广州大学附属中学2019年12月2019年度高考模拟试题-高考模拟
2019年普通高等学校招生全国统一考试广东省模拟试卷含答案
2019年普通高等学校招生全国统一考试广东省模拟试卷(一)语文试题注意事项:1.答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上。
2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
3.非选择题的作答:用签字笔直接答在答题卡上对应的答题区域。
写在试题卷、草稿纸和答题卡上的非答题区域均无效。
4.考试结束后,请将本试卷和答题卡一并上交。
一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。
道德的本质不是远离“得”,而是要学会如何在处理现实复杂利益关系中获得正当性;道德的完满也不是不要“得”,而是能够自如地运用符合“德”的方式去“得”,儒家“孝”伦理发展到“德”的阶段,便在个体自身内部完成了“孝”的内化,但这只是抽象地完成。
儒家“孝”伦理的意义与价值,决不仅仅是精神的自我完成,而是“外化为他物”。
这种现实外化就是“得”,就是使儒家“孝”伦理能够更有效地干预现实社会生活。
“得”是儒家“孝”伦理逻辑运行的目的。
但“得”的实现与获取也不能偏离伦理的逻辑。
在儒家“孝”伦理中,“德”与“得”互相投射,形成了具有丰富内涵的逻辑结构。
第一,“得”必颓有“德”。
在中国传统社会,因为孝行而获得社会广泛认可的孝子不乏其人,这种认可包括物质和精神两方面的嘉奖:在物质上能够获得上层的封赏,比如对孝子实行赦免赋税的优惠等;在精神上获得社会的广泛赞誉,孝子们被旌表门闾、载入史书,其而能够因为孝行被选入官。
反之,如果有不孝者,则被除名削爵,永世不得续用。
第二,“德”必然能“得”。
舜因何能贵为天子,因为舜是大孝之人,德行高远。
而且这种大德能使老百姓受益,自然就会受到上天的保佑,所以大德之人必然会“得”。
“德”不以“得”为目的,但“德”却必然有“得”的报答。
父子是血亲相连的天伦关系,如果孝敬双亲是为了赢得孝子的美名和求得功利,则损害了亲亲之情,使人失去最基本的情感依托。
【市级联考】广东省广州市2019届高三第二次模拟考试数学(理)试题(解析版)
2019年广州市普通高中毕业班综合测试(二)理科数学一、选择题.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.)A. B. D.【答案】B【解析】【分析】根据复数的几何意义建立不等式关系即可.若复数在复平面内对应的点在第三象限,所以的取值范围是故选B.【点睛】该题考查的是有关复数在复平面内对应的点的问题,属于简单题目.2.)A. 或【答案】D【解析】【分析】先解分式不等式求集合A,再由补集的定义直接求解即可.【详解】解:由,则R故选:D.【点睛】本题主要考查集合的基本运算,比较基础.3.的样本,若样本中8辆,则 )A. 96B. 72C. 48D. 36【答案】B 【解析】 【分析】根据分层比例列式求解.B.【点睛】本题考查分层抽样,考查基本分析求解能力,属基础题.4.)A. 21B. 22C. 23D. 24【答案】B 【解析】试题分析:运行第一次,,,;运行第二次,,,,,停止运行,所以输出的B .考点:程序框图.5. )A.B.D.【答案】D 【解析】 【分析】根据对称列式求解.D.【点睛】本题考查关于直线对称点问题,考查基本分析求解能力,属基础题.6.从某班6名学生(其中男生4人,女生2人)中任选3人参加学校组织的社会实践活动.设所选3人中女)A. B. 1 D. 2【答案】B【解析】【分析】先列随机变量,再分别求解对应概率,最后根据数学期望公式求结果.,所以,选B.【点睛】本题考查数学期望,考查基本分析求解能力,属基础题.7.)A. B. D.【答案】D【解析】【分析】再根据二倍角正切公式得结果.【详解】因,且,因为,从而 D.【点睛】本题考查同角三角函数关系以及二倍角正切公式,考查基本分析求解能力,属基础题.8.的左焦点,则双曲线的离心率为()A. B. D.【答案】A【解析】【分析】再根据切线得OE.,所以PF,PF,A.【点睛】本题考查双曲线定义以及离心率,考查基本分析求解能力,属中档题.9.,且点)A. B. D.【答案】C【解析】【分析】设A(s,t),求得函数y的导数可得切线的斜率,解方程可得切点A,代入直线方程,再由基本不等式可得所求最小值.【详解】解:设A(s,t),y=x3﹣2x2+2的导数为y′=3x2﹣4x,可得切线的斜率为3s2﹣4s,切线方程为y=4x﹣6,可得3s2﹣4s=4,t=4s﹣6,解得s=2,t=2或由点A在直线mx+ny﹣l=0(其中m>0,n>0),可得2m+2n=1成立,(s,2m+2n))=2(32(当且仅当n时,取得最小值6+4,故选:C.【点睛】本题考查导数的运用:求切线斜率,以及基本不等式的运用:求最值,考查运算能力,属于基础题.10.的图像的一条对称轴为()A. B. D.【答案】C【解析】【分析】.,选C.【点睛】本题考查由图象求函数解析式、三角函数图象变换以及正弦函数性质,考查基本分析求解能力,属中档题.11.已知点在直线上,的中点为)A.B.D.【答案】B 【解析】【分析】.M 在直线AB,,因此的取值范围为选B.【点睛】本题考查线性规划求范围,考查基本分析求解能力,属中档题.12.的)A. B. D.【答案】D【解析】【分析】先设切点B.【详解】A为圆心,B,则在B点处切线的斜率为,选D.【点睛】本题考查利用导数求函数最值,考查综合分析求解能力,属难题.二、填空题.13.是夹角为.【答案】【解析】【分析】,;.故答案为:【点睛】考查单位向量的概念,向量的数量积运算及计算公式,向量长度的求法.14.80________.【答案】2【解析】解:(ax-1)5的展开式中x3的系数C53(ax)3•(-1)2=10a3x3=80x3,则实数a的值是2,15.秦九韶是我国南宋著名数学家,在他的著作《数书九章》中有己知三边求三角形面积的方法:“以小斜幂并大斜幂减中斜幂,余半之,自乘于上.以小斜幂乘大斜幂减上,余四约之,为实.一为从隅,开平方得,,,的对边为.1,成等差数列,则________.【解析】【分析】再根据余弦定理化简得1成等差数列,所以的最大值为.【点睛】本题考查正余弦定理以及二次函数性质,考查基本分析求解能力,属中档题.16.________.【解析】【分析】.正四面体外接球恰为圆锥内切球,所以【点睛】本题考查圆锥内切球以及正四面体外接球,考查基本分析求解能力,属中档题.三、解答题.解答应写出文字说明、证明过程和演算步骤.17.,(1)求数列(2,求数列【答案】 (2)【解析】【分析】(1)解法1:运用等比数列的通项公式,解方程可得首项和公比,即可得到所求通项公式;解法2:运用等比数列的性质建立方程.(2,利用错位相减求和.【详解】解法1:(1的公比为,是递增的等比数列,所以数列解法2:(1,是递增的等比数列,(2)由(1①-所以【点睛】本题考查等比数列的通项公式的运用,考查数列的错位相减求和,以及化简整理的运算能力,属于基础题.18.科研人员在对人体脂肪含量和年龄之间关系的研究中,获得了一些年龄和脂肪含量的简单随机样本数据,如下表: (年龄(脂肪根据上表的数据得到如下的散点图.(1)根据上表中的样本数据及其散点图: (i(i )计算样本相关系数(精确到0.01),并刻画它们的相关程度. (20.01),并根据回归方程估计年龄为50岁时人体的脂肪含量.【答案】(1) (ⅰ)47 (ⅱ)见解析;%.【解析】【分析】(1)(i)根据上表中的样本数据,利用平均数的公式求得结果;(ii 以推断人体脂肪含量和年龄的相关程度很强.(2结果.【详解】(1)根据上表中的样本数据及其散点图:.因为,(2.的线性回归方程为.【点睛】该题考查的是有关回归分析的问题,涉及到的知识点有平均值的计算,根据相关系数r的大小判断相关性,回归直线的性质,属于简单题目.19.(1(2.【答案】(1)见解析;(2【解析】【分析】(1)先根据计算得线线线线垂直,再根据线面垂直判定定理以及面面垂直判定定理得结论,(2)建立空间直角坐标系,利用空间向量求二面角.【详解】(1为中,中,,,,所以平面(2由(1设平面的法向量为设二面角为,由于的余弦值为.【点睛】本题考查线面垂直判定定理、面面垂直判定定理以及利用空间向量求二面角,考查基本分析论证与求解能力,属中档题.20.(1(2并说明理由.【答案】(1;(2)相离.【解析】【分析】(1)根据直接法求轨迹方程,(2离与半径大小进行判断.【详解】(1,整理得所以动点的轨迹的方程(2的直线为轴时,显然不合题意.因为,.的中点坐标为.到直线的距离为.与以线段为直径的圆相离.【点睛】本题考查直接法求轨迹方程以及直线与圆位置关系,考查基本分析求解能力,属中档题.21.(1)讨论函数的单调性;(2【答案】(1)见解析;(2)见证明【解析】【分析】(1)函数f(x)的定义域为(0,+∞),f′(x x>0,利用分类讨论思想,结合导数性质能讨论函数f(x)的单调性.(2)先求k f(﹣2k)=ln(﹣2k.然后证明x1+x2≥)(1+t)2<﹣8lnt,即证8lnt+(1+t)2<0,(t>0).设h(t)=8lnt+(1+t)2,t>1.则h(t)=8t>1.由此能证明x1+x2>【详解】(1,函数.时,,,时,函数时,函数(2方法1:由(1要使函数有两个零点,首先,,则因为,所以在上单调递增,的取值范围是.方法2:,则,且:方法1:,即,即证.,所以即证,.所以.在上单调递减,.方法2:,即,需证.,所以即证所以在上单调递减,.方法3:因为,是函数,需证.只需证.,所以,所以.方法4:因为,是函数,即证明,则.所以在上单调递增,所以.,.方法5:,所以在上单调递减.在上恒成立.【点睛】本题考查函数单调性的讨论,考查不等式的性质,考查导数性质、函数的单调性、最值等基础知识,考查运算求解能力,考查化归与转化思想,是难题.22.(.在以坐标原点为极点,.(1(2.【答案】(1;(2【解析】【分析】(1的普通方程,根据2)利用直线参数方程几何意义求解.【详解】(1,.因为(2)解法1的直角坐标方程为,可设该方程的两个根为整理得,因为,所以综上所述,直线的倾斜角为解法2,两点,且的,整理得.综上所述,直线【点睛】本题考查参数方程化普通方程、极坐标方程化直角坐标方程以及直线参数方程应用,考查综合分析求解能力,属中档题.23.[选修4-5:不等式选讲](1)时,解不等式(2)若存在实数x a的取值范围.【答案】(1;(2【解析】【分析】(1)根据绝对值定义转化为两个不等式组,解可得,(2)根据绝对值定义转化为分段函数,根据函数最值可得结果.【详解】(1综上可知,不等式的解集为(2..所以实数的取值范围为.【点睛】本题考查含绝对值不等式,考查基本分析求解能力,属基本题.。
2019-2020学年广州市大学附属中学高三英语模拟试卷及答案解析
2019-2020学年广州市大学附属中学高三英语模拟试卷及答案解析第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项AWhen the weather is bad or when the flu breaks out, we can let the kids do some fun things at home, which can be beneficial to kids.Reading out loudIf your children are young enough, don't forget to read books to them out loud! Few children dislikehaving a good book read to them, and it's great for the development of their brains. However, if your children are a bit older and have moved onto more advanced books, there is always the choice of listening to an audiobook. This can also be done while they're doing something else.Playing board gamesMaybe your children's table is full of board games, which have been forgotten for a long time. It's a good time to bring them out when playing outside is no longer a choice. Surely, playing board games is a great way to connect with children. In addition, many board games are designed to get children thinking!Having a dance partyConsidering that all you need is a speaker or maybe just a phone, you can have a dance party wherever you are! This is a great way to get kids’ bodies moving when they are inside. Play some of your children's favorite music and let them dance to it. Not only is it good exercise, but it will help your children feel time is flying!Doing jigsaw (拼图) puzzlesFor most people that have children, it's common to have at least one jigsaw puzzle at home. Jigsaw puzzles are great because everyone can do them on their own time. Besides, your whole family will have a sense of achievement when everyone is smiling over the finished product.1. What do reading out loud and playing board games have in common?A. They both develop children's team spirit.B. They both improve children's listening ability.C. They both do good to children's thinking ability.D. They both focus on interaction between children.2. Which of the following combines exercise and music?A. Reading out loud.B. Playing board games.C. Doing jigsaw puzzles.D. Having a dance party.3. What is the purpose of the text?A. To list four interesting children's parties.B. To recommend four children's favorite books.C. To introduce some activities for children inside.D. To show some funny things for children outside.BTo stay healthy and fit, Chinesestudents do group exercises every day at school. Most of you probably do the same set of exercises. But some school exercises have grown popular online due to their local and innovative designs.Singing in Sichuan dialects with energetic movements and unique mask-changing is not just a Sichuan Opera performance. It’s the routine exercise for students of Mianyang Foreign Languages Experimental School in Sichuan province.“Sichuan Opera is a local opera, and it is now facing a gap in inheritance (传承). Therefore, we cooperated with Mianyang Intangible Cultural Heritage Center to create a simple and easy-to-learn Sichuan Opera exercise,” said Shen Junhua, who is in charge of organizing the school’s exercise between classes.According to Shen, this new type of exercise has been practiced since 2017 and has been popular among students. When students enroll (入学), they will spend several weeks practicing it. At present, almost all of the students and teachers have mastered it.“In fact, we had hardly heard of Sichuan Opera before teachers taught us how to do the Sichuan Opera exercise,” said Li Yangwenwen, 14, an eighth grade student who also joined the school’s Sichuan Opera club out of interest. “It’s very different from normal exercises. After practicing it, we found it very beautiful and became interested in it. Now, almost all of the students look forward to our daily exercise time and feel excited to do it.”