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高二2017-2018学年度期中测试题

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昌黎汇文二中2017-2018学年度第一学期期中考试高二英语命题人：白艳第一卷（选择题）第一部分：听力（共两节，满分30分）第一节（每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳答案，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What do we know about the woman?A. She is richB. She is looking for a jobC. She is out of condition2. What will the woman do?A. Buy a carB. Take a testC. Have a driving lesson3. Where will the speakers go?A. A restaurantB. A libraryC. A park4. What does the man want to do?A. Book a room for a partyB. Hold a birthday party for his childC. Have a meeting with two adults5. Why is the man worried?A. He did poorly in the final examB. He’s not prepared for the coming examC. He has to hand in his homework in three days第二节（每小题1.5分，满分22.5分）听下面5段对话或独白。

每段对话后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听每段对话前，你将有时间阅读各个小题，每小题5秒钟；听完后，各个小题将给出5秒钟的作答时间。

每段对话读两遍。

听下面一段对话，回答第6和第7两个小题。

6. What does the woman like about the first skirt?A Its material B. Its color C. Its design7. What does the woman buy at last?A. A pale yellow skirt and a red coat.B. A pale yellow coat and a white skirt.C. A white shirt and a pale yellow coat. 听下面一段对话，回答第8和第9两个小题。

8. When did the woman learn a lesson?A. When she attended a concertB. When she went to an openingC. When she had a test9. What does the man want to drink?A. A large Coke.B. A medium CokeC. A small Coke听下面一段对话，回答第10至12三个小题。

10. What was the weather like on Monday?A. SunnyB. WindyC. Rainy.11. What did the man do on Thursday?A. He went boatingB. He went swimmingC. He went shopping12. How did the man feel on Friday?A. TiredB. Excited.C. Disappointed听下面一段对话，回答第13至16四个小题。

13. When should the man leave the dorm?A. On MondayB. On TuesdayC. On Thursday14. Why is the man calling?A. To make a complaint.B. To cancel the courseC. To find a place to live in15. How will the woman help the man?A. Look for a private house for himB. Give him a list of private housesC. Find another dorm for him16. What can we learn from the conversation?A. The man has been in the dorm for 10 weeksB. The man began the course at the end of JulyC. The woman will call the man several days later听下面一段独白，回答第17至20四个小题。

17. Wh ere will this year’s Motor Show be held?A．On the Park School FieldB. On the Three Tree Show GroundC. On the Park School Show Ground18. What troubled visitors in the Motor Show last year?A. Parking carsB. Buying ticketsC. Finding the place19. How much is it for two adults and two children?A. £.18B. £ 24C. £ 3020. What does the speaker advise people to do?A. Take cameras with themB. Buy a family ticketC. Get tickets in advance第二部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C、D）中，选出最佳选项。

