山东省聊城一中2020-2021学年高三4月份线上模拟试题

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山东省聊城市第一中学2020-2021学年度高三下学期地区联考模拟试题语文答案

高三年级一模检测题（一）语文参考答案一、现代文阅读（35分）（一）现代文阅读I（本题共5小题，19.分）1.C（A“当下社会发生的特定事实基础……不能产生重要的影响”错。

B“向群体和社会层面展开的特点是由社会记忆的底色或基调决定的”错。

D“我们”与“他们”的说法，与“我群和他群”的概念，都是为区别自己所在群体和其他群体的表达，有相似的地方；但“我们”与“他们”具体指“中华民族”和“外族”，在这一点上含义又有不同）2.A（“个体记忆促成社会分类”，只是形成社会认同感的微观层面的路径，此外还有其他路径）3.A（材料一第四段的核心观点是权力或各方面的利益关系会借助对社会记忆的建构来实现对社会中人们思想的引导，形成社会认同。

A项内容符合中心论点的意思。

B谈的是了解能促成团体的稳定。

C.说的是机构、组织自身的改变。

D没有重新建构的意思）4.材料—侧重从理论上论述社会认同形成的一般规律。

材料二侧重从实践层面论述如何建构中国国家认同。

（4分。

每点2分，言之成理亦可）5.武汉网友泪奔，是个体记忆鲜活性和冲击力的体现，视频被大量转发说明其影响并形成了社会认同。

“90后”的心声体现了担当的社会记忆，符合外部期望和本身需要，收获大量点赞说明实现了社会认同。

央视频道推出“抗疫”纪实，是国家层面对人们进行思想引导，形成共同的社会记忆，引发广泛共鸣体现国民认同。

（6分。

每点2分，意思对即可）（二）现代文阅读（16分）6.B（“说明王德贵比刘强对此次运送任务有着更清醒的认识”，理解错误）7.B（“形象刻画出王德贵恼羞成怒的心理”，理解错误）8.突出体现志愿军对朝鲜人民的体贴与关爱。

集中寄托对未来生活的期盼和希望。

有力控诉侵略战争的残酷与罪恶。

（4分。

每点2分，答出两点即可）9.历史真实小说写志愿军战士护送朝鲜妇女到达后方家园，展现军民情深，不回避战争的残酷，这符合历史真实。

（2分）审美感受志愿军战士对朝鲜妇孺关心备至；朝鲜妇女天真淳朴，心怀感激，都展现了人性美。

2020届山东省聊城一中高三4月线上模拟英语试题(解析版)

聊城一中高三加强训练（一）英语强试题阅读（共两节，每题2.5分；满分97.5分）第一节Laura Sides was a psychology major at the University of Nottingham in 2004. She first noticed signs of her dad’s developing dementia(痴呆) when she moved to Nottingham. She said, "Dad was a doctor, so he knew exactly what had happened to him, but people try to hide it when they are ill. Then, I came home for my 21st birthday and arranged to meet him, but he never showed up as he’d forgotten. That’s when I knew something serious had happened."So, aged 21, she decided to leave university and look after him herself. She lived close by, popping in every day to make sure he was eating, and that the house was tidy, before heading off to her work.Besides challenging moments, there was a time when looking after her dad was a pure joy. "We’d wake up, I’d ask what he wanted to do that day, and however ridiculous the adventure is, off we’d go."Sadly, in 2009, 5 years later, Laura lost her father. Before he died, Laura went to a hospital appointment with him, where doctors mentioned that his form of Alzheimer’s disease was genetic meaning there was a fifty-fifty chance that she had inherited it. For several years Laura agonised over whether to be tested, finally finding out in August 2017 that she has the APP gene, meaning that, like him, she will develop the condition within a decade.At first, she struggled, feeling as if her life lacked purpose. Then, during a sleepless night in the summer of 2018, she decided at around 2 a.m. to enter the 2019 London Marathon sponsored b y the charities Alzheimer’s Society and Alzheimer’s Research UK.She hoped to start the conversation around early-onset Alzheimer’s disease and to encourage people to talk about it more openly. "I remember when Dad was ill, people wouldn’t know how to reac t, but I want to be honest and open," she added. "The more information we can get, the less of a taboo(忌讳) we will feel. That said, the support I’ve received so far after going public has been amazing —that’s what is carrying me through."1. Laura noticed her father’s dementia when .A. her father told her his condition in personB. people nearby informed her of his father’s conditionC. her father forgot his own birthday partyD. her father forgot to attend her 21-year-old birthday party2. The underlined word "agonised" in Paragraph 4 probably means .A. excitedB. struggledC. shockedD. delighted3. Laura started the open talk in the hope of .A. getting people to talk about Alzheimer’s disease openlyB. earning some money to help treat her Alzhei mer’s diseaseC. making herself stronger to fight against Alzheimer’s diseaseD. raising funds for charities Alzheimer’s Society and Alzheimer’s disease Research UK4. Which words can best describe Laura?A. Caring and positive.B. Careful and honest.C. Patient and cautious.D. Devoted and modest.【答案】1. D 2. B 3. A 4. A【解析】【分析】本文是一篇记叙文。

山东省聊城一中高三4月线上模拟英语试题

2020届山东省聊城一中高三4月线上模拟英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读选择Laura Sides was a psychology major at the University of Nottingham in 2004. She first noticed signs of her dad’s developing dementia(痴呆) when she moved to Nottingham. She said, "Dad was a doctor, so he knew exactly what had happened to him, but people try to hide it when they are ill. Then, I came home for my 21st birthday and arranged to meet him, but he never showed up as he’d forgotten. That’s when I knew something serious had happened."So, aged 21, she decided to leave university and look after him herself. She lived close by, popping in every day to make sure he was eating, and that the house was tidy, before heading off to her work.Besides challenging moments, there was a time when looking after her dad was a pure joy. "We’d wake up, I’d ask what he wanted to do that day, and however ridiculous the adventure is, off we’d go."Sadly, in 2009, 5 years later, Laura lost her father. Before he died, Laura went to a hospital appointment with him, where doctors mentioned that his form of Alzheimer’s disease was genetic meaning there was a fifty-fifty chance that she had inherited it. For several years Laura agonised over whether to be tested, finally finding out in August 2017 that she has the APP gene, meaning that, like him, she will develop the condition within a decade.At first, she struggled, feeling as if her life lacked purpose. Then, during a sleepless night in the summer of 2018, she decided at around 2 a.m. to enter the 2019 London Marathon sponsored by the charities Alzheimer’s Society and Alzheimer’s Research UK.She hoped to start the conversation around early-onset Alzheimer’s disease and to encourage people to talk about it more openly. "I rememb er when Dad was ill, people wouldn’t know how to react, but I want to be honest and open," she added. "The more information we can get, the less of a taboo(忌讳) we will feel. That said, the support I’ve received so far after going public has been amazing —that’s what is carrying me through."1．Laura noticed her father’s dementia when .A．her father told her his condition in personB．people nearby informed her of his father’s conditionC．her father forgot his own birthday partyD．her father forgot to attend her 21-year-old birthday party2．The underlined word "agonised" in Paragraph 4 probably means .A．excited B．struggledC．shocked D．delighted3．Laura started the open talk in the hope of .A．getting people to talk about Alzheimer’s disease openlyB．ear ning some money to help treat her Alzheimer’s diseaseC．making herself stronger to fight against Alzheimer’s diseaseD．raising funds for charities Alzheimer’s Society and Alzheimer’s disease Research UK 4．Which words can best describe Laura?A．Caring and positive. B．Careful and honest.C．Patient and cautious. D．Devoted and modest.Life on the street is a constant struggle for homeless people. In times of extreme weather conditions, that struggle becomes even more difficult. Recently, homeless people across Chicago faced freezing to death if they couldn't find shelter for the night.Thankfully, one local woman refused to let that happen. On January 30，2018, 34-year-old Candice Payne, a local managing broker, was lucky enough to have shelter from the dangerous conditions.“It was - 20℃, and I knew they were going to be sleeping on ice and I had to do something,” said Payne. Payne started brainstorming different ways she could possibly help. Finally，she decided to see if there were any rooms available at local inns and hotels that she could get to help those stuck on the street.For Payne, her mission was personal. According to Payne, her husband, Carlos Callahan，had lived on the streets at one point in his life. Based on his experiences, Payne knew that the homeless people still out on the street desperately needed help and that if she didn't step up to help, no one likely would.However, when Payne explained what she was trying to do，many of the local hotels refused to allow her to pay for the rooms as they didn't want homeless people to stay in their rooms. “No one wanted them, but one hotel, the Amber Inn, was nice enough to allow me to buy the rooms,” said Payne.Payne's selfless act made news across the country. However, she insisted she had neverdo ne it for attention. “I am a regular person, ” said Payne, who spent thousands of dollars of her own money to help complete strangers.“It all sounded like a rich person did this, but I’m just a little black girl from the South Side. ”5．What does the under lined word “ that” in Paragraph 2 refer to?A．Locals living on the street‘B．The homeless freezing to death.C．The extreme weather.D．The shelter for the homeless.6．How did Candice Payne help the homeless people?A．By drawing public attention.B．By giving them money directly.C．By taking them to her own house.D．By buying hotel rooms for them.7．What may contribute to Payne’s nice act?A．Her husband's past experiences.B．The requests from the homeless.C．Her desire to become famous.D．The coldness of local hotels.8．Which of the following can be the best title for the text?A．A Black Fighter Changes the WorldB．A Woman’s Curiosity Brings a RewardC．A Regular Woman Makes a DifferenceD．A Couple's Brave Act Moves the CountryBeing "young is associated with all the good things in life - beauty, hope, and energy. But youth also has negative associations - impulsiveness, trouble -making, and irresponsibility. This negative side seems to be what society focuses on more, which is why young people have mostly been considered as idle and difficult.But when it comes to Generation Z - those born between 1996 and 2010 - this stereotype doesn't seem to apply anymore.In Japan, for example, Gen Z-ers are less likely to buy on impulse, but take into consideration more a product's true value. They' re looking at the companies, not just theproducts," Masahiko Uotani, CEO of Japanese cosmetics company Shiseido, told Bloomberg. They're asking, 'Are they really delivering value to the society? Are they promoting diversity and inclusion?"Gen Z-ers are also more grounded than we' ve expected them to be. According to a recent survey by Bank of America, more than half of young adults aged between 18 and 23 said they were planning to buy a house within five years. And they' re not just saying it - they are willing to make sacrifices for it, including getting a second job and saving money for down payment instead of spending it on a vacation."Despite their young age, this group is pragmatic and actively planning for their future," D．Steve Boland, head of Consumer Lending at Bank of America, told USA Today. "They have a clear vision how they are willing to help themselves in order to make it happen.Social issues are also at the center of concern of Gen Z-ers, who take themselves as a changing force of the world. In India, for example, young people who have just reached the voting age are eager to vote for a new leader who is capable of solving problems that matter the most to them, including pollution, unemployment and women' s safety.As a Gen-Zer yourself, what is your plan for the future?9．What do the underlined words this stereotype" in paragraph2 refer to ?A．Being young is good.B．Gen Z-ers are born after 1996.C．The traditional poor impressions on the youth.D．The associations with young people.10．We can infer from the third paragraph that______________.A．Gen Z-ers in Japan are pickyB．The Gen Z-ers are self-centeredC．The Gen Z-ers care little about productsD．The Gen Z-ers are wise when shopping11．What's Steve Boland's attitude to the Gen Z-ers ?A．Approving. B．Negative.C．Indifferent. D．Critical.12．Which of the following words may best describe the Gen Z-era?A．Confident and independent. B．Visionary and responsible. C．Persistent and down-to-earth. D．Active and creative.Self-driving cars have been backed by the hope that they will save lives by getting involved in fewer crashes with fewer injuries and deaths than human-driven cars. But so far，most comparisons between human drivers and automated vehicles have been unfair.Crash statistics for human-driven cars are gathered from all sorts of driving situations，and on all types of roads. However，most of the data on self-driving cars' safety have been recorded often in good weather and on highways，where the most important tasks are staying in the car's own lane and not getting too close to the vehicle ahead. Automated cars are good at those tasks，but so are humans.It is true that self-driving cars don't get tired，angry，frustrated or drunk. But neither can they yet react to uncertain situations with the same skill or anticipation of an attentive human driver. Nor do they possess the foresight to avoid potential perils. They largely drive from moment to moment，rather than think ahead to possible events literally down the road.To a self-driving car，a bus full of people might appear quite similar to an uninhabited corn field. Indeed，deciding what action to take in an emergency is difficult for humans，but drivers have sacrificed themselves for the greater go od of others. An automated system’s limited understanding of the world means it will almost never evaluate(评估) a situation the same way a human would. And machines can't be programmed in advance to handle every imaginable set of events.Some people may argue that the promise of simply reducing the number of injuries and deaths is enough to support driverless cars. But experience from aviation(航空) shows that as new automated systems are introduced，there is often an increase in the rate of disasters.Therefore comparisons between humans and automated vehicles have to be performed carefully. To fairly evaluate driverless cars on how well they fulfill their promise of improved safety，it's important to ensure the data being presented actually provide a true comparison. After all，choosing to replace humans with automation has more effects than simply aone-for-one exchange.13．What makes the comparison between self-driving cars and human-driven cars unfair? A．Self-driving cars never get tired. B．Statistics are collected differently. C．Machines can make decisions faster. D．Self-driving cars know the world better. 14．What does the underlined word “perils” in Paragraph 3 most probably mean? A．Dangers. B．Self-driving cars.C．Pedestrians. D．Human-driven cars.15．In which aspect can self-driving cars beat human-driven cars?A．Driving steadily. B．Climbing steep slopes. C．Evaluating the cost of loss. D．Making complex decisions.16．Why does the author write this text?A．To support human-driven cars.B．To show his doubt about self-driving cars.C．To call for exact evaluation of self-driving cars.D．To stress the importance of reducing car accidents.Empathy (同理心)is one of those strange qualities—something almost everyone wants, but few know how to truly give or receive it. In a world where self-satisfaction is emphasized, it is in short supply but high demand. This is all the more reason to teach the next generation what it means to have empathy for those around them.What Is Empathy?— Many people confuse sympathy and empathy, but they are two distinct values. Empathy is not just the ability to understand someone's feelings; criminals often take advantage of people by appearing to understand their feelings and subsequently gaining their trust. Empathy is more than that. Not only is it the ability to recognize how someone feels, but it also values and respects the feelings of another person. It means treating others with kindness, dignity, and understanding.Kids Need To See Adults Show Empathy—While some children are gifted with naturally kind hearts, in most cases kids need to see empathy modeled by the adults around them. It begins with the way parents relate to their children. Parents who show an interest in the things that matter to their kids and respond to emotions in a positive and caring way are teaching the skill of empathy.Meet Emotional Needs—When children have their emotional needs met, two things happen. They learn how to meet the emotional needs of others and they are anchored in what they are receiving, meaning that they are secure enough to give to others when the need arises but first they need to receive. An empty jug cannot fill a cup.It's a good idea to talk to kids about emotions and how other people experience them. Give their emotions names (for example, jealousy, anger, and love) and teach them that these are normal. Talk to them about how to handle emotions in a positive way and point out situationswhere other people are experiencing emotions. Teach them about respecting the emotions of others and show them how to act in a situation where a response is required.17．Why is it that the next generation are taught to have empathy?A．Because people tend to center themselves. B．Because everyone lacks empathy. C．Because empathy is a strange quality. D．Because it's better to give than to receive. 18．Which situation can empathy be used in?A．When a mathematician is calculating the area of a farmland.B．When a teacher is comforting a student about his failure in exams.C．When a criminal is cheating a victim.D．When a dancer is dancing to music.19．What does the underlined sentence "An empty jug cannot fill a cup" mean?A．An empty jug is too small to hold a cup.B．It's a must to talk to kids about emotions.C．Adults should set an example to kids.D．Kids give empathy with their emotional needs met first.20．What is the main topic of the passage?A．How to train kids to have empathy.B．How to distinguish sympathy and empathy.C．How to help kids finish empathy-related tasks.D．Whether kids can be trained to be more empathetic.A large body of research has been developed in recent years to explain many aspects of willpower. Most of the researchers exploring self-control do so with an obvious goal in mind: How can willpower be strengthened? If willpower is truly a limited resource, as the research suggests, what can be done to make it stay strong?Avoiding temptation (诱惑) is an effective method for maintaining self-control, which is called the “out of sight, out of mind” principle. One recent study, for inst ance, found office workers less attracted to candy in the desk drawer than that on top of their desks, in plain sight.The research suggesting that we possess a limited reservoir of self-control raises a troubling question. When we face too many temptations, are we to fail? Not necessarily. Researchers don’t believe that one’s willpower is ever completely exhausted. Rather, people appear to hold some willpower in reserve, saved for future demands. The right motivationallows us to tap into those reserves, allowing us to carry on even when our self-control strength has been run down. High motivation might help overcome weakened willpower-at least to a point.Willpower may also be made less vulnerable（脆弱）to being exhausted in the first place. Researchers who study self-control often describe it as being like a muscle that gets tired with heavy use. But there is another aspect to the muscle comparison, they say. While muscles become exhausted by exercise in the short term, they are strengthened by regular exercise in the long term. Similarly, regular practices of self-control may improve willpower strength.The evidence from willpower-exhaustion studies also suggests that making a list of resolutions on New Year’s Eve is the worst possible approach. Being exhaust ed in one area can reduce willpower in other areas, so it makes more sense to focus on a single goal at a time. In other words, don’t try to quit smoking, adopt a healthy diet and start a new exercise plan at the same time. Taking goals one by one is a better approach. Once a good habit is in place, Baumeister says, you’ll no longer need to draw on your willpower to maintain the behavior. Eventually healthy habits will become routine, and won’t require making decisions at all. 21．From the studies in the passage we learn that .A．people have unlimited self-controlB．high motivation ensures one’s successC．willpower is hardly completely exhaustedD．too many temptations often lead to failure22．The underlined phrase “tap into” in Paragraph 3 most pro bably means . A．make use of B．run out ofC．build D．increase23．The author compares self –control to muscles .A．to prove the long-term effect of willpowerB．to show the significance of regular exerciseC．to argue that self-control can he easily used upD．to explain the benefits of practicing self-control24．To develop a good habit, which of the following does the author prefer?A．“I will give up dessert and do exercise.”B．“I will set three goals this new semester.”C．“I will read an English novel every month.”D．“I will keep myself from any temptation.”二、七选五We are surrounded by messages trying to get our attention. Advertisers, politicians and other groups all try to get us to agree with them. 25．Propaganda is not always a bad thing, but it often hides the truth about a topic. Its very similar to advertising, whose goal is to get people to buy something, while propaganda is to change what people believe.Propaganda TechniquesTo create propaganda, a common technique is name-calling. A politician might call someone a traitor or liar, which makes the other person look bad. Another is the trend technique. People want to be on the side in the favor. 26．，Some propaganda is based on fear. It scares people into choosing a certain side or taking action, which might not be bad. Anti—smoking campaigns are one example. They may scare people into thinking that if they smoke, they will get cancer.27．Propaganda is used by many groups, such as businesses, politicians and the government .It is spread through posters, television and radio. The Internet makes it easy to send messages to the whole world.Types of PropagandaPolitical propaganda has been around as long as there have been politicians 28．People have used it for many years. War propaganda began during World War I, more than 100 years ago.To see if propaganda is saying the truth, people need to do extra work. For example, ads might always sound like they're true. But before you believe them, try to think about who created and paid for them. Find out why they did it. 29．A．Propaganda TargetsB．Propaganda MediumsC．Another word for this is propagandaD．So most people will agree with those politiciansE.Then, you can decide if they're being honest or notF.However, religious propaganda was actually the first official oneG.People might join someone who claims to be popular or winningChoosing a major is a very important thing in our life. However, it can be extremely challenging. So it’s necessary to l earn some simple tips that much easier. 30．. And how does that choice affect your future career? We can help guide you in choosing a major.Your major is your primary field of study in your undergraduate (大学的）program. It is the subject in which you earn your bachelor’s degree. After completing a list of general education courses, you spend the rest of the time studying one subject of your choice. 31．.For many students, it's best to try out different classes to figure out a major you’ll like, but you need to have a couple of solid choices. 32．, you should begin taking courses you are required to complete before graduation, so it will decrease your chances of needing to switch majors later because you already know what to expect.33．, you should begin thinking about how your field will lead to a job as soon as possible.Consider clubs, research programs and part-time jobs related to potential career choices for your major.For more career-specific majors like nursing and engineering, the majority of your studies will include skills that can be used in a future job. 34．. The sooner you begin working in a lab, on research projects or helping as a teaching assistant for a professor, the better you'll be able to figure out your career.A．You have to think of ways to choose a majorB．Be sure that you choose a major that suits your interestsC．Once you have a better idea of which major you like bestD．While your major may or may not lead directly to a careerE.Do some research to find out which majors can help you get that jobF.Roughly one-third to one-half of your courses will be related to your majorG.However, you should still consider doing activities outside of the classroom to better prepare youOne of my earliest memories was watching my mom talking on our old phone. I was fascinated that she could talk to someone who wasn't actually in the room with her. 35．I was wondering how she managed to talk with someone she couldn't see.36．Later, we had mobile phones that could be carried around the room. Then came computers and smart phones. These days I can send an e-mail around the world in a second. My daughter's smart phone has a hundred apps and a dozen social media accounts. She stores all ofher information in a mysterious place known as "the cloud", where she can take it out at any time. It seems that we are more connected in this world than ever before.37．Few take time to talk face to face. Instead of looking into eyes of our beloved ones, we stare at our screens. It seems as if the most connected generation is also the loneliest.Don't let technology take our time and ruin our life. Make the time to meet and to communicate with each other. Take a walk on the beach with a friend. Have a long conversation with the phones off and the hearts on. 38．Remember that we are here to love each other, help each other and make this world a better place. 39．But when it can't, turn it off.A．That was a long time ago.B．Connect offline as well as online.C．We should communicate with each other with phones off.D．Smart phones have both advantages and disadvantages.E.When your technology can help to do these things, then use it.F.However, what bothers me is that we rarely associate with each other nowadays.G.When she left the room, I slowly walked over to the phone and stared at it for a while.三、用单词的适当形式完成短文语法填空Traditional Chinese medicine (TCM) isn't 40．( wide ) supported by most Westerners. 41．, this is starting to change. TCM will be included in the new version of the International Classification of Diseases (ICD), which 42．(bring) out by the WHO in the near future. This is the 43．(one) time for TCM to be included in the ICD, serving as the international standard for diseases and health conditions.China has been making efforts 44．(promote) TCM overseas. Twenty-six TCM centers were set up overseas over the past three years, according to a report 45．(publish) by Xinhua News Agency. In 2015, Chinese scientist Tu Youyou 46．(win) the Nobel Prize in Physiology or Medicine for her discovery of a drug that can cure malaria(疟疾). She said she was inspired by traditional Chinese medicine.The inclusion of TCM in the ICD is a mainstream acceptance that will have significant influence around the world. TCM has seen some growth in other countries for all these years. It is reported 47．a number of famous foreign people use it. For example, during the 2016 RioSummer Olympics, US 48．(swim) Michael Phelps was seen using cupping. Cupping is a traditional Chinese medical practice which has been around 49．more than 2,000 years.语法填空Paper cutting was recognized as our national intangible cultural heritage in 2006. According to experts, paper cutting conveys the culture shared between China 50．foreign countries to wish for family reunions and maintain links with loved ones, alive or dead.In the movie Coco, for instance, the 12-year-old Miguel and his family dance and sing to celebrate a 51．(tradition) festival in Mexico, when colorful paper cuttings are hung on the street. Chinese people also cut images of small figures 52．(commemorate) their family who have passed away. The difference is that most Chinese paper cuttings are red, while those in other countries 53．(be) often made in many other colors.The fairy tale writer Hans Christian Andersen liked to cut 54．(character), such as princesses, out of paper while 55．(tell) stories to children. In China, female friends and family members used to chat and make paper cuttings together.“56．the patterns and colors may be different, paper cuttings share the same function of maintaining emotional ties among people,” says Yang Huizi, 57．art teacher at Beijing Union University.Yang 58．(study) and performed the art for over a decade. Besides routine university courses, she also organizes nonprofit paper cutting activities 59．are open to the public in Beijing to promote basic knowledge of paper cutting.参考答案1．D2．B3．A4．A【分析】本文是一篇记叙文。

