2018年广二模答案
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2020年高考历史一轮复习课时精练12【国共十年对峙时期】一、选择题1.(2018四川资阳二模,28)1919年5月7日《晨报》刊登了一篇题为《国民外交协会宣言》的文章,文中指出青岛、山东是“齐鲁名邦”“孔孟之圣迹”“文明吐露之源泉”。
由此可见当时的部分文化精英希望()A.借传统文化深化爱国意识B.促使民主思想进一步扩散C.重塑儒家思想的主流地位D.借文化圣地博得列强的理解2.(2018北京海淀期末,1)有学者认为五四运动期间“新知识分子从象牙塔中走出来,到十字街头,大庭广众之间……与商人、店员、都市工人、工业家以及职业政客相往来”。
这说明五四运() A.促进新文化运动兴起 B.知识分子是运动的主力军C.具有广泛的群众基础D.目标是推翻资产阶级政权3.(2018河南洛阳期末,16)1921年中共一大通过的《中国共产党纲领》规定:“以无产阶级革命军队推翻资产阶级,由劳动阶级重建国家,直至消灭阶级差别……废除资本私有制,没收一切生产资料,如机器、土地、厂房、半成品等,归社会所有。
”这些规定说明中国共产党()A.未能正确认识中国国情B.认识到人民军队的重要性C.坚持共产主义基本原则D.照搬俄国的民主革命道路4.(2018河北邯郸摸底,10)下表是中国早期的铁路工人斗争情况。
这种变化可以用来表明()时间结果1912—1920年只是一部分人参加的斗争,大都是在被逼无奈或被教唆利用的情况下发生的,没有其他产业工人的声援1921—1923年罢工持续时间长,牵涉方面广、斗争复杂、组织严密、斗争策略娴熟和胜利成果(经济、政治诉求)全面A.工人斗争由自发阶段走向自觉阶段B.工人阶级逐步成为独立政治力量C.工人阶级和资产阶级矛盾日趋尖锐D.工人阶级的民主意识在迅速提高5.(2018天津部分地区一模,6)胡适指出:民国十三年以后,国民党一度成为中国“一个簇新的社会重心……这四五年来,又逐渐失去做社会重心的资格了”。
国民党成为中国“一个簇新的社会重心”的主要原因是()A.北伐战争的胜利B.国民党的改组C.黄埔军校的建立D.共产国际帮助6.(2018广西南宁二模,29)1924年后,共产党一方面推动国民党中央先后设立了农民部、工人部、青年部、妇女部和商人部,工运、农运计划及相关政策、法规也相继出台;另一方面积极从事国民党不愿意做的下层民众动员工作,这说明()A.共产党主导群众运动B.国共合作在基层未实现C.国民党不愿发动民众D.中国共产党掌握了革命领导权7.(2018重庆调研卷,29)1927年7月毛泽东提出农民武装可以“上山”,以造成军事势力的基础;同月,中共中央决定组成中央前敌委员会负责领导武装暴动。
2018届上海市青浦区高三二模考试语文试题(解析版)一、积累运用1. 按要求填空。
(1)__________,水面清圆,一一风荷举。
(周邦彦《苏幕遮》)(2)锦江春色来天地,__________。
(_________)《登楼》)(3)李白的《登金陵凤凰台》中抒情的句子是“________,________。
”【答案】(1). 叶上初阳干宿雨(2). 玉垒浮云变古今(3). 杜甫(4). 总为浮云能蔽日(5). 长安不见使人愁【解析】试题分析:名句默写分为两大类,一是给出语境的理解性默写,二是给出上句写下句的直接默写。
默写要注意字形,而字形与字义分不开,学生应借助字义来识记字形。
注意重点字的写法。
如“垒”“蔽”要理解字义去记忆。
学生记忆的时候应该结合诗句的意思。
2. 按要求选择。
(1)好同学赴国外深造,离别之后对他表达思念之情,以下句子最不合适的一项A.但愿人长久,千里共婵娟。
B.桃花潭水深千尺,不及汪伦送我情。
C.离恨恰如春草,更行更远还深。
D.故人何在,烟水茫茫。
(2)3月5日学雷锋,学校组织义卖活动,面对前来的老师同学,以下表达最得体的一项是A.欢迎同学、老师光临惠顾。
B.一分钱一分货,您看着好就多给点。
C.爱心播种希望,奉献点亮青春。
D.走过路过,不要错过。
【答案】⑴B;⑵A【解析】(1)试题分析:本题主要考查情境理解,需要注意结合诗句内容和题干要求。
“好同学赴国外深造,离别之后对他表达思念之情”,需要注意“思念”之情。
B项,表达了对汪伦深情相送的感激。
用“深千尺”的潭水比喻送别之深情,生动而形象,而又加“不及”二字,更增强了诗句的动人力量。
这首有明显的民歌风味的诗词自然质朴,清新流畅。
诗人用眼前普通的景物作比喻,写出了与友人的真挚情意。
(2)试题分析:本题主要考查语言表达。
B项,“多给点”不符合要求;C项,“义卖”活动,“爱心”“奉献”不符合要求;D项,太口语化,不符合学校环境要求。
本文部分内容来自网络整理,本司不为其真实性负责,如有异议或侵权请及时联系,本司将立即删除!== 本文为word格式,下载后可方便编辑和修改! ==疼痛银行阅读答案篇一:201X年广州二中初中毕业班二模语文试题201X年广州市二中初中毕业班二模语文试题第一部分积累与运用(共34分)一、(6小题,19分)1、下列词语中,加点字读音正确的一项是()(3分)..A、襁褓(qiǎng)勾当(g?u)咫尺(zhǐ)盘虬卧龙(qiú)....B、倔强(jua)倒坍(tān)妖娆(yāo)群蚁排衙(yá)....C、镶嵌(qiàn)睥睨(pì)脑髓(suǐ)惟妙惟肖(qiào)....D、哂笑(shěn)瘠薄(jí)凛冽(lǐng)鳞次栉比(zhì)....2、下列词语中,没有错别字的一项是()(3分)A、鞠躬尽瘁连声喏喏进退维谷格物至知B、众目睽睽诚惶诚恐影影绰绰五彩斑斓C、更胜一畴猝然长逝蹑手蹑脚囊莹映雪D、廓然无累冥思暇想水波鳞鳞长吁短叹3、下列语段中,加点词语运用不恰当的一项是()(3分)...4月25日14时11分,尼泊尔发生了骇人听闻的8.1级地震,震源深度20千米。
得知情况后,中国....义无反顾地派出救援队前往灾区,以实际行动与灾区人民相濡以沫。
这次地震让许多尼泊尔百姓失去亲人........与家园,每当回想起地震一幕,幸存者依然心有余悸。
....A、骇人听闻B、义无反顾C、相濡以沫D、心有余悸4、下列句子中,没有语病的一项是()(3分)A、本着“客户至上,服务第一”为原则,中国工商银行推出一系列更人性化的服务。
B、随着李克强总理敦促提高网速和降低网费,对于广大网民来说,无疑是一件好事。
C、“一带一路”建设秉持的是共商、共建、共享原则,不是封闭的,而是开放包容的。
D、在现今激烈的市场竞争中,我们所缺乏的:一是勇气不足,二是谋略不当。
2018北京各区初三二模语文试卷分类汇编(名著阅读)答案版朝阳区6、古典小说中,人物的绰号往往能体现其外貌特征或性格特点。
智多星吴用、豹子头林冲、鼓上蚤时迁……他们都出自《》,作者是。
(2分)6.答案:水浒传施耐庵三、名著阅读(共6分)16. 孝,是一种传统美德。
请结合下面几则语录中的任意一则,说说你在生活中如何尽孝。
(3分)①子曰:“父母唯其疾之忧。
”(《论语·为政》)②子曰:“父母在,不远游,游必有方。
”(《论语·里仁》)③子曰:“色难。
有事,弟子服其劳;有酒食,先生馔。
曾是以为孝乎?”(《论语·为政》)④子曰:“父母之年,不可不知也;一则以喜,一则以惧。
”(《论语·里仁》)⑤子曰:“今之孝者,是谓能养。
至于犬马,皆能有养;不敬,何以别乎?”(《论语·为政》)答:17.小说中人物间的矛盾和冲突最能突显人物性格。
请结合你读过的名著,选择相关情节,说说其中体现的人物性格。
(3分)答:16.答案示例:选择③句。
孔子认为孝最难得的是对父母保持敬爱和悦的态度。
我总是能够和颜悦色地和父母沟通交流,从不对父母发脾气,这就是我对父母的孝。
(评分标准:共3分。
选择并理解2分,做法1分)17.答案示例:鲁达听闻金氏父女被郑屠欺侮,感到愤愤不平。
他见义勇为,三拳打死镇关西,表现出他嫉恶如仇、侠肝义胆的性格特点。
(评分标准:共3分。
情节2分,性格1分)丰台区4.关山难越,更衬托出人的信念坚定,意志坚强。
①替父从军,“万里赴戎机,关山度若飞”;关羽为了追寻兄长,②,成为千古佳话。
无论古今中外,攻坚克难、勇闯难关,都是精神力量的体现。
《海底两万里》中的尼摩船长在海底航行时多次遇险,他都能沉着机变地应对,比如③;《红岩》中的江姐、许云峰等共产党员不畏敌人的酷刑,在狱中坚持斗争,体现了④的“红岩精神”。
(4分)4.(4分)①花木兰②过五关斩六将③潜艇在去往南极途中被困冰下,艇内缺氧,船长指挥大家砸冰自救,最终脱险④英勇无畏评分说明:每空1分。
2018届高三广州二模关于蚯蚓甲虫小鸟的寓言故事的语文作文题目及12篇范文一片曾经丰饶的土地逐渐变得贫瘠,许多动物陆续离开了这里。
小鸟对蚯蚓说:“我也要走了,你不走吗?”蚯蚓说:“我还是想留在这里。
”小鸟拍拍翅膀,依依不舍地飞走了。
蚯蚓看了看小鸟的背影,继续埋头松土。
甲虫从远方赶来,和蚯蚓一起忙碌起来。
几年之后,小鸟回来了,带着一群伙伴,它们嘴里衔着各种各样的种子。
小鸟惊讶地发现,这片土地已经变得松软,还长出了小草。
蚯蚓说:“你离开的这段时间,甲虫和我一起翻土堆肥,这里渐渐变好了。
”小鸟和伙伴们把种子播撒在大地上。
大地上开满了鲜花,蝴蝶和蜜蜂也来了。
根据上面的材料,结合自己的感受和思考,写一篇文章。
要求选好角度,明确文体,自拟标题;不要套作,不得抄袭;不少于800字。
审题立意:一、关于题意理解。
(一)对于资料的理解:题目提供的是动物们通过努力让瘠薄的土地变得丰饶美丽的寓言资料。
资料中虽然只有“小鸟”“蚯蚓”和“甲虫”三个主要角色,情节和对话也非常简单,却包含着深刻的社会意义,也为考生提供了较多的写作角度。
材料叙述的内容可分为三个层次。
第一层,动物们陆续离开逐渐贫瘠的土地。
这是寓言的背景,也为下文小鸟、蚯蚓和甲虫的行为提供了动机和原因。
第二层,蚯蚓留下来辛苦耕耘,甲虫从远方赶来帮忙,小鸟离开几年后带回种子。
这是寓言的核心内容,小鸟、蚯蚓和甲虫方式不同,但都在为这片土地劳心尽力。
第三层,蚯蚓和甲虫的劳作初见成效,小鸟带回的种子让大地开满鲜花。
这是寓言的结局,大家的努力让这片土地重新变得丰饶美丽。
这样的叙述使材料有了丰厚的现实意义和多样的思考维度。
热爱乡梓之情、建设家园之心,历来为人所称道。
资料中的“土地”内涵丰富,可以让考生遐想到“乡村”“故乡”甚至“故国”等。
资料并未提及这片“土地”由“丰饶”变得“瘠薄”的原因,考生不应将此作为行文的重点,而应将思考的重心放在小鸟、蚯蚓和甲虫让瘠薄的土地变得丰饶美丽的不同方式上。
2018年广东省梅州市高考语文二模试卷一、现代文阅读1.阅读下面的文字,完成下列小题。
《管子•牧民》曰:“上服度则六亲固,四维张则君令行。
”其中,“四维”指的是礼、义、廉、耻。
“四维”是国家运行的重要支柱,礼绝则国倾,义绝则国危,廉绝则国覆,耻绝则国灭。
国家顺应礼制法度才能使父母兄弟妻子各得其所,才能使社会关系稳固。
抛弃礼,社会发展将失去平衡。
近些年来,社交礼仪、家礼、开笔礼、婚礼等越来越受重视,尤其是婚礼,中西合璧,花样繁多。
然而,遗憾的是我国并没有一部现行的官修礼仪文本如《汉仪》《大唐开元礼》般让人们参考或效仿,以至于很多礼仪活动失去了原有的含义,甚至成了娱乐、哗众取宠的噱头。
婚礼,是人生中最重要的礼仪。
《礼记》将之视为“礼之本”。
古代国君在迎娶夫人的求婚辞中写道:“请君之玉女,与寡人共有敝邑,事宗庙社稷。
”除传宗接代以外,婚礼更多表达的是夫妇一体、相扶相携、尊卑等同的含义。
婚礼仪式中“御布对席”“合卺而饮”“共牢而食”传递的就是这个意思。
另外,古代婚礼中还有“拜见公婆”这一礼节,即新娘要在婚后的第二天向公公婆婆见礼,以表达对长辈的尊重。
由此可见,婚礼绝不是一个简单的形式,而是一场正视夫妻关系的启蒙教育。
然而,在现代社会,如此神圣庄严的时刻竟被一些人视为儿戏,甚至演绎为一场闹剧。
尤其表现在那些荒唐而野蛮的闹婚陋习方面,闹公婆,闹新郎,闹新娘,手段粗俗卑劣,甚至威胁到当事人生命安全。
礼仪等同于文明。
中国庞大的礼仪系统是人类社会由野蛮迈向文明的一块历史丰碑。
每一种礼仪都暗含着对道德的敬畏,对生命的敬畏。
周代时始制定礼制,这一时期制定了天子之礼、诸侯之礼、卿相之礼,到后来才延伸出士人之礼。
从孔子以仁释礼开始,经过汉、唐、宋、明、清等多个朝代,政府和官员几番做出“礼下庶人”的尝试,专门制定了“庶人之礼”供民间通用,但均未得到普及。
主要原因是社会存在严格的等级制度,贵族阶层予以阻挠。
具体而言,礼仪和礼制是门阀贵族彰显身份、维护特权的工具,根据等级品阶的不同,他们所穿的服装的样式、所用的器物、乘坐的车辇等各不相同,庶人用礼,损害到贵族阶层的威仪和利益。
2018年普通高等学校招生全国统一考试广东省语文模拟试卷(二)注意事项:1.答题前,考生须认真核对条形码上的姓名、考生号、考场号和座位号,并将其贴在指定位置,然后用0.5毫米黑色字迹签字笔将自己所在的县(市、区)、学校以及自己的姓名、考生号、考场号和座位号填写在答题卡和试卷的指定位置,并用2B铅笔在答题卡的“考生号”处填涂考生号。
2.回答选择题时,选出每小题答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其他答案标号。
回答非选择题时,用0.5毫米黑色字迹签字笔将答案写在答题卡上。
写在本试卷上无效。
3.考试结束后,将本试卷和答题卡一并交回。
一、现代文阅读(一)论述类文本阅读阅读下面的文字,完成下列小题。
草原文化精神从根本上说是一种动态的、积极的、以个体奋斗为主要内容的进取性文化。
北方草原具有独特性,它既有深远悠长的民族牧歌情调式的生活方式,又有严酷杀伐的沧桑、厚重的历史,从而导致了草原民族对于人与自然关系方面的特殊认识。
一方面,人与自然的关系极为直接、密切,人对自然的依赖性突出,因而对自然的亲近之感、迫近之念、忧虑之情更为突出、深挚;它不似江南山水对人的细膩轻抚,更不像士子优游于小桥流水,流溢着别致而精微的艺术感触,而是将民族生存与自然环境直接联系起来,把对草原的关切与民族的整体性发展联结起来,发出了“草原不在、民族何存”的呐喊,汉代流传的《匈奴歌》就是其中代表。
另一方面,人与自然架构起初步的生态美学桥梁。
蒙古民族第一部史传文学《蒙古秘史》当中就显现着极为浓郁的生态观念。
由于生存环境的严酷、险恶、千变万化,草原民族“逐水草而居”的动态生存观念异常鲜明。
草原民族生存的“动态”和“变化”的特点使人的适应能力和冲破自然环境束缚的能力分外突出。
草原民族强调人在强大的自然环境和社会面前的力量、智慧、意志力,从而形成一种对英雄的崇拜的观念。
《史记·匈奴列传》中的冒顿单于和《蒙古秘史》中的札木合就是鲜活的例证,只不过后者还显现着强烈的崇信重义之美。
普通高等学校招生全国统一考试文科综合地理部分文科综合共300分,考试用时150分钟。
地理试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,第Ⅰ卷1至4页,第Ⅱ卷5至8页,共100分。
答卷前,考生务必将自己的姓名填写在答题卡上。
答卷时,考生务必将第Ⅰ卷及第Ⅱ卷的答案都涂写在答题卡上,答在试卷上的无效。
祝各位考生考试顺利!第Ⅰ卷注意事项:1.每题选出答案后,用铅笔将答题卡上对应题目的答案标号涂黑。
如需改动,用橡皮擦干净后,再选涂其他答案标号。
2.本卷共11题,每题4分,共44分。
在每题列出的四个选项中,只有一项是最符合题目要求的。
和煦春日,共享单车风靡津城。
一辆辆色彩明快的单车穿梭在都市,徜徉于乡野。
据此回答1~3题。
1. 越来越多的津城市民选择共享单车作为出行方式。
其行为能更好彰显社会进步价值理念的是A.近自然,享春光 B.出行便,灵活畅C.循环用,零排放 D.时间省,性价强2. 下列能够体现天津城市发展历程的骑行线路是甲乙丙丁图1A.甲—丙—丁 B.乙—丁—甲C.丙—甲—丁 D.丁—丙—甲3. 途经渤海湾沿岸连片的工地,看似铺天盖地的候鸟飞翔场景,实则是由于候鸟A. 遭违法猎捕已绝迹B. 繁殖期间数量增加C. 经人工繁育后放飞D. 所栖息空间被挤占读亚洲东部某区域海平面等压线分布图(单位:hPa),回答4~5题。
图24. 假如气压变化只与气温有关,图中两条纬线和三条经线的交点中,气温日较差最小的是A.(30°N,120°E) B.(40°N,140°E)C.(40°N,130°E) D.(30°N,130°E)5. 该时间段内,M点①偏南风转西北风②风力减弱③天气晴朗④气温降低A.①②③ B.②③④ C.①③④ D.①②④鲜切水果是为了满足消费者的即食需求,对新鲜水果进行处理,使产品保持生鲜状态的鲜切制品。
鲜果切从加工到送至客户手中不超过6小时,以确保顾客能食用到最新鲜的果切。
2018上海各区二模作文解析及范文【奉贤2018年高三二模作文题】阅读以下材料,自选角度,自拟题目,写一篇不少于800字的文章。
冲突是指发生在两个或两个以上事物之间的互相对抗。
达伦多夫在《社会冲突论》中认为,冲突也是解决问题的有效方式。
课代表说:诚如题干所说,冲突是指发生在两个或两个以上事物之间的互相对抗,很多时候这样的对抗不可避免地会带来破坏,但是破坏之后就是新生。
通过对抗,“幸存”下来的一方也许就是解决方法。
素材推荐:1、辛亥革命,结束了中国两千余年的封建帝制,开创了亚洲第一个共和体制。
随着封建帝制的废除,依附于封建帝制的种种丑恶制度也被次第扫除。
民族平等、民权自由、民生幸福的呼声在中华大地回荡。
空前的民主气象,竞办实业的浪潮,形成了生机勃勃的局面。
【2018虹口高三二模】网络流行词“佛系”,最早来源于2014年日本某杂志,指一种怎么都行、无须走心、看淡一切的活法,一切随缘。
该词一出迅速流行,更有不少人自称佛系,有佛系追星、佛系购物、佛系恋爱、佛系员工、佛系学生等种种说法。
请写一篇文章,谈谈你对此的思考解析:“佛系”一词自流行以来人们对其一直有两种态度,有人认同“佛系”的洒脱和平和,也有人认为“佛系”其实就是一种逃避的借口,而无论是哪一种都有话可说,有例可证。
1、宠辱不惊,闲看庭前花开花落;去留无意,漫随天外云卷云舒——陈继儒《小窗幽记》2、你拥有青春的时候,就要感受它。
不要虚掷你的黄金时代,不要去倾听枯燥乏味的东西,不要设法挽留无望的失败,不要把你的生命献给无知、平庸和低速。
这些都是我们时代病态的目标,虚假的理想。
活着,把你宝贵的内在生命活出来。
什么都别错过。
——奥斯卡·王尔德《道林·格雷的画像》少年法盛唐中国人自古以来面对困厄的出路大抵有两条:一条是唐人的,是用蓬勃的朝气去改造人的处境;另条是晋人的,是用隐逸的志趣去修养人的内在。
没想到,在这个活力四射的盛世,一些自称“佛系”的年轻人似乎悄然转向了第二条路。
