四川省绵阳市2020届高三英语第三次诊断性测试试题
四川省绵阳市高三英语第三次诊断性考试试题(扫描版)

绵阳市高中2013级第三次诊断性考试英语试题参考答案及评分标准第Ⅰ卷(选择题,共100分)第一部分:听力(共二节,满分30分)第一节 (共5小题;每小题1.5分,满分7.5分)1-5 CBCCB第二节 (共15小题;每小题1.5分,满分22.5分)6-10 BABAA 11-15 CBBAB 16-20 BBCAB第二部分:阅读理解(共二节,满分40分)第一节(共15小题:每小题2分,满分30分)21-25 ADBDA 26-30 CDABD 31-35 BCDDB第二节(共5小题:每小题2分,满分10分)36-40 FEABG第三部分英语知识运用(共二节,满分45分)第一节完形填空(共20小题;每小题1.5分,满分30分)41-45 BCDBA 46-50 BCACD 51-55 DAACB 56-60 DBDBC第Ⅱ卷(非选择题,共50分)第二节(共10小题:每小题1.5分,满分15分)61. exists 62. to understand 63. the 64. from/against 65.which66. Fortunately 67. be treated 68. their 69. faced 70. effective评分标准:有任何错误,包括用词错误、单词拼写错误(含大小写)或语法形式错误,均不给分。
第四部分写作(共两节,满分35分)第一节短文改错(共10小题,每小题1分,满分10分)As is known to us all, birthday is important to everybody. It was Jerry’s birthday on yesterday.Some of his classmates held ^birthday party to congratulate him. They bring a lot of nice gifts.a broughtTheir English teacher, Mary, was also present at the party with his best wishes forJerry. He was sohermoving that he couldn’t help expressing his thank to them. For the whole afternoon they sang andmoved thanksdanced happily without think about their school work. In the evening they decidedto study eventhinkinghard to make up for it. What happy they all were! I really wish I had join them. harder How joined评分标准:有任何错误,包括用词、修改及标号的位置、单词拼写错误(含大小写)或语法形式错误,均不给分。
四川省绵阳市2019届高三第三次诊断性考试英语

四川省绵阳市高中2019 级第二次诊断性考试英语试题本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,满分 注意事项:1.答题前,考生务必将自己的学校、班级、姓名、考号用0.5 毫米的黑色签字笔填写在答题卡上,并检查条表码粘贴是否正确。
2.选择题( 1— 65)使用 2B 铅笔填涂在答题卡对应题目标号的位置上,非选择题用 0.5 毫米黑色签字笔书写在答题卡的对应题框内,超出答题区域书写的答案无效;在草稿纸、试题卷上答 题无效。
3.考试结束以后,将答题卡收回。
第Ⅰ卷(选择题,共 100 分)第一部分:英语知识运用(共两节,满分 50 分)第一节:语法和词汇知识(共 20 小题;每小题 1分,满分 20 分)从 A 、B 、C 、D 四个选项中,选出可以填入空白处的最佳选项,并在答题卡上将该项涂黑。
1.—I can 't stand you! I 'm leaving.— . See if I care.A .No problemB .Come onC . No doubtD .Go ahead 2. Remember, better healthin feet, Walk more, feel the difference!A . measuresB .has measuredC . is measuredD .was measured3. Memorial Day is a US national holidayto remember those who gave their lives in the service ofthe nation. 4.— Do you ask why I am learning English?— Yes, I mean, what are you going to do with it? Are you going to be a teacher of English or somethinglike ? 5.You are welcome to ask whatever questions. Don ' t6. It is easy to set the goal of our production for next year, butto achieve it takes a lot of trouble.7. Is this the reason he gave us for the delay of the project?A .howB .whatC . why·1·8. Being a tour guide offers me a chance to work for money and meanwhile travel for.A . pleasureB .honourC . leisureD .cheer150 分,考试时间 120 分钟。
2022届成都和绵阳一二三诊英语听力试题

1.成都市2019级高中毕业班第一次诊断性检测—英语听力试卷第一节1.What will the woman do this weekend?A.Walk in the mountain.B.Finish her paper.C.Attend a wedding.2.How does the man feel now?A.Annoyed.B.Surprised.C.Pitiful.3.Where does the conversation probably take place?A.At home.B.At the agency.C.In the office.4.How much should the man pay?A.$2.B.$4.C.$6.5.What are the speakers mainly talking about?A.The man's son.B.A writer.C.A book.第二节听第6段材料,回答第6至7题。
6.What is the possible relationship between the speakers?A.Strangers.B.Friends.C.Classmates.7.What contributes most to the effect of the man's stories?A.The content.B.The words.C.The ending.听第7段材料,回答第8至9题。
8.How will the man go to the destination?A.By shuttle.B.By taxi.C.By subway.9.Where does the conversation probably take place?A.At the airport.B.In the hotel.C.In the museum.听第8段材料,回答第10至12题。
THUSSAT中学生标准学术能力2023-2024学年高三上学期9月诊断性测试英语试卷(Word版含

THUSSAT中学生标准学术能力2023-2024学年高三上学期9月诊断性测试英语试卷(Word版含答案,无听力音频及听力原文)中学生标准学术能力诊断性测试2023年9月测试英语参考答案1-5 BCDCD 6-10 BCBDA 11-15 CBDCA 16-20 FACGE21-25 BACDC 26-30 DABDC 31-35 BBCAD36.born 37.to 38.to study 39.striking 40.has had 41.unforgettable 42.whose 43.difficulty 44.was published 45.it作文参考:Dear Ben,It is a pleasure to know that you have finished the journey in China. You mentioned which gift is most suitableto bring back to your country, and I strongly recommend Chinese tea, especially the famous Yunnan Pu-erh tea. Dueto the reason that it has a rich history in China and is a great choice if you want to explore the depths of Chinese teaculture. What’s more, it has the function of help you