2016年广州市普通高中毕业班综合测试
广州市普通高中毕业班综合测试(广一模)含答案word版
2016年广州市普通高中毕业班综合测试(一)物理试卷第I 卷二、选择题(本题共8小题,每小题6分。
在每小题给出的四个选项中,第14-18题只有一项符合题意要求,第19-21题有多项符合题意要求。
全部选对得6分,选对但不全的得3分,有选错的得0分)14.物理学史上是哪位科学家、由于哪项贡献而人们称为“能称出地球质量的人”A .阿基米德,发现了杠杆原理B .牛顿,发现了万有引力定律C .伽利略,测出了重力加速度的值D .卡文迪许,测出了万有引力常量 15.不计重力的两个带电粒子M 和N 沿同一方向经小孔S 垂直进入匀强磁场,在磁场中的径迹如图。
分别用v M 与v N , t M 与t N ,M Mm q 与NN m q 表示它们的速率、在磁场中运动的时间、荷质比,则A .如果M M m q =NNm q ,则v M > v N B .如果M M m q =NNm q ,则t M < t N C .如果v M = v N ,则M Mm q >N N m q D .如果t M = t N ,则M Mm q >NN m q 16.如图a ,理想变压器原、副线圈的匝数比为2∶1,与副线圈相连的两个灯泡完全相 同、电表都为理想电表。
原线圈接上如图b 所示的正弦交流电,电路正常工作。
闭合开关后,A .电压表示数增大B .电流表示数增大C .变压器的输入功率增大D .经过灯泡的电流频率为25 Hz 17.如图,窗子上、下沿间的高度H=1.6m ,墙的厚度d=0.4m ,某人在离墙壁距离L=1.4m 、距窗子上沿h=0.2m 处的P 点,将可视为质点的小物件以v 的速度水平抛出,小物件直接穿过窗口并落在水平地面上,取g=10m/s 2。
则v 的取值范围是A .7>v m/sB .32.v <m/sC .7m/s m/s 3<<vD .3m/s m/s 32<<v .18.电梯经过启动、匀速运行和制动三个过程,从低楼层到达高楼层,启动和制动可看作是匀变速直线运动。
广东省广州市2016年普通高中毕业班综合测试(一)理综试题(Word)
2016 年广州市普通高中毕业班综合测试(一)理科综合2016.3.17第I卷一、选择题:本题包括13 小题,每小题 6 分,共78 分。
在每小题给出的四个选项中,只有一个选项符合题目要求。
1.下列关于生物膜的叙述,正确的是A .生物膜是对生物体内所有膜结构的统称B •组成生物膜的脂质中脂肪最丰富C•生物膜都是选择透过性膜D •细胞器膜是细菌进行代谢的结构基础2.下列有关酶的叙述,正确的是A •酶的合成都要经过转录和翻译两个过程B •酶可提供化学反应所需的活化能C . ATP合成酶都存在于线粒体中D .比较过氧化氨酶和Fe3+的催化效率可证明酶具有高效性3.下列有关育种的叙述,正确的是A .杂交育种所选双亲必须是纯合子B .诱变所得植株若不出现预期性状就应丢弃C.多倍体育种时可用秋水仙索处理幼苗D •单倍体育种依据的遗传学原理是基因突变4.下列有关免疫的说法,正确的是A .机体主要依赖免疫调节维持内环境稳态B .效应T 细可诱导靶细胞凋亡C .人在初次接触过敏原时会发生过敏反应D •再次接触同种抗原,记忆细胞分化成浆细胞释放淋巴因子5•用32P或35S标记T2噬菌体并分别与无标记的细菌混合培养,保温一定时间后经搅拌、离心得到上清液和沉淀物,并测量放射性。
对此实验的叙述,不正确的是.A .实验目的是研究遗传物质是DNA 还是蛋白质B-保温时间过长会使32P标记组上清液的放射性偏低C・搅拌不充分会使35S标记组沉淀物的放射性偏高D •实验所获得的子代噬菌体不含,35S而部分可含有32P6.下列有关湿地生态系统的叙述,不正确的是A •湿地蓄洪防旱的作用体现了生物多样性的直接价值B •在群落水平上研究湿地,需要确定其范围和边界C.弃耕水稻田中的群落演替属于次生演替D •保护湿地生物多样性有利于维持该生态系统的稳定性7.下列叙述正确的是A .顺-2 -丁烯和反-2-丁烯的加氢产物不同B •甲醛、氯乙烯和乙二醇均可作为合成聚合物的单体C.ABS 树脂、光导纤维及碳纤维都是有机高分子材料D •酸性条件下,C2H5C018OC2H5的水解产物是C2H5CO18OH和C2H5OH&设N A为阿伏加德罗常数,下列叙述中正确的是A . 78 g Na2O2中存在的共价键总数为N AB. 0. 1mol9038Sr原子中含中子数为3.8N AC. 氢氧燃料电池负极消耗 2. 24 L气体时,电路中转移的电子数为0.1N AD. 0.1 mol氯化铁溶于1L水中,所得溶液中Fe3+的数目为0. 1N A9•三聚磷酸可视为三个磷酸分子(磷酸结构式见右图)之间脱去两个水分子的产物,三聚磷酸钠(俗称“五钠”)是常用的水处理剂。
2016年广州市普通高中毕业班综合测试试题(理科)
2016 年广州市普通高中毕业班综合测试(一)
理科数学
注意事项: 1.本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。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,并用铅笔在答题卡上的相应位置填涂考生号。 2.回答第Ⅰ卷时,选出每小题答案后,用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,再选涂其它答案标号。写在本试卷上无效。 3.回答第Ⅱ卷时,将答案写在答题卡上。写在本试卷上无效。 4.考试结束后,将本试卷和答题卡一并交回。
(Ⅰ)求 AD 的长; (Ⅱ)求△ ABC 的面积. (18) (本小题满分 12 分) 从某企业生产的某种产品中抽取 100 件, 测量这些产品的质量指标值, 由测量结果得到如图所示的频 率分布直方图,质量指标值落在区间 55, 65 , 65, 75 , 75,85 内的频率之比为 4 :2:1 . (Ⅰ)求这些产品质量指标值落在区间
x2 y 2 1 a 0, b 0 的左顶点为 A , 右焦点为 F , 点 B 0, b , 且 BA BF 0 , a 2 b2
.
3
则双曲线 C 的离心率为
(15) x x 2 的展开式中, x 的系数为
2
4
. (用数字填写答案) 则函数 g x 2 f x 2 的零点个数为
A1
C1 B1
D O A B
C
(20) (本小题满分 12 分)
0 ,点 B 2, 2 在 已知椭圆 C 的中心在坐标原点,焦点在 x 轴上,左顶点为 A ,左焦点为 F1 2,
椭圆 C 上, 直线 y kx k 0 与椭圆 C 交于 E ,F 两点, 直线 AE ,AF 分别与 y 轴交于点 M ,N . (Ⅰ)求椭圆 C 的方程; (Ⅱ)以 MN 为直径的圆是否经过定点?若经过,求出定点的坐标;若不经过,请说明理由.
2016年广州市普通高中毕业班综合测试
2016年广州市普通高中毕业班综合测试(一)英语2016.3本试卷分第1卷(选择题)和第Ⅱ卷(非选择题)两部分。
考试结束后,将本试卷和答题卡一并交回。
第I卷第一部分阅读理解(共两节,满分40分)第一节 (共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
A1.Who is organizing the Festival of Books?A. Book TV.B. The Los Angeles Times.C. Natural History Museum.D. Los Angeles Public Library.2. How are the best book prizes chosen?A. By a group of experts.B. By a vote of the authors.C. By the reading public.D. By festival sponsors.3. What is true about tickets to the festival?A. They are only for published authors.B. They include transport to each event.C. They can only be applied for onlineD. They can be purchased at the event location.BIn 1800, only three percent of the world's population lived in cities. Only one city —Beijing —had a population of over a million. Most people lived in rural areas, and never saw a city in their lives. In 1900, just a hundred years later, roughly 150 million people lived in cities, and the world’s ten largest cities all had populations exceeding one million. By 2000, the number of city dwellers exceeded three billion; and in 2008, the world's population crossed a tipping point — more than half of the people on Earth lived in cities. By 2050, that could increase to over two-thirds. Clearly, humans have become an urban species.In the 19th and early 20th centuries, many people viewed cities negatively — crowded, dirty environments full of disease and crime. They feared that as cities got bigger, living conditions would worsen. In recent decades, however, attitudes have changed. Many experts now think urbanization (城市化) is good news, offering solutions to the problems of Earth’s growing population.Harvard economist Edward Glaeser, author of The Victory of the Cir, is one such person. Glaeser argues that cities are very productive because “the absence of space between people” reduces the cost of transporting goods, people, and ideas. While the flow of goods has always been important to cities, what is most important today is the flow of ideas. Successful cities enable people to learn from each other easily, and attract and reward smart people with higher wages.Another urbanization supporter is environmentalist Stewart Brand. Brand believes cities help the environment because they allow haft of the world’s population to live on about four percent of the land. T his leaves more space for open country, such as farmland. City dwellers also have less impact per person on the environment than people in the countryside. Their roads, sewers, and power lines need fewer resources to build and operate. City apartments require less energy to heat, cool, and light. Most importantly, people in cities drive less so they produce fewer greenhouse gases per person.So it’s a mistake to see urbanization as evil; it’s a natural part of development. The challenge is how to manage the growth.4. What is Paragraph 1 mainly about?A. The history of modern cities.B. Changes taking place within cities.C. How cities have grown over time.D. Why modern cities are changing.5. How have experts’ attitudes towards cities changed in recent decades?A. They now view the weaknesses as strengths.B. They no longer see city-riving as attractive.C. They accept city life in spite of its problems.D. They think city-riving provides more benefits.6. Which of the following would Edward Glaeser agree with?A. Cities provide more economic opportunities.B. City people get along better with each other.C. Over-crowded cities result in problems.D. Cities limit the flow of ideas.7. According to Paragraph 4, what would be the result of moving people out of cities?A. Economic production would be reduced.B. There would be less farmland available.C. People would travel less frequently.D. House values would fall greatly.CIt was December 25, 1914, only 5 months into World War I. German, British, and French soldiers, already sick and tired of the senseless killing, disobeyed their superiors and started socialising with “the enemy” along two-thirds of the Western Front.German troops held up Christmas trees with signs, “Merry Christmas”. Thousands of troops ran across the battlefields covered with dead bodies. They sang Christmas songs, exchanged photographs of loved ones back home, shared food and even played football. Soldiers hugged men they had been trying to kill a few short hours before. They agreed to warn each other if their officers forced them to fire their weapons, and to aim high.Fear ran through the military leaders on both sides. They felt that their power was being challenged: soldiers declaring their brotherhood with each other and refusing to fight. Generals declared this unexpected peacemaking illegal and said that participating soldiers would face a military court. Those found guilty would be imprisoned or even shot. By March 1915 the socialising movement had been destroyed and the killing machine was back in full operation. Over the next three years more than fifteen million people died in the war.Not many people have heard the story of the Christmas Truce (休战). On Christmas Day, 1988, a local radio host in Bos ton played “Christmas in the Trenches”, a song about the Christmas Truce, several times and was stunned by his listeners’ response. Thousands of people called in, praising the song, with many moved to tears by the amazing events it described.You can probably guess why the callers were in tears. The Christmas Truce story goes against most of what we have been taught about people. It lets us see the world as it can be and says, “This really happened once.” It shows us the potential we have as humans, and contradicts all of those TV and newspaper stories that tell us how mean and heartless people are. It is like hearing that our deepest wishes really are true: the world really can be different.8. The soldiers began socialising with the enemy because ________.A. they couldn't bear the meaningless killingB. it was the best