2018届黄浦区⾼考数学⼆模试卷(附答案)黄浦区2018年⾼考模拟考数学试卷(完卷时间:120分钟满分:150分) 2018.4考⽣注意:1.每位考⽣应同时收到试卷和答题卷两份材料,解答必须在答题卷上进⾏,写在试卷上的解答⼀律⽆效; 2.答卷前,考⽣务必将姓名等相关信息在答题卷上填写清楚,并在规定的区域贴上条形码; 3.本试卷共21道试题,满分150分;考试时间120分钟.⼀、填空题(本⼤题共有12题,满分54分)考⽣应在答题纸相应编号的空格内直接填写结果,每个空格填对前6题得4分、后6题得5分,否则⼀律得零分.1.已知集合{}{}1,2,31,A B m ==,,若3m A -∈,则⾮零实数m 的数值是. 2.不等式|1|1x ->的解集是.3.若函数()f x =是偶函数,则该函数的定义域是.4.已知ABC ?的三内⾓A B C 、、所对的边长分别为a b c 、、,若2222sin a b c bc A =+-,则内⾓A 的⼤⼩是.5.已知向量a 在向量b ⽅向上的投影为2-,且3b =,则a b ? = .(结果⽤数值表⽰) 6.⽅程33log (325)log (41)0x x ?+-+=的解x = .7.已知函数2sin cos 2()1cos x xf x x-=,则函数()f x 的单调递增区间是.8.已知α是实系数⼀元⼆次⽅程22(21)10x m x m --++=的⼀个虚数根,且||2α≤,则实数m 的取值范围是.9.已知某市A 社区35岁⾄45岁的居民有450⼈,46岁⾄55岁的居民有750⼈,56岁⾄65岁的居民有900⼈.为了解该社区35岁⾄65岁居民的⾝体健康状况,社区负责⼈采⽤分层抽样技术抽取若⼲⼈进⾏体检调查,若从46岁⾄55岁的居民中随机抽取了50⼈,试问这次抽样调查抽取的⼈数是⼈.10.将⼀枚质地均匀的硬币连续抛掷5次,则恰好有3次出现正⾯向上的概率是.(结果⽤数值表⽰) 11.已知数列{}n a 是共有k 个项的有限数列,且满⾜11(2,,1)n n nna a n k a +-=-=- ,若1224,51,0k a a a ===,则k = .12.已知函数2()(02)f x ax bx c a b =++<<对任意R x ∈恒有()0f x ≥成⽴,则代数式(1)(0)(1)f f f --的13.在空间中,“直线m ⊥平⾯α”是“直线m 与平⾯α内⽆穷多条直线都垂直 ”的答( ).(A )充分⾮必要条件 (B )必要⾮充分条件 (C )充要条件 (D )⾮充分⾮必要条件14.⼆项式40的展开式中,其中是有理项的项数共有答( ). (A ) 4项 (B ) 7项 (C ) 5项 (D ) 6项15.实数x y 、满⾜线性约束条件3,0,0,10,x y x y x y +≤??≥≥??-+≥?则⽬标函数23w x y =+-的最⼤值是答( ).(A ) 0 (B ) 1 (C ) 2- (D ) 316.在给出的下列命题中,是假命题的是答( ).(A )设O A B C 、、、是同⼀平⾯上的四个不同的点,若(1)(R)OA m OB m OC m =?+-?∈,则点A B C 、、必共线(B )若向量a b 和是平⾯α上的两个不平⾏的向量,则平⾯α上的任⼀向量c都可以表⽰为(R)c a b λµµλ=+∈、,且表⽰⽅法是唯⼀的(C )已知平⾯向量OA OB OC、、满⾜||||(0)OA OB OC r r ==> |=|,且0OA OB OC ++= ,则ABC ?是等边三⾓形(D )在平⾯α上的所有向量中,不存在这样的四个互不相等的⾮零向量ab c d、、、,使得其中任意两个向量的和向量与余下两个向量的和向量相互垂直三、解答题(本⼤题满分76分)本⼤题共有5题,解答下列各题必须在答题卷的相应编号规定区域内写出必要的步骤.17.(本题满分14分)本题共有2个⼩题,第1⼩题满分4分,第2⼩题满分10分.在四棱锥P ABCD-中,P A A B ⊥平⾯,,,1,AB AD BC AD BC ⊥=045CD CDA =∠=.(1)画出四棱锥P ABCD -的主视图;(2)若PA BC =,求直线PB 与平⾯PCD 所成⾓的⼤⼩.(结果⽤反三⾓函数值表⽰)18.(本题满分14分)本题共有2个⼩题,第1⼩题满分6分,第2⼩题满分8分.某企业欲做⼀个介绍企业发展史的铭牌,铭牌的截⾯形状是如图所⽰的扇形环⾯(由扇形OAD 挖去扇形OBC 后构成的).已知10,(010)OA OB x x ==<<⽶⽶,线段BA CD 、线段与弧BC 、弧AD 的长度之和为30⽶,圆⼼⾓为θ弧度. (1)求θ关于x 的函数解析式;(2)记铭牌的截⾯⾯积为y ,试问x 取何值时,y 的值最⼤?并求出最⼤值.19.(本题满分14分)本题共有2个⼩题,第1⼩题满分6分,第2⼩题满分8分.已知动点(,)M x y 到点(2,0)F 的距离为1d ,动点(,)M x y 到直线3x =的距离为2d,且12d d =. (1)求动点(,)M x y 的轨迹C 的⽅程; (2)过点F 作直线:(2)(0)l y k x k =-≠交曲线C 于P Q 、两点,若OPQ ?的⾯积OPQ S ?=(O 是20.(本题满分16分)本题共有2个⼩题,第1⼩题满分4分,第2⼩题满分6分,第3⼩题满分6分.已知函数22, 10,()=1, 0 1.x x f x x x --≤(1) 求函数()f x 的反函数1()f x -;(2)试问:函数()f x 的图像上是否存在关于坐标原点对称的点,若存在,求出这些点的坐标;若不存在,说明理由; (3)若⽅程()|()240f x f x ax +---=的三个实数根123x x x 、、满⾜:123x x x <<,且32212()x x x x -=-,求实数a 的值.21.(本题满分18分)本题共有3个⼩题,第1⼩题满分3分,第2⼩题满分6分,第3⼩题满分9分.定义:若数列{}n c 和{}n d 满⾜*10,0,N nn n c d n +>>=∈且c ,则称数列{}n d 是数列{}n c 的“伴随数列”.已知数列{}n b 是数列{}n a 的伴随数列,试解答下列问题: (1)若*(N )nn b a n =∈,1b {}n a 的通项公式n a ;(2)若*11(N )n n n b b n a +=+∈,11b a 为常数,求证:数列2n n b a ???????? ???????是等差数列; (3)若*1N )n nb n +=∈,数列{}n a 是等⽐数列,求11a b 、的数值.黄浦区2018年⾼考模拟考数学试卷参考答案和评分标准2018.4说明:1.本解答仅列出试题的⼀种解法,如果考⽣的解法与所列解答不同,可参考解答中的评分精神进⾏评分.2.评阅试卷,应坚持每题评阅到底,不要因为考⽣的解答中出现错误⽽中断对该题的评阅,当考⽣的解答在某⼀步出现错误,影响了后继部分,但该步以后的解答未改变这⼀题的内容和难度时,可视影响程度决定后⾯部分的给分,这时原则上不应4π5.6- 6.27.3[,],Z 88k k k ππππ-+∈ 8.3(4- 9.140 10.516 11.50 12.3.⼆、选择题.13.()A 14.()B 15.()D 16.()D三、解答题. 17.(本题满分14分)本题共有2个⼩题,第1⼩题满分4分,第2⼩题满分10分.解 (1)主视图如下:(2) 根据题意,可算得1,2AB AD ==. ⼜1PA BC ==,按如图所⽰建⽴空间直⾓坐标系,可得,(0,0,0),(1,0,0),(1,1,0),(0,2,0),(0,0,1)A B C D P .于是,有(1,0,1),(1,1,0),(0,2,1)PB CD PD =-=-=-.设平⾯PCD 的法向量为(,,)n x y z =,则0,0,n CD n PD ??==?? 即0,20.x y y z -+=??-=? 令2z =,可得1,1y x ==,故平⾯PCD 的⼀个法向量为(1,1,2)n =.设直线PB 与平⾯PCD 所成⾓的⼤⼩为θ,则||sin ||||n PB n PB θ?== .所以直线PB 与平⾯PCD所成⾓的⼤⼩为arcsin 6.18.(本题满分14分)本题共有2个⼩题,第1⼩题满分6分,第2⼩题满分8分.解 (1)根据题意,可算得弧BC x θ=?(m ),弧10AD θ=(m ). ⼜30BA CD BC CD +++=弧弧,于是,10101030x x x θθ-+-+?+=,所以,210(010)10x x x θ+=<<+.(2) 依据题意,可知2化简,得2550y xx =-++25225()24x =--+. 于是,当52x =(满⾜条件010x <<)时,max 2254y =(2m ).答所以当52x =⽶时铭牌的⾯积最⼤,且最⼤⾯积为2254平⽅⽶.19.(本题满分14分)本题共有2个⼩题,第1⼩题满分6分,第2⼩题满分8分.解 (1)结合题意,可得12|3|d d x ==-.⼜12d d == 22162x y +=. 因此,所求动点(,)M x y 的轨迹C 的⽅程是22162x y +=. (2) 联⽴⽅程组221,62(2),x y y k x ?+==-?得222设点1122(,)(,)P x y Q x y 、,则2122212212,13126,130.k x x k k x x k ?+=?+?-?=?+?>于是,弦||PQ=点O到直线l的距离d=.由OPQS==42210k k-+=,解得1k=±,且满⾜0>,即1k=±都符合题意.因此,所求直线的⽅程为2020x y x y--=+-=或.20.(本题满分16分)本题共有3个⼩题,第1⼩题满分4分,第2⼩题满分6分,第3⼩题满分6分.解(1) 2x x--≤<-≤≤∴当10x-≤<时,()2,0()2 f x x f x=-<≤且.由2y x=-,得12x y=-,互换x y 与,可得11()(02)2f x x x-=-<≤.当01x≤≤时,2()1,()0f x x f x,得x=x y与,可得1()10)f x x-=-≤≤.11, 0<2,2()10.x xf xx--≤∴=-≤≤(2) 答函数图像上存在两点关于原点对称.设点00000(,)(01)(,)A x y xB x y<≤--+-=,即200120x x-+=,解得001(1,)x x=舍去,且满⾜01x<≤.因此,函数图像上存在点1,2(12) A B-和关于原点对称.(3) 考察函数()y f x=与函数y=当12x-≤≤-x ax---=,解得2+2x a =-,且由21+2a -≤-≤,得02a ≤≤.当12x -<≤时,有()f x <240ax -=,化简得 22(4)40a x ax ++=,解得2 4=0+4a x x a =-,或(当02a ≤≤时,24024aa -<-<+). 于是,123224,,024ax x x a a =-=-=++. 由32212()x x x x -=-,得22442=2(+)+442a a a a a -++,解得32a -=.因为1a =不符合题意,舍去;02a <=<,满⾜条件.因此,所求实数a =21.(本题满分18分)本题共有3个⼩题,第1⼩题满分3分,第2⼩题满分6分,第3⼩题满分9分.解 (1)根据题意,有*10,0,N n n n a b a n +>>=∈且.由*(N )nn b a n =∈,1b =111n a a b +===,*N n ∈.所以n a =*N n ∈.证明 (2) *11(N )n n n b b n a +=+∈,*10,0,N n n n a b a n +>>=∈且,∴11n n b a ++==n n b a ++=*N n ∈.∴22111n n n n b b a a ++-= ? ?,*N n ∈.∴数列2n n b a?是⾸项为211b a ?? ???、公差为1的等差数列.解(3)*1N )n n b n +=∈,*10,0,N n n n a b a n +>>=∈且,*N n n a b n <+≤∈,得11n a +<≤.{}n a 是等⽐数列,且0n a >,设公⽐为(0)r r >,则1*1(N )n n a a r n -=∈. ∴当1r >,即lim n n a →∞→+∞,与11n a +<⽭盾.因此,1r >不成⽴.当01r <<,即lim 0n n a →∞→,与11n a +<≤01r <<不成⽴.∴ 1r =,即数列{}n a 是常数列,于是,1n a a =*11(N )n n b n +∴=∈. 100n b b >∴> ,,数列{}n b 也是等⽐数列,设公⽐为(0)q q >,有11n n b b q +=.2n a +∴=可化为222221111111(1)2(1)0(1n n b a q a b q a a a --+-=<≤,*N n ∈.2222422111111111(1)0,20,(1)0,4(2)0b a a b a a a b a ->≠->?=-≥,∴关于x 的⼀元⼆次⽅程22222111111(1)2(1)0b a x a b x a a --+-=有且仅有两个⾮负实数根.⼀⽅⾯,n q (*N n ∈)是⽅程22222111111(1)2(1)0b a x a b x a a --+-=的根;另⼀⽅⾯,若1(0)q q ≠>,则⽆穷多个互不相等的234,,,,,,n q q q q q 都是该⼆次⽅程的根.这与该⼆次⽅程有且仅有两个⾮负实数根⽭盾!1q ∴=,即数列{}n b 也是常数列,于是,1n b b =,*N n ∈.∴由*1N )n nb n +=∈,得1a =把1a =1n a +=解得1b11a b ?=?∴?。
2018年4月佛山市高三生物教学质量检测(二)理综生物试题一、选择题1.下表中人体不同细胞的寿命和分裂能力不同,以下说法错误..的是A.白细胞的凋亡比红细胞快,这与白细胞吞噬病原体有关B.癌细胞的无限增殖是由正常基因突变为原癌基因引起的C.通常情况下,细胞的分化程度越高,细胞分裂能力越弱D.小肠上皮细胞寿命最短,这与基因控制的衰老和凋亡有关2.某同学进行了下列有关酶的实验:甲组: 淀粉溶液+新鲜唾液+斐林试剂→ 有砖红色沉淀乙组: 蔗糖溶液+新鲜唾液+斐林试剂→ 无砖红色沉淀丙组: 蔗糖溶液+蔗糖酶溶液+斐林试剂→ ?下列叙述正确的是A.丙组的实验结果是“无砖红色沉淀”B.加入斐林试剂后通常还应沸水浴加热C.该同学的实验目的是验证酶的专一性D.为省去水浴加热步骤,可用碘液代替斐林试剂3.鸡霍乱病原菌易致鸡死亡。
1880年,巴斯德用久置的鸡霍乱病原菌对鸡群进行注射,意外发现全部鸡存活。
再次培养新鲜病原菌,并在第一次注射鸡群的基础上扩大注射范围,结果仅有部分鸡存活。
下列分析正确的是A.久置的霍乱病原菌已经失去了抗原性B.第二次注射存活的鸡中,绝大多数接受过第一次注射C.如果继续对存活的鸡注射新鲜病原菌,结果是仅有少部分鸡存活D.第一次注射时,鸡体内没有相应的记忆细胞,因此只能发生非特异性免疫4.将DNA双链都被32P标记的某一雄性动物细胞(染色体数为2N)置于不含32P的培养基中,该细胞经过两次连续分裂形成4个大小形态相似的子细胞。
下列分析错误..的是A.若子细胞的染色体数为N,则子细胞的DNA均含32PB.若子细胞的染色体数为2N,则子细胞的DNA可能均含32PC.若子细胞中的染色体都含32P,则细胞分裂过程中会发生基因重组D.若子细胞中有的染色体不含32P,则这是同源染色体彼此分离的结果5.对板栗园内的栗瘿蜂和长尾小蜂的数量进行连续多年的监测,结果见下图。
下列说法正确的是A.栗瘿蜂在与长尾小蜂的种间竞争中占优势B.利用性信息素来防止果园害虫属于化学防治C.调查栗瘿蜂和长尾小蜂的种群密度用样方法D.栗树同化的能量部分储存于有机物中,其余以热能散失6.红绿色盲是一种常见的伴X染色体隐性遗传病。
2018年上海初三二模汇编—议论文2018年初三二模奉贤区(一)阅读下文,完成第14—18题(18分)在阅读中拥有“心力量”①我们看到的世界,总是受到许多因素的制约。
而阅读,就是那副帮助我们清晰看世界的眼镜。
②人的生活,包括物质生活与精神生活两个方面。
伟大的书,本身就同时拥有安顿灵魂和激发创造的力量。
物质的发展永远会受到各种制约,精神的蓬勃才是不竭的源泉。
毫无疑问,阅读正是为了让我们的生活更精彩、更有条理、更有方向、更有智慧。
更重要的是,阅读让我们的心灵有一个安身立命之所。
③社会需要诚信,人们需要信仰。
读书会让人知道世界的深奥,会让人明白自己的无知与渺小,会让人产生敬畏之心。
而敬畏之心是建立信仰的重要基础。
阅读不是为了帮助我们逃避生活,逃避与人打交道,而是为了帮助我们更积极地面对生活,拥有更和谐的人际关系。
共读共行,才能同心同行。
因此,我们需要用阅读来重建信仰。
④我们需要用阅读来创新创造。
好的书会让我们更年轻、更有活力。
读书,应该是一种挑战,向未知世界、同时向自己智力的一种挑战。
虽然批评式的阅读本身也是一种重要的读书方法,不过读书不是为了批评书中的缺点与错误,而是为了创新创造。
尤其对于研究问题与建构理论而言,阅读应该以学其所长为主。
⑤我们需要用阅读来探索世界。
我们看到的世界,总是受到许多因素的制约,比如“近视”“老花”等原因,或者“粗心”“马虎”等心理,让我们无法清晰把握这个世界。
而阅读,能够帮助我们清晰看世界。
⑥我们需要用阅读来反思自我。
“一千个读者就有一千个哈姆雷特”,同样一本书里,每个读者所读到的、所汲取的,都带有个人色彩。
每个真正用心阅读的人,总是能够在书中2018年初三二模金山区(一)阅读下文,完成第15—19题(20分)涵养几分静气①一则故事,读来耐人寻味。
父亲丢了一块表,抱怨着四处寻找,可怎么也找不到;等他出去后,儿子悄悄进屋,不一会儿就找到了。
父亲很好奇,儿子则回答:我就安静地坐着,听到滴答滴答的声音,表就找到了。
