【经典汇编】2019-2020年广东省广州市三模：广州市2019届高三第三次模拟考试数学(理)试题-含答案
广东省广州市2019-2020学年高考数学三模试卷含解析
广东省广州市2019-2020学年高考数学三模试卷一、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．已知复数为纯虚数(为虚数单位)，则实数（ ） A ．-1 B ．1C ．0D ．2【答案】B 【解析】 【分析】 化简得到，根据纯虚数概念计算得到答案.【详解】为纯虚数，故且，即.故选：. 【点睛】本题考查了根据复数类型求参数，意在考查学生的计算能力.2．已知实数,x y 满足,10,1,x y x y y ≥⎧⎪+-≤⎨⎪≥-⎩则2z x y =+的最大值为（ ）A ．2 B．32C ．1D ．0【答案】B 【解析】 【分析】作出可行域，平移目标直线即可求解. 【详解】 解：作出可行域：由2z x y =+得，1122y x z =-+ 由图形知，1122y x z =-+经过点时，其截距最大，此z 时最大10y x x y =⎧⎨+-=⎩得1212x y ⎧=⎪⎪⎨⎪=⎪⎩，11,22C ⎛⎫ ⎪⎝⎭ 当1212x y ⎧=⎪⎪⎨⎪=⎪⎩时，max 1232222z =+⨯=故选：B 【点睛】考查线性规划，是基础题.3．设曲线(1)ln y a x x =--在点()1,0处的切线方程为33y x =-，则a =（ ） A ．1 B ．2 C ．3 D ．4【答案】D 【解析】 【分析】利用导数的几何意义得直线的斜率，列出a 的方程即可求解 【详解】 因为1y a x'=-，且在点()1,0处的切线的斜率为3，所以13a -=，即4a =. 故选：D 【点睛】本题考查导数的几何意义，考查运算求解能力，是基础题4．已知数列{}n a 是公比为2的正项等比数列，若m a 、n a 满足21024n m n a a a <<，则()21m n -+的最小值为（ ） A ．3 B ．5C ．6D ．10【答案】B 【解析】 【分析】利用等比数列的通项公式和指数幂的运算法则、指数函数的单调性求得110m n <-<再根据此范围求()21m n -+的最小值．【详解】Q 数列{}n a 是公比为2的正项等比数列，m a 、n a 满足21024n m n a a a <<，由等比数列的通项公式得11111122210242n m n a a a ---⋅<⋅<⋅，即19222n m n -+<<，10222m n -∴<<，可得110m n <-<，且m 、n 都是正整数，求()21m n -+的最小值即求在110m n <-<，且m 、n 都是正整数范围下求1m -最小值和n 的最小值，讨论m 、n 取值.∴当3m =且1n =时，()21m n -+的最小值为()23115-+=.故选：B ． 【点睛】本题考查等比数列的通项公式和指数幂的运算法则、指数函数性质等基础知识，考查数学运算求解能力和分类讨论思想，是中等题．5．在ABC ∆中,,A B C ∠∠∠所对的边分别是,,a b c ，若3,4,120a b C ︒==∠=，则c =（ ）A ．37B ．13C D【答案】D 【解析】 【分析】直接根据余弦定理求解即可． 【详解】解：∵3,4,120a b C ︒==∠=，∴2222cos 9161237c a b ab C =+-=++=，∴c = 故选：D ． 【点睛】本题主要考查余弦定理解三角形，属于基础题． 6．已知集合A {}0,1,2=，B={}(2)0x x x -<,则A∩B= A ．{}1 B ．{}0,1C ．{}1,2D ．{}0,1,2【答案】A 【解析】 【分析】先解A 、B 集合，再取交集。
广东省广州市2019届高三地理第三次模拟考试试题(含解析)
广东省广州市2019届高三地理第三次模拟考试试题（含解析）一、单项选择题东莞市某家具企业于2010年起从意大利进口价值3亿元人民币的机器人，由机器手组成的自动化生产线替代原来的工人，帮助企业完成锯木、切块、涂漆等工作。
该企业还从非洲和欧洲进口环保木材，生产的家具以高价在北上广深等一线城市销售。
2014年后，该企业通过一系列改革措施促进内销。
据此完成下列各题。
1. 采用机器人进行生产对该企业所起作用较小的是A. 提高生产效率B. 节约生产成本C. 保证产品质量D. 降低销售费用2. 该家具企业从非洲和欧洲进口环保木材的主要原因是A. 满足高端消费需求B. 降低生产成本C. 提高出口产品质量D. 开拓国外市场3. 2014年后，该企业促进内销的最重要措施是A. 扩大自动化生产规模B. 改善交通运输条件C. 拓宽二三线城市市场D. 降低进口原料成本【答案】1. D 2. A 3. C【解析】【1题详解】根据材料采用机器人生产能够使用自动化生产线代替工人，节省劳动力投入，能够提高生产效率，节约生产成本，故AB错误；机器人操作精度比人力更高，产品质量更好，故C错误；机器人替代的是生产环节，不影响营销环节的费用，故该题选D。
【2题详解】根据材料“该企业还从非洲和欧洲进口环保木材，生产的家具以高价在北上广深等一线城市销售”说明进口环保木材的生产成本较高，故B错误；其销售市场主要在北上广深等一线城市，故CD错误；实用环保木材的家具，价格较高，市场主要面向高收入群体，满足高端消费需求，故该题选A。
【3题详解】扩大自动化生产规模能够提高产量和提升产品的品质，不属于扩大内销的措施，故A错误；改善交通运输条件，能够提高产品的运输能力，对其扩大内销影响较小，故B错误；降低进口原料成本，能降低生产成本，能够增大北上广深等城市的销量，对其整体内销影响不大，故D错误；我国二三线城市众多，经济增长速度较快，市场需求量大，所以扩大其内销的最主要措施应拓宽二三线城市的市场，故该题选C。
2019-2020学年广州市第三中学高三英语三模试卷及参考答案
2019-2020学年广州市第三中学高三英语三模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AWhen the sun shines brightly, it provides a great chance to get outdoor things done. Like making hay! At least, that is what farmers from the past would say. ―Make hay while the sun shines.This idiom is very old, dating back to Medieval times. Rain would often ruin the process of making hay. So, farmers had no choice but to make hay when the sun was shining.Today, we all use this expression, not just farmers. When conditions are perfect to get something done, we can say, ―It’s a good idea to make hay while the sun shines.In other words, you are taking advantage of a good situation or of good conditions. You are making the most of your opportunities. These all mean ―making hay while the sun shines.And sometimes we use this expression to mean we beat someone to the punch, or we got ahead of someone else. And other times you make hay while the sun shines to make good use of the chance to do something while it lasts. You are being opportunistic – taking advantage of a good opportunity. For example, my friend Ozzy was sick for a week and could not go to work. So, his co-worker Sarah -- who doesn’t like him -- took advantage of his illness and stole his project! Talk about making hay while the sun shines.Sometimes when you make hay while the sun shines you are staying ahead of a problem – like in this example:Hey, do you want to go hiking with me and my friends this weekend? The weather is going to be beautiful! I wish I could. But I have to finish my taxes. It’s the last weekend before they’re due.Oh, that’s too bad.Wait. What about your taxes?My taxes are done. I was off from work a couple of weeks ago and made hay while the sun shined. I got all of it done!I wish I would have taken advantage of my time off last week___1___All I did was lay around thehouse.And that’s all the time we have for these Words and Their Stories. But join us again next week. You can listen while you’re making dinner or riding to work. Yeah, make hay while the sun shines.1.Which of the following best matches ―make hay whilethe sun shines in paragraph 2?A.Sow nothing, reap nothing.B.Sharp tools make good work.C.Strike while the iron is hot.D.One swallow doesn’t make a summer.2.According to the underlined sentence, what feeling does the speaker express?A.AdmirableB.RegretfulC.AnnoyedD.Indifferent3.Where is the passage probably taken from?A.A radio programB.A magazineC.A brochureD.A novelBI had very good parents. My mother came toAmericafromScotlandby herself when she was 11, and she didn’t have much education. My dad was kind of a street kid, and he eventually went into the insurance business, selling nickel policies door to door.One day, my dad asked his boss, “What's the toughest market to sell?” and the insurance guy replied “Well, black people. They don’t buy insurance.” My dad thought, but they have kids; they have families. Why wouldn’t they buy insurance? So he said, “Give meHarlem.”When my dad died in 1994, I talked about him onThe Tonight Show. I told the story of how he worked in Harlem and how he always taught us to be open-minded and not to say or think things of racism (种族主义). Then one day, I got a letter from a woman who was about 75 years old.She wrote that when she was a little girl, a man used to come to her house to collect policies. She said this man was the only white person who had ever come to dinner at their house. The man was very kind to her, she said, and his name was Angelo—was this my father?The letter made me cry. I called her up and said yes, that was in fact my dad, and she told me how kind he had been to her family. Her whole attitude toward white people was based on that one nice man she met in her childhood, who always treated her with kindness and respect and always gave her a piece of candy. From this experience, I learned a valuable life lesson: never judge people and be open-minded and kind to others.4. What did my father do after knowing what was the toughest market to sell?A. He asked his boss to give him some insurance.B. He went toScotlandto improve his education.C. He specially went to white families with kids.D. He choseHarlemto face the toughest challenge.5. What can we learn from the third paragraph?A. It was rare that a businessman had dinner in his customer's house.B. Angelo was the only white person to sell insurance inHarlem.C. The little girl admired Angelo very much.D. Racism was a serious problem inAmericaat that time.6. Which of the following can best describe the author’s father?A. Stubborn and generous.B. Patient and intelligent.C. Determined and open-minded.D. Confident and romantic.7. What can be the best title of the passage?A. Memories from a TV Show.B. A Letter from an Old Lady.C. Life Lessons from My Father.D. My Father's Experience inHarlem.C“One person’s trash is another person’s treasure.” That’s a common expression, but the next time you throw something away, think about a twist on the old saying. What if your trash could become your own treasure? Many creative, thrifty, and environmentally minded people have come up with a way to makethathappen. It’s called upcycling. Our world would be a better place if everyone would begin upcycling.Upcycling is the practice of taking an unwanted item and turning it into something useful. For example, how about that pair of jeans with a hole in one knee? It could become a new pillow for your bedroom.Upcycling is not the same as recycling. Upcycling is actually much better for the environment. Recycling takes an item made of glass, paper, metal, or plastic, breaks it down to its base material, and then uses that material to make another product. This requires a great deal of energy. On the other hand, when you choose to upcycle, the only energy you use is your own. And upcycling not only reduces the amount of trash that goes into our landfills, but it also protects natural resources, such as oil and gas. Recycling is good for the environment, but upcycling is even better.Upcycling also makes a family’s budget stretch further. Of course, the idea of reusing items to save money is not new. During the Great Depression in the 1930s, many families lived on a tight budget. People had to use what they already had in order to meet their needs.As responsible citizens, we should all be concerned with protecting our environment and budgeting our resources. Upcycling is a fun and creative way to help. The next time you go to toss something into the trash can, stop and think about what it could become. Chances are, there’s a brand-new item in your hand just waiting to be upcycled.8. Why does the author mention an old saying in the first paragraph?A. To arise reader’s awareness of upcycling.B. To stress the importance of upcycling.C. To lead in the topic of upcycling.D. To show the idea of upcycling.9. Which one below belongs to upcycling?A. An old ladder is transformed into a bookshelf.B. Old tin cans are transported to landfill.C. A broken wooden door is chopped up.D. Old cloth is made into a paper bed.10. What is the difference between recycling and upcycling?A. Upcycling is much more creative.B. Recycling is much easier to achieve.C. Recycling is much more cost-saving.D. Upcycling is much more energy-efficient.11. What can be inferred from the text?A. Upcycling is popular at present.B. Upcycling is replacing recycling.C. Upcycling is worth recommending.D. Upcycling is a tradition in daily life.DTrees are “social creature” that communicate with each other in cooperative ways that hold lessons for humans, too, ecologist Suzanne Simard says. Simard grew up in Canadian forests as a child of loggers before becoming an ecologist. She's now a professor of forest ecology at theUniversityofBritish Columbia.Trees are linked to neighboring trees by a network of fungi below the surface of the earth that resembles the nervous networks in the brain, she explains. In one study, Simard watched as a Douglas fir tree that had been injured by insects appeared to send chemical warning signals to a pine nearby. The pine tree then produced defense enzymes to protect against the insect.