2019-2020学年宁夏青铜峡市高级中学(吴忠中学青铜峡分校)高一下学期期中考试英语试题
青铜峡市高级中学吴忠中学青铜峡分校高一数学下学期期中试题含解析
【详解】 故选:B
【点睛】本题考查了平面向量基本定理的应用,考查了平面向量加法的几何意义,考查了共线向量的性质,属于基础题。
8。已知 ,则 的值为( )
A. B. C. D.
【答案】B
【解析】
【分析】
直接利用余弦的二倍角公式可得结果.
宁夏青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020学年高一数学下学期期中试题(含解析)
一、选择题:(共12小题,每小题5分,共60分)
1.数列1, , , , ,…的一个通项公式 是( )
A. B。 C. D.
【答案】D
【解析】
【分析】
通过观察数列 分子和分母,猜想出数列的通项公式.
【详解】由于数列的分母是奇数列,分子 是自然数列,故通项公式为 。故选D.
【点睛】本小题考查观察数列给定的项,猜想数列的通项公式.根据分子和分母的规律,易得出正确的选项。属于基础题.
2。 的值是
A。 B。 C. D。
【答案】C
【解析】
分析:首先观察题中所给的式子,结合正弦的倍角公式,可以确定其为 的值,借助于特殊角的三角函数值求得结果。
详解:根据正弦的倍角公式可得 ,
故选C.
21.已知等差数列 的公差 ,且 。
(1)求 及 ;
(2)若等比数列 满足 , ,求数列 的前n项和 .
【答案】(1) , ;(2) 。
【解析】
【分析】
(1)直接利用等差数列公式计算得到答案.
(2)计算 ,再利用分组求和法计算得到答案。
【详解】(1)由 ,得 ,又 ,∴ ,∴ ;
(2)由题意 , ,即 ,∴ ,于是 ,
宁夏青铜峡市高中(吴忠中学青铜峡分校)2019-2020学年高一下学期期中考试化学试题 Word版含解析
2019-2020学年第二学期高一年级化学期中试卷可能用到的相对原子质量:H-1 C-12 N-14 O-16 Na-23 S-32 Cl-35.5 Fe-56一、选择题1.我国著名的化学家、教育家徐光宪先胜因在稀土金属等研究领域做出杰出贡献,荣获了2008年度“国家最高科学技术奖”。
14058Ce是地壳中含量最高的稀土金属饰元素。
下列关于14058Ce的说法错误的是A. 质量数为140B. 中子数为82C. 质子数为58D. 核外电子数为198【答案】D【解析】【分析】14058Ce中质子数为58,质量数为140,根据质量数A=质子数Z+中子数N,可求出中子数,根据核外电子数=质子数=原子序数=核电荷数可以知道核外电子数。
【详解】A、14058Ce中Ce的左上角代表质量数,因此14058Ce的质量数为140,所以A选项是正确的;B、14058Ce中Ce的左上角代表质量数为140,左下角为质子数为58,根据质量数A=质子数Z+中子数N可以知道:中子数N=140-58=82,所以B选项是正确的;C、14058Ce中Ce的左下角为质子数,因此14058Ce的质子数为58,所以C选项是正确的;D、根据核外电子数=质子数可以知道,核外电子数为58,故D错误;综上所述,本题正确选项D。
2.元素的下列性质,随原子序数的递增不是周期性变化的是( )A. 相对原子质量B. 化合价C. 原子半径D. 元素的化学性质【答案】A【解析】【分析】【详解】随着原子序数的递增,核外电子排布、元素的化合价、原子半径、金属性和非金属性呈现周期性变化,但由于相对原子质量的决定因素是原子中质子数、中子数及同一元素的同位素原子在自然界中的物质的量分数,所以相对原子质量不呈周期性变化,A选项符合题意;3.下列元素原子半径最大的是A. LiB. FC. NaD. Cl【答案】C【解析】【详解】同周期自左向右原子半径逐渐减小,同主族自上而下原子半径逐渐增大,则原子半径大小顺序是Na>Cl、Li>F、Na>Li、Cl>F,因此原子半径最大的是Na,答案选C。
宁夏青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020学年高一下学期空中课堂在线考试英语试题
高级中学2019---2020学年(二)网课阶段测试卷试卷满分100分,测试时间60分钟。
出卷教师:2020-3-11一:阅读理解(每小题2分,共计40分)[A]The World Around UsChannel : C8A study of the ancient Egyptian pyramids and the area around the River Nile in Egypt. The filming of this programme is a work of art as it is so thoughtfully done. As well as the obvious camels, there are also many other desert animals and plant life.Summer HolidaysChannel: M6Some of the best summer holidays are on offer this year. Tonight’s progr amme shows a weekend in Disneyland in Paris, cheap sailing holidays in the Mediterranean and a shopping and sightseeing trip to New York.The Creative MindChannel: C4One of the most popular programmes on TV at the moment, The Creative Mind has different artistic themes from exhibition reviews, information about major and smaller museums, and interviews with artists, writers, actors and musicians.Prepared for Special MomentsChannel: TF1The fun cookery programme that offers lots of exciting ideas from ch ildren’s birthday parties to that candlelit dinner for the boss and his wife. Easy to follow step by step instructions that show you how to make your dinner party a little bit special.Born to RunChannel: W9An interesting story of a young man with learning difficulties who overcame (克服) the problems in his life, through his great talent (天赋) for athletics. This is a true story of how one person made the most of his life and also helped many other people with similar problems.1. Which channel should you turn to if you are very interested in wildlife?A. C4.B. M6.C. C8.D. W9.2. Which programme is suitable for those who like watching TV interviews?A. The World Around Us.B. Summer Holidays.C. The Creative Mind.D. Prepared for Special Moments.3. Which of the following words can best describe the hero of Born to Run?A. Creative.B. Dishonest.C. Self-centered.D. Strong-minded.[B]Actually,we never live well well without music. About 50 years ago,Bob Dylan played at the NewPort Fork Festival and was widely booed(一片嘘声). The audience may have been unhappy,but Dylan’s performance helped change the direction of music and culture in the US.The mid-1960s were a time of great change. One such place of change was the world of folk music. Bob Dylan became a symbol of change when he moved from acoustic(原声的) to electric guitar.Rock music historian Elijah Wald has written the new book, called Dylan Goes Electric about the challenge.“There was a moment in the early 1960s when you could look at the Billboard charts,and 7 of the top 10 albums were folk records. And Joan Baez, Peter Paul and Mary,the Kingston Trio all had huge,huge,huge number one records.”And then this happened. T he “British I nvasion” introduced the world to the Bea tles and grew a huge fan based for rock music. That worried many folk musicians, says Elijah Wald. He hoped that Dylan would come to the rescue,however,a new Dylan sound came from the stage. Dylan,had gone electric, and the followers of folk music were not pleased.“When Dylan went electric,I think one of the issues was the feeling that-wait a minute,he’s g one over to the enemy.”Folk lovers had looked to Dylan to save their movement for rock and roll,but Wald says Dylan felt differently about the music.Dylan had always liked rock and roll,and Dylan didn't think of rock and roll as stupid Music. In fact,Dylan was Beatles fan. He later said that from the first time he heard the Beatles,he knew “they were pointing to the direction w here music had to go.”Beyond the music, Dylan’s performance that night also marked a turn in American culture.4.How did the audience feel when Dylan played at the NewPort Fork Festival?A: excited B: unsatisfied C: respectful D: bored5. Which is True about American music world in the early 1960s,according to Elijah Wald? A:Great changes have taken place in folk music.B: People had grown sincere love for rock music.C: Folk music was once a huge mainstream pop trend .D:Dylan was the best American musicians at that time.6.What did folk musicians and followers expect Dylan to do?A: Change folk music for better. B: Recover the glory of folk music.C: Introduce the world to the Beatles. D: Play rock and roll against the Beatles.7.What will the author most probably talk about next?A: Dylan’s deep love for the Beatles,rock and roll music.B: Dylan’s beliefs in the new direction music should go.C: Dylan’s music as a great contribution to American folk music.D: Dylan’s performance as a mark of a change in American culture.[C]What do you usually do when a flying insect lands on your arm? Most likely you blow it away. But what if that bug were a ladybug (瓢虫)? Then chances are you would let it stay. What is it about ladybugs that we like?Farmers once thought ladybugs were a good luck sign. A ladybug in the field meant that the crops would be successful and the weather would be good. The farmers may have been exaggerating (夸大) about what a very small bug can do, but in fact ladybugs do help out with the crops. They eat harmful insects.Actually, it is not the adult ladybug that eats the other insects. It’s the ladybug larvae (幼虫). Ladybugs lay their eggs on leaves that are covered with aphids or other insects. When the larvae come out of the eggs, they are very hungry and will eat mites, aphids, mealybugs, green flies, and other crop-destroyers.A single ladybug larva can eat over 1,000 aphids in one day. In the 1800s, ladybugs were brought all the way from Australia to California and set free among orange trees that were dying from being eaten by insects. The little red and black “eating machines” completely saved all the trees.Ladybugs go about their daily business without worrying much about anything. Very few birds or insects will eat a ladybug because it gives off something that is dangerous to eat. Ladybugs also protect themselves by playing dead. They fall off leaves and remain very still until their enemy goes away.So next time you see a ladybug, think about what it may be up to. Maybe it is off to save an orange tree. Maybe its next move will be to fall off a leaf and play dead or stand up to a bird about 100 times its size. Maybe it’s letting you know that the weather will be good tomorrow. Who knows? Ladybugs are special that way.8. What do we know about mites and aphids?A. They do harm to the crops.B. They eat more than they should.C. They are food for adult ladybugs.D. They are the enemy of ladybug larvae.9. Why were ladybugs brought from Australia to California in the 1800s?A. To report the weather for farmers.B. To increase their number.C. To do research on them.D. To save orange trees.10. What does the underlined part “stand up to” in the last paragraph probably mean?A. Play with.B. Worry about.C. Get along with.D. Fight back against.11. What does the author think of the ladybug? A. It is a crop-destroyer.B. It is a good luck sign.C. It is a little flying helper.D. It is a big eating machine.[D]As you research music,you will find music that is familiar to you. You will find music,which tells of interesting places and exciting things to do. You will find music,which expresses feelings that are often your own.Music is an expression of the people. As you research,you will find music of people at work and play. You will find music expressing love of the country,love of nature and love of home.Music is also an expression of the comp osers. The composer expresses his own musical ideas. he studies the materials of music and discovers ways of using them. He looks for new kinds of musical expression.Music can suggest actions and feelings which we all share. we can enjoy playing and singing,music,dancing and listening to the music of the people and the composers of different times and places。
宁夏青铜峡市高级中学吴忠中学青铜峡分校2019-2020学年高一数学下学期期中试题【含答案】
2.
的值是
A.
B.
C.
D.
3.sin70。cos20。+cos70。sin 20。( )
A.0 B.1
C.-1
D.
4.
中,若
,则
A.锐角三角形 B.直角三角形 C.等腰三角形
的形状为 D.等边三角形
5.若三点
、
、
共线,则 的值为( )
A.
B.
C.
D.
6.若 | a | 2,| b | 2,且(a - b)⊥ a,则a与b的夹角是
宁夏青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020 学年高一数学 下学期期中试题
1、选择题:(共 12 小题,每小题 5 分,共 60 分)
2345
1.数列1, 3
, 5
, 7
, 9
,...的一个通项公式an是()
A. n 2n +1
n B.
2n + 3
n C.
2n -1
D. n 2n - 3
A.
B.
C.
D.
7.在ABC中,BD 1 DC,则AD
2
2 AB 1 AC
A. 3
3
1 AB 3 AC
B. 4
4
1 AB 2 AC
C. 3
3
D.
2 AB - 1 AC 33
8.已知sin 1 ,则cos2的值为() 8
A.
