A-1 A答案
第一章至第七章阶段测验A卷参考答案(李泽亮等同学整理)
《材料成形原理》阶段测验（第一章）班级：姓名：学号成绩：1、下图中偶分布函数g(r)，液体g(r)为c图，晶态固体g(r)为a图，气体g(r)为b图。
（a）（b）（c）2、液态金属是由大量不停“游动”着的原子团簇组成，团簇内为某种有序结构，团簇周围是一些散乱无序的原子。
由于“能量起伏”，一部分金属原子（离子）从某个团簇中分化出去，同时又会有另一些原子组合到该团簇中，此起彼伏，不断发生着这样的涨落过程，似乎原子团簇本身在“游动”一样，团簇的尺寸及其内部原子数量都随时间和空间发生着改变，这种现象称为结构起伏。
3、对于液态合金，若同种元素的原子间结合力F(A-A、B-B) 大于异类元素的原子间结合力F(A-B)，则形成富A及富B的原子团簇，具有这样的原子团簇的液体仅有“拓扑短程序”；若熔体的异类组元具有负的混合热，往往F(A -B)＞F(A-A、B-B)，则在液体中形成具有A-B化学键的原子团簇，具有这样的原子团簇的液体同时还有“化学短程序”。
4、液体的原子之间结合力（或原子间结合能U）越大，则内摩擦阻力越大，粘度也就越大。
液体粘度η随原子间结合能U按指数关系增加，即（公式）：⎪⎪⎭⎫⎝⎛=TUTBBkexpk23τδη。
5、加入价电子多的溶质元素，由于造成合金表面双电层的电荷密度大，从而造成对表面压力大，而使整个系统的表面张力增大。
6、铸件的浇注系统静压头H越大，液态金属密度1ρ及比热1C、合金的结晶潜热H∆越大，浇注温度浇T、铸型温度T型越高，充型能力越强。
7、两相质点间结合力越大，界面能越小，界面张力就越小。
两相间的界面张力越大，则润湿角越大，表示两相间润湿性越差。
8、铸件的浇注系统静压头H越大，液态金属密度1ρ及比热1C、合金的结晶潜热H∆越小，浇注温度浇T、铸型温度T型越高，充型能力越强。
9、右图为碱金属液态的径向分布函数RDF，请在图中标注液态K的平均原子间距r1的位置，并以积分面积（涂剖面线）表达液态K的配位数N1的求法。
七年级英语《一本》答案
七年级英语《一本》答案【篇一：七年级上册英语参考答案1】=txt>uint 1一、a) 1.what’s2. card 3. last 4. seven/six 5. phoneb) 1-5 acbccc) 1. her 2. answer 3. first 4. your 5. i 6. eight7. to meet8. is 9. tony 10. its二、1-5 bcccc 6-10acacc 11-15ccacb三、1. telephone number 2. last/family name 3. in english 4. three; two5. an ; his四、1. how are 2. i am 3. it is4. what is family/last 5. what color五、dacbe六、1-5 abbac 6-10aabaa七、a. 1. girl 2. 89325613. black 4. red5. amyb. 1. linda 2. cruisfe 3. 12 4. four 5. 473-2106八、 my name is alice smith. alice is my first name. smith is my family name. i am 13 yearsold. and i like red . i think it’s nice. my telephone number is 3258673.unit 2二、b) 1.your 2.me 3.her 4. that 5.keys 6. hisc) 1.is2.are3.spell 4.is5. isn’t 6. call二、1-5cbbbc 6-10cabac 11-15 cacac三、1.baseball；isn’t2. not his 3. how ；spell 4. is ；an 5. excuse me四、1. is that 2. is not 3. what’s this 4. it’s not 5. call； at五、daceb六、1-5accbc6-10 dbcad七、cacda八、 lost founda set of keys a backpack. some booksmy name is joae. and a pencilcase are in it.please call 83602388 call mr. green at 46300029unit 3一、a) 1.sister 2. grandmother 3. photo/picture 4. cousin 5. uncle 6. parentss 6. this 7. his 8. picturesc) 1.are; is 2. look 3.isn’t 4. thanks 5. to meet二、1-5 bcaaa6-10 bcbcc 11-15 cbaab三、1. this; son ; it’s2. good friend 3. family photo 4. are those5. these; cousins 四、1. is ; your 2. are not 3. are these 4.they a ren’t 5. that ; a五、gcfad六、1-5 bcdaa 6-10 cdcdc七、acdcd八、 hello! i am a student. there are 6 people in my family. this is my father. he is a worker.this is my mother. she is a teacher. those are my grandfather and grandmother. look! that boy is my brother. he is 6 yearsold , and i am 12. welcome to my home.unit 4一、a) 1. where2. under 3. tape 4. bring 5. chair 6. floor 7. math8. knowb) 1. keys 2. things3. those4. my, your 5. cdsc) 1. is 2. meet3. are 4. are; aren’t 5. take二、1-5 bdbac6-10 bcbcd11-15 ccdca三、1. where’s my 2. need；are they 3. i don’t know 4.her plant 5. is四、1. where is 2.yes,they 3. where are 4. no; isn’t5. what is五、adceb六、1-5 abccb 6-10 cbacb七、badac八、this is a picture of my room. it’s very nice. my bed is under the window. my hat is on it.near the bed, there is is a desk. my book is on the desk andmy pen is on the desk, too. under the desk, it’s my backpack. where is my chair? oh, it’s next to th e bed.unit 5一、a）1. basketball 2. has 3. volleyball 4. difficult 5. only 6. every7. collection 8. manyb) 1. us 2. interesting 3. balls 4. them 5. clubsc) 1. go 2. dont have 3. sounds 4. does; like 5. plays 6. has7. bring二、1-5 accad 6-10 baaba 11-15 bbabc三、1. great sports collection 2. racket 3.sounds good 4. lets play soccer 5.run 四、1. do; have 2. doesnt have 3. what does ; have 4. let’s play 5. she doesn’t五、1-5 cdebf六、1-5 badad 6-10 bbbca七、a)1-5 bbaac b) 1-5 ftfft八、参考范文：my good friendi have a good friend . his name is tim. he likes sports. and he likes collecting sports things. he has 5 baseballs, 8 volleyballs, 4 soccer balls and 3 basketballs. he plays sport every day.unit 6一、a）1. hamburgers 2. cream 3. likes 4. bananas; oranges5. strawberries6. lunch7. starb) 1. tomatoes 2. fries 3. salad(s) 4. well 5. healthy 6. chicken 7. runnerc) 1.playing 2.are 3. likes 4. make 5. doesn’t like二、1-5 accbc 6-10 cbccc 11-15 dbcac三、1. healthy food 2. do; like3. for; likes4. for dinner 5. many students四、1. are boys 2. doesn’t like 3. does; play4. no, they don’t 5. what do五、1-5 cdbea六、1-5 acbda 6-10 dbcad七、1-5 bdbad 6-10 ccdbc八、参考范文：breakfast lunch dinner1. eggs1. hamburgers 1. chicken2. bananas2. salad 2. tomatoes3. apples 3. pears 3. french friesevery day, i eat lots of healthy food. for breakfast, i like eggs, bananas, and apples. for lunch, i like hamburgers, salad and pears. and for dinner, i have chicken, tomatoes and french fries.unit 7一、a）1. white 2. much 3. color 4. shorts; long 5. store 6. sale 7. sweater 8.price；dressb) 1. dollars 2. yuan 3. shoes 4. our 5. yourselfc) 1. are 2. do 3. sells4. afford5. take二、1-5 dddcd 6-10 cbcad 11-15 abdbc三、1. how much 2. here; are 3. afford 4. on sale 5. have a look四、1. how much 2. these are ; hats 3. buy 4. doesn’t sell clothes 5. short五、1-5 dfabe六、1-5 abadd 6----10 bcbac七、1-5 baccb6.yellow 7.shoes 8. white 9. socks 10. pants八、参考范文：come and buy your clothes at jennifer’s clothing sale. we have sweaters in black, white and blue for only 40 yuan each. andfor girls, we have hats in all colors-only 20 yuan. for boys, we have t-shirts for 25 yuan. come and see for yourself. you can choose your favorite clothes.unit 8一、a）1. party 2. contest 3. music 4. date 5. festival 6. twelfth7. october8. ageb) 1.chinese 2. first 3.days 4. second 5. trips 6. thirty-first 7. months 8. his 二、1-5 badcb6-10bacac 11-15 bdabc,三、1. when; birthday 2. how old3. years old4. march eighth 5. school trip 四、1. when; is2. do; have 3. how old 4. it isn’t5. my birthday is june 20 th五、1-5 bcfda六、1-5 bcbac 6-10 cacab七、1-5 acadd 6-10 aacdc八、参考范文：hello! this is my friend. his name is robert. he is 13 years old. his birthday is april 15th. he likes basketball and music. for food, he likes hamburgers, chicken and fruits. he thinks they are healthy. he has three good friends---helen, kate and mona.c) 1. help 2. have 3. is 4. likeunit 9一、a) 1. sad 2. favorite 3. weekends 4. history 5. about 6. students 7. boring 8. scaryb) 1. actor 2. successful 3. really 4. funny 5. chinese 6.actionc) 1. has 2. to drink 3. doesn’t like 4. thinks 5. to see 6. learn二、1-5 cdbcb6-10 babcb11-15 dbdcc三、1. goes to see 2. learn about 3. kind of songs 4. don’t like四、1. what; does 2. do; want 3. my favorite 4. they; movies 5. and comedies五、1-5 ecadg六、1-5 cbadb6-10 aaadc七、(a): dcbbb (b): tfftt八、（one possible version）my family’s hobbiesi have a happy family. my father likes reading newspapers. he likes documentaries, because he thinks they are helpful. he likes playing soccer. my mother likes reading books, and she likes seeing beijing opera., she thinks it’s interesting. but i don’t like it. i think it’s boring. i like singing, dancing andseeing comedies. i also like playing ping-pong. we have our hobbies. we are happy every day.unit10一、a) 1. chess 2. show 3. draw 4. address 5. why 6. sundayb) 1. musician 2. kids 3. our 4. pianos 5. english 6. wellc) 1. do 2. painting 3. aren’t/ are not 4. swimming5. to join6. swimming7. dance8. singing二、1-5 ddbdb6-10 cbdbc11-15 daddb三、1. want; for 2. be in 3. are; good with 4. need help for四、1. what club 2. can; play; can’t 3. can’t make 4. what can; do五、1-5 dfcae六、1-5 adcab6-10 caaad七、(a): 1-5 dccdb(b): 1-5 ffttf八、（one possible version）students wanted for our school partywe want some students for our school celebration party. can you sing? can you dance? can you play the piano or guitar?can you play the violin or erhu? can you give skit performance?can you perform cross-talk? if you can do any of these, you can make our school party more successful. please call mr zhang at 3254968 .come and join the school party.unit 11一、a) 1. around/about 2. shower 3. homework 4. soon 5. saturday 6. letterb) 1. teeth 2. usually 3. me 4. hours 5. jobs6. workers 7. wishes 8. funnyc) 1. brushes 2. starts 3. telling 4. listens 5. get 6. to know二、1-5 babcd6-10 bcaba11-15 acbaa三、1. write; tell; about 2. his ;job; is 3. does; go 4. takes; work四、1. do you; my 2. takes; shower 3. what time does 4. doesn’t eat【篇二：初一英语基本翻译练习加答案1】>1.是回家/玩电子游戏/上学的时候了。
