2022-2023学年湖南省长沙市第一中学高一上学期第三次月考英语试题
长沙市雅礼中学2022-2023学年高三下学期月考试卷(八)英语试题(原卷版)

雅礼中学2023届高三月考试卷(八)英语注意事项:1.答卷前生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。
2.回答选择题时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,强用橡度擦千净后,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。
第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
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个选项中选出最佳选项。
听完每段对话后,你都10秒钟的时间来回答有关小题和阅读下小题。
每段对话仅读一遍。
例:How much is the shirt?A.£19.15.B.£9.18.C.£9.15.答案是C。
1 What is the worst part of Joe's story?A.The grammar.B.The handwriting.C.The spelling2.What do the mams class want to do this Sunday?A.Go fora swim.B.Do some sunbathing.C.Collect the rubbish.3.Who probably picked the woman's roses?A.Her son.B.The man.C.Her neighbor.4.What are the speakers mainly discussing?A.Holiday plans.B.Work problems.C.Familv members.5.How did Cynthia do in her history test?A.She failed it.B.She barely passed.C.She did very well.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。
湖南省长沙市长郡双语实验中学2022-2023学年九年级上学期第一次月考英语试卷(含答案)

2022-2023-1长郡双语九年级上第一次月考第二部分阅读(共三节,满分50分)第一节(共15小题每小题2分,满分30分)阅读下列材料,从每题所给的A、B、C三个选项中,选出最佳选项。
Abetter choose _______.A. Better English CenterB. Three-tree Language SchoolC. The International Horse of English22. How much will John pay if he takes part in the evening program in the International Horse of English?A. 1200 yuan.B.600 yuan.C.700 yuan.BMid-Autumn Festival, which fell on Sept 10 this year, is one of the most important traditional Chinese festivals. A cultural show. “2022 Adventures on Mid-Autumn Festival” staged(上演)by Henan TV led viewers on a journey to the palace of the moon at 7:30 pm, Sept 9! Here are three fantastic programs in it.Missing You like Seeing You, a dance, was from the show. Performers brought the audience a love story between a craftsman and a girl during the Mid-Autumn Festival, while showing the beauty of Jun ware(钧瓷)at the same time. Jun ware, from Yuzhou, Henan, became popular in the Song Dynasty.Fly up High with the Wind told the story of “FlightPioneer” Wan Hoo, a scholar(学者)of the Ming Dynasty.He was lost in the “dream of flight”. Space flight is ashared dream of all humans. Follow the performanceAdventures of the Mid-Autumn Festival 2022to feel theromance of this unique festival.In this multilingual song Our Souls Together HeavenwardFly from the performance, children from differentcountries and ethnicities(种族)sang a song as one,expressed reunion and enjoyed the moon together.23. People enjoyed the cultural show at _______.A. 7:30am,Sept 9B. 7:30pm, Sept 10C. 7:30pm, Sept 924. The dance Missing You like Seeing You showed us _______.A. the dream of space flightB. a love story and the beauty of Jun wareC. friendship among different countries25. Which of the following is True according to the passage?A. A cultural show “2022 Adventures on Mid-Autumn Festival”was staged by Hunan TV.B. The song Our Souls Together Heavenward Fly was sung only by Chinese children.C. Wan Hoo was the famous flight pioneer of the Ming Dynasty.CNot many people know that at the Children’s Hospital of Fudan University in Shanghai, there is a learning difficulties clinic(门诊)Zhu Daqian is the doctor of the hospital’s psychology department(心理科), who is also the leader of the clinic. She told China Daily that the clinic was set up in September 2020 because of a growing number of unfriendly relationships between parents and children.Facing their children’s terrible school grades, most parents think that their child is short of study goals or has a low IQ. Few of them consider that their child might have a learning disability, said Zhu.Similar clinics in other cities such as Beijing, Nanjing and Wuhan, have seen a great rise in the number of asking for help, according to local reports.Learning disabilities can be caused by many things, including a child’s neural(神经的)development, mental problems and family fights, Zhu said. There are also different kinds of learning disabilities, such as dyslexia(阅读障碍)and attention deficit hyperactivity disorder(ADHD, 注意缺陷多动障碍). The key to dealing with these problems is to pay attention to them earlier, she added.What’s more, different learning disabilities need different treatments. For example, for children with ADHD, medicine can help. Modern technology can also help treat children’s mental problems.Although it is good to hear that many children do well in their studies after treatment, Zhu warned that medicine and intervention(干预)can’t help children get good grades. “We’re here to help them find their own ways to learn,” said Zhu.26. Why was the learning difficulties clinic set up?A. Because many parents have unfriendly relationships with their children.B. Because modern technology can help children with mental problems.C. Because many parents think that more children have a low IQ.27. What can we learn from Paragraph 3?.A. Most parents think that their child might have a learning disability.B. Most children with terrible grades have no study goals.C. Parents should consider whether the children have a learning disability.28. What does the underlined word “treatment” mean in Paragraph 6?A. 对待方式B. 治疗方法C.款待29. Which of the following is True according to the passage?A. Similar clinics in Hunan are also popular now.B. Family fights may cause learning disabilities.C. The key to dealing with dyslexia is to take the medicine.30. What is the purpose of writing this passage?A. To warn parents not to ask their children to study too hard.B. To tell people what learning disabilities are.C. To help parents deal with children's learning disabilities.DIn ancient times, poems(诗歌)were importantways for people to express their feelings, and record thesuccess and difficulties in their lives. Now, they stillplay a big role in people’s lives. Today, we are going tolearn about a famous scholar who has closely connected her life with poems. She is Ye Jiaying.Professor Ye has spent all her life teaching poetry. And her life is just like poems. Her childhood and girlhood were a happy and joyful poem. She was born into a well-off family in Beijing. By 15, she could write great poems. She graduated from Beijing Normal University with the highest score for her year in 1945.However, her midlife was a poem of sadness and difficulties. She and her daughter lost their house after her husband was thrown into prison in Taiwan. And she had to work hard to support the family by teaching poetry. Several years later, she moved to Canada with her family. Then she got another blow in 1976 when her eldestdaughter and son-in-law both lost their lives in a traffic accident.When living abroad, Ye missed the Chinese mainland very much. So she gave up all her achievements in Canada and came back to China in 1979. Then she taught poetry at Nankai University. This period of Ye’s life was a poem that was filled with flowers and honor. But Ye herself doesn’t care much about it. She thinks it is her mission(使命)in life to pass down classical Chinese poems to the young since they are so beautiful and valuable. She even gave away all her collections of books and paintings along with personal savings(存款)to Nankai University. Although life gives Ye lots of difficulties, she repays it with the beauty of poetry.31. Who is Ye Jiaying according to the passage?A. A famous scholar who has connected her life with poems.B. A poet who just stayed at home to take care of her family.C. A teacher who taught English at Nankai University.32. Which of the following is True according to the passage?A. Her family was too poor to help her study poems.B. She lived a happy and joyful life in her childhood.C. She graduated from Nankai university in 1945.33. How many times did Ye experience sadness and difficulties during her midlife?A. OnceB. TwiceC. Three times34. From the last paragraph,we can infer that _______.A. Her life was wonderful when she taught poetry at Nankai University.B. She cares much about her great achievements in Canada.C. It is her mission to give away all her collections of books and paintings.35. What’s the best title of this passage?A. Ye Jiaying’s Life with PoetryB. Ye Jiaying’s Achievements of PoetryC. Ye Jiaying’s Sadness and Happiness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人空白处的最佳选项。
湖南省长沙市雅礼中学2022-2023学年高二下学期3月考试英语试卷PDF版含答案

长沙市雅礼中学2023年上学期高二月考英语试卷(本试卷满分150分,考试时间:120分钟)注意事项:1.答卷前:先将自己的姓名、准考证号填写在试卷和答题卡上,并将准考证条码粘贴在答题卡上指定位置。
2.选择题,每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑。
3.非选择题,用0.5mm黑色签字笔写在答题卡上对应的答题区域,写在非答题区域无效。
第一部分听力(共两节,满分30分)第一节听下面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。
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 will the speakers do next?A.Visit a friend.B.Pick up Billy.C.Buy some beans.2.What does Andy Clarkes do?A.A public librarian.B.A TV actor.C.A famous lawyer.3.What are the speakers mainly talking about?A.Gifts for Jason.B.A baseball game.C.The woman’s retirement.4.What went on at Cooper’s last night?A.A movie show.B.A birthday party.C.A sales promotion.5.What problem do the speakers have?A.They are late for work.B.They are stuck in a traffic jam.C.They are lost on the way.第二节听下面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。
2022-2023学年湖南省长沙市雅礼中学高一下学期期末考试英语试题

2022-2023学年湖南省长沙市雅礼中学高一下学期期末考试英语试题 What to See in Langkawi Underwater World Langkawi Set along the Pantai Cenang beach town, Underwater World Langkawi houses more than 500 species of sea creatures including harbour seals, rockhopper penguins, seahorses, as well as mandarin ducks. One of the highlights is a 15-metre walkthrough underwater tunnel, where you can enjoy close-up views of sharks, giant stingrays, and green turtles.
