2014潍坊市二模

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年高考模拟训练试题理科综合(二)本试卷分第I 卷(选择题)和第Ⅱ卷(非选择题)两部分，共15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I 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：略二、选择题(本题包括7道小题．共42分．每小题给出的四个选项中．有的只有一个选项正确，有的多个选项正确。

全部选对的得6分．选对但不全的得3分．有选错的得0分)14．下列说法正确的是A ．功是矢量B ．A 、B 两点间的电势差U AB =-5.0V 中“-”号表示方向C ．法拉第发现了电磁感应现象，并制作了世界上第一台发电机D ．用国际单位制中的基本单位表示磁感应强度的单位应为kg ／(A ·s 2)15．如图所示，底面粗糙、碗口水平的半球形容器放置于水平地面上，将一劲度系数为k 的轻弹簧一端固定在内壁光滑的半球形容器底部O'处(O 为球心)，弹簧另一端与质量为m 的小球相连，小球静止于P 点，OP 与水平方向的夹角为θ=30°．容器的质量为M ，半径为R ．重力加速度为g ，下列说法正确的是A ．轻弹簧对小球的作用力大小为mg 23 B ．弹簧原长为kmg R + C ．容器与水平地面之间无摩擦力D ．容器对水平地面的压力小于(m+M)g16．质量为m 的人造地球卫星与地心的距离为r 时，引力势能可表示为rGMm E p -=，其中G 为引力常量，M 为地球质量。

该卫星原来在半径为R 1的轨道上绕地球做匀速圆周运动，由于受到极稀薄空气的摩擦作用，飞行一段时间后其圆周运动的半径变为R 2，此过程中因摩擦而产生的热量为A ．)11(12R R GMm -B ．)11(21R R GMm -C ．)11(212R R GMm -D ．)11(221R R GMm - 17．如图所示，圆心在O 点、半径为R 的光滑圆弧轨道ABC 竖直固定在水平桌面上，OC 与OA 的夹角为60°，轨道最低点A 与桌面相切．一足够长的轻绳两端分别系着质量为m 1和m 2的两小球(均可视为质点)，挂在圆弧轨道光滑边缘C 的两边，开始时m 1位于C 点，然后从静止释放。

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 =________。

保密★启用前 试卷类型：A潍坊实验中学2014届高三第二次单元过关检测物理试题 2013.12注意事项：1．本试卷分第Ⅰ卷（选择题）、第Ⅱ卷（非选择题）两部分，共100分。

考试时间90分钟。

2．答第Ⅰ卷前，考生务必将自己的姓名、准考证号、考试科目、试卷类型用铅笔涂写在答题卡上。

3．每小题选出答案后，用铅笔把答题卡上对应题目的答案涂黑，如需改动，用橡皮擦干净后，再选涂其他答案，答在试卷上无效。

山东中学联盟 4．考试结束，将第Ⅱ卷和答题卡一并交回。

第Ⅰ部分（选择题48分）一、选择题：（本题共12小题，每小题4分，共48分。

在每小题给出的四个选项中，有一个或多个选项正确，全部选对得4分，选对但不全得2分，选错或不选得0分）1. 把一条导线平行地放在如图所示磁针的上方附近，当导线中有电流时，磁针会发生偏转。

首先观察到这个实验现象的物理学家是 ( )A ．奥斯特B ．法拉第C ．欧姆D ．洛伦兹2.若带正电荷的小球只受到电场力作用，则它在任意一段时间内 ( ) A ．一定沿电场线由高电势处向低电势处运动 B ．一定沿电场线由低电势处向高电势处运动C ．不一定沿电场线运动，但一定由高电势处向低电势处运动D ．不一定沿电场线运动，也不一定由高电势处向低电势处运动I3．如图所示，实线为电场线，虚线为等势面，且AB=BC 。

电场中 A 、 B 、C 三点的场强分别为EA 、EB 、EC 电势分别为 A B C ϕϕϕ、、，AB 、BC 间的电势差分别为U AB U BC 则下列关系式中正确的是（）A ．ABC ϕϕϕ>> B ．E C >E B >E A C ．U AB <U BCD ．U AB =U BC4.小灯泡通电后其电流I 随所加电压U 变化的图线如图所示，P 为图线上一点，PN 为图线的切线，PQ 为U 轴的垂线，PM 为I 轴的垂线。

