江苏省G4(苏州中学、盐城中学、扬州中学、常州中学)2021届高三上学期期末调研数学试题含答案
2021届江苏省扬州中学高三语文上学期期末试卷及答案
2021届江苏省扬州中学高三语文上学期期末试卷及答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下列小题。
房东庐隐（1）当我们坐着山兜，停在一座山坡上时，兜夫“哎哟”的舒了一口气，意思是说“这可到了”。
那里有一所三楼三底的中国式洋房。
在这所房子的对面，是峙立着无数的山峦，当晨曦窥云的时候，我们睡在床上，可以看见万道霞光，从山背后冉冉而升。
跟着雾散云开，露出艳丽的阳光。
再加着晨气清凉，稍带冷意的微风，吹着我们不曾掠梳的散发，真有些感觉得环境的松软。
这种幽丽的地方，我们城市里熏惯了煤烟气的人住着，真是有些自惭形秽，虽然我们的外表强于他们乡下人，但是他们乡下人至少要比我们离大自然近得多，他们的心要比我们干净得多。
就是我那老房东，虽然她的样子特别的朴质，然而她却比我们这些好像知道什么似的人，更知道些自然的趣味。
（2）她已经五十八岁了，她的老伴比她小一岁，可是他俩所做的工作，真不像年纪这么大的人做的。
他们的儿媳妇一天到晚不在家，早上五点钟就到田地里去做工，到黄昏的时候，她有时肩上挑着几十斤重的柴就来家了。
在他们家里，从不预备什么钟，他们每一个人的手上也永没有戴什么手表，然而他们看见日头正照在头顶上便知道午时到了，除非是阴雨的天气，他们有时见了我们，或者要问一声：师姑，现在十二点了罢！据他们的习惯，对于做工时间的长短也总有个准儿。
（3）住在城市里的人每天都能在五点钟左右起来，恐怕是绝无仅有，然而在这岭里的人，确没有一个人能睡到八点钟起来。
说也奇怪，我也喜欢上了早起，朝旭未出将出的天容和阳光未普照的山景，实在别有一种情趣。
到了晚上，大家同坐在院子里讲家常，我们从楼上的栏杆望下去，老女房东便笑嘻嘻地说：“师姑！晚上如果怕热，就把门开着睡。
”我说：“那怪怕的，倘若来个贼呢？……这院子又只是一片石头叠就的短墙，又没个门!”“呵哟师姑！真真的不碍事，我们这里从来没有过贼，我们往常洗了衣服，晒在院子里，有时被风吹了掉在院子外头，也从没有人给拾走。
江苏省G4(苏州中学、盐城中学、扬州中学、常熟中学)2021届高三期末调研卷英语解析
苏高中，常熟省中，盐城中学，扬州中学四校联考解析A篇是一篇应用文，首先说明在生意场上喝酒的不可避免性，再从“喝什么”、“勇气、数量、质量”和“一般的礼节”三个方面介绍中国的饮酒文化。
文章理解容易，比较简单。
B篇是一篇议论文，阐述了在现代化喧嚣的城市中独处的必要性和独处的难点，作者呼吁人们多一点独处。
文章需要一定的理解力，难度中等。
C篇阅读讲的是某国接种疫苗的举措，且数据证实了有效性。
主要考察段落理解，较容易定位，难度中等。
D篇文章从随身听的起源和优缺点谈起，引申出关于公共场合缓解孤独的建议，呼吁人们适当的和陌生人进行交流。
总体考察文章大意的理解，难度中等偏上。
龚露老师详细解析：A篇第21题C 推断题。
啤酒是清凉的，清爽的淡啤酒可以冷却喝白酒的灼热感，所以啤酒是辅助性饮料，辅助白酒，A是开胃菜，B是主要酒水，C是辅助酒水，D是饭后酒水，故选C。
第22题D 推断题。
由文章第五段可知，与人喝酒时底线是，最好喝他们给的任何东西，这叫做“勇气”，即是“酒胆”，底线可以推断出重要的，而且在该段最后明确提出了该点是最重要的，故选D。
第23题C 细节题。
由文章最后一段可知，碰杯时，年轻人应该比年长者举的杯更低，而且如果年轻人比那些迟到的人迟到，那就被认为是极其严重的酒后惩罚，junior对应a young adult，故选C。
B篇第24题D 推断题。
文章第一段提出了一个情景，一个人的桌子，这样的一顿饭总是让人“刮目相看”，但是人们通常都是和他的伙伴一起吃饭，独处是一种奢华的选择，一个人的桌子通常会招致不赞同的目光，D正确，A项all the time无法体现。
第25题A 主旨大意题。
文章第二段主要介绍了现代化生活中我们的时间被各种社交充斥，我们失去了独处的时间，与此对比，阐述了独处的好处—用自己的方式来处理，故选A。
第26题B 细节题。
由文章第三段可知，性格内向者通过与个人思想的相处反思补充个人的能量释放个性特质，sit alone with my thoughts对应reflect on their own thoughts。
2021年江苏省扬州中学高三英语上学期期末考试试卷及参考答案
2021年江苏省扬州中学高三英语上学期期末考试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AThe Origins of Famous BrandsOur lives are full of brand names and trademarked products that we use every day. Although many brand names are simple acronyms(首字母缩略词) or versions of their founders names, some of the companies we trust every day actually have fascinating and surprising back stories.StarbucksIt seems fitting that the most famous coffee brand in the world would take its name from one of the world’s greatest works of literature. The inspiration for the name of the coffeehouse came from Herman Melville’sMoby Dick. The founders’ original idea was to name the company after the Captain Ahab’s ship, but they eventually decided that Pequot wasn’t a great name for coffee, so they chose Ahab’s first mate, Starbucks, as the name instead.GoogleGoogle was originally called Backrub, for it searched for links in every corner of the Web. In 1997, when the founders of the company were searching for a new name showing a huge amount of data for their rapidly improving search technology, a friend suggested the word “googol”. When a friend tried to register the new domain (域) name, he misspelled “googol” as “google”.NikeOriginally founded as a distributor for Japanese running shoes, the company was originally named BRS, or Blue Ribbon Sports. In 1971, BRS introduced its own soccer shoe, a model called Nike, which is alsothe name for the Greek goddess of victory. In 1978, the company officially renamed itself as Nike, Inc.The right name is essential to a company’s success, and a great origin story is just as important as a great product. An attractive origin story is one more thing that keeps customers guessing, wondering, and buying its products.1. What is the name of the Captain Ahab’s ship?A. Moby Dick.B. Starbucks.C. Pequot.D. Herman Melville.2. Why did the founders of the Google want to change its name?A. They mistook their name.B. They wanted new customers.C. The company’s original name was too long.D. The company’s search technology was improving rapidly.3. Where does the importance of the origin story of one company lie in?A. It can change the company’s image.B. It can add myth to the company.C. It explains the development of the company to customers.D. It makes customers imagine and purchase its goods.BIn recent years,people have been focusing on the quality of food that children are fed in schools. Former First Lady Michelle Obama worked hard to make school lunches healthier, resulting in new menus that featured less fat and salt, more fruits and vegetables.But high-quality nutrients count for little when there is no time to eat them. Amy Ettinger reports, "There is no national standard on how much time kids get to eat that meal. " And with schools being occupied with test scores, teachers are using every available minute for lesson time, which often leaves kids without enough eating time.This is a problem because the length of the school lunch period is a key factor (因素) in how much nutrition children actually gel. Research has found that having less than 20 minutes for lunch results in children consuming much less of their lunch than those with more than 20 minutes.This is really terrible. For many low-income kids, that cafeteria lunch can represent half their daily energy intake. There's also another terrible message that it's acceptable to wolf down food as fast as possible before rushing off to your next class. Cafeteria time should be a chance to interact with friends, to learn important social skills, to observe and share varieties of food. It should be a break