“By combining opera with daily exercise, the daily class activity allows students to perceive and understand Sichuan Opera’s culture”, Shen said. “After years of continuous effort to spread the seeds of traditional culture, the younger generation is finally catching on.”4. What do students in Shen’s school do during the group exercise?A. They do normal exercise .B. They sing pop songs in Sichuan dialects.C. They do mask-changing in a Sichuan Opera performance.D. They combine group exercise with Sichuan Opera.5. Why do they adopt the new type of exercise?A. To attract new students to the school.B. To inherit local culture.C. To create an easy-to-learn exercise.D. To make the school’s group exercise popular.6. How do teachers and students react to the group exercise?A. Calm.B. Indifferent.C. Enthusiastic.D. Uninterested.7. What can we learn from the last paragraph?A. Shen’s continuous effort is highly praised.B. The younger generation will have a stronger body.C. The students can better understand their local culture.D. Students help to spread the seeds of traditional culture toyounger generation.C“My P.E. teacher taught me maths.”It has been a common joke for years but when a P.E. teacher applied for the head-teacher job, many parents worried that “the joke could come true”. Some parents asked, “Does the P.E. teacher know maths and English?If not, how is he supposed to tutor the students?” But there were still parents who believed P.E. teachers had more time to discipline the students since they didn’t have many classes.This concern is actually a “subject bias”, that is, choosing a P.E. teacher as the head-teacher is not good for the students’ grades since he doesn’t know Chinese, maths, or English. It is acceptable for Chinese teachers, maths teachers and English teachers to be head-teachers because these subjects are important to entrance exams and scores. Such a concern reflects parents’ anxiety in the current educational environment, which tends to link the head-teacher’s responsible subject with the facts whether the school cares about the class and the children’s performances in the subject.Many people care about their kids’ academic performances only. They don’t care about their kids’ P.E. performance at all. Even if kids have P.E. classes, parents care little. P.E. teachers are in humble position and their classes are often occupied by other teachers. Of course, if students, P.E. performance is related to grades and entrance exams, parents won’t mind “a P.E. teacher being the head-teacher”.At the moment, P.E. is gradually included in entrance exams.In terms of high school entrance examination,Guangzhouwill launch a new high school entrance exam which includes P.E. performance and Health examination in 2021, raising the score to 70 points.In terms of the college entrance examination, universities having the right of independent enrollments added P.E. tests to their entrance exams in 2019. This practice is seen as an important signal that the assessment of physical fitness and athletic ability, which are important aspects of a student’s overall quality, may be included in the college entrance examination in the future.We hope that it’s a trend for P.E. teachers, music teachers and art teachers to become head-teachers.8. By saying “My P.E. teacher taught me maths”, what does the author intend to show us?A. It is just a joke that seldom happened in reality.B. P.E. teachers hardly assist students in maths.C. P.E. teachers are good at teaching maths..D. Parents doubt the ability of P.E. teacher.9. We can infer from paragraph 2 that a P.E. teacher .A. is of little benefit for students’ academic performanceB. is humbler than Chinese, math or English teachersC. reflects whether the school cares about the classD. gets unfairly judged due to the current educational system10. Why doesGuangzhouraise proportion of P.E. performance in high school entrance examination?A. To raise students’ awareness of physical health.B. To call on parents to pay attention to P.E. teachers.C. To test the overall ability of high school students.D. To make P.E. teachers equal with other teachers.11. What opinion does the author hold towards P.E. teachers working as head-teachers?A. Neutral.B. Unexpected.C. Supportive.D. Critical.DA trip to thelibrary was like a great journey to a different country. To get there, we had to walk a mile. But our weekly journeys to the library were a piece of perfection. I had around me at one time all the people I loved best-my father and mother and brothers and sister--and all the things I loved best- quiet, space and books.I read a lot of books about science: not the spaceships my brothers preferred, but the birds and the bees--literally. I brought home a book of birds and searched the trees for anything other than robins (知更鸟). I went through a phrase of loving books with practical science experiments and used up a whole bottle of white vinegar by pouring it on the sides of our apartment building to prove that it was constructed of limestone (石灰石).One Saturday, as I wandered through the young adult section, I saw a title: Little Women, by Lousia May Alcott. I had learned from experience that titles weren’t everything. A book that sounded great on the shelf could be dull once you got it home. So I sat in a chair near the shelves to skim the first paragraphs.I read and read and read Little Women until it was time to walk home, and, except for a few essential interruptions like sleeping and eating, I did not put it down until the end. Even the freedom to watch weekend television held no appeal for me in the wake of Alcott' s story. It was about girls, for one thing, girls who could almost be like me, especially Jo. I had found someone who thought and felt the way I did.12. What can we say about the author’s family?A. They enjoyed traveling abroad.B. They were library frequenters.C. They were very fond of walking.D. They led a perfectly quiet life.13. What does the author mainly want to show in paragraph 2?A. Her different hobbies from her brothers.B. How she conducted science experiments.C. Why she loved books about the birds and the bees.D. Her reading interests during a particular period of time.14. What opinion does the author hold on books?A. Book titles can sometimes be misleading.B. Science books are as interesting as novels.C. The first few paragraphs of a book are attractive.D. Books seem duller when read in libraries than at home.15. How would the author describe Little Women?A. It helped her to discover her true character.B. It made her forget about food and sleep.C. It inspired confidence in her.D. It kept her absorbed.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
2019年广州市高考模拟试卷语文试题及答案
2019届广州市高三模拟试卷语文一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。
“天下”是中国传统文化对世界秩序的一种原初想象,“天下主义”是以“天下”理念为核心,由具有普遍性和开放性的世界秩序、价值规范与理想人格构成的思想体系。
在当今全球化语境之下,中国文化理念和文化战略的自觉自信体现的正是“天下主义”的精神内核。
文化自信首先是一种“以天下观天下”的世界观的自信。
《道德经》有云:“修之于天下,其德乃普。
故以身观身,以家观家,以乡观乡,以国观国,以天下观天下。
”“天下”是中国文化特有的思维尺度,是一个最宏大、最完备的分析单位,具有最广阔的容纳力。
面对差异性的多元文化格局,西方文化多以民族国家为基本单位,文化视域限于国家、民族内部,虽然也有关于世界的思考,比如斯多葛学派的“世界主义”等,但其思考方式是“以国家观世界”,与中国文化的立足点和尺度不同。
中国文化的“天下”蕴含了“天下无外”的理想,各个民族的历史文化在世界内部是平等共存的;中国文化依循“修身、齐家、治国、平天下”的进路,从“身-家-国”逻辑同构的角度,最终达到“天下大同”的境界。
与西方文化相较,“天下”的世界观更具有开放性和包容性。
文化自信的核心是“以天下为一家”的价值观的自信。
中国文化是以儒家文化为代表的伦理型文化,梁漱溟认为“中国伦理始于家庭而不止于家庭”,中国文化重视家庭生活,整个社会关系是依照家庭关系推广发挥的。
“以天下为一家”的价值观实质上是一种关系性伦理,把“自我”和“他者”看成一体共生的关系。
“天下”是一个最大的家,家庭利益的最大化就是个体利益的最大化,共同体的善与个体的善是统一的。
文化自信最终体现为一种人格自信。
无论是“以天下观天下”的世界观,还是“以天下为一家”的价值观,最终都沉淀为个体的精神品格。
中国文化倡导和推崇的理想人格是“君子”,这是由中国人独特的精神气质所决定的。
2019年广州市高三一模考试试卷
2019年广州市高三一模考试试卷各位读友大家好,此文档由网络收集而来,欢迎您下载,谢谢本试卷分选择题和非选择题两部分,共10页,满分为150分。
考试用时150分钟。
注慧事项: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考生号填写在答题卡上,用2B铅笔将试卷类型填涂在答题卡上。
2、选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑;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宇迹钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内的相应位置上;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。
不按以上要求作答无效。
4、考生必须保持答题卡的整洁,考试结束后,将答题卡交回。
第一部分选择题一、1、下列词语中加点的字的读音,全都不相同的一组是A、驾驭翁妪熨贴誉满全球与会人员B、把持对峙侍候有恃无恐严阵以待c、落榜落枕落色落在家里落叶归根D、帐簿怅然涨潮张皇失措为虎作伥2、下列词语中没有错别字的一组是A、演绎事必躬亲忸怩作态有志者事竟成B、缉捕声名雀起沁人心脾事实胜于雄辩c、噩耗意气风发鞭辟入理多行不义必自毙D、清澈坐收渔利乔装打扮文武之道一张一驰3、下列标点符号使用有错误的一项是A、卓别林完成他的电影《独裁者》后,一出版商要求卓别林给他25000美元的转让费,理由是电影借用了他《独裁者》一书的书名。
B、有人断言,先生这样的天才,“在号称有四千年文明史的中国才出现一个,恐怕能跟他伦比的一个也没有”。
这话说得未免夸大。
c、牡丹、吊钟、水仙、大丽、梅花、菊花、山茶、兰花、芍药、桃花……等等,春秋冬三季的鲜花都一起挤到了广州的迎春花市上。
D、第一对敌人要狠,要压倒它,要消灭它;第二对自己人、对人民、对官长、对部下要和,要团结:这是我们军队一向的方针。
4、依次填入下列各句横线处的词语,最恰当的一组是①英国是一个议会制的君主立宪国家,它的君主是世袭的,但议会并没给他过多的________,国家的一切决定都要经过漫长的协商。
2019届广东省广州市高三第二次模拟考试数学(理)试卷及解析
2019届广州市高三第二次模拟考试数学(理)试卷一、选择题.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.已知复数在复平面内对应的点在第三象限,则实数的取值范围是()A. B. C. D.【答案】B【解析】【分析】根据复数的几何意义建立不等式关系即可.【详解】,若复数在复平面内对应的点在第三象限,则,解得,所以的取值范围是,故选B.2.已如集合,则()A. 或B. 或C. 或D. 或【答案】D【解析】【分析】先解分式不等式求集合A,再由补集的定义直接求解即可.【详解】解:由10,即0,即解得,即,则R故选:D.3.某公司生产,,三种不同型号的轿车,产量之比依次为,为检验该公司的产品质量,用分层抽样的方法抽取一个容量为的样本,若样本中种型号的轿车比种型号的轿车少8辆,则()A. 96B. 72C. 48D. 36 【答案】B【解析】【分析】根据分层比例列式求解.【详解】由题意得选B.4.执行如图所示的程序框图,则输出的值是()A. 21B. 22C. 23D. 24 【答案】B【解析】试题分析:运行第一次,,,;运行第二次,,,;运行第三次,,;运行第四次,,不满足,停止运行,所以输出的的值是,故选B.5.已知点与点关于直线对称,则点的坐标为()A. B. C. D.【答案】D【解析】【分析】根据对称列式求解.【详解】设,则,选D.6.从某班6名学生(其中男生4人,女生2人)中任选3人参加学校组织的社会实践活动.设所选3人中女生人数为,则数学期望()A. B. 1 C. D. 2【答案】B【解析】【分析】先列随机变量,再分别求解对应概率,最后根据数学期望公式求结果.【详解】因为,所以因此,选B.7.已知,其中,则()A. B. C. D.【答案】D【解析】【分析】先根据同角三角函数关系求得,再根据二倍角正切公式得结果.【详解】因,且,所以,因为,所以,因此,从而,,选D.8.过双曲线的左焦点作圆的切线,切点为,延长交双曲线右支于点,若,则双曲线的离心率为()A. B. C. D.【答案】A【解析】【分析】先根据条件得,再根据切线得OE,结合双曲线定义列等式,解得离心率.【详解】设右焦点,因为,所以,因为,所以, 由双曲线定义得,因为⊥PF,所以⊥PF,因此,选A.9.若曲线在点处的切线方程为,且点在直线(其中,)上,则的最小值为()A. B. C. D.【答案】C【解析】【分析】设A(s,t),求得函数y的导数可得切线的斜率,解方程可得切点A,代入直线方程,再由基本不等式可得所求最小值.【详解】解:设A(s,t),y=x3﹣2x2+2的导数为y′=3x2﹣4x,可得切线的斜率为3s2﹣4s,切线方程为y=4x﹣6,可得3s2﹣4s=4,t=4s﹣6,解得s=2,t=2或s,t,由点A在直线mx+ny﹣l=0(其中m>0,n>0),可得2m+2n=1成立,(s,t,舍去),则(2m+2n)()=2(3)≥2(3+2)=6+4,当且仅当n m时,取得最小值6+4,故选:C.10.函数的部分图像如图所示,先把函数图像上各点的横坐标缩短到原来的倍,纵坐标不变,再把得到的图像向右平移个单位长度,得到函数的图像,则函数的图像的一条对称轴为()A. B. C. D.【答案】C【解析】【分析】先根据图象求,再根据图象变换得,最后根据正弦函数性质求对称轴.【详解】由图得,从而,,,选C.11.已知点在直线上,点在直线上,的中点为,且,则的取值范围为()A. B. C. D.【答案】B【解析】【分析】先确定所在直线,再根据,得轨迹为一条线段,最后根据斜率公式求结果.【详解】因为点在直线上,点在直线上,所以M在直线上,即,因为,所以轨迹为一条线段AB,其中,因此的取值范围为,选B.12.若点与曲线上点的距离的最小值为,则实数的值为( )A.B.C.D.【答案】D 【解析】 【分析】先设切点B ,再根据导数几何意义以及最值列式解得实数的值. 【详解】因为,所以由题意得以A 为圆心,为半径的圆与曲线相切于点B,设,则在B 点处切线的斜率为,所以,选D.二、填空题. 13.若,是夹角为的两个单位向量,向量,则________.【答案】 【解析】 【分析】 根据条件即可求出,,从而可以求出,进而得出.【详解】解:,;∴;∴.故答案为:. 14.若的展开式中的系数是80,则实数的值是________.【答案】2 【解析】解:(ax-1)5的展开式中x 3的系数C 53(ax )3•(-1)2=10a 3x 3=80x 3,则实数a的值是2,15.秦九韶是我国南宋著名数学家,在他的著作《数书九章》中有己知三边求三角形面积的方法:“以小斜幂并大斜幂减中斜幂,余半之,自乘于上.以小斜幂乘大斜幂减上,余四约之,为实.一为从隅,开平方得积.”如果把以上这段文字写成公式就是,共中,,是的内角,,的对边为.若,且,1,成等差数列,则面积的最大值为________. 【答案】【解析】【分析】先根据正弦定理得,再根据余弦定理化简得【详解】因为,所以,因此,因为,1,成等差数列,所以+=2,因此,即面积的最大值为.16.