江西省宜春第一中学高一2015届英语下学期第三次模拟(理)试卷

2017届高一年级下学期期末考试英语试卷命题人：刘红瑛第一部分听力（共两节，满分30分）第一节(共5小题；每小题1。

5分，满分7。

5分）1 What is the woman going to do next？A. Have a meeting。

B. See the manager。

C。

Do some translation.2 How did the woman feel about her performance？A。

Satisfied. B. Disappointed。

C. Worried.3 What did the woman do today？A。

She went to a multicultural fair. B。

She had dinner in a Thai restaurant.C. She learned a native American dance。

4 Where are the speakers?A. At a library.B. At a computer lab. C。

At a print shop。

5 Why does the man come to the woman?A。

To take a picture of her. B。

To ask for a new ID card。

C。

To fill out a form.第二节（共15小题；每小题1.5分,满分22.5分）听第6 段材料，回答第6、7 题。

6 What is the relationship between the speakers?A。

Teacher and student. B. Classmates. C。

Strangers。

7 Where is the Language Arts building？A. On the right of the bridge。

B. At the end of Campus Center Walk。

奉新一中2015届高三模拟考试英语试卷2015.5.25.本试卷分第I卷（选择题）和第II卷（非选择题）两部分，满分150分，考试用时120分钟。

第I卷第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5分）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选项，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.How much will the man pay for the watch?A. $75.B. $60.C. $15.2.What are the speakers mainly talking about?A.The menu.B. The lunch.C. The school website.3.Where will the speakers meet?A.At the cinema.B. At the bar.C. At the bus stop.4.What will the man do this winter?A. Work in the garden.B. Get a job in a store.C. Do building work outside.5.What kind of room might the man take tonight?A.A standard room.B. A twin room.C. A single room.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

每段对话或独白后有几个小题，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。

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

每段对话或独白读两遍。

听第6段材料，回答第6和第7题。

于对市爱美阳光实验学校高一英语下学期第三次考试试题第一： 听力(共两节, 总分值30分) 第一节(共5小题；每题分, 总分值分)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 三个选项中选出最正确选项, 并标在试卷的相位置。

听完每段对话后, 你都有10秒钟的时间来答复有关小题和阅读下一小题。

每段对话仅读一遍 1. How much money does the man probably have?A. 20 dollars.B. 30 dollars.C. 50 dollars.2. What will the woman do this afternoon?A. Read a book.B. go shopping.C. visit Ann.3. What does the woman think of the video?A. Disappointing.B. violent.C. funny.4. What does the man suggest?A. Staying at home.B. going fishing.C. going to the cinema.5. How long does the woman usually sleep?A. For about 4 hours.B. for about 6 hours.C. for about 8 hours.第二节〔共15小题；每题分，总分值2分〕听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A 、B 、C 三个选项中选出最正确选项，并标在试卷的相位置。

听每段对话或独白前，你将有时间阅读各个小题，每题5秒钟；听完后，各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料，答复第6至7题。

6. What was the woman before probably?A. A secretary.B. a computerprogrammer. C. a factory worker. 7. What do we know about the woman?A. She gets along well with her colleaguesB. She worksovertime every day. C. she gets a bad salary. 听第7段材料， 答复第8至9题。

高一上学期第三次月考英语试卷命题人：王婷2015.12第一部分：听力(共两节，满分30分)第一节(共5小题；每小题1.5分，满分7.5分)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选项，并标在答题卷的相应位置。

听完每段对话后，你都有l0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. How much did the man tip the woman?A. $1.B. $10.C. $100.2. How will the woman go home?A. Walk.B. Run.C. Drive.3. What meal will the speakers eat?A. Breakfast.B. Lunch.C. Dinner.4. Who was at the door?A. The postman.B. The man’s friend.C. The woman’s brother.5. When will the tomatoes be ready?A. Right now.B. In three weeks.C. In two months.第二节(共15小题，每小题1.5分，满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在答题卷的相应位置。

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

每段对话或独白读两遍。

听第6段材料, 回答第6、7题。

6. Where are the speakers?A. At a hotel.B. At a store.C. At a restaurant.7. How does the woman feel in the end?A. Excited.B. Sick.C. Scared.听第7段材料，回答第8、9题。

江西省宜春市中学高一英语联考试题含解析一、选择题1. Over the past 20 years, the Internet has helped change our world in_____ way or another for the better .A oneB any C. every D. either参考答案：A2. —The teacher said，“Let's have a rest.”—What did the teacher say?—________.A．The teacher said to us to have a restB．The teacher asked to have a restC．The teacher suggested having a restD．The teacher advised to have a rest参考答案：C[带有let的祈使句变为间接引语时，可用“suggest doing或suggest＋从句”的形式。

句意：老师建议休息一下。

]3. This is the first time that I ________at the meeting.A. had spokenB. have spokenC. amD. was参考答案：B4. ----Where did you get to know each other?---- It was on the farm ______ we worked during the last summer vocation.A. whereB. thereC. whichD. that参考答案：A5. How did the movie finally ______? I couldn’t wait to see the ending.A. come outB. come acrossC. come offD. come away参考答案：A6. –My teacher punished me this morning.--Why? You’ve done nothing wrong, ______ I can see.A. as long asB. as far asC. as well asD. as soon as参考答案：B7. Many Chinese are ______ bringing down the price of housing, thinking it is too high.A. in favour ofB. in honour ofC. in search ofD. in charge of参考答案：A8. —Is ______ America ______ American or______ European country? —Sorry, I have no idea.A.an; an; anB. the ;/ ;/C. /; an/aD./; an; an参考答案：C略9. This atlas(地图集) _______ 40 maps, ________that of Great Britain.A. contains; includingB. includes; containingC. contains; containingD. includes; including参考答案：A10. –Why didn’t you ask the teacher the question?-- She had left the office ________ I had time to do it．A．as B．till C．after D．before参考答案：D11. The poor girl had to _____ her job to look after her elderly mother.A. gave in toB. gave offC. gave outD. gave up参考答案：d略12. I was ________ asking him what that work might be, but something in his manner showed me that the question would be an unwelcome one.A. at the edge ofB. on the point ofC. in the case ofD. at the mercy of参考答案：B【详解】考查介词短语词义辨析。

江西省宜春市宜丰县宜丰中学2019-2020学年高一英语下学期质量检测试题〔考试时间：120分钟总分为150分〕温馨提示：请将全部答案填写在答题卡上。

第一局部听力〔共两节，共30分〕第一节〔共5小题；每一小题1.5分，共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三个选项中选出最优选项，并标在试卷的相应位置。

听完每段对话后，你将有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. When did the speakers last see each other?A. Seven years ago.B. Eight years ago.C. Ten years ago.2. Why can’t the man go to the show?A. He isn’t fine.B. He has to work.C. He slepttoo late.3. What is the probable relationship between the speakers?A. Classmates.B. Friends.C. Cousins.4. Where is the woman going?A. The taxi company.B. The XinHua Hotel.C. The HiltonHotel.5. What does the man seem to care about most?A. The price.B. The bedroom.C. The size.第二节〔共15小题；每一小题1.5分，共22.5分〕听下面5段对话或独白，每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最优选项，并标在试卷的相应位置。

听每段对话或独白前，你将有时间阅读各个小题，每一小题5秒钟；听完后，每一小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

