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2022年高考物理复习重点专练33 机械振动(解析版)

2022年高考物理复习重点专练33 机械振动(解析版)

高三物理复习专题专练(力学部分)专题33 机械振动【典例专练】一、简谐运动1.如图甲所示,弹簧振子以O 点为平衡位置,在光滑水平面上的A 、B 两点之间做简谐运动。

取水平向右为正方向,振子的位移x 随时间t 的变化如图乙所示,下列说法正确的是( )A .0.6s t =时,振子在O 点右侧6cm 处B .振子在0.2s t =时和 1.0s t =时的速度相同C .6s t =时,振子的加速度方向水平向右D . 1.0s t =到 1.4s t =的时间内,振子的加速度和速度都逐渐增大 【答案】C【解析】A .由题图乙可知,该振动的振幅、周期为12cm A =, 1.6s T =所以2π1.25πrad/s Tω== 结合振动图像可知,该振动的方程为12sin(1.25π)cm x t =在0.6s t =时,振子的位移1x =12sin(1.25π0.6)cm ⨯=故A 错误;B .由振动图像可知,t =0.2s 时振子从平衡位置向右运动, 1.0s t =时振子从平衡位置向左运动,速度方向不同,故B 错误;C .6s t =时,振子的位移2x =12sin(1.25π6)cm 12cm ⨯=-处在A 点,故此时加速度方向向右,故C 正确;D . 1.0s t =到 1.2s t =的时间内振子向最大位移处运动,速度减小,加速度增大, 1.2s t =到 1.4s t =时间内振子从最大位移处向平衡位置运动,则速度增大,加速度减小,故D 错误。

故选C 。

2.如图所示,一底端有挡板的斜面体固定在水平面上,其斜面光滑,倾角为θ。

一个劲度系数为k 的轻弹簧下端固定在挡板上,上端与物块A 连接在一起,物块B 紧挨着物块A 静止在斜面上。

某时刻将B 迅速移开,A 将在斜面上做简谐运动。

已知物块A 、B 的质量分别为A m 、B m ,若取沿斜面向上为正方向,移开B 的时刻为计时起点,则A 的振动位移随时间变化的图像是( )A .B .C .D .【答案】B【解析】将A 、B 作为一个整体,平衡时,设弹簧压缩量为AB x ,有()sin A B AB m m g kx θ+=移开B 后,A 平衡时,设弹簧压缩量为A x ,有sin A A m g kx θ=设A 振动的振幅为0A ,则0sin B AB A x A m g x kθ-==因取沿斜面向上为正方向,0t =时刻物块A 处于负的最大位移处,故其位移随时间变化的图像应是B 。

中考总复习专项训练题(三)(改错专练)

中考总复习专项训练题(三)(改错专练)

Jim was too frightened not to move.The snake 8.________
A B C D
6.You can sing in English,and so does he.
A B C D
7.Why not ask for help when you are with trouble?
A B C D
8.Ann didn't know how work out the problem in class.
the school,opened the door and went in the 7.________
classroom.That was nice and warm there 8.________
but Miss Jones was happy.But a small boy looked at her for a few 9.________
C D
25.How proudly they were when they heard the good news!
A B C D
26.It took the boy one and a half hour to fall asleep last night.
A B C D
A B C D
23.I hope my son to get on well with his classmates.
A B C D
24.—How did you come here this morning?
A B
—I came here by Mr Smith's car.

期末复习专练:完形填空(三)2020-2021学年人教版下册八年级英语

期末复习专练:完形填空(三)2020-2021学年人教版下册八年级英语

八年级英语期末复习专练:完形填空(三)There was once a small country. It hadn’t rained for a long time in the country. People there were starting to go 1 because of the bad harvests(收成).One day, a group of musicians traveled to the country. They tried to perform for the people.2 with so many problems in the land, no one wanted to3 music. So the musicians tried to find out4 it didn’t rain. They found the sky was overcast(阴沉的).“It has been 5 like this for many months, but not a drop of rain has fallen,” people told them.“Don’t worry. We’ll bring 6 to the country,” the musicians said, and then they began to play music at the top of the highest mountain.The 7 flew out of their instruments and rose into the clouds. And the notes(音符) started to 8 with the clouds. They ran here and there, up and 9 . Before long, the clouds cried with laughter and began to 10 their tears. The musicians brought a happy rain to the country.1.A.bad B.tired C.hungry2.A.When B.Since C.But3.A.listen to B.learn from C.find out4.A.when B.why C.how5.A.snowy B.rainy C.cloudy6.A.rain B.food C.joy7.A.music B.laughter C.cry8.A.help B.fight C.play9.A.in B.out C.down10.A.see B.drop C.stopI've lived in large cities in my whole life, and was used to the fast pace(节奏)of a city atmosphere(气氛). So 11 I was placed in a small town in France for my year abroad, I was more than a little nervous. However, now that I've lived here for a week, I've 12 some veryuseful and interesting things.While I was in a big city, I was 13 in a hurry. I had no time to just 14 or slow down. But in a small town I've noticed that time works in a 15 way. Instead of noisy cars in the morning, I can enjoy 16 the beautiful view brings to me on my way home.The people here are clearly less stressed than those in the nearest city. While I was 17 , I stopped to sit on a bench in the town square. And the next moment an elderly man sat next to me and he started a 18 about the new iPhone 11. This 19 me a great chance to test my French and I know that if I were in a big city, I wouldn't have this experience.Now I enjoy 20 in the small town. And I'm glad to see another way of French life that I have never experienced before.11.A.unless B.when C.because D.until 12.A.missed B.enjoyed C.learnt D.changed 13.A.never B.sometimes C.seldom D.always 14.A.stop B.work C.travel D.reply 15.A.difficult B.important C.funny D.different 16.A.how B.what C.why D.which 17.A.running out B.handing out C.working out D.hanging out 18.A.conversation B.suggestion C.question D.lesson19.A.got B.covered C.gave D.refused 20.A.itself B.himself C.ourselves D.myselfThere was a boy called Tony. He was shy but good at singing. He could sing many songs21 he learned and understood things slowly.One day he sat in a corner of the classroom with tears in his eyes. Mrs. Black, his music teacher came and asked, “What makes you so 22 ?” Tony answered, “We will have a singing competition. I am afraid I would lose it and other boys 23 me and call me ‘Slow Tony' again.” In a kind 24 , Mrs. Black said, “Look at the snail(蜗牛) on that wall. As far as we know, it is slow, right? When you 25 heart, think of the snail. It can give you strength(力量). ” Tony thought he could run a race with the snail. So he decided to practice his songs until the snail got to the 26 of the wall.At last, the day came. Tony tried his best to sing his songs in the competition. He sang very well and won the 27 . His monitor asked, “How could you learn the songs so well?” “The snail on the wall taught me 28 to do it.” said Tony, “It didn't stop, nor turn back, but went on. And I thought I would finish my task in the same way. By the time the snail got to the top of the wall, I had 29 it all.” “Well done, Tony!” said the tall boy happily, “Now, class, 30 should cheer for Tony and the snail on the wall.”21.A.unless B.until C.although D.because 22.A.proud B.bored C.excited D.sad23.A.look at B.laugh at C.fall behind D.believe in 24.A.voice B.smile C.face D.joke25.A.open B.give C.lose D.have26.A.top B.middle C.bottom D.side27.A.price B.prize C.chance D.gift28.A.how B.when C.where D.why 29.A.learned B.noticed C.left D.changed 30.A.he B.you C.they D.we“I hate my parents. They always tell me I should do this, and I should not do that. That sometimes makes me angry.” said Zhang Yang, a middle school student in Xi'an. Do you have31 same problem? Perhaps your parents also had a similar problem when they were your age long ago. Why does it seem that some parents are not so friendly in their 32 eyes?The truth is that one likes worrying about things when he or she becomes a parent. Take your parents as an example. They worry about 33 about you. They worry about your choice of friends, your work at school, how much sleep you get, and so on. They even worry about 34 . All these things are a part of your life. They want you to grow up healthily.So how can you make things a little 35 ?Just make sure that your parents know what you are doing. Make sure that your homework 36 on time so that you won't get them into trouble. Get them to know your friends. Call them if you 37 somewhere else so that they won't be worried. Say sorry to them if you do something wrong. Take responsibility for what you38 and done. Talk about your ideas with them. Then they may talk about 39 with you.Most of all, try to think about why your parents do this or do that. Someday, when you become a parent, it 40 be able to help you know how to get on with your children.31.A.a B.an C.the D./32.A.friends B.friends'C.children D.children's 33.A.something B.some things C.everything D.nothing 34.A.what do you eat B.how do you eatC.which you eat D.what you eat35.A.easy B.easier C.easily D.more easily 36.A.is finished B.is finishing C.finished D.finishes 37.A.will stay B.stay C.stays D.stayed 38.A.said B.have said C.say D.had said 39.A.themselves B.them C.theirs D.they40.A.need B.would C.may D.shouldOnce upon a time a teacher and his student lay down under a tree near the grass area. The student asked the teacher, “Sir, I really don’t know how we 41 our friends?”42 for a few seconds, the teacher then answered, “It’s a hard and an 43 question.” The teacher continued, “Look! There is a lot of grass. Why don’t you walk there? Please don’t walk backwards; walk straight ahead. On your way, try to find a blade (片) of beautiful grass, pick it and give it to me. But just one.”The student said, “Well. OK…Wait for me…” and walked straight ahead to the field.A few minutes later, he came back.The teacher asked, “Well, I don’t see a beautiful blade of grass in your hand.”The student said, “On my journey, I found a few beautiful blades of grass, but I thought I would find a 44 one, so I didn’t pick them. But I didn’t realize I was at the 45 of the field, and I hadn’t picked any. I didn’t go back 46 you told me not to.”The teacher said, “That’s what will 47 in real life.”In the story, grass is the people 48 you; the beautiful blades of grass are the people that attract you and the grassy (茂盛的) field is 49 .In looking for your friends, don’t always compare (比较) and hope that there will be a better one. By doing that, you’ll 50 your lifetime; remember that “Time never goes back.”41.A.find B.like C.think D.feel42.A.Calm B.Afraid C.Silent D.Ready 43.A.easy B.interesting C.exciting D.amazing 44.A.good B.nice C.best D.better45.A.top B.beginning C.end D.middle 46.A.after B.until C.because D.while 47.A.happen B.form C.achieve D.miss48.A.for B.around C.against D.among 49.A.power B.knowledge C.confidence D.time50.A.limit B.waste C.bet D.reduceSam was very nervous. It was his first day in a new school, and lunch was only thirty51 away. For most students, lunch is the best part of the school day. It is time when you chat52 your friends, get to know what interesting things everyone is doing, and, of course, eat. Sam, however, didn't know anyone or anything, like where to sit or with whom to sit at lunch.Sam's mother was in the army, so his family 53 a lot. He had really liked his old school and hated leaving his two 54 friends of all. They still talked to each other through phone calls and emails, but he couldn't see them every day. He wondered if their friendship would continue. He didn't want to lose them, but he knew it 55 be hard for them to stay close.But Sam didn't want his mother to feel sorry for him 56 they had no choice. She had always been ready to defend(保卫) not only the family but also the country. Sam loved her so much. He had wanted to mean it, but there was still always worry in him when they moved.The lunch time came 57 . As Sam was walking to the dining room, his name 58 . “Excuse me, Sam.” Sam turned around and saw five friendly faces.“Would you like 59 us?” asked one of them.The unexpected invitation was exactly what 60 . Sam nodded yes happily and joined them. He was sure about his future school life now.51.A.second B.seconds C.minute D.minutes 52.A.with B.about C.around D.for 53.A.moves B.moved C.will move D.has moved 54.A.good B.best C.poor D.poorest 55.A.could B.must C.would D.should 56.A.unless B.after C.because D.if57.A.slow B.slowly C.quick D.quickly 58.A.called B.is called C.was called D.is calling 59.A.to join B.join C.joins D.joined 60.A.does he need B.he needs C.did he need D.he neededA few years ago the company I work for sent my wife and me to live in New York for a year. I’ve always loved jogging, so I was really 61 when I found out our flat was next to Central Park. This 62 that every morning I could go for a run before I went to work.Because a lot of people had told me to be 63 of thieves(窃贼) in the park, I didn’t usually take anything with me. What could they 64 from me if I didn’t have anything? But one morning my wife asked me to buy some 65 on the way home so I took a $10 note out of my wallet.While I was running through the park, another jogger 66 into me. He said sorry and continued running. I thought it was a bit 67 so I checked my pocket and found that the money was missing. I started to run after him right away. I 68 caught him by his arm. I started shouting and told him to give me the $10. I’m not usually a 69 person but I really got very angry. This seemed to frighten(惊吓) him and he quickly put his hand in his pocket and gave me the money. Then he ran away as fast as he could.I bought the bread and went home. As soon as I got there, I began to tell my wife my story. “You won’t believe what happened to me,” I 70 . She stopped me, “I know. You left the money for the bread on the kitchen table.”61.A.angry B.happy C.guilty D.curious 62.A.made B.explained C.meant D.proved63.A.careful B.sure C.proud D.afraid64.A.hide B.borrow C.steal D.receive 65.A.coffee B.milk C.fruit D.bread 66.A.bumped B.looked C.turned D.moved 67.A.crazy B.natural C.strange D.rude 68.A.suddenly B.finally C.politely D.slowly 69.A.cold-heart B.warm-heart C.cool-headed D.hot-headed 70.A.started B.added C.insisted D.doubtedIt was a cold windy winter day and I was walking home from work. My head pointed down against the wind when I noticed an old lady. She was 71 on the sidewalk in her house slippers(拖鞋). As I passed her, I saw that she was well into her eighties, and she was holding some72 in her hands.I walked on and arrived at the traffic light. Waiting there, I saw the post box across the street. The light turned green, 73 I pretended(假装)to check my phone and waited for her to catch up with me. In this neighborhood, even crossing with the green light was 74 . Drivers often drove wildly through red lights and stop signs.I stayed there until she 75 me. I turned to her and touched her gently on the arm. She smiled and took my hand in hers and held it. Her small hand was soft and warm. Hand-in-hand, we were waiting for the 76 to change.When the light turned green, we walked together slowly across the street, 77 the letters in the mailbox. Then I took her back across the street.As I walked the rest of the way home, I could still feel the 78 of her hand in my own.People say the first language of all humans is 79 . A caring touch helps to make a bad situation 80 . And even now, every time I think of the old lady, I can still feel the power of touch.71.A.coming B.walking C.running D.riding 72.A.letters B.pictures C.books D.flowers73.A.but B.and C.so D.because74.A.strange B.crazy C.dangerous D.safe 75.A.called B.reached C.found D.caught 76.A.way B.light C.ride D.traffic 77.A.kept B.covered C.dropped D.locked 78.A.warmth B.trust C.care D.love 79.A.help B.kindness C.touch D.action 80.A.real B.necessary C.popular D.hopeful参考答案1.C 2.C 3.A 4.B 5.C 6.A 7.A 8.C 9.C 10.B11.B 12.C 13.D 14.A 15.D 16.B 17.D 18.A 19.C 20.D21.C 22.D 23.B 24.A 25.C 26.A 27.B 28.A29.A30.D31.C32.D33.C34.D35.B36.A37.B38.B39.C40.C41.A42.C43.A44.D45.C46.C47.A48.B49.D50.B51.D52.A53.B54.B55.C56.C57.D58.C59.A60.D61.B62.C63.A64.C65.D66.A67.C68.B69.D70.A71.B72.A73.A74.C75.B76.B77.C78.A79.C80.D。

2020届高考英语二轮复习系列之疯狂专练三 词性转化及词性变化单句填空+语法填空(word版含答案)

2020届高考英语二轮复习系列之疯狂专练三 词性转化及词性变化单句填空+语法填空(word版含答案)

当空格处所需词类与括号中所给词的词类不同时,就需要词类转化。

我们可据以下3条规则顺利解题:(1)作主语或宾语用名词形式;(2)作定语、表语或补足语用形容词形式;(3)修饰动词、形容词或另一副词,作状语,用副词形式。

具体解题技巧如下:第一步:分析结构,确定要填的词在句中充当哪种句子成分。

在名词前作定语、在系动词后作表语、作主语和宾语的补足语,一般要用形容词;修饰动词、形容词或副词,或修饰整个句子,作状语,用副词;作主语或宾语用名词,或者在冠词、形容词性物主代词或名词所有格后,用名词。

第二步:根据构词法将括号中的词变成所需要的词类。

注意:1. 有时不但要注意词性转换,而且还要考虑用表示相反意义的前缀或后缀, 其逻辑意义才通顺;2. 当所给词的词性与空格处所需词的词性相同时, 无需改变词性, 就可能是加只改变词义但不改变词性的前缀了。

形容词→副词wide→widely 形容词→形容词比较级/最高级wide→wider/widest 形容词→名词wide→width 形容词→动词wide→ widen 动词→名词 instruct → instruction (s)特殊变性 happy →happily, simple →simply ,true →truly, arrange →arrangement ,judge →judgment1.【2019·全国 II 卷】Her years of hard work have _____(final)been acknowledged after a customer nominated(提名)her to be Cheshire's Woman Of The Year.疯狂专练三 词性转化单句填空+语法填空 技巧点拨常考考点 小题狂练2【2019·全国卷I】It is difficult to figure out a global population of polar bears as much of the range has been _____ (poor) studied.3. 【2019·全国卷I改编】It leads to a ______ (believe) that populations are increasing4. 【2019·浙江卷】Other American studies showed no _____ (connect) between uniforms and school performance.5.【2019·浙江卷】School uniforms are _____ (tradition) in Britain, but some schools are starting to get rid of them.6.【2018·全国卷I I】According to the World Bank, China accounts for about 30 percent of total _____ (globe) fertilizer consumption.7.【2018·全国卷III】I'm a _____ (science) who studies animals such as apes and monkeys.8. 【2018·全国卷I】Running is cheap, easy and it’s always (energy).9.【2017·全国卷I I】This development was only possible with the _____ (introduce) of electric-powered engines and lifts.10.【2019·浙江卷改编】When the children are walking to school on dark mornings, car drivers can ______ (easy) see them.11. She took the job for ______(variety)reasons.12. It (eventual) became China’s primary faith, more widespread than the original Chinese religions of Confucianism and Taoism.13. When they were free from work,they invited us to local events and let us know of an interesting ________(compete) to watch,together with the story behind it.14. The obvious one is money; eating out once or twice a week may be (afford) but doing this most days adds up.15. He explained that he tried to ________ (simple) the show and ensured the piglet is unforgettable.16.The _______ (produce) made his fortune thanks to the young peppa pig.17.Now that you are able to take up arms and protect your motherland,we are all________(pride)of you.18. I suggest we should find new ways to protect the ______(endanger)animals.19. In the course of ______(evolve),some birds have lost the power of flight.20. In order to protect the environment people shouldn’t destroy ______(nature)balance.直击考题passage1体裁主题字数建议用时议论文空气污染的影响215字9分钟The silent killer — air pollution causes seven million premature (过早的) ___1___ (death) a year, not just in___2___ (develop) countries but also in UK and the USA as well. In China,people are well aware of the health problems it brings.I check every part every day. If it has heavy air pollution, I will prepare masks for my family. I dislike ___3___ when the air is bad. Because bad air makes it difficult for me to breathe and I think it does influence my ___4___ (perform) at work. A new research in China has also found a link between air quality and levels of intelligence. In the study, 25,000 people living across China __5___ (test) in language and math skills last year. They found the results of those ___6___ lived in more polluted areas were ___7___ (negative) impacted, especially in languages. It adds to a growing body of evidence that air pollution has an effect not just on the lungs and heart ___8___ on the brain.This research shows that the longer we are exposed to air pollution, the ___9___ (many) problems we’re storing up for later life. Campaigners hope this study will persuade the British government ___10___ (think) hard before it publishes its clean air strategy next year.passage2【晋冀鲁豫中原名校2019届高三第三次联考】Stand in line at any grocery store or sit in any hospital waiting room, and you see people staring at their phones, probably catching up on news or just relieving their boredom. This seems harmless enough. But could being on our phones affect our___1___(able) to connect with the people around us?__2___new study aimed to answer that question.Researcher Kostadin Kushlev and his colleagues asked pairs of college students,___3___were strangers to each other___4___(come) into a small lab waiting room-either with or without their phones. They were told that the researchers were running a bit late and they needed to wait. While waiting, their faces were___5___(secret) videotaped. Afterwards, the students___6___(report) how they felt and how much they interacted(互动)with the other participants.The researchers studied videotapes of the faces of airs who interacted,___7___(measure) how often they smiled. The___8___(result)?People with phones exhibited fewer smiles overall. They spent 30% less of the time smiling___9___people without phones, signaling less interest in connecting with others. What's more, thirty-two participants with phones didn't interact at all in the waiting room. These findings show that using phones in public _____10_____(affect) one's interaction with others. Consequently,cell phones should be used wisely.passage3【江西省名校(临川一中、南昌二中)2019届高三5月联合考试】Louis Cha, better known by his pen name Jin Yong, was a Chinese wuxia novelist and essayist, ___1___co-founded the Hong Kong daily newspaper Ming Bao in 1959 and ___2___ (serve) as the first editor-in-chief. He was Hong Kong’s most famous writer.His wuxia novels have ___3___ widespread following in Chinese communities worldwide. His 15 works ___4___ (write) between 1955 and 1972 earned him a reputation as one of the greatest and most popular wuxia writers ever. By the time of his ___5___ (die) he was the best-selling Chinese author, and over 100 million copies of his works have been sold worldwide. His works have the unusual ability ___6___ (go) beyond geographical and ideological barriers separating Chinese communities of the world, achieving ___7___ (great) success than any other contemporary writer.Most of his works ____8____ (translate) into many languages so far, including English, French, etc. Even the asteroid (小行星) 10930 Jinyong (1998 CR2) is named ____9____ him. We cannot thank him more for ____10____ he has done in novel literature.答案与解析小题狂练1.【答案】finally【解析】考查副词用法。