2020-2021学年度山东省聊城市高考一模考试数学（理）试题及答案

2020-2021学年度山东省聊城市高考一模考试数学（理）试题及答案高考模拟试题理科数学（一）第Ⅰ卷（选择题共60分）一、选择题（本大题共12个小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的.）1.已知集合2{|1}A x x =<，{|lg(1)0}B x x =+≥，则 A B =I （）A ．[0,1)B ．(1,)-+∞C ．(0,1)D ．(1,0]-2.设复数2(1)1i z i-=+，则z =（）A ．4B ．2C ．2D ．13.设等差数列{}n a 的前n 项和为n S ，若13104S =，65a =，则数列{}n a 的公差为（） A ．2 B ．3 C ．4 D ．54.我国三国时期的数学家赵爽为了证明勾股定理创制了一幅“勾股圆方图”，该图是由四个全等的直角三角形组成，它们共同围成了一个如图所示的大正方形和一个小正方形.设直角三角形中一个锐角的正切值为3.在大正方形内随机取一点，则此点取自小正方形内的概率是（）A ．110B ．15 C ．310 D ．255.设等比数列{}n a 的各项均为正数，其n 前项和为n S ，则“1921202S S S +>”是“数列{}n a 是递增数列”的（）A ．充分不必要条件B ．必要不充分条件C ．充要条件D ．既不充分也不必要条件6.已知直线l 与抛物线C ：24y x =相交于A ，B 两点，若线段AB 的中点为(2,1)，则直线l 的方程为（）A ．1y x =-B ．25y x =-+C ．3y x =-+D ．23y x =-7.已知函数()(1010)xxf x x -=-，不等式(12)(3)0f x f -+>的解集为（）A ．(,2)-∞B ．(2,)+∞C ．(,1)-∞D ．(1,)+∞8.已知双曲线C ：22221(0,0)x y a b a b-=>>的右焦点2F 到渐近线的距离为4，且在双曲线C 上到2F 的距离为2的点有且仅有1个，则这个点到双曲线C 的左焦点1F 的距离为（） A ．2 B ．4 C ．6 D ．8 9.执行如图所示的程序框图，若输出的结果为1.5，则输入k 的值应为（）A ．4.5B ．6C ．7.5 D．910.在ABC ?中，BC 边上的中线AD 的长为2，点P 是ABC ?所在平面上的任意一点，则PA PB PA PC ?+?u u u r u u u r u u u r u u u r的最小值为（）A ．1B ．2C ．-2D ．-111.如图是某几何体的三视图，其中俯视图为等边三角形，正视图为等腰直角三角形，若该几何体的各个顶点都在同一个球面上，则这个球的体积与该几何体的体积的比为（）A ．73πB ．289πC ．1479πD ．43π12.已知函数3,21(),20x x a x x f x a e x x ?--≤-??+=??--<<??恰有3个零点，则实数a 的取值范围为（）A ．11,3e ??-- B ．211,e e ??--C ．221,3e ??--D ．21,33??--第Ⅱ卷（非选择题共90分）二、填空题（本大题共4个小题，每小题5分，共20分）13.设x ，y 满足约束条件102020x y x y x y -+≥??-≤??+≤?，则12()16x yz =的最大值为．14.某工厂从生产的一批产品中随机抽出一部分，对这些产品的一项质量指标进行了检测，整理检测结果得到如下频率分布表：15.2922()y x x ++的展开式中常数项为． 16.若函数()sin()4f x m x π=+x 在开区间7(0,)6π内，既有最大值又有最小值，则正实数m 的取值范围为．三、解答题：共70分.解答应写出文字说明、证明过程或演算步骤.第17～21题为必考题，每个试题考生都必须作答.第22、23题为选考题，考生根据要求作答. （一）必考题：共60分17.已知数列{}n a 满足12a =-，124n n a a +=+. （Ⅰ）证明：{4}n a +是等比数列；（Ⅱ）求数列{}n a 的前n 项和n S .18.某教育培训中心共有25名教师，他们全部在校外住宿.为完全起见，学校派专车接送教师们上下班.这个接送任务承包给了司机王师傅，正常情况下王师傅用34座的大客车接送教师.由于每次乘车人数不尽相同，为了解教师们的乘车情况，王师傅连续记录了100次的乘车人数，统计结果如下：（Ⅰ）若随机抽查两次教师们的乘车情况，求这两次中至少有一次乘车人数超过18的概率；（Ⅱ）有一次，王师傅的大客车出现了故障，于是王师傅准备租一辆小客车来临时送一次需要乘车的教师.可供选择的小客车只有20座的A 型车和22座的B 型车两种，A 型车一次租金为80元，B 型车一次租金为90元.若本次乘车教师的人数超过了所租小客车的座位数，王师傅还要付给多出的人每人20元钱供他们乘出租车.以王师傅本次付出的总费用的期望值为依据，判断王师傅租哪种车较合算？19.如图，四棱锥P ABCD -中，PAD ?为等边三角形，且平面PAD ⊥平面ABCD ，22AD BC ==，AB AD ⊥，AB BC ⊥.（Ⅰ）证明：PC BC ⊥；（Ⅱ）若直线PC 与平面ABCD 所成角为60o，求二面角B PC D --的余弦值.20.已知圆224x y +=经过椭圆C ：22221(0)x y a b a b+=>>的两个焦点和两个顶点，点(0,4)A ，M ，N 是椭圆C 上的两点，它们在y 轴两侧，且MAN ∠的平分线在y 轴上，AM AN ≠. （Ⅰ）求椭圆C 的方程；（Ⅱ）证明：直线MN 过定点. 21.已知函数()22xf x e kx =--.（Ⅰ）讨论函数()f x 在(0,)+∞内的单调性；（Ⅱ）若存在正数m ，对于任意的(0,)x m ∈，不等式()2f x x >恒成立，求正实数k 的取值范围.（二）选考题：共10分.请考生在22、23两题中任选一题作答，如果多做，则按所做的第一题记分.22.选修4-4：坐标系与参数方程在直角坐标系xOy 中，圆C 的普通方程为2246120x y x y +--+=.在以坐标原点为极点，x 轴正半轴为极轴的极坐标系中，直线l 的极坐标方程为sin()24πρθ+=（Ⅰ）写出圆C 的参数方程和直线l 的直角坐标方程；（Ⅱ）设直线l 与x 轴和y 轴的交点分别为A 、B ，P 为圆C 上的任意一点，求PA PB ?u u u r u u u r的取值范围.23.选修4-5：不等式选讲已知函数()22f x x a a =++，a R ∈.（Ⅰ）若对于任意x R ∈，()f x 都满足()(3)f x f x =-，求a 的值；（Ⅱ）若存在x R ∈，使得()21f x x a ≤--+成立，求实数a 的取值范围.高考模拟理科数学（一）答案一、选择题1-5: ACBDC 6-10: DADBC 11、12：CA二、填空题13. 4 14. 144 15. 67216. 23m <<+三、解答题17.解：（Ⅰ）∵12a =-，∴142a +=，∵124n n a a +=+，∴1428n n a a ++=+2(4)n a =+，∴1424n n a a ++=+，∴{4}n a +是以2为首项，2为公比的等比数列.（Ⅱ）由（Ⅰ），可知42n n a +=，∴24n n a =-. ∴12n n S a aa =+++2(24)(24)=-+-(24)n++-2(222)4nn =+++-2(12)412n n -=--1224n n +=--.∴1242n n S n +=--.18.解：（Ⅰ）由题意得，在一次接送中，乘车人数超过18的概率为0.8. 记“抽查的两次中至少有一次乘车人数超过18”为事件A ，则()1(10.8)P A =--(10.8)0.96-=.即抽查的两次中至少有一次乘车人数超过18的概率为0.96.（Ⅱ）设X 表示租用A 型车的总费用（单位：元），则X 的分布列为.设Y 表示租用B 型车的总费用（单位：元），则Y 的分布列为.因此以王师傅本次付出的总费用的期望值为依据，租B 型车较合算. 19.证明：（Ⅰ）取AD 的中点为O ，连接PO ，CO ，∵PAD ?为等边三角形，∴PO AD ⊥.底面ABCD 中，可得四边形ABCO 为矩形，∴CO AD ⊥，∵PO CO O =I ，∴AD ⊥平面POC ，∵PC ?平面POC ，∴AD PC ⊥. 又//AD BC ，所以BC PC ⊥.（Ⅱ）由面PAD ⊥面ABCD ，PO AD ⊥，∴PO ⊥平面ABCD ，可得OP ，OD ，OC 两两垂直，又直线PC 与平面ABCD 所成角为60o，即60PCO ∠=o，由2AD =，知3PO =，得1CO =.建立如图所示的空间直角坐标系O xyz -，则(0,0,3)P ，(0,1,0)D ，(1,0,0)C ，(1,1,0)B -，(0,1,0)BC =u u u r ，(1,0,3)PC =-u u u r ，(1,1,0)CD =-u u u r ，设平面PBC 的一个法向量为(,,)n x y z =r. ∴030y x z =-=??，令1z =，则(3,0,1)n =r ，设平面PDC 的一个法向量为(',',')m x y z =u r，∴''0'3'0x y x z -=-=??，令'1z =，则(3,3,1)m =u r ， cos ,m n <>u r r m n m n ?=u r ru r r 27727==，∵二面角B PC D --为钝角，∴二面角B PC D --的余弦值为27 -.20.解：（Ⅰ）圆224x y +=与x 轴交点(2,0)±即为椭圆的焦点，圆224x y +=与y 轴交点(0,2)±即为椭圆的上下两顶点，所以2c =，2b =.从而22a =因此椭圆C 的方程为：22184x y +=. （Ⅱ）设直线MN 的方程为y kx m =+.由22184y kx m x y =++=??，消去y 得222(21)4280k x kmx m +++-=.设11(,)M x y ，22(,)N x y ，则122421kmx x k +=-+，21222821m x x k -=+. 直线AM 的斜率1114y k x -=14m k x -=+；直线AN 的斜率2224y k x -=24m k x -=+. 12k k +=1212(4)()2m x x k x x -++2(4)(4)228m km k m --=+-216(1)28k m m -=-. 由MAN ∠的平分线在y 轴上，得120k k +=.又因为AM AN ≠，所以0k ≠，所以1m =.因此，直线MN 过定点(0,1).21.解：（Ⅰ）'()2xf x e k =-，(0,)x ∈+∞，当2k ≤时，因为22xe >，所以'()0f x >，这时()f x 在(0,)+∞内单调递增.当2k >时，令'()0f x >得ln2k x >；令'()0f x <得0ln 2kx <<. 这时()f x 在(0,ln )2k 内单调递减，在(ln,)2k+∞内单调递增. 综上，当2k ≤时，()f x 在(0,)+∞内单调递增，当2k >时，()f x 在(0,ln )2k 内单调递减，在(ln,)2k+∞内单调递增. （Ⅱ）①当02k <≤时，因为()f x 在(0,)+∞内单调递增，且(0)0f =，所以对于任意的(0,)x m ∈，()0f x >.这时()2f x x >可化为()2f x x >，即2(2)20x e k x -+->.设()2(2)2xg x e k x =-+-，则'()2(2)xg x e k =-+，令'()0g x =，得2ln 2k x +=，因为2ln02k +>，所以()g x 在2(0,ln )2k +单调递减.又因为(0)0g =，所以当2(0,ln)2k x +∈时，()0g x <，不符合题意. ②当2k >时，因为()f x 在(0,ln )2k 内单调递减，且(0)0f =，所以存在00x >，使得对于任意的0(0,)x x ∈都有()0f x <.这时()2f x x >可化为()2f x x ->，即2(2)20xe k x -+-+>.设()2(2)2xh x e k x =-+-+，则'()2(2)xh x e k =-+-.（i ）若24k <≤，则'()0h x <在(0,)+∞上恒成立，这时()h x 在(0,)+∞内单调递减，又因为(0)0h =，所以对于任意的0(0,)x x ∈都有()0h x <，不符合题意. （ii ）若4k >，令'()0h x >，得2ln 2k x -<，这时()h x 在2(0,ln)2k -内单调递增，又因为(0)0h =，所以对于任意的2(0,ln )2k x -∈，都有()0h x >，此时取02min{,ln}2k m x -=，对于任意的(0,)x m ∈，不等式()2f x x >恒成立. 综上，k 的取值范围为(4,)+∞. 22.解：（Ⅰ）圆C 的参数方程为2cos 3sin x y θθ=+??=+?（θ为参数）.直线l 的直角坐标方程为20x y +-=.（Ⅱ）由直线l 的方程20x y +-=可得点(2,0)A ，点(0,2)B .设点(,)P x y ，则PA PB ?u u u r u u u r(2,)(,2)x y x y =--?--.2222x y x y =+--2412x y =+-.由（Ⅰ）知2cos 3sin x y θθ=+??=+?，则PA PB ?u u u r u u u r 4sin 2cos 4θθ=++)4θ?=++.因为R θ∈，所以44PA PB -≤?≤+u u u r u u u r23.解：（Ⅰ）因为()(3)f x f x =-，x R ∈，所以()f x 的图象关于32x =对称. 又()2||22a f x x a =++的图象关于2a x =-对称，所以322a -=，所以3a =-. （Ⅱ）()21f x x a ≤--+等价于2210x a x a ++-+≤. 设()g x =221x a x a ++-+，则min ()(2)(21)g x x a x a =+--+1a a =++. 由题意min ()0g x ≤，即10a a ++≤. 当1a ≥-时，10a a ++≤，12a ≤-，所以112a -≤≤-；当1a <-时，(1)0a a -++≤，10-≤，所以1a <-，综上12a ≤-.。

聊城一中2020届高三4月份线上模拟化学试题(word无答案)

聊城一中2020届高三4月份线上模拟化学试题一、单选题(★★) 1 . 生活离不开化学。

某种金属制成的器皿，放置于空气中，其表面会逐渐变黑，如将表面变黑的上述器皿放入盛有食盐水的铝制容器中浸泡，一段时间后，黑色完全褪去。

下列成语与该金属有关的是（）A．衣紫腰银B．点石成金C．铜鸵荆棘D．铁柞成针(★) 2 . 下图有机物的正确命名为 ( )A．2-乙基-3,3-二甲基戊烷B．3,3-二甲基-4-乙基戊烷C．3,3,4-三甲基己烷D．3,4,4-三甲基己烷(★★) 3 . 下列说法不正确的是A．利用盐析可分离提纯蛋白质B．用碘水检验淀粉是否发生水解C．棉花的主要成分为纤维素，属于天然纤维D．油脂的主要成分为高级脂肪酸甘油酯，可用于制取肥皂(★★★★) 4 . 下列离子方程式书写正确的是A．金属Na加入水中：Na+ 2H2O="=" Na+ + 2OH－+ H2↑B．NaOH溶液吸收Cl2：Cl2 + 2OH－="=" Cl－+ ClO－+ H2OC．Al2(SO4)3溶液和过量的NaOH溶液反应：Al3+＋3OH－="=" Al(OH)3↓D．过量的Fe和稀硝酸反应：Fe + 4H+ + NO3—="=" Fe3++ NO↑+ 2H2O(★★★★) 5 . 下列叙述正确的是A．向混有苯酚的苯中加入浓溴水，过滤，可除去其中的苯酚B．向酸性KMnO4溶液中滴加维生素C溶液，酸性KMnO4溶液褪色，说明维生素C具有还原性C．向淀粉溶液中加稀硫酸，加热，加新制Cu(OH)2悬浊液，加热未见红色沉淀，说明淀粉未水解D．向NaOH溶液中加入溴乙烷，加热，再加入AgNO3溶液，产生沉淀，说明溴乙烷发生了水解(★★) 6 . 下列方法中，可以使0.10 mol·L − 1 氨水中NH 3·H 2O的电离程度减小的是A．加入少量0.10 mol·L− 1盐酸B．加水稀释C．加入少量0.10 mol·L− 1NaOH溶液D．加入少量NaCl固体(★) 7 . 关于晶体的叙述中，正确的是（）A．原子晶体中，共价键的键能越大，熔、沸点越高B．分子晶体中，分子间的作用力越大，该分子越稳定C．分子晶体中，共价键的键能越大，熔、沸点越高D．某晶体溶于水后，可电离出自由移动的离子，该晶体一定是离子晶体(★★) 8 . 下列说法正确的是A．1 mol NH3中含有的质子数约为6.02×1023B．0.5 mol·L− 1 NaCl溶液中含Cl－的物质的量为0.5 molC．8 g S在足量O2中完全燃烧转移的电子数约为9.03×1023D．标准状况下，22.4 L SO2和CO2的混合气体所含原子数约为1.806×1024(★★) 9 . 下列事实不能用元素周期律解释的是A．酸性：HClO4 > H2SO3B．碱性：NaOH> Mg(OH)2C．气态氢化物的稳定性：H2O > H2S D．Cl2从NaBr溶液中置换出Br2(★★) 10 . 自然界中的许多植物中含有醛，其中有些具有特殊香味，可作为植物香料使用，例如桂皮含肉桂醛( CH=CH－CHO)，杏仁含苯甲醛( CHO)。

