重点名校高考考前模拟——2014学年高三第一学期期中考试试卷.
百强重点名校高考备考_2014届武汉市一中高三第一次模拟考试语文试题(完美整理版)

2014届武汉市一中高三第一次模拟考试语文试题未0420 17582014届武汉市一中高三第一次模拟考试语文试题一、(15分,每小题3分)1.下列各组词语中,加点字的读音全都正确的一组是A.脊梁(jǐ)炮制(páo)白炽灯(zhì)鳞次栉比(zhì)B.混沌(hǜn)悭吝(jiān)一沓纸(dá)寡廉鲜耻(xiǎn)、c.伺机(sì)噱头(xuã)刽子手(guì)畏葸不前(xǐ)D.昵称(nì)攻讦(jiã)撂挑子(1iào)喟然长叹(wâi)2.下列各句中没有错别字的一项是A.把汉语修辞方式融入花鸟画创作中的实例不胜梅举,屡屡被众多画家广泛使用。
B.面对浩如烟海的有字书,尽管我们浩首穷经、读破万卷,也只能涉猎沧海一粟。
C.晚清时期,随着诸多脍炙人口的反满革命文章的流播,不几年,革命论胜行于中国。
D.普京来自底层,其亲属中没有政治上的显赫者,也没有在其他领域中出类拔萃的人。
3.下列各句中加点的词语使用不恰当的一句是A.社会生活好比一本大书,需要慢慢地去品读,才能读懂其中蕴含的道理,才能从中获得人生真谛。
B.要把爱心转化成行动,我们需要看问题,找办法,察效果,虽然错综复杂的事态总在阻挠这三步的实施。
C.毛泽东诗词是独具风神的艺术品,更是中国现代史上吐纳风云、评说春秋、飞扬情志的文化现象。
D.李零在其新作中指出,孔子是一个四处游说、替统治者操心、与虎谋皮、拼命劝他们改邪归正的人。
4.下列各句中没有语病的一项是A.美国研究人员最近报告说:电击以醋和废水为养分的细菌,可以制造出清洁的氢燃料能够替代汽油给车辆提供动力。
B.人的全面发展并不是个孤立的历史发展过程,而是始终如何处理好人与自然、人与社会、人与人的关系网络中。
C.尽管西方节日对中国近代节日的多样化产生了影响,但中国近代节日的民族特性并没消失,、其内容、样式基本保留。
浙江省宁波重点中学2014届高三上学期期中考试英语试题 Word版含答案

2013-2014学年宁波重点中学高三期中考(英语)【注意事项】:本试题卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,共9页,全卷满分120分,考试时间120分钟.第Ⅰ卷(选择题)第一部分:英语知识运用(共两节,满分30分)第一节:单项填空(共20小题;每小题0.5分,满分10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,并在答题纸上将该选项标号涂黑。
1. The movie Transformers 3 brought in 1.2 billion dollars in 2011, which was ________ newheight in ________ director’s career.A. a; aB. a; theC. the; aD. the; the2. Judging by the ________ he keeps, Mark must be an extremely wealthy man.A. cooperationB. characteristicC. motivationD. company3. —Why, my new computer is rubbish! Five hours has passed and it still hasn’t finished thedownloading.—Don’t worry. You know, sometimes the Internet ________ be very slow.A. canB. mayC. shouldD. must4. Young children are often advised not to spoil their appetite by eating sweets ________ the three meals.A. amongB. duringC. betweenD. through5. According to the original design, the plant would be able to _______ at least 8,000 pianos a year.A. bring outB. take outC. put outD. work out6. Security was such a major concern at the G8 Summit that no journalists could approach themeeting zone without ________ the special pass.A. producingB. issuingC. commandingD. involving7. Jessica has always been honest and straightforward, and it doesn’t mat ter ________ thatshe’s talking to.A. who is itB. who it isC. it is whoD. it is whom8. Joan stood outside a shopping mall for a long time, as if ________ somebody.A. to wait forB. waiting forC. waited forD. having waited for9. —Dad, dad! My head hurts so much. I think I’m dying.—________, darling! The doctor says you will feel better soon.A. That will doB. There, thereC. Hi, thereD. So long10. Researchers have found clear and convincing evidence ________ happiness is the key tobetter health and longer lives.A. whereB. whetherC. howD. that11. When Steven suddenly remembered that he had left his mobile in the office, the train wasalready ________.A. on the goB. on the riseC. on the runD. on the move12. In spite of the fact that soccer is the most popular international sport, it isn’t what most girls ________.A. go in forB. make up forC. do away withD. put up with13. Mr. Smith doesn’t like all that empty talk between Susan and Joan, which he thinks willdefinitely get them ________.A. anywhereB. anythingC. nowhereD. nothing14. What the educational expert wants to stress is that ________ from worry and anxiety candefinitely contributes to a child’s mental and physical health.A. responsibilityB. freedomC. sorrowD. comfort15. —I remember I left my laptop in the study last night. Why ________ here?—Oh, I thought you might need it so I brought it here for you.A. does it comeB. has it comeC. had it comeD. will it come16. —Jenny, I hate to say it, but you really must go!—________.A. All right, if you insistB. Not likelyC. Mind your own businessD. You must be kidding17. Chemical pollution will bring about a threat to agriculture and food chain, and ________ tohuman health.A. frequentlyB. significantlyC. steadilyD. consequently18. The weather that day may not be as good as expected, in ________ case we will have to putoff the event.A. whatB. thatC. thisD. which19. To his credit, his translation manages to be ________ to the spirit of the original.A. faithfulB. admirableC. automaticD. considerate20. The new model ________ the market in three months, all the shops are now selling the oldmodel at a discount.A. hittingB. hitC. having hitD. to hit第二节:完型填空(共20小题;每小题1分,满分20分)阅读下面短文,掌握其大意,然后从21—40题所给的四个选项(A、B、C、D)中,选出最佳选项,并在答题纸上将该选项标号涂黑。
2014学年第一学期高三年级期中考试文科综合试卷

2014-2015学年度第一学期高三级文科综合期中考试试卷本试卷分单项选择题和综合题两部分,共19页,满分为300分。
考试用时150分钟。
注意事项:1、答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名和学号填写在答题卡和答卷密封线内相应的位置上,用2B铅笔将自己的学号填涂在答题卡上。
2、选择题每小题选出答案后,有2B铅笔把答题卡上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其他答案;不能答在试卷上。
3、非选择题必须用黑色字迹的钢笔或签字笔在答卷纸上作答,答案必须写在答卷纸各题目指定区域内的相应位置上,超出指定区域的答案无效;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液。
不按以上要求作答的答案无效。
4、考生必须保持答题卡的整洁和平整。
一、选择题:本大题共35小题,每小题4分,共140分。
在每小题列出的四个选项中,只有一项是符合题目要求的。
2014年8月11日凌晨2时前后,广大天文爱好者欣赏到了本年度的“超级月亮”(指本年度“最大最圆”的月亮)。
“超级月亮”上一次发生在2013年6月23日,下一次将出现在2015年9月28日。
1.“超级月亮”出现时,下列说法正确的是()A.地球公转速度最快B.出现的时间具有规律性C.月亮、地球、太阳三者的位置不在一条直线上D.会诱发强烈地震2.读“中亚地壳西藏至准噶尔(沿85 ºE)的地形与莫霍面剖面图”,下列说法正确的是()A.莫霍面是一条水平直线B.莫霍面是地壳和岩石圈的分界线C.地壳厚度的变化取决于地表形态的起伏D.地形高度与莫霍面深度呈现“倒影”的关系3.当东半球恰好全部为夜半球时()A.太阳直射在(110°W , 0°)B.地球上新一天的范围比旧一天的范围大C.南半球获得的太阳辐射比北半球多D.东半球的平均气温比西半球高下左图为东半球示意图,下右图为90°E附近二分二至日中的某一节气海平面气压图(单位:hPa),读图回答问题:4.由气压值可推断,此时()A.夏威夷高压势力强盛B.气压带、风带处于偏北位置C.大陆上等温线向高纬凸出D.长江中下游盛行西北风近年来石家庄深陷雾霾污染之中。
2014届高三上学期期中考试化学试卷(含答案)

2014届高三上学期期中考试化学试卷一、单项选择题:本题包括10小题,每小题2分,共计20分.每小题只有一个选项符合题意.1.(2分)化学科学对提高人类生活质量和促进社会发展具有重要作用,下列说法正确的是()①用高效半导体照明灯取代白炽灯,有利于节能减排②研发安全特效药物,有利于抵御人类疾病③规范食品添加剂生产和科学使用,有利于保障人体健康和生命安全④推广城市生活垃圾处理的减量化、资源化、无害化技术,有利于环境保护和资源的充分利用分子的电子式:铝离子结构示意图:比例模型电子式为,=0.09mol8.(2分)短周期元素W、X、Y和Z的原子序数依次增大.金属W是制备一种高效电池的重要材料,X原子的最外层电子数是内层电子数的2倍,元素Y是地壳中含量最丰富的10.(2分)类比(比较)是研究物质性质的常用方法之一,可预测许多物质的性质.但类二、不定项选择题:本题包括5小题,每小题4分,共计20分.每小题只有一个或两个选项符合题意.若正确答案只包括一个选项,多选时,该题得0分;若正确答案包括两个选项时,只选一个且正确的得2分,选两个且都正确的得满分,但只要选错一个,该小题就得0分.12.(4分)A~E是中学常见的5种化合物,A、B是氧化物,它们之间的转化关系如图所示.则下列说法正确的是()+2Fe+2Al13.(4分)锑(Sb)在自然界一般以硫化物的形式存在,我国锑的蕴藏量为世界第一.从硫化物中提取单质锑一般是先在高温下将硫化物转化为氧化物,再用碳还原:①2Sb2S3+3O2+6Fe Sb4O6+6FeS ②Sb4O6+6C4Sb+6CO↑14.(4分)(2011•广东)下列实验现象预测正确的是()15.(4分)某同学设计如下实验测量m g铜银合金样品中铜的质量分数:下列说法中不正确的是()铜的质量分数为:,所以合金中银的质量为×三、非选择题16.(16分)(2012•浙江模拟)大气中SO2和NO x是形成酸雨的主要物质.某地酸雨中可能含有下列离子:Na+、Mg2+、NH4+、Cl﹣、SO32﹣、SO42﹣、NO3﹣和NO2﹣等.某研究小组取该地一定量的酸雨,浓缩后将所得试液分成4份,进行如下实验:第一份滴加适量的淀粉KI溶液,呈蓝色;第二份滴加用盐酸酸化的BaCl2溶液,有白色沉淀析出;第三份滴加NaOH 溶液,加热,产生的气体能使湿润的红色石蕊试纸变蓝;第四份加入足量硝酸酸化的AgNO3溶液,有沉淀产生,静置,在上层清液中滴加酸性KMnO4溶液,不褪色.已知:K sp(Ag2SO4)=1.20×10﹣5请回答下列问题:(1)该酸雨中肯定存在的离子有SO42﹣、NO3﹣、NH4+;肯定不存在的离子有SO32﹣、NO2﹣,说明其不存在的理由:SO32具有较强的还原性,酸性条件下,与NO3不能共存;若有NO2﹣,能使酸性KMnO4溶液褪色.(2)写出试液中滴加淀粉KI溶液所发生反应的离子方程式:6I﹣+2NO3﹣+8H+=3I2+2NO↑+4H2O.(3)设计实验方案,检验该试液中是否存在Cl﹣:取少量试液,滴加足量的Ba(NO3)2溶液,静置;取上层清液,滴加硝酸酸化的AgNO3溶液,若有白色沉淀产生,则存在Cl.(4)该研究小组为了探究NO参与的硫酸型酸雨的形成,在烧瓶中充入含有少量NO的SO2气体,慢慢通入O2,该过程中发生的化学反应有2NO+O2=2NO2、NO2+SO2=SO3+NO,再喷射适量蒸馏水即得硫酸型酸雨.说明NO的作用:催化剂.17.(12分)(2011•镇江模拟)工业上常用铁质容器盛装冷浓硫酸.