高二下学期期末测试
上海市闵行区2023-2024学年高二下学期6月期末英语试题

2023学年第二学期高二年级学业质量调研英语试卷(考试时间105分钟,试卷满分115分)考生注意:1.本次考试设试卷和答题纸两部分。
所有答题必须涂(选择题)或写(非选择题)在答题纸上,做在试卷上一律不得分。
2.答题前,务必在答题纸规定的地方张贴条形码并填写考生信息。
II.Grammar and VocabularySection ADirections: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 last seven years, most states have banned texting by drivers. Public service campaigns have also tried a wide range of methods to persuade people to put down their phones (21) ______ they are behind the wheel.Yet the problem, by just about any measure, appears to be getting worse. Americans are still texting while driving, as well as using social networks and taking photos. Road accidents, which (22) ______ (fall) for years, are now rising sharply.That is partly (23) ______ people are driving more, but Mark Rosekind, the chief of the National Highway Traffic Safety Administration, said inattentive driving was “only increasing, unfortunately.”“Big change requires big ideas,” he said in a speech last month, (24) ______ (refer) broadly to the need to improve road safety. So (25) ______ (change) a distinctly modern behavior, lawmakers and public health experts are reaching back to an old approach: They want to treat (26) ______ (distract) driving like driving after consuming alcohol.An idea (27) ______ lawmakers in New York is to give police officers a new device called the Textalyzer. It would work like this: An officer arriving at the scene of a crash could ask for the phones of the drivers and use the Textalyzer to check in the operating system for recent activity. The technology could determine (28) ______ a driver had just texted, emailed or done anything else not allowed under New York’s hands-free driving laws.“We need something on the books (29) ______ can change people’s behavior,” said Félix W. Ortiz, who pushed for the state’s ban on hand-held devices by drivers. “If the Textalyzer bill becomes law,” he said, “people are going to be (30) ______ (afraid) to put their hands on the cell phone.”Section BDirections: 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.A.focused B.labels C.contributes D.accident E.duly F.present G.rarely H.tricks I.available J.equipped K.attemptThis Is Why Most Grocery Stores Lack WindowsWhen you’re grocery shopping, have you noticed that supermarkets often lack windows, and if they have them, they’re only at the front of the store? You may 31 pay attention to the architectural features of the building because you’re there to shop for food, not admire the layout.But that’s the point. If grocery stores had windows, would it be easier to ignore your main shopping task? There are all sorts of supermarket 32 grocery stores have to keep you shopping longer. It’s no 33 that grocery stores often lack windows and there are a few reasons behind this trend.One of them is a (n) 34 to keep people inside longer. Stores want to create a separate environmentwithin their store where the outside world doesn’t exist. In spite of the rain or sunshine, your attention remains 35 . The technique also prevents shoppers from noticing it’s getting dark out. It really 36 to an immersive (沉浸式的) shopping experience, for better or worse.Keeping daylight out of stores can help preserve the products, as some fresh produce can go bad faster in direct sunlight. Too much sun exposure can even cause packaging 37 to fade. Having windows in stores would also reduce the space 38 to display products. Not to mention, 39 with strong structural supports, outside walls can hold the heavier items on the shelves of those walls.Due to the high cost of constructing windows and storefronts, along with the potential security risks they probably 40 , retailers (零售商) choose to minimize the number of entry points into their space in order to cut expenses and improve safety.III.Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Social media can be a convenient way to extend your network, staying in touch with your grandma or sharing photos of your new puppy. 41 , social media can be terrible for your health. It’s our 42 of it that’s out of control.British evolutionary psychologist Robin Dunbar developed the idea that 150 people is the maximum number of meaningful connections anyone can have. You may have 800 friends on Wechat, but you’re not 43 them in person. Your QQ connections may be vast, but how many of them do you have 44 interactions with?If you’re using social media to feel more connected, a recent study published in the American Journal of Health Promotion suggests that it’s not 45 , and it also brought about unfavorable connections and even depression. Positive interactions on social media don’t help people feel 46 . Negative interactions, on the other hand, bring more feelings of sadness. The same goes for 47 , which social media encourages. Another study, published by the American Psychological Association, shows that comparing yourself to others through social media also produces 48 effects, leading to symptoms of depression.Reaching for your cell phone as a mental break is also a (n) 49 idea. Research by Rutgers University compared participants