青骄第二课堂回家答案
青骄第二课堂回家答案1、禁毒微动漫：X任务第一集1、动画中猪爸爸车里藏的是哪一种毒品？A、冰毒B、大麻C、海洛因D、鸦片答案：A2、设卡盘查时缉毒犬们依靠哪一种感官来查找毒品？A、触觉B、听觉C、嗅觉D、视觉答案：C3、猪爸爸为什么会撞上刺猬妈妈?A、醉酒后驾车B、开车拨打手机，反应驰援C、毒瘾发作产生幻觉，未发现刺猬妈妈D、故意谋杀刺猬妈妈答案：C4、 X探长是哪一种动物形象的拟人化？A、猪B、刺猬C、鸭子D、狗答案：D5、毒品的危害主要表现为（）A、毁灭自己B、祸及家庭C、危害社会D、危及后代答案：ABCD2、禁毒微动漫：X任务第二集1、狱警是哪一种动物形象的拟人化？A、猪B、刺猬C、鸭子D、狗答案：C2、毒枭大猩猩贩卖的是哪一种毒品？A、病毒B、大麻C、海洛因D、摇头丸答案：D3、舞厅中梅花鹿小姐为何被毒贩强行喂入毒品？A、为了杀害梅花鹿小姐B、为了让梅花鹿小姐对毒品上瘾，以便贩卖毒品C、为了绑架梅花鹿小姐D、为梅花鹿小姐治病答案：B4、X探长选择成为一名缉毒警察的主要原因是什么？A、因为警察威风帅气B、为了打击毒品犯罪，让和X探长母亲一样的悲剧不再重演C、为了抓捕毒枭黑猩猩D、为打击地下制毒工厂答案：B5、X探长的母亲注射的可能是哪一种毒品？B、大麻C、海洛因D、鸦片答案：AC3、禁毒预防教育片1、下列物品中不属于毒品的是？A、鸦片B、海洛因C、酒精D、杜冷丁答案：C2、（）问题已经成为当今世界公害？A、吸烟B、赌博C、毒品D、贪污答案：C3、下列物质中，属于毒品的是？A、罂粟壳B、香烟C、茶叶D、酒精答案：A4、冰毒属于（）？A、致幻剂B、抑制剂C、兴奋剂D、麻醉剂答案：C5、长期吸食鸦片会使人丧失（）？B、嗅觉C、味觉D、免疫力答案：D4、如何拒绝毒品1、有这么一句话“一日吸毒，十年戒毒，终生想毒”用以表明毒品的（）？A、生理依赖B、躯体状况C、心瘾D、外在表现答案：C2、冰毒在（）比较常见？A、家里B、歌舞厅C、餐厅D、学校答案：B3、以下不属于毒品的基本特征的是（）？A、依懒性B、安全性C、非法性答案：B4、吸毒成瘾一般具有的特征？A、药物耐受性B、生理依赖性C、心理依赖性D、以上三项均正确答案：D5、毒品的危害有（）A、危害个人B、危害家庭C、危害社会D、ABC选项都正确答案：D5、远离毒品从预防药物滥用开始1、药物滥用造成的药物依赖性包括A、身体依赖性和精神依赖性B、药物耐受性和病人依从性C、精神依赖性和病人依从性D、中枢兴奋性和精神依赖性答案：A2、下面哪些药物出于非医疗目的大量滥用会依赖成瘾？A、杜冷丁B、曲马多C、布洛芬D、复方甘草片答案：ABD3、关于预防药物滥用说法正确的有A、遵循医嘱，谨慎用药B、及早发现，及早干预C、一旦成瘾，系统治疗D、保险起见，药物停产答案：ABC6、毒品，那么远，那么近-X侦探的少年往事1、大家在日常生活中要对毒品提高警惕，那么下面哪种方法是正确的呢？A、不进入酒吧、迪厅等场所B、不食用来路不明的零食及饮料C、吸食毒品很刺激，想要尝试D、不结交涉毒“损友”答案：ABD2、吸食毒品后人可能出现的状况是A、极度亢奋B、萎靡不振C、休克失去意识D、越来越聪明、睿智E、产生幻想幻觉答案：ABCE7、X侦探：做识毒辨毒的明眼人1、以下哪几种毒品可能是“暗黑四大天王之一”合成大麻素的乔装？A、大麻B、 K粉C、邮票D、小树枝答案：CD2、合成大麻素家族，时而会以什么样的形象出现？A、小树枝B、邮票C、冰糖D、卡通贴纸答案：ABD3、以下属于毒品原植物的是？A、大麻B、罂粟C、古柯D、虞美人答案：ABC8、毒品大拿海洛因，对它说“NO”下决心1、吸食海洛因后，会出现以下哪些症状？（多选）A、心跳加速、产生幻觉B、精力充沛、神清气爽C、恶心呕吐、无精打采D、皮肤溃烂、牙齿脱落正确答案：A、C、D2、4号海洛因又被叫做什么？A、冰粉B、白粉C、K粉D、米粉正确答案：B1、禁毒微动漫：X任务第三集之疯狂的暗夜1、动画中X探长在两起案件现场发现了什么关键线索?A、指纹B、血迹C、脚印D、毛发答案：毛发2、牛大婶发狂咬人是因为吸食了哪种毒品?A、摇头丸B、神仙水C、冰毒D、氯胺酮答案：B3、动画中牛大婶毒瘾发作后引发了什么症状?A、流涕流泪B、腹痛、腹泻C、四肢抽搐D、精神发狂答案：D4、我国禁毒工作的治本之策是?A、预防教育B、依法严厉打击C、切断毒品来源D、戒断治疗答案：A5、要拒绝毒品，我们除了要知道什么是毒品、知道毒品极易成瘾、知道毒品的危害以外，还要怎么做?A、树立正确的人生观B、养成良好行为习惯C、拒绝不良诱惑D、出入情况复杂的娱乐场所答案：ABC2、禁毒微动漫：X任务第四集之吞噬1、动画中小袋鼠宝宝为什么突然在公园的长椅上昏迷?A、毒瘾发作B、害怕兔小姐C、想要找妈妈D、身体不好生病了答案：A2、袋鼠妈妈为什么每天都去买同一款包装独特的蛋糕?A、袋鼠宝宝喜欢吃这款蛋糕B、这款蛋糕里藏着毒品C、袋鼠妈妈喜欢吃这款蛋糕D、这款蛋糕包装精美答案：B3、羊奶奶蛋糕店的地下室里有什么?A、制毒原料B、面粉C、水果D、牛奶答案：A4、袋鼠宝宝是如何染上毒瘾的?A、袋鼠宝宝吃了羊奶奶的蛋糕B、袋鼠妈妈让袋鼠宝宝吃了藏有毒品的蛋糕C、袋鼠宝宝吸食毒品D、因为袋鼠妈妈在怀孕的时候吸食毒品答案：D5、中学生应该怎样做才能远离毒品、拒绝毒品?A、了解毒品的特征B、认识到毒品的危害C、学习西方大麻文化D、能够辨别毒品答案：ABD3、毒品危害家庭典型案例1、当陌生人主动给你饮料和水时，你应该怎么做?A、果断的接住B、礼貌的拒绝C、大声呵斥D、不理不睬答案：B2、“一人吸毒，全家遭殃”说明了毒品的什么危害?A、个人危害B、家庭危害C、社会危害D、身体危害答案：B3、以下不属于吸毒违法行为的是( )?A、服食摇头丸B、注射海洛因C、服用安眠药D、吸食大麻答案：C4、如何加强自我保护意识( )?A、不区别的结交朋友B、树立防毒意识C、对毒品有好奇心D、追求刺激答案：B5、吸毒行为产生的环境因素有很多，而这里的环境指的是( )?A、社区环境B、人际环境C、家庭环境D、以上A、B、C均是答案：D4、他们怎么了?请远离毒品!1、当你的朋友或毒贩逼你吸毒时，你应该怎么办?A、秘密打110报案B、坚决反抗C、吸一次试试看答案：A2、当自己的好朋友吸毒后，不正确的做法是什么?A、自己也吸毒，然后陪好朋友一起戒毒B、劝说好朋友戒毒C、渐渐远离这类朋友D、直接断绝来往答案：A3、吸食方式中危害最大的是( )?A、吸入式B、口服式C、注射式D、静脉注射答案：D4、禁毒是全社会的( )?A、共同责任B、义务C、权利答案：A5、下列不是毒品的是( )?A、罂粟B、大麻C、跳舞草答案：C5、新武松打虎1、《新武松打虎》中“打虎”打的是A、黑恶势力B、毒品犯罪C、暴力犯罪D、电信诈骗答案：B2、下面表述正确、完整的是A、禁毒是全社会的共同责任B、禁毒是人民政府的责任C、禁毒是公安机关的责任D、禁毒是公民自身的责任答案：A3、下列属于毒品犯罪的是A、走私、贩卖、运输、制造毒品B、非法持有毒品C、包庇毒品犯罪分子D、容留他人吸毒答案：ABCD4、2018年6月，总书记就禁毒工作作出重要指示强调，禁毒工作事关____、_____、_____，毒品一日不除，禁毒斗争就一日不能松懈。
大学英语(二)第一阶段a标准答案
大学英语(二)第一阶段a答案————————————————————————————————作者：————————————————————————————————日期：大学英语（二）20171203第一阶段a交卷时间：2017-09-25 15:56:25一、词汇与语法1.(5分)Dress at that party was ________, with no neckties or fancy dresses.• A. formal• B. informal• C. serious• D. fashionable纠错得分： 52.(5分)There is no money ________ for this project.• A. workable• B. capable• C. available• D. durable纠错得分： 53.(5分)He was very upset when Stanford ________ his application for schol arship.• A. granted• B. accepted• C. considered• D. rejected纠错得分： 54.(5分)The country is ________ and the people live in peace.• A. prosperous• B. needy• C. messy• D. violent纠错得分： 55.(5分)I simply couldn't ________ his intention.• A. figure out• B. bear out• C. get out• D. bring out纠错得分： 5二、交际英语1.(5分)Let me introduce myself. I'm steward．- _________• A. What a pleasure.• B. Pleased to meet you.• C. I don't know.• D. Thanks a lot.纠错得分： 52.(5分)Hello, could I speak to Mike please? - _________• A. Who are you?• B. Who is speaking?• C. What's wrong?• D. Why?纠错得分： 53.(5分)Well done and_________.- Thank you very much!• A. Not at all• B. Congratulations• C. That's right• D. You are welcome纠错得分： 54.(5分)Would you please show me your bankbook? - _________• A. Sorry, I have no idea.• B. Here you are.• C. Come with me.• D. Yes, I'd like to.纠错得分： 55.(5分)Can I ask you a few questions?- _________• A. What do you do?• B. It's a good idea.• C. No, thanks.• D. Certainly.纠错得分： 5三、阅读理解1.Community service is an important component of education here at ou r university. We encourage all students to volunteer for at least one co mmunity activity before they graduate. A new community program call ed "One On One" helps elementary students who've fallen behind. Your education majors might be especially interested in it because it offer s the opportunity to do some teaching, that is, tutoring in math and E nglish.You'd have to volunteer two hours a week for one semester. You can choose to help a child with math, English, or both. Half-hour lessons a re fine, so you could do a half hour of each subject two days a week. Professor Dodge will act as a mentor to the tutors---he'll be available to help you with lesson plans or to offer suggestions for activities. He has office hours every Tuesday and Thursday afternoon. You can sign up for the program with him and begin the tutoring next week.I'm sure you'll enjoy this community service…and you'll gain valuable e xperience at the same time. It looks good on your resume, too, showi ng that you've had experience with children and that you care about y our community. If you'd like to sign up, or if you have any questions, stop by Professor Dodge's office this week.纠错(1)(5分)What is the purpose of the talk? ____________• A. To explain a new requirement for graduation.• B. To interest students in a new community program.• C. To discuss the problems of elementary school students.• D. To recruit elementary school teachers for a special program. 得分： 5(2)(5分)What is the purpose of the program that the dean describes? ____ ______• A. To find jobs for graduating students.• B. To help education majors prepare for final exams.• C. To offer tutorials to elementary school students.• D. To provide funding for a community service project.得分： 5(3)(5分)What does Professor Dodge do? ____________• A. He advises students to participate in the special program.• B. He teaches part-time in an elementary school.• C. He observes elementary school students in the classroom• D. He helps students prepare their resumes.得分： 5(4)(5分)What should students interested in the tutorials do? __________• A. Contact the elementary school.• B. Sign up for a special class.• C. Submit a resume to the dean.• D. Talk to Professor Dodge.得分： 5(5)(5分)Whom do you think the speaker addresses to? _________• A. Faculty• B. Students• C. Freshman• D. Graduating students of the university.得分： 5四、选词填空1.Man can't go on increasing his number at the present rate. In the nex t 30 years man will face a period of crisis. 1 experts believe that there will be a widespread food shortage. Other experts think this is too pes simistic（悲观的）, and man can prevent things 2 worse than they are n ow.One thing that man can do is to limit the number of babies born. The need for this is obvious, 3 it is not easy to achieve. People have to be persuaded to limit their families. In the countries of the population explosion, many people like big families. The parents think that this 4 a b igger income for the family and ensures there will be someone in the family who will look after them in old age.Several governments have 5 birth control policies in recent years. Amon g them are Japan, China. India and Egypt. In some cases the results h ave not been successful. Japan has been an exception. People were en couraged to limit their families in the 1950's, which has effectively red uced the birth rate in Japan.纠错(1)(5分)1• A. From getting• B. Some• C. But• D. Adopted• E. Brings得分： 5(2)(5分)2• A. From getting• B. Some• C. But• D. Adopted• E. Brings得分： 5(3)(5分)3• A. From getting• B. Some• C. But• D. Adopted• E. Brings得分： 5(4)(5分)4• A. From getting• B. Some• C. But• D. Adopted• E. Brings得分： 5(5)(5分)5• A. From getting• B. Some• C. But• D. Adopted• E. Brings得分： 5。
