高二下期3月份月考
山东省东营市利津县高级中学2023-2024学年高二下学期3月月考生物试题(原卷版)
C.卵裂过程进行有丝分裂，胚胎中DNA总量不断增加
D.卵裂需要消耗能量，有机物总量和种类都不断减少
20.下表为几种限制酶的识别序列及其切割位点，由此推断，下列说法正确的是（）
限制酶名称
识别序列和切割位点
限制酶名称
识别序列和切割位点
BamH I
5'-G↓GATCC -3'
C.电泳凝胶中DNA分子的迁移速率与凝胶的浓度、DNA分子大小和构象等有关
D.设计引物从图1中完整DNA片段中扩增出完整X，至少需要3次PCR
二、选择题：本题共5小题，每小题有一个或多个选项符合题目要求。
16.某生物活动小组为探究当地农田土壤中分解尿素的细菌的数量，进行了取样、系列梯度稀释、涂布平板、培养、计数等步骤，实验操作过程如下，下列说法正确的是（）
2023-2024学年第一学期3月份教学质量检测生物试题
一、选择题：本题共15小题，每小题只有一个选项符合题目要求。
1.下列有关传统发酵技术及其应用的叙述，正确的是（）
A.酿酒过程中密封的时间越长，酵母菌产生的酒精量就越多
B.制作果醋时，在液体培养基的表面将会形成单菌落
C.制作泡菜时，乳酸菌可以将葡萄糖分解成乳酸和CO2
A.a中放置的一般是幼龄动物的器官或组织，原因是其细胞分裂能力强
B c过程属于原代培养，d过程属于传代培养
C.c、d瓶中的培养液需要加入血清等天然成分
D.将培养的细胞置于95%O2和5%CO2混合气体的CO2培养箱中进行培养
19.下列关于卵裂的叙述，错误的是（ ）
A.卵裂时期细胞不断分裂，胚胎体积不断增大
A.受精卵发育为桑葚胚的过程是在透明带内完成的，孵化后进入囊胚期阶段
上海市进才中学2023-2024学年高二下学期3月月考英语试题
上海市进才中学2023-2024学年高二下学期3月月考英语试题一、语法填空Directions: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.Over the past decade or so, biologists have shown that we are filled with microbiomes (微生物组), inside and out. This microbiome, 1 (compose) of bacteria, fungi and viruses, profoundly influences our health and fitness and sometimes is even linked to our emotional state.The oral microbiome gets far less attention, but we 2 (know) of it for a long time. In 1891, US dentist Willoughby D．Miller first proposed that bacteria could leave the mouth, travel to other parts of the body and cause disease. We only began to get supporting evidence in 1989, when researchers noticed that people who had experienced a heart attack were rated as having oral health that was about twice as poor as 3 of a control group. Even when age, social class and smoking habits 4 (account) possible factors, the results remained almost the same. A solid link seemed to be there.More recently, thanks to DNA sequencing technology, by cataloguing the microbes in our mouths, we are now finding that the types of bacteria people have 5 (live) there seem to be associated with a growing number of conditions like cancer. Perhaps the most striking example is Alzheimer’s disease. It can be found that people with gum (牙龈) disease are 6 increased risk of developing this condition, which slowly robs people of their memories, personalities and cognitive function. However, until recently, it was unclear 7 poor oral health was a contributing cause of Alzheimer’s or a consequence of it.Then, in 2019, scientists discovered some species of bacteria known 8 (cause) gum disease — including one called Porphyromonas gingivalis — which are inside the brains of people who died of Alzheimer’s disease. If the mouth bacteria were getting into the brain, that lent weight to the idea that they could be a cause of Alzheimer’s.Researchers are still trying to grasp how this could happen. 9 the mechanism is, they note that there may be a way you can protect yourself in advance. Dental scientists at theUniversity of Melbourne, Australia, is developing a vaccine against gum disease 10 you can reduce the risks of gum diseases significantly.二、选词填空Directions: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.Luxury brands used to speak in monologues. However, nowadays, influencers are becoming the ambassadors of those luxurious brands.In the age of social media, the buyers are having a voice in products, in particular, the influencers. These individuals have won large followings by 11 and attacking occasionally a variety of products. Their fame stems from the clever use of Instagram, Snapchat or TikTok. Their posts seem trivial. Their business isn’t.For consumers, influencers are at once a walking advert and a trusted friend. For the brands, they are a(n) 12 . And for regulators, they are the subject of ever closer review. On March 29th, news reports 13 that China’s authorities were planning new restrictions concerning livestreaming platforms. The limitations 14 how much money internet users can spend tipping their favorite influencers, how much those influencers can earn from fans and what they are allowed to post.There hardly exists any 15 of the size of the influencer industry. One in 2020 from the National Bureau of Statistics in China, where influencers gained prominence earlier than in the West, assessed its contribution to the economy at $210bn, equivalent to 1.4% of GDP. As with many things digital, the pandemic seems to have given it a(n) 16 as more people were glued to their smartphones more of the time. The influencer ecosystem is challenging the 17 principle of luxury-brand management. Apart from being one-directional, campaigns have tendedto be 18 , unchanging and expensive. The same smile from the same photograph of the same Hollywood star would induce passers-by to purchase an item for many years. Such star-led campaigns can be unappealing to teenagers and 20-somethings 19 authenticity over timeless glamour. And influencers, with their girl-or boy-next-door charm, offer this for a small portion of the fee of a big-name star. The best ones are able to repackage a brand’s message in a way that is 20 with their voice, their followers’ tastes and their platform of choice. So to be a top-ten brand, you have to know how to play the digital game. If you don’t, you are not going to be top ten for very long.