AEvery year, TIME selects the best inventions that are making the world better, smarter and—in some cases—a little more fun. Here are the ones on last year’s list.The Hello SenseIt’s hard to believe that an alarm clock, called Hello Sense, can not only be beautiful but also improve your sleep. It can control the temperature, light and even air quality in your bedroom to help you have a perfect sleep environment. It can wa ke you up when you’re in the lightest part of your sleep cycle and ready to wake. You can also speak to Sense to find out how well you slept.The Folding Bike Helmet (头盔)People refuse to wear a helmet because it’s thick and heavy. The Morpher, a bike helmet made from interweaved plastics is just as strong as its traditional ones, but flexible enough to fold almost totally flat, making it easier to carry.Shoes That Tie ThemselvesAlmost everyone who sees the movie Back to the Future wants a pair of shoes that tie themselves. Now, thanks to Nike, the shoe dream is a reality. When you press a button near the tongue, the sneakers themselves tighten and loosen around your feet. Simplified shoe fastening could be helpful for the athletes during the competition, and it’s especially useful for people with the impaired (损坏的)motor nerve.Sweet Potatoes That Could Save LivesIn sub-Saharan Africa, the lack of vitamin (维他命) A puts children at risk of blindness, and other diseases. Plant scientists are helping some countries grow their own solutions in the form of sweet potatoes. And they fight against droughts. Last year, the orange-colored sweet potatoes won the World Food Prize and have been seen as a healthy crop for all.21.What do we know about Hello Sense?A.It shows you ways of sleeping wellB.It can serve as a light in your bedroom.C.It tries its best to improve your quality of sleep.D.It can wake you up at any time every day.22.Who are more suitable to wear shoes that tie themselves?A.Ones who enjoy seeing Back to the Future.B.Ones who practice running.C.Ones who are fans of Nike.D.Ones who have trouble with the motor nerve23.Which column is the passage probably taken from?A. historyB. travelC. scienceD. cultureBLast week, I went to Montpellier in the south of France for an exam. So this trip was for business, not for pleasure. However, I was looking forward to seeing some of my friends, who studied in Paul Valery University in Montpellier.I was quite happy to go to Montpellier because it was a part of France that I hadn’tseen before. Upon arriving there, I was greeted at the train station by one of my friends. We decided to go for a walk to Place de la Comedie before heading to our exam. Then, we received a text message from our lecturer, saying that the exam had been cancelled. I was angry because I had travelled three hours from Toulouse for no purpose.After venting our anger, my friend and I decided to go to the nearest shopping centre--- the Odysseum. I enjoyed my time there, especially since my friends took me to the aquarium (水族馆) there, and I picked up even more French words for sea creatures like asea lion. So my journey to Montpellier wasn’t totally without purpose. After this, we went to Pizza Pino for lunch in the Odysseum where food and good service made me impressed.I asked my friends to show me some of the sites, like Place de la Comedie. My favoritesite of the whole trip must have been the Arc de Triomphe, because behind the Arc de Triomphe is a park area with an area to stand. From here, my friends and I watched the sunset before I had to go back to Toulouse.In a word, my trip wasn’t a total w aste. I had a great day in Montpellier with my friends and learnt some new words in French.24. The author went to Montpellier to __________.A. take part in an examB. get together with his friendsC. enjoy places of French interestD. study further in Paul Valery University25. What did the author like best in Montpellier?A. Having a walk to Place de la ComedieB. Going shopping in the OdysseumC. Delicious food offered by Pizza PinoD. Enjoying the sunset from the Arc de Triomphe26. Why did the author think the journey to Montpellier wasn’t totally withoutpurpose?A. The author found real pleasure in MontpellierB. The author learned French from his friendsC. The author could avoid stress from the examD. The author didn’t expect t he good service he received27. What can we infer from the author’s trip to Montpellier?A. The trip to Montpellier is worthwhileB. We should make good preparations in advanceC. Trips are sometimes a waste of time and moneyD. Trips and culture are hand in handC．