke to him. 24 the time the son was 18 years old, in his last year of school, he’d become a very good football player. When his team qualified for the championship, he 25 his father to finally come to see him play this one time; so his dad 26 he'd be there.On the day of the 27 , the son was on the field warming up just 28 the game when he saw his dad come into the stadium with two other men. Then shortly after the game started, his dad 29 with his friends and never returned. It's now been 40 years 30 that day, yet this 58-year-old son says the painful 31 he felt as a teenager is still very real. When his father recently died at age 83, he stood alone next to his dad's coffin at the funeral home and said: “Dad, we could have 32 so much love and good times together, 33 I never knew you.”A father has great power and 34 in the lives of his children. Studies have shown that the No. 135 of troubled boys and young men is that they had fathers who didn’t 36 care. They didn't spend time with their sons so they weren’t 37 to them. There are millions of invisible 38 like this. But “No man stands as tall as when he bends down to help his son.” How about you? Do you need to give more time and 39 to your children? It's never too late to 40 .21. A. family B. work C. hobby D. housework22. A. often B. ever C. even D. never23. A. In fact B. As a result C. Of course D. Above all24. A. At B. In C. By D. For25. A. begged B. advised C. permitted D. ordered26. A. admitted B. pretended C. suspected D. promised27. A. ceremony B. graduation C. championship D. concert28. A. after B. on C. before D. during29. A. stood B. left C. chatted D. sat30. A. since B. until C. after D. before31. A. relief B. disappointment C. fear D. anxiety32. A. lacked B. shared C. received D. mixed33. A. and B. but C. so D. then34. A. trust B. favor C. influence D. courage35. A. reason B. case C. cause D. concern36. A. simply B. only C. initially D. really37. A. close B. similar C. polite D. known38. A. mothers B. sons C. fathers D. players39. A. money B. freedom C. attention D. award40. A. learn B. change C. develop D. adapt第二部分阅读理解（共25小题；每小题2分，满分50分）AIf you didn’t look at them, they weren’t there, right? On seeing those in rags on the corner, I only wanted the light to turn green fast enough so that I wouldn't have to keep pretending not to see them.Then, one day, as I was holding the hand of my best friend Jane who was young but dying of breast cancer, she told me that she made every moment count by slowing down and by seeing everything. I held her hand for five years and then she passed away. She did teach me something. It took a while for her words to really sink in. I can be a slow learner.I started by seeing everything, and focused on their presence. One day I saw a Vietnam veteran(老兵). I asked him what would make his day.“A hot cup of coffee,” he told me.I bought him a cup of coffee a stack of pancakes, some eggs, and so on. After our meal together, I asked him if there was anything else that would make his life a bit easier.“A new pair of socks,” he said.Socks, really? I actually happened to be wearing a nice pair of wool hiking socks at that very moment. I told him that I wanted to give him the pair I was wearing, if he would accept it. Finally, he agreed. We sat down on a bench, and he started to first take off his boots, then remove the black socks that had once been white off his feet. I think a layer or two of skin might have come off with them. Taking his new pair of socks, he held the socks up to his cheeks and said they were warm and smelled as good as me, pools of tears in his eyes.Such a simple luxury I used to take for granted. Now, I always have an extra pair with me in my car. They are always my best pair, just waiting to be given away.41. On seeing the homeless, the author used to_______.A. hide in the cornerB. show mercy to themC. act as if they were not thereD. make fun of them42. What can we learn about the author from Paragraph 2?A. He was slow in learning things.B. His best friend was once a teacher.C. He was sad that his friend was dying.D. He didn't understand Jane's words at first.43. What does the underlined word “them” in the last paragraph but one refer to?A. The veteran's boots.B. The veteran’s feet.C. The wool hiking socks.D. The veteran's old socks.44. Why does the author always have an extra pair of socks in his car?A. To decorate his car.B. To sell it to others.C. To keep it for a change.D. To donate it to a needy person.45. What does the author want to convey to us?A. All men are born equal.B. A beggar’s purse is bottomless.C. Don’t miss doing any good thing.D. Don’t trouble until trouble troubles you.BAmerican high school students are terrible writers, and one education reform group thinks it has an answer: robots. Or, more accurately, robot-readers-computers programmed to scan student essays and spit out a grade.Mark Shermis, professor of the College of Education at the University of Akron, is helping to hold a contest, set up by the William and Flora Hewlett Foundation ( WFHF), that promises $ 100,000 in prize money to programmers who write the best automated grading software. “If you’re a high school teacher and you give a writing task, you're walking home with 150 essays,” Shermis said. “You're going to need some help.”Automated essay grading was first proposed in the 1960s, but computers back then were not up to the task. In the late 1990s, as technology improved, several textbook and testing companies jumpedinto the field. Today, computers are used to grade essays on South Dakota’s student writing assessments and a handful of other exams, including the TOEFL test of English fluency, taken by foreign students.The Hewlett contest aims to show that computers can grade as well as English teachers- only much more quickly and without all that depressing red ink. Automated essay scoring is “objective,”Shermis said. “And it can be done immediately. If students finish an essay at l0 pm, they get a result at 10:0l pm.”Take, for instance, the Intelligent Essay Assessor, a web-based tool marketed by Pearson Education, Inc. Within seconds it can analyze an essay for spelling, grammar, organization, and help students to make revisions. The program scans for key words and analyzes semantic(语义的) patterns, and Pearson claims that it can understand the meaning of text much the same as a human reader.46. The text is written to introduce________.A. robot-readersB. education reform in AmericaC. Hewlett contestD. William and Flora Hewlett Foundation47. What does the underlined phrase "spit out "in Paragraph I probably mean?A. Give.B. Organize.C. Analyze.D. Check.48. From Paragraph 3, we know that in the 1960s _______.A. computers were not easy to getB. automated grading software was popularC. people refused automated essay gradingD. computers couldn't grade essays automatically49. What does Paragraph 4 focus on?A. The prize of Hewlett contest.B. The advantages of automated essay scoring.C. The application of automated essay scoring.D. Teachers' opinions about Hewlett contest.50. The Intelligent Essay Assessor can________.A. rewrite essaysB. underline the mistakes in red inkC. understand the meaning of textD. correct key words and patternsCIn order to increase their job chances after college, Chinese students are turning to a special practice —Eiffel Tower nose jobs (鼻整形术). The latest trend in plastic surgery promises to create a nose that is classic, slim and sloping, similar to the sweeping curve of the Eiffel Tower.Surgeon Wang Xuming said: “We are influenced by the beauty of the Eiffel Tower, we are not content to just add something to the nose, we reconstruct it.” The surgery costs about US $ 10,000 and involves the enlarging of the nose using tissue from the forehead.Hundreds of posters advertising the procedure are put up all over Chongqing city, where surgeon Xuming runs his practice. They show a Western-looking woman with an almost-too- perfect nose, against an outline of the Eiffel Tower.Interestingly, many young women in China are eager to achieve a western appearance, as theybelieve it will give them an advantage in the highly competitive job market. “Some students face a lot of employment pressure after graduation. If their facial features are good, they’ll have more chances of finding a job,”said surgeon Xuming. “We've had students getting the Eiffel Tower nose; it’s helped them a lot.”Apparently, Chinese employers are quite particular about appearances and prefer attractive candidates. Some of them even go as far as putting height and weight requirements in their employment ads. Plastic surgeons across the country are reporting an increase in the number of students choosing beauty “improvement”.According to a Mr. Li, hospital manager at surgeon Xuming’s clinic, most of their customers are female and the bill is taken care of by the family. “They usually come in with their mothers, and tend to be from well-off backgrounds,” he said.51. Chinese college students choose to have a nose operation to_______.A. marry