山东省聊城市2021届新高考数学四月模拟试卷含解析

山东省聊城市2021届新高考数学四月模拟试卷一、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．已知数列{}n a 满足：12125 1,6n n n a a a a n -≤⎧=⎨-⎩L …()*n N ∈)若正整数()5k k ≥使得2221212k k a a a a a a ++⋯+=⋯成立，则k =（）A ．16B ．17C ．18D ．19【答案】B 【解析】【分析】计算2226716...5n n a a a a a n ++++=-+-，故2221211...161k k k a a a a k a +++++=+-=+，解得答案.【详解】当6n ≥时，()1211111n n n n n a a a a a a a +--==+-L ，即211n n n a a a +=-+，且631a =.故()()()222677687116......55n n n n a a a a a a a a a n a a n +++++=-+-++-+-=-+-，2221211...161k k k a a a a k a +++++=+-=+，故17k =.故选：B . 【点睛】本题考查了数列的相关计算，意在考查学生的计算能力和对于数列公式方法的综合应用. 2．已知集合A={x|–1<x<2}，B={x|x>1}，则A ∪B= A ．（–1，1） B ．（1，2）C ．（–1，+∞）D ．（1，+∞）【答案】C 【解析】【分析】根据并集的求法直接求出结果. 【详解】∵{|12},{|1}A x x B x =-<<=> ， ∴(1,)A B =-+∞U ，故选C. 【点睛】考查并集的求法，属于基础题.3．空间点到平面的距离定义如下：过空间一点作平面的垂线，这个点和垂足之间的距离叫做这个点到这个平面的距离．已知平面α，β，λ两两互相垂直，点A α∈，点A 到β，γ的距离都是3，点P 是α上的动点，满足P 到β的距离与P 到点A 的距离相等，则点P 的轨迹上的点到β的距离的最小值是（） A ．33- B ．3C ．33- D ．32【答案】D 【解析】【分析】建立平面直角坐标系，将问题转化为点P 的轨迹上的点到x 轴的距离的最小值，利用P 到x 轴的距离等于P 到点A 的距离得到P 点轨迹方程，得到()26399y x =-+≥，进而得到所求最小值.【详解】如图，原题等价于在直角坐标系xOy 中，点()3,3A ，P 是第一象限内的动点，满足P 到x 轴的距离等于点P 到点A 的距离，求点P 的轨迹上的点到x 轴的距离的最小值．设(),P x y ，则()()2233y x y =-+-，化简得：()23690x y --+=，则()26399y x =-+≥，解得：32y ≥，即点P 的轨迹上的点到β的距离的最小值是32. 故选：D . 【点睛】本题考查立体几何中点面距离最值的求解，关键是能够准确求得动点轨迹方程，进而根据轨迹方程构造不等关系求得最值.4．百年双中的校训是“仁”、“智”、“雅”、“和”.在2019年5月18日的高三趣味运动会中有这样的一个小游戏.袋子中有大小、形状完全相同的四个小球，分别写有“仁”、“智”、“雅”、“和”四个字，有放回地从中任意摸出一个小球，直到“仁”、“智”两个字都摸到就停止摸球.小明同学用随机模拟的方法恰好在第三次停止摸球的概率.利用电脑随机产生1到4之间（含1和4）取整数值的随机数，分别用1，2，3，4代表“仁”、“智”、“雅”、“和”这四个字，以每三个随机数为一组，表示摸球三次的结果，经随机模拟产生了以下20组随机数：141 432 341 342 234 142 243 331 112 322342 241 244 431 233 214 344 142 134 412由此可以估计，恰好第三次就停止摸球的概率为（） A ．14B ．15C ．25D ．35【答案】A 【解析】【分析】由题意找出满足恰好第三次就停止摸球的情况，用满足恰好第三次就停止摸球的情况数比20即可得解. 【详解】由题意可知当1，2同时出现时即停止摸球，则满足恰好第三次就停止摸球的情况共有五种：142，112，241，142，412.则恰好第三次就停止摸球的概率为51204p ==. 故选：A. 【点睛】本题考查了简单随机抽样中随机数的应用和古典概型概率的计算，属于基础题. 5．已知点(m,8)在幂函数()(1)n f x m x =-的图象上，设,(ln ),()m a f b f c f n n π⎛⎫=== ⎪⎝⎭，则（） A ．b ＜a ＜c B ．a ＜b ＜cC ．b ＜c ＜aD ．a ＜c ＜b【答案】B 【解析】【分析】先利用幂函数的定义求出m 的值，得到幂函数解析式为f （x ）＝x 3，在R 上单调递增，再利用幂函数f （x ）的单调性，即可得到a ，b ，c 的大小关系. 【详解】由幂函数的定义可知，m ﹣1＝1，∴m ＝2， ∴点（2，8）在幂函数f （x ）＝x n 上， ∴2n ＝8，∴n ＝3，∴幂函数解析式为f （x ）＝x 3，在R 上单调递增，∵23m n =，1＜lnπ＜3，n ＝3， ∴mln n nπ＜＜， ∴a ＜b ＜c ，故选：B. 【点睛】本题主要考查了幂函数的性质，以及利用函数的单调性比较函数值大小，属于中档题.6．已知抛物线220y x ＝的焦点与双曲线()222210,0x y a b a b-=>>的一个焦点重合，且抛物线的准线被双曲线截得的线段长为92，那么该双曲线的离心率为（）A ．54 B ．53C ．52D 【答案】A 【解析】【分析】由抛物线220y x ＝的焦点(5,0)得双曲线()222210,0x y a b a b-=>>的焦点(5,0)±，求出5c ＝，由抛物线准线方程5x =-被曲线截得的线段长为92，由焦半径公式2292b a =，联立求解.【详解】解：由抛物线220y x ＝，可得220p ＝，则10p ＝，故其准线方程为5x =-， Q 抛物线220y x ＝的准线过双曲线()222210,0x y a b a b-=>>的左焦点， 5c ∴＝．Q 抛物线220y x ＝的准线被双曲线截得的线段长为92， 2292b a ∴=，又22225c a b +＝＝，4,3a b ∴＝＝，则双曲线的离心率为54c e a ==．故选：A ．【点睛】本题考查抛物线的性质及利用过双曲线的焦点的弦长求离心率. 弦过焦点时，可结合焦半径公式求解弦长．7．已知复数21iz i =-，则z 的虚部为（） A ．－1B ．i -C ．1D ．i【解析】【分析】分子分母同乘分母的共轭复数即可. 【详解】2i 2i(i 1)22i 1i i 1(i 1)(i+1)2z +-+====----，故z 的虚部为1-. 故选：A. 【点睛】本题考查复数的除法运算，考查学生运算能力，是一道容易题.8．若x 、y 满足约束条件220100x y x y y --≤⎧⎪-+≥⎨⎪≤⎩，则32z x y =+的最大值为（）A ．5B ．9C ．6D ．12【答案】C 【解析】【分析】作出不等式组所表示的可行域，平移直线32z x y =+，找出直线在y 轴上的截距最大时对应的最优解，代入目标函数计算即可. 【详解】作出满足约束条件220100x y x y y --≤⎧⎪-+≥⎨⎪≤⎩的可行域如图阴影部分（包括边界）所示．由32z x y =+，得322z y x =-+，平移直线322z y x =-+，当直线322zy x =-+经过点()2,0时，该直线在y 轴上的截距最大，此时z 取最大值，即max 32206z =⨯+⨯=. 故选：C.本题考查简单的线性规划问题，考查线性目标函数的最值，一般利用平移直线的方法找到最优解，考查数形结合思想的应用，属于基础题.9．已知函22()(sin cos )2cos f x x x x =++，,44x ππ⎡⎤∈-⎢⎥⎣⎦，则()f x 的最小值为（）A ．2-B ．1C ．0D ．【答案】B 【解析】【分析】())2,4f x x π=++,44x ππ⎡⎤∈-⎢⎥⎣⎦，32444x πππ-≤+≤利用整体换元法求最小值.【详解】由已知，2()12sin cos 2cos sin 2cos22f x x x x x x =++=++)2,4x π=++又44x ππ-≤≤，32444x πππ∴-≤+≤，故当244x ππ+=-，即4πx =-时，min ()1f x =.故选：B. 【点睛】本题考查整体换元法求正弦型函数的最值，涉及到二倍角公式的应用，是一道中档题. 10． “1cos 22α=-”是“3k παπ=+，k Z ∈”的（） A ．充分不必要条件 B ．必要不充分条件 C ．充要条件 D ．既不充分又不必要条件【答案】B 【解析】【分析】先求出满足1cos 22α=-的α值，然后根据充分必要条件的定义判断．【详解】由1cos 22α=-得2223k παπ=±，即3k παπ=±，k Z ∈ ，因此“1cos 22α=-”是“3k παπ=+，k Z ∈”的必要不充分条件．故选：B ．【点睛】本题考查充分必要条件，掌握充分必要条件的定义是解题基础．解题时可根据条件与结论中参数的取值范围进行判断．11．如图所示的程序框图，当其运行结果为31时，则图中判断框①处应填入的是（）A ．3?i ≤B ．4?i ≤C ．5?i ≤D ．6?i ≤【答案】C 【解析】【分析】根据程序框图的运行，循环算出当31S =时，结束运行，总结分析即可得出答案. 【详解】由题可知，程序框图的运行结果为31，当1S =时，9i =；当1910S =+=时，8i =；当19818S =++=时，7i =；当198725S =+++=时，6i =；当1987631S =++++=时，5i =. 此时输出31S =. 故选：C. 【点睛】本题考查根据程序框图的循环结构，已知输出结果求条件框，属于基础题.12．已知函数()32,0log ,0x x f x x x ⎧≤=⎨>⎩，则3=f f ⎛⎫ ⎪ ⎪⎝⎭⎝⎭（） A 2B ．12C ．3log 2-D ．3log 2【答案】A 【解析】【分析】根据分段函数解析式，先求得3f ⎝⎭的值，再求得33f f ⎛⎫⎛ ⎪ ⎪⎝⎭⎝⎭的值. 【详解】依题意1233331log log 32f -===-⎝⎭，12312222f f f -⎛⎫⎛⎫=-== ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭. 故选：A 【点睛】本小题主要考查根据分段函数解析式求函数值，属于基础题. 二、填空题：本题共4小题，每小题5分，共20分。

2020届山东省聊城市一中高三4月份线上模拟语文试题解析

聊城一中高三加强训练（一）语文试题一、现代文阅读（57分）阅读下面的文字，完成下面小题。

材料一：近年来，宋代文学研究的突破与创新多集中于诗、文、笔记等以往关注较少的文体领域，而传统的词学研究则相对较为沉寂。

在此背景下，马里扬《内美的镶边——宋词的文本形态与历史考证》一书的出版，可谓为宋词研究的拓展与深化提供了珍贵的尝试。

作者对其研究有着非常自觉的反省，以“内美的镶边’这一巧妙譬喻为其工作定位。

所谓“内美”，盖指词之为词的文体特性。

深入阐明这一点固为词学研究的根本要旨，然而作者自谓本书的研究并不直接探索“内美”，而将先从“镶边”的工作做起。

所谓“镶边”，作者借用高友工《美典：中国文学研究论集》一书中的“外缘研究”概念加以阐发，谓其为对与宋词文体特质相关的外缘因素之考察，但又与以文献、历史考证本身为目的的“外部研究”有所不同:“外部研究,侧重于文献的整理与作家生平及作品背景的查考，而外缘研究则是在文学本体研究之内的，也可以说是文学的历史与文化的批评，不是单纯的文献与历史的研究”,其意图在于“借助文献学或历史学的方法，要来进入对文本的文学特性的研究”。

虽以对宋词“内美”的关怀为底色，然而占据本书主体并最能体现作者研究功力的部分，仍是大量扎实、细腻甚至趋于烦琐的实证性考据工作。

书中所呈现的判断与观点，皆建立在对诸如“犯曲”结构与文辞格式、王安石文集编撰、苏轼与杨绘之交往、晏几道歌词“投赠"事件等具体问题的辨析之上。

在直接材料有限、史实面目不清的情况下,作者在茫茫史料间钩沉爬梳,如农民耕种般对一手文献材料一寸一寸地耐心耕耘，遂使史料间的隐晦联系逐渐显影，模糊的历史事件有了较为清晰的轮廓。

尤令人敬佩的是,虽以深厚的考证功力见长，作者呈现于书中的学术探索却不止于此。

在作者看来，实证性的考据工作，应当通向对文学本质问题的揭示与阐明:“我们认为，古典文学研究当中的‘考证’本身，恐怕不只是一种态度、方式与基础，或者应该本就是一种批评。

2020-2021学年聊城市第一中学高三生物月考试卷及答案

2020-2021学年聊城市第一中学高三生物月考试卷及答案一、选择题：本题共15小题，每小题2分，共30分。

每小题只有一个选项符合题目要求。

1. 下列关于细胞学说及其建立的叙述，错误的是（）A. 细胞学说主要是由施莱登和施旺提出的B. 细胞学说认同细胞是一个相对独立的单位C. 细胞学说认为细胞分为真核细胞和原核细胞D. 细胞学说的重要内容之一是动物和植物都是由细胞发育而来的2. 如图是细胞核模型图，下列有关叙述正确的是（）A.细胞的遗传信息库主要与4密切相关B.人体成熟的红细胞中2很小，因此红细胞代谢较弱C.5由四层磷脂分子组成，与核质的信息交流有关D.3的存在实现了细胞核是遗传和代谢的中心3. 下列有关种间关系的叙述，不正确的是（）A. 马蛔虫和马属于寄生B. 食性相同的异种鸟类分布在树林的不同位置能有效减小种间竞争C. 同一区域的两种动物之间没有共同的食物，不存在竞争关系D. 组成地衣的藻类和真菌之间属于互利共生的关系4. 下列与生活相联系的生物学知识中，说法正确的是（）A.胆固醇是动物细胞膜的重要成分，参与血液中脂质的运输，过多摄入有益无害B.糖尿病患者饮食虽然受到严格限制，但不具甜味的米饭、馒头等可随意食用C.患急性肠炎的病人脱水时，需要及时补水，同时也需要补充体内丢失的无机盐D.鸡蛋煮熟后，蛋白质发生了变性，不容易被蛋白酶水解，因此，吃熟鸡蛋难消化5. 某研究小组探究两种生长素类似物对月季插条生根的影响，得到右图所示实验结果。

相关判断错误的是A.实验自变量是生长素类似物的种类和浓度B.实验中“对照组”插条处理溶液可能是蒸馏水C.结果显示等浓度的IBA对生根促进作用比NAA强D.结果表明NAA、IBA对生根的作用具有两重性6. 与肌肉注射相比，静脉点滴因能将大剂量药物迅速送到全身细胞而疗效显著。

图中a、b、c为内环境的相关组成（其中b为血浆）。

下列叙述不正确的是（）A. 图中a为组织液，是体内绝大多数细胞直接生活的环境B. 正常情况下，a大量被毛细血管吸收进入b，少量被毛细淋巴管吸收成为cC. 静脉点滴的葡萄糖进入人体后到达组织细胞内至少需穿过5层细胞膜D. 静脉点滴一定浓度的血浆蛋白溶液有助于缓解因营养不良引起的组织水肿7. 下列有关细胞器的说法，不正确的是（）A. 所有细胞器都含有蛋白质B. 根尖细胞中含有核酸的细胞器有线粒体、核糖体C. 每种生物体内都含有一定结构和功能的细胞器D. 合成并分泌抗体的浆细胞中粗而内质网发达8. 研究发现调节性T细胞具有抑制免疫反应的功能，防止免疫反应过度损伤自身。

山东2020┄2021届高三模拟考试理综试卷4Word版含解析

理科综合能力测试注意事项：1、本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。

答题前，考生务必将自己的姓名、考生号填写在答题卡上。

2、回答第Ⅰ卷时，选出每小题的答案后，用铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号。

写在试卷上无效。

3、回答第Ⅱ卷时，将答案填写在答题卡上，写在试卷上无效。

4、考试结束，将本试卷和答题卡一并交回。

可能用到的相对原子质量：H-1C-12 N-14 O-16 Na-23 Cl-35.5Ca-40 Ti-48 Co-59第Ⅰ卷一、选择题：本大题共13小题，每小题6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．下列关于真核细胞结构和功能的叙述，正确的是A．内质网上核糖体合成的性激素与第二性征的维持有关B．纤维素组成的细胞骨架与细胞形态的维持有关C．线粒体与有氧呼吸过程中CO2和H2O的产生有关D．吞噬细胞中的溶酶体可合成、储存各种水解酶，用于分解侵入细胞的多种病菌【解析】性激素为固醇类物质，在卵巢和睾丸相应细胞的内质网上合成，不在核糖体上合成，性激素有促进动物生殖器官发育、维持第二性征等作用，A错误；真核细胞中有维持细胞形态、保持细胞内部结构有序性的细胞骨架，细胞骨架是由蛋白质纤维组成的网架结构，B错误；有氧呼吸第二阶段是丙酮酸和水彻底分解成二氧化碳和[H]，并释放少量能量，此过程发生在线粒体基质中，有氧呼吸第三阶段是有氧呼吸第一阶段和第二阶段产生的[H]，经过一系列反应与氧结合形成水，同时放出大量能量，此过程发生在线粒体内膜上，C正确；吞噬细胞中的溶酶体储存着多种水解酶，用于分解侵入细胞的多种病菌和吞入的细胞碎片等，但这些酶不是在溶酶体内合成，而是在核糖体上合成，D错误。

【答案】C2．下列关于双链DNA的结构和复制的叙述，正确的是A．DNA分子中的磷酸、碱基、脱氧核糖交替排列构成基本骨架B．DNA分子中碱基间的氢键使DNA分子具有较强的特异性C．DNA分子复制时，解旋酶与DNA聚合酶不能同时发挥作用D．噬菌体遗传物质DNA的复制所需要的原料全部由宿主细胞提供【解析】DNA分子中的磷酸和脱氧核糖交替排列构成基本骨架，A错误；DNA分子的特异性与组成DNA的碱基数目和排列顺序有关，与碱基间氢键无关，B错误；DNA分子是边解旋边复制，即解旋酶与DNA聚合酶能同时发挥作用，C错误；噬菌体是病毒，没有细胞结构，其遗传物质DNA的复制所需要的原料全部由宿主细胞提供，D正确。