小学六年级语文毕业复习句子专项练习题三篇

小学六年级语文毕业复习句子专项练习题三篇

【导语】亲爱的同学们,在愉快⽽紧张的六年学习中,我们⼀直期待着丰收,马上就要毕业考试了,准备了《⼩学六年级语⽂毕业复习句⼦专项练习题三篇》,快来练习⼀下吧!篇⼀ (⼀)发挥想象,补充句⼦。

(4分) ①春姑娘迈着轻盈的步⼦⾛来了。

她提着神奇的⼩花篮,把五彩的鲜花洒向⼭坡,撒向⽥野;她_______________________________________________________ ②A是⼀座⾦字塔,C是未满的⽉⽛⼉,D是⼀把竖琴,Z是__________,F是 _____________ (⼆)选择适当的关联词语填空。

(4分) 只有……才…… 不但……⽽且…… 不管……都…… 虽然……但是…… 李悦是个优秀的少先队员,他( )学习勤奋,( )体育优秀。

( )有多忙,他( )坚持晨跑,( )他的⾝体⼀直很棒,( )他经常说:“( )做⼀个全⾯发展的学⽣,将来( )能为祖国的现代化建设作贡献。

” (三)指出下列各句⽤了哪⼀种修辞⼿法。

(3分) 1、⼀阵春风过后,⼩草跳起了欢乐的舞蹈,有时舒展双臂,有时弯腰触地,有时左右摇摆……( ) 2、黄河是我们中华民族的摇篮。

( ) 3、漓江的⽔真静啊,静得让你感觉不到它在流动;漓江的⽔真清啊,清得让你可以看见江底的沙⽯;漓江的⽔真绿啊,绿得仿佛那是⼀块⽆瑕的翡翠。

( )篇⼆ ⼀、选关联词填空:4分 因为……所以…… 不但……⽽且…… 只要……就…… 虽然……但是…… 如果……就…… 即使……也…… A.⼤明( )刻苦学习,( )热爱劳动。

B.⼤明( )刻苦学习,( )成绩很好。

C.⼤明( )刻苦学习,( )不注意锻炼⾝体。

D.⼤明( )刻苦学习,( )能提⾼成绩。

⼆、修改病句(改在原句上)。

6分 1.⾛进拉萨古城,我们就看到极富特⾊的藏民居和藏戏唱腔。

2.“马踏飞燕”的作者是东汉时期的艺术珍品。

3.早晨,当阳光照耀在校园时,教室⾥传出了琅琅的⼀阵阵读书声。

2023中考数学一轮复习专题3

2023中考数学一轮复习专题3

专题3.2 平面直角坐标系与一次函数、反比例函数(基础篇)(真题专练)一、单选题1.(2021·黑龙江牡丹江·中考真题)如图,在平面直角坐标系中A (﹣1,1)B (﹣1,﹣2),C (3,﹣2),D (3,1),一只瓢虫从点A 出发以2个单位长度/秒的速度沿A →B →C →D →A 循环爬行,问第2021秒瓢虫在( )处.A .(3,1)B .(﹣1,﹣2)C .(1,﹣2)D .(3,﹣2)2.(2021·山东济南·中考真题)反比例函数()0ky k x=≠图象的两个分支分别位于第一、三象限,则一次函数y kx k =-的图象大致是( )A .B .C .D .3.(2021·四川德阳·中考真题)下列函数中,y 随x 增大而增大的是( ) A .y =﹣2x B .y =﹣2x +3C .y 2x=(x <0) D .y =﹣x 2+4x +3(x <2)4.(2021·内蒙古呼和浩特·中考真题)在平面直角坐标系中,点()3,0A ,()0,4B .以AB 为一边在第一象限作正方形ABCD ,则对角线BD 所在直线的解析式为( ) A .147y x =-+B .144y x =-+C .142y x =-+D .4y =5.(2021·湖南娄底·中考真题)如图,直线y x b =+和4y kx =+与x 轴分别相交于点(4,0)A -,点(2,0)B ,则040x b kx +>⎧⎨+>⎩解集为( )A .42x -<<B .4x <-C .2x >D .4x <-或2x >6.(2021·黑龙江大庆·中考真题)已知反比例函数ky x=,当0x <时,y 随x 的增大而减小,那么一次的数y kx k =-+的图像经过第( ) A .一,二,三象限 B .一,二,四象限 C .一,三,四象限D .二,三,四象限7.(2021·福建·中考真题)如图,一次函数()0y kx b k =+>的图象过点()1,0-,则不等式()10k x b -+>的解集是( )A .2x >-B .1x >-C .0x >D .1x >8.(2021·辽宁朝阳·中考真题)如图,O 是坐标原点,点B 在x 轴上,在OAB 中,AO =AB =5,OB =6,点A 在反比例函数y =kx(k ≠0)图象上,则k 的值( )A .﹣12B .﹣15C .﹣20D .﹣309.(2021·湖南湘西·中考真题)如图所示,小英同学根据学习函数的经验,自主尝试在平面直角坐标系中画出了一个解析式为21y x 的函数图象.根据这个函数的图象,下列说法正确的是( )A .图象与x 轴没有交点B .当0x >时0y >C .图象与y 轴的交点是1(0,)2- D .y 随x 的增大而减小10.(2021·四川达州·中考真题)在反比例函数21k y x+=(k 为常数)上有三点()11,A x y ,()22,B x y ,()33,C x y ,若1230x x x <<<,则1y ,2y ,3y 的大小关系为( )A .123y y y <<B .213y y y <<C .132y y y <<D .321y y y <<11.(2021·浙江杭州·中考真题)已知1y 和2y 均是以x 为自变量的函数,当x m =时,函数值分别为1M 和2M ,若存在实数m ,使得120M M +=,则称函数1y 和2y 具有性质P .以下函数1y 和2y 具有性质P 的是( )A .212y x x =+和21y x =--B .212y x x =+和21y x =-+C .11y x =-和21y x =--D .11y x=-和21y x =-+二、填空题12.(2021·青海西宁·中考真题)在平面直角坐标系xOy 中,点A 的坐标是(–2)1-,,若//AB y轴,且9AB =,则点B 的坐标是________.13.(2021·广西河池·中考真题)从﹣2,4,5这3个数中,任取两个数作为点P 的坐标,则点P 在第四象限的概率是__________.14.(2021·辽宁丹东·中考真题)在函数y =中,自变量x 的取值范围_________. 15.(2021·湖北黄石·中考真题)将直线1y x =-+向左平移m (0m >)个单位后,经过点(1,−3),则m 的值为______.16.(2021·内蒙古呼和浩特·中考真题)正比例函数1y k x =与反比例函数2k y x=的图象交于A ,B 两点,若A 点坐标为-,则12k k +=__________.17.(2021·四川眉山·中考真题)一次函数()232y a x =++的值随x 值的增大而减少,则常数a 的取值范围是______.18.(2021·江苏苏州·中考真题)若21x y +=,且01y <<,则x 的取值范围为______. 19.(2021·山东青岛·中考真题)列车从甲地驶往乙地.行完全程所需的时间()h t 与行驶的平均速度()km/h v 之间的反比例函数关系如图所示.若列车要在2.5h 内到达,则速度至少需要提高到__________km/h .20.(2021·江苏徐州·中考真题)如图,点,A D 分别在函数36,y y x x-==的图像上,点,B C 在x 轴上.若四边形ABCD 为正方形,点D 在第一象限,则D 的坐标是_____________.21.(2021·北京·中考真题)在平面直角坐标系xOy 中,若反比例函数(0)ky k x =≠的图象经过点()1,2A 和点()1,B m -,则m 的值为______________.22.(2021·湖南邵阳·中考真题)已知点()11,A y ,()22,B y 为反比例函数3y x=图象上的两点,则1y 与2y 的大小关系是1y ______2y .(填“>”“=”或“<”)23.(2021·广西河池·中考真题)在平面直角坐标系中,一次函数2y x =与反比例函数()0ky k x=≠的图象交于()11,A x y ,()22,B x y 两点,则12y y +的值是____________.24.(2021·江苏淮安·中考真题)如图(1),△ABC 和△A ′B ′C ′是两个边长不相等的等边三角形,点B ′、C ′、B 、C 都在直线l 上,△ABC 固定不动,将△A ′B ′C ′在直线l 上自左向右平移.开始时,点C ′与点B 重合,当点B ′移动到与点C 重合时停止.设△A ′B ′C ′移动的距离为x ,两个三角形重叠部分的面积为y ,y 与x 之间的函数关系如图(2)所示,则△ABC 的边长是___.三、解答题25.(2021·甘肃兰州·中考真题)小军到某景区游玩,他从景区入口处步行到达小憩屋,休息片刻后继续前行,此时观光车从景区入口处出发的沿相同路线先后到达观景点,如图,1l ,2l 分别表示小军与观光车所行的路程()m y 与时间()min x 之间的关系. 根据图象解决下列问题:(1)观光车出发______分钟追上小军; (2)求2l 所在直线对应的函数表达式;(3)观光车比小军早几分钟到达观景点?请说明理由.26.(2021·河南·中考真题)猕猴嬉戏是王屋山景区的一大特色,猕猴玩偶非常畅销.小李在某网店选中A ,B 两款猕猴玩偶,决定从该网店进货并销售.两款玩偶的进货价和销售价如下表:(1)第一次小李用1100元购进了A ,B 两款玩偶共30个,求两款玩偶各购进多少个; (2)第二次小李进货时,网店规定A 款玩偶进货数量不得超过B 款玩偶进货数量的一半.小李计划购进两款玩偶共30个,应如何设计进货方案才能获得最大利润,最大利润是多少? (3)小李第二次进货时采取了(2)中设计的方案,并且两次购进的玩偶全部售出,请从利润率的角度分析,对于小李来说哪一次更合算? (注:利润率100%=⨯利润成本)27.(2021·山东淄博·中考真题)如图,在平面直角坐标系中,直线11y k x b =+与双曲线22k y x=相交于()()2,3,,2A B m --两点. (1)求12,y y 对应的函数表达式;(2)过点B 作//BP x 轴交y 轴于点P ,求ABP △的面积; (3)根据函数图象,直接写出关于x 的不等式21k k x b x+<的解集.参考答案1.A【分析】根据点的坐标求出四边形ABCD 的周长,然后求出第2021秒是爬了第几圈后的第几个单位长度,从而确定答案.解: A (﹣1,1)B (﹣1,﹣2),C (3,﹣2),D (3,1)∴ 四边形ABCD 是矩形()1--2=1+2=3AB ∴=()=3--1=4BC343414AB BC CD AD ∴+++=+++=∴瓢虫转一周,需要的时间是14=72秒 2021=2887+5⨯ ,∴ 按A →B →C →D →A 顺序循环爬行,第2021秒相当于从A 点出发爬了5秒,路程是:52=10⨯个单位,10=3+4+3,所以在D 点()3,1 .故答案为:A【点拨】本题考查了点的变化规律,根据点的坐标求出四边形ABCD 一周的长度,从而确定2021秒瓢虫爬完了多少个整圈的矩形,不成一圈的路程在第几圈第几个单位长度的位置是解题的关键. 2.D【分析】根据题意可得0k >,进而根据一次函数图像的性质可得y kx k =-的图象的大致情况.解:反比例函数()0ky k x=≠图象的两个分支分别位于第一、三象限, 0k ∴>△一次函数y kx k =-的图象与y 轴交于负半轴,且经过第一、三、四象限. 观察选项只有D 选项符合. 故选D【点拨】本题考查了反比例函数的性质,一次函数图像的性质,根据已知求得0k >是解题的关键. 3.D【分析】一次函数当a >0时,函数值y 总是随自变量x 增大而增大,反比例函数当k >0时,在每一个象限内,y 随自变量x 增大而增大,二次函数根据对称轴及开口方向判断增减性.解:A .一次函数y =-2x 中的a =-2<0,y 随x 的增大而减小,故不符合题意. B .一次函数y =-2x +3中的a =-2<0,y 随自变量x 增大而减小,故不符合题意.C .反比例函数y =2x (x <0)中的k =2>0,在第三象限,y 随x 的增大而减小,故不符合题意.D .二次函数y =-x 2+4x +3(x <2),对称轴x =2ba-=2,开口向下,当x <2时,y 随x 的增大而增大,故符合题意. 故选:D .【点拨】本题考查了一次函数、反比例函数、二次函数的增减性;熟练掌握一次函数、二次函数、反比例函数的性质是关键. 4.A【分析】过点D 作DE x ⊥轴于点E ,先证明()ABO DAE AAS ≅,再由全等三角形对应边相等的性质解得(7,3)D ,最后由待定系数法求解即可. 解:正方形ABCD 中,过点D 作DE x ⊥轴于点E , 90ABO BAO BAO DAE ∠+∠=∠+∠=︒ABO DAE ∴∠=∠90,BOA AED AB AD ∠=∠=︒= ()ABO DAE AAS ∴≅ 3,4AO DE OB AE ∴==== (7,3)D ∴设直线BD 所在的直线解析式为(0)y kx b k =+≠, 代入()0,4B ,(7,3)D 得473b k b =⎧⎨+=⎩ 174k b ⎧=-⎪∴⎨⎪=⎩ 147y x ∴=-+,故选:A .【点拨】本题考查待定系数法求一次函数的解析式,涉及正方形性质、全等三角形的判定与性质等知识,是重要考点,难度较易,掌握相关知识是解题关键. 5.A【分析】根据图像以及两交点(4,0)A -,点(2,0)B 的坐标得出即可. 解:△直线y x b =+和4y kx =+与x 轴分别相交于点(4,0)A -,点(2,0)B ,△观察图像可知040x b kx +>⎧⎨+>⎩解集为42x -<<,故选:A .【点拨】本题考查了一次函数与一元一次不等式组,能根据图像和交点坐标得出答案是解此题的关键. 6.B【分析】根据反比例函数的增减性得到0k >,再利用一次函数的图象与性质即可求解. 解:△反比例函数ky x=,当0x <时,y 随x 的增大而减小, △0k >,△y kx k =-+的图像经过第一,二,四象限, 故选:B .【点拨】本题考查反比例函数和一次函数的图象与性质,掌握反比例函数和一次函数的图象与性质是解题的关键. 7.C【分析】先平移该一次函数图像,得到一次函数()()10y k x b k =-+>的图像,再由图像即可以判断出 ()10k x b -+>的解集.解:如图所示,将直线()0y kx b k =+>向右平移1个单位得到 ()()10y k x b k =-+>,该图像经过原点,由图像可知,在y 轴右侧,直线位于x 轴上方,即y >0, 因此,当x >0时,()10k x b -+>, 故选:C .【点拨】本题综合考查了函数图像的平移和利用一次函数图像求对应一元一次不等式的解集等,解决本题的关键是牢记一次函数的图像与一元一次不等式之间的关系,能从图像中得到对应部分的解集,本题蕴含了数形结合的思想方法等. 8.A【分析】过A 点作AC △OB ,利用等腰三角形的性质求出点A 的坐标即可解决问题. 解:过A 点作AC △OB ,△AO =AB ,AC △OB ,OB =6, △OC =BC =3,在Rt △AOC 中,OA =5,△AC 4==,△A (﹣3,4),把A (﹣3,4)代入y =k x,可得k =﹣12 故选:A .【点拨】本题考查反比例函数图象上的点的性质,等腰三角形的性质,勾股定理等知识,解题的关键是熟练掌握基本知识,属于中考常考题型.9.A【分析】根据函数图象可直接进行排除选项.解:由图象可得:10x -≠,即1x ≠,A 、图象与x 轴没有交点,正确,故符合题意;B 、当01x <<时,0y <,错误,故不符合题意;C 、图象与y 轴的交点是()0,2-,错误,故不符合题意;D 、当1x <时,y 随x 的增大而减小,且y 的值永远小于0,当1x >时,y 随x 的增大而减小,且y 的值永远大于0,错误,故不符合题意;故选A .【点拨】本题主要考查反比例函数的图象与性质,熟练掌握反比例函数的图象与性质是解题的关键.10.C【分析】根据k >0判断出反比例函数的增减性,再根据其坐标特点解答即可.解:△210k +>,△反比例函数图象的两个分支在第一、三象限,且在每个象限内y 随x 的增大而减小, △B (x 2,y 2),C (x 3,y 3)是双曲线k y x=上的两点,且320x x >>, △点B 、C 在第一象限,0<y 3<y 2,△A (x 1,y 1)在第三象限,△y 1<0,△132y y y <<.故选:C .【点拨】本题考查了由反比例函数图象的性质判断函数图象上点的坐标特征,理解基本性质是解题关键.11.A【分析】根据题中所给定义及一元二次方程根的判别式可直接进行排除选项.解:当x m =时,函数值分别为1M 和2M ,若存在实数m ,使得120M M +=,对于A 选项则有210m m +-=,由一元二次方程根的判别式可得:241450b ac -=+=>,所以存在实数m ,故符合题意;对于B 选项则有210m m ++=,由一元二次方程根的判别式可得:241430b ac -=-=-<,所以不存在实数m ,故不符合题意;对于C 选项则有110m m---=,化简得:210m m ++=,由一元二次方程根的判别式可得:241430b ac -=-=-<,所以不存在实数m ,故不符合题意;对于D 选项则有110m m--+=,化简得:210m m -+=,由一元二次方程根的判别式可得:241430b ac -=-=-<,所以不存在实数m ,故不符合题意;故选A .【点拨】本题主要考查一元二次方程根的判别式、二次函数与反比例函数的性质,熟练掌握一元二次方程根的判别式、二次函数与反比例函数的性质是解题的关键.12.(2,8)-或(2,10)--【分析】由题意,设点B 的坐标为(-2,y ),则由AB =9可得(1)9y --=,解方程即可求得y 的值,从而可得点B 的坐标.解:△//AB y 轴△设点B 的坐标为(-2,y )△AB =9 △(1)9y --=解得:y =8或y =-10△点B 的坐标为(2,8)-或(2,10)--故答案为:(2,8)-或(2,10)--【点拨】本题考查了平面直角坐标系求点的坐标,解含绝对值方程,关键是抓住平行于坐标轴的线段长度只与两点的横坐标或纵坐标有关,易错点则是考虑不周,忽略其中一种情况.13.13【分析】先画树状图展示所有6种等可能的结果,利用第四象限点的坐标特征确定点P 在第四象限的结果数,然后根据概率公式计算,即可求解.解:画出树状图为:共有6种等可能的结果,它们是:(-2,4),(-2,5),(4,-2),(4,5),(5,4),(5,-2), 其中点P 在第四象限的结果数为2,即(4,-2),(5,-2),所以点P 在第四象限的概率为:2163= . 故答案为:13 . 【点拨】本题考查了列表法与树状图法求概率和点的坐标特征,通过列表法或树状图法列举出所有可能的结果求出n ,再从中选出符合事件A 或B 的结果数目m ,求出概率是解题的关键.14.3x ≥【分析】根据被开方数大于等于0,分母不等于0列式计算即可得解.解:根据题意得:3020x x -≥⎧⎨-≠⎩,解得3x ≥ △自变量x 的取值范围是3x ≥.故答案为:3x ≥.【点拨】本题考查了函数自变量的范围,一般从三个方面考虑:(1)当函数表达式是整式时,自变量可取全体实数;(2)当函数表达式是分式时,考虑分式的分母不能为0;(3)当函数表达式是二次根式时,被开方数非负.15.3【分析】根据平移的规律得到平移后的解析式为()1y x m =-++,然后把点(1,−3)的坐标代入求值即可.解:将一次函数y =-x +1的图象沿x 轴向左平移m (m ≥0)个单位后得到()1y x m =-++, 把(1,−3)代入,得到:()311m -=-++,解得m =3.故答案为:3.【点拨】本题主要考查了一次函数图象与几何变换,用平移规律“左加右减,上加下减”直接代入函数解析式求得平移后的函数解析式是解题的关键.16.8-【分析】将A 点坐标为-分别代入正比例函数1y k x =与反比例函数2k y x =的解析式中即可求解.解:1y k x =和2k y x=过点A -12k ==-2(6k -=-12(2)(6)8k k +=-+-=-故答案为8-.【点拨】本题考查了待定系数法求正比例函数和反比例函数的解析式,有理数的加法运算,正确的实用待定系数法求解析式是解题的关键.17.32a <- 【分析】由题意,先根据一次函数的性质得出关于a 的不等式230a +<,再解不等式即可.解:一次函数()232y a x =++的值随x 值的增大而减少,230a ∴+<, 解得:32a <-, 故答案是:32a <-. 【点拨】本题考查了一次函数的图象与系数的关系,解题的关键是:熟知一次函数的增减性.18.102x << 【分析】根据21x y +=可得y =﹣2x+1,k =﹣2<0进而得出,当y =0时,x 取得最大值,当y =1时,x 取得最小值,将y =0和y =1代入解析式,可得答案.解:根据21x y +=可得y =﹣2x+1,△k =﹣2<0△01y <<,△当y =0时,x 取得最大值,且最大值为12, 当y =1时,x 取得最小值,且最小值为0, △102x << 故答案为:102x <<. 【点拨】此题考查了一次函数的性质,熟练掌握一次函数的性质是解题的关键. 19.240 【分析】由设,k t v=再利用待定系数法求解反比例函数解析式,把 2.5t =h 代入函数解析式求解v 的值,结合图象上点的坐标含义可得答案. 解:由题意设,k t v= 把()200,3代入得:2003600,k tv ==⨯=600,t v∴= 当 2.5t =h 时,6002402.5v ==km/h , 所以列车要在2.5h 内到达,则速度至少需要提高到240km/h ,故答案为:240km/h .【点拨】本题考查的是反比例函数的应用,掌握利用待定系数法求解反比例函数的解析式是解题的关键.20.(2,3)【分析】根据正方形和反比例函数图像上点的坐标特征,设D 点坐标为(m ,6m),则A 点坐标为(2m - ,6m ),进而列出方程求解. 解:△四边形ABCD 为正方形,△设D 点坐标为(m ,6m ),则A 点坐标为(2m - ,6m ), △m -(2m -)=6m ,解得:m =±2(负值舍去), 经检验,m =2是方程的解,△D 点坐标为(2,3),故答案是:(2,3).【点拨】本题主要考查反比例函数与平面几何的综合,掌握反比例函数图像上点的坐标特征,是解题的关键.21.2-【分析】由题意易得2k =,然后再利用反比例函数的意义可进行求解问题.解:把点()1,2A 代入反比例函数()0k y k x=≠得:2k =, △12m -⨯=,解得:2m =-,故答案为-2.【点拨】本题主要考查反比例函数的图象与性质,熟练掌握反比例函数的图象与性质是解题的关键.22.>【分析】根据反比例函数的性质,当反比例系数k >0,在每一象限内y 随x 的增大而减小可得答案. 解:△ 反比例函数的解析式为3y x =,k >0,△ 在每个象限内y 随x 的增大而减小,△ 1<2,△1y >2y .故答案为:>.【点拨】本题主要考查了反比例函数的性质,掌握反比例函数的性质是解题的关键. 23.0【分析】根据正比例函数和反比例函数的图像关于原点对称,则交点也关于原点对称,即可求得12y y +解:一次函数2y x =与反比例函数()0k y k x =≠的图象交于()11,A x y ,()22,B x y 两点, 一次函数2y x =与反比例函数()0k y k x=≠的图象关于原点对称, ∴12y y +0= 故答案为:0【点拨】本题考查了正比例函数和反比例函数图像的性质,掌握以上性质是解题的关键. 24.5【分析】在点B '到达B 之前,重叠部分的面积在增大,当点B '到达B 点以后,且点C '到达C 以前,重叠部分的面积不变,之后在B '到达C 之前,重叠部分的面积开始变小,由此可得出B 'C '的长度为a ,BC 的长度为a +3,再根据△ABC 的面积即可列出关于a 的方程,求出a 即可.解:当点B '移动到点B 时,重叠部分的面积不再变化,根据图象可知B 'C '=a ,A B C S '''∆=过点A '作A 'H △B 'C ',则A 'H 为△A 'B 'C '的高,△△A 'B 'C '是等边三角形,△△A 'B 'H =60°,△sin60°=A H A B '''=△A 'H ,△12A B C S a '''∆=⋅2= 解得a =﹣2(舍)或a =2,当点C '移动到点C 时,重叠部分的面积开始变小,根据图像可知BC =a +3=2+3=5,△△ABC 的边长是5,故答案为5.【点拨】本题主要考查动点问题的函数图象和三角函数,关键是要分析清楚移动过程可分为哪几个阶段,每个阶段都是如何变化的,先是点B '到达B 之前是一个阶段,然后点C '到达C 是一个阶段,最后B '到达C 又是一个阶段,分清楚阶段,根据图象信息列出方程即可. 25.(1)6;(2)300-4500y x =;(3)观光车比小军早8分钟到达观景点,理由见解析.【分析】(1)由图像可知,1l ,2l 的交点,即为两者到达同一位置,所以在21分钟时观光车追上小军,而观光车是在15分钟时出发的,所以观光车出发6分钟后追上小军;(2)设2l 所在直线对应的函数表达式为y kx b =+,将经过两点(15,0)和(21,1800)带入表达式y kx b =+,得300-4500y x =;(3)由图像可知,到达观景点需要3000m 的路程,小军到达观景点的时间为33min ,通过2l 所在直线对应的函数表达式300-4500y x =,可知,观光车到达观景点的时间为25min x =,因此观光车比小军早33min 25min 8min -=到达观景点.解:(1)由图像可知,在21min 时,1l ,2l 相交于一点,表示在21min 时,小军和观光车到达了同一高度,此时观光车追上了小军, 观光车是在15min 时出发,△21min-15min=6min ,△观光车出发6分钟后追上小军;(2)设2l 所在直线对应的函数表达式为y kx b =+,由图像可知,直线2l 分别经过(15,0)和(21,1800)两点,将两点带入2l 函数表达式y kx b =+得:150211800k b k b +=⎧⎨+=⎩ 解得:3004500k b =⎧⎨=-⎩△2l 函数表达式为300-4500y x =;(3)由图像可知,到达观景点需要3000m 的路程,小军到达观景点的时间为33min ,△观光车2l 函数表达式为300-4500y x =,△将=3000y 带入300-4500y x =,可知观光车到达观景点所需时间为=25min x , △33min-25min=8min ,△观光车比小军早8分钟到达观景点.答:(1)观光车出发6分钟追上小军;(2)2l 所在直线对应的函数表达式为300-4500y x =;(3)观光车比小军早8分钟到达观景点,理由见解析.【点拨】本题考查了一次函数的应用,熟练掌握待定系数法求出函数解析式是解答本题的关键.26.(1)A 款20个,B 款10个;(2)A 款10个,B 款20个,最大利润是460元;(3)第二次更合算.理由见解析【分析】(1)根据题意列二元一次方程组,解方程组即可;(2)根据条件求得利润的解析式,再判断最大利润即可;(3)分别求出第一次和第二次的利润率,比较之后即可知道哪一次更合算.解:(1)设A ,B 两款玩偶分别为,x y 个,根据题意得:30{4030=1100x y x y +=+ 解得:2010x y =⎧⎨=⎩ 答:两款玩偶,A 款购进20个,B 款购进10个.(2)设购进A 款玩偶a 个,则购进B 款(30)a -个,设利润为y 元则(5640)(4530)(30)y a a =-+--=1615(30)a a +-=450+a (元) A 款玩偶进货数量不得超过B 款玩偶进货数量的一半1(30)2a a ∴≤- 10a ∴≤,又0,a ≥010,a ∴≤≤ 且a 为整数,10-<∴当10a =时,y 有最大值max 460.y ∴=(元)∴A 款10个,B 款20个,最大利润是460元.(3)第一次利润20(5640)10(4530)=470⨯-+⨯-(元)∴第一次利润率为:470100%=42.7%1100⨯ 第二次利润率为:460100%=46%1040+2030⨯⨯⨯ 42.7%46%<∴第二次的利润率大,即第二次更划算.【点拨】本题考查了二元一次方程组的应用,最大利润方案问题,利润率求解等问题,一次函数最值问题,理解题意,根据题意列出方程组是解题的关键.27.(1)11y x =-+,26y x =-;(2)152ABP S =;(3)20x -<<或3x > 【分析】(1)由题意先求出2y ,然后得到点B 的坐标,进而问题可求解;(2)由(1)可得ABP △以PB 为底,点A 到PB 的距离为高,即为点A 、B 之间的纵坐标之差的绝对值,进而问题可求解;(3)根据函数图象可直接进行求解.解:(1)把点()2,3A -代入反比例函数解析式得:6k =-, △26y x=-, △点B 在反比例函数图象上,△26m -=-,解得:3m =,△()3,2B -,把点A 、B 作代入直线解析式得:112332k b k b -+=⎧⎨+=-⎩,解得:111k b =-⎧⎨=⎩, △11y x =-+;(2)由(1)可得:()2,3A -,()3,2B -,△//BP x 轴,△3BP =,△点A 到PB 的距离为()325--=, △1153522ABP S =⨯⨯=; (3)由(1)及图象可得:当21k k x b x +<时,x 的取值范围为20x -<<或3x >. 【点拨】本题主要考查反比例函数与一次函数的综合,熟练掌握反比例函数与一次函数的图象与性质是解题的关键.。