wellB. look coolC. have a good jobD. look like westerners52. What is mainly talked about in Paragraph 3?A. The posters.B. A nose job.C. The Eiffel Tower.D. A plastic surgery procedure.53. It can be inferred from Paragraph 5 that________.A. employers are potential customersB. ads promote the plastic surgeryC. appearance is as important as height and weightD. attractive appearance seems to increase job chances54. From what Mr. Li said we can learn that ________.A. most families can't afford the surgeonB. the number of plastic surgeons is increasingC. patients can be well looked after at the clinicD. their customers are usually from wealthy families55.What’s the author's attitude towards Eiffel Tower nose jobs?A. Skeptical.B. Objective.C. Critical.D. Worried.DLaw enforcement agencies across the United States are using cameras to take pictures of automobile license plates. The idea is to build a computerized collection of information detailing the daily travel of millions of Americans.Detective Mohammed Tabibi uses a license plate reader, also known as a LPR to look for stolen vehicles. “It has somewhat paid. We have caught some people involved in some serious crimes because of the LPR,” said Tabibi.The use of LPRs is growing across the United States. Some are fixed to poles along Roadsides, others are placed in law-enforcement vehicles. Privacy groups are concerned about the growing use ofthese devices. Jay Stanley is with the American Civil Liberties Union (ACLU). “What they are also doing is storing everybody's time, place, and location. And many police departments are holding that information indefinitely. You know in our society, the government doesn’t follow you and invade your privacy and track you unless it has a specific reason that you are involved in wrongdoing,” said Stanley.Kevin Rearden served as Captain of the Arlington county Police, who also headed the county’s Homeland Security Department before he retired. He said county policy called for the LPR information to be kept for six months. “We originally had a two-month period, and the detectives requested the chief extend it to six months because they found in so many investigations, keeping it for two months wasn't long enough,” said Rearden, “but other law enforcement agencies that use the information may store it for unlimited periods of time.”Supporters of privacy rights say they have no problem with police departments taking pictures of license plates to investigate crimes. But Jay Stanley says they're against storing the information for long periods of time.However, Captain Rearden disagrees, “They keep saying the word ‘tracking’. I would be lucky if I could find you in a few places in Arlington at a particular time. By no stretch of the imagination wouldI be able to track you.”56. What does the text mainly tell us?A. The use of LPRs is controversial.B. Privacy protection is important.C. LPRs help investigate crimes.D. ACLU is involved in wrongdoing.57. The function of a LPR itself is to_______.A. give directionsB. take picturesC. identify locationD. analyze information58. Jay Stanley agrees that________.A. tracking people all the time costs too muchB. using LPRs is likely to invade one's privacyC. police are not allowed to store private informationD. every citizen is supposed to support the government59. How long was the LPR information kept in the beginning?A. Two months.B. Six months.C. Two years.D. Unlimited periods of time.60. According to the last paragraph, Kevin Rearden thinks that_______.A. tracking is almost impossibleB. imagination never endsC. he is lucky to live in ArlingtonD. time and place are both important in trackingEItalians were reminded to slow down and relax on World Slow Day, an annual event celebrating life's simple pleasures.“Let’s take this day to stop and think about all the things we miss while we're rushing through our lives,” said Bruno Contigiani, the President of the Art of Living Slowly Association. Contigiani, 62, aone-time high-powered manager, is now an ambassador for the slow life movement around the world. He started the first World Slow Day in 2007 to encourage the values of living and working at a more natural pace, and to make people rethink their daily lifestyle.Contigiani’s association suggests “14 commandments (诫条)” for living better, such as waking up five minutes earlier to enjoy breakfast without rushing. Others include walking whenever possible, and reading in the evenings instead of watching television.This year, Contigiani left Italy where the event has spread around the country for Shanghai, one of the fastest moving cities in the world. The slow-living supporter said he wandered around the busiest streets of the commercial center for an entire afternoon, inviting people to “slow down”. “Among the 14 commandments, the one about waking up five minutes earlier was the most popular by far,”Contigiani said.Back in Italy, the now well-known event hosts a lot of activities, such as reminding people to stop and smell the roses. In central Milan: you would probably be fined if you walk too fast. In parks and public spaces, free yoga and Tai Chi (太极拳) lessons are important parts of the events.Italian farmers' union noted that Italians spent less and less time preparing meals, a habit connected to rising levels of obesity. Therefore, the group said World Slow Day was a good opportunity to remind Italians to take extra time at meals in particular.World Slow Day is by now an international event. A total of 90 “Slow Cities”in 11 countries inspired by the “live well” philosophy are supporting the day.61. What's the best title of the text?A. Slow Life SupportersB. Lifestyle of ItalyC. World Slow DayD. Living Well62. World Slow Day is intended to__________.A. tell people to walk slowlyB. encourage people to enjoy pleasant thingsC. advise people to adjust to modern lifestyleD. remind people to live and work at a natural pace63. Which of the following goes against the “commandments”?A. Walking whenever possible.B. Slowing down to smell flowers.C. Watching TV in the evenings instead of reading.D. Waking up five minutes earlier to enjoy breakfast.64. We know from Paragraph 6 that .A. obesity had little to do with eating quicklyB. free yoga in particular is popular with ItaliansC. Italian farmers' union supported World Slow DayD. Italians spent more and more time preparing meals65. In which section of a newspaper is the text probably put?A. Advertisement.B. Culture.C. Business.D. Entertainment.第Ⅱ卷（共45分）第一节阅读表达（共5小题；每小题3分，满分15分）阅读下面短文并回答问题，然后将答案写到答案纸相应的位置上（请注意题后的词数要求）[1] Calling a taxi is easy ---- you stand in the street and stretch out your hand. But have you tried to call a taxi using a mobile application (软件)? It’s not hard ---- you download a taxi-calling application, input your location and destination, click “calli ng” and wait for a taxi to pick you up. Didi taxi and Kuaidi Taxi are two of the most popular taxi apps.[2] However, what appeals to people about the apps is not only convenience but also potential prizes. In early January, Didi announced that for each successful ride arranged over the app, both passengers and drivers get 10 yuan subsidies (补贴). A few days later, Kuaidi decided to offer 10 yuan for passengers and 15 yuan for drivers per ride.[3] However popular the taxi-calling apps may be, questions have arisen. One big concern is safety. In order to claim as many subsidies as possible, some drivers constantly check the apps to grab more taxi call s, even while they're driving. “It greatly distracts drivers’ attention, thus causing a threat to road safety,” Beijing Review commented. Meanwhile, like many other digital devices, apps may be quite handy for tech-savvy young people, but it could bring a challenge for senior citizens who don't know how to use them, commented China Daily. “As drivers prefer taking taxi orders from apps, they will overlook those p eople standing on the street,” Beijing Review continued.[4] “Taxi-calling apps are a new thing and we should observe them carefully,” said Liu Xiaohua, deputy director of Guangdong Transport Department. “Authorities are expected to regulate the market when they cause saf ety issues or market disorder.”[5] Some cities________________________. Beijing said last month that each taxi is only allowed to be linked with one taxi-calling app, China Daily reported. In Shanghai, drivers will be banned from using these apps during rush hour starting on March l. The city also decided to include these taxi apps in its own taxi-calling service platform.66. What's the main idea of the text? ( no more than 15 words)67. Complete the following sentence using a proper word from Paragraph 2.Apart from saving us money, calling a taxi using a mobile app also brings us__________________. 68. What is Paragraph 3 mainly about? ( no more than 5 words)69. What does the underlined word "they" in Paragraph 4 refer to?70. Fill in the blank in Paragraph 5 with proper words. ( no more than 5 words)第二节写作（满分30分）假设你是李华，你的英国笔友Mike即将完成学业，不久前来信就是否到中国工作询问你的意见。