in day, a chance to relax before heading into the afternoon.As Ettinger explains,some parents are hoping the National Parent Teacher Association will address this issue. This, in turn, would help parents push their kids' schools for better lunch time standards. Meanwhile, if you have a kid in this situation, you can help by packing a healthy lunch to spare them the cafeteria lineup. Make the foods easy to eat, provide non-messy snacks that can be eaten in class, put great effort into serving a hearty breakfast, and sit down as a family for dinner whenever possible.4. What did Michelle Obama make efforts to improve?A. The quality of school lunches.B. The performance of school kids.C. The school lunch time kids have.D. The eating habits of school kids.5. What happens to children in American schools?A. They are occupied with many tests.B. They fail to get along with each other.C. They consume more meat than before.D. They have less lunch time than before.6. How are low-income kids influenced by the problem at school?A. They can't go to classes on time.B. They can't have enough energy.C. They can't share different kinds of food.D. They can't hold a positive attitude toward life.7.What can parents do to solve the problem?A. Prepare a better lunch for their kids.B. Stop their kids going to the cafeteria.C. Force schools to make adjustments to lunch.D. Guide their kids on how to pack their own lunch.CEarthquakes are a natural disaster—except when they're man-made. The oil and gas industry has forcefully used the technique known as hydraulic fracturing (水力压裂法) to destroy sub-surface rock and liberate the oil and gas hiding there. But the process results in large amounts of chemical-filled waste water. Horizontal drilling (水平钻孔) for oil can also produce large amount of natural, unwanted salt water. The industry deals with this waste water by pumping it into deep wells.On Monday, the US Geological Survey published for the first time an earthquake disaster map covering both natural and “induced” quakes. The map and a report show that parts of the central United States now face a ground-shaking disaster equal to the famously unstable terrain (不稳定地形) of California.Some 7 million people live in places easily attacked by these man-made quakes, the USGS said The list of places at highest risk of man-made earthquakes includes Oklahoma, Kansas, Texas, Arkansas, Colorado, New Mexico, Ohio and Alabama. Most of these earthquakes are ly small, in the range of magnitude (震级) 3, but some have been more powerful, including a magnitude 5.6 earthquake in 2011 in Oklahoma that was connected to waste water filling.Scientists said they do not know ifthere is an upper limit on the magnitude of man-made earthquakes; this is an area of active research Oklahoma has had prehistoric earthquakes as powerful as magnitude 7.It's not immediately clear whether this new research will change industry practices, or even whether it will surprise anyone in the areas of newly supposed danger. In Oklahoma, for example, the natural rate of earthquakes is only one or two a year, but there have been hundreds since hydraulic fracturing and horizontal drilling, with the waste water filling, became common in the last ten years.8. What kind of human activities can cause earthquakes?A. The man-made produced waste water in the factories.B. The process of digging deep wells in those poor areas.C. The advanced techniques used to deal with waste water.D. The oil or gas industry's work connected with the earth.9. What does the underlined word “induced” in paragraph 2 mean?A. Man-made.B. Reduced.C. Newly-built.D. Controlled.10. How much magnitude can man-made earthquakes reach?A. It's been said as small as magnitude 3.B. It has been said as high as magnitude 7.C. It's being studied without a final conclusion.D. It has risen by an average of magnitude 5. 6.11. What is the best title for the text?A. Natural Earthquakes in America Are Disappearing NowB. 7 Million Americans at Risk of Man-Made EarthquakesC. Time for Oil and Gas Industry Change Their Working PracticeD. More Often Earthquakes as Powerful as Magnitude 7 in AmericaD“Heavy hearts, like heavy clouds in the sky, are best relieved by the letting of a little water, the French writer Antoine de Rivarol wrote. This love letter to the cleansing beauty of a good cry is a comforting thought at atime when the continuing stress of the COVID-19 has added heaviness to each of our lives.Scientifically, de Rivarol's poetic image doesn't, if you'll forgive the words used in the poem, hold water. There's limited research on crying, partly because of the difficulty of copying the behavior of real crying in a lab.But even within the previous studies, there's little evidence to suggest that crying provides a physiological cleansing of poisons in people's body.Psychologists believe the relief of a good cry connects with a different emotional process. “It seems that crying occurs just after the peak of the emotional experience, and crying is associated with this return to homeostasis: the process of maintaining a stable psychological state,” said Lauren Bylsma. He also said holding back tears can have negative physical consequences, including headaches and muscle tension. Such restriction can also limit our experiences of joy, gratitude and other positive emotions if we avoid acknowledging our feelings.For me crying has been easier said than done during the COVID-19. Psychologists say it's normal to feel stopped up by the stresses of the