日照实验高中2019级模块考试(必修5)
山东日照实验高中高二上学期期末数学复习(必修5+选修2-1)理科练习三
山东日照实验高中高二上学期期末数学复习理科练习三数 学（理） 第Ⅰ卷 (选择题 共60分)一、选择题：本小题共12小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的.请把答案填涂在答题纸的相应位置. 1. △ABC 的内角A 、B 、C 的对边分别为a 、b 、c ，若c ＝2，b ＝6，B ＝120°，则a 等于( )A. 6 B ． 2 C. 3 D. 22. 已知平面α的法向量是()2,3,1-，平面β的法向量是()4,,2λ-，若//αβ， 则λ的值是（ ） A ．103-B ．6-C ．6D ．1033．已知, , a b c 满足c b a <<，且0ac <，那么下列选项中一定成立的是( ) A. ab ac > B. ()0c b a -< C. 22cb ab < D. ()0ac a c ->4. 等差数列}{n a 中，已知前15项的和9015=S ，则8a 等于（ ） A ．245 B ．12 C ．445D ．6 5. 下列有关命题的说法正确的是( )A ．命题“若21x =，则1=x ”的否命题为：“若21x =，则1x ≠”． B ．“1x =-”是“2560x x --=”的必要不充分条件．C ．命题“x R ∃∈，使得210x x ++<”的否定是：“x R ∀∈， 均有210x x ++<”． D ．命题“若x y =，则sin sin x y =”的逆否命题为真命题6. (2010年浙江)设S n 为等比数列{a n }的前n 项和，8a 2＋a 5＝0，则S 5S 2＝( )A ．11B ．5C ．－8D ．－117. 若椭圆的两焦点为（－2，0）和（2，0），且椭圆过点)23,25(-，则椭圆方程是 （ ）A ．14822=+x y B ．161022=+x y C ．18422=+x y D ．161022=+y x8. 若ABC ∆的内角,,A B C 所对的边,,a b c 满足22()4a b c +-=，且060C =，则a b +的最小值为( )A B ．C ．43D ．8-9. 已知正方体1111D C B A ABCD -中，E 为11D C 的中点，则异面直线AE 与BC 所成角的余弦值为A. 0B.21 C. 32 D. 32- 10.若不等式ax 2＋8ax ＋21<0的解集是{x |－7<x <－1}，那么a 的值是( ) A ．1 B ．2 C ．3 D ．411．若双曲线22221(0,0)x y a b a b-=>>的右焦点为F ，若过F 且倾斜角为︒60的直线与双曲线的右支有且只有一个交点，则此双曲线离心率e 的取值范围是（ ） A ．[]2,1B ．()2,1C ．()+∞,2D ． [)+∞,212.若抛物线24y x =的焦点是F ,准线是l ,则经过点F 、M （4，4）且与l 相切的圆共有（ ）．A.4个B.2个C.1个D.0个第2卷（非选择题 共100分）二、填空题:本大题共4小题，每小题4分，满分16分.请把答案填在答题纸的相应位置. 13．等差数列{}n a 中，若34512,a a a ++=则71a a += .14. 已知向量)0,1,1(=→a ，)2,0,1(-=→b ，且→→+b a k 与→→-b a 2互相垂直，则k 的值是 15. 设0>x ，0>y ，且1116x y+=，则x y +的最小值为 ． 16. 点P 是抛物线x y 42=上一动点，则点P 到点)1,0(-A 的距离与P 到直线1-=x 的距离和的最小值是 .三、解答题:本大题共6小题，共74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 的前n 项和为n S ，且n a 是n S 与2的等差中项，⑴求12,a a 的值；⑵求数列{}n a 的通项公式。
2019-2020学年日照市实验高级中学高三英语一模试题及参考答案
2019-2020学年日照市实验高级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AFour Best Hikes in the WorldThere's nothing like getting out and getting some fresh air on a hike. No matter whether your idea of a hike is a leisure walk or climbing the highest mountain on Earth, we've got you covered. Below are four best hikes inthe world.Torres del Paine W CircuitLocation (位置): Patagonia. ChileDistance: 37 + milesTime: 5~6 daysBest time to go: October to JanuaryThe W Circuit is one of the most recommended hikes you'll find. Not only will you appreciate the diverse landscapes and striking granite pillars (花岗岩柱子), but you'll probably meet some new friends along the way.Grand Canyon Rim - to - Rim HikeLocation: Arizona, the United StatesDistance: 48 milesTime: 1~3 daysBest time to go: May to June, September to OctoberThere's no better way to experience one of the greatest wonders in the world. Located in one of the USA's most beautiful parks, the views are ly appealing. Just make sure you're prepared for the challenge.Trek to PetraLocation: JordanDistance: 47 milesTime: 5~ 6 daysBest time to go: October to AprilTake the road less traveled through the Kingdom of Jordan and experience one of the seven wonders of the world. Hike through canyons, gorges and ridges, and see tombs and temples along the way all while avoidingcrowds of tourists.Yosemite Grand TraverseLocation: California, the United StatesDistance: 60 milesTime: 6~7 daysBest time to go: July to SeptemberKnown for some of the best hiking in the world, Yosemite National Park is famous for its views and huge sequoia (红杉) trees. Praised byNational Geographic, the Yosemite Grand Traverse will take you through waterfalls and green mountaintops.1.Which of the following is the best time for the hike in Patagonia, Chile?A.AprilB.MayC.AugustD.December2.Where should you go for a less crowded hike?A.JordanB.Patagonia, ChileC.Arizona, the United StatesD.California, the United States3.What can you do along the Yosemite Grand Traverse?A.Plant sequoia treesB.Appreciate waterfallsC.Visit local templesD.Climb granite pillarsBIn the past, most people received their news from newspapers, magazines, radio and TV. But now, almost anyone can report and publish on the Internet and share it as news through social media. But the problemis that not all of the information is true and not all of the reporting is trustworthy.Howard Schneider, a former editor of the newspapersNewsday,started the Center for News Literacy (素养) at Stony Book University in 2007. The center has multiple projects, but the most famous one is a course to teach news literacy. The course trains students to look for various details that may indicate the truth of the story.Michelle Sheng is a third-year student at theUniversityofMichigan. Sheng finds that students either just stop reading the news or only take news from one source that they trust. "A lot of people are tired of the news. People are too busy to keep up with the news, and it is really easy to take whatever news is given to you because you don’t have the time to figure it out yourself,“ she says.For her part, Sheng recently created a digital exhibit for the university library of images to educate students on steps they can take to better analyze the news.It is important to educate an even larger audience, beyond American university students. The Center for News Literacy has developed teaching resources, as well as a free online news literacy course. It is also trying to reach a younger audience. It has partnered with several secondary schools in the American state ofNew Yorkto teach news literacy.People should research and confirm what they read online. However, to change human behavior is a difficult thing, but that really is the only thing that is going to help. The biggest problem is not getting people to be able to recognize bad journalism or false news, but getting people to want to recognize it. Our brains are wired to the Internet to seek out information that agrees with our current beliefs.4. What’s purpose of the course “News Literacy”？A. To get rid of false information on the Internet.B. To make people realize the risk on the Internet.C. To train students to tell true information from the false.D. To teach students good habits of using information online.5. Why do students have difficulty judging the truth of news?A. They are too lazy.B. They are bored with news.C. They lack news resources.D. They lack time to check its realness.6. What does the Center try to do besides teaching university students?A. Educate the public.B. Improve the service online.C. Prevent children going online.D. Set up several secondary schools.7. What did the author suggest doing to solve the problem of false news?A. Believing whatever you see.B. Changing human behaviors.C. Questioning all the news online.D. Only trusting reliable information.CThe Gata used to look annoyed when they received power bills that routinely topped $200. Last Septemberthe couple moved into a 1,500-square-foot home in Premier Gardens, an area of 95 “zero-energy homes” (ZEH) just outside town. Now they're actually eager to see their electricity bills. The grand total over the 10 months they've lived in the three-bedroom house: $75. For the past two months, they haven’t paid a cent.ZEH communities are the leading edge of technologies that might someday create houses that produce as much energy as they consume. Premier Gardens is one of a half-dozen subdivisions (住宅开发项目) in California where every home cuts power consumption by 50%, mostly by using low power appliances and solar panels.Aside from the panels on the roof, Premier Gardens looks like a community of traditional homes. But inside, special windows cut power bills by blocking solar heat in summer and keeping indoor warmth winter.The rest of the energy savings comes from the solar units. They don't just feed the home they serve. If they generate more power than the home is using, the excess flows into the utility's power grid(电网). The residents are billed by “net metering”: they pay for the amount of power that they get from the grid, minus the kilowatts(千瓦) they feed into it. If a home generates more power than it uses, the bill is zero.That sounds like a bad deal for the power company, but it's not. Solar homes produce the most power on the hot sunny afternoons when everyone rushes home to turn up the air conditioner. "It helps us lower usage at peak power times," says solar expert Mike Keesee. “That lets us avoid building costly plants or buying expensive power at peak usage time.”What’s not to like? Mostly the costs. The special features can add $25,000 or more to the purchase price of a house. Tax breaks bring the cost down, especially in California, but in many states ZEHs can be extremely expensive. For the consumer, it's a matter of paying now for the hardware to save later on the power bill.8. Why is the Gata eager to see their electricity bills now?A. They want to cut down their utility' expenses.B. They want to know if they are able to pay.C. They want to see how much they have saved.D. They want to avoid being overcharged.9. What is special about the ZEH communities?A. They are built in harmony with the environment.B. They have created cutting edge technologies.C. They are subdivided into half a dozen sections.D. They aim to be independent in power supply.10. What does the "net metering" practice mean to the power company?A. More pressure at peak timeB. Reduced operational costs.C. Increased electricity output.D. Less profits in the short term.11. The author believes that buying a house in a ZEH community __________.A. is a worthy investment in the long runB. is but a dream for average consumersC. gives the owner great tax benefitsD. contributes toenvironmental protectionDBertha von Suttner received the Nobel Peace Prize in 1905—she was the first woman to receive it, and also the inspiration for the creation of the Nobel Prize.She met Alfred Nobel, a rich millionaire, by answering hisnewspaper ad for a secretary. Although she only worked for him for a few weeks, she remained good friends with Alfred Nobel for the next 20 years. When she became involved in the peace movement inEurope, she promised to keep Nobel informed of its progress. When Alfred Nobel died in 1896, his will included the establishment of a peace prize, thanks to Bertha von Suttner’s influence.Bertha von Suttner was born in an aristocratic (贵族) military family, but she spent the second half of her life working for peace. She wrote books, attended peace conferences, gave lectures and helped organize peace societies inAustria,GermanyandHungary, as well as the International Peace Bureau inSwitzerland. Her