way to avoid being killedC. they feared that they would be caughtD. their enemies were from similar backgrounds9. How did the generals finally put an end to the soldiers’ socialisi ng?A. They sent the soldiers’ loved ones to prison.B. They moved the two groups of soldiers further apart.C. They increased the number of officers to control the soldiers.D. They warned the soldiers that they would face severe punishment.10. The u nderlined word “stunned” in Paragraph 4 most probably means ________.A. satisfiedB. shockedC. amusedD. confused11. The author uses the story of the soldiers to imply that human beings ________.A. are not trustworthy under stressB. are naturally aggressive and war-likeC. are basically caring and kindheartedD. will always do what is in their own self-interestDEarlier this year, the social media website Facebook announced that it would work with several news organisations — including The New York Times, The Guardian, and the BBC — to place news stories directly into users’ personal Facebook webpage.Stories published using Facebook Instant will load more quickly and keep the style of the original publisher, who will keep all the advertising income the stories earn — at least for now. The deal shows how important social, media has become to news organisations, and is a clear sign of how the world of news is changing — and has been for a while.When Google News began in 2002, many saw it as the death of the newspaper. It had no human editor. Instead Google used, and still uses, a secret computer program that selects and displays news stories according to the reader’s personal interests. More recently, Associated Press and Yahoo! have bee n punishing computer-written articles. Both use special software to automatically produce stories about company financial results and sports reports — areas where the quality of writing is felt to be of secondary importance to the accuracy of the data.Should we be worried about such developments? I think we should. One concern is that facebook, Google and other social media websites see journalism as a sideline, a way of putting people in front of advertisements. It isn’t their primary function — so if it stops making them lots of money, they're likely to stop doing it.There’s also a concern that computer-written articles are not actually journalism at all, because what a human news team produces is actually quite complex. A well-written news story puts information in context, offers a voiceto each side of an argument and brings the public new knowledge.Though economics and speed of delivery mean readers will probably choose a computer-written story over a carefully shaped article — at least for daily news — I don't think the computers will be writing any in-depth articles for a while yet.12. What is the main purpose of the article?A. To report on a new computer service offered by Facebook..B. To advise readers against reading computer-written news.C. To express concern about recent trends in online news.D. To describe the process of online news reporting.13. Computer-written news reports have so far focused on sports and finance because ________.A. these are the most popular topics for online readersB. there are fewer journalists specialising in these areasC. information on these topics is more easily availableD. writing style is less important than accuracy in these areas14. What does the underlined word “It” in Paragraph 3 refer to?A. Journalism.B. Advertising.C. Facebook.D. Business.15. In Paragraph 4, which of the following is mentioned as a characteristic of a well-written news article?A. The information presented is up-to-date.B. The author's opinion is clear.C. Different views on the topic are presented.D. The language used is vivid.第二节 (共5小题,每小题2分,满分l0分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
2016年广州市普通高中毕业班综合测试(二)文科综合试题
试卷类型:A 2016年广州市普通高中毕业班综合测试(二)文科综合2016.4注意事项:1.本试卷分第I卷(选择题)和;第H卷(非选择题)两部分。
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,并用铅笔在答题卡上的相应位置填涂考生号。
2.回答第I卷时,选出每小题答案后,用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,再选涂其它答案标号。
写在本试卷上无效。
3.回答第H卷时,将答案写在答题卡上。
写在本试卷上无效。
4.考试结束后,将本试卷和答题卡一并交回。
,.第I卷本卷共35个小题,每小题4分,共140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。
中国在建造空间站的同时,将发送一个单独的"光学舱",搭载一架在功能上类似"哈勃"的太空望远镜。
哈勃望远镜距离地面约547千米,挠地球一周耗时约97分钟。
运用所学知识,完成1-3题。
1.哈勃太空望远镜能够全天候观测星空,这是因为A.没有经历昼夜交替B.没有受大气影响C.没有绕太阳公转D.位于地球同步轨道上2.哈勃望远镜观测地球为一个美丽的"蓝色"星球,这是因为A.海洋水体为蓝色B.海水藻类呈蓝色C.大气散射太阳光D.望远镜镜头色散3.在下列四个波段的观测中,太空望远镜比地面望远镜更具优势的是A.红光B.黄光C.蓝光D.紫外光谢贝利河发源于高原山地,其上、中、下游分别流经山地、高原、平原,读下图完成4 -6题。
4.该河流的流量特点是A.自上游向下游增大B.自上游向下游减小C.中游最大,向上、下游减小D.中游最小,向上、下游增大5.该河流的补给水源最主要是A.雨水B.冰雪融水C.湖泊水D.季节性积雪融水6.该河流下游无支流,其最主要成因是A.干温季分明B.地形平坦C.地上河D.植被稀少某发达国家劳动力数量呈下降趋势。
读该国劳动力年龄结构变化示意图,完成7-9题。
2016届广州市普通高中毕业班综合测试(一)(文数)
2016届广州市普通高中毕业班综合测试(一)数学(文科)注意事项:1.本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
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,并用铅笔在答题卡上的相应位置填涂考生号。
2.回答第Ⅰ卷时,选出每小题答案后,用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,再选涂其它答案标号。
写在本试卷上无效。
3.回答第Ⅱ卷时,将答案写在答题卡上。
写在本试卷上无效。
4.考试结束后,将本试卷和答题卡一并交回。
第Ⅰ卷一.选择题:本大题共12小题,每小题5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{}11A x x =-≤≤,{}220B x x x =-≤,则A B =(A ){}12x x -≤≤ (B ){}10x x -≤≤ (C ){}12x x ≤≤ (D ){}01x x ≤≤ (2)已知复数3i1iz +=+,其中i 为虚数单位,则复数z 所对应的点在 (A )第一象限 (B )第二象限 (C )第三象限 (D )第四象限(3)已知函数()2,1,1,1,1x x x f x x x⎧-≤⎪=⎨>⎪-⎩则()()2f f -的值为(A )12 (B )15 (C )15- (D )12-(4)设P 是△ABC 所在平面内的一点,且2CP PA =,则△PAB 与△PBC 的面积之比是(A )13 (B )12 (C )23 (D )34(5)如果函数()cos 4f x x ωπ⎛⎫=+⎪⎝⎭()0ω>的相邻两个零点之间的距离为6π,则ω的值为 (A )3 (B )6 (C )12 (D )24(6)执行如图所示的程序框图,如果输入3x=,则输出k的值为(A)6 (B)8 (C)10 (D)12(7)在平面区域(){},0112x y x y≤≤≤≤,内随机投入一点P,则点P的坐标(),x y满足2y x≤的概率为(A)14(B)12(C)23(D)34(8)已知()sin6f x xπ⎛⎫=+⎪⎝⎭,若3sin5α=2πα⎛⎫<<π⎪⎝⎭,则12fαπ⎛⎫+=⎪⎝⎭(A)10-(B)10-(C)10(D)10(9)如果1P,2P,…,nP是抛物线C:24y x=上的点,它们的横坐标依次为1x,2x,…,nx,F是抛物线C的焦点,若1210nx x x+++=,则12nPF P F P F+++=(A)10n+(B)20n+(C)210n+(D)220n+(10)一个六棱柱的底面是正六边形,侧棱垂直于底面,所有棱的长都为1,顶点都在同一个球面上,则该球的体积为(A)20π(B(C)5π(D(11)已知下列四个命题:1p:若直线l和平面α内的无数条直线垂直,则lα⊥;2p:若()22x xf x-=-,则x∀∈R,()()f x f x-=-;3p:若()11f x xx=++,则()0,x∃∈+∞,()01f x=;4p:在△ABC中,若A B>,则sin sinA B>.其中真命题的个数是(A)1 (B)2 (C)3 (D)4(12)如图,网格纸上小正方形的边长为1,粗线画出的是某个四面体的三视图,则该四面体的表面积为(A)8+(B)8+(C)2+(D)1224++第Ⅱ卷本卷包括必考题和选考题两部分.第13题~第21题为必考题,每个试题考生都必须做答.第22题~第24题为选考题,考生根据要求做答. 二.填空题:本大题共4小题,每小题5分. (13)函数()33f x x x =-的极小值为 .(14)设实数x ,y 满足约束条件230,230,3x y x y x --≤⎧⎪+-≤⎨⎪≥-⎩, 则23z x y =-+的取值范围是 .(15)已知双曲线C :22221x y a b-=()0,0a b >>的左顶点为A ,右焦点为F ,点()0,B b ,且0=⋅BF BA ,则双曲线C 的离心率为 .(16)在△ABC 中,点D 在边AB 上,CD BC ⊥,AC =5CD =,2BD AD =,则AD 的长为 .三.解答题:解答应写出文字说明,证明过程或演算步骤. (17)(本小题满分12分)已知数列{}n a 是等比数列,24a =,32a +是2a 和4a 的等差中项. (Ⅰ)求数列{}n a 的通项公式;(Ⅱ)设22log 1n n b a =-,求数列{}n n a b 的前n 项和n T .(18)(本小题满分12分)从某企业生产的某种产品中抽取100件, 测量这些产品的质量指标值,由测量结果 得到如图所示的频率分布直方图,质量指标值落在区间[)55,65,[)65,75,[]75,85 内的频率之比为4:2:1.(Ⅰ)求这些产品质量指标值落在区间[]75,85内的频率;(Ⅱ)用分层抽样的方法在区间[)45,75内抽取一个容量为6的样本,将该样本看成一个总体,从中任意抽取2件产品,求这2件产品都在区间[)45,65内的概率.(19)(本小题满分12分)如图,四棱柱1111ABCD A BC D -的底面ABCDAC BD O = ,1AO ⊥底面ABCD ,1==AA AB (Ⅰ)证明:BD ⊥平面1ACO ; (Ⅱ)若60BAD ∠=,求点C 到平面1OBB(20)(本小题满分12分)已知椭圆C 的中心在坐标原点,焦点在x 轴上,左顶点为A ,左焦点为()120F -,,点(B 在椭圆C 上,直线()0y kx k =≠与椭圆C 交于E ,F 两点,直线AE ,AF 分别与y 轴交于点M ,N .(Ⅰ)求椭圆C 的方程;(Ⅱ)在x 轴上是否存在点P ,使得无论非零实数k 怎样变化,总有MPN ∠为直角?若存在,求出点P 的坐标;若不存在,请说明理由.(21)(本小题满分12分)已知函数()e ln 1x f x m x =--.(Ⅰ)当1m =时,求曲线()y f x =在点()()11f ,处的切线方程; (Ⅱ)当1m ≥时,证明:()1f x >.请考生在第22、23、24题中任选一题做答,如果多做,则按所做的第一题计分.做答时请写清题号.(22)(本小题满分10分)选修4-1:几何证明选讲如图所示,△ABC 内接于⊙O ,直线AD 与⊙O 相切于点 A ,交BC 的延长线于点D ,过点D 作CA DE //交BA 的延长线于点E . (Ⅰ)求证:BE AE DE ⋅=2;(Ⅱ)若直线EF 与⊙O 相切于点F ,且4EF =,2EA =,求线段AC 的长.(23)(本小题满分10分)选修4-4:坐标系与参数方程在平面直角坐标系xOy 中,以坐标原点O 为极点,x 轴正半轴为极轴建立极坐标系,曲线C 的极坐标方程为θρsin 2=,[)0,2θ∈π. (Ⅰ)求曲线C 的直角坐标方程;(Ⅱ)在曲线C 上求一点D ,使它到直线l:32x y t ⎧=⎪⎨=-+⎪⎩(t 为参数,t ∈R )的距离最短,并求出点D 的直角坐标.(24)(本小题满分10分)选修4-5:不等式选讲设函数()f x x x = (Ⅰ)当1a =时,求不等式()12f x ≥的解集; (Ⅱ)若对任意[]0,1a ∈,不等式()f x b ≥的解集为空集,求实数b 的取值范围.数学(文科)参考答案评分说明: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(1)D (2)D (3)C (4)B (5)B (6)C (7)A (8)B(9)A(10)D(11)B(12)A二.填空题(13)2-(14)[]6,15- (15 (16)5三.解答题(17)解:(Ⅰ)设数列{}n a 的公比为q ,因为24a =,所以34a q =,244a q =.…………………………………………1分因为32a +是2a 和4a 的等差中项,所以()32422a a a +=+.……………………2分 即()224244q q +=+,化简得220q q -=.因为公比0q ≠,所以2q =.………………………………………………………4分 所以222422n n n n a a q --==⨯=(*n ∈N ).…………………………………………5分 (Ⅱ)因为2n na =,所以22log 121n nb a n =-=-.所以()212nn n a b n =-.……………………………………………………………7分 则()()231123252232212n n n T n n -=⨯+⨯+⨯+⋅⋅⋅+-+-, ①()()23412123252232212n n n T n n +=⨯+⨯+⨯+⋅⋅⋅+-+-. ②………………9分①-②得,()2312222222212n n n T n +-=+⨯+⨯+⋅⋅⋅+⨯--……………………………………10分()()()11142221262321212n n n n n ++-=+⨯--=-----,所以()16232n n T n +=+-.……………………………………………………………12分(18)解:(Ⅰ)设区间[]75,85内的频率为x ,则区间[)55,65,[)65,75内的频率分别为4x 和2x .…………………………1分 依题意得()0.0040.0120.0190.03010421x x x +++⨯+++=,……………3分 解得0.05x =.所以区间[]75,85内的频率为0.05.………………………………………………4分 (Ⅱ)由(Ⅰ)得,区间[)45,55,[)55,65,[)65,75内的频率依次为0.3,0.2,0.1.