广东省广州执信中学2019学年高二第二学期期末考试试卷(深圳外国语学校命题)

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执信中学高二第语文二学期期末考试

执信中学高二第语文二学期期末考试执信中学高二第语文二学期期末考试本试卷分选择题和非选择题两部分，共10页，满分为150分。

考试用时150分钟。

注意事项：1、答卷前，考生务必用黑色字迹的钢笔或签字笔将自己的姓名和学号填写在答题卡和答卷密封线内相应的位置上，用2B铅笔将自己的学号填涂在答题卡上。

2、选择题每小题选出答案后，有2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案；不能答在试卷上。

3、非选择题必须用黑色字迹的钢笔或签字笔在答卷纸上作答，答案必须写在答卷纸各题目指定区域内的相应位置上，超出指定区域的答案无效；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

4、考生必须保持答题卡的整洁和平整。

第一部分选择题(共9题，27分)一、基础知识部分（15分）1．下列加点字的读音全部正确的一项是（）A．狡黠x iá画帖tiè胴体tònɡ翘首以待qiáoB．巨擘bò饿殍fú虎穴xué顺蔓摸瓜wànC．忸怩ní踮脚diǎn奢靡mǐ力能扛鼎ɡānɡD．沏茶qī市侩kuài攻讦jié泾渭分明jīnɡ2．依次填入下列各句横线处的词语，最恰当的一项是（）因为文人只是文人，他们或许能写出不少感动人的故事，自己却很少有这种故事。

有时也出现这种故事了，但这又与文化史关系不大。

他们在做这些事情的时候，不是以文人的身份，是以忠臣或守将的身份，进入了政治史和军事史，从而着文化史。

A．毕竟似乎但推进B．到底似乎而推展C．毕竟好像而推进D．到底好像但推展3．下列各句中加点的成语使用恰当的一句是（）A．在这次竞选总统的演讲中，奥巴马以幽默的语言，独到的见解与深刻的思想赢得了选民热烈的掌声，其言辞真如空谷足音，令人耳目一新。

B．广州市教育局近日组织了一次中学生“学会感恩”演讲比赛，参赛的选手们个个广开言路，酣畅淋漓地发挥自己的口才。

广东省广州执信中学2019学年高二第二学期期末考试试卷(深圳外国语学校命题)

2019学年第二学期高二期末考试试卷物理本试卷分选择题和非选择题两部分，共5页，满分100分，考试用时75分钟注意事项1.答题卡前，考生务必用黑色字迹的钢笔或签字笔将自己的校名，姓名、考号、座位号等相关信息填写在答题卡指定区域2.选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑：如需改动，用橡皮擦干净后，再选涂其它答案：不能答在试卷上3.非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上如需改动，先划掉原来的答案，然后再写上新的答案：不准使用铅笔和涂改液不按以上要求作答的答案无效4.考生必须保持答题卡的整洁第一部分选择题（共52分）一、选择题：本题共8小，每小题4分，共32分，在每小题给出的四个选项中，只有一项符合题目要求。

1.以下关于物理学史，物理概念和物理思维的说法正确的是A.法拉弟通过十余年的研究，最终发现了电磁感应定律B爱因斯坦最早提出量子的概念，并成功地解释了光电效应现象C.楞次首先测出阴极射线的比荷，并把它命名为电子D.卢瑟福用α粒子轰击氮原子核而发现质子是人类第一次实现的原子核的人工转变2．新冠肺炎防控中有一个重要环节是对外来人员进行体温检测，检测用的体温枪工作原理就是黑体辐射定律。

黑体辐射的实验规律如图所示，由图可知下列描述正确的是A．随着温度升高，各种波长的辐射强度都有增加B．温度降低，可能部分波长的辐射强度会减小C．随温度升高，辐射强度的极大值向频率较小的方向移动D．随温度降低，辐射强度的极大值向波长较短的方向移动3．已知23290Th发生衰变的半衰期为24天，关于衰变下列说法正确的是A．衰变放出γ射线是23290Th从高能级向低能级跃迁时放出的B．23290Th发生β衰变产生的新核有143个中子，91个电子C．现在有80个23290Th，经过96天后，未衰变的23290Th原子数为5个D ．β衰变是弱相互作用引起的，β射线是核内一个中子转化成一个质子时放出的4.图甲是浇花的一种喷壶，图乙是喷壶的切面简化图，假设喷壶中装有水但未装满，里面有一部分压强为P 0的空气。

广州市执信中学高二英语期末考试真题

广州市执信中学高二英语期末考试真题1.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。

ASwimming with DolphinsIf you have ever dreamed of swimming with dolphins, then Dolphin Cove in Jamaica is the perfect place. Our 25,000-square-foot award-winning facility offers a number of exciting opportunities and programs.Interacting with dolphins can be a dream come true. Experience the thrill of “riding’” a dolphin belly ride, where two dolphins push the palms of your feet and take you for a ride through the water. Or, choose a dolphindorsal(背鳍) tow，where you are pulled through the water by holding onto the dorsal fin of the dolphin as he swims backward. Many people feel the dorsal tow is the most exciting activity offered.The deep-water swim offers the opportunity to experience the power and speed of a dolphin. You wear a dive mask(潜水面具) and hold onto the dolphin’s dorsal fin as he swims through the water. On the swim, the dolphin will kindly show you some of his favourite activities like handshakes, kisses and even a great dolphin song.Age Requirement1.With a Dolphin Encounter Program, children aged 1 to 5 must be accompanied by a paying adult.2.The Dolphin Swim Program is only available to persons 8 years of age or older.3.Children aged 1 to 7 can participate in the Encounter Program free of charge with a paying adult.Dolphin Cove VideoVideo cameras may not be taken into the water during any of the swim programs.Photographers will be present at all swim programs to take the best photos for you. You can also bring your underwater cameras if you like.（Leading Journals）1. The text is mainly intended for______.A. tourists who want to spend a holiday in JamaicaB. people who are interested in different kinds of dolphinsC. tourists planning a visit to Dolphin Cove in JamaicaD. people who have ever dreamed of swimming with dolphins2. Which of the following activities is considered to be the most exciting?A. The deep-water swim.B. The dolphin dorsal tow.C. The dolphin belly ride.D. The Dolphin Swim Program.3. Children aged from 1 to 5 can attend the Dolphin EncounterProgram______.A. aloneB. with an adultC. for freeD. with a swimming coach4. What should visitors bring with them if they want to take photos while swimming with dolphins?A. Video cameras.B. Dive masks.C. Underwater cameras.D. Photographers.【答案】CDBDBAfter a day with dolphins at Dolphin Cove, a visit to Dunn’s River Falls Jamaica is another must-do experience.Dunn’s River Falls is a famous waterfall near Ocho Rios, Jamaica. It is about 183 meters high and 55 meters wide. Over time，the falls have been known as a living natural museum of history and adventure. Here are some reasons why visiting the falls is a great opportunity.First, it is a truly beautiful place. As you walk up the falls with your tour group, you will see crystal-clear water glasses trickling down the rocks, the sound of flowing water can be heard throughout the climb. Take a swim in the refreshing pools of water on the way up but don’t forget to hold handsand create a human rope with your fellow visitors as a sign of togetherness.Second, it is an amazing adventure. The waterfall climbs starts at the bottom of the falls and continues up about 183 meters. Visitors can climb on their own or pyramid(金字塔) style in a human rope. Tours at the falls offer guides to assist visitors in case they need help; however, many people like to experience climbing the falls unassisted, which is also very popular.Finally, it is a great site for photography. You will want your camera with you to capture the memories of one of the most amazing adventures of your life. Rental cameras are available, but using your own, with water protection if necessary, is better.So come, and enjoy all that Dunn’s River Falls has to offer. Your visit will be unforgettable!5. What do we know about Dunn’s River Falls?A. It is located in Ocho Rios, Jamaica.B. It is the highest waterfall in the world.C. It was made by human hands.D. It has a pool for visitors to have a swim.6. What can visitors do on the way up Dunn’s River Falls?A. Walk in the crystal-clear water.B. Take a sleep on the rocks.C. Enjoy a beautiful landscape.D. Create a human rope to support others.7. If you need help while climbing the falls,______.A. you can ask the tour guide for assistanceB. you can get a map at the entrance of itC. you can climb with a partnerD. you can use the rope assist8. What does the text suggest visitors do at Dunn’s River Falls?A. Photograph the moments of climbing the falls.B. Rent a camera for taking photos.C. Take a refreshing swim in the falls.D. Use your camera underwater for better photos.【答案】ACADCVisitors often tell Dolphin Cove staff that swimming with dolphins is high on their “bucket lists”. Here is what a few visitors have said.-They made me feel comfortable and at ease in the water.-I was able to create many great memories.-This is an experience my family and I will never forget.-The trainers were knowledgeable and fun to be around.-Dolphin Cove is definitely the place to come if swimming with dolphins is your dream.Visiting Dolphin Cove proved to be one of the best experiences of my life! I can’t wait to come back!9. Dolphin Cove is a place where visitors can______.A. relax in the waterB. learn about the trainersC. talk about their experiencesD. get professional dolphin training10. The comment in paragraph 3 suggests that swimming with dolphins______.A. helps create lifelong memoriesB. provides an educational experienceC. is suitable for visitors of all agesD. requires professional swimming skills11. How do visitors generally feel about visiting Dolphin Cove?A. They feel relaxed and at ease.B. They find it educational and fun.C. They are highly satisfied with the experience.D. They believe more trainers should be available.【答案】CADDDear all,There is a talk being held at our school on Monday，April 25th, at 3:00 p.m. The speaker is Mark Jenkins, a famous explorer and writer. He will be talking about his latest book, “Into the Hollow Earth”，and sharing his personal experiences in caves and underground systems. The talk will be held in the school auditorium, and it is open to all students and teachers.In his book, Mark Jenkins writes about his amazing journey into the deepest caves on earth. He will talk about the difficulties he faced underground, the fascinating discoveries he made, and the wonderful underground landscapes he explored. By sharing his experiences, Mark hopes to encourage everyone to explore the world and pursue their own dreams.This is a great opportunity for all of us to learn from a renowned explorer. Don’t miss out on this incredible journey into the Hollow Earth! We look forward to seeing you there!12. Who is the speaker of the talk?A. A school teacher.B. A famous writer.C. An explorer.D. A student.13. What will Mark Jenkins talk about in his book?A. His journey into the earth’s core.B. His exploration of caves in the world.C. His experience in the school auditorium.D. His wonderful underground landscapes.14. What does Mark hope to achieve by sharing his experiences?A. To encourage people to travel more.B. To explore the world’s famous caves.C. To persuade students to write more books.D. To encourage people to pursue their dreams.15. Who is invited to attend the talk?A. Students and teachers.B. Teachers only.C. Explorers and writers.D. Famous speakers.【答案】CABD。

广州执信中学数学高二下期末测试(含答案解析)

