高一10月月考测试题
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2024级“高中”10月高一年级新高考月考测试语文(答案在最后)(考试时间:150分钟满分:150分)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。
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:“姜还是老的老”“我上次听到这句话,还是在上次”……这类毫无意义却又挑不出毛病的语句,被人们戏称为“废话文学”。
在信息获取和观点表达日益便捷的今天,这类对文本符号的二次改造与“语言玩梗”大行其道,不禁让人好奇,互联网语境下,“不好好讲话”的魔力究竟在哪儿?最初的“废话文学”,内含着人们戏谑、反讽与抵抗意味的阴阳怪气。
过去“路遥车马慢”,信息承载量与传播速度都很低,寄信隔着万水千山,发短信要字斟句酌,人们恨不得一句话把所有事情讲完。
互联网时代,看完某篇形式主义言辞空洞的文章,围观一个长达几分钟但内涵不超过两句话的无效视频,大家惊觉,原来互联网时代,早已不是“听君一席话,胜读十年书”。
出于对低信息增量内容的不满,人们开始用语言予以反击。
这一届年轻人偏爱“躺平”,消解无用信息的方式,就是用同样的“无意义”进行风险对冲。
于是,“废话”式名言警句被改造出来,并活学活用于各类语境。
“废话文学”的生产十分简便,把单位从分钟变成秒、用同义词西红柿替换番茄……实在不行,还可以向鲁迅先生学习,把说过的句子再说一遍,“你说了两句话,一句是废话,另一句也是废话”。
由此一来,“不好好说话”便演变成某种社交策略,“废话文学”在本身语义之外,逐渐衍生出各场景独有的意义和情绪表达。
面对朋友空洞的长篇大论,一句“除了内容,你说得都挺好”,就能以戏谑的口吻表达出某种不满,含蓄中夹着杀伤力,对方也不至于下不来台。
2024级“贵百河—武鸣高中”10月高一年级新高考月考测试数 学(考试时间:120分钟 满分:150分)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。
3.回答非选择题时,将答案写在答题卡上,写在试卷上无效。
一、单选题:本题共8小题,每小题5分,共40分。
在每小题给出的选项中,只有一项是符合题目要求的。
1.已知集合,集合,则图中阴影部分表示的集合为()A . B.C .D .2.已知命题,则是( )A .B .C .D .3.已知集合,则“”是“集合M 仅有1个真子集”的( )A .必要不充分条件B .充分不必要条件C .充要条件D .既不充分又不必要条件4.已知函数的对应关系如下表,函数的图象如图,则的值为()A .3B .0C .1D .25.给出下列结论:①两个实数a ,b 之间,有且只有a ﹥b ,a =b ,a <b 三种关系中的一种;②若,则a ﹥b ;③若,;④已知,则.其中正确结论的个数为( )A .1B .2C .3D .4x123230{32}A x x =-<<{05}B x x =<<{35}x x -<<{02}x x <<{30}x x -<≤{3025}x x x -<≤≤<或2:1,1p x x ∀<->p ⌝21,1x x ∃≤-≤21,1x x ∃<-≤21,1x x ∀<->21,1x x ∀≥->{}()210R M x ax x a =-+=∈14a =)(x f y =)(x g y =()1f g ⎡⎤⎣⎦1>ab0a b >>0a bc d d c >>⇒>0ab >11a b a b>⇔<()f x6.已知函数的定义域是,则的定义域为()A .B .C .D .7.已知函数,若对于任意的实数与至少有一个为正数,则实数m 的取值范围是( )A .B .C .D .8.已知正实数a ,b ,记,则M 的最小值为()AB .2C .1D .二、多选题:本题共3小题,每小题6分,共18分。
山东省菏泽市第一中学2024-2025学年高一上学期10月月考英语试题一、阅读理解When you are about to go to a boarding school (寄宿制学校) in England, there are many different questions that may come to mind. But once you look at them with some perspective (视角), you will certainly feel easy. Here is a normal boarding day.Early Morning:Usually boarders get up around 7:00 am and have around an hour to take a shower and put on their uniform before breakfast.Lessons:Classes start at 9: 00 am and every lesson lasts for 50 minutes. After two lessons, at 10: 40 am, you’ll have a short break. The next period of classics will include two more lessons.Lunch:Lunch is usually held around 12:30 pm at the dining hall, where you’ll join your friends to enjoy a hot dish. After an hour of lunch, you’ll have three or four more lessons to attend.Dinner:At 5:00 pm you will have finished your school lessons for the day. Most boarding schools in the UK offer their full boarders different kinds of hot meals to choose from.Activities/ Sports:All boarding schools in the UK provide many different kinds of activities for their boarders, such as football, swimming, golf or art.Prep:An important part of boarding school life is the supervised homework session known as “prop”. Although “prep” might sound stressful, it’s a great way for you to keep up with your studies.Free time:Once you have done all your classes and activities, it’s time to relax.Bedtime:In most boarding schools, the lights go out around 10:30 pm.Being nervous just before you go to a boarding school is completely normal and understandable. Hopefully, knowing the usual routine (常规) can help you. Once you are there, you will also see how exciting life in a boarding school in England can be.1.What can help you keep up with your studies in the boarding school?A.Doing activitıes.B.Having lessons.C.Enjoying free time.D.Supervised homework sessions.2.What do we know about boarding schools in England from the text?A.Classes usually start at 8:30 am.B.Students can have a short break after four lessons.C.They don’t give students any free time.D.They turn off the lights around 10:30 pm. 3.What is the purpose of the text?A.To help students know about boarders’ life.B.To attract more students to boarding schools.C.To introduce a new school life.D.To advertise for boarding schools.I want to share with you a story from 28 years ago. My dad was a used car salesman. Every Thursday night, he would head off to Shreveport, LA for an auction (拍卖会). Most of the time, I drove a car over there for him so he could sell it at the auction.One day I was riding with my dad when he noticed a hitch-hiker (搭便车的人) with a backpack. Without hesitation, he pulled the car over and offered him a ride. Dad asked him his name, and proceeded to talk to him about all sorts of things. Dad asked him where he was going. The hitch-hiker told him that he was heading for the west. I can’t recall why but he told Dad a lot of things that had occurred to him and that persuaded him to make that decision. He talked about the tragic events that occurred to him several years before. He was in low spirits, but I could see that the hitch-hiker’s attitude was changing as someone was really listening to him.We drove 45 minutes before the hitch-hiker got off. We pulled over and Dad told him to keep his head up and things would start looking up for him soon. He reached into his pocket and handed the hitch- hiker a twenty-dollar bill. The guy smiled. He nearly lit up right there on the cold, dark highway.We drove on and my dad did not say a single thing. I was still completely surprised by what I had just witnessed. I was always told by everyone never to pick up a hitch-hiker and yet my dad did it every single time he saw one. While reflecting upon that story I came to understand that just one single kind act could change someone’s life. And I am sure that my father’s deed made that poor man’s day.4.What made the hitch-hiker become less upset?A.The writer’s father talking to him about all sorts of things.B.The writer’s father offering him a free ride.C.The writer’s father really listening to him.D.The writer’s father agreeing to drive him to his destination.5.When his father helped the hitch-hiker, the writer ___________.A.was deeply moved B.strongly disagreedC.admired his father D.couldn’t understand6.Which of the following words can be used to describe the father?A.Wealthy.B.Warm-hearted.C.Far-sighted.D.Adventurous.7.What’s the writer’s main purpose of writing this text?A.To tell a story of his father.B.To show his respect for his father.C.To prove his father is the best teacher.D.To advise people to learn from the hitch-hiker.What you do after studying could have a big effect on how well you learn and remember. Today on Education Tips, we will explore two no-cost ways that can help you improve your learning: wakeful rest and sleep.When many students finish studying, they often go straight to another activity. Perhaps they look at their phone or computer. They might even play a video game or watch television. But research suggests that resting after your study may help you remember what you have studied.The basic idea is this: by stopping your activity after the study, your brain gets a chance to rest. Resting is difficult when you are playing computer games.While this might sound unusual to you, many studies have explored the benefits of resting after learning—what is called “wakeful rest”.A 2019 study found that both young and older adults were able to better remember information they learned after doing wakeful rest.If you want to give “wakeful rest” a try, here are a few simple things to do. Rest quietly for five to ten minutes. Do not look at your phone, read stories or play games. It is really that easy!Besides wakeful rest, sleep is also important for learning. The Division of Sleep Medicine at Harvard Medical School (HMS) notes that sleep helps people to learn in two ways. A report on the school’s website explains that “a sleep-deprived person cannot focus attention well and therefore cannot learn well”. It also says, “Sleep itself helps us to memorize and