于对市爱美阳光实验学校一中2021-高一级第三次阶段测试英语试题第一英语知识运用〔共两节，总分值35分〕第一节单项填空〔共15小题；每题1分，总分值15分〕1. —What is your plan for the holiday, Paul?—I’ll go on a trip to Yunnan Province with my family. —_____. A. Have a nice time B. Good luck C. Come back soon D. Congratulations2. —Have you finished your experiment report, Jane?—Oh, my God. I’ve _____ forgotten all about that.A. stronglyB. extremelyC. entirelyD.freely3. It was the first time that they _____to think about what use could be _____ the waste things.A. have begun; made upB. had begun; made ofC. had begun; made fromD. began; made out4. Here are Daphne and David; _____ is my brother.A. the laterB. laterC. the latterD. latter5. Ladies and gentlemen, please fasten your seat belts. The plane _____.A. takes offB. is taking offC. has taken offD. took off 6. She is the most _____ one of my friends; you can just depend on her.A. reliableB. beautifulC. fortunateD. stubborn7. I know you are very _____ ice-cream, but you will gain weight if you eat too much.A. afraid ofB. tired ofC. full ofD. fondof8. Mr. Black has been teaching English in a middle school _____ he came to China.A. for everB. ever sinceC. since thenD. longbefore9. When they knew there was a panda in the zoo, the children could____wait to see it.A. almostB. nearlyC. hardD. hardly10. Little Tony insisted that his mother _____ him to school.A. sendB. sentC. sendsD. sending11. He is one of the most excellent students in our class, and he always _____ himself in his studies. A. buries B. hidesC. coversD. leaves12. At the sight of the _____ scene, he was almost _____ to death.A. frightened；frighteningB. frightened；frightenedC. frightening；frighteningD. frightening；frightened13. _____ seemed as if nobody knew anything about the matter.A. ThisB. ThatC. ItD. One34. Here is my classmate _____ bicycle was lost yesterday.A. hisB. thatC. whoseD. which15. Anything _____ could be found has been used to repair the damaged bridge.A. whichB. thatC. whatD. 不填第二节：完形填空〔共20小题，总分值20分〕Nowadays it is found that school students seldom pay much attentionto sports. Is it because they have no 16 in sports? It may not be the fact. They often say they have other 17 important things to do.18 are these important things? Examinations! They have to 19 themselves for the most important School Certificates〔毕业证〕of Education Examination, and also to 20 with the test and the examsin school. So many of them will 21 bookworms (书虫，书呆子) .In the summer holidays of the former years they could do 22 they liked, 23 in the summer holidays of the present years, they have to give all 24 time to the preparation. So studies have prevented them from going in 25 .Because of the pressure 26 their parents and teachers, they haveto work harder and spend most of their time on books. Many parents 27 their children to pass this examination in order to get a stepping stonefor higher studies or better jobs. As for the 28 themselves, they want to get good results 29 they can further their studies in the universities. So it is necessary that they 30 up their school activities, especially sports.Indeed, a complete education cannot go 31 physical training, for a quick mind seldom 32 along with a weak body. It is well said, “All work no play 33 Jack a dull〔迟钝的〕boy.〞 In one word, without a 34 body, you can never achieve anything, let alone 〔更不用说〕 a great success in 35 .16. A. lesson B. time C. interest D. place17. A. very B. more C. few D. much18. A. Where B. How C. Which D. What19. A. prepare B. get C. make D. ask20. A. agree B. fight C. spend D. deal21. A. have B. become C. change D.find22. A. something B. when C. anything D. which23. A. so B. and C. however D.but24. A. my B. our C. your D. their25. A. games B. music C. work D. sports26. A. from B. of C. in D. to27. A. have B. want C. make D. let28. A. students B. teachers C. parentsD. examinations29. A. while B. instead of C. so thatD. even if30. A. keep B. put C. take D. give31. A. for B. after C. with D. without32. A. goes B. stands C. lives D. lies33. A. turns B. makes C. gets D. brings34. A. weak B. strong C. useful D. helpful 35. A. sports B. writing C. training D. life第二阅读理解(共15小题，每题2分，总分值30分)AThe best way of learning a language is using it. The best way of learning English is talking in English as much as possible. Sometimes you'll get your words mixed up (混合) and people will not understand you. Sometimes people will say things too quickly and you can't understand them. But if you keep your sense of humor , you can always have a good laugh at the mistakes you make. Don't be unhappy if people seem to be laughing at your mistakes. It's better for people to laugh at your mistakes than to be angry with you, because they don't understand what you are saying. The most important thing for learning English is: "Don't be afraid of making mistakes because everyone makes mistakes."36. The writer thinks that the most important thing for you to learna language is___________.A. readingB. practisingC. talking about itD. listening37. What should you do in learning English?A. Try to make some mistakes.B. Avoid making any mistakes.C. Remember as many new words as you can.D. Use it as often as you cam38. If people laugh at the mistakes you make, you should ___________.A. be angry with themB. be angry with yourselfC. not careD. believe you are right39. When you make a mistake, you should___________.A. never make any mistakes againB. tell others not to make the same mistakeC. punish yourself for making itD. keep your sense of humor40. The story tells us:" ___________."A. It is normal (正常的) that we make some mistakes in learning EnglishB. Everyone must make mistakesC. We can avoid making mistakes in learning a languageD. Laughing can help one learn English wellBFor thousands of years, man has enjoyed the taste of apples. Apples, which are about 85 percent water, grow almost everywhere in the world, both the hottest and coldest areas. Among the leading countries in apple production are China, France and the United States.There are various kinds of apples, but a very few make up the majority of those grown for sale. The three most common kinds grown in the United States are Delicious, Golden Delicious, and McIntosh.Apples are different in colour, size and taste. The colour of the skin may be red, green, or yellow. They have various sizes, with Delicious apples being among the largest. The taste may be sweet or tart(酸的). Generally, sweet apples are eaten fresh while tart apples are used to make applesauce(苹果酱).Apple trees may grow as tall as twelve meters. They do best in areasthat have very cold winters. Although no fruit is produced during the winter, this cold period is good for the tree.41. It can be learned from the text that Delicious apples areA. grown in FranceB. sold everywhereC. very bigD. quite sweet42. Cold winter weather is good for __________.A. producing large applesB. the growthof apple treesC. improving the taste of applesD. the increase ofwater in apples43. China, France and the United States are considered to __________.A. be large producers of applesB. be large producers of applesauceC. have the longest history in apple productionD. have the coldest winter among apple producing countries44. According to the third paragraph, what can we learn?A. the history of the production of applesB. the differences of apples in size, taste and places of originC. the different ways of eating applesD. the different usages of different kinds of applesCDear Editor,I’m a Senior I student in a middle school. This term, my favorite teacher, Miss Mao, nolonger teaches us. I want to see her, but I’m afraid that she no longer likes me and I don’t want totrouble her. I really miss her.What should I do?Tian YanDear Tian Yan,It’s bad luck that you have lost your favorite teacher, But if she is still in your school ,nothing can stop you going to see her. When she isn’t busy, ask her if she minds having a quickchat. You can then tell her she was your favorite teacher------ everyone is happy to know that theyare liked!If she has left the school, it will be more difficult to meet her. It will be hard, but rememberpeople always come and go in their lives. We can’t rely on them to be with us all the time.You may be sad to say goodbye to her, but we can remember and learn from her. Think ofher best qualities.You could also try looking for similar qualities in your other teachers. Study hard and giveyour new teacher a chance.In time, you might start to see that he or she has different qualities tolearn from.Finally, you can not completely rely on other people to get you through your studies , oreven your life.Editor45 The letters are from________.A bookB newspaperC posterD advertisement46 Which of the following does the editor advise Tuan Yan NOT to do?A Find a time to see Miss MaoB Have a short chat with Miss MaoC Tell Miss Mao she likes herD Keep quiet about the thing47. According to the passage, students can make progress in theirstudies _______.A completely by their favorite teachersB completely by their favorite classmatesC mainly by their parentsD mainly by themselvesDA young man was walking through a supermarket to pick up a few things when he noticed an old lady following him around. Thinking nothing of it,he ignored her and continued on. Finally, he went to the checkout line, but she got in front of him. “Pardon me,〞 she said, “ I’m sorry if my staring at you has made you feel uncomfortable. It’s just that you look just like my son who just died recently.〞“I’m very sorry,〞replied the young man, “Is there anything I can do for you?〞“Yes,〞she said, “As I’m leaving, can you say ‘Goodbye mother’? It would make me feel much better〞. “Sure,〞answered the young man.As the old lady was leaving the counter, he called out from behind her, “Goodbye mother!〞 As he stepped up to the checkout counter and put what he bought on the counter, he saw that his total was $12. “How can that be?〞he asked, “I only bought a few things!〞“ Your mother said that you would pay for her.〞 said the clerk.48 The underlined word ‘ignore’ most probably means _______.A. not to pay attention toB. to put on a smile atC. to watch out forD. to look down upon49. Which of the following is the right order of the events in the story?a. the man said “Goodbye mother!〞b. The woman left the supermarket.c. The woman spoke to the man.d. The man came to the counter.e. The woman got in front of the man.f. The woman followed the man.A. e,f,c,b,a,dB. c,e,f,a,b,dC.f,e,c,a,b,d D. d,e,c,f,b,a50. Which of the following would be the best title?A. An Unexpected Meeting at the SupermarketB. A Foolish MotherC. A Total of $ 12 at the SupermarketD. A Mother and Her Son第三阅读及写作〔共两节，总分值35分)第一小节〔共5小题；每题2分,总分值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正确选项。