2018届高三英语二模汇编——六选四1、2018黄浦二模Directions:Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.A.The sensors on a fully self -driving car require special care.B.Orduña wouldn’t reveal exactly how they’re washing the vehicles.C.The most advanced cars on the planet require an old-fashioned handwashing.D.Meanwhile, some companies, such as Cruise, are building sensor cleaning equipment into their vehicles.E.There are a range of problems with putting a self-driving vehicle through a traditional car wash, experts say.F. A self-driving vehicle’s external needs to be cleaned even more frequently than a typical car because thesensors must remain free of obstacles.For Self-driving Cars, Car Washes are a NightmareCar washes have been automated for decades, but companies developing fully self-driving vehicles must rely on a human touch to keep their cars and trucks in working condition.__67__ For example, soap remainder or water spots could effectively “blind” a self-driving car. A traditional car wash’s heavy brushes could jar the vehicle’s sensors, disturbing their standardization and accuracy. Even worse, sensors, which can cost over $100,000, could be broken. __68__ Dirt, dead bugs, bird droppings or water spots can impact the vehicle’s ability to drive safely.Avis, which has years of experience managing rental cars, has been tasked with cleaning and refueling the self-driving vans of Waymo, the self-driving arm of Google’s parent company. Avis chief innovation officer Arthur Orduña told CNN, “There are special processes that definitely require a lot more care and focus, and you have to clean the vans quite often.”__69__ But other self-driving car companies such as Toyota, Aptiv, Drive, AI and Uber described to CNN that they use microfiber (微纤维) cloths along with rubbing alcohol, water or glass cleaner for manual cleanings.__70__ This should ease some need for manual cleaning. But because self-driving vehicles can have dozens of sensors, Seeva CEO Diane Lansinger doesn’t imagine products like this will be able to clean every camera, radar or LIDAR, a laser sensor that most experts see as essential for self-driving vehicles.答案:67-70 EFBD2、2018普陀二模A. In daily life, imitation can hurt us if we subconsciously hold poor role models.B. Creative people have an endless resource of ideas.C. It is how to use imagination creatively that troubles us.D. Why follow someone else’s way of cooking when I could create my own?E. But if you begin to enter this field, imitation proves useful.F. If you are going to follow someone, focus on their talent, not their bad character or unacceptablebehaviors.Blind imitation is self-destruction. To those who do not recognize their unique worth, imitation appears attractive; to those who know their strength, imitation is unacceptable.In the early stages of skill or character development, imitation is helpful. When I first learned to cook, I used recipes and turned out some tasty dishes. But soon I grew bored. ______67______Imitating role models is like using training whe els on a child’s bicycle; they help you get going, but once you find your own balance, you fly faster and farther without relying on them.______68______ If, as a child, you observed people whose lives were bad, you may have accepted their fear and pain as normal and gone on to follow what they did. If you do not make strong choices for yourself, you will get the results of the weak choices of others.In the field of entertainment, our culture glorifies celebrities. Those stars look great on screen. But when they step off screen, their personal lives may be disastrous. ______69______Blessed is the person willing to act on their sudden desire to create something unique. Think of the movies, books, teachers, and friends that have affected you most deeply. They touched you because their creations were motivated by inspiration, not desperation. The world is changed not by those who do what has been done before them, but by those who do what has been done inside them. ______70______ The problem a creator faces is not running out of material; it is what to do with the material knocking at the door of imagination.Study your role models, accept the gifts they have given, and leave behind what does not serve you. Then you can say, “I stand on the shoulders of my ancestors’ tragedies and declare victory, and know that they are cheering me on.”答案:67-70 D A F B3、2018徐汇二模A.Research shows that co-rumination is a double-edged sword.B.In that way, a problem shared can really be a problem halved.C.Girls are more likely than boys to co-ruminate with their close friends.D.Indeed, having close friends to trust may protect you against poor mental health.E.This human tendency to work things out in one’s mind, is common.F.How you co-ruminate matters too.People discuss their pro blems with friends in the hope that they’ll gain some insight into how to solve them. And even if they don’t find a way to solve their problems, it feels good to let off some steam.(67)_______________ How problems are discussed, though, can be the difference between halving a problem or doubling it.The term psychologists use for negative problem sharing is “co-rumination”. Co-rumination is the mutual encouragement to discuss problems repeatedly going over the same problems, anticipating future problems and focusing on negative feelings.(68)________________ In a study involving children aged seven to 15 years of age, researchers found thatco-rumination in both boys and girls is associated with “high-quality” and close friendships. However, in girls, it was also associated with anxiety and depression (the same association was not found with the boys). And studies suggest that co-rumination isn’t just a problem for girls. Co-rumination with work colleagues can increase the risk of stress and burn out, one study suggests.(69)________________ In a group of adults, the effects of co-rumination was compared between face-to-face contact, telephone contact, texting and social media. The positive effects of co-rumination were found in face-to-face contact, telephone contact and texting, but not in social media. The negative aspects of co-rumination (anxiety) was found in face-to-face communication and telephone contact, but not texting or social media. Verbal forms of communication seem to enhance both the positive and negative aspects of co-rumination more than non-verbal communication.Discussing problems with friends doesn’t always have to lead to worsening mental health, as long as the discussion involves finding solutions and the person with the problem acts on those solutions. Then, relationships can be positive and beneficial to both parties. (70)________________答案:67-70 DAFB4、2018杨浦二模A. We could see increased tensions and conflict as a result of pressure on scarce resources.B. The report shows that with proper efforts the number of climate migrants could be reduceddramatically, from 140 million to 40 million by 2050.C. Some measures that could help helpless people adapt to climate change were on the agenda.D. However, it noted that cities must take the time to plan for the possibility for a flood of people.E. Countries that are becoming more restrictive on migration are those who are largely responsible forglobal warming.F. It found that unless urgent action is taken, the regions will likely have to deal with tens of millions ofso-called climate migrants.A new report by the World Bank shows that the effects of climate change could force 140 million people to move within their countries by 2050.The report looked at three developing regions of the world—sub-Saharan Africa, South Asia and Latin America. __________67__________Climate migrants are people who are forced to move within their country because of water scarcity, crop failure, rising sea levels and storm surges due to climate change.__________68__________ It is important to help people make good decisions about whether to stay where they are or move to new locations where they are less vulnerable.The report noted that the effects of climate change will often force people to move from rural areas suffering from droughts or crop failures to cities where there are different opportunities. __________69__________“Without the right planning and support, people migrating from rural areas into cities could be facing new and even more dangerous risks,” said Kanta Kumari Rigaud, the report’s team lead. “__________70__________” Rigaud added.The report recommends key actions to help prevent wide-scale climate migration: cutting global greenhouse gas emissions; improving development planning at the local level for climate migration; and investing in data to better understand climate migration trends in each country.The report notes that any rise in climate migration willbe in addition to millions of other migrants within countries, moving for economic, social, political or other reasons.