2014年山东省潍坊市高三考点回扣即高考模拟训练（四）英语试题第I卷(共105分)第一部分英语知识应用(共两节，满分55分)第一节单项填空(共10小题；每小题1．5分，满分15分)从A、B、C、D四个选项中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。

1．—Oh，my God! I've just broken the vase that my mother likes best．—______________．It can’t be helped．A．That’s fine B．All right C．Never mind D．Not at all2．Paul fell down and broke his wrist，____________ was a pity．A．which B．what C．that D．it3. I must admit you _____________ but I'm afraid the answer is still no．A．will try B．try C．have tried D．had tried4．Lora，you ____________ leave all your books on the floor like this!A．wouldn’t B．mustn’t C．needn’t D．may not5．Mr．Black is willing to do ______________ Mrs．Black tells him to do．A．whichever B．however C．whatever D．whenever6．While reading an interesting story book in my room，______________.A．the light suddenly went out B．the doorbell rangC．someone was shouting outside D．I heard someone shouting outside7．—Is Tom still smoking?—No．By next Saturday he _____________ for a whole month without smoking a single cigarette.A．will go B．will have gone C．goes D．has been going8．Shanghai is the first city in the world _____________ a high-speed maglev train，from the city to Pudong Airport．A．building B．having built C．to have built D. built9．Mother was busy washing dishes ________________ Father was watching TV．A．until B．if C．since D．while10．—I was worried about my math，but Mr．Brown gave me an A.A．Don’t worry about it． B．Congratulations! That’s a difficult course．C．Mr．Brown is very good．D．Good luck to you!第二节完形填空(共两篇，满分40分)阅读下面短文，从短文后各题所给的四个选项(A、B、C和D)中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。

第I 卷(选择题，共50分)一、选择题：本大题共10小题．每小题5分。

共50分．把正确答案涂在答题卡上.1.已知集合{}{}(){}*2,4124,,,,log x A B C x y x A y B y N ===∈∈∈，，，且，则C 元素个数是A.2B.3C.4D.5 2.已知()():230p x a x x p q -<4;-->⌝⌝，若是的充分不必要条件，则实数a 的取值范围A. 16a a <->或B. 16a a <-≥或C. 16a -≤≤D. 16a -<<3.已知向量()()cos ,2,sin ,1//tan 4a b a b πααα⎛⎫=-=-⎪⎝⎭，且，则等于A.3B. 3-C. 13D. 13- 4.执行右图的程序框图，任意输入一次()()0101x x y y ≤≤≤≤与，则能输出数对(),x y 的概率为 A. 14B.13 C. 23 D. 345.下列说法正确的个数是①“在ABC ∆中，若sin sin A B >>，则A B ”的逆命题是真命题；②“1m =-”是“直线()2110mx m y +-+=和直线320x my ++=垂直”的充要条件；③“三个数,,a b c 成等比数列”是“b =④命题“32,10x R x x ∀∈-+≤”的否定是“33000,10x R x x ∃∈-+>”A.1B.2C.3D.46.已知各项均不为零的数列{}n a ，定义向量()()*1,,,1,n n n n c a a b n n n N +==+∈，则下列命题中是真命题的是A.若对任意的*n N ∈，都有//n n c b 成立，则数列{}n a 是等差数列B.若对任意的*n N ∈，都有//n n c b 成立，则数列{}n a 是等差数列C.若对任意的*n N ∈，都有n n c b ⊥成立，则数列{}n a 是等差数列D.若对任意的*n N ∈，都有n n c b ⊥成立，则数列{}n a 是等比数列7.已知非零向量AB AC与满足102AB AC AB AC BC AB AC AB AC ⎛⎫ ⎪+⋅=⋅= ⎪⎝⎭，且，则ABC ∆为 A.等腰非等边三角形 B.等边三角形C.三边均不相等的三角形D.直角三角形 8.一个篮球运动员投篮一次得3分的概率为a ，得2分的概率为b ，不得分的概率为(),,0,1c a b c ∈⎡⎤⎣⎦，，已知他投篮一次得分的期望是2，则213a b+的最小值为 A. 323 B. 283 C. 143 D. 1639.设不等式组4,010x y x x y +≤⎧⎪-≥⎨⎪-≥⎩表示的平面区域为 D.若圆()()()222:110C x y r r +++=>经过区域D 上的点，则r的取值范围是A. ⎡⎣B.⎡⎣ C. (0, D. ( 10.设()f x 是定义在R 上的偶函数，对x R ∈，都有()()[]22,2,0f x f x x -=+∈-且当时，()112x f x ⎛⎫=- ⎪⎝⎭，若在区间(]2,6-内关于x 的方程()()()log 201a f x xa -+=>恰有3个不同的实数根，则a 的取值范围是A. ()1,2B. ()2,+∞C. (D. )2 第II 卷（非选择题，共100分）二、填空题：本大题共5小题，每小题5分，共25分.把正确答案填在答题卡相应的位置上.11.复数2a i i+-在复平面内所对应的点在实轴上，那么实数a =___________.12.若()5224100125321x a a x a x a x a +=+++⋅⋅⋅+，则的值为____________.13.函数()tan 0y x y a ωω=>=与直线相交于A ，B 两点，且AB 最小值为π，则函数()cos f x x x ωω=-的单调增区间是___________.14.如图，12,F F 是椭圆221:14x C y +=与双曲线2C 的公共焦点，A ，B 分别是12,C C 在第二、四象限的公共点.若四边形12AF BF 为矩形，则2C 的离心率是_________.15.关于函数()()21lg 0x f x x x+=≠，有下列命题： ①其图象关于y 轴对称；②当()0x f x >时，是增函数；当()0x f x <时，是减函数；③()f x 的最小值是lg 2；④()f x 在区间()()1,02,-+∞、上是增函数；⑤()f x 无最大值，也无最小值.其中所有正确结论的序号是_____________.三、解答题：本大题共6小题，共75分.解答应写出必要的文字说明、证明过程或演算步骤.16.（本小题满分12分）已知锐角ABC ∆中内角A 、B 、C 的对边分别为2226cos ,sin 2sin sin a b c a b ab C C A B +==、、，且.（I ）求角C 的值；（II ）设函数()()sin cos 06f x x x πωωω⎛⎫=--> ⎪⎝⎭，且()f x 图象上相邻两最高点间的距离为π，求()f A 的取值范围.17.（本小题满分12分）李先生家住H 小区，他工作在C 科技园区，从家开车到公司上班路上有12L L 、两条路线（如图），1L 路线上有123A A A 、、三个路口，各路口遇到红灯的概率均为12；2L 路线上有12B B 、两个路口，各路口遇到红灯的概率依次为3345，.（I ）若走1L 路线，求最多遇到1次红灯的概率；（II ）若走2L 路线，求遇到红灯次数的X 的数学期望；（III ）按照“平均遇到红灯次数最少”的要求，请你帮助李先生从上述两条路线中选择一条最好的上班路线，并说明理由.18.（本小题满分12分）如图，在底面是正方形的四棱锥P ABCD PA -⊥中，面ABCD ，BD 交AC 于点E ，F 是PC 中点，G 为AC 上一点.（I ）求证：BD FG ⊥；（II ）确定点G 在线段AC 上的位置，使FG//平面PBD ，并说明理由；（III ）当二面角B PC D --的大小为23π时，求PC 与底面ABCD 所成角的正切值.19.（本小题满分12分）已知数列{}n a 是首项为111,44a q ==公比的等比数列，设()*1423log n n b a n N +=∈，数列{}n c 满足n n n c a b =⋅.（I ）求数列{}n c 的前n 项和n S ；（II ）若2114n c m m ≤+-对一切正整数n 恒成立，求实数m 的取值范围. 20.（本小题满分12分）以椭圆()2222:10x y C a b a b+=>>的中心O 为半径的圆称为该椭圆的“准圆”.设椭圆C 的左顶点为P ，左焦点为F ，上顶点为Q ，且满足2,OFQ PQ S OPQ S ∆∆==. （I ）求椭圆C 及其“准圆”的方程；（II ）若椭圆C 的“准圆”的一个弦ED （不与坐标轴垂直）与椭圆C 交于M 、N 两点，试证明：当0OM ON ⋅=时，试问弦ED 的长是否为定值，若是，求出该定值；若不是，请说明理由.21.（本小题满分12分）已知函数()()()211,ln .f x a x x g x x =-+-=（I ）若()()()()1,0a F x g x f x ==-+∞求在，上的最大值； （II ）证明：对任意的正整数n ，不等式()23412ln 149n n n ++++⋅⋅⋅+>+都成立； （III ）是否存在实数()0a a >，使得方程()()()21141,g x f x a e x e ⎛⎫'=+-- ⎪⎝⎭在区间内有且只有两个不相等的实数根？若存在，请求出a 的取值范围；若不存在，请说明理由.。