三、完形填空A cancer diagnosis will force King Charles III out of the public eye for the foreseeable future. For a highly 21 royal family that has cultivated its public image through countless appearances a year — ribbon-cuttings, ship launchings, gala benefits and so on — the marginalization of Charles may finally force the royals to rethink how they 22 themselves in a social-media age.The king’s illness is the latest 23 to the British royal family, which has seen its ranks thinning by death, scandal (Prince Andrew) and self-exile (流放) (Prince Harry and his wife, Meghan). Regardless of what is predicted, it appears that the king’s cancer presses the royal family into 24 territory.The answer to it, royal watchers argue, may be 25 . During the coronavirus pandemic, Elizabeth conducted meeting s via Zoo m calls,becoming 26 enough with it that she made jokes with the digitally distorted faces on her computer screen.Social media can also be employed to enhance the 27 exposure of family members. The royal family’s Instagram account 28 more than 13 million followers. But for young people, who spend whole day following their favorite celebrities online, a royal turning up to dedicate a new primary school may not 29 as much as it did to their previous generations.Prince Harry, the king’s younger son, fell out bitterly with his family after his 30 from royal duties and relocation to California in 2020. Undoubtedly, the greatest burden falls onthe mere remaining 41-year-old heir (继承人), William, who has been recognized as a qualified successor for 31 a role on issues from climate change to homelessness. Meanwhile, he has jealously guarded his family’s 32 , demanding his office release no photographs of three young children visiting their mother in the hospital. That approach stood 33 his father, who approve the disclosure of an unusual amount of detail about his recent cancer diagnosis. The scrutiny (审查) of William will 34 increase, experts said, as he occupies a more central place in the Windsor family hierarchy.Queen Elizabeth viewed assuming kingly duties as so 35 that she steeled herself, two days before her death at 96, to meet with the outgoing prime minister in Scotland, masking her own condition. Charles, though, has departed from long family practice “in the hope it may assist public understanding for all those around the world who are affected by cancer.”21．A．apparent B．controversial C．visible D．generous 22．A．project B．market C．illustrate D．propose 23．A．witness B．blow C．solution D．disloyalty 24．A．uncharted B．unnoticed C．indefensible D．inexcusable 25．A．human-initiated B．strategy-focused C．technology-drivenD．goal-oriented26．A．frustrated B．preoccupied C．content D．comfortable 27．A．in-depth B．in-person C．in-between D．in-built 28．A．claims B．calculates C．confirms D．clarifies 29．A．matter B．contribute C．relate D．bring 30．A．dismissal B．survival C．withdrawal D．renewal 31．A．carving out B．applying for C．identifying with D．reflecting on 32．A．connection B．priority C．presence D．privacy 33．A．in parallel to B．as opposed to C．on behalf of D．in honor of 34．A．scarcely B．effortlessly C．approximately D．inevitably 35．A．grave B．distinct C．exceptional D．progressive四、阅读理解First, I must get settled into school. My classes begin today at the PRIVET! Russian Academy of Language Studies, where I will be attending class five days a week, four hours a day.I know I am such a shameless student. I laid my clothes out last night, just like I did before my first day of first grade, with my patent leather shoes and my new lunch box.The last thing I want is to end up in a Level One class, which would be so humiliating for me. Given that I already took a whole entire semester of Russian at my Night School for Divorced Ladies in New York, and that I spent the summer memorizing flash cards. The thing is, I don’t even know how many levels this school has, but the me re mention of “level” sparks a resolve within me to aim for Level Two—at least.It’s such a hard test! I can’t get through even a tenth of it!In the end, it’s OK, though.So it’s hammering down rain today, but I show up early, wander about the school and smugly walk past all those Level One students (who must be cookies, really) and enter my first class. Here I am with my peers. But it becomes swiftly evident that these are not my peers and that I have no business being here. I feel like I’m swimming, but barely. Like I’m taking in water with every breath. The teacher, a skinny guy (Why are the teachers so skinny here? I don’t trust skinny Russians.), is going way too fast, is skipping over whole chapters of the textbook, saying, “You already know this, you already know that…” and keeping up a rapid-fire conversation with my apparently fluent classmates. My stomach is gripped in horror and I’m gasping for air and praying he won’t call on me. Just as soon as the break comes, I run out of that classroom on wobbling legs and I scurry all the way over to the administrative office almost in tears, where I beg in very clear English if they could please move me down to a Level One class. And so they do. And now I am here.36．What do we know about the writer from the first two paragraphs?A．She is a newly-admitted student majoring in language teaching.B．She has no knowledge of this foreign language and is put to shame.C．She is full of anticipation and readiness for new educational pursuit.D．She attends the same school to build upon prior academic achievements.37．What does the underlined word smugly in paragraph 5 mean?A．arrogantly B．furiously C．secretly D．nobly38．Why does the writer withdraw from Level Two class in the end?A．The teacher singles her out for her inability to converse smoothly in foreign language.B．The age and culture divide leads to an overwhelming sense of alienation with classmates.C．The unexpected discomfort in her stomach prompts a pause in regular learning activities.D．The unthinkable difficulty in catching up with the teaching rhythm destroys confidence. 39．What might be the best title of the passage?A．Well Begun, Half Done.B．An Idle Youth, A Needy Age.C．More Haste, Less Speed.D．