理工英语1-国家开放大学电大机考网考题目答案
理工英语1一、交际用语1、A: Could we make a tentative appointment for, say, this Saturday? - -B: （）答案：That's fine for me.2、- -A: Hello, Bill Burton speaking. （）- B: Hello, Mr, Burton this is Jenny Jen kins of Bradford and Sons returning your call,答案：What can I do for you?3.A: I can put you down for eleven o'clock. Is that OK?- -B:( )答案：How about 3 o'clock in the afternoon?4、A: I have an appointment with Mr. Xie at about 10 o'clock tomorrow morning, but would you please postpone it to sometime next week?- B:答案：All right.5、A: I have many questions about study ing abroad. - B:Would you like to make an appointment? ()答案：Let's say Wednesday afternoon at two?6、- A: () are the most popular things people buy online? - B:Definitely, clothes.答案：What do you think7. - -Are you good at Building Materials, Rose?- （）. But I will try to study it well this term.Not very8、-Can I help you,sir?-I'dliketohave100().I want my students to draw pictures.答案：pieces of paper9、- Could you please tell me something about the two（）?- （）,They are exchange students of No.1 Middle School.答案：Germans; All right10.- -Do you like reading books?- Yes. Each ofus （）to do more reading in and after class.答案：is encouraged11、- -Don't forget to come to our party this w eekend!答案：Sure. See you.12、- -Excuse me, can you tell me where the nearest bus stop is?- -I'm sorry, 1 have no idea.I （）.答案：am a stranger here13、- Excuse me, fficer. 1 can't find the subway entrance. （）-See those glass doors? Go in and you will see it on your right.答案：Can you help me?14、- -Excuse me,can you tell me w here the nearest bus stopis?- -I'm sorry,I have no idea.I （）.答案：am a stranger here15、- -Excuse me. Can you tell me how to get to the Linden Street?答案：Sorry, I'm new here.16、-Excuse me. You must be Mr. Liu Hua from China?- （）答案：Yes, nice to meet you.17、- Forgotten something? 1 can keep an eye on your kids if you want to go and get it.一（）Thank you all the same.答案：Well, I can do without it.18、- Goodbye. everyone. -Bye, Sally! Don't forget to write.答案：Stay in touch.19、一-Goodbye, everyone. （）- -Bye, Sally! Don't forget to write.答案：Stay in touch.20-Have you had a nice journey?- （）答案：Yes. we have.21.- Haven't seen you for ages, Mike.（）-Pretty good. Everything goes well.答案：How is it going?22- Hello, I'm Harry Potter.- Glad to meet you. I'm Miller. But（）答案：call me Paul23、- -His parents are present there （）， but their minds are somewhere else.答案：physically24、- -I mean the best way is using your 0w n mobile phone or computer to operate it.答案：What do you mean?25、- -I'm going to Hainan for my holiday.答案：Have a nice trip!26、- I'm so excited to meet you. May I introduce myself to you?答案：Sure.27、- I'm so excited to meet you. May 1 introduce myself to you?一（）答案：Sure please! .29、-Is there anything I can do?- （）答案：Thanks, but I can manage.30、一-It is easy to get the software we need （）the market is small.答案：although31、- -I'm going to Hainan for my holiday.- （）答案：Have a nice trip32、- -I'm so excited to meet you. May I introduce myself to you?- （）答案：Sure.33、 - Let's meet at 7:30 outside the gate of （）。
2015年高考英语全国卷1-答案
2015普通高等学校招生全国统一考试（全国新课标卷1）英语答案解析第Ⅰ卷第一部分听力第一节1.【答案】A【解析】原文：W: What time is your train leaving?M: It leaves at ten. I've got fifty minutes 1eft.W: You'd better hurry. Or you won't be able to cat hit.2.【答案】C【解析】原文：M: Nice weather we're having. Don't you think?W: No.I1is too cold.M: I think it is just right.W: I'd prefer it a few degrees warmer.3.【答案】A【解析】原文：M: Now, let's stop talking and get going. I need to be in my office in fifteen minutes, or I'll be late for a meeting.W: Okay, bye.4.【答案】B【解析】原文：M: This course is really difficult.W: I don't think it’s all that bad. And we' 11 benefit a 1ot from it.M: So, you're taking it too?W: That's true.5.【答案】C【解析】原文：W: Could you turn that off? I can't hear myself think.M: What?W: The radio.M: Oh! Sorry.第二节6.【答案】B7.【答案】A【解析】原文：W: Hi Michael! I heard you just came back from a holiday.M: Yes. I stayed for a week in China, and five days in India.W: You do travel a 1ot, don't you? Last year, you went to Norway, right?M: Well, I've been to quite some countries, but not yet to Norway Last summer. I toured Russia for two weeks.8.【答案】B9.【答案】C【解析】原文：M: Sally, do you like seafood?W: Yes, of course.M: Is there anything you especially like?W: Well, I really don't know. I can never remember the names.M: Olay. Is there any food you don't eat?W: Well, I don't eat chicken. And I don't like eggs, either. But I like all kinds of fish and vegetables.M: Then 1et's 1ook at the menu and see what they've got for us.10.【答案】B11.【答案】A12.【答案】B【解析】原文：M: You look pale, Stephanie. What's wrong?W: I don't feel good. I have a bad headache. In fact, I haven't got much sleep this past week, and I feel really tired. M: Why don't you go to see a doctor?W: Yeah, I think I should. But I have a report due tomorrow. Ms. Jenkins needs it for the board meeting next Monday.M: Well, it's Wednesday today. Why don't you talk to Ms. Jenkins, and ask if you can hand it in on Friday morning? W: Maybe I should try. I guess I just need a good sleep. Thanks, George.M: If you need any help with the report just let me know.13.【答案】C14.【答案】A15.【答案】B16.【答案】C【解析】原文：W: Anything interesting in the paper today, dear?M: Well, yeah. There are a few here that might interest us. Here's one for just four hundred dollars. It only has one bedroom. But it sounds nice near Lake Street.W: Yeah. Let me see what the cheapest two-bedroom apartment is. Oh, here's one on Market Street. It's a real bargain. Only three hundred and fifty dollars. But it doesn't have any furniture.M: Well, it costs a lot to buy all the furniture.W: Oh, here's another one for just over four hundred dollars. T his sounds very interesting! It’s on South Street. That's a nice area.M: Yes, it's quiet. Did you say two bedrooms?W: Yes, at four hundred and fifteen dollars.M: Why don't we go and have a 1ook?W: Okay, I'll give them a call.17.【答案】B18.【答案】A19.【答案】A20.【答案】C【解析】原文：Look at this picture. It's the London Tea Trade Centre. As you can see, it is on the north bank of the river Thames. It is the center of an important industry in the everyday lives of the British people. Tea is the British national drink. Every man, woman, and child over ten years of age has on average over four cups a day. Or someone thousand, five hundred cups annually. About thirty percent of the world's exports of tea makes its way to London. And Britain is by far the largest importer of tea in the world. Now, in the second picture, you can see how tea is tasted in the Tea Trade Centre before it is sold. Here, different types of tea are tasted by skilled tea tasters. Before they're sold at each week's tea sale. It' s amazing to see them at work. Over a hundred kinds of tea are laid out in a line on 1ong tables. The tasters generally taste teas with milk, since that is how the majority of British people drink their tea. The tasters move down the line with surprising speed, tasting from a spoon and deciding what is a fair price for each tea. After that, they...第二部分阅读理解第一节21.【答案】C【解析】根据文章第二段第二句James Brindley is recognized as one of the leading early canal engineers.所对应的日期November 7th。
