寿宁二中高三月考二

寿宁县高级中学2018-2019学年高三上学期11月月考数学试卷含答案一、选择题1． 已知2a =3b =m ，ab ≠0且a ，ab ，b 成等差数列，则m=（ ）A ．B ．C ．D ．62． 如图，网格纸上的正方形的边长为1，粗线画出的是某几何体的三视图，则这个几何体的体积为（）A ．30B ．50C ．75D ．1503． 直径为6的球的表面积和体积分别是（ ）A ．B ．C ．D ．144,144ππ144,36ππ36,144ππ36,36ππ4． 在函数y=中，若f （x ）=1，则x 的值是（）A ．1B ．1或C ．±1D ．5． 已知集合M={0，1，2}，则下列关系式正确的是（ ）A ．{0}∈MB ．{0}MC ．0∈MD ．0M∉⊆6． 在△ABC 中，已知A=30°，C=45°，a=2，则△ABC 的面积等于（ ）A ．B ．C ．D ．7． α是第四象限角，，则sin α=（）A ．B ．C ．D ．8． 函数y=2|x|的图象是（）A ．B ．C ．D ．9． 已知回归直线的斜率的估计值是1.23，样本点的中心为（4，5），则回归直线的方程是（）班级_______________ 座号______ 姓名_______________ 分数__________________________________________________________________________________________________________________A ． =1.23x+4B ． =1.23x ﹣0.08C ． =1.23x+0.8D ． =1.23x+0.0810．下列命题中的假命题是（）A ．∀x ∈R ，2x ﹣1＞0B ．∃x ∈R ，lgx ＜1C ．∀x ∈N +，（x ﹣1）2＞0D ．∃x ∈R ，tanx=211．若集合A={x|﹣2＜x ＜1}，B={x|0＜x ＜2}，则集合A ∩B=（）A ．{x|﹣1＜x ＜1}B ．{x|﹣2＜x ＜1}C ．{x|﹣2＜x ＜2}D ．{x|0＜x ＜1}12．如图，一隧道截面由一个长方形和抛物线构成现欲在随道抛物线拱顶上安装交通信息采集装置若位置C 对隧道底AB 的张角θ最大时采集效果最好，则采集效果最好时位置C 到AB 的距离是（）A ．2mB ．2mC ．4 mD ．6 m二、填空题13．抛物线y 2=6x ，过点P （4，1）引一条弦，使它恰好被P 点平分，则该弦所在的直线方程为 ．14．函数的单调递增区间是 ．15．设某总体是由编号为的20个个体组成，利用下面的随机数表选取个个体，选取方01,02,…,19,206法是从随机数表第1行的第3列数字开始从左到右依次选取两个数字，则选出来的第6个个体编号为________．【命题意图】本题考查抽样方法等基础知识，意在考查统计的思想．16．在△ABC 中，角A ，B ，C 所对的边分别为a ，b ，c ，若△ABC 不是直角三角形，则下列命题正确的是 （写出所有正确命题的编号）①tanA •tanB •tanC=tanA+tanB+tanC ②tanA+tanB+tanC 的最小值为3③tanA ，tanB ，tanC 中存在两个数互为倒数④若tanA ：tanB ：tanC=1：2：3，则A=45°⑤当tanB ﹣1=时，则sin 2C ≥sinA •sinB ．17．“黑白配”游戏，是小朋友最普及的一种游戏，很多时候被当成决定优先权的一种方式．它需要参与游戏的人（三人或三人以上）同时出示手势，以手心（白）、手背（黑）来决定胜负，当其中一个人出示的手势与其它人都不一样时，则这个人胜出，其他情况，则不分胜负．现在甲乙丙三人一起玩“黑白配”游戏．设甲乙丙三人每次都随机出“手心（白）、手背（黑）”中的某一个手势，则一次游戏中甲胜出的概率是 ．　18．已知函数f （x ）的定义域为[﹣1，5]，部分对应值如下表，f （x ）的导函数y=f ′（x ）的图象如图示． x ﹣1045f （x ）12211818 0792 4544 1716 5809 7983 86196206 7650 0310 5523 6405 0526 6238下列关于f （x ）的命题：①函数f （x ）的极大值点为0，4；②函数f （x ）在[0，2]上是减函数；③如果当x ∈[﹣1，t]时，f （x ）的最大值是2，那么t 的最大值为4；④当1＜a ＜2时，函数y=f （x ）﹣a 有4个零点；⑤函数y=f （x ）﹣a 的零点个数可能为0、1、2、3、4个．其中正确命题的序号是 ．三、解答题19．（本小题满分10分）选修4—4：坐标系与参数方程以坐标原点为极点，以轴的非负半轴为极轴建立极坐标系，已知曲线的参数方程为（x C ⎪⎩⎪⎨⎧==θθsin 2cos 2y x θ为参数，），直线的参数方程为（为参数）．],0[πθ∈l 2cos 2sin x t y t ì=+ïí=+ïîaat （I ）点在曲线上，且曲线在点处的切线与直线垂直，求点的极坐标；D C C D +2=0x y +D （II ）设直线与曲线有两个不同的交点，求直线的斜率的取值范围．l C l 【命题意图】本题考查圆的参数方程、直线参数方程、直线和圆位置关系等基础知识，意在考查数形结合思想、转化思想和基本运算能力．20．已知函数，（其中常数m ＞0）（1）当m=2时，求f （x ）的极大值；（2）试讨论f （x ）在区间（0，1）上的单调性；（3）当m ∈[3，+∞）时，曲线y=f （x ）上总存在相异两点P （x 1，f （x 1））、Q （x 2，f （x 2）），使得曲线y=f （x ）在点P 、Q 处的切线互相平行，求x 1+x 2的取值范围．21．在直角坐标系xOy中，圆C的参数方程（φ为参数）．以O为极点，x轴的非负半轴为极轴建立极坐标系．（Ⅰ）求圆C的极坐标方程；（Ⅱ）直线l的极坐标方程是ρ（sinθ+）=3，射线OM：θ=与圆C的交点为O，P，与直线l的交点为Q，求线段PQ的长．22．甲、乙两位同学参加数学竞赛培训，在培训期间他们参加5次预赛，成绩如下：甲：78 76 74 90 82乙：90 70 75 85 80（Ⅰ）用茎叶图表示这两组数据；（Ⅱ）现要从中选派一人参加数学竞赛，你认为选派哪位学生参加合适？说明理由．23．如图，在底面是矩形的四棱锥P﹣ABCD中，PA⊥平面ABCD，PA=AB=2，BC=2，E是PD的中点．（1）求证：平面PDC⊥平面PAD；（2）求二面角E﹣AC﹣D所成平面角的余弦值．24．（本小题满分12分）111]在如图所示的几何体中，是的中点，.D AC DB EF //（1）已知，，求证：平面； BC AB =CF AF =⊥AC BEF （2）已知分别是和的中点，求证： 平面.H G 、EC FB //GH ABC寿宁县高级中学2018-2019学年高三上学期11月月考数学试卷含答案（参考答案）一、选择题题号12345678910答案C．B D C C B B B D C题号1112答案D A二、填空题13．　3x﹣y﹣11=0　．14．　[2，3）　．15．1916．　①④⑤　17． ．18．　①②⑤　．三、解答题19．20．21．22．23．24．（1）详见解析；（2）详见解析.。