Anyone who has had a long-term illness knows that recovering at home can be lonely. This is especially true of children, who may feel cut from their friends and classmates.Now, these children may have a high-tech friend to help feel less lonely. A robot called A V1 goes to school for a child who is homebound while he is recovering from a long-term illness. And the child’s school friends help carry the robot between classes and place the robot on the child’s desk.A company called No Isolation created the robot. The founder Dolva explains how the robot A V1 works. She says from home, the child uses a tablet or a phone to start the robot. Then they use the same device to control the robot’s movements.At school, the robot sits on the child’s desk in the classroom and the child uses a tablet or a phone to start it, control its movements with touch, and talk through it. So it’s the eyes and the ears and the voice at school. The student can take part in classroom activities from wherever they are recovering. The robot is equipped with speakers, microphones and cameras that make communicating easy.A V1 is large and looks like a human for a reason. Dolva says this is important because the robot is supposed to be a friend to the children. It can’t be just a tiny camera because the other kids can pick it up and take it with them. And robots are, quite simply, cool. The robot just became available to the public. Hopefully A V1 will help some children feel less lonely while they are absent from classes.28. When can the robot A V1 be brought to the classroom?A. When a child feels lonely and cut off at school.B When the teacher is ill and can’t attend classesC. When a child is sick at home or in hospitalD. When a child wants to help with his lessons29. What will the robot A V1 do in class?A. Join in class activities through the control of the childB. Record the lessons for the child to learn back at home.C. Present recorded information about the child’s illnessD. Record daily activities of the classmates for the child30. Why is the robot A V1 designed to be large and look like a human?A. To make the child less lonely at home or in hospitalB. To make sure that it won’t break down easily in classC. To make it store more school information for the childD. To let the child’s classmates feel it’s their friend.31. Which of the following is closest in meaning to the underlined word “homebound”?A. Cut away from the worldB. Forced to stay indoorsC. Having the family responsibilityD. Losing control of everythingDYoga is a way to bring strength and calm to a person’s mind and body. People use breathing to try to calm and control their minds. Laughter Yoga comes from these same ideas. It uses breathing exercise and laughter to strengthen the mind and body. For example, people doing Laughter Yoga perform simple laughter exercises together in a group. While they are laughing, they concentrate on breathing in a particular way. They also move and stretch their bodies.Doctor Madan Kataria in Mumbai, India invented Laughter Yoga. Through his research of laughter he found people did not know the difference between forced or false laughter and real laughter. Forced laughter can lead to a feeling of happiness. And then Kataria had the idea for a group of people who would laugh together.Kataria gathered a few of his friends in a public park in 1995 in Mumbai. His friends formed the first Laughter Club. That small group grew and grew! And today there are thousands of laughter yoga clubs all over the world.Laughter Club meetings are similar all around the world. People in a club usually perform a series of exercises, including body movement, correct breathing, and of course, laughter! At the beginning of the meeting, people may have to force themselves to laugh. But in the end, everyone is usually laughing in a real way.Kataria claims that laughter is actually a very good form of exercise. It helps to improve the immune system(免疫系统)，lower blood pressure and help lift depression.Laughter Yoga is good for everyone because everyone can laugh. Experts explain that laughter crosses all barriers. It does not matter how old you are, where you are from, what language you speak, or what abilities you have. Everyone likes to laugh.32. What do Yoga and Laughter Yoga have in common?A. They use physical exercise to make the mind and body strong.B. They focus on the importance of laughing in our daily lifeC. They require people to form groups to do exercise.D. They make people feel what a happy life they live33. What made Madan Kataria come up with Laughter Club?A. His friends’ requirementsB. His research about laughterC. His feeling of happinessD. His wonder of real laughter34. According to the experts, we can infer that _________.A. they approve of Laughter Yoga very muchB. they advise everyone to laugh as much as possibleC. they think everyone should be treated equallyD. they deal with their problems through laughing35. What’s the main idea of the passage?A. Tell the meaning of Laughter in our lifeB. Compare Yoga with Laughter YogaC. Show the exercise ---- Laughter YogaD. How to keep us happy all the time第二节（共5小题；每小题2分,满分10分）根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。