Opening Hours: Monday-Friday 09:30-18:30, Saturday-Sunday 09:30-22:30 Price Range:RM40(adults)and RM30(children aged 3 to 12 years old) Langkawi Wildlife Park & Bird Paradise Great for families and animal lovers, Langkawi Wildlife Park & Bird Paradise houses over 2,500 unique birds from 150 species such as canaries, parakeets, owls, eagles, toucans, hornbills and flamingos. The park is fitted with a walk-in aviary(鸟舍)and a 15-metre-tall man-made waterfall, where some of the birds are free to roam about and interact(互动)with visitors.
湖南省长沙市长郡中学2023-2024学年高一上学期期末考试英语试题

4.How did the children react when they were given the paper map?
A.The teaching focus in today's classroom.
B.The situations where paper maps are used.
C.The necessity of digital maps in the modem world.
D.The benefit of developing paper map skills for kids.
D.encourage students to participate in hands-on learning
2.What are students supposed to do to earn college credits?
A.Attend various courses.B.Carry out STEM research.
Your hard work will be combined with social events and fun activities. And you’ll still have time for your own adventures on campus! Email the SPP office at******************.
Academic life
SPP invites you to join other highly motivated teens from 87 countries in our summer programs for high school graduates. You can take college courses alongside undergraduates either on campus or online. And you can also earn up to eight college credits by conducting in-depth STEM research with individual instructors or as part of a group project.
2022-2023学年湖南省长沙市湖南师范大学附属中学高一上学期第一次月考化学试题(解析版)

湖南省长沙市湖南师范大学附属中学2022-2023学年高一上学期第一次月考化学试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.成语是中华民族灿烂文化中的瑰宝,许多成语中蕴含着丰富的化学原理。
下列成语原意涉及氧化还原反应的是A.滴水成冰B.铁杵成针C.钻木取火D.木已成舟【答案】C【详解】A.滴水成冰是物质状态的改变,属于物理变化,故A不符合题意;B.铁杵成针只是物质形状的变化,没有新的物质生成,则没有涉及化学变化,故B不符合题意;C.钻木取火有燃烧的发生,氧气参与反应生成化合物,同时氧元素的化合价发生变化,属于氧化还原反应,故C符合题意;D.木已成舟只是物质形状的变化,没有新物质生成,属于物理变化,故D不符合题意;答案为C。
2.下列有关物质的分类合理的是A.氧化物:Fe3O4、NO2、SO3B.碱:NaOH、KOH、Na2CO3C.铵盐:NH4Cl、NH4NO3、NH3·H2O D.混合物:空气、矿泉水、金属钠【答案】A【分析】碱是电离出的阴离子都是氢氧根离子的化合物;盐是由金属阳离子或铵根离子和酸根离子构成的化合物;氧化物是含有两种元素一种为氧元素的化合物;混合物是由多种物质组成的;【详解】A.Fe3O4、NO2、SO3是含有两种元素一种为氧元素的化合物,属于氧化物,A正确;B.Na2CO3是由钠离子和碳酸根离子构成的化合物,属于盐,B错误;C.NH3·H2O是电离出的阴离子都是氢氧根离子的化合物,为碱,C错误;D.金属钠是金属单质,D错误;故选A。
3.有如下物质:①NaCl溶液;①熔融MgCl2;①CuSO4固体;①NaOH固体;①金属铜;A.属于非电解质的有①①①B.不能导电的有①①①C.属于电解质的有①①①①D.属于混合物的有①①①①【答案】B【分析】电解质是溶于水或在熔融状态下能够导电的化合物;非电解质是溶于水和在熔融状态下都不能够导电的化合物;溶液导电的原因是存在自由移动的离子,金属导电的原因是存在自由移动的电子;【详解】A.因①NaCl溶液和①金属铜都不属于化合物,既不是电解质也不是非电解质,A错误;B.因①CuSO4固体和①NaOH 固体中离子不能自由移动,①蔗糖中没有自由移动的离子,则不能导电,选项B正确;C.①NaCl溶液为混合物,既不是电解质也不是非电解质;①熔融MgCl2中有自由移动的离子,能导电,①熔融MgCl2①CuSO4·5H2O固体① NaOH固体都是在熔融状态下或水溶液中能导电的化合物,都属于电解质,C错误;D.①熔融的MgCl2、①CuSO4·5H2O固体和①蔗糖都是只有一种物质组成,则属于纯净物,D错误。
2023-2024学年湖南省长沙市第一中学高二上学期入学考试英语试题(含听力)

2023-2024学年湖南省长沙市第一中学高二上学期入学考试英语试题(含听力)1. What time is it now?A.7:30 p.m. B.8:00 p.m. C.8:30 p.m.2. What are the speakers talking about?A.How to preserve strawberries.B.Where to buy strawberries.C.How to wash strawberries.3. What was Becky doing when a car crashed into her house?A.Doing the gardening. B.Eating dinner. C.Doing the cleaning.4. What caused Matthew’s cough?A.The flu. B.The season. C.The cold wind. 5. What does the woman mean?A.She has no time for dinner.B.She forgot the wedding anniversary.C.She wants to eat something different.听下面一段较长对话,回答以下小题。
6. Why is Dylan displeased with Ellie?A.She didn’t finish her work.B.She uses too much perfume.C.She speaks too loud in the office.7. What does the man think of the manager’s solution?A.Disapproving. B.Sympathetic. C.Understanding.听下面一段较长对话,回答以下小题。
湖南省长沙市第一中学2022-2023学年高三下学期月考卷(六)数学试题及答案

长沙市一中2023届高三月考试卷(六)数学时量:120分钟 满分:150分一、单项选择题(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,,则( ) {}32,Z M x x n n ==-∈{}2,1,0,1,2N =--M N ⋂=A. B. C.D. {}2,1-{}1,2-{}1,1-{}2,0,2-2. 已知复数满足,为虚数单位,则( )z ()1i 1i z -=+i z =A. B. C. D. i 11i 22+1i +3. 已知,,,一束光线从点出发经AC 反射后,再经BC 上点D 反射,()30A -,()3,0B ()0,3C ()1,0F -落到点上.则点D 的坐标为( )()1,0E A. B. C. D. 15,22⎛⎫ ⎪⎝⎭33,22⎛⎫ ⎪⎝⎭()1,2()2,14. 若,且,则( ) ππ,24α⎛⎫∈-- ⎪⎝⎭23π1cos cos 222αα⎛⎫++=- ⎪⎝⎭tan α=A. B. C. D. 2-3--5. 据一组样本数据,求得经验回归方程为,且.现发现()(()1122,,,,,,n n x y x y x y ⋅⋅⋅ 1.20.4y x =+3x =这组样本数据中有两个样本点和误差较大,去除后重新求得的经验回归直线的斜率为()1.2,0.5()4.8,7.5l 1.1,则( )A. 去除两个误差较大的样本点后,的估计值增加速度变快y B. 去除两个误差较大的样本点后,重新求得的回归方程对应直线一定过点()3,5C. 去除两个误差较大的样本点后,重新求得的回归方程为1.10.7y x =+D. 去除两个误差较大的样本点后,相应于样本点的残差为0.1()2,2.76. 在四面体中,,,,,则该四面体的PABC PA AB ⊥PA AC ⊥120BAC ∠=︒2AB AC AP ===外接球的表面积为( )A. B. C. D.12π16π18π20π7. 已知圆O 的半径为1,A 为圆内一点,,B ,C 为圆O 上任意两点,则的最小值是12OA =AC BC ⋅( )A. B. C. D. 18-116-116188. 设是定义在上的函数,若是奇函数,是偶函数,函数()f x R ()2f x x +()f x x -,若对任意的,恒成立,则实数的最大值为()()[]()(),0,121,1,f x x g x g x x ∞⎧∈⎪=⎨-∈+⎪⎩[]0,x m ∈()3g x ≤m ( ) A. B. C. D. 133********二、多项选择题(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. 