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2018年高三第二次月考数学试卷（文）一、选择题（每小题5分,共60分） 1。

已知全集集合则 ( ）A 。

B. C. D 。

2。

如果角,那么的值是( )A. B. C 。

D 。

3。

已知函数，则该函数图象（ ) A 。

关于点对称 B 。

关于点对称 C 。

关于直线对称 D 。

关于直线对称4. 函数的零点所在的大致区间是( ) A 。

（1,2）） C 。

D.5.( ) A. B. C. D 。

6。

函数满足,那么函数的图象大致是（ ) A 。

B. C 。

D.7。

的图像可以看作是把的图像作下列平移而得到( ) A 。

向左平移个单位长度 B.向右平移个单位长度C 。

向左平移个单位长度。

向右平移个单位长度8.已知（ )A 。

B. C. D 。

9.设函数在上可导，其导函数的图象如图所示,则的极大值点是（ ),U R ={|l g 1},{|22},xA x xB x =<=≤=A B (,1)-∞()0,1(,1]-∞(0,1]θ12⎛⎫- ⎪ ⎪⎝⎭tan θ12--()c o s 23f x x π⎛⎫=+ ⎪⎝⎭,03π⎛⎫ ⎪⎝⎭,04π⎛⎫ ⎪⎝⎭3x π=4x π=2()ln f x x x =-(3,4)(,)e +∞,45,=oA B C a B A ∆=则o30o60o 30150o 或o60120o 或()af x x =()24f =()()1a gxl o g x =+3s in 23y x π⎛⎫=+ ⎪⎝⎭3s i n 2y x =6π6π3π3πa 为第二象限角=1-3-13f(x)R'()f x f(x)A 。

2020年寿宁二中高三英语二模试题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AThere have been many great painters in the rich history of Chinese art. Here are four of the greatest painters from China.Li Cheng (919—967, Five Dynasties and early Song Dynasty)Li Cheng contributed greatly to one of the golden ages of landscape paintings in world history. During his time, he was considered the best landscape painter ever. He is remembered especially for the winter landscapes he created and for simple compositions of tall, old evergreens set against a dry landscape. Several of his paintings are in thin ink which gives them a foggy appearance.Fan Kuan (990—1020 , Song Dynasty)Fan Kuan began his career by modeling Li Cheng's work but later created his own style, claiming that the only true teacher was nature. His finest workTravelers among Mountains and Streamsis a masterpiece of landscape painting and many future artists turned to it for inspiration.Qi Baishi (1864-1957)One of the greatest contemporary Chinese painters, Qi Baishi is known for not being influenced by Western styles like most painters of his time. He can be considered as the last great traditional painter of China. He painted almost everything from insects to landscapes. He is regarded highly in Chinese art for the freshness that he brought to the familiar types of birds and flowers, insects and grass.Wu Guanzhong (1919—2010)Widely considered as the founder of modern Chinese painting , Wu Guanzhong has painted various aspects of China, like its architecture, plants, animals, people and landscapes. Wu went on to combine Western and Chinese styles to create a unique form of modem art. In 1992, he became the first living Chinese artist whose work was exhibited at the British Museum.1.What do we know about Li Cheng?A.He loved landscape paintings.B.He copied many artists' work.C.His work gained worldwide recognition.D.He was considered as Fan Kuan's teacher.2.What is the main feature of Qi Baishi's paintings?A.They have foggy appearances.B.They lack diversity in the theme.C.They come under Western influence.D.They show advanced traditional painting skills.3.What did the four Chinese painters have in common?A.They were all modern painters.B.They all created landscape paintings.C.They were all impacted by Western art.D.They were all pioneers intraditional art history.BHaley Curfman, 25, of Blackwell, Oklahoma, is a teacher at Blackwell Public School and last yearshe bought a plain(朴素的) white dress, which she set up a station in her classroom for her students to decorate, encouraging them to go and draw on the dress whenever they had free time. Haley set up a station at a table with the dress and some color1 ed markers so that her students could draw on it, having first seen the idea on Pinterest.After the kids had finished the design, she then surprised them all by wearing the dress to class, sharing pictures of herself in it on Facebook in a post that has since been shared over 200, 000 times.For teachers looking to do something similar for their own students, Haley said that she bought the dress off Amazon for less than $ 20 about six months ago and pre-washed the dress before she started the project. She used markers from Walmart. But she said these things can be bought in other places. She said, “To do the project, I set up a station at a table with the dress and markers. It takes anywhere from two weeks to a month to complete as we just work on it here and there when time allows. You'd better give the students enough time so they don't have to hurry.”“Teachers have been sharing their dresses, T-shirts, etc. with me that they've been creating since the 1950s with the same idea. It is amazing, and I love the fact that you are sharing them with me! Thank you all so much for your kindness and support,” she wrote on Facebook.Asked why she came up with the idea, Haley told Scary Mommy. “We don't have art in our school, so, I always try to do little creative projects when possible.”4. What did Haley use the plain white dress to do?A. To teach her students painting.B. To prepare for her presentation.C. To ask her students to draw on it.D. To help her students with their homework.5. What do we know about the finished dress?A. It took Haley by surprise.B. It is kept by Haley at home.C. It is the product of a new idea.D. It has enjoyed popularity online.6. What should a teacher do if he or she wants to follow Haley's example?A. Buy the same dress online.B. Give the kids enough time.C. Buy the same markers in Walmart.D. Leave the finished work untouched.7. Why did Haley carry out the project?A. To teach art to her students.B. To show her idea on Pinterest.C. To share it with other teachers.D. To exciteher students' creativity.CI had just delivered a memorable speech, and I was about to learn how the judges decided my performance. The audience leaned forward and a period of silence fell across the room. I felt the drum rolled in my heart.The third-place winner was announced. The name was not mine. Then the second-place winner, still not me. At last, the moment of truth came. I was about to either enjoy the warmth of victory or regret the months’ preparation. My heart felt closer to the latter.Losing is a part of life, and I have dealt with it on more than one occasion. However, it was an indescribable feeling to drive a 200-mile round trip, get up very early on a freezing Saturday morning, and yet still finish fourth out of four competitors in my group. After Lincoln lost the 1858 Illinois Senate race, he said, “I felt like the 12-year-old boy who kicked his toe. I was too big to cry and it hurt too bad to laugh.” Oh yeah, I could relate.I had spent many hours in front of a computer and in libraries doing research for the Lincoln Bicentennial Speech Contest. After not placing in the first year of the contest, I really wanted to compete again. Lincoln had many failures, but he never allowed them to defeat his spirit or ambition, so I was not going to give up on a second contest! I reworked my speech for the following year, but again I did not place.I couldn’t accept the fact that I failed twice in something that I had worked so hard on, until I thought about my hero. Never mind the lost prize money and praise—through learning stories about Lincoln, I discovered that I can fail successfully.8. How did the author feel after finishing his speech?A. Delighted.B. Annoyed.C. Thrilled.D. Nervous.9. What can be inferred from Paragraph 3?A. He was regretful about his not being fully prepared.B. He felt upset for getting up early on a chilly morning.C. He once kicked and hurt his toe when he was 12 years old.D. He turned out to be the last one of his group in the contest.10. Why did the author decide to enter the second contest?A. He was eager to prove himself to be the best contestant.B. He was inspired by the never-give-up spirit of Lincoln.C. He was willing to enjoy the warmth and joy of victory.D. He was determined to win the prize money and praise.11. Which of the following can be the best title for the text?A. A memorable hero in my lifeB. Never mind others’ judgmentsC. Losing is an indescribable feelingD. Stand up from where we tripped overDLearning to say “yes, and”When I first heard about the improvisation (即兴交流) class, I was hesitating. As a quiet and shy girl, I feared improvising in front of strangers. However,I knew I wanted to work as a science communicator after finishing my Ph.D., so it seemed like a perfect opportunity to learn how to speak and communicate with others effectively. I signed up, knowing the experience would give me help.During our first class, we learned an important concept of improvisation: “yes, and.” It means that, as improvisers, we’d better accept what fellow performers say. If someone says that rhinos (犀牛) are librarians, for example, then rhinos are librarians. We do not question the logic; we say “yes” and then continue with the scene as if nothing is wrong.The first few scenes were hard, but as weeks turned into months, I became more comfortable andeven started to enjoy our classes. I became better at listening, relating to my conversation partners, and communicating clearly in the moment. Once when I was giving a presentation about my science, an audience member surprised me with a question that didn’t grow out of the information I’d presented. Instead of getting confused and nervous, I took the “yes, and” approach—accepting the question and letting my mind focus on why it was asked. Thathelped me find an appropriate answer. I got pretty excited about it.The benefits of improvisation go beyond communication. Before attending the class, I would get stuck when my experiments produced unexpected data, thinking that I had made a mistake. But now, instead of getting discouraged, I will stay open to the possibility that the results are real, keep exploring the data and end up identifying a new type of cell—one that isn’t behaving as expected.I think all scientists can benefit from this lesson. If the data say rhinos are librarians, then it’s worth findingout whether rhinos are, in fact, librarians. As scientists, our job isn’t to challenge data that support a preconceived (先入为主的) story, but to say “yes, and.”12. Why did the author attend the improvisation class?A. To get a different experience.B. To finish her Ph.D. at university.C. To give up her job as a science communicator.D. To improve her speaking and communicating ability.13. What was the author’s change after attending the improvisation class?A. She formed her own idea quickly.B. She came up with lots of creative responses.C. She paid more attention to the logic of answers.D. She became a good listener before giving an opinion.14. The author mentions applying the “yes, and” approach to her scientific experiments to ______.A. explain the process of using the methodB. prove the benefits of the improvisation classC. share her own research experiences with readersD. attract fellow scientists to attend the improvisation class15. What can be inferred about scientists from the last paragraph?A. They should attend the improvisation class.B. They should question all preconceived ideas.C. They should carry on research by admitting earlier data.D. They should try to improve their professional knowledge.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。

