二轮复习综合试题1

郑州第一中学2024届高三第二轮复习质量检测试题数学试题注意事项：1．答卷前，考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。

用2B 铅笔将试卷类型（B ）填涂在答题卡相应位置上。

将条形码粘贴在答题卡右上角"条形码粘贴处"。

2．作答选择题时，选出每小题答案后，用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案。

答案不能答在试题卷上。

3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液。

不按以上要求作答无效。

4．考生必须保证答题卡的整洁。

考试结束后，请将本试卷和答题卡一并交回。

一、选择题：本题共12小题，每小题5分，共60分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1．如图，正方体的底面与正四面体的底面在同一平面α上，且//AB CD ，若正方体的六个面所在的平面与直线CE EF ，相交的平面个数分别记为m n ，，则下列结论正确的是（ ）A ．m n =B ．2m n =+C ．m n <D ．8m n +<2．已知椭圆C 的中心为原点O ，(25,0)F -为C 的左焦点，P 为C 上一点，满足||||OP OF =且||4PF =，则椭圆C 的方程为（ ）A ．221255x y +=B ．2213616x y +=C ．2213010x y += D ．2214525x y += 3．如图，设P 为ABC ∆内一点，且1134AP AB AC =+，则ABP ∆与ABC ∆的面积之比为A ．14 B ．13 C ．23D ．164．若实数,x y 满足不等式组2,36,0,x y x y x y +≥⎧⎪-≤⎨⎪-≥⎩则3x y +的最小值等于（ ）A ．4B ．5C ．6D ．75．已知ABC 是边长为3的正三角形，若13BD BC =，则AD BC ⋅=A ．32- B ．152 C ．32D ．152-6．做抛掷一枚骰子的试验，当出现1点或2点时，就说这次试验成功，假设骰子是质地均匀的.则在3次这样的试验中成功次数X 的期望为（ ） A ．B ．C ．1D ．27．已知等差数列{}n a 的公差为2-，前n 项和为n S ，1a ，2a ，3a 为某三角形的三边长，且该三角形有一个内角为120︒，若n m S S ≤对任意的*n ∈N 恒成立，则实数m =（ ）. A ．6B ．5C ．4D ．38．执行如图所示的程序框图，则输出的S 的值是（ ）A ．8B ．32C ．64D ．1289．函数()()()sin 0,0f x x ωϕωϕπ=+><<的图象如图所示，为了得到()cos g x x ω=的图象，可将()f x 的图象（ ）A ．向右平移6π个单位 B ．向右平移12π个单位C ．向左平移12π个单位D ．向左平移6π个单位 10．已知函数()cos f x x m x =+，其图象关于直线3x π=对称，为了得到函数()g x x =的图象，只需将函数()f x 的图象上的所有点（ ） A ．先向左平移6π个单位长度，再把所得各点横坐标伸长为原来的2倍，纵坐标保持不变 B ．先向右平移6π个单位长度，再把所得各点横坐标缩短为原来的12，纵坐标保持不变 C ．先向右平移3π个单位长度，再把所得各点横坐标伸长为原来的2倍，纵坐标保持不变 D ．先向左平移3π个单位长度，再把所得各点横坐标缩短为原来的12，纵坐标保持不变 11．已知命题p ：1m =“”是“直线0x my -=和直线0x my +=互相垂直”的充要条件；命题q ：对任意()2,∈=+a R f x x a 都有零点；则下列命题为真命题的是（ ）A ．()()p q ⌝∧⌝B ．()p q ∧⌝C ．p q ∨D ．p q ∧12．下列函数中，值域为R 的偶函数是（ ） A ．21y x =+B ．x x y e e -=-C ．lg y x =D．y 二、填空题：本题共4小题，每小题5分，共20分。

2023届高三二轮复习联考（一）新高考卷语文试题注意事项：1.答题前，考生务必将自己的姓名、考场号、座位号、准考证号填写在答题卡上。

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，将本试卷和答题卡一并交回。

考试时间150分钟，满分150分一、现代文阅读（35分）（一）现代文阅读I（本题共5小题，17分）阅读下面的文字，完成1～5题。

材料一：妙夺天工而又不露人巧地反映生活，是《红楼梦》艺术上的一大特色，也是世界文学宝库里的罕见贡献。

正是这个特色，使我们在《红楼梦》里几乎看不见虚构和夸张，好像一切都是“追踪蹑迹”地照实摹写下来的。

实际上，艺术总是离不开虚构和夸张。

"饰穷其要"的夸张，不仅不排斥作品的真实性，反而是更加有助于突出事物的本质。

由于《红楼梦》是以所谓“亲睹亲闻”的生活经验作为创作的基础，而又经过艺术上的反复锤炼，所以它能把夸张、虚构这些为创作所不可缺少的艺术手法运用到天衣无缝的境地，以至只见它的自然逼真，而不见人工斧凿之痕。

像《红楼梦》这样自然逼真到去尽夸饰的作品，在中外文学史上都是十分罕见的。

譬如《三国演义》这部小说，在艺术上也有其不可忽视的成就，但它常常是采取特异的行动和令人惊奇的情节来加强故事性或传奇性。

夸张，成为它的一个主要艺术手法，即使在许多细节描写上也莫不如此。

如关羽使的那把重八十二斤的青龙偃月刀，不仅于史无据，事实上这样重的刀也不适于作战，违背了古代兵器锋利轻便的要求。

但是，《三国演义》的作者通过艺术上的夸张，生动地渲染出关羽的勇武绝伦的性格。

然而，夸张的手法，如果运用不当或过度，又容易使读者感到那是“小说家言”了。

而《红楼梦》却绝少令人产生“小说家言”的感觉。

相反地，它却使一些“红学家”以为那是一部“实录自传”，甚至是一部真人真事、按年按月的记录。

高三第二轮复习模拟测试卷英语试题(一)本试卷6页，满分120分。

考试用时120分钟注意事项：1.答卷前，考生务必用黑色字迹的钢笔或签字笔将自己的校名、姓名、考号、座位号等相关信息填写在答题卡指定区域内。

2.选择题每小题选出答案后，请用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案；不能答在试卷上。

.3.非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液，不按以上要求作答的答案无效。

4. 考生必须保证答题卡的整洁。

第二部分阅读理解(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，从每题所给的四个选项(A、B、C和D)中，选出最佳选项，并在答题卡上将该项涂。