已知函数在区间上有且仅有3条对称轴,给出下列四个结论,正确()()πsin 04f x x ωω⎛⎫=+> ⎪⎝⎭[]0,π的是( )A. 的取值范围是 ω913,44⎡⎫⎪⎢⎣⎭B. 在区间上有且仅有3个不同的零点()f x ()0,πC. 的最小正周期可能是 ()f x 4π5D. 在区间()f x π0,15⎛⎫ ⎪⎝⎭10. 已知抛物线C :的焦点为F ,准线为,A ,B 是C 上异于点O 的两点,O 为坐标原点,则22x y =l ( )A. 的方程为 l 12x =-B. 若,则 32AF =AOF AC. 若,则0OA OB ⋅= 9OA OB ⋅≥D. 若,过AB 的中点D 作于点E ,则的最小值为 120AFB ∠=︒DE l ⊥AB DE11. 如图,正方体中,顶点在平面内,其余顶点在的同侧,顶点到1111ABCD A B C D -A αα1,,B C A α的距离分别为,则( )1,2,3A. 平面BD A αB. 平面平面1A AC ⊥αC. 直线与所成角比直线与所成角大1AB α1AA αD.12. 已知,为正实数,且,则( )a b 26ab a b ++=A. 的最大值为2B. 的最小值为5 ab 2a b +C. 的最小值为D. 1211a b +++98()0,3a b -∈三、填空题(本题共4小题,每小题5分,共20分.)13. 设直线是曲线的一条切线,则_________.10x y ++=ln y a x =-=a 14. 楼道里有8盏灯,为了节约用电,需关掉3盏互不相邻的灯,则关灯方案有_________种.15. 过双曲线:右焦点作直线,且直线与双曲线的一条渐近线垂直,C ()222210x y a b a b-=>>F l l C 垂足为A ,直线与另一条渐近线交于点B .且点A ,B 位于x 轴的异侧,O 为坐标原点,若的内切l OAB A 圆的半径为,则双曲线C 的离心率为__________. 23b 16. 小说《三体》中,一个“水滴”摧毁了人类整个太空舰队,当全世界第一次看到“水滴”的影像时,所有人都陶醉于它那绝美的外形.这东西真的是太美了,像梦之海中跃出的一只镜面海豚,仿佛每时每刻都在宇宙之夜中没有尽头地滴落着.有科幻爱好者为“水滴”的轴截面设计了二维数学图形,已知集合.由集合中所有的点组成的图形如图中阴影部分()()(){}22,cos sin 4,0P x y x y θθθπ=-++=≤≤P所示,中间白色部分就如美丽的“水滴”.则图中“水滴”外部阴影部分的面积为_________.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. 记为正项数列的前项和,已知是4与的等比中项.n S {}n a n 1n a +n S (1)求的通项分式;{}n a (2)证明:. 2222123111154n a a a a +++⋅⋅⋅+<18. 已知a ,b ,c 分别为三个内角A ,B ,C 的对边,且.ABCA cos sin a C C b c +=+(1)求A ;(2)已知M 为BC 的中点,且,的平分线交BC 于N ,求线ABC A AM =BAC ∠段AN 的长度.19. 近日,某芯片研发团队表示已自主研发成功多维先进封装技术XDFOI ,可以实现4nm 手机SOC 芯片的封装,这是中国芯片技术的又一个重大突破,对中国芯片的发展具有极为重要的意义.可以说国产4nm 先进封装技术的突破,激发了中国芯片的潜力,证明了知名院士倪光南所说的先进技术是买不来的、求不来的,自主研发才是最终的出路.研发团队准备在国内某著名大学招募人才,准备了3道测试题,答对两道就可以被录用,甲、乙两人报名参加测试,他们通过每道试题的概率均为,且相互独立,若()01p p <<甲选择了全部3道试题,乙随机选择了其中2道试题,试回答下列问题.(所选的题全部答完后再判断是否被录用)(1)求甲和乙各自被录用的概率;(2)设甲和乙中被录用的人数为,请判断是否存在唯一的值,使得?并说明理由. ξp 0p () 1.5E ξ=20. 如图,四棱锥的底面是边长为2的正方形,. P ABCD -ABCD 2PA PB ==(1)证明:;PAD PBC ∠=∠(2)当直线PA 与平面PCD 所成角的正弦值最大时,求此时二面角的大小.P AB C --21. 已知,D 是圆C :上的任意一点,线段DF 的垂直平分线交DC 于点P . ()1,0F -()22116x y -+=(1)求动点P 的轨迹的方程:Γ(2)过点的直线与曲线相交于A ,B 两点,点B 关于轴的对称点为,直线交轴于(),0M t l Γx B 'AB 'x 点,证明:为定值.N OM ON ⋅ 22. 已知函数,. ()1e ln axf x x x-=+a ∈R (1)当时,求函数的最小值;1a =()f x x -(2)若函数的最小值为,求的最大值.()f x xa a长沙市一中2023届高三月考试卷(六)数学时量:120分钟 满分:150分一、单项选择题(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,,则( ) {}32,Z M x x n n ==-∈{}2,1,0,1,2N =--M N ⋂=A.B. C. D. {}2,1-{}1,2-{}1,1-{}2,0,2-【答案】A【解析】【分析】利用列举法及交集的定义即可求解.【详解】,{}}{32,Z ...,5,2,1,4,7,M x x n n ==-∈=-- 所以.{}2,1M N =- 故选:A.2. 已知复数满足,为虚数单位,则( )z ()1i 1i z -=+i z =A.B. C. D. i 11i 22+1i +【答案】B【解析】【分析】根据向量的除法和向量模的求法,变形的,即可求解. 1i 1i z +==-【详解】, 1i 1i z +===+-故选:B3. 已知,,,一束光线从点出发经AC 反射后,再经BC 上点D 反射,()30A -,()3,0B ()0,3C ()1,0F -落到点上.则点D 的坐标为( )()1,0E A. B. C. D.15,22⎛⎫ ⎪⎝⎭33,22⎛⎫ ⎪⎝⎭()1,2()2,1【答案】C【解析】【分析】根据入射光线与反射光线的性质可知方程,由与的交点可得D ,求坐标即可.GH GH BC【详解】根据入射光线与反射光线关系可知,分别作出关于的对称点,,F E ,AC BC ,G H 连接,交于,则D 点即为所求,如图,GH BCD因为所在直线方程为,,设,AC 3y x =+(1,0)F -()G x y ,则,解得,即, 132211y x y x -⎧=+⎪⎪⎨⎪=-⎪+⎩3,2x y =-=(3,2)G -由所在直线方程为,,同理可得,BC 3y x =-+(1,0)E (3,2)H 所以直线方程为,由解得, GH 2y =32y x y =-+⎧⎨=⎩(1,2)D 故选:C4. 若,且,则( ) ππ,24α⎛⎫∈-- ⎪⎝⎭23π1cos cos 222αα⎛⎫++=- ⎪⎝⎭tan α=A.B. C. D. 2-3--【答案】C【解析】【分析】利用三角函数的诱导公式及二倍角的正弦公式,结合三角函数的齐次式法即可求解.【详解】因为,所以, ππ,24α⎛⎫∈-- ⎪⎝⎭tan 1α<-由,得,即, 23π1cos cos 222αα⎛⎫++=- ⎪⎝⎭21cos sin 22αα+=-222cos 2sin cos 1cos sin 2ααααα+=-+所以,即,解得 212tan 11tan 2αα+=-+2tan 4tan 30αα++=或(舍).tan 3α=-tan 1α=-故选:C.5. 据一组样本数据,求得经验回归方程为,且.现发现()()()1122,,,,,,n n x y x y x y ⋅⋅⋅ 1.20.4y x =+3x =这组样本数据中有两个样本点和误差较大,去除后重新求得的经验回归直线的斜率为()1.2,0.5()4.8,7.5l1.1,则( )A. 去除两个误差较大的样本点后,的估计值增加速度变快y B. 去除两个误差较大的样本点后,重新求得的回归方程对应直线一定过点()3,5C. 去除两个误差较大的样本点后,重新求得的回归方程为1.10.7y x =+D. 去除两个误差较大的样本点后,相应于样本点的残差为0.1()2,2.7【答案】C【解析】【分析】根据直线的斜率大小判断A ;求出判断B ;再求出经验回归方程判断C ;计算残差判断D 作l y 答.【详解】对于A ,因为去除两个误差较大的样本点后,经验回归直线的斜率变小,则的估计值增加速l y 度变慢,A 错误;对于B ,由及得:,因为去除的两个样本点和, 1.20.4y x =+3x =4y =()1.2,0.5()4.8,7.5并且,因此去除两个样本点后,样本的中心点仍为, 1.2 4.80.57.53,422++==(3,4)因此重新求得的回归方程对应直线一定过点,B 错误;(3,4)对于C ,设去除后重新求得的经验回归直线的方程为,由选项B 知,,解得l ˆ1.1y x a=+ˆ4 1.13a =⨯+, ˆ0.7a=所以重新求得的回归方程为,C 正确;1.10.7y x =+对于D ,由选项C 知,,当时,,则, 1.10.7y x =+2x = 1.120.72.9y =⨯+= 2.7 2.90.2-=-因此去除两个误差较大的样本点后,相应于样本点的残差为,D 错误.()2,2.70.2-故选:C6. 在四面体中,,,,,则该四面体的PABC PA AB ⊥PA AC ⊥120BAC ∠=︒2AB AC AP ===外接球的表面积为( )A.B. C. D. 12π16π18π20π【答案】D【解析】【分析】由线面垂直的判定定理可得平面,设底面的外心为,外接球的球心为,PA ⊥ABC ABC A G O 为的中点,可得四边形为平行四边形,所以,在中,由余弦定理及正弦定理D PA ODAG 1OG =ABC 可求,故可求外接球的半径,根据球的表面积公式即可求解.AG 【详解】因为,,平面,PA AB ⊥PA AC ⊥,,AB AC A AB AC =⊂ ABC 所以平面.PA ⊥ABC设底面的外心为,外接球的球心为,则平面,所以. ABC A G O OG ⊥ABC //PA OG 设为的中点,DPA因为,所以.OP OA =DO PA ⊥因为平面,平面,PA ⊥ABC AG ⊂ABC 所以,所以.PA ⊥AG //OD AG 因此四边形为平行四边形,所以. ODAG 112OG AD PA ===因为,,120BAC ∠=︒2AB AC ==所以,BC ===由正弦定理,得. 242AG AG ==⇒=所以该外接球的半径满足,R )()2225R OG AG =+=故该外接球的表面积为.24π20πS R ==故选:D.7. 