则下列说法中正确的是（ ）A ．随着所加电压的增大，小灯泡的电阻增大B ．对应P 点，小灯泡的电阻为R ＝U 1I 2C ．对应P 点，小灯泡的电阻为R ＝U 1I 2－I 1D ．对应P 点，小灯泡的功率为图中矩形PQOM 所围的面积5. 在阴极射线管中电子流方向由左向右，其上方有一根通有如图所示电流的直导线，导线与阴极射线管平行，则阴极射线将 ( )A ．向上偏转B ．向下偏转C ．向纸里偏转D ．向纸外偏转6.如图所示，在光滑水平面上一轻质弹簧将挡板和一条形磁铁连接起来，此时磁铁对水平面的压力为N 1，现在磁铁左上方位置固定一导体棒，当导体棒中通以垂直纸面向里的电流后，磁铁对水平面的压力为N 2 ，则以下说法正确的是（ ） A ．弹簧长度将变长 B ．弹簧长度将变短C ．N 1＞N 2D ．N 1＜N 2NSI 2 M P I 1 NQ17.如图所示,一束电子以大小不同的速率沿图示方向飞入横截面为一正方形的匀强磁场区,在从ab边离开磁场的电子中，下列判断正确的是（）A.从b点离开的电子速度最大山东中学联盟B.从b点离开的电子在磁场中运动时间最长C.从b点离开的电子速度偏转角最大D.在磁场中运动时间相同的电子,其轨迹线一定重合8. 如图表示一交流电的电流随时间而变化的图象，此交流电流的有效值是（）A．3.5A B．25.3 AC．25 A D．5A9.矩形导线框abcd固定在匀强磁场中，磁感线的方向与导线框所在平面垂直，规定磁场的正方向垂直低面向里，磁感应强度B随时间变化的规律如图所示.若规定顺时针方向为感应电流I的正方向，下列各图中正确的是 ( )10. (2011·新课标全国卷)如图，一理想变压器原、副线圈的匝数比为1∶2；副线圈电路中接有灯泡，灯泡的额定电压为220 V，额定功率为22 W；原线圈电路中接有电压表和电流表。

山东省潍坊市二模理科综合2014.5本试卷分第I卷(选择题)和第Ⅱ卷(非选择题)两部分，共16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。

中学联盟网2．第I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：H 1 C 12 N 14 O 16 Na 23 S 32 C1 35．5 Ti 48Fe 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)1．细胞内某些生物膜上的蛋白质被糖基化形成糖蛋白，起到识别信息分子、保护自身免受破坏等作用，下列哪种细胞器膜上最可能存在较多的糖蛋白A．内质网B．溶酶体C．高尔基体D．线粒体2．血浆中的各类蛋白质分子统称为血浆蛋白，下列关于血浆蛋白的说法错误的是A．有的能维持血浆渗透压B．有的能催化肽聚糖的水解C．有的可以参与免疫调节D．有的可以引起肌肉细胞收缩3．下列关于呼吸作用中间产物丙酮酸的叙述，正确的是A．丙酮酸氧化分解释放的能量少量储存在ATP中B．丙酮酸→酒精的阶段只能释放少量能量C．催化丙酮酸氧化分解的酶只分布在线粒体基质中D．丙酮酸彻底氧化分解的产物是CO2、水、A TP4．右图表示某肽链片段合成的过程，下列有关叙述正确的是A．①和②过程都需要能量和DNA聚合酶B．③过程的模板和场所含有不同的五碳糖C．图中DNA模板链上的G突变为C，蛋白质的结构不会改变D．若仅a链发生突变，A变为C，则该片段复制3次共需游离的腺嘌呤脱氧核苷酸18个5．酒精是生物学实验中常用的试剂，下列有关酒精及作用的叙述，不正确的是A．观察生物组织细胞中的脂肪颗粒时需要用50％的酒精溶液洗去浮色B．观察植物根尖细胞有丝分裂时需要用盐酸和酒精的等体积混合液解离根尖C．观察DNA和RNA在细胞中的分布时需要用8％的酒精水解细胞D．可用70％的酒精保存从土壤中采集的小动物6．下列关于遗传变异与进化的分析中，正确的是A．染色体结构变异必然导致染色体上基因数目的变化B．一般来说，种群内频率高的基因所控制的性状更适应环境C．基因突变产生的有利变异决定了生物进化的方向D．金鱼能与野生鲫鱼杂交并产生可育后代，说明二者之间不存在生殖隔离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。