novel Lay Down your Arms, was one of the most influential anti-war books of all time, and helped to make her a leader of the peace movement in Europe. Its end to war theme was both the ambition (抱负) and the most important goal in the life of this great woman.Bertha von Suttner worked so hard for peace because she believed that a terrible war would break out inEuropeif nations didn’t work hard to establish lasting peace institutions. She made many major achievements for a more peaceful world, but two months after she died, World War I broke out. A hundred years after she won the Nobel Peace Prize, nations still seem to view war as a choice to work out their problems. But like Bertha von Suttner did, many today are working hard around the world to help strengthen peace institutions and spread the idea that it’s time to put an end to war.12. Which of the following is true about Bertha von Suttner?A. She worked for Alfred Nobel for 20 years.B. She helped Alfred Nobel draw up his will.C. She persuaded Alfred Nobel to join the peace movement.D. She inspired Alfred Nobel to establish the Nobel Peace Prize.13. Paragraph 3 is mainly about Bertha von Suttner’s _____________.A. efforts and contributions to the peace movement.B. family background and work experiences.C. writing career and life experiences.D. ambition and goals in life.14. What do we know aboutLay Doun Your Arms?A. It was based on a true story.B. It recorded Bertha von Suttner’s daily life.C. It was about an aristocratic military family.D. It showed Bertha von Suttner’s wish for peace.15. What can we infer about Bertha von Suttner from the last paragraph?A. Her fight for peace is still shared by many.B. She failed to found peace institutions.C. She successfully predicted awar.D. She lost her life in World War I.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。
山东省日照市2019-2020学年高考数学五模试卷含解析
山东省日照市2019-2020学年高考数学五模试卷一、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x ，y 满足约束条件020x y x y y -≥⎧⎪+≤⎨⎪≥⎩，则2z x y =+的最大值为A ．1B ．2C ．3D ．4【答案】D【解析】【分析】作出不等式组对应的平面区域，利用目标函数的几何意义，利用数形结合即可得到结论．【详解】作出不等式组表示的平面区域如下图中阴影部分所示，2z x y =+等价于2y x z =-+，作直线2y x =-，向上平移，易知当直线经过点()2,0时z 最大，所以max 2204z =⨯+=，故选D ．【点睛】本题主要考查线性规划的应用，利用目标函数的几何意义，结合数形结合的数学思想是解决此类问题的基本方法．2．博览会安排了分别标有序号为“1号”“2号”“3号”的三辆车，等可能随机顺序前往酒店接嘉宾．某嘉宾突发奇想，设计两种乘车方案．方案一：不乘坐第一辆车，若第二辆车的车序号大于第一辆车的车序号，就乘坐此车，否则乘坐第三辆车；方案二：直接乘坐第一辆车．记方案一与方案二坐到“3号”车的概率分别为P 1，P 2，则（ ）A ．P 1•P 2＝14 B ．P 1＝P 2＝13 C ．P 1+P 2＝56 D ．P 1＜P 2【答案】C【解析】【分析】将三辆车的出车可能顺序一一列出，找出符合条件的即可.【详解】三辆车的出车顺序可能为：123、132、213、231、312、321方案一坐车可能：132、213、231，所以，P1＝3 6；方案二坐车可能：312、321，所以，P1＝26；所以P1+P2＝56故选C.【点睛】本题考查了古典概型的概率的求法，常用列举法得到各种情况下基本事件的个数，属于基础题.3．设x，y满足约束条件34100640280x yx yx y-+≥⎧⎪+-≥⎨⎪+-≤⎩，则2z x y=+的最大值是（）A．4B．6C．8D．10【答案】D【解析】【分析】作出不等式对应的平面区域，由目标函数的几何意义，通过平移即可求z的最大值．【详解】作出不等式组的可行域，如图阴影部分，作直线0l：20x y+=在可行域内平移当过点A时，2z x y=+取得最大值.由34100280x yx y-+≥⎧⎨+-≤⎩得：()2,4A，max10z∴=故选：D【点睛】本题主要考查线性规划的应用，利用数形结合是解决线性规划题目的常用方法，属于基础题.4．设x∈R，则“|1|2x-<“是“2x x<”的（）A．充分而不必要条件B．必要而不充分条件C ．充要条件D ．既不充分也不必条件【答案】B【解析】【分析】 解出两个不等式的解集，根据充分条件和必要条件的定义，即可得到本题答案.【详解】由|1|2x -<，得13x -<<，又由2x x <，得01x <<，因为集合{|01}{|13}x x x x <<⊂-<<，所以“|1|2x -<”是“2x x <”的必要不充分条件.故选：B【点睛】本题主要考查必要不充分条件的判断，其中涉及到绝对值不等式和一元二次不等式的解法.5．已知正项等比数列{}n a 满足76523a a a =+，若存在两项m a ，n a ，使得219m n a a a ⋅=，则19m n+的最小值为（ ）.A ．16B ．283C ．5D ．4【答案】D【解析】【分析】由76523a a a =+，可得3q =，由219m n a a a ⋅=，可得4m n +=，再利用“1”的妙用即可求出所求式子的最小值.【详解】设等比数列公比为(0)q q >，由已知，525523a a q a q =+，即223q q =+，解得3q =或1q =-（舍），又219m n a a a ⋅=，所以211111339m n a a a --⋅=， 即2233m n +-=，故4m n +=，所以1914m n +=1919()()(10)4n m m n m n m n++=++ 1(1044≥+=，当且仅当1,3m n ==时，等号成立. 故选：D.【点睛】本题考查利用基本不等式求式子和的最小值问题，涉及到等比数列的知识，是一道中档题.6．若()()613x a x -+的展开式中3x 的系数为-45，则实数a 的值为（ ）A ．23B ．2C ．14D ．13【答案】D【解析】【分析】将多项式的乘法式展开，结合二项式定理展开式通项，即可求得a 的值.【详解】∵()()()()666131313x a x x x a x -+=+-+所以展开式中3x 的系数为2233663313554045C aC a -=-=-， ∴解得13a =. 故选：D.【点睛】本题考查了二项式定理展开式通项的简单应用，指定项系数的求法，属于基础题.7．已知(1,3),(2,2),(,1)a b c n ===-r r r ，若()a c b -⊥r r r，则n 等于（ ）A ．3B ．4C ．5D ．6 【答案】C【解析】【分析】 先求出(1,4)a c n -=-r r ，再由()a c b -⊥r r r ，利用向量数量积等于0，从而求得n .【详解】由题可知(1,4)a c n -=-r r ，因为()a c b -⊥r r r，所以有()12240n -⨯+⨯=，得5n =，故选：C.【点睛】该题考查的是有关向量的问题，涉及到的知识点有向量的减法坐标运算公式，向量垂直的坐标表示，属于基础题目. 8．定义在[]22-,上的函数()f x 与其导函数()f x '的图象如图所示，设O 为坐标原点，A 、B 、C 、D 四点的横坐标依次为12-、16-、1、43，则函数()x f x y e=的单调递减区间是（ ）A ．14,63⎛⎫- ⎪⎝⎭B ．1,12⎛⎫- ⎪⎝⎭C ．11,26--⎛⎫ ⎪⎝⎭D ．()1,2【答案】B【解析】【分析】先辨别出图象中实线部分为函数()y f x =的图象，虚线部分为其导函数的图象，求出函数()x f x y e =的导数为()()x f x f x y e '='-，由0y '<，得出()()f x f x '<，只需在图中找出满足不等式()()f x f x '<对应的x 的取值范围即可.【详解】若虚线部分为函数()y f x =的图象，则该函数只有一个极值点，但其导函数图象（实线）与x 轴有三个交点，不合乎题意；若实线部分为函数()y f x =的图象，则该函数有两个极值点，则其导函数图象（虚线）与x 轴恰好也只有两个交点，合乎题意.对函数()x f x y e =求导得()()x f x f x y e '='-，由0y '<得()()f x f x '<，由图象可知，满足不等式()()f x f x '<的x 的取值范围是1,12⎛⎫- ⎪⎝⎭， 因此，函数()x f x y e =的单调递减区间为1,12⎛⎫- ⎪⎝⎭. 故选：B.【点睛】本题考查利用图象求函数的单调区间，同时也考查了利用图象辨别函数与其导函数的图象，考查推理能力，属于中等题.9．已知抛物线22(0)y px p =>上的点M 到其焦点F 的距离比点M 到y 轴的距离大12，则抛物线的标准方程为（ ）A ．2y x =B ．22y x =C ．24y x =D ．28y x = 【答案】B【解析】【分析】由抛物线的定义转化，列出方程求出p ，即可得到抛物线方程．【详解】由抛物线y 2＝2px （p ＞0）上的点M 到其焦点F 的距离比点M 到y 轴的距离大12，根据抛物线的定义可得122p =，1p ∴=，所以抛物线的标准方程为：y 2＝2x ． 故选B ．【点睛】本题考查了抛物线的简单性质的应用，抛物线方程的求法，属于基础题．10．如下的程序框图的算法思路源于我国古代数学名著《九章算术》中的“更相减损术”．执行该程序框图，若输入的a ，b 分别为176，320，则输出的a 为（ ）A ．16B ．18C ．20D ．15【答案】A【解析】【分析】 根据题意可知最后计算的结果为a b ，的最大公约数.【详解】输入的a ，b 分别为176,320,根据流程图可知最后计算的结果为a b ，的最大公约数，按流程图计算320-176=144,176-144=32,144-32=112,112-32=80,80-32=48,48-32=16,32-16=16,易得176和320的最大公约数为16，故选:A.【点睛】本题考查的是利用更相减损术求两个数的最大公约数,难度较易.11．已知函数()cos 221f x x x =++，则下列判断错误的是（ ）A ．()f x 的最小正周期为πB ．()f x 的值域为[1,3]-C ．()f x 的图象关于直线6x π=对称 D ．()f x 的图象关于点,04π⎛⎫- ⎪⎝⎭对称 【答案】D【解析】【分析】先将函数()cos 221f x x x =++化为()2sin 216f x x π⎛⎫=++ ⎪⎝⎭，再由三角函数的性质，逐项判断，即可得出结果.【详解】Q ()cos 221f x x x =++可得1()2cos 2sin 212sin 21226f x x x x π⎛⎫⎛⎫=⋅++=++ ⎪ ⎪ ⎪⎝⎭⎝⎭对于A ，()f x 的最小正周期为22||2T πππω===,故A 正确； 对于B ，由1sin 216x π⎛⎫-≤+≤ ⎪⎝⎭，可得1()3f x -≤≤，故B 正确； 对于C ，Q 正弦函数对称轴可得：()02,62x k k Z πππ+=+∈ 解得：()0,612x k k Z ππ=+∈， 当0k =，06x π=，故C 正确； 对于D ，Q 正弦函数对称中心的横坐标为：()02,6x k k Z ππ+=∈ 解得：()01,212x k k Z ππ=+∈ 若图象关于点,04π⎛⎫-⎪⎝⎭对称，则12124k πππ+=-解得：23k =-，故D 错误； 故选：D.【点睛】 本题考查三角恒等变换，三角函数的性质，熟记三角函数基本公式和基本性质，考查了分析能力和计算能力，属于基础题.12．已知{}n a 为等差数列，若2321a a =+，4327a a =+，则5a =( )A ．1B ．2C ．3D ．6【答案】B【解析】【分析】利用等差数列的通项公式列出方程组，求出首项和公差，由此能求出5a ．【详解】∵{a n }为等差数列，2343a 2a 1,a 2a 7=+=+, ∴()()1111a d 2a 2d 1a 3d 2a 2d 7⎧+=++⎪⎨+=++⎪⎩， 解得1a ＝﹣10，d ＝3，∴5a ＝1a +4d ＝﹣10+11＝1．故选：B ．【点睛】本题考查等差数列通项公式求法，考查等差数列的性质等基础知识，考查运算求解能力，是基础题．二、填空题：本题共4小题，每小题5分，共20分。
山东省日照市2019届高三第五次模拟考试理科综合物理试卷
山东省日照市2019届高三第五次模拟考试理科综合物理试卷本试题卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共16页，38题（含选考题）。
全卷满分300分。
考试用时150分钟。
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二、选择题：本题共8小题，每小题6分，共48分。
在每小题给出的四个选项中，第14～17题只有一项符合题目要求，第18～21题有多项符合题目要求。
全部选对的得6分，选对但不全的得3分，有选错的得0分。
1.氢原了能级如图，一群氢原子处于n=4能级上。
当氢原子从n=4能级跃迁到n=3能级时，辐射光的波长为1884nm，下列判断正确的是A. 氢原子向低能级跃迁时，最多产生4种谱线B. 从高能级向低能级跃迁时，氢原子核一定向外放出能量C. 氢原子从n=3能级跃迁到n=2能级时，辐射光的波长大于1884nmD. 用从能级n=2跃迁到n=1辐射的光照射的铂，能发生光电效应【答案】D【解析】【分析】本题涉及氢原子的能级公式和跃迁，光子的发射，光子能量的计算，光电效应等知识点，涉及面较广，只有入射光子的能量大于金属的逸出功才会发生光电效应；【详解】A、根据知，一群处于能级上的氢原子向低能级跃迁时最多产生6种谱线，故A错误；B、由高能级向低能级跃迁，氢原子向外辐射能量，不是原子核辐射能量，故B错误；C 、从和的能级差大于和的能级差，则从能级跃迁到能级比从能级跃迁到能级辐射出电磁波的频率大，波长短，即辐射光的波长小于，故C错误；D 、从能级跃迁到能级辐射出的光子的能量为：，而使金属发生光电效应的条件是光子的能量大于电子的逸出功，故可以发生光电效应，故D正确。
山东省日照市2019-2020学年高考数学五模考试卷含解析
山东省日照市2019-2020学年高考数学五模考试卷一、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．已知函数()5sin 12f x x π⎛⎫=+ ⎪⎝⎭，要得到函数()cos g x x =的图象，只需将()y f x =的图象( ) A ．向左平移12π个单位长度B ．向右平移12π个单位长度C ．向左平移512π个单位长度 D ．向右平移512π个单位长度 【答案】A 【解析】 【分析】根据函数图像平移原则，即可容易求得结果. 【详解】因为sin cos 122f x x x ππ⎛⎫⎛⎫+=+= ⎪ ⎪⎝⎭⎝⎭，故要得到()g x ，只需将()f x 向左平移12π个单位长度.故选：A. 【点睛】本题考查函数图像平移前后解析式的变化，属基础题.2．若复数z 满足1(120)z i -=,则复数z 在复平面内对应的点在( ) A ．第一象限 B ．第二象限C ．第三象限D ．第四象限【答案】A 【解析】 【分析】化简复数，求得24z i =+，得到复数在复平面对应点的坐标，即可求解. 【详解】由题意，复数z 满足1(120)z i -=，可得()()()10121024121212i z i i i i +===+--+， 所以复数z 在复平面内对应点的坐标为(2,4)位于第一象限 故选：A. 【点睛】本题主要考查了复数的运算，以及复数的几何表示方法，其中解答中熟记复数的运算法则，结合复数的表示方法求解是解答的关键，着重考查了推理与计算能力，属于基础题.3．由曲线3,y x y ==围成的封闭图形的面积为（ ）A ．512 B ．13C ．14D ．12【答案】A 【解析】 【分析】先计算出两个图像的交点分别为()()0,0,1,1，再利用定积分算两个图形围成的面积. 【详解】封闭图形的面积为)1331412000215||3412x dx x x =-=⎰.选A. 【点睛】本题考察定积分的应用，属于基础题.解题时注意积分区间和被积函数的选取.4．根据最小二乘法由一组样本点(),i i x y （其中1,2,,300i =L ），求得的回归方程是ˆˆˆybx a =+，则下列说法正确的是( )A ．至少有一个样本点落在回归直线ˆˆˆybx a =+上 B ．若所有样本点都在回归直线ˆˆˆybx a =+上，则变量同的相关系数为1 C ．对所有的解释变量i x （1,2,,300i =L ），ˆˆibx a +的值一定与i y 有误差 D ．若回归直线ˆˆˆybx a =+的斜率ˆ0b >，则变量x 与y 正相关 【答案】D 【解析】 【分析】对每一个选项逐一分析判断得解. 【详解】回归直线必过样本数据中心点，但样本点可能全部不在回归直线上﹐故A 错误；所有样本点都在回归直线ˆˆˆybx a =+上，则变量间的相关系数为1±，故B 错误； 若所有的样本点都在回归直线ˆˆˆy bx a =+上，则ˆˆbx a +的值与y i 相等，故C 错误；相关系数r 与ˆb符号相同，若回归直线ˆˆˆy bx a =+的斜率ˆ0b >，则0r >，样本点分布应从左到右是上升的，则变量x 与y 正相关，故D 正确． 故选D ． 【点睛】本题主要考查线性回归方程的性质，意在考查学生对该知识的理解掌握水平和分析推理能力.5．设集合{}2320M x x x =++>，集合1{|()4}2xN x =≤ ，则 M N ⋃=（ ）A ．{}2x x ≥- B ．{}1x x >-C ．{}2x x ≤-D ．R【答案】D 【解析】试题分析：由题{}{}2320|21M x x x x x x =++=--或，{}2111|()4|()|2222x x N x x N x x -⎧⎫⎪⎪⎧⎫⎛⎫=≤=≤==≥-⎨⎬⎨⎬ ⎪⎩⎭⎝⎭⎪⎪⎩⎭，M N R ∴⋃=，选D考点：集合的运算6．已知正方体1111ABCD A B C D -的棱长为2，E ，F ，G 分别是棱AD ，1CC ，11C D 的中点，给出下列四个命题: ①1EF B C ⊥;② 直线FG 与直线1A D 所成角为60︒;③ 过E ，F ，G 三点的平面截该正方体所得的截面为六边形; ④ 三棱锥B EFG -的体积为56. 其中，正确命题的个数为（ ） A ．1 B ．2C ．3D ．4【答案】C 【解析】 【分析】画出几何体的图形，然后转化判断四个命题的真假即可． 【详解】 如图；连接相关点的线段，O 为BC 的中点，连接EFO ，因为F 是中点，可知1B C OF ⊥，1EO B C ⊥，可知1B C ⊥平面EFO ，即可证明1B C EF ⊥，所以①正确；直线FG 与直线1A D 所成角就是直线1A B 与直线1A D 所成角为60︒；正确； 过E ，F ，G 三点的平面截该正方体所得的截面为五边形；如图：是五边形EHFGI ．所以③不正确； 如图：三棱锥B EFG -的体积为： 由条件易知F 是GM 中点， 所以B EFG B EFM F BEM V V V ---==, 而=2311522131=2222BEM ABE EDM ABMD S S S S ∆∆+⨯-⨯⨯-⨯-⨯=-梯形， 1551326F EBMV -=⨯⨯=．所以三棱锥B EFG -的体积为56，④正确； 故选：C ． 【点睛】本题考查命题的真假的判断与应用，涉及空间几何体的体积，直线与平面的位置关系的应用，平面的基本性质，是中档题． 7．已知函数()2ln 2xx f x ex a x=-+-（其中e 为自然对数的底数）有两个零点，则实数a 的取值范围是（ ）A ．21,e e⎛⎤-∞+ ⎥⎝⎦B ．21,e e ⎛⎫-∞+⎪⎝⎭ C ．21,e e⎡⎫-+∞⎪⎢⎣⎭D ．21,e e⎛⎫-+∞ ⎪⎝⎭【答案】B【解析】 【分析】求出导函数()f x '，确定函数的单调性，确定函数的最值，根据零点存在定理可确定参数范围． 【详解】21ln ()2()xf x x e x-'=--，当(0,)x e ∈时，()0f x '>，()f x 单调递增，当(,)x e ∈+∞时，()0f x '<，()f x 单调递减，∴在(0,)+∞上()f x 只有一个极大值也是最大值21()f e e a e=+-，显然0x →时，()f x →-∞，x →+∞时，()f x →-∞，因此要使函数有两个零点，则21()0f e e a e =+->，∴21a e e<+． 故选：B ． 【点睛】本题考查函数的零点，考查用导数研究函数的最值，根据零点存在定理确定参数范围．8．已知直线30x y m -+=过双曲线C ：22221(0,0)x y a b a b-=>>的左焦点F ，且与双曲线C 在第二象限交于点A ，若||||FA FO =（O 为坐标原点），则双曲线C 的离心率为 A ．2 B ．31+C ．5D ．51-【答案】B 【解析】 【分析】 【详解】直线30x y m -+=的倾斜角为π3，易得||||FA FO c ==．设双曲线C 的右焦点为E ，可得AFE △中，90FAE ∠=o ，则||3AE c =，所以双曲线C 的离心率为313e c c==+-.故选B ．9．一个几何体的三视图如图所示，则这个几何体的体积为（ ）A ．63π+ B ．6πC 163πD ．163π+【答案】B 【解析】 【分析】还原几何体可知原几何体为半个圆柱和一个四棱锥组成的组合体，分别求解两个部分的体积，加和得到结果. 【详解】由三视图还原可知，原几何体下半部分为半个圆柱，上半部分为一个四棱锥半个圆柱体积为：2211123622V r h πππ==⨯⨯=四棱锥体积为：2114333V Sh ==⨯⨯⨯=原几何体体积为：126V V V π=+= 本题正确选项：B 【点睛】本题考查三视图的还原、组合体体积的求解问题，关键在于能够准确还原几何体，从而分别求解各部分的体积.10．在等差数列{}n a 中，若n S 为前n 项和，911212a a =+，则13S 的值是（ ） A ．156 B ．124C ．136D ．180【答案】A 【解析】 【分析】因为711911212a a a a +==+，可得712a =，根据等差数列前n 项和，即可求得答案. 【详解】Q 711911212a a a a +==+，∴712a =， ∴()113137131313121562a a S a +===⨯=.故选：A. 【点睛】本题主要考查了求等差数列前n 项和，解题关键是掌握等差中项定义和等差数列前n 项和公式，考查了分析能力和计算能力，属于基础题.11．一个正三角形的三个顶点都在双曲线221x ay +=的右支上，且其中一个顶点在双曲线的右顶点，则实数a 的取值范围是（ ） A ．()3,+∞ B．)+∞C．(,-∞D ．(),3-∞-【答案】D 【解析】 【分析】因为双曲线分左右支，所以0a <，根据双曲线和正三角形的对称性可知：第一象限的顶点坐标为(1t +，)(0)t >，将其代入双曲线可解得． 【详解】因为双曲线分左右支，所以0a <，根据双曲线和正三角形的对称性可知：第一象限的顶点坐标为(1t +)(0)t >，将其代入双曲线方程得：22(1))1t a ++=， 即2113t a -=+，由0t >得3a <-．故选：D ． 【点睛】本题考查了双曲线的性质，意在考查学生对这些知识的理解掌握水平． 12．已知实数,x y 满足约束条件11220220x y x y x y ≥-⎧⎪≥-⎪⎨-+≥⎪⎪--≤⎩，则23x y -的最小值是A ．2-B ．72-C ．1D ．4【答案】B 【解析】 【分析】 【详解】作出该不等式组表示的平面区域，如下图中阴影部分所示， 设23z x y =-，则2133y x z =-，易知当直线2133y x z =-经过点D 时，z 取得最小值，由1220xx y=-⎧⎨-+=⎩，解得112xy=-⎧⎪⎨=⎪⎩，所以1(1,)2D-，所以min172(1)322z=⨯--⨯=-，故选B．二、填空题：本题共4小题，每小题5分，共20分。