No Pain, No Gain.I guess I was a little bored. For the past hour, I’d been on the phone with Daniele, the head of my office in Italy, reviewing our latest purchases of Italian gold, Murano glass and Italian-made shoes and handbags.“Daniele,” I said, “What is the hottest jewelry in Italy right now?” His reply? Woven gold bracelets studded with gems. He texted me some photos and I knew immediately that this was jewelry that Raffinato just had to have.RAFFINATO IS ONE OF AMERICA’S LARGEST RETAILERS OF ITALIAN-MADEJEWELRYPresenting the Italiano Fantasia Bracelets, two designs that are prime examples of Italy’sfinest artisanship. Each of these bracelets includes more than 20 brilliant cut gems of Diamond Aura®, our Ultimate Diamond Alternative®, in a setting finished with 18 karat Italian gold.What is DiamondAura®? It’s a sparkling marvel that rivals even the finest diamonds (D Flawless) with its transparent color and clarity: Both are so hard they can cut glass. Don’t believe me? The book “Jewelry and Gems: The Buying Guide,” praised the technique used in our diamond alternative :“The best diamond simulation to date, and even some jewelers have mistaken these stones for mined diamonds,” it raved.The best part about these bracelets? The price. Because of our longstanding connections in Arezzo, the mecca of Italian goldsmithing, we can offer both bracelets together for just $99, a fraction of the price you’ll pay anywhere else for similar jewelry.Order today. These bracelets are one of our hottest sellers this year, and with disruptions in the supply chain, we can only guarantee that we have 1,279 861 of these bracelets on hand for this ad.Jewelry Specifications:• Made in Arezzo, Italy. 18k gold finish• Diamond Aura®, the Ultimate Diamond Alternative®• Fit wrists up to 7 ¼"Italian Fantasia Bracelet CollectionA．X Bracelet (¼ ctw) $399 $59* Save $340B．Woven Bracelet (⅓ ctw) $299 $69* Save $230*special price only for customers using the offer code*an extra $50 refund for any purchase exceeding 6 pieces.40．Which of the descriptions is true about the advertised products?A．The bracelets boast original Italian design and craftsmanship.B．There is no restriction on wrist size due to its adjustable design.C．A supply chain disruption leads to products being out of stock.D．the jewelry is framed with real diamonds of the finest quality.41．Raffinato is mainly in charge of _________.A．invention B．production C．distribution D．exhibition42．If a local adolescent wants to buy 2 Woven Bracelets and 3 Sets of Both as souvenir forroommates, the minimum price he/she has to pay is ______.A．$295B．$385C．$415D．$435At Cleveland Bridge, in Bath, a long line of traffic is building up. Although the Georgian structure was praised for its handsome Greek Revival style by Nikolaus Pevsner, an architectural master, it was built for horses, not cars. Repairs will close the bridge for several months, causing bigger jams and more pollution in a city where air quality is already a cause of concern.Bath is an extreme example of a tradeoff faced by much of the country. Britain has the oldest housing stock in Europe, with one in five homes more than 100 years old. Period features are valued and often protected by law. Yet as efforts to cut carbon emissions intensify, they are contradicting attempts to preserve heritage. It is a “delicate balance” says Wera Hobhouse, Bath’s Member of Parliament (议员) “What is the public benefit of dealing with the climate emergency, versus protecting a heritage asset?”Two years ago, Bath was among the first British cities to declare a “climate emergency” when it also promised to go carbon-neutral by 2030. Yet Bath also wants—and is legally required to preserve its heritage. With Roman remains and Georgian streets that spread across the Avon Valley in shades of honey and butter, the city is designated a world heritage site by UNESCO. About 60% of it is further protected by the government as a conservation area, more than 5,000 of Bath’s buildings—nearly 10% of the total—are listed as being of special architectural or historical interest, making it a criminal offence to alter them without permission.Many of the features that make Bath’s Georgian buildings so delicate also make them leaky. Buildings of traditional construction make up 30% of Bath’s housing stock but take up 40% of domestic carbon emissions, according to the Centre for Sustainable Energy, a charity. British homes are rated for energy efficiency on a scale from A to G; most traditional buildings in the city are an F or G.Transport is another area where climate and heritage clash. Bath’s 17th-century streets lack room for bike lanes. Joanna Wright was recently relieved of her role as Bath’s climate chief after proposing that North Road, which leads to the university, should be closed to traffic. In two years she was unable to install any on-street electric-vehicle charging points, partly because of the “nightmare” of getting permission to dig up old pavements.All this means going carbon neutral by 2030 looks hard, but the city is at least beginning to make compromises. In March it launched the first “clean-air zone” outside London, charging drivers to enter central Bath. A trial has made 160 electric motoreycles available to hire. And local opinion seems to be shifting in favour of sustainability. “The discussion has moved dramatically towards considering the climate emergency,” says Ms. Hobhouse.43．What problem does Bath face?A．It is getting more and more crowded due to the maintenance work underway.B．Its housing stock with a long history are being altered without authorization.C．Its historical significance makes it hard to be reformed into an eco-friendly city.D．Its promise to go carbon-neutral by 2030 is greeted with doubt from its citizens. 