(完整)2018年高考真题汇编-函数,推荐文档

2018年高考真题汇编(函数与导数)

函数与导数1 .【2018年浙江卷】函数【解析】分析:先研究函数的奇偶性』再研究雷数在G"）上的符号，即可判断选择详解；令= 2圍血滋，因为^ e =刃*血2（—x） = —2罔血Zx = —fG（）p所以fOO = 2團血2耳力奇画数’排除选项止出因为工匸$町时『f@） < 0,所以曲穩选项J选D点睛：有关函数图象的识别问题的常见题型及解题思路：（1）由函数的定义域，判断图象的左、右位置，由函数的值域，判断图象的上、下位置；（2）由函数的单调性，判断图象的变化趋势；（3）由函数的奇偶性，判断图象的对称性；（4）由函数的周期性，判断图象的循环往复.c = b眉2. 【2018年理天津卷】已知il=lo^^in2, 2 ,则a, b, c的大小关系为A. u > b>cB.b>u> e C c> b> a D.c> a> b【答案】D【解析】分析：由题意结合对数函数的性质整理计算即可求得最终结果b = ln2 = -^―e （0A）c= 3詰=和g* > Sg声详解：由题意结合对数函数的性质可知: "忆吆>1, 5慾, 2据此可得：•本题选择D选项.点睛：对于指数幕的大小的比较，我们通常都是运用指数函数的单调性，但很多时候，因幕的底数或指数不相同，不能直接利用函数的单调性进行比较•这就必须掌握一些特殊方法•在进行指数幕的大小比较时，若底数不同，则首先考虑将其转化成同底数，然后再根据指数函数的单调性进行判断•对于不同底而同指数的指数幕的大小的比较，利用图象法求解，既快捷，又准确.龙兰0*3. 【2018年理新课标I卷】已知函数I曲乩北〉心饥巧二“/) + +a .若g (x)存在2个零点，则a的取值范围是A. [ - 1, 0)B. [0 , +R)C. [ - 1 , +R)D. [1 , +R)【答案】C【解析】分析；首先根据存在2个零点，得到方程f CO十""哨两个亀将其转化为金〉二-覽-口有两个解，即直线y =-第-诣曲^二fCO有两个交点”根据題中所给的函数解析式，画出函数f何的團像(将町4掉A再画出直绳=-补并将其上下移动』从图中可臥发现走丄时/龊7=-電-口与曲线y=f^＞有两个玄点'从而求得结果.详解：画出函数的图像，7■-了在y轴右侧的去掉，再画出直线卜:讨，之后上下移动，可以发现当直线过点A时，直线与函数图像有两个交点，并且向下可以无限移动，都可以保证直线与函数的图像有两个交点，即方程■有两个解，也就是函数有两个零点，此时满足，即• ，故选C.点睛：该题考查的是有关已知函数零点个数求有关参数的取值范围问题，在求解的过程中，解题的思路是将函数零点个数问题转化为方程解的个数问题，将式子移项变形，转化为两条曲线交点的问题，画出函数的图像以及相应的直线，在直线移动的过程中，利用数形结合思想，求得相应的结果4. 【2018年理新课标I卷】设函数兀心--，若$叩为奇函数，则曲线：在点’ 处的切线方程为A.卜「阙B. H" - '■ - -IC."划D.【答案】D【解析】分析；利用奇函数偶此项系数为零求得"X进而得到的解析式，再对“）求导得出桩戋的斜率©进而求得切线方程.详解；因豹画数雇苛函数J 解得"二4所以』⑴二卯1,门>）二阪y 所臥厂◎二九代町二g所汰曲线y二厲刃在点（啦处的切线方程为y-m））二比建简可得y二知故选D点睛：该题考查的是有关曲线卜在某个点凤煮強；；|处的切线方程的问题，在求解的过程中，首先需要确定函数解析式，此时利用到结论多项式函数中，奇函数不存在偶次项，偶函数不存在奇次项，从而求得相应的参数值，之后利用求导公式求得帀，借助于导数的几何意义，结合直线方程的点斜式求得结果•5. 【2018年全国卷川理】设“=』0目仇2°収，方=衍的帖，贝UA. N + bunbcOB.C. u + bcOca/iD. kb<OCQ +市【答案】B1 i I 11【解析】分析：求出-= io^^ 2t-=lo^.32,得到- +二的范围，进而可得结果。

(完整word版)2018全国高考英语完型填空真题汇编(含答案)(2),推荐文档

During my second year at the city college, I was told that the education department was offering a “free” course, called Thinking Chess, for three credits. I41 the idea of taking the class because, after all, who doesn’t want to42 a few dollars? More than that, I’d always wanted to learn chess. And, even if I weren’t43 enough about free credits, news about our 44 was appealing enough to me. He was an international grand master, which 45 I would be learning from one of the game’s46 I could hardly wait to 47 him.Maurice Ashley was kind and smart, a former graduate returning to teach, and this 48 was no game for him; he meant business. In his introduction, he made it 49 that our credits would be hard-earned. In order to 50 the class among other criteria, we had to write a paper on how we plan to 51 what we would learn in class to our future professions and 52 .to our lives.I managed to get an A in that 53 and leaned life lessons that have served me well beyond the54 .Ten years after my chess class with Ashley, I' m still putting to use what he 55 me：“the absolute most important 56 that you learn when you play chess is how to make good 57 . On every single move you have to 58 a situation, process what your opponent (对手) is doing and 59 the best move from among all your options.” These words s till ring true today in my 60 as a journalist.41. A put forward B jumped at C tried out D turned down42. A waste B earn C save D pay43. A excited B worried C moved D tired44. A title B competitor C textbook D instructor45. A urged B demanded C held D meant46. A fastest B easiest C best D rarest47. A interview B meet C challenge D beat48. A chance B qualification C honor D job49. A real B perfect C clear D possible50. A attend B pass C skip D observe51. A add B expose C apply D compare52. A eventually B naturally C directly D normally53. A game B presentation C course D experiment54. A criterion B classroom C department D situation55. A taught B wrote C questioned D promised56. A fact B step C manner D skill57. A grades B decisions C impressions D comments58. A analyze B describe C rebuild D control59. A announce B signal C block D evaluate60. A role B desire C concern D behaviorTwo weeks earlier, my son, Ben, had got in touch. He’d moved to England with his mum when he was three and it had been 13 years since I’d __41__ seen him. So imagine my __42__ when he emailed me saying he wanted to come to visit me.I was __43__! I arrived early at Byron Bay where we were supposed to __44__. The bay was __45__ in sunshine, and there was a group of kayakers around 150m off the shore. Getting a little __46__, I realized one kayak（皮划艇）was in __47__. ＂Something’s not __48__!＂I took off my T-shirt and __49__ into the water. I saw there were two instructors on board and a man lying across the middle. He was __50__ violently. Linking arms with one of the instructors, I helped __51__ the young man out of the water. He was unconscious and as I looked at his face, something __52_ to me. Those brown eyes were very __53__. ＂What’s his name?＂I asked the instructor. ＂Ben,＂he replied, and immediately I __54__. That stranger was my son!The instructors called for an ambulance. __55__, after a brief stay in hospital, Ben was well enough to be allowed to __56__ and later the family met up for dinner. We chatted about everything and then Ben __57__ to me. ＂I just want to say thank you,＂he said. ＂You __58__ my life!＂I still can’t believe what a __59__ it was. I’m just so glad I was there __60__ to help my son.41. A. also B. often C. even D. last42. A. delight B. relief C. anger D. worry43. A. scared B. shocked C. thrilled D. ashamed44. A. talk B. stay C. meet D. settle45. A. bathed B. clean C. deep D. formed46. A. faster B. closer C. heavier D. wiser47. A. trouble B. advance C. question D. battle48. A. real B. right C. fair D. fit49. A. stared B. sank C. dived D. fell50. A. arguing B. fighting C. shouting D. shaking51. A. lead B. persuade C. carry D. keep52. A. happened B. occurred C. applied D. appealed53. A. sharp B. pleasant C. attractive D. familiar54. A. agreed B. hesitated C. doubted D. knew55. A. Fortunately B. Frankly C. Sadly D. Suddenly56. A. return B. relax C. speak D. leave57. A. joked B. turned C. listened D. pointed58. A. created B. honored C. saved D. guided59. A. coincidence B. change C. pity D. pain60. A. on board B. in time C. for sure D. on purposeWhen most of us get a text message on our cell phone from an unknown person, we usually say ＂sorry, __41__ number!＂and move on. But when Dennis Williams __42__ a text that clearly wasn’t intended for him, he did something __43__.On March 19, Dennis got a group text __44__ him that a couple he didn’t know were at the hospital, waiting for the __45__ of a baby.＂Congratulations! But I think someone was mistaken,＂Dennis __46__. The baby was born and update texts were __47__ quickly from the overjoyed grandmother, Teresa. In her __48__, she didn’t seem to realize that she was __49__ the baby’s photos with a complete stranger. ＂Well, I don’t __50__ you all but I will get there to take pictures with the baby,＂replied Dennis before asking which room the new __51__ were in.Much to the family’s surprise, Dennis stuck to his __52__! He turned up at the hospital __53__ gifts for the new mother Lindsey and her baby boy. Lindsey’s husband was totally __54__ by the unexpected visit. ＂I don’t think we would have randomly invited him over but we __55__ it and the gifts.＂Teresa __56__ a photo of the chance meeting on a social networking website __57__ by the touching words: ＂What a __58__ this young man was to our family! He was so __59__ and kind to do this.＂The post has since gained the __60__ of social media users all over the world, receiving more than 184,000 shares and 61,500 likes in just three days.41.A. unlucky B. secret C. new D. wrong42.A. received B. translated C. copied D. printed43.A. reasonable B. special C. necessary D. practical44.A. convincing B. reminding C. informing D. warning45.A. wake-up B. recovery C. growth D. arrival46.A. responded B. interrupted C. predicted D. repeated47.A. coming in B. setting out C. passing down D. moving around48.A. opinion B. anxiety C. excitement D. effort49.A. comparing B. exchanging C. discussing D. sharing50.A. accept B. know C. believe D. bother51.A. parents B. doctors C. patients D. visitors52.A. dream B. promise C. agenda D. principle53.A. bearing B. collecting C. opening D. making54.A. discouraged B. relaxed C. astonished D. defeated55.A. admit B. need C. appreciate D. expect56.A. found B. selected C. developed D. posted57.A. confirmed B. simplified C. clarified D. accompanied58 .A. pity B. blessing C. relief D. problem59.A. smart B. calm C. sweet D. fair60.A. sympathy B. attention C. control D. trustThe Homeless HeroFor many, finding an unattended wallet filled with £400 in cash would be a source（来源）of temptation（诱惑）. But the __16__ would no doubt be greater if you were living on the streets with little food and money. All of this makes the actions of the homeless Tom Smith __17__more remarkable.After spotting a __18__ on the front seat inside a parked car with its window down, he stood guard in the rain for about two hours waiting for the __19__ to return.After hours in the cold and wet, he __20___ inside and pulled the wallet out hoping to find some ID so he could contact（联系）the driver, only to __21__ it contained £400 in notes, with another £50 in spare change beside it.He then took the wallet to a nearby police station after __22__ a note behind to let the owner know it was safe. When the car’s owner John Anderson and his colleague Carol Lawrence returned to the car—which was itself worth £35, 000—in Glasgow city centre, they were __23__to find two policemen standing next to it. The policemen told them what Mr. Smith did and that the wallet was __24__.The pair were later able to thank Mr. Smith for his __25__.Mr. Anderson said: ＂I couldn’t believe that the guy never took a penny. To think he is sleeping on the streets tonight __26__ he could have stolen the money and paid for a place to stay in. This guy has nothing and __27__ he didn’t take the wallet for himself；he thought about others __28__. It’s unbelievable. It just proves there are __29__ guys out there.＂Mr. Smith’s act __30__ much of the public’s attention. He also won praise from social media users after Mr. Anderson __31__ about the act of kindness on Facebook.Now Mr. Anderson has set up an online campaign to __32__ money for Mr. Smith and other homeless people in the area, which by yesterday had received £8,000. ＂I think the faith that everyone has shown __33__ him has touched him. People have been approaching him in the street; he’s had job __34__ and all sorts,＂Mr. Anderson commented.For Mr. Smith, this is a possible life-changing __35__. The story once again tells us that one good turn deserves another.16. A. hope B. aim C. urge D. effort17. A. still B. even C. ever D. once18. A. wallet B. bag C. box D. parcel19. A. partner B. colleague C. owner D. policeman20. A. turned B. hid C. stepped D. reached21. A. discover B. collect C. check D. believe22. A. taking B. leaving C. reading D. writing23. A. satisfied B. excited C. amused D. shocked24. A. safe B. missing C. found D. seen25. A. service B. support C. kindness D. encouragement26. A. when B. if C. where D. because27. A. rather B. yet C. already D. just28. A. too B. though C. again D. instead29. A. honest B. polite C. rich D. generous30. A. gave B. paid C. cast D. drew31. A. learned B. posted C. cared D. heard32. A. borrow B. raise C. save D. earn33. A. of B. at C. for D. in34. A. details B. changes C. offers D. applications35. A. lesson B. adventure C. chance D. challengeRaynor Winn and her husband Moth became homeless due to their wrong investment. Their savings had been36 to pay lawyers’ fees. To make matters worse, Moth was diagnosed（诊断）with a37 disease. There was no 38 , only pain relief.Failing to find any other way out, they decided to make a 39 journey, as they caught sight of an old hikers’(徒步旅行者）guide.This was a long journey of unaccustomed hardship and 40 recovery. When leaving home, Raynor andMoth had just ￡320 in the bank. They planned to keep the 41 low by living on boiled noodles, with the 42 hamburger shop treat.Wild camping is 43 in England. To avoid being caught, the Winns had to get their tent up 44 and packed it away early in the morning. The Winns soon discovered that daily hiking in their 50s is a lot 45 than they remember it was in their 20s. Raynor 46 all over and desired a bath. Moth, meanwhile, after an initial 47 , found his symptoms were strangely 48 by their daily tiring journey.49 , the couple found that their bodies turned for the better, with re-found strong muscles that they thought had 50 forever. "Our hair was fried and falling out, nails broken, clothes 51 to a thread, but we were alive."During the journey, Raynor began a career as a nature writer. She writes, " 52 had taken every material thing from me and left me torn bare, an empty page at the end of a(n) 53 written book. It had also given me a 54 , either to leave that page 55 or to keep writing the story with hope. I chose hope.”36. A. drawn up B. used up C. backed up D. kept up37. A. mild B. common C. preventable D. serious38. A. cure B. luck C. care D. promise39. A. business B. walking C. bus D. rail40. A. expected B. frightening C. disappointing D. surprising41. A. budget B. revenue C. compensation D. allowance42. A. frequent B. occasional C. abundant D. constant43. A. unpopular B. lawful C. attractive D. illegal44. A. soon B. early C. late D. slowly45. A. harder B. easier C. cheaper D. funnier46. A. rolled B. bled C. ached D. trembled47. A. struggle B. progress C. excitement D. research48. A. developed B. controlled C. reduced D. increased49. A. Initially B. Eventually C. Temporarily D. Consequently50. A. gained B. kept C. wounded D. lost51. A. sewn B. washed C. worn D. ironed52. A. Doctors B. Hiking C. Lawyers D. Homelessness53. A. well B. partly C. neatly D. originally54. A. choice B. reward C. promise D. break55. A. loose B. full C. blank D. missingNo one is born a winner. People make themselves into winners by their own __16__.I learned this lesson from a(n) __17__ many years ago. I took the head __18__ job at a school in Baxley, Georgia. It was a small school with a weak football program.It was a tradition for the school’s old team to play agains t the __19__ team at the end of spring practice. The old team had no coach, and they didn’t even practice to __20__ the game. Being the coach of the new team, I was excited because I knew we were going to win, but to my disappointment we were defeated. I c ouldn’t __21__ I had got into such a situation. Thinking hard about it, I came to __22__ that my team might not be the number one team in Georgia, but they were __23__ me. I had to change my __24__about their ability and potential.I started doing anything I could to help them build a little __25__. Most important, I began to treat them like __26__. That summer, When the other teams enjoyed their __27___, we met every day and __28__passing and kicking the football.Six months after suffering our __29__on the spring practice field, we won our first game and our second, and continued to __30__. Finally, we faced the number one team in the state. I felt that it would be a __31__for us even if we lost the game. But that wasn’t what happened. My boys beat the best team in Georgia, giving me one of the greatest __32__of my life!From the experience I learnt a lot about how the attitude of the leader can __33__ the members of a team. Instead of seeing my boys as losers, I pushed and__34__them. I helped them to see themselves __35__, and they built themselves into winners.Winners are made, but born.16. A. luck B. tests C. efforts D. nature17. A. experiment B. experience C. visit D. show18. A. operating B. editing C. consulting D. coaching19. A. successful B. excellent C. strong D. new20. A. cheer for B. prepare for C. help with D. finish with21. A. believe B. agree C. describe D. regret22. A. realize B. claim C. permit D. demand23. A. reacting to B. looking for C. depending on D. caring about24. A. decision B. attitude C. conclusion D. intention25. A. pride B. culture C. fortune D. relationship26. A. leaders B. partners C. winners D. learners27. A. rewards B. vacations C. health D. honor28. A. risked B. missed C. considered D. practiced29. A. defeat B. decline C. accident D. mistake30. A. relax B. improve C. expand D. defend31. A. shame B. burden C. victory D. favor32. A. chances B. thrills C. concerns D. offers33. A. surprise B. serve C. interest D. affect34. A. encouraged B. observed C. protected D. impressed35. A. honestly B. individually C. calmly D. differentlyA young English teacher saved the lives of 30 students when he took 36 of a bus after its driver suffered a serious heart attack. Guy Harvold, 24, had 37 the students and three course leaders from Gatwick airport, and they were travelling to Bourmemouth to 38 their host families. They were going to 39 a course at the ABC Language School in Bournemouth where Harvold works as a 40 .Harvold, who has not 41 his driving test, said, “I realized the bus was out of control when I was 42 the students.”The bus ran into trees at the side of the road and he 43 the driver was slumped (倒伏) over the wheel. The driver didn’t 44 . He was unconscious. The bus45 a lamp post and it broke the glass on the front door before Harvold 46 to bring the bus toa stop. Police 47 the young teacher’s quick thinking. If he hadn’t48 quickly, there could have been a terrible 49 .The bus driver never regained consciousness and died at Easy Surrey Hospital. He had worked regularly with the 50 and was very well regarded by the teachers and students. Harvold said, “I was51 that no one else was hurt, but I hoped that the driver would 52 .The head of the language school told the local newspaper that the school is going to send Harvold on a weekend 53 to Dublin with a friend, thanking him for his 54 . A local driving school has also offered him six 55 driving lessons.36. A. control B. care C. advantage D. note37. A. taken in B. picked up C. tracked down D. helped out38. A. greet B. thank C. invite D. meet39. A. present B. introduce C. take D. organize40. A. drive B. doctor C. librarian D. teacher41. A. given B. marked C. passed D. conducted42. A. speaking to B. waiting for C. returning to D. looking for43. A. learned B. noticed C. mentioned D. doubted44. A. sleep B. cry C. move D. recover45. A. ran over B. went by C. carried D. hit46. A. remembered B. continued C. prepared D. managed47. A. witnessed B. recorded C. praised D. understood48. A. appeared B. reacted C. escaped D. interrupted49. A. delay B. accident C. mistake D. experience50. A. airport B. hospital C. school D. police51. A. happy B. fortunate C. touched D. sorry52. A. survive B. retire C. relax D. succeed53. A. project B. trip C. dinner D. duty54. A. bravery B. skill C. quality D. knowledge55. A. necessary B. easy C. different D. freeWe have all heard how time is more valuable than money, but is it __36__ to have too much?I__37__ back in high school I spent most of my day at school since I also __38__a team sport. By the time I got home, I only had a few hours to do my homework, and I had to do it __39__.When I got into college, things __40__. I suddenly found myself out of class before noon time. Because of all this __41__ there was no sense of __42__ to do my school work immediately.I was performing this action of waiting until it later became__43__.Once that happened, I just kept__44__my studying further and further back in my day. Then I got to the point where I was__45__really late at night to get my work alone.One day I __46__a former classmate of mine who was __47__ a lot of money running a sideline(副业).Since his regular job was __48__,I asked him why he just didn't do his sideline full-time. He said without the job, he would __49__ have too much time and would just do what I did back in__50__.He said that if he __51__the job, he would lose his __52__ to work and succeed.So, try __53__your tine with other work. This is why there is a __54__ that if you want something done, ask a __55__person to do it.36.A. true B. fair C. strange D. possible37.A. remember B. admit C. understand D. expect38.A. watched B. loved. C. Coached D. played39.A. al last B. right away C. of course D. mattered40.A. happened B. repeated C. changed D. mattered41.A. extra B. difficult C. valuable D. limited42.A. duty B. achievement C. urgency D. direction43.A. burden B. relief C. risk D. habit44.A. pushing B. taking C. setting D. calling45.A. hanging out B. staying up C. jogging round D. showing off46.A. met B. helped C. treated D. hired47.A. raising B. wasting C. demanding D. making48.A. safe B. important C. boring D. rewarding49.A. luckily B. hardly C. hopefully D. simply50.A. childhood B. college C. town D. business51.A. quit B. found C. accepted D. kept52.A. heart B. chance C. drive D. way53.A. saving B. filling up C. giving up D. trading54.A. message B. story C. saying D. fact55.A. careful B. busy C. reliable D. kind2018新课标I卷BCADD CBDCB CACBA DBADA 2018新课标II卷DACCA BABCD CBDDA DBCAB 2018新课标III卷 DABDC AACDB ABACC DDBCB 2018北京卷CBACD ABDAC ABDAD BBDCC 2018江苏卷BDABD ABDCA CACBD CDBAC 2018天津卷CBDDB AACBA CBDAB CBDAD 2017.11浙江卷ABDCD CABCD DCBBC AABAD 2018浙江卷DADBC ACDAB ADCDB ACBCB。