past year. We should find opportunities to release and process our emotions.Watching a tear-jerking movie, having an emotional conversation with a close friend, and writing in a journal are healthy ways toelicita cry. Physical activity like light-footed walking or even dancing can also signal our bodies to release some emotional tightness. We can then open up to the flow of feelings that leave us feeling lighter and refreshed—like a clear sky after a soaking rain.12. What is the weakness of the studies ever clone on crying?A. They were clone in a laboratory setting.B. They cared little about different forms of crying.C. They were always concentrated on people's daily life.D. They showed little about the positive physical effect of crying.13. What is the function of crying according to Lauren Bylsma?A. Curing people of their diseases.B. Keeping emotionally balanced.C. Producing negative mental results.D. Expanding people's experience of joy.14. What does the underlined word “elicit” in the last paragraph mean?A. Produce.B. Postpone.C. Control.D. Repeat.15. What are people advised to do according to the text?A. Learn to hold back their tears wisely.B. Share their emotion with their colleagues.C. Have a good cry when necessary.D. Try to avoid admitting our feelings.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。
2021届江苏省常州市四校高三上学期期末联考数学试题
绝密★启用前江苏省常州市四校联考2021届高三上学期期末测试数学试题注意事项：1、答题前填写好自己的姓名、班级、考号等信息 2、请将答案正确填写在答题卡上一、单项选择题（本大题共8小题，每小题5分，共计40分．在每小题给出的四个选项中，只有一个是符合题目要求的，请把答案添涂在答题卡相应位置上）1．已知集合A ＝{﹣2，﹣1，0，1，2}，B ＝{}21A y y x x =+∈，，则AB ＝A ．∅B ．{1，2}C ．{﹣2，0，2}D ．{﹣2，﹣1，1，2} 2．当复数20211i ()i 1i=a a -+时，实数a 的值可以为3附：2()()()()K a b c d a c b d =++++，其中n a b c d =+++；与年龄段无关”B ．在犯错误的概率不超过0.1%的前提下，认为“能接种与年龄段有关”C ．有99%以上的把握认为“能接种与年龄段无关”D ．有99%以上的把握认为“能接种与年龄段有关”4．函数()x xf x -=ABCD5．设随机变量N ξ(μ，1)，函数2()2f x x x ξ=+-没有零点的概率是0.5，则P(0＜ξ≤1)＝ 附：若N ξ(μ，2σ)，则P(μσ-＜X ≤μσ+)≈0.6826，P(2μσ-＜X ≤2μσ+)≈0.9544．A ．0.1587B ．0.1359C ．0.2718D ．0.34136．在探索系数A ，ω，ϕ，b 对函数Asin()y x b ωϕ=++(A ＞0，ω＞0)图象的影响时，我们发现，系数A 对其影响是图象上所有点的纵坐标伸长或缩短，通常称为“振幅变换”；系数ω对其影响是图象上所有点的横坐标伸长或缩短，通常称为“周期变换”；系数ϕ对其影响是图象上所有点向左或向右平移，通常称为“左右平移变换”；系数b 对其影响是图象上所有点向上或向下平移，通常称为“上下平移变换”．运用上述四种变换，若函数()sin f x x =的图象经过四步变换得到函数()2sin(2)13g x x π=-+的图象，且已知其中有一步是向右平移3π个单位，则变换的方法共有 A ．6种B ．12种C ．16种D ．24种 7．俄国著名飞机设计师埃格•西科斯基设计了世界上第一架四引擎飞机和第一种投入生产的直升机，当代著名的“黑鹰”直升机就是由西科斯基公司生产的．1992年，为了远程性和安全性上与美国波音747竞争，欧洲空中客车公司设计并制造了A340，是一种有四台发动机的远程双过道宽体客机，取代只有两台发动机的A310．假设每一架飞机的引擎在飞行中出现故障率为1﹣p ，且各引擎是否有故障是独立的，已知A340飞机至少有3个引擎正常运行，飞机就可成功飞行；A310飞机需要2个引擎全部正常运行，飞机才能成功飞行．若要使A340飞机比A310飞机更安全，则飞机引擎的故障率应控制的范围是 A ．(23，1)B ．(13，1)C ．(0，23)D ．(0，13) 8．已知数列{}n a 满足211n n n a a a +=-+(*N n ∈)，设12111n n S a a a =+++，且10910231a S a -=-，则数列{}n a 的首项1a 的值为A ．23B ．1C ．32D ．2 二、多项选择题（本大题共4小题，每小题5分，共计20分．在每小题给出的四个选项中，至少有两个是符合题目要求的，请把答案添涂在答题卡相应位置上）9．2020年的“金九银十”变成“铜九铁十”，全国各地房价“跳水”严重，但某地二手房交易却“逆市”而行．下图是该地某小区2019年12月至2020年12月间，当月在售二手房均价（单位：万元/平方米）的散点图．（图中月份代码1~13分别对应2019年12月~2020年12月）根据散点图选择y a =+和ln y c d x =+两个模型进行拟合，经过数据处理得到的两个回归方程分别为0.9369y =+0.95540.0306ln y x =+，并得到以下一些统计量的值：注：x y A ．当月在售二手房均价y 与月份代码x 呈负相关关系B ．由0.9369y =+2021年3月在售二手房均价约为1.0509万元/平方米C ．曲线0.9369y =+0.95540.0306ln y x =+都经过点(x ，y )D ．模型0.95540.0306ln y x =+回归曲线的拟合效果比模型0.9369y =+ 10．若22012(1)+(1)++(1)n n n x x x a a x a x a x +++=++++，且121125n a a a n -+++=-，则下列结论正确的是 A ．n ＝6B ．(12)n x +展开式中二项式系数和为729C ．2(1)+(1)++(1)n x x x +++展开式中所有项系数和为126D ．12323321n a a a na ++++=11．已知抛物线C ：22(0)y px p =>的焦点为F ，过F 的直线l 交抛物线C 于点A ，B ，且A(4p，a )，3AF 2=.下列结论正确的是A ．p ＝4B ．a =C ．BF ＝3D ．△AOB 12．若函数()f x 是连续的平滑曲线，且在[a ，b]上恒非负，则其图象与直线x ＝a ，x ＝b ，x轴围成的封闭图形面积称为()f x 在[a ，b]上的“围面积”．根据牛顿—莱布尼兹公式，计算围面积时，若存在函数()F x 满足()()F x f x '=，则()()F b F a -的值为()f x 在[a ，b]上的围面积．下列围面积计算正确的有A ．函数3()f x x =在[0，1]上的围面积为14B ．函数()2x f x =在[0，2]上的围面积为2ln 2C ．函数2()cos f x x =在[0，4π]上的围面积为148π+D ．函数()ln f x x =在[e ，e 2]上的围面积为e 2三、填空题（本大题共4小题，每小题5分，共计20分．请把答案填写在答题卡相应位置上）13．在四边形ABCD 中，AB ＝8，若31DA CA CB 44=+，则AB CD ⋅＝ ．14．在平面直角坐标系xOy 中，设双曲线C ：22221x y a b-=(a ＞0，b ＞0)的右焦点为F ，若双曲线的右支上存在一点P ，使得△OPF 是以P 为直角顶点的等腰直角三角形，则双曲线C 的离心率为 ．15．在△ABC 中，已知AC ＝1，∠A 的平分线交BC 于D ，且AD ＝1，BD，则△ABC 的面积为 ．16．矩形ABCD 中，ABBC ＝1，现将△ACD 沿对角线AC 向上翻折，得到四面体D —ABC ，则该四面体外接球的体积为 ；设二面角D —AC —B 的平面角为θ，当θ在[3π，2π]内变化时，BD 的范围为 ．（第一空2分，第二空3分） 四、解答题（本大题共6小题，共计70分．请在答题卡指定区域内作答．解答时应写出文字说明、证明过程或演算步骤） 17．（本小题满分10分）在△ABC 中，a ，b ，c 分别为角A ，B ，C 所对的边．在①(2)cosB cosC a c b -=；②ABC BC=2S ⋅△；③sin B sin(B )3π++=这三个条件中任选一个，作出解答．（1）求角B 的值；（2）若△ABC 为锐角三角形，且b ＝1，求△ABC 的面积的取值范围． 18．（本小题满分12分）某公司在市场调查中，发现某产品的单位定价x （单位：万元/吨）对月销售量y （单位：吨）有影响．对不同定价i x 和月销售量1, 2,8()i y i =数据作了初步处理：表中x z =．经过分析发现可以用xa y +=来拟合y 与x 的关系． （1）求y 关于x 的回归方程；（2）若生产1吨产品的成本为1.6万元，那么预计价格定位多少时，该产品的月利润取最大值，求此时的月利润．附：对于一组数据(1w ，1v )，(2w ，2v )，…，(n w ，n v )，其回归直线v αβω=+的斜率和截距的最小二乘估计分别为：^1122211()()()n niii ii i nniii i v n v v vn ωωωωβωωωω====---⋅==--⋅∑∑∑∑，v αβω=-．19．（本小题满分12分）已知等差数列{}n a 和等比数列{}n b 满足13a =，12b =，2221a b =-，333a b =+． （1）求{}n a 和{}n b 的通项公式；（2）将{}n a 和{}n b 中的所有项按从小到大的顺序排列组成新数列{}n c ，求数列{}n c 的前100项和100S ．20．（本小题满分12分）在多面体ABCDE 中，平面ACDE ⊥平面ABC ，四边形ACDE 为直角梯形，CD ∥AE ，AC ⊥AE ，AB ⊥BC ，CD ＝1，AE ＝AC ＝2，F 为DE 的中点，且点E 满足EB 4EG =．（1）证明：GF ∥平面ABC ；（2）当多面体ABCDE 的体积最大时，求二面角A —BE —D 的余弦值．21．（本小题满分12分）已知函数()(0)ln axf x a x=>． （1）当函数()f x 在1ex =处的切线斜率为﹣2时，求()f x 的单调减区间；（2）当x ＞1时，ln()e ln x x xa f x x≥⋅，求a 的取值范围．22．（本小题满分12分）已知椭圆C ：22221(0)x y a b a b +=>>的离心率为12，且过点A(2，3)，右顶点为B ．（1）求椭圆C 的标准方程；（2）过点A 作两条直线分别交椭圆于点M ，N ，满足直线AM ，AN 的斜率之和为﹣3，求点B 到直线MN 距离的最大值．江苏省常州市四校联考2021届高三上学期期末测试数学试题2021．01一、单项选择题（本大题共8小题，每小题5分，共计40分．在每小题给出的四个选项中，只有一个是符合题目要求的，请把答案添涂在答题卡相应位置上）1．已知集合A ＝{﹣2，﹣1，0，1，2}，B ＝{}21A y y x x =+∈，，则AB ＝A ．∅B ．