日照实验高中级模块考试(必修4).doc
日照实验高中级模块考试（必修4） 20XX.6一、选择题（本大题共10小题，每小题5分，共50分.每小题有四个选项，其中只有一项是正确的，请把说选答案填在下表中.） 1.下列命题正确的是A.第一象限角是锐角B.钝角是第二象限角C.终边相同的角一定相等D.不相等的角，它们终边必不相同 2.函数12sin()24y x π=-+的周期，振幅，初相分别是A.4π，2，4π B. 4π，2-，4π- C. 4π，2，4π D. 2π，2，4π3.如果1cos()2A π+=-，那么sin()2A π+=A.12B.12C.12D.124.函数2005sin(2004)2y x π=-是A.奇函数B.偶函数C.非奇非偶函数D.既是奇函数又是偶函数 5.给出命题（1）零向量的长度为零，方向是任意的. （2）若a ，b 都是单位向量，则a ＝b . （3）向量AB 与向量BA 相等.（4）若非零向量AB 与CD 是共线向量，则A ，B ，C ，D 四点共线. 以上命题中，正确命题序号是A.（1）B.（2）C.（1）和（3）D.（1）和（4） 6.如果点(sin 2P θ，cos 2)θ位于第三象限，那么角θ所在象限是 A.第一象限 B.第二象限 C.第三象限 D.第四象限7.在四边形ABCD 中，如果0AB CD =，AB DC =，那么四边形ABCD 的形状是 A.矩形 B.菱形 C.正方形 D.直角梯形 8.若α是第一象限角，则sin cos αα+的值与1的大小关系是A.sin cos 1αα+>B.sin cos 1αα+=C.sin cos 1αα+<D.不能确定 9.在△ABC 中，若sin 2cos sin C A B =，则此三角形必是A.等腰三角形B.正三角形C.直角三角形D.等腰直角三角形 10.如图，在△ABC 中，AD 、BE 、CF 分别是BC 、点G ，则下列各等式中不正确的是A.23BG BE =B.2CG GF =C.12DG AG =D.121332DA FC BC +=二、填空题（本大题共4小题，每小题5分，共20分）11.设扇形的周长为8cm ，面积为24cm ，则扇形的圆心角的弧度数是 .12.已知tan 2α=，3tan()5αβ-=-，则tan β= . 13.已知(3a =，1)，(sin b α=，cos )α，且a ∥b ，则4sin 2cos 5cos 3sin αααα-+＝ .14.给出命题：（1）在平行四边形ABCD 中，AB AD AC +=.（2）在△ABC 中，若0AB AC <，则△ABC 是钝角三角形. （3）在空间四边形ABCD 中，,E F 分别是,BC DA 的中点，则1()2FE AB DC =+. 以上命题中，正确的命题序号是 .三、解答题（本大题共6小题，共80分，解答应写出文字说明，证明过程或演算步骤） 15.（本小题满分13分）已知3sin 25α=，53[,]42αππ∈. （1）求cos2α及cos α的值；（2）求满足条件sin()sin()2cos x x ααα--++=的锐角x . 16.（本小题满分13分）已知函数()sin22x xf x =+，x R ∈. （1）求函数()f x 的最小正周期，并求函数()f x 在[2,2]x ππ∈-上的单调递增区间； （2）函数()sin ()f x x x R =∈的图象经过怎样的平移和伸缩变换可以得到函数()f x 的图象.17.（本小题满分13分）已知电流I 与时间t 的关系式为sin()I A t ωϕ=+. （1）下图是sin()I A t ωϕ=+(0,)2πωϕ><求sin()I A t ωϕ=+的解析式； （2）如果t 在任意一段1150秒的时间内，电流 sin()I A t ωϕ=+都能取得最大值和最小值，1900- 1180那么ω的最小正整数值是多少？18.（本小题满分13分）已知向量(3,4)OA =-，(6,3)OB =-，(5,3)OC m m =---. （1）若点,,A B C 能够成三角形，求实数m 应满足的条件； （2）若△ABC 为直角三角形，且A ∠为直角，求实数m 的值.19.（本小题满分13分）设平面内的向量(1,7)OA =，(5,1)OB =，(2,1)OM =，点P 是直线OM 上的一个 动点，且8PA PB =-，求OP 的坐标及APB ∠的余弦值.20.（本小题满分13分） 已知向量33(cos,sin )22x x a =，(cos ,sin )22x x b =-，且[,]2x ππ∈. （1）求a b 及a b +；（2）求函数()f x a b a b =++的最大值，并求使函数取得最大值时x 的值.日照实验高中级模块考试（必修4） 20XX.6一、选择题BCBBA BAAAC二、填空题11. 2 12. -13 13. 5714. （1）（2）（3） 三、解答题15.解：（1）因为5342παπ<<，所以5232παπ<<. ………………………（2分）因此4cos 25α==-. ………………………………（4分）由2cos 22cos 1αα=-，得cos 10α=-. ……………………（8分） （2）因为sin()sin()2cos 10x x ααα--++=-，所以2cos (1sin )10x α-=-，所以1sin 2x =. ………………………（11分） 因为x 为锐角，所以6x π=. ………………………………………………（13分）16.解：sin2sin()2223x x x y π=+=+. （1）最小正周期2412T ππ==. ……………………………………………（3分）令123z x π=+，函数sin y z =单调递增区间是[2,2]()22k k k Z ππππ-++∈.由 1222232k x k πππππ-+≤+≤+，得 544,33k x k k Z ππππ-+≤≤+∈. ………………………………（5分）取0k =，得533x ππ-≤≤，而5[,]33ππ-⊂[2,2]ππ-，所以，函数sin 22x x y =，[2,2]x ππ∈-得单调递增区间是5[,]33ππ-.…………………………………………………………………………（8分） （2）把函数sin y x =图象向左平移3π，得到函数sin()3y x π=+的图象，…（10分）再把函数sin()3y x π=+的图象上每个点的横坐标变为原来的2倍，纵坐标不变，得到函数sin()23x y π=+的图象， …………………………………（11分）然后再把每个点的纵坐标变为原来的2倍，横坐标不变，即可得到函数2sin()23x y π=+的图象. …………………………………………………（13分） 17.解：（1）由图可知300A =，设11900t =-，21180t =， ……………………（2分）则周期211112()2()18090075T t t =-=+=， …………………………（4分） ∴2150T πωπ==. ………………………………………………………（6分）1900t =-时，0I =，即1sin[150()]0900πϕ⋅-+=，sin()06πϕ-=.而2πϕ<， ∴6πϕ=.故所求的解析式为300sin(150)6I t ππ=+. ……………………………（8分） （2）依题意，周期1150T ≤，即21150πω≤，(0)ω>， …………………（10分） ∴300942ωπ≥>，又*N ω∈，故最小正整数943ω=. ……………（13分） 18.解：（1）已知向量(3,4)OA =-，(6,3)OB =-，(5,3)OC m m =---，若点,,A B C 能构成三角形，则这三点不共线，即AB 与BC 不共线. ……（4分） (3,1)AB =，(2,1)AC m m =--， 故知3(1)2m m -≠-， ∴实数12m ≠时，满足条件. …………………………………………………（8分） （若根据点,,A B C 能构成三角形，必须任意两边长的和大于第三边的长，即由ABBC CA +>去解答，相应给分）（2）若△ABC 为直角三角形，且A ∠为直角，则AB AC ⊥， …………（10分） ∴3(2)(1)0m m -+-=， 解得74m =. …………………………………………………………………（13分） 19.解：设(,)OP x y =. ∵点P 在直线OM 上，∴OP 与OM 共线，而OM (2,1)=，∴20x y -=，即2x y =，有(2,)OP y y =. ………………………………（2分） ∵(12,7)PA OA OP y y =-=--，(52,1)PB OB OP y y =-=--，……（4分） ∴(12)(52)(7)(1)PA PB y y y y =--+--，即252012PA PB y y =-+. …………………………………………………（6分）又8PA PB =-， ∴2520128y y -+=-，所以2y =，4x =，此时(4,2)OP =. ……………………………………（8分） (3,5),(1,1)PA PB =-=-. 于是34,2,8PA PB PA PB ===-. …………………………………（10分）∴cos 34PA PB APB PA PB∠===⋅. ………………………（13分）20.解：（1）33coscos sin sin cos 22222x x x xa b x =-=， ……………………（3分） (cosa b += ………………………（4分）=2cos x == …………………………………………（7分） ∵[,]2x ππ∈， ∴cos 0x <.∴2cos a b x +=-. …………………………………………………………（9分） （2）2()cos 22cos 2cos 2cos 1f x a b a b x x x x =++=-=-- 2132(cos )22x =-- …………………………………………………（11分） ∵[,]2x ππ∈， ∴1cos 0x -≤≤， ……………………………………（13分）∴当cos 1x =-，即x π=时max ()3f x =. ………………………………（15分）。
2019-2020学年日照市实验高级中学高三英语上学期期末考试试题及答案解析
2019-2020学年日照市实验高级中学高三英语上学期期末考试试题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ACitisport inNewportWe at Citisport aim to improve sports training and facilities inNewport, giving you more opportunities to try both new and traditional sports.GolfWe are pleased to be able to offer lessons at Kingsway Golf Centre just outsideNewport. These are run by experienced golf professionals, and are held on an all-weather practice area. The adult lessons are open to anyone aged 13 and over, and are suitable for all levels from beginners upwards. These take place on Wednesdays from 3:00 to 4:00 pm over a period of six weeks. Children’s lessons for 7-12 year old are held from 2:00 to 3:00 pm on Saturdays during term time.TennisThe Citisport tennis courses provide an opportunity for local people to develop their skills on the brand-new indoor tennis court at Newport Leisure Centre. All equipment can be provided, but please feel free to use your own racket (球拍) if you prefer. Our Starter course is held on Mondays from 7:00 to 8:00 pm, and is for beginners of 12 years and over. Our Improver course, which takes place on Tuesdays from 8:00 to 9:00 pm, is for players with some experience.Football for girlsBy popular request, Citisport is holding another one-day girls-only football course. This aims to give local girls the chance to learn essential skills and develop more advanced ones. The course will take place on Saturday, 9th November from 9:00 am to 5:00 pm, and is open to all girls aged 10-14 years living in theNewportarea.GymnasticsThis course is for beginners aged 8-14 and will provide an introduction to basic skills. There is a maximum of six pupils per coach in each class. At the end of the course there is a demonstration for friends and family of all the skills learnt there. The course will take place on Thursdays from 6:00 to 7:00 pm.1. What can we know about the Citisport golf lessons?A. You can take lessons at Kingsway Golf Centre insideNewport.B. The golf lessons can take place only in good weather.C. Teenagers can attend golf lessons on Wednesday afternoons.D. Children’s lessons usually last 2 or 3 hours on Saturday afternoons.2. Which of the following statements is true about the Citisport tennis courses?A. You can get the skills in an open-air court.B. You must take your own racket during the course.C. You can take the Monday course if you are a green hand.D. You can become an experienced player after the courses.3. Which course lasts only one day according to the text?A. Golf.B. Tennis.C. Football for girls.D. Gymnastics.BAlex Wong, a junior atMarkKeppelHigh SchoolinAlhambra,California, is working hard on his application to a top college. His resume shows off his nearly straight A’s in difficult classes, experience at a summer program atStanfordUniversity, Eagle Scout project and time on the soccer team as well as the school choir. But his steady progress stopped unexpectedly this year. Aiming to open access to college-level Advanced Placement (大学预科) courses, his schoolbegan using a computer-based lottery to give out spaces. Alex got shut out of all three of the courses he requested.The new system caused anger among families whose children failed to get into AP courses, which many consider important to develop advanced skills, improve grade-point averages and allow students to earn college credit, saving them and their families tuition dollars. Students and parents wrote to administrators to complain, circulated a petition (请愿) and launched a Facebook group for trading classes. “I’M DESPERATE! I’LL GIVE YOU FREE FOOD,” one student, Kirk Hum, posted on the 210-member AP Flea Market Facebook group.AP classes have long been held dear by the most talented and ambitious students.But now they are seen as positive for all students who are willing to push themselves – and schools are increasingly viewing access to them as a basic educational right. But this change has brought challenges.Miracle Vitangcol, a junior atDowntownMagnetsHigh Schoolwith average grades and test scores, is failing her AP US history class. She said she can’t handle the rapid pace and volume of material she needs to remember. But she said she intends to stick it out because the class is teaching her to manage her time, take good notes and work hard. “I’m struggling to adjust,” she said. “But I keep telling myself: ‘It’s OK. You can do it. Just push yourself’.”Some critics worry that the open-access movement is pushing too many unprepared students into AP classes, as shown by higher exam failure rates over the last decade. They also fear that open enrollment (录取) policies are encouraging teachers to weaken courses and give out high grades to students who don’t deserve them. “While expanding access is generally a good thing, we need to make sure we’re not watering down the experience for the high achievers,” said Michael Petrilli, executive vice president of the Thomas B. Fordham Institute, a Washington-based educational policy organization.4. The purpose of the new AP courses system at Alex Wong’s