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福建省厦门第一中学2017-2018学年高二上学期期中考试数学理试题

福建省厦门第一中学2017-2018学年高二上学期期中考试数学理试题

福建省厦门第一中学2017—2018学年度上期中考高二年理科数学试卷满分为150分，考试时间120分钟．一、选择题：本大题共12个小题，每小题5分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的．在答题卷上的相应题目的答题区域内作答．1．如果0a b <<，那么下列各式一定成立的是 （ ）A ．0>-b aB ．bc ac <C ．22b a >D ．ba 11<2．已知命题p ：“若ab ＝1，则a ＋b ≥2”，则下列说法正确的是 ( ) A ．命题p 的逆命题是“若ab ≠1，则a ＋b <2” B ．命题p 的逆命题是“若a ＋b <2，则ab ≠1” C ．命题p 的否命题是“若ab ≠1，则a ＋b <2” D ．命题p 的否命题是“若a ＋b ≥2，则ab ＝1”3.已知数列{}n a 满足：11112n n a a ++=+，且22a =，则4a 等于 （ ）A. B. 11 C. 12 D. 234. {}n a 是公差不为0的等差数列，满足22224567a a a a +=+，则该数列的前10项和10S =（ ）A.-10B. -5C. 0D. 55. 如图，一艘船自西向东匀速航行，上午10时到达一座灯塔P 的南偏西75°距塔68海里的M 处，下午2时到达这座灯塔的东南方向的N 处，则这艘船航行的速度为 ( )A.1762海里/时 B ．346海里/时C.1722海里/时D ．342海里/时6. ABC ∆的内角A 、B 、C 的对边分别为.,,c b a 若c b a ,,成等比数列，且c=2a ，则cosB=（ ）A.41 B.43C.42D.327.已知命题p ：x 2＋2x －3>0；命题q ：x >a ，且⌝q 的一个充分不必要条件是⌝p ，则a 的取值范围是( )A ．[1，＋∞)B ．(－∞，1]C ．[－1，＋∞)D ．(－∞，－3]8．已知函数f (x )＝x 2＋mx －1，若对于任意x ∈[m ，m ＋1]，都有f (x )＜0成立，则实数m 的取值 范围是 ( )A. 2(,0)2-B. 3(,0)2- C. 32(,)22-- D. 22(,)22-9. 已知()20{,|20}360x y D x y x y x y +-≤⎧⎪=-+≤⎨⎪-+≥⎩，给出下列四个命题：()1:,,0;P x y D x y ∀∈+≥ ()2,,210;P x y D x y ∀∈-+≤：()31:,,4;1y P x y D x +∃∈≤-- ()224,,2;P x y D x y ∃∈+≥： 其中真命题的是 ( )A. 12,P PB. 23,P PC. 34,P PD. 24,P P10.在△ABC 中，角A 、B 、C 所对的边分别为a 、b 、c ，且BC ，则c bb c+的最大值是 ( )D. 411.已知等差数列{}n a 满足20152017201620170,0a a a a +>+<,12323412n n n n T a a a a a a a a a ++=+++,若对任意正整数n ,恒有n k T T ≤,则正整数k 的值是 ( )A ．2014B ．2015C ．2016D ．201712．已知各项都为整数的数列{}n a 中， 12a =，且对任意的*N n ∈，满足1122n n n a a +-<+， 2n n a a +- 321n >⨯-，则2017a = ( )A. 201732⋅B. 20172+2 C. 20172+1 D. 20172二、填空题：本大题共4小题，每小题5分，共20分．在答题卷上的相应题目的答题区域内作答．13. 命题p 的否定是“对∀x ∈(0，＋∞)，x >x ＋1”，则命题p 是 ． 14. 用一根长为12的钢筋焊接一个正三棱柱形状的广告牌支架，则该三棱柱的侧面积的最大值是__________．15．在△ABC 中，B ＝60°，AC ，则AB ＋2BC 的最大值为 ．16．已知数列{}n a 的前n 项和为n S ，若11a =， 2n n a n a =-， 211n n a a +=+，则100S =__________．（用数字作答）三、解答题：本大题共6小题，共70分，解答题应写出文字说明，证明过程或演算步骤，在答题卷上相应题目的答题区域内作答．17.已知2()2f x ax bx =++,关于x 的不等式()0f x >的解集为()1,2-. （1）求函数()f x 的解析式；（2）若0m >，解关于x 的不等式23(1)2()m m x m f x -+-++<18. 已知a R ∈，命题[]2:1,2,-0p x x a ∀∈≥，命题2q :22,-0x R x ax a ∃∈++=.（1）若命题“p q ∧”为真命题，求实数a 的取值范围；（2）若命题“p q ∨”为真命题，命题“p q ∧”为假命题，求实数a 的取值范围.19. 若ABC ∆的内角,,A B C 所对的边分别为,,a b c ，且满足sin cos 0a B A =（1）求A ；（2）当2a b ==时， 求ABC ∆的面积．20. 设ABC ∆的内角,,A B C 的对边分别为,,a b c ，且满sin sin [cos cos()]sin A B A B C π+=--⋅.（1）试判断ABC ∆的形状，并说明理由；（2）若1a b c ++=ABC ∆面积的最大值.21. 已知数列{}n a 的前n 项和21n n S a =-.{}n b 是公差不为0的等差数列，其前三项和为3，且3b 是25,b b 的等比中项.