(完整word版)2018全国高考英语完型填空真题汇编(含答案),推荐文档

During my second year at the city college, I was told that the education department was offering a “free” course, called Thinking Chess, for three credits. I41 the idea of taking the class because, after all, who doesn’t want to42 a few dollars? More than that, I’d always wanted to learn chess. And, even if I weren’t43 enough about free credits, news about our 44 was appealing enough to me. He was an international grand master, which 45 I would be learning from one of the game’s46 I could hardly wait to 47 him.Maurice Ashley was kind and smart, a former graduate returning to teach, and this 48 was no game for him; he meant business. In his introduction, he made it 49 that our credits would be hard-earned. In order to 50 the class among other criteria, we had to write a paper on how we plan to 51 what we would learn in class to our future professions and 52 .to our lives.I managed to get an A in that 53 and leaned life lessons that have served me well beyond the54 .Ten years after my chess class with Ashley, I' m still putting to use what he 55 me：“the absolute most important 56 that you learn when you play chess is how to make good 57 . On every single move you have to 58 a situation, process what your opponent (对手) is doing and 59 the best move from among all your options.” These words s till ring true today in my 60 as a journalist.41. A put forward B jumped at C tried out D turned down42. A waste B earn C save D pay43. A excited B worried C moved D tired44. A title B competitor C textbook D instructor45. A urged B demanded C held D meant46. A fastest B easiest C best D rarest47. A interview B meet C challenge D beat48. A chance B qualification C honor D job49. A real B perfect C clear D possible50. A attend B pass C skip D observe51. A add B expose C apply D compare52. A eventually B naturally C directly D normally53. A game B presentation C course D experiment54. A criterion B classroom C department D situation55. A taught B wrote C questioned D promised56. A fact B step C manner D skill57. A grades B decisions C impressions D comments58. A analyze B describe C rebuild D control59. A announce B signal C block D evaluate60. A role B desire C concern D behaviorTwo weeks earlier, my son, Ben, had got in touch. He’d moved to England with his mum when he was three and it had been 13 years since I’d __41__ seen him. So imagine my __42__ when he emailed me saying he wanted to come to visit me.I was __43__! I arrived early at Byron Bay where we were supposed to __44__. The bay was __45__ in sunshine, and there was a group of kayakers around 150m off the shore. Getting a little __46__, I realized one kayak（皮划艇）was in __47__. ＂Something’s not __48__!＂I took off my T-shirt and __49__ into the water. I saw there were two instructors on board and a man lying across the middle. He was __50__ violently. Linking arms with one of the instructors, I helped __51__ the young man out of the water. He was unconscious and as I looked at his face, something __52_ to me. Those brown eyes were very __53__. ＂What’s his name?＂I asked the instructor. ＂Ben,＂he replied, and immediately I __54__. That stranger was my son!The instructors called for an ambulance. __55__, after a brief stay in hospital, Ben was well enough to be allowed to __56__ and later the family met up for dinner. We chatted about everything and then Ben __57__ to me. ＂I just want to say thank you,＂he said. ＂You __58__ my life!＂I still can’t believe what a __59__ it was. I’m just so glad I was there __60__ to help my son.41. A. also B. often C. even D. last42. A. delight B. relief C. anger D. worry43. A. scared B. shocked C. thrilled D. ashamed44. A. talk B. stay C. meet D. settle45. A. bathed B. clean C. deep D. formed46. A. faster B. closer C. heavier D. wiser47. A. trouble B. advance C. question D. battle48. A. real B. right C. fair D. fit49. A. stared B. sank C. dived D. fell50. A. arguing B. fighting C. shouting D. shaking51. A. lead B. persuade C. carry D. keep52. A. happened B. occurred C. applied D. appealed53. A. sharp B. pleasant C. attractive D. familiar54. A. agreed B. hesitated C. doubted D. knew55. A. Fortunately B. Frankly C. Sadly D. Suddenly56. A. return B. relax C. speak D. leave57. A. joked B. turned C. listened D. pointed58. A. created B. honored C. saved D. guided59. A. coincidence B. change C. pity D. pain60. A. on board B. in time C. for sure D. on purposeWhen most of us get a text message on our cell phone from an unknown person, we usually say ＂sorry, __41__ number!＂and move on. But when Dennis Williams __42__ a text that clearly wasn’t intended for him, he did something __43__.On March 19, Dennis got a group text __44__ him that a couple he didn’t know were at the hospital, waiting for the __45__ of a baby.＂Congratulations! But I think someone was mistaken,＂Dennis __46__. The baby was born and update texts were __47__ quickly from the overjoyed grandmother, Teresa. In her __48__, she didn’t seem to realize that she was __49__ the baby’s photos with a complete stranger. ＂Well, I don’t __50__ you all but I will get there to take pictures with the baby,＂replied Dennis before asking which room the new __51__ were in.Much to the family’s surprise, Dennis stuck to his __52__! He turned up at the hospital __53__ gifts for the new mother Lindsey and her baby boy. Lindsey’s husband was totally __54__ by the unexpected visit. ＂I don’t think we would have randomly invited him over but we __55__ it and the gifts.＂Teresa __56__ a photo of the chance meeting on a social networking website __57__ by the touching words: ＂What a __58__ this young man was to our family! He was so __59__ and kind to do this.＂The post has since gained the __60__ of social media users all over the world, receiving more than 184,000 shares and 61,500 likes in just three days.41.A. unlucky B. secret C. new D. wrong42.A. received B. translated C. copied D. printed43.A. reasonable B. special C. necessary D. practical44.A. convincing B. reminding C. informing D. warning45.A. wake-up B. recovery C. growth D. arrival46.A. responded B. interrupted C. predicted D. repeated47.A. coming in B. setting out C. passing down D. moving around48.A. opinion B. anxiety C. excitement D. effort49.A. comparing B. exchanging C. discussing D. sharing50.A. accept B. know C. believe D. bother51.A. parents B. doctors C. patients D. visitors52.A. dream B. promise C. agenda D. principle53.A. bearing B. collecting C. opening D. making54.A. discouraged B. relaxed C. astonished D. defeated55.A. admit B. need C. appreciate D. expect56.A. found B. selected C. developed D. posted57.A. confirmed B. simplified C. clarified D. accompanied58 .A. pity B. blessing C. relief D. problem59.A. smart B. calm C. sweet D. fair60.A. sympathy B. attention C. control D. trustThe Homeless HeroFor many, finding an unattended wallet filled with £400 in cash would be a source（来源）of temptation（诱惑）. But the __16__ would no doubt be greater if you were living on the streets with little food and money. All of this makes the actions of the homeless Tom Smith __17__more remarkable.After spotting a __18__ on the front seat inside a parked car with its window down, he stood guard in the rain for about two hours waiting for the __19__ to return.After hours in the cold and wet, he __20___ inside and pulled the wallet out hoping to find some ID so he could contact（联系）the driver, only to __21__ it contained £400 in notes, with another £50 in spare change beside it.He then took the wallet to a nearby police station after __22__ a note behind to let the owner know it was safe. When the car’s owner John Anderson and his colleague Carol Lawrence returned to the car—which was itself worth £35, 000—in Glasgow city centre, they were __23__to find two policemen standing next to it. The policemen told them what Mr. Smith did and that the wallet was __24__.The pair were later able to thank Mr. Smith for his __25__.Mr. Anderson said: ＂I couldn’t believe that the guy never took a penny. To think he is sleeping on the streets tonight __26__ he could have stolen the money and paid for a place to stay in. This guy has nothing and __27__ he didn’t take the wallet for himself；he thought about others __28__. It’s unbelievable. It just proves there are __29__ guys out there.＂Mr. Smith’s act __30__ much of the public’s attention. He also won praise from social media users after Mr. Anderson __31__ about the act of kindness on Facebook.Now Mr. Anderson has set up an online campaign to __32__ money for Mr. Smith and other homeless people in the area, which by yesterday had received £8,000. ＂I think the faith that everyone has shown __33__ him has touched him. People have been approaching him in the street; he’s had job __34__ and all sorts,＂Mr. Anderson commented.For Mr. Smith, this is a possible life-changing __35__. The story once again tells us that one good turn deserves another.16. A. hope B. aim C. urge D. effort17. A. still B. even C. ever D. once18. A. wallet B. bag C. box D. parcel19. A. partner B. colleague C. owner D. policeman20. A. turned B. hid C. stepped D. reached21. A. discover B. collect C. check D. believe22. A. taking B. leaving C. reading D. writing23. A. satisfied B. excited C. amused D. shocked24. A. safe B. missing C. found D. seen25. A. service B. support C. kindness D. encouragement26. A. when B. if C. where D. because27. A. rather B. yet C. already D. just28. A. too B. though C. again D. instead29. A. honest B. polite C. rich D. generous30. A. gave B. paid C. cast D. drew31. A. learned B. posted C. cared D. heard32. A. borrow B. raise C. save D. earn33. A. of B. at C. for D. in34. A. details B. changes C. offers D. applications35. A. lesson B. adventure C. chance D. challengeRaynor Winn and her husband Moth became homeless due to their wrong investment. Their savings had been36 to pay lawyers’ fees. To make matters worse, Moth was diagnosed（诊断）with a37 disease. There was no 38 , only pain relief.Failing to find any other way out, they decided to make a 39 journey, as they caught sight of an old hikers’(徒步旅行者）guide.This was a long journey of unaccustomed hardship and 40 recovery. When leaving home, Raynor andMoth had just ￡320 in the bank. They planned to keep the 41 low by living on boiled noodles, with the 42 hamburger shop treat.Wild camping is 43 in England. To avoid being caught, the Winns had to get their tent up 44 and packed it away early in the morning. The Winns soon discovered that daily hiking in their 50s is a lot 45 than they remember it was in their 20s. Raynor 46 all over and desired a bath. Moth, meanwhile, after an initial 47 , found his symptoms were strangely 48 by their daily tiring journey.49 , the couple found that their bodies turned for the better, with re-found strong muscles that they thought had 50 forever. "Our hair was fried and falling out, nails broken, clothes 51 to a thread, but we were alive."During the journey, Raynor began a career as a nature writer. She writes, " 52 had taken every material thing from me and left me torn bare, an empty page at the end of a(n) 53 written book. It had also given me a 54 , either to leave that page 55 or to keep writing the story with hope. I chose hope.”36. A. drawn up B. used up C. backed up D. kept up37. A. mild B. common C. preventable D. serious38. A. cure B. luck C. care D. promise39. A. business B. walking C. bus D. rail40. A. expected B. frightening C. disappointing D. surprising41. A. budget B. revenue C. compensation D. allowance42. A. frequent B. occasional C. abundant D. constant43. A. unpopular B. lawful C. attractive D. illegal44. A. soon B. early C. late D. slowly45. A. harder B. easier C. cheaper D. funnier46. A. rolled B. bled C. ached D. trembled47. A. struggle B. progress C. excitement D. research48. A. developed B. controlled C. reduced D. increased49. A. Initially B. Eventually C. Temporarily D. Consequently50. A. gained B. kept C. wounded D. lost51. A. sewn B. washed C. worn D. ironed52. A. Doctors B. Hiking C. Lawyers D. Homelessness53. A. well B. partly C. neatly D. originally54. A. choice B. reward C. promise D. break55. A. loose B. full C. blank D. missingNo one is born a winner. People make themselves into winners by their own __16__.I learned this lesson from a(n) __17__ many years ago. I took the head __18__ job at a school in Baxley, Georgia. It was a small school with a weak football program.It was a tradition for the school’s old team to play agains t the __19__ team at the end of spring practice. The old team had no coach, and they didn’t even practice to __20__ the game. Being the coach of the new team, I was excited because I knew we were going to win, but to my disappointment we were defeated. I c ouldn’t __21__ I had got into such a situation. Thinking hard about it, I came to __22__ that my team might not be the number one team in Georgia, but they were __23__ me. I had to change my __24__about their ability and potential.I started doing anything I could to help them build a little __25__. Most important, I began to treat them like __26__. That summer, When the other teams enjoyed their __27___, we met every day and __28__passing and kicking the football.Six months after suffering our __29__on the spring practice field, we won our first game and our second, and continued to __30__. Finally, we faced the number one team in the state. I felt that it would be a __31__for us even if we lost the game. But that wasn’t what happened. My boys beat the best team in Georgia, giving me one of the greatest __32__of my life!From the experience I learnt a lot about how the attitude of the leader can __33__ the members of a team. Instead of seeing my boys as losers, I pushed and__34__them. I helped them to see themselves __35__, and they built themselves into winners.Winners are made, but born.16. A. luck B. tests C. efforts D. nature17. A. experiment B. experience C. visit D. show18. A. operating B. editing C. consulting D. coaching19. A. successful B. excellent C. strong D. new20. A. cheer for B. prepare for C. help with D. finish with21. A. believe B. agree C. describe D. regret22. A. realize B. claim C. permit D. demand23. A. reacting to B. looking for C. depending on D. caring about24. A. decision B. attitude C. conclusion D. intention25. A. pride B. culture C. fortune D. relationship26. A. leaders B. partners C. winners D. learners27. A. rewards B. vacations C. health D. honor28. A. risked B. missed C. considered D. practiced29. A. defeat B. decline C. accident D. mistake30. A. relax B. improve C. expand D. defend31. A. shame B. burden C. victory D. favor32. A. chances B. thrills C. concerns D. offers33. A. surprise B. serve C. interest D. affect34. A. encouraged B. observed C. protected D. impressed35. A. honestly B. individually C. calmly D. differentlyA young English teacher saved the lives of 30 students when he took 36 of a bus after its driver suffered a serious heart attack. Guy Harvold, 24, had 37 the students and three course leaders from Gatwick airport, and they were travelling to Bourmemouth to 38 their host families. They were going to 39 a course at the ABC Language School in Bournemouth where Harvold works as a 40 .Harvold, who has not 41 his driving test, said, “I realized the bus was out of control when I was 42 the students.”The bus ran into trees at the side of the road and he 43 the driver was slumped (倒伏) over the wheel. The driver didn’t 44 . He was unconscious. The bus45 a lamp post and it broke the glass on the front door before Harvold 46 to bring the bus toa stop. Police 47 the young teacher’s quick thinking. If he hadn’t48 quickly, there could have been a terrible 49 .The bus driver never regained consciousness and died at Easy Surrey Hospital. He had worked regularly with the 50 and was very well regarded by the teachers and students. Harvold said, “I was51 that no one else was hurt, but I hoped that the driver would 52 .The head of the language school told the local newspaper that the school is going to send Harvold on a weekend 53 to Dublin with a friend, thanking him for his 54 . A local driving school has also offered him six 55 driving lessons.36. A. control B. care C. advantage D. note37. A. taken in B. picked up C. tracked down D. helped out38. A. greet B. thank C. invite D. meet39. A. present B. introduce C. take D. organize40. A. drive B. doctor C. librarian D. teacher41. A. given B. marked C. passed D. conducted42. A. speaking to B. waiting for C. returning to D. looking for43. A. learned B. noticed C. mentioned D. doubted44. A. sleep B. cry C. move D. recover45. A. ran over B. went by C. carried D. hit46. A. remembered B. continued C. prepared D. managed47. A. witnessed B. recorded C. praised D. understood48. A. appeared B. reacted C. escaped D. interrupted49. A. delay B. accident C. mistake D. experience50. A. airport B. hospital C. school D. police51. A. happy B. fortunate C. touched D. sorry52. A. survive B. retire C. relax D. succeed53. A. project B. trip C. dinner D. duty54. A. bravery B. skill C. quality D. knowledge55. A. necessary B. easy C. different D. freeWe have all heard how time is more valuable than money, but is it __36__ to have too much?I__37__ back in high school I spent most of my day at school since I also __38__a team sport. By the time I got home, I only had a few hours to do my homework, and I had to do it __39__.When I got into college, things __40__. I suddenly found myself out of class before noon time. Because of all this __41__ there was no sense of __42__ to do my school work immediately.I was performing this action of waiting until it later became__43__.Once that happened, I just kept__44__my studying further and further back in my day. Then I got to the point where I was__45__really late at night to get my work alone.One day I __46__a former classmate of mine who was __47__ a lot of money running a sideline(副业).Since his regular job was __48__,I asked him why he just didn't do his sideline full-time. He said without the job, he would __49__ have too much time and would just do what I did back in__50__.He said that if he __51__the job, he would lose his __52__ to work and succeed.So, try __53__your tine with other work. This is why there is a __54__ that if you want something done, ask a __55__person to do it.36.A. true B. fair C. strange D. possible37.A. remember B. admit C. understand D. expect38.A. watched B. loved. C. Coached D. played39.A. al last B. right away C. of course D. mattered40.A. happened B. repeated C. changed D. mattered41.A. extra B. difficult C. valuable D. limited42.A. duty B. achievement C. urgency D. direction43.A. burden B. relief C. risk D. habit44.A. pushing B. taking C. setting D. calling45.A. hanging out B. staying up C. jogging round D. showing off46.A. met B. helped C. treated D. hired47.A. raising B. wasting C. demanding D. making48.A. safe B. important C. boring D. rewarding49.A. luckily B. hardly C. hopefully D. simply50.A. childhood B. college C. town D. business51.A. quit B. found C. accepted D. kept52.A. heart B. chance C. drive D. way53.A. saving B. filling up C. giving up D. trading54.A. message B. story C. saying D. fact55.A. careful B. busy C. reliable D. kind2018新课标I卷BCADD CBDCB CACBA DBADA 2018新课标II卷DACCA BABCD CBDDA DBCAB 2018新课标III卷 DABDC AACDB ABACC DDBCB 2018北京卷CBACD ABDAC ABDAD BBDCC 2018江苏卷BDABD ABDCA CACBD CDBAC 2018天津卷CBDDB AACBA CBDAB CBDAD 2017.11浙江卷ABDCD CABCD DCBBC AABAD 2018浙江卷DADBC ACDAB ADCDB ACBCB。

