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

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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。

2021届江苏省扬州中学高三英语上学期期末考试试题及答案解析

2021届江苏省扬州中学高三英语上学期期末考试试题及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项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 novelBWolves have a certain undeserved reputation: fierce, dangerous, good forhunting down deer and farmers’ livestock. However, wolves have a softer, more social side, one that has been embraced by a heart-warming new initiative.In a bid to save some of Europe’s last wolves, scientists have explored the willingness of these supposedly fierce creatures to help others of their kind. Female wolves, the scientists have discovered, make excellent foster parents to wolf cubs that are not their own. The study, published in Zoo Biology, suggests that captive-bred wolfcubs(幼兽)could be placed with wild wolf families, boosting the wild population.The gray wolf was once the world’s most widely distributed mammal, but it became extinct as a result of widespread habitat destruction and the deliberate killing of wolves suspectedof preying on livestock. Fear and hatred of the wolf have since become culturally rooted, fuelled by myths, fables and stories.In Scandinavia, the gray wolf is endangered, the remaining population found by just five animals. As a result, European wolves are severely inbred and have little geneticvariability（变异性）, making them vulnerable to threats, such as outbreaks of disease that they can’t adapt to quickly. So Inger Scharis and Mats Amundin of Linkoping University, in Sweden, started Europe’s first gray wolf-fostering program. They worked with wolves keptat seven zoos across Scandinavia. Eight wolf cubs between four and six days old were removed from their natural parents and placed with other wolf packs in other zoos. The foster mothers accepted the new cubs placed in their midst.The welfare of the foster cubs and the wolves’ natural behavior were monitored using a system of surveillance cameras. The foster cubs had a similar growth rate as their step siblings in the recipient litter, as well as their biological siblings in the source litter. The foster cubs had a better overall survival rate, with 73% surviving until 33 weeks, than their biological siblings left behind, of which 63% survived. That rate of survival is similar tothat seen in wild wolf cubs. Scientists believe that wolves can recognize their young, but this study suggests they can only do so once cubs are somewhere between three to seven weeks of age.If captive-bred cubs can be placed with wild-living families, which already have cubs of a similar age, not only will they have a good chance of survival, but they could help dramatically increase the diversity of the wild population, say the researchers. Just like the wild wolves they would join, these foster cubs would need protection from hunting. Their arrival could help preserve the future of one of nature’s most iconic and polarizing animals.4. What’s the theme of the passage?A. Giving wolf cubs a new lifeB. Foster wolf parents and foster cubsC. The fate of wild wolvesD. Changing diversity of wild wolves5. Which of the following flow chart best demonstrates the relationship between the wolves?A. B.C. D.6. Which of the following statements is true?A. Female wolves are willing to raise wolf cubs of 3 to 7 weeks old.B. Foster cubs are accepted by foster parents and are well bred.C. Man’s hostile attitude towards wolves roots in myths, fables and stories.D. Foster cubs and their biological siblings have similar growth rate and survival rate.7. What’s the purpose of the research?A. To help wolves survive various threatsB. To improve wolves’ habitat and stop deliberate killingC. To save endangered wolves by increasing their populationD. To raise man’s awareness of protecting wolvesCHenry Cavill: Bring Superman to LifeHenry Cavill knew that he wanted to be a star at 16 years of age, after a chance meeting with movie star Russell Crowe who inspired hispassion for acting. But for the British-born actor, the bright lights and attraction ofHollywoodwere a long way away. Supported by his secretary mother and stockbroker father, he decided to study drama during high school. His journey to super star began.Before gaining the international recognition he has now, Cavill tried out for roles in the Harry Potter and Twilight series but failed to get either. He would have to keep waiting for his big chance.Determined as ever, Cavill took any acting jobs he could get his hands on and appeared in several low-budget horror movies and TV shows in hopes of getting noticed. It almost worked. In the early 2000s, at just 22 years old, he narrowly missed out on becoming the new James Bond. Finally, in 2007, his hard work paid off. He won a leading role as the first Duke of Suffolk in the period showThe Tudors. The TV show was very popular and helped to raise Cavill's popularity inAmerica.In 2011, Cavil landed his breakout role, playing Superman in the DC Extended Universe. He hasn't looked back and has since starred in many hit films, such asMission: Impossible- Fallout.More recently, he stepped back on to the small screen. Since 2019, he has starred in the popular seriesThe Witcher, adapted from the book series and video games of the same name. In the TV show, Cavill played a brave monster hunter named Geralt of Rivia, which was the perfect role for Cavill because he was a fan of the video games. Cavill also got a chance to play a classic English character — master detective Sherlock Holmes — in 2020'sEnola Holmes.However, Cavill isn't just a good guy on screen. His charity work also makes him a real-life hero. In 2014, he took part in the Ice Bucket Challenge while wearing his full Superman suit to support the ALS Association. Currently, he is an ambassador for the UK's Royal Marines Charity, which supports war veterans (退伍军人). Why does he do it? He love to make people feel good and bring smiles to people' faces. Indeed, Henry Cavill in living proof that you don't always need to wear a cape (斗篷) to act like a hero.8. Why did Cavil act in low-budget film and TV works early in his career?A. He was too polite to refuse.B. He was hoping to get noticed.C. He was encouraged to do so by his parents.D. He was friends with the directors of the projects.9. The role of the monster hunter was the perfect for Cavill because ________ .A. he had experienced hunting monstersB. he had played the same role in a movieC. he knew the writer of the books personallyD. he enjoyed the video games that the show was rooted in10. Which of the following words can best describe Cavill?A. Modest and friendly.B. Determined and kind.C. Talented and faithful.D. Honest and considerate.11. What made Cavill a real-life hero?A. Being a successful actor.B. Playing Superman on screen.C. Devoting to charities.D. Wearing a cape to take part in activities.DCanadaIs Our NeighbourCanada and the United States are neighbours.They are on the same land.They share the same long boundary(国界).These two nations are similar in many ways.Canada buys many goods from the United States.Cars and clothes are two examples.The United States also buys goods from Canada.Much of the paper used in the United States comes from Canada.Some of the oilweuse comes from Canada,too.Americans travel toCanadaon holiday.And Canadians often visit the United States.It is easy for the people of one country to go to the other country.Canadians read about the United States in newspapers and magazines.Many Americans watch Canadian baseball and hockey (曲棍球)matches on Sundays.However,there are important differences between theUnited Statesand Canada.The United States has more people.Because the population is smaller,there are more open places in Canada.There is much unused land.This is another important difference.12.Canadabuys from theUnited States.A.oil and paperB.nothingC.many thingsD.everything13.In the first paragraph “we” means ________.A.CanadiansB.AmericansC.ChineseD.students14.The people in theUnited Stateslike Canadian ________.A.baseballB.basketballC.newspapersD.oil15.Which of the following statements is WRONG?A.Canada has less people than theUSA.B.Canada has not used all the land.C.Canada is connected withAmerica.D.Canadians don’t like hockey.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G4(苏州中学、盐城中学、扬州中学、常熟中学)高三教学情况期末调研卷英语(解析版)

2021年G4学校高三教学情况期末调研英语 2021.01注意事项：1.答题前，考生务必将自己的姓名、考试号填写在答题卡上。

2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

回答非选择题时，将答案写在答题卡上，写在本试卷上无效。

3.考试结束后，将答题卡交回，试卷学生自行保管好。

第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. Where is the conversation probably taking place?A.:At school.B. At a cafe.C. At the man's house.2. How will the woman probably get paid?A. In cash.B. Through WeChat.C.By credit card.3. What are the speakers likely to do?A. Hide from the bears.B. Scare the bears away.C. Feed the bears.4. How is the man probably feeling?A.Relaxed.B.Determined.C. Uncertain.5. What is the conversation mainly about?A. Teammates discussing a game.B.A young child talking to her hero.C. A girl asking for advice from her brother.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

专题29 空间向量与立体几何(解答题)(新高考地区专用)(解析版)