B.
C.
D.
9.已知cosα = - 3(α ∈(π ,π)),则tan(α + π ) = ( )
(1)
,且
,
,-------2 分
于是 (2)
宁夏青铜峡市高级中学吴忠中学青铜峡分校2019-2020学年高一英语下学期期中试题【含答案】
宁夏青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020学年高一英语下学期期中试题第Ⅰ卷听力(30分)第一节(共5小题;每小题1.5分,满分7.5分)请听下面5段对话,选出最佳选项。
第一部分听力(共两节,满分30分)请听下面5段对话,选出最佳选项。
I. How many classes does the woman have on Wednesday?A. 2.B. 4.C. 5.2. Where did Jane grow up?A. In Sweden .B. In Switzerland.C. In the US.3. What time is the man’s flight?A. At 9:00 a.m.B. At 9:50 a.m.C. At 10:40 a.m.4.What is the woman hunting for?A. A good movie .B. A nice magazine 7C. A fortune list5. What did the woman do last night?A. She attended a lecture.B. She went to the hospital .C. She made a speech on education.第二节(共15小题;每小题1.5分,满分22.5分)请听下面5段对话或独白,选出最佳选项。
请听第6段材料,回答第6至8题6. What is the probable relationship between the two speakers?A. Sister and brother.B. Teacher and student.C. Husband and wife.7.How many children do they have?A. 3B. 4C.58. How does the man feel?A. TiredB. FreeC. Bad请听第7段材料,回答第9至12题。
宁夏吴忠市青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020高一下学期期中试题(含解析)
A. 三大改造的完成
B. 人民公社的建立
C. 一五计划的实施
D. 土地改革的进行
【答案】A
【解析】
【详解】1956年底三大改造的完成,私营企业被改造成社会主义公有制的一部分,直接导致私营企业“集体消亡”,故A正确;1958年人民公社建立时,私营企业早已消亡,故B错误;一五计划包括工业化和三大改造,不够准确,故C错误;土地改革进行的时间是1950年,时间与题目不符,故D错误.
17。邓小平曾说:后来苏联的模式僵化了。计划经济部等于社会主义,社会主义也有市场,计划和市场都是经济手段。为此,中共十四大明确提出我国经济体制改革的目标是
A。 全面推进社会主义新农村建设B. 建立现代企业制度
C。 开拓对外开放新格局D。 建立社会主义市场经济体制
【答案】D
【解析】
【详解】结合所学可知,中共十四大明确要坚持党的基本路线不动摇,关键是坚持以经济建设为中心不动摇;要抓住机遇,加快改革开放和现代化建设的步伐;提出我国经济体制改革的目标是建立社会主义市场经济体制,故D项正确;全面推进社会主义新农村建设提出于党的十六届五中全会,排除A项;党的十四届三中全会明确提出建立现代企业制度,排除B项;十四大未涉及对开开放格局的扩大,排除C项。故选D
13.中共七届二中全会决议指出,民主革命胜利以后,党的基本任务之一是“使中国稳步地由农业国转变为工业国".新中国的工业化建设始于
A。 土地改革的推进B。 三大改造的完成
C.C
【解析】
土地改革的推进是要完成民主革命的任务,排除A;1956年底,三大改造完成标志着我国进入社会主义初级阶段,排除B;1957年,一五计划超额完成,实现了国民经济的快速增长,奠定了我国工业化的基础,故选C;中共八大是要探索社会主义建设道路,排除D.
2019-2020学年宁夏青铜峡市高级中学高一下学期期中考试地理试题
★祝考试顺利★注意事项:1、答题前,请先将自己的姓名、准考证号用0.5毫米黑色签字笔填写在试题卷和答题卡上的相应位置,并将准考证号条形码粘贴在答题卡上的指定位置。
用2B 铅笔将答题卡上试卷类型A 后的方框涂黑。
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答案用0.5毫米黑色签字笔写在答题卡上对应的答题区域内,写在试题卷、草稿纸和答题卡上的非选修题答题区域的答案一律无效。
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一、单选题。
(共30小题,每小题2分,共60分)1.下列因素中长江中下游平原地区由开发早期的限制因素转变为现在有利因素的是( )A .稠密的水系B .黏重的土壤C .便利的交通D .平坦的地形2.聚落的建筑风格与自然环境密切相关。
下列地区和当地民居搭配不相符的是( )A .陕北延安——窑洞B .西双版纳——竹楼C .云贵高原——蒙古包D .江南地区——尖顶屋 2019-2020学年第二学期 高一年级地理期中试卷青铜峡市高级中学吴忠中学青铜峡分校3.2011年3月日本大地震引发海啸,当时为快速了解沿岸地区的受灾范围,应采用的技术手段是A.遥感B.地理信息系统C.全球定位系统D.人工实地调查4.在罗布泊进行科考,可为科技工作者全天候提供地理位置信息的技术是( )A.遥感B.地理信息系统C.全球定位系统D.管理信息系统5.由于灌溉措施不当,在西北干旱气候条件下,最容易出现的环境问题是( )A.水土流失B.土壤沙化C.水体富营养化D.土壤次生盐渍化黄土高原地区开发历史悠久,在全国经济发展中具有重要地位,但近些年来人地矛盾十分尖锐。
宁夏吴忠市青铜峡市高级中学2019-2020学年高一第一学期期中考试试题 数学(含解析)
宁夏吴忠市青铜峡市高级中学2019-2020学年高一第一学期期中考试试题 数学一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.设全集为R ,集合A {}|33x x =-<<,{}|15B x x =-<≤,则()R A C B ⋂=( ) A. (]3,1-- B. (3,1)-- C. (3,0)- D. (3,3)-【答案】A 【解析】试题分析:由{}|15B x x =-<≤的{}51U C B x x x =≤-或,所以()R A C B ⋂={}|31x x -<≤-,选A . 考点:集合的运算2.设函数f (x )=21,1,2,1,x x x x⎧+≤⎪⎨>⎪⎩则f (f (3))=( )A.15B. 3C.23D.139【答案】D 【解析】 【详解】()231,33f >∴=, 22213((3))()()1339f f f ==+=,故选D.【此处有视频,请去附件查看】3.函数()()2ln 1f x x x =+-的定义域为( )A. [)2,1- B. (]2,1-C. []2,1-D. ()1,+∞【答案】A 【解析】依题意有2010x x +≥⎧⎨->⎩,解得[)2,1x ∈-.4.下列函数中,在区间(0,)+∞上是增函数的是( )A. 2y x =- B. 1y x=C. 12xy ⎛⎫= ⎪⎝⎭D. 2log y x =【答案】D 【解析】【详解】试题分析:2y x =-在(0,)+∞上是减函数,故A 不对;1y x=在(0,)+∞上是减函数,故B 不对;12xy ⎛⎫= ⎪⎝⎭在(0,)+∞上是减函数,故C 不对.;2log y x =在(0,)+∞上是增函数,故D 对考点:函数的单调性.5.已知幂函数()y f x =的图象过点122⎛ ⎝⎭,则()4f 的值为( )A.14B. 2C. 4D.116【答案】B 【解析】 【分析】根据幂函数的定义和待定系数法,求出幂函数的表达式,即可求值. 【详解】设幂函数为()f x x α=,()y f x =的图象过点12,22⎛ ⎝⎭,1212()222αα--∴===12α∴=.()12f x x ∴=,()124442f ∴===,故选B .【点睛】本题主要考查了利用待定系数法求函数解析式,同时考查了幂函数的概念,属于基础题. 6.满足关系{}1{1,2,3,4}B ⊆⊆的集合B 的个数( ) A. 5个 B. 6个C. 7个D. 8个【答案】D 【解析】 【分析】根据题意得,B 是{1,2,3,4}的一个包含元素1子集,一共有8个.【详解】满足关系式{1}⊆B ⊆{1,2,3,4}的集合B 有{1},{1,3},{1,2},{1,4},{1,2,3},{1,{1,3,4},{1,2,3,4}一共有8个. 故选D .【点睛】本题考查元素与集合关系的判断和子集的应用,属于基本题. 7.若2x =3,则x 等于( ) A. 3log 2 B. lg2lg3-C.lg2lg3D.lg3lg2【答案】D 【解析】 【分析】化指数式为对数式,再由换底公式得答案. 【详解】由2x=3,得x 2332lg log lg ==. 故选D .【点睛】本题考查指数式与对数式的互化,考查换底公式的应用,是基础题.8.已知2(1)5f x x x +=+,那么()f x =( )A. 234x x ++B. 234x x +-C. 23x x +D. 25x x +【答案】B 【解析】 【分析】先令1t x =+,则22()(1)5(1)34f t t t t t =-+-=+-,即可求得函数解析式. 【详解】解:设1t x =+,则1x t =-, 则22()(1)5(1)34f t t t t t =-+-=+-, 即函数解析式为()f x =234x x +-, 故选:B.【点睛】本题考查了利用换元法求函数解析式,属基础题.9.已知32121=0.3log 22a b c -⎛⎫== ⎪⎝⎭,,,则a ,b ,c 的大小关系( )A. a b c >>B. a c b >>C. c b a >>D. b a c >>【解析】 【分析】利用指数函数的单调性与1作比较可以得出a 与b 的大小关系,通过对数函数的图像性质可以得到0c <,得到最终的结果.【详解】由指数函数和对数函数图像可知:32121(0,1),0.31,log 202a b c -⎛⎫=∈=>=< ⎪⎝⎭,则a b c ,,的大小关系是:b a c >>. 故选D .【点睛】本题考查了指数函数与对数函数的单调性,考查了推理能力与计算能力,属于基础题. 10.当01a <<时,在同一坐标系中xy a =与log a y x =的图像大致是( )A. B.C. D.【答案】B 【解析】【详解】解析过程略11.如果奇函数()f x 在区间[]3,7上是增函数,且最小值为5,那么()f x 在区间[]7,3--上是( ) A. 增函数且最小值为5- B. 增函数且最大值为5- C. 减函数且最小值为5- D. 减函数且最大值为5-【答案】B 【解析】 【分析】根据奇偶性和函数在[]3,7上的单调性可知()f x 在[]7,3--上为增函数,由()35f =可知()35f -=-,由单调性确定()3f -为最大值. 【详解】()f x 为奇函数 ()f x ∴图象关于原点对称()f x 在[]3,7上为增函数 ()f x ∴在[]7,3--上为增函数()f x ∴在[]7,3--上的最小值为()7f -;最大值为()3f -又()f x 在[]3,7上最小值为()35f = ()()335f f ∴-=-=- 即()f x 在[]7,3--上为增函数且最大值为5- 本题正确选项:B【点睛】本题考查根据函数的奇偶性和单调性求解函数值的问题,关键是能够通过奇偶性得到对称区间内的单调性,从而确定最值点.12.若()f x 是偶函数,且对任意12,x x ∈(0,)+∞且12x x ≠,都有()()21210-f x f x x x -<,则下列关系式中成立的是( )A. 123()()()234f f f >->B. 132()()()243f f f >->C. 312()()()423f f f >->D. 321()()()432f f f ->>【答案】A 【解析】 分析】由于对任意的x 1,x 2∈(0,+∞),都有()()21210-f x f x x x -<,可得函数f (x )在(0,+∞)上单调递减,即可得出.【详解】∵对任意的x 1,x 2∈(0,+∞),都有()()21210-f x f x x x -<, ∴函数f (x )在(0,+∞)上单调递减, 又∵123234<<, ∴123f f f ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪>>,又∵f (x )是偶函数,∴f (﹣23)=f (23). ∴123234f f f ⎛⎫⎛⎫⎛⎫-⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭>>. 故选A .【点睛】本题考查了函数的奇偶性、单调性的应用,属于基础题. 二、填空题(每小题5分,共20分) 13.已知函数()f x 是定义在R 上的奇函数,当(,0)x ∈-∞时,32()2f x x x =+,则(2)f =__________.【答案】12 【解析】 【分析】由函数的奇偶性可知()()22f f =--,代入函数解析式即可求出结果. 【详解】函数()f x 是定义在上的奇函数,()()f x f x -=-,则()()f x f x =--,()()()()322222212f f ⎡⎤=--=-⨯-+-=⎣⎦.【点睛】本题主要考查函数的奇偶性,属于基础题型. 14.若指数函数()()1xf x a a =>在区间[]0,2上的最大值和最小值之和为10,则a 的值为__【答案】3 【解析】 【分析】先由当1a >时,指数函数()xf x a =为增函数,则在区间[]0,2上,()2max f x a =,()0min 1f x a ==,再结合已知条件运算即可得解.【详解】解:因为当1a >时,指数函数()x f x a =为增函数, 则在区间[]0,2上,()2max f x a =,()0min 1f x a ==,又指数函数()()1xf x aa =>在区间[]0,2上的最大值和最小值之和为10,则2110a +=,即29a =, 又1a >,即3a =, 故答案为:3.15.二次函数22y x ax b =++在[1,)-+∞上单调递增,则实数a 的取值范是____.【答案】[1,+∞) 【解析】 【分析】二次函数的开口向上,在[1,)-+∞上单调递增,所以对称轴要在区间的左边. 