lose weight as well as keep healthy. Hope my recommendationcan help you. By the way, welcome to China back again. Wish you a nicejourney.Yours,Li Hua短文改错:Running, especially outside, is good for ease anxiety. While it can definitely be hardly to get up and run the firsteasing hardcouple of days, the benefits start flowing in. Throughout the day you may feel energetic, less anxious, and happiermoreon account to the release of some waste in your body. A run a day do not scare the anxiety away for everyone, and itof does butis great to do everything actively. And if you run social, your anxiety can be lessened and your social time can besociallyincreased. Try running when you have a big test which is came up. Try running the next time when you need to sitcomingdown to study for a couple of hour.hours续写参考:My dad told the other two to keep moving; we’d follow shortly. He took my hands in his and squeezed. “I’mnot leaving you. It’s almost dark. We need to keep moving. I’ll help you.” “I can’t,” I said as the tears pooled in myeyes. “You can. Just take one tiny step at a time. Don’t look at the mountain. Keep your eyes focused on your feetand on me.” Dad never let go of my hands as he continued to encourage me.“Just one tiny step at a time,” Dad repeated as we climbed together. Some of my steps were so minuscule thatwe hardly moved at all, but I clung to the power in his words. Eventually, we did make it over the ridge, grateful tofind that it was the last formidable obstacle in our way. Years have passed since that hike, and Dad’s words havecontinued to be a lifeline for me. It might take me a little longer to conquer my mountains, but one tiny step at a timewill eventually get me where I need to be.第1 页共1 页{#{ABAIIUoQogoigAgAABQJAAJ AAgQCQgCFQiCFEiCGgQOkQBCk ACAAIoCGIoBGFRAAIAAEAoBAwABBFAAQBFAAAB=A}#A}=}#}中学生标准学术能力诊断性测试2023 年9 月测试英语试卷本试卷共150 分,考试时间100 分钟。
四川省达州市2024届高三下学期二模考试 英语含答案

达州市普通高中2024届第二次诊断性测试英语试题(答案在最后)该试卷由四部分组成。
第一、二部分和第三部分的第一节为选择题。
第三部分的第二节和第四部分为非选择题。
满分150分,考试用时120分钟。
注意事项:1.答题前,考生务必将自己的学校、班级、姓名、考号用0.5毫米的黑色签字笔填写在答题卡上,并检查条形码粘贴是否正确。
2.选择题(1-60)使用2B铅笔填涂在答题卡对应题目标号的位置上,非选择题用0.5毫米的黑色签字笔书写在答题卡的对应题框内,超出答题区域书写的答案无效;在草稿纸、试题卷上答题无效。
3.考试结束以后,将答题卡收回。
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。
例:How much is the shirt?A.£19.15.B.£9.15.C.£9.18.答案是B.1.What’s the man worried about?A.A coming lecture.B.The students’behavior.C.Customs in a foreign country.2.Where will the woman stay?A.At a hotel.B.At a guesthouse.C.At a backpacker hostel.3.How will the man go home?A.He will take a bus.B.His mother will pick him up.C.Jane’s mother will drive him home.4.When should the man take the cake out?A.In15minutes.B.In30minutes.C.In45minutes.5.What is the weather like in the woman’s city?A.Changeable.B.Cold.C.Windy.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。
四川省绵阳市高中2024届高三突击班第一次诊断性考试模拟测试英语答案

YTUN理科突击班高中2021级第一次诊断性考试参考答案英语第一部分听力(共两节,满分30分)第一节1-5ACABB第二节6-10ABBAC11-15ABCCC16-20BCABA第二部分阅读理解21.D22.A23.B【来源】江苏省2023-2024学年高三第一届“七夕杯”高中英语能力检测试题(原创模拟试题)(含听力)【导语】本文是应用文。
文章介绍了四位杰出女性。
21.细节理解题。
根据Maria Sibylla Merian部分中的“Merian was fascinated by insects,and shebegan collecting,studying,and drawing them when she was as young as13.(梅里安对昆虫非常着迷,她在13岁时就开始收集、研究和绘制昆虫。
)”可知,Maria SibyllaMerian酷爱研究昆虫。
故选D。
22.推理判断题。
根据Madam CJ.Walker部分中的“Madam C.J.Walker developed a conditionthat caused her to lose her hair,and with it,an interest in hair care.(C.J.沃克夫人患上了一种疾病,导致她失去了头发,同时也使她对护发产生了兴趣。
)”可知,Madam C.J.Walker掉发严重,因此对护发产生了兴趣。
故选A。
23.细节理解题。
根据Hedy Lamarr部分中的“But her film career is far from her only achievement.she was also a brilliant inventor.(但她的电影生涯远不是她唯一的成就。
她也是一位杰出的发明家。
)”可知,Hedy Lamarr不仅是电影明星而且是发明家。
故选B。
24.A25.C26.A27.B【来源】2023届重庆市七校高三三诊考试英语试题【导语】本文是一篇记叙文。
四川省绵阳市高中2024届高三突击班第一次诊断性考试模拟测试 语文答案

YTUN理科突击班高中2021级第一次诊断性考试参考答案语文1.D2.A3.D【来源】四川省绵阳市绵阳中学2022-2023学年高三上学期综合质量检测语文试题【解析】1.本题考查学生理解文章内容,筛选并整合文中信息的能力。
A.“由于人有着不同于自然界生物的长久记忆,使得记忆对于人类历史的建构异常重要”原因分析错误,原文第一段“与自然界的生物不同,人不仅有长久的记忆,而且还能通过语言、文字、图像等媒介不断强化、延伸这一能力。
就此而言,记忆对于人类历史的建构具有异乎寻常的重要性”中的“此”既指“长久的记忆”,又指“还能通过语言、文字、图像等媒介不断强化、延伸这一能力”。
选项以偏概全。
B.“古器物”错,原文第三段“历史文物作为家庭陈设或文人雅集的谈资”说的是“历史文物”,另外,原因的概括也不全面,原文说的是“它不但更便于唤醒人的历史记忆,而且让人‘身生后世,眼对前朝’,在古与今的交相辉映中有效延展了记忆的张力”。
C.“都采用‘托史言志’”错误,原文第三段“使‘托史言志’成为历代士人最重要的话语方式”说的是“最重要”的方式,而不是“都采用”。