专题29 空间向量与立体几何（解答题）1．如图，在三棱锥P ABC -中，平面PAC ⊥平面ABC ，PC AC ⊥，BC AC ⊥，2AC PC ==，4CB =，M 是PA 的中点．（1）求证：PA ⊥平面MBC ；（2）设点N 是PB 的中点，求二面角N MC B --的余弦值．【试题来源】陕西省咸阳市2020-2021学年高三上学期高考模拟检测（一）（理）【答案】（1）证明见解析；（2）3．【解析】（1）平面PAC ⊥平面ABC ，平面PAC平面ABC =AC ，BC ⊂平面ABC ，BC AC ⊥，所以BC ⊥平面PAC ，因为PA ⊂平面PAC ，所以BC PA ⊥，因为AC PC =，M 是PA 的中点，所以CM PA ⊥，因为CMBC C =，,CM BC ⊂平面MBC ，所以PA ⊥平面MBC ．（2）因为平面PAC ⊥平面ABC ，平面PAC平面ABC =AC ，PC ⊂平面PAC ，PC AC ⊥，所以PC ⊥平面ABC ，因为BC ⊂平面ABC ，所以PC BC ⊥，以C 为原点，CA ，CB ，CP 为x ，y ，z 轴正方向，建立如图所示的空间直角坐标系，(2,0,0)A ，(0,4,0)B ，(0,0,0)C ，(0,0,2)P ，(1,0,1)M ，(0,2,1)N ，则(1,0,1)CM =，(0,2,1)CN =，(2,0,2)PA =-，由（1）知(2,0,2)PA =-是平面MBC 的一个法向量，设(,,)n x y z =是平面MNC 的法向量，则有00CM n CN n ⎧⋅=⎨⋅=⎩，即020x z y z +=⎧⎨+=⎩，令1y =，则2z =-，2x =，所以(2,1,2)n =-，设二面角N MC B --所成角为θ，由图可得θ为锐角，则2cos cos ,||||PA n PA n PA n θ⋅⨯=<>===【名师点睛】解题的关键是熟练掌握面面垂直的性质定理，线面垂直的判定和性质定理，并灵活应用，处理二面角或点到平面距离时，常用向量法求解，建立适当的坐标系，求得所需点的坐标及向量坐标，求得法向量坐标，代入夹角或距离公式，即可求得答案． 2．在四棱锥P ABCD -中，PAB △为直角三角形，90APB ∠=︒且12PA AB CD ==，四边形ABCD 为直角梯形，//AB CD 且DAB ∠为直角，E 为AB 的中点，F 为PE 的四等分点且14EF EP =，M 为AC 中点且MF PE ⊥．（1）证明：AD ⊥平面ABP ；（2）设二面角A PC E --的大小为α，求α的取值范围．【试题来源】山东省德州市2020-2021学年高三上学期期末【答案】（1）证明见解析；（2）,32ππα【解析】（1）取PE 的中点N ，连接AN ，DN ，CE ，如图所示：因为12AE AB =，12AP AB =，所以AP AE =，AN PE ⊥．因为四边形ABCD 为直角梯形，且90DAB ∠=︒，12CD AB =，所以四边形AECD 为正方形，即M 为DE 的中点．因为14EF EP =，N 为PE 的中点，所以F 为EN 的中点．所以//MF DN ．因为MF PE ⊥，所以DN PE ⊥．所以PE DN PE ANPE DN AN N ⊥⎧⎪⊥⇒⊥⎨⎪⋂=⎩平面ADN ．因为DA ⊂平面ADN ，所以PE DA ⊥．所以DA AB DA PEDA PE AB E ⊥⎧⎪⊥⇒⊥⎨⎪⋂=⎩平面ABP ．（2）以A 为原点，AB ，AD 分别为y ，z 轴，垂直AB 的直线为x 轴，建立空间直角坐标系，如图所示：设AD a =，1PA CD ==，2AB =，则()0,0,0A，1,02P ⎫⎪⎪⎝⎭，()0,1,0E ，()0,1,C a ． 31,02AP ⎛⎫= ⎪ ⎪⎝⎭，()0,1,AC a =，1,02PE ⎛⎫=- ⎪ ⎪⎝⎭，()0,0,CE a =-．设平面PAC 的法向量()111,,n x y z =，则1111310220AP n x yAC n y az ⎧⋅=+=⎪⎨⎪⋅=+=⎩，令1y =，解得11x =，1z =，故1,3,n⎛=- ⎝⎭．设平面PEC 的法向量()222,,m x y z =，则222310220PE mx y CE m az ⎧⋅=-+=⎪⎨⎪⋅=-=⎩，令2y =21x =，20z =，故()1,3,0m =．由图知，二面角A PC E --的平面角α为锐角，所以11cos 0,2α-⎛⎫==⎪⎝⎭．故,32ππα．3．如图，在四棱锥P ABCD -中，底面ABCD 为直角梯形，AD BC ∥，112BC AD ==且CD =E 为AD 的中点，F 是棱PA 的中点，2PA =，PE ⊥底面ABCD ．AD CD ⊥（1）证明：//BF平面PCD ；（2）求二面角P BD F --的正弦值；（3）在线段PC （不含端点）上是否存在一点M ，使得直线BM 和平面BDF 所成角的正弦值为13？若存在，求出此时PM 的长；若不存在，说明理由．【试题来源】天津市滨海七校2020-2021学年高三上学期期末联考【答案】（1）证明见解析；（2（3）存在，7PM = 【解析】（1）由题意得//BC DE ，=BC DE ，90ADC ∠=︒，所以四边形BCDE 为矩形，又PE ⊥面ABCD ，如图建立空间直角坐标系E xyz -，则()0,0,0E ，()1,0,0A，()B ，()1,0,0D -，(P ，()C -，1,0,22F ⎛ ⎝⎭，设平面PCD的法向量为(),,m x y z=，()0,DC =，(DP =则00DC m DP m ⎧⋅=⎨⋅=⎩，则0x ==⎪⎩，则0y =，不妨设x =1z =，可得()3,0,1m =-，又1,22BF ⎛⎫= ⎪ ⎪⎝⎭，可得0BF m ⋅=，因为直线BF ⊄平面BCD ，所以//BF 平面BCD ．（2）设平面PBD 的法向量为()1111,,x n y z =，()1,DB =，(0,BP =，则1100DB n BP n ⎧⋅=⎪⎨⋅=⎪⎩，即111100x ⎧+=⎪⎨+=⎪⎩，不妨设x =()13,1,1n =--，设平面BDF 的法向量为()2222,,n xy z =，32DF ⎛= ⎝⎭，则2200DB n DF n ⎧⋅=⎪⎨⋅=⎪⎩，即222203022x x z ⎧+=⎪⎨+=⎪⎩，不妨设2x =，可得()2n =-，因此有121212cos ,65n n n n n n ⋅==-⋅，（注：结果正负取决于法向量方向）于是21212465sin ,1cos ,n n n n =-=，所以二面角P BD F --．（3）设((),PM PC λλλ==-=-，()0,1λ∈(),BM BP PM λ=+=-，由（2）可知平面BDF 的法向量为()23,1,3n =-，2223cos ,BM n BM n BM n⋅===⋅，有23410λλ-+=，解得1λ=（舍）或13λ=，可得1,333PM ⎛=-- ⎝⎭，所以73PM =． 4．在四棱锥P ABCD -中，PA ⊥平面ABCD ，PA =//DC AB ，90DAB ∠=︒，3AB =，2AD CD ==，M 是棱PD 的中点．（1）求异面直线DP 与BC 所成的角的余弦值；（2）求AM 与平面PBC 所成的角的大小；（3）在棱PB 上是否存在点Q ，使得平面QAD 与平面ABCD 所成的锐二面角的大小为60°？若存在，求出AQ 的长；若不存在，说明理由．【试题来源】天津市南开中学2020-2021学年高三上学期第四次月考【答案】（1；（2）45︒；（3）125．【解析】如图，以,,AD AB AP 所在直线分别为,,x y z 轴建立如图所示空间直角坐标系，则(()()()()(,0,0,0,3,0,0,2,2,0,0,2,0,P A B C D M ，（1）(0,DP =-，()1,2,0BC =-，所以cos,DP BC==，即异面直线DP与BC（2）(AM=，(3,0,PB=-，()1,2,0BC=-设平面PBC的法向量(),,m x y z=，则mPBm BC⎧⋅=⎨⋅=⎩，3020xx y⎧-=⎪⎨-+=⎪⎩，所以可取(m=，设AM与平面PBC所成的角为θ，则sin cos,AM mθ===，所以AM与平面PBC所成的角为45︒；（3）平面ABCD的法向量可取()10,0,1n=，设(()3,0,3,0,PQ PBλλλ==-=-，则()3Qλ，所以()3AQλ=，()0,2,0AD =，设平面QAD的法向量为()2222,,n x y z=，则22nAQn AD⎧⋅=⎪⎨⋅=⎪⎩，()2223020x zyλ⎧+=⎪⎨=⎪⎩，可取()223,0,3nλ=-，因为平面QAD与平面ABCD所成的锐二面角的大小为60°.所以121cos,2n n=，12=，解得25λ=或2λ=-（舍）所以6,0,55AQ⎛=⎝⎭，所以61255AQ⎛==5．如图，在正四面体A BCD-中，点E，F分别是,ABBC的中点，点G，H分别在,CD AD 上，且14DH AD=，14DG CD=．（1）求证：直线,EH FG 必相交于一点，且这个交点在直线BD 上；（2）求直线AB 与平面EFGH 所成角的正弦值．【试题来源】陕西省榆林市2020-2021学年高三上学期第一次高考模拟测试（理）【答案】（1）证明见解析；（2．【解析】（1）因为//,//EF AC GH AC ，11=,=24EF AC GH AC ，所以//GH EF 且12GH EF =，故E ，F ，G ，H 四点共面，且直线,EH FG 必相交于一点，设=EH FG M ，因为,∈M EH EH平面ABD ，所以M ∈平面ABD ，同理：M ∈平面BCD ，而平面ABD ⋂平面BCD BD =，故M ∈平面BCD ，即直线,EH FG 必相交于一点，且这个交点在直线BD 上；（2）取BD 的中点O ，则,⊥⊥BD OA BD OC ，所以BD ⊥平面AOC ，不妨设OD =，则BD AC ==12AO CO ==，所以1441441921cos 212123+-∠==⨯⨯AOC ，以O 为坐标原点建立如图所示的空间直角坐标系，则(0,(12,0,0),(6,--A B C F G ，故=BA ，(=-FG ，(8,0,=-AC ，(4,0,=-EF ，设平面EFGH 的法向量为(,,)n x y z =，由00n EF n FG ⎧⋅=⎨⋅=⎩可得50y x ⎧+=⎪⎨-=⎪⎩，令x =，则(52,=n ，则182cos ,3||||92⋅<>===⨯BA n BA n BA n ，故直线AB 与平面EFGH ． 6．如图，已知四边形ABCD 为菱形，对角线AC 与BD 相交于O ，60BAD ∠=︒，平面ADEF平面BCEF =直线EF ，FO ⊥平面ABCD ，22BC CE DE EF ====（1）求证：//EF DA ；（2）求二面角A EF B --的余弦值．【试题来源】江西省五市九校协作体2021届高三第一次联考【答案】（1）证明见解析；（2）35．【解析】（1）因为四边形ABCD 为菱形，所以//AD BC ，AD ⊄平面BCEF ，BC ⊂平面BCEF ，//AD ∴平面BCEF ，因为平面ADEF平面BCEF =直线,EF AD ⊂平面ADEF ，所以//EF AD ；（2）因为四边形ABCD 为菱形，所以AC BD ⊥，因为OF ⊥平面ABCD ，所以以O 为坐标原点、OA ，OB ，OF 为x ，y ，z 轴建立空间直角坐标系，取CD 中点M ，连EM ，OM ，60BAD ︒∠=，21BC OA OC OB OD =∴====，2BC CD CE DE CDE ====∴为正三角形，EM =11//,=,//,=22OM BC OM BC EF BC EF BC，//,=//,=EF OM EF OM OF EM OF EM∴∴，从而1(0,1,0),((0,1,0),(22A B C D E---，设平面ADEF一个法向量为(,,)m x y z=，则m DAm DE⎧⋅=⎨⋅=⎩，即12yx y⎧+=⎪⎨+=⎪⎩，令11,(1,x y z m=∴===-，设平面BCEF一个法向量为(,,)n x y z=，则n BCn EC⎧⋅=⎨⋅=⎩，即122yx y⎧-=⎪⎨-+-=⎪⎩，令11,(1,3,1)x y z n=∴==-=--，3cos,5|||,|m nm nm n⋅∴<>==，因此二面角A EF B--的余弦值为35．7．如图，在四棱锥P ABCD-中，90BAD∠=，//AD BC，PA AD⊥，PA AB⊥，122PA AB BC AD====．（1）求证：//BC平面PAD；（2）求平面PAB与平面PCD所成锐二面角的余弦值．【试题来源】北京房山区2021届高三上学期数学期末试题【答案】（1）证明见解析；（2【解析】（1）解法1．因为//BC AD，BC⊄平面PAD，AD⊂平面PAD，所以//BC平面PAD，解法2．因为PA AD⊥，PA AB⊥，AD AB⊥，所以以A为坐标原点，,,AB AD AP所在直线分别为x轴、y轴、z轴，建立如图所示空间直角坐标系A xyz-，则(0,0,0),(2,0,0),(0,4,0),(0,0,2),(2,2,0)A B D P C ，平面PAD 的法向量为(1,0,0)t ， (0,2,0)BC = ，因为 0120000t BC ⋅=⨯+⨯+⨯= ，BC ⊄平面PAD ，所以//BC 平面PAD ；（2）因为PA AD ⊥，PA AB ⊥AD AB ⊥，所以以A 为坐标原点，,,AB AD AP 所在直线分别为x 轴、y 轴、z 轴，建立如图所示空间直角坐标系A xyz -，则(0,0,0),(2,0,0),(0,4,0),(0,0,2),(2,2,0)A B D P C所以平面PAB 的法向量为(0,1,0)n = ，设平面PCD 的法向量为(,,)m x y z =， (2,2,2)PC =-，(0,4,2)PD =- ，所以2220042020x y z x y m PC m PC y z z y m PD m PD ⎧⎧+-==⎧⎧⊥⋅=⇒⇒⇒⎨⎨⎨⎨-==⊥⋅=⎩⎩⎩⎩，令1(1,1,2)y m ==得，cos ,1n mn m n m ⋅<>===⨯，设平面PAB 与平面PCD 所成角为θθ，为锐角，所以cos θ=． 8．如图，在四棱锥P ABCD -中，底面ABCD 为菱形，平面PAD ⊥平面ABCD ，PA PD ⊥，PA PD =，3BAD π∠=，E 是线段AD 的中点，连结BE ．（1）求证：BE PA ⊥；（2）求二面角A PD C --的余弦值；（3）在线段PB 上是否存在点F ，使得//EF 平面PCD ？若存在，求出PF PB 的值；若不存在，说明理由．【试题来源】北京市朝阳区2021届高三上学期期末数学质量检测试题【答案】（1）证明见解析；（2）7-；（3）存在；12PF PB =．【解析】（1）因为四边形ABCD 为菱形，所以AB AD =．因为3BAD π∠=，E 为AD 的中点，所以BE AD ⊥．因为平面PAD ⊥平面ABCD ，平面PAD平面ABCD AD =，所以BE ⊥平面PAD ．因为PA ⊂平面PAD ，所以BE PA ⊥．（2）连结PE ．因为PA PD =，E 为AD 的中点，所以PE AD ⊥．由（1）可知BE ⊥平面PAD ，所以BE AD ⊥，PE BE ⊥．设2AD a =，则PE a =．如图，建立空间直角坐标系E xyz -．所以(,0,0),,0),(2,0),(,0,0),(0,0,)A a B C a D a P a --．所以),0(D C a =-，(,0,)D a P a =．因为BE ⊥平面PAD ，所以(0,,0)EB =是平面PAD 的一个法向量．设平面PCD 的法向量为(,,)x y z =n ，则00n DC n DP ⎧⋅=⎨⋅=⎩，即00ax ax az ⎧-+=⎪⎨+=⎪⎩，所以,.x x z ⎧=⎪⎨=-⎪⎩令3x =，则1y =，z =(3,1,n =．所以cos ,||||7n EB n EB n EB ⋅===．由题知，二面角A PD C --为钝角，所以其余弦值为- （3）当点F 是线段PB 的中点时，//EF 平面PCD ．理由如下：因为点E ∈/平面PCD ，所以在线段PB 上存在点F 使得//EF 平面PCD 等价于0EF ⋅=n ．假设线段PB 上存在点F 使得//EF 平面PCD ．设([0,1])PF PBλλ=∈，则PF PB λ=．所以(0,0,),),)EF EP PF EP PB a a a a a λλλ=+=+=+-=-．由)0EF a a a λ⋅=-=n ，得12λ=．所以当点F 是线段PB 的中点时，//EF 平面PCD ，且12PF PB =． 9．如图，在四棱锥P ABCD -中，PD ⊥平面ABCD ，4PD =，底面ABCD 是边长为2的正方形，E ，F 分别为PB ，PC 的中点．