潍坊二模试题及答案一、选择题（每题3分，共30分）1. 下列关于潍坊的描述中，哪一项是不正确的？A. 潍坊是中国山东省的一个地级市B. 潍坊以风筝闻名于世C. 潍坊位于中国的南部D. 潍坊是中国历史文化名城答案：C2. 潍坊二模试题中，选择题的总分是多少？A. 20分B. 25分C. 30分D. 35分答案：C3. 下列哪一项不是潍坊二模试题的特点？A. 覆盖面广B. 难度适中C. 题目新颖D. 答案唯一答案：D4. 潍坊二模试题的命题人是谁？A. 张老师B. 李老师C. 王老师D. 赵老师答案：B5. 潍坊二模试题的考试时间是多久？A. 90分钟B. 120分钟C. 150分钟D. 180分钟答案：B6. 潍坊二模试题的总分是多少？A. 100分B. 120分C. 150分D. 180分答案：C7. 潍坊二模试题的考试形式是什么？A. 闭卷考试B. 开卷考试C. 口试D. 笔试答案：D8. 潍坊二模试题的考试地点在哪里？A. 潍坊一中B. 潍坊二中C. 潍坊三中D. 潍坊四中答案：A9. 潍坊二模试题的考试日期是什么时候？A. 2024年4月1日B. 2024年5月1日C. 2024年6月1日D. 2024年7月1日答案：C10. 潍坊二模试题的考试目的是什么？A. 选拔优秀学生B. 检验学习成果C. 促进学生交流D. 以上都是答案：D二、填空题（每题2分，共20分）1. 潍坊二模试题的命题人是______。

答案：李老师2. 潍坊二模试题的考试时间是______分钟。

答案：1203. 潍坊二模试题的总分是______分。

答案：1504. 潍坊二模试题的考试形式是______。

答案：笔试5. 潍坊二模试题的考试地点是______。

答案：潍坊一中6. 潍坊二模试题的考试日期是______月______日。

答案：6月1日7. 潍坊二模试题的考试目的之一是______。

答案：检验学习成果8. 潍坊二模试题的选择题总分是______分。

理 科 综 合二、选择题（本题包括7道小题，共42分．每小题给出的四个选项中，有的只有一个选项正确，有的多个选项正确，全部选对的得6分，选对但不全的得3分，有选错的得0分）14．物理学在揭示现象本质的过程中不断发展，下列说法正确的是A ．通电导线受到的安培力，实质上是导体内运动电荷受到洛仑兹力的宏观表现B ．穿过闭合电路的磁场发生变化时电路中产生感应电流，是因为变化磁场在周围 产生了电场使电荷定向移动C ．磁铁周围存在磁场，是因为磁铁内有取向基本一致的分子电流D ．踢出去的足球最终要停下来，说明力是维持物体运动的原因15.已知地球的平均密度为1ρ，火星的平均密度为2ρ，设绕地球做圆周运动的卫星最小运行周期为T 1,绕火星做圆周运动的卫星最小运行周期为T 2，则为12T T16.如图所示，理想变压器原副线圈匝数比为10：1，电源电压u=220sin100πtV ，通过电阻0R 接在变压器原线圈两端，开关闭合后，电压表示数为12V ，电流表的示数为10A.以下说法正确的是A ．0R 的阻值是100ΩB ．电源的功率是120WC ．t=0.0ls 时刻，电阻R 中电流最大D ．若将开关断开，电压表的示数仍然是12V17.如图所示，在两个等量异种电荷形成的电场中，D 、E 、F 是两电荷连线上等间距的三个点，三点的电势关系是，K 、M 、L 是过这三个点的等势线，其中等势线L 与两电荷连线垂直．带电粒子从a 点射人电场后运动轨迹与三条等势线的交点是a 、b 、c ，粒子在a 、b 、c 三点的电势能分别是,以 下判断正确的是A ．带电粒子带正电B ．C.D.18．如图所示，质量为m1、m2。

的两个物块，通过定滑轮和动滑轮以及轻绳悬挂在天花板的M、N两点，平衡时悬挂于M点的绳与竖直方向的夹角为α，悬挂于N点的绳与竖直方向的夹角为β。

现将悬点M向左缓慢移动一小段距离后固定，再次平衡后，关于,αβ以及M、N两点所受作用力的变化情况，以下判断正确的是（滑轮重力和绳与滑轮之间的摩擦力不计）A．α和β均变大，且α=βB．α和β均不变，且α=2βC．绳对M、N的作用力不变D．绳对M、N的作用力变大1 9．如图所示，轻弹簧上端通过一轻绳固定，下端拴一小球，小球与光滑的三角形斜面接触，弹簧处于竖直状态。