44．The percentage mentioned in the underlined sentence (Paragraph 4) is intended to _________.A．highlight the long history of Bath’s building complexB．show the low energy efficiency of the historic buildingsC．illustrate the delicate features that Bath’s buildings shareD．challenge the validity and credibility of the A-G scale45．Why was Joanna Wright removed from her position?A．She proposed the North Road not be accessible to the public.B．She failed to solve the conflict between transport and heritage.C．She stood right in the way of the heritage preservation efforts.D．She struggled to get authorized to dig up those old pavements.46．What can be inferred from the passage?A．Wera Hobhouse keeps a balance between public benefit and heritage protection.B．Traditional buildings in Bath are not available to visitors not having permits.C．Bath has been setting the pace for the construction of a carbon-neutral city.D．Local people in Bath used to favor heritage conservation over sustainability.Immersive Art Draws People InWith bold, swirling brushstrokes (绘画技巧) and vivid colors, Vincent van Gogh’s stirring Starry Night brings to life a turbulent (汹涌的) sky. It’s one of the most recognizable paintings in the world. And gazing at the scenic canvas can make museum visitors feel starstruck.But seeing this masterpiece on a gallery wall isn’t the only way art fans can experience its impact 47 . They find themselves surrounded by shimmering colors that dance before their eyes and ripple (涟漪) at their feet. These exhibitions digitally project moving images onto walls, floors, and sometimes onto viewers themselves. They are examples of immersive art.48 . While it can be hard to characterize, it’s generally a multisensory, an interactive event that engages viewers and lets them feel like part of the artwork. One thing is certain — these exhibitions have been wildly popular — selling out tickets in cities worldwide.Van Gogh gained fame only after his death. In fact, the 19th-century painter sold just one painting during his lifetime. But now he is immersive art’s biggest superstar. His work has been showcased in various exhibitions featuring immense images. 49 . One show, Van Gogh Alive, has appeared in 65 countries since 2011. It even features a signature scent for visitors to sniff. Shows of other artists — including Monet, Renoir, and Chagall — have lit up venues, too.The popularity of immersive art has been powered party by social media. As visitors post selfies featuring van Gogh’s art or videos of friends stepping into a fantastical fridge, these experiences draw bigger and bigger crowds. 50 . That’s because many curators and creators share a common goal — to help more people get into art!A．Immersive art doesn’t simply mean sitting in a glass case or fitting in a frame.B．The art work is animated and accompanied by music, voices, and background sound.C．In fact, some exhibits give people a chance to be enveloped by van Gogh’s celebrated painting. D．Meanwhile, traditional museums are following the trend and applying immersive technologies E．Critics once described his work as a multidimensional mystery house leading to secret passages.F．More important is the ability to bridge the gap between traditional art spaces and modern audiences.五、书面表达51．Directions:Read the following passage. Summarize the main idea and the main point (s) of the passage in no more than 60 words. Use your own words as far as possible.Preventing ResistanceThe development of drug-resistant pathogens (病原体) can be prevented in at least three ways. First, sufficiently high concentrations of the drug can be maintained in a patient’s body for a long enough time to kill all sensitive cells and hold others long enough for the body’s defenses to defeat them. Discontinuing a drug before all of the pathogens have been neutralized promotes the development of resistance. For this reason, it is important that patients finish their entire antimicrobial prescription (抗菌药物) and resist the temptation to “save some for another day.”A second way to prevent resistance is to use antimicrobial agents in combination so that pathogens resistant to one drug will be killed by the second, and vice versa. Additionally, one drug sometimes enhances the effect of a second drug in a process called synergism. Enhanced effect can also result from combining an antimicrobial drug and a chemical.A third way to reduce the development of resistance is to limit the use of antimicrobials to necessary cases. Unfortunately, many antimicrobial agents are used indiscriminately, in both developed countries and in lessdeveloped regions where many are available without a physician’s prescription. In the United States, an estimated 50 percent of prescriptions for antibacterial agents to treat sore throats and 30 percent of prescriptions for ear infections, are inappropriate because the diseases are viral in nature. Likewise, because antibacterial drugs have no effect on cold and flu viruses, 100 percent of antibacterial prescriptions for treating diseases are unnecessary. As discussed previously, the use of antimicrobial agents encourages the reproduction of resistant bacteria by limiting the growth of sensitive cells; therefore, inappropriate use of such drugs increases the likelihood that resistance of bacteria will multiply.In order to maintain the effectiveness of antimicrobial treatments, responsible practices must be implemented._______________________________________________________________________________ _______________________________________________________________________________ _______________________________________________________________________________ _______________________________________________________________________________ _______________________________________________________________________________ ___________________________________________________________六、翻译52．你要认识到：如果感觉困难，放平心态，那是因为它真的很难。