有一个底面半径为,轴截面为正三角形的圆锥纸盒,在该纸盒内放一个棱长均为的四面体,并且四面体在纸盒内可以任意转动,则的最大值为________. 【答案】【解析】【分析】先求圆锥内切球半径,再根据取最大值时,四面体外接球恰为圆锥内切球,解得结果.【详解】设圆锥内切球半径为,则,所以,因为取最大值时,正四面体外接球恰为圆锥内切球,所以,解得.三、解答题.解答应写出文字说明、证明过程和演算步骤.17.已知是递增的等比数列,,.(1)求数列的通项公式;(2)令,求数列的前项和.【答案】(1) (2)【解析】【分析】(1)解法1:运用等比数列的通项公式,解方程可得首项和公比,即可得到所求通项公式;解法2:运用等比数列的性质建立方程.(2)的通项公式是等差数列与等比数列的乘积,利用错位相减求和.【详解】解法1:(1)设等比数列的公比为,因为,,所以解得或因为是递增的等比数列,所以,.所以数列的通项公式为.解法2:(1)设等比数列的公比为,因为,,所以,是方程的两个根.解得或因为是递增的等比数列,所以,,则.所以数列的通项公式为.(2)由(1)知.则,①在①式两边同时乘以得,,②①-②得,即, 所以.18.科研人员在对人体脂肪含量和年龄之间关系的研究中,获得了一些年龄和脂肪含量的简单随机样本数据,如下表: (年(脂根据上表的数据得到如下的散点图.(1)根据上表中的样本数据及其散点图:(i)求;(i)计算样本相关系数(精确到0.01),并刻画它们的相关程度.(2)若关于的线性回归方程为,求的值(精确到0.01),并根据回归方程估计年龄为50岁时人体的脂肪含量.附:参考数据:,,,,,,参考公式:相关系数回归方程中斜率和截距的最小二乘估计公式分别为,.【答案】(1) (ⅰ)47 (ⅱ)见解析;(2) ;%.【解析】【分析】(1)(i)根据上表中的样本数据,利用平均数的公式求得结果;(ii)利用公式求得相关系数的值,从而可以推断人体脂肪含量和年龄的相关程度很强.(2)利用回归直线过样本中心点,求得,得到回归直线的方程,再将代入回归直线方程求得结果.【详解】(1)根据上表中的样本数据及其散点图:(ⅰ).(ⅱ).因为,,所以.由样本相关系数,可以推断人体脂肪含量和年龄的相关程度很强.(2)因为回归方程为,即.所以.【或利用】所以关于的线性回归方程为.将代入线性回归方程得.所以根据回归方程估计年龄为岁时人体的脂肪含量为%.【点睛】该题考查的是有关回归分析的问题,涉及到的知识点有平均值的计算,根据相关系数r的大小判断相关性,回归直线的性质,属于简单题目.19.如图,在四棱锥中,底面为菱形,,,且.(1)求证:平面平面;(2)若,求二面角的余弦值.【答案】(1)见解析;(2).【解析】【分析】(1)先根据计算得线线线线垂直,再根据线面垂直判定定理以及面面垂直判定定理得结论,(2)建立空间直角坐标系,利用空间向量求二面角.【详解】(1)证明:取中点,连结,,,因为底面为菱形,,所以.因为为的中点,所以.在△中,,为的中点,所以.设,则,,因为,所以.在△中,,为的中点,所以.在△ 和△ 中,因为,,,所以△ △ .所以.所以.因为,平面,平面,所以平面.因为平面,所以平面平面.(2)因为,,,平面,平面,所以平面.所以.由(1)得,,所以,,所在的直线两两互相垂直.以为坐标原点,分别以所在直线为轴,轴,轴建立如图所示的空间直角坐标系.设,则,,,,所以,,,设平面的法向量为,则令,则,,所以.设平面的法向量为,则令,则,,所以.设二面角为,由于为锐角,所以.所以二面角的余弦值为.【点睛】本题考查线面垂直判定定理、面面垂直判定定理以及利用空间向量求二面角,考查基本分析论证与求解能力,属中档题.20.在平面直角坐标系中,动点分别与两个定点,的连线的斜率之积为.(1)求动点的轨迹的方程;(2)设过点的直线与轨迹交于,两点,判断直线与以线段为直径的圆的位置关系,并说明理由.【答案】(1);(2)相离.【解析】【分析】(1)根据直接法求轨迹方程,(2)先用坐标表示以线段为直径的圆方程,再根据圆心到直线距离与半径大小进行判断.【详解】(1)设动点的坐标为,因为,,所以,整理得.所以动点的轨迹的方程.(2)过点的直线为轴时,显然不合题意.所以可设过点的直线方程为,设直线与轨迹的交点坐标为,,由得.因为,由韦达定理得=,=.注意到=.所以的中点坐标为.因为.点到直线的距离为.因为,即,所以直线与以线段为直径的圆相离.【点睛】本题考查直接法求轨迹方程以及直线与圆位置关系,考查基本分析求解能力,属中档题.21.己知函数.(1)讨论函数的单调性;(2)若函数有两个零点,,求的取值范围,并证明. 【答案】(1)见解析;(2)见证明【解析】【分析】(1)函数f(x)的定义域为(0,+∞),f′(x),x>0,利用分类讨论思想,结合导数性质能讨论函数f(x)的单调性.(2)先求k的取值范围是,再证明f(﹣2k)=ln(﹣2k)0.然后证明x1+x2≥2,即证(1)(1+t)2<﹣8lnt,即证8lnt+()(1+t)2<0,(t>0).设h(t)=8lnt+()(1+t)2,t>1.则h(t)=8lnt﹣t2﹣2t,t>1.由此能证明x1+x2>2.【详解】(1)解:因为,函数的定义域为,所以.当时,,所以函数在上单调递增.当时,由,得(负根舍去),当时,,当时,,所以函数在上单调递减;在上单调递增.综上所述,当时,函数在上单调递增;当时,函数在上单调递减,在上单调递增(2)先求的取值范围:方法1:由(1)知,当时,在上单调递增,不可能有两个零点,不满足条件.当时,函数在上单调递减,在上单调递增,所以,要使函数有两个零点,首先,解得.因为,且,下面证明.设,则.因为,所以.所以在上单调递增,所以.所以的取值范围是.方法2:由,得到.设,则.当时,,当时,,所以函数在上单调递减,在上单调递增.所以由.因为时,,且,要使函数有两个零点,必有.所以的取值范围是.再证明:方法1:因为,是函数的两个零点,不妨设,令,则.所以即.所以,即,,.要证,即证.即证,即证.因为,所以即证,或证.设,.即,.所以.所以在上单调递减,所以.所以.方法2:因为,是函数有两个零点,不妨设,令,则.所以即.所以,即,,.要证,需证.即证,即证.因为,所以即证.设,则,.所以在上单调递减,所以.所以.方法3:因为,是函数有两个零点,不妨设,令,则.所以即.要证,需证.只需证.即证,即证.即证.因为,所以,即.所以.而,所以成立.所以.方法4:因为,是函数有两个零点,不妨设,令,则.由已知得即.先证明,即证明.设,则.所以在上单调递增,所以,所证不等式成立.所以有.即.因为(),所以,即.所以.方法5:要证,其中,,即证.利用函数的单调性,只需证明.因为,所以只要证明,其中.构造函数,,则.因为(利用均值不等式),所以在上单调递减.所以.所以在上恒成立.所以要证的不等式成立.【点睛】本题考查函数单调性的讨论,考查不等式的性质,考查导数性质、函数的单调性、最值等基础知识,考查运算求解能力,考查化归与转化思想,是难题.22.在直角坐标系中,倾斜角为的直线的参数方程为(为参数).在以坐标原点为极点,轴正半轴为极轴的极坐标系中,曲线的极坐标方程为.(1)求直线的普通方程与曲线的直角坐标方程;(2)若直线与曲线交于,两点,且,求直线的倾斜角.【答案】(1);(2)或.【解析】【分析】(1)根据平方关系消参数得直线的普通方程,根据得曲线的直角坐标方程(2)利用直线参数方程几何意义求解.【详解】(1)因为直线的参数方程为(为参数),当时,直线的直角坐标方程为.当时,直线的直角坐标方程为.因为,因为,所以. 所以的直角坐标方程为.(2)解法1:曲线的直角坐标方程为,将直线的参数方程代入曲线的方程整理,得.因为,可设该方程的两个根为,,则 ,.所以.整理得, 故. 因为,所以或,解得或综上所述,直线的倾斜角为或. 解法2:直线与圆交于,两点,且, 故圆心到直线的距离.①当时,直线的直角坐标方程为,符合题意.②当时,直线的方程为. 所以,整理得.解得.综上所述,直线的倾斜角为或.【点睛】本题考查参数方程化普通方程、极坐标方程化直角坐标方程以及直线参数方程应用,考查综合分析求解能力,属中档题.23.[选修4-5:不等式选讲] 己知函数.(1)当时,解不等式;(2)若存在实数x ,使得成立,求实数a 的取值范围.【答案】(1);(2) .【解析】【分析】(1)根据绝对值定义转化为两个不等式组,解可得,(2)根据绝对值定义转化为分段函数,根据函数最值可得结果.【详解】(1)当时,由,得.当时,,解得.当时,,解得.综上可知,不等式的解集为.(2)由,得.则.令,则问题等价于因为.所以实数的取值范围为.【点睛】本题考查含绝对值不等式,考查基本分析求解能力,属基本题.2019届广东省广州市高三第二次模拟考试数学(理)试卷。
2019年广州市高考数学模拟测试(后附满分答案)
2019年广州市高考数学模拟测试(附满分答案)第Ⅰ卷 (选择题 共4 0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. 若集合{}21|21|3,0,3x A x x B xx ⎧+⎫=-<=<⎨⎬-⎩⎭则A ∩B 是( ) A .11232x x x ⎧⎫-<<-<<⎨⎬⎩⎭或 B.{}23x x << C.122x x ⎧⎫-<<⎨⎬⎩⎭ D .112x x ⎧⎫-<<-⎨⎬⎩⎭2.如图,在复平面内,复数1z ,2z 对应的向量分别是OA ,OB ,则复数12z z 对应的点位于( ) A.第一象限 B.第二象限C.第三象限D.第四象限3.已知等差数列{}n a 中,25a = ,411a =,则前10项和=10S ( )A . 55B . 155C . 350D . 400 4. 函数f(x)=1+log x 2与g(x)=2x-1在同一直角坐标系下的图象大致是 ( )5.平面四边形ABCD 中0AB CD +=,()0AB AD AC -=⋅,则四边形ABCD 是 ( ) A .矩形 B .梯形 C .正方形 D .菱形6. 一个四棱锥的三视图如图所示,其中主视图是腰长为1的等腰 直角三角形,则这个几何体的体积是 A .21 B .1 C .23D .27.下列命题:①函数22()sin cos f x x x =-的最小正周期是π;②函数()(1f x x =-是偶函数; ③若111(1)adx a x=>⎰,则a e =; ④椭圆)0(3222>=+m m y x 的离心率不确定。
其中所有的真命题是( )A.①②B.③④C.②④D.①③8.定义在R 上的函数()f x ,如果存在函数()(,g x kx b k b =+为常数),使得()()f x g x ≥对一切实数x 都成立,则称()g x 为函数()f x 的一个“承托函数”.现有如下命题:①对给定的函数()f x ,其承托函数可能不存在,也可能有无数个;②()2g x x =为函数()2x f x =的一个承托函数;③定义域和值域都是R 的函数()f x 不存在承托函数.其中正确的命题是( ).A.①B.②C.①③D.②③第Ⅱ卷(非选择题 共110分)二、填空题:本大题共7小题,考生作答6小题,每小题5分,满分30分. (一)必做题(9~13题)(一)必做题(9~13题)9. 已知()πϕϕπ<<=+0,23)2sin(,则ϕtan =________ .10.若52345012345(12),x a a x a x a x a x a x +=+++++则a 3= 。
广州中山大学附属中学2019年高三数学模拟真题及答案
广州中山大学附属中学2019年高三数学模拟真题及答案【试题】一、选择题(本大题有15小题,每小题2分,共30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。
请将你认为正确选项的字母写在答题卷上。
1. 一次函数 y = kx + b 的图象是一条直线,已知点(2, 3)在此直线上,且当 x = 5 时,函数值为12,那么 k 的值为()A. 2B. 3C. 4D. 52. 设正数 a,b 满足 a^2 + b^2 = 1,那么 a + b 的最大值是(注:结果要写成准确的小数,如√2+√3要写成√2+√3,根号内不能有小数)A. 1B. √2C. √3D. 23. 定义序列 {a_n} 为:a_1 = 2,a_{n+1} = 4a_n - 3(n ≧ 1),则 a_1+ a_2 + a_3 + … + a_{10} 的值为()A. 16335B. 16339C. 16343D. 163474. 在直角坐标系中,点 A 的坐标是(1, 2),点 B 在 x 轴正半轴上,若线段 AB 的中点坐标为 (-3, 1),那么 B 的坐标为()A. (-5, 0)B. (-2, 0)C. (0, 0)D. (2, 0)5. 已知集合 A = {1, 2, 3},集合 B = {x | x^2 + px + q = 0 的根},若A ∩B = ∅,则 p + q 的取值范围是(注:结果要写成一个范围,如 [1,3] )A. [2, 6)B. [1, 4)C. (0, 5)D. (1, 3]6. 函数 f(x) = x^2 - 4ax + 2 在区间 (2, 5) 上是减函数,则 a 的取值范围是()A. (1/2, 2)B. (1, 2)C. (3/2, 2)D. (2, 4)7. 已知arctan (1/a) + arctan (1/b) = π/4(其中:a > 0,b > 0),那么a +b 的最小正整数值是()A. 2B. 4C. 6D. 108. 将九个不同的数两两互不相同地放入一个九宫格中,填满此九宫格,那么九个数的和最大为()A. 65B. 67C. 72D. 909. 已知动点 M 在直线 6x + 8y = 64 上,点 P 在直线 x - y = 0 上,点N 在直线 4x + 3y = 12 上,若等腰三角形 MPN 的周长最小,则点 M 的坐标为()A. (2, 4)B. (3, 6)C. (6, 3)D. (8, 2)10. 若集合 A = {(x, y) | x^2 + y^2 = 1},集合 B = {(x, y) | x = t, y = t - 1},那么集合A ∩ B 为()A. 一个点B. 两个点C. 无穷多个点D. 无交集11. 已知四阶行列式 D = | k k k k || k k^2 k^3 k^4 || k k^3 k^5 k^7 || k k^4 k^7 k^11 |,则 D 的值为(注:结果要写成 k 的多项式)A. k^9B. k^5C. k^13D. k^1512. 设函数 f(x) = a^{3x^2 + 2x - 1} 是 (1, 2) 上的减函数,则 a 的取值范围是()A. (0, 1)B. (0, 1]C. (1, ∞)D. (1, +∞)13. 若点 P(2, a) 在直线 3x - 4y = 10 上,则 a 的取值范围是()A. [-2, 2)B. (-2, 2]C. (-∞, -2) ∪ (2, +∞)D. (-∞, -2] ∪ (2, +∞)14. 函数f(x) = ax^2 + bx + c (a ≠ 0) 的图象过点 (1, 2),且点 (2, 1) 在其右侧关于直线 x = -1 的对称点处,那么 f(x) 的一个零点是()A. -1B. 0C. 1D. 215. 设对于所有的 x,函数 f(x) 满足 f(x + 1) = 2f(x) + x + 1,若 f(0) = -1,那么 f(2019) 的值为()A. 2^{2019} - 2018B. 2^{2019} - 2017C. 2^{2020} - 2019D. 2^{2020} - 2020【试题答案】1. C2. B3. D4. B5. D6. D7. B8. C9. C 10. C 11. C12. C 13. C 14. C 15. A【解析】1. 题目给出一次函数的表达式 y = kx + b,再根据已知条件可得到方程组:2 = 2k + b12 = 5k + b解方程组得 k = 4,因此答案为 C。
2019-2020学年广州大学附属中学高三英语一模试题及参考答案
2019-2020学年广州大学附属中学高三英语一模试题及参考答案第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项AWelcome to join our Summer Youth Language Program to improve English language skills, make new friends worldwide. and have a good time here! The program capacity is limited, so if you are interested, you should register as soon as possible.Dates andCostsAll programs require a $ 100 nonrefundable deposit (不退还的押金)to reserve a spot in the program.June 15—July 16 5-week program:( $ 1920)July 20—August 20 5-week program:( $ 1920)June 15—August 20 10-week program:( $ 3620)You can also study with us for shorter periods. 4-week programs cost $ 1580 tuition and 3-week programs cost $ 1240.DiscountsThere is a 10% discount for each additional family member!Appropriate AgesRecommended Ages:14 and olderRestrictions:Students who are younger than 16 must have a parent or guardian with them.Program ScheduleIn the morning, you will join the all-aged Intensive English classes from 9:00 am to 2:00 pm, Monday to Thursday, where they can meet other students from worldwide. In the afternoonfrom 2 to 3 pm, we will have fun after-school activities, like soccer in the Park, visit toScienceMuseumand story writing competition.Items Students Should BringClothes:Shirts, a jacket, long pants ,a swimsuit and comfortable shoes, etc.Other personal items:Camera phone? plug adapter, photos of friends/family.Study materials:Notebooks, pens, and pencils.Airport Safe Items:Don't bring foods and snacks, or they'll be charged before you board the airport.1. How much will be charged if you and your brother join in a 3-week program?A. $ 1920,B. $ 2356.C. $3002.D. $ 3620.2. What will participants do at 2:30 pm on Monday?municate with foreign students.B. Attend intensive English classes.C. Join in some interesting activities.D. Talk with their parents on the phone.3. Why should students avoid bringing foods and snacks?A. To save space for their luggage.B. To protect the environment.C. To avoid any unwanted fees.D. To follow the rules of the airport.B“Snowplow(扫雪机) parenting” is the newest parenting style that can include parents booking their adult children haircuts, calling their college kids to wake them up so that they don’t sleep through a test, and even calling their kids’ employers.