四川省达州市大竹县文星中学2016届高三12月月考数学试题
四川省达州市大竹县文星中学高三物理12月月考试题
四川省大竹县文星中学2016届高三12月月考物理试题一、单选题1.为了验证拉住月球使它围绕地球运动的力与拉着苹果下落的力以及地球、众行星与太阳之间的作用力是同一性质的力,同样遵从平方反比定律,牛顿进行了著名的“月地检验”。
已知月地之间的距离为60R(R 为地球半径),月球围绕地球公转的周期为T,引力常量为G.则下列说法中正确的是A.物体在月球轨道上受到的地球引力是其在地面附近受到的地球引力的B.由题中信息可以计算出地球的密度为C.物体在月球轨道上绕地球公转的向心加速度是其在地面附近自由下落时的加速度的D.由题中信息可以计算出月球绕地球公转的线速度为2.如图所示,一水平的浅色长传送带上放置一质量为m的煤块(可视为质点),煤块与传送带之间的动摩擦因数为μ。
初始时,传送带与煤块都是静止的。
现让传送带以恒定的加速度a开始运动,当其速度达到v后,便以此速度做匀速运动。
经过一段时间,煤块在传送带上留下了一段黑色痕迹后,煤块相对于传送带不再滑动,关于上述过程,以下判断正确的是(重力加速度为g)A.μ与a之间一定满足关系B.黑色痕迹的长度为C.煤块从开始运动到相对于传送带静止经历的时间为D.煤块与传送带由于摩擦而产生的热量为3.如图所示为四分之一圆柱体OAB的竖直截面,半径为R,在B点上方的C点水平抛出一个小球,小球轨迹恰好在D点与圆柱体相切,OD与O B的夹角为60°,则C点到B点的距离为A. B. C. D.R4.有a、b、c、d四颗地球卫星,a还未发射,在赤道表面上随地球一起转动,b是近地轨道卫星,c是地球同步卫星,d是高空探测卫星,它们均做匀速圆周运动,各卫星排列位置如图所示,则A.a 的向心加速度等于重力加速度gB.在相同时间内b 转过的弧长最长C.c 在2小时内转过的圆心角是D.d 的运动周期有可能是20小时5.如图所示,两根材料相同的均匀导体柱m 和n ,m 长为l ,n 长为2l ,串联在电路中时,沿x 轴方向电势变化如图像φ–x 所示,选取x =3l 处电势为零,则导体柱m 、n 的横截面积之比为A.31B.21C.41D.52 6.如图所示为一种获得高能粒子的装置,环形区域内存在垂直纸面向外、大小可调节的匀强磁场,质量为m 、电荷量为+q 的粒子在环中做半径为R 的圆周运动,A 、B 为两块中心开有小孔的极板,原来电势都为零,每当粒子顺时针飞经A 板时,A 板电势升高为φ,B 板电势仍保持为零,粒子在两板间电场中得到加速,每当粒子离开B 板时,A 板电势又降为零,粒子在电场中一次次加速下动能不断增大,而绕行半径不变A.粒子从A 板小孔处由静止开始在电场作用下加速,绕n 圈后回到A 板时获得的总动能为2nq φB.在粒子绕行的整个过程中,A 板电势可以始终保持为φC.在粒子绕行的整个过程中,每一圈的周期不变D.为使粒子始终保持在半径为R的圆轨道上运动,磁场必须周期性递增,则粒子绕行第n圈时的磁感应强度为7.如图所示,有5000个质量均为m的小球,将它们用长度相等的轻绳依次连接,再将其左端用细绳固定在天花板上,右端施加一水平力使全部小球静止。
四川省达州市大竹县文星中学2015~2016学年度高一数学12月月考试题
四川省大竹县文星中学2015-2016学年高一12月月考数学试题一、单选题1.已知集合,集合,则A. B. C. D.2.函数的定义域为A. B. C. D.3.已知函数,其定义域是,则下列说法正确的是A.有最大值,无最小值B.有最大值,最小值C.有最大值,无最小值D.有最大值,最小值4.设,则二次函数的图象可能是A. B.C. D.5.已知函数为偶函数,那么在上是A.单调递增函数B.单调递减函数C.先减后增函数D.先增后减函数6.偶函数在区间[0,4]上单调递减,则有A. B.C. D.7.若,,则A. B. C. D.8.已知函数和均为奇函数,在区间上有最大值5,那么在上的最小值为A.-5B.-9C.-7D.-19.下列哪组中的函数与是同一函数A.B.C.D.10.若,则不等式的解集是A. B. C. D.11.函数的大致图象是12.已知函数与函数的图象关于直线对称,函数的图象与的图象关于轴对称,若,则实数的值为A. B. C. D.二、填空题13.已知定义在R 上的奇函数,当0x >时,,那么0x <时,.14.设为常数且,是定义在上的奇函数,当时,,若对一切都成立,则的取值范围为_____________________.15.已知奇函数在区间上是单调递增函数,且在区间上的最大值为8,最小值为,则16.设全集集合则___________.三、解答题17.计算下列各式的值: (1);(2).18.已知函数.(1)判断的奇偶性,并证明;(2)求使的的取值范围.19.某省两相近重要城市之间人员交流频繁,为了缓解交通压力,特修一条专用铁路,用一列火车作为交通车,已知该车每次拖4节车厢,一天能往返16次,如果每次拖7节车厢,则每天能往返10次.(注明:往、返各算一次)(1)若每天往返的次数是车头每次拖挂车厢节数的一次函数,求此一次函数解析式;(2)在(1)的条件下,每节车厢能载乘客110人,问这列火车每天往返多少次才能使运营人数最多?并求出每天最多运营人数.20.设函数满足:①对任意实数都有;②对任意,都有恒成立;③不恒为0,且当时,.(1)求,的值;(2)判断函数的奇偶性,并给出你的证明;(3)定义: “若存在非零常数T,使得对函数定义域中的任意一个,均有,则称为以T为周期的周期函数”.试证明:函数为周期函数,并求出的值.21.已知.(1)当时,求;(2)若,求实数的取值范围.22.已知函数.(1)若为奇函数,求的值;(2)若在内有意义,求的取值范围;(3)在(2)的条件下,判断并证明的单调性.参考答案1-5 BBADA 6-10 ADBCA 11-12 BC 13.21x x -++ 14.15.-15 16.17.(1)原式314242314113[()]1(3)813.2(2)22-=-+-=++-=- (2)原式239log (28)323 1.32=⨯⨯-=-=- 18.(1)由,得.故的定义域为.∵∴是奇函数.(2)当时,由,得,所以,当时,由,得,所以故当时,的取值范围是;当时,的取值范围是.19.(1)设每天往返y 次,每次挂x 节车厢,由题意y =kx +b ,当x =4时,y =16,当x =7时,y =10,得到16=4k +b ,10=7k +b . 解得:k =-2,b =24,∴y =-2x +24(2)设每天往返y 次,每次挂x 节车厢,由题意知,每天挂车厢最多时,运营人数最多,设每天运营S 节车厢,则S =xy =x (-2x +24)=-2x 2+24x =-2(x -6)2+72,所以当x =6时,S ma x =72,此时y =12,则每日最多运营人数为110×72=7920(人).答:这列火车每天往返12次,才能使运营人数最多,每天最多运营人数为7920人.20.解:(1)由于不恒为0,故存在,使,令,则,所以,令,由并令得:,结合以上结果可得又令 (因为)所以,,故;(2)令,得:,以及有即有,即有为偶函数;(3)由并取得,又为偶函数, 则,即是以2为周期的周期函数;令,再令.而,解得,,由得,,所以又由于是以2为周期的周期函数,21(1)当时,; 因为,所以.(2)因为所以因为, 所以或.当时,,所以; 当时,或; 所以综上,或.22.(1)∵为奇函数,∴,∴a=1.(2)若在内恒有意义,则在上恒成立;而x+1>0,∴>0,∴在上恒成立,∴a>5;(3)当a>5时,在定义域上为减函数;由得定义域为(﹣1,a).令﹣1<<<a,由于﹣=,∵﹣1<<<a,∴a﹣>a﹣>0,1+>1+>0,∴, 即,即;所以﹣>0,即;∴在(﹣1,a)为减函数.。
四川省大竹县文星中学2016届高三历史12月月考试卷(含解析)
四川省大竹县文星中学2016届高三12月月考历史试卷注意事项:1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上第I卷(选择题)一、单选题:共12题每题4分共48分1.《史记》载:“德厚者位尊,禄重者宠荣……朱弦洞越,大羹玄酒,所以防其淫侈,救其雕敝。
是以君臣朝廷尊卑贵贱之序,下黎庶车舆衣服宫室饮食嫁娶丧祭之分,事有宜适,物有节文。
”材料现象反映出A.重义轻利的义利观念B.节用尚贤的墨家思想C.贵贱有序的礼乐制度D.尊卑有别的宗法观念【答案】C【解析】本题主要考查古代中国的政治制度。
重义轻利的义利观念材料无从体现,A项错误,故排除;由材料“是以君臣朝廷尊卑贵贱之序,下黎庶车舆衣服宫室饮食嫁娶丧祭之分,事有宜适,物有节文。
”可知材料强调等级秩序下的合理行事方式,与节俭没有关联,故B项错误;根据材料“是以君臣朝廷尊卑贵贱之序,下黎庶车舆衣服宫室饮食嫁娶丧祭之分,事有宜适,物有节文。
”森严的等级制度下强调社会各阶层必须遵守贵贱有序的礼乐制度,故C项正确;尊卑有别的宗法观念不能反映材料的本质,故D项错误。
2.自汉武帝以来,汉代政坛曾出现“世为边郡守”、“世为二千石”的现象,这主要是因为A.中央集权的加强B.君主专制的减弱C.儒学地位的提高D.察举制促使世家大族开始出现【答案】C【解析】本题主要考查儒家成为正统思想。
题干信息只是强调累世公卿现象,无法体现中央集权的加强,故A项错误;汉武帝倡导儒学是为了巩固专制主义中央集权,君主专制并没有减弱,故B项错误;结合所学知识自汉武帝尊崇儒学,特别是元帝以后,经学兴盛,著名的儒者以传经为业,通过经学入仕,形成了一些累世公卿的家族,故C项正确;D项中世家大族开始出现说法错误,故排除。
3.钱穆指出,起于元代的行省制度与中央集权不同,“实近似于一种变相的封建,乃是一种分权统御制也”,“此种制度在平时足以障碍地方政事之推进,而增加地方与中央之隔阂;而待一旦中央政权削弱,各行省转易成为反抗中央,分区割据之凭藉。
四川省达州市大竹县文星中学2016届高三12月月考英语试题
四川省大竹县文星中学2016届高三12月月考英语试题第I卷(选择题)一、单项选择1.It is generally believed that communication skills are becoming ______ it takes to be a good doctor.A.whetherB.thatC.howD.What2.He stood at the window, thinking where he his camera last.A.sawB.has seenC.had seenD.would see3.Mr. Black, as well as the professor who________ from Beijing University, ________ to attend our school meeting.e; ises; arees; ise; are4.The disease is so terrible that it made Jack’s life difficult, one problem leading to .A.the otherB.othersC.anotherD.the others5.I feel that one of my main duties a teacher is to help the students to become better learners.A.forB.likeC.asD.with6.A new study suggests that yelling at children may have consequences that go beyond of beating them.A.onesB.theseC.thoseD.that7.Take care! Accidents _________ happen along this part of the road.A.willB.wouldC.mustD.Shall8.The world has lost one of its most respected statesmen (政治家) —Nelson Man dela, ______once said, “I learned that courage was not the _______ of fear, but the triumph over it.”A.who; absenceB.whom; presenceC.that; absenceD.whom; present9.It makes no difference you connect to the company server from the next room or the next country. A.that B.which C.when D.whether10.The electronic red-packet has been so ______ that AliPay and Tencent Wechat compete against each other openly and secretly since the beginning of this year.A.popularB.convenientC.favorableD.arbitrary11.The manager’s confident words have our doubts about how the plan will be carried out.A.gotten upB.taken upC.cleared upD.given up12.In an economy income per head used to rise by barely 1% a year, current growth rates feel like a miracle.A.whenB.whoseC.thatD.where13.— Oh, my God,I left my portable computer in the taxi!—______. Let’s call the taxi company first.A.Pray for itB.Cheer upC.Forget itD.Calm down14.As a famous public figure, you have a duty to _______ yourself, especially in public places.A.focusB.behaveC.guideD.operate15."Talking with others in the real world always me embarrassed and my heart beats quickly. I never dare to look in anyone’s eyes when ," the 22-year-old said, describing her anxiety.A.make; speakingB.makes; speakingC.makes; speakD.make; speak二、完形填空Imagine waking up in a hospital bed. And the entire left side of your body isn’t movable. This was the 16I found myself in after my crash. I 17so severely for my head and neck were twisted to one side and stuck in that disturbing 18.The prognosis(预断)for my 19was not good. The doctor told my family,“I 20to say this, but he’ll be lucky to survive the next 48 hours. ”21as my family was at the news, the one thing that had been holding them together was the 22that with the proper medical treatment, I could recover. The doctor’s prognosis hit them like a hammer, 23any hope. The 24that I might die touched each of them deeply.For the next two days, my parents kept waiting outside my room. They could 25eat or sleep. The doctor’s prognosis 26heavily on their hearts. Yet with each passing hour they become slightly more hopeful that my chances of 