【详解】二次函数22y x ax b =++的对称轴为x a =-, ∵()f x 在[1,)-+∞上单调递增, ∴1a --,即1a ≥.【点睛】研究二次函数的单调性时,要注意开口方向及对称轴与区间的位置关系.16.已知函数()f x 是定义在R 上的偶函数,当[0,)x ∈+∞时,()f x 是增函数,且(1)0f -=,则不等式()0f x <的解集为___________【答案】{}|11x x -<< 【解析】 【分析】根据函数的奇偶性和单调性之间的关系,将不等式进行转化,即可得到不等式的解集. 【详解】∵偶函数f (x )在[0,+∞)上增函数,f (﹣1)=0,∴f(﹣1)=f (1)=0, 则函数f (x )对应的图象如图:则f (x )<0的解为﹣1<x <1, 即不等式的解集为(﹣1,1), 故答案为{}|11x x -<<.查函数性质的应用.三、解答题(本大题共6小题,共70分,解答时应写出必要的文字说明、证明过程或演算步骤) 17.计算:(1)11232012720.148π-⎛⎫⎛⎫+-+ ⎪ ⎪⎝⎭⎝⎭(2)2lg 25lg 44log ++ 【答案】(1)101 (2)4 【解析】 【分析】(1)由分数指数幂的运算性质()m nmna a=运算即可得解;(2)由对数的运算性质log log log a a a m n mn +=运算即可得解.【详解】解:(1)11232012720.148π-⎛⎫⎛⎫+-+ ⎪ ⎪⎝⎭⎝⎭1132(1)(2)323333()10()110011012222⨯⨯-⨯-=+-+=+-+=; (2)2lg 25lg 4log 4++2lg100log 4224=+=+=. 【点睛】本题考查了分数指数幂的运算及对数的运算,属基础题. 18.已知集合A ={x |2≤x <7},B ={x |3<x <10},C ={x |x a ≤}. (1)求A ∪B ,(∁R A )∩B ;(2)若A ∩C ≠∅,求a 的取值范围.【答案】(1) {x |2≤x <10}, {x |7≤x <10};(2) 2a ≥ 【解析】 【分析】(1)根据交、并、补集的运算分别求出A∪B,(∁R A )∩B;(2)根据题意和A∩C≠∅,即可得到a 的取值范围.【详解】解:(1)因为A ={x |2≤x <7},B ={x |3<x <10}, 所以A ∪B ={x |2≤x <10}. 因为A ={x |2≤x <7}, 所以∁R A ={x |x <2,或x ≥7}, 则(∁R A )∩B ={x |7≤x <10}.(2)因为A ={x |2≤x <7},C ={x |x a ≤},且A ∩C ≠∅,所以2a ≥【点睛】高考对集合知识的考查要求较低,均是以小题的形式进行考查,一般难度不大,要求考生熟练掌握与集合有关的基础知识.纵观近几年的高考试题,主要考查以下两个方面:一是考查具体集合的关系判断和集合的运算.解决这类问题的关键在于正确理解集合中元素所具有属性的含义,弄清集合中元素所具有的形式以及集合中含有哪些元素.二是考查抽象集合的关系判断以及运算. 19.已知函数f (x )=211x x ++, (1)判断函数在区间[1,+∞)上的单调性,并用定义证明你的结论. (2)求该函数在区间[1,4]上的最大值与最小值. 【答案】(1)增函数,证明见解析 (2)min 3()2f x =,max 9()5f x =【解析】 【分析】(1)设121x x ≤<,再利用作差法判断12(),()f x f x 的大小关系即可得证; (2)利用函数在区间[]1,4上为增函数即可求得函数的最值. 【详解】解:(1)函数f (x )=211x x ++在区间[1,+∞)上为增函数, 证明如下:设121x x ≤<, 则12121212122121()()011(1)(1)x x x x f x f x x x x x ++--=-=<++++, 即12()()f x f x <,故函数f (x )=211x x ++在区间[1,+∞)上为增函数; (2)由(1)可得:函数f (x )=211x x ++在区间[]1,4上为增函数,则min 2113()(1)112f x f ⨯+===+,max 2419()(4)415f x f ⨯+===+, 故函数f (x )在区间[]1,4上的最小值为32,最大值为95. 【点睛】本题考查了利用定义法证明函数的单调性及利用函数单调性求函数的最值,属基础题.20.已知函数()()()22f x log 3x log 3x =+--,()1求()f 1.()2判断并证明函数()f x 的奇偶性;()3已知()2f lga log 5=,求a 的值.【答案】(1)1; (2)()3,3-; (3)100 【解析】 【分析】()1将x=1代入计算即可;()2先求定义域并判断是否关于原点对称,然后用奇偶性定义判断;()3先计算f (lga ),再解方程可得. 【详解】()()()()221f 1log 31log 31211=+--=-=;()2要使函数()()()22f x log 3x log 3x =+--有意义,则{3x 03x 0+>->,解得3x 3-<<,∴函数()f x 的定义域为()3,3-;()()()()22f x log 3x log 3x f x -=--+=-,∴函数()f x 奇函数.()()()()2223f lga log 3lga log 3lga log 5=+--=,3lga53lga+∴=-,且3lga 3-<<,解得a 100=.a 100=.【点睛】本题考查了函数奇偶性定义证明及对数的运算性质,属基础题.21.已知定义在R 上的奇函数()f x ,当0x <时2(1)2f x x x =++.(1)求函数()f x 的表达式; (2)请画出函数()f x 的图象;【答案】(1)2221,0()0,021,0x x x f x x x x x ⎧-+->⎪==⎨⎪++<⎩【解析】【分析】(1)先设0x >,则0x -<,再结合函数的奇偶性求函数解析式即可;(2)结合函数解析式作图像即可得解.【详解】解:(1)设0x >,则0x -<,又函数()f x 为奇函数,则22()()[()2()1]21f x f x x x x x =--=--+-+=-+-,又函数()f x 为R 上的奇函数,则(0)0f =,故2221,0()0,021,0x x x f x x x x x ⎧-+->⎪==⎨⎪++<⎩;(2)由(1)可得:函数()f x 的图象如图所示:【点睛】本题考查了利用函数的奇偶性求函数解析式,重点考查了函数图像的作法,属基础题.22.已知二次函数f (x )满足条件f (0)=1,及f (x +1)﹣f (x )=2x .(1)求函数f (x )的解析式;(2)在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方,试确定实数m 的取值范围.【答案】(1)()21f x x x =-+(2)m <﹣1 【解析】【分析】(1)根据二次函数f (x )满足条件f (0)=1,及f (x +1)﹣f (x )=2x ,可求f (1)=1,f (﹣1)=3,从而可求函数f (x )的解析式;(2)在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方,等价于x 2﹣x +1>2x +m 在[﹣1,1]上恒成立,等价于x 2﹣3x +1>m 在[﹣1,1]上恒成立,求出左边函数的最小值,即可求得实数m 的取值范围.【详解】解:(1)令x =0,则∵f (x +1)﹣f (x )=2x ,∴f (1)﹣f (0)=0,∴f (1)=f (0)∵f (0)=1∴f (1)=1, ∴二次函数图象的对称轴为12x =. ∴可令二次函数的解析式为f (x )21()2y a x h ==-+.令x =﹣1,则∵f (x +1)﹣f (x )=2x ,∴f (0)﹣f (﹣1)=﹣2∵f (0)=1∴f (﹣1)=3, ∴114934a h a h ⎧+=⎪⎪⎨⎪+=⎪⎩ ∴a =1,34h = ∴二次函数的解析式为()2213()124y f x x x x ==-+=-+ (2)∵在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方∴x 2﹣x +1>2x +m 在[﹣1,1]上恒成立∴x 2﹣3x +1>m 在[﹣1,1]上恒成立令g (x )=x 2﹣3x +1,则g (x )=(x 32-)254- ∴g (x )=x 2﹣3x +1在[﹣1,1]上单调递减,∴g (x )min =g (1)=﹣1,∴m <﹣1.【点睛】本题重点考查二次函数解析式的求解,考查恒成立问题的处理,解题的关键是将在区间[﹣1,1]上,y=f(x)的图象恒在y=2x+m的图象上方,转化为x2﹣3x+1>m在[﹣1,1]上恒成立.。
2019-2020学年宁夏吴忠市青铜峡市高级中学高一上学期期中数学试题(解析版)
2019-2020学年宁夏吴忠市青铜峡市高级中学高一上学期期中数学试题一、单选题1.设全集为R ,集合A {}|33x x =-<<,{}|15B x x =-<≤,则()R A C B ⋂=( ) A .(]3,1-- B .(3,1)-- C .(3,0)- D .(3,3)-【答案】A【解析】试题分析:由{}|15B x x =-<≤的{}51U C B x x x =≤-或,所以()R A C B ⋂={}|31x x -<≤-,选A .【考点】集合的运算2.设函数f (x )=21,1,2,1,x x x x⎧+≤⎪⎨>⎪⎩则f (f (3))=( )A .15B .3C .23D .139【答案】D 【解析】【详解】()231,33f >∴=Q , 22213((3))()()1339f f f ==+=,故选D.3.函数()()ln 1f x x -的定义域为( )A .[)2,1- B .(]2,1-C .[]2,1-D .()1,+∞【答案】A【解析】依题意有2010x x +≥⎧⎨->⎩,解得[)2,1x ∈-.4.下列函数中,在区间(0,)+∞上是增函数的是( )A .2y x =- B .1y x= C .12xy ⎛⎫= ⎪⎝⎭ D .2log y x =【答案】D【解析】试题分析:2y x =-在(0,)+∞上是减函数,故A 不对;1y x=在(0,)+∞上是减函数,故B 不对;12xy ⎛⎫= ⎪⎝⎭在(0,)+∞上是减函数,故C 不对.考点:函数的单调性.5.已知幂函数()y f x =的图象过点1,22⎛ ⎝⎭,则()4f 的值为( )A .14B .2C .4D .116【答案】B【解析】根据幂函数的定义和待定系数法,求出幂函数的表达式,即可求值. 【详解】设幂函数为()f x x α=,()y f x =Q 的图象过点1,22⎛⎫ ⎪ ⎪⎝⎭,121()2222αα--∴=== 12α∴=.()12f x x ∴=,()12442f ∴===,故选:B . 【点睛】本题主要考查了利用待定系数法求函数解析式,同时考查了幂函数的概念,属于基础题. 6.满足关系{}1{1,2,3,4}B ⊆⊆的集合B 的个数( ) A .5个 B .6个C .7个D .8个【答案】D【解析】根据题意得,B 是{1,2,3,4}的一个包含元素1子集,一共有8个. 【详解】满足关系式{1}⊆B ⊆{1,2,3,4}的集合B 有{1},{1,3},{1,2},{1,4},{1,2,3},{1,2,4},{1,3,4},{1,2,3,4}一共有8个. 故选:D . 【点睛】本题考查元素与集合关系的判断和子集的应用,属于基本题. 7.若2x =3,则x 等于( ) A .3log 2 B .lg2lg3-C .lg2lg3D .lg3lg2【解析】化指数式为对数式,再由换底公式得答案. 【详解】由2x =3,得x 2332lg log lg ==. 故选D . 【点睛】本题考查指数式与对数式的互化,考查换底公式的应用,是基础题. 8.已知2(1)5f x x x +=+,那么()f x =( ) A .234x x ++ B .234x x +- C .23x x + D .25x x +【答案】B【解析】先令1t x =+,则22()(1)5(1)34f t t t t t =-+-=+-,即可求得函数解析式. 【详解】解:设1t x =+,则1x t =-,则22()(1)5(1)34f t t t t t =-+-=+-, 即函数解析式为()f x =234x x +-, 故选:B. 【点睛】本题考查了利用换元法求函数解析式,属基础题.9.已知32121=0.3log 22a b c -⎛⎫== ⎪⎝⎭,,,则a ,b ,c 的大小关系( )A .a b c >>B .a c b >>C .c b a >>D .b a c >>【答案】D【解析】利用指数函数的单调性与1作比较可以得出a 与b 的大小关系,通过对数函数的图像性质可以得到0c <,得到最终的结果. 【详解】由指数函数和对数函数图像可知:32121(0,1),0.31,log 202a b c -⎛⎫=∈=>=< ⎪⎝⎭,则a b c ,,的大小关系是:b a c >>.【点睛】本题考查了指数函数与对数函数的单调性,考查了推理能力与计算能力,属于基础题. 10.当01a <<时,在同一坐标系中x y a =与log a y x =的图像大致是( )A .B .C .D .【答案】B 【解析】【详解】 解析过程略11.如果奇函数()f x 在区间[]3,7上是增函数,且最小值为5,那么()f x 在区间[]7,3--上是( )A .增函数且最小值为5-B .增函数且最大值为5-C .减函数且最小值为5-D .减函数且最大值为5-【答案】B【解析】根据奇偶性和函数在[]3,7上的单调性可知()f x 在[]7,3--上为增函数,由()35f =可知()35f -=-,由单调性确定()3f -为最大值.【详解】()f x Q 为奇函数 ()f x ∴图象关于原点对称()f x Q 在[]3,7上为增函数 ()f x ∴在[]7,3--上为增函数()f x ∴在[]7,3--上的最小值为()7f -;最大值为()3f -又()f x 在[]3,7上最小值为()35f = ()()335f f ∴-=-=-本题正确选项:B 【点睛】本题考查根据函数的奇偶性和单调性求解函数值的问题,关键是能够通过奇偶性得到对称区间内的单调性,从而确定最值点.12.若()f x 是偶函数,且对任意12,x x ∈(0,)+∞且12x x ≠,都有()()21210-f x f x x x -<,则下列关系式中成立的是( )A .123()()()234f f f >->B .132()()()243f f f >->C .312()()()423f f f >->D .321()()()432f f f ->>【答案】A【解析】由于对任意的x 1,x 2∈(0,+∞),都有()()21210-f x f x x x -<,可得函数f (x )在(0,+∞)上单调递减,即可得出. 