湖北省枣阳一中2014-2015学年高二下学期第三次月检考试文数试题 Word版含答案

枣阳一中2014-2015学年度高二下学期第三次月检测二、阅读理解（共20小题, 每小题2分，满分40分）AIn the summer of 1978 an English farmer was driving his tractor through a field of wheat when he discovered that some of his wheat was lying flat on the ground. The flattened (变平的) wheat formed a circle about six meters across. Around this circle were four smaller circles of flattened wheat. The five circles were in a formation like five dots (点). During the following years, farmers in England found the strange circles in their fields more and more often.The circles are called “crop circles” because they appear in the fields of grain—usually wheat or corn. The grain in the circles lies flat on the ground but is never broken; it continues to grow, and farmers can later harvest it. Farmers always discover the crop circles in the morning, so the circles probably form at night. They appear only in the months from May to September.At first, people thought that the circles were a hoax. Probably young people were making them as a joke, or farmers were making them to attract tourists. To prove that the circles were a hoax, people tried to make circles exactly like the ones that farmers had found. They could not do it. They couldn’t enter a field of grain without leaving tracks, and they couldn’t flatt en the grain without breaking it.Many people believe that beings from outer space are making the circles to communicate（交流）with us from far away and that the crop circles are messages from them.Scientists who have studied the crop circles suggested several possibilities. Some scientists say that a downward rush of wind leads to the formation of the circles—the same downward rush of air that sometimes causes an airplane to crash (坠毁). Other scientists say that forces within the earth cause the circles to appear. There is one problem with all these scientific explanations: crop circles often appear in formations, like the five-dot formation. It is hard to believe that any natural force could form those.1．In the summer of 1978, an English farmer discovered in his field that ________.A. some of his wheat had been damagedB. his grain was growing up in circlesC. his grain was moved into several circlesD. some of his wheat had fallen onto the ground2．According to the text, the underlined part “a hoax” (line 1, para. 3 ) probably means ________.A. an action made to fool peopleB. a special way to plant cropsC. a research on the force of windsD. an experiment for the protection of crops3．Which of the following may prove that the crop circles are not made by man？A. The farmers couldn’t step out of the field.B. The farmers couldn’t make the circles round.C. The farmers coul dn’t leave without footprints.D. The farmers couldn’t keep the wheat straight up.BWith their weakening bodies, advanced age or increasing pressure of work or study, people have been advised and usually persuaded to have health care products. Do you remember your first time to take a nutritional supplement (营养补充品)？What was your first impression? Madeline on Dec. 24，2010 7: 25 PMYears ago, as a young mom with two small children, I struggled to keep up with the demands of a busy life. One day, I woke up feeling so tired that I knew something had to change!A friend gave me some Royal Jelly—an amazing substance (物质) from the beehive (蜂房). I was doubtful, but desperate... so I tried it! In time, I began to experience renewed energy and vitality (活力) like never before! Now, over two decades later, I travel all round the country, sharing my life-changing experience. I'm grateful for the energy to keep up and love to share this wonderful discovery with others.Blanca on Dec. 25，2010 7: 02 PM Ten years ago, when I was 73, my mind wanted to be busy and useful, but my body didn't. I became tired and lethargic (无精打采的). That's when my daughter Madeline tried to persuade me to try her special Royal Jelly. It made such a dramatic (巨大的) difference in her life, so she was sure it could help me too! Hesitating for 2 days, I gave in. She was right! I no longer felt worn out. I had a new, youthful zest (热情) for life and I've been OK —thanks to Royal Jelly!Lori on Dec. 25，2010 10: 27 PMWhen my mom Madeline was introduced to Royal Jelly, it affected my life, too! Mom gave me some of her fantastic Royal Jelly. Boy, things did turn around! I had more energy and stamina (耐力) and was finally able to keep up with high school and, eventually, college activities. Today, I'm a busy mom in my mid-thirties with two active boys and a new business! I have enough energy to do it all! I'm glad I took my mom's advice.4．From the text, we learn that __________.A. Madeline became more than willing to take Royal Jelly at the very beginningB. Lori was grateful partly because Royal Jelly helped him to improve his studiesC. Royal Jelly is a product that can help renew a person's energy and vitalityD. Amy has benefited from Royal Jelly so she posted a topic about the product5．Which of the following is TRUE of Blanca?A. She replied to the topic on Christmas Eve.B. She tried Royal Jelly without hesitation.C. She is now supposed to be in her seventies.D. She was the second one to reply to the topic.6．What's the relationship between the Royal Jelly takers?A. Lori and Blanca are Madeline's grandchildren.B. Madeline is Blancas daughter and Lori's mother.C. Blanca is Lori and Madeline's great grandmother.D. They have no blood relationship but friendship.7．Where can we most probably read this text?A. In a sports club.B. On an Internet page.C. In a fiction.D. In a travel magazine.CIn my profession as an educator and health care provider, I have worked with numerous children infected with the virus that causes AIDS. The relationships that I have had with these special kids have been gifts in my life. They have taught me so many things, but I haveespecially learned that great courage can be found in the smallest of packages. Let me tell you about Tyler.Tyler was born infected with HIV: his mother was also infected. From the very beginning of his life, he was dependent on medications to enable him to survive. When he was five, he had a tube surgically inserted in a vein in his chest. This tube was connected to a pump, which he carried in a small backpack on his back. Medications were hooked up to this pump and were continuously supplied through this tube to his bloodstream. At times, he also needed supplemented oxygen to support his breathing.Tyler wasn't willing to give up one single moment of his childhood to this deadly disease. It was not unusual to find him playing and racing around his backyard, wearing his medicine-laden backpack and dragging his tank of oxygen behind him in his little wagon. All of us who knew Tyler marveled at his pure joy in being alive and the energy it gave him. Tyler's mom often teased him by telling him that he moved so fast that she needed to dress him in red. That way, when she peered through the window to check on him playing in the yard, she could quickly spot him.This dreaded disease eventually wore down even the likes of a little dynamo like Tyler. He grew quite ill and, unfortunately, so did his HIV-infected mother. When it became apparent that he wasn't going to survive, Tyler's mom talked to him about death. She comforted him by telling Tyler that she was dying too, and that she would be with him soon in heaven.A few days before his death, Tyler beckoned me over to his hospital bed and whispered, "l might die soon. I'm not scared. When I die, please dress me in red. Mom promised she's coming to heaven, too. I'll be playing when she gets there, and I want to make sure she can find me." 8．What is the boy Tyler's attitude towards death?A. Pessimistic.B. Optimistic.C. Sorrowful.D. Fearful.9．Tyler requested the writer to dress him in red when he died simply because __________.A. red is a lucky colorB. red might help to cure himC. his mom could spot him easilyD. he could find more mates by wearing red10．Which of the following might serve as a possible title for this passage?A. My Unusual ProfessionB. A Caring MotherC. Mother and SonD. Dying in Red11．The underlined word "dynamo" in the fourth paragraph here means "__________".A. a promising and helpful youthB. an extremely energetic personC. a rare and beautiful flowerD. a magic and understanding superstarD“When your mother scolds you, you can look elsewhere and think about other things. Just ignore her words. But remember: such a tough attitude cannot be used often.”These words are from a series of cartoons which outline skills for children to fight against their mothers. The images have created heated debate among Chinese netizens.Regarded as “a book for children aged 6 to 12 who are always scolded by their parents”，the cartoons, drawn by two 10-year-old Beijing girls, list over 20 skills which children can use to deal with their mothers' anger such as crying, pretending to be ashamed, fleeing into the toilet and pleasing her afterwards.Each skill is described with vivid pictures and humorous notes. The creativity of the young girls has amazed netizens, the Yangtse Evening Post reported on Thursday.According to one of the girls' mothers, her daughter once received a poor mark in an exam, and the mother blamed her and compared her performance with another classmate. The daughter's feelings led to her creating the cartoons.The girl's father, who first posted the pictures on his Sohu Microblog on Monday, said he hopes parents pay close attention to the pictures, allow children to feel free to develop their own characteristics and try not to criticize them so often.The father said the cartoons aren't finished yet; his daughter will continue with them when she has time.“The cartoons, although an individual case, reflec t a modern phenomenon and some of the problems within Chinese family education，” said Yu Qinfang, an expert on family education. According to a survey of 104 children and their parents, Yu discovered that as many as 51.9 percent of primary school students hate being urged to do things by their mothers.“Not giving children enough ti me and hurrying them to do things seems to be a very tiny detail within family life, but it is potentially a huge problem which can easily be ignored by parents. A mother's blame may lead to negative feelings within her child's heart，” Yu said. “Parents sh ould learn to blame less and be more patient.”12．What amazed Chinese netizens according to the text?A．The girls' poor marks.B．The girls' skills against their mothers.C．The girls' creativity.D．The girls' tough attitude to their mothers.13．What do we know about the book?A．It tells us how to control mothers' feelings.B．It includes a number of vivid pictures and humorous notes.C．It is intended for primary school teachers.D．It is finished by children aged 6 to 12.14．By posting the cartoons on his microblog, the father wanted to ________.A．show off his daughter's clevernessB．blame his daughter for drawing these cartoonsC．encourage his daughter to make greater progressD．get other parents to draw a lesson from the cartoons15．According to a survey, Yu Qinfang found that primary school pupils ________.A．dislike being forced to do thingsB．like drawing cartoons in classC．seldom do well in examsD．enjoy being compared with others16．We can conclude from the text that ________.A．parents often ignore their children's hobbiesB．parents often speak highly of their childrenC．how to develop children's gifts puzzles parentsD．how to blame children needs parents' wisdomEFew people would question the value of taking part in sports for young people. With proper training, supervision, protective equipment and techniques, and a proper emphasis on winning, sports can develop a healthy body and spirit and a life-long interest in being active and fit. Without such measures, childhood sports can lead to injuries and even paralysis or death. Even in the best conditions, no activity can be risk-free. But most serious hazards are preventable. Cyclists and football players can reduce their risks by wearing helmets; hockey players by wearing masks; basketball and tennis players by wearing eye guards; baseball players by wearing batting pelmets.Besides, risks to individual players can often be found, and thus prevented, through a properly performed medical exam before a child plays. For accidents that may not be preventable, having an emergency plan and first-aid equipment, and someone trained to use the equipment, can be life-saving.Still, each year, according to the American College of Sports Medicine, more than 775, 000 children under 14 are treated in emergency rooms for sports injuries, nearly half of them preventable. An estimated 300, 000 athletes experience exercise-related head illnesses each year, and almost all of them should have been avoided.Further, from half to three-fourths of sports-related concussions (脑震荡) are never even diagnosed; the injured are often sent back to play too soon and put at risk of another more serious brain-damaging concussion. To help reduce these risks, the National Center for Sports Safety, with the National Athletic Trainers, Association, offers a three-hour online safety course for coaches for $ 28 at www. SportsSafety. org.17．Which of the following is TRUE according to the passage?A. All the accidents can be prevented.B. All the accidents cannot be prevented.C. Lives can be saved so long as there is proper equipment.D. Lives cannot be saved even if there is proper equipment.18．What does the underlined word "hazards" (in Paragraph 2) mean?A. Mistakes.B. Diseases.C. Dangers.D. Situations.19．It is implied in the passage that __________.A. prevention of injuries is not paid enough attention toB. children under 14 are more easily hurt in sportsC. most head illnesses are related to exerciseD. none of the head illnesses should have happened20．What can coaches mainly learn from the online safety course?A. How to cure brain-damaging concussion.B. How to diagnose brain-damaging concussion.C. How to predict the possibility of brain damage.D. How to deal with the injured properly.三、完形填空（共20小题, 每小题1.5分，满分30分）Children model themselves largely on their parents. They do so mainly through identification. Children identify a parent when they believe they have the qualities and feelings that are of that parent. The things parents do and say---and the they do and say to them--therefore strongly influence a child's . However, parents must consistently behave like the type of they want their child to become.A parent's actions affect the self-image that a child forms identification. Children who see mainly positive qualities in their will likely learn to see themselves in a positive way. Children who observe chiefly qualities in their parents will have difficulty positive qualities in themselves. Children may their self-image, however, as they become increasingly by peers groups standards before they reach 13.Isolated(孤立的) events, dramatic(突然的) ones, do not necessarily have a permanent on a child's behavior. Children interact such events according to their established attitudes andprevious training. Children who know they are loved can, , accept the divorce of their parents' or a parent's early . But if children feel unloved, they may interpret such events a sign of rejection or punishment.In the same way, all children are not influenced by toys and games, reading matter, and television programs. in the case of a dramatic change in family relations, the of an activity or experience depends on how the child interprets it.21．A. to B. with C. around D. for22．A. informed B. characteristic C. conceived D. indicative 23．A. gesture B. expression C. way D. Extent24．A. behavior B. words C. mood D. reactions25．A. person B. humans C. creatures D. adult26．A. in turn B. nevertheless C. also D. as a result 27．A. before B. besides C. with D. through28．A. eyes B. parents C. peers D. behaviors29．A. negative B. cheerful C. various D. complex30．A. see B. seeing C. to see D. to seeing31．A. modify B. copy C. give up D. continue32．A. mature B. influenced C. unique D. independent 33．A. not B. besides C. even D. finally34．A. idea B. wonder C. stamp D. effect35．A. luckily B. for example C. at most D. theoretically 36．A. death B. rewards C. advice D. teaching37．A. as B. being C. of D. For38．A. even B. at all C. alike D. as a whole 39．A. Even B. Since C. Right D. As40．A. result B. effect C. scale D. Cause四、阅读填空（共10小题；每小题1分，满分10分）A listener from China has written for advice about how to lose weight. Michael in Shanghai says he is 26 and has battled obesity for most of his life.Obesity, a severe weight problem, is a complex condition. A doctor may advise taking medicine along with changing one's behaviors. Experts say that the most successful weight-loss plans include a wellbalanced diet and exercise.People who want to avoid weight gain have to balance the number of calories they eat with the number of calories they use. To lose weight, you can reduce the number of calories you take in, increase the number you use, or both.A recent study looked at four of the most popular dieting plans in the US. Researchers at Stanford University in California studied over 300 overweight women, mostly in their thirties and forties. Each woman went on one of the four plans: Atkins, The Zone, Ornish or LEARN. The women attended diet classes and received written information about the food plans.At the end of the year, the women on the Atkins diet had lost the most, more than four and one-half kilograms on average. They also did better on tests for cholesterol levels and blood pressure.Christopher Gardner, who led the study, says the Atkins diet may be more successful because of its simple message to eat less sugar. He also says that the advice to increase protein in the diet leads to more satisfying meals. He says that there was not enough money to study men, but that men would probably have similar results.Last week, another report based on thirty-one studies suggested that only a small minority of people have long-term success with dieting. Most dieters regained their lost weight within five years and often they become more overweight. But those who kept the weight off generally were the ones who exercised.Title: 41．to lose weight五、短文改错（共10小题；每小题1分，满分10分）51．假定英语课上老师要求同桌之间交换修改作文，请你修改你同桌写的以下作文。