健康中国行动答题答案
健康中国行动答题答案1. 下列哪项不是引起心力衰竭常见的诱因?A.骨折(正确答案)B.心律失常C.妊娠D.感染2. 精神（心理）性职业紧张属A.生产过程中的有害因素B.劳动条件中的有害因素C.生产过程中的有害因素D.劳动过程中的有害因素(正确答案)3. 《全民健身条例》规定每年为全民健身日。
A.6月8日B.6月18日C.8月8日(正确答案)D.8月18日4. 关于诺如病毒感染引起的胃肠炎，说法不正确的是_。
A.具有发病急、传播速度快、涉及范围广等特点B.症状是恶心、呕吐、腹部痉挛性腹泻C.症状通常持续10-15天(正确答案)D.症状一般摄入病毒后24-48小时出现，但是暴露后12小时也可能出现症状5. 儿童哪个症状常可能是儿童支气管哮喘的唯一症状?A.胸闷B.喘息C.慢性或反复咳嗽(正确答案)D.呼吸困难6. 为巩固毛囊，一般可在植发1-2周后开始使用米诺地尔，建议使用_。
A.1个月B.3个月C.至少6个月(正确答案)D.至少12个月7. 关于药品，以下说法错误的是?A.药品说明书用于指导医生和患者选择使用(正确答案)药品，具有法律效力B.大部分药品属于处方药，如注射剂、毒麻药品等C.非处方药不需要凭执业医师处方即可自行购买使用D.处方药不可擅自使用、停用或增减剂量8. 胰岛素因保存不当遇到低温环境，以下选项表述不正确的是_。
A.不会有不良影响(正确答案)B.冷冻结冰后，会导致胰岛素变性从而失效C.冷冻之后，即使解冻，胰岛素也不能再使用D.一旦发现胰岛素结冰，则应妥善丢弃，换用新的胰岛素9. 导致婴儿营养不良的主要原因是_。
A.喂养不当(正确答案)B.疾病因素C.母乳不足D.小儿偏食10. 下列产品中哪个不会促使女性衰老_。
A.咖啡(正确答案)B.白酒C.烟雾尘埃D.铝制品11. 婴幼儿高热惊厥发病率较高，多见于_的婴幼儿。
A.3月龄-3岁B.6月龄-2岁C.6月龄-3岁(正确答案)D.6月龄-6岁12. 中国国家节水标志由_构成。
2023年数学高考复习真题演练(2021-2022年高考真题)10 对数与对数函数 (含详解)
专题10 对数与对数函数【考点预测】 1.对数式的运算(1)对数的定义：一般地，如果(0x a N a =>且1)a ≠，那么数x 叫做以a 为底N 的对数，记作log a x N =，读作以a 为底N 的对数，其中a 叫做对数的底数，N 叫做真数．(2)常见对数：①一般对数：以(0a a >且1)a ≠为底，记为log N a ，读作以a 为底N 的对数；②常用对数：以10为底，记为lg N ； ③自然对数：以e 为底，记为ln N ； (3) 对数的性质和运算法则：①1log 0a =；log 1a a =；其中0a >且1a ≠；②log Na a N =(其中0a >且1a ≠，0N >)； ③对数换底公式：log log log c a c bb a=； ④log ()log log a a a MN M N =+； ⑤log log log aa a MM N N=-； ⑥log log (m na a nb b m m=，)n R ∈； ⑦log a b a b =和log b a a b =； ⑧1log log a b b a=； 2.对数函数的定义及图像(1)对数函数的定义：函数 log a y x =(0a >且1)a ≠叫做对数函数． 对数函数的图象【方法技巧与总结】 1.对数函数常用技巧在同一坐标系内，当1a >时，随a 的增大，对数函数的图象愈靠近x 轴；当01a <<时，对数函数的图象随a 的增大而远离x 轴．(见下图)【题型归纳目录】题型一：对数运算及对数方程、对数不等式 题型二：对数函数的图像题型三：对数函数的性质（单调性、最值（值域）） 题型四：对数函数中的恒成立问题 题型五：对数函数的综合问题 【典例例题】题型一：对数运算及对数方程、对数不等式例1．（2022·全国·高三专题练习）（1）计算331log 2327lg 50lg 2+++； （2）已知()23log log lg 1x ⎡⎤=⎣⎦，求实数x 的值； （3）若185a =，18log 9b =，用a ，b ，表示36log 45． 例2．（2022·全国·高三专题练习）（1）求23151log log 8log 2725⋅⋅的值. （2）已知9log 5=a ，37b =，试用a ，b 表示21log 35例3．（2022·全国·高三专题练习）（1）已知a ，b ，c 均为正数，且3a ＝4b ＝6c ，求证：212ab c+=；（2）若60a ＝3，60b ＝5，求12(1)12a bb ---的值．例4．（2022·全国·模拟预测）若e 4a =，e 25b =，则（ ） A ．a ＋b ＝100B ．b －a ＝ea 增大a 增大C ．28ln 2ab <D ．ln6b a ->例5．（2022·全国·模拟预测）已知实数x ，y 满足0x >，0y >，1x ≠，1y ≠，y x x y =，log 4y xx y+=，则x y +=（ ） A ．2 B ．4 C ．6 D ．8例6．（2022·北京昌平·二模）已知函数2()42(0)f x ax ax a =-+<，则关于x 的不等式2()log f x x >的解集是（ ）A ．(,4)-∞B ．(0,1)C ．(0,4)D ．(4,)+∞例7．（2022·全国·江西师大附中模拟预测（文））已知函数()122log ,1,1,1,x x f x x x >⎧⎪=⎨⎪-≤⎩则不等式()(1)f x f x <-的解集为______．例8．（2022·辽宁·东北育才学校二模）若函数()f x 满足：（1）1x ∀，()20,x ∈+∞且12x x ≠，都有()()21210f x f x x x -<-；（2）()()1122x f f x f x x ⎛⎫=- ⎪⎝⎭，则()f x =___________.（写出满足这些条件的一个函数即可）例9．（2022·全国·高三专题练习）设函数()log m f x x =（0m >且1m ≠）的图像经过点()3,1.（1）解关于x 的方程()()22(1)10f x m f x m +-+-=；（2）不等式()()10f x a f x +⋅->⎡⎤⎡⎤⎣⎦⎣⎦的解集是1,93⎛⎫⎪⎝⎭，试求实数a 的值.【方法技巧与总结】对数的有关运算问题要注意公式的顺用、逆用、变形用等.对数方程或对数不等式问题是要将其化为同底，利用对数单调性去掉对数符号，转化为不含对数的问题，但这里必须注意对数的真数为正.题型二：对数函数的图像例10．（2022·山东潍坊·二模）已知函数()()log a f x x b =-（0a >且1a ≠）的图像如图所示，则以下说法正确的是（ ）A ．0a b +<B ．1ab <- C ．01b a << D ．log 0a b >例11．（2022·江苏省高邮中学高三阶段练习）函数log (3)1(0a y x a =+->且1)a ≠的图象恒过定点A ，若点A 在直线10mx ny ++=上，其中0mn >，则11+m n的最小值为（ ） A ．3-B ．1C ． 3+D ．2+（多选题）例12．（2022·福建·莆田二中模拟预测）已知函数()()log a g x x k =+（0a >且1a ≠）的图象如下所示．函数()()1x x f x k a a -=--的图象上有两个不同的点()11,A x y ，()22,B x y ，则（ ）A ．1a >，2k >B ．()f x 在R 上是奇函数C ．()f x 在R 上是单调递增函数D ．当0x ≥时，()()22f x f x ≤例13．（2022·全国·高三专题练习）已知223,20(){1ln ,021x x x f x x x -+-≤<=≤≤+，若()()g x f x ax a =--的图象与x 轴有3个不同的交点，则实数a 的取值范围为______.【方法技巧与总结】研究和讨论题中所涉及的函数图像是解决有关函数问题最重要的思路和方法.图像问题是数和形结合的护体解释.它为研究函数问题提供了思维方向.题型三：对数函数的性质（单调性、最值（值域））例14．（2022·陕西·榆林市第十中学高二期中（文））函数()22log 43y x x =+-的一个单调增区间是（ ）A ．3,2⎛⎫-∞ ⎪⎝⎭B ．3,2∞⎡⎫+⎪⎢⎣⎭C ．31,2⎛⎫- ⎪⎝⎭D ．3,42⎡⎫⎪⎢⎣⎭例15．（2022·天津·南开中学二模）已知函数()21,14log 1,1a ax x x f x x x ⎧--≤⎪=⎨⎪->⎩是R 上的单调函数，则实数a 的取值范围为（ ） A ．11,42⎡⎫⎪⎢⎣⎭B ．11,42⎡⎤⎢⎥⎣⎦C ．10,2⎛⎤ ⎥⎝⎦D ．1,12⎛⎫ ⎪⎝⎭例16．（2022·浙江·模拟预测）己知实数,(1,)∈+∞a b ，且33log log 3log log 4b a a b +=+，则（ ） Ab a <<B．b a <Ca b <D．a b <<例17．（2022·全国·高三专题练习（理））函数f (x )＝log ax (0＜a ＜1)在[a 2，a ]上的最大值是（ ） A ．0 B ．1C ．2D ．a 例18．（2022·重庆·模拟预测）若函数()2()log 341a f x x ax =-+-有最小值，则实数a 的取值范围是（ ）A．⎫⎪⎪⎝⎭B．C．⎛ ⎝⎭D．)+∞【方法技巧与总结】研究和讨论题中所涉及的函数性质是解决有关函数问题最重要的思路和方法.性质问题是数和形结合的护体解释.它为研究函数问题提供了思维方向.题型四：对数函数中的恒成立问题例19．（2022·北京·高三专题练习）若不等式2log 0a x x -<在10,2⎛⎫ ⎪⎝⎭内恒成立，则a 的取值范围是（ ） A ．1116a ≤< B ．1116a << C ．1016a <≤D ．1016a <<例20．（2022·江苏·高三专题练习）已知函数22414ax x y -+⎛⎫= ⎪⎝⎭的值域为10,16⎛⎤⎥⎝⎦，若不等式()()log 4log 2x a x a t t ⋅<-在[]1,2x ∈上恒成立，则t 的取值范围是（ ） A ．2,25⎛⎫ ⎪⎝⎭B ．2,5⎛⎫+∞ ⎪⎝⎭C ．(,2)-∞D ．()0,2例21．（2022·浙江·高三阶段练习）已知函数()29x f x x+=，()2log g x x a =+，若存在[]13,4x ∈，任意[]24,8x ∈，使得()()12f x g x ≥，则实数a 的取值范围是___________. 例22．（2022·全国·高三专题练习）已知函数()ln f x x x =-，已知实数0a >，若2()e ln 0x f x a a ++≥在()0+∞，上恒成立，求实数a 的取值范围. 例23．（2022·全国·高三专题练习）已知函数()log (0,1)x a f x a x a a =+>≠在[1,2]上的最大值与最小值之和为6log 2a +． （1）求实数a 的值；（2）对于任意的[2,)x ∈+∞，不等式()10kf x -≥恒成立，求实数k 的取值范围．例24．（2022·陕西安康·高三期末（文））已知函数()()()2log 2log 30,1a a f x x x a a =++>≠.(1)若()32f =，求a 的值；(2)若对任意的[]8,12x ∈，()6f x >恒成立，求a 的取值范围.例25．（2022·上海·高三专题练习）已知2()32log f x x =-，2()log g x x =. （1）当[]1,4x ∈时，求函数[]()1()y f x g x =+⋅的值域；（2）对任意12,2n n x +⎡⎤∈⎣⎦，其中常数n N ∈，不等式()2()f x f kg x ⋅>恒成立，求实数k的取值范围.【方法技巧与总结】（1）利用数形结合思想，结合对数函数的图像求解；（2）分离自变量与参变量，利用等价转化思想，转化为函数的最值问题.（3）涉及不等式恒成立问题，将给定不等式等价转化，借助同构思想构造函数，利用导数探求函数单调性、最值是解决问题的关键.题型五：对数函数的综合问题例26．（2022·河北·张家口市第一中学高三阶段练习）已知定义域为()0,∞+的单调递增函数()f x 满足：()0,x ∀∈+∞，有()()ln 1f f x x -=，则方程()242f x x x =-+-的解的个数为（ ）A ．3B ．2C ．1D ．0例27．（2022·四川雅安·三模（文））设()f x 是定义在R 上的偶函数，对任意R x ∈，都有()()4f x f x +=，且当[]2,0x ∈-时，()163xf x ⎛⎫=- ⎪⎝⎭．若在区间(]2,6-内关于x 的方程()()()log 201a f x x a -+=>恰有3个不同的实数根，则a 的取值范围是（ ）．A ．()1,2B ．()2,+∞C ．(D ．)2例28．（2022·广西柳州·高一期中）已知0a b >>，且1a b +=，则（ ）A．sin sin a b > B ．11a b> C ．22a b +>D ．lg lg 0a b +=例29．（2022·河北保定·二模）已知函数2332xxy =-在()0,∞+上先增后减，函数3443xxy =-在()0,∞+上先增后减.若()231log log x =()321log log 0x a =>，()()242422log log log log x x b ==，()()343433log log log log 0x x c ==>，则（ ） A ．a c <B ．b a <C ．c a <D ．a b <例30．（2022·广东·三模）已知,R a b ∈，e 是自然对数的底，若e ln b b a a +=+，则a b的取值可以是（ ） A ．1B ．2C ．3D ．4例31．（2022·全国·高三专题练习）已知0x 是函数()22e ln 2x f x x x -=+-的零点，则020e ln xx -+=_______．【过关测试】一、单选题 1．（2022·辽宁辽阳·二模）区块链作为一种新型的技术，被应用于许多领域.在区块链技术中，某个密码的长度设定为512B ，则密码一共有5122种可能，为了破解该密码，在最坏的情况下，需要进行5122次运算.现在有一台计算机，每秒能进行142.510⨯次运算，那么在最坏的情况下，这台计算机破译该密码所需的时间大约为（参考数据lg20.3≈ 1.58≈）（ ） A ．1393.1610s ⨯ B ．1391.5810s ⨯ C ．1401.5810s ⨯D ．1403.1610s ⨯2．（2022·山东·肥城市教学研究中心模拟预测）已知1log 3m p =，9p n =，其中0m >且1m ≠，0n >且1n ≠，若20m n -=，则p 的值为（ ） A ．3log 2B ．2log 3C ．2D ．33．（2022·河南安阳·模拟预测（文））已知正实数x ，y ，z 满足(34zx y ==，则（ ） A ．111x y z+=B ．111y z x+= C ．112x y z += D ．112x z y+=4．（2022·河南·南阳中学高三阶段练习（文））已知函数()()()ln 22ln 33f x x x =++-，则()f x （ ）A ．是奇函数，且在0,1上单调递增B ．是奇函数，且在0,1上单调递减C ．是偶函数，且在0,1上单调递增D ．是偶函数，且在0,1上单调递减5．（2022·全国·高三专题练习）函数()log (1)2a f x x =-+的图象恒过定点 A ．（2，2）B ．（2，1）C ．（3，2）D ．（2，0）6．（2022·安徽六安·一模（文））设函数()2f x =()()2ln 41g x ax x =-+，若对任意的1R x ∈，都存在实数2x ，使得()()12f x g x =成立，则实数a 的取值范围为（ ） A ．(],4-∞B ．(]0,4C ．[]0,4D ．(]0,27．（2022·湖北·荆门市龙泉中学二模）设0a >且1a ≠，sin cos a x x x >+对(0,)4x π∈恒成立，则a 的取值范围是（ ） A ．(0,)4πB ．(0,]4πC ．(,1)(1,)42ππ⋃D ．[,1)4π8．（2022·浙江·模拟预测）己知实数,(1,)∈+∞a b ，且33log log 3log log 4b a a b +=+，则（ ）A b a <<B ．b a <C a b <D ．a b <<二、多选题9．（2022·重庆市天星桥中学一模）已知0,0a b >>，且1a b +=，则下列结论正确的是（ ） A ．11a b+的最小值是4 B ．1ab ab+的最小值是2C ．22a b +的最小值是D ．22log log a b +的最小值是2-10．（2022·广东汕头·二模）设a ，b ，c 都是正数，且469a b c ==，则下列结论正确的是（ ） A ．2ab bc ac +=B ．ab bc ac +=C ．4949b b a c ⋅=⋅D ．121c b a=-11．（2022·河北·高三阶段练习）下列函数中，存在实数a ，使函数()f x 为奇函数的是（ ）A ．()(lg f x x =B ．()2f x x ax =+C ．()21xaf x e =-- D ．()()2ln 2xx f x x e a =+-12．（2022·江苏·南京师大附中高三开学考试）当102x <≤时，4log xa x ≤，则a 的值可以为（ ）AB C D三、填空题13．