“‘Helicopter (直升机) parenting’ means monitoring their kids’ every activity,which is out of date.” Claire Cain Miller and Jonah Engel Bromwich wrote in The New York Times. “Some rich mothers and fathers now are more like snowplows: clearing any problems in their children’s path to success so that they don’t have to meet failure or lose opportunities.”There is a mother who started a charity in her son’s name to try to raise his chances of being accepted to the college. Another parents spent years helping their daughter avoid foods with sauce, which she didn’t like. Once she got to college, she had problems with the food in her school because it was all covered in sauce.A survey says that three-quarters of parents of children between the ages of 18 and 28 ask for doctor visits or haircuts for their children, and 11% say they would call their kids’ bosses whether their children are having an issue at work.As reported, wealthy parents try to get their children into top colleges by giving a large amount of money to a school, such as paying for a building. This parenting has become the most popular way to raise children, whatever the income, education, or race is.Julie, a teacher at Stanford, told the Times that “snowplow parenting” is not a reasonable approach. “The parents should prepare the kid for the road, instead of preparing the road for the kid,” she said.4. How does Julie like “snowplow parenting”?A. It is unreasonable.B. It is advanced.C. It is accepted by teachers.D. It is refused by rich people.5. What is the character of “helicopter parenting”?A. Parents make kids popular.B. Parents provide little money for kids.C. Parents ask kids to care for themselves.D. Parents watch over kids’ every activity.6. What should parents do according to Julie?A. Do as wealthy parents do.B. Make kids be prepared.C. Make roads be prepared.D. Do as little as possible.7. What’s the best title for the text?A. Helicopter Parenting.B. The Similarity in Parenting.C. A Research on Parenting.D. A New Kind of Parenting.CAt Aizo Chuo Hospital in Japan, employees greet newcomers, guide patients to and from the surgery area, and print out maps of the hospital for confused visitors. They don’t take lunch breaks or even get paid. Why? They’re robots!Robots have long worked in factories, helping to build cars and electronic appliances. But today’s robots don’t just do the jobs of people-they actually look and act a lot like people.Kansei, arobot from Japan, has a plastic face covering 19 movable parts. The robot can make 36 facial expressions in response to different words. Kansei shakes in fear at the word “war” and smiles when it hears the word “dinner”.Researchers in Europe are going even further with iCub, a “baby” robot. They are teaching it to speak and hold conversations.The ability to interact is crucial for robots that will one day work closely with humans says robotics professor ChrisAtkeson. “ This will require robots to understand what you say and how you are feeling and respond with appropriate emotions, ” he told WR News.Japanese scientist Minoru Asada agrees. He is building a robot called CB2 that acts like a real baby. “ Right now, it only goes, ̒Ah, ah. ̓But as we develop its learning function, it will start saying more complex sentences and moving on its own, ” Asada says. “ Next-generation robots need to be able to learn and develop by themselves.”Intelligent robot will become more important in the future, as populations age and the number of human workers declines in many countries. “ We’re going to have many more old people and not enough young people to care for them,” says robot researcher Matthew Mason. “ Technology can help the old people live at home longer, instead of going to nursing homes.”8. According to the passage, what jobs have robots already performed?A. Giving advice, answering customer questions and planning events.B. Producing factory goods, building cars and greeting customers.C. Greeting customers, producing factory goods and performing surgery.D. Building cars, driving passengers and providing directions.9. The second paragraph in this passage is mainly about?A. To explain how a robot works.B. To define what a robot is.C. To describe the functions of modern robots.D. To predict the future uses of robots.10. How does the Kansei robot react on the word “fire”?A. Use languages to warn nearby humans.B Back up its memory files.C. Activate an automatic fire alarm.D. Produce a worried look on his face.11. In Asada’s opinion, the next step for robots will be to develop_______.A. the ability to learn independentlyB. the ability to understand human commandsC. the capacity to interact with humansD. the willingness to work togetherDLarry was on another of his underwater expeditions(探险)but this time, it was different. He decided to take his daughter along with him. She was only ten years old. This would be her first trip with her father on what he had always been famous for.Larry first began diving when he was his daughter’s age. Similarly, his father had taken him along on one of his expeditions. Since then, he had never looked back. Larry started out by renting diving suits from the small diving shop just along the shore. He had hated them. They were either too big or too small. Then, there was the instructor. He gave him a short lesson before allowing him into the water with his father. He had made an exception. Larry would never have been able to go down without at least five hours of theory and another similar number of hours on practical lessons with a guide. Children of his age were not even allowed to dive.After the first expedition, Larry’s later diving adventures only got better and better. There was never a dull moment. In his black and blue suit and with an oxygen tank fastened on his back, Larry dived from boats into the middle of the ocean. Dangerous areas did not prevent him from continuing his search. Sometimes, he was limited to a cage underwater but that did not bother him. At least, he was still able to take photographs of the underwater creatures.Larry’s first expedition without his father was in the Cayman Islands. There were numerous diving spots in the area and Larry was determined to visit all of them .Fortunately for him, a man offered to take him around the different spots for rry didn’t even know what the time was, how many spots he dived into or how many photographs he had taken.The diving spots afforded such a wide range of fish and sea creatures that Larry saw more than thirty varieties of creatures.Larry looked at his daughter. She looked as excited as he had been when he was her age. He hoped she would be able to continue the family tradition. Already, she looked like she was much braver than had been then. This was the key to a successful underwater expedition.12. In what way was this expedition different for Larry?A. His daughter had grown up.B. He had become a famous diver.C. His father would dive with him.D. His daughter would dive with him.13. What can be inferred from Paragraph 2?A. Larry had some special right.B. Larry liked the rented diving suits.C. Divers had to buy diving equipment.D. Ten-year-old children were permitted to dive.14. What can be learned from the underlined sentence?A. Larry didn’t wear a watch.B. Larry was not good at math.C. Larry had a poor memory.D. Larry enjoyed the adventure.15. What did Larry expect his daughter to do?A. Become a successful diver.B. Make a good diving guide.C. Take a lot of photo underwater.D. Have longer hours of training.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
广东省广州市2019届高三第二次模拟考试数学(文)试题含答案
2019年广州市普通高中毕业班综合测试(二)文科数学2019.4本试卷共6页,23小题,满分150分。
考试用时120分钟。
注意事项:1.答卷前,考生务必将自己的姓名和考生号、试室号、座位号填写在答题卡上,用2B铅笔在答题卡的相应位置填涂考生号,并将试卷类型(B)填涂在答题卡相应位置上。
2.作答选择题时,选出每小题答案后,用2B铅笔在答题卡上对应题目选项的答案信息点涂黑;如需改动,用橡皮擦干净后,再选涂其他答案。
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作答,答案必须写在答题卡各题目指定区域内的相应位置上;如需改动,先划掉原来的答案,然后再写上新答案;不准使用铅笔和涂改液。
不按以上要求作答无效。
4.考生必须保证答题卡的整洁。
考试结束后,将试卷和答题卡一并交回。
一、选择题:本题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合A={x∈N|0<x<6} , B={2, 4, 6, 8} ,则 A∩B=A.{0,1,3,5}B.{0,2,4,6}C. {1,3,5}D.{2,4,6}2.已知复数z=m(3+i)-(2+i)在复平面内对应的点在第三象限,则实数m的取值范围是A. B. C. D.3.某公司生产A,B,C三种不同型号的轿车,产量之比依次为2:3:4,为检验该公司的产品质量,用分层抽样的方法抽取一个容量为n的样本,若样本中A种型号的轿车比B种型号的轿车少8辆,则n=A. 96B. 72C. 48D. 364.执行如图所示的程序框图,则输出z的值是A. 21B. 22C. 23D. 245.从某班5名学生(其中男生3人,女生2人)中任选3人参加学校组织的社会实践活动,则所选3人中至少有1名女生的概率为A. B. C. D.6.函数y=的部分图像如图所示,则函数的解析式为A. B.C. D.7.设等比数列{a n}的前n项和为S n,则下列等式中一定成立的是A. S n+S2n=S3nB. S22n=S n S3nC. S22n=S n+S2n- S3nD. S2n + S22n=S n (S2n+S3n)8.已知双曲线拘渐近线方程为5x±3y=0,则此双曲线的离心率为A. B. C. D.9.一个圆锥的体积为,当这个圆锥的侧面积最小时,其母线与底面所成角的正切值为A. B. C. D.10.设a≥b≥c,且1是一元二次方程ax2+ bx+c=0的一个实根,则的取值范围为A.[-2,0] B.C.D.11.在三棱锥P-ABC中,PA=PB=PC=2,AB=AC=I,BC=,则该三棱锥的外接球的表面积为A. B. C. D.12.己知函数与的图像上存在关于x轴对称的点,则实数a的取值范围为A.B.C.D.二、填空题:本题共4小题,每小题5分,共20分.13.已知向量,向量,则=14. 《莱茵德纸草书》是世界上最古老的数学著作之一.书中有一道这样的题目:把100个面包分给5个人,使每人所得份量成等差数列,且较大的三份之和的是较小的两份之和,则最小一份的量为.15.若函数f(x)=x2 -x+l+ alnx在(0,+∞)上单调递增,则实数a的取值范围是.16.己知点P在直线x+2y-l=0上,点Q在直线x+2y+3=O E,PQ的中点为M(x0,y0),且-1≤y0 -x0≤7,则的取值范围是____.三、解答题:共70分,解答应写出文字说明、证明过程和演算步骤,第17~21题为必考题,每个试题考生都必须做答.第22、23题为选考题,考生根据要求做答.(一)必考题:共60分.17. (本小题满分12分)△ABC中角A,B,C的对边分别为a,b,c,已知(1)求的值;(2)若c=2,求△ABC的面积.18. (本小题满分12分)如图,在四棱锥P-ABCD中,底面ABCD是边长为2的菱形,∠BAD=60°,∠APD=90°,且PA=PD,AD=PB.(1)求证:AD⊥PB;(2)求点A到平面PBC的距离.19. (本小题满分12分)科研人员在对人体脂肪含量和年龄之间关系的研究中,获得了一些年龄和脂肪含量的简单随机样本数据,如下表:根据上表的数据得到如下的散点图.(1)根据上表中的样本数据及其散点图:(i)求;(ii)计算样本相关系数(精确到0.01),并刻画它们的相关程度.(2)若y关于x的线性回归方程为,求的值(精确到0.01),并根据回归方程估计年龄为50岁时人体的脂肪含量.附:参考数据:参考公式:相关系数回归方程中斜率和截距的最小二乘估计公式分别为20. (本小题满分12分)从抛物线y 2=36x 上任意一点P 向x 轴作垂线段,垂足为Q ,点M 是线段PQ 上的一点,且满足(1)求点M 的轨迹C 的方程;(2)设直线x=my+1(m ∈R)与轨迹c 交于A ,B 两点,T 为C 上异于A ,B 的任意一点,直线AT ,BT 分别与直线x=-1交于D ,E 两点,以DE 为直径的圆是否过x 轴上的定点?若过定点,求出符合条件的定点坐标;若不过定点,请说明理由.21. (本小题满分12分)已知函数f(x)=(x+2)lnx+ax 2- 4x+ 7a . (1)若a=,求函数f(x)的所有零点; (2)若a ≥,证明函数f(x)不存在极值.(二)选考题:共10分.请考生在第22、23题中任选一题作答,如果多做,则按所做的第一题计分. 22.[选修4 -4:坐标系与参数方程](本小题满分10分) 在直角坐标系xOy 中,倾斜角为α的直线l 的参数方程为⎩⎨⎧+=+=ααsin 3,cos 2t y t x (t 为参数).