2024年江西省宜春市第一中学高三下学期第三次模拟考试数学试卷（新高考）本试卷满分150分，考试时间120分钟注意事项：1．答卷前，考生务必将自己的姓名､准考证号填写在答题卡上．2．回答选择题时，选出每小题答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑．如需改动，用橡皮擦干净后，再选涂其他答案标号．回答非选择题时，将答案写在答题卡上，写在本试卷上无效．3．考试结束后，将本试卷和答题卡一并交回．一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1. 已知集合{|lg 0}M x x =>，{|N x y ==，则M N ⋂=（ ） A. （1，2]B. (4,)+∞C. (1,2)[4,)+∞D. (1,2][4,)+∞2. 已知复数z 满足2(1i)|34i |(1i)z -=++，则z 的虚部是（ ）A. -25B. -5C. 1D. 53. 下列说法不正确的是（ ）A. 一组数据1，4，14，6，13，10，17，19的25％分位数为5B. 一组数据m ，3，2，5，7中位数为3，则m 的取值范围是(,3]-∞C. 若随机变量1~(4,)3X B ，则方差(31)4D X +=D. 若随机变量2~(1,)X N σ，且(01)0.4<<=P X ，则(2)0.1P X >= 4. 设等差数列{}n a 的前n 项和为n S ，且1248S S -=，则133S S -=（ ） A. 10B. 12C. 14D. 165. 已知,m n 是空间中两条不同的直线，α，β是两个不同的平面，则下列说法错误的是（ ） A. 若m α⊥、//n α，则m n ⊥ B. 若m α⊥，//m n ，则n α⊥ C. 若//m n ，n β⊥，m α⊥，则//αβD. 若m α⊥，m n ⊥，则//n α的6. 已知0a >，且1a ≠，若函数1()(ln )x f x a x a -=-在(1,)+∞上单调递减，则a 的取值范围是（ ） A. 1(0,]eB. 1[,1)eC. (1,e]D. [e,)+∞7. 已知抛物线C ：24y x =焦点为F ，动直线l 与抛物线C 交于异于原点O 的A ，B 两点，以线段OA ，OB 为邻边作平行四边形OAPB ，若点()4,P m （0m >），则当||||AF BF +取最小值时，m =（ ） A. 2B.C. 3D.8.已知a =，b =，ln 44c =，其中e 2.71828= 为自然对数的底数，则（ ） A. b a c <<B. b c a <<C. a b c <<D. c b a <<二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9. 同时抛出两枚质地均匀骰子甲、乙，记事件A ：甲骰子点数为奇数，事件B ：乙骰子点数为偶数，事件C ：甲、乙骰子点数相同．下列说法正确的有（ ） A. 事件A 与事件B 对立 B. 事件A 与事件B 相互独立 C. 事件A 与事件C 相互独立D. ()()P C P AB =10. 古希腊数学家阿波罗尼斯的著作《圆锥曲线论》中给出了阿波罗尼斯圆的定义：在平面内，已知两定点A ，B 之间的距离为a （非零常数），动点M 到A ，B 的距离之比为常数λ（0λ>，且1λ≠），则点M 的轨迹是圆，简称为阿氏圆．在平面直角坐标系xOy 中，已知()()4,0,2,0A B -，点M 满足||2||MA MB =，则下列说法正确的是（ ）A. AMB 面积的最大值为12B. MA MB ⋅的最大值为72C. 若()88Q ,，则||2||MA MQ +的最小值为10D. 当点M 不在x 轴上时，MO 始终平分AMB ∠11. 设椭圆C ：22184x y +=的左、右焦点分别为1F ，2F ，坐标原点为O ．若椭圆C 上存在一点P ，使得||OP = ）A. 123cos 5F PF ∠=B. 125PF PF ⋅=C. 12F PF △的面积为2D. 12F PF △1-的的12. 如图，正方体1111ABCD A B C D -的棱长为2，设P 是棱1CC 的中点，Q 是线段1C P 上的动点（含端点），M 是正方形11BCC B 内（含边界）的动点，且1//A M 平面1D AP ，则下列结论正确的是（ ）A. 存在满足条件的点M ，使11A M AD ⊥B. 当点Q 在线段1C P 上移动时，必存在点M ，使1A M BQ ⊥C. 三棱锥11C A PM -的体积存在最大值和最小值D. 直线1A M 与平面11BCC B 所成角的余弦值的取值范围是11[,]32三、填空题：本题共4小题，每小题5分，共20分.13. 已知a ，b 均为非零向量，若|2|||2||a b b a -== ，则a 与b的夹角为________．14. 已知π0θ4<<，且πtan 2θtan(θ)44⋅+=，则cos 2θ1sin 2θ=-________． 15. 已知0x >，0y >，且满足2249630x y xy ++-=，则23x y +最大值为________． 16. 已知方程ln e 1xx mx+=+在(0,1)上有两个不相等的实数根，则实数m 的取值范围是________． 四、解答题：本题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.17. 在ABC 中，设角A ，B ，C 所对的边分别为a ，b ，c ．已知120C =︒，ABC 的周长为15，面积为． （1）求ABC 的外接圆面积；（2）设D 是边AB 上一点，在①CD 是边AB 上的中线；②CD 是ACB ∠的角平分线这两个条件中任选一个，求线段CD 的长．18. 在正项数列{}n a 中，已知11a =，且11(1)1n n n nna n a a a +++-=． （1）求数列{}n a 的通项公式；的（2）求证：2(1)3nn a ≤+<．19. 如图，在四棱锥P-ABCD 中，已知底面ABCD 为菱形，平面PAB ⊥底面ABCD ，M 为棱BC 上异于点C 的一点，O 为棱AB 的中点，且PA PB AB ==，60ABC ∠=︒．（1）若BD PM ⊥，求证：M 为BC 中点；（2）若平面POM 与平面PAC，求BM BC 的值．20. 据教育部统计，2024届全国高校毕业生规模预计达1179万，同比增加21万，岗位竞争激烈．为落实国务院关于高校毕业生就业工作的决策部署，搭建高校毕业生和用人单位求职招聘的双向对接通道，促进高校毕业生高质量充分就业，某市人社局联合市内高校开展2024届高校毕业生就业服务活动系列招聘会．参加招聘会的小王打算依次去甲、乙、丙三家公司应聘．假设小王通过某公司的专业测试就能与该公司签约，享受对应的薪资待遇，且不去下一家公司应聘，或者放弃签约并参加下一家公司的应聘；若未通过测试，则不能签约，也不再选择下一家公司．已知甲、乙、丙三家公司提供的年薪分别为10万元、12万元、18万元，小王通过甲、乙、丙三家公司测试的概率分别为23，12，13，通过甲公司的测试后选择签约的概率为34，通过乙公司的测试后选择签约的概率为35，通过丙公司的测试后一定签约．每次是否通过测试、是否签约均互不影响．（1）求小王通过甲公司的测试但未与任何公司签约的概率；（2）设小王获得的年薪为X （单位：万元），求X 的分布列及其数学期望． 21. 已知函数()e l R n (e ),x x f x x x a a --∈=，．（1）若()0f x ≤对(0,)∀∈+∞x 恒成立，求实数a 的取值范围；（2）若曲线()y f x =与x 轴交于A ，B 两点，且线段AB 的中点为00(),M x ，求证：01x >．22. 已知以点M 为圆心的动圆经过点1(3,0)F -，且与圆心为2F 的圆22(3)12x y -+=相切，记点M 的轨迹为曲线C ．（1）求曲线C 的方程；的（2）若动直线l 与曲线C 交于11(,)A x y ，22(,)B x y 两点（其中120y y >），点A 关于x 轴对称的点为A'，且直线BA'经过点()1,0P -． （ⅰ）求证：直线l 过定点；（ⅱ）若||||PA PB +=，求直线l 的方程．参考答案一、选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1. 已知集合{|lg 0}M x x =>，{|N x y ==，则M N ⋂=（ ） A. （1，2] B. (4,)+∞ C. (1,2)[4,)+∞ D. (1,2][4,)+∞【答案】D 【解析】【分析】分别化简集合,M N ，取交集即可.【详解】{|lg 0}{|1}M x x x x =>=>，{|{|4N x y x x ===≥或2}x ≤，所以{}1{|4M N x x x x ⋂=⋂≥或][()2}1,24,x ∞≤=⋃+. 故选:D .2. 已知复数z 满足2(1i)|34i |(1i)z -=++，则z 的虚部是（ ） A. -25 B. -5C. 1D. 5【答案】B 【解析】【分析】由复数的模定义求得|34i |+，利用复数的四则运算求得z ，再由共轭复数定义得55i z =--即可得出结论. 【详解】由2(1i)|34i |(1i)z -=++，得(1i)52iz -=， 所以10i 10i(1i)55i 1i 2z +===-+-，所以55i z =--． 故选：B ．3. 下列说法不正确的是（ ）A. 一组数据1，4，14，6，13，10，17，19的25％分位数为5B. 一组数据m ，3，2，5，7的中位数为3，则m 的取值范围是(,3]-∞C. 若随机变量1~(4,)3X B ，则方差(31)4D X +=D. 