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2020-2021学年海南省华侨中学高三英语月考试题及参考答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项ACitisport inNewportWe at Citisport aim to improve sports training and facilities inNewport, giving you more opportunities to try both new and traditional sports.GolfWe are pleased to be able to offer lessons at Kingsway Golf Centre just outsideNewport. These are run by experienced golf professionals, and are held on an all-weather practice area. The adult lessons are open to anyone aged 13 and over, and are suitable for all levels from beginners upwards. These take place on Wednesdays from 3:00 to 4:00 pm over a period of six weeks. Children’s lessons for 7-12 year old are held from 2:00 to 3:00 pm on Saturdays during term time.TennisThe Citisport tennis courses provide an opportunity for local people to develop their skills on the brand-new indoor tennis court at Newport Leisure Centre. All equipment can be provided, but please feel free to use your own racket (球拍) if you prefer. Our Starter course is held on Mondays from 7:00 to 8:00 pm, and is for beginners of 12 years and over. Our Improver course, which takes place on Tuesdays from 8:00 to 9:00 pm, is for players with some experience.Football for girlsBy popular request, Citisport is holding another one-day girls-only football course. This aims to give local girls the chance to learn essential skills and develop more advanced ones. The course will take place on Saturday, 9th November from 9:00 am to 5:00 pm, and is open to all girls aged 10-14 years living in theNewportarea.GymnasticsThis course is for beginners aged 8-14 and will provide an introduction to basic skills. There is a maximum of six pupils per coach in each class. At the end of the course there is a demonstration for friends and family of all the skills learnt there. The course will take place on Thursdays from 6:00 to 7:00 pm.1. What can we know about the Citisport golf lessons?A. You can take lessons at Kingsway Golf Centre insideNewport.B. The golf lessons can take place only in good weather.C. Teenagers can attend golf lessons on Wednesday afternoons.D. Children’s lessons usually last 2 or 3 hours on Saturday afternoons.2. Which of the following statements is true about the Citisport tennis courses?A. You can get the skills in an open-air court.B. You must take your own racket during the course.C. You can take the Monday course if you are a green hand.D. You can become an experienced player after the courses.3. Which course lasts only one day according to the text?A. Golf.B. Tennis.C. Football for girls.D. Gymnastics.BSalad plants have already been grown in old sheltersand tunnels. Urban farming is a regular topic of interest at places like the World Economic Forum (WEF) in Davos, where leaders consider whether the world's food system, blamed for causing both obesity and malnutrition, can be fixed. There are already plenty of urban farming projects around the world, particularly in the US, Japan and the Netherlands, from urban fish and plant farms to vertical farming.“It's becoming an expanding industry,” said Richard Ballard, one of the founders of the farm Growing Underground. “There're several other businesses starting up in London in containers, and there are other vertical farms around the country now.”Growing Underground is not a standard farm. The rows of crops could be in almost any tunnel, but these plants are 100 feet below Clapham High Street and show that urban agriculture is, in some cases at least, nota fad. The underground farm has occupied a part of the Second World War air-raid shelters for nearly five years, and Ballard is planning to expand into the rest of the space later this year.Growing Underground supplies herb and salad mixes to grocery shops, supermarkets and restaurants. Being in London creates an advantage, Ballard says, as they can harvest and deliver in an hour.He adds other advantages. Being underground means temperatures never go below 15℃surface greenhouses need to be heated. They can do more harvests: 60 crops a year, compared with about seven in a traditional farm. Electricity to power the lights is a major cost, but the company believes renewable energy will become cheaper.Similar British companies include the Jones Food Company in Lincolnshire, while in the US AeroFarms has several projects in New Jersey, and Edenworks in Brooklyn.4. What do we know about urban farming?A. It leads to a healthier lifestyle.B. It is rarely discussed at the WEF.C. Different farming methods are used.D. Local governments pay efforts to develop it.5. Which of the following best explains "a fad" underlined in Paragraph 3?A. A dream that's easy to realize.B. A field controlled for a long time.C. An approach to a serious problem.D. A fashion that’s popular for a short time.6. What can we learn about the underground farm?A. It is more productive than a traditional farm.B. It provides food directly to the customers.C. Its major products are herbs and salads.D. It uses less energy than a greenhouse.7. What can be a suitable title for the text?A. Current food system causes health problemsB. Growing Underground attracts more peopleC. Traditional farming will be replaced soonD. Urban farming isstill thought costly and time-consumingCIn a study published in Nature Machine Intelligence, researchers at Ohio State University show how artificial intelligence（AI）can follow clinical trials to identify drugs for repurposing, a solution that can help advance innovative treatments.Repurposing drugs is legal and not unusual. When doctors prescribe(开处方)drugs that have been approved by the Food and Drug Administration(FDA)for purposes different from what is printed on the labels, the drugs are being used “off-label” Just because a drug is FDA-approved for a specific type of disease does not prevent it from having possible benefits for other purposes.For example, Metformin, a drug that is FDA-approved for treating type 2 diabetes, is also used to treat PCOS(a disease of women), and other diseases. Trazodone, an anti-depressant with FDA-approval to treat depression, is also prescribed by doctors to help treat patients with sleep issues.The Ohio State University research team created an AI deep learning model for predicting treatment probability with patient data including the treatment, outcomes, and potential confounders(干扰因素).Confounders are related to the exposure and outcome. For example, a connection is identified between music festivals and increases in skin rashes(红疹). Music festivals do not directly cause skin rashes. In this case, one possible confounding factor between the two may be outdoor heat, as music festivals tend to run outdoorswhen the temperature is high, and heat is a known cause for rashes. When working with real-world data, confounders could number in the thousands. AI deep learning is well-suited to find patterns in the complexity of potentially thousands of confounders.The researcher team used confounders including population data and co-prescribed drugs. With this proof-of-concept, now clinicians have a powerful AI tool to rapidly discover new treatments by repurposing existing medications.8. What do we know about a drug used off-label?A. It is sold without a label.B. It is available at a low price.C. Its uses extend beyond the original ones.D. Its clinical trials are rejected by doctors.9. Metformin and Trazodone are similar as both of them________.A. are used off-labelB. treat rare diseasesC. result in sleep issuesD. are medical breakthroughs10. What