湖北省枣阳市第一中学2015-2016学年度下学期高二年级3月月考数学（文科）试题本试卷两大题22个小题，满分150分，考试时间120分钟★ 祝考试顺利 ★第I 卷（选择题共60分）一、选择题（本大题12小题，每小题5分，共60分） 1．已知命题:,25xp x R ∀∈=，则p ⌝为( ) A 、,25x x R ∀∉= B 、,25xx R ∀∈≠ C 、00,25x x R ∃∈= D 、00,25x x R ∃∈≠2．“30α=o ”是 “1sin 2α=”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3．已知命题p ：200,10x R mx ∃∈+≤，命题q ：2,10.x R x mx ∀∈++>若q p ∨为假命题，则实数m 的取值范围为( )A ．22≤≤-mB ．2-≤m 或2≥mC ．2-≤mD ．2≥m 4．（5分）（2011•天津）设集合A={x ∈R|x ﹣2＞0}，B={x ∈R|x ＜0}，C={x ∈R|x （x ﹣2）＞0}，则“x∈A ∪B”是“x∈C”的（ ） A.充分而不必要条件 B.必要而不充分条件 C.充分必要条件 D.即不充分也不必要条件 5．已知椭圆的一个焦点为F(0，1)，离心率12e =，则该椭圆的标准方程为 A ．22134x y += B ．22143x y += C ．2212x y += D ．2212y x += 6．椭圆的焦距为 ( )A.10B.5C.D.7．若椭圆经过原点，且焦点分别为120103F F （，），（，）， 则该椭圆的短轴长为（ ） A 3、23、2 D 、48．椭圆216x 错误!未指定书签。