school is to ______.A. make sure all students get access to the AP courses they desire.B. ensure that students have a fair chance to get access to AP courses.C. improve the academic performance of students in AP courses.D. separate high achievers from average students through the new courses.5. According to the article, the AP Flea Market Facebook group is a place where ______.A. students’ parents send their complaints to school administrators.B. students share tips about saving money for college.C. students offer items to trade for the AP courses they need.D. students can find support and guidance on their AP study.6. Which of the following statements would Michael Petrilli agree with?A. Opening AP courses to all students is a bad idea.B. School administrators should maintain high academic standards for AP courses.C. High schools should stop charging students for taking AP courses.D. Access to AP courses is necessary for students applying for top American colleges.7. The author used Miracle Vitangcol’s example to show that ______.A. students need to remember too much in their AP courses.B.AP courses pose a big challenge to unprepared students.C. the secret to success in AP courses is to keep pushing yourself.D. average students don’t deserve their places in AP courses.COne billion people in the world are short of water. How can this problem be solved. Some suggestions have been to desalinate ocean water or to build enormous water pipelines from areas where water is abundant. (Suggestions such as these prove extremely expensive when they are actually used.) One possibility that scientistsare considering is pulling icebergs from either the North Pole or the South Pole to parts of the world with a water shortage. Although many questions must be answered before such a project could be tried, moving icebergs seems a reasonable possibility in the future.Engineers, mathematicians, and glaciologists from a dozen countries have been considering the iceberg as a future source of water. Saudi Arabia is particularly interested in this project because it has a great water shortage. Scientists estimate that it would take 128 days to transport a large iceberg (about 1/2 square mile) to Saudi Arabia. Yet the iceberg would be completely melted by the 104th day. Therefore, insulation would be essential, but how to insulate the iceberg remains an unsolved problem.The problems in transporting an iceberg are numerous. The first problem is choosing the iceberg to pull. The icebergs that form in the North Pole are quite difficult to handle because of their shape. Only a small portion extends above the water — most of the iceberg is below the surface, which would make it difficult to pull. South Pole icebergs, on the other hand, are flat and float like table tops. Thus they would be much easier to move.How can a 200-million-ton iceberg be moved. No ship is strong enough to pull such enormous weight through the water. Perhaps several ships could be used. Attaching ropes to an iceberg this size is also an enormous problem. Engineers think that large nails or long metal rods could be driven into the ice. What would happen if the iceberg splits into several pieces during the pulling. Even if an iceberg with very few cracks were chosen, how could it be pulled through stormy waters. Furthermore, once the iceberg reached its destination, very few ports would be deep enough to store it.All of these problems must be solved before icebergs can become a reasonable source of water. Yet scientists estimate that it will be possible to transport them in the near future. Each year, enough icebergs form to supply the whole world with fresh water for a full year. In addition, icebergs are free and nonpolluting. As a solution to the world’s water problems, icebergs may be a workable possibility.8. What is a problem in transporting iceberg?A. The size of the iceberg.B. The colour of the iceberg.C. The salt in the iceberg.D. The movement of air and water.9. What is the author’ attitude towards transporting iceberg?A. Pessimistic.B. Objective.C. Optimistic.D. Unconcerned.10. What does the last paragraph mainly tell us?A. It is hard to use iceberg.B. Iceberg are a good choice.C. There are problems with iceberg.D. Man finds no other ways to solve water shortage.11. What can be a suitable title for the text?A. Shortage of water.B. Icebergs for water.C. Scientists and icebergs.D. Iceberg—scientists headache.DYou’re in a crowd of people who are all asking for the same thing. How do you make your voice heard above the rest? Be different. Don’t shout. Lisa, 25, was waiting to board a plane flying fromLondontoAustriafor Christmaswhen the flight was cancelled.“There were about a hundred of us unable to leave,” she says. “Everyone else was shouting at the airport staff. Instead of joining in, I walked up to the man behind the ticket desk very quietly and said, ‘This must be so awful for you! I don’t know how you deal with these situations—it’s not even your fault. I could never handle it as well as you are.’ Without my even asking, he found me a seat on another airline with an upgrade to first class. He was happy to do a favor forsomeone who was appreciative instead of unfriendliness.”Flattery (恭维) is an essential element of the sweet-talk strategy. “It’s human psychology that stroking a person’s ego (自我) with a few well-directed praises makes them want to prove you right,” says apsychologist. “Tell someone they’re pretty and they’ll instantly fix their hair; praise their sense of humor and they’ll tell a joke.”You need help and there’s ly no reason that the person will want to lend a hand. Allison, 26. a lawyer, realized she’d made a huge mistake on a batch of documents. “The only way I could fix the problem was to get the help of a colleague who I knew didn’t like me,” she said.Allison then went to the woman’s office and explained her problem. “As I was saying to the boss the other day you’re the only person who would know how to handle a situation like this, what would you suggest I do?” “Feeling pumped up (鼓励), she set about helping me and we finished the job on time, and she was happy to help.” Allison said.12. Whatwould have happened at the airport according to paragraph 1?A. The departure hall was filled with noise.B. Someone screamed just lo be different.C. The passengers waited on board patiently.D. The airport stuff were rude to the passengers.13. Why did the man put Lisa on another airline?A. He admired Lisa’s beauty.B. He appreciated her attitude.C. He was ready to help others.D. He was blamed for the cancellation.14. What is the third paragraph mainly about?A. The potential benefits of ego.B. The strategy to start small talk.C. The great importance of flattery.D. The value of humor in daily life.15. What can we learn about Allison’s colleague?A. She was a popular lawyer.B. She was always ready to help others.C. She always got praise from Allison.D. She did a great favor for Allison eventually.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年日照市高中必修五数学上期末第一次模拟试卷(带答案)
2019年日照市高中必修五数学上期末第一次模拟试卷(带答案)一、选择题1．若正实数x ，y 满足141x y +=，且234yx a a +>-恒成立，则实数a 的取值范围为（ ） A ．[]1,4-B ．()1,4-C ．[]4,1-D ．()4,1-2．已知等比数列{}n a 的公比为正数，且239522,1a a a a ⋅==，则1a = ( )A ．12B ．2C ．2D ．223．正项等比数列中，的等比中项为，令，则（ ） A ．6B ．16C ．32D ．644．在ABC ∆中，2AC =,22BC =,135ACB ∠=o ，过C 作CD AB ⊥交AB 于D ，则CD =( ) A ．25B ．2C ．3D ．55．已知数列{}n a 的前n 项和为n S ，点(,3)n n S +*()n N ∈在函数32xy =⨯的图象上，等比数列{}n b 满足1n n n b b a ++=*()n N ∈，其前n 项和为n T ，则下列结论正确的是（ ）A ．2n n S T =B ．21n n T b =+C ．n n T a >D ．1n n T b +<6．已知数列{}n a 的前n 项和为n S ，1112n n a S a +＝，＝， 则n S ＝( )A ．12n -B ．13()2n -C ．12()3n - D ．112n - 7．若直线()10,0x ya b a b+=>>过点(1,1)，则4a b +的最小值为（ ） A ．6B ．8C ．9D ．108．设变量,x y 、满足约束条件236y xx y y x ≤⎧⎪+≥⎨⎪≥-⎩，则目标函数2z x y =+的最大值为（ ）A ．2B ．3C ．4D ．99．已知等差数列{}n a ,前n 项和为n S ,5628a a +=,则10S =( ) A ．140 B ．280C ．168D ．5610．在△ABC 中，角A ，B ，C 所对的边长分别为a ，b ，c ，若∠C=120°，c=a ，则A ．a ＞bB ．a ＜bC ．a ＝bD ．a 与b 的大小关系不能确定11．变量,x y 满足条件1011x y y x -+≤⎧⎪≤⎨⎪>-⎩，则22(2)x y -+的最小值为（ ） A ．32B ．5C ．5D ．9212．如图，为了测量山坡上灯塔CD 的高度，某人从高为=40h 的楼AB 的底部A 处和楼顶B 处分别测得仰角为=60βo，=30αo ，若山坡高为=35a ，则灯塔高度是（ ）A ．15B ．25C ．40D ．60二、填空题13．若首项为1a ，公比为q （1q ≠）的等比数列{}n a 满足21123lim()2n n a q a a →∞-=+，则1a 的取值范围是________.14．已知0a >，0b >，当()214a b ab++取得最小值时，b =__________． 15．在ABC ∆中，角A ，B ，C 的对边分别为a ，b ，c .2C A π-=，1sin 3A =，3a =，则b =______.16．已知等差数列{}n a 的公差为()d d 0≠，前n 项和为n S ，且数列{}n S n +也为公差为d 的等差数列，则d =______．17．设无穷等比数列{}n a 的公比为q ，若1345a a a a =+++…，则q =__________________．18．已知二次函数f （x ）=ax 2+2x+c （x ∈R ）的值域为[0，+∞），则11a c c a+++的最小值为_____．19．在数列{}n a 中，11a =，且{}n a 是公比为13的等比数列．设13521T n n a a a a L -=++++，则lim n n T →∞=__________．(*n ∈N ) 20．若无穷等比数列{}n a 的各项和为2，则首项1a 的取值范围为______.三、解答题21．在()f x 中，角,,A B C 的对边分别为,,a b c ，满足(2)cos cos b c A a C -=． (1)求角A 的大小(2)若3a =，求ABC △的周长最大值． 22．如图，测量河对岸的塔高AB 时，可以选与塔底B 在同一水平面内的两个测点C 与D . 现测得BCD α∠=，BDC β∠=，CD s =，并在点C 测得塔顶A 的仰角为θ，求塔高AB .23．已知等差数列{}n a 的所有项和为150，且该数列前10项和为10，最后10项的和为50.（1）求数列{}n a 的项数； （2）求212230a a a ++⋅⋅⋅+的值.24．已知数列{}n a 的前n 项和为n S ，且221n n n S na a =+-． （1）求数列{}n a 的通项公式； （2）若数列21n a ⎧⎫⎨⎬⎩⎭的前n 项和为n T ，证明：4nT <． 25．已知a ，b ，c 分别为△ABC 三个内角A ，B ，C 的对边，且acos C ＋3asin C －b －c ＝0.(1)求A ；(2)若AD 为BC 边上的中线，cos B ＝17，AD ＝1292，求△ABC 的面积． 26．已知点（1，2）是函数()(0,1)xf x a a a =>≠的图象上一点，数列{}n a 的前n 项和是()1n S f n =-.（1）求数列{}n a 的通项公式；（2）若1log n a n b a +=，求数列{}n n a b •的前n 项和n T【参考答案】***试卷处理标记，请不要删除一、选择题 1．B 解析：B 【解析】 【分析】 根据1444y y x x x y ⎛⎫⎛⎫+=++ ⎪ ⎪⎝⎭⎝⎭，结合基本不等式可求得44yx +≥，从而得到关于a 的不等式，解不等式求得结果. 【详解】 由题意知：1442444y y x yx x x y y x⎛⎫⎛⎫+=++=++ ⎪ ⎪⎝⎭⎝⎭ 0x Q >，0y > 40x y ∴>，04yx>424x y y x ∴+≥=（当且仅当44x y y x =，即4x y =时取等号） 44yx ∴+≥ 234a a ∴-<，解得：()1,4a ∈- 本题正确选项：B 【点睛】本题考查利用基本不等式求解和的最小值问题，关键是配凑出符合基本不等式的形式，从而求得最值.2．D解析：D 【解析】设公比为q ，由已知得()22841112a q a q a q ⋅=，即22q=，又因为等比数列{}n a 的公比为正数，所以q 212a a q ===，故选D. 3．D解析：D 【解析】因为，即，又，所以.本题选择D 选项.4．A解析：A 【解析】 【分析】先由余弦定理得到AB 边的长度，再由等面积法可得到结果. 【详解】根据余弦定理得到222222AC BC AB AC BC +-=-⨯⨯将2AC =,22BC =,代入等式得到AB=5 再由等面积法得到112252522222CD CD ⨯=⨯⇒=故答案为A. 【点睛】这个题目考查了解三角形的应用问题，涉及正余弦定理，面积公式的应用，在解与三角形有关的问题时，正弦定理、余弦定理是两个主要依据.解三角形时，有时可用正弦定理，有时也可用余弦定理，应注意用哪一个定理更方便、简捷一般来说 ,当条件中同时出现ab 及2b 、2a 时，往往用余弦定理，而题设中如果边和正弦、余弦函数交叉出现时，往往运用正弦定理将边化为正弦函数再结合和、差、倍角的正余弦公式进行解答.5．D解析：D 【解析】 【分析】 【详解】由题意可得：332,323n nn n S S +=⨯=⨯- ，由等比数列前n 项和的特点可得数列{}n a 是首项为3，公比为2的等比数列，数列的通项公式：132n n a -=⨯ ，设11n nb b q -= ，则：111132n n n b q b q --+=⨯ ，解得：11,2b q == ，数列{}n b 的通项公式12n nb -= ，由等比数列求和公式有：21nn T =- ，考查所给的选项：13,21,,n n n n n n n n S T T b T a T b +==-<< .本题选择D 选项.6．