（1）求,n n a b ； （2）若()112222n n a b a b a b n t +++≥-+，求实数t 的取值范围.22. 已知数列{}n a 与{}n b 的前n 项和分别为n A 和n B ，且对任意n *∈N ，112()n n n n a a b b ++-=-恒成立． （1）若21,2n A n b ==，求n B ； （2）若对任意n *∈N ，都有n n a B =及3124122334113n n n b b b b a a a a a a a a ++++++<成立，求正实数1b 的取值范围； （3）若12,a =2n n b =，是否存在两个互不相等的整数,s t (1)s t <<，使11,,s ts tA A AB B B 成等差数列？若存在，求出,s t 的值；若不存在，请说明理由．福建省厦门第一中学2017—2018学年度上期中考高二年理科数学试卷答 案 卷一、选择题：1-5．CCBCA 6-10.BAADD 11-12.CD211sin 22S bc A ===，即2sin a A =，222222cos 2cos 4sin()4,63c b b c a a a A A A A A b c bc bc ππ+-++==+=+=+≤=取等.11. 由20152017201620a a a +=>得20160a >,由201620170a a +<得20170a <,所以等差数列{}n a 的公差0d <,故2016n ≤时0n a >,2017n ≥时0n a <,所以2014n ≤时120n n n a a a ++>, 2015201620170a a a <,2016201720180a a a >,当2017n ≥时120n n n a a a ++<,又()2015201620172016201720182016201720152018a a a a a a a a a a +=+()2016201720162017a a a a =+>0,所以2016n =时n T 最大,12. 12211112232122n n n n n n n n n a a a a a a +++++--<+++=⋅=-++，即 2321321n n n n a a +⋅<-<⋅+-，又2n n Z a a +-∈，则有232n n n a a +=-⋅.则320152017201713120172015()()23(222)2a a a a a a =+-++-=++++=二、填空题13. 00(0,1x x ∃∈+∞>+ 14. 615. 16. 130615. 由正弦定理可知，sin(120),sin ,sin sin AC ACAB A BC A B B=-= 则有AB ＋2BC =2sin(120)4sin 5sin )A A A A A ϕ-+=+=+≤.16. 由题设可得2211n n a a n ++=+，取1,2,3,,49n =⋅⋅⋅可得23456798992,3,4,,50a a a a a a a a +=+=+=⋅⋅⋅+=，将以上49个等式两边分别相加可得23456798992504912742a a a a a a a a +++++++⋅⋅⋅++=⨯=；又3163126251250251005012,31,65,16,2519,5031a a a a a a a a a a a a =+==-==-==+==-==-=，所以10011274311306S =++=.三、解答题17. 解：（1）根据题意得220ax bx ++=的两根为2,121=-=x x ，且0a < 由根与系数的关系可求得1,1a b =-=所以2()2f x x x =-++． （2）原不等式可化为23(1)2()m m x m f x -++++<，即223()0x m m x m -++<，即2()()0x m x m --<，又0m >，所以当2m m <，即01m <<时，2m x m <<； 当2=m m ，即1m =时，原不等式的解集为∅； 当2m m >，即1m >时，2m x m <<．综上所述，当01m <<时，原不等式的解集为{}2x m x m <<，当1m =时，原不等式的解集为∅，当1m >时，原不等式的解集为{}2x m x m <<．18.解：（1）命题p 为真命题时：令()2-f x x a =，根据题意，只要[]1,2x ∈时，()min 0f x ≥即可，也就是1-01a a ≥⇒≤；命题q 为真命题时，()24420a a ∆=--≥，解得2a ≤-或1a ≥；“p q ∧”为真命题，即,p q 都为真命题，则有(,2]{1}21a a a a ≤⎧⇒∈-∞-⎨≤-≥⎩1或. （2）由（1）可知，当命题p 为真命题时，1a ≤，因为命题“p q ∨”为真命题，命题“p q ∧”为假命题，所以命题p 与q 一真一假，当命题p 为真，命题q 为假时，12121a a a ≤⎧⇒-<<⎨-<<⎩,当命题p为假，命题q 为真时，1121或a a a a >⎧⇒>⎨≤-≥⎩.综上：(2,1)(1,)a ∈-⋃+∞19.解：(1)由正弦定理可得：sin cos 0sin sin cos 0a B A A B B A =⇔=，又sin 0B >，则有sin 0A A =，即tan A =又(0,),A π∈则有3A π=.（2）由余弦定理，得2222cos a b c bc A =+-，而2a b ==， 3A π=，得2742c c =+-，即2230c c --=，因为三角形的边0c >，所以3c =，则ABC S ∆=1sin 2bc A =．20.解：（1）依题意得sin sin (cos cos )sin A B A B C +=+法一：由正余弦定理可得：222222()22b c a a c b a b c bc ac+-+-+=+.