已知圆O 的半径为1,A 为圆内一点,,B ,C 为圆O 上任意两点,则的最小值是12OA =AC BC ⋅ ( )A.B. C. D. 18-116-11618【答案】A【解析】 【详解】首先设与所成角为,根据题意得到OA BC θ,再根据()1cos cos 2AC BC OC OA BC OC BC OA BC BC BCO BC θ⋅=-⋅=⋅-⋅=∠- 求解即可. 221111cos 2222BC BC BC BC θ-≥-【点睛】如图所示:设与所成角为,OA BCθ因为, ()1cos cos 2AC BC OC OA BC OC BC OA BC BC BCO BC θ⋅=-⋅=⋅-⋅=∠- 因为,112cos 2BC BCO BC OC ∠== 所以 211cos 22AC BC BC BC θ⋅=- 因为,当时,等号成立. 221111cos 2222BC BC BC BC θ-≥- 0θ= 因为,所以当时,取得最小值为, 02BC ≤≤ 12BC = 21122BC BC - 18-所以当时,取得最小值为. 12BC = AC BC ⋅ 18-故选:A8. 设是定义在上的函数,若是奇函数,是偶函数,函数()f x R ()2f x x +()f x x -,若对任意的,恒成立,则实数的最大值为()()[]()(),0,121,1,f x xg x g x x ∞⎧∈⎪=⎨-∈+⎪⎩[]0,x m ∈()3g x ≤m ( )A. B. C. D. 133********【答案】B【解析】【分析】由是奇函数,是偶函数,求出,再根据()2f x x +()f x x -()2f x x x =-,作出函数的图象即可求解. ()()[]()(),0,121,1,f x xg x g x x ∞⎧∈⎪=⎨-∈+⎪⎩()g x【详解】因为是奇函数,是偶函数, ()2f x x +()f x x -所以,解得,()()()()()22f x x f x x f x x f x x⎧-+-=--⎪⎨-+=-⎪⎩()2f x x x =-由, ()()[]()(),0,121,1,f x x g x g x x ∞⎧∈⎪=⎨-∈+⎪⎩当时,则,所以, ()1,2x ∈()10,1x -∈()()()2121gx g x f x =-=-同理:当时,,()2,3x ∈()()()()214242g x g x g x f x =-=-=-以此类推,可以得到的图象如下:()gx由此可得,当时,,()4,5x ∈()()164g x f x =-由,得,解得或, ()3g x ≤()()16453x x --≤174x ≤194x ≥又因为对任意的,恒成立,[]0,x m ∈(3g x ≤所以,所以实数的最大值为. 1704m <≤m 174故选:B.【点睛】本题考查了奇函数与偶函数的性质,抽象函数的周期性,通过递推关系分析出每一个区间的解析式是本题的关键,数形结合是解题中必须熟练掌握一种数学思想,将抽象转化为形象,有助于分析解决抽象函数的相关问题. 二、多项选择题(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. 已知函数在区间上有且仅有3条对称轴,给出下列四个结论,正确()()πsin 04f x x ωω⎛⎫=+> ⎪⎝⎭[]0,π的是( )A. 的取值范围是 ω913,44⎡⎫⎪⎢⎣⎭B. 在区间上有且仅有3个不同的零点()f x ()0,πC. 的最小正周期可能是 ()f x 4π5D. 在区间上单调递增 ()f x π0,15⎛⎫ ⎪⎝⎭【答案】ACD【解析】【分析】由,得,再根据函数在区间上有且仅有条对称轴,[]0,πx ∈πππ,π444x ωω⎡⎤+∈+⎢⎥⎣⎦()f x []0,π3可得,可求出的取值范围判断A ,再利用三角函数的性质可依次判断BCD . 5ππ7ππ242ω≤+<ω【详解】由,得, []0,πx ∈πππ,π444x ωω⎡⎤+∈+⎢⎥⎣⎦因为函数在区间上有且仅有条对称轴,()f x []0,π3所以,解得,故A 正确; 5ππ7ππ242ω≤+<91344ω≤<对于B ,,, (0,π)x ∈ ∴πππ,π444x ωω⎛⎫+∈+ ⎪⎝⎭, ∴π5π7ππ,422ω⎛⎫+∈ ⎪⎝⎭当时,在区间上有且仅有个不同的零点; π5π,3π42x ω⎛⎤+∈ ⎥⎝⎦()f x (0,π)2当时,在区间上有且仅有个不同的零点,故B 错误; π7π3π,42x ω⎛⎫+∈ ⎪⎝⎭()f x (0,π)3对于C ,周期,由,则, 2πT ω=91344ω≤<414139ω<≤, ∴8π8π139T <≤又,所以的最小正周期可能是,故C 正确; 84ππ58π,139⎛⎤∈ ⎥⎝⎦()f x 4π5对于D ,,, π0,15x ⎛⎫∈ ⎪⎝⎭∴ππππ,44154x ωω⎛⎫+∈+ ⎪⎝⎭又,, 91344ω≤<∴ππ2π7ππ,0,1545152ω⎡⎫⎛⎫+∈⊆⎪ ⎪⎢⎣⎭⎝⎭所以在区间上一定单调递增,故D 正确. ()f x π0,15⎛⎫ ⎪⎝⎭故选:ACD.10. 已知抛物线C :的焦点为F ,准线为,A ,B 是C 上异于点O 的两点,O 为坐标原点,则22x y =l ( )A. 的方程为 l 12x =-B. 若,则 32AF =AOF AC. 若,则0OA OB ⋅= 9OA OB ⋅≥D. 若,过AB 的中点D 作于点E ,则的最小值为 120AFB ∠=︒DE l ⊥AB DE【答案】BD【解析】【分析】A 选项,由抛物线方程得到准线方程,A 错误;由焦半径公式得到,进而求出1A y =A x =从而得到的面积,B 正确;由得到,,表达出AOF A 0OA OB ⋅=4A B x x =-4A B y y =,结合基本不等式求出最值,C 错误;作出辅助线,设()2222232A B A B OA OB x y y x ⋅=++,由焦半径公式得到,结合余弦定理,基本不等式得到的最小值. ,AF a BF b ==2a b DE +=AB DE【详解】的焦点为,准线方程为,故A 错误; 22x y =F ⎛ ⎝12y =-由焦半径公式可知:,解得, 1322A AF y =+=1A y =故,故 222A A x y ==A x =所以的面积为,B 正确; AOF A 111222A OF x ⋅=⨯=若,则,即,解得:, 0OA OB ⋅= 0A B A B x x y y +=22104A B A B x x x x +=4A B x x =-则,4A B y y =故 ()()()2222222223232A A B B AB A B OA OB x y x y x y y x ⋅=++=++≥+,32264A B A B x x y y =+⋅=故,当且仅当时,等号成立,C 错误;8OA OB ⋅≥A B A B x y y x =过点作⊥l 于点,过点B 作⊥l 于点,A 1AA 1A 1BB 1B设,所以, ,AF a BF b ==2a b DE +=因为()2222222cos AB a b ab AFB a b ab a b ab =+-∠=++=+-, ()()22223342a b a b a b DE ++⎛⎫≥+-== ⎪⎝⎭所以. AB ≥故选:BD【点睛】圆锥曲线中最值或范围问题的常见解法:(1)几何法,若题目的条件和结论能明显体现几何特征和意义,则考虑利用几何法来解决;(2)代数法,若题目的条件和结论能体现某种明确的函数关系,则可首先建立目标函数,再求这个函数的最值或范围.11. 如图,正方体中,顶点在平面内,其余顶点在的同侧,顶点到1111ABCD A B C D -A αα1,,B C A α的距离分别为,则( )1,2,3A. 平面BD A αB. 平面平面1A AC ⊥αC. 直线与所成角比直线与所成角大1AB α1AA αD.【答案】ABD【解析】【分析】根据点到面的距离的性质,结合线面垂直的判定定理、线面角的定义、面面相交的性质进行求解判断即可.【详解】解:设的交点为,显然是、的中点,,AC BD O O AC BD 因为平面,到平面的距离为,所以到平面的距离为,ABCD A α= C α2O α1又到平面的距离为,B α1所以平面,即平面,即A 正确;//BO α//BD α设平面,ABCD l α= 所以,//BD l 因为是正方形,所以,ABCD AC BD ⊥又因为平面,平面,1AA ⊥ABCD BD ⊂ABCD 所以,因为平面,1AA BD ⊥11,,AA AC A AA AC ⋂=⊂1A AC 所以平面,因此有平面,而,BD ⊥1A AC l ⊥1A AC l ⊂α所以平面平面,因此选项B 正确;1A AC ⊥α设到平面的距离为,1B αd 因为平面,是正方形,点,B 到的距离分别为,1,11AA B B A α= 11AA B B 1A α3所以有, 31422d d +=⇒=设正方体的棱长为,1111ABCD A B CD -a设直线与所成角为,所以, 1AB αβ14sin AB β===设直线与所成角为,所以, 1AA αγ133sin AA aγ==因为,因此选项C 不正确;3>sin sin βγβγ<⇒<因为平面平面,平面平面,1A AC ⊥α1A AC ⋂A α=所以在平面的射影与共线,1,C A α,E F A显然,如图所示:1112,3,,,CE A F AC AA a AA AC ====⊥由,11ECA CAE CAE A AF ECA A AF ∠+∠=∠+∠⇒∠=∠, 111cos ,sin A F CE ECA A AF AC AA ∠=∠=由, 2212249cos sin 112ECA A AF a a a ∠+∠=⇒+=⇒=因此选项D 正确,故选:ABD 12. 已知,为正实数,且,则( )a b 26ab a b ++=A. 的最大值为2B. 的最小值为5 ab 2a b +C. 的最小值为D. 1211a b +++98()0,3a b -∈【答案】AC【解析】【分析】由已知条件结合基本不等式及相关结论分别检验各选项即可求解.【详解】依题意,对于A :因为,26ab a b ++=所以,62ab a b ab =++≥+当且仅当时取等号,2a b =令,则有,0t =>260t +-≤解得,又因为, t -≤≤0t =>所以,即0t <≤0<≤的最大值为2,故A 选项正确;ab 对于B :因为,26ab a b ++=所以, ()221162222224a b ab a b ab a b a b +=++=⨯++≤⨯++当且仅当时取等号,2a b =令,则有,20t a b =+>28480t t +-≥解得或(舍去),4t ≥t 12≤-即,所以的最小值为4,24a b +≥2a b +故B 选项错误;对于C :因为,26ab a b ++=所以, 12111888b b a ++==++所以,81221119888111a b b b +++≥=+++=++当且仅当,即时等式成立, 2118b b +=+3b =所以的最小值为,故C 选项正确; 1211a b +++98对于D :当,时,, 14a =225b =()4.150,3a b -=∉所以D 选项错误;故选:AC.三、填空题(本题共4小题,每小题5分,共20分.)13. 