2020届聊城市第一中学高三英语模拟试卷及参考答案

2020届聊城市第一中学高三英语模拟试卷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AAs a nation, we are getting bigger and eating more. But there are effective ways to control your appetite and eat only as much as you need.Keep away from low-nutrition snacksThat means ice cream, sweets, chips biscuits, cakes and any other salty orsugary snacks you eat between meals. Although we have a tendency to eat them, you can learn to live without these unhealthy-and-fattening-additions to your diet. Try to make it a habit to eat them only when offered at social events or as a special treat.Leave half an hour between main course and dessertHaving a break between courses gives your brain time to receive the fullness signal and make you more likely to refuse the sweet stuff. And, in fact, as soon as you feel the first signals of fullness, remove your plate from the table. That will tell your brain that food time is over.Make yours a small helpingPut an end to super-sized portions. You won’t be missing out—today’s small was the medium or large of a few decade ago. Select or serve yourself a modest portion and eat it slowly enjoying the flavors. Before you know it, small will feel just right. What’s more, ordering the smaller size leads to wearing the smaller size.Distract yourselfWhen you find yourself hunting down food, even though you’re even hungry, do something else for 20 minutes. Drink a large glass of water as thirst is often confused with a desire for food. Choose something that engages your brain as well as your hands, such as writing a letter or listening to a song. You could also go for a short walk or do something that you enjoy. If you think you really are hungry, set an alarm for 20 minutes’ time and if you still want to eat when it rings, fine. If not, the urge will have passed.1. Which way suits you better if you tend to order a large portion of food?A. Distract yourself.B. Make yours a small helping.C. Keep away from low-nutrition snacks.D. Leave half an hour between main course and dessert.2. Why should you have a break between main course and dessert?A. To give people time to chat.B. To have a good appetite for sweet stuff.C. To reduce appetite for dessert.D. To give the host time to remove your plate.3. When you find yourself pursuing for food, what should you do?A. Eat some biscuits.B. Eat some sugary snacks.C. Have some soft drinks.D. Listen to a lovely melody.BThe world's largest iceberg is floating toward South Georgia in the southernAtlantic Ocean. Scientists fear the iceberg could crash into the island and block major feeding areas for a large population of penguins and seals. If the iceberg hits the island, it could prevent the penguins and seals from reaching food supplies.The huge iceberg is named A68a. It broke away fromAntarctica's Larsen C Ice Shelf in 2017. Satellite images show the iceberg has remained in one piece. It is estimated to be about 150 kilometers long and 48 kilometers wide. It is traveling at one kilometer per hour and is on a path to hitSouth Georgiain around 30 days.This is the time of year when seals and penguins spend a lot of time caring for their young. The distance that parents have to travel to find food is crucial. That means they have to go a lot further or go around the iceberg to find sources of food.Ecologists say an iceberg crash would also disturb materials settled on the seabed, possibly polluting the surrounding seas. As the iceberg melts, it would also release large amounts of fresh water into the ocean. This could affect krill（磷虾）populations that are a major source of food for the island's wildlife. The iceberg could remain for up to 10 years and change the area's whole ecosystem. These are globally significant populations of these species. If these species fail in this particular area, then the numbers globally are going to go down quite dramatically.Professor Geraint Tarling, an ecologist at the British Antarctic Survey, said, "The breaking off of icebergs fromAntarcticais a natural process. But the process is changing with climate change. What we're seeing with models and some observations now is that this is happening at an increasing rate. And so, this might become more of a usual thing in the future."4. Why are the scientists worried about the coming iceberg crash?A. It will bring extremely cold weather.B. It will destroy the feeding areas of the animals.C. It will put wildlife on the island at risk of starving.D. It will prevent animals from moving to other places.5. What is paragraph 2 mainly about?A. The characteristic of the iceberg.B. The importance of the iceberg.C. The traveling speed of the iceberg.D. The forming process of the iceberg.6. What damage can an iceberg crash bring according to paragraph 4?A. Using up much fresh water.B. Polluting the surrounding farms.C. Changing the world's ecosystem.D. Affecting the number of certain species.7. How does Tarling think of the breaking off of icebergs fromAntarctica?A. It may slow down in the near future.B. It may become common in the future.C. It has a great influence on the climate.D. It helps scientists conduct a sea study.CAs an eco-minimalist, Su Yige has maintained an environmentally friendly and sustainable lifestyle for the past three years while studying in Canada. She takes her own bag when she goes grocery shopping and uses second-hand items as often as possible. She avoids almost all paper-related products unless she has to use a public toilet while away from home.Diligence and thrift are time-honored traditional Chinese virtues. Su's family is a good example of this, according to the native of Weihai, Shandong province. Although they have little formal “green” knowledge, her parents lead a very environmentally friendly life.For example, the family has used the same hair dryer for more than a decade, and Su remembers many of her mother's clothes from as far back as kindergarten. “As long as something can still be used, my mother will not replace it with a new item,” she said.“I frequently asked my father to bring the plastic bag back home after dumping our waste in the trash bin. He was unhappy, and argued that instead of making that request, I should go downstairs to dump the waste myself” she said. In another move, her father criticized her for doing too much shopping online. Eventually, they both made steps toward becoming better environmentalists. Her father brings the bag back for reuse and she has only bought two pieces of clothing online in the past six months.Back in Canada, Su is looking forward to finding a job related to sustainable development in China after she graduates as a computer science major in the summer.8. Which of the following best describes Su Yige?A. Conservative.B. Nostalgic.C. Economical.D. Productive.9. What can be found about Su's mother according to the third paragraph?A. She has a lot of formal green knowledge.B. She regards using the same items as a lifestyle.C. She always wears old clothes due to lack of money.D. She will not replace the old items until they are out of style.10. What can be inferred about Su and her father?A. Both of them like to criticize each other.B. Su's father is particular about her clothes.C. The relationship between them is very tense.D. They urge each other to become more environmentally friendly.11. What can we learn about the author soon after she graduates in Canada?A. She will stay there to look for a job.B. She will put effort into computer science.C. She will devote herself to her motherland's future.D. She will come back to China to stay with her parents.DYou’ve heard that plastic is polluting the oceans — between 4.8 and 12.7 million tonnes enter ocean ecosystems every year. But does one plastic straw or cup really make a difference? Artist Benjamin Von Wong wants you to know that it does. He builds massive sculptures out of plastic garbage, forcing viewers to re-examine their relationship to single-use plastic products.At the beginning of the year, the artist built a piece called “Strawpocalypse,” a pair of 10-foot-tall plasticwaves, frozen mid-crash. Made of 168,000 plastic straws collected from several volunteer beach cleanups, the sculpture made its first appearance at the Estella Place shopping center in Ho Chi Minh City, Vietnam.Just 9% of global plastic waste is recycled. Plastic straws are by no means the biggest source（来源）of plastic pollution, but they’ve recently come under fire because most people don’t need them to drink with and, because of their small size and weight, they cannot be recycled. Every straw that’s part ofVon Wong’s artwork likely came from a drink that someone used for only a few minutes. Once the drink is gone, the straw will take centuries to disappear.In a piece from 2018, Von Wong wanted to illustrate（说明）a specific statistic: Every 60 seconds, a truckload’s worth of plastic enters the ocean. For this work, titled “Truckload of Plastic,” Von Wong and a group of volunteers collected more than 10,000 pieces of plastic, which were then tied together to look like they’d been dumped（倾倒）from a truck all atonce.Von Wong hopes that his work will also help pressure big companies to reduce their plastic footprint.12. What are Von Wong’s artworks intended for?A. Beautifying the city he lives in.B. Introducing eco-friendly products.C. Drawing public attention to plastic waste.D. Reducing garbage on the beach.13. Why does the author discuss plastic straws in paragraph 3?A. To show the difficulty of their recycling.B. To explain why they are useful.C. To voice his views on modern art.D. To find a substitute for them.14. What effect would “Truckload of Plastic” have on viewers?A. Calming.B. Disturbing.C. Refreshing.D. Challenging.15. Which of the following can be the best title for the text?A. Artists’ Opinions on Plastic SafetyB. Media Interest in Contemporary ArtC. Responsibility Demanded of Big CompaniesD. Ocean Plastics Transformed into Sculptures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。

山东省聊城市2020届高三高考模拟(一)物理试题

聊城市2020年普通高中学业水平等级考试模拟卷物理试题（一）1.答题前，考生先将自己的姓名、考生号、座号填写在相应位置，认真核对条形码上的姓名、考生号和座号，并将条形码粘贴在指定位置上。

2.选择题答案必须使用2B 铅笔（按填涂样例）正确填涂；非选择题答案必须使用0.5毫米黑色签字笔书写，绘图时，可用2B 铅笔作答，字体工整、笔迹清楚。

3.请按照题号在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。

保持卡面清洁，不折叠、不破损。

一、单项选择题：本题共8小题。

每小题3分，共24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.下列说法中正确的是（）A.在光电效应实验中，对于同一种金属而言，相同颜色入射光的强度越大，飞出的光电子的最大初动能就越大B.结合能越大的原子核越稳定C.一个氢原子从4n =的能级向低能级跃迁时，最多可以辐射6种不同频率的光D.23892U 衰变成能20682Pb 要经过6次β衰变和8次α衰变2.真空中，两个相距L 的固定点电荷P 、Q 所带的电荷量的绝对值分别为P Q 和Q Q ，在它们共同形成的电场中，有一条电场线如图中实线所示，实线上的箭头表示电场线的方向，电场线上标出了M 、N 两点，其中过N 点的切线与P 、Q 连线平行，且NPQ NQP ∠>∠，则（）A.P 带负电，Q 带正电B.点电荷P 、Q 所带电荷量的关系为P Q Q Q <C.在M 点由静止释放一带正电的检验电荷，该电荷仅在电场力的作用下有可能沿电场线运动到N 点D.带负电的检验电荷在M 点的电势能大于在N 点的电势能3.如图所示，固定在地面上圆盘的上表面粗糙水平，一小物块从圆盘边缘上的P 点，以大小相同的初速度在圆盘上沿与直径PQ 成不同角θ开始滑动，小物块运动到圆盘另一边缘时的速度大小为v ，则2v θ-图像应为图中的（）A. B. C. D.4.下列说法正确的是（）A.同种元素的固体，可能由于原子（或分子）的排列方式不同而成为不同的晶体B.悬浮微粒越大，在某一瞬间撞击它的液体分子数越多，布朗运动越明显C.系统不可能从单一热源吸热全部用于做功D.布料做成的雨伞，纤维间虽然有缝隙，但是不漏雨，原因是雨水和布料浸润5.如图所示，理想变压器原线圈连接灯C后与120V的交流电源相连，副线圈并联两个灯泡A和B，灯泡A、C的额定功率皆为30W，正常发光时电阻皆为30Ω，已知三灯泡均正常发光，流过原线圈的电流为1.0A，则下列说法中正确的是（）A.原、副线圈匝数比为2：1B.原、副线圈匝数比为4：1C.流过灯泡B的电流为2AD.正常发光时B灯电阻为30Ω6.如图所示，ad、bd、cd是竖直面内三根固定的光滑细杆，a、b、c、d位于同一圆周上，a点为圆周的最高点，d点为最低点。

山东省聊城一中2020届高三4月线上模拟英语试题

聊城一中高三加强训练（一）英语强试题阅读（共两节，每题2.5分；满分97.5分）第一节Laura Sides was a psychology major at the University of Nottingham in 2004. She first noticed signs of her dad’s developing dementia(痴呆) when she moved to Nottingham. She said, "Dad was a doctor, so he knew exactly what had happened to him, but people try to hide it when they are ill. Then, I came home for my 21st birthday and arranged to meet him, but he never showed up as he’d forgotten. That’s when I knew something serious had happened."So, aged 21, she decided to leave university and look after him herself. She lived close by, popping in every day to make sure he was eating, and that the house was tidy, before heading off to her work.Besides challenging moments, there was a time when looking after her dad was a pure joy. "We’d wake up, I’d ask what he wanted to do that day, and however ridiculous the adventure is, off we’d go."Sadly, in 2009, 5 years later, Laura lost her father. Before he died, Laura went to a hospital appointment with him, where doctors mentioned that his form of Alzheimer’s disease was genetic meaning there was a fifty-fifty chance that she had inherited it. For several years Laura agonised over whether to be tested, finally finding out in August 2017 that she has the APP gene, meaning that, like him, she will develop the condition within a decade.At first, she struggled, feeling as if her life lacked purpose. Then, during a sleepless night in the summer of 2018, she decided at around 2 a.m. to enter the 2019 London Marathon sponsored by the charities Alzheimer’s Society and Alzheimer’s Research UK.She hoped to start the conversation around early-onset Alzheimer’s disease and to encourage people to talk about it more openly. "I remember when Dad was ill, people wouldn’t know how to reac t, but I want to be honest and open," she added. "The more information we can get, the less of a taboo(忌讳) we will feel. That said, the support I’ve received so far after going public has been amazing —that’s what is carrying me through."1. Laura noticed her father’s dementia when .A. her father told her his condition in personB. people nearby informed her of his father’s conditionC. her father forgot his own birthday partyD. her father forgot to attend her 21-year-old birthday party2. The underlined word "agonised" in Paragraph 4 probably means .A. excitedB. struggledC. shockedD. delighted3. Laura started the open talk in the hope of .A. getting people to talk about Alzheimer’s disease openlyB. earning some money to help treat her Alzhei mer’s diseaseC. making herself stronger to fight against Alzheimer’s diseaseD. raising funds for charities Alzheimer’s Society and Alzheimer’s disease Research UK4. Which words can best describe Laura?A. Caring and positive.B. Careful and honest.C. Patient and cautious.D. Devoted and modest.【答案】1. D 2. B 3. A 4. A【解析】【分析】本文是一篇记叙文。

山东省聊城市2021届新高考物理四月模拟试卷含解析

山东省聊城市2021届新高考物理四月模拟试卷一、单项选择题：本题共6小题，每小题5分，共30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．如图所示，细线的一端固定于O点，另一端系一小球，在水平拉力作用下，小球以恒定速率在竖直平面内由A点运动到B点，下列说法中正确的是( )A．水平拉力先增大后减小B．水平拉力先减小后增大C．水平拉力的瞬时功率先减小后增大D．水平拉力的瞬时功率逐渐增大【答案】D【解析】【分析】【详解】AB．小球是以恒定速率运动，即做匀速圆周运动，小球受到的重力G、水平拉力F、绳子拉力T，三者的合力必是沿绳子指向O点，设绳子与竖直方向夹角是θ，F与G的合力必与绳子拉力在同一直线上，则有=(1)tanF mgθ球由A点运动到B点θ 增大，说明水平拉力逐渐增大，故AB错误；CD．由几何关系可知拉力F的方向与速度v的夹角也是θ，所以水平力F 的瞬时功率是=(2)p Fvθcos联立两式可得=p mgvθsin从A 到B 的过程中，θ是不断增大的，所以水平拉力F 的瞬时功率是一直增大的．故C 错误，D 正确。