若随机变量2~(1,)X N σ，且(01)0.4<<=P X ，则(2)0.1P X >= 【答案】C 【解析】【分析】对于A ，先把数据从小到大排列，利用百分位定义计算即可；对于B ，根据中位数的定义讨论即可；对于C ，根据二项分布的方差公式计算即可；对于D ，根据正态分布的对称性求解.【详解】对于A ，该组数据共8个，且825%2⨯=，所以25％分位数为从小到大排列后第2个数和第3个数的平均数，即为4652+=，故A 正确； 对于B ，若5m ≥，则这组数据由小到大排列依次为2，3，5，m ，7或2，3，5，7，m ，中位数为5，不合题意；若35m <<，则这组数据由小到大排列依次为2，3，m ，5，7，中位数为3m ≠，不合题意； 若3m ≤，则这组数据由小到大排列依次为2，m ，3，5，7或m ，2，3，5，7，中位数为3，故实数m 的取值范围是(,3]-∞，故B 正确；对于C ，若随机变量1~(4,3X B ，则118()4(1)339D X =⨯⨯-=，所以28(31)3()989D X D X +==⨯=，故C 错误；对于D ，若随机变量2~(1,)X N σ，且(01)0.4<<=P X ，则(2)0.5(12)0.5(01)0.1P X P X P X >=-<<=-<<=，故D 正确．故选：C ．4. 设等差数列{}n a 的前n 项和为n S ，且1248S S -=，则133S S -=（ ） A. 10 B. 12C. 14D. 16【答案】A 【解析】【分析】利用等差数列求和公式化简1248S S -=可得12152a d +=，将133S S -化简可得13315(215)S S a d -=+，计算可得结果.【详解】设{}n a 的公差为d ，由1248S S -=，得1112114312(4)822a d a d ⨯⨯+-+=， 化简为12152a d +=， 所以1331111312321335(215)521022S S a d a d a d ⨯⨯⎛⎫-=+-+=+=⨯= ⎪⎝⎭． 故选：A ．5. 已知,m n 是空间中两条不同的直线，α，β是两个不同的平面，则下列说法错误的是（ ） A. 若m α⊥、//n α，则m n ⊥ B. 若m α⊥，//m n ，则n α⊥ C. 若//m n ，n β⊥，m α⊥，则//αβ D. 若m α⊥，m n ⊥，则//n α【答案】D 【解析】【分析】对于A ，可过n 作平面β，使l βα⋂=，则//n l ，即可判断；对于B ，由线面垂直的性质即可判断；对于C ，由条件，可得m β⊥，又m α⊥，则//αβ，即可判断；对于D ，要考虑n 可能在平面α内，即可判断．【详解】对于A ，当//n α时，过n 作平面β，使l βα⋂=，则//n l ，因为m α⊥，l ⊂α，所以m l ⊥，所以m n ⊥，故A 正确；对于B ，当m α⊥，//m n ，由线面垂直的性质可得n α⊥，故B 正确；对于C ，因为//m n ，n β⊥，所以m β⊥，又m α⊥，所以//αβ，故C 正确； 对于D ，当m α⊥，m n ⊥时，n 可能在平面α内，故D 错误． 故选：D ．6. 已知0a >，且1a ≠，若函数1()(ln )x f x a x a -=-在(1,)+∞上单调递减，则a 的取值范围是（ ） A. 1(0,]eB. 1[,1)eC. (1,e]D. [e,)+∞【答案】D 【解析】【分析】根据题意，转化为()0f x '≤在(1,)+∞上恒成立，令()()g x f x '=，利用导数求得函数()g x 单调递减，得到()(1)ln g x g a a a <=-，得出ln 0a a a -≤，即可求解. 【详解】由函数1()(ln )x f x a x a-=-，可得()ln x af x a a x-'=因为()f x 在(1,)+∞上单调递减，所以()0f x '≤在(1,)+∞上恒成立， 令()()ln x a g x f x a a x '==-，则22()(ln )0x ag x a a x'=--<， 所以()g x 在(1,)+∞上单调递减，所以()(1)ln g x g a a a <=-，即()ln f x a a a '<-， 则ln 0a a a -≤，解得e a ≥，即实数a 的取值范围是[e,)+∞． 故选：D ．7. 已知抛物线C ：24y x =的焦点为F ，动直线l 与抛物线C 交于异于原点O 的A ，B 两点，以线段OA ，OB 为邻边作平行四边形OAPB ，若点()4,P m （0m >），则当||||AF BF +取最小值时，m =（ ） A. 2B.C. 3D.【答案】B 【解析】【分析】根据题意，由抛物线的方程可得焦点坐标以及准线方程，然后分别过A 、B 、M 向准线作垂线，||||AF BF +取最小值即直线AB 过焦点()1,0F 时，再结合点差法代入计算，即可得到结果.【详解】由题可知焦点()1,0F ，准线=1x -，设线段AB 中点为00(,)M x y ，即为OP 中点， 则0422x ==，02my =．分别过A 、B 、M 向准线作垂线，垂足分别为1A ，1B ，1M ，如图所示．则||||||AF BF AB +≥，当直线AB 过焦点()1,0F 时取等号，此时10||2||2|1|426AB MM x ==+=+=．的设11(,)A x y 、22(,)B x y ，直线AB 的斜率为k ，由21122244y x y x ⎧=⎨=⎩，两式相减，得2212124()y y x x -=-，所以12121222y y y y x x +-⋅=-， 即02y k =，得00221y y ⋅=-，所以22y =，又0m >，所以02m y == 故选：B ． 8.已知a =，b =，ln 44c =，其中e 2.71828= 为自然对数的底数，则（ ） A. b a c << B. b c a <<C. a b c <<D. c b a <<【答案】A 【解析】【分析】首先将,,a b c 化成统一形式，构造函数()ln xf x x=()0x >，研究单调性进而比较大小即可.【详解】由题意得a ==，b ==，ln 42ln 2ln 2442c ===； 设()ln x f x x =，则21ln ()xf x x -'=， 当0e x <<时，()0f x '>，所以()f x单调递增，又02e <<<<，所以(2)f f f <<ln 22<<，所以b a c <<. 故选：A .二、选择题：本题共4小题，每小题5分，共20分.在每小题给出的选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9. 同时抛出两枚质地均匀的骰子甲、乙，记事件A ：甲骰子点数为奇数，事件B ：乙骰子点数为偶数，事件C ：甲、乙骰子点数相同．下列说法正确的有（ ） A. 事件A 与事件B 对立 B. 事件A 与事件B 相互独立 C. 事件A 与事件C 相互独立 D. ()()P C P AB =【答案】BC 【解析】【分析】对于A ，甲骰子点数为奇数，乙骰子点数为偶数，事件可以同时发生，由对立事件的概念可判断；对于B ，计算出()()()P A P B P AB ，，根据()()()P AB P A P B =可以判定两个事件是否相互独立；对于C ，计算出()()()P A P C P AC ，，根据()()()P AC P A P C =可以判定两个事件是否相互独立；对于D ，由前面可知()()P C P AB ，，即可判断是否相等.【详解】由题意，得1()2P A =，1()2P B =，61()366P C ==， 对于A ，当甲为奇数点，且乙为偶数点时，事件可以同时发生，所以事件A 与事件B 不互斥，故事件A 与事件B 不对立，故A 错误；对于B ，由题意知11331166C C 1()C C 4P AB ==，又111()()()224P A P B P AB =⨯==，故事件A 与事件B 相互独立，故B 正确； 对于C ，31()3612P AC ==，又111()()()2612P A P C P AC =⨯==，故事件A 与事件C 相互独立，故C 正确；对于D ，由上知，11()()64P C P AB =<=，故D 错误． 故选：BC ．10. 古希腊数学家阿波罗尼斯的著作《圆锥曲线论》中给出了阿波罗尼斯圆的定义：在平面内，已知两定点A ，B 之间的距离为a （非零常数），动点M 到A ，B 的距离之比为常数λ（0λ>，且1λ≠），则点M 的轨迹是圆，简称为阿氏圆．在平面直角坐标系xOy 中，已知()()4,0,2,0A B -，点M 满足||2||MA MB =，则下列说法正确的是（ ）A. AMB 面积的最大值为12B. MA MB ⋅的最大值为72C. 若()88Q ,，则||2||MA MQ +的最小值为10D. 当点M 不在x 轴上时，MO 始终平分AMB ∠【答案】ABD 【解析】【分析】设点(,)M x y ，由条件可得点M 的轨迹方程，即可判断A ，由向量数量积的运算律代入计算，即可判断B ，由点与圆的位置关系，即可判断C ，由角平分线定理即可判断D【详解】对于A ，设点(,)M x y ，由||2||MA MB ==， 化为22(4)16x y -+=，所以点M 的轨迹是以点()4,0为圆心、4为半径的圆，所以AMB 面积的最大值为11||641222AB r =⨯⨯=，故A 正确； 对于B ，设线段AB 的中点为N ，2222()()||(81)(14)72MA MB MN NA MN NB MN NA ⋅=+⋅+=-≤+--+= ，当点M 的坐标为()8,0时取等号，故MA MB ⋅的最大值为72，故B 正确； 对于C ，显然点()8,8Q 在圆外，点()2,0B 在圆内，()2222220MA MQ MB MQ MB MQ BQ +=+=+≥==，当B ，M ，Q 三点共线且点M 在线段BQ 之间时，min (||2||)20MA MQ +=，故C 错误； 对于D ，由||4OA =，||2OB =，有||||2||||OA MA OB MB ==，当点M 不在x 轴上时， 由三角形内角平分线分线段成比例定理的逆定理知，MO 是AMB 中AMB ∠的平分线，故D 正确．