{1，2}C ．{﹣2，0，2}D ．{﹣2，﹣1，1，2} 答案：D 2．当复数20211i ()i 1i=a a -+时，实数a 的值可以为 A ．0B ．1C ．﹣1D ．±1 答案：C3附：2()()()()()n ad bc K a b c d a c b d -=++++，其中n a b c d =+++；与年龄段无关”B ．在犯错误的概率不超过0.1%的前提下，认为“能接种与年龄段有关”C ．有99%以上的把握认为“能接种与年龄段无关”D ．有99%以上的把握认为“能接种与年龄段有关” 答案：D 4．函数()x x f x -=ABCD 答案：A 5．设随机变量N ξ(μ，1)，函数2()2f x x x ξ=+-没有零点的概率是0.5，则P(0＜ξ≤1)＝ 附：若N ξ(μ，2σ)，则P(μσ-＜X ≤μσ+)≈0.6826，P(2μσ-＜X ≤2μσ+)≈0.9544．A ．0.1587B ．0.1359C ．0.2718D ．0.3413 答案：B6．在探索系数A ，ω，ϕ，b 对函数Asin()y x b ωϕ=++(A ＞0，ω＞0)图象的影响时，我们发现，系数A 对其影响是图象上所有点的纵坐标伸长或缩短，通常称为“振幅变换”；系数ω对其影响是图象上所有点的横坐标伸长或缩短，通常称为“周期变换”；系数ϕ对其影响是图象上所有点向左或向右平移，通常称为“左右平移变换”；系数b 对其影响是图象上所有点向上或向下平移，通常称为“上下平移变换”．运用上述四种变换，若函数()sin f x x =的图象经过四步变换得到函数()2sin(2)13g x x π=-+的图象，且已知其中有一步是向右平移3π个单位，则变换的方法共有 A ．6种B ．12种C ．16种D ．24种 答案：B 7．俄国著名飞机设计师埃格•西科斯基设计了世界上第一架四引擎飞机和第一种投入生产的直升机，当代著名的“黑鹰”直升机就是由西科斯基公司生产的．1992年，为了远程性和安全性上与美国波音747竞争，欧洲空中客车公司设计并制造了A340，是一种有四台发动机的远程双过道宽体客机，取代只有两台发动机的A310．假设每一架飞机的引擎在飞行中出现故障率为1﹣p ，且各引擎是否有故障是独立的，已知A340飞机至少有3个引擎正常运行，飞机就可成功飞行；A310飞机需要2个引擎全部正常运行，飞机才能成功飞行．若要使A340飞机比A310飞机更安全，则飞机引擎的故障率应控制的范围是A ．(23，1)B ．(13，1)C ．(0，23)D ．(0，13) 答案：C8．已知数列{}n a 满足211n n n a a a +=-+(*N n ∈)，设12111n n S a a a =+++，且10910231a S a -=-，则数列{}n a 的首项1a 的值为A ．23B ．1C ．32D ．2 答案：C二、多项选择题（本大题共4小题，每小题5分，共计20分．在每小题给出的四个选项中，至少有两个是符合题目要求的，请把答案添涂在答题卡相应位置上）9．2020年的“金九银十”变成“铜九铁十”，全国各地房价“跳水”严重，但某地二手房交易却“逆市”而行．下图是该地某小区2019年12月至2020年12月间，当月在售二手房均价（单位：万元/平方米）的散点图．（图中月份代码1~13分别对应2019年12月~2020年12月）根据散点图选择y a =+和ln y c d x =+两个模型进行拟合，经过数据处理得到的两个回归方程分别为0.9369y =+0.95540.0306ln y x =+，并得到以下一些统计量的值：注：x y A ．当月在售二手房均价y 与月份代码x 呈负相关关系B ．由0.9369y =+2021年3月在售二手房均价约为1.0509万元/平方米C ．曲线0.9369y =+0.95540.0306ln y x =+都经过点(x ，y )D ．模型0.95540.0306ln y x =+回归曲线的拟合效果比模型0.9369y =+ 答案：BD10．若22012(1)+(1)++(1)n n n x x x a a x a x a x +++=++++，且121125n a a a n -+++=-，则下列结论正确的是 A ．n ＝6B ．(12)n x +展开式中二项式系数和为729C ．2(1)+(1)++(1)n x x x +++展开式中所有项系数和为126D ．12323321n a a a na ++++=答案：ACD11．已知抛物线C ：22(0)y px p =>的焦点为F ，过F 的直线l 交抛物线C 于点A ，B ，且A(4p，a)，3AF 2=.下列结论正确的是A ．p ＝4B ．a =C ．BF ＝3D ．△AOB 答案：BCD12．若函数()f x 是连续的平滑曲线，且在[a ，b]上恒非负，则其图象与直线x ＝a ，x ＝b ，x轴围成的封闭图形面积称为()f x 在[a ，b]上的“围面积”．根据牛顿—莱布尼兹公式，计算围面积时，若存在函数()F x 满足()()F x f x '=，则()()F b F a -的值为()f x 在[a ，b]上的围面积．下列围面积计算正确的有A ．函数3()f x x =在[0，1]上的围面积为14B ．函数()2x f x =在[0，2]上的围面积为2ln 2C ．函数2()cos f x x =在[0，4π]上的围面积为148π+D ．函数()ln f x x =在[e ，e 2]上的围面积为e 2答案：ACD三、填空题（本大题共4小题，每小题5分，共计20分．请把答案填写在答题卡相应位置上）13．在四边形ABCD 中，AB ＝8，若31DA CA CB 44=+，则AB CD ⋅＝ ．答案：﹣16解析：根据题意可知四边形ABCD 是梯形，且1CD AB 4=-，所以21AB CD AB 164⋅=-=-．14．在平面直角坐标系xOy 中，设双曲线C ：22221x y a b-=(a ＞0，b ＞0)的右焦点为F ，若双曲线的右支上存在一点P ，使得△OPF 是以P 为直角顶点的等腰直角三角形，则双曲线C 的离心率为 ．15．在△ABC 中，已知AC ＝1，∠A 的平分线交BC 于D ，且AD ＝1，BD ，则△ABC 的面积为 ．16．矩形ABCD 中，AB BC ＝1，现将△ACD 沿对角线AC 向上翻折，得到四面体D —ABC ，则该四面体外接球的体积为 ；设二面角D —AC —B 的平面角为θ，当θ在[3π，2π]内变化时，BD 的范围为 ．（第一空2分，第二空3分）答案：43π，]四、解答题（本大题共6小题，共计70分．请在答题卡指定区域内作答．解答时应写出文字说明、证明过程或演算步骤） 17．（本小题满分10分）在△ABC 中，a ，b ，c 分别为角A ，B ，C 所对的边．在①(2)cosB cosC a c b -=；②ABC BC=2S ⋅△；③sin B sin(B )3π++=这三个条件中任选一个，作出解答．（1）求角B 的值；（2）若△ABC 为锐角三角形，且b ＝1，求△ABC 的面积的取值范围． 解：（1）选①由正弦定理得：2sin cos sin cos sin cosC A B C B B -= 2sin cos sin A B A ∴=()0,sin 0A A π∈∴>1cos 2B ∴=()0,3B B ππ∈∴=选②32ABC BA BC S⋅=1cos 2sin 2B ac B =⋅sin B B ∴=()0,sin 0cos 0B B B π∈∴>∴>3B π∴=选③1sin sin 22B B B ++=1cos 12B B +=sin 16B π⎛⎫∴+= ⎪⎝⎭()70,,666B B ππππ⎛⎫∈∴+∈ ⎪⎝⎭623B B πππ∴+=∴=（2）由正弦定理得：sin sin sin a b c A B C ===,a A c C ∴==12sin sin23S ac B A A π⎛⎫∴==- ⎪⎝⎭26A π⎛⎫=- ⎪⎝⎭ 锐角三角形ABC02262032A A C A πππππ⎧<<⎪⎪∴⇒<<⎨⎪<=-<⎪⎩52,666A πππ⎛⎫∴-∈ ⎪⎝⎭64S ⎛∴∈ ⎝⎦18．（本小题满分12分） 某公司在市场调查中，发现某产品的单位定价x （单位：万元/吨）对月销售量y （单位：吨）有影响．对不同定价i x 和月销售量1, 2,8()i y i =数据作了初步处理：表中x z =．经过分析发现可以用xa y +=来拟合y 与x 的关系． （1）求y 关于x 的回归方程；（2）若生产1吨产品的成本为1.6万元，那么预计价格定位多少时，该产品的月利润取最大值，求此时的月利润．附：对于一组数据(1w ，1v )，(2w ，2v )，…，(n w ，n v )，其回归直线v αβω=+的斜率和截距的最小二乘估计分别为：^1122211()()()n niii ii i nni i i i v n v v vn ωωωωβωωωω====---⋅==--⋅∑∑∑∑，v αβω=-．解：（1）令1z x=，则y a b z =+⋅ 则8^1822123956894358208988i ii i i z y z yb z z ==-⋅⋅===-⋅--∑∑ ^^2a y b z =-⋅=- ^52y x∴=-+（2）月利润()()^58 1.62 1.68.22y x x x x x T ⎛⎫⎛⎫⋅-=--=-+⎪ ⎝⎭⎝=⎪⎭8.20.2≤-=（当且仅当82x x=即2x =时取等号） 答：（1）y 关于x 的回归方程为^52y x=-+； （2）预计价格定位2万元/吨时，该产品的月利润取最大值，最大值为0.2万元 19．（本小题满分12分）已知等差数列{}n a 和等比数列{}n b 满足13a =，12b =，2221a b =-，333a b =+． （1）求{}n a 和{}n b 的通项公式；（2）将{}n a 和{}n b 中的所有项按从小到大的顺序排列组成新数列{}n c ，求数列{}n c 的前100项和100S ．解：（1）由223221443223d q d q d q a q+=⋅-=-⎧⎧⇒⎨⎨+=⋅+=⎩⎩ 2,4q d ∴==41,2n n n a n b ∴=-=（2）当{}n c 的前100项中含有{}n b 的前7项时，令841225664.25n n -<=⇒< 此时至多有64771+=项（不符）当{}n c 的前100项中含有{}n b 的前8项时，令9412512128.25n n -<=⇒< 则{}n c 的前100项中含有{}n b 的前8项且含有{}n a 的前92项()8100212929192341702051017530212S -⨯⎛⎫∴=⨯+⨯+=+= ⎪-⎝⎭20．（本小题满分12分）在多面体ABCDE 中，平面ACDE ⊥平面ABC ，四边形ACDE 为直角梯形，CD ∥AE ，AC ⊥AE ，AB ⊥BC ，CD ＝1，AE ＝AC ＝2，F 为DE 的中点，且点E 满足EB 4EG =．（1）证明：GF ∥平面ABC ；（2）当多面体ABCDE 的体积最大时，求二面角A —BE —D 的余弦值．解：（1）分别取EB AB ,中点N M ,，连结ND MN CM ,,. 在梯形ACDE 中，EA DC //且EA DC 21=，且N M ,分别为BE BA ,中点EA MN EA MN 21,//=∴CD MN CD MN =∴,//∴四边形CDNM 是平行四边形DN CM //∴又EB EG 41=，N 为EB 中点，G ∴为EN 中点，又F 为ED 中点DN GF //∴ CM GF //∴又⊂CM 平面ABC ，⊄GF 平面ABC //GF ∴平面ABC（2）在平面ABC 内，过B 作AC BH ⊥交AC 于H .平面⊥ACDE 平面ABC ，平面 ACDE 平面AC ABC =， ⊂BH 平面ABC ，AC BH ⊥， ∴⊥BH 平面ACDEBH ∴即为四棱锥ACDE B -的高，又底面ACDE 面积确定，所以要使多面体ABCDE 体积最大，即BH 最大，此时2AB BC ==过点H 作AE ，易知HP HC HB ,,两两垂直，以{}HP HC HB ,,为正交基底建立如图所示的平面直角坐标系xyz H -则)1,1,0(),2,1,0(),0,0,1(),0,1,0(D E B A --)1,2,0(),2,1,1(),0,1,1(-=--==DE BE AB设),,(1111z y x n =为平面ABE 的一个法向量，则⎪⎩⎪⎨⎧=⋅=⋅011BE n AB n ，所以⎩⎨⎧=+--=+02011111z y x y x ，取)0,1,1(1-=n 设),,(2222z y x n =为平面DBE 的一个法向量，则⎪⎩⎪⎨⎧=⋅=⋅022BE n DE n ，所以⎩⎨⎧=+--=+-020222222z y x z y ，取)2,1,3(2=n 所以77||||,cos 212121=⋅>=<n n n n n n ， 由图，二面角D BE A --为钝二面角，所以二面角D BE A --的余弦值为77- 21．