27were a little bit better.With the 2848 hours passing, although I was once 29to the limit of my life, I 30 to quit my life, so they felt somewhat 31. Maybe the doctor had made a(n) 32. After all, doctors aren’t necessarily right. Bit by bit, hope began to return to them. 33, they still had no idea what the future held for me.My survival surprised everyone. It would not have been 34if not for my determination not to quit my life. 35will always happen around you as long as you don’t give up.16.A.despair B.situation C.place D.pain17.A.choked B.trembled C.enjoyed D.suffered18.A.position B.. action C.behavior D.height19.A.spirit B.disability C.recovery D.treatment20.A.regret B.decide C.tend D.wish21.A.Impatient B.Angry C.Confused D.Upset22.A.desire B.intention C.hope D.lie23.A.offering B.destroying C.raising D.abandoning24.A.exception B.choice C.fact D.acceptance25.A.barely B.gradually C.hopelessly D.slightly26.A.reacted B.weighed C.froze D.expanded27.A.promotion B.escape C.improvementD.survival28.A.convincing B.disappointing C.exciting D.frightening29.A.directed B.pushed C.ordered D.guided30.A.refused B.pretended C.struggled D.attempted31.A.confident B.satisfied C.relieved D.stressful32.A.joke B.guess C.apology D.mistake33.A.However B.Therefore C.Instead D.Besides34.A.alarming B.possible C.reasonable D.pessimistic35.A.Accidents B.Coincidences C.Miracles D.Successes三、阅读理解:共20题每题2分共40分Anyone can try to lead a group, but not every individual is cut out for leadership. The better leaders possess a few qualities that can mean the difference between the success and failure of the group. These are the qualities the leader of higher rank will look for when choosing a leader for a group, or when evaluating the performance of a leader. They're also the qualities team members want in a group leader, and appreciate when they find them. Take ResponsibilityGroup leaders might share tasks around a group as necessary, but eventually a group leader needs to be able to accept that responsibility lies on her shoulders.That means that if things go wrong in a group project, she's the one who must accept the consequences and work out what mistakes were made. The group leader won't always have the power to control everything group members do, but she should be ready to admit any mistakes the group has made as a result of her leadership.Concern for MembersThe group leader has a commitment to the task or project at hand, but perhaps more importantly, he has a real concern for each and every person who is part of his group. This means getting to know the strengths, weaknesses and goals of team members, as well as making time to build the group through collective activities. The group leader should make sure that everyone is included, even if an individual is new to a group.Good ListenerThe group leader needs to be able to listen to the suggestions, complaints (抱怨) and ideas of group members. Not only will this allow complaints to be addressed and potentially suitable ideas to be put into practice, but a leader who listens will also encourage group members to share their concerns and thoughts, creating an atmosphere of free speech and productivity.36.The passage is mainly about____________.A.choosing a good leader for a group.B.assessing the performance of a leader.C.qualities of a good group leader.D.team members’ appr eciation of a leader.37.A good leader should always be able to_______.A.tell when things go wrong in a group project.B.control everything group members do.C.make group members admit their mistakes.D.recognize any mistakes as his own.38.According to“Concerns for members”, which of the following is a good quality of a leader?A.fairB.reliableC.determinedD.generous39.Which of the following is not the benefit of being a good listener?A.Offer a chance for group members to express themselves freely.B.Allow the leader to put all his ideas into practice easily.C.Make it possible to deal with group members’ complaints.D.Make group members productive by sharing their thoughts.In her new book, “The Smartest Kids in the World,” Amanda Ripley, an investi gative journalist, tells the story of Tom, a high-school student from Gettysburg, Pennsylvania, who decides to spend his senior year in Warsaw, Poland. Poland is a surprising educational success story: in the past decade, the country raised students’ test scores from significantly below average to well above it. Polish kids have now outscored(超过……分数) American kids in math and scien ce, even though Poland spends, on average, less than half as much per student as the United States does. One of the most striking differences between the high school Tom attended in Gettysburg and the one he ends up at in Warsaw is that the latter has no football team, or, for that matter, teams of any kind.Those American High Schools lavish more time and money on sports than on math is an old complaint. This is not a matter of how any given student who plays sports does in school, but of the culture and its priorities. This December, when the latest Program for International Student Assessment(PISA) results are announced, it’s sa fe to predict that American high-school students will once again display their limited skills in math and reading, outscored not just by students in Poland but also by stud ents in places like South Korea, Belgium, the Netherlands, Finland, Singapore, and Japan. Meanwhile, they will have played some very exciting football games, which will have been breathlessly written up in their hometown papers.Why does this situation continue? Well, for one thing, kids like it. And for another, according to Ripley, parents seem to like the arrangement, too. She describes a tour she took of a school in Washington D.C., which costs thirty thousand dollars a year. The tour leader—a mother with three children in the school—was asked about the school’s flaws(暇疵). When she said that the math program was weak, none of the parents taking the tour reacted. When she said that the football program was weak, the parents suddenly became concerned. “Really?” one of them asked worriedly, “What do you mean?”One of the ironies (讽刺) of the situation is that sports reveal what is possible. American kids’ performance on the field shows just how well they can do when expectations are high. It’s too bad that their test scores show the same thing.40.Tom decides to spend his senior year in Poland because _______.A.there are striking differences between the 2 countriesB.Polish kids are better at learningC.sports are not supported at schools in GettysburgD.he intends to improve his scores41.According to Paragraph 2, we know that _______.A.too much importance is placed on sports in AmericaB.little time is spent on sports in Japanese schoolsC.American high schools complain about sports timeD.PISA plays a very important role in America42.The underlined sentence in the last paragraph means _______.A.American students’ academic performance worries their parents a lotB.high expectations push up American students’ academic performanceC.low expectations result in American students’ poor PISA performancecking practice contributes to Ameri can students’ average performance43.The purpose of this article is to _______.A.draw public attention to a weakness in American school traditionB.call on American