高三数学(文)2014．04本试卷共5页，分第I 卷(选择题)和第Ⅱ卷(非选择题)两部分．共150分．考试时间120分钟．第I 卷(选择题 共50分)注意事项：1．答卷前，考生务必用0．5毫米黑色签字笔将自己的、号、考试科目填写在规定的位置上。

2．第I 卷每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号。

3．第Ⅱ卷必须用0．5毫米黑色签字笔作答，答案必须写在答题卡各题目指定区域相应的位置；如需改动，先划掉原来的答案，然后再写上新的答案，不得使用涂改液，胶带纸、修正带和其他笔。

4．不按以上要求作答以及将答案写在试题卷上的，答案无效。

一、选择题：本大题共10小题。

每小题5分，共50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.若复数z 满足()1i z i z +=，则的虚部为 A.2i - B.12- C.2i D.122.已知集合{}(){}2210,l 10,A x x B x ox A B g =-≤=-≤⋂=则 A.[]0,2 B.(]0,2 C.(]1,2D.()1,2 3.下列结论正确的是A.若向量a//b ，则存在唯一的实数a b λλ=使B.已知向量,a b 为非零向量，则“,a b 的夹角为钝角”的充要条件是“0a b •<”C.“若3πθ=，则1cos 2θ=”的否命题为“若132πθθ≠≠，则cos ” D.若命题22:,10:,10p x R x x p x R x x ∃∈-+<⌝∀∈-+>，则4.为了调查学生携带手机的情况，学校对高一、高二、高三三个年级的学生进行分层抽样调查.已知高一有学生1000人、高二有1200人；三个年级总共抽取了66人，其中高一抽取了20人，则高三年级的全部学生数为A.1000B.1100C.1200D.13004.已知()()()21sin ,42f x x x f x f x π⎛⎫'=++ ⎪⎝⎭为的导函数，则()'y f x =图象大致是6.已知,αβ表示平面，,m n 表示直线，,m βαβ⊥⊥，给出下列四个结论；①,n n αβ∀⊂⊥；②,n m n β∀⊂⊥；③,//n m n α∀⊂；④,n m n α∃⊂⊥. 则上述结论中正确的个数为A.1B.2C.3D.47.已知函数()2f x x x =+，执行右边的程序框图，若输出的结果是3132，则判断框中的条件应是A. 30n ≤B. 31n ≤C. 32n ≤D. 33n ≤ 8.已知双曲线()2222:10x y C a b a b-=>0,>的左、右焦点分别是12F F 、，过2F 垂直x 轴的直线与双曲线C 的两渐近线的交点分别是M 、N ，若1MF ∆N 为正三角形，则该双曲线的离心率为A.21B.3C.13D.23+9.某几何体的三视图如图所示，则该几何体外接球的表面积为A.43π B.323π C.4π D.16π10.已知定义在R 上的函数()y f x =对任意的x 满足()()1,11f x f x x +=--≤<当时，()3f x x =.函数()1,0,1,0a og x x g x x x⎧>⎪=⎨-<⎪⎩，若函数()()()[)6h x f x g x =--+∞在，上有6个零点，则实数a 的取值围是A.()1077⎛⎫⋃+∞ ⎪⎝⎭，，B.(]117997⎡⎫⋃⎪⎢⎣⎭，，C.(]11199⎡⎫⋃⎪⎢⎣⎭，，D.[)117997⎛⎤⋃ ⎥⎝⎦，， 第II 卷（非选择题 共100分）注意事项：将第II 卷答案用0.5mm 的黑色签字笔答在答题卡的相应位置上.二、填空题：本大题共5小题，每小题5分，共25分.11.已知12,e e 是夹角为60的两个单位向量.若向量1232a e e =+，则a =________。

山东省潍坊市2014届高三数学考点回扣即模拟训练试题 理〔四〕第I 卷(选择题，共50分)一、选择题：本大题共10小题．每一小题5分。

共50分．把正确答案涂在答题卡上. 1.集合{}{}(){}*2,4124,,,,logxA B C x y x A y B y N ===∈∈∈，，，且，如此C 元素个数是A.2B.3C.4D.52.()():230p x a x x p q -<4;-->⌝⌝，若是的充分不必要条件，如此实数a 的取值范围 A.16a a <->或 B.16a a <-≥或 C.16a -≤≤ D.16a -<<3.向量()()cos ,2,sin ,1//tan 4a b a b πααα⎛⎫=-=- ⎪⎝⎭，且，则等于 A.3B.3-C.13D.13- 4.执行右图的程序框图，任意输入一次()()0101x x y y ≤≤≤≤与，如此能输出数对(),x y 的概率为A.14B.13 C.23D.345.如下说法正确的个数是①“在ABC ∆中，假设sin sin A B >>，则A B 〞的逆命题是真命题；②“1m =-〞是“直线()2110mx m y +-+=和直线320x my ++=垂直〞的充要条件； ③“三个数,,a b c 成等比数列〞是“b ac =〞的既不充分也不必要条件；④命题“32,10x R x x ∀∈-+≤〞的否认是“33000,10x R x x ∃∈-+>〞A.1B.2C.3D.46.各项均不为零的数列{}n a ，定义向量()()*1,,,1,n n n n c a a b n n n N +==+∈，如此如下命题中是真命题的是A.假设对任意的*n N ∈，都有//n n c b 成立，如此数列{}n a 是等差数列 B.假设对任意的*n N ∈，都有//n n c b 成立，如此数列{}n a 是等差数列C.假设对任意的*n N ∈，都有n n c b ⊥成立，如此数列{}n a 是等差数列 D.假设对任意的*n N ∈，都有n n c b ⊥成立，如此数列{}n a 是等比数列7.非零向量AB AC 与满足102AB AC AB AC BC AB AC AB AC ⎛⎫ ⎪+⋅=⋅= ⎪⎝⎭，且，如此ABC ∆为A.等腰非等边三角形B.等边三角形C.三边均不相等的三角形D.直角三角形8.一个篮球运动员投篮一次得3分的概率为a ，得2分的概率为b ，不得分的概率为(),,0,1c a b c ∈⎡⎤⎣⎦，，他投篮一次得分的期望是2，如此213a b+的最小值为 A.323 B.283 C.143D.1639.设不等式组4,010x y x x y +≤⎧⎪-≥⎨⎪-≥⎩表示的平面区域为D.假设圆()()()222:110C xy r r +++=>经过区域D 上的点，如此r 的取值范围是A.⎡⎣B.⎡⎣C.(0, D. (10.设()f x 是定义在R 上的偶函数，对x R ∈，都有()()[]22,2,0f x f x x -=+∈-且当时，()112xf x ⎛⎫=- ⎪⎝⎭，假设在区间(]2,6-内关于x 的方程()()()log 201a f x x a-+=>恰有3个不同的实数根，如此a 的取值范围是 A.()1,2 B.()2,+∞ C.( D.)2第II 卷〔非选择题，共100分〕二、填空题：本大题共5小题，每一小题5分，共25分.把正确答案填在答题卡相应的位置上. 11.复数2a ii+-在复平面内所对应的点在实轴上，那么实数a =___________. 12.假设()5224100125321x a a x a x a x a +=+++⋅⋅⋅+，则的值为____________.13.函数()tan 0y x y a ωω=>=与直线相交于A ，B 两点，且AB 最小值为π，如此函数()3sin cos f x x x ωω=-的单调增区间是___________.14.如图，12,F F 是椭圆221:14x C y +=与双曲线2C 的公共焦点，A ，B 分别是12,C C 在第二、四象限的公共点.假设四边形12AF BF 为矩形，如此2C 的离心率是_________.15.关于函数()()21lg 0x f x x x+=≠，有如下命题：①其图象关于y 轴对称；②当()0x f x >时，是增函数；当()0x f x <时，是减函数； ③()f x 的最小值是lg 2；④()f x 在区间()()1,02,-+∞、上是增函数； ⑤()f x 无最大值，也无最小值.其中所有正确结论的序号是_____________.三、解答题：本大题共6小题，共75分.解答应写出必要的文字说明、证明过程或演算步骤. 16.〔本小题总分为12分〕锐角ABC ∆中内角A 、B 、C 的对边分别为2226cos ,sin 2sin sin a b c a b ab C C A B +==、、，且. 〔I 〕求角C 的值；〔II 〕设函数()()sin cos 06f x x x πωωω⎛⎫=--> ⎪⎝⎭，且()f x 图象上相邻两最高点间的距离为π，求()f A 的取值范围.17.〔本小题总分为12分〕李先生家住H 小区，他工作在C 科技园区，从家开车到公司上班路上有12L L 、两条路线〔如图〕，1L 路线上有123A A A 、、三个路口，各路口遇到红灯的概率均为12；2L 路线上有12B B 、两个路口，各路口遇到红灯的概率依次为3345，.〔I 〕假设走1L 路线，求最多遇到1次红灯的概率； 〔II 〕假设走2L 路线，求遇到红灯次数的X 的数学期望；〔III 〕按照“平均遇到红灯次数最少〞的要求，请你帮助李先生从上述两条路线中选择一条最好的上班路线，并说明理由.18.〔本小题总分为12分〕如图，在底面是正方形的四棱锥P ABCD PA -⊥中，面ABCD ，BD 交AC 于点E ，F 是PC 中点，G 为AC 上一点.〔I 〕求证：BD FG ⊥；〔II 〕确定点G 在线段AC 上的位置，使FG//平面PBD ，并说明理由； 〔III 〕当二面角B PC D --的大小为23π时，求PC 与底面ABCD 所成角的正切值. 19.〔本小题总分为12分〕 数列{}n a 是首项为111,44a q ==公比的等比数列，设()*1423log n n b a n N +=∈，数列{}n c 满足n n n c a b =⋅.〔I 〕求数列{}n c 的前n 项和n S ； 〔II 〕假设2114n c m m ≤+-对一切正整数n 恒成立，求实数m 的取值范围. 20.〔本小题总分为12分〕以椭圆()2222:10x y C a b a b+=>>的中心O .设椭圆C 的左顶点为P ，左焦点为F ，上顶点为Q ，且满足2,OFQ PQ S OPQ ∆∆==. 〔I 〕求椭圆C 与其“准圆〞的方程；〔II 〕假设椭圆C 的“准圆〞的一个弦ED 〔不与坐标轴垂直〕与椭圆C 交于M 、N 两点，试证明：当0OM ON ⋅=时，试问弦ED 的长是否为定值，假设是，求出该定值；假设不是，请说明理由.21.〔本小题总分为12分〕函数()()()211,ln .f x a x x g x x =-+-=〔I 〕假设()()()()1,0a F x g x f x ==-+∞求在，上的最大值； 〔II 〕证明：对任意的正整数n ，不等式()23412ln 149n n n++++⋅⋅⋅+>+都成立； 〔III 〕是否存在实数()0a a >，使得方程()()()21141,g x f x a e x e ⎛⎫'=+-- ⎪⎝⎭在区间内有且只有两个不相等的实数根？假设存在，请求出a 的取值范围；假设不存在，请说明理由.。