江苏省南京市秦淮区2023-2024学年高二下学期3月月考数学试卷
江苏省南京市秦淮区2023-2024学年高二下学期3月月考数学试卷一、单选题1．已知点()3,1,0A -，若向量()2,5,3AB =-u u u r，则点B 的坐标是（ ）. A ．()1,6,3-B ．()5,4,3-C ．()1,6,3--D ．()2,5,3-2．设()ln f x x x =，()f x '为()f x 的导函数，若()02f x '=，则0x =（ ） A ．2eB ．ln 2C ．eD ．ln 223．用0，1，2，…，5这6个数字组成无重复数字的三位数的个数是（ ） A ．36AB ．3365A A -C ．255AD ．35A4．()(2)(3)(4)(15)N ,15x x x x x x +----∈>L 可表示为（ ） A ．132A x - B ．142A x - C ．1315A x -D ．1415A x -5．在三棱柱111ABC A B C -中，记1AA a =u u u r r ，AB b =u u u r r ，AC c =u u u r r ，点P 满足12BP PC =uu r uuu r ，则AP =u u u r（ ）A ．113332a b c -+r r rB ．121333a b c ++r r rC ．223331a b c +-r r rD ．212333a b c ++r r r6．已知双曲线()22210x y a a-=>的右焦点与抛物线28y x =的焦点重合，则此双曲线的渐近线方程是（ ）A ．y =B ．y =C ．y =D ．y = 7．数学对于一个国家的发展至关重要，发达国家常常把保持数学领先地位作为他们的战略需求.现某大学为提高数学系学生的数学素养，特开设了“古今数学思想”，“世界数学通史”，“几何原本”，“什么是数学”四门选修课程，要求数学系每位同学每学年至多选3门，大一到大三三学年必须将四门选修课程选完，则每位同学的不同选修方式有（ ） A ．60种B ．78种C ．84种D ．144种8．已知EF 是棱长为8的正方体的一条体对角线，空间一点M 满足40ME MF ⋅=-u u u r u u u r，AB 是正方体的一条棱，则AM AB ⋅u u u u r u u u r的最小值为（ ）A ．)164B ．(162C ．(164D ．)162二、多选题9．设数列{}n a 的前n 项和为n S ，已知11a =，13n n a S +=，*n ∈N ，则（ ） A ．24S =B ．6416a a =C ．数列{}n a 是等比数列D ．数列{}n S 是等比数列10．平面α经过三点()1,0,1A -，()0,1,0B ，()1,2,0C -，向量()1,,n u t =r是平面α的法向量，则下列四个选项中正确的是（ ）A ．直线AB 的一个方向向量为()2,2,2a =-rB ．线段AB 的长度为3C ．平面α的法向量n r中1u t +=D ．向量AB u u u r 与向量BC u u u r11．甲，乙，丙，丁四名志愿者主动到,,A B C 三所山区学校参加支教活动，要求每个学校至少安排一名志愿者，下列结论正确的是（ ）A ．共有72种安排方法B ．若甲被安排在A 学校，则有12安排方法C ．若A 学校需要两名志愿者，则有12种安排方法D ．若甲、乙不能在同一所学校，则有6种安排方法三、填空题12．已知1121n n C -+=，那么n =； 13．如图，已知空间四边形ABCD 中，2AB =-u u u r r r a c ，568CD =+-u u u r r r ra b c ，对角线AC ，BD 的中点分别为E ，F ，则EF u u u r＝.(用向量,,a b c r r r 表示)14．已知i r ，j r ，k r 是不共面向量，a v =2i r -j r +3k r ，b v =-i r +4j r -2k r ，c v =7i r +5j r +λk r ，若a v，b v ，c v三个向量共面，则实数λ等于.四、解答题15．甲组有3名男生．3名女生；乙组有4名男生，2名女生．(1)从这些学生中选出3人参加活动，至少有1名女生的不同选法有多少种？(2)从甲、乙两组中各选出2名学生，选出的4人中恰有1名女生的不同选法有多少种？ (3)将这些学生排成两排，两组的女生站第一排，两组的男生站第二排，且同组学生均相邻，共有多少种不同的排法？16．记等差数列{}n a 的前n 项和为n S ，735a a +=，126a =．设2n n n a b n ⋅=．(1)求16S 的值；(2)记2n K 为数列{}n b 的前2n 项和，n T 为数列{}2n b 的前n 项和，且2n n K tT =，求实数t 的值．17．已知椭圆C ：()222210+=>>x y a b a b (P ．(1)求椭圆C 的方程；(2)过右焦点F 的直线l 与椭圆C 交于A ，B 两点，若3AF FB =u u u r u u u r，求PAB V 的面积．18．已知函数()()2e 2e x xf x a a x =+--．(1)若函数()f x 在点()()0,0f 处的切线与直线30x y +=垂直，求a 的值； (2)讨论函数()f x 的单调性；(3)若()f x 有两个零点，求a 的取值范围．19．如图，在四棱锥P ABCD -中，底面ABCD 是边长为2的菱形，PAD △为等边三角形，平面PAD ⊥平面ABCD ，PB BC ⊥．点E 在线段PC 上．(1)若3PE EC，在PB上找一点F，使得E F A D、、、四点共面，并说明理由；(2)求点A到平面PBC的距离；(3)若直线AE与平面ABCD ADE与平面ABCD夹角的余弦值．。
湖北省天门市2023-2024学年高二下学期3月月考数学试题含答案
湖北省天门2023-2024学年度高二下学期三月月考数学试题（答案在最后）考试内容：选修一第一章——选修三第六章6.1考试时间：2024年3月31日出题人：审题人：一、单选题（共40分）1.某圆锥的侧面积为16π，其侧面展开图为一个半圆，则该圆锥的底面半径长为（）A.2B.4C. D.【答案】C 【解析】【分析】设圆锥的母线长为l ，底面半径为r ，由题意得到2ππr l =求解.【详解】设圆锥的母线长为l ，底面半径为r ，即侧面展开图的半径为l ，侧面展开图的弧长为πl .又圆锥的底面周长为2πr ，所以2ππr l =，即圆锥的母线长2l r =.所以圆锥的侧面积为2π2π16πrl r ==，解得r =故选：C.2.若直线1l ：2(1)40x m y +++=与直线2l ：320mx y +-=平行，则m 的值为（）A.2B.3- C.2或3- D.2-或3-【答案】C 【解析】【分析】依题意可得23(1)0m m ⨯-+=，求出m 的值，再检验即可.【详解】直线1l ：2(1)40x m y +++=与直线2l ：320mx y +-=平行，则23(1)0m m ⨯-+=，解得3m =-或2m =，当3m =-时，此时直线1l ：2240x y -+=与直线2l ：3320x y -+-=平行，当2m =时，此时直线1l ：2340x y ++=与直线2l ：2320x y +-=平行，故3m =-或 2.m =故选：C3.等比数列{}n a 的各项均为正数，且564718a a a a +=，则3132310log log log a a a ++⋅⋅⋅+=（）A.12B.10C.5D.32log 5【答案】B 【解析】【分析】利用等比数列的性质，结合对数的运算法则即可得解.【详解】因为{}n a 是各项均为正数的等比数列，564718a a a a +=，所以564756218a a a a a a +==，即569a a =，则11029569a a a a a a ==== 记3132310log log log S a a a =++⋅⋅⋅+，则3103931log log log S a a a =+⋅+⋅⋅+，两式相加得()()()3110329310132log log log 10log 920S a a a a a a =++⋅⋅⋅+=⨯=，所以10S =，即3132310log log log 10a a a ++⋅⋅⋅+=.故选：B.4.已知函数()()()ln 2ln 4f x x x =-+-，则()f x 的单调递增区间为（）A.()2,3 B.()3,4 C.(),3-∞ D.()3,+∞【答案】A 【解析】【分析】根据对数真数大于零可构造不等式组求得函数定义域；利用导数可求得函数单调递增区间．【详解】由2040x x ->⎧⎨->⎩得：24x <<，即()f x 的定义域为()2,4；()()()()23112424x f x x x x x -'=-=---- ，∴当()2,3x ∈时，()0f x ¢>；当()3,4x ∈时，()0f x '<；()f x \的单调递增区间为()2,3．故选：A ．5.已知函数()2xf x =，则函数()f x 的图象在点()()0,0f 处的切线方程为（）A.10x y --=B.10x y -+=C.ln 210x y ⋅--=D.ln 210x y ⋅-+=【答案】D【分析】求出函数()f x 的导数，再利用导数的几何意义求出切线方程.【详解】函数()2xf x =，求导得()2ln 2x fx '=，则(0)ln 2f '=，而(0)1f =，所以所求切线方程为1ln 2(0)y x -=⋅-，即ln 210x y ⋅-+=.故选：D6.在平面直角坐标系xOy 中，点()()1,0,2,3A B -，向量OC mOA nOB =+，且40m n --=.若P 为椭圆2217y x +=上一点，则PC 的最小值为（）A.B.C.D.【答案】A 【解析】【分析】根据给定条件，求出点C 的轨迹，再借助三角代换及点到直线距离公式求出最小值.【详解】设点(,)C x y ，由()()1,0,2,3A B -及OC mOA nOB =+，得(,)(2,3)x y m n n =-+，即23x m ny n=-+⎧⎨=⎩，而40m n --=，消去,m n 得：3120x y -+=，设椭圆2217y x +=上的点(cos ),R P θθθ∈，则点P 到直线3120x y -+=的距离d =，其中锐角ϕ由tanϕ=确定，当sin()1θϕ+=时，min d =PC d ≥ ，所以PC 的故选：A【点睛】思路点睛：求出椭圆上的点与其相离的直线上点的距离最小值，可转化为求椭圆上的点到直线距离有最小值解决.7.5人排一个5天的值日表，每天排一人值日，每人可以排多天或不排，但相邻两天不能排同一人，值日表排法的总数为（）A.120B.324C.720D.1280【分析】利用分步乘法计数原理计算即可.【详解】第一天可以排5个人中的任意一个，有5种排法；第二天可以排另外4个人中任意一个，有4种排法；第三天同上，有4种排法；第四天同上，有4种排法；第五天同上，有4种排法．根据分步乘法计数原理得所有的排法总数为544441280⨯⨯⨯⨯=．故选:D ．8.函数32()(1)f x x a x x b =+--+为R 上的奇函数，过点1,12P ⎛⎫- ⎪⎝⎭作曲线()y f x =的切线，可作切线条数为（）A.1B.2C.3D.不确定【答案】A 【解析】【分析】根据奇函数确定3()f x x x =-，求导得到导函数，设出切点，根据切线方程公式计算01x =-，计算切线得到答案.【详解】()3232()(1)(1)f x x a x x b f x x a x x b -=-+-+=-=--++--，故1a =，0b =，3()f x x x =-，2()31x f x '=-，设切点为()00,Mxy ，则2000012()311y f x x x '-=+=-，且30000()f x x x y -==，整理得到()()20001410x x x +-+=，解得01x =-，(1)2f '-=，故切线方程为22y x =+，故选：A二、多选题（共18分）9.公差为d 的等差数列{}n a ，其前n 项和为n S ，110S >，120S <，下列说法正确的有（）A.0d < B.70a > C.{}n S 中5S 最大D.49a a <【分析】利用等差数列性质结合给定条件可得60a >，670a a +<，再逐项分析判断作答.【详解】由()111116111102a a S a +==>，得60a >，又()()112126712602a a S a a +==+<，得，670a a +<，所以60a >，70a <，数列{}n a 是递减数列，其前6项为正，从第7项起均为负数，等差数列{}n a ，公差0d <，A 选项正确；70a <，B 选项错误；前6项和最大，C 选项错误；由40a >，90a <，有4949670a a a a a a -=+=+<，则49a a <，D 选项正确.