在以坐标原点为极点,x轴正半轴为极轴的极坐标系中,曲线C 的极坐标方程为ρ2= 2p cos θ+8. (1)求直线l 的普通方程与曲线C 的直角坐标方程; (2)若直线l 与曲线C 交于A ,B 两点,且求直线l 的倾斜角.23.[选修4-5:不等式选讲](本小题满分10分) 己知函数f(x) =|2x-l|-a . (1)当a=l 时,解不等式f(x)>x+1;(2)若存在实数x,使得f(x)< f(x+1)成立,求实数a的取值范围.绝密 ★ 启用前2019年广州市普通高中毕业班综合测试(二)文科数学试题答案及评分参考评分说明:1.本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分参考制订相应的评分细则.2.对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应得分数的一半;如果后继部分的解答有较严重的错误,就不再给分.3.解答右端所注分数,表示考生正确做到这一步应得的累加分数. 4.只给整数分数.选择题不给中间分.一、选择题二、填空题13 14.53 15.1,8⎡⎫+∞⎪⎢⎣⎭16.2,05⎡⎤-⎢⎥⎣⎦三、解答题17.解:(1)因为tan tan 2(tan tan )cos cos A BA B B A+=+, 所以sin sin sin sinB 2cos cos cos cos cos cos A B A A B A B A B ⎛⎫+=+⎪⎝⎭.………………………………………………1分 化简得()2sin cos cos sin sin sin A B A B A B +=+.………………………………………………2分即()2sin sin sin A B A B +=+.………………………………………………………………………3分 因在ABC ∆中,A B C ++=π,则()()sin sin sin A B C C π+=-=.……………………………4分 从而sin sin 2sin A B C +=.…………………………………………………………………………… 5分 由正弦定理,得2a b c +=. 所以=2a bc+.……………………………………………………………………………………………6分 (2)由(1)知2a bc +=,且2c =,所以4a b +=.……………………………………………………7分 因为=3C π,所以()222222cos 22a b ab ca b c C ab ab+--+-==.……………………………………9分 即122cos32ab abπ-=. 所以4ab =.……………………………………………………………………………………………10分所以11sin 4sin 223ABC S ab C ∆π==⨯⨯= 所以△ABC12分18.(1)证明:取AD 的中点O ,连结OP ,OB ,BD ,因为底面ABCD 为菱形,60BAD ∠=,所以AD AB BD ==.…………………………………1分 因为O 为AD 的中点,所以BO AD ⊥. ……………2分 在△PAD 中,PA PD =,O 为AD 的中点,所以PO AD ⊥. ………………………………………3分 因为BOPO O =,所以AD ⊥平面POB .………4分因为PB ⊂平面POB ,所以AD PB ⊥.………………………………………………………………5分 (2)解法1:在Rt △ PAD 中,2AD =,所以1PO =.因为底面ABCD 是边长为2的菱形,60BAD ∠=,所以BO =.……………………………6分D CBAPO在△PBO 中,1PO =,BO =2PB BC ==,因为222PO BO PB +=,所以PO BO ⊥.……………………………………………………………7分 【6-7分段另证:在△APD 中,90APD ∠=,O 为AD 的中点,所以12PO AD AO ==. 在△ BOP 和△ BOA 中,因为PO AO =, PB AD AB ==,BO BO =,所以△ BOP ≅△ BOA . 所以90BOP BOA ∠=∠=.所以OP OB ⊥.】 由(1)有PO AD ⊥,且ADBO O =,AD ⊂平面ABCD ,BO ⊂平面ABCD ,所以PO ⊥平面ABCD .…………………………………………………………………………………8分 在△PBC 中,由(1)证得AD PB ⊥,且//BC AD ,所以BC PB ⊥.因为2PB BC ==,所以2PBC S ∆=.…………………………………………………………………9分 在△ABC 中,2AB BC ==,120ABC ∠=, 所以1sin 2ABC S AB BC ABC =⨯⨯⨯∠=.………………………………………………………10分 设点A 到平面PBC 的距离为h , 因为A PBC P ABC V V --=,即1133PBC ABC S h S PO ∆=.……………………………………………………11分 所以122ABC PBC S PO h S ∆===.所以点A 到平面PBC .…………………………………………………………………12分 解法2:因为//AD BC ,BC ⊂平面PBC ,AD ⊄平面PBC , 所以//AD 平面PBC .所以点A 到平面PBC 的距离等于点O 到平面PBC 的距离.………………………………………6分 过点O 作OH PB ⊥于点.…………………………7分因为OH ⊂平面POB ,所以BC ⊥OH . 因为PBBC B =,PB ⊂平面PBC ,BC ⊂平面PBC ,所以OH ⊥平面PBC .…………………………………8分 在Rt △ PAD 中,2AD =,所以1PO =.因为底面ABCD 是边长为2的菱形,60BAD ∠=,所以BO =.……………………………9分 在△PBO 中,1PO =,BO =2PB BC ==,因为222PO BO PB +=,所以PO BO ⊥.…………………………………………………………10分 【9-10分段另证:在△APD 中,90APD ∠=,O 为AD 的中点,所以12PO AD AO ==. 在△ BOP 和△ BOA 中,因为PO AO =,PB AD AB ==,BO BO =,所以△ BOP ≅△ BOA . 所以90BOP BOA ∠=∠=.所以OP OB ⊥.】在△PBO 中,根据等面积关系得PB OH PO OB ⨯=⨯.…………………………………………11分所以PO OB OH PB ⨯===. 所以点A 到平面PBC的距离为2.…………………………………………………………………12分19.解:(1)根据上表中的样本数据及其散点图:(ⅰ)262739414953565860614710x +++++++++==.…………………………………2分(ⅱ)rni ix y nx y-=∑=…………3分==…………………………………4分=5分6.56≈54.18≈,所以0.98r ≈.……………………………………………………………………………………………6分 由样本相关系数0.98r ≈,可以推断人体脂肪含量和年龄的相关程度很强.………………………7分(2)因为回归方程为ˆˆ 1.56ybx =+,即ˆ 1.56a =. 所以ˆ27 1.56ˆ0.5447y abx--==≈.【或利用()()()121ˆn iii nii x x y y bx x ==--=-∑∑()1221ni ii nii x y nx yxn x==-=-∑∑837.80.541548=≈】……………………………10分 所以y 关于x 的线性回归方程为ˆ0.54 1.56yx =+. 将50x =代入线性回归方程得ˆ0.5450 1.5628.56y=⨯+=.……………………………………11分 所以根据回归方程估计年龄为50岁时人体的脂肪含量为28.56%.………………………………12分 【结论没写28.56%扣1分】20.解:(1)设(),M x y ,()00,P x y ,则点Q 的坐标为()0,0x .因为2PM MQ =,所以()()000,2,x x y y x x y --=--,………………………………………………………………………1分即00,3.x x y y =⎧⎨=⎩ ………………………………………………………………………………………………2分因为点P 在抛物线236y x =上,所以20036y x =,即()2336y x =.………………………………………………………………………3分所以点M 的轨迹C 的方程为24y x =.…………………………………………………………………4分(2)解法1:设直线1x my =+与曲线C 的交点坐标为A 211,4y y ⎛⎫ ⎪⎝⎭,222,4y B y ⎛⎫⎪⎝⎭,由韦达定理得+1y 2y =4m ,1y 2y =4-.……………………………………………………………5分设点200,4y T y ⎛⎫ ⎪⎝⎭,则10220101444AT y y k y y y y -==+-.………………………………………………………6分 所以直线AT 的方程为2000144y y y x y y ⎛⎫-=- ⎪+⎝⎭.令1x =-,得点D 的坐标为010141,y y y y ⎛⎫-- ⎪+⎝⎭.…………………………………………………………7分同理可得点E 的坐标为020241,y y y y ⎛⎫-- ⎪+⎝⎭.………………………………………………………………8分如果以DE 为直径的圆过x 轴某一定点(),0N n ,则满足0ND NE ∙=.…………………………9分因为010********,1,y y y y ND NE n n y y y y ⎛⎫⎛⎫--∙=--∙-- ⎪ ⎪++⎝⎭⎝⎭()()()2212001220012124161++y y y y y y n y y y y y y -++=+++. 所以()2200200416161++044y my n y my --+=+-.………………………………………………………………10分 即()2140n +-=,解得1n =或3n =-.……………………………………………………………11分 故以DE 为直径的圆过x 轴上的定点()1,0和()3,0-.………………………………………………12分 解法2:直线1x =与曲线C 的交点坐标为()1,2A ',()1,2B '-,若取()0,0T ',则A T '',B T ''与直线1x =-的交点坐标为()1,2D '--,()1,2E '-, 所以以D E ''为直径的圆的方程为()2214x y ++=.该圆与x 轴的交点坐标为()1,0和()3,0-.所以符合题意的定点只能是()11,0N 或()23,0N -.…………………………………………………6分设直线1x my =+与曲线C 的交点坐标为A 211,4y y ⎛⎫ ⎪⎝⎭,222,4y B y ⎛⎫⎪⎝⎭,由韦达定理得+1y 2y =4m ,1y 2y =4-.……………………………………………………………7分设点200,4y T y ⎛⎫ ⎪⎝⎭,则10220101444AT y y k y y y y -==+-.………………………………………………………8分 所以直线AT 的方程为2000144y y y x y y ⎛⎫-=- ⎪+⎝⎭.令1x =-,得点D 的坐标为010141,y y y y ⎛⎫-- ⎪+⎝⎭.…………………………………………………………9分同理可得点E 的坐标为020241,y y y y ⎛⎫-- ⎪+⎝⎭.………………………………………………………………10分若点()11,0N 满足要求,则满足110N D N E ∙=.因为0102110102442,2,y y y y N D N E y y y y ⎛⎫⎛⎫--∙=-∙- ⎪ ⎪++⎝⎭⎝⎭()()212001220012124164+y y y y y y y y y y y y -++=+++20020041616=4+044y my y my --+=+-.……11分 所以点()11,0N 满足题意. 同理可证点()23,0N -也满足题意.故以DE 为直径的圆过x 轴上的定点()1,0和()3,0-.………………………………………………12分 21.(1)解:当21=a 时,217()(2)ln 422f x x x x x =++-+, 函数)(x f 的定义域为),0(+∞,…………………………………………………………………………1分 且()2ln 3f x x x x'=++-.……………………………………………………………………………2分 设()2ln 3g x x x x=++-, 则()()222221122()1x x x x g x x x x x+-+-'=-+==()0x >.当01x <<时,()0g x '<;当1x >时,()0g x '>,即函数()g x 在()0,1上单调递减,在()1,+∞上单调递增,…………………………………………3分 所以当0x >时,()()10g x g ≥=(当且仅当1=x 时取等号).…………………………………4分 即当0x >时,()0f x '≥(当且仅当1=x 时取等号).所以函数()f x 在),0(+∞单调递增,至多有一个零点. ………………………………………………5分 因为(1)0f =,1=x 是函数)(x f 唯一的零点. 所以若21=a ,则函数()f x 的所有零点只有1=x .…………………………………………………6分 (2)证法1:因为2()(2)ln 47f x x x ax x a =++-+,函数)(x f 的定义域为),0(+∞,且2()ln 24x f x x ax x+'=++-.…………………………………7分 当12a ≥时,()2ln 3f x x x x'≥++-,………………………………………………………………9分 由(1)知032ln ≥-++x xx .………………………………………………………………………10分 即当0x >时()0f x '≥,所以()f x 在()0,+∞上单调递增.……………………………………………………………………11分 所以)(x f 不存在极值.…………………………………………………………………………………12分 证法2:因为2()(2)ln 47f x x x ax x a =++-+, 函数)(x f 的定义域为),0(+∞,且2()ln 24x f x x ax x+'=++-.…………………………………7分 设2()ln 24x m x x ax x+=++-, 则2221222()2ax x m x a x x x +-'=-+=()0x >.设)0( 22)(2>-+=x x ax x h ,则()m x '与)(x h 同号.当21≥a 时,由2()220h x ax x =+-=,解得10x =<,20x =>.……………………………………………8分可知当20x x <<时,()0h x <,即()0m x '<,当2 x x >时,()0h x >,即()0m x '>,所以()f x '在()20,x 上单调递减,在()2,x +∞上单调递增.…………………………………………9分 由(1)知032ln ≥-++x xx .………………………………………………………………………10分 则2222222()ln 3(21)(21)0f x x x a x a x x '=++-+-≥-≥. 所以2()()0f x f x ''≥≥,即()f x 在定义域上单调递增.…………………………………………11分 所以)(x f 不存在极值.…………………………………………………………………………………12分22.(1)解法1:因为直线l 的参数方程为⎩⎨⎧+=+=ααsin 3,cos 2t y t x (t 为参数),当=2απ时,直线l 的直角坐标方程为2x =.…………………………………………………………1分 当2απ≠时,直线l的直角坐标方程为()tan 2y x α=-.……………………………………3分 因为222,cos x y x ρρθ=+=,…………………………………………………………………………4分因为8cos 22+=θρρ,所以2228x y x +=+.所以C 的直角坐标方程为08222=--+x y x .………………………………………………………5分解法2:因为直线l 的参数方程为⎩⎨⎧+=+=ααsin 3,cos 2t y t x (t 为参数),则有sin 2sin sin cos ,cos sin cos ,x t y t αααααααα=+⎧⎪⎨=+⎪⎩……………………………………………………………2分所以直线l的直角坐标方程为()sin cos 2sin 0x y αααα--= .………………………3分因为222,cos x y x ρρθ=+=,…………………………………………………………………………4分因为8cos 22+=θρρ,所以2228x y x +=+.所以C 的直角坐标方程为08222=--+x y x .………………………………………………………5分 (2)解法1:曲线C 的直角坐标方程为08222=--+x y x ,将直线l 的参数方程代入曲线C 的方程整理,得05)cos 2sin 32(2=-++t t αα.……………6分 因为020)cos 2sin 32(2>++=∆αα,可设该方程的两个根为1t ,2t ,则()122cos t t αα+=-+ ,125t t =-.……………………………………………………7分所以12AB t t =-===8分整理得)2cos 3αα+=,故2sin 6απ⎛⎫+= ⎪⎝⎭9分 因为0α≤<π,所以63αππ+=或263αππ+=, 解得6απ=或2απ= 综上所述,直线l 的倾斜角为6π或2π.…………………………………………………………………10分解法2:直线l 与圆C 交于A ,B 两点,且AB =,故圆心)0,1(C 到直线l 的距离1)22(92=-=d .…………………………………………………6分 ①当2απ=时,直线l 的直角坐标方程为2=x ,符合题意.…………………………………………7分 ②当0,,22αππ⎡⎫⎛⎫∈π⎪ ⎪⎢⎣⎭⎝⎭时,直线l 的方程为0tan 23tan =-+-ααy x .所以1tan 1|tan 230tan |2=+-+-=αααd ,………………………………………………………………8分tan α-=解得6απ=.………………………………………………………………………………………………9分 综上所述,直线l 的倾斜角为6π或2π.…………………………………………………………………10分23.(1)解:当1a =时,由()f x x >,得2111x x -->+.…………………………………………1分当12x ≥时,2111x x -->+, 解得3x >. 当12x <时,1211x x -->+,解得13x <-.…………………………………………………………4分 综上可知,不等式()1f x x >+的解集为 133x x x ⎧⎫><-⎨⎬⎩⎭或.……………………………………5分(2)解法1:由1()(1)2f x f x <+,得1212122a x a x --<+-. 则22121a x x >--+.…………………………………………………………………………………6分 令()22121g x x x =--+, 则问题等价于min (())a g x >因为123,,211()61,,22123,,2x x g x x x x x ⎧-+<-⎪⎪⎪=-+-≤≤⎨⎪⎪->⎪⎩……………………………………………………………………9分min 1()22g x g ⎛⎫==- ⎪⎝⎭.所以实数a 的取值范围为(2,)-+∞.…………………………………………………………………10分 解法2:因为2121(21)(21)x x x x --+≤--+,………………………………………………6分即221212x x -≤--+≤,则21212x x --+≥-.……………………………………………7分 所以()2121212212g x x x x x =--++-≥-+-≥-,…………………………………………8分 当且仅当12x =时等号成立.……………………………………………………………………………9分 所以min ()2g x =-.所以实数a 的取值范围为(2,)-+∞.…………………………………………………………………10分。
2019-2020学年广州大学附属中学高三英语月考试卷及参考答案