【详解】∵对任意的x 1,x 2∈(0,+∞),都有()()21210-f x f x x x -<,∴函数f (x )在(0,+∞)上单调递减, 又∵123234<<, ∴123234f f f ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭>>, 又∵f (x )是偶函数,∴f (﹣23)=f (23). ∴123234f f f ⎛⎫⎛⎫⎛⎫-⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭>>. 故选:A . 【点睛】本题考查了函数的奇偶性、单调性的应用,属于基础题.二、填空题13.已知函数()f x 是定义在R 上的奇函数,当(,0)x ∈-∞时,32()2f x x x =+,则【答案】12【解析】由函数的奇偶性可知()()22f f =--,代入函数解析式即可求出结果. 【详解】函数()f x 是定义在上的奇函数,()()f x f x -=-,则()()f x f x =--,()()()()322222212f f ⎡⎤=--=-⨯-+-=⎣⎦.【点睛】本题主要考查函数的奇偶性,属于基础题型. 14.若指数函数()()1xf x a a =>在区间[]0,2上的最大值和最小值之和为10,则a 的值为__ 【答案】3【解析】先由当1a >时,指数函数()xf x a =为增函数,则在区间[]0,2上,()2max f x a =,()0min 1f x a ==,再结合已知条件运算即可得解.【详解】解:因为当1a >时,指数函数()xf x a =为增函数,则在区间[]0,2上,()2max f x a =,()0min 1f x a ==,又指数函数()()1xf x aa =>在区间[]0,2上的最大值和最小值之和为10,则2110a +=,即29a =, 又1a >,即3a =, 故答案为:3. 【点睛】本题考查了指数函数的单调性及最值的求法,属基础题.15.二次函数22y x ax b =++在[1,)-+∞上单调递增,则实数a 的取值范是____. 【答案】[1,+∞)【解析】二次函数的开口向上,在[1,)-+∞上单调递增,所以对称轴要在区间的左边. 【详解】二次函数22y x ax b =++的对称轴为x a =-,∴1a --…,即1a ≥. 【点睛】研究二次函数的单调性时,要注意开口方向及对称轴与区间的位置关系.16.已知函数()f x 是定义在R 上的偶函数,当[0,)x ∈+∞时,()f x 是增函数,且(1)0f -=,则不等式()0f x <的解集为___________【答案】{}|11x x -<<【解析】根据函数的奇偶性和单调性之间的关系,将不等式进行转化,即可得到不等式的解集. 【详解】∵偶函数f (x )在[0,+∞)上为增函数,f (﹣1)=0, ∴f (﹣1)=f (1)=0, 则函数f (x )对应的图象如图:则f (x )<0的解为﹣1<x <1, 即不等式的解集为(﹣1,1), 故答案为:{}|11x x -<<. 【点睛】本题主要考查不等式的解法,利用函数的奇偶性和单调性之间的关系是解决本题的关键,综合考查函数性质的应用.三、解答题17.计算:(1)11232012720.148π-⎛⎫⎛⎫+-+ ⎪ ⎪⎝⎭⎝⎭(2)2lg 25lg 44log ++ 【答案】(1)101 (2)4(2)由对数的运算性质log log log a a a m n mn +=运算即可得解. 【详解】 解:(1)11232012720.148π-⎛⎫⎛⎫+-+ ⎪ ⎪⎝⎭⎝⎭1132(1)(2)323333()10()110011012222⨯⨯-⨯-=+-+=+-+=; (2)2lg 25lg 4log 4++2lg100log 4224=+=+=. 【点睛】本题考查了分数指数幂的运算及对数的运算,属基础题. 18.已知集合A ={x |2≤x <7},B ={x |3<x <10},C ={x |x a ≤}. (1)求A ∪B ,(∁R A )∩B ;(2)若A ∩C ≠∅,求a 的取值范围.【答案】(1) {x |2≤x <10}, {x |7≤x <10};(2) 2a ≥【解析】(1)根据交、并、补集的运算分别求出A ∪B ,(∁R A )∩B ;(2)根据题意和A∩C≠∅,即可得到a 的取值范围. 【详解】解:(1)因为A ={x |2≤x <7},B ={x |3<x <10}, 所以A ∪B ={x |2≤x <10}. 因为A ={x |2≤x <7}, 所以∁R A ={x |x <2,或x ≥7}, 则(∁R A )∩B ={x |7≤x <10}.(2)因为A ={x |2≤x <7},C ={x |x a ≤},且A ∩C ≠∅,所以2a ≥ 所以a 的取值范围为2a ≥. 【点睛】高考对集合知识的考查要求较低,均是以小题的形式进行考查,一般难度不大,要求考生熟练掌握与集合有关的基础知识.纵观近几年的高考试题,主要考查以下两个方面:一是考查具体集合的关系判断和集合的运算.解决这类问题的关键在于正确理解集合中元素所具有属性的含义,弄清集合中元素所具有的形式以及集合中含有哪些元素.二是考查抽象集合的关系判断以及运算.(1)判断函数在区间[1,+∞)上的单调性,并用定义证明你的结论. (2)求该函数在区间[1,4]上的最大值与最小值. 【答案】(1)增函数,证明见解析 (2)min 3()2f x =,max 9()5f x =【解析】(1)设121x x ≤<,再利用作差法判断12(),()f x f x 的大小关系即可得证; (2)利用函数在区间[]1,4上为增函数即可求得函数的最值. 【详解】 解:(1)函数f (x )=211x x ++在区间[1,+∞)上为增函数, 证明如下:设121x x ≤<, 则12121212122121()()011(1)(1)x x x x f x f x x x x x ++--=-=<++++, 即12()()f x f x <,故函数f (x )=211x x ++在区间[1,+∞)上为增函数; (2)由(1)可得:函数f (x )=211x x ++在区间[]1,4上为增函数,则min 2113()(1)112f x f ⨯+===+,max 2419()(4)415f x f ⨯+===+, 故函数f (x )在区间[]1,4上的最小值为32,最大值为95. 【点睛】本题考查了利用定义法证明函数的单调性及利用函数单调性求函数的最值,属基础题.20.已知函数()()()22f x log 3x log 3x =+--,()1求()f 1.()2判断并证明函数()f x 的奇偶性; ()3已知()2f lga log 5=,求a 的值.【答案】(1)1; (2)()3,3-; (3)100. 【解析】()1将x=1代入计算即可;()2先求定义域并判断是否关于原点对称,然后用奇偶性定义判断;【详解】()()()()221f 1log 31log 31211=+--=-=; ()2要使函数()()()22f x log 3x log 3x =+--有意义,则{3x 03x 0+>->,解得3x 3-<<,∴函数()f x 的定义域为()3,3-;()()()()22f x log 3x log 3x f x -=--+=-Q ,∴函数()f x 为奇函数.()()()()2223f lga log 3lga log 3lga log 5=+--=Q ,3lga53lga+∴=-,且3lga 3-<<,解得a 100=.a 100=.【点睛】本题考查了函数奇偶性定义证明及对数的运算性质,属基础题.21.已知定义在R 上的奇函数()f x ,当0x <时2(1)2f x x x =++. (1)求函数()f x 的表达式; (2)请画出函数()f x 的图象;【答案】(1)2221,0()0,021,0x x x f x x x x x ⎧-+->⎪==⎨⎪++<⎩(2)函数()f x 的图像见解析【解析】(1)先设0x >,则0x -<,再结合函数的奇偶性求函数解析式即可; (2)结合函数解析式作图像即可得解. 【详解】解:(1)设0x >,则0x -<, 又函数()f x 为奇函数,则22()()[()2()1]21f x f x x x x x =--=--+-+=-+-, 又函数()f x 为R 上的奇函数,故2221,0()0,021,0x x x f x x x x x ⎧-+->⎪==⎨⎪++<⎩; (2)由(1)可得:函数()f x 的图象如图所示:【点睛】本题考查了利用函数的奇偶性求函数解析式,重点考查了函数图像的作法,属基础题. 22.已知二次函数f (x )满足条件f (0)=1,及f (x +1)﹣f (x )=2x . (1)求函数f (x )的解析式;(2)在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方,试确定实数m 的取值范围.【答案】(1)()21f x x x =-+(2)m <﹣1 【解析】(1)根据二次函数f (x )满足条件f (0)=1,及f (x +1)﹣f (x )=2x ,可求f (1)=1,f (﹣1)=3,从而可求函数f (x )的解析式;(2)在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方,等价于x 2﹣x +1>2x +m 在[﹣1,1]上恒成立,等价于x 2﹣3x +1>m 在[﹣1,1]上恒成立,求出左边函数的最小值,即可求得实数m 的取值范围.【详解】解:(1)令x =0,则∵f (x +1)﹣f (x )=2x ,∴f (1)﹣f (0)=0,∴f (1)=f (0)∵f (0)=1∴f (1)=1,∴二次函数图象的对称轴为12x =. ∴可令二次函数的解析式为f (x )21()2y a x h ==-+.令x =﹣1,则∵f (x +1)﹣f (x )=2x ,∴f (0)﹣f (﹣1)=﹣2∵f (0)=1∴f (﹣1)=3, ∴114934a h a h ⎧+=⎪⎪⎨⎪+=⎪⎩ ∴a =1,34h = ∴二次函数的解析式为()2213()124y f x x x x ==-+=-+ (2)∵在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方∴x 2﹣x +1>2x +m 在[﹣1,1]上恒成立∴x 2﹣3x +1>m 在[﹣1,1]上恒成立令g (x )=x 2﹣3x +1,则g (x )=(x 32-)254- ∴g (x )=x 2﹣3x +1在[﹣1,1]上单调递减,∴g (x )min =g (1)=﹣1,∴m <﹣1.【点睛】本题重点考查二次函数解析式的求解,考查恒成立问题的处理,解题的关键是将在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方,转化为x 2﹣3x +1>m 在[﹣1,1]上恒成立.。
2019-2020学年宁夏青铜峡市高级中学(吴忠中学青铜峡分校)高二下学期期中考试数学(理)试题 Word版
一、选择题(本大题共12个小题,每小题5分,共60分)1.某校的书法绘画,乐器演奏,武术爱好三个兴趣小组的人数分别为600,400,300,若用分层抽样方法抽取n名学生参加某项活动,已知从武术小组中抽取了6名学生,则n的值为()A.20 B.22 C.23 D.262.某校8位学生的本次月考成绩恰好都比上一次的月考成绩高出50分,则以该8位学生这两次的月考成绩各自组成样本,则这两个样本不变的数字特征是()A.方差B.中位数C.众数D.平均数3.某学校调查了200名学生每周的自习时间(单位:小时),制成了如图所示的频率分布直方图,其中自习时间的范围是[17.5,30],样本数据分组为[17.5,20),[20,22.5),[22.5,25),[25,27.5),[27.5,30].根据直方图,这200名学生中每周的自习时间不少于22.5小时的人数是()A.56 B.60 C.140 D.1204.某公司从A、B两个部门中各选出6名员工参加本季度的笔试考核,他们取得的成绩(满分100分)的茎叶图如图,其中A部门员工成绩的中位数是83,B部门员工成绩的平均数是85,则x y+的值为()A.2 B.3 C.4 D.55.若101022110)2(xaxaxaax++++=-Λ, 则( ) A.1 B.-1 C.1023 D.-10232019-2020学年第二学期高二年级数学(理)期中试卷命题人:青铜峡市高级中学吴忠中学青铜峡分校6.5名学生中有且只有3名同学会颠足球,从中任意选取2人,则这2人都会颠足球的概率为()A.310B.25C.12D.357.我国明朝数学家程大位著的《算法统宗》里有一道闻名世界的题目:“一百馒头一百僧,大僧三个更无争.小僧三人分一个,大小和尚各几丁?”.如右图所示的程序框图反映了对此问题的一个求解算法,则输出n的值为()A.20B.25C.30D.358.现采用随机模拟的方法估计某运动员射击4次,至少击中3次的概率;先由计算器给出0到9之间取整数值的随机数,指定0、1、2表示没有击中目标,3、4、5、6、7、8、9表示击中目标,以4个随机数为一组,代表射击4次的结果,经随机模拟产生了20随机数:7527,0293,7140,9857,0347,4373,8636,6947,1417,46980371,6233,2616,8045,6011,3661,9597,7424,7610,4281根据以上数据估计该射击运动员射击4次至少击中3次的概率为()A.0.55 B.0.6 C.0.65 D.0.79.在高三下学期初,某校开展教师对学生的家庭学习问卷调查活动,已知现有3名教师对4名学生家庭问卷调查,若这3名教师每位至少到一名学生家中问卷调查,又这4名学生的家庭都能且只能得到一名教师的问卷调查,那么不同的问卷调查方案的种数为()A.72 B.48 C.36 D.2410.函数()21ln 4f x x x =-的大致图象是( ) A . B . C . D .11.若函数2()x f x x e a =-恰有3个零点,则实数a 的取值范围是( ) A .24(,)e+∞ B .24(0,)eC .2(0,4)eD .(0,)+∞12.若函数()2ln f x x x bx =+-在[)1,+∞是增函数,则b 的最大值是( ) A .3B .22C .2D 2二、填空题(本大题共4个小题,每小题5分,共20分)13.