can be inferred about “confounders”?A. They are possible treatments.B. They are environmental factors.C. They can be easily recognized in real-world data.D. They should be taken into serious consideration.11. What is the main idea of the text?A. AI examines benefits of existing drugs.B. AI identifies off-label uses for drugs.C. AI finds new drugs for common diseases.D. AI proves the power of drug research.DWhen you walk on a sandy beach, it takes more energy than striding down a sidewalk — because the weight of your body pushes into the sand. Turns out, the same thing is true for vehicles driving on roads. The weight of the vehicles creates a very shallow indentation (凹陷) in the pavement (路面) — and it makes it such that it’scontinuously driving up a very shallow hill.Jeremy Gregory, a sustainability scientist at M.I.T. and histeam modeled how much energy could be saved — and green-house gases avoided — by simply stiffening (硬化) the nation’s roads and highways. And they found that stiffening 10 percent of the nation’s roads every year could prevent 440 megatons of carbon dioxide equivalent emissions over the next five decades — enough to offset half a percent of projected transportation sector emissions over that time period. To put those emissions savings into context — that amount is equivalent to how much CO2 you’d spare the planet by keeping a billion barrels of oil in the ground — or by growing seven billion trees — for a decade.The results are in the Transportation Research Record.As for how to stiffen roads? Gregory says you could mix small amounts of synthetic fibers orcarbon nanotubes into paving materials. Or you could pave with cement-based concrete, which is stiffer than asphalt (沥青).This system could also be a way to shave carbon emissions without some of the usual hurdles. Usually, when it comes to reducing emissions in the transportation sector, you’re talking about changing policies related to vehicles and also driver behavior, which involves millions and millions of people — as opposed to changing the way we design and maintain our pavements. That’s just on the order of thousands of people who are working in transportation agencies. And when it comes to retrofitting (翻新) our streets and highways —those agencies are where the rubber meets the road.12. Why does the author mention “walk on a sandy beach” in paragraph 1?A. To present a fact.B. To make a contrast.C. To explain a rule.D. To share an experience.13. What suggestion does the author give to reduce CO2 emissions?A. Hardening the road.B. Keeping oil in the ground.C. Growing trees for decades.D. Improving the transportation.14. What is the advantage of this suggestion?A. Gaining more support.B. Consuming less money.C. Involving more people.D. Facing fewer usual obstacles.15. What does the underlined part mean in the last paragraph?A.Those agencies are likely to make more rules.B. Those agencies will change some related policies.C. Those agenciesmight put more rubber tires on the roads.D. Those agencies will play a key role in making this happen.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020届海口市琼山华侨中学高三英语月考试题及答案解析第一部分阅读（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A、B、C、D四个选项中选出最佳选项AEast Yorkshire has typical unpredictable British weather. So here are some ideas to keep everybody happy when the weather is not the most ideal.William's Den, North CaveThe outdoor and indoor areas are suitable for children of all ages to have fun.There are nests to explore, rope bridges to cross, a tree-house and a slide. The attached Kitchen provides fresh food made from locally sourced ingredients serving a selection of treats.East Riding Leisure CentresKnown for a fun learner pool alongside an incredible fun zone with two slides as well, it is perfect for kids to find their feet in the water, have fun and explore. Its 6 climbing walls offer a different challenge on each. This place is suitable for anyone over the age of 4 and you can refuel at cafe with fresh food, snacks and cakes.Sewerby Hall and GardensWhen the weather’s not sure, take cover in the Hall and learn how life was in the early 1900’s for the residents and workers of the house. Then explore the zoo and meet the pigs, parrots and penguins! Kids of all ages are welcome.Withernsea LighthouseThere’s no limitation to the age of kids to climb Withernsea Lighthouse, which is 144 steps to the top, with full views of the East Yorkshire Coast at the top of it. Enjoy the museum on the ground floor and learn what life is like working and living in a lighthouse. The souvenir shop provides attractive gifts for visitors at a fair price.1.Which one is unsuitable for kids of all ages?A.William’s Den, North Cave.B.East Riding Leisure Centres.C.Sewerby Hall and Gardens.D.Withernsea Lighthouse.2.Where can kids enjoy food?A.In William’s Den, North Cave and Sewerby Hall and Gardens.B.In East Riding Leisure Centres and Withernsea LighthouseC.In William’s Den, North Cave and East RidingLeisure Centres.D.In Sewerby Hall and Gardens and Withernsea Lighthouse.3.Where does this passage probably come from?A.A geography textbook.B.A science report.C.A finance magazine.D.A travel brochure.BOne rainy afternoon, I was on a crosstown bus when ayoung woman jumped on. She had a child with her who must have been about 3 or 4 years old.The bus was full, bumpy, and it soon got noisy as her kid began crying because he couldn’t sit next to his mother. There were a couple of open seats, but they weren’t together. She wasflusteredand looked embarrassed.Then another woman, a little older, stood up and moved so that the mother and child could sit together. The mom smiled as a thank-you. And then three words came out of the older woman’s mouth that elevated the entire energy of that bus ride: “I’ve been there”.Simple, undramatic and honest. In that moment, it seemed to unite people. Why? Because almost all experiences are shared human experiences. We forget that, as we forge (前进) through life, focused onour own troubles and needs—which are actually less unique than we think. How can these three words create more connection in your life? Ask yourself: “Where am I holding back?One thing I know for sure is this: Healing others helps heal yourself. I noticed this recently with my friend, Tracy, who took a new friend who had suffered a miscarriage under her wing. Tracy had three of them before having her daughter two years ago. Our intellect needs a doctor to explain the medical side of things, yes. But our souls need human connection to help us along. No one can do that better than someone who has been exactly where you are.Can the essence of these three words help you make a small difference right now? It can be as simple as volunteering your seat, sharing some helpful advice or even lightening the mood with a joke when you notice that someone’s uncomfortable—because we’re all in this together.4. The underlined word “flustered” in the second paragraph is closest in meaning to _______.A. angryB. anxiousC.scaredD. upset5. What does the woman mean by saying “ I’ve been there”in the third paragraph？