schools to learn from the Polish modelpare Polish schools with those in AmericaD.explain what is wrong with American schools and provide solutionsLoch Ness, the largest freshwater lake in the British Isles, is twenty four miles long and, at one point, one mile wide. It has an average depth of four hundred and fifty feet and at times drops close to a thousand. It is cold and murky(浑浊的), with dangerous currents. In short, it is the perfect place to hide a monster from even the sharpest eyes of science.The Loch Ness Monster, also called Nessie, is supposedly living in this area. The earliest recorded sighting of the Loch Ness Monster was in the biography of Life of St. Columba by Adamnan in the year AD 565. The monster apparently attacked a man who was swimming in the River Ness.The monster didn’t make headlines again until August 27, 1930, when 3 fishermen reported s eeing a creature with 20 feet long approaching their boat, throwing water in the air. In 1933, after a new road was built along the edge of the Loch, the number of reports rose suddenly. Early in 1934, Author Grant, a young student, was out on his motorcycle one evening when he almost ran into the monster as it crossed the road. Grant’s description of the thing — small head, long thin neck and tail with a big body, seemed to match the appearance of the plesiosaur(蛇颈龙), an aquatic(水生的) type of dinosaur that has been extinct(已灭绝的)for 65 million years.The Loch Ness Investigation Bureau was formed in 1962 to act as a research organization for information about the creature. Even now, efforts have continue d to find the monster. A great deal of information was discovered about the Loch, but they haven’t yet to produce any specific evidence of a monster.Skeptics(怀疑论者) argue that the water in the Loch is too cold for a plesiosaur to live in. They also argue that an air-breathing animal, like a whale or seal, would spend much more time on the surface than the creature seems to, and would be sp otted more often.Some scientists have wondered if the sightings might be caused by an underwater wave which is known to sometimes occur in deep, long, and cold lakes, like Loch Ness. Such a wave might push debris(废弃物)to the surface that might look like a strange animal.However, none of these is identified.44.According to the skeptics, which of the following is TRUE?A.It is impossible for a monster to live in cold water.B.The Loch Ness Monster often stays under the water.C.The Loch Ness Monster is an air-breathing animal.D.There is no so-called monster in Loch Ness.45.The purpose of setting up the Loch Ness Investigation Bureau is to .A.research the plesiosaur in the Loch NessB.protect the Nessie in the lakeC.collect some information about the NessieD.catch the Loch Ness Monster46.Which of the following is the correct order for the things that happened in the passage?a. A young student met with a monster crossing the road.b. A swimmer was attacked by a monster in Loch Ness.c. A new road was built along the edge of the Loch.d. The Loch Ness Investigation Bureau was set up.e. Three fishermen saw a creature swimming towards their boat.A.b, e, c, a, dB.a,b, e, d, cC.b, d, a, c, eD.d, c, e, b, a47.We can infer from the passage that .A.Nessie is an aquatic type of dinosaurB.Nessie has a 20-foot-long bodyC.Nessie is an underwater waveD.Nessie is still a mystery48.What does this passage mainly talk about?A.The natural scenery of Loch Ness.B.The Nessie.C.Skeptics’ opinions on Loch Ness Monster.D.The Loch Ness Investigation Bureau’s research results.Wikipedia is a free-access, free content Internet encyclopedia(百科全书), supported and hosted by the non-profit Wikipedia Foundation. Wikipedia is ranked among the ten most popular websites and is considered the Internet’s largest and most popular general reference book. Now, Wikipedia is becoming Wookiepedia as scientists hope the informative website will help us reach out to intelligent life forms. So aliens can learn about the human race. Astronomers would like to beam (播送) the entire contents of Wikipedia into space in the hope of contacting aliens. They want to send messages to hundreds of star systems and planets 20 light years away using radio telescopes. The Search for Extraterrestrial Intelligence Institute, in California, wants to use powerful radio telescopes to try to reach Chewbacca and his mates in a galaxy(银河系)far away.The plans will be discussed by astronomers at the weekend as some scientists fear the reply from ET might not turn out to be friendly. Institute scientist David Black said, “One question is if there are dangerous creatures we might be drawing their attention to ourselves. Another is if we go ahead, what message should be sent? There could be many civilizations out there,but if they are all listening and no one is broadcasting or responding, then nothing will happen.”Professor Stephen Hawking, who has warned that intelligent aliens probably exist and we should keep a safe distance from them, is among those. “If aliens visit us, the outcome would be much as when Columbus landed in America which didn’t turn out well for the Native Americans,” he said.Signals from Earth’s radio and TV bro adcasts have been heading out in space for some 60 years reaching around 5,000 stars.49.What does the underlined word those in the fourth paragraph refer to?A.aliens from spaceB.astronomers sending signalsC.scientists for the plansD.scientists against the plans50.What has Prof. Stephen Hawking warned?A.intelligent aliens would probably land in America.B.the plans will be in vain and nothing will happen.C.none of the civilizations would reply.D.aliens would bring disasters to the human race.51.Which can be the best title?A.Wikipedia, the most popular websiteB.Wikipedia to be beamed into spaceC.Wikipedia, aliens’ best friendD.Wikipedia to result in a disasterA great loss—Shirley Temple dies at 85February 12,2014NG,Associated PressShirley Temple Black, who died on February 10th at age 85, wasn’t just a child star. She was THE child star—the sweet little girl whose shining smile helped illumine some of the darkest days the US has known during the Great Depression.It’s hard today to imagine the super star Shirley was once “America’s Little Darling”. She sang and danced her way to the top of the box office in such films as Bright Eyes, Curly Top and Heidi. By 1940, she had appeared in 43 films. Temple teamed with Bill Robison in four movies, and their dance on the stairs in The Little Colonel is still a legendary film moment.In the 1930s, her name on a movie introduction assured(保证) a packed house. She inspired dolls, dresses, dishes—even a drink (alcohol-free, of course).US Pres ident Franklin D. Roosevelt once famously said that “as long as our country has Shirley Temple, we will be all right.’’Unlike so many of today’s child stars, Temple didn’t end up with her name appearing across the headlines for bad behaviors. Instead of getting her photos on front pages or struggling with drugs and alcohol, Temple went on to a second career in diplomacy (外交), including presidential appointments as ambassador to Ghana.She surprised a lot of people who doubted her with her grace, knowledge and eagerness to serve. In fact, her career in public service (20 years) was longer than her career in movies (19). The role she valued most, however, was as wife, mother, grandmother and great-grandmother.The world has lost a treasured Hollywood legend. But her movies will allow that little dynamic figure to continue charming audiences for a very long time.52.The word “illumine” in Paragraph 1 means _______.Shorten B. Sweeten C. strengthen D. brighten53.Temple, as a child movie star, can best be described as _______.A.sweet and livelyB.gentle and kindC.smart and knowledgeableD.shy and attractive54.What part did she regard as the most important in her life?A.A top movie star.B.A businesswoman.C.Her family role.D.Her diplomacy career.55.Where does this passage possibly come from?A.A biography.B.A newspaper.C.A magazine.D.A poster.