高考语文复习:材料作文专练三元思辨类

高考语文复习:材料作文专练三元思辨类

材料作文专练-------三元思辨类1.阅读下面的材料,根据要求写作。

古人认为书法临帖可以分为三个阶段:对临、背临和意临。

对临就是将字帖摆在眼前,照着帖上的运笔和结构写,追求形似;背临意谓脱离碑帖而临,是靠记忆来书写,力求与原作形神兼似;而意临指在继承古人用笔与精神上能有所取舍,自发创造,加之自己的理解和风格特点。

“临书画固贵逼真,尤贵能避其熟。

”临摹书画的过程中做到对临,追求逼真形似很重要,但更重要的是做到在背临的基础上走向意临,在学习古人法度的过程中不被习惯和法度束缚。

以上材料对我们颇具启示意义。

请结合材料写一篇文章,体现你的感悟与思考。

要求:选准角度,确定立意,明确文体,自拟标题;不要套作,不得抄袭;不得泄露个人信息;不少于800字。

2.阅读下面材料,根据要求写作。

三千年前,我们的先祖观察尺蠖软且细长的身体屈伸而行,触发灵感,留下了“尺蠖之屈,以求信(通‘伸’)也”的哲思。

生活在21世纪的中国科技人员,创造性地设计了“祝融号”火星车的主动悬挂系统,模拟尺蠖的运动,提高了火星车在复杂地形自主脱困的能力。

最近有信息传出,美国宇航局最新设计的月球车借鉴了中国火星车的设计,将以一种像尺蠖爬行一样的协调方式移动轮子,中国科学家回应:欢迎共享。

以上材料对我们颇具启示,“尺蠖”这小小的自然之物,给古人以灵感,给今人以创造,让科技在世界共享。

复兴中学将在五四青年节组织以“灵感·创造·共享”为主题的演讲活动,请结合以上材料写一篇演讲稿,体现你的认识与思考。

3.阅读下面的材料,根据要求写作。

,鲁迅先生说,青年“所多的是生力,遇见深林,可以辟成平地的;遇见旷野,可以栽种树木的;遇见沙漠,可以开掘井泉的”。

在实现中华民族伟大复兴的新征程上,应对重大挑战、抵御重大风险、克服重大阻力、解决重大矛盾,迫切需要迎难而上、挺身而出、勠力同心的担当精神。

一切视探索尝试为畏途、一切把负重前行当吃亏、一切“躲进小楼成一统”逃避责任的思想和行为,都是要不得的,都是成不了事的,也是难以真正获得人生快乐的。

2023届高考语文复习默写专练理解性默写附答案

2023届高考语文复习默写专练理解性默写附答案

2023届高考语文复习默写专练理解性默写附答案2023届高考语文复习小题集训理解性默写(一)1.韩愈在《师说》中指出,无论“他”生在“我”的前面还是后面,“我”2.《登高》中,用落叶和江水抒发时光易逝、壮志难酬的感伤的句子是3.陶渊明在《桃花源记》中,写桃花源中老人孩子过着幸福生活的语句是对偶句描写了秦人对从六国剽掠而来的珍宝不知珍惜,生活奢侈浪费无度。

愁绪的形象描绘。

6.黄鹂俗称黄莺,常常被赋予美好欢快的象征意义,是中国古诗中常见的意句则运用自然环境描写,侧面表现琵琶女弹奏技艺高超。

8.《劝学》中以马为喻,从反面强调积累的重要性的两句是:9.在《氓》中,用“淇水、湿地”的有界来反衬男子的变化无常的句子是明人们开始都能做得很好,但很少有能坚持到最后的。

11.《子路、曾皙、冉有、公西华侍坐》中,孔子“哂由”是因为12.古代诗人非常喜欢用比喻的手法来描绘声音,如:13.“绿水青山就是金山银山”,我们要学会敬畏自然,就像苏轼《赤壁赋》14.《阿房宫赋》中,作者运用夸张手法描写阿房宫占地之广,建筑之高的句15.在《谏逐客书》中,作者先后列举泰山高大和河海深邃的例子,目的是说16.面对落花,多情而敏感的诗人们常常为之感叹,并借以表达自己的情感,两句,运用夸张、想象的手法,借助物和人的反应,把洞箫那种悲咽低回的哀音表现得十分形象、真切。

再弹一曲。

(《琵琶行并序》)19.《荀子·劝学》中,与王之涣《登鹳雀楼》中“欲穷千里目,更上一层楼”写的是周瑜的儒将装束。

参考答案1.其闻道也固先乎吾余嘉其能行古道2.无边落木萧萧下不尽长江滚滚来3.黄发垂髫并怡然自乐4.鼎铛玉石金块珠砾5.小楼昨夜又东风恰似一江春水向东流6.两个黄鹂鸣翠柳自在娇莺恰恰啼(或“千里莺啼绿映红”“隔叶黄鹂空好音”“阴阴夏木啭黄鹂”“几处早莺争暖树”)7.轻拢慢捻抹复挑唯见江心秋月白8.骐骥一跃不能十步9.淇则有岸隰则有泮10.有善始者实繁能克终者盖寡11.其言不让吾与点也12.大弦嘈嘈如急雨小弦切切如私语13.物各有主虽一毫而莫取14.覆压三百余里隔离天日15.王者不却众庶故能明其德16.无可奈何花落去似曾相识燕归来(落花人独立微雨燕双飞)17.舞幽壑之潜蛟泣孤舟之嫠妇18.呕哑嘲哳难为听如听仙乐耳暂明19.吾尝跂而望矣不如登高之博见也20.雄姿英发羽扇纶巾2023届高考语文复习小题集训理解性默写(二)2.《念奴娇·过洞庭》一词中,描写词人要舀尽西江的水,用北斗星做酒器招3.苏洵在《六国论》中比较了秦获得土地和诸侯失去土地的方式后,得出了在于战的结论。

2023新教材高考化学二轮专题复习 专练17 非选择题提分练(三)

2023新教材高考化学二轮专题复习 专练17 非选择题提分练(三)

专练17 非选择题提分练(三)1.[2022·海南卷]以Cu2O、ZnO等半导体材料制作的传感器和芯片具有能耗低、效率高的优势。

回答问题:(1)基态O原子的电子排布式为,其中未成对电子有个。

(2)Cu、Zn等金属具有良好的导电性,从金属键的理论看,原因是________________________________________________________________________ ________________________________________________________________________。

(3)酞菁的铜、锌配合物在光电传感器方面有着重要的应用价值。

酞菁分子结构如下图,分子中所有原子共平面,所有N原子的杂化轨道类型相同,均采取杂化。

邻苯二甲酸酐()和邻苯二甲酰亚胺()都是合成酞菁的原料,后者熔点高于前者,主要原因是________________________________________________________________________ ________________________________________________________________________。

(4)金属Zn能溶于氨水,生成以氨为配体,配位数为4的配离子,Zn与氨水反应的离子方程式为___________________________________________________________________ ________________________________________________________________________。

(5)ZnO晶体中部分O原子被N原子替代后可以改善半导体的性能,Zn—N键中离子键成分的百分数小于Zn—O键,原因是________________________________________________________________________ ________________________________________________________________________。

2021届高考英语一轮语法复习 专题12 非谓语动词专练(三)(含解析)

2021届高考英语一轮语法复习 专题12 非谓语动词专练(三)(含解析)

2021高考英语一轮复习语法考点非谓语动词专练(三)一、基础达标测试(本题共20小题,每题1分,共20分)1.I heard a passenger behind me shouting to the driver, but he refused _____________ (stop) until we reached the next stop。

【答案】to stop【解析】考查动词不定式。

句意:我听到后面一位乘客向司机喊叫,但他拒绝停车,直到我们到达下一站.refuse to do sth。

是固定搭配,意思是“拒绝干某事”,故填to stop.2.I regret ________ (inform) you that they are unable to come to your wedding tomorrow.【答案】to inform【解析】考查固定短语。

句意:我很遗憾地通知你,他们明天不能来参加你的婚礼了。

regret to do sth.很遗憾做某事,固定短语。

故填to inform。

3.My teacher is always the first person ________ (get) to the office。

【答案】to get【解析】考查动词不定式。

句意:我的老师总是第一个到达办公室的人。

当名词被序数词修饰时,用动词不定式作后置定语。

该句中名词person前被序数词the first修饰,故填to get.4。

She became the first black woman ________(elect) to the Senate.【答案】to be elected【解析】考查不定式的被动语态。

此处是the+序数词+n+to do结构,因the first black woman与elect之间是被动关系,需要使用不定式被动结构to be done形式。

故填to be elected。

高考语文二轮复习考点题型变形专练3词语组合题型含解析

高考语文二轮复习考点题型变形专练3词语组合题型含解析

词语组合题型1、在下面一段话的空缺处依次填入词语,最恰当的一组是( )成为文化世家的必要条件是家庭有着___________的文化底蕴。

北宋最著名的文化世家非苏氏一门莫属,父亲苏洵发奋苦读、勤于笔耕的美名___________,而苏轼和苏辙两兄弟在文坛的影响更是超过了父亲,真可谓___________。