（1）求证：平面ADE ⊥平面PCD ；（2）求直线BF 与平面ADE 所成角的正弦值．【试题来源】北京市东城区2021届高三上学期期末考试【答案】（1）证明见解析；（2）15．【解析】（1）因为PD ⊥平面ABCD ，所以PD AD ⊥．因为底面ABCD 是正方形，所以AD CD ⊥．因为PD CD D ⋂=，所以AD ⊥平面PCD ．因为AD ⊂平面ADE ，所以平面ADE ⊥平面PCD ．（2）因为PD ⊥底面ABCD ，所以PD AD ⊥，PD CD ⊥．因为底面ABCD 是正方形，所以AD CD ⊥．如图建立空间直角坐标系D xyz -．因为4PD =，底面ABCD 为边长为2的正方形，所以()0,0,4P ，()2,0,0A ，()2,2,0B ，()0,2,0C ，()0,0,0D ，()1,1,2E ，()0,1,2F ．则()2,0,0DA =，()1,1,2DE =，()2,1,2BF =--．设平面ADE 的法向量(),,m x y z =，由00m DA m DE ⎧⋅=⎨⋅=⎩，可得2020x x y z =⎧⎨++=⎩．令1z =-，则0x =，2y =．所以()0,2,1m =-．设直线BF 与平面ADE 所成角为θ，则,sincos ,9BF mBF m BF m θ====．所以直线BF 与平面ADE ．【名师点睛】本题考查了面面垂直的判定，核心是要求面面垂直，先考虑线面垂直；同时也考查了线面角的计算方法，核心是要求正弦值，先求余弦值．10．如图，已知11ABB A 是圆柱1OO 的轴截面，O 、1O 分别是两底面的圆心，C 是弧AB 上的一点，30ABC ∠=，圆柱的体积和侧面积均为4π．（1）求证：平面1ACA ⊥平面1BCB ；（2）求二面角11B A B C --的大小．【试题来源】江西省吉安市2021届高三大联考数学（理）（3-2）试题【答案】（1）证明见解析；（2）60 ．【解析】（1）因为1AA 是圆柱的母线，所以1AA ⊥平面ABC ，因为BC ⊂平面ABC ，所以1AA BC ⊥，又C 是弧AB 上的一点，且AB 是圆O 的直径，所以AC BC ⊥，因为1AA AC A =，所以BC ⊥平面1ACA ，又BC ⊂平面1BCB ，所以平面1ACA ⊥平面1BCB ；（2）设圆柱的底面半径为r ，母线长为l ，因为圆柱的体积和侧面积均为4π，所以2244rl r l ππππ=⎧⎨=⎩，解得，2r ，1l =，即4AB =，11AA =，因为30ABC ∠=，所以2AC =，BC =设圆柱过C 点的母线为CD ，以C 为原点，CA ，CB ，CD 所在直线分别为x 轴､y 轴､z 轴建立空间直角坐标系C xyz -，如图所示；则()0,0,0C ，()B ，()12,0,1A ，()1B ；所以()12,0,1CA =，()10,CB =，()12,BA =-，()10,0,1BB = 设平面11CA B 的法向量为(),,n x y z =，由1120000x z n CA n CB z ⎧+=⎧⋅=⎪⎪⇒⎨⎨⋅=+=⎪⎪⎩⎩，取z =x =1y =-，所以平面11CA B的一个法向量为(3,n =--，设平面11BA B 的法向量为(),,m a b c=，由1102000m BA a c m BB c ⎧⎧⋅=-+=⎪⎪⇒⎨⎨⋅==⎪⎪⎩⎩，取1b =，则a =0c ，所以平面11BA B 的一个法向量为()3,1,0m =，所以1cos ,23n mm n n m ⋅===-+⋅，由图中可看出二面角11B A B C --是锐角，故二面角11B A B C --的值为60．【名师点睛】证明面面垂直的方法：（1）利用面面垂直的判定定理，先找到其中一个平面的一条垂线，再证明这条垂线在另外一个平面内或与另外一个平面内的一条直线平行即可；（2）利用性质：//,αββγαγ⊥⇒⊥（客观题常用）；（3）面面垂直的定义（不常用）；（4）向量方法：证明两个平面的法向量垂直，即法向量数量积等于0．11．如图1，正方形ABCD ，边长为a，,E F 分别为,AD CD 中点，现将正方形沿对角线AC 折起，折起过程中D 点位置记为T ，如图2．（1）求证：EF TB ⊥；（2）当60TAB ︒∠=时，求平面ABC 与平面BEF 所成二面角的余弦值．【试题来源】安徽省黄山市2020-2021学年高三上学期第一次质量检测（理）【答案】（1）证明见解析；（2．【解析】（1）取AC 中点O ，连,,OT OB BT ，因为ABCD 为正方形，所以,AC OT AC OB ⊥⊥，又OT OB O ⋂=，所以AC ⊥平面OBT ，而TB ⊂平面OBT ，所以AC TB ⊥．又,E F 分别为,AD CD 中点，所以//EF AC ，所以EF TB ⊥；（2）因为60TAB ︒∠=，所以TAB △为等边三角形，TB a =，又2OT OB a ==，所以222TB OB OT =+，即OT OB ⊥．如图建立空间直角坐标系O xyz -，则,0,0,0,,B E F ⎫⎛⎛⎪ ⎝⎭⎝⎭⎝⎭，220,,0,,,2244EF a EB a ⎛⎫⎛⎫==- ⎪ ⎪⎝⎭⎝⎭，平面ABC 法向量(0,0,1)m =设平面BEF 法向量(,,1)x n y =，由00n EF n EB ⎧⋅=⎨⋅=⎩，00244y ay =⎧+-=⎩，012y x =⎧⎪⎨=⎪⎩，1,0,1,cos ,2||||11mn n m n m n ⋅⎛⎫=<>=== ⎪⋅⎝⎭⋅，记平面ABC 与平面BEF 所成二面角为θ，则θ为锐角，所以cos 5θ=即平面ABC 与平面BEF ． 12．如图所示，四棱柱1111ABCD A B C D -的底面是菱形，侧棱垂直于底面，点E ，F 分别在棱1AA ，1CC 上，且满足113AE AA =，113CF CC =，平面BEF 与平面ABC 的交线为l ．（1）证明：直线l ⊥平面1BDD ；（2）已知2EF =，14BD =，设BF 与平面1BDD 所成的角为θ，求sin θ的取值范围．【试题来源】海南省2021届高三年级第二次模拟考试【答案】（1）证明见解析；（2）35⎫⎪⎪⎝⎭．【解析】（1）如图，连接AC ，与BD 交于点O ．由条件可知//AE CF ，且AE CF =，所以//AC EF ，因为EF ⊂平面BEF ，所以//AC 平面BEF ．因为平面BEF 平面ABC l =，所以//AC l ．因为四棱柱1111ABCD A B C D -的底面是菱形，且侧棱垂直于底面，所以AC BD ⊥，1AC BB ⊥，又1BD BB B ⋂=，所以AC ⊥平面1BDD ，所以l ⊥平面1BDD ．（2）如图所示，以O 为坐标原点，分别以OB ，OC 的方向为x ，y 轴的正方向建立空间直角坐标系．设2BD a =，因为1BD BD <，所以02a <<．则OB a =，1DD ==所以(,0,0)B a ，(0,1,0)C，F ⎛ ⎝．由（1）可知(0,1,0)OC =是平面1BDD的一个法向量，而BF a ⎛=- ⎝，所以sin cos ,OC BF OC BF OC BF θ⋅=<>===当02a <<35<<，即3sin 5θ⎫∈⎪⎪⎝⎭．【名师点睛】求空间角的常用方法：（1）定义法，由异面直线所成角、线面角、二面角的定义，结合图形，作出所求空间角，再结合题中条件，解对应三角形，即可求出结果；（2）向量法：建立适当的空间直角坐标系，通过计算向量夹角（直线方向向量与直线方向向量、直线方向向量与平面法向量，平面法向量与平面法向量）余弦值，即可求出结果．13．在三棱柱111ABC A B C -中，1AB AC ==，1AA =AB AC ⊥，1B C ⊥平面ABC ，E 是1B C 的中点．（1）求证：平面1AB C ⊥平面11ABB A ；（2）求直线AE 与平面11AAC C 所成角的正弦值．【试题来源】浙江省宁波市2020-2021学年高三上学期期末【答案】（1）证明见解析；（2【解析】（1）由1B C ⊥平面ABC ，AB平面ABC ，得1AB B C ⊥，又AB AC ⊥，1CB AC C =，故AB ⊥平面1AB C ， AB 平面11ABB A ，故平面11ABB A ⊥平面1AB C ．（2）以C 为原点，CA 为x 轴，1CB 为z 轴，建立如图所示空间直角坐标系，则()0,0,0C ，()1,0,0A ，()1,1,0B，又BC =11BB AA == 故11CB =，()10,0,1B ，10,0,2E ⎛⎫ ⎪⎝⎭，()1,0,0CA =， ()111,1,1AA BB ==--，11,0,2AE ⎛⎫=- ⎪⎝⎭，设平面11AAC C 的一个法向量为(),,n x y z =，则100n CA n AA ⎧⋅=⎪⎨⋅=⎪⎩，即00x x y z =⎧⎨--+=⎩，令1y =，则1z =， ()0,1,1n =，设直线AE 与平面11AAC C 所成的角为θ，故1sin 102nAEn AE θ⋅===，即直线AE 与平面11AAC C14．如图，在平面四边形PABC 中，PA AC ⊥，AB BC ⊥，PA AB ==，2AC =，现把PAC △沿AC 折起，使P 在平面ABC 上的射影为O ，连接OA 、OB ，且OB//AC ．（1）证明：OB ⊥平面PAO ；（2）求二面角O PB C --的余弦值．【试题来源】安徽省六安市示范高中2020-2021学年高三上学期教学质量检测（理）【答案】（1）证明见解析；（2）【解析】（1）PO ⊥平面ABC ，AC ⊂平面ABC ，PO AC ∴⊥，又PA AC ⊥，PAPO P =，所以AC ⊥平面PAO ， //OB AC ，所以OB ⊥平面PAO ；（2）在Rt ABC 中，AB =2AC =，则1BC ==，30BAC ∴∠=，在Rt OAB 中，903060OAB ∠=-=，所以12OA AB ==，32OB =，Rt PAO 中，PA =AO =32OP ∴==，以点O 为坐标原点，OB 、OA 、OP 所在直线分别为x 、y 、z 轴建立空间直角坐标系O xyz -，则0,,02A ⎛⎫ ⎪ ⎪⎝⎭、,02C ⎛⎫ ⎪ ⎪⎝⎭、3,0,02B ⎛⎫ ⎪⎝⎭、30,0,2P ⎛⎫ ⎪⎝⎭，所以33,0,22PB ⎛⎫=- ⎪⎝⎭，32PC ⎛⎫=- ⎪ ⎪⎝⎭，由（1）可知()0,1,0m =为平面POB 的一个法向量，设平面平PBC 的法向量为(),,n x y z =，则有330223202x z x y z ⎧-=⎪⎪⎨⎪-=⎪⎩y x z x ⎧=⎪⇒⎨⎪=⎩，取x =(3,n =-，cos ,717m n m n m n ⋅===-⋅⨯，由图可知，二面角O PB C --为钝角，所以，二面角O PB C --的余弦值为7-． 15．在四棱锥P ABCD -中，平面PAD ⊥平面ABCD ，底面ABCD 为直角梯形，//,90BC AD ADC ∠=︒，11,2BC CD AD E ===为线段AD 的中点，过BE 的平面与线段,PD PC 分别交于点,G F ．（1）求证：GF ⊥平面PAD ；（2）若PA PD ==G为PD 的中点，求平面PAB 与平面BEGF所成锐二面角的余弦值．【试题来源】安徽省名校2020-2021学年高三上学期期末联考（理）【答案】（1）证明见解析；（2．【解析】证明：（1）因为12BC AD =，且E 为线段AD 的中点，所以BC DE =，因为//BC AD ，所以四边形BCDE 为平行四边形，所以//BE CD ，因为CD ⊂平面,PCD BE ⊂/平面PCD ，所以//BE 平面PCD ，又平面BEGF ⋂平面PCD GF =，所以//BE GF ，又BE AD ⊥，且平面PAD ⊥平面ABCD ，平面PAD平面ABCD AD =，所以BE ⊥平面PAD ，所以GF ⊥平面PAD ；（2）因为,PA PD E =为线段AD 的中点，所以PE AD ⊥，‘’因为平面PAD ⊥平面ABCD ，所以PE ⊥平面ABCD ，以E 为坐标原点，EA 的方向为x 轴正方向建立如图所示的空间直角坐标系E xyz -；则11(0,0,1),(1,0,0),(0,1,0),(0,0,0),(1,0,0),,0,22P A B E D G ⎛⎫--⎪⎝⎭，则11(1,0,1),(0,1,1),(0,1,0),(1,0,1),,0,22PA PB BE DP EG ⎛⎫=-=-=-==- ⎪⎝⎭，设平面PAB 的法向量为()111,,m x y z =，则0{0PA m PB m ⋅=⋅=，，，即11110,0x z y z -=⎧⎨-=⎩，不妨令11x =，可得(1,1,1)n =为平面BEGF 的一个法向量，设平面BEGF 的法向量为()222,,n x y z =，则0{0BE n EG n ⋅=⋅=，，，即222011022y x z =⎧⎪⎨-+=⎪⎩，，不妨令21x =，可得(1,0,1)n =为平面BEGF 的一个法向量，设平面PAB 与平面BEGF 所成的锐二面角为α，于是有2cos |cos ,|32m n α=〈〉==；所以平面PAB 与平面BEGF ．16．如图所示，在四棱锥S ABCD -中，底面ABCD 是正方形，对角线AC 与BD 交于点F ，侧面SBC 是边长为2的等边三角形，点E 在棱BS 上．（1）若//SD 平面AEC ，求SE EB的值；（2）若平面SBC ⊥平面ABCD ，求二面角B AS C --的余弦值．【试题来源】江苏省G4（苏州中学、常州中学、盐城中学、扬州中学）2020-2021学年高三上学期期末联考【答案】（1）1；（2．【解析】（1）连结EF ，因为//SD 平面AEC ，SD ⊂平面BSD ，平面BSD ⋂平面AEC EF =，所以//SD EF ．因为底面ABCD 是正方形，F 为AC 中点，所以EF 是SD 的中位线，则1SE EB=．（2）取BC 的中点为O ，AD 的中点为M ，连结MO ，则MO BC ⊥，因为平面SBC ⊥平面ABCD ，平面SBC平面ABCD BC =，OM ⊂平面ABCD ，所以OM ⊥平面SBC ．又OS BC ⊥，所以O 为坐标原点．以{},,OS OC OM 为正交基底建立空间直角坐标系O xyz -．则()0,1,2A -，()010B -,,，()0,1,0C，)S，1,022E ⎛⎫- ⎪ ⎪⎝⎭，从而()SC =-，()0,2,2AC =-，()0,0,2AB =-，()3,1,2AS =-．设平面ASC 的法向量为(),,m x y z =，则0,0.m SC m AC ⎧⋅=⎪⎨⋅=⎪⎩，即0,0.y y z ⎧+=⎪⎨-=⎪⎩取1x =，则y =z = 所以平面ASC的一个法向量为(1,3,m =．设平面ASB 的法向量为(),,n x y z =，则0,0.n AB n AS ⎧⋅=⎪⎨⋅=⎪⎩，即20,20.z y z -=⎧⎪+-=取y =1x =-，0z =．所以平面ASB 的一个法向量为()1,3,0n =-．所以7cos ,7m n m n m n ⋅〈〉==．因为二面角B AS C --的平面角为锐角，所以二面角B AS C --的余弦值为7．【名师点睛】本题的核心在考查空间向量的应用，需要注意以下问题：（1）求解本题要注意两点：一是两平面的法向量的夹角不一定是所求的二面角，二是利用方程思想进行向量运算，要认真细心，准确计算．（2）设,m n 分别为平面α，β的法向量，则二面角θ与,m n <>互补或相等．求解时一定要注意结合实际图形判断所求角是锐角还是钝角．17．在三棱锥P ABC -中，底面ABC 为正三角形，平面PBC ⊥平面,1,ABC PB PC D ==为AP 上一点，2,AD DP O =为三角形ABC 的中心．（1）求证：AC ⊥平面OBD ；（2）若直线PA 与平面ABC 所成的角为45︒，求二面角A BD O --的余弦值．【试题来源】山东省威海市2020-2021学年高三上学期期末【答案】（1）证明见解析；（2）5．【解析】（1）证明：连接AO 并延长BC 交于点E ，则E 为BC 中点，连接PE ．如图所示：因为О为正三角形ABC 的中心，所以2,AO OE =又2AD DP =，所以//,DO PE 因为PB PC =，E 为BC 中点，所以,PE BC ⊥ 又平面PBC ⊥平面ABC ，平面PBC 平面ABC BC =，所以PE ⊥平面,ABC 所以DO ⊥平面,ABC AC ⊂平面PBC ，所以,DO AC ⊥又,AC BO DO BO O ⊥⋂=，所以AC ⊥平面OBD ．（2）由PE ⊥平面ABC 知，所以45PAE ∠=︒ ，所以,PE AE =所以,ABE PBE ≌ 所以1AB PB BC AC ====，由（1）知，,,EA EB EP 两两互相垂直，所以分别以,,EA EB EP 的方向为,,x y z 轴正方向，建立如图所示空间直角坐标系，则1,0,,0,0,0,,22263A B P D ⎛⎫⎛⎫⎛⎛⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎝⎭⎝⎭⎝⎭⎝⎭，10,,02C ⎛⎫- ⎪⎝⎭所以31,,0,2231,,623AB BD ⎛⎫-⎛= ⎪ ⎪⎝⎭=-⎝⎭，设平面ABD 的法向量为(),,n x y z =，则302302x y nBD z y n AB x ⎧⋅=-=⎪⎪⎨⎪⋅=-+=⎪⎩，令1,x =可得1y z ==，则()1,3,1n =．