in the process of completing a task who took a break with their cell phones, with paper and pencils, and who took no break at all. Those who used their cell phones during their break solved 22% fewer problems and took 19% longer to complete their tasks than those under the other two 50 .So how should you 51 your social media usage?First, know your time and 52 of use. Use the tracking function on your device to find out how much you’re using it and what you’re using it for. After this, get away from your device unless you have to use it. Tough as it might be, turn off and tune into the world around you.Then, be in charge of your 53 , rather than let it be in charge of you. Your phone’s rings or vibrations (震动) don’t mean you must respond to them. Remind yourself that you’re in charge, not your device or the people on the other end of it who’ve just contacted you.Finally, consider using your device as a (n) 54 builder. Use your apps to find your friends and make your dinner reservation so you can catch up face-to-face in your favorite restaurant, where you’ll keep your phone out of sight.When you’re in control of your 55 , social media becomes a tool to enrich your life but not a distraction that makes you feel miserable.()41.A.Moreover B.Therefore C.Nevertheless D.Likewise()42.A.use B.reform C.ignorance D.range()43.A.commenting on B.cooperating with C.competing with D.connecting with()44.A.virtual B.meaningful C.constant D.complex()45.A.automatic B.productive C.working D.appealing()46.A.more confident B.more satisfied C.calmer D.happier()47.A.comparison B.sharing C.connection D.variety()48.A.instructive B.restrictive C.negative D.active()49.A.widespread B.bad C.creative D.undervalued()50.A.conditions B.assumptions C.influences D.developments()51.A.reduce B.analyze C.manage D.track()52.A.strategies B.boundaries C.efficiency D.purposes()53.A.device B.schedule C.decision D.emotion()54.A.knowledge B.character C.relationship D.confidence()55.A.digital consumption B.personal interaction C.daily routine D.working habit Section BDirections:Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have read.(A)As midnight approaches in Svalbard, a young polar bear climbs onto an iceberg (冰山) and carves himself a temporary bed before falling asleep. This peaceful moment, capturing the essence of Earth’s delicacy, was taken by Nima Sarikhani, who won Wildlife Photographer of the Year People’s Choice Award, which is decided by public vote.Sarikhani, from the UK, spent three days searching Norway’s Svalbard islands in the hope of catching sight of these symbolic Arctic (北极的) animals. He was finally rewarded with the sight of both an older and younger male shortly after his ship changed course. Sarikhani seized the chance to take a photo of the younger bear sleeping on the iceberg—a scene that not only awakens the bond between an animal and its habitat, but also the need to act on global challenges like climate change and habitat loss.These days, the sight of a lone polar bear on melting ice is a familiar symbol of the impact of climate change. But that is for a good reason: global warming is having a particularly rapid effect on the Arctic, which is heating up at a rate three times as high as that of the average around the world. Among those hardest hit are these bears, which are increasingly threatened by the reduction in sea ice cover that they rely on to hunt and raise young.Though his photo is intended to arouse emotion in those who see it, Sarikhani is optimistic that with the right actions, it isn’t too late for polar bears. The image will be displayed in an exhibition at the Natural History Museum, London, along with the competition’s four runners-up, until 30 June.()56.Sarikhani’s journey in Svalbard can be described as ______.A.purposeful B.leisurely C.profit-driven D.research-led()57.It can be inferred from paragraph 3 that ______.A.polar bears are likely to die outB.the warming of the Arctic just beginsC.sea ice cover is vital for polar bears’ survivalD.polar bears are adapting well to climate change()58.According to the last paragraph, what’s Sarikhani’s hope for the impact of his photo?A.To arouse people’s sadness for the scene.B.To motivate people to protect polar bears.C.To attract more visitors to the Arctic.D.To boost the popularity of the exhibition.()59.Which of the following might be the best title of the passage?A.Svalbard’s quietness: Polar bears’ comfortable zoneB.Arctic adventure: Search for a dramatic landscapeC.Bearing witness: Polar bears’ shelter on melting iceD.Ice and isolation: A faraway natural wonder(B)Four Books Worth ReadingEntangled Life (Illustrated) by Merlin SheldrakeFollowing 2020’s real hit Entangled Life, biologist Sheldrake returns in time with an impressive illustrated edition. At over 240 pages, his passion and knowledge leaps off every spread. From the microscopic to the splendid, the colour images create an entertaining and charming experience.Anna Atkins: Cyanotypes by Peter WaltherOften copied, seldom improved on, the elegant prints of Anna Atkins are timeless. Atkins uses light-sensitive