故选：ABD ． 11. 设椭圆C ：22184x y +=的左、右焦点分别为1F ，2F ，坐标原点为O ．若椭圆C 上存在一点P ，使得||OP = ）A. 123cos 5F PF ∠=B. 125PF PF ⋅=C. 12F PF △的面积为2D. 12F PF △1-【答案】ACD 【解析】【分析】根据已知求出P 点坐标，根据两点间距离公式分布求出12,PF PF ，在12F PF △中利用余弦定理可判定A ，利用向量数量积公式可判定B ，三角形面积公式可判定C ，根据等面积法可判定D.【详解】法1：由题意得a =12||24F F c ===，则1(2,0)F -，2(2,0)F ．由对称性可设00(,)P x y （00x >，00y >），1||PF m =，2||PF n =，12F PF θ∠=，由2200184x y ⎧+=⎪=，解得001x y ⎧=⎪⎨=⎪⎩1(2,0)F -，2(2,0)F ，所以m ==，n ==，所以5mn ===．由椭圆的定义得2m n a +==12F PF △中，由余弦定理，得22212||2cos F F m n mn θ=+-，即2224()22cos 2525cos m n mn mn θθ=+--=-⨯-⨯， 解得3cos 5θ=，故A 正确； 123cos 535PF PF mn θ⋅==⨯= ，故B 错误；12F PF △的面积为1211sin 5222F PF S mn θ==⨯= ，故C 正确； 设12F PF △的内切圆半径为r ，由12F PF △的面积相等，得12121(||)2F PF S m n F F r =++△，即124)2r =+，解得1r =-，故D 正确． 故选：ACD ．法2：设1||PF m =，2||PF n =，12F PF θ∠=．易知a =，2c ==，由极化恒等式，得22121||||743PF PF OP OF ⋅=-=-=，故B 错误；由中线长定理得222212(||||)22m n OP OF +=+=，由椭圆定义得2m n a +== 所以222()222232m n m n mn mn +=++=+=，所以5mn =，所以123cos 5PF PF mn θ⋅== ，故A 正确；由3cos 5θ=，得4sin 5θ==，所以12114sin 52225F PF S mn θ==⨯⨯= ，故C 正确； 设12F PF △的内切圆半径为r ，由12F PF △的面积相等，得12121(||)2F PF S m n F F r =++△，在即124)2r =+，解得1r =-，故D 正确． 故选：ACD ．12. 如图，正方体1111ABCD A B C D -的棱长为2，设P 是棱1CC 的中点，Q 是线段1C P 上的动点（含端点），M 是正方形11BCC B 内（含边界）的动点，且1//A M 平面1D AP ，则下列结论正确的是（ ）A. 存在满足条件的点M ，使11A M AD ⊥B. 当点Q 在线段1C P 上移动时，必存在点M ，使1A M BQ ⊥C. 三棱锥11C A PM -的体积存在最大值和最小值D. 直线1A M 与平面11BCC B 所成角的余弦值的取值范围是11[,]32【答案】ABC 【解析】【分析】由已知，取11B C 的中点E ，1BB 的中点F ，并连接，可得点M 的轨迹为线段EF ．对于A ，连接1AC ，1B C 交1BC 于点O ，可得1BC ⊥平面11A B C ，当M 为线段EF 中点时，11BC A M ⊥，又11//BC AD ，则可判断：对于B ，分别以向量DA ，DC ，1DD的方向为x ，y ，z 轴的正方向建立空间直角坐标系，由空间向量坐标运算可得存在10A M BQ ⋅=，即可判断；对于C ，设点M 到1C P 的距离为h ，可知当M 与E 重合时，min 11h EC ==，当M 与F 重合时，max 2h FP ==，即可求出三棱锥11C A PM -的体积存在最大值和最小值，则可判断；对于D ，由11A B ⊥平面11BCC B 知，11A MB Ð即为直线1A M 与平面11BCC B 所成的角，在1B EF 中，可得11B M ≤≤，则得2tan θ≤≤进而得1cos 3θ≤≤.【详解】取11B C 的中点E ，1BB 的中点F ，连接1A E ，1A F ，EF ，1BC ，如图所示．易知11////EF BC AD ，11//A F D P ，因为EF ⊂平面1A EF ，EF ⊄平面1D AP ，所以//EF 平面1D AP ， 同理，1//A F 平面1D AP ，又1A F EF F ⋂=，又1,EF A F ⊂平面1A EF ， 所以平面1//A EF 平面1D AP ，又1//A M 平面1D AP ， 所以1A M ⊂平面1A EF ，故点M 的轨迹为线段EF ．对于A ，连接1AC ，1B C 交1BC 于点O ，如图所示．则11BC B C ⊥，又111A B BC ⊥，1111A B B C B = ，111A B B C ⊂、平面11A B C ， 所以1BC ⊥平面11A B C ，当M 为线段EF 中点时，11BC A M ⊥， 因11//BC AD ，所以11A M AD ⊥，故A 正确；对于B ，分别以向量DA ，DC ，1DD的方向为x ，y ，z 轴的正方向建立空间直角坐标系，如图所示．为则1(2,0,2)A ，()()()()2,2,0,1,2,2,2,2,1,0,2,B E F Q m ()12m ≤≤，设EM EF λ=（01λ≤≤），得(1,2,2)M λλ+-，从而1(1,2)A M λλ=-- ， 又(2,0,)BQ m =-，令10A M BQ ⋅= ，得2(1)0m λλ---=，当0λ=时，显然不合题意； 当01λ<≤时，由2212m λλ-≤=≤，解得1223λ≤≤， 即当点Q 在线段1C P 上移动时，均存在点M ，使1A M BQ ⊥，故B 正确； 对于C ，设点M 到1C P 的距离为h ， 则三棱锥11C A PM -的体积为111111111111113323C A PM A C PM C PM V V S A B C P h A B h --==⨯=⨯⨯⨯=△，当M 与E 重合时，min 11h EC ==，得11min 1()3C A PM V -=；当M 与F 重合时，max 2h FP ==，得11max 2()3C A PM V -=，故C 正确；对于D ，设直线1A M 与平面11BCC B 所成的角为θ、连接1B M ，如图所示．由11A B ⊥平面11BCC B 知，11A MB θ∠=，在1B EF 中，11111B E B F B M B F EF ⨯=≤≤=，得2tan θ≤≤， 所以2222sin 1cos 48cos cos θθθθ-≤=≤，所以1cos 3θ≤≤D 错误． 故选：ABC ．【点睛】关键点点睛，本题关键是先找到点M 的轨迹，对于B 选项，通过设出向量的含参坐标，借助参数的范围满足条件，得到答案；对于C 选项，利用等积转化，转化成棱锥高取得最值，可得体积最值；对于D 选项，关键是找到线面角正切的范围，进而得到余弦的范围.三、填空题：本题共4小题，每小题5分，共20分.13. 已知a ，b 均为非零向量，若|2|||2||a b b a -== ，则a 与b的夹角为________．【答案】π3【解析】【分析】根据题意，求得2||a b a ⋅= ，结合||2||b a =，利用向量的夹角公式，即可求解. 【详解】由|2|||a b b -= ，可得22|2|||a b b -= ，即2224||4||||a a b b b -⋅+= ，解得2||a b a ⋅= ， 因为||2||b a = ，所以22||1cos ,2||||2||a b a a b a b a ⋅=== ， 又因为0,πa b ≤≤ ，所以π,3a b = ．故答案为：π3.14. 已知π0θ4<<，且πtan 2θtan(θ)44⋅+=，则cos 2θ1sin 2θ=-________． 【答案】3 【解析】【分析】先结合二倍角的正切与两角和的正切公式及角θ的取值范围，得到tan θ，再利用倍角公式把cos 2θ1sin 2θ-转化为齐次式求解.【详解】由πtan 2θtan(θ44⋅+=，得22tan θtan θ141tan θ1tan θ+⋅=--， 即22tan θ5tan θ20-+=，又π0θ4<<，所以1tan θ2=，从而2222cos 2θcos θsin θ1sin 2θsin θcos θ2sin θcos θ-=-+-2(cos θsin θ)(cos θsin θ)(cos θsin θ)+-=-cos θsin θcos θsin θ+=-111tan θ2311tan θ12++===--． 故答案为：315. 已知0x >，0y >，且满足2249630x y xy ++-=，则23x y +的最大值为________． 【答案】2 【解析】【分析】解法1、根据题意，得到22491236x y xy xy ++=+，结合基本不等式求得23(23)34x y +≤，进而求得23x y +的最大值；解法2、根据题意，得到222(96)33x y xy x +++=，利用权方和不等式得24(23)x y +≥，进而求得23x y +的最大值．