某奶茶店的日销售收入y (单位:百元)与当天平均气温x (单位:C ︒)之间的关系如下:x 2-1-0 1 2 y5m221通过上面的五组数据得到了x 与y 之间的线性回归方程:ˆ 2.8yx =-+; 但现在丢失了一个数据m ,该数据m 应为____________14.已知函数,()cos sin f x x x =+, 则()4f π'= .15.某兴趣小组有2名女生和3名男生,现从中任选2名学生去参加活动,则至多有一名男生的概率为_____________.16.731x x ⎛⎫⎪⎝⎭x _____.(用数字填写答案)三、解答题:本大题共6小题,满分70分.解答须写出文字说明、证明过程和演算步骤. 17.(本题10分) 设x =1与x =2是函数f (x )=a ln x +bx 2+x 的两个极值点. (1)试确定常数a 和b 的值;(2)判断x =1,x =2是函数f (x )的极大值点还是极小值点,并说明理由.18.(本题12分) 选修4-4: 参数方程与极坐标在直角坐标系xOy 中,圆C 的参数方程为6cos 1sin x ty t =-+⎧⎨=-+⎩(t 为参数),在以坐标原点O 为极点,x 轴非负半轴为极轴建立的极坐标系中,直线l 的极坐标方程为sin 04πρθ⎛⎫--= ⎪⎝⎭.(1)求圆C 的普通方程和直线l 的直角坐标方程;(2)设点P 是圆C 上任一点,求点P 到直线l 距离的最小值.19.(本题12分) 选修4-5:不等式选讲 已知函数()2221f x x x =-++. (1)求不等式()7f x >的解集;(2)若()f x 的最小值为(0,0)a b a b a b +>>且、为实数,求22a b +的最小值,并指出此时a b 、的值.20.(本题12分) 某地区某农产品近几年的产量统计如表:(I )根据表中数据,建立关于t 的线性回归方程$$y bta =+$; (Ⅱ)根据线性回归方程预测2019年该地区该农产品的年产量.附:对于一组数据()11,t y ,()22,t y ,…, (),n n t y ,其回归直线$$y bta =+$的斜率和截距的最小二乘估计分别为:()()()121ˆˆˆ,niii ni i t t y y bay bt t t ==--==--∑∑. (参考数据:()()612.8ii i tty y =--=∑,计算结果保留小数点后两位)21.(本题12分) 某校从参加某次知识竞赛的同学中,选取60名同学将其成绩(百分制,均为整数)分成[4050),,[5060),,[6070),,[7080),,[8090),,[90100],六组后,得到部分频率分布直方图(如图),观察图形中的信息,回答下列问题:(1)求分数[7080),内的频率,并补全这个频率分布直方图; (2)从频率分布直方图中,估计本次考试成绩的中位数;(3)若从第1组和第6组两组学生中,随机抽取2人,求所抽取2人成绩之差的绝对值大于10的概率.22.(本题12分) 设函数()ln f x x =,()1g x ax =+,a R ∈,记()()()F x f x g x =-. (1)求曲线()y f x =在x e =处的切线方程; (2)求函数()F x 的单调区间;(3)当0a >时,若函数()F x 没有零点,求a 的取值范围.高二数学理科期中答案13. 4 14. 0 15. 1016. 3517. (1)∵f (x )=a ln x +bx 2+x ,∴f ′(x )=ax+2bx +1. 由极值点的必要条件可知:f ′(1)=f ′(2)=0,∴a +2b +1=0且2a+4b +1=0,解方程组得,a =23- ,b =16- .(2)由(1)可知f (x )=23-ln x 16-x 2+x ,且函数f (x )=23-ln x 16-x 2+x 的定义域是(0,+∞),f ′(x )=23-x -131-x +1=(1)(2)3x x x--- .当x ∈(0,1)时,f ′(x )<0;当x ∈(1,2)时,f ′(x )>0;当x ∈(2,+∞)时,f ′(x )<0; 所以,x =1是函数f (x )的极小值点, x =2是函数f (x )的极大值点.18.(1)由6cos 1sin x t y t=-+⎧⎨=-+⎩消去参数t ,得()()22611x y +++=,所以圆C 的普通方程为()()22611x y +++=. 由sin 04πρθ⎛⎫--= ⎪⎝⎭,得sin cos 2ρθρθ-=,所以直线l 的直角坐标方程为20x y -+=. (2)设点P 的坐标为()6cos ,1sin t t -+-+, 则点P 到直线l 的距离为d ==sin 24t π⎛⎫=+- ⎪⎝⎭, 当sin 14t π⎛⎫-=- ⎪⎝⎭时,d 取最小值,min 12d =-.19.(1)原不等式等价于1111224-1741737x x x x x ⎧⎧≥<--≤<⎧⎪⎪⎨⎨⎨>⎩⎪⎪-+>>⎩⎩或或分别解得3,22x x -无解,,综上所述,不等式()7f x >的解集为()3,2,2⎛⎫-∞-⋃+∞ ⎪⎝⎭.(2)依题意,可知3a b +=,()()22222222222a b a b ab a b a b a b +=++≤+++=+2293,22a b a b +≥==当且仅当等号成立20.(1)由题意可知:1234563.56t +++++==,6.6 6.777.17.27.476y +++++==,()()()()6222222212.5 1.50.50.5 1.5 2.517.5i i t t =-=-+-+-+++=∑,∴()()()12112.80.1617.ˆ5niii ni t t y y bt t ==--===-∑∑,又70.16ˆˆ 3.5 6.44ay bt =-=-⨯=, ∴y 关于t 的线性回归方程为0.1664ˆ.4yt =+. (2)由(1)可得,当年份为2019年时,年份代码8t =,此时0.168 6.4.ˆ4772y=⨯+=, 所以,可预测2019年该地区该农产品的年产量约为7.72万吨.21.(1)设分数在[)70,80内的频率为x ,根据频率分布直方图,则有()0.010.01520.0250.005101x +⨯++⨯+=,可得0.3x =, 所以频率分布直方图为:(2)以中位数为准做一条垂直于横轴的直线,这条直线把频率分布直方图分成面积相等的两个部分,由频率分布直方图知中位数要把最高的小长方形三等分, 所以中位数是1170107333+⨯=,所以估计本次考试成绩的中位数为1733(3)设所抽取2人成绩之差的绝对值大于10为事件M , 第1组学生数:600.16⨯=人(设为1,2,3,4,5,6) 第6组学生数:600.053⨯=人(设为,,A B C )所有基本事件有:12,13,14,15,16,1,1,1A B C ,23,24,25,26,2A ,2B ,2C ,34,35,36,3A ,3B ,3C ,45,46,4A ,4B ,4C ,56,5A ,5B ,5C ,6A ,6B ,6C ,AB ,AC ,BC 共有35种,事件M 包括的基本事件有:1,1,1A B C ,2A ,2B ,2C ,3A ,3B ,3C ,4A ,4B ,4C ,5A ,5B ,5C ,6A ,6B ,6C 共有18种所以()181362P M ==.22.(1)1()f x x '=,则函数()f x 在x e =处的切线的斜率为1k e=.又()1f e =, 所以函数()f x 在x e =处的切线方程为11()y x e e-=-,即1y x e =(2)()ln 1F x x ax =--,11()axF x a x x-=-=',(0x >).①当0a ≤时,()0F x '>,()F x 在区间(0,)+∞上单调递增;②当0a >时,令()0F x '<,解得1x a>;令()0F x '>,解得10x a <<.综上所述,当0a ≤时,函数()F x 的增区间是(0,)+∞; 当0a >时,函数()F x 的增区间是1(0,)a ,减区间是1(,)a+∞. (3)依题意,函数()F x 没有零点,即()ln 10F x x ax =--=无解.由(2)知,当0a >时,函数()F x 在区间1(0,)a 上为增函数,区间1(,)a+∞上为减函数,由于(1)10F a =--<,只需111()ln 1ln 20F a a aa a =-⋅-=--<, 解得2a e ->.所以实数a 的取值范围为21(,)e+∞.。
2019-2020学年宁夏吴忠市青铜峡市高级中学高一上学期期中考试数学试题
高级中学2019-2020学年(一)期中考试高一年级数学试卷★祝考试顺利★ 注意事项:1、答题前,请先将自己的姓名、准考证号用0.5毫米黑色签字笔填写在试题卷和答题卡上的相应位置,并将准考证号条形码粘贴在答题卡上的指定位置。
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一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.设全集为R ,集合A {}|33x x =-<<,{}|15B x x =-<≤,则()R A C B ⋂=( ) A. (]3,1-- B. (3,1)-- C. (3,0)- D. (3,3)-【答案】A 【解析】试题分析:由{}|15B x x =-<≤的{}51U C B x x x =≤-或,所以()R A C B ⋂={}|31x x -<≤-,选A .考点:集合的运算2.设函数f (x )=21,1,2,1,x x x x⎧+≤⎪⎨>⎪⎩则f (f (3))=( )A.15B. 3C.23D.139【答案】D 【解析】【详解】()231,33f >∴=Q , 22213((3))()()1339f f f ==+=,故选D.【此处有视频,请去附件查看】3.函数()()ln 1f x x =-的定义域为( )A. [)2,1- B. (]2,1-C. []2,1-D. ()1,+∞【答案】A 【解析】依题意有2010x x +≥⎧⎨->⎩,解得[)2,1x ∈-.4.下列函数中,在区间(0,)+∞上是增函数的是( )A. 2y x =-B. 1y x=C. 12xy ⎛⎫= ⎪⎝⎭D.2log y x =【答案】D 【解析】【详解】试题分析:2y x =-在(0,)+∞上是减函数,故A 不对;1y x=在(0,)+∞上是减函数,故B 不对;12xy ⎛⎫= ⎪⎝⎭在(0,)+∞上是减函数,故C 不对.;2log y x =在(0,)+∞上是增函数,故D 对考点:函数的单调性.5.已知幂函数()y f x =的图象过点1,22⎛⎫⎪ ⎪⎝⎭,则()4f 的值为( )A.14B. 2C. 4D.116【答案】B 【解析】 【分析】根据幂函数的定义和待定系数法,求出幂函数的表达式,即可求值.【详解】设幂函数为()f x xα=,()y f x =Q 的图象过点1,22⎛⎝⎭,121()2222αα--∴===12α∴=.()12f x x ∴=,()12442f ∴===,故选B .【点睛】本题主要考查了利用待定系数法求函数解析式,同时考查了幂函数的概念,属于基础题.6.满足关系{}1{1,2,3,4}B ⊆⊆的集合B 的个数( ) A. 5个 B. 6个C. 7个D. 8个【答案】D 【解析】 【分析】根据题意得,B 是{1,2,3,4}的一个包含元素1子集,一共有8个.【详解】满足关系式{1}⊆B ⊆{1,2,3,4}的集合B 有{1},{1,3},{1,2},{1,4},{1,2,3},{1,2,4},{1,3,4},{1,2,3,4}一共有8个. 故选D .【点睛】本题考查元素与集合关系的判断和子集的应用,属于基本题. 7.若2x =3,则x 等于( ) A. 3log 2 B. lg2lg3- C. lg2lg3D. lg3lg2【答案】D 【解析】 【分析】化指数式为对数式,再由换底公式得答案. 【详解】由2x =3,得x 2332lg log lg ==.故选D .【点睛】本题考查指数式与对数式的互化,考查换底公式的应用,是基础题. 8.已知2(1)5f x x x +=+,那么()f x =( ) A. 234x x ++ B. 234x x +- C. 23x x + D. 25x x +【答案】B 【解析】 【分析】先令1t x =+,则22()(1)5(1)34f t t t t t =-+-=+-,即可求得函数解析式. 【详解】解:设1t x =+,则1x t =-, 则22()(1)5(1)34f t t t t t =-+-=+-, 即函数解析式为()f x =234x x +-, 故选:B.【点睛】本题考查了利用换元法求函数解析式,属基础题.9.已知32121=0.3log 22a b c -⎛⎫== ⎪⎝⎭,,,则a ,b ,c 的大小关系( )A. a b c >>B. a c b >>C. c b a >>D. b a c >>【答案】D 【解析】 【分析】利用指数函数的单调性与1作比较可以得出a 与b 的大小关系,通过对数函数的图像性质可以得到0c <,得到最终的结果.【详解】由指数函数和对数函数图像可知:32121(0,1),0.31,log 202a b c -⎛⎫=∈=>=< ⎪⎝⎭,则a b c ,,的大小关系是:b a c >>. 故选D .【点睛】本题考查了指数函数与对数函数的单调性,考查了推理能力与计算能力,属于基础题.10.当01a <<时,在同一坐标系中x y a =与log a y x =的图像大致是( )A. B.C. D.【答案】B 【解析】【详解】解析过程略11.如果奇函数()f x 在区间[]3,7上是增函数,且最小值为5,那么()f x 在区间[]7,3--上是( )A. 增函数且最小值为5-B. 增函数且最大值为5-C. 减函数且最小值为5-D. 减函数且最大值为5-【答案】B 【解析】 【分析】根据奇偶性和函数在[]3,7上的单调性可知()f x 在[]7,3--上为增函数,由()35f =可知()35f -=-,由单调性确定()3f -为最大值.【详解】()f x Q 为奇函数 ()f x ∴图象关于原点对称()f x Q 在[]3,7上为增函数 ()f x ∴在[]7,3--上为增函数()f x ∴在[]7,3--上的最小值为()7f -;最大值为()3f -又()f x 在[]3,7上最小值为()35f = ()()335f f ∴-=-=- 即()f x 在[]7,3--上为增函数且最大值为5-本题正确选项:B【点睛】本题考查根据函数的奇偶性和单调性求解函数值的问题,关键是能够通过奇偶性得到对称区间内的单调性,从而确定最值点.12.