（本小题满分12分）已知函数()(0)ln axf x a x=>． （1）当函数()f x 在1ex =处的切线斜率为﹣2时，求()f x 的单调减区间；（2）当x ＞1时，ln()e ln x x xa f x x≥⋅，求a 的取值范围．解：（1）()ln axf x x=定义域为()()0,11,+∞．因为()''ln ax f x x ⎛⎫== ⎪⎝⎭所以()f x 在1x e=处的切线斜率为2a -． 所以1a =．所以()ln x f x x =，()''ln x f x x ⎛⎫== ⎪⎝⎭令()'0fx =，则x e =（2）由题()lnln x xa f x e x≥⋅对任意),1(+∞∈x 恒成立所以ln ln xae x a ≥-对任意),1(+∞∈x 恒成立方法一：所以()ln ln ln a xe a x x x +++≥+对任意),1(+∞∈x 恒成立所以()ln ln ln ln a xx ea x e x +++≥+对任意),1(+∞∈x 恒成立令()xg x e x =+则()()ln ln g a x g x +≥对任意),1(+∞∈x 恒成立 因为()'10xg x e =+>所以()g x 在R 上单调增所以ln ln a x x +≥对任意),1(+∞∈x 恒成立 所以()()max ln ln 1a x x x ≥-> 令()()ln 1h x x x x =->因为()'1110x g x x x-=-=< 所以()g x 在(1,)+∞上单调减所以()()11g x g <=-所以ln 1a ≥-即1a e ≥方法二：设)1(ln ln )(>+-=x a x ae x h x，则01)(''1)('2>+=-=xae x h x ae x h x x ，，所以)('x h 在),1(+∞单调递增，又1)1('-=ae h若ea 1≥，则0)1('≥h ，所以0)('≥x h 恒成立，所以)('x h 在),1(+∞单调递增，又011ln )1(=-≥+=a ae h ，所以0)(≥x h 恒成立，符合题意.若ea 10<<，则011ln )1(=-<+=a ae h ，不符合题意，舍去. 综上所述，ea 1≥.22．（本小题满分12分）已知椭圆C ：22221(0)x y a b a b +=>>的离心率为12，且过点A(2，3)，右顶点为B ．（1）求椭圆C 的标准方程；（2）过点A 作两条直线分别交椭圆于点M ，N ，满足直线AM ，AN 的斜率之和为﹣3，求点B 到直线MN 距离的最大值．解：（1）由题222224122491b c a a c b a c a b ⎧⎪+==⎧⎪⎪⎪=⇒=⎨⎨⎪⎪=⎩⎪+=⎪⎩ 所以C 的标准方程为1121622=+y x （2）若直线MN 斜率不存在，设),(),,(0000y x N y x M -，则⎩⎨⎧==⇒⎪⎪⎩⎪⎪⎨⎧-=---+--=+0432323112160000002020y x x y x y y x ，此时N M ,重合，舍去. 若直线MN 斜率存在，设),(),,(2211y x N y x M t kx y MN ，：+=，联立⎪⎩⎪⎨⎧+==+t kx y y x 1121622得04848)34(222=-+++t ktx x k ，所以34484,3482221221+-=+-=+k t x x k kt x x由题323232211-=--+--x y x y ，即323232211-=--++--+x t kx x t kx化简得.0244))(92()32(2121=+-+--++t x x k t x x k因此.0244)348)(92(34484)32(222=+-+---++-+t k ktk t k t k 化简得0686822=---++t k t kt k 即0)24)(32(=++-+t k t k若032=-+t k ，则32+-=k t ，直线MN 过点)3,2(A ，舍去， 所以024=++t k ，即24--=k t ，因此直线MN 过点)2,4(-P . 又点)0,4(B ，所以点B 到直线MN 距离最大值即2=BP , 此时2-=y MN ：，符合题意.所以点B 到直线MN 距离最大值为2。
2021年G4学校高三教学情况期末调研 语文
盐城中学·苏州中学·扬州中学·常州中学2021年G4学校高三教学情况期末调研语文2021.01一、现代文阅读（一）现代文阅读Ⅰ（本题共5小题，19分）阅读下面的文字，完成1-5题。
材料一：高尔基说：“一般说来，神话乃是自然现象，对自然的斗争，以及社会生活在广大的艺术概括中的反映。
”这就说明了神话的产生，是基于现实生活，而并不是出于人类头脑里的空想。
所以当我们研究神话的起源，古代每一时期的神话所包含的特定意义等诸如此类的问题的时候，都不能离开当时人类的现实生活、劳动和斗争而作凭空的推想。
中国神话的“源”，求诸古籍记载，自然最早莫过于属于巫书性质的《山海经》。
它实际上是从战国初年到汉代初年这一段长时间内众多无名氏的作品，初步推断可能是楚地和巴地的人所作，有巫师和文人参与其事。
但是追本溯源，还应当推寻到传说中夏禹、伯益那个历史时代。
好些神话故事经由那个时代的酋长而兼巫师身份的人物，口头直接传承下来乃是大有可能的。
根据我的研究，万物有灵论时期已是神话的初步发展阶段，《山海经》所记载的神话，大都属此阶段。
但在前万物有灵论时期，即已有萌芽状态的神话产生了。
这个时期相当于马克思在《摩尔根﹤古代社会﹥一书摘要》中所说的蒙昧时期的中级阶段，亦即以生产方式为分期的旧石器时期的中期。
这个时期产生的神话，多以动植物为主要描述的对象，尤其着重叙写的是动物，性质和后世的童话、寓言相近。
我称这个时期的神话为活物论神话，以别于万物有灵论时期的神话。
那时候的人们，刚从动物分离出来不久，还存在着物我混同的原始思维的心理状态，视眼前的万物，不论是动物植物，或山川日月星辰风雨云霞等，都认为是和自己一样有生命有意志的活物，由此而在集体无意识中产生的叙写它们之间或它们与人类交往的故事，就是最早时期的神话——活物论神话。
原始的宗教思想萌芽于此，图腾主义也由此而来。
但《山海经》保留这种神话已经不多了，只还有两三个残片遗存其中，较多的是保留在先秦时代的寓言里。
江苏省G4(苏州中学、盐城中学、扬州中学、常州中学)2021届高三上学期期末调研化学卷
+19 2 88 1 2021届盐城中学、常州中学、苏州中学、扬州中学四校联考高三化学可能用到的相对原子质量：H 1 C 12 O 16 S 32 Fe 56一、单项选择题：共13题，每题3分，共39分。
每小题只有一个选项最符合题意。
1．现代社会的发展与进步离不开材料，下列有关材料的说法不正确．．．的是 A ．“中国天眼”球面射电望远镜的钢铁“眼眶”属于新型无机非金属材料 B ．“天宫二号”的硅太阳能电池板可将光能直接转换为电能 C ．北京大兴国际机场航站楼的多面体玻璃属于硅酸盐材料 D ．国庆阅兵仪式上坦克喷涂的聚氨酯涂料属于有机高分子材料2．某酒精检验器的工作原理为2K 2Cr 2O 7＋3C 2H 5OH ＋8H 2SO 4＝3CH 3COOH ＋2Cr 2(SO 4)3＋2K 2SO 4+11H 2O 。
下列说法正确的是A ．Cr 元素基态原子的核外电子排布式为[Ar]3d 44s 2B ．固态C 2H 5OH 是分子晶体C ．H 2O 的电子式为H ＋[∶··O ··∶]2－H ＋D ．K +的结构示意图为3．下列有关物质性质与用途具有对应关系的是A ．二氧化硫具有漂白性，可用作制溴工业中溴的吸收剂B ．苯的密度比H 2O 小，可用于萃取碘水中的碘C ．Na 具有强还原性，可用于和TiCl 4溶液反应制备TiD ．Mg 2Si 3O 8·nH 2O 能与酸反应，可用于制胃酸中和剂阅读下列材料，完成4-6题：2021年1月20日中国科学院和中国工程院评选出2020年世界十大科技进展，排在第四位的是一种可借助光将二氧化碳转化为甲烷的新型催化转化方法：CO 2+4H 2＝CH 4+2H 2O ，这是迄今最接近人造光合作用的方法。
4．下列有关CO 2、CH 4的说法正确的是 A ．CO 2的空间构型是V 形B ．电负性由大到小的顺序是O>C>HC ．CH 4是极性分子D ．CO 2转化为CH 4利用了CO 2的还原性5．CO 2加氢制CH 4的一种催化机理如图，下列说法正确的是 A ．反应中La 2O 3是中间产物B ．反应中La 2O 2CO 3可以释放出带负电荷的CO 2·C ．H 2经过Ni 活性中心裂解产生活化态H ·的过程中ΔS>0D ．使用TiO 2作催化剂可以降低反应的焓变，从而提高化学反应速率6．某光电催化反应器如图所示，A 电极是Pt/CNT ，B 电极是TiO 2。
江苏省G4联盟(苏州中学、扬州中学、盐城中学、常州中学)22-23学年高三12月联考数学试题 附答案
G4联盟—苏州中学、扬州中学、常州中学、盐城中学2022－2023学年第一学期12月联合调研高三数学一、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．设集合A ＝{－1，0}，B ＝{x |－2＜x ＜0}，则A ∩B ＝ A ．{－1}B ．{－1，0}C ．{x |－2＜x ＜0}D ．{x |－2＜x ≤0}2．若复数z 的共轭复数z 满足i ⋅z ＝4＋3i （其中i 为虚数单位），则z z ⋅的值为AB ．5C ．7D ．253．下图是近十年来全国城镇人口、乡村人口的折线图（数据来自国家统计局）．根据该折线图，下列说法错误的是 A ．城镇人口与年份星现正相关B ．乡村人口与年份的相关系数r 接近1C ．城镇人口逐年增长率大致相同D ．可预测乡村人口仍呈现下降趋势4．函数y ＝2x 2－e |x |在[－2，2]的图象大致为A ．B ．C ．D ．5．若椭圆的焦点为F 1，F 2，过F 1的最短弦PQ 的长为10，△PF 2Q 的周长为36，则此椭圆的离心率为A ．13B ．3C ．23D ．36．南宋时期，秦九韶就创立了精密测算雨量、雨雪的方法，他在《数学九章》载有“天池盆测雨”题，使用一个圆台形的天池盆接雨水．观察发现体积一半时的水深大于盆高的一半，体积一半时的水面面积大于盆高一半时的水面面积，若盆口半径为a ，盆地半径为b （0＜b ＜a ），根据如上事实，可以抽象出的不等关系为A <B <C ．22222a b a b ++⎛⎫<⎪⎝⎭D ．33322a b a b ++⎛⎫<⎪⎝⎭7．在数列{a n }中，()()111sin sin 10n n n n a a a a ++-⋅+=，则该数列项数的最大值为 A ．9B ．10C ．11D ．128．在△ABC 中，AB ＝4，BC ＝3，CA ＝2，点P 在该三角形的内切圆上运动，若AP mAB nAC =+（m ，n 为实数），则m ＋n 的最小值为 A ．518B ．13C ．718D ．49二、选择题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分．9．已知a ＞0，b ＞0，a ＋b ＝1，则A ．114a b+≤B ．22a b+≥C ．log 2a ＋log 2b ≤－2D ．1sin sin 2sin2a b +≤ 10．已知函数()x a a x f x e e --=+，()x a a x g x e e --=-，则 A ．函数y ＝g （x ）有且仅有一个零点B ．f ′（x ）＝g （x ）且g ′（x ）＝f （x ）C ．函数y ＝f （x ）g （x ）的图象是轴对称图形D ．函数()()g x y f x =在R 上单调递增 11．乒乓球（tabletennis ），被称为中国的“国球”，是一种世界流行的球类体育项目，是推动外交的体育项目，被誉为“小球推动大球”．某次比赛采用五局三胜制，当参赛甲、乙两位中有一位赢得三局比赛时，就由该选手晋级而比赛结束．每局比赛皆须分出胜负，且每局比赛的胜负不受之前已赛结果影响．假设甲在任一局赢球的概率为p （0≤p ≤1），实际比赛局数的期望值记为f （p ），下列说法正确的是 A ．三局就结束比赛的概率为p 3＋（1－p ）3B ．f （p ）的常数项为3C ．1435f f ⎛⎫⎛⎫< ⎪⎪⎝⎭⎝⎭D ．13328f ⎛⎫=⎪⎝⎭ 12．在四棱锥P －ABCD 中，底面ABCD 为正方形，P A ⊥底面ABCD ，P A ＝AB ＝1．G 为PC 的中点，M 为平面PBD 上一点下列说法正确的是A ．MGB ．若MA ＋MG ＝1，则点M 的轨迹是椭圆C．若MA =M 的轨迹围成图形的面积为12π D ．存在点M ，使得直线BM 与CD 所成角为30°三、填空题：本题共4小题，每小题5分，共20分．13．在6x ⎛⎝的展开式中，常数项为 ．14．