寿宁县第二中学2018-2019学年上学期高三数学10月月考试题班级__________ 座号_____ 姓名__________ 分数__________一、选择题1． 复数是虚数单位）的虚部为（ ）i iiz (21+=A ．B ．C ．D ．1-i -i 22【命题意图】本题考查复数的运算和概念等基础知识，意在考查基本运算能力．2． 如图F 1、F 2是椭圆C 1：+y 2=1与双曲线C 2的公共焦点，A 、B 分别是C 1、C 2在第二、四象限的公共点，若四边形AF 1BF 2为矩形，则C 2的离心率是（）A ．B ．C ．D ．3． 在数列{a n }中，a 1=3，a n+1a n +2=2a n+1+2a n （n ∈N +），则该数列的前2015项的和是（）A ．7049B ．7052C ．14098D ．141014． 函数（，）的部分图象如图所示，则 f （0）的值为（ ）()2cos()f x x ωϕ=+0ω>0ϕ-π<<A. B. C. D. 32-1-【命题意图】本题考查诱导公式，三角函数的图象和性质，数形结合思想的灵活应用.5． 不等式≤0的解集是（）A ．（﹣∞，﹣1）∪（﹣1，2）B ．[﹣1，2]C ．（﹣∞，﹣1）∪[2，+∞）D ．（﹣1，2]6． 若,则的值为（ ）()()()()2,106,10x x f x f f x x -≥⎧⎪=⎨+<⎡⎤⎪⎣⎦⎩()5f A ． B ． C.D ．101112137． 复数的虚部为（ ）A ．﹣2B ．﹣2iC ．2D ．2i8． 已知抛物线C ：的焦点为F ，准线为，P 是上一点，Q 是直线PF 与C 的一个交点，若y x 82=l l ，则（ ）FQ PF 2==QF A ．6B ．3C ．D ．3834第Ⅱ卷（非选择题，共100分）9． 函数f （x ）＝sin （ωx ＋φ）（ω＞0，－≤φ≤）的部分图象如图所示，则的值为（）π2π2φωA.B ．1814C. D ．11210．已知角的终边经过点()3P x ，()0x <且cos θ=，则等于（ ）A ．1-B ．13-C．3-D ．11．为了得到函数的图象，只需把函数y=sin3x 的图象（）A ．向右平移个单位长度B ．向左平移个单位长度C ．向右平移个单位长度D ．向左平移个单位长度12．“”是“圆关于直线成轴对称图形”的（ ）3<-b a 056222=++-+a y x y x b x y 2+=A ．充分不必要条件 B ．必要不充分条件C ．充分必要条件 D ．既不充分也不必要条件【命题意图】本题考查圆的一般方程、圆的几何性质、常用逻辑等知识，有一定的综合性，突出化归能力的考查，属于中等难度．二、填空题13．已知△的面积为，三内角，，的对边分别为，，．若，ABC S A B C 2224S a b c +=+则取最大值时．sin cos(4C B π-+C =14．若非零向量，满足|+|=|﹣|，则与所成角的大小为 ．15．已知点E 、F 分别在正方体的棱上，且, ,则面AEF 与面ABC 所成的二面角的正切值等于 .16．【徐州市2018届高三上学期期中】已知函数（为自然对数的底数），若，则实数 的取值范围为______．17．直线l 1和l 2是圆x 2+y 2=2的两条切线，若l 1与l 2的交点为（1，3），则l 1与l 2的夹角的正切值等于　_________　。