理综化学参考答案选择题（本题包括7小题，每小题分，共分。

每小题只有一个选项符合题意。

） 7．B 8．9．B 10． 11．C 12． 13．C 28．（1分）（1）2Al +2OH－+6H2O=2 [Al(OH)4]－ + 3H2↑（2分）（）（2分）（）4Al (s)+ 3MnO2 (s) ＝3Mn(s) +2Al2O3(s) ΔH＝1789 kJ/mol（3分）（）Si3N4 +16HF=3SiF4 + 4NH4F （2分） （）0.024 （3分） 减小（2分） （）c(Ce4+)?[c(OH－)]4 （分） ；8 （2分） 29．（1分）（1） 氧化（2分） CO2 + 2e-==CO+O2-（2分） （2）SiO2、Cu（各1分）Fe2+ （2分） 2Fe3+ + Cu=Cu2+ + 2Fe2+（2分） 5.2≤pH＜5.4 （2分） b （2分） 5：11 （3分） ．（1分）．（）CO2在水中存在平衡：CO2+H2OH2CO3HCO3－+H+，有H2SO4存在时，可使上述平衡向左移动，从而减少CO2在水中的溶解，同时吸收挥发出来的HCl气体（分）HCl气体CO2在水中的溶解H2SO4电离出的H+使平衡左移 （）2HCO3－+Ba2+=BaCO3↓+ CO2↑+H2O（分）．（1）碱石灰或生石灰或氢氧化钠固体（任意填一种）（2分） （2）Cu和Cu2O（3分）（）偏小（2分）（）、（分） （）防止F中水蒸气进入D中，并吸收氨气（2分） . （1分）（1）醛基 （1分） （2）（分）（3）G：CH2＝CHOOCCH3 （分） M： （分） （4） +3NaOHCH3COONa++NaBr+ H2O（3分） （5）（分） 24．（12分）（1） RNA聚合酶、DNA解旋酶 核孔 从a到b TCAGCC （2）与蛋白质的加工与运输有关 D （2分） （3） 阻止GA-MYB基因的转录（2分） 启动a-淀粉酶基因的转录（2分） （4）催化大麦种子中储藏的淀粉水解，为种子萌发提供可利用的小分子物质 25．（13分） （1）叶绿体 叶绿素a （2）逐渐下降 28 （3）淀粉和蔗糖的含量 （4）减少（降低） 在叶片中积累 （5）a、c和e （2分） （6）150（2分） C（2分） 26．（9分）（1）光合作用 同化作用 （2）三个 捕食和竞争（2分） （3）次生演替 （4）调节种间关系，维持生态系统相对稳定 （5）y=25x-20ax（2分） 27．（14分） （1）12 （2）紫花腋生 紫花腋生：蓝花腋生：白花腋生=12：3：1（2分）（3）Aabb和aaBb（2分） （4）3：0：1（2分） （5）实验一：出现紫花腋生：紫花顶生：白花腋生：白花顶生=9：3：3：1（或出现四种表现型） 出现紫花顶生：紫花腋生：白花腋生=1：2：1（或出现三种表现型） 实验二：aaBbEe（2分） 出现蓝花腋生：蓝花顶生：白花腋生：白花顶生=9：3：3：1（或出现四种表现型）蓝花腋生：白花顶生=3：1（或出现两种表现型） 35．（12分）（1）促性腺 同期发情 （2）获能 减数第二次分裂中（2分） （3）显微注射 抗凝血酶（2分） （4）桑椹胚或囊胚 胚胎分割 （5）早期胚胎在相同生理环境条件下空间位置的转移（2分） ： ：。