如图，将绘有函数()sin 2f x M πϕ⎛⎫=+⎪⎝⎭（M ＞0，0＜φ＜π）部分图象的纸片沿x 轴折成直二面角，此时A ，Bφ＝ ．15．我们利用“错位相减”的方法可求等比数列的前n 项和，进而可利用该法求数列{（2n －1）⋅3n }的前n 项和S n ，其操作步骤如下：由于S n ＝1×31＋3×32＋…＋（2n －1）⋅3n ，()23131333213n n S n +=⨯+⨯++-⋅，从而()()21232323213n n n S n +=--⨯++⨯+-⋅，所以()1133n n S n +=-⋅+，始比如上方法可求数列{n 2⋅3n }的前n 项和T n ，则2T n ＋3＝ ．16．已知函数f （x ）是定义在R 上的偶函数，且当x ≥0时，f （x ）＝2x ．若对任意x ∈[1，3]，不等式f （x ＋a ）≤f 2（x ）恒成立，则实数a 的取值范围是 ．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．（本小题满分10分）在数列{a n }中，a ＝1，其前n 项和S n 满足2S n ＝（n ＋1）a n ，n ∈N *． （1）求数列{a n }的通项公式a n ；（2）若m 为正整数，记集合22n nn a a m a ⎧⎫⎪⎪+⎨⎬⎪⎪⎩⎭≤的元素个数为b m ，求数列{b m }的前20项和． 18．（本小题满分12分）在轴截面为正方形ABCD 的圆柱中，M ，N 分别为弧AD ，弧BC 的中点，且在平面ABCD 的两侧．（1）求证：四边形ANCM 是矩形； （2）求二面角B －MN －C 的余弦值．19．（本小题满分12分）文化月活动中，某班级在宣传栏贴出标语“学好数学好”，可以不同断句产生不同意思，“学/好数学/好”指要学好的数学，“学好/数学/好”强调数学学习的重要性，假设一段时间后，随机有N 个字脱落． （1）若N ＝3，用随机变量X 表示脱落的字中“学”的个数，求随机变量X 的分布列及期望； （2）若N ＝2，假设某同学检起后随机贴回，求标语恢复原样的概率． 20．（本小题满分12分）记△ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，已知b ＝1，c ＝2． （1）若2CD DB =，2AD CB ⋅=，求A ； （2）若23C B π-=，求△ABC 的面积． 21．（本小题满分12分）在平面直角坐标系xOy 中，已知点P 在抛物线C 1：y 2＝4x 上，圆C 2：（x －2）2＋y 2＝r 2（0＜r ＜2）． （1）若r ＝1，Q 为圆C 2上的动点，求线段PQ 长度的最小值；（2）若点P 的纵坐标为4，过P 的直线m ，n 与圆C 2相切，分别交抛物线C 1于A ，B （异于点P ），求证：直线AB 过定点．22．（本小题满分12分）若对实数x 0，函数f （x ），g （x ）满足f （x 0）＝g （x 0）且f ′（x 0）＝g ′（x 0），则称()()()0,,f x x x F x g x x x <⎧⎪=⎨⎪⎩≥为“平滑函数”，x 0为该函数的“平滑点”．已知()323122x f x ax x x =-+，g （x ）＝bx ln x ． （1）若1是平滑函数F （x ）的“平滑点”， （ⅰ）求实数a ，b 的值；（ⅱ）若过点P （2，t ）可作三条不同的直线与函数y ＝F （x ）的图象相切，求实数t 的取值范围； （2）对任意b ＞0，判断是否存在a ≥1，使得函数F （x ）存在正的“平滑点”，并说明理由．G4联盟—苏州中学、扬州中学、常州中学、盐城中学2022-2023学年第一学期12月联合调研高三数学答案及其解析一、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．【答案】A 2．【答案】D【解析】4334i z i z i ⋅=+⇒=-，所以25z z ⋅= 3．【答案】B【解析】因为乡村人口与年份望负线性相关关系，所以r 接近－1，故选B 4．【答案】D 5．【答案】C【解析】由题意得22245109436b b a a a ⎧⎧==⎪⇒⎨⎨=⎩⎪=⎩，所以6c ==，故椭圆离心率为23c e a == 6．【答案】D 7．【答案】C【解析】()()()()()()11112111cos cos sin sin sin 2n n n n n n n n n n n n n a a a a a a a a a a a a a +++++++--+--++⎡⎤⎡⎤⎣⎦⎣⎦-⋅+==-21sin 10n a =，所以{}2sin n a 为等差数列，公差为110，所以()2211sin sin 1110n a a n =+-⨯≤，所以110n -211sin 111a n -⇒≤≤≤，故选C8．【答案】B【解析】()m n AP mAB nAC m n AB AC m n m n ⎛⎫=+=++⎪++⎝⎭，由P 在内切圆上，故APm n m n AB AC m n m n +=⎛⎫+⎪++⎝⎭，则11cos 16A =，所以BC 边上高为2h =6r =，故由平行线等比关系，可得213h r m n h -+=≥，故选B 二、选择题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分．9．【答案】BCD 【解析】选项A ，应该是114a b +≥，B ：22221a ba b+++≥，B 正确；C ：222log log 2log 22a b a b ++=-≤，C 正确；D ：1sin sin 2sin cos 2sin 222a b a b a b +-+=⋅≤，D 正确；答案为BCD 10．【答案】ABD【解析】AB 正确，因为()f x 关于x a =轴对称，()g x 关于(),0a 中心对称，故()()f x g x 为中心对称图形，C 错误：而()()()()()220'g x f x q x f x B x ⎡⎤-=>⎢⎥∠⎣⎦或根据一般得分离常数变形可知D 正确；答案为：ABD 11．【答案】ABD 【解析】 显然A 正确；()()()()()323131223343141151f p p p C p p C p p C p p ⎡⎤⎡⎤=+-+-+-+⨯-⎣⎦⎣⎦()03f =，13328f B ⎛⎫=⇒ ⎪⎝⎭，D 正确； 求导或根据()f p 关于12对称，且p 越极端，越可能快结束，有11412352--≤，得1435f f ⎛⎫⎛⎫> ⎪ ⎪⎝⎭⎝⎭， 故答案为：ABD 12．【答案】ABC 【解析】A 选项判断：应用等体积法，可()()min min 1122MG AG =≥A 正确；B 选项：因为面PBD 不与AG 垂直，也不平行，故轨迹不可能时圆，即为椭圆，B 正确；C 选项判断：设MH ⊥面PBD ，H ∈面PBD ，2112MA HM =⇒=，故C 正确； D 选项判断：由于CD 与面PBD 夹角θ满足1sin2θ=>，故[],6πθπθ∉-，D 错误；综上所述，答案为ABC三、填空题：本题共4小题，每小题5分，共20分．13．【答案】15【解析】展开式的通项为()()36621661rr r r Tr Cx C x --+⎛==- ⎝，当31602r -=，4r =时，为常数项15 14．【答案】56π【解析】如图，因为()f x 的周期为242T ππ==，所以22T CD ==，22TCD ==，所以AB ===解得M =所以()2f x x πϕ⎛⎫=+ ⎪⎝⎭，所以()0f ϕ==，1sin 2ϕ=，因为0ϕπ<<，所以6πϕ=或56π，又因为函数()f x 在y 轴右侧单调递减，所以56πϕ=． 15．【答案】()2113n n n +-+⋅【解析】2122213233n n T n =⨯+⨯+⋯+⋅① 222321313233n n T n +=⨯+⨯+⨯+⋅②②－①()()()222222322123123233133n n n T n n n +⎡⎤=-+-⋅+-⋅++--⋅+⋅⎣⎦()()3321333532133n n n n +=--⋅+⋅++-⋅+⋅()()212112333313n n n n n S n S n n n +++=---+⋅=-+⋅=-+⋅所以()212313n n T n n ++=-+⋅16．【答案】[]3,1-． 【解析】()()()()[]2221,3f x a fx f x f x x +==⇒∀∈≤，[]23,1x a a +⇒∈-≤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．（本小题满分10分） 解析：（1）()()()()()111212221212nn n n n n n n n n a S n a a S S n a na n n a na n n---=+⇒=-=+-⇒-=⇒=≥≥11111n n a a a n n -===⇒=-（2）2214222n n a n m m n m a n n ⎛⎫+⇒+⇒-+ ⎪⎝⎭≤≤≤， 因为1422n n ⎛⎫+ ⎪⎝⎭≥，当且仅当2n =时成立， 所以10b =，21b =，当3n ≥，35b =，47b =，59b =，611b =，…，2339b = 所以{}m b 的前20项和为()135739378+++++=．18．（本小题满分12分） 【解析】（1）设轴截面正方形ABCD 边长为2a ，取弧BC 另一侧的中点Q ， 则BC 与NQ 垂直平分，且2BC NQ a ==， 所以四边形BNCQ为正方形，BQ NC ==，因为M 为弧AD 中点，所以MQ AB ∥，四边形ABQM 为矩形， 所以AM BQ ∥，所以AM CN ∥，所以四边形AMCN 为平行四边形，因为AN ==，MN ==，所以22228AM AN MN a +==，所以AM ⊥AN ，所以四边形ANCM 为矩形； （2）由（1）知，MB MC ===，BN CN ==，MN =，所以2MNB MNC π∠=∠=所以MNB MNC ∆∆≌，Rt △MBN 斜边MN上的高2h a ==， 作BP ⊥MN ，则CP ⊥MN ，∠BPC 即为二面角B -MN -C的平面角，2BP CP ==，2BC a =， 在△BPC 中，由余弦定理得222222341cos 233BP CP BC a a BPC BP CP a +--∠===-⨯，二面角B -MN -C 的余弦值为13- 19．（本小题满分12分） 【解析】（1）随机变量X 的可能取值为0，1，2，12C()33351010C P X C ===，()1223356110C C P X C ===，()2123353210C C P X C ===，随机变量X 的分布列如下表：随机变量X 的期望为()163012 1.2101010E X =⨯+⨯+⨯= 法二：随机变量X 服从超几何分布X ～H （3，2，5），所以()26355E X =⨯= （2）设脱落一个“学”为事件A ，脱落一个“好”为事件B ，脱落一个“数”为事件C ，事件M 为脱落两个字M AA BB AB AC BC =++++，()2225110C P AA C ==，()2225110C P BB C ==，()112225410C C P AB C ⋅==，()112125210C C P AC C ⋅==，()112125210C C P BC C ⋅==， 所以某同学捡起后随机贴回，标语恢复原样的概率为()()()()()()()11413125525P P AA P BB P AB P AC P BC =+⨯+++⨯=+⨯=，法二：掉下的两个字不同的概率为1020.810p -==， 所以标语恢复原样的概率为()110.62p p -+=． 20．（本小题满分12分） 解：（1）()112123333CD DB AD AB BD AB BC AB AC AB AB AC =⇒=+=+=+-=+ 所以()22212118112cos 233333333AD CB AB AC AB AC AB AC AB AC A ⎛⎫⋅=+-=--⋅=--⨯⨯=⇒⎪⎝⎭1cos 2A =，因为()0,A π∈，所以3A π=（2）法一： 因为23C B π-=，所以562A C π=-，62AB π=-， 因为2c b =，sin 2sinC B =， 则5sin 2sin 6262A A ππ⎛⎫⎛⎫-=-⎪ ⎪⎝⎭⎝⎭化简整理得tan 29A =，所以22tan2sin 1tan 2AA A ==+故面积为1sin 214S bc A == 法二：因为2sin 2sin c b C B =⇒=， 因为23C B π-=，所以2sin 2sin sin 3B B B B π⎛⎫+=⇒=⎪⎝⎭①， 联立22sin cos 1B B +=②解得sin cos B B ⎧=⎪⎪⎨⎪=⎪⎩，所以sin 2sin C B ==，232C B ππ=+> 所以cos 0C <，则cos C ==所以()sin sin sin cos cos sin 14A B C B C B C =+=+= 所以△ABC的面积为1sin 214ABC S bc A ∆==． 