that memorizing is very important for learning new information.”The National Sleep Foundation (NSF) suggests that adults between the ages of 20 and 64 get between 7 and 9 hours of sleep per day. Teenagers may need a little bit more, and people over the age of 65 may need a little less.Try to get some “wakeful rest” after a study. Then try to get a good night’s sleep. That’s it—a simple, no-cost way to help you remember what you learn.8.What is paragraph 3 mainly about?A.The importance of sleep.B.The good ways to get a rest.C.The reason for having a rest.D.The types of resting activities.9.Which activity might be recommended shortly after studying?A.Reading a newspaper.B.Playing a video game.C.Sitting there doing nothing.D.Having a nice sleep at night.10.How long should teenagers sleep each day according to NSF?A.Less than 7 hours.B.Just 7 hours.C.No more than 9 hours.D.Around 9 hours.11.What can be a suitable title for the text?A.Low-cost Tips on Working B.Good Things to Do After Your StudyC.Why Do We Need to Have a Good Sleep?D.How Long Should We Rest AfterLearning?Exams are brain-burning. The silence of the hall; the beating of the clock; the watchful eye of the teacher; the proud expression of the person who has finished 15 minutes early. So it is not surprising that those who feel anxious in the exam do worse than those who do not. But different from the existing(现存的) point of view, the recent study by Maria Theobald shows that the pressure of revision affects(影响) your exam performance.To test this idea, Dr Theobald worked with 309 German students who were preparing for their final exam. During the 100 days before the real exam, these students used an online learning app which showed them old exam questions and recorded their performance. And 40 days before the real exam, they also took a mock(模拟) exam. Their levels of anxiety were also be recorded and assessed(评定), every day for 40 days before the real exam and also on the day of that exam.The result was not what she expected. Anxiety on the day of the test did not predict(预测) exam performance at all. It was the level of knowledge in the mock exam and the earlier practices that predicted it. Those who performed well in the mock exams also did well in the real test, no matter how anxious they were on the day.It turned out that what really affected students were high levels of anxiety during the weeks before the exam. The greater a student’s anxiety in the days before the exam, the lower his or her knowledge gain was. To reduce this anxiety, she gives some advice for students when they revise. First, they can raise their confidence in their own abilities by reminding(提醒) themselves of how much they know. Second, they can diminish the importance of the test by reminding themselves that, though it is important, it is not a life or death situation. It really isn’t.12.What influences exam results according to the existing view?A.Exam levels.B.Exam schedule.C.Pressure from the exam hall.D.High expectations(期望) from teachers. 13.How did Dr Theobald go on the study?A.By asking students some questions.B.By collecting the data(数据).C.By asking experts` advice.D.By following students’ performances. 14.What does the underlined word “diminish” probably mean in the last paragraph?A.Reduce.B.Influence.C.Realize.D.Understand. 15.Which of the following is the suitable title for the text?A.Never Too Late to Revise B.Never Put Off Revision till TomorrowC.Anxiety in Revision, Failure in Exam D.The More Pressure, The Better ResultsPeer PressurePeer pressure is a part of life. 16 This peer group may be of similar age but can also be defined by other commonalities (共同点), including motherhood, professional affiliations, and your local neighborhood.Peer pressure is not always negative. Trying to fit into a healthy social group, for example, of peers getting good grades, joining sports teams, and making plans for their futures, is positive.17 Peer influence can show you there is support, encouragement, and community available to you. By seeing someone else do something positive, you may reflect on your own life choices, goals, and where you spend your time.18 Being pressured by peer, whether it happens in person or online, may shake your sense of identity and self-confidence. For example, you may feel pressure to do unsafe things that have risks you may not fully know.19 You can also work on surrounding yourself with more positive influences. Some ways of coping with peer pressure include:● Not spending time with people who pressure you to do things that feel wrong or dangerous.● Learning to say “no” and practicing leaving situations that feel unsafe or uncomfortable.● Talking to a trusted peer or professional (teacher, counselor) if you have problems saying “no” or are feeling pressured to change something about yourself.● 20In the journey of life, peers influence our steps. If you embrace positive guidance and stand firm against negativity, you can become the architects of your own story.A.Some refer to this type of peer “pressure” as peer “influence”.B.It can be direct or indirect, positive or negative.C.It is any type of influence that comes from a peer group.D.Getting over peer pressure means not accepting the negative influence of others.E.While peer influence can improve your life, peer pressure can cause problems.F.Feeling pressured to do things that may make you feel bad about yourself is unhealthy.G.Befriending people who have a positive influence.二、完形填空Jenna had graduated from middle school and was ready for new 21 in high school.22 , high school was different. In the first week, Jenna tried out for a cheerleading team. She 23 very excellent girls, and she knew it would be difficult for her to be chosen. Two hours later, the judge(裁判)24 a list of the girls for the second try-out. Her name wasn’t on the list. Feeling 25 , she walked home carrying her schoolbag full of homework.26 home, she started with math. She had always been a good math student, but now she was 27 . She moved on to English and history, and was 28 to find that she didn’t have any trouble with those subjects. Feeling better, she decided not to 29 math for the time being.The next day in30 class, Jenna spent most of her time working out the problems that had given her so much 31 . By the end of class, she finally got the answers. As she gathered (收拢)her books, Jenna 32 she’d continue to try to fit in at her new school. She wasn’t sure if she’d succeed, but she knew she had to 33 . High school was just as her mom had said, "You will feel like a small fish in a big pond 34 a big fish in a small pond. The challenge is to become the 35 fish you can be."21.A.courses B.decisions C.challenges D.exercises 22.A.So B.However C.Therefore D.Besides 23.A.fought B.connected C.beat D.encouraged 24.A.pronounced B.forgot C.saw D.heard 25.A.strange B.happy C.sad D.lonely 26.A.Arriving B.Going C.Staying D.Leaving 27.A.struggling B.improving C.working D.continuing 28.A.anxious B.disappointed C.scared D.relaxed 29.A.work with B.prepare for C.worry about D.give up 30.A.physics B.history C.English D.math 31.A.pleasure B.hope C.trouble D.courage 32.A.decided B.accepted C.refused D.felt33.A.swim B.try C.ask D.travel34.A.in exchange for B.in case of C.in terms ofD.instead of35.A.thinnest B.smallest C.best D.gentlest三、语法填空阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式。