一、选择题1．已知关于x 的方程20ax bx c ++=，其中,,a b c 都是非零向量，且,a b 不共线，则该方程的解的情况是（） A ．至少有一个解 B ．至多有一个解 C ．至多有两个解D ．可能有无数个解2．函数()sin()(0,0,)2f x A x A πωφωφ=+>><的部分图象如图所示,若将()f x 图象向左平移4π个单位后得到()g x 图象,则()g x 的解析式为( )A ．2()2sin(2)3g x x π=+ B ．5()2sin(2)6g x x π=- C ．()2sin(2)6g x x π=+D ．()2sin(2)3g x x π=-3．12sin(2)cos(2)ππ+-⋅-( ) A ．sin2cos2+ B ．cos2sin 2- C ．sin2cos2-D ．cos2sin2±-4．将函数()sin 3f x x πω⎛⎫=+⎪⎝⎭的图象向左平移(0)ϕϕ>个单位长度后得到函数()cos2g x x =的图象，则ϕ的最小值为（）A ．3π B ．6π C ．12πD ．24π5．平面向量(1,2)a =，(4,2)b =，c ma b =+（m R ∈），且c 与a 的夹角等于c 与b 的夹角，则m =（） A ．2-B ．1-C ．1D ．26．已知向量a 、b 、c 满足a b c +=，且::1:1:2a b c =a 、b 夹角为（） A ．4π B ．34π C ．2π D ．23π 7．在锐角ABC 中，4sin 3cos 5,4cos 3sin 3A B A B +=+=C 等于（）A ．150B ．120C ．60D ．308．若动直线x a =与函数()sin f x x =和()cos g x x =的图像分别交于M N ，两点，则MN 的最大值为（）A ．1BC D ．29．在给出的下列命题中，是假命题的是（） A ．设O A B C 、、、是同一平面上的四个不同的点，若(1)(R)OA m OB m OC m =⋅+-⋅∈，则点、、A B C 必共线B ．若向量,a b 是平面α上的两个不平行的向量，则平面α上的任一向量c 都可以表示为(R)c a b λμμλ=+∈、，且表示方法是唯一的C ．已知平面向量OA OB OC 、、满足|(0)OA OB OC r r ===，且0OA OB OC ++=，则ABC ∆是等边三角形D ．在平面α上的所有向量中，不存在这样的四个互不相等的非零向量a b c d 、、、，使得其中任意两个向量的和向量与余下两个向量的和向量相互垂直 10．已知函数()(0,0)y sin x ωθθω=+<为偶函数，其图象与直线1y =的某两个交点横坐标为1x 、2x ，若21x x -的最小值为π，则（） A ．2,2πωθ==B ．1,22==πωθ C ．1,24==πωθ D ．2,4==πωθ11．已知角α的终边经过点()2,1P -，则sin cos sin cos αααα-=+（）A ．4-B ．3-C ．12D ．3412．已知函数()2sin(2)(0)f x x ϕϕπ=+<<，若将函数()f x 的图象向右平移6π个单位后关于y 轴对称，则下列结论中不正确．．．的是 A ．56πϕ=B ．(,0)12π是()f x 图象的一个对称中心C ．()2f ϕ=-D ．6x π=-是()f x 图象的一条对称轴13．若O 为ABC ∆所在平面内一点，()()20OB OC OB OC OA -⋅+-=，则ABC ∆形状是（）． A ．等腰三角形 B ．直角三角形 C ．正三角形 D ．以上答案均错14．已知角6πα-的顶点在原点，始边与x 轴正半轴重合，终边过点()5,12P -，则7cos 12πα⎛⎫+= ⎪⎝⎭( )A ．26-B ．26-C ．26D ．2615．设0002012tan15cos 2sin 2,,221tan 15a b c =-==+，则有（） A ．c a b <<B ．a b c <<C ．b c a <<D ．a c b <<二、填空题16．已知ABC ∆是顶点为A 腰长为2的等腰直角三角形，P 为平面ABC 内一点，则()PA PB PC ⋅+的最小值是__________．17．设tan α、tan β是方程2320x x -+=的两个根，则()tan αβ+=________________. 18．空间四点,,,A B C D 满足3AB =，=7BC ，||=11CD ，||=9DA ，则·AC BD =_______．19．实数x ，y 满足223412x y +=，则23x y +的最大值______．20．已知平面向量a ，b 满足|a |=1，|b |=2，|a ﹣b a 在b 方向上的投影是__________．21．已知函数()2cos sin 2=-f x x x ，则()f x 的最大值是__________． 22．在ABC ∆中，角,,A B C 所对的边分别为,,a b c ，1a =，3B π=，当ABC ∆的面积等tan C =__________．23．已知ABC ∆，4AB AC ==，2BC =，点D 为AB 延长线上一点，2BD =，连结CD ，则cos BDC ∠=__________．24．已知0>ω，在函数sin y x ω=与cos y x ω=的图象的交点中，距离最短的两个交点，则ω值为__________．25．设G 是ABC ∆的重心（即三条中线的交点），AB a =，AC b =，试用a 、b 表示AG =________. 三、解答题26．已知a ，b ，c 分别为ABC ∆内角A ，B ，C 的对边，222sin 2cos 22B Aa b b c +=+．（1）求B ；（2）若6c =，[2,6]a ∈，求sin C 的取值范围．27．设函数()sin 1f x x x =+.（1）求函数()f x 的值域和函数的的单调递增区间；（2）当()135f α=，且263ππα<<时，求2sin 23πα⎛⎫+⎪⎝⎭的值. 28．已知数列{}n a 是公差不为零的等差数列，1a ＝1，且139,,a a a 成等比数列． (1)求数列{}n a 的通项；(2)设2n an b =，求数列{}n b 的前n 项和S n .29．已知:4,(1,3)a b ==- （1）若//a b ，求a 的坐标；（2）若a 与b 的夹角为120°，求a b -.30．如图，在△ABC 中，∠ABC ＝90°，AB ＝3，BC ＝1，P 为△ABC 内一点，∠BPC ＝90°. (1)若PB ＝12，求PA ； (2)若∠APB ＝150°，求tan ∠PBA ．【参考答案】2016-2017年度第*次考试试卷参考答案**科目模拟测试一、选择题 1．B 2．C 3．C 4．C 5．D 6．C7．D8．B9．D10．A11．B12．C13．A14．B15．A二、填空题16．【解析】【分析】以所在直线为轴建立坐标系设运用向量的坐标运算和向量数量积的坐标表示得出关于的表达式配方即可得出结论【详解】以所在直线为轴以边上的高为轴建立坐标系是直角边为2的等腰直角三角形且为直角顶17．【解析】【分析】利用二次方程根与系数的关系得出和的值然后利用两角和的正切公式计算可求出的值【详解】由二次方程根与系数的关系得出因此故答案为【点睛】本题考查两角和的正切公式的应用同时也考查了二次方程根18．0【解析】【分析】由代入再由代入进一步化简整理即可【详解】因为故答案为0【点睛】本题主要考查向量的数量积运算灵活运用数量积的运算公式即可属于常考题型19．【解析】分析：根据题意设则有进而分析可得由三角函数的性质分析可得答案详解：根据题意实数xy满足即设则又由则即的最大值5；故答案为：5点睛：本题考查三角函数的化简求值关键是用三角函数表示xy20．【解析】分析：根据向量的模求出•=1再根据投影的定义即可求出详解：∵||=1||=2|﹣|=∴||2+||2﹣2•=3解得•=1∴在方向上的投影是=故答案为点睛：本题考查了平面向量的数量积运算和投影21．【解析】分析：对函数求导研究函数的单调性得到函数的单调区间进而得到函数的最值详解：函数设函数在故当t=时函数取得最大值此时故答案为：点睛：这个题目考查了函数最值的求法较为简单求函数的值域或者最值常用22．【解析】由题意即则所以由余弦定理所以所以应填答案点睛：解答本题的思路是先借助三角形的面积公式求出边进而运用余弦定理求出边然后再运用余弦定理求出进而求出最后求出23．【解析】取中点中点由题意中又所以故答案为24．【解析】由题意令则所以即当；当如图所示由勾股定理得解得25．【解析】【分析】延长交于点利用重心的性质得出以及中线向量可求出的表达式【详解】延长交于点则点为线段的中点由平面向量加法的平行四边形法则可知则为的重心因此故答案为【点睛】本题考查向量的基底分解解题的关三、解答题 26． 27． 28． 29． 30．2016-2017年度第*次考试试卷参考解析【参考解析】**科目模拟测试一、选择题 1．B 解析：B 【解析】【分析】根据平面向量基本定理可知(),c a b R λμλμ=+∈，从而将方程整理为()()20x a x b λμ+++=，由,a b 不共线可得200x x λμ⎧+=⎨+=⎩，从而可知方程组至多有一个解，从而得到结果.【详解】由平面向量基本定理可得：(),c a b R λμλμ=+∈则方程20ax bx c ++=可变为：20ax bx a b λμ+++= 即：()()20xa xb λμ+++=,a b 不共线 200x x λμ⎧+=∴⎨+=⎩可知方程组可能无解，也可能有一个解∴方程20ax bx c ++=至多有一个解本题正确选项：B 【点睛】本题考查平面向量基本定理的应用，关键是能够利用定理将方程进行转化，利用向量和为零和向量不共线可得方程组，从而确定方程解的个数.2．C解析：C 【解析】【分析】根据函数的图象求出函数()f x 的解析式，再根据图象的平移变换得到()g x 的解析式即可. 【详解】由图象可知，A =2,541264T πππ=-=, 2T ππω∴==, 2ω∴=,又当512x π=时，52sin(2)212πφ⨯+=，即5sin()16πφ+=， 2πφ<， 3πφ∴=-，故()sin()f x x π=-223，将()f x 图象向左平移4π个单位后得到()g x ，∴ ()2sin[2()]2sin(2)436g x x x πππ=+-=+，故选：C 【点睛】本题主要考查了正弦型函数的图象与性质，图象的变换，属于中档题.3．C解析：C 【解析】【分析】先利用诱导公式化简角，然后利用正弦的二倍角公式和完全平方式结合角在各个象限中的符号化简即可得到答案. 【详解】==，∵22ππ<<，∴sin2cos20->.∴原式sin2cos2=-. 故选C. 【点睛】本题考查诱导公式和二倍角公式以及三角函数在各个象限中的符号的应用，属于基础题.4．C解析：C 【解析】【分析】根据题意得到变换后的函数解析式，利用诱导公式求得结果【详解】由题，向左平移(0)ϕϕ>不改变周期，故2ω=，∴平移得到()sin 2sin 22cos 233x x x ππϕϕ⎡⎤⎡⎤⎛⎫++=++= ⎪⎢⎥⎢⎥⎣⎦⎝⎭⎣⎦ 2+=+232k ππϕπ∴，12k πϕπ∴=+0ϕ>，∴当0k =时，min 12πϕ=，故选C【点睛】本题考查函数()sin y A x ωϕ=+的图象变换规律，利用诱导公式完成正、余弦型函数的转化5．D解析：D 【解析】【分析】【详解】()()4,22,422258c m m a c m m m =++⋅=+++=+,()()44222820b c m m m ⋅=+++=+,5,2025a b ===,c 与a 的夹角等于c 与b 的夹角 ,c a c bc a c b ⋅⋅=⋅⋅,=,解得2m =, 故选D. 【考点定位】向量的夹角及向量的坐标运算.6．C解析：C 【解析】【分析】对等式a b c +=两边平方，利用平面向量数量积的运算律和定义得出0a b ⋅=，由此可求出a 、b 的夹角. 【详解】等式a b c +=两边平方得2222a a b b c +⋅+=，即2222cos a b b c a θ+⋅+=，又::1:1:a b c =0a b ⋅=，a b ∴⊥，因此，a 、b 夹角为2π，故选：C. 【点睛】本题考查平面向量夹角的计算，同时也考查平面向量数量积的运算律以及平面向量数量积的定义，考查计算能力，属于中等题.7．D解析：D 【解析】【分析】由题：()()224sin 3cos 25,4cos 3sin 12A B A B +=+=，两式相加即可求出sin()A B +，进而求出A B +，角C 得解.【详解】由题：()()224sin 3cos 25,4cos 3sin 12A B A B +=+=，2216sin 24sin cos 9cos 25A A B B ++=，2216cos 24cos sin 9sin 12A A B B ++=，两式相加得：()1624sin cos cos sin 937A B A B +++=，1sin()2A B +=，所以1sin sin(())2C A B π=-+=，且C 为锐角，所以30C =. 故选：D 【点睛】此题考查同角三角函数基本关系与三角恒等变换综合应用，考查对基本公式的掌握和常见问题的处理方法.8．B解析：B 【解析】【分析】【详解】构造函数，根据辅助角公式，对函数的解析式进行化简，再根据正弦函数求出其最值，即可得到答案．则可知2()sin cos sin 4F x x x x π⎛⎫=-=- ⎪⎝⎭，F（x 2,故|MN|2,故选B9．D解析：D 【解析】【分析】【详解】由()()1OA m OB m OC OA OC m OB OC CA mCB =⋅+-⋅⇒-=⋅-⇒=⋅ 则点、、A B C 必共线，故A 正确；由平面向量基本定理可知B 正确；由 (0)OA OB OC r r ===>可知O 为ABC ∆的外心，由0OA OB OC ++=可知O 为ABC ∆的重心，故O 为ABC ∆的中心，即ABC ∆是等边三角形，故C 正确；存在四个向量（1，0），（0，1），（2，0），（0，-2）其中任意两个向量的和向量与余下两个向量的和向量相互垂直,D 错误故选D.10．A解析：A 【解析】分析：首先根据12x x -的最小值是函数的最小正周期，求得ω的值，根据函数是偶函数，求得θ的值，从而求得正确的选项.详解：由已知函数sin()(0)y x ωθθπ=+<<为偶函数，可得2πθ=，因为函数sin()(0)y x ωθθπ=+<<的最大值为1，所以21x x -的最小值为函数的一个周期，所以其周期为T π=，即2=ππω，所以=2ω，故选A.点睛：该题考查的是有关三角函数的有关问题，涉及到的知识点有函数的最小正周期的求法，偶函数的定义，诱导公式的应用，正确使用公式是解题的关键，属于简单题目.11．B解析：B 【解析】【分析】根据角的终边上一点的坐标，求得tan α的值，对所求表达式分子分母同时除以cos α，转化为只含tan α的形式，由此求得表达式的值. 【详解】依题意可知1tan 2α=-，11sin cos tan 1231sin sin tan 112αααααα----===-++-+．故选B. 【点睛】本小题主要考查三角函数的定义，考查齐次方程的计算，属于基础题. 12．