A. The woman was on the bus and saw what had happened to the boy.B. The woman got to her destination and was ready to get off the bus.C. The woman once had the similar experience with that mother.D. The woman took the exact seat that the boy was on just now.6. Which of the following statements is TRUE according to the passage?A. Everyone has his or her own unique problem that is difficult to solve.B. Doctors can help us get through when we have mental or physical problems.C. The author’s friend Tracy felt better after she was comforted by her new friend.D. One can indeed make a difference to those in need of help by doing simple things.7. The passage isintended to _______.A. show a harmonious world by telling some touching storiesB. praise those who are willing to help others in emergenciesC. appeal to readers to give timely help to those in needD. illustrate some ways of helping others in detailCIf you ever find yourself trapped in the wilderness without food, you'llhave to figure out how to feed yourself. Many plants in the wild areedible, but many are also poisonous. So it is necessary to learn how to determine whether the plants you find can be eaten safely.Avoid using this method without careful planning. Some plants can be deadly, and even if you follow these guidelines perfectly, there is always a chance that a plant will make you seriously ill. Prepare yourself for wilderness outings by learning about the local plants, and carry a guidebook to help you identify plants. Even if you are unprepared and cannot find food you know to be safe, remember that, depending on your activity level, the human body can go for days without food, and you’re better off being hungry than being poisoned.Testing the plant in your mouth is dangerous, so go forward very slowly and carefully. First, hold a small portion of the prepared plant part against your lip for 3 minutes. Do not put the plant in your mouth. If you notice any burning, tingling (刺痛), or other reactions, discontinue testing. Second, place another small portion of the plant part on your tongue. Hold the plant on your tongue without chewing for 15 minutes. Discontinue testing if you notice any reaction. Third, chew the plant and holdit in your mouth for 15 minutes. Chew the plant well, and do not swallow. Discontinue testing if you notice any reaction. Fourth, swallow the small portion of the plant. Wait 8 hours. Do not eat or drink anything during this period except purified water. If you feel sick, immediately throw up what you eat and drink plenty of water. If activated charcoal (活性炭) is available, take that with the water.8. What’s the meaning of the underlined world “edible” in paragraph 1?A. Suitable for using as food.B. Widely spread.C. Existing in large quantities.D. Not widely known.9. What can we know from paragraph 2?A. Planning is unnecessary when using the method.B. Not all plants in the wild can serve as food generally.C. Suffering hunger can be more dangerous than testing plants.D. Following the method perfectly can ensure safety.10. Which is the correct order of testing plants in the mouth?① wait and see ② chew it in the mouth③ put it on the tongue④ put it against lips ⑤ swallow itA. ③④②①⑤B. ④③②①⑤C. ③④②⑤①D.④③②⑤①11. Where might the passage come from?A. A student’s diary.B. A science report.C. A guide book for camping.D. A doctor’ s notebook.DAccording to a survey published by the American Institutes for Research last year, a total of 57 colleges were operating some form of CBE programs and about 85 percent of all the higher education officials said they were either designing a CBE program at their school or were considering doing so.Students in a CBE program choose a central field of study, just as they would at a traditional college or university. Yet instead of attending a series of classes led by professors or teaching assistants at schools, the students study online and direct themselves.CBE programs require students to show their understanding of a given set of sills Students must prove their mastery of skills that relate to their field of choice by taking related exams. Once they have met all the requirements of their study programs, the students will get their degrees.CBE programs have made use of many new technologies, especially internet and online media. This helps reduce barriers for nontraditional and other students by bringing higher education to them. And programs that permit students to work at their own speed may save students' money by reducing the time it takes for them to earn a degree.But some educators have concerns about the value of the education that CBE programs offer. Johann NeematWesternWashingtonUniversityargues that the purpose of higher education is not simply to help students master certain skills. It should teach students how to think critically (批判性地) understand the subjects they are studying more deeply and see how they are connected to other subjects. Only that way can they put the knowledge to better use.He said, “You need to explore, think .. get shaken, have a conversation and struggle. Andthose things take time.”Instead of supporting CBE, he adds, policy makers and educators should look for ways to improve access and reduce costs for traditional higher education.12. How are CBE programs different from traditional college education?A. They require students to choose their subjects.B. They offer shorter curricula and are less expensive.C. They heavily rely on the information technologies.D. They allow students to take easier examinations.13. What can we learn from Johann Neem's words?A Free access to traditional education should be provided.B. Higher education just focuses on critical thinking skills.C. Students should spend longer time completing the degree courses.D. College students should be challenged to explore around their subjects.14. How does Johann Neem's attitude toward CBE programs?A. Supportive.B. Disapproving.C. Sympathetic.D. Uncaring.15. What is the author's purpose in writing the text?A. To press policy-makers to provide more affordable education.B. To show the disadvantages of the traditional college education.C. To introduce a new controversial trend in the higher education.D. To encourage educators to improve the quality of CBE programs.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020-2021学年海南省高三5月段考物理（解析版）姓名:_____________ 年级:____________ 学号:______________题型选择题填空题解答题判断题计算题附加题总分得分1. （知识点：粒子在有界磁场中运动）如图所示，两金属板正对并水平放置，分别与平行金属导轨连接，Ⅰ、Ⅱ、Ⅲ区域有垂直导轨所在平面的匀强磁场．金属杆ab与导轨垂直且接触良好，并一直向右匀速运动．某时刻ab进入Ⅰ区域，同时一带电小球从O点沿板间中轴线水平射入两板间．ab在Ⅰ区域运动时，小球匀速运动；ab从Ⅲ区域右边离开磁场时，小球恰好从金属板的边缘离开．已知板间距为4d，导轨间距为L，Ⅰ、Ⅱ、Ⅲ区域的磁感应强度大小相等、宽度均为d．带电小球质量为m，电荷量为q，ab 运动的速度为v0，重力加速度为g．求：（1）小球带何种电荷及磁感应强度B的大小；（2）ab在Ⅱ区域运动时，小球的加速度a大小；（3）要使小球恰好从金属板的边缘离开，ab运动的速度v0要满足什么条件。
【答案】（1）正电；（2）2g （3）【解析】试题分析：（1）小球带正电ab在磁场区域运动时，产生的感应电动势大小为：①金属板间产生的场强大小为：②ab在Ⅰ磁场区域运动时，带电小球匀速运动，有③联立①②③得：④（2）ab在Ⅱ磁场区域运动时，设小球的加速度a，依题意，有⑤评卷人得分联立③⑤得：⑥（3）依题意，ab分别在Ⅰ、Ⅱ、Ⅲ磁场区域运动时，小球在电场中分别做匀速、类平抛和匀速运动，设发生的竖直分位移分别为SⅠ、SⅡ、SⅢ；ab进入Ⅲ磁场区域时，小球的竖直分速度为vⅢ．则：SⅠ=0⑦SⅡ=⑧SⅢ=vⅢ⑨vⅢ=⑩又：SⅠ+SⅡ+SⅢ=2d⑾联立可得：⑿考点：法拉第电磁感应定律；带电粒子在复合场中的运动.如图所示，导热良好的薄壁气缸放在水平面上，用横截面积为S＝1.0×10－2m2的光滑薄活塞将一定质量的理想气体封闭在气缸内，活塞杆的另一端固定在墙上．此时活塞杆与墙刚好无挤压。