重庆市第一中学2024-2025学年高一上学期10月月考数学试题一、单选题1.已知集合{}{}432A B x x ==,,则A B =I ( )A .2163x x ⎧⎫<≤⎨⎬⎩⎭ B .{}316x x ≤<C .223x x ⎧⎫<≤⎨⎬⎩⎭D .{}02x x ≤≤2.命题.“230,1x x x ∃<+>”的否定是( ) A .230,1x x x ∀≥+≤ B .230,1x x x ∀<+≤ C .230,1x x x ∃<+≤D .230,1x x x ∃≥+≤3.已知函数()2f x +的定义域为()3,4-,则函数()1f xg x +的定义域为( )A .()4,3-B .()2,5-C .1,33⎛⎫⎪⎝⎭D .1,53⎛⎫ ⎪⎝⎭4.使得“[]21,2,0x x x a ∀∈+-≤”为真命题的一个充分不必要条件是( )A .2a ≥B .2a >C .6a >D .6a ≥5.若正实数,x y 满足3x y +=,且不等式22823m m x y+>-+恒成立,则实数m 的取值范围是( )A .{31}mm -<<∣ B .{3m m <-∣或1}m > C .{13}mm -<<∣D .{1mm <-∣或3}m > 6.函数()()()245,2231,2x a x x f x a x x ⎧-++<⎪=⎨-+≥⎪⎩满足对12,R x x ∀∈且12x x ≠,都有()()()12120f x f x x x --<⎡⎤⎣⎦,则实数a 的取值范围是( ) A .30,2⎛⎫⎪⎝⎭B .30,2⎡⎫⎪⎢⎣⎭C .()0,1D .[]0,17.已知,a b 均为正实数,且1a b +=,则下列选项错误的是( )AB .34a a b ++的最小值为7+C .()()11a b ++的最大值为94D .2232a b a b +++的最小值为16 8.含有有限个元素的数集,定义其“交替和”如下:把集合中的数按从小到大的顺序排列,然后从最大的数开始交替地加减各数,例如{}4,6,9的“交替和”是9647-+=;而{}5的交替和是5,则集合{}Z 54M x x =∈-≤≤∣的所有非空子集的“交替和”的总和为( ) A .2048B .2024C .1024D .512二、多选题9.已知,,a b c ∈R ;则下列不等式一定成立的有( ) A .若0ab ≠且a b <,则11a b> B .若0a b >>,则20242024b b a a +<+ C .若,a bcd >>,则ac bd >D .()221222a b a b ++≥--10.下列说法正确的是( )A .若p 是q 的必要不充分条件,p 是r 的充要条件,则q 是r 的充分不必要条件B .若关于x 的不等式2430kx kx k -++≥的解集为R ,则实数k 的取值范围是01k <≤C .若不等式()()30x ax b x c-+≤-的解集为[)[)2,13,∞-⋃+,则不等式2320ax ax b --≥的解集为[]1,4-D .“[]()21,3,2130a ax a x a ∃∈---+-<”为假命题的充要条件为[]51,0,43x ⎡⎤∈-⋃⎢⎥⎣⎦11.已知函数()f x 的定义域为[)0,+∞,且满足当[)0,2x ∈时,()22f x x x =-+,当2x ≥时,恒有()()2f x f x λ=-,且λ为非零常数,则下列说法正确的有( )A .()()101320272024f f λ+=B .当12λ=时,反比例函数()1g x x =与()f x 在()0,2024x ∈上的图象有且仅有6个交点C .当0λ<时,()f x 在区间[]2024,2025上单调递减D .当1λ<-时,()f x 在[]()*0,4n n ∈N 上的值域为2122,n n λλ--⎡⎤⎣⎦三、填空题12.已知集合{}210A xx =-=∣,则集合A 有个子集. 13.已知集合[]()(){}1,4,10A B xx a ax ==+-≤∣,若A B B =U 且0a ≥,则实数a 的取值范围是.14.若正实数x ,y 满足()()332331423x y x y -+-=--,则2346y x x x y++的最小值为.四、解答题15.已知函数()21,122,1x x f x x x ⎧->-⎪=⎨⎪--≤-⎩.(1)若()01f x =,求0x 的值;(2)若()3f a a <+,求实数a 的取值范围. 16.已知函数()f x =A ,集合{}321B xx =->∣. (1)求A B U ;(2)集合{}321M xa x a =-≤≤-∣,若M ()R A ð,求实数a 的取值范围. 17.已知二次函数()f x 的图象过原点()0,0,且对任意x ∈R ,恒有()26231x f x x --≤≤+.(1)求()1f -的值; (2)求函数()f x 的解析式;(3)记函数()g x m x =-,若对任意(]11,6x ∈,均存在[]26,10x ∈,使得()()12f x g x >,求实数m 的取值范围.18.教材中的基本不等式可以推广到n 阶:n 个正数的算数平均数不小于它们的几何平均数.也即:若12,,,0n a a a >L,则有*12,2n a a a n n n+++∈≥N L ,当且仅当12n a a a ===L 时取等.利用此结论解决下列问题:(1)若,,0x y z >,求24y z x x y z++的最小值;(2)若10,2x ⎛⎫∈ ⎪⎝⎭,求()312x x -的最大值,并求取得最大值时的x 的值;(3)对任意*k ∈N ,判断11k k ⎛⎫+ ⎪⎝⎭与1111k k +⎛⎫+ ⎪+⎝⎭的大小关系并加以严格证明.19.已知定义在11,,22⎛⎫⎛⎫-∞-⋃+∞ ⎪ ⎪⎝⎭⎝⎭上的函数()f x 同时满足下列四个条件:①512f ⎛⎫=- ⎪⎝⎭;②对任意12x >,恒有()()0f x f x -+=; ③对任意32x >,恒有()0f x <; ④对任意,0a b >,恒有111222f a f b f ab ⎛⎫⎛⎫⎛⎫+++=+ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭.(1)求32f ⎛⎫- ⎪⎝⎭的值;(2)判断()f x 在1,2⎛⎫+∞ ⎪⎝⎭上的单调性,并用定义法证明;(3)若对任意[]1,1t ∈-,恒有()()21232f t k t k -+-+≤,求实数k 的取值范围.。
邢台市第一中学2023-2024学年高一上学期10月月考数学试卷 学校:___________姓名:___________班级:___________考号:___________ 一、选择题1、设集合{2,3,5,7}A =,{1,2,3,5,8}B =则A B =( )A.{1,3,5,7}B.{2,3}C.{2,3,5}D.{1,2,3,5,7,8} 2、已知0a >,函数2()f x ax bx c =++,若0x 满足关于x 的方程20ax b +=,则下列选项的命题中为假命题的是( )A.x ∃∈R ,0()()f x f x ≤B.x ∃∈R ,0()()f x f x ≥C.x ∀∈R ,0()()f x f x ≤D.x ∀∈R ,0()()f x f x ≥3、若a b >,则下列正确的是( )A.b c a c -<-B.22a b >C.ac >1b < 4、若a ,0b >,且3ab a b =++,则ab 的取值范围是( )A.1ab ≤B.9ab ≥C.3ab ≥D.19ab ≤≤5、已知全集U =R ,集合{|08,}A x x x =<<∈R 和{|35,}B x x x =-<<∈Z 关系的韦恩图如图所示,则阴影部分所表示集合中的元素共有( )A.3个B.4个C.5个D.无数个6、已知x ∈R ,则“(1)(2)0x x --≤成立”是“|1||2|1x x -+-=成立”的_____条件( )A.充分不必要B.必要不充分C.充分必要D.既不充分也不必要 7、已知[1,1]a ∈-,不等式2(4)420x a x a +-+->恒成立,则x 的取值范围为( )A.(,2)(3,)-∞+∞B.(,1(2,)-∞+∞C.(,1)(3,)-∞+∞D.(1,3)8、下列结论中正确的个数是( )①命题“有些平行四边形是矩形”是存在量词命题;②命题“{|x y y ∃∈是无理数},2x 是有理数”是真命题;③命题“a b >是22ac bc >的必要不充分条件”是真命题;④命题“0x ∃>,使2210x x ++≤”的否定为“0x ∀≤,都有2210x x ++>”;⑤若x ∈R ,则函数y =+A.3B.2C.1D.0若a b ≠,则( ) A.甲先到达终点 B.乙先到达终点C.甲乙同时到达终点D.无法确定谁先到达终点10、若集合A 满足x ∈A ,则称集合A 为自倒关系集合.在集合111,0,,,1,2,3,423M ⎧⎫=-⎨⎬⎩⎭的所有非空子集中,具有自倒关系的集合的个数为( ) A.7 B.8 C.16 D.15二、多项选择题11、设集合{}1,9,A m =,{}2,1B m =若A B B =,则满足条件的实数m 的值是( )A.0B.1C.3D.-312、下列命题正确的是( )A.若0a b >>,m >B.若正数a 、b 满足a b +=11b +≥+C.若0x >,则23x --D.若()2x x y =-,0x >,0y >,则2x y +的最小值是9;13、已知函数2()4f x x ax =-+,a ∈R 则下列叙述正确的是( )A.若对x ∀∈R 都有()0f x ≥成立,则44a -≤≤B.若[1,2]x ∃∈使得()0f x ≥有解,则4a ≤C.若1x ∃,20x >且12x x ≠使得()()120f x f x ==,则4a >D.若()0f x ≤的解集是[1,4],则5a =14、设集合{}61,M x x k k ==+∈Z ,{}64,N x x k k ==+∈Z ,{}32,P x x k k ==-∈Z ,则下列说法中正确的是( )A.M N P ≠=⊂B.()M P N ≠⊂C.M N =∅D.P M N =15、若0a >,0b >,且21a b +=,则下列说法正确的是( )22a b + 16、若01,22x ⎡⎤∃∈⎢⎥⎣⎦,使得200210x x λ-+<成立是假命题,则实数λ可能取值是( )1≤的解集为____________. 18、某班级共有50名学生做物理、化学两种实验,已知物理实验做得正确的有40人,化学实验做得正确的有31人,两种实验都做错的有4人,则这两种实验都做正确的有____________人.19、一元二次不等式23208kx kx +-<对一切实数x 都成立,则实数k 的取值范围为___________.20、已知实数a ,b ,c 满足222425a b c ++=,则23ab c +的最大值为__________.四、解答题21、已知0m >,0n >,关于x 的不等式2200x mx --<的解集为{|2}x x n -<<. (1)求m ,n 的值;(2)正实数a ,b 满足na mb +=22、已知{}2|40A x x x =+=,(){}22|2110B x x a x a =+++-=.(1)若A 是B 的子集,求实数a 的值;(2)若B 是A 的子集,求实数a 的取值范围.23、如图所示,设矩形()ABCD AB BC >的周长为24,把它沿AC 翻折,翻折后AB '交DC 于点P ,设AB x =.(1)用x 表示DP ,并求出x 的取值范围;(2)求ADP △面积的最大值及此时x 的值.24、已知函数2()(1)2f x mx m x =-++,m ∈R .(1)设m <()f x mx <;(2)设0m >,若当[2,)x ∈+∞时()f x参考答案1、答案:D解析:因为{2,3,5,7}A =,{1,2,3,5,8}B =所以{1,2,3,5,7,8}A B =,故选:D.2、答案:C解析:因为,0x 满足关于x 的方程20ax b +=,所以,0x =2()f x ax bx c =++取得最小值,因此,x ∀∈R ,0()()f x f x ≤是假命题,选C.3、答案:A解析:由a b >,两边同时减去c ,有a c b c ->- ,故选项A 正确;1a =,2b =-时,22a b > 不成立,排除B 选项;当0c <时,由a b >得ac bc <,排除C 选项;1a =,b =-<故选:A.4、答案:B 解析:因为3ab a b =++,a ,0b >,所以33,ab a b =++≥即230-≥,当且仅当3a b ==时,等号成立; 解得9ab ≥或1ab ≤-(舍). 故选:B.5、答案:A解析:由题意知,集合{2,1,0,1,2,34}B =--,,因为集合{|08,}A x x x =<<∈R , 由集合的交运算可得,{1,2,3,4}A B =,故阴影部分所表示集合为(){}2,1,0B A B =--,其中的元素共有三个.故选:A.6、答案:C解析:充分性:若(1)(2)0x x --≤,则12x ≤≤,所以|1||2|121x x x x -+-=-+-=; 必要性:根据绝对值的性质:若||||||a b a b +=-,则0ab ≤,若|1||2|1x x -+-=,且|(1)(2)|1x x ---=,则有(1)(2)0x x --≤.所以“(1)(2)0x x --≤成立”是“|1||2|1x x -+-=成立”的充要条件.故选:C.7、答案:C解析:令()2(2)44f a x a x x =-+-+,则不等式2(4)420x a x a +-+->恒成立转化为()0f a >在[1,1]a ∈-上恒成立.∴有(1)0(1)0f f ->⎧⎨>⎩,即22(2)4402440x x x x x x ⎧--+-+>⎨-+-+>⎩, 整理得:22560320x x x x ⎧-+>⎨-+>⎩, 解得:1x <或3x >.∴x 的取值范围为()(),13,-∞+∞.故选:C.8、答案:A解析:对于①: 含有存在量词的命题 ,叫做特称命题(存在性命题), ”有些”为存在量词,故①正确;对于②:当x =对于③:若a b >成立,当0c =时22ac bc >不成立,若22ac bc >成立,则0c ≠,所以a b >成立, 故a b >是22ac bc >的必要不充分条件,所以③成立;对于④: 命题“0x ∃>,使2210x x ++≤”的否定为“0x ∀>,都有2210x x ++>”,故④错误;对于⑤:2y =≥≠综上只有①②③正确;故选:A.9、答案:A2T b S +=,则2S T a b =+设选手乙总共用时T2S T b '=,则2Sb Sa T ab +'= ()()()()()()22244202222S ab a b Sab S a b S a b S Sb Sa T T a b ab ab a b ab a b ab a b ⎡⎤-+-+--+⎣⎦'-=-===<++++即T T '<,即甲先到达终点故选:A. 10、答案:D解析:根据自倒关系集合的定义可知,当x =-1=-;当x =1x=1=;当x ==x ==x =14=不存在;所以{}1,12,2⎧⎫⎨⎬⎩⎭,13,3⎧⎫⎨⎬⎩⎭,{}1-必须分别在一起,可以把它们看作一个元素,所以自倒关系集合的个数为42115-=. 故选D.11、答案:ACD 解析:因为A B B =,所以B A ⊆,若29m =,则3m =±,满足题意,若2m m =,则0m =或1m =,1m =不合题意,0m =满足题意.故选:ACD.12、答案:BC++a m b m 因为0a b >>,0m >,所以0a b ->,()()0a b m bb m ->+a mb m +->+>对于选项B,因为1a b +=,所以113a b +++=,()111111114112113113113b a a b a b a b a b ++⎛⎫⎛⎫+=++++=++≥ ⎪ ⎪++++++⎝⎭⎝⎭=b ==对于选项C,因为0x >,43x x +≥=x ==所以4232x x--≤-对于选项D,因为()2x x=-21x=, 所以()124224x y x y x y y x y x⎛⎫+=++=+≥= ⎪⎝⎭=4x =,2y =时,等号成立,所以2x y +的最小值是8,故D 错误. 故选:BC.13、答案:ACD解析:对于A 项,由已知可得,0∆≤,即2160a -≤,解得44a -≤≤,故A 项正确; 对于B 项,由已知可得[1,2]x ∃∈使得2()40f x x ax =-+≥有解, 即a x ≤+[1,2]x ∈上有解,只需max 4a x x ⎛⎫≤+ ⎪⎝⎭即可. 设()g x x=1x ∀,2[1,2]x ∈且12x x <,则()()()1212121244g x g x x x x x x x -=+--=-因为1x ,2[1,2]x ∈且12x x <,所以1214x x <<,且120x x -<,所以,()()120g x g x ->,()()12g x g x >.所以,()g x x =[1,2]x ∈上单调递减, 所以,()max 145g x =+=,所以5a ≤,故B 错误;对于C 项,由已知可得,2()40f x x ax =-+=有两个不相等正实根, 则1212204Δ160x x a x x a +=>⎧⎪=⎨⎪=->⎩,所以124x x a +=>,故C 项正确; 对于D 项,由已知可得,1和4是方程240x ax -+=的两个根,则214016440160a a a -+=⎧⎪-+=⎨⎪∆=->⎩,解得5a =,故D 项正确.故选:ACD.14、答案:CD 解析:{}(){}61,3212,M x x k k x x k k ==+∈==⋅+-∈Z Z , {}(){}64,3222,N x x k k x x k k ==+∈==⋅+-∈Z Z , 当k ∈Z 时,21k +为奇数,22k +为偶数, 则M N ≠,MN P =,M N =∅,P M N =.故选:CD.15、答案:ABC解析:21122222a b ab a b +⎛⎫=⨯⋅≤⨯= ⎪⎝⎭a b=,即a ==22112a b=++=+≤+=,当且仅当a ==成立,1222a a ba b a a b a b a b++=+=++≥==b ==()2221424142a b a bab ab +=+-=-≥,当且仅当a ==22a b +的故选:ABC.16、答案:AB解析:由条件可知1,22x ⎡⎤∀∈⎢⎥⎣⎦,2210x x λ-+≥是真命题,即2212x x x λ+≤=min 12x x ⎛⎫≤+ ⎪⎝⎭,1,22x ⎡⎤∈⎢⎥⎣⎦ 设()12f x x x =+≥=1,22⎡⎤∈⎢⎥⎣⎦等号成立的条件是112,22x x x ⎡⎤=⇒=⎢⎥⎣⎦,所以()f x的最小值是即λ≤故选:AB.17、答案:[)2,2-≤10-≤, 0≤,等价于()()22020x x x ⎧+-≤⎨-≠⎩, 解得22x -≤<;1≤的解集为[)2,2- 故答案为:[)2,2-. 