2014年山东省潍坊市高考数学二模试卷（理科）学校:___________姓名：___________班级：___________考号：___________一、选择题(本大题共10小题，共50.0分)1.若复数z满足（1+i）•z=i，则z的虚部为（）A.-B.-C.D.【答案】D【解析】解：∵（1+i）•z=i，∴z===，∴z的虚部为，故选：D．由题意可得z=，再利用两个复数代数形式的乘除法法则，虚数单位i的幂运算性质，计算求得结果．本题主要考查两个复数代数形式的乘除法法则，虚数单位i的幂运算性质，属于基础题．2.设集合A={x||2x-1|≤3}，B={x|y=lg（x-1）}，则A∩B=（）A.（1，2）B.[1，2]C.（1，2]D.[1，2）【答案】C【解析】解：由A中不等式得：-3≤2x-1≤3，解得：-1≤x≤2，即A=[-1，2]，由B中y=lg（x-1），得到x-1＞0，即x＞1，∴B=（1，+∞），则A∩B=（1，2]．故选：C．求出A中不等式的解集确定出A，求出B中x的范围确定出B，找出A与B的交集即可．此题考查了交集及其运算，熟练掌握交集的定义是解本题的关键．3.下列结论正确的是（）A.若向量∥，则存在唯一的实数λ使=λB.已知向量，为非零向量，则“，的夹角为钝角”的充要条件是“•＜0’’C.“若θ=，则cosθ=”的否命题为“若θ≠，则cosθ≠”D.若命题p：∃x∈R，x2-x+1＜0，则￢p：∀x∈R，x2-x+1＞0【答案】C【解析】解：若向量∥，≠，则存在唯一的实数λ使=λ，故A不正确；已知向量，为非零向量，则“，的夹角为钝角”的充要条件是“•＜0，且向量，不共线”，故不正确；条件否定，结论否定，可知C正确；若命题p：∃x∈R，x2-x+1＜0，则￢p：∀x∈R，x2-x+1≤0，故D不正确．故选：C．根据向量共线定理判断A，向量，为非零向量，则“，的夹角为钝角”的充要条件是“•＜0，且向量，不共线”，可判断B，条件否定，结论否定，可判断C；命题p：∃x∈R，x2-x+1＜0，则￢p：∀x∈R，x2-x+1≤0，可判断D．本题考查命题的真假判断与应用，考查学生分析解决问题的能力，知识综合性强．4.已知f（x）=x2+sin，f′（x）为f（x）的导函数，则f′（x）的图象是（）A. B. C. D.【答案】A【解析】解：由f（x）=x2+sin=x2+cosx，∴f′（x）=x-sinx，它是一个奇函数，其图象关于原点对称，故排除B，D．又f″（x）=-cosx，当-＜x＜时，cosx＞，∴f″（x）＜0，故函数y=f′（x）在区间（-，）上单调递减，故排除C．故选：A．先化简f（x）=x2+sin=x2+cosx，再求其导数，得出导函数是奇函数，排除B，D．再根据导函数的导函数小于0的x的范围，确定导函数在（-，）上单调递减，从而排除C，即可得出正确答案．本题主要考查函数的单调性与其导函数的正负之间的关系，即当导函数大于0时原函数单调递增，当导函数小于0时原函数单调递减．5.已知α，β表示平面，m，n表示直线，m⊥β，α⊥β，给出下列四个结论：①∀n⊂α，n⊥β；②∀n⊂β，m⊥n；③∀n⊂α，m∥n；④∃n⊂α，m⊥n，则上述结论中正确的个数为（）A.1B.2C.3D.4【答案】B【解析】解：由α，β表示平面，m，n表示直线，m⊥β，α⊥β，知：①∀n⊂α，则n∥β或n⊂β或n与β相交，故①错误；②∀n⊂β，由直线与平面垂直的性质，知m⊥n，故②正确；③∀n⊂α，则m与n相交、平行或异面，故③错误；④由m⊥β，α⊥β知，在平面α中至少有一条直线与m垂直，∴∃n⊂α，m⊥n，故④正确．故选：B．利用空间中线线、线面、面面间的位置关系求解．本题考查命题真假的判断，是基础题，解题时要注意空间思维能力的培养．6.已知函数f（x）=x2+x，执行如图所示的程序框图，若输出的结果是，则判断框中的条件应是（）A.n≤30B.n≤31C.n≤32D.n≤33【答案】B【解析】∴解：∵函数f（x）=x2+x，∴f（n）=n（n+1），由程序框图知：算法的功能是求S=++…+=1-的值，∵输出的结果是，∴跳出循环的n值为32，∴判断框内的条件应填：n＜32或n≤31．故选：B．算法的功能是求S=++…+的值，根据输出的结果判断跳出循环的n值，从而确定判断框内应填的条件．本题考查了当型循环结构的程序框图，根据框图的流程判断算法的功能是解答本题的关键．7.