故选D。
2.本题考查学生分析论点、论据和论证方法的能力。
A.“与民间传统的记忆呈现方式做了对比”错误,原文第三段“与民间记忆总是游离于真实历史不同,士人的相关记忆则更多奠基于实际的历史知识训练”说的是两者形成记忆的方式不同,而不是呈现记忆的方式不同。
故选A。
3.本题考查学生分析概括作者在文中的观点态度的能力。
D.“只有当人们面临的现实境遇残酷时才会被美化,成为个人情感寄托的载体”错误,由原文第一段“几乎一切过往均会成为诱发诗情记忆的对象”可知。
故选D。
4.D5.B6.①懂得人与人相处的基本礼仪。
采访伊始,ChatGPT与记者互致问候,并表示很高兴与记者交流。
②知道谦虚礼让,说话留有余地。
以没有感知能力巧妙回避火了的话题,说不能知道自己的知名度。
说自己的回答不是100%正确,告诉学生仍然需要他们思考和判断。
2020届四川省绵阳市高三第三次诊断性测试数学(理)试题及答案

绝密★启用前2020届四川省绵阳市高三第三次诊断性测试数学(理)试题注意事项:1、答题前填写好自己的姓名、班级、考号等信息2、请将答案正确填写在答题卡上一、单选题1.设集合A={(x,y)|x2+y2=1},B={(x,y)|x+y=1},则A∩B中元素的个数是()A.0 B.1 C.2 D.3答案:C可画出圆x2+y2=1和直线x+y=1的图象,从而可看出它们交点的个数,从而得出A∩B 中的元素个数.解:画出x2+y2=1和x+y=1的图象如下:可看出圆x2+y2=1和直线x+y=1有两个交点,∴A∩B的元素个数为2.故选:C.点评:考查了描述法的定义,交集的定义及运算,数形结合解题的方法,考查了计算能力,属于容易题.2.已知复数z满足(1﹣i)•z=3i|,则z=()A.1﹣i B.1+i C.2﹣2i D.2+2i答案:B利用复数的运算法则、模的计算公式即可得出. 解:(1﹣i )•z =|3+i|,∴(1+i )(1﹣i )•z =2(1+i ),则z =1+i . 故选:B . 点评:本题考查了复数的运算法则、模的计算公式,考查了推理能力与计算能力,属于容易题. 3.已知x •log 32=1,则4x =() A .4 B .6C .432logD .9答案:D利用对数的性质和运算法则及换底公式求解. 解:∵x •log 32=1, ∴x =log 23,∴4x 243944log log ===9, 故选:D . 点评:本题考查对数值的求法,是基础题,解题时要认真审题,注意对数的性质、运算法则及换底公式的合理运用,属于容易题.4.有报道称,据南方科技大学、上海交大等8家单位的最新研究显示:A 、B 、O 、AB 血型与COVID ﹣19易感性存在关联,具体调查数据统计如图:根据以上调查数据,则下列说法错误的是()A .与非O 型血相比,O 型血人群对COVID ﹣19相对不易感,风险较低B .与非A 型血相比,A 型血人群对COVID ﹣19相对易感,风险较高C .与O 型血相比,B 型、AB 型血人群对COVID ﹣19的易感性要高 D .与A 型血相比,非A 型血人群对COVID ﹣19都不易感,没有风险答案:D根据频率分布直方图,利用频率、频数与样本容量的关系,患者占有比例即可解答. 解:根据A 、B 、O 、AB 血型与COVID ﹣19易感性存在关联,患者占有比例可知: A 型37.75%最高,所以风险最大值,比其它血型相对易感; 故而D 选项明显不对. 故选:D . 点评:本题考查由频数直方图,看频数、频率,判断问题的关联性,属于中档题5.在二项式2()nx x-的展开式中,仅第四项的二项式系数最大,则展开式中常数项为() A .﹣360 B .﹣160 C .160 D .360答案:B根据展开式二项式系数最大,求出n =6,然后利用展开式的通项公式进行求解即可. 解:∵展开式中,仅第四项的二项式系数最大, ∴展开式共有7项,则n =6, 则展开式的通项公式为T k+1=C 6kx 6﹣k (2x-)k =(﹣2)k C 6kx 6﹣2k , 由6﹣2k =0得k =3,即常数项为T 4=(﹣2)3C 36=-160, 故选:B . 点评:本题主要考查二项展开式的应用,求出n 的值,结合展开式的通项公式是解决本题的关键.属于中档题.6.在△ABC 中,已知sin 2sin cos C A B =,则△ABC 一定是() A .等腰直角三角形 B .等腰三角形 C .直角三角形D .等边三角形答案:B根据三角形内角和定理以及诱导公式,将sin 2sin cos C A B =化为sin()2sin cos A B A B +=,再根据两角和的正弦公式和两角差的正弦公式的逆用公式化为in 0()s A B -=,最后根据,A B 的范围,可得A B =.解:在△ABC 中,因为sin 2sin cos C A B =, 所以sin[()]2sin cos A B A B π-+=, 所以sin()2sin cos A B A B +=所以sin cos cos sin 2sin cos A B A B A B +=, 所以sin cos cos sin 0A B A B -=, 所以in 0()s A B -=, 所以,A B k k Z π-=∈, 因为0,0A B ππ<<<<, 所以0,k A B ==,所以△ABC 一定是等腰三角形. 故选:B 点评:本题考查了三角形的内角和定理,考查了诱导公式,考查了两角和与差的正弦公式,属于基础题.7.已知两个单位向量,a b →→的夹角为120°,若向量c →═2a b →→-,则a →•c →=() A .52B .32C .2D .3答案:A根据平面向量的数量积定义,计算即可. 解:由题意知|a →|=|b →|=1,且a →•b →=1×1×cos120°12=-,又向量c →═2a b →→-,所以a →•c →=22a a →→-•b →=2×1﹣(12-)52=.故选:A . 点评:本题考查了平面向量的数量积运算问题,是基础题.8.数学与建筑的结合造就建筑艺术品,2018年南非双曲线大教堂面世便惊艳世界,如图.若将此大教堂外形弧线的一段近似看成焦点在y 轴上的双曲线()222210>,>0-=y x a b a b 上支的一部分,且上焦点到上顶点的距离为2,到渐近线距离为22,则此双曲线的离心率为( )A .2B .3C .22D .3答案:B利用已知条件求出方程组,得到a ,c ,即可求解双曲线的离心率. 解:双曲线22221(0y x a b a b-=>,>0)的上焦点到上顶点的距离为2,到渐近线距离为22可得:22222222c a bca b c a b -=⎧=+=+⎩,解得a =1,c =3,b =2, 所以双曲线的离心率为:e ca==3. 故选:B . 点评:本题考查双曲线的简单性质的应用,双曲线的离心率的求法,是基本知识的考查,属于中档题.9.设函数f (x )210210x x x x -⎧+=⎨--⎩,>,<则下列结论错误的是()A .函数f (x )的值域为RB .函数f (|x|)为偶函数C .函数f (x )为奇函数D .函数f (x )是定义域上的单调函数答案:A根据题意,依次分析选项是否正确,综合即可得答案. 解:根据题意,依次分析选项:对于A ,函数f (x )210210x x x x -⎧+=⎨--⎩,>,<,当x >0时,f (x )=2x +1>2,当x <0时,f(x )=﹣2﹣x﹣1=﹣(2﹣x+1)<﹣2,其值域不是R ,A 错误;对于B ,函数f (|x|),其定义域为{x|x ≠0},有f (|﹣x|)=f (|x|),函数f (|x|)为偶函数,B 正确;对于C ,函数f (x )210210x x x x -⎧+=⎨--⎩,>,<,当x >0时,﹣x <0,有f (x )=2x +1,f (﹣x )=﹣f (x )=﹣2﹣x﹣1,反之当x <0时,﹣x >0,有f (x )=﹣2x﹣1,f (﹣x )=﹣f (x )=2x +1,综合可得:f (﹣x )=﹣f (x )成立,函数f (x )为奇函数,C 正确;对于D ,函数f (x )210210x x x x -⎧+=⎨--⎩,>,<,当x >0时,f (x )=2x+1>2,f (x )在(0,+∞)为增函数,当x <0时,f (x )=﹣2﹣x﹣1<﹣2,f (x )在(﹣∞,0)上为增函数,故f (x )是定义域上的单调函数; 故选:A . 点评:本题考查分段函数的性质,涉及函数的值域、奇偶性、单调性的分析,属于中档题. 10.已知函数f (x )=sin (ωx+φ)(ω>0,02πϕ<<)的最小正周期为π,且关于08π⎛⎫-⎪⎝⎭,中心对称,则下列结论正确的是() A .f (1)<f (0)<f (2) B .f (0)<f (2)<f (1) C .f (2)<f (0)<f (1) D .f (2)<f (1)<f (0)答案:D根据条件求出函数的解析式,结合函数的单调性的性质进行转化判断即可. 解:∵函数的最小周期是π, ∴2πω=π,得ω=2,则f (x )=sin (2x+φ), ∵f (x )关于08π⎛⎫-⎪⎝⎭,中心对称,∴2×(8π-)+φ=k π,k ∈Z , 即φ=k π4π+,k ∈Z ,∵02πϕ<<,∴当k =0时,φ4π=,即f (x )=sin (2x 4π+),则函数在[8π-,8π]上递增,在[8π,58π]上递减,f (0)=f (4π), ∵4π<1<2,∴f (4π)>f (1)>f (2), 即f (2)<f (1)<f (0), 故选:D . 点评:本题主要考查三角函数值的大小比较,根据条件求出函数的解析式,利用三角函数的单调性进行判断是解决本题的关键,属于中档题.11.