18、答案:25 解析:根据题意可设:全班的学生组成的集合为U 做对物理实验的学生组成的集合为A做对化学实验的学生组成的集合为B并将两种实验都做对的学生记为x 人,则可用文氏图将其关系表示如下: 结合文氏图及题意知: ()()4031450x x x -++-+=,解之得:25x = 故两种实验都做对的学生为25人.故答案为:25.19、答案:(3,0)-解析:由不等式23208kx kx +-<对一切实数都成立, 可得202034208k k k k ⎧⎪≠⎪⎪<⎨⎪⎛⎫⎪-⨯⨯-< ⎪⎪⎝⎭⎩即30k -<<.故答案为:(3,0)-.解析:因为22ab a b =⋅≤2b =时取到等号,所以2252332232c c c ab c -⎛⎫≤+=-- ⎪⎝⎭+由225c≤可知c =3abc +≤21、答案:(1)10n =,8m =;(2)9.解析:(1)根据题意,不等式2200x mx --<的解集为{|2}x x n -<<, 即方程2200x mx --=的两根为-2和n ,则有()()2220n m n -+=⎧⎪⎨-⨯=-⎪⎩,, 解可得10n =,8m =. 2()正实数a ,b 满足2na mb +=,即1082a b +=,变形有541a b +=, ()1114554145955b a a b b a b a b ⎛⎫+=++=+++≥+= ⎪⎝⎭, 当且仅当a ==∴15a +22、答案:(1)1a =(2)1a ≤-或1a =解析:(1)因为{}{}2|404,0A x x x =-=+=, 若A 是B 的子集,则{}4,0B A ==-,所以2Δ0402(1)401a a >⎧⎪-+=-+⎨⎪-⨯=-⎩,解得1a =.(2)若B 是A 的子集,则B A ⊆.①若B 为空集,则()22Δ4(1)41880a a a =+--=+<,解得1a <-; ②若B 为单元素集合,则()22Δ4(1)41880a a a =+--=+=,解得1a =-. 将1a =-代入方程()222110x a x a +++-=,得20x =,解得0x =,所以{}0B =,符合要求; ③若B 为双元素集合,{}4,0B A ==-,则1a =. 综上所述,1a ≤-或1a =.23、答案:(1)()7212612DP x x=-<< (2)最大值为108-=解析:(1)由矩形()ABCD AB BC >的周长为24,且AB x =,可得12AD x =-, 在ACP △中,易知PAC PCA ∠=∠,所以可得AP PC =,因此DP PB '=; 所以AP AB PB AB DP x DP ''=-=-=-, 在Rt ADP △中,由勾股定理可得()()22212x DP x DP -+=-,整理可得12DP =-又AB BC >,即12x x >-,依题意解得612x <<, 即可得()7212612DP x x =-<< (2)在Rt ADP △中,()()11724321212108661222ADP S AD DP x x x x x ⎛⎫⎛⎫=⋅=--=-+<< ⎪ ⎪⎝⎭⎝⎭△; 又612x <<,所以4326x x +≥=当且仅当4326x x=,即x =所以4321086108ADP S x x ⎛⎫=-+≤- ⎪⎝⎭△即当x =ADP 面积的最大值为108-24、答案:(1)答案见解析解析:(1)不等式即2(21)20mx m x -++<,即(2)(1)0x mx --< 当0m =时,即20x ->,解得2x >,当0m ≠时,由(2)(1)0x mx --=得,122,x x ==(i )若m <2<,原不等式解得x <2x >,(ii )若0m <<2>,原不等式解得2x <<综上,当0m <时,解集为{x x <∣2}>; 当0m =时,解集为{2}x x >∣;当0m <<12x x m ⎫<<⎬⎭∣.(2)由0m >知()f x 开口向上,对称轴0x =当02x ≤,即m ≥函数()f x 在[2,)x ∈+∞上单调递增,最小值为(2)2f m ==38=;当02x >,即0m <<函数()f x 在[)02,x 单调递减,在[)0,x +∞上单调递增,最小值为()20614m m f x m --===(舍),.。
2024-2025学年度上学期第一次月考高一语文试题本试卷共24题,共150分,共5页。
考试时间为150分钟。
考试结束后,只交答题卡。
一、现代文阅读(35分)(一)信息类文本阅读(本题共5小题,19分)阅读下面的文字,完成下面小题。
赵艳(以下简称赵):很多读者对你的认识是从《哦,香雪》开始的,这是一篇意境和文字都非常优美的文章,可以说是当代文学中诗化小说的代表作之一。
这篇文章读起来很有青春的诗意和激情,一气呵成,像是从生命中活泼泼地流淌出来的东西,很感性,也很朴实。
你自己如何评价这篇作品在你所有创作中的地位和价值?铁凝(以下简称铁):它在我整个创作中不是以价值高或低,分量轻或重来衡量的,而是它本身具有一种不可替代性,我后来的小说在叙事上更成熟,更像一个自觉作家的写作,但《哦,香雪》焕发出来的对人生、对情感、对生活、对希望那种透明的激情是不可替代的。
一个年轻的作家,如果有本领,写出那样的作品并不难,但是岁月过去了,我们经历了很多,如果还保有那样的爱和希望,还拥有那种透明的心境,那就真是不容易的。
我想,《哦,香雪》对我的意义就在这里。
现在,香雪的时代好像已经过去了,我歌颂的也的确不是封闭的处于蒙昧状态的大山里的人和事。
生活在变,生活里的人也在变,但是我们依然需要香雪。
孙犁说过:“女孩子们心中,埋藏着人类原始的多种美德。
”我认为,发掘我们内心的多种原始美德是任何作家在任何时代都不应该放弃的,哪怕在经历了人生的苦难之后,外在的形式变了,内部那个坚硬的核应该还在。
赵:无论是美还是丑,无论外壳怎样变换,内在的核始终如一地朝向对人类美好的情感、珍贵的品性的追求。
铁:是的。
我的作品中前后的人物和故事都有鲜明的不同,中间还有一些起伏。
文学可以而且应该有多种途径,有时是这个故事触动了你,有时是那个人使你有倾诉的愿望,但最终我将这些人和事融合在一起,通过他们,我想要实现的是文学温暖世界的功能。
汪曾祺说过“人是孤儿”,人真的是很孤独的,所以人需要文学的世界、希望的世界,虽然也许那最美好的境界,我们永远到达不了,但愿望在那儿,我们就会有向前走的勇气,对生活有所期待,这就够了。
2023北京首都师大附高一10月月考数 学第Ⅰ卷(共50分)一、选择题(本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,选出符合题目要求的一项)1. 下列各式:①{}10,1,2∈;②{}0,1,2∅⊆;③{}{}10,1,2∈;④{}{}0,1,22,0,1=,其中错误的个数是( ) A. 1个B. 2个C. 3个D. 4个2. 命题“2x ∃<,220x x −<”的否定是( ) A. 2x ∃≤,220x x −≥ B. 2x ∀≥,02x << C. 2x ∃<,220x x −≥D. 2x ∀<,0x ≤或2x ≥3. 将下列多项式因式分解,结果中不含因式()2x +的是( ) A. 224x x + B. 2312x −C. 26x x +−D. ()()228216x x −+−+4. 若集合{}{3},21,Z A xx B x x n n =<==+∈∣∣,则A B =( )A. ()1,1−B. ()3,3−C. {}1,1−D. {}3,1,1,3−−5. 如图,I 是全集,M 、P 、S 是I 的3个子集,则阴影部分所表示的集合是( )A. ()M P SB. ()M P SC. ()M P SD. ()M P S6. 已知p :111x <+,q :()10x x +<,则p 是q 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7. 下列结论成立的是( ) A. 若ac bc <,则a b > B. 若a b >,则22a b > C. 若a b >,则11a b< D. 若110a b<<,则0b a <<8. 设集合11,Z ,,Z 3663k k M x x k N x x k ⎧⎫⎧⎫==+∈==+∈⎨⎬⎨⎬⎩⎭⎩⎭||,则( ) A. MNB. M NC. N MD. M N ⋂=∅9. 已知,,A B C 是三个集合,若A B B C ⋃=⋂,则一定有( ) A. A C ⊆B. C A ⊆C. C A ≠D. A =∅10. 设()C M 表示非空集合M 中元素的个数,已知非空集合,A B .定义()(),()()()(),()()C A C B C A C B A B C B C A C A C B −≥⎧⊗=⎨−<⎩,若{}1,2A =,()(){}2220B x x ax x ax =+++=且1A B ⊗=,则实数a 的所有取值为( )A. 0B. 0,−C. 0,D. −,0,第Ⅱ卷(共70分)二、填空题(本大题共5小题,每小题5分,共25分)11. 方程组322327x y x y +=⎧⎨−=⎩的解集用列举法表示为______________.12. 若“25x m >−”是“|x |<1”的必要不充分条件,则实数m 的取值范围是___________ 13. 设a ,b ∈R ,集合{}2,0,1{,,0}a a b −=,则a b +的值是______.14. 已知集合{}|3A x a x =≤≤,{}|0B x x =<,若A B ⋂=∅,则实数a 的取值范围是______. 15. 当两个集合中有一个集合为另一个集合的子集时,称两个集合之间构成“全食”;当两个集合有公共元素,但互不为对方子集时,称两个集合之间构成“偏食”,对于集合11,,12A ⎧⎫=−⎨⎬⎩⎭,{}2B x x a ==|.若A 与B 构成“全食”,则a 的取值范围是______;若A 与B 构成“偏食”,则a 的取值范围是______.三、解答题(本大题共4小题,共45分.解答应写出文字说明,演算步骤或证明过程)16. 已知全集U =R ,集合{R |211}A x x =∈−≤,集合{R |12}B x x =∈−<≤. (1)求集合A B ⋂及()UA B ⋃;(2)若集合{|2,0}=∈≤<>C x R a x a a ,且C B ⊆,求实数a 的取值范围. 17. 已知关于x 的一元二次方程()22230x m x m +−+=有两个实数根1x ,2x .(1)求实数m 的取值范围; (2)若12126x x x x +=−,求m 的值.18. 已知全集U =R ,812x A xx ⎧⎫+=>⎨⎬−⎩⎭,{}22240B x x mx m =−+−<,{}14C x x =−<<,在①Ux A ∈;②x A C ∈;③x A C ∈⋃;这三个条件中任选一个补充到下列问题中并作答.问题:设p :______,q :x B ∈,是否存在实数m ,使得p 是q 的必要不充分条件?若实数m 存在,求m 的取值范围;若实数m 不存在,说明理由.19. 已知集合{}1,2,,A n =⋅⋅⋅(3n ≥),表示集合A 中的元素个数,当集合A 的子集i A 满足2i A =时,称i A 为集合A 的二元子集,若对集合A 的任意m 个不同的二元子集1A ,2A ,…,m A ,均存在对应的集合B 满足:①BA ⊆;②B m =;③1i BA ≤(1i m ≤≤),则称集合A 具有性质J .(1)当3n =时,若集合A 具有性质J ,请直接写出集合A 的所有二元子集以及m 的一个取值; (2)当6n =,4m =时,判断集合A 是否具有性质J ?并说明理由.参考答案第Ⅰ卷(共50分)一、选择题(本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,选出符合题目要求的一项)1. 【答案】A【分析】根据集合与集合的关系,元素与集合的关系即可求解.【详解】由元素与集合的关系可知{}10,1,2∈正确,{}{}10,1,2∈不正确, 由集合之间的关系知{}0,1,2∅⊆正确, 由集合中元素的无序性知{}{}0,1,22,0,1=正确, 故错误的个数为1, 故选:A【点睛】本题主要考查了元素与集合的关系,集合的子集,集合的相等,属于容易题. 2. 【答案】D【分析】根据存在量词命题的否定是全称量词命题即可得到结果. 【详解】命题“2x ∃<,220x x −<”是存在量词命题, 又22002x x x −<⇒<<,所以其否定为全称量词命题,即为“2x ∀<,0x ≤或2x ≥”. 故选:D. 3. 【答案】C【分析】利用提取公因式法判断A ,利用公式法判断B ,利用十字相乘法判断C 、D. 【详解】对于A.原式()22x x =+,不符合题意;对于B.原式()()()234322x x x =−=+−,不符合题意;对于C 原式()()23x x =−+,符合题意; 对于D.原式()()22242x x =−+=+,不符合题意. 故选:C. 4. 【答案】C【分析】解绝对值不等式得A ,根据交集的定义计算即可.【详解】解3x <得33x −<<,即()3,3A =−,B 为奇数集,故{}1,1A B =−.故选:C 5. 【答案】C【分析】根据Venn 图表示的集合运算作答.【详解】阴影部分在集合,M P 的公共部分,但不在集合S 内,表示为()⋂⋂M P S , 故选:C . 6. 【答案】D【分析】分别求出,p q ,再分析出,p q 的推导关系. 【详解】()11110010111x x x x x x −<⇒−<⇒<⇒+>+++, 所以:0p x >或1x <−,而:10q x −<<,所以p 是q 的既不充分也不必要条件, 故选:D 7. 【答案】D【分析】根据不等式的性质或举出反例对各选项逐一判断即可.【详解】选项A :当0c >时,若ac bc <,则a b <,当0c <时,若ac bc <,则a b >,故A 说法错误; 选项B :若1a =,2b =−满足a b >,此时22a b <,故B 说法错误; 选项C :当0a b >>或0a b >>时, 11a b <,当0a b >>时, 11a b>,故C 说法错误;选项D :当110a b<<时,0ab >,所以不等式同乘ab 可得0b a <<,故D 说法正确; 故选:D 8. 【答案】B【分析】根据集合,M N 的表达式,可求出集合M 是16的奇数倍,N 是16的整数倍,即可得出,M N 的关系.【详解】由()11,Z 21,Z 366k M x x k x x k k ⎧⎫⎧⎫==+∈==+∈⎨⎬⎨⎬⎩⎭⎩⎭||可知,集合M 表示的是16的奇数倍; 由()11,Z 2,Z 636k N x x k x x k k ⎧⎫⎧⎫==+∈==+∈⎨⎬⎨⎬⎩⎭⎩⎭||可知,集合N 表示的是16的整数倍; 即可知M 是N 的真子集,即M N . 故选:B 9. 【答案】A 【分析】根据()B C B ⋂⊆,以及()B C C ⋂⊆,结合已知条件,即可判断集合之间的关系. 【详解】因为()B C B ⋂⊆,又A B B C ⋃=⋂, 故可得()A B B ⋃⊆,则A B ⊆; 因为()B C C ⋂⊆,又A B B C ⋃=⋂,故可得()A B C ⋃⊆,则B C ⊆; 综上所述:A B C ⊆⊆. 故选:A.【点睛】本题考查由集合的运算结果,求集合之间的关系,属基础题. 10. 【答案】D【分析】由题意可得集合B 中的元素个数为1个或3个,分集合B 中的元素个数为1和集合B 中的元素个数为3两种情况,再结合一元次方程根的个数求解即可. 【详解】解:由2220xax x ax 可得20x ax或220x ax ++=,又因为{}1,2A =,1A B ⊗=, 所以集合B 中的元素个数为1个或3个, 当集合B 中的元素个数为1时,则20x ax有两相等的实数根,且220x ax ++=无解,所以22080a a ⎧=⎨−<⎩,解得0a =;当集合B 中的元素个数为3时,则20x ax有两不相等的实数根,且220x ax ++=有两个相等且异于方程20x ax 的根的解,所以20Δ80a a ≠⎧⎨=−=⎩,解得a =a =−综上所述,0a =或a =a =− 故选:D.【点睛】关键点睛:本题的关键是根据题意得出集合B 中的元素个数为1个或3个.第Ⅱ卷(共70分)二、填空题(本大题共5小题,每小题5分,共25分)11. 【答案】(){}3,7−【分析】首先根据方程组求出其解,然后运用列举法表示出对应的解集即可(以有序数对(),a b 的形式表示元素).【详解】因为322327x y x y +=⎧⎨−=⎩,所以37x y =⎧⎨=−⎩,所以列举法表示解集为:(){}3,7−.故答案为(){}3,7−.【点睛】本题考查二元一次方程组解集的列举法表示,难度较易.二元一次方程组的解用列举法表示时,可将元素表示成有序数的形式:(),x y . 12. 【答案】(],2−∞【分析】根据题意得到(1,1)− (25,+)m −∞,再利用数轴得到不等式,解出不等式即可. 【详解】||<1,1<<1x x ∴−>25x m −是||1x <的必要不充分条件,(1,1)∴− (25,+)m −∞,251,2m m ∴−≤−∴≤, ∴实数m 的取值范围是(,2]−∞,故答案为: (,2]−∞. 13. 【答案】0【分析】由集合相等的含义,分类讨论元素对应关系即可. 【详解】由集合元素互异性:0a ≠,又{}2,0,1{,,0}a a b −=,则21a a b ⎧=⎨=−⎩或21a ba ⎧=⎨=−⎩,解得11a b =⎧⎨=−⎩或11a b =−⎧⎨=⎩,故0a b += 故答案为:0 14. 【答案】0a ≥【分析】分别讨论A =∅和A ≠∅两种情况求解.【详解】因为A B ⋂=∅, 若3a >,则A =∅,满足题意;若3a ≤,则应满足0a ≥,所以03a ≤≤, 综上,0a ≥. 故答案为:0a ≥.15. 【答案】 ①. {|0a a <或}1a = ②. 14⎧⎫⎨⎬⎩⎭【分析】分情况解集合B ,再根据“全食”与“偏食”的概念分析集合中元素满足的关系列式求解即可. 【详解】由{}2B x x a ==|可知,当a<0时,B =∅,此时B A ⊆; 当0a =时,{}0B =,此时A B ⋂=∅,当0a >时,{B =; 又11,,12A ⎧⎫=−⎨⎬⎩⎭,若A 与B 构成“全食”,则B A ⊆, 当a<0时,满足题意;当0a =时,不合题意;当0a >时,要使B A ⊆,则{}1,1B =−1=,解得1a =; 综上,A 与B 构成“全食”时,a 的取值范围是{|0a a <或}1a =; 若A 与B 构成“偏食”时,显然0a ≤时,不满足题意,当0a >时,由A B ⋂≠∅,所以11,22B ⎧⎫=−⎨⎬⎩⎭12=,解得14a =,此时a 的取值范围是14⎧⎫⎨⎬⎩⎭.