故选D 。

2．研究光电效应现象的实验电路如图所示，A 、K 为光电管的两个电极，电压表V 、电流计G 均为理想电表。

已知该光电管阴极K 的极限频率为ν0，元电荷电量为e ，普朗克常量为h ，开始时滑片P 、P ＇上下对齐。

现用频率为ν的光照射阴极K （ν>ν0），则下列说法错误的是A ．该光电管阴极材料的逸出功为hν0B ．若加在光电管两端的正向电压为U ，则到达阳极A 的光电子的最大动能为hv-hv 0+eUC ．若将滑片P 向右滑动，则电流计G 的示数一定会不断增大D ．若将滑片P ＇向右滑动，则当滑片P 、P ＇间的电压为0hv hv e-时，电流计G 的示数恰好为0 【答案】C 【解析】【详解】A ．由极限频率为ν0，故金属的逸出功为W 0= hν0，A 正确；B ．由光电效应方程可知，电子飞出时的最大动能为 0k E hv W =-由于加的正向电压，由动能定理kk eU E E '=- 解得0kE hv hv eU '=-+ 故B 正确；C ．若将滑片P 向右滑动时，若电流达到饱和电流，则电流不在发生变化，故C 错误；D ．P ＇向右滑动时，所加电压为反向电压，由k eU E =可得hv hv U e-=则反向电压达到遏止电压后，动能最大的光电子刚好不能参与导电，则光电流为零，故D 正确；故选C 。

山东省聊城市2020┄2021届高三4月第二次模拟考试试题英语

第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一个小题。

每段对话仅读一遍。

例：How much is the shirt？A．￡19．15．B．￡9．18．C．￡9．15．答案是C。

1．What is the man doing？A．Reading a magazine．B．Checking his email． C．Typing a report．2．What time is it now？A．12：30．B．1：00．C．1：30．3．Why does the woman often eat at restaurants？A．She likes the food there．B．She doesn’t like cooking．C．She’s too busy to cook．4．What are the speakers discussing？A．Which bus to take．B．Which way to go．C．Which stop to get off．5．Where are the speakers？A．In a clothes store．B．In a car shop．C．In a parking lot．第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料，回答第6、7题。

6．What sport made the man injured？A．Football．B．Basketball．C．Baseball．7．How soon can the man probably return to court？A．In a month．B．In two months．C．In five months．听第7段材料，回答第8、9题。

2020年山东省聊城市高考数学一模(4月份)试卷(Word解析版)