“This was a breakthrough,” Simard says. The trees were sharing “information that actually is important to the health of the whole forest.”In addition to warning each other of danger, Simard says that trees have been known to share nutrients at critical times to keep each other healthy. She says the trees in a forest are often linked to each other via an older tree she calls a “mother” or “hub” tree.“In connecting with all the trees of different ages, the mother trees can actually ease the growth of these young trees,” she says. “The young trees will link into the network of the old trees and benefit from that huge resource capacity. And the old trees would also pass a little bit of carbon and nutrients and water to the young trees, at crucial times in their lives, that actually help them survive.”The study of trees took on a new resonance for Simard when she suffered from breast cancer. During her treatment, she learned that one of the medicines she relied on was actually obtained from what some trees produce for their own mutual defense. She explains her research on cooperation in the forest, and shares her personal story in the new bookFinding the Mother Tree: Discovering the Wisdom of theForest.12. How could a Douglas fir tree send chemical warning signals to a pine nearby?A. By an underground network of fungi.B. By the nervous networks in the brain.C. By making cooperation with each other.D. By holding lessons to it as human beings.13. According to Simard, what was a breakthrough?A. Simard was-brought up in Canadian forests.B. She became a professor of forest ecology.C. The pine tree produced defense enzymes.D. Vital information was shared among trees.14. What helped Simard understand trees further?A. Her rich knowledge of trees.B. Her childhood in the forest.C. Her medicine gained from trees.D. Her research on cooperation.15. In Simard's book we may discover the wisdom of the forest except .A. communicating cooperativelyB. warning each other of dangerC. sharing nutrientsat critical timesD. sacrificing mother trees for survival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。
广州市2019年高三毕业班第三次模拟考试 语文试卷
广州市2019年高三毕业班第三次模拟考试语文本试卷共10页，满分为150分。
考试时间150分钟。
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一、现代文阅读（36分）（一）论述类文本阅读（本题共3小题，9分）阅读下面的文字，完成1-3题日本、印度茶业虽然在制度的构建主体、组织形式等方面呈现出不同，但有着根本的共同之处，即实现全体茶业参与者的利益均衡而非仅仅一部分茶业参与者的既得利益，构建有利于整个茶业发展的有效秩序。
由印、日的成功可见，他们的整个社会能够建立一个完整有效的生产、销售、组织制度体系，这是其成功的关键。
反观中国情形，正如1891年湖北盐茶牙厘局针对华茶为何衰落进行的调查所指出的那样：华茶在生产、收集以及加工过程中，都普遍存在着资金短缺，而资金短缺的部分原因是体制的松散结构。
这种结构不仅导致了产品质量的下降，而且还由于茶叶运抵汉口出售之前要换好几手，层层加码使其价格抬升，其标价就比其竞争对手高的多。
总之，数百年来在国内贸易中运行得很好的、由收集代理人与中间人组成的、精致的网络，一旦面对新的体制外竞争形势，却被证明是笨拙的、无能为力的了。
为什么中国不能构建印度、日本等国有效的茶业制度呢？在近代中国，特别是在晚清和北洋政府时期，政府干预经济的能力相当弱。
政府不能够为市场交易提供必要的秩序，也不可能为市场的运作提供具体的规则，同时由于单个企业力量是有限的，那么市场交易规则的构建主体谁来承担？杜恂诚教授认为：“商会和同业公会责无旁贷地肩负起市场操作层面的创建和完善制度秩序的责任”。
广东省广州市2019届高三第三次模拟考试数学(文)试卷
广东省广州市2019届高三第三次模拟考试数学（文）试题本试题卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共8页，23题（含选考题）。
全卷满分150分。
考试用时120分钟。
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6、保持卡面清洁，不折叠，不破损，不得使用涂改液、胶带纸、修正带等。
7、考试结束后，请将本试题卷、答题卡、草稿纸一并依序排列上交。
第Ⅰ卷（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，，则（）A. B. C. D.【答案】C【解析】则，选C.2.已知复数的实部和虚部相等，则（）A. 2B. 3C.D.【答案】D【解析】,，，选D.3.已知向量满足，则（）A. －12B. －20C.D.【答案】A【解析】试题分析：由题意得，，由，解得得，则，故选A．考点：1、向量的坐标运算；2、平面向量的数量积公式．4.数列是公差不为零的等差数列，并且是等比数列的相邻三项，若，则等于（）A. B. C. D.【答案】B【解析】设等差数列首项为，公差为，，则，，，；设等比数列公比为，，，选B.5.函数的部分图象如图所示，为了得到的图象，只需将函数的图象（）A. 向左平移个单位长度B. 向左平移个单位长度C. 向右平移个单位长度D. 向右平移个单位长度【答案】B【解析】解析，将代入得，故可将函数的图象向左平移个单位长度得到的图象.6.公元263年左右，我国数学家刘徽发现当圆内接正多边形的边数无限增加时，多边形面积可无限逼近圆的面积，并创立了“割圆术”．利用“割圆术”刘徽得到了圆周率精确到小数点后两位的近似值3.14，这就是著名的“徽率”．如图是利用刘徽的“割圆术”思想设计的一个程序框图，则输出n的值为(参考数据：)A. 12B. 24C. 36D. 48【答案】B【解析】试题分析：第一次循环：；第二次循环：；第三次循环：，满足条件，跳出循环，输出.故选B.考点：1.数学文化；2.程序框图.【名师点睛】本题考查数学文化与程序框图，属中档题；数学文化是高考新增内容，程序框图是第年高考的必考内容，掌握循环程序的运行方法，框图以赋值框和条件框为主，按照框图箭线方向和每个框的指令要求运行，注意条件框的要求是否满足，运行程序时要准确. 7.若三个正实数满足，且不等式有解，则实数的取值范围是（）A. B. C. D.【答案】C【解析】,则，不等式有解，则，解得或,选C.8.在上随机地取两个实数，则事件“直线与圆相交”发生的概率为（）A. B. C. D.【答案】A【解析】直线与圆相交,只需圆心到直线的距离小于半径，即，，，由于，在直角坐标系下画出满足条件的正方形区域，其面积为16，满足的面积为，根据几何概型求概率公式得：，选A.9.某三棱锥的三视图如图所示，其侧（左）视图为直角三角形，则该三棱锥最长的棱长等于（）A. B. C. D.【答案】C【解析】根据三视图恢复为原几何体，原几何体为三棱锥，其中底面，底面为直角三角形，，其中，，计算，最长棱为.10.已知抛物线的焦点到双曲线渐近线的距离为，点是抛物线上的一动点，到双曲线的上焦点的距离与到直线的距离之和的最小值为3，则该双曲线的方程为（）A. B. C. D.【答案】B【解析】抛物线的焦点,双曲线的渐近线方程为，任取一条渐近线，焦点到渐近线的距离，为抛物线的准线，到准线的距离等于到焦点的距离，到双曲线的上焦点的距离与到直线的距离之和的最小值为，则，选B.11.在等腰直角中，在边上且满足：，若，则的值为（）A. B. C. D.【答案】A【解析】根据题意，在线段上，过作，垂足为，作，垂足为，若设，由于，得，根据题意；，即，，故选A.12.定义在R上的奇函数，当时，，则关于的函数的所有零点之和为（）A. B. C. D.【答案】B【解析】由已知可得时，，当时，，令，，此时无解；当时，由，得；当，由，得；根据函数为奇函数，所以当时，，同样当时，的解依次为，，；所以，选B.第Ⅱ卷（共90分）二、填空题（每题5分，满分20分，将答案填在答题纸上）13.抛物线的焦点坐标为__________．【答案】【解析】抛物线方程化为，，抛物线的焦点坐标为.14.设函数则__________．【答案】-1【解析】,.15.现有一半球形原料，若通过切削将该原料加工成一正方体工件，则所得工件体积与原料体积之比的最大值为__________．【答案】【解析】【分析】设球半径为R，正方体边长为a，由题意得当正方体体积最大时：R2，由此能求出所得工件体积与原料体积之比的最大值．【详解】设球半径为R，正方体边长为a，由题意得当正方体体积最大时：R2，∴R，∴所得工件体积与原料体积之比的最大值为：．故答案为【点睛】本题考查空间几何体的体积，，是中档题，关键是确定正方体体积最大时体积之比为最大值.16.已知函数的图像为曲线，若曲线存在与直线垂直的切线，则实数的取值范围为．【答案】【解析】试题分析：，；若存在与垂直的切线，则有解；即有解，．考点：1．导数的几何意义；2．两直线的位置关系．三、解答题（本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.）17.已知函数，当时，的最小值为.（1）求的值；（2）在中，已知，延长至，使，且，求的面积.【答案】（1）;（2）.【解析】试题分析：利用两角和公式、降幂公式及辅助角公式把函数解析式化为标准形式，根据的范围求出的范围，根据的最小值为，求出的值；利用，求出角，在根据正弦定理、余弦定理及面积公式解题.试题解析：（Ⅰ），当时，，∴∴由（Ⅰ）知，又,∴，又∴，故∴在中，由余弦定理可得：解得:∴在中，又∴，18.某大学高等数学这学期分别用两种不同的数学方式试验甲、乙两个大一新班（人数均为人，入学数学平均分和优秀率都相同；勤奋程度和自觉性都一样）.现随机抽取甲、乙两班各名的高等数学期末考试成绩，得到茎叶图：（1）学校规定：成绩不得低于85分的为优秀，请填写下面的列联表，并判断“能否在犯错误率的概率不超过0.025的前提下认为成绩优异与教学方式有关？”下面临界值表仅供参考：（参考方式：，其中）（2）现从甲班高等数学成绩不得低于80分的同学中随机抽取两名同学，求成绩为86分的同学至少有一个被抽中的概率.【答案】（1）见解析；（2）.【解析】试题分析：根据茎叶图所提供的数据，填写列联表，根据独立性检验方法先计算随机变量观测值，计算要准确，保留3位小数，根据临界值表发现，因此在犯错误的概率不超过0.025的前提下，可以认为成绩优秀与数学方式有关；甲班不低于80分有6人，随机抽取两人，用列举法列出15种情况，至少有1名86分的情况有9种，求出概率值.试题解析：（1），因此在犯错误的概率不超过0.025的前提下，可以认为成绩优秀与数学方式有关.（2）甲班不低于80分有6人，随机抽取两人，用列举法列出15种情况，至少有1名86分的情况有9种，19.如图，正方形的边长等于2，平面平面.（1）求证：平面；（2）求三棱锥的体积.【答案】（1）见解析；（2）.【解析】试题分析：根据线面平行的判定定理证明线面平行，连接交于，取中点，连接，借助中位线定理和平行四边形的判定与性质证明线线平行，进而得证；根据线面平行，可以转化三棱锥的体积，又，利用平面，求出体积.试题解析：（1）证明：连接，记，取的中点，连接∵点分别是的中点，∴，又∴∴四边形为平行四边形∴，即，又面∴面（2）在面内，过点作，交于点，由已知条件可得，在梯形中，∴，即，从而∴∴∵面面，面面∴面∵面∴点到平面的距离等于点到平面的距离∴.【点睛】求三棱锥的体积要灵活运用转化思想，一是灵活选用顶点，方便利用体高的数值，方便求底面面积；二是灵活使用平行转化、对称转化、比例转化，使所求的四棱锥的体积的底面积与高计算简单准确.20.已知椭圆的一个焦点为，左右顶点分别为，经过点的直线与椭圆交于两点.（1）求椭圆方程；（2）记与的面积分别为和，求的最大值.【答案】（1）；（2）．【解析】试题分析：（1）根据条件建立参数，，所满足的方程，解方程组即可求解；（2）建立的函数表达式，求函数最值即可求解.试题解析：（1）∵点为椭圆的一个焦点，∴，又∵，∴，∴椭圆方程为；（2）当直线斜率不存在时，直线方程为，此时，，与的面积相等，，当直线斜率存在时，设直线方程为（），设，显然，异号，由得，显然，方程有实根，且，，此时，由可得，当且仅当时等号成立，∴的最大值为.考点：1.椭圆的标准方程；2.椭圆中的最值问题.【方法点睛】求解范围问题的常见求法：（1）利用判别式来构造不等关系，从而确定参数的取值范围；（2）利用已知参数的范围，求新参数的范围，解这类问题的核心是在两个参数之间建立等量关系；（3）利用隐含或已知的不等关系建立不等式，从而求出参数的取值范围；（4）利用基本不等式求出参数的取值范围；（5）利用函数的值域的求法，确定参数的取值范围．21.设函数，的图象在点处的切线与直线平行．（1）求的值；（2）若函数，且在区间上是单调函数，求实数的取值范围．【答案】（1）；（2）.【解析】试题分析：（1）由题意知，曲线y=f（x）的图象在点（1，f（1））处的切线斜率为3，求导数，代入计算，即可得出结论；（2）求导数，分类讨论，即可求实数a的取值范围．试题解析：（1）由题意知，曲线的图象在点处的切线斜率为3，所以，又，即，所以．（2）由（1）知，所以，①若在区间（0，+∞）上为单调递减函数，则在（0，+∞）上恒成立，即，所以．令，则，由，得，由，得，故在（0，1]上是减函数，在[1，+∞）上是增函数，则，无最大值，在（0，+∞）上不恒成立，故在（0，+∞）不可能是单调减函数②若在（0，+∞）上为单调递增函数，则在（0，+∞）上恒成立，即，所以，由前面推理知，的最小值为，∴，故a的取值范围是.点睛：已知函数单调性求参即可转化为导数恒大于等于或恒小于等于0问题，即为恒成立问题.（1）根据参变分离，转化为不含参数的函数的最值问题；（2）若就可讨论参数不同取值下的函数的单调性和极值以及最值，最终转化为，若恒成立；（3）若恒成立，可转化为请考生在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平面直角坐标系中，直线的参数方程为（其中为参数）.现以坐标原点为极点，轴的正半轴为极轴建立极坐标系，曲线的极坐标方程为. （1）写出直线普通方程和曲线的直角坐标方程；（2）过点，且与直线平行的直线交于两点，求.【答案】（1）；（2）.【解析】试题分析：利用两式相减削去参数，把直线的参数方程化为普通方程，再利用公式和把极坐标方程化为直角坐标方程，涉及弦长问题常用直线的参数方程解决，写出过点与直线平行的直线的参数方程，把直线的参数方程化为代入到圆的方程，利用直线的参数方程的几何意义，把表示为，再利用求出 .试题解析：（1）由，消去参数，得直线的普通方程为.又由得，由得曲线的直角坐标方程为.（2）过点且与直线平行的直线的参数方程为将其代入得，则，知，所以23.选修4-5：不等式选讲已知函数（其中）.（1）若时，求不等式的解集；（2）若不等式对任意实数恒成立，求的取值范围.【答案】（1）；（2）.【解析】试题分析：利用零点分区间讨论法解绝对值不等式，得出解集；不等式对任意实数恒成立，根据绝对值不等式的性质得出，只需满足解不等式求出的取值范围.（1）当时即.①当时，得，解得；②当时，得，不成立，此时；③当时，得，解得.综上，不等式的解集为（2）因为，由题意，即或，解得或，即的取值范围是。
广东省广州市广东实验中学2019-2020学年高三第三次阶段考试理科数学试题(解析版)