已知双曲线C：-=1（a＞0，b＞0）的左、右焦点分别是F1、F2过F2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△MF1N为正三角形，则该双曲线的离心率为（）A. B. C. D.2+【答案】A【解析】解：双曲线C：-=1（a＞0，b＞0）的渐近线方程为bx±ay=0，x=c时，y=±，∵△MF1N为正三角形，∴2c=×，∴a=b，∴c=b，∴e==．故选：A．求出双曲线C的两渐近线方程，利用△MF1N为正三角形，建立三角形，即可求出该双曲线的离心率．本题考查双曲线的简单性质，考查学生的计算能力，比较基础．8.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.πB.πC.4πD.16π【答案】D【解析】解：由三视图知：几何体为圆锥，圆锥的高为1，底面半径为，设外接球的半径为R，如图：则（R-1）2+3=R2⇒R=2．∴外接球的表面积S=4π×22=16π．故选：D．几何体为圆锥，根据三视图判断圆锥的高与底面半径，设外接球的半径为R，结合图形求得R，代入球的表面积公式计算．本题考查了由三视图求几何体的外接球的表面积，结合图形的求得外接球的半径是解答本题的关键．9.在区间[-3，3]上任取两数x，y，使x2-y-1＜0成立的概率为（）A. B. C. D.【答案】A【解析】解：由题意可得，区间[-3，3]上任取两数x，y，区域为边长为6的正方形，面积为36，x2-y-1＜0的区域是图中阴影区域以外的部分，其面积S==，∴在区间[-3，3]上任取两数x，y，使x2-y-1＜0成立的概率为=．故选：A．该题涉及两个变量，故是与面积有关的几何概型，分别表示出满足条件的面积和整个区域的面积，最后利用概率公式解之即可．本题主要考查了与面积有关的几何概率的求解，解题的关键是准确求出区域的面积，属于中档题．10.已知定义在R上的函数y=f（x）对任意的x满足f（x+1）=-f（x），当-1≤x＜1时，f（x）=x3．函数g（x）=，＞，＜若函数h（x）=f（x）-g（x）在[-6，+∞）上有6个零点，则实数a的取值范围是（）A.（0，）∪（7，+∞）B.[，）∪（7，9]C.[，1）∪（1，9]D.（，]∪[7，9）【答案】B【解析】解：∵对任意的x满足f（x+1）=-f（x），∴f（x+2）=-f（x+1）=f（x），即函数f（x）是以2为最小正周期的函数，画出函数f（x）、g（x）在[-6，+∞）的图象，由图象可知：在y轴的左侧有2个交点，只要在左侧有4个交点即可．则＜即有＞或＜＜＜或＜，故7＜a≤9或≤a＜．故选：B．f（x）=x3．函数g（x）=[-6，+∞）上有6个零点，即函数f（x）与g（x）的交点的个数，由函数图象的变换，分别做出y=f（x）与y=g（x）的图象，由此求得a的取值范围．本题考查函数图象的变化与运用，涉及函数的周期性，对数函数的图象等知识点，关键是作出函数的图象，由此分析两个函数图象交点的个数．二、填空题(本大题共5小题，共25.0分)11.已知，是夹角为60°的两个单位向量，若向量=3+2，则||= ______ ．【答案】【解析】解：∵=1，=°=．∴=+4+12=9+4+12×=19．∴=故答案为：．利用数量积的运算和性质即可得出．本题考查了数量积的运算和性质，属于基础题．12.现将如图所示的5个小正方形涂上红、黄两种颜色，其中3个涂红色，2个涂黄色，若恰有两个相邻的小正方形涂红色，则不同的涂法种数共有______ ．（用数字作答）【答案】6【解析】解：当涂红色两个相邻的小正方形在两端时是有=4，当涂红色两个相邻的小正方形在不在两端时是有=2，则不同的涂法种数共有4+2=6种．故答案为：6．根据涂红色两个相邻的小正方形的位置进行分类，利用分类计数原理即可解得．本题主要考查了分类计数原理，本题的关键是根据涂红色两个相邻的小正方形位置进行分类．13.已知抛物线C：y2=2px（p＞0）上一点P（2，m）（m＞0），若P到焦点F的距离为4，则以P为圆心且与抛物线C的准线相切的圆的标准方程为______ ．【答案】（x-2）2+（y-4）2=16【解析】解：由题意结合抛物线的定义可得P到准线的距离为4，∴2-（-）=4，求得p=4，∴抛物线C：y2=8x．点P（2，m）代入抛物线C：y2=8x，结合m＞0，可得m=4．再根据题意可得圆的半径为4，故所求的圆的标准方程为（x-2）2+（y-4）2=16，故答案为：（x-2）2+（y-4）2=16．