2020年高考（4月份）数学一模试卷一、选择题.1．已知集合A＝{x∈N*|x＜4}，B＝{x|x（x﹣2）≤0}，则集合A∩B中元素的个数为（）A．1B．2C．3D．42．已知复数z满足z（1+2i）＝|3+4i|（i是虚数单位），则z的共轭复数z=（）A．1+2i B．1﹣2i C．﹣1+2i D．﹣1﹣2i3．“a＜2”是“∀x∈R，a≤x2+1’为真命题”的（）A．充分不必要条件B．必要不充分条件C．充要条件D．既不充分也不必要条件4．已知cos(α−π6)=35，则sin(α+π3)=（）A．35B．−35C．45D．−455．将某校高一3班全体学生分成三个小组分别到三个不同的地方参加植树活动，若每个学生被分到三个小组的概率都相等，则这个班的甲，乙两同学分到同一个小组的概率为（）A．23B．12C．13D．196．数列1，6，15，28，45，…中的每一项都可用如图所示的六边形表示出来，故称它们为六边形数，那么第10个六边形数为（）A．153B．190C．231D．2767．正方体ABCD﹣A1B1C1D1的棱长为1，点M是棱CC1的中点，点A，B，D，M都在球O的球面上，则球O的表面积为（）A．32πB．3πC．94πD．9π8．高斯是德国著名的数学家，近代数学奠基者之一，享有“数学王子“的称号，为了纪念数学家高斯，人们把函数y＝[x]，x∈R称为高斯函数，其中[x]表示不超过x的最大整数．设{x}＝x﹣[x]，则函数f（x）＝2x{x}﹣x﹣1的所有零点之和为（）A．﹣1B．0C．1D．2二、多项选择题：本题共4小题，每小题5分，共20分.在每小题给出的四个选项中有多项符合题目要求.全部选对的得5分，部分选对的得3分，有选错的得0分.9．下列说法正确的是（）A ．回归直线一定经过样本点的中心（x ，y ）B ．若两个具有线性相关关系的变量的相关性越强，则线性相关系数r 的值越接近于1C ．在残差图中，残差点分布的水平带状区域越窄，说明模型的拟合精度越高D ．在线性回归模型中，相关指数R 2越接近于1，说明回归的效果越好10．若双曲线C ：x 2a 2−y 2b 2=1(a ＞0，b ＞0)的实轴长为6，焦距为10，右焦点为F ，则下列结论正确的是（）A ．C 的渐近线上的点到F 距离的最小值为4B ．C 的离心率为54 C ．C 上的点到F 距离的最小值为2D ．过F 的最短的弦长为32311．已知直线l ：2kx ﹣2y ﹣kp ＝0与抛物线C ：y 2＝2px （p ＞0）相交于A ，B 两点，点M（﹣1，﹣1）是抛物线C 的准线与以AB 为直径的圆的公共点，则下列结论正确的是（）A ．p ＝2B ．k ＝﹣2C ．|AB |＝5D ．△MAB 的面积为5√512．若实数a ≥2，则下列不等式中一定成立的是（）A ．（a +1）a +2＞（a +2）a +1B ．log a （a +1）＜log a +1（a +2）C ．log a (a +1)＜a+1aD ．log a+1(a +2)＜a+2a+1三、填空题：本题共4小题，每小题5分，共20分.13．已知(ax −2x 2)5的展开式中x ﹣1的系数为﹣40，则实数a ＝． 14．若函数f （x ）＝sin x +cos x 在[0，a ]上单调递增，则a 的取值范围为． 15．已知a =（cos α，sin α），b =（sin β，cos β），且α+β＝100°，则向量a 与b 的夹角θ＝．16．点M ，N 分别为三棱柱ABC ﹣A 1B 1C 1的棱BC ，BB 1的中点，设△A 1MN 的面积为S 1，平面A 1MN 截三棱柱ABC ﹣A 1B 1C 1所得截面面积为S ，五棱锥A 1﹣CC 1B 1NM 的体积为V 1，三棱柱ABC ﹣A 1B 1C 1的体积为V ，则V 1V = ，S 1S = ．四、解答题：本题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤. 17．①a 5＝b 3+b 5，②S 3＝87③a 9﹣a 10＝b 1+b 2这三个条件中任选一个，补充在下面问题中，并给出解答．设等差数列{a n }的前n 项和为S n ，数列{b n }的前n 项和为T n ，________，a 1＝b 6，若对于任意n ∈N *都有T n ＝2b n ﹣1，且S n ≤S k （k 为常数），求正整数k 的值．注：如果选择多个条件分别解答，那么按第一个解答计分．18．在平面四边形ABCD 中，AB =√2，AD =√17，∠ABD =45°．（1）求△ABD 的面积；（2）设M 为BD 的中点，且MC ＝MB ，求四边形ABCD 周长的最大值．19．如图，在四边形ABCD 中，BC ＝CD ，BC ⊥CD ，AD ⊥BD ，以BD 为折痕把△ABD 折起，使点A 到达点P 的位置，且PC ⊥BC ．（1）证明：PD ⊥平面BCD ；（2）若M 为PB 的中点，二面角P ﹣BC ﹣D 等于60°，求直线PC 与平面MCD 所成角的正弦值．20．已知椭圆C ：x 2a 2+y 2b 2=1(a ＞b ＞0)的长轴长为4，右焦点为F ，且椭圆C 上的点到点F 的距离的最小值与最大值的积为1，圆O ：x 2+y 2＝1与x 轴交于A ，B 两点．（1）求椭圆C的方程；（2）动直线l：y＝kx+m与椭圆C交于P，Q两点，且直线l与圆O相切，求△APQ的面积与△BPQ的面积乘积的取值范围．21．2020年春节期间，武汉市爆发了新型冠状病毒肺炎疫情，在党中央的坚强领导下，全国人民团结一心，众志成城，共同抗击疫情．某中学寒假开学后，为了普及传染病知识，增强学生的防范意识，提高自身保护能力，校委会在全校学生范围内，组织了一次传染病及个人卫生相关知识有奖竞赛（满分100分），竞赛奖励规则如下，得分在[70，80）内的学生获三等奖，得分在[80，90）内的学生获二等奖，得分在[90，100]内的学生获一等奖，其他学生不得奖．教务处为了解学生对相关知识的掌握情况，随机抽取了100名学生的竞赛成绩，并以此为样本绘制了如下样本频率分布直方图．（1）现从该样本中随机抽取两名学生的竞赛成绩，求这两名学生中恰有一名学生获奖的概率；（2）若该校所有参赛学生的成绩X近似服从正态分布N（μ，σ2），其中σ≈15，μ为样本平均数的估计值，利用所得正态分布模型解决以下问题：（i）若该校共有10000名学生参加了竞赛，试估计参赛学生中成绩超过79分的学生数（结果四舍五人到整数）；（ii）若从所有参赛学生中（参赛学生数大于10000）随机抽取3名学生进行座谈，设其中竞赛成绩在64分以上的学生数为ξ，求随机变量ξ的分布列和均值．附：若随机变量X服从正态分布N（μ，σ2），则P（μ﹣σ＜X≤μ+σ）≈0.6827，P （μ﹣2σ＜X≤μ+2σ）≈0.9544，P（μ﹣3σ＜X≤μ+3σ）≈0.9973．22．已知函数f（x）＝ax+x2lnx．（1）证明：当a≤0时，函数f（x）有唯一的极值点；（2）设a为正整数，若不等式f（x）＜e x在（0，+∞）内恒成立，求a的最大值．参考答案一、单项选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1．已知集合A＝{x∈N*|x＜4}，B＝{x|x（x﹣2）≤0}，则集合A∩B中元素的个数为（）A．1B．2C．3D．4【分析】可以求出集合A，B，然后进行交集的运算求出A∩B，进而可得出集合A∩B 中元素的个数．解：∵A＝{1，2，3}，B＝{x|0≤x≤2}，∴A∩B＝{1，2}，∴集合A∩B中元素的个数为2．故选：B．【点评】本题考查了描述法、列举法的定义，一元二次不等式的解法，交集的运算，集合元素的定义，考查了计算能力，属于基础题．2．已知复数z满足z（1+2i）＝|3+4i|（i是虚数单位），则z的共轭复数z=（）A．1+2i B．1﹣2i C．﹣1+2i D．﹣1﹣2i【分析】把已知等式变形，再由复数代数形式的乘除运算化简得答案．解：由z（1+2i）＝|3+4i|＝5，得z=51+2i=5(1−2i)(1+2i)(1−2i)=1−2i，∴z=1+2i．故选：A．【点评】本题考查复数代数形式的乘除运算化简，考查复数的基本概念，是基础题．3．“a＜2”是“∀x∈R，a≤x2+1’为真命题”的（）A．充分不必要条件B．必要不充分条件C．充要条件D．既不充分也不必要条件【分析】先化简，再判断单调性．解：“∀x∈R，a≤x2+1’为真命题”，则a≤1，则a＜2是a≤1的必要不充分条件，故选：B．【点评】本题考查解不等式，以及充要性，属于基础题．4．已知cos(α−π6)=35，则sin(α+π3)=（）A．35B．−35C．45D．−45【分析】由题意利用诱导公式求得要求式子的值．解：已知cos(α−π6)=35，则sin(α+π3)=cos[π2−（α+π3）]＝cos（π6−α）＝cos（α−π6）=35，故选：A．【点评】本题主要考查诱导公式的应用，属于基础题．5．将某校高一3班全体学生分成三个小组分别到三个不同的地方参加植树活动，若每个学生被分到三个小组的概率都相等，则这个班的甲，乙两同学分到同一个小组的概率为（）A．23B．12C．13D．19【分析】这个班的甲，乙两同学分配的基本事件总数n＝3×3＝9，这个班的甲，乙两同学分到同一个小组包含的基本事件个数m＝3，由此能求出这个班的甲，乙两同学分到同一个小组的概率．解：将某校高一3班全体学生分成三个小组分别到三个不同的地方参加植树活动，每个学生被分到三个小组的概率都相等，这个班的甲，乙两同学分配的基本事件总数n＝3×3＝9，这个班的甲，乙两同学分到同一个小组包含的基本事件个数m＝3，则这个班的甲，乙两同学分到同一个小组的概率为p=mn=39=13．故选：C．【点评】本题考查概率的求法，考查古典概型等基础知识，考查运算求解能力，是基础题．6．数列1，6，15，28，45，…中的每一项都可用如图所示的六边形表示出来，故称它们为六边形数，那么第10个六边形数为（）A．153B．190C．231D．276【分析】根据已知数，求出其规律即可求解结论．解：因为：1，6＝1+5，15＝1+5+9，28＝1+5+9+13，45＝1+5+9+13+19；即这些六边形数是由首项为1，公差为4的等差数列的和组成的；所以：c n＝1•n+n(n−1)2×4＝2n2﹣n；∴第10个六边形数为：2×102﹣10＝190．故选：B．【点评】本题主要考查归纳推理，归纳推理的一般步骤是：（1）通过观察个别情况发现某些相同性质；（2）从已知的相同性质中推出一个明确表达的一般性命题（猜想）．7．正方体ABCD﹣A1B1C1D1的棱长为1，点M是棱CC1的中点，点A，B，D，M都在球O的球面上，则球O的表面积为（）A．32πB．3πC．94πD．9π【分析】由题意可得底面外接圆的圆心为对角线BD的中点E，过E做底面的垂线在垂线上取O使OM＝OB，则O为外接球的球心，画图可得（详见解答）在两个三角形中求出外接球的半径，进而求出外接球的表面积．解：取BD的中点E，则E为△ADB的外心；由题意可得外接球的球心直线EO上，设球心为O，连接MO，OB可得都是外接球的半径，过O作OH⊥MC与H；则OB2＝OE2+EB2＝OE2+(√22)2①OM2＝OH2+MH2＝CE2+MH2=(√22)2+（12−OE)22②由①②可得：R2＝OB2=916；∴则球O的表面积为：4πR2=9π4；故选：C．【点评】本题考查三棱锥的外接球的半径与棱长的关系，及球的的表面积公式，属于中档题，解题时要认真审题，注意空间思维能力的培养．8．高斯是德国著名的数学家，近代数学奠基者之一，享有“数学王子“的称号，为了纪念数学家高斯，人们把函数y＝[x]，x∈R称为高斯函数，其中[x]表示不超过x的最大整数．设{x}＝x﹣[x]，则函数f（x）＝2x{x}﹣x﹣1的所有零点之和为（）A．﹣1B．0C．1D．2【分析】由题意，函数f（x）＝2x{x}﹣x﹣1的零点，即方程2x（x﹣[x]）﹣x﹣1＝0的根，也就是2x（x﹣[x]）＝x+1的根，x＝0显然不是上面方程的根；当x≠0时，方程化为2（x﹣[x]）=1+1x，作出两函数y＝2（x﹣[x]）与y＝1+1x的图象，数形结合得答案．解：∵{x}＝x﹣[x]，∴f（x）＝2x{x}﹣x﹣1＝2x（x﹣[x]）﹣x﹣1，函数f（x）＝2x{x}﹣x﹣1的零点，即方程2x（x﹣[x]）﹣x﹣1＝0的根，也就是2x（x﹣[x]）＝x+1的根．x＝0显然不是上面方程的根；当x≠0时，方程化为2（x﹣[x]）=1+1 x．作出两函数y＝2（x﹣[x]）与y＝1+1x的图象如图：由图可知，两函数的交点除（﹣1，0）之外，其余的交点关于（0，1）中心对称，则函数f（x）＝2x{x}﹣x﹣1的所有零点之和为﹣1．故选：A．【点评】本题考查函数零点与方程根的关系，考查数学转化思想方法与数形结合的解题思想方法，是中档题．二、多项选择题：本题共4小题，每小题5分，共20分.在每小题给出的四个选项中有多项符合题目要求.全部选对的得5分，部分选对的得3分，有选错的得0分.9．下列说法正确的是（）A．回归直线一定经过样本点的中心（x，y）B．若两个具有线性相关关系的变量的相关性越强，则线性相关系数r的值越接近于1C．在残差图中，残差点分布的水平带状区域越窄，说明模型的拟合精度越高D．在线性回归模型中，相关指数R2越接近于1，说明回归的效果越好【分析】根据学过的知识点进行判断．解：回归直线一定经过样本点的中心（x，y），A对；若两个具有线性相关关系的变量的相关性越强，则线性相关系数r的值越接近于1或﹣1，B错；在残差图中，残差点分布的水平带状区域越窄，说明模型的拟合精度越高，C对；在线性回归模型中，相关指数R2越接近于1，说明回归的效果越好，D对，故选：ACD．【点评】本题考查回归方程的性质，属于基础题．10．若双曲线C：x2a2−y2b2=1(a＞0，b＞0)的实轴长为6，焦距为10，右焦点为F，则下列结论正确的是（）A．C的渐近线上的点到F距离的最小值为4B ．C 的离心率为54 C ．C 上的点到F 距离的最小值为2D ．过F 的最短的弦长为323【分析】由题意求出a ，b ，c 的值，进而可得所给命题的真假．解：由题意可得2a ＝6，2c ＝10，所以a ＝3，c ＝5，b =√c 2−a 2=4，右焦点F （5，0），渐近线的方程为4x ﹣3y ＝0，所以C 的渐近线上的点到F 距离的最小值F 到渐近线的距离d =bc c=b ＝4，所以A 正确，离心率e =c a =53，所以B 不正确；双曲线上的点为顶点到焦点的距离最小，5﹣3＝2，所以C 正确；过焦点最短的弦长为垂直与x 轴的直线与双曲线的弦长，2b 2a =323，故D 正确，故选：ACD ．【点评】本题考查双曲线的性质，属于中档题．11．已知直线l ：2kx ﹣2y ﹣kp ＝0与抛物线C ：y 2＝2px （p ＞0）相交于A ，B 两点，点M（﹣1，﹣1）是抛物线C 的准线与以AB 为直径的圆的公共点，则下列结论正确的是（）A ．p ＝2B ．k ＝﹣2C ．|AB |＝5D ．△MAB 的面积为5√5 【分析】将直线l 的方程整理可得恒过定点即为抛物线的焦点，由M 在抛物线的准线上求出p 的值，进而可得AB 的方程，与抛物线联立求出两根之和及两根之积，再由以AB 为直径的圆过M 点可得可得MA →⋅MB →=0可得k 的值，进而求出弦长AB 的值，再求M 到直线AB 的距离可得三角形MAB 的面积，可判断所给命题的真假．解：直线l ：2kx ﹣2y ﹣kp ＝0整理可得k （2x ﹣p ）﹣2y ＝0，恒过（p 2，0），即过抛物线的焦点F ，所以抛物准线方程为x =−p 2，点M （﹣1，﹣1）是抛物线C 的准线与以AB 为直径的圆的公共点，M 在抛物线的准线上，所以−p 2=−1，解得p ＝2，所以A 正确，焦点坐标为（1，0），直线l 整理可得y ＝k （x ﹣1），设A （x 1，y 1），B （x 2，y 2），联立直线与抛物线的方程{y =k(x −1)y 2=4x ，整理可得：k 2x 2﹣（2k 2+4）x +k 2＝0，x 1x 2＝1，x 1+x 2=2k 2+4k2，y 1+y 2＝k （x 1+x 2﹣2）=4k，y 1y 2＝﹣4√x 1x 2=−4，由题意可得MA →⋅MB →=0，即（x 1+1，y 1+1）•（x 2+1，y 2+1）＝0 整理可得x 1x 2+（x 1+x 2）+1+y 1y 2+（y 1+y 2）+1＝0，代入可得1+2k 2+4k2+1﹣4+4k +1＝0，解得：4k +4k+1＝0，解得k ＝﹣2，所以B 正确，所以x 1x 2＝1，x 1+x 2＝3，所以弦长|AB |=√1+k 2√(x 1+x 2)2−4x 1x 2=√5⋅√5=5，所以C 正确；直线AB 的方程为：y ＝﹣2（x ﹣1），即2x +y ﹣2＝0，所以M 到直线AB 的距离d =5=√5，所以S △MAB =12|AB |•d =12⋅5⋅√5=5√52，所以D 不正确，故选：ABC ．【点评】本题考查直线恒过的定点的坐标及直线与抛物线的综合，面积的求法，属于中档题．12．若实数a ≥2，则下列不等式中一定成立的是（） A ．（a +1）a +2＞（a +2）a +1 B ．log a （a +1）＜log a +1（a +2） C ．log a (a +1)＜a+1a D ．log a+1(a +2)＜a+2a+1 【分析】令f （x ）=lnxx ，可得f ′（x ）=1−lnx x 2，利用导数研究其单调性即可判断出ACD 的大小关系．对于B ．令g （x ）＝log x （x +1）（x ≥2）．利用导数研究其单调性即可判断出B 的大小关系．解：令f （x ）=lnxx ，则f ′（x ）=1−lnx x 2，可得函数f （x ）在（0，e ）上单调递增，在（e ，+∞）上单调递减， ∵实数a ≥2，∴a +1＞e ， ∴ln(a+1)a+1＞ln(a+2)a+2，∴（a +1）a +2＞（a +2）a +1．（a +2）lna ＞（a +1）ln （a +2），可得：log a +1（a +2）＜a+2a+1．ln(a+1) a+1与lnaa的大小关系不确定．可得：AD正确，C不正确．对于B．令g（x）＝log x（x+1）（x≥2）．g′（x）=xlnx−(x+1)ln(x+1)ln2x＜0，∴函数g（x）在[2，+∞）上单调递减，∴log a（a+1）＞log（a+1）（a+2），综上可得：只有AD正确．故选：AD．【点评】本题考查了利用导数研究函数的单调性极值与最值、数形结合方法、等价转化方法、对数运算性质，考查了推理能力与计算能力，属于中档题．三、填空题：本题共4小题，每小题5分，共20分.13．已知(ax−2x2)5的展开式中x﹣1的系数为﹣40，则实数a＝﹣1．【分析】利用通项公式即可得出．解：(ax−2x2)5的展开式中通项公式：T k+1=∁5k（ax）5﹣k(−2x2)k=（﹣2）k•∁5k•a5﹣k•x5﹣3k，令5﹣3k＝﹣1，解得k＝2．∵x﹣1的系数为﹣40，∴（﹣2）2∁52•a3＝﹣40，解得a＝﹣1．故答案为：﹣1．【点评】本题考查了二项式定理的通项公式．考查考生的计算能力，属于基础题．14．若函数f（x）＝sin x+cos x在[0，a]上单调递增，则a的取值范围为（0，π4]．【分析】由题意利用正弦函数的单调性，求得a的范围．