（2022·天津·二模）已知()4log 41log x y +=+2x y +的最小值为__________.14．（2022·全国·高三专题练习）已知23e ln 3x x x -+=，则3e ln x x -+=__________．15．（2022·河南·模拟预测（文））已知函数()241,1log ,1x x f x x x ⎧-≤=⎨>⎩，若1()2f a <≤，则实数a的取值范围为___________.16．（2022·河南·开封高中模拟预测（文））已知函数()y f x =为奇函数，且对定义域内的任意x 都有()()11f x f x +=--．当()1,2x ∈时，()21log f x x =-．给出以下4个结论： ①函数()y f x =的图象关于点()(),0k k ∈Z 成中心对称；②函数()y f x =是以2为周期的周期函数；③当()0,1x ∈时，()()2log 21f x x =--； ④函数()y f x =在()(),1k k k +∈Z 上单调递减． 其中所有正确结论的序号为______． 四、解答题17．（2022·北京·高三专题练习）已知函数()log (0),1)a f x x a a =>≠且，设1a >，函数log a y x =的定义域为[m ，n ] (m <n )，值域为[0，1]，定义“区间[m ，n ]的长度等于n －m ”，若区间[m ，n ]长度的最小值．．．为5,6求实数a 的值；18．（2022·全国·高三专题练习（理））已知函数f (x )＝log a (x ＋1)－log a (1－x )，a ＞0且a ≠1． (1)求f (x )的定义域；(2)判断f (x )的奇偶性并予以证明； (3)当a ＞1时，求使f (x )＞0的x 的解集．19．（2022·北京·高三专题练习）已知函数()log (0)1)a f x x a a =>≠且，作出|()|y f x =的大致图像并写出它的单调性；20．（2022·全国·高三专题练习）已知函数()()44log 3log 4f x x x =-⋅.当1,164x ⎡⎤∈⎢⎥⎣⎦时，求该函数的值域；21．（2022·全国·高三专题练习）已知：函数()0.51log 1axf x x -=-在其定义域上是奇函数，a 为常数． (1)求a 的值．(2)证明：()f x 在()1,+∞上是增函数．(3)若对于[]3,4上的每一个x 的值，不等式()12xf x m ⎛⎫>+ ⎪⎝⎭恒成立，求实数m 的取值范围．22．（2022·北京东城·高三期末）曲线ln y x =在点(,ln )A t t 处的切线l 交x 轴于点M . (1)当t e =时，求切线l 的方程；(2)O为坐标原点，记AMO的面积为S，求面积S以t为自变量的函数解析式，写出其定义域，并求单调增区间.专题10 对数与对数函数【考点预测】 1.对数式的运算(1)对数的定义：一般地，如果(0x a N a =>且1)a ≠，那么数x 叫做以a 为底N 的对数，记作log a x N =，读作以a 为底N 的对数，其中a 叫做对数的底数，N 叫做真数．(2)常见对数：①一般对数：以(0a a >且1)a ≠为底，记为log N a ，读作以a 为底N 的对数；②常用对数：以10为底，记为lg N ； ③自然对数：以e 为底，记为ln N ； (3) 对数的性质和运算法则：①1log 0a =；log 1a a =；其中0a >且1a ≠；②log Na a N =(其中0a >且1a ≠，0N >)； ③对数换底公式：log log log c a c bb a=； ④log ()log log a a a MN M N =+； ⑤log log log aa a MM N N=-； ⑥log log (m na a nb b m m=，)n R ∈； ⑦log a b a b =和log b a a b =； ⑧1log log a b b a=； 2.对数函数的定义及图像(1)对数函数的定义：函数 log a y x =(0a >且1)a ≠叫做对数函数． 对数函数的图象【方法技巧与总结】 1.对数函数常用技巧在同一坐标系内，当1a >时，随a 的增大，对数函数的图象愈靠近x 轴；当01a <<时，对数函数的图象随a 的增大而远离x 轴．(见下图)【题型归纳目录】题型一：对数运算及对数方程、对数不等式 题型二：对数函数的图像题型三：对数函数的性质（单调性、最值（值域）） 题型四：对数函数中的恒成立问题 题型五：对数函数的综合问题 【典例例题】题型一：对数运算及对数方程、对数不等式例1．（2022·全国·高三专题练习）（1）计算331log 2327lg 50lg 2+++； （2）已知()23log log lg 1x ⎡⎤=⎣⎦，求实数x 的值； （3）若185a =，18log 9b =，用a ，b ，表示36log 45． 【答案】（1）7；（2）109；（3）2a bb+-． 【解析】（1）利用对数恒等式和对数的运算法则计算即可； （2）利用指对互化可得实数x 的值；（3）先求出a ，再利用换底公式结合对数的运算法则求得结果．【详解】（1）原式=()23lg 510lg25lg51lg26lg5lg26lg107++⨯+=+++=++=+=；（2）因为()23log log lg 1x ⎡⎤=⎣⎦，所以()3log lg 2x =，所以2lg 39x ==，所以x =109；a 增大a 增大（3）因为185a =，所以18log 5a =，所以()()()181818183618181818log 59log 45log 5log 9log 45log 36log 182log 18log 189⨯+====⨯+÷1818181818log 5log 9log 18log 18log 92a bb++=+--．例2．（2022·全国·高三专题练习）（1）求23151log log 8log 2725⋅⋅的值. （2）已知9log 5=a ，37b =，试用a ，b 表示21log 35 【答案】（1）18；（2）21a bb ++. 【解析】 【分析】（1）首先根据题意得到原式()()()2352log 53log 23log 3=-⋅⋅-，再利用换底公式化简即可得到答案.（2）首先根据题意得到3log 7b =，3log 52=a ，再利用换底公式化简即可得到答案. 【详解】（1）原式()()()1233232355log 5log 2log 32log 53log 23log 3--=⋅⋅=-⋅⋅-lg5lg 2lg31818lg 2lg3lg5=⋅⋅⋅=（2）由37b =得到3log 7b =， 由9log 5=a ，得到31log 52=a ，即3log 52=a . 33321333log 35log 5log 72log 35log 21log 7log 31a bb ++===++.【点睛】本题主要考查对数的换底公式，同时考查指数、对数的互化公式，属于中档题.例3．（2022·全国·高三专题练习）（1）已知a ，b ，c 均为正数，且3a ＝4b ＝6c ，求证：212a b c+=；（2）若60a ＝3，60b ＝5，求12(1)12a bb ---的值． 【答案】（1）详见解析；（2）2. 【解析】【分析】（1）设3461a b c k ===>，应用指对数的互化有346log ,log ,log a k b k c k ===，进而应用换底公式及对数的运算性质分别求21a b +、2c，即可证结论；（2）应用指对数互化有6060log 3,log 5a b ==，应用对数的运算性质求12(1)a bb ---，进而可求12(1)12a b b ---的值.【详解】（1）设346a b c k ===，则1k >. ∴346log ,log ,log a k b k c k ===，∴3421212log 3log 4log 9log 4log 362log 6log log k k k k k k a b k k+=+=+=+==， 而6222log 6log k c k==， ∴212a b c+=. （2）由题设知：6060log 3,log 5a b ==，得606011log 5log 12b -=-=，60606011log 3log 5log 4a b --=--=， ∴60121260log 42log 21log 22(1)2log 122a b b --===-， 则121log 22(1)12122a b b ---==.例4．（2022·全国·模拟预测）若e 4a =，e 25b =，则（ ） A ．a ＋b ＝100 B ．b －a ＝e C ．28ln 2ab < D ．ln6b a ->【答案】D 【解析】 【分析】利用指数和对数互化，得到a ，b 后逐项判断. 【详解】对于A ，由e 4a =，e 25b =，得ln 4a =，ln 25b =，所以ln 4ln 25ln100a b +=+=，故A 错误；对于B ，25ln 25ln 4ln4b a -=-=，故B 错误； 对于C ，2ln 4ln 252ln 2ln168ln 2ab =⨯>⨯=，故C 错误；对于D ，25ln 25ln 4lnln 64b a -=-=>，故D 正确． 故选：D ．例5．（2022·全国·模拟预测）已知实数x ，y 满足0x >，0y >，1x ≠，1y ≠，y x x y =，log 4y xx y+=，则x y +=（ ） A ．2 B ．4 C ．6 D ．8【答案】C 【解析】 【分析】 根据y x x y =得到lg lg x xy y =，再利用换底公式得到2x y=，利用lg 2lg x y =，即2x y =，求出4x =，2y =，所以6x y +=.【详解】由y x x y =，得lg lg y x x y =，lg lg x xy y=. 由log 4y x x y +=，lg log lg y x x y =，所以lg 4lg x x y y+=， 所以4x xy y +=，解得：2x y=，则lg 2lg x y =，即2x y =， 所以4x =，2y =，所以6x y +=， 故选：C.例6．（2022·北京昌平·二模）已知函数2()42(0)f x ax ax a =-+<，则关于x 的不等式2()log f x x >的解集是（ ）A ．(,4)-∞B ．(0,1)C ．(0,4)D ．(4,)+∞【答案】C 【解析】 【分析】由二次函数的性质判断()f x 区间单调性，根据解析式知()f x 恒过(4,2)且(0)2f =，进而确定区间值域，再由对数函数性质求2log y x =的对应区间值域，即可得不等式解集. 【详解】由题设，()f x 对称轴为2x =且图象开口向下，则()f x 在(0,2)上递增，(2,)+∞上递减， 由2()42(4)2f x ax ax ax x =-+=-+，即()f x 恒过(4,2)且(0)2f =， 所以(0,4)上()2f x >，(4,)+∞上()2f x ，而2log y x =在(0,)+∞上递增，且(0,4)上2y <，(4,)+∞上2y >，所以2()log f x x >的解集为(0,4). 故选：C例7．（2022·全国·江西师大附中模拟预测（文））已知函数()122log ,1,1,1,x x f x x x >⎧⎪=⎨⎪-≤⎩则不等式()(1)f x f x <-的解集为______．【答案】12x x ⎧⎫>⎨⎬⎩⎭【解析】 【分析】 分1x ≤、12x <≤和2x >，依次解不等式，再取并集即可.【详解】当1x ≤时，不等式()(1)f x f x <-为2211(1)x x -<--，解得112x <≤； 当12x <≤时，不等式()(1)f x f x <-为212log 1(1)x x <--，易知21122log log 10,1(1)0x x <=--≥，解得12x <≤；当2x >时，不等式()(1)f x f x <-为1122log log (1)x x <-，解得2x >；综上，解集为：12x x ⎧⎫>⎨⎬⎩⎭.故答案为：12x x ⎧⎫>⎨⎬⎩⎭.例8．（2022·辽宁·东北育才学校二模）若函数()f x 满足：（1）1x ∀，()20,x ∈+∞且12x x ≠，都有()()21210f x f x x x -<-；（2）()()1122x f f x f x x ⎛⎫=- ⎪⎝⎭，则()f x =___________.（写出满足这些条件的一个函数即可） 【答案】12log x，（log a x ，(0<a <1)都对）【解析】 【分析】满足第一个条件，表示函数是单调递减函数，第二个条件正好是符合对数的运算性质； 【详解】对于条件①，不妨设12x x <，则210x x ->，∵()()21210f x f x x x -<-，∴()()210f x f x -<∴12()()f x f x >，∴()f x 为()0,+∞上的单调递增函数，对于条件②，刚好符合对数的运算性质，故这样的函数可以是一个单调递减的对数函数. 故答案为:12log x.（log ax ，(0<a <1)都对）例9．（2022·全国·高三专题练习）设函数()log m f x x =（0m >且1m ≠）的图像经过点()3,1.（1）解关于x 的方程()()22(1)10f x m f x m +-+-=；（2）不等式()()10f x a f x +⋅->⎡⎤⎡⎤⎣⎦⎣⎦的解集是1,93⎛⎫⎪⎝⎭，试求实数a 的值. 【答案】（1）9x =或181x =；（2）2a =. 【解析】 【分析】(1)根据给定条件求出m 值，并代入方程，再解方程即得.(2)由给定解集借助对数函数单调性求出()f x 范围，换元借助一元二次不等式即可得解. 【详解】（1）由已知得()31f =，即log 31m =，则3m =，于是得()3log f x x =， 方程222()(1)()10()2()80f x m f x m f x f x +-+-=⇔+-=， 从而得()2f x =或()4f x =-，即3log 2x =或3log 4x =-，9x =或181x =， 所以原方程的根为9x =或181x =； （2）依题意，函数()3log f x x =中，1,93x ⎛⎫∈ ⎪⎝⎭，从而得()3log 1,2x ∈-.又()()()()3310log 1log 0f x a f x x x a +⋅->⇔+⋅-<⎡⎤⎡⎤⎣⎦⎣⎦，令3log x t =， 即一元二次不等式()()10t t a +⋅-<的解集为()1,2-，因此有-1，2是关于t 的方程()()10t t a +⋅-=的两根，则2a =， 所以实数a 的值为2.【方法技巧与总结】对数的有关运算问题要注意公式的顺用、逆用、变形用等.对数方程或对数不等式问题是要将其化为同底，利用对数单调性去掉对数符号，转化为不含对数的问题，但这里必须注意对数的真数为正.题型二：对数函数的图像例10．（2022·山东潍坊·二模）已知函数()()log a f x x b =-（0a >且1a ≠）的图像如图所示，则以下说法正确的是（ ）A ．0a b +<B ．1ab <-C ．01b a <<D ．log 0a b >【答案】C 【解析】 【分析】结合函数()f x 的图象可得1a >和10b -<<，然后逐项分析即可求出结果. 【详解】由图象可知()f x 在定义域内单调递增，所以1a >，令()()log 0a f x x b =-=，即1x b =+，所以函数()f x 的零点为1b +，结合函数图象可知011b <+<，所以10b -<<，因此0a b +>，故A 错误；0-<<a ab ，又因为1a >，所以1a -<-，因此1ab <-不一定成立，故B 错误；因为10b a a a -<<，即11b a a <<，且101a<<，所以01b a <<，故C 正确； 因为01b <<，所以log log 1a a b <，即log 0a b <，故D 错误， 故选：C.例11．（2022·江苏省高邮中学高三阶段练习）函数log (3)1(0a y x a =+->且1)a ≠的图象恒过定点A ，若点A 在直线10mx ny ++=上，其中0mn >，则11+m n的最小值为（ ） A．3-B ．1C ． 3+D ．2+【答案】C 【解析】 【分析】由对数函数的性质，可得()2,1A --，可得21m n +=，再根据基本不等式“1”的用法，即可求出结果.【详解】解：因为函数log (3)1(0a y x a =+->且1)a ≠的图象恒过定点()2,1A --，所以210m n --+=，即21m n +=， 所以()1111223n m m n m n m n m n⎛⎫+=++=++ ⎪⎝⎭， 又0mn >，所以0,0n mm n>>所以2333n m m n ++≥=，当且仅当2n m m n =，即1n =时取等号.故选：C.（多选题）例12．（2022·福建·莆田二中模拟预测）已知函数()()log a g x x k =+（0a >且1a ≠）的图象如下所示．函数()()1x xf x k a a -=--的图象上有两个不同的点()11,A x y ，()22,B x y ，则（ ）A ．1a >，2k >B ．()f x 在R 上是奇函数C ．()f x 在R 上是单调递增函数D ．当0x ≥时，()()22f x f x ≤【答案】BCD 【解析】 【分析】对于A 结合对数型函数图像相关知识求解；对于B 运用定义法判断()f x 是否在R 上是奇函数；对于C 运用定义法判断函数单调性；对于D 通过作差法并对式子变形即可判断. 【详解】对于A ，由图像可知，函数()()log a g x x k =+（0a >且1a ≠）在()2,-+∞上单调递增，所以1a >，因为()g x 经过()1,0-，所以()()1log 10a g k -=-+=，所以01a k =-+，2k =，故A 错误.对于B ，()x x f x a a -=-，定义域R 关于原点对称，()()x xf x a a f x --=-=-，所以()f x 在R 上是奇函数，故B 正确.