为研究铁质材料与热浓硫酸的反应,某学习小组进行了以下探究活动:(1)将已去除表面氧化物的铁钉(碳素钢)放入冷浓硫酸中,10分钟后移入硫酸铜溶液中,片刻后取出观察,铁钉表面无明显变化,其原因是铁表面被钝化.(2)另称取铁钉6.0g放入15.0ml浓硫酸中,加热,充分应后得到溶液X并收集到气体Y.①甲同学认为X中除Fe3+外还可能含有Fe2+.写出生成Fe2+所有可能的离子反应方程式:Fe+2H+=Fe2++H2↑;Fe+2Fe3+=3Fe2+.若要确认其中有Fe2+,应选用d(选填序号).a.KSCN溶液和氯水b.铁粉和KSCN溶液c.浓氨水d.酸性KMnO4溶液②乙同学取336ml(标准状况)气体Y通入足量溴水中,发生反应的化学方程式为:SO2+Br2+2H2O=2HBr+H2SO4.然后加入足量BaCl2溶液,经适当操作后得干燥固体2.33g.由于此推知气体Y中SO2的体积分数为66.7%.=的体积分数:18.(12分)实验室里可用如图一所示的装置制取氯酸钾、次氯酸钠,并验证氯水的性质.图一中①为氯气发生装置;②的试管中盛有15mL 30%的KOH溶液,并置于水浴中;③的试管中盛有15mL 8%的NaOH溶液,并置于冰水浴中;④的试管中加有紫色石蕊试液;⑤为尾气吸收装置.(1)制取氯气时,在烧瓶中先加入一定量的二氧化锰固体,再通过分液漏斗(填写仪器名称)向烧瓶中加入适量的浓盐酸(填写试剂名称).(2)为除去氯气中混有的氯化氢气体,可在①和②之间安装盛有b(选填字母编号)的净化装置.a.碱石灰b.饱和食盐水c.浓硫酸d.饱和碳酸氢钠溶液(3)本实验制取次氯酸钠的离子方程式是Cl2+2OH﹣=Cl﹣+ClO﹣+H2O.(4)比较制取氯酸钾和次氯酸钠的条件,两者的差异是反应的温度不同、反应物浓度不同.(5)反应完毕经冷却后,②的试管中有大量晶体析出,图二中符合该晶体溶解度随温度变化规律的曲线是M(选填字母);从②的试管中分离该晶体的操作是过滤(填写实验操作名称).(6)实验中可观察到④的试管中溶液颜色会发生变化:最初溶液由紫色变为红色,原因是氯气与水反应生成酸使石蕊显红色;接着溶液逐渐变为无色,是因为氯气与水反应生成次氯酸将石蕊氧化为无色.19.(14分)(2011•安徽)MnO2是一种重要的无机功能材料,粗MnO2的提纯是工业生产的重要环节.某研究性学习小组设计了将粗MnO2(含有较多的MnO和MnCO3)样品转化为纯MnO2实验,其流程如下(1)第①步加稀H2SO4时,粗MnO2样品中的MnO和MnCO3(写化学式)转化为可溶性物质.(2)第②步反应的离子方程式:5Mn2++2ClO3﹣+4H2O=5MnO2+Cl2↑+8H+.(3)第③步蒸发操作必需的仪器有铁架台(含铁圈)、酒精灯、蒸发皿、玻璃棒,已知蒸发得到的固体中有NaClO3和NaOH,则一定还有含有NaCl(写化学式).(4)若粗MnO2样品的质量为12.69g,第①步反应后,经过滤得到8.7g MnO2,并收集到0.224LCO2(标准状况下载),则在第②步反应中至少需要0.02mol mol NaClO3.20.(12分)焦亚硫酸钠(Na2S2O5)常用作食品漂白剂.其制备工艺流程如下:已知:反应Ⅱ包含2NaHSO3Na2S2O5+H2O等多步反应.(1)反应I的化学方程式为CO2+NH3+NaCl+H2O=NaHCO3↓+NH4Cl,反应Ⅰ进行时应先通入NH3气体.(2)硫磺燃烧前先加热成液态并通过喷嘴喷入焚硫炉中,目的是增大硫与空气的接触面积,使硫磺充分燃烧,加快反应速率.在上述工艺流程中可循环使用的物质是CO2、H2O.(3)反应Ⅱ中需控制参加反应的气体与固体的物质的量之比接近2:1,若气体量不足,则会导致Na2S2O5产品中Na2SO3、Na2CO3等杂质含量增加,Na2S2O5产率降低.(4)Na2S2O5与稀硫酸反应放出SO2,其离子方程式为S2O5+2H=2SO2↑+H2O.3321.(14分)X、Y、Z、M、G五种元素分属三个短周期,且原子序数依次增大.X、Z同主族,可形成离子化合物ZX;Y、M同主族,可形成MY2、MY3两种分子.请回答下列问题:(1)Y在元素周期表中的位置为第二周期第ⅥA族.(2)上述元素的最高价氧化物对应的水化物酸性最强的是HClO4(写化学式),非金属气态氢化物还原性最强的是H2S(写化学式).(3)Y、G的单质或两元素之间形成的化合物可作水消毒剂的有O3、Cl2(写出其中两种物质的化学式).(4)X2M的燃烧热△H=﹣a kJ•mol﹣1,写出X2M燃烧反应的热化学方程式:2H2S(g)+3O2(g)=2SO2(g)+2H2O(l),△H=﹣2aKJ•mol﹣1.(5)ZX的电子式为;ZX与水反应放出气体的化学方程式为NaH+H2O=NaOH+H2↑.故答案为:;。
2014-2015学年山东省潍坊市重点中学高三(上)期中数学试卷和答案(文科)

2014-2015学年山东省潍坊市重点中学高三(上)期中数学试卷(文科)一、选择题:本题共10小题,每小题5分,共50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.(5分)集合A={0,2,a},B={1,a2},若A∪B={0,1,2,4,16},则a的值为()A.0 B.1 C.2 D.42.(5分)已知函数f(x)=ax5﹣bx3+cx,f(﹣3)=2,则f(3)的值为()A..2 B.﹣2 C.6 D.﹣63.(5分)设α是第二象限角,p(x,4)为其终边上的一点,且cosα=x,则sinα=()A.B.﹣ C.D.﹣4.(5分)已知向量=(2,3),=(﹣1,2),若与共线,则m的值为()A.B.2 C.D.﹣25.(5分)定义在R上的函数y=f(x)满足,(x﹣)f′(x)>0,任意的x1<x2,都有f(x1)>f(x2)是x1+x2<5的()A.充分不必要条件 B.必要不充分条件C.充分必要条件D.既不充分也不必要条件6.(5分)已知函数f(x)=,则f(log27)=()A.B.C.D.7.(5分)在△ABC中,若b=2,A=120°,三角形的面积S=,则三角形外接圆的半径为()A.B.2 C.2 D.48.(5分)已知f(x)=,若f(0)是f(x)的最小值,则t的取值范围为()A.[﹣1,2]B.[﹣1,0]C.[1,2]D.[0,2]9.(5分)已知f(x)=x2+cosx,f′(x)为f(x)的导函数,则y=f′(x)的图象大致是()A.B.C.D.10.(5分)已知x∈R,符号[x]表示不超过x的最大整数,若关于x的方程(a为常数)有且仅有3个不等的实根,则a的取值范围是()A.B.C.D.二、填空题:本大题共5小题,每小题5分,共25分,把答案填在答案纸的相应位置上.11.(5分)如果过曲线y=x4﹣x上点P处的切线平行于直线y=3x+2,那么点P 的坐标为.12.(5分)将函数y=3sin(3x+)的图象向右平移个单位后得到函数的图象.13.(5分)已知向量=(λ,2),=(﹣3,5),且向量与的夹角为锐角,则λ的取值范围是.14.(5分)已知f(x)=,定义f1(x)=f′(x),f2(x)=[f1(x)]′,…,f n+1(x)=[f n(x)]′,n∈N*.经计算f1(x)=,f2(x)=,f3(x)=,…,照此规律,则f n(x)=.15.(5分)下图展示了一个由区间(0,1)到实数集R的映射过程:区间(0,1)中的实数m对应数轴上的点M,如图1;将线段AB围成一个圆,使两端点A、B恰好重合,如图2;再将这个圆放在平面直角坐标系中,使其圆心在y轴上,点A的坐标为(0,1),如图3.图3中直线AM与x轴交与点N(n,0),则m 的象就是n,记作f(m)=n下列说法中正确的命题的序号是(填出所有正确命题的序号).①;②f(x)是奇函数;③f(x)在定义域上单调递增;④f(x)的图象关于点(,0)对称.三、解答题:本大题共6小题,共75分,解答应写出文字说明,证明过程或演算步骤.16.(12分)已知集合A={x|x2﹣3x+2≤0},集合B为函数y=x2﹣2x+a的值域,集合C={x|x2﹣ax﹣4≤0},命题p:A∩B≠∅;命题q:A⊆C.(1)若命题p为假命题,求实数a的取值范围;(2)若命题p∧q为真命题,求实数a的取值范围.17.(12分)已知函数f(x)=sin(x﹣),f′(x)是f(x)的导函数.(1)求函数F(x)=[f′(x)]2﹣f(x)f′(x)的最小值和相应的x值.(2)若f(x)=2f′(x),求的值.18.(12分)已知f(x)为定义在[﹣1,1]上的奇函数,当x∈[﹣1,0]时,函数解析式为f(x)=﹣(b∈R).(1)求b的值,并求出f(x)在[0,1]上的解析式.(2)求f(x)在[﹣1,1]上的值域.19.(12分)设函数f(x)=sin(ωx+)﹣2sin2x+1(ω>0),直线y=﹣与函数f(x)图象相邻两交点的距离为π.(1)求ω的值.(2)在△ABC中,角A、B、C所对的边分别是a、b、c,若点(B,0)是函数y=f(x)图象的一个对称中心,且b=3,求△ABC面积的最大值.20.(13分)5A级景区沂山为提高经济效益,现对某一景点进行改造升级,提高旅游增加值,经过市场调查,旅游增加值y万元与投入x(x≥10)万元之间满足:y=f(x)=ax2+x﹣bln,a、b为常数,当x=10万元,y=19.2万元;当x=50万元,y=74.4万元.(参考数据:In2=0.7,In3=1.1,In5=1.6)(1)求f(x)的解析式.(2)求该景点改造升级后旅游利润T(x)的最大值.(利润=旅游增加值﹣投入)21.(14分)已知函数f(x)=e x﹣x2+a,x∈R的图象在点x=0处的切线为y=bx.(e ≈2.71828).(Ⅰ)求函数f(x)的解析式;(Ⅱ)当x∈R时,求证:f(x)≥﹣x2+x;(Ⅲ)若f(x)>kx对任意的x∈(0,+∞)恒成立,求实数k的取值范围.2014-2015学年山东省潍坊市重点中学高三(上)期中数学试卷(文科)参考答案与试题解析一、选择题:本题共10小题,每小题5分,共50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.(5分)集合A={0,2,a},B={1,a2},若A∪B={0,1,2,4,16},则a的值为()A.0 B.1 C.2 D.4【解答】解:∵A={0,2,a},B={1,a2},A∪B={0,1,2,4,16}∴∴a=4,故选:D.2.(5分)已知函数f(x)=ax5﹣bx3+cx,f(﹣3)=2,则f(3)的值为()A..2 B.﹣2 C.6 D.﹣6【解答】解:∵函数f(x)=ax5﹣bx3+cx,∴f(﹣x)=﹣f(x)∵f(﹣3)=2,∴f(3)=﹣2,故选:B.3.(5分)设α是第二象限角,p(x,4)为其终边上的一点,且cosα=x,则sinα=()A.B.﹣ C.D.﹣【解答】解:由题意|PO|=,所以cosα==x,因为α是第二象限角,解得:x=﹣3,∴sinα=.故选:A.4.(5分)已知向量=(2,3),=(﹣1,2),若与共线,则m的值为()A.B.2 C.D.﹣2【解答】解:由题意可知=m(2,3)+4(﹣1,2)=(2m﹣4,3m+8)=(2,3)﹣2(﹣1,2)=(4,﹣1)∵与共线∴(2m﹣4)×(﹣1)=(3m+8)×4∴m=﹣2故选:D.5.(5分)定义在R上的函数y=f(x)满足,(x﹣)f′(x)>0,任意的x1<x2,都有f(x1)>f(x2)是x1+x2<5的()A.充分不必要条件 B.必要不充分条件C.充分必要条件D.既不充分也不必要条件【解答】解:∵,∴f(x)=f(5﹣x),即函数y=f(x)的图象关于直线x=对称.又因,故函数y=f(x)在(,+∞)上是增函数.再由对称性可得,函数y=f(x)在(﹣∞,)上是减函数.∵任意的x1<x2,都有f(x1)>f(x2),故x1和x2在区间(﹣∞,)上,∴x1+x2<5.反之,若x1+x2<5,则有x2 ﹣<﹣x1,故x1离对称轴较远,x2 离对称轴较近,由函数的图象的对称性和单调性,可得f(x1)>f(x2).综上可得,“任意的x1<x2,都有f(x1)>f(x2)”是“x1+x2<5”的充要条件,故选:C.6.(5分)已知函数f(x)=,则f(log27)=()A.B.C.D.【解答】解:因为,所以=.而,所以.故选:C.7.(5分)在△ABC中,若b=2,A=120°,三角形的面积S=,则三角形外接圆的半径为()A.B.2 C.2 D.4【解答】解:△ABC中,∵b=2,A=120°,三角形的面积S==bc•sinA=c•,∴c=2=b,故B=(180°﹣A)=30°.再由正弦定理可得=2R==4,∴三角形外接圆的半径R=2,故选:B.8.(5分)已知f(x)=,若f(0)是f(x)的最小值,则t的取值范围为()A.[﹣1,2]B.[﹣1,0]C.[1,2]D.[0,2]【解答】解:法一:排除法.当t=0时,结论成立,排除C;当t=﹣1时,f(0)不是最小值,排除A、B,选D.法二:直接法.由于当x>0时,f(x)=x++t在x=1时取得最小值为2+t,由题意当x≤0时,f(x)=(x﹣t)2,若t≥0,此时最小值为f(0)=t2,故t2≤t+2,即t2﹣t﹣2≤0,解得﹣1≤t≤2,此时0≤t≤2,若t<0,则f(t)<f(0),条件不成立,选D.9.(5分)已知f(x)=x2+cosx,f′(x)为f(x)的导函数,则y=f′(x)的图象大致是()A.B.C.D.【解答】解:∵f(x)=x2+cosx,∴f′(x)=x﹣sinx,为奇函数,关于原点对称,排除B,D,设g(x)=f′(x)=x﹣sinx,则g(x)=0,得x=sinx,由图象可知方程有三个根,在图象A正确,故选:A.10.(5分)已知x∈R,符号[x]表示不超过x的最大整数,若关于x的方程(a为常数)有且仅有3个不等的实根,则a的取值范围是()A.B.C.D.【解答】解:由f(x)=﹣a=0得=a,①若x>0,设g(x)=,则当0<x<1,[x]=0,此时g(x)=0,当1≤x<2,[x]=1,此时g(x)=,此时<g(x)≤1,当2≤x<3,[x]=2,此时g(x)=,此时<g(x)≤1,当3≤x<4,[x]=3,此时g(x)=,此时<g(x)≤1,当4≤x<5,[x]=4,此时g(x)=,此时<g(x)≤1,作出函数g(x)的图象,要使f(x)=﹣a有且仅有三个零点,即函数g(x)=a有且仅有三个零点,则由图象可知<a≤,②若x<0,设g(x)=,则当﹣1≤x<0,[x]=﹣1,此时g(x)=﹣,此时g(x)≥1,当﹣2≤x<﹣1,[x]=﹣2,此时g(x)=﹣,此时1≤g(x)<2,当﹣3≤x<﹣2,[x]=﹣3,此时g(x)=﹣,此时1≤g(x)<,当﹣4≤x<﹣3,[x]=﹣4,此时g(x)=﹣,此时1≤g(x)<,当﹣5≤x<﹣4,[x]=﹣5,此时g(x)=﹣,此时1≤g(x)<,作出函数g(x)的图象,要使f(x)=﹣a有且仅有三个零点,即函数g(x)=a有且仅有三个零点,则由图象可知≤a<,综上:<a≤或≤a<,故选:B.二、填空题:本大题共5小题,每小题5分,共25分,把答案填在答案纸的相应位置上.11.(5分)如果过曲线y=x4﹣x上点P处的切线平行于直线y=3x+2,那么点P 的坐标为(1,0).【解答】解:由y=x4﹣x,得到y′=4x3﹣1,又直线y=3x+2的斜率为3,则4x3﹣1=3,解得x=1,把x=1代入曲线方程得:y=0,所以点P的坐标为(1,0).故答案为:(1,0).12.(5分)将函数y=3sin(3x+)的图象向右平移个单位后得到函数y=3sin3x 的图象.【解答】解:将函数y=3sin(3x+)的图象向右平移个单位,所得图象对应的函数解析式为:y=3sin[3(x﹣)+]=3sin3x.故答案为:y=3sin3x.13.(5分)已知向量=(λ,2),=(﹣3,5),且向量与的夹角为锐角,则λ的取值范围是.【解答】解:由题意可得>0,且不共线,即﹣3λ+10>0,且,解得λ∈,故答案为:.14.(5分)已知f(x)=,定义f1(x)=f′(x),f2(x)=[f1(x)]′,…,f n+1(x)=[f n(x)]′,n∈N*.经计算f1(x)=,f2(x)=,f3(x)=,…,照此规律,则f n(x)=.【解答】解:∵f1(x)==,f2(x)==,f3(x)==,…,由此归纳可得:f n(x)=,故答案为:15.