【详解】解法1、由2249630x y xy ++-=，可得22491236x y xy xy ++=+， 由基本不等式得2223(23)3233()2x y x y x y ++=+⋅≤+，可得23(23)34x y +≤， 所以232x y +≤，当且仅当23x y =时取等号，联立方程组222349630x y x y xy =⎧⎨++-=⎩，解得12x =，13y =，故23x y +的最大值为2． 解法2、由2249630x y xy ++-=，可得222(96)33x y xy x +++=，因为0,0x y >>，由权方和不等式得222(3)(3)111133x y x x y x +++++≥，即24(23)x y +≥， 所以232x y +≤，当且仅当3113x y x+=，即23x y =时取等号， 联立方程组222349630x y x y xy =⎧⎨++-=⎩，解得12x =，13y =，故23x y +的最大值为2． 故答案为：2.16. 已知方程ln e 1xx mx+=+在(0,1)上有两个不相等的实数根，则实数m 的取值范围是________． 【答案】(1,e 1)- 【解析】【分析】先分离参数m ，构造函数()e ln xf x x x x =--，将问题转化为函数()y f x =的图象与直线y m =在(0,1)上有两个交点，再将()f x 变形，构造函数()ln h t t t =-，0e t <<，通过导数研究函数()h t 进而求出m 的取值范围.【详解】由ln e 1xx mx+=+，得e ln x m x x x =--，令()e ln x f x x x x =--， 则函数()y f x =的图象与直线y m =在(0,1)上有两个交点， 而()e (ln e ln )e ln(e )x x x xf x x x x x =-+=-；令()e x g x x =，(0,1)x ∈，则()(1)e 0x g x x '=+>恒成立，故()g x 在(0,1)上单调递增，故(0)()(1)g g x g <<，即0e e x x <<， 令e x t x =，函数()ln h t t t =-，0e t <<，则函数()y h t =的图象与直线y m =有两个交点，由11()1t h t t t-'=-=， 则当(0,1)t ∈时，()0h t '<，当(1,e)t ∈时，()0h t '>，故()h t 在(0,1)上单调递减，在(1,e)上单调递增，所以函数()h t 在1t =处有极小值， 且e 1(e)()(1)1ln11h h t h -=>=-=≥，当0t >且0t →时，()h t →+∞， 所以1e 1m <<-，即实数m 的取值范围是(1,e 1)-. 故答案为：(1,e 1)-.【点睛】方法点睛：()()()()e ln e ln e ln e ln e 0xxxxxf x x x x x x x x x =--=-+=->，将函数化成相同整体变量进而构造函数解决参数取值范围是解决导数问题的常用方法.四、解答题：本题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.17. 在ABC 中，设角A ，B ，C 所对的边分别为a ，b ，c ．已知120C =︒，ABC 的周长为15，面积为．（1）求ABC 的外接圆面积；（2）设D 是边AB 上一点，在①CD 是边AB 上中线；②CD 是ACB ∠的角平分线这两个条件中任选一个，求线段CD 的长． 【答案】（1）49π3（2）答案见解析 【解析】【分析】（1）由ABC，求得15ab =，再由ABC 的周长为15，得到15a b c +=-，结合余弦定理，求得7c =，再由正弦定理，求得外接圆半径即可求解； （2）若选择①：法1：由1()2CD CA CB =+ ，结合向量的运算法则，即可求解； 法2：设b a >，列出方程组求得3,5a b ==，结合cos cos 0ADC CDB ∠+∠=，列出方程，即可求解； 若选择②，设b a >，求得3,5a b ==，根据ABC ACD BCD S S S =+△△△，列出方程，即可求解； 法2：由sin sin sin ACB BCD ACDCD AC BC∠∠∠=+，列出方程，即可求解.【小问1详解】 解：由ABC，可得1sin1202ABC S ab =︒=△15ab =， 又由ABC 的周长为15，可得15a b c ++=，即15a b c +=-， 由余弦定理得22222cos ()22cos120c a b ab C a b ab ab =+-=+--︒21(15)215215()2c =--⨯-⨯⨯-，解得7c =，设外接圆半径为R ，由正弦定理得72sin120R =︒，所以R =，所以ABC 的外接圆面积为249ππ3R =． 【小问2详解】 解：若选择①：法1：由（1）知，158a b c +=-=及15ab =，由1()2CD CA CB =+ ，可得222211||()(2cos120)44CD CA CB b a ab =+=++︒的221119[()3](8315)444a b ab =+-=⨯-⨯=，所以||CD =CD =． 法2：不妨设b a >，由158a b c +=-=及15ab =，解得3,5a b ==， 在ACD 和BCD △中，可得cos cos 0ADC CDB ∠+∠=，由余弦定理得22222277()5()3220772222CD CD CD CD +-+-+=⨯⨯⨯⨯，解得CD =． 若选择②，不妨设b a >，由158a b c +=-=及15ab =，解得3,5a b ==， 法1：由ABC ACD BCD S S S =+△△△，115sin 603sin 6022CD CD =⨯⨯︒+⨯⨯︒，解得158CD =． 法2：由张角定理，得sin sin sin ACB BCD ACDCD AC BC∠∠∠=+，即sin120sin 60sin 6053CD ︒︒︒=+，解得158CD =， 18. 在正项数列{}n a 中，已知11a =，且11(1)1n n n nna n a a a +++-=． （1）求数列{}n a 的通项公式； （2）求证：2(1)3nn a ≤+<． 【答案】（1）1n a n= （2）证明见解析 【解析】【分析】（1）根据题意，化简得到1(1)0n n n a na ++-=，得到数列{}n na 为等差数列，进而求得数列的通项公式；（2）由（1）知1n a n =，结合二项式定理，得到111(1)(1)1C n n n n a n n+=+≥+⋅，再结合1C 11!2k nk k n k -≤≤，结合等比数列的求和公式，即可得证. 【小问1详解】解：由11(1)1n n n nna n a a a +++-=，可得2211(1)0n n n n n a a a na ++++-=， 即11()[(1)]0n n n n a a n a na ++++-=，因为10,0n n a a +>>，所以1(1)0n n n a na ++-=， 所以数列{}n na 是首项为1，公差为0的等差数列，又因为11a =，所以1n na =，所以数列{}n a 的通项公式为1n a n=． 【小问2详解】 解：由（1）知1n a n=， 则121211111(1)(11C C 1C 2nnn n n n n a n n n n n+=+=+⋅+⋅+++⋅= ≥，当1n =时，取等号， 因为1C (1)(2)(1)11!!2k nk k k n n n n k n n k k ----+=≤≤ ， 所以12221111111111121C C 111331222212n n n n n n n n n ---+⋅+⋅++≤+++++=+=-<- ， 所以2(1)3nn a ≤+<．19. 如图，在四棱锥P-ABCD 中，已知底面ABCD 为菱形，平面PAB ⊥底面ABCD ，M 为棱BC 上异于点C 的一点，O 为棱AB 的中点，且PA PB AB ==，60ABC ∠=︒．（1）若BD PM ⊥，求证：M 为BC 的中点；（2）若平面POM 与平面PAC，求BM BC 的值．【答案】（1）证明见解析（2）14BM BC = 【解析】【分析】（1）根据题意，证得PO ⊥平面ABCD ，得到BD PO ⊥，再证得BD ⊥平面POM ，得到BD OM ⊥，进而得到//OM AC ，即可得到M 为BC 的中点；（2）以O 为原点，建立空间直角坐标系， 设BM BC λ=，分别求得平面PAC 和平面POM 的法向量，结合向量的夹角公式，列出方程求得λ的值，即可求解. 【小问1详解】证明：因为PA PB =，O 为AB 的中点，所以PO AB ⊥，又平面PAB ⊥底面ABCD ，平面PAB ⋂底面ABCD AB =，PO ⊂平面PAB ， 所以PO ⊥平面ABCD ，因为BD ⊂平面ABCD ，所以BD PO ⊥，因为BD PM ⊥，且PO PM P = ，,PO PM ⊂平面POM ，所以BD ⊥平面POM ， 又因为OM ⊂平面POM ，所以BD OM ⊥，因为BD AC ⊥，且,OM AC ⊂底面ABCD ，所以//OM AC ， 又因为O 为AB 的中点，所以M 为BC 的中点． 