已知x 为实数,[x]表示不超过x 的最大整数,若函数f (x )=x ﹣[x],则函数x xg x f x e=+()()的零点个数为() A .1 B .2C .3D .4答案:B函数x x g x f x e =+()()的零点个数,即方程xxf x e =-()的零点个数,也就是两函数y=f (x )与y x xe=-的图象的交点个数,画出图象,数形结合得答案. 解:函数x x g x f x e =+()()的零点个数,即方程xxf x e =-()的零点个数,也就是两函数y =f (x )与y x xe =-的交点个数.由y x x e =-,得y ′21x x xx e xe x e e--=-=. 可知当x <1时,y ′<0,函数单调递减,当x >1时,y ′>0,函数单调递增.作出两函数y =f (x )与y xxe =-的图象如图:由图可知,函数xxg x f x e =+()()的零点个数为2个. 故选:B . 点评:本题考查函数的零点与方程根的关系,考查数形结合的解题思想方法,训练了利用导数研究函数的单调性,是中档题.12.在△ABC 中,∠C =90°,AB =2,3AC =,D 为AC 上的一点(不含端点),将△BCD 沿直线BD 折起,使点C 在平面ABD 上的射影O 在线段AB 上,则线段OB 的取值范围是() A .(12,1) B .(12,32) C .(32,1) D .(0,32) 答案:A由题意,OC ⊥平面ABD ,根据三余弦定理,线线角的余弦值等于线面角的余弦值与射影角余弦值的积,从而求解. 解:由题意,OC ⊥平面ABD , 如图:设∠CBD =θ,∠CBO =θ1,则∠ABD =60°-θ;则cos θ=cos θ1×cos (60°﹣θ) 所以cos θ1()6013cos cos tan θθθ==︒-+∵θ∈(30°,60°); ∴OB =cos θ1∈(12,1). 故选:A .本题考查△ABC 的折叠和三余弦定理(最小角定理),要求熟悉余弦定理,属于中档题. 二、填空题13.已知cossin22αα-=,则sin α=_____. 答案:45将已知等式两边平方,利用同角三角函数基本关系式,二倍角的正弦函数公式即可求解. 解:∵225cossinαα-=, ∴两边平方可得:cos 22α+sin 22α-2cos 1sin 225αα=,可得1﹣sin α15=, ∴sin α45=. 故答案为:45.点评:本题主要考查了同角三角函数基本关系式,二倍角的正弦函数公式在三角函数化简求值中的应用,考查了转化思想,属于容易题.14.若曲线f (x )=e x cosx ﹣mx ,在点(0,f (0))处的切线的倾斜角为34π,则实数m =_____. 答案:2对函数求导,然后得f ′(0)314tan π==-,由此求出m 的值. 解:f ′(x )=e x(cosx ﹣sinx )﹣m .∴3'0114f m tan π=-==-(). ∴m =2. 故答案为:2 点评:本题考查导数的几何意义以及切线问题.抓住切点处的导数为切线斜率列方程是本题的基本思路.属于容易题.15.已知F 1,F 2是椭圆C :()222210x y a b a b+=>>的两个焦点,P 是椭圆C .上的一点,∠F 1PF 2=120°,且△F 1PF 2的面积为43,则b =_____. 答案:2根据正余弦定理可得PF 1•PF 2=16且4c 2=(2a )2﹣16,解出b 即可. 解:△F 1PF 2的面积12=PF 1•PF 2sin120°34=PF 1•PF 2=43,则PF 1•PF 2=16, 又根据余弦定理可得cos120°2221212122PF PF F F PF PF +-=⋅,即4c 2=PF 12+PF 22+16=(2a )2﹣32+16,所以4b 2=16,解得b =2, 故答案为:2. 点评:本题考查椭圆性质,考查正、余弦定理的应用,属于中档题.16.在一个半径为2的钢球内放置一个用来盛特殊液体的正四棱柱容器,要使该容器所盛液体尽可能多,则该容器的高应为_____. 答案:433设正四棱柱的高为h ,底面边长为a ,用h 表示出a ,写出正四棱柱容器的容积,利用导数求出V 取最大值时对应的h 值. 解:设正四棱柱的高为h ,底面边长为a ,如图所示;则h 2+2a 2=(2×2)2, 所以a 2=812-h 2,所以正四棱柱容器的容积为V =a 2h =(812-h 2)h 12=-h 3+8h ,h ∈(0,4);求导数得V ′32=-h 2+8,令V ′=0,解得h 3=,所以h ∈(0,3)时,V ′>0,V (h )单调递增;h ,4)时,V ′<0,V (h )单调递减;所以h =时,V 取得最大值.故答案为:3. 点评:本题考查了球内接正四棱柱的体积的最值问题,也考查了利用导数求函数的最值问题,是中档题. 三、解答题17.若数列{a n }的前n 项和为S n ,已知a 1=1,a n+123n S =. (1)求S n ; (2)设b n 1n s =,求证:b 1+b 2+b 3+…+b n 52<. 答案:(1)S n =(53)n ﹣1;(2)详见解析. (1)由数列的递推式:a n+1=S n+1﹣S n ,结合等比数列的定义和通项公式,可得所求; (2)求得b n 1n s ==(35)n ﹣1,由等比数列的求和公式和不等式的性质,即可得证. 解: (1)a n+123n S =,可得a n+1=S n+1﹣S n 23=S n , 由a 1=1,可得S 1=1,即S n+153=S n ,可得数列{S n }是首项为1,公比为53的等比数列, 则S n =(53)n ﹣1;(2)证明:b n 1n s ==(35)n ﹣1, 则b 1+b 2+b 3+…+b n 31()55532215n-==--•(35)n 52<.点评:本题考查数列的递推式和等比数列的通项公式和求和公式的运用,考查定义法和运算能力、推理能力,属于中档题.18.如图,已知点S 为正方形ABCD 所在平面外一点,△SBC 是边长为2的等边三角形,点E 为线段SB 的中点.(1)证明:SD//平面AEC ;(2)若侧面SBC ⊥底面ABCD ,求平面ACE 与平面SCD 所成锐二面角的余弦值. 答案:(1)详见解析;(215. (1)连接BD 交AC 于F ,连接EF ,由已知结合三角形的中位线定理可得EF ∥SD ,再由直线与平面平行的判定可得SD ∥平面AEC ;(2)取BC 的中点O ,连接OF 并延长,可知OF ⊥OC ,利用线面垂直的判定定理与性质定理可得:OS ⊥OF ,OS ⊥OC ,建立空间直角坐标系,分别求出平面CDS 与平面ACE 的一个法向量,由两法向量所成角的余弦值可得平面ACE 与平面SCD 所成锐二面角的余弦值. 解:(1)证明:连接BD 交AC 于F ,连接EF ,∵ABCD 为正方形,F 为BD 的中点,且E 为BS 的中点, ∴EF ∥SD .又SD ⊄平面AEC ,EF ⊂平面AEC , ∴SD ∥平面AEC ;(2)取BC 的中点O ,连接OF 并延长,可知OF ⊥OC ,在等边三角形SBC 中,可得SO ⊥BC ,∵侧面SBC ⊥底面ABCD ,且侧面SBC ∩底面ABCD =BC , ∴SO ⊥平面ABCD ,得OS ⊥OF ,OS ⊥OC .以O 为坐标原点,分别以OF ,OC ,OS 所在直线为x ,y ,z 轴建立空间直角坐标系,得:A (2,﹣1,0),C (0,1,0),E (0,12-3,D (2,1,0),S (0,03. ()200CD →=,,,(03CS →=-,,,()220AC →=-,,,1322AE →⎛=- ⎝⎭,,. 设平面CDS 与平面ACE 的一个法向量分别为()n x y z ,,=,()111m x y z =,,.由2030n CD x n CS y z ⎧⋅==⎪⎨⋅=-+=⎪⎩,取z =1,得()031n →=,,; 由1111122013202m AC x y m AE x y ⎧⋅=-+=⎪⎨⋅=-++=⎪⎩,取x 1=1,得(3m →=,,. ∴cos231525m nm n m n→→→→→→⋅===⋅<,>. ∴平面ACE 与平面SCD 15. 点评:本题考查直线与平面平行与垂直的判定、法向量与数量积的应用、空间角,考查空间想象能力与思维能力、计算能力,属中档题.19.2020年3月,各行各业开始复工复产,生活逐步恢复常态,某物流公司承担从甲地到乙地的蔬菜运输业务.已知该公司统计了往年同期200天内每天配送的蔬菜量X (40≤X <200,单位:件.注:蔬菜全部用统一规格的包装箱包装),并分组统计得到表格如表:若将频率视为概率,试解答如下问题:(1)该物流公司负责人决定随机抽出3天的数据来分析配送的蔬菜量的情况,求这3天配送的蔬菜量中至多有2天小于120件的概率;(2)该物流公司拟一次性租赁一批货车专门运营从甲地到乙地的蔬菜运输.已知一辆货车每天只能运营一趟,每辆货车每趟最多可装载40件,满载才发车,否则不发车.若发车,则每辆货车每趟可获利2000元;若未发车,则每辆货车每天平均亏损400元.为使该物流公司此项业务的营业利润最大,该物流公司应一次性租赁几辆货车? 答案:(1)485512;(2)3. (1)记事件A 为“在200天随机抽取1天,其蔬菜量小于120件”,则P (A )38=,由此能求出随机抽取的3天中配送的蔬菜量中至多有2天的蔬菜量小于120件的概率. (2)由题意得每天配送蔬菜量X 在[40,80),[80,120),[120,160),[160,200)的概率分别为11118428,,,,设物流公司每天的营业利润为Y ,若租赁1辆车,则Y 的值为2000元,若租赁2辆车,则Y 的可能取值为4000,1600,若租赁3辆车,则Y 的可能取值为6000,3600,1200,若租赁4辆车,则Y 的可能取值为8000,5600,3200,800,分别求出相应的数学期望,推导出为使该物流公司此项业务的营业利润最大,该物流公司应一次性租赁3辆货车. 