+28y 错误!未指定书签。

=1的离心率为( )(A)13错误!未指定书签。

(B) 12错误!未指定书签。

(C) 33错误!未指定书签。

(D)22错误!未指定书签。

枣阳一中2014-2015学年度高二下学期第三次月检测历史试题学校:___________姓名：___________班级：___________考号：___________一、选择题（30×2=60分）1．“人人自有定盘针，万化根源总在心。

却笑从前颠倒见，枝枝叶叶外头寻。

”这首诗反映了（）A．孟子的“仁政” B．朱熹的“格物致知”C．王阳明的“心学” D．顾炎武的“经世致用”2．朱熹是继孔孟之后最重要的儒学思想家，是宋朝理学的集大成者。

他提出“存天理，灭人欲”，其中“天理”主要是指（）A.世界万物的本原B. 封建道德规范和等级秩序C.社会发展的规律D.人的情感感受3．“国家将有失道之败，而天乃先出灾害以潜告之，不知自省，又出怪异警告之，尚不知变，而伤败乃至。

”这反映董仲舒的（）A．大一统思想 B．天人感应思想C．君权神授思想 D．无为而治思想4．和谐是中国传统文化的基本价值取向。

下列体现人与自然和谐思想的语句是（）①已所不欲，勿施于人。

②天地与我并生，而万物与我为一。

③万物并育而不相害，道并行而不相悖。

④德不忧者不能怀远，才不大者不能博见。

A.①②B. ②③C.②④D. ③④5．“无论什么人，不管他是教皇、主教、传教士或是修士、修女，世俗权力都有权来管他……”对上述材料理解正确的是A、强调世俗权力高于教权B、主张国家权力为神所授C、主张政教合一D、教皇权力高于王权6．日内瓦之所以被称为“新教的罗马”，主要是因为A．加尔文教主张比较激进B．加尔文在瑞士日内瓦进行的宗教改革影响最大，传播到欧洲许多地区C．加尔文教倡导《圣经》是最高权威D．加尔文严厉排除异己7．中世纪时天主教会成为斗争的焦点，爆发了宗教“异端”运动。

对宗教“异端”运动理解最为准确的是A、是指人们反对教会的斗争B、是指人们利用宗教思想发动的反对封建王权的斗争C、是指人们常常利用与教会说教不同的“异端”宗教思想发动民众的反封建斗争D、是指反对天主教会的反封建斗争8．下列宗教改革的内容，属于马丁•路德和加尔文所共有的是：①信仰得救②因信称义③简化仪式④制定严格的道德标准A．①②③ B．②③④ C．①③ D．②④9．明代中叶，昆曲兴起并逐渐成为戏剧主流。