设直线是曲线的一条切线,则_________.10x y ++=ln y a x =-=a 【答案】2-【解析】【分析】设切点为,根据导数的几何意义求出切点的横坐标,再根据切点即在曲线上又在切线上()00,x y 即可得解.【详解】设切点为,()00,x y , 1y x '=-则,所以, 0011x x y x ==-=-'01x =所以切点为,()1,a 又切线为,10x y ++=所以,解得.110a ++=2a =-故答案为:.2-14. 楼道里有8盏灯,为了节约用电,需关掉3盏互不相邻的灯,则关灯方案有_________种.【答案】20【解析】【分析】根据题意,原问题等价于在5盏亮灯的6个空隙中插入3盏不亮的灯,由组合公式计算即可求解.【详解】依题意,原问题等价于在5盏亮灯的6个空隙中插入3盏不亮的灯,则有种方案.36C 20=故答案为:20. 15. 过双曲线:右焦点作直线,且直线与双曲线的一条渐近线垂直,C ()222210,0x y a b a b-=>>F l l C 垂足为A ,直线与另一条渐近线交于点B .且点A ,B 位于x 轴的异侧,O 为坐标原点,若的内切l OAB A 圆的半径为,则双曲线C 的离心率为__________. 23b【解析】 【分析】作出图象,设的内切圆的圆心为,易知在的平分线上,过分别作OAB A M M AOB ∠Ox M 于,于,则有四边形为正方形,则,MN OA ⊥N MT AB ⊥T MTAN 2||||3b NA MN ==2||3b ON a =-,由,可得,由斜率公式即可得答案. tan MNb AOF ON a∠==2a b =【详解】解:如图所示:设A 在第一象限,由题意可知,其中为点到渐近线的距离,, AF d b ===d (c,0)F b y x a =||OF c =所以, ||OA a ===设的内切圆的圆心为,OAB A M 则在的平分线上,M AOB ∠Ox 过分别作于,于,M MN OA ⊥N MT AB ⊥T 又因为于,FA OA ⊥A 所以四边形为正方形,MTAN所以, 2||||3b NA MN ==所以, 2||||||3b ON OA NA a =-=-又因为, 2||3tan 2||3bMN b AOF b ON aa ∠===-所以, 2233a b a =-,2a b =所以,22225c a b b =+=所以, c =所以. c e a ===. 16. 小说《三体》中,一个“水滴”摧毁了人类整个太空舰队,当全世界第一次看到“水滴”的影像时,所有人都陶醉于它那绝美的外形.这东西真的是太美了,像梦之海中跃出的一只镜面海豚,仿佛每时每刻都在宇宙之夜中没有尽头地滴落着.有科幻爱好者为“水滴”的轴截面设计了二维数学图形,已知集合.由集合中所有的点组成的图形如图中阴影部分()()(){}22,cos sin 4,0P x y x y θθθπ=-++=≤≤P 所示,中间白色部分就如美丽的“水滴”.则图中“水滴”外部阴影部分的面积为_________.【答案】 16π3+【解析】【分析】根据图形与,建立直角坐标系,画出图形,()()(){}22,cos sin 4,0πP x y x y θθθ=-+-=≤≤求出相应的坐标,先求第一、二象限的阴影面积,再求第三象限的阴影面积,再求和即可求解.【详解】根据题意,建立直角坐标系,如图所示:在方程,中, ()()22cos sin 4x y θθ-+-=0πθ≤≤令,则有, 0x =222cos 2sin sin 4y y θθθ+-+=所以,其中, 12sin y yθ=-0πθ≤≤所以,所以, []sin 0,1θ∈[]12sin 0,2y y θ=-∈解得,1y ⎡⎤⎤∈-⎣⎦⎦所以,,,, (A ()0,3E ()0,1G -(0,D 令,则有,0θ=()2214x y -+=所以,,()1,0C ()3,0N 令,则有πθ=()2214x y ++=所以,. ()1,0B -()3,0M -由,,易得与线段()3,0M -()3,0N ()0,3E A MEN MN 组成的图形为的上半圆,229x y +=由此可知,在第一、第二象限中的阴影面积是由 的上半圆减去上半圆 229x y +=()2214x y -+=与上半圆相交的部分形成, ()2214x y ++=即与线段组成的面积,设为. A BACBC S 水滴上部由,,三点易得 (A ()1,0B -()1,0C 为边长为2的等边三角形,ABC A所以 212ππ263ABC AnC S S =⨯⨯-=-A 弓形所以,4π23ABC AnC S S S =+=A 弓形水滴上部设第一、二象限的阴影面积为, 1S 则. 19π9π4π19π2236S S =-=-+=+水滴上部由,,易得与线段 ()1,0B -()1,0C ()0,1G -A BGCBC 组成的图形为的下半圆, 221x y +=设在第三象限中的阴影面积为, 2S 则有, 2π4MOD MpD S S S =+-A 弓形由图知11322MOD S MO OD =⨯⨯=⨯=A ,,11222MBD S MB OD =⨯⨯=⨯=A 2π3MBD ∠=所以,214ππ233MBD MpD S S =⨯⨯-=A 弓形所以,2π4ππ13π43412MOD MpD S S S =+-=+-=A 弓形所以图中“水滴”外部阴影部分的面积为:. 1219π13π16π226123S S S ⎛=+=⨯=+ ⎝故答案为:. 16π3+【点睛】本题考查了圆与三角函数综合的知识点,可以根据图形的对称性建立直角坐标系,将图形转化为实际的数据,割补法是求阴影面积常用的方法,需要考生有一定的分析转化能力.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. 记为正项数列的前项和,已知是4与的等比中项. n S {}n a n 1n a +n S (1)求的通项分式;{}n a (2)证明:. 2222123111154n a a a a +++⋅⋅⋅+<【答案】(1)21n a n =-(2)证明见解析 【解析】【分析】(1)由等比中项得,进而由递推式计算出,并得到,得数列()214n n a S +=11a =12n n a a --=是等差数列,进而可求解;{}n a (2)由,从第二项开始放缩即可证明. ()22111114121n a n n n ⎛⎫=<- ⎪-⎝⎭-【小问1详解】∵是4与的等比中项,∴①. 1n a +n S ()214n n a S +=当时,,∴. 1n =()2111144a S a +==11a =当时,②,2n ≥()21114n n a S --+=由①-②得,, ()()()22111144n n n n n a a S S a --+-+=-=∴, ()()1120n n n n a a a a ----+=∵,∴,0n a >12n n a a --=∴数列是首项为l ,公差为2的等差数列, {}n a ∴的通项公式. {}n a 21n a n =-【小问2详解】由(1)得,2111a =当时,,2n ≥()222111111444121n a n n n n n ⎛⎫==<=- ⎪--⎝⎭-∴ 22222221232311111111n na a a a a a a +++⋅⋅⋅+=+++⋅⋅⋅+1111111115151114122314444n n n n ⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫<+-+-+⋅⋅⋅+-=+-=-< ⎪ ⎪ ⎪ ⎪⎢⎥-⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦18. 已知a ,b ,c 分别为三个内角A ,B ,C 的对边,且.ABC A cos sin a C C b c +=+(1)求A ;(2)已知M 为BC 的中点,且,的平分线交BC 于N ,求线ABC A AM =BAC ∠段AN 的长度. 【答案】(1) π3A =(2) AN =【解析】【分析】(1)根据题意,由正弦定理的边角互化将原式化简,再结合三角恒等变换即可求得结果; (2)根据题意,可得,再结合三角形()22222242AMAB ACAB AB AC AC c b bc =+=+⋅+=++的面积公式,代入计算,即可得到结果. 【小问1详解】由题意知中,, ABC A cos sin a C C b c +=+由正弦定理边角关系得:则sin cos sin A C A C,()sin sin sin sin sin cos cos sin sin B C A C C A C A C C =+=++=++, sin cos sin sin A C A C C =+∵,()0,πC ∈∴, sin 0C ≠cos 1A A -=∴,∴, π2sin 16A ⎛⎫-= ⎪⎝⎭π1sin 62A ⎛⎫-= ⎪⎝⎭又,, ()0,πA ∈ππ5π,666A ⎛⎫-∈- ⎪⎝⎭所以,即. ππ=66A -π3A =【小问2详解】如下图所示,在中,为中线,ABC A AM∴, 2AM AB AC =+∴,()22222242AMAB ACAB AB AC AC c b bc =+=+⋅+=++ ∴. 2212b c bc ++=∵, ABC S =△1sin 2bc A ==3bc =∴,b c +==∵, ABC ABN ACN S S S =+△△△,∴. ()1πsin 26b c AN AN =+=AN =19. 近日,某芯片研发团队表示已自主研发成功多维先进封装技术XDFOI ,可以实现4nm 手机SOC 芯片的封装,这是中国芯片技术的又一个重大突破,对中国芯片的发展具有极为重要的意义.可以说国产4nm 先进封装技术的突破,激发了中国芯片的潜力,证明了知名院士倪光南所说的先进技术是买不来的、求不来的,自主研发才是最终的出路.研发团队准备在国内某著名大学招募人才,准备了3道测试题,答对两道就可以被录用,甲、乙两人报名参加测试,他们通过每道试题的概率均为,且相互独立,若()01p p <<甲选择了全部3道试题,乙随机选择了其中2道试题,试回答下列问题.(所选的题全部答完后再判断是否被录用)(1)求甲和乙各自被录用的概率;(2)设甲和乙中被录用的人数为,请判断是否存在唯一的值,使得?并说明理由. ξp 0p () 1.5E ξ=【答案】(1)甲被录用的概率为,乙被录用的概率为2332p p -2333p p -(2)不存在;理由见解析 【解析】【分析】(1)分析已知,甲被录用符合二项分布,乙被录用符合组合排列,分别利用对应求概率公式计算即可.(2)先分析的可能取值,然后分别求解对应概率,再利用离散型数学期望的公式表示出数学期望,然后构ξ造函数,利用求导分析函数单调性,进而判断即可. 