对于C ，对于()x xf x a a -=-，由题意不妨令1212,,x x x R x R >∈∈，则()()()()()121212121212121212111x x x x x x x x x x x x x x x x a a a a a f x f x a a a a a a a a ++++--⎛⎫⎛⎫-=---=-+=⎪ ⎪⎝⎭⎝⎭，因为1212,,x x x R x R >∈∈，1a >，所以12121210,0,0x x x x x x a a a a +++>>->，即()()12f x f x >，所以()f x 在R 上是单调递增函数，故C 正确.对于D ，()()()()()()()()()2222222x x x x x x x x x x x x x x a a a a a a a a a a a a a x f a f x --------=---=---+--=-()()()()22322221111112x x x x x x xx xxxa a a a a a a a a aa----+-⎛⎫⎛⎫--=⎪-==⎪⎝⎭⎝⎭，因为1a >，0x ≥，所以()3210,010,xxxa a a +≥>->，所以()()23101x x xa a a-+-≤，当且仅当0x =时等号成立，即当0x ≥时，()()22f x f x ≤成立，故D 正确.故选：BCD例13．（2022·全国·高三专题练习）已知223,20(){1ln ,021x x x f x x x -+-≤<=≤≤+，若()()g x f x ax a =--的图象与x 轴有3个不同的交点，则实数a 的取值范围为______. 【答案】ln 31[,)3e【解析】 【分析】由分段函数解析式，结合导数研究|()|f x 的性质，再将问题转化为|()|f x 与(1)y a x =+有3个不同交点，应用数形结合的思想有(1)y a x =+与|()|f x 在02x ≤≤上至少有2个交点，最后由导数求它们相切或(1)y a x =+过(2,ln 3)时参数a 的值，即可知a 的取值范围. 【详解】由题设，20x -≤<上239()2()48f x x =--+，故值域为[14,0]-且单调递增；02x ≤≤上()f x '=101x -<+，故()f x 值域为[ln 3,0]-且单调递减； ∴|()|f x 在20x -≤<上值域为[0,14]且单调递减；在02x ≤≤上值域为[0,ln 3]且单调递增； 要使()g x 与x 轴有3个不同的交点，即|()|f x 与(1)y a x =+有3个不同交点，它们的图象如下：∴由图知：要使函数图象有3个交点，则(1)y a x =+与|()|f x 在02x ≤≤上至少有2个交点， 由02x ≤≤，1()|()|ln1g x f x x ==-+，则1()|()|1g x f x x '==+，此时，若|()|f x 与(1)y a x =+相切时，切点为(,(1))m a m +， ∴111ln (1)1a m a m m ⎧=⎪⎪+⎨⎪-=+⎪+⎩，可得1e a =，当(1)y a x =+过(2,ln 3)时，有3ln3a =，得ln 33a =， ∴ln 313ea ≤<. 故答案为：ln 31[,)3e【点睛】关键点点睛：根据已知研究|()|f x 的性质，并将问题转化为|()|f x 与(1)y a x =+的交点问题，应用导数的几何意义、数形结合的思想求参数范围.【方法技巧与总结】研究和讨论题中所涉及的函数图像是解决有关函数问题最重要的思路和方法.图像问题是数和形结合的护体解释.它为研究函数问题提供了思维方向.题型三：对数函数的性质（单调性、最值（值域））例14．（2022·陕西·榆林市第十中学高二期中（文））函数()22log 43y x x =+-的一个单调增区间是（ ） A ．3,2⎛⎫-∞ ⎪⎝⎭B ．3,2∞⎡⎫+⎪⎢⎣⎭C ．31,2⎛⎫- ⎪⎝⎭D ．3,42⎡⎫⎪⎢⎣⎭【答案】C 【解析】【分析】先求出函数的定义域，再利用复合函数单调性法则“同增异减”即可求解.【详解】函数()22log 43y x x=+-的定义域为()1,4-.要求函数()22log 43y x x =+-的一个单调增区间，只需求243y x x =+-的增区间，只需32x <. 所以312x -<<. 所以函数()22log 43y x x =+-的一个单调增区间是31,2⎛⎫- ⎪⎝⎭. 故选：C例15．（2022·天津·南开中学二模）已知函数()21,14log 1,1a ax x x f x x x ⎧--≤⎪=⎨⎪->⎩是R 上的单调函数，则实数a 的取值范围为（ ） A ．11,42⎡⎫⎪⎢⎣⎭B ．11,42⎡⎤⎢⎥⎣⎦C ．10,2⎛⎤ ⎥⎝⎦D ．1,12⎛⎫ ⎪⎝⎭【答案】B 【解析】 【分析】分函数()f x 在R 上的单调递减和单调递增求解. 【详解】当函数()21,14log 1,1a ax x x f x x x ⎧--≤⎪=⎨⎪->⎩是R 上的单调递减函数，所以01112514a aa ⎧⎪<<⎪⎪≥⎨⎪⎪-≥-⎪⎩，解得1142a ≤≤，因为0a >且1a ≠，所以当1x ≤时，()f x 不可能是增函数，所以函数()f x 在R 上不可能是增函数，综上：实数a 的取值范围为11,42⎡⎤⎢⎥⎣⎦，故选：B例16．（2022·浙江·模拟预测）己知实数,(1,)∈+∞a b ，且33log log 3log log 4b a a b +=+，则（ ） Ab a << B．b a < Ca b < D．a b <<【答案】A 【解析】 【分析】对33log log 4log log 3a b a b -=-利用换底公式等价变形，得333311log log log log -<-b a b a，结合1y x x=-的单调性判断b a <，同理利用换底公式得343411log log log log b a b a ->-，即34log log b a >，再根据对数运算性质得4log log log a =>3log y x =单调性，b >解. 【详解】由33log log 4log log 3a b a b -=-可得333343111log log log log log log b a a b a a-=-<-， 因为1y x x=-在(,0),(0,)-∞+∞上单调递增，且3log a ，3log (0,)b ∈+∞，所以33log log b a <，即b a <， 其次，343411log log log log b a b a->-，所以34log log b a >，又因为4log log log a =>3log y x =单调递增，所以由3log log b >b >b a <． 故选：A例17．（2022·全国·高三专题练习（理））函数f (x )＝log ax (0＜a ＜1)在[a 2，a ]上的最大值是（ ） A ．0 B ．1 C ．2 D ．a【答案】C 【解析】【分析】根据对数函数的单调性可求出结果. 【详解】∵0＜a ＜1，∴f (x )＝log ax 在[a 2，a ]上是减函数， ∴f (x )max ＝f (a 2)＝log aa 2＝2． 故选：C例18．（2022·重庆·模拟预测）若函数()2()log 341a f x x ax =-+-有最小值，则实数a 的取值范围是（ ）A ．⎫⎪⎪⎝⎭B ．C ．⎛ ⎝⎭D ．)+∞【答案】A 【解析】 【分析】根据对数函数的性质可得()()0,11,a ∈+∞且23410x ax -+->，则0∆>，即可求出a 的大致范围，再令23410x ax -+-=的根为1x 、2x 且12x x <，()2341u x x ax =-+-，log a y u =，对a 分两种情况讨论，结合二次函数、对数函数的单调性判断即可； 【详解】解：依题意()()0,11,a ∈+∞且23410x ax -+->，所以216120a ∆=->，解得a >a <()1,a ⎫∈+∞⎪⎪⎝⎭，令23410x ax -+-=的根为1x 、2x 且12x x <，()2341u x x ax =-+-，log a y u =，若()1,a ∈+∞，则log a y u =在定义域上单调递增，()2341u x x ax =-+-在12,3a x ⎛⎫ ⎪⎝⎭上单调递增，在22,3a x ⎛⎫⎪⎝⎭上单调递减，根据复合函数的单调性可知，()2()log 341a f x x ax =-+-在12,3a x ⎛⎫ ⎪⎝⎭上单调递增，在22,3a x ⎛⎫ ⎪⎝⎭上单调递减，函数不存在最小值，故舍去；若a ⎫∈⎪⎪⎝⎭，则log a y u =在定义域上单调递减，()2341u x x ax =-+-在12,3a x ⎛⎫⎪⎝⎭上单调递增，在22,3a x ⎛⎫ ⎪⎝⎭上单调递减，根据复合函数的单调性可知，()2()log 341a f x x ax =-+-在12,3a x ⎛⎫⎪⎝⎭上单调递减，在22,3a x ⎛⎫ ⎪⎝⎭上单调递增，所以函数在23a x =取得最小值，所以a ⎫∈⎪⎪⎝⎭； 故选：A【方法技巧与总结】研究和讨论题中所涉及的函数性质是解决有关函数问题最重要的思路和方法.性质问题是数和形结合的护体解释.它为研究函数问题提供了思维方向.题型四：对数函数中的恒成立问题例19．（2022·北京·高三专题练习）若不等式2log 0a x x -<在10,2⎛⎫ ⎪⎝⎭内恒成立，则a 的取值范围是（ ） A ．1116a ≤< B ．1116a << C ．1016a <≤D ．1016a <<【答案】A 【解析】 【分析】根据对数函数的图象与性质，分1a >和01a <<两种情况分类讨论，结合函数的单调性，列出不等式，即可求解. 【详解】当1a >时，由1(0,)2x ∈，可得log 0a x <，则log 0a x ->，又由20x >，此时不等式2log 0a x x -<不成立，不合题意；当01a <<时，函数log a y x =在1(0,)2上单调递减，此时函数log a y x =-在1(0,)2上单调递增，又由2yx 在1(0,)2上单调递增，要使得不等式2log 0a x x -<在1(0,)2内恒成立，可得211()log 022a -≤，解得1116a ≤<.故选：A.例20．（2022·江苏·高三专题练习）已知函数22414ax x y -+⎛⎫= ⎪⎝⎭的值域为10,16⎛⎤⎥⎝⎦，若不等式()()log 4log 2x a x a t t ⋅<-在[]1,2x ∈上恒成立，则t 的取值范围是（ ） A ．2,25⎛⎫ ⎪⎝⎭B ．2,5⎛⎫+∞ ⎪⎝⎭C ．(,2)-∞D ．()0,2【答案】A 【解析】根据题意，先求得12a =，把不等式()()1122log 4log 2x x t t ⋅<-在[]1,2x ∈上恒成立，转化为402042x xx x t t t t ⎧⋅>⎪->⎨⎪⋅>-⎩在[]1,2x ∈上恒成立，结合指数幂的运算性质，即可求解. 【详解】由题意，函数22414ax x y -+⎛⎫= ⎪⎝⎭的值域为10,16⎛⎤⎥⎝⎦，可得函数y 的最大值为116，当0a =时，函数2414x y -+⎛⎫= ⎪⎝⎭显然不存在最大值；当0a >时，函数22414ax x y -+⎛⎫= ⎪⎝⎭在1,x a ⎛⎫∈-∞ ⎪⎝⎭上单调递增，在1,x a ⎛⎫∈+∞ ⎪⎝⎭上单调递减，当1x a =时，函数y 有最大值，即12411416a a -+⎛⎫= ⎪⎝⎭，解得12a =； 当0a <时，22414ax x y -+⎛⎫= ⎪⎝⎭在1,x a ⎛⎫∈-∞ ⎪⎝⎭上单调递减，在1,x a ⎛⎫∈+∞ ⎪⎝⎭上单调递增，此时函数y 无最大值，所以()()1122log 4log 2x xt t ⋅<-在[]1,2x ∈上恒成立， 即402042x xx x t t t t ⎧⋅>⎪->⎨⎪⋅>-⎩在[]1,2x ∈上恒成立， 由40x t ⋅>在[]1,2x ∈上恒成立，可得0t >；由20x t ->在[]1,2x ∈上恒成立，即2x t <在[]1,2上恒成立，可得2t <； 由42x x t t ⋅>-在[]1,2x ∈上恒成立，即2114122x x x xt >=++在[]1,2上恒成立，令()122xxf x =+，可得函数()f x 在[]1,2上单调递增，所以()()min512f x f ==，即25t >， 综上可得225t <<，即实数t 的取值范围是2,25⎛⎫⎪⎝⎭.故选：A. 例21．（2022·浙江·高三阶段练习）已知函数()29x f x x+=，()2log g x x a =+，若存在[]13,4x ∈，任意[]24,8x ∈，使得()()12f x g x ≥，则实数a 的取值范围是___________. 【答案】13,4∞⎛⎤- ⎥⎝⎦【解析】 【分析】将问题转化为在对应区间上max max ()()f x g x ≥，结合对勾函数、对数函数的性质求()f x 、()g x 的区间最值，即可求a 的范围. 【详解】若()f x 在[3,4]上的最大值max ()f x ，()g x 在[4,8]上的最大值max ()g x ， 由题设，只需max max ()()f x g x ≥即可.在[3,4]上，9()6f x x x =+≥=当且仅当3x =时等号成立， 由对勾函数的性质：()f x 在[3,4]上递增，故max 25()4f x =. 在[4,8]上，()g x 单调递增，则max ()3g x a =+， 所以2534a ≥+，可得134a ≤.故答案为：13,4∞⎛⎤- ⎥⎝⎦.例22．（2022·全国·高三专题练习）已知函数()ln f x x x =-，已知实数0a >，若2()e ln 0x f x a a ++≥在()0+∞，上恒成立，求实数a 的取值范围. 【答案】12ea ≥. 【解析】 【分析】把不等式作等价变形，构造函数()ln g x x x =+，借助其单调性可得2e x a x ≥，分离参数构造函数并求出最大值作答. 【详解】函数()ln f x x x =-定义域为(0,)+∞，则(0,)∀∈+∞x ：222()e ln 0e ln l 2n e ln ln x x x f x a a a a x a a x x x x++≥⇔+≥⇔+≥+++22e e )n ln(l x x a a x x ⇔≥++，令()ln g x x x =+，函数()g x 在(0,)+∞上单调递增，则有原不等式等价于()()2e xg a g x ≥22e e x xx a x a ⇔≥⇔≥， 令2()e x x h x =，0x >，求导得：212()exx h x -'=，当102x <<时，()0h x '>，当12x >时，()0h x '<， 因此，函数()h x 在1(0,)2上单调递增，在1(,)2+∞上单调递减，当12x =时，max 11()()22eh x h ==，则12ea ≥， 所以实数a 的取值范围是12ea ≥. 【点睛】关键点睛：涉及不等式恒成立问题，将给定不等式等价转化，借助同构思想构造函数，利用导数探求函数单调性、最值是解决问题的关键.例23．（2022·全国·高三专题练习）已知函数()log (0,1)xa f x a x a a =+>≠在[1,2]上的最大值与最小值之和为6log 2a +． （1）求实数a 的值；（2）对于任意的[2,)x ∈+∞，不等式()10kf x -≥恒成立，求实数k 的取值范围． 【答案】（1）2；（2）1,5⎡⎫+∞⎪⎢⎣⎭【解析】 【分析】（1）根据指对数函数的单调性得函数()log (0,1)xa f x a x a a =+>≠在[1,2]上是单调函数，进而得260+-=a a ，解方程得2a =；（2）根据题意，将问题转化为对于任意的[2,)x ∈+∞，1()k f x ≥恒成立，进而求函数的最值即可. 【详解】解：（1）因为函数,log (0,1)xa y a y x a a ==>≠在[1,2]上的单调性相同， 所以函数()log (0,1)xa f x a x a a =+>≠在[1,2]上是单调函数，所以函数()f x 在[1,2]上的最大值与最小值之和为2log 26log 2a a a a ++=+，所以260+-=a a ，解得2a =和3a =-（舍） 所以实数a 的值为2.（2）由（1）得2()2log x f x x =+，因为对于任意的[2,)x ∈+∞，不等式()10kf x -≥恒成立，所以对于任意的[2,)x ∈+∞，1()k f x ≥恒成立， 当[2,)x ∈+∞时，2()2log x f x x =+为单调递增函数， 所以()()25f x f ≥=，所以11()5f x ≤，即15k ≥ 所以实数k 的取值范围1,5⎡⎫+∞⎪⎢⎣⎭【点睛】本题考查指对数函数的性质，不等式恒成立求参数范围，考查运算求解能力，回归转化思想，是中档题.本题第二问解题的关键在于根据题意，将问题转化为任意的[2,)x ∈+∞，1()k f x ≥恒成立求解.例24．（2022·陕西安康·高三期末（文））已知函数()()()2log 2log 30,1a a f x x x a a =++>≠. (1)若()32f =，求a 的值；(2)若对任意的[]8,12x ∈，()6f x >恒成立，求a 的取值范围.【答案】(1)13a =；(2)()1,11,82⎛⎫⋃ ⎪⎝⎭. 【解析】 【分析】（1）由()32f =可求得log 3a 的值，进而可求得实数a 的值；（2）由()6f x >可得出log 3a x <-或log 1>a x ，分01a <<、1a >两种情况讨论，可得出关于实数a 的不等式，由此可解得实数a 的取值范围. (1)解：因为()32f =，所以()2log 32log 332a a ++=，所以()2log 310a +=，所以log 31a =-，解得13a =.(2)解：由()6f x >，得()2log 2log 30a a x x +->，即()()log 3log 10a a x x +->，即log 3a x <-或log 1>a x .当01a <<时，log 12log log 8a a a x ≤≤，则log 83a <-或log 121a >，因为log 12log 10a a <=，则log 121a >不成立，由log 83a <-可得318a ⎛⎫< ⎪⎝⎭，得112a <<；当1a >时，log 8log log 12a a a x ≤≤，则log 123a <-或log 81a >，因为log 12log 10a a >=，则log 123a <-不成立，所以log 81a >，解得18a <<. 综上，a 的取值范围是()1,11,82⎛⎫⋃ ⎪⎝⎭.例25．（2022·上海·高三专题练习）已知2()32log f x x =-，2()log g x x =. （1）当[]1,4x ∈时，求函数[]()1()y f x g x =+⋅的值域；。