用分层抽样的方法在区间[)45,75内抽取一个容量为6的样本,则在区间[)45,55内应抽取0.3630.30.20.1⨯=++件,记为1A ,2A ,3A .在区间[)55,65内应抽取0.2620.30.20.1⨯=++件,记为1B ,2B . 在区间[)65,75内应抽取0.1610.30.20.1⨯=++件,记为C .…………………6分 设“从样本中任意抽取2件产品,这2件产品都在区间[)45,65内”为事件M ,则所有的基本事件有:{}12,A A ,{}13,A A ,{}11,A B ,{}12,AB ,{}1,AC ,{}23,A A , {}21,A B ,{}22,A B ,{}2,A C ,{}31,A B ,{}32,A B ,{}3,A C ,{}12,B B ,{}1,B C ,{}2,B C ,共15种.…………………………………………………………………8分事件M 包含的基本事件有:{}12,A A ,{}13,A A ,{}11,A B ,{}12,AB ,{}23,A A , {}21,A B ,{}22,A B ,{}31,A B ,{}32,A B ,{}12,B B ,共10种.…………10分所以这2件产品都在区间[)45,65内的概率为102153=.………………………12分 (19)(Ⅰ)证明:因为1AO ⊥平面ABCD ,BD ⊂平面ABCD , 所以1AO ⊥BD .……………………………………………………………………1分 因为ABCD 是菱形,所以CO ⊥BD .……………………………………………2分因为1AO CO O = ,1AO ,CO ⊂平面1ACO , 所以BD ⊥平面1ACO .……………………………………………………………3分 (Ⅱ)解法一:因为底面ABCD 是菱形,AC BD O = ,21==AA AB ,60BAD ∠=,所以1OB OD ==,OA OC ==……………………………………………4分所以OBC ∆的面积为112212OBC S OB OC ∆==⨯=⨯⨯…………………5分 因为1AO ⊥平面ABCD ,AO ⊂平面ABCD , 所以1AO AO ⊥,11AO ==.………………………………………6分因为ABCD B A 平面//11,所以点1B 到平面ABCD 的距离等于点1A 到平面ABCD 的距离1AO .…………7分 由(Ⅰ)得,BD ⊥平面1A AC .因为1A A ⊂平面1AAC ,所以BD ⊥1A A . 因为B B A A 11//,所以BD ⊥1B B .………………………………………………8分 所以△1OBB 的面积为111121212OBB S OB BB ∆=⨯⨯==⨯⨯.……………………9分 设点C 到平面1OBB 的距离为d , 因为11C OBB B OBC V V --=,所以O A S d S OBC OBB 131311⋅=⋅∆∆.………………………………………………10分 所以111212OBC OBBS AO d S ∆∆⋅===所以点C 到平面1OBB 解法二:由(Ⅰ)知BD ⊥因为BD ⊂平面11BB D D ,所以平面1A CO ⊥平面11BB D D .…4分 连接11AC 与11B D 交于点1O , 连接1CO ,1OO ,因为11AA CC =,11//AA CC ,所以11CAAC 为平行四边形. 又O ,1O 分别是AC ,11AC 的中点,所以11OAO C 为平行四边形. 所以111OC OA ==.…………………………………………………………………6分 因为平面11OAO C 与平面11BB D D 交线为1OO , 过点C 作1CH OO ⊥于H ,则CH ⊥平面11BB D D .………………………………8分 因为,O A C O 11//1AO ⊥平面ABCD ,所以·1O C ⊥平面ABCD . 因为OC ⊂平面ABCD ,所以·1O C ⊥OC ,即△1OCO 为直角三角形.………10分所以1122O C OC CH OO ⋅===.所以点C 到平面1OBB2……………………………………………12分(20)(Ⅰ)解法一:设椭圆C 的方程为22221(0)x y a b a b+=>>,因为椭圆的左焦点为()120F -,,所以224a b -=.……………………………1分 设椭圆的右焦点为()220F ,,已知点(2B 在椭圆C 上, 由椭圆的定义知122BF BF a +=,所以2a =.………………………………………………………2分所以a =2b =.………………………………………………………3分所以椭圆C 的方程为22184x y +=.………………………………………………4分 解法二:设椭圆C 的方程为22221(0)x y a b a b+=>>,因为椭圆的左焦点为()120F -,,所以224a b -=. ①…………………1分因为点(2B 在椭圆C 上,所以22421a b+=. ②…………………2分由①②解得,a =2b =.…………………………………………………3分所以椭圆C 的方程为22184x y +=.………………………………………………4分 (Ⅱ)解法一:因为椭圆C 的左顶点为A ,则点A的坐标为()-.…………5分因为直线(0)y kx k =≠与椭圆22184x y +=交于两点E ,F , 设点()00,E x y (不妨设00x >),则点()00,F x y --.联立方程组22,184y kx x y =⎧⎪⎨+=⎪⎩消去y 得22812x k =+.所以0x =0y =………………………………………………6分所以直线AE的方程为y x =+.……………………………7分因为直线AE 与y 轴交于点M ,令0x =得y =M ⎛⎫⎝.……………………8分同理可得点N ⎛⎫ ⎝.…………………………………………………9分 假设在x 轴上存在点(,0)P t ,使得MPN ∠为直角,则0MP NP ⋅=.………10分即20t =,即240t -=.………………………11分解得2t =或2t =-.故存在点()2,0P 或()2,0P -,无论非零实数k 怎样变化,总有MPN ∠为直角. ………………………………12分 解法二: 因为椭圆C 的左端点为A ,则点A的坐标为()-.……………5分因为直线(0)y kx k =≠与椭圆22184x y +=交于两点E ,F , 设点00(,)E x y ,则点00(,)F x y --.所以直线AE的方程为y x =+.………………………………6分 因为直线AE 与y 轴交于点M ,令0x =得y =M ⎛⎫⎝.……………………………7分同理可得点N ⎛⎫⎝.……………………………………………………8分假设在x 轴上存在点(),0P t ,使得MPN ∠为直角,则0MP NP ⋅=.即20t +=,即2220808y t x +=-. (※)…………9分 因为点00(,)E x y 在椭圆C 上,所以2200184x y +=,即220082x y -=.……………………………………………10分 将220082x y -=代入(※)得240t -=.………………………………………11分解得2t =或2t =-.故存在点()2,0P 或()2,0P -,无论非零实数k 怎样变化,总有MPN ∠为直角. ………………………………12分 解法三:因为椭圆C 的左顶点为A ,则点A的坐标为()-.……………5分因为直线(0)y kx k =≠与椭圆22184x y +=交于两点E ,F ,设点(),2sin E θθ(0θ<<π),则点(),2sin F θθ--.……6分 所以直线AE的方程为y x =+.………………………7分因为直线AE 与y 轴交于点M ,令0x =得2sin cos 1y θθ=+,即点2sin 0,cos 1M θθ⎛⎫⎪+⎝⎭.………………………………8分同理可得点2sin 0,cos 1N θθ⎛⎫⎪-⎝⎭.………………………………………………………9分假设在x 轴上存在点(,0)P t ,使得MPN ∠为直角,则0MP NP ⋅=.………10分即22sin 2sin 0cos 1cos 1t θθθθ--+⨯=+-,即240t -=.…………………………………11分解得2t =或2t =-.故存在点()2,0P 或()2,0P -,无论非零实数k 怎样变化,总有MPN ∠为直角. ………………………………12分(21)(Ⅰ)解:当1m =时,()e ln 1x f x x =--,所以1()e xf x x'=-.………………………………………………………………1分 所以(1)e 1f =-,(1)e 1f '=-. …………………………………………………2分 所以曲线()y f x =在点()()11f ,处的切线方程为(e 1)(e 1)(1)y x --=--. 即()e 1y x =-.………………………………………………………………………3分 (Ⅱ)证法一:当1m ≥时,()e ln 1e ln 1x xf x m x x =--≥--.要证明()1f x >,只需证明e ln 20xx -->.……………………………………4分 以下给出三种思路证明e ln 20xx -->.思路1:设()e ln 2xg x x =--,则1()e x g x x'=-. 设1()e xh x x =-,则21()e 0xh x x'=+>, 所以函数()h x =1()e xg x x'=-在0+∞(,)上单调递增.…………………………6分 因为121e 202g ⎛⎫'=-< ⎪⎝⎭,(1)e 10g '=->,所以函数1()e xg x x '=-在0+∞(,)上有唯一零点0x ,且01,12x ⎛⎫∈ ⎪⎝⎭.…………8分 因为0()0g x '=时,所以01ex x =,即00ln x x =-.………………………………9分 当()00,x x ∈时,()0g x '<;当()0,x x ∈+∞时,()0g x '>.所以当0x x =时,()g x 取得最小值()0g x .……………………………………10分故()000001()=e ln 220xg x g x x x x ≥--=+->. 综上可知,当1m ≥时,()1f x >.………………………………………………12分 思路2:先证明e 1xx ≥+()x ∈R .………………………………………………5分 设()e 1x h x x =--,则()e 1x h x '=-.因为当0x <时,()0h x '<,当0x >时,()0h x '>,所以当0x <时,函数()h x 单调递减,当0x >时,函数()h x 单调递增. 所以()()00h x h ≥=.所以e 1xx ≥+(当且仅当0x =时取等号).………………………………………7分 所以要证明e ln 20xx -->,只需证明()1ln 20x x +-->.……………………………………………………8分 下面证明ln 10x x --≥. 设()ln 1p x x x =--,则()111x p x x x-'=-=. 当01x <<时,()0p x '<,当1x >时,()0p x '>,所以当01x <<时,函数()p x 单调递减,当1x >时,函数()p x 单调递增. 所以()()10p x p ≥=.所以ln 10x x --≥(当且仅当1x =时取等号).………………………………10分由于取等号的条件不同, 所以e ln 20xx -->.综上可知,当1m ≥时,()1f x >.………………………………………………12分 (若考生先放缩ln x ,或e x、ln x 同时放缩,请参考此思路给分!) 思路3:先证明e ln 2xx ->.因为曲线e x y =与曲线ln y x =的图像关于直线y x =对称,设直线x t =()0t >与曲线e xy =,ln y x =分别交于点A ,B ,点A ,B 到直线y x =的距离分别为1d ,2d ,则)12AB d d +. 其中1t d =2d =()0t >.①设()e t h t t =-()0t >,则()e 1t h t '=-. 因为0t >,所以()e 10t h t '=->.所以()h t 在()0,+∞上单调递增,则()()01h t h >=.所以1t d =>. ②设()ln g t t t =-()0t >,则()111t g t t t -'=-=.因为当01t <<时,()0g t '<;当1t >时,()0g t '>,所以当01t <<时,()ln g t t t =-单调递减;当1t >时,()ln g t t t =-单调递增. 所以()()11g t g ≥=.所以2d =≥.所以)122AB d d =+>=⎭. 综上可知,当1m ≥时,()1f x >.………………………………………………12分证法二:因为()e ln 1xf x m x =--,要证明()1f x >,只需证明e ln 20xm x -->.…………………………………4分以下给出两种思路证明e ln 20xm x -->.思路1:设()e ln 2xg x m x =--,则1()e x g x m x'=-. 设1()e xh x m x =-,则21()e 0xh x m x'=+>. 所以函数()h x =()1e xg x m x'=-在()0+∞,上单调递增.……………………6分因为11221e 2e 202mm g m m m m ⎛⎫⎛⎫'=-=-< ⎪ ⎪⎝⎭⎝⎭,()1e 10g m '=->,所以函数1()e xg x m x '=-在()0+∞,上有唯一零点0x ,且01,12x m ⎛⎫∈⎪⎝⎭.……8分 因为()00g x '=,所以01ex m x =,即00ln ln x x m =--.……………………9分 当()00,x x ∈时,()0g x '<;当()0,x x ∈+∞时,()0g x '>.所以当0x x =时,()g x 取得最小值()0g x .……………………………………10分 故()()000001e ln 2ln 20xg x g x m x x m x ≥=--=++->. 综上可知,当1m ≥时,()1f x >.………………………………………………12分 思路2:先证明e 1()xx x ≥+∈R ,且ln 1(0)x x x ≤+>.……………………5分 设()e 1xF x x =--,则()e 1x F x '=-.因为当0x <时,()0F x '<;当0x >时,()0F x '>, 所以()F x 在(,0)-∞上单调递减,在(0,)+∞上单调递增. 所以当0x =时,()F x 取得最小值(0)0F =.所以()(0)0F x F ≥=,即e 1xx ≥+(当且仅当0x =时取等号).……………7分由e 1()x x x ≥+∈R ,得1ex x -≥(当且仅当1x =时取等号).………………8分所以ln 1(0)x x x ≤->(当且仅当1x =时取等号).……………………………9分 再证明e ln 20xm x -->.因为0x >,1m ≥,且e 1xx ≥+与ln 1x x ≤-不同时取等号,所以()()e ln 2112x m x m x x -->+---()()11m x =-+0≥.综上可知,当1m ≥时,()1f x >.………………………………………………12分(22)(Ⅰ)证明:因为AD 是⊙O 的切线,所以DAC B ∠=∠(弦切角定理).………………1分因为CA DE //,所以DAC EDA ∠=∠.……………………………2分 所以EDA B ∠=∠.因为AED D EB ∠=∠(公共角),所以△AED ∽△DEB .……………………………………………………………3分 所以DE AE BEDE=.即2DE AE BE = .…………………………………………………………………4分 (Ⅱ)解:因为EF 是⊙O 的切线,EAB 是⊙O 的割线,所以EB EA EF ⋅=2(切割线定理).……………………………………………5分 因为4EF =,2EA =,所以8EB =,6AB EB EA =-=.…………………7分 由(Ⅰ)知BE AE DE ⋅=2,所以4DE =.………………………………………8分 因为CA DE //,所以△BAC ∽△BED . ………………………………………9分 所以BA ACBEED =.所以6438BA EDAC BE⋅⨯===. …………………………………………………10分(23)(Ⅰ)解:由θρsin 2=,[)0,2θ∈π,可得22sin ρρθ=.…………………………………………………………………1分因为222x y ρ=+,sin y ρθ=,…………………………………………………2分 所以曲线C 的普通方程为2220x y y +-=(或()2211x y +-=). …………4分(Ⅱ)解法一:因为直线的参数方程为32x y t ⎧=⎪⎨=-+⎪⎩(t 为参数,t ∈R ),消去t 得直线l 的普通方程为5y =+. ……………………………………5分因为曲线C :()2211x y +-=是以G ()1,0为圆心,1为半径的圆,设点()00,D x y ,且点D 到直线l :5y =+的距离最短,所以曲线C 在点D 处的切线与直线l:5y =+平行. 即直线GD 与l 的斜率的乘积等于1-,即(0011y x -⨯=-.………………7分 因为()220011x y +-=,解得02x =-或0x = 所以点D 的坐标为12⎛⎫ ⎪ ⎪⎝⎭,或322⎛⎫ ⎪ ⎪⎝⎭,.……………………………………9分由于点D到直线5y =+的距离最短,所以点D的坐标为322⎛⎫⎪ ⎪⎝⎭,.……………………………………………………10分 解法二:因为直线l的参数方程为32x y t ⎧=⎪⎨=-+⎪⎩(t 为参数,t ∈R ),消去t 得直线l50y +-=.……………………………………5分因为曲线C ()2211x y +-=是以G ()1,0为圆心,1为半径的圆,因为点D 在曲线C 上,所以可设点D ()cos ,1sin ϕϕ+[)()0,2ϕ∈π.………7分 所以点D 到直线l的距离为d =2sin 3ϕπ⎛⎫=-+ ⎪⎝⎭.………………………………8分 因为[)0,2ϕ∈π,所以当6ϕπ=时,min 1d =.…………………………………9分 此时D 322⎛⎫ ⎪ ⎪⎝⎭,,所以点D的坐标为322⎛⎫⎪ ⎪⎝⎭,.……………………………10分(24)(Ⅰ)解:当1a =时,()12f x ≥等价于112x x +-≥.……………………1分 ①当1x ≤-时,不等式化为112x x --+≥,无解;②当10x -<<时,不等式化为112x x ++≥,解得104x -≤<; ③当0x ≥时,不等式化为112x x +-≥,解得0x ≥.…………………………3分 综上所述,不等式()1≥x f 的解集为1,4⎡⎫-+∞⎪⎢⎣⎭.………………………………4分 (Ⅱ)因为不等式()f x b ≥的解集为空集,所以()max b f x >⎡⎤⎣⎦.…………………5分以下给出两种思路求()f x 的最大值.思路1:因为()f x x x =+--()01a ≤≤,当x ≤()f x x x =-=0<.当x <<时,()f x x x =+2x =≤=当x ≥()f x x x ==所以()max f x ⎡⎤⎣⎦=……………………………………………………7分思路2:因为 ()f x x x =+--x x ≤=当且仅当x所以()max f x ⎡⎤⎣⎦=……………………………………………………7分因为对任意[]0,1a ∈,不等式()f x b ≥的解集为空集,所以max b >.………………………………………………………8分以下给出三种思路求()g a =.思路1:令()g a =所以()21g a =+2212≤++=.=12a =时等号成立. 所以()max g a =⎡⎤⎣⎦所以b的取值范围为)∞.…………………………………………………10分思路2:令()g a =因为01a ≤≤,所以可设2cos a θ= 02θπ⎛⎫≤≤⎪⎝⎭, 则()g a=cos sin 4θθθπ⎛⎫=+=+≤ ⎪⎝⎭当且仅当4θπ=时等号成立. 所以b的取值范围为)∞.…………………………………………………10分思路3:令()g a =因为01a ≤≤,设⎪⎩⎪⎨⎧-==ay ax 1,则221x y +=,10,10≤≤≤≤y x .问题转化为在221x y +=,10,10≤≤≤≤y x 的条件下, 求z x y =+的最大值.利用数形结合的方法容易求得z此时2x y ==.所以b的取值范围为)∞.…………………………………………………10分。
2016年广州市普通高中毕业班综合测试