湖北省襄阳市枣阳高中2014-2015学年高二（下）期末数学试卷（文科）一、选择题（本大题共10题，每题5分，共计50分）1．如图，已知双曲线C：﹣=1（a＞0，b＞0）的右顶点为A，O为坐标原点，以A为圆心的圆与双曲线C的某渐近线交于两点P、Q，若∠PAQ=60°且=3，则双曲线C的离心率为（）A．B．C．D．2．已知条件p：x2﹣4≤0，条件q：≥0，则￢p是q的（）A．充分不必要条件B．必要不充分条件C．充要条件D．既非充分也非必要条件3．曲线y=x2在点M（）的切线的倾斜角的大小是（）A．30°B．45°C．60°D．90°4．有如下四个结论：①分别在两个平面内的两条直线一定是异面直线；②过平面α的一条斜线有一个平面与平面α垂直；③“x＞0”是“x＞1”的必要条件；④命题“∀x∈R，x2﹣x+1＞0”的否定是“∀x∈R，x2﹣x+1≤0”．其中正确结论的个数为（）A．4 B．3 C．2 D．15．如果命题“￢（p∧q）”是真命题，则（）A．命题p、q均为假命题B．命题p、q均为真命题C．命题p、q中至少有一个是真命题D．命题p、q中至多有一个是真命题6．若函数y=x2+（2a﹣1）x+1在区间（﹣∞，2]上是减函数，则实数a的取值范围是（）A．[﹣，+∞）B．（﹣∞，﹣]C．[，+∞）D．（﹣∞，]7．已知p：x≥k，q：＜1，如果p是q的充分不必要条件，则实数k的取值范围是（）A．[2，+∞）B．（2，+∞）C．[1，+∞）D．（﹣∞，﹣1）8．命题“所有能被5整除的数都是偶数”否定形式是（）A．所有不能被5整除的数都是偶数B．所有能被5整除的数都不是偶数C．存在一个不能被5整除的数都是偶数D．存在一个能被5整除的数不是偶数9．以椭圆+=1的长轴端点为焦点、以椭圆焦点为顶点的双曲线方程为（）A．﹣=1 B．﹣=1 C．﹣y2=1 D．﹣y2=110．如图，设抛物线y2=4x的焦点为F，不经过焦点的直线上有三个不同的点A，B，C，其中点A，B在抛物线上，点C在y轴上，则△BCF与△ACF的面积之比是（）A．B．C．D．二、填空题（本大题共5题，每题5分，共计25分）11．“p：x∈{x|x2﹣x﹣2≥0}”，“q：x∈{x|x＜a}”，若￢p是q的充分不必要条件，则a的取值范围是．12．已知双曲线﹣=1（a＞0，b＞0）的左顶点与抛物线y2=2px（p＞0）的焦点的距离为4，且双曲线的一条渐近线与抛物线的准线的交点坐标为（﹣2，﹣1），则双曲线的焦距为．13．做一个无盖的圆柱形水桶，若要使体积是27π，且用料最省，则圆柱的底面半径为．14．过抛物线y2=4x焦点的直线l的倾斜角为，且l与抛物线相交于A、B两点，O为原点，那么△AOB的面积为．15．椭圆=1（a＞b＞0）的左、右焦点分别为F1，F2，若椭圆上存在点P使线段PF1与以椭圆短轴为直径的圆相切，切点恰为线段PF1的中点，则该椭圆的离心率为．三、解答题（75分）16．已知椭圆C：+=1（a＞b＞0）的左．右焦点为F1、F2，离心率为e．直线l：y=ex+a 与x轴．y轴分别交于点A、B，M是直线l与椭圆C的一个公共点，P是点F1关于直线l的对称点，设=λ．（Ⅰ）证明：λ=1﹣e2；（Ⅱ）确定λ的值，使得△PF1F2是等腰三角形．17．已知椭圆C：+=1（a＞b＞0）的离心率为，F是椭圆的焦点，点A（0，﹣2），直线AF的斜率为，O为坐标原点．（1）求椭圆C的方程；（2）设过点A的直线与C相交于P、Q两点，当△OPQ的面积最大时，求l的方程．18．已知F1，F2分别为椭圆的上、下焦点，F1是抛物线C1：x2=4y的焦点，点M是C1与C2在第二象限的交点，且|MF1|=（1）求椭圆C1的方程；（2）与圆x2+（y+1）2=1相切的直线l：y=k（x+t），kt≠0交椭圆C1于A，B，若椭圆C1上一点P满足+=λ，求实数λ的取值范围．19．已知椭圆C 1：+x2=1（a＞1）与抛物线C：x2=4y有相同焦点F1．（Ⅰ）求椭圆C1的标准方程；（Ⅱ）已知直线l1过椭圆C1的另一焦点F2，且与抛物线C2相切于第一象限的点A，设平行l1的直线l交椭圆C1于B，C两点，当△OBC面积最大时，求直线l的方程．20．已知F1、F2分别是椭圆+=1的左、右焦点，曲线C是坐标原点为顶点，以F2为焦点的抛物线，过点F1的直线l交曲线C于x轴上方两个不同点P、Q，点P关于x轴的对称点为M，设=（Ⅰ）若λ∈[2，4]，求直线L的斜率k的取值范围；（Ⅱ）求证：直线MQ过定点．21．已知函数f（x）=+ax，x＞1．（Ⅰ）若f（x）在（1，+∞）上单调递减，求实数a的取值范围；（Ⅱ）若a=2，求函数f（x）的极小值；（Ⅲ）若存在实数a使f（x）在区间（）（n∈N*，且n＞1）上有两个不同的极值点，求n的最小值．湖北省襄阳市枣阳高中2014-2015学年高二（下）期末数学试卷（文科）参考答案与试题解析一、选择题（本大题共10题，每题5分，共计50分）1．如图，已知双曲线C：﹣=1（a＞0，b＞0）的右顶点为A，O为坐标原点，以A为圆心的圆与双曲线C的某渐近线交于两点P、Q，若∠PAQ=60°且=3，则双曲线C的离心率为（）A．B．C．D．考点：双曲线的简单性质．专题：计算题；圆锥曲线的定义、性质与方程．分析：确定△QAP为等边三角形，设AQ=2R，则OP=R，利用勾股定理，结合余弦定理，即可得出结论．解答：解：因为∠PAQ=60°且=3，所以△QAP为等边三角形，设AQ=2R，则OP=R，渐近线方程为y=x，A（a，0），取PQ的中点M，则AM=由勾股定理可得（2R）2﹣R2=（）2，所以（ab）2=3R2（a2+b2）①在△OQA中，=，所以7R2=a2②①②结合c2=a2+b2，可得=．故选：B．点评：本题考查双曲线的性质，考查余弦定理、勾股定理，考查学生的计算能力，属于中档题．2．已知条件p：x2﹣4≤0，条件q：≥0，则￢p是q的（）A．充分不必要条件B．必要不充分条件C．充要条件D．既非充分也非必要条件考点：必要条件、充分条件与充要条件的判断．专题：简易逻辑．分析：求出满足条件￢p的x的范围，和满足条件q的x的范围，判断两个范围的包含关系，进而可用集合法判断出￢p与q的充要关系．解答：解：∵条件p：x2﹣4≤0，∴条件￢p：x2﹣4＞0，即x∈（﹣∞，﹣2）∪（2，+∞）；∵条件q：≥0，即x∈（﹣∞，﹣2]∪（2，+∞）；且（﹣∞，﹣2）∪（2，+∞）⊊（﹣∞，﹣2]∪（2，+∞）；故￢p是q的充分不必要条件，故选：A点评：判断充要条件的方法是：①若p⇒q为真命题且q⇒p为假命题，则命题p是命题q的充分不必要条件；②若p⇒q为假命题且q⇒p为真命题，则命题p是命题q的必要不充分条件；③若p⇒q为真命题且q⇒p为真命题，则命题p是命题q的充要条件；④若p⇒q为假命题且q⇒p为假命题，则命题p是命题q的即不充分也不必要条件．⑤判断命题p与命题q所表示的范围，再根据“谁大谁必要，谁小谁充分”的原则，判断命题p与命题q的关系．3．曲线y=x2在点M（）的切线的倾斜角的大小是（）A．30°B．45°C．60°D．90°考点：利用导数研究曲线上某点切线方程；直线的倾斜角．专题：计算题．分析：欲判别切线的倾斜角的大小，只须求出其斜率的值即可，故先利用导数求出在x=处的导函数值，再结合导数的几何意义即可求出切线的斜率．从而问题解决．解答：解：y'=2x∴当x=时，y'=1，得切线的斜率为1，所以k=；∴1=tanα，∴α=450，故选B．点评：本小题主要考查直线的斜率、导数的几何意义、利用导数研究曲线上某点切线方程等基础知识，考查运算求解能力．属于基础题．4．有如下四个结论：①分别在两个平面内的两条直线一定是异面直线；②过平面α的一条斜线有一个平面与平面α垂直；③“x＞0”是“x＞1”的必要条件；④命题“∀x∈R，x2﹣x+1＞0”的否定是“∀x∈R，x2﹣x+1≤0”．其中正确结论的个数为（）A．4 B．3 C．2 D．1考点：命题的真假判断与应用．专题：简易逻辑．分析：利用两个平面内的两条直线的位置关系可判断①；利用面面垂直的判定定理可判断②；利用充分条件与必要条件的概念可判断③；利用全称命题与特称命题的关系可判断④．解答：解：①分别在两个平面内的两条直线可能平行，也可能相交、异面，故①错误；②过平面α外斜线上一点P作PO⊥α，则斜线与PO确定的平面β⊥α，故过平面α的一条斜线有一个平面与平面α垂直，正确；③“x＞0”不能⇒“x＞1”，充分性不成立，反之“x＞1”⇒是“x＞0”，即必要性成立，故③正确；④命题“∀x∈R，x2﹣x+1＞0”的否定是“∃x∈R，x2﹣x+1≤0”，故④错误；综上所述，其中正确结论的个数为2个．故选：C．点评：本题考查命题的真假判断与应用，着重考查充分条件与必要条件的概念、全称命题与特称命题的关系及空间直线与平面的位置关系，属于中档题．5．如果命题“￢（p∧q）”是真命题，则（）A．命题p、q均为假命题B．命题p、q均为真命题C．命题p、q中至少有一个是真命题D．命题p、q中至多有一个是真命题考点：复合命题的真假．专题：计算题．分析：可知p∧q是假命题，由复合命题的真假可知：命题p，q中至少有一个是假命题，进而可得答案．解答：解：由题意可知：“￢（p∧q）”是真命题，∴p∧q是假命题，由复合命题的真假可知：命题p，q中至少有一个是假命题，即命题p，q中至多有一个是真命题，故选D点评：本题考查复合命题的真假，属基础题．6．若函数y=x2+（2a﹣1）x+1在区间（﹣∞，2]上是减函数，则实数a的取值范围是（）A．[﹣，+∞）B．（﹣∞，﹣]C．[，+∞）D．（﹣∞，]考点：函数单调性的性质．专题：计算题．分析：由已知中函数的解析式，结合二次函数的图象和性质，可以判断出函数y=x2+（2a﹣1）x+1图象的形状，分析区间端点与函数图象对称轴的关键，即可得到答案．解答：解：∵函数y=x2+（2a﹣1）x+1的图象是方向朝上，以直线x=为对称轴的抛物线又∵函数在区间（﹣∞，2]上是减函数，故2≤解得a≤﹣故选B．