iron salt solution and exposure to sunlight to create beautiful Cyanotypes, pictures in a specific dark blue color. Walther’s book is comprehensive and perfect for Atkins fans or anyone keen to learn.The Earth in Our Hands by Thomas PesquetPesquet’s breathtaking collection of photographs captured from the International Space Station (ISS) follows in the footsteps of astronaut Don Pettit’s Spaceborne. Pesquet took more than 245,000 images over two missions, with his book including 200 of these. It is the closest most of us will get to being on the ISS—a deeply engaging read.Looking at Trees by Sophie HowarthThis absorbing book features 26 of the world’s leading photographers. From a photo of Hollywood juniper (杜松) in California to a dreamlike image of Halfway Gardens in South Africa, Howarth asks us to value trees at a time of environmental challenge. It’s a book you will return to.()60.The four books can be classified into ______.A.art magazines B.photo books C.historical novels D.instruction books()61.Which book is a remade version of a popular book?A.Entangled Life (Illustrated). B.Anna Atkins: Cyanotypes.C.The Earth in Our Hands. D.Looking at Trees.()62.What can be learnt from the introduction of the four books?A.Looking at Trees mainly focuses on valuable trees.B.Don Pettit has joined space missions with Pesquet.C.Cyanotypes of Anna Atkins enjoy a high reputation.D.All the books introduced care about the environment.(C)Conventionally, being overweight is bad for health. This may not always be the case. The latest evidence comes from a study in which people classed as overweight, but not extremely fat, had a lower death rate within a certain period than people with a supposedly ideal weight. This suggests that the threshold (阈值) for classifying individuals as overweight may have been set too low.It is uncontroversial that being very heavy is bad for health, but it is unclear at what point health risks begin. Doctors usually ad vise people to lose weight if their Body Mass Index (BMI) is high, which is calculated by dividing a person’s weight in kilograms by the square of their height in meters. In most countries, a healthy weight is defined as a BMI between 18.5 and 24.9.Having a BMI between 25 and 29.9 is classed as overweight and 30 and above asextremely fat.Previous research made waves when it found that people whose BMI was somewhat over the “healthy” threshold of 25 may have a slightly lower death rate than slimmer individuals. But many of the studies are fairly old and were done when people were mostly slimmer, and subjects weren’t racially diverse, says Aayush Visaria, a researcher in New Jersey.To address those issues, Visaria tracked the survival of about 500,000 racially diverse US adults of known height and weight for up to 20 years. Having a BMI between 25 and 27.4 carried a 5 percent lower risk of death in this period than a BMI within the healthy category of 22.5 to 24.9. A slightly higher BMI, of 27.5 to 29.9, seemed even better, linked with a 7 percent lower risk of death.One criticism is that the apparent benefit of being overweight could be a misconception, as people who lose weight due to illness are more likely to die. However, in the new research, the pattern persisted even if people who died within two years of entering the study were ruled out from the figures.Visaria says it’s too early to conclude that having an “overweight” BMI outperforms being in the healthy category, because population studies may be prejudiced and lead to misinterpretation of the results. “We aren’t certain if this is truly interpretable,” he says. “A more appropriate message is that BMI isn’t a good indicator of death risk—other factors, like body fat distribution, also play a role.”()63.The purpose of the first paragraph is to ______.A.put forward a controversial issueB.discuss reasons for a lower death rateC.reveal a finding that questions a common beliefD.justify the traditional “overweight” threshold()64.According to paragraph 3 and 4, Visaria resolved the limitations of previous research by ______. A.arriving at a more specific findingB.tracking the research for another 20 yearsC.calming down a wave of criticism from the crowdD.adopting a large, diverse sample over an extended period()65.According to the passage, “the pattern” in paragraph 5 refers to ______.A.the misconception about the benefit of being overweightB.the increased likelihood of death for individuals losing weight due to illnessC.the changing perception of BMI classification over timeD.the lower risk of death for slightly heavier persons in a set time compared to slimmer ones()66.Which of the following can be inferred from the passage?A.Weight and blood pressure are two primary factors related to BMI.B.To accurately reflect someone’s health risk, a broader view is necessary.C.Visaria will continue to figure out a more scientific BMI threshold.D.Visaria is confident of the reliability of his research.Section CDirections: Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.A.For example, they usually point at a box and say “open”.B.You should be mindful of the potential impact of gestures.C.Make fake gestures on certain occasions if necessary.D.Here are some tips that can help you take advantage of gestures.E.This can give you a way of seeing what others are thinking but not saying.F.Gesturing while learning will also help them solve the same problem in a new way.Tuning in to GesturesGesture gets its power in part from the fact that it is seldom noticed by the speaker or listener yet is easily understood and included in our conversations. The challenge is to use it to good effect. 