C解析：C 【解析】函数()()2sin 2f x x ϕ=+的图象向右平移6π个单位，可得()2sin 23g x x πϕ⎛⎫=-+ ⎪⎝⎭,() 2sin 23g x x πϕ⎛⎫=-+ ⎪⎝⎭的图象关于y 轴对称，所以32k ππϕπ-+=+, 0k =时可得5=6πϕ，故5()2sin(2)6f x x π=+，555()=2sin()2sin 2362f πππϕ+==，()2f ϕ=-不正确，故选C. 13．A解析：A 【解析】【分析】根据向量的减法运算可化简已知等式为()0CB AB AC ⋅+=，从而得到三角形的中线和底边垂直，从而得到三角形形状. 【详解】()()()20OB OC OB OC OA CB AB AC -⋅+-=⋅+= ()CB AB AC ∴⊥+∴三角形的中线和底边垂直 ABC ∆∴是等腰三角形本题正确选项：A 【点睛】本题考查求解三角形形状的问题，关键是能够通过向量的线性运算得到数量积关系，根据数量积为零求得垂直关系.14．B解析：B 【解析】分析：利用三角函数的定义求得66cos sin ππαα⎛⎫⎛⎫-- ⎪ ⎪⎝⎭⎝⎭，结果，进而利用两角和的余弦函数公式即可计算得解．详解：由三角函数的定义可得512,613613cos sin ππαα⎛⎫⎛⎫-=--= ⎪ ⎪⎝⎭⎝⎭，则773cos cos cos 12661264ππππππααα⎛⎫⎛⎫⎛⎫+=-++=-+ ⎪⎪⎪⎝⎭⎝⎭⎝⎭33=cos cos sin sin 6464ππππαα⎛⎫⎛⎫⎛⎫⎛⎫--- ⎪⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎝⎭512=13213226⎛⎛⎫---⋅=- ⎪ ⎝⎭⎝⎭点睛：本题考查任意角的三角函数的定义，两角和与差的余弦函数公式，考查了计算能力和转化思想，属于基础题．15．A解析：A 【解析】【分析】利用两角差的正弦公式化简a ，分子分母同乘以2cos 15结合二倍角的正弦公式化简b ，利用降幂公式化简c ，从而可得结果. 【详解】()sin 302sin28a =︒-︒=︒ ，222sin15cos15sin 30cos 15cos 15b ==+sin28a >=sin25sin28,c a b a c ==︒<︒=∴>>，故选A.【点睛】本题主要考查二倍角的正弦公式、二倍角的余弦公式，两角差的正弦公式，意在考查综合运用所学知识解答问题的能力，属于中档题.二、填空题16．【解析】【分析】以所在直线为轴建立坐标系设运用向量的坐标运算和向量数量积的坐标表示得出关于的表达式配方即可得出结论【详解】以所在直线为轴以边上的高为轴建立坐标系是直角边为2的等腰直角三角形且为直角顶解析：1-【解析】【分析】以BC 所在直线为x 轴建立坐标系，设P x y （，），运用向量的坐标运算和向量数量积的坐标表示，得出()PA PB PC ⋅+关于x y ，的表达式，配方即可得出结论．【详解】以BC 所在直线为x 轴，以BC 边上的高为y 轴建立坐标系，ABC ∆是直角边为2的等腰直角三角形，且A 为直角顶点，斜边22BC =，则022020A B C -（，），（，），（，），设P x y （，），则2222PB PC PO x y PA x y （，），（，），+==--=-∴()22222 22222212PA PB PC x y x y ⋅+=+-=+--（，∴当202x y ==，时，()PA PB PC ⋅+取得最小值-1．故答案为：-1．【点睛】本题考查了平面向量的数量积运算，运用坐标法解题是关键，属于中档题．17．【解析】【分析】利用二次方程根与系数的关系得出和的值然后利用两角和的正切公式计算可求出的值【详解】由二次方程根与系数的关系得出因此故答案为【点睛】本题考查两角和的正切公式的应用同时也考查了二次方程根解析：3-. 【解析】【分析】利用二次方程根与系数的关系得出tan tan αβ+和tan tan αβ的值，然后利用两角和的正切公式计算可求出()tan αβ+的值. 【详解】由二次方程根与系数的关系得出tan tan 3αβ+=，tan tan 2αβ=，因此，()tan tan 3tan 31tan tan 12αβαβαβ++===---，故答案为3-.【点睛】本题考查两角和的正切公式的应用，同时也考查了二次方程根与系数的关系，考查运算求解能力，属于中等题.18．0【解析】【分析】由代入再由代入进一步化简整理即可【详解】因为故答案为0【点睛】本题主要考查向量的数量积运算灵活运用数量积的运算公式即可属于常考题型解析：0 【解析】【分析】由BD AD AB =-代入·AC BD ，再由AC AD DC AC AB BC ，=+=+代入进一步化简整理即可. 【详解】因为()()()······AC BD AC AD AB AC AD AC AB AD DC AD AB BC =-=-=+-+()()222222211··22AB AD DC AD AB BC AB AD DC AD DC AD AB =+--=++-+--()()()2222222221111122222BC AB BC AB AD AC DC AD AB AC +++=+-+--+ ()()()222222111811219490222BC AB AD DC AB BC ++=--+=--+=. 故答案为0 【点睛】本题主要考查向量的数量积运算，灵活运用数量积的运算公式即可，属于常考题型.19．【解析】分析：根据题意设则有进而分析可得由三角函数的性质分析可得答案详解：根据题意实数xy 满足即设则又由则即的最大值5；故答案为：5点睛：本题考查三角函数的化简求值关键是用三角函数表示xy解析：【解析】分析：根据题意，设2cos x θ=，y θ=，则有24cos 3sin x θθ+=+，进而分析可得()25sin x θα+=+，由三角函数的性质分析可得答案．详解：根据题意，实数x ，y 满足223412x y +=，即22143x y +=，设2cos x θ=，y θ=，则()24cos 3sin 5sin x θθθα=+=+，3tan 4α⎛⎫= ⎪⎝⎭，又由()15sin 1θα-≤+≤，则525x -≤≤，即2x +的最大值5；故答案为：5．点睛：本题考查三角函数的化简求值，关键是用三角函数表示x 、y ．20．【解析】分析：根据向量的模求出•=1再根据投影的定义即可求出详解：∵||=1||=2|﹣|=∴||2+||2﹣2•=3解得•=1∴在方向上的投影是=故答案为点睛：本题考查了平面向量的数量积运算和投影解析：12【解析】分析：根据向量的模求出a •b =1，再根据投影的定义即可求出．详解：∵|a |=1，|b |=2，|a ﹣b ∴|a |2+|b |2﹣2a •b =3，解得a •b =1， ∴a 在b 方向上的投影是a b b⋅=12，故答案为12点睛：本题考查了平面向量的数量积运算和投影的定义，属于中档题．21．【解析】分析：对函数求导研究函数的单调性得到函数的单调区间进而得到函数的最值详解：函数设函数在故当t=时函数取得最大值此时故答案为：点睛：这个题目考查了函数最值的求法较为简单求函数的值域或者最值常用解析：2【解析】分析：对函数求导，研究函数的单调性，得到函数的单调区间，进而得到函数的最值.详解：函数()2cos sin2f x x x =-，()22sin 2cos24sin 2sin 2,f x x x x x =----'=设()()[]2sin ,422,1,1t x f x g t t t t ===--∈-'，函数在11-1-122⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭，，故当t=12-时函数取得最大值，此时,66x f ππ⎛⎫=--= ⎪⎝⎭故答案为：2. 点睛：这个题目考查了函数最值的求法，较为简单，求函数的值域或者最值常用的方法有：求导研究单调性，或者直接研究函数的单调性，或者应用均值不等式求最值.22．【解析】由题意即则所以由余弦定理所以所以应填答案点睛：解答本题的思路是先借助三角形的面积公式求出边进而运用余弦定理求出边然后再运用余弦定理求出进而求出最后求出解析：-【解析】由题意1sin 23ac π=4c =⇒=，则b ==，所以由余弦定理cosC ==sin C ==tan (C ==-- 点睛：解答本题的思路是先借助三角形的面积公式求出边4c =，进而运用余弦定理求出边b ==，然后再运用余弦定理求出cosC ==，进而求出sin C ==tan (C ==- 23．【解析】取中点中点由题意中又所以故答案为【解析】取BC 中点,E DC 中点F ，由题意,AE BC BF CD ⊥⊥，cos BDC sin DBF ∠=∠,ABE ∆中，1cos 4BE ABC AB ∠==，1cos 4DBC ∴∠=-，又21cos 12sin ,sin 4DBC DBF DBF ∴∠=-∠=-∴∠=，所以cos BDC ∠=，24．【解析】由题意令则所以即当；当如图所示由勾股定理得解得解析：π【解析】由题意，令sin cos x x ωω=， sin cos 0x x ωω-=，则sin 04x πω⎛⎫-= ⎪⎝⎭，所以4x k πωπ-=， k Z ∈，即14x k ππω⎛⎫=⋅+ ⎪⎝⎭，当10,4k x πω==， 12y =；当251,4k x πω==， 222y =-，如图所示，由勾股定理得()()()22221213y y x x -+-=，解得ωπ=.25．【解析】【分析】延长交于点利用重心的性质得出以及中线向量可求出的表达式【详解】延长交于点则点为线段的中点由平面向量加法的平行四边形法则可知则为的重心因此故答案为【点睛】本题考查向量的基底分解解题的关解析：1133a b +. 【解析】【分析】延长AG 交BC 于点D ，利用重心的性质得出23AG AD =以及中线向量 ()12AD AB AC =+可求出AG 的表达式. 【详解】延长AG 交BC 于点D ，则点D 为线段BC 的中点，由平面向量加法的平行四边形法则可知2AD AB AC a b =+=+，则1122AD a b =+， G 为ABC ∆的重心，因此，221111332233AG AD a b a b ⎛⎫==⨯+=+ ⎪⎝⎭，故答案为1133a b +. 【点睛】本题考查向量的基底分解，解题的关键就是三角形重心的性质和中线向量的应用，考查分析问题和解决问题的能力，属于中等题.三、解答题 26．（1）3B π=；（2）32⎤⎥⎣⎦. 【解析】【分析】（1）利用二倍角公式和正弦定理以及两角和与差的正弦公式进行化简，求解出cos B 的值后即可求出B 的值；（2）根据余弦定理先求解出b 的取值范围，然后根据sin sin c BC b=求解sin C 的取值范围. 【详解】（1）已知得2(1cos )12cos2A a B c b ⎛⎫-=+- ⎪⎝⎭，由正弦定理得sin sin cos sin sin cos A A B C B A -=-，即sin sin sin()sin()A C A B A B =+-=++sin()2sin cos A B A B -=， ∴1cos 2B =，解得3B π=．（2）由余弦定理得222222cos 636(3)27b a c ac B a a a =+-=-+=-+，∵[2,6]a ∈，∴b ∈，sin sin 2c B C b ⎤=∈⎥⎣⎦．【点睛】本题考查解三角形的综合应用，难度一般.（1）解三角形的边角化简过程中要注意隐含条件A B C π++=的使用；（2）求解正弦值的范围时，如果余弦值的范围容易确定也可以从余弦值方面入手，若余弦值不容易考虑则可以通过正弦定理将问题转化为求解边与角的正弦的比值范围.27．（1）值域是[]1,3-，单调递增区间为52+266k k ππππ⎡⎤-+⎢⎥⎣⎦，；（2）2425-.【解析】【分析】（1）根据三角函数的关系式，即可求求函数f （x ）的值域和函数的单调递增区间．（2）根据三角函数的诱导公式即可得到结论．【详解】（1）依题意()sin 1f x x x =+ 2sin 13x π⎛⎫=++ ⎪⎝⎭. 因为22sin 23x π⎛⎫-≤+≤ ⎪⎝⎭，则12sin 133x π⎛⎫-≤++≤ ⎪⎝⎭. 即函数()f x 的值域是[]1,3-. 令32222k x k πππππ-+≤+≤+，Z k ∈，解得52+266k x k ππππ-+≤≤，Z k ∈，所以函数()f x 的单调递增区间为52+266k k ππππ⎡⎤-+⎢⎥⎣⎦，，Z k ∈.（2）由()132sin 135f παα⎛⎫=++= ⎪⎝⎭，得4sin 35πα⎛⎫+= ⎪⎝⎭. 因为263ππα<<，所以23ππαπ<+<时，得3cos 35πα⎛⎫+=- ⎪⎝⎭.所以2sin 2sin233ππαα⎛⎫⎛⎫+=+=⎪ ⎪⎝⎭⎝⎭ 2sin cos 33ππαα⎛⎫⎛⎫++= ⎪ ⎪⎝⎭⎝⎭ 432425525-⨯⨯=-. 【点睛】三角函数求值的类型如下：（1）给角求值：关键是正确选用公式，以便把非特殊角的三角函数转化为特殊角的三角函数.（2）给值求值：关键是找出已知式与待求式之间的联系及函数的差异. ①一般可以适当变换已知式，求得另外函数式的值，以备应用； ②变换待求式，便于将已知式求得的函数值代入，从而达到解题的目的.（3）给值求角：实质是转化为“给值求值”，先求角的某一函数值，再求角的范围，确定角.28．(1)a n ＝n . (2)S n ＝2n ＋1－2. 【解析】【分析】【详解】(1)由题设知公差d ≠0，由a 1＝1，a 1，a 3，a 9成等比数列得121d +＝1812dd++，解得d ＝1，d ＝0(舍去)，故{a n }的通项a n ＝1＋(n －1)×1＝n . (2)由(1)知2=2n an nb =，由等比数列前n 项和公式得 S n ＝2＋22＋23＋…＋2n ＝()21212n --＝2n ＋1－2.点评：掌握等差、等比数列的概念及前n 项和公式是此类问题的关键．29．（1）(2,-或(2,-.（2）27a b -= 【解析】试题分析：（1）利用向量共线定理、数量积运算性质即可得出．（2）利用数量积运算性质即可的．试题解析：（1）∵(1,3b =-，∴2b =，与b 共线的单位向量为1,22bc b ⎛=±=±- ⎝⎭.∵4,//a a b =，∴()2,23a a c ==-或()2,23-. （2）∵04,2,,b 120a b a ===，∴b cos ,b 4a b a a ⋅==-， ∴()222228a ba ab b -=-⋅+=，∴27a b -=.点睛：平面向量中涉及有关模长的问题时，常用到的通法是将模长进行平方，利用向量数量积的知识进行解答，很快就能得出答案；另外，向量是一个工具型的知识，具备代数和几何特征，在做这类问题时可以使用数形结合的思想，会加快解题速度.30．（1）72（2）34【解析】试题分析:（1）在三角形中，两边和一角知道，该三角形是确定的，其解是唯一的，利用余弦定理求第三边.（2）利用同角三角函数的基本关系求角的正切值.（3）若是已知两边和一边的对角，该三角形具有不唯一性，通常根据大边对大角进行判断.（4）在三角兴中，注意这个隐含条件的使用.试题解析:解：（1）由已知得∠PBC ＝60°，所以∠PBA ＝30°. 在△PBA 中，由余弦定理得PA 2＝.故PA 7. 5分（2）设∠PBA ＝α，由已知得PB ＝sin α. 在△PBA 中，由正弦定理得3sin sin150sin(30)αα=︒︒-， 3＝4sin α. 所以tan α3tan ∠PBA 3分考点：（1）在三角形中正余弦定理的应用.（2）求角的三角函数.。