根据题意可得2-（-）=4，求得p=4，可得抛物线C：y2=8x．把点P（2，m）代入抛物线的方程，求得m的值，可得圆心和半径，从而得到所求的圆的标准方程．本题主要考查抛物线的定义和标准方程的应用，求圆的标准方程的方法，求出m的值，是解题的关键，属于中档题．14.曲线y=xsinx在点A（，），B（-，））处的切线分别为l1，l2，设l1，l2及直线x-2y+2=0围成的区域为D（包括边界）．设点P（x，y）是区域D内任意一点，则x+2y 的最大值为______ ．【答案】6【解析】解：∵y=xsinx，∴y′=sinx+xcosx，x=，y′=1；x=-，y′=-1，∴l1：y-=x-，即y=x；l2：y-=-（x-），即y=-x，l1，l2及直线x-2y+2=0围成的区域为D（包括边界），如图所示，交点坐标分别为（0，0）、（2，2）、（-，），∴在（2，2）处，x+2y的最大值为6．故答案为：6．求出函数的切线方程，作出对应的平面区域，利用线性规划的知识进行求解即可得到结论．本题主要考查导数的几何意义的应用，以及线性规划的有关知识，利用数形结合是解决本题的关键．15.如图所示，位于东海某岛的雷达观测站A，发现其北偏东45°，与观测站A距离20海里的B处有一货船正匀速直线行驶，半小时后，又测得该货船位于观测站A东偏北θ（0°＜θ＜45°）的C处，且cosθ=，已知A、C两处的距离为10海里，则该货船的船速为______ 海里/小时．【答案】4【解析】解：∵cosθ=，∴sin=，由题意得∠BAC=45°-θ，即cos∠BAC=cos（45°-θ）=，∵AB=20，AC=10，∴由余弦定理得BC2=AB2+AC2-2AB•AC cos∠BAC，即BC2=（20）2+102-2×20×10×=800+100-560=340，即BC=，设船速为x，则=2，∴x=4（海里/小时），故答案为：4根据余弦定理求出BC的长度即可得到结论．本题主要考查解三角形的应用，根据条件求出cos∠BAC，以及利用余弦定理求出BC的长度是解决本题的关键．三、解答题(本大题共6小题，共75.0分)16.已知函数f（x）=A sin（ωx+）（A＞0，ω＞0）的振幅为2，其图象的相邻两个对称中心之间的距离为．（Ⅰ）若f（α+）=，0＜α＜π，求sinα；（Ⅱ）将函数y=f（x）的图象向右平移个单位得到y=g（x）的图象，若函数y=g（x）-k是在[0，π]上有零点，求实数k的取值范围．【答案】解：（Ⅰ）依题意，A=2，T==，∴ω=3，∴f（x）=2sin（3x+）…2分又f（α+）=2sin[3（+）+]=2sin（2α+）=2cos2α=，∴cos2α=…4分∴sin2α==，又0＜α＜π，∴sinα=…6分（Ⅱ）将函数y=f（x）的图象向右平移个单位得到y=g（x）=2sin[3（x-）+]=2sin（3x-）的图象，…8分则函数y=g（x）-k=2sin（3x-）-k，∵x∈[0，π]，∴3x-∈[-，]，∴-≤2sin（3x-）≤2…11分∵函数y=g（x）-k在[0，π]上有零点，∴y=g（x）与y=k在[0，π]上有交点，∴实数k的取值范围是[-，2]…12分【解析】（Ⅰ）利用函数y=A sin（ωx+φ）的图象性质可求得A=2，T=，解得ω=3，于是可得函数y=f（x）的解析式，从而可由f（α+）=，0＜α＜π，求得sinα；（Ⅱ）利用函数y=A sin（ωx+φ）的图象变换，可求得g（x）=2sin（3x-），利用正弦函数的单调性与最值可求得x∈[0，π]时该函数的值域，利用y=g（x）与y=k在[0，π]上有交点，即可求得实数k的取值范围．本题考查函数y=A sin（ωx+φ）的图象性质与图象变换，考查正弦函数的单调性与最值，考查等价转化思想与运算求解能力，属于中档题．17.直三棱柱ABC-A1B1C1中，AB⊥BC，BC=，BB1=2，AC1与A1C交于一点P，延长B1B到D，使得BD=AB，连接DC，DA，得到如图所示几何体．（Ⅰ）若AB=1，求证：BP∥平面ACD，（Ⅱ）若直线CA1与平面BCC1B1所成的角为30°，求二面角D-AC-C1的余弦值．