解：∵函数f （x ）＝sin x +cos x =√2sin （x +π4）在[0，a ]上单调递增，而x +π4∈[π4，a +π4]， ∴a +π4≤π2，即 0＜a ≤π4，故答案为：（0，π4]．【点评】本题主要考查正弦函数的单调性，属于基础题．15．已知a =（cos α，sin α），=（sin β，cos β），且α+β＝100°，则向量a 与的夹角θ＝ 10° ．【分析】根据题意，由a 、b 的向量坐标计算分析可得|a |、|b |以及a →•b →的值，结合向量数量积的计算公式计算可得答案．解：根据题意，a =（cos α，sin α），b =（sin β，cos β），则|a →|=√cos 2α+sin 2α=1，|b →|=√sin 2β+cos 2β=1，则a →•b →=cos αsin β+sin αcos β＝sin （α+β）＝sin100°，则cos θ=a →⋅b→|a →||b →|=sin100°，又由0°≤θ≤100°，则θ＝10°；故答案为：10°【点评】本题考查向量数量积的坐标计算，涉及三角函数的和角公式，属于基础题． 16．点M ，N 分别为三棱柱ABC ﹣A 1B 1C 1的棱BC ，BB 1的中点，设△A 1MN 的面积为S 1，平面A 1MN 截三棱柱ABC ﹣A 1B 1C 1所得截面面积为S ，五棱锥A 1﹣CC 1B 1NM 的体积为V 1，三棱柱ABC ﹣A 1B 1C 1的体积为V ，则V 1V=712，S 1S=35．【分析】如图所示，延长NM 交直线C 1C 于点P ，连接PA 1交AC 于点Q ，连接QM ．可得截面为四边形A 1NMQ ．由BB 1∥CC 1，M 为BC 的中点，可得△PCM ≌△NBM ．可得△A 1MN 的面积S 1=12S △A 1NP ，由QC ∥A 1C 1，利用平行线的性质可得S1S．△BMN 的面积与四边形B 1C 1BC 面积的关系，五棱锥A 1﹣CC 1B 1NM 的体积与四棱锥A 1﹣B 1C 1BC 的关系，而三棱锥A 1﹣ABC 的体积=23V ，即可得出V1V．解：如图所示，延长NM 交直线C 1C 于点P ，连接PA 1交AC 于点Q ，连接QM ．平面A 1MN 截三棱柱ABC ﹣A 1B 1C 1所得截面为四边形A 1NMQ ．∵BB 1∥CC 1，M 为BC 的中点，则△PCM ≌△NBM ．点M 为PN 的中点．∴△A 1MN 的面积S 1=12S △A 1NP ，∵QC ∥A 1C 1，PC PC 1=13=PQ PA 1，∴△A 1QM 的面积=23S △A 1PM ，∴S 1S=35．∵△BMN 的面积=18S 四边形B 1C 1BC ，∴五棱锥A 1﹣CC 1B 1NM 的体积为V 1=78V 四棱锥A 1−B 1C 1BC ，而三棱锥A 1﹣ABC 的体积=23V ， ∴V 1V=78×23V V =712．故答案为：712，35．【点评】本题考查了空间位置关系、三角形面积之比、三棱锥与四棱锥及其三棱柱的体积．考查了推理能力与计算能力，属于中档题．四、解答题：本题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤. 17．①a 5＝b 3+b 5，②S 3＝87③a 9﹣a 10＝b 1+b 2这三个条件中任选一个，补充在下面问题中，并给出解答．设等差数列{a n }的前n 项和为S n ，数列{b n }的前n 项和为T n ，________，a 1＝b 6，若对于任意n ∈N *都有T n ＝2b n ﹣1，且S n ≤S k （k 为常数），求正整数k 的值．注：如果选择多个条件分别解答，那么按第一个解答计分．【分析】由T n＝2b n﹣1，结合数列递推式；n＝1时，b1＝T1，n≥2时，b n＝T n﹣T n﹣1，结合等比数列的定义和通项公式可得b n，然后选三个条件中一个，结合等差数列的通项公式可得a n，再讨论{a n}的正负项，即可得到所求值．解：由T n＝2b n﹣1，可得n＝1时，b1＝1；n≥2时，T n﹣1＝2b n﹣1﹣1，相减可得b n＝2b n ﹣2b n﹣1，即b n＝2b n﹣1，由此可得{b n}为首项为1，公比为2的等比数列，故b n＝2n﹣1，①当a5＝b3+b5，a1＝b6＝32，a5＝4+16＝20，设{a n}的公差为d，则20＝32+4d，解得d＝﹣3，所以a n＝32﹣3（n﹣1）＝35﹣3n．因为当n≤11时，a n＞0，当n＞11时，a n＜0，所以当n＝11时，S n取得最大值，因此正整数k的值为11．②当S3＝87时，a1＝32，3a2＝87，设{a n}的公差为d，则3（32+d）＝87，解得d＝﹣3，所以a n＝32﹣3（n﹣1）＝35﹣3n．因为当n≤11时，a n＞0，当n＞11时，a n＜0，所以当n＝11时，S n取得最大值，因此正整数k的值为11．③当a9﹣a10＝b1+b2时，a1＝32，a9﹣a10＝3，设{a n}的公差为d，则d＝﹣3，所以a n＝32﹣3（n﹣1）＝35﹣3n．因为当n≤11时，a n＞0，当n＞11时，a n＜0，所以当n＝11时，S n取得最大值，因此正整数k的值为11．【点评】本题考查等差数列和等比数列的定义和通项公式的运用，以及数列递推式的运用，考查等差数列的前n项和的最值情况，考查方程思想和运算能力、推理能力，属于中档题．18．在平面四边形ABCD中，AB=√2，AD=√17，∠ABD=45°．（1）求△ABD的面积；（2）设M为BD的中点，且MC＝MB，求四边形ABCD周长的最大值．【分析】（1）连接BD，在△ABD中，设BD＝x，由余弦定理可得x2﹣2x﹣15＝0，解得x，可得BD的值，进而根据三角形的面积公式即可求解．（2）由M为BD的中点，可得MC为△BCD的边BD的中线，又MC＝MB，可得∠BCD＝90°，利用勾股定理可求BC2+CD2＝BD2＝25，根据基本不等式可求BC+CD≤5√2，即可求解四边形ABCD的周长的最大值．解：（1）连接BD，在△ABD中，由余弦定理可得AD2＝AB2+BD2﹣2AB•BD•cos∠ABD，设BD＝x，则17＝2+x2﹣2√2x cos45°，即x2﹣2x﹣15＝0，解得x＝5或x＝﹣3（舍去），所以BD＝5，所以S△ABD=12AB•BD•sin∠ABD=12×√2×5×√22=52．（2）由M为BD的中点，可得MC为△BCD的边BD的中线，又MC＝MB，可得MC=12BD，所以∠BCD＝90°，所以BC2+CD2＝BD2＝25，又（BC+CD）2＝BC2+CD2+2BC•CD≤2（BC2+CD2）＝50，所以BC+CD≤5√2，当且仅当BC＝CD时等号成立，所以AB+AD+BC+CD≤6√2+√17，即四边形ABCD的周长的最大值为6√2+√17．【点评】本题主要考查了余弦定理，三角形的面积公式，勾股定理，基本不等式在解三角形中的综合应用，考查了转化思想和数形结合思想的应用，属于中档题．19．如图，在四边形ABCD中，BC＝CD，BC⊥CD，AD⊥BD，以BD为折痕把△ABD折起，使点A到达点P的位置，且PC⊥BC．（1）证明：PD⊥平面BCD；（2）若M为PB的中点，二面角P﹣BC﹣D等于60°，求直线PC与平面MCD所成角的正弦值．【分析】（1）由已知可得BC ⊥平面PCD ，得到BC ⊥PD ，再由PD ⊥BD ，利用直线与平面垂直的判定可得PD ⊥平面BCD ；（2）由PC ⊥BC ，CD ⊥BC ，可得∠PCD 是二面角P ﹣BC ﹣D 的平面角，则∠PCD ＝60°，取BD 的中点O ，连接OM ，OC ，可得OM ，OC ，OD 两两互相垂直，以O 为坐标原点，分别以OC ，OD ，OM 所在直线为x ，y ，z 轴建立空间直角坐标系，分别求出平面MCD 的一个法向量n →与CP →的坐标，由两向量所成角的余弦值可得直线PC 与平面MCD 所成角的正弦值．【解答】（1）证明：∵BC ⊥CD ，BC ⊥PC ，且PC ∩CD ＝C ， ∴BC ⊥平面PCD ，又∵PD ⊂平面PCD ，∴BC ⊥PD ． ∵PD ⊥BD ，BD ∩BC ＝B ， ∴PD ⊥平面BCD ；（2）解：∵PC ⊥BC ，CD ⊥BC ，∴∠PCD 是二面角P ﹣BC ﹣D 的平面角，则∠PCD ＝60°，因此PD ＝CD •tan60°=√3CD ，取BD 的中点O ，连接OM ，OC ，由已知可得OM ，OC ，OD 两两互相垂直，以O 为坐标原点，分别以OC ，OD ，OM 所在直线为x ，y ，z 轴建立空间直角坐标系，设OB ＝1，则P （0，1，√6），C （1，0，0），D （0，1，0），M （0，0，√62），CP →=(−1，1，√6)，CD →=(−1，1，0)，CM →=(−1，0，√62)．设平面MCD 的一个法向量为n →=(x ，y ，z)，由{n →⋅CD →=−x +y =0n →⋅CM →=−x +√62z =0，取z =√2，得n →=(√3，√3，√2)．∴cos ＜n →，CP →＞=CP →⋅n→|CP →|⋅|n →|=√34．故直线PC 与平面MCD 所成角的正弦值为√34．【点评】本题考查直线与平面垂直的判定，考查空间想象能力与思维能力，训练了利用空间向量求解空间角，是中档题． 20．已知椭圆C ：x 2a 2+y 2b 2=1(a ＞b ＞0)的长轴长为4，右焦点为F ，且椭圆C 上的点到点F 的距离的最小值与最大值的积为1，圆O ：x 2+y 2＝1与x 轴交于A ，B 两点．（1）求椭圆C 的方程；（2）动直线l ：y ＝kx +m 与椭圆C 交于P ，Q 两点，且直线l 与圆O 相切，求△APQ 的面积与△BPQ 的面积乘积的取值范围．【分析】（1）根据题意可得2a ＝4，（a ﹣c ）（a +c ）＝1，解得a ，b ，进而得椭圆C 的方程．（2）联立直线l 与椭圆的方程，△＞0且设P （x 1，y 1），Q （x 2，y 2），则x 1+x 2=−8km 1+4k2，x 1x 2=4m 2−41+4k2，还能得到A ，B 两点到直线l 得距离分别为d 1，d 2，S △APQ •S △BPQ =1216k 2+1k2+8，由基本不等式可得出答案．解：（1）设椭圆C 的焦距为2c ，则由已知得2a ＝4，（a ﹣c ）（a +c ）＝1，解得a ＝2，b ＝1，所以椭圆C 的方程为x 24+y 2=1．（2）由{y =kx +mx 24+y 2=1，得（1+4k 2）x 2+8kmx +4m 2﹣4＝0， △＝（8km ）2﹣4（1+4k 2）（4m 2﹣4）＝16（1+4k 2﹣m 2）＞0，设P （x 1，y 1），Q （x 2，y 2），则x 1+x 2=−8km 1+4k2，x 1x 2=4m 2−41+4k2，所以|PQ |=√1+k 2⋅√(x 1+x 2)2−4x 1x 2=√1+k 2⋅√(−8km 1+4k2)2−4⋅4m 2−41+4k2=√1+k 2⋅√16(1+4k 2−m 2)(1+4k 2)2．因为直线l 与O 相切，所以点O 到直线l 得距离d =√1+k=1，即1+k 2＝m 2．所以△＝48k 2，由△＞0，得k 2＞0，又A ，B 两点到直线l 得距离分别为d 1=√1+k，d 2=√1+k，所以△APQ 的面积与△BPQ 的面积乘积为S △APQ •S △BPQ =14（√1+k 2⋅√16(1+4k 2−m 2)(1+4k 2)2）•√1+k 2•√1+k 2=4(1+4k 2−m 2)|k 2−m 2|(1+4k 2)2=4(1+4k 2−1−k 2)|k 2−1−k 2|(1+4k 2)2=12k2(1+4k 2)2=1216k 2+1k2+8，因为k 2＞0，所以16k 2+1k2+8≥16•116k 2+1k2+8∈（0，116]，S △APQ •S △BPQ ∈（0，34]，所以△APQ 的面积与△BPQ 的面积乘积的取值范围为（0，34]．【点评】本题考查椭圆的标准方程，直线与椭圆的相交问题，属于中档题．21．2020年春节期间，武汉市爆发了新型冠状病毒肺炎疫情，在党中央的坚强领导下，全国人民团结一心，众志成城，共同抗击疫情．某中学寒假开学后，为了普及传染病知识，增强学生的防范意识，提高自身保护能力，校委会在全校学生范围内，组织了一次传染病及个人卫生相关知识有奖竞赛（满分100分），竞赛奖励规则如下，得分在[70，80）内的学生获三等奖，得分在[80，90）内的学生获二等奖，得分在[90，100]内的学生获一等奖，其他学生不得奖．教务处为了解学生对相关知识的掌握情况，随机抽取了100名学生的竞赛成绩，并以此为样本绘制了如下样本频率分布直方图．（1）现从该样本中随机抽取两名学生的竞赛成绩，求这两名学生中恰有一名学生获奖的概率；（2）若该校所有参赛学生的成绩X 近似服从正态分布N （μ，σ2），其中σ≈15，μ为样本平均数的估计值，利用所得正态分布模型解决以下问题：（i）若该校共有10000名学生参加了竞赛，试估计参赛学生中成绩超过79分的学生数（结果四舍五人到整数）；（ii）若从所有参赛学生中（参赛学生数大于10000）随机抽取3名学生进行座谈，设其中竞赛成绩在64分以上的学生数为ξ，求随机变量ξ的分布列和均值．附：若随机变量X服从正态分布N（μ，σ2），则P（μ﹣σ＜X≤μ+σ）≈0.6827，P （μ﹣2σ＜X≤μ+2σ）≈0.9544，P（μ﹣3σ＜X≤μ+3σ）≈0.9973．【分析】（1）由样本频率分布直方图得样本中获一等奖的6人，获二等奖的8人，获三等奖的16人，共30人获奖，70人没有获奖，从该样本中随机抽取两名学生的竞赛成绩，基本事件总数为n=C1002，设“抽取的两名学生中恰有一名学生获奖”为事件A，则事件A包含的基本事件的个数为m=C701C301，由此能求出这两名学生中恰有一名学生获奖的概率．（2）（i）由样本频率分布直方图得样本平均数的估计值，得到所有参赛学生的成绩X 近似值服从正态分布N（64，152），由此能估计参赛学生中成绩超过79分的学生数．（ii）由μ＝64，得P（X＞64）=12，得到随机变量ξ～B（3，12），由此能求出ξ的分布列和E（ξ）．解：（1）由样本频率分布直方图得样本中获一等奖的6人，获二等奖的8人，获三等奖的16人，共30人获奖，70人没有获奖，从该样本中随机抽取两名学生的竞赛成绩，基本事件总数为n=C1002，设“抽取的两名学生中恰有一名学生获奖”为事件A，则事件A包含的基本事件的个数为m=C701C301，∴这两名学生中恰有一名学生获奖的概率p（A）=C701C301C1002=1433．（2）由样本频率分布直方图得样本平均数的估计值：μ＝35×0.006×10+45×0.012×10+55×0.018×10+65×0.034×10+75×0.016×10+85×0.008×10+95×0.006×10＝64，所有参赛学生的成绩X 近似值服从正态分布N （64，152），（i ）∵μ+σ＝79，∴P （X ＞79）≈1−0.68272=0.15865， ∴估计参赛学生中成绩超过79分的学生数为： 0.15868×10000≈1587（人）．（ii ）由μ＝64，得P （X ＞64）=12，即从所有参赛学生中随机抽取1名学生，该生竞赛成绩在64分以上的概率为12，∴随机变量ξ～B （3，12），P （ξ＝0）=C 30(12)3=18， P （ξ＝1）=C 31(12)(12)2=38， P （ξ＝2）=C 32(12)2(12)=38， P （ξ＝3）=C 33(12)3=18，∴ξ的分布列为：ξ 0 1 2 3 P18383818∴E （ξ）=0×18+1×38+2×38+3×18=32．【点评】本题考查概率、离散型随机变量的分布列、数学期望的求法，考查频率分布直方图、二项分布的性质等基础知识，考查运算求解能力，是中档题． 22．已知函数f （x ）＝ax +x 2lnx ．（1）证明：当a ≤0时，函数f （x ）有唯一的极值点；（2）设a 为正整数，若不等式f （x ）＜e x 在（0，+∞）内恒成立，求a 的最大值．【分析】（1）由于f ′（x ）＝a +x +2xlnx ，设g （x ）＝a +x +2xlnx ，则g ′（x ）＝2lnx +3，分x ∈（e −32，+∞）与x ∈（0，e −32）两类讨论，可证得当a ≤0时，函数f （x ）有唯一的极值点；（2）当x ∈（0，+∞）时，不等式f （x ）＜e x，等价于e xx 2−lnx −a x ＞0，设h （x ）=e x x 2−lnx −2x ，利用导数，可求得（x ）min ＝h （2）=e 24−ln 2﹣1＞0，从而可求得a 的最大值．【解答】（1）证明：函数f （x ）的定义域为（0，+∞），f ′（x ）＝a +x +2xlnx ，设g （x ）＝a +x +2xlnx ，则g ′（x ）＝2lnx +3，…2分①当x ∈（e −32，+∞）时，因为g ′（x ）＞0，所以g （x ）在（e −32，+∞）内单调递增，又因为g （e −32）＝a +e −32+2e −32ln e −32=a ﹣2e −32＜0，g （e ﹣a ）＝a +e ﹣a +2e ﹣a lne ﹣a ＝a +e ﹣a ﹣2ae ﹣a ＝e ﹣a +a （1﹣2e ﹣a ）＞0，所以存在x 0∈（e −32，e ﹣a ），使g （x 0）＝0，对于x ∈（e −32，x 0），都有g （x ）＜0，对于x ∈（x 0，+∞），都有g （x ）＞0；…4分②当x ∈（0，e −32）时，g （x ）＝a +x （1+2lnx ）＜a ﹣2x ＜0，…5分因为g ′（x ）＞0，所以g （x ）在（e −32，+∞）内单调递增，又因为g （e −32）＝a +e −32+2e −32ln e −32=a ﹣2e −32＜0，综上可得，f ′（x 0）＝0，当∈（0，x 0）时，f ′（x ）＜0，当x ∈（x 0，+∞）时，f ′（x ）＞0，因此，当a ≤0时，函数f （x ）有唯一的极值点；…6分（2）解：当x ∈（0，+∞）时，不等式f （x ）＜e x，等价于e x x 2−lnx −ax ＞0，令x ＝1，得a ＜e ，又因为a 为正整数，所以a ＝1或2，当a ＝2，…7分不等式f （x ）＜e x ，即e x x 2−lnx −2x ＞0，设h （x ）=e x x 2−lnx −2x ，则h ′（x ）=x 2e x−2xe x 4−1x +22=(x−2)(e x−x)3，…8分设k （x ）＝e x ﹣x ，则k ′（x ）＝e x ﹣1，因为当x ＞0时，k ′（x ）＞0，所以函数k （x ）在[0，+∞）上单调递增，又因为k （0）＝1＞0，所以当x ＞0时，k （x ）＞0，即e x ﹣x＞0，…9分令h ′（x ）＝0，得x ＝2，因为e x ﹣x ＞0，所以当x ∈（0，2）时，h ′（x ）＜0，当x ∈（2，+∞）时，h ′（x ）＞0，时，所以h （x ）min ＝h （2）=e 24−ln 2﹣1，又因为h （2）＞0，所以h （x ）＞0，因此，当x ＞0时，h （x ）＞0恒成立．也就是说当a ＝2时，不等式f （x ）＜e x 在（0，+∞）内恒成立…11分故的最大值为2…12分【点评】本题考查利用导数研究函数的极值与最值，考查等价转化思想、分类与整合思想，函数与方程思想的综合运用，考查逻辑推理与数学运算能力，属于难题．。