故答案为:{|0a a <或}1a =;14⎧⎫⎨⎬⎩⎭三、解答题(本大题共4小题,共45分.解答应写出文字说明,演算步骤或证明过程)16. 【答案】(1)(1,1]A B ⋂=−,(1,)UA B ⋃=−+∞;(2)(0,1]【分析】(1)解一元一次不等式求集合A ,再应用集合的交并补运算求A B ⋂及()UA B ⋃.(2)由集合的包含关系可得2a ≤2,结合已知即可得a 的取值范围. 【小问1详解】由211x −≤得:1x ≤,所以(,1]A ∞=−,则(1,)UA =+∞,由(1,2]B =−,所以(1,1]A B ⋂=−,(1,)UA B ⋃=−+∞.【小问2详解】 因为C B ⊆且0a >, 所以2a ≤2,解得1a ≤. 所以a 的取值范围是(0,1]. 17. 【答案】(1)34m ≤ (2)1m =−【分析】(1)根据根的判别式列不等式,然后解不等式即可;(2)根据韦达定理得到1223x x m +=−+,212x x m =,然后代入求解即可.【小问1详解】因为有两个实根,所以()222341290m m m ∆=−−=−+≥,解得34m ≤. 【小问2详解】由题意得()122323x x m m +=−−=−+,212x x m =,所以2236m m −+=−,整理得 ()()310m m −+=,解得3m =或-1,因为34m ≤,所以1m =−. 18. 【答案】答案见解析【分析】分别求解集合,A B ,并求解三个条件的集合,再根据必要不充分条件,转化为集合的包含关系,即可列式求解. 【详解】不等式8831100222x x x x x x +++>⇔−>⇔<−−−,即()()320x x +−<, 解得:32x −<<,即{}32A x x =−<<,()()22240220x mx m x m x m −+−<⇔−−−+<⎡⎤⎡⎤⎣⎦⎣⎦,解得:22m x m −<<+,即{}22B x m x m =−<<+, 若选①,{3UA x x =≤−或2}x ≥,:p {3U x A x x ∈=≤−或2}x ≥,{}:22q x B x m x m ∈=−<<+,若p 是q 的必要不充分条件,则BUA ,即23m +≤−或22m −≥,解得:5m ≤−或4m ≥;所以存在实数m ,使得p 是q 的必要不充分条件,m 的范围为5m ≤−或4m ≥; 若选②,{}12A C x x ⋂=−<<,:p {}12x A C x x ∈⋂=−<<,{}:22q x B x m x m ∈=−<<+,若p 是q 的必要不充分条件,则B ()A C ,则2122m m −≥−⎧⎨+≤⎩,解集为∅;所以不存在实数m ,使得p 是q 的必要不充分条件; 若选③,{}34A C x x ⋃=−<<,:p {}34x A C x x ∈⋃=−<<,{}:22q x B x m x m ∈=−<<+,若p 是q 的必要不充分条件,则B ()A C ,则2324m m −≥−⎧⎨+≤⎩,解得:12m −≤≤;所以存在实数m ,使得p 是q 的必要不充分条件,m 的取值范围为12m −≤≤; 19. 【答案】(1)答案见解析 (2)不具有,理由见解析【分析】(1)根据集合A 具有性质J 的定义即可得出答案;(2)当6n =,4m =时,利用反证法即可得出结论. 【小问1详解】当3n =时,{}1,2,3A =,集合A 的所有二元子集为{}{}{}1,2,1,3,2,3,则满足题意得集合B 可以是{}1或{}2或{}3,此时1m =, 或者也可以是{}1,2或{}1,3或{}2,3,此时2m =; 【小问2详解】当6n =,4m =时,{}1,2,3,4,5,6A =,假设存在集合B ,即对任意的()1234,,,,4,114i A A A A B B A i =⋂≤≤≤,则取{}{}{}{}12341,2,3,4,5,6,2,3A A A A ====,(4A 任意构造,符合题意即可) 此时由于4B =,由抽屉原理可知,必有()223i B A i ⋂=≤≤, 与题设矛盾,假设不成立, 所以集合A 是不具有性质J .【点睛】关键点点睛:此题对学生的抽象思维能力要求较高,特别是对数的分析,在解题时注意对新概念的理解与把握是解题的关键.。
长沙第一中学2024—2025学年度高一第一学期阶段性检测化学时量:60分钟 满分:100分 得分:______可能用到的相对原子质量:H~1 O~16 Cl~35.5 Mn~55一、选择题(本题共12小题,每小题4分,共48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.下列有关说法正确的是( )A .属于纯净物B .金属氧化物均为碱性氧化物,非金属氧化物均为酸性氧化物C .由一种元素组成的物质是纯净物D .酸性氧化物均能溶于水生成相应的酸2.下列水溶液中的各组离子因发生氧化还原反应而不能大量共存的是( )A .、、、B .、、、C .、、、D .、、、3.下列对有关物质的分类不正确的是( )选项物质分类不同类物质A 干冰、白酒、加碘盐、食醋混合物干冰B 液氨、纯硫酸、液态氧化钠、生石灰不导电液态氧化钠C 云、烟、雾、淀粉溶液胶体淀粉溶液D铝、铁、锌、氧气还原剂氧气4.下列各组中的反应可以用同一个离子方程式表示的是( )A .氧化钠与稀盐酸混合;氧化铜与稀硫酸B .氢氧化铜、氢氧化钡分别与盐酸反应C .Zn 分别与稀盐酸和稀硝酸反应D .和分别与氢氧化钠溶液反应5.下列几个反应的导电性变化,不符合图像的是()A .向饱和石灰水中不断通入B .向稀硫酸中滴加溶液C .向稀溶液中滴加溶液D .向稀盐酸中滴加溶液6.下列离子方程式书写正确的是()42FeSO 7H O ⋅Na +2Ba+Cl -24SO -2Ca+3HCO -Cl -K+4MnO -K +I -H+H +Cl -Na +23CO -24H SO 4NaHSO 2CO ()2Ba OH ()2Ba OH 4CuSO 3AgNOA .用醋酸除去水垢中的:B .铜与稀硫酸反应:C .和NaOH 溶液混合:D .与稀盐酸反应:7.某溶液中含有较大量的、、三种阴离子,如果只取一次该溶液就能够分别将3种阴离子依次检验出来。
下列实验操作的操作顺序中,正确的是()①滴加溶液 ②过滤 ③滴加溶液 ④滴加溶液A .①②④②③B .④②③②①C .①②③②④D .④②①②③8.利用粗(含有杂质MnO 和)制取纯的流程如图。
2024学年第一学期海宁市静安高级中学10月月考测试高—数学试题卷注意事项:1.本试题共4页,满分150分,考试时间120分钟。
2.答题前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名、试室号、座位号填写在答题卷上。
3.答题必须使用黑色字迹的钢笔或签字笔作答,答案必须写在答题卷上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.考生必须保持答题卷整洁,考试结束后,将答题卷交回,试卷自己保存。
第一部分(选择题共58分)一、选择题:本题共8小题,每小题5分,共40分。
在每小题给出的四个选项中,只有一项是符合题目要求的。
1.集合用列举法表示为()A. B. C. D.2.不等式的解集为( )A. B. C. D.或3.已知二次函数的图象与轴有交点,则的取值范围是()A. B. C.且 D.且4.下列四组函数中,与不相等的是()A.与B.与C.与D.与5.对于,用表示不大于的最大整数,例如:,,则“”是“”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{}N |21x x+∈-≤{0,1,2,3}{1,2,3}{0,1,2,3,4}{1,2,3,4}305x x+<-{|5}x x >{|3}x x <-{|35}x x -<<{|3x x <-5}x >2(3)21y k x x =-++x k 4k <4k ≤4k <3k ≠4k ≤3k ≠()f x ()g x ()||f x x =()g x =2()1f x x =+2()1g t t =+||()x f x x =1,0()1,0x g x x >⎧=⎨-<⎩()f x =()g x =x ∀∈R []x x [π]3=[ 2.1]3-=-[][]x y >x y >6.已知,则的最小值为( )A. B. C. D.7.已知函数的定义域是,则函数的定义域是( )A. B.C. D.8.若两个正实数,满足,且不等式恒成立,则实数的取值范围( )A. B.或C. D.或二、多项选择题(本题共3小题,每小题6分,共18分.在每小题给出的四个选项中,有多项符合题目要求。
中华中学2023—2024学年度第一学期学情调研(二)高一数学本卷调研时间:120分钟总分:150分一、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,只有一项是符合衣有限公司在暑假期间加班生产提供(](0,20)x x ∈(万元)的专项补贴.该制衣有限公司在收到市政府x (万元)补贴后,产量将增加到(3)t x =+(万件).同时该制衣有限公司生产t (万件)产品需要投入成本为36(73)t x t ++(万元),并以每件42(8)t+元的价格将其生产的产品全部售出.注:收益=销售金额+政府专项补贴-成本.(1)求该制衣有限公司暑假期间,加班生产所获收益y (万元)关于专项补贴x (万元)的表达式,并求当加班生产所获收益不低于35万元时,实数x 的取值范围;(2)南京市政府的专项补贴为多少万元时,该制衣有限公司假期间加班生产所获收益y (万元)最大?【解析】(1)4236873y t x t x t t ⎛⎫⎛⎫=+⋅+-++ ⎪⎪⎝⎭⎝⎭36422t x t =+--.因为3t x =+,所以363634224533y x x x x x =++--=--++.................................................3分由35y ≥,得3645353x x --+≥,即2760x x -+≤,所以16x ≤≤,又020x <≤,所以实数x 的取值范围是[1,6]..........................................6分(2)因为36453y x x =--+()363483x x ⎡⎤=-+++⎢⎥+⎣⎦.(020x <≤)..........................8分又因为(]0,20x ∈,所以3630,03x x +>>+,所以()363123x x ++≥=+(当且仅当36333x x x +==+即时取“=”)所以124836y ≤-+=,即当3x =万元时,y 取最大值36万元............................................11分答:南京市政府的专项补贴为3万元时,该制衣有限公司假期间加班生产所获收益最大....12分22.(12分)已知函数2()3f x x ax =++,Ra ∈(1)若函数)(1x f y =的定义域为R ,求实数a 的取值范围;(2)若当[]2,2x ∈-时,函数a x f y -=)(有意义,求实数a 的取值范围.(3)若函数a x a x f x g +--=)2()()(,函数)]([x g g y =的最小值是5,求实数a 的值.【解析】由)(1x f y =定义域为R ,则2()3f x x ax =++的值域大于0,所以2120a ∆=-<,所以(a ∈-........................................2分(2)由[2,2],x y ∈-=有意义,即()0f x a -≥恒成立,令2()()3,[2,2]h x f x a x ax a x =-=++-∈-最小值非负,221()(3,[2,2].24a h x x a a x =+--+∈-①当22a-<-即4a >时,()h x 在[2,2]-单调递增,min ()(2)73h x h a =-=-,所以4477303a a a a >⎧>⎧⎪⇒⎨⎨-≤≤⎩⎪⎩,所以a φ∈;................................4分②当222a-≤-≤即44a -≤≤时,()h x 在[2,2]-先单调递减后递增,2min1()()324a h x h a a =-=--+,所以224444441623041204a a a a a a a a -≤≤⎧-≤≤-≤≤⎧⎧⎪⇒⇒⎨⎨⎨-≤≤--+≥+-≤⎩⎩⎪⎩,所以[4,2]a ∈-;......6分③当22a->即4a <时,()h x 在[2,2]-单调递减,min ()(2)7h x h a ==+,所以44707a a a a <-<-⎧⎧⇒⎨⎨+≤≥-⎩⎩,所以[7,4)a ∈--综上:[7,2]a ∈-...............................................................8分(3)222()3(2)23(1)22g x x ax a x a x x a x a a =++--+=+++=+++≥+.令22()2,[()]23(1)2t g x a y g g x t t a t a =≥+==+++=+++....................9分①当21a +<-,即3a <-,min 25y a =+=,所以25333a a a a +==⎧⎧⇒⎨⎨<-<-⎩⎩无解;.....10分②当21a +≥-,即3a ≥-,2min (2)2(2)35y a a a =+++++=,所以231(2)3(2)40a a a a ≥-⎧⇒=-⎨+++-=⎩;.....................................11分综上: 1.a =-...............................................................12分。
2024~2025学年度高中同步月考测试卷(一)高一数学测试模块:必修第一册考生注意:1.本试卷分选择题和非选择题两部分.满分150分,考试时间120分钟.2.答题前,考生务必用直径0.5毫米黑色墨水签字笔将密封线内项目填写清楚.3.考生作答时,请将答案答在答题卡上.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑;非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答,超出答题区域书写的答案无效,在试题卷、草稿纸上作答无效.4.本试卷主要命题范围:北师大版必修第一册第一章.一、选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合,则集合的子集个数为( )A .4B .8C .10D .162.不等式的解集为( )A . B . C . D .3.已知集合,若,则实数a 的值为( )A .B .3C .3或D .64.已知实数a ,b ,c ,d 满足,则下列结论正确的是( )A .B .C .D .5.已知关于x 的不等式的解集为,其中a ,b ,c 为常数,则不等式的解集是( )A .B .C .D .6.某校高一年级组织趣味运动会,有跳远球类跑步三项比赛,共有24人参加比赛,其中有12人参加跳远比赛,有11人参加球类比赛,有16人参加跑步比赛,同时参加跳远和球类比赛的有4人,同时参加球类和跑步比赛的有5人,没有人同时参加三项比赛,则( )A .同时参加跳远和跑步比赛的有4人B .仅参加跳远比赛的有3人{2,3,4},{0,1}A B =={,,}C z z x y x A y B ==+∈∈∣342x ≤-1124x x ⎧⎫<≤⎨⎬⎩⎭,2114x x x ⎧⎫≥<⎨⎬⎩⎭或1124x x ⎧⎫≤≤⎨⎬⎩⎭11,24x x x ⎧⎫≥≤⎨⎬⎩⎭或{,||,3}A a a a =-3A ∈3-3-0a b c d >>>>a d b c ->-ab cd >a c b d ->-ac bd>20ax bx c ++>{27}xx -<<∣20cx bx a ++≤211,7x x x ⎧⎫≤-≥⎨⎬⎩⎭或11,27x x x ⎧⎫≤-≥⎨⎬⎩⎭或1127x x ⎧⎫-≤≤⎨⎬⎩⎭1172x x ⎧⎫-≤≤⎨⎬⎩⎭C .仅参加跑步比赛的有5人D .同时参加两项比赛的有16人7.已知全集U ,集合M ,N 满足,则( )A . B .C .D .8.已知实数x 满足,则的最小值为( )A .9B .18C .27D .36二、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.下列结论中正确的是( )A .B .C .D .10.已知,若q 是的充分条件,则q 可以是( )A .B .C .D .11.定义,则下列说法正确的是( )A .B .对任意的且C .若对任意实数恒成立,则实数a 的取值范围是D .若存在,使不等式成立,则实数a 的取值范围是三、填空题:本题共3小题,每小题5分,共15分.12.命题“”的否定是_________.13.已知集合,若,则实数m 的最大值为__________.14.已知实数a ,b 满足,且,则的最小值为____________.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.(本小题满分13分)已知集合.(1)若成立的一个必要条件是,求实数m 的取值范围;(2)若,求实数m 的取值范围.16.