广东省北京师范大学东莞石竹附属学校2016届高三上学期第二次月考港澳台学生英语试题 Word版含答案.doc

2015-2016学年第一学期港澳台学生高三试题（四）英语（2015.12）注意事项：1.答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

2.将答案涂写在答题卡上，写在本试卷上无效。

3.考试结束后，将本试卷和答题卡一并交回。

I.听力（共20 小题，每小题1.5 分；满分30 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个小题。

每段对话仅读一遍。

1.What does the man mean?A．He doesn’t like the journey. B．He has the same journey in the past.C．He likes the journey very much.2.Where does the conversation probably take place?A．In a clothes shop. B．In a tailor’s shop. C．In an exhibition hall. 3.How many times has the woman been late for school?A．Twice B．Four times. C．Five times.4.When should the man hand in his paper?A．Today B．Tomorrow C．The day after tomorrow.5. Who called the police station?A．A passer-by B．The thief. C. The car driver.听下面5段对话或独白。

每段对话或独白后有几个小题，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。

听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟。

2015～2016学年度第一学期第二次月考试卷 高三数学（理科） [答卷时长（120）分钟 总分：150分] 命题人：张培利一、 选择题：（本大题共12小题，每小题5分，共60分） 1、 设集合{}|24x A x =≤，集合{}|lg(1)B x y x ==-，则A B 等于( )A ．(1,2)B ． (1,2]C ． [1,2)D ． [1,2]2、当2x ππ-≤≤时,函数()sin f x x x =+的（ ）A ．最大值是1，最小值是．最大值是1，最小值是1- C ．最大值是2，最小值是 D ．最大值是2，最小值是1- 3、 下列命题中的假命题是（ ）．A ．2,log 0x R x ∃∈=B ．0,2>∈∀x R xC ．0tan ,=∈∃x R xD ．,30x x R ∀∈> 4、 已知1122log log a b <，则下列不等式一定成立的是（ ）．A ．11a b > B ．11()()43a b < C ．ln()0a b -> D ．31a b -<5、一个三棱锥的三视图是三个直角三角形，如图所示，则该三棱锥的外接球表面积( ) A ．29π B ．30π C ．292πD ．216π 6、已知函数()f x 的图像是连续不断的，有如下的x ，()f x 的对应表则函数f x 存在零点的区间有( ) A ．区间[][]1,22,3和B ．区间[][]2,33,4和C ．区间[][][]2,33,44,5、和D ．区间[][][]3,44,55,6、和 7、函数()()sin f x A x ωϕ=+（其中0,2A πϕ><）的图像如图所示，为了得到俯视图主视图侧视图π7πx()cos 22g x x π⎛⎫=- ⎪⎝⎭的图像，只需将()f x 的图像( )A ．向左平移3π个长度单位 B ．向右平移3π个长度单位．260y +=垂直,则a =（ ）3⎩( )A ．52,2⎡⎤⎢⎥⎣⎦B ．55,42⎡⎤⎢⎥⎣⎦ C ．45,52⎡⎤⎢⎥⎣⎦ D ．5,24⎡⎤⎢⎥⎣⎦11、如图，从气球A 上测得正前方的河流的两岸B ，C 的俯角分别为075，030，此时气球的高是60m ，则河流的宽度BC 等于( ) A ．1)m - B ．1)m - C ．1)m - D ．1)m +12、若偶函数()f x 的图像关于1x =对称，且当[0,1]x ∈时，()f x x =，则函数()y f x =的图象与函数lg y x =的图象的交点个数为（ ）A. 14B. 16C. 18D. 20二、填空题： （每小题5分，共20分）13、已知函数，⎪⎩⎪⎨⎧<+≥=3)1(3)21()(x x f x x f x则(l)f = 。

2018--2019学年度上学期高三10月月考英语试题*本试题卷共5页。

全卷满分120分，折合分135分，考试用时120分钟。

第二部分阅读理解 (共两节, 满分40分)第一节 (共15小题; 每小题2分, 满分30分)阅读下面短文, 从每题所给的A、B、C、和D四个选项中, 选出最佳选项, 并在答题卡上将该项涂黑。