解:(1)记事件A 为“在200天随机抽取1天,其蔬菜量小于120件”, 则P (A )38=, ∴随机抽取的3天中配送的蔬菜量中至多有2天的蔬菜量小于120件的概率为:p 22120333335355485()()()88888512C C C ⎛⎫⎛⎫=++= ⎪⎪⎝⎭⎝⎭. (2)由题意得每天配送蔬菜量X 在[40,80),[80,120),[120,160),[160,200)的概率分别为11118428,,,, 设物流公司每天的营业利润为Y , 若租赁1辆车,则Y 的值为2000元,若租赁2辆车,则Y的可能取值为4000,1600,P(Y=4000)78=,P(Y=1600)18=,∴Y的分布列为:∴E(Y)=4000160086⨯+⨯=3700元.若租赁3辆车,则Y的可能取值为6000,3600,1200,P(Y=6000)58 =,P(Y=3600)14 =,P(Y=1200)18 =,∴Y的分布列为:∴E(Y)600036001200848=⨯+⨯+⨯=4800元,若租赁4辆车,则Y的可能取值为8000,5600,3200,800,P(Y=8000)18 =,P(Y=5600)12 =,P(Y=3200)14 =,P(Y=800)18 =,∴Y的分布列为:∴E(Y)8000560032008008248=⨯+⨯+⨯+⨯=4700,∵4800>4700>3700>2000,∴为使该物流公司此项业务的营业利润最大,该物流公司应一次性租赁3辆货车. 点评:本题考查概率、离散型随机变量的分布列、数学期望的求法,考查频数分布表、古典概型等基础知识,考查运算求解能力,是中档题. 20.已知函数f (x )=ax ﹣(a+2)lnx 2x-+2,其中a ∈R . (1)当a =4时,求函数f (x )的极值;(2)试讨论函数f (x )在(1,e )上的零点个数.答案:(1)极大值6ln2,极小值4;(2)分类讨论,详见解析.(1)把a =4代入后对函数求导,然后结合导数可求函数的单调性,进而可求极值; (2)先对函数求导,然后结合导数与单调性关系对a 进行分类讨论,确定导数符号,然后结合导数与函数的性质可求. 解:(1)当a =4时,f (x )=4x ﹣6lnx 2x -+2,()()22221162'4x x f x x x x--=-+=(),x >0,易得f (x )在(0,12),(1,+∞)上单调递增,在(112,)上单调递减, 故当x 12=时,函数取得极大值f (12)=6ln2,当x =1时,函数取得极小值f (1)=4,(2)()()222122'ax x a f x a x x x--+=-+=(), 当a ≤0时,f (x )在(1,e )上单调递减,f (x )<f (1)=a ≤0,此时函数在(1,e )上没有零点;当a ≥2时,f (x )在(1,e )上单调递增,f (x )>f (1)=a ≥2,此时函数在(1,e )上没有零点; 当02a e≤<即2e a ≥时,f (x )在(1,e )上单调递减,由题意可得,1020f a f e ae a e =⎧⎪⎨=--⎪⎩()>()<, 解可得,0()21a e e -<<,当22a e <<即21e a<<时,f (x )在(1,2a )上单调递减,在(2e a ,)上单调递增, 由于f (1)=a >0,f (e )=a (e ﹣1)()2224120e e e e e ---=->>,令g (a )=f (2a )=2﹣(a+2)ln 2a-a+2=(a+2)lna ﹣(1+ln2)a+4﹣2ln2,令h (a )2'2g a lna ln a ==+-(),则22'a h a a-=()<0, 所以h (a )在(22e ,)上递减,h (a )>h (2)=1>0,即g ′(a )>0, 所以g (a )在(22e ,)上递增,g (a )>g (2e )=240e->, 即f (2a)>0,所以f (x )在(1,e )上没有零点, 综上,当0<a ()21e e -<时,f (x )在(1,e )上有唯一零点,当a ≤0或a ()21e e ≥-时,f (x )在(1,e )上没有零点.点评:本题综合考查了导数与函数性质的应用,体现了转化思想与分类讨论思想的应用,属于难题.21.已知动直线l 过抛物线C :y 2=4x 的焦点F ,且与抛物线C 交于M ,N 两点,且点M 在x 轴上方.(1)若线段MN 的垂直平分线交x 轴于点Q ,若|FQ|=8,求直线l 的斜率; (2)设点P (x 0,0),若点M 恒在以FP 为直径的圆外,求x 0的取值范围.答案:(1)3±;(2)x 0∈[0,1)∪(1,9). (1)由题意可得直线l 的斜率存在且不为0,设l 的方程与抛物线联立,求出两根之和及两根之积,进而可得MN 的中点坐标,进而可得MN 的中垂线方程,令y =0可得Q 的坐标,进而求出|QF|的值,由题意可得直线l 的斜率;(2)由题意可得∠FMP 为锐角,等价于MF MP →→⋅>0,求出MF MP →→⋅的表达式,换元等价于h (t )=t 2+(3﹣x 0)4+x 0,t >0恒成立,分两种情况求出x 0取值范围. 解:(1)由题意可得直线l 的斜率存在且不为0,设直线l 的方程为:x =ty+1,设M (x 1,y 1),N (x 2,y 2),线段MN 的最大E (x 0,y 0),联立直线与抛物线的方程可得:214x ty y x =+⎧⎨=⎩,整理可得y 2﹣4ty ﹣4=0, 所以y 1+y 2=4t ,y 1y 2=﹣4,所以y 0=2t ,x 0=ty 0+1=2t 2+1,即E (2t 2+1,2t ), 故线段MN 的中垂线方程为:y ﹣2t =﹣t (x ﹣2t 2﹣1), 令y =0,则Q (2t 2+3,0), 所以|FQ|=|22+3﹣1|=8, 解得t =,所以直线l 的斜率k 1t ==; (2)点M 恒在以FP 为直径的圆外,则∠FMP 为锐角,等价于MF MP →→⋅>0,设M (214y ,y 1),F (1,0),P (x 0,0),则MP →=(x 0214y -,﹣y 1),MF →=(1214y -,﹣y 1),故MF MP →→⋅=(x 0214y -)(1214y -)+y 1242113164y y =++(1214y -)x 0>0恒成立, 令t 214y =,t >0,原式等价于t 2+3t+(1﹣t )x 0>0对任意t >0恒成立,即t 2+(3﹣x 0)4+x 0>0对任意t >0恒成立, 令h (t )=t 2+(3﹣x 0)4+x 0,t >0, ①△=(3﹣x 0)2﹣4x 0<0,即1<x 0<9,②0030200x h ∆≥⎧⎪-⎪≤⎨⎪≥⎪⎩(),解得0≤x 0≤1,又因为x 0≠1,故x 0∈[0,1), 综上所述x 0∈[0,1)∪(1,9). 点评:本题考查抛物线的性质及直线与抛物线的综合及点在圆外的性质,属于中难题. 22.如图,在极坐标系中,曲线C 1是以C 1(4,0)为圆心的半圆,曲线C 2是以22C π⎫⎪⎭,为圆心的圆,曲线C 1、C 2都过极点O .(1)分别写出半圆C 1,C 2的极坐标方程; (2)直线l :()3R πθρ=∈与曲线C 1,C 2分别交于M 、N 两点(异于极点O ),P 为C 2上的动点,求△PMN 面积的最大值. 答案:(1)1:C 802cos πρθθ⎛⎫=≤≤⎪⎝⎭;2:C ()230sin ρθθπ=≤≤;(2)334. (1)直接利用转换关系的应用,把参数方程极坐标方程和直角坐标方程之间进行转换. (2)利用三角函数关系式的变换和三角形的面积的公式的应用求出结果. 解:(1)曲线C 1是以C 1(4,0)为圆心的半圆, 所以半圆的极坐标方程为802cos πρθθ⎛⎫=≤≤⎪⎝⎭, 曲线C 2是以232C π⎛⎫⎪⎝⎭,为圆心的圆,转换为极坐标方程为()230sin ρθθπ=≤≤.(2)由(1)得:|MN|=|823|133M N cossinππρρ-=-=.显然当点P 到直线MN 的距离最大时,△PMN 的面积最大. 此时点P 为过C 2且与直线MN 垂直的直线与C 2的一个交点, 设PC 2与直线MN 垂直于点H , 如图所示:在Rt △OHC 2中,|223|6HC OC sinπ==所以点P 到直线MN 的最大距离d 22333||3C HC r =+==, 所以113333122PMNSMN d =⨯⋅=⨯=点评:本题考查的知识要点:参数方程极坐标方程和直角坐标方程之间的转换,三角函数关系式的恒等变换,三角形的面积公式的应用,主要考查学生的运算能力和转换能力及思维能力,属于中档题.23.已知函数f (x )=|x ﹣2|+|x+1|. (1)解关于x 的不等式f (x )≤5;(2)若函数f (x )的最小值记为m ,设a ,b ,c 均为正实数,且a+4b+9c =m ,求11149a b c++的最小值.答案:(1){x|﹣2≤x ≤3};(2)3.(1)将f (x )写为分段函数的形式,然后根据f (x )≤5,利用零点分段法解不等式即可;(2)利用绝对值三角不等式求出f (x )的最小值m ,然后由a+4b+9c =m ,根据111111149349a b c a b c ⎛⎫++=++ ⎪⎝⎭(a+4b+9c ),利用基本不等式求出11149a b c++的最小值. 解:(1)f (x )=|x ﹣2|+|x+1|212312211x x x x x -⎧⎪=-≤≤⎨⎪-+-⎩,>,,<. ∵f (x )≤5, ∴2152x x -≤⎧⎨⎩>或﹣1≤x ≤2或2151x x -+≤⎧⎨-⎩<,∴﹣2≤x ≤3,∴不等式的解集为{x|﹣2≤x ≤3}.(2)∵f (x )=|x ﹣2|+|x+1|⩾|(x ﹣2)﹣(x+1)|=1 ∴f (x )的最小值为1,即m =3, ∴a+4b+9c =3.()11111114949349a b c a b c a b c ⎛⎫++=++++ ⎪⎝⎭ 14499334949b a b c c a a b c b a c ⎛⎫=++++++ ⎪⎝⎭21 1323⎛+= ⎝3, 当且仅当1493a b c ===时等号成立, ∴11149a b c++最小值为3. 点评:本题考查了绝对值不等式的解法,绝对值三角不等式和利用基本不等式求最值,考查了分类讨论思想和转化思想,属中档题.。