2014年山东省潍坊市高考数学模拟试卷（四）（理科）一、选择题：本大题共10小题．每小题5分．共50分．把正确答案涂在答题卡上. 1. 已知集合A ={2, 4}，B ={1, 2, 4}，C ={(x, y)|x ∈A, y ∈B, 且log x y ∈N ∗}，则C 元素个数是（ ）A 2B 3C 4D 52. 已知：p:|x −a|<4，q ：(x −2)(3−x)>0，若￢p 是￢q 的充分不必要条件，则实数a 的取值范围是（ ）A a <−1或a >6B a ≤−1或a ≥6C −1≤a ≤6D −1<a <6 3. 已知向量a →=(cosα, −2)，b →=(sinα, 1)，且a → // b →，则tan(α−π4)等于( ) A 3 B −3 C 13D −134.如图，阅读程序框图，任意输入一次x(0≤x ≤1)与y(0≤y ≤1)，则能输出数对(x, y)的概率为（ ） A 13B 23C 14D 345. 下列说法正确的个数是①“在△ABC 中，若sinA >sinB ，则A >B”的逆命题是真命题；②“m =−1”是“直线mx +(2m −1)y +1=0和直线3x +my +2=0垂直”的充要条件； ③“三个数a ，b ，c 成等比数列”是“b =√ac”的既不充分也不必要条件；④命题“∀x ∈R ，x 3−x 2+1≤0”的否定是“∃x 0∈R ，x 03−x 03+1>0”（ ） A 1 B 2 C 3 D 46. 已知各项均不为零的数列{a n }，定义向量c n→=(a n ,a n+1)，b n →=(n,n +1)，n ∈N ∗．下列命题中真命题是（ ） A 若∀n ∈N∗总有c n → // b n →成立，则数列{a n }是等差数列 B 若∀n∈N∗总有c n → // b n →成立，则数列{a n }是等比数列 C 若∀n ∈N ∗总有c n →⊥b n →成立，则数列{a n }是等差数列 D 若∀n ∈N∗总有c n→⊥b n →成立，则数列{a n }是等比数列7. 已知非零向量AB →与AC →满足(AB →|AB|→+AC →|AC|→)⋅BC →=0，且AB →|AB|→⋅AC →|AC|→=−12，则△ABC 为（ ）A 等腰非等边三角形B 等边三角形C 三边均不相等的三角形D 直角三角形 8. 一个篮球运动员投篮一次得3分的概率为a ，得2分的概率为b ，不得分的概率为c（a, b, c ∈(0, 1)），已知他投篮一次得分的均值为2，2a+13b的最小值为（ ）A 323B 283C 143D 1639. 设不等式组{x +y ≤4y −x ≥0x −1≥0，表示的平面区域为D ，若圆C ：(x +1)2+(y +1)2=r 2(r >0)经过区域D 上的点，则r 的取值范围是（ ）A [2√2, 2√5]B (2√2, 3√2]C (3√2, 2√5]D (0, 2√2)∪(2√5, +∞)10. 设f(x)是定义在R 上的偶函数，对于任意的x ∈R ，都有f(x −2)＝f(2+x)，且当x ∈[−2, 0]时，f(x)=(12)x −1，若在区间(−2, 6]内关于x 的方程f(x)−log a (x +2)＝0恰有3个不同的实数解，则a 的取值范围是（ ）A (1, 2)B (2, +∞)C (1, √43) D (√43, 2)二、填空题：本大题共5小题，每小题5分，共25分.把正确答案填在答题卡相应的位置上. 11. 复数a+i 2−i在复平面内所对应的点在实轴上，那么实数a =________．12. 若(2x 2+1)5=a 0+a 1x 2+a 2x 4+...+a 5x 10，则a 3的值为________．13. 已知函数y =tanωx(ω>0)与直线y =a 相交于A 、B 两点，且|AB|最小值为π，则函数f(x)=√3sinωx −cosωx 的单调增区间是________．14.如图，F 1，F 2是椭圆C 1:x 24+y 2=1与双曲线C 2的公共焦点，A ，B 分别是C 1，C 2在第二、四象限的公共点．若四边形AF 1BF 2为矩形，则C 2的离心率是________．15. 关于函数f(x)=lgx 2+1|x|(x ≠0)，有下列命题①其图象关于y 轴对称；②当x >0时，f(x)是增函数；当x <0时，f(x)是减函数； ③f(x)的最小值是lg2；④f(x)在区间(−1, 0)、(2, +∞)上是增函数； ⑤f(x)无最大值，也无最小值其中所有正确结论的序号是________．三、解答题：本大题共6小题，共75分.解答应写出必要的文字说明、证明过程或演算步骤.16. 已知锐角三角形ABC 中内角A 、B 、C 的对边分别为a ，b ，c ，a 2+b 2=6abcosC ，且sin 2C =2sinAsinB ．（1）求角C 的值；（2）设函数f(x)=sin(ωx −π6)−cosωx(ω>0)，且f(x)图象上相邻两最高点间的距离为π，求f(A)的取值范围．17. 如图，李先生家住H 小区，他工作在C 科技园区，从家开车到公司上班路上有L 1、L 2两条路线，L 1路线上有A 1、A 2、A 3三个路口，各路口遇到红灯的概率均为12；L 2路线上有B 1、B 2两个路口，各路口遇到红灯的概率依次为34，35．（1）若走L 1路线，求最多遇到1次红灯的概率； （2）若走L 2路线，求遇到红灯次数X 的数学期望；（3）按照“平均遇到红灯次数最少”的要求，请你帮助李先生从上述两条路线中选择一条最好的上班路线，并说明理由．18. 如图，在底面是正方形的四棱锥P −ABCD 中，PA ⊥面ABCD ，BD 交AC 于点E ，F 是PC 中点，G 为AC 上一点． （1）求证：BD ⊥FG ；（2）确定点G 在线段AC 上的位置，使FG // 平面PBD ，并说明理由；（3）当二面角B −PC −D 的大小为2π3时，求PC 与底面ABCD 所成角的正切值．19. 已知数列{a n }是首项为a 1=14，公比q =14的等比数列，设b n +2=3log 14a n (n ∈N ∗)，数列{c n }满足c n =a n ⋅b n ． （1）求{b n }的通项公式；（2）若c n ≤14m 2+m −1对一切正整数n 恒成立，求实数m 的取值范围． 20. 以椭圆C:x 2a 2+y 2b 2=1(a >b >0)的中心O 为圆心，√a 2+b 2为半径的圆称为该椭圆的“准圆”．设椭圆C 的左顶点为P ，左焦点为F ，上顶点为Q ，且满足|PQ|=2，S △OPQ =√62S △OFQ． （1）求椭圆C 及其“准圆”的方程；（2）若椭圆C 的“准圆”的一个弦ED （不与坐标轴垂直）与椭圆C 交于M 、N 两点，试证明：当OM →⋅ON →=0时，试问弦ED 的长是否为定值，若是，求出该定值；若不是，请说明理由． 21. 已知函数f(x)=a(x −1)2+x −1，g(x)=lnx ．（1）若a =1，求F(x)=g(x)−f(x)在(0, +∞)上的最小值；（2）证明：对任意的正整数n，不等式2+34+49+...+n+1n>ln(n+1)都成立；（3）是否存在实数a(a>0)，使得方程2g(x)x =f′(x+1)−(4a−1)在区间(1e, e)内有且只有两个不相等的实数根？若存在，请求出a的取值范围；若不存在，请说明理由．2014年山东省潍坊市高考数学模拟试卷（四）（理科）答案1. B2. C3. B4. A5. C6. A7. A8. D9. A10. D11. −212. 8013. [−π3+2kπ,2π3+2kπ](k∈Z)14. √6215. ①③④16. 解：（1）∵ sin2C=2sinAsinB，∴ 由正弦定理有：c2=2ab，由余弦定理有：a2+b2=c2+2abcosC=c2(1+cosC)①又a2+b2=6abcosC=3c2cosC②由①②得1+cosC=3cosC，∴ cosC=12，又0<C<π，∴ C=π3；（2）f(x)=sin(ωx−π6)−cosωx=√3sin(ωx−π3)∵ f(x)图象上相邻两最高点间的距离为π，∴ T=π∴ 2πω=π∴ ω=2∴ f(x)=√3sin(2x−π3)∴ f(A)=√3sin(2A−π3)∵ π6<A <π2，∴ 0<2A −π3<2π3∴ 0<sin(2A −π3)≤1∴ 0<f(A)≤√3．