21．（本小题满分12分）【解析】 （1）设()2,2P t t ，则211PQ PC -=≥，当()0,0P ，Q 为2PC 线段与圆2C 的交点时，min 1PQ =（2）题意可知()4,4P ，过P 点直线()44y k x -=-与圆2C 相切，r =，即()222416160r k k r --+-=，①设直线AB 为：()()441m x n y -+-=，则与抛物线C 的交点方程可化为：()()()()()()24844444(4)4y y m x n y x m x n y -+--+-=--+-⎡⎤⎡⎤⎣⎦⎣⎦， 令44y z x -=-，则：()()2188440n z m n z m ++--=，② 题意有，①②方程同解，故有()()()[]()2233164164818444y r r m n m n -⎡⎤⎣=---+⨯=--+-⎦-， 即：2111m n -=，故：直线AB 恒过()6,7-．22．（本小题满分12分）【解析】（1）（ⅰ）()21'332f x ax a =-+，()[]'1lng x b x =+， 由题意可知10a -=，且532a b -=， 故解得：1a =，12b =， （ⅱ）进一步()323,122ln ,12x x x x F x x x ⎧-+<⎪⎪=⎨⎪⎪⎩≥，过点()2,P t 作()F x 的切线，切点()(),x F x 满足方程：()()()2F x t F x x -=-，故题意等价于方程：()()()'2t F x F x x =--有3个不同根，()()()()'2p x F x F x x =--，()()()''2p x F x x =--， 代入得1,2x ⎛⎫∈-∞ ⎪⎝⎭时， ()p x 单调递减，1,22x ⎡⎫∈⎪⎢⎣⎭时，()p x 单调递增，[)2,x ∈+∞时，()p x 单调递减， 故()13,2,ln 228t p x x ⎧⎫⎛⎫⎛⎫∈∈=-⎨⎬ ⎪ ⎪⎝⎭⎝⎭⎩⎭ （2）题意等价于：0b ∀>，是否1a ∃≥，使得[]3223ln 221331ln 2x ax x bx x ax x b x ⎧-+=⎪⎪⎨⎪-+=+⎪⎩有解 消a 有：()313212ln 122ln 1x x b x b x ---=-⇒=-，其中由0b >，可得23x ⎛∈ ⎝，故题意进一步化简23x ⎛∀∈ ⎝，是否1a ∃≥，使得()3ln 3122ln 1x x x a x x -+=-成立，23x ⎛⇔∀∈ ⎝，()23ln 3122ln 1x x x x x -+-≤是否恒成立 设()()2243ln 231q x x x x x x =--+-，()()'83ln q x x x =-，故2,13x ⎛⎫∈ ⎪⎝⎭时，单调递减，(x ∈，()q x 单调递增，故：()()10q x q =≥得证，即0b ∀>，31a ≥，使得()F x 存在的“平滑点”．。
2021届江苏省盐城中学高三英语上学期期末试题及参考答案
2021届江苏省盐城中学高三英语上学期期末试题及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AFor some people, there’s no better companion than mans best friend-a dog. This four-legged pet can bring comfort and joy and provide much- needed exercise for you when it needs walkies! This probably explains why dog ownership increased last year because people spent more time at home during he CovID-I9 lockdown.However, as demand for a new dog increased, so did the price tag. Popular breeds, such as Cockapoos and Cocker Spaniels, saw even sharper price increases, and puppies have been selling for $3,000 or more.Animal welfare charities fearthat high prices could encourage puppy farming, smuggling (走私) or dog theft. An investigation found some breeders have been selling puppies and kittens on social media sites--something charities have called “extremely irresponsible”.But despite some new owners purchasing a dog legally, maybe from a rescue center or registered breeder, they’ve proved to be ill-prepared for life with a new pet, and the pet itself has found it hard tocome to terms withlife in a new home.Looking to the future, there are concerns about the welfare of these much-loved pets. Lan Alkin manager of the Oxfordshire Animal Sanct uary in the UK, notes: “At the moment, the dogs are having a great time, but separation anxiety could still surface when people go back to work.” And Cliare Calder from the UKs Dogs Trust rescue charity says, “The economic situation also means that some people may find they can’t afford to look aftera dog.” The message is not to buy a dog in haste and to pick one that fits into our lifestyle.1. The greater demand for dogs can cause the following problems except ________.A. illegal trade of dogsB. less dog farmingC. high prices of dogsD. online sale of dogs2. What does the underlined phrase"come to terms with"in paragraph 4 mean?A. Fit in withB. Go in forC. Make up for.D. End up with3. What can we learn from the last paragraph?A. Despite the problems, dogs are living happily.B. The writer has a positive attitude towards dogs future.C. Experts are worried that dogs will be unaffordable to people.D. The writer advises people to think twice before keeping dogs as pets.BWhile the arts can' t stop the COVID-19 virus or the social unrest we see in the world today, they can give us insight into the choices we make when moving through crises and chaos. The arts invite everyone to think in new ways.We often experience works of art as something that's pleasing to our senses without a full understanding of the creative effort. Great art often shows us contradictions and crises, and we can learn a great deal from their resolutions(解决). Through our understanding of art, we can gain a deeper understanding of how we might overcome our own challenges. In understanding extremes of contrast, we can see the beauty in art with themes that are not simply pleasing for their magnificent features or qualities.Beethoven offers a wonderful example of moving artfully through crises and chaos. He composed his Symphony No. 9 as his hearing loss became more and more pronounced. The opening of the symphony seems to come out of nowhere, from near silence in the opening to a full expression of what many consider to be the joy of freedom and universal brotherhood with Schiller’s Ode to joy（欢乐颂). Beethoven appears to have created a work of art that not only freed him from his personal struggles, but one that also speaks to the joy of living together in peace and harmony.Have a dialogue between the two opposing parts and you will find that they always start out fighting each other until we come to an appreciation of difference—a oneness of the two opposingforces.The arts offer many lessons that can help us gain the knowledge we need to move more confidently in today’ s competitive and uncertain environment. An openness to arts-based solutions will give you more control over your future.4. What value does art have beyond pleasing people's senses?A. It brings people inner peace.B. It contributes to problem-solving.C. It reduces the possibility of crises.D. It deepens understanding of music.5. What can we learn