AI recently heard a story about a famous research scientist who had made several very important medical achievements. A newspaper reporter interviewed him why he was able to be so much more creative than theaverage person. What set him so far apart from othersHe responded that it all came from an experience with his mother that occurred when he was about two years old. He had been trying to remove a bottle of milk from the refrigerator when he dropped the slippery bottle, spilling its contents all over the kitchen floor—a real sea of milk!When his mother came into the kitchen，instead of shouting at him, giving him a lecture or punishing him, she said, “Robert, what a great and wonderful mess you have made! I have rarely seen such a huge pool of milk. Well, the damage has already been done. Would you like to get down and play in the milk for a few minutes before we clean it up？”Indeed, the boy did. After a few minutes, his mother said, “You know, Robert, whenever you make a mess like this, eventually you have to clean it up and restore everything to its proper order. How would you like to do that We could use a sponge(海绵)，a towel or a mop. Which do you prefer？” He chose the sponge and together they cleaned up the spilled milk.His mother then said，“You know，what we have here is a failed experiment in how to effectively carry a big milk bottle with two tiny hands. Let's go out in the back yard and fill the bottle with water and see if you can find a way to carry it without dropping it. ”The little boy learned that if he grasped the bottle at the top near the lip with bothhands, he could make it. What a wonderful lesson!This famous scientist then said that it was at that moment that he knew he didn't need to be afraid to make mistakes.21.When the mother found the kitchen floor covered with milk, she ________.A. gave the boy a lecture instead of shouting at himB. praised the boy instead of punishing himC. felt satisfied with the boy's trying to help himselfD. calmed down the boy and helped him find a way to clean the floor22.The child's experience resulted in the following BUT ________.A. offering the boy a chance to grasp the bottleB. benefiting the boy all his lifeC. helping the boy be more creativeD. making the boy realize the mistake is of value23. According to the passage, the way the scientist's mother used is ________.A. instructiveB. strictC. formalD. strangeBToday's buzz words, which include global warming and zero emissions, are causing everyday people (not just celebrities) to look for ways to reduce their carbon footprint. Based on this trend, entire districts,known as eco-communities, are being designed with green initiatives in mind. Dockside Green in Victoria, British Columbia, Canada is one of these communities. If everything goes according to plan, Dockside Green will be a self-sufficient community. The plans describe a variety of environmental features, some of which have never been seen.Builders of Dockside Green have the environment in mind with every choice they make. They ensure proper ventilation (通风), and guarantee residents 100% fresh indoor air. Interior and exterior building materials, such as paints and wood, are natural and non-toxic (无毒的). Eco-conscious builders use bamboo wherever possible because it is durable and does not require pesticides to grow.Energy efficiency is one of the top priorities in eco-communities, such as Dockside Green. Not only do energy-efficient appliances and light fixtures reduce the environmental impact of heating and hot water, they also save residents and business owners money. Dockside Green claims that home owners will use 55% less energy than average residents in Canada. Studies show that people use approximately 20% less energy when they are billed for exactly what they use. In addition, water is treated at Dockside Green and reused on site for flushing toilets. Planners of eco-communities such as Dockside Green must take the future into account. Dockside Green plans on reusing 90% of its construction waste. They also plan to continue using local suppliers forall of their transport and maintenance needs. This is a great way to reduce emissions. Finally, plans are in the works for a high-tech heating system that will use renewable biomass instead of fossil fuels.The first two stages of development at Dockside Green have already been completed, and many other plans are in the works. Other smaller eco-villages can be found around the world. Residents of these communities aim to reduce their carbon footprint and be models for sustainable living.24. The purpose of the article is to __________.A. introduce eco-communitiesB. describe Dockside Green’s plansC. advertise Dockside GreenD. illustrate sustainable living25. According to the article, what percentage of energy saving will Dockside Green residents saveA. 100%.B. 90%.C. 55%.D. 20%.26. Bamboo is an environmentally friendly choice for builders because it __________.A. provides good air circulationB. keeps pests out of housesC. is toxic substances freeD. can be used for a long time27. What will Dockside Green try to do in the futureA. To find fossil fuel alternatives.B. To develop an interior transportation system.C. To use most of construction waste again.D. To cooperate with other eco-villages.CAs you move out of your teens and into your 20s and 30s, your musical tastes start to solidify and you likely quit keeping up with popular music. Now, research has found the average age at which your music library is unlikely to change: 33.The study's author reached this conclusion by analyzing data on U. S. Spotify users and comparing it to artist popularity data from music intelligence company The Echo Nest.To determine the age at which we stop seeking out new tunes, the author gathered self-reported age data from Spotify and looked at users' " Taste Profiles," which tracks how many times listeners have streamed individual artists. These artists were then matched to their popularity rank on The Echo Nest.The average teen listens almost only to music among the Billboard, the 200 highest-ranking albums in the country, but this music represents a smaller part of their streaming as they age. In their teens, they listen to a lot of the same music, over and over again. Frank T. McAndrew, a professor of psychology at Knox College, explains that's the "mere exposure effect" at work, and it basically means, the more we're exposed to something, the more we like it.There are many reasons why a person may stop streaming the currenttop musical hits, and the author looked into one by identifying Spotify listeners with large amounts of children's music and nursery rhymes in their libraries. In other words, when users may have become parents. But the study concludes with some good news for parents: “If you're getting older and can't find yourself staying as relevant as you usedto, have nofear — just wait for your kidsto becometeenagers, and you'll get exposed to all thepopularmusic of the day once again!”28. What happens after teens grow olderA. They quit listening to music.B. Their interest in music fades.C. They change their music libraries.D. Their musical tastes become stable.29. What is the way of the researchA. Interviews.B. Analyzing data.C. Observation.D. Tracking popularity.30. How do teens develop their musical tastesA. They listen to what they like repeatedly.B. They act like their favorite popular stars.C. They experience different styles of music.D. They like high-ranking music worldwide.31. What's the good news for parents in the last paragraphA. Parents will regain their faith in music.B. Parents will influence their teens in music.C. Parents will hear their favorite music again.D. Parents will enjoy the music for teens again.DThey still bite, but new research shows lab-grown mosquitoes are fighting dengue fever — a dangerous disease that they normally would spread. Dengue infections appear to be dropping fast in communities in Indonesia, Vietnam, Brazil and Australia that are filled with the specially grown mosquitoes.Researchers first injected （注射）mosquito eggs with Wolbachia bacteria that's common in insects and harmless to people in a lab. Infected females then pass the bacteria on through their eggs. Releasing enough Wolbachia carriers, both the females that bite and the males that don't, allows mating（交配）to spread the bacteria through a local mosquito population.Rather than using chemicals to wipe out pests, “this is really about transforming the mosquito," said Cameron Simmons of the nonprofit World Mosquito Program, which is conducting the research.The first success came from Australia. Mosquitoes carrying Wolbachia were released in parts of North Queensland starting in 2011, and gradually spread through the local mosquito population. Dengue is spread when a mosquito bites someone who is infected, and then bites another person, but somehow Wolbachia blocks that — and local spread has nearly disappeared in those North Queensland Communities,Simmons said.The studies are continuing in other countries. But the findings, presented at a meeting of the American Society for Tropical Medicine and Hygiene, suggest it's possible to turn at least some mosquitoes from a public health threat into annoying biters.The work marks “exciting progress," said Michigan State University professor Zhiyong Xi, who wasn't involved with the project but has long studied how Wolbachia can turn mosquitoes against themselves.More research is needed, specialists cautioned. "The results are pretty exciting — strong levels of reductions — but there clearly are going to be things to be learned from the areas where the reductions are not as great," said Penn State University professor Elizabeth McGraw.32. What can we learn about the lab-grown mosquitoesA. They lose the ability to bite people.B. They become harmful to human beings.C. They carry dangerous Wolbachia bacteria.D. They spread Wolbachia bacteria by mating.33. What does the underlined word "that" refer to in Paragraph 4A. The bite from a mosquito.B. The local community.C. The spread of Dengue.D. The infected person.34. What do the last two paragraphs suggestA. This research proves a complete success.B. Scientists agree on this research.C. There is still room for improvement.D. Mosquitoes turn themselves against harmful ones.35. What is the best title for the textA. Bacteria- infected Mosquitoes Bite Deadly Dengue.B. Transform Mosquitoes into Annoying BitersC. Fight against Mosquitoes with WolbachiaD. Harmless Mosquitoes Are on Their Way第二节(共5小题:每小题2分，满分10分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