答案:67-70 F B D A5、2018崇明二模A.Age really should be treated as just a number.B.Felt age might play a role in more than just how you feel.C.Feeling young is about maintaining vitality as you get older.D.Self-felt age has the potential to change, so interventions (干预) may be possible.E.And other studies suggest that there could be more benefits to thinking yourself younger,besides a longer life.F.People who feel younger than the number of years they have on the clock get more pleasurethan people who feel their age.Feel Young at Heart and You’ll Enjoy a Longer LifeAge-liars and birthday-deniers... you’d best learn a thing or two from those who are young at heart. People who feel younger than their actual age may live longer than those who feel older than they truly are, a new study says.67 Results from the study, published in the Journal of the American Medical Association, suggest that people who feel a year or more older than they truly are could have around 41 percent greater risk for death.Researchers looked at nearly 6,500 older adults, with an average age of 65.8 for the study. Around 70 percent of them felt younger than they were, about a quarter felt their precise age and just under 5 percent felt a year or more older than they were, when asked “How old do you feel you are?”Those who felt older than they were had a higher death rate after a follow-up period of 99 months. While just 14.3 and 18.5 percent of people who felt younger or felt their age, respectively, died during those 99 months, 24.6 percent of those who felt aged beyond their years had died.The authors say more research is needed on the topic, but suggest it could be that those who feel “young at heart” have healthier behaviors and a stronger will to live. “ 68 Individuals who feel older could be targeted with health messages promoting positive health behaviors and attitudes toward aging,” the authors write in the study.The good news is that you can change your feeling of how young you are. 69 One recent study found that helping participants have positive feelings toward age, by showing them positive word associations, helped older adults improve in physical tasks like balancing and getting up out of a chair, in as little as four weeks. Another study found that negative feeling of aging and poor memory can make older adults feel up to five years older, regardless of their actual mental abilities.There you have it. 70答案:67. B 68. D 69. E 70. A6、2018长宁二模A.Some experts think that the wealth gained from trade allowed for such leisure in sports and the arts.B. Experts believe that a king and mostly a female priest ruled the government and controlled trade.C. In ancient Greek myths, Minoan society was quite prosperous and highly civilized.D. The palace that Evans unearthed a century ago was the first proof of Minoan culture.E. Minoan culture didn’t exist before Arthur Evans discovered the palace under the earth.F.Although not yet decoded, written script on clay tablets appears to list trade accounts.The Minoans: A Forgotten PeopleThe first advanced culture in ancient Greece was the Minoan culture. For thousands of years, knowledge of these people survived only in Greek myths. In the late 19th century, archaeologists began to unearth ruins. This inspired Arthur Evans to begin digging on the island of Crete near mainland Greece. On a dig in Kbossos, Evans found an ancient palace. Experts think that it was the palace of King Minos, acentral figure in many Greek myths.67 With his team, he uncovered a vast structure, varied works of art, and many hieroglyphic records. These finds, together with later finds, comprise all that experts know about Minoan culture.From the evidence experts gathered, it is clear that the Minoans were ahead of their time.The palace at Knossos was five floors high with hundreds of rooms. Buildings throughout the ancient city had plumbing and flush toilets. Stone pavement lined the surfaces of the roads. In addition, the Minoans possessed a highly developed naval fleet for long-distance trade. 68 These records confirm the central role of commerce in culture.Their analysis of the evidence also offers insight into some aspects of Minoan society.69 Ruins and artwork suggest that people of all classes enjoyed a high degree of social and gender equality. Religious icons(图符) show that Minoans worshiped bulls, the natural world, and many female gods.An unusual feature of Minoans culture was the pursuit of leisure interests. Sport and visual arts were central to Minoan life. Boxing and bull jumping, a sport in which players jumped over live bulls, were popular. Although bull jumping may have served some ritual purpose, experts believe that it was done mostly for fun. Similarly, although some works of art showed political and religious themes, other works served only as pleasant décor(装饰品). 70The Minoans met their demise after a series of natural disasters. Experts believe that group from the Greek mainland capitalized on these events and looked over the island.答案:67—70 DFBA7、2018松江(闵行)二模A. The rhythm of life is a slight beat.B. Experts say every aspect of human bodies is influenced by daily rhythms.C. Jenny Hope claims to have tested her body clocks to the limit.D. This is a key reason why shift work can cause so many problems.E. Many people study the state of their biorhythms before making their daily plans.F. Such a dangerous combination never gave their bodies’ natural rhythms a chance to settledown.Rhythm of LifeChoosing the right time to sleep, the correct moment to make decisions, the best hour to eat—and even go into hospital—could be your key to perfect health.Centuries after man discovered the rhythms of the planets and the cycles of crops, scientists have learned that we too live by precise rhythms that govern everything from our basic bodily functions to mental skills. Man is a prisoner of time.But it’s not just the experts who are switching on to the way our bodies work. 67Prince Charles consults a chart which tells him when he will be at his peak on a physical, emotional and intellectual level. Boxer Frank Bruno is another who charts his biorhythms to plan for big fights.68 Sleep, blood pressure, hormone levels and heartbeat all follow their own clocks, which may bear only slight relation to our manmade 24hour cycle.Research shows that in laboratory experiments when social signals and, most importantly, light indicators such as dawn are taken away, people lose touch with the 24hour clock and sleeping patterns change. Temperature and heartbeat cycles lengthen and settle into “days” lasting about 25 hours.In the real world, light and dark keep adjusting internal clock to the 24hour day. But the best indicator of performance is body temperature. As it falls from a 10 p.m. high of 37.2°C to a predawn low of 36.1°C, mental functions fall too. 69The most famous example is the nuclear accident at Three Mile Island in the US. The three operators in the control room worked alternating weeks of day, evening and night shifts.70 Investigators believe this caused the workers to overlook a warning light and fail to close an open valve.Finding the secret of what makes us tick has long fascinated scientists and work done over the last decade has yielded important clues. The aim is to help us become more efficient. For example, the time we eat may be important if we want to maximize intellectual or sporting performance. There is already evidence suggesting that the time when medicine is given to patients affects how well it works.答案:67-70 EBDF8、2018宝山二模A.They thus focused on two elements that are typical for contemporary work environments.B.The researchers used online questionnaires to obtain data.C.The notion that obtaining external feedback about one's ideas is essential for increasingcreativity is deeply rooted in society.D.The most important is that when an organization stimulates feedback seeking, it needs toensure that the work environment is excellent enough to use the benefits of feedback.E.The question, however, is whether these beneficial effects always occur.F. The creative achievements were assessed by their direct managers.Seeking feedback not always sufficient for stimulating creativity It is widely believed that seeking feedback from colleagues, managers, friends and family enhances employees' creativity. But is this always the case? No, a positive effect depends on the work environment. This isthe conclusion of new joint research study led by UvA work and organizational psychologist Roy Sijbom.