【答案】（Ⅰ）证明：取AC的中点E，连接PE，DE…1分则PE，∵BD=AB=1，BB1=2，∴BD=BB1=CC1，又∵BD∥CC1，∴BD CC1，∴PE BD，∴四边形DBPE为平行四边形，∴BP∥DE， (3)分∵BP⊄面ACD，DE⊂面ACD，…4分∴BP∥平面ACD，…5分（Ⅱ）解：由题意知，AB⊥BC，AB⊥BB1，∴AB⊥面BC1，∴A1B1⊥面BC1连接B1C，则∠A1CB1为直线CA1与平面BCC1B1所成的角，则∠A1CB1=30°，…6分在R t△A1B1C中，B1C=，tan A1CB1．∴A1B1=…7分以B为原点，分别以BC，BB1，AA1为x、y、z轴建立如图所示的空间直角坐标系，则A（0，0，），C（，0，0），D（0，-，0），∴=（，0，-），=（0，-，-），…8分设面ACD的法向量为=（x，y，z），则即，取z=1，则=（1，-1，1）…9分在平面ABC内取面AC1的一个法向量=（x，0，z），则=x-z=0，取x=1，则z=1，∴=（1，0，1）…10分∴cos＜，＞==，…11分由图知二面角D-AC-C1为钝角，二面角D-AC-C1的余弦值为-…12分【解析】（Ⅰ）取AC的中点E，连接PE，DE，证明四边形DBPE为平行四边形，从而BP∥平面ACD；（Ⅱ）轴建立空间直角坐标系，用向量法解决．空间直角坐标系本题考查线面平行，考查面面角，考查向量知识的运用，解题的关键是正确建立坐标系，属于中档题．18.某超市制定“五一”期间促销方案，当天一次性购物消费额满1000元的顾客可参加“摸球抽奖赢代金券”活动，规则如下：①每位参与抽奖的顾客从一个装有2个红球和4个白球的箱子中逐次随机摸球，一次只摸出一个球；②若摸出白球，将其放回箱中，并再次摸球；若摸出红球则不放回，工作人员往箱中补放一白球后，再次摸球；③如果连续两次摸出白球或两个红球全被摸出，则停止摸球．停止摸球后根据摸出的红球个数领取代金券，代金券数额Y与摸出的红球个数x满足如下关系：Y=144+72x（单位：元）．（Ⅰ）求一位参与抽奖顾客恰好摸球三次即停止摸球的概率；（Ⅱ）求随机变量Y的分布列与期望．【答案】解：（Ⅰ）恰好摸球三次即停止摸球包含三种情况：①红白红；②白红红；③红白白，∴所求事件的概率为：p==．（Ⅱ）x的可能取值为0，1，2，对应随机变量Y的可能取值为144，216，288，则P（Y=144）=，P（Y=216）=，P（Y=288）=1-=，∴Y的分布列为：【解析】（Ⅰ）恰好摸球三次即停止摸球包含三种情况：①红白红；②白红红；③红白白，由此能求出一位参与抽奖顾客恰好摸球三次即停止摸球的概率．（Ⅱ）x的可能取值为0，1，2，对应随机变量Y的可能取值为144，216，288，分别求出相应的概率，由此能求出随机变量Y的分布列与期望．本题考查概率的求法，考查离散型随机变量的分布列和数期望的求法，解题时要认真审题，是中档题．19.已知等差数列{a n}，a1+a3+a5=42，a4+a6+a8=69；等比数列{b n}，b1=2，log2（b1b2b3）=6．（Ⅰ）求数列{a n}和数列{b n}的通项公式；（Ⅱ）设c n=a n-b n，求数列{|c n|}的前n项和T n．【答案】解：（Ⅰ）设等差数列{a n}的公差为d，∵a1+a3+a5=3a3=42，∴a3=14，a4+a6+a8=3a6=69，∴a6=23，∴d==3．a n=a3+（n-3）d=14+（n-3）•3=3n+5．设等比数列{b n}的公比为q，由log2（b1b2b3）=6，得b1b2b3=26，即，∴b2=4，则q==2，∴．（Ⅱ）c n=a n-b n=（3n+5）-2n，c n+1-c n=[3（n+1）+5]-2n+1-（3n+5）+2n=3-2n，当n=1时，c2-c1=1＞0，c2＞c1，当n≥2时，3-2n＜0，c n+1＜c n，又c1=6，c2=7，c3=6，c4=1，c5=-12，…∴{c n}的前4项为正，从第5项开始往后各项为负，设数列{c n}的前n项和为S n，S n=（a1-b1）+（a2-b2）+…+（a n-b n）=（a1+a2+…+a n）-（b1+b2+…+b n）=（2n+1-2），∴当n≤4时，T n=|c1|+|c2|+…+|c n|=c1+c2+…+c n=S n=+2；当n≥5时，T n=c1+c2+c3+c4-（c5+c6+…+c n）=S4-（S n-S4）=2S4-S n∴，，．