67 Gesture more when you speak. It will help you learn and understand, and also think in a more abstract way. If you gesture while talking, you will remember more of what you have said. Do it while you are multitasking and it will lighten your mental load. What’s more, when you gesture, the people around you tend to do so too. 68 Encourage gestures in your children, students and anyone else you are trying to teach. This will help them understand the material you are conveying. 69 That is important because being able to generalise what we learn is essential to acquiring new knowledge.Pay attention to other people’s gestures. These offer a window into the thoughts that speakers have but don’t express in their words. These thoughts are often at the cutting-edge of their knowledge or address issues that are uppermost in their minds. Noticing and responding to such gestures will improve your interactions at home and at work.Observe the gesturing of infants (婴儿). While learning to talk, children typically convey sentence-like meanings in a combination of gesture and speech before using words alone. 70 If a child fails to produce these gesture-word combinations, it may be a sign that their spoken language development will be delayed, allowing you to intervene and help.IV.Summary WritingDirections: 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.What Is a Superiority Complex?A superiority complex is a behavior that suggests a person believes they are somehow superior to others. People with this complex often have overstated opinions of themselves. They may believe their abilities and achievements are better than those of others. However, the complex is believed to be a defense system for feelings of inadequacy that we all struggle with.It is unclear why a person develops a superiority complex. Multiple incidents may be the root cause. For example, it may be the result of many failures. A person tries to complete a specific goal or achieve a desired outcome, but they don’t succeed. They learn to handle the anxiety and stress of the failure by pretending to be above it. If they feel protected from their failures in this way, they may repeat this behavior in the future.These behaviors can begin at an early age. When a child is learning to cope with challenges, they may learn to control feelings of inadequacy or fear. A superiority complex may develop. Likewise, it may also happen later in life. As teens and adults, a person has many opportunities to try new things among new people. If these situations are not successfully coped with, a person may develop a superiority complex to overcome feeling isolated.People with a superiority complex are unlikely to be a threat to anyone’s physical health. However, the continuous lies and overstatements can become annoying to others and may negatively affect relationships. It can push away other people in your life and shrink your social circle. If you are in a relationship with a person who you think has the issue, encourage the person to seek help. For example, never hesitate to reach out to mental health professionals. They can find healthier ways to deal with hidden feelings.V.TranslationDirections: Translate the following sentences into English, using the words given in the brackets.72.检查过程中未发现任何机器损坏。
2023-2024学年云南省昭通一中教研联盟高二下学期期末质量检测物理试卷 A卷

2023-2024学年云南省昭通一中教研联盟高二下学期期末质量检测物理试卷 A卷1.气流流动的过程中都会发出噪声,如图1所示的消声器可以用来削弱高速气流产生的噪声。
波长为频率为的声波沿水平管道自左向右传播,在声波到达处时,分成上下两束波,这两束声波在处相遇时可削弱噪声,则()A.该消声器在处可削弱噪声,是因为上、下两束波到达处的波速不同B.该消声器在处可削弱噪声,是因为上、下两束波在处的振幅不同C.该消声器在处削弱噪声时,上、下两束波从到的路程差可能为D.该消声器在处削弱噪声时,上、下两束波从到的时间差可能为2.在一条平直公路上,甲、乙两辆汽车从时刻开始的位移一时间图像如图所示。
甲的图像为抛物线,乙的图像为倾斜直线。
已知甲的加速度大小为,再根据图中所提供的其他信息,下列说法正确的是()A.时刻甲的速度等于乙的速度B.甲的图像在时刻的切线与乙的图像平行C.甲的初速度可能不为0 D.甲、乙在时刻处在同一地点3.水面下深处有一点光源,发出两种不同颜色的光和,光在水面上形成了如图所示的一个有光线射出的圆形区域,该区域的中间为由两种单色光所构成的复色光圆形区域,周围为光构成的圆环。
若光的折射率为,则下列说法正确的是()A.光从水中进入空气中传播速度变大,波长变短B.复色光圆形区域的面积为C.在水中,光发生全反射的临界角比光的小D.用同一装置做双缝干涉实验,光的干涉条纹比光的窄4.氢原子能级图如图甲所示。
某群处于基态的氢原子受某光子照射后可辐射出三种不同频率的光,其中有两种能使乙图中逸出功为的极钾金属发生光电效应,通过乙图实验装置得到这两种光分别实验时的电流和电压读数,绘出丙图中①、②两根曲线,则下列说法正确的是()A.三种光中波长最长的光是从跃迁到时产生的B.乙图中当滑片向左端移动时,电流表示数不断增大到饱和电流后保持不变C.丙图中①曲线对应的入射光光子能使钾金属产出最大初动能为的光电子D.丙图中5.智能寻迹小车目前被应用于物流配送等多个领域,为测试不同寻迹小车的刹车性能,让它们在图甲中A点获得相同的速度,并沿直线AB刹车,最终得分为刹车停止时越过的最后一条分值线对应的分数,每相邻分值线间距离为0.5m。
浙江省杭州市2023-2024学年高二下学期数学期末检测试卷(含解析)