点评：本题考查的知识点是函数单调性的性质，其中熟练掌握二次函数的图象和性质是解答本题的关键．7．已知p：x≥k，q：＜1，如果p是q的充分不必要条件，则实数k的取值范围是（）A．[2，+∞）B．（2，+∞）C．[1，+∞）D．（﹣∞，﹣1）考点：必要条件、充分条件与充要条件的判断．专题：简易逻辑．分析：求出不等式q的等价条件，根据充分条件和必要条件的定义即可得到结论．解答：解：∵＜1，∴﹣1=＜0，即（x﹣2）（x+1）＞0，∴x＞2或x＜﹣1，∵p是q的充分不必要条件，∴k＞2，故选：B．点评：本题主要考查充分条件和必要条件的应用，利用不等式之间的关系是解决本题的关键，比较基础．8．命题“所有能被5整除的数都是偶数”否定形式是（）A．所有不能被5整除的数都是偶数B．所有能被5整除的数都不是偶数C．存在一个不能被5整除的数都是偶数D．存在一个能被5整除的数不是偶数考点：命题的否定；全称命题．专题：阅读型．分析：本题中所给的命题是一个全称命题，书写其否定要注意它的格式的变化，即量词的变化，写出它的否定命题，再对比四个选项得出正确选项解答：解：∵全称命题“所有被5整除的整数都是偶数”∴全称命题“所有被5整除的整数都是偶数”的否定是“存在一个被5整除的整数不是偶数”，对比四个选项知，D选项是正确的故选D点评：本题考查命题的否定，解答本题关键是正解全称命题的否定命题的书写格式，结论要否定，还要把全称量词变为存在量词．9．以椭圆+=1的长轴端点为焦点、以椭圆焦点为顶点的双曲线方程为（）A．﹣=1 B．﹣=1 C．﹣y2=1 D．﹣y2=1考点：椭圆的简单性质．专题：计算题；圆锥曲线的定义、性质与方程．分析：确定椭圆的焦点、顶点坐标，可得双曲线的顶点、焦点坐标，即可求出双曲线的方程．解答：解：椭+=1的焦点坐标为（±，0），两个顶点为（±2，0），∴双曲线的顶点为（±，0），焦点坐标为（±2，0），∴双曲线的方程为﹣=1．故选：A．点评：本题考查双曲线的方程与性质，考查学生的计算能力，比较基础．10．如图，设抛物线y2=4x的焦点为F，不经过焦点的直线上有三个不同的点A，B，C，其中点A，B在抛物线上，点C在y轴上，则△BCF与△ACF的面积之比是（）A．B．C．D．考点：直线与圆锥曲线的关系．专题：圆锥曲线的定义、性质与方程．分析：根据抛物线的定义，将三角形的面积关系转化为的关系进行求解即可．解答：解：如图所示，抛物线的准线DE的方程为x=﹣1，过A，B分别作AE⊥DE于E，交y轴于N，BD⊥DE于E，交y轴于M，由抛物线的定义知BF=BD，AF=AE，则|BM|=|BD|﹣1=|BF|﹣1，|AN|=|AE|﹣1=|AF|﹣1，则===，故选：A点评：本题主要考查三角形的面积关系，利用抛物线的定义进行转化是解决本题的关键．二、填空题（本大题共5题，每题5分，共计25分）11．“p：x∈{x|x2﹣x﹣2≥0}”，“q：x∈{x|x＜a}”，若￢p是q的充分不必要条件，则a的取值范围是a≥2．考点：必要条件、充分条件与充要条件的判断．专题：简易逻辑．分析：求出p的等价条件，利用充分不必要条件的定义建立，建立条件关系即可求实数a的取值范围．解答：解：由x2﹣x﹣2≥0得x≥2或x≤﹣1，即p：x≥2或x≤﹣1，￢p：﹣1＜x＜2．若￢p是q的充分不必要条件，则{x|﹣1＜x＜2}⊊{x|x＜a}，即a≥2，故答案为：a≥2．点评：本题主要考查充分条件和必要条件的应用，考查学生的推理能力．利用不等式的性质是解决本题的关键．12．已知双曲线﹣=1（a＞0，b＞0）的左顶点与抛物线y2=2px（p＞0）的焦点的距离为4，且双曲线的一条渐近线与抛物线的准线的交点坐标为（﹣2，﹣1），则双曲线的焦距为2．考点：双曲线的简单性质．专题：圆锥曲线的定义、性质与方程．分析：由已知方程即可得出双曲线的左顶点、一条渐近线方程与抛物线的焦点、准线的方程，再根据数量关系即可列出方程，解出即可．解答：解：∵双曲线﹣=1（a＞0，b＞0）的左顶点（﹣a，0）与抛物线y2=2px（p＞0）的焦点F的距离为4，∴；又双曲线的一条渐近线与抛物线的准线的交点坐标为（﹣2，﹣1），∴渐近线的方程应是，而抛物线的准线方程为，因此，，联立得，解得，∴=2．故双曲线的焦距为．故答案为．点评：熟练掌握圆锥曲线的图象与性质是解题的关键．13．做一个无盖的圆柱形水桶，若要使体积是27π，且用料最省，则圆柱的底面半径为3．考点：函数最值的应用．专题：应用题．分析：设圆柱的高为h，半径为r则由圆柱的体积公式可得，πr2h=27π，即，要使用料最省即求全面积的最小值，而S全面积=πr2+2πrh==（法一）令S=f（r），结合导数可判断函数f（r）的单调性，进而可求函数取得最小值时的半径（法二）：S全面积=πr2+2πrh==，利用基本不等式可求用料最小时的r解答：解：设圆柱的高为h，半径为r则由圆柱的体积公式可得，πr2h=27πS全面积=πr2+2πrh==（法一）令S=f（r），（r＞0）=令f′（r）≥0可得r≥3，令f′（r）＜0可得0＜r＜3∴f（r）在（0，3）单调递减，在[3，+∞）单调递增，则f（r）在r=3时取得最小值（法二）：S全面积=πr2+2πrh====27π当且仅当即r=3时取等号当半径为3时，S最小即用料最省故答案为：3点评：本题主要考查了圆柱的体积公式及表面积的最值的求解，解答应用试题的关键是要把实际问题转化为数学问题，根据已学知识进行解决．14．过抛物线y2=4x焦点的直线l的倾斜角为，且l与抛物线相交于A、B两点，O为原点，那么△AOB的面积为．考点：抛物线的应用．专题：计算题．分析：S△AOB=，其中d为l到AB的距离，或者把△AOB分成△OFA与OFB，设A （x1，y1），B（x2，y2），则S△AOB=OF|y1﹣y2|．解答：解：抛物线y2=4x焦点F（1，0），l的方程为y=tan（x﹣1），即y=（x﹣1），与抛物线方程y2=4x联立消去x得y2﹣y﹣4=0，得y2﹣﹣4=0，则S△AOB=S△OFA+S△OFB=OF|y1﹣y2|=OF=×1×=．故答案为：．点评：本题三角形借助于抛物线这一特殊背景出现，因此若考虑到抛物线的定义，便会得出如上的解答过程．当然用S△AOB=，其中d为l到AB的距离也完全可以．15．椭圆=1（a＞b＞0）的左、右焦点分别为F1，F2，若椭圆上存在点P使线段PF1与以椭圆短轴为直径的圆相切，切点恰为线段PF1的中点，则该椭圆的离心率为．考点：椭圆的简单性质．专题：计算题；直线与圆；圆锥曲线的定义、性质与方程．分析：设线段PF1的中点为M，另一个焦点F2，利用OM是△F1PF2的中位线，以及椭圆的定义求出直角三角形OMF1的三边之长，使用勾股定理求离心率．解答：解：设线段PF1的中点为M，另一个焦点F2，由题意知，OM=b，又OM是△F1PF2的中位线，∴OM=PF2=b，PF2=2b，由椭圆的定义知PF1=2a﹣PF2=2a﹣2b，又MF1=PF1=（2a﹣2b）=a﹣b，又OF1=c，直角三角形OMF1中，由勾股定理得：（a﹣b）2+b2=c2，又a2﹣b2=c2，可得2a=3b，故有4a2=9b2=9（a2﹣c2），由此可求得离心率e==，故答案为：．点评：本题考查椭圆的定义、方程和性质，考查直线和圆相切的条件，考查运算能力，属于中档题．三、解答题（75分）16．已知椭圆C：+=1（a＞b＞0）的左．右焦点为F1、F2，离心率为e．直线l：y=ex+a 与x轴．y轴分别交于点A、B，M是直线l与椭圆C的一个公共点，P是点F1关于直线l的对称点，设=λ．（Ⅰ）证明：λ=1﹣e2；（Ⅱ）确定λ的值，使得△PF1F2是等腰三角形．考点：直线与圆锥曲线的综合问题．专题：证明题；综合题．分析：（Ⅰ）因为A、B分别是直线l：y=ex+a与x轴、y轴的交点，所以A、B的坐标分别是（﹣，0）（0，a）．由题设知点M的坐标是（﹣c，）．由=λ得（﹣c+，）=λ（，a）．从而解得λ=1﹣e2．（Ⅱ）因为PF1⊥l，所以∠PF1F2=90°+∠BAF1为钝角，要使△PF1F2为等腰三角形，必有|PF1|=c．由题设知当λ=时，△PF1F2为等腰三角形．解答：解：（Ⅰ）因为A、B分别是直线l：y=ex+a与x轴、y轴的交点，所以A、B的坐标分别是（﹣，0）（0，a）．由得．这里c=．所以点M的坐标是（﹣c，）．由=λ得（﹣c+，）=λ（，a）．即．解得λ=1﹣e2．（Ⅱ）因为PF1⊥l，所以∠PF1F2=90°+∠BAF1为钝角，要使△PF1F2为等腰三角形，必有|PF1|=|F1F2|，即|PF1|=c．设点F1到l的距离为d，由|PF1|═d===c，得=e．所以e2=，于是λ=1﹣e2=．即当λ=时，△PF1F2为等腰三角形．点评：本题考查直线和圆锥曲线的综合问题，解题时要认真审题，仔细求解，合理地运用公式．17．已知椭圆C：+=1（a＞b＞0）的离心率为，F是椭圆的焦点，点A（0，﹣2），直线AF的斜率为，O为坐标原点．（1）求椭圆C的方程；（2）设过点A的直线与C相交于P、Q两点，当△OPQ的面积最大时，求l的方程．考点：直线与圆锥曲线的综合问题；椭圆的标准方程．专题：圆锥曲线的定义、性质与方程．分析：（1）利用椭圆的离心率以及直线的斜率，求出椭圆的几何量，然后求椭圆C的方程；（2）由设直线的斜率为k，方程为y=kx﹣2，联立直线与椭圆方程，通过△=16（4k2﹣3）＞0，求出k的范围，设P（x1，y1），Q（x2，y2），利用韦达定理，求出|PQ|，坐标原点O到直线的距离，得到S△OPQ的表达式，利用换元法以及基本不等式，通过面积的最大值，求出k的值，得到直线方程．