(5分)下图展示了一个由区间(0,1)到实数集R的映射过程:区间(0,1)中的实数m对应数轴上的点M,如图1;将线段AB围成一个圆,使两端点A、B恰好重合,如图2;再将这个圆放在平面直角坐标系中,使其圆心在y轴上,点A的坐标为(0,1),如图3.图3中直线AM与x轴交与点N(n,0),则m 的象就是n,记作f(m)=n下列说法中正确的命题的序号是③④(填出所有正确命题的序号).①;②f(x)是奇函数;③f(x)在定义域上单调递增;④f(x)的图象关于点(,0)对称.【解答】解:如图,因为在以为圆心,为半径的圆上运动,对于①当=时.的坐标为(﹣,1﹣),直线AM的方程为所以点N的坐标为(﹣1,0),故f()=﹣1,即①错对于②,因为实数m所在区间(0,1)不关于原点对称,所以f(x)不存在奇偶性.故②错.对于③,当实数m越来越大时,如图直线AM与x轴的交点N(n,0)也越来越往右,即n也越来越大,所以f(x)在定义域上单调递增,即③对.对于④当实数m=时,对应的点在点A的正下方,此时点N(0,0),所以f()=0,再由图形可知f(x)的图象关于点(,0)对称,即④对.故答案为③④.三、解答题:本大题共6小题,共75分,解答应写出文字说明,证明过程或演算步骤.16.(12分)已知集合A={x|x2﹣3x+2≤0},集合B为函数y=x2﹣2x+a的值域,集合C={x|x2﹣ax﹣4≤0},命题p:A∩B≠∅;命题q:A⊆C.(1)若命题p为假命题,求实数a的取值范围;(2)若命题p∧q为真命题,求实数a的取值范围.【解答】解:∵y=x2﹣2x+a=(x﹣1)2+a﹣1≥a﹣1∴A={x|x2﹣3x+2≤0}={x|1≤x≤2},B={y|y≥a﹣1},C={x|x2﹣ax﹣4≤0},(1)由命题p为假命题可得A∩B=∅∴a﹣1>2∴a>3(2)∵命题p∧q为真命题命题∴p,q都为真命题即A∩B≠∅且A⊆C.∴解可得0≤a≤317.(12分)已知函数f(x)=sin(x﹣),f′(x)是f(x)的导函数.(1)求函数F(x)=[f′(x)]2﹣f(x)f′(x)的最小值和相应的x值.(2)若f(x)=2f′(x),求的值.【解答】解:(1)∵f(x)=sin(x﹣)=sinx﹣cosx∴f′(x)=cosx+sinx∵F(x)=[f′(x)]2﹣f(x)f′(x),∴F(x)=(cosx+sinx)2﹣(cosx+sinx)(sinx﹣cosx)=cos2x+sin2x+1=sin(2x+)+1,其最小值为1﹣,此时x=kπ﹣,k∈Z,(2)∵f(x)=2f′(x),∴sinx﹣cosx=2(cosx+sinx),∴tanx=﹣3∴===18.(12分)已知f(x)为定义在[﹣1,1]上的奇函数,当x∈[﹣1,0]时,函数解析式为f(x)=﹣(b∈R).(1)求b的值,并求出f(x)在[0,1]上的解析式.(2)求f(x)在[﹣1,1]上的值域.【解答】解:(1)∵f(x)为定义在[﹣1,1]上的奇函数,且f(x)在x=0处有意义,∴f(0)=0,即f(0)=1﹣b,∴b=1.设x∈[0,1],则﹣x∈[﹣1,0]∴f(﹣x)=,f(x)=2x﹣4x,.所以f(x)=2x﹣4x在[0,1]上的解析式为f(x)=2x﹣4x,(2)当x∈[0,1],f(x)=2x﹣4x=2x﹣(2x)2,∴设t=2x(t>0),则g(t)=﹣t2+t,∵x∈[0,1],t∈[1,2]当t=1时,最大值为1﹣1=0,当t=0时,取最小值﹣2,∴函数在[0,1]上取最小值﹣2,最大值为0,∵f(x)为定义在[﹣1,1]上的奇函数,∴函数在[﹣1,0]上取最小值0,最大值为2,所以f(x)在[﹣1,1]上的值域[﹣2,2]19.(12分)设函数f(x)=sin(ωx+)﹣2sin2x+1(ω>0),直线y=﹣与函数f(x)图象相邻两交点的距离为π.(1)求ω的值.(2)在△ABC中,角A、B、C所对的边分别是a、b、c,若点(B,0)是函数y=f(x)图象的一个对称中心,且b=3,求△ABC面积的最大值.【解答】解:(1)函数f(x)=sin(ωx+)﹣2sin2x+1(ω>0)=sinωxcos+cosωxsin+cosωx=sinωx+cosωx=sin(ωx+),∵函数的最大值为,最小值为﹣,直线y=﹣与函数f(x)图象相邻两交点的距离为π,可得函数的最小正周期为=π,求得ω=2.(2)由于f(x)=sin(2x+),故有f(B)=sin(2B+)=0,∴B=,或B=.若B=,则cosB==,化简可得ac=a2+c2﹣9≥2ac﹣9,∴ac≤9,故△ABC面积ac•sinB的最大值为×=.若B=,则cosB=﹣=,化简可得﹣ac=a2+c2﹣9≥2ac﹣9,∴ac≤9(2﹣),故△ABC面积ac•sinB的最大值为×(2﹣)×=.20.(13分)5A级景区沂山为提高经济效益,现对某一景点进行改造升级,提高旅游增加值,经过市场调查,旅游增加值y万元与投入x(x≥10)万元之间满足:y=f(x)=ax2+x﹣bln,a、b为常数,当x=10万元,y=19.2万元;当x=50万元,y=74.4万元.(参考数据:In2=0.7,In3=1.1,In5=1.6)(1)求f(x)的解析式.(2)求该景点改造升级后旅游利润T(x)的最大值.(利润=旅游增加值﹣投入)【解答】解:(1)由条件可得,解得a=﹣,b=1.则f(x)=﹣+x﹣ln(x≥10).(2)由T(x)=f(x)﹣x=﹣+x﹣ln(x≥10),则T′(x)=﹣+﹣=﹣,令T'(x)=0,则x=1(舍)或x=50,当x∈(10,50)时,T'(x)>0,因此T(x)在(10,50)上是增函数;当x>50时,T'(x)<0,因此T(x)在(50,+∞)上是减函数,故x=50为T(x)的极大值点,也是最大值点,且最大值为24.4万元.即该景点改造升级后旅游利润T(x)的最大值为T(50)=24.4万元.21.(14分)已知函数f(x)=e x﹣x2+a,x∈R的图象在点x=0处的切线为y=bx.(e ≈2.71828).(Ⅰ)求函数f(x)的解析式;(Ⅱ)当x∈R时,求证:f(x)≥﹣x2+x;(Ⅲ)若f(x)>kx对任意的x∈(0,+∞)恒成立,求实数k的取值范围.【解答】解:(Ⅰ)f(x)=e x﹣x2+a,f'(x)=e x﹣2x.由已知,f(x)=e x﹣x2﹣1.…(4分)(Ⅱ)令φ(x)=f(x)+x2﹣x=e x﹣x﹣1,φ'(x)=e x﹣1,由φ'(x)=0,得x=0,当x∈(﹣∞,0)时,φ'(x)<0,φ(x)单调递减;当x∈(0,+∞)时,φ'(x)>0,φ(x)单调递增.∴φ(x)min=φ(0)=0,从而f(x)≥﹣x2+x.…(8分)(Ⅲ)f(x)>kx对任意的x∈(0,+∞)恒成立对任意的x∈(0,+∞)恒成立,令,∴.由(Ⅱ)可知当x∈(0,+∞)时,e x﹣x﹣1>0恒成立,…(10分)令g'(x)>0,得x>1;g'(x)<0,得0<x<1.∴g(x)的增区间为(1,+∞),减区间为(0,1).g(x)min=g(1)=0.∴k<g(x)min=g(1)=e﹣2,∴实数k的取值范围为(﹣∞,e﹣2).…(14分)。
北京市第14中学2014届高三上学期期中考试数学(理)试题 Word版含答案

北京市第14中学2013-2014学年度第一学期期中测试 高三数学(理)试题本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,第Ⅰ卷第1 页至第2 页;第Ⅱ卷第3页至第5页,答题纸第7页至第12 页。
共150分,考试时间120分钟。
请在答题纸第7、9、11 页左侧密封线内书写班级、姓名、准考证号。
考试结束后,将本试卷的答题纸和答题卡一并交回。
第Ⅰ卷(选择题 共40分)一、选择题(本大题共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S = R ,}2|2||{},032|{2<-=≤--=x x B x x x A ,那么集合)(B A C S 等于( )A .}30|{≤<x xB .}21|{<≤-x xC .{|0,3}x x x ≤>或D .}2,1|{≥-<x x x 或2.下列说法错误的是( )A .“1x >”是“1x >”的充分不必要条件B .若p 且q 为假命题,则p q 、均为假命题C .命题“若2430x x -+=,则3x =”的逆否命题是:“若3x ≠,则2430x x -+≠”D .命题p :“x R ∃∈,使得210x x ++<”,则p ⌝:“x R ∀∈,均有210x x ++≥”3.若向量a 、b 满足a +b =(2,-1),a =(1,2),则向量a 与b 的夹角等于( )A .︒135 B . ︒120 C .︒60 D .︒454. 下列函数中,周期为1的奇函数是( ) A.212sin y x π=- B. sin cos y x x ππ=C.tan2y x π= D. sin 23y x ππ=+() 5.若定义在R 上的偶函数)(x f 满足),()2(x f x f =+ 且当]1,0[∈x 时,,)(x x f =则方程0||log )(3=-x x f 的根的个数是( )A .2B .3C .4D .66.设函数ax x x f m +=)(的导函数'()21f x x =+,则数列⎭⎬⎫⎩⎨⎧)(1n f 的前n 项和为( ) A.1n n + B. 12++n n C.1-n nD.nn 1+ 7.已知△ABC 中,︒=∠30A ,AB ,BC 分别是23+,23-的等差中项与等比中项,则△ABC 的面积等于( )A .23 B .43 C .23或3 D .23或438. 对于下列命题:①已知i 是虚数单位,函数1,(0)()1,(0)x ii x f x ia a x +⎧⋅>⎪=-⎨⎪-≤⎩在R 上连续,则实数a=2.②五本书排成一排,若A 、B 、C 三本书左右顺序一定(不一定相邻),那么不同排法有3333A A ⋅③如图,⊙O 中的弦AB 与直径CD 相交于点p ,M 为DC 延长线上一点,MN 为⊙O 的切线,N 为切点,若AP =8, PB =6, PD =4, MC =6,则MN 的长为332④在极坐标系(ρ,θ)(0 ≤θ <2π)中,曲线ρ=2sin θ 与1cos -=θρ交点的极坐标为3)4π⑤设2014cos ,()n n xdx x xπ=-⎰则二项式的展开式的常数项为6其中假命题的序号是( ) A .②⑤ B . ②③ C . ② D . ①④第Ⅱ卷(共110分)二、填空题:(本大题每小题5分,满分30分) 9.若33sin()25-=πα,且α的终边过点(),2P x ,则x = ;tan()πα+= . 10.已知数列}{n a 是等差数列,其前n 项和为S n ,12,2344==S a . 则数列}{n a 的通项公式=n a ;n=时,S n 最大.11.函数2sin cos cos sin ++=ϕωϕωx A x A y)20,0,0(πϕω<<>>A 的图象如右,则ω=______,ϕ=______.12.函数())1,0(13log ≠>-+=a a x y a 的图象恒过定点A ,且点A 在直线01=++ny mx 上,其中0>mn ,则nm 21+的最小值为 . 13.在正方形ABCD 中,已知AB =2,M 为BC 的中点,若N 为正方形内(含边界)任意一点,则AM ·AN 的最大值为 .14. 已知函数2,(0)()21,(0)x e x f x ax x -⎧-≤=⎨->(a 是常数且0>a ).对于下列命题:三、解答题(本大题共6小题,共80分.) 15. (本小题满分13分)在数列{}n a 中,13a =,121n n a a n -=--+ *(2)n n ≥∈N ,且 . (Ⅰ)证明:数列{}n a n +是等比数列; (Ⅱ)求{}n a 的通项公式; (Ⅲ)求数列{}n a 的前n 项和n S .16.(本小题满分13分)盒内有大小相同的9个球,其中2个红色球,3个白色球,4个黑色球.规定取出1个红色球得1分,取出1个白色球得0分,取出1个黑色球得1-分.现从盒内一次性取3个球. (Ⅰ)求取出的3个球得分之和恰为1分的概率;(Ⅱ)设ξ为取出的3个球中白色球的个数,求ξ的分布列和数学期望. 17.(本小题共13分)已知向量(sin , cos )x x =a ,(cos ,sin 2cos )x x x =-b ,24ππ<<-x .(Ⅰ)若a b ∥,求x ;(Ⅱ)设()f x =⋅a b ,求()f x 的单调减区间;(Ⅲ)函数()f x 经过平移后所得的图象对应的函数是否能成为奇函数?如果是,说出平移方案;如果否,说明理由.18. (本小题共13分)已知函数.)2ln()(2c bx x x x f ++-+=(Ⅰ)若函数 f (x )在点x=1处的切线与直线0273=++y x 垂直,且f (-1)=0,求函数f (x )在区间[0,3]上的最小值;(Ⅱ)若f (x )在区间[0,1]上为单调减函数,求b 的取值范围. 19.(本小题共14分)设函数2()(1)2ln(1)f x x x =+-+(Ⅰ)若在定义域内存在0x ,而使得不等式0()0f x m -≤能成立,求实数m 的最小值; (Ⅱ)若函数2()()g x f x x x a =---在区间[]0,2上恰有两个不同的零点,求实数a 的取值范围20. (本小题共14分)已知)(x f 是定义在R 上的函数,1)1(=f ,且∀R x x ∈21,,总有1)()()(2121++=+x f x f x x f 恒成立.(Ⅰ)记()()1g x f x =+,求证:()g x 是奇函数; (Ⅱ)对∀*N n ∈,有)(1n f a n =,1)21(1+=+n n f b ,记n n nb c a =,求{}n c 的前n 项和n S ;(Ⅲ)求n n n a a a n F 221)(+++=++ ),2(N n n ∈≥的最小值.高三数学期中测试答案及评分标准(理科)一、选择题:本大题每小题5分,满分40分. CBAB CADC 二、填空题:本大题每小题5分,满分30分.9. 32-,43- 10. n a n -=211;n=5 11.ω=3,ϕ=3π12. 8 13. 6 14.①③④ 三、解答题:本大题6小题,满分80分15.(13分)解: (Ⅰ)11121111-=-++--=-++---n a n a n a n a n n n n ,由定义知数列{}na n +是等比数列;…5分(Ⅱ)因为数列{}n a n +是等比数列,公比为-1,首项为4, 则4)1(1⋅-+-=-n n n a *∈N n …….8分(Ⅲ) (1),(2,)2(1)4,(21,)2n n nn k k N S n n n k k N**+⎧-=∈⎪⎪=⎨+⎪-+=-∈⎪⎩ …13分17. ( 13分)解:(I )若a b ∥,则2sin (sin 2cos )cos ,x x x x ⋅-=……1分sin 2cos 2,x x -=即tan 21x ∴=-…………2分又∵24ππ<<-x , ∴ππ<<-x 22,∴42π-=x 或43π, 8π-=x 或83π………4分(II )2()2sin cos 2cos sin2cos21=2sin(2)14f x x π=⋅⋅--a b =x x -x =x -x -2()2sin cos 2cos sin2cos2)14x x π=⋅⋅--a b =x x -x =x -x -………7分令Z k k x k ∈+≤-≤+,2234222πππππ得,Z k k x k ∈+≤≤+,8783ππππ,又24ππ≤≤-x ∴)8,4(ππ--和)2,83(ππ是()f x 的单调减区间………11分 (Ⅲ)是,将函数()f x 的图象向上平移1个单位,再向左平移,8k k N +∈ππ个单位或向右平移7,8k k N +∈ππ个单位,即得函数()2g x x =的图象,而()g x 为奇函数………13分18. (13分)解:(1).221)(b x x x f +-+=' (2分) 因为与直线0273=++y x 垂直的直线的斜率为4,37)1(,37=='b f 得令又f (-1)=ln (2-1)-1-4+c =0,所以c =5 f (x )=ln (x +2)-x 2+4x -5,4221)(+-+='x x x f (6分) 由223,0)(=='x x f 得 当]223,0[∈x 时,f ′(x )≥0,f (x )单调递增 当]3,223[∈x 时,f ′(x )≤0,f (x )单调递减(8分) 又f (0)=ln2+5,f (3)=ln5+8,所以f (x )在[0,3]最小值为ln2+5 (10分) (Ⅱ)因为f (x )是减函数所以]1,0[2120221)(∈+-≤≤+-+='x x x b b x x x f 对即恒成立(12分) 因为212+-x x 在[0,1]上单调递增 所以(2x -21+x )min =-21所以当b ≤-21时,f (x )在区间[0,1]上单调递减(13分)19. (14分)解:(Ⅰ)要使得不等式0()0f x m -≤能成立,只需min ()m f x ≥。