浙江省杭州市2023-2024学年高二下学期数学期末检测试卷考生注意:1.答题前,考生务必将自己的姓名、考生号填写在试卷和答题卡上,并将考生号条形码粘贴在答题卡上的指定位置.2.回答选择题时,选出每小题答案后,用铅笔把答题卡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.一、单项选择题: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.已知集合,则( ){}{}31,1e M x x N x x =-<=<≤M N ⋂=A .B .C .D .{}23x x <≤{}24x x <<{}2e x x <≤{}1e x x <≤2.已知复数,则在复平面内对应的点位于( )i 31i z -=-z A .第一象限B .第二象限C .第三象限D .第四象限3.样本数据的中位数和平均数分别为( )27,30,28,34,35,35,43,40A .34,35B .34,34C .34.5,35D .34.5,344.已知直线与圆有公共点,则的可能取值为( )30kx y k --=22:1O x y +=k A .1B .C .D .131-2-5.在中,角的对边分别是,且,则ABC ,,A B C ,,a b c ()()2sin 2sin 2sin a A b c B c b C=+++( )cos A =A .B .C .D .12-1312236.已知正方体的棱长为为棱的中点,则四面体的体积为1111ABCD A B C D -2,P 1BB 1ACPD ( )A .2B C .D .837.已知,则( )4sin25α=-tan2πtan 4αα=⎛⎫+ ⎪⎝⎭A .4B .2C .D .2-4-8.已知双曲线的上焦点为,圆的圆心位于,且与的22:1C y x -=F A x C 上支交于两点,则的最小值为( ),BD BF DF+A.B CD21-二、多项选择题:本题共3小题,每小题6分,共18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6分,部分选对的得部分分,有选错的得0分.9.已知分别是定义域为的偶函数和奇函数,且,设函数()(),f x g x R ()()e xf xg x +=,则( )()()()g x G x f x =()G x A .是奇函数B .是偶函数C .在上单调递减D .在上单调递增R R 10.将函数的图象向左平移个单位长度后,所得的图象关于轴()πsin (0)3f x x ωω⎛⎫=+> ⎪⎝⎭π3y 对称,则( )A .的图象关于直线对称B .的最小值为()f x π3x =ω12C .的最小正周期可以为D .的图象关于原点对称()f x 4π52π3f x ⎛⎫- ⎪⎝⎭11.如图,有一个棱台形的容器(上底面无盖),其四条侧棱均相1111ABCD A B C D -1111D C B A 等,底面为矩形,,容器的深度为,容器壁的厚度忽略11111111m 224AB BC A B B C====1m不计,则下列说法正确的是( )A .1AA =B .该四棱台的侧面积为(2mC .若将一个半径为的球放入该容器中,则球可以接触到容器的底面0.9m D .若一只蚂蚁从点出发沿着容器外壁爬到点A 1C 三、填空题:本题共3小题,每小题5分,共15分.12.的展开式中的系数为 .(用数字作答)712x x ⎛⎫+ ⎪⎝⎭3x 13.已知椭圆的左、右焦点分别为为上一动点,则的取22224:1(0)3x y C a a a +=>12,,F F A C 12AF AF 值范围是.14.已知两个不同的正数满足,则的取值范围是.,a b 33(1)(1)a b a b ++=ab 四、解答题:本题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.15.已知函数()1e 4xf x =(1)求曲线在点处的切线在轴上的截距;()y f x=()()1,1f l y (2)探究的零点个数.()f x 16.如图,在直三棱柱中,为棱上一点,111ABC A BC -12,1,AB BC AC AA M ====1CC 且.1AM BA ⊥(1)证明:平面平面;AMB ⊥1A BC (2)求二面角的大小.B AM C --17.设数列满足,且.{}n a ()122n n na n a +=+14a=(1)求的通项公式;{}n a(2)求的前项和.{}n a n n S 18.在机器学习中,精确率、召回率、卡帕系数是衡量算法性能的重要指标.科研机Q R k 构为了测试某型号扫雷机器人的检测效果,将模拟战场分为100个位点,并在部分位点部署地雷.扫雷机器人依次对每个位点进行检测,表示事件“选到的位点实际有雷”,表示事A B 件“选到的位点检测到有雷”,定义:精确率,召回率,卡帕系数()Q P A B =()R P B A =,其中.1o ee p p k p -=-()()()()()(),o e p P AB P AB p P A P B P A P B =+=+(1)若某次测试的结果如下表所示,求该扫雷机器人的精确率和召回率.Q R 实际有雷实际无雷总计检测到有雷402464检测到无雷102636总计5050100(2)对任意一次测试,证明:.()212Q R QR k Q R P AB +-=-+-(3)若,则认为机器人的检测效果良好;若,则认为检测效果一般;若0.61k <≤0.20.6k <≤,则认为检测效果差.根据卡帕系数评价(1)中机器人的检测效果.00.2k ≤≤k 19.已知抛物线的焦点为,以点为圆心作圆,该圆与轴的正、负半轴分别2:4C y x =F F x 交于点,与在第一象限的交点为.,H G C P (1)证明:直线与相切.PG C (2)若直线与的另一交点分别为,直线与直线交于点.,PH PF C ,M N MN PG T (ⅰ)证明:;4TM TN=(ⅱ)求的面积的最小值.PNT【分析】求得集合,可求{}24M x x =<<M N⋂【详解】因为,{}{}{}3124,1e M x x x x N x x =-<=<<=<≤所以.{}2e M N x x ⋂=<≤故选:C .2.B【分析】根据复数的四则运算和共轭复数的概念,以及复数的几何意义即可求解.【详解】因为,()()()()3i 1i i 342i 2i 1i 1i 1i 2z -++---====----+所以,2i z =-+故在复平面内对应的点为位于第二象限.z (2,1)-故选:B.3.D【分析】先将样本数据按从小到大进行排列,再根据样本数据的中位数、平均数概念公式进行计算即可.【详解】将样本数据按照从小到大的顺序排列可得,27,28,30,34,35,35,40,43故中位数为,343534.52+=平均数为.()12728303435354043348⨯+++++++=故选:D.4.B,求解即可.1≤【详解】由直线与圆有公共点,30kx y k --=22:1O x y +=可得圆心到直线的距离为,()0,0O 30kx y k--=1d =≤解得,所以的取值范围为.k ≤≤k ⎡⎢⎣故选:B.【分析】根据题意,利用正弦定理化简得,结合余弦定理,即可求解.222b c a bc +-=-【详解】因为,()()2sin 2sin 2sin a A b c B c b C =+++由正弦定理得,即,()()2222a b c b c b c=+++222b c a bc +-=-又由余弦定理得.2221cos 22b c a A bc +-==-故选:C.6.A【分析】设与交于点,证得平面,得到,且AC BD O AC ⊥11BDD B 113OPD V S AC =⨯中,结合,即可求解.AC =11BDD B 111111BDD B BOP B OP D P D ODD S S S S S =--- 【详解】设与交于点,在正方形中,,AC BD O ABCD AC BD ⊥又由正方体中,平面,1111ABCD A B C D -1DD ⊥ABCD 因为平面,可得,AC ⊂ABCD 1AC DD ⊥又因为且平面,所以平面,1BD DD D = 1,BD DD ⊂11BDD B AC ⊥11BDD B所以四面体的体积为,且,1ACPD 113OPD V S AC =⨯ AC =在对角面中,可得,11BDD B 111111BDD B BOP B D P OPD ODD S S S S S =-=--所以四面体的体积为.1ACPD 123V =⨯=故选:A.7.D【分析】由已知可得,利用,可求值.251tan tan 2αα+=-tan2tan 4απα⎛⎫+ ⎪⎝⎭22tan 1tan 2tan ααα=++【详解】因为,所以,2222sin cos 2tan 4sin2sin cos tan 15ααααααα===-++251tan tan 2αα+=-所以.2tan22tan 1tan tan 4ααπαα=⨯-⎛⎫+ ⎪⎝⎭221tan 2tan 2tan 41tan (1tan )1tan 2tan ααααααα-===-++++故选:D.8.B【分析】设出圆的方程与双曲线方程联立,可得,进而可得,利用两点1212,x x xx +22121x x +=间距离公式求出,并利用不等式方法求出其最小值.BF DF+【详解】由题可知.设圆,,.(F 22:()2A x a y -+=()11,B x y ()22,D x y 联立,得,则,22221()2y x x a y ⎧-=⎨-+=⎩222210x ax a -+-=212121,2a x x a x x -+==因此,故.()22212121221x x x x x x +=+-=222222121212112213y y x x x x +=+++=++=+=因为,所以,同理可得22111y x -=11BF===-.21DF =-故.)122BF DF yy +=+-又,且,故,从而22123y y +=12,1yy≥1y =≤=2y=≤=.())22121y y -≤所以)122BF DF y y +=+-2=2=2=2≥2==当时,有,,此时1a =()0,1B (D 11BF DF +=-+=所以的最小值是BF DF+故选:B.关键点睛:本题解题关键是由圆的方程与双曲线方程联立得到,再用不等式方法求22121x x +=其最小值.9.AD【分析】根据奇、偶性得到方程组求出、的解析式,从而得到的解析式,再()f x ()g x ()G x 由奇偶性的定义判断的奇偶性,利用导数判断函数的单调性.()G x 【详解】因为①,所以,()()e xf xg x +=()()e xf xg x --+-=即②,联立①②,解得,()()e xf xg x --=()()e e e e ,22x x x xf xg x --+-==所以,定义域为,又,()e e e e x x x x G x ---=+R ()()e e e e x xx xG x G x ----==-+所以是奇函数,又,()G x ()()()()()2222ee e e 40eeeexx x x xx xx G x ----+--=+'=>+所以在上单调递增,故A ,D 正确,B 、C 错误.()G x R 故选:AD10.ABD【分析】根据图象平移判断A ,根据关于直线对称可得判断B ,由周π3x =()132k k ω=+∈Z 期计算可判断C ,可先证明函数关于点对称,再由图象平移判断D.