寿宁县二中2018-2019学年高三上学期11月月考数学试卷含答案一、选择题1． 已知一个算法的程序框图如图所示，当输出的结果为21时，则输入的值为（ ）A ．2B ．1-C ．1-或2D ．1-或102． 对于区间[a ，b]上有意义的两个函数f （x ）与g （x ），如果对于区间[a ，b]中的任意数x 均有|f （x ）﹣g（x ）|≤1，则称函数f （x ）与g （x ）在区间[a ，b]上是密切函数，[a ，b]称为密切区间．若m （x ）=x 2﹣3x+4与n （x ）=2x ﹣3在某个区间上是“密切函数”，则它的一个密切区间可能是（ ）A ．[3，4]B ．[2，4]C ．[1，4]D ．[2，3]3． 函数y=f （x ）在[1，3]上单调递减，且函数f （x+3）是偶函数，则下列结论成立的是（ ） A ．f （2）＜f （π）＜f （5） B ．f （π）＜f （2）＜f （5）C ．f （2）＜f （5）＜f （π）D ．f （5）＜f （π）＜f （2）4． 如图所示，函数y=|2x ﹣2|的图象是（ ）A． B． C． D．5． 已知U=R ，函数y=ln （1﹣x ）的定义域为M ，集合N={x|x 2﹣x ＜0}．则下列结论正确的是（ ） A ．M ∩N=N B ．M ∩（∁U N ）=∅ C ．M ∪N=U D ．M ⊆（∁U N ）6． 已知直线x ﹣y+a=0与圆心为C 的圆x 2+y 2+2x ﹣4y+7=0相交于A ，B两点，且•=4，则实数a的值为（ ） A．或﹣B．或3C．或5D ．3或57． 阅读如图所示的程序框图，运行相应的程序，若输出的的值等于126，则判断框中的①可以是（ ）班级_______________ 座号______ 姓名_______________ 分数__________________________________________________________________________________________________________________A．i＞4？B．i＞5？C．i＞6？D．i＞7？8．某校在暑假组织社会实践活动，将8名高一年级学生，平均分配甲、乙两家公司，其中两名英语成绩优秀学生不能分给同一个公司；另三名电脑特长学生也不能分给同一个公司，则不同的分配方案有（）A．36种B．38种C．108种D．114种9．在等比数列{a n}中，已知a1=3，公比q=2，则a2和a8的等比中项为（）A．48 B．±48 C．96 D．±9610．设复数z满足z（1+i）=2（i为虚数单位），则z=（）A．1﹣i B．1+i C．﹣1﹣i D．﹣1+i11．函数f（x）=x3﹣3x2+5的单调减区间是（）A．（0，2）B．（0，3）C．（0，1） D．（0，5）12．已知f（x）是定义在R上的奇函数，且f（x﹣2）=f（x+2），当0＜x＜2时，f（x）=1﹣log2（x+1），则当0＜x＜4时，不等式（x﹣2）f（x）＞0的解集是（）A．（0，1）∪（2，3）B．（0，1）∪（3，4）C．（1，2）∪（3，4）D．（1，2）∪（2，3）二、填空题13．一个圆柱和一个圆锥的母线相等，底面半径也相等，则侧面积之比是．14．已知a=（cosx﹣sinx）dx，则二项式（x2﹣）6展开式中的常数项是．15．计算：×5﹣1=．16．设复数z满足z（2﹣3i）=6+4i（i为虚数单位），则z的模为．17．若圆与双曲线C：的渐近线相切，则_____；双曲线C的渐近线方程是____．18．椭圆+=1上的点到直线l：x﹣2y﹣12=0的最大距离为．三、解答题19．【无锡市2018届高三上期中基础性检测】已知函数()()2ln 1.f x x mx m R =--∈ （1）当1m =时，求()f x 的单调区间；（2）令()()g x xf x =，区间1522,D e e -⎛⎫= ⎪⎝⎭，e 为自然对数的底数。