由（1）知AC ⊥平面,DBO 故1,02AC ⎛⎫=-- ⎪ ⎪⎝⎭为平面DBO 的法向量，所以2cos ,5nAC n AC n AC -⋅===-，由图可知二面角A BD O --的为锐二面角，所以二面角A BDO --的余弦值为5． 18．如图，在几何体ABCDEF 中，四边形ABCD 为等腰梯形，且22AB CD ==，60ABC ∠=︒，四边形ACFE 为矩形，且FB =，M ，N 分别为EF ，AB 的中点．（1）求证：//MN 平面FCB；（2）若直线AF 与平面FCB 所成的角为60°，求平面MAB 与平面MAC 所成锐二面角的余弦值．【试题来源】山西省运城市2021届高三上学期期末（理）【答案】（1）证明见解析；（2．【解析】（1）取BC 的中点Q ，连接NQ ，FQ ，则1//2NQ AC ，且12NQ AC =，又1//2MF AC ，且12MF AC = ，所以//MF NQ 且MF NQ =，所以四边形MNQF 为平行四边形，所以//MN FQ ，因为FQ ⊂平面FCB ，MN ⊄平面FCB ，所以//MN 平面FCB ；（2）由四边形ABCD 为等腰梯形，且22AB CD ==，60ABC ∠=︒，可得1BC =，AC =90ACB ∠=︒，所以AC BC ⊥．因为四边形ACFE 为矩形，所以AC CF ⊥，所以AC ⊥平面FCB ，所以AFC ∠为直线AF 与平面FCB 所成的角，即60AFC ∠=︒，所以1FC =．因为FB =，所以222FB FC CB =+，所以FC BC ⊥．则可建立如图所示的空间直角坐标系C xyz -，3(3,0,0),(0,1,0),,0,12A B M ⎛⎫ ⎪⎝⎭，所以3,0,1,(3,1,0)2MA AB ⎛⎫=-=- ⎪⎝⎭，设(,,)m x y z =为平面MAB 的法向量，则00MA m AB m ⎧⋅=⎨⋅=⎩，即30230x z x y ⎧-=⎪⎨⎪-+=⎩，取23x =，则(23,6,3)m =为平面MAB 的一个法向量，又(0,1,0)n =为平面MAC的一个法向量，所以657257cos 571||m n mn m n ⋅〈〉====⨯∣∣，故平面MAB 与平面MAC 所成锐二面角的余弦值为5719． 19．如图，该多面体由底面为正方形ABCD 的直四棱柱被截面AEFG 所截而成，其中正方形ABCD 的边长为4，H 是线段EF 上（不含端点）的动点，36==FC EB ．（1）若H 为EF 的中点，证明：//GH 平面ABCD ；（2）若14=EH EF ，求直线CH 与平面ACG 所成角的正弦值．【试题来源】河南省驻马店市2020-2021学年高三上学期期末考试（理）【答案】（1）证明见解析；（26．【解析】（1）证明：取BC 的中点M ，连接HM ，DM ．因为该多面体由底面为正方形ABCD 的直四棱柱被截面AEFG 所截而成，所以截面AEFG 是平行四边形，则4=-=DG CF EB ．因为36==FC EB ，所以1(26)42=⨯+=HM ，且DG//HM ，所以四边形DGHM 是平行四边形，所以GH //DM ．因为DM ⊂平面ABCD ，GH ⊄平面ABCD ，所以//GH 平面ABCD ．（2）解：如图，以D 为原点，分别以DA ，DC ，DG 的方向为x 轴、y 轴、z 轴的正方向，建立空间直角坐标系D xyz -，则(4,0,0)A ，(0,4,0)C ，(0,0,4)G ，(3,4,3)H ，(4,4,0)=-AC ，(4,0,4)=-AG ，(3,0,3)=CH ．设平面ACG 的法向量为(,,)n x y z =，则440440AC n x y AG n x z ⎧⋅=-+=⎨⋅=-+=⎩，令1x =，得(1,1,1)n =．因为cos ,3||||32⋅〈〉===⨯CH n C n n CH H ，所以直线CH 与平面ACG 所成角的正弦值为3．【名师点睛】本题考查了立体几何中的线面平行的判定和线面角的求解问题，意在考查学生的空间想象能力和逻辑推理能力；解答本题关键在于能利用直线与直线、直线与平面关系的相互转化，通过严密推理证明线线平行从而得线面平行，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．20．如图，已知四边形ABCD 和BCEG 均为直角梯形，//AD BC ，//CE BG ，且2BCD BCE π∠=∠=，120ECD ∠=︒．22BC CD CE AD BG ====．（1）求证：//AG 平面BDE ；（2）求二面角E BD C --的余弦值．【试题来源】安徽省蚌埠市2020-2021学年高三上学期第二次教学质量检查（理）【答案】（1）证明见解析；（2 【解析】（1）证明：在平面BCEG 中，过G 作GN CE ⊥于N ，交BE 于M ，连DM ，由题意知，MG MN =，////MN BC DA 且12MN AD BC ==，因为//MG AD ，MG AD =，故四边形ADMG 为平行四边形，所以//AG DM ，又DM ⊂平面BDE ，AG ⊂/平面BDE ，故//AG 平面BDE ．（2）由题意知BC ⊥平面ECD ，在平面ECD 内过C 点作CF CD ⊥交DE 于F ，以C 为原点，CD ，CB ，CF 的方向为x ，y ，z 轴的正方向建立空间直角坐标系，不妨设1AD =，则22BC CD CE BG ====．且()0,0,0C ，()2,0,0D ，()0,2,0B ，(E -，设平面EBD 的法向量(),,n x y z =，则由0,0,DE n BD n ⎧⋅=⎨⋅=⎩得30,220,x x y ⎧-=⎪⎨-=⎪⎩ 取1y =，得(1,1,3n =，易知平面BCD 的一个法向量为()0,0,1m =，3cos ,51m nm n m n ⋅==⋅=⋅E BD C --． 21．如图，在四棱锥P ABCD -中，底面ABCD 是边长为2的正方形，M 为PC 的中点．（1）求证：//AP 平面BDM ；（2）若PB PC ==CD PC ⊥，求二面角C DM B --的余弦值．【试题来源】河南省湘豫名校2020-2021学年高三上学期1月月考（理）【答案】（1）证明见解析；（2．【解析】（1）连接AC 交BD 于E ，连接EM ，则E 为AC 中点，所以EM 为APC △的中位线，所以//EM AP ，因为EM ⊂平面BDM ，AP ⊄平面BDM ，所以//AP 平面BDM ．（2）在PBC 中，因为2224PB PC BC +==，所以PB PC ⊥，取BC 中点O ，AD 中点F ，连接PO ，OF ，则PO BC ⊥，1PO =，因为BC CD ⊥，CD PC ⊥，BC 、PC ⊂平面PBC ，BC PC C ⋂=，所以CD ⊥平面PBC ，因为PO ⊂平面PBC ，所以CD PO ⊥，因为PO BC ⊥，BC CD C ⋂=，BC 、CD ⊂平面ABCD ，所以PO ⊥平面ABCD ，因为OF ⊂平面ABCD ，所以PO OF ⊥，所以PO ，OF ，OB 两两垂直，如图所示，以O 为原点，OF ，OB ，OP 分别为x 轴，y 轴，z 轴建立空间直角坐标系，则(2,1,0)D -，(0,0,1)P ，(0,1,0)B ，(0,1,0)C -，所以110,,22M ⎛⎫- ⎪⎝⎭，可得112,,22DM ⎛⎫=- ⎪⎝⎭，(2,2,0)BD =-，(2,0,0)CD =．设平面BDM 的法向量为()111,,m x y z =，则0 0m BD m DM ⎧⋅=⎨⋅=⎩，即11111220112022x y x y z -=⎧⎪⎨-++=⎪⎩，取(1,1,3)m =，设平面CDM 的法向量为()222,,n x y z =，则00n CD n DM ⎧⋅=⎨⋅=⎩，即222220112022x x y z =⎧⎪⎨-++=⎪⎩，取(0,1,1)n =-，所以222cos ,11||||112m n m nm n ⋅〈〉===⋅⨯，所以二面角C DM B --的余弦值为11．22．如图所示，矩形ABCD 和梯形BEFC 所在平面互相垂直， //BE CF ，BCF CEF ∠=∠=90°，AD =EF =（1）求证：EF ⊥平面DCE（2）当AB 的长为何值时，二面角A EF C --的大小为60°．【试题来源】山东省菏泽市2020-2021学年高三上学期期末【答案】（1）证明见解析；（2）60°．【解析】（1）因为平面ABCD ⊥平面BEFC ，平面ABCD 平面BEFC BC =，CD BC ⊥，CD ⊂平面ABCD ，所以CD ⊥平面BEFC ，EF ⊂平面BEFC ，从而CD EF ⊥．因为EF CE ⊥，CD CE C =，,CD CE ⊂平面CDE ，所以EF ⊥平面CDE ．（2）如图所示，以点C 为坐标原点，以CB 、CF 和CD 所在直线分别为x 轴、y 轴和z 轴建立空间直角坐标系．过点E 作EG CF ⊥于点G ．在Rt EFG中，EG AD ==EF =1FG =．因为CE EF ⊥，则90EFC ECF BCE ∠=︒-∠=∠，所以Rt EFG Rt ECB △△，EG GF EF BE BC EC==，所以2,BE CE == 所以2CG =，所以3CF =．设AB a ，则()0,0,0C，)A a，)E ，()0,3,0F ．()0,2,AE a =-，()EF =-，()2,2,0CE =，设平面AEF 的法向量(),,n x y z =．则00n AE n EF ⎧⋅=⎨⋅=⎩，即200y az y -=⎧⎪⎨+=⎪⎩，令2z=，得,2n a ⎫=⎪⎭．因为CD ⊥平面EFC ，()0,0,CD a =，所以1cos ,2n CD ==，解得a =所以当AB =A EF C --的大小为60°．【名师点睛】本题考查空间向量法求二面角．求空间角的方法：（1）几何法（定义法）：根据定义作出二面角的平面角并证明，然后解三角形得出结论；（2）空间向量法：建立空间直角坐标系，写出各点为坐标，求出平面的法向量，由两个平面法向量的夹角得二面角（它们相等或互补）．23．如图，四棱锥E ABCD -中，底面ABCD 为直角梯形，其中AB BC ⊥，//CD AB ，面ABE ⊥面ABCD ，且224AB AE BE BC CD =====，点M 在棱AE 上．（1）证明：当2MA EM =时，直线//CE 平面BDM ；（2）当AE ⊥平面MBC 时，求二面角E BD M --的余弦值．【试题来源】内蒙古赤峰市2021届高三模拟考试（理）【答案】（1）证明见解析；（2．【解析】（1）连结BD 与AC 交于点N ，连结MN ，//AB CD ，24AB CD ==， CND ANB ∴△∽△，12CD CN AB AN ∴==， 12EM MA =，EM CN MA AN∴=，MN //EC ∴，又MN ⊂面BDM ，CE ⊂面BDM ，//CE ∴平面BDM ．（2）AE 平面MBC ，AE BM ∴⊥，M ∴是AE 的中点，取AB 的中点为O ， OE ∴⊥平面ABCD ，以OD ，OA ，OE 所在的直线为x ，y ，z 轴建立空间直角坐标系O xyz -，则(0,2,0)B-，E ，(2,0,0)D ，(0,2,0)A ，M ，设平面EBD 的法向量为()1111,,x n y z=，则1111112200020x y n BD n BE y ⎧+=⎧⋅=⎪⎪⇒⎨⎨⋅=+=⎪⎪⎩⎩，令11z =，则1y=1x =1(3,3,1)n ∴=-，设平面BDM 的法向量为()2222,,n x y z =，则2222222200030x y n BD n BM y ⎧+=⎧⋅=⎪⎪⇒⎨⎨⋅==⎪⎪⎩⎩，令2z 21y =-，21x =，1(1,13)n ∴=-， 1212123105cos ,||n n n n n n ⋅∴<>===⋅ ∴二面角E BD M --的余弦值为35． 24．已知正方体1111ABCD A B C D -，棱长为2，M 为棱CD 的中点，N 为面对角线1BC 的中点，如图．（1）求证：ND AN ⊥；（2）求平面1AMD 与平面11AAC C 所成锐二面角的余弦值．【试题来源】安徽省池州市2020-2021学年高三上学期期末（理）【答案】（1）证明见解析；（2 【解析】（1）取BC 的中点分别为F ，连接NF ，DF ，因为N ，F 分别为1BC ，BC 的中点，1111ABCD A B C D -是正方体，易得NF ⊥平面ABCD ，所以NF AM ⊥；因为FC MD =，AD DC =，FCD MDA ∠=∠，所以FCD MDA ≌△△，所以CFD DMA ∠=∠，所以90FDC DMA ∠+∠=︒，所以FD AM ⊥，因为NF FD F =，NF ⊂平面NFD ，FD ⊂平面NFD ，所以AM ⊥平面NFD ，又DN ⊂平面NFD ，所以ND AM ⊥；（2）以A 为原点，分别以AB 、AD 、1AA 方向为x 轴、y 轴、z 轴正方向，建立如下图所示空间直角坐标系：连接BD ，1C D ，在正方体1111ABCD A B C D -中，易知1BD C D =，且N 为1BC 中点，所以1DN BC ⊥．又11//BC AD ，所以1AD DN ⊥．因为1AD AM A =，1AD ⊂平面1AMD ，AM ⊂平面1AMD ，所以ND ⊥平面1AMD ，故ND 为平面1AMD 的一个法向量；由1111ABCD A B C D -是正方体，得BD ⊥平面11AAC C ，故BD 为平面11AAC C 的一个法向量，因为()2,0,0B ，()0,2,0D ，()2,1,1N ，所以()2,1,1ND =--，()2,2,0BD =-，所以(cos ,ND BDND BD ND BD -⋅<>===⋅则平面1AMD 与平面11AAC C25．如图，正方形ADEF 与梯形ABCD 所在的平面互相垂直，AD CD ⊥，AB ∥CD ，122AB AD CD ===，点M 在线段EC 上．（1）当点M 为EC 中点时，求证：BM ∥平面ADEF ；（2）当平面BDM 与平面ABFM 在线段EC 上的位置．【试题来源】宁夏固原市第五中学2021届高三年级期末考试（理）【答案】（1）证明见解析；（2）点M 为EC 中点．【解析】（1）以直线DA 、DC 、DE 分别为x 轴、y 轴、z 轴建立空间直角坐标系，则(2,0,0)A ，(2,2,0)B ，(0,4,0)C ，(0,0,2)E ，所以(0,2,1)M ．所以(2,0,1)BM =-，又(0,4,0)DC =是平面ADEF 的一个法向量．因为0BM DC ⋅=即BM DC ⊥，BM ⊄平面ADEF ，所以BM ∥平面ADEF ；（2）设(,,)M x y z ，则(,,2)EM x y z =-，又(0,4,2)EC =-，设()01EM EC λλ=≤≤，则0,4,22x y z λλ===-，即(0,4,22)M λλ-．设111(,,)n x y z =是平面BDM 的一个法向量，则11112204(22)0DB n x y DM n y z λλ⎧⋅=+=⎪⎨⋅=+-=⎪⎩，取11x =得11y =-，此时显然1λ=时不符合，则121z λλ=-，即2(1,1,)1n λλ=--，又由题设，(2,0,0)DA =是平面ABF 的一个法向量，所以cos ,622DA n DA n DA n ⋅===⋅，解得12λ=，即点M 为EC 中点．【名师点睛】利用法向量求解空间面面角的关键在于“四破”：第一，破“建系关”，构建恰当的空间直角坐标系；第二，破“求坐标关”，准确求解相关点的坐标；第三，破“求法向量关”，求出平面的法向量；第四，破“应用公式关”．26．如图所示，在多面体ABCDEF 中，//AB CD ，AB BC ⊥，22AB BC CD ==，四边形ADEF 为矩形，平面ADEF ⊥平面ABCD ，AF AB λ=．（1）证明：//DF 平面BCE ；（2）若二面角C BE F --λ的值．【试题来源】江西宜春市2021届高三上学期数学（理）期末试题【答案】（1）证明见解析；（2）1．【解析】（1）取AB 的中点为M ，连接FM CM DM ，，，因为//AM CD 且AM CD =，四边形AMCD 为平行四边形，所以//AD MC 且AD MC =，因为四边形ADEF 为矩形，所以//FE MC 且=FEMC ，所以四边形EFMC 是平行四边形，所以//FM EC ，且EC ⊂平面BEC ，FM ⊄平面BEC ，。