AAND ENJOY A FREE * FLIGHT TO ANY DESTINATION IN ASI A!With a registration fee of just $50 per child, children under the age of 12 can join Eagle Airways' FLY* Child must be accompanied by two paying adults.** Terms and conditions apply.21. One of the benefits mentioned in the advertisement is ____.A. a free flight to any destination in the worldB.30% off any book purchased at Ruby BookstoreC. a free bowl of dessert at any restaurant at the airportD. a discount on any course at Tanya Language School22. Which of the following bookings may receive the most benefits?23. Which of the following is TRUE according to the advertisement?A. You need to pay $50 to sign up a child for the club.B. Club members enjoy free travel insurance for any flight.C. The advertisement is intended for students of all ages.D. Any child must be accompanied by at least one paying adult.BI've loved my mother's desk since I was just tall enough to sit above the top of it. Mother sat writing letters. Standing by her chair, looking at the ink bottle, pens, and white paper, I decided that the act of writing must be a most wonderful thing in the world.Years later, during her final illness, Mother kept different things for mysisterand brother. "But the desk," she said again, "is for Elizabeth."I never saw her angry, never saw her cry. I knew she loved me; she showed in action. But as a young girl, I wanted to have heart-to-heart talks between mother and daughter.They never happened. And a gulf opened between us. I was "too emotional". But she lived "on the surface".As years passed and I had my own family. I loved my mother and thanked her for our happy family. I wrote to her in careful words and asked her to let me know in any way she chose that she did forgive me.My hope turned to disappointment, then little interest and, finally, peace - it seemed that nothing happened. I couldn't be sure that the letter had even got to Mother.I only knew that I had written it, and I could stop trying to make her into someone she was not.But the present of her desk told me, as she'd never been able to, that she was pleased that writing was my chosen work. I cleaned the desk carefully and found some papers inside - a photo of my father and a one-paper letter, folded and refolded many times. It was my letter."In any way you choose, Mother, you always chose the act that speaks louder than words."24. When did the writer begin to love her mother's desk?A. After Mother died.B. Before she became a writer.C. When she was a child.D. When Mother gave it to her.25. What does the passage want to show?A. Mother was actually kind in her heart to her daughter.B. Mother was too serious about her daughter in words.C. Mother wrote to her daughter in careless words.D. Mother wrote to her daughter in careful words.26. What does the underlined word "gulf" in the passage mean?A. Deep understanding between the old and the young.B. Different ideas between mother and daughter.C. Free talks between mother and daughter.D. Part of the sea going far in land.27. What did Mother do with her daughter's letter asking for forgiveness?A. She had never received the letter.B. For years, she often talked about the letter.C. She read the letter again and again till she died.D. She didn't forgive her daughter at all in all her life.CThe production of coffee beans is a huge, profitable business, but, unfortunately, full-sun production is taking over the industry and bringing about a lot of damage. The change in how coffee is grown from shade-grown production to full-sun production endangers the very existence of certain animals and birds, and even disturbs the worl d’s ecological balance.On a local level, the damage of the forest required by full-sun fields affects the area’s birds and animals. The shade of the forest trees provides a home for birds and other species that depend on the trees’ flowers and fruits. Ful l-sun coffee growers destroy this forest home. As a result, many species are quickly dying out.On a more global level, the destruction of the rainforest for full-sun coffee fields also threatens (威胁) human life. Medical research often makes use of the forests' plant and animal life, and the destruction