ω()f x 2π,03⎛⎫- ⎪⎝⎭【详解】对于A ,将的图象向左平移个单位长度后,关于轴对称,所以的图()f x π3y ()f x 象关于直线对称,故A 正确;π3x =对于B ,由题可知,解得,又,所以的最小()ππππ332k k ω+=+∈Z ()132k k ω=+∈Z 0ω>ω值为,故B 正确;12对于C ,若最小正周期,则,由B 项可知,不存在满足条件的,故C 错4π5T =2π52T ω==ω误;对于D ,因为,代入,得2π2ππsin 333f ω⎛⎫⎛⎫-=-+ ⎪ ⎪⎝⎭⎝⎭()132k k ω=+∈Z ,()2πsin 2π03f k ⎛⎫-=-= ⎪⎝⎭所以的图象关于点对称,将的图象向右平移个单位长度可以得到()f x 2π,03⎛⎫- ⎪⎝⎭()f x 2π3的图象,2π3f x ⎛⎫- ⎪⎝⎭则对称中心对应平移到坐标原点,故的图象关于原点对称,故D 正确.2π,03⎛⎫-⎪⎝⎭2π3f x ⎛⎫- ⎪⎝⎭故选:ABD 11.BD【分析】由勾股定理即可判断A ,由梯形的面积公式代入计算,即可判断B ,做出轴截面图形代入计算,即可判断C ,将四棱台展开,然后代入计算,即可判断D 【详解】对于A ,由题意可得,故A错误;132AA ==对于B ,梯形11ADD A =所以梯形的面积为11ADD A 242+=梯形,11ABB A=所以梯形的面积为,11ABB A 122+=故该四棱台的侧面积为,故B正确;2⨯=对于C ,若放入容器内的球可以接触到容器的底面,则当球的半径最大时,球恰好与面、面、面均相切,11ADD A 11BCC B ABCD 过三个切点的截面如图(1)所示,由题意可知棱台的截面为等腰梯形,较长的底边上的底角的正切值为,则,12212=-tan 2MPN ∠=-由于互补,故,,MPN MON ∠∠tan 2MON ∠=则,所以,从而球的半径为22tan 21tan MOPMOP ∠=-∠tanMOP ∠=,0.9=<所以将半径为的球放入该容器中不能接触到容器的底面,故C 错误;0.9cm对于D ,将平面与平面展开至同一平面,ABCD 11DCC D 如图(2),则,1AC ==将平面与平面展开至同一平面,如图(3),ABCD 11BCC B 则,145333044AC ⎛=+=< ⎝D 正确.故选:BD难点点睛:解答本题的难点在于选项D 的判断,解答时要将空间问题转化为平面问题,将几何体侧面展开,将折线长转化为线段长,即可求解.12.672【分析】利用二项式定理,求得二项展开式中的通项,把含x 的进行幂运算合并,然后令指数等于3,即可求解.【详解】因为通项为,令,得,712x x ⎛⎫+ ⎪⎝⎭77721771C (2)2C rr r r r rr T x x x ---+⎛⎫== ⎪⎝⎭72r 3-=2r =所以的系数为.3x 72272C 672-=故672.13.1,33⎡⎤⎢⎥⎣⎦【分析】先根据椭圆、、之间的关系,求出,再根据椭圆的定义,把换成a b c 12c a=1AF ,最后根据,代入即可.22a AF -[]2,AF a c a c ∈-+【详解】设椭圆的半焦距为,则,C (0)c c >12c a==,12222221AF a AF aAF AF AF -==-因为,即,[]2,AF a c a c ∈-+213,22AF a a ⎡⎤∈⎢⎥⎣⎦所以,即.2211,33a AF ⎡⎤-∈⎢⎥⎣⎦121,33AF AF ⎡⎤∈⎢⎥⎣⎦故答案为.1,33⎡⎤⎢⎥⎣⎦14.10,4⎛⎫⎪⎝⎭【分析】本题将条件式化简后结合基本不等式得出关于ab 的不等式,再构造函数并利用函数的单调性求解即可.【详解】将两边展开,33(1)(1)a b a b ++=得到,22113333a a b b a b +++=+++从而,()()221130ab a b a b ⎛⎫-+-+-= ⎪⎝⎭故,而,()130a b a b ab ⎛⎫-++-= ⎪⎝⎭a b¹故,又,130a b ab ++-=00a b >,>故,133a b ab =++>从而.321+<设函数,则,()3223g x x x=+112gg ⎛⎫<= ⎪⎝⎭观察易得在,()g x ()0,∞+12<又,所以.0,0a b >>104ab <<故答案为.10,4⎛⎫ ⎪⎝⎭关键点点睛:本题考查函数与不等式的综合,其关键是利用均值不等式构造关于ab 的不等式,再构造函数并利用函数的单调性解决问题.321+<()3223g x x x =+15.(1)12-(2)有两个零点()f x【分析】(1)求得,,利用导数的几何意()1e 4x f x '=()e 1142f ='-()e 114f =-义,求得切线方程,进而求得其在轴上的截距;y(2)得到在上递增,结合,得到,()1e 4x f x '=()0,∞+()10,104f f ⎛⎫ ⎪⎝⎭''01,14x ⎛⎫∃∈ ⎪⎝⎭使得,进而求得单调性,结合零点的存在性定理,即可求解.()00f x '=()f x【详解】(1)解析:由函数,可得,()1e 4x f x =()1e 4x f x '=()e 1142f ='-又,所以的方程为,即,()e 114f =-l ()e 1e 11424y x ⎛⎫=--+- ⎪⎝⎭e 11422y x ⎛⎫=-- ⎪⎝⎭令,可得,所以直线在轴上的截距为.0x =12y =-l y 12-(2)解:因为和上均单调递增,1e 4x y =y =()0,∞+所以在上单调递增,()1e 4x f x '=()0,∞+又因为,所以,使得,()141111e 10,1e 04442f f ⎛⎫=-=''- ⎪⎝⎭01,14x ⎛⎫∃∈ ⎪⎝⎭()00f x '=所以,当时,,在单调递减;()00,x x ∈()0f x '<()f x ()00,x 当时,,在单调递增,()0,x x ∞∈+()0f x '>()f x ()0,x ∞+又因为,()()14100111e 1e 0,110,4e 2010041044f f f ⎛⎫=->=-=- ⎪⎝⎭所以有两个零点.()f x 方法点睛:已知函数零点(方程根)的个数,求参数的取值范围问题的三种常用方法:1、直接法,直接根据题设条件构建关于参数的不等式(组),再通过解不等式(组)确定参数的取值范围2、分离参数法,先分离参数,将问题转化成求函数值域问题加以解决;3、数形结合法,先对解析式变形,在同一平面直角坐标系中作出函数的图象,然后数形结合求解.结论拓展:与和相关的常见同构模型e xln x①,构造函数或;e ln e ln e ln a a a a b b b b ≤⇔≤()lnf x x x =()e xg x x =②,构造函数或;e e ln ln e ln a a a b b a b b <⇔<()ln x f x x =()e x g x x =③,构造函数或.e ln e ln e ln a a a a b b b b ±>±⇔±>±()lnf x x x =±()e xg x x =±16.(1)证明见解析(2)4π【分析】(1)由线面垂直得到,结合勾股定理逆定理得到,证明出1AA BC ⊥BC AC ⊥平面,得到,结合题目条件证明出平面,得到面面垂直;BC⊥11AA C C AMBC ⊥AM ⊥1A BC (2)建立空间直角坐标系,设点,根据向量垂直得到方程,求出()0,0,M a ,进而求出平面的法向量,得到二面角的余弦值,得到答案.a M ⎛=⎝【详解】(1)在直三棱柱中,平面,111ABC A B C -1AA ⊥ABC ∵平面,BC ⊂ABC ∴,1AA BC ⊥∵2,1,AB BC AC ===∴,222AB AC BC =+∴,BC AC ⊥,平面,1AC AA A⋂=1,AC AA ⊂11AA C C ∴平面.BC ⊥11AA C C 平面,AM ⊂ 11AA C C ∴,AM BC ⊥,平面,11,AM A B A B BC B ⊥= 1,A B BC ⊂1A BC ∴平面.AM ⊥1A BC 又平面,AM ⊂AMB平面平面.∴AMB ⊥1A BC (2)由(1)可知两两垂直,1,,CA CB CC 如图,以点为坐标原点,所在直线分别为轴、轴、轴建立空间直角坐标C 1,,CA CB CC x y z 系,Cxyz 则.())()10,0,0,,,0,1,0C AAB设点,()0,0,M a 则.()()()1,,0,1,0,AM a BA CB AB ==-==,解得.11,30AM BA AM BA ⊥∴⋅=-+=a M ⎛=∴ ⎝设平面的法向量为,AMB (),,m x y z =则可取.0,0,m AM z m AB y ⎧⋅==⎪⎨⎪⋅=+=⎩(m = 易知为平面的一个法向量.()0,1,0n CB ==AMCcos ,m n m n m n ⋅〈〉===⋅故由图可知二面角的大小为.B AM C --4π17.(1)()12nn a n n =+⋅(2)()21224+=-+⋅-n n S n n【分析】(1)由已知可得,累乘法可求的通项公式;()122n n n a a n ++={}n a (2)由(1)可得,利用错位相减法可求的前()1212223212nn S n n =⨯⨯+⨯⨯+++⋅ {}n a 项和.n n S 【详解】(1)由题易知,且,0n a ≠()122n n n a a n ++=所以,()2341231212324251231n n n a a a a a a a a n -+⨯⨯⨯⨯⨯⨯⨯=⨯⨯⨯⨯- 所以,()()121121212n n n n n a n n a --+⋅==+⋅⨯所以也满足该式,()112,n n a n n a =+⋅所以.()12nn a n n =+⋅(2),①()1212223212nn S n n =⨯⨯+⨯⨯+++⋅ ,②()()2121221212n n n S n n n n +=⨯⨯++-⋅++⋅ ②-①,得.()()11212212222n n n S n n n +=+⋅-⨯⨯+⨯++⋅ 设,③1212222nn T n =⨯+⨯++⋅ 则,④()23121222122n n n T n n +=⨯+⨯++-⋅+⋅ ④-③,得,()()()1121112222222122n n n n n n T n n n ++++=⋅-+++=⋅--=-+ 所以.()()()1121122124224n n n n S n n n n n +++=+⋅--⋅-=-+⋅-18.(1);.0.625=Q 0.8R =(2)证明见解析(3)0.32【分析】(1)利用条件概率的计算公式计算即可;(2)由条件概率与互斥事件的概率公式证明即可;(3)由(2)计算出的值,判断机器人的检测效果即可.k 【详解】(1),()()()400.62564P AB Q P A B P B ====.()()()400.850P AB R P B A P A ====(2),()()()()()()1111111o e oe e P AB P AB p p p k p p P A P B P A P B ----==-=-----要证明,()212Q R QR k Q R P AB +-=-+-需证明.()()()()()()()1221P AB P AB Q R QR Q R P AB P A P B P A P B --+-=+---等式右边:()()()()()()()()||2||22||2P A B P B A P A B P B A Q R QR Q R P AB P A B P B A P AB +-+-=+-+-.()()()()()()()()()()()()()22P AB P AB P AB P AB P B P A P B P A P AB P AB P AB P B P A +-⨯⨯=+-()()()()()()()22P A P B P AB P A P B P A P B +-=+-等式左边:因为,()()()()()1P A B P AB P A P B P AB ⋃=-=+-所以()()()()()()()()()()()()()121111P AB P AB P A P B P AB P A P B P A P B P A P B P A P B --+-=⎡⎤⎡⎤------⎣⎦⎣⎦.()()()()()()()22P A P B P AB P A P B P A P B +-=+-等式左右两边相等,因此成立.()212Q R QRk Q R P AB +-=-+-(3)由(2)得,因为,0.6250.820.6250.810.320.6250.820.4k +-⨯⨯=-=+-⨯0.20.320.6<<所以(1)中机器人的检测效果一般.19.(1)证明见解析(2)(ⅰ)证明见解析;(ⅱ)163【分析】(1)根据题意,表示出直线的方程,然后与抛物线方程联立,由即可证明;PG Δ0=(2)(ⅰ)根据题意,设直线的方程为,与抛物线方程联立,即可得到点的PF 1x ty =+,N H 坐标,从而得到直线的方程,再与抛物线方程联立,即可得到点的坐标,再结合相似PH M 三角形即可证明;(ⅱ)由条件可得,再由代入计算,即可43PNTPNES S =△△12PNES EP EN = 证明.