(本小题满分15分)M N U ⊆⊆()()U U M N =∅ ððM N M = ()U M N M = ð()()U U M N M= ðð103x <<11213x x+-0∈∅{0}=∅{}∅∈∅{0}∅⊆:2p x ≥p ⌝3x ≥1x ≤2x >0x <*(1)(1)x y x y =+-1*33*2=2x >-111,*112x x x≠-=++,(1)*(23)33x x a x a ----≥--{13}aa -<<∣2x ≥(1)*(23)33x a x a ----≤--27a a ⎧⎫≥⎨⎬⎩⎭23,430x x x ∈++=R {3,2,0,2,3},{}M N xx m =--=≥∣M N M = 11a b -<<<2a b +=1311aa b ++-{26},{22}A xx B x m x m =-<<=-<<+∣∣x B ∈x A ∈A B =∅记全集,集合,.(1)若,求;(2)若,求a 的取值范围;(3)若,求a 的取值范围.17.(本小题满分15分)已知实数a ,b 满足.(1)求实数a ,b 的取值范围;(2)求的取值范围.18.(本小题满分17分)如图所示,为宣传某运动会,某公益广告公司拟在一张矩形海报纸上设计大小相等的左右两个矩形宣传栏,宣传栏的面积之和为,为了美观,要求海报上四周空白的宽度均为,两个宣传栏之间的空隙的宽度为,设海报纸的长和宽分别为.(1)求y 关于x 的函数表达式;(2)为节约成本,应如何选择海报纸的尺寸,可使用纸量最少?19.(本小题满分17分)已知:,q :关于x 的方程的两根均大于1.(1)若p 为真命题,求实数a 的取值范围;(2)若p 和q 中一个为真命题一个为假命题,求实数a的取值范围.U =R {221,}A xa x a a =-≤≤+∈R ∣{3,7}B x x x =≤≥∣或4a =()U A B ðA B =R A B A = 18,34a b a b ≤+≤≤-≤25a b -2700dm 2dm 3dm dm,dm x y 2:1,30p x x ax a ∀≥---+≥2260 x ax a -+-=2024~2025学年度高中同步月考测试卷(一)·高一数学参考答案、提示及评分细则1.D ,故其子集的个数为16.故选D .2.B 不等式,即,等价于解得或,所以原不等式的解集为.故选B .3.A 由,,则,不符合集合元素的互异性;若,则或(舍),,此时符合集合元素的特征;若,即,则不符合集合元素的互异性.故.故选A .4.A 对于A ,,所以,则,故A 正确;对于BCD ,取,,,,满足,显然,,故BCD 错误.故选A .5.C 关于x 的一元二次不等式的解集为,则,且,7是一元二次方程的两根,于是解得则不等式化为,即,解得,所以不等式的解集是.故选C .6.C 设同时参加跳远和跑步比赛的有x 人,由题意画出韦恩图,如图,则,解得,故A 错误;仅参加跳远比赛的人数为,故B 错误;仅参加跑步比赛的人数为,故C 正确;同时参加两项比赛的人数为,故D 错误.故选C .{}2,3,4,5C =342x ≤-11402x x -≤-(114)(2)0,20,x x x --≤⎧⎨-≠⎩114x ≥2x <11,24x x x ⎧⎫≥<⎨⎬⎩⎭或3A ∈3a =||3a =||3a =3a =-3a =36a -=-{3,3,6}A =--33a -=6a =||6a =3a =-0a b c d >>>>0d c ->->a d b c ->-2a =1b =2c =-4d =-0a b c d >>>>28,45ab cd a c b d =<=-=<=-4ac bd =-=20ax bx c ++>{27}xx -<<∣0a <2-20ax bx c ++=0,27,27,a b a c a ⎧⎪<⎪⎪-+=-⎨⎪⎪-⨯=⎪⎩5,14,0,b a c a a =-⎧⎪=-⎨⎪<⎩20cx bx a ++≤1450ax ax a --+≤2214510x x +-≤1127x -≤≤20cx bx a ++≤1127x x ⎧⎫-≤≤⎨⎬⎩⎭84251124x x x -+++++-=6x =862-=1165-=46515++=7.B 全集U ,集合M ,N 满足,绘制图,如图:对于A:,故A 错误;对于B:,故B 正确;对于C:,故C 错误;对于D:,故D 错误.故选B .8.C 因为,所以,又因为,所以(当且仅当,即时等号成立).故选C .9.CD 是不含任何元素的集合,所以是指元素为的集合,所以,故AB 错误,C 正确;是任何集合的子集,所以,故D 正确.故选CD .10.BD 因为条件,所以,对于A ,因为不能推出,所以不是的充分条件,故A 错误;对于B ,因为能推出,所以是的充分条件,故B 正确;对于C ,因为不能推出,所以不是的充分条件,故C 错误;对于D ,因为能推出,所以是的充分条件,故D 正确.故选BD .M N U ⊆⊆Venn ()()U U U M N N = ðððM N M = ()U M N =∅ ð()()U U U M N M = ððð103x <<0131x <-<3(13)1x x +-=1123123121336[3(13)]1515271331331313x x x x x x x x x x x x -⎛⎫+=+=+-⨯+=++≥+= ⎪----⎝⎭133613x x x x -=-19x =∅0,{}∉∅∅∅{}∅∈∅∅{0}∅⊆:2p x ≥:2p x <3x ≥2x <3x ≥2x <1x ≤2x <1x ≤2x <2x >2x <2x >2x <0x <2x <0x <2x <11.ABD 对于A ,,即,故A 正确;对于B ,,故B 正确;对于C , 恒成立,即恒成立,则,解得,故C 错误;对于D ,由题可知存在,使得成立,即成立,又,得a 的取值范围是,故D 正确.故选ABD .12. 由特称量词命题的否定为全称量词命题得,命题“”的否定为“”.13. 因为且,所以,则,所以m 的最大值为.14由题易得,则,又,即时等号成立,的最小值为.15.解:(1)是的一个必要条件,,显然,,且,解得,即m 的取值范围为. 6分(2)若,,或,解得,或,即m 的取值范围为,或. 13分16.解:(1)因为,所以,所以,或, 2分1*3(11)(13)4,3*2(13)(12)4=+⨯-=-=+⨯-=-1*33*2=111121*111121212x x x x x x x x++⎛⎫⎛⎫=+-=⋅= ⎪⎪++++++⎝⎭⎝⎭(1)*(23)(11)x a x x a ----=+--2[1(23)]()(33)3(33)333x x a x x a x a a ---=-+=+--≥--2(1)10x a x +-+≥2(1)40a ∆=--≤13a -≤≤2x ≥2(1)10x a x +-+≤11a x x ≥++min 1712x x ⎛⎫++= ⎪⎝⎭72a a ⎧⎫≥⎨⎬⎩⎭2,430x x x ∀∈++≠R 2,430x x x ∃∈++=R 2,430x x x ∀∈++≠R 3-{3,2,0,2,3},{}M N xx m =--=≥∣M N M = M N ⊆3m ≤-3-1-2a b =-13163133111111a b a b a b a b -+=+=+-+-+-+-133(1)1[(1)(1)]13441111a b a b a b b a +-⎛⎫++-+=+++≥+=+ ⎪+--+⎝⎭13211a b ∴+≥++-3(1)111a b b a +-=-+2,4a b ==1311aa b ∴++-231+=x A ∈ x B ∈B A ∴⊆B ≠∅26m ∴+≤22m -≥-04m ≤≤{04}mm ≤≤∣A B =∅ 26m ∴-≥22m +≤-8m ≥4m ≤-{4m m ≤-∣8}m ≥4a ={29}A xx =≤≤∣U {2A xx =<∣ð9}x >所以,或. 4分(2)因为,所以6分解得,故a 的取值范围为. 8分(3)因为,所以,9分①当,即时,,显然满足,符合题意;11分②当,即时,,因为,所以,或,所以,或,14分综上所述,,或,即a 的取值范围为,或. 15分17.解:(1),①,②①②两式相加,得,.3分,③ 5分∴①③两式相加,得, 7分的取值范围为的取值范围为. 8分(2)令,,9分,10分,11分又,,12分, 14分的取值范围为.15分18.解:(1)由题知,两个矩形宣传栏的长为,宽为, 2分U (){2A B x x =< ∣ð9}x >A B =R 23,217,a a -≤⎧⎨+≥⎩35a ≤≤{35}aa ≤≤∣A B A = A B ⊆221a a ->+3a <-A =∅A B ⊆221a a -≤+3a ≥-A ≠∅A B ⊆27a -≥213a +≤9a ≥31a -≤≤1a ≤9a ≥{1aa ≤∣9}x ≥18ab ≤+≤ 34a b ≤-≤4212a ≤≤26a ∴≤≤34,43a b b a ≤-≤∴-≤-≤- 35325,22b b -≤≤∴-≤≤a ∴{26},aa b ≤≤∣3522b b ⎧⎫-≤≤⎨⎬⎩⎭,x a b y a b =+=-,22x y x ya b +-∴==737325()()2222a b y x a b a b ∴-=-=--+21734,()1422a b a b ≤-≤∴≤-≤ 18,8()1a b a b ≤+≤∴-≤-+≤-3312()22a b ∴-≤-+≤-37325()()2222a b a b ∴-≤--+≤25a b ∴-325252522a b a b -⎧⎫⎨-≤≤⎩-⎬⎭72x -4y -, 6分整理得.8分(2)由(1)知,即,10分,∴由基本不等式可得,12分令,解得(舍去)或.14分,当且仅当即时等号成立, 16分∴海报长,宽时,用纸量最少,最少用纸量为. 17分19.解:(1)若p 为真命题,即为真命题,当时,成立,此时;2分当时,,所以在内恒成立, 4分令,则,所以,当且仅当,即时等号成立. 7分所以,故实数a 的取值范围为, 8分(2)设关于x 的方程的两根分别为,则且,所以即11分解得,即实数a 的取值范围为.13分因为p 和q 中一个为真命题一个为假命题,所以p 真q 假,或p 假q 真,当p 真q 假时,所以,15分72(4)7002x y -∴⨯⨯-=7004(7)7y x x =+>-(7)(4)700x y --=47672xy x y =++7,4x y >> 47672672xy x y =++≥+t =26720t --≥t ≤-t ≥1008xy ∴≥47,47672,x y xy x y =⎧⎨=++⎩42,24x y ==42dm 24dm 21008dm 21,30x x ax a ∀≥---+≥1x =-2(1)(11)30a ---++≥a ∈R 1x >-10x +>231x a x +≤+{1}xx >-∣1x t +=1(0)x t t =->2223(1)34242221x t t t t x t t t +-++-===+-≥-=+4t t=2(1)t x ==2a ≤{2}aa ≤∣2260x ax a -+-=12,x x 11x >212121,2,6x x x a x x a >+==-()()()()21212(2)4(6)0,110,110,a a x x x x ⎧---≥⎪-+->⎨⎪-->⎩260,22,6210,a a a a a ⎧+-≥⎪>⎨⎪--+>⎩723a ≤<723a a ⎧⎫≤<⎨⎬⎩⎭2,72,,3a a a ≤⎧⎪⎨<≥⎪⎩或2a <当p 假q 真时,所以,所以实数a 的取值范围为. 17分2,72,3a a >⎧⎪⎨≤<⎪⎩723a <<72,23a a a ⎧⎫<<<⎨⎬⎩⎭∣或。
辽宁省普通高中2024-2025学年度上学期10月月考模拟试题高一物理注意事项: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.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。
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、选择题:本题共10小题,共46分。
在每小题给出的四个选项中,第1~7题中只有一项符合题目要求,每小题4分;第8~10题有多项符合题目要求,每小题6分,全部选对的得6分,选对但不全的得3分,有选错的得0分。
1.在2018年11月6日上午9时开幕的第十二届航展上,中国航天科工集团有限公司正在研发的高速飞行列车仿真模型首次亮相。
该飞行列车利用磁悬浮技术及近真空管道线路减小阻力,未来项目落地时最大运行速度可达4000km/h ,从郑州到北京的路程为693km ,只要12min 就可到达,真是“嗖”的一声,人就到了。
根据以上信息判断,下列说法正确的是( ) A .“2018年11月6日上午9时”和“12min”都是时刻 B .从郑州到北京的路程为693km ,“路程”是矢量 C .飞行列车从郑州到北京的平均速率为3465km/hD .若研究飞行列车经过某一路标所用的时间,可将列车看成质点2.将一弹力球从某一高度处由静止释放,弹力球从地面弹起的最大高度不到释放高度的一半,不计球与地面相互作用的时间和空气阻力,以向下为正方向,下列关于弹力球运动的v t −和a t −图像,可能正确的是( )A .B .C .D .3.如图所示,t =0时刻,一个物体以08m /s v =的初速度沿光滑斜面向上滑动,加速度的大小为2m/s 2,运动到最高点之后,又以相同的加速度往回运动,则物体( ) A .第2s 末的速度大小为4m/s B .前3s 内的位移是9m C .第4s 末的加速度为零 D .前5s 内的路程是15m4.如图所示,一列“和谐号”动车,每节车厢的长度均为l ,列车启动过程中可视为匀加速直线运动,列车员站在列车一侧的站台上,已知第3节车厢经过列车员的时间为t 1,第4节车厢经过列车员的时间为t 2,则列车的加速度为( )A .l (t 1-t 2)t 1t 2(t 1+t 2)B .l (t 1+t 2)t 1t 2(t 1-t 2)C .2l (t 1+t 2)t 1t 2(t 1-t 2)D .2l (t 1-t 2)t 1t 2(t 1+t 2)5.一长为L 的金属管从地面以0v 的速率竖直上抛,管口正上方高()h h L >处有一小球同时自由下落,金属管落地前小球从管中穿过.已知重力加速度为g ,不计空气阻力.关于该运动过程说法正确的是( )A .小球穿过管所用时间大于LB .若小球在管上升阶段穿过管,则0>vC 0v<<D .小球不可能在管上升阶段穿过管6.某人驾驶一辆汽车甲正在平直的公路上以某一速度匀速运动,突然发现前方50m 处停着一辆乙车,立即刹车,刹车后做匀减速直线运动。
河南省许昌高级中学2024-2025学年高一上学期10月月考物理试题一、单选题1.高速公路的测速有两种:一种是定点测速,定点测速不超过限速的10%不会被处罚;另一种是区间测速,区间测速由两个测速点组成,分别是区间的起点位置和终点位置,在起点记录通过的时刻,在终点再记录一次通过的时刻,根据区间的里程和前后两次时间间隔,确定该路段是否超速行驶。
某段高速路上的定点测速提示牌如图甲所示,该路段的区间测速提示牌如图乙所示,下列说法正确的是()A.该路段小车的区间限速为120m/sB.图乙中的100是指该区间大车行驶的最大平均速度C.若小车某时刻瞬时速度达到35m/s,则该小车一定会受到处罚D.若小车通过该区间测速路段的时间为7min,则该小车未违反区间测速的限速规定2.如图所示,将弹性小球以10m/s的速度从距地面2m处的A点竖直向下抛出,小球落地后竖直向上反弹,经过距地面1.5m高的B点时,向上的速度为7m/s,从A到B过程,小球共用时0.3s,则此过程中()A.小球的位移大小为0.5m,方向竖直向上B.小球速度变化量的大小为3m/s,方向竖直向下C.小球平均速度的大小为8.5m/s,方向竖直向下D.小球平均加速度的大小约为256.7m/s,方向竖直向上3.新情境扫地机器人因操作简单、使用方便,已逐渐走进了人们的生活。
某次清扫过程中,主人在A处启动扫地机器人,在B处完成清扫工作,其规划清扫路线如图所示,完成清扫任务用时180s。
数据表明:机器人清扫路线的总长度为36m,A,B两点间的距离为9m。
下列说法正确的是()A.机器人在该过程中的位移大小为36mB.机器人在该过程中的平均速率为0.05m/sC.机器人在该过程中的平均速度大小为0.05m/sD.机器人在B点的瞬时速率大小一定为0.2m/s4.交警通常用超声波来测量高速路上的汽车是否超速,已知发射装置间隔2s的时间发出一个超声波。
假设一辆汽车在平直的高速公路上匀速行驶,0时刻发出第一个超声波,经过2s 的时间发射装置接收到汽车反射回来的超声波,同时发射第二个超声波,再经过1.6s的时间发射装置接收到汽车反射回来的第二个超声波,已知声速为330m/s,该路段的限速为120km/h。