B解析：B 【解析】 【分析】利用公式1n n n a S S -=-计算得到11323,2n n n n S S S S ++==，得到答案. 【详解】由已知1112n n a S a +==，，1n n n a S S -=- 得()12n n n S S S -=-，即11323,2n n n n S S S S ++==， 而111S a ==，所以13()2n n S -=.故选B. 【点睛】本题考查了数列前N 项和公式的求法，利用公式1n n n a S S -=-是解题的关键.7．C解析：C 【解析】 【详解】 因为直线()10,0x y a b a b+=>>过点()1,1,所以11+1a b = ,因此114(4)(+)5+59b a a b a b a b +=+≥+= ,当且仅当23b a ==时取等号,所以选C.点睛：在利用基本不等式求最值时，要特别注意“拆、拼、凑”等技巧，使其满足基本不等式中“正”(即条件要求中字母为正数)、“定”(不等式的另一边必须为定值)、“等”(等号取得的条件)的条件才能应用，否则会出现错误.8．D解析：D 【解析】 【分析】由约束条件作出可行域，化目标函数为直线方程的斜截式，数形结合得到最优解，联立方程组求得最优解的坐标，把最优解的坐标代入目标函数得结论. 【详解】画出满足约束条件236y xx y y x ≤⎧⎪+≥⎨⎪≥-⎩的可行域，如图，画出可行域ABC ∆，(2,0)A ，(1,1)B ，(3,3)C ， 平移直线2z x y =+，由图可知，直线2z x y =+经过(3,3)C 时 目标函数2z x y =+有最大值，2z x y =+的最大值为9.故选D. 【点睛】本题主要考查线性规划中，利用可行域求目标函数的最值，属于简单题.求目标函数最值的一般步骤是“一画、二移、三求”：（1）作出可行域（一定要注意是实线还是虚线）；（2）找到目标函数对应的最优解对应点（在可行域内平移变形后的目标函数，最先通过或最后通过的顶点就是最优解）；（3）将最优解坐标代入目标函数求出最值.9．A解析：A 【解析】由等差数列的性质得，5611028a a a a +==+，∴其前10项之和为()11010102814022a a +⨯==，故选A. 10．A解析：A 【解析】 【分析】由余弦定理可知c 2＝a 2+b 2﹣2ab cos C ，进而求得a ﹣b 的表达式，根据表达式与0的大小，即可判断出a 与b 的大小关系． 【详解】解：∵∠C ＝120°，ca ，∴由余弦定理可知c 2＝a 2+b 2﹣2ab cos C ，（）2＝a 2+b 2+ab ．∴a 2﹣b 2＝ab ，a ﹣b ，∵a ＞0，b ＞0， ∴a ﹣b ，∴a ＞b 故选A ． 【点睛】本题考查余弦定理，特殊角的三角函数值，不等式的性质，比较法，属中档题．11．C解析：C 【解析】由约束条件画出可行域，如下图，可知当过A(0,1)点时，目标函数取最小值5，选C.12．B解析：B 【解析】 【分析】过点B 作BE DC ⊥于点E ，过点A 作AF DC ⊥于点F ，在ABD ∆中由正弦定理求得AD ，在Rt ADF ∆中求得DF ，从而求得灯塔CD 的高度． 【详解】过点B 作BE DC ⊥于点E ，过点A 作AF DC ⊥于点F ，如图所示，在ABD ∆中，由正弦定理得，sin sin AB ADADB ABD=∠∠，即sin[90(90)]sin(90)h ADαβα=︒--︒-︒+，cos sin()h AD αβα∴=-，在Rt ADF ∆中，cos sin sin sin()h DF AD αβββα==-，又山高为a ，则灯塔CD 的高度是3340cos sin 22356035251sin()2h CD DF EF a αββα⨯⨯=-=-=-=-=-． 故选B ．【点睛】本题考查了解三角形的应用和正弦定理，考查了转化思想，属中档题．二、填空题13．【解析】【分析】由题意可得且即且化简可得由不等式的性质可得的取值范围【详解】解：故有且化简可得且即故答案为：【点睛】本题考查数列极限以及不等式的性质属于中档题解析：33(0,)(,3)22U【解析】 【分析】由题意可得1q <且0q ≠，即11q -<<且0q ≠，211232a a a =+，化简可得13322a q =+由不等式的性质可得1a 的取值范围. 【详解】解：21123lim()2n n a q a a →∞-=+Q 21123lim 2n a a a →∞∴=+，lim 0nn q →∞= 故有11q -<<且0q ≠，211232a a a =+化简可得13322a q =+ 103a ∴<<且132a ≠即133(0,)(,3)22a ∈U 故答案为：33(0,)(,3)22U 【点睛】本题考查数列极限以及不等式的性质，属于中档题.14．【解析】【分析】根据均值不等式知即再由即可求解注意等号成立的条件【详解】（当且仅当等号成立）（当且仅当等号成立）（当且仅当等号成立）故答案为【点睛】本题主要考查了均值不等式不等式等号成立的条件属于中 解析：14【解析】 【分析】根据均值不等式知，4a b +≥=()2416a b ab +≥，再由41684ab a b +≥=⋅即可求解，注意等号成立的条件. 【详解】4a b +≥=Q （当且仅当4a b =等号成立），()2416a b ab ∴+≥（当且仅当4a b =等号成立），()2444a b a b ∴++≥⋅8=（当且仅当4a b =等号成立）， ()224281a a a∴+=⇒=. 故答案为14b =. 【点睛】本题主要考查了均值不等式，不等式等号成立的条件，属于中档题.15．7【解析】【分析】先求出再利用正弦定理求最后利用余弦定理可求【详解】因为所以故且为锐角则故由正弦定理可得故由余弦定理可得故即或因为为钝角故故故答案为：7【点睛】三角形中共有七个几何量（三边三角以及外解析：7 【解析】 【分析】先求出sin 3C =，再利用正弦定理求c ，最后利用余弦定理可求b . 【详解】 因为2C A π-=，所以2C A π=+，故sin sin cos 2C A A π⎛⎫=+= ⎪⎝⎭， 且A为锐角，则cos A =sin 3C =. 由正弦定理可得sin sin a c A C =，故3sin 31sin 3a Cc A⨯=== 由余弦定理可得2222cos a b c bc A =+-，故297223b b =+-⨯即7b =或9b =， 因为C 为钝角，故c b >，故7b =. 故答案为：7. 【点睛】三角形中共有七个几何量（三边三角以及外接圆的半径），一般地，知道其中的三个量（除三个角外），可以求得其余的四个量. （1）如果知道三边或两边及其夹角，用余弦定理；（2）如果知道两边即一边所对的角，用正弦定理（也可以用余弦定理求第三条边）； （3）如果知道两角及一边，用正弦定理.16．【解析】【分析】表示出再表示出整理并观察等式列方程组即可求解【详解】等差数列的公差为前项和为设其首项为则=又数列也为公差为的等差数列首项为所以=即：整理得：上式对任意正整数n 成立则解得：【点睛】本题 解析：12【解析】 【分析】表示出n S【详解】等差数列{}n a 的公差为()0d d ≠，前n 项和为n S ，设其首项为1a ， 则n S =()112n n na d -+，又数列也为公差为d=()1n d-()1n d=-=上式对任意正整数n成立，则)212122dddda d d⎧=⎪=⎪-+=⎪⎩，解得：12d=，134a=-【点睛】本题主要考查了等差数列的前n项和及通项公式，考查了方程思想及转化思想、观察能力，属于中档题．17．【解析】【分析】由可知算出用表示的极限再利用性质计算得出即可【详解】显然公比不为1所以公比为的等比数列求和公式且故此时当时求和极限为所以故所以故又故故答案为：【点睛】本题主要考查等比数列求和公式当时【解析】【分析】由1345a a a a=+++…可知1q<,算出345a a a+++…用1a表示的极限,再利用性质计算得出q即可.【详解】显然公比不为1,所以公比为q的等比数列{}n a求和公式1(1)1-=-nna qSq,且1345a a a a=+++…,故01q<<.此时1(1)1-=-nna qSq当n→∞时,求和极限为11aq-,所以3345...1aa a aq+++=-,故2311345...=11a a qa a a aq q=+++=--,所以2211101a qa qqq=⇒+-=-,故12q-=,又01q<<,故12q=..【点睛】本题主要考查等比数列求和公式1(1)1-=-nna qSq,当01q<<时1lim1nnaSq→∞=-.18．4【解析】【分析】先判断是正数且把所求的式子变形使用基本不等式求最小值【详解】由题意知则当且仅当时取等号∴的最小值为4【点睛】】本题考查函数的值域及基本不等式的应用属中档题解析：4 【解析】 【分析】先判断a c 、是正数，且1ac =，把所求的式子变形使用基本不等式求最小值． 【详解】由题意知，044010a ac ac c =-=∴=V ＞，，，＞，则111111 2224a c a c a c c a c c a a c a c a +++=+++=+++≥+=+=（）（），当且仅当1a c ==时取等号．∴11a c c a +++的最小值为4． 【点睛】】本题考查函数的值域及基本不等式的应用．属中档题.19．【解析】【分析】构造新数列计算前n 项和计算极限即可【详解】构造新数列该数列首项为1公比为则而故【点睛】本道题考查了极限计算方法和等比数列前n 项和属于中等难度的题目解析：9lim 8n n T →∞=【解析】 【分析】构造新数列{}21n a -，计算前n 项和，计算极限，即可。
日照实验高中数学必修5模块考试(理科)
日照实验高中数学必修5模块考试（理科）第Ⅰ卷 选择题（共60分）一、选择题：本大题共12小题，每小题5分，共60分. 在每小题给出的四个选项中，只有一项是符合题目要求的. 1.若1a ＜1b＜ 0 ，则下列结论不正确的是 A. 2a ＜b 2 B. a b ＜b 2 C.b aa b+＞2 D. b a b a +>+ 2.在△ABC 中，已知8=a ，B=060，C=075，则b 等于A ．64B ．54C ．34D ．322 3.已知等差数列{}n a 的公差为2,若431,,a a a 成等比数列, 则2a 等于 A –4 B –6 C –8 D –10 4.在△ABC 中，已知0120C ,6b ,4a ===，则sinA 的值是A1957B 721C 383D 1957-5.在△ABC 中，AB=3，BC=13，AC=4，则边AC 上的高为A ．223 B ．233 C ．23D ．33 6.不等式035322>--x x 的解集是A ⎭⎬⎫⎩⎨⎧>-<527x x x 或 B ⎭⎬⎫⎩⎨⎧><<5270x x x 或 C ⎭⎬⎫⎩⎨⎧>-<<-7525x x x 或 D {}55>-<x x x 或 7.已知等差数列||||,}{93a a a n =中，公差0<d ，则使前n 项和n S 取最大值的正整数n 的值是A ．4或5B ．5或6C ．6或7D ．8或98.在△ABC 中，2,32,300===AC AB B ，则△ABC 的面积是A 32B 3C 32或34D 3或329.等比数列{}n a 前n 项的积为n T ，若3618a a a 是一个确定的常数，那么数列10T ，13T ，17T ，25T 中也是常数的项是A ．10T B ．13T C ．17T D ．25T 10.若实数x y ，满足1000x y x y x ⎧-+⎪+⎨⎪⎩，，，≥≥≤则23x yz +=的最小值是A ．0B ．1CD ．911.已知数列{}n a 中，++∈++==N n a n na a n n ,2)1(,211 ,则11a 等于A ．36B ．38C ．40D ．4212.若不等式n)1(2a )1(1n n+-+<-对任意正整数n 恒成立。
2019-2020学年日照市实验高级中学高三英语期末试卷及答案解析
2019-2020学年日照市实验高级中学高三英语期末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AOlder adults who sleep six hours or fewer a night may have elevated risk for dementia(痴呆症) and other cognitive (认知的) issues, a new study finds.Researchers at Stanford University measured seniors' (ages 65 to 85) dementia risk and cognitive abilities, finding higher risk in those patients who regularly slept six or fewer hours compared to those who slept seven or eight hours. Those seniors who slept nine or more hours also had lower cognitive functions and other health issues, but the researchers didn't find the same high dementia risk in this group.The findings demonstrate how important it is for adults to maintain a healthy sleep cycle, especially as they get older.As adults age, it's common for their sleep patterns to change or becomedisrupted— leading to longer, shorter, or more irregular sleep. This disruption may be linked to Alzheimer's and other forms of dementia, impacting seniors' ability to remember information, problem-solve, and go through everyday behaviors. Sleep disruption can also be caused by or heighten depression, cardiovascular disease（冠心病）, and other conditions.A recommended sleep time for seniors is seven to eight hours, the researchers said. Six or fewer hours corresponded to short sleep, while nine or more hours corresponded to long sleep.The Stanford researchers measured levels of beta amyloid, a protein in the brain that is typically found in high levels when a patient develops Alzheimer's. In addition, the researchers used several tests for memory, attention, spatial skills, and executive function to identify patients' cognitive abilities. Those patients sleeping for six hours or fewer a night were more likely to develop dementia, the researchers found. The low-sleep patients had higher levels of beta amyloid.The Stanford researchers found that patients with lower sleep also performed worse on memory tests, while those with higher sleep (more hours) performed worse on executive function tests，which measure the brain's ability to switch between different tasks.―The main takeaway is that it is important to maintain healthy sleep late in life, Winer told CNN.1. What does the underlined word “disrupted” in paragraph 4 mean?A. difficultB. disorderedC. dissolvedD. different2. According to the findings, which of the following is NOT related to the disrupted sleep?A. It is more likely to cause old people to have bad memories over issues.B. It may contribute to dementia, cardiovascular disease and other illnesses.C. Some daily behaviors perhaps differ from those whose sleeping is normal.D It tends to bring all the old people to undergo brain scans and cognitive tests.3. What can we infer from the study?A. A proper sleep time for seniors is seven to eight hours.B. Low and high sleep patients were both poor at memory tests.C. Executive function test is applied to measure the capacity of brain.D. Keeping a healthy sleep for older adults late in life is crucial.BThese days, football is one of the most popular sports in the world. Given that Neil Armstrong wanted to take a football to the Moon, we could even say that it is also the most popular sport out of this world! The history of the game goes back over two thousand years to Ancient China. It was then known as cuju (kick ball), a game using a ball of animal skins with hair inside. Goals were hung in the air. Football as we know it today started inGreat Britain, where the game was given new rules.That football is such a simple game to play is perhaps the basis of its popularity. It is also a game that is very cheap to play. You don’t need expensive equipment; even the ball doesn’t have to cost much money. All over the world you can see kids playing to their hearts’ content with a ball made of plastic bags.Another factor behind football’s global popularity is the creativity and excitement on the field. It is fun enough to attract millions of people. You do not have to be a fan to recognize the skill of professional players or to feel the excitement of a game ending with a surprising twist.What’s more, football has become one of the best ways for people to communicate: it does not require words, but everyone understands it. It breaks down walls and brings people together on and off the field.“Some people believe football is a matter of life and death, ...” said Bill Shankly, the famous footballer and manager. “I can tell you with certainty it is much, much more important than that.” This might sound funny, but one only has to think about the Earth to realize that our planet is shaped like a football.4. What can we know from paragraph one?A. Some people like to play football on the Moon.B. The game called cuju was given new rules today.C. Cuju is different from football as we know it today.D. Many people like playing a ball made of plastic bags.5. According to the author, there are ________ reasons why football became so popular in the world.A. 3B. 4C. 5D. 66. What can be inferred from the last paragraph?A. Football is round.B. Football is more than just a sport.C. Our planet is shaped like a football.D. What Bill Shankly said sounds funny.7. What’s the author’s purpose in writing the passage?A. To talk about the history of football.B. To express his/her love of football.C. To explain why football is such a popular game.D. To prove that he/she is a professional football fan.CIsraeli Paintings—Israeli artist Menashe Kadishman will hold a personal show named "Flock of Sheep" from November 26 to December 20 at the China National Art Museum On show are 550 colourful oil paintings of sheep heads.His works have been on show inthe Metropolitan Museum in New York and Tate Gallery in London over the past 30 years.Time: 9: 00 a.m.