英语中级听力2听写答案1-10课
听写答案Lesson OnePassage 1: The KnowledgeBecoming a London taxi driver isn‟t easy. In order to obtain a licence to drive a taxi in London, candidates have to pass a detailed examination. They have to learn not only the streets, landmarks and hotels, but also the quickest way to get there. This is called “The Knowledge” by London Cab drivers and it can take years of study and practice to get …The Knowledge‟. Candidates are examined not only on the quickest routes but also on the quickest routes at different times of the day. People who want to pass the examination spend much of their free time driving or even cycling around London, studying maps and learning the huge street directory by heart.Passage 2: The UndergroundTravelling on the London underground (the …tube‟) presents few difficulties for visitors because of the clear colour- coded maps. It is always useful to have plenty of spare change with you because there are often long queues at the larger stations. If you have enough change you can buy your ticket from a machine. Y ou will find signs which list the stations in alphabetical order, with the correct fares, near the machines. There are automatic barriers which are operated by the tickets. Y ou should keep the ticket, because it is checked at the destination.Lesson 2The Foolish FrogOnce upon a time a big, fat frog lived in a tiny shallow pond. He knew every plant and stone in it, and he could swim across it easily. He was the biggest creature in the pond, so he was very important. When he croaked, the water-snails listened politely. And the water-beetles always swam behind him. He was very happy there.One day, while he was catching flies, a pretty dragon-fly passed by. …Y ou‟re a very fine frog,‟she sang, …but why don‟t you live in a bigger pond? Come to my pond. Y ou‟ll find a lot of frogs there. Y ou‟ll meet some fine fish, and you‟ 11 see the dangerous ducks. And you must see our lovely water-lilies. Life in a large pond is wonderful!‟…Perhaps it is rather dull here,‟ thought the foolish frog. S o he hopped after the dragon-fly.But he didn‟t like the big, deep pond. It was full of strange plants. The water-snails were rude to him, and he was afraid of the ducks. The fish didn‟t like him, and he was the smallest frog there. He was lonely and unhappy.He sat on a water-lily leaf and croaked sadly to himself, …I don‟t like it here. I think I‟ll go home tomorrow.‟But a hungry heron flew down and swallowed him up for supper.Lesson 3SleepIt‟s clear that everyone needs to sleep. Most people rarely think about how and why they sleep, however. We know that if we sleep well, we feel rested. If we don‟t sleep enough, we often feel tired and irritable. It seems there are two purposes of sleep: physical rest and emotional and psychological rest: We need to rest our bodies and our minds. Both are important in order for us to be healthy. Each night we alternate between two kinds of sleep: active sleep and passive sleep. The passive sleep gives our body the rest that‟s needed and prepares us for active sleep, in which dreaming occurs.Throughout the night, people alternate between passive and active sleep. The brain rests, then it becomes active, then dreaming occurs. The cycle is repeated: the brain rests, then it becomes active, then dreaming occurs. This cycle is repeated several times throughout the night~. During eight hours of sleep, people dream for a total d one and half hours on the average.Lesson 4Books Belong to the PastSir,I visited my old school yesterday. It hasn‟t changed in thirty years. The pupils were sitting in the same desks and reading the same books. When are schools going to move into the modern world? Books belong to the past. In our homes radio and television bring us knowledge of the world. We can see and hear the truth for ourselves. If we want entertainment most of us prefer a modern film to a classical novel. In the business world computers store information, so that we no longer need encyclopaedias and dictionaries. But in the schools teachers and pupils still use books. There should be a radio and television set in every classroom, and a library of tapes and records in every school. The children of today will rarely open a book when they leave school. The children of tomorrow won‟t need to read and Write at all.M. P. MillerLondonLesson 5The School Holidays Are T oo LongToday the children of this country have at last returned to work. After two months‟ holiday pupils have started a new term. How many adults get such long holidays? Two to four weeks in the summer and public holidays--that‟s all the working man gets. As for the average woman, she‟slucky to get a holiday at all. Children don‟t need such long holidays. In term-time they start work later and finish earlier than anyone else.In the holidays most of them get bored, and some get into trouble. What a waste! If their overworked parents were given more free time instead, everyone would be happier.This isn‟t just a national problem either--it‟s worldwide. Dates may be different from country to country, but the patt ern‟s the same. Why should children do half as much work and get twice as much holiday as their parents?Lesson 6Sign LanguageDeaf people, people who can‟t hear, are still able to communicate quite well with a special language. It‟s called sign langua ge. The speaker of sign language uses hand gestures in order to communicate. Basic sign language has been used for a long, long time, but sign language wasn‟t really developed until about 250 years ago. In the middle of the 1700s a Frenchman named Epée developed sign language. Epée was able to speak and hear, but he worked during most of his life as a teacher of deaf people in France. Epée developed a large number of vocabulary words for sign language. Epée taught these words to his deaf students. Epée‟s sy stem used mostly picture :image signs. We call them picture image signs because the signs create a picture. For example, the sign for sleep is to put both hands together, and then to place the hands flat against the right side of your face, and then to lower your head slightly to the right. This action was meant to show the position of sleep. So we call it a picture image sign.Lesson 7Credit CardsMany businesses, such as department stores, restaurants, hotels and airline companies, use a credit system for selling their products and services. In a credit system, the seller agrees to sell something to the buyer without immediately receiving cash. The buyer receives the goods or services immediately and promises to pay for them later. This “buy- now- pay- later” credit system is quite old. People have been buying things on credit for centuries. But nowadays people use credit cards. There are two types of credit cards. One type is issued directly by a Store to a customer. Many large department stores issue credit cards to their customers. The store credit card can be used to make purchases only at a particular store. The other kind of credit card is issued by a credit company. Credit cards from credit companies can be used to buy things almost anywhere. If you have a major credit card, you can buy airplane tickets, stay at hotels, and eat at restaurants with it. Most large credit companies are connected to large banks. So if you want a credit card from a credit company, you generally have to make an application at a bank. After an applicant receives a credit card, he or she can make purchases, using the card.Lesson 8The Boy Who Cried WolfOnce upon a time there was a very naughty shepherd-boy. He often fell asleep while he was watching his sheep. And he told lies. The villagers shook their heads and said, …That boy will come to a bad end.