四、七选五:Seeing with SoundAccording to a British news report,some blind people may finally get a chance to “see”. 56Dr. Peter Meijer,a scientist at Philips Research Laboratories in the Netherlands,has developed a new system called the vOICe.The three middle letters in vOICe stand for “Oh,I see.” Meijer’s groundbreaking technology may change the lives of visually impaired people all over the world.A wearable setup of vOICe system consists of a head-mounted camera,stereo headphones and a notebook PC. The system translates visual images from a camera into complex sounds. 57The program is based on the theory that people can hear certain sounds and learn to translate them into meaningful mental images. Everything has its own unique sound. 58The height of an object or person could be determined by pitch(音调).And a built-in color identifier speaks out color names when it is turned on. What the v OICe users have to lear n is which sound goes with which object. Meijer says that he is counting on the brain’s ability to adapt.59Within two weeks,most people who experiment with the vOICe system are able to identify objects such as walls and doors. They are also able to identify certain situations,including whether the lights in a room are on or off. 60Meijer thinks that translating will eventually become automatic for many users of the vOICe,bringing a form of vision to them for the very first time.A. Brighter areas sound louder than darker areas.B. It then sends these sounds to a person through headphones.C. However, they are not learning to see with their eyes;they are learning to see with their ears.D. The vOICe system is aiming to treat blindness by translating images from a camera into audio signals.E. Meijer supposes that the brain is interested in the information “content”,but not the information “carrier”.F. To evaluate the new system’s effectiveness, a number of visually impaired people were chosen to test thev0ICe.G. Over time, some users have even learned to “watch” television or “recognize” the outlines of buil dings as they walk.第II卷(非选择题)五、语法填空:Many teens in high school want to try out for sports because they think it will make them more popular. But there are lots of other benefits 61sports that teens may overlook.With TV, movies, computers, and video games 62(become)more and more popular, it has become so much 63(easy) for teens to be by themselves rather than going out with friends. Kids used to hang out at the mall or drive around the town; now they just sit at home. Getting teens into a sport gives them 64opportunity to go out and socialize. 65they may not find a new best friend, they will learn how to interact and work as a team, s omething they’ll find 66(use) later in life.More and more kids are becoming overweight. If teens see that their physical condition is causing them to perform 67(bad), they may 68(motivate) to do other activities to get healthy. By the time your child 69(reach) their teenager year, part of good parenting will be providing them with direction and 70 (encourage) and continuing to help them develop a healthy style of living.六、书面表达:71.下面的漫画反映了一种社会现象,请你根据对这幅漫画的理解,用英语写一篇短文, 描述一下他们的不同情况,并就如何解决此问题发表你的看法。
【解析】四川省大竹县文星中学2016届高三12月月考数学试卷 Word版含解析[ 高考]
四川省大竹县文星中学2016届高三12月月考数学试题一、单选题1.已知集合A={x|x2-x-2<0},集合,则下列结论正确的是A.A=BB.ABC.BAD.A∩B=【答案】C【解析】本题考查集合的基本运算.因为,,所以B A.选C.【备注】集合的基本运算为高考常考题型,要求熟练掌握.2.已知函数,命题p:∀x∈[0,+∞),f(x)≤1,则A.p是假命题,p:x0∈[0,+∞),f(x0)>1B.p是假命题,p:x∈[0,+∞),f(x)≥1C.p是真命题,p:x0∈[0,+∞),f(x0)>1D.p是真命题,p:x∈[0,+∞),f(x)≥1【答案】C【解析】本题考查命题及其关系,全称量词与特称量词.函数,命题p:∀x∈[0,+∞),f(x)≤1,p是真命题, p:x0∈[0,+∞),f(x0)>1.选C.【备注】全称命题的否定是特称命题.3.“函数在区间(0,+∞)上为增函数”是“=3”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【答案】B【解析】本题考查充分条件与必要条件的判定、对数函数的单调性,意在考查考生的分析理解能力.由函数在区间(0,+∞)上为增函数,则,故是“=3”的必要不充分条件.故本题正确答案为B.4.若抛物线的焦点与双曲线的一个焦点重合,则的值为A.1B.-1C.2D.4【答案】A【解析】本题考查抛物线和双曲线的简单几何性质,意在考查考生的运算求解能力.由抛物线的焦点,双曲线的右焦点为,得,.故本题正确答案为A.5.已知函数定义域是,则y=f(2|x|-1)的定义域是A. B.[-1,4] C. D.【答案】C【解析】本题考查复合函数的定义域的求法,意在考查考生的运算求解能力. 函数定义域是,则,即,故函数定义域为,则得.故本题正确答案为C.6.若函数A.5B.4C.3D.2【答案】B【解析】本题考查三角函数的图像与性质.设函数的周期为,由图知;因为,所以.选B.【备注】“知图求式”.7.如图,在边长为1的正三角形中,分别为边上的动点,且满足,,其中分别是的中点,则的最小值为A. B. C. D.【答案】C【解析】本题主要考查向量的基本运算.==,(,当取得最小值,所以故选C.【备注】平面向量基本定理是重点.8.已知双曲线的右焦点为,直线与一条渐近线交于点A,若的面积为 (O为原点),则抛物线的准线方程为A. B. C. D.【答案】A【解析】本题主要考查双曲线和抛物线的基本性质.不妨设直线与渐近线y=交于点A,则A(,所以的面积为,解得a=b,所以抛物线的准线方程为x=-=-1,故选A.【备注】抛物线的考查重点是定义.9.某几何体的三视图(单位:)如图所示,其中侧视图是一个边长为2的正三角形,则这个几何体的体积是A. B. C. D.【答案】B【解析】本题主要考查简单几何体的三视图.由三视图可知,该几何体是底面为直角梯形的直四棱锥,且底面直角梯形的上底为1,下底为2,高为2,棱锥的高为=【备注】高考常考题,需熟练掌握.10.已知函数若互不相等,且,则的取值范围是A. B. C. D.【答案】C【解析】本题考查分段函数的图像、函数的零点,意在考查考生的数形结合思想及分析理解能力.作出函数图像,设,故利用三角函数的对称性得,由得,故.故本题正确答案为C.11.下列命题中正确的是A.函数是奇函数B.函数在区间上是单调递增的C.函数的最小值是D.函数是最小正周期为2的奇函数【答案】C【解析】本题主要考查本题主考查三角函数的奇偶性、单调性、周期与最值,考查了分析问题与解决问题的能力.区间不关于原点对称,所以A错误;函数在区间上是单调递减,所以B错误;因为,所以函数的最小值是,则C正确;函数是最小正周期为1的奇函数,故D错误.12.如图,设D是图中边长分别为1和2的矩形区域,E是D内位于函数图象下方的区域(阴影部分),从D内随机取一个点M,则点M取自E内的概率为A. B. C. D.【答案】C【解析】本题主要考查利用积分求面积以及几何概型概率的求法,考查了分析问题与解决问题的能力.图中矩形的面积为2,利用积分求出阴影的面积为==,利用几何概率公式可得点M 取自E内的概率为.二、填空题13.已知集合,N=,若,则的值是_______【答案】【解析】本题考查集合的运算,意在考查考生的运算求解能力. 集合,若,得集合N可能为,由N=,当得,适合条件,若,则得,若,则得,综上,a的值为.故本题正确答案为.14.已知角θ的顶点为坐标原点,始边为x轴非负半轴,若P(4,y)是角θ终边上一点,且sin θ=,则y=________【答案】-8【解析】本题考查任意角的三角函数,意在考查考生的分析理解能力.依题意,sin θ=,且点P(4,y),得,则得.故本题正确答案为-8.15.若是偶函数,则___________.【答案】【解析】本题考查函数的性质,对数函数.因为是偶函数,所以,即,解得.【备注】特殊值代入,事半功倍.16.某程序框图如图所示,则输出的S的值是______________.【答案】【解析】本题主要考查程序框图.第一次循环结束:S=1,k=2;第二次循环结束:S=,k=3;第三次循环结束:S=,k=4;第四次循环结束:S=,k=5;此时k<5不成立,循环结束,输出【备注】高考常考题,需要熟练掌握.三、解答题17.已知向量,,设函数.(Ⅰ)求的单调递增区间;(Ⅱ)求在上的最大值和最小值.【答案】(Ⅰ)===.当时,解得,的单调递增区间为.(Ⅱ)当时,,由标准函数在上的图像知==.所以在上的最大值和最小值分别为.【解析】本题考查平面向量数量积、三角恒等变换、三角函数单调性、函数在区间上的最值,意在考查考生的运算求解能力.(Ⅰ)利用平面向量数量积公式结合三角恒等变换求得,从而求得单调区间.(Ⅱ)利用整体思想求得函数的最值.18.已知数列的前项和为满足且.(1) 令证明:;(2) 求的通项公式.【答案】(1),,.(2) , ,累加得,经检验,符合【解析】本题主要考查数列的递推公式、、累加法,考查了分析问题与解决问题的能力.(1)由可得,又因为则结论得证;(2)根据利用累加法可得,则可以求得,利用即可求得的通项公式.19.函数(且)是定义在实数集上的奇函数.(1)若,试求不等式的解集;(2)若且在上的最小值为,求的值.【答案】(1)是定义在R上的奇函数,,又且易知f(x)在R上单调递增,原不等式化为:,即,不等式的解集为.(2),即(舍去)令∵x≥1,∴t≥f(1)=,∴g(t)=-2mt+2=+2-当时,当时,当时,当时,,解得,舍去.综上可知.【解析】本题考查函数的奇偶性,函数的单调性,二次函数的最值,意在考查考生的分类讨论思想及分析问题与解决问题的能力及运算求解能力.(1)利用函数为奇函数及增函数将原不等式化为,从而求得不等式的解集.(2)利用求得a的值,令,得g(t)=-2mt+2=+2-,分类讨论求得其最小值,从而求得m的值.20.某批发市场对某种商品的日销售量(单位:吨)进行统计,最近50天的统计结果如下:若以上表中频率作为概率,且每天的销售量相互独立.(Ⅰ)求5天中该种商品恰好有两天的销售量为1.5吨的概率;(Ⅱ)已知每吨该商品的销售利润为2千元,表示该种商品某两天销售利润的和(单位:千元),求的分布列和数学期望.【答案】解:(Ⅰ),,依题意,随机选取一天,销售量为吨的概率,设5天中该种商品有天的销售量为1.5吨,则,.(Ⅱ)的可能取值为,则:,,,,,4 5 6 7 80.04 0.2 0.37 0.3 0.09所以的分布列为:的数学期望==.【解析】本题考查二项分布、离散型随机变量的分布列和数学期望.(Ⅰ)先求得销售量为吨的概率,然后利用二项分布求得其概率.(Ⅱ)的可能取值为,分别求得其概率,写出分布列和数学期望.21.如图,已知的两条角平分线和相交于在上,且.(1)证明:四点共圆:(2)证明:平分.【答案】(1)在中,因为,所以,因为是角平分线,所以,故,于是,即,所以四点共圆.(2)连结,则为的平分线,得,由(1)知四点共圆,所以.又,又由,且平分,可得,可得,所以平分.【解析】本题考查四点共圆,圆周角定理.【备注】三角形相似,圆周角定理,弦切角定理,切割线定理等. 22.已知,求证:(1);(2).【答案】证明:(1),,,,,,.专业文档,,,.【解析】本题考查基本不等式的应用,意在考查考生的推理论证能力.(1),从而问题得证;(2)由,得,从而问题得证.珍贵文档。