【小问2详解】解：连接OC ，因为,60AB BC ABC =∠=︒，所以ABC 为正三角形，所以OC AB ⊥，以O 为原点，分别以向量OB ，OC ，OP的方向为x ，y ，z 轴正方向建立空间直角坐标系，如图所示，设4AB =，则(2,0,0),(2,0,0),(0,(0,0,A B C P -，所以(0,0,OP =，(BC =-，AC =，(2,0,AP =， 设1111(,,)n x y z = 是平面PAC的一个法向量，则1111112020n AC x n AP x ⎧⋅=+=⎪⎨⋅=+=⎪⎩，取1x =111y z ==-，所以11,1)n =--，设(01)BM BC λλ=≤<，则(22,,0)M λ-，即(22,,0)OM λ=- ， 设2222(,,)n x y z = 是平面POM的一个法向量，则22222(22)0n OM x y n OP λ⎧⋅=-+=⎪⎨⋅==⎪⎩， 取21x =，可得220y z ==，所以2(1,n = ， 因为平面POM 与平面PAC所以121212cos ,n n n n n n ⋅===整理得24510λλ-+=，解得14λ=，或1λ=（舍去），所以14BM BC =．20. 据教育部统计，2024届全国高校毕业生规模预计达1179万，同比增加21万，岗位竞争激烈．为落实国务院关于高校毕业生就业工作的决策部署，搭建高校毕业生和用人单位求职招聘的双向对接通道，促进高校毕业生高质量充分就业，某市人社局联合市内高校开展2024届高校毕业生就业服务活动系列招聘会．参加招聘会的小王打算依次去甲、乙、丙三家公司应聘．假设小王通过某公司的专业测试就能与该公司签约，享受对应的薪资待遇，且不去下一家公司应聘，或者放弃签约并参加下一家公司的应聘；若未通过测试，则不能签约，也不再选择下一家公司．已知甲、乙、丙三家公司提供的年薪分别为10万元、12万元、18万元，小王通过甲、乙、丙三家公司测试的概率分别为23，12，13，通过甲公司的测试后选择签约的概率为34，通过乙公司的测试后选择签约的概率为35，通过丙公司的测试后一定签约．每次是否通过测试、是否签约均互不影响．（1）求小王通过甲公司的测试但未与任何公司签约的概率；（2）设小王获得的年薪为X （单位：万元），求X 的分布列及其数学期望． 【答案】（1）19180（2）分布列见解析，295【解析】【分析】（1）记事件A ：小王通过甲公司的测试，但未通过乙公司的测试，记事件B ：小王通过甲、乙公司的测试，但未通过丙公司的测试，根据相互独立事件及互斥事件的概率公式计算可得；（2）依题意X 的可能取值为0，10，12，18，求出所对应的概率，即可得到分布列与数学期望.【小问1详解】记事件A ：小王通过甲公司的测试，但未通过乙公司的测试， 记事件B ：小王通过甲、乙公司的测试，但未通过丙公司的测试， 则2311()(1(1)34212P A =⨯-⨯-=，231311()(1)(1)(13425345P B =⨯-⨯⨯-⨯-=， 显然A 与B 互斥，所以小王通过甲公司的测试但未与任何公司签约的概率1119()()1245180P P A P B =+=+=． 【小问2详解】依题意X 的可能取值为0，10，12，18， 则11979(0)3180180P X ==+=，231(10)342P X ==⨯=， 21131(12)342520P X ==⨯⨯⨯=，211211(18)3425390P X ==⨯⨯⨯⨯=，则X 的分布列如下表：X 0 10 12 18P7918012120 190故7911129()0101218180220905E X =⨯+⨯+⨯+⨯=． 21. 已知函数()e l R n (e ),x x f x x x a a --∈=，．（1）若()0f x ≤对(0,)∀∈+∞x 恒成立，求实数a 的取值范围；（2）若曲线()y f x =与x 轴交于A ，B 两点，且线段AB 的中点为00(),M x ，求证：01x >． 【答案】（1）(,e]-∞（2）证明见解析 【解析】【分析】（1）根据题意，把不等式()0f x ≤转化为e ln e x xx a x ≤-恒成立，令e ln ()e x xx h x x=-，求得2(1)(1ln )()e xx x x h x x-+-'=，设()1ln g x x x =+-，利用导数求得()g x 的单调性，结合()0g x >，进而得到()h x 单调性和()min e h x =，即可求解；（2）根据题意，转化为()a h x =有两个实根，设1201x x <<<，因为ln ()(ln )e x x h x x x -=-，转化为1122ln ln 1122(ln )e (ln )e x x x x x x x x ---=-，构造()e x H x x =，利用导数得到()H x 在[1,)+∞递增，得到12121ln ln x x x x -=-，转化为证1211222(1)ln 1x x x x x x ->+，令2(1)()ln 1x G x x x -=-+在利用导数求得函数()G x 的单调性，得到()0G x >，取12(0,1)x x x =∈，即可得证. 【小问1详解】解：因为函数()e ln (e )xxf x x x a =--，可得其定义域为(0,)+∞，由()0f x ≤，即e ln (e )0xxx x a --≤，化为e ln e x xxa x≤-，因为()0f x ≤对(0,)∀∈+∞x 恒成立，即e ln e x xxa x≤-恒成立，令e ln ()e x xx h x x =-，则min [()]a h x ≤，可得2221(1)ln (1)(1ln )()e e x xx x x x x x h x x x-+--+-'==， 设()1ln g x x x =+-，则11()1x g x x x-'=-=， 当(0,1)x ∈时，()0g x '<，当(1,)x ∈+∞时，()0g x '>，故()g x 在区间(0,1)内单调递减，在区间(1,)+∞内单调递增，所以()(1)20g x g ≥=>， 当(0,1)x ∈时，()0h x '<，当(1,)x ∈+∞时，()0h x '>，故()h x 在区间(0,1)内单调递减，在区间(1,)+∞内单调递增，则()()min 1e h x h ==， 所以实数a 的取值范围是(,e]-∞． 【小问2详解】 证明：令()0f x =，由（1）知方程()a h x =有两个不等实根，且一根小于1，另一根大于1，不妨设1201x x <<<，因为ln e ln ()e (ln )e x xx x xh x x x x-=-=-，所以1122ln ln 1122(ln )e (ln )e x x x x x x x x ---=-，又因为ln 1x x -≥，构造函数()e x H x x =，[1,)x ∈+∞，则()(1)e 0x H x x '=+>， 得()H x 在[1,)+∞单调递增，12()()H x H x =，即1122ln ln x x x x -=-，即1212ln ln x x x x -=-，即12121ln ln x x x x -=-， 要证01x >，即证1212x x +>， 即证1212122ln ln x x x x x x +->-，即证1211222(1)ln 1x x xx x x ->+， 构造函数2(1)()ln ,(0,1)1x G x x x x -=-∈+， 则22222414(1)(1)()0(1)(1)(1)x x x G x x x x x x x -+-'=-==-<+++， 故()G x 在区间(0,1)内单调递减，则()(1)0>=G x G ，即2(1)ln 01x x x -->+， 取12(0,1)x x x =∈，则有1211222(1)ln 01x x xx x x -->+，即12121212ln ln x x x x x x +->=-，故01x >． 【点睛】方法点睛：对于利用导数研究不等式的恒成立与有解问题的求解策略： 1、合理转化，根据题意转化为两个函数的最值之间的比较，列出不等式关系式求解； 2、构造新函数，利用导数研究函数的单调性，求出最值，从而求出参数的取值范围； 3、利用可分离变量，构造新函数，直接把问题转化为函数的最值问题．4、根据恒成立或有解求解参数的取值时，一般涉及分离参数法，但压轴试题中很少碰到分离参数后构造的新函数能直接求出最值点的情况，进行求解，若参变分离不易求解问题，就要考虑利用分类讨论法和放缩法，注意恒成立与存在性问题的区别．22. 已知以点M 为圆心的动圆经过点1(3,0)F -，且与圆心为2F 的圆22(3)12x y -+=相切，记点M 的轨迹为曲线C ．（1）求曲线C 方程；的（2）若动直线l 与曲线C 交于11(,)A x y ，22(,)B x y 两点（其中120y y >），点A 关于x 轴对称的点为A'，且直线BA'经过点()1,0P -． （ⅰ）求证：直线l 过定点；（ⅱ）若||||PA PB +=，求直线l 的方程．【答案】（1）22136x y -=（2）（ⅰ）证明见解析；（ⅱ）30x y ±+= 【解析】【分析】（1）根据动圆M 与圆2F 相切，由1212||||||||6MF MF F F -=<=，利用双曲线的定义求解；（2）（ⅰ）设直线l 的方程为x my t =+（显然l 与x 轴不平行），与22136x y-=联立，由0AP BP k k +=求解；（ⅱ）由（ⅰ）知，当3t =-时，1221221m y y m +=-，12212021y y m =>-，然后由||||||||||PA PB PA PB A B ''+=+=求解.【小问1详解】圆22(3)12x y -+=的圆心坐标为2(3,0)F ，半径r =． 动圆M 与圆2F 相切有两种情况，即内切或外切，所以1212||||||||6MF MF F F -=<=，所以点M 在以1F ，2F 为焦点的双曲线上，且该双曲线的实轴长为2a =，26c =， 所以2226b c a =-=，所以曲线C 的方程是22136x y -=．【小问2详解】（ⅰ）设直线l 的方程为x my t =+（显然l 与x 轴不平行），与22136x y -=联立，得222(21)4260m y mty t -++-=，。