2018年各地高考数学文科分类汇编——函数

（全国1卷12）答案：（全国1卷13）答案：（全国2卷3）函数2e e ()x xf x x --=的图象大致为答案：B（全国2卷12）已知()f x 是定义域为(,)-∞+∞的奇函数，满足(1)(1)f x f x -=+．若(1)2f =，则(1)(2)(3)(50)f f f f ++++=A ．50-B ．0C ．2D ．50答案：C（全国3卷7）答案：B（全国3卷16）答案：-2（北京卷8）设集合，则（A ）对任意实数a ，（2,1）（B ）对任意实数a ，（2,1）（C ）当且仅当a 0时,（2,1）（D ）当且仅当a时,（2,1）答案：D（天津卷5）已知13313711log ,(),log 245===a b c ，则,,a b c 的大小关系为（A ）>>a b c （B ）>>b a c （C ）>>c b a （D ）>>c a b答案：D 解析：37log 2=a ，1331log =log 55=c ，又3log x 在+（0，）∞单调递增， 3371log log 522∴<<<，即12∴<<<a c ，131()4=b ，函数1()4=x y 的底数小于1，1()4是定义域内单调递减的函数∴=xy ，10311b ()()144∴=<=b 12∴<<<<ac ，即b <<a c（天津卷14）)[)2122,0,,()3,,22,0,x x a x a R f x x x x a x ⎧++-≤∈=∈-+∞⎨-+->⎩已知函数若对任意 ()a f x x ≤恒成立，则的取值范围是答案：1[,2]8解析：当2[3,0],22x x x a x 恒成立2min(32)2ax x当2(0,),22x x x a x 恒成立2max1()28x x a综上，1[,2]8a。

2018年度高考文科数学分类汇编专栏三函数与导数

《2018年高考文科数学分类汇编》第三篇:函数与导数一、选择题1.【2018全国一卷6】设函数()()321f x x a x ax =+-+．假设()f x 为奇函数，那么曲线()y f x =在点()00，处的切线方程为 A ．2y x =-B ．y x =-C ．2y x =D ．y x =2.【2018全国二卷10】若()cos sin f x x x =-在[,]a a -是减函数，那么a 的最大值是A ．π4B ．π2C ．3π4D ．π3.【2018全国三卷9】函数422y x x =-++的图像大致为4.【2018浙江卷5】函数y =||2x sin2x 的图象可能是A ．B ．C ．D ．二、填空题1.【2018全国二卷13】曲线2ln y x =在点(1,0)处的切线方程为__________．2.【2018天津卷10】已知函数f (x )=e x ln x ，f ′(x )为f (x )的导函数，那么f ′（1）的值为__________．3.【2018江苏卷11】假设函数32()21()f x x ax a =-+∈R 在(0,)+∞内有且只有一个零点，那么()f x 在[1,1]-上的最大值与最小值的和为．三．解答题1.【2018全国一卷21】已知函数()e ln 1xf x a x =--．（1）设2x =是()f x 的极值点．求a ，并求()f x 的单调区间；（2）证明：当1ea ≥时，()0f x ≥．2.【2018全国二卷21】已知函数()()32113f x x a x x =-++．（1）假设3a =，求()f x 的单调区间；（2）证明：()f x 只有一个零点．3．【2018全国三卷21】已知函数21()exax x f x +-=．（1）求曲线()y f x =在点(0,1)-处的切线方程；（2）证明：当1a ≥时，()e 0f x +≥．4.【2018北京卷19】设函数2()[(31)32]e xf x ax a x a =-+++.（Ⅰ）假设曲线()y f x =在点(2,(2))f 处的切线斜率为0，求a ；（Ⅱ）假设()f x 在1x =处取得极小值，求a 的取值范围.5.【2018天津卷20】设函数123()=()()()f x x t x t x t ---，其中123,,t t t ∈R ,且123,,t t t 是公差为d 的等差数列．（I ）假设20,1,t d ==求曲线()y f x =在点(0,(0))f 处的切线方程；（II ）假设3d =，求()f x 的极值；（III ）假设曲线()y f x =与直线2()63y x t =---有三个互异的公共点，求d 的取值范围．6．【2018江苏卷17】某农场有一块农田，如下图，它的边界由圆O 的一段圆弧MPN （P 为此圆弧的中点）和线段MN 组成．已知圆O 的半径为40米，点P 到MN 的距离为50米．现计划在此农田上修建两个温室大棚，大棚Ⅰ内的地块形状为矩形ABCD ，大棚Ⅱ内的地块形状为CDP △，要求,A B 均在线段MN 上，,C D 均在圆弧上．设OC 与MN 所成的角为θ．（1）用θ别离表示矩形ABCD 和CDP △的面积，并确信sin θ的取值范围；（2）假设大棚Ⅰ内种植甲种蔬菜，大棚Ⅱ内种植乙种蔬菜，且甲、θ为何值时，能乙两种蔬菜的单位面累年产值之比为43∶．求当使甲、乙两种蔬菜的年总产值最大．7.【2018江苏卷19】（本小题总分值16分）记(),()f x g x ''别离为函数(),()f x g x 的导函数．假设存在0x ∈R ，知足00()()f x g x =且00()()f x g x ''=，那么称0x 为函数()f x 与()g x 的一个“S 点”．（1）证明：函数()f x x =与2()22g x x x =+-不存在“S 点”；（2）假设函数2()1f x ax =-与()ln g x x =存在“S 点”，求实数a 的值；8.【2018浙江卷22】已知函数f (x x ln x ．（Ⅰ）若f (x )在x =x 1，x 2(x 1≠x 2)处导数相等，证明：f (x 1)+f (x 2)>8−8ln2；（Ⅱ）若a ≤3−4ln2，证明：关于任意k >0，直线y =kx +a 与曲线y =f (x )有唯一公共点．9.【2018上海卷19】（此题总分值14分，第1小题总分值6分，第2小题总分值8分）某群体的人均通勤时刻，是指单日内该群体中成员从居住地到工作地的平均历时，某地上班族S 中的成员仅以自驾或公交方式通勤，分析显示：当S 中()%0100x x <<的成员自驾时，自驾群体的人均通勤时刻为⎪⎩⎪⎨⎧<<-+≤<=10030,9018002300,30)(x x x x x f （单位：分钟），而公交群体的人均通勤时刻不受x 阻碍，恒为40分钟，试依照上述分析结果回答以下问题： I ）当x 在什么范围内时，公交群体的人均通勤时刻少于自驾群体的人均通勤时刻？II ）求该地上班族S 的人均通勤时刻g x （）的表达式；讨论g x （）的单调性，并说明其实际意义.参考答案一、选择题1.D2.A3.D4.D 二、填空题1. 22-=x y2.e 3．3-三．解答题1．解：（1）f （x ）的概念域为(0)+∞，，f ′（x ）=a e x –1x．由题设知，f ′（2）=0，因此a =212e ．从而f （x ）=21e ln 12e x x --，f ′（x ）=211e 2e x x-．当0<x <2时，f ′（x ）<0；当x >2时，f ′（x ）>0．因此f （x ）在（0，2）单调递减，在（2，+∞）单调递增．（2）当a ≥1e 时，f （x ）≥e ln 1exx --．设g （x ）=e ln 1e x x --，则e 1()e x g x x'=-．当0<x <1时，g ′（x ）<0；当x >1时，g ′（x ）>0．因此x =1是g （x ）的最小值点．故当x >0时，g （x ）≥g （1）=0．因此，当1ea ≥时，()0f x ≥．2．解：（1）当a =3时，f （x ）=3213333x x x ---，f ′（x ）=263x x --．令f ′（x ）=0解得x =3-x =3+当x ∈（–∞，3-3++∞）时，f ′（x ）>0；当x ∈（3-3+ f ′（x ）<0．故f （x ）在（–∞，3-3++∞）单调递增，在（3-3+（2）由于210x x ++>，因此()0f x =等价于32301x a x x -=++．设()g x =3231x a x x -++，那么g ′（x ）=2222(23)(1)x x x x x ++++≥0，仅当x =0时g ′（x ）=0，因此g （x ）在（–∞，+∞）单调递增．故g （x ）最多有一个零点，从而f （x ）最多有一个零点．又f （3a –1）=22111626()0366a a a -+-=---<，f （3a +1）=103>，故f （x ）有一个零点．综上，f （x ）只有一个零点．3.解：（1）2(21)2()e xax a x f x -+-+'=，(0)2f '=．因此曲线()y f x =在点(0,1)-处的切线方程是210x y --=．（2）当1a ≥时，21()e (1e )e x x f x x x +-+≥+-+．令21()1e x g x x x +≥+-+，那么1()21e x g x x +'≥++．当1x <-时，()0g x '<，()g x 单调递减；当1x >-时，()0g x '>，()g x 单调递增；因此()g x (1)=0g ≥-．因此()e 0f x +≥．4.解：（Ⅰ）因为2()[(31)32]e xf x ax a x a =-+++，因此2()[(1)1]e xf x ax a x '=-++.2(2)(21)e f a '=-，由题设知(2)0f '=，即2(21)e 0a -=，解得12a =. （Ⅱ）方式一：由（Ⅰ）得2()[(1)1]e (1)(1)e xxf x ax a x ax x '=-++=--. 若a >1，那么当1(,1)x a∈时，()0f x '<；当(1,)x ∈+∞时，()0f x '>. 因此()f x 在x =1处取得极小值.若1a ≤，那么当(0,1)x ∈时，110ax x -≤-<，因此()0f x '>.因此1不是()f x 的极小值点. 综上可知，a 的取值范围是(1,)+∞.方式二：()(1)(1)e xf x ax x '=--.（1）当a =0时，令()0f x '=得x =1.(),()f x f x '随x 的转变情形如下表：∴()f x 在x =1处取得极大值，不合题意. （2）当a >0时，令()0f x '=得121,1ax x ==. ①当12x x =，即a =1时，2()(1)e 0xf x x '=-≥，∴()f x 在R 上单调递增， ∴()f x 无极值，不合题意.②当12x x >，即0<a <1时，(),()f x f x '随x 的转变情形如下表：∴()f x 在x =1处取得极大值，不合题意.③当12x x <，即a >1时，(),()f x f x '随x 的转变情形如下表：∴()f x 在x =1处取得极小值，即a >1知足题意. （3）当a <0时，令()0f x '=得121,1ax x ==. (),()f x f x '随x 的转变情形如下表：∴()f x 在x =1处取得极大值，不合题意. 综上所述，a 的取值范围为(1,)+∞.5.解：（I ）由已知，可得f (x )=x (x −1)(x +1)=x 3−x ，故()f x '=3x 2−1，因此f (0)=0，(0)f '=−1，又因为曲线y =f (x )在点(0，f (0))处的切线方程为y −f (0)=(0)f '(x −0)，故所求切线方程为x +y =0．（Ⅱ）由已知可得f (x )=(x −t 2+3)(x −t 2)(x −t 2−3)=(x −t 2)3−9(x −t 2)=x 3−3t 2x 2+(3t 22−9)x −t 23+9t 2．故()f x '=3x 2−6t 2x +3t 22−9．令()f x '=0，解得x =t 2，或x =t 2 当x 转变时，()f x '，f (x )的转变如下表：因此函数f (x )的极大值为f (t 2)3−9×(f (x )的极小值为f (t 2+3−)=−（Ⅲ）解：曲线y =f (x )与直线y =−(x −t 2)−x 的方程(x −t 2+d )(x −t 2)(x −t 2−d )+(x −t 2有三个互异的实数解，令u =x −t 2，可得u 3+(1−d 2)u ．设函数g (x )=x 3+(1−d 2)x ，那么曲线y =f (x )与直线y =−(x −t 2)−y =g (x )有三个零点．()g'x =3x 3+(1−d 2)．当d 2≤1时，()g'x ≥0，这时()g x 在R 上单调递增，不合题意．当d 2>1时，()g'x =0，解得x 1=，x 2．易患，g (x )在(−∞，x 1)上单调递增，在[x 1，x 2]上单调递减，在(x 2，+∞)上单调递增．g (x )的极大值g (x 1)=g (+．g (x )的极小值g (x 2)=g)=+ 若g (x 2)≥0，由g (x )的单调性可知函数y =g (x )最多有两个零点，不合题意．若2()0,g x <即322(1)27d ->，也确实是||d >现在2||d x >，(||)||630,g d d =+>且312||,(2||)6||2||636210630d x g d d d -<-=--+<-+<，从而由()g x 的单调性，可知函数()y g x =在区间1122(2||,),(,),(,||)d x x x x d -内各有一个零点，符合题意．因此，d 的取值范围是(,10)(10,)-∞-+∞．6．解：（1）连结PO 并延长交MN 于H ，那么PH ⊥MN ，因此OH =10．过O 作OE ⊥BC 于E ，那么OE ∥MN ，因此∠COE =θ，故OE =40cos θ，EC =40sin θ，那么矩形ABCD 的面积为2×40cos θ（40sin θ+10）=800（4sin θcos θ+cos θ）， △CDP 的面积为12×2×40cos θ（40–40sin θ）=1600（cos θ–sin θcos θ）．过N 作GN ⊥MN ，别离交圆弧和OE 的延长线于G 和K ，那么GK =KN =10．令∠GOK =θ0，那么si n θ0=14，θ0∈（0，π6）．当θ∈[θ0，π2）时，才能作出知足条件的矩形ABCD ，因此sin θ的取值范围是[14，1）．答：矩形ABCD 的面积为800（4sin θcos θ+cos θ）平方米，△CDP 的面积为 1600（cos θ–sin θcos θ），sin θ的取值范围是[14，1）．（2）因为甲、乙两种蔬菜的单位面累年产值之比为4∶3，设甲的单位面积的年产值为4k ，乙的单位面积的年产值为3k （k >0），那么年总产值为4k ×800（4sin θcos θ+cos θ）+3k ×1600（cos θ–sin θcos θ） =8000k （sin θcos θ+cos θ），θ∈[θ0，π2）．设f （θ）=sin θcos θ+cos θ，θ∈[θ0，π2），则222()cos sin sin (2sin sin 1)(2sin 1)(sin 1)f θθθθθθθθ=--=-+-=--+′．令()=0f θ′，得θ=π6，当θ∈（θ0，π6）时，()>0f θ′，因此f （θ）为增函数；当θ∈（π6，π2）时，()<0f θ′，因此f （θ）为减函数，因此，当θ=π6时，f （θ）取到最大值．答：当θ=π6时，能使甲、乙两种蔬菜的年总产值最大．7.解：（1）函数f （x ）=x ，g （x ）=x 2+2x -2，那么f ′（x ）=1，g ′（x ）=2x +2．由f （x ）=g （x ）且f ′（x ）= g ′（x ），得 222122x x x x ⎧=+-⎨=+⎩，此方程组无解，因此，f （x ）与g （x ）不存在“S ”点．（2）函数21f x ax =-（），()ln g x x =，则12f x ax g x x'='=（），（）．设x 0为f （x ）与g （x ）的“S ”点，由f （x 0）=g （x 0）且f ′（x 0）=g ′（x 0），得200001ln 12ax x ax x ⎧-=⎪⎨=⎪⎩，即200201ln 21ax x ax ⎧-=⎪⎨=⎪⎩，（*）得01ln 2x =-，即120e x -=，那么1221e 22(e )a -==．当e2a =时，120e x -=知足方程组（*），即0x 为f （x ）与g （x ）的“S ”点．因此，a 的值为e2．（3）对任意a >0，设32()3h x x x ax a =--+．因为(0)0(1)1320h a h a a =>=--+=-<，，且h （x ）的图象是不中断的，因此存在0x ∈（0，1），使得0()0h x =．令03002e (1)x x b x =-，那么b >0．函数2e ()()xb f x x a g x x=-+=，，则2e (1)()2()x b x f x x g x x -=-=′，′．由f （x ）=g （x ）且f ′（x ）=g ′（x ），得22e e (1)2xx b x a x b x x x ⎧-+=⎪⎪⎨-⎪-=⎪⎩，即00320030202e e (1)2e (1)2e (1)x x xx x x a x x x x x x x ⎧-+=⋅⎪-⎪⎨-⎪-=⋅⎪-⎩，（**）现在，0x 知足方程组（**），即0x 是函数f （x ）与g （x ）在区间（0，1）内的一个“S 点”．因此，对任意a >0，存在b >0，使函数f （x ）与g （x ）在区间（0，+∞）内存在“S 点”．8.解：（Ⅰ）函数f （x ）的导函数11()2f x x x '=-，由12()()f x f x ''=得1212111122x x x x -=-，因为12x x ≠，因此121112x x +=．由大体不等式得4121212122x x x x x x =+≥．因为12x x ≠，因此12256x x >．由题意得12112212121()()ln ln ln()2f x f x x x x x x x x x +=-+-=-．设1()ln 2g x x x =-，那么1()(4)4g x x x '=-，因此x（0，16） 16 （16，+∞） ()g x '− 0 + ()g x 2−4ln2因此g （x ）在[256，+∞）上单调递增，故12()(256)88ln 2g x x g >=-，即12()()88ln 2f x f x +>-．（Ⅱ）令m =()e a k -+，n =21()1a k++，那么 f （m ）–km –a >|a |+k –k –a ≥0，f （n ）–kn –a <()a n k n n --≤()n k n<0，因此，存在x 0∈（m ，n ）使f （x 0）=kx 0+a ，因此，关于任意的a ∈R 及k ∈（0，+∞），直线y =kx +a 与曲线y =f （x ）有公共点．由f （x ）=kx +a 得ln x x a k --=设h （x ）ln x x a -- 则h ′（x ）=22ln 1()12x x a g x a x x --+--+=，其中g （x ）ln x x ．由（Ⅰ）可知g （x ）≥g （16），又a ≤3–4ln2，故–g （x ）–1+a ≤–g （16）–1+a =–3+4ln2+a ≤0，因此h ′（x ）≤0，即函数h （x ）在（0，+∞）上单调递减，因此方程f （x ）–kx –a =0最多1个实根．综上，当a ≤3–4ln2时，关于任意k >0，直线y =kx +a 与曲线y =f （x ）有唯一公共点．9.解（1）①当300≤<x 时，自驾群体人均通勤时刻为30分钟，公交群体人均通勤时刻为40分钟，现在公交群体人均通勤时刻大于自驾群体人均通勤时刻。