—4: 00 p.m., November 26—December 20.Place: China National Art Museum, 1 Wusi Dajie, Dongcheng District, Beijing.Tel: 6401-2252Russian Ballet—The Kremlin Ballet from Russia will perform two immortal classical ballets—"Swan Lake" and "The Nutcracker"—at the Beijing Beizhan Theatre. Set up in 1990, the theatre has a number of first-class ballet dancers. Most of their performances are classical.Time: 7: 15 p.m., December 5 and 6 ("Swan Lake"); 7: 15 p.m., December 7 ("The Nutcracker")Place: Beizhan Theatre, Xiwai Dajie, Xicheng District, Beijing.Tel: 6605-3388Folk Concert—The Central Conservatory of Music will hold a folk concert in memory of the late musician Situ Huacheng.On the programme are many popular folks such as "Moon Night on the Bamboo Tower", "Celebrating Harvest", "Deep and Lasting Friendship", "Golden Snake Dances Wildly" and "Children's Holiday".Time: 7: 30 p.m., November 25.Place: Beijing Concert Hall, 1 Beixinhuajie, Xicheng District, Beijing.Tel: 6605-58128.If a child is very fond of dancing, his parents should take him to ________.A.China National Art MuseumB.Beizhan TheatreC.Beijing Concert HallD.1 Wusi Dajie9.Menashe Kadishman is well known for painting ________.A.deerB.birdsC.sheepD.flowers10.Which of the following is TRUE?A.The folk concert will last three days.B.The ballet "The Nutcracker" will be put on once.C.The Israeli paintings will be on show for a month.D.China National Art Museum lies in Xicheng District.11.If you dial the telephone number 6605-3388 on Dec. 8, you can ________.A.go to the folk concertB.visit the Art MuseumC.watch the balletD.none of the aboveDThe regular world presented to us by our five senses — you could call it reality 1.0 — is not always the most user-friendly. We get lost in unfamiliar cities; we meet people whose language we don’t understand. So why not try the improved version: augmented reality（AR）or reality 2.0 ? AR technology adds computer-produced images on the real world via a mobile phone camera or special video glasses.Early forms of AR are already here — smart phones can deliver information about nearby ATMs and restaurants and other points of interest. But that’s just the beginning. A few years from now the quantity of information available will have increased hugely. You will not only see that there’s a Chinese restaurant on the next block, but you will be able to see the menu and read reviews of it.This is where the next revolution in computing will take place: in the interface（界面）between the real world and the information brought to us via the Internet. Imagine bubbles floating before your eyes, filled with cool information about anything and everything that you see in front of you.Let’s jump ahead to ten years from now. A person trying to fix a car won’t be reading a book with pictures; he will be wearing a device that projects animated 3D computer graphics onto the equipment under repair, labelling parts and giving step-by-step guidance.The window onto the AR world can be a smart phone or special video glasses. But in ten years’ time these will have been replaced by contact lenses(隐形眼镜) with tiny LEDs, which present something at a readable distance in front of eyes. So a deaf person wearing these lenses will be able to see what people are saying.The question is, while we are all absorbed in our new augmented reality world, how willwe be communicating with each other?12. What is the text mainly about?A. The relationship between reality 1.0 and reality 2.0.B. Different forms of the AR technology.C. The next information technology revolution.D. The popularity of the AR technology.13. Which of the following will AR technology support according to the text?A. To pay for things online conveniently.B. To play online games merrily.C. To offer information efficiently.D. To communicate with others socially.14. What are Contact lenses with tiny LEDs used for?A. Show texts and images.B. Protect people’s eyes.C. Help deaf people communicate.D. Replace smart phone.15. What’s the author’s attitude towards the AR technology?A. Indifferent.B. Critical.C. Concerned.D. Favourable.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年日照市高中必修五数学上期中第一次模拟试卷(带答案)
2019年日照市高中必修五数学上期中第一次模拟试卷(带答案)一、选择题1．如果111A B C ∆的三个内角的余弦值分别等于222A B C ∆的三个内角的正弦值，则A ．111ABC ∆和222A B C ∆都是锐角三角形 B ．111A B C ∆和222A B C ∆都是钝角三角形C ．111A B C ∆是钝角三角形，222A B C ∆是锐角三角形D ．111A B C ∆是锐角三角形，222A B C ∆是钝角三角形2．设ABC ∆的三个内角, , A B C 成等差数列，sin A 、sin B 、sin C 成等比数列，则这个三角形的形状是 （ ） A ．直角三角形B ．等边三角形C ．等腰直角三角形D ．钝角三角形3．定义在()(),00,-∞⋃+∞上的函数()f x ,如果对于任意给定的等比数列{}n a ,若(){}nf a 仍是比数列，则称()f x 为“保等比数列函数”.现有定义在()(),00,-∞⋃+∞上的如下函数： ①()3f x x =；②()xf x e =；③()f x =④()ln f x x =则其中是“保等比数列函数”的()f x 的序号为（ ） A ．①②B ．③④C ．①③D ．②④4．已知数列{a n } 满足a 1=1，且111()(233n n n a a n -=+≥，且n ∈N*），则数列{a n }的通项公式为（ ）A ．32nn a n =+B ．23n n n a +=C ．a n =n+2D ．a n =（ n+2）·3n5．当()1,2x ∈时，不等式220x mx ++≥恒成立，则m 的取值范围是（ ） A ．()3,-+∞B．()-+∞C ．[)3,-+∞D．)⎡-+∞⎣6．,x y 满足约束条件362000x y x y x y -≤⎧⎪-+≥⎪⎨≥⎪⎪≥⎩,若目标函数(0,0)z ax by a b =+>>的最大值为12，则23a b+的最小值为 ( )A ．256B ．25C ．253D ．57．在ABC ∆中，角A ，B ，C 所对的边分别是a ，b ，c ，60A =︒，a=4b =，则B =（ ） A ．30B =︒或150B =︒ B ．150B =︒ C ．30B =︒D ．60B =︒8．已知数列{}n a 中，3=2a ，7=1a ．若数列1{}na 为等差数列，则9=a ( ) A ．12B ．54C ．45D ．45-9．设等差数列{a n }的前n 项和为S n ，已知(a 4－1)3＋2 016(a 4－1)＝1，(a 2 013－1)3＋2 016·(a 2 013－1)＝－1，则下列结论正确的是( ) A ．S 2 016＝－2 016，a 2 013>a 4 B ．S 2 016＝2 016，a 2 013>a 4 C ．S 2 016＝－2 016，a 2 013<a 4 D ．S 2 016＝2 016，a 2 013<a 410．在等差数列{}n a 中，如果123440,60a a a a +=+=，那么78a a +=（ ） A ．95B ．100C ．135D ．8011．设{}n a 是首项为1a ，公差为-2的等差数列，n S 为其前n 项和，若1S ，2S ，4S 成等比数列，则1a = ( ) A ．8B ．-8C ．1D ．-112．若正数,x y 满足40x y xy +-=，则3x y+的最大值为 A ．13B ．38C ．37D ．1二、填空题13．若变量x ，y 满足22390x y x y x +≤⎧⎪-≤⎨⎪≥⎩，则z ＝2x +y 的最大值是_____．14．已知数列111112123123n+++++++L L L ，，，，，，则其前n 项的和等于______． 15．设不等式组30,{230,1x y x y x +-<--≤≥表示的平面区域为1Ω，平面区域2Ω与1Ω关于直线20x y +=对称，对于任意的12,C D ∈Ω∈Ω，则CD 的最小值为__________．16．已知120,0,2a b a b>>+=，2+a b 的最小值为_______________.17．已知关于x 的一元二次不等式ax 2+2x+b ＞0的解集为{x|x≠c}，则227a b a c+++（其中a+c≠0）的取值范围为_____．18．已知在△ABC 中，角,,A B C 的对边分别为,,a b c ，若2a b c +=，则C ∠的取值范围为________19．设a ＞0,b ＞0． 若关于x,y 的方程组1,{1ax y x by +=+=无解，则+a b 的取值范围是 ．20．设等差数列{}na 的前n 项和为n S ．若35a =，且1S ，5S ，7S 成等差数列，则数列{}n a 的通项公式n a =____．三、解答题21．已知等差数列{}n a 的前n 项和为n S ，各项为正的等比数列{}n b 的前n 项和为n T ，11a =-，11b =，222a b +=.（1）若335a b +=，求{}n b 的通项公式； （2）若321T =，求3S 22．已知数列{}n a 的首项123a =，且当2n ≥时，满足1231312n n a a a a a -++++=-L ． （1）求数列{}n a 的通项公式； （2）若2n n nb a =，n T 为数列{}n b 的前n 项和，求n T ． 23．设数列{}n a 满足113,23nn n a a a +=-=⋅．（Ⅰ）求数列{}n a 的通项公式n a ；（Ⅱ）若n n b na =，求数列{}n b 的前n 项和n S ．24．已知函数()sin (0)f x m x x m =+>的最大值为2. （Ⅰ）求函数()f x 在[0,]π上的单调递减区间； （Ⅱ）ABC ∆中,()()sin 44f A f B A B ππ-+-=,角,,A B C 所对的边分别是,,a b c ，且060,3C c ==，求ABC ∆的面积．25．已知等差数列{}n a 的前n 项和为n S ，且211a =，7161S =． （1）求数列{}n a 的通项公式；（2）若6512n n S a n >--，求n 的取值范围； （3）若11n n n b a a +=，求数列{}n b 的前n 项和n T ． 26．数列{}n a 中，11a = ，当2n ≥时，其前n 项和n S 满足21()2n n n S a S =⋅-．（1）求n S 的表达式； (2)设n b ＝21nS n +,求数列{}n b 的前n 项和n T ．【参考答案】***试卷处理标记，请不要删除一、选择题 1．D 解析：D 【解析】 【分析】 【详解】111A B C ∆的三个内角的余弦值均大于0，则111A B C ∆是锐角三角形，若222A B C ∆是锐角三角形，由，得2121212{22A AB BC C πππ=-=-=-，那么，2222A B C π++=，矛盾，所以222A B C ∆是钝角三角形，故选D.2．B解析：B 【解析】 【分析】先由ABC ∆的三个内角, , A B C 成等差数列，得出2,33B AC ππ=+=,又因为sin A 、sin B 、sin C 成等比数列，所以23sin sin sin 4B AC =⋅=，整理计算即可得出答案.【详解】因为ABC ∆的三个内角, , A B C 成等差数列，所以2,33B AC ππ=+=, 又因为sin A 、sin B 、sin C 成等比数列， 所以23sin sin sin 4B AC =⋅=所以222sin sin sin sin cos sin cos333A A A A A πππ⎛⎫⎛⎫⋅-=⋅-⎪ ⎪⎝⎭⎝⎭21111132sin 2cos 2sin 2424442344A A A A A π⎛⎫=+=-+=-+= ⎪⎝⎭ 即sin 213A π⎛⎫-= ⎪⎝⎭又因为203A π<< 所以3A π=故选B 【点睛】本题考查数列与三角函数的综合，关键在于求得2,33B AC ππ=+=，再利用三角公式转化，属于中档题.3．C解析：C 【解析】 【分析】设等比数列{}n a 的公比为q ，验证()()1n n f a f a +是否为非零常数，由此可得出正确选项. 【详解】设等比数列{}n a 的公比为q ，则1n na q a +=. 对于①中的函数()3f x x =，()()3313112n n n n n n f a a a q f a a a +++⎛⎫=== ⎪⎝⎭，该函数为“保等比数列函数”；对于②中的函数()xf x e =，()()111n n n n a a a n a n f a e e f a e++-+==不是非零常数，该函数不是“保等比数列函数”； 对于③中的函数()f x =()()1n n f a f a +===，该函数为“保等比数列函数”；对于④中的函数()ln f x x =，()()11ln ln n n n na f a f a a ++=不是常数，该函数不是“保等比数列函数”.故选：C.【点睛】本题考查等比数列的定义，着重考查对题中定义的理解，考查分析问题和解决问题的能力，属于中等题.4．B解析：B 【解析】试题分析：由题可知，将111()(233n n n a a n -=+≥，两边同时除以，得出，运用累加法，解得，整理得23n n n a +=； 考点：累加法求数列通项公式5．D解析：D 【解析】由()1,2x ∈时，220x mx ++≥恒成立得2m x x ⎛⎫≥-+⎪⎝⎭对任意()1,2x ∈恒成立，即max 2,m x x ⎡⎤⎛⎫≥-+ ⎪⎢⎥⎝⎭⎣⎦Q 当2x 时，2x x ⎛⎫-+ ⎪⎝⎭取得最大值22,22m -∴≥-，m 的取值范围是)22,⎡-+∞⎣，故选D.【易错点晴】本题主要考查利用基本不等式求最值以及不等式恒成立问题，属于中档题. 利用基本不等式求最值时，一定要正确理解和掌握“一正，二定，三相等”的内涵：一正是，首先要判断参数是否为正；二定是，其次要看和或积是否为定值（和定积最大，积定和最小）；三相等是，最后一定要验证等号能否成立（主要注意两点，一是相等时参数否在定义域内，二是多次用≥或≤时等号能否同时成立）.6．A解析：A 【解析】 【分析】先画不等式组表示的平面区域，由图可得目标函数(0,0)z ax by a b =+>>何时取最大值，进而找到a b ，之间的关系式236,a b +=然后可得23123()(23)6a b a b a b+=++，化简变形用基本不等式即可求解。
山东省日照市日照实验高级中学2019_2020学年高一数学上学期期末考试试题含解析
又 ， ，
，解得 或 ，
所以 或 .
【点睛】本题考查了向量坐标的加法和数乘运算，平行向量的坐标关系，根据向量的坐标求向量长度的方法，考查了计算能力，属于基础题．
20.某中学为了丰富学生的业余生活，开展了一系列文体活动，其中一项是同学们最感兴趣的 对 篮球对抗赛，现有甲乙两队进行比赛，甲队每场获胜的概率为 .且各场比赛互不影响.
8.函数 大致图象是（ ）
A. B.
C. D.
【答案】C
【解析】
【分析】
根据特殊位置的 所对应的 的值，排除错误选项，得到答案.
【详解】因为
所以当 时， ，故排除A、D选项，
而 ，
所以
即 是奇函数，其图象关于原点对称，故排除B项，
故选C项.
【点睛】本题考查根据函数的解析式判断函数图象，属于简单题.
9.若 ，则 的最小值是（ ）．
①当 时，方程 有两不相等的实数根，设为 ， ，
则 ，
， 均为负数，由函数图象知 ，故不存在实数使得 ，即方程 无实数根；
②当 时，方程 有两不相等的实数根，设为 ， ，
则 ，
， 均为正数且 ，
设 则 ，由图可知，存在 个不同 实数使得 ，
存在 个不同的实数使得 ，
即方程 有 个不同的实数根；
（3）当 时，方程无解，则方程 无实数根；
综上可得正确的有①④，错误的有②③
故选：
【点睛】本题考查了分段函数，以及函数与方程的思想，数形结合的思想，属于难题．
二、填空题
13.已知: ，用 表示 __________．
【答案】
【解析】
【分析】
根据对数的运算法则及对数的性质计算可得.
2019-2020学年日照市实验高级中学高三英语上学期期末考试试题及参考答案