江西省宜春市第一中学高一英语月考试题含解析一、选择题1. On Sunday when I was a child, my father and I _____ get up early and go fishing.A. couldB. wouldC. mightD. should参考答案：B略2. He went to the hospital for a medical examination.And that is ________ he was absent from the meeting.A．why B．when C．whetherD．what参考答案：A[句意：他去医院体检了，那就是他缺席会议的原因。

why表原因，引导表语从句，在从句中作状语。

]3. Either we or he _______considering buying the old house to be pulled down soon.A. areB. isC. wereD. was参考答案：B4. Only when you grow up _______appreciate the importance of responsibility.A. you will B you can C. will you D. do you参考答案：C5. —What do you think of their marriage?—________，if I were Kate, I wouldn't marry him.A．Generally speaking B．I have no ideaC．To be honest D．To my surpriseC解析考查插入语。

句意：——你对于他们的婚姻有什么看法？——说实话，如果我是肯特的话，我就不会嫁给他了。

根据答语可知，答话人并不看好这桩婚事。

To be honest“实话实说”，符合语境。

2021年宜春一中高三英语第三次联考试题及参考答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项AIn theUnited States, the word "holiday" is synonymous with celebration. The following tenholidaysper year are proclaimed by the federal government.Independence DayIndependence Day is annually celebrated on July 4 and is often known as "the Fourthof July”. It is the anniversary of the publication of the declaration of independence fromGreat Britainin 1776. Now it is celebrated in all the states. The army marks the occasion by firing a 13-gun salute every year. Ceremonies may include parades, official speeches, visits to historic monuments and fireworks displays.Memorial DayThis holiday, on the fourth Monday of every May, is a day on which Americans honor the dead. Originally a day on which flags and flowers were placed on graves of soldiers who died in the American Civil War, now it has become a day on which the dead of all wars and all other dead are remembered the same way.Veterans DayVeterans Day was established to honor Americans who had served in World War I. It falls on November 11, the day when that war ended in 1918, but it now honors veterans of all wars in which the United States has fought Veterans' organizations hold parades or other special ceremonies, and the US president customarily places a wreath on the Tomb of the Unknowns at Arlington National.ThanksgivingThanksgiving Day is celebrated on the fourth Thursday in November. It has been an annual tradition in theUnited Statessince 1863. Today, people celebrate Thanksgiving to remember these early days. The most important part of the celebration is a traditional dinner. Thanksgiving dinner almost always includes some of the foods served at the first feast: roast turkey, cranberry sauce potatoes pumpkin pies. Before the meal begins, families often pause to give thanks.1. When isIndependence Day?A. May 14，B. July 13.C. July 14.D. July 4.2. Which holiday honors dead soliders?A. Independence Day.B. Memorial Day.C. Veterans Day.D. Thanksgiving.3. What will Americans do on Thanksgiving Day?A. They say thanks.B. They havefriend gatherings.C. They go on holiday.D. They buy many cards.BGrowing up as kids we are told to share our toys and notto be selfish. We also live in an age when discussing our feelings is encouraged. But when does it all become too much? With new crazes trending all the time, such as dance challenges and wearing a carpet as a dress, the question is: when can sharing become oversharing on social media?“Oversharing” has become associated with social media, but it isn'texclusiveto this platform. Imagine you head to a party and meet x k w someone. Within five minutes they have revealed private details about their life. While some of us may try to escape these people, according to marriage advisor Carolyn Cole, this form of oversharing could come from a strong desire to connect with someone. But how does this translate to social media?Dr. Christopher Hand, a lecturer in cyberpsychology (网络心理学），says the more details people disclose, the less sympathy we express when things go wrong. It seems that searching for sympathy by oversharing is generally considered as negative rather than the cry for help it could really be.However, Dr. Hand's research also seems to suggest that the more we post on a platform, the more socially attractive we become-provided that the posts that we bang out are positive. Even back in 2015, Gwendolyn Seidman PhD said that we should avoid complaining and being negative online. We should also avoid showing off, especially about our love lives. It makes sense-if your date is going “that well", would you really have time to share a photo with text?So, how can you know if you are oversharing? Well, why not ask your friends in real life. They would probably be happy to tell you if your posts about your breakfast or your complaints about your lack of money really are too much.4. What does the underlined word “exclusive" in paragraph 2 mean?A. Unique.B. Similar.C. Relevant.D. Fundamental.5. Why do some people prefer oversharing at parties?A. To draw others' attention.B. To satisfy others' curiosity.C. To remove negative feelings.D. To develop good relationships.6. Which of the following may Dr. Hand agree with?A. Sharing more details online can attract more sympathy.B. Oversharing negative experiences is equal to crying for help.C. Sharing negative posts can't help one become socially attractive.D. Oversharing isn't likely to happen online when things go wrong.7. According to the text, what should be avoided for online sharing?A. Reflecting on past bad manners.B. Showing a great many expensive goods.C. Writing a recipe for a balanced breakfast.D. Recording unforgettable moments with friends.CWhy do you check social media? Is it to keep up with everything that your friends and family are doing? Is it to find new trendy spots to eat?Regardless of the reason, you may find yourself with different degrees of envy or discomfort after a quick look at your phone. Then you might be suffering from a phenomenon known as “Fear of Missing Out (FOMO). ”While the phenomenon of FOMO can be traced back for centuries, it had never been the issue as it is today, causing widespread discussion and research. This rise in checking social media is naturally connected to the increasing leading position that social media holds over our lives. Every time someone opens their WeChat Moments, Facebook, Twitter... etc. , they are bombarded with the highlight reels of other peoples' lives. A sunny beach, delicious-looking food, a super cute kitten-they are all uplifting photos, yet they're very likely to bring about more unhappiness than joy. You see, the first thought to come out of your mind may be “Wow, that's so cool/delicious/cute”, but then it takes a hard U-turn. You're thinking: “I wish I were there” or “my life is so boring compared to his/hers.” The more you see, the more likely you are to have these negative feelings.What's worse is the habit many people have of turning to social media in search of happiness when they arefeeling down, not realizing that they are just going to end up in a negative cycle of endless disappointment. Montesquieu once said: “If one only wished to be happy, this could be easily accomplished; but we wish to be happier than other people, and this is always difficult, for we believe others to be happier than they are.”FOMO will go hand in hand with dissatisfaction and envy. Appreciate what you already have, because someone else out there in the world would gladly give everything to be you.8. What is the purpose of the questions in paragraph 1?A. To introduce the topic of the passage.B. To explain the function of FOMO.C. To describe the features of FOMO.D. To give the reasons for checking social media.9. How might people feel seeing other's perfect life through social media?A. Joyful.B. Admiring.C. Comfortable.D. Envious.10. What can we conclude from Montesquieu's words in paragraph 3?A. We could turn to social media for happiness.B. We couldn't realise our dream without hard work.C. We couldn't harvest happiness through comparison.D. We could live better than others by showing ourselves online.11. What is the author's attitude towards FOMO?A. Indifferent.B. Objective.C. Doubtful.D. Hopeful.DMove over, helicopter parents. “Snowplow (扫雪机) parents” are the newest reflection of an intensive (强化的) parenting style that can include parents booking their adult children haircuts, texting their college kids to wake them up so they don’t sleep through a test, and even calling their kids’ employers.Helicopter parenting the practice of wandering anxiously near one’s children, monitoring their every activity, is so 20th century. Some rich mothers and fathers now are more like snowplows: machines moving ahead, clearing any difficulties in their children’s path to success, so they don’t have to suffer failure, frustration (挫折) or lose opportunities.It starts early, when parents get on wait lists for excellent preschools before their babies are born and try to make sure their kids never do anything that may frustrate them. It gets more intense when school starts: running forgotten homework to school or calling a coach to request that their children make the team.Rich parents may have more time and money to devote to making sure their children don’t ever meet with failure, but it’s not only rich parents practicing snowplow parenting. This intensive parenting has become the mostwelcome way to raise children, regardless of income, education, or race.Yes, it’s a parent’s job to support the children, and to use their adult wisdom to prepare for the future when their children aren’t mature enough to do so. That’s why parents hide certain toys from babies to avoid getting angry or take away a teenager’s car keys until he finishes his college applications.But snowplow parents can take it too far, some experts say. If children have never faced a difficulty, what happens when they get into the real world?“Solving problems, taking risks and overcoming frustration are key life skills,” many child development experts say, “and if parents don’t let their children experience failure, the children don’t acquire them.”12. What do we know about snowplow parenting?A. It appeared before helicopter parenting.B. It costs parents less than helicopter parenting.C. It was a typical phenomenon of the 20th century.D. It provides more than enough services for children.13. What is mainly discussed about snowplow parenting in Paragraph 4?A. Its cost.B. Its benefits.C. Its popularity.D. Its ending.14. Why does the author mention parents’ taking away car keys?A. To show teenagers are no better than babies.B. To advise teenagers not to treat their cars as toys.C. To advise parents not to buy cars for their teenagers.D. To show it’s appropriate to help children when necessary.15. What’s the possible result of snowplow parenting according to the experts?A. Children lacking problem-solving ability in reality.B. Children mastering more key life skills than parents.C. Children gaining great success in every aspect of life.D. Children meeting no problems or frustration after growing up.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。