about Beethoven's Symphony No. 9?A. It celebrates freedom and unity.B. It aims to show crises and chaos.C. It opens with Schiller's Ode to Joy.D. It is unfinished due to his hearing loss.6. What is the author's suggestion on dealing with conflicting forces?A. Leaving things as they are.B. Making a choice between them.C. Separating them from each other.D. Engaging them in a conversation.7. Which of the following can be the best title for the text?A. How COVID-19 changes artB. Essentials of Symphony No. 9C. Moving artfully through crisesD. Joy in the eyes of BeethovenCDid you know people who live in different parts ofChinahave different habits and preferences? For example, people from southernChinaprefer to eat vegetables, while people from northChinalike to eat meat. According to a new study in a journal, gene variations (变异) might be responsible for these differences. Researchers fromChina’s BGI collected genetic information from 141,431 Chinese women, who came from 31 provinces and consisted of 36 ethnic minority groups.They found that natural selection has played an important role in the ways that people living in different regions of China have developed, affecting their food preferences, immunities (免疫力) to illness and physical features.A variation of the gene FADS2 is more commonly found in northern people. It helps people metabolize (新陈代谢) fatty acids, which suggests a diet that is rich in flesh. This is due to climate differences.Northern Chinais at a higher latitude. This weather is difficult to grow vegetables in. Therefore, northerners tend to eat more meat.The study also found differences in the immune systems of both groups. Most people in southernChinacarry the gene CR1, which protects against malaria. Malaria was once quite common in southernChina. In order to survive, the genes of people in the south evolved to fight against this disease. However, people in the south are also more sensitive to certain illnesses, as they lack the genes to stop them.Genes can also cause physical differences between northerners and southerners. Most northerners have the ABCC11 gene, which causes dry earwax, less body smell and fewer sweats. These physical differences are also more beneficial to living in cold environments. Southerners are less likely to have this gene, as it did not develop in their population.8. What did the new study focus on?A. Regions.B. Eating habits.C. Gene variations.D. Ethnic minority groups.9. What is the main function of the gene FADS2?A. It helps store fat.B. It helps digest meat.C. It helps gain weight.D. It helps treat an illness.10. According to the study, most northerners ________.A. sweat less frequentlyB. are immune to malariaC. prefer vegetables to meatD. are more sensitive to climates11. How many differences did the study find related to genes?A. Two.B. Three.C. Four.D. Five.DDogs are often called as “man's best friend”, MacKenzie, a four-pound Chihuahua（吉娃娃）, was named winner of the 2020 American Hero Dog Competition on October 19, 2020.In its tenth year in 2020 the annual contest is the brainchild of American Humane, the country's first national charitable organization founded for the safety and well-being of animals. Often called the “Oscars for dogs”, the award recognizes dogs who make extremely great contributions to society.The competition of 2020 attracted over 400 entries（参赛者）from across the country. These heroic dogs have gone above the call of duty, saving lives, comforting the ill and aged and reminding us of the powerful, age-old ties between animals and people. While all were impressive, it was tiny MacKenzie who wonthe judges' hearts.MacKenzie's growth was not easy. Born with a mouth disability, she had to be fed through a tube（管子）for the first year of her life. Despite her own struggles, she always seemed to think more of other animals in need. “Never have I seen such a will to live. Though sick, she carefully looked after the baby animals at the rescue（救助）center,” said her caretaker.A life-saving operation performed in 2014 gave MacKenzie the ability to eat independently. The seven-year-old chihuahua is now working for the Mia Foundation, an organization that rescues and nurses animals with inborn disabilities. The chihuahua does an excellent job and has raised various animals. She plays nurse, cleans, comforts and hugs them, acting as their mother and teaching them how to socialize, play and have good manners.In addition to her role as an animal caretaker, MacKenzie also visits schools to educate kids about theimportance of accepting physical differences in both animals and people. Her heartwarming and inspiring story makes MacKenzie a worthy receiver ofAmerica's top dog honor.12. What can we infer about the American Hero Dog Competition?A. It was first held in 2010B. It was held to honor caretakers of dogs.C. It takes place every ten years.D. It was started by a charitable organization.13. With what quality did MacKenzie win the award?A. Talent and braveryB. Friendliness and care.C. Courage and selflessness.D. Confidence and independence.14. In which aspect can students benefit from MacKenzie's visits?A. Learning from failures.B. Understanding the disabled.C. Valuing physical health.D. Developing practical ability.15. What's the best title for the text?A. Dogs Are Man's Best Friends.B. Treat Dogs the Way We Want to Be Treated.C. Touching Stories between MacKenzie and PeopleD. 2020 American Hero Dog: A TinyChihuahua.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。