2016年广州市普通高中毕业班综合测试(二)文科综合2016.4第I卷本卷共35个小题,每小题4分,共140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。
中国在建造空间站的同时,将发送一个单独的"光学舱",搭载一架在功能上类似"哈勃"的太空望远镜。
哈勃望远镜距离地面约547千米,挠地球一周耗时约97分钟。
运用所学知识,完成1-3题。
1.哈勃太空望远镜能够全天候观测星空,这是因为A.没有经历昼夜交替B.没有受大气影响C.没有绕太阳公转D.位于地球同步轨道上2.哈勃望远镜观测地球为一个美丽的"蓝色"星球,这是因为A.海洋水体为蓝色B.海水藻类呈蓝色C.大气散射太阳光D.望远镜镜头色散3.在下列四个波段的观测中,太空望远镜比地面望远镜更具优势的是A.红光B.黄光C.蓝光D.紫外光谢贝利河发源于高原山地,其上、中、下游分别流经山地、高原、平原,读下图完成4-6题。
4.该河流的流量特点是A.自上游向下游增大B.自上游向下游减小C.中游最大,向上、下游减小D.中游最小,向上、下游增大5.该河流的补给水源最主要是A.雨水B.冰雪融水C.湖泊水D.季节性积雪融水6.该河流下游无支流,其最主要成因是A.干温季分明B.地形平坦C.地上河D.植被稀少某发达国家劳动力数量呈下降趋势。
读该国劳动力年龄结构变化示意图,完成7-9题。
7.10年来,该国15-29岁年龄段劳动力比重保持稳定的最主要原因是A.降低就业年龄B.鼓励妇女就业C.人口迁移D.出生率稳定8.10年来,该国人口A.快速增长B.缓慢增长C.零增长D.负增长9.引起该国劳动力年龄结构变化的最主要原因是A.老龄化严重B.死亡率较低C.人口迁移D.出生率过低南安大略地区在该国的农业生产中占有重要地位,读图完成10-11题。
10.该地区的农业地域类型是A.商品谷物农业B.混合农业C.大牧场放牧业D.乳畜业11.引起该地区农作物播种面积变化的最主要因素是A.国际市场B.劳动力价格C.气候变化D.城市化.(22分)阅读图文材料,完成下列要求。
广东省广州市普通高中2016届高三下学期综合测试(一)文
2016年广州市普通高中毕业班综合测试(一)文科数学一、选择题(本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.)1.已知集合{}11A x x =-≤≤,{}220B x x x =-≤,则AB =( )(A ){}12x x -≤≤ (B ){}10x x -≤≤ (C ){}12x x ≤≤ (D ){}01x x ≤≤ 【答案】D 【解析】试题分析:{}{}22002x x x x x B =-≤=≤≤,所以{}01x x A B =≤≤,故选D .考点:1、一元二次不等式;2、集合的交集. 2.已知复数3i1iz +=+,其中i 为虚数单位,则复数z 所对应的点在( ) (A )第一象限 (B )第二象限 (C )第三象限 (D )第四象限 【答案】D 【解析】试题分析:()()()()23133321112i i i i i i z i i i i +-+-+-====-++-,所以复数z 所对应的点()2,1Z -,在第四象限,故选D .考点:1、复数的除法运算;2、复数的几何意义.3.已知函数()2,1,1,1,1x x x f x x x⎧-≤⎪=⎨>⎪-⎩则()()2f f -的值为( )(A )12 (B )15 (C )15- (D )12-【答案】C 【解析】试题分析:()()()22226f -=---=,所以()()()1126165f f f -===--,故选C . 考点:分段函数求值.4.设P 是△ABC 所在平面内的一点,且2CP PA =,则△PAB 与△PBC 的面积之比是( )(A )13 (B )12 (C )23 (D )34【答案】B 【解析】试题分析:依题意,得C 2P =PA ,设点P 到C A 的距离为h ,所以∆PAB 与C ∆PB 的面积之比是C1121C 2C 2hS S h ∆PAB∆PB PA⋅PA ===P P ⋅,故选B .考点:三角形的面积. 5.如果函数()cos 4f x x ωπ⎛⎫=+ ⎪⎝⎭()0ω>的相邻两个零点之间的距离为6π,则ω的值为( )(A )3 (B )6 (C )12 (D )24 【答案】B 【解析】试题分析:由题意得:26πT =,解得:3πT =,因为23ππωT ==,所以6ω=,故选B . 考点:三角函数的性质.6.执行如图所示的程序框图,如果输入3x =,则输出k 的值为( )(A )6 (B )8 (C )10 (D )12【答案】C 【解析】试题分析:第1次运行:2339x =⨯+=,2k =,9100>,否;第2次运行:29321x =⨯+=,4k =,21100>,否;第3次运行:221345x =⨯+=,6k =,45100>,否;第4次运行:245393x =⨯+=,8k =,93100>,否;第5次运行:2933189x =⨯+=,10k =,189100>,是,所以输出10k =.故选C .考点:程序框图. 7.在平面区域(){},0112x y x y ≤≤≤≤,内随机投入一点P ,则点P 的坐标(),x y 满足2y x ≤的概率为( ) (A )14 (B )12 (C )23 (D )34【答案】A 【解析】试题分析:作出平面区域,如图所示,其中阴影部分符合2y x ≤,其面积为11111224S =⨯⨯=,正方形的面积为111S =⨯=,所以点P 的坐标(),x y 满足2y x ≤的概率是114S S =,故选A .考点:1、线性规划;2、几何概型. 8.已知()sin 6f x x π⎛⎫=+⎪⎝⎭,若3sin 5α=2πα⎛⎫<<π ⎪⎝⎭,则12f απ⎛⎫+= ⎪⎝⎭( )(A )(B ) (C (D【答案】B 【解析】试题分析:因为3sin 5α=,2παπ<<,所以4c o s 5α==-,所以sin 124f ππαα⎛⎫⎛⎫+=+ ⎪ ⎪⎝⎭⎝⎭34sin coscos sin44525210ππαα=+=⨯-⨯=-,故选B . 考点:1、同角三角函数的基本关系;2、两角和的正弦公式.9.如果1P ,2P ,…,n P 是抛物线C :24y x =上的点,它们的横坐标依次为1x ,2x ,…,n x ,F是抛物线C 的焦点,若1210n x x x +++=,则12n PF P F P F +++=( )(A )10n + (B )20n + (C )210n + (D )220n +【答案】A 【解析】试题分析:抛物线C 的焦点()F 1,0,准线方程是1x =-,由抛物线的定义得:11F 1x P =+,22F 1x P =+,⋅⋅⋅,F 1n n x P =+,所以1212F F F 1nn x x x n n P +P+⋅⋅⋅+P =++⋅⋅⋅++=+,故选A .考点:抛物线的定义.10.一个六棱柱的底面是正六边形,侧棱垂直于底面,所有棱的长都为1,顶点都在同一个球面上,则该球的体积为( )(A )20π (B (C )5π (D 【答案】D 【解析】=R =,所以该球的体积为34V R 3π=343π=⨯=⎝⎭D . 考点:1、六棱柱的外接球;2、球的体积. 11.已知下列四个命题:1p :若直线l 和平面α内的无数条直线垂直,则l α⊥; 2p :若()22x x f x -=-,则x ∀∈R ,()()f x f x -=-;3p :若()11f x x x =++,则()00,x ∃∈+∞,()01f x =; 4p :在△ABC 中,若A B >,则sin sin A B >.其中真命题的个数是( )(A )1 (B )2 (C )3 (D )4【答案】B 【解析】试题分析:若直线l 和平面α内的无数条直线垂直,则l α⊥或//l α,所以1p 是假命题;()22x x f x --=-()()22x x f x -=--=-,所以2p 是真命题;由111x x +=+得:0x =,所以3p 是假命题;a b A >B ⇒>2R sin 2R sin sin sin ⇒A >B ⇒A >B ,所以4p 是真命题.故选B .考点:1、直线与平面的位置关系;2、函数的奇偶性;3、全称命题与特称命题;4、正弦定理.12.如图,网格纸上小正方形的边长为1,粗线画出的是某个四面体的三视图,则该四面体的表面积为( )(A )8+(B )8+(C )2+(D )1224+【答案】A 【解析】试题分析:该四面体是如图中的三棱锥D C -AB ,D B =AB =,1C A =D AB 的底边D A =的表面积是11242422S =⨯⨯+⨯⨯114822+⨯+⨯+A .考点:1、三视图;2、空间几何体的表面积.第Ⅱ卷(非选择题共90分)二、填空题(本大题共4小题,每题5分,满分20分.) 13.函数()33f x x x =-的极小值为 . 【答案】2- 【解析】试题分析:()233f x x '=-,令()0f x '=得1x =±,当1x <-或1x >时,()0f x '>,当11x -<<时,()0f x '<,所以当1x =时,函数()f x 有极小值,且极小值是()311312f =-⨯=-.考点:导数研究函数的极值.14.设实数x ,y 满足约束条件230,230,3x y x y x --≤⎧⎪+-≤⎨⎪≥-⎩, 则23z x y =-+的取值范围是 .【答案】[]6,15- 【解析】试题分析:作出可行域如图所示:由23z x y =-+可得233z y x =+表示的是斜率为23,截距为3z 的平行直线系.当截距最大时,z 最大,当截距最小时,z 最小.当过直线230x y --=与直线230x y +-=的交点()3,0A 时,截距最小,min 2306z =-⨯+=-,当过直线230x y +-=与直线3x =-的交点()3,3B -时,截距最大,()max 233315z =-⨯-+⨯=,所以23z x y =-+的取值范围是[]6,15-. 考点:线性规划.15.已知双曲线C :22221x y a b-=()0,0a b >>的左顶点为A ,右焦点为F ,点()0,B b ,且0BA BF =,则双曲线C 的离心率为 .【答案】12【解析】试题分析:由题意得:(),0a A -,()F ,0c ,所以(),a b BA =--,()F ,c b B =-,因为F 0BA⋅B =,所以20b ac -=,因为222b c a =-,所以220c ac a --=,两边同除以2a ,得210e e --=,解得:e =(舍去)或e =. 考点:1、双曲线的简单几何性质;2、平面向量的坐标运算.16.在△ABC 中,点D 在边AB 上,CD BC ⊥,AC =5CD =,2BD AD =,则AD 的长为 . 【答案】5考点:余弦定理.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.(本小题满分12分)已知数列{}n a 是等比数列,24a =,32a +是2a 和4a 的等差中项. (Ⅰ)求数列{}n a 的通项公式;(Ⅱ)设22log 1n n b a =-,求数列{}n n a b 的前n 项和n T . 【答案】(I )2n n a =;(II )()16232n n n +T =+-⨯.【解析】试题分析:(I )设数列{}n a 的公比,由题意列出关于q 的方程,解出q ,进而可得数列{}n a 的通项公式;(II )先求出数列{}n b 的通项公式,再利用错位相减法可得数列{}n n a b 的前n 项和n T .试题解析:(Ⅰ)设数列{}n a 的公比为q , 因为24a =,所以34a q =,244a q =.…………………………………………1分因为32a +是2a 和4a 的等差中项,所以()32422a a a +=+.……………………2分 即()224244q q +=+,化简得220q q -=.因为公比0q ≠,所以2q =.………………………………………………………4分 所以222422n n n n a a q --==⨯=(*n ∈N ).…………………………………………5分(Ⅱ)因为2n na =,所以22log 121n nb a n =-=-.所以()212nn n a b n =-.……………………………………………………………7分 则()()231123252232212n n n T n n -=⨯+⨯+⨯+⋅⋅⋅+-+-, ①()()23412123252232212n n n T n n +=⨯+⨯+⨯+⋅⋅⋅+-+-. ②………………9分①-②得,()2312222222212n n n T n +-=+⨯+⨯+⋅⋅⋅+⨯--……………………………………10分()()()11142221262321212n n n n n ++-=+⨯--=-----,所以()16232n n T n +=+-.……………………………………………………………12分 考点:1、等比数列的通项公式;2、数列求和. 18.(本小题满分12分)从某企业生产的某种产品中抽取100件,测量这些产品的质量指标值,由测量结果得到如图所示的频率分布直方图,质量指标值落在区间[)55,65,[)65,75,[]75,85内的频率之比为4:2:1. (Ⅰ)求这些产品质量指标值落在区间[]75,85内的频率;(Ⅱ)用分层抽样的方法在区间[)45,75内抽取一个容量为6的样本,将该样本看成一个总体,从中任意抽取2件产品,求这2件产品都在区间[)45,65内的概率.【答案】(I )0.05;(II )23. 【解析】试题分析:(I )利用频率分布直方图中所有频率之和等于1可得这些产品质量指标值落在区间[]75,85内的频率;(II )先算出落在区间[)45,55,[)55,65,[)65,75内的产品件数,再列举出从6件产品中任意抽取2件产品的基本事件和这2件产品都在区间[)45,65内的基本事件,进而利用古典概型公式可得这2件产品都在区间[)45,65内的概率. 试题解析:(Ⅰ)设区间[]75,85内的频率为x ,则区间[)55,65,[)65,75内的频率分别为4x 和2x .…………………………1分 依题意得()0.0040.0120.0190.03010421x x x +++⨯+++=,……………3分 解得0.05x =.所以区间[]75,85内的频率为0.05.………………………………………………4分 (Ⅱ)由(Ⅰ)得,区间[)45,55,[)55,65,[)65,75内的频率依次为0.3,0.2,0.1. 用分层抽样的方法在区间[)45,75内抽取一个容量为6的样本,则在区间[)45,55内应抽取0.3630.30.20.1⨯=++件,记为1A ,2A ,3A .在区间[)55,65内应抽取0.2620.30.20.1⨯=++件,记为1B ,2B . 在区间[)65,75内应抽取0.1610.30.20.1⨯=++件,记为C .…………………6分 设“从样本中任意抽取2件产品,这2件产品都在区间[)45,65内”为事件M ,则所有的基本事件有:{}12,A A ,{}13,A A ,{}11,A B ,{}12,A B ,{}1,A C ,{}23,A A ,{}21,A B ,{}22,A B ,{}2,A C ,{}31,A B ,{}32,A B ,{}3,A C ,{}12,B B ,{}1,B C ,{}2,B C ,共15种.…………………………………………………………………8分事件M 包含的基本事件有:{}12,A A ,{}13,A A ,{}11,A B ,{}12,A B ,{}23,A A ,{}21,A B ,{}22,A B ,{}31,A B ,{}32,A B ,{}12,B B ,共10种.…………10分所以这2件产品都在区间[)45,65内的概率为102153=.………………………12分考点:1、频率分布直方图;2、古典概型;3、分层抽样. 19.(本小题满分12分)如图,四棱柱1111ABCD A BC D -的底面ABCD 是菱形,ACBD O =,1AO ⊥底面ABCD ,21==AA AB .(Ⅰ)证明:BD ⊥平面1ACO ; (Ⅱ)若60BAD ∠=,求点C 到平面1OBB 的距离.【答案】(I )证明见解析;(II . 【解析】试题分析:(I )由题意可证1D A O ⊥B ,C D O ⊥B ,进而可证D B ⊥平面1C A O ;(II)先将点1B 到平面CD AB 的距离转化为点1A 到平面CD AB 的距离,再利用等积法可得点C 到平面1OBB 的距离.试题解析:(Ⅰ)证明:因为1AO ⊥平面ABCD ,BD ⊂平面ABCD , 所以1AO ⊥BD .……………………………………………………………………1分 因为ABCD 是菱形,所以CO ⊥BD .……………………………………………2分 因为1AO CO O =,1AO ,CO ⊂平面1ACO , 所以BD ⊥平面1A CO .……………………………………………………………3分(Ⅱ)解法一:因为底面ABCD 是菱形,ACBD O =,21==AA AB ,60BAD ∠=,所以1OB OD ==,OA OC =4分所以OBC ∆的面积为112212OBC S OB OC ∆==⨯=⨯⨯5分 因为1AO ⊥平面ABCD ,AO ⊂平面ABCD , 所以1AO AO ⊥,11AO ==.………………………………………6分因为11A B 平面ABCD ,所以点1B 到平面ABCD 的距离等于点1A 到平面ABCD 的距离1AO .…………7分 由(Ⅰ)得,BD ⊥平面1A AC .因为1A A ⊂平面1AAC ,所以BD ⊥1A A . 因为11A AB B ,所以BD ⊥1B B .………………………………………………8分所以△1OBB 的面积为111121212OBB S OB BB ∆=⨯⨯==⨯⨯.……………………9分 设点C 到平面1OBB 的距离为d , 因为11C OBB B OBC V V --=,所以111133OBB OBC S d S A O D D =gg .………………………………………………10分所以111212OBC OBBS AO d S ∆∆⋅===所以点C 到平面1OBB的距离为2.……………………………………………12分 解法二:由(Ⅰ)知BD ⊥平面1ACO , 因为BD ⊂平面11BB D D ,所以平面1ACO ⊥平面11BB D D .…4分 连接11AC 与11B D 交于点1O ,连接1CO ,1OO ,因为11AA CC =,11//AA CC ,所以11CAAC 为平行四边形. 又O ,1O 分别是AC ,11AC 的中点,所以11OAO C 为平行四边形. 所以111OC OA ==.…………………………………………………………………6分 因为平面11OAO C 与平面11BB D D 交线为1OO , 过点C 作1CH OO ⊥于H ,则CH ⊥平面11BB D D .………………………………8分 因为11O CA O ,1AO ⊥平面ABCD ,所以·1O C ⊥平面ABCD . 因为OC ⊂平面ABCD ,所以·1O C ⊥OC ,即△1OCO 为直角三角形.………10分所以1122O C OC CH OO ⋅===.所以点C 到平面1OBB212分考点:1、线面垂直;2、点到平面的距离. 20.(本小题满分12分)已知椭圆C 的中心在坐标原点,焦点在x 轴上,左顶点为A ,左焦点为()120F -,,点(B 在椭圆C 上,直线()0y kx k =≠与椭圆C 交于E ,F 两点,直线AE ,AF 分别与y 轴交于点M ,N .(Ⅰ)求椭圆C 的方程;(Ⅱ)在x 轴上是否存在点P ,使得无论非零实数k 怎样变化,总有MPN ∠为直角?若存在,求出点P 的坐标;若不存在,请说明理由.【答案】(I )22184x y +=;(II)()2,0或()2,0-.解法二:设椭圆C 的方程为22221(0)x y a b a b+=>>,因为椭圆的左焦点为()120F -,,所以224a b -=. ①…………………1分因为点(2B 在椭圆C 上,所以22421a b +=. ②…………………2分由①②解得,a =2b =.…………………………………………………3分所以椭圆C 的方程为22184x y +=.………………………………………………4分(Ⅱ)解法一:因为椭圆C 的左顶点为A ,则点A 的坐标为()-.…………5分因为直线(0)y kx k =≠与椭圆22184x y +=交于两点E ,F , 设点()00,E x y (不妨设00x >),则点()00,F x y --.