of such species could prevent researchers from finding cures for certain diseases. In addition, new coffee-growing techniques are poisoning the water locally, and eventually the world's groundwater.Both locally and globally, the continued spread of full-sun coffee plantations (种植园) could mean the destruction of the rainforest ecology. The loss of shade trees is already causing a slight change in the world's climate, and studies show that loss of oxygen-giving trees also leads to air pollution and global warming. Moreover, the new growing techniques are contributing to acidic (酸性的) soil conditions.It is obvious that the way much coffee is grown affects many aspects of life, from the local environment to the global ecology. But consumers do have a choice.They can purchase shade-grown coffee whenever possible, although at a higher cost. The future health of the planet and mankind is surely worth more than an inexpensive cup of coffee.28. What can we learn about full-sun coffee production from Paragraph 4?A. It limits the spread of new growing techniques.B. It leads to air pollution and global warming.C. It slows down the loss of shade trees.D. It improves local soil conditions.29. The purpose of the text is to________.A. entertainB. advertiseC. instructD. persuade30. Where does this text probably come from?A. An agricultural magazine.B. A medical journal.C. An engineering textbook.D. A tourist guide.31. Which of the following shows the structure of the whole text?(P：Paragraph)DEvery year more people recognize that it is wrong to kill wildlife for "sport." Progress in this direction is slow because shooting is not a sport for watching, and only those few who take part realize the cruelty and destruction.The number of gunners, however, grows rapidly. Children too young to develop proper judgments through independent thought are led a long way away by their gunning parents. They are subjected to advertisements of gun producers who describe shooting as good for their health and gun carrying as a way of putting redder blood in the veins (血管). They are persuaded by gunner magazines with stories honoring the chase and the kill. In school they view motion pictures which are supposedly meant to teach them how to deal with arms safely but which are actually designed to stimulate a desire to own a gun. Wildlife is disappearing because of shooting and because of the loss of wild land habitat. Habitat loss will continue with our increasing population, but can we slow the loss of wildlife caused by shooting? There doesn't seem to be any chance if the serious condition of our birds is not improved.Wildlife belongs to everyone and not to the gunners alone. Although most people do not shoot, they seem to forgive shooting for sport because they know little or nothing about it. The only answer, then, is to bring the truth about sport shooting to the great majority of people.Now, it is time to realize that animals have the same right to life as we do and that there is nothing fair or right about a person with a gun shooting the harmless and beautiful creatures. The gunners like to describe what they do as character-building, but we know that to wound an animal and watch it go through the agony of dying can make nobody happy. If, as they would have you believe, gun-carrying and killing improve human-character, then perhaps we should encourage war.32. According to the text, the films children watch at school actually_______.A. teach them how to deal with guns safelyB. praise hunting as character-huntingC. describe hunting as exerciseD. encourage them to have guns of their own33. Most people do not seem to be against hunting because_____.A. it helps to build human characterB. they have little knowledge of itC. it is too costly to stop killing wildlifeD. they want to keep wildlife under control34. The underlined word “agony” in the last paragraph probably means_____.A. formB. conditionC. painD. sadness35. It can be inferred from the text that the author seems to ______.A. worry about the existence of wildlifeB. blame the majority of peopleC. be in favor of warD. be in support of hunting第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