17. 设“走L 1路线最多遇到1次红灯”为事件A ，包括没有遇到红灯和只遇到红灯一次两种情况．则P(A)=C 30×(12)3+C 31×12×(12)2=12，所以走L 1路线，最多遇到1次红灯的概率为12． 依题意，X 的可能取值为0，1，2． P(X =0)=(1−34)×(1−35)=110，P(X =1)=34×(1−35)+(1−34)×35=920，P(X =2)=34×35=920．随机变量X 的分布列为：所以EX =110×0+920×1+920×2=2720．设选择L 1路线遇到红灯次数为Y ，随机变量Y 服从二项分布Y ∼B(3,12)，所以EY =3×12=32．因为EX <EY ， 18. 证明：（1）∵ PA ⊥面ABCD ，四边形ABCD 是正方形，其对角线BD ，AC 交于点E ， ∴ PA ⊥BD ，AC ⊥BD ， ∴ BD ⊥平面PAC ， ∵ FG ⊂平面PAC ， ∴ BD ⊥FG 解（2）：当G 为EC 中点，即AG =34AC 时，FG // 平面PBD ， 理由如下：连接PE ，由F 为PC 中点，G 为EC 中点，知FG // PE ， 而FG Ë平面PBD ，PE ⊂平面PBD ， 故FG // 平面PBD ．解（3）：作BH ⊥PC 于H ，连接DH ，∵ PA ⊥面ABCD ，四边形ABCD 是正方形， ∴ PB =PD ，又∵ BC =DC ，PC =PC ， ∴ △PCB ≅△PCD ，∴ DH ⊥PC ，且DH =BH ，∴ ∠BHD 就是二面角B −PC −D 的平面角， 即∠BHD =2π3，∵ PA ⊥面ABCD ，∴ ∠PCA 就是PC 与底面ABCD 所成的角 连接EH ，则EH ⊥BD ，∠BHE =π3，EH ⊥PC ， ∴ tan∠BHE =BE EH=√3，而BE =EC ，∴EC EH=√3，∴ sin∠PCA =EH EC=√33，∴ tan∠PCA =√22， ∴ PC 与底面ABCD 所成角的正切值是√22或用向量方法：解：以A 为原点，AB ，AD ，AP 所在的直线分别为x ，y ，z 轴建立空间直角坐标系如图所示，设正方形ABCD 的边长为1，则A(0, 0, 0)，B(1, 0, 0)，C(1, 1, 0)，D(0, 1, 0)，P(0, 0, a)(a >0)，E(12,12,0)，F(12,12,a2)，G(m, m, 0)(0<m <√2)(1)BD →=(−1, 1, 0)，FG →=(m −12,m −12,−a2)，BD →×FG →=−m +12+m −12+0=0， ∴ BD ⊥FG（2）要使FG // 平面PBD ，只需FG // EP ，而EP →=(12,12,−a)，由FG →=12EP →可得{m −12=12λ−a2=−aλ，解得l =1，m =34，∴ G(34, 34, 0)，∴ AG →=34AC →， 故当AG =34AC 时，FG // 平面PBD（3）设平面PBC 的一个法向量为u →=(x, y, z)，则{u →⋅BC →=0˙，而PC →=(1,1,−a)，BC →=(0,1,0)，∴ {x +y −az =0y =0，取z =1，得u →=(a, 0, 1)，同理可得平面PDC 的一个法向量为v →=(0, a, 1)，设u →，v →所成的角为β，则|cosβ|=|cos2π3|=12，即|u →||v →|˙=12，∴√a 2+1⋅√a 2+1=12，∴ a =1∵ PA ⊥面ABCD ，∴ ∠PCA 就是PC 与底面ABCD 所成的角， ∴ tan∠PCA =PA AC=√2=√2219. 解：（1）由题意知，a n =(14)n (n ∈N ∗)， 易得b n =3log 14a n −2=3n −2；（2）c n =a n ⋅b n =(3n −2)⋅(14)n ，∴ c n+1−c n =(3n +1)⋅(14)n+1−(3n −2)⋅(14)n =9(1−n)⋅(14)n+1(n ∈N ∗)， ∴ 当n =1时，c 2=c 1=14，当n ≥2时，c n+1<c n ，即c 1=c 2>c 3>c 4>...>c n ，∴ 当n =1时，c n 取最大值是14，又c n ≤14m 2+m −1对一切正整数n 恒成立， ∴ 14m 2+m −1≥14，即m 2+4m −5≥0，解得：m ≥1或m ≤−5． 20. 解：（1）设椭圆的左焦点F(−c, 0)(c >0)， 由S △OPQ =√62S △OFQ 得12ab =√62⋅12bc ，化为a =√62c ． 由|PQ|=2可得√a 2+b 2=2，联立{a =√62c √a 2+b 2=2a 2=b 2+c 2，解得a 2=3，b 2=1，c 2=2．∴ 椭圆C 的标准方程为x 23+y 2=1，椭圆C 的“准圆”的方程为x 2+y 2=4．（2）设直线ED 的方程为y =kx +t ，与椭圆的交点为M(x 1, y 1)，N(x 2, y 2)， 联立{y =kx +tx 23+y 2=1，化为(1+3k 2)x 2+6ktx +3t 2−3=0，∴ x 1+x 2=−6kt1+3k 2，x 1x 2=3t 2−31+3k 2，可得y 1y 2=(kx 1+t)(kx 2+t)=t 2−3k 21+3k 2，由OM →⋅ON →=0，得x 1x 2+y 1y 2=0，即3t 2−31+3k 2+t 2−3k 21+3k 2=4t 2−3k 2−31+3k 2=0，∴ t 2=34(k 2+1)，此时满足△=36k 2t 2−4(1+3k 2)(3t 2−3)=27k 2+3>0成立．则点O 到弦ED 的距离d =2=√t 21+k 2=√34=√32， ∴ |ED|=2√4−34=√13是定值． 21. （1）解：a =1时，f(x)=x 2−x ， F(x)=lnx −x 2+x ，F′(x)=−(2x+1)(x−1)x，由F′(x)=0得x 1=−12，x 2=1，∵ x ∈(0, +∞)，∴ x ∈(0, 1)时，F(x)递减，x ∈(1, +∞)时，F(x)递增， 则x =1为极小值点，也为最小值点， 故F(x)min =F(1)=0．（2）证明：由（1）知lnx ≤x 2−x ，令x =i+1i>1，lni+1i<(i+1i)2−i+1i=i+1i 2，即ln(i +1)−lni <i+1i 2，故∑[n i=1ln(i +1)−lni]<2+34+49+...+n+1n ，即有不等式2+34+49+...+n+1n>ln(n +1)恒成立；（3）解：假设存在实数a(a >0)，使得方程2g(x)x=f′(x +1)−(4a −1)在区间(1e , e)内有且只有两个不相等的实数根． 将方程2g(x)x=f′(x +1)−(4a −1)整理得ax 2+(1−2a)x −lnx =0，设H(x)=ax 2+(1−2a)x −lnx ，由已知可转化为H(x)在(1e , e)内有且只有两个零点， 所以H′(x)=2ax +1−2a −1x =(2ax+1)(x−1)x令H′(x)=0，a >0，解得x 1=1，x 2=−12a（舍去），当x ∈(0, 1)，H′(x)<0，H(x)单调递减，x ∈(1, +∞)，H(x)单调递增． 故H(x)在(1e , e)内有且只有两个零点，只需{H(1e )>0H(1)<0H(e)>0即{a e 2+1−2ae +1>0a +1−2a <0ae 2+(1−2a)e −1>0 则{a <e 2+e2e−1a >1a >1−ee −2e解得1<a <e 2+e 2e−1，即存在a >0，且a 的取值范围是(1, e 2+e2e−1)．。