联立方程组22,184y kx x y =⎧⎪⎨+=⎪⎩消去y 得22812x k =+.所以0x =0y =6分所以直线AE的方程为y x =+.……………………………7分因为直线AE 与y 轴交于点M ,令0x =得y =M ⎛ ⎝.……………………8分同理可得点N ⎛ ⎝.…………………………………………………9分 假设在x 轴上存在点(,0)P t ,使得MPN ∠为直角,则0MP NP ⋅=.………10分即20t =,即240t -=.………………………11分解得2t =或2t =-.故存在点()2,0P 或()2,0P -,无论非零实数k 怎样变化,总有MPN ∠为直角.…………12分解法二: 因为椭圆C 的左端点为A ,则点A的坐标为()-.……………5分因为直线(0)y kx k =≠与椭圆22184x y +=交于两点E ,F , 设点00(,)E x y ,则点00(,)F x y --. 所以直线AE的方程为y x =+.………………………………6分 因为直线AE 与y 轴交于点M ,令0x =得y =,即点M ⎛⎫⎝.……………………………7分同理可得点N ⎛ ⎝.……………………………………………………8分 假设在x 轴上存在点(),0P t ,使得MPN ∠为直角,则0MP NP ⋅=.即20t =,即222808y t x +=-. (※)…………9分 因为点00(,)E x y 在椭圆C 上,所以2200184x y +=,即220082x y -=.……………………………………………10分将220082x y -=代入(※)得240t -=.………………………………………11分解得2t =或2t =-.故存在点()2,0P 或()2,0P -,无论非零实数k 怎样变化,总有MPN ∠为直角.………………12分解法三:因为椭圆C 的左顶点为A ,则点A的坐标为()-.……………5分因为直线(0)y kx k =≠与椭圆22184x y +=交于两点E ,F ,设点(),2sin E θθ(0θ<<π),则点(),2sin F θθ--.……6分所以直线AE的方程为y x =+.………………………7分因为直线AE 与y 轴交于点M ,令0x =得2sin cos 1y θθ=+,即点2sin 0,cos 1M θθ⎛⎫⎪+⎝⎭.………………………………8分同理可得点2sin 0,cos 1N θθ⎛⎫⎪-⎝⎭.………………………………………………………9分假设在x 轴上存在点(,0)P t ,使得MPN ∠为直角,则0MP NP ⋅=.………10分 即22sin 2sin 0cos 1cos 1t θθθθ--+⨯=+-,即240t -=.…………………………………11分解得2t =或2t =-.故存在点()2,0P 或()2,0P -,无论非零实数k 怎样变化,总有MPN ∠为直角.………12分考点:1、椭圆的标准方程;2、椭圆的简单几何性质;3、直线与圆锥曲线的位置关系. 21.(本小题满分12分)已知函数()e ln 1x f x m x =--.(Ⅰ)当1m =时,求曲线()y f x =在点()()11f ,处的切线方程; (Ⅱ)当1m ≥时,证明:()1f x >. 【答案】(I )()1y e x =-(II )证明见解析. 【解析】试题分析:(I )先代入1m =,对()f x 求导数,再算出()1f ',()1f ,进而可得曲线()y f x =在点()()1,1f 处的切线方程;(II )先构造函数()ln 2xg x e x =--,再利用导数可得()g x 的最小值,,进而可证当1m ≥时,()1f x >.试题解析:(Ⅰ)解:当1m =时,()e ln 1x f x x =--,所以1()e x f x x'=-.………………………………………………………………1分 所以(1)e 1f =-,(1)e 1f '=-. …………………………………………………2分 所以曲线()y f x =在点()()11f ,处的切线方程为(e 1)(e 1)(1)y x --=--. 即()e 1y x =-.………………………………………………………………………3分 (Ⅱ)证法一:当1m ≥时,()e ln 1e ln 1x xf x m x x =--≥--.要证明()1f x >,只需证明e ln 20xx -->.……………………………………4分 以下给出三种思路证明e ln 20xx -->.思路1:设()e ln 2xg x x =--,则1()e x g x x'=-. 设1()e xh x x =-,则21()e 0xh x x'=+>, 所以函数()h x =1()e xg x x'=-在0+∞(,)上单调递增.…………………………6分因为121e 202g ⎛⎫'=-< ⎪⎝⎭,(1)e 10g '=->,所以函数1()e xg x x '=-在0+∞(,)上有唯一零点0x ,且01,12x ⎛⎫∈ ⎪⎝⎭.…………8分 因为0()0g x '=时,所以01ex x =,即00ln x x =-.………………………………9分 当()00,x x ∈时,()0g x '<;当()0,x x ∈+∞时,()0g x '>.所以当0x x =时,()g x 取得最小值()0g x .……………………………………10分 故()000001()=e ln 220xg x g x x x x ≥--=+->. 综上可知,当1m ≥时,()1f x >.………………………………………………12分 思路2:先证明e 1xx ≥+()x ∈R .………………………………………………5分 设()e 1xh x x =--,则()e 1xh x '=-.因为当0x <时,()0h x '<,当0x >时,()0h x '>,所以当0x <时,函数()h x 单调递减,当0x >时,函数()h x 单调递增. 所以()()00h x h ≥=.所以e 1xx ≥+(当且仅当0x =时取等号).………………………………………7分 所以要证明e ln 20xx -->,只需证明()1ln 20x x +-->.……………………………………………………8分 下面证明ln 10x x --≥. 设()ln 1p x x x =--,则()111x p x x x-'=-=. 当01x <<时,()0p x '<,当1x >时,()0p x '>,所以当01x <<时,函数()p x 单调递减,当1x >时,函数()p x 单调递增. 所以()()10p x p ≥=.所以ln 10x x --≥(当且仅当1x =时取等号).………………………………10分由于取等号的条件不同, 所以e ln 20xx -->.综上可知,当1m ≥时,()1f x >.………………………………………………12分 (若考生先放缩ln x ,或e x、ln x 同时放缩,请参考此思路给分!) 思路3:先证明e ln 2xx ->.因为曲线e x y =与曲线ln y x =的图像关于直线y x =对称,设直线x t =()0t >与曲线e xy =,ln y x =分别交于点A ,B ,点A ,B 到直线y x = 的距离分别为1d ,2d ,则)12AB d d +. 其中1t d =2d =()0t >.①设()e t h t t =-()0t >,则()e 1t h t '=-. 因为0t >,所以()e 10t h t '=->.所以()h t 在()0,+∞上单调递增,则()()01h t h >=.所以1t d =>. ②设()ln g t t t =-()0t >,则()111t g t t t -'=-=.因为当01t <<时,()0g t '<;当1t >时,()0g t '>,所以当01t <<时,()ln g t t t =-单调递减;当1t >时,()ln g t t t =-单调递增. 所以()()11g t g ≥=.所以2d =≥所以)122AB d d +>=⎭. 综上可知,当1m ≥时,()1f x >.………………………………………………12分证法二:因为()e ln 1x f x m x =--,要证明()1f x >,只需证明e ln 20xm x -->.…………………………………4分以下给出两种思路证明e ln 20xm x -->.思路1:设()e ln 2x g x m x =--,则1()e xg x m x'=-. 设1()e xh x m x =-,则21()e 0xh x m x'=+>. 所以函数()h x =()1e xg x m x'=-在()0+∞,上单调递增.……………………6分因为11221e 2e 202m mg m m m m ⎛⎫⎛⎫'=-=-< ⎪ ⎪⎝⎭⎝⎭,()1e 10g m '=->,所以函数1()e xg x m x '=-在()0+∞,上有唯一零点0x ,且01,12x m ⎛⎫∈⎪⎝⎭.……8分 因为()00g x '=,所以01ex m x =,即00ln ln x x m =--.……………………9分 当()00,x x ∈时,()0g x '<;当()0,x x ∈+∞时,()0g x '>.所以当0x x =时,()g x 取得最小值()0g x .……………………………………10分 故()()000001e ln 2ln 20xg x g x m x x m x ≥=--=++->. 综上可知,当1m ≥时,()1f x >.………………………………………………12分 思路2:先证明e 1()xx x ≥+∈R ,且ln 1(0)x x x ≤+>.……………………5分 设()e 1xF x x =--,则()e 1x F x '=-.因为当0x <时,()0F x '<;当0x >时,()0F x '>,所以()F x 在(,0)-∞上单调递减,在(0,)+∞上单调递增. 所以当0x =时,()F x 取得最小值(0)0F =.所以()(0)0F x F ≥=,即e 1xx ≥+(当且仅当0x =时取等号).……………7分 由e 1()x x x ≥+∈R ,得1ex x -≥(当且仅当1x =时取等号).………………8分所以ln 1(0)x x x ≤->(当且仅当1x =时取等号).……………………………9分 再证明e ln 20xm x -->.因为0x >,1m ≥,且e 1xx ≥+与ln 1x x ≤-不同时取等号,所以()()e ln 2112x m x m x x -->+---()()11m x =-+0≥.综上可知,当1m ≥时,()1f x >.………………………………………………12分 考点:1、导数的几何意义;2、利用导数研究函数的单调性;3、利用导数研究函数的最值;4、不等式的证明.请考生在第22、23、24三题中任选一题作答,如果多做,则按所做的第一题记分.解答时请写清题号.22.(本小题满分10分)选修4-1:几何证明选讲如图所示,△ABC 内接于⊙O ,直线AD 与⊙O 相切于点A ,交BC 的延长线于点D ,过点D 作DE CA 交BA 的延长线于点E .(Ⅰ)求证:2DE AE BE =;(Ⅱ)若直线EF 与⊙O 相切于点F ,且4EF =,2EA =,求线段AC 的长.【答案】(I )证明见解析;(II )3.【解析】(I )利用弦切角定理和D //C E A 证D ∆AE ∽D ∆EB ,进而可证2D E =AE⋅BE ;(II )先利用切割线定理可得EB 和AB ,利用(I )的结论可得D E ,再由D //C E A 可得C ∆BA ∽D ∆EB ,进而可得C A .试题分析:试题解析:(Ⅰ)证明:因为AD 是⊙O 的切线, 所以DAC B ∠=∠(弦切角定理).………………1分 因为DECA ,所以DAC EDA ∠=∠.……………………………2分 所以EDA B ∠=∠.因为AED DEB ∠=∠(公共角),所以△AED ∽△DEB .……………………………………………………………3分 所以DE AE BEDE=.即2DE AE BE =.…………………………………………………………………4分 (Ⅱ)解:因为EF 是⊙O 的切线,EAB 是⊙O 的割线,所以2EF EA EB = (切割线定理).……………………………………………5分 因为4EF =,2EA =,所以8EB =,6AB EB EA =-=.…………………7分 由(Ⅰ)知2DE AE BE =,所以4DE =.………………………………………8分 因为DE CA ,所以△BAC ∽△BED . ………………………………………9分所以BA ACBEED =.所以6438BA EDAC BE⋅⨯===. …………………………………………………10分考点:1、相似三角形的判定定理;2、弦切角定理;3、切割线定理. 23.(本小题满分10分)选修4-4:坐标系与参数方程在平面直角坐标系xOy 中,以坐标原点O 为极点,x 轴正半轴为极轴建立极坐标系,曲线C 的极坐标方程为θρsin 2=,[)0,2θ∈π.(Ⅰ)求曲线C 的直角坐标方程;(Ⅱ)在曲线C 上求一点D ,使它到直线l:32x y t ⎧=⎪⎨=-+⎪⎩(t 为参数,t ∈R )的距离最短,并求出点D 的直角坐标.【答案】(I )2220x y y +-=(或()2211x y +-=);(II)32⎫⎪⎪⎝⎭. 【解析】试题分析:(I )先两边同乘ρ得22sin ρρθ=,再利用222x y ρ=+,sin y ρθ=可得曲线C 的直角坐标方程;(II )先消去t 可得直线l 的普通方程,再设点D 的坐标,利用垂直可得0x ,进而检验可得点D 的坐标.试题解析:(Ⅰ)解:由θρsin 2=,[)0,2θ∈π,可得22sin ρρθ=.…………………………………………………………………1分 因为222x y ρ=+,sin y ρθ=,…………………………………………………2分所以曲线C 的普通方程为2220x y y +-=(或()2211x y +-=). …………4分(Ⅱ)解法一:因为直线的参数方程为32x y t ⎧=⎪⎨=-+⎪⎩(t 为参数,t ∈R ),消去t 得直线l的普通方程为5y =+. ……………………………………5分因为曲线C :()2211x y +-=是以G ()1,0为圆心,1为半径的圆,设点()00,D x y ,且点D 到直线l:5y =+的距离最短, 所以曲线C 在点D 处的切线与直线l:5y =+平行. 即直线GD 与l 的斜率的乘积等于1-,即(0011y x -⨯=-.………………7分 因为()220011x y +-=,解得02x =-或02x =.所以点D 的坐标为12⎛⎫ ⎪ ⎪⎝⎭,或32⎫⎪⎪⎝⎭,.……………………………………9分由于点D 到直线5y =+的距离最短,所以点D 的坐标为32⎫⎪⎪⎝⎭,.……………………………………………………10分解法二:因为直线l 的参数方程为32x y t ⎧=+⎪⎨=-+⎪⎩(t 为参数,t ∈R ),消去t 得直线l 50y +-=.……………………………………5分因为曲线C ()2211x y +-=是以G ()1,0为圆心,1为半径的圆,因为点D 在曲线C 上,所以可设点D ()cos ,1sin ϕϕ+[)()0,2ϕ∈π.………7分所以点D 到直线l 的距离为d =2sin 3ϕπ⎛⎫=-+ ⎪⎝⎭.………………………………8分 因为[)0,2ϕ∈π,所以当6ϕπ=时,min 1d =.…………………………………9分此时D 32⎫⎪⎪⎝⎭,,所以点D 的坐标为32⎫⎪⎪⎝⎭,.……………………………10分 考点:1、极坐标方程与直角坐标方程的互化;2、参数方程与普通方程的互化. 24.(本小题满分10分)选修4-5:不等式选讲设函数()f x x x =+-. (Ⅰ)当1a =时,求不等式()12f x ≥的解集; (Ⅱ)若对任意[]0,1a ∈,不等式()f x b ≥的解集为空集,求实数b 的取值范围.【答案】(I )1,4⎡⎫-+∞⎪⎢⎣⎭;(II ))+∞.【解析】试题分析:(I )先代入1a =得()1f x x x =+-,写出分段函数,再求解()12f x ≥,进而可得实数x 的取值范围;(II)先由已知条件得()max b f x >⎡⎤⎣⎦,再利用绝对值不等式可得()f x 的最大值,进而利用基本不等式可得实数b 的取值范围. 试题解析:(Ⅰ)解:当1a =时,()12f x ≥等价于112x x +-≥.……………………1分 ①当1x ≤-时,不等式化为112x x --+≥,无解; ②当10x -<<时,不等式化为112x x ++≥,解得104x -≤<;③当0x ≥时,不等式化为112x x +-≥,解得0x ≥.…………………………3分综上所述,不等式()1≥x f 的解集为1,4⎡⎫-+∞⎪⎢⎣⎭.………………………………4分 (Ⅱ)因为不等式()f x b ≥的解集为空集,所以()max b f x >⎡⎤⎣⎦.…………………5分 以下给出两种思路求()f x 的最大值.思路1:因为()f x x x =+- ()01a ≤≤,当x ≤()f x x x =-=0<.当x <时,()f x x x =2x =≤=当x ≥()f x x x =+=所以()max f x ⎡⎤⎣⎦=7分思路2:因为 ()f x x x =x x ≤+==当且仅当x ≥ 所以()max f x ⎡⎤⎣⎦=7分因为对任意[]0,1a ∈,不等式()f x b ≥的解集为空集,所以max b >.………………………………………………………8分以下给出三种思路求()g a =.思路1:令()g a = 所以()21ga =+2212≤++=.=12a =时等号成立. 所以()max g a =⎡⎤⎣⎦所以b的取值范围为)∞.…………………………………………………10分思路2:令()g a =因为01a ≤≤,所以可设2cos a θ= 02θπ⎛⎫≤≤⎪⎝⎭, 则()g a=cos sin 4θθθπ⎛⎫+=+≤ ⎪⎝⎭当且仅当4θπ=时等号成立. 所以b的取值范围为)∞.…………………………………………………10分思路3:令()g a =因为01a ≤≤,设x y ìï=ïíï=ïî则221x y +=()01,01x y##.问题转化为在221x y +=()01,01xy ##的条件下,求z x y =+的最大值.利用数形结合的方法容易求得z此时2x y ==.所以b的取值范围为)∞.…………………………………………………10分考点:1、绝对值不等式;2、恒成立问题;3、基本不等式.。
2016届广州市普通高中毕业班综合测试(二)理综试题
2016届广州市普通高中毕业班综合测试(二)理科综合注意事项:1. 本试卷分第I卷(选择题)和第II卷(非选择题)两部分。
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,并用铅笔在答题卡上的相应位置填涂考生号。
2. 回答第I卷时,选出每小题答案后,用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。
写在本试卷上无效。
3. 回答第II卷时,将答案写在答题卡上。
写在本试卷上无效。
4. 考试结束后,将本试卷和答题卡一并交回。
可能用到的相对原子质量:H-1 C-12 O-16 Mg-24 Cl-35.5 K-39 V-51 Ni-59第I卷一、选择题:本题包括13小题,每小题6分,共78分。
在每小题给出的四个选项中,只有一个选项符合题目要求。
1.下列物质进入细胞需要载体蛋白的是A.酒精进人胃粘膜细胞B.CO2进入叶肉细胞C.钾离子进入丽藻细胞D.母乳中的抗体进入婴儿小肠上皮细胞2.下列关于人体细胞衰老的叙述,正确的是A.细胞衰老只发生在老年期B.衰老细胞的形态、结构和功能发生改变C.衰老细胞的新陈代谢速率加快D.“自由基学说”认为DNA随着细胞分裂会周期性变短3.下列关于人体内环境稳态的叙述中,错误..的是A.血浆渗透压与蛋白质、无机盐含量有关B.甲状腺激素和尿素均可存在于内环境中C.某些激素可以影响神经系统的发育和功能D.内环境是细胞代谢的主要场所4.下列关于生物进化的叙述,错误..的是A.捕食者的存在有利于增加物种多样性B.外来物种入侵可改变生物进化的方向和速度C.自然选择会导致种群基因频率定向改变D.共同进化就是不同物种之同在相互影响中不断进化和发展的过程5.某种植物的株高由三对独立遗传的基因控制,每个显性基因对植物株高的增加效应相同且能叠加。
已知隐性纯合子和显性纯合子的株高分别为10cm和82cm。