专题15 复数的四则运算(解析版)

专题15 复数的四则运算一、单选题1．若复数Z 满足()·1 2z i i -=(i 是虚数部位)，则下列说法正确的是 A ．z 的虚部是-i B ．Z 是实数C ．z =D ．2z z i +=【试题来源】江苏省盐城市滨海中学2020-2021学年高三上学期迎八省联考考前热身【答案】C【分析】首先根据题意化简得到1z i =-，再依次判断选项即可．【解析】()()()22122211112i i i i iz i i i i ++====---+-．对选项A ，z 的虚部是1-，故A 错误．对选项B ，1z i =-为虚数，故B 错误．对选项C ，z ==C 正确．对选项D ，112z z i i +=-++=，故D 错误．故选C 2．已知复数1z i =+（i 为虚数单位），则1z在复平面内对应的点在 A ．第一象限 B ．第二象限 C ．第三象限D ．第四象限【试题来源】安徽省六安市示范高中2020-2021学年高三上学期教学质量检测（文）【答案】D【分析】由复数的运算化简1z，再判断复平面内对应的点所在象限．【解析】因为()()11111122i i z i i -==-+-，所以1z 在复平面内对应的点11 ,22⎛⎫- ⎪⎝⎭在第四象限．故选D3．已知复数1z i =+（i 为虚数单位），则1z在复平面内对应的点在 A ．第一象限 B ．第二象限 C ．第三象限D ．第四象限【试题来源】安徽省六安市示范高中2020-2021学年高三上学期教学质量检测（理）【答案】D 【分析】化简复数1z，利用复数的几何意义可得出结论．【解析】因为()()11111112i i z i i i --===++-，所以1z在复平面内对应的点的坐标为11,22⎛⎫- ⎪⎝⎭，在第四象限．故选D ． 4．设复数z 满足11zi z+=-，则z = A ．i B ．i - C ．1D ．1i +【试题来源】山东省威海市2020-2021学年高三上学期期末【答案】B【分析】利用除法法则求出z ，再求出其共轭复数即可【解析】11zi z+=-得()11z i z +=-，即()()()()111111i i i z i i i i ---===++-，z i =-，故选B. 5．(1)(4)i i -+= A ．35i + B ．35i - C ．53i +D ．53i -【试题来源】安徽省皖西南联盟2020-2021学年高三上学期期末（文）【答案】D【分析】根据复数的乘法公式，计算结果．【解析】2(1)(4)4453i i i i i i -+=-+-=-．故选D 6．设复数z 满足()11z i i -=+，则z 的虚部为． A ．1- B ．1 C ．iD ．i -【试题来源】安徽省芜湖市2020-2021学年高三上学期期末（文）【答案】B【分析】利用复数的除法化简复数z ，由此可得出复数z 的虚部．【解析】()11z i i -=+，()()()211111i iz i i i i ++∴===--+，因此，复数z 的虚部为1．故选B ． 7．若复数z 满足21zi i=+，则z = A ．22i + B ．22i - C ．22i --D ．22i -+【试题来源】安徽省芜湖市2020-2021学年高三上学期期末（理）【答案】C【分析】求出()2122z i i i =+=-+，再求解z 即可．【解析】()2122z i i i =+=-+，故22z i =--，故选C. 8．将下列各式的运算结果在复平面中表示，在第四象限的为A ．1ii + B ．1ii +- C ．1i i-D ．1i i--【试题来源】河南省湘豫名校2020-2021学年高三上学期1月月考（文）【答案】A【分析】对A 、B 、C 、D 四个选项分别化简，可得．【解析】由11ii i+=-在第四象限．故选A ．【名师点睛】（1）复数的代数形式的运算主要有加、减、乘、除及求低次方根；（2）复数除法实际上是分母实数化的过程．9．若复数z 满足()z 1i i +=- (其中i 为虚数单位)则复数z 的虚部为A ．12-B ．12C ．12i -D ．12i【试题来源】安徽省马鞍山市2020-2021学年高三上学期第一次教学质量监测（文）【答案】A【分析】先由已知条件利用复数的除法运算求出复数z ，再求其虚部即可．【解析】由()z 1i i +=-可得()()()111111222i i i z i i i ----===--+-，所以复数z 的虚部为12-，故选A 10．复数z 满足()212()z i i -⋅+=（i 为虚数单位），则复数z 在复平面内对应的点在 A ．第一象限 B ．第二象限 C ．第三象限D ．第四象限【试题来源】宁夏吴忠市2021届高三一轮联考（文）【答案】D【分析】先计算复数221z i i=++，再求其共轭复数，即可求出共轭复数对应的点，进而可得在复平面内对应的点所在的象限．【解析】由()()212z i i -⋅+=得()()()()21212211112i i z i i i i i ---====-++-，所以1z i =+，1z i =-．所以复数z 在复平面内对应的点为()1,1-，位于第四象限，故选D ．11．已知复数z 满足(2)z i i -=(i 为虚数单位)，则z = A ．125i-+ B ．125i-- C ．125i- D ．125i+ 【试题来源】安徽省名校2020-2021学年高三上学期期末联考（文）【答案】A【分析】由已知可得2iz i=-，再根据复数的除法运算可得答案．【解析】因为(2)z i i -=，所以()()()2122225i i i i z i i i +-+===--+．故选A ． 12．已知复数3iz i-=，则z =A ．4 BCD ．2【试题来源】江西省吉安市“省重点中学五校协作体”2021届高三第一次联考（文）【答案】B【分析】利用复数代数形式的乘除运算化简，再由复数模的计算公式求解．【解析】因为()()()3331131i i i i z i i i i -⋅----====--⋅-，所以z ==B ．【名师点睛】本题考查复数代数形式的乘除运算，考查复数模的求法，属于基础题． 13．复数z 满足：()11i z i -=+，其中i 为虚数单位，则z 的共轭复数在复平面对应的点的坐标为 A ．0,1 B ．0,1 C ．1,0D ．()1,0【试题来源】江西宜春市2021届高三上学期数学（理）期末试题【答案】A【分析】先由()11i z i -=+求出复数z ，从而可求出其共轭复数，进而可得答案【解析】由()11i z i -=+，得21i (1i)2ii 1i (1i)(1+i)2z ++====--，所以z i =-，所以其在复平面对应的点为0,1，故选A 14．已知复数312iz i+=-，则z =A ．1 BCD ．2【试题来源】湖南省岳阳市平江县第一中学2020-2021学年高二上学期1月阶段性检测【答案】B【分析】利用复数的除法法则化简复数z ，利用复数的模长公式可求得z ．【解析】()()()()2312337217121212555i i i i i z i i i i +++++====+--+，因此，z ==B ． 15．设复1iz i=+（其中i 为虚数单位），则复数z 在复平面内对应的点位于A ．第一象限B ．第二象限C ．第三象限D ．第四象限【试题来源】江苏省南通市如皋市2020-2021学年高三上学期期末【答案】A【分析】利用复数的除法化简复数z ，利用复数的几何意义可得出结论．【解析】()()()1111111222i i i i z i i i i -+====+++-，因此，复数z 在复平面内对应的点位于第一象限．故选A ．16．已知(1)35z i i +=-，则z = A ．14i - B ．14i -- C ．14i -+D ．14i +【试题来源】江苏省盐城市一中、大丰高级中学等四校2020-2021学年高二上学期期末联考【答案】B【分析】由复数的除法求解．【解析】由题意235(35)(1)3355141(1)(1)2i i i i i i z i i i i -----+====--++-．故选B 17．复数(2)i i +的实部为 A ．1- B ．1 C ．2-D ．2【试题来源】浙江省绍兴市上虞区2020-2021学年高三上学期期末【答案】A【分析】将(2)i i +化简即可求解．【解析】(2)12i i i +=-+的实部为1-，故选A ．18．已知i 是虚数单位，(1)2z i i +=，则复数z 所对应的点位于 A ．第一象限 B ．第二象限 C ．第三象限D ．第四象限【试题来源】山东省德州市2019-2020学年高一下学期期末【答案】D【分析】利用复数的运算法则求解复数z ，再利用共轭复数的性质求z ，进而确定z 所对应的点的位置．【解析】由(1)2z i i +=，得()()()()2121211112i i i i z i i i i -+====+++-，所以1z i =-，所以复数z 所对应的点为()1,1-，在第四象限，故选D ．【名师点睛】对于复数的乘法，类似于多项式的四则运算，可将含有虚数单位i 的看作一类同类项，不含i 的看作另一类同类项，分别合并即可；对于复数的除法，关键是分子分母同乘以分母的共轭复数，解题中要注意把i 的幂写成最简形式． 19．若复数2iz i=+，其中i 为虚数单位，则z =A B C ．25D ．15【试题来源】重庆市南开中学2020-2021学年高二上学期期末【答案】B【分析】先利用复数的除法运算法则化简复数2iz i=+，再利用复数模的公式求解即可．【解析】因为()()()21212222555i i i i z i i i i -+====+++-，所以z ==，故选B ． 20．52i i-= A ．152i--B ．52i-- C ．152i- D ．152i+ 【试题来源】江西省吉安市2021届高三上学期期末（文）【答案】A【分析】根据复数的除法的运算法则，准确运算，即可求解．【解析】由复数的运算法则，可得()5515222i i i ii i i ----==⨯．故选A ．21．设复数z 满足()1z i i R +-∈，则z 的虚部为 A ．1 B ．-1 C ．iD ．i -【试题来源】湖北省2020-2021学年高三上学期高考模拟演练【答案】B【分析】根据复数的运算，化简得到()11(1)z i i a b i +-=+++，根据题意，求得1b =-，即可求得z 的虚部，得到答案．【解析】设复数,(,)z a bi a b R =+∈，则()11(1)z i i a b i +-=+++，因为()1z i i R +-∈，可得10b +=，解得1b =-，所以复数z 的虚部为1-．故选B ． 22．若复数151iz i-+=+，其中i 为虚数单位，则z 的虚部是 A ．3 B ．3- C ．2D ．2-【试题来源】安徽省淮南市2020-2021学年高三上学期第一次模拟（文）【答案】A【分析】先利用复数的除法运算，化简复数z ，再利用复数的概念求解．【解析】因为复数()()()()1511523111i i i z i i i i -+--+===+++-，所以z 的虚部是3，故选A. 23．若m n R ∈、且4334im ni i+=+-（其中i 为虚数单位），则m n -= A ．125- B ．1- C ．1D ．0【试题来源】湖北省部分重点中学2020-2021学年高三上学期期末联考【答案】B【分析】对已知进行化简，根据复数相等可得答案．【解析】因为()()()()433443121225343434916i i i ii m ni i i i +++-+====+--++，根据复数相等，所以0,1m n ==，所以011m n -=-=-．故选B ．24．若复数z满足()36z =-（i 是虚数单位），则复数z =A．32-B．32- C．322+D．322-- 【试题来源】湖北省荆州中学2020-2021学年高二上学期期末【答案】A【分析】由()36z =-，得z =，利用复数除法运算法则即可得到结果．【解析】复数z满足()36z +=-，6332z --=====-∴+，故选A ．25．若复数2i()2i+=∈-R a z a 是纯虚数，则z = A ．2i - B ．2i C ．i -D ．i【试题来源】河南省驻马店市2020-2021学年高三上学期期末考试（理）【答案】D【分析】由复数的除法运算和复数的分类可得结果．【解析】因为2i (2i)(2i)22(4)i2i (2i)(2i)5+++-++===-+-a a a a z 是纯虚数，所以22040a a -=⎧⎨+≠⎩，则1a =，i =z ．故选D ．26．复数12z i =+，213z i =-，其中i 为虚数单位，则12z z z =⋅在复平面内的对应点位于 A ．第一象限 B ．第二象限 C ．第三象限D ．第四象限【试题来源】江苏省G4（苏州中学、常州中学、盐城中学、扬州中学）2020-2021学年高三上学期期末联考【答案】D【分析】根据复数的乘法法则，求得55z i =-，即可求得答案．【解析】由题意得122(2)(13)25355i i i i i z z z =+-=-==--⋅，所以12z z z =⋅在复平面内的对应点为（5，-5）位于第四象限，故选D27．复数2()2+∈-R a ia i 的虚部为 A ．225+aB ．45a - C ．225a -D ．45a +【试题来源】河南省驻马店市2020-2021学年高三上学期期末考试（文）【答案】D【分析】由得数除法运算化为代数形式后可得．【解析】因为2i (2i)(2i)22(4)i 2i (2i)(2i)5+++-++==-+-a a a a ，所以其虚部为45a +．故选D ． 28．复数z 满足()12z i i ⋅+=，则2z i -=ABCD ．2【试题来源】安徽省蚌埠市2020-2021学年高三上学期第二次教学质量检查（文）【答案】A【分析】先利用除法化简计算z ，然后代入模长公式计算．【解析】()1i 2i z ⋅+=变形得22222221112-+====++-i i i i z i i i ，所以2121-=+-=-==z i i i i A ．29．i 是虚数单位，若()17,2ia bi ab R i-=+∈+，则ab 的值是 A ．15- B ．3- C ．3D ．15【试题来源】山东省菏泽市2020-2021学年高三上学期期末【答案】C【分析】根据复数除法法则化简得数后，由复数相等的定义得出,a b ，即可得结论．【解析】17(17)(2)2147132(2)(2)5i i i i i i i i i ------===--++-，所以1,3a b =-=-，3ab =．故选C ． 30．复数3121iz i -=+的虚部为 A ．12i -B ．12i C ．12-D ．12【试题来源】江西省赣州市2021届高三上学期期末考试（理）【答案】C【分析】由复数的乘除法运算法则化简为代数形式，然后可得虚部．【解析】231212(12)(1)1223111(1)(1)222i i i i i i i z i i i i i ---++--=====-+--+，虚部为12-．故选C ． 31．若复数z 满足(1)2i z i -=，i 是虚数单位，则z z ⋅=AB ．2C ．12D ．2【试题来源】内蒙古赤峰市2021届高三模拟考试（理）【答案】B【分析】由除法法则求出z ，再由乘法法则计算．【解析】由题意222(1)2()11(1)(1)2i i i i i z i i i i ++====-+--+，所以(1)(1)2z z i i ⋅=-+--=．故选B ． 32．若23z z i +=-，则||z =A ．1 BCD ．2【试题来源】河南省（天一）大联考2020-2021学年高三上学期期末考试（理）【答案】B【分析】设(,)z a bi a b R =+∈，代入已知等式求得,a b 后再由得数的模的定义计算．【解析】设(,)z a bi a b R =+∈，则22()33z z a bi a bi a bi i +=++-=-=-，所以以331a b =⎧⎨-=-⎩，解得11a b =⎧⎨=⎩，所以==z B ．33．复数z 满足(2)(1)2z i i -⋅+=（i 为虚数单位），则z = A ．1 B ．2CD 【试题来源】宁夏吴忠市2021届高三一轮联考（理）【答案】C【分析】先将复数化成z a bi =+形式，再求模．【解析】由(2)(1)2z i i -⋅+=得2211z i i i-==-+，所以1z i =+，z ==C ．34．已知a R ∈，若()()224ai a i i +-=-（i 为虚数单位），则a = A ．-1 B ．0 C ．1D ．2【试题来源】浙江省杭州市2020-2021学年高三上学期期末教学质量检测【答案】B【分析】将()()22ai a i +-展开可得答案．【解析】()()()222444ai a i a a i i +-=+-=-，所以0a =，故选B.35．已知i 为虚数单位，且复数3412ii z+=-，则复数z 的共轭复数为 A ．12i -+ B ．12i -- C ．12i +D ．1 2i -【试题来源】湖北省孝感市应城市第一高级中学2020-2021学年高二上学期期末【答案】D【分析】根据复数模的计算公式，以及复数的除法运算，求出z ，即可得出其共轭复数．【解析】因为3412i i z+=-，所以512z i =-，则()()()512512121212i z i i i i +===+--+，因此复数z 的共轭复数为1 2i -．故选D ． 36．已知复数i()1ia z a +=∈+R 是纯虚数，则z 的值为 A ．1 B ．2 C ．12D ．-1【试题来源】江西省赣州市2021届高三上学期期末考试（文）【答案】A【分析】根据复数除法运算化简z ，根据纯虚数定义求得a ，再求模长．【解析】()()()()11121122a i i a i a a z i i i i +-++-===+++-是纯虚数，102102a a +⎧=⎪⎪∴⎨-⎪≠⎪⎩，解得1a =-，所以z i ，1z =．故选A ． 37．设复数11iz i，那么在复平面内复数31z -对应的点位于 A ．第一象限 B ．第二象限 C ．第三象限D ．第四象限【试题来源】陕西省咸阳市2020-2021学年高三上学期高考模拟检测（一）（理）【答案】C【分析】利用复数的除法法则化简复数z ，再将复数31z -化为一般形式，即可得出结论．【解析】()()()21121112i ii z i i i i ---====-++-，3113z i ∴-=--，因此，复数31z -在复平面内对应的点位于第三象限．故选C ． 38．已知复数13iz i-=+（i 为虚数单位），则z 在复平面内对应的点位于 A ．第一象限B ．第二象限C ．第三象限D ．第四象限【试题来源】江西省南昌市新建区第一中学2020-2021学年高二上学期期末考试（理）【答案】D【分析】将复数化简成z a bi =+形式，则在复平面内对应的点的坐标为(),a b ，从而得到答案．【解析】因为1(1)(3)24123(3)(3)1055i i i i z i i i i ----====-++-，所以z 在复平面内对应的点12(,)55-位于第四象限，故选D.39．若复数2(1)34i z i+=+，则z =A ．45 B ．35C ．25D 【试题来源】成都市蓉城名校联盟2020-2021学年高三上学期（2018级）第二次联考【答案】C 【分析】先求出8625iz -=，再求出||z 得解．【解析】由题得()()()()212342863434343425i i i i iz i i i i +-+====+++-，所以102255z ===．故选C. 40．设复数11iz i，那么在复平面内复数1z -对应的点位于 A ．第一象限 B ．第二象限 C ．第三象限D ．第四象限【试题来源】陕西省咸阳市2020-2021学年高三上学期高考模拟检测（一）（文）【答案】C【分析】先求出z i =-，11z i -=--，即得解．【解析】由题得21(1)21(1)(1)2i i iz i i i i ---====-++-，所以11z i -=--，它对应的点的坐标为(1,1)--，所以在复平面内复数1z -对应的点位于第三象限．故选C. 二、多选题1．已知m ∈R ，若6()64m mi i +=-，则m =A ．B ．1-CD ．1【试题来源】2021年高考一轮数学（理）单元复习一遍过【答案】AC【分析】将6()m mi +直接展开运算即可．【解析】因为()()66661864m mi m i im i +=+=-=-，所以68m =，所以m =故选AC ． 2．设复数z 满足1z i z+=，则下列说法错误的是 A ．z 为纯虚数B ．z 的虚部为12i -C ．在复平面内，z 对应的点位于第三象限D ．2z = 【试题来源】2021年新高考数学一轮复习学与练【答案】AB【分析】先由复数除法运算可得1122z i =--，再逐一分析选项，即可得答案．【解析】由题意得1z zi +=，即111122z i i -==---，所以z 不是纯虚数，故A 错误；复数z 的虚部为12-，故B 错误；在复平面内，z 对应的点为11(,)22--，在第三象限，故C 正确；2z ==，故D 正确．故选AB 【名师点睛】本题考查复数的除法运算，纯虚数、虚部的概念，复平面内点所在象限、复数求模的运算等知识，考查计算求值的能力，属基础题．3．已知复数122z =-，则下列结论正确的有 A ．1z z ⋅=B ．2z z =C ．31z =-D ．202012z =-+ 【试题来源】山东新高考质量测评联盟2020-2021学年高三上学期10月联考【答案】ACD【分析】分别计算各选项的值，然后判断是否正确，计算D 选项的时候注意利用复数乘方的性质．【解析】因为111312244z z ⎛⎫⎛⎫-+=+= ⎪⎪⎪⎪⎝⎭⎭=⎝⋅，所以A 正确；因为221122z ⎛⎫=-⎪⎪⎝⎭=，122z =+，所以2z z ≠，所以B 错误；因为3211122z z z ⎛⎫⎛⎫=⋅=-=- ⎪⎪ ⎪⎪⎝⎭⎝⎭，所以C 正确；因为6331z z z =⋅=，所以()202063364431112222zzz z z ⨯+⎛⎫===⋅=-⋅-=-+ ⎪ ⎪⎝⎭，所以D 正确，故选ACD ．