化简整理可得：222()()()a b a b a b c ++=+，又0a b +>，则22290a b c C +=⇒=︒，即为直角三角形.法二：由正弦定理知：sin()sin()cos sin cos sin B C A C A C B C +++=+，展开化简得(sin sin )cos 0A B C +=，又sin sin 0A B +>，则cos 090C C =⇒=︒，即为直角三角形.（2）1a b c a b =++=+≥，当且仅当a b =时取等，≤1124ABC S ab ∆=≤，即ABC ∆面积的最大值为14，当且仅当a b =时取等.21.解：（1）1n =时，1111211a S a a ==-⇒=，1n >时111222n n n n n n n a S S a a a a ---=-=-⇒=，所以{}n a 是以1为首项，2为公比的等比数列，即12n n a -=.设{}n b 的公差为0d ≠，依题意有1231333b b b b d ++=+=，2253,b b b ⋅=即21111()(4)(2)0b d b d b d b d +⋅+=+⇒=，解得10,1b d ==，即1n b n =-.（2）由（1），可知， 12,1n n n a b n -==-，从而()112n n n a b n -=-⨯，令1122n n n T a b a b a b =+++，即()()122112222212n n n T n n --=⨯+⨯++-⨯+-⨯，③×2，得()()231212222212n n n T n n -=⨯+⨯++-⨯+-⨯，④ -④，得()231222212n n n T n --=++++--⨯()()221222212n n n n n -=--⨯=--⨯--， 即(2)22nn T n =-+，故题设不等式可化为()22(2)nt n n -≥-，（*）当1n =时，不等式（*）可化为2t -≥-，解得2t ≥； 当2n =时，不等式（*）可化为00≥，此时t R ∈；当3n ≥时，不等式（*）可化为2n t ≤，因为数列{}2n 是递增数列，所以8t ≤， 综上， t 的取值范围是[]2,8.22.解：（1）1n =时，111a S ==，1n >时121n n n a S S n -=-=-，所以*21()n a n n N =-∈. 则有11n n b b +-=，即{}n b 是以2为首项，1为公比的等差数列，即1n b n =+，(1)(3)222n n n n n B n -+=+=. （2）依题意得112(),n n n n B B b b ++-=-即112()n n n b b b ++=-，即12n nb b +=，且111b B a == {}n b 是以1b 为首项，2为公比的等比数列，112n n b b -=, 11(12)(21)12n n n b B b -==--, 所以111111111n n n n n n n n n n n n b b B B a a B B B B B B +++++++-===-， 则31211223112231111111111111111(1)21n n n n n n n b b b a a a a a a B B B B B B B B b b ++++++++=-+-++-=-=-<-， 则111,3b ≤即13b ≥（3）由112()n n n n a a b b ++-=-得：112n n n a a ++-=，所以当2n ≥时，11232211()()()()n n n n n a a a a a a a a a a ---=-+-++-+-+132********n n n -+=+++++=-，当1n =时，上式也成立，则21242,22n n n n A n B ++=--=-，所以2124222221n n n n n A n nB ++--==---. 法一：假设存在两个互不相等的整数,s t (1)s t <<，使11,,s ts tA A AB B B 成等差数列，即 111122212121212121t t s s s t t s A A A t s s tB B B +=⇔+=⇔=+-----. 又有2112121s t s t =+>--，即2120s s --<，设*()221,2,s f s s s s N =--≥∈.则有(1)()220sf s f s +-=->，即数列{()}f s 单调递增， 又(2)10f =-<，(3)10f =>，则有()0f s <⇒ 2.s =当2s =时，21121213t s t s =-=--，即2310,3t t t --=≥.同理可证当3t ≥数列{231}tt --递增，当3t =时2312t t --=-舍去，当4t ≥时4231212130t t --≥--=>，即2310tt --=无解，综上所述，不存在两个互不相等的整数,s t (1)s t <<，使11,,s ts tA A AB B B 成等差数列.法二：11111(1)2102121(21)(21)n n n n n n n n n A A n n n B B +++++-+-=-=>----，即数列{}n n A B 单调递增. 2[1,2)21n n n A nB =-∈-，又111123()222s t s t A A A B B B +=+<=， 又123312431131,,3272A A A B B B ==<=>，则2,s =所以11523t s t s A A A B B B =-= 又3434115265,73153A AB B =<=>，34t ⇒<<，则这样的t 不存在. 综上所述，不存在两个互不相等的整数,s t (1)s t <<，使11,,s ts tA A AB B B 成等差数列.。