山东省济南一中等四校2014届高三上学期期中联考 文科数学 含答案
济南市高三部分学校调研考试(11月)数学(文科)本试卷共4页,分第I 卷(选择题)和第Ⅱ卷(非选择题)两部分.共150分,考试时间120分钟,第I 卷(选择题共60分)注意事项:l .答第1卷前,考生务必将自己的姓名、准考证号、考试科目用铅笔涂写在答题卡上.2.每题选出答案后,用2B 铅笔把答题卡对应题目的答案标号涂黑.如需改动,用橡皮擦干净后,再改涂其他答案标号.一、选择题(本题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.已知全集U={0,1,2,3,4),集合A={1,2,3),B={2,4},则()U C A B 为A.{1,2,4) B 。
{2,3,4) C 。
{0,2,4) D.{0,2,3,4)2.设z ∈R,则x=l 是21x =的A .充分不必要条件B .必要不充分条件C .充要条件D .既不充分也不必要条件3.已知函数log3,0()2,0x x f x x x >⎧=⎨≤⎩,则1(())9f f = A.4 B .14 C .一4 D .14-4.设平面向量(1,2),(3,1)a b ==-,则2a b +=A .5B .6 C 。
17 D. 34 5.已知数列{}n a 的前n 项和为n S ,且221n Sn =-,则3a 等于 A .-10 B .6 C .10 D .14 6.函数ln x xy x =的图像可能是7.为了得到函数sin 2y x =的图象,只需把函数sin(2)6y x π=+的图象 A. 向左平移6π个单位 B .向左平移12π个单位 C .向右平移6π个单位 D .向右平移12π个单位 8.已知两点(1,0),(1,3)A B -,向量(21,2)a k =-,若AB a ⊥,则实数k 的值为A. —2 B .-l C .1 D .29.等差数列{}n a 公差为2,若134,,a a a 成等比数列,则2a 等于 A .-4 B .-6 C .-8 D .-1010.设357log 6,log 10,log 14a b c ===,则A. c>b 〉aB.b 〉c 〉a C 。
百强重点名校高考备考_浙江省杭州市2014届高三上学期期中七校联考(完美整理版)
浙江省杭州市2014届高三上学期期中七校联考未1203 1633浙江省杭州市2014届高三上学期期中七校联考语文试题一.语言文字运用(共24分,其中选择题每小题3分)1、下列词语中加点的字,注音有错误的一组是()A.跻身(jī)粗糙(cāo)锲而不舍(qiâ)未雨绸缪(móu)B.整饬(chì)分泌(bì)孜孜不倦(zī)穷形尽相(xiàng)C.伺候(cì)对峙(zhì)胼手胝足(pián)外强中干(gān)D.修葺(qì)城垣(yuán)栉风沐雨(zhì)沆瀣一气(xiâ)2、下列各项中,没有错别字的一组是()A.文身九州人情是故再接再厉B.安装杀戮罄竹难书金榜提名C.诡谲调剂洁白无瑕炙手可热D.桥礅坐镇世外桃源天翻地覆3、下列各句中,加点的词语运用错误的一项是()A.杭州要解决交通拥堵问题难度非常大,但有关部门仍应勉力为之。
B.按照新课改精神,教师应当以培养学生的综合能力为根本,当学生遇到疑难问题时,教师要善于启发诱导,不要越俎代庖。
C.当今中国,各行各业对外语的要求已经到了令人啼笑皆非的地步,外语考试更是五花八门。
D.刘欢自幼喜爱音乐,他勤学苦练,朝歌夜弦,终于获得成功,奥运开幕式上的一曲《我和你》更是让人回味无穷。
4、下列各句中,没有语病的一句是()A.早在2002年,我国就明令禁止“瘦肉精”使用于养殖业,然而这个餐桌上的“毒瘤”多年来未被根除,频繁出现的“瘦肉精”事件,令人们再添食品安全之忧。
B.浙江财智女人会曾经接了万阜乡的23个“留守孩子”到杭州游玩,圆了山里娃去山外看世界的梦想;现在,她们又赶到山乡万阜,不仅是爱心的不断延续,更是一次“回访”。
C.英国伦敦高等法院日前裁决,《达·芬奇密码》的丹·布朗被指控抄袭另一畅销书著作权的罪名不能成立。
湖北省部分重点中学2014届高三上学期期中联考数学(理)试题 Word版含答案
2013年秋季湖北省部分重点中学期中联考高三数学试卷(理科)命题学校:红安一中 命题教师:黄孝银 审题教师: 刘中帅考试时间:2013年11月7日下午2:30~4:30 试卷满分:150分一.选择题:本大题共10小题,每小题5分,满分50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.若全集{}{}0,1,2,32U U C A ==且,则集合A 的真子集共有 A .3个B .5个C .7个D .8个 2. cos 48sin108cos 42cos 72+=A .B .12C .sin114︒D .cos114︒ 3.下列各组命题中的假命题是A .1,20x x R -∀∈>B .2,(1)0x N x +∀∈->C .,lg 1x R x ∃∈<D .,tan 2x R x ∃∈= 4.右图是函数sin()(0,0,)2y A x A πωφωφ=+>><在区间5,66ππ⎡⎤-⎢⎥⎣⎦上的图象,为了得到这个函数的图象,只要将sin ()y x x R =∈的图象上所有的点A .向左平移3π个单位长度,再把所得各点的横坐标缩短到 原来的12倍,纵坐标不变 B .向左平移3π个单位长度,再把所得各点的横坐标伸长到原来的2倍,纵坐标不变 C .向左平移6π个单位长度,再把所得各点的横坐标缩短到原来的12倍,纵坐标不变D .向左平移6π个单位长度,再把所得各点的横坐标伸长到原来的2倍,纵坐标不变5.已知等比数列{}n a 中,公比1q >,且168a a +=,3412a a =,则116aa =A .2B .3C .6D .3或66.若a b c <<,则函数()()()()()()()f x x a x b x b x c x c x a =--+--+--的两个零点分别位于区间A.(),a b 和(),b c 内B.(),a -∞和(),a b 内C.(),b c 和(),c +∞内D.(),a -∞和(),c +∞内6π-56π7.设a ,b ,c 均为正数,且122log aa =, 121log 2b b ⎛⎫= ⎪⎝⎭, 21log 2cc ⎛⎫= ⎪⎝⎭则A .a <b <cB .c <b <aC .c <a <bD .b <a <c8.P 是ABC ∆所在平面上的一点,满足02=++PC PB PA ,若ABC ∆的面积为1,则ABP ∆的面积为 A. 1 B.2 C.12 D.139.从1开始的自然数按如图所示的规则排列,现有一个三角形框架在图中上下或左右移动,使每次恰有九个数在此三角形内,则这九个数的和可以为A .2097B .2264C .2111D .201210.我们把形如()()x y f x ϕ=的函数称为幂指函数,幂指函数在求导时,可以利用对数法:在函数解析式两边求对数得ln ()ln ()y x f x ϕ=,两边求导数,得()()ln ()()()y f x x f x x y f x ϕϕ'''=+,于是()()()()ln ()()()x f x y f x x f x x f x ϕϕϕ'⎡⎤''=+⎢⎥⎣⎦,运用此方法可以探求得函数1xy x =的一个单调递增区间是 A .(),4e B .11,e e e e ⎛⎫-+ ⎪⎝⎭C .(1,1)e e -+D . (0,)e 二. 填空题:本大题共5小题,每小题5分,满分25分.把答案填在答题卡的横线上. 11.若()f x '为()f x 的导函数,且()f x =,则11()f x dx -'=⎰ ▲ .12.已知tan()3πα-=-22sin cos 3cos 2sin αααα-= ▲ .13.如右图,测量河对岸的塔高AB 时,可以选与塔底B 在同一水平面内的两个测点C 与D ,测得15BCD ︒∠=,30BDC ︒∠=,30CD =米,并在点C 测得塔顶A 的仰角为60︒,则塔高AB = ▲ 米.14.已知函数f (x )=|x +11x-|,则关于x 的方程2()6()0f x f x c -+= (c ∈R)有6个不同实数解的充要条件是 ▲ .15. (1)若指数函数xy a =的图象与直线y x =相切,则a = ▲ ;(2)如果函数()log xa f x a x =-不存在零点,则a 的取值范围为 ▲ .三.解答题:本大题共6小题,共75分.解答应写出文字说明、证明过程或演算步骤. 16. (本小题满分12分)1 2 3 4 5 6 7 8910 11 12 13 14 15 16 1718 19 20 21 22 23 24 25 26 2728 29 30 31 32 33 34 35 36 37 38 39 40…已知集合231{|1,[,2]},{|||1}22A y y x x xB x x m ==-+∈-=-≥;命题:p x A ∈,命题:q x B ∈,并且命题p 是命题q 的充分条件,求实得数m 的取值范围.17.(本小题满分12分)已知sin 2()sin xf x x x=+(1)求()f x 的最大值及取得最大值时x 的取值的集合; (2)在△ABC 中,a b c 、、分别是角A ,B ,C所对的边,若a =,且对()f x 的定义域内的每一个x ,都有()()f x f A ≤恒成立,求AB AC ⋅的最大值.18.(本小题满分12分)已知数列{}n a 满足:121,(0).a a a a ==>数列{}n b 满足1(*)n n n b a a n N +=∈. (1)若{}n a 是等差数列,且312b =,求a 的值及{}n a 的通项公式; (2)若{}n a 是等比数列,求{}n b 的前项和n S ;(3)当{}n b 是公比为1a -的等比数列时,{}n a 能否为等比数列?若能,求出a 的值;若不能,请说明理由.19.(本小题满分12分)某校学生社团心理学研究小组在对学生上课注意力集中情况的调查研究中,发现其注意力指数p 与听课时间t 之间的关系满足如图所示的曲线.当(0,14]t ∈时,曲线是二次函数图象的一部分,当[14,40]t ∈时,曲线是函数log (5)83(0a y x a =-+>且1)a ≠图象的一部分.根据专家研究,当注意力指数p ≥80时,听课效果最佳. (1) 试求()p f t =的函数关系式;(2) 老师在什么时段内安排核心内容能使得学生听课效果最佳?请说明理由.。
2014-2015年河南省名校高三(上)期中数学试卷及参考答案(理科)
2014-2015学年河南省名校高三(上)期中数学试卷(理科)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.请把答案填涂在答题纸的相应位置.1.(5分)在复平面内,复数Z=+i2015对应的点位于()A.第四象限B.第三象限C.第二象限D.第一象限2.(5分)已知集合M={x|y=lg},N={y|y=x2+2x+3},则(∁R M)∩N=()A.{x|0<x<1}B.{x|x>1}C.{x|x≥2}D.{x|1<x<2}3.(5分)已知sin2α=﹣,α∈(﹣,0),则sinα+cosα=()A.B.﹣ C.﹣ D.4.(5分)设f(x)是定义在R上的奇函数,当x<0时,f(x)=x﹣e﹣x(e为自然数的底数),则f(ln6)的值为()A.ln6+6 B.ln6﹣6 C.﹣ln6+6 D.﹣ln6﹣65.(5分)已知向量+=(2,﹣8),﹣=(﹣8,16),则与夹角的余弦值为()A.B.C.D.6.(5分)执行如图所示的程序框图,会输出一列数,则这个数列的第3项是()A.870 B.30 C.6 D.37.(5分)函数f(x)=sin(2x+φ)(|φ|<π)的图象向左平移个单位后关于原点对称,则函数f(x)在[0,]上的最小值为()A.﹣B.﹣ C.D.8.(5分)某几何体的三视图如图所示,且该几何体的体积是3,则正视图中的x的值是()A.2 B.C.D.39.(5分)已知数列{a n}为等差数列,{b n}为等比数列,且满足:a1003+a1013=π,b6•b9=2,则tan=()A.1 B.﹣1 C.D.10.(5分)如图,把圆周长为1的圆的圆心C放在y轴上,顶点A(0,1),一动点M从A开始逆时针绕圆运动一周,记=x,直线AM与x轴交于点N(t,0),则函数t=f(x)的图象大致为()A.B. C.D.11.(5分)已知函数f(x)=,若a、b、c互不相等,且f(a)=f(b)=f(c),则a+b+c的取值范围是()A.(1,2014)B.(1,2015)C.(2,2015)D.[2,2015]12.(5分)已知定义在R上的函数f(x)满足f(x+1)=f(1﹣x)且在[1,+∞)上是增函数,不等式f(ax+2)≤f(x﹣1)对任意x∈[,1]恒成立,则实数a 的取值范围是()A.[﹣3,﹣1]B.[﹣2,0]C.[﹣5,﹣1]D.[﹣2,1]二、填空题:本大题共4小题,每小题5分,满分20分.请把答案填在答题纸的相应位置.13.(5分)已知tan(θ﹣π)=2,则sin2θ+sinθcosθ﹣2cos2θ+3的值为.14.(5分)图中阴影部分的面积等于.15.(5分)设正实数x,y,z满足x2﹣3xy+4y2﹣z=0,则当取得最大值时,+﹣的最大值为.16.(5分)设f(x)是定义在R上的偶函数,且对于∀x∈R恒有f(x+1)=f(x ﹣1),已知当X∈[0,1]时,f(x)=()1﹣x,则(1)f(x)的周期是2;(2)f(x)在(1,2)上递减,在(2,3)上递增;(3)f(x)的最大值是1,最小值是0;(4)当x∈(3,4)时,f(x)=()x﹣3其中正确的命题的序号是.三、解答题:本大题共5小题,共70分.解答应写出文字说明,证明过程或演算步骤.17.(12分)设函数.(Ⅰ)求f(x)的最大值,并写出使f(x)取最大值是x的集合;(Ⅱ)已知△ABC中,角A,B,C的对边分别为a,b,c.若.求a的最小值.18.(12分)已知数列{a n}的前n项和为S n,S n=2a n﹣2.(1)求数列{a n}的通项公式;(2)设b n=log2a n,c n=,记数列{c n}的前n项和T n,若对n∈N*,T n≤k (n+4)恒成立,求实数k的取值范围.19.(12分)如图,在斜三棱柱ABC﹣A1B1C1中,O是AC的中点,A1O⊥平面ABC,∠BCA=90°,AA1=AC=BC.(Ⅰ)求证:A1B⊥AC1;(Ⅱ)求二面角A﹣BB1﹣C的余弦值.20.(12分)设椭圆+=1(a>b>0)的左、右焦点分别为F1、F2,右顶点为A,上顶点为B,已知|AB|=|F1F2|.(Ⅰ)求椭圆的离心率;(Ⅱ)设P为椭圆上异于其顶点的一点,以线段PB为直径的圆经过点F1,经过原点O的直线l与该圆相切,求直线l的斜率.21.(12分)已知函数f(x)=x2+a(x+lnx),x>0,a∈R是常数.(1)求函数y=f(x)的图象在点(1,f(1))处的切线方程;(2)若函数y=f(x)图象上的点都在第一象限,试求常数a的取值范围;(3)证明:∀a∈R,存在ξ∈(1,e),使f′(ξ)=.四、请考生在第22、23、24三题中任选一题作答,如果多做,则按所做的第22题计分.【选修4-1:几何证明选讲】22.