2020年海口市琼山华侨中学高三英语月考试卷及答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项AInformation on school visits to Kew GardensEnjoy yourselves in a wonderland of science with over 50,000 living plants and a variety of educational events or amusing activities. Here is essential information about planning a school visit to Kew.Educational course pricesYou can plan a self-led visit or book one of our educational courses. Students will take part in the educational courses in groups of 15. Prices vary according to different situations.EYFS (Early Years Foundation Stage) to Key Stage 4:45-minute course: 35/group 90-minute course: 70/groupKey Stage 5:Half day (one course): 80/group Full day (two courses): 160/groupTeachers and adults:Up to required key stage proportions (比例): FreeAdults needed for 1:1 special educational needs support: FreeAdults above the required proportions: 11/personThe payment will due within 28 calendar days of making the booking.Health and safetyRequired supervising (监护) adult-student proportions:Key Stage 1: 1:5 Key Stage 2: 1:8Key stage 3: 1:10 Key Stage 4: 1:12Key Stage 5: 1:12The group sizes should be controlled if you are visiting potentially busy areas such as the glasshouse and other attractions. The maximum number of students visiting the glasshouses is 15 per group and each group to Kew shops should include no more than 10 students.If there is an emergency, please contact the nearest Kew staff member or call Constabulary on 0208 32 3333 for direct and quick support. Please do not call 999.Planning your visitYour tickets and two planning passes will be sent to you upon receipt of your payment. You can complete your risk assessment with the passes, ensure you bring your tickets and the receipt document and show them to the staff members at the gate on the day of your visit.Recommended timingsThe Kew Gardens opens at 10 am. You are recommended to spend at least three to five hours on your visit. The closing time varies throughout the year. But the earliest is 3:30 pm. We have a fixed schedule for educational courses, which is from 10:30 am to 2:20 pm.1.How much should a group of 15 Key Stage I students and 4 teachers pay for a 45-minute course?A.35B.46C.57D.812.What should one do in an emergency?A.Check the risk assessment.B.Call 999 immediately.C.Ask adults or teachers for help.D.Seek help from the staff member nearby.3.What is the purpose of the text?A.To introduce Kew Gardens.B.To give tips on visiting Kew Gardens.C.To attract potential visitors to Kew Gardens.D.To inform coming activities in Kew Gardens.BI waschecking out at the supermarket counter on Wednesday night, ready to pay for my bananas, when all ofa sudden, fear came upon me. My wallet was gone. And I could only have left it one place: the G9 bus, from which I had gotten off minutes earlier and which was now speeding to some stops. The moment of realizing it was gone was followed by mental math. How much time and money would it cost to replace the credit cards, the driver's license, the expensive lipstick ($ 55!).Two hours after I was back at my house, I heard a knock on the door. My husband answered while I sat in the dining room on the phone with a credit card company. "Does Jennifer live here?" I heard someone say. In her hand was my wallet, without a penny missing. She left before I could offer my gratitude to her.After I posted the story, I heard from her boyfriend, who identified the good citizen as Erin Ball, a 26-year-oldgirl working for a trade organization.Once I figured out her, I called to thank her. She said she spotted my wallet and thought that it's more dangerous to go to a stranger's house than leaving the wallet with the driver, but she still decided to take the chance. "If I were in that situation, I would want someone to try to find me," she said. Ball doesn't find her actions particularly excellent. She added, "It's not hard to do small things for people."After Ball found my wallet, she decided to post a picture of my driver's license online before going to my house, trying to see if anyone knew me. No sooner had she left my doorstep than I got emails from two neighbors who recognized my face, both offering to help me find my missing property.Ball found my house on a bitterly cold night,for which I was extremely grateful. Looking back, I'm not surprised someone had wanted to help a stranger. A warm current of honesty and harmony is running through this town.4. What do we know about the author according to paragraph 1?A. She missed the G9 bus.B. She paid for her bananas.C. She replaced the credit cards.D. She found she had left her wallet on the bus.5. Who helped the author find Ball?A. The G9 driver.B. The girl's boyfriend.C. The author's neighbors.D. The author's husband.6. What did Ball do first after finding the wallet?A. Ball called the author.B. Ball went to the author's house.C. Ball gave the wallet to the bus driver.D. Ball posted a photo of the author's driving license.7. Which of the following best describes Erin Ball?A. Humorous and kind.B. Generous and demanding.C. Honest and warm-hearted.D. Caring and outgoing.CIt's a popular belief that a fish's memory lasts for only seven seconds. It may seem sad to think that they don't remember what they've eaten or where they’ve been, and they don't identify you or any of their friends--every moment intheir life would be like seeing the world for the first time.But don't be so quick to feel sorry for them. A new study has found that fish have a much better memory than we used to think. In fact, certain species of fish can even remember events from as long as 12 days ago. In the study, researchers from Mac Ewan University in Canada trained a kind of fish called African cichlids to go to a certain area of their tank to get food.They then waited for 12 days before putting them back in the tank again. Researchers used computer software to monitor the fish’s movements.They found that after such a long break the fish still went to the same place where they first got food. This suggested that they could remember their past experiences.In fact. scientists had been thinking for a long time that African cichlids might have a good memory. An earlier study showed that they behaved aggressively(挑衅地) in front of certain fish, perhaps because they remembered their past "fights".But until the latest findings, there was no clear evidence.Just as a good memory can make our lives easier, it also plays an important part when a fish is trying to survivein the wild. "If fish are able to remember that a certain area contains safe food, they will be able to go back to that area without putting their lives at risk,"lead researcher Trevor Hamilton told Live Science.For a long time, fish were placed far below chimpanzees, dolphins and mice on the list of smart animals.But this study has given scientists a new understanding of their intelligence.8. According to the text, people commonly believe that ______.A. fish don't recognize any of their friendsB. a fish's memory lasts for only seven minutesC. fish can only remember part of their past experiencesD. fish can remember things that happened long ago9. How can fish benefit most from a good memory?A. They can remember their enemies and fight.B. They can remember where to get food and survive.C. They can remember their friends and help each other.D. They can remember where to go when in danger.10. What can we learn from the text?A. Only African cichlids have a good memory.B. African cichlids can remember things for 12 days.C. African cichlids always treat other fish aggressively.D. African cichlids don't belong to the list of smart animals.11. What is the text mainly about?A. What we can learn from fish.B. Fish having a very bad memory.C. How fish improve their memory.D. Fish being smarter than we thought.DWith their tiny brains and excellent ability to memorize nectar locations, honeybees are a favorite model organism for studying learning and memory. Such research has indicated that to form long-term memories—ones that last a day or more—the insects need to repeat a training experience at least three times. By contrast, short-and mid-term memories that last seconds to minutes and minutes to hours, respectively, need only a single learning experience.Exceptions to this rule have been observed, however. For example, in some studies, bees formed long-lasting memories after a single learning event. Such