现将三对基因均杂合的两植株杂交,F1中株高为70cm的个体所占比例为A.1/32 B. 3/32 C. 3/16 D. 31/326.下列关于植物激素的叙述,错误..的是A.植物激素直接参与细胞代谢发挥调节作用B.一定浓度的赤霉素可促进植物种子萌发C.顶端优势的现象体现了生长素作用的两重性D.多种植物激素相互作用共同调节植物的生长7.合成氨是人类科学技术上的一项重大突破,曾三次获诺贝尔化学奖。
2016年广州市普通高中毕业班综合测试(一)(地理)含答案
2016年广州市普通高中毕业班综合测试(一)文科综合第І卷从“南粮北运”到“北粮南运“,中国农业经历千年形成的粮食产销格局在最近的30年间被彻底改变了,下图为中国粮食产销格局示意图,读图完成1-3题。
1.引起“南粮北运”向“北粮南运”转变的最主要原因是A.全球气候变化B.城市化地区差异C.人口分布变化D.土地生产力差异2.影响东北地区成为我国最大商品粮基地的最主要原因是A.人均耕地多B.高速铁路的建设C.全球变暖、热量条件改善D.新品种、新技术广泛推广3.当前,制约“北粮南运”的最主要因素是A.气候条件B.铁路运输C.生产水平D.市场需求一对英国夫妇1月3日从佛得角群岛(约17°N,23°W)出发,驾驶游艇前往安提瓜岛(约17°02′N,61°50′W),如下图所示,1月9日航程近半时,游艇发生动力故障,只能随水漂流,30天后在M地被救起。
运用所学知识,完成4-6题。
4.佛得角群岛上的植被主要是A.热带雨林B.常绿硬叶林C.热带荒漠D.热带季雨林5.出发地与计划目的地之间的最短距离约为A.2100公里B.4100公里C.6100公里D.8100公里6.M地最可能位于A.安提瓜岛B.亚马孙河口C.直布罗陀海峡D.百慕大群岛下图为2010年上海人口年龄结构示意图,读图完成7-9题。
7.上海流入人口比例最大的年龄段是A.20~24岁B.25~29岁C.30~34岁D.35~39岁8.上海是我国人口老龄化程度下降最快的城市之一,主要原因是A.出生率大幅下降B.死亡率迅速上升C.大量年轻劳力流入D.大量老龄人口迁出9.目前,上海市流动人口正处在A.劳动力个体流动阶段B.老人跟着孩子流动阶段C.夫妻带着孩子流动阶段D.农村流动人口回流阶段下图是北德维纳河流域示意图,该流域的地形是宽阔的、多丘陵的平原,据此完成10-11题。
10.影响该流域冬季降水的海洋气流主要来自A.南方B.北方C.东方D.西方11.该河流主要的补给水源是①雨水②季节性积雪融水③冰川融水④湖泊水A.①②B.③④C.①③D.②④第ІІ卷36.(22分)阅读图文材料,完成下列要求。
广州市2016届普通高中毕业班综合测试(一)英语试题
2016年广州市普通高中毕业班综合测试(一)英语2016.3 本试卷分第1卷(选择题)和第Ⅱ卷(非选择题)两部分。
考试结束后,将本试卷和答题卡一并交回。
第I卷第一部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
A1. Who is organizing the Festival of Books?A. Book TV.B. The Los Angeles Times.C. Natural History Museum.D. Los Angeles Public Library.2. How are the best book prizes chosen?A. By a group of experts.B. By a vote of the authors,C. By the reading public.D. By festival sponsors.3. What is true about tickets to the festival?A. They are only for published authors.B. They include transport to eachevent.C. They can only be applied for onlineD. They can be purchased at the eventlocation.BIn 1800, only three percent of the world's population lived in cities. Only one city —Beijing —had a population of over a million. Most people lived in rural areas, and never saw a city in their lives. In 1900, just a hundred years later, roughly 150 million people lived in cities, and the world’s ten largest cities all had populations exceeding one million. By 2000, the number of city dwellers exceeded three billion; and in 2008, the world's population crossed a tipping point —more than half of the people on Earth lived in cities. By 2050, that could increase to over two-thirds. Clearly, humans have become an urban species.In the 19th and early 20th centuries, many people viewed cities negatively —crowded, dirty environments full of disease and crime. They feared that as cities got bigger, living conditions would worsen. In recent decades, however, attitudes have changed. Many experts now think urbanization (城市化) is good news, offering solutions to the problems of Earth’s growing population.Harvard economist Edward Glaeser, author of The Victory of the Cir, is one such person. Glaeser argues that cities are very productive because “the absence of space between people” reduces the cost of transporting goods, people, and ideas. While the flow of goods has always been important to cities, what is most important today is the flow of ideas. Successful cities enable people to learn from each other easily, and attract and reward smart people with higher wages.Another urbanization supporter is environmentalist Stewart Brand. Brand believes cities help the environment because they allow haft of the world’s population to live on about four percent of the land. This leaves more space for open country, such as farmland. City dwellers also have less impact per person on the environment than people in the countryside. Their roads, sewers, and power lines need fewer resources to build and operate. City apartments require less energy to heat, cool, and light. Most importantly, people in cities drive less so they produce fewer greenhouse gases per person.So it’s a mistake to see urbanization as evil; it’s a natural part of development. The challenge is how to manage the growth.4. What is Paragraph 1 mainly about?A. The history of modern cities.B. Changes taking place within cities.C. How cities have grown over time.D. Why modern cities are changing.5. How have experts’ attitudes towards cities changed in recent decades?A. They now view the weaknesses as strengths.B. They no longer see city-riving as attractive.C. They accept city life in spite of its problems.D. They think city-riving provides more benefits.6. Which of the following would Edward Glaeser agree with?A. Cities provide more economic opportunities.B. City people get along better with each other.C. Over-crowded cities result in problems.D. Cities limit the flow of ideas.7. According to Paragraph 4, what would be the result of moving people out of cities?A. Economic production would be reduced.B. There would be less farmland available.C. People would travel less frequently.D. House values would fall greatly.CIt was December 25, 1914, only 5 months into World War I. German, British, and French soldiers, already sick and tired of the senseless killing, disobeyed their superiors and started socialising with “the enemy”along two-thirds of the Western Front.German troops held up Christmas trees with signs, “Merry Christmas”. Thousands of troops ran across the battlefields covered with dead bodies. They sang Christmas songs, exchanged photographs of loved ones back home, shared food and even played football. Soldiers hugged men they had been trying to kill a few short hours before. They agreed to warn each other if their officers forced them to fire their weapons, and to aim high.Fear ran through the military leaders on both sides. They felt that their power was being challenged: soldiers declaring their brotherhood with each other and refusing to fight. Generals declared this unexpected peacemaking illegal and said that participating soldiers would face a military court. Those found guilty would be imprisoned or even shot. By March 1915 the socialising movement had been destroyed and the killing machine was back in full operation. Over the next three years more than fifteen million people died in the war.Not many people have heard the story of the Christmas Truce (休战). On Christmas Day, 1988, a local radio host in Boston played “Christmas in the Trenches”, a song about the Christmas Truce, several times and was stunned by his listeners’response. Thousands of people called in, praising the song, with many moved to tears by the amazing events it described.You can probably guess why the callers were in tears. The Christmas Truce story goes against most of what we have been taught about people. It lets us see the worldas it can be and says, “This really happened once.” It shows us the potential we have as humans, and contradicts all of those TV and newspaper stories that tell us how mean and heartless people are. It is like hearing that our deepest wishes really are true: the world really can be different.8. The soldiers began socialising with the enemy because ________.A. they couldn't bear the meaningless killingB. it was the best way to avoid being killedC. they feared that they would be caughtD. their enemies were from similar backgrounds9. How did the generals finally put an end to the soldiers’ socialising?A. They sent the soldiers’ loved ones to prison.B. They moved the two groups of soldiers further apart.C. They increased the number of officers to control the soldiers.D. They warned the soldiers that they would face severe punishment.10. The underlined word “stunned” in Paragraph 4 most probably means ________.A. satisfiedB. shockedC. amusedD. confused11. The author uses the story of the soldiers to imply that human beings ________.A. are not trustworthy under stressB. are naturally aggressive and war-likeC. are basically caring and kindheartedD. will always do what is in their own self-interestDEarlier this year, the social media website Facebook announced that it would work with several news organisations —including The New York Times, The Guardian, and the BBC—to place news stories directly into users’personal Facebook webpage. Stories published using Facebook Instant will load more quickly and keep the style of the original publisher, who will keep all the advertising income the stories earn —at least for now. The deal shows how important social, media has become to news organisations, and is a clear sign of how the world of news is changing —and has been for a while.When Google News began in 2002, many saw it as the death of the newspaper. It had no human editor. Instead Google used, and still uses, a secret computer program that selects and displays news stories according to the reader’s personal interests. More recently, Associated Press and Yahoo!have been punishing computer-written articles. Both use special software to automatically produce stories about company financial results and sports reports —areas where the quality of writing is felt to be of secondary importance to the accuracy