四川省大竹县文星中学高一12月月考数学试题 Word版含答案
大竹县文星中学2014-2015学年高一12月月考数学试卷(时间:120分钟 满分:150分)一、(本大题共12个小题,每小题5分,共60分,每小题给出的四个备选答案中,有且仅有一个是正确的)1.设集合U ={x |0<x <10,x ∈N +},若A ∩B ={2,3},A ∩(∁U B )={1,5,7},(∁U A )∩(∁U B )={9},则集合B =( )A .{2,3,4}B .{2,3,4,6}C .{2,4,6,8}D .{2,3,4,6,8}[答案] D[解析] ∵U ={1,2,3,4,5,6,7,8,9}, ∵A ∩B ={2,3},∴2∈B,3∈B . ∵A ∩(∁U B )={1,5,7}, ∴1∈A,5∈A,7∈A,1∉B,5∉B,7∉B . ∵(∁U A )∩(∁U B )={9}∴9∉A,9∉B , ∴A ={1,2,3,5,7},B ={2,3,4,6,8}.2.若集合P ={x |2≤x <4},Q ={x |x ≥3},则P ∩Q 等于( ) A .{x |3≤x <4} B .{x |3<x <4} C .{x |2≤x <3} D .{x |2≤x ≤3}[答案] A[解析] P ∩Q ={x |2≤x <4}∩{x |x ≥3}={x |3≤x <4}.3.函数f (x )=3x 21-x +3x +1的定义域是( )A .(-13,+∞)B .(-13,1)C .[-13,1)D .[0,1)[答案] C[解析] 要使函数有意义,应满足⎩⎪⎨⎪⎧1-x >03x +1≥0,∴-13≤x <1,故选C.4.设函数f (x )=⎩⎪⎨⎪⎧1,x >00,x =0-1,x <0,g (x )=⎩⎪⎨⎪⎧1,x ∈Q 0,x ∈∁R Q ,则f [g (π)]的值为( ) A .1 B .0 C .-1 D .π[答案] B[解析] g (π)=0,∴f [g (π)]=f (0)=0.5.设(x ,y )在映射f 下的象是(2x +y ,x -2y ),则在f 下,象(2,1)的原象是( ) A .(12,32)B .(1,0)C .(1,2)D .(3,2)[答案] B[解析] 由⎩⎪⎨⎪⎧ 2x +y =2x -2y =1,得⎩⎪⎨⎪⎧x =1y =0,故选B.6.函数g (x )=2x +5x 的零点所在的一个区间是( ) A .(0,1) B .(1,2) C .(-1,0) D .(-2,-1)[答案] C[解析] 本题考查函数零点存在区间的判断,只要计算函数在区间两个端点处的值是否异号即可,因为g (-1)=2-1-5<0,g (0)=20=1>0,故选C.7.由于被墨水污染,一道数学题仅能见到如下文字:已知二次函数y =x 2+bx +c 的图象经过(1,0),…,求证这个二次函数的图象关于直线x =2对称.根据已知信息,题中二次函数图象不具有的性质是( ) A .过点(3,0) B .顶点(2,-2)C .在x 轴上截线段长是2D .与y 轴交点是(0,3) [答案] B[解析] ∵二次函数y =x 2+bx +c 的图象经过点(1,0), ∴1+b +c =0,又二次函数的图象关于直线x =2对称, ∴b =-4,∴c =3.∴y =x 2-4x +3,其顶点坐标为(2,-1),故选B.8.已知a =21.2,b =(12)-0.8,c =2log 52,则a ,b ,c 的大小关系为( )A .c <b <aB .c <a <bC .b <c <aD .b <a <c[答案] A[解析] 本题考查基本函数的性质.a =21.2,b =(12)-0.8=20.8,c =2log 52=log 522=log 54,因为21.2>20.8>1,所以a >b >1,c =log 54<1,所以a ,b ,c 的大小关系为a >b >c ,故选A.9.已知偶函数f (x )在(-∞,-2]上是增函数,则下列关系式中成立的是( ) A .f (-72)<f (-3)<f (4)B .f (-3)<f (-72)<f (4)C .f (4)<f (-3)<f (-72)D .f (4)<f (-72)<f (-3)[答案] D[解析] ∵f (x )在(-∞,-2]上是增函数, 又-4<-72<-3,∴f (4)=f (-4)<f (-72)<f (-3).10.设函数y =x 3与y =22-x 的图象的交点为(x 0,y 0),则x 0所在的区间是( )A .(0,1)B .(1,2)C .(2,3)D .(3,4)[答案] B[解析] 令f (x )=x 3-22-x ,由题意知x 0是函数f (x )的零点,又f (1)=1-2=-1<0,f (2)=8-1=7>0,故选B.11.设a =60.5,b =0.56,c =log 60.5,则a ,b ,c 的大小关系为( ) A .a >b >c B .b >a >c C .c >b >a D .a >c >b[答案] A[解析] a =60.5>60=1,b =0.56<0,50=1, 又0.56>0,∴0<0.56<1, c =log 60.5<log 61=0,∴a >b >c .12.对实数a 和b ,定义运算“⊗”:a ⊗b =⎩⎪⎨⎪⎧a ,a -b ≤1b ,a -b >1,设函数f (x )=(x 2-2)⊗(x -1),x ∈R .若函数y =f (x )-c 的图象与x 轴恰有两个公共点,则实数c 的取值范围是( )A .(-1,1]∪(2,+∞)B .(-2,-1]∪(1,2]C .(-∞,-2)∪(1,2]D .[-2,-1][答案] B[解析] 依题意可得f (x )=⎩⎪⎨⎪⎧x 2-2,-1≤x ≤2x -1,x <-1或x >2作出其示意图如图所示.由数形结合知,实数c 需有1<c ≤2或-2<c ≤-1.二、填空题(本大题共4个小题,每空4分,共16分,把正确答案填在题中横线上) 13.已知函数f (x +1)=3x +4,则f (x )的解析式为________________. [答案] f (x )=3x +1[解析] 设x +1=t ,∴x =t -1, ∴f (t )=3(t -1)+4=3t +1,∴f (x )=3x +1. 14.3log 925+log 2-1(2+1)的值为__________.[答案] 4[解析] 3 log 925+log2-1(2+1)=3 log 35+log2-1(2-1)-1=5-1=4.15.定义域为R 的函数y =f (x )的值域是[a ,b ],则函数y =f (x +a )的值域是________. [答案] [a ,b ][解析] 函数f (x +a )的图象只是由f (x )的图象向左或向右平移得到,函数值y 没有变化. 16.对于定义域在R 上的函数f (x ),若实数x 0满足f (x 0)=x 0,则称x 0是函数f (x )的一个不动点.若函数f (x )=x 2+ax +1没有不动点,则实数a 的取值范围是__________.[答案] (-1,3)[解析] 由题意,得方程x 2+ax +1=x ,即 x 2+(a -1)x +1=0无实根, ∴Δ=(a -1)2-4=a 2-2a -3<0, ∴-1<a <3.三、解答题(本大题共6个小题,共74分,解答应写出文字说明,证明过程或演算步骤) 17.(本小题满分12分)已知函数f (x )=log 2(x -1)的定义域为集合A ,函数g (x )=(12)x (-1≤x ≤0)的值域为集合B .(1)求A ∩B ;(2)若C ={x |a ≤x ≤2a -1},且C ⊆B ,求实数a 的取值范围. [解析] (1)要使函数f (x )有意义,应满足log 2(x -1)≥0, ∴x -1≥1,∴x ≥2. ∴A ={x |x ≥2}.∴g (x )=(12)x (-1≤x ≤0)是减函数,∴当x =-1时,g (x )取最大值2, 当x =0时,g (x )取最小值1, ∴B ={x |1≤x ≤2},∴A ∩B ={2}. (2)∵C ⊆B ,①当C =∅时满足题意,即a >2a -1,解得a <1;②当C ≠∅时,则有⎩⎪⎨⎪⎧a ≥12a -1≤2,解得1≤a ≤32.综上实数a 的取值范围是(-∞,32].18.(本小题满分12分)设a ,b ,c 为正数,且满足a 2+b 2=c 2. (1)求证:log 2(1+b +c a )+log 2(1+a -cb)=1;(2)若log 4(1+b +c a )=1,log 8(a +b -c )=23,求a ,b ,c 的值.[解析] (1)log 2(1+b +c a )+log 2(1+a -cb )=log 2a +b +c a +log 2a +b -cb=log 2(a +b )2-c 2ab=log 2(a 2+b 2-c 2)+2ab ab=log 22=1.(2)由log 4(1+b +c a )=1,log 8(a +b +c )=23,得1+b +ca=4,a +b -c =4,又a 2+b 2=c 2,整理可得⎩⎪⎨⎪⎧b +c =3a a +b -c =4a 2+b 2=c 2,解得a =6,b =8,c =10.19.(本小题满分12分)2009年某个体企业受金融危机和国家政策调整的影响,经历了从亏损到盈利的过程,下面的二次函数图象(部分)刻画了该公司年初以来的累积利润S (万元)与时间t (月)之间的关系(即前t 个月的利润总和S 与t 之间的关系,0≤t ≤12).请根据图象提供的信息解答下列问题:(1)求累积利润S (万元)与时间t (月)之间的函数关系式; (2)截止到第几月末公司累积利润可达到9万元? (3)该企业第四季度所获利润是多少? [解析]设S (t )=at 2+bt +c , 将点(0,0),(6,0),(3,-3)代入得 ⎩⎪⎨⎪⎧36a +6b =09a +3b =-3c =0,解得⎩⎪⎨⎪⎧a =13b =-2c =0.∴函数关系式S (t )=13t 2-2t (0≤t ≤12).(2)令S =9即13t 2-2t =9,解得t =9或t =-3(舍),∴截止到9月末公司累积利润可达到9万元. (3)S (12)=13×144-2×12=24(万元),S (9)=13×81-2×9=9(万元),∴第四季度获利S (12)-S (9)=24-9=15(万元). 答:第四季度所获利润为15万元.20.(本小题满分12分)若关于x 的方程x 2+mx +m -1=0有一个正根和一个负根,且负根的绝对值较大,求实数m 的取值范围.[解析] 根据题意,画出f (x )=x 2+mx +m -1的图象,如图所示.图象的对称轴为直线x =-m2.因为方程x 2+mx +m -1=0有一个正根和一个负根, 则函数f (x )有两个零点x 1,x 2, 由题意不妨设x 1>0,x 2<0,且|x 1|<|x 2|. 由题意,有⎩⎪⎨⎪⎧f (0)<0-m 2<0,故⎩⎪⎨⎪⎧m -1<0m >0.∴ 0<m <1.即所求的取值范围为(0,1).21.(本小题满分12分)已知定义在R 上的函数f (x )满足f (log 2x )=x +ax ,a 为常数.(1)求函数f (x )的表达式; (2)如果f (x )为偶函数,求a 的值;(3)如果f (x )为偶函数,用函数单调性的定义讨论f (x )的单调性. [解析] (1)令log 2x =t ,则x =2t . ∴f (t )=2t +a2t .∴f (x )=2x +a2x (x ∈R ).(2)由f (-x )=f (x ),则2-x +a 2-x =2x+a 2x , ∴(2x -2-x )(1-a )=0对x ∈R 均成立. ∴1-a =0,即a =1. (3)当a =1时,f (x )=2x +12x ,设0≤x 1<x 2,则 f (x 1)-f (x 2)=2x 1+12x 1-(2 x2+12x 2) =(2 x 1-2 x 2)(1-12 x 1+x 2),∵2 x 1-2 x 2<0,1-12 x 1+x 2>0,∴f (x 1)-f (x 2)<0.即f (x 1)<f (x 2).因此f (x )在区间[0,+∞)上是增函数. 同理当x 1<x 2<0时, f (x 1)-f (x 2)>0,∴f (x )在区间(-∞,0)上是减函数.22.(本小题满分14分)已知函数f (x )=lg(m x -2x )(0<m <1). (1)当m =12时,求f (x )的定义域;(2)试判断函数f (x )在区间(-∞,0)上的单调性并给出证明; (3)若f (x )在(-∞,-1]上恒取正值,求m 的取值范围.[解析] (1)当m =12时,要使f (x )有意义,须(12)x -2x >0,即2-x >2x ,可得:-x >x ,∴x <0∴函数f (x )的定义域为{x |x <0}.(2)设x 2<0,x 1<0,且x 2>x 1,则Δ=x 2-x 1>0 令g (x )=m x -2x ,则g (x 2)-g (x 1)=m x 2-2 x 2-m x 1+2 x 1 =m x 2-m x 1+2 x 1-2 x 2 ∵0<m <1,x 1<x 2<0, ∴m x 2-m x 1<0,2 x 1-2 x 2<0 g (x 2)-g (x 1)<0,∴g (x 2)<g (x 1) ∴lg[g (x 2)]<lg[g (x 1)], ∴Δy =lg(g (x 2))-lg(g (x 1))<0, ∴f (x )在(-∞,0)上是减函数.(3)由(2)知:f (x )在(-∞,0)上是减函数, ∴f (x )在(-∞,-1]上也为减函数,∴f (x )在(-∞,-1]上的最小值为f (-1)=lg(m -1-2-1)所以要使f (x )在(-∞,-1]上恒取正值, 只需f (-1)=lg(m -1-2-1)>0,即m -1-2-1>1,∴1m >1+12=32,∵0<m <1,∴0<m <23.。
四川省大竹县文星中学高一数学12月月考试题
学年高一12月月考大竹县文星中学2014-2015 数学试卷)满分:150分(时间:120分钟分,每小题给出的四个备选答案中,有且5分,共60一、(本大题共12个小题,每小题)仅有一个是正确的BBAxxABAUx),(?{2,3},?∩(?1.设集合={)∩(|0<)<10,=∈N},若{1,5,7}∩=UUU+B)( {9},则集合==B.{2,3,4,6} A.{2,3,4}D.{2,4,6,8} {2,3,4,6,8}C.D[答案]U,=∵[解析] {1,2,3,4,5,6,7,8,9}BB,AB. ∈,∴2∵∈∩3={2,3}BA {1,5,7}?,)∵=∩(U BB,A,A,A,B,. ∴5∈77∈?1∈1??5BA,AB (??)∩(?9)={9}∴?,9∵UU BA=∴{2,3,4,6,8}={1,2,3,5,7},.QPxPxxQx) ∩等于={2.若集合|={|2≤( <4},≥3},则xxxx<4} .{|3≤<4} |3<BA.{xxxx≤3}<3} { |2≤|2≤D.C.{A] [答案xQPxxxxx |3≤[解析] ∩<4}={|2≤.<4}∩{|{≥3}=2x3xfx+1的定义域是( +()3= ) 3.函数x-111A.(-,+∞) ,B.(-1)331C.[.D[0,1)-,1)3[答案] Cx>0-1??,解析] 要使函数有意义,应满足[?x+1≥03??1x<1,故选C.∴-≤3- 1 -x>01,??x Q,∈1??x?00,=xfxg, 4.设函数,())=(=?x Q0,?∈???R?x<0,-1πfg) [( )]则的值为(B.1 .0 AD .π 1 C.-B[答案]gfgf(0)=[0.[解析] (π)]=(π)=0,∴xyfxyxyf下,象(2,1)的原象是( ,-5.设(2,) )在映射)下的象是(2,则在+13A.(,)B.(1,0) 22D(1,2) .(3,2)C.B[答案]xxy122+==????,故选由B.,得[解析] ??yxy0=12-=????x xxg) 的零点所在的一个区间是2+56.函数( ()=B.(0,1) (1,2) A.D.(-(C.-1,0) 2,C[答案]本题考查函数零点存在区间的判断,只要计算函数在区间两个端点处的值是否[] 解析01-gg C. ,故选2==,=-1)2-5<01>0(0)异号即可,因为( .由于被墨水污染,一道数学题仅能见到如下文字:72xcbxyx,…,求证这个二次函数的图象关于直线的图象经过+(1,0)已知二次函数+==2对称.) 根据已知信息,题中二次函数图象不具有的性质是((3,0) A.过点2) (2,-.顶点B x2 C.在轴上截线段长是y(0,3) .