(10分)选修4﹣1:几何证明选讲如图,已知圆上的,过C点的圆的切线与BA的延长线交于E点.(Ⅰ)证明:∠ACE=∠BCD;(Ⅱ)若BE=9,CD=1,求BC的长.【选修4-4:坐标系与参数方程】23.已知直线l:(t为参数)经过椭圆C:(φ为参数)的右焦点F.(Ⅰ)求m的值;(Ⅱ)设直线l与椭圆C交于A,B两点,求|FA|•|FB|的最大值与最小值.【选修4-5:不等式选讲】24.已知函数f(x)=|2x+1|+|2x﹣3|(1)求不等式f(x)≤6的解集;(2)若关于x的不等式f(x)≤|a﹣2|的解集非空,求实数a的取值范围.2014-2015学年河南省名校高三(上)期中数学试卷(理科)参考答案与试题解析一、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.请把答案填涂在答题纸的相应位置.1.(5分)在复平面内,复数Z=+i2015对应的点位于()A.第四象限B.第三象限C.第二象限D.第一象限【解答】解:复数Z=+i2015=﹣i=﹣i=﹣.复数对应点的坐标(),在第四象限.故选:A.2.(5分)已知集合M={x|y=lg},N={y|y=x2+2x+3},则(∁R M)∩N=()A.{x|0<x<1}B.{x|x>1}C.{x|x≥2}D.{x|1<x<2}【解答】解:集合M={x|y=lg},,解得:0<x<1,M={x|0<x<1},∴∁R M={x|x≤0或x≥1}N={y|y=x2+2x+3}={y|y≥2},(∁R M)∩N=[2,+∞)故选:C.3.(5分)已知sin2α=﹣,α∈(﹣,0),则sinα+cosα=()A.B.﹣ C.﹣ D.【解答】解:∵α∈(﹣,0),∴sinα+cosα>0,∴(sinα+cosα)2=1+sin2α=,∴sinα+cosα=,故选:A.4.(5分)设f(x)是定义在R上的奇函数,当x<0时,f(x)=x﹣e﹣x(e为自然数的底数),则f(ln6)的值为()A.ln6+6 B.ln6﹣6 C.﹣ln6+6 D.﹣ln6﹣6【解答】解:∵当x<0时,f (x)=x﹣e﹣x,∴f(﹣ln6)=﹣ln6﹣e ln6=﹣ln6﹣6,又∵f (x)是定义在R上的奇函数,∴f(ln6)=﹣f(﹣ln6)=ln6+6故选:A.5.(5分)已知向量+=(2,﹣8),﹣=(﹣8,16),则与夹角的余弦值为()A.B.C.D.【解答】解:由向量,,得=(﹣3,4),=(5,﹣12),所以||=5,||=13,=﹣63,即与夹角的余弦值cosθ==.故选:B.6.(5分)执行如图所示的程序框图,会输出一列数,则这个数列的第3项是()A.870 B.30 C.6 D.3【解答】解:当N=1时,A=3,故数列的第1项为3,N=2,满足继续循环的条件,A=3×2=6;当N=2时,A=6,故数列的第2项为6,N=3,满足继续循环的条件,A=6×5=30;当N=3时,A=30,故数列的第3项为30,故选:B.7.(5分)函数f(x)=sin(2x+φ)(|φ|<π)的图象向左平移个单位后关于原点对称,则函数f(x)在[0,]上的最小值为()A.﹣B.﹣ C.D.【解答】解:函数f(x)=sin(2x+φ)图象向左平移个单位得,由于函数图象关于原点对称,∴函数为奇函数,又|φ|<π,∴,得,∴,由于,∴0≤2x≤π,∴,当,即x=0时,.故选:A.8.(5分)某几何体的三视图如图所示,且该几何体的体积是3,则正视图中的x的值是()A.2 B.C.D.3【解答】解:根据三视图判断几何体为四棱锥,其直观图是:V==3⇒x=3.故选:D.9.(5分)已知数列{a n}为等差数列,{b n}为等比数列,且满足:a1003+a1013=π,b6•b9=2,则tan=()A.1 B.﹣1 C.D.【解答】解:数列{a n}为等差数列,{b n}为等比数列,且满足:a1003+a1013=π,b6•b9=2,所以a1+a2015=a1003+a1013=π,b7•b8=b6•b9=2,所以tan=tan=.故选:D.10.(5分)如图,把圆周长为1的圆的圆心C放在y轴上,顶点A(0,1),一动点M从A开始逆时针绕圆运动一周,记=x,直线AM与x轴交于点N(t,0),则函数t=f(x)的图象大致为()A.B. C.D.【解答】解:当x由0→时,t从﹣∞→0,且单调递增,由→1时,t从0→+∞,且单调递增,∴排除A,B,C,故选:D.11.(5分)已知函数f(x)=,若a、b、c互不相等,且f(a)=f(b)=f(c),则a+b+c的取值范围是()A.(1,2014)B.(1,2015)C.(2,2015)D.[2,2015]【解答】解:作出函数的图象如图,直线y=m交函数图象于如图,不妨设a<b<c,由正弦曲线的对称性,可得(a,m)与(b,m)关于直线x=对称,因此a+b=1,当直线y=m=1时,由log2014x=1,解得x=2014,即x=2014,∴若满足f(a)=f(b)=f(c),(a、b、c互不相等),由a<b<c可得1<c<2014,因此可得2<a+b+c<2015,即a+b+c∈(2,2015).故选:C.12.(5分)已知定义在R上的函数f(x)满足f(x+1)=f(1﹣x)且在[1,+∞)上是增函数,不等式f(ax+2)≤f(x﹣1)对任意x∈[,1]恒成立,则实数a 的取值范围是()A.[﹣3,﹣1]B.[﹣2,0]C.[﹣5,﹣1]D.[﹣2,1]【解答】解:定义在R上的函数f(x)满足f(x+1)=f(1﹣x)且在[1,+∞)上是增函数,可得出函数图象关于x=1对称,且函数在(﹣∞,1)上减,由此得出自变量离1越近,函数值越小,综合考虑四个选项,四个选项中的集合中都有﹣1,0不存在于A,C两个选项的集合中,B中集合是D中集合的子集,故可通过验证a的值取0与1时两种情况得出正确选项.当a=0时,不等式f(ax+2)≤f(x﹣1)变为f(2)≤f(x﹣1),有函数f(x)图象特征可得出|2﹣1|≤|x﹣1﹣1|,解得x≥3或x≤1,满足,不等式f(ax+2)≤f(x﹣1)对任意x∈[,1]恒成立,由此排除A,C两个选项.当a=1时,不等式f(ax+2)≤f(x﹣1)变为f(x+2)≤f(x﹣1),有函数f(x)图象特征可得出|x+2﹣1|≤|x﹣1﹣1|,解得x≤,不满足不等式f(ax+2)≤f (x﹣1)对任意x∈[,1]恒成立,由此排除D选项.综上可知,B选项是正确的.故选:B.二、填空题:本大题共4小题,每小题5分,满分20分.请把答案填在答题纸的相应位置.13.(5分)已知tan(θ﹣π)=2,则sin2θ+sinθcosθ﹣2cos2θ+3的值为.【解答】解:∵已知tan(θ﹣π)=2=tanθ,则sin2θ+sinθcosθ﹣2cos2θ+3=+3=+3=+3=,故答案为.14.(5分)图中阴影部分的面积等于1.【解答】解:根据题意,该阴影部分的面积为=x3=(13﹣03)=1故答案为:115.(5分)设正实数x,y,z满足x2﹣3xy+4y2﹣z=0,则当取得最大值时,+﹣的最大值为1.【解答】解:由正实数x,y,z满足x2﹣3xy+4y2﹣z=0,∴z=x2﹣3xy+4y2.∴===1,当且仅当x=2y>0时取等号,此时z=2y2.∴+﹣==≤1,当且仅当y=1时取等号,即+﹣的最大值是1.故答案为1.16.(5分)设f(x)是定义在R上的偶函数,且对于∀x∈R恒有f(x+1)=f(x ﹣1),已知当X∈[0,1]时,f(x)=()1﹣x,则(1)f(x)的周期是2;(2)f(x)在(1,2)上递减,在(2,3)上递增;(3)f(x)的最大值是1,最小值是0;(4)当x∈(3,4)时,f(x)=()x﹣3其中正确的命题的序号是(1)(2)(4).【解答】解:(1)∵对任意的x∈R恒有f(x+1)=f(x﹣1),∴f(x+2)=f[(x+1)﹣1]=f(x),即2是f(x)的周期,(1)正确;(2)∵x∈[0,1]时,f(x)=()1﹣x=2x﹣1为增函数,又f(x)是定义在R上的偶函数,∴f(x)在区间[﹣1,0]上单调递减,又其周期T=2,∴f(x)在(1,2)上递减,在(2,3)上递增,(2)正确;(3)由(2)x∈[0,1]时,f(x)=()1﹣x=2x﹣1为增函数,f(x)在区间[﹣1,0]上单调递减,且其周期为2可知,f (x )max =f (1)=21﹣1=20=1,f (x )min =f (0)=20﹣1=,故(3)错误; (4)当x ∈(3,4)时,x ﹣4∈(﹣1,0),4﹣x ∈(0,1), ∴f (4﹣x )=()1﹣(4﹣x )=,又f (x )是周期为2的偶函数,∴f (4﹣x )=f (x )=,(4)正确.综上所述,正确的命题的序号是(1)(2)(4), 故答案为:(1)(2)(4).三、解答题:本大题共5小题,共70分.解答应写出文字说明,证明过程或演算步骤.17.(12分)设函数.(Ⅰ)求f (x )的最大值,并写出使f (x )取最大值是x 的集合; (Ⅱ)已知△ABC 中,角A ,B ,C 的对边分别为a ,b ,c .若.求a 的最小值.【解答】解:(Ⅰ)f (x )=cos (2x ﹣)+2cos 2x=(cos2xcos +sin2xsin)+(1+cos2x )=cos2x ﹣sin2x +1=cos (2x +)+1,(3分)∵﹣1≤cos (2x +)≤1,即cos (2x +)最大值为1,∴f (x )的最大值为2,(4分) 要使f (x )取最大值,cos (2x +)=1,即2x +=2kπ(k ∈Z ),解得:x=kπ﹣(k ∈Z ),则x 的集合为{x |x=kπ﹣(k ∈Z )};(6分)(Ⅱ)由题意,f (B +C )=cos [2(B +C )+]+1=,即cos (2π﹣2A +)=,化简得:cos (2A ﹣)=,(8分)∵A ∈(0,π),∴2A ﹣∈(﹣,),则有2A﹣=,即A=,(10分)在△ABC中,b+c=2,cosA=,由余弦定理,a2=b2+c2﹣2bccos=(b+c)2﹣3bc=4﹣3bc,(12分)由b+c=2知:bc≤=1,当且仅当b=c=1时取等号,∴a2≥4﹣3=1,则a取最小值1.(14分)18.(12分)已知数列{a n}的前n项和为S n,S n=2a n﹣2.(1)求数列{a n}的通项公式;(2)设b n=log2a n,c n=,记数列{c n}的前n项和T n,若对n∈N*,T n≤k (n+4)恒成立,求实数k的取值范围.【解答】解:(1)当n=1时,a1=S1=2a1﹣2,解得a1=2.当n≥2时,a n=S n﹣S n﹣1=2a n﹣2﹣(2a n﹣1﹣2)=2a n﹣2a n﹣1,化为a n=2a n﹣1,∴数列{a n}是以2为公比的等比数列,∴.(2)∵b n=log2a n==n,∴c n==.∴数列{c n}的前n项和T n=+…+==.∵对n∈N*,T n≤k(n+4)恒成立,∴,化为=.∵n++5=9,当且仅当n=2时取等号.∴,∴.∴实数k的取值范围是.19.(12分)如图,在斜三棱柱ABC﹣A1B1C1中,O是AC的中点,A1O⊥平面ABC,∠BCA=90°,AA1=AC=BC.(Ⅰ)求证:A1B⊥AC1;(Ⅱ)求二面角A﹣BB1﹣C的余弦值.【解答】解:(Ⅰ)因为A1O⊥平面ABC,所以A1O⊥BC.又BC⊥AC,所以BC⊥平面A1ACC1,所以AC1⊥BC.…(2分)因为AA1=AC,所以四边形A1ACC1是菱形,所以AC1⊥A1C.所以AC1⊥平面A1BC,所以A1B⊥AC1.…(5分)(Ⅱ)以OC为单位长度,建立如图所示的空间直角坐标系O﹣xyz,则A(0,﹣1,0),B(2,1,0),C(0,1,0),C1(0,2,).=(2,2,0),=(0,1,),设=(x,y,z)是面ABB1的一个法向量,则•=0,•=0,即,取x=,得=(,﹣,1).同理面CBB1的一个法向量为=(0,﹣,1).…(10分)因为cos<>=.二面角A﹣BB 1﹣C是锐二面角,所以二面角A﹣BB1﹣C的余弦值.…(12分)20.(12分)设椭圆+=1(a>b>0)的左、右焦点分别为F1、F2,右顶点为A,上顶点为B,已知|AB|=|F1F2|.(Ⅰ)求椭圆的离心率;(Ⅱ)设P为椭圆上异于其顶点的一点,以线段PB为直径的圆经过点F1,经过原点O的直线l与该圆相切,求直线l的斜率.【解答】解:(Ⅰ)设椭圆的右焦点为F2(c,0),由|AB|=|F1F2|,可得,化为a2+b2=3c2.又b2=a2﹣c2,∴a2=2c2.∴e=.(Ⅱ)由(Ⅰ)可得b2=c2.因此椭圆方程为.设P(x0,y0),由F1(﹣c,0),B(0,c),可得=(x0+c,y0),=(c,c).∵,∴=c(x0+c)+cy0=0,∴x0+y0+c=0,∵点P在椭圆上,∴.联立,化为=0,∵x0≠0,∴,代入x0+y0+c=0,可得.∴P.设圆心为T(x1,y1),则=﹣,=.∴T,∴圆的半径r==.设直线l的斜率为k,则直线l的方程为:y=kx.∵直线l与圆相切,∴,整理得k2﹣8k+1=0,解得.∴直线l的斜率为.21.(12分)已知函数f(x)=x2+a(x+lnx),x>0,a∈R是常数.(1)求函数y=f(x)的图象在点(1,f(1))处的切线方程;(2)若函数y=f(x)图象上的点都在第一象限,试求常数a的取值范围;(3)证明:∀a∈R,存在ξ∈(1,e),使f′(ξ)=.【解答】(1)解:函数f(x)=x2+a(x+lnx)的导数f′(x)=2x+a(1+),f(1)=1+a,f′(1)=2+2a,则函数y=f(x)的图象在点(1,f(1))处的切线为y﹣(1+a)=(2+2a)(x﹣1),即y=(1+a)(2x﹣1);(2)解:①a=0时,f(x)=x2,因为x>0,所以点(x,x2)在第一象限,依题意,f(x)=x2+a(x+lnx)>0;②a>0时,由对数函数性质知,x∈(0,1)时,lnx∈(﹣∞,0),alnx∈(﹣∞,0),从而“∀x>0,f(x)=x2+a(x+lnx)>0”不成立;③a<0时,由f(x)=x2+a(x+lnx)>0得,设,g′(x)=+,则g(x)≥g(1)=﹣1,从而,﹣1<a<0;综上所述,常数a的取值范围﹣1<a≤0.(3)证明:直接计算知,设函数g(x)=f′(x)﹣=2x﹣(e+1)+﹣,,,当a>e(e﹣1)2或时,<0,因为y=g(x)的图象是一条连续不断的曲线,所以存在ξ∈(1,e),使g(ξ)=0,即ξ∈(1,e),使f′(ξ)=;当时,g(1)、g(e)≥0,而且g(1)、g(e)之中至少一个为正,由均值不等式知,,等号当且仅当时成立,所以g(x)有最小值,且,此时存在ξ∈(1,e)(或),使g(ξ)=0.综上所述,∀a∈R,存在ξ∈(1,e),使f′(ξ)=.四、请考生在第22、23、24三题中任选一题作答,如果多做,则按所做的第22题计分.【选修4-1:几何证明选讲】22.(10分)选修4﹣1:几何证明选讲如图,已知圆上的,过C点的圆的切线与BA的延长线交于E点.(Ⅰ)证明:∠ACE=∠BCD;(Ⅱ)若BE=9,CD=1,求BC的长.【解答】(Ⅰ)证明:∵,∴∠ABC=∠BCD.又∵EC为圆的切线,∴∠ACE=∠ABC,∴∠ACE=∠BCD.(Ⅱ)∵EC为圆的切线,∴∠CDB=∠BCE,由(Ⅰ)可得∠BCD=∠ABC.∴△BEC∽△CBD,∴,∴BC2=CD•EB=1×9=9,解得BC=3.【选修4-4:坐标系与参数方程】23.已知直线l:(t为参数)经过椭圆C:(φ为参数)的右焦点F.(Ⅰ)求m的值;(Ⅱ)设直线l与椭圆C交于A,B两点,求|FA|•|FB|的最大值与最小值.【解答】解:(Ⅰ)椭圆的参数方程化为普通方程,得,∴a=5,b=3,c=4,则点F的坐标为(4,0).∵直线l经过点(m,0),∴m=4.…(4分)(Ⅱ)将直线l的参数方程代入椭圆C的普通方程,并整理得:(9cos2α+25sin2α)t2+72tcosα﹣81=0.设点A,B在直线参数方程中对应的参数分别为t1,t2,则|FA|•|FB|=|t1t2|=.…(8分)当sinα=0时,|FA|•|FB|取最大值9;当sinα=±1时,|FA|•|FB|取最小值.…(10分)【选修4-5:不等式选讲】24.已知函数f(x)=|2x+1|+|2x﹣3|(1)求不等式f(x)≤6的解集;(2)若关于x的不等式f(x)≤|a﹣2|的解集非空,求实数a的取值范围.【解答】解:(1)∵函数f(x)=|2x+1|+|2x﹣3|,∴不等式f(x)≤6 等价于①,或②,或③.解①求得﹣1≤x<﹣;解②求得﹣≤x≤;解③求得<x≤2.综合可得,原不等式的解集为[﹣1,2].(2)∵f(x)=|2x+1|+|2x﹣3|≥|2x+1﹣(2x﹣3)|=4,则f(x)的最小值为4.若关于x的不等式f(x)≤|a﹣2|的解集非空,则|a﹣2|≥4,a﹣2≥4,或a ﹣2≤﹣4,求得a≥6,或a≤﹣2,故a的范围为{a|a≥6,或a≤﹣2 }.。