(67)________For example, managers are encouraged to engage customers in order to confirm whether their business model is workable and scholars attend conferences to obtain feedback on their research results. A definite assumption is that individuals who have obtained feedback will also actually (be able to) use it.“The idea is simple: seeking feedback from different sources - also known as feedback source variety -- benefits one's creativity since it leads to a greater diversity of viewpoints”, says Sijbom. “And the more diverse the viewpoints, the more it benefits one's creativity because by combining all the different viewpoints new perspectives will emerge that in turn will result in more creativity. (68)________ ”The researchers examined how specific characteristics of the immediate work environment influence the relationship between feedback source variety and creative performance. (69)________ One is the recognized rate of change of performance standards. The other is the extent to which employees feel they have sufficient time to develop creative ideas at work (experienced creative time pressure). “We discovered a growing relationship between the search for input from a variety of feedback sources and creativity, but only when performance standards within an organization are changing and when a re latively low creative time pressure is experienced”, says Sijbom.Sijbom offers several recommendations: “(70)________In a more concrete sense, organizations can, for example, consider using feedback workshops in which employees are encouraged to reflect on diverse feedback and equipped with techniques and strategies on how to absorb feedback in their daily work. In addition, managers should not only stimulate their employees to actively cultivate relationships with potential feedback sources within and outside the organization, but also provide sufficient time to process the feedback obtained from these relationships.”答案:67--70 CEAD9、2018奉贤二模A. For OpenClassrooms, the company will certainly make profits as a technology provider.B. And it can explain why more and more people tend to apply for online courses.C. It’s the exact same degree that you would get at IESA, except that you won’t see any teachers.D.That’s why every week, you will get to video chat with a teacher.E. For OpenClassrooms, the company will surely enjoy a reputation for its technology.F. OpenClassrooms lets you work and study at the same time, and pay a lot less.You won’t have any excuse to skip class anymore. French startup OpenClassrooms is offering the first State-recognized bachelor degree in France that uses only MOOCs (massive open online courses). The startup partnered with IESA Multimédia to create this program.There are three learning paths in engineering, design and marketing. Students will have to complete all the courses and required projects in order to get their degree. 67 IESA is already working on 40 different MOOCs for this program.On average, it will take a year of hard work in order to complete all the classes. As always, it’s hard to keep going when you sign up for a MOOC. 68This kind of degree has many key advantages. For IESA, it gives the school more students. IESA is a privateschool, and its end goal is to make as much money as possible. So with these new MOOC students, IESA will be able to get more money per teacher on average.69 The startup already offers a course for €20 per month, but you need to pay €300 per month for the Premium Plus offering to use the state-recognized program. It’s unclear how much OpenClassrooms will keep, but it should be more than €20 per month.For students, it’s a cheaper way to get a degree. Maybe you can’t afford to study for three years at IESA and pay €6,950 per year. 70 Sure, it’s probably a less enjoyable experience than going to your school and spending time with other students and teachers, but it makes sense for some students.It’s an interesting new direction for OpenClassrooms, and I can’t wait to see whether other schools will start working with the startup to provide online courses. It will be interesting to see whether the first students are satisfied with this kind of degree as well.答案:67-70 C D A F10、2018嘉定二模题目与长宁区相同11、2018浦东二模A. Being simple might be another reason.B. It was the only affordable way to play them.C. We should have admiration for this old technology.D. The current trend for old games shows no sign of slowing.E. Newer consoles and their games are incredibly expensive.F. So it seems like it’s not ‘game over’ for old-school technology.Retro GamingThere’s no doubt that in today’s digital world, computer games are extremely sophisticated and capable of creating virtual reality experiences that were unimaginable only a few years ago. So I am interested to see that the simplistic games that I grew up with, are making a revival. But Why?In the 1970s, the original place to play a computer game was at an arcade. Here, you and your mates could try out the new big names in games such as Space Invaders and Pacman. _____67_____ And because of the technology involved, the gaming machines were too big to fit into your house.But in the 1980s and 90s, gaming arrived in our homes and people like me were addicted. The sound of beeping became a familiar sound emanating from bedrooms across the land! Names such as Tetris, Sonic and Street Fighter became popular language in the playground -- and now they are being talked about -- and played again. One of the reasons is the low cost. The BBC spoke to gamer, Gemma Wood, who says that: _____68_____ I understand that a lot of hard work has gone into the design etc., but how can anyone justify £50 to £60 for a game that you might not even enjoy?_____69_____ The graphics on old games may not compare with the detail and definition of modern gamesbut they are fun and easy to use by children and adults alike. And of course, nostalgia plays its part. Some people want to relive their childhood while for others, it is a chance to show their children the computer games they grew up with.Technology journalist, KG Orphanides, says “it's important to recognize how well-designed many of those classic games are... the developers had so little space to work with -- your average Sega Mega Drive or SNES cartridge had a maximum capacity of just 4mb -- and limited graphics and sound capabilities.” This compares to an average capacity of 40G in today's games. _____70_____This craze for using retro hardware and grabbing an old joystick is certainly catching on. And to persuade those of us who are not sure about downgrading the gaming experience, manufacturers such as Nintendo, are bringing back some of their older consoles in new style casing.答案:67-70 B E A C12、2018静安二模A. The appropriate feeding and caring styles were greatly rewarded in all aspects.B. It was claimed that simply feeding and changing them would be adequate.C. Some, however, demonstrate little influence with roughly ordinary behavior pattern.a> D. Maternal attachment plays a fundamental role in shaping who we are.b> E. Necessities are not just the availability of food and water.c> F. And they simply didn’t grow like normal infants.early maltreatment can derail a child’s development requires careful study.In a famous research, Harry Harlow had demonstrated that proper psychological and physical development of infants requires nurturing and attention from a parent. ___67___ In that research, socially isolated monkey babies that were removed from their mothers were found to be clinging to a cloth-covered surrogate(替代的) mother for comfort.Such experimentations sound cruel. They, however, have been critical in helping change policies in human orphanages( 孤儿院) in the U.S. For centuries some orphanages treated infants equally inhumanely. Despite early evidence that orphanage infants were far more likely to die than others, supporters argued that it didn’t matter whether children had “parents” specially devoted to them at the orphanage. ___68___ Orphans were supposed to be in positive mental and physical health until adoptive parents were found. Babies, they said, couldn’t remember anyway.The harrowing consequences of these theories were most vividly brought to light in Romania in the 1980s and ‘90s. A ban on abortion(流产) led to a surge in orphanage babies. Simply being fed and changed without individualized affection, some babies present serious problems. Many developed violent behaviors, repetitively rocking or banging their heads. Some were cold and withdrawn or indiscriminately affectionate. ___69___ Their head sizes were especially small. They even had problems with attention and comprehension. The longer these children were left alone, the more damage was seen.The lack of a secure attachment relationship in the early years has destructive consequences for both physical and mental health later in life, with long-lasting effects. The persistence of these effects emphasizes the need to intervene early in life. The Nobel-prize-winning economist James Heckman, has long argued that investing in early childhood education provides a greater return for society than virtually any other type of spending. It is obviously reflected in increased educational success and productivity,. The reduced crime, addiction, distressand disorder point to the same theory. Early life conditions critically affect adult life. ___70___ Remove it and the harm is great.