故选：AD.10.已知函数()()322R x x a a f x x =-++∈的图像为曲线C ，下列说法正确的有（）A.R a ∀∈，()f x 都有两个极值点B.R a ∀∈，()f x 都有零点C.R a ∀∈，曲线C 都有对称中心D.R a ∃∈，使得曲线C 有对称轴【答案】ABC 【解析】【分析】根据函数极值的定义、零点的定义，结合函数的对称性的性质逐一判断即可.【详解】A ：()()()()3222341311x x x a f x x x x x f x '=-++⇒=-+=--，当1x >时，()()0,f x f x '>单调递增，当113x <<时，()()0,f x f x '<单调递减，当13x <时，()()0,f x f x '>单调递增，因此13x =是函数的极大值点，1x =是函数的极小值点，因此本选项正确；B ：当x →+∞时，()f x →+∞，当x →-∞时，()f x →-∞，而函数()f x 是连续不断的曲线，所以一定存在0R x ∈，使得()0f x =，因此本选项正确；C ：假设曲线C 的对称中心为(),b c ，则有()()()()()()32322222,f b x f b x c b x b x b x a b x b x b x a c ++-=⇒+-+++++---+-+=化简，得()232322b x c a b b b -=---+，因为x ∈R ，所以有322320320227b b c a b b b c a ⎧=⎪-=⎧⎪⇒⎨⎨---+=⎩⎪-=⎪⎩，因此给定a 一个实数，一定存在唯一的一个实数c 与之对应，因此假设成立，所以本选项说法正确；D ：由上可知当x →+∞时，()f x →+∞，当x →-∞时，()f x →-∞，所以该函数不可能是关于直线对称，因此本选项说法不正确，故选：ABC11.已知正方体1111ABCD A B C D -的棱长为1，下列四个结论中正确的是（）A.直线1B C 与直线1AD 所成的角为90B.直线1B C 与平面1ACD 所成角的余弦值为33C.1B D ⊥平面1ACD D.点1B 到平面1ACD 的距离为32【答案】ABC 【解析】【分析】如图建立空间直角坐标系，求出1B C 和1AD uuu r的坐标，由110AD B C ⋅= 可判断A ；证明10AC B D ⋅= ，110AD B D ⋅=，再由线面垂直的判定定理可判断C ；计算11cos ,B D B C 的值可得线面角的正弦值，再求出夹角的余弦值可判断B ；利用向量求出点A 到平面11D B C 的距离可判断D.【详解】如图以D 为原点，分别以1,,DA DC DD 所在的直线为,,x y z 轴建立空间直角坐标系，则()0,0,0D ，()1,0,0A ，()0,1,0C ，()10,0,1D ，()11,1,1B ，对于A ：()11,0,1B C =-- ，()11,0,1AD =-，因为()()()111100110B C AD ⋅=-⨯-+⨯+-⨯= ，所以11AD B C ⊥ ，即11B C AD ⊥，直线1B C 与直线1AD 所成的角为90 ，故选项A 正确；对于C ：因为()1,1,0AC =- ，()11,0,1AD =- ，()11,1,1B D =---，所以11100AC B D ⋅=-+= ，111010AD B D ⋅=+-= ，所以1AC B D ⊥ ，11AD B D ⊥uuur uuu r ，因为1AC AD A =I ，1,AC AD ⊂平面A 1，所以1B D ⊥平面1ACD ，故选项C 正确；对于B ：由选项C 知：1B D ⊥平面1ACD ，所以平面1ACD 的一个法向量()11,1,1B D =---，因为()11,0,1B C =-- ，所以111111cos ,B D B C B D B C B D B C⋅=== 即直线1B C 与平面1ACD 所成，所以直线1B C 与平面1ACD33=，故选项B 正确；对于D ：因为()11,0,1B C =-- ，平面1ACD 的一个法向量()11,1,1B D =---，所以点1B 到平面1ACD的距离为1113B D B C d B D⋅=== ，故选项D 不正确.故选：ABC.三、填空题（共15分）12.若抛物线22y px =-过点()1,2-，则该抛物线的焦点为________．【答案】()1,0-【解析】【分析】根据题意，代入求得2p =，结合抛物线的几何性质，即可求解.【详解】解：将()1,2-代入抛物线方程22y px =-，可得2p =，即24y x =-，所以抛物线24y x =-的焦点为()1,0-．故答案为：()1,0-.13.已知等比数列{}n a 的前n 项和为n S ，且满足122n n S λ+=+，则实数λ的值是_____．【答案】-2【解析】【分析】由已知推得1q ≠，继而结合等比数列的前n 项和的特点及已知即可求解.【详解】等比数列{}n a 中，由122n n S λ+=+可得122n n S λ=+，则11122a S λ==+，若公比1q =，则2211224,02S a λλλ=+==+∴=，则13323S a =≠，故1q ≠，则等比数列的前n 项和()1111111n nn a q a S qa q a a--=⋅--=-，（1q ≠），故令112λ=-，即2λ=-，故答案为：2-14.若e e e e ()cos 22x x x xf x x x ---+=+，则不等式(sin )(cos )0f x f x +>的解集是________．【答案】π3π|2π2π,44x k x k k ⎧⎫-<<+∈⎨⎬⎩⎭Z 【解析】【分析】根据奇偶性的定义和导数分析可知()f x 在[]1,1-内单调递增，且为奇函数，进而可得sin cos x x >-，利用辅助角公式结合正弦函数运算求解.【详解】取()f x 的定义域为[]1,1-，关于原点对称，且()()()e e e e e e e e ()cos cos sin 2222x x x x x x x xf x x x x x f x -----+-+-=-+-=--=-，所以()f x 为定义在[]1,1-上的奇函数，因为()e e e e e e e e ()cos sin sin cos e e cos 2222x x x x x x x xx x f x x x x x x ------+-+'=-++=+，若[]1,1x ∈-，则e 0,e cos 00,x x x ->>>，可得()()e e cos 0x xf x x -'=+>，可知()f x 在[]1,1-内单调递增，对于不等式(sin )(cos )0f x f x +>，则(sin )(cos )(cos )f x f x f x >-=-，且[][]sin 1,1,cos 1,1x x ∈--∈-，可得sin cos x x >-，整理得πsin cos 04x x x ⎛⎫+=+> ⎪⎝⎭，令π2π2ππ,4k x k k <+<+∈Z ，解得π3π2π2π,44k x k k -<<+∈Z ，所以不等式(sin )(cos )0f x f x +>的解集是π3π|2π2π,44x k x k k ⎧⎫-<<+∈⎨⎬⎩⎭Z .故答案为：π3π|2π2π,44x k x k k ⎧⎫-<<+∈⎨⎬⎩⎭Z .四、解答题（共77分）15.已知函数()ln 1f x x ax =++.（1）当1a =-时，求()f x 的最大值.（2）讨论函数()f x 的单调性.【答案】（1）0（2）答案见解析【解析】【分析】（1）利用导数求解函数最值即可.（2）含参讨论函数单调性即可.【小问1详解】当1a =-时，()ln 1f x x x =-+，由0x >，所以()111x f x x x-=-='，当01x <<时，()0f x '>，所以函数()f x 在()0,1上单调递增；当1x >时，()0f x '<，所以函数()f x 在()1,∞+上单调递减；故()()max 1ln1110f x f ==-+=；【小问2详解】定义域为(0,)+∞，()1f x a x'=+，当0a ≥时，()10f x a x+'=>，()f x 在(0,)+∞上递增；当a<0时，令()10f x a x +'=>，解得10,x a ⎛⎫∈- ⎪⎝⎭，令()10f x a x +'=<，解得1,x a ∞⎛⎫∈-+ ⎪⎝⎭.于是()f x 在10,a ⎛⎫-⎪⎝⎭上单调递增；在1,a ⎛⎫-+∞ ⎪⎝⎭上单调递减.16.如图，在底面为菱形的直四棱柱1111ABCD A B C D -中，12π,23BAD AA AB ∠===，,,E F G 分别是111,,BB CC DD 的中点．（1）求证：1A E GC ∥；（2）求平面1A EF 与平面ABCD 所成夹角的大小．【答案】（1）证明见解析（2）π6【解析】【分析】（1）建立空间直角坐标系，利用向量的坐标运算即可求解，（2）根据法向量的夹角即可求解.【小问1详解】取BC 中点H ，连接AH因为底面ABCD 为菱形，2π3BAD ∠=，所以AH AD ⊥以A 为原点，1,,AH AD AA 所在直线分别为x 轴，y 轴，z 轴，建立如图所示的空间直角坐标系，则()()()10,0,2,3,1,1,0,2,1A E G -，()()3,1,0,3,1,1C F ))13,1,1,3,1,1A E GC =--=-- 1A E GC∴ ∥1A E GC∴∥【小问2详解】设平面1A EF 的法向量为(),,n x y z =又()0,2,0EF = 所以100n A E n EF ⎧⋅=⎪⎨⋅=⎪⎩ 即3020y z y --==⎪⎩取1x =，则0,3y z ==(3n = ()10,0,2AA = 为平面ABCD 的法向量，设平面1A EF 与平面ABCD 的夹角为θ，则11233cos 222AA n AA nθ⋅===⨯ π6θ∴=∴平面1A EF 与平面ABCD 的夹角为π617.已知数列{}n a 的前n 项和n S 满足()1122n n S n +=-+．（1）求{}n a 的通项公式；（2）求数列12·1n n a n ++⎧⎫⎨⎬+⎩⎭的前n 项和n T ．【答案】（1）2nn a n =⨯（2）()2124n n T n +=+⨯-【解析】【分析】（1）由已知结合数列的和与项的递推关系即可求解；（2）先求数列121n n a n ++⎧⎫⎨⎬+⎩⎭的通项公式，然后利用错位相减求和即可求解．【小问1详解】当1n =时，112a S ==，当2n ≥时，由()1122n n S n +=-+，得()1222n n S n -=-+，则()()1112222n n n n n n a S S n n n +-=-=---=⨯，因为11212a ==⨯，所以2n n a n =⨯；【小问2详解】由（1）可知，()112·221n n n a n n +++=+⨯+，则()234132425222n n T n +=⨯+⨯+⨯+⋯++⨯，则()3452232425222n n T n +=⨯+⨯+⨯+⋯++⨯，则()234123222222n n n T n ++-=⨯+++⋯+-+⨯()()12812122212n n n -+-=+-+⨯-()22122822n n n ++=+--+⨯()2412n n +=-+⨯，所以()2124n n T n +=+⨯-．18.在平面直角坐标系xOy 中，已知椭圆2222:1x y C a b +=（0a b >>过点(2,1)P，且离心率2e =．（1）求椭圆C 的方程；（2）直线l 的斜率为12，直线l 与椭圆C 交于A 、B 两点，求PAB 的面积的最大值．【答案】（1）22182x y +=（2）2【解析】【分析】（1）利用222c e a =，可得22234a b a -=，再将点P 坐标代入方程，解方程组求得,a b 从而可得椭圆的方程；（2）设直线l 的方程为1,2y x m =+，代入椭圆方程中整理得222240x mx m ++-=，借助根的判别式可得||2m <，结合根与系数的关系可得AB ==直线的距离公式可求出点P 到直线的距离d ，再利用三角形面积公式1||2PAB S d AB =⋅ 和基本不等式进行求解，即可解决问题.【小问1详解】因为22222234c a b e a a -===，所以224a b =，①因为椭圆C 过点(2,1)P ，所以22411a b +=，②由①②解得228,2a b ==，所以椭圆的方程为22182x y +=．【小问2详解】设直线l 的方程为()()11221,,,,2y x m A x y B x y =+，联立2212182y x m x y ⎧=+⎪⎪⎨⎪+=⎪⎩，得222240x mx m ++-=，所以212122,24x x m x x m +=-=-，又直线l 与椭圆相交，所以2248160m m =-+> ，解得||2m <，则AB ==P 到直线l的距离d ==，所以221142222PAB m m S d AB +-=⋅==≤= ，当且仅当22m =，即m =时，PAB 的面积取得最大值为2．19.已知函数()2e e x x f x a x =-+，其中0a >.