广东实验中学2020届高三级第三次阶段考试数 学（理科）注意事项：1．答卷前，考生务必用黑色字迹的钢笔或签字笔将自己的姓名、考号填写在答题卡上. 2．选择题每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其它答案；不能答在试卷上.3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在另发的答题卷各题目指定区 域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液，不按以上要求作答的答案无效.4．考生必须保持答题卡的整洁，考试结束后，将答题卷和答题卡一并收回.第Ⅰ卷（共60分）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.集合{}260A x x x =--<，集合{}2|log 1B x x =<，则AB =（ ）A. ()2,3-B. (),3-∞C. ()2,2-D. ()0,2【答案】A 【解析】 【分析】先由二次不等式的解法得{}|23A x x =-<<，由对数不等式的解法得{}|02B x x =<<，再结合集合并集的运算即可得解.【详解】解不等式260x x --<,解得23x -<<,则{}|23A x x =-<<,解不等式2log 1x <,解得02x <<,即{}|02B x x =<<, 即AB =()2,3-,故选:A.【点睛】本题考查了二次不等式的解法及对数不等式的解法，重点考查了集合并集的运算，属基础题.2.己知i 是虚数单位，复数z 满足1zi z=-，则z 的模是( ) A. 1B.12C.2【答案】C 【解析】 【分析】利用复数的运算法则和模的计算公式即可得出． 【详解】1zz=-i ， ∴z ＝i -zi ， ∴z 1(1)11222i i i i i ===++-， ∴|z|2==， 故选：C ．【点睛】本题考查了复数的运算法则和模的计算公式，属于基础题．3.若2,a ln =125b -=，201cos 2c xdx π=⎰，则,,a b c 的大小关系（ ）A. a b c <<B. b a c <<C. c b a <<D. b c a <<【答案】D 【解析】 【分析】利用对数函数的性质，以及微积分定理与12比较即可.【详解】12,2a ln =>=121,25b -=<== ()02111cos sin 22220c xdx x ππ=⎰=⨯=，故选：D【点睛】本题考查实数大小的比较，考查对数函数的性质，微积分定理，考查利用中间量比较大小，属于常考题型.4.若2sin cos 12x x π⎛⎫-+= ⎪⎝⎭，则cos2x =（ ）A. 89-B. 79-C.79D. -1【答案】C 【解析】 【分析】利用诱导公式化简得到sin x ，再结合二倍角的余弦公式即可求解. 【详解】2sin sin 1x x +=，即1sin 3x =所以22cos 212sin 1799x x =-=-= 故选C【点睛】本题主要考查了三角函数的化简和求值，属于基础题.5.(,2)m ∈-∞-是方程222156x y m m m +=---表示的图形为双曲线的（ ） A. 充分不必要条件 B. 必要不充分条件 C. 充要条 D. 既不充分也不必要条件【答案】A 【解析】 分析】方程表示双曲线，可得()()()5320m m m --+<，解得m 范围即可判断出结论，解得m 范围即可判断出结论．【详解】由方程222156x y m m m +=---表示的图形为双曲线， 可得()()2560m m m ---<，即()()()5320m m m --+<即2m <-，或35m <<， ∴ (,2)m ∈-∞-是方程222156x y m m m +=---表示的图形为双曲线的充分不必要条件，故选：A【点睛】本题考查了双曲线的标准方程、不等式的解法、简易逻辑的判定方法，考查了推理能力与计算能力，属于中档题．6.点P 是ABC △所在平面上一点，若2355AP AB AC =+，则ABP △与ACP △的面积之比是（ ） 【A.35B.52C.32D.23【答案】C 【解析】 【分析】由向量的线性运算可得32=BP PC ，即点P 在线段AB 上，且32=BP PC ,由三角形面积公式可得:ABP S ∆APC S ∆:3:2BP PC ==，得解.【详解】解：因为点P 是ABC △所在平面上一点，又2355AP AB AC =+， 所以2233-=-5555AP AB AC AP ,即23=55BP PC ,即32=BP PC ， 则点P 在线段BC 上，且32=BP PC , 又1sin 2APC S AP PC APC ∆=∠，1sin 2ABP S AP BP APB ∆=∠, 又APB APC π∠+∠=，即sin sin APC APB ∠=∠， 所以点P 在线段BC 上，且32=BP PC , :ABP S ∆APC S ∆1sin :2AP BP APB =∠1sin 2AP PC APC ∠:3:2BP PC ==， 故选C.【点睛】本题考查了向量的线性运算及三角形的面积公式，重点考查了运算能力，属中档题.7.已知()121sin 221x xf x x x -⎛⎫=-⋅ ⎪+⎝⎭，则函数()y f x =的图象大致为（）A. B.C. D.【答案】D 【解析】 【分析】由函数解析式可得()()f x f x =-，则函数()y f x =为偶函数，其图像关于y 轴对称，再取特殊变量4π得04f π⎛⎫< ⎪⎝⎭，即可得在()0,∞+存在变量使得()0f x <，再观察图像即可. 【详解】解：因为()121sin 221xxf x x x -⎛⎫=-⋅ ⎪+⎝⎭， 则()121sin 221x x f x x x ---⎛⎫-=-+⋅ ⎪+⎝⎭=121sin 221xx x x -⎛⎫-⋅ ⎪+⎝⎭，即()()f x f x =-，则函数()y f x =为偶函数，其图像关于y 轴对称，不妨取4x π=，则 ()4421(08221f x πππ-=-<+，即在()0,∞+存在变量使得()0f x <， 故选D.【点睛】本题考查了函数奇偶性的判断及函数的图像，重点考查了函数的思想，属中档题.8.某班上午有五节课，分別安排语文，数学，英语，物理，化学各一节课．要求语文与化学相邻，数学与物理不相邻，且数学课不排第一节，则不同排课法的种数是 A. 24 B. 16 C. 8 D. 12【答案】B 【解析】 【分析】根据题意，可分三步进行分析：（1）要求语文与化学相邻，将语文与化学看成一个整体，考虑其顺序；（2）将这个整体与英语全排列，排好后，有3个空位；（3）数学课不排第一行，有2个空位可选，在剩下的2个空位中任选1个，得数学、物理的安排方法，最后利用分步计数原理，即可求解． 【详解】根据题意，可分三步进行分析：（1）要求语文与化学相邻，将语文与化学看成一个整体，考虑其顺序，有222A =种情况；（2）将这个整体与英语全排列，有222A =中顺序，排好后，有3个空位；（3）数学课不排第一行，有2个空位可选，在剩下的2个空位中任选1个， 安排物理，有2中情况，则数学、物理的安排方法有224⨯=种， 所以不同的排课方法的种数是22416⨯⨯=种，故选B ．【点睛】本题主要考查了排列、组合的综合应用，其中解答红注意特殊问题和相邻问题与不能相邻问题的处理方法是解答的关键，着重考查了分析问题和解答问题的能力，属于中档试题．9.已知函数22()2sin cos ()sin (0)24x f x x x ωπωωω=-->在区间25[,]36ππ-上是增函数，且在区间[0,]π上恰好取得一次最大值，则ω的范围是（ ）A. 3(0,]5B. 13[,]25C. 13[,]24D. 15[,)22【答案】B【解析】 【分析】先化简()f x ，再根据正弦函数性质列方程与不等式，解得结果.【详解】222()2sin cos ()sin sin (1cos())sin 422x f x x x x x x ωππωωωωω=--=+-- 2sin (1sin )sin sin x x x x ωωωω=+-=因为()f x 在区间25[,]36ππ-上是增函数，且在区间[0,]π上恰好取得一次最大值， 所以255,,236222ππωπωπππωπ-≤-≤≤<，即13[,]25ω∈ 故选B点睛】本题考查二倍角余弦公式、辅助角公式以及正弦函数性质，考查综合分析与求解能力，属中档题.10.设变量y 满足约束条件342y xx y x ≥⎧⎪+≤⎨⎪≥-⎩则z ＝|x －3y |的最大值为( )A. 8B. 4C. 2D.5【答案】A 【解析】由题意作出满足条件的可行域如图中阴影部分，【则对于目标函数z=|x ﹣3y |，平移直线y=13x 可知， 当直线经过点A （﹣2，2）时，z=|x ﹣3y |取得最大值， 代值计算可得z max =|﹣2﹣3×2|=8． 故选A ．11.AOB 中，OA a OB b ==，，满足||2a b a b ⋅=-=，则AOB ∆的面积的最大值为（ ）A. B. 2C.D. 【答案】A 【解析】 【分析】利用数量积公式以及平方关系计算得到sin AOB ∠，利用模长公式以及基本不等式得到||||4a b ≤，结合三角形面积公式化简即可求解.【详解】||||cos 2a b a b AOB ⋅=∠=,即2cos ||||AOB a b ∠=2(||||)4sin |||||||a b AOB a b a b -∴∠==⎪⎭22||||2||2a b a a b b -=-⋅+= ，即228||||2||||a b a b =+≥所以||||4a b ≤ 所以22(||||)41111||||sin ||||=(||||)4164=3222|||AOBa b S a b AOB a b a b a b ∆-=∠=-≤-故选A【点睛】本题主要考查了平面向量的数量积公式以及模长公式的应用，属于中档题.12.椭圆22221(0)x y a b a b +=>>上有一点P ，1F ，2F 分别为椭圆的左、右焦点，椭圆内一点Q 在线段2PF 的延长线上，且1,QF QP ⊥15sin 13F PQ ∠=，则该椭圆离心率的取值范围是（ ）A. ⎫⎪⎪⎝⎭B. 15⎛ ⎝⎭C. 1,52⎛⎫ ⎪ ⎪⎝⎭D. 2⎝⎭【答案】D 【解析】 【分析】要求出离心率的取值范围，得列出不等关系，解出e 的取值范围；首先满足QF 1⊥QP ，点Q 在椭圆的内部，故点Q 轨迹在以F 1F 2为直径，原点为圆心的圆上，且圆在椭圆的内部，圆半径c ＜椭圆短半轴b ，由a 2﹣c 2＝b 2，可解得e 的一个范围；其次由sin ∠F 1QF 2，可求得cos ∠F 1QF 2．在△PF 1F 2中，而|F 1F 2|＝2c ，|PF 1|+|PF 2|＝2a 是定值，由基本不等式可得PF 1|•|PF 2|122()2PF PF +≤；由余弦定理得4c2＝|PF 1|2+|PF 2|2﹣2|PF 1||PF 2|cos ∠F 1QF 2，结合不等关系即可解出e 的取值范围．【详解】解：∵QF 1⊥QP ，∴点Q 在以F 1F 2为直径，原点为圆心的圆上，∵点Q 在椭圆的内部，∴以F 1F 2为直径的圆在椭圆内，∴c ＜b ；∴c 2＜a 2﹣c 2，∴212e ＜，故0＜e 2∵sin ∠F 1PQ 513=，∴cos ∠F 1PQ 1213=； 设|PF 1|＝m ，则|PF 2|＝n ，而|F 1F 2|＝2c ，|PF 1|+|PF 2|＝m +n ＝2a ， 在△PF 1F 2中，由余弦定理得 4c 22212213m n mn =+-⋅． ∴4c 2＝（m +n ）2﹣2mn ﹣2mn •1213； 即4c 2＝4a 25013-mn ；∴mn ()222625a c =-； 由基本不等式得：mn 2()2m n +≤=a 2， 当且仅当m ＝n 时取等号；由题意知：QF 1⊥QP ，∴m ≠n ，∴mn 2()2m n +=＜a 2， ∴()222625a c -＜a 2∴a 2＜26c 2；故2126e ＞，∴e综上可得：26e 2． 故选：D ．【点睛】本题考查了椭圆的性质、圆的性质，余弦定理、基本不等式的应用，考查了推理能力与计算能力，属于难题．第Ⅱ卷（共90分）二、填空题（每题5分，满分20分，将答案填在答题纸上）13.设函数()3ln 2f x x x x =+，则曲线()y f x =在点()1,2处的切线方程是___________．【答案】750x y --= 【解析】 【分析】先求函数()f x 的导函数()'fx ，再由导数的几何意义，求()'17f =，则曲线()y f x =在点()1,2处的切线的斜率为7，再由直线的点斜式方程求解即可.【详解】解：因为()3ln 2f x x x x =+，所以()'2ln 16fx x x =++，则()'21ln11617f =++⨯=，即曲线()y f x =在点()1,2处的切线方程是27(1)y x -=-，即750x y --=， 故答案为750x y --=.【点睛】本题考查了导数的几何意义、直线的点斜式方程，重点考查了导数的应用及运算能力，属基础题.14.()422x x --的展开式中，3x 的系数为 ． （用数字填写答案） 【答案】40- 【解析】试题分析：()422x x --()422x x ⎡⎤=-+⎣⎦展开后只有()42x +与()33242C x x -+中含3x 项其系数和为133124432240C C C ⨯-⨯⨯=-，故答案为40-.考点：二项展开式定理.15.己知函数sin ()xx af x e-=有极值，则实数a 的取值范围为_____________【答案】( 【解析】 【分析】求出函数的导函数，则cos sin ()xx x af x e -+'=有可变零点，求三角函数的值域得到结果.【详解】由sin ()x x a f x e -=可得：cos sin ()xx x af x e -+'=，∵函数sin ()xx af x e-=有极值， ∴cos sin ()xx x af x e-+'=有可变零点，∴cos sin 0x x a -+=，即sin cos 4a x x x π⎛⎫=-=-⎪⎝⎭，∴(a ∈故答案为：( 【点睛】本题考查函数存在极值条件，考查三角函数的值域问题，考查转化思想，属于中档题.16.点D 是直角ABC ∆斜边AB 上一动点，5,AC =4,BC =将直角ABC ∆沿着CD 翻折，使B DC '∆与ADC ∆构成直二面角，则翻折后AB '的最小值是_______．【解析】 【分析】过点B ′作B ′E ⊥CD 于E ，连结BE ，AE ，设∠BCD ＝∠B ′CD ＝α，则有B ′E ＝4sin α，CE ＝4cos α，2ACE πα∠=-，由此利用余弦定理、勾股定理能求出当4πα=时，AB．【详解】解：过点B ′作B ′E ⊥CD 于E ，连结BE ，AE ， 设∠BCD ＝∠B ′CD ＝α，则有B ′E ＝4sin α，CE ＝4cos α，2ACE πα∠=-，在△AEC 中，由余弦定理得：222516402AE cos cos cos πααα⎛⎫=+-- ⎪⎝⎭＝25+16cos 2α﹣40sin αcos α， 在Rt △AEB ′中，由勾股定理得：AB '2＝AE 2+B ′E 2＝25+16cos 2α﹣40sin αcos α+16sin 2α＝41﹣20sin2α， ∴当4πα=时，AB的【点睛】本题考查线段长的最小值的求法，考查余弦定理、勾股定理、直二面角等基础知识，运算求解能力，考查函数与方程思想，是中档题．三、解答题：共70分．解答应写出文字说明、证明过程和演算步骤．第17～21题为必考题，每个试题考生都必须做答．第22、23题为选考题，考生根据要求做答． （一）必考题：共60分．17.设等差数列{}n a 的前n 项和为n S ，公比是正数的等比数列{}n b 的前n 项和为n T ，已知．1122331,3,8,15a b a b T S ==+=-= (Ⅰ)求{}{},n n a b 的通项公式； (Ⅱ)若数列{}n c 满足11211222n n n n a c a c a c n +--+++=--对任意*n N ∈都成立；求证：数列{}n c 是等比数列．【答案】（1）1,32n n n a n b -==⋅；（2）证明见解析．【解析】(Ⅰ)设数列{}n a 的公差为d ，数列{}n b 的公比为(0)q q >2375d q q q d +=+-=由题意得……………………………………………………………2分2375d q q q d +=+-=解得………………………………………………………5分(Ⅱ)由知两式相减：………………………………8分…………………………………………………………………10分当时，，适合上式即是等比数列…………………………18.如图，在梯形ABCD中，AB∥CD，AD＝DC＝CB＝1，∠BCD＝120°，四边形BFED为矩形，平面BFED⊥平面ABCD，BF＝1.(1)求证：AD⊥平面BFED；(2)点P在线段EF上运动，设平面PAB与平面ADE所成锐二面角为θ，试求θ的最小值．【答案】（1）证明见解析（2）θ最小值为60°【解析】【分析】（1）在梯形ABCD 中，利用勾股定理，得到AD ⊥BD ，再结合面面垂直的判定，证得DE ⊥平面ABCD ，即可证得AD ⊥平面BFED ；（2）以D 为原点，直线DA ，DB ，DE 分别为x 轴，y 轴，z 轴建立如图所示的空间直角坐标系，求得平面PAB 与平面ADE 法向量，利用向量的夹角公式，即可求解． 【详解】（1）证明：在梯形ABCD 中，∵AB ∥CD ，AD ＝DC ＝CB ＝1，∠BCD ＝120°，∴AB ＝2. ∴BD 2＝AB 2＋AD 2－2AB ·AD ·cos 60°＝3. ∴AB 2＝AD 2＋BD 2，∴AD ⊥BD .∵平面BFED ⊥平面ABCD ，平面BFED ∩平面ABCD ＝BD ， DE ⊂平面BFED ，DE ⊥DB ，∴DE ⊥平面ABCD ， ∴DE ⊥AD ，又DE ∩BD ＝D ，∴AD ⊥平面BFED .（1）由(1)知，直线AD ，BD ，ED 两两垂直，故以D 为原点，直线DA ，DB ，DE 分别为x 轴，y 轴，z 轴建立如图所示的空间直角坐标系，令EP ＝λ(0≤λ，则D (0,0,0)，A (1,0,0)，B (00)，P (0，λ，1)， 所以AB ＝(－10)，BP ＝(0，λ1)． 设n 1＝(x ，y ，z )为平面PAB 的法向量，由1100n AB n BP ⎧⋅=⎪⎨⋅=⎪⎩得00x y z λ⎧-=⎪⎨+=⎪⎩，取y ＝1，则n 1＝1λ)．因为n 2＝(0,1,0)是平面ADE 的一个法向量，所以cos θ＝1212n n n n ⋅＝214+. 因为0≤λλ时，cos θ有最大值12，所以θ的最小值为60°.【点睛】本题考查了线面垂直关系的判定与证明，以及空间角的求解问题，意在考查学生的空间想象能力和逻辑推理能力，解答中熟记线面位置关系的判定定理和性质定理，通过严密推理是线面位置关系判定的关键，同时对于立体几何中角的计算问题，往往可以利用空间向量法，通过求解平面的法向量，利用向量的夹角公式求解.19.已知椭圆C 的中心在坐标原点，焦点在x 轴上，左顶点为A ，左焦点为()12,0F -，点(B 在椭圆C 上，直线()0y kx k =≠与椭圆C 交于E ，F 两点，直线AE ，AF 分别与y 轴交于点M ，N . （1）求椭圆C 的方程；（2）以MN 为直径的圆是否经过定点？若是，求出定点的坐标；若不经过，请说明理由．【答案】（Ⅰ）22184x y +=；（Ⅱ）经过两定点()12,0P ，()22,0P -. 【解析】试题分析：（Ⅰ）椭圆的左焦点为()120F -，，所以224a b -=．由点(2,B 在椭圆C 上，得22421a b +=，进而解出,a b 得到椭圆C 的方程；（Ⅱ）直线(0)y kx k =≠与椭圆22184x y +=联立，解得,E F 的坐标（用k 表示），设出AE ，AF 的方程，解出,M N 的坐标，圆方程用k 表示，最后可求得MN 为直径的圆经过两定点.试题解析：（Ⅰ） 设椭圆C 的方程为22221(0)x y a b a b+=>>，因为椭圆的左焦点为()120F -，，所以224a b -=．因为点(2,B 在椭圆C 上，所以22421a b+=． 由①②解得，a =2b =．所以椭圆C 的方程为22184x y +=．（Ⅱ）因为椭圆C 的左顶点为A ，则点A的坐标为()-．因为直线(0)y kx k =≠与椭圆22184x y +=交于两点E ，F ， 设点()00,E x y （不妨设00x >），则点()00,F x y --．联立方程组22,{184y kx x y =+=消去y 得22812x k =+．所以0x =，则0y =．所以直线AE的方程为y x =+．因为直线AE ，AF 分别与y 轴交于点M ，N ，令0x =得y =，即点M ⎛⎫⎝．同理可得点N ⎛⎫ ⎝．所以MN ==设MN 的中点为P ，则点P的坐标为0,P ⎛ ⎝⎭．则以MN为直径的圆的方程为22x y k ⎛⎫++= ⎪ ⎪⎝⎭2， 即224x y y k++=． 令0y =，得24x =，即2x =或2x =-．故以MN 为直径的圆经过两定点()12,0P ，()22,0P -． 考点：1、 待定系数法求椭圆；2、圆的方程及几何意义.20.某汽车公司最近研发了一款新能源汽车，并在出厂前对100辆汽车进行了单次最大续航里程的测试．现对测试数据进行分析，得到如图所示的频率分布直方图：（1）估计这100辆汽车的单次最大续航里程的平均值（同一组中的数据用该组区间的中点值代表）. （2）根据大量的汽车测试数据，可以认为这款汽车的单次最大续航里程X 近似地服从正态分布())2,N μσ，经计算第（1）问中样本标准差s 的近似值为50．用样本平均数x 作为μ的近似值，用样本标准差s 作为σ的估计值，现任取一辆汽车，求它的单次最大续航里程恰在250千米到400千米之间的概率.参考数据：若随机变量服从正态分布()2N μσ，，则()0.6827P μσξμσ-<+≈…，(33)0.9973P μσξμσ-<+≈…，(22)0.9545P μσξμσ-<+≈….（3）某汽车销售公司为推广此款新能源汽车，现面向意向客户推出“玩游戏，送大奖”活动，客户可根据抛掷硬币的结果，操控微型遥控车在方格图上行进，若遥控车最终停在“胜利大本营”，则可获得购车优惠券3万元．已知硬币出现正、反面的概率都是0.5方格图上标有第0格、第1格、第2格、…、第20格．遥控车开始在第0格，客户每掷一次硬币，遥控车向前移动一次．若掷出正面，遥控车向前移动一格（从k 到1k +）若掷出反面遥控车向前移动两格（从k 到2k +），直到遥控车移到第19格胜利大本营）或第20格（失败大本营）时，游戏结束．设遥控车移到第1(1)9n n 剟格的概率为P 试证明{}1n n P P --是等比数列，并求参与游戏一次的顾客获得优惠券金额的期望值．【答案】（1）300；（2）0.8186；（3）证明见解析，期望值为201212⎡⎤⎛⎫-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦，约2万元.【解析】 【分析】 0000(1)利用每组中点值乘以其频率,再求和即可得到平均值； (2)由(1)可知300μ=，利用0.95450.6827(250400)0.95452P X -<≤≈-求解即可；(3)根据题意可知：得出移到第n 格两种方式①遥控车先到第2n -格，又掷出反面；②遥控车先到第1n -格，又掷出正面，由此得到211122n n n P P P --=+，利用定义证明其为等比数列，结合累加法得出n P 的表达式，由此得到19P ，20P ，根据题意得出参与游戏一次的顾客获得优惠券金额为X 万元，3X =或0，分别求出3X =或0的概率，然后求出期望即可.【详解】（1）0.002502050.004502550.00950305x =⨯⨯+⨯⨯+⨯⨯0.004503550.00150405300+⨯⨯+⨯⨯=（千米）（2）因为X 服从正态分布2(300,50)N所以0.95450.6827(250400)0.95450.81862P X -<≤≈-=（3）遥控车开始在第0格为必然事件，01P =，第一次掷硬币出现正面，遥控车移到第一格，其概率为12,即112P =．遥控车移到第n （219n 剟）格的情况是下列两种，而且也只有两种． ①遥控车先到第2n -格，又掷出反面，其概率为212n P - ②遥控车先到第1n -格，又掷出正面，其概率为112n P - 所以211122n n n P P P --=+，1121()2n n n n P P P P ---∴-=--∴当119n 剟时，数列1{}n n P P --是公比为12-的等比数列 2312132111111,(),(),()2222nn n P P P P P P P -∴-=--=--=-⋅⋅⋅-=- 以上各式相加，得2311111()()()()2222nn P -=-+-+-+⋅⋅⋅+-=11()1()32n ⎡⎤---⎢⎥⎣⎦1211()32n n P +⎡⎤∴=--⎢⎥⎣⎦（0,1,2,,19n =⋅⋅⋅）， ∴获胜的概率2019211()32P ⎡⎤=--⎢⎥⎣⎦失败的概率1920181111232P P ⎡⎤==+⎢⎥⎣⎦（） ∴设参与游戏一次的顾客获得优惠券金额为X 万元，3X =或0∴X 的期望2019202111131()01()21()32322EX ⎡⎤⎡⎤⎡⎤=⋅-+⋅+=-⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦∴参与游戏一次的顾客获得优惠券金额的期望值为201212⎡⎤⎛⎫-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦，约2万元.【点睛】本题主要考查了频率分布直方图求平均数、正态分布求概率，等比数列的证明以及数学期望的求法，题目较为综合，考查面较广，属于难题. 21.已知函数sin ()()cos sin xf xg x x x x x==⋅-，.（1）判断函数()g x 在区间(0)3π，上零点的个数； （2）函数()f x 在区间(0)+∞，上的极值点从小到大分别为1234nx x x x x ，，，，，证明：（Ⅰ）()()120f x f x +<；（Ⅱ）对一切()()()()*1230n n N f x f x f x f x ∈++++<，成立.