results are often regarded as circumstantial anomalies, says Martin Giurfa of the University of Toulouse. But the anomalous findings, together with research showing that fruit flies and ants can form long-term memories after single experiences, aroused Giurfa’s curiosity. Was it possible that honeybees could reliably do the same? Giurfa reasoned that the ability to form long-term memories might depend on the particular type of bee and the experience. Within a honeybee colony, there are nurses, who clean the hive and feed the young; guards, who patrol and protect the hive; and foragers, who search for nectar.While previous studies have tested bees as a whole, Ciurfa and his colleagues focused on foragers, tasking them with remembering an experience relevant to their role: an odor associated with a sugary reward.The researchers observed that a single exposure to a reward-paired odor was enough for most forager bees to remember that specific odor the following day. Many foragers could even remember the odor three day later.The results do not mean that all prior research was wrong, says André Fiala of the University of Göttingen. “People have done the experiments in a different way.” Still, the new results do show that “the commonly held belief that one needs multiple training trials to achieve long-term memory is not always true,” he says, and this“really advances the field.”12. What does the author want to tell us through Paragraph 1?A. A model for memory research.B. The classification of memories.C. New research on learning and memory.D. Previous findings on memory formation.13. Which factor might influence a bee’s memory of an experience, according to Giurfa’s research?A. Whether the bee's role is related.B. Whether the bee is introduced or native.C. How often the bee repeats the experience.D. How long the bee is exposed to the reward.14. What is Andre Fiala’s attitude towards the new results?A. Doubtful.B. Favorable.C. Intolerant.D. Unclear.15. Which of the following is a suitable title for the text?A. Learning and Memory: How Honeybees RememberB. Honeybee Memory: Honeybee Knows What to DoC. Honeybees Remember after Just One LessonD. Honeybees Use Memory for Communication第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020届海南省华侨中学2017级高三第五次月考数学试卷★祝考试顺利★(解析版)一、单选题（单选题每个小题只有一个正确选项,每小题5分,共计40分）1.已知复数224(1)+=-iz i （i 为虚数单位）,则z 的模||z 为（ ）A. C. 5 【答案】B【解析】化简得到2z i =-+,再计算z 得到答案.【详解】224242(12)i i z i i i ++===--+-,故z =故选B2.设集合{}1,0,1,2,3,4A =-,{}|2B x x A x A =∈∈,,则集合B 中元素的个数为（）A. 1B. 2C. 3D. 4【答案】C【解析】先求出集合B ,再确定元素个数.【详解】因为{}1,0,1,2,3,4A =-,{}|2B x x A x A =∈∈,,所以{}0,1,2B =,所以集合B 中有3个元素,故选:C.3.在等比数列{}n a 中,若435,,a a a 成等差数列,则数列{}n a 的公比为（ ）A. －1或－2B. 1或－2C. 1或2D. －2【答案】B【解析】由等差中项的性质可得3452a a a =+,从而有220q q +-=,进而可得解.【详解】因为在等比数列{}n a 中,435,,a a a 成等差数列,所以345332322a a a a a a q q ⇒=++⋅⋅=,又0n a ≠,所以220q q +-=,解得1q =或2q =-,故选:B.4.设21log 3a =,432b =,2313c ⎛⎫= ⎪⎝⎭,则（ ） A . a b c <<B. c a b <<C. b c a <<D. a c b << 【答案】D【解析】根据指数,对数函数的单调性分别比较,,a b c 与0,1的大小关系即可. 【详解】221log log 103a =<=, 41322=2b =>02311133c <⎛⎫⎛⎫= ⎪ ⎪⎝⎝⎭=⎭,故01c <<,所以a c b <<,故选:D.5.在正方体1111ABCD A B C D -中,E 为棱1CC 的中点,则异面直线AE 与CD 所成角的正切值为【答案】C【解析】利用正方体1111ABCD A B C D -中,//CD AB ,将问题转化为求共面直线AB 与AE 所成角的正切值,在ABE ∆中进行计算即可.。
海南省海口市华侨中学2025届高三第五次模拟考试数学试卷考生请注意：1．答题前请将考场、试室号、座位号、考生号、姓名写在试卷密封线内，不得在试卷上作任何标记。
2．第一部分选择题每小题选出答案后，需将答案写在试卷指定的括号内，第二部分非选择题答案写在试卷题目指定的位置上。
3．考生必须保证答题卡的整洁。
考试结束后，请将本试卷和答题卡一并交回。
一、选择题：本题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1．已知实数集R ，集合{|13}A x x =<<，集合1|2B x y x ⎧⎫==⎨⎬-⎩⎭，则()R A C B ⋂=（ ） A ．{|12}x x <≤ B ．{|13}x x << C ．{|23}x x ≤<D ．{|12}x x <<2．在等差数列{}n a 中，若n S 为前n 项和，911212a a =+，则13S 的值是（ ） A ．156B ．124C ．136D ．1803．已知六棱锥P ABCDEF -各顶点都在同一个球（记为球O ）的球面上，且底面ABCDEF 为正六边形，顶点P 在底面上的射影是正六边形ABCDEF 的中心G ，若6PA =，2AB =，则球O 的表面积为（ ）A ．163πB ．94π C ．6πD ．9π4．若，则（ ） A ．B ．C ．D ．5．已知集合2{|1}A x x =<，{|ln 1}B x x =<，则 A ．{|0e}A B x x =<< B ．{|e}A B x x =< C ．{|0e}A B x x =<<D ．{|1e}AB x x =-<<6．()2523(2)x x x --+的展开式中，5x 项的系数为（ ） A ．－23B ．17C ．20D ．637．在平面直角坐标系中，经过点(22,2)P ，渐近线方程为2y x =的双曲线的标准方程为（ ）A ．22142-=x yB ．221714x y -=C ．22136x y -=D ．221147y x -=8．已知集合{}2|320M x x x =-+≤，{}|N x y x a ==-若M N M ⋂=，则实数a 的取值范围为（ ）A ．(,1]-∞B ．(,1)-∞C ．(1,)+∞D ．[1,)+∞9．已知函数()xf x e b =+的一条切线为(1)y a x =+，则ab 的最小值为（ ） A ．12e-B ．14e-C ．1e-D ．2e-10．已知甲、乙两人独立出行，各租用共享单车一次（假定费用只可能为1、2、3元）．甲、乙租车费用为1元的概率分别是0.5、0.2，甲、乙租车费用为2元的概率分别是0.2、0.4，则甲、乙两人所扣租车费用相同的概率为（ ） A ．0.18B ．0.3C ．0.24D ．0.3611．如图，正方体1111ABCD A B C D -的棱长为1，动点E 在线段11A C 上，F 、M 分别是AD 、CD 的中点，则下列结论中错误的是（ ）A ．11//FM AC ，B ．存在点E ，使得平面//BEF 平面11CCD D C ．BM ⊥平面1CC FD ．三棱锥B CEF -的体积为定值12．已知某几何体的三视图如右图所示，则该几何体的体积为（ ）A ．3B ．103C ．113D ．83二、填空题：本题共4小题，每小题5分，共20分。
海南省2020届高三年级第五次模拟考试数学试题一、选择题1．设复数2z m i =++，若z 的实部与虚部相等，则实数m 的值为（ ）A ．-3B ．-1C ．1D ．3【答案】B【解析】因为复数2z m i =++的实部与虚部相等，所以21+=m ，解得1m =-．故选：B. 2．sin1050︒=（ ）A ．-1B ．12- C ．0 D 【答案】B【解析】()()1sin1050sin 336030sin 302︒=⨯︒-︒=-︒=-．故选:B 3．设集合(){}211A x x =-≤，(){}20B x x x =+≤，则AB =（ ）A ．[]1,1-B ．[]0,2C ．[]22-,D ．[]2,1-【答案】C【解析】∵(){}{}21102A x x x x =-≤=≤≤，(){}{}2020B x x x x x =+≤=-≤≤，∴[]2,2AB =-，故选：C ．4．已知函数()2e ,0,,0,x x f x ax x ⎧-≥=⎨<⎩若()()01f f =，则a 的值为（ ）A ．1B ．0C ．-1D ．2【答案】A 【解析】()()()()()20e 111ff f f a =-=-=-=，所以a 的值为1．故选：A.5．统计与人类活动息息相关，我国从古代就形成了一套关于统计和整理数据的方法．据宋元时代学者马端临所著的《文献通考》记载，宋神宗熙宁年间（公元1068－1077年），天下诸州商税岁额：四十万贯以上者三，二十万贯以上者五，十万贯以上者十九……五千贯以下者七十三，共计三百十一．由这段内容我们可以得到如下的统计表格：则宋神宗熙宁年间各州商税岁额（单位：万贯）的中位数大约为（ ） A ．0.5 B ．2C ．5D ．10【答案】B【解析】总频数为311，则中位数是所有数据从小到大第156个数据，156733548--=，中位数大约在区间[)1,3的中点处，所以中位数大约为2．故选：B6．已知等差数列{}n a 的前n 项和为n S ，若316214S a a -+=，则9S =（ ）A ．7B ．10C ．63D ．18【答案】C【解析】等差数列{}n a 的首项为1a ，公差为d ∴311323332S a d a d ⨯=+=+，615a a d =+，∴111133252814a d a a d a d +-++=+=，∴147a d +=，即57a =∴()199599632a a S a+⨯===．故选：C7．函数()()224log log 44xf x x =⋅的最小值为（ ） A ．94-B ．2-C ．32-D ．0【答案】A【解析】由题意知()f x 的定义域为(0,)+∞．所以，()()()()2224224log log 4log log 41log 4xf x x x x =⋅=-⋅+，()()()()22222221992log 1log log log 2log 244f x x x x x x ⎛⎫=-++=--=--≥- ⎪⎝⎭，故选：A ．8．从某个角度观察篮球（如图1），可以得到一个对称的平面图形，如图2所示，篮球的外轮席为圆O ，将篮球表面的粘合线看成坐标轴和双曲线，若坐标轴和双曲线与圆O 的交点将圆O 的周长八等分，且AB BC CD ==，则该双曲线的离心率为（ ）A 2B 3C ．355D ．477【答案】D【解析】设双曲线的方程为()222210,0x y a b a b-=>>，则OC a =．因为AB BC CD ==，所以2CD OC =，所以33OD OC a ==．因为坐标轴和双曲线与圆O 的交点将圆O 的周长八等分，所以点22a ⎫⎪⎭在双曲线上，代入双曲线方程得2299122a b -=，解得2297b a =．所以双曲线的离心率为229471177c b e a a==+=+=D ． 二、多选题9．已知正方形ABCD 的边长为2，向量a ，b 满足2AB a =，2AD a b =+，则（ ）A ．22b =B ．a b ⊥C ．2a b ⋅=D ．()4a b b +⊥【答案】AD【解析】由条件可b AD AB BD =-=，所以22b BD ==，A 正确；12a AB =，与BD 不垂直，B 错误；122a b AB BD ⋅=⋅=-，C 错误；4a b AB AD AC +=+=，根据正方形的性质有AC BD ⊥，所以()4a b b +⊥，D 正确．故答案为：AD .10．设α和β是两个不同的平面，m ，n 是两条不同的直线，则下列说法正确的是（ ）A ．若//m α，βn//，//m n ，则//αβB ．若m α⊥，n β⊂，//αβ，则m n ⊥C ．若m α⊥，n β⊥，m n ⊥，则αβ⊥D ．若m α⊥，n β⊥，//αβ，则//m n【答案】BCD【解析】//m α，βn//，//m n ，并不能推出//αβ，这时α和β还可能相交，故A 错误；若m α⊥，//αβ，则m β⊥，又n β⊂，则m n ⊥，B 正确；若m α⊥，m n ⊥，则//n α或n ⊂α，又n β⊥，则αβ⊥，C 正确；若m α⊥,//αβ，中m β⊥，又n β⊥，则//m n ，D 正确. 11．函数()()π6sin 0f x x ωω⎛⎫ ⎪⎝⎭=+>的最小正周期为π2，则（ ） A ．ω的值为4B ．()f x 图象的一条对称轴为直线π6x = C ．π6f x ⎛⎫+⎪⎝⎭是偶函数 D ．函数()f x 在区间ππ,412⎡⎤--⎢⎥⎣⎦上的最大值为12【答案】BC【解析】对A ，因为ππ2T ω==，所以ω的值为2，A 错误； 对B ，()πsin 26f x x ⎛⎫=+ ⎪⎝⎭，当π6x =时，()1f x =，所以π6x =是函数()f x 图象的一条对称轴，B 正确； 对C ，πππsin 2666f x x ⎡⎤⎛⎫⎛⎫+=++= ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦πsin 2cos 22x x ⎛⎫+= ⎪⎝⎭，所以π6f x ⎛⎫+ ⎪⎝⎭是偶函数，C 正确； 对D ，当ππ,412x ⎡⎤∈--⎢⎥⎣⎦时，ππ2,063x ⎡⎤+∈-⎢⎥⎣⎦，所以函数πsin 26y x ⎛⎫=+ ⎪⎝⎭的取值范围是2⎡⎤-⎢⎥⎣⎦，所以函数()f x 在区间ππ,412⎡⎤--⎢⎥⎣⎦D 错误． 故选：BC12．设椭圆22193x y +=的右焦点为F ，直线(0y m m =<<与椭圆交于A ，B 两点，则（ ）A ．AF BF +为定值B ．ABF 的周长的取值范围是[]6,12C ．当2m =时，ABF 为直角三角形 D ．