与D轴交点是B答案] [2cbxxy,(1,0)++的图象经过点] [解析∵二次函数=xcb∴2,又二次函数的图象关于直线0++==对称,1cb3.∴,∴4=-=- 2 -2xyx B.1),故选3,其顶点坐标为=(2-4,-+∴11.2-0.8cababc)( ,=2log2,则8.已知=2,,=(),的大小关系为52cabcba<<B .A.<<baacbc <DC..<<<A ] [答案120.8-0.81.2cab,log4log2[解析] 本题考查基本函数的性质.=2,==()=2,==2log2cbabacabc A.55520.81.2>的大小关系为,>,>2>1,所以>,故选>1,4<1=log,所以因为25xf) (2])在(-∞,-上是增函数,则下列关系式中成立的是( 9.已知偶函数7fff(4)3)<((-)<-.A27fff(4)-(-3)<)<B.(27fff)((-3)<C.(4)<-27fff3)-)<D.-(4)<((2D[答案]xf2]∵解析] -∞,-(上是增函数,)在([7 3,又-4<-<-27ffff(--)<∴3)(4)=.(-4)<(2x-32xyyyxx所在的区间是( ),则=2的图象的交点为(.设函数10,=与)..(0,1) (1,2) AD.C(2,3) .(3,4)000BB[答案]x-32xfxfxxff(2),2=-1<0是函数-()的零点,又(1)=] [解析令()=-21由题意知,0B.=17>0,故选-=860.5ccabab) 0.5log0.5611.设=,=,=,则,,的大小关系为 ( 6bbaacc A.B>.>> >abccba.D .C>>>>A答案[]000.56ba=<0,50.5=,=>66=] 解析[11,66 0.5又0<0.5,∴>0,<1- 3 -cacb.0,∴>>log=0.5<log1=66baa≤1,-??2xxabxfab-?=(,设函数-”:(2))=12.对实数(和?,定义运算“??bab>1-,??cxcxyfx)R=( (轴恰有两个公共点,则实数).若函数-的图象与1),的取值范围是∈(1,2] ,-1]∪B..(-1,1]∪(2,+∞) (-2A1]-2,-D.[∪C.(-∞,-2)(1,2]B] [答案2xx≤2,-1≤-2??xf (依题意可得=)] [解析?xxx>21<--1,或??作出其示意图如图所由数形结合知,ccc1.实数需有1<≤-≤2或-2<) 16分,把正确答案填在题中横线上二、填空题(本大题共4个小题,每空4分,共xxffx+4,则(.13.已知函数)(的解析式为+1)=3________________xxf1)=] 3(+[答案txxt=-,∴1[解析] 设,+1=xftttfx1. 3+3(4-1)+=3)+1,∴=∴(()=25log.(214.3+1)+的值为log__________912-4 ] [答案-5 log251 log4.1=1)3=+5log-([解析] 32+-log2(+1)=391-2-21yfxabyfxa)的值域是________],则函数+=15.定义域为R的函数=.(()的值域是[,ab]] [,[答案fxafxy没有变的图象向左或向右平移得到,函数值+)的图象只是由[解析] 函数)((化.fxxfxxxfx)是函数满足(的一)=,则称16.对于定义域在R上的函数(,若实数)(00002aaxxxf__________1没有不动点,则实数个不动点.若函数.()=+的取值范围是+1,3)-答案] ([2xxax 1] [解析由题意,得方程++=,即- 4 -2xax+11)0无实根,+(=-22aaa 0∴Δ=(1)--4=,-2-3<a3.<∴-1<) 个小题,共三、解答题(本大题共674分,解答应写出文字说明,证明过程或演算步骤xxAfxg==)log-,函数1217.(本小题满分分)已知函数((1)的定义域为集合21x Bx. )≤0)的值域为集合((-1≤2BA∩;(1)求aBaCCxax 1}={,且|,求实数≤?≤2的取值范围.(2)若-xxf ((-1)≥0,)有意义,应满足[解析] (1)要使函数log2xx-1≥1,∴∴≥2.xxA |={≥2}.∴1x xxg )≤0)是减函数,∴((-1≤)=(2xxg 2(,∴当)=-1时,取最大值xxg 1(,当)=0时,取最小值BxABx{2}={∩|1≤.≤2},∴∴=BC∵,?(2)aaaC,解得;?时满足题意,即<1>21①当-=a≥1?3?aC.≤,解得②当1≤≠?时,则有?2a-1≤22??3a.]综上实数(的取值范围是-∞,2222cabcab. ,=为正数,且满足分)设,+.18(本小题满分12ccab-+ 1;(1)+log+)log(1)求证:(1+=22bacb2+cbaabc,求+,)(1+=1,log,-)=若(2)log的值.(84a3cabc-+) (1解析] (1)log++)+log(1[22bacbbcaa-+++ log=log+22ba22cab-+=log2ab222abbac2++-=log2ab-5 -1. ==log22cb2+cab)(2)由log(1=1,log+(=+,+)84a3cb+222cbcaab=4,又-=,得1+=4++,a abc3+=??cab?4+=-,整理可得?222?cba=+cab10.,6,==8解得=经历了从)2009年某个体企业受金融危机和国家政策调整的影响,1219.(本小题满分分S)(万元亏损到盈利的过程,下面的二次函数图象(部分)刻画了该公司年初以来的累积利润tttSt请根据图象之间的关系,个月的利润总和0≤与与时间(月)之间的关系(≤12).即前提供的信息解答下列问题:tS之间的函数关系式;月万元)与时间)(1)求累积利润((万元?截止到第几月末公司累积利润可达到9(2)(3)该企业第四季度所获利润是多少?2catbttS+,[解析]设(+)= 3)代入得(6,0),(3,-将点(0,0),1?baa0636=+=??3??ba?39+3=-.,解得b2=-???c?0=c0=12Stttt≤12).(0≤-∴函数关系式2( )=312Stt=9,=9即-2(2)令3tt=-3(舍),解得9=或∴截止到9月末公司累积利润可达到9万元.1S(12)=×144-2×12=24(万元(3)), 3- 6 -1S=(9),×81-2×9=9(万元)3SS(9)=24-9=15(万元)∴第四季度获利.(12)-答:第四季度所获利润为15万元.2mmxxx有一个正根和一个负根,且负=+1+.20(本小题满分12分)若关于0的方程-m的取值范围.根的绝对值较大,求实数2mxmxfx1[解析+] 根据题意,画出的图象,如图所示.(=)-+mx .=-图象的对称轴为直线22mmxx +有一个正根和一个负根,-+1=因为方程0xxfx ,(,)则函数有两个零点21xxxx |. |<|<0,且|由题意不妨设>0,2121 f <00? ?m 1<0-???.,故由题意,有m ?m >0<0-????2m <1.∴ 0< 即所求的取值范围为(0,1).aaxfxfx ,)R )已知定义在上的函数=(为常数.)满足+.21(本小题满分12分(log 2xxf 求函数)(的表达式;(1)afx ()为偶函数,求(2)如果的值;xffx ((3)如果的单调性.())为偶函数,用函数单调性的定义讨论t xxt . (1)令log ==2,则[解析] 2a t tf . )=(2∴+t 2a x xfx .()=2+(R ∈)∴x 2aa xx -xffx (2)由(-=),(),则2+=2+xx -22xx -ax ∈R 均成立. )(1-)=0对2(2∴-aa =1.01∴-=,即- 7 -1x xfa ,+)==1时,2( (3)当x 2xx ,则<设0≤ 2111x x xffx +(2) -+(-)(=2)21 xx 2122211 x x ), )(1=(2--221 x x +2211 x x >0, <0,1∵2--221x x +221fxfx )<0. )∴-((21fxfx ).即(( )<21fx )在区间[0,+∞)上是增函数.因此 (xx <0时,< 同理当21fxfx )>0,(( )-21fx )在区间(-∞,∴(0)上是减函数.xx mxmf <1).)(0<-14分)已知函数2( )=lg(22.(本小题满分1fxm )的定义域;=(时,求(1)当 2fx )在区间(-∞,(0)上的单调性并给出证明;(2)试判断函数fxm 的取值范围.上恒取正值,求 在(-∞,-(3)若1]()11xxxx -fmx )有意义,须()-2[解析] (1)当=>0(,即2>2, 时,要使 22xxx <0>,∴可得:-xxxf <0}|(.)的定义域为{∴函数xxxx Δxx >0 >=,且设(2),则<0,-<0112122xx mgx -)=2令,( x xx x mmxgxg 2+-2)-(-则)(=122112 x x x x mm =+2-2-2211xmx <0<∵0<,<1,21 x x x x mm <0 --∴2<0,22121xggxxggx )((()-)<()<0,∴1221xxgg ,(∴lg[()])]<lg[12xg Δygx ))<0))-lg(,∴=lg(((12xf 上是减函数.(-∞,∴0)()在xf 上是减函数,-∞,在(2)(3)由知:()(0) - 8 -fx )在(-∞,-1]∴上也为减函数,(-1-1mffx ) -=lg(2()在(-∞,-1]上的最小值为(-1)∴fx )在(-∞,-(1]上恒取正值,所以要使-1-1mf )>02-1)=lg(, 只需-(1131--1m , 即>1+-2=>1,∴ m 222mm .,∴0<<0<∵<1 3- 9 -。
四川省达州市大竹县文星中学高二12月月考数学试题
四川省大竹县文星中学2015-2016学年高二12月月考数学试题一、单选题1.若下列不等式成立的是A. B. C. D.2.某同学在一次综合性测试中语文、数学、英语、科学、社会5门学科的名次在其所在班级里都在前三名(记第一名为1,第二名为2,第三名为3,依此类推且没有并列名次情况),则称该同学为超级学霸,现根据不同班级的甲、乙、丙、丁四位同学对一次综合性测试名次数据的描述,一定可以推断是超级学霸的是A.甲同学:平均数为2,中位数为2B.乙同学:中位数为2,唯一的众数为2C.丙同学:平均数为2,标准差为2D.丁同学:平均数为2,唯一的众数为23.已知直线与圆有公共点,则A. B.C. D.4.在同一坐标系中,方程与的曲线大致是5.在△ABC中,若,则△ABC的形状是A.锐角三角形B.直角三角形C.钝角三角形D.不能确定6.在等差数列中,其前n项和是S n,若,则在中最大的是A. B. C. D.7.下列说法正确的是A.命题“∀x∈R,e x>0”的否定是“∃x∈R,e x>0”B.命题“已知x、y∈R,若x+y≠3,则x≠2或y≠1”是真命题C.“x2+2x≥ax在x∈上恒成立”⇔“(x2+2x)min≥(ax)max在x∈上恒成立”D.命题“若a=-1,则函数f(x)=ax2+2x-1只有一个零点”的逆命题为真命题8.已知命题,使;命题当时,的最小值为4.下列命题是真命题的是A. B. C. D.9.在数列中,已知对任意,则等于A. B. C. D.10.已知函数,则下列说法正确的为A.函数的最小正周期为2πB.的最大值为C.的图象关于直线x=﹣对称D.将的图象向右平移,再向下平移个单位长度后会得到一个奇函数的图象11.设圆(x+1)2+y2=25的圆心为C,A(1,0)是圆内一定点,Q为圆周上任一点,线段AQ的垂直平分线与CQ的连线交于点,则的轨迹方程为A. B. C. D.12.已知实数a,b,c满足,且,若实数是函数的一个零点,那么下列不等式中,不可能成立的是A. B. C. D.二、填空题13.某班级有50名学生,现要采取系统抽样的方法在这50名学生中抽出10名学生,将这50名学生随机编号1~50号并分组,第一组1~5号,第二组6~10号,,第十组46~50号,若在第三组中抽得号码为12的学生,则在第八组中抽得号码为________的学生.14.已知直线l:x﹣y+3=0被圆C:(x﹣a)2+(y﹣2)2=4截得的弦长为2,则a的值为. 15.直线经过的定点坐标为.16.如果三棱锥A-BCD的底面BCD是正三角形,顶点A在底面BCD上的射影是△BCD的中心,则这样的三棱锥称为正三棱锥.给出下列结论:①正三棱锥A-BCD中必有AB⊥CD,BC⊥AD,AC⊥BD;②正三棱锥A-BCD所有相对棱中点连线必交于一点;③当正三棱锥A-BCD所有棱长都相等时,该棱锥内切球和外接球半径之比为1:2;④若正三棱锥A-BCD的侧棱长均为2,侧面三角形的顶角为40°,过点B的平面分别交侧棱AC,AD于M,N,则△BMN周长的最小值等于.以上结论正确的是.(写出所有正确命题的序号).三、解答题17.等差数列中,=2,=2.(I)求的通项公式;(II)设求数列的前项和18.已知圆C的方程为,点是坐标原点,直线与圆C交于两点. (1)求的取值范围;(2)设是线段上的点,且,请将表示为的函数,并求其定义域.19.设的内角A ,B ,C.所对的边分别为a ,b ,c ,已知1,b =2,.(Ⅰ)求的周长;(Ⅱ)若求的值.20.如图是正方形,是正方形的中心,底面是的中点.求证:(1)//平面; (2)平面平面;(3)若,求四棱锥P -ABCD 的体积.21.实数满足圆的标准方程.(Ⅰ)求的最小值;(Ⅱ)求定点到圆上点的距离的最大值.22.已知等差数列满足;数列的前n 项和为,且满足.(Ⅰ)分别求数列的通项公式;C PABD OE(Ⅱ)若对任意的恒成立,求实数k的取值范围.参考答案1-5 CDDDC 6-10 BBABD 11-12 DD13.3714. 1或-315.16.①②④17. (Ⅰ)设等差数列的公差为d,则解得,.所以的通项公式为.(Ⅱ),所以.18. (1)将y=kx代入x2+(y-4)2=4中,得(1+k2)x2-8kx+12=0.(*) 由Δ=(-8k)2-4(1+k2)×12>0,得k2>3.所以,k的取值范围是(-∞,)∪(,+∞).(2)因为M,N在直线l上,可设点M,N的坐标分别为(x1,kx1),(x2,kx2), 则|OM|2=(1+k2)x12,|ON|2=(1+k2)x22,又|OQ|2=m2+n2=(1+k2)m2.由,得,即.由(*)式可知,x1+x2=,x1x2=, 所以.因为点Q在直线y=kx上,所以,代入中并化简,得5n2-3m2=36.由及k2>3,可知0<m2<3,即m∈(,0)∪(0,). 根据题意,点Q在圆C内,则n>0,所以.于是,n与m的函数关系为(m∈(,0)∪(0,)).19. (Ⅰ)∵∴∴△ABC周长为a+b+c=1+2+2=5(Ⅱ)∵∴20. (1)连接AC,OE,AC BD=O,在△PAC中,∵E为PC中点,O为AC中点.∴PA//EO,又∵EO平面EBD ,PA平面EBD,∴PA //面BD E.(2)∵PO底面ABCD,∴PO B D.又∵BD AC,∴BD平面PA C.又BD平面BDE,∴平面PAC平面BD E.(3)∵ABCD是正方形的中心,∴∵PO底面ABCD.21. (1)令k=,即y=k(x-4),结合图形可知当直线y=k(x-4)与圆相切时k最小,即最小,由=2得k=或k=0(舍去),故k=得最小值为(2)由得点在圆外故到圆上点的距离的最大值为+2=2+2.22.(1)由.∴ .由,得,即.又(n)∴{是等比数列,其中首相为,公比为3,所以(2)所以原不等式可转化为(对n恒成立,∴对n恒成立.令.当n3时,即; 当n3时,.∴当n=3时有最大值,最大值为所以。
四川省达州市大竹县文星中学高三数学下学期期初试卷 理(含解析)