寿宁县第二高级中学2018-2019学年高三上学期11月月考数学试卷含答案一、选择题1． （﹣6≤a ≤3）的最大值为（ ）A ．9B ．C ．3D ．2． 两条平行直线3x ﹣4y+12=0与3x ﹣4y ﹣13=0间的距离为（ ）A ．B ．C ．D ．53． 若f （x ）=sin （2x+θ），则“f （x ）的图象关于x=对称”是“θ=﹣”的（）A ．充分不必要条件B ．必要不充分条件C ．充要条件D ．既不充分又不必要条件4． 已知函数，其中，对任意的都成立，在122()32f x x ax a =+-(0,3]a ∈()0f x ≤[]1,1x ∈-和两数间插入2015个数，使之与1，构成等比数列，设插入的这2015个数的成绩为，则（ ）T T =A ．B ．C ．D ．20152201532015232015225． 设等比数列{a n }的公比q=2，前n 项和为S n ，则=（）A ．2B ．4C ．D ．6． 在△ABC 中，已知a=2，b=6，A=30°，则B=（）A ．60°B ．120°C ．120°或60°D ．45°7． 若函数y=x 2+bx+3在[0，+∞）上是单调函数，则有（）A ．b ≥0B ．b ≤0C ．b ＞0D ．b ＜08． 已知集合A={y|y=x 2+2x ﹣3}，，则有（）A ．A ⊆B B ．B ⊆AC ．A=BD ．A ∩B=φ9． 正方体的内切球与外接球的半径之比为（ ）A ．B ．C ．D ．10．函数f （x ）=1﹣xlnx 的零点所在区间是（）A ．（0，）B ．（，1）C ．（1，2）D ．（2，3）11．如图，空间四边形OABC 中，，，，点M 在OA 上，且，点N 为BC 中点，则等于（）班级_______________ 座号______ 姓名_______________ 分数__________________________________________________________________________________________________________________A ．B ．C ．D ．12．的展开式中，常数项是（ ）62)21(x x -A ．B ．C ．D ．45-451615-1615二、填空题13．已知函数f （x ）=sinx ﹣cosx ，则= ．14．在（1+2x ）10的展开式中，x 2项的系数为 （结果用数值表示）．15．已知正方体ABCD ﹣A 1B 1C 1D 1的一个面A 1B 1C 1D 1在半径为的半球底面上，A 、B 、C 、D 四个顶点都在此半球面上，则正方体ABCD ﹣A 1B 1C 1D 1的体积为 ．16．已知函数，且，则，的大小关系()f x 23(2)5x =-+12|2||2|x x ->-1()f x 2()f x 是．17．用1，2，3，4，5组成不含重复数字的五位数，要求数字4不出现在首位和末位，数字1，3，5中有且仅有两个数字相邻，则满足条件的不同五位数的个数是 .（注：结果请用数字作答）【命题意图】本题考查计数原理、排列与组合的应用，同时也渗透了分类讨论的思想，本题综合性强，难度较大.18．设满足条件，若有最小值，则的取值范围为．,x y ,1,x y a x y +≥⎧⎨-≤-⎩z ax y =-a 三、解答题19．已知函数f （x ）=x 2﹣ax+（a ﹣1）lnx （a ＞1）．（Ⅰ） 讨论函数f （x ）的单调性；（Ⅱ） 若a=2，数列{a n }满足a n+1=f （a n ）．（1）若首项a 1=10，证明数列{a n }为递增数列；（2）若首项为正整数，且数列{a n }为递增数列，求首项a 1的最小值．20．已知三次函数f（x）的导函数f′（x）=3x2﹣3ax，f（0）=b，a、b为实数．（1）若曲线y=f（x）在点（a+1，f（a+1））处切线的斜率为12，求a的值；（2）若f（x）在区间[﹣1，1]上的最小值、最大值分别为﹣2、1，且1＜a＜2，求函数f（x）的解析式．21．某运动员射击一次所得环数X的分布如下：X0～678910P00.20.30.30.2现进行两次射击，以该运动员两次射击中最高环数作为他的成绩，记为ξ．（I）求该运动员两次都命中7环的概率；（Ⅱ）求ξ的数学期望Eξ．22．设函数f（x）=ax2+bx+c（a≠0）为奇函数，其图象在点（1，f（1））处的切线与直线x﹣6y﹣7=0垂直，导函数f′（x）的最小值为﹣12．（1）求a，b，c的值；（2）求函数f（x）的单调递增区间，并求函数f（x）在[﹣1，3]上的最大值和最小值．23．设函数f（x）是定义在R上的奇函数，且对任意实数x，恒有f（x+2）=﹣f（x），当x∈[0，2]时，f（x）=2x﹣x2．（1）求证：f（x）是周期函数；（2）当x∈[2，4]时，求f（x）的解析式；（3）求f（0）+f（1）+f（2）+…+f（2015）的值．24．已知定义在区间（0，+∞）上的函数f（x）满足f（）=f（x1）﹣f（x2）．（1）求f（1）的值；（2）若当x＞1时，有f（x）＜0．求证：f（x）为单调递减函数；（3）在（2）的条件下，若f（5）=﹣1，求f（x）在[3，25]上的最小值．寿宁县第二高级中学2018-2019学年高三上学期11月月考数学试卷含答案（参考答案）一、选择题1． 【答案】B【解析】解：令f （a ）=（3﹣a ）（a+6）=﹣+，而且﹣6≤a ≤3，由此可得函数f（a ）的最大值为，故（﹣6≤a ≤3）的最大值为=，故选B ．【点评】本题主要考查二次函数的性质应用，体现了转化的数学思想，属于中档题．　2． 【答案】D【解析】解：两条平行直线3x ﹣4y+12=0与3x ﹣4y ﹣13=0间的距离为： =3．故选：D ．【点评】本题考查平行线之间的距离公式的求法，考查计算能力．　3． 【答案】B【解析】解：若f （x ）的图象关于x=对称，则2×+θ=+k π，解得θ=﹣+k π，k ∈Z ，此时θ=﹣不一定成立，反之成立，即“f （x ）的图象关于x=对称”是“θ=﹣”的必要不充分条件，故选：B【点评】本题主要考查充分条件和必要条件的判断，结合三角函数的对称性是解决本题的关键．　4． 【答案】C 【解析】试题分析：因为函数，对任意的都成立，所以，解得22()32f x x ax a =+-()0f x ≤[]1,1x ∈-()()1010f f -≤⎧⎪⎨≤⎪⎩或，又因为，所以，在和两数间插入共个数，使之与，构成等3a ≥1a ≤-(0,3]a ∈3a =122015,...a a a 2015比数列，，，两式相乘，根据等比数列的性质得，T 122015...a a a =g 201521...T a a a =g ()()2015201521201513T a a ==⨯，故选C.T =201523考点：1、不等式恒成立问题；2、等比数列的性质及倒序相乘的应用.5． 【答案】C【解析】解：由于q=2，∴∴；故选：C．6．【答案】C【解析】解：∵a=2，b=6，A=30°，∴由正弦定理可得：sinB===，∵B∈（0°，180°），∴B=120°或60°．故选：C．7．【答案】A【解析】解：抛物线f（x）=x2+bx+3开口向上，以直线x=﹣为对称轴，若函数y=x2+bx+3在[0，+∞）上单调递增函数，则﹣≤0，解得：b≥0，故选：A．【点评】本题考查二次函数的性质和应用，是基础题．解题时要认真审题，仔细解答．8．【答案】B【解析】解：∵y=x2+2x﹣3=（x+1）2﹣4，∴y≥﹣4．则A={y|y≥﹣4}．∵x＞0，∴x+≥2=2（当x=，即x=1时取“=”），∴B={y|y≥2}，∴B⊆A．故选：B．【点评】本题考查子集与真子集，求解本题，关键是将两个集合进行化简，由子集的定义得出两个集合之间的关系，再对比选项得出正确选项．9．【答案】C【解析】解：正方体的内切球的直径为，正方体的棱长，外接球的直径为，正方体的对角线长，设正方体的棱长为：2a ，所以内切球的半径为：a ；外接球的直径为2a ，半径为： a ，所以，正方体的内切球与外接球的半径之比为：故选C 10．【答案】C【解析】解：∵f （1）=1＞0，f （2）=1﹣2ln2=ln ＜0，∴函数f （x ）=1﹣xlnx 的零点所在区间是（1，2）．故选：C ．【点评】本题主要考查函数零点区间的判断，判断的主要方法是利用根的存在性定理，判断函数在给定区间端点处的符号是否相反．　11．【答案】B【解析】解： ===；又，，，∴．故选B ．【点评】本题考查了向量加法的几何意义，是基础题．　