‟One day, when he was feeling very bored, the boy decided to play a practical joke on the villagers. He ran down the hill. …Wolf,wolf!‟ he cried. …Help, come quickly. Wolf!‟All the villagers seized their spears and ran to help him. But there was no wolf. …He heard you,‟ the naughty boy lied, …and ran away.‟ When everyone had gone, he started to laugh.Three weeks later, when he was feeling very bored indeed, he decided to play the same trick again. …Wolf, wolf!‟ he shouted.…Help, come quickly. Wolf!‟ Most of the villagers hurried to help him. This time the boy laughed at them. …Ha, ha… There wasn‟t a wolf,‟ he said. …What a good joke!‟ The villagers were very angry.…Lies are not jokes,‟ they said.Two days later the boy woke up suddenly. He had fallen asleep in the afternoon sun. What was that big dark animal coming towards his flock? Suddenly it seized a lamb. …Wolf!‟ screamed the boy. …Wolf. Help, come quickly. Wolf!‟ But none of the villagers came to help him. He screamed again. The wolf heard him and licked its lips. …I like lamb,‟ it thought, …but shepherd-boy tastes much nicer.‟When the shepherd-boy didn‟t come home that night, some of the villagers went to look for him. They found a few bones.Lesson 9Passage 1: How to Make WineThis is how wine is made in our winery. After the grapes are picked in late summer, they are pressed so that all the juice runs out. Then the juice is separated from the skins and pips and it is put into large containers and left to ferment. Later, it is put into smaller containers. Then it is left for about a year when it is put into bottles. If it is a good wine, the bottles are kept for several years but the cheaper wines are sold immediately.Passage 2: Alan SimpsonThe mystery of the man found wandering in the city centre has now been solved. The man, whose name is now known to be Alan Simpson, is a medical student. Mr. Simpson was taking part in an experiment conducted by the university department of psychology, when he walked away, unnoticed by the staff supervising the experiment. He has now regained his memory, and has left hospital. Several people, including his sister, April Simpson, telephoned the police to identify Mr. Simpson after seeing his picture in the press.Lesson 10Voice AnalysisIf we want to measure voice features very accurately, we can use a voice analyser. A voiceanalyser can show four characteristics of a speaker‟s voice. No two speakers‟ voices are alike. To get a voice sample, you have to speak into the voice analyser. The voice analyser is connected to a computer. From just a few sentences of normal speech, the computer can show four types of information about your voice. It will show nasalization, loudness, frequency and length of articulation. The first element, nasalization, refers to how much air normally goes through your nose when you talk. The second feature of voice difference is loudness. Loudness is measured in decibels. The number of decibels in speaking is determined by the force of air that comes from the lungs. The third feature of voice variation is frequency. By frequency we mean the highness or lowness of sounds. The frequency of sound waves is measured in cycles per second. Each sound of a language will produce a different frequency. The final point of voice analysis concerns the length of articulation for each sound. This time length is measured in small fractions for each second. From all four of these voice features —length of articulation, frequency, loudness and nasalization — the voice analyser can give an exact picture of a person‟s voice.。
初中数学北师大版七年级下册第一章 整式的乘除1.6完全平方公式-章节测试习题(1)
章节测试题1.【题文】化简求值．（）求的值，其中．（）若，求的值．【答案】(1)22;(2)6【分析】（1）根据平方差公式，单项式乘多项式的运算法则，进行运算，然后和合并同类项后把的值代入进行计算即可得解；根据完全平方公式，单项式乘多项式的运算法则进行运算，然后和合并同类项后,把已知式子的值整体代入即可得解；【解答】解：（），，，∵，∴原式，，．（），，，∵，∴，∴原式．2.【题文】我们知道对于一个图形，通过不同的方法计算图形的面积可以得到一个数学等式．例如：由图1可得到(a+b)²=a²+2ab+b²．图1 图2 图3（1）写出由图2所表示的数学等式：_____________________；写出由图3所表示的数学等式：_____________________；（2）利用上述结论，解决下面问题：已知a+b+c=11，bc+ac+ab=38，求a²+b²+c²的值．【答案】（a+b+c）2=a2+b2+c2+2ab+2ac+2bc （a-b-c）2=a2+b2+c2-2ab-2ac+2bc 45【分析】（1）根据数据表示出矩形的长与宽，再根据矩形的面积公式写出等式的左边，再表示出每一小部分的矩形的面积，然后根据面积相等即可写出等式．（2）根据利用（1）中所得到的结论，将a+b+c=11，bc+ac+ab=38，作为整式代入即可求出．【解答】解：(1)根据题意,大矩形的面积为：小矩形的面积为：(2)由（1）得3.【题文】已知，求：（1）的值；（2）的值；（3）的值.【答案】（1）-30；（2）；（3）【分析】（1）提公因式，然后将a+b=5和ab=-6整体代入求值；（2）将原式利用配方法转化为两根的和与两根的积来解答；（3）将原式利用配方法转化为两根的和与两根的积来解答．【解答】解：（1）∵，∴；（2）；（3），故.4.【题文】利用图形中面积的等量关系可以得到某些数学公式．例如，根据图甲，我们可以得到两数和的平方公式：（a+b）2=a2+2ab+b2．你根据图乙能得到的数学公式是怎样的？写出得到公式的过程．【答案】（a﹣b）2=a2﹣2ab+b2．【分析】根据图形，左上角正方形的面积等于大正方形的面积减去两个矩形的面积，然后加上多减去的右下角的小正方形的面积．【解答】解：∵大正方形的面积= a2还可以表示为5.【题文】先化简，再求值：(1)(9x3y－12xy3＋3xy2)÷(－3xy)－(2y＋x)(2y－x)，其中x＝1，y＝－2；(2)(m－n)(m＋n)＋(m＋n)2－2m2，其中m、n满足方程组【答案】(1) －2x2－y，0;(2) 2mn，-6.【分析】（1）根据多项式除以单项式和平方差公式化简，然后代入求值；（2）根据完全平方公式和平方差公式化简，然后解方程组求出m、n的值后再代入求值.【解答】解：(1)原式＝－3x2＋4y2－y－4y2＋x2＝－2x2－y.当x＝1，y＝－2时，原式＝－2＋2＝0.(2)①＋②，得4m＝12，解得m＝3.将m＝3代入①，得3＋2n＝1，解得n＝－1.故方程组的解是(m－n)(m＋n)＋(m＋n)2－2m2＝m2－n2＋m2＋2mn＋n2－2m2＝2mn，当m＝3，n＝－1时，原式＝2×3×(－1)＝－6.6.【题文】已知a2+b2=1,a-b=,求a2b2与(a+b)4的值.【答案】【分析】把目标代数式化成包含已知代数式的形式.【解答】解：因为a2+b2=1,a-b=,所以(a-b)2=a2+b2-2ab.所以ab=- [(a-b)2-(a2+b2)]=.所以a2b2=(ab)2=.因为(a+b)2=(a-b)2+4ab.=,所以(a+b)4=[(a+b)2]2=.7.【题文】请认真观察图形，解答下列问题：（1）根据图中条件，用两种方法表示两个阴影图形的面积的和（只需表示，不必化简）；并由此得到怎样的等量关系？请用等式表示；（2）如果图中的a，b（a＞b）满足a2+b2=53，ab=14，求：①a+b的值；②a－b 的值．【答案】（1）a2+b2=(a+b)2-2ab;(2)①9;②5.【分析】（1）两个阴影部分的面积可以用阴影部分面积相加和用总面积减去非阴影部分面积来表示。
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1
2006~2007学年第一学期化工原理A-1试卷A标准答案及评分标准
制定教师： 樊丽华 审核人： 梁英华 制定时间： 2007.1.4
一、 填空题（每题2分，共20分）（一题多空的每空的分数平均分配）
1、14.3
2、16，32
3、16
4、湍流
5、出口阀，旁路（支路）
6、加热蒸汽温度，空气侧的对流传热系数
7、冷凝水未排出，不凝气未排出
8、160
9、阻止，低
10、49％
二、选择题（每题2分，共24分）
1 2 3 4 5 6
A D C A A B
7 8 9 10 11 12
C C B A B A
三、计算题（共22分）
解：（1）管路的特性方程：
取贮槽液面为1-1截面，高位槽液面为2-2截面，在两截面之间列柏努利方程，
取贮槽液面为基准水平面， （1分）
feHgpguzHgpguz
2222121
1
22
（3分）
已知：,01zmz102，021uu，01p，kPap1002（表压）
所以，当压力表读数为100kPa时管路的特性方程为：
2232
2.2081.910001010010eeeeBQBQBQgpZH
（1） （3分）
离心泵的特性曲线：24100.740QH （2）
2
联立（1）、（2），并将sLQ/10代入，得：
224
)01.0(2.20)01.0(100.740BH
e
解得：525/1028.1msB （2分）
则当压力表读数为100kPa时管路特性方程为：
25
1028.12.20eeQH
（3） （1分）
（2）当压力表读数为80kPa时，由于流动在阻力平方区，故（3）式中B值不
变，而常数将有所减小，则：
mK15.1881.910001080103
（2分）
于是：251028.115.18eeQH （4） （1分）
（3）输水体积流量：
联立泵的特性方程及式（4），可求得输水体积流量，即：
2524
1028.115.18100.740eeQQ
（3分）
解得：smQe/1005.132 （1分）
（4）离心泵的有效功率：
mQHee28.32100.74024
（2分）
gHQNe
（2分）
W332581.910000105.028.32
（1分）
四、计算题（共13分）
解：（1）框数和框厚
根据滤液体积用恒压过滤方程计算过滤面积，进而确定框数。由滤饼的体积
计算框厚。
3
1004.0/4.0/mvVVe
（1分）
由于忽略介质阻力，恒压过滤方程简化为：
22
KAV
（2分）
即：25236001073.210A
解得：29.31mA （1分）
3
又nA281.081.0 （1分）
则：n＝24.31，取n＝25个 （1分）
所需滤框厚度为：mb0244.0)81.025/(4.02，实取框厚25mm （2分）
（2）操作压强
实际过滤面积为：2281.3225281.0mA （1分）
最佳工况时的过滤时间为45min，则：
'22
604581.3210K
/sm1044.325'K
（2分）
kPa5.1852.1471073.21044.355'p
（2分）
五、计算题（共21分）
解：（1）校核该换热器是否合用：
机油放出的热量为：
kJ/h280000)100200(0.21400)(21TTcWQ
phh
（2分）
冷却水的出口温度为：
6.802.412002800002512pcccWQtt
℃ （2分）
平均温度差为：
5.95251006.80200ln251006.80200ln12211221tTtTtTtTtm℃ （2分）
总传热系数为：
6.175201800252001111000iiddK W/(m2.℃) （3分）
所需的传热面积为：
22
3
00464.45.956.175360010280000mmtKQSm
＝
需
（3分）
即换热器不适用。
（2）采取的措施：
4
（在Q及K0一定的前提下，若增大冷却水用量，先从热量衡算方程看：
t
必减小，而t1不变，则冷却水出口温度t2下降；再从传热速率方程看：mt必增
加，使S0下降，从而使原换热器面积够用。）所以采取的措施为增大冷却水用量。
（2分）
因Q一定，K0不变，204mS＝，则用原换热器面积操作时的平均温度差为：
7.11046.175360010280000300SKQt
m
℃ （2分）
假定平均温度差可用算术均值计算，则：
7.110225100200221221ttTtTtm℃
则冷却水出口温度为：6.532t℃ （2分）
校核：2952.1251006.5320012tt （1分）
冷却水用量为：
kg/h2331256.532.428000012ttcQWpcc （2分）
（注：对于计算题，由于前一步的结果错误而影响后面各步的计算结果的，后面
步骤解题思路正确者，不扣分。）