【名师点睛】本题考查复数乘法与乘方的计算，其中还涉及到了共轭复数的计算，难度较易． 4．下面是关于复数21iz =-+的四个命题，其中真命题是A ．||z =B ．22z i =C ．z 的共轭复数为1i -+D ．z 的虚部为1-【试题来源】福建省龙海市第二中学2019-2020学年高二下学期期末考试【答案】ABCD【分析】先根据复数的除法运算计算出z ，再依次判断各选项．【解析】()()()2121111i z i i i i --===---+-+--，z ∴==，故A 正确；()2212z i i =--=，故B 正确；z 的共轭复数为1i -+，故C 正确；z 的虚部为1-，故D 正确；故选ABCD ．【名师点睛】本题考查复数的除法运算，以及对复数概念的理解，属于基础题． 5．若复数351iz i-=-，则A ．z =B ．z 的实部与虚部之差为3C ．4z i =+D ．z 在复平面内对应的点位于第四象限【试题来源】2021年新高考数学一轮复习学与练【答案】AD【分析】根据复数的运算先求出复数z ，再根据定义、模、几何意义即可求出．【解析】()()()()351358241112i i i iz i i i i -+--====---+，z ∴==，z 的实部为4，虚部为1-，则相差5，z 对应的坐标为()41-,，故z 在复平面内对应的点位于第四象限，所以AD 正确，故选AD ．6．已知复数202011i z i+=-(i 为虚数单位)，则下列说法错误的是A ．z 的实部为2B ．z 的虚部为1C ．z i =D ．||z =【试题来源】2021年新高考数学一轮复习学与练【答案】AC【分析】根据复数的运算及复数的概念即可求解．【解析】因为复数2020450511()22(1)11112i i i z i i i i +++=====+---，所以z 的虚部为1，||z =，故AC 错误，BD 正确．故选AC. 7．已知复数cos sin 22z i ππθθθ⎛⎫=+-<< ⎪⎝⎭（其中i 为虚数单位）下列说法正确的是A ．复数z 在复平面上对应的点可能落在第二象限B ．z 可能为实数C ．1z =D ．1z的虚部为sin θ 【试题来源】湖北省六校（恩施高中、郧阳中学、沙市中学、十堰一中、随州二中、襄阳三中）2020-2021学年高三上学期11月联考【答案】BC【分析】分02θπ-<<、0θ=、02πθ<<三种情况讨论，可判断AB 选项的正误；利用复数的模长公式可判断C 选项的正误；化简复数1z，利用复数的概念可判断D 选项的正误．【解析】对于AB 选项，当02θπ-<<时，cos 0θ>，sin 0θ<，此时复数z 在复平面内的点在第四象限；当0θ=时，1z R =-∈；当02πθ<<时，cos 0θ>，sin 0θ>，此时复数z 在复平面内的点在第一象限．A 选项错误，B 选项正确；对于C 选项，22cos sin 1z θθ=+=，C 选项正确；对于D 选项，()()11cos sin cos sin cos sin cos sin cos sin i i z i i i θθθθθθθθθθ-===-++⋅-，所以，复数1z的虚部为sin θ-，D 选项错误．故选BC ． 8．已知非零复数1z ，2z 满足12z z R ∈，则下列判断一定正确的是 A ．12z z R +∈B ．12z z R ∈C ．12z R z ∈D ．12z R z ∈【试题来源】重庆市南开中学2020-2021学年高二上学期期中【答案】BD【分析】设12,(,,,)z a bi z c di a b c d R =+=+∈，结合选项逐个计算、判定，即可求解．【解析】设12,(,,,)z a bi z c di a b c d R =+=+∈，则()()12()()z z a bi c di ac bd ad bc i =++=-++，则0ad bc +=，对于A 中，12()()z z a bi c di a c b d i +=+++=+++，则12z z R +∈不一定成立，所以不正确；对于B 中，12()()ac bd ad bc z R i z =-+∈-一定成立，所以B 正确；对于C 中，()()()()2122()()a bi c di a bi ac bd ad bc i R c di c di c z di z c d+-++--==∈++-+=不一定成立，所以不正确；对于D 中，()()()()2122()()a bi c di a bi ac bd ad bc iR c di c di c z di z c d ++++++==∈--++=一定成立，所以正确．故选BD ．9．已知复数()()()32=-+∈z a i i a R 的实部为1-，则下列说法正确的是 A ．复数z 的虚部为5- B ．复数z 的共轭复数15=-z i C．z =D ．z 在复平面内对应的点位于第三象限【试题来源】辽宁省六校2020-2021学年高三上学期期中联考【答案】ACD【分析】首先化简复数z ，根据实部为-1，求a ，再根据复数的概念，判断选项．【解析】()()()()23232323223z a i i a ai i i a a i =-+=+--=++-，因为复数的实部是-1，所以321a +=-，解得1a =-，所以15z i =--，A ．复数z 的虚部是-5，正确；B ．复数z 的共轭复数15z i =-+，不正确；C ．z ==D ．z 在复平面内对应的点是()1,5--，位于第三象限，正确．故选ACD 10．已知复数cos sin 22z i ππθθθ⎛⎫=+-<< ⎪⎝⎭（其中i 为虚数单位），下列说法正确的是（） A ．复数z 在复平面上对应的点可能落在第二象限 B ．cos z θ=C ．1z z ⋅=D ．1z z+为实数【试题来源】山东省菏泽市2021届第一学期高三期中考试数学（B ）试题【答案】CD【分析】利用复数对应点，结合三角函数值的范围判断A ；复数的模判断B ；复数的乘法判断C ；复数的解法与除法，判断D ．【解析】复数cos sin ()22z i ππθθθ=+-<<（其中i 为虚数单位），复数z 在复平面上对应的点(cos ,sin )θθ不可能落在第二象限，所以A 不正确；1z ==，所以B 不正确；22·(cos sin )(cos sin )cos sin 1z z i i θθθθθθ=+-=+=．所以C 正确；11cos sin cos sin cos()sin()2cos cos sin z i i i z i θθθθθθθθθ+=++=++-+-=+为实数，所以D 正确；故选CD11．已知i 为虚数单位，下面四个命题中是真命题的是 A ．342i i +>+B ．24(2)()a a i a R -++∈为纯虚数的充要条件为2a =C ．()2(1)12z i i =++的共轭复数对应的点为第三象限内的点D ．12i z i +=+的虚部为15i 【试题来源】2020-2021年新高考高中数学一轮复习对点练【答案】BC【分析】根据复数的相关概念可判断A ，B 是否正确，将()2(1)12z i i =++展开化简可判断C 选项是否正确；利用复数的除法法则化简12iz i+=+，判断D 选项是否正确．【解析】对于A ，因为虚数不能比较大小，故A 错误；对于B ，若()242a a i ++-为纯虚数，则24020a a ⎧-=⎨+≠⎩，解得2a =，故B 正确；对于C ，()()()211221242z i i i i i =++=+=-+，所以42z i =--对应的点为()4,2--位于第三象限内，故C 正确；对于D ，()()()()12132225i i i i z i i i +-++===++-，虚部为15，故D 错误．故选BC ． 12．已知复数(12)5z i i +=，则下列结论正确的是A ．|z |B ．复数z 在复平面内对应的点在第二象限C ．2z i =-+D ．234z i =+【试题来源】河北省邯郸市2021届高三上学期期末质量检测【答案】AD【分析】利用复数的四则运算可得2z i =+，再由复数的几何意义以及复数模的运算即可求解．【解析】5512122121212()()()()i i i z i i i i i i -===-=+++-，22,||34z i z z i =-==+ 复数z 在复平面内对应的点在第一象限，故AD 正确．故选AD13．已知i 是虚数单位，复数12i z i -=(z 的共轭复数为z )，则下列说法中正确的是 A ．z 的虚部为1B ．3z z ⋅=C ．z =D ．4z z +=【试题来源】山东省山东师大附中2019-2020学年高一下学期5月月考【答案】AC 【分析】利用复数的乘法运算求出122i z i i-==--，再根据复数的概念、复数的运算以及复数模的求法即可求解．【解析】()()()12122i i i z i i i i ---===---，所以2z i =-+，对于A ，z 的虚部为1，故A 正确；对于B ，()2225z z i ⋅=--=，故B 不正确；对于C ，z =C 正确；对于D ，4z z +=-，故D 不正确．故选AC14．早在古巴比伦时期，人们就会解一元二次方程．16世纪上半叶，数学家得到了一元三次、一元四次方程的解法．此后数学家发现一元n 次方程有n 个复数根（重根按重数计）．下列选项中属于方程310z -=的根的是A．12 B．12-+ C．122-- D ．1【试题来源】江苏省苏州市2020-2021学年高二上学期1月学业质量阳光指标调研【答案】BCD【分析】逐项代入验证是否满足310z -=即可．【解析】对A，当122z =+时， 31z -31122i ⎛⎫+- ⎪ ⎪⎭=⎝21112222⎛⎫⎛⎫+⋅+- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭=21121344i ⎛⎫=++⋅ ⎪⎛⎫+- ⎪ ⎝ ⎭⎭⎪⎪⎝12112⎛⎫=-+⋅⎛⎫+- ⎪ ⎪⎝⎭⎪ ⎪⎝⎭2114⎫=-+-⎪⎪⎝⎭ 13144=--- 2=-，故3120z -=-≠，A 错误；对B，当12z =-时，31z -3112⎛⎫-+- ⎪ ⎪⎝⎭=211122⎛⎫⎛⎫-⋅-- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭=2113124242i ⎛⎫=-+⋅ ⎪ ⎪⎛⎫-+- ⎪ ⎪⎝⎭⎝⎭1221122⎛⎫-⎛⎫=--⋅ ⎪+ - ⎪ ⎪⎝⎭⎪⎝⎭21142⎛⎫=-- ⎪ ⎪⎝⎭ 13144=+- 0=，故310z -=，B 正确；对C，当12z =-时，31z-31122⎛⎫--- ⎪ ⎪⎝⎭=21112222⎛⎫⎛⎫--⋅--- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭=21131442i ⎛⎫=++⋅ ⎪ ⎪⎛⎫--- ⎪ ⎪⎝⎭⎝⎭12112⎛⎫-⎛⎫=-+⋅ ⎪- - ⎪ ⎪⎝⎭⎪⎝⎭2114⎫=--⎪⎪⎝⎭13144=+-0=，故310z -=，C 正确；对D ，显然1z =时，满足31z =，故D 正确．故选BCD ．15．已知复数()()122z i i =+-，z 为z 的共轭复数，则下列结论正确的是A ．z 的虚部为3iB ．5z =C ．4z -为纯虚数D ．z 在复平面上对应的点在第四象限【试题来源】湖南师范大学附属中学2020-2021学年高二上学期期末【答案】BCD【分析】先根据复数的乘法运算计算出z ，然后进行逐项判断即可．【解析】因为()()12243z i i i =+-=+，则z 的虚部为3，5z z ===，43z i -=为纯虚数，z 对应的点()4,3-在第四象限，故选BCD ．三、填空题1．已知复数z 满足(1)1z i i ⋅-=+(i 为虚数单位)，则z =_________．【试题来源】上海市松江区2021届高三上学期期末（一模）【答案】1【分析】把已知等式变形，利用复数代数形式的乘除运算化简，再由复数模的计算公式求解．【解析】由(1)1z i i ⋅-=+，得21(1)1(1)(1)i i z i i i i ++===--+，所以1z =．故答案为1． 2．i 是虚数单位，复数1312i i-+=+_________．【试题来源】天津市七校2020-2021学年高三上学期期末联考【答案】1i +【分析】分子分母同时乘以分母的共轭复数12i -，再利用乘法运算法则计算即可．【解析】()()()()22131213156551121212145i i i i i i i i i i i -+--+-+-+====+++--．故答案为1i +． 3．若复数z 满足方程240z +=，则z =_________．【试题来源】上海市复旦大学附属中学2020-2021学年高二上学期期末【答案】2i ±【分析】首先设z a bi =+，再计算2z ，根据实部和虚部的数值，列式求复数．．【解析】设z a bi =+，则22224z a b abi =-+=-，则2240a b ab ⎧-=-⎨=⎩，解得02a b =⎧⎨=±⎩，所以2z i =±，故答案为2i ±. 4．复数21i-的虚部为_________．【试题来源】上海市上海交通大学附属中学2020-2021学年高二上学期期末【答案】1【分析】根据分母实数化，将分子分母同乘以分母的共轭复数1i +，然后即可判断出复数的虚部．【解析】因为()()()2121111i i i i i +==+--+，所以复数的虚部为1，故答案为1． 5．若复数z 满足(12)1i z i +=-，则复数z 的虚部为_________．【试题来源】山东省山东师大附中2019-2020学年高一下学期5月月考【答案】35【分析】根据复数的除法运算法则，求出z ，即可得出结果．【解析】因为(12)1i z i +=-，所以()()()()112113213121212555i i i i z i i i i -----====--++-，因此其虚部为35．故答案为35． 6．复数34i i+=_________．【试题来源】北京市东城区2021届高三上学期期末考试【答案】43i -【分析】分子和分母同乘以分母的共轭复数，整理后得到最简形式即可．【解析】由复数除法运算法则可得， ()343434431i i i i i i i i +⋅+-===-⋅-，故答案为43i -． 7．已知复数(1)z i i =⋅+，则||z =_________．【试题来源】北京市西城区2020-2021学年高二上学期期末考试【分析】根据复数的运算法则，化简复数为1z i =-+，进而求得复数的模，得到答案．【解析】由题意，复数(1)1z i i i =⋅+=-+，所以z == 8．i 是虚数单位，复数73i i-=+_________．【试题来源】宁夏银川一中2020-2021学年高二上学期期末考试（文）【答案】2i -【分析】根据复数除法运算法则直接计算即可．【解析】()()()()27372110233310i i i i i i i i i ----+===-++-．故答案为2i -． 9．设复数z 的共轭复数是z ，若复数143i z i -+=，2z t i =+，且12z z ⋅为实数，则实数t 的值为_________．【试题来源】宁夏银川一中2020-2021学年高二上学期期末考试（理）【答案】34【分析】先求出12,z z ，再计算12z z ⋅即得解．【解析】由题得14334i z i i-+==+，2z t i =-，所以12(34)()34(43)z z i t i t t i ⋅=+-=++-为实数，所以3430,4t t -=∴=．故答案为34【名师点睛】复数(,)a bi a b R +∈等价于0b =，不需要限制a ．10．函数()n nf x i i -=⋅（n N ∈，i 是虚数单位）的值域可用集合表示为_________．【试题来源】上海市上海中学2020-2021学年高二上学期期末【答案】{}1【分析】根据复数的运算性质可函数的值域．【解析】()()1111nn n n n n n n f x i i i i i i i i --⎛⎫=⋅⋅⋅⋅= ⎪⎝=⎭==，故答案为{}1． 11．已知()20212i z i +=（i 为虚数单位），则z =_________．【试题来源】河南省豫南九校2021届高三11月联考教学指导卷二（理）【分析】由i n 的周期性，计算出2021i i =，再求出z ，求出z ．【解析】因为41i =，所以2021i i =，所以i 12i 2i 55z ==++，所以z z == 【名师点睛】复数的计算常见题型：（1）复数的四则运算直接利用四则运算法则；（2）求共轭复数是实部不变，虚部相反；（3）复数的模的计算直接根据模的定义即可．12．若31z i =-（i 为虚数单位），则z 的虚部为_________．【试题来源】江西省上饶市2021届高三第一次高考模拟考试（文）【答案】32-【分析】利用复数的除法化简复数z ，由此可得出复数z 的虚部．【解析】()()()313333111122i z i i i i i +==-=-=-----+，因此，复数z 的虚部为32-．故答案为32-． 13．设i 为虚数单位，若复数z 满足()21z i -⋅=，则z =_________．【试题来源】江西省上饶市2020-2021学年高二上学期期末（文）【答案】2i +【分析】利用复数的四则运算可求得z ，利用共轭复数的定义可求得复数z ．【解析】()21z i -⋅=，122z i i ∴=+=-，因此，2z i =+．故答案为2i +． 14．已知i 是虚数单位，则11i i+=-_________．【试题来源】湖北省宜昌市2020-2021学年高三上学期2月联考【答案】1【分析】利用复数的除法法则化简复数11i i +-，利用复数的模长公式可求得结果．【解析】()()()21121112i i i i i i i ++===--+，因此，111i i i +==-．故答案为1． 15．i 是虚数单位，复数103i i=+____________．【试题来源】天津市南开中学2020-2021学年高三上学期第四次月考【答案】13i +【分析】根据复数的除法运算算出答案即可．【解析】()()()()10310313333i i i i i i i i i -==-=+++-，故答案为13i +. 16．在复平面内，复数()z i a i =+对应的点在直线0x y +=上，则实数a =_________．【试题来源】北京市丰台区2021届高三上学期期末练习【答案】1【分析】由复数的运算法则和复数的几何意义直接计算即可得解．【解析】2()1z i a i ai i ai =+=+=-+，其在复平面内对应点的坐标为()1,a -，由题意有：10a -+=，则1a =．故答案为1．17．已知复数z 满足()1234i z i +=+（i 为虚数单位），则复数z 的模为_________．【试题来源】江苏省苏州市2020-2021学年高二上学期1月学业质量阳光指标调研【分析】求出z 后可得复数z 的模．【解析】()()3412341121255i i i i z i +-+-===+，5z == 18．复数1i i-(i 是虚数单位)的虚部是_________．【试题来源】北京通州区2021届高三上学期数学摸底（期末）考试【答案】1-【分析】先化简复数得1i 1i i-=--，进而得虚部是1-【解析】因为()()221i i 1i i i 1i i i--==--=--，所以复数1i i-(i 是虚数单位)的虚部是1-．故答案为1-. 19．已知i 是虚数单位，复数11z i i =+-，则z =_________．【试题来源】山东省青岛市2020-2021学年高三上学期期末【答案】2【分析】根据复数的除法运算，化简复数为1122z i =-+，再结合复数模的计算公式，即可求解．【解析】由题意，复数()()111111122i z i i i i i i --=+=+=-+----，所以2z ==．故答案为2． 20．计算12z ==_______．【试题来源】2021年高考一轮数学（理）单元复习一遍过【答案】-511【分析】利用复数的运算公式，化简求值．【解析】原式1212369100121511()i ==+=-+=--．【名师点睛】本题考查复数的n次幂的运算，注意31122⎛⎫-+= ⎪ ⎪⎝⎭，()212i i +=，以及()()612211i i ⎡⎤+=+⎣⎦，等公式化简求值．四、双空题1．设32i i 1ia b =++(其中i 为虚数单位，a ，b ∈R )，则a =_________，b =_________．【试题来源】浙江省绍兴市嵊州市2020-2021学年高三上学期期末【答案】1- 1- 【分析】利用复数的除法运算化简32i 1i 1i=--+，利用复数相等的定义得到a ，b 的值，即得解．【解析】322(1)2211(1)(1)2i i i i i a bi i i i ----===--=+++-，1,1a b ∴=-=-. 故答案为－1；－1.2．已知k ∈Z ， i 为虚数单位，复数z 满足：21k i z i =-，则当k 为奇数时，z =_________；当k ∈Z 时，|z +1+i |=_________．【试题来源】2020-2021学年【补习教材寒假作业】高二数学（苏教版）【答案】1i -+ 2【分析】由复数的运算及模的定义即可得解．【解析】当k 为奇数时，()()2211k k k i i ==-=-，所以1z i -=-即1z i =-+，122z i i ++==；当k 为偶数时，()()2211k k k i i ==-=，所以1z i =-，122z i ++==；所以12z i ++=．故答案为1i -+；2．3．若复数()211z m m i =-++为纯虚数，则实数m =_________，11z=+_________．【试题来源】浙江省金华市义乌市2020-2021学年高三上学期第一次模拟考试【答案】1 1255i - 【分析】由题可得21010m m ⎧-=⎨+≠⎩，即可求出m ，再由复数的除法运算即可求出．【解析】复数()211z m m i =-++为纯虚数，21010m m ⎧-=∴⎨+≠⎩，解得1m =，。