of the data.Should we be worried about such developments? I think we should. One concern is that facebook, Google and other social media websites see journalism as a sideline, a way of putting people in front of advertisements. It isn’t their primary function —so if it stops making them lots of money, they're likely to stop doing it.There’s also a concern that computer-written articles are not actually journalismat all, because what a human news team produces is actually quite complex. A well-written news story puts information in context, offers a voice to each side of an argument and brings the public new knowledge.Though economics and speed of delivery mean readers will probably choose a computer-written story over a carefully shaped article —at least for daily news —I don't think the computers will be writing any in-depth articles for a while yet.12. What is the main purpose of the article?A. To report on a new computer service offered by Facebook..B. To advise readers against reading computer-written news.C. To express concern about recent trends in online news.D. To describe the process of online news reporting.13. Computer-written news reports have so far focused on sports and finance because ________.A. these are the most popular topics for online readersB. there are fewer journalists specialising in these areasC. information on these topics is more easily availableD. writing style is less important than accuracy in these areas14. What does the underlined word “It” in Paragraph 3 refer to?A. Journalism.B. Advertising.C. Facebook.D. Business.15. In Paragraph 4, which of the following is mentioned as a characteristic of a well-written news article?A. The information presented is up-to-date.B. The author's opinion is clear.C. Different views on the topic are presented.D. The language used is vivid.第二节(共5小题,每小题2分,满分l0分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
2016年广州市普通高中毕业班综合测试
2016年广州市普通高中毕业班综合测试(二)数学(文科)第Ⅰ卷一. 选择题:本大题共12小题,每小题5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{}0,1,2M =,{11,N x x =-≤≤x ∈Z}, 则(A)M N ⊆ (B) N M ⊆ (C) {}0,1M N = (D) M N N = (2) 已知()1i i +=a b +i (,a b ∈R ),其中i 为虚数单位,则a b +的值为 (A) 1-(B)0 (C)1 (D) 2 (3) 已知等比数列{}n a 的公比为12-, 则135246a a a a a a ++++的值是 (A)2- (B) 12-(C) 12(D) 2 (4) 从数字1,2,3,4,5中任取2的两位数,则这个两位数大于30的概率是(A) 15 (B)25 (C)35 (D)45(5) 执行如图的程序框图,若程序运行中输出的一组数是(),12x -,则x 的值为(A) 27 (B) 81(C) 243 (D) 729(6) 不等式组0,2,22x y x y x y -≤⎧⎪+≥-⎨⎪-≥-⎩的解集记为D , 若(),a b D ∈, 则(A) 1 (B) 4 (C) 1- (D) 4- (7) 已知函数()sin 24f x x π⎛⎫=+⎪⎝⎭,则下列结论中正确的是 (A)函数()f x 的最小正周期为2π(B) 函数()f x 的图象关于点,04π⎛⎫⎪⎝⎭对称 (C) 由函数()f x 的图象向右平移8π个单位长度可以得到函数sin 2y x =的图象 (D) 函数()f x 在区间5,88ππ⎛⎫⎪⎝⎭上单调递增 (8) 已知1F , 2F 分别是椭圆C ()2222:10x y a b a b +=>>的左, 右焦点,点1,2A ⎛ ⎝⎭在椭 圆C 上, 124AF AF +=, 则椭圆C 的离心率是 (A)12(B) 4 (C) 23(D) 2(9) 已知球O 的半径为R ,,,A B C 三点在球O 的球面上,球心O 到平面ABC 的距离为12R ,2AB AC ==,120BAC ︒∠=, 则球O 的表面积为 (A)169π (B)163π (C)649π (D)643π (10) 已知命题p :x ∀∈N *, 1123x x⎛⎫⎛⎫≥ ⎪ ⎪⎝⎭⎝⎭,命题q :x ∃∈R, 122x x-+=命题中为真命题的是(A) p q ∧ (B) ()p q ⌝∧ (11) 如图, 网格纸上的小正方形的边长为1, 的是某几何体的三视图, (A)86+π (B)46+π (C)412+π (D)812+π(12) 设函数()f x 的定义域为R , ()f x -=()3f x x =, 则函数()()cos g x x π=-22⎢⎥⎣⎦(A) 4 (B) 3 (C) 2 (D) 1第Ⅱ卷本卷包括必考题和选考题两部分。
广州市2016届高三普通高中毕业班综合测试(一模)语文试卷 含答案
2016年广州市普通高中毕业班综合测试(一)语文2016。
3 注意事项:1.本试卷共8页,18题,满分为150分,考试用时150分钟;分第Ⅰ卷(阅读题)和第Ⅱ卷(表达题)两部分;答卷前,考生务必将自己的姓名、准考证号、试室号、座位号填写在答题卡上,并用铅笔在答题卡上的相应位置填涂考生号.2.作答时,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.第I卷阅读题甲必考题一、现代文阅读(9分,每小题3分)国内外学界在讨论中国古代对外关系时,首先就会想到古代的朝贡制度以及中国中心主义,由此关注起中国儒家的天下观。
不过,天下观只是儒家的世界秩序学说,在涉外实践中如何实现理想的世界秩序,儒家学者们还提出了一套涉外理念与之配合。
儒家的天下观旨在构建理想的“天下”秩序。
要使“天下”有序,就要“协和万邦"。
《尚书〃尧典》中记载,帝尧家族内部关系融洽后,又先后在各部落以及更远的外邦间次第推行德治与仁爱,使各部落以及外部邦国和平有序。
儒家把“和”看作是处理中国与外国关系的思想准则。
“和也者,天下之达道也。
"即“和"是天下通行的道理,是天下各安其所的交往方式。
而“和”的本质是“仁",“仁”不仅适用于人际关系,也适合于国家间的关系.《左传》将这种“仁”推及国家间的关系,认为“亲仁善邻,国之宝也".儒家崇尚和谐,强调“和为贵”。
如何达到“和为贵”?其中的重要途径是“和而不同"。
孔子说“君子和而不同,小人同而不和”。
“和而不同",在处理国家间的关系上,就是要允许、承认并尊重各国间的文化、信仰和制度的不同,求同存异,从而达到和谐共处.尤其值得指出的是,尽管儒家主张“天下一家”,但后世儒者却多主张“王者不治夷狄"。
他们旨在阐述,“中国"帝王不要去直接统治“夷狄”,应与“夷狄”各守其境,和平相处。
所以,即使是明朝派遣郑和率领庞大水军巡游东南亚以及印度洋诸国时,他们也是秉持着“圣王”对于“夷狄之邦,则以不治治之"的坚定理念,来处理与这些国家间的政治关系.亲仁善邻,协和万邦,还要处理好“王道"与“霸道”之间的关系。
2016广州一模文综试题(word版,标准答案)解析
2016年广州市普通高中毕业班综合测试(一)文科综合2016,3.17第I卷本卷共35个小题,每小题4分,共140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。
从“南粮北运”到“北粮南运”,中国农业经历千年形成的粮食产销格局在最近的30年间被彻底改变了。
下图为中国粮食产销格局示意图,读图完成l一3题。
1.引起“南粮北运”向“北粮南运”转变的最主要原因是A.全球气候变化B.城市化地区差异C.人口分布变化D.土地生产力差异2.影响东北地区成为我国最大商品粮基地的最主要原因是A.人均耕地多B.高速铁路的建设C.全球变暖、热量条件改善D.新品种、新技术广泛应用3-当前,制约“北粮南运”的最主要因素是A.气候条件B.铁路运输C.生产水平D.市场需求一对英国夫妇1月3日从佛得角群岛(约17°N,23°W)出发,驾驶游艇前往安提瓜岛(约17°02'N,61°50'W),如下图所示。
1月9日航程近半时,游艇发生动力故障,只能随水漂流,30天后在M地被救起。
运用所学知识,完成4~6题:4.佛得角群岛上的植被主要是A.热带雨林B.常绿硬叶林C.热带荒漠D.热带季雨林5.出发地与计划目的地之间的最短距离约为A. 2100公里B.4100公里C.6100公里D.8100公里6.M地最可能位于A.安提瓜岛B.亚马孙河口C.直布罗陀海峡D.百慕大群岛下图为2010年上海人口年龄结构示意图,读图完成7—9题。
7.上海流人人口比例最大的年龄段是A. 20N24岁B.25 ~29岁C.30—34岁D.35—39岁8.上海是我国人口老龄化程度下降最快的城市之一,主要原因是A.出生率大幅下降B.死亡率迅速上升C.太量年轻劳力流入D.大量老龄人口迁出9.目前上海市流动人口正处在A.劳动力个体流动阶段 B.老人跟着孩子流动阶段C.夫妻带着孩子流动阶段D.农村流动人口回流阶段下图是北德维纳河流域示意图,该流域的地形是宽阔的、多丘陵的平原,据此完成10 —11题。
2016 年广州市普通高中毕业班综合测试衣答案
2016年广州市普通高中毕业班综合测试(二)语文参考答案及评分标准1.【内容理解】(3分)C (原文表述是“融合了新闻报道的现实性与细致观察及小说的技巧与道德眼光”,选项表述为“运用小说家的直觉洞察力”,与原文不符。
)2.【理解分析】(3分)D (原文表述是“非虚构文学……总是容易被和虚构文学对立起来谈”,选项表述为“非虚构文学与虚构文学是对立的”,与作者观点不符;“非虚构文学与虚构文学是对立的”与后面的结果没有因果关系。
3.【概括分析】(3分)C (原文是“非虚构写作只能尽最大努力接近‘真实’”,不可能达到真正的“真实”,选项表述与原文不符。
)4.【文言理解】(3分)C (“扫除之职固当”作“以为”的宾语,“尔”是句末语气助词,“所以”表结果,因此A、B选项错误。
“力士”应做下句主语,结合上句意思,“逐”的不是力士,因此D 选项错误)5.【实词理解】(3分)A (“特指男女十五岁成年”理解不恰当,应该是特指“女子十五岁成年”。
)6.【概括分析】(3分)B (牟子才作记不是“以致”后面内容的直接原因;“从旁阻挠”的是其他“宦者”,不是董宋臣。
)7.【文言翻译】(10分)(1)于是亲自写信给他说:“我想见你很久了,所以有这个任命,你千万不要有疑虑,为了我(你就)勉为其难出来做官吧。
”[5分。
“赐御笔”1分,“是命”1分,“起”1分,大意2分。
](2)当时士大夫稍微触犯当权者,就被流放,子才向皇上请求重罪的轻罚,轻罪的免除处罚。
[5分。
“迕”1分,“窜流”1分,“放还”1分,大意2分。
]参考译文:牟子才,字荐叟,井研人。
他的八世祖牟允良生下来一岁时,遇上淳化年间强盗作乱,全家在此难中灭绝,仅有一个未成年的姑姑幸免,她用缸盖住允良,允良得以辛免。
牟子才小时候跟着父亲寄居在陈咸家,陈咸大奏音乐大摆宴席,牟子才闭门读书好像没听到看到,见到他的人都觉得他与众不同。
嘉定十六年牟子才考中进士。
当时丁大全与董宋臣内外勾结败坏朝政,牟子才多次上疏请求辞职还乡。
2016届广州市普通高中毕业班综合测试(理综)
2016届广州市普通高中毕业班综合测试(一)理科综合注意事项:1. 本试卷分第I卷(选择题)和第II卷(非选择题)两部分。
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,并用铅笔在答题卡上的相应位置填涂考生号。
2. 回答第I卷时,选出每小题答案后,用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。
写在本试卷上无效。
3. 回答第II卷时,将答案写在答题卡上。
写在本试卷上无效。
4. 考试结束后,将本试卷和答题卡一并交回。
可能用到的相对原子质量:H-1 C-12 O-16 Na-23 S-32 Fe-56 Cu-64第I卷一、选择题:本题包括13小题,每小题6分,共78分。
在每小题给出的四个选项中,只有一个选项符合题目要求。
7.下列叙述正确的是A.顺-2 -丁烯和反-2-丁烯的加氢产物不同B.甲醛、氯乙烯和乙二醇均可作为合成聚合物的单体C.ABS树脂、光导纤维及碳纤维都是有机高分子材料D.酸性条件下,C2H5CO18OC2H5的水解产物是C2H5CO18OH和C2H5OH8.设N A为阿伏加德罗常数,下列叙述中正确的是A.78 g Na2O2中存在的共价键总数为N A原子中含中子数为B.9038C.氢氧燃料电池负极消耗2. 24 L气体时,电路中转移的电子数为D.mol氯化铁溶于1L水中,所得溶液中Fe3+的数目为0. 1N A9.三聚磷酸可视为三个磷酸分子(磷酸结构式见右图)之间脱去两个水分子的产物,三聚磷酸钠(俗称“五钠”)是常用的水处理剂。
下列说法错误..的是A.三聚磷酸中P的化合价为+5B.三聚磷酸钠的化学式为Na3P3O10C.以磷酸钠为原料通过化合反应也能生成三聚磷酸钠D.多聚磷酸的结构可表示为10.W、X、Y、Z均为短周期主族元素,原子序数依次增加,且互不同族,其中只有两种为金属元素,W原子的最外层电子数与次外层电子数相等,W与Z、X与Y这两对原子的最外层电子数之和均为9,单质X与Z都可与NaOH溶液反应。
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2016年广州市普通高中毕业班综合测试(一)英语第I卷第一部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C、和D)中,选出最佳选项,并在答题卡上将该项涂黑。
ADear Reader,As a proud member of this community, the Los Angeles Times is committed to more than just reporting the news. We believe in developing connections with our neighbors . An important part of that mission is our annual Festival of Books, which has grown to become an essential piece of the L.A. cultural scene.This year marks our 21st anniversary, and we're working to make the 2016 festival the best one yet. Over two wonderful days, you’ll have a chance to meet your favorite authors, hear live music, see original art works and listen to poetry.The festival will conclude with the Book Prizes Ceremony, where the best book prizes — as judged by our panel of experts — will be presented. The People's Choice Award , as chosen by you, the reading public, will also be announced.All festival events are free but, due to limited seating at our event locations, you will need a ticket to enter. The free tickets can only be applied for here at our website. This year we are also organizing free buses to the most popular events.The Festival of Books is our gift to the city and will be here soon. I look forward to seeing you there. Allen PetersonLos Angeles Times — Chief Executive1. Who is organizing the Festival of Books?A. Book TV.B. The Los Angeles Times.C. Natural History Museum.D. Los Angeles Public Library.2. How are the best book prizes chosen ?A. By a group of experts.B. By a vote of the authors.C. By the reading public.D. By festival sponsors.3. What is true about tickets to the festival?A. They are only for published authors.B. They include transport to each event.C. They can only be applied for online.D. They can be purchased at the event loca tion.BIn 1800, only three percent of the world’s population lived in cities. Only one city —Beijing —had a population of over a million. Most people lived in rural areas, and never saw a city in their lives. In 1900, just a hundred years