【小问1详解】由题意,设甲答对题目的个数为,得, X ()~3,X B p 则甲被录用的概率为,()2232313C 132P pp p p p =-+=-乙被录用的概率为. ()222332C 133P p p p p =-=-【小问2详解】的可能取值为0,1,2,ξ则,()()()12011P P P ξ==--, ()()()1212111P P P P P ξ==-+-,()122P PP ξ==∴ ()()()()()121212*********E P P P P P P PPξ=⨯--+⨯-+-+⨯⎡⎤⎣⎦,23232312323365 1.5P P p p p p p p =+=-+-=-=,32101230p p ∴-+=设,()()321101230f p p p p +=<<-则.()23024f p p p '=-∴当时,单调递减,405p <<()f p 当时,单调递增,415p <<()f p 又,,,()03f =()11f =4110525f ⎛⎫=> ⎪⎝⎭所以不存在的值,使得.p 0p ()00f p =20. 如图,四棱锥的底面是边长为2的正方形,.P ABCD -ABCD 2PA PB ==(1)证明:;PAD PBC ∠=∠(2)当直线PA 与平面PCD 所成角的正弦值最大时,求此时二面角的大小. P AB C --【答案】(1)证明见解析(2)6π【解析】【分析】(1) 分别取,的中点,,连接,,,证明出,可得AB CD E F PE EF PF PC PD =,由此可证得结论成立;PAD PBC ≌△△(2)先根据条件推出为二面角的平面角,设,建立空间直角坐标系,利用PEF ∠P AB C --PEF α∠=空间向量法结合基本不等式求出直线与平面所成角的正弦值的最大值,求出对应的角的值,即PA PCD 可求解. 【小问1详解】分别取,的中点,,连接,,, AB CD E F PE EF PF ∵,为的中点,∴.PA PB =E AB PE AB ⊥∵四边形为正方形,则且,∴. ABCD AB CD ∥AB CD =CD PE ⊥∵,分别为,的中点,∴,∴,E F AB CD EF AD ∥EF CD ⊥∵,∴平面.EF PE E ⋂=CD ⊥PEF∵平面,∴. PF ⊂PEF CD PF ⊥在中,PCD A ∵为的中点,,∴. F CD CD PF ⊥PC PD =又∵,,∴, PA PB =AD BC =PAD PBC ≌△△从而可得. PAD PBC ∠=∠【小问2详解】由(1)可知,,PE AB ⊥EF AB ⊥∴为二面角的平面角,且,PEF ∠P AB C --PE ==以点为坐标原点,,所在直线分别为x ,轴建立如下图所示的空间直角坐标系,E EB EFy设,其中,PEF α∠=0απ<<则,,,,,,()1,0,0A -()1,0,0B ()1,2,0C ()1,2,0D -()0,2,0F ()P αα,,.()AP αα= ()2,0,0DC =u u ur()FP αα=- 设平面的法向量为,PCD (),n x y z =由,即,取, 00n DC n FP ⎧⋅=⎪⎨⋅=⎪⎩202)0x y z αα=⎧⎪⎨-⋅=⎪⎩y α=则,,∴,2z α=-0x=(),2n αα=-cos ,n AP n AP n AP⋅<>==⋅==令,(77t α-=∈-+则, cos α=则,cos ,n AP <>==≤=当且仅当时,即当时,等号成立.1t =cos α=6πα=所以当直线与平面所成角的正弦值最大时,二面角为.PA PCD P AB C --6π21. 已知,D 是圆C :上的任意一点,线段DF 的垂直平分线交DC 于点P . ()1,0F -()22116x y -+=(1)求动点P 的轨迹的方程:Γ(2)过点的直线与曲线相交于A ,B 两点,点B 关于轴的对称点为,直线交轴于(),0M t l Γx B 'AB 'x 点,证明:为定值. N OM ON ⋅【答案】(1)22143x y +=(2)证明见解析 【解析】【分析】(1)由中垂线性质,可知,得动点P 的轨迹以,F 42PC PF PC PD DC FC +=+==>=C 为焦点的椭圆;(2)将直线与曲线方程联立,利用韦达定理及题目条件表示出点N 坐标,后可得答案. l Γ【小问1详解】圆:,圆心为,半径为4,C ()22116x y -+=)1,0因为线段DF 的垂直平分线交DC 于P 点,所以, PD PF =所以, 42PC PF PC PD DC FC +=+==>=所以由椭圆定义知,P 的轨迹是以,F 为焦点的椭圆, C 则,,.242a a =⇒=221c c =⇒=2223b a c =-=故轨迹方程为:.22143x y +=【小问2详解】依题意,直线不垂直于坐标轴,设直线的方程为,将其与方程联立:l l ()0x my t m =+≠Γ,消去x 得. 22143x my tx y =+⎧⎪⎨+=⎪⎩()2223463120m y mty t +++-=方程判别式,设,,则,()2248430m t+->()11,A x y ()22,B x y ()22,B x y '-由韦达定理有,,122634mt y y m -+=+212231234t y y m -=+则直线的方程为,AB '()121112y y y y x x x x +-=--令()1212211212N 121212202my y t y y x y x y y yy x m t y y y y y y +++=⇒===⋅++++,则,得.2312426t m t mt t -=⋅+=-40,N t ⎛⎫ ⎪⎝⎭()400,,,OM t ON t ⎛⎫== ⎪⎝⎭∴.即为定值4.44OM ON t t ⋅=⋅= OM ON ⋅ 22. 已知函数,.()1e ln axf x x x-=+a ∈R (1)当时,求函数的最小值; 1a =()f x x -(2)若函数的最小值为,求的最大值. ()f x xa a 【答案】(1)0(2)1【分析】(1)当时,令,求得,根据在不同区间1a =()()F x f x x =-()()()121e x x x x F x --=-'()F x '的符号判断的单调性,由单调性即可求出的最小值;()F x ()()F x f x x =-(2)将等价变换为,借助第(1)问中判断的符号时()≥f x a x ()0f x ax -≥()()()121e x x xx F x --=-'构造的在时取最小值,取,将问题转化为有解问题即可.()1ex g x x -=-1x =()ln g ax x -ln 1ax x -=【小问1详解】当时,令,,1a =()()1e ln x x x F xf x x x-+=--=()0,x ∈∞则,()()()()()11112221e e 11e e 11x x x x x x x x x x x x F xx x ------+-'==-⋅-+-=令,,则,()1ex g x x -=-x ∈R ()1e 1x g x -'=-易知在上单调递增,且,()g x 'R ()10g '=∴当时,,在区间上单调递减,且,()0,1x ∈()0g x '<()g x ()0,1()()110e x g x x g -=->=当时,,在区间上单调递增,且,()1,x ∈+∞()0g x '>()g x ()1,+∞()()110e x g x x g -=->=∴当时,,在区间上单调递减, ()0,1x ∈()()()121e 0x x x F x x --'=-<()F x ()0,1当时,,在区间上单调递增,()1,x ∈+∞()()()121e 0x x x F xx --'=->()F x ()1,+∞当时,取得极小值,也是最小值,,1x =()F x ()()11mine 1ln1101F x F -==+-=∴当时,函数的最小值为. 1a =()f x x -0【小问2详解】由已知,的定义域为, ()f x ()0,∞+若函数的最小值为,则有,∴,, ()f x x a ()≥f x a x()f x ax ≥()0f x ax -≥令,即的最小值为,()()h x f x ax =-()()1e ln ax x ax h x x ax xf -+=--=0由第(1)问知,当且仅当时,取最小值,1x =()1ex g x x -=-()10g =∴当且仅当时,取得最小值, ln 1ax x -=()ln g ax x -0又∵,()()()l 1l 1n 1n n e e ln ln ln ee ax ax ax x x g ax x ax x x ax x ax h x x-----=--=+-=+-=∴只需令有解,即有解, ln 1ax x -=ln 1x a x+=令,,则, ()ln 1x H x x+=()0,x ∈+∞()()221ln 1ln x x x x H x x x ⋅-+'==-当时,,在区间上单调递增, ()0,1x ∈()2ln 0xH x x '=->()H x ()0,1当时,,在区间上单调递减, ()1,x ∈+∞()2ln 0xH x x'=-<()H x ()1,+∞∴, ()()ln 111x a H x H x+==≤=综上所述,若函数的最小值为,则的最大值为. ()f x xa a 1【点睛】在导数压轴题中,常常会使用前问的结论或某一步构造的函数,解决后面的问题.本题第(2)问中直接求导分析的单调性较为困难,这里使用了换元思想,借助第()()1e ln ax x ax h x x ax xf -+=--=(1)问构造的,使,以达到简化运算的目的.()1ex g x x -=-()()ln g ax x h x -=。
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一中2022-2023年高一第一学期第三次月考英语第二部分阅读(共两节,满分50 分)第一节(共15小题;每小题2.5分,满分37.5)阅读下列短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。