2017—2018学年度高二上学期期中考试

2017—2018学年度高二上学期期中考试

2017—2018学年度高二上学期期中考试第一部分：听力（共两节，满分20分）第一节（共5小题；每小题1分，满分5分）1. Where does the conversation probably take place?A. In a restaurant.B. In a furniture store.C. I n the woman's company.2. What does the man want to do first?A. See the elephants.B. Have a cup of tea.C. Watch t he dolphin show.3. What is the woman probably going to do?A. Make a list.B. Do some shopping.C. Make a chocolate pie.4. Why does the woman look upset?A. She experienced a theft.B. She was given a parking ticket.C. She couldn't find a parking space.5. How will the woman go to the city church?A. By bike.B. By bus.C. By car.第二节（共15小题；每小题1分，满分15分）听第6段材料，回答第6至7。

6. What activity is the man organizing?A. A school reunionB. A birthday partyC. A farewell banquet7.Why does the woman recommend the Omni Hotel?A. The price is quite reasonable.B. She likes the party r ooms there.C. Her brother is a staff member there.听第7段材料，回答第8至9题。

2017—2018学年度第二学期期中考试

2017—2018学年度第二学期期中考试

2017—2018学年度第二学期期中考试高二数学（文）试题（导数、选修1－2）第Ⅰ卷（选择题 共50分）一、选择题（本大题共10个小题，每小题5分，共50分。

在每小题给出的4个选项中，只有一项是符合题目要求的） 1．211+)i复数（的值是（ ）A ． 2iB ．－2iC ． 2D ． －22．若)(,sin cos 2)('ααf x x f 则-=等于（ B ） A ．αsin -B ．αcos - Cααcos sin 2-- Dαcos 3-3．按流程图的程序计算，若开始输入的值为3x =，则输出的x 的值是 （ ）A ．6B ．21C ．156D ．2314．在独立性检验中，统计量有两个临界值：3.841和6.635；当2K ＞3.841时，有95%的把握说明两个事件有关，当＞6.635时，有99%的把握说明两个事件有关，当 3.841时，认为两个事件无关．在一项打鼾与患心脏病的调查中，共调查了2000人，经计算的=20.87，2K 2K 2K ≤2K 输入x计算(1)2x x x +=的值 100x >输出结果x是否根据这一数据分析，认为打鼾与患心脏病之间（ ） A ．有95%的把握认为两者有关 B ．约有95%的打鼾者患心脏病C ．有99%的把握认为两者有关D ．约有99%的打鼾者患心脏病5．如图是函数32()f x x bx cx d =+++的大致图象，则等于（）A ．169B ．109C ．89D ．2896．有下列关系：①正方体的体积与棱长；②曲线上的点与该点的坐标之间的关系；③苹果的产量与气候之间的关系；④森林中的同一种树木，其横断面直径与高度之间的关系，其中有相关关系的是（ ） A ．①②③B ．①②C ．②③D ．③④7．用反证法证明命题：“一个三角形中不能有两个直角”的过程归纳为以下三个步骤：①9090180A B C C ++=︒+︒+>︒，这与三角形内角和为180︒相矛盾，90A B ==︒不成立；②所以一个三角形中不能有两个直角；③假设三角形的三个内角A 、B 、C 中有两个直角，不妨设90A B ==︒，正确顺序的序号为（ ） A ．①②③B ．③①②C ．①③②D ．②③①8．设函数322()3(1)1f x kx k x k =+--+在区间（0,4）上是减函数，则的取值2221x x +k范围是（ ） A ．13k <B ．103k <≤C ．103k ≤≤D ．13k ≤9．如图所示是()y f x =的导数图象，则正确的判断是（ ）①()f x 在(－3,1)上是增函数；②x ＝－1是()f x 的极小值点；③()f x 在(2,4)上是减函数，在(－1,2)上是增函数；④x ＝2是()f x 的极小值点．A ．①②③B ．②③C ．③④D ．①③④10．设△ABC 的三边长分别为a ，b ，c ，△ABC 的面积为S ，内切圆半径为r ，则2Sr ab c=++，类比这个结论可知：四面体S －ABC 的四个面的面积分别为S 1，S 2，S 3，S 4，内切球半径为R ，四面体S －ABC 的体积为V ，则R ＝（ ） A ．1234VS S S S +++ B ．12342VS S S S +++ C ．12343VS S S S +++D ．12344VS S S S +++第Ⅱ卷（非选择题，共100分）二．填空题（本大题共5个小题，每小题5分，共25分，把答案填在答题卡中横线上） 11． 求垂直于2610x y -+=且与3235y x x =+-相切的直线方程__________________12．已知,,2x y R x y ∈+<则y x ,中至多有一个大于1，在用反证法证明时，假设应为________． 13．若函数24()1xf x x =+在区间(,21)m m +上单调递增，则实数m 的取值范围是 ．14．将全体正整数排成一个三角形数阵：1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 … … … … … …根据以上排列规律，数阵中第n (n ≥3)行的从左至右的第3个数是__________．15．已知f (x )＝(2x －x 2)e x ，给出以下几个结论：①f (x )>0的解集是{x |0<x <2}；②f (－2)是极小值，f (2)是极大值；③f (x )没有最小值，也没有最大值；④f (x )有最大值，没有最小值． 其中判断正确的是________．三、解答题（本大题共6个小题，满分75分，解答应写出必要的文字说明、证明过程和演算步骤） 16．（本小题满分12分）已知复数(1)3(1)2i i z i-++=-，若21z az b i ++=-，（1）求z ；（2）求实数a , b 的值．17．（本小题满分12分）某人酷爱买彩票，一次他购买了1000注的彩票，共有50注中奖，于是他回到家对彩票的号码进行了分析，分析后又去买了1500注的彩票，有75注中奖．请分析他对号码的研究是否对中奖产生了大的影响．18．（本小题满分12分）某种产品的广告费支出x与销售额y(单位：百万元)之间有如下对应数据：x 24568y 3040605070（1）画出散点图；（2）求回归直线方程；（3）试预测广告支出为10百万元时，销售额多大？(注：b＝∑i＝1nx i y i－n x－y－∑i＝1nx2i－n x－2，a＝y－－b x－)．19．（本小题满分12分）已知正数a,b,c,d满足a+b=c+d，且a<c≤d<b，求证:a b c d+<+．20．（本小题满分13分）已知函数32f x x x=-+()23 3.（1）求曲线()=在点2y f xx=处的切线方程；（2）若关于x的方程()0+=有三个不同的实根，求实数m的取f x m值范围．