中国古代史综合检测一、选择题(本大题16小题,每题3分,共计48分,每题只有一个选项是正确的)1.(2022天津河东一模)兴隆洼遗址,发现于内蒙古赤峰敖汉旗,是距今8 000多年前的原始村落遗址。

这里发掘出170余座半穴居房址,最大的房址面积达140平方米。

由此判断,当时兴隆洼地区()A.雨水稀少,没有森林B.地势低洼,气候潮湿C.遵循礼制,等级森严D.地势较高,聚落而居2.(2022辽宁沈阳二模)《礼记·王制》云:“天地寒暖燥湿,广谷大川异制。

民生其间者异俗”,西周封建着眼于天、地、人,尊“地之义”“民之行”“修其教,不易其俗;齐其政,不易其宜”,以求“藏天下于地方,由地方以见天下”。

据此可知,西周封建()A.消弭了地域间的文化差异B.源于自然风土民情C.秉持天下一家的治国理念D.以神权政治为核心3.(2022四川绵阳模拟)老子认为“民之饥,以其上食税之多”;孔子叹“苛政猛于虎”;孟子斥责梁惠王之流“庖有肥肉,厩有肥马,民有饥色,野有饿莩”是“率兽而食人”;庄子则揭露国君们“轻用民死”。

这反映出,当时他们()A.吸收各家学说走向融合B.具有强烈的社会责任感C.得到统治者的积极支持D.迎合地主阶级政治诉求4.(2022山东潍坊一模)据睡虎地秦墓出土的竹简记载:擅兴奇祠,赀(注:罚)二甲。

可(何)如为奇?王室所当祠固有矣,擅有鬼立(位)殹为奇,它不为。

这一规定意在()A.加强思想统一B.打击地方祭祀C.节省财政开支D.完善法律体系5.(2022山东德州二模)南朝宋孝武帝任用出身贫寒的戴法兴、巢尚之等人担任中枢机构要职,负责协助皇帝起草诏令决策国家大事,从而出现了“寒人掌机要”的局面。

这一现象反映出()A.门阀士族政治结束B.庶族地主掌握权力中枢C.君主权力有所强化D.九品中正制已遭到废弃6.(2022山东临沂二模)唐朝诗人王建在《凉州行》中写道:“蕃人旧日不耕犁,相学如今种禾黍。

驱羊亦著锦为衣,为惜毡裘防斗时。

2023届高三二轮复习联考(一)山东卷地理试题注意事项：1.答卷前，考生务必将自己的姓名、考场号、座位号、准考证号填写在答题卡上。

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，将本试卷和答题卡一并交回。

考试时间为90分钟，满分100分一、选择题：本题共15小题，每小题3分，共45分。

在每小题给出的四个选项中只有一项是最符合题目要求的。

某校地理研学小组在有关人员协助下，测量了春分～夏至的日出方位角(即正东方向偏北或偏南的角度)。

表1为30 °N 和60 °N 日出方位角测量数据统计表。

据此完成1～2题。

1.根据表中数据可以推知A.天文四季的划分标准B.日出与日落时间早晚C.地转偏向力的纬度差异D.五带具有的天文现象2.家住俄罗斯某城市(约60 °N) 的某同学在立秋日(太阳直射点纬度约为16.3°)看到太阳的升起方向约A.东偏北13°B.东偏南13°C.东偏北34°D.东偏南34°我国某河流下游流域分布众多湖泊。