AUrban(乡村的) gardens are valuable places to communities. They provide green spaces to grow sustainable food, build community cohesion(凝聚力), make new friends, connect with the earth, and much more. So, let's check out our list of 4 inspiring urban gardens in the US.Gotham GreensWhere: New York & ChicagoWhat: Gotham Greens first started in Brooklyn and now has four locations in New York City and Chicago. Their flagship farm in Brooklyn produces over 100,000 pounds of greens per year. But it doesn't just produce healthy local vegetables. It is using high-tech greenhouses with solar panels(太阳能电池板) to make sure the food grown is healthy and sustainable.Baltimore Urban Gardening with StudentsWhere: Baltimore, MarylandWhat: The Baltimore Urban Gardening with Students (BUGS) program encourages students to get their hands dirty and plant vegetables through their after-school and summer programs. Many of these kids don't have access to green spaces, and have never had the opportunity to grow food.ReVision Urban FarmWhere: Boston, MassachusettsWhat: ReVision Urban Farm in Boston works in partnership with the ReVision Family Home a shelter for 22 homeless parents and their kids. The farm provides these families with information on healthy eating, and access to the farm's fresh vegetables. The organization also provides job training to help these families escape the cycle of poverty.SwaleWhere: New YorkWhat: Swale, a floating food forest located on a large boat, is a project meant to inspire citizens to rethink the relationship between our cities and our food. This urban garden serves as both a living art exhibit(展览品) and an educational farm. Food forests are sustainable gardens that include vegetables, fruit, nut trees, bushes, herbs, and vines- each one complementing(完美) the other in a symbiotic(共生的) relationship.21. What does the BUGS program mainly do?A. Provide job training for students.B. Use high-tech greenhouses to grow healthy food.C. Create a sustainable garden on a large boat.D. Offer students the opportunity to grow vegetables.22. Which urban garden helps homeless people get out of poverty?A. Gotham Greens.B. Baltimore Urban Gardening with Students.C. ReVision Urban Farm.D. Swale.23. Where can citizens go to see a food forest?A. Chicago.B. Baltimore.C. Boston.D. New YorkBDo you like to keep fit? We're always told that regular exercise is good for our body and mind. More and more people are taking up activities that improve their fitness. But is there a risk that some of us might get obsessed (着迷的) and overdo it?Well, for some people, fitness has become an obsession as they aim for perfection. And fitness trackers (追踪器) and apps can add to this addiction (癒), especially if someone is driven by achievement and perfectionism. And sharing data on social media means exercising becomes public and competitive, which could cause problems in someone who is vulnerable(脆弱的).Experts say this can lead to a medical condition called orthorexia nervosa, or addiction to healthy eating and over-exercise. Untreated, it can lead to malnutrition and mental health complications.Too much exercise can also take its toll on someone's physical health as well. Symptoms of over-exercising include injuries such as a broken leg and a low immune system. So how much exercise is too much? Researchers found the ideal pace to jog was about eight kilometers per hour and that it was best to jog no more than three times a week or for 2.5 hours in total, showing that moderate jogging is possibly more beneficial than being inactive or undertaking strenuous (剧烈的) jogging.If you're more of a couch potato than a quick runner, this might sound like good news. But for amateur (业余的) athletes who can't help but push their bodies to the limit the advice from Martin Turner. a sports and exercise psychologist(心理学家),is, “ It's all about letting go, not being obsessed, learning not to controleverything, saying, “You don't need to be perfect.”24. What contributes most to people's addiction to over-exercise according to paragraph 2?A. Their urge for social support.B. Their pursuit of perfectionism.C. Their addiction to fitness apps.D. Their concern over health issues.25. What does the underlined phrase “take its toll on” in paragraph 3 mean?A. Have a bad effect on.B. Go hand in hand with.C. Make little difference to.D. Play an important role in.26. What is the text mainly about?A. The risk of fitness obsession.B. The importance of mental healthC. The benefits of moderate jogging.D. The symptoms of over-exercising.27. In which section of a newspaper may this text appear?A. Science.B. Opinion.C. Culture.D. LifestyleCA woman held her phone tightly to her heart the way a churchgoer might hold a Bible. She was anxious to take a picture of an impressive bunch of flowers that sat not so far away, but first she had to get through a crowd of others pushing their way to do the same. The cause of this was Bouquets to Art, one of the most popular events at the de Young Museum in San Francisco.Flower sellers were asked to create flower arrangements that respond to pieces of art on display, from ancient carvings to contemporary sculptures. It's extremely attractive and also memorable, to the point that it has become a problem.In recent years, the de Young