A.深厚家喻户晓长江后浪推前浪B.深厚不言而喻这山望着那山高C.深重不言而喻长江后浪推前浪D.深重家喻户晓这山望着那山高2、在下面一段话空缺处依次填入词语,最恰当的一组是( )台上一分钟,台下十年功。

接地气的草根戏班,________的地方唱腔,着实令戏迷们为之________。

台上是帝王将相才子佳人,台下是男女老少芸芸众生,演戏人精益求精,演得________;看戏人津津有味,笑得前仰后合。

台上演员声泪俱下,台下观众也跟着哭鼻子抹眼泪:人们就这样莫名其妙地融进了悲剧的氛围之中。

A.字正腔圆迷倒惟妙惟肖B.原汁原味倾倒惟妙惟肖C.原汁原味迷倒栩栩如生D.字正腔圆倾倒栩栩如生3、在下面一段话的空缺处依次填入词语,最恰当的一组是庄子于学无所不窥,但真正令人无法__________的是他的天才和洒脱。

谁能像他那样用微笑来面对丑恶?而这微笑,只是轻微的一丝,不易察觉地掠过他的脸,便如炎阳照雪,那些丑陋悄然__________。

他只微微一笑,显出大智慧在面对丑恶世界时的从容与最使人_______的平淡。

A.望其项背溶化喜不自胜B.望尘莫及融化喜不自胜C.望尘莫及溶化忍俊不禁D.望其项背融化忍俊不禁4、依次填入下面横线处的词语,最恰当的一项是( )公益与每个人_________。

在互联网时代成长起来的“微公益”,真正实现了公益的平民化、常态化。

_________你没有亿万身价,没有强大的社会影响力,_________不妨碍你从事公益事业。

捐赠一本书,提供一份午餐,甚至转发一条微博,爱心就在你我之间_________。

2020年中考英语总复习专练三:改错专练(无答案)

2020年中考英语总复习专练三:改错专练(无答案)

2020年中考英语总复习专练三:改错专练(无答案)以下各题中均有一处错误,请指出并改正。

1.They flying kites in the park now.A B C D2.Would you please not to make any noise?The baby has just fallen asleep.A B C D3.Few of them can drive,can't they?A B C D4.His father had lots of moneys,so he could buy everything for him.A B C D5.What a bad weather!It's raining hard.A B C D6.You can sing in English,and so does he.A B C D7.Why not ask for help when you are with trouble?A B C D8.Ann didn't know how work out the problem in class.A B C D9.The Shutes are enjoying to live in the country.A B C D10.The children will go to the zoo if it won't rain tomorrow.A B C D11.Mary felt happily because her aunt gave her a present.A B C D12.Could you guess where did I put the ball?A B C D13.What about to play cards now?A B C D14.At first the child was sad.But he stopped to cry when he saw his mother.A B C D15.Is there new anything in today's newspaper?A B C D16.I have already bought this computer for two years.A B C D17.The teachers want to make a contribution to help the poor children return to school.A B C D18.Her parents used to taking a walk after supper when they were in that city.A B C D19.We had to wait for our teacher for the other two hours.A B C D20.The Greens have never gone to the Great Wall.A B C D21.I don't think play basketball is a lot of fun.A B C D22.The only letter which I have received is still on the shelf.A B C D23.I hope my son to get on well with his classmates.A B C D24.—How did you come here this morning?A B—I came here by Mr Smith's car.C D25.How proudly they were when they heard the good news!A B C D26.It took the boy one and a half hour to fall asleep last night.A B C D27.He asked me if my father enjoy fishing at weekends.A B C D28.Could you tell me how much did he pay for the new jeans?A B C D29.She wants to go and stay with her aunt for sometimes in her summer holiday.A B C D30.Who can tell me how to do?I need some help.A B C D短文改错专练〔A〕There are usually three school terms in Britain:autumn,1.________spring and summer terms.The schools usually have a five-day holiday halfway in each term.Sometime the schools 2.________take their pupils in trips at half-term.Holidays can 3.________be different in different place.The schools usually have 4.________ten days in Christmas,ten days at Easter and six weeks 5.________in summer from the end of July to the begin of September.6.________Students can have the lunch in the school dining room.In 7.________these years more and most students have brought their 8.________own lunch〔packed lunch〕.They don't like eat in the dining 9.________room.All pupils enjoy talking how bad school food is.10.________〔B〕Once there was a businessman.Every day he worked too much.So later he found he could not asleep at night,but kept 1.________on going to sleep in the day.He became very worried that he went 2.________to see his doctor.〝Can you help me,doctor?〞he asked.〝I slept very good before,3.________but recently〔近来〕I haven't been able sleep for more than two 4.________hours a night.〞The doctor looked him up carefully.At last he gave the 5.________man some medicine,and told him to work less hard.Then 6.________he would soon be better.So the businessman grew worse and worse.He slept 7.________even less than ago at night,and was always falling 8.________asleep in his office.He visited to his doctor very often,9.________and it spent the doctor a long time to find 10.________the reason.The businessman's wife gave him the wrong medicine.She was giving him the sleeping medicine in the morning,and the one to keep him awake at night.〔C〕Miss Jones was a teacher.His home was far 1.________to her school,and she always walked there in the 2.________morning.All the pupils were very young.Miss Jones 3.________walked to school in a very cold and windy morning 4.________in October,and the cold wind goes into her eyes,and 5.________big tears began running out of them.She reached to 6.________the school,opened the door and went in the 7.________classroom.That was nice and warm there 8.________but Miss Jones was happy.But a small boy looked at her for a few 9.________seconds,put his arms round her and said kind,10.________〝Don't cry,Miss.Our school isn't very bad.〞〔D〕By midday,the sun was very strong.Jim was very tired 1.________to walk.There were not some trees near the road,so he 2.________rested by a big stone.After drink some water,he took off 3.________his shirt,lied down on the ground and fell asleep at once.4.________He was much tired that he didn't wake up until the evening.5.________He was just about to jump when he felt something move 6.________near his feet.He looked up and saw a long black snake.7.________Jim was too frightened not to move.The snake 8.________began to crawl〔爬〕cross his legs.It crawled on and on 9.________until it disappeared under the rock.Jim jumped to his feet,pick up his shirt and ran off down the road.10.________〔E〕Uncle Dick sent John and Rose a big vase〔花瓶〕.And their room was so big that they didn't have 1.________enough room for it.So they give it to a 2.________friend of them.The vase went from one house 3.________to another.At the end a woman sold it to Mr Hill.4.________One day John and Rose heard a letter from Uncle 5.________Dick.He was came to visit them.Of course,6.________he is going to see his own present in their 7.________house.They went to the shop,and Mr Hill had 8.________already sold the vase.When they got home,they 9.________found the same vase again,because Uncle Dickhad bought it from Mr Hill and sent it them 10.________。

2023年高考复习专项练习一轮语文第三板块 专题一 练案一 正确使用词语(包括熟语)

2023年高考复习专项练习一轮语文第三板块 专题一 练案一 正确使用词语(包括熟语)

第三板块语言文字运用专题一基于情境运用的语言策略练案一正确使用词语(包括熟语)1.在下面文段的横线处分别填入词语,最恰当的一项是(3分)()文学形象是文学理论中具有普遍意义的重要范畴。

在黑格尔的美学体系中,形象具有多义性。

一般认为,文学形象能够使读者更加集中地欣赏自然美、生活美和艺术美,作品里悲伤与喜悦的感情,其中的美好和丑恶,获得美的享受。

A.五花八门领会判定从而B.包罗万象领会判别从而C.五花八门感受判别因而D.包罗万象感受判定因而2.在下面一段话的横线处依次填入词语,最恰当的一项是(3分)()中国文学具有源远流长、的特点。

若论文学的悠久,只有古希腊文学、古印度文学可以与中国文学比肩;若论文学传统的不断,任何别的国家和民族的文学都不能与中国文学。

中国文学一直以丰富的作品、持久的魅力着全世界。

A.博大精深绵延相提并论滋养B.博古通今绵延齐头并进滋养C.博古通今绵亘相提并论滋补D.博大精深绵亘齐头并进滋补3.依次填入下列横线处的词语,最恰当的一项是(3分)()孙犁在给贾平凹的散文集《月迹》作的序中,更是地指出:文艺之途正如人生之途,过早的金榜、骏马、高官、高楼,过多的花红热闹,鼓噪喧腾,并不一定是好事。

人之一生,或是作家一生,要能经受得和寂寞,忍受得污蔑和凌辱。

要之,在这条道路上,冷也能安得,热也能处得,风里也来得,雨里也去得。

在历史上,到头来退却的,或者说是的,常常不是坚定的战士,而是那些跳梁的小丑。

这样的文字,为人为文的至理名言。

A.一语破的清贫偃旗息鼓竟是B.一针见血清苦销声匿迹不啻C.一语破的清苦偃旗息鼓竟是D.一针见血清贫销声匿迹不啻4.在下列横线处依次填入词语,最恰当的一项是(3分)()阐释学的观点是,不再追求一个阐释的终点。

阐释不是,千方百计地搜索某一种定音的标准答案,从而结束漫长的理论。

即使某个时代的读者达成了评价一部作品的,另一个时代的阅读又可能不同的观点。

A.披沙拣金跋涉共识催生B.拨云见日旅行共识衍生C.披沙拣金旅行见解衍生D.拨云见日跋涉见解催生5.在下面一段话的横线处依次填入词语,最恰当的一项是(3分)()我看过许多民国时期的文人论战,地说,鲁迅跟林语堂斗嘴,比鲁迅跟叶灵凤斗嘴,好看多了。

2020版高考英语大二轮专题复习新方略专练:语言知识运用练(三) Word版含解析

2020版高考英语大二轮专题复习新方略专练:语言知识运用练(三) Word版含解析

语言知识运用练(三)语言知识运用(共两节,满分45分)第一节完形填空(共20小题;每小题1.5分,满分30分)阅读下面短文,从短文后各题所给的A、B、C和D四个选项中,选出可以填入空白处的最佳选项。

[2019·大连高三一模] Being jobless for the last three months, I am upset nowadays. My son has lost his __1__ from a good school and he might have to go to a less qualified school. I have moved from a spacious (宽敞的) apartment to a small one in order to __2__ my living expenses. My wife has built extra stress. On top of that, my father-in-law __3__ a week ago, which has added fuel to fire, __4__ our family into a more terrible state.In these circumstances I have two mental __5__:either to feel upset and keep losing my peace of mind, or __6__ my negative thoughts with super mental powers about self-confidence and consistency and __7__ applying for new jobs with positive attitude.I choose the second option because I believe that the pain I am __8__ today will build up my __9__. My strengths will then increase my confidence and make me a (n) __10__ man one day. I understand that good and bad stages are part of __11__. While good times make me happy, bad times __12__ the “inner” me.Two months later, __13__,I have finally landed a job which is very exciting and offers a nice salary. Time has healed my wife's mental stress __14__ the death of her father. My son has already been accepted by another good school because the admissions for new academic year were still __15__!As I am back on the track of normal life, I can __16__ say that the most important things that helped me __17__ the crisis (危机) were my consistent character of patience, and __18__ attitude towards life. So, while I wish you all the best life can __19__ you, I would highly recommend taking the same attitude towards life __20__ you are trapped in any of such situations.体裁:记叙文题材:个人经历主题:积极应对困境【语篇解读】本文是一篇记叙文,讲述了作者成功应对困境的故事启迪读者要一直保持积极的生活态度。