2015-2016学年广东省北师大东莞石竹附中高三（上）期中数学试卷（理科）一、选择题：（本大题共12小题，每小题5分，共60分）1．若{1}⊆A⊆{1，2，3}，则这样的集合A有( )A．1个B．2个C．3个D．4个2．下列有关命题的说法正确的是( )A．命题“若x2=1，则x=1”的否命题为：“若x2=1，则x≠1”B．命题“若x=y，则sinx=siny”的逆否命题为真命题C．命题“存在x∈R，使得x2+x+1＜0”的否定是：“对任意x∈R，均有x2+x+1＜0”D．“x=﹣1”是“x2﹣5x﹣6=0”的必要不充分条件3．设变量x，y满足约束条件，则目标函数z=x+6y的最大值为( ) A．3 B．4 C．18 D．404．设函数f（x）=，则f（）的值为( )A．B．﹣C．D．185．“a=2”是“函数f（x）=|x﹣a|在区间11．已知函数f（x）是定义在R上的偶函数，且在区间B．C．D．（0，2]12．用C（A）表示非空集合A中的元素个数，定义A*B=，若A={1，2}，B={x||x2+ax+1|=1}，且A*B=1，由a的所有可能值构成的集合是S，那么C（S）等于( )A．4 B．3 C．2 D．1二、填空题：（每小题5分，共20分）13．已知全集U={1，2，3，4，5，6，7}，A={3，4，5}，B={1，3，6}，则A∩（∁U B）等于__________．14．已知函数f（x）=a x+b（a＞0，a≠1）的定义域和值域都是，则a+b=__________．15．已知x＞0，y＞0，若不等式恒成立，则m的最大值为__________．16．如图，互不相同的点A1，A2，…，A n，…和B1，B2，…，B n，…分别在角O的两条边上，所有A n B n相互平行，且所有梯形A n B n B n+1A n+1的面积均相等，设OA n=a n，若a1=1，a2=2，则数列{a n}的通项公式是__________．三、解答题：本大题共6小题，共80分．解答写出文字说明、演算步骤或推证过程．17．已知二次函数f（ x ）=x2+ax+b关于x=1对称，且其图象经过原点．（1）求这个函数的解析式；（2）求函数在x∈（0，3]的值域．18．已知p：方程x2+mx+1=0有两个不等的负实根，q：方程4x2+4（m﹣2）x+1=0无实根．若“p或q”为真，“p且q”为假．求实数m的取值范围．19．如题图，三棱锥P﹣ABC中，PC⊥平面ABC，PC=3，∠ACB=．D，E分别为线段AB，BC 上的点，且CD=DE=，CE=2EB=2．（Ⅰ）证明：DE⊥平面PCD（Ⅱ）求二面角A﹣PD﹣C的余弦值．20．某纺纱厂生产甲、乙两种棉纱，已知生产甲种棉纱1吨需耗一级子棉2吨、二级子棉1吨；生产乙种棉纱需耗一级子棉1吨、二级子棉2吨，每1吨甲种棉纱的利润是600元，每1吨乙种棉纱的利润是900元，工厂在生产这两种棉纱的计划中要求消耗一级子棉不超过300吨、二级子棉不超过250吨．甲、乙两种棉纱应各生产多少，能使利润总额最大？21．设S n为数列{a n}前n项和，对任意的n∈N*，都有S n=2﹣a n，数列{b n}满足，b1=2a1，（1）求证：数列{a n}是等比数列，并求{a n}的通项公式；（2）求数列{b n}的通项公式；（3）求数列的前n项和T n．22．（14分）已知函数f（x）=e x+e﹣x，其中e是自然对数的底数．（1）证明：f（x）是R上的偶函数；（2）若关于x的不等式mf（x）≤e﹣x+m﹣1在（0，+∞）上恒成立，求实数m的取值范围；（3）已知正数a满足：存在x0∈x＞0时，f（x）=x（1+x3），即有f（﹣x）=﹣x（1﹣x3），又函数f（x）为奇函数，则f（﹣x）=﹣f（x），即有f（x）=x（1﹣x3）（x＜0），故选A．【点评】本题考查函数的奇偶性及运用：求函数的解析式，注意运用定义，考查运算能力，属于基础题．8．已知S n为等差数列{a n}的前n项和，若S1=1，，则的值为( )A．B．C．D．4【考点】等差数列的性质．【专题】计算题．