received more than a thousand complaints from people who felt that cell phones had spoiled their experience of the exhibit. Institutions of fine art around the world face similar problems as the desire to take photographs becomes a huge attraction for museums, as well as something thatupsets some of their patrons(资助人). So the de Young responded with a kind of compromise: carving out "photo free" hours during the exhibition's six-day run.One common complaint about the effect of social media on museum culture is thatpeople seem to be missing out on experiences because they are so busy collecting evidence of them. A study recently published in the journal Psychological Science suggests there is some truth to this. It finds that people who keep taking photos of an exhibit and posting them on social media rather than simply observing it, have a hard time remembering what they see. But the issue is complex for the professionals running museums. Linda Butler, the de Young's head of marketing and visitor experience, acknowledges that not everyone wants a museum to be "a photo-taking playland". Yet a lot of people do and she believes that the de Young is in no position to judge one reason for buying a $ 28 ticket to be more valid than another. “If we removed social media and photography,” she says,“we would risk becoming irrelevant.”28. What was the woman eager to do according to paragraph 1?A. To get her phone.B. To escape the crowdC. To take a photo.D. To push ahead.29. How did the de Young respond to the complaints from visitors?A. By setting periods without photo-taking.B. By making the exhibition free of charge.C. By compromising with the government.D. By extending the free exhibition hours.30. The recent study finds that the use of social media in museums may _________.A. uncover the truthB. cause irrelevancyC. bring more pleasureD. play a negative role31. Which of the following may Linda Butler support?A. Reducing admission prices.B. Meeting different needs of the visitors.C. Reserving judgement in public.D. Banning social media and photography.DCuckoos(布谷鸟) are masters of cheating. When it comes to raising young, they don't spend the energy building a nest(鸟巢), protecting eggs or feeding children. Instead, the female passes these roles on to other birds. They don't raise their own young. Usually, they lay their eggs in the nests of other birds, fooling other birds into thinking the cuckoo eggs are theirs.To succeed in doing this, & female cuckoo watches over her chosen nest to observe feeding times. When the host parent leaves its nest in search of food, the cuckoo quickly lays her eggs among those already in the nest. Sometimes, she will even destroy and remove one of the host's eggs to make room for her own.Cuckoos are medium-sized birds with long tails, and often have gray or brown backs. When they hatch(孵化) and begin to grow in a host's nest, the difference between the two can be obvious to an onlooker. Often the cuckoo is twice the size of its foster parents(养父母),but still continues to receive food from them.The cuckoo imposter(冒名顶替者)is usually the only baby bird that the host parent has to care for. This is because when the cuckoo hatches after around 11 days, it gets rid of all the other eggs in the nest. It will lift each egg onto its back before throwing them one by one over the edge of the nest. Even then, the nonbiological parent will continue to treat it as one of its own.Also, cuckoos have developed to produce eggs that are similar in color to their main hosts'. This reduces the chances of eggs being attacked. Female cuckoos have been known to take host birds' attention away after laying their eggs by producing a noise similar to Eurasian sparrowhawks, scaring birds away from returning to the nest and allowing time for the cuckoo to make her escape unnoticed.32. What do mother cuckoos usually do when it comes to raising their young?A. Attend to eggs.B. Build a new nest.C. Learn feeding skills.D. Rely on other birds.33. Wh at does the underlined word “them” in paragraph 3 refer to?A. Other cuckoos.B. Baby cuckoos.C. Cuckoo's host parents.D. Cuckoo's birth parents.34. What can be inferred about the cuckoo imposter?A. It usually hatches out earlier than the other eggs in the nest.B. It throws all the other eggs out of nests with its mother's help.C. It often makes a noise to scare other birds away from the nest.D. It looks much larger than other eggs in the nest before hatching.35. Which can be the best title for the text?A. How cuckoos protect their childrenB. How cuckoos fool other birdsC. How cuckoos destroy others' nestsD. How cuckoos produce eggs第二节(共5小题;每小题2.5分,满分12.5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,选项中有两项为多余选项。