2023中考数学一轮复习专题3

2023中考数学一轮复习专题3

专题3.3 平面直角坐标系与一次函数、反比例函数(巩固篇)(真题专练)一、单选题1.(2021·四川自贡·中考真题)如图,()8,0A ,()2,0C -,以点A 为圆心,AC 长为半径画弧,交y 轴正半轴于点B ,则点B 的坐标为( )A .()0,5B .()5,0C .()6,0D .()0,62.(2021·内蒙古鄂尔多斯·中考真题)已知:AOCD 的顶点()0,0O ,点C 在x 轴的正半轴上,按以下步骤作图:①以点O 为圆心,适当长为半径画弧,分别交OA 于点M ,交OC 于点N .①分别以点M ,N 为圆心,大于12MN 的长为半径画弧,两弧在AOC ∠内相交于点E .①画射线OE ,交AD 于点()2,3F ,则点A 的坐标为( )A .5,34⎛⎫- ⎪⎝⎭B .(3C .4,35⎛⎫- ⎪⎝⎭D .(23.(2021·辽宁锦州·中考真题)如图,在四边形DEFG 中,①E =①F =90°,①DGF =45°,DE =1,FG =3,Rt ①ABC 的直角顶点C 与点G 重合,另一个顶点B (在点C 左侧)在射线FG 上,且BC =1,AC =2,将①ABC 沿GF 方向平移,点C 与点F 重合时停止.设CG 的长为x ,①ABC 在平移过程中与四边形DEFG 重叠部分的面积为y ,则下列图象能正确反映y 与x 函数关系的是( )A .B .C .D .4.(2021·辽宁营口·中考真题)已知一次函数y kx k =-过点()1,4-,则下列结论正确的是( )A .y 随x 增大而增大B .2k =C .直线过点()1,0D .与坐标轴围成的三角形面积为25.(2021·贵州安顺·中考真题)小星在“趣味数学”社团活动中探究了直线交点个数的问题.现有7条不同的直线()1,2,3,4,5,6,7n n y k x b n =+=,其中12345,k k b b b ===,则他探究这7条直线的交点个数最多是( ) A .17个B .18个C .19个D .21个6.(2021·山东滨州·中考真题)如图,在OAB 中,45BOA ∠=︒,点C 为边AB 上一点,且2BC AC =.如果函数()90y x x=>的图象经过点B 和点C ,那么用下列坐标表示的点,在直线BC 上的是( )A .(-2019,674)B .(-2020,675)C .(2021,-669)D .(2022,-670)7.(2021·湖北荆门·中考真题)在同一直角坐标系中,函数y kx k =-与(0)||ky k x =≠的大致图象是( )A .①①B .①①C .①①D .①①8.(2021·辽宁丹东·中考真题)如图,点A 在曲线到12(0)y x x=>上,点B 在双曲线2(0)ky x x=<上,//AB x 轴,点C 是x 轴上一点,连接AC 、BC ,若ABC 的面积是6,则k 的值( )A .6-B .8-C .10-D .12-9.(2021·山东淄博·中考真题)如图,在平面直角坐标系中,四边形AOBD 的边OB 与x 轴的正半轴重合,//AD OB ,DB x ⊥轴,对角线,AB OD 交于点M .已知:2:3,AD OB AMD =的面积为4.若反比例函数ky x=的图象恰好经过点M ,则k 的值为( )A .275B .545C .585D .1210.(2021·山东威海·中考真题)一次函数()1110y k x b k =+≠与反比例函数()2220k y k x=≠的图象交于点(1,2)A --,点(2,1)B .当12y y <时,x 的取值范围是( ) A .1x <- B .10x -<<或2x > C .02x <<D .02x <<或1x <-11.(2021·内蒙古呼伦贝尔·中考真题)点()()()1235,,3,,3,y y y --都在反比例函数()0ky k x=>的图像上,则( ) A .312y y y >> B .123y y y >>C .132y y y >>D .213y y y >>二、填空题12.(2021·辽宁盘锦·中考真题)如图,在平面直角坐标系xOy 中,点A 在x 轴负半轴上,点B 在y 轴正半轴上,①D 经过A ,B ,O ,C 四点,①ACO =120°,AB =4,则圆心点D 的坐标是________13.(2021·山东潍坊·中考真题)在直角坐标系中,点A 1从原点出发,沿如图所示的方向运动,到达位置的坐标依次为:A 2(1,0),A 3(1,1),A 4(﹣1,1),A 5(﹣1,﹣1),A 6(2,﹣1),A 7(2,2),….若到达终点A n (506,﹣505),则n 的值为 _______.14.(2021·广西梧州·中考真题)如图,在同一平面直角坐标系中,直线l 1:y 14=x 12+与直线l 2:y =kx +3相交于点A ,则方程组11423y x y kx ⎧=+⎪⎨⎪=+⎩的解为 ___.15.(2021·贵州毕节·中考真题)如图,在平面直角坐标系中,点()11,1N 在直线:l y x =上,过点1N 作11N M l ⊥,交x 轴于点1M ;过点1M 作12M N x ⊥轴,交直线l 于点2N ;过点2N 作22N M l ⊥,交x 轴于点2M ;过点2M 作23M N x ⊥轴,交直线l 于点3N ;…;按此作法进行下去,则点2021M 的坐标为_____________.16.(2021·广西贺州·中考真题)如图,一次函数4y x =+与坐标轴分别交于A ,B 两点,点P ,C 分别是线段AB ,OB 上的点,且45OPC ∠=︒,PC PO =,则点P 的标为________.17.(2021·山东日照·中考真题)如图,在平面直角坐标系xOy 中,正方形OABC 的边OC 、OA 分别在x 轴和y 轴上,10OA =,点D 是边AB 上靠近点A 的三等分点,将OAD △沿直线OD 折叠后得到'OA D △,若反比例函数()0ky k x=≠的图象经过'A 点,则k 的值为_______.18.(2021·辽宁鞍山·中考真题)如图,ABC 的顶点B 在反比例函数(0)ky x x=>的图象上,顶点C 在x 轴负半轴上,//AB x 轴,AB ,BC 分别交y 轴于点D ,E .若32BE CO CE AD ==,13ABCS =,则k =_____.19.(2021·四川巴中·中考真题)如图,平行于y 轴的直线与函数y 1k x =(x >0)和y 22x=(x>0)的图象分别交于A 、B 两点,OA 交双曲线y 22x=于点C ,连接CD ,若OCD 的面积为2,则k =_______.20.(2021·湖北荆门·中考真题)如图,在平面直角坐标系中,Rt OAB 斜边上的高为1,30AOB ∠=︒,将Rt OAB 绕原点顺时针旋转90︒得到Rt OCD △,点A 的对应点C 恰好在函数(0)k y k x =≠的图象上,若在ky x =的图象上另有一点M 使得30MOC ∠=︒,则点M 的坐标为_________.21.(2021·黑龙江齐齐哈尔·中考真题)如图,点A 是反比例函数1(0)k y x x=<图象上一点,AC x ⊥轴于点C 且与反比例函数2(0)k y x x=<的图象交于点B ,3AB BC = ,连接OA ,OB ,若OAB 的面积为6,则12k k +=_________.22.(2021·内蒙古通辽·中考真题)如图,11OA B ,122A A B ,233A A B △…,1n n n A A B -都是斜边在x 轴上的等腰直角三角形,点1A ,2A ,3A ,…,n A 都在x 轴上,点1B ,2B ,3B ,…,n B 都在反比例函数()10y x x=>的图象上,则点n B 的坐标为__________.(用含有正整数n 的式子表示)23.(2021·山东潍坊·中考真题)如图,在直角坐标系中,O 为坐标原点a y x =与by x=(a >b >0)在第一象限的图象分别为曲线C 1,C 2,点P 为曲线C 1上的任意一点,过点P 作y 轴的垂线交C 2于点A ,作x 轴的垂线交C 2于点B ,则阴影部分的面积S ①AOB =_______.(结果用a ,b 表示)24.(2021·黑龙江绥化·中考真题)如图,在平面直角坐标系中,O 为坐标原点,MN 垂直于x 轴,以MN 为对称轴作ODE 的轴对称图形,对称轴MN 与线段DE 相交于点F ,点D 的对应点B 恰好落在(0,0)ky k x x=≠<的双曲线上.点O E 、的对应点分别是点C A 、.若点A 为OE 的中点,且1AEF S =△,则k 的值为____.25.(2021·广西柳州·中考真题)如图,一次函数2y x =与反比例数()0ky k x=>的图像交于A ,B 两点,点M 在以()2,0C 为圆心,半径为1的C 上,N 是AM 的中点,已知ON 长的最大值为32,则k 的值是_______.三、解答题26.(2021·山东青岛·中考真题)某超市经销甲、乙两种品牌的洗衣液,进货时发现,甲品牌洗衣液每瓶的进价比乙品牌高6元,用1800元购进甲品牌洗衣液的数量是用1800元购进乙品牌洗衣液数量的45.销售时,甲品牌洗衣液的售价为36元/瓶,乙品牌洗衣液的售价为28元/瓶.(1)求两种品牌洗衣液的进价;(2)若超市需要购进甲、乙两种品牌的洗衣液共120瓶,且购进两种洗衣液的总成本不超过3120元,超市应购进甲、乙两种品牌洗衣液各多少瓶,才能在两种洗衣液完全售出后所获利润最大?最大利润是多少元?27.(2021·辽宁沈阳·中考真题)如图,平面直角坐标系中,O 是坐标原点,直线15(0)y kx k =+≠经过点()3,6C ,与x 轴交于点A ,与y 轴交于点B .线段CD 平行于x 轴,交直线34y x =于点D ,连接OC ,AD .(1)填空:k = __________.点A 的坐标是(__________,__________); (2)求证:四边形OADC 是平行四边形;(3)动点P 从点O 出发,沿对角线OD 以每秒1个单位长度的速度向点D 运动,直到点D 为止;动点Q 同时从点D 出发,沿对角线OD 以每秒1个单位长度的速度向点O 运动,直到点O 为止.设两个点的运动时间均为t 秒. ①当1t =时,CPQ 的面积是__________.①当点P ,Q 运动至四边形CPAQ 为矩形时,请直接写出此时t 的值.28.(2021·甘肃兰州·中考真题)如图,一次函数12y x b =-+与反比例函数()100y x x =-<,()0ky x x=>图象分别交于()2,A m -,()4,B n ,与y 轴交于点C ,连接OA ,OB .(1)求反比例函数()0ky x x =>和一次函数12y x b =-+的表达式;(2)求AOB 的面积.29.(2021·山东济南·中考真题)如图,直线32y x =与双曲线()0k y k x=≠交于A ,B 两点,点A 的坐标为(),3m -,点C 是双曲线第一象限分支上的一点,连接BC 并延长交x 轴于点D ,且2BC CD =.(1)求k 的值并直接写出....点B 的坐标; (2)点G 是y 轴上的动点,连接GB ,GC ,求GB GC +的最小值;(3)P 是坐标轴上的点,Q 是平面内一点,是否存在点P ,Q ,使得四边形ABPQ 是矩形?若存在,请求出所有符合条件的点P 的坐标;若不存在,请说明理由.参考答案1.D 【分析】先根据题意得出OA =8,OC =2,再根据勾股定理计算即可 【详解】解:由题意可知:AC =AB ①()8,0A ,()2,0C - ①OA =8,OC =2 ①AC =AB =10在Rt ①OAB 中,6OB = ①B (0,6) 故选:D【点拨】本题考查勾股定理、正确写出点的坐标,圆的半径相等、熟练进行勾股定理的计算是关键 2.A 【分析】由题意得:OE 平分①AOC ,结合AD ①OC ,可得AO=AF ,设AH =m ,则AO =AF =2+m ,根据勾股定理,列出方程,即可求解. 【详解】解:由作图痕迹可知:OE 平分①AOC ,①①AOF =①COF ,①在AOCD 中,AD ①OC , ①①COF =①AFO , ①①AOF =①AFO ,①AO=AF , ①()2,3F , ①FH =2,OH =3,设AH =m ,则AO =AF =2+m , ①在Rt AOH 中,AH 2+OH 2=AO 2, ①m 2+32=(2+m ) 2,解得:54m =, ①A 5,34⎛⎫- ⎪⎝⎭,故选A .【点拨】本题主要考查平行四边形的性质,尺规作角平分线,勾股定理,等腰三角形的判定和性质,推出AO=AF ,利用勾股定理列出方程,是解题的关键. 3.B 【分析】根据移动过程分三个阶段讨论,第一个是点B 到达点G 之前,即0<x <1时,求出y 和x 的关系式,确定图象,第二个是点C 到达点H 之前,即1<x <2时,求出y 和x 的关系式,确定图象,第三个是点C 到达点F 之前,即2<x <3时,求出y 和x 的关系式,确定图象,即可确定选项. 【详解】解:过点D 作DH ①EF ,①①DGF =45°,DE =1,FG =3, ①EH =2,DH =EF =2,当0<x <1时,重叠部分为等腰直角三角形,且直角边长为x , ①y =212x ,①102>, ①该部分图象开口向上,当1<x <2时,如图,设A 'B '与DG 交与点N ,A 'C '与DG 交与点M , 则S 重叠=S ①GMC '﹣S ①GNB ', 设B 'K =a ,则NK =2a , ①GC '=x ,B 'C '=1, ①GB '=x ﹣1,①①GKN 是等腰直角三角形, ①GK =NK , ①x ﹣1+a =2a , ①a =x ﹣1, ①NK =2x ﹣2,①21(1)(22)212GNB S x x x x '∆=--=-+,①212GMC S x '∆=, ①S 重叠=212x ﹣(x 2﹣2x +1)=21212x x -+-,①102-<, ①该部分图象开口向下,当2<x <3时,重叠部分的面积为S ①ABC ,是固定值, ①该部分图象是平行x 轴的线段, 故选:B .【点拨】本题主要考查动点问题的函数图象,关键是要把移动过程分成几个阶段,然后根据每个阶段的情况单独讨论,确定y 和x 之间的函数关系式,从而确定图象. 4.C 【分析】将点()1,4-代入一次函数解析式,求出k 的值,利用一次函数的图象与性质逐一判断即可.解:①一次函数y kx k =-过点()1,4-, ①4k k =--,解得2k =-,①一次函数为22y x =-+,y 随x 增大而减小,故A 和B 错误; 当1x =时,0y =,故C 正确;该一次函数与x 轴交于点()1,0,与y 轴交于点()0,2, ①与坐标轴围成的三角形面积为11212⨯⨯=,故D 错误;故选:C .【点拨】本题考查一次函数的图象与性质,利用待定系数法求出一次函数解析式是解题的关键. 5.B 【分析】因为题中已知12345,k k b b b ===,可知:第1、2条直线相互平行没有交点,第3、4、5条直线交于一点,由此即可求解此题. 【详解】解:①直线()1,2,3,4,5,6,7n n y k x b n =+=,其中12345,k k b b b === ①第1、2条直线相互平行没有交点,第3、4、5条直线交于一点, ①这5条直线最多有7个交点,第6条直线,与前面5条直线的交点数最多有5个, 第7条直线,与前面6条直线的交点数最多有6个, ①得出交点最多就是7+5+6=18条, 故选:B .【点拨】本题考查了两条直线相交或平行问题,做题关键在于分析得出两条平行直线,三条直线相交于一点. 6.D 【分析】根据反比例函数图象上点的坐标特征,求出B 、C 点的坐标,再写出BC 解析式,再判断点在BC 上.解:作BD OA ⊥,CE OA ⊥,45BOA ∠=︒,BD OD ∴=,设(,)B a a ,∴9a a=, 3a ∴=或3a =-(舍去), 3BD OD ∴==,(3,3)B , 2BC AC =.3ABAC ,BD OA ⊥,CE OA ⊥,//BD CE ∴,.ABD ACE ∴∆∆∽3BD ABCE AC==, ∴33CE=, 1CE ∴=,图象经过点C ,∴91x=, 9x ∴=,(9,1)C设BC 的解析式为y kx b =+,3319k b k b=+⎧⎨=+⎩,解得134k b ⎧=-⎪⎨⎪=⎩, ∴143y x =-+,当2019x =-时,677y =, 当2020x =-时,16773y =, 当2021x =时,26693y =-, 当2022x =时,670y =-, 故选:D .【点拨】本题考查反比例函数图象上的点的性质,能求出BC 的解析式是解题的关键. 7.B 【分析】根据k 的取值范围,分别讨论k >0和k <0时的情况,然后根据一次函数和反比例函数图象的特点进行选择正确答案. 【详解】 解:当k >0时,一次函数y=kx -k 经过一、三、四象限, 函数的(0)||ky k x =≠(k≠0)的图象在一、二象限, 故选项①的图象符合要求. 当k <0时,一次函数y=kx -k 经过一、二、四象限, 函数的(0)||ky k x =≠(k≠0)的图象经过三、四象限, 故选项①的图象符合要求. 故选:B .【点拨】此题考查反比例函数的图象问题;用到的知识点为:反比例函数与一次函数的k 值相同,则两个函数图象必有交点;一次函数与y 轴的交点与一次函数的常数项相关. 8.C 【分析】根据//AB x 轴可以得到6ABCAOBS S==,转换成反比例函数面积问题即可解题.【详解】连接OA 、OB ,设AB 与y 轴交点为M ,①//AB x 轴 ①AB ①y 轴,6ABCAOBS S==①12BOMS k =,1212AOMS =⨯= ①6ABCAOBBOMAOMS S SS==+=①1162k += 解得10k =± ①点B 在双曲线2(0)ky x x=<上,且B 在第二象限 ①0k < ①10k =- 故选C【点拨】本题考查反比例函数问题,熟记反比例函数面积与k 的关系是解题的关键. 9.B 【分析】过点M 作ME ①x 轴于点E ,则有ME ①BD ,2MEOk S=,进而可得ADM BOM ∽、OME ODB ∽,然后根据相似三角形的面积比与相似比的关系可进行求解.【详解】解:过点M 作ME ①x 轴于点E ,如图所示:①DB x⊥轴,①ME①BD,①//AD OB,①ADM BOM∽,①:2:3AD OB=,①249 ADMBOMS ADS OB⎛⎫==⎪⎝⎭,①AMD的面积为4,①9BOMS=,①:2:3AD OB=,①:3:5OM OD=,由题可知①OMB、①OBD的高是相同的,则有35BOM OBDS S=,①453OBDS=,①ME①BD,①OME ODB∽,①2925 OMEODBS OMS OD⎛⎫==⎪⎝⎭,①275OMES=,由反比例函数k 的几何意义可得:2MEOk S =,①0k >, ①545k =; 故选B .【点拨】本题主要考查反比例函数k 的几何意义及相似三角形的性质与判定,熟练掌握反比例函数k 的几何意义及相似三角形的性质与判定是解题的关键. 10.D 【分析】先确定一次函数和反比例函数解析式,然后画出图象,再根据图象确定x 的取值范围即可. 【详解】解:①两函数图象交于点(1,2)A --,点(2,1)B①112=12k b k b --+⎧⎨=+⎩,221k -=-,解得:1=11k b ⎧⎨=-⎩,k 2=2 ①11y x =-,22y x=画出函数图象如下图:由函数图象可得12y y <的解集为:0<x <2或x <-1.故填D .【点拨】本题主要考查了运用待定系数法求函数解析式以及根据函数图象确定不等式的解集,根据题意确定函数解析式成为解答本题的关键. 11.A 【分析】根据反比例函数的增减性解答即可. 【详解】 解:①()0ky k x=>, ①在每个象限内,y 随着x 的增大而减小, ①-5<-3<0<3, ①312y y y >>, 故选:A .【点拨】此题考查反比例函数的增减性:当k >0时,在每个象限内,y 随着x 的增大而减小;当k <0,在每个象限内,y 随着x 的增大而增大.12.D (1) 【分析】先利用圆内接四边形的性质得到①ABO =60°,再根据圆周角定理得到AB 为①D 的直径,则D 点为AB 的中点,接着利用含30度的直角三角形三边的关系得到OB =2,OA =以A (0),B (0,2),然后利用线段的中点坐标公式得到D 点坐标. 【详解】解:①四边形ABOC 为圆的内接四边形, ①①ABO +①ACO =180°, ①①ABO =180°−120°=60°, ①①AOB =90°, ①AB 为①D 的直径, ①D 点为AB 的中点,在Rt①ABO 中,①①ABO =60°,①OB =12AB =2,①OA =①A (0),B (0,2),①D 点坐标为(1).故答案为(1).【点拨】本题考查了圆周角定理:在同圆或等圆中,同弧或等弧所对的圆周角相等,都等于这条弧所对的圆心角的一半.半圆(或直径)所对的圆周角是直角,90°的圆周角所对的弦是直径.也考查了坐标与图形性质.13.2022【分析】 终点()506505n A -,在第四象限,寻找序号与坐标之间的关系可求n 的值. 【详解】解:①()506505-,是第四象限的点, ①()506505n A -,落在第四象限. ①在第四象限的点为()()()()61014213243506505n A A A A ---⋯-,,,,,,,,. ①64121042214432=⨯-+=⨯-+=⨯-+,,,18442=⨯-+⋯,, ①450522022n =⨯-+=.故答案为:2022【点拨】本题考查了点坐标的位置及坐标变化规律的知识点,善于观察并寻找题目中蕴含的规律是解题的关键.14.21x y =⎧⎨=⎩【分析】由题意,两直线的交点坐标就是这两条直线组成的方程组的解,即可得到答案.【详解】解:根据题意,①直线l 1:y 14=x 12+与直线l 2:y =kx +3相交于点A (2,1), ①方程组11423y x y kx ⎧=+⎪⎨⎪=+⎩的解为21x y =⎧⎨=⎩; 故答案为:21x y =⎧⎨=⎩. 【点拨】本题考查了一次函数与二元一次方程组的关系,解题的关键是掌握两直线的交点坐标就是这两条直线组成的方程组的解.15.(20212,0).【分析】根据题目所给的解析式,求出对应的1M 坐标,然后根据规律求出n M 的坐标,最后根据题目要求求出最后答案即可.【详解】解:如图,过点N 作NM ①x 轴于M将1x =代入直线解析式y x =中得1y =①1OM MN ==,MON ∠=45°①1ONM =∠90°①1ON NM =①1ON NM ⊥①11OM MM ==①1M 的坐标为(2,0)同理可以求出2M 的坐标为(4,0)同理可以求出3M 的坐标为(8,0)同理可以求出n M 的坐标为(2n ,0)①2021M 的坐标为(20212,0)故答案为:(20212,0).【点拨】本题主要考查了直线与坐标轴之间的关系,解题的关键在于能够发现规律.16.(--【分析】过P 作PD ①OC 于D ,先求出A ,B 的坐标,得①ABO =①OAB =45°,再证明①PCB ①①OP A ,从而求出BD =OD =【详解】如图所示,过P 作PD ①OC 于D ,①一次函数4y x =+与坐标轴分别交于A ,B 两点,①A (-4,0),B (0,4),即:OA =OB ,①①ABO =①OAB =45°,①①BDP 是等腰直角三角形,①①PBC =①CPO =①OAP =45°,①①PCB +①BPC =135°=①OP A +①BPC ,①①PCB =①OP A ,又①PC =OP ,①①PCB ①①OP A (AAS ),①AO =BP =4,①Rt ①BDP 中,BD =PD =BP=①OD =OB −BD =,①P (-).故答案是:P (-,.【点拨】本题主要考查了一次函数图象上点的坐标特征以及等腰三角形的性质,结合等腰三角形的性质,判定全等三角形是解决问题的关键.17.48【分析】过A '作EF OC ⊥于F ,交AB 于E ,设(,)A m n ',OF m =,A F n '=,通过证得①A OF '∽①DA E ',得到310103m n n m ==--,解方程组求得m 、n 的值,即可得到A '的坐标,代入(0)k y k x =≠即可求得k 的值.【详解】解:过A '作EF OC ⊥于F ,交AB 于E ,90OA D ∠'=︒,90OA F DA E ∴∠'+∠'=︒,90OA F AOF ∠'+∠'=︒,DA E AOF ∴∠'=∠',A FO DEA ∠'=∠',∴①A OF '∽①DA E ', ∴OF A F OA A E DE A D''=='',设(,)A m n ',OF m ∴=,A F n '=,正方形OABC 的边OC 、OA 分别在x 轴和y 轴上,10OA =,点D 是边AB 上靠近点A 的三等分点,103DE m ∴=-,10A E n '=-, ∴310103m n n m ==--, 解得6m =,8n =,(6,8)A ∴', 反比例函数(0)k y k x=≠的图象经过A '点, 6848k ∴=⨯=,故答案为48.【点拨】本题考查了正方形的性质,反比例函数图象上点的坐标特征,三角形相似的判定和性质,求得A '的坐标是解题的关键.18.18【分析】过点B 作BF x ⊥轴于点F ,通过设参数表示出①ABC 的面积,从而求出参数的值,再利用①ABC 与矩形ODBF 的关系求出矩形面积,即可求得 k 的值.【详解】解:如图,过点B 作BF x ⊥轴于点F .//AB x 轴,DBE COE ∴∽,DB BE DE CO CE EO∴==,32BE CO CE AD ==, 32DB DE BE CO CO EO CE AD ∴====, 设3CO a =,3DE b =,则2AD a =,2OE b =,332DB a ∴=,5OD b =, 92a BD ∴=, 132a AB AD DB ∴=+=, 1113513222ABC a S AB OD b =⋅⋅=⨯⨯=, 45ab ∴=, 94551822ODBF a ab S BD OD b ⋅=⋅===矩形, 又反比例函数图象在第一象限,18k ∴=,故答案为18.【点拨】此题考查反比例函数知识,涉及三角形相似及利用相似求长度,矩形面积公式等,难度一般.19.8【分析】设A (m ,k m ),则B (m ,2m ),D (m ,0),C (n ,k n ),由112=222OCD C m S OD y m n n ===△得出12n m =,再根据()1122OCD OAD ACD k S S S k m n m=-=--△△△求解即可得到答案. 【详解】解:设A (m ,k m ),则B (m ,2m ),D (m ,0),C (n ,k n ), ①112=222OCD C m S OD y m n n ===△, ①12n m =, 又①()1122OCD OAD ACD k S S S k m n m=-=--△△△ 112m n k m -⎛⎫=- ⎪⎝⎭12n k m =14k = ①124k = 解得8k故答案为:8.【点拨】本题主要考查了反比例函数与一次函数的交点问题,反比例函数比例系数的几何意义,函数图像上点的坐标特征,三角形的面积,解题的关键在于能够熟练掌握相关知识进行求解.20. 【分析】利用30的正切可以求出C 点坐标,再利用C 、M 在(0)k y k x =≠上,设M 的坐标,最后通过30MOF ∠=︒可以求出M 点的坐标.【详解】解:如图,过点C 作CE y ⊥轴,过点M 作MF x ⊥轴,由题意可知30EOC MOF ∠=∠=︒,1CE =则tan 30CE OE ==︒C 在(0)k y k x=≠上,k ∴=设)M m (0)m > 30MOF ∠=︒tan MOF ∴∠=解得1,1m m ==-(不符合题意,舍去)所以M故答案为:.【点拨】本题考查了直角三角形的性质,特殊角的锐角三角函数,反比例函数性质,正确理解题意,求出C 点的坐标是解决问题的关键.21.20【分析】利用反比例函数比例系数k 的几何意义得到S ①AOC =12|1k |=-112k ,S ①BOC =12|2k |=-212k ,利用AB =3BC 得到S ①ABO =3S ①OBC =6,所以-212k =2,解得2k =-4,再利用-112k =6+2得1k =-16,然后计算1k +2k 的值.【详解】解:①AC ①x 轴于点C ,与反比例函数y =2k x (x <0)图象交于点B , 而1k <0,2k <0,①S ①AOC =12|1k |=-112k ,S ①BOC =12|2k |=-212k , ①AB =3BC ,①S ①ABO =3S ①OBC =6,即-212k =2,解得2k =-4, ①-112k =6+2,解得1k =-16, ①1k +2k =-16-4=-20.故答案为:-20.【点拨】本题考查了反比例函数比例系数k 的几何意义:在反比例函数的图象上任意一点向坐标轴作垂线,这一点和垂足以及坐标原点所构成的三角形的面积是12|k |,且保持不变.22. 【分析】根据等腰直角三角形的性质,得到1B 的横,纵坐标相等,在结合反比例函数解析式求得该点的坐标,再根据等腰三角形的性质和反比例函数的解析式首先求得各个点的坐标,发现其中的规律,从而得到答案.【详解】11OB A △为等腰三角形∴直线1OB 的解析式为y x = 由题意得:1y x y x =⎧⎪⎨=⎪⎩解得1x =()111B ∴,1OB ∴=112OA ∴==()12,0A ∴122A A B △为等腰三角形∴设直线12A B 的解析式为y x b =+02b ∴=+,解得2b =-∴直线12A B 的解析式为2y x =- ∴21y x y x =-⎧⎪⎨=⎪⎩解得1x =)21B ∴21222B A A y ∴==∴点2A ()233A A B △为等腰三角形∴设直线23A B 的解析式为1y x b =+∴10b =解得1b =-∴直线23A B的解析式为y x =-1y x y x ⎧=-⎪⎨=⎪⎩解得x =∴3B 综上可得:点()111B ,,点)21B,点3B 总结规律可得n B坐标为:故答案为: 【点拨】本题综合考查了等腰直角三角形的性质以及结合反比例函数的解析式求得点的坐标,解答本题的关键是找出其中的规律求出坐标.23.12a 22b a- 【分析】设B (m ,b m ),A (b n,n ),则P (m ,n ),阴影部分的面积S ①AOB =矩形的面积﹣三个直角三角形的面积可得结论.【详解】解:设B (m ,b m ),A (b n,n ),则P (m ,n ), ①点P 为曲线C 1上的任意一点,①mn =a ,①阴影部分的面积S ①AOB =mn 12-b 12-b 12-(m b n -)(n b m-) =mn ﹣b 12-(mn ﹣b ﹣b 2b mn+)=mn ﹣b 12-mn +b 22b mn- 12=a 22b a-. 故答案为:12a 22b a-. 【点拨】本题考查了反比例函数的系数k 的几何意义,矩形的面积,反比例函数图象上点的坐标特征等知识,本题利用参数表示三角形和矩形的面积并结合mn =a 可解决问题. 24.24-【分析】先利用轴对称和中点的定义,确定EG 和EO 之间的关系,再利用平行线分线段成比例定理及推论,得到FG 和OD 之间的关系,设EG =x ,FG =y ,用它们表示出D 点坐标,接着得到B 点坐标,利用1AEF S =△,得到1xy =,再利用反比例函数的定义,计算出B 点横纵坐标的积,即为所求k 的值.【详解】解:如图所示,由轴对称的性质可知:GE =GA ,CG =OG ,BC =OD ,①点A 为OE 的中点,①AE =OA , ①1244EG EG EG OE AE EG ===, ①MN ①y 轴, ①14FG EG OD EO ==, ①=4OD FG ,①1AEF S =△, ①112AE FG ⋅=, ①1212EG FG ⨯⋅=, ①1EG FG ⋅=,设EG =x ,FG =y ,则OG =3x ,OD =4y ,①()0,4D y ,因为D 点和B 点关于MN 对称,①()6,4B x y -①1EG FG ⋅=,①1xy =①6424x y -⋅=-,①点B 恰好落在(0,0)k y k x x=≠<的双曲线上, ①24k =-,故答案为:24-.【点拨】本题考查了轴对称的性质、中点的定义、平行线分线段成比例定理的推论、反比例函数的定义等内容,解决本题的关键是牢记相关定义与性质,能根据题意在图形中找到对应关系,能挖掘图形中的隐含信息等,本题蕴含了数形结合的思想方法等.25.3225【分析】根据题意得出ON 是ABM 的中位线,所以ON 取到最大值时,BM 也取到最大值,就转化为研究BM 也取到最大值时k 的值,根据,,B C M 三点共线时,BM 取得最大值,解出B 的坐标代入反比例函数即可求解.【详解】解:连接BM ,如下图:在ABM 中,,O N 分别是,AB AM 的中点,ON ∴是ABM 的中位线,12ON BM ∴=, 已知ON 长的最大值为32, 此时的3BM =,显然当,,B C M 三点共线时,取到最大值:3BM =,13BM BC CM BC =+=+=,2BC ∴=,设(,2)B t t ,由两点间的距离公式:2BC ==,22(2)44t t ∴-+=, 解得:124,05t t ==(取舍), 48(,)55B ∴, 将48(,)55B 代入()0k y k x=>, 解得:3225k =, 故答案是:3225.【点拨】本题考查了一次函数、反比例函数、三角形的中位线、圆,研究动点问题中线段最大值问题,解题的关键是:根据中位线的性质,利用转化思想,研究BM 取最大值时k 的值. 26.(1)甲品牌洗衣液进价为30元/瓶,乙品牌洗衣液进价为24元/瓶;(2)购进甲品牌洗衣液40瓶,乙品牌洗衣液80瓶时所获利润最大,最大利润是560元【分析】(1)设甲品牌洗衣液每瓶的进价是x 元,则乙品牌洗衣液每瓶的进价是(x -6)元,根据数量=总价÷单价,结合用1800元购进乙品牌洗衣液数量的45,即可得出关于x 的分式方程,解之经检验后即可得出结论;(2)设可以购买m 瓶乙品牌洗手液,则可以购买(100-m )瓶甲品牌洗手液,根据总价=单价×数量,结合总费用不超过1645元,即可得出关于m 的一元一次不等式,解之即可得出m 的取值范围,再取其中的最大整数值即可得出结论.【详解】解:(1)设甲品牌洗衣液进价为x 元/瓶,则乙品牌洗衣液进价为()6x -元/瓶, 由题意可得,180********x x =⋅-, 解得30x =,经检验30x =是原方程的解.答:甲品牌洗衣液进价为30元/瓶,乙品牌洗衣液进价为24元/瓶.(2)设利润为y 元,购进甲品牌洗衣液m 瓶,则购进乙品牌洗衣液()120m -瓶,由题意可得,()30241203120m m +-≤,解得40m ≤,由题意可得,()()()363028*********y m m m =-+--=+,①20k =>,①y 随m 的增大而增大,①当40m =时,y 取最大值,240480560y =⨯+=最大值.答:购进甲品牌洗衣液40瓶,乙品牌洗衣液80瓶时所获利润最大,最大利润是560元.【点拨】本题考查分式方程的应用,一次函数的应用,一元一次不等式的应用,解题的关键是灵活运用所学知识解决问题.27.(1)3-,5,0;(2)见解析;(3)①12;①55+【分析】(1)代入C 点坐标即可得出k 值确定直线的解析式,进而求出A 点坐标即可; (2)求出AD 点坐标,根据CD OA =,//CD OA ,即可证四边形OADC 是平行四边形; (3)①作CH OD ⊥于H ,设出H 点的坐标,根据勾股定理计算出CH 的长度,根据运动时间求出PQ 的长度即可确定CPQ ∆的面积;①根据对角线相等确定PQ 的长度,再根据P 、Q 的位置分情况计算出t 值即可.【详解】解:(1)直线15(0)y kx k =+≠经过点(3,6)C ,3156k ∴+=,解得3k =-,即直线的解析式为315y x =-+,当0y =时,5x =,(5.0)A ∴,(2)线段CD 平行于x 轴,D ∴点的纵坐标与C 点一样,又D 点在直线34y x =上, 当6y =时,8x =,即(8,6)D ,835CD ∴=-=,5OA =,OA CD ∴=,又//OA CD ,∴四边形OADC 是平行四边形;(3)①作CH OD ⊥于H ,H 点在直线34y x =上,∴设H 点的坐标为3(,)4m m , 2223(3)(6)4CH m m ∴=-+-,2223(8)(6)4DH m m =-+-, 由勾股定理,得222CH DH CD +=, 即2222233(3)(6)(8)(6)544m m m m -+-+-+-=, 整理得245=m 或8(舍去), 3CH ∴=,810OD =,∴当1t =时,10118PQ OD t t =--=--=,11831222CPQ S PQ CH ∆∴=⋅=⨯⨯=, ①10OD =,当05t 时,102PQ t =-,当510t 时,210PQ t =-,当点P ,Q 运动至四边形CPAQ 为矩形时,PQ AC =,(5AC ==当05t 时,102t -=,解得5t =当510t 时,210t -=解得5t =综上,当点P ,Q 运动至四边形CPAQ 为矩形时t 的值为55+【点拨】本题主要考查一次函数的性质,熟练掌握待定系数法求解析式,平行四边形的性质和矩形的性质是解题的关键.28.(1)()80y x x =>,142y x =-+;(2)12. 【分析】(1)把点A 的坐标代入()100y x x =-<m 的值,得出A 的坐标代入12y x b =-+,求出一次函数的解析式,进而求得点B 的坐标,利用B 点的坐标求得()0ky x x =>的解析式;(2)根据一次函数解析式求得点C 的坐标,再将y 轴作为分割线,求得①AOB 的面积;【详解】解:(1)①()2,A m -,在函数()100y x x=-<的图象上, ①m =5,①A (-2,5),把A (-2,5)代入12y x b =-+得:15(2)2b =-⨯-+, ①b =4,①一次函数12y x b =-+的表达式为:142y x =-+, ①()4,B n 在函数142y x =-+的图象上, ①n =2,①()4,2B ,把()4,2B 代入()0k y x x =>得:2=4k ,①k =8, ①反比例函数的解析式为:()80y x x=>; (2)①C 是直线AB 与y 轴的交点,直线AB :142y x =-+, ①当x =0时,y =4,①点C (0,4),即OC =4,①A (-2,5),()4,2B ,①AOB AOC BOC S S S =+△△△=12×4×2+12×4×4=12;【点拨】本题考查了反比例函数与一次函数的交点问题,用待定系数法求一次函数与反比例函数的解析式,根据题意求出C 点坐标是解题的关键.29.(1)6k =,B (2,3);(2)(3)P (132,0)或(0,133). 【分析】(1)根据直线32y x =经过点A (),3m -,可求出点A (-2,-3),因为点A 在()0k y k x =≠图象上,可求出k ,根据点A 和点B 关于原点对称,即可求出点B ;(2)先根据2BC CD =利用相似三角形的性质求出点C ,再根据对称性求出点B 关于y 轴的对称点B ’,连接B ’C ,即B ’C 的长度是GB GC +的最小值;(3)先作出图形,分情况讨论,利用相似三角形的性质求解即可.【详解】(1)解:因为直线32y x =经过点A (),3m -, 所以332m -=⨯, 所以m =-2,所以点A (-2,-3),因为点A 在()0k y k x=≠图象上, 所以()236k =-⨯-=, 因为32y x =与双曲线()0k y k x=≠交于A ,B 两点, 所以点A 和点B 关于原点对称,所以点B (2,3);(2)过点B ,C 分别作BE ①x 轴,CF ①x 轴,作B 关于y 轴对称点B’,连接B’C ,因为BE ①x 轴,CF ①x 轴,所以BE //CF ,所以BED CFD , 所以BE BD CF CD=, 因为2BC CD =, 所以31BE BD CF CD ==, 因为B (2,3),所以BE =3,所以CF =1,所以C 点纵坐标是1,将1C y =代入6y x=可得:x =6, 所以点C (6,1),又因为点B’是点B 关于y 轴对称的点,所以点B’(-2,3),所以B’C ==,即GB GC +的最小值是(3)解:①当点P 在x 轴上时,当①ABP =90°,四边形ABPQ 是矩形时,过点B 作BH ①x 轴,因为①OBP =90°,BH ①OP ,所以OHBBHP ,。