12．【答案】D【解析】，2612316611()()()22r r r r r r r T C x C x x --+=-=-令，解得．1230r -=4r =∴常数项为．446115()216C -=二、填空题13．【答案】 ．【解析】解：∵函数f （x ）=sinx ﹣cosx=sin （x ﹣），则=sin （﹣）=﹣=﹣，故答案为：﹣．【点评】本题主要考查两角差的正弦公式，属于基础题．　14．【答案】　180　【解析】解：由二项式定理的通项公式T r+1=C n r a n ﹣r b r 可设含x 2项的项是T r+1=C 7r （2x ）r 可知r=2，所以系数为C 102×4=180，故答案为：180．【点评】本题主要考查二项式定理中通项公式的应用，属于基础题型，难度系数0.9．一般地通项公式主要应用有求常数项，有理项，求系数，二项式系数等．　15．【答案】　2　．【解析】解：如图所示，连接A 1C 1，B 1D 1，相交于点O ．则点O 为球心，OA=．设正方体的边长为x ，则A 1O=x ．在Rt △OAA 1中，由勾股定理可得： +x 2=，解得x=．∴正方体ABCD ﹣A 1B 1C 1D 1的体积V==2．故答案为：2．16．【答案】]12()()f x f x 【解析】考点：不等式，比较大小．【思路点晴】本题主要考查二次函数与一元二次方程及一元二次不等式三者的综合应用. 分析二次函数的图象，主要有两个要点：一个是看二次项系数的符号，它确定二次函数图象的开口方向；二是看对称轴和最值，它确定二次函数的具体位置．对于函数图象判断类似题要会根据图象上的一些特殊点进行判断，如函数图象与正半轴的交点，函数图象的最高点与最低点等．17．【答案】48【解析】18．【答案】[1,)+∞【解析】解析：不等式表示的平面区域如图所示，由得，当,1,x y a x y +≥⎧⎨-≤-⎩z ax y =-y ax z =-01a ≤<时，平移直线可知，既没有最大值，也没有最小值；当时，平移直线可知，在点A 处取得最小1l z 1a ≥2l z 值；当时，平移直线可知，既没有最大值，也没有最小值；当时，平移直线可知，10a -<<3l z 1a ≤-4l 在点A 处取得最大值，综上所述，．1a ≥三、解答题19．【答案】 【解析】解：（Ⅰ）∵，∴（x ＞0），当a=2时，则在（0，+∞）上恒成立，当1＜a ＜2时，若x ∈（a ﹣1，1），则f ′（x ）＜0，若x ∈（0，a ﹣1）或x ∈（1，+∞），则f ′（x ）＞0，当a ＞2时，若x ∈（1，a ﹣1），则f ′（x ）＜0，若x ∈（0，1）或x ∈（a ﹣1，+∞），则f ′（x ）＞0，综上所述：当1＜a ＜2时，函数f （x ）在区间（a ﹣1，1）上单调递减，在区间（0，a ﹣1）和（1，+∞）上单调递增；当a=2时，函数（0，+∞）在（0，+∞）上单调递增；当a ＞2时，函数f （x ）在区间（0，1）上单调递减，在区间（0，1）和（a ﹣1，+∞）上单调递增．（Ⅱ）若a=2，则，由（Ⅰ）知函数f （x ）在区间（0，+∞）上单调递增，（1）因为a 1=10，所以a 2=f （a 1）=f （10）=30+ln10，可知a 2＞a 1＞0，假设0＜a k ＜a k+1（k ≥1），因为函数f （x ）在区间（0，+∞）上单调递增，∴f （a k+1）＞f （a k ），即得a k+2＞a k+1＞0，由数学归纳法原理知，a n+1＞a n对于一切正整数n都成立，∴数列{a n}为递增数列．（2）由（1）知：当且仅当0＜a1＜a2，数列{a n}为递增数列，∴f（a1）＞a1，即（a1为正整数），设（x≥1），则，∴函数g（x）在区间上递增，由于，g（6）=ln6＞0，又a1为正整数，∴首项a1的最小值为6．【点评】本题考查导数的运用：求单调区间，同时考查函数的零点存在定理和数学归纳法的运用，考查运算能力，属于中档题．选做题：本题设有（1）（2）（3）三个选考题，每题7分，请考生任选2题作答，满分7分．如果多做，则按所做的前两题计分．【选修4-2：矩阵与变换】20．【答案】【解析】解：（1）由导数的几何意义f′（a+1）=12∴3（a+1）2﹣3a（a+1）=12∴3a=9∴a=3（2）∵f′（x）=3x2﹣3ax，f（0）=b∴由f′（x）=3x（x﹣a）=0得x1=0，x2=a∵x∈[﹣1，1]，1＜a＜2∴当x∈[﹣1，0）时，f′（x）＞0，f（x）递增；当x∈（0，1]时，f′（x）＜0，f（x）递减．∴f（x）在区间[﹣1，1]上的最大值为f（0）∵f（0）=b，∴b=1∵，∴f（﹣1）＜f（1）∴f（﹣1）是函数f（x）的最小值，∴∴∴f（x）=x3﹣2x2+1【点评】曲线在切点处的导数值为曲线的切线斜率；求函数的最值，一定要注意导数为0的根与定义域的关系．21．【答案】【解析】解：（1）设A=“该运动员两次都命中7环”，则P（A）=0.2×0.2=0.04．（2）依题意ξ在可能取值为：7、8、9、10且P（ξ=7）=0.04，P（ξ=8）=2×0.2×0.3+0.32=0.21，P（ξ=9）=2×0.2×0.3+2×0.3×0.3×0.32=0.39，P（ξ=10）=2×0.2×0.2+2×0.3×0.2+2×0.3×0.2+0.22=0.36，∴ξ的分布列为：ξ78910P0.040.210.390.36ξ的期望为Eξ=7×0.04+8×0.21+9×0.39+10×0.36=9.07．【点评】本题考查概率的求法，考查离散型随机变量的数学期望的求法，是中档题，解题时要认真审题，注意相互独立事件概率乘法公式的合理运用．22．【答案】【解析】解：（1）∵f（x）为奇函数，∴f（﹣x）=﹣f（x），即﹣ax3﹣bx+c=﹣ax3﹣bx﹣c，∴c=0．∵f′（x）=3ax2+b的最小值为﹣12，∴b=﹣12．又直线x﹣6y﹣7=0的斜率为，则f′（1）=3a+b=﹣6，得a=2，∴a=2，b=﹣12，c=0；（2）由（1）知f（x）=2x3﹣12x，∴f′（x）=6x2﹣12=6（x+）（x﹣），列表如下：x （﹣∞，﹣）﹣（﹣，（，+∞））f′（x）+0﹣0+f（x）增极大减极小增所以函数f（x）的单调增区间是（﹣∞，﹣）和（，+∞）．∵f（﹣1）=10，f（）=﹣8，f（3）=18，∴f（x）在[﹣1，3]上的最大值是f（3）=18，最小值是f（）=﹣8．23．【答案】【解析】（1）证明：∵f（x+2）=﹣f（x），∴f（x+4）=f[（x+2）+2]=﹣f（x+2）=f（x），∴y=f（x）是周期函数，且T=4是其一个周期．（2）令x∈[﹣2，0]，则﹣x∈[0，2]，∴f（﹣x）=﹣2x﹣x2，又f（﹣x）=﹣f（x），∴在x∈[﹣2，0]，f（x）=2x+x2，∴x∈[2，4]，那么x﹣4∈[﹣2，0]，那么f（x﹣4）=2（x﹣4）+（x﹣4）2=x2﹣6x+8，由于f（x）的周期是4，所以f（x）=f（x﹣4）=x2﹣6x+8，∴当x∈[2，4]时，f（x）=x2﹣6x+8．（3）当x∈[0，2]时，f（x）=2x﹣x2．∴f（0）=0，f（1）=1，当x∈[2，4]时，f（x）=x2﹣6x+8，∴f（2）=0，f（3）=﹣1，f（4）=0∴f（1）+f（2）+f（3）+f（4）=1+0﹣1+0=0，∵y=f（x）是周期函数，且T=4是其一个周期．∴2016=4×504∴f（0）+f（1）+f（2）+…+f（2015）=504×[f（0）+f（1）+f（2）+f（3）]=504×0=0，即求f（0）+f（1）+f（2）+…+f（2015）=0．【点评】本题主要考查函数周期性的判断，函数奇偶性的应用，综合考查函数性质的应用．24．【答案】【解析】解：（1）令x1=x2＞0，代入得f（1）=f（x1）﹣f（x1）=0，故f（1）=0．…（4分）（2）证明：任取x1，x2∈（0，+∞），且x1＞x2，则＞1，由于当x＞1时，f（x）＜0，所以f（）＜0，即f（x1）﹣f（x2）＜0，因此f（x1）＜f（x2），所以函数f（x）在区间（0，+∞）上是单调递减函数．…（8分）（3）因为f（x）在（0，+∞）上是单调递减函数，所以f（x）在[3，25]上的最小值为f（25）．由f（）=f（x1）﹣f（x2）得，f（5）=f（）=f（25）﹣f（5），而f（5）=﹣1，所以f（25）=﹣2．即f（x）在[3，25]上的最小值为﹣2．…（12分）【点评】本题主要考查抽象函数的应用，利用赋值法以及函数单调性的定义是解决本题的关键．　。