图1 为该河流下游与某湖泊一年内的水位变化图。

据此完成3～5题。

3.该河流下游流域的气候特点是A.雨季较长，降水较多B.冬季寒冷，结冰期长C.冬季积雪厚度大D.地下水埋藏较浅4.与河流相比，湖泊水位季节变化较小，说明湖泊A.由多种水源补给B.夏季蒸发旺盛C.涵养水源功能显著D.冬季无结冰期5.夏季河流与湖泊之间的径流方向是A.湖泊水流向河流B.河流水流向湖泊C.白天，湖泊水流向河流D.夜晚，河流水流向湖泊近年来，康养产业已经成为发展前景广阔的朝阳产业。

根据消费需求、资源差异，又衍生出森林康养、气候康养、海洋康养、温泉康养、田园康养等产业类型。

乡村旅居+田园康养是康养产业发展的必然趋势。

2024届高三二轮复习联考(一)地理试题注意事项：1.答卷前，考生务必将自己的姓名、考场号、座位号、准考证号填写在答题卡上。

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，将本试卷和答题卡一并交回。

考试时间75 分钟，满分100分一、选择题：本题共16 小题，每小题3分，共 48 分。

在每小题给出的四个选项中只有一项是最符合题目要求的。

偏岩古镇位于我国亚热带季风区，常年多雨多雾，是高湿地区。

古镇内部主街呈南北走向，依凭周围山川拱抱，迎纳阳光，形成良好生态小气候。

下图示意偏岩古镇与现代聚落选址(图中实线表示等高线，颜色越深表示海拔越高)。

据此完成1~3 题。

1.推测偏岩古镇建筑最可能朝向A.东B.南C.西D.北2.影响偏岩古镇主街走向的主导因素是A.夏季风风向B.冬季风风向C.最小风频风向①.最大风频风向3.与现代聚落相比，偏岩古镇选址更加关注A.资源状况 B 气候适应C.交通条件D.建设成本受地形与气流运动影响，我国西南部某山区突发性降水多发。

下图示意该区域5月某次降水产生时沿降水中心附近104°E的垂直经向环流剖面图(图中箭头代表气流运动方向，箭头长短代表风速大小)。

据此完成4~6题。

4.此次降水的特点为A.降水时间长B.集中在39°N附近C.降水强度大D.受准静止锋影响大5.与近地面气流甲相比，气流乙A.运动速度快B.为偏北风C.上升强烈D.空气湿度大6.与此次降水成因相似的降水过程是喜马拉雅冷杉广泛分布于喜马拉雅山脉中部地区海拔2300 ~4000 米范围内的尼泊尔境内。

近几十年来，喜马拉雅地区温度持续升高，低海拔地区与高海拔地区(林线附近)的冷杉生长状况受水热条件影响差异明显。

据此完成7~9题。

7.喜马拉雅冷杉生长区的气候特征是A.炎热湿润B.凉爽湿润C.炎热干燥D.凉爽干燥8.最有利于加快喜马拉雅冷杉在高海拔地区生长速率的条件是A.冬季温度上升B.夏季温度上升C.冬季降水增多D.夏季降水增多9.与高海拔地区相比，喜马拉雅冷杉在低海拔地区A.受土壤条件影响大B.受热量条件影响大C.受光照条件影响大D.受水分条件影响大伊索普斯草原保护区位于美国哈德逊河口(潮汐河口)，自2007 年开始实施岸线修复工程，拆除砖石岸线(图甲)，放置石块作为基底，在石基上方放置软质石笼并用土工织物包裹土環和岩石，在软石笼之间混种多种天然灌木物种。

2023届高考历史大二轮专题复习试题第一部分古代中国和古代世界专题3Word版含答案第一部分专题三一、选择题1．(2023·高考历史模拟卷一)《礼记》中载：“殷人尊神，率民以事神，先鬼而后礼”。

周代统治者提出了“惟命于不常”(《康诰》)、“皇天无亲，惟德是辅”(《尚书》)和“天视自我民视，天听自我民听”(《泰誓》)等思想。

殷周变革中天命观的发展实质上体现了(D) A．政权神秘化色彩不断加强B．早期民主政治的特征C．对于天命神意的批判否定D．一定程度的理性色彩【解析】试题分析：本题主要考查学生结合所学知识分析认知历史事物的能力。

由材料信息“殷人尊神，率民以事神，先鬼而后礼”，反映出殷人将鬼神作为万物主宰的天命观。

由“惟命于不常”“惟德是辅”“天视自我民视，天听自我民听”，周代则认为命运无常，只有品德高尚、按人民意愿办事，治国有方的人才是天命的延续。

从天命观的这一发展，可以看出天命观彰显出一定理性色彩，D项符合题意；A项与材料的描述相悖，明显错误；B项属于无关项；C项不符合题干“实质上”要求；综上，本题正确答案选D。

2．(2023·乌鲁木齐)古代儒家学者认为：“修己以安人，修己以安百姓。

”这表明儒者(C)A．不能适应现实政治C．具有经世济民情怀B．反对进行社会变革D．以复古为政治目标【解析】“修己以安人，修己以安百姓”是指通过自身素养和道德水平的提高，回报他人和社会，体现的是对于社会的责任感，因此这种主张表明儒者具有经世济民的情怀，故答案为C项；A项结论无法得出，排除；材料信息没有涉及对于社会变革的态度，排除B项；D项“复古”材料没有体现，排除。