2023新教材高考化学二轮专题复习 专练3 元素推断专项集训

2023新教材高考化学二轮专题复习 专练3 元素推断专项集训

专练3 元素推断专项集训一、单项选择题1.[2022·江苏省苏、锡、常、镇四市一调]短周期主族元素X、Y、Z、W的原子序数依次增大,X与Z处于同一主族,Y是短周期中电负性最大的元素,Z是同周期基态原子中未成对电子数最多的元素,W的族序数是周期序数的2倍。

下列有关说法正确的是( )A.元素Z在周期表中位于第3周期第ⅤA族B.原子半径:r(X)<r(Y)C.Z的第一电离能比W的小D.X的简单气态氢化物的热稳定性比Z的弱2.[2022·福建四地市第一次检测]某种净水剂由原子序数依次增大的R、W、X、Y、Z 五种元素组成。

五种元素分处三个短周期,包含地壳中含量前三的元素。

基态Z原子的成对电子占据的轨道数与未成对电子占据的轨道数之比为7∶2。

下列说法不正确的是( )A.简单离子半径:Z>X>WB.简单氢化物稳定性:W>Z>YC.X与Z形成的化合物在水中会生成沉淀和气体D.第一电离能:Z>Y>X3.[2022·湖南省邵阳市一模]物质A可用作抗氧化增效剂等,其结构式如图所示。

物质A的组成元素X、Y、Z、Q为原子序数依次增大的短周期主族元素,且Y的一种单质是天然存在的最硬的物质。

下列说法错误的是( )A.物质A与稀硫酸反应生成的有机物能发生缩聚反应B.X、Y、Z、Q四种元素原子半径的大小顺序为Q>Z>Y>XC.Y与Z分别形成的简单氢化物的沸点:Z>YD.Q、Z两种元素形成的化合物中可能存在共价键4.[2022·广东省韶关市一模]一种高效电解质的结构如图所示,W、Y、X、Z、Q均为短周期元素,且原子序数依次增大,X与Q同族,Y和Z的原子序数之和与Q相等。

下列说法正确的是( )A.X与Q的最高化合价均为+6B.化合物QX2、YX2均能与NaOH溶液反应C.简单氢化物的沸点:Q>Z>XD.W单质在空气中燃烧的产物是W2O25.[2022·广东省茂名市一模]化合物W可用于农药生产,其结构如图所示,其中X、Y、Z、M、N是原子序数依次增大的短周期主族元素,Y与N同主族。

2023中考数学一轮复习专题3

2023中考数学一轮复习专题3

专题3.10 “设参求值”解决函数动点问题(真题专练)中考中“设参求值”是解题中常用的方法,其解题步骤为:设参数-表示点的坐标-表示线段长-建立等量关系-建立方程-解方程消参。