【答案】（1）两个零点；（2）（I ）见解析；（Ⅱ）见解析 【解析】 【分析】(1)对()g x 求导，利用导数得出函数()g x 的单调性，结合零点存在性定理即可得出零点的个数； (2) （Ⅰ）对函数()f x 求导，由(1)得出12x x ，的范围，进而得到21x x π>+，利用诱导公式即可得出()()120f x f x +<; （Ⅱ）由（Ⅰ）得出22n ππ+>221n n x x π->+>2n π，结合cos y x =的单调性确定221221()()cos cos 0n n n n f x f x x x --+=+<，且221()0,()0n n f x f x -><，对n 为偶数和奇数进行分类讨论，即可得出对一切()()()()*1230n n N f x f x f x f x ∈++++<，成立.【详解】（1）()cos sin cos sin g x x x x x x x '=--=- 当(]0x π∈，时，sin 0()0x g x '>∴<，()g x 在0π（，）上单调递减，()(0)0g x g <=，()g x ∴在(]0π，上无零点 当(],2x ππ∈时，sin 0()0x g x '<∴>，()g x 2ππ（，）上单调递增，()0,(2)20,g g ππππ=-<=>()g x ∴在(]2ππ，上有唯一零点当(]2,3x ππ∈时，sin 0()0x g x '>∴<，()2,3)g x ππ在（上单调递减(2)0,(3)0g g ππ><，(]()2,3g x ππ∴在上有唯一零点综上，函数()g x 在区间()03π，上有两个零点． （2）cos sin ()2x x x f x x-'=（I ）由（1）知()f x 在(]0x π∈，无极值点；在(],2x ππ∈有极小值点，即为1x ； 在(]2,3x ππ∈有极大值点，即为2x ，同理可得，在(]3,4ππ有极小值点3x ， 在(],(1)n n ππ+有极值点n x .由cos sin 0n n n x x x -=得tan n n x x =21211tan tan tan(),x x x x x π>∴>=+35()0,()10,(2)0,()022g g g g ππππ<=-<>< 1235(,),(2,)22x x ππππ∴∈∈，215,(2,)2x x πππ+∈，由函数tan y x =在52,2ππ⎛⎫⎪⎝⎭单调递增， 得21x x π>+，12121212sin sin ()()cos cos x x f x f x x x x x ∴+=+=+， 由cos y x =在52,2ππ⎛⎫ ⎪⎝⎭单调递减得211cos cos()cos x x x π<+=-∴12()()0f x f x +<；（Ⅱ）同理212((21),2),(2,2)22n n x n n x n n πππππ-∈--∈+，22n ππ+ >221n n x x π->+>2n π由cos y x =在2,2()2n n n N πππ⎛⎫+∈ ⎪⎝⎭上单调递减得221cos cos n n x x -<-221221()()cos cos 0n n n n f x f x x x --∴+=+<，且221()0,()0n n f x f x -><当n 为偶数时，从1()f x 开始相邻两项配对，每组和均为负值，即[][][]12341()(()()()()0n n f x f x f x f x f x f x -++++⋅⋅⋅++<），结论成立；当n 为奇数时，从1()f x 开始相邻两项配对，每组和均为负值，还多出最后一项也是负值，即[][][]123421()()()()()()()0n n n f x f x f x f x f x f x f x --++++⋅⋅⋅+++<，结论也成立．综上，对一切n N +∈，123()()()()0n f x f x f x f x +++⋅⋅⋅+<成立.【点睛】本题主要考查了导数在研究函数性质的应用、零点存在性定理、余弦函数的单调性，考查面较广，属于难题.（二）选考题：请考生在第22、23题中任选一题做答，如果多做，则按所做的第一题计分．22.在平面直角坐标系xOy 中，以坐标原点为极点，x 轴正半轴为极轴建立极坐标系，曲线1C 的极坐标方程为24cos 30ρρθ-+=，[)0,2θ∈π. （1）求1C 的直角坐标方程；（2）曲线2C 的参数方程为cos 6sin6x t y t ππ⎧=⎪⎪⎨⎪=⎪⎩（t 为参数），求1C 与2C 的公共点的极坐标.【答案】(1) 22(2)1x y -+=(2) 6π⎫⎪⎭【解析】【详解】（1）将222{cos x y xρρθ=+=代入24cos 30ρρθ-+=得：()2221x y -+=. （2）由题设可知，2C 是过坐标原点，倾斜角为6π的直线， 因此2C 的极坐标方程为6πθ=或7,06πθρ=>， 将6πθ=代入21:30C ρ-+=，解得：ρ=将76θπ=代入1C得ρ=12,C C 公共点极坐标为6π⎫⎪⎭．23.设()f x x 1x 1=-++ . (1)求()f x x 2≤+ 的解集;(2)若不等式()a 12a 1f x a +--≥,对任意实数a 0≠恒成立,求实数x 的取值范围.【答案】(1) []0,2(233)22x x ≤-≥或. 【解析】 试题分析：(1)分情况讨论去绝对值求解即可；(2)整理,再结合绝对值三角不等式可得121111112123a a aa a a a+--=+--≤++-=，再解不等式113x x -++≥即可. 试题解析：(1)由()f x x 2≤+有201112x x x x x +≥⎧⎪≤-⎨⎪---≤+⎩或2011112x x x x x +≥⎧⎪-<<⎨⎪-++≤+⎩或201112x x x x x +≥⎧⎪≥⎨⎪-++≤+⎩解得0x 2≤≤,∴所求解集为[]0,2.(2a 12a 1)a+--=111112123a a a a+--≤++-=, 当且仅当11120a a ⎛⎫⎛⎫+-≤ ⎪⎪⎝⎭⎝⎭时取等号.由不等式()a 12a 1f x a+--≥对任意实数a 0≠恒成立,可得x 1x 13-++≥,解得33x x 22≤-≥或.。
2019-2020学年广州市第三中学高三英语三模试题及答案
2019-2020学年广州市第三中学高三英语三模试题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AA medical capsule robot is a small,often pill-sized device that can do planned movement inside the body after being swallowed or surgically inserted. Most models use wireless electronics or magnets or a combination of the two to control the movement of the capsule. Such devices have been equipped with cameras to allow observation and diagnosis, with sensors that “feel,” and even with mechanical needles that administer drugs.But in practice, Biomechatronics engineer Pietro Valdastri has found that developing capsule models from scratch (从头开始) is costly, time-consuming and requires advanced skills. “The problem was we had to do them from scratch every time,” said Valdastri in an interview. “And other research groups were redeveloping those same modules from scratch, which didn’t make sense.”Since most of the capsules have the same parts of components: a microprocessor, communication submodules, an energy source, sensors, and actuators (致动器), Valdastri and his team made the modular platform in which the pieceswork in concertand can be interchanged with ease. They also developed a flexible board on which the component parts are snapped in like Legos. The board can be folded to fit the body of the capsule, down to about 14 mm. Additionally, they compiled (编译) a library of components that designers could choose from, enabling hundreds of different combinations. They arranged it all in a free online system. Designers can take the available designs or adapt them to their specific needs.“Instead of redeveloping all the modules from scratch, people with limited technological experience can use our modules to build their own capsule robots in clinical use and focus on their innovation,” Valdastri said.Now, the team has designed a capsule equipped with a surgical clip to stop internal bleeding. Researchers at Scotland’s Royal Infirmary of Edinburg have also expressed interest in using the system to make a crawling capsule that takes images of the colon(结肠). One research group, led by professors at the Institute of Digestive Disease of the Chinese University of HongKong, is making a swimming capsule equipped with a camera that pushes itself through the stomach.One limitation of Valdastri’s system is that it’s only for designing models. Researchers can confirm their hypotheses (假设) and do first design using the platform, but will need to move to a custom approach to develop their capsules further and make them practical for clinical use.1. According to the passage, Valdastri and his team created the platform to ________.A. adopt the latest technologiesB. make their robots dream come trueC. help build specialized capsule robotsD. do preciser observation and diagnosis2. What does the underlined phrase “work in concert” mean in Para.3?A. Perform live.B. Run independently.C. Act in a cooperative way.D. Carry on step by step.3. What can be learnt from the passage?A. Valdastri’s system can’t provide a complete capsule creation.B. The modular platform is more useful than a custom approach.C. The capsules can move in human’s body automatically.D. It costs more to module the capsules on the board.BYou’re in a crowd of people who are all asking for the same thing. How do you make your voice heard above the rest? Be different. Don’t shout. Lisa, 25, was waiting to board a plane flying fromLondontoAustriafor Christmaswhen the flight was cancelled.“There were about a hundred of us unable to leave,” she says. “Everyone else was shouting at the airport staff. Instead of joining in, I walked up to the man behind the ticket desk very quietly and said, ‘This must be so awful for you! I don’t know how you deal with these situations—it’s not even your fault. I could never handle it as well as you are.’ Without my even asking, he found me a seat on another airline with an upgrade to first class. He was happy to do a favor forsomeone who was appreciative instead of unfriendliness.”Flattery (恭维) is an essential element of the sweet-talk strategy. “It’s human psychology that stroking a person’s ego (自我) with a few well-directed praises makes them want to prove you right,” says apsychologist. “Tell someone they’re pretty and they’ll instantly fix their hair; praise their sense of humor and they’ll tell a joke.”You need help and there’s ly no reason that the person will want to lend a hand. Allison, 26. a lawyer, realized she’d made a huge mistake on a batch of documents. “The only way I could fix the problem was to get the help of a colleague who I knew didn’t like me,” she said.Allison then went to the woman’s office and explained her problem. “As I was saying to the boss the other day you’re the only person who would know how to handle a situation like this, what would you suggest I do?”“Feeling pumped up (鼓励), she set about helping me and we finished the job on time, and she was happy to help.” Allison said.4. Whatwould have happened at the airport according to paragraph 1?A. The departure hall was filled with noise.B. Someone screamed just lo be different.C. The passengers waited on board patiently.D. The airport stuff were rude to the passengers.5. Why did the man put Lisa on another airline?A. He admired Lisa’s beauty.B. He appreciated her attitude.C. He was ready to help others.D. He was blamed for the cancellation.6. What is the third paragraph mainly about?A. The potential benefits of ego.B. The strategy to start small talk.C. The great importance of flattery.D. The value of humor in daily life.7. What can we learn about Allison’s colleague?A. She was a popular lawyer.B. She was always ready to help others.C. She always got praise from Allison.D. She did a great favor for Allison eventually.CIdentifying the chemical makeup of pigment (色素) used in ancient documents, paintings, and watercolor1 s is criticalto restoring and conserving the precious artworks. However, despite numerous efforts, scientists had been unable to determine the source of folium, a popular blue dye used to color1 manuscripts (手稿) in Europe during the middle ages — from the 5th to the 15th century. Now, a team of researchers fromPortugalhas finally uncovered the mysterious ingredient responsible for the gorgeous blueish-purple color1 that helped bring ancient illustrations and texts to life.The research team began byporing overinstructions penned by European dye makers from the 12th, 14th, and 15th centuries. They found what they were seeking in a 15th-century text entitledThe Book on How to Make All the Color Paints for Illuminating Books. However, translating the instructions was no easy task. It was written in the now extinct Judaeo-Portuguese language, and though the source of the dye was traced back to a plant, no name was mentioned.However, by piecing together suggestions from the text, the scientists were able to determine that the dye was made from the bluish-green berries of the chrozophora tinctoria plant. After an extensive search, the team found a few varieties of the plant growing along the roadside near the town ofMonsarazin southPortugal.The detailed instructions gave the researchers critical clues — including the best time to pick the berries. “You need to squeeze the fruits, being careful not to