【详解】(1)由题意知,()1,0F 设,则,()2,2(0)P n n n >21PF n =+所以,所以,21GF FH n ==+()2,0G n -所以直线的斜率为,方程为.PG 1n ()21y x n n =+联立方程得,()221,4,y x n n y x ⎧=+⎪⎨⎪=⎩22440y ny n-+=因为,所以直线与相切.Δ0=PG C (2)(ⅰ)设直线的方程为,PF 1x ty =+由可得,则,又因为,所以.24,1,y x x ty ⎧=⎨=+⎩2440y ty --=4P N y y =-()2,2P n n 212,N n n ⎛⎫- ⎪⎝⎭由(1)知,点,直线的斜率为,方程为,()22,0H n +PH n -()22y n x n=---由得,由,()224,2,y x y n x n ⎧=⎪⎨=---⎪⎩224480y y n n +--=248P M y y n =--得.22444,2M n n n n ⎛⎫++-- ⎪⎝⎭作,垂足为,则,直线的方程为,NE PG ⊥E EN PM ∥EN 212y n x n n ⎛⎫=---⎪⎝⎭将直线与的方程联立,得解得.EN PG ()2212,1,y n x n n y x n n ⎧⎛⎫=--- ⎪⎪⎪⎝⎭⎨⎪=+⎪⎩11,E n n ⎛⎫-- ⎪⎝⎭所以,所以,2211441,,4,4EN n PM n n n n n ⎛⎫⎛⎫=+--=+-- ⎪ ⎪⎝⎭⎝⎭ 4PM EN =由相似三角形的性质可得.4TM TN=(ⅱ)由(ⅰ)知,所以,故,4TM TN=4TP TE=43PNT PNES S =△△因为,221111,,1,EP n n EN n n n n ⎛⎫⎛⎫=++=+-- ⎪ ⎪⎝⎭⎝⎭ 所以(当且仅当时等号成立),()323311114222PNEn S EP EN n n n +⎛⎫===+≥ ⎪⎝⎭ 1n =故,即的面积的最小值为.41633PNT PNES S =≥△△PNT 163方法点睛:利用韦达定理法解决直线与圆锥曲线相交问题的基本步骤如下:(1)设直线方程,设交点坐标为;()()1122,,,x y x y (2)联立直线与圆锥曲线的方程,得到关于(或)的一元二次方程,必要时计算;x y ∆(3)列出韦达定理;(4)将所求问题或题中的关系转化为、(或、)的形式;12x x +12x x 12y y +12y y (5)代入韦达定理求解.。
江苏省扬州市2023-2024学年高二下学期6月期末考试语文试题(解析版)

2023-2024学年第二学期期末检测高二语文2024.06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.试卷共8页;满分为150分,考试时间为150分钟。
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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,完成下面小题。
①在一般人看,美是物所固有的。
有些人物生来就美,有些人物生来就丑。
比如称赞一个美人,你说她像一朵鲜花,像一颗明星,像一只轻燕,你决不说她像一个布袋,像一条犀牛或是像一只癞蛤蟆。
这就分明承认鲜花、明星和轻燕一类事物原来是美的,布袋、犀牛和癞蛤蟆一类事物原来是丑的。
美人的美也是如此,和看者无关。
②这种见解并不限于一般人,许多哲学家和科学家也是如此想。
所以他们费许多心力去实验最美的颜色是红色还是蓝色,最美的形体是曲线还是直线,最美的音调是G调还是F调。
但是,如果美本来是物的属性,则凡是长眼睛的人们应该都可以看到,应该都承认它美,好比一个人的高矮,有尺可量,是高大家就要都说高,是矮大家就要都说矮。
可是,美的估定就没有一个公认的标准。
假如你说一个人美,我说她不美,你用什么方法可以说服我呢?有些人欢喜辛稼轩而讨厌温飞卿,有些人欢喜温飞卿而讨厌辛稼轩,这究竟谁是谁非呢?同是一个对象,有人说美,有人说丑,从此可知美本在物之说有些不妥了。
③因此,有一派哲学家说美是心的产品。
美如何是心的产品,他们的说法却不一致。
康德以为美感判断是主观的而却有普遍性,因为人心的构造彼此相同。
黑格尔以为美是在个别事物上见出概念或理想。
比如你觉得峨眉山美,由于它表现“庄严”“厚重”的概念。
山东济宁市2024年高二下学期7月期末考试物理试题(解析版)

2023—2024学年度第二学期质量检测高二物理试题2024.07注意事项:1.答卷前,考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置。
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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。
一、单项选择题:本题共8小题,每小题3分,共24分。
每小题只有一个选项符合题目要求。
1. 甲图为水黾停在水面上的情景;乙图为分子间作用力与分子间距离的关系;丙图为金刚石微观结构示意图;丁图中的每条折线表示一个炭粒每隔相同时间位置的连线。
下列说法正确的是()A. 水黾能停在水面上是因为其受到了水的表面张力作用B. 当分子间距离为0r时,分子间作用力为0,分子间没有引力和斥力C. 金刚石晶体中,原子按照一定规则排列,具有空间上的周期性D. 炭粒沿着笔直的折线运动,说明水分子在短时间内的运动是规则的【答案】C【解析】【详解】A.水黾能停在水面上是因为液体表面存在张力,而不是受到张力作用,故A错误;B.图(乙)中0r处,分子引力和斥力的合力为零,但引力和斥力均不为零,故B错误;C.晶体中原子(或分子、离子)都按照一定规则排列,具有空间上的周期性,故C正确;D.图(丁)是每隔一定时间把水中炭粒的位置记录下来,然后用线段把这些位置按时间顺序依次连接起来,而炭粒本身并不是沿折线运动,该图说明炭粒的运动(布朗运动)是不规则的,从而反映了水分子运动的不规则性,故D错误。
故选C 。
2. 太阳黑子群13679在北京时间2024年05月24日10时49分爆发了一个C4.9级小耀斑,耀斑从太阳的日冕抛射出高能的电子、离子和原子云气团,其中的一部分射向了地球,使地球上许多高纬地区出现了美丽壮观的极光。
如图所示,科学家发现并证实,这些高能带电粒子流向两极做螺旋运动,旋转半径不断减小,原因可能是( )A. 重力对粒子做正功,使其动能增大B. 越接近两极,地磁场的磁感应强度越大C. 洛伦兹力对粒子做负功,使其动能减小D. 空气分子与粒子碰撞,使粒子的带电量减少【答案】B 【解析】【详解】带电粒子在磁场中做圆周运动,根据2v Bvq m r=可得mv r Bq=A .若动能增大,则粒子速度增大, 粒子做圆周运动的半径会增大,A 错误;B .越接近两极,地磁场的磁感应强度B 越大,则半径减小,B 正确;C .洛伦兹力不做功,C 错误;D .若粒子带电量减小,则半径增大,D 错误。
山西省长治市2023-2024学年高二下学期6月期末考试语文试题(含答案)

长治市2023-2024学年高二下学期6月期末考试语文试卷注意事项: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。
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,将本试题卷和答题卡一并交回。
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,完成下面小题。
作为唐代最优秀的诗人之一,李白一直受到后代读者的喜爱。
对李白诗歌的注释、评点、和韵、拟作等数量很多,形成丰富的接受史。
但也有一些读者发现李白诗歌存在不完美的地方,因此对其进行批评,甚至删改。
对李白诗歌的删改有两种方式。
第一,改动诗歌的字词。
如《静夜思》宋刻本作:“床前看月光,疑是地上霜。
举头望山月,低头思故乡。
”但在明清时期的选本中,文字逐渐发生变动。
“看月光”在李攀龙《唐诗选》、王士祯《唐人万首绝句选》、沈德潜《唐诗别裁集》等选本中均为“明月光”,“望山月”在李攀龙《古今诗删》、乾隆《唐宋诗醇》等选本中均为“望明月”。
经过众多名家名选的接力,以及《唐诗三百首》等普及性较大的蒙学著作的推动,《静夜思》今天通行的版本变为:“……举头望明月,低头思故乡。
”又如《早发白帝城》:“……两岸猿声啼不尽,轻舟已过万重山。
”“啼不尽”,在高棅《唐诗品汇》、王士祯《唐人万首绝句选》、沈德潜《唐诗别裁集》、乾隆《唐宋诗醇》、孙洙《唐诗三百首》等选本中,均改为“啼不住”。
这些对诗歌字词的改动,并没有版本依据,是后人从审美角度出发作出的主观判断,却得到诸多选本的赞同。
第二,删去部分诗句。
这种删改的方式最为常见,有时删去部分诗句,会改变诗歌的体式。
如《关山月》:“明月出天山,苍茫云海间。
……戍客望边邑,思归多苦颜。
高楼当此夜,叹息未应闲。
”是乐府诗,陆深删去此诗后四句,刘大櫆《历朝诗约选》亦然,吴昌祺《删订唐诗解》也认为:“去后四句,竟似五言律矣。
下学期高二语文期末试题汇编:文学类文本阅读

文学类文本阅读广西百色市2023—2024学年高二下学期期末教学质量调研测试语文试卷(二)现代文阅读Ⅱ(本题共4小题,17分)阅读下面的文字,完成下面小题。
串门去(节选)李娟进入二月后,白天越来越长,天气一天比一天暖和。
当我们迎来冬牧场上第一拨正式拜访的客人阿孜拉和她的妈妈后,也憋不住了。
在当天的晚餐桌上大家商量着列出了一份计划表,开始挑选合适的日子陆续出门拜访邻居。
先由年轻人开始,然后是嫂子,接着是居麻。
其实早在一月,我和加玛就说好了二月一起出去串门的事。
为什么非要二月不可呢?因为,若是十二月和一月的话,白昼一晃而过,哪怕去最近的人家拜访,也未必能当天去当天回。
又不能走夜路,夜里有狼。
我很早就开始期待这趟行程了。
期待的同时却发愁没有像样的外套。
都脏得不像样子,脏得洗都没法洗……随着二月一天天来临,大家也替我着急起来。
我决定穿我的皮大衣,虽然臃肿不合身,毕竟是干净的。
大家都说:“豁切①,又不是去放羊。
”太邋遢了。
我又想穿我的短羽绒衣。
因为它是深色的,不太显脏,而且合身又利落。
大家说:“不行,太冷!”我说:“天不是已经热了嘛。
”加玛用汉语说:“去,热。
回来,冷。
”——回来的话,太阳就西斜了,温度迅速降了下来。
居麻大方地说:“行啦行啦,我的衣服借给你吧!”我伤心地说:“豁切!”终于到了出门那一天。
一大早加玛就提醒我,一定要穿刚洗过的那条裤子!这天她洗脸的时间格外漫长,然后又足足打扮了半个小时。
我坚持穿我相对体面的那件但大家七嘴八舌地表示反对。
我只好妥协,只好把肮脏的长羽绒衣套在短羽绒衣外面。
依大家说的,一到地方就赶紧脱下来塞在马鞍后。
本来这天打算去北面牧场拜访加玛的一个同学家,但因一时找不到散养在外的坐骑,耽搁了些时候。
等备好了马要出发时,突然东北面沙丘上出现了两个骑马的人。
走近一看,正是那个同学和她的妈妈!真不愧是最适合串门的一个日子啊。
两人真不愧是好朋友,想到一起了。
于是我们又卸了马鞍,脱去衣服。
山东省潍坊市2023-2024学年高二下学期期末考试语文试题(含答案)

山东省潍坊市2023-2024学年下学期期末考试高二语文试题2024.7注意事项:1.答卷前,考生务必将自己的姓名、考生号和座号填写在答题卡和试卷指定位置上2.回答选择题时,选出每小题答案后,用铅笔把答题卡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。
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。
材料一:一个作家如何选择和书写细节?首先必须准确。
美国诗人庞德说过:“写作的道德标准只有一个,那就是它的表达在根本上是否准确。