江苏省苏州中学扬州中学盐城中学常州中学2022-2023学年高三上学期12月G4联考数学试卷解析版

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}【答案】A2．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A ．7B ．5C ．7D ．25【答案】D【解析】i ⋅z -＝4＋3i ⇒z -＝3－4i ，所以z ⋅z -＝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【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．函数y ＝2x 2－e |x |在[－2，2]的图象大致为【答案】D5．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 ．33C ．23D ．63【答案】C6．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 ．3a ＋b 2＜3a ＋3b2B ．a ＋b 2＜a ＋b2C ．(a ＋b 2)2＜a 2＋b 22D ．(a ＋b 2)3＜a 3＋b 32【答案】D7．在数列{a n }中，sin(a n ＋1－a n ) sin(a n ＋1＋a n )＝110，则该数列项数的最大值为A ．9B ．10C ．11D ．12【答案】C 【解析】8．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 ＝m →AB ＋n →AC (m ，n 为实数)，则m ＋n 的最小值为A ．518B ．13C ．718D ．49【答案】B【解析】→AP ＝m →AB ＋n →AC ＝(m ＋n )(m m ＋n →AB ＋n m ＋n →AC )，由P 在内切圆上，故m ＋n ＝|→AP ||(m m ＋n →AB ＋n m ＋n→AC )|，则cos A ＝1116，所以BC 边上高为h ＝152，内切圆半径r ＝156，故由平行线等比关系，可得m ＋n ≥h －2r h＝13，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．已知a ＞0，b ＞0，a ＋b ＝1，则A ．1a ＋1b ≤4B ．2a ＋2b ≥22C ．log 2a ＋log 2b ≤－2D ．sin a ＋sin b ≤2sin12【答案】BCD【解析】选项A ，应该是1a ＋1b ≥4，B ：2a ＋2b ≥2a ＋b2＋1，B 正确；C ：log 2a ＋log 2b ≤2log 2a ＋b 2＝－2，C 正确；D ：sin a ＋sin b ＝2sin a ＋b 2·cos a －b 2≤2sin 12，D 正确；答案为BCD10．已知函数f (x )＝ex －a＋ea －x，g (x )＝ex －a－ea －x，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y ＝g (x )f (x )在R 上单调递增【答案】ABD【解析】AB 正确，因为f (x )关于x ＝a 轴对称，g (x )关于(a ，0)中心对称，故f (x )g (x )为中心对称图形，C 错误：而[g (x )f (x )]'＝f (x )2－q (x )2∠B (x )＞0或根据一般得分离常数变形可知D 正确；答案为：ABD11．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 ．f (13)＜f (45)D ．f (12)＝338【答案】ABD【解析】显然A 正确：f (p )＝3[p 3＋(1－p )]＋4[C 13p 3(1－p )＋C 13p (1－p )3]＋5×C 24p 2(1－p )2f (0)＝3,f (12)＝338⇒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12－13≤45－12，得f (13)＞f (45)，故答案为：ABD12．在四棱锥P －ABCD 中，底面ABCD 为正方形，PA ⊥底面ABCD ，PA ＝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 ．MG 的最小值为36B ．若MA ＋MG ＝1，则点M 的轨迹是椭圆C ．若MA ＝156，则点M 的轨迹围成图形的面积为π12D ．存在点M ，使得直线BM 与CD 所成角为30°【答案】ABC【解析】A 选项判断：应用等体积法，可(MG )min ≥12(AG )min ＝12 13，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 ,MA ＝156⇒HM 2＝112，故C 正确；D 选项判断：由于CD 与面PBD 夹角θ满足sin θ＝13＞12,故π6[θ,π－θ]，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．在(x －1x)6的展开式中，常数项为．【答案】15【解析】展开式的通项为Tr ＋1＝C r 6(x )6－r(－1xr )＝(－1)r C r 6x 6－32r ，当16－32r ＝0，r ＝4时，为常数项1514．如图，将绘有函数f (x )＝M sin(π2＋φ)(M ＞0，0＜φ＜π)部分图象的纸片沿x 轴折成直二面角，此时A ，B 之间的距离为10，则φ＝．【答案】5π6【解析】如图，因为f (x )的周期为T ＝2ππ2＝4,所以CD ＝T 2＝2,BC ＝M 2＋4,所以AB ＝AC 2＋BC 2＝2M 2＋4＝10，解得M ＝3，所以f (x )＝3sin(π2x ＋φ)，所以f (0)＝3sin φ＝32，sin φ＝12，因为0＜φ＜π，所以φ＝π6或5π6，又因为函数f (x )在y 轴右侧单调递减，所以φ＝5π6．15．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 ，3S n ＝1×32＋3×33＋…＋(2n －1)⋅3n ＋1，从而2S n ＝－3－(2×32＋…＋2×3n )＋(2n －1)⋅3n ＋1，所以S n ＝(n －1)⋅3n ＋1＋3，始比如上方法可求数列{n 2⋅3n }的前n 项和T n ，则2T n ＋3＝．【答案】(n 2－n ＋1)·3n ＋1【解析】T n ＝12×31＋22×32＋…＋n 2·3n①3T n ＝12×32＋22×33＋×＋n 2·3n ＋1②②－①2T n ＝－3＋(12－22)⋅32＋(22－32)·33＋…＋[(n －1)2－n 2]·3n＋n 2·3n ＋1＝－3－(3·33＋5·33＋…＋(2n －1)·3n)＋n 2⋅3n ＋1＝－3－(S n －3)＋n 2·3n ＋1＝－S n ＋n 2·3n ＋1＝(n 2－n ＋1)⋅3n ＋1所以2T n ＋3＝(n 2－n ＋1)⋅3n ＋116．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)≤f2(x)恒成立，则实数a的取值范围是．【答案】[－3，1]．．．．【解析】f(x＋a)≤f2(x)＝f2(|x|)＝f(2|x|)⇒∀x∈[1，3]，|x＋a|≤|2|⇒a∈[－3，1]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{a n|a n2＋2a n≤m}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．【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BC＝NQ＝2a，所以四边形BNCQ为正方形，BQ＝NC＝2a，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＝AB2＋BN2＝6a，MN＝MQ2＋QN a＝22a，所以AM2＋AN2＝MN2＝8a2，所以AM⊥AN，所以四边形ANCM为矩形；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．【解析】(1)随机变量X的可能取值为0，1，2，C12P(X＝0)＝C33C35＝110，P(X＝1)＝C12C23C35＝610，P(X＝2)＝C22C13C35＝310，随机变量X的分布列如下表：X012P 110610310随机变量X的期望为E(X)＝0×110＋1×610＋2×310＝1.2法二：随机变量X服从超几何分布X~H(3，2，5)，所以E(x)＝3×25＝65(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，P(AA)＝C22C25＝110，P(BB)＝C22C25＝110，P(AB)＝C12·C12C25＝410，P(AC)＝C12·C11C25＝210，P(BC)＝C12·C11C25＝210,所以某同学捡起后随机贴回，标语恢复原样的概率为P ＝(P (AA )＋P (BB ))×1＋(P (AB )＋P (AC )＋P (BC ))×12＝15＋45×12＝35，法二：掉下的两个字不同的概率为p ＝10－210＝0.8,所以标语恢复原样的概率为(1－p )＋12p ＝0.6．20．(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→CD ＝2→DB ，→AD ·→CB ＝2，求A ；(2)若C －B ＝2π3，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．【解析】(1)设P (t 2，2t )，则PQ ≥PC 2－1＝(t 2－2)2＋4t 2－1≥1,当P (0，0)，Q 为PC 2线段与圆C 2的交点时,PQ min ＝1(2)题意可知P (4，4)，过P 点直线y －4＝k (x －4)与圆C 2相切，则|2k －4|1＋k2＝r ，即(4－r 2)k 2－16k ＋16－r 2＝0，①设直线AB 为：m (x －4)＋n (y －4)＝1，则与抛物线C 的交点方程可化为：(y －4)2＋8(y －4)[m (x －4)＋n (y －4)]＝4(x －4)[m (x －4)＋n (y －4)]，令z ＝y －4x －4,则：(1＋8n )z 2＋(8m －4n )z －4m ＝0，②题意有，①②方程同解，故有y ＝[(16－r 2)－(4－r 2)]＋34×(－16)＝[－4m －8n －1]＋34(8m －4n ),即：2m －11n ＝1，故：直线AB 恒过(6，－7)．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′(x0)，则称F (x )x )，x ＜x 0(x )，x ≥x 0为“平滑函数”，x 0为该函数的“平滑点”．已知f (x )＝ax 3－32x 2＋12x ，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．【解析】(1)(i)f '(x )＝3ax 2－3a ＋12，g ′(x )＝b [1＋ln x ]，由题意可知a －1＝0，且3a －52＝b ,故解得：a ＝1,b ＝12,(ii)进一步F (x )2＋x2，x ＜1x ≥1，过点P (2，t )作F (x )的切线，切点(x ，F (x ))满足方程：F (x )－t ＝F (x )(x －2)，故题意等价于方程：t ＝F (x )－F ′(x )(x －2)有3个不同根，p (x )＝F (x )－F ′(x )(x －2)，p '(x )＝－F ′(x )(x －2)，代入得x ∈(－∞,12)时，p (x )单调递减，x ∈[12,2)时，p (x )单调递增，x ∈[2，＋∞)时，p (x )单调递减，故t ∈{p (x )|x ∈(12,2)}＝(－38,ln2)(2)题意等价于：∀b＞0，是否 a≥13－32x2＋x2＝bx ln xax2－3x＋12＝b[1＋ln x]有解消a有：1－－32x＝b(2ln x－1)⇒b＝1－32x2ln x－1，其中由b＞0，可得x∈(23,e),故题意进一步化简∀x∈(23,e),是否∃a≥1，使得a＝3x ln x－3x＋12x(2ln x－1)成立，⇔∀x∈(23,e)，3x ln x－3x＋1≤2x2(2ln x－1)是否恒成立设q(x)＝(4x2－3x)ln x－2x2＋3x－1，q'(x)＝(8x－3)ln x，故x∈(23，1)时，单调递减，x∈(1,e),q(x)单调递增，故：q(x)≥q(1)＝0得证，即∀b＞0，3a≥1，使得F(x)存在的“平滑点”．。

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生物试题含答案

江苏省盐城中学·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生物试题2021.02注意事项考生在答题前请认真阅读本注意事项及各题答题要求1.本试卷共8 页，满分为100分，考试时间为75分钟。

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一、单项选择题：共15题，每题2 分，共30分。

每题只有一个选项最符合题意。

1 .下列关于细胞中化合物的叙述，错误的是A .糖原、纤维素酶和脂肪酸的组成元素都主要是C、H、OB.细胞中合成淀粉、蛋白质及核酸的过程都会产生水C.dATP可为DNA分子复制提供原料和能量D .通过“食盐补碘”可以有效预防“大脖子病”的发生2.下列有关利用传统发酵技术制作果酒、果醋及泡菜的叙述，正确的是A.发酵原料和发酵装置都需进行灭菌，以防止杂菌污染B.发酵用菌种的细胞呼吸类型及生物膜系统的组成都是相同的C.三者的发酵液都主要是因为产生了大量二氧化碳而呈酸性D .制作泡菜时加入一定量的陈泡菜汁能加快发酵进程3.下图是细胞中内质网腔驻留蛋白逃逸及回收的示意图，其中途径1由内质网运输至高尔基体，途径2则由高尔基体运输至内质网，通过KDEL受体的作用回收从内质网逃逸的驻留蛋白。

下列相关叙述正确的是A.在内质网中,KDEL序列能与KDEL受体必异性结合B.细胞合成的蛋白质除部分经过途径2运输外.其他都要经过途径1运输C.胰高血糖素、抗体、消化酶等分泌蛋白上一般不存在KDEL 序列D.除图中所示囊泡的定向运输外，细胞中其他囊泡的运输方向则是随机的4 .生物体结构与功能相统一的观点，既体现在细胞等生命系统水平上，也体现在分子水平上。

2021年盐城中学高三语文上学期期末试题及答案

2021年盐城中学高三语文上学期期末试题及答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

黄昏【英国】萨基诺尔曼·葛尔特茨比坐在海德公园的长凳上。

这是三月初的一个傍晚，暮色苍茫，笼罩大地，只有那微弱的月光和点点星星的亮光冲淡着昏暗的夜幕。

马路和人行道都空落落的。

然而，就在这样的夜色中仍有不少被人们遗忘的小人物在活动着。

他们有的荡来荡去，无声无息：有的把自己点缀在长凳和木椅上，毫不显眼，在昏暗中，他们的身影已无法辨认清楚。

葛尔特茨比此时觉得眼前的景色与他的心情完全和谐。

黄昏，在他看来，是失意者的时刻。

经过奋斗仍不免遭到惨败的男男女女，在这日薄西山的时候纷纷出来活动，躲避着好奇者的寻根问底。

长凳另一端，就在他身旁，坐着一位老先生。

从他的神态里，可以看出他正在和社会抗衡，但是他的气概已趋衰退。

坐了一会儿，老人起身离去，慢慢消失在黑暗中。

空出来的位子几乎立刻就被一个年轻人占据了。

但是他面部的神情并不比那位老人开朗，嘴里还狠狠地骂了一声，好像是要强调：在这个世界上，没有一件事能使他称心如意。

“看来您心情不好啊。

”葛尔特茨比说道，心想他这番表演准是为了引起自己的注意。

年轻人转过身来，脸上的神情非常坦然。

但是葛尔特茨比反而因此一下子警觉了起来。

“要是陷入我的困境，您的心情也好不了，”他回答说，“我干了一件有生以来最傻的事。

”“是吗？”葛尔特茨比不动声色地问道。

“我今天下午到的伦敦，本打算在伯塔刚尼安饭店落脚，”年轻人接着说道，“可是到了那儿我才发现，饭店已经被拆掉了，我只好去了另一家旅店。

到了我的住处，就出去买香皂了——我讨厌旅店里的香皂，可自己又忘记准备了。

我在街上溜达一会儿，在酒吧喝了杯酒，又逛了逛商店，然后转身回旅馆。

就在这时候，忽然意识到，我根本没记住旅馆叫什么，更不知道它在哪条街上。

这多么尴尬！我在伦敦又举目无亲。

我出来的时候，身上只带了一先令。

买了块香皂，喝了杯酒，也就花得差不多了，只怕要落得个流浪街头，无处栖身了。

2021届江苏省扬州中学高三生物上学期期末试卷及参考答案

2021届江苏省扬州中学高三生物上学期期末试卷及参考答案一、选择题：本题共15小题，每小题2分，共30分。

每小题只有一个选项符合题目要求。

1. 中枢神经系统包括（）A. 脑和脊髓B. 脑和脑神经C. 大脑和小脑D. 脑神经和脊神经2. 内环境稳态的维持是细胞进行正常生命活动的必要条件。

下列生命现象中，有利于内环境稳态维持的是A. 健康人体中，组织液、血浆和淋巴中各种化学成分保持恒定不变B. 在寒冷环境中，机体通过神经体液调节使散热量低于炎热环境中的散热量C. 在缺水状态下，下丘脑合成的抗利尿激素减少，机体尿量也随之减少D. 处于饥饿状态下，血液中的葡萄糖浓度下降，胰岛素的浓度也会下降3. 下列过程能双向进行的是（）A. 植物生长素的极性运输B. 能量在生态系统中的流动C. 在反射活动中，兴奋在神经纤维上的传导D. 兔子与狼之间的信息传递4. 下列关于内环境及稳态的叙述，不正确的是（）A. 人体的每一个细胞都生活在血浆、组织液和淋巴中B. 内环境与外环境的物质交换需要各器官、系统的协调活动C.缓冲物质H2CO3/NaHCO3对血液中pH的调节需要肺和肾脏的参与D. “血钙过高会引起肌无力”证明稳态是机体进行正常生命活动的必要条件5. 昆虫的保护色越来越逼真，它们的天敌的视觉也越来越发达，结果双方都没有取得明显的优势，这说明A.自然选择不起作用B.生物为生存而进化C.双方在斗争中用进废退D.双方相互选择，共同进化6. 2019年7月1日《上海市生活垃圾管理条例》正式实施，这意味着垃圾分类进入了强制时代。