河南省南阳市寿宁县高级中学高三数学文月考试卷含解析一、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. 已知定义在R上的函数满足，若方程有5个实根，则正实数的取值范围是A． B．C． D．参考答案：2. 对于函数f（x）=，设f2（x）=f[f（x）]，f3（x）=f[f2（x）]，…，f n+1（x）=f[f n（x）]（n∈N*，且n≥2），令集合M={x|f2015（x）=﹣x，x∈R}，则集合M为（）A．空集B．实数集C．单元素集D．二元素集参考答案：A【考点】元素与集合关系的判断；集合的表示法．【专题】集合．【分析】根据条件可分别求出f2（x），f3（x），f4（x），f5（x），f6（x），f7（x），会得出f7（x）=f3（x），从而从f3（x）开始每4项便重复出现f3（x），而2015=3+4×503，从而有f2015（x）=f3（x），这样即可解出方程f2015（x）=﹣x，这便可得到集合M所含元素的情况，从而找出正确选项．【解答】解：，，，，f6（x）=﹣x，；∴f7（x）=f3（x）；∴从f3（x）开始组成了一个以f3（x）为首项，以周期为4重复出现，且2015=3+4×503；∴；∴；整理得x2=﹣1；M=?．故选：A．【点评】考查已知f（x）求f[g（x）]的方法，周期的概念，以及描述法表示集合，空集的概念．3. 如图，是双曲线：的左、右焦点，过的直线与的左、右两支分别交于两点．若为等边三角形，则双曲线的离心率为（）A. B. C. D.参考答案：D略4. 等比数列中，，，则（）A．－4 B．4 C．±4 D．－5参考答案：A5. 已知圆x2+y2+2x﹣6y+5=0，将直线y=2x+λ向上平移2个单位与之相切，则实数λ的值为（）A．﹣7或3 B．﹣2或8 C．﹣4或4 D．0或6参考答案：B【考点】直线与圆的位置关系．【分析】根据直线平移的规律，由直线y=2x+λ向上平移2个单位得到平移后直线的方程，然后因为此直线与圆相切得到圆心到直线的距离等于半径，利用点到直线的距离公式列出关于λ的方程，求出方程的解即可得到λ的值．【解答】解：由题意知：直线2x﹣y+λ=0平移后方程为2x﹣y+λ+2=0．圆x2+y2+2x﹣6y+5=0的圆心坐标为（﹣1，3），半径为又直线与圆相切，则圆心到直线的距离等于圆的半径，即=，得λ=﹣2或8，故选B．【点评】此题考查学生掌握平移的规律及直线与圆相切时所满足的条件，灵活运用点到直线的距离公式化简求值，是一道中档题．6. 下列说法正确的是( )A．a2＞b2是a＞b的必要条件B．“若a∈（0，1），则关于x的不等式ax2+2ax+1＞0解集为R”的逆命题为真C．“若a，b不都是偶数，则a+b不是偶数”的否命题为假D．“已知a，b∈R，若a+b≠3，则a≠2或b≠1”的逆否命题为真参考答案：D【考点】命题的真假判断与应用．【专题】综合题；简易逻辑．【分析】对4个选项分别进行判断，即可得出结论．【解答】解：A，当a=﹣2，b=1时，a2＞b2成立，但a＞b不成立，即“a2＞b2”是“a＞b”的不充分条件；当a=1，b=﹣1时，a＞b成立，但a2＞b2不成立，即“a2＞b2”是“a＞b”的不必要条件，故“a2＞b2”是“a＞b”的既不充分也不必要条件，故不正确；B，由命题p：不等式ax2+2ax+1＞0的解集为R可得a＞0 且 4a2﹣4a＜0，或者a=0，解得0≤a＜1，故不正确；C，命题“若a，b不都是偶数，则a+b不是偶数”的否命题为：若a，b都是偶数，则a+b是偶数，正确；D，“若a+b≠3，则a≠1或b≠2”的逆否命题是：“若a=1且b=2，则a+b=3”是真命题，正确．故选：D．【点评】本题考查命题真假的判断，四种命题的关系，以及原命题与它的逆否命题真假性相同的应用，属于中档题．7. 已知函数是R上的奇函数，且当时，设函数,若，则实数的取值范围是A． B．C．(1,2) D．参考答案：D8. 设分别是椭圆的左右焦点,若在其右准线上存在点,使为等腰三角形,则椭圆的离心率的取值范围是（）A．B．C．D．参考答案：C9. 在椭圆上有两个动点P ，Q ，E （3,0）为定点，EP⊥EQ，则最小值为（ ）A. 6B.C. 9D.参考答案：A设，则有，因为EP⊥EQ，所以，即，因为，所以当时，取得最小值6，故选择A 。