当1m =时，ABF 【答案】ACD【解析】设椭圆的左焦点为F '，则AF BF '=，∴=6AF BF AF AF '+=+为定值，A 正确；ABF 的周长为AB AF BF ++，因为AF BF +为定值6，∴AB 的范围是()0,6，∴ABF 的周长的范围是()6,12，B 错误；将y =与椭圆方程联立，可解得A ⎛ ⎝⎭，B ⎝⎭，又∵)F ，∴260222AF BF ⎛⎛⋅=+-+=⎭⎝⎭，∴ABF 为直角三角形，C 正确；将1y =与椭圆方程联立，解得()A ，)B ，∴112ABFS=⨯=D 正确． 故选：ACD 三、填空题13．能够说明“*x ∀∈N ，22x x ≥”是假命题的一个x 值为__________．【答案】3【解析】因为*3x =∈N ，而3223<，∴说明“*x ∀∈N ，22x x ≥”是假命题．14．为了给国外新冠肺炎疫情严重的地区提供援助，国内某机构计划派出由5人组成的专家指导小组，其中甲、乙、丙3人通晓英语，丁、戊2人通晓法语，现从中随机选出通晓英语、法语的专家各1名作为领队，则甲和丁至少有1人被选中的概率为__________． 【答案】23【解析】从5人中选出通晓英语、法语的专家各1名的可能结果为（甲，丁），（甲，戊），（乙，丁），（乙，戊），（丙，丁），（丙，戊），共6种情况．甲和丁至少有1人被选中的有（甲，丁），（甲，戊），（乙，丁），（丙，丁），共4种情况． 甲和丁至少有1人被选中的概率为42==63P ． 15．一个底面半径为r ，高为h 的圆柱内接于半径为R 的球O 中，若h=R ，则rR=__________．【解析】做出该圆柱内接于球O 的轴截面如图所示，则OA R =，22h ROB ==，AB r =，在OAB 中，2222322R AB OA OB R R r ⎛⎫=-=-== ⎪⎝⎭，所以32r R =．16．设()f x '是奇函数()f x 的导函数，()23f -=-，且对任意x ∈R 都有()2f x '<，则()2f =_________，使得()e 2e 1x x f <-成立的x 的取值范围是_________．【答案】3 ()ln 2,+∞【解析】∵()f x 是奇函数，∴()()223f f =--=，设()()2g x f x x =-，则()()22g f =-41=-，()()20g x f x ''=-<， ∴()g x 在R 上单调递减， 由()e2e1xxf <-得()e e 21x x f -<-，即()()2e xg g <，∴e 2x >，得ln 2x >， 四、解答题17．设33M a =-，22N a =，4T a =，给出以下四种排序：①M ，N ，T ；②M ，T ，N ；③N ，T ，M ；④T ，N ，M ．从中任选一个，补充在下面的问题中，解答相应的问题．已知等比数列{}n a 中的各项都为正数，11a =，且__________依次成等差数列． （Ⅰ）求{}n a 的通项公式；（Ⅱ）设,01,{1,1,n n n n na ab a a <≤=>数列{}n b 的前n 项和为n S ，求满足100n n S b >的最小正整数n ．注：若选择多种排序分别解答，按第一个解答计分． 【解析】（解答一）选②或③：（Ⅰ）设{}n a 的公比为q ，则0q >．由条件得423223a a a =-，又因为11a =，所以32223q q q =-，即22320q q +-=，解得12q =（负值舍去）．所以112n n a -=．（Ⅱ）由题意得112n n b -=，则1112121212n nn n S ---==-．由100n n S b >得 112110022n n n --->，即2101>n，又因为*n ∈N ，所以n 的最小值为7． （解答二）选①或④：（Ⅰ）设{}n a 的公比为q ，则0q >．由条件得24343a a a =-，又因为11a =，所以3243q q q =-，即2340q q --=，解得4q =（负值舍去）．所以14n n a -=．（Ⅱ）由题意得114n n b -=，则11141413414n n n n S ---==⨯-．由100n n S b >得 1141100344n n n --->⨯，即4301n >，又因为*n ∈N ，所以n 的最小值为5． 18．设ABC 的内角A ，B ，C 所对的边分别为a ，b ，c ．sin A ，cos C 分别为方程212530x x +-=的两根．（Ⅰ）求sin B ；（Ⅱ）若2a =，求ABC 的面积． 【解析】（1）解方程212530x x +-=得113x =，234x =-．因为(),0,πA C ∈，所以1sin 3A =，3cos 4C =-，所以cos 3A ==，sin C == 因为πA B C ++=，所以()sin sin sin cos cos sin B A C A C A C =+=+13134346⎛⎫=⨯-+=-+⎪⎝⎭．（结果写成312也对）（2）由正弦定理sin sin c a C A =，所以2sin 41sin 3a Cc A⨯===， 所以1sin 2ABC S ac B =△11224⎛=⨯⨯-+=+ ⎝⎭．也对） 19．甲、乙两人进行围棋比赛，比赛要求双方下满五盘棋，开始时甲每盘棋赢的概率为34，由于心态不稳，甲一旦输一盘棋，他随后每盘棋赢的概率就变为12．假设比赛没有和棋，且已知前两盘棋都是甲赢． （Ⅰ）求第四盘棋甲赢的概率；（Ⅱ）求比赛结束时，甲恰好赢三盘棋的概率．【解析】（Ⅰ）设事件A 为“第四盘棋甲赢”，若第四盘棋甲赢，分两种情况： 若第三盘棋和第四盘棋都是甲赢，概率13394416P =⨯=， 若第三盘棋乙赢，第四盘棋甲赢，概率2111428P =⨯=， ∴()12911116816P A P P =+=+=； （Ⅱ）设事件B 为“比赛结束时，甲恰好赢三盘棋”，若甲恰好赢三盘棋，则他在后三盘棋中只赢一盘，分三种情况：若甲第三盘赢，概率33113144232P ⎛⎫=⨯⨯-= ⎪⎝⎭， 若甲第四盘赢，概率41111142216P ⎛⎫=⨯⨯-= ⎪⎝⎭， 若甲第五盘赢，概率51111142216P ⎛⎫=⨯-⨯= ⎪⎝⎭， ∴()345311732161632P B P P P =++=++=． 20．如图，在三棱柱111ABC A B C -中，BC ⊥平面11ABB A ，四边形11ABB A 为菱形．（Ⅰ）证明：1AB ⊥平面1A BC ；（Ⅱ）若160ABB ∠=︒，4AB =，二面角11C A B A --的余弦值为217，求三棱锥1C ABB -的体积．【解析】（Ⅰ）因为四边形11ABB A 为菱形，所以11AB A B ⊥． 因为BC ⊥平面11ABB A ，1AB ⊂平面11ABB A ，所以1AB BC ⊥． 又因为1A BBC B =，1A B ⊂平面1A BC ，BC ⊂平面1A BC ，所以1AB ⊥平面1A BC ．（Ⅱ）以B 为坐标原点，分别以1BB ，BC 所在的直线为x 轴和z 轴， 以过B 点垂直平面11BB C C 的直线为y 轴，建立空间直角坐标系如图所示．设()0BC h h =>，则()0,0,0B ，()14,0,0B ，()16,23,0A ，()0,0,C h ．所以()14,0,B C h =-，()112,23,0B A =．设平面11CA B 的法向量为(),,n x y z =，则1110,0,n B C n B A ⎧⋅=⎪⎨⋅=⎪⎩即40,2230,x hz x -+=⎧⎪⎨+=⎪⎩令1x =，得341,3n h ⎛⎫=- ⎪ ⎪⎝⎭． 由条件知()0,0,BC h =为平面11AA B 的一个法向量．设二面角11C A B A --的平面角为θ，易知θ为锐角．则cos 7θ==，解得4h =．所以11111444sin 603323C A BB B A B V BC S -=⨯︒=⨯=⨯⨯⨯⨯⨯． 21．已知抛物线2:4C y x =的焦点为F ，过F 的直线交抛物线C 于()11,A x y ，()22,B x y 两点．（Ⅰ）当14y =时，求2y 的值；（Ⅱ）过点A 作抛物线准线的垂线，垂足为E ，过点B 作EF 的垂线，交抛物线于另一点D ，求ABD △面积的最小值．【解析】（Ⅰ）由题意知()1,0F ，设直线AB 的方程为1x ty =+，联立21,4,x ty y x =+⎧⎨=⎩消去x 得2440y ty --=． 由根与系数的关系得124y y =-．当14y =时，21y =-．（Ⅱ）设()00,D x y ，()20,4m m m A ⎛⎫⎪⎝≠⎭，则()1,E m -， 由（Ⅰ）知24y m =-，所以244,B m m⎛⎫- ⎪⎝⎭．因为BD EF ⊥，0112EF m m k -==---，所以2BD k m=． 所以直线BD 的方程为2424y x m m m⎛⎫+=- ⎪⎝⎭，即28240x my m ---=． 联立方程组得228240,4,x my my x ⎧---=⎪⎨⎪=⎩消去x 得2216280y my m ---=， 所以202y y m +=，202168y y m=--． ()2222020202644432y y y y y y m m -=+-=++，所以20BD y y =-=试卷第11页，总12页 设点A 到BD 的距离为d，则22168m d ++==．所以332222111618816244ABD S BD d m m ⎛⎫⎛⎫=⋅=++≥= ⎪ ⎪ ⎪⎝⎭⎝⎭， 当且仅当2m =±时等号成立，所以ABD △面积的最小值为16． 22．已知0a >，函数()()21ln f x a x x =--，()111ex g x x -=-． （Ⅰ）求()f x 的单调区间；（Ⅱ）证明：当1x >时，()0g x >；（Ⅲ）若()()f x g x >在区间()1,+∞上恒成立，求a 的取值范围．【解析】（Ⅰ）()()212120ax f x ax x x x-'=-=>． 因为0a >，由()0f x '=，得x = 由()0f x '>，得x >()0f x '<，得x <． 所以()f x的单调递增区间为⎫+∞⎪⎭，单调递减区间为⎛ ⎝． （Ⅱ）证明：设()1e x x x ϕ-=-，则()1e 1x x ϕ-'=-． 当1x >时，()0x ϕ'>，所以()x ϕ在()1,+∞上单调递增， 所以()()10x ϕϕ>=，即1e x x ->，所以11e 1x x-<， 所以当1x >时，()0g x >． （Ⅲ）当102a <<1>，由（Ⅰ）知，()10f f <=，而0g >，此时()()f x g x >在区间()1,+∞上不恒成立． 当12a ≥时，设()()()()1h x f x g x x =-≥．试卷第12页，总12页 当1x >时，()21221111112121e x h x ax x x x x x x x x -'=-+->-+->-+ 22210x x x -+=>， 所以()h x 在()1,+∞上单调递增，所以()()10h x h >=， 即此时()()f x g x >恒成立．综上所述，a 的取值范围是1,2⎡⎫+∞⎪⎢⎣⎭．。
海南省海口市达标名校2020年高考五月质量检测化学试题一、单选题（本题包括15个小题，每小题4分，共60分．每小题只有一个选项符合题意）1．下列有关物质性质与用途具有对应关系的是A．Si02熔点很高，可用于制造坩埚B．NaOH能与盐酸反应，可用作制胃酸中和剂C．Al(OH)3是两性氢氧化物，氢氧化铝胶体可用于净水D．HCHO可以使蛋白质变性，可用于人体皮肤伤口消毒2．设N A为阿伏加德罗常数的值，下列叙述正确的是A．所含共价键数均为0.4N A的白磷(P4)和甲烷的物质的量相等B．1 mol Na与O2反应，生成Na2O和Na2O2的混合物共失去N A个电子C．1mol Na2O2固体中含有离子总数为4N AD．25℃时，pH=13的氢氧化钠溶液中约含有N A个氢氧根离子3．聚苯胺是一种在充放电过程中具有更优异可逆性的电极材料。
Zn一聚苯胺二次电池的结构示意图如图所示，设N A为阿伏加德罗常数的值。
下列说法错误的是（）A．放电时，外电路每通过0.1N A个电子时，锌片的质量减少3.25gB．充电时，聚苯胺电极的电势低于锌片的电势C．放电时，混合液中的Cl-向负极移动D．充电时，聚苯胺电极接电源的正极，发生氧化反应4．测定硫酸铜晶体中结晶水含量实验，经计算相对误差为+0.4%，则下列对实验过程的相关判断合理的为（）A．所用晶体中有受热不挥发的杂质B．用玻璃棒搅拌时沾去少量的药品C．未将热的坩埚放在干燥器中冷却D．在实验结束时没有进行恒重操作5．一定条件下，不能与SiO2反应的物质是（）A．CaCO3B．NaOH C．盐酸D．焦炭6．已知：25℃时，K sp[Ni(OH)2]=2.0×10-15，K sp[Fe(OH)3]=4.0×10-38。
将含Fe2O3、Ag、Ni的某型废催化剂溶于盐酸，过滤，滤渣为Ag，所得溶液中c(Ni2+)=c(Fe3+)=0.4mol/L。
向该溶液中滴加一定浓度的NaOHB．加入NaOH溶液时，先产生Ni(OH)2沉淀C．当滴定到溶液pH=5时，溶液中lg() ()2+3+NiFecc约为10D．当滴定到溶液呈中性时，Ni2+已沉淀完全7．短周期元素W、X、Y、Z的原子序数依次增加，W原子的核外电子数等于电子层数，X2-和Y+的核外电子排布相同，X与Z同族。
海南侨中2014届高三第5次数学（理科）测试卷第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.)1.复数i iz 2143++=的共轭复数z =（C ） A.i 52511- B.i 51152- C.i 52511+ D.i 51152+ 2.不等式组⎪⎩⎪⎨⎧≤-+≥--020222x x x x 的解集用数轴表示为（B ）3.如右图所示的程序框图.若两次输入x 的值分别为π和3π-，则两次运行程序输出的b 值分别为（A ）A.π，23-B.1，23C.,023D.π-，23- 4.双曲线12222=-by a x )0,0(>>b a 的一个焦点F 到它的一条渐近线距离x 满足a x a 3≤≤，则该双曲线的离心率的取值范围为（D ）A.),2[+∞B.)10,1(C.)10,2[D.)10,2[5.已知m ，n 为两条不同的直线,α，β为两个不同的平面,下列命题中正确的是（D ）A ．l m ⊥,l n ⊥,且,m n α⊂,则l α⊥B ．若平面α内有不共线的三点到平面β的距离相等,则βα//C ．若n m m ⊥⊥,α,则α//nD ．若α⊥n n m ,//,则α⊥m 6.若锐角α满足3cos 32sin 2=+αα，则)322tan(πα+的值是（B ） A.73- B.73 C.773-D.7737.如图是一台微波炉的操作界面.若一个两岁小孩触碰E D C B A 、、、、五个按钮是等可能的，则他不超过两次按钮启动微波炉的概率为（B ） A.257 B.259 C.258 D.25118.下列命题中真命题的个数为（B ）①R x ∈∃0,使得2cos sin =+x x .②锐角ABC ∆中，恒有1tan tan >B A . ③R x ∈∀,不等式012<--ax ax 成立的充要条件为：04<<-a A.0B.1C.2D.39.（理）二项式nx )1(+展开式的二项式系数之和为64，则nx )1(-展开式第四项的系数为(C) A.15B.20C.20- D.15-10.平行四边形ABCD 中，点E 为AD 中点，连接AC BE 、且交于点F .若AE y AB x AF +=)(R y x ∈、，则=y x :（C ）A.3:1B.3:2C.2:1D.4:311.已知集合},,20,20|),{(R c a c a c a A ∈<<<<=，则任取(,)a c A ∈，关于x 的方程022=++c x ax 无实根的概率（D ）A ．22ln 1+B ．42ln 21+C ．22ln 1-D ．42ln 23- 12.(理)某几何体的三视图如右所示，若该几何体的外接球的表面积为π3， 则正视图中=a (A) A.2B.23C.2D.π 第Ⅱ卷二、填空题：(本大题共4小题，每小题5分，把答案填在答题卡中的指定位置）13．对于n N +∈的命题，下面四个判断：①若2()1222nf n =++++L ，则(1)1f =；②若21()1222n f n -=++++L ，则(1)12f =+；③若111()12321f n n =+++++L ，则(1)f 11123=++；④若111()1231f n n n n =++++++L ，则1111(1)()3233341f k f k k k k k +=+++-++++其中正确命题的序号为___③④__________．15在ABC ∆中，角,,A B C 的对边分别为a ，b ，c.已知5sin 13B =，且a ，b ，c 成等比数列. 则11tan tan A C+= 513.15．已知实数,x y 满足0024x y x y s y x ≥⎧⎪≥⎪⎨+≤⎪⎪+≤⎩，当23s ≤≤时，目标函数32z x y =+的最大值函数()f s 的最小值为_____6________．16．将正奇数1,3,5,7,L 按右表的方式进行排列，记ij a 表示第i 行 第j 列的数，若2013ij a =，则i j +的值为 254 .三、解答题（本大题共5小题，共60分.解答题应写出文字说明、证明过程或演算步骤.请将大体的过程写在答题卷中指定的位置）17.（本小题满分12分）已知数列{}n a 的前n 项和为n S ，且10a =，对任意n ∈N *，都有()11n n na S n n +=++. （1）求数列{}n a 的通项公式；（2）若数列{}n b 满足22log log n n a n b +=，求数列{}n b 的前n 项和n T .