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2014学年第一学期高三英语期中考试试卷第I卷(共103分)I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. 145 minutes B. 140 minutes C. 130 minutes D. 160 minutes2. A. In a bank B. At a hotel C. At a tourist agency D. In a supermarket.3. A. A travel agency clerk. B. An office secretaryC. An emergency room nurseD. A hotel receptionist4. A. A manager B. A teacher C. A secretary D. jobless5. A. Only four courses are offered next year.B. Five courses may be too many.C. It would be wise to take one more course.D. A sensible decision must be made as soon as possible.6. A. In a restaurant across the street. B. In a garden near the restaurant.C. In a driving club not far away.D. In a car on the way.7. A. Betty will start working in this company in four years.B. Betty will be working in this company for at least four more years.C. Betty began to work in this company more than four years ago.D. Betty left her company less than four years ago.8. A. David was too tired and depressed to studyB. David didn‘t take the woman‘s advice.C. David didn‘t do well in any of his exams.D. David failed in his exam because he was depressed.9. A. Lisa is still in the kitchen. B. Lisa didn‘t drive her car that day.C. Lisa doesn‘t like her new car.D. The key on the table was a special one.10. A. The battery is not correctly positioned.B. The woman doesn‘t know how the calculator works.C. The calculator needs a new battery to work properly.D. The man should put down the different batteries.Section BDirections: In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Political system B. Religion C. Working language D. Race12. A. Discuss current issues. B. Celebrate their friendship.C. Join in a writing competition.D. Attend an arts and crafts competition.13. A. The Common Wealth Games. B. An Important Holiday.C. The Common Wealth members.D. An international association.Questions 14 through 16 are based on the following passage.14. A. A particular spider in U S. B. Habitats of all Spiders in North America.C. How to defend spiders.D. People‘s fear of spiders and powerful poison.15. A. Most spiders will not bite even when handled by people.B. Most spiders are poisonous and dangerous although they are shy.C. Most spiders tend to attack people directly with their silk.D. Most spiders have sharp eyes and do not like being disturbed.16. A. Because she is too frightening to find a husband for herself.B. Because the female spider is larger than the male one.C. Because the female spider often kills her husband.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)English is ___25____ (widely) used language in the history of our planet. One in every seven human beings ___26____ speak it. More than half of the world‘s books and three quarters of international mail are in English. Of all languages, English has the largest vocabulary —perhaps as many as two million words.However, let‘s face it: English is a crazy l anguage. There is no egg in ____27____ eggplant, neither pine nor apple in a pineapple and no ham in a hamburger. Sweet-meats are candy, while sweetbreads, ___28____ aren‘t sweet, are meat.We take English ___29____ granted. But when we explore its paradoxes (矛盾), we find that quicksand can work slowly, boxing rings are square, public bathrooms have no baths in them.And why is it ___30____ a writer writes, but fingers don‘t fing, grocers don‘t groce, and hammers don‘t ham? If the plural of tooth is teeth, shouldn‘t the plural of booth be beeth?How can a slim chance and a fat chance be the same, while a wise man and a wise guy are opposites? How can overlook and oversee be opposites, while quite a lot and quite a few are alike? How can the weather be hot as hell one day and cold as hell the next?So far English, ___31____ (invent) by people not computers, ___32____ (reflect) the creativity of human beings.(B)About 18 months ago, my father was in hospital recovering __33__ a major lung operation. My mother had recently passed away, and my father had taken the loss of his partner of 55 years and had lost interest in life.__34___(t ry) to get him to eat each day was quite a chore as he didn‘t want anything. The one thing, however, that he would ask us to bring him was ice-cream.One evening, to our surprise, he refused ___35___(eat ) the ice-cream, so I placed __36__ in a staffroom freezer. A little while later, my son decided he wanted it, so I fetched it for him.As I passed another ward, a woman asked, ‗Are there more ice-cream?‖ .__37__ I had explained the situation, she apologized. She then said the she had cancer and could eat very little, other than the occasional ice-cream.The next morning, I decided to buy two ice-creams. On the way to Dad‘s room, I stopped in at the sick woman‘s room and of fered her the ice-cream I‘d bought for h er. She was totally surprised at what I __38__ (think) for her, and accepted the gift with tears in her eyes. I spoke with her for a few minutes, explaining__39_ was happening in my family and listened to her similar story of pain and suffering. It was clear that she didn‘t have many visitors, and the ice-cream and our short chat meant a great deal to her.I repeated the gesture a few days later, and this time, I __40__(reward) with a big hug.I never even thought to ask her name, never saw her again, but it made me realize that an act of kindness can be more rewarding when you give it, rather than receive it.Section BDirections: Complete the passage with the words given in the table. Each word can be used only once. There is an extra one that you will find no use for.terrifying night in the high country, they were wandering 41 –cold, wet and hungry. ―They‘ll never find us,‖ one girl said between sobs. ―We‘re all going to die.‖ Then, an 11-year-old, Evancell Towne, stepped forward. ―I‘m not going to die,‖ she said 42 . ―I‘ve heard that if you follow a little stream, it empties into a bigger stream and finally you come to a town. I‘m going to follow that str eam we saw. The rest of you can come if you want.‖Evancell43 resolutely into the woods toward a tiny stream. The others followed. For more than five hours they 44 through bushes along ever-larger stream beds. Finally, they heard voices, and their shouts brought the rescue party. The 45 youngster had led the group to safety.