在函数中常常用此方法解决动点问题,设参数可以一个或两个,据题特征而定。

一、单选题1.(2021·山东滨州·中考真题)如图,在OAB 中,45BOA ∠=︒,点C 为边AB 上一点,且2BC AC =.如果函数()90y x x=>的图象经过点B 和点C ,那么用下列坐标表示的点,在直线BC 上的是( )A .(-2019,674)B .(-2020,675)C .(2021,-669)D .(2022,-670)2.(2017·山东滨州·中考真题)在平面直角坐标系内,直线AB 垂直于x 轴于点C (点C 在原点的右侧),并分别与直线y =x 和双曲线y =相交于点A 、B ,且AC +BC =4,则△OAB的面积为 A .2+3或2-3B .+1或-1C .2-3D .-13.(2019·四川眉山·中考真题)如图,一束光线从点()4,4A 出发,经y 轴上的点C 反射后经过点()10B ,,则点C 的坐标是( )A .10,2⎛⎫ ⎪⎝⎭B .40,5⎛⎫ ⎪⎝⎭C .()0,1D .()0,24.(2021·四川乐山·中考真题)如图,已知直线1:24l y x =-+与坐标轴分别交于A 、B 两点,那么过原点O 且将AOB 的面积平分的直线2l 的解析式为( )A .12y x =B .y x =C .32y x =D .2y x =5.(2021·广东广州·中考真题)在平面直角坐标系xOy 中,矩形OABC 的点A 在函数()10y x x =>的图象上,点C 在函数()40y x x=-<的图象上,若点B 的横坐标为72-,则点A 的坐标为( )A .1,22⎛⎫⎪⎝⎭B .⎝C .12,2⎛⎫⎪⎝⎭D .⎭6.(2021·辽宁营口·中考真题)如图,在平面直角坐标系中,菱形ABCD 的边BC 与x 轴平行,A ,B 两点纵坐标分别为4,2,反比例函数ky x=经过A ,B 两点,若菱形ABCD 面积为8,则k 值为( )A .-B .-C .8-D .-7.(2021·黑龙江·中考真题)如图,在平面直角坐标系中,菱形ABCD 的边AD y ⊥轴,垂足为E ,顶点A 在第二象限,顶点B 在y 轴正半轴上,反比例函数(0ky k x=≠,0)x >的图象同时经过顶点C D 、.若点C 的横坐标为5,2BE DE =,则k 的值为( )A .403 B .52C .54D .2038.(2019·广西玉林·中考真题)已知抛物线21:(1)12C y x =--,顶点为D ,将C 沿水平方向向右(或向左)平移m 个单位,得到抛物线1C ,顶点为1D ,C 与1C 相交于点Q ,若160DQD ︒∠=,则m 等于( )A .±B .±C .﹣2或D .﹣4或9.(2020·江苏宿迁·中考真题)如图,在平面直角坐标系中,Q 是直线y=﹣12x+2上的一个动点,将Q 绕点P(1,0)顺时针旋转90°,得到点Q ',连接OQ ',则OQ '的最小值为( )A B C D 10.(2017·湖北荆门·中考真题)已知:如图,在平面直角坐标系中,等边的边长为6,点在边上,点在边上,且.反比例函数的图象恰好经过点和点.则的值为 ( )A.B.C.D.11.(2021·湖南怀化·中考真题)如图,菱形ABCD的四个顶点均在坐标轴上,对角线AC、BD交于原点O,AE BC⊥于E点,交BD于M点,反比例函数0)y x=>的图象经过线段DC的中点N,若4BD=,则ME的长为()A.53ME=B.43=MEC.1ME=D.23 ME=12.(2017·广西·中考真题)如图,垂直于x轴的直线AB分别与抛物线:(x≥0)和抛物线:(x≥0)交于A,B两点,过点A作CD△x轴分别与y轴和抛物线C2交于点C,D,过点B作EF△x轴分别与y轴和抛物线C1交于点E,F,则的值为()A.B.C.D.二、填空题13.(2020·江苏泰州·中考真题)如图,点P在反比例函数3yx=的图像上且横坐标为1,过点P作两条坐标轴的平行线,与反比例函数kyx=()0k<的图像相交于点A、B,则直线AB与x轴所夹锐角的正切值为______.14.(2019·四川乐山·中考真题)如图,点P是双曲线C:4yx=(0x>)上的一点,过点P作x轴的垂线交直线AB:122y x=-于点Q,连结OP,OQ.当点P在曲线C上运动,且点P在Q的上方时,△POQ面积的最大值是______.15.(2014·黑龙江牡丹江·中考真题)如图,在平面直角坐标系中,点A(0,4),B(3,0),连接AB,将△AOB沿过点B的直线折叠,使点A落在x轴上的点A′处,折痕所在的直线交y轴正半轴于点C,则直线BC的解析式为_____.16.(2018·山东东营·中考真题)如图,在平面直角坐标系中,点A 1,A 2,A 3,…和B 1,B 2,B 3,…分别在直线y=15x+b 和x 轴上.△OA 1B 1,△B 1A 2B 2,△B 2A 3B 3,…都是等腰直角三角形.如果点A 1(1,1),那么点A 2018的纵坐标是_____.17.(2018·黑龙江大庆·中考真题)已知直线y=kx (k≠0)经过点(12,﹣5),将直线向上平移m (m >0)个单位,若平移后得到的直线与半径为6的△O 相交(点O 为坐标原点),则m 的取值范围为_____.18.(2021·江苏南通·中考真题)平面直角坐标系xOy 中,已知点()2,39P m n -,且实数m ,n 满足240m n -+=,则点P 到原点O 的距离的最小值为___________.19.(2021·江苏无锡·中考真题)如图,在平面直角坐标系中,O 为坐标原点,点C 为y 轴正半轴上的一个动点,过点C 的直线与二次函数2yx 的图象交于A 、B 两点,且3CBAC ,P 为CB 的中点,设点P 的坐标为(,)(0)P x y x >,写出y 关于x 的函数表达式为:________.20.(2011·广西钦州·中考真题)如图,一次函数y=-2x 的图象与二次函数y=-x 2+3x 图象的对称轴交于点B .(1)写出点B 的坐标 ;(2)已知点P 是二次函数y=-x 2+3x 图象在y 轴右侧部分上的一个动点,将直线y=-2x 沿y 轴向上平移,分别交x 轴、y 轴于C 、D 两点. 若以CD 为直角边的PCD 与OCD 相似,则点P 的坐标为 .21.(2015·浙江湖州·中考真题)如图,已知抛物线C 1:y=a 1x 2+b 1x+c 1和C 2:y=a 2x 2+b 2x+c 2都经过原点,顶点分别为A ,B ,与x 轴的另一个交点分别为M 、N ,如果点A 与点B ,点M 与点N 都关于原点O 成中心对称,则抛物线C 1和C 2为姐妹抛物线,请你写出一对姐妹抛物线C 1和C 2,使四边形ANBM 恰好是矩形,你所写的一对抛物线解析式是___________22.(2021·广西柳州·中考真题)如图,一次函数2y x =与反比例数()0ky k x=>的图像交于A ,B 两点,点M 在以()2,0C 为圆心,半径为1的C 上,N 是AM 的中点,已知ON 长的最大值为32,则k 的值是_______.三、解答题23.(2019·四川乐山·中考真题)如图,已知过点(1,0)B 的直线1l 与直线2l :24y x =+相交于点(1,)P a -.(1)求直线1l 的解析式; (2)求四边形PAOC 的面积.24.(2014·江苏苏州·中考真题)如图,已知函数12y x b =-+的图象与x 轴、y 轴分别交于点A ,B ,与函数y =x 的图象交于点M ,点M 的横坐标为2.在x 轴上有一点P (a ,0)(其中a>2),过点P 作x 轴的垂线,分别交函数12y x b =-+和y =x 的图象于点C ,D(1)求点A 的坐标; (2)若OB =CD ,求a 的值.25.(2015·江苏盐城·中考真题)如图,在平面直角坐标系xOy 中,已知正比例函数34y x =与一次函数7y x =-+的图像交于点A , (1)求点A 的坐标;(2)设x 轴上一点P (a ,0),过点P 作x 轴的垂线(垂线位于点A 的右侧),分别交34y x =和7y x =-+的图像于点B 、C ,连接OC ,若BC=75OA ,求△OBC 的面积.26.(2017·浙江台州·中考真题)如图,直线:与直线:相交于点.(1)求的值;(2)垂直于轴的直线与直线,分别交于点,若线段长为2,求的值.27.(2017·江苏无锡·中考真题)(2017江苏省无锡市)操作:“如图1,P 是平面直角坐标系中一点(x 轴上的点除外),过点P 作PC △x 轴于点C ,点C 绕点P 逆时针旋转60°得到点Q .”我们将此由点P 得到点Q 的操作称为点的T 变换.(1)点P (a ,b )经过T 变换后得到的点Q 的坐标为 ;若点M 经过T 变换后得到点N (6,,则点M 的坐标为 .(2)A 是函数y =图象上异于原点O 的任意一点,经过T 变换后得到点B . △求经过点O ,点B 的直线的函数表达式;△如图2,直线AB 交y 轴于点D ,求△OAB 的面积与△OAD 的面积之比.28.(2017·湖北荆州·中考真题)如图在平面直角坐标系中,直线334y x =-+与x 轴、y 轴分别交于A 、B 两点,点P 、Q 同时从点A 出发,运动时间为t 秒.其中点P 沿射线AB 运动,速度为每秒4个单位长度,点Q 沿射线AO 运动,速度为每秒5个单位长度.以点Q 为圆心,PQ 长为半径作△Q .(1)求证:直线AB 是△Q 的切线;(2)过点A 左侧x 轴上的任意一点C (m ,0),作直线AB 的垂线CM ,垂足为M .若CM 与△Q 相切于点D ,求m 与t 的函数关系式(不需写出自变量的取值范围);(3)在(2)的条件下,是否存在点C ,直线AB 、CM 、y 轴与△Q 同时相切?若存在,请直接写出此时点C 的坐标;若不存在,请说明理由.29.(2019·山东东营·中考真题)已知抛物线24y ax bx+=﹣经过点()()20,40A B ,-,,与y 轴交于点C.()求这条抛物线的解析式;1()如图1,点P是第三象限内抛物线上的一个动点,当四边形ABPC的面积最大时,求点P 2的坐标;()如图2,线段AC的垂直平分线交x轴于点E,垂足为,D M为抛物线的顶点,在直线DE 3上是否存在一点G,使CMG的周长最小?若存在,求出点G的坐标;若不存在,请说明理由.参考答案【解析】【分析】根据反比例函数图象上点的坐标特征,求出B 、C 点的坐标,再写出BC 解析式,再判断点在BC 上.解:作BD OA ⊥,CE OA ⊥,45BOA ∠=︒,BD OD ∴=,设(,)B a a ,∴9a a=, 3a ∴=或3a =-(舍去), 3BD OD ∴==,(3,3)B , 2BC AC =.3ABAC ,BD OA ⊥,CE OA ⊥,//BD CE ∴,.ABD ACE ∴∆∆∽3BD ABCE AC==, ∴33CE=, 1CE ∴=,图象经过点C ,∴91x=, 9x ∴=,设BC 的解析式为y kx b =+,3319k b k b=+⎧⎨=+⎩, 解得134k b ⎧=-⎪⎨⎪=⎩, ∴143y x =-+,当2019x =-时,677y =, 当2020x =-时,16773y =, 当2021x =时,26693y =-, 当2022x =时,670y =-, 故选:D .【点拨】本题考查反比例函数图象上的点的性质,能求出BC 的解析式是解题的关键. 2.A 【解析】如图,分线段AB 在双曲线和直线y=x 交点的左右两侧两种情况,设点C 的坐标为(m ,0),则点A 的坐标为(m ,m ),点B 的坐标为(m , ),因AC+BC=4,所以m+=4,解得m=2± ,当m=2-时,即线段AB 在双曲线和直线y=x 交点的左侧,求得AC=2-,BC=2+,所以AB=(2+)-(2-)=2,即可求得△OAB 的面积为;当m=2+时,即线段AB 在双曲线和直线y=x 交点的右侧,求得AC=2+,BC=2-,所以AB=(2+)-(2-)=2,即可求得△OAB的面积为,故选A.3.B 【解析】【分析】延长AC 交x 轴于点D ,利用反射定律,可得1OCB ∠=∠,利用ASA 可证()COD COB ASA ∆≅∆,已知点B 坐标,从而得点D 坐标,利用A ,D 两点坐标,求出直线AD 的解析式,即可求得点C 坐标.如图所示,延长AC 交x 轴于点D .设()0,C c△这束光线从点()4,4A 出发,经y 轴上的点C 反射后经过点()10B ,, △由反射定律可知,1OCB ∠=∠, △△1=△OCD , △OCB OCD ∠=∠, △CO DB ⊥于O , △COD COB ∠=∠=90°,在COD ∆和COB ∆中OCD OCBOC OC COD COB ∠=∠⎧⎪=⎨⎪∠=∠⎩,△()COD COB ASA ∆≅∆, △1OD OB ==, △()1,0D -,设直线AD 的解析式为y kx b =+,△将点()4,4A ,点()1,0D -代入得:440k bk b =+⎧⎨=-+⎩,解得:4545k b ⎧=⎪⎪⎨⎪=⎪⎩,△直线AD 的解析式为:4455y x =+, △点C 坐标为40,5⎛⎫⎪⎝⎭.故选B .【点拨】本题考查了反射定律、全等三角形的判定与性质、待定系数法求一次函数解析式等知识点,综合性较强,难度略大. 4.D 【解析】【分析】根据已知解析式求出点A 、B 的坐标,根据过原点O 且将AOB 的面积平分列式计算即可; 如图所示,当0y =时,240x -+=, 解得:2x =, △()2,0A ,当0x =时,4y =, △()0,4B , △C 在直线AB 上, 设(),24C m m -+,△12OBC C S OB x =⨯⨯△,12OCA C S OA y =⨯⨯△,△2l 且将AOB 的面积平分, △OBC OCA S S =△△, △y C C OB x OA ⨯=⨯, △()4224m m =⨯-+, 解得1m =, △()1,2C ,设直线2l 的解析式为y kx =, 则2k =, △2y x =; 故答案选D .【点拨】本题主要考查了一次函数的应用,准确计算是解题的关键. 5.A 【解析】【分析】构造K 字形相似,由面积比得出相似比为2,从而得出A 点坐标与C 点坐标关系,而P 是矩形对角线交点,故P 是AC 、BO 的中点,由坐标中点公式列方程即可求解. 解:过C 点作CE △x 轴,过A 点作AF △x 轴,△点A 在函数()10y x x =>的图象上,点C 在函数()40y x x=-<的图象上, △2OCE S =△,12OAF S =△, △CE △x 轴,△90CEO ∠=︒,90OCE COE ∠+∠=︒, △在矩形OABC 中,90AOC ∠=︒, △90AOF COE ∠+∠=︒, △OCE AOF ∠=∠, △OCE AOF △△,△2CE OEOF AF ==, △2CE OF =,2OE AF =,设点A 坐标为1(,)x x,则点B 坐标为2(,2,)x x -,连接AC 、BO 交于点P ,则P 为AC 、BO 的中点, △27()2x x +-=-,解得:112x =,24x =-(不合题意,舍去), △点A 坐标为1(,2)2,故选A .【点拨】本题考查了反比例函数与几何图形的综合,关键是构造相似三角形,根据反比例函数的系数k 的几何意义,由面积比得到相似三角形的相似比,从而确定点A 与点C 的坐标关系. 6.A 【解析】【分析】过点A 作AE BC ⊥,设,44k A ⎛⎫⎪⎝⎭,,22k B ⎛⎫ ⎪⎝⎭,根据菱形的面积得到AB 的长度,在Rt ABE △中应用勾股定理即可求解.解:过点A 作AE BC ⊥,△A ,B 两点纵坐标分别为4,2,反比例函数ky x=经过A ,B 两点, △设,44k A ⎛⎫⎪⎝⎭,,22k B ⎛⎫ ⎪⎝⎭,△2AE =,244kk k BE =-+=-, △菱形ABCD 面积为8, △8BC AE ⋅=,解得4BC =, △4AB BC ==,在Rt ABE △中,222AB AE BE =+,即22242BE =+,解得BE = △k =- 故选:A .【点拨】本题考查反比例函数图象上点的坐标特征、菱形的性质等内容,根据提示做出辅助线是解题的关键. 7.A 【解析】【分析】由题意易得5,AB BC CD AD AD//BC ====,则设DE =x ,BE =2x ,然后可由勾股定理得()225425x x -+=,求解x ,进而可得点5,5k C ⎛⎫ ⎪⎝⎭,则2,45k D ⎛⎫+ ⎪⎝⎭,最后根据反比例函数的性质可求解. 解:△四边形ABCD 是菱形, △,AB BC CD AD AD//BC ===,△AD y ⊥轴,△90DEB AEB ∠=∠=︒, △90DEB CBO ∠=∠=︒, △点C 的横坐标为5,△点5,5k C ⎛⎫⎪⎝⎭,5AB BC CD AD ====,△2BE DE =,△设DE =x ,BE =2x ,则5AE x =-,△在Rt △AEB 中,由勾股定理得:()225425x x -+=, 解得:122,0x x ==(舍去), △2,4DE BE ==, △点2,45k D ⎛⎫+ ⎪⎝⎭,△245k k ⎛⎫⨯+= ⎪⎝⎭,解得:403k =; 故选A .【点拨】本题主要考查菱形的性质及反比例函数与几何的综合,熟练掌握菱形的性质及反比例函数与几何的综合是解题的关键. 8.A 【解析】【分析】先表示出平移后的函数为21(1)12y x m =---,得到(1,1)D -,1(1,1)D m +-,求出Q点的横坐标为:22m +,代入21(1)12y x =--求得22,128m m Q ⎛⎫+- ⎪⎝⎭,再根据等腰直角三角形的性质得到2222211128m mm ⎛⎫+⎛⎫--+= ⎪ ⎪⎝⎭⎝⎭+,解出m 即可求解. 抛物线21:(1)12CC y x =--沿水平方向向右(或向左)平移m 个单位得到21(1)12y x m =---△(1,1)D -,1(1,1)D m +-,△Q 点的横坐标为:22m +, 代入21(1)12y x =--求得22,128m m Q ⎛⎫+- ⎪⎝⎭, 若160DQD ︒∠=,则1DQD ∆是等边三角形, △1||QD DD m ==,由勾股定理得,2222211128m mm ⎛⎫+⎛⎫--+= ⎪ ⎪⎝⎭⎝⎭+,解得m =± 故选A .【点拨】此题主要考查二次函数与几何,解题的关键是熟知二次函数的性质及直角三角形的性质. 9.B 【解析】【分析】利用等腰直角三角形构造全等三角形,求出旋转后Q′的坐标,然后根据勾股定理并利用二次函数的性质即可解决问题. 解:作QM△x 轴于点M ,Q′N△x 轴于N ,设Q(m ,122m -+),则PM=1m﹣,QM=122m -+, △△PMQ=△PNQ′=△QPQ′=90°, △△QPM+△NPQ′=△PQ′N+△NPQ′,△△QPM=△PQ′N , 在△PQM 和△Q′PN 中, '90''PMQ PNQ QPM PQ N PQ Q P ∠=∠=︒⎧⎪∠=∠⎨⎪=⎩, △△PQM△△Q′PN(AAS),△PN=QM=122m -+,Q′N=PM=1m ﹣,△ON=1+PN=132m -,△Q′(132m -,1m ﹣),△OQ′2=(132m -)2+(1m ﹣)2=54m 2﹣5m+10=54(m ﹣2)2+5,当m=2时,OQ′2有最小值为5, △OQ′故选:B .【点拨】本题考查了一次函数图象上点的坐标特征,一次函数的性质,三角形全等的判定和性质,坐标与图形的变换-旋转,二次函数的性质,勾股定理,表示出点的坐标是解题的关键. 10.A 【解析】试题分析:过点C 作CE△x 轴于点E ,过点D 作DF△x 轴于点F ,设BD=a ,则OC=3a ,根据等边三角形的性质结合解含30度角的直角三角形,可找出点C 、D 的坐标,再利用反比例函数图象上点的坐标特征即可求出a 、k 的值,此题得解. 过点C 作CE△x 轴于点E ,过点D 作DF△x 轴于点F ,如图所示. 设BD=a ,则OC=3a .△△AOB 为边长为6的等边三角形,△△COE=△DBF=60°,OB=6. 在Rt△COE 中,△COE=60°,△CEO=90°,OC=3a , △△OCE=30°,△OE=a ,CE=,△点C (,).同理,可求出点D 的坐标为(6﹣a ,a ).△反比例函数(k≠0)的图象恰好经过点C 和点D , △k=×a=(6﹣a )×a ,△a=,k=.故选A .考点:反比例函数图象上点的坐标特征;等边三角形的性质;含30度角的直角三角形. 11.D 【解析】【分析】根据菱形的性质得出D 点的坐标,利用反比例函数0)y x >的图象经过线段DC 的中点N ,求出C 点的坐标,进而得出30ODC ∠=︒;根据菱形的性质可得260ABC ADC ODC ∠=∠=∠=︒,AB BC =,可判定ABC 是等边三角形;最后找到ME 、AM 、AE 、OB 之间的数量关系求解. △菱形ABCD ,4BD = △2OD OB ==△D 点的坐标为(0,2) 设C 点坐标为(c x ,0) △线段DC 的中点N △设N 点坐标为(2cx ,1)又△反比例函数0)y x =>的图象经过线段DC 的中点N132c =⋅,解得c x即C 0),OC =在Rt ODC 中,3tan 2OC ODC OD ∠===△30ODC ∠=︒△菱形ABCD△260ABC ADC ODC ∠=∠=∠=︒,AB BC =,30OBC ODC ∠=∠=︒ △ABC 是等边三角形又△AE BC ⊥于E 点,BO OC ⊥于O 点 △2AE OB ==,AO BE =△AO BE =,90AOB AEB ∠=∠=︒,AMO BME ∠=∠ △()AOM BEM AAS ≅ △AM BM = 又△在Rt BME 中,sin 30MEBM=︒ △1sin 30=2ME AM =︒ △1122333ME AE ==⨯= 故选:D .【点拨】本题考查菱形的性质、等边三角形的判定和特殊角30的三角函数.菱形的性质,四边相等,对角相等,对角线互相垂直且平分一组对角.等边三角形的判定,有一个角为60︒角的等腰三角形是等边三角形.特殊角30的三角函数,1sin 30=2︒,cos30︒tan 30︒. 12.D 【解析】试题分析:设点A 、B 横坐标为a ,则点A 纵坐标为,点B 的纵坐标为,△BE△x 轴,△点F 纵坐标为,△点F 是抛物线上的点,△点F 横坐标为x==,△CD△x 轴,△点D 纵坐标为,△点D 是抛物线上的点,△点D 横坐标为x==2a ,△AD=a ,BF=,CE=,OE=,△则= ==,故选D .考点:二次函数图象上点的坐标特征;综合题. 13.3【解析】【分析】由题意,先求出点P 的坐标,然后表示出点A 和点B 的坐标,即可求出答案. 解:△点P 在反比例函数3y x=的图像上且横坐标为1, △点P 的坐标为:(1,3),如图,AP△x 轴,BP△y 轴, △点A 、B 在反比例函数ky x=()0k <的图像上, △点A 为(,33k),点B 为(1,k ),△直线AB 与x 轴所夹锐角的正切值为: 3tan 313kk α-==-; 故答案为:3.【点拨】本题考查了反比例函数与一次函数的综合,解直角三角形的应用,解题的关键是掌握反比例函数的性质与一次函数的性质进行解题. 14.3 【解析】【分析】令PQ 与x 轴的交点为E ,根据双曲线的解析式可求得点A 、B 的坐标,由于点P 在双曲线上,由双曲线解析式中k 的几何意义可知△OPE 的面积恒为2,故当△OEQ 面积最大时△POQ 的面积最大.设Q (a ,122a -)则S △OEQ =12 ×a×(122a -)=214-a a =21(1)12-+a ,可知当a=2时S △OEQ 最大为1,即当Q 为AB 中点时△OEQ 为1,则求得△POQ 面积的最大值是是3.△122y x=-交x轴为B点,交y轴于点A,△A(0,-2),B(4,0)即OB=4,OA=2令PQ与x轴的交点为E△P在曲线C上△△OPE的面积恒为2△当△OEQ面积最大时△POQ的面积最大设Q(a, 122a-)则S△OEQ=12×a×(122a-)=214-a a=21(1)12-+a当a=2时S△OEQ最大为1即当Q为AB中点时△OEQ为1故△POQ面积的最大值是是3.【点拨】本题考查了反比例函数与一次函数几何图形面积问题,二次函数求最大值,解本题的关键是掌握反比例函数中k的几何意义,并且建立二次函数模型求最大值.15.y=﹣12x+32【解析】【分析】在Rt△OAB中,OA=4,OB=3,用勾股定理计算出AB=5,再根据折叠的性质得BA′=BA=5,CA′=CA,则OA′=BA′﹣OB=2,设OC=t,则CA=CA′=4﹣t,在Rt△OA′C中,根据勾股定理得到t2+22=(4﹣t)2,解得t=32,则C点坐标为(0,32),然后利用待定系数法确定直线BC的解析式解:△A(0,4),B(3,0),△OA=4,OB=3,在Rt△OAB中,,△△AOB 沿过点B 的直线折叠,使点A 落在x 轴上的点A′处, △BA′=BA=5,CA′=CA , △OA′=BA′﹣OB=5﹣3=2, 设OC=t ,则CA=CA′=4﹣t , 在Rt △OA′C 中, △OC 2+OA′2=CA′2,△t 2+22=(4﹣t )2,解得t=32,△C 点坐标为(0,32),设直线BC 的解析式为y=kx+b ,把B (3,0)、C (0,32)代入得3k+b=03b=2⎧⎪⎨⎪⎩,解得1k=-23b=2⎧⎪⎪⎨⎪⎪⎩△直线BC 的解析式为y=﹣12x+32故答案为y=﹣12x+32.【考点】翻折变换(折叠问题);待定系数法求一次函数解析式. 16.20173()2【解析】分析:因为每个A 点为等腰直角三角形的直角顶点,则每个点A 的纵坐标为对应等腰直角三角形的斜边一半.故先设出各点A 的纵坐标,可以表示A 的横坐标,代入解析式可求点A 的纵坐标,规律可求.详解:分别过点A 1,A 2,A 3,…向x 轴作垂线,垂足为C 1,C 2,C 3,…△点A 1(1,1)在直线y=15x+b 上△代入求得:b=4 5△y=15x+45△△OA1B1为等腰直角三角形△OB1=2设点A2坐标为(a,b)△△B1A2B2为等腰直角三角形△A2C2=B1C2=b△a=OC2=OB1+B1C2=2+b把A2(2+b,b)代入y=15x+45解得b=3 2△OB2=5同理设点A3坐标为(a,b)△△B2A3B3为等腰直角三角形△A3C3=B2C3=b△a=OC3=OB2+B2C3=5+b把A2(5+b,b)代入y=15x+45解得b=9 4以此类推,发现每个A的纵坐标依次是前一个的32倍则A2018的纵坐标是(32)2017故答案为(32)2017点睛:本题为一次函数图象背景下的规律探究题,结合了等腰直角三角形的性质,解答过程中注意对比每个点A的纵坐标变化规律.17.0<m<13 2【解析】【分析】利用待定系数法得出直线解析式,再得出平移后得到的直线,求与坐标轴交点的坐标,转化为直角三角形中的问题,再由直线与圆的位置关系的判定解答.【详解】把点(12,﹣5)代入直线y=kx得,﹣5=12k,△k=﹣5 12;由y=﹣512x平移m(m>0)个单位后得到的直线l所对应的函数关系式为y=﹣512x+m(m>0),设直线l与x轴、y轴分别交于点A、B,(如图所示)当x=0时,y=m;当y=0时,x=125m,△A(125m,0),B(0,m),即OA=125m,OB=m,在Rt△OAB中,135m ==,过点O作OD△AB于D,△S△ABO=12OD•AB=12OA•OB,△12OD•135m=12×125m×m,△m>0,解得OD=1213m,由直线与圆的位置关系可知1213m <6,解得m<132,故答案为0<m<13 2.【点睛】本题考查了直线的平移、直线与圆的位置关系等,能用含m的式子表示出原点到平移后的直线的距离是解题的关键.本题有一定的难度,利用数形结合思想进行解答比较直观明了.18【解析】【分析】由已知得到点P 的坐标为(m ,33m +),求得PO =利用二次函数的性质求解即可. 解:△240m n -+=,△24n m =+,则23933n m -=+, △点P 的坐标为(m ,33m +),△PO = △100>,△210189m m ++当1892010m =-=-时,有最小值, 且最小值为910,△PO【点拨】本题考查了点的坐标,二次函数的图象和性质,熟练掌握二次函数的性质是解决本题的关键. 19.283y x =【解析】【分析】过点A 作AN △y 轴,过点B 作BM 垂直y 轴,则BM △AN ,13AN AC BM CB ==,设A (-a ,a 2),则B (3a ,9a 2),求出C (0,3a 2),从而得P (32a ,26a ),进而即可得到答案.解:过点A 作AN △y 轴,过点B 作BM 垂直y 轴,则BM △AN , △CBM CAN ∽, △3CB AC ,△13AN AC BM CB ==, 设A (-a ,a 2),则B (3a ,9a 2), 设直线AB 的解析式为:y =kx +b ,则2293a ka b a ka b ⎧=-+⎨=+⎩,解得:223k a b a =⎧⎨=⎩,△直线AB 的解析式为:y =2ax +3a 2, △C (0,3a 2), △P 为CB 的中点,△P (32a ,26a ),△2326x ay a⎧=⎪⎨⎪=⎩,即:283y x =,故答案是:283y x =.【点拨】本特纳主要考查二次函数与一次函数的综合,相似三角形的判定和性质,掌握函数图像上点的坐标特征,是解题的关键.20.(1)3(,)23-;(2)(2,2),15(,)24,1111(,)416,1326(,)525【解析】(1)△抛物线y=-x 2+3x 的对称轴为x=332(1)2-=⨯-,△当x=32时,y=-2x=-3,即B 点(32,-3);(2)设D (0,2a ),则直线CD 解析式为y=-2x+2a ,可知C (a ,0),即OC :OD=1:2, 则OD=2a ,OC=a ,根据勾股定理可得:. 以CD 为直角边的△PCD 与△OCD 相似,当△CDP=90°时,若PD :DC=OC :OD=1:2,则a ,设P 的横坐标是x ,则P 点纵坐标是-x 2+3x ,根据题意得:222222222(32))(3)()x x x a x x x a +-+-=+=-++-,解得:1212{x a ==,则P 的坐标是:(15,24),若DC :PD=OC :OD=1:2,同理可以求得P (2,2),当△DCP=90°时,若PC :DC=OC :OD=1:2,则P (1111,416), 若DC :PD=OC :OD=1:2,则P (1326,525).21.2y =+,2y =+(答案不唯一,只要符合条件即可).【解析】试题分析:因点A 与点B ,点M 与点N 都关于原点O 成中心对称,所以把抛物线C 2看成抛物线C 1以点O 为旋转中心旋转180°得到的,由此即可知a 1,a 2互为相反数,抛物线C 1和C 2的对称轴直线关于y 轴对称,由此可得出b 1=b 2.抛物线C 1和C 2都经过原点,可得c 1=c 2,设点A (m ,n ),由题意可知B (-m ,-n ),由勾股定理可得ABMN=︱4m ︱,又因四边形ANBM 是矩形,所以AB=MN,4m =,解得223,m n m n ==即,设抛物线的表达式为2()y a x m n =-+,任意确定m 的一个值,根据m n =n 的值,抛物线过原点代入即可求得表达式,然后在确定另一个表达式即可.l例如,当m=1时,抛物线的表达式为2(1)y a x =-把x=0,y=0代入解得a=即2y =+,所以另一条抛物线的表达式为2y =+.考点:旋转、矩形、二次函数综合题.22.3225【解析】【分析】根据题意得出ON 是ABM 的中位线,所以ON 取到最大值时,BM 也取到最大值,就转化为研究BM 也取到最大值时k 的值,根据,,B C M 三点共线时,BM 取得最大值,解出B 的坐标代入反比例函数即可求解.解:连接BM ,如下图:在ABM 中,,O N 分别是,AB AM 的中点,ON ∴是ABM 的中位线,12ON BM ∴=, 已知ON 长的最大值为32, 此时的3BM =,显然当,,B C M 三点共线时,取到最大值:3BM =,13BM BC CM BC =+=+=,2BC ∴=,设(,2)B t t ,由两点间的距离公式:2BC ==,22(2)44t t ∴-+=, 解得:124,05t t ==(取舍), 48(,)55B ∴, 将48(,)55B 代入()0k y k x=>, 解得:3225k =, 故答案是:3225. 【点拨】本题考查了一次函数、反比例函数、三角形的中位线、圆,研究动点问题中线段最大值问题,解题的关键是:根据中位线的性质,利用转化思想,研究BM 取最大值时k 的值.23.(1)1y x =-+;(2)52【解析】【分析】(1)根据P 点是两直线交点,可求得点P 的纵坐标,再利用待定系数法将点B 、点P 的坐标代入直线l 1解析式,得到二元一次方程组,求解即可.(2)根据解析式可求得点啊(-2,0),点C (0,1),由四边形∆∆=-PAB BOC PAOC S S S 可求得四边形PAOC 的面积解:(1)△点P 是两直线的交点,将点P (1,a )代入24y x =+得2(1)4⨯-+=a ,即2a =则P 的坐标为(1,2)-,设直线1l 的解析式为:y kx b =+(0)k ≠,那么02k b k b +=⎧⎨-+=⎩, 解得:11k b =-⎧⎨=⎩. 1l ∴的解析式为:1y x =-+.(2)直线1l 与y 轴相交于点C ,直线2l 与x 轴相交于点A∴C 的坐标为(0,1),A 点的坐标为(2,0)-则3AB =,而四边形∆∆=-PAB BOC PAOC S S S ,∴PAOC S 四边形1153211222=⨯⨯-⨯⨯= 【点拨】本题考查了一次函数求解析式,求一次函数与坐标轴围成的图形面积,解本题的关键是求得各交点坐标求得线段长度,将不规则图形转化为规则图形求面积.24.(1)(6,0);(2)4.【解析】试题分析:(1)先利用直线y=x上的点的坐标特征得到点M的坐标为(2,2),再把M(2,2)代入y=﹣12x+b可计算出b=3,得到一次函数的解析式为y=﹣12x+3,然后根据x轴上点的坐标特征可确定A点坐标为(6,0);(2)先确定B点坐标为(0,3),则OB=CD=3,再表示出C点坐标为(a,﹣12a+3),D点坐标为(a,a),所以a﹣(﹣12a+3)=3,然后解方程即可.试题解析:解:(1)△点M在直线y=x的图象上,且点M的横坐标为2,△点M的坐标为(2,2),把M(2,2)代入y=﹣12x+b得﹣1+b=2,解得b=3,△一次函数的解析式为y=﹣12x+3,把y=0代入y=﹣12x+3得﹣12x+3=0,解得x=6,△A点坐标为(6,0);(2)把x=0代入y=﹣12x+3得y=3,△B点坐标为(0,3),△CD=OB,△CD=3,△PC△x轴,△C点坐标为(a,﹣12a+3),D点坐标为(a,a)△a﹣(﹣12a+3)=3,△a=4.考点:两条直线相交或平行问题.25.(1)A(4,3);(2)28.【解析】【分析】(1)点A是正比例函数34y x=与一次函数y=-x+7图像的交点坐标,把34y x=与y=-x+7联立组成方程组,方程组的解就是点A的横纵坐标;(2)过点A作x轴的垂线,在Rt△OAD中,由勾股定理求得OA的长,再由BC=75OA求得OB的长,用点P的横坐标a表示出点B、C的坐标,利用BC的长求得a值,根据12OBCS BC OP∆=⋅即可求得△OBC的面积.解:(1)由题意得:347y xy x⎧=⎪⎨⎪=-+⎩,解得43xy=⎧⎨=⎩,△点A的坐标为(4,3).(2)过点A作x轴的垂线,垂足为D,在Rt△OAD中,由勾股定理得,5OA△775755BC OA==⨯=.△P(a,0),△B(a,34a),C(a,-a+7),△BC=37(7)744a a a--+=-,△7774a-=,解得a=8.△11782822OBCS BC OP∆=⋅=⨯⨯=.26.(1)m-1(2)a=或a=【解析】试题分析:(1)把点P(1,b)分别代入l1和l2,得到b和m的值.(2)将直线x=a分别与直线l1、l2联立求出C和D的坐标,根据CD=2,列出关于a的方程求出a的值即可.试题解析:(1)把点P(1,b)代入y=2x+1,得b=2+1=3,把点P(1,3)代入y=mx+4,得m+4=3,△m=-1.(2)直线x=a与直线l1的交点C为(a,2a+1),与直线l2的交点D为(a,-a+4).△CD=2,△|2a+1-(-a+4)|=2,即|3a -3|=2,△3a -3=2或3a -3=-2,△a=或a=.考点:1、待定系数法求一次函数解析式,2、两条直线相交或平行问题27.(1)(a +,12b );(9,-;(2)△y =;△34. 【解析】【分析】(1)连接CQ 可知△PCQ 为等边三角形,过Q 作QD△PC ,利用等边三角形的性质可求得CD 和QD 的长,则可求得Q 点坐标;设出M 点的坐标,利用P 、Q 坐标之间的关系可得到点M 的方程,可求得M 点的坐标;(2)△可取A (2,利用T 变换可求得B 点坐标,利用待定系数示可求得直线OB 的函数表达式;△由待定系数示可求得直线AB 的解析式,可求得D 点坐标,则可求得AB 、AD 的长,可求得△OAB 的面积与△OAD 的面积之比.解:(1)如图1,连接CQ ,过Q 作QD△PC 于点D ,由旋转的性质可得PC=PQ ,且△CPQ=60°,△△PCQ 为等边三角形,△P (a ,b ),△OC=a ,PC=b ,△CD=12PC=12b ,,△Q (,12b );设M (x ,y ),则N 点坐标为(,12y ),△N (6,△612x y y ⎧=⎪⎪⎨⎪=⎪⎩,解得9x y =⎧⎪⎨=-⎪⎩ △M (9,﹣;(2)△△A 是函数图象上异于原点O 的任意一点, △可取A (2,72,12△B (72, 设直线OB 的函数表达式为y=kx ,则72△直线OB 的函数表达式为; △设直线AB 解析式为y=k′x+b ,把A 、B坐标代入可得2'7'2k b k b ⎧+=⎪⎨+⎪⎩k b '⎧=⎪⎪⎨⎪=⎪⎩, △直线AB 解析式为y=, △D (0),且A (2,B (72,△34OAB OAD S AB S AD ===△△.28.(1)证明见解析;(2)m =4﹣354t 或m =4﹣54t ;(3)(﹣38,0)或(278,0)或(﹣272,0)或(32,0). 【解析】试题分析:(1)只要证明△PAQ△△BAO ,即可推出△APQ=△AOB=90°,推出QP△AB ,推出AB 是△O 的切线;(2)分两种情形求解即可:△如图2中,当直线CM 在△O 的左侧与△Q 相切时,设切点为D,则四边形PQDM是正方形.△如图3中,当直线CM在△O的右侧与△Q相切时,设切点为D,则四边形PQDM是正方形.分别列出方程即可解决问题.(3)分两种情形讨论即可,一共有四个点满足条件.试题解析:(1)如图1中,连接QP.在Rt△AOB中,OA=4,OB=3,△AB==5,△AP=4t,AQ=5t,△,△△PAQ=△BAO,△△PAQ△△BAO,△△APQ=△AOB=90°,△QP△AB,△AB是△O的切线.(2)△如图2中,当直线CM在△O的左侧与△Q相切时,设切点为D,则四边形PQDM 是正方形.易知PQ=DQ=3t,CQ=•3t=,△OC+CQ+AQ=4,△m+t+5t=4,△m=4﹣t.△如图3中,当直线CM在△O的右侧与△Q相切时,设切点为D,则四边形PQDM是正方形.△OC+AQ﹣CQ=4,△m+5t﹣t=4,△m=4﹣t.(3)存在.理由如下:如图4中,当△Q在y则的右侧与y轴相切时,3t+5t=4,t=,由(2)可知,m=﹣或.如图5中,当△Q在y则的左侧与y轴相切时,5t﹣3t=4,t=2,由(2)可知,m=﹣或.综上所述,满足条件的点C 的坐标为(﹣,0)或(,0)或(﹣,0)或(,0).考点:一次函数综合题29.(1) 2142y x x +-=;(2)点P 的坐标为()2,4--;(3)315,48G ⎛⎫- ⎪⎝⎭ 【解析】【分析】(1) 用待定系数法即可得到答案;(2)连接OP ,设点21,42P x x x ⎛⎫+- ⎪⎝⎭,由题意得到AOC OCP OBPS S S S ∴++=()2216x ++=.即可得到答案. (3)用待定系数法求解析式,再结合勾股定理即可得到答案.解:1()抛物线4y ax bx +-=经过点()()2,0,40A B -,, 424016440a b a b +-=⎧∴⎨--=⎩, 解得1,21a b ⎧=⎪⎨⎪=⎩ ∴抛物线解析式为2142y x x +-=; 2()如图1,连接OP ,设点21,42P x x x ⎛⎫+- ⎪⎝⎭,其中40x -<<,四边形ABPC 的面积为S ,由题意得0,4C -(),AOC OCP OBP S S S S ∴++=()1124422x =⨯⨯+⨯⨯-2114422x x ⎛⎫+⨯⨯--+ ⎪⎝⎭, 24228x x x ---+=,2412x x -+=-,。