高一年级10月月考数学试题考试时长:60分钟 满分100分一.选择题(每小题5分,共60分)1.命题“2,220x x x ∃∈++≤R ”的否定是( )A .2,220x R x x ∀∈++>B .2,220x R x x ∀∈++≤ C .2,220x R x x ∃∈++> D .2,220x R x x ∃∈++≥ 2.已知全集U R =,集合{}2|230A x x x =-->,集合{|||2}B x x =≤.则下图的阴影部分表示的集合为( )A .{}12x x -≤<B .{}23x x -<≤C .{}23x x <≤D .{}13x x -≤≤ 3.已知13a -≤≤,24b ≤≤,则2a b -的取值范围是( ).A .624a b -≤-≤B .0210a b ≤-≤C .422a b -≤-≤D .521a b -≤-≤ 4.设集合22{,},{0,,}A a b B a b ==-,若A B ⊆,则a b -=( )A .2-B .2C .2-或2D .05.设0,10a b <-<<,则下列选项中正确的是( )2.A a ab ab >> 2.B ab ab a >> 2.C ab ab a >>2.D ab a ab >> 6.在下列四个命题中,①若p 是q 的充分不必要条件,则q 是p 的必要不充分条件;②若0,0a b d c >><<,则ac bd >;③“2430x x -+≥”是“2x >”的必要不充分条件; x x的最小值为2 正确的个数为( )A .1B .2C .3D .47.若不等式22221463x mx m x x ++<++对一切实数x 恒成立,则实数m 的取值范围是( ) A .{}13m m << B .{}3m m < C .{|1m m <或}2m > D .R8.己知命题p : “关于x 的方程240x x a -+=有实根”,若非p 为真命题的充分不必要条件为31a m >+,则实数m 的取值范围是( ) A. {}1x x ≥ B {}1x x >. C. {}1x x < D. {}1x x ≤9.关于x 的不等式22280(0)x ax a a --<>的解集为12(,)x x ,且:2115x x -=,则a=( )A .52 B .72 C .154 D .152 10.若110a b <<,则下列不等式:(1),(2),(3),(4)2b a a b ab a b a b a b +<<<+>中, 正确不等式的序号是( )A .(1)(2) B .(2)(3) C .(3)(4) D .(1)(2)(4)11.已知不等式220x ax a -+->的解集为{}12x x x x x <>或,其中120x x <<,则 121211x x x x +++的最大值为( ) A .32 B .32- C .2 D .0 12.设,,,a b c d 均为大于零的实数,且1abcd =,令()()m a b c d b c d cd =+++++, 则22a b m ++的最小值( ).8.4.5A B C D ++二.填空题(每小题4分,共16分)13.2a 与1a -的大小关系为________.14.已知二次函数2(0)y ax bx c a =++≠,且不等式20(0)ax bx c a ++>≠的解集为{}12x x <<,若二次函数的最大值等于1,则a 的值为________.15.设0,0x y >>,且2116()y x y x -=,则当1x y +取最小值时,221x y+=________.16.若关于x 的不等式22(21)x ax -<的解集中恰好有3个整数,则实数a 的取值范围是________.三.解答题17.(12分)若关于x 的不等式()22210xa x a a -+++≤的解集为A ,不等式322x -≥的解集为B .(1)求集合A ;(2)已知x B ∈是x A ∈的必要不充分条件,求实数a 的取值范围.18.(12分)食品加工厂要用某一侧原有墙体建造一间墙高为3米,底面积为12平方米,且背面靠墙的长方体形状的保管员室,由于此保管员室的后面靠墙,无需建造费用,因此甲工程队给出的报价为:屋子前面新建墙体的报价为每平方米400元,左右两面新建墙体报价为每平方米150元,屋顶和底面以及其它报价共计7200元,设屋子的左右两侧墙的长度均为x 米,(26)x ≤≤(1)当左右两面墙的长度为多少时,甲工程队报价最低?(2)现有乙工程队也参与此保管室建造竞标,其给出的整体报价为900(1)a x x+元(0a >),若无论左右两面墙的长度为多少米,乙工程队都能竞标成功,试求a 的取值范围.石家庄二中高一年级10月开学测数学试题答案一.选择题1-12 ACACC BABAD DB二.填空题14. 4- 13. 21a a >- 15. 12 16.2549916a <≤ 三.解答题17.(1)原不等式可化为:()()10x a x a --+≤⎡⎤⎣⎦,解得1a x a ≤≤+, 所以集合{}|1A x a x a =≤≤+;(2)不等式322x -≥可化为:321222x x x--=≥--0, 等价于()()212020x x x --≥⎧⎪⎨-≠⎪⎩,解得122x ≤<,所以集合1|22B x x ⎧⎫=≤<⎨⎬⎩⎭, 因为x B ∈是x A ∈的必要不充分条件,所以A B ,故1212a a ⎧≥⎪⎨⎪+<⎩,解得112a ≤<. 18.。
河南省青桐鸣大联考2024-2025学年高一上学期10月月考数学试卷一、单选题 1.命题“0x ∃>,21x x<”的否定为( ) A .0x ∀≤,21x x < B .0x ∀>,21x x ≥ C .0x ∀≤,21x x ≥ D .0x ∃≤,21x x ≥ 2.已知集合{}|24A x x =∈-<<Z ,{}|32B x x =∈-<≤N ,则A B =I ( ) A .{}|23x x -<< B .{}1,0,1,2-, C .{}0,1,2D .{}1,23.在下列函数中,与函数y )A .321x xy x +=+B .21x xy x +=-C .y =D .3xy x=4.已知0x ≥,0y ≥,且2a y =-,b x =a ,b 的大小关系为( ) A .a b >B .a b <C .a b ≥D .a b =5.已知a ,b ,c 为实数,则“>是“a c b c ->-”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6.已知关于x 的不等式240mx mx --≥的解集为∅,则实数m 的取值范围为( ) A .()4,0-B .(]4,0-C .(]16,0-D .[)16,0-7.某体育局为调查学生观看第33届巴黎奥运会的情况,统计了某高中高一(1)班55名学生的观看情况:55名学生观看比赛项目都集中在球类比赛、水上运动比赛、田径比赛这三类,其中8名学生只观看了球类比赛,5名学生只观看了水上运动比赛,6名学生只观看了田径比赛,既观看过球类比赛又观看过水上运动比赛的学生有24名,既观看过球类比赛又观看过田径比赛的学生有20名,既观看过水上运动比赛又观看过田径比赛的学生有18名,则该班这三类比赛都观看过的学生人数为( ) A .10B .11C .12D .138.函数()1,4,11x x f x x x x ⎧≤⎪=⎨+>⎪-⎩的值域为( ) A .[)5,5,4⎛⎤-∞+∞ ⎥⎝⎦UB .5,54⎡⎤⎢⎥⎣⎦C .[)3,4,4⎛⎤-∞+∞ ⎥⎝⎦UD .3,44⎡⎤⎢⎥⎣⎦二、多选题9.下列命题中是真命题的有( ) A .0x ∀≥,21x x >- B .0x ∃<,2x x =C .“1x <”是“2560x x -+>”的充分不必要条件D .“四边形为菱形”是“四边形为正方形”的充分不必要条件10.已知0a >,0b >,且2a b ab +=,则下列结论一定正确的是( )A .1a >B .3ab ≥C .210a b +≥D .1129216a b ⎛⎫+≤ ⎪⎝⎭11.已知非空集合A ,B ,定义{|A B x x A -=∈且}x B ∉,{|A B x x A B ⊗=∈U 且}x A B ∉I ,则下列结论一定正确的是( )A .()A AB B -=ðB .()()A B A B B A ⊗=--UC .当A B B A ⊗=-时A B ⊆D .当A B B A -=-时,A B ⊗=∅三、填空题12.已知函数()22,12, 1.x x f x x x -+≤⎧=⎨->⎩则12f f ⎡⎤⎛⎫= ⎪⎢⎥⎝⎭⎣⎦. 13.已知函数()f x 满足()1122f f x x x ⎛⎫+= ⎪⎝⎭,则()1f =.14.设{}m a x ,,,abcd 表示a ,b ,c ,d 中最大的数,已知x ,y 均为正数,则12max ,,2,x y x y ⎧⎫⎨⎬⎩⎭的最小值为.四、解答题15.已知全集{}1,2,3,4,5,6U =,集合{},4,5A a =,{}1,3,4,6B =. (1)若(){}2,3,4,5U A B =U ð,求实数a 的值;(2)若()()U U A B ⋂=∅痧,求出集合A 的所有真子集. 16.按要求完成下面问题.(1)已知21a -<<-,46b <<,求ba的取值范围;(2)已知,a b ∈R ,证明“0ab ≤”是“a b a b -=+”的充要条件.17.某厂共投资6百万元建设A ,B 两条高端精密产品的生产线,经预测,A 生产线投资x百万元,其年利润()A G x B 生产线投资()6x -百万元,其年利润()B G x .已知06x <<,当1x =时,A 生产线年利润为2百万元;当5x =时,B 生产线年利润为8百万元;当4x =时,两条生产线年利润之和为(10+百万元.设两条生产线年利润之和()()()A B f x G x G x =+. (1)求()f x 的解析式;(2)如何投资才能使()f x 取得最大值?求此时()f x 的最大值. 18.已知函数()f x =的定义域为A ,集合{}|24B x x =≤≤,且A B =∅I ,A B ⋃=R(1)求实数a ,b 的值.(2)设集合{}|224C x m x m =-<<+. ①若C A ,求正数m 的最小值;②若A C A ⋃≠,且A C I 中只含有两个正整数元素,求实数m 的取值范围. 19.已知0a >,0b >. (1)若22a b +=,证明2274a b +≥; (2)若2a b +=,不等式()4402mm a b+≥>+恒成立,求m 的取值范围;c>的最大值.(3)若0。
高一10月月考英语测试题本试题卷共4 页,4大题。
全卷满分150分。
考试用时120 分钟。
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第I卷第一部分:听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)1. What is the woman going to do this evening?A. Go to the dinner party.B. Go to the park.C. Visit her sister.2. What time is it now?A. 8:45B. 8:30C. 8:153. What does the man mean?A. He had a good time at the party.B. He doesn’t want to see his friends again.C. He didn’t enjoy the party at all.4. What does the woman suggest the man do?A. Go on a diet.B. Talk to her sister.C. Introduce himself first.5. Where is Mr. Black now?A. In the office.B. Out for lunch.C. At the Friendship Hotel.第二节(共15小题;每小题1.5分,满分22.5分)听第6段材料,回答第6至8题。
6. What are the two speakers doing?A. They are talking in the man’s office.B. They are talking over the phone.C. They are talking at a university.7. How did the woman learn about the job?A. A friend told her about it.B. She saw the advertisement on TV.C. She read the advertisement in newspaper.8. Why is the woman interested in the job?A. Because it paid well.B. Because it is in a university.C. Because it is near her home.听第7段材料,回答第9至11题。
9. Where is the man going?A. He is going to the Palace Museum.B. He is going to the Palace Hotel.C. He is going to the railway station.10. How is he going?A. By bus.B. By subway.C. By train.11. How long does it take?A. Half an hour.B. 13 minutes.C. 40 minutes. 听第8段材料,回答第12至14题。
12. What will happen on Saturday?A. There will be a football match.B. There will be a baseball match.C. There will be a basketball match.13. Who will they play against?A. A team from Iowa.B. A team from Michigan.C. A team from Chicago.14. What will the man do next?A. He will go to have a good rest in bed.B. He will go on practicing.C. He will go to see some films of games.听第9段材料,回答第15至17题。
15. Why can’t the man find Susan’s home?A. Because he hasn’t been told how to get there.B. Because he forgot to bring the map Susan gave him.C. Because he is so forgetful that he can’t remember her address.16. Where is the bank?A. It is next to the bus stop.B. It is on the corner beside the bus stop.C. It is on the corner across the street.17. Where is Susan’s house?A. It is the seventh house from the corner on the man’s right.B. It is the seventeenth house from the corner on the man’s left.C. It is the seventeenth house from the corner on the man’s right.听第10段材料,回答第18至20题。
18. How many countries did Mr. Bell visit last summer?A. Only one.B. Three.C. Four.19. Why did Mr. Bell feel very tired after returning home?A. He visited too many places in a hurry.B. His holiday was not long enough for him to travel in Europe.C. He was on the train or on a bus all day long during his holiday.20. What will be a good idea to spend a two-week summer holiday?A. Visiting only one city by the sea with a friend.B. Visiting only a few cities near the sea with some friends.C. Visiting some museums and famous places in different cities.