2014-2015学年四川省达州市大竹县文星中学高三(下)期初数学试卷(理科)一、选择题:共12题每题5分共60分1.设全集U={1,3,5,7},M={1,|a﹣5|},M⊆U,∁U M={5,7},则a的值为()A. 2或﹣8 B.﹣8或﹣2 C.﹣2或8 D. 2或82.已知i是虚数单位,则=()A.﹣i B.+i C.+i D.﹣i3.命题“对任意x∈R都有x2≥1”的否定是()A.对任意x∈R,都有x2<1 B.不存在x∈R,使得x2<1C.存在x0∈R,使得x02≥1D.存在x0∈R,使得x02<14.函数f(x)=的图象大致为()A.B.C.D.5.定义在R上的奇函数f(x)满足f(x+1)=f(﹣x),当x∈(0,]时,f(x)=(1﹣x),则f(x)在区间(1,)内是()A.减函数且f(x)>0 B.减函数且f(x)<0 C.增函数且f(x)>0 D.增函数且f(x)<06.已知向量=(λ+1,1),=(λ+2,2),若(+)⊥(﹣),则λ=()A.﹣4 B.﹣3 C.﹣2 D.﹣17.已知实数x,y满足约束条件,则z=x+3y的最大值等于()A. 9 B. 12 C. 27 D. 368.已知两条直线l1:y=m和l2:y=(m>0,m≠),l1与函数y=|log2x|的图象从左至右相交于点A、B,l2与函数y=|log2x|的图象从左至右相交于点C、D.记线段AC和BD在x轴上的投影长度分别为a,b,当m变化时,的最小值为()A. 16 B. 8 C. 4 D. 29.已知某四棱锥的三视图(单位:cm)如图所示,则该四棱锥的体积是()A.B.C.D.10.阅读如图的程序框图,若运行相应的程序,则输出的S的值是()A. 39 B. 21 C. 81 D. 10211.设a=log37,b=21.1,c=0.83.1,则()A. b<a<c B. c<a<b C. c<b<a D. a<c<b 12.下列说法正确的是()A.若a>b,则<B.函数f(x)=e x﹣2的零点落在区间(0,1)内C.函数f(x)=x+的最小值为2D.若m=4,则直线2x+my+1=0与直线mx+8y+2=0互相平行二、填空题:共4题每题4分共16分13.已知集合A={x|0<x<3},B={x|x﹣1<0},则A∩B=.14.已知函数f(x)=asin2x+cos(2x+)的最大值为1,则a= .15.已知与的夹角为120°,若(+)⊥(﹣2)且||=2,则在上的投影为.16.如图展示了一个由区间(0,1)到实数集R的映射过程:区间(0,1)中的实数m对应数轴上的点M,如图1;将线段AB围成一个圆,使两端点A,B恰好重合(点M从点A按逆时针方向运动至点B),如图2;再将这个圆放在平面直角坐标系中,使其圆心在y轴上,点A的坐标为(0,1),如图3.图3中直线AM与x轴交于点N(n,0),则m的象就是n,记作f(m)=n.下列说法中正确命题的序号是.(填出所有正确命题的序号)①f()=1;②f(x)在定义域上单调递增;③方程f(x)=0的解是x=;④f(x)是奇函数;⑤f(x)的图象关于点(,0)对称.三、解答题:共6题每题12分共74分17.在△ABC中,内角A,B,C对边的边长分别是a,b,c.已知c=2,C=.(Ⅰ)若△ABC的面积等于,试判断△ABC的形状,并说明理由;(Ⅱ)若sinC+sin(B﹣A)=2sin2A,求△ABC的面积.18.己知数列{a n}满足a1=1,a n+1=(n∈N*),(Ⅰ)证明数列{ }是等差数列;(Ⅱ)求数列{a n)的通项公式;(Ⅲ)设b n=n(n+1)a n求数列{b n}的前n项和S n.19.如图,四边形ABCD是正方形,EA⊥平面ABCD,EA∥PD,AD=PD=2EA,F,G,H分别为PB,EB,PC的中点.(1)求证:FG∥平面PED;(2)求平面FGH与平面PBC所成锐二面角的大小.20.甲、乙两人参加一次英语口语考试,已知在备选的10道试题中,甲能答对其中的6题,乙能答对其中的8题.规定每次考试都从备选题中随机抽出3题进行测试,至少答对2题才算合格.(Ⅰ)分别求甲、乙两人考试合格的概率;(Ⅱ)求甲、乙两人至少有一人考试合格的概率.21.已知抛物线C:y=mx2(m>0),焦点为F,直线2x﹣y+2=0交抛物线C于A、B两点,P 是线段AB的中点,过P作x轴的垂线交抛物线C于点Q,(1)若抛物线C上有一点R(x R,2)到焦点F的距离为3,求此时m的值;(2)是否存在实数m,使△ABQ是以Q为直角顶点的直角三角形?若存在,求出m的值;若不存在,说明理由.22.定义在R上的函数y=f(x),f(0)≠0,当x>0时,f(x)>1,且对任意的a、b∈R,有f(a+b)=f(a)•f(b).(1)求证:f(0)=1;(2)求证:对任意的x∈R,恒有f(x)>0;(3)求证:f(x)是R上的增函数;(4)若f(x)•f(2x﹣x2)>1,求x的取值范围.2014-2015学年四川省达州市大竹县文星中学高三(下)期初数学试卷(理科)参考答案与试题解析一、选择题:共12题每题5分共60分1.设全集U={1,3,5,7},M={1,|a ﹣5|},M ⊆U ,∁U M={5,7},则a 的值为( ) A . 2或﹣8 B . ﹣8或﹣2 C . ﹣2或8 D . 2或8考点: 补集及其运算;集合的确定性、互异性、无序性. 分析: 由C U M={5,7},得5∉M ,7∉M ,由M ⊆U ,可得|a ﹣5|=3,解出a 即可. 解答: 解:由题意|a ﹣5|=3,a=2或8 故选D 点评: 本题考查集合的基本运算,属基本题.2.已知i 是虚数单位,则=( )A .﹣iB .+iC .+iD .﹣i考点: 复数代数形式的乘除运算. 专题: 数系的扩充和复数. 分析: 直接利用复数代数形式的除法运算化简求值. 解答: 解:=.故选:B . 点评: 本题考查了复数代数形式的除法运算,是基础的计算题.3.命题“对任意x ∈R 都有x 2≥1”的否定是( )A . 对任意x ∈R ,都有x 2<1B . 不存在x ∈R ,使得x 2<1C . 存在x 0∈R ,使得x 02≥1D . 存在x 0∈R ,使得x 02<1考点: 全称命题;命题的否定. 专题: 规律型. 分析: 利用汽车媒体的否定是特称命题写出结果判断即可. 解答: 解:因为全称命题的否定是特称命题,所以命题“对任意x ∈R 都有x 2≥1”的否定是:存在x 0∈R ,使得.故选:D . 点评: 本题考查全称命题的否定,注意量词以及形式的改变,基本知识的考查.4.函数f (x )=的图象大致为( )A.B.C.D.考点:函数的图象.专题:函数的性质及应用.分析:先研究函数的性质,可以发现它是一个奇函数,再研究函数在原点附近的函数值的符号,从而即可得出正确选项.解答:解:此函数是一个奇函数,故可排除B,D两个选项;又当自变量从原点左侧趋近于原点时,函数值为负,图象在X轴下方,当自变量从原点右侧趋近于原点时,函数值为正,图象在x轴上方,故可排除B,A选项符合,故选A.点评:本题考查由函数的性质确定函数图象,其研究规律一般是先研究单调性与奇偶性,再研究某些特殊值.5.定义在R上的奇函数f(x)满足f(x+1)=f(﹣x),当x∈(0,]时,f(x)=(1﹣x),则f(x)在区间(1,)内是()A.减函数且f(x)>0 B.减函数且f(x)<0 C.增函数且f(x)>0 D.增函数且f(x)<0考点:奇偶性与单调性的综合.专题:函数的性质及应用.分析:根据条件推出函数的周期性,利用函数的周期性得:f(x)在(1,)上图象和在(﹣1,﹣)上的图象相同,利用条件、奇偶性、对数函数单调性之间的关系即可得到结论.解答:解;因为定义在R上的奇函数满足f(x+1)=f(﹣x),所以f(x+1)=﹣f(x),即f(x+2)=﹣f(x+1)=f(x),所以函数的周期是2,则f(x)在(1,)上图象和在(﹣1,﹣)上的图象相同,设x∈(﹣1,﹣),则x+1∈(0,),又当x∈(0,]时,f(x)=(1﹣x),所以f(x+1)=(﹣x),由f(x+1)=f(﹣x)得,f(﹣x)=(﹣x),所以f(x)=﹣f(﹣x)=﹣(﹣x),由x∈(﹣1,﹣)得,f(x)=﹣(﹣x)在(﹣1,﹣)上是减函数,且f(x)<f(﹣1)=0,所以则f(x)在区间(1,)内是减函数且f(x)<0,故选:B.点评:本题主要考查函数奇偶性和单调性的应用,利用条件推出函数的周期性是解决本题的关键,综合考查函数性质的综合应用,考查了转化思想.6.已知向量=(λ+1,1),=(λ+2,2),若(+)⊥(﹣),则λ=()A.﹣4 B.﹣3 C.﹣2 D.﹣1考点:数量积判断两个平面向量的垂直关系.专题:平面向量及应用.分析:利用向量的运算法则、向量垂直与数量积的关系即可得出.解答:解:∵,.∴=(2λ+3,3),.∵,∴=0,∴﹣(2λ+3)﹣3=0,解得λ=﹣3.故选B.点评:熟练掌握向量的运算法则、向量垂直与数量积的关系是解题的关键.7.已知实数x,y满足约束条件,则z=x+3y的最大值等于()A. 9 B. 12 C. 27 D. 36考点:简单线性规划.专题:不等式的解法及应用.分析:由约束条件作出可行域,数形结合得到最优解,联立方程组求出最优解的坐标,代入目标函数得答案.解答:解:由约束条件作出可行域如图,联立,解得:A(3,3),化目标函数z=x+3y为,由图可知,当直线过A时,直线在y轴上的截距最大,z最大.此时z=3+3×3=12.故选:B.点评:本题考查了简单的线性规划,考查了数形结合的解题思想方法,是中档题.8.已知两条直线l1:y=m和l2:y=(m>0,m≠),l1与函数y=|log2x|的图象从左至右相交于点A、B,l2与函数y=|log2x|的图象从左至右相交于点C、D.记线段AC和BD在x轴上的投影长度分别为a,b,当m变化时,的最小值为()A. 16 B. 8 C. 4 D. 2考点:对数函数的图像与性质.专题:函数的性质及应用.分析:由题意设A,B,C,D各点的横坐标分别为x A,x B,x C,x D,依题意可求得为x A,x B,x C,x D的值,a=|x A﹣x C|,b=|x B﹣x D|,下面利用基本不等式可求最小值解答:解:设A,B,C,D各点的横坐标分别为x A,x B,x C,x D,则﹣log2x A=m,log2x B=m;﹣log2x C=,log2x D=;∴x A=2﹣m,x B=2m,x C=,x D=.∴a=|x A﹣x C|,b=|x B﹣x D|,∴==又m>0,∴m+=m+1+﹣1≥2﹣1=4﹣1=3,当且仅当m=1时取“=”号,∴≥23=8,故选:B.点评:本题考查对数函数图象与性质的综合应用,理解投影的概念并能把问题转化为基本不等式求最值是解决问题的关键,属中档题.9.已知某四棱锥的三视图(单位:cm)如图所示,则该四棱锥的体积是()A.B.C.D.考点:由三视图求面积、体积.专题:空间位置关系与距离.分析:由已知中的三视图我们要以判断出几何体为一个四棱锥,且由图中标识的数据,可以判断出几何体的棱长,高等几何量值,代入棱锥体积公式,可得答案.解答:解:由已知中的三视图可得该几何体是一个以正视图为底的四棱锥底面面积S=4×(1+1)=8高h=故该四棱锥的体积V=Sh=故选C点评:本题考查的知识点是由三视图求体积,其中根据已知条件判断出几何体的几何形状及棱长,高等几何量值,是解答的关键.10.阅读如图的程序框图,若运行相应的程序,则输出的S的值是()A. 39 B. 21 C. 81 D. 102考点:循环结构.专题:图表型.分析:用列举法,通过循环过程直接得出S与n的值,得到n=4时退出循环,即可.解答:解:第一次循环,S=3,n=2;第二次循环,S=3+2×32=21,n=3;第三次循环,S=21+3×33=102,n=4;第四次循环,不满足条件,输出S=21+3×33=102,故选D.点评:本题考查循环结构,判断框中n=4退出循环是解题的关键,考查计算能力.11.设a=log37,b=21.1,c=0.83.1,则()A. b<a<c B. c<a<b C. c<b<a D. a<c<b考点:对数值大小的比较.专题:函数的性质及应用.分析:分别讨论a,b,c的取值范围,即可比较大小.解答:解:1<log37<2,b=21.1>2,c=0.83.1<1,则c<a<b,故选:B.点评:本题主要考查函数值的大小比较,根据指数和对数的性质即可得到结论.12.下列说法正确的是()A.若a>b,则<B.函数f(x)=e x﹣2的零点落在区间(0,1)内C.函数f(x)=x+的最小值为2D.若m=4,则直线2x+my+1=0与直线mx+8y+2=0互相平行考点:命题的真假判断与应用.专题:综合题.分析:A中取特值,a正b负即可判断;B中由根的存在性定理只需判断f(0)f(1)的符号;C中注意检验基本不等式求最值时等号成立的条件;D中可先求出“直线2x+my+1=0与直线mx+8y+2=0互相平行”的充要条件.解答:解:若a=1,b=﹣1,不等式不成立,排除A;f(0)•f(1)=﹣2(e﹣2)<0,而且函数f(x)在区间(0,1)内单增,所以f(x)在区间(0,1)内存在唯一零点,B正确;令x=﹣1,则f(x)=﹣2,不满足题意,C错;若m=4,则直线重合,D错;故选:B.点评:本题考查不等式性质、基本不等式求最值、函数的零点问题、充要条件的判断等知识,考查知识点较多,属于中档题.二、填空题:共4题每题4分共16分13.已知集合A={x|0<x<3},B={x|x﹣1<0},则A∩B={x|0<x<1} .考点:交集及其运算.专题:集合.分析:根据题意和交集的运算求出A∩B.解答:解:因为集合A={x|0<x<3},B={x|x﹣1<0}={x|x<1},所以A∩B={x|0<x<1},故答案为:{x|0<x<1}.点评:本题考查交集及其运算,属于基础题.14.已知函数f(x)=asin2x+cos(2x+)的最大值为1,则a= 0或.考点:三角函数中的恒等变换应用.专题:三角函数的图像与性质.分析:首先,借助于两角和的余弦公式,展开cos(2x+),然后,利用辅助角公式,得到f(x)=sin(2x+θ),再利用最值关系式,求解a的取值.解答:解:∵函数f(x)=asin2x+cos(2x+)=asin2x+cos2xcos﹣sin2xsin=asin2x+cos2x﹣sin2x=(a﹣)sin2x+cos2x=sin(2x+θ),∴f(x)max═∴=1,两边平方,得(a﹣)2+=1,∴|a﹣|=,∴a=0或.故答案为:0或.点评:本题综合考查了两角和与差的三角公式及其灵活运用,辅助角公式,三角函数的最值等知识,考查比较综合,属于中档题.15.已知与的夹角为120°,若(+)⊥(﹣2)且||=2,则在上的投影为﹣.考点:平面向量数量积的运算.专题:平面向量及应用.分析:因为向量与的夹角为120°,所以在上的投影为cos120°=﹣,问题转化为求.解答:解:∵与的夹角为120°,若(+)⊥(﹣2)且||=2,∴(+)•(﹣2)=0,即﹣﹣22=0,∴4+﹣22=0,解得=,∴在上的投影为cos120°=﹣=﹣×=﹣.故答案为:﹣.点评:本题考查在上的投影的求法,是基础题,解题时要认真审题,注意向量垂直的性质的合理运用.16.如图展示了一个由区间(0,1)到实数集R的映射过程:区间(0,1)中的实数m对应数轴上的点M,如图1;将线段AB围成一个圆,使两端点A,B恰好重合(点M从点A按逆时针方向运动至点B),如图2;再将这个圆放在平面直角坐标系中,使其圆心在y轴上,点A的坐标为(0,1),如图3.图3中直线AM与x轴交于点N(n,0),则m的象就是n,记作f(m)=n.下列说法中正确命题的序号是②③⑤.(填出所有正确命题的序号)①f()=1;②f(x)在定义域上单调递增;③方程f(x)=0的解是x=;④f(x)是奇函数;⑤f(x)的图象关于点(,0)对称.考点:进行简单的合情推理.专题:阅读型;函数的性质及应用;推理和证明.分析:由题中对映射运算描述,对五个命题逐一判断其真伪,①m=此时M恰好处在左半圆弧的中点上,求出直线AM的方程后易得N的横坐标,即可判断;②可由图3,由M的运动规律观察出函数值的变化,得出单调性,即可判断;③可由②的单调性,结合图3即可判断;④可由奇偶函数的定义域关于原点对称来确定正误;④可由图3中圆关于y轴的对称判断出正误.解答:解:对于①,因为当m=,此时M恰好处在左半圆弧的中点上,此时直线AM的方程为y=x+1,即f()=﹣1,故①错;对于②,当x从0→1变化时,点N从左边向右边移动,其对应的坐标值渐渐增大,故f(x)在定义域上单调递增,故②正确.对于③,由②f(x)在定义域上单调递增,则M运动到AB的中点,即有直线AM为x=0,即有f()=0,故③正确;对于④,由于函数f(x)的定义域为(0,1),不关于原点对称,则函数f(x)是非奇非偶函数,故④错.对于⑤,由图3可以看出,当M点的位置离中间位置相等时,N点关于y轴对称,即此时函数值互为相反数,故可知f(x)的图象关于点(,0)对称,故⑤正确.故答案为:②③⑤.点评:本题考查映射的概念,解答本题关键是理解题设中所给的对应关系,正确认识三个图象的意义,由此对五个命题的正误作出判断,本题题型新颖,寓数于形,是一个考查理解能力的题,对题设中所给的关系进行探究,方可得出正确答案,本题易因为理解不了题意而导致无法下手,题目较抽象.三、解答题:共6题每题12分共74分17.在△ABC中,内角A,B,C对边的边长分别是a,b,c.已知c=2,C=.(Ⅰ)若△ABC的面积等于,试判断△ABC的形状,并说明理由;(Ⅱ)若sinC+sin(B﹣A)=2sin2A,求△ABC的面积.考点:正弦定理.专题:解三角形.分析:(Ⅰ)△ABC为等边三角形,理由为:利用余弦定理列出关系式,把c,cosC的值代入得到关系式,再由△ABC的面积等于,利用三角形面积公式列出关系式,两式联立求出a与b的值,即可对于△ABC的形状做出判断;(Ⅱ)已知等式利用诱导公式及二倍角的正弦函数公式化简,再利用和差化积公式变形,由cosA为0与cosA不为0两种情况,分别求出三角形ABC面积即可.解答:解:(Ⅰ)△ABC为等边三角形,理由为:∵c=2,C=,∴由余弦定理得:c2=a2+b2﹣2abcosC,即a2+b2﹣ab=4①,∵△ABC的面积等于②,∴absinC=,即ab=4,联立①②解得:a=b=2,则△ABC为等边三角形;(Ⅱ)由sinC+sin(B﹣A)=2sin2A,变形得:sin(B+A)+sin(B﹣A)=4sinAcosA,即sinBcosA=2sinAcosA,若cosA=0,即A=,由c=2,C=,得b=,此时△ABC面积S=bc=;若cosA≠0,可得sinB=2sinA,由正弦定理得:b=2a③,联立①③得:a=,b=,此时△ABC面积为S=absinC=.