―A born leader,‖ people say when I tell this story. In their 46 , people like Evancell are47 at birth to lead, while others are destined to follow. But my work with athletes, students, Air Force pilots and hotel managers has convinced me that leaders aren‘t born –they‘re made. Men and women who run organizations, lead the community and captain sports teams are the 48 of strict parents who follow rules that help develop a leadership mentality –strength of mind and independent thinking. Their children don‘t easily give in under pressure; they follow their own 49 . A leadership mentality pays off tomorrow as well as today. Childhood leadership in class and after-school activities is more accurate a predictor of adult50 for both sexes than intelligence or academic grades.III. Reading ComprehensionSection ADirections:For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Napping for a while at daytime is a very smart and healthy move. The Mayo Clinic says naps _51 relaxation, better mood and alertness, and a sharper working 52 . A 2008 British study found that compared to getting more nighttime sleep, a mid-day nap was the best way to 53 the mid-afternoon sleepiness.According to the Harvard Health Letter, several studies have shown that people 54 new information better when they take a nap shortly after learning it. And, most 55 , a 2007 study of nearly 24,000 Greek adults in the Archives of Internal Medicine found that people who napped 56 had a 37 percent reduced risk of dying from heart disease com pared to people who didn‘t nap.Of course, napping isn‘t57 for everyone. If you‘re suffering from inability to sleep, naps that are too long or taken too late in the day can 58 with your ability to fall or stay asleep at night.But for most, naps can make you feel sharper and happier. Naps provide different benefits 59 on how long they are. A 20-minute nap will boost alertness and concentration; a 90-minute snooze(小睡)can 60 creativity.According to , you 61 a natural nap in body temperature between 1 p.m. and 3 p.m. A short nap at this time can boost alertness for several hours and, for most people, shouldn‘t 62 being able to fall asleep at night.Pick a dark, cozy place that‘s not too warm or too chilly. 63 napping on the couch instead of in bed, so you‘re less64 to snooze for too long.Surprisingly, the best place to take a nap may be a hammock(吊床)if you have one. A Swiss study 65 last year found that people fell asleep faster and had deeper sleep when they napped in a hammock than in a bed. That same rocking motion that puts babies to sleep works wonders for grown-ups, too.51. A. relieve B. promote C. operate D. support52. A. feeling B. frame C. smell D. mind53. A. cope with B. put aside C. talk about D. carry upon54. A. remark B. consider C. remember D. concern55. A. reportedly B. unbelievably C. constantly D. frankly56. A. regularly B. enormously C. heavily D. strongly57. A. exact B. wrong C. right D. accurate58. A. connect B. deal C. compete D. interfere59. A. focusing B. depending C. force D. basing60. A. enlarge B. engage C. enhance D. enroll61. A. explore B. experience C. exercise D. implement62. A. produce B. handle C. affect D. urge63. A. postpones B. discourages C. acknowledges D. recommends64. A. obliged B. tempted C. adopted D. attracted65. A. pronounced B. published C. discovered D. cultivatedDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)On a hill 600 feet above the surrounding land, we watch the lines of rain move across the scene, the moon rise over the hills, and the stars appear in the sky. The views invite a long look from a comfortable chair in front of the wooden house.Every window in our wooden house has a view, and the forest and lakes seldom look the same as the hour before. Each look reminds us where we are.There is space for our three boys to play outside, to shoot arrows, collect tree seeds, build earth houses and climb trees.Our kids have learned the names of the trees, and with the names have come familiarity and appreciation. As they tell all who show even a passing interest, maple(枫树)makes the bestfighting sticks and white pines are the best climbing trees.The air is clean and fresh. The water from the well has a pleasant taste, and it is perhaps the healthiest water our kids will ever drink. Though they have one glass a day of juice and the rest is water, they never say anything against that.The seasons change just outside the door. We watch the maples turn every shade of yellow and red in the fall and note the poplars‘(杨树)putting out the first green leaves of spring. The rainbow smelt fills the local steam as the ice gradually disappears, and the wood frogs start to sing in pools after being frozen for the winter. A family of birds rules our skies and flies over the lake.66. What can be learned from Paragraph 2?A. The scenes are colorful and changeable.B. There are many windows in the wooden house.C. The views remind us that we are in a wooden house.D. The lakes outside the windows are quite different in color.67. By mentioning the names of the trees, the author aims to show that ______ .A. the kids like playing in treesB. the kids are going to study in natureC. the kids have learned much knowledgeD. the kids find trees useful learning tools68. What does the underlined sentence in the last paragraph mean?A. The change of seasons is easily felt.B. The seasons make the scenes change.C. The weather often changes in the forest.D. The door is a good position to enjoy changing seasons.69. What is the main purpose of the author writing the text?A. To describe the beauty of the scene around the house.B. To introduce her children‘s happy life in the forest.C. To show that living in the forest is healthful.D. To share the joy of living in the nature.(B)One of the greatest contributions to the first Oxford English Dictionary was also one of its most unusual. In 1879, Oxford University in England asked Prof. James Murray to serve as editor for what was to be the most ambitious dictionary in the history of the English language. It would include every English word possible and would give not only the definition but also the history of the word and quotations (引文)showing how it was used.This was a huge task.,so Murrary had to find volunteers from Britain, the United States, and the British colonies to search every newspaper, magazine, and book ever written in English. Hundreds of volunteers responded, including William Chester Minor. Dr. Minor was an American surgeon who had served in the Civil War and was now living in England. He gave his address as ―Broadmoor, Crowthorne, Berkshire,‖‖ 50 miles from Oxford.Minor joined the army of volunteers sending words and quotations to Murray. Over the next 17 years, he became one of the staff‘s most valued contributors.But he was also a mystery. In spite of many invitations, he would always decline to visitOxford. So in 1897,Murray finally decided to travel to Crowthorne himself. When he arrived, he found Minor locked in a book-lined cell at the Broadmoor Asylum(精神病院)for the Criminally insane.Murray and Minor became friends, sharing their love of words. Minor continued contributing to the dictionary, sending in more than 10,000 submissions in 20 years. Murray continued to visit Minor regularly, sometimes taking walks with him around the asylum grounds.In 