广东省广州市执信中学2018-2019学年高二下学期期末考试理数试题PDF版含答案

2018-2019学年高二级数学科期末考试试卷第一部分选择题（共60分）最新试卷十年寒窗苦，踏上高考路，心态放平和，信心要十足，面对考试卷，下笔如有神，短信送祝福，愿你能高中，马到功自成，金最新试卷多少汗水曾洒下，多少期待曾播种，终是在高考交卷的一刹尘埃落地，多少记忆梦中惦记，多少青春付与流水，人生，总有一次这样的成败，才算长大。

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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2{20}A x x x，{0}B x x ，则集合AB 等于（）A ．{2}xx B ．{21}xx C ．{1}xx D ．{01}xx 2.已知复数(1)23i z i （i 为虚数单位），则z 在复平面内对应的点位于（）A ．第一象限B ．第二象限 C．第三象限 D．第四象限3.等差数列n a 中，79416,1a a a ，则16a 的值是（）A ．22B ．16 C．15 D ．184.已知角的顶点与原点重合，始边与x 轴的非负半轴重合，终边在直线2y x 上，则22cossin 等于（）A ．45B．35C．35D ．455.如下图所示，1OA ，在以O 为圆心，以OA 为半径的半圆弧上随机取一点B ，则AOB的面积小于14的概率为（）A ．13B ．14C．12D ．166.某几何体的三视图如下图所示，则此几何体的体积是（）A ．203B．6C ．103D ．1637.函数22xy x 的图象大致是（）8.执行下面的程序框图，如果输入的10N，那么输出的S ( )A ．109B ．169C ．95D．20119.设01,1b a c ，则（）A ．2abbbc B．log log b a a c b c C．ccabba D．log log a b c c10.直三棱柱111ABCA B C 中，90BCA，,M N 分别是1111,A B AC 的中点，1BCCA CC ，则BM 与AN 所成的角的余弦值为（）A ．110B ．25C ．3010D ．2211.过抛物线22(0)ypx p 的焦点F 的直线l 依次交抛物线及其准线于点,,A B C ，若2BCBF ，且3AF ，则抛物线的方程为（）A ．2yx B．22yx C ．23y x D ．24yx12.我们把形如(0,0)b yabxa的函数称为“莫言函数”，其图象与y 轴的交点关于原点的对称点称为“莫言点”，以“莫言点”为圆心且与“莫言函数”的图象有公共点的圆称为“莫言圆”，当1a b 时，“莫言圆”的面积的最小值是（）A ．2B．52C．e D．3第二部分非选择题（共90分）二、填空题：本大题共4小题，每小题5分，共20分.13.设向量(,1),(1,3)a m b ，且()0a a b ，则m ___________.14.已知ABC 的面积为32，2AC ，3BAC，则ACB___________.15.已知54(12)(1)x ax 的展开式中2x 的系数为26，则实数a 的值为___________.16.已知实数,x y 满足不等式组204025xy x yxy ，则2()xy xy的取值范围是___________. 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）设数列n a 的前n 项和为n S ，已知213nnS a ，*n N .（1）求通项公式n a 及n S ；（2）设10nn b a ，求数列n b 的前n 项和nT 18.（本小题满分12分）在如图所示的几何体中，四边形ABCD 为平行四边形，90ABD，EB面ABCD ，//EF AB ，2AB ，3EB ，1,13EF BC ，且M 是BD 的中点.（1）求证: //EM 平面ADF ；（2）求二面角DAF B 的大小.19.（本小题满分12分）为及时了解适龄公务员对开放生育二胎政策的态度，某部门随机调查了90位30岁到40岁的公务员，得到情况如下表：（1）完成表格，并判断是否有99%以上的把握认为“生二胎意愿与性别有关”，并说明理由；（2）现把以上频率当作概率，若从社会上随机独立抽取三位30岁到40岁的男公力员访问，求这三人中至少有一人有意愿生二胎的概率.（2）已知15位有意愿生二胎的女性公务员中有两位来自省妇联，该部门打算从这15位有意愿生二胎的女性公务员中随机邀请两位来参加座谈，设邀请的2人中来自省女联的人数为X ，求X 的公布列及数学期望()E X .男性公务员女性公务员总计有意愿生二胎30 15 无意愿生二胎2025总计附：22()()()()()n ad bc ka b c d a c b d 20.（本小题满分12分）设椭圆2222:1(3)x y C aab的右焦点为F ，右顶点为M ，且113eOF OM FM，（其中O 为原点，e 为椭圆的离心率.20()P kk 0.050 0.010 0.001 0k 3.8416.63510.828（1）求椭圆C 方程；（2）若过点F 的直线l 与C 相交于,A B 两点，在x 轴上是否存在点N ，使得NA NB 为定值？如果有，求出点N 的坐标及相应定值；如果没有，请说明理由.21.（本小题满分12分）已知函数()()xf x ekx k k R .（1）试讨论函数()y f x 的单调性；（2）若该函数有两个不同的零点12,x x ，试求：（i ）实数k 的取值范围；（ii ）证明：124x x .请考生在22、23题中任选一题作答，如果多做，则按所做的第一题记分.22.（本小题满分10分）选修4-4：坐标系与参数方程曲线C 的极坐标方程是4sin()6，直线l 的参数方程是32545xt yt（t 为参数）.（1）将曲线C 的极坐标方程转化为直角坐标方程；（2）设直线l 与x 轴的交点是,M N 为曲线C 上一动点，求MN 的取值范围.23. （本小题满分10分）选修4-5：不等式选讲已知函数()3f x m x ，不等式()1f x 的解集为(1,5)；（1）求实数m 的值；（2）若关于x 的不等式()x af x 恒成立，求实数a 的取值范围.2015-2016学年度第二学期高二级数学科期末考试答案一、选择题 DCABA CACDC CD。

广东省执信中学高二下学期期末考试(英语)(1)

广东省执信中学高二下学期期末考试英语试卷本试卷分选择题和非选择题两部分，共14页，满分为150分。

考试用时1。

2、选择题每小题选出答案后，有2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案；不能答在试卷上。

不按以上要求作答的答案无效。

4、考生必须保持答题卡的整洁和平整。

第I卷(共80分)I、听力：第一节: 听力理解(2段共6小题；每小题1分，满分6分)每段播放两遍。

各段后有几个小题，各段播放前每小题有5秒钟的阅题时间。

请根据各段播放内容及其相关小题，在5秒钟内从题中所给的A、B、C选项中，选出最佳选项，并在答题卡上将该项涂黑。

听第一段对话, 回答第1- 3题。

1.What does the woman do in the afternoons?A.Go sightseeing.B. Do homework.C. Take classes.2.Why does the woman want to improve her English?A.To get a promotion soon.B.To make friends with English people.C.To communicate with her customers better.3.What do we know about the woman?A.She knew little English before she came to London.B.She has never been to London before.C.She practises English with her roommates.听第二段对话, 回答第4- 6题。

广州市2019-2020年高二下学期期末考试英语试题

下学期期末教学质量测试高二英语试卷本试卷共四部分。

第一、二部分和第三部分的第一节为选择题。

第三部分的第二节和第四部分为非选择题。

试卷共12页。

考试结束，将本试卷和答题卡一并交回。

第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.音频Which sport does Jane like best?A. Tennis.B. V olleyball.C. Basketball.【答案】A【解析】【分析】此题为听力题，解析略。

【详解】此题为听力题，解析略。

2.音频What will the woman do?A. Deliver the man’s baby.B. Attend the man’s wife.C. Take care of the man’s dog.【答案】C【解析】【分析】此题为听力题，解析略。

【详解】此题为听力题，解析略。

3.音频What did the man do?A. He played in the park.B. He removed the rubbish.C. He fixed the shower.【答案】B【分析】此题为听力题，解析略。

【详解】此题为听力题，解析略。

4.音频When will the woman probably arrive at Beijing Hotel?A. 11:20.B. 11:30.C. 11:50.【答案】C【解析】【分析】此题为听力题，解析略。

【详解】此题为听力题，解析略。

5.音频What are the two speakers mainly talking about?A. Where to spend the weekend.B. What to buy as birthday gifts.C. Whether to hold a birthday party.【答案】B【解析】【分析】此题为听力题，解析略。

2018-2019学年广东省深圳外国语学校高二(下)期末数学试卷(理科)