解答：解：（1）设F（c，0），由题意k AF=，∴c=，又∵离心率=，∴a=2，∴b==1，椭圆C的方程为；﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（4分）（2）由题意知，直线的斜率存在，设直线的斜率为k，方程为y=kx﹣2，联立直线与椭圆方程：，化简得：（1+4k2）x2﹣16kx+12=0，由△=16（4k2﹣3）＞0，∴k2＞，设P（x1，y1），Q（x2，y2），则x1+x2=，x1x2=，﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（6分）∴|PQ|==，坐标原点O到直线的距离为d=，S△OPQ=••=，﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（8分）令t=（t＞0），则S△OPQ==，∵t+，当且仅当t=，即t=2时等号成立，∴S△OPQ≤1，故当t=2，即，k2=＞，∴k=±时，△OPQ的面积最大，﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（10分）此时直线的方程为：y=±x﹣2．﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（12分）点评：本题考查椭圆的方程的求法，直线与椭圆的方程的综合应用，考查分析问题解决问题的能力．18．已知F1，F2分别为椭圆的上、下焦点，F1是抛物线C1：x2=4y的焦点，点M是C1与C2在第二象限的交点，且|MF1|=（1）求椭圆C1的方程；（2）与圆x2+（y+1）2=1相切的直线l：y=k（x+t），kt≠0交椭圆C1于A，B，若椭圆C1上一点P满足+=λ，求实数λ的取值范围．考点：直线与圆锥曲线的综合问题；椭圆的简单性质．专题：直线与圆；圆锥曲线的定义、性质与方程．分析：（1）利用抛物线的方程和定义即可求出点M的坐标，再利用椭圆的定义即可求出；（2）根据直线与圆相切则圆心到直线距离等于半径，可得k=，联立直线与椭圆方程，结合椭圆上一点P满足+=λ，可得到λ2的表达式，进而求出实数λ的取值范围．解答：解：（Ⅰ）由题知F1（0，1），所以a2﹣b2=1，又由抛物线定义可知MF1=y M+1=，得y M=，于是易知M（﹣，），从而MF1==，由椭圆定义知2a=MF1+MF2=4，得a=2，故b2=3，从而椭圆的方程为+=1；（Ⅱ）设A（x1，y1），B（x2，y2），P（x0，y0），则由+=λ知，x1+x2=λx0，y1+y2=λy0，且+=1，①又直线l：y=k（x+t），kt≠0与圆x2+（y+1）2=1相切，所以有=1，由k≠0，可得k=（t≠±1，t≠0）②又联立消去y得（4+3k2）x2+6k2tx+3k2t2﹣12=0，且△＞0恒成立，且x1+x2=﹣，x1x2=，所以y1+y2=k（x1+x2）+2kt=，所以得P（，），代入①式得+=1，所以λ2=，又将②式代入得，λ2=，t≠0，t≠±1，易知（）2++1＞1，且（）2++1≠3，所以λ2∈（0，）∪（，4），所以λ的取值范围为{λ|﹣2＜λ＜2且λ≠0，且λ≠±}．点评：熟练掌握圆锥曲线的定义和性质、向量相等、直线与圆锥曲线的相交问题及根与系数的关系是解题的关键．本题需要较强的计算能力，注意分类讨论的思想方法应用．19．已知椭圆C 1：+x2=1（a＞1）与抛物线C：x2=4y有相同焦点F1．（Ⅰ）求椭圆C1的标准方程；（Ⅱ）已知直线l1过椭圆C1的另一焦点F2，且与抛物线C2相切于第一象限的点A，设平行l1的直线l交椭圆C1于B，C两点，当△OBC面积最大时，求直线l的方程．考点：直线与圆锥曲线的综合问题；椭圆的标准方程；直线与圆锥曲线的关系．专题：圆锥曲线的定义、性质与方程．分析：（Ⅰ）求出抛物线的F1（0，1），利用椭圆的离心率，求出a、b即可求解椭圆方程．（Ⅱ）F2（0，﹣1），由已知可知直线l1的斜率必存在，联立方程组，利用相切求出k，然后利用直线的平行，设直线l的方程为y=x+m联立方程组，通过弦长公式点到直线的距离求解三角形的面积，然后得到所求直线l的方程．解答：解：（Ⅰ）∵抛物线x2=4y的焦点为F1（0，1），∴c=1，又b2=1，∴∴椭圆方程为：+x2=1．…（4分）（Ⅱ）F2（0，﹣1），由已知可知直线l1的斜率必存在，设直线l1：y=kx﹣1由消去y并化简得x2﹣4kx+4=0∵直线l1与抛物线C2相切于点A．∴△=（﹣4k）2﹣4×4=0，得k=±1．…（5分）∵切点A在第一象限．∴k=1…（6分）∵l∥l1∴设直线l的方程为y=x+m由，消去y整理得3x2+2mx+m2﹣2=0，…（7分）△=（2m）2﹣12（m2﹣2）＞0，解得．设B（x1，y1），C（x2，y2），则，．…（8分）又直线l交y轴于D（0，m）∴…（10分）=当，即时，．…（11分）所以，所求直线l的方程为．…（12分）点评：本题主要考查椭圆、抛物线的有关计算、性质，考查直线与圆锥曲线的位置关系，考查运算求解能力及数形结合和化归与转化思想．20．已知F1、F2分别是椭圆+=1的左、右焦点，曲线C是坐标原点为顶点，以F2为焦点的抛物线，过点F1的直线l交曲线C于x轴上方两个不同点P、Q，点P关于x轴的对称点为M，设=（Ⅰ）若λ∈[2，4]，求直线L的斜率k的取值范围；（Ⅱ）求证：直线MQ过定点．考点：三点共线；圆锥曲线的综合．专题：计算题．分析：（I）求出曲线C的方程，把PQ的方程x=my﹣1 （m＞0）代入曲线C的方程化简可得y2﹣4my+4=0，利用根与系数的关系及=，可得=λ++2=4m2，据λ∈[2，4]，求得直线L的斜率的范围．（II）根据﹣=0，可得M、Q、F 2三点共线，故直线MQ过定点F2 （1，0 ）．解答：解：（I）令P（x1，y1），Q（x2，y2），由题意，可设抛物线方程为y2=2px由椭圆的方程可得F1（﹣1，0），F2 （1，0 ）故p=2，曲线C的方程为y2=4x，由题意，可设PQ的方程x=my﹣1 （m＞0）．把PQ的方程代入曲线C的方程化简可得y2﹣4my+4=0，∴y1+y2=4m，y1y2=4．又=，∴x1+1=λ（x2+1），y1=λy2，又=λ++2=4m2．λ∈[2，4]，∴2+≤λ+≤4+，≤m2≤，∴≤≤∴直线L的斜率k的取值范围为[，]．（II）由于P，M关于X轴对称，故M（x1，﹣y1），∵﹣=+==0，∴M、Q、F2三点共线，故直线MQ过定点F2 （1，0 ）．点评：本题考查椭圆、抛物线的标准方程、简单性质，三点共线的条件，根据题意，得到2+≤λ+≤4+，是解题的关键．21．已知函数f（x）=+ax，x＞1．（Ⅰ）若f（x）在（1，+∞）上单调递减，求实数a的取值范围；（Ⅱ）若a=2，求函数f（x）的极小值；（Ⅲ）若存在实数a使f（x）在区间（）（n∈N*，且n＞1）上有两个不同的极值点，求n的最小值．考点：利用导数研究函数的极值；利用导数研究函数的单调性．专题：导数的综合应用．分析：（Ⅰ）求出函数的导数，利用f′（x）≤0在x∈（1，+∞）上恒成立，得到a的表达式，利用函数的最小值求出a的范围．（Ⅱ）通过a=2，化简函数的解析式，求出函数的导数，利用导数的符号，判断函数的单调性，求出极小值．（Ⅲ）判断aln2x+lnx﹣1=0在上有两个不等实根，法一：构造函数，推出，求出n的最小值．法二：利用，推出a的表达式，列出然后求解n的最小值．解答：（本小题满分13分）解：（Ⅰ），由题意可得f′（x）≤0在x∈（1，+∞）上恒成立；﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（1分）∴，﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（2分）∵x∈（1，+∞），∴lnx∈（0，+∞），﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（3分）∴时函数t=的最小值为，∴﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（4分）（Ⅱ）当a=2时，﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（5分）令f′（x）=0得2ln2x+lnx﹣1=0，解得或lnx=﹣1（舍），即﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（7分）当时，f′（x）＜0，当时，f′（x）＞0∴f（x）的极小值为﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（8分）（Ⅲ）原题等价于f′（x）=0在，且n＞1）上有两个不等的实数根；由题意可知﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（9分）即aln2x+lnx﹣1=0在上有两个不等实根．﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（10分）法一：令，g（u）=au2+u﹣1∵g（0）=﹣1＜0，根据图象可知：，整理得﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（11分）即，解得n＞2，∴n的最小值为3．﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（13分）法二：令，﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（11分）由题意可知解得解得n＞2，∴n的最小值为3．﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣（13分）点评：本题考查函数的单调性以及函数的极值，构造法的应用，考查转化思想以及计算能力．。