(完整版)2018年各地高考数学文科分类汇编——三角函数,推荐文档

答案：（全国 3 卷 4）
答案：B （全国 3 卷 6）
答案：C
（北京卷 7）在平面坐标系中，
,
,
（如图），点 P 在其中一段上，角以 O
，则 P 所在的圆弧是
（A）
（B）
（C）
（D）答案：C
,
是圆
上的四段弧
（北京卷 16）已知函数
+
.
（Ⅰ）求
的最小正周期
（Ⅱ）若答案：
在区间
上的最大值为，求的最小值.
（全国 1 卷 8）答案：（全国 1 卷 11）答案：
（全国 2 卷 10）若 f (x) cos x sin x 在[0, a] 是减函数，则 a 的最大值是
A. π 4
答案：C
B. π 2
C. 3π
4
D. π
5π 1
（全国 2 卷 15）已知 tan α
4
，(2x
) 的图象向右平移个单位长度，所得图象对
5
10
应的函数
（A）在区间[
上单调递增（B）在区间[
上单调递减
（C）在区间 4 , 4上] 单调递增（D）在区间
4
, 0] ] 上单调递减
[,] 4
[,
2
2
答案：A
解析： y sin(2x ) 向右移动个单位长度得到
5
10
y sin[（2 ）x - ，] 即 y sin 2x ， 10 5
单增区间为： +2k 2x 2k (k Z )
+k x
2 k (kZ)
2
当4k 0 时，函4数 y sin(2x 在区间[
)
,]

2018年高考真题汇编(函数与导数)

函数与导数1．【2018年浙江卷】函数y=sin2x的图象可能是A. B.C. D.【答案】D点睛：有关函数图象的识别问题的常见题型及解题思路：（1）由函数的定义域，判断图象的左、右位置，由函数的值域，判断图象的上、下位置；（2）由函数的单调性，判断图象的变化趋势；（3）由函数的奇偶性，判断图象的对称性；（4）由函数的周期性，判断图象的循环往复．2．【2018年理天津卷】已知，，，则a，b，c的大小关系为A. B. C. D.【答案】D【解析】分析：由题意结合对数函数的性质整理计算即可求得最终结果.详解：由题意结合对数函数的性质可知：，，，据此可得：.本题选择D选项.点睛：对于指数幂的大小的比较，我们通常都是运用指数函数的单调性，但很多时候，因幂的底数或指数不相同，不能直接利用函数的单调性进行比较．这就必须掌握一些特殊方法．在进行指数幂的大小比较时，若底数不同，则首先考虑将其转化成同底数，然后再根据指数函数的单调性进行判断．对于不同底而同指数的指数幂的大小的比较，利用图象法求解，既快捷，又准确．3．【2018年理新课标I卷】已知函数．若g（x）存在2个零点，则a的取值范围是A. [–1，0）B. [0，+∞）C. [–1，+∞）D. [1，+∞）【答案】C详解：画出函数的图像，在y轴右侧的去掉，再画出直线，之后上下移动，可以发现当直线过点A时，直线与函数图像有两个交点，并且向下可以无限移动，都可以保证直线与函数的图像有两个交点，即方程有两个解，也就是函数有两个零点，此时满足，即，故选C.点睛：该题考查的是有关已知函数零点个数求有关参数的取值范围问题，在求解的过程中，解题的思路是将函数零点个数问题转化为方程解的个数问题，将式子移项变形，转化为两条曲线交点的问题，画出函数的图像以及相应的直线，在直线移动的过程中，利用数形结合思想，求得相应的结果.4．【2018年理新课标I卷】设函数，若为奇函数，则曲线在点处的切线方程为A. B. C. D.【答案】D点睛：该题考查的是有关曲线在某个点处的切线方程的问题，在求解的过程中，首先需要确定函数解析式，此时利用到结论多项式函数中，奇函数不存在偶次项，偶函数不存在奇次项，从而求得相应的参数值，之后利用求导公式求得，借助于导数的几何意义，结合直线方程的点斜式求得结果. 5．【2018年全国卷Ⅲ理】设，，则A. B. C. D.【答案】B【解析】分析：求出，得到的范围，进而可得结果。

2018年高考数学-函数含答案

2018年高考复习专题-函数一．函数1、函数的概念：（1）定义：设A 、B 是非空的数集，如果按照某个确定的对应关系f ，使对于集合A 中的任意一个数x ，在集合B 中都有唯一确定的数)(x f 和它对应，那么就称f ：A →B 为从集合A 到集合B 的一个函数．记作：y =)(x f ，x ∈A ．其中，x 叫做自变量，x 的取值范围A 叫做函数的定义域；与x 的值相对应的y 值叫做函数值，函数值的集合{)(x f | x ∈A }叫做函数的值域．（2）函数的三要素：定义域、值域、对应法则（3）相同函数的判断方法：①表达式相同（与表示自变量和函数值的字母无关）；②定义域一致 (两点必须同时具备)2、定义域：（1）定义域定义：函数)(x f 的自变量x 的取值范围。

（2）确定函数定义域的原则：使这个函数有意义的实数的全体构成的集合。

（3）确定函数定义域的常见方法：①若)(x f 是整式，则定义域为全体实数②若)(x f 是分式，则定义域为使分母不为零的全体实数例:求函数xy 111+=的定义域。

③若)(x f 是偶次根式，则定义域为使被开方数不小于零的全体实数例1．求函数 ()2143432-+--=x x xy 的定义域。

例2．求函数()02112++-=x x y 的定义域。

④对数函数的真数必须大于零⑤指数、对数式的底必须大于零且不等于1⑥若)(x f 为复合函数，则定义域由其中各基本函数的定义域组成的不等式组来确定⑦指数为零底不可以等于零，如)0(10≠=x x⑧实际问题中的函数的定义域还要保证实际问题有意义. （4）求抽象函数（复合函数）的定义域已知函数)(x f 的定义域为[0,1]求)(2x f 的定义域已知函数)12(-x f 的定义域为[0,1）求)31(x f -的定义域3、值域 :（1）值域的定义：与x 相对应的y 值叫做函数值，函数值的集合叫做函数的值域。

（2）确定值域的原则：先求定义域（3）常见基本初等函数值域：一次函数、二次函数、反比例函数、指数函数、对数函数、三角函数（正余弦、正切）（4）确定函数值域的常见方法：①直接法：从自变量x 的范围出发，推出()y f x =的取值范围。

2018年全国各地高考数学试题及解答分类汇编大全(04 导数及其应用)

x
则曲线 y 2ln x 在点 1,0 处的切线的斜率为 k f 1 2 ，则所求切线方程为 y 0 2 x 1 ，即 y 2x 2 ．
4．（2018 全国新课标Ⅱ理）曲线 y 2 ln(x 1) 在点 (0, 0) 处的切线方程为__________．
4．【答案】 y 2x
x
,1
1
1,1a
f x
0
f x
Z
极大值
]
f x 在 x 1处取得极大值，不合题意． ③当 x1 x2 ，即 a 1时， f x ， f x 随 x 的变化情况如下表：
x
,1 a
1 a
1 a
,1
f x
0
1 a 0 极小值
1 0
1 a
,
Z
1,
f x
Z
极大值
]
极小值
Z
f x 在 x 1处取得极小值，即 a 1满足题意．
1 x
1)2 k 1
4
16
0 ，得 h(x) 有两个极值点 x1, x2 (x1 x2 ) ，
∴
1 x1
1 4
，∴ 0
x1
16 .
可知 h(x) 在 (0, x1) 递增， (x1, x2 ) 递减， (x2 , ) 递增，
∴ h(x1) kx1
x1
ln x1
a
( 2
1 x1
1) x1
（1）证明：函数 f (x) x 与 g(x) x2 2x 2 不存在“S 点”；
（2）若函数 f (x) ax2 1与 g(x) ln x 存在“S 点”，求实数 a 的值；（3）已知函数 f (x) x2 a ，g(x) bex ．对任意 a 0 ，判断是否存在 b 0 ，使函数 f (x) 与 g(x)

2018年各地高考真题分类汇编(文)-三角函数---教师版(可编辑修改word版)

2 3 330 三角函数和解三角形1.（2018 年全国 1 文科·8）已知函数 f ( x ) = 2 cos 2 x - sin 2x + 2 ，则 BA. f ( x ) 的最小正周期为 π，最大值为 3B. f ( x ) 的最小正周期为 π，最大值为 4C. f (x ) 的最小正周期为2π ，最大值为 3D. f (x ) 的最小正周期为2π ，最大值为 42.（2018 年全国 1 文科·11）已知角的顶点为坐标原点，始边与 x 轴的非负半轴重合，终边上有两点 A (1，a ) ， B (2 ，b ) ，且cos 2= 2，则 a - b = B 3A.15 B. 5C. 25 5D ．13.（ 2018 年全国 1 文科· 16） △ ABC 的内角 A ，B ，C 的对边分别为 a ，b ，c ，已知b s in C +c sin B = 4a sin B sin C ， b 2 + c 2 - a 2 = 8 ，则△ABC 的面积为．4. （2018 年全国 2 文科·7）．在△ABC 中， cos C = 5 ， BC = 1 ， AC = 5 ，则 AB = AA. 4 2 5B. C ． D ．25.（2018 年全国 2 文科·10）若 f (x ) = cos x - sin x 在[0, a ] 是减函数，则 a 的最大值是 CA.π4B.π 2C. 3π4D. π6．（2018 年全国 2 文科·15）已知 tan(α -5π) = 1，则tan α = 3．4 527.（2018 年全国 3 文科·4）若sin= 1，则cos 2= B3A.89B.79C. - 79 D. - 89229 58.（2018 年全国 3 文科·6）函数 f (x) =tan x1+ tan2x的最小正周期为CA．πB．πC．πD．2π 4 29.（2018 年全国3 文科·11）△ABC 的内角A ，B ，C 的对边分别为a ，b ，c ．若a2 +b2 -c2△ABC 的面积为4，则C =CππA.B．2 3ππ C．D．4 610.（2018 年北京文科·7）在平面直角坐标系中， AB, C D, E F , G H 是圆x2+y2= 1上的四段弧（如图），点P 在其中一段上，角以O为始边，OP 为终边，若tan< cos< sin，则P 所在的圆弧是C（A） AB （B）C D（C）E F （D）G H11.（2018 年北京文科·14）若△ABC 的面积为cB=60°；的取值范围是（2，+∞）.a3(a2 +c2 -b2 ) ,且∠C 为钝角，则412.（2018 年天津文科·6）将函数y = sin(2x +图象对应的函数A ππ) 的图象向右平移个单位长度，所得5 107 （A ）在区间[- π π, ] 上单调递增（B ）在区间[- 4 4 π , 0] 上单调递减4π ππ（C ）在区间[ , ] 上单调递增（D ）在区间[ , π] 上单调递减4 2213.（2018 年江苏·7）．已知函数 y = sin(2x +)(- π << π) 的图象关于直线 x = π对称，则的值是．2 2 314. （2018 年江苏·13）在△ABC 中，角 A , B , C 所对的边分别为 a , b , c ， ∠ABC = 120︒ ，∠ABC 的平分线交 AC 于点 D ，且 BD = 1，则4a + c 的最小值为 9 ．15.（2018 年浙江·13）在△ABC 中，角 A ，B ，C 所对的边分别为 a ，b ，c ．若 a = ，b =2，A =60°，则 sin B =217 ，c = 3 ．16.（2018 年北京文科·16）（本小题 13 分）已知函数 f (x ) = sin 2 x + 3 sin x cos x .（Ⅰ）求 f (x ) 的最小正周期；（Ⅱ）若 f (x ) 在区间[- π , m ] 上的最大值为 3，求m 的最小值.3216.（共 13 分）解：（Ⅰ）f (x ) = 1- cos 2x +3 sin 2x = 3 sin 2x - 1 cos 2x + 1 = sin(2x - π) + 1 ，2 2 2 2 2 6 2所以 f (x ) 的最小正周期为T =2π = π .2（Ⅱ）由（Ⅰ）知 f (x ) = sin(2x - π) + 1.6 2π π 5π π因为 x ∈[- , m ]，所以2x - ∈[- , 2m - ] .3 6 6 67 π π 要使得 f (x ) 在[- π , m ] 上的最大值为 3 ，即sin(2x - π) 在[- π, m ] 上的最大值为 1.所以2m - ≥ 6 2 3 ，即 m ≥π 2 6 3π .学科&网 3所以m 的最小值为 .317.（2018 年天津文科·16）（本小题满分 13 分）在△ABC 中，内角 A ，B ，C 所对的边分别为 a ，b ，c ．已知 b sin A =a cos(B – π)．6（Ⅰ）求角 B 的大小；（Ⅱ）设 a =2，c =3，求 b 和 sin(2A –B )的值．（16）本小题主要考查同角三角函数的基本关系，两角差的正弦与余弦公式，二倍角的正弦与余弦公式，以及正弦定理、余弦定理等基础知识，考查运算求解能力．满分 13 分．（ Ⅰ ）解：在△ ABC 中，由正弦定理 a = sin A bsin B，可得 b sin A = a sin B ，又由 b sin A = a cos(B - π) ，得 a sin B = a cos(B - π) ，即sin B = cos(B - π) ，可得tan B = 6 6 6．又因为 B ∈(0 ，π) ，可得 B = π．3（Ⅱ）解：在△ABC 中，由余弦定理及 a =2，c =3，B = π，有b 2 = a 2 + c 2 - 2ac cos B = 7 ，3故 b = ．由 b s in A = a cos(B - π) ，可得 6sin A =．因为 a <c ，故cos A =．因此sin 2 A = 2sin A cos A =4 3 ， cos 2 A = 2 cos 2 A - 1 = 177所以， sin(2 A - B ) = sin 2 A cos B - cos 2 A sin B =4 3 ⨯ 1 - 1⨯ 3 = 3 3 7 2 7 2 1418.（2018 年江苏·16）（本小题满分 14 分）33 727已知,为锐角，tan=4，cos(+) =-5．3 5（1）求cos 2的值；（2）求tan(-)的值．16．本小题主要考查同角三角函数关系、两角和（差）及二倍角的三角函数，考查运算求解能力．满分14 分．解：（1）因为tan=4 ，tan=sin，所以sin=4 cos．3 cos 3因为sin2+c os2=1，所以cos2=9，25因此，cos 2= 2 cos2- 1 =-7 ．25（2）因为,为锐角，所以+∈(0,π)．又因为cos(+)=-5，所以sin(+)=5=2 5，5因此tan(+)=-2．因为tan=4，所以tan 2=32 tan1 -tan2=-24，7因此，tan(-) = tan[2- (+)] =tan 2- tan(+)=-2．1+ t an 2tan(+) 1119.（2018 年浙江·18）（本题满分14 分）已知角α 的顶点与原点O 重合，始边与x 轴的非负半轴重合，它的终边过点P（-3，-4）．5 5（Ⅰ）求sin（α+π）的值；（Ⅱ）若角β 满足sin（α+β）= 5，求cosβ 的值．1318.本题主要考查三角函数及其恒等变换等基础知识，同时考查运算求解能力。