【解析】（Ⅰ）设等差数列{a n}的公差为d，由等差数列的性质及已知可分别求得a3=14，a6=23，进而可求d，由通项公式可得a n；设等比数列{b n}的公比为q，由log2（b1b2b3）=6，得b1b2b3=26，由等比数列的性质可得b2=4，则q==2，由通项公式可得b n；（Ⅱ）易求c n=a n-b n=（3n+5）-2n，由c n+1-c n=[3（n+1）+5]-2n+1-（3n+5）+2n=3-2n 的符号可判断{c n}的前4项为正，从第5项开始往后各项为负，设数列{c n}的前n项和为S n，利用等差、等比数列的求和公式可求S n=（a1-b1）+（a2-b2）+…+（a n-b n）=（a1+a2+…+a n）-（b1+b2+…+b n），然后分n≤4，n≥5两种情况讨论可求T n．本题考查等差、等比数列的通项公式、求和公式，考查分类讨论思想，考查学生的运算求解能力，属中档题．20.如图，椭圆C：+=1（a＞b＞0）的短轴长为2，点P为上顶点，圆O：x2+y2=b2将椭圆C的长轴三等分，直线l：y=mx-（m≠0）与椭圆C交于A、B两点，PA、PB与圆O交于M、N两点．（Ⅰ）求椭圆C的方程；（Ⅱ）求证△APB为直角三角形；（Ⅲ）设直线MN的斜率为n，求证：为定值．【答案】（Ⅰ）解：∵椭圆C：+=1（a＞b＞0）的短轴长为2，∴2b=2，解得b=1，∵圆O将椭圆的长轴三等分，∴2b=，∴a=3b=3，∴椭圆C的方程为．（Ⅱ）证明：由，消去y得（1+9m2）x2-，设A（x1，y1），B（x2，y2），则，，又P（0，1），∴，，===（1+m2）•-==0∴PA⊥PB，∴△PAB为直角三角形．（Ⅲ）证明：由（Ⅱ）知PA⊥PB，由题意知PA，PB的斜率存在且不为0，设直线PA的斜率为k，k＞0，则PA：y=kx+1，由，得或，∴，，又直线l过点（0，-），则m==，由，得，或，∴M（，），又∵PM⊥PN，∴MN为⊙O的直径，∴MN过原点，∴n=，又∵m≠0，∴k2-1≠0，∴n≠0，∴=，∴为定值．【解析】（Ⅰ）由椭圆C：+=1（a＞b＞0）的短轴长为2，解得b=1，由圆O将椭圆的长轴三等分，得a=3b=3，由此能求出椭圆C的方程．（Ⅱ）由，得（1+9m2）x2-，由此推导出，从而能证明△PAB为直角三角形．（Ⅲ）设直线PA的斜率为k，k＞0，则PA：y=kx+1，由，得，，又直线l过点（0，-），则m=，由，得M（，），MN过原点，n=，由此能证明为定值．本题考查椭圆方程的求法，考查三角形为直角三角形的证明，考查两数比值为定值的证明，解题时要认真审题，注意函数与方程思想的合理运用．21.已知函数f（x）=a x+x2-xlna（a＞0且a≠1）．（Ⅰ）求函数f（x）的单调区间；（Ⅱ）a＞l，证明：当x∈（0，+∞）时，f（x）＞f（-x）；（Ⅲ）若对任意x1，x2，x1≠x2，且当f（x1）=f（x2）时，有x1+x2＜0，求a的取值范围．【答案】解：（Ⅰ）f′（x）=a x•lna+2x-lna，令g（x）=f′（x），∴g′（x）=a x（lna）2+2＞0，∴g（x）是（-∞，+∞）上的增函数，∵g（0）=0，∴x＞0时，g（x）＞g（0）=0，此时f′（x）＞0，x＜0时，g（x）＜g（0）=0，此时f′（x）＜0，∴f（x）在（0，+∞）单调递增，在（-∞，0）单调递减．（Ⅱ）设h（x）=f（x）-f（-x）=a x-a-x-2xlna，∴h′（x）=（a x+a-x）lna-2lna，∵a＞1，故lna＞0，∴h′（x）≥2-2lna=2lna-2lna=0，∴h（x）在（0，+∞）单调递增；∴h（x）＞h（0）=0，即x∈（0，+∞）时，f（x）＞f（-x）．（Ⅲ）由于x1≠x2，且f（x1）=fx2），由（Ⅰ）知x1，x2异号，不妨设x1＜0，x2＞0，则x1，-x2∈（-∞，0），由（Ⅱ）知：当a＞1时，f（x1）=f（x2）＞f（-x2），∵x∈（-∞，0）时，f（x）单调递减，故x1＜-x2，∴x1+x2＜0，即a＞1适合题意；当0＜a＜1时，lna＜0，由（Ⅱ）h（x）=a x-a-x-2xlnah′（x）=（a x+a-x）lna-2lna≤2lna-2lna=0，∴h（x）在（0，+∞）单调递减，h（x）＜h（0）=0，即f（x）＜f（-x），故f（x1）=f（x2）＜f（-x2），∵x∈（-∞，0）时，f（x）单调递减，x1＞-x2，x1+x2＞0，即0＜a＜1不合题意，综上：a＞1．【解析】（Ⅰ）通过对函数求导确定单调区间，（Ⅱ）设出新函数，通过对新函数求导找到单调区间，确定最小值，从而问题得解，（Ⅲ）对a进行讨论，由前两问综合得出．本题考察了导数的综合应用，函数的单调性，分类讨论思想，是一道综合题．。