聊城一中2020届高三4月份线上模拟数学

17.（10
分）已知
f
x
2cosx
sin
x
6
3sinx cosx sin2 x
（1）求函数 y f x 的单调递增区间；
（2）设
ABC
的内角
A
满足
f
A
2
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AB
AC
3 ，求边 BC 的最小值．
18.（12
分）已知数列an 的前 n
项和为
Sn
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3 4
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Sn1
an1
1 2
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P（，﹣1），且△PF1F2 的面积为 2 （Ⅰ）求椭圆 C 的标准方程（Ⅱ）设斜率为 1 的直线 l 与以原点为圆心，半径为的圆交于 A，B 两点，与椭圆 C 交于 C，D 两点，且|CD|＝λ|AB|（λ∈R），当λ取得最小值时，求直线 l 的方程
3
21．（12 分）某公司即将推车一款新型智能手机，为了更好地对产品进行宣传，需预估市民购买该款手机是否与年龄有关，现随机抽取了 50 名市民进行购买意愿的问卷调查，若得分低于 60 分，说明购买意愿弱；若得分不低于 60 分，说明购买意愿强，调查结果用茎叶图表示如图所示．
且
n
2
），数列 bn 满足：
b1
37 4
，且
3bn
bn1
n
1
（
n
N
*
且
n
2
）.
（1）求证：数列bn an 为等比数列；
（2）求数列bn 的前 n 项和的最小值.
盟
19.（12 分）如图，在三棱锥 S ABC 中，SA 底面 ABC, AC AB SA 2, AC AB, D, E 分别是 AC, BC

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