（1）当1a =时，求函数()f x 在0x =处的切线方程；（2）讨论函数()f x 的极值点的个数；（3）若对任意的0a >，关于x 的方程()f x m =仅有一个实数根，求实数m 的取值范围.【答案】（1）20x y -=（2）见解析（3）3ln 2,2⎡⎫-++∞⎪⎢⎣⎭【解析】【分析】（1）求导得斜率，再利用点斜式求直线方程；（2）求导，讨论判别式与0的关系得单调性即可求解极值点个数；（3）构造新函数()2ee x x g x a x m =-+-，判单调性，得到()()120,ln 2,ln 2,x x ∞∈∈+，结合()10g x <或()20g x >即可求解.【小问1详解】当1a =时，()()22e e ,2e e 1x x x x f x x f x '=-+=-+，()02f '=，()00f =，所以函数()f x 在0x =处的切线方程为()020y x -=-，即20x y -=.【小问2详解】()22e e 1x x f x a '=-+，令()0,e x f x t ='=，得2210at t -+=，则18a ∆=-.当18a ≥时，0∆≤，此时()0f x '≥，故函数()f x 在(),∞∞-+上单调递增，没有极值点；当108a <<时，0∆>，令()0f x '=，则1e 4x a =，则1211ln ln 44x x a a-+==，则当()1,x x ∞∈-时，()0f x '>，当()12,x x x ∈时，()0f x '<，当()2,x x ∞∈+时，()0f x '>，则()f x 在()()12,,,x x ∞∞-+单调递增，在()12,x x 单调递减，此时函数()f x 有两个极值点.综上所述，当18a ≥时，函数()f x 没有极值点；当108a <<时，函数()f x 有两个极值点.【小问3详解】依题意，2e e x x a x m -+=，记()2e e x x g x a x m =-+-，()()g x f x '='.（i ）由（2）知当18a ≥时，()0g x '≥，则函数()g x 在(),∞∞-+上单调递增；可知当x →-∞时，()g x ∞→-，当x →+∞时，()g x ∞→+，故当18a ≥时，函数()g x 恰有一个零点，方程()f x m =仅有一个实数根，此时R m ∈.（ii ）当108a <<时，()g x 在()1,x ∞-上单调递增，在()12,x x 上单调递减，在()2,x ∞+单调递增，()()112222122e e 12e e 10x x x x g x a g x a ''=-+==-+=，则121222e 1e 12e 2ex x x x a --==，所以()()1112111e 1ee 22x x x g x g x a x m x m ==-+-=-+--极大值，()()2222222e 1e e 22x x x g x g x a x m x m ==-+-=-+--极小值，因为当(),x g x ∞∞→-→-，当(),x g x ∞∞→+→+，故只需()10g x <或()20g x >，令()e 122x h x x =-+-，则()e 12xh x '=-+，故当(),ln 2x ∞∈-时，()0h x '>，当()ln 2,x ∞∈+时，()0h x '<，则()h x 在(),ln 2∞-单调递增，在()ln 2,∞+单调递减；又121ln ln ln4x x a -===又108a <<，故()0,1，则()()120,ln 2,ln 2,x x ∞∈∈+，所以()()12331,ln 2,,ln 222h x h x ∞⎛⎫⎛⎫∈--+∈--+ ⎪ ⎪⎝⎭⎝⎭，故3ln 22m ≥-+.综上所述，实数m 的取值范围为3ln 2,2∞⎡⎫-++⎪⎢⎣⎭.【点睛】关键点点睛：本题考查函数极值点及零点个数问题，解决问题关键是利用第二问单调性解决第三问零点问题，并利用构造函数法求函数值域。
江西师范大学附属中学2023-2024学年高二下学期3月月考数学试题参考答案
第一次月考一、单选题1. 等差数列{}n a 中，1239a a a ++=，4516a a +=，则6a =（ ） A. 9 B. 10 C. 11 D. 12【答案】C【解析】因为1231339a a a a d ++=+=，4512716+=+=a a a d ， 所以可解得1a 1,d 2，所以61511011a a d =+=+=，故选：C2．在正项等比数列{}n a 中，n S 为其前n 项和，若1010S =，2030S =，则30S 的值为（ ） A ．50 B ．70 C ．90 D ．110【答案】B【解析】由等比数列的片段和性质得10S ，1200S S −，3020S S −成等比数列 所以()()22010103020S S S S S −=− 所以()()23030101030S −=−， 解得3070S =. 故选：B.3．用数学归纳法证明“1111112331n n n n ++++>++++”时，假设n k =时命题成立，则当1n k =+时，左端增加的项为（ ） A ．134k + B ．11341k k −++ C ．111323334k k k +++++ D ．11232343(1)k k k +−+++ 131k +++111+31323k k k ++++111+31331111233123k k k k k k k ⎫++−⎪+++⎭⎫+++⎪++++⎭故选：D4．已知数列{}n a 为等差数列，首项10a >，若101210131a a <−，则使得0n S >的n 的最大值为（ ） A ．2022 B ．2023C ．2024D ．20255. 已知数列{}n a 为正项递增等比数列，123212a a a ++=，12311176a a a ++=，则该等比数列的公比q =（ ）A. 2B. 3C. 4D. 5【答案】A【解析】由题意10,1a q >>， 由123212a a a ++=，1312321231322111716a a a a a a a a a a a a +++++==+=， 得2221726a =，所以23a =（23a =−舍去），所以132********q a a q =−=++=， 整理得22520q q −+=，解得2q （12q =舍去）， 所以2q.故选：A.6．近几年，我国在电动汽车领域有了长足的发展，电动汽车的核心技术是动力总成，而动力总成的核心技术是电机和控制器，我国永磁电机的技术已处于国际领先水平.某公司计划今年年初用196万元引进一条永磁电机生产线，第一年需要安装､人工等费用24万元，从第二年起，包括人工､维修等费用每年所需费用比上一年增加8万元，该生产线每年年产值保持在100万元.则引进该生产线后总盈利的最大值为（ ） A ．204万元 B ．220万元C ．304万元D ．320万元7. 已知数列{}n a 的前n 项和为n S ，且满足12cos 3n n n a a a ++++=，11a =，则2023S =（ ） A. 0 B.12C. lD. 32【答案】C【解析】解：()()()20231234567202120222023S a a a a a a a a a a =++++++++++2π5π1coscos 33=++++2018π2021πcoscos33+ 2π5π1337cos cos 133⎛⎫=+⨯+= ⎪⎝⎭．故选:C ．8．已知数列{}n a 的前n 项和为n S ，数列{}n b 的前n 项和为n T ，且111,1,1n n n n a S n a b a +=+==+，则使得n T M <恒成立的实数M 的最小值为（ ）A ．1B ．32C ．76D ．2【答案】C【解析】数列{}n a 中，11a =，1n n a S n +=+，当2n ≥时，11n n a S n −=+−，两式相减得11n n n a a a +−=+，二、多选题9．在等比数列{}n a 中，11a =，427a =，则（ ） A ．{}1n n a a +的公比为9 B ．{}31log n a +的前20项和为210C ．{}n a 的前20项积为2003D ．()111()231nn k k k a a −+=+=−∑2020++=，n a 的前201919033⨯⨯=，因为()1313n n a −++}1n n a a ++的前)13213n −=−10．下列命题中正确的是（ ）A ．已知随机变量16,3XB ⎛⎫⎪⎝⎭，则()3212D X += B ．若随机事件A ，B 满足：()12P A =，()23P B =，()56P A B ⋃=，则事件A 与B 相互独立C ．若事件A 与B 相互独立，且()()01P A P B <<，则()()P A B P A =D ．若残差平方和越大，则回归模型对一组数据()11,x y ，()22,x y ，…，(),n n x y 的拟合效果越好11. 已知数列1C ：0，2，0，2，0，现在对该数列进行一种变换，规则f ：每个0都变为“2，0，2”，每个2都变为“0，2，0”，得到一个新数列，记数列()1k k C f C +=，1,2,3,k =，且n C 的所有项的和为n S ，则以下判断正确的是（ ）A. n C 的项数为153n −⋅B. 4136S =C. 5C 中0的个数为203D. 1531n n S −=⋅−【答案】ABC【解析】设数列{}n C 的项数为一个数列{}n a ，因为1C 中有5项，即15a =， 根据题意：在f 作用下，每个0都变为“2,0,2”，每个2都变为“0,2,0”， 所以有()13Nn n a a n *+=∈，由此可知数列{}n a 为首相15a =，公比3q =的等比数列， 所以n C 的项数为153n n a −=⋅，故A 正确；根据变换规则，若数列的各项中，2与0的个数相同， 则与之相邻的下一个数列中2与0的个数也相同；若2比0多n 个，则与之相邻的下一个数列中2比0的个数少n 个， 若2比0少n 个，则与之相邻的下一个数列中2比0的个数多n 个，因为1C 中有5项，其中2个2，3个0，2比0少1个， 所以2C 的15项中，2比0的个数多1个，以此类推，若n 为奇数，则数列的各项中2比0少1个， 若n 为偶数，则数列的各项中2比0多1个，4C 中4n =，项数为353135⋅=个，n 为偶数，所以2的个数为1351682+=， 所以4682136S =⨯=，所以B 正确；5C 中共有453405⋅=项，其中5n =为奇数，所以数列中有40512032+=个0，所以C 正确； D 选项，n S 的值与n 的奇偶有关()()11531531n n n n S n −−⎧⋅−⎪=⎨⋅+⎪⎩为奇数为偶数，所以D 错误. 故选：ABC.【点睛】方法点睛：学生在理解相关新概念、新法则 (公式)之后，运用学过的知识，结合已掌握的技能，通过推理、运算等解决问题.在新环境下研究“旧”性质.主要是将新性质应用在“旧”性质上，创造性地证明更新的性质，落脚点仍然是数列求通项或求和. 三、填空题12．已知等差数列{}n a 中，24a =，616a =，若在数列{}n a 每相邻两项之间插入三个数，使得新数列也是一个等差数列，则新数列的第41项为___． 【答案】31【解析】设等差数列{}n a 的公差为d ，则62123624a a d −===−， 在数列{}n a 每相邻两项之间插入三个数，则新的等差数列{}n b 的公差为344d =， 故新数列的首项为431−=，故通项公式为()33111444n b n n =+−=+， 故4131413144b =⨯+=. 故答案为：3113．箱子中装有5个大小相同的小球，其中3个红球、2个白球.从中随机抽出2个球，在已知抽到红球的条件下，则2个球都是红球的概率为 ．14．已知n S 是各项均为正实数的数列{}n a 的前n 项和，221111,60n n n n a a a a a ++=−−=，若*,2270n n n n S a ma ∀∈−+≥N ，则实数m 的取值范围是 ．(2)记n n n b a c ⋅=，n T 为n c 的前n 项和，求n T .【解】（1）解：由已知可得32112127a b a b d q d q =++=++=+①， ()()22231122212a b a d b q d q −=+−=+−=②，联立①②，得()()26320q q q q +−=+−=，解得3q =−或2q，2q，代入①式可得在曲线()y f x =上(1)3f '⇒−=，21a a ++−(1n ⋅++=，)1+；()(1nn −−⋅，)()(1212233445212222k k k k k ⎡+++⋅⋅−⋅+⋅−⋅++−⋅−⋅+⎣[]12224222k +⋅−⋅−⋅−−⋅()()222224221k k k k k k k k =+−+++=+−+=，即T 2n =n 2.18．已知数列{}n a 的前 n 项和为n S ，()*∈−=N n S a n n 2.（1）求数列{}n a 的通项公式； （2）是否存在实数λ ,使数列⎭⎬⎫⎩⎨⎧++n n n S 2λλ为等差数列？若存在， 求出λ的值； 若不存在，请说明理由； （3）已知数列{}n b ，()()1121++=+−n n nn a a b ，其前 n 项和为n T ，求使得442m T m n<<−对所有*N n ∈都成立的自然数m 的值.的一动点，PAB 面积的最大值为C 交于,D 两点，记ODE 的面积为,DN EN 的斜率分别为12,k k .联立221431x y x my ⎧+=⎪⎨⎪=+⎩，消去x 可得()234m y +所以()()222Δ3636341441m m m =++=+且12122269,3434m y y y y m m +=−=−++， ODES=1,t t =≥2631t t =+试卷第11页，共11页。