2019-2020学年广州大学附属中学高三英语月考试卷及参考答案第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项AOn the 100th anniversary of the Communist Party of China (CPC), red tourism has gained popularity among tourists who flood in to visit historic sites with a modern revolutionary heritage.JinggangshanThis is one of the most crucial and splendid chapters of history of establishing Red China as well as a unique and wonderful ecosystem, which is covered with rich forest, rugged peaks and several memorials to the Red Army. The best time to visit is between April and October, with the most temperature timing April and May when the large azaleas (杜鹃花) bloom.Open: 8:00-17:00 (Feb. 16-Nov. 15). 8:00-16:30 (Nov. 16-Feb. 15)XibaipoIt is an old revolutionary base where the leadership of the Communist Party of China was stationed, drawing up the blueprint for a new country. A memorial hall was built to honor the memory of this site. The lake and the hill here add brilliance and beauty to each other and form pleasant scenery.Open: Tuesdays to Sundays 9:30-17:00 (Xibaipo Memorial Hall)The Nanhu Revolutionary Memorial HallA new exhibition is held with updated display approaches, including phantom imaging (全息影像) and oil painting, which are used to improve visitors' experiences. The exhibition shows four stages of the CPC from its establishment to its achievements.Open: Tuesdays to Sundays 8:30-18:00 (closed on Mondays)Former Site of the Editorial Department ofNew YouthNew Youthstarted the New Culture Movement and spread the influence of the May Fourth Movement. The site was briefly based in Beijing but moved back to Shanghai in 1920 and also served as the office for the Communist Party of China Central Committee in the 1920s.Open: Thursdays to Tuesdays 9:00 - 11:30, 13:30 - 16:30 (closed on Wednesdays)1. Where would visitors learn more about the history of the Red Army?A. Jinggangshan.B. Xibaipo.C. The Nanhu Revolutionary Memorial Hall.D. Former Site of the Editorial Department ofNew Youth.2. What do we know about the Nanhu Revolutionary Memorial Hall?A. It focuses on Chinese achievements in art.B. It mainly advertises the coming anniversary.C. It applies modernized methods to the exhibition.D. It briefly introduces the rise and fall of Nanhu.3. When can tourists visit Former Site of the Editorial Department ofNew Youth?A. At 1:00 p.m. on Mondays.B. At 9:00 a.m. on Wednesdays.C. At 2:00 p.m. on Fridays.D. At 5:00 p.m on Sundays.BAccording to statistics published by the BPI (Buying Power Index) a couple of months ago, digital streaming (流媒体) now accounts for 80 percent of the music consumption in the UK. Despite the incredible growth of online streaming platforms like iTunes, Apple Music and Tidal over the past 15 years, a more traditional medium has also seen a return of interest and sales in the music industry. In 2020, almost one in five of all albums purchased in the UK is vinyl (黑胶唱片), and it has once again become the most popular physical musical medium.With digital streaming so easy and convenient, why are so many peopledrawn to traditional records? Some experts claim that vinyl is a physical medium for experiencing music, something tangible (有形的) to hold and own. For most people, having something tangible and interacting with it gives depth to the experience of music. Listening to an album and touching it the way the artist intended can make them feel more connected to the music and the artist. Records are physical products that can be not only displayed but also gifted, shared, traded and passed down through generations.Sound quality is another hot topic. A lot of music lovers feel that the analogue sound (模拟声音) vinyl offers is superior to modern digital audio, particularly with regards to the compressed formats streaming platforms use. There’s a common belief that old-school analogue audio has a warmer, fuller sound than digitised music. For vinyl followers, the very defect traditional recorders often have, such as the familiar crackle (劈啪作响) when the record starts, bring the music to life in a different way.There’s aritualisticaspect to vinyl that a lot of people are drawn to, too. The act of putting a record on—carefully removing the record from the sleeve, placing it on the record player and gently dropping the needleon the right groove (凹槽)—is a more assiduous (一丝不苟的), mindful way of engaging with music. When you’re listening to vinyl, you can’t tap abutton and go about your day while the streaming service provides hours of music. You need to stay close to the record player to move the needle and flip the record over.It’s clear that the vinyl interest is well underway, and vinyl records are truly making a comeback. In an increasingly digital society, there’s something to be said for analogue experiences. Perhaps one of the great things about being alive in the 21st century is our ability to have the best of both worlds—the timeless appeal of physical records alongside the easy access to vast music libraries that streaming offers.4. What are the statistics published by the BPI used to show?A. An increase in music consumption.B. The recovery of music industry.C. A comeback of a physical medium.D. The acceptance of online streaming.5. According to some experts, why does vinyl interest many people?A. It attracts people by its realistic feel.B. It offers simple access to different music.C. It shares a new way to enjoy music.D. It provides people with perfect sound effect.6. The underlined word “ritualistic” in Para.4 means something ______.A. Overlooked by society.B. Updated very frequently.C. Performed as part of a ceremony.D. Kept for a long time without changing.7. How does the writer feel about the future development of music medium?A. Traditional records will get underway.B. The analogue experiences may matter more.C. Vinyl sales will boom with technological advance.D. There should be a good mix of old and new.CIt’s 13:30 and 28-year-old Marten Pella 's smart phone starts pinging, a signal that it’s time for us to stop working around his living room table and instead start our workout routine together. A cartoon character wearingbright red shorts on video begins instructing us to do star-jumps and sit-ups around his apartment.Pella, a research assistant at Stockholm University, is part of the Hoffice movement, which invites workers-freelancers(自由职业者)or full-time employees who can do their jobs remotely—to work at each other’s homes to increase productivity and enjoy an active social life.Those attending Hoffice events advertised on Facebook are typically asked to work silently in 45-minute blocks, before taking short breaks together to exercise, or simply chatting over a coffee. In addition, each participant shares daily objectives with the rest of the group upon arrival, and is invited to report back on whether or not they have achieved them at the end of the day.“Often when I am alone, I can work focused for a couple of hours but then I’m easily distracted(分心).The help of others makes me so much more disciplined.” says Pella, who attends Hoffice events as both a guest and a host. Lunches mean networking and connecting with new contacts. “People are coming from really different areas and have different professions so there can be really interesting discussions,” he says.The Hoffice movement has grown quickly since it was founded in 2014 by Swedish psychologist Christofer Franzen, now 37. He had been giving lectures on the benefits of collective(集体的)intelligence, but realised he was spending most of his own time working alone at his kitchen table. “I wanted to test more structured home co-working with friends in similar situations,” he says.Franzen says that holding events in houses and apartments creates a unique atmosphere, because there’s a sense of community and desire to contribute. He’s looking for ways to expand the social value of Hoffice, by matching up members with relevant skills to share and even encouraging jobseekers to join its gatherings.8. Where is Pella when his smart phone starts pinging?A. In his own home.B. In his office.C. In another person’s home.D. At Stockholm University.9. What do people attending Hoffice events do first when they meet?A. Watch an exercise video.B. Work silently for 45 minutes.C. Tell each other their daily plans.D. Report what they have achieved.10. What does Franzen really mean by saying “friends in similar situations”?A. They usually work alone.B. They often give lectures.C. They study collective intelligence.D. They have to work at a kitchen table.11. What is the best title for the text?A. Sharing Comfortable WorkplacesB. A New Way to Make New FriendsC. Benefits of Collective IntelligenceD. Working from Others’ HomesDWhen I was trying to find a place where to spend my December holidays, I met by chance some cheap flights to Iceland. After checking just a few winter pictures of Iceland, I realized that the country, known as the land of fire and ice, during the cold months of the year could offer me experiences I had never had before.For sure you can’t miss the chance to go to Iceland in winter if your traveling wish list includes at least one of the crazy experiences Iceland can offer. Iceland in the North Atlantic Ocean is a paradise (乐园) for all those who want to see the northern lights, experience cold weather conditions and put themselves in geothermal (地热的) baths while the snow is falling on their head.The best way to move around Iceland is with a rental car. Distances are huge and public transport in winter is not really common out of the major towns. As we wanted to be even more convenient we decided to rent a small camper (野营车). Sleeping and cooking in a camper saved us a lot of driving, money and gave us the chance tobe always in the right place at the right time.There were also no locals and in many cases no tourist facilities (设备). For us, as we slept in a camper, it was easier. But for tourists traveling by normal cars it is necessary to check the opening times ofhotels and restaurants as many of them run just from June to September.It is amazing to experience how the weather is changing in Iceland. However, Icelanders prefer to stay inside their houses. They have even no time to complain about the weather in December. All they care about is Christmas. They love to decorate their houses, sing Christmas songs and eat typical Christmas food.12. Why is Iceland famous as the land of fire and ice?A. Because tourists would like to play with fire on the ice.B. Because it is too dry to easily cause fire to happen.C. Because it is hot inside a house and cold outside.D. Because there exist hot springs and freezing ice.13. What did the author think of the rented camper?A. It was not only practical but also economical.B. It was convenient but cost them more money.C. It provided the best chance to see the new country.D. It was much faster than other public transport.14. What does the last paragraph imply?A. The Icelanders prefer to live with their family.B. The joy of Christmas drives the freezing weather away.C. December is the coldest month of the year.D. The Icelanders are always positive and stay outside.15. What does this passage most probably come from?A. A textbookB. A scientific reportC. A travel magazineD. A news report第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