解：（1）21n n S na n n +=--Q …①212(1)(1)(1)n n S n a n n ++∴=+-+-+…②由②-①得：121(1)22n n n a n a na n +++=+---，21(1)(1)2(1)n n n a n a n +++=+++212n n a a ++=+，即：212n n a a ++-=…③;又21122a S a =+=+，即：212a a -=…④综合③、④可得：对*n N ∈,有12n n a a +-=成立.∴数列{}n a 是以10a =为首项，公差2d =的等差数列.所以数列{}n a 的通项公式为：22n a n =-.（2）Q 数列{}n b 满足22log log n n a n b +=，∴2222log log n n n b -+=，2log 22nb n n∴=-，14n n b n -∴=⋅.01221142434(1)44 0n n n T n n --∴=⋅+⋅+⋅++-⋅+⋅+L …⑤ 12140 1424 (1)44n n n T n n -∴=+⋅+⋅++-⋅+⋅L …⑥由⑤-⑥可得：0121344444n nn T n --=++++-⋅L 41441n n n -=-⋅-，441(31)41399n n n n n n T ⋅--⋅+∴=-=18．某校学生会组织部分同学，用“10分制”随机调查“阳光”社区人们的幸福度.现从调查人群中随机抽取12名，如图所示的茎叶图记录了他们的幸福度分数(以小数点前的一位数字为茎，小数点后的一位数字为叶)：(1)指出这组数据的众数和中位数；(2)若幸福度不低于9.5分，则称该人的幸福度为“极幸福”.求从这12人中随机选取3人，至多有1人是“极幸福”的概率； (3)以这12人的样本数据来估计整个社区的总体数据，若从该社区(人数很多)任选2人，记ξ表示抽到“极幸福”的人数，求ξ的分布列及数学期望． 18.解：（1）众数：8.6；中位数：8.7；……………………………2分（2）设i A 表示所取3人中有i 个人是“极幸福”，至多有1人是“极幸福”记为事件A ，则3129390133121248()()()55C C C P A P A P A C C =+=+=；…………………6分（3）ξ的可能取值为0，1，2.239(0)()416P ξ===；12133(1)448P C ξ==⨯⨯=；211(2)()416P ξ===；…….10分所以ξ的分布列为：9311()012168162E ξ=⨯+⨯+⨯=………..……….…12分另解：ξ的可能取值为0，1，2.则1~(2,)4B ξ，2213()()()(0,1,2)44k k kP k C k ξ-===其中ξ 0 12 P916 38116所以11()242E ξ=⨯=. 19.已知四边形ABCD 是菱形,2==DB DA ,ABCD DD 面⊥1，点P 为线段1OD 上的任一点.(1)若21=DD ,1OD DP ⊥,求OD 与面1D AC 所成角的正切值；(2)若二面角D AD C --1的平面角的余弦值为515，求线段1DD 的长. 解析：（1）Q AC BD 、为四边形ABCD 的两条对角线，AC BD ∴⊥. 又ABCD DD 面⊥1，AC ABCD ⊂面,1AC DD ∴⊥.且1111,,DD DB D DD D DB DB D DB ⋂=⊂⊂Q 面面，1AC D DB ∴⊥面. 再1DP D DB ⊂Q 面,DP AC ∴⊥,且1OD DP ⊥，1DP D AC ∴⊥面.OD ∴与面1D AC 所成角为DOP ∠.由条件21=DD ，1DO =,1tan 2DD DOP DO∴∠== （2）如图建立空间直角坐标系oxyz ，则)0,0,3(A ，)0,1,0(-D ,)2,1,0(1-D ,易求得面DA D 1的一个法向量)0,3,1(1-=n .设线段1DD 的长为0z ，),1,0(01z D -∴，),1,3(01z AD --=，)0,0,32(-=AC ,设面C AD 1的一个法向量),,(2z y x n =.由⎪⎩⎪⎨⎧=⋅=⋅00221n AC n AD ，可得：⎩⎨⎧==-+0030x z z y x ，由0=x ，z z y 0=，令1=z ，可得：0z y = )1,,0(02z n =∴,由（2）已知面面DA D 1的一个法向量)0,3,1(1-=n ，再因二面角D AD C --1的平面角的余弦值为515，515123||||||2002121=+=⋅⋅∴z z n n n n ，可解得：20=z ，即：线段1DD 的长为2.20.（本小题满分12分）已知椭圆C ：)0(12222>>=+b a by a x 经过点)23,1(P ，离心率21=e ，直线l 的方程为4=x .(1)求椭圆C 的方程；(2)AB 是经过右焦点F 的任一弦（不经过点P ），设直线AB 与l 相交于点M ，记PA ,PB ,PM 的斜率分别为321,,k k k ，问：是否存在常数λ，使得321k k k λ=+？若存在，求出λ的值，若不存在，说明理由。
2020届海南华侨中学高三第五次月考数学试题一、单选题 1．已知复数224(1)+=-iz i （i 为虚数单位），则z 的模||z 为（ ）A ．BC ．5D【答案】B【解析】化简得到2z i =-+，再计算z 得到答案. 【详解】224242(12)i iz i i i++===--+-，故z =故选B 【点睛】本题考查了复数模的计算，意在考查学生的计算能力.2．设集合{}1,0,1,2,3,4A =-，{}|2B x x A x A =∈∈,，则集合B 中元素的个数为（ ） A ．1 B ．2C ．3D ．4【答案】C【解析】先求出集合B ,再确定元素个数. 【详解】因为{}1,0,1,2,3,4A =-,{}|2B x x A x A =∈∈,, 所以{}0,1,2B =, 所以集合B 中有3个元素, 故选:C. 【点睛】本题考查集合,属于简单题.3．在等比数列{}n a 中，若435,,a a a 成等差数列，则数列{}n a 的公比为（ ） A ．－1或－2 B ．1或－2C ．1或2D ．－2【解析】由等差中项的性质可得3452a a a =+,从而有220q q +-=,进而可得解. 【详解】因为在等比数列{}n a 中,435,,a a a 成等差数列,所以345332322a a a a a a q q ⇒=++⋅⋅=, 又0n a ≠,所以220q q +-=,解得1q =或2q =-, 故选:B. 【点睛】本题主要考查等差中项的性质运用,考查等比数列和计算能力,难度不大.4．设21log 3a =，432b =，2313c ⎛⎫= ⎪⎝⎭，则（ ）A ．a b c <<B ．c a b <<C ．b c a <<D ．a c b <<【答案】D【解析】根据指数,对数函数的单调性分别比较,,a b c 与0,1的大小关系即可. 【详解】221log log 103a =<=, 41322=2b =>2311133c <⎛⎫⎛⎫= ⎪ ⎪⎝⎝⎭=⎭,故01c <<,所以a c b <<, 故选:D. 【点睛】本题考查指数,对数式的大小比较,属于基础题.5．在正方体1111ABCD A B C D -中，E 为棱1CC 的中点，则异面直线AE 与CD 所成角的正切值为A ．2B ．2C D ．2【答案】CAE 所成角的正切值，在ABE ∆中进行计算即可.【详解】在正方体1111ABCD A B C D -中，//CD AB ，所以异面直线AE 与CD 所成角为EAB ∠， 设正方体边长为2a ，则由E 为棱1CC 的中点，可得CE a =，所以5BE a =，则55tan 22BE a EAB AB a ∠===.故选C.【点睛】求异面直线所成角主要有以下两种方法：（1）几何法：①平移两直线中的一条或两条，到一个平面中；②利用边角关系，找到（或构造）所求角所在的三角形；③求出三边或三边比例关系，用余弦定理求角； （2）向量法：①求两直线的方向向量；②求两向量夹角的余弦；③因为直线夹角为锐角，所以②对应的余弦取绝对值即为直线所成角的余弦值.6．唐朝著名的凤鸟花卉浮雕银杯（如图1所示），它的盛酒部分可以近似地看做是半球与圆柱的组合体（如图2），当这种酒杯内壁表面积固定时（假设内壁表面光滑，表面积为S 平方厘米，半球的半径为R 厘米），要使酒杯容积不大于半球体积的两倍，则R 的取值范围为（ ）3S ⎛3S ⎫3S SD ． 【答案】D【解析】根据题意,酒杯内壁表面积为圆柱与半球的表面积,列出S 的表达式,再求出体积V ,解不等式即可.【详解】设圆柱的高度与半球的半径分别为h ,R , 则表面积222S R Rh ππ=+,故22SRh R ππ=-, 所以酒杯的容积323233224()332323S S V R R h R R R R R R ππππππ=+=+-=-+„,所以2523S R π„,又202SR π->,所以22523S R R ππ<„,R <, 故选:D. 【点睛】本题考查了组合体的体积和表面积的计算,难度不大.7．设a r ，b r 是非零向量，“a b a b ⋅=r r r r ”是“//a b r r ”的（ ）A ．充分而不必要条件B ．必要而不充分条件C ．充分必要条件D ．既不充分也不必要条件【答案】A【解析】cos ,a b a b a b ⋅=⋅r r r r r r ，由已知得cos ,1a b =r r ，即,0a b =r r ，//a b r r .而当//a b r r 时，,a b r r 还可能是π，此时a b a b ⋅=-r r r r ，故“a b a b ⋅=r r r r ”是“//a br r ”的充分而不必要条件，故选A.【考点】充分必要条件、向量共线.8．在三棱锥V ABC -中，面VAC ⊥面ABC ，2VA AC ==，VA AC ⊥，BA BC ⊥则三棱锥V ABC -的外接球的表面积是（ ）【答案】D【解析】设AC 边的中点为D ,VC 边的中点为O ,则由题意可推出VA ⊥面ABC ,又因为BA BC ⊥,则点D 为ABC ∆的外接圆圆心,从而点O 为V ABC -的外接球球心,最后代入数据求解即可. 【详解】如图所示,设AC 边的中点为D ,因为BA BC ⊥,则点D 为ABC ∆的外接圆圆心, 因此三棱锥V ABC -的外接球球心在过点D 的垂线上, 因为面VAC ⊥面ABC ,面VAC I 面ABC AC =,VA AC ⊥, 所以VA ⊥面ABC ,设VC 边的中点为O ,则//VA OD ,即V ABC -的外接球球心在直线OD 上, 又VA AC ⊥,则VO AO =,则点O 即为V ABC -的外接球球心,因为2VA AC ==,所以V ABC -的外接球半径122R VO VC ===, 因此三棱锥V ABC -的外接球的表面积为248R ππ=, 故选:D.【点睛】本题考查三棱锥外接球表面积的求法,需要学生具备一定的空间思维与想象能力,属于中档题.二、多选题9．关于函数2sin 314y x π⎛⎫=++ ⎪⎝⎭，下列叙述正确的是（ ）A ．函数的最小正周期为23π π⎛⎫C ．其图象关于直线4πx =-对称 D ．其图象可由2sin 14y x π⎛⎫=++ ⎪⎝⎭图象上所有点的横坐标变为原来的3倍得到 【答案】AC【解析】利用三角函数的图像及性质一一判断选项正误即可. 【详解】2sin 314y x π⎛⎫=++ ⎪⎝⎭,其最小正周期23T π=,故选项A 正确; 当4x π=时,32sin()12sin 1144y πππ=++=+=,其关于,14π⎛⎫⎪⎝⎭对称,故选项B 错误; 当4πx =-时,334442x ππππ+=-+=,故选项C 正确; 2sin 14y x π⎛⎫=++ ⎪⎝⎭图象上所有点的横坐标变为原来的3倍得到函数12sin 134y x π⎛⎫=++ ⎪⎝⎭,故选项D 错误;故选:AC 【点睛】本题考查三角函数图像､性质的应用,难度不大.10．已知函数21()21x x f x +=-，()2g x x =，则下列结论正确的是（ ）A ．()()f x g x 为奇函数B ．()()f x g x 为偶函数C ．()()f x g x +为奇函数D ．()()f x g x +为非奇非偶函数【答案】BC【解析】先判断函数(),()f x g x 的奇偶性,再利用函数奇偶性的性质判断选项正误. 【详解】21()21x xf x +=-,其定义域为(,0)(0,)-∞+∞U ,21(21)212()()21(21)212x x x xx x x xf x f x ----++⋅+-====---⋅-, 故函数()f x 为奇函数, 又()2g x x =为奇函数,根据函数奇偶性的性质可知:()()f x g x 为偶函数,()()f x g x +为奇函数, 故选:BC. 【点睛】本题考查函数奇偶性的判断及其性质应用,难度不大.11．如图，等边三角形ABC 的中线AF 与中位线DE 相交于G ，已知A ED '∆是ADE ∆绕DE 旋转过程中的一个图形，下列命题中，正确的是（ ）A ．动点A '在平面ABC 上的射影在线段AF 上B ．恒有平面AGF '⊥平面BCDEC ．三棱锥A EFD '-的体积有最大值D ．旋转过程中二面角A DE C '--的平面角始终为A GF '∠ 【答案】ABCD【解析】由斜线的射影定理可判断A 正确;由面面垂直的判定定理,可判断B 正确;由三棱锥的体积公式,可判断C 正确;由二面角的平面角定义可判断D 正确. 【详解】A D A E ''=Q ,ABC ∆是正三角形, ,A G DE GF DE '⊥⊥， DE ⊥平面A GF '，因为DE ⊂平面BCED ，所以平面AGF '⊥平面BCEDA '∴在平面ABC 上的射影在线段AF 上,故A 正确;由A 知, DE ⊥平面A GF ',DE ⊂平面BCED∴恒有平面AGF '⊥平面BCED ,故B 正确;三棱锥A FED ¢-的底面积是定值,体积由高即A '到底面的距离决定,平面A DE 'I 平面CDE DE =,且,A G DE GF DE '⊥⊥,则二面角A DE C '--的平面角为A GF '∠,故D 正确; 故选:ABCD. 【点睛】本题考查了线面､面面垂直的判定定理及性质定理的运用,考查了二面角的平面角的概念,需要学生具备一定的空间想象能力.12．已知函数()2x f x =，2()g x x ax =+（其中a R ∈）.对于不相等的实数1x ，2x ，设()()1212f x f x m x x -=-，()()1212g x g x n x x -=-下列说法正确的是（ ）A ．对于任意不相等的实数1x ，2x ，都有0m >；B ．对于任意的a 及任意不相等的实数1x ，2x ，都有0n >；C ．对于任意的a ，存在不相等的实数1x ，2x ，使得m n =；D ．对于任意的a ，存在不相等的实数1x ，2x ，使得m n =-. 【答案】AD【解析】运用指数函数的单调性,即可判断A;由二次函数的单调性,即可判断B;通过函数2()2x h x x ax =+-,求出导数判断单调性,即可判断C;通过函数2()2x h x x ax =++,求出导数判断单调性,即可判断D. 【详解】对于A,由指数函数的单调性可得()f x 在R 上递增,即有0m >,则A 正确; 对于B,由二次函数的单调性可得()g x 在(,)2a -∞-递减,在(2a-,)+∞递增,则0n >不恒成立,则B 错误;对于C,若m n =,可得1212()()()()f x f x g x g x -=-,即为1122()()()()g x f x g x f x -=-, 设2()2x h x x ax =+-,则应有12()()h x h x =,而()22ln 2x h x x a '=+-,当a →-∞,()h x '小于0,()h x 单调递减,则C 错误;对于D,若m n =-,可得1212()()[()()]f x f x g x g x -=--,即为1122()()()()f x g x f x g x +==+而()22ln 2x h x x a '=++,对于任意的a ,()h x '不恒大于0或小于0, 即()h x 在定义域上有增有减,则D 正确. 故选:AD. 【点睛】本题考查函数的单调性及运用,运用指数函数和二次函数的单调性,以及导数判断单调性是解题的关键.三、填空题13．已知两个单位向量,a b v v 满足||3||a b b +=rr r ，则,a b v v 的夹角为__________．【答案】3π 【解析】将已知等式两边平方后,利用向量的夹角公式可解得. 【详解】因为a r ,b r 是单位向量,所以||||1a b ==r r,因为||3||a b b +=rr r,所以22()3||a b b +=r r r , 所以22223||a b a b b ++⋅=r rr r r ,所以222||||2||||cos ,3||a b a b a b b ++<>=rr rrrr r , 因为||||1a b ==rr,所以3111cos ,2112a b --<>==⨯⨯rr , 又,[0,]a b π<>∈rr ,所以,3a b π<>=rr .故答案为:3π 【点睛】本题考查了向量的数量积和向量夹角公式,属于基础题. 14．已知等比数列中，，数列是等差数列，且，则_______．【解析】根据等比数列的性质得到再由等差数列的中项的性质得到：.【详解】根据等比数列的性质得到：，∴(舍去)，由等差数列的中项的性质得到：，∴.故答案为：8. 【点睛】对于等差等比数列的小题，常用到的方法，其一是化为基本量即首项和公差，其二是观察各项间的脚码关系，即利用数列的基本性质.15．