later, roughly 150 million people lived in cities, and the world's ten largest cities all had populations exceeding one million. By 2000, the number of city dwellers exceeded three billion; and in 2008, the world’s population crossed a tipping point — more than half of the people on Earth lived in cities. By 2050, that could increase to over two- thirds. Clearly, humans have become an urban species.In the 19th and early 20th centuries, many people viewed cities negatively - - crowded, dirty environments full of disease and crime. They feared that as cities got bigger, living conditions would worsen. In recent decades, however, attitudes have changed. Many experts now think urbanization(城市化)is good news, offering solutions to the problems of Earth’s growing population. Harvard economist Edward Glaeser, author of The Victory of the City, is one such person. Glaeser argues that cities are very productive because “the absence of space between people” reduces the cost of transporting goods, people, and ideas. While the flow of goods has always been important to cities, what is most important today is the flow of ideas. Successful cities enable people to learn from each other easily, and attract and reward smart people with higher wages.Another urbanization supporter is environmentalist Stewart Brand. Brand believes cities help the environment because they allow half of the world’s population to live on about four percent of the land. This leaves more space for open country, such as far mland. City dwellers also have less impact per person on the environment than people in the countryside. Their roads, sewers, and power lines need fewer resources to build and operate. City apartments require less energy to heat cool, and light. Most importantly, people in cities drive less so they produce fewer greenhouse gases per person. So it’s a mistake to see urbanization as evil; it’s a natural part of development. The challenge is how to manage the growth.4. What is Paragraph 1 mainly about?A. The history of modem cities.B. Changes taking place within cities.C. How cities have grown over time.D. Why modem cities are changing.5. How have experts* attitudes towards cities changed in recent decades?A. They now view the weaknesses as strengths.B. They no longer see city-living as attractive.C. They accept city life in spite of its problems.D. They think city-living provides more benefits.6. Which of the following would Edward Glaeser agree with?A. Cities provide more economic opportunities.B. City people get along better with each other.C. Over-crowded cities result in problems.D. Cities limit the flow of ideas.7. According to Paragraph 4, what would be the result of moving people out of cities?A. Economic production would be reduced.B. There would be less farmland available.C. People would travel less frequently.D. House values would fall greatly.CIt was December 25, 1914, only 5 months into World War I. German, British, and French soldiers, already sick and tired of the senseless killing, disobeyed their superiors and started socialising with “the enemy” along two-thirds of the Western Front. German troops held up Christmas trees with signs, “Merry Christmas”. Thousands of troops ran across the battlefields covered with dead bodies. They sang Christmas songs, exchanged photographs of loved ones back home, shared food andeven played football. Soldiers hugged men they had been trying to kill a few shorthours before. They agreed to warn each other if their officers forced them to fire their weapons, and to aim high. Fear ran through the military leaders on both sides. They felt that their power was being challenged: soldiers declaring their brotherhood with each other and refusing to fight. Generals declared this unexpected peacemaking illegal and said that participating soldiers would face a military court. Those found guilty would be imprisoned or even shot. By March 1915 the socialising movement had been destroyed and the killing machine was back in full operation Over the next three years more than fifteen million people died in the war.Not many people have heard the story of the Christmas Truce(休战). On Christmas Day, 1988, a local radio host in Boston played “Christmas in the Trenches”, a song about the Christmas Truce, several times and was stunned by his listeners’ response. Thousands of people called in, praising the song, with many moved to tears by the amazing events it described.You can probably guess why the callers were in tears. The Christmas I nice story goes against most of what we have been taught about people. It lets us see the world as it can be and says, “This really happened once.” It shows us the potential we have as humans, and contradicts all of those TV and newspaper stories that tell us how mean and heartless people are. It is like hearing that our deepest wishes really are true: the world really can be different.8. The soldiers began socialising with the enemy because ________.A. they couldn't bear the meaningless killingB. it was the best way to avoid being killedC. they feared that they would be caughtD. their enemies were from similar backgrounds9. How did the generals finally put an end to the soldiers' socialising?A. They sent the soldiers' loved ones to prison.B. They moved the two groups of soldiers further apart.C. They increased the number of officers to control the soldiers.D. They warned the soldiers that they would face severe punishment.10. The underlined word “stunned” in Paragraph 4 most probably means________ .A. satisfiedB. shockedC. amusedD. confused11. The author uses the story of the soldiers to imply that human beings ______.A. are not trustworthy under stressB. are naturally aggressive and war-likeC. are basically caring and kindheartedD. will always do what is in their own self-interestDEarlier this year, the social media website Facebook announced that it would work with several news organizations — including The New York Times, The Guardian, an d the BBC — to place news stories directly into users’ personal Facebook webpage. Stories published using Facebook Instant will load more quickly and keep the style of the original publisher, who will keep all the advertising income the stories earn — at least for now. The deal shows how important social media has become to news organizations, and is a clear sign of how the world of news is changing — and hasbeen for a while. When Google News began in 20021 many saw it as the death of the newspaper. It had no human editor. Instead Google used, and still uses, a secret compu ter program that selects and displays news stories according to the reader^ personal interests. Morerecently, Associated Press and Yahoo! have been publishing computer-written articles. Both use special software to automatically produce stories about company financial results and sports reports — areas where the quality of writing is felt to be of secondary importance to the accuracy of the data.Should we be worried about such developments? I think we should. One concern is that Facebook, Google and other social media websites see journalism as a sideline, a way of putting people in front of advertisements. It isn't their primary function — so if it stops making them lots of money, they’re likely to stop doing it.There's also a concern that computer-written articles are not actually journalism at all, because what a human news team produces is actually quite complex. A well-written news story puts information in context, offers a voice to each side of an argument and brings the public new knowledge.Though economics and speed of delivery mean readers will probably choose a computer- written story over a carefully shaped article — at least for daily news —I don’t think the computers will be writing any in-depth articles for a while yet.12. What is the main purpose of the article?A. To report on a new computer service offered by Facebook.B. To advise readers against reading computer-written news.C. To express concern about recent trends in online news.D. To describe the process of online news reporting.13. Computer-written news reports have so far focused on sports and finance because ________.A. these are the most popular topics for online readersB. there are fewer journalists specializing in these areasC. information on these topics is more easily availableD. writing style is less important than accuracy in these areas14. What does the underlined word “It” in Paragraph 3 refer to?A. Journalism.B. Advertising.C. Facebook.D. Business15. In Paragraph 4, which of the following is mentioned as a characteristic ofwell-written news article?A. The information presented is up-to-date.B. The author’s opinion is clearC. Different views on the topic are presented.D. The language used is vivid 第二节(共5小题,每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项,选项中有两项为多余选项。