【分析】根据首项等于S1，得到首项的值，利用等差数列的前n项和公式化简，即可求出公差d的值，然后再利用等差数列的前n项和公式化简所求的式子，把求出的首项和公差代入即可求出值．【解答】解：由S1=a1=1，，得到=4，解得d=2，则===．故选A【点评】此题考查学生灵活运用等差数列的前n项和公式化简求值，掌握等差数列的性质，是一道基础题．9．函数f（x）=ln（x2+1）的图象大致是( )A． B．C．D．【考点】函数的图象．【专题】函数的性质及应用．【分析】∵x2+1≥1，又y=lnx在（0，+∞）单调递增，∴y=ln（x2+1）≥ln1=0，函数的图象应在x轴的上方，在令x取特殊值，选出答案．【解答】解：∵x2+1≥1，又y=lnx在（0，+∞）单调递增，∴y=ln（x2+1）≥ln1=0，∴函数的图象应在x轴的上方，又f（0）=ln（0+1）=ln1=0，∴图象过原点，综上只有A符合．故选：A【点评】对于函数的选择题，从特殊值、函数的性质入手，往往事半功倍，本题属于低档题．10．定义在R上的函数f（x）满足f（x+6）=f（x）．当﹣3≤x＜﹣1时，f（x）=﹣（x+2）2；当﹣1≤x＜3时，f（x）=x．则f（1）+f（2）+f（3）+…+f=( )A．335 B．336 C．338 D．2 016【考点】函数的周期性；函数的值．【专题】规律型；整体思想；函数的性质及应用．【分析】可得函数的周期为6，计算f（1）+f（2）+f（3）+f（4）+f（5）+f（6）的值，结合规律可得．【解答】解：∵定义在R上的函数f（x）满足f（x+6）=f（x），∴函数f（x）为周期为6的周期函数，∵当﹣3≤x＜﹣1时，f（x）=﹣（x+2）2，当﹣1≤x＜3时，f（x）=x，∴f（1）=1，f（2）=2，f（3）=f（﹣3）=﹣1，f（4）=f（﹣2）=0，f（5）=f（﹣1）=﹣1，f（6）=f（0）=0，∴f（1）+f（2）+f（3）+f（4）+f（5）+f（6）=1f（1）+f（2）+f（3）+…+f=336×1=336故选：B．【点评】本题考查函数的周期性，属基础题．11．已知函数f（x）是定义在R上的偶函数，且在区间B．C．D．（0，2]【考点】奇偶性与单调性的综合．【专题】函数的性质及应用．【分析】根据偶函数的定义将所给的式子化为：f（|log2a|）≤f（1），再利用偶函数的单调性列出关于a的不等式求解．【解答】解：∵f（x）是定义在R上的偶函数，∴，∴可变为f（log2a）≤f（1），即f（|log2a|）≤f（1），又∵在区间，则a+b=﹣．【考点】函数的值域．【专题】函数的性质及应用．【分析】对a进行分类讨论，分别题意和指数函数的单调性列出方程组，【解答】解：当a＞1时，函数f（x）=a x+b在定义域上是增函数，所以，解得b=﹣1，=0不符合题意舍去；当0＜a＜1时，函数f（x）=a x+b在定义域上是减函数，所以解得b=﹣2，a=综上a+b=，故答案为；﹣【点评】本题考查指数函数的单调性的应用，以及分类讨论思想，属于基础题15．已知x＞0，y＞0，若不等式恒成立，则m的最大值为12．【考点】基本不等式．【专题】转化思想；整体思想；不等式．【分析】题目转化为m≤（+）（x+3y）恒成立，由基本不等式求（+）（x+3y）的最小值可得．【解答】解：∵x＞0，y＞0，不等式恒成立，∴m≤（+）（x+3y）恒成立，又（+）（x+3y）=6++≥6+2=12当且仅当=即x=3y时取等号，∴（+）（x+3y）的最小值为12，由恒成立可得m≤12，即m的最大值为12，故答案为：12．【点评】本题考查基本不等式求最值，涉及恒成立问题，属基础题．16．如图，互不相同的点A1，A2，…，A n，…和B1，B2，…，B n，…分别在角O的两条边上，所有A n B n相互平行，且所有梯形A n B n B n+1A n+1的面积均相等，设OA n=a n，若a1=1，a2=2，则数列{a n}的通项公式是．【考点】数列的应用；数列的函数特性．【专题】压轴题；等差数列与等比数列．