2019-2020学年日照市实验高级中学高三英语上学期期末考试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AMany workers have had no choice but to adapt to working from home in recent months since offices shut down due to the COVID-19 pandemic (新冠疫情). And the noisy situation and endless housework may result in a terrible emotion. A new option is waiting foryou. That is WFH: work from a hotel.Hotel FigueroA special program titled Work Perks aims to reposition some of 94-year-old Hotel Figuero’s 268 rooms as day-use offices.According to Managing Director Connie Wang, the set-up launched in June and is a great opportunity to get out of their houses with high-speed Wi-Fi, unlimited printing privileges and free parking. The 350-square-foot rooms sell for $ 129 per day, with an option to extend to an overnight stay for an additional $ 20.The WytheA boutique hotel in Brooklyn. The hotel recently announced a partnership with co-working office space company Industrious through which it is recycling 13 second-story guest rooms to serve as offices for up to four people.Each of the rooms has a small outdoor platform, and dogs are welcome. Pricing starts at $ 200 and goes up to $ 275, depending on how many people use the space.The SawyerThe Sawyer, in Sacramento, California, is offering pool cabanas (更衣室) for use as outdoor offices, complete with fast Wi-Fi, free parking and catered lunch for $ 150 per day.HotelsByDayYannis Moati founded HotelsByDay back in 2015. That company has grown to include more than 1,500 hotels, and has seen a significant increase in the number of inquiries for day-use bookings lately.Moati said the current situation will force hotels to upgrade themselves to stay alive, and he predicted that offering rooms for day-use only is one of the directions they will go.1.How much should one pay for a 24-hour stay in Hotel Figuero?A.$ 129.B.$ 149.C.$ 150.D.$ 200.2.Which hotel allows pets in?A.The Wythe.B.The Sawyer.C.HotelsByDay.D.Hotel Figuero.3.What do we know about Yannis Moati?A.He started a program titledWork Perks.B.He has upgraded at least 1,500 rooms.C.He usually predicts everything correctly.D.He is optimistic about the WFH trend.BChinese paleontologists (古生物学家) have determined that, about 47 million years ago, subtropical forests once existed on the high-altitude Qinghai-Tibet Plateau.The conclusion, which appears in a paper published on Tuesday, was drawn based on the large number of fossils found in theBaingoinBasinat an altitude of nearly 5,000 meters during the second comprehensive scientific expedition to the plateau.A joint team from theXishuangbannaTropicalBotanical Gardenconducted the research on the fossils. By combining the findings and models, the team recreated the climate and altitude that existed 47 million years ago, showing that the central plateau had an altitude of just 1,500 meters and an annual average temperature of 19℃, says Su Tao, a researcher from the tropical botanical garden and first author of the paper.“It was covered by thick forest and was rich in water and grass. It is fair tocall it the ‘ShangriLa’ of ancient times,” Su adds.The researchers have also found over 70 plant fossils, the majority of which are most closely related to plant life in today's subtropical or tropical regions.“This is enough to show that the central part of the now high-altitude, freezing Qinghai-Tibet Plateau had flourishing subtropical plants 47 million years ago,” Su says.The findings provide new evidence for the study of the evolutionary history of biodiversity and the evolution of the plateau's landscape, according to Zhou Zhekun, the paper's corresponding author and a researcher at the tropical botanical garden.Chinalaunched the second comprehensive scientific expedition to the Qinghai-Tibet Plateau in June 2017, 40 years after the first. Lasting up to 10 years, the expedition will conduct a series of studies focusing on the plateau'sglaciers, its biodiversity and ecological changes, and will also monitor the changes in climate.4. How did the paper come to the conclusion?A. Through the observation of the Baingoin basin.B. Through the fossils found in scientific expedition.C. Through the drawing of a large number of fossils.D. Through the adventure on the Qinghai-Tibetan Plateau.5. What can be inferred according to Su Tao?A. The average altitude of the plateau was 1,500 meters.B. “Shangrila”means a place with abundant water and grass.C. The flourishing subtropical plants have covered the plateau.D. The fossils found by researchers are tropical or subtropical plants now.6. Where might the passage come from?A. The Times.B. The Wall Street Journal.C. Chinese National Geography.D. The Economist.7. What is the purpose of the passage?A. To instruct.B. To educate.C. To persuade.D. To inform.CPreparations for the Tokyo Olympics have suffered another challenge after a survey found that 60% of people in Japan want them to be cancelled,less than three months before the Games are scheduled to open.Japan has extended a state of emergency in Tokyo and several other regions until the end of May as it struggles to control a fast increase in COVID-19 cases caused by new, more catching variants（变异体）with medical staff warning that health services in some areas are on the edge of breaking down.The Olympics, which were delayed by a year due to the pandemic, are set to open on 23 July, with the International Olympic Committee（IOC）and organizers insisting that measures will be put in place to ensure the safety of athletes and other visitors, as well as a nervous Japanese public.The survey, conducted between 7 and 9 May by the conservative Yomiuri Shimbun, showed 60% wanted the Games cancelled as opposed to 39% who said they should be held. “Postponement” — an option abandoned by the IOC — was not offered as a choice.Of those who said the Olympics should go ahead, 23% said they should take place without audience. Foreignaudience have been banned but a final decision on native attendance will be made in June.Another poll conducted at the weekend by TBS News found 65% wanted the Games cancelled or postponed again, with 37% voting to give up the event altogether and 28% calling for another delay. A similar poll in April conducted by Kyodo news agency found 70% wanted the Olympics cancelled or postponed.The IOC's vice president, John Coates, said that while Japanese sentiment about the Games “was a concern”, he could foresee no situation under which the sporting events would not go ahead.8. How many Japanese wish the Olympics would not be held in Tokyo according to the survey?A. 60%.B. 28%.C. 37%.D. 70%.9. What should be put into consideration if the Olympics open?A. The economic crisis.B. The urban transport.C. The safety of athletes.D. The health condition of citizens.10. What is some people's attitude towards foreign spectators in Paragraph 5?A. Welcome.B. Unfriendly.C. Cold.D. Unsupported.11. What can we conclude from John Coates'words?A. The Olympics will be stopped this year.B. The Olympics will be put off.C. The Olympics will be held normally.D. The Olympics will take place in other place.DThe Great Barrier Reef's outlook remains “very poor” despite coral (珊瑚) recovery over the past year, Australian government scientistssaid Monday, just days before a UNESCO ruling on the site's world heritage (遗产) status.The United Nations cultural agency recommended last month that the world's largest reef (珊瑚礁) system be placed on its endangered list because of damage to the corals largely caused by climate change.The Australian Institute of Marine Science (AIMS) said the corals were now in a “recovery window” after a decade of harmful heat stress and cyclones (旋风). But such opportunities were becoming rarer due to the influence ofclimate change, the government agency, which has monitored the reef for 35 years, said in its annual report released today. “The increasing emergence of climate-related extreme weather events and starfish outbreaks is causing more severe and frequent pressures, giving the reef fewer opportunities like this to recover,”CEO Paul Hardisty said. The scientists surveyed 127 reef sites in 2021 and found hard coral cover hadincreased at 69 of the 81 locations surveyed in the past two years.Separate scientific research released last October found the 2, 300-kilometre (1, 400 miles) system had lost half its corals since 1995, with a series of ocean heatwaves causing mass coral death.Britta Schaffelke, research program director at AIMS, said the latest findings provided a slight hope that the reef still has the power of recovering. But she added that its future is still very poor because of the dangers of climate change and other factors that are affecting the reef.UNESCO has urged Australia to take urgent climate action but the government has long resisted calls to commit to net zero emissions (排放) by 2050. The government has said it hopes to meet the target “as soon as possible” without harming its economy, insisting dealing with climate change requires a global effort. The reef was worth about US $4. 8 billion a year in tourism for the Australian economy and there are fears that an “in danger” listing could weaken its tourist appeal.12. What is the major cause of the damage to the corals?A. The climate change.B. Lack of money.C. Over development.D. Too many tourists.13. What is mainly talked about in Paragraph 3?A. The result of the survey.B. The efforts AIMS has made.C. The slight chance of the recovery.D. The terrible situation of the climate.14. What is Britta Schaffelke's attitude towards the future of the reef?A. Unclear.B. Positive.C Intolerant. D. Anxious.15. What can we infer from the last paragraph?A. Australia wants to put the reef on the endangered list.B. The Australian government has ignored UNESCO's demand.C. Australia hopes to keep a balance between emission target and its economy.D. The Australian government refuses to take its share of responsibility of climate change.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学年日照市实验高级中学高三英语下学期期末考试试题及答案解析