3．(2023·高考押题卷一)战国申不害重“术”，其思想以加强君主专制为核心；王夫之却将其学说斥为“洪水猛兽”，主张在政府内部分权，各级官吏都要有自己的权力，君主不能过多的干涉。

造成这种认识差异的决定性因素是(C)A．阶级立场根本不同C．社会发展阶段不同B．受外来思想影响不同D．学术指导思想的不同【解析】申不害是法家代表人物，他主张加强君主专制，要求建立君主专制的中央集权的国家，在我国由奴隶社会向封建社会的过渡时期，他的主张符合新兴封建地主阶级的要求。

2.复数、平面向量考向1 复数的概念、运算及几何意义1.(2022·河南开封一模)设(1+i 4n+3)z=i,n ∈Z ,则在复平面内,复数z 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2.(2022·全国甲·理1)若z=-1+√3i,则zz -1=( )A.-1+√3iB.-1-√3iC.-13+√33iD.-13−√33i3.(2022·全国乙·理2)已知z=1-2i,且z+a z +b=0,其中a ,b 为实数,则( ) A.a=1,b=-2 B.a=-1,b=2 C.a=1,b=2 D.a=-1,b=-24.(2022·山东潍坊一模)已知复数z 满足z+3=4z +5i,则在复平面内复数z 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 5.(2022·新高考Ⅰ·2)若i(1-z )=1,则z+z =( ) A.-2B.-1C.1D.2考向2 平面向量的概念及线性运算6. (2022·河南名校联盟一模)如图,在△ABC 中,点M 是AB 上的点且满足AM ⃗⃗⃗⃗⃗⃗ =3MB ⃗⃗⃗⃗⃗⃗ ,P 是CM 上的点,且MP ⃗⃗⃗⃗⃗⃗ =15MC ⃗⃗⃗⃗⃗⃗ ,设AB ⃗⃗⃗⃗⃗ =a ,AC ⃗⃗⃗⃗⃗ =b ,则AP ⃗⃗⃗⃗⃗ =( )A.12a +14b B.35a +15b C.14a +12bD.310a +35b7.(2022·河南名校联盟一模)下列关于平面向量的说法正确的是( ) A.若AB ⃗⃗⃗⃗⃗ ,CD ⃗⃗⃗⃗⃗ 共线,则点A ,B ,C ,D 必在同一直线上 B.若a ∥b 且b ∥c ,则a ∥cC.若G 为△ABC 的外心,则GA ⃗⃗⃗⃗⃗ +GB⃗⃗⃗⃗⃗ +GC ⃗⃗⃗⃗⃗ =0 D.若O 为△ABC 的垂心,则OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ =OB ⃗⃗⃗⃗⃗ ·OC⃗⃗⃗⃗⃗ =OC ⃗⃗⃗⃗⃗ ·OA ⃗⃗⃗⃗⃗ 8.(2022·新高考Ⅰ·3)在△ABC 中,点D 在边AB 上,BD=2DA.记CA ⃗⃗⃗⃗⃗ =m ,CD ⃗⃗⃗⃗⃗ =n ,则CB ⃗⃗⃗⃗⃗ =( ) A.3m -2nB.-2m +3nC.3m +2nD.2m +3n9.(2022·河南许昌质检)正方形ABCD 中,P ,Q 分别是边BC ,CD 的中点,AP ⃗⃗⃗⃗⃗ =x AC ⃗⃗⃗⃗⃗ +y BQ ⃗⃗⃗⃗⃗ ,则x=( ) A.1113B.65C.56D.3210.(2022·河南名校联盟一模)如图,在同一个平面内,向量OA ⃗⃗⃗⃗⃗ 与OC ⃗⃗⃗⃗⃗ 的夹角为α,且tan α=7,向量OB ⃗⃗⃗⃗⃗ 与OC ⃗⃗⃗⃗⃗ 的夹角为45°,且|OA ⃗⃗⃗⃗⃗ |=|OB ⃗⃗⃗⃗⃗ |=1,|OC ⃗⃗⃗⃗⃗ |=√2.若OC ⃗⃗⃗⃗⃗ =m OA ⃗⃗⃗⃗⃗ +n OB ⃗⃗⃗⃗⃗ (m ∈R ,n ∈R ),则n-m= . 考向3 平面向量的数量积11.(2022·新高考Ⅱ·4)已知向量a =(3,4),b =(1,0),c =a +t b ,若<a ,c >=<b ,c >,则实数t=( ) A.-6 B.-5C.5D.612. (2022·新高考八省第二次T8联考)如图,在同一平面内沿平行四边形ABCD 两边AB ,AD 向外分别作正方形ABEF ,正方形ADMN ,其中AB=2,AD=1,∠BAD=π4,则AC ⃗⃗⃗⃗⃗ ·FN ⃗⃗⃗⃗⃗=( )A.