”我读许多小说,时间一长,会忘掉情节,甚至是主人公的名字,但能清晰地记得其中的一些细节。
我常常折服于优秀作家对日常生活中细节的准确把握和扎实表达,他们目光如炬,不耍花招,像技艺高超的渔夫,一叉下去就是一条鱼,鱼身上水淋淋的,散发出腥气,鱼还在挣扎,但已经被捕获了。
许多人描写细节喜欢用大量的形容词和花哨的比喻,看上去很美,让人感觉作者有才华,却华而不实,词不达意,有一种无力的感觉。
也有人描写的是日常生活中随处可见的细节,我们翻开许多杂志一眼就能看到,这样的细节准确但无聊、无神。
好作家不这样写细节,他们描述的细节往往是真实的细节,生活中确实存在,还能打动你。
一般作家就捕捉不到,因为它们太平常、太普通了,看上去不够美,也不够感人。
但正是因为有了这些细节,小说的现场感才更强,让读者感觉写什么都像是真的。
这类细节,是优秀小说的基础,使小说具有了真实性。
好作家描述的细节除了来源于真实生活,还有一种是在生活的基础上通过想象来呈现。
意大利作家卡尔维诺的《我们的祖先》三部曲写的内容都是假的,《不存在的骑士》中的骑士没有身体;《树上的男爵》中的男爵永远待在树上不下来;《分成两半的子爵》中的子爵被打成两半活了下来,一半代表善良,一半代表邪恶。
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高二下学期期末测试高二下学期期末测试高二语文(满分160分,考试用时150分钟)一、语言文字运用(15分)1.下列词语中加点的字,每对读音都不相同的一组是(3分)A.请帖/字帖当真/安步当车模板/装模作样B.载体/下载连累/罪行累累稽首/无稽之谈C.对峙/仗恃憧憬/灯影幢幢拜谒/断碑残碣D.联袂/抉择提防/金榜题名疏浚/提前竣工2.下列各句中,没有语病的一项是(3分)A.日前,《丹阳日报》新开辟的“质量丹阳”专栏是宣传质量技术监督工作的新窗口,强化质量技术监督职能的新载体,加强质量技术监督管理和服务水平的新平台。
B.外交部发言人表示,任何人也不能动摇中国政府和人民维护国家主权与领土完整的坚定意志,任何人也阻挡不了两岸推进交流合作,开创两岸关系和平发展的历史潮流。
C.导演陆川坦言拍《南京!南京!》这部电影的一个很大的目的就是为了让包括日本在内的全世界观众看到这部电影,目前,这部影片的日本商业发行已有突破。
D.卫生部5月2日零时许通报:已要求沪京粤等地卫生行政部门立即对AM098航班上的所有乘客按密切接触者进行隔离和为期7天的医学观察。
3.阅读下面的文字,概括“沙面文物建筑”所面临的三个方面的问题。
(每一个方面概括不超过8个字)(3分)广州的沙面是广州近代文物建筑最多的地方,由于地处珠江边,被称为“广州的外滩”。
但是,沙面要想真正成为“外滩”,却面临着考验。
广州沙面的建筑大多是十九世纪末、二十世纪初建造的砖木、砖土混合及钢筋混凝土结构建筑物。
由于年代久远,导致材料老化,混凝土风化,天花板脱落,墙、楼板和瓦面破损,管网陈旧漏水。
但保护沙面建筑群的手续繁琐,一砖一瓦的改变都必须上报国家文物局审批。
这导致了文物修缮审批手续繁琐,使沙面的保护利用工作非常被动,停滞不前。
因此,做实做细整体批复沙面古建筑的规划设计方案,迫在眉睫。
建筑文物的修缮与保护所涉及到的居民住户的搬迁,建筑群的维护、加固、内部装修等,需要的费用也是巨大的。
而这些费用如果完全依靠沙面所在的荔湾区文物部门负责,是不可能得到保证的。
吸引民间资本进入,允许文物建筑土地使用权的置换,都是有效的集资方式。
第一方面:第二方面:第三方面:4.某地一个跟甲型H1N1疑似患者同乘一辆汽车的民工回到家里,探望妻子和小孩,当地疾病防疫部门登门进行住点(住在需医学观察者家)隔离观察,需先跟其妻子说明情况,取得家人的配合与支持。
他们该怎么说呢?请将下面设计的内容补充完整。
(不超过80字)(6分)您的爱人回家途中跟甲型流感疑似患者乘坐的是同一辆汽车,这种流感传染性很强,二、文言文阅读(19分)阅读下面的文言文,完成5—8题。
郑遨,字云叟,滑州白马人也。
唐明宗祖庙讳遨,故世行其字。
遨少好学,敏于文辞。
唐昭宗时,举进士不中,见天下已乱,有拂衣远去之意,欲携其妻、子与俱隐,其妻不从,遨乃入少室山为道士。
其妻数以书劝遨还家,辄投之于火,后闻其妻、子卒,一恸而止。
遨与李振故善,振后事梁贵显,欲以禄遨,遨不顾,后振得罪南窜,遨徒步千里往省之,由是闻者益高其行。
其后,遨闻华山有五粒松,脂沦入地,千岁化为药,能去三尸,因徙居华阴,欲求之。
与道士李道殷、罗隐之友善,世目以为三高士。
遨种田,隐之卖药以自给,道殷有钓鱼术,钩而不饵,又能化石为金,遨尝验其信然,而不之求也。
节度使刘遂凝数以宝货遗之,遨一不受。
唐明宗时以左拾遗、晋高祖时以谏议大夫召之,皆不起,即赐号为逍遥先生。
天福四年卒,年七十四。
遨之节高矣,遭乱世不污于荣利,至弃妻、子不顾而去,岂非与世自绝而笃爱其身者欤?然遨好饮酒弈棋,时时为诗章落人间,人间多写以缣素,相赠遗以为宝,至或图写其形,玩于屋壁,其迹虽远而其名愈彰,与乎石门之徒异矣。
与遨同时张荐明者,燕人也。
少以儒学游河朔,后去为道士,通老子、庄周之说。
高祖召见,问“道家可以治国乎?”对曰:“道者,妙万物而为言,得其极者,安居衽席之间可以治天地也。
”高祖大其言,延入内殿讲《道德经》,拜以为师。
荐明闻宫中奉时鼓,曰:“陛下闻鼓乎?其声一而已。
五音十二律,鼓无一焉,然和之者鼓也。
夫一,万事之本也,能守一者可以治天下。
”高祖善之,赐号通玄先生,后不知其所终。
(选自欧阳修《新五代史隐逸传》)5.对下列句子中加点词的解释,不正确的一项是(3分)A.后闻其妻、子卒,一恸而止恸:极度悲哀,痛哭B.因徙居华阴,欲求之因:因为C.至或图写其形,玩于壁屋写:描摹D.与乎石门之徒异矣徒:同一类人6.下列句子中,均能说明郑遨“节高”的一组是(3分)①遨少好学,敏于文辞②后闻妻子卒,一恸而止③欲以禄遨,遨不顾④节度使刘遂凝数以宝货遗之,遨⑤遭乱世不污于荣利⑥然遨好饮酒弈棋,时时为诗章落人间A.③④⑤B.①④⑤C.①②⑥D.②③⑥7.下列对原文有关内容的分析和概括,不正确的一项是(3分)A.郑遨年轻时好学能文,后来考进士却落第了,又恰逢社会动荡,于是隐居山林,当了道士。
B.郑遨与道友李道殷、罗隐交情深厚,三人虽屡受朝廷征召,但均未出仕,被世人看作是三位高洁之士。
C.郑遨隐居山林,但经常有诗稿散落民间,人们对这些诗稿珍爱有加,甚至把它们誊写在丝绢上,作为馈赠佳品。
D.张荐明与郑遨同为当时著名的隐士,他曾喜好儒学,后来则一心崇奉道家学说,他认为道家学说对治国大有裨益。
8.翻译下列句子。
(10分)(1)其妻数以书劝遨还家,辄投之于火。
(3分)▲(2)遨尝验其信然,而不之求也。
(4分)▲(3)高祖大其言,延入内殿讲《道德经》,拜以为师。
▲三、古诗词鉴赏(10分)9.阅读下面这首古诗,然后回答问题。
除夜宿石头驿戴叔伦旅馆谁相问?寒灯独可亲。
一年将尽夜,万里未归人。
寥落悲前事,支离笑此身。
愁颜与衰鬓,明日又逢春。
(1)诗歌的首联运用了什么写法?这样写有什么作用?(3分)▲(2)“一年将尽夜,万里未归人”这一联历来为人激赏,称其为“客中除夕夜之绝唱”(《诗薮》),请你说说这一联的妙处。
(3分)▲(3)“寥落悲前事,支离笑此身”一句中既写到“悲”又提及“笑”,这样写是否矛盾?请结合诗歌内容作简要说明。
(4分)▲四、名句名篇默写(8分)10.补写出下列名句名篇中的空缺部分。
(1)众里寻他千百度,▲,那人却在,灯火阑珊处。
(辛弃疾《青玉案·元夕》)(2)零落成泥碾作尘,▲。
(苏轼《卜算子·咏梅》)(3)▲,帘卷西风,人比黄花瘦。
(李清照《醉花阴》)(4)▲,不思量,自难忘。
(苏轼《江城子》)(5)相见时难别亦难,▲。
▲,蜡炬成灰泪始干。
(李商隐《无题》)(6)▲,月涌大江流。
(杜甫《旅夜书怀》)(7)高山仰止,▲。
(《诗经》)五、现代文阅读(一)(23分)阅读下面的文字,完成11—14题。
斑驳都江镇张劲①21世纪的皮鞋叩击在18世纪的石板路上,踏响的并非都是历史的回声。
这取决于历史对于现实的触发,以及来访者对于历史的兴趣。
②我们在三都县都江镇上流连。
装饰一新的音响店里,正播放着爱死爱活的流行歌曲;缤纷的时装铺面,正叫卖着绿肥红瘦的女裙男衫;延伸出去的街面上,正斑驳着汽车、拖拉机和摩托车留下的车辙、油迹……古镇已兴奋在时尚之中。
当时尚在历史上面趾高气扬的时候,历史索性埋下头去,闭目养起神来。
③穿过古旧的城门洞,步入老城的上江街,我们才听到了历史的短叹长吟。
街身已经凹陷,一如老人瘪皱的脸,五官歪斜,但鳞次栉比的木房、砖房仍顽强而艰难地撑持着,关闭了近三个世纪的嘴角,默守着心中的万语千言。
雍正九年(1731年),实行“改土归流”的清政府,曾在这里设立都江厅,建筑城垣,驻扎绿营兵镇守。
厅城初为土城,为防止苗民造反,两年后改为石城。
石城周长三里七分,设城门、城楼各四,炮台八座,城内诸般官廨齐备。
都江厅所辖地界“广七十里,袤一百八十里”。
至乾隆年间,城内再建文昌阁、关帝庙、城隍庙、万寿宫等,时“商旅出于途,汉苗杂于市”,厅城已成为都柳江上游一带的军事重镇和商贸重镇。
④如今,270多年时光流去,古镇只剩下东西南北四座城门和一小段城墙遗迹了。
立于城楼之上,四顾苍茫。
迎接我们这些隔世目光的,除了山坡下的都柳江水,四面的层峦叠嶂,剩下的便是秋阳与秋风了。
秋阳下的断墙残垣,像一截干枯的蛇蜕,蜷伏在丛莽之中,任凭着野草、荆棘的蔓延与承包。
浩荡秋风频频入怀,它在向来客表示礼节性问候的同时,也在反反复复地叙说着一个主题——战争与和平的故事。
⑤清咸丰五年(1855年),本地苗族首领罗光明等人在太平天国运动影响下,组织“斋教”义军,反抗清政府的暴虐统治,其根据地便设在附近大坪山上。
义军数次攻克都江厅城及周边州、县,拉锯战持续了十六年之久,至同治十年(1871年)方告失败。
我们脚下便是当年的古战场,昔日帐前鸣战马,今朝城下泣秋虫,那些纷纭杀气,淋漓鲜血,早都交与冷月秋风了。
据《三都县志》等资料记载,咸丰、同治年间上江城内常住人口约1800余户,6000余人,有地方官吏、绿营兵士及其家属,也有经商、务农和从事手工业生产的市民。
绿营兵士、商贩,多为江西、湖广等外籍汉人,家属和普通市民则多为本地汉人和世居的少数民族同胞。
战乱结束后,只见“田地荒芜,人烟萧索,枯骨枕野,惨不堪言”,于是便有了后继官员的治理,有了“城乡义冢”即“万人坟”的诞生。
⑥“万人坟”是战争创伤结的痂。
它不高大,却甚有名,圆形土丘掩映在松柏丛中,站立成南门外的一道特别的风景。
坟墓为同治十二年(1873年)地方官吏配合当地士绅所建。
据碑文显示,当时“城厢内外,尸骸遍野”,十余年无人拾捡,墓内所葬,乃收集来的“枯骨数十担”。
引起我兴趣的是,这数十担枯骨,有阵亡的绿营兵官军,有无辜市民,也有不少起义军将士,而墓碑上皆以“万古流芳”颂之,以“祭扫维新”、“体魄长存”悼之。
这种不分敌我、不问民族、不管阶级、不计尊卑贵贱,无论生前怎样,死后都被定格在同一座坟墓里的做法,是因为白骨缠野,确实难以区分所致,还是因为本就不想区分,一律皆视之为大清子民,同墓合葬,以显示胜利者的大度仁心所致……这就只好交给历史学家去研究了。
⑦“万人坟”不远的冈峦上,有民族小学一所,小学就建在当年都江厅通判衙门的旧址上。
它三重石阶引路,一对石狮把门,居高而立,面对碧水青山,气势仍颇为雄壮。
学校前身原为厅城义学,开办于清雍正十年(1732年)。
当时贵州巡抚奏准办“苗疆义学”,计划吸收苗、水等少数民族子弟入学,除武力征服外,再施以文治教化。
但义学所收实际只为城内绿营官兵子弟和外籍商贩子弟,与原来的设想大相径庭。
到咸丰五年,罗光明义军反清战事起后,义学也就停办了。
至战乱结束,“万人坟”修建之时,也恰是义学恢复之日。
因此,前者恰似一部旧书的后记,而后者,则如一册新著的序言了。
此时义学的学生,有厅城官兵、居民的子弟,也有附近村寨的少数民族子弟。
直到后来义学被新式小学取代,学校历史也与都江古镇一样地久长。