break the seeds, and then to put them on linen (亚麻).” The scientist says the detail was important since broken seeds polluted the pigment, producing an inferior quality ink. The dyed linen, which was left to dry, was an efficient way to store and transport the pigment during ancient times. When needed, the artist would simply cut off a piece of the cloth and dip it with water to squeeze out the blue color1 .Once the key ingredient had been identified, the researchers began to determine the dye’s molecular structure. To their surprise, they found that folium was not like any other known permanent blue dyes — it was an entirely new class of color1 , one they named chrozophoridin. “Chrozophoridin was used in ancient times to make a beautiful blue dye for painting.” the team wrote in the study. “Thus, we believe that this will not be our final word on this amazing plant and its story and that further discoveries will follow soon.”8. The primary purpose of the study is to ________.A. restore and conserve ancient precious artworksB. determine the substance making up the foliumC. prove the ancient dye-making technique was organicD. identify which class of color1 folium belongs to9. The underlined phrase “poring over”in the second paragraph means ________.A. discussing publiclyB. testing repeatedlyC. passing directlyD. reading carefully10. What can be learned about the blue dye folium?A. It was essentially an inferior type of ink.B. It was the only kind made from wild berries.C. It could be carried and used easily.D. It was carefully squeezed from broken seeds.11. The article is mainly about _________.A. how the mystery ofa thousand-year-old blue dye was solvedB. why the researchers took the trouble to recreate the dyeC. what needs to be done to make an organic dye from a plantD. when and where the discovery of the dye was madeDI was at the hardware store the other day and overheard a woman tell Ed., the manager, that fall was her favorite time of year. Ed., because he liked to keep his customers happy, agreed that fall was a wonderful season, but I could tell he was lying.I was going through my mind recently, trying to find sweet memories of fall. I failed. I met my wife in the summer and married her two summers later. My sons were born in the winter and summer, my granddaughter in the winter. I’ve been fried twice in my life, both in fall. One October, a truck carrying tofu ran a red light and hit me, destroying my favorite car, combining the three things I most hate - trucks, tofu and October.I'm not saying fall is without its attractions. The leavesare beautiful. But fall's vacillation (立场摇摆)is troubling, its effort (努力)to please everyone, its continuous search for the middle ground to be all things to all people. Say what you will about summer and winter, at least they have the courage to keep their opinions strong, even if they kill us with extreme heat or cold.I recently read a story of a man coming out of a three-month coma (昏迷). It started in early fall and ended just as winter came. I hope if I were ever in a coma I would be just as lucky as the man.Upon my awakening, one of my families who stood around my bed would ask. "Don't you remember anything from the past three months?""Not the first thing," I would happily report.If I ever have enough money. I'm going to buy a second home inAustralia, so that when fall starts here, I can move there for three months, just when spring is starting.12. What did Ed think of the customers words according to the author?A. Ed understood them and supported the customer completely.B. Ed might hold a different opinion on the topic.C. Ed believed the customer wasn't telling the truth.D. Ed thought they stood for most peoples' opinion.13. What can we infer from Paragraph 2?A.The author has a big and loving family.B. The author is having a hard time at work.C. It is important to follow the rules of the road.D. Nothing good has happened to the author in fall.14. Why does the author say the man in Paragraph 4 was lucky?A. Because he slept fall away.B. Because he ho sweet memories of fall.C. Because hedreamed of fill many times.D. Because he was met by his family when waking up.15. Which of the following does the author most want to do?A. Drivetracks.B. Eat tofu dishes.C. Watch leaves falling in fall.D. Move toAustraliain October.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。
广州市2019-2020年高三调研测试试题 数学(文)
高中毕业班第三次调研测试数学（文科）本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共24小题，共150分，考试时间120分钟。
注意事项： 1．答题前，考生先将自己的姓名、准考证号码填写清楚，将条形码准确粘贴在条形码区域内；2．选择题必须用2B 铅笔填涂；非选择题必须使用0.5毫米的黑色字迹的签字笔书写，字体工整、笔迹清楚；3．请按照题号顺序在各题的答题区域内作答，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；4．作图可先使用铅笔画出，确定后必须用黑色字迹的签字笔描黑；5．保持卡面清洁，不要折叠，不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。
第Ⅰ卷一、选择题：本大题共12题，每小题5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. 设全集={1,2,3,4,5,6,7,8}U ，集合{1,2,3,5}A =，={2,4,6}B ，则()U BA =ðA ．{2}B ．{4,6}C ．{1,3,5}D ．{4,6,7,8}2．复数3ii-= A ．13i +B ．13i --C ．13i -+D ．13i -3．设2()2f x ax bx =++是定义在[1,1]a +上的偶函数，则2a b +=A ．0B ．2C ．2-D ．124．已知(1,2)a =-，(2,)b m =，若a b ⊥，则=||bA ．12B ．1C ．3D ．55．下列有关命题的说法正确的是A ．“(0)0f =”是“函数()f x 是奇函数”的充要条件；B ．若:p 2000,10x R x x ∃∈-->.则:p ⌝2,10x R x x ∀∈--<；C ．若p q ∧为假命题，则,p q 均为假命题；D ．“若3πα=，则1cos 2α=”的否命题是“若3πα≠，则1cos 2α≠”． 6．已知,x y 满足14210x x y x y ≥⎧⎪+≤⎨⎪--≤⎩，则2z x y =+的最大值为A ．3B ．4C ．6D ．77．已知双曲线2222:1(0,0)x y C a b a b-=>>的离心率为52，则双曲线C 的渐近线方程为A ．14y x =±B ．13y x =±C ．12y x =±D．y x =±8．执行如图所示的程序框图，输出的T = A ．29B ．44C ．52D ．629．一个几何体的三视图如图所示，其中正视图与侧视图都是斜边长为2的直角三角形，俯视图是半径为1的四分之一圆周和两条半径，则这个几何体的体积为A ．312πB ．36πC ．34πD ．33π 10．若函数2()sin3sin sin 2f x x x x πωωω⎛⎫=++⎪⎝⎭（0ω>） 的最小正周期为π，则()f x 在区间203，π⎡⎤⎢⎥⎣⎦上的值域为A ．3[0]2，B ．13[]22，-C ．1[1]2，-D ．31[]22，-开始结束S = 3,n = 1,T = 2T > 2S?S = S + 3n = n + 1T = T + 3n输出T 是否8题图9题图正视图俯视图侧视图11．对于问题：“已知关于x 的不等式20ax bx c ++>的解集为(1,2)-，解关于x 的不等式20a x b xc -+>”，给出如下一种解法： 解：由20a x b x c ++>的解集为(1,2)-，得2()()0a xb xc -+-+>的解集为 (2,1)-，即关于x 的不等式20a x b xc -+>的解集为(2,1)-． 参考上述解法，若关于x 的不等式0k x b x a x c ++<++的解集为11(1,)(,1)32--， 则关于x 的不等式1011kx bx ax cx ++<++的解集为 A ．()()2,21,3- B ．()()3,11,2--C ．()(),,2311-- D ．()(),,3112--12．若函数()f x 满足1()1(1)f x f x +=+，当[0,1]x ∈时，()f x x =，若在区间(1,1]-上，()()2g x f x m x m =--有两个零点，则实数m 的取值范围是A ．103m <≤B ．102m <<C ．112m <≤ D ．113m <<第Ⅱ卷二、填空题：本大题共4个小题,每小题5分。
2019-2020学年广州市第三中学高三英语第三次联考试卷及答案解析
2019-2020学年广州市第三中学高三英语第三次联考试卷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AHottest Travel DestinationsSometimes figuring out the best place to go can be difficult. Here are some hottest travel destinations that offer some of the most beautiful, artistic, and fashionable places on the globe.Ibiza, SpainWhile Ibiza is knownas a party city, there is far more to do in this historic town than simply drink and dance under the stars. Built mainly in the second century, it’s a world heritage site with architecture dating back to as early as the 7th century. Don’t miss Charo Ruiz, Ibiza’s most famous fashion brand. Plus, the Ibiza Fashion Festival takes place every June.Tangier, MoroccoKnown as the “door to Africa”, Tangier has a rich and complex history dating back thirty centuries. It has all the beauty of the natural world. After spending a morning sunbathing by the Mediterranean Sea, get in some amazing shopping and discover great deals on everything. Before you head home, stop off in Marakesh to visit some of the top Moroccan designers.Havana, CubaStepping onto the streets of Havana feels like stepping back in time. For a day of sightseeing, check out the remaining architecture of Old Havana, which was built ten centuries ago. You can easily do it with one of the area’s many walking tours. Or visit the Museum of Rum for a taste of the island’s most popular wine. You might come across a clothing shop offering some classic finds!Melbourne AustraliaBuilt largely during the 1850s gold rush, Melbourne remains as alive as ever. Make sure to check out the hottest Australian brands. Moreover, visit the Block Arcade in Collins Street to see some of the 19th century architectural details the world has to offer. And, if you want to catch the largest consumer fashion festival in the world, grab tickets for the yearly Virgin Australia Melbourne Fashion Festival.1.What can visitors do in both Ibiza and Havana?A.Taste local wine.B.Visit modern Museums.C.Appreciate ancient buildings.D.Enjoy parties under the stars.2.To attend the globally largest consumer fashion festival, you have to go to ________.A.SpainB.MoroccoC.Havana.D.Australia3.Which of the following cities is the oldest?A.Ibiza.B.Tangier.C.Havana.D.Melbourne.BIt's the near future.Animal populations have fallen sharply and 80% of species are extinct.The forests are so rare that you need to make a booking to visit one. Birds also face extinction.The Arctic terns,a species evolved to fly across the world on4000kmannual journeys，are on their last migration (迁徙) to Antarctica.The Last Migrationby the Sydney-based writer Charlotte McConaghy is a different sort of climate novel，one in which the heroine's(女主人公)damaged soul is as much a story as the damaged environment This is McConaghy's first work of literary fiction,after a history publishing in science fiction and a romantic fantasy series."I wanted to try and engage with the climate crisis closely,"she said."It's hard to nail down where he book came from.But I had Toni Morrison's words in my head:'If there's a book you really want to read,but i hasn't been written yet,then you must write it.'I love that. It really speaks to me.""I wanted to write about the way the natural world is disappearing but I didn't know a way in."The way in”, she says, was to"go travelling.I went to Ireland and Iceland,and thought about these incredible journeys of the terns and these people who study hes journeys."The book became a story of a double journey: the migration of the birds,and a broken woman's travelling to the end of the earth.Much of the book is told in flashbacks, the action