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单词拼写. 根据句意和首字母完成单词。

(五)1. T___________ of Japanese people lost their lives in the earthquake in 2011. Chinese people try their best to help them.2. Everyone went to the zoo e__________ Wei Ming last Tuesday. Because he had to look after his little brother.3. Liu Wei is a disabled boy. He learned the piano by h__________.4. Could you p__________ the books for me on your way to school, Amy?5. Television was invented by John Baird and it is one of the greatest i___________ in the world.6. He wants to be a bus d___________ like his grandfather.7. You should choose two w__________ teachers to take part in the speech.8. It s________ that they had finished it at the same time.9. Please pay a___________ to your handwriting.10. I won’t be free u________ 8:00 this evening.11. The story-book is quite p________________ with the students.12. It’s quite d____________ for a child to stay alone at home.13. Nothing is more important than s______________ for the pupils.14. Our school is as beautiful as t___________ .15. As a citizen, we mustn’t do anything a _____________ the law.16. Her family was so p____________ that she couldn’t go to school.17. It’s very important for us to s_____________ the environment.18. Sorry. I have lost the dictionary you l______________ me last week.19. For some people, drinking green t__________ is good for their health.20. He is very l___________ to be alive after that accident.21. The twins are going to be singers when they g________ up.22. –What are you doing this afternoon, Jack?-I’m going to the library to b________ some books.23. My d______________ is to be a reporter.24. Listening to light music makes us feel r_____________.24. Do you know who i____________ the first computer.综合填空(共10小题,计10分)阅读短文,根据短文内容和已给出的首字母,在空白处填入适当的单词。

将完整的单词写在短文后面的横线上。

Once there was a poor little girl living near a forest. She had no family and no one to love her. So she often f sad and lonely.One day, when she was walking in the forest, she found that a small b was trapped unluckily in a bush. The butterfly tried to fly away b failed. The kind little girl saved the butterfly with great care. Instead of flying away, the butterfly turned into a beautiful fairy (仙女). The little girl was very s .―Thank you for s me. You are so kind. I will make any of your dreams come true.‖ said the fairy.The little girl thought for a moment and then said, ―I want to be h!‖The fairy said, ―Very well. I will help you.‖ And she said something in the little girl’s ear. Then the fairy disappeared.As the kind little girl grew up, she was always ready to help people in need and was popular among the villagers. No one in the village was as happy as she was. Everyone asked her the s of her happiness. She always smiled and answered ―The secret of my happiness is that I listened to a kind when I was a little girl.‖When the kind girl became a very old woman and was dying, the neighbors in the village all gathered (聚拢) around her bed because they were afraid that her secret of happiness would die with her. They asked, ―Please tell us what the kind fairy said.‖The lovely old woman still s and said, ―She told me that everyone needed me, no matter how safe they seemed, no matter how rich or poor, no matter how old or y . She said that helping others would make me happy all my life.‖根据汉语意思用英语完成句子,将答案写在横线上。

1. 告诉我你要到哪儿去。

Tell me ___ you _ _ __ .2. 对我来说,回答这个问题太容易。

too me answer this question .3. 她对英语越来越感兴趣。

She is becoming more and __ _ ___ ___ ____ English.4. 外面太冷了,我宁愿呆在家里,而不愿出去。

It’s too cold outside. I ____ ___ stay at home than __ ____ out.5. 我六点钟出了家门,以便赶上火车。

I left home at 6:00 _____ __ _____ __ I could catch the train.6. 有一个健康的生活方式很重要。

7.吉姆忘了给他的朋友打电话。

Jim forgot __ __ __ his friend.8. 我独自一人,但我不感到寂寞。

I was ___ , but I didn’t ___ ___ .阅读表达(共5小题,计10分)阅读短文,根据要求完成下面各小题。

83 They do the same things as us, but they do them differently.On Earth, we put food on a plate and water in a glass. Gravity (引力) holds the food down and keeps the water in the glass. But in space, there is almost no gravity. So food can float(漂浮) away, and astronauts must eat and drink carefully.Sleeping is very difficult in space. Some astronauts like to float in the air, but most like to be in a sleeping bag. They tie(拴)the bag to a wall so that they don’t float away in the night.Washing is difficult. There is no shower or bath, so astronauts must use a sponge(海绵), They brush their teeth normally: they don’t want wet toothpaste to float around the spacecraft!Exercise is very important in space. On Earth, your legs carry your body, but in space, astronauts float, so they do not use their legs. 84 This is very bad for their legs, so they must exercise for thirty minutes every day. They often use an exercise bicycle for this.After a day’s work, astronauts relax. They may listen to music, read, watch films, play games, or talk to their friends and families by radio. On the International Space Station, astronauts sometimes race from one end of the station to the other. The most popular pastime is looking out of the window, looking at the space and watching the Earth.(一)根据短文内容回答下列问题。

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