-2√2B.2√2C.0D.-1 13.(2022·山东威海期末)已知向量a ,b 满足|a |=|b |=2,且a -b 在a 上的投影为2+√3,则<a ,b >=( )A.π6 B.π3C.2π3D.5π614.(2022·山东潍坊期末)已知正方形ABCD 的边长为2,MN 是它的内切圆的一条弦,点P 为正方形四条边上的动点,当弦MN 的长度最大时,PM ⃗⃗⃗⃗⃗⃗ ·PN ⃗⃗⃗⃗⃗⃗ 的取值范围是( ) A.[0,1]B.[0,√2]C.[1,2]D.[-1,1]15.(2022·山东济宁一模)等边三角形ABC 的外接圆的半径为2,点P 是该圆上的动点,则PA⃗⃗⃗⃗⃗ ·PB ⃗⃗⃗⃗⃗ +PB ⃗⃗⃗⃗⃗ ·PC ⃗⃗⃗⃗⃗ 的最大值为( ) A.4 B.7C.8D.111 3,且|a|=1,|b|=3,则(2a+b)·b=.16.(2022·全国甲·理13)设向量a,b的夹角的余弦值为2.复数、平面向量1.B 解析: ∵i 4n+3=i 4n ·i 3=-i, ∴(1+i 4n+3)z=(1-i)z=i, ∴z=i1-i =i (1+i )(1-i )(1+i )=-12+12i,∴复数z 在复平面内对应的点为-12,12位于第二象限. 故选B . 2.C 解析: zz -1=√3i(-1+√3i )(-1-√3i )-1=√3i(-1)2+(√3)2-1=-13+√33i,故选C .3.A 解析: ∵z=1-2i, ∴z =1+2i,∴z+a z +b=1-2i +a (1+2i)+b=a+b+1+(2a-2)i =0, ∴{a +b +1=0,2a -2=0, 解得{a =1,b =-2.故选A .4.A 解析: 设z=x+y i,x ,y ∈R ,则z =x-y i,由z+3=4z +5i 得(x+y i)+3=4(x-y i)+5i,即(x+3)+y i =4x+(5-4y )i,于是得{x +3=4x ,y =5-4y ,解得x=y=1,则有z=1+i 对应的点为(1,1),所以在复平面内复数z 对应的点在第一象限. 故选A .5.D 解析: ∵i(1-z )=1, ∴z=i -1i=1+i, ∴z =1-i . ∴z+z =2. 故选D .6.B 解析: AP ⃗⃗⃗⃗⃗ =AM ⃗⃗⃗⃗⃗⃗ +MP ⃗⃗⃗⃗⃗⃗ =AM ⃗⃗⃗⃗⃗⃗ +15MC ⃗⃗⃗⃗⃗⃗ =AM ⃗⃗⃗⃗⃗⃗ +15(AC ⃗⃗⃗⃗⃗ −AM ⃗⃗⃗⃗⃗⃗ )=45AM ⃗⃗⃗⃗⃗⃗ +15AC ⃗⃗⃗⃗⃗ =45×34AB ⃗⃗⃗⃗⃗ +15AC ⃗⃗⃗⃗⃗ =35a +15b .7.D 解析: 若AB ⃗⃗⃗⃗⃗ ,CD ⃗⃗⃗⃗⃗ 共线,则直线AB 与CD 平行或重合,∴点A ,B ,C ,D 不一定在同一直线上,A 错;当b =0时,满足a ∥b 且b ∥c ,不能得出a ∥c ,B 错; 当G 为△ABC 的重心,则GA⃗⃗⃗⃗⃗ +GB ⃗⃗⃗⃗⃗ +GC ⃗⃗⃗⃗⃗ =0,C 错; 若O 为△ABC 的垂心,则OB ⃗⃗⃗⃗⃗ ⊥AC ⃗⃗⃗⃗⃗ ,∴OB ⃗⃗⃗⃗⃗ ·AC ⃗⃗⃗⃗⃗ =0, 即OB ⃗⃗⃗⃗⃗ ·(OC ⃗⃗⃗⃗⃗ −OA ⃗⃗⃗⃗⃗ )=0,∴OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ =OB ⃗⃗⃗⃗⃗ ·OC ⃗⃗⃗⃗⃗ ,同理OB ⃗⃗⃗⃗⃗ ·OC ⃗⃗⃗⃗⃗ =OC ⃗⃗⃗⃗⃗ ·OA ⃗⃗⃗⃗⃗ ,∴D 正确,故选D . 8.B解析: 如图.∵BD=2DA ,∴AB ⃗⃗⃗⃗⃗ =3AD ⃗⃗⃗⃗⃗ ,∴CB ⃗⃗⃗⃗⃗ =CA ⃗⃗⃗⃗⃗ +AB ⃗⃗⃗⃗⃗ =CA ⃗⃗⃗⃗⃗ +3AD ⃗⃗⃗⃗⃗ =CA ⃗⃗⃗⃗⃗ +3(CD ⃗⃗⃗⃗⃗ −CA ⃗⃗⃗⃗⃗ )=-2CA ⃗⃗⃗⃗⃗ +3CD ⃗⃗⃗⃗⃗ . 又CA ⃗⃗⃗⃗⃗ =m ,CD ⃗⃗⃗⃗⃗ =n ,所以CB ⃗⃗⃗⃗⃗ =-2m +3n . 