jumping between the south coast of New South Wales to the west coast of Ireland and to Greenland."I've always been fascinated with Ireland: the landscape, the people and the poetry and music.I was fascinated with writing a character from there. It was a way to connect more with the place."McConaghy says she also wanted to have a character who was"of two places"."I had lived in 21 houses by the time I was 21, as a result I definitely know how it feels to feel as if you are not sure where you belong and feeling as if you are between two worlds."4. How isThe Last Migrationdifferent from other climate novels?A. It forecasts environmental destruction.B. It features a bird's cross-continental migration.C It combines science fiction well with romantic fantasy.D. It attaches equal importance to the heroine's broken soul.5. What did McConaghy think of Toni Morrison's words?A. Inspirational.B. One-sided.C. Authoritative.D. Casual.6. Why did McConaghy go travelling?A. To appreciate the landscape of Ireland.B. To follow the migration of the birds.C. To get away from her tiresome life.D. To find ideas for her new book.7. How might a character "of two places"feel?A. Content and carefree.B. More connected with nature.C. Lacking in a sense of belonging.D. Knowledgeable about the world.CThere is an old army joke about an officer who asks some soldiers whether any of them are interested in music. When four hands go up, the officer says, “Right, men. You can carry this grand piano down to the officers mess.”Job recruitment has become more complicated since that story first did the rounds. Today's careers require a lot more than just raw music but that sometimes makes jobs hard to define. An unfortunate result is a form of “adjective inflation” in recruitment ads as employers attempt to make routine tasks sound exciting.Candidates must sometimes wonder whether they are applying for a 9-to-5-role or to become a member of the Marvel “Avengers”. On Indeed, a job-bunting website, a bar was recently looking for "bartenders who are people focused, quality-driven, and have superhero hospitality powers". The ability to give customers the correct change was not mentioned.Another British company advertised for “a call-centre ninja, a superhero in people", a Job description which sounds a little over-the top for what was in fact a role at an insurance company in Isleworth. In case you think thatad was not typical. Indeed also had jobs demanding “ninja-like attention to detail". Short of turning up for the interview dressed head-to-toe in black, and then sneaking up behind the managing director at his desk, it is hard to see how candidates could show their ninja qualities.Not all companies require candidates to possess the qualities of a ninja, of course. Some require applicants to be passionate. The Bluewater shopping mall in southeastEnglandwas looking for “passionate sales-driven brand ambassadors” while “passionate crew members” were needed at a bakery in westLondonfor a wage of just £8. 23 an hour. In fact, passion is pretty hard to keep consistently for 40 hours a week, month after month. Job applicants should find some information from the kind of ads that companies place. If a job ad talks about passion or superheroes, run away faster than a speeding Batmobile. Being a ninja should be reserved for teenage mutant ninja turtles.8. Why does the author mention the joke in Paragraph 1?A. To show that officers enjoy playing tricks on soldiers.B. To introduce the topic of overstated job ads.C. To explain the origin of complex job ads.D. To describe soldiers' everyday life.9. Which of the following offers a job demanding ninja qualities?A. The bar.B. The bakery.C. The insurance company.D. The Bluewater shopping mall.10. What does the author say about the ads requiring passion?A. They are unrealistic.B. They are typical of want ads.C. They are appealing to applicants.D. They are uncommon on Indeed.11. How does the author sound when talking about today’s job ads?A. Curious.B. Hesitant.C. Humorous.D. Sympathetic.DAs I was walking home，a boy seemed to be singing a song in a very low voice，walking infront of me and carrying some of the same high school books I had with me.I caught up to him，and said “Hello”．I could tell he wasstartled，and dropped one of his books.I picked it up and as I handed it to him，he said “Thank you.” in a strange way，like witha strong accent.We soon began walking together and talking.It turns out he was in my science class，and I didn’t even knowit!His name was Ahmad，and his family had just moved here this school year.He invited me into his house，and his family treated me like a special guest，which made me feel very welcome.He said hospitality (好客) is very basic to his culture，and strangers were always treated very kindly.His mom brought out some great food，and offered me some tea.Ahmad’s father and two sisters wanted to hearall about me and my family，and my schooling.Ahmad’s whole family had to leave their lifelong home because war had broken out，and it wasn’t safe there anymore.They left with only what they could carry.His family was happy to feel safe，and they got used toAmerican culture quickly.They seemed pleased that Ahmad had brought a friend home to meet them.And as my friendship with Ahmad has continued，I now realize that the world is so much bigger than I thought！I also realize that your friends don’t have to be just like you.Differences make the world go round.12. What does the underlined word “startled” in Paragraph 1 mean?A. Excited.B. Moved.C. Surprised.D. Interested.13. Why was the writer treated like a special guest by Ahmad’s family?A. They liked strangers who came to their home for help.B. They wanted to know more information about the writer.C. Strangers were always treated very kindly in their culture.D. The writer was an American and knew much about their culture.14. Where does the writer probably come from?A. Canada.B. America.C. China.D. Australia.15. What can we learn from the passage?A. People from different countries can be good friends.B. Friends need different cultures and different accents.C. Your friends should have a lot of things the same with you.D. You can’t make friends with people from other countries.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。
广州市2019年高三毕业班第三次模拟考试 文科数学试题
3C ． 13D ．34．已知函数 f ( x ) = ( ) x - 2 x ，则 f ( x )5. 已知曲线 C 1： y = sin x ，C 2： y = sin(2 x -2πA ．把 C 1 上各点的横坐标伸长到原来的 2 倍，纵坐标不变，再把得到的曲线向右平移 个单位长度，得B ．把C 1 上各点的横坐标伸长到原来的 2 倍，纵坐标不变，再把得到的曲线向左平移 个单位长度，得广州市 2019 年高三毕业班第三次模拟考试数学（文科）本试卷共 23 题，共 150 分，共 4 页，考试结束后将本试卷和答题卡一并收回．注意事项：1.答题前，考生先将自己的姓名、准考证号码填写清楚.2.选择题必须使用 2B 铅笔填涂；非选择题必须使用 0.5 毫米黑色字迹的签字笔书写，字体工整、笔迹清楚.3.请按照题目的顺序在答题卡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、试卷上答题无效.4.作图可先使用铅笔画出，确定后必须用黑色字迹的签字笔描黑.5.保持卡面清洁，不要折叠、不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀.一、选择题：本题共12小题，每小题5分，共60分．在每个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 | y = lg( x - 2)} ， B = (-2, 3) ，则 AB =A ． (-2, 2)(2,3） B ． (-2, 2) C ． (2, 3) D ． [2, 3)2．已知 a ∈ R , i 是虚数单位，若 z = 3 + ai ， | z |= 2 ，则 a =A ． 7或- 7B ．1 或-1C ． 2D ． -23．已知向量 a = (1,2), b = (2, -1), c = (1, λ ) ，若 (a + b ) ⊥ c ，则 λ 的值为A ． -3B ． -112A. 是奇函数，且在 R 上是增函数B. 是偶函数，且在 R 上是增函数C. 是奇函数，且在 R 上是减函数D. 是偶函数，且在 R 上是减函数3) ，则下面结论正确的是π3到曲线 C 2π 3．．．8012 8 C ． 1A ． 1 2到曲线 C 2C ．把 C 1 上各点的横坐标缩短到原来的到曲线 C 2D ．把 C 1 上各点的横坐标缩短到原来的1 π倍，纵坐标不变，再把得到的曲线向右平移 个单位长度，得2 31 π倍，纵坐标不变，再把得到的曲线向左平移 个单位长度，得2 3到曲线 C 26. 已知数列{a } 满足 (n + 1)a = na （ n ∈ N * ），a = 2 ，等比数列{b } 满足 b = a ，b = a ，则{b }nn n +1 2 n 1 1 2 2 n的前 6 项和为A ． -64B ． 63C ． 64D ．1267. 某工厂为提高生产效率，开展技术创新活动，提出了完成某项生产任务的两种新的生产方式．为比较两种生产方式的效率，选取 40 名工人，将他们随机分成两组，每组 20 人，第一组工人用第一种生产方式，第二组工人用第二种生产方式．根据工人完成生产任务的工作时间（单位：min ）绘制了如右茎叶图：则下列结论中表述不正确的是A. 第一种生产方式的工人中，有 75%的工人完成生产任务所需要的时间至少 80 分钟B. 第二种生产方式比第一种生产方式的效率更高C. 这 40 名工人完成任务所需时间的中位数为 80D. 无论哪种生产方式的工人完成生产任务平均所需要的时间都是 分钟.8．右图为中国古代刘徽的《九章算术注》中研究“勾股容方”问题的图形，图中△ABC 为直角三角形，四边形 DEFC 为它的内接正方形，已知 BC=2，AD EAC=4，在△ABC 上任取一点，则此点取自正方形 DEFC 的概率为2 451A ．B ．C ．D ．99 9 29．如图，网格纸上虚线小正方形的边长为 1，实线画出的是某几何体的三视图，则该几何体上下两部分的体积比为1 1B ．D ．64CF B10. 过双曲线 x 2 y2 - a b 2= 1(a > 0,b > 0) 两焦点且与 x 轴垂直的直线与双曲线的四个交点组成一个正方形，则该双曲线的离心率为2C ． 5 - 1n 4=.6 ，点 D 在 BC 上，A ．2B ． 3D ． 5 + 1211.已知圆锥的顶点为 S ，底面圆周上的两点 A 、 B 满足 ∆SAB 为等边三角形，且面积为 4 3 ，又知 SA 与圆锥底面所成的角为 45°，则圆锥的表面积为A ． 8 2πB ． 4( 2 + 2)πC ． 8( 2 + 1)πD ． 8( 2 + 2)π12. 已知点 P 在直线 x + 2 y - 1 = 0 上，点 Q 在直线 x + 2 y + 3 = 0 上， M ( x , y ) 为 PQ 的中点，且y > 2x + 1 ，则 0 0 y0 的取值范围是 x1A ． [ , +∞)B ． (- 3 1 1 1 1 1, ) C ． (-∞,0) (0, ) D ． (- ,0) (0, ]2 3 3 2 3二、填空题：本题共 4 小题，每小题 5 分，共 20 分．13．命题“对 ∀x ∈[-1,1], x 2 + 3x - 1 > 0 ”的否定是 _______.14．在曲线 f ( x ) = x 3 - 4x 的所有切线中，斜率最小的切线方程为 .1 15．若圆 x2 + y 2 = 1与圆 x 2 + y 2 - 6 x - 8 y - m = 0 相切，则 m 的值为. 416. 如图，给出一个直角三角形数阵，满足每一列的数成等差数列，从第三12 ，14行起，每一行的数成等比数列，且每一行的公比相等，记第 i 行第 j 列的数为 a (i ≥ j , i 、j ∈ Z + ) ,则 aij3 3 3 ， ，4 8 16 ......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 题～第 21 题为必考题，每个试题考生都必须做答．第 22 题～第 23 题为选考题，考生根据要求做答．（一）必考题：共 60 分17.（12 分）在△ABC 中， AC = 4 2 ， ∠C =π1cos ∠ADC = - .3（1）求 AD 的长；（△2）若 ABD 的面积为 2 2 ，求 AB 的长；B18. （12 分）如图，在四边形 ABED 中，AB//DE ，AB ⊥ BE ，点 C 在 AB 上，且 AB ⊥ CD ，AC=BC=CD=2，现将△ACD 沿 CD 折起，使点 A到达点 P 的位置，且 PE = 2 2 .（1）求证：平面 PBC ⊥ 平面 DEBC ；（2）求三棱锥 P-EBC 的体积.19.（12 分）某地种植常规稻 A 和杂交稻 B ，常规稻 A 的亩产稳定为 500 公斤，统计近年来数据得到每年常规稻 A 的单价比当年杂交稻 B 的单价高 50%．统计杂交稻 B 的亩产数据，得到亩产的频率分布直方图如下；统计近10年来杂交稻 B 的单价（单位：元 /公斤）与种植亩数（单位：万亩）的关系，得到的10 组数据记为( x , y )(i = 1,2, 10) ，并得到散点图如下，参考数据见下．ii（1）求出频率分布直方图中m 的值，若各组的取值按中间值来计算，求杂交稻的亩产平均值；（2）判断杂交稻 B 的单价 y （单位：元/公斤）与种植亩数 x （单位：万亩）是否线性相关，若相关，试根据以下统计的参考数据求出 y 关于 x 的线性回归方程；（3）调查得到明年此地杂交稻 B 的种植亩数预计为 2 万亩，估计明年常规稻 A 的单价，若在常规稻 A和杂交稻 B 中选择，明年种植哪种水稻收入更高？统计参考数据： x = 1.60 ， y = 2.82 ， ∑( x - x )( y - y ) = -0.52 ， ∑ ( x - x ) ∑ ( x - x )( y - y )n．已知椭圆 C : + = 1 ，直线 l : y = x + m （ m ∈ R ）与椭圆 C 交于不同的两点 A 、 B .（2）若0 < a < 1， x 是函数 f ( x ) 的零点， f '(x ) 是 f ( x ) 的导函数，求证： f '( ) < f '(x ) < f '(3) ．2（ B B附：线性回归方程 y = bx + a ， b =ni =110 10i i ii =1 i =1i i∑ ( x - x )2ii =12 = 0.65 ，20.