第二部分阅读理解(共两节,满分40分)ADiscoverNewsmagazine of science devoted to the wonders and stories of modern science, written for the educated general reader. Published by Disney Magazine Publishing Co., Discover tells many of the same stories professionals (专业人员) read in Scientific American. A truly delightful family science magazine, each issue (每期) brings to light new and newsworthy topics to make dinnertime and water-cooler conversations interesting.Cover Price: $59.88Price: $19.95 ($1.66/issue)You Save: $39.93 (67%)Issues: 12 issues/12 monthsSelfPublished by Conde Nast Publications Inc., Self is a handbook devoted to women’s overall physical (身体的) and mental health. Every issue contains usable articles such as “Style Lab”, in which wearable clothes are mixed and matched on non-models and the “Eat-right Road Map”, with tips on how to eat properly.Cover Price: $35.86Price: $15.00 ($2.5/issue)You Save: $20.86 (58%)Issues: 6 issues/12 monthsInStyleInStyle is a guide (手册) to the lives and lifestyles of th e world’s famous people. The magazine covers the choices people make about their homes, their clothes and their free time activities. With photos and articles, it opens the door to these people’s homes, families, parties and weddings, offering ideas about beauty, fitness and in general, lifestyles. Publisher: The Time Inc. Magazine Company.Cover Price: $47.88Price: $23.88 ($2.38/issue)You Save: $24.00 (50%)Issues: 10 issues/12 monthsWiredThis magazine is designed for leaders in the field of information engineering including top managers and professionals in the computer, business, design and education industries. Published by Conde Nast Publications Inc., Wired often carries articles on how technology changes people’s lives.Cover Price: $59.40Price: $10.00 ($1.00/issue)You Save: $49.40 (83%)Issues: 10 issues/12 months21. Which two magazines are published by the same publisher?A. Wired and InStyle.B. Discover and InStyle.C. Self and Discover.D. Self and Wired.22. Which magazine offers the biggest price cut?A. InStyle.B. Wired.C. Discover.D. Self.23. The “Style Lab” in Self provides readers with articles which .A. offer advice to ordinary women on clothesB. show how a woman can become famousC. introduce places with the best foodD. discuss ways of training models 24. Those who are interested in management and the use of high technology would probably choose .A. InStyle.B. Self.C. Wired.D. Discover.BYears ago, I lived in a building in a large city. The next building was only a few feet away from mine. There was a woman living there, and I had never met her, yet I could see she sat by her window each afternoon, sewing or reading.After several months had gone by, I began to notice that her window was dirty. Everything was unclear through the dirty window. I would say to myself, “I wonder why that woman doesn’t wash her window. It really looks terrible.”One bright morning I decided to clean my flat, including washing the window on the inside.Late in the afternoon when I finished the cleaning, I sat down by the window with a cup of coffee for a rest. What a surprise! Across the way, the woman sitting by her window was clearly visible (可见的). Her window was clean!Then it dawned on me. I had been criticizing (批评) her dirty window, but all the time I was watching hers through my own dirty window.That was quite an important lesson for me. How often had I looked at and criticized others through the dirty window of my heart, through my own shortcomings?Since then, whenever I wanted to judge someone, I asked myself first, “Am I looking at him through my own dirty window?” I try to clean the window of my own world so that I may see the world abo ut me more clearly.25. “It dawned on me” in paragraph5 probably means “_______”.A. I began to understand itB. it cheered me upC. I knew it grew lightD. it began to get dark26. It’s clear that ________.A. the writer had never met the woman beforeB. the writer often washed the windowC. they both worked as cleanersD. they lived in a small town27. From the passage, we can learn _______.A. one shouldn’t criticize others very oftenB. one should often make his windows cleanC. one must judge himself before he judges othersD. one must look at others through his dirty windowsCBooks are for reading, but man must bring to their reading a desire to learn and a power of absorbing. Reading should be active, not passive.When students first go to a library, they may be at a loss as to what to read of all the different subjects. Well, Bacon tells you to “look at weak places in your armor(盔甲)”, and shows you how to fill up the blanks in your knowledge. On the other hand, it is no good just trying to fill your mind with knowledge. Knowledge in itself is often useless. A mind filled with too much knowledge is like a room too full of furniture; a man cannot walk about freely in it, and look out of the windows. It is much better to concentrate on a few subjects which interest you and to deal lightly with the others than to march heavily through the whole range(范围) of learning, like a silly tourist going through a museum and not missing a single object. If you try to master every subject you may become very wise, but you will be very lonely and you will probably lose all your friends. So you must learn to pick and choose, and you must also learn to look here and there in a library likea camel eating grass on the grass land. If you watch it eating, you will see that although it is supposed to be one of the most stupid animals in the world, it has at least one of the qualities of the cultured man, the power to pick and choose. A student looking for mental food in a library should take the camel as his model.28. The writer thinks that one must ______.A. read as many books as he canB. read books on the subjects that interest himC. read and absorb a lotD. try to read books on all the different subjects29. A cultured man is similar to a camel because_______.A. neither of them is interested in knowledgeB. both of them have the ability to selectC. neither of them can be considered wiseD. the man reads books as much as a camel eats grass30. The term “mental food” refers to _______.A. grasslandB. subjectsC. booksD. brain31. In the second paragraph the writer mainly discusses_______.A. how to compare furniture with booksB. why books must be absorbedC. how to select reading materialsD. why students go to a libraryDGoing green seems to be fad(时尚)for a lot of people these days. Whether that is good or bad, we can’t really say, but for the two of us, going green is not a fad but a lifestyle.On April 22, 2011, we decided to go green every single day for an entire year. This meant doing 365 different green things, and it also meant challenging ourselves to go green beyond easy things. Rather than recycle and reduce our energy, we had to think of 365 different green things to do and this was no easy task.With the idea of going green every single day for a year, Our Green Year started. My wife and I decided to educate people about how they could go green in their lives and hoped we could show people all the green things that could be done to help the environment. We wanted to push the message that every little bit helps.Over the course of Our Green Year, we completely changed our lifestyle. We now shop at organic(有机的)stores. We consume less meat, choosing green food. We have greatly reduced our buying we don’t need. We have given away half of what we owned through websites. Our home is kept clean by vinegar and lemon juice, with no chemical cleaners. We make our own butter, enjoying the smell of home-made fresh bread. In our home office, anyone caught doing something ungreen might be punished.Our minds have been changed by Our Green Year. We are grateful for the chance to have been able to go green and educate others. We believe that we do have the power to change things and help our planet.32. What might be the best title for the passage?A. Going Green.B. Protecting the Planet.C. Keeping Open-MindedD. Celebrating Our Green Year.33. It was difficult for the couple to live a green life for the whole year because _________.A. they were expected to follow the green fadB. they didn’t know how to educate other peopleC. they were unwilling to reduce their energyD. they needed to perform different green tasks34. What did the couple do over the course of Our Green Year?A. They tried to get out of their ungreen habits.B. They ignore others’ ungreen behavior.C. They chose better chemical cleaners.D. They sold their home-made food. 35. What can we infer form the last paragraph?A. The government will give support to the green people.B. The couple may continue their project in the future.C. Some people disagree with the couple’s green ideas.D. Our Green Year is becoming a national campaign.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。