枣阳一中2014-2015学年度高二下学期第三次月检测语文试题学校:___________姓名：___________班级：___________考号：___________一、选择题（共5小题，每小题3分，共15分）1．下列加点字的注音完全正确的一项是（）A.帷．幕（wéi）屋脊．（jī）翘．起（qiào）翘．首以盼（qiáo）B.埋．怨（mán）接榫．（shǔn）厦．门（xià）高楼大厦．（xià）C.点缀．（zhuì）辍．学（chuò）角．色（jiǎo）凤毛麟角．（jiǎo）D.穹．窿（qióng）哺．育（bǔ）门槛．（kǎn）直栏横槛．（jiàn）2．下列词语书写全都正确的一项是（）A.羡妒斟酌婉蜓曲折卓有成效B.援例尺牍光采夺目顽固凶残C.叫嚣蕴藉长途拔涉杀一敬百D.游离门栓矫揉造作残羹冷炙3．依次填入下列各句横线中的词语，恰当．．的一项是（）（3分）①群众有等候____公开行刑的习惯，所以并没有表现出十分不耐烦的样子。

②当人们能够从各方面看见他被人用绳子和皮条绑在刑台的轮盘上的时候，场内_____了一阵笑声和喊声，人们认出他就是伽西莫多。

③人们看见他眼睛冒火，筋脉鼓起，四肢____，一下子就把皮条和绳子都挣开了。

④他打破了一直固执地保持着的____，用又嘶哑又愤怒的声音吼叫。

A、观赏爆发蜷曲缄默B、观赏暴发蜷缩缄默C、观看暴发蜷缩沉默D、观看爆发蜷曲沉默4．下列句子中，没有．．语病的一项是（）（3分）A．当前各级政府要完善公务消费、预算编制和审计、绩效考核以及监督问责等全方位的制度建设，切实遏制公款消费的各种违纪违规现象。

B．在过去的一个星期里，大家对教研室赵主任起草的教学大纲从多角度提出质疑，经过几轮认真的讨论和修改，最终达成共识。

C．国人热议的“例外”服饰在借“第一夫人效应”拓展市场后，国货品牌自身应向高品质方向发展，其当务之急是树立自身形象。

枣阳一中2014-2015学年度高二下学期第三次月检测英语试题学校:___________姓名：___________班级：___________考号：___________一、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秒钟的时间回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. How does the man come here?A. By bus.B. By taxi.C. By car.2. Why isn’t Helen present?A. She forgot to come.B. She changed her decision.C. She wasn’t invited.3. What’s the probable relationship between the two speakers?A. Husband and wife.B. Mother and son.C. Doctor and patient.4. What’s the man’s job?A. A shop assistant.B. A tailor.C. A salesman.5. What does the man mean?A. He can’t go to the cinema.B. He can go to the cinema on Saturday morning.C. He can go to the cinema on Saturday evening.第二节（共15小题；每小题1.5分,满分22.5分）请听下面5段对话或独白,选出最佳选项。

请听第6段材料,回答第6、7题。

枣阳二中2014—2015学年度下学期高二期中考试 化学 命题人 李长江 评卷人 得分 一、选择题（本大题共18题，每题3分，共54分。

在每题给出的四个选项中只有一个是符合题目要求的）1．下列有关晶体的说法正确的是A．具有规则几何外形的固体均为晶体 B．晶体具有自范性，非晶体没有自范性C．没有固定熔点 D．将玻璃加工成规则的固体即变成晶体2．下列各组元素中，属于同一周期的是A．碳、氧、硫 B．硅、磷、氯 C．锂、钠、钾 D．氟、氯、溴3．X、Y、Z、R是1～18号元素中的四种元素，它们的原子结构有以下特点：①元素X原子的M层比L层少3个电子；②元素Y的2价阴离子的核外电子排布与氖原子相同；③元素Z原子的L层比M层多5个电子；④元素R原子的L层比K层多3个电子，其中金属性最强的是A．X B．Y C．Z D．R4．在离子 RO3n- 中, 共有 x 个核外电子,原子的质数为, 则 R 原子核内含有的子数目是A. Ax+n+48 B. Ax+n+24 C. x-n24 D. X+n24 5．应用元素周期律分析下列推断，其中正确的组合是①第三周期金属元素最高价氧化物对应的水化物，其碱性随原子序数的增大而减弱②砹（At）是第ⅦA族元素，其氢化物的稳定性大于HCl③第二周期非金属元素的气态氢化物溶于水后，水溶液均为酸性④铊（Tl）与铝同主族，其单质既能与盐酸反应，又能与氢氧化钠溶液反应A．①②④ B．③④C．①③④ D．①6．下列有关比较中正确的是A．酸性： H3PO4 ＞ HNO3 B．稳定性： HBr ＞ HCl C．金属性：Na ＞ K D．非金属性： Cl ＞ S 7．A、B、C为短周期元素，在周期表中如图所示：A、C两种元素的原子核外电子数之和等于B原子的核电荷数。

有关说法错误的是　A．A的氧化物有多种形式B．B最高价氧化物对应的水化物是一种弱酸C．C的氢化物比HCl稳定D．D元素形成的单质不止一种A.在常温下，Na2SiO3+H2O+CO2=Na2CO3+H2SiO3↓，说明H2CO3酸性强于H2SiO3；在高温下，Na2CO3+SiO2 Na2SiO3+CO2↑，说明H2SiO3酸性强于H2CO3 B.c（NH4+）相等的（NH4）2SO4、（NH4）2CO3、（NH4）2Fe（SO4）2溶液，则 c[（NH4）2SO4]＞c[（NH4）2CO3] ＞c[（NH4）2Fe（SO4）2] C.向1L0.3mol/L的NaOH溶液中缓慢通入CO2气体至溶液增重8.8g时，则溶液中 2c（Na+）=3[c（HCO3-）+c（CO32-）+c（H2CO3）] D.在海带中提取碘的实验中既可以用四氯化碳作萃取剂也可以用酒精作萃取剂 9．下列叙述正确的是（ ） A．13C和14C属于同一种元素，它们互为同位素? B．2H和3H是不同的核素，它们的质子数不相等? C．14C和14N的质量数相等，它们的中子数相等 D．6Li和7Li的电子数相等，中子数相等 10．元素周期律和元素周期表是学习化学的重要工具，下列说法中，不正确的是A．HF、HCI、HBr、HI的还原性依次增强，热稳定性依次减弱B．P、S、CI得电子能力和最高价氧化物对应水化物的酸性均依次增强C．IA族的金属单质与水反应比同周期IIA族的金属单质剧烈D．除稀有气体外，第三周期元素的原子半径和离子半径随原子序数的增加而减小11．短周期元素X、Y、Z在周期表中的位置如下图所示，则下列说法中错误的是 A．X、Y、Z 形成的单质中，X的单质最稳定B．Y所在周期中所有主族元素，Y的原子半径最小C．Z所在族中元素的最高正价均为+6价D．能发生 Cl2+H2Z=Z+2HCl 的反应A. Li、Be、BB. Li,Na,K的金属性依次增强 C. P、S、C1 D. N、0、F下列说法中正确的是 （ ）A．NH4+和H3O+两个离子的电子数、质子数中子数相等B．1H2和2H2互为同位素，它们的质子数相等C．14C和14N的质量数相等，它们是同素异形体D．14C16O和12C18O两个分子所含质子数、中子数和原子个数均相等15． D、广泛使用高效节能的荧光电灯可以减少二氧化碳的排放量 16．下列叙述中不正确的是A．任一主族元素均由非金属元素和金属元素组成B．第n周期第n主族的元素可能是金属元素，也可能是非金属元素C．第n周期有(8-n)种非金属元素(1 HBrO4 > HIO4B．碱性： NaOH > Mg(OH)2 >Al(OH)3C．稳定性： PH3 > H2S > HClD．非金属性：F > O > S18．现有短周期元素X、Y、Z、M，X、Y位于同主族，Z、M位于同主族，Y、Z位于同周期，X与Z、M都不在同一周期，Z的核电荷数是M的2倍。