2019-2020学年广州市大学附属中学高三语文上学期期末考试试题及答案
2019-2020学年广州市大学附属中学高三语文上学期期末考试试题及答案一、现代文阅读(36分)(一)现代文阅读I(9分)阅读下面的文章,完成各题。
幸会,妈妈张春我妈年轻的时候是一名会计,在食品站工作。
那个年代的屠夫看不起坐办公室的,男人看不起女人,双重歧视。
我妈妈不服,就学会了杀猪。
一个二十来岁的女孩,穿着黑色的皮围裙,按倒一头猪,手起刀落,干脆利落。
后来我妈走到哪儿,那帮屠夫叔叔们就跟到哪儿。
她的本职业务也顶呱呱,现在已经60多岁,对数字依然非常敏感,心算精确到个位数。
我们小时候爱吃手指,把手指甲都啃坏了。
她就给我和哥哥在胸前吊了一粒甘草片。
因为甘草片比手指头好吃,我们就不吃手指头了。
我4岁的时候,看到其他小孩子在高楼外的屋檐上追跑嬉闹,也想跟上去。
妈妈没有打我骂我,而是去买了一个大西瓜,带我们站到那个楼顶上,瞅着下面没人,把西瓜扔了下去,然后说:“你们看,摔下去就是这个样子!”还有一次,在家里看电视剧《哪吒闹海》,看到哪吒自杀的时候,我一边伤心地大哭,一边去上学。
走出好远,后边远远传来妈妈的声音,她边喊边跑:“哪吒没有死——被他师父救活了——不要哭了!”她起码追了200米。
我不到8岁的时候,妈妈就和我说,不要让男人和你太亲密,更不要让男人碰你。
洗澡上厕所,就算是爸爸、哥哥也不能看。
上小学四年级时,一次我和另外两个小女孩走在路上,一个20多岁的男人来和我们说话,然后挨个儿抱我们,说要掂掂有多重。
我看到他抱起一个女孩,撩起了她的衣服,突然觉得不对,大喊一声:快跑!——很难想象,如果没有妈妈的早期教导,当时会发生什么事。
我初中的时候第一次收到情书,非常忧心,试探地拿给妈妈看。
妈妈仔细地看完,然后笑眯眯地叠起来还给我,说:“青春真好,还有人给你写这样的信。
”我后来听说很多女孩子不再对妈妈说心事,就是从第一封类似的书信开始的,而我却松了一口气,好像今后没有什么事不能和妈妈说的了。
但我们之间也不都是美好时光。
广东省广州大学附属中学2019-2020学年高三下学期第三次线上测试数学(文)试题
2020届广大附中高三文科数学试题(0419)一、选择题(本大题共12小题,每小题5分,共60分.)1.若22i 1iz =+-,则z z ⋅=( ) A. 2C. 10【答案】C 【解析】 【分析】先将复数22i 1iz =+-化简,然后与其共轭复数相乘可得结果. 【详解】因为22i=1+i+2i=1+3i 1iz =+-,所以z z ⋅=221310+=, 故选:C .【点睛】此题考查复数及其共轭复数的运算,属于基础题. 2.已知集合{3,2,1,0,1,2,3}A =---,1{1}2B x x =∈≤--Z ,则A B =( )A .{1}B. {1,1}-C. {1,2}D. {3,2,1,0,1}---【答案】A 【解析】 【分析】先求出集合B 中的元素,然后求两集合,A B 的公共部分 【详解】因为1{|1}{|120}{|12}{1}2B x x x x x x =∈≤-=∈-≤-<=∈≤<=-Z Z Z ,所以{}1A B ⋂=,故选:A .【点睛】此题考查集合的交集运算,属于基础题3.已知角α的终边经过点()1,2P -,则()cos πα-=( )B.D. 【答案】B 【解析】 【分析】根据角α的终边经过点()1,2P -,利用三角函数的定义求得()25cos 12α==+-,再利用诱导公式求()cos πα-.【详解】因为角α的终边经过点()1,2P -, 所以()25cos 12α==+-, 所以()5cos cos παα-=-=-. 故选:B【点睛】本题主要考查三角函数的定义及诱导公式,还考查了运算求解的能力,属于基础题. 4.执行如图所示的程序框图,则当输入的x 分别为3和6时,输出的值的和为( )A. 45B. 35C. 147D. 75【答案】D 【解析】 【分析】根据循环终止条件,分别求得输入3和6的结果,再求和. 【详解】当输入的x 为3时,27544y =-=. 当输入的x 为6时,26531y =-=. 所以输出的值的和为75. 故选:D【点睛】本题主要考查程序框图中的循环结构,还考查了逻辑推理的能力,属于基础题.5.据国家统计局发布的数据,2019年11月全国CPI(居民消费价格指数),同比上涨4.5%,CPI上涨的主要因素是猪肉价格的上涨,猪肉加上其他畜肉影响CPI上涨3.27个百分点.下图是2019年11月CPI一篮子商品权重,根据该图,下列四个结论正确的有______.①CPI一篮子商品中权重最大的是居住②CPI一篮子商品中吃穿住所占权重超过50%③猪肉在CPI一篮子商品中权重 2.5%④猪肉与其他禽肉在CPI一篮子商品中权重约为0.18%【答案】①②③【解析】【分析】结合两个图,对四个结论逐个分析可得出答案.【详解】对于①,CPI一篮子商品中居住占23%,所占权重最大,故①正确;++=,权重超过50%,故②正确;对于②,CPI一篮子商品中吃穿住所占19.9%8%23%50.9%对于③,由第二个图可知,猪肉在CPI一篮子商品中权重为2.5%,故③正确;+=,故④错对于④,由第二个图可知,猪肉与其他禽肉在CPI一篮子商品中权重约为2.5% 2.1% 4.6%误.故答案为:①②③.【点睛】本题考查统计图的识别和应用,考查学生的分析问题、解决问题的能力,属于基础题.6.刘徽是我国魏晋时期伟大的数学家,他在《九章算术》中对勾股定理的证明如图所示.“勾自乘为朱方,股自乘为青方,令出入相补,各从其类,因就其余不移动也.合成弦方之幂,开方除之,即弦也”.已知图中网格纸上小正方形的边长为1,其中“正方形ABCD为朱方,正方形BEFG为青方”,则在五边形AGFID 内随机取一个点,此点取自朱方的概率为()A.1637B.949C.937D.311【答案】C 【解析】 【分析】首先明确这是一个几何概型面积类型,然后求得总事件的面积和所研究事件的面积,代入概率公式求解. 【详解】因为正方形ABCD 为朱方,其面积为9,五边形AGFID 的面积为37ABCD BGFE DCI IEF S S S S ∆∆+++=, 所以此点取自朱方的概率为937. 故选:C【点睛】本题主要考查了几何概型的概率求法,还考查了数形结合的思想和运算求解的能力,属于基础题. 7.已知圆224210x yx y +-++=关于双曲线()2222:10,0x y C a b a b-=>>的一条渐近线对称,则双曲线C 的离心率为( ) 5 B. 55 D.54【答案】C 【解析】 【分析】将圆224210x y x y +-++=,化为标准方程为,求得圆心为()21-,.根据圆224210x y x y +-++=关于双曲线()2222:10,0x y C a b a b-=>>的一条渐近线对称,则圆心在渐近线上,12b a =.再根据21c b e a a ⎛⎫==+ ⎪⎝⎭求解.【详解】已知圆224210x y x y +-++=, 所以其标准方程为:()()22214x y -++=,所以圆心为()21-,. 因为双曲线()2222:10,0x y C a b a b-=>>, 所以其渐近线方程为by x a=±, 又因为圆224210x yx y +-++=关于双曲线()2222:10,0x y C a b a b-=>>的一条渐近线对称, 则圆心在渐近线上, 所以12b a =.所以c e a ===故选:C【点睛】本题主要考查圆的方程及对称性,还有双曲线的几何性质 ,还考查了运算求解的能力,属于中档题.8.在ABC 中,AD AB ⊥,3,BC BD =||1AD =,则AC AD ⋅的值为( ) A. 1 B. 2C. 3D. 4【答案】C 【解析】 【分析】由题意转化(3)AC AD AB BD AD ⋅=+⋅,利用数量积的分配律即得解. 【详解】AD AB ⊥,3,BC BD =||1AD =,()(3)AC AD AB BC AD AB BD AD ∴⋅=+⋅=+⋅ 2333AB AD BD AD AD =⋅+⋅==故选:C【点睛】本题考查了平面向量基本定理和向量数量积综合,考查了学生综合分析,转化划归,数学运算能力,属于中档题.9.函数()()()22lg 101x f x x x =+-+在[]22-,上的图象大致为( )A. B. C. D.【答案】A 【解析】 【分析】根据函数的特点,结合选项的图象特征,利用特殊值进行验证排除确定.【详解】因为()()0lg 101lg 20f =+=>,排除B ,D.又因为()()4461012lg 1016lg 010f ⎛⎫+=+-=< ⎪⎝⎭,排除C. 故选:A【点睛】本题主要考查函数的图象,还考查了理解辨析,特殊法应用的能力,属于中档题. 10.在ABC ∆中,22223sin a b c ab C ++=,则ABC ∆的形状是 ( ) A. 锐角三角形 B. 直角三角形C. 钝角三角形D. 等边三角形【答案】D 【解析】 【分析】由余弦定理可知2222cos a b c ab C +-=,与已知条件相加,得到cos 3C π⎛⎫-⎪⎝⎭的表达式,利用基本不等式得到范围,结合其本身范围,得到cos 13C π⎛⎫-= ⎪⎝⎭,从而得到C 的大小,判断出ABC ∆的形状,得到答案. 【详解】由余弦定理可知2222cos a b c ab C +-=,22223sin a b c ab C ++=两式相加,得到()22cos 32cos 3a b ab C C ab C π⎛⎫+=+=-⎪⎝⎭所以222cos 1322a b ab C ab ab π+⎛⎫-== ⎪⎝⎭≥,当且仅当a b =时,等号成立, 而[]cos 1,13C π⎛⎫-∈- ⎪⎝⎭所以cos 13C π⎛⎫-= ⎪⎝⎭, 因为()0,C π∈,所以2,333C πππ⎛⎫-∈- ⎪⎝⎭所以03C π-=,即3C π=,又a b =,所以ABC ∆是等边三角形, 故选D 项.【点睛】本题考查余弦定理解三角形,基本不等式,余弦型函数的性质,判断三角形的形状,属于中档题. 11.已知正方体1111ABCD A BC D -的棱长为2,点P 在线段1CB 上,且12B P PC =,平面α经过点1,,A P C ,则正方体1111ABCD A BC D -被平面α截得的截面面积为( )A. 36B. 26C. 5D.534【答案】B 【解析】 【分析】先根据平面的基本性质确定平面,然后利用面面平行的性质定理,得到截面的形状再求解. 【详解】如图所示:1,,A P C 确定一个平面α,因为平面11//AA DD 平面11BB CC , 所以1//AQ EC ,同理1//AE QC , 所以四边形1AEC Q 是平行四边形. 即正方体被平面截的截面. 因为12B P PC =, 所以112C B CE =, 即1EC EB ==所以11AE EC AC ===由余弦定理得:22211111cos 25AE EC AC AEC AE EC +-∠==⨯所以1sin AEC ∠= 所以S四边形1AEQC 1112sin 2AE EC AEC =⨯⨯⨯∠=故选:B【点睛】本题主要考查平面的基本性质,面面平行的性质定理及截面面积的求法,还考查了空间想象和运算求解的能力,属于中档题.12.定义:()(){}N f x g x ⊗表示()()f x g x <的解集中整数的个数.若()()()22log ,11f x x g x a x ==++,且()(){}1N f x g x ⊗=,则实数a 的取值范围是( )A. 1,04⎡⎤-⎢⎥⎣⎦B. 1,04⎛⎤- ⎥⎝⎦C. (],0-∞D. 11,4⎡⎫--⎪⎢⎣⎭【答案】B 【解析】 【分析】根据函数图象,结合()(){}1N f x g x ⊗=,则有(1)410(2)911g a g a =+>⎧⎨=+≤⎩求解. 【详解】因为()(){}1N f x g x ⊗=则有(1)410(2)911g a g a =+>⎧⎨=+≤⎩解得:104a -<≤ 故选:B【点睛】本题主要考查函数与不等式问题,还考查了数形结合的思想和运算求解的能力,属于中档题.二、填空题(本大题共4小题,每小题5分,共20分)13.已知a =(4,﹣1),b =(2,t 2﹣1),若a b ⋅=5,则t =_________. 【答案】2± 【解析】 【分析】结合已知,直接利用向量数量积的坐标表示代入即可求解t .【详解】∵a =(4,﹣1),b =(2,t 2﹣1),∴a •b =4×2﹣(t 2﹣1)=5,t 2=4,则t =±2. 故答案为:±2.【点睛】本题主要考查了向量数量积的坐标表示的简单应用是,属于基础试题.14.函数()f x 是定义在R 上的奇函数,且满足(1)(1)f x f x +=-+.当01x <≤时,2020()log f x x =-,则1()(2019)(2020)2020f f f ++=__________.【解析】 【分析】由函数()f x 是定义在R 上的奇函数,可得(0)0f =,再结合(1)(1)f x f x +=-+可得()f x 的周期为4,然后利用函数的性质将自变量化简到(0,1]上进行求解【详解】因为()f x 是定义在R 上的奇函数,所以()()f x f x -=-,且(0)0f =. 又因为(1)(1)f x f x +=-+,所以(2)()f x f x +=-,所以(2)()f x f x +=-, 可得(4)()f x f x +=,所以奇函数()f x 的周期为4, 所以202011()(2019)(2020)log (1)(0)20202020f f f f f ++=-+-+ 20201(1)(0)1log 101f f =-+=++=.故答案为:1.【点睛】此题考查函数的奇偶性、周期性,考查运算能力,属于中档题15.在三棱锥A BCD -中,,2,AB AD AB AD BC CD ⊥====当三棱锥A BCD -的体积最大时,三棱锥A BCD -外接球的体积与三棱锥A BCD -的体积之比为__________.【答案】8π【解析】 【分析】根据题意,当面BCD ⊥面ABD 时,三棱锥A BCD -的体积最大.此时取BD 的中点O ,由,2,AB AD AB AD ⊥==,得4BD =,OA=2,同理根据BC CD ==,且222BC CD BD +=,由直角三角形中线定理可得2OC =,从而得到外接圆半径R =2,再分别利用体积公式求解. 【详解】如图所示:当面BCD ⊥面ABD 时,三棱锥A BCD -的体积最大. 取BD 的中点O ,因为,2,23AB AD AB AD ⊥==, 所以4BD =,OA=2, 22BC CD ==,222BC CD BD +=, 2OC =,外接圆半径R =2, V 球343233R ππ==,11432232323A BCD V -=⨯⨯⨯=, 三棱锥A BCD -外接球的体积与三棱锥A BCD -的体积之比为83π故答案为:83π【点睛】本题主要考查组合体的体积问题,还考查了逻辑推理和运算求解的能力,属于中档题.16.已知双曲线C :()222210,0x y a b a b-=>>的左右焦点分别为1F ,2F ,过1F 的直线l 与圆222x y a +=相切于点T ,且直线l 与双曲线C 的右支交于点P ,若114F P FT =,则双曲线C 的离心率为______. 【答案】53【解析】 【分析】根据题意,作出图形,结合双曲线第一定义,再将所有边长关系转化到直角三角形2MPF 中,化简求值即可【详解】如图,由题可知12OF OF c ==,OT a =,则1FT b =, 又114F P FT =,3TP b ∴=,14F P b ∴=, 又122PF PF a -=,242PF b a ∴=-作2//F M OT ,可得22F M a =,TM b =,则2PM b = 在2MPF ∆,22222PM MF PF +=,即()222c b a =-,2b a c =+又222c a b =+,化简可得223250c ac a --=,同除以2a ,得23250e e --=解得53e =双曲线的离心率为53【点睛】本题考查了利用双曲线的基本性质求解离心率的问题,利用双曲线的第一定义和中位线定理将所有边长关系转化到直角三角形2MPF 中是解题关键,一般遇到此类题型,还是建议结合图形来进行求解,更直观更具体三、解答题(本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤)17.2019年12月16日,公安部联合阿里巴巴推出的“钱盾反诈机器人”正式上线,当普通民众接到电信网络诈骗电话,公安部钱盾反诈预警系统预警到这一信息后,钱盾反诈机器人即自动拨打潜在受害人的电话予以提醒,来电信息显示为“公安反诈专号”.某法制自媒体通过自媒体调查民众对这一信息的了解程度,从5000多参与调查者中随机抽取200个样本进行统计,得到如下数据:男性不了解这一信息的有50人,了解这一信息的有80人,女性了解这一信息的有40人.(1)完成下列22⨯列联表,问:能否在犯错误的概率不超过0.01的前提下,认为200个参与调查者是否了解这一信息与性别有关?了解 不了合计(2)该自媒体对200个样本中了解这一信息的调查者按照性别分组,用分层抽样的方法抽取6人,再从这6人中随机抽取3人给予一等奖,另外3人给予二等奖,求一等奖与二等奖获得者都有女性的概率.附:22(),()()()()n ad bcK n a b c da b a c c d b d-==+++ ++++【答案】(1)能在犯错误的概率不超过0.01的前提下,认为200个参与调查者是否了解这一信息与性别有关.(2)3 5【解析】【分析】(1)男性不了解这一信息的有50人,了解这一信息的有80人,女性了解这一信息的有40人,补全列联表.再根据22⨯列联表,代入求临界值的公式,求观测值,利用观测值临界表进行比较.(2)根据了解这一信息的男女比例,确定抽取6人中,男女的人数,然后列举从6人中任取3人的基本事件的总数,再从中找出含有一名女性的基本事件的个数,再代入古典概型概率公式求解.【详解】(1)由随机抽取200个样本进行统计,男性不了解这一信息的有50人,了解这一信息的有80人,女性了解这一信息的有40人.得22⨯列联表如下,222()200(30804050)0.3663 6.635()()()()1307080120n ad bc K a b a c c d b d -⨯-⨯===<++++⨯⨯⨯所以能在犯错误的概率不超过0.01的前提下,认为200个参与调查者是否了解这一信息与性别有关. (2)从了解这一信息的调查者按照性别分组,用分层抽样的方法抽取6人中,男性有8064120⨯=人,女性有2人,设男生编号为1,2,3,4,女性编号分别为5,6,则“从这6人中任选3人”的基本事件有; (1,2,3),(1,2,4),(1,2,5),(1,2,6),(1,3,4),(1,3,5),(1,3,6),(1,4,5),(1,4,6) ,(1,5,6),(2,3,4),(2,3,5) (2,3,6),(2,4,5),(2,4,6), (2,5,6),(3,4,5) (3,4,6) ,(3,5,6),(4,5,6)共20个 其中事件A “一等奖与二等奖获得者都有女性”的基本事件有(1,2,5),(1,2,6),(1,3,5),(1,3,6),(1,4,5),(1,4,6),(2,3,5) (2,3,6),(2,4,5),(2,4,6),(3,4,5) (3,4,6)共12个 所以一等奖与二等奖获得者都有女性的概率为35【点睛】本题主要考查独性检验和古典概型概率的求法,还考查了数据处理和运算求解的能力,属于中档题.18.已知函数()f x 满足()1111f x x +=++,数列{}n a 满足()*1112,n n a a f n N a +⎛⎫==∈ ⎪⎝⎭. (1)求证:数列{}n a 是等差数列; (2)若112,2n n n n n b a T b b b -=⋅=+++,求n T .【答案】(1)证明见解析;(2)2n n T n =⨯ 【解析】 【分析】(1)根据函数()f x 满足()1111f x x +=++,得()11f x x=+.