2018年高考真题理科数学分类汇编(解析版)：函数及答案

2018年高考真题理科数学分类汇编（解析版）
函数
1、（2018年高考（安徽卷））函数=()y f x 的图像如图所示，在区间[],a b 上可找到(2)n n ≥ 个不同的数12,...,,n x x x 使得1212()()()==,n n
f x f x f x x x x 则n 的取值范围是（A ）{}3,4 （B ）{}2,3,4
（C ） {}3,4,5 （D ）{}2,3
【答案】B
【解析】由题知，过原点的直线与曲线相交的个数即n 的取值.用尺规作图，交点可取2,3,4. 所以选B
2、（2018年高考（北京卷））函数f (x )的图象向右平移一个单位长度，所得图象与y =e x 关于y 轴对称，则f (x )=
A.1e x +
B. 1e x -
C. 1e x -+
D. 1e x --
3、（2018年高考（广东卷））定义域为R 的四个函数3y x =,2x y =,2
1y x =+,2sin y x =中,奇函数的个数是( )
A . 4
B ．3
C ．2
D ．【解析】C ；考查基本初等函数和奇函数的概念,是奇函数的为3y x =与2sin y x =,故选C ．
4、（2018年高考（全国（广西）卷））已知函数()()()-1,021f x f x -的定义域为，则函数的定义域为
（A ）()1,1- （B ）11,2⎛⎫- ⎪⎝⎭ （C ）()-1,0 （D ）1,12⎛⎫
⎪⎝⎭
【答案】B 【解析】由题意可知 1210,x -<+<，则1
12x -<<-。