2018-2019学年广东省深圳外国语学校高二（下）期末数学试卷（理科）1.（单选题，5分）设复数z满足z（1+3i）=2-i（其中i为虚数单位），则z在复平面内对应的点落在（）A.第一象限B.第二象限C.第三象限D.第四象限2.（单选题，5分）已知集合A={x|y= √−x2+4x }，B={x|x2+x-6≥0}，则A∩∁R B=（）A.[0，2）B.（-3，4]C.（-3，0]D.[2，4]3.（单选题，5分）图中为截止2019年3月末，我国的外汇储备近1年的变化折线图，由此得到以下说法，其中叙述正确的是（）A.近1年以来，我国外汇储备月增长量最大的月份是2019年3月B.2018年4月至10月，我国外汇储备连续下降C.2018年底，我国外汇储备降至近年来最低D.截止2019年3月末，我国外汇储备连续第五个月上升4.（单选题，5分）如图，已知幂函数y=x a的图象过点P（2，4），则图中阴影部分的面积等于（）A. 163B. 83 C. 43 D. 235.（单选题，5分）已知双曲线C ： x 2a 2 - y 2b 2 =1的离心率e= 54 ，且其右焦点为F 2（5，0），则双曲线C 的方程为（） A. x 24 - y 23 =1B. x 29 - y 216 =1 C. x 216 - y 29 =1 D. x 23 - y 24 =1 6.（单选题，5分）已知a ＞0，且a≠1，函数f （x ）=log a （6-ax ），则“1＜a ＜3”是“f （x ）在（1，2）上单调递减”的（） A.充分不必要条件 B.必要不充分条件 C.充要条件D.既不充分也不必要条件7.（单选题，5分）《孙子算经》是中国古代重要的数学著作．其中的一道题“今有木，方三尺，高三尺，欲方五寸作枕一枚．问：得几何？”意思是：“有一块棱长为3尺的正方体方木，要把它作成边长为5寸的正方体枕头，可作多少个？”现有这样的一个正方体木料，其外周已涂上油漆，则从切割后的正方体枕头中任取一块，恰有一面涂上油漆的概率为（） A. 125216 B. 827 C. 49 D. 148.（单选题，5分）已知函数f （x ）=x 3-2xf′（1）+a-2，若f （x ）为奇函数，则曲线y=f （x ）在点（a ，f （a ））处的切线方程为（）A.2x-y=0B.y=0C.10x-y-16=0D.x-y+2=09.（单选题，5分）某学校共有2000名学生，各年级男、女生人数如表：方法从全校学生中抽取80名学生，则三年级应抽取的学生人数为（）A.16B.20C.30D.40（x∈R），若关于x的方程f2（x）-tf（x）+t-1=0恰好10.（单选题，5分）已知f（x）= |x|e x有4个不相等的实数根，则实数t的取值范围为（），2）∪（2，e）A.（1eB.（1，1）e+1）C.（1，1eD.（1，e）e11.（单选题，5分）点M（3，2）到抛物线C：y=ax2（a＞0）准线的距离为4，F为抛物线的最小值为（）的焦点，点N（1，1），当点P在直线l：x-y=2上运动时，|PN|−1|PF|A. 3−2√28B. 2−√24C. 5−2√28D. 5−2√2412.（单选题，5分）点P为棱长是2的正方体ABCD-A1B1C1D1的内切球O球面上的动点，点M为B1C1的中点，若满足DP⊥BM，则动点P的轨迹的长度为（）A. √5π5B. 2√5π5C. 4√5π5D. 8√5π513.（填空题，5分）定义运算法则如下：a ⊗b=a12+b−13，a⊕b=lga2−lgb12，M=21 4⊗8125，N=√2⊕125，则M+N=___ ．14.（填空题，5分）若随机变量X～N（2，32），且P（X≤1）=P（X≥a），则（ax- 1√x）5展开式x2项的系数是 ___15.（填空题，5分）《聊斋志异》中有这样一首诗：“挑水砍柴不堪苦，请归但求穿墙术．得诀自诩无所阻，额上坟起终不悟．”在这里，我们称形如以下形式的等式具有“穿墙术”：2 √23=√223，3√38=√338，4√415=√4415，5√524=√5524，•……则按照以上规律，若8 √8n =√88n具有“穿墙术”，则n=___ ．16.（填空题，5分）历史上，许多人研究过圆锥的截口曲线如图，在圆锥中，母线与旋转轴夹角为30°，现有一截面与圆锥的一条母线垂直，与旋转轴的交点O距离圆锥顶点M长度为1，对于所得截口曲线给出如下命题：① 曲线形状为椭圆；② 点O为该曲线上任意两点最长距离的三等分点；③ 该曲线上任意两点最长距离其为32，最短距离为23√3；④ 该曲线的离心率为√33．其中正确命题的序号为___ ．17.（问答题，12分）已知m＞0，设命题p：不等式x+|x-2m|＞1的解集为R，命题q：函数f（x）= −12x2+4x-3lnx在区间[m，m+1]上不单调，若命题p∨q为真命题，P∧q为假命题，求实数m的取值范围．18.（问答题，12分）如图，已知四边形ABCD为梯形，AB || CD，∠DAB=90°，BDD1B1为矩形，平面BDD1B1⊥平面ABCD，又AB=AD=BB1=1，CD=2．（1）证明：CB1⊥AD1；（2）求二面角B1-AD1-C的余弦值．19.（问答题，12分）已知椭圆E：x2a2+y2b2=1(a＞b＞0)，A、B为椭圆的左右顶点，F（√3，0）为其右焦点，点P为椭圆上位于第一象限内的点，且直线PA与直线PB的斜率之积为−14．（1）求椭圆E的方程；（2）设O为坐标原点，直线y=kx+2交椭圆E与M、N两点，求△MON面积的最大值．20.（问答题，12分）近年来，随着网络的普及，数码产品早已走进千家万户的生活，为了节约资源，促进资源循环利用，折旧产品回收行业得到迅猛发展，电脑使用时间越长，回收价值越低，某二手电脑交易市场对2018年回收的折旧电脑交易前使用的时间进行了统计，得到如图所示的频率分布直方图，在如图对时间使用的分组中，将使用时间落入各组的频率视为概率．（1）若在该市场随机选取3个2018年成交的二手电脑，求至少有2个使用时间在（4，8]上的概率；（2）根据电脑交易市场往年的数据，得到如图所示的散点图，其中x（单位：年）表示折旧电脑的使用时间，y（单位：百元）表示相应的折旧电脑的平均交易价格．（ⅰ）由散点图判断，可采用y=e a+bx作为该交易市场折旧电脑平均交易价格与使用年限x的回归方程，若t=lny i，t=110∑t i10i=1，选用如下参考数据，求y关于x的回归方程收购1000台折旧电脑所需的费用附：参考公式：对于一组数据（u i ，v i ）（i=1，2，…，n ），其回归直线 v ̂ = α̂ +βu 的斜率和截距的最小二乘估计分别为： β̂=∑u i v i −nuvni=1∑u i 2−nu2n i=1 ，α̂=v ‾−β̂u ‾ ．参考数据：e 3.25≈26，e 2.65≈14，e 2.05≈7.8，e 1.45≈4.3，e 0.85≈2.3．21.（问答题，12分）已知函数 f (x )=−alnx −e x x+ax ，a ∈R ．（Ⅰ）当a ＜0时，讨论函数f （x ）的单调性；（Ⅱ）当a=1时， F (x )=f (x )+(x +1x )e x −bx ，对任意x∈（0，+∞），都有F （x ）≥1恒成立，求实数b 的取值范围．22.（问答题，10分）在平面直角坐标系中，直线l 的参数方程为 {x =tcosαy =tsinα （t 为参数，0≤α＜π）．以坐标原点为极点，x 轴正半轴为极轴建立极坐标系，曲线C 的极坐标方程为ρ2-4=4ρcosθ-2ρsinθ．（Ⅰ）写出曲线C 的直角坐标方程；（Ⅱ）若直线l 与曲线C 交于A ，B 两点，且AB 的长度为2 √5 ，求直线l 的普通方程．23.（问答题，0分）设函数f （x ）=|x-1|-|2x+1|的最大值为m ．（1）作出函数f （x ）的图象；（2）若a 2+2c 2+3b 2=m ，求ab+2bc 的最大值．。

学年广东省广州市执信中学高二下学期期末考试英语试卷精选文档

学年广东省广州市执信中学高二下学期期末考试英语试卷精选文档 TTMS system office room 【TTMS16H-TTMS2A-TTMS8Q8-2015-2016学年度第二学期高二级英语科期末考试试卷本试卷分选择题和非选择题两部分，共14页，满分140分，考试用时120分钟。

注意事项：1．答题前，考生务必用黑色字迹的钢笔或签字笔将自己的姓名、考号、座位号等相关信息填写在答题卡指定区域内。

2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其它答案；不能答在试卷上。

3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

4．考生必须保持答题卡的整洁。

第Ⅰ卷第一部分阅读理解（共两节，满分40分）第一节阅读理解（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。

（A）Experience the newly opened Grand Canyon (大峡谷) West Skywalk in Colorado. Departing from Grand Canyon’s South Rim by Airplane to Grand Canyon’s West Rim, you will land and take a ground tour to the Skywalk! Walk on air for 70 feet over the edge of Grand Canyon West.This Skywalk has been open since March 28, 2007. Daily visitorship to the Skywalk has been over 4,000 people. Please be patient to enjoy your moment on the Skywalk.After you have experienced the one and only Grand Canyon Skywalk Glass Bridge, you will return to the Grand Canyon West Airport and take your Airplane for a flight back to the South Rim of the Canyon. This is a tour never to be forgotten as you will have walked on air over the Grand Canyon.Tour Itinerary (行程)1. This advertisement is for __________.A. Grand Canyon WestB. Grand Canyon SouthC. Grand CanyonD. the Skywalk2. The package fee does NOT cover the cost of _______.3. According to the Tour Itinerary, the route is ______.A. South Rim→Guano Point→West Airport→Eagle Point→West Airport→South RimB. South Rim→West Airport→Guano Point→Eagle Point→West Airport→South RimC. South Rim→West Airport→Eagle Point→Guano Point→West Airport→South RimD. South Rim→West Airport→Eagle Point→West Airport→Guano Point→South Rim（B）Alex Pang’s amusing new book The Distraction Addiction addresses those of us who feel panic without a cellphone or computer. And that, he claims, is pretty much all of us. When we’re not online, where we spend four months annually, we’re engaged in the stressful work of trying to get online.The Distraction Addiction is not framed as a self-help book. It’s a thoughtful examination of of how technological advances change consciousness. A “professional futurist”, Pang urges an a pproach which he calls “contemplative (沉思的) computing.” He asks that you pay full attention to “how your mind and body interact with computers and how your attention and creativity are influenced by technology.”Pang’s first job is to free you from the common misconception that doing two things at once allows you to get more done. What is commonly called multitasking is, in fact, switch-tasking, and its harmful effects on productivity are well documented. Pang doesn't advocate returning to a pre-Internet world. Instead, he asks you to “take a more ecological view of your relationships with technologies. Modern devices or media may be making specific tasks easier or faster but at the same timema king your work and life harder.”The Distraction Addiction is particularly fascinating on how technologies have changed certain fields of labor—often for the worse. For architects, computer-aided design has become essential but in some ways has cheapened the design process. As one architect puts it, “Architecture is first and foremost about thinking...and drawing is a more productive way of thinking” than computer-aided design. Somewhat less amusing are Pang’s solutions for kicking the Internet habit. Herecommends the usual behavior-controlling approaches, familiar to anyone who has completed a quit-smoking program. Keep logs to study your online profile and decide what you can knock out, download a program like Freedom that locks you out of your browser, or take a “digital Sabbath (安息日)”: “Unless you’re a reporter or emergency-department doctor, you'll discover that your world doesn’t fall apart when you go offline.”4. Alex Pang’s new book is aimed for readers who ______.A. find their work online too stressfulB. go online mainly for entertainmentC. are fearful about using the cellphone or computerD. can hardly tear themselves away from the Internet5. What does Alex Pang try to do in his new book?A. Offer advice on how to use the Internet effectively.B. Warn people of the possible dangers of Internet use.C. Predict the trend of future technological development.D. Examine the influence of technology on the human mind.6. What is the common view on multitasking?A. It enables people to work more efficiently.B. It is in a way quite similar to switch-tasking.C. It makes people’s work and life even harder.D. It distracts people’s attention from useful work.7. What is Alex Pang’s recommendation for Internet users?A. They use the Internet as little as possible.B. They keep a record of their computer use time.C. They exercise self-control over their time online.D. They entertain themselves online on off-days only.（C）I live in the land of Disney, Hollywood and year-round sun. You may think people in such a glamorous, fun-filled place are happier than others. If so, you have some mistaken ideas about the nature of happiness.Many intelligent people still equate happiness with fun. The truth is that fun and happiness have little or nothing in common. Fun is whatwe experience during an act. Happiness is what we experience after an act. It is a deeper, more abiding emotion.Going to an amusement park or ball game, watching a movie or television, are fun activities that help us relax, temporarily forget our problems and maybe even laugh. But they do not bring happiness, because their positive effects end when the fun ends.I have often thought that if Hollywood stars have a role to play, it is to teach us that happiness has nothing to do with fun. These rich, beautiful individual have constant access to glamorous parties, fancy cars, expensive homes, everything that spells “happiness”. But in memoir after memoir, celebrities reveal that unhappiness hidden beneath all their fun: depression, alcoholism, drug addiction, broken marriages, troubled children and profound loneliness.Ask a bachelor why he resists marriage even though he finds datingto be less and less satisfying. If he’s honest, he will tell you that he is afraid of making a commitment. For commitment is in fact quite painful. The single life is filled with fun, adventure and excitement. Marriage has such moments, but they are not its most distinguishing features.Similarly, couples that choose not to have children are deciding in favor of painless fun over painful happiness. They can dine out wheneverthey want and sleep as late as they want. Couples with babies are lucky to get a whole night’s sleep or a three-day vacation. I don’t know any parent who would choose the word fun to describe raising children. However, they all feel blessed to have children.Understanding and accepting that true happiness has nothing to do with fun is one of the most liberating realizations we can ever come to. It liberates time: now we can devote more hours to activities that can genuinely increase our happiness. It liberates money: buying that new car or those fancy clothes that will do nothing to increase our happiness now seems pointless. And it liberates us from envy: we now understand that all those rich and glamorous people we were so sure are happy because they are always having so much fun actually may not be happy at all.8. Which of the following can be inferred from the passage?A. Fun creates long-lasting satisfaction.B. Fun provides enjoyment while pain leads to happiness.C. Happiness is enduring whereas fun is short-lived.D. Fun that is long-standing may lead to happiness.9. To the author, Hollywood stars all have an important role to play that is to __________.A. write memoir after memoir about their happiness.B. tell the public that happiness has nothing to do with fun.C. teach people how to enjoy their lives.D. bring happiness to the public instead of going to glamorous parties.10. Couples having babies ____________.A. are lucky since they can have a whole night’s sleep.B. find fun in tucking them into bed at night.C. find more time to play and joke with them.D. gain happiness from their endeavor.11. If one gets the meaning of the true sense of happiness, he will ____________.A. stop playing games and joking with othersB. make the best use of his time increasing happinessC. give a free hand to moneyD. keep himself with his family（D）If you think a high-factor sunscreen (防晒霜) keeps you safe from harmful rays, you may be wrong. Research in this week’s Nature showsthat while factor 50 reduces the number of melanomas (黒瘤) and delays their occurrence, it can’t prevent them. Melanomas are the most aggressive skin cancers. You have a higher risk if you have red or blond hair, fair skin, blue or green eyes, or sunburn easily, or if a close relative has had one. Melanomas are more common if you have periodic intense exposure to the sun. Other skin cancers are increasingly likely with long-term exposure.There is continuing debate as to how effective sunscreen is in reducing melanomas--- the evidence is weaker than it is for preventing other types of skin cancer. A 2011 Australian study of 1,621 peoplefound that people randomly selected to apply sunscreen daily had halfthe rate of melanomas of people who used cream as needed. A second study, comparing 1,167 people with melanomas to 1,101 who didn't have the cancer, found that using sunscreen routinely, alongside other protection such as hats, long sleeves or staying in the shade, did give some protection. This study said other forms of sun protection ---not sunscreen--- seemed most beneficial. The study relied on peopleremembering what they had done over each decade of their lives, so it's not entirely reliable. But it seems reasonable to think sunscreen gives people a false sense of security in the sun.Many people also don't use sunscreen properly ---applyinginsufficient amounts, failing to reapply after a couple of hours and staying in the sun too long. It is sunburn that is most worrying--- recent research shows five episodes of sunburn in the teenage years increases the risk of all skin cancers.The good news is that a combination of sunscreen and covering up can reduce melanoma rates, as shown by Australian figures from their slip-slop-slap campaign. So if there is a heat wave this summer, it would be best for us, too, to slip on a shirt, slop on (抹上) sunscreen and slap on a hat.12. What does the research in Nature say about a high-factor sunscreen?A. It is ineffective in preventing melanomas.B. It is ineffective in case of intense sunlight.C. It is ineffective with long-term exposure.D. It is ineffective for people with fair skin.13. What do we learn from the 2011 Australian study of 1,621 people?A. Sunscreen should be applied alongside other protection measures.B. High-risk people benefit the most from the application of sunscreen.C. Irregular application of sunscreen does women more harm than good.D. Daily application of sunscreen helps reduce the incidence of melanomas.14. What does the author say about the second Australian study?A. It misleads people to rely on sunscreen for protection.B. It helps people to select the most effective sunscreen.C. It is not based on direct observation of the subjects.D. It confirms the results of the first Australian study.15. What does the author suggest to reduce melanoma rates?A. Using both covering up and sunscreen.B. Staying in the shade whenever possible.C. Using covering up instead of sunscreen.D. Applying the right amount of sunscreen.第二节阅读填空（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。