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四川省绵阳市2020届高三英语第三次诊断性测试试题第一部分听力(共两节,满分30分)回答听力部分时,先将答案标在试卷上。
听力部分结束前,你将有两分钟的时间将你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腰,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腰井阅读下一小题。
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:C.1.What will the woman do tonight?A.Eat out.B.Stay at home.C.See a movie.2.How will the man go to the Central Park?A.By bus.B.By underground.C.On foot.3.Who is David?A.The man's cousin.B.The man's brother.C.The man's uncle.4.What is Andy doing?A.Doing homework.B.Tidying his bedroom.C.Washing dishes.5.What is the weather like now?A.Windy.B.Sunny.C.Rainy.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、7题。
6.What is the relationship between the speakers?A.Classmates.B.Teacher and student.C.Boss and secretary.7.What is the woman going to write about?A.An imaginary person.B.A famous model.C.Her grandfather.听第7段材料,回答第8至9题。
8.What does the woman want Tom to do?A.Fix her computer.B.Go to the library.C.Hang out with friends.9.What is the woman's phone number?A.678-3566.B.678-3356.C.678-3556.听第8段材料,回答第10至12题。
10.Why is the woman nervous?A.She is new in her school.B.She is not ready for tomorrow's lessons.C.She is not good at making friends.11.When does morning reading begin?A.At8:00.B.At 7:40.C.At 7:30.12.What can students do if hungry?A.Eat something after the second class.B.Eat something in class.C.Ask teachers for some cookies.听第9段材料,回答第13至16题。
13.Where does the conversation probably take place?A.At home.B.In a store.C.In a restaurant.14.Why does the woman want to exchange the dress?A.It is expensive.B.It doesn't fit her.C.Its color doesn't suit her.15.What is on sale today?A.Shoes.B.Skirts.C.Blouses.16.What does the woman decide to do?A.Have her money back.B.Wait for the sale next weekend.C.Take a pair of shoes.听第10段材料,回答第17至20题。
17.What is the aim of the speaker's website?A.To help people with their spoken English.B.To make the website more popular.C.To advise people to learn English18.What can be learned everyday?A.Important grammar.B.Different accents of English pronunciation.C.New English expressions.19.What is the feature of the online lessons?A.Long but easy.B.Short but difficult.C.Short but interesting.20.What does the speaker think is the key to learning English?A.Learn a little and remember clearly daily.B.Learn as much new vocabulary as possible.C.Talk to others as much as possible.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和D)中,选出最佳选项。
井在答题卡上将该选项涂黑。
AThe recent few weeks have been the perfect time for a good book and here are some good ones that allow readers to get completely lost in another world.Watch Me DisappearIf you like mysteries with family drama,Janelle Brown's best-selling thriller is for you,The story follows a wealthy wife and mom who goes on a hike and never returns.Her lonely husband and teenage daughter,Olive(who's dealing with her own problems),are confused with sorrow while trying to find out what happens.The Boy from the WoodsIf you pick up one of Harlan Coben's books,you won't be able to put itdown.His misty stories keep readers fascinated.The Boy from the Woods,is sure to satisfy fans of his twisty,heart-racing plots filled with interesting and exciting atmosphere. This is the book to read if you love masterful surprises.How to Walk AwayKatherine Center's best-selling novel follows a young woman,Margaret,with her perfect fiance,a pilot who takes her on a upsetting flight that changes her life forever.Readers adore Margaret's wisdom and humor as she faces a changed world and figures out what love really looks like.The Last One LeftThis novel,first published in 1967,still attracts readers decadeslater.Novelist Dean Koontz wrote the introduction to the recent edition,where he describes reading each of John D.MacDonald's novels“at least three times,some of them twice that often."This is your next read if you want to be swept away by