答案:67-70 EBFD13、2018青浦二模A. I truly express my respect for Shan and his team for their contributions.B. With regard to cultural heritage restoration, Shan said the museum opened a restoration hospital atthe end of 2016.C. A total of 600 people from all walks of life, including over 100 foreign guests, participated in theactivity.D. I will learn more about Chinese culture from the magnificent ancient objects.E. Our design teams often study consumer demands and create cultural items that are nice to look atand practical to use.F. Traditional craftsmanship is combined with modern methods, and the lives of ancient culturalobjects will be lengthened by the so-called doctors.Make traditional treasures come aliveThe Palace Museum Director Shan Jixiang delivered a cultural heritage speech on Feb 27 in Beijing, which was co-organized by the Beijing Diplomatic Service Bureau and Beijing Housing Service Corporation for Diplomatic Missions. 67On the theme The World of the Palace Museum and the Palace Museum of the World, the 64-year-old director shared his ideas about how to make traditional treasures come alive again. During the speech, which lasted two and a half hours, Shan touched on topics including upgrading museum infrastructure(基础设施), restoring cultural sites, digitalizing online museums, setting up restoration hospitals, providing better visitor experiences and promoting the Palace Museum’s cultural items.“The abundant collection of cultural objects at the Palace Museum is the inspiration for the creative souvenirs and cultural items available,” Shan said. “ 68 ” Throughout 2017, the total sales of Palace Museum’s cultural items have been more than 1 billion yuan ($158million). Explaining the huge success of Palace Museum’s cultural souvenirs, Shan said: “The museum opened a shop on the e-commerce website Taobao in 2008, but sales remained neither high nor low for years, as more than 80 percent of the souvenirs sold in stores in the past were not related to our museum.” “Therefore, I wanted to change the situation. Now, souvenirs from the Palace Museum cover almost every aspect of life. After all, what matters to a museum is not how many visitors they have, but how close they are to people’s daily lives.”69 Around 200 “doctors” are employed to analyze, examine, detect flaws or damage in ancient objects and restore them using more than 100 pieces of specialized equipment, including 3-D printers and scanners. The restoration hospital covers 13,000 square meters and boasts the nation’s most advanced restoration workshops.John Aquilina, Malta’s ambassador to China said that Shan’s speech showed a totally different Palace Museum to foreign people. “China enjoys a long and profound culture and many of the national treasures have been preserved at the Palace Museum. It is no easy task to preserve them well. 70 ”。
2018年中考数学二模试卷一、.选择题:本大题共12小题,在每小题给出的四个选项中,只有一项是正确的,请把正确的选项选出来.每小题选对得3分,选错、不选或选出的答案超过一个均记零分1.计算(ab2)3的结果是()A.ab5B.ab6C.a3b5D.a3b62.下列各式中,不成立的是()A.|﹣3|=3 B.﹣|3|=﹣3 C.|﹣3|=|3| D.﹣|﹣3|=33.在实数﹣,0,,,,中,无理数有()A.1个B.2个C.3个D.4个4.如图,AB是⊙O直径,∠AOC=130°,则∠D=()A.65°B.25°C.15°D.35°5.如图是由四个小正方体叠成的一个立体图形,那么它的主视图是()A.B.C.D.6.已知抛物线y=x2﹣x﹣1与x轴的一个交点为(m,0),则代数式m2﹣m+2014的值为()A.2012 B.2013 C.2014 D.20157.如图,在△ABC中,已知∠C=90°,BC=3,AC=4,⊙O是内切圆,E,F,D分别为切点,则tan∠OBD=()A.B.C.D.8.如图,在▱ABCD中,AC与BD交于点O,点E是BC边的中点,OE=1,则AB的长是()A.1 B.2 C.D.49.某快餐店用米饭加不同炒菜配制了一批盒饭,配土豆丝炒肉的有25盒,配芹菜炒肉丝的有30盒,配辣椒炒鸡蛋的有10盒,配芸豆炒肉片的有15盒.每盒盒饭的大小、外形都相同,从中任选一盒,不含辣椒的概率是()A.B.C.D.10.定义:如果一元二次方程ax2+bx+c=0(a≠0)满足a+b+c=0,那么我们称这个方程为“凤凰”方程.已知ax2+bx+c=0(a≠0)是“凤凰”方程,且有两个相等的实数根,则下列结论正确的是()A.a=c B.a=b C.b=c D.a=b=c11.如图,已知△ABC中,∠ABC=90°,AB=BC,三角形的顶点在相互平行的三条直线l1,l2,l3上,且l1,l2之间的距离为2,l2,l3之间的距离为3,则AC的长是()A.B. C. D.712.如图,抛物线y=ax2+bx+c与x轴交于点A(﹣1,0),顶点坐标为(1,n),与y轴的交点在(0,2)、(0,3)之间(包含端点),则下列结论:①当x>3时,y<0;②3a+b>0;③﹣1≤a≤﹣;④3≤n≤4中,正确的是()A.①②B.③④C.①④D.①③二、填空题:本大题共6小题,共24分,只要求填写最后结果,每小题填对得4分.13.因式分解:x2﹣2xy+y2=.14.将三角板(不是等腰的)顶点放置在直线AB上的O点处,使AB∥CD,则∠2的余弦值是.15.如图,△ABC中,AB=AC,∠A=30°,DE垂直平分AC,则∠BCD的度数为.16.方程x2﹣2x﹣1=0的解是.17.如图,点E在正方形ABCD内,满足∠AEB=90°,AE=6,BE=8,则阴影部分的面积是.18.猜数字游戏中,小明写出如下一组数:,,,,…,小亮猜想出第六个数字是,根据此规律,第n个数是.三、选修题、本小题满分6分,请在下列两个小题中,任选其一完成即可19.(1)解方程组:(2)解不等式组:.四、解答题:本大题共7个小题,满分54分.解答时请写出必要的演推过程.20.计算﹣2sin45°+(﹣2)﹣3+()0.21.为了解学生的课余生活情况,某中学在全校范围内随机抽取部分学生进行问卷调查.问卷中请学生选择最喜欢的课余生活种类(2007•台州)如图,△ABC内接于⊙O,点D在半径OB的延长线上,∠BCD=∠A=30°.(1)试判断直线CD与⊙O的位置关系,并说明理由;(2)若⊙O的半径长为1,求由弧BC、线段CD和BD所围成的阴影部分面积.(结果保留π和根号)23.海丰塔是无棣灿烂文化的象征(如图①),喜爱数学实践活动的小伟查资料得知:海丰塔,史称唐塔,原名大觉寺塔,始建于唐贞观十三年(公元639年),碑记为“尉迟敬德监建”,距今已1300多年,被誉为冀鲁三胜之一.小伟决定用自己所学习的知识测量海丰塔的高度.如图②,他利用测角仪站在B处测得海丰塔最高点P的仰角为45°,又前进了18米到达A处,在A处测得P的仰角为60°.请你帮助小伟算算海丰塔的高度.(测角仪高度忽略不计,≈1.7,结果保留整数).24.如图,在平行四边形ABCD中,AE⊥BC于E,AF⊥CD于F,BD与AE、AF分别相交于G、H.(1)求证:△ABE∽△ADF;(2)若AG=AH,求证:四边形ABCD是菱形.25.我们把一个半圆与抛物线的一部分合成的封闭图形称为“蛋圆”,如果一条直线与“蛋圆”只有一个交点,那么这条直线叫做“蛋圆”的切线.如图,点A、B、C、D分别是“蛋圆”与坐标轴的交点,AB为半圆的直径,点M为圆心,A点坐标为(﹣2,0),B点坐标为(4,0),D点的坐标为(0,﹣4).(1)你能求出经过点C的“蛋圆”切线的解析式吗?试试看;(2)请你求出“蛋圆”抛物线部分的解析式,并写出自变量x的取值范围.(3)你能求出经过点D的“蛋圆”切线的解析式吗?能,请写出过程,不能,请说明理由.参考答案与试题解析一、.选择题:本大题共12小题,在每小题给出的四个选项中,只有一项是正确的,请把正确的选项选出来.每小题选对得3分,选错、不选或选出的答案超过一个均记零分1.计算(ab2)3的结果是()A.ab5B.ab6C.a3b5D.a3b6【考点】幂的乘方与积的乘方.【分析】根据积的乘方的性质进行计算,然后直接选取答案即可.【解答】解:(ab2)3=a3•(b2)3=a3b6.故选D.【点评】本题考查积的乘方,把积中的每一个因式分别乘方,再把所得的幂相乘.2.下列各式中,不成立的是()A.|﹣3|=3 B.﹣|3|=﹣3 C.|﹣3|=|3| D.﹣|﹣3|=3【考点】绝对值.【分析】根据绝对值的意义选择.【解答】解:A中|﹣3|=3,正确;B中﹣|3|=﹣3,正确;C中|﹣3|=|3|=3,正确;D中﹣|﹣3|=﹣3,不成立.故选D.【点评】本题考查绝对值的化简:一个正数的绝对值是它本身;一个负数的绝对值是它的相反数;0的绝对值是0.3.在实数﹣,0,,,,中,无理数有()A.1个B.2个C.3个D.4个【考点】无理数.【分析】根据无理数的三种形式求解.【解答】解:=3,=﹣2,无理数有:,,共2个.故选B.【点评】本题考查了无理数的知识,解答本题的关键是掌握无理数的三种形式:①开方开不尽的数,②无限不循环小数,③含有π的数.4.如图,AB是⊙O直径,∠AOC=130°,则∠D=()A.65°B.25°C.15°D.35°【考点】圆周角定理.【专题】压轴题.【分析】先根据邻补角的定义求出∠BOC,再利用圆周角定理求解.【解答】解:∵∠AOC=130°,∴∠BOC=180°﹣∠AOC=180°﹣130°=50°,∴∠D=×50°=25°.故选B.【点评】本题利用了圆周角定理和邻补角的概念求解.5.如图是由四个小正方体叠成的一个立体图形,那么它的主视图是()A.B.C.D.【考点】简单组合体的三视图.【分析】找到从正面看所得到的图形即可,注意所有的看到的棱都应表现在主视图中.【解答】解:从正面看易得第一层有3个正方形,第二层中间有1个正方形.故选C.【点评】本题考查了三视图的知识,主视图是从物体的正面看得到的视图.6.已知抛物线y=x2﹣x﹣1与x轴的一个交点为(m,0),则代数式m2﹣m+2014的值为()A.2012 B.2013 C.2014 D.2015【考点】抛物线与x轴的交点.【分析】把x=m代入方程x2﹣x﹣1=0求得m2﹣m=1,然后将其整体代入代数式m2﹣m+2014,并求值.【解答】解:∵抛物线y=x2﹣x﹣1与x轴的一个交点为(m,0),∴m2﹣m﹣1=0,解得m2﹣m=1.∴m2﹣m+2014=1+2014=2015.故选:D.【点评】本题考查了抛物线与x轴的交点.解题时,注意“整体代入”数学思想的应用,减少了计算量.7.如图,在△ABC中,已知∠C=90°,BC=3,AC=4,⊙O是内切圆,E,F,D分别为切点,则tan∠OBD=()A.B.C.D.【考点】三角形的内切圆与内心;切线长定理.【专题】压轴题.【分析】首先根据切线的性质和切线长定理证得四边形OECD是正方形,那么AC+BC﹣AB即为2R(⊙O 的半径R)的值,由此可得到OD、CD的值,进而可在Rt△OBD中求出∠OBD的正切值.【解答】解:∵BC、AC、AB都是⊙O的切线,∴CD=CE、AE=AF、BF=BD,且OD⊥BC、OE⊥AC;易证得四边形OECD是矩形,由OE=OD可证得四边形OECD是正方形;设OD=OE=CD=R,则:AC+BC﹣AB=AE+R+BD+R﹣AF﹣BF=2R,即R=(AC+BC﹣AB)=1,∴BD=BC﹣CD=3﹣1=2;在Rt△OBD中,tan∠OBD==.故选C.【点评】此题考查的是三角形的外切圆,切线长定理以及锐角三角形函数的定义,难度适中.8.如图,在▱ABCD中,AC与BD交于点O,点E是BC边的中点,OE=1,则AB的长是()A.1 B.2 C.D.4【考点】平行四边形的性质;三角形中位线定理.【分析】由四边形ABCD是平行四边形,根据平行四边形的对角线互相平分,即可求得OC=OA,又由点E 是BC边的中点,根据三角形中位线的性质,即可求得AB的长.【解答】解:∵四边形ABCD是平行四边形,∴OC=OA,∵点E是BC边的中点,即BE=CE,∴OE=AB,∵OE=1,∴AB=2.故选B.【点评】此题考查了平行四边形的性质与三角形中位线的性质.注意平行四边形的对角线互相平分,三角形的中位线平行于三角形的第三边且等于第三边的一半.9.某快餐店用米饭加不同炒菜配制了一批盒饭,配土豆丝炒肉的有25盒,配芹菜炒肉丝的有30盒,配辣椒炒鸡蛋的有10盒,配芸豆炒肉片的有15盒.每盒盒饭的大小、外形都相同,从中任选一盒,不含辣椒的概率是()A.B.C.D.【考点】概率公式.【分析】让不含辣椒的盒饭数除以总盒饭数即为从中任选一盒,不含辣椒的概率.