若()f x 是偶函数,且对任意12,x x ∈(0,)+∞且12x x ≠,都有()()21210-f x f x x x -<,则下列关系式中成立的是( )A. 123()()()234f f f >->B. 132()()()243f f f >->C. 312()()()423f f f >->D. 321()()()432f f f ->>【答案】A 【解析】 【分析】由于对任意的x 1,x 2∈(0,+∞),都有()()21210-f x f x x x -<,可得函数f (x )在(0,+∞)上单调递减,即可得出. 【详解】∵对任意的x 1,x 2∈(0,+∞),都有()()21210-f x f x x x -<,∴函数f (x )在(0,+∞)上单调递减,又∵123234<<, ∴123234f f f ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭>>, 又∵f (x )是偶函数,∴f (﹣23)=f (23). ∴123234f f f ⎛⎫⎛⎫⎛⎫-⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭>>. 故选A .【点睛】本题考查了函数的奇偶性、单调性的应用,属于基础题.二、填空题(每小题5分,共20分)13.已知函数()f x 是定义在R 上奇函数,当(,0)x ∈-∞时,32()2f x x x =+,则(2)f =__________.【答案】12 【解析】 【分析】由函数的奇偶性可知()()22f f =--,代入函数解析式即可求出结果. 【详解】函数()f x 是定义在上的奇函数,()()f x f x -=-,则()()f x f x =--,()()()()322222212f f ⎡⎤=--=-⨯-+-=⎣⎦.【点睛】本题主要考查函数的奇偶性,属于基础题型. 14.若指数函数()()1xf x a a =>在区间[]0,2上的最大值和最小值之和为10,则a 的值为__【答案】3 【解析】 【分析】先由当1a >时,指数函数()xf x a =为增函数,则在区间[]0,2上,()2max f x a =,()0min 1f x a ==,再结合已知条件运算即可得解.【详解】解:因为当1a >时,指数函数()xf x a =为增函数,则在区间[]0,2上,()2max f x a =,()0min 1f x a ==,又指数函数()()1xf x aa =>在区间[]0,2上的最大值和最小值之和为10,则2110a +=,即29a =, 又1a >,即3a =, 故答案为:3.【点睛】本题考查了指数函数的单调性及最值的求法,属基础题.15.二次函数22y x ax b =++在[1,)-+∞上单调递增,则实数a 的取值范是____. 【答案】[1,+∞) 【解析】 【分析】二次函数的开口向上,在[1,)-+∞上单调递增,所以对称轴要在区间的左边.【详解】二次函数22y x ax b =++的对称轴为x a =-, ∵()f x 在[1,)-+∞上单调递增, ∴1a --…,即1a ≥.【点睛】研究二次函数的单调性时,要注意开口方向及对称轴与区间的位置关系.16.已知函数()f x 是定义在R 上的偶函数,当[0,)x ∈+∞时,()f x 是增函数,且(1)0f -=,则不等式()0f x <的解集为___________ 【答案】{}|11x x -<< 【解析】 【分析】根据函数的奇偶性和单调性之间的关系,将不等式进行转化,即可得到不等式的解集. 【详解】∵偶函数f (x )在[0,+∞)上增函数,f (﹣1)=0,∴f (﹣1)=f (1)=0,则函数f (x )对应的图象如图:则f (x )<0的解为﹣1<x <1, 即不等式的解集为(﹣1,1), 故答案为{}|11x x -<<.【点睛】本题主要考查不等式的解法,利用函数的奇偶性和单调性之间的关系是解决本题的关键,综合考查函数性质的应用.三、解答题(本大题共6小题,共70分,解答时应写出必要的文字说明、证明过程或演算步骤)17.计算:(1)11232012720.148π-⎛⎫⎛⎫+-+ ⎪ ⎪⎝⎭⎝⎭(2)2lg 25lg 44log ++ 【答案】(1)101 (2)4 【解析】 【分析】(1)由分数指数幂的运算性质()m nmna a=运算即可得解;(2)由对数的运算性质log log log a a a m n mn +=运算即可得解. 【详解】解:(1)11232012720.148π-⎛⎫⎛⎫+-+ ⎪ ⎪⎝⎭⎝⎭1132(1)(2)323333()10()110011012222⨯⨯-⨯-=+-+=+-+=; (2)2lg 25lg 4log 4++2lg100log 4224=+=+=. 【点睛】本题考查了分数指数幂的运算及对数的运算,属基础题. 18.已知集合A ={x |2≤x <7},B ={x |3<x <10},C ={x |x a ≤}. (1)求A ∪B ,(∁R A )∩B ;(2)若A ∩C ≠∅,求a 的取值范围. 【答案】(1) {x |2≤x <10}, {x |7≤x <10};(2) 2a ≥ 【解析】 【分析】(1)根据交、并、补集的运算分别求出A ∪B ,(∁R A )∩B ;(2)根据题意和A∩C≠∅,即可得到a 的取值范围.【详解】解:(1)因为A ={x |2≤x <7},B ={x |3<x <10}, 所以A ∪B ={x |2≤x <10}. 因为A ={x |2≤x <7}, 所以∁R A ={x |x <2,或x ≥7}, 则(∁R A )∩B ={x |7≤x <10}.(2)因为A ={x |2≤x <7},C ={x |x a ≤},且A ∩C ≠∅,所以2a ≥ 所以a 的取值范围为2a ≥.【点睛】高考对集合知识的考查要求较低,均是以小题的形式进行考查,一般难度不大,要求考生熟练掌握与集合有关的基础知识.纵观近几年的高考试题,主要考查以下两个方面:一是考查具体集合的关系判断和集合的运算.解决这类问题的关键在于正确理解集合中元素所具有属性的含义,弄清集合中元素所具有的形式以及集合中含有哪些元素.二是考查抽象集合的关系判断以及运算. 19.已知函数f (x )=211x x ++, (1)判断函数在区间[1,+∞)上的单调性,并用定义证明你的结论. (2)求该函数在区间[1,4]上的最大值与最小值. 【答案】(1)增函数,证明见解析 (2)min 3()2f x =,max 9()5f x =【解析】 【分析】(1)设121x x ≤<,再利用作差法判断12(),()f x f x 的大小关系即可得证; (2)利用函数在区间[]1,4上为增函数即可求得函数的最值. 【详解】解:(1)函数f (x )=211x x ++在区间[1,+∞)上为增函数, 证明如下:设121x x ≤<, 则12121212122121()()011(1)(1)x x x x f x f x x x x x ++--=-=<++++, 即12()()f x f x <,故函数f (x )=211x x ++在区间[1,+∞)上为增函数; (2)由(1)可得:函数f (x )=211x x ++在区间[]1,4上为增函数,则min 2113()(1)112f x f ⨯+===+,max 2419()(4)415f x f ⨯+===+, 故函数f (x )在区间[]1,4上的最小值为32,最大值为95. 【点睛】本题考查了利用定义法证明函数的单调性及利用函数单调性求函数的最值,属基础题.20.已知函数()()()22f x log 3x log 3x =+--,()1求()f 1.()2判断并证明函数()f x 的奇偶性;()3已知()2f lga log 5=,求a 的值.【答案】(1)1; (2)()3,3-; (3)100【解析】【分析】()1将x=1代入计算即可;()2先求定义域并判断是否关于原点对称,然后用奇偶性定义判断;()3先计算f (lga ),再解方程可得. 【详解】()()()()221f 1log 31log 31211=+--=-=;()2要使函数()()()22f x log 3x log 3x =+--有意义,则{3x 03x 0+>->, 解得3x 3-<<,∴函数()f x 的定义域为()3,3-;()()()()22f x log 3x log 3x f x -=--+=-Q ,∴函数()f x 奇函数. ()()()()2223f lga log 3lga log 3lga log 5=+--=Q ,3lga 53lga+∴=-,且3lga 3-<<, 解得a 100=.a 100=.【点睛】本题考查了函数奇偶性定义证明及对数的运算性质,属基础题.21.已知定义在R 上的奇函数()f x ,当0x <时2(1)2f x x x =++.(1)求函数()f x 的表达式;(2)请画出函数()f x 的图象;【答案】(1)2221,0()0,021,0x x x f x x x x x ⎧-+->⎪==⎨⎪++<⎩(2)函数()f x 的图像见解析【解析】【分析】(1)先设0x >,则0x -<,再结合函数的奇偶性求函数解析式即可;(2)结合函数解析式作图像即可得解.【详解】解:(1)设0x >,则0x -<,又函数()f x 为奇函数,则22()()[()2()1]21f x f x x x x x =--=--+-+=-+-,又函数()f x 为R 上的奇函数,则(0)0f =,故2221,0()0,021,0x x x f x x x x x ⎧-+->⎪==⎨⎪++<⎩;(2)由(1)可得:函数()f x 的图象如图所示:【点睛】本题考查了利用函数的奇偶性求函数解析式,重点考查了函数图像的作法,属基础题.22.已知二次函数f (x )满足条件f (0)=1,及f (x +1)﹣f (x )=2x .(1)求函数f (x )的解析式;(2)在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方,试确定实数m 的取值范围.【答案】(1)()21f x x x =-+(2)m <﹣1 【解析】【分析】(1)根据二次函数f (x )满足条件f (0)=1,及f (x +1)﹣f (x )=2x ,可求f (1)=1,f (﹣1)=3,从而可求函数f (x )的解析式;(2)在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方,等价于x 2﹣x +1>2x +m 在[﹣1,1]上恒成立,等价于x 2﹣3x +1>m 在[﹣1,1]上恒成立,求出左边函数的最小值,即可求得实数m 的取值范围.【详解】解:(1)令x =0,则∵f (x +1)﹣f (x )=2x ,∴f (1)﹣f (0)=0,∴f (1)=f (0)∵f (0)=1∴f (1)=1, ∴二次函数图象的对称轴为12x =. ∴可令二次函数的解析式为f (x )21()2y a x h ==-+.令x =﹣1,则∵f (x +1)﹣f (x )=2x ,∴f (0)﹣f (﹣1)=﹣2∵f (0)=1∴f (﹣1)=3, ∴114934a h a h ⎧+=⎪⎪⎨⎪+=⎪⎩ ∴a =1,34h = ∴二次函数的解析式为()2213()124y f x x x x ==-+=-+ (2)∵在区间[﹣1,1]上,y =f (x )的图象恒在y =2x +m 的图象上方∴x2﹣x+1>2x+m在[﹣1,1]上恒成立∴x2﹣3x+1>m在[﹣1,1]上恒成立令g(x)=x2﹣3x+1,则g(x)=(x32-)254-∴g(x)=x2﹣3x+1在[﹣1,1]上单调递减,∴g(x)min=g(1)=﹣1,∴m<﹣1.【点睛】本题重点考查二次函数解析式的求解,考查恒成立问题的处理,解题的关键是将在区间[﹣1,1]上,y=f(x)的图象恒在y=2x+m的图象上方,转化为x2﹣3x+1>m在[﹣1,1]上恒成立.。
2019-2020学年宁夏青铜峡市高级中学(吴忠中学青铜峡分校)高一下学期期中考试政治试题、
青铜峡市高级中学2019——2020年(二)期中考试高一年级政治测试卷出题:一选择题(每小题2分共60分)1.《中华人民共和国宪法》第41条规定:“中华人民共和国公民对于任何国家机关和国家工作人员,有提出批评和建议的权利。
”宪法赋予公民监督权的根本原因是( )A.我国是人民当家作主的社会主义国家B.监督权是公民的基本政治权利C.人民民主是社会主义的生命和基石D.人民民主具有广泛性和真实性2.网络给人们的信息交流提供了便利的平台,同时,借助这一平台,谣言也得以迅速传播。
创造健康和谐的网络舆论环境,必须( )①提升网络参与者的素养②提高政府的行政管理水平③防止网络信息传播扩散④增强公民的民主管理能力A.①②B.①④C.②③D.③④3.所谓晒客,指的是那些热衷于用文字、照片和视频等方式将私人物件以及生活经历放在网上曝光、与人分享的网友。
晒客族的一个口号就是“只有不想晒的,没有不能晒的”。
这种观点( )A.是正确的,因为网络是一片无拘无束的天空B.是正确的,因为公民不是网络世界的权利主体C.是错误的,因为公民的自由是在法律范围内的自由D.是错误的,因为真正的自由是不存在的4.网络日渐成为公民表达诉求、进行民主监督的重要渠道。
然而,也存在着某些人利用网络散布谣言、侵犯他人隐私等问题。
为此,作为公民应该( )①充分利用网络表达意愿②充分利用网络实现自身权益③坚持权利与义务统一④坚持理性表达,有序参与A.①②B.②③C.①④D.③④5.为进一步推进基层民主政治的发展,某省人大常委会修订了《村民委员会选举办法》,改无目标的“海选”为有组织的直选,“自荐直选”被列为选举方式之一。
据统计,全省88.2%的村采用“自荐直选”方式,受到村民的欢迎。
这反映了( )①通过创新选举方式,扩大公民的民主权利②通过改革选举方式等途径,健全基层民主③我国公民享有的选举权和被选举权日益制度化、规范化④通过完善基层民主制度,保证人民依法直接行使民主权利A.①③B.①④C.②③D.②④6.2015年10月16日,湖北省通山县洪港镇洪港社区党群中心内,社区党员干部通过远教平台,收看社区换届选举法律知识宣传片。
宁夏青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020学年高一语文下学期期中试题
宁夏青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020学年高一语文下学期期中试题一、现代文阅读(27分)(一)文学类文本阅读(本题共3小题,15分)阅读下面文章,完成1~3小题。
献给艾青的红玫瑰曹积三①十多年过去了,那束红玫瑰,依稀仍在眼前灿灿地绽放着。
按说,送仙逝者驾鹤西去,应献白色或黄色的菊花,可我觉得,艾青是浪漫的诗人,更是世俗的反叛者,他一生追逐太阳,为光明战斗,为光明歌唱,献一束充盈着太阳颜色的红玫瑰更为合适.于是,那天,我捧着一束如火的红玫瑰,赶到东四十三条97号艾老的府上送他远行。