21．（本小题满分14分） 已知函数()2a f x x x=+，()ln g x x x =+，其中0a >．（1）若1x =是函数()()()h x f x g x =+的极值点，求实数a 的值； （2）若对任意的[]12,1x x e ∈，（e 为自然对数的底数）都有()1f x ≥()2g x 成立，求实数a 的取值范围．高二数学（文）试题（A ）参考答案（导数、选修1-2）1—10 BBDC Ａ DBDBC11．063=++y x 12．y x ,均大于１ 13．(]0,1-14．262+-n n 15．①②④16．解：（1）z=1+i （2）⎩⎨⎧=-=43b a17．解：根据题意可知购买1000注的彩票，中奖50注，未中奖的有950注；购买1500注彩票，中奖75注，未中奖的有1425注．列出对应的2×2列联表如下：中奖注数 未中奖注数总计 未分析 50 950 1000 分析后 75 1425 1500 总计12523752500假设H 0：对彩票号码的研究与中奖无关． 由表中数据，得K 2的观测值为 k ＝－21000×1500×125×2375＝0.因为0<2.706，所以没有足够的证据说明对彩票号码的分析与中奖有关． 18．解：（1）根据表中所列数据可得散点图如下：（2）列出下表，并用科学计算器进行有关计算i12345x i 2 4 5 6 8 y i 30 40 60 50 70 x i y i60160300300560因此，x ＝255＝5，y ＝2505＝50∑i ＝15x 2i ＝145，∑i ＝15y 2i ＝13500， ∑i ＝15x i y i ＝1380， 于是可得b ＝∑i ＝15x i y i －5x y∑i ＝15x 2i －5x2＝1380－5×5×50145－5×52＝6.5； a ＝y －b x ＝50－6.5×5＝17.5.因此，所求回归直线方程为y ^＝6.5x ＋17.5.（3）据上面求得的回归直线方程，当广告费支出为10百万元时，y ^＝6.5×10＋17.5＝82.5（百万元）即这种产品的销售收入大约为82.5百万元． 19．证明：要证明a b c d +<+，需证明2()a b c d 2+)<(+， 需证明a +b+2ab <c+d+2cd , 因为a +b =c +d ,所以只需证明ab <cd , 需证明ab -bc <cd -bc , 需证明b (a -c )<c(d -b ), 考虑a +b =c +d ,即a -c =d -b , 需证明(a -c )(b -c )<0, 考虑a -c <0,需证明b -c >0, 而b -c >0显然成立,所以a b c d +<+成立.20. 解（1）2()66,(2)12,(2)7,f x x x f f ''=-==∴曲线()y f x =在2x =处的切线方程为712(2)y x -=-，即12170x y --=； （2）记322()233,()666(1)g x x x m g x x x x x '=-++=-=- 令()0,0g x x '==或1.则,(),()x g x g x '的变化情况如下表x(,0)-∞0 （0,1） 1(1,)+∞()g x ' +0 -0 +()g x递增极大递减极小递增当0,()x g x =有极大值3;1,()m x g x +=有极小值2m +. 由()g x 的简图知，当且仅当(0)0,(1)0g g >⎧⎨<⎩即30,3220m m m +>⎧-<<-⎨+<⎩时，函数()g x 有三个不同零点，过点A 可作三条不同切线.所以若过点A 可作曲线()y f x =的三条不同切线，m 的范围是(3,2)--.21．解：（1）∵()22ln a h x x x x=++，其定义域为()0 +∞，，∴()2212a h x x x'=-+．∵1x =是函数()h x 的极值点，∴()10h '=，即230a -=．∵0a >，∴3a =． 经检验当3a =时，1x =是函数()h x 的极值点，∴3a =．（2）对任意的[]12,1x x e ∈，都有()1f x ≥()2g x 成立等价于对任意的[]12,1x x e ∈，都有()mi n f x ⎡⎤⎣⎦≥()max g x ⎡⎤⎣⎦．当x ∈［1，e ］时，()110g x x'=+>．∴函数()ln g x x x =+在[]1e ，上是增函数．∴()()max 1g x g e e ==+⎡⎤⎣⎦． ∵()()()2221x a x a a f x x x +-'=-=，且[]1,x e ∈，0a >．①当01a <<且x ∈［1，e ］时，()()()20x a x a f x x +-'=>，∴函数()2a f x x x=+在［1，e ］上是增函数，∴()()2min 11f x f a ==+⎡⎤⎣⎦.由21a +≥1e +，得a ≥e ，又01a <<，∴a 不合题意．②当1≤a ≤e 时，若1≤x ＜a ，则()()()2x a x a f x x +-'=<，若a ＜x ≤e ，则()()()2x a x a f x x +-'=>．∴函数()2a f x x x=+在[)1,a 上是减函数，在(]a e ，上是增函数．∴()()min 2f x f a a ==⎡⎤⎣⎦.由2a ≥1e +，得a ≥12e +， 又1≤a ≤e ，∴12e +≤a ≤e ． ③当a e >且x ∈［1，e ］时，()()()20x a x a f x x +-'=<， ∴函数()2a f x x x=+在[]1e ，上是减函数． ∴()()2min a f x f e e e ==+⎡⎤⎣⎦. 由2a e e+≥1e +，得a ≥e ， 又a e >，∴a e >．综上所述，a 的取值范围为1,2e +⎡⎫+∞⎪⎢⎣⎭．。

2017—2018学年度高二上学期期中考试_1

2017—2018学年度高二上学期期中考试_1

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C. Introducing a company.
2. What are the speakers mainly talking about?
A. A university. B. A lab. C. A physicist.
3. What does the woman want to do?
A. Ask the man for advice.
7. How long will the woman stay in Shanghai this time?
A. Three weeks. B. Two weeks.C. One week.
请听第7段材料，回答第8~9题。
8. What does the woman want the man to do for the party?
请听第9段材料，回答第13~16题。
13. What’s the relationship between the speakers?
A. Husband and wife. B. Brother and sister. C. Mother and son.
14. Where is the man most probably?
Dawn rushed home to get her husband to help and called 911. Thetwo then returned with some blankets to cover Ms. St. Onge, while they waited for the ambulance (救护车) to arrive. Though still in hospital, the elderly woman is recovering well and forever grateful to Danny, for following up on his hunch!