枣阳一中2014-2015学年度高二下学期第三次月检测生物试题学校:___________姓名：___________班级：___________考号：___________一、选择题（30×2=60分）1．下列有关监测和预防遗传病的做法，正确的是( )A.为了倡导婚姻自由，允许近亲结婚B.为了保护个人隐私，严禁基因诊断C.为了减少遗传病的发生，尽量减少运动D.为了优生优育，进行遗传咨询和产前诊断2．我国婚姻法规定禁止近亲结婚的理论依据是（）A．近亲结婚后代必患遗传病B．近亲结婚不符合我国某些地区的风俗习惯C．近亲结婚违反社会伦理道德D．近亲结婚会使后代患遗传病的机会增加3．蜜蜂的雄蜂是由未受精的卵细胞发育而成的，雌蜂是受精卵发育成的。

蜜蜂的体色，褐色相对于黑色为显性。

现有褐色雄蜂与黑色蜂王杂交，则F2的体色将是A．全部是褐色 B．蜂王和工蜂是褐色，雄蜂都是黑色C．褐色︰黑色＝1︰1 D．蜂王和工蜂是黑色，雄蜂都是褐色4．有翅昆虫有时会出现残翅和无翅的突变类型，这类昆虫在正常情况下很难生存下来，但在经常遭到暴风雨袭击的岛屿，这种突变类型因不能飞行，而未被大风吹到海里淹死，这一事实说明（）A．突变多数是有害的 B．突变多数是有利的C．突变的有利和有害并不是绝对的 D．突变的有利和有害是绝对的5．下图所示为某家族的遗传系谱图，下列分析正确的是A．由1号、2号、4号和5号，可推知此病为显性遗传病B．由5号、6号与9号，可推知致病基因为显性基因C．由图可以推知2号为显性纯合子、5号为杂合子D．7号和8号婚配，后代患病概率为1/36．如图为利用纯合高秆(D)抗病(E)水稻和纯合矮秆(d)易染病(e)水稻快速培育优良品种纯合矮秆抗病水稻(ddEE)的示意图，下列有关叙述正确的是( )。

A．①过程的主要目的是让控制不同优良性状的基因组合到一起B．②过程中非同源染色体的自由组合发生在减数第一次分裂前期C．③过程获得的植株为单倍体，单倍体就是指体细胞中含有一个染色体组的个体D．获得无子番茄所用原理与④过程相同7．—个处在有丝分裂后期的洋葱根尖细胞内有32条染色体，则洋葱体细胞中来自父方的染色体数和同源染色体的对数分别是()A．16和16B．32和16 C．8和16D．8和8 8．下列细胞中只能有一个染色体组的是（）A．人的精原细胞B．人的初级精母细胞C．人的次级精母细胞D．人的精细胞9．染色体组型是描述一个生物体内所有染色体的大小、形状和数量的图像。

枣阳一中2014-2015学年度高二下学期第三次月检测化学试题学校:___________姓名：___________班级：___________考号：___________一、选择题（30×2=60分）1．下列操作能达实验目的且符合实验室安全措施的是A．稀释浓硫酸，应将水慢慢注入盛有浓硫酸的烧杯中并搅拌B．酒精灯洒出的酒精万一在桌上着火，应立即用湿抹布扑盖C．碱溶液沾到皮肤上，应尽快用抹布擦拭干净，再涂上稀盐酸D．将金属钠在研钵中研成粉末，使钠与水反应的实验更安全2．石英沙（主要成分SiO2）是一种常见工业原料，其中常混有少量碳酸钙。

要除去石英沙中的少量碳酸钙，应采用的方法是A．用稀硫酸溶解后过滤分离 B．用酒精灯加热C．用水溶解后过滤分离 D．用稀盐酸溶解后过滤分离3．下列关于分散系的说法不正确的是A．石灰乳不是分散系B．分散剂一定是液态C．一种分散系里只有分散质和分散剂两种物质D．Fe(OH)3胶体是一种液态分散系4．下列物质混合发生化学反应，且反应属于离子反应的是：A．H2和O2反应生成水B．锌片投入稀硫酸中C．KClO3(固体)和MnO2(固体)混合加热制O2D．NaOH溶液和K2SO4溶液混合5．某些补铁剂的成分中含有硫酸亚铁，长期放置会因氧化而变质。

检验硫酸亚铁是否变质的试剂是A．稀盐酸 B．KSCN溶液 C．氯水 D．铁粉6．N A代表阿伏加德罗常数的数值。

下列说法中正确的是A．在1L0.1mol／L碳酸钠溶液中，阴离子总数大于0.1 N AB．1molCaC2中含有阴离子数目为2N AC．120gNaHSO4在水溶液和熔融状态下电离出来的离子总数均为3N AD．2mol SO2和1 mol O2混合气体在V2O5存在的条件下于密闭容器中加热反应后容器内物质分子数小于2N A7．月壤中的3He是核聚变发电最清洁最安全的理想物质，地球上氦元素主要以4He形式存在，3He仅有15 t左右；而月球上的3He有数百万吨之多，可供全世界开采500年。

..枣阳一中2014-2015学年度高二下学期第三次月检测理科数学试题学校:___________姓名：___________班级：___________考号：___________一、选择题（10小题，每小题5分，共50分） 1．三角形的面积为()c b a r c b a S ,,,21⋅++=为三角形的边长，r 为三角形内切圆的半径，利用类比推理，可得出四面体的体积为（ ）A.abc V 31=B.Sh V 31=C.()r S S S S V 432131+++=（4321,,,S S S S 分别为四面体的四个面的面积，r 为四面体内切球的半径）D.)(,)(31为四面体的高h h ac bc ab V ++=2．已知函数()()x f x y x R e=∈满足'()()f x f x >，则(1)f 与(0)ef 大小关系是（ ）A 、(1)(0)f ef <B 、(1)(0)f ef >C 、(1)(0)f ef =D 、不能确定 3．下列几种推理中是演绎推理的序号为（ ）A 、由2022<，2132<， 2242<猜想21)1(2+<-n n （+∈N n ）B 、半径为r 的圆的面积2r s π=，单位圆的面积π=sC 、猜想数列211⨯、321⨯、 431⨯的通项为)1(1+=n n a n （+∈N n ）D 、由平面直角坐标系中，圆的方程为222)()(r b y a x =-+-推测空间直角坐标系中球的方程为2222)()()(r c z b y a x =-+-+-4．已知抛物线x y 42=的焦点F ，该抛物线上的一点A 到y 轴的距离为3，则=AF A ．4 B ．5 C ．6 D ．75．从6,5,4,3,2,1这六个数字中,任取三个组成无重复数字的三位数,但当三个数字中有2 和3时,2必须排在3前面(不一定相邻),这样的三位数有A.108个B.102个C.98个D.96个6．抛物线28x y =-的准线方程是 （ ）117．已知复数z 满足(33)3i z i +=，则z 为A ．3322i －B. 3344i －C. 3322i ＋D. 3344i ＋ 8．抛物线2y ax =的准线方程是2y =，则a 的值为 （ ）A ．18 B ．18- C ．8 D ．-89．已知双曲线22221(0,0)x y a b a b-=>>的一条渐近线与直线310x y ++=垂直，则双曲线的离心率等于 62310310．已知点A 为双曲线x 2y 2=1的左顶点，点B 和点C 在双曲线的右分支上，△ABC 是等边三角形，则△ABC 的面积是 （ ） (A)33 (B) 233 (C) 33 (D) 63二、填空题（5小题，每小题5分，共25分）11．现有7件互不相同的产品，其中有4件次品，3件正品，每次从中任取一件测试，直到4件次品全被测出为止，则第三件次品恰好在第4次被测出的所有检测方法有_____种.12．若椭圆221x my +=的离心率为32，则它的长半轴长为_______________ 13．已知命题a x x p 3|1||1:|≥++-恒成立，命题xa y q )12(:-=为减函数，若“q p ∧”为真命题，则a 的取值范围是 . 14．三、解答题（本大题共有3个小题，共40分。