如图所示，在四棱锥P ABCD -中，PA ⊥底面ABCD ，且底面各边都相等，M 是PC 上的一动点，当点M 满足条件①BM DM ⊥，②DM PC ⊥，③BM PC ⊥中的______时，平面MBD ⊥平面PCD （只要填写一个你认为是正确的条件序号即可）.【答案】②（或③）【解析】推出BD PC ⊥,则要得到平面MBD ⊥平面PCD ,即要得到PC ⊥平面MBD ,故只需PC 垂直平面MBD 内的一条与BD 相交的直线即可. 【详解】PA ⊥Q 底面ABCD ,PA BD ∴⊥,Q 底面各边都相等,AC BD ∴⊥,PA AC A =Q I ,BD ∴⊥平面PAC ,BD PC ∴⊥,∴当DM PC ⊥(或)BM PC ⊥时,即有PC ⊥平面MBD ,而PC ⊂平面PCD ,∴平面MBD ⊥平面PCD . 故答案为:②(或③).本题考查线面､面面垂直的判定与性质应用,需要学生具备一定的空间想象能力与逻辑思维能力.16．设函数()f x 的定义域为R ，满足(2)2()f x f x -=，且当(0,2)x ∈时，2()2f x x x =-，则52f ⎛⎫= ⎪⎝⎭______，若16()()log g x f x x =-，则()g x 有______个零点.【答案】383【解析】由题可得1()(2)2f x f x =-,故52f ⎛⎫= ⎪⎝⎭1122f ⎛⎫⎪⎝⎭,再将()g x 的零点问题转换为函数()f x 与16()log h x x =的图象交点问题求解. 【详解】因为(2)2()f x f x -=,所以1()(2)2f x f x =-, 又当(0,2)x ∈时,2()2f x x x =-,所以52f ⎛⎫=⎪⎝⎭1511113212222248f f ⎛⎫⎛⎫⎛⎫-==⨯-= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭, 画出16(),()log f xh x x =的图象如下图所示:161611(3)(3)log 3,(5)(5)log 5,24f g f g =>==<= 因此两函数图象有3个交点,即()g x 有3个零点, 故答案为:38;3.【点睛】本题考查函数性质的应用,考查数形结合法解决函数零点问题,属于中档题.四、解答题 17．如图，直三棱柱中，AC BC ⊥,1AC BC ==,12CC =,点M 是11A B 的中点.（1）求证：1B C //平面1AC M ； （2）求三棱锥11A AMC -的体积. 【答案】(1)证明见解析；（2）16. 【解析】（1）连接1A C 交1AC 与N ，则N 为1A C 的中点，利用三角形中位线定理可得1//MN B C ，再由线面平行的判定定理可得结果；（2）由等积变换可得11A AMC V -11A A C M V -=，再利用棱锥的体积公式可得结果.【详解】（1）连接1A C 交1AC 与N ，则N 为1A C 的中点， 又M Q 为11A B 的中点，1//MN B C ∴，又因为MN ⊂平面1AC M ，1B C ⊄平面1AC M ， 1//B C ∴平面1AC M ；（2）因为，直三棱柱111A B C ABC -中，AC BC ⊥,1AC BC ==,12CC =，且点M 是11A B 的中点 所以11A AMC V -11A A C M V -=11113A C M S AA ∆=⨯11111132A C B S AA ∆=⨯⨯ 11111123226=⨯⨯⨯⨯⨯=. 【点睛】本题主要考查线面平行的判定定理、利用等积变换求三棱锥体积，属于中档题.证明线面平行的常用方法：①利用线面平行的判定定理，使用这个定理的关键是设法在平面内找到一条与已知直线平行的直线，可利用几何体的特征，合理利用中位线定理、线面平行的性质或者构造平行四边形、寻找比例式证明两直线平行.②利用面面平行的性质，即两平面平行，在其中一平面内的直线平行于另一平面.18．已知数列{}n a ，n S 为其前n 项和，22n S n n =+.（1）求{}n a 的通项公式： （2）若21n n n b a a +=，记n T 为数列{}n b 的前n 项和，求n T .【答案】（1）n a n =；（2）32342(1)(2)n n n +-++ 【解析】(1)根据22n S n n =+,由1n n n a S S -=-即可求出{}n a 的通项公式:(2)利用裂项相消法即可求出答案. 【详解】(1)当1n =时,11222S a ==,即11a =,由22n S n n =+得212(1)1(1)n S n n n -=-+->,两式相减得:22(1)n a n n =>, 即(1)n a n n =>, 又1n =时上式也成立, 故n a n =; (2)由(1)知,1111(2)22n b n n n n ⎛⎫==- ⎪++⎝⎭,则1111111112324352n T n n ⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫=-+-+-++- ⎪ ⎪ ⎪ ⎪⎢⎥+⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦L 111112212n n ⎡⎤=+--⎢⎥++⎣⎦32342(1)(2)n n n +=-++. 【点睛】本题考查由数列的求和公式求通项公式,考查裂项相消法求和,难度不大.在由数列的求和公式求通项公式时,需注意n 的取值范围.19．2019年，海南等8省公布了高考改革综合方案将采取“312++”模式，即语文、数学、英语必考，然后考生先在物理、历史中选择1门，再在思想政治、地理、化学、生物中选择2门为了更好进行生涯规划，甲同学对高一一年来的七次考试成绩进行统计分析，其中物理、历史成绩的茎叶图如图所示.（1）若甲同学随机选择3门功课，求他选到物理、地理两门功课的概率； （2）试根据茎叶图分析甲同学的物理和历史哪一学科成绩更稳定.（不需计算） （3）甲同学发现，其物理考试成绩y （分）与班级平均分x （分）具有线性相关关系，统计数据如下表所示，试求当班级平均分为50分时，其物理考试成绩.（计算$a，ˆb 时精确到0.01）x （分）57 61 65 72 74 77 84 y （分）76828285879093参考数据：71490ii x==∑，71595i i y ==∑，72134840i i x ==∑，72150767i i y ==∑，7141964i ii x y==∑，()()71314i i i x x y y =--=∑.参考公式：()()()1122211ˆn niii ii i nni ii i x x y y x y n x ybx x xn x ====---⋅⋅==--⋅∑∑∑∑，ˆˆay b x =-⋅ 【答案】（1）14；（2）物理；（3）73 【解析】(1)直接利用枚举法与古典概型概率计算公式求解;(2)由茎叶图可知物理成绩的方差s 2物理<历史成绩的方差s 2历史,故物理成绩更稳定; (3)由表格数据先求,x y ,再利用公式求出回归方程,进而得解. 【详解】(1)记物理､历史分别为1A ,2A ,思想政治､地理､化学､生物分别为1B ,2B ,3B ,4B , 由题意可知考生选择的情形有{}112,,A B B ,{}113,,A B B ,{}114,,A B B ,{}123,,A B B ,{}124,,A B B ,{}134,,A B B ,{}212,,A B B ,{}213,,A B B ,{}214,,A B B ,{}223,,A B B ,{}224,,A B B ,{}234,,A B B ,共12种,他选到物理､地理两门功课的满情形有{}112,,A B B {}123,,A B B {}124,,A B B ,共3种, ∴甲同学选到物理､地理两门功课的概率为31124P ==; (2)由茎叶图可知物理成绩数据更集中, 故物理成绩的方差2s <物理历史成绩的方差2s 物理,故物理成绩更稳定;(3)57616572747784707x ++++++==,85y =,∴717222174196477085314ˆ0.58348407705407i ii ii x y x y bx x==-⋅⋅-⨯⨯===≈-⨯-⋅∑∑,$ˆ850.587044.40ay b x =-⋅=-⨯≈, ∴y 关于x 的回归方程为0.5844.40y x =+, 当50x =时,0.585044.4073y =⨯+≈. 【点睛】本题考查古典概型,考查茎叶图以及回归方程,属于中档题.在解决古典概型问题时,常利用枚举法进行答题.20．ABC ∆的内角A ，B ，C 的对边分别为a ，b ，c ，已知2cos 2c A b a =-（1）求角C ；（2）若D 是边BC 的中点，5,21AC AD ==求AB 的长；【答案】（1）3C π=；（2197；【解析】（1）首先根据正弦定理边角互化，得到2sin cos 2sin sin C A B A =-，由()sin sin B A C =+，代入化简，最后得到1cos 2C =求角C ；（2）首先在ACD ∆中，根据余弦定理求CD ，然后在ABC ∆中再利用余弦定理求边AB . 【详解】（1）2cos 2c A b a =-Q ，∴由正弦定理得2sin cos 2sin sin C A B A =-，2sin cos 2sin sin C A A C A =+（）-∴，2sin cos 2sin cos 2cos sin sin C A A C A C A =+-∴，2sin cos in ,sin 0A C s A A =≠∴，1cos 2C ∴=， (),3C C ππ∈=Q 0，∴，（2）在ACD ∆中，由余弦定理得2222cos AD AC CD AC CD C =+-⋅⋅ 221255CD CD =+-∴ 2540CD CD -+=，1CD =∴或4CD =，当1CD =时，2BC =ABC ∆中，由余弦定理得2222cos AB AC BC AC BC C =+-⋅⋅1254252219=+-⨯⨯⨯=19AB =∴，当4CD =时，8BC =2222cos AB AC BC AC BC C =+-⋅⋅12564258492=+-⨯⨯⨯= 7AB =∴19AB =∴或7AB =.【点睛】本题考查正余弦定理解三角形，属于基础题型，一般在含有边和角的等式中，可根据正弦定理的边角互化公式转化为三角函数恒等变形问题.21．如图1，在直角梯形ABCD 中，//AB CD ，90A ∠=︒，2AB =，6CD =，3AD =，E 为CD 上一点，且4DE =，过E 作//EF AD 交BC 于F ，现将CEF ∆沿EF 折到PEF ∆，使60PED ∠=︒，如图2.（1）求证：PE ⊥平面ADP（2）在线段PF 上是否存在一点M ，使DM 与平面ADP 所成的角为30°？若存在，确定点M 的位置；若不存在，请说明理由.【答案】（1）证明见解析；（2）不存在，理由见解析【解析】(1)解法一:由EF PE ⊥,EF DE ⊥,推出EF ⊥平面PDE ,即有AD ⊥平面PDE ,故AD PE ⊥,结合PE PD ⊥即可推出PE ⊥平面APD ;解法二:建立空间直角坐标系,利用向量推出结论;(2)由(1)知AD ⊥平面PDE ,故以DA 所在的直线为x 轴,以DE 所在的直线为y 轴,在平面DPE 内过D 作DE 的垂线,以垂线所在直线为z 轴,建立空间直角坐标系,设M 是线段PF 上一点,则存在01λ≤≤,使PM PF λ=u u u u r u u u r,再利用向量,结合线面角公式列式求解即可. 【详解】 (1)解法一:∵4DE =,2PE =,60PED ∠=︒,由余弦定理得22212232DE PE PD DE PE PD +-=⋅⋅⋅⇒=, ∵22216PD PE DE +==,∴PE PD ⊥, 又直角梯形ABCD 中,//EF AD , ∴EF PE ⊥,EF DE ⊥,PE DE E =I , 则EF ⊥平面PDE ,又∵//EF AD ,∴AD ⊥平面PDE ,∴AD PE ⊥,又因为直线AD ,PD 在平面APD 内,且相交于D ,∴PE ⊥平面APD . 解法二:以为EF PE ⊥,EF DE ⊥,且PE DE E =I , 则EF ⊥平面PDE ,所以平面DEF ⊥平面PDE ,以DA 所在的直线为x 轴,以DE 所在的直线为y 轴,在平面DPE 内过D 作DE 的垂线,以垂线所在直线为z 轴,建立空间直角坐标系,如图所示:则(0,0,0)D ,(3,0,0)A ,(3P ,(0,4,0)E ,∴(3,0,0)DA =u u u r,(3DP =u u u r ,(0,3EP =-u u u r ,∴0DA EP ⋅=u u u r u u u r ,0DP EP ⋅=u u u r u u u r,∴DA EP ⊥u u u r u u u r ,DP EP ⊥u u u r u u u r∴DA EP ⊥,DP EP ⊥,∵DA ,DP 是平面ADP 内的相交直线, ∴PE ⊥平面APD .(2)由(1)知AD ⊥平面PDE ,∴平面ADE ⊥平面PDE ,以DA 所在的直线为x 轴,以DE 所在的直线为y 轴,在平面DPE 内过D 作DE 的垂线,以垂线所在直线为z 轴,建立空间直角坐标系,如图所示:则(0,0,0)D ,(3P ,(0,4,0)E ,3,4,02F ⎛⎫⎪⎝⎭, 则(0,3EP =-u u u r ,3,1,32PF ⎛=- ⎝u u u r ,∵PE ⊥平面ADP ,∴平面ADP 的一个法向量为(0,3n EP ==-r u u u r,设M 是线段PF 上一点,则存在01λ≤≤,使PM PF λ=u u u u r u u u r,∴(33,1,32DM DP PM λ⎛=+-+ ⎝u u u u r u u u r u u u u r 3,3,332λλλ⎛⎫=+-+ ⎪⎝⎭,2cos ,2548n DM n DM n DM λ⋅==⨯+r u u u u rr u u u u r r u u u u r , 如果直线DM 与平面ADC 所成的角为30°,那么cos ,sin 30n DM ︒=r u u u u r,2122548λ=±+,解得21613λ=,此方程在[]0,1内无解, 所以在线段PF 上不存在一点M ,使DM 与平在ADP 所成的角为30°. 【点睛】本题考查线面垂直的判定及应用,考查空间向量在线面角上的应用,需要学生具备一定的空间思维及想象能力,属于中档题.22．已知函数2()ln (21)(1)f x x ax a x a =+-+++. （1）若12a =，分析()f x 的单调性.（2）若对1x ∀>，都有()0f x >恒成立，求a 的取值范围；（3）证明：2222222212n n n k n nn n n n++++⋅⋅⋯⋅⋅⋯⋅>对任意正整数n 均成立，其中e 为自然对数的底数.【答案】（1）单调增区间为(0,)+∞，无减区间；（2）1,2⎡⎫+∞⎪⎢⎣⎭；（3）证明见解析 【解析】(1)直接对函数求导,利用导数研究其单调性即可; (2)对()f x 求导后,再根据a 的取值进行分情况讨论即可;(3)题目可变形为证明不等式22222222121ln ln ln ln 2n n n k n n n n n n ++++++⋅⋅⋅++⋅⋅⋅+>恒成立,又由(1)可得1ln (1)1(1)2x x x ⎡⎤>---⎢⎥⎣⎦在(1,)+∞恒成立,则令21k x n =+,即有2224221ln 122k k k k n n nn n ⎛⎫+>-≥- ⎪⎝⎭,据此即可推出结论.【详解】(1)12a =,213()ln 222f x x x x =+-+,2(1)()x f x x-'=,(0,)x ∈+∞,故()0f x '>在(0,)+∞上恒成立,所以()f x 的单调增区间为(0,)+∞,无减区间.(2)1()2(21)f x ax a x '=+-+22(21)1(21)(1)ax a x ax x x x-++--==. ∵1x >,∴10x ->,故:①当0a ≤时,()0f x '≤,()f x 在(1,)+∞上单调递减,而(1)0f =,∴()0f x <,不符合题意;②当12a ≥时,即112a≤,()f x 在(1,)+∞上单调递增, 而()(1)0f x f >=,∴符合题意;③当102a <<时,11,2x a ⎛⎫∈ ⎪⎝⎭,()0f x '<,()f x 在11,2a ⎛⎫ ⎪⎝⎭上单调递减,而(1)0f =,∴此时()0f x <,不符合题意;第 21 页 共 21 页 综上所述,a 的取值范围为1,2⎡⎫+∞⎪⎢⎣⎭. (3)证明:要证明2222222212n n n k n n n n n n++++⋅⋅⋯⋅⋅⋯⋅>, 等价于证明22222222121ln ln ln ln 2n n n k n n n n n n ++++++⋅⋅⋅++⋅⋅⋅+>, 由(1)可得1ln (1)1(1)2x x x ⎡⎤>---⎢⎥⎣⎦在(1,)+∞恒成立, 令21k x n =+,1,2,3,,k n =⋅⋅⋅,则221k n≤, ∴2224221ln 122k k k k n n nn n ⎛⎫+>-≥- ⎪⎝⎭, ∴2222222212ln ln ln ln n n n k n n n n n n++++++⋅⋅⋅++⋅⋅⋅+22121122n n n n ++⋅⋅⋅+>-⨯= ∴22222222121ln ln ln ln 2n n n k n n n n n n ++++++⋅⋅⋅++⋅⋅⋅+>成立, ∴()()()()22222123nn n n n n n +⋅+⋅+⋅⋯⋅+>成立.【点睛】本题考查利用导数研究函数的单调性,解决恒成立问题以及不等式证明问题,难度较大.。