【分析】设，利用已知可得A1B1是三角形OA2B2的中位线，得到==，梯形A1B1B2A2的面积=3S．由已知可得梯形A n B n B n+1A n+1的面积=3S．利用相似三角形的性质面积的比等于相似比的平方可得：，，，…，已知，，可得，…．因此数列{}是一个首项为1，公差为3等差数列，即可得到a n．【解答】解：设，∵OA1=a1=1，OA2=a2=2，A1B1∥A2B2，∴A1B1是三角形OA2B2的中位线，∴==，∴梯形A1B1B2A2的面积=3S．故梯形A n B n B n+1A n+1的面积=3S．∵所有A n B n相互平行，∴所有△OA n B n（n∈N*）都相似，∴，，，…，∵，∴，，…．∴数列{}是一个等差数列，其公差d=3，故=1+（n﹣1）×3=3n﹣2．∴．因此数列{a n}的通项公式是．故答案为．【点评】本题综合考查了三角形的中位线定理、相似三角形的性质、等差数列的通项公式等基础知识和基本技能，考查了推理能力和计算能力．三、解答题：本大题共6小题，共80分．解答写出文字说明、演算步骤或推证过程．17．已知二次函数f（ x ）=x2+ax+b关于x=1对称，且其图象经过原点．（1）求这个函数的解析式；（2）求函数在x∈（0，3]的值域．【考点】二次函数的性质．【专题】计算题．【分析】（1）由已知条件列方程，即可得解（2）根据二次函数对称轴与区间的位置关系，确定原函数在（0，3]上的单调性，由单调性求值域【解答】解：（1）二次函数f（x）关于x=1对称∴∴a=﹣2又f（x）的图象经过原点∴b=0∴f（x）的解析式为f（x）=x2﹣2x（2）∵对称轴x=1落在区间（0，3]内，且抛物线开口向上∴函数在（0，1]上单调递减，在上单调递增∴x=1时，f（x）有最小值，最小值为f（1）=1﹣2=﹣1；x=3时，f（x）有最大值，最大值为f（3）=9﹣6=3∴f（x）的值域是【点评】本题考查用待定系数法求二次函数的解析式和求二次函数的最值问题，需注意区间与对称轴的位置关系．属简单题18．已知p：方程x2+mx+1=0有两个不等的负实根，q：方程4x2+4（m﹣2）x+1=0无实根．若“p或q”为真，“p且q”为假．求实数m的取值范围．【考点】复合命题的真假；一元二次方程的根的分布与系数的关系．【专题】分类讨论．【分析】根据题意，首先求得p、q为真时m的取值范围，再由题意p，q中有且仅有一为真，一为假，分p假q真与p真q假两种情况分别讨论，最后综合可得答案．【解答】解：由题意p，q中有且仅有一为真，一为假，若p为真，则其等价于，解可得，m＞2；若q为真，则其等价于△＜0，即可得1＜m＜3，若p假q真，则，解可得1＜m≤2；若p真q假，则，解可得m≥3；综上所述：m∈（1，2]∪故可取=（2，1，1），由（Ⅰ）知DE⊥平面PCD，故平面PCD的法向量可取=（1，﹣1，0），∴两法向量夹角的余弦值cos＜，＞==∴二面角A﹣PD﹣C的余弦值为．【点评】本题考查二面角，涉及直线与平面垂直的判定，建系化归为平面法向量的夹角是解决问题的关键，属难题．20．某纺纱厂生产甲、乙两种棉纱，已知生产甲种棉纱1吨需耗一级子棉2吨、二级子棉1吨；生产乙种棉纱需耗一级子棉1吨、二级子棉2吨，每1吨甲种棉纱的利润是600元，每1吨乙种棉纱的利润是900元，工厂在生产这两种棉纱的计划中要求消耗一级子棉不超过300吨、二级子棉不超过250吨．甲、乙两种棉纱应各生产多少，能使利润总额最大？【考点】简单线性规划的应用．【专题】数形结合；不等式的解法及应用．【分析】利用线性规划知识求解，建立约束条件，作出可行域，再根据目标函数z=600x+900y，利用截距模型，平移直线找到最优解，即可．【解答】解：设生产甲、乙两种棉纱分别为x吨、y吨，利润总额为z元，则目标函数为z=600x+900y．作出以上不等式组所表示的平面区域（如图），即可行域．作直线l：600x+900y=0，即直线l：2x+3y=0，把直线l向右上方平移至l1的位置时，直线经过可行域上的点M，且与原点距离最大，此时z=600x+900y取最大值．