高瑛大姐欣喜地对我说:“艾青喜欢红玫瑰.”我感到一阵宽慰.后来得知,艾老走后,赶来献花的人络绎不绝,所献之花,满满地拉了两卡车,还没有运完.②戴红领巾的时候,我因读到《大堰河》,而晓得了艾青。
岂料这位可敬的诗人命途多舛。
他的诗集一度被图书馆“禁借”,为此,我曾备感困惑和惆怅,但不管有人怎么往他身上大泼污水,那些美丽的诗行仍令我的心头发烫。
值得庆幸的是,波诡云谲里,虽经七灾八难,他终于挺了过来,如同涅槃的凤凰.然而,直到1988年的初秋,我才得以见到心仪多年的诗人。
③那是一个下午,北影有个剧组开机,约我去写点什么。
并告诉我说,拍摄地点在东城丰收胡同21号四合院。
到了那里才知道,那是艾青的家。
处于京华一隅的这个小院,像是远离尘嚣,别有一番天地,它静谧、安适,特别是那斑红灿烂的花木和葡萄架上缀满的果实,更平添了几分温馨.拍戏,是个扰人的活儿,把原本整洁的院子弄得很凌乱.艾老的夫人高瑛大姐古道热肠,对此毫不介意,还替剧组找东拿西,不停地张罗着……小憩时,她从葡萄架上,剪下葡萄,请大家尝鲜。
就在这时,我望见艾老透过书房的窗子,正笑吟吟地望着我们.我急忙去向他问好。
④那天,艾老身着中山装,笑容可掬,热诚得如同相识多年的朋友。
我们聊起了电影,他告诉我,他很喜欢电影,在延安的时候,放的是几部苏联的原版影片,一边放,一边由萧三翻译,人们看得津津有味。
宁夏青铜峡市高级中学吴忠中学青铜峡分校2019_2020学年高一化学下学期期中试题
宁夏青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020学年高一化学下学期期中试题可能用到的相对原子质量:H-1 C-12 N-14 O-16 Na-23 S-32 Cl-35.5 Fe-56一、选择题(本大题共25个小题,每小题2分,共50分,每小题只有一个选项是正确的)1.我国著名的化学家、教育家徐光宪先生因在稀土金属等研究领域做出杰出贡献,荣获了“国家最高科学技术奖”。
是地壳中含量最高的稀土金属元素。
下列关于的说法错误的是()A.质量数为140 B.中子数为82 C.质子数为58 D.核外电子数为1982.元素的下列性质,随原子序数的递增不是周期性变化的是()A.相对原子质量B.化合价C.原子半径D.元素的化学性质3.下列元素原子半径最大的是()A.LiB.FC.NaD.Cl4.下列物质中,只含有非极性共价键的是A. MgCl2B. Ba(OH)2C. O2D. H2O25.下列有关周期表的说法正确的是()A.短周期是第一、二、三、四周期B.元素周期表有18个族C.第ⅠA族的元素全部是金属元素D.元素周期表含元素最多的族是第ⅢB族6.下列各组中的性质比较中,不正确的是A.酸性 HClO4>HBrO4>HIO4B.碱性 Ba(OH)2>Ca(OH)2>Mg(OH)2C.还原性 F->Cl->Br-D.稳定性 HCl>H2S>PH37.下列原子序数所对应的元素组中,两者可形成离子键的是()A. 12和9B. 1和17C. 14和6D. 15和88.下列关于卤族元素由上到下性质递变规律的叙述,正确的是( )①单质的氧化性减弱②单质的颜色加深③气态氢化物的稳定性增强A.①②③B.①②C.②③D.①③9.下列化学用语书写正确的是( )A. 氯离子的结构示意图:B. 作为相对原子质量测定标准的碳核素:C. 氯化镁的电子式:D. 用电子式表示氯化氢分子的形成过程:10.某元素原子M层电子数是K层电子数的3.5倍,则该元素最高价氧化物对应水化物的化学式为A. HNO3 B.H3PO4 C.H2SO4 D.HClO411.在元素周期表中,金属元素与非金属元素分界线附近,能找到A.制半导体材料的元素B.制农药的元素C.制催化剂的元素D.制耐高温合金的元素12.可用来判断金属性强弱的依据是A.原子电子层数的多少B.最外层电子数的多少C.最高价氧化物的水化物的碱性强弱D.等物质的量的金属置换氢气的多少13.同一周期X、Y、Z三种元素,已知最高价氧化物对应水化物的酸性是HXO4>H2YO4>H3ZO4,则下列判断错误的是 ( )A.阴离子的还原性Z3->Y2->X-B.气态氢化物的稳定性HX>H2Y>ZH3C.非金属性X>Y>ZD.原子半径X>Y>Z14. 短周期中三元素a、b、c在周期表中的位置如图,下列有关这三种元素的叙述正确的是:A.a是一种活泼的非金属元素B.b的氢化物很稳定C.c的最高价氧化物的水化物是一种弱酸D.b元素的最高化合物为+7价。
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宁夏青铜峡市高级中学(吴忠中学青铜峡分校)2019-2020学年高一下学期期中考试英语试题第Ⅰ卷听力(30分)第一节(共5小题;每小题1.5分,满分7.5分)请听下面5段对话,选出最佳选项。
第一部分听力(共两节,满分30分)请听下面5段对话,选出最佳选项。
I. How many classes does the woman have on Wednesday?A. 2.B. 4.C. 5.2. Where did Jane grow up?A. In Sweden.B. In Switzerland.C. In the US.3. What time is the man’s flight?A. At 9:00 a.m.B. At 9:50 a.m.C. At 10:40 a.m.4.What is the woman hunting for?A. A good movie.B. A nice magazine 7C. A fortune list5. What did the woman do last night?A. She attended a lecture.B. She went to the hospital.C. She made a speech on education.第二节(共15小题;每小题1.5分,满分22.5分)请听下面5段对话或独白,选出最佳选项。
请听第6段材料,回答第6至8题6. What is the probable relationship between the two speakers?A. Sister and brother.B. Teacher and student.C. Husband and wife.7.How many children do they have?A. 3B. 4C.58. How does the man feel?A. TiredB. FreeC. Bad请听第7段材料,回答第9至12题。
9. At what time does the early flight get to Paris?A. At 7:50B. At 11:05.C. At 9:00.10.At what time can the woman get to the center if she takes the early plane?A. 9:40.B. At 10:10.C. At 11:45.11. Which of the following is true?. ?A.There are two flights to Paris every morning.B. There is a flight to Paris every half hour.C. There are two flights to Paris expect Friday.12. How long does it take the bus to get to the city center?A. 14 minutes.B. 40 minutesC. 4 minutes请听第8段材料,回答第13、14题13.When did the conversation take place?A. In the morning.B. In the afternoonC. In the evening.14. What should the man like before eating ?A. Some orange.B. Some coffee.C. Some ham.请听第9段材料,回答第15至17题。
15. What happened to the woman?A. She lost her jobB. The boss allowed her to go on a holiday.C. She was given a chance to study abroad.16. How did the man feel about the woman’s boss’s decision?A.He understood it well.B. It was completely wrong.C. He did ‘t understand it.17. What should the woman do in the future?A. She should go to work on time.B. She should learn new things and work hard.C. Both A and B.请听第10段材料,回答第18至20题。
18.To many British people. what is an ideal way to relax for a few minutes?A. A cup of tea.B.A cup of coffeeC.A glass of beer.19. When did the English custom of drinking tea begin to form?A. In the 14 century.B. In the 15 centuryC. In the 17th and 18th centuries.20 Which Coffee House first offered tea?A. Lynn’s Coffee House.B. Tom’s Coffee HouseC. Twining’s Coffee House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。
AWhen I was a child, I often dreamed of the time when I could leave home and escape to the city. We lived on a farm, in the winter especially, we were quite cut off from the outside world. As soon as I left school, I packed my bags and moved to the capital. However, I soon discovered that my life has its problems too.One big disadvantage is money. It costs so much to go out, not to mention basics like food and housing. Another disadvantage is pollution. I suffer from asthma(哮喘), and the air is so bad that I am afraid to go outside. Then there is the problem of traveling round. Although I have a car, I seldom use it because of the traffic jams. One choice is to go by bicycle, but that can be quite dangerous.Of course there are advantages. First, there is so much to do in the city, whatever you taste in culture or entertainment(娱乐活动). Besides, there are wonderful jobs and greater chances of moving to a more important job or position. Finally, if you like shopping, the variety of goods is very surprising --- and. what is more, shops are often only a short walk away.Is life better then, in the city? Perhaps it is. when you are in your teens or twenties. However, as you get older, and especially if you have small children, the peace of the countryside may seem preferable. I certainly hope to move back there soon.21. What was the writer always thinking about when he was a child?A. Staying on the farm.B. Leaving home for the city.C. Moving to the countryside.D. Running away from the school.22. Which of the following is true about the writer?A. He is very old now.B. He is in good health.C. He prefers driving a car.D. He lives in the city now.23. In the passage, the writer tries to __________.A. describe his life in the countrysideB. express his opinions about way of lifeC. an interest in the outside worldD. persuade the reader to live in the cityBSome people think that as more and more people have televisions in their homes, fewer and fewer people will buy books and newspapers. Why read an article in the newspaper, when the TV