广东省广州市执信中学高二下学期期末考试英语试题

第一卷一．听力：请根据录音和所给中文，用英语写出三个问题和回答五个问题。

（共8小题，满分12分）In this part, you are required to act as a role and complete three communicative tasks: listen to the speaker, ask the speaker three questions and then answer five questions.情景介绍角色：你是学生。

任务：（1）在电视节目中向牛津大学的国际部主任提问；（2）根据谈话内容回答另一同学的提问。

请听完对话后，根据下面中文提示请你用英语提出三个问题（请把英语问题写在答卷上）Now please ask the speaker three questions. You have 20 seconds to prepare the question.1、你们拥有多少国际学生呢？2、去牛津学习是不是几乎不可能呢？3、那边的教育费用如何呢？Now please get ready to answer five questions in English. You’re allowed 10 seconds to prepare the answer. （电脑提问两遍，请把英语回答写在答卷上）二. 单项选择（共20小题，每小题1分，满分20分）1. As there is less and less coal and oil, scientists are exploring new ways of making use of ________ energy, such as sunlight, wind and water for power and fuel.A. antiqueB. alternativeC. arbitraryD. authentic2. ________ wise or eminent, judges are human and can make mistakes.A. EitherB. No matter howC. How muchD. Whether3. Every possible means ________ to prevent the air pollution, but the sky is still notclear.A. has been usedB. are usedC. is usedD. have been used4. After that, he knew he could ________ any emergency by doing what he could do tothe best of his ability.A. get away withB. get on withC. get throughD. get across5. Though ________ to see us, the professor gave us a warm welcome.A. was surprisedB. surpriseC. being surprisedD. surprised6. Teachers have to constantly update their knowledge in order to maintain theirprofessional ________.A. consequenceB. independenceC. competenceD. intelligence7. Ideally ________ for Broadway theaters and Fifth Avenue, the New York Park is afavorite with many guests.A. locatedB. locatingC. being locatedD. having been located8. He was thought to have been killed in action because his unit was lost and ________had been destroyed.A. each of his recordsB. all records of hisC. every record of hisD. all records of him9. Bats are surprisingly long-lived creatures, some________ a life span of around 20years.A. havingB. hadC. haveD. to have10. When Alice came to herself, she didn’t know how long she ________ there.A .has lain B. had been lyingC. has been lyingD. was lying11. People in Guangzhou are proud of ________ we have achieved in the past tenyears.A. thatB. whatC. whichD. how12. Even a mighty express train can be held up by signals. Which of the following wordscan be used to replace “held up”?A. delayedB. slowed upC. haltedD. retarded13.________ life pace continues to speed up, we are quickly losing the art ofenjoyment.A. whileB. WithC. AsD. When14. Until recently historians assumed that calendars ________ with the advent ofagriculture.A. came to existenceB. came to existC. were existedD. were existential15. There was a chance in a million that she might be right. Which of the followingphrases can be used to replace “a chance in a million”?A. the scarcest chanceB. the greatest chanceC. the remotest chanceD. the lightest chance16. I first met Lisa three years ago. She ________ at a radio shop at the time.A. was workingB. has workedC. had workedD. had been working17. The poor boy, ________ with extraordinary strength, ________ from the ground.A. filling; risingB. full; raisedC. filled; risenD. filled; rose18. There is no need to worry too much. In fact we shall solve this problem ________law and education.A. by means ofB. in terms ofC. in case ofD. in control of19. The dealer gave a brief ________ and told him that it was worth $50.A. eyeB. regardC. glanceD. peer20. When you are home, give a call to let me know you ________ safely.A. are arrivingB. have arrivedC. had arrivedD. will arrive三. 完形填空：（共15小题，每小题1分，满分15分）When traveling in the Canadian Rockies by car, I often notice a road sign that says, “A fed bear is a dead bear.” 21 , I did not get it. Why is a fed bear a dead one? According to a friend, many travelers used to throw their food from their cars for the bears. 22 , the bears turned to the roadside for food and slowly lost their ability to take care of themselves. When winter came, fewer travelers took to the mountains, which 23 less food for the bears, some of whom starved or frozen to death. So the Canadian government 24 warning signs along the road, essentially advising people not to feed the bears.This reminded me of a scientific experiment. Some white mice were 25 into two groups. One group spent their days only eating and sleeping, 26 the other, fed only with half the amount of food 27 , had to search for food. Half a year later, scientists found that the mice that had to search for their own food were 28 , while the fully fed ones were either ill or dead. It was 29 that the underfed white mice, in the 30 of searching for their food, had kept healthy by exercising in finding food, being adaptable, and 31 their immunity.Many over-concerned parents are feeding their children like bears or white mice. At present, children are only good at reading and studying, and are not 32 to think independently and act for themselves. 33 placed in strange environments, they are lost, confused, and helpless. Parents do not understand what it takes to 34 their children’s long-term success. They forget the most important thing---and that is how to cultivate their children into 35 adults, so that they can bravely undertake challenges and succeed in the future.21. A. First of all B. At the beginning C. After all D. In the end22. A. Normally B. Gradually C. Generally D. Actually23. A. proved B. explained C. meant D. advocated24. A. make up B. take up C. keep up D. put up25. A. separated B. divided C. arranged D. determined26. A. since B. when C. while D. if27. A. available B. regular C. delicious D. necessary28. A. alive B. popular C. healthy D. lazy29. A. evident B. reliable C. funny D. curious30. A. stage B. process C. aim D. period31. A. practicing B. affecting C. reducing D. improving32. A. forced B. advised C. allowed D. encouraged33. A. Until B. Unless C. Once D. As34. A. indicate B. ensure C. demand D. predict35. A. independent B. skilful C. helpful D. successful四. 阅读:第一节：阅读下面短文，然后从每题中所给的四个选项（A、B、C和D）中，选出最佳选项。

广东省广州市2019年高二下学期数学期末考试试卷A卷

广东省广州市2019年高二下学期数学期末考试试卷A卷姓名:________ 班级:________ 成绩:________一、双空题 (共4题；共4分)1. （1分）若幂函数的图象过函数的图象所经过的定点，则 ________，________.2. （1分） (2019高二下·镇海期末) 已知f（x）＝x2﹣4x+2，数列{an}是等差数列且单调递减，a1＝f（a+1），a2＝0，a3＝f（a﹣1），则数列{an}的公差为________，数列{an}的通项公式为________．3. （1分） (2019高二下·镇海期末) 若函数为偶函数，则k＝________，f（0）＝________．4. （1分） (2019高二下·镇海期末) 若2x+3y＝2，则4x+1+9y的最小值为________；的最小值为________．二、填空题 (共3题；共3分)5. （1分）和﹣终边相同的角为________．6. （1分）(2018·南阳模拟) 在中，角所对的边分别为，若且，则面积的最大值为________．7. （1分） (2019高二下·镇海期末) 若对于任意x∈[1，4]，不等式0≤ax2+bx+4a≤4x恒成立，|a|+|a+b+25|的范围为________．三、解答题 (共5题；共25分)8. （5分） (2017高二下·雅安期末) 已知p：﹣x2+4x+12≥0，q：x2﹣2x+1﹣m2≤0（m＞0）．（Ⅰ）若p是q充分不必要条件，求实数m的取值范围；（Ⅱ）若“￢p”是“￢q”的充分条件，求实数m的取值范围．9. （5分） (2016高一上·济南期中) 求下列各式的值：（Ⅰ）（Ⅱ）log3 ﹣ln1．10. （5分）已知a>0,设p：函数在R上是增函数；q：不等式对∀x∈R恒成立.若“p∨q”为真,“p∧q”为假,求实数a的取值范围.11. （5分） (2019高二下·镇海期末) 已知函数．（Ⅰ）若函数f（x）的最小值为8，求实数a的值；（Ⅱ）若函数g（x）＝|f（x）|+f（x）﹣16有4个零点，求实数a的取值范围．12. （5分） (2019高二下·镇海期末) 已知正实数列a1 ， a2 ，…满足对于每个正整数k，均有，证明：（Ⅰ）a1+a2≥2；（Ⅱ）对于每个正整数n≥2，均有a1+a2+…+an≥n．参考答案一、双空题 (共4题；共4分)1-1、2-1、3-1、4-1、二、填空题 (共3题；共3分)5-1、6-1、7-1、三、解答题 (共5题；共25分)8-1、9-1、10-1、11-1、12-1、。