（12 分）x 2y 263 2 3（1）若 | AB |= 5 33，求 m 的值；（2）试求 || OA |2 - | OB |2 | （其中 O 为坐标原点）的最大值. 21．(12 分)已知函数 f ( x ) = x - a - 1ee x + a ln x - x （ a < 1 ，e 是自然对数的底， e ≈ 2.72 ）（1）讨论 f ( x ) 的单调性；30 0（二）选考题：共 10 分.请考生在第 22、23 题中任选一题作答，如果多做，则按所做的第一题计分．22. [选修 4-4：坐标系与参数方程] （10 分）以原点 O 为极点，x 轴的非负半轴为极轴建立极坐标系，已知曲线 C 的极坐标方程为 ρ 2 cos 2θ = a 2（ a ∈ R ,a 为常数）），过点 P (2,1) 、倾斜角为 30︒ 的直线 l 的参数方程满足 x = 2 +3 2t ， t 为参数）．（1）求曲线 C 的普通方程和直线 l 的参数方程；（2）若直线 l 与曲线 C 相交于 A 、 两点（点 P 在 A 、 之间），且 | P A | ⋅ | PB |= 2 ，求 a 和 || PA | - | PB ||的值．23. [选修 4 - 5：不等式选讲] (10 分）已知函数 f ( x ) =| x + 1| - | x - 1| ，（1）求函数 f ( x ) 的值域；（2）若 x ∈ [-2,1] 时， f ( x ) ≤ 3x + a ，求实数 a 的取值范围．解析: 8.设 CD = x ，则由 x⨯ 4 ⨯1⨯ 4 l 4 5 x广州市 2019 年高三毕业班第三次模拟考试（文科）参考答案及评分说明一、本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制订相应的评分细则．二、对计算题当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的内容和难度，可视影响的程度决定给分，但不得超过该部分正确解答应得分数的一半；如果后续部分的解答有较严重的错误，就不再给分．三、解答右端所注分数，表示考生正确做到这一步应得的累加分数．四、只给整数分数．一、选择题题序答案1C 2B 3A 4C 5C 6B 7D 8B 9C 10D 11C 12B4 - x 4 4 1 4=，得 x = , p = ( )2 ⨯ = 选 B 24 3 3 4 99. 由三视图知该几何体是下方为长方体上方为一直三棱柱的组合体，其上下体积比为121 = 42 ⨯3 6.10 ．将 x = c 代入双曲线的方程得 y 2= b 4 b 2 b 2 1 ⇒ y =± ，则 c = ⇒ ac = c 2 - a 2 ⇒ e - =1 ，解得 a 2 a a ee = 5 + 1.211. 设圆锥母线长为 ,由 ∆SAB 为等边三角形，且面积为 3 得34l 2 = 4 3 ⇒ l = 4 ，设圆锥底面半径为r ，则又 S A 与圆锥底面所成的角为45°得 r = 2 2 ，故 S =π r l + π r 2 = 8( 2 + 1)π .表12. 因直线 x + 2 y - 1 = 0 与 x + 2 y + 3 = 0 平行，故点 M 的轨迹为与两直线距离相等且平行于两直线的直线，其方程为 x + 2 y + 1 = 0 ，即点 M ( x , y ) 满足 x + 2 y + 1 = 0 ，而满足不等式 y > 2x + 1 的点在0 0 0 0 0 03 1直线 y = 2 x + 1 的上方，易得直线 x + 2 y + 1 = 0 与 y = 2 x + 1 的交点为 (- , - ) ，故问题转化为求射5 53 y线（不含端点）x + 2 y + 1 = 0 （ x < - ）上的点 M ( x , y ) 与坐标原点 (0,0) 连线斜率、即 0 的取0 0 0 0 0值范围, 故y= kxOM 1 1∈ (- , ) . 2 3651n 4 = ⨯ ( )3 =2二、填空题题序1314 15 16答案∃x ∈[-1,1],x 2+ 3x -1 ≤ 0 y = -4 x （或 4 x + y = 0 ） -9或11n32解 析 ： 15. 圆 x 2 + y 2 - 6 x - 8 y - m = 0 的 圆 心 为 (3, 4) ， 半 径 r =25 + m ， 若 两 圆 外 切 ， 则2 5+ m + 1 = ， m = -9 ，若两圆内切，则 25 + m -1 = 5 解得 m = 11.16. a1，每一行的公比均为 ，由等差数列的通项公式知第 n 4 21 1 n n 1 n 行第一个数为 + (n - 1)⨯ = ，故 a 4 4 4 42 32 三、解答题.17.解：（1）∵ cos ∠ADC = - 1 3,且 0 < ∠ADC < π∴ sin ∠ADC = 1 - (1 )2 = 32 2 3，--------------------------2 分正弦定理有 AD AC AC sin ∠C 1 3= ，得 AD = = 4 2 ⨯ ⨯ = 3 ；-----5 分sin ∠C sin ∠ADC sin ∠ADC 2 2 2（2）∵ sin ∠ADB = sin(π - ∠ADC ) = sin ∠ADC = 2 23， ---------------------6 分S ∆ABD = 1AD ⋅ BD ⋅ sin ∠ADB = 2BD ，∴ 2BD = 2 2 ，得 BD = 2 ，--------------------------------------------------8 分1又∵ cos ∠ADB = cos(π - ∠ADC ) = - cos ∠ADC = ，-------------------------9 分31由余弦定理得 AB 2 = 32 + 22 - 2 ⨯ 3 ⨯ 2 ⨯ = 9 ，3∴ AB = 3 ．----------------------------------------------------12 分18. 解：（1）证明：∵AB⊥BE，AB⊥CD，∴BE//CD，-----1 分∵AC⊥CD，∴PC⊥CD，∴PC⊥BE，---------2 分又 BC⊥BE，PC∩BC=C，∴EB⊥平面 PBC ，-------------------4 分又∵EB ⊂ 平面 DEBC ，∴平面 PBC ⊥ 平面 DEBC ；------------------6 分E - P BC = S1（2）解法 1：∵AB//DE，结合 CD//EB 得 BE=CD=2，------------7 分由（1）知 EB⊥平面 PBC ，∴EB⊥PB，由 PE = 2 2得 PB = PE 2 - EB 2 = 2 ,------------------------8 分∴△PBC 为等边三角形，∴ S∆PBC = 3 4⨯ 22= 3 ,----------------10 分∴VP - E BC= V ⋅ EB = ⨯ 3 ⨯ 2 1 13∆PBC3= 2 33.---------------------------------------12 分【 解 法 2 ： ∵A B//DE ， 结 合 CD//EB 得 BE=CD=2 ，-------------------------7 分由（1）知 EB⊥平面 PBC ，∴EB⊥PB，由 PE = 2 2 ，得 PB = PE 2 - EB 2 = 2 ,-------------------------------------------------8 分∴△PBC 为等边三角形，取 BC 的中点 O ，连结 OP ，则 PO = 3 ,-----------------------------10 分∵PO⊥BC，∴PO⊥平面 EBCD ，∴ VS3 ∆EBC1 12 3⋅ PO = ⨯ ⨯ 22 ⨯ 3 = .--------------------------12 分】3 2 319.解：（1）由 m ⨯ 30 + 0.01⨯ 20 + 0.02 ⨯ 20 + 0.025 ⨯10 = 1，解得 m = 0.005 ．---------------------------------------------------------2 分过程一：杂交稻 B 的亩产平均值为：[(730 + 790 + 800) ⨯ 0.005 + (740 + 780) ⨯ 0.01 + (750 + 770) ⨯ 0.02 + 760 ⨯ 0.025] ⨯10= 116 + 152 + 304 + 190 = 762 ．-------------------------------------------------5 分【过程二：设杂交稻 B 的亩产数据为 n 个，则杂交稻 B 的亩产平均值为：[(730 + 790 + 800) ⨯ 0.05n + (740 + 780) ⨯ 0.1n + (750 + 770) ⨯ 0.2n + 760 ⨯ 0.25n ]⨯1n= 116 + 152 + 304 + 190 = 762 ．--------------------------------------------5 分】（2）因为散点图中各点大致分布在一条直线附近，所以可以判断杂交稻 B 的单价 y 与种植亩数 x 线性相关，--------------------------------6 分由题目提供的数据得： b =-0.52 = -0.8 ，0.65由 y = bx + a 得 a = y - bx = 2.82 + 0.8 ⨯1.60 = 4.10 ，ˆ ˆ 2 433 1 3 2 3 1 3 1所以线性回归方程为 y= -0.8 x + 4.10 ．----------------------------------------8 分（3）明年杂交稻 B 的单价估计为 y = -0.8 ⨯ 2 + 4.10 = 2.50 元／公斤，明年常规稻 A 的单价估计为 2.50 ⨯ (1+ 50%) = 3.75 元／公斤；----------------------------10 分明年常规稻 A 的每亩平均收入估计为 500 ⨯ 3.75 = 1875 元／亩，明年杂交稻 B 的每亩平均收入估计为 762 ⨯ 2.50 = 1905 元／亩，因 1905>1875，所以明年选择种杂交稻 B 收入更高．----------------------------------12 分⎧2 x 2 + 3 y 2 = 6,⎪20.解：（1）由⎨ 6⎪ y =x + m . ⎩ 3消去 y 并整理得 4 x 2 + 2 6mx + 3(m 2 - 2) = 0 ，----------2 分∵直线 l 与椭圆 C 交于不同的两点 A 、 B ，∴ ∆ = (2 6m )2 - 48(m 2 - 2) > 0 ，即 -2 < m < 2 ， ----------------------------3 分设 A ( x , y ), B ( x , y ) ，则 x + x = - 1 1 2 2 1 2 6m 3(m 2 - 2), x x =1 2 ，---------------------------4 分| AB |2= ( x - x )2+ ( y - y )2= [1+ ( 1 2 1 2 6 3 2)2 ]( x - x )2 = (1+ )[( x + x )2 - 4 x x ]1 2 1 2 1 2即 5 3 ⨯ [(- 6m 25 )2 - 3(m 2 - 2)] = 2 3 6，解得 m =± .--------------------------------6 分3（2）∵ | OA |2 - | OB |2 = ( x 2 + y 2 ) - ( x 1122 + y 2 )21 1 1 1= x 2 + 2(1- x 2 ) - [ x 2 + 2(1- x 2 )] = ( x 2 - x 2 ) = ( x + x )( x - x ) -----------7 分 1 2 2 2 1 2又 | x - x |= ( x + x )2- 4 x x =12121 23 3 6m 2 - 3(m 2 - 2) = 6 - m 2 = 4 - m 2 ---8 分 2 2 21 6 | m | 6 4 - m2 1∴ || OA |2 - | OB |2 | = ⨯ ⨯ = m 2 (4 - m 2 )3 2 2 2∵ m 2 (4 - m 2 ) ≤ (m 2 + 4 - m 22)2 = 4 ，---------------------------10 分∴ || OA |2- | OB |2| = 1 1m 2 (4 - m 2 ) ≤ ⨯ 4 = 1 ，2 2即 || OA |2 - | OB |2 | 的最大值为 1.（当且仅当 m = ± 2 时，取得最大值）---------------12 分e x a e x 121.解：（1） f '(x ) = ( x - a ) + - 1 = ( x - a )( - ) ( x > 0) ，--------------------------1 分e x e x设 g ( x ) = e x 1- ( x > 0) ，e x2 2e x1过程一：由y=和y=-在(0,+∞)上单调递增，e xe x1【过程二：由x>0得g'(x)=+e x2>0】可知g(x)在(0,+∞)上单调递增，又g(1)=0，所以当x∈(0,1)时，g(x)<0，当x∈(1,+∞)时，g(x)>0，①当a≤0时，x-a>0，当x∈(0,1)时，f'(x)<0；当x∈(1,+∞)时，f'(x)>0．-----------------------3分②当0<a<1时，由f'(x)=0得x=a或x＝1，当x∈(0,a)时，x-a<0，g(x)<0，f'(x)>0；当x∈(a,1)时，f'(x)<0；当x∈(1,+∞)时，f'(x)>0．-----------------------5分综上所述：当a≤0时，f(x)在(0,1)上单调递减，在(1,+∞)上单调递增；当0<a<1时，f(x)在(0,a)单调递增，在(a,1)上单调递减，在(1,+∞)上单调递增．----------------------------------------6分（2）方法一（分析法）：当0<a<1时，由（1）知f(x)在(0,1]上的最大值为f(a)，可知f(a)=-e a-1+a ln a-a<0，所以f(x)在(0,1]上无零点．若x是函数f(x)的零点，则x>1，------------------------------------------------7分00e x1∵f'(x)=(x-a)(-)(x>1)，e x过程一：由y=x-a和y=e x1e x1-在(1,+∞)上单调递增，且->0、x-a>0，e x e xe x1e x1【过程二：设h(x)=f'(x)，则h'(x)=(-)+(x-a)(+e x e x2)，e x1e x1由x>1得->0，(x-a)(+e x e x2)>0，所以h'(x)>0，】可知f'(x)在(1,+∞)上单调递增，-------------------------------------------8分要证f'(3)<f'(x)<f'(3)，只需证32<x<3，----------------------------9分由（1）知f(x)在(1,+∞)上单调递增，只需证f(3)<f(x)<f(3)，又f(x)=0，------------------------------10分00由 ln 3 而 f ( ) = ( - a ) e + a ln 3 由 ln 3 < ln e = 1 ， e > 1 ，得 ln - e < 0 ，又 - < 0 ，所以 f ( ) < 0 ； 所以 f ( ) < 0 < f (3) ，又 f ( x ) = 0 ，即 f ( ) < f ( x ) < f (3) ，--------------------------9 分 2 22 只需证 f (3 ) < 0 且 f (3) > 0 ． 23 1 3 3 3 e 3 f ( ) = ( - a ) e + a ln - = (ln - e )a + - ， 2 2 2 2 2 2 23 e 33 < ln e = 1 ， e > 1 ，得 ln - e < 0 ，又 - < 0 ，所以 f ( ) < 0 ； 2 2 2 2 2f (3) = (2 - a )e 2 + a ln3 - 3 ，由 2 - a > 1得 f (3) > e 2 + a ln3 - 3 > 0 ，综上所述，得证．---------------------------------------------------------------------12分方法二（综合法）：当 0 < a < 1时，由（1）知 f ( x ) 在 (0,1] 上的最大值为 f (a ) ，可知 f (a ) = -e a -1 + a ln a - a < 0 ，所以 f ( x ) 在 (0,1] 上无零点．若 x 是函数 f ( x ) 的零点，则 x > 1 ，--------------------------------------------------7 分 02 2 13 3 3 e 3 - = (ln - e )a + - ， 2 2 2 2 2 3 e 3 3 2 2 2 2 2f (3) = (2 - a )e 2 + a ln3 - 3 ，由 2 - a > 1得 f (3) > e 2 + a ln3 - 3 > 0 ，3 3 0 0 由（1）知 f ( x ) 在 (1, + ∞) 上单调递增，所以e x 1 而f '(x ) = ( x - a )( - ) ( x > 1) ， e x3 2< x < 3 ，----------------------------10 分 0由 y = x - a 和 y = e x 1 e x 1 - 在 (1, + ∞) 上单调递增，且 - > 0 、 x - a > 0 ， e x e x可知 f '(x ) 在 (1, + ∞) 上单调递增，---------------------------------------------11 分3 所以 f '( ) < f '(x ) < f '(3) ，得证．---------------------------------------12 分 022.解：（1）由 ρ 2 cos 2θ = a 2 得 ρ 2 (cos 2 θ - sin 2 θ ) = a 2 ，-----------------------------1 分又 x = ρ cos θ ， y = ρ sin θ ，得 x 2 - y 2 = a 2 ，∴C 的普通方程为 x 2 - y 2 = a 2 ，--------------------------------------------------2 分∵过点 P (2,1) 、倾斜角为 30︒ 的直线 l 的普通方程为 y =3 3( x - 2) + 1 ，----------------3 分； 法二： f ( x ) = ⎨2 x , -1 ≤ x < 1 ，得 -2 ≤ f ( x ) ≤ 2 ， ⎪ 2, x ≥ 1由 x = 2 + 3 2 1 t 得 y = 1 + t 2⎧ 3 ⎪⎪ x = 2 + 2 ∴直线 l 的参数方程为 ⎨ ⎪ y = 1 + t ⎪⎩ 2 t （t 为参数） -------------------------------5 分⎧ 3 ⎪⎪ x = 2 + 2 （2）将 ⎨ ⎪ y = 1 + t ⎪⎩ 2t代入 x 2 - y 2 = a 2 ，得 t 2 + 2(2 3 - 1)t + 2(3 - a 2 ) = 0 ， --------------------------------------------6 分依题意知 ∆ = [2(2 3 - 1)]2 - 8(3 - a 2 ) > 0则上方程的根 t 、 t 就是交点 A 、B 对应的参数，∵ t ⋅ t = 2(3 - a 2 ) ， 1 2 1 2由参数 t 的几何意义知| P A | ⋅ | PB |=| t | ⋅ | t |=| t ⋅ t | ，得 | t ⋅ t |= 2 ，1 2 1 2 1 2∵点 P 在 A 、B 之间，∴ t ⋅ t < 0 ，1 2 ∴ t ⋅ t = -2 ，即 2(3 - a 2 ) = -2 ，解得 a 2 = 4 （满足 ∆ > 0 ），∴ a = ±2 ，------------8 分 1 2 ∵ || P A | - | PB ||=|| t | - | t ||=| t + t | ，又 t + t = -2(2 3 -1)， 12 1 2 1 2∴ || P A | - | PB ||= 4 3 - 2 ． ------------------------------------------------------10 分23.解：（1）法一： | f ( x ) |=|| x + 1| - | x - 1|| ≤| ( x + 1) - ( x - 1) |= 2 ，∴ -2 ≤ f ( x ) ≤ 2 ， f ( x ) 的值域为[-2, 2]；---------------------------------4 分⎧ -2, x < -1 ⎪ ⎩∴ f ( x ) 的值域为[-2, 2]；----------------------------------------------------4 分（2）由 f ( x ) ≤ 3x + a 得 a ≥| x + 1| - | x - 1| -3x ，由 x ∈ [-2,1] 得 x - 1 ≤ 0 ，∴ a ≥| x + 1| + x - 1 - 3x =| x + 1| -2 x - 1 ，---------------------------------5 分设 g ( x ) =| x + 1| -2 x - 1 (-2 ≤ x ≤ 1) ，① 当 -2 ≤ x ≤ -1 时， x + 1 ≤ 0 ， g ( x ) = -( x + 1) - 2 x - 1 = -3x - 2 ，∴ g ( x )max = g (-2) = 4 ；----------------------------------------------------7 分② 当 -1 < x ≤ 1 时， x + 1 > 0 ， g ( x ) = x + 1 - 2 x - 1 = - x ，∴g(x)<g(-1)=1；---------------------------------------------------------9分综上知，g(x)max=4，由a≥g(x)恒成立，得a≥4，即a的取值范围是[4,+∞)．---------------10分。
2019-2020学年广州中学高三英语第三次联考试卷及参考答案