又因为数列{}n a 满足()*11n n a f n N a +⎛⎫=∈ ⎪⎝⎭,化简利用等差数列的定义证明.(2)由(1)知:()111n a a n d n =+-=+,则()112n n b n -=+⋅,符合等差数列与等比数列相应项之积构成的数列,用错位相减法求和.【详解】(1)因为函数()f x 满足()1111f x x +=++, 令1x t ,所以()11f t t=+, 即()11f x x=+. 因为数列{}n a 满足()*11n n a f n N a +⎛⎫=∈⎪⎝⎭, 所以11n n a a +-=,又12,a =所以{}n a 是以2为首项,以1为公差的等差数列. (2)由(1)知:()11n a a n d n =+-=+1. 所以()112n n b n -=+⋅.()0121223242...12n n T n -=⨯+⨯+⨯+++⨯, ()1232223242...12n n T n =⨯+⨯+⨯+++⨯,两式相减得:()1232222...12n n T n -=++++-+⨯,()1211212nn n -=+-+⨯-,2n n =-⨯,所以2n n T n =⨯.【点睛】本题主要考查等差数列的证明和错位相减法求和,还考查了运算求解的能力,属于中档题. 19.如图,在四棱锥P ABCD -中,PAD ∆是等边三角形,O 是AD 上一点,平面PAD ⊥平面,ABCD //,,1,2,3AB CD AB AD AB CD BC ⊥===.(1)若O是AD的中点,求证:OB⊥平面POC;(2)设=ODOAλ=,当λ取何值时,三棱锥B POC-3【答案】(1)证明见解析;(2)1λ=.【解析】【分析】(1)在平面ABCD中,由勾股定理证明OB OC⊥.在空间中,由平面PAD⊥平面,ABCD得到PO⊥平面,ABCD从而有PO⊥OB,再利用线面垂直的判定定理证明. (2)设OD OAλ=,所以0OD Aλ=,则有22OD OA OA OAλ+=+=PAD∆是等边三角形,平面PAD⊥平面,ABCD得到点P到平面ABCD的距离,即为四棱锥P ABCD-的高,且6h=体积法转化133B POC P BOC BOC V V S h--∆===()11132222BOC S AB CD AD AB OA AD OD∆=+⨯-⨯-⨯=()1232OAλ+=. 【详解】(1)因为//,,1,2,3AB CD AB AD AB CD BC⊥===,所以()2222AD BC CD AB=--=因为O是AD的中点,所以2OA OD==223,6OB OC==,所以222OB OC BC+=,所以OB OC⊥.又因为平面PAD ⊥平面,ABCD 所以PO ⊥平面,ABCD 所以PO ⊥,0OB PO OC ⋂=, 所以OB ⊥平面POC . (2)设ODOAλ=, 所以0OD A λ=,因为PAD ∆是等边三角形,平面PAD ⊥平面,ABCD点P 到平面ABCD 的距离,即为四棱锥P ABCD -的高,且h =因为13B POC P BOC POC V V S h --∆==所以()1112222BOC S AB CD AD AB OA AD OD ∆=+⨯-⨯-⨯=整理得:()12OA λ+=又因为OD OA OA OA λ+=+=解得1λ=【点睛】本题主要考查面面垂直的性质定理,线面垂直的判定定理以及几何体体积的求法,还考查了转化化归的思想和逻辑推理的能力,属于中档题.20.已知抛物线2:4C y x =与过点(2,0)的直线l 交于,M N 两点.(1)若MN =l 的方程; (2)若12MP MN =,PQ y ⊥轴,垂足为Q ,探究:以PQ 为直径的圆是否过定点?若是,求出该定点的坐标;若不是,请说明理由.【答案】(1)20x -=或20x -=;(2)过定点,(2,0) 【解析】 【分析】(1)设出直线l 的方程2()x my m =+∈R ,联立直线与抛物线方程,利用根与系数的关系及弦长公式计算即可;(2)设以PQ 为直径的圆经过点()00,A x y ,()20022,2AP m x m y =+--,()00,2AQ x m y =--,利用0AP AQ ⋅=得()2220000042420x m y m x y x --++-=,令00220004204020x y x y x -=⎧⎪=⎨⎪+-=⎩解方程组即可.【详解】(1)由题可知,直线l 的斜率不为0,设其方程为2()x my m =+∈R , 将2x my =+代入24y x =,消去x 可得2480y my --=,显然216320m ∆=+>,设()11,M x y ,()22,N x y ,则124y y m +=,128y y =-,所以12||MN y y =-==因为||MN ==m =,所以直线l 的方程为20x--=或20x -=. (2)因为12MP MN =,所以P 是线段MN 的中点, 设(),P P P x y ,则由(1)可得()2121242222P m y y x x x m +++===+,1222P y y y m +==,所以()222,2P m m +,又PQ y ⊥轴,垂足为Q ,所以(0,2)Q m ,设以PQ 为直径的圆经过点()00,A x y ,则()20022,2AP m x m y =+--,()00,2AQ x m y =--,所以0AP AQ ⋅=,即()()220002220x m x m y -+-+-=,化简可得()2220000042420x m y m x y x --++-=①,令00220004204020x y x y x -=⎧⎪=⎨⎪+-=⎩,可得0020x y =⎧⎨=⎩,所以当02x =,00y =时,对任意的m ∈R ,①式恒成立, 所以以PQ 为直径的圆过定点,该定点的坐标为(2,0).【点睛】本题考查直线与抛物线的位置关系,涉及到抛物线中的定点问题,考查学生的计算能力,是一道中档题.21.已知函数()()ln a xf x x a R x=+∈. (1)若函数()f x 的图象在2x e =处的切线与y x =平行,求实数a 的值;(2)设()()()201,221a g x xf x x a x <≤=-+-.求证:()g x 至多有一个零点.【答案】(1)0a =;(2)证明见解析. 【解析】 【分析】(1)由函数()()ln a x f x x a R x =+∈,求导()()21ln 1a x f x x-'=+,根据函数()f x 的图象在2x e =处的切线与y x =平行,则有()2411af e e'=-+=求解.(2)根据()()()()22221ln 21g x xf x x a x a x x a x =-+-=-+-,求导()()()()()222121221x a x a x x a ag x x a x x x ---+-'=-+-=-=-,易知当0,()0x a g x '<<>,当,()0x a g x '><,当x a =时,()()max ln 1g x a a a =+-,只要论证()max 0g x ≤即可.【详解】(1)已知函数()()ln a xf x x a R x=+∈, 所以()()21ln 1a x f x x-'=+, 所以()()()222421ln 11a e af e ee -'=+=-+, 因为函数()f x 的图象在2x e =处的切线与y x =平行,所以()2411af e e '=-+=, 解得0a =.(2)因为()()()()22221ln 21g x xf x x a x a x x a x =-+-=-+-,所以()()()()()222121221x a x a x x a ag x x a x x x---+-'=-+-=-=-, 当0,()0x a g x '<<>,当,()0x a g x '><, 所以当x a =时,()()max ln 1g x a a a =+-, 令ln 1t a a =+-,所以110t a'=+>, 所以t 在()01a ∈,上是增函数.所以0t ≤,即()0g x ≤. 所以()g x 至多有一个零点.【点睛】本题主要考查导数的几何意义以及导数在函数零点中的应用,还考查了转化化归的思想和运算求解的能力,属于难题.请考生在第22、23两题中任选一题作答.注意:只能做所选定的题目.如果多做,则按所做的第一个题目计分.22.在平面直角坐标系xOy 中,曲线C 的参数方程为1cos sin x y ϕϕ=+⎧⎨=⎩(ϕ为参数).在以坐标原点为极点,x轴的正半轴为极轴的极坐标系中,直线l 的极坐标方程为sin 36πρθ⎛⎫-= ⎪⎝⎭. (1)求曲线C 的普通方程及直线l 的直角坐标方程; (2)求曲线C 上的点到直线l 的距离的最大值与最小值.【答案】(1)()()22:110C x y y -+=≥,,:60l x +=(2)最大值52,最小值1 【解析】 【分析】(1)由曲线C 的参数方程1cos sin x y ϕϕ=+⎧⎨=⎩,得cos 1,sin x y ϕϕ=-=两式平方相加求解,根据直线l 的极坐标方程sin 36πρθ⎛⎫-= ⎪⎝⎭,展开有1sin cos 322ρθρθ-=,再根据sin ,cos y x ρθρθ==求解.(2)因为曲线C 是一个半圆,利用数形结合,圆心到直线的距离减半径即为最小值,最大值点由图可知.【详解】(1)因为曲线C 的参数方程为1cos sin x y ϕϕ=+⎧⎨=⎩所以cos 1,sin x y ϕϕ=-=两式平方相加得:()()22110x y y -+=≥,因为直线l 的极坐标方程为sin 36πρθ⎛⎫-= ⎪⎝⎭. 所以31sin cos 32ρθρθ-= 所以3132y x -= 即360x y -+=(2)如图所示:圆心C 到直线的距离为:1322d +'== 所以圆上的点到直线的最小值为:min 1d d r '=-=则点M (2,0)到直线的距离为最大值:max 23522d +== 【点睛】本题主要考查参数方程,普通方程及极坐标方程的转化和直线与圆的位置关系,还考查了数形结合的思想和运算求解的能力,属于中档题.23.已知()|||2|f x x x =+-.(1)求不等式|4|()x f x x>的解集; (2)若()f x 的最小值为M ,且22(,,)a b c M a b c ++=∈R ,求证:22249a b c ++≥. 【答案】(1)(,0)(3,)-∞⋃+∞;(2)证明见解析【解析】【分析】(1)分0x <、02x <≤和2x >三种情况,分别解不等式,进而可得出答案;(2)先求出()f x 的最小值,可求出的M 的值,再结合柯西不等式,可证明结论.【详解】(1)当0x <时,|4|()x f x x>等价于|||2|4x x +->-,该不等式恒成立; 当02x <≤时,()|||2|2f x x x =+-=,则|4|()x f x x >等价于24>,该不等式不成立; 当2x >时,()|||2|22f x x x x =+-=-,则|4|()x f x x >等价于2224x x >⎧⎨->⎩,解得3x >, 所以不等式|4|()x f x x>的解集为:(,0)(3,)-∞⋃+∞. (2)因为()|||2||(2)|2f x x x x x =+-≥--=,当02x ≤≤时取等号,所以2M =,222a b c ++=, 由柯西不等式可得22222222224(22)(122)()9()a b c a b c a b c =++≤++++=++, 当且仅当244,,999a b c ===时等号成立,所以22249a b c ++≥. 【点睛】本题考查绝对值不等式的解法,考查不等式的证明,考查分类讨论的数学思想的应用,考查学生的推理论证能力,属于基础题.。
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广州大学附属中学2019年12月2019年度高考模拟试题-高考模拟各位读友大家好,此文档由网络收集而来,欢迎您下载,谢谢第Ⅰ卷(选择题共42分)一、(18分,每小题3分)1.下列词语中加点的字的读音,全部不相同的一组是()A.花圃胸脯店铺精心哺育惊魂甫定B.恪守炮烙格斗络绎不绝洛阳纸贵c.荟萃市侩烩饭绘声绘色脍炙人口D.逡巡唆使梭镖怙恶不悛崇山峻岭2.下列各组词语中,有错别字的一组是()A.成功计日程功抱怨以德报怨B.不凡要言不烦颜色察颜观色c.清秀山清水秀锤炼千锤百炼D.勾通理解沟通广义集思广益3.依次填入下列各句横线处的词语,最恰当的一组是()①前年是毛泽东同志____一百周年,不少出版社出了不少纪念的书。
②全新的载体可以____优秀传统文化的因子,使之释放出夺目的光辉。
③一位七十岁的老大娘激动地说:“改革开放后,日子好了,也不用田里劳动了,在家带孙子,____着呢!”A.诞辰激活悠闲B.诞辰激发悠闲c.诞生激发自在D.诞生激活自在4.下列各句中加点的成语的使用,恰当的是()A.假如我们的作家肯肝胆相照地和人民打成一片,那么一定会写出更好的作品来。
B.今后的国家科研计划,将一改过去的单位为中心的老面孔,摇身一变为以课题为中心。
c.虽然王郅治上场时间不长,但他的弱点和长处都已暴露无遗。
D.他想:“这次调入新的单位,如果锋芒毕露,就会招致一些麻烦,我不能操之过急。
”5.下列句子,没有语病的一项是()A.“扬州八怪”并非都是扬州人,其实指的是形成于扬州的八位画家的某些共同特征。
B.虽然国家教委早有规定,不能把考试作为奖惩学校教师的唯一依据,但某些地区县市的政府、教育主管部门却将考试依然作为奖惩的硬性指标。
c.当客观世界的种种困难在知识面前迎刃而解的时候,知识的力量也就十分鲜明地显露出来。
D.由于部分国有企业产权不明晰,管理制度不健全,使企业缺乏创新的动力和相应的实力,以致难以适应市场竞争。
6.依次填入下面一段文字中横线处的语句,与上下文衔接最恰当的一组是()以苏东坡之才,治国安邦都会有独特的建树,他任杭州太守期间的政绩就是明证。
可是他太富于诗人气质了,禁不住________________,结果总是得罪人。
他的诗名________________,但他的五尺之躯却难以见容当权派。
________________________________,他都照例不受欢迎。
①不平则鸣,有感而发②有感而发,不平则鸣③冠绝一时,流芳百世④流芳百世,冠绝一时⑤无论同党秉政,还是政敌当道⑥无论政敌当道,还是同党秉政A.①④⑤B.②③⑥c.②③⑤D.①④⑥二、阅读下文,完成7-10题。
(12分,每小题3分)宣德青花瓷陈润民明代景德镇御窑厂烧造的宣德青花瓷器,以其古朴、典雅的造型,晶莹艳丽的釉色,多姿多彩的纹饰而闻名于世,被称为中国青花瓷器的典范。
宣德青花在烧造数量上是空前的,据《大明会典》记载,宣德八年(1433),朝廷一次就向景德镇下达了烧造龙凤瓷器四十四万三千五百件的任务,其中青花占大多数。
产品不仅供宫廷之需,而且也作为商品大量行销海外,以及对国外入贡者的答赠,成为东西文化交流的见证。
当时宣德皇帝和王皇后对艺术品具有浓厚的兴趣,很大程度上促进了手工艺术方面的发展。
除瓷器外,宣德年间的铜器、雕漆及织绣等,俱能成就辉煌。
宣德青花造型多种多样,富于变化。
常见的有盘、碗、瓶、罐、花盆、壶、洗、钵、盒、三足炉、八方烛台等。
受当时外来文化的影响,还有许多造型摹仿西亚地区的金银器和陶器,如花浇、鱼篓樽、盘座、折沿盆等。
整体风格是雄伟浑厚,庄重古朴。
大件器皿增多,均为分段粘接而成,胎体厚,制作非常规整,比例协调,没有变形现象。
小件器物精致细巧,厚薄适度,具有典雅秀美的艺术风采。
青花用料有进口、国产两种,往往是根据纹饰的不同而决定使用哪种钴料描绘哪个部位的纹饰。
从宫中传世品来看,以进口料描绘为主。
这种进口青料来自波斯,是郑和下西洋带回来的。
与国产料不同的是,其含铁量特别高,含锰量低,所以呈色深沉浓艳,有着自然晕散的艺术效果,料色溶融在釉中,出现银黑色结晶斑点,在一定光线下有锡光色,而且呈三角形结晶,用手抚摸釉面凹凸不平。
用这种青料描绘的纹饰具有中国画的水墨韵味,被视为无法模仿的特色。
宣德青花纹饰上具有突出的时代特征,改变了元代青花层次繁密的布局风格及粗犷的画法,装饰上渐趋疏朗,规矩中富于变化。
注重从自然界选取素材。
植物纹有:枇杷纹、栀子花、百合、灵芝、葡萄等。
最常见的是以茶花、菊花、牡丹、莲花组合描绘在一起。
动物纹有:龙、凤、鱼、麻雀、狮子等。
人物纹有:仕女、吹箫引凤、婴戏图等。
海水纹在宣德青花中大量出现,也最为擅长,多是用作辅助纹饰,采用写实手法描绘,海水起伏翻腾,汹涌澎湃,有气势。
另外,梵文、藏文也开始出现在青花瓷器上。
综观宣德青花纹饰,取材范围广泛,绘画讲究气势壮阔,具有很强的艺术感染力。
宣德青花款识曾有满器身的说法,根据器物造型的不同在口沿、耳、颈、碗心、足底、腹部均有书写。
主要是以足内中心的青花双圈六字楷书书款“大明宣德年制”居多。
模仿晋唐小楷的笔法,笔划精细适中,笔法遒劲有力。
已故古陶瓷鉴定专家孙瀛洲总结宣德款识,作有歌诀:“宣德年款遍器身,楷刻印篆暗阳阴,横竖花四双单圈,晋唐小楷最出群。
”当时写篆书款极少,双圈往往有深浅,字体清晰,六字有的大小不一,起落笔处呈尖状,“大”字多撇短捺长,“德”字“心”上无一横,此为宣德年款的鉴定要点。
(选自《文史知识》xxxx年第1期,有删改)7.下列各项不属于作者所认为的宣德青花得到极大发展的原因的一项是()A.当时的宣德皇帝和王皇后对艺术品具有浓厚的兴趣。
B.外来文化的影响,特别是西亚地区文化的影响。
c.从波斯进口的青料的使用。
D.宣德青瓷大量行销海外,成为东西文化交流的见证。
8.以下不属于作者认为宣德青花是“中国青花瓷器的典范”的一项是()A.宣德青花烧造的数量空前,朝廷曾一次就下达了烧造龙凤瓷器四十四万三千五百件的任务。
B.宣德青花造型多种多样,富于变化,具有雄伟深厚、庄重古朴、典雅秀美的艺术风采。
c.宫中传世有青花釉色,呈色深沉浓艳,有着自然晕散的艺术效果,描绘的纹饰具有中国画的水墨韵味。
D.宣德青花装饰疏朗,规矩中富于变化,取材广泛,绘画气势壮阔。
9.对原文最后一段内容理解不正确的一项是()A.宣德青花的款识,根据器物造型的不同,分别在口沿、耳、颈、碗心、足底、腹部等处。
B.宣德青花的款识,主要模仿晋唐小楷笔法,笔法遒劲有力。
c.宣德青花也有少量篆书款识,双圈往往有深浅,字体清晰。
D.鉴定宣德年款篆书款的要点是“大明宣德年制”六字小楷,其大小不一,起落笔处呈尖状,“大”字撇短捺长,“德”字“心”上无一横。
10.根据原文所给的信息,下列推断不正确的一项是()A.“中国”一词在英语中用“china”表示,可能就是因为宣德青花瓷器奠定了中国瓷器在世界的地位。
B.元代青花的特点是布局风格层次繁密,画法粗犷,装饰上不疏朗,规矩而少变化。
c.宣德青花如果不用进口青料,就不会成为我国瓷器的名品之一。
D.海水纹在宣德青花中大量出现,是明政府扩大其在海外的政治影响,与各国进行贸易往来的体现。
三、阅读下面一段文言文,完成11~15题及第Ⅱ卷16题。
(选择题12分,每小题3分)应侯谓昭王曰:“亦闻恒思有神丛①与?恒思(地名)有悍少年,请与神博,曰:‘吾胜丛,丛藉我神三日;不胜丛,丛困我。
’乃左手为丛投,右手自为投,胜丛。
丛藉其神三日,丛往求之,遂弗归。
五日而丛枯,七日而丛亡。
今国者,王之丛;势者,王之神。
藉人如此,得无危乎?臣未尝闻指大于臂,臂大于股,若有此,则病必甚矣。
百人舆瓢而趋,不如一人持而走疾。
百人诚舆瓢,瓢必裂。
今秦国,华阳用之,穣侯用之,太后用之,王亦用之。
不称瓢为器则已,称瓢为器,国必裂矣。
臣闻之也:‘木实繁者枝必披,枝之披者伤其心。
都大者危其国,臣强者危其主。
’且今邑中自斗食以上至尉、内史乃王左右,有非相国之人者乎?国无事则已,国有事,臣必见王独立于庭也。
臣窃为王恐,恐万世之后,有国者非王子孙也。
“臣闻古之善为政也,其威内扶,其辅外布,而治政不乱不逆,使者直道而行,不敢为非。
今太后使者分裂诸侯而符布天下操大国之势强征兵伐诸侯战胜攻取利尽归于陶(即穣侯);国之币帛,竭入太后之家;境内之利,分移华阳。
古之所谓‘危主灭国之道’必从此起。
三贵竭国以自安,然则令何得从王出,权何得毋分,是王果处三分之一也。
”(选自《战国策·秦策》)【注】①:神丛,叫“丛”的神,实际上就是托身于草木茂盛的地方的神灵。
11.下列加点词语的解释,不正确的一项是A.恒思有悍少年,请与神博博:打赌B.百人诚舆瓢诚:果真c.木实繁者枝必披,披:裂开D.古之所谓‘危主灭国之道’必从此起危:危险12.下列各组句子中,加点词语的意义和用法不相同的一组是A.乃左手为丛投B.使杞子、杨孙、逢孙戍之,乃还今国者,王之丛今者殊不欲食c.百人舆瓢而趋抱明月而长终D.三贵竭国以自安宁许以负秦曲13.下列有关应侯论述的秦国主要危机,阐述正确的一项是A.民风狡猾,尤其是青年,不遵守承诺,导致信用缺失。
B.一个工作往往许多人去做,导致人浮于事,效率低下。
c.权臣当政,任用亲信控制经济、外交,国君权力被削弱。
D.国君师心自用,有大事时不善于听取谋臣的意见。
14.下面是文中未标点的一段话,标点正确的一项是A.今,太后使者分裂,诸侯而符布天下,操大国之势强,征兵伐诸侯,战胜攻取利,尽归于陶B.今太后使者分裂诸侯,而符布天下,操大国之势,强征兵,伐诸侯,战胜攻取,利尽归于陶c.今太后使者,分裂诸侯而符,布天下,操大国,之势强,征兵伐诸侯,战,胜攻取利,尽归于陶D.今太后使者分裂诸侯,而符布天下操大国之,势强征兵,伐诸侯,战胜攻取利,尽归于陶第Ⅱ卷15.将文中划横线的句子翻译成现代汉语。
(8分)(1)丛藉其神三日,丛往求之,遂弗归。
(2分)译文:__________________________________ __________________________(2)且今邑中自斗食以上至尉、内史乃王左右,有非相国之人者乎?(3分)译文:____________________________________________________________(3)(3)臣闻古之善为政也,其威内扶,其辅外布,而治政不乱不逆,使者直道而行,不敢为非。
(3分)译文:__________________________________ __________________________16.阅读下面一首宋词,然后回答问题(6分)玉楼春欧阳修别后不知君远近,触目凄凉多少闷。
渐行渐远渐无书,水阔鱼沉何处问。