a master of mystery and excitement.21.What is Watch Me Disappear probably about?A.A family outing.B.Teenager problems.C.Sorrow at a dead mom.D.Misty missing of a woman.22.Which book is for romance lovers?A.How to Walk Away.B.The Boy from the Woods.C.Watch Me Disappear.D.The Last One Left.23.Who is the author of The Last One Left?A.Dean Koontz.B.John D.MacDonald.C.Katherine Center.D.Harlan Coben.BThe past few years found me working long hours at my marketing job while my husband,Steven,put in equally fall days as a physics tutor.The weekends were spent photographing weddings and portraits.While we were able to save some money,we were both burned out from the constant tiring work.We felt the need for both renewal and vacation experiences.So,in January,we set out on a seven-month,cross-country road trip to visit all 61 National Parks in the United States.Steven was ready to dive headlong into the unknown,while I,someone who enjoys more planned fun,had to adapt.I was amazed at how easily I turned to be a more relaxed version who wasn't worried where to park or sleep.We both became so accustomed to sleeping in the van,despite the foreign sound outside,which we eventually let fade into white noise.Over the course of our seven-month trip,several interruptions changed our route, including the government shutdown,which left the first 10 parks inaccessible. Another time,I fell ill and lost my voice right as we were heading to the Virgin Islands National Park.Positively,while we've done photography professionally for years.this trip really helped us grow in our craft.From rapidly trying to shoot dolphins to lining up panoramic(全景的))shots,this trip was a photographer's dream.I never daredout into complete darkness to photograph nights capes,but the views of the Milky Way we managed to shoot were well worth it.One of our big goals on this trip was to come away from each park with one spectacular image,something that pushed us both to become better artists in the process.That shared goal placed us on the same team:It brought us together,gave us more to talk about,and encouraged us to motivate each other along the way.24.Why did the couple take such a trip?A.They were both out of work.B.They needed some photos of parks.C.They wanted a getaway from exhausting work.D.They wanted to make money by photographing.25.What surprised the author on the trip?A.The trip was full of unexpected difficulties.B.The trip was nothing like her planned fun.C.She enjoyed his husband's way of travel.D.She had several fights with her husband.26.What failed them to visit the first 10 parks?A.Bad sleep in the van.B.The author's illness.C.The change of route.D.The government shutdown.27.What have the couple gained from the trip?A.They've caught some dolphins.B.They've became famous artists.C.They've bettered their relationship.D.They've got their professional photographer certificates.CThe slogan for the Cultural Heritage and Sustainable Development Fund is"In Love With China"-and this special message has bcen shown in its own logo.The logo uses the characters for"wind"and"phoenix"(风凰)in ancient oracle bone script(甲骨文)on a circular Chinese fan.Madam Kang Jiaqi,the executive director of CHSDF, IN LOVE WITH CHINA explains the many layers of meaning behind the logo:“The logo adopts the writing style of oracle bone script-dating back over 3,000 years-carved onto either turtle shell or ox bones."The Chinese character for*wind' and the character for"phoenix' are almost the same,with just one stroke difference.In China's ancient past,the character for"phoenix' was widely considered to bring peace and happiness and represents good fortune.The character"wind' has a long history and is also known as a name representing culture and spiritual power.""Both the characters"wind' and 'phoenix 'share the same component in their characters,which is also used in the slogan 'In Love With China'.This extraordinary combination represents how experts lead fashion,which in tum leads culture-and that culture is the basis of creativity.Therefore,it represents the importance of traditional Chinese culture in global creativity."Taking the Chinese oracle bone character of"wind 'and' phoenix' as