解方程组，解得M的坐标为（）因此，当x=，y=时，z取得最大值．此时zmax=600×+900×=130000．答：应生产甲种棉纱吨，乙种棉纱吨，能使利润总额达到最大，最大利润总额为13万元．【点评】本题考查用线性规划解决实际问题中的最值问题，解题的关键是确定约束条件，作出可行域，利用目标函数的类型，找到最优解，属中档题．21．设S n为数列{a n}前n项和，对任意的n∈N*，都有S n=2﹣a n，数列{b n}满足，b1=2a1，（1）求证：数列{a n}是等比数列，并求{a n}的通项公式；（2）求数列{b n}的通项公式；（3）求数列的前n项和T n．【考点】数列的求和．【专题】计算题．【分析】（1）当n=1时，由a1=S1=2﹣a1，可求a1，n≥2时，由a n=S n﹣S n﹣1，可得a n=与a n﹣1之间的递推关系，结合等比数列的通项公式可求a n（2）由，可得，结合等差数列的通项公式可求，进而可求b n（3）由（1）（2）可求，利用错位相减求和即可求解【解答】（本小题满分14分）证明：（1）当n=1时，a1=S1=2﹣a1，解得a1=1．…当n≥2时，a n=S n﹣S n﹣1=a n﹣1﹣a n，即2a n=a n﹣1．∴．…∴数列{a n}是首项为1，公比为的等比数列，即．…解：（2）b1=2a1=2．…∵，∴，即．…∴是首项为，公差为1的等差数列．…∴，…（3）∵，则．…所以，…即，①…则，②…②﹣①得，…（13分）故．…（14分）【点评】本题主要考查了利用数列的递推公式求解数列的通项公式、等差数列与等比数列的通项公式的应用，还考查了错位相减求和方法的应用22．（14分）已知函数f（x）=e x+e﹣x，其中e是自然对数的底数．（1）证明：f（x）是R上的偶函数；（2）若关于x的不等式mf（x）≤e﹣x+m﹣1在（0，+∞）上恒成立，求实数m的取值范围；（3）已知正数a满足：存在x0∈（2）利用参数分离法，将不等式mf（x）≤e﹣x+m﹣1在（0，+∞）上恒成立，进行转化求最值问题即可求实数m的取值范围；（3）构u造函数，利用函数的单调性，最值与单调性之间的关系，分别进行讨论即可得到结论．【解答】解：（1）∵f（x）=e x+e﹣x，∴f（﹣x）=e﹣x+e x=f（x），即函数：f（x）是R上的偶函数；（2）若关于x的不等式mf（x）≤e﹣x+m﹣1在（0，+∞）上恒成立，即m（e x+e﹣x﹣1）≤e﹣x﹣1，∵x＞0，∴e x+e﹣x﹣1＞0，即m≤在（0，+∞）上恒成立，设t=e x，（t＞1），则m≤在（1，+∞）上恒成立，∵=﹣=﹣，当且仅当t=2时等号成立，∴m．（3）令g（x）=e x+e﹣x﹣a（﹣x3+3x），则g′（x）=e x﹣e﹣x+3a（x2﹣1），当x＞1，g′（x）＞0，即函数g（x）在[1，+∞）上单调递增，故此时g（x）的最小值g（1）=e+﹣2a，由于存在x0∈[1，+∞），使得f（x0）＜a（﹣x03+3x0）成立，故e+﹣2a＜0，即a＞（e+），令h（x）=x﹣（e﹣1）lnx﹣1，则h′（x）=1﹣，由h′（x）=1﹣=0，解得x=e﹣1，当0＜x＜e﹣1时，h′（x）＜0，此时函数单调递减，当x＞e﹣1时，h′（x）＞0，此时函数单调递增，∴h（x）在（0，+∞）上的最小值为h（e﹣1），注意到h（1）=h（e）=0，∴当x∈（1，e﹣1）⊆（0，e﹣1）时，h（e﹣1）≤h（x）＜h（1）=0，当x∈（e﹣1，e）⊆（e﹣1，+∞）时，h（x）＜h（e）=0，∴h（x）＜0，对任意的x∈（1，e）成立．①a∈（（e+），e）⊆（1，e）时，h（a）＜0，即a﹣1＜（e﹣1）lna，从而e a﹣1＜a e﹣1，②当a=e时，a e﹣1=e a﹣1，③当a∈（e，+∞）⊆（e﹣1，+∞）时，当a＞e﹣1时，h（a）＞h（e）=0，即a﹣1＞（e ﹣1）lna，从而e a﹣1＞a e﹣1．【点评】本题主要考查函数奇偶性的判定，函数单调性和最值的应用，利用导数是解决本题的关键，综合性较强，运算量较大．。