故选B。

2018年高考数学分类汇编函数及答案解析

2018年高考数学分类汇编函数一．选择题1、（2018年高考全国卷I文科5）（5分）设函数f（x）=x3+（a﹣1）x2+ax．若f（x）为奇函数，则曲线y=f（x）在点（0，0）处的切线方程为（）A．y=﹣2x B．y=﹣x C．y=2x D．y=x【解答】解：函数f（x）=x3+（a﹣1）x2+ax，若f（x）为奇函数，可得a=1，所以函数f（x）=x3+x，可得f′（x）=3x2+1，曲线y=f（x）在点（0，0）处的切线的斜率为：1，则曲线y=f（x）在点（0，0）处的切线方程为：y=x．故选：D．2、（2018年高考全国卷1文科9）（5分）已知函数f（x）=，g（x）=f（x）+x+a．若g（x）存在2个零点，则a的取值范围是（）A．[﹣1，0）B．[0，+∞）C．[﹣1，+∞）D．[1，+∞）【解答】解：由g（x）=0得f（x）=﹣x﹣a，作出函数f（x）和y=﹣x﹣a的图象如图：当直线y=﹣x﹣a的截距﹣a≤1，即a≥﹣1时，两个函数的图象都有2个交点，即函数g（x）存在2个零点，故实数a的取值范围是[﹣1，+∞），故选：C．3、（2018年高考全国卷1理科2）（5分）已知集合A={x|x2﹣x﹣2＞0}，则∁R A=（）A．{x|﹣1＜x＜2}B．{x|﹣1≤x≤2}C．{x|x＜﹣1}∪{x|x＞2}D．{x|x≤﹣1}∪{x|x≥2}【解答】解：集合A={x|x2﹣x﹣2＞0}，可得A={x|x＜﹣1或x＞2}，则：∁R A={x|﹣1≤x≤2}．故选：B．4、（2018年高考全国卷1理科5）（5分）设函数f（x）=x3+（a﹣1）x2+ax．若f（x）为奇函数，则曲线y=f（x）在点（0，0）处的切线方程为（）A．y=﹣2x B．y=﹣x C．y=2x D．y=x【解答】解：函数f（x）=x3+（a﹣1）x2+ax，若f（x）为奇函数，可得a=1，所以函数f（x）=x3+x，可得f′（x）=3x2+1，曲线y=f（x）在点（0，0）处的切线的斜率为：1，则曲线y=f（x）在点（0，0）处的切线方程为：y=x．故选：D．5、（2018年高考全国卷1理科9）（5分）已知函数f（x）=，g（x）=f（x）+x+a．若g（x）存在2个零点，则a的取值范围是（）A．[﹣1，0）B．[0，+∞）C．[﹣1，+∞）D．[1，+∞）【解答】解：由g（x）=0得f（x）=﹣x﹣a，作出函数f（x）和y=﹣x﹣a的图象如图：当直线y=﹣x﹣a的截距﹣a≤1，即a≥﹣1时，两个函数的图象都有2个交点，即函数g（x）存在2个零点，故实数a的取值范围是[﹣1，+∞），故选：C．6、（2018年高考全国卷2文科3）（5分）函数f（x）=的图象大致为（）A．B．CD．【解答】解：函数f（﹣x）==﹣=﹣f（x），则函数f（x）为奇函数，图象关于原点对称，排除A，当x=1时，f（1）=e﹣＞0，排除D．当x→+∞时，f（x）→+∞，排除C，故选：B．7、（2018年高考全国卷2文科12）（5分）已知f（x）是定义域为（﹣∞，+∞）的奇函数，满足f（1﹣x）=f（1+x），若f（1）=2，则f（1）+f（2）+f（3）+…+f（50）=（）A．﹣50 B．0 C．2 D．50【解答】解：∵f（x）是奇函数，且f（1﹣x）=f（1+x），∴f（1﹣x）=f（1+x）=﹣f（x﹣1），f（0）=0，则f（x+2）=﹣f（x），则f（x+4）=﹣f（x+2）=f（x），即函数f（x）是周期为4的周期函数，∵f（1）=2，∴f（2）=f（0）=0，f（3）=f（1﹣2）=f（﹣1）=﹣f（1）=﹣2，f（4）=f（0）=0，则f（1）+f（2）+f（3）+f（4）=2+0﹣2+0=0，则f（1）+f（2）+f（3）+…+f（50）=12[f（1）+f（2）+f（3）+f（4）]+f（49）+f（50）=f（1）+f（2）=2+0=2，故选：C．8、（2018年高考全国卷2理科3）（5分）函数f（x）=的图象大致为（）A．B．C．D．【解答】解：函数f（﹣x）==﹣=﹣f（x），则函数f（x）为奇函数，图象关于原点对称，排除A，当x=1时，f（1）=e﹣＞0，排除D．当x→+∞时，f（x）→+∞，排除C，故选：B．9.（2018年高考全国卷2理科11）（5分）已知f（x）是定义域为（﹣∞，+∞）的奇函数，满足f（1﹣x）=f（1+x），若f（1）=2，则f（1）+f（2）+f（3）+…+f（50）=（）A．﹣50 B．0 C．2 D．50【解答】解：∵f（x）是奇函数，且f（1﹣x）=f（1+x），∴f（1﹣x）=f（1+x）=﹣f（x﹣1），f（0）=0，则f（x+2）=﹣f（x），则f（x+4）=﹣f（x+2）=f（x），即函数f（x）是周期为4的周期函数，∵f（1）=2，∴f（2）=f（0）=0，f（3）=f（1﹣2）=f（﹣1）=﹣f（1）=﹣2，f（4）=f（0）=0，则f（1）+f（2）+f（3）+f（4）=2+0﹣2+0=0，则f（1）+f（2）+f（3）+…+f（50）=12[f（1）+f（2）+f（3）+f（4）]+f（49）+f（50）=f（1）+f（2）=2+0=2，故选：C．10.（2018年高考全国卷3文科7）（5分）下列函数中，其图象与函数y=lnx的图象关于直线x=1对称的是（）A．y=ln（1﹣x）B．y=ln（2﹣x）C．y=ln（1+x） D．y=ln（2+x）【解答】解：首先根据函数y=lnx的图象，则：函数y=lnx的图象与y=ln（﹣x）的图象关于y轴对称．由于函数y=lnx的图象关于直线x=1对称．则：把函数y=ln（﹣x）的图象向右平移2个单位即可得到：y=ln（2﹣x）．即所求得解析式为：y=ln（2﹣x）．故选：B．11.（2018年高考全国卷3文科9）（5分）函数y=﹣x4+x2+2的图象大致为（）A．B．C．D．【解答】解：函数过定点（0，2），排除A，B．函数的导数f′（x）=﹣4x3+2x=﹣2x（2x2﹣1），由f′（x）＞0得2x（2x2﹣1）＜0，得x＜﹣或0＜x＜，此时函数单调递增，排除C，故选：D．12.（2018年高考全国卷3理科7）（5分）函数y=﹣x4+x2+2的图象大致为（）A．B．C．D．【解答】解：函数过定点（0，2），排除A，B．函数的导数f′（x）=﹣4x3+2x=﹣2x（2x2﹣1），由f′（x）＞0得2x（2x2﹣1）＜0，得x＜﹣或0＜x＜，此时函数单调递增，排除C，故选：D．13.（2018年高考全国卷3理科12）（5分）设a=log0.20.3，b=log20.3，则（）A．a+b＜ab＜0 B．ab＜a+b＜0 C．a+b＜0＜ab D．ab＜0＜a+b【解答】解：∵a=log0.20.3=，b=log20.3=，∴=，，∵，，∴ab＜a+b＜0．故选：B．14.（2018年北京市高考数学试卷文科8）（5分）设集合A={（x，y）|x﹣y≥1，ax+y＞4，x﹣ay≤2}，则（）A．对任意实数a，（2，1）∈A B．对任意实数a，（2，1）∉AC．当且仅当a＜0时，（2，1）∉A D．当且仅当a≤时，（2，1）∉A【解答】解：当a=﹣1时，集合A={（x，y）|x﹣y≥1，ax+y＞4，x﹣ay≤2}={（x，y）|x﹣y ≥1，﹣x+y＞4，x+y≤2}，显然（2，1）不满足，﹣x+y＞4，x+y≤2，所以A，C不正确；当a=4，集合A={（x，y）|x﹣y≥1，ax+y＞4，x﹣ay≤2}={（x，y）|x﹣y≥1，4x+y＞4，x ﹣4y≤2}，显然（2，1）在可行域内，满足不等式，所以B不正确；故选：D．15.（2018年北京市高考数学试卷理科8）（5分）设集合A={（x，y）|x﹣y≥1，ax+y＞4，x﹣ay ≤2}，则（）A．对任意实数a，（2，1）∈A B．对任意实数a，（2，1）∉AC．当且仅当a＜0时，（2，1）∉A D．当且仅当a≤时，（2，1）∉A【解答】解：当a=﹣1时，集合A={（x，y）|x﹣y≥1，ax+y＞4，x﹣ay≤2}={（x，y）|x﹣y ≥1，﹣x+y＞4，x+y≤2}，显然（2，1）不满足，﹣x+y＞4，x+y≤2，所以A，C不正确；当a=4，集合A={（x，y）|x﹣y≥1，ax+y＞4，x﹣ay≤2}={（x，y）|x﹣y≥1，4x+y＞4，x ﹣4y≤2}，显然（2，1）在可行域内，满足不等式，所以B不正确；故选：D．16.（2018年上海市高考数学试卷16）（5分）设D是含数1的有限实数集，f（x）是定义在D上的函数，若f（x）的图象绕原点逆时针旋转后与原图象重合，则在以下各项中，f（1）的可能取值只能是（）A．B．C．D．0【解答】解：设D是含数1的有限实数集，f（x）是定义在D上的函数，若f（x）的图象绕原点逆时针旋转后与原图象重合，故f（1）=cos=，故选：B．17.（2018年浙江省高考数学试卷5）（4分）函数y=2|x|sin2x的图象可能是（）A．B．C．D．【解答】解：根据函数的解析式y=2|x|sin2x，得到：函数的图象为奇函数，故排除A和B．当x=时，函数的值也为0，故排除C．故选：D．18.（2018年天津市高考数学试卷文科3）（5分）设x∈R，则“x3＞8”是“|x|＞2”的（）A．充分而不必要条件B．必要而不充分条件C．充要条件D．既不充分也不必要条件【解答】解：由x3＞8，得x＞2，则|x|＞2，反之，由|x|＞2，得x＜﹣2或x＞2，则x3＜﹣8或x3＞8．即“x3＞8”是“|x|＞2”的充分不必要条件．故选：A．19.（2018年天津市高考数学试卷文科5）（5分）已知a=log3，b=（），c=log，则a，b，c的大小关系为（）A．a＞b＞c B．b＞a＞c C．c＞b＞a D．c＞a＞b【解答】解：∵a=log 3，c=log=log35，且5，∴，则b=（）＜，∴c＞a＞b．故选：D．20.（2018年天津市高考数学试卷理科5）（5分）已知a=log2e，b=ln2，c=log，则a，b，c的大小关系为（）A．a＞b＞c B．b＞a＞c C．c＞b＞a D．c＞a＞b【解答】解：a=log 2e＞1，0＜b=ln2＜1，c=log=log23＞log2e=a，则a，b，c的大小关系c＞a＞b，故选：D．二．填空题1.（2018年高考全国卷I文科13）（5分）已知函数f（x）=log2（x2+a），若f（3）=1，则a=．【解答】解：函数f（x）=log2（x2+a），若f（3）=1，可得：log2（9+a）=1，可得a=﹣7．故答案为：﹣7．2.（2018年高考全国卷2文科13）（5分）曲线y=2lnx在点（1，0）处的切线方程为．【解答】解：∵y=2lnx，∴y′=，当x=1时，y′=2∴曲线y=2lnx在点（1，0）处的切线方程为y=2x﹣2．故答案为：y=2x﹣2．3.（2018年高考全国卷2理科13）（5分）曲线y=2ln（x+1）在点（0，0）处的切线方程为．【解答】解：∵y=2ln（x+1），∴y′=，当x=0时，y′=2，∴曲线y=2ln（x+1）在点（0，0）处的切线方程为y=2x．故答案为：y=2x．4.（2018年高考全国卷3文科16）（5分）已知函数f（x）=ln（﹣x）+1，f（a）=4，则f（﹣a）=．【解答】解：函数g（x）=ln（﹣x）满足g（﹣x）=ln（+x）==﹣ln（﹣x）=﹣g（x），所以g（x）是奇函数．函数f（x）=ln（﹣x）+1，f（a）=4，可得f（a）=4=ln（﹣a）+1，可得ln（﹣a）=3，则f（﹣a）=﹣ln（﹣a）+1=﹣3+1=﹣2．故答案为：﹣2．5.（2018年高考全国卷3理科14）（5分）曲线y=（ax+1）e x在点（0，1）处的切线的斜率为﹣2，则a=．【解答】解：曲线y=（ax+1）e x，可得y′=ae x+（ax+1）e x，曲线y=（ax+1）e x在点（0，1）处的切线的斜率为﹣2，可得：a+1=﹣2，解得a=﹣3．故答案为：﹣3．6.（2018年北京市高考数学试卷理科13）（5分）能说明“若f（x）＞f（0）对任意的x∈（0，2]都成立，则f（x）在[0，2]上是增函数”为假命题的一个函数是．【解答】解：例如f（x）=sinx，尽管f（x）＞f（0）对任意的x∈（0，2]都成立，当x∈[0，）上为增函数，在（，2]为减函数，故答案为：f（x）=sinx．7.（2018年江苏省高考数学试卷5）（5分）函数f（x）=的定义域为．【解答】解：由题意得：≥1，解得：x≥2，∴函数f（x）的定义域是[2，+∞）．故答案为：[2，+∞）．8.（2018年江苏省高考数学试卷9）（5分）函数f（x）满足f（x+4）=f（x）（x∈R），且在区间（﹣2，2]上，f（x）=，则f（f（15））的值为．【解答】解：由f（x+4）=f（x）得函数是周期为4的周期函数，则f（15）=f（16﹣1）=f（﹣1）=|﹣1+|=，f（）=cos（）=cos=，即f（f（15））=，故答案为：9.（2018年江苏省高考数学试卷11）（5分）若函数f（x）=2x3﹣ax2+1（a∈R）在（0，+∞）内有且只有一个零点，则f（x）在[﹣1，1]上的最大值与最小值的和为．【解答】解：∵函数f（x）=2x3﹣ax2+1（a∈R）在（0，+∞）内有且只有一个零点，∴f′（x）=2x（3x﹣a），x∈（0，+∞），①当a≤0时，f′（x）=2x（3x﹣a）＞0，函数f（x）在（0，+∞）上单调递增，f（0）=1，f（x）在（0，+∞）上没有零点，舍去；②当a＞0时，f′（x）=2x（3x﹣a）＞0的解为x＞，∴f（x）在（0，）上递减，在（，+∞）递增，又f（x）只有一个零点，∴f（）=﹣+1=0，解得a=3，f（x）=2x3﹣3x2+1，f′（x）=6x（x﹣1），x∈[﹣1，1]，f′（x）＞0的解集为（﹣1，0），f（x）在（﹣1，0）上递增，在（0，1）上递减，f（﹣1）=﹣4，f（0）=1，f（1）=0，∴f（x）min=f（﹣1）=﹣4，f（x）max=f（0）=1，∴f（x）在[﹣1，1]上的最大值与最小值的和为：f（x）max+f（x）min=﹣4+1=﹣3．10（2018年上海市高考数学试卷4)（4分）设常数a∈R，函数f（x）=1og2（x+a）．若f（x）的反函数的图象经过点（3，1），则a=．【解答】解：∵常数a∈R，函数f（x）=1og2（x+a）．f（x）的反函数的图象经过点（3，1），∴函数f（x）=1og2（x+a）的图象经过点（1，3），∴log2（1+a）=3，解得a=7．故答案为：7．10.（2018年上海市高考数学试卷7)（5分）已知α∈{﹣2，﹣1，﹣，1，2，3}，若幂函数f（x）=xα为奇函数，且在（0，+∞）上递减，则α=．【解答】解：∵α∈{﹣2，﹣1，，1，2，3}，幂函数f（x）=xα为奇函数，且在（0，+∞）上递减，∴a是奇数，且a＜0，∴a=﹣1．故答案为：﹣1．11.（2018年上海市高考数学试卷11)（5分）已知常数a＞0，函数f（x）=的图象经过点P（p，），Q（q，）．若2p+q=36pq，则a=．【解答】解：函数f（x）=的图象经过点P（p，），Q（q，）．则：，整理得：=1，解得：2p+q=a2pq，由于：2p+q=36pq，所以：a2=36，由于a＞0，故：a=6．故答案为：612.（2018年上海市高考数学试卷12)（5分）已知实数x1、x2、y1、y2满足：x12+y12=1，x22+y22=1，x1x2+y1y2=，则+的最大值为．【解答】解：设A（x1，y1），B（x2，y2），=（x1，y1），=（x2，y2），由x12+y12=1，x22+y22=1，x1x2+y1y2=，可得A，B两点在圆x2+y2=1上，且•=1×1×cos∠AOB=，即有∠AOB=60°，即三角形OAB为等边三角形，AB=1，+的几何意义为点A，B两点到直线x+y﹣1=0的距离d1与d2之和，显然d1+d2≤AB=1，即+的最大值为1，故答案为：1．13.（2018年天津市高考数学试卷文科10）（5分）已知函数f（x）=e x lnx，f′（x）为f（x）的导函数，则f′（1）的值为．【解答】解：函数f（x）=e x lnx，则f′（x）=e x lnx+•e x；∴f′（1）=e•ln1+1•e=e．故答案为：e．14.（2018年天津市高考数学试卷文科13）（5分）已知a，b∈R，且a﹣3b+6=0，则2a+的最小值为．【解答】解：a，b∈R，且a﹣3b+6=0，可得：3b=a+6，则2a+==≥2=，当且仅当2a=．即a=﹣3时取等号．函数的最小值为：．故答案为：．15.（2018年天津市高考数学试卷文科14）（5分）己知a∈R，函数f（x）=．若对任意x∈[﹣3，+∞），f（x）≤|x|恒成立，则a的取值范围是．【解答】解：当x≤0时，函数f（x）=x2+2x+a﹣2的对称轴为x=﹣1，抛物线开口向上，要使x≤0时，对任意x∈[﹣3，+∞），f（x）≤|x|恒成立，则只需要f（﹣3）≤|﹣3|=3，即9﹣6+a﹣2≤3，得a≤2，当x＞0时，要使f（x）≤|x|恒成立，即f（x）=﹣x2+2x﹣2a，则直线y=x的下方或在y=x上，由﹣x2+2x﹣2a=x，即x2﹣x+2a=0，由判别式△=1﹣8a≤0，得a≥，综上≤a≤2，故答案为：[，2]．16.（2018年天津市高考数学试卷理科13）（5分）已知a，b∈R，且a﹣3b+6=0，则2a+的最小值为．【解答】解：a，b∈R，且a﹣3b+6=0，可得：3b=a+6，则2a+==≥2=，当且仅当2a=．即a=﹣3时取等号．函数的最小值为：．故答案为：．17.（2018年天津市高考数学试卷理科14）（5分）已知a＞0，函数f（x）=．若关于x的方程f（x）=ax恰有2个互异的实数解，则a的取值范围是．【解答】解：当x≤0时，由f（x）=ax得x2+2ax+a=ax，得x2+ax+a=0，得a（x+1）=﹣x2，得a=﹣，设g（x）=﹣，则g′（x）=﹣=﹣，由g（x）＞0得﹣2＜x＜﹣1或﹣1＜x＜0，此时递增，由g（x）＜0得x＜﹣2，此时递减，即当x=﹣2时，g（x）取得极小值为g（﹣2）=4，当x＞0时，由f（x）=ax得﹣x2+2ax﹣2a=ax，得x2﹣ax+2a=0，得a（x﹣2）=x2，当x=2时，方程不成立，当x≠2时，a=设h（x）=，则h′（x）==，由h（x）＞0得x＞4，此时递增，由h（x）＜0得0＜x＜2或2＜x＜4，此时递减，即当x=4时，h（x）取得极小值为h（4）=8，要使f（x）=ax恰有2个互异的实数解，则由图象知4＜a＜8，故答案为：（4，8）19（2018年浙江省高考数学试卷15）（6分）已知λ∈R，函数f（x）=，当λ=2时，不等式f（x）＜0的解集是．若函数f（x）恰有2个零点，则λ的取值范围是．【解答】解：当λ=2时函数f（x）=，显然x≥2时，不等式x﹣4＜0的解集：{x|2≤x＜4}；x＜2时，不等式f（x）＜0化为：x2﹣4x+3＜0，解得1＜x＜2，综上，不等式的解集为：{x|1＜x＜4}．函数f（x）恰有2个零点，函数f（x）=的草图如图：函数f（x）恰有2个零点，则λ∈（1，3]．故答案为：{x|1＜x＜4}；（1，3]．三.解答题1.（2018年高考全国卷I文科21）（12分）已知函数f（x）=ae x﹣lnx﹣1．（1）设x=2是f（x）的极值点，求a，并求f（x）的单调区间；（2）证明：当a≥时，f（x）≥0．【解答】解：（1）∵函数f（x）=ae x﹣lnx﹣1．∴x＞0，f′（x）=ae x﹣，∵x=2是f（x）的极值点，∴f′（2）=ae2﹣=0，解得a=，∴f（x）=e x﹣lnx﹣1，∴f′（x）=，当0＜x＜2时，f′（x）＜0，当x＞2时，f′（x）＞0，∴f（x）在（0，2）单调递减，在（2，+∞）单调递增．证明：（2）当a≥时，f（x）≥﹣lnx﹣1，设g（x）=﹣lnx﹣1，则﹣，当0＜x＜1时，g′（x）＜0，当x＞1时，g′（x）＞0，∴x=1是g（x）的最小值点，故当x＞0时，g（x）≥g（1）=0，∴当a≥时，f（x）≥0．2.（2018年高考全国卷I理科21）（12分）已知函数f（x）=﹣x+alnx．（1）讨论f（x）的单调性；（2）若f（x）存在两个极值点x1，x2，证明：＜a﹣2．【解答】解：（1）函数的定义域为（0，+∞），函数的导数f′（x）=﹣﹣1+=﹣，设g（x）=x2﹣ax+1，当a≤0时，g（x）＞0恒成立，即f′（x）＜0恒成立，此时函数f（x）在（0，+∞）上是减函数，当a＞0时，判别式△=a2﹣4，①当0＜a≤4时，△≤0，即g（x）＞0，即f′（x）＜0恒成立，此时函数f（x）在（0，+∞）上是减函数，②当a＞2时，x，f′（x），f（x）的变化如下表：x（0，）（，）（，+∞）f′（x）﹣0+0﹣f（x）递减递增递减综上当a≤2时，f（x）在（0，+∞）上是减函数，当a＞2时，在（0，），和（，+∞）上是减函数，则（，）上是增函数．（2）由（1）知a＞2，0＜x1＜1＜x2，x1x2=1，则f（x1）﹣f（x2）=（x2﹣x1）（1+）+a（lnx1﹣lnx2）=2（x2﹣x1）+a（lnx1﹣lnx2），则=﹣2+，则问题转为证明＜1即可，即证明lnx1﹣lnx2＞x1﹣x2，即证2lnx1＞x1﹣在（0，1）上恒成立，设h（x）=2lnx﹣x +，（0＜x＜1），其中h（1）=0，求导得h′（x）=﹣1﹣=﹣=﹣＜0，则h（x）在（0，1）上单调递减，∴h（x）＞h（1），即2lnx﹣x +＞0，故2lnx＞x﹣，则＜a﹣2成立．3.（2018年高考全国卷2文科21）（12分）已知函数f（x）=x3﹣a（x2+x+1）．（1）若a=3，求f（x）的单调区间；（2）证明：f（x）只有一个零点．【解答】解：（1）当a=3时，f（x）=x3﹣a（x2+x+1），所以f′（x）=x2﹣6x﹣3时，令f′（x）=0解得x=3，当x∈（﹣∞，3﹣2），x∈（3﹣2，+∞）时，f′（x）＞0，函数是增函数，当x∈（3﹣2时，f′（x）＜0，函数是单调递减，综上，f（x）在（﹣∞，3﹣2），（3﹣2，+∞），上是增函数，在（3﹣2上递减．（2）证明：因为x2+x+1=（x+）2+，所以f（x）=0等价于，令，则，所以g（x）在R上是增函数；取x=max{9a，1}，则有=，取x=min{9a，﹣1}，则有=，所以g（x）在（min{9a，﹣1}，max{9a，1}）上有一个零点，由单调性则可知，f（x）只有一个零点．4.（2018年高考全国卷2理科21）（12分）已知函数f（x）=e x﹣ax2．（1）若a=1，证明：当x≥0时，f（x）≥1；（2）若f（x）在（0，+∞）只有一个零点，求a．【解答】证明：（1）当a=1时，函数f（x）=e x﹣x2．则f′（x）=e x﹣2x，令g（x）=e x﹣2x，则g′（x）=e x﹣2，令g′（x）=0，得x=ln2．当∈（0，ln2）时，h′（x）＜0，当∈（ln2，+∞）时，h′（x）＞0，∴h（x）≥h（ln2）=e ln2﹣2•ln2=2﹣2ln2＞0，∴f（x）在[0，+∞）单调递增，∴f（x）≥f（0）=1，解：（2），f（x）在（0，+∞）只有一个零点⇔方程e x﹣ax2=0在（0，+∞）只有一个根，⇔a=在（0，+∞）只有一个根，即函数y=a与G（x）=的图象在（0，+∞）只有一个交点．G，当x∈（0，2）时，G′（x）＜0，当∈（2，+∞）时，G′（x）＞0，∴G（x）在（0，2）递增，在（2，+∞）递增，当→0时，G（x）→+∞，当→+∞时，G（x）→+∞，∴f（x）在（0，+∞）只有一个零点时，a=G（2）=．5.（2018年高考全国卷3文科21）（12分）已知函数f（x）=．（1）求曲线y=f（x）在点（0，﹣1）处的切线方程；（2）证明：当a≥1时，f（x）+e≥0．【解答】解：（1）=﹣．∴f′（0）=2，即曲线y=f（x）在点（0，﹣1）处的切线斜率k=2，∴曲线y=f（x）在点（0，﹣1）处的切线方程方程为y﹣（﹣1）=2x．即2x﹣y﹣1=0为所求．（2）证明：函数f（x）的定义域为：R，可得=﹣．令f′（x）=0，可得，当x时，f′（x）＜0，x时，f′（x）＞0，x∈（2，+∞）时，f′（x）＜0．∴f（x）在（﹣），（2，+∞）递减，在（﹣，2）递增，注意到a≥1时，函数g（x）=ax2+x﹣1在（2，+∞）单调递增，且g（@）=4a+1＞0函数g（x）的图象如下：∵a≥1，∴，则≥﹣e，∴f（x）≥﹣e，∴当a≥1时，f（x）+e≥0．6.（2018年高考全国卷3理科21）（12分）已知函数f（x）=（2+x+ax2）ln（1+x）﹣2x．（1）若a=0，证明：当﹣1＜x＜0时，f（x）＜0；当x＞0时，f（x）＞0；（2）若x=0是f（x）的极大值点，求a．【解答】（1）证明：当a=0时，f（x）=（2+x）ln（1+x）﹣2x，（x＞﹣1）．，，可得x∈（﹣1，0）时，f″（x）≤0，x∈（0，+∞）时，f″（x）≥0∴f′（x）在（﹣1，0）递减，在（0，+∞）递增，∴f′（x）≥f′（0）=0，∴f（x）=（2+x）ln（1+x）﹣2x在（﹣1，+∞）上单调递增，又f（0）=0．∴当﹣1＜x＜0时，f（x）＜0；当x＞0时，f（x）＞0．（2）解：由f（x）=（2+x+ax2）ln（1+x）﹣2x，得f′（x）=（1+2ax）ln（1+x）+﹣2=，令h（x）=ax2﹣x+（1+2ax）（1+x）ln（x+1），h′（x）=4ax+（4ax+2a+1）ln（x+1）．当a≥0，x＞0时，h′（x）＞0，h（x）单调递增，∴h（x）＞h（0）=0，即f′（x）＞0，∴f（x）在（0，+∞）上单调递增，故x=0不是f（x）的极大值点，不符合题意．当a＜0时，h″（x）=8a+4aln（x+1）+，显然h″（x）单调递减，①令h″（0）=0，解得a=﹣．∴当﹣1＜x＜0时，h″（x）＞0，当x＞0时，h″（x）＜0，∴h′（x）在（﹣1，0）上单调递增，在（0，+∞）上单调递减，∴h′（x）≤h′（0）=0，∴h（x）单调递减，又h（0）=0，∴当﹣1＜x＜0时，h（x）＞0，即f′（x）＞0，当x＞0时，h（x）＜0，即f′（x）＜0，∴f（x）在（﹣1，0）上单调递增，在（0，+∞）上单调递减，∴x=0是f（x）的极大值点，符合题意；②若﹣＜a＜0，则h″（0）=1+6a＞0，h″（e﹣1）=（2a﹣1）（1﹣e）＜0，∴h″（x）=0在（0，+∞）上有唯一一个零点，设为x0，∴当0＜x＜x0时，h″（x）＞0，h′（x）单调递增，∴h′（x）＞h′（0）=0，即f′（x）＞0，∴f（x）在（0，x0）上单调递增，不符合题意；③若a＜﹣，则h″（0）=1+6a＜0，h″（﹣1）=（1﹣2a）e2＞0，∴h″（x）=0在（﹣1，0）上有唯一一个零点，设为x1，∴当x1＜x＜0时，h″（x）＜0，h′（x）单调递减，∴h′（x）＞h′（0）=0，∴h（x）单调递增，∴h（x）＜h（0）=0，即f′（x）＜0，∴f（x）在（x1，0）上单调递减，不符合题意．综上，a=﹣．7.（2018年北京市高考数学试卷文科19)（13分）设函数f（x）=[ax2﹣（3a+1）x+3a+2]e x．（Ⅰ）若曲线y=f（x）在点（2，f（2））处的切线斜率为0，求a；（Ⅱ）若f（x）在x=1处取得极小值，求a的取值范围．【解答】解：（Ⅰ）函数f（x）=[ax2﹣（3a+1）x+3a+2]e x的导数为f′（x）=[ax2﹣（a+1）x+1]e x．曲线y=f（x）在点（2，f（2））处的切线斜率为0，可得（4a﹣2a﹣2+1）e2=0，解得a=；（Ⅱ）f（x）的导数为f′（x）=[ax2﹣（a+1）x+1]e x=（x﹣1）（ax﹣1）e x，若a=0则x＜1时，f′（x）＞0，f（x）递增；x＞1，f′（x）＜0，f（x）递减．x=1处f（x）取得极大值，不符题意；若a＞0，且a=1，则f′（x）=（x﹣1）2e x≥0，f（x）递增，无极值；若a＞1，则＜1，f（x）在（，1）递减；在（1，+∞），（﹣∞，）递增，可得f（x）在x=1处取得极小值；若0＜a＜1，则＞1，f（x）在（1，）递减；在（，+∞），（﹣∞，1）递增，可得f（x）在x=1处取得极大值，不符题意；若a＜0，则＜1，f（x）在（，1）递增；在（1，+∞），（﹣∞，）递减，可得f（x）在x=1处取得极大值，不符题意．综上可得，a的范围是（1，+∞）．8.（2018年北京市高考数学试卷理科18)（13分）设函数f（x）=[ax2﹣（4a+1）x+4a+3]e x．（Ⅰ）若曲线y=f（x）在点（1，f（1））处的切线与x轴平行，求a；（Ⅱ）若f（x）在x=2处取得极小值，求a的取值范围．【解答】解：（Ⅰ）函数f（x）=[ax2﹣（4a+1）x+4a+3]e x的导数为f′（x）=[ax2﹣（2a+1）x+2]e x．由题意可得曲线y=f（x）在点（1，f（1））处的切线斜率为0，可得（a﹣2a﹣1+2）e=0，解得a=1；（Ⅱ）f（x）的导数为f′（x）=[ax2﹣（2a+1）x+2]e x=（x﹣2）（ax﹣1）e x，若a=0则x＜2时，f′（x）＞0，f（x）递增；x＞2，f′（x）＜0，f（x）递减．x=2处f（x）取得极大值，不符题意；若a＞0，且a=，则f′（x）=（x﹣2）2e x≥0，f（x）递增，无极值；若a＞，则＜2，f（x）在（，2）递减；在（2，+∞），（﹣∞，）递增，可得f（x）在x=2处取得极小值；若0＜a＜，则＞2，f（x）在（2，）递减；在（，+∞），（﹣∞，2）递增，可得f（x）在x=2处取得极大值，不符题意；若a＜0，则＜2，f（x）在（，2）递增；在（2，+∞），（﹣∞，）递减，可得f（x）在x=2处取得极大值，不符题意．综上可得，a的范围是（，+∞）．9.（2018年江苏省高考数学试卷19）（16分）记f′（x），g′（x）分别为函数f（x），g（x）的导函数．若存在x0∈R，满足f（x0）=g（x0）且f′（x0）=g′（x0），则称x0为函数f（x）与g （x）的一个“S点”．（1）证明：函数f（x）=x与g（x）=x2+2x﹣2不存在“S点”；（2）若函数f（x）=ax2﹣1与g（x）=lnx存在“S点”，求实数a的值；（3）已知函数f（x）=﹣x2+a，g（x）=．对任意a＞0，判断是否存在b＞0，使函数f（x）与g（x）在区间（0，+∞）内存在“S点”，并说明理由．【解答】解：（1）证明：f′（x）=1，g′（x）=2x+2，则由定义得，得方程无解，则f（x）=x与g（x）=x2+2x﹣2不存在“S点”；（2）f′（x）=2ax，g′（x）=，x＞0，由f′（x）=g′（x）得=2ax，得x=，f（）=﹣=g（）=﹣lna2，得a=；（3）f′（x）=﹣2x，g′（x）=，（x≠0），由f′（x0）=g′（x0），得b=﹣＞0，得0＜x0＜1，由f（x0）=g（x0），得﹣x02+a==﹣，得a=x02﹣，令h（x）=x2﹣﹣a=，（a＞0，0＜x＜1），设m（x）=﹣x3+3x2+ax﹣a，（a＞0，0＜x＜1），则m（0）=﹣a＜0，m（1）=2＞0，得m（0）m（1）＜0，又m（x）的图象在（0，1）上连续不断，则m（x）在（0，1）上有零点，则h（x）在（0，1）上有零点，则f（x）与g（x）在区间（0，+∞）内存在“S”点．10.（2018年天津市高考数学试卷文科20）（14分）设函数f（x）=（x﹣t1）（x﹣t2）（x﹣t3），其中t1，t2，t3∈R，且t1，t2，t3是公差为d的等差数列．（Ⅰ）若t2=0，d=1，求曲线y=f（x）在点（0，f（0））处的切线方程；（Ⅱ）若d=3，求f（x）的极值；（Ⅲ）若曲线y=f（x）与直线y=﹣（x﹣t2）﹣6有三个互异的公共点，求d的取值范围．【解答】解：（Ⅰ）函数f（x）=（x﹣t1）（x﹣t2）（x﹣t3），t2=0，d=1时，f（x）=x（x+1）（x﹣1）=x3﹣x，∴f′（x）=3x2﹣1，f（0）=0，f′（0）=﹣1，∴y=f（x）在点（0，f（0））处的切线方程为y﹣0=﹣1×（x﹣0），即x+y=0；（Ⅱ）d=3时，f（x）=（x﹣t2+3）（x﹣t2）（x﹣t2﹣3）=﹣9（x﹣t2）=x3﹣3t2x2+（3﹣9）x ﹣+9t2；∴f′（x）=3x2﹣6t2x+3﹣9，令f′（x）=0，解得x=t2﹣或x=t2+；当x变化时，f′（x），f（x）的变化情况如下表；x（﹣∞，t2﹣）t2﹣（t2﹣，t2+）t2+（t2+，+∞）f′（x）+0﹣0+f（x）单调增极大值单调减极小值单调增∴f（x）的极大值为f（t2﹣）=﹣9×（﹣）=6，极小值为f（t2+）=﹣9×=﹣6；（Ⅲ）曲线y=f（x）与直线y=﹣（x﹣t2）﹣6有三个互异的公共点，等价于关于x的方程（x﹣t2+d）（x﹣t2）（x﹣t2﹣d）+（x﹣t2）﹣6=0有三个互异的实数根，令u=x﹣t2，可得u3+（1﹣d2）u+6=0；设函数g（x）=x3+（1﹣d2）x+6，则曲线y=f（x）与直线y=﹣（x﹣t2）﹣6有3个互异的公共点，等价于函数y=g（x）有三个不同的零点；又g′（x）=3x2+（1﹣d2），当d2≤1时，g′（x）≥0恒成立，此时g（x）在R上单调递增，不合题意；当d2＞1时，令g′（x）=0，解得x1=﹣，x2=；∴g（x）在（﹣∞，x1）上单调递增，在（x1，x2）上单调递减，在（x2，+∞）上也单调递增；∴g（x）的极大值为g（x1）=g （﹣）=+6＞0；极小值为g（x2）=g （）=﹣+6；若g（x2）≥0，由g（x）的单调性可知，函数g（x）至多有两个零点，不合题意；若g（x2）＜0，即＞27，解得|d|＞，此时|d|＞x2，g（|d|）=|d|+6＞0，且﹣2|d|＜x1；g（﹣2|d|）=﹣6|d|3﹣2|d|+6＜0，从而由g（x）的单调性可知，函数y=g（x）在区间（﹣2|d|，x1），（x1，x2），（x2，|d|）内各有一个零点，符合题意；∴d的取值范围是（﹣∞，﹣）∪（，+∞）．11.（2018年天津市高考数学试卷理科20）（14分）已知函数f（x）=a x，g（x）=log a x，其中a＞1．（Ⅰ）求函数h（x）=f（x）﹣xlna的单调区间；（Ⅱ）若曲线y=f（x）在点（x1，f（x1））处的切线与曲线y=g（x）在点（x2，g（x2））处的切线平行，证明x1+g（x2）=；（Ⅲ）证明当a≥e时，存在直线l，使l是曲线y=f（x）的切线，也是曲线y=g（x）的切线．【解答】（Ⅰ）解：由已知，h（x）=a x﹣xlna，有h′（x）=a x lna﹣lna，令h′（x）=0，解得x=0．由a＞1，可知当x变化时，h′（x），h（x）的变化情况如下表：x（﹣∞，0）0（0，+∞）h′（x）﹣0+h（x）↓极小值↑∴函数h（x）的单调减区间为（﹣∞，0），单调递增区间为（0，+∞）；（Ⅱ）证明：由f′（x）=a x lna，可得曲线y=f（x）在点（x1，f（x1））处的切线的斜率为lna．由g′（x）=，可得曲线y=g（x）在点（x2，g（x2））处的切线的斜率为．∵这两条切线平行，故有，即，两边取以a为底数的对数，得log a x2+x1+2log a lna=0，∴x1+g（x2）=；（Ⅲ）证明：曲线y=f（x）在点（）处的切线l1：，曲线y=g（x）在点（x2，log a x2）处的切线l2：．要证明当a≥时，存在直线l，使l是曲线y=f（x）的切线，也是曲线y=g（x）的切线，只需证明当a≥时，存在x1∈（﹣∞，+∞），x2∈（0，+∞）使得l1与l2重合，即只需证明当a≥时，方程组由①得，代入②得：，③因此，只需证明当a≥时，关于x1的方程③存在实数解．设函数u（x）=，既要证明当a≥时，函数y=u（x）存在零点．u′（x）=1﹣（lna）2xa x，可知x∈（﹣∞，0）时，u′（x）＞0；x∈（0，+∞）时，u′（x）单调递减，又u′（0）=1＞0，u′=＜0，故存在唯一的x0，且x0＞0，使得u′（x0）=0，即．由此可得，u（x）在（﹣∞，x0）上单调递增，在（x0，+∞）上单调递减，u（x）在x=x0处取得极大值u（x0）．∵，故lnlna≥﹣1．∴=．下面证明存在实数t，使得u（t）＜0，由（Ⅰ）可得a x≥1+xlna，当时，有u（x）≤=．∴存在实数t，使得u（t）＜0．因此，当a≥时，存在x1∈（﹣∞，+∞），使得u（x1）=0．∴当a≥时，存在直线l，使l是曲线y=f（x）的切线，也是曲线y=g（x）的切线．12.（2018年浙江省高考数学试卷22）（15分）已知函数f（x）=﹣lnx．（Ⅰ）若f（x）在x=x1，x2（x1≠x2）处导数相等，证明：f（x1）+f（x2）＞8﹣8ln2；（Ⅱ）若a≤3﹣4ln2，证明：对于任意k＞0，直线y=kx+a与曲线y=f（x）有唯一公共点．【解答】证明：（Ⅰ）∵函数f（x）=﹣lnx，∴x＞0，f′（x）=﹣，∵f（x）在x=x1，x2（x1≠x2）处导数相等，∴=﹣，∵x1≠x2，∴+=，由基本不等式得：=≥，∵x1≠x2，∴x1x2＞256，由题意得f（x 1）+f（x2）==﹣ln（x1x2），设g（x）=，则，∴列表讨论：x（0，16）16（16，+∞）g′（x）﹣0+g（x）↓2﹣4ln2↑∴g（x）在[256，+∞）上单调递增，∴g（x1x2）＞g（256）=8﹣8ln2，∴f（x1）+f（x2）＞8﹣8ln2．（Ⅱ）令m=e﹣（|a|+k），n=（）2+1，则f（m）﹣km﹣a＞|a|+k﹣k﹣a≥0，f（n）﹣kn﹣a＜n（﹣﹣k）≤n（﹣k）＜0，∴存在x0∈（m，n），使f（x0）=kx0+a，∴对于任意的a∈R及k∈（0，+∞），直线y=kx+a与曲线y=f（x）有公共点，由f（x）=kx+a，得k=，设h（x）=，则h′（x）==，其中g（x）=﹣lnx，由（1）知g（x）≥g（16），又a≤3﹣4ln2，∴﹣g（x）﹣1+a≤﹣g（16）﹣1+a=﹣3+4ln2+a≤0，∴h′（x）≤0，即函数h（x）在（0，+∞）上单调递减，∴方程f（x）﹣kx﹣a=0至多有一个实根，综上，a≤3﹣4ln2时，对于任意k＞0，直线y=kx+a与曲线y=f（x）有唯一公共点．312018高考数学汇编函数第页共页。