1910, Minor left Broadmoor for an asylum in his native America. Murray was at the port to wave goodbye to his remarkable friend.Minor died in 1920, seven years before the first edition of the Oxford English Dictionary was completed. The 12 volumes defined 414,825 words, and thousands of them were contributions from a very scholarly and devoted asylum patient.70. According to the text, the first Oxford English Dictionary _________.A. came out before minor diedB. was edited by an American volunteerC. included the English words invented by MurrayD. was intended to be the most ambitious English dictionary71. How did Dr. Minor contributed to the dictionary?A. He helped Murray to find hundreds of volunteers.B. He sent newspapers, magazines and books to Murray.C. He provided a great number of words and quotationsD. he went to England to work with Murray.72. Why did Dr. Minor refuse to visit Oxford?A. He was shut in an asylumB. He lived far from OxfordC. He was busy writing a bookD. He disliked traveling73. Prof. Murray and Dr. Minor became friends mainly because __________.A. they both served in the Civil War.B. They had a common interest in wordsC. Minor recovered with the help of MurrayD. Murray went to America regularly to visit Minor(C)Last night I was driving from Harrisburg to Lewisburg,a distance of about eighty miles. It was late. Several times I got stuck behind a slow-moving truck on a narrow road with a solid white line on my left, and I became increasingly impatient.At one point along an open road, I came to a crossing with a traffic light. I was alone on the road by now, but as I drove near the light, it turned red and I made a stop. I looked left, right and behind me. Nothing. Not a car, no suggestion of car lamps, but there I sat, waiting for the light to change, the only human being for at least a mile in any direction.I started wondering why 1 refused to run the light I was not afraid of being caught, because there was clearly no policeman around, and there certainly would have been no danger in going through it.Much later that night, the question of why I'd stopped for that light came back to me. I think I stopped because it's part of a contract(契约)we all have with each other. It's not only the law, but itis an agreement we have, and we trust each other to honor it: we don't go through red lights.Trust is our first inclination(倾向).Doubting others does not seem to be natural to us. The whole construction of our society depends on mutual(相互)trust, not distrust .We do what we say we'll do; we show up when we say we'll show up; and we pay when we say we'll pay. We trust each other in these matters, and we're angry or disappointed with the person or organization that breaks the trust we have in them.I was so proud of myself for stopping for the red light that night.74. Why did the author get impatient while driving?A. He was lonely on the road.B. He was slowed down by a truck.C. He got tired of driving too long.D. He came across too many traffic lights.75. What was the author's immediate action when the traffic light turned red?A. Stopping still.B. Driving through it.C. Looking around for other cars.D. Checking out for traffic police.76. The event made the author strongly believe thatA. traffic rules may be unnecessaryB. doubting others is human natureC. patience is important to driversD. a society needs mutual trust77. Why was the author proud of himself?A. He kept his promise.B. He held back his anger.C. He followed his inclination.D. He made a right decision.Section CDirections: Read the passage carefully. Then answer the questions or complete the statements in no more than 12 words..Scientists have tried to come up with biological explanations for the difference between boys and girls.However, none were believable enough to explain the general picture. As one scientist points out, ―There are slight genetic differences between the sexes at birth which may affect the subjects boys and girls choose. But the difficulty is that by the time children reach school age, there are so many other effects that it is almost impossible to tell whether girls are worse at science and maths, or whether they‘ve been brought up to think of these subjects as boys‘ ‗territory‘‖.Statistics show that in mathematics, at least, girls are equal to boys. A recent report suggests that girls only stop studying mathematics because of social attitudes. One of the reports‘ authors says, ―While it is socially unacceptable for people not to be able to read and write, it is still acceptable for women to say that they are ‗hope-less‘ at maths. Our research shows that, althoughgirls get marks which are as good as the boys, they have not been encouraged to do so.‖The explanation for the difference, which is very clear during the teenage years, goes as far back as early childhood experiences. From their first days in nursery school, girls are not encouraged to work on their own or to complete tasks, although boys are. For example, boys and not girls, are often asked to ‗help‘ with repair work. This encouragement leads to a way of learning how to solve problems later on in life. Evidence shows that exceptional mathematicians and scientists did not have teachers who supplied answers; they had to find out for themselves.A further report on maths teaching shows that teachers seem to give more attention to boys than to girls.Most teachers who took part in the study admitted that they expect their male students to do better at mathematics and science subjects than their female students. All of this tends to encourage boys to work harder in these subjects, gives them confidence and makes them believe that they can succeed.Interestingly, both boys and girls tend to regard such ‗male‘ subjects like mathematics and science as difficult. Yet it has been suggested that girls avoid mathematics courses, not because they are difficult, but for social reasons.Mathematics and science are mainly male subjects, and therefore, as girls become teenagers, they are less likely to take them up. Girls do not seem to want to be in open competition with boys. Neither do they want to do better than boys because they are afraid to appear less female and so, less attractive.78. According to scientific studies, girls can learn maths as well as boys if___________________ ______________________________________________________________________________.79. Those who made extraordinary contributions in mathematics and science_________________ ______________________________________________________________________________.80. Why are boys favored over girls when it comes to mathematics and science?______________________________________________________________________________.81. What caused girls to give up mathematics courses?______________________________________________________________________________.第II卷(共47分)I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1.他已经养成了每天晚饭以后散步的好习惯。