2019-2020学年广州中学高三英语第三次联考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ALocated besideLake Geneva, the Olympic Museum houses more than 10,000 artificial objects and hours of interactive contents highlighting some of the best moments during the Olympics. Here are some of the museum’s most moving moments.The Olympic ParkThe journey through the Olympic Museum begins in the Olympic Park, an 8,000-square-meter outdoor area in front of the museum overlooking Lake Geneva and theAlps. The park contains artwork and sculptures that show respect to the world of sport.The first Olympic symbolThe “Olympic Rings” flag was designed by Coubertin in 1913. The rings represent the five continents that participate in the Olympics: Africa, Asia,America,AustraliaandEurope. The six color1 s include at least one color1 that is represented on the flag of every country.The stadiumsThe stadiums that host the Olympic Games are as much of a celebration of design as the games are a celebration of sportsmanship. Guests can explore plans and models of Olympic stadiums’ past and present, including one of the games’ most attractive stadiums, the Bird’s Nest from Beijing 2008 Olympics.The Olympic medalsHave you ever wondered what an Olympic medal looks like? The Olympic Museum has a room that houses every bronze, silver, and gold medal from every Olympic Games dating back to the first modern Olympics of 1896. Each medal design is a unique representation of the year and location in which the games were held.1.Which moment do you see first when exploring the Olympic Museum?A.The Olympic Park.B.The first Olympic symbol.C.The stadiums.D.The Olympic medals.2.What do you know from The first Olympic symbol?A.The first modern Olympics took place inGreece.B.There are six color1 s on the flag of every country.C.Australia used to be the largest continent on earth.D.The “Olympic Rings” flag was created in 1913.3.What can you do in the section of The stadiums?A.Admire the view ofLake Geneva.B.Meet some famous designers.C.Enjoy the model of the Bird’s Nest.D.Talk with guests of honour.BMy first week working in a restaurant, one of the servers said something that stuck with me: Everyone should work in a restaurant for at least a year. At the time, I didn't get it, but I took the advice to heart and worked in restaurants on and off for the next eight years. Before realizing it, I mastered many important skills, one of which is communication skill.When I was little, I was so shy that I used to hide behind my mom whenever someone spoke to me. And when I first started in restaurants, I had two personalities: Restaurant Lizzy and Home Lizzy. It was easier to pretend to be a different person while at work, since it was so different from who I actually was. But gradually, the skills I learned working in restaurants helped Home Lizzy come out of her shell in the real world.When you work in a restaurant, you don't have the luxury of hiding behind your parents to avoid talking to people. I'm still 110% an introvert, but restaurant work helped me communicate. Working in a restaurant not only helped me speak clearly, deliberately and directly but also taught me how to talk about almost everything. Some guests don't want their servers to interact too much with them, and that's fine. But some sit at the bar simply to chat with you. You learn how to judge your guests' level of interest in communicating with you, and how to exit a conversation at the appropriate time.My restaurant work is something that I'm most proud of and I know I wouldn't be the person I am today without those eight years of experience. If you're still on the fence about working in a restaurant for that long, start with one year. I doubt that you'll look back.4. What did the writer think of the server's words?A. Impressive.B. Ridiculous.C. Amusing.D. Logical.5. What do we infer from Paragraph 2?A. The writer tried different jobs.B. The writer became more sociable.C. The writer used an invented name.D. The writer had a hard time at work.6. Which of the following best describes the writer's restaurant work?A. Boring.B. Relaxing.C. Worthwhile.D. Unchallenging.7. What message does the writer try to convey in this passage?A. A strong-willed soul can reach his goal.B. Things are difficult before they are easy.C. Communication skills advance your career.D. Restaurantwork helps to achieve a better self.CIn sportthe sexes are separate. Women and men do not run or swim in the same races. Women are less strong than men.That at least is what people say.Women are called the weaker sex, or, if men want to please them, the fair sex. But boys and girls are taught together at schools and universities. There are women who are famous Prime Ministers, scientists and writers. And women live longer than men. A European woman can expect to live until the age of 74, a man only until he is 68. Are women’s bodies really weaker?The fastest men can run a mile in under 4 minutes. The best women need 4.5 minutes. Women’s time is always slower than men’s, but some facts are a surprise. Some of the fastest women swimmers today are teenage girls. One of them swam 400 meters in 4 minutes 21.2 seconds when she was only 16. The first ‘Tartan’ in film was an Olympic swimmer, Johnny Weissmuller. His fastest 400 meters was 4 minutes 49.1 seconds, which is 37.9 seconds slower than a girl 50 years later! This does not mean that women are catching men up. Conditions are very different now and sport is much more serious. It is so serious that some women athletes are given hormone injections. At the Olympics a doctor has to check whether the women athletes are really women or not. It seems sad that sport has such problems. Life can be very complicated when there are two separate sexes!8. Women are called the weaker sex because _________.A. women do as much as menB. people think women are weaker than menC. sport is easier for men than for womenD. in sport the two sexes are always together9. Which of the following is true?A. Boys and girls study separately everywhere.B Women do not run or swim in races with men.C. Famous Prime Ministers are women.D. Men can expect to live longer than women in Europe.10. The underlined part “That at least is what people say” means people _________.A. say other things, tooB. don’t say this muchC. say this but may not think soD. only think this11. In this passage the author implies that _________.A. womenare weaker than men, but fasterB. women are slower than men, but strongerC. men are not always stronger and faster than womenD. men are faster and stronger than womenDJIANLI, Hubei Province, June 2 (Xinhua) — A cruise ship carrying more than 450 people sank in the Yangtze River overnight, which could be China’s worst sinking disaster in decades. As of 6 p.m. Tuesday, 14 peoplehad been rescued from the capsized vessel, with five others confirmed dead. The rest are still missing, although rescuers said there could be more survivors. Strong winds and heavy rain are hampering rescue efforts.The Eastern Star sank in only 15 meters of water “within one or two minutes” of being caught in freak weather in Jianli, according to the ship’s captain and chief engineer who survived the incident. The ship left the eastern Chinese city ofNanjingon May 28 bound forChongqingMunicipality.The ship was carrying 403 passengers, five tour guides, and 46 crew, rather than the previously reported 47. Most passengers were tourists fromShanghaiand its neighboringprovinceofJiangsu, aged between 3 and 83, with most in their 60s and 70s.The 76.5-meter-long and 11-meter-wide vessel has been in service for nearly 20 years and can carry up to 534 people. It is owned by Chongqing Dongfang Shipping. Waterway officials said they have no record of the company being involved in any previous sinking incidents.According to weather forecast, most of the Yangtze basin will be subject to downpours over the next 10 days, with heavy rain expected in the area where the search is underway.Police, waterway authorities and fire departments have sent more than 150 boats and over 4,000 personnel to the scene. The Chinese Navy has sent diving forces to search for the missing. The team is composed of soldiers from the navy’s fleets in the North Sea, East China Sea, andSouth China Sea, as well as students of the Naval University of Engineering. In addition, five helicopters were dispatched fromBeijingandHubei’s provincial capital ofWuhanon Tuesday morning, along with an IL-76 transport plane.12. We can infer from the text that ________.A. the incident was very suddenB. the weather benefited the rescueC. the ship had bad safety recordsD. the ship’s captain was drowned13. Which of the following statements is TRUE about the Eastern Star?A. It was overloaded.B. It’s been launched recently.C. It had 454 people on board.D. Its owner met similar incidents.14. What’s the main idea of the last paragraph?A. How people came to the rescue.B. Why the rescue was quite difficult.C. What the rescue force consisted of.D. How the rescue operation went on.15. The author’s purpose of writing the text is most likely to ________.A. informB. explainC. describeD. entertain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。
广东省广州市2019届高三第三次模拟考试理科综合物理试题
广东省广州市2019届高三第三次模拟考试理科综合物理试题本试题卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共16页，38题（含选考题）。
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一、选择题1.“复兴号”动车组在京沪高铁率先实现350公里时速运营，我国成为世界上高铁商业运营速度最高的国家。
一列“复兴号”正在匀加速直线行驶途中，某乘客在车厢里相对车厢以一定的速度竖直向上抛出一个小球，则小球（）A. 在最高点对地速度为零B. 在最高点对地速度最大C. 落点位置与抛出时车厢的速度大小无关D. 抛出时车厢速度越大，落点位置离乘客越远【答案】C【解析】【分析】小球竖直抛出后，水平方向有与车厢相同的初速度，竖直方向做竖直上抛运动，结合运动的合成知识进行解答.【详解】在匀加速直线行驶的车厢中竖直向上抛出的小球具有水平速度，则在最高点对地速度不为零，选项A 错误；小球在运动过程中，竖直速度逐渐减小，水平速度不变，则在最低点对地速度最大，选项B 错误；小球抛出时竖直向上的初速度一定时，在空中运动的时间t 一定，则设此时车厢的速度为v ，则落地时的位置与抛出时的位置间距为：221122x vt at vt at D =+-=，即落点位置与抛出时车厢的速度大小无关，选项C 正确，D 错误；故选C.2.一跳伞运动员从悬停的直升飞机上跳下，2s 时开启降落伞，运动员跳伞过程中的v -t 图象如图所示，根据图象可知运动员（ ）A. 在2~6s 内速度方向先向上后向下B. 在2~6s 内加速度方向先向上后向下C. 在2~6s 内先处于失重状态后处于超重状态D. 在0~20s 内先匀加速再匀减速最终匀速直线运动【答案】C【解析】【分析】v-t 图像的斜率等于加速度，斜率的符号反映加速度的方向；加速度向下为失重，加速度向上为超重。