故选B .9.C 解析: ∵P ,Q 分别是正方形边BC ,CD 的中点,∴AC ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +A D ⃗⃗⃗⃗⃗⃗ ,AP ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +12AD ⃗⃗⃗⃗⃗ ,BQ ⃗⃗⃗⃗⃗ =-12AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ ,∴AP ⃗⃗⃗⃗⃗ =x AC ⃗⃗⃗⃗⃗ +y BQ ⃗⃗⃗⃗⃗ =x (AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ )+y -12AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ =x-12y AB ⃗⃗⃗⃗⃗ +(x+y )AD ⃗⃗⃗⃗⃗ =AB ⃗⃗⃗⃗⃗ +12AD ⃗⃗⃗⃗⃗ ,∴{x -12y =1,x +y =12,∴{x =56,y =-13,故选C . 10.12解析: 由题意在题图中以O 为原点,OA ⃗⃗⃗⃗⃗ 方向为x 轴非负半轴,过O 与OA 垂直向上为y 轴正方向建立平面直角坐标系(图略),则A (1,0),∵向量OA ⃗⃗⃗⃗⃗ 与OC ⃗⃗⃗⃗⃗ 的夹角为α, tan α=7,∴cos α=√210,sin α=7√210, 又|OC ⃗⃗⃗⃗⃗ |=√2,∴C15,75,cos(α+45°)=-35,sin(α+45°)=45,∴B -35,45, ∵OC ⃗⃗⃗⃗⃗ =m OA ⃗⃗⃗⃗⃗ +n OB ⃗⃗⃗⃗⃗ ,∴15,75=m (1,0)+n -35,45,∴{m -35n =15,45n =75,解得{m =54,n =74,∴n-m=12. 11.C 解析: 由题意得c =(3+t ,4),cos <a ,c >=cos <b ,c >,故9+3t+16|c |×5=3+t|c |×1,解得t=5.故选C .12.C 解析: AC ⃗⃗⃗⃗⃗ ·FN ⃗⃗⃗⃗⃗ =(AB ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ )·(FA ⃗⃗⃗⃗⃗ +AN ⃗⃗⃗⃗⃗⃗ )=AB ⃗⃗⃗⃗⃗ ·FA ⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ ·FA ⃗⃗⃗⃗⃗ +A A ⃗⃗⃗⃗⃗⃗ ·AN ⃗⃗⃗⃗⃗⃗ +AD ⃗⃗⃗⃗⃗ ·AN ⃗⃗⃗⃗⃗⃗ =0+|AD ⃗⃗⃗⃗⃗ ||FA ⃗⃗⃗⃗⃗ |cos π4+|AB ⃗⃗⃗⃗⃗ ||AN⃗⃗⃗⃗⃗⃗ |cos 3π4+0=√2−√2=0.选C . 13.D 解析: (a -b )·a =|a -b ||a |cos <a -b ,a >=(2+√3)·2, 即a 2-a ·b =4+2√3,a ·b =-2√3.所以|a ||b |cos <a ,b >=-2√3,cos <a ,b >=-√32,<a ,b >=5π6.14.A 解析: 由题当弦MN 长度最大时,即MN 为直径,设弦MN 的中点为O ,由题意,PM ⃗⃗⃗⃗⃗⃗ ·PN ⃗⃗⃗⃗⃗⃗ =(PO ⃗⃗⃗⃗⃗ +OM ⃗⃗⃗⃗⃗⃗ )·(PO ⃗⃗⃗⃗⃗ +ON ⃗⃗⃗⃗⃗⃗ )=PO ⃗⃗⃗⃗⃗ 2−14MN ⃗⃗⃗⃗⃗⃗⃗ 2=PO ⃗⃗⃗⃗⃗ 2-1=|PO ⃗⃗⃗⃗⃗ 2|-1,由1≤|PO ⃗⃗⃗⃗⃗ |≤√2,得PM ⃗⃗⃗⃗⃗⃗ ·PN⃗⃗⃗⃗⃗⃗ 的取值范围是[0,1]. 15.C解析: 如图所示,建立平面直角坐标系,设△ABC 的边长为a ,则asinA =2R=4(R 为△ABC 外接圆半径),所以a=2√3,A (0,3),B (-√3,0),C (√3,0),△ABC 的外接圆的方程为x 2+(y-1)2=4,设P 点坐标为(2cos θ,1+2sin θ),θ∈R ,PA ⃗⃗⃗⃗⃗ ·PB ⃗⃗⃗⃗⃗ +PB ⃗⃗⃗⃗⃗ ·PC ⃗⃗⃗⃗⃗ =PB ⃗⃗⃗⃗⃗ (PA ⃗⃗⃗⃗⃗ +PC ⃗⃗⃗⃗⃗ )=4+2√3cos θ+2sin θ=4+4cos θ-π6≤8,当cos θ-π6=1时,等号成立.故选C .。