湖南省永州市2023-2024学年高二下学期3月月考化学试题含答案
2024年上期永州市高二第一次月考卷化学2024.3出题人：（答案在最后）温馨提示：1．本试卷满分100分，考试时间75分钟。
2．答题前，考生务必将自己的姓名、准考证号填写在答题卡上，并按规定贴好条形码。
3．请将全部答案填写在答题卡上。
可能用到的相对原子质量：H-1C-12O-16S-32Cl-35.5Zn-65Ga-70As-75一、单选题（共42分，每小题只有一个选项符合题意，每小题3分）1．化学与人类生活、社会可持续发展密切相关，下列说法错误的是A．75%酒精灭活细菌，利用了乙醇使蛋白质变性的功能B．长沙马王堆出土的素纱禪衣(丝绸)，其主要成分是纤维素C．神舟飞船的轨道舱壳体采用的铝合金具有密度小、抗腐蚀能力强等特性D．华为Mate60高清镜头中使用的COC/COP(环状聚烯烃)是高分子化合物，也是混合物2．下列有机物的命名正确的是A．二甲苯B．4-甲基-2-乙基-1-戊烯C．3-甲基-3-乙基-1-丁炔D．2，4，4-三甲基戊烷3．下列化学用语错误的是A.CH3+的电子式：B.反-2-丁烯的结构简式：C.基态Fe2+的价电子轨道表示式：D.空间填充模型，可表示CH4分子，也可表示CCl4分子4．下列各项括号内为杂质，后面为除杂质操作，其中正确的是A．甲醛(乙酸)，蒸馏B．乙酸乙酯(乙酸)，分液C．苯(四氯化碳)，加水振荡、过滤D．乙烷(乙烯)，通过足量的酸性KMnO4溶液，洗气5．设N A为阿伏加德罗常数的值。
下列说法错误的是A．1mol乙烯中含有的σ键数目为5N AB．23g CH3CH2OH中sp3杂化的原子数为1.5N AC.标准状况下，22.4L CHCl3中含有氯原子数目为3N AD．48g正丁烷和10g异丁烷的混合物中共价键数目为13N A6．分子式为C5H10O2的有机物属于酯的同分异构体的种数为A．6种B．7种C．8种D．9种7．某芳香烃的相对分子质量为106，分子中有3中化学环境不同的氢原子，该芳香烃是A. B. C. D.8．科学家以可再生碳资源木质素为原料合成姜油酮的过程如图所示，下列说法正确的是A．香兰素的分子式为C7H8O3B．香兰素与姜油酮互为同系物C．脱氢姜酮中所有碳原子可能共平面D．姜油酮中的官能团有4种9．在探索苯分子结构的过程中，人们写出了符合分子式“C6H6”的多种可能结构（如图所示），下列说法正确的是A．c、e的一氯代物均有2种B．五种物质均能与氢气发生加成反应C．a、b、c、e能使溴的四氯化碳溶液褪色D．五种物质中，只有a、e分子的所有原子处于同一平面10．从海带中提取精品甘露醇(C 6H 14O 6)的流程如图所示。
四川省成都市第七中学2022-2023学年高二下学期3月月考英语试卷及答案
成都七中高2024届高二下期3月阶段性考试试卷英语考试时间：120分钟试题满分：150分注意事项：1. 本试卷分第I卷（选择题）和第II卷（非选择题）两部分。
2. 考生务必将自己的姓名、考生号填写在答题卡上。
3. 作答时，将答案写在答题卡上。
写在本试卷上无效。
4. 考试结束后．将答题卡交回。
第I卷(100分)第一部分听力（共两节，满分30分）第一节（共5个小题；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，从每题所给的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.What happened to the man?A.He lost his horse.B.He was bitten by a dog.C.He was bitten by a horse.2.What does the woman mean?A.John doesn’t put his ideas into practice.B.John doesn’t like dreaming.C.John has too few dreams.3.What does the man advise the woman to do?A.Go to the biology department.B.Teach herself the courses.C.Wait for his help.4.What will the man probably do?A.See a doctor.B. Attend a meeting.C. Visit the woman.5.Why should the man apologize to Sonia?A. He did her hair badly.B. He didn’t notice her new hairstyle.C. He made fun of her new hairstyle.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。
重庆市重点中学2023-2024学年高二下学期3月月考试题 语文含答案
重庆市重点中学高2025届高二下期3月联考语文试题（答案在最后）注意事项：1.本试卷满分为150分，考试时间为150分钟。
2.答卷前，考生务必将自己的姓名、班级、准考证号准确填写在答题卡上。
3.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。
4.考试结束后，将答题卡交回。
一、现代文阅读（35分）（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。
①“自得”字面意思即是自己得到，论学问时就是指自己领会知识的意思。
孟子说：“君子深造之以道，欲其自得之也。
自得之则居之安，居之安则资之深，资之深则取之左右逢其原，故君子欲其自得之也。
”孟子认为学问致道的关键在于“自得”，只有“自得”的知识才会取用自如、受益匪浅。
“自得”作为普遍意义的论学观念，后来也被有些人用来讨论诗学。
对于诗人而言，“自得”是他们在生活中与前人经验相融汇的顿然晓悟。
如嵇康“目送归鸿，手挥五弦。
俯仰自得，游心太玄”是领会老庄游心自然的意趣，陶渊明“屡空既有人，春兴岂自免”是深契于箪食瓢饮而不改其乐的颜回。
对于读者而言，当我们面对一首诗时，随文解义是通文字者的自然反应。
如朱熹曾举林逋“疏影横斜水清浅，暗香浮动月黄昏”，欧阳修、苏轼等名公大家都道此句好，但它到底好在哪里呢?这一句的文义并不难理解。
然而它的“言外之意”却是要读者自身去体会。
朱熹认为，在理解诗语文义的基础上能情不自已，获得一种会心的快乐，如此便是“自得”。
“自得”并非某种天启式的领悟，而与读者个人的知识积累与精神体验密切相关。
②先说“自得”与知识的关系。
林逋的论梅诗句“疏影横斜水清浅，暗香浮动月黄昏”，苏轼曾与友人议论其妙处。
有人提出林逋此句用以描述桃、李、杏同样没有问题，苏轼幽默地回答：“可则可，但恐杏与桃、李不敢承当。
”引得在座众人大笑。
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高2020届高二下期第一次月考地理试题
一．选择题
读“两半岛图”，回答1-2题
1.A地位于B地的
A.西北方
B.西南方
C.东北方
D.东南方
2.两图的比例尺
A.甲图大于乙图
B.甲图小于乙图
C.甲图等于乙图
D.不能确定大小
下图为某区域等高线地形图，已知乙村在甲村的西北方向。
读图回答下列3-5题。
3.图示河流的干流流向是
A.自西北流向东南
B.自西南流向东北
C.自东北流向西南
D.自西向东流
4．关于图中规划公路的叙述，正确的是
A.高差可能为280米
B.长度约为6千米
C.穿越河谷地带
D.穿越鞍部地区
5．沿上图中M—N剖面线绘制的地形剖面图是
读等高线图(图中等高距为100米，直线处有一条小河)，回答6-7题。
6.下列关于图中河流的叙述，不正确的是
A.该河在A处可见瀑布景观
B.该河自西北流向东南
C.该河夏季水量大、冬季水量小
D.该河各段的流速有较大变化
7.图中B处有一条闭合的等高线，下列有关叙述正确的是
A.B处的海拔一定大于500米
B.站在B处一定能见到A处的瀑布
C.B处的地形可能是个小洼地
D.B处的降水量一定是图中最丰富的
下图中M、N两地位于同一经线上，相距600千米，N地年太阳辐射总量少于M地。
读图完成8-9题。
8．一年当中，当N地降水明显多于M地时( )
A．M 地温和多雨B．M 地白昼比N 地长
C．N 地受副高控制 D．N 地受西北风影响
9．关于图示区域的说法，正确的是( )
A．图中山脉的形成与太平洋板块有关
B．图中河流径流季节变化小
C．图中M、N 附近自然带相同
D．图中陆地上等温线发生弯曲，主要是受地形影响
读下面某区域示意图,回答10-11题。
10.2015年2月该区域部分地区遭受了罕见的雪灾，其中受灾较严重的城市可能有( ) A．①④B．①②C．②③D．③④
11.关于乙岛屿的叙述，正确的是( )
A．虚线框内东南部地势较平坦，西北部地势较陡峻
B．该岛因火山喷发而形成
C．自然植被为常绿阔叶林
D．沿岸海可能有不冻港
二．综合题
36.下图为我国东南沿海某地等高线图,图示区域最近准备在P处拦水筑坝,同时修建一条连接城镇的公路。
读图完成下列问题。
(18分)
(1) 分析水库大坝建在P处的原因。
(4分)
(2)试从气候、河流水文特征两方面分析该区域建水库的原因。
（6分）
（3）根据图上信息，分析图中公路选线的主要区位条件（4分）
（4）某同学通过对该地区考察发现B地植被长势比A地好，试分析产生此现象的原因（4分）
37.（16分）阅读非洲肯尼亚相关图文材料，据此回答下列小题。
肯尼亚境内多高原，平均海拔1500米，东非大裂谷东支纵切高原南北，大裂谷谷底分布着深浅不等的湖泊，并屹立着许多火山。
农业是肯尼亚的支柱产业之一，近年来鲜花产业发展较快，出口鲜花约占欧盟市场的1／3。
肯尼亚航空业较为发达，与30多个欧美国家和地区通航。
（1）简述肯尼亚的地理位置特征。
（6分）
(2)说出肯尼亚境内的主要气候类型及其成因。
（6分）
(3)分析肯尼亚发展鲜花产业有利的区位条件（4分）
(4)用板块构造学说解释东非大裂谷谷底多湖泊、火山的成因。
（4分）
38.（22分）阅读图文资料，回答下列小题。
下图中左图为世界某半岛及附近地区等高线图，R地盛产地中海栓皮栎，生长缓慢，环境适应性强，其外层树皮是理想的软木塞原料。
栓皮栎栎炭火力强而耐久，是良好的薪炭材；皮质坚硬不易燃烧；是防火林的优良树种。
我国云南也有少量栓皮栎分布，欲大量引种地中海栓皮栎。
右图为波尔图和巴塞罗那的年降水量资料（单位：毫米）。
（1）根据材料判断R地地形类型并说明理由。
（6分）
(2)比较波尔图和巴塞罗那的降水差异并分析原因。
（6分）
(3)评价大量引种地中海栓皮栎对云南省的生态影响。
（6分）
高2020届高二下期第一次月考地理参考答案
1-5ABCDC 6-11CCDDBD
36.（1）位于峡谷，建坝工程量小；上游小盆地利于蓄水；集水区域较广（4分）
(2)地处季风气候区，降水量大；降水季节、年际变化大；河流流量大；流量变化大。
（6分）
（3）公路经过地形平坦，建设工程量小，难度小，投资成本小
公路经过多个城镇，方便出行，促进经济发展
（4）B地与A地比较：B地位于阳坡，A地位于阴坡，B地光照、热量条件好；
B地处于夏季风的迎风坡，降水丰富
37. （1）位置：肯尼亚位于非洲东部（2分）；赤道横贯中部，全境位于热带（2分）；东南濒临印度洋
（2分）；北与埃塞俄比亚、东北与索马里交界，西与乌干达、西南与坦桑尼亚接壤（1分）；意义：地
处赤道附近，水热充足，有利于农作物生长；与多国接壤利于与邻国的交往和合作；东南临海，有利于
发展海洋事业（任答2点得2分）。
（2）气候类型：热带稀树草原气候（2分）；成因：境内多高原，地势高（2分）；对流不旺盛（2分）。
（3）肯尼亚地处赤道的东非高原上，全年气候温和，降水适中，有利于种植鲜花肯尼亚航空业发达，有多条航线连接欧美，利于鲜花出口
肯尼亚为发展中国家，劳动力土地廉价，种植鲜花成本低
国家政策支持，利于壮大鲜花产业
（4）东非大裂谷为板块的张裂地区，地壳不稳定，多地层陷落湖及火山（4分）。
38. （1）高原（2分）；理由：R地等高线稀疏，（2分）地形起伏小；海拔较高，东部沿海等高线密集，1000米等高线呈闭合曲线（2分）。
（2）差异：波尔图年降水量较巴塞罗那多，雨季较巴塞罗那长（2分）。
原因：波尔图位于迎风坡，受西风影响明显，且时间长，降水丰富且雨季较长（或巴塞罗那位于西风背
风坡，降水较少）（4分）。
（3）栎炭火力强而耐久，有利于缓解能源压力，减轻对森林的过度砍伐；皮质坚硬不易燃烧，有利于防治森林火灾；云南省属于喀斯特地貌区，土壤贫瘠，栓皮栎环境适应性强，存活率高；外来物种入侵，破坏原有的生态环境。
（任答3点得6分）。