2018年高考真题汇编(函数与导数)

2018年高考真题汇编(函数与导数)函数与导数1．【2018年浙江卷】函数y=sin2x的图象可能是A. B.C. D.【答案】D点睛：有关函数图象的识别问题的常见题型及解题思路：（1）由函数的定义域，判断图象的左、右位置，由函数的值域，判断图象的上、下位置；（2）由函数的单调性，判断图象的变化趋势；（3）由函数的奇偶性，判断图象的对称性；（4）由函数的周期性，判断图象的循环往复．2．【2018年理天津卷】已知，，，则a，b，c的大小关系为A. B. C. D.【答案】D【解析】分析：由题意结合对数函数的性质整理计算即可求得最终结果.详解：由题意结合对数函数的性质可知：，，，据此可得：.本题选择D选项.点睛：对于指数幂的大小的比较，我们通常都是运用指数函数的单调性，但很多时候，因幂的底数或指数不相同，不能直接利用函数的单调性进行比较．这就必须掌握一些特殊方法．在进行指数幂的大小比较时，若底数不同，则首先考虑将其转化成同底数，然后再根据指数函数的单调性进行判断．对于不同底而同指数的指数幂的大小的比较，利用图象法求解，既快捷，又准确．3．【2018年理新课标I卷】已知函数．若g（x）存在2个零点，则a的取值范围是A. [–1，0） B. [0，+∞） C. [–1，+∞） D. [1，+∞）【答案】C详解：画出函数的图像，在y轴右侧的去掉，再画出直线，之后上下移动，可以发现当直线过点A时，直线与函数图像有两个交点，并且向下可以无限移动，都可以保证直线与函数的图像有两个交点，即方程有两个解，也就是函数有两个零点，此时满足，即，故选C.点睛：该题考查的是有关已知函数零点个数求有关参数的取值范围问题，在求解的过程中，解题的思路是将函数零点个数问题转化为方程解的个数问题，将式子移项变形，转化为两条曲线交点的问题，画出函数的图像以及相应的直线，在直线移动的过程中，利用数形结合思想，求得相应的结果.4．【2018年理新课标I卷】设函数，若为奇函数，则曲线在点处的切线方程为A. B. C. D.【答案】D点睛：该题考查的是有关曲线在某个点处的切线方程的问题，在求解的过程中，首先需要确定函数解析式，此时利用到结论多项式函数中，奇函数不存在偶次项，偶函数不存在奇次项，从而求得相应的参数值，之后利用求导公式求得，借助于导数的几何意义，结合直线方程的点斜式求得结果.5．【2018年全国卷Ⅲ理】设，，则A. B. C. D.【答案】B【解析】分析：求出，得到的范围，进而可得结果。