2022届高三二轮复习联考（一）新高考卷II英语试卷注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

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1.What do the speakers need to buy?A.Two chairs.B.A sofa bed.C.A TV set.2. Where is the woman's husband now?A.In London.B.In Tokyo.C.In Paris.3.What does the woman think of Greg?A.He is interesting.B. He is handsome.C.He is very tall.4.How will the man get to Paris?A.By plane.B.By train.C.By bus.5. What are the speakers mainly talking about?A.A party.B. A necklace.C.A dress.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

1.集合、常用逻辑用语、不等式考向1 集合的概念及运算1.(2022·全国甲·理3)设全集U={-2,-1,0,1,2,3},集合A={-1,2},B={x|x 2-4x+3=0},则∁U (A ∪B )=( ) A.{1,3} B.{0,3} C.{-2,1} D.{-2,0}2.(2022·全国乙·理1)设全集U={1,2,3,4,5},集合M 满足∁U M={1,3},则( )A.2∈MB.3∈MC.4∉MD.5∉M3.(2022·新高考八省第二次T8联考)设集合A={x|log 2(x-1)<2},B={x|x<5},则( )A.A=BB.B ⊆AC.A ⊆BD.A ∩B=⌀ 4.(2022·安徽蚌埠质检三)设集合M={x|x=C 5m ,m ∈N *,m ≤5},则M 的子集个数为( )A.8B.16C.32D.64考向2 充分条件、必要条件与充要条件5.(2022·浙江·4)设x ∈R ,则“sin x=1”是“cos x=0”的( ) A.充分不必要条件 B.必要不充分条件 C.充分必要条件D.既不充分也不必要条件6.(2022·河南濮阳一模)“b ≤1”是“函数f (x )={bx +2,x >0,log 2(x +2)+b ,-2<x ≤0是在(-2,+∞)上的单调函数”的( )A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件7.若x ,y ∈R ,则“x<|y|”是“x 2<y 2”的( ) A.充分不必要条件 B.必要不充分条件 C.充要条件D.既不充分也不必要条件8.(2022·河南许昌质检)若(x-a )2<4成立的一个充分不必要条件是1+12-x ≤0,则实数a 的取值范围为( ) A.(-∞,4] B.[1,4] C.(1,4)D.(1,4]考向3 常用逻辑用语9.(2022·河南郑州质检)已知命题p :∃x 0∈R ,3sin x 0+4cos x 0=4√2;命题q :∀x ∈R ,1e |x|≤1.则下列命题中为真命题的是 ( )A.p ∧qB.(¬p )∧qC.p ∨(¬q )D.¬(p ∨q )10.(2022·河南焦作一模)已知命题p :∃x 0∈N *,lg x 0<0,q :∀x ∈R ,cos x ≤1,则下列命题是真命题的是( ) A.p ∧q B.(¬p )∧q C.p ∧(¬q )D.¬(p ∨q )11.(2022·河南洛阳一模)已知命题p :"x ∈R ,x 2+x+1>0;命题q :若a>b ,则1a<1b.下列命题为真命题的是( ) A.(¬p )∨q B.(¬p )∧(¬q ) C.p ∧qD.p ∨q12.若“∃x 0∈12,2,使得2x 02-λx 0+1<0成立”是假命题,则实数λ的取值范围为 .考向4 不等关系及线性规划13.(2022·河南许昌质检)已知a>b>0,且a+b=1,则下列结论正确的是( ) A.ln(a-b )>0 B.√a +√b >2 C.b a >a bD.1a +1b >414.(2022·河南焦作二模)已知x ,y 满足约束条 件{2x -3y +6≥0,2x +y +2≥0,4x -y -8≤0,则3x-2y 的最大值为 ( )A.1B.4C.7D.1115.(2022·浙江·3)若实数x ,y 满足约束条件{x -2≥0,2x +y -7≤0,x -y -2≤0,则z=3x+4y 的最大值是( )A.20B.18C.13D.616.(2022·河南濮阳一模)设x ,y 满足约束条件{y ≥2x ,y ≥-x ,y ≤2,则z=y-x 的最大值是 .1.集合、常用逻辑用语、不等式1.D 解析: 由题意知B={1,3},则A ∪B={-1,1,2,3}, 所以∁U (A ∪B )={-2,0}, 故选D .2.A 解析: ∵U={1,2,3,4,5},∁U M={1,3}, ∴M={2,4,5},∴2∈M ,3∉M ,4∈M ,5∈M. 故选A .3.C 解析: log 2(x-1)<2⇔0<x-1<4⇔1<x<5,∴A={x|log 2(x-1)<2}={x|1<x<5},即A ⊆B ,故选C .4.A 解析: 因为C 51=C 54,C 52=C 53,所以集合中含有3个元素,则M 的子集个数为23=8,故选A .5.A 解析: 由sin x=1,得x=2k π+π2,k ∈Z ,此时cos x=0;由cos x=0,得x=k π+π2,k ∈Z ,此时sin x=±1,故选A .6.B 解析: 依题意,函数f (x )是在(-2,+∞)上的单调函数, ∵y=log 2(x+2)+b 在(-2,0]上单调递增, ∴f (x )在(-2,+∞)上单调递增, 需b>0且1+b ≤2,即0<b ≤1. 故选B .7.B 解析: 由x<|y|推不出x 2<y 2,如x=-3,y=1;由x 2<y 2得|x|<|y|,又因为x ≤|x|,所以x ≤|x|<|y|,所以x 2<y 2⇒x<|y|. 故选B .8.D 解析: 根据题意,(x-a )2<4⇔-2<x-a<2⇔a-2<x<a+2,不等式的解集为(a-2,a+2); 1+12-x ≤0⇔3-x2-x ≤0⇔(x-3)(x-2)≤0且x ≠2,解得2<x ≤3,不等式的解集为(2,3]; 若(x-a )2<4成立的一个充分不必要条件是1+12-x ≤0,则(2,3]⫋(a-2,a+2);则有{a -2≤2,a +2>3,解得1<a ≤4,即a 的取值范围为(1,4]. 故选D .9.B 解析: ∵3sin x+4cos x=5sin(x+θ)∈[-5,5],tan θ=43,4√2>5,∴命题p 为假命题.∵|x|≥0,∴1e|x|≤1e=1,∴命题q 为真命题,∴p ∧q 为假命题;(¬p )∧q 为真命题;p ∨(¬q )为假命题;¬(p ∨q )为假命题.故选B .10.B 解析: 因为∀x ∈N *,lg x ≥0,所以命题p 为假命题,¬p 为真命题.因为∀x ∈R ,cos x ≤1成立,所以命题q 为真命题,所以(¬p )∧q 为真命题.11.D 解析: 对命题p ,因为x 2+x+1=x+122+34>0恒成立,故命题p 为真命题.对命题q ,当a 为正数,b 为负数时,命题不成立,故命题q 为假命题,故只有选项D 为真命题,故选D .12.(-∞,2√2] 解析: 由题意得,“∀x ∈12,2,2x 2-λx+1≥0”为真命题,即λ≤2x+1x .因为2x+1x≥2√2x ·1x=2√2,当且仅当2x=1x,即x=√22时,等号成立,所以实数λ的取值范围为(-∞,2√2].13.D 解析: ∵a>b>0,且a+b=1,∴12<a<1,0<b<12, ∴0<a-b<1,ln (a-b )<0,故A 错误;∵1>a>b>0,∴√a +√b <1+1=2,故B 错误; 令f (x )=lnxx (0<x<1),则f'(x )=1-lnxx 2>0,故f (x )在(0,1)上单调递增,故lna a>lnb b,即b ln a>a ln b ,即ln a b >ln b a ,∴a b >b a ,故C 错误; ∵a>b>0,∴1a +1b =a+b a +a+b b =2+b a +a b ≥2+2√b a ·ab=4,当且仅当a=b 时,等号成立,∴1a +1b >4,故D正确.14. D 解析: 不等式组{2x -3y +6≥0,2x +y +2≥0,4x -y -8≤0表示的平面区域如图中阴影部分所示,联立方程组{2x +y +2=0,4x -y -8=0,解得{x =1,y =-4,即B (1,-4),平移直线3x-2y=0至经过点B 时目标函数u=3x-2y 取得最大值,即u max =3×1-2×(-4)=11.15. B 解析: 根据约束条件画出可行域.可知当直线y=-34x+z4过点(2,3)时,z 取到最大值,为18,故选B .16.4 解析: 画出可行域如图所示,化目标函数为斜截式方程y=x+z ,则当直线y=x+z 在y 轴上截距最大时,z 取得最大值,联立{y =2,y =-x , 解得{x =-2,y =2,。