2019-2020学年日照市实验高级中学高三英语下学期期末考试试题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ASilicon Valley VolunteersDo you want to give back to your community while making new friends? Silicon Valley Volunteers is an organization that includes both wishes. The organization is designed to help people find opportunities to volunteer in theirlocal community while meeting others with similar interests. Silicon Valley Volunteers focuses on providing opportunities that fit into the busy schedules (日程安排) of professionals. Many opportunities listed on the site are in the evenings or weekends.Please browse (浏览) the website to learn more about the organization. In addition to volunteer opportunities, there are other opportunities for becoming a leader as well. For any questions about the group, ***************************.Discussion about the groupCourtney H: Hey everyone. I’m new to this group and would love to start meeting up with you guys. I’m a little unclear on how this group works. When you guys go ahead and have an event can someone message me the infomation? Thanks so much.Former member: How about an online discussion for ways we can still volunteer. I just checked in here and thought there would be online activities!Maria O: We need volunteers on Friday, November 8th for The Silicon Valley Philanthropy Day! Volunteers would act as greeters, check-in registrars, and ushers (接待员). The event is from 11 am to 1:30 pm on Nov. 8th @ **********************************************************************************.Firstcome,first serve!1.What is one of the purposes of the organization?A.To help find job opportunities.B.To help make new friends.C.To help develop a new hobby.D.To help experience the joy of sharing.2.Where can you learn more about the organization?A.In the magazine.B.In the newspaper.C.On the Internet.D.On TV.3.What kind of volunteers are needed from Maria O?A.Baby-sitters.B.Translators.C.Teachers.D.Greeters.BBeing an Olympian (奥运会选手) demands focus, determination, and a competitive spirit. Plus, representing your country is a lot of pressure. However, two athletes recently showed the world another quality that is definitely worth championing.Qatar's Mutaz Essa Barshim and Italy's Gianmarco Tamberi were competing in the high jump on Sunday when they reached a stalemate (僵局). Both men had managed to jump over a surprising 2.37 meters with no faults along the way. However, after three attempts neither managed the next level of 2.39 meters.An Olympic official suggested ajump-offbetween the two friends and rivals (竞争对手) to determine who would get the gold medal. But Barshim had another plan to reward their efforts.“Can we have two golds?” he asked the official.The official agreed and the two men jumped for joy. This was the first time a gold medal had been shared since 1912. “He is one of my best friends, not only on the track, but outside the track. We work together. This is a dream come true.” shared Barshim.The decision to share the medal was particularly meaningful to Tamberi. The Italian had suffered an ankle injury that prevented him from competing in the Rio Olympics in 2016, and it nearly put an end to his career altogether. So this year he brought along his cast to this year's Olympics with “Road to Tokyo 2021” to inspire him along the way.For Barshim, the gold has topped off his already impressive medal collection, having received a bronze and a silver medal in 2012 and 2016 respectively.Despite all the glory of receiving a gold medal for their countries, their achievement means so much more. These two individuals, trying to do their very best for their countries, have provided a wonderful example to all those competing in sports. They've summed up exactly what it means to take part in a global event with a generous and compassionate (有同情心的) spirit.4. What happened to Barshim and Tamberi in the competition?A. They ended in a tie.B. They quit the competition.C. They set a new record.D. They ran out of strength.5. What does the underlined word “jump-off” in Paragraph 3 refer to?A. Debate.B. Vote.C. Celebration.D. Extra round.6. Why was the gold medal particularly meaningful to Tamberi?A. It could bring him a lot of money.B. It may make up for his regret in 2016.C. It was a glory for his country.D. It could complete his medal collection.7. What does the author mainly want to convey in the text?A. The importance of sharing.B. The glory of winning gold medals.C. The valuable and special team spirit.D. The considerate and sharing Olympic spirit.CThereare two days that set you on your path in life: the day you’re born, and the day you realize why you were born.Growing up south of Chicago in Harvey, Illinois, most people just had their heads down trying to make it from point A to point B. I was the same way, just going with the flow. I played basketball in high school because I was good at it and because other people thought I should until I discovered my talent.I give up basketball and started doing speeches. It wasn’t a popular decision but my grandfather told me to do what made me happy. I fell in love with comedy and performing. And when I discovered the passion, I realized why I was born.I knew I had something to offer —I knew that not only am I powerful, but I can make a difference.I realized a long time ago that my dream is not to be famous or rich. My talent is to entertain. But it’s more than that. I have the chance to reach people, to brighten days, to bring laughter and positive energy into lives and inspire. And I am grateful forit.Acting putting myself out there and having doors closed on me time and time again has taught me a lot about myself. I have learned to trust what I have to offer the world over momentary doubt. I’ve learned to put my faith over my feelings. And I've grown a tough skin. More importantly, I have learned there is a long way towards our goals and that when we put our talents and passion to work, we determine our value.Like a lot of places across the country, there’s poverty, crime, violence and unemployment in Harvey. And growing up there, a lot of people have tragically low expectations for life. But I know that with the rightopportunity and with help along the way, everyone can find their passion and go after it. My life is proof.8. What was the author born to do according to the text?A. Be a basketball player.B. Act and perform.C. Make speeches.D. Teach people.9. What does the underlined word “it” in Paragraph 5 refer to?A. Chance.B. Energy.C. Days.D. Laughter.10. What is the author’s purpose of writing this text?A. To help others find their talents.B. To prove his decision was right.C. To inspire people to follow their dreams.D. To encourage people to set a goal.11. What can be the best tile for the text?A. Success Lies in Hard Work.B. How to Achieve the Dream Is Important.C. The Two Important Days in Life.D. The Day I Realized What I Was Born to Do.DCanadaIs Our NeighbourCanada and the United States are neighbours.They are on the same land.They share the same long boundary(国界).These two nations are similar in many ways.Canada buys many goods from the United States.Cars and clothes are two examples.The United States also buys goods from Canada.Much of the paper used in the United States comes from Canada.Some of the oilweuse comes from Canada,too.Americans travel toCanadaon holiday.And Canadians often visit the United States.It is easy for the people of one country to go to the other country.Canadians read about the United States in newspapers and magazines.Many Americans watch Canadian baseball and hockey (曲棍球)matches on Sundays.However,there are important differences between theUnited Statesand Canada.The United States has more people.Because the population is smaller,there are more open places in Canada.There is much unused land.This is another important difference.12.Canadabuys from theUnited States.A.oil and paperB.nothingC.many thingsD.everything13.In the first paragraph “we” means ________.A.CanadiansB.AmericansC.ChineseD.students14.The people in theUnited Stateslike Canadian ________.A.baseballB.basketballC.newspapersD.oil15.Which of the following statements is WRONG?A.Canada has less people than theUSA.B.Canada has not used all the land.C.Canada is connected withAmerica.D.Canadians don’t like hockey.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。
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日照实验高中2019级模块考试（必修5）
一、选择题：（每小题5分，共60分）
1、ΔABC 中,a =1,b =3, A =30°,则B 等于 A ．60° B ．60°或120° C ．30°或150° D ．120°
2、两灯塔A,B 与海洋观察站C 的距离都等于a (km), 灯塔A 在C 北偏东30°,B 在C 南偏东60°,则A,B 之间相距
A ．a (km)
B ．3a (km)
C ．2a (km)
D ．2a (km)
3、等差数列{a n }中，已知a 1＝1
3
，a 2+a 5＝4，a n ＝33，则n 为
A ．50
B ．49
C ．48
D ．47 4、已知等比数列{a n }的公比为2, 前4项的和是1, 则前8项的和为 A ．15. B ．17. C ．19. D ．21
5、等差数列{a n }中，a 1+a 2+…+a 50＝200，a 51+a 52+…+a 100＝2700，则a 1等于 A ．－1221 B ．－21．5 C ．－20．5 D ．－20
6、设集合y x y x y x A --=1,,|),{(是三角形的三边长}，则A 所表示的平面区域（不含边界的阴影部分）是
．
A
B ．
C ．
D ．
7、已知-9,a 1,a 2,-1四个实数成等差数列，-9,b 1,
b 2,b 3,-1五个实数成等比数列，
则b 2(a 2-a 1)= （）
A.8
B.-8
C.±8
D.
8、目标函数y x z +=2，变量y x ,满足43035251x y x y x -+<⎧⎪
+≤⎨⎪≥⎩，则有
A ．3,12min max ==z z
B ．,12max =z z 无最小值
C ．z z ,3min =无最大值
D ．z 既无最大值，也无最小值
9、在三角形ABC 中，如果()()3a b c b c a bc +++-=，那么A 等于
8
9
A ．030
B ．060
C ．0120
D ．0150 10、已知数列{}n a 的前n 项和()21n S n n =+则5a 的值为
A ．80
B ．40
C ．20
D ．10 11、f x ax ax ()=+-21在R 上满足f x ()<0，则a 的取值范围是 A ．a ≤0 B ．a <-4 C ．-<<40a D ．-<≤40a 12．若实数a 、b 满足a +b =2，则3a +3b 的最小值是
A ．18
B ．6
C ．23
D ．243 二、填空题：（每小题4分，共16分，答案写在第二卷上）
13、在△ABC 中，sin A =2cos B sin C ，则三角形为 三角形
14、不等式
21
131
x x ->+的解集是 ． 15、已知数列{ a n }满足条件a 1 = –2 , a n + 1 =2 +
n
n
a 1a 2-, 则a 5 = ． 16、若关于x 的不等式m x x ≥-42对任意]1,0[∈x 恒成立，则实数m 的取值范围是 ．
日照实验高中2004级模块考试（必修5）
一、填空题答案：
1 3、 14、
15、 16、 三、解答题： 17、（12分）三个数成等比数列,其积为512,如果第一个数与第三个数各减2,则成等
差数列，求这三个数．
18、（12分）解关于x 的不等式ax 2－(a ＋1)x ＋1＜0．
19、（12分）如图，在四边形ABCD 中，已知AD ⊥CD, AD=10, AB=14, ∠BDA=60︒,
∠BCD=135︒ 求BC 的长．
20、（12分）在某海滨城市附近海面有一台风，据测，当前台风中心位于城市O （如图）的东偏南)10
2(cos =θθ方向300km 的海面P 处,并以20km/h 的速度向西
偏北45°方向移动，台风侵袭的范围为圆形区域，当前半径为60km ，并以10km/h
的速度不断增大，问几小时后该城市开始受到台风的侵袭？受到台风的侵袭的时间有多少小时？
O P θ45°东西
北东
21、（12分）某工厂用两种原料A 、B 配成甲、乙两种药品，每生产一箱甲药品使用4kg 的A 原料，耗时1小时，每生产一箱乙药品使用4kg 的B 原料，耗时2小时，该厂每天最多可从原料厂获取16kg 的A 原料和12kg 的B 原料，每天只能有8小时的合成生产时间，该厂生产一箱甲药品获得3万元，生产一箱乙药品获得1万元，怎样安排生产才能获利最大？最大利润是多少？ 22、（14分）设,4,221==a a 数列}{n b 满足：,1n n n a a b -=+ 122n n b b +=+，
(1) 求证：数列}2{+n b 是等比数列(要指出首项与公比)，
(2) 求数列}{n a 的通项公式．
参考答案：
一、选择题
1-5BCABC 6-10ABDBC 11-12DB 二、填空题
13、等腰14、 1|23x x ⎧
⎫-<<-⎨⎬⎩
⎭ 15、107 16、(,3]-∞-
三、解答题
17、解:设三数为.,,aq a q a ⎪⎩
⎪⎨⎧⎩⎨
⎧==⇒=-+⎪⎪⎭⎫ ⎝⎛-=∴282)2(25123q a a aq q a a 或⎪⎩⎪⎨⎧==.218q a 则三数为,4,816或,168,.4
18、解: 16．解：当a ＝0时，不等式的解为x ＞1；当a ≠0时，分解因式a (x －a
1)(x
－1)＜0 当a ＜0时，原不等式等价于(x －a
1)(x －1)＞0，不等式的解为x ＞1或x ＜a
1；
当0＜a ＜1时，1＜a
1，不等式的解为1＜x ＜a
1；
当a ＞1时，a
1＜1，不等式的解为a
1＜x ＜1；
当a ＝1时，不等式的解为 。
19、解：在△ABD 中，设BD=x
则BDA AD BD AD BD BA ∠⋅⋅-+=cos 2222 即 60cos 1021014222⋅⋅-+=x x 整理得：096102=--x x
解之：161=x 62-=x （舍去）
由余弦定理： BCD BD CDB BC ∠=∠sin sin ∴2830sin 135
sin 16
=⋅=
BC 20、解：设经过t 小时台风中心移动到Q 点时，台风边沿恰经过O 城， 由题意可得：OP=300，PQ=20t ，OQ=r(t)=60+10t 因为102cos =
θ，α=θ-45°,所以10
2
7sin =θ，54cos =α
由余弦定理可得：OQ 2=OP 2+PQ 2-2·OP ·PQ ·αcos
即 (60+10t)2=3002+(20t)2-2·300·20t ·5
4
即0288362=+-t t ， 解得121=t ,242=t
-2t 121=t
答：12小时后该城市开始受到台风气侵袭，受到台风的侵袭的时间有12小时？
21、解、参考新课标必修五第98页例题题。
22解：(1)),2(222211
+=+⇒+=++n n n n b b b b ,22
2
1=+++n n b b
又42121=-=+a a b , ∴ 数列}2{+n b 是首项为4,公比为2的等比数列. (2)2224211-=⇒⋅=+∴+-n n n n b b .
.221-=-∴-n n n a a
令),1(,,2,1-=n n 叠加得)1(2)222(232--+++=-n a n n ,
22)2222(32+-++++=∴n a n n .22221
2)12(21n n n n -=+---=+。