山东省潍坊市2014届高三第二次模拟考试英语试题本试卷分第1卷（选择题）和第Ⅱ卷（非选择题）两部分。

第1卷1至10页，第Ⅱ卷11至12页。

满分为150分。

考试用时为120分钟。

第I卷（共105分）注意事项：1．答第1卷前，考生务必将自己的姓名、准考证号、考试科目涂写在答题卡上。

2．每小题选出答案后，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。

不能答在试卷上。

第一部分英语知识运用（共两节，满分55分）第一节单项填空（共10小题；每小题1.5分，满分15分）从每题所给的A、B、C、D四个选项中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。

1. If you want to speak English without _______foreigner's accent, you have to learn ________correct pronunciation of those words.A．the；a B．a；the C．a；不填D．the；不填2.The MH370 from Malaysia crashed in the southern Indian Ocean,________global concern about flight safety.A. to causeB.causedC.causingD.having caused3. A1l the candidates are required to be at the Conference Hall at 2:00 pm________ otherwiseinformed．A. unlessB.ifC.whileD.though4. Although faced with stress from constant exams, she lost________ of her enthusiasmforstudy.A. someB.neitherC.noneD.all5. -________you go to the movies with me tonight?- I'd love to．A. WillB.MustC.MightD.Shall6. It is not quite clear_______our local government will do with air pollution.A. whetherB. howC. whatD. that7. - How long have you been here , Alice?- Only a few minutes. Susan_______me here.A. walksB. walkedC. has walkedD. had walked8. Left_______in the reading room alone quite by accident ,the boy felt nervous.A. lockingB. having lockedC. to lockD. locked9. I'd like to place things_______it is convenient to get them.A. in whichB. whatC. whereD. that10.- I guess you find the speech boring.- _______. I like it.A. Not in the leastB. Don't mention itC. No doubtD. I couldn't agree more .第二节完形填空 (共 30小题;11-20题每小题1分,21-40小题 1.5分, 满分40分) 阅列短文 ,掌握其大意，从每题所给的 A .B .C .D四个选项中 ,选出最佳项,并在答题卡上将该项涂黑。

山东省潍坊市2014届高三理综（化学部分）4月模拟考试（潍坊市二模，无答案）本试卷分第I卷（选择题）和第Ⅱ卷（非选择题）两部分，共12页，满分300分，考试用时150分钟。

考试结束后，将本试卷和答题卡一并交回。

答卷前，考生务必将自己的姓名、准考证号、考试科目填涂在答题卡规定的地方。

第I卷（必做题，共107分）注意事项：1．每小题选出答案后，用2B铅笔把答题卡上对应的答案标号涂黑。

如需改动，用橡皮擦干净以后，再涂写其他答案标号。

只答在试卷上不得分。

2．第1卷共20道小题，1-13题每小题5分，14-20题每小题6分，共107分。

以下数据可供答题时参考：相对原子质量：H1 C 12 N 14 0 1 6 Na 23 S 32 Cl 35.5 Ti 48 Fe 56 Cu 64 Zn 65 Ba 137一、选择题（本题包括13道小题，每小题只有一个选项符合题意）7．下列说法正确的是A．淀粉和纤维素的最终水解产物都是葡萄糖，二者互为同分异构体B．高纯度的二氧化硅是制备光导纤维、太阳能电池板的主要材料C．石油的分馏、裂化、裂解都属于化学变化D．合金材料中可能含有非金属元素8．短周期元素R、T、Q、W、G在元素周期表中的相对位置如右图所示，其中Q是无机非金属材料的主角。

下列判断正确的是A．离子半径：T< W<GB．等物质的量的W、G单质分别与足量铁粉反应，后者消耗的铁粉多C．最简单气态氢化物的热稳定性：Q<RD．T和G组成的化合物溶于水，其水溶液呈中性9．下列两种物质间发生反应，消耗等物质的量的酸，产生气体最多的是A．木炭与浓硝酸 B．铜与稀硝酸 C．锌与稀硫酸 D．木炭与浓硫酸10．已知 一紫罗兰酮的结构简式是，关于该有机物的说法正确的是A.其分子式为C13H18OB．能使溴的四氯化碳溶液、酸性高锰酸钾溶液退色C．属于芳香族化合物D．能发生加成反应、取代反应、水解反应11．根据下列操作及现象，所得结论正确的是12．下列溶液中有关物质的浓度关系正确的是A.c （NH 4＋）相等的NH 4HCO 3、NH 4HSO 4、NH 4Cl 溶液中：B ．向醋酸钠溶液中加入适量醋酸，得到的酸性混合溶液中：C ．1．0mol 1L -⋅Na 2CO 3 溶液中：D ．某二元弱酸的酸式盐NaHA 溶液中：13．kJ 1mol -⋅。

数学（理）参考答案及评分标准一、选择题（每小题5分，共50分）CADDA DCBBC二、填空题（每小题5分，共25分）11．12．13．14．15．三、解答题：16．（本小题满分12分）解：（Ⅰ），…………2分由得：，∴…………………4分∴………………5分由得：，∴的单调递减区间为：．…………………………6分（Ⅱ）∵，即，由正弦定理得：，……………8分，，∴,………………10分∵△锐角三角形，∴，…………………11分∴的取值范围为．………………………………12分17．（本小题满分12分）（Ⅰ）证明：因为四边形是菱形，所以. …………… 1分因为平面平面，且四边形是矩形，所以平面，…………… 2分又因为平面，所以. ……………… 3分因为，所以平面. …………… 4分（Ⅱ）解：设，取的中点，连接，因为四边形是矩形，分别为的中点，所以，又因为平面，所以平面，由，得两两垂直，………………5分所以以为原点，所在直线分别为x轴，y轴，z轴，如图建立空间直角坐标系.………… 6分因为底面是边长为2的菱形，，，所以，，，.得，.………………7分设平面的法向量为，所以即令，得. …………9分由平面，得平面的法向量为，…………10分则. ……………11分由图可知二面角为锐角，所以二面角的大小为.……………12分18.（本小题满分12分）解：（Ⅰ）甲同学先从三个项目中随机抽取两项，共有种方法，则恰好抽取跳、掷两个项目的概率为.……………… 2分（Ⅱ）经过两项测试就能达标的概率是：.……………… 4分（Ⅲ）的取值分别为，……………… 5分当，甲参加随机抽取的两项测试全部合格或者全不合格，此时，…………… 8分当，甲参加随机抽取的两项测试应该是一项合格另一项不合格，必须进行第三次测试，此时，（解法二：）……………… 10分则的分布列是……………… 11分数学期望.……………… 12分19．（本小题满分12分）解：（Ⅰ）因为，，①所以当时，．………………………………………………………1分当时，，②…………………2分①－②得，．………………………………………………………………4分所以．…………………………………………………………………………5分因为，适合上式，所以．……………6分（Ⅱ）由（Ⅰ）得．所以……………………………8分．…………………………………………………………10分所以．…12分20．（本小题满分13分）解：（Ⅰ）由题意知：，………………………………………1分椭圆上的点满足，且，．，．，……………………………………………2分又，………………………………………………3分椭圆的方程为．…………………………………………………4分（Ⅱ）由题意知、，（1）当直线与轴垂直时，、，则的方程是：，的方程是：，直线与直线的交点为，∴点在直线上. ……………………………………………………………………6分（2）当直线不与轴垂直时，设直线的方程为，、，由得∴，…………………………………………………7分，，共线，∴………………8分又，，需证明共线，需证明，只需证明若，显然成立，若，即证明∵成立，……………………………………………11分∴共线，即点总在直线上. ………………………………………12分21．（本小题满分14分）解：（Ⅰ）当时，，所以，…………………1分，，………………………………………2分又因为切线过，所以切线方程为.…………………4分（Ⅱ）的定义域为，…………………5分令，其判别式，…………………6分①当，故上单调递增，…5分②当，的两根都小于0，在上，，故上单调递增．…………………6分③当，设的两根为，，………………7分当时，；当时，；当时，，故分别在上单调递增，在上单调递减．………………………………………8分（Ⅲ）由（Ⅱ）可知：当在上有两个极值点…9分因为所以，………………10分由（Ⅱ）可知：，于是，若存在，使得，则，即，亦即，……………………11分设函数，当时，，…………12分所以在上单调递增，………………………………………13分而，所以，这与式矛盾．故不存在，使得．…………………………14分第11 页共11 页。