广东省广州市执信中学2019-2020学年高二数学理下学期期末试卷含解析

广东省广州市执信中学2019-2020学年高二数学理下学期期末试卷含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 已知a＞0，b＞0，则的最小值是（）A.2B.C. 4D. 5参考答案：C略2. 已知函数y=f（x）在定义域内可导，其图象如图，记y=f（x）的导函数为y=f′（x），则不等式f′（x）≥0的解集为参考答案：[-4,-4/3]U[1,11/3]【考点】函数的单调性与导数的关系．【分析】根据导数与函数的单调性的关系，f′（x）≥0，f（x）为增函数，f′（x）≤0，f（x）为减函数，利用此性质来求f′（x）≥0的解集；【解答】解：如图f（x）在与上为增函数，可得f′（x）≥0，故[-4,-4/3]U[1,11/3]．【点评】此题考查函数的单调性与导数的关系，此题出的比较新颖，是一道基础题．3. 如图，长方形的四个顶点为，曲线经过点．现将一质点随机投入长方形中，则质点落在图中阴影区域的概率是（）A． B． C． D．参考答案：D4. 已知x＞0，由不等式x+≥2=2，x+=≥3=3，…，可以推出结论：x+≥n+1（n∈N*），则a=（）A． 2n B．3n C．n2 D．nn参考答案：C略5. 已知某等差数列共有10项，其奇数项之和为15，偶数项之和为30，则其公差为（）A．5 B．4 C．3 D．2参考答案：C【分析】写出数列的第一、三、五、七、九项的和即5a1+（2d+4d+6d+8d），写出数列的第二、四、六、八、十项的和即5a1+（d+3d+5d+7d+9d），都用首项和公差表示，两式相减，得到结果．【解答】解：，故选C．【点评】等差数列的奇数项和和偶数项和的问题也可以这样解，让每一个偶数项减去前一奇数项，有几对得到几个公差，让偶数项和减去奇数项和的差除以公差的系数．6. 定义在(－∞,0)∪(0,+∞)上的函数f(x),如果对于任意给定的等比数列{a n}, {f(a n)}仍是等比数列,则称f(x)为“保等比数列函数”. 现有定义在(－∞,0)∪(0,+∞)上的如下函数:①f(x)=x2; ②f(x)=2x; ③f(x)=; ④f(x)=ln|x|.则其中是“保等比数列函数”的f(x)的序号为（）A．①②B．③④C．①③D．②④参考答案：C7. 直三棱柱ABC-A1B1C1中，各侧棱和底面的边长均为a，点D是CC1上任意一点，连接A1B,BD, A1D,AD,则三棱锥A- A1BD的体积为（）A. B. C. D.参考答案：B8. 命题函数的单调增区间是，命题函数的值域为，下列命题是真命题的为（）A． B . C. D.参考答案：B略9. 某校有高一学生n名，其中男生数与女生数之比为6:5，为了解学生的视力情况，现要求按分层抽样的方法抽取一个样本容量为的样本，若样本中男生比女生多12人，则n=（）A. 990B. 1320C. 1430D. 1560参考答案：B【分析】根据题意得出样本中男生和女生所占的比例分别为和，于是得出样本中男生与女生人数之差为，于此可求出的值。

执信中学2019-2019高二下学期期末考试物理共8页文档

执信中学2019-2019高二下学期期末考试物理试卷本试卷分选择题和非选择题两部分，共10页，满分为100分。

考试用时90分钟。

2、选择题每小题选出答案后，有2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案；不能答在试卷上。

不按以上要求作答的答案无效。

4、考生必须保持答题卡的整洁和平整。

第一部分选择题(共68分)一、单选题（本题12小题，只有一个选项正确，每小题3分，共36分。

）1．在物理学理论建立的过程中，有许多伟大的科学家做出了贡献。

关于科学家和他们的贡献，下列说法中正确的是A．牛顿最早提出了“力不是维持物体运动的原因”的观点B．卡文迪许首先通过天平和弹簧秤测出了万有引力的常量C．安培提出了分子电流假说D．汤姆生通过对阴极射线的研究发现了电子，并提出了原子的核式结构学说2．卢瑟福α粒子散射实验的结果表明A．可以用人工方法研究原子核的结构B．原子的全部正电荷都集中在原子中央很小的核内C．质子比电子轻D．原子核内存在着中子3．下列四个方程中，表示α衰变的是A．+→ThU2349023892He42B．eNgNa0124122411-+→C．3141569236123592++→+BaKrnU n10D．nHeHH1423121+→+4．如图所示，木块放在表面光滑的小车上并随小车一起沿桌面向左做匀速直线运动．当小车遇障碍物而突然停止运动时，车上的木块将A．立即停下来B．立即向前倒下C．立即向后倒下D．仍继续向左做匀速直线运动5.运动员跳伞将经历加速下降和减速下降两个过程，将人和伞看成一个系统，在这两个过程中，下列说法中正确的是A．阻力对系统始终做负功B．系统受到的合外力始终向下C．重力做功使系统的重力势能增加D．任意相等的时间内重力做的功相等6.关于天然放射现象，下列说法中正确的是A．放射性元素的原子核内的核子有半数发生变化所需的时间就是半衰期B．放射性物质放出的射线中，α粒子动能很大，因此贯穿物质的本领很强C．当放射性元素的原子的核外电子具有较高能量时，将发生β衰变D．放射性元素的原子核发生衰变时，将会伴随着γ射线的辐射7.子弹水平射入一个置于光滑水平面上的木块中，则有A．子弹对木块的冲量必大于木块对子弹的冲量B．子弹受到的冲量和木块受到的冲量大小相等、方向相反C．当子弹与木块以同一速度运动后，子弹与木块的动量一定相等D．子弹与木块的动量变化的方向相反，大小不一定相等8. 某个带电粒子仅在电场力作用下由A点运动到B点，电场线、粒子在A点的初速度，以及运动轨迹如图所示。

2019-2020学年广州市执信中学高三英语下学期期末考试试卷及答案

2019-2020学年广州市执信中学高三英语下学期期末考试试卷及答案第一部分阅读（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AGetting your kid to bed at night is seriously one of the most challenging things you'll ever have to do. Most kids are just so full of energy that they'll tire you out before they're halfway through their store of energy. An easy thing to calm down your child to get into bed is giving in and allowing some iPad screen time. However, it's really not a great idea, just like you thought.Researchers at theArizonaStateUniversityconducted a study with 547 kids between the ages of 7 to 9. Their parents tracked how much screen time the kids were allowed along with their sleep patterns. The study found that kids who did not engage in screen time before bed slept for 23 more minutes every week and also went to sleep about 34 minutes earlier than those playing with iPad. Although that might not seem like so much more time, quality of sleep is vastly important in Children's development.The CDC's (美国疾病控制中心)2018 National Youth Risk Survey outlines that good quality sleep can impact a child's life in many ways, including affecting grades and also weight gain. Students with an "A" average slept for 30 or more minutes per night than those with a "D" or"F" average.A 2018PennsylvaniaStateUniversitystudy showed that children with irregular bedtimes had a higher risk of having increased body weight. Those with consistent and age-appropriate bedtimes when they were 9 years old had a healthier BMI (体质指数)at age 15 than those with irregular bedtimes.Hard as it is, it's really important not to give in and hand over an iPad to your child who is about to go to bed. Just like it's important for adults to go to sleep without any distractions, it's even more important for kids.1. What do the findings of the researchers at theArizonaStateUniversitysuggest?A. More sleep is necessary for children's development.B. Enough sleep helps improve academic performances.C. Screen time before bed leads to later and less sleep.D. Children sleeping irregularly are easy to gain weight.2. What is the text mainly about?A. How is screen time affecting teenagers?B. What are negative effects of irregular bedtimes?C. When should you get your kid to bed at night?D. Why is screen time before bed a bad idea for kids?3. Who is the text intended for?A. Parents.B. Children.C. Teachers.D. Researchers.BLast year, 138,000San Franciscoresidents used Airbnb, a popular app designed to connect home renters and travelers. It’s a striking number for a city with a population of about 850,000, and it was enough for Airbnb to win a major victory in local elections, asSan Franciscovoters struck down a debatable rule that would have placed time restrictions and other regulations on short-term rental services.The company fiercely opposed the measure, Proposition F, with a nearly $10 million advertising campaign. It also contacted its San Franciscan users with messages urging them to vote against Proposition F.Most people think of Airbnb as a kind of couch-surfing app. The service works for one-night stays on road trips and longer stays in cities, and it often has more competitive pricing than hotels. It’s a textbook example of the “sharing economy”, but not everyone is a fan.The app has had unintended consequences inSan Francisco. As the San Francisco Chronicle reported last year, a significant amount of renting on Airbnb is not in line with the company’s image: middle-class families putting up a spare room to help make ends meet. Some users have taken advantage of the service, using it to turn their multiple properties into vacation rentals or even full-time rentals. Backers of Proposition F argued that this trend takes spaces off the conventional, better-regulated housing market and contributes to rising costs.“The fact is, widespread abuse of short-term rentals is taking much needed housing off the market and harming our neighborhoods,” said ShareBetter SF, a group that supported Proposition F. Hotel unions have protested the company’s practices inSan Franciscoand other cities, saying that it creates an illegal hotel system.San Franciscois in the middle of a long-term, deeply rooted housing crisis that has seen the cost of living explode. Actually, explode is a generous term. The average monthly rent for an apartment is around $4, 000. Located on a narrow outcropping of land overlooking the bay,San Franciscosimply doesn’t have enough space to accommodate the massive inflow of young, high-salaried tech employees flocking toSilicon Valley.As the Los Angeles Times reported, someSan Franciscoresidents supported the measure simply because it seemed like a way to check a big corporation. Opponents of Proposition F countered that the housing crisis runs much deeper, and that passing the rule would have discouraged a popular service while doing little to solve thecity’s existing problems.4. The intention of Proposition F is to ________.A. place time limits in local election.B. set limits on short-term rental.C. strike down a controversial rule.D. urge users to vote against Airbnb.5. What is the negative consequence of Airbnb onSan Francisco?A. It shrinks the living space of middle-class families.B. Users are taken advantage of by the service financially.C. It makes the house market more competitive.D. It indirectly leads to high house rental price.6. The housing crisis inSan Franciscoresults from ________.A. explosion of the living costB. its geographic characteristicsC. generosity of local enterprisesD. inflow of migrant population7. Theauthor’s attitude toward Proposition F is ________.A. objectiveB. supportiveC. negativeD. indifferentCCuckoos don’t bother building their own nests—they just lay eggs that perfectly imitate those of other birds and take over their nests. But other birds are wishing up, evolving some seriously impressive tricks to spot the cuckoo eggs.Cuckoos are often know asparasites, meaning that they hide their eggs in the nest of other species. To avoid detection, the cuckoos have evolved so that eggs seem reproduction of those of their preferred targets. If the host bird doesn’t notice the strange egg in its nest, the little cuckoo will actually take the entire nest for itself after it comes out, taking the other eggs on its back and dropping them out of the nest.To avoid this unpleasant fate for their young, the other birds have evolved a few smart ways to spot the fakes, which we’re only now beginning to fully understand. One of the most amazing finds is that birds have an extra colour-sensitive cell in their eyes, which makes them far more sensitive to ultraviolet wavelengths and allows them to see a far greater range of colours than humans can. This allows cautious birds to detect a fake egg which might be exactly the same to our eyes.Fascinatingly, we’re actually able to observe different bird species at very different points in their evolutionary war with the cuckoos. For instance, some cuckoos lay their eggs in the nests of the redstarts. Theblue eggs these cuckoos lay are practically alike to those of the redstarts, and yet they are still sometimes rejected. Compare that with cuckoos who target dunnocks. While those birds lay perfectly blue eggs, their cuckoo invaders just lay white eggs with brown irregular shaped spots. And yet dunnocks barely ever seem to notice the obvious trick.Biologists suspect these more easily fooled species like the dunnocks are on the same evolutionary path as the redstarts, but they have a long way to go until they evolve the same levels of suspicion. What’s remarkable is that the dunnocks fakes are so bad and the redstart ones so good, and yet cuckoos are still more successful with the former than the latter.It speaks to just how thoroughly a species’ behavior can be changed by the pressures of natural selection, or it might just be a bit of strategic cooperation on the part of the dunnocks. Biologists have suggested that these birds are willing to tolerate a parasite every so often because they don’t want to risk accidentally getting rid of one of their own eggs.8. This passage can be most likely found in a ________.A. science surveyB. nature magazineC. zoo advertisementD. travel journal9. What does the underlined word “parasite” in paragraph 2 most probably refer to?A. Animals that work together to raise young.B. Small harmful animals such as worms or mice.C. Animals that can adapt to changing environments.D. Animals which live on or inside other host animals.10. Which of the following is TRUE about the dunnock according to the passage?A. It is colour-blind and therefore cannot identify foreign eggs in the nest.B. It can easily remove cuckoo eggs from the nest because fakes are so bad.C. It is a host bird that is more likely to raise a cuckoo chick than the redstart.D. It is unable to evolve and hence accepts cuckoo eggs that appear in the nest.11. Which of the following can be inferred from the passage?A. Dunnocks may eventually learn to recognise foreign eggs.B. Redstarts seem to be less suspicious compared to dunnocks.C. Cuckoo birds are good at taking responsibility for their own young.D. It is very easy for cuckoos to imitate the colouring of the dunnock’s egg.DIn June, 2021, a group of students from eight high schools in Winnipeg, the capital of Canada’s Manitoba province, will begin test-launching (试发射) a satellite the size of a Rubik’s cube.The one-kilogram Win-Cube satellite, named for its home city and its shape, will be put into low orbit. Once in space, it can perform for a few months or up to several years, communicating information that could help find the signs of earthquakes.There are 80 similar satellite projects worldwide, but this is the first high-school based program of its kind in Canada. 30 Manitoba high school students are having a hand in designing and building the satellite, in cooperation with aerospace (航空航天的) experts and 10 students from the University of Manitoba, and with support from two other organizations.The Win-Cube project is not something that goes on a piece of paper; it is real-world engineering, allowing high school students to have an opportunity to learn more about the exciting world of engineering through their participation in this challenging program. It is also taken as a wonderful example of the unique partnerships within Manitoba. Designing, building and launching a satellite with high-school participation will bring this world-class educational project into reality and Manitoba closer to space“These Manitoba high school students deserve congratulations for their enthusiasm, innovation (创新), and a strong love for discovery,” said Education, Citizenship and Youth Minister Peter Bjomson. “We want to make science more relevant, interesting and attractive to high school students by showing them how classroom studies can relate to practical experience in the workplace or, in this case, in space,” Bjomson added.The Win-Cube program is mainly aimed at inspiring a strong desire for discovery on the part of the students. It also shows Manitoba’s devotion to research and innovation and the development of a skilled workforce—all important drivers of knowledge-based economic growth.12. What can we learn from Mr. Bjomson? .A. Those Manitoba high school students are worth praising.B. The study of space can be practically made in classrooms.C. Manitoba high schools are famous for the study of space.D. Scientific research is too far away from high school students.13. What is the primary purpose of the project ? .A. To find the early signs of earthquakes.B. To relate studies to practical.C. To help high school students study real-world engineering.D. To inspire a strong desire for discovery among the students.14. According to the passage, what can we know about the Win-Cube satellite? .A. It is named after Manitoba and its shape.B. It is intended for international communication.C. It is designed like a Rubik’s cube both in shape and size.D. It is challenged by university students around the world.15. What may be the best title for the passage?A. Manitoba SchoolB. Win-Cube ProgramC. Space Co-operationD. Satellite Launching第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