垃圾分类是改善城市生态环境、促进经济社会可持续发展的一项重要内容。

保护生态环境，也是当前世界各国关注的焦点。

回答下列问题：（1）城市中的垃圾经过分解者处理可减少对环境的污染，分解者在物质循环方面的主要功能是___________。

某研究人员利用分解者处理一些生活垃圾时发现效果不理想，可能的原因是___________。

2021年江苏扬中高级中学高三英语上学期期末试卷及答案解析

2021年江苏扬中高级中学高三英语上学期期末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AOne day when I was 12, my mother gave me an order: I was to walk to the public library, and borrow at least one book for the summer. This was one more weapon for her to defeat my strange problem inability to read.In the library,I found my way into the "Children's Room." I sat down on the floor and pulled a few books off the shelf at random. The cover of a book caught my eye. It presented a picture of a beagle. I had recently had a beagle, the first and only animal companion I ever had as a child. He was my secret sharer, but one morning, he was gone, given away to someone who had the space and the money to care for him. I never forgot my beagle. Without opening the book—Amos, the Beagle with a Plan ,1 borrowed it from the library for the summer.Under the shade of a bush, I started to read about Amos. I read very, very slowly with difficulty. Though pages were turned slowly, I got the main idea of the story about a dog who, like mine, had been separated from his family and who finally found his way back home. That dog was my dog, and I was the little boy in the book. At the end of the story, my mind continued the final scene of reunion, on and on, until my own lost dog and I were, in my mind, running together.My mother's call returned me to the real world. I suddenly realized something: I had read a book, and I had loved reading that book.I never told my mother about my “miraculous” experience that summer, but she saw a slow but remarkable improvement in my classroom performance during the next year. And years later ,she was proud that her son had read thousands of books, was awarded a PhD in literature, and authored his own books, articles, poetry and fiction. The power of the words has held.1. The author's mother told him to borrow a book in order to ________.A. let him spend a meaningful summerB. encourage him to do more walkingC. help cure him of his reading problemD. make him learn more about weapons2. The book caught the author's eye because .A. it reminded him of his own dogB. he found its title easy to understandC. it contained pretty pictures of animalsD. he liked children's stories very much3. Which one could be the best title of the passage?A. Mum's Strict Order.B. My Passion forReading.C. Reunion with My Beagle.D. The Charm of a Book.BA 25-year-old American with a university degree can expect to livea decade longer than a peer who dropped out of high school. Although researchers have long known that the rich live longer than the poor, this education gap is less well documented. And although the average American’s expected span(预期寿命) has been smooth in recent year—and, shockingly, even fell between 2015 and 2017—that of the one-third with a bachelor’s degree has continued to lengthen.This gap in life expectancy is growing, according to new research published in the report of the National Academy of Sciences. Anne Case and Angus Deation ofPrincetonUniversityfound that the lifespans of those with and without a bachelor’s degree started to become different in the 1990s and 2000s. This gap grew even wider in the 2010s.What is the link between schooling and longevity(长寿)? Some argue that better-educated people develop healthier lifestyles: each additional year of study reduces the chances of being a smoker and of being overweight. The better-educated earn more, which in turn is associated with greaterhealth.Ms Case and Mr Deaton argue that changes in labor markets, including the rise of automation and increased demand for highly-educated workers, coupled with the rising costs of employer-provided health care, have decreased the supply of well-paid jobs for those without a degree. This may be contributing to higher rates of alcohol and drug use, suicide and other “deaths of despair”.The authors argued that the educational gap in mortality(致死率) will widenin the wake ofthe covid-19 pandemic. ForAmerica’s overall life expectancy to start climbing again, improvements will be needed across all social groups, not just among the privileged few.4. When did the lifespans of people with and without a degree vary greatly?A. In the 1990s.B. In the 2020sC. In the 2000sD. In the 2010s5. According to the article, changes in labor markets reduce jobs for those without a degree. Which change is NOT included?A. The rising spending of employer-provided health care.B. The gap in life expectancy.C. Raised request for better-educated workers.D. The development of automation.6. What does the underlined phrase “in the wake of” probably mean ?A. afterB. untilC. beforeD. while7. What is the best title for the text?A. Changes in labor market.B. Quit bad habits by Further studyC. Educated Americans live longer.D. Highly-educated people develop healthier lifestyles.CIf you ever get the impression that your dog can "tell" whether you look delighted or annoyed, you may be onto something. Dogs may indeed be able to distinguish between happy and angry human faces, according to a new studyResearchers trained a group of 11 dogs to distinguish between images(图像)of the same person making either a happy or an angry face. During the training stage, each dog was shown only the upper half or the lower half of the person's face. The researchers then tested the dogs' ability to distinguish between human facial expressions by showing them the other half of the person's face on images totally different from the ones used in training. The researchers found that the dogs were able to pick the angry or happy face by touching a picture of it with their noses more often than one would expect by random chance.The study showed the animals had figured out how to apply what they learned about human faces during training to new faces in the testing stage. "We can rule out that the dogs simply distinguish between the pictures based on a simple cue, such as the sight of teeth," said study author Corsin Muller. "Instead, our results suggest that the successful dogs realized that a smiling mouth means the same thing as smiling eyes, and the same rule applies to an angry mouth having the same meaning as angry eyes.""With our study, we think we can now confidently conclude that at least some dogs can distinguish human facial expressions," Muller toldLive Science.At this point, it is not clear why dogs seem to be equipped with the ability to recognize different facial expressions in humans. "To us, the most likely explanation appears to be that the basis lies in their living with humans, which gives them a lot of exposure to human facial expressions and this exposure has provided them with many chances to learn to distinguish between them." Muller said.8. The new study focused on whether dogs can_________.A. distinguish shapesB. make sense of human facesC. feel happy or angryD. communicate with each other9. What can we learn about the study from paragraph 2?A. Researchers tested the dogs in random order.B. Diverse methods were adopted during training.C. Pictures used in the two stages were differentD. The dogs were photographed before the lest.10. What is the last paragraph mainly about?A. A suggestion for future studies.B. A possible reason for the study findings.C. A major limitation of the studyD. An explanation of the research method.11. In which section is the text most likely to be found in a newspaper ?A. EntertainmentB. EconomyC. ScienceD. NatureDIf you’ve ever had a dog, you know just how deep a connection you can develop with “man’s best friend”. But a dog has a much shorter life span — about 12 to 15 years long — than humans, which means every dog owner has to go through the heartbreaking moment when their loving pet passes away.Why not make a clone of that dog then? This is the solution offered by a South Korean company, Sooam Biotech Research Foundation. The company has successfully cloned at least 400 dogs, mostly for US customers, ever since it pioneered the technique in 2005. Now, Sooam Biotech is planning to introduce their business toUKdog owners, offering them dogs that look just like their lost ones.Meanwhile, another dog is selected to supply an egg.Researchers then replace the DNA in the egg with that from the skin cell and implant the egg into the womb (子宫) of a female dog. The egg grows into a puppy over the following two months. To clone a dog, researchers first need to take a skin cell from a living dog or one that has just died.The whole process takes less than a day, but it comes at a shockingly high price — around £63,000 (614,000 yuan). But if you can’t afford it now, you can also save the cells in a laboratory and access them at a later date. Just like identical twins of humans, they share the exact same DNA but there will still be small differences betweenthem. “The spots on a Dalmatian clone will be different, for example,” Insung Hwang, head of Sooam Biotech, toldThe Guardian. However, as magical as cloning might sound, there is no guarantee that the cloned dog will be a perfect replica of the original one.Dog owners will also have to accept the fact that personality is not “clone-able”. Apart from genes, personality is also determined by upbringing and environment, which are both “random elements [that] cloning technologies simply cannot overcome”, Professor Tom Kirkwood atNewcastle University,UK, toldThe Telegraph.Perhaps bringing our dogs back with cloning is not the best way to remember them after all.Kirkwood, a dog owner himself, pointed out: “An important aspect of our relationship with them is coming to terms with the pain of letting go.”12. According to the article, Sooam Biotech Research Foundation is ______.A. working on plans to help dog owners enjoy their pets longerB. offering a way to help dogs give birth to more puppiesC. providing a service that will make copies of pet dogsD. introducing a completely new technique to clone dogs13. Which of the following statements about dog cloning is TRUE according to the article?A. Dog cloning technology hadn’t been put into practice until recently.B. Dog cloning is very expensive and usually takes several months to complete.C. Dog cloning is very popular among US andUKpet owners.D. Cloned dogs might develop different habits and characteristics even though they look very similar.14. Which of the following shows the correct order of the dog cloning process?a. an egg is taken from another dogb. a skin cell is taken from the pet dog and saved in a laboratoryc. the egg is placed in the womb of a female dogd. the DNA of the egg is replaced by the DNA from the skin celle. the egg grows into a puppy in two monthsA. acbde.B. adbce.C. bacde.D. badce.15. We can learn from the article thatKirkwood______ dog cloning.A. disapproves ofB. supportsC. is afraid ofD. is curious about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年江苏扬中高级中学高三英语上学期期末考试试卷及参考答案

2021年江苏扬中高级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AGet up to 19% off the cover pricePlus, get digital access with your paid print subscription●Up-to-date news that touches your lifeFrom money-saving tips and quick reports on the latest healthcare, to inspiring articles on world events, you'll discover hundreds of ideas for living a richer, more satisfying life.●Read it anytime, anywhereGet a l-year-print subscription ofReader's Digestmagazine today and you'll also get free digital instantly. With digital access, you can read the latest issue ofReader's Digestanytime, anywhere! Plus, you can quickly access your past issues online, too.●Continuous renewal serviceYour subscription will automatically renew at the end of each term until you cancel. You authorize us to charge you credit/debit cardat the discounted rate on the renewal service unless you cancel. You may cancel at anytime by visiting Customer Care and receive a refund on all unmailed issues.●Other informationThe cover price ofReader's Digestis $3.99 per issue and it is currently published 10 times annually. Please check the confirmation page and your mailbox to download detailed instructions.1. What is the annual fee for subscription?A. $32. 3.B. $39. 9.C. $40.D. $47. 9.2. Which of the following words best describes the content inReader's Digest?A. Touching and amusing.B. Inspiring and practical.C. Amazing and entertaining.D. Educational and theoretical.3. If you subscribe toReader’s Digest, you can ________.A. have as many issues as possible every yearB. renew your subscription at the original rateC. get back your money for the issues not mailedD. obtain all the past issues online anytime, anywhereBIn a world simultaneously on fire and underwater thanks to climate change, scientists have announced some good news: Several important tuna (金枪鱼) species have stepped back from the edge of extinction.The unexpectedly fast recovery speaks to the success of efforts over the past decade to end overfishing. But tuna are not the only species scientists are discussing at the 2021 World Conservation Congress in Marseille, France, which is organized by the International Union for Conservation of Nature (IUCN). Researchers caution that many other marine species remainimperiled. For instance, more than a third of the world's sharks remain threatened with extinction due to overfishing, habitat loss, and climate change.“I think the good news is that sustainable fisheries are possible,” says Beth Polidoro, a marine biologist at Arizona State University. “We can eat fish in a proper way and without driving the population to the point where it is on the road to collapse or extinction."At the same time, she warned that the changes in status should not be an reason to catch as many fish as we want.The IUCN, which ranks the world's most endangered species on its Red List of Threatened Species and is backed by 16,000 experts across the globe, also announced at the meeting that some animals are moving in the other direction, onto the Red List. One notable example is the Komodo dragon, an island-living lizard at particular risk from climate change.For the better part of two decades, Polidoro has been part of a specialist group tasked with assessing the statuses of more than 60 species of tuna and billfishes for the IUCN.Her team announced its first comprehensive findings in 2011, mentioning that a number of commercially fished tuna species were dangerously close to disappearing.According to the new data, the Atlantic bluefin tuna (Thunnus thynnus), once listed as endangered, now qualifies for a status of least concern. As does the yellowfin tuna (Thunnus albacares) and albacore tuna (Thunnus alalunga), which were both considered near-threatened the last time they were assessed.4. What does the underlined word “imperiled” in paragraph 2 mean?A. EndangeredB. ConservedC. ExtinctD. Safe5. What can we infer from Polidoro's words?A. Too many fish are being eaten by human beings.B. Eating fish does not necessarily lead to its extinction.C. Fish species are on the edge of dying out if no action is taken.D. The situation of underwater species are changing for the better.6. Which of following statement is true according to the passage?A. Some Tuna species are wiped out by overfishing.B. Tuna are ranked as the world's most endangered species.C. Climate change poses a threat to most species in water and on land.D. Three tuna species have been saved from extinction according to the data.7. What's the main idea of the passage?A. Some tuna species are reported endangered recently.B. IUCN has helped saved a great many marine species.C. Improvement has been made in saving marine species.D. Great efforts should be made to conserve species underwater.CWe touch our faces all the time, and it had never seemed to be a big problem—until COVID-19 arrived. Touching our faces—the "T-zone" of our eyes, nose and mouth in particular—can mean giving ourselves the deadly virus.This is why organizations like the Centers for Disease Control and Prevention (CDC) have suggested that we avoid touching our faces. "Just stop this simple behavior," William Sawyer, founder of Henry the Hand, a nonprofit organization that promotes hand hygiene (卫生), told The Washington Post. "It's the one behavior that would be better than any vaccine (疫苗) ever created."Yet, stopping this "simple" behavior might be harder than you think because it's already hardwired (固定存在于) into our system.Some face touchingis an automatic reflex (反射) —like when there is an itch (痒) on your nose, you'll scratch (挠) it without thinking. According to CNBC, a 2014 study found that touching your face also helps to reduce stress and regulate emotions. For example, you're more likely to do it when you feel awkward or embarrassed. According to Dacher Keltner, a psychologist at the University of California Berkeley, US, this action may also come with a social function: When you are talking to someone and want to change the subject, for example, touching your face is like "the curtains on a stage, closing up one act of the social drama, ushering (引导) in the next," Keltner told the BBC.Moreover, face-touching in almost all of these occasions is subconscious, which means it's very hard to change "because you don't even know you're doing it", said Sawyer. But you're not alone. In a 2015 study, whereagroup of medical students were filmed in class, it was found that they touched their faces an average of 23 times an hour—with 44 percent of the touches being in the "T-zones". That was particularly surprising since medical students were supposed to know better.Since it's so hard to shake the habit, maybe the easiest way is to wash our hands more often. This way, we can be sure that our hands are free from the novel coronavirus.8. What do the first two paragraphs talk about?A. The best way to fight COVID-19B. How organizations are fighting COVID-19.C. Typical hand hygiene problems in the fight against COVID-19.D. The necessity to avoid touching our faces to fight COVID-19.9. Why is it hard for people to stop touching their faces?A. It makes people feel more confident.B. Many are unaware of this behavior's risks.C. They usually do it automatically and subconsciouslyD. Many think the action helps them express their emotions.10. Which of the following is a social function of touching faces, according to the text?A. Using it as a sign to change the topicB. Bringing a conversation to an end.C. Showing an interest in the ongoing subject.D. Making others feel relaxed while talking.11. What is the author's purpose in mentioning a 2015 study on a group of medical students?A. To give tips on how to stop touching faces.B. To prove that it is common for people to touch their faces.C. To show it is impossible to shake the habit of touching your face.D. To show how hygiene awareness helps people avoid touching their faces.DWith graduation days being celebrated all over the country, a student who has to use a wheelchair honored his mother on his graduation day in a special way. Easley High School graduate, Alex Mays surprised people present when he got up and walked across the stage at Clemson's Littlejohn Coliseum.“I was really happy—it made me feel good,” Alex said.Alex was not given a chance to live right from his birth. He was born at 25 weeks and weighed just 1 pound, 10 ounces at birth. When he was very young, he had a disease and lost the ability to walk. After his mother's death in 2013, Alex had several other difficult life changes until he came to live with his grandparents, Dousay and her husband, Dewayne. Dousay said that when Alex came to live with them, they decided to bring him up in the best possible way they could.Last fall, Alex said that he would walk across the stage to get his diploma to honor his late mother. He practiced hard and worked with a physical therapist for 9 months to complete his plan.The only help Alex got was from his mom's best friend, Tonya Johnson, who pushed his wheelchair to the stage wearing one of his mother's favorite shirts. “I had support from my family. I couldn't have done it without them,” Alex said.“Alex made everyone in the building feel encouraged that day” Pickens County School District public information specialist John Eby said. “The school teachers knew he was going to get up to get his diploma, but the distance he walked was a surprise, even to them,” Eby said.“Some of life's most important tests aren’t given in a classroom; Alex tested himself and passed with flying color1 s,” Eby added.12. In what way did Alex honor his late mother on his graduation day?A. By dressing like her.B. By saying sorry to her.C. By inviting her best friend.D. By walking to get his diploma.13. What can we learn from Paragraph 3?A. Alex was born healthy.B. Alex went through a lot.C. Alex had a purpose in life as a child.D. Alex has lived with his grandparents all the time.14. What did Alex also express on his graduation day?A. His big regret in life.B. His feelings for hisschool.C. His thanks for his family.D. His will to complete his study.15. Which of the following words can best describe Alex?A. Strong-minded.B. Warm-hearted.C. Cool-headed.D. Easy-going.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。

江苏省G4(苏州中学、盐城中学、扬州中学、常熟中学)2021届高三教学情况期末调研英语试题及参考答案

2021年G4学校高三教学情况期末调研英语参考答案2021.01听力1-5ABACB6-10BCCAB11-15BCACB16-20CCBBC阅读21-23CDC24-27DABD28-31CDAC32-35BDAC 7选5阅读36-40CFDBG完型41-45CABBD46-50ACDBC51-55DABAC语法填空56.expression57.so/and58.was called59.universally60.when 61.has been used62.categories63.those64.made65.a 1.应用文写作Dear Sir/Madam,I am a student from Jiangsu,China.I saw the ads on your website and would like to sign up as a part-time bird watcher for your program.I am living near a lake,where lots of migrant birds spend the winter.I have been fascinated with birds since I was a little kid and I can identify at least thirty kinds of birds.I have a telescope and a camera with telephoto lens,with which I watch and photograph birds.I would be happy to join your program and do my part in protecting those beautiful species.YoursLi Hua 2.读后续写Curious,Annie got up and looked out of her window.There,sitting by his mother’s feet,was a boy playing with a Rubik’s cube,his face twisted with each turn of the cube and his mismatched eyes were fixed on the colourful object. Despite his deformity,there was something angelic about this boy when he showed his mother the finished cube.The mother smiled at him as if he were the most handsome boy in the world.1Annie turned her eyes away from the scene and saw her own figure in the mirror.Yes,she was plump and her face was chubby.But she was healthy and her skin rosy.With loving starry eyes and curly hair,she looked almost like some lady from a classic portrait.Mum once said when she played Beethoven on the piano in her purple dress,she looked like a real princess.She smiled to her reflection,first tentatively and then with more confidence.She made a new decision.She would throw her bikini into the dustbin.After all,a girl doesn’t have to fit in a bikini to be beautiful.2。