2018年高考数学分类汇编三角函数及答案详解

2018年高考数学分类汇编三角函数1、（2018年高考全国卷1理科）16．（5分）已知函数f（x）=2sinx+sin2x，则f （x）的最小值是．【解答】解：由题意可得T=2π是f（x）=2sinx+sin2x的一个周期，故只需考虑f（x）=2sinx+sin2x在[0，2π）上的值域，先来求该函数在[0，2π）上的极值点，求导数可得f′（x）=2cosx+2cos2x=2cosx+2（2cos2x﹣1）=2（2cosx﹣1）（cosx+1），令f′（x）=0可解得cosx=或cosx=﹣1，可得此时x=，π或；∴y=2sinx+sin2x的最小值只能在点x=，π或和边界点x=0中取到，计算可得f（）=，f（π）=0，f（）=﹣，f（0）=0，∴函数的最小值为﹣，故答案为：．2、（2018年高考全国卷1理科）17．（12分）在平面四边形ABCD中，∠ADC=90°，∠A=45°，AB=2，BD=5．（1）求cos∠ADB；（2）若DC=2，求BC．【解答】解：（1）∵∠ADC=90°，∠A=45°，AB=2，BD=5．∴由正弦定理得：=，即=，∴sin∠ADB==，∵AB＜BD，∴∠ADB＜∠A，∴cos∠ADB==．（2）∵∠ADC=90°，∴cos∠BDC=sin∠ADB=，∵DC=2，∴BC===5．3、（2018年高考全国卷1文科）8．（5分）已知函数f（x）=2cos2x﹣sin2x+2，则（）A．f（x）的最小正周期为π，最大值为3B．f（x）的最小正周期为π，最大值为4C．f（x）的最小正周期为2π，最大值为3D．f（x）的最小正周期为2π，最大值为4【解答】解：函数f（x）=2cos2x﹣sin2x+2，=2cos2x﹣sin2x+2sin2x+2cos2x，=4cos2x+sin2x，=3cos2x+1，=，=，故函数的最小正周期为π，函数的最大值为，故选：B．4、（2018年高考全国卷1文科）11．（5分）已知角α的顶点为坐标原点，始边与x轴的非负半轴重合，终边上有两点A（1，a），B（2，b），且cos2α=，则|a﹣b|=（）A．B．C．D．1【解答】解：∵角α的顶点为坐标原点，始边与x轴的非负半轴重合，终边上有两点A（1，a），B（2，b），且cos2α=，∴cos2α=2cos2α﹣1=，解得cos2α=，∴|cosα|=，∴|sinα|==，|tanα|=||=|a﹣b|===．故选：B．5、（2018年高考全国卷1文科）16．（5分）△ABC的内角A，B，C的对边分别为a，b，c．已知bsinC+csinB=4asinBsinC，b2+c2﹣a2=8，则△ABC的面积为．【解答】解：△ABC的内角A，B，C的对边分别为a，b，c．bsinC+csinB=4asinBsinC，利用正弦定理可得sinBsinC+sinCsinB=4sinAsinBsinC，由于sinBsinC≠0，所以sinA=，则A=由于b2+c2﹣a2=8，则：，①当A=时，，解得：bc=，所以：．②当A=时，，解得：bc=﹣（不合题意），舍去．故：．故答案为：．6、（2018年高考全国卷2理科）6．（5分）在△ABC中，cos=，BC=1，AC=5，则AB=（）A．4B．C．D．2【解答】解：在△ABC中，cos=，cosC=2×=﹣，BC=1，AC=5，则AB====4．故选：A．7、（2018年高考全国卷2理科）10．（5分）若f（x）=cosx﹣sinx在[﹣a，a]是减函数，则a的最大值是（）A．B．C．D．π【解答】解：f（x）=cosx﹣sinx=﹣（sinx﹣cosx）=，由，k∈Z，得，k∈Z，取k=0，得f（x）的一个减区间为[，]，由f（x）在[﹣a，a]是减函数，得，∴．则a的最大值是．故选：A．8、（2018年高考全国卷2理科）15．（5分）已知sinα+cosβ=l，cosα+sinβ=0，则sin（α+β）=．【解答】解：sinα+cosβ=l，两边平方可得：sin2α+2sinαcosβ+cos2β=1，①，cosα+sinβ=0，两边平方可得：cos2α+2cosαsinβ+sin2β=0，②，由①+②得：2+2（sinαcosβ+cosαsinβ）=1，即2+2sin（α+β）=1，∴2sin（α+β）=﹣1．∴sin（α+β）=．故答案为：．9、（2018年高考全国卷2文科）7．（5分）在△ABC中，cos=，BC=1，AC=5，则AB=（）A．4B．C．D．2【解答】解：在△ABC中，cos=，cosC=2×=﹣，BC=1，AC=5，则AB====4．故选：A．10、（2018年高考全国卷2文科）10．（5分）若f（x）=cosx﹣sinx在[0，a]是减函数，则a的最大值是（）A．B．C．D．π【解答】解：f（x）=cosx﹣sinx=﹣（sinx﹣cosx）=﹣sin（x﹣），由﹣+2kπ≤x﹣≤+2kπ，k∈Z，得﹣+2kπ≤x≤+2kπ，k∈Z，取k=0，得f（x）的一个减区间为[﹣，]，由f（x）在[0，a]是减函数，得a≤．则a的最大值是．故选：C11、（2018年高考全国卷2文科）15．（5分）已知tan（α﹣）=，则tanα=．【解答】解：∵tan（α﹣）=，∴tan（α）=，则tanα=tan（α+）=====，故答案为：．12、（2018年高考全国卷3理科）4．（5分）若sinα=，则cos2α=（）A．B．C．﹣D．﹣【解答】解：∵sinα=，∴cos2α=1﹣2sin2α=1﹣2×=．故选：B．13、（2018年高考全国卷3理科）9．（5分）△ABC的内角A，B，C的对边分别为a，b，c．若△ABC的面积为，则C=（）A．B．C．D．【解答】解：∵△ABC的内角A，B，C的对边分别为a，b，c．△ABC的面积为，==，∴S△ABC∴sinC==cosC，∵0＜C＜π，∴C=．故选：C．14、（2018年高考全国卷3理科）15．（5分）函数f（x）=cos（3x+）在[0，π]的零点个数为3．【解答】解：∵f（x）=cos（3x+）=0，∴3x+=+kπ，k∈Z，∴x=+kπ，k∈Z，当k=0时，x=，当k=1时，x=π，当k=2时，x=π，当k=3时，x=π，∵x∈[0，π]，∴x=，或x=π，或x=π，故零点的个数为3，故答案为：315、（2018年高考全国卷3文科）4．（5分）若sinα=，则cos2α=（）A．B．C．﹣D．﹣【解答】解：∵sinα=，∴cos2α=1﹣2sin2α=1﹣2×=．故选：B．16、（2018年高考全国卷3文科）6．（5分）函数f（x）=的最小正周期为（）A．B．C．πD．2π【解答】解：函数f（x）===sin2x的最小正周期为=π，故选：C．17、（2018年高考全国卷3文科）11．（5分）△ABC的内角A，B，C的对边分别为a，b，c．若△ABC的面积为，则C=（）A．B．C．D．【解答】解：∵△ABC的内角A，B，C的对边分别为a，b，c．△ABC的面积为，∴S==，△ABC∴sinC==cosC，∵0＜C＜π，∴C=．故选：C．18、（2018年高考北京卷理科）15．（13分）在△ABC中，a=7，b=8，cosB=﹣．（Ⅰ）求∠A；（Ⅱ）求AC边上的高．【解答】解：（Ⅰ）∵a＜b，∴A＜B，即A是锐角，∵cosB=﹣，∴sinB===，由正弦定理得=得sinA===，则A=．（Ⅱ）由余弦定理得b2=a2+c2﹣2accosB，即64=49+c2+2×7×c×，即c2+2c﹣15=0，得（c﹣3）（c+5）=0，得c=3或c=﹣5（舍），则AC边上的高h=csinA=3×=．19、（2018年高考北京卷理科）7．（5分）在平面直角坐标系中，记d为点P（cosθ，sinθ）到直线x﹣my﹣2=0的距离．当θ、m变化时，d的最大值为（）A．1 B．2 C．3 D．4【解答】解：由题意d==，tanα=﹣，∴当sin（θ+α）=﹣1时，d max=1+≤3．∴d的最大值为3．故选：C．20、（2018年高考北京卷理科）11．（5分）设函数f（x）=cos（ωx﹣）（ω＞0），若f（x）≤f（）对任意的实数x都成立，则ω的最小值为．【解答】解：函数f（x）=cos（ωx﹣）（ω＞0），若f（x）≤f（）对任意的实数x都成立，可得：，k∈Z，解得ω=，k∈Z，ω＞0则ω的最小值为：．故答案为：．21、（2018年高考北京卷文科）7．（5分）在平面直角坐标系中，，，，是圆x2+y2=1上的四段弧（如图），点P其中一段上，角α以Ox为始边，OP 为终边．若tanα＜cosα＜sinα，则P所在的圆弧是（）A．B．C．D．【解答】解：A．在AB段，正弦线小于余弦线，即cosα＜sinα不成立，故A不满足条件．B．在CD段正切线最大，则cosα＜sinα＜tanα，故B不满足条件．C．在EF段，正切线，余弦线为负值，正弦线为正，满足tanα＜cosα＜sinα，D．在GH段，正切线为正值，正弦线和余弦线为负值，满足cosα＜sinα＜tanα不满足tanα＜cosα＜sinα．故选：C．22、（2018年高考北京卷文科）14．（5分）若△ABC的面积为（a2+c2﹣b2），且∠C为钝角，则∠B=；的取值范围是（2，+∞）．【解答】解：△ABC的面积为（a2+c2﹣b2），可得：（a2+c2﹣b2）=acsinB，，可得：tanB=，所以B=，∠C为钝角，A∈（0，），cotA∈（，+∞）．===cosB+cotAsinB=cotA∈（2，+∞）．故答案为：；（2，+∞）．23、（2018年高考北京卷文科）16．（13分）已知函数f（x）=sin2x+sinxcosx．（Ⅰ）求f（x）的最小正周期；（Ⅱ）若f（x）在区间[﹣，m]上的最大值为，求m的最小值．【解答】解：（I）函数f（x）=sin2x+sinxcosx=+sin2x=sin（2x﹣）+，f（x）的最小正周期为T==π；（Ⅱ）若f（x）在区间[﹣，m]上的最大值为，可得2x﹣∈[﹣，2m﹣]，即有2m﹣≥，解得m≥，则m的最小值为．24、（2018年高考天津卷理科）6．（5分）将函数y=sin（2x+）的图象向右平移个单位长度，所得图象对应的函数（）A．在区间[，]上单调递增B．在区间[，π]上单调递减C．在区间[，]上单调递增D．在区间[，2π]上单调递减【解答】解：将函数y=sin（2x+）的图象向右平移个单位长度，得到的函数为：y=sin2x，增区间满足：﹣+2kπ≤2x≤，k∈Z，减区间满足：≤2x≤，k∈Z，∴增区间为[﹣+kπ，+kπ]，k∈Z，减区间为[+kπ，+kπ]，k∈Z，∴将函数y=sin（2x+）的图象向右平移个单位长度，所得图象对应的函数在区间[，]上单调递增．故选：A．25、（2018年高考天津卷理科）15．（13分）在△ABC中，内角A，B，C所对的边分别为a，b，c．已知bsinA=acos（B﹣）．（Ⅰ）求角B的大小；（Ⅱ）设a=2，c=3，求b和sin（2A﹣B）的值．【解答】解：（Ⅰ）在△ABC中，由正弦定理得，得bsinA=asinB，又bsinA=acos（B﹣）．∴asinB=acos（B﹣），即sinB=cos（B﹣）=cosBcos+sinBsin=cosB+，∴tanB=，又B∈（0，π），∴B=．（Ⅱ）在△ABC中，a=2，c=3，B=，由余弦定理得b==，由bsinA=acos（B﹣），得sinA=，∵a＜c，∴cosA=，∴sin2A=2sinAcosA=，cos2A=2cos2A﹣1=，∴sin（2A﹣B）=sin2AcosB﹣cos2AsinB==．26、（2018年高考天津卷文科）6．（5分）将函数y=sin（2x+）的图象向右平移个单位长度，所得图象对应的函数（）A．在区间[]上单调递增B．在区间[﹣，0]上单调递减C．在区间[]上单调递增D．在区间[，π]上单调递减【解答】解：将函数y=sin（2x+）的图象向右平移个单位长度，所得图象对应的函数解析式为y=sin[2（x﹣）+]=sin2x．当x∈[]时，2x∈[，]，函数单调递增；当x∈[，]时，2x∈[，π]，函数单调递减；当x∈[﹣，0]时，2x∈[﹣，0]，函数单调递增；当x∈[，π]时，2x∈[π，2π]，函数先减后增．故选：A．27、（2018年高考天津卷文科）16．（13分）在△ABC中，内角A，B，C所对的边分别为a，b，c．已知bsinA=acos（B﹣）．（Ⅰ）求角B的大小；（Ⅱ）设a=2，c=3，求b和sin（2A﹣B）的值．【解答】解：（Ⅰ）在△ABC中，由正弦定理得，得bsinA=asinB，又bsinA=acos（B﹣）．∴asinB=acos（B﹣），即sinB=cos（B﹣）=cosBcos+sinBsin=cosB+，∴tanB=，又B∈（0，π），∴B=．（Ⅱ）在△ABC中，a=2，c=3，B=，由余弦定理得b==，由bsinA=acos（B﹣），得sinA=，∵a＜c，∴cosA=，∴sin2A=2sinAcosA=，cos2A=2cos2A﹣1=，∴sin（2A﹣B）=sin2AcosB﹣cos2AsinB==．24、（2018年高考浙江卷）18．（14分）已知角α的顶点与原点O重合，始边与x轴的非负半轴重合，它的终边过点P（﹣，﹣）．（Ⅰ）求sin（α+π）的值；（Ⅱ）若角β满足sin（α+β）=，求cosβ的值．【解答】解：（Ⅰ）∵角α的顶点与原点O重合，始边与x轴非负半轴重合，终边过点P（﹣，﹣）．∴x=﹣，y=，r=|OP|=，∴sin（α+π）=﹣sinα=；（Ⅱ）由x=﹣，y=，r=|OP|=1，得，，又由sin（α+β）=，得=，则cosβ=cos[（α+β）﹣α]=cos（α+β）cosα+sin（α+β）sinα=，或cosβ=cos[（α+β）﹣α]=cos（α+β）cosα+sin（α+β）sinα=．∴cosβ的值为或．。

2018年高考试题分类汇编(三角函数)

2018年高考试题分类汇编(三角函数) 2018年高考试题分类汇编（三角函数）考点1：任意角的三角函数考法1：三角函数的定义已知角$\alpha$的顶点与坐标原点重合，始边与$x$轴的非负半轴重合，终边上两点$A(1,a)$，$B(2,b)$，且$\cos2\alpha=\frac{1}{3}$，则$a-b=5\sqrt{3}$。

考法2：三角函数的图像与性质1.（2018·全国卷Ⅲ理）函数$f(x)=\cos(3x+\frac{\pi}{6})$在$[0,\pi]$的零点的个数为6.2.（2018·江苏）已知函数$y=\sin(2x+\varphi)$，($-\frac{\pi}{2}<\varphi<\frac{\pi}{2}$)关于直线$x=\frac{\pi}{4}$对称，则$\varphi$的值是$-\frac{\pi}{4}$。

3.（2018·天津文科）将函数$y=\sin(2x+\frac{\pi}{3})$的图象向右平移$\frac{1}{2}$个单位长度，所得图象关于直线$x=\frac{\pi}{3}$对称的图象对应的函数为$y=\sin(2x+\frac{5\pi}{6})$。

4.（2018·天津理科）将函数$y=\sin(2x+\frac{\pi}{3})$的图象向右平移$\frac{1}{2}$个单位长度，所得图象对应的函数在区间$[\frac{3\pi}{4},2\pi]$上单调递减。

5.（2018·北京理科）设函数$f(x)=\cos(\omega x-\frac{\pi}{4})$，若$f(x)\leq f(\frac{\pi}{4})$对任意的实数$x$都成立，则$\omega$的最小值为$\frac{2}{\pi}$。

6.（2018·全国卷Ⅱ文科）若函数$f(x)=\cos x-\sin x$在$[0,a]$是减函数，则$a$的最大值为$\frac{3\pi}{4}$。

2018年高考真题汇编--函数

2018年高考真题汇编--函数一、单选题1.（2018•卷Ⅰ）设函数,则满足f(x+1)<f(2x)的x的取值范围是（）A. (-∞,-1]B. (0,+∞)C. (-1,0)D. (-∞,0)2.（2018•卷Ⅰ）已知函数，.若存在2个零点，则a的取值范围是( )A. B. C. D.3.（2018•卷Ⅱ）已知是定义域为的奇函数，满足。

若，则（）A.-50B.0C.2D.504.（2018•卷Ⅱ）函数的图像大致为( )A. B. C. D.5.（2018•卷Ⅲ）函数的图像大致为( )A. B.C. D.6.（2018•卷Ⅲ）下列函数中，其图像与函数的图像关于直线对称的是（）A. B. C. D.7.（2018•卷Ⅲ）设，，则（）A. B. C. D.8.（2018•天津）已知，，，则a，b，c的大小关系为（）A. B. C. D.9.（2018•卷Ⅰ）设函数，若为奇函数，则曲线y=f(x)在点（0,0）处的切线方程为（）A.y=-2xB.y=-xC.y=2xD.y=x二、填空题（共14题；共15分）10.（2018•卷Ⅰ）已知函数f(x)=log2(x2+a).若f(3)=1,则a=________.11.（2018•卷Ⅲ）已知函数，，则________。

12.（2018•天津）已知a，b∈R，且a–3b+6=0，则2a+ 的最小值为________．13.（2018•天津）已知a∈R，函数若对任意x∈［–3，+ ），f(x)≤ 恒成立，则a的取值范围是________．14.（2018•天津）已知，函数若关于的方程恰有2个互异的实数解，则的取值范围是________.15.（2018•上海）已知，若幂函数为奇函数，且在上递减，则α=________16.（2018•上海）设常数，函数，若的反函数的图像经过点，则a=________。

17.（2018•浙江）已知λ∈R，函数f(x)= ，当λ=2时，不等式f(x)<0的解集是________．若函数f(x)恰有2个零点，则λ的取值范围是________．18.（2018•江苏）函数的定义域为________.19.（2018•卷Ⅲ）曲线在点处的切线的斜率为，则________．20.（2018•卷Ⅱ）曲线在点处的切线方程为________.21.（2018•卷Ⅱ）曲线在点处的切线方程为________.22.（2018•天津）已知函数f(x)=e x ln x，f ′(x)为f(x)的导函数，则f ′（1）的值为________．23.（2018•江苏）若函数在内有且只有一个零点，则在上的最大值与最小值的和为________三、解答题（共8题；共70分）24.（2018•卷Ⅰ）已知函数（1）讨论的单调性；（2）若存在两个极值点，证明：25.（2018•卷Ⅰ）已知函数f(x)=ae x-lnx-1（1）设x=2是f(x)的极值点,求a,并求f(x)的单调区间（2）证明:当a≥ 时,f(x)≥026.（2018•卷Ⅱ）已知函数（1）若a=1,证明：当时，（2）若在只有一个零点，求.27.（2018•卷Ⅱ）已知函数（1）若a=3，求的单调区间（2）证明：只有一个零点28.（2018•卷Ⅲ）已知函数（1）求函数在点处的切线方程（2）证明:当时，29.（2018•卷Ⅲ）已知函数．（1）若，证明：当时，；当时，；（2）若是的极大值点，求．30.（2018•北京）设函数=[ -(4a+1)x+4a+3] .（I）若曲线y= f（x）在点(1, )处的切线与X轴平行,求a：(II)若在x=2处取得极小值，求a的取值范围。

2018年高考数学—函数(解答+答案)

3
8.（18 北京文（19）（本小题 13 分））
设函数 f (x) [ax2 (3a 1)x 3a 2]ex . （Ⅰ）若曲线 y f (x) 在点 (2, f (2)) 处的切线斜率为 0，求 a；（Ⅱ）若 f (x) 在 x 1处取得极小值，求 a 的取值范围.
9.（18 全国二文 21．（12 分））
当 1 x 0 时，g(x) 0 ；当 x 0 时，g(x) 0 .故当 x 1 时，g(x) g(0) 0 ，
且仅当 x 0 时， g(x) 0 ，从而 f (x) 0 ，且仅当 x 0 时， f (x) 0 .
所以 f (x) 在 (1, ) 单调递增.学#科网
又 f (0) 0 ，故当 1 x 0 时， f (x) 0；当 x 0 时， f (x) 0 .
f (x) 在 (0, ) 只有一个零点当且仅当 h(x) 在 (0, ) 只有一个零点．
（i）当 a 0 时， h(x) 0 ， h(x) 没有零点；（ii）当 a 0 时， h'(x) ax(x 2)ex ．
当 x (0, 2) 时， h'(x) 0 ；当 x (2, ) 时， h'(x) 0 ．
7
综上， f (x) 在 (0, ) 只有一个零点时， a e2 ． 4
4.解：（1）当 a 0 时， f (x) (2 x) ln(1 x) 2x ， f (x) ln(1 x) x . 1 x
设函数 g(x)
f
(x)
ln(1 x) x 1 x
，则 g(x)
x (1 x)2
.
（1）若 a 0 ，证明：当 1 x 0时， f x 0 ；当 x 0 时， f x 0 ；（2）若 x 0 是 f x 的极大值点，求 a ．