广东执信中学18-19学度高二下年末试题--数学文

广东执信中学18-19学度高二下年末试题--数学文高二级数学科(文科)期末考试试卷本试卷分选择题和非选择题两部分共4页，总分值为150分.考试用时120分钟.考前须知：1、答卷前，考生务必用黑色字迹旳钢笔或签字笔将自己旳姓名和学号填写在答题卡和答卷密封 _ 线内相应旳位置上，用 2B 铅笔将自己旳学号填涂在答题卡上.2、选择题每题选岀【答案】后，用2B 铅笔把答题卡上对应题目旳【答案】标号涂黑；如需改动，用橡皮擦洁净后，再选涂其他【答案】；不能答在试卷上.3、非选择题必须用黑色字迹旳钢笔或签字笔在答卷纸上作答，【答案】必须写在答卷纸各题目指定区域内旳相应位置上，超岀指定区域旳【答案】无效；如需改动，先划掉原来旳【答案】，然后再写上新旳【答案】；不准使用铅笔和涂改液.不按以上要求作答旳【答案】无效 .4、考生必须保持答题卡旳整洁和平坦 .第一部分选择题(共50分)【一】选择题：本大题共 10小题，每题5分，共50分、在每题给出旳四个选项中，只有一项为哪一项符合题目要求旳、1、全集U=R 那么正确表示集合 M={— 1，0，1 }和N={ Xx 2+1 = 0 }关系旳韦恩〔Venn 〕4、以下函数中，在其定义域内既是奇函数又是增函数旳是既非充分也非必要条件1 bi2 i 是纯虚数〔是虚数单位，b 为实数〕，那么b =C.D.-22 13.假设函数 f(x)二sin x (x ・ R)，A.最小正周期为-■旳奇函数；B.那么f (x)是最小正周期为二旳奇函数;2 C.最小正周期为2二旳偶函数；D. 最小正周期为A. y =2xB. y =log 2X A.充分非必要条件B.c. y =x |x | D . y 二 sinx必要非充分条件C.充要条件D.①假设a //M ，b//M, 那么a//b 2.假设复数②假设a//M, b丄M那么b丄aI③假设a M b M,且 c ±a , c ±b,那么c 丄M ④假设a 丄M, a 〃N ，那么 Ml NA. 510B. 1022C. 254D. 2568 一个四面体旳顶点在空间直角坐系O-xyz 中旳坐标分别是〔1, 0, 1〕，〔 1, 1, 0〕，〔 0, 1,1〕，〔0, 0, 0〕，画该四面体三视图中旳正视图时，以 zOx 平面为投影面，那么得到旳正视图可为③当1 ■ a 2时，函数y = f (x) - a 有4个零点. A 0个 B 、3个 C. 2 个 D. 1个第二部分非选择题(共100分)【二】填空题：本大题共 4小题，每题5分，总分值20分、本大题分为必做题和选做题 . 〔一〕必做题：第 11、12、13题是必做题，每道试题考生都必须做答、9.在平面直角坐标系 xOy 中，己知圆C 在x 轴上截得线段长为 2 一，在y 轴上截得线段长为2匚圆心C 旳轨迹方程是 A x 2 y 2 =1y 2「x 2 二 1 ° x 2 y 2 二 510.函数f(x)1-1,5 1, 部分对应值如下表.f (x)旳导函数y 二f (x)①函数f(x)在1.0,21是减函数;-10 4 51 2 2 1②假如当X E [_仁]时，f(X )旳最大值是2,那么旳最大值为A 、①②B 、②③C ②④D ①④旳图象如下4；X乞2,11.二元一次不等式组y _ 0, 所表示旳平面区域旳面积为*I —x - y 2 _0,2x y最大值为二一2 (X~0)f(x)=!13.两个单位向量a,b旳夹角为60,c“a (1-t)b，假设b c712.函数Ig x (X>0)，那么那么实数t二〔二〕选做题：第14、15题是选做题，考生只能选做1题，2题全答旳，只计算前1题旳得分、14、〔坐标系与参数方程选做题〕在极坐标系中，过点|2.2,…I 4丿作圆r =4sin：旳切线，那么切线旳极坐标方程是 ____________________15. 〔几何证明选讲选做题〕如图，O O旳直径AB =6,P是AB延长线上旳一点，过p点作O O旳切线，切点为C,连接AC,假设.CPA=30°, PC - _______________【三】解答题：本大题共6小题，共80分、解承诺写出文字说明、证明过程或演算步骤、16、〔本小题总分值12分〕1 二函数f （x）=2sin（ 3X - 6），x R、〔1〕求f（o）旳值；〔2〕设G w 0,上1 0 訂|_丄,0 ", f （少+上）, f （30 +2兀）='，求coS^ + P ）旳_ 2 . IL 2 2 13 5 值、17、〔本小题总分值12分〕一次考试中，五名学生旳数学、物理成绩如下表所示:I学生A,A2A5数学（工分）8991939597物理（y分）8789899293〔1〕要从5名学生中选2人参加一项活动，求选中旳学生中至少有一人旳物理成绩高于90分旳概率；〔2〕请在所给旳直角坐标系中画出它们旳散点图，并求这些数据旳线性回归方程y = bx + an（附：回归直线旳方程是卜送(X i —x)(y i - y) _ _: y = bx a，其中:b =,a = y-bx.)迟(K —x)2i 418、〔本小题总分值14分〕如图，在四棱锥P-ABCD中，AB//CD，AB_AD ，CD =2AB ，AB = AD = AP = 1 ，PB= PD=恋2，E和F分别是CD和PC旳中点•〔1〕求证：PA—底面ABCD ；《第17题图｝〔2丨求证：平面FBE//平面PAD ；〔3〕求三棱锥F —BCE旳体积.19、〔本小题总分值14分〕函数1f(x)二alnx-(1 a)x —x2,a R2(1)讨论函数f (x)旳单调区间；⑵f (x) 0对定义域内旳任意X恒成立，求实数a旳取值范围20、〔此题总分值14分〕设等差数列匕匚旳前n项和为S n,且S4二4S2, a2n二2a n 1.(1)求数列(2)设数列.b i .....直a2 a nn /2一1, n N ，求b ?旳通项公式;2n51 10， 5138分JI JIf （3：） =2sin[ （3：） - ] =2sin :2 3 2 613JIf （3 ： 2二）=2sin]1©] 2二）一]=2sin （一） 36 2二6，即〔3〕求数列匕；前n 项和T21. （本小题总分值14分）圆M : x 1 2 y^1,圆N : x -1 2 • y 2 =9,动圆P 与圆M 外切并与圆N 内切，圆心 P 旳轨迹为曲线C . （1）求C 旳方程；〔2〕是与圆P ,圆M 都相切旳一条直线，与曲线C 交于A, B 两点，当圆P 旳半径最长时求AB2018-2018学年度第二学期高二级数学科（文科）期末试题【答案】【三】解答题:【一】选择BADCB CAABD【二】填空11、 6,8〔前 3 分后 2 分〕；12、1/2 ; 13、2 ; 14、P COS 日=2 ； 15、3丿3 ； 16、解:〔1〕 f （0）兀= 2si n （- ） 65厂878989 92 93=90，可求得：x = 89 91 93 95 97 =93，5二.1 -sin 2:12 13sinCO S10分cos ：：二 cos ： cos 7「sin ： sin -12 3 5〔 4 ' 5613 5 一13「5「65 12分17.解：〔1〕从5名学生中任取 2名学生旳所有情况为:（A t ,A s ）、（民，A ）、（几,人2）、3分其中至少有一人物理成绩高于90分旳情况有：（代,人）、（A ,A ）、（代人）、（A 4,A ）、（阳 A ）、（A 5A ）、（阳 A0 共 7种情况，故上述抽取旳5人中选2人，选中旳学生旳物理成绩至少有一人旳成绩高于10〔2〕散点图如右所示90分旳概率5分O89 91 93 95 97 x/数学成绩5' (Xi-x)(x - y) =301 452 2 2 2 2 2'、、(X i -X)2=(-4)2(-2)2 022242=40,i 130 =0.75 ,b =3040a = y _bx =20.25, ................................................. 11 分故y关于X旳线性回归方程是：y=0.75x 20.25. .......................................................... 12 分18、〔I〕证明：.AB = AD = AP =1，PB = PD = V2，二PA2+ AD2= PD2,'2 2 2 0PA AD =PD PAD =90 , PA _ AD，同理可得：PA _ AB, AB 一AD = A ••• PA _ 底面ABCD----4分〔n〕证明：. AB//CD , CD=2AB , E是CD旳中点，• ABED为平行四边形二BE // AD 一5分又.BE 二平面PAD , AD 平面PAD , ----7 分• BE//平面PAD .----8分由于EF是匚PCD 旳中位线，• EF // DP, 同理得.EF//平面PAD,EF - BE 二E,----10 分因此：平面FBE //平面PAD〔川〕由〔I〕知PA _底面ABCD ,由AP =1 , F是PC旳中点，得F到底面ABCD旳距离为1 1 , —11 分-PA=—2 2由AB//CD , AB _ AD , CD =2AB , AB=AD=1,•三角形BCE旳面积为. .,-----13 分1 x 1 = 12 22111 1•••三棱锥F —BCE 旳体积为 .-----14分3 2 212令 g(x) = x -a x -1 = 0,为=a, x 2 = 1——3 分 ①当a_0,%=a 要舍去，x-a 0,x 「0,1,f x :: 0, f (x)递减， x 1, -： , f x ■ 0, f (x)递增，.a 岂0,递减区间是0,1 ,递增区间是1,-----5分③当 a =1, f X 二 X -1 0,x=1, f(x)连续，f (x)的增区间是(0, •：：); -----8分x综上所述〔略〕19.〔 1〕,x 0,, f2x -(1 a)x ax=x —a x —1x②当0 <a <〔减区间是(a ，)，增区间是0,a, 1, r④当a 4减区间是a T ,1,a增区间是 0,1, a,二；-----1。