2021年寿宁二中高三英语二模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ALocated in the beautiful Sichuan Basin, Chongqing is a magical 8D city. The natural history and cultural scenery of the area provide children with learning opportunities because they can enjoy the many wonders of this area.Fengjie Tiankeng Ground JointTiankeng Diqiao Scenic Area is located in the southern mountainous area of Fengjie County. The Tiankeng pit is 666 meters deep and is currently the deepest tiankeng in the world. The scenic spot is divided into ten areas including Xiaozhai Tiankeng, Tianjingxia Ground, Labyrinth River, and Longqiao River. There are many and weird karst cave shafts, and countless legends haunt them.Youyang Peach GardenYouyang Taohuayuan Scenic Area is a national forest park, a national 5A-level scenic spot, and a national outdoor sports training base. Located in the hinterland of Wuling Mountain. The Fuxi Cave in the scenic spot is about 3,000 meters long, with winding corridors, deep underground rivers, and color1 ful stalactites. The landscape is beautiful.Jinyun Mountain National Nature ReserveJinyun Mountain is located in Beibei District of Chongqing City, about 45 kilometers away from the Central District of Chongqing City. The nine peaks of Jinyun Mountain stand upright and rise from the ground. The ancient trees on the mountain are towering, the green bamboos form the forest, the environment is quiet, and the scenery is beautiful, so it is called "Little Emei". Among them, Yujian Peak is the highest, 1050 meters above sea level; Lion Peak is the most precipitous and spectacular, and the other peaks are also unique.Chongqing People's SquareChongqing's Great Hall of the People, one of the landmarks of Chongqing, gives people the deepest impression than its magnificent appearance resembling the Temple of Heaven. It also uses the traditional method of central axis symmetry, with colonnade-style double wings and a tower ending, plus a large green glazed roof, large red pillars, white railings, double-eave bucket arches, and painted carved beams.1.How deep is the Tiankeng Ground Joint?A.666mB.3,000mC.45kmD.1050m2.Which of the following rocks can you see in Youyang Peach Garden?A.LimestoneB.StalactiteC.MarbleD.Quartzite3.Which attraction is closest to downtown Chongqing?A.Fengjie Tiankeng Ground JointB.Jinyun Mountain National Nature ReserveC.Chongqing People's SquareD.Youyang Peach GardenBRemember when your mom told you not to eat too many candy bars or sweets because they can cause tooth decay (蛀牙)? However, it turns out that chocolate can be moresalutaryto your teeth than you might expect. Recent studies show that chocolate can effectively fight against tooth decay, as if we need another excuse to eat chocolate.Chocolate offers protection like fluoride, a main ingredient in most household toothpastes. Not only does chocolate protect our teeth, but it can do so very effectively. Studies show that chocolate has compounds that provide strong protection for teeth. One of the compounds in chocolate, CBH, is shown to protect even more effectively than fluoride.Tooth decay happens when bacteria work to turn sugar into acids in our mouth. This is why eating foods with high sugar content can lead to more tooth decay. The compounds in chocolate, however, are anti-bacteria and can fight against bacteria in your mouth. The CBH compound in particular also works to strengthen tooth enamel (牙釉质), andprotects against tooth decay.Does this mean you can cat as much chocolate as you want without worrying about your teeth? It depends on the types of chocolate that you like. The protective effect of chocolate is most effective when you chew on cocoa beans. Of course, this option is not very appealing to; most people. A more tasty option is to choose dark chocolate with little sugar content, ideally no more than 6 to 8 grams per serving. For other types of chocolate with higher sugar content, the effect will be lessened. However, because of the protective compounds, it is still better for your teeth than other sweets and desserts containing the same amount of sugar.4. The word “salutary” in paragraph 1 means?A. Beneficial.B. Harmful.C. Familiar.D. Useless.5. What can we know about the compound CBH in chocolate?A. It can help chocolate cure tooth decay.B. It can effectively stop teeth from decaying.C. It may protect teeth better than toothpastes do.D. It may soon replace most household toothpastes.6. How does chocolate fight tooth decay?A. By breaking down acids.B. By building up compounds.C. By fixing up tooth enamel.D. By fighting against bacteria.7. What's the main idea of the text?A. Chocolate plays the role of toothpaste.B. Chocolate protects against tooth decay.C. Chocolate is the best choice for teeth protection.D. Chocolate is healthier to teeth than other sweets.CSpain's tourism industry is looking to Chinese tourists for its high-endmarket, according to Rafael Cascales, president of the Spain-China Tourism Association (ATEC). “It is the kind of tourism that is not only interested in the sun, beach and the “all-included” culture. They enjoy culture, wine, history and nature, and the new Chinese tourists would also want to spend more money in Spain," said Cascales in a recent interview with Xinhua.“They are younger, more women travel and they are more cosmopolitan (见多识广的).They also travel on their own or in couples or in smaller groups. The old-fashioned large groups of visitors have not disappeared, but this new form of traveling is becoming more important,55he said.Speaking of the consumption pattern of the new kind of Chinese tourists, Cascales said, “The money they spend is distributed better because they will book one flight with one airline, the hotel with another company and the restaurant with another.” In his eyes, “Chinese tourists are very important because they combine two things: there are a large number of them and they spend more money than anyone else — almost four times more than tourists from other countries." They not only travel abroad in the summer months when Spain has to compete with the sun and beaches in countries such as Turkey and Egypt, but also travel in the off-peak seasons of a year, according to Cascales.In 2017, Spain is the second most popular tourist destination in the world, only after France. It attracted about 82 million visitors, 700,000 of them from China, a number which is estimated to rise to about 2.2 million by 2022.“We are ready; we have the infrastructure (基础设施) at every level, especially in hotel capacity. Here those visitors can find what they are looking for, including the luxury items which distinguish them,” Cascales noted.8. What are the features of the new Chinese tourists according to Cascales?A. They are cautious about spending money in Spain.B. They are likely to travel in smaller groups now.C. They are only interested in the sun and beach.D. They are mainly male visitors of middle age.9. What can we learn about the consumption pattern of new Chinese tourists?A. They will reserve flights and hotels with different companies.B. They will spend less money than tourists from other countries.C. They will travel abroad only during the off-peak seasons.D. They will spend the money in different cities.10. What is done to meet the demands of Chinese tourists?A. Local cultures in Spain are promoted.B. Well-furnished hotels are provided.C. Best and expensive items are offered for free.D. More shopping sites are constructed.11. What is the purpose of this text?A. To introduce the tourism industry of Spain.B. To show Spain's desire to attract Chinese tourists.C. To describe the features of Chinese tourists.D. To advertise Spain as a top tourist destination.DScientists have long sought to prevent sharp memories from dulling with age, but the problem remains unsettled. Now research published in Scientific Reports suggests virtual reality might help older people recall facts and events based on specific details.The study involved 42 healthy older adults from the San Francisco Bay Area. Half spent a dozen hours over four weeks playing a virtual-reality game called Labyrinth; they wore headsets and walked in place, walking virtual neighborhoods while completing small tasks. The other half, in the control group, used electronic tablets to play games that did not require recalling details. After 15 sessions (期), the latter performed roughly the same asbefore on a long-term memory test. But the Labyrinth players gain an improvement in memory through the VR game. A scientist Peter Wais of the University of California said the improvements brought them up to the level of another group of younger adults taking the same memory tests.Meredith Thompson, an education researcher, studies learning through VR games but was not involved in the new study. It would be great to actually follow people over time and see what this type of game does for long-term memory. She says, adding VR can provide greater involvement than other games. Wais's team is now investigating how long the observed effects last and which elements of the training have the most impact.A cognitive (认知)psychologist, Daniel Simons, who was also not involved in the study, notes experiments with other games that claim to train the brain have often failed to evaluate this. And it remains unclear how test performance in a laboratorysetting might translate to real-world situations. The outcome, Simons notes, “needs to be repeated, ideally with a much larger group, before it’s treated as a strong finding.”For now, Wais says, the team hopes its studies with similar-sized groups will help draw funding to test the game in a larger pool of participants.12. What is the passage mainly about?A. People's memory gradually fails as they age.B. People of different ages should play VR games.C. Virtual reality changes people's memory.D. Virtual reality improves older adults' memory.13. What is Meredith Thompson's attitude toward the research?A. satisfied.B. enthusiastic.C. cautious.D. concerned.14. According to the scientists, the research needs to be improved due to ________.A. the lack of financial support.B. the limited pool of participants.C. the unsatisfying test performance.D. the impractical application in real world.15. Where may the passage come from?A. A novel.B. A review.C. A magazine.D. A guidebook.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。