【解答】解:配土豆丝炒肉的有25盒,配芹菜炒肉丝的有30盒,配辣椒炒鸡蛋的有10盒,配芸豆炒肉片的有15盒,全部是80盒,不含辣椒的有70盒,所以从中任选一盒,不含辣椒的概率是=.故选A .【点评】本题比较容易,考查等可能条件下的概率.用到的知识点为:概率=所求情况数与总情况数之比.10.定义:如果一元二次方程ax 2+bx+c=0(a ≠0)满足a+b+c=0,那么我们称这个方程为“凤凰”方程.已知ax 2+bx+c=0(a ≠0)是“凤凰”方程,且有两个相等的实数根,则下列结论正确的是( ) A .a=c B .a=b C .b=c D .a=b=c 【考点】根的判别式. 【专题】压轴题;新定义.【分析】因为方程有两个相等的实数根,所以根的判别式△=b 2﹣4ac=0,又a+b+c=0,即b=﹣a ﹣c ,代入b 2﹣4ac=0得(﹣a ﹣c )2﹣4ac=0,化简即可得到a 与c 的关系.【解答】解:∵一元二次方程ax 2+bx+c=0(a ≠0)有两个相等的实数根, ∴△=b 2﹣4ac=0,又a+b+c=0,即b=﹣a ﹣c ,代入b 2﹣4ac=0得(﹣a ﹣c )2﹣4ac=0,即(a+c )2﹣4ac=a 2+2ac+c 2﹣4ac=a 2﹣2ac+c 2=(a ﹣c )2=0, ∴a=c . 故选A【点评】一元二次方程根的情况与判别式△的关系: (1)△>0⇔方程有两个不相等的实数根; (2)△=0⇔方程有两个相等的实数根; (3)△<0⇔方程没有实数根.11.如图,已知△ABC 中,∠ABC=90°,AB=BC ,三角形的顶点在相互平行的三条直线l 1,l 2,l 3上,且l 1,l 2之间的距离为2,l 2,l 3之间的距离为3,则AC 的长是( )A.B. C. D.7【考点】勾股定理;全等三角形的性质;全等三角形的判定.【专题】计算题;压轴题.【分析】过A、C点作l3的垂线构造出直角三角形,根据三角形全等和勾股定理求出BC的长,再利用勾股定理即可求出.【解答】解:作AD⊥l3于D,作CE⊥l3于E,∵∠ABC=90°,∴∠ABD+∠CBE=90°又∠DAB+∠ABD=90°∴∠BAD=∠CBE,,∴△ABD≌△BCE∴BE=AD=3在Rt△BCE中,根据勾股定理,得BC==,在Rt△ABC中,根据勾股定理,得AC=×=2;故选A.【点评】此题要作出平行线间的距离,构造直角三角形.运用全等三角形的判定和性质以及勾股定理进行计算.12.如图,抛物线y=ax2+bx+c与x轴交于点A(﹣1,0),顶点坐标为(1,n),与y轴的交点在(0,2)、(0,3)之间(包含端点),则下列结论:①当x>3时,y<0;②3a+b>0;③﹣1≤a≤﹣;④3≤n≤4中,正确的是()A.①②B.③④C.①④D.①③【考点】二次函数图象与系数的关系.【专题】计算题;压轴题.【分析】①由抛物线的对称轴为直线x=1,一个交点A(﹣1,0),得到另一个交点坐标,利用图象即可对于选项①作出判断;②根据抛物线开口方向判定a的符号,由对称轴方程求得b与a的关系是b=﹣2a,将其代入(3a+b),并判定其符号;③根据两根之积=﹣3,得到a=﹣,然后根据c的取值范围利用不等式的性质来求a的取值范围;④把顶点坐标代入函数解析式得到n=a+b+c=c,利用c的取值范围可以求得n的取值范围.【解答】解:①∵抛物线y=ax2+bx+c与x轴交于点A(﹣1,0),对称轴直线是x=1,∴该抛物线与x轴的另一个交点的坐标是(3,0),∴根据图示知,当x>3时,y<0.故①正确;②根据图示知,抛物线开口方向向下,则a<0.∵对称轴x=﹣=1,∴b=﹣2a,∴3a+b=3a﹣2a=a<0,即3a+b<0.故②错误;③∵抛物线与x轴的两个交点坐标分别是(﹣1,0),(3,0),∴﹣1×3=﹣3,∴=﹣3,则a=﹣.∵抛物线与y轴的交点在(0,2)、(0,3)之间(包含端点),∴2≤c≤3,∴﹣1≤﹣≤﹣,即﹣1≤a≤﹣.故③正确;④根据题意知,a=﹣,﹣=1,∴b=﹣2a=,∴n=a+b+c=c.∵2≤c≤3,∴≤c≤4,即≤n≤4.故④错误.综上所述,正确的说法有①③.故选D.【点评】本题考查了二次函数图象与系数的关系.二次函数y=ax2+bx+c系数符号由抛物线开口方向、对称轴、抛物线与y轴的交点抛物线与x轴交点的个数确定.二、填空题:本大题共6小题,共24分,只要求填写最后结果,每小题填对得4分.13.因式分解:x2﹣2xy+y2=(x﹣y)2.【考点】因式分解-运用公式法.【专题】计算题.【分析】根据完全平方公式直接解答即可.【解答】解:原式=(x﹣y)2.故答案为(x﹣y)2.【点评】本题考查了因式分解﹣﹣运用公式法,熟悉因式分解是解题的关键.14.将三角板(不是等腰的)顶点放置在直线AB上的O点处,使AB∥CD,则∠2的余弦值是.【考点】特殊角的三角函数值;平行线的性质.【专题】探究型.【分析】先根据平行线的性质及直角三角板的特点求出∠2的度数,再根据特殊角的三角函数值进行解答即可.【解答】解:由三角板的特点可知,∠D=60°,∵AB∥CD,∴∠D=∠2=60°,∴cos∠2=cos60°=.故答案为:.【点评】本题考查的是直角三角板的特点及平行线的性质、特殊角的三角函数值,熟记特殊角的三角函数值是解答此题的关键.15.如图,△ABC中,AB=AC,∠A=30°,DE垂直平分AC,则∠BCD的度数为45°.【考点】线段垂直平分线的性质.【专题】计算题.【分析】首先利用线段垂直平分线的性质推出∠DAC=∠DCA,根据等腰三角形的性质可求出∠ABC=∠ACB,易求∠BCD的度数.【解答】解:∵AB=AC,∠A=30°(已知)∴∠ABC=∠ACB==75°∵DE垂直平分AC,∴AD=CD;∴∠A=∠ACD=30°,∴∠BCD=∠ACB﹣∠ACD,∴∠BCD=45°;故答案为:45°.【点评】本题主要考查了线段垂直平分线的性质以及等腰三角形的性质,难度一般.16.方程x2﹣2x﹣1=0的解是x1=1+,x2=1﹣.【考点】解一元二次方程-配方法.【分析】首先把常数项2移项后,然后在左右两边同时加上一次项系数﹣2的一半的平方,然后开方即可求得答案.【解答】解:∵x2﹣2x﹣1=0,∴x2﹣2x=1,∴x2﹣2x+1=2,∴(x﹣1)2=2,∴x=1±,∴原方程的解为:x1=1+,x2=1﹣.故答案为:x1=1+,x2=1﹣.【点评】此题考查了配方法解一元二次方程.解题时注意配方法的一般步骤:(1)把常数项移到等号的右边;(2)把二次项的系数化为1;(3)等式两边同时加上一次项系数一半的平方.选择用配方法解一元二次方程时,最好使方程的二次项的系数为1,一次项的系数是2的倍数.17.如图,点E在正方形ABCD内,满足∠AEB=90°,AE=6,BE=8,则阴影部分的面积是76.【考点】勾股定理;正方形的性质.【分析】根据勾股定理求出AB,分别求出△AEB和正方形ABCD的面积,即可求出答案.【解答】解:∵在Rt△AEB中,∠AEB=90°,AE=6,BE=8,∴由勾股定理得:AB==10,∴正方形的面积是10×10=100,∵△AEB的面积是AE×BE=×6×8=24,∴阴影部分的面积是100﹣24=76,故答案是:76.【点评】本题考查了正方形的性质,三角形的面积,勾股定理的应用,主要考查学生的计算能力和推理能力.18.猜数字游戏中,小明写出如下一组数:,,,,…,小亮猜想出第六个数字是,根据此规律,第n个数是.【考点】规律型:数字的变化类.【分析】根据分数的分子是2n,分母是2n+3,进而得出答案即可.【解答】解:∵分数的分子分别是:2 2=4,23=8,24=16,…分数的分母分别是:2 2+3=7,23+3=11,24+3=19,…∴第n个数是.故答案为:.【点评】此题主要考查了数字变化规律,根据已知得出分子与分母的变化规律是解题关键.三、选修题、本小题满分6分,请在下列两个小题中,任选其一完成即可19.(1)解方程组:(2)解不等式组:.【考点】解二元一次方程组;解一元一次不等式组.【专题】计算题.【分析】(1)方程组利用加减消元法求出解即可;(2)求出不等式组中两不等式的解集,找出解集的公共部分即可.【解答】解:(1)①+②得:4x=20,即x=5,把x=5代入①得:y=1,则方程组的解为;(2),由①得:x<﹣1,由②得:x≤2,则不等式组的解集为x<﹣1.【点评】此题考查了解二元一次方程组,以及解一元一次不等式组,熟练掌握运算法则是解本题的关键.四、解答题:本大题共7个小题,满分54分.解答时请写出必要的演推过程.20.计算﹣2sin45°+(﹣2)﹣3+()0.【考点】实数的运算;零指数幂;负整数指数幂;特殊角的三角函数值.【专题】计算题.【分析】原式第一项利用二次根式性质化简,第二项利用特殊角的三角函数值计算,第三项利用负整数指数幂法则计算,最后一项利用零指数幂法则计算即可得到结果.【解答】解:原式=﹣1﹣2×﹣+1=﹣.【点评】此题考查了实数的运算,熟练掌握运算法则是解本题的关键.21.为了解学生的课余生活情况,某中学在全校范围内随机抽取部分学生进行问卷调查.问卷中请学生选择最喜欢的课余生活种类(2)易知选择音乐类的有4人,选择美术类的有3人.记选择音乐类的4人分别是A1,A2,A,小丁;选择美术类的3人分别是B1,B2,小李.可画出树状图如下:由树状图可知共有12种选取方法,小丁和小李都被选中的情况仅有1种,所以小丁和小李恰好都被选中的概率是或列表:由表可知共有12中选取方法,小丁和小李都被选中的情况仅有1种,所以小丁和小李恰好都被选中的概率是;(3)由(1)可知问卷中最喜欢体育运动的学生占40%,由样本估计总体得得500×40%=200名.所以该年级中最喜欢体育运动的学生约有200名.【点评】本题考查的是条形统计图和扇形统计图及用样本估计总体等知识的综合运用.读懂统计图,从不同的统计图中得到必要的信息是解决问题的关键.条形统计图能清楚地表示出每个项目的数据;扇形统计图直接反映部分占总体的百分比大小.22.如图,△ABC内接于⊙O,点D在半径OB的延长线上,∠BCD=∠A=30°.(1)试判断直线CD与⊙O的位置关系,并说明理由;(2)若⊙O的半径长为1,求由弧BC、线段CD和BD所围成的阴影部分面积.(结果保留π和根号)【考点】切线的判定;扇形面积的计算.【专题】几何综合题.【分析】(1)由已知可证得OC⊥CD,OC为圆的半径所以直线CD与⊙O相切;(2)根据已知可求得OC,CD的长,则利用S阴影=S△COD﹣S扇形OCB求得阴影部分的面积.【解答】解:(1)直线CD 与⊙O 相切, ∵在⊙O 中,∠COB=2∠CAB=2×30°=60°, 又∵OB=OC , ∴△OBC 是正三角形, ∴∠OCB=60°, 又∵∠BCD=30°, ∴∠OCD=60°+30°=90°, ∴OC ⊥CD , 又∵OC 是半径, ∴直线CD 与⊙O 相切.(2)由(1)得△OCD 是Rt △,∠COB=60°, ∵OC=1, ∴CD=,∴S △COD =OC •CD=,又∵S 扇形OCB =,∴S 阴影=S △COD ﹣S 扇形OCB =.【点评】此题主要考查学生对切线的性质及扇形的面积公式的理解及运用.23.海丰塔是无棣灿烂文化的象征(如图①),喜爱数学实践活动的小伟查资料得知:海丰塔,史称唐塔,原名大觉寺塔,始建于唐贞观十三年(公元639年),碑记为“尉迟敬德监建”,距今已1300多年,被誉为冀鲁三胜之一.小伟决定用自己所学习的知识测量海丰塔的高度.如图②,他利用测角仪站在B 处测得海丰塔最高点P 的仰角为45°,又前进了18米到达A 处,在A 处测得P 的仰角为60°.请你帮助小伟算算海丰塔的高度.(测角仪高度忽略不计,≈1.7,结果保留整数).【考点】解直角三角形的应用-仰角俯角问题.【分析】设海丰塔的高OP=x,在Rt△POB中表示出OB,在Rt△POA中表示出OA,再由AB=18米,可得出方程,解出即可得出答案.【解答】解:设海丰塔的高OP=x,在Rt△POB中,∠OBP=45°,则OB=OP=x,在Rt△POA中,∠OAP=60°,则OA==x,由题意得,AB=OB﹣OA=18m,即x﹣x=18,解得:x=27+9,故海丰塔的高度OP=27+9≈42米.答:海丰塔的高度约为42米.【点评】本题考查了解直角三角形的应用,要求学生能借助仰角构造直角三角形并解直角三角形,注意方程思想的运用.24.如图,在平行四边形ABCD中,AE⊥BC于E,AF⊥CD于F,BD与AE、AF分别相交于G、H.(1)求证:△ABE∽△ADF;(2)若AG=AH,求证:四边形ABCD是菱形.【考点】菱形的判定;全等三角形的判定与性质;平行四边形的性质;相似三角形的判定与性质.【专题】证明题.【分析】(1)利用两角对应相等可证出△ABE∽△ADF;(2)利用(1)的结论,先证出△ABG≌△ADH,得到AB=AD,那么平行四边形ABCD是菱形.【解答】证明:(1)∵AE⊥BC,AF⊥CD,∴∠AEB=∠AFD=90度.∵四边形ABCD是平行四边形,∴∠ABE=∠ADF.∴△ABE∽△ADF.(2)∵△ABE∽△ADF,∴∠BAG=∠DAH.∵AG=AH,∴∠AGH=∠AHG,从而∠AGB=∠AHD,∴△ABG≌△ADH,∴AB=AD.∵四边形ABCD是平行四边形,∴四边形ABCD是菱形.【点评】本题利用了相似三角形的判定和性质,全等三角形的判定和性质以及菱形的判定.25.我们把一个半圆与抛物线的一部分合成的封闭图形称为“蛋圆”,如果一条直线与“蛋圆”只有一个交点,那么这条直线叫做“蛋圆”的切线.如图,点A、B、C、D分别是“蛋圆”与坐标轴的交点,AB为半圆的直径,点M为圆心,A点坐标为(﹣2,0),B点坐标为(4,0),D点的坐标为(0,﹣4).(1)你能求出经过点C的“蛋圆”切线的解析式吗?试试看;(2)请你求出“蛋圆”抛物线部分的解析式,并写出自变量x的取值范围.(3)你能求出经过点D的“蛋圆”切线的解析式吗?能,请写出过程,不能,请说明理由.【考点】二次函数综合题.【分析】(1)易得点A、B的坐标,用交点式设出二次函数解析式,把D坐标代入即可.自变量的取值范围是点A、B之间的数.(2)先设出切线与x轴交于点E.利用直角三角形相应的三角函数求得EM的长,进而求得点E坐标,把C、E坐标代入一次函数解析式即可求得所求的解析式.(3)设出所求函数解析式,让它与二次函数组成方程组,消除y,让跟的判别式为0,即可求得一次函数的比例系数k.【解答】解:(1)如图,设经过点C“蛋圆”的切线CE交x轴于点E,连结CM,∴CM⊥CE,又∵A点坐标为(﹣2,0),B点坐标为(4,0),AB为半圆的直径,点M为圆心,∴M点的坐标为(1,0),∴AO=2,BO=4,OM=1.又因为CO⊥x轴,所以CO2=AO•OB,解得:CO=2,又∵CM⊥CE,CO⊥x轴,∴CO2=EO•OM,解之得:EO=8,∴E点的坐标是(﹣8,0),∴切线CE的解析式为:y=x+2;(2)根据题意可得:A(﹣2,0),B(4,0);则设抛物线的解析式为y=a(x+2)(x﹣4)(a≠0),又∵点D(0,﹣4)在抛物线上,∴a=;∴y=x2﹣x﹣4自变量取值范围:﹣2≤x≤4;(3)设过点D(0,﹣4),“蛋圆”切线的解析式为:y=kx﹣4(k≠0),由题意可知方程组只有一组解.即kx﹣4=x2﹣x﹣4有两个相等实根,∴k=﹣1,∴过点D“蛋圆”切线的解析式y=﹣x﹣4;【点评】本题以半圆与抛物线合成的封闭图形“蛋圆”为背景,考查一次函数、二次函数有关性质,解题过程中涉及解一元一次方程、一元二次方程、方程组相关知识与技能,是一道综合性很强的试题.。
2018年广州二模理科数学试题(含详细答案)2018年广州市普通高中毕业班综合测试(二)理科数学试卷,共5页,23小题,满分150分,考试用时120分钟。
注意事项:1.在答题卡上填写姓名、考生号、试室号和座位号,并用2B铅笔填涂考生号。
2.选择题用2B铅笔在答题卡上填涂,填涂错误需用橡皮擦干净。
3.填空题和解答题必须使用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡指定区域内,不得使用铅笔和涂改液。
4.必须保持答题卡整洁,考试结束后将试卷和答题卡一并交回。
一、选择题:共12小题,每小题5分,共60分。
1.已知z1=1+2i,z2=1-i,则z1z2=6.2.已知集合M={x|x≤2,x∈Z},N={x|x-2x-3<0},则M=[-1,2]。
3.执行如图所示的程序框图,若输出y=3,则输入x的值为2.4.已知C: (x^2/a^2)-(y^2/b^2)=1(a>0,b>0)的渐近线与圆(x-2)^2+y^2=1相切,则C的渐近线方程为y=±(x/3)。
5.根据图表,结论B“2010年以来我国实际利用外资规模逐年增加”是正确的。
6.已知cos(α)+cos(β)=1/2,sin(α)+sin(β)=√3/2,则α-β=π/3.7.已知椭圆C: (x^2/16)+(y^2/9)=1,点P(4,1)在C上,则点P关于x轴的对称点P'的坐标为(4,-1)。
二、填空题:共6小题,每小题5分,共30分。
8.已知函数f(x)=x^3+ax^2+bx+c,当x=1时,f(x)=0,f'(1)=0,f''(1)=2,则a=-3,b=3,c=-1.9.已知向量a=2i+j,b=i+2j,则|a-b|=√10.10.已知函数f(x)在区间[0,1]上具有二阶连续导数,且f(0)=f(1)=0,f''(x)+2f'(x)+f(x)=0,则f(x)=e^(-x)(x^2-2x)。