news can bring you the information in a few minutes and with pictures? Why read the life story of a famous man, when a short television program can tell you all that you want to know?Television has not killed reading, however. Today, newspapers sell in very large numbers. And books of every kind are sold more than ever before. Books are still a cheap way to get information and enjoyment. Although some books with hard covers are expensive, many books are printed today as paperbacks (平装本), which are quite cheap. A paperback collection of short stories, for example, is always cheaper than an evening at the cinema or the theater, and you can keep a book forever and read it many times.Books are a wonderful provider of knowledge and pleasure and some types of books should be in every home. Every home should have a good dictionary. A good encyclopedia (百科全书), though expensive, is useful, too, because you can find information on any subject. Besides, you can have such books as history books., science textbooks, cookbooks, and collections of stories and poems. Then from time to time you can take a book of poems off your shelves and read the thoughts and feelings of your favorite poets.24. It can be inferred from the passage that ________.A. TV programs are a chief provider of knowledgeB. cinemas are the best choice in getting informationC. newspapers are an expensive way to enjoy oneselfD. reading is a cheap way of learning and having fun25. What does the sentence “Television has not killed reading, however ”underlined in the second paragraph suggest?A. People only need reading, though.B. Reading is more fun than television.C. Reading is still necessary today.D. Watching television doesn’t help reading.26. What can we learn from the passage?A. Fewer and fewer people will buy books.B. A good dictionary should be kept in every home.C. Books with hard covers sell better than paperbacks.D. More people like TV programs about famous men.CIn the summer of 1885, nine-year-old Joseph Meister was very ill. He had been wounded by a sick dog that had rabies, a very dangerous disease. His parents were told that there was probably only one man who could save Joseph's life —Louis Pasteur.When Pasteur was a young boy in France, he spent many hours every day with the chemist (药剂师) who lived in his small town. At that time, the chemist had to make all the medicines himself. Young Louis enjoyed watching the chemist as he worked and helped those people who came to him each day.As a schoolboy, Pasteur worked slowly and carefully. At first, his teachers thought that young Louis might be a slow learner. Through elementary school, high school, and college, Pasteur worked in the same thoughtful way. Then he became a college professor (教授) and a scientist, and he continued to work very carefully.Pasteur was studying about the germs (细菌) that cause rabies when Joseph Meister became ill. In fact, Pasteur believed he had a medical treatment for rabies, but he had never given it to a person before. At first, Pasteur was afraid to treat Joseph, but the poor child was dying. Pasteur gave Joseph an inoculation (预防接种) every day for ten days. Slowly, the child became better.During his lifetime, Pasteur studied germs and learned how they cause diseases in animals and people. He developed vaccinations (疫苗) that prevent many of these diseases. On September28, 1895, Louis Pasteur passed away, at the age of 72. The work of this great man has been of great help to modern medicine.27. The story of Joseph Meister is given to _____.A. express the author's sadness.B. introduce the subject of the textC. show some common diseases in 1885.D. warn children to stay away from dogs28. According to the text, young Louis _____.A. was once badly hurt by a dog.B. was very interested in medicineC. made a living by working for a chemist.D. had been thought highly of by his teachers29. We can learn from Paragraph 3 that Louis Pasteur _____.A. was always patient.B. was clever but proud.C. was a slow learner.D. was a humorous professor.30. What would be the best title for the text ?A. Germs and diseases.B. Rabies: a terrible disease.C. The earliest chemist in France.D. Louis Pasteur: a great scientist.DIt doesn’t matter when or how much a person sleeps, but everyone needs some rest to stay alive. That’s what all doctors thought, until they heard about AI Herpin. AI Herpin, it was said, never slept, Could this be true? The doctors decided to see this strange man themselves.AI Herpin was 90 years old when the doctors came to his home in New Jersey. They thought for sure that he got some sleep of some kind. So they stayed with him and watched every movement he made. But they were surprised. Though they watched him hour after hour and day after day, they never saw Herpin sleeping. In fact, he did not even own a bed. He never needed one.The only rest that Herpin sometimes got was sitting in a comfortable chair and reading newspapers. The doctors were puzzled by this strange continuous sleeplessness. They found only one answer that might explain his condition. Herpin remembered some talk about his mother having been injured several days before he was born. But that was all. Was this the real reason? No one could be sure.Herpin died at the age of 94.31. The main idea of this passage is that _______.A. large numbers of people do not need sleepB. a person was found who actually didn’t need any sleepC. everyone needs some sleep to stay aliveD. people can live longer by trying not to sleep32. The doctors came to visit Herpin. expecting to ______.A. cure him of his sleeplessnessB. find that his sleeplessness was not really trueC. find a way to free people from the need of sleepingD. find out why some old people didn’t need any sleep33. After watching him closely, the doctors came to believe that AL Herpin ________.A. was too old to need any sleepB. often slept in a chairC. needed no sleep at allD. needed some kind of sleep34. One reason that might explain Herpin’s sleeplessness was ________.A. that he hadn’t got a bedB. that he had gradually got rid of the sleeping habitC. his mother’s injury before he was bornD. his magnificent physical condition35. AI Herpin’s condition could be regarded as ______.A. a common oneB. a rare oneC. very healthyD. one that could be cured第二节根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。