our logo represents the beginning of Chinese civilization,which suggests that it can pass on Chinese culture through the cooperation between the CHSDF and the cooperative platform of global designers,"explains Jiaqi.Logo artist Sam Chung similarly explains her design process,"The decision of the oracle bone script of' wind' comes from the root of the characteritself.Originated from the character shape of 'phoenix,''wind's 'oracle bonescript still presents wonder, despite its more circular shape.The three tassels(流苏)at the end of the moon-shaped fan further give a feeling of lightness while hinting at both Eastern and Western flavors."28.What do we know about the logo?A.It's carved on turtle shells.B.It's drawn on circular Chinese fans.C.It's designed in a new writing style.D.It's a combination of two similar characters.29.Why are' wind' and 'phoenix' chosen?A.They are easily written.B.They symbolize good fortune.C.They represent Chinese cultural and spiritual power.D.They promote global creativity based on Chinese culture.30.What can be inferred from the last paragraph?A.Sam Chung is the actual designer of the logo.B.The idea for the logo comes from abroad.C.The director has little say in the decision.D.Tassels are symbol of western culture.31.What's the passage mainly about?A.Chinese influence on the world.B.The making of the logo for CHSDF.C.Chinese culture and civilization.D.Chinese characters 'wind' and' phoenix'.DAs data leak and identity theft become more and more common,the market is growing for fingerprint or iris scans(虹工膜扫描)-to keep others out of private e-spaces.They're still expensive,though,and some people are unwilling to have personal identifiers taken and kept by a third party.Researchers say they have come up with a low-cost device-a smart keyboard. It precisely measures the sound with which one types and the pressure fingers apply to each key.These patterns are unique to each individual,says Jun Chen,a doctoral engineering student.By measuring how somebody types a password(码),he says, the keyboard can determine people's identities,and thus,by extension,whether they should be granted access to the computer it's connected to-regardless of whether someone gets the password right.It doesn't require a new type of technology that people aren't already familiar with."Everybody uses a keyboard...and everybody types differently,"Chen says.The device powers itself by generating electricity when a person's fingertips touch the keys-multi-layer plastic materials,press down,and lift again,which completes an electric circuit with the keyboard.The keyboard could offer a stronger layer of security by analyzing things like the force of a user's typing and the time between key presses.This phenomenon,called"contact electrification,"is the same process that creates static electricity,Chen says:"lt's like when you run your hand across a wool blanket and see sparks(电火花)in the darkness."In a study describing the technology published in the journal ACS Nano,the researchers had 100 volunteers type the word touch four times using the keyboard. Data gleaned from the device could be used to identify individual participants based on how they typed,with very low error rates,Chen says.So far,there is just one such keyboard.But,Chen says,it should be pretty straightforward to commercialize and is mostly made of inexpensive,plastic-like parts. The team hopes it could make it to market in about five years.32.What is the feature of the smart device?A.It's inexpensive.B.It's a new type of technology.C.It's inaccessible without a password.D.It can recognize people's fingerprint.33.How is the smart device powered?A.By fingers touching it.B.By replaceable batteries.C.By producing power of its own.D.By being connected to electricity.34.What does the underlined word"glean"probably mean in Paragraph 6?A.Collect.B.Involve.C.Contain.D.Leak.35.Which can be the best title for this passage?A.Smart Keyboard Can Produce Electricity ItselfB.Smart Keyboard Can identify You by How You TypeC.Smart Keyboard Can Protect Personal Data PrivacyD.Smart Keyboard Can Measure Typing Forces and Time第二节(共5小:每小圈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。