2018年河南省信阳市高考数学二模试卷(文科)一、选择题:本大题共12小题,每小题5分,满分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.(5分)复数的实部与虚部分别为()A.7,﹣3 B.7,﹣3i C.﹣7,3 D.﹣7,3i2.(5分)已知全集U=R,集合A={x|x<﹣或x>1},B={x|﹣1≤x≤2,x∈Z},则图中阴影部分所表示的集合等于()A.{﹣1,2}B.{﹣1,0}C.{0,1}D.{1,2}3.(5分)已知命题p:∃x∈R,e x﹣x﹣1≤0,则¬p为()A.∀x∉R,e x﹣x﹣1>0 B.∀x∈R,e x﹣x﹣1≥0C.∀x∈R,e x﹣x﹣1>0 D.∃x∈R,e x﹣x﹣1>04.(5分)已知双曲线﹣=1(a>0,b>0)的一条渐近线经过点(3,),则双曲线的离心率为()A.B.2 C.或2 D.或25.(5分)某同学先后投掷一枚骰子两次,第一次向上的点数记为x,第二次向上的点数记为y,在直角坐标系xoy中,以(x,y)为坐标的点落在直线2x﹣y=1上的概率为()A.B.C.D.6.(5分)如果f(x)是定义在R上的奇函数,那么下列函数中,一定为偶函数的是()A.y=x+f(x)B.y=xf(x)C.y=x2+f(x)D.y=x2f(x)7.(5分)已知等比数列{a n}的前n项和为S n,a1+a3=,且a2+a4=,则等于()A.4n﹣1 B.4n﹣1 C.2n﹣1 D.2n﹣18.(5分)执行如图所示的程序框图,若输出的y的值为5,则判断框中可以填入的条件是()A.i<6?B.i<5?C.i<4?D.i<3?9.(5分)一名法官在审理一起珍宝盗窃案时,四名嫌疑人甲、乙、丙、丁的供词如下,甲说:“罪犯在乙、丙、丁三人之中”:乙说:“我没有作案,是丙偷的”:丙说:“甲、乙两人中有一人是小偷”:丁说:“乙说的是事实”.经过调查核实,四人中有两人说的是真话,另外两人说的是假话,且这四人中只有一人是罪犯,由此可判断罪犯是()A.甲B.乙C.丙D.丁10.(5分)在△ABC中,内角A、B、C所对的边分别为a、b、c.若asinBcosC+csinBcosA=b,且a>b,则B等于()A. B.C. D.11.(5分)已知x=﹣与x0分别是函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<)的一条对称轴和零点,且|x0+|min=,则φ等于()A.﹣B.﹣C.D.12.(5分)已知函数f(x)(x∈R)满足f(﹣x)=8﹣f(4+x),函数g(x)=,若函数f(x)与g(x)的图象共有168个交点,记作P i(x i,y i)(i=1,2,…,168),则(x1+y1)+(x2+y2)+…+(x168+y168)的值为()A.2018 B.2017 C.2016 D.1008二、填空题:本大题共4小题,每小题5分,共20分.把答案填在答题卡的相应位置.13.(5分)设向量=(1,﹣2),+=(0,2),则|﹣2|=.14.(5分)已知椭圆C:x2+my2=1的焦点在y 轴上,长轴长是短轴长的两倍,则m=.15.(5分)直线ax+by+c=0与圆C:x2﹣2x+y2+4y=0相交于A,B两点,且||=,则•=.16.(5分)某化肥厂生产甲、乙两种肥料,生产一车皮甲种肥料需要磷酸盐4吨、硝酸盐18吨;生产一车皮乙种肥料需要磷酸盐1吨、硝酸盐15吨.已知生产一车皮甲种肥料产生的利润是10万元,生产一车皮乙种肥料产生的利润是5万元.现库存磷酸盐10吨、硝酸盐66吨.如果该厂合理安排生产计划,则可以获得的最大利润是.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.第17~21题为必考题,每个试题考生都必须作答.第22、23题为选考题,考生根据要求作答.17.(12分)已知△ABC的内角A,B,C的对边分别为a,b,c,且acosB+bsinA=0.(Ⅰ)求角B的大小;(Ⅱ)若a+c=4,b=,求△ABC的面积.18.(12分)已知等差数列{a n}的前n项和为S n,且S5=45,S6=60.(Ⅰ)求数列{a n}的通项公式a n;(Ⅱ)若数列{b n}满足b n+1﹣b n=a n(n∈N*),且b1=3,求的前n项和T n.19.(12分)某二手车交易市场对某型号的二手汽车的使用年数x(0<x≤10)与销售价格y(单位:万元/辆)进行整理,得到如下的对应数据:246810使用年数售价16139.57 4.5(1)试求y关于x的回归直线方程:(参考公式:=,=﹣.)(2)已知每辆该型号汽车的收购价格为ω=0.05x2﹣1.75x+17.2万元,根据(1)中所求的回归方程,预测x为何值时,销售一辆该型号汽车所获得的利润z最大?20.(12分)已知抛物线C:x2=2py(p>0)的焦点为F,直线l过点F交抛物线C于A、B两点.且以AB为直径的圆M与直线y=﹣1相切于点N.(1)求C的方程;(2)若圆M与直线x=﹣相切于点Q,求直线l的方程和圆M的方程.21.(12分)已知函数f(x)=4x2+﹣a,g(x)=f(x)+b,其中a,b为常数.(1)若x=1是函数y=xf(x)的一个极值点,求曲线y=f(x)在点(1,f(1))处的切线方程;(2)若函数f(x)有2个零点,f(g(x))有6个零点,求a+b的取值范围.选考题:共10分.请考生从第22、23题中任选一题作答,并用2B铅笔将答题卡上所选题目对应的题号右侧方框涂黑,按所涂题号进行评分;多涂、多答,按所涂的首题进行评分;不涂,按本选考题的首题进行评分.[选修4-4:坐标系与参数方程]22.(10分)已知直线l的参数方程为(其中t为参数),曲线C1:ρ2cos2θ+3ρ2sin2θ﹣3=0,以坐标原点为极点,x轴正半轴为极轴,建立极坐标系,两种坐标系中取相同长度单位.(Ⅰ)求直线l的普通方程及曲线C1的直角坐标方程;(Ⅱ)在曲线C1上是否存在一点P,使点P到直线l的距离最大?若存在,求出距离的最大值及点P的直角坐标;若不存在,请说明理由.[选修4-5:不等式选讲]23.已知函数f(x)=|x﹣5|﹣|x﹣2|.(Ⅰ)若∃x∈R,使得f(x)≤m成立,求实数m的取值范围;(Ⅱ)解不等式x2﹣8x+15+f(x)≤0.2018年河南省信阳市高考数学二模试卷(文科)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,满分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.(5分)复数的实部与虚部分别为()A.7,﹣3 B.7,﹣3i C.﹣7,3 D.﹣7,3i【解答】解:=,∴z的实部与虚部分别为7,﹣3.故选:A.2.(5分)已知全集U=R,集合A={x|x<﹣或x>1},B={x|﹣1≤x≤2,x∈Z},则图中阴影部分所表示的集合等于()A.{﹣1,2}B.{﹣1,0}C.{0,1}D.{1,2}【解答】解:∵A={x|x<﹣或x>1},全集U=R,∴∁U A={x|﹣≤x≤1},∵B={﹣1,0,1,2},∴由图象可知阴影部分对应的集合为B∩(∁U A)={0,1}.故选:C.3.(5分)已知命题p:∃x∈R,e x﹣x﹣1≤0,则¬p为()A.∀x∉R,e x﹣x﹣1>0 B.∀x∈R,e x﹣x﹣1≥0C.∀x∈R,e x﹣x﹣1>0 D.∃x∈R,e x﹣x﹣1>0【解答】解:因为特称命题的否定是全称命题,所以命题p:∃x∈R,e x﹣x﹣1≤0,则¬p为∀x∈R,e x﹣x﹣1>0.故选:C.4.(5分)已知双曲线﹣=1(a>0,b>0)的一条渐近线经过点(3,),则双曲线的离心率为()A.B.2 C.或2 D.或2【解答】解:双曲线﹣=1(a>0,b>0)的一条渐近线经过点(3,),可得,即,可得,解得e=.故选:A.5.(5分)某同学先后投掷一枚骰子两次,第一次向上的点数记为x,第二次向上的点数记为y,在直角坐标系xoy中,以(x,y)为坐标的点落在直线2x﹣y=1上的概率为()A.B.C.D.【解答】解:由题意知本题是一个古典概型,∵试验发生包含的事件是先后掷两次骰子,共有6×6=36种结果,满足条件的事件是(x,y)为坐标的点落在直线2x﹣y=1上,当x=1,y=1,x=2,y=3;x=3,y=5,共有3种结果,∴根据古典概型的概率公式得到以(x,y)为坐标的点落在直线2x﹣y=1上的概率:P=.故选:A.6.(5分)如果f(x)是定义在R上的奇函数,那么下列函数中,一定为偶函数的是()A.y=x+f(x)B.y=xf(x)C.y=x2+f(x)D.y=x2f(x)【解答】解:∵f(x)是奇函数,∴f(﹣x)=﹣f(x).对于A,g(﹣x)=﹣x+f(﹣x)=﹣x﹣f(x)=﹣g(x),∴y=x+f(x)是奇函数.对于B,g(﹣x)=﹣xf(﹣x)=xf(x)=g(x),∴y=xf(x)是偶函数.对于C,g(﹣x)=(﹣x)2+f(﹣x)=x2﹣f(x),∴y=x2+f(x)为非奇非偶函数,对于D,g(﹣x)=(﹣x)2f(﹣x)=﹣x2f(x)=﹣g(x),∴y=x2f(x)是奇函数.故选:B.7.(5分)已知等比数列{a n}的前n项和为S n,a1+a3=,且a2+a4=,则等于()A.4n﹣1 B.4n﹣1 C.2n﹣1 D.2n﹣1【解答】解:∵等比数列{a n}的前n项和S n,且a1+a3=,a2+a4=,∴两式相除可得公比q=,∴a1=2,∴a n==,S n==4(1﹣),∴=2n﹣1,故选:D.8.(5分)执行如图所示的程序框图,若输出的y的值为5,则判断框中可以填入的条件是()A.i<6?B.i<5?C.i<4?D.i<3?【解答】解:模拟执行程序,可得x=1,y=1,i=1满足条件,执行循环体,y=2,x=1,i=2满足条件,执行循环体,y=3,x=2,i=3满足条件,执行循环体,y=5,x=3,i=4由题意,此时应该不满足条件,退出循环,输出y的值为5.故判断框中可填入的条件是i≤3?或i<4?.故选:C.9.(5分)一名法官在审理一起珍宝盗窃案时,四名嫌疑人甲、乙、丙、丁的供词如下,甲说:“罪犯在乙、丙、丁三人之中”:乙说:“我没有作案,是丙偷的”:丙说:“甲、乙两人中有一人是小偷”:丁说:“乙说的是事实”.经过调查核实,四人中有两人说的是真话,另外两人说的是假话,且这四人中只有一人是罪犯,由此可判断罪犯是()A.甲B.乙C.丙D.丁【解答】解:在甲、乙、丙、丁四人的供词不达意中,可以看出乙、丁两人的观点是一致的,因此乙、丁两人的供词应该是同真或同假(即都是真话或者都是假话,不会出现一真一假的情况);假设乙、丁两人说的是真话,那么甲、丙两人说的是假话,由乙说真话推出丙是罪犯的结论;由甲说假话,推出乙、丙、丁三人不是罪犯的结论;显然这两个结论是相互矛盾的;所以乙、丁两人说的是假话,而甲、丙两人说的是真话;由甲、丙的供述内容可以断定乙是罪犯,乙、丙、丁中有一人是罪犯,由丁说假说,丙说真话,推出乙是罪犯.故选:B.10.(5分)在△ABC中,内角A、B、C所对的边分别为a、b、c.若asinBcosC+csinBcosA=b,且a>b,则B等于()A. B.C. D.【解答】解:△ABC中,asinBcosC+csinBcosA=b,由正弦定理得:sinAsinBcosC+sinCsinBcosA=sinB,且sinB≠0,∴sinAcosC+sinCcosA=,∴sin(A+C)=;又A+B+C=π,∴sin(A+C)=sin(π﹣B)=sinB=;又a>b,∴B=.故选:D.11.(5分)已知x=﹣与x0分别是函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<)的一条对称轴和零点,且|x0+|min=,则φ等于()A.﹣B.﹣C.D.【解答】解:x=﹣与x0分别是函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<)的一条对称轴和零点,则:①,ω•x0+φ=kπ②,②﹣①得:ω|ω•x0+φ|=,由于:|x0+|min=,解得:ω=2.故:(k∈Z),解得:φ=(k∈Z),当k=﹣1时,.故选:B.12.(5分)已知函数f(x)(x∈R)满足f(﹣x)=8﹣f(4+x),函数g(x)=,若函数f(x)与g(x)的图象共有168个交点,记作P i(x i,y i)(i=1,2,…,168),则(x1+y1)+(x2+y2)+…+(x168+y168)的值为()A.2018 B.2017 C.2016 D.1008【解答】解:函数f(x)(x∈R)满足f(﹣x)=8﹣f(4+x),可得:f(﹣x)+f(4+x)=8,即函数f(x)关于点(2,4)对称,函数g(x)===4+可知图象关于(2,4)对称;∴函数f(x)与g(x)的图象共有168个交点即在(2,4)两边各有84个交点.而每个对称点都有:x1+x2=4,y1+y2=8,∵有168个交点,即有84组.故得:(x1+y1)+(x2+y2)+…+(x168+y168)=(4+8)×84=1008.故选:D.二、填空题:本大题共4小题,每小题5分,共20分.把答案填在答题卡的相应位置.13.(5分)设向量=(1,﹣2),+=(0,2),则|﹣2|=.【解答】解:根据题意,向量=(1,﹣2),+=(0,2),则=+﹣=(﹣1,4),则﹣2=(3,﹣10),则|﹣2|==;故答案为:.14.(5分)已知椭圆C:x2+my2=1的焦点在y轴上,长轴长是短轴长的两倍,则m=.【解答】解:根据题意,椭圆C:x2+my2=1的焦点在y轴上,则其标准方程为:+=1,其中a=,b=1,若长轴长是短轴长的两倍,则a=2b,则有=2,解可得m=;故答案为:.15.(5分)直线ax+by+c=0与圆C:x2﹣2x+y2+4y=0相交于A,B两点,且||=,则•=﹣.【解答】解:圆C:x2﹣2x+y2+4y=0⇔(x﹣1)2+(y+2)2=5,如图,过C作CD⊥AB于D,AB=2AD=2AC•sin∠CAD,∴,∴∠CAD=30°,∴∠ACB=120°,则•==﹣.故答案为:﹣.16.(5分)某化肥厂生产甲、乙两种肥料,生产一车皮甲种肥料需要磷酸盐4吨、硝酸盐18吨;生产一车皮乙种肥料需要磷酸盐1吨、硝酸盐15吨.已知生产一车皮甲种肥料产生的利润是10万元,生产一车皮乙种肥料产生的利润是5万元.现库存磷酸盐10吨、硝酸盐66吨.如果该厂合理安排生产计划,则可以获得的最大利润是30万元.【解答】解:设x、y分别为计划生产甲、乙两种混合肥料的车皮数,于是满足以下条件:再设分别生产甲、乙两种肥料各x、y车皮产生的利润为z=10000x+5000y=5000(2x+y),由得两直线的交点M(2,2).令t=2x+y,当直线L:y=﹣2x+t经过点M(2,2)时,它在y轴上的截距有最大值为6,此时z=30000.故分别生产甲、乙两种肥料各2车皮时产生的利润最大为30万元.故答案为:30万元.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.第17~21题为必考题,每个试题考生都必须作答.第22、23题为选考题,考生根据要求作答.17.(12分)已知△ABC的内角A,B,C的对边分别为a,b,c,且acosB+bsinA=0.(Ⅰ)求角B的大小;(Ⅱ)若a+c=4,b=,求△ABC的面积.【解答】解:(Ⅰ)∵且acosB+bsinA=0.由正弦定理可得:sinAcosB+sinAsinB=0∵sinA≠0,∴sinB=﹣cosB,∴tanB=﹣.∴B=120°.(Ⅱ)由余弦定理得b2=a2+c2﹣2accosB=(a+c)2﹣ac,∵a+c=4,b=,∴ac=3.∴△ABC的面积S=.18.(12分)已知等差数列{a n}的前n项和为S n,且S5=45,S6=60.(Ⅰ)求数列{a n}的通项公式a n;(Ⅱ)若数列{b n}满足b n+1﹣b n=a n(n∈N*),且b1=3,求的前n项和T n.【解答】解:(I)设等差数列{a n}的公差为d,∵S5=45,S6=60.∴=45,d=60,解得a1=5,d=2.∴a n=5+2(n﹣1)=2n+3.﹣b n=a n=2n+3,且b1=3,(II)b n+1∴b n=(b n﹣b n﹣1)+(b n﹣1﹣b n﹣2)+…+(b2﹣b1)+b1=2(n﹣1)+3+2(n﹣2)+3+…+2×1+3+3==n2+2n.∴==.∴的前n项和T n=+…+==.19.(12分)某二手车交易市场对某型号的二手汽车的使用年数x(0<x≤10)与销售价格y(单位:万元/辆)进行整理,得到如下的对应数据:246810使用年数售价16139.57 4.5(1)试求y关于x的回归直线方程:(参考公式:=,=﹣.)(2)已知每辆该型号汽车的收购价格为ω=0.05x2﹣1.75x+17.2万元,根据(1)中所求的回归方程,预测x为何值时,销售一辆该型号汽车所获得的利润z最大?【解答】解:(1)由已知:,则,所以回归直线的方程为.(2)z=﹣1.45x+18.7﹣(0.05x2﹣1.75x+17.2)=﹣0.052x2+0.3x+1.5=﹣0.05(x﹣3)2+1.95,所以预测当x=3时,销售利润z取得最大值.20.(12分)已知抛物线C:x2=2py(p>0)的焦点为F,直线l过点F交抛物线C于A、B两点.且以AB为直径的圆M与直线y=﹣1相切于点N.(1)求C的方程;(2)若圆M与直线x=﹣相切于点Q,求直线l的方程和圆M的方程.【解答】解:(1)设A(x1,y1),B(x2,y2),则|AB|=y1+y2+p=2p,又∵以AB为直径的圆M与直线y=﹣1相切,∴|FN|=|AB|=+1,即|AB|=p+2,∴p=2,∴抛物线C的方程为x2=4y.(2)设直线l的方程为y=kx+1,代入x2=4y中,化简整理得x2﹣4kx﹣4=0,∴x1+x2=4k,x1x2=﹣4,∴,∴圆心的坐标为M(2k,2k2+1),∵圆M与直线相切于点Q,∴|MQ|=|MN|,∴,解得,此时直线l的方程为,即x﹣2y+2=0,圆心,半径,∴圆M的方程为.21.(12分)已知函数f(x)=4x2+﹣a,g(x)=f(x)+b,其中a,b为常数.(1)若x=1是函数y=xf(x)的一个极值点,求曲线y=f(x)在点(1,f(1))处的切线方程;(2)若函数f(x)有2个零点,f(g(x))有6个零点,求a+b的取值范围.【解答】解:(1)函数f(x)=4x2+﹣a,则y=xf(x)=4x3+1﹣ax的导数为y′=12x2﹣a,由题意可得12﹣a=0,解得a=12,即有f(x)=4x2+﹣12,f′(x)=8x﹣,可得曲线在点(1,f(1))处的切线斜率为7,切点为(1,﹣7),即有曲线y=f(x)在点(1,f(1))处的切线方程为y+7=7(x﹣1),即为y=7x﹣14;(2)由f(x)=4x2+﹣a,导数f′(x)=8x﹣,当x>时,f′(x)>0,f(x)递增;当x<0或0<x<时,f′(x)<0,f(x)递减.可得x=处取得极小值,且为3﹣a,由f(x)有两个零点,可得3﹣a=0,即a=3,零点分别为﹣1,.令t=g(x),即有f(t)=0,可得t=﹣1或,则f(x)=﹣1﹣b或f(x)=﹣b,由题意可得f(x)=﹣1﹣b或f(x)=﹣b都有3个实数解,则﹣1﹣b>0,且﹣b>0,即b<﹣1且b<,可得b<﹣1,即有a+b<2.则a+b的范围是(﹣∞,2).选考题:共10分.请考生从第22、23题中任选一题作答,并用2B铅笔将答题卡上所选题目对应的题号右侧方框涂黑,按所涂题号进行评分;多涂、多答,按所涂的首题进行评分;不涂,按本选考题的首题进行评分.[选修4-4:坐标系与参数方程]22.(10分)已知直线l的参数方程为(其中t为参数),曲线C1:ρ2cos2θ+3ρ2sin2θ﹣3=0,以坐标原点为极点,x轴正半轴为极轴,建立极坐标系,两种坐标系中取相同长度单位.(Ⅰ)求直线l的普通方程及曲线C1的直角坐标方程;(Ⅱ)在曲线C1上是否存在一点P,使点P到直线l的距离最大?若存在,求出距离的最大值及点P的直角坐标;若不存在,请说明理由.【解答】解:(Ⅰ)直线l的参数方程为(其中t为参数),转化为直角坐标方程为:x﹣y+1=0.曲线C1:ρ2cos2θ+3ρ2sin2θ﹣3=0,转化为直角坐标方程为:.(Ⅱ)由(Ⅰ)知:C1的参数方程为:(θ为参数).所以:点P到直线l的距离d==,则:,此时:cos()=1,解得:(k∈Z).所以:,故P()到直线l的距离最大.[选修4-5:不等式选讲]23.已知函数f(x)=|x﹣5|﹣|x﹣2|.(Ⅰ)若∃x∈R,使得f(x)≤m成立,求实数m的取值范围;(Ⅱ)解不等式x2﹣8x+15+f(x)≤0.【解答】解:(Ⅰ)f(x)=|x﹣5|﹣|x﹣2|=,当2<x<5时,﹣3<7﹣2x<3,所以﹣3≤f(x)≤3,∴m≥﹣3;(Ⅱ)不等式x2﹣8x+15+f(x)≤0,即﹣f(x)≥x2﹣8x+15由(1)可知,当x≤2时,﹣f(x)≥x2﹣8x+15的解集为空集;当2<x<5时,﹣f(x)≥x2﹣8x+15,即x2﹣10x+22≤0,∴5﹣≤x<5;当x≥5时,﹣f(x)≥x2﹣8x+15,即x2﹣8x+12≤0,∴5≤x≤6;综上,原不等式的解集为{x|5﹣≤x≤6}.。