点评:此题考查了正弦定理,余弦定理,以及三角形面积公式,熟练掌握正弦定理是解本题的关键.18.己知数列{a n}满足a1=1,a n+1=(n∈N*),(Ⅰ)证明数列{ }是等差数列;(Ⅱ)求数列{a n)的通项公式;(Ⅲ)设b n=n(n+1)a n求数列{b n}的前n项和S n.考点:数列的求和;等差关系的确定.专题:等差数列与等比数列.分析:(Ⅰ)由a n+1=(n∈N*)变形两边取倒数即可得出;(Ⅱ)由(I)利用等差数列的通项公式即可得出;(Ⅲ)由(Ⅱ)知,b n=n(n+1)a n=n•2n,利用“错位相减法”和等比数列的前n项和公式即可得出.解答:解:(Ⅰ)∵数列{a n}满足a1=1,a n+1=(n∈N*),∴,即,∴数列是公差为1的等差数列.(Ⅱ)由(Ⅰ)可得=n+1,∴.(Ⅲ)由(Ⅱ)知,b n=n(n+1)a n=n•2n,∴S n=1×2+2×22+3×23+…+n•2n,2S n=22+2×23+…+(n﹣1)•2n+n•2n+1,两式相减得:﹣S n=2+22+…+2n﹣n•2n+1=﹣n•2n+1=(1﹣n)•2n+1﹣2,∴S n=(n﹣1)•2n+1+2.点评:本题考查了等差数列与等比数列的通项公式及其前n项和公式、“错位相减法”,考查了变形的能力,考查了推理能力与计算能力,属于难题.19.如图,四边形ABCD是正方形,EA⊥平面ABCD,EA∥PD,AD=PD=2EA,F,G,H分别为PB,EB,PC的中点.(1)求证:FG∥平面PED;(2)求平面FGH与平面PBC所成锐二面角的大小.考点:用空间向量求平面间的夹角;直线与平面平行的判定;与二面角有关的立体几何综合题.专题:综合题;空间位置关系与距离;空间角;空间向量及应用.分析:(1)利用三角形的中位线的性质证明FG∥PE,再根据直线和平面平行的判定定理证得结论;(2)建立空间直角坐标系,根据两个平面的法向量所成的角与二面角相等或互补,由两个平面法向量所成的角求解二面角的大小解答:(1)证明:∵F,G分别为PB,BE的中点,∴FG∥PE,∵FG⊄平面PED,PE⊂平面PED,∴FG∥平面PED;(2)解:∵EA⊥平面ABCD,EA∥PD,∴PD⊥平面ABCD,∵AD,CD⊂平面ABCD,∴PD⊥AD,PD⊥CD.∵四边形ABCD是正方形,∴AD⊥CD.以D为原点,建立如图所示的空间直角坐标系,设EA=1∵AD=PD=2EA,∴D(0,0,0),P(0,0,2),A(2,0,0),C(0,2,0),B(2,2,0),E(2,0,1),∴=(2,2,﹣2),=(0,2,﹣2).∵F,G,H分别为PB,EB,PC的中点,∴F(1,1,1),G(2,1,0.5),H(0,1,1),∴=(﹣1,0,0.5),=(﹣2,0,0.5)设=(x,y,z)为平面FGH的一个法向量,则,得=(0,1,0)同理可得平面PBC的一个法向量为=(0,1,1),∴cos<,>=||=,∴平面FGH与平面PBC所成锐二面角的大小为45°.点评:本题考查了线面平行的判定,考查了面面角,训练了利用平面法向量求解二面角的大小,解答此类问题的关键是正确建系,准确求用到的点的坐标,此题是中档题.20.甲、乙两人参加一次英语口语考试,已知在备选的10道试题中,甲能答对其中的6题,乙能答对其中的8题.规定每次考试都从备选题中随机抽出3题进行测试,至少答对2题才算合格.(Ⅰ)分别求甲、乙两人考试合格的概率;(Ⅱ)求甲、乙两人至少有一人考试合格的概率.考点:相互独立事件的概率乘法公式;古典概型及其概率计算公式.专题:计算题.分析:(Ⅰ)设甲乙两人考试合格分别为事件A、B;根据题意,由排列、组合公式,易得答案,(Ⅱ)因为事件A、B相互独立,先计算“甲、乙两人考试均不合格的概率”,由“甲、乙两人考试均不合格”与“甲、乙两人至少有一人考试合格”为对立事件,根据独立事件的概率公式,计算可得答案.解答:解:(Ⅰ)设甲乙两人考试合格分别为事件A、B,则P(A)===,P(B)===;答:甲乙两人考试合格的概率分别为和;(Ⅱ)因为事件A、B相互独立,所以甲、乙两人考试均不合格的概率为P(•)=P()•P()=(1﹣)(1﹣)=,甲乙两人至少有一人考试合格的概率为P=1﹣P(•)=1﹣=;答:甲、乙两人至少有一人考试合格的概率为.点评:本题考查对立事件、相互独立事件的概率计算,为了简化计算,一般把“至少”、“最多”一类的问题转化为对立事件,由其公式,计算可得答案.21.已知抛物线C:y=mx2(m>0),焦点为F,直线2x﹣y+2=0交抛物线C于A、B两点,P 是线段AB的中点,过P作x轴的垂线交抛物线C于点Q,(1)若抛物线C上有一点R(x R,2)到焦点F的距离为3,求此时m的值;(2)是否存在实数m,使△ABQ是以Q为直角顶点的直角三角形?若存在,求出m的值;若不存在,说明理由.考点:直线与圆锥曲线的综合问题.专题:计算题;压轴题.分析:(1)先求出焦点坐标,再利用抛物线的定义把焦点F的距离为3转化为到准线的距离为3即可求m的值;(也可以直接利用两点间的距离公式求解.)(2)△ABQ是以Q为直角顶点的直角三角形即是,把直线方程和抛物线方程联立,可以得到A,B两点的坐标进而求得P以及Q的坐标,代入,即可求出m的值.解答:解:(1)∵抛物线C的焦点,∴,得.(2)联立方程,消去y得mx2﹣2x﹣2=0,设A(x1,mx12),B(x2,mx22),则(*),∵P是线段AB的中点,∴,即,∴,得,若存在实数m,使△ABQ是以Q为直角顶点的直角三角形,则,即,结合(*)化简得,即2m2﹣3m﹣2=0,∴m=2或(舍去),∴存在实数m=2,使△ABQ是以Q为直角顶点的直角三角形.点评:本题考查抛物线的应用以及直线与抛物线的综合问题.解决本题的关键是看清题中给出的条件,灵活运用韦达定理,中点坐标公式进行求解.22.定义在R上的函数y=f(x),f(0)≠0,当x>0时,f(x)>1,且对任意的a、b∈R,有f(a+b)=f(a)•f(b).(1)求证:f(0)=1;(2)求证:对任意的x∈R,恒有f(x)>0;(3)求证:f(x)是R上的增函数;(4)若f(x)•f(2x﹣x2)>1,求x的取值范围.考点:抽象函数及其应用;函数单调性的判断与证明.专题:计算题;证明题.分析:(1)利用赋值法解决,令x=y=0即得;(2)利用条件:“当x>0时,f(x)>1”,只须证明当x≤0时,f(x)>0即可;(3)利用单调函数的定义证明,设x1<x2,将f(x2)写成f[(x2﹣x1)+x1]的形式后展开,结合(2)的结论即可证得;(4)由f(x)•f(2x﹣x2)>f(0)得f(3x﹣x2)>f(0).结合f(x)的单调性去掉符号“f”后,转化成一元二次不等式解决即可.解答:(1)证明:令a=b=0,则f(0)=f2(0).又f(0)≠0,∴f(0)=1.(2)证明:当x≤0时,﹣x>0,∴f(0)=f(x)•f(﹣x)=1.∴f(﹣x)=>0.又x>0时f(x)≥1>0,∴x∈R时,恒有f(x)>0.(3)证明:设x1<x2,则x2﹣x1>0.∴f(x2)=f(x2﹣x1+x1)=f(x2﹣x1)•f(x1).∵x2﹣x1>0,∴f(x2﹣x1)>1.又f(x1)>0,∴f(x2﹣x1)•f(x1)>f(x1).∴f(x2)>f(x1).∴f(x)是R上的增函数.(4)解:由f(x)•f(2x﹣x2)>1,f(0)=1得f(3x﹣x2)>f(0).又f(x)是R上的增函数,∴3x﹣x2>0,∴0<x<3.点评:本题主要考查抽象函数及其应用、函数单调性的判断与证明.解本题的关键是灵活应用题目条件,尤其是(3)中“f(x2)=f[(x2﹣x1)+x1]”是证明单调性的关键,这里体现了向条件化归的策略.。
四川省达州市大竹县文星中学高三数学下学期开学调研考试试题 文
四川省达州市大竹县文星中学2015届高三数学下学期开学调研考试试题 文考试时间:120分钟;满分150分 第I 卷(选择题)一、选择题:共12题 每题5分 共60分 1.设P={x ︱x <4},Q={x ︱2x <4},则 A.Q P ⊆ B.P Q ⊆C.Q C P R ⊆D.P C Q R ⊆【答案】B【解析】本题考查集合间的基本关系。
Q={x ︱22x -<<},所以P Q ⊆。
选B 。
2.|(3+2i)-(4-i)|等于( )A.58 B .10 C .2 D .-1+3i【答案】 B【解析】 原式=|-1+3i|=-12+32=10.3.命题“对任意x R ∈都有21x ≥”的否定是 A.对任意x R ∈,都有21x < B.不存在x R ∈,使得21x < C.存在0x R ∈,使得201x ≥ D.存在0x R ∈,使得201x <【答案】D【解析】本题考查本题考查全称量词与存在量词。
根据全称命题的否定是特称命题,所以命题“对任意x ∈R 都有21x ≥”的否定是:存在,使得.所以选D.4.函数()2sin 1xf x x =+的图象大致为【答案】A【解析】本题考查三角函数的图像和奇函数的图像性质。
首先由()f x 为奇函数,得()2sin 1xf x x =+的图象关于原点对称,排除C 、D ,又由0πx <<时,()0f x >知,所以选A.5.定义在R 上的奇函数()f x 满足(1)()f x f x +=-,当1(0,]2x ∈时,12()log (1)f x x =-,则()f x 在区间3(1,)2内是 A.减函数且()0f x > B.减函数且()0f x < C.增函数且()0f x > D.增函数且()0f x <【答案】B【解析】本题主要考查函数的奇偶性和单调性。
由此可知函数的周期为2,根据复合函数判断可知函数利用函数和周期性可知B 正确.6.已知向量m =(λ+1,1),n =(λ+2,2),若(m +n )⊥(m -n ),则λ= A.-4 B.-3C.-2D.- 1【答案】B【解析】本题考查平面向量的数量积。
四川省大竹县文星中学高三数学12月月考试卷(含解析)
四川省大竹县文星中学2016届高三12月月考数学试题一、单选题1.已知集合A={x|x2-x-2<0},集合,则下列结论正确的是A.A=BB.ABC.BAD.A∩B=【答案】C【解析】本题考查集合的基本运算.因为,,所以B A.选C.【备注】集合的基本运算为高考常考题型,要求熟练掌握.2.已知函数,命题p:∀x∈[0,+∞),f(x)≤1,则A.p是假命题,p:x0∈[0,+∞),f(x0)>1B.p是假命题,p:x∈[0,+∞),f(x)≥1C.p是真命题,p:x0∈[0,+∞),f(x0)>1D.p是真命题,p:x∈[0,+∞),f(x)≥1【答案】C【解析】本题考查命题及其关系,全称量词与特称量词.函数,命题p:∀x∈[0,+∞),f(x)≤1,p是真命题, p:x0∈[0,+∞),f(x0)>1.选C.【备注】全称命题的否定是特称命题.3.“函数在区间(0,+∞)上为增函数”是“=3”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【答案】B【解析】本题考查充分条件与必要条件的判定、对数函数的单调性,意在考查考生的分析理解能力.由函数在区间(0,+∞)上为增函数,则,故是“=3”的必要不充分条件.故本题正确答案为B.4.若抛物线的焦点与双曲线的一个焦点重合,则的值为A.1B.-1C.2D.4【答案】A【解析】本题考查抛物线和双曲线的简单几何性质,意在考查考生的运算求解能力.由抛物线的焦点,双曲线的右焦点为,得,.故本题正确答案为A.5.已知函数定义域是,则y=f(2|x|-1)的定义域是A. B.[-1,4] C. D.【答案】C【解析】本题考查复合函数的定义域的求法,意在考查考生的运算求解能力. 函数定义域是,则,即,故函数定义域为,则得.故本题正确答案为C.6.若函数A.5B.4C.3D.2【答案】B【解析】本题考查三角函数的图像与性质.设函数的周期为,由图知;因为,所以.选B.【备注】“知图求式”.7.如图,在边长为1的正三角形中,分别为边上的动点,且满足 ,,其中分别是的中点,则的最小值为A. B. C. D.【答案】C【解析】本题主要考查向量的基本运算.==,(,当取得最小值,所以故选C.【备注】平面向量基本定理是重点.8.已知双曲线的右焦点为,直线与一条渐近线交于点A,若的面积为(O为原点),则抛物线的准线方程为A. B. C. D.【答案】A【解析】本题主要考查双曲线和抛物线的基本性质.不妨设直线与渐近线y=交于点A,则A(,所以的面积为,解得a=b,所以抛物线的准线方程为x=-=-1,故选A.【备注】抛物线的考查重点是定义.9.某几何体的三视图(单位:)如图所示,其中侧视图是一个边长为2的正三角形,则这个几何体的体积是A. B. C. D.【答案】B【解析】本题主要考查简单几何体的三视图.由三视图可知,该几何体是底面为直角梯形的直四棱锥,且底面直角梯形的上底为1,下底为2,高为2,棱锥的高为=【备注】高考常考题,需熟练掌握.10.已知函数若互不相等,且,则的取值范围是A. B. C. D.【答案】C【解析】本题考查分段函数的图像、函数的零点,意在考查考生的数形结合思想及分析理解能力.作出函数图像,设,故利用三角函数的对称性得,由得,故.故本题正确答案为C.11.下列命题中正确的是A.函数是奇函数B.函数在区间上是单调递增的C.函数的最小值是D.函数是最小正周期为2的奇函数【答案】C【解析】本题主要考查本题主考查三角函数的奇偶性、单调性、周期与最值,考查了分析问题与解决问题的能力.区间不关于原点对称,所以A错误;函数在区间上是单调递减,所以B错误;因为,所以函数的最小值是,则C正确;函数是最小正周期为1的奇函数,故D错误.12.如图,设D是图中边长分别为1和2的矩形区域,E是D内位于函数图象下方的区域(阴影部分),从D内随机取一个点M,则点M取自E内的概率为A. B. C. D.【答案】C【解析】本题主要考查利用积分求面积以及几何概型概率的求法,考查了分析问题与解决问题的能力.图中矩形的面积为2,利用积分求出阴影的面积为==,利用几何概率公式可得点M 取自E内的概率为.二、填空题13.已知集合,N=,若,则的值是_______【答案】【解析】本题考查集合的运算,意在考查考生的运算求解能力. 集合,若,得集合N可能为,由N=,当得,适合条件,若,则得,若,则得,综上,a的值为.故本题正确答案为.14.已知角θ的顶点为坐标原点,始边为x轴非负半轴,若P(4,y)是角θ终边上一点,且sin θ=,则y=________【答案】-8【解析】本题考查任意角的三角函数,意在考查考生的分析理解能力.依题意,sin θ=,且点P(4,y),得,则得.故本题正确答案为-8.15.若是偶函数,则___________.【答案】【解析】本题考查函数的性质,对数函数.因为是偶函数,所以,即,解得.【备注】特殊值代入,事半功倍.16.某程序框图如图所示,则输出的S的值是______________.【答案】【解析】本题主要考查程序框图.第一次循环结束:S=1,k=2;第二次循环结束:S=,k=3;第三次循环结束:S=,k=4;第四次循环结束:S=,k=5;此时k<5不成立,循环结束,输出【备注】高考常考题,需要熟练掌握.三、解答题17.已知向量,,设函数.(Ⅰ)求的单调递增区间;(Ⅱ)求在上的最大值和最小值.【答案】(Ⅰ)===.当时,解得,的单调递增区间为.(Ⅱ)当时,,由标准函数在上的图像知==.所以在上的最大值和最小值分别为.【解析】本题考查平面向量数量积、三角恒等变换、三角函数单调性、函数在区间上的最值,意在考查考生的运算求解能力.(Ⅰ)利用平面向量数量积公式结合三角恒等变换求得,从而求得单调区间.(Ⅱ)利用整体思想求得函数的最值.18.已知数列的前项和为满足且.(1) 令证明:;(2) 求的通项公式.【答案】(1),,.(2) , ,累加得,经检验,符合【解析】本题主要考查数列的递推公式、、累加法,考查了分析问题与解决问题的能力.(1)由可得,又因为则结论得证;(2)根据利用累加法可得,则可以求得,利用即可求得的通项公式.19.函数(且)是定义在实数集上的奇函数.(1)若,试求不等式的解集;(2)若且在上的最小值为,求的值.【答案】(1)是定义在R上的奇函数,,又且易知f(x)在R上单调递增,原不等式化为:,即,不等式的解集为.(2),即(舍去)令∵x≥1,∴t≥f(1)=,∴g(t)=-2mt+2=+2-当时,当时,当时,当时,,解得,舍去.综上可知.【解析】本题考查函数的奇偶性,函数的单调性,二次函数的最值,意在考查考生的分类讨论思想及分析问题与解决问题的能力及运算求解能力.(1)利用函数为奇函数及增函数将原不等式化为,从而求得不等式的解集.(2)利用求得a的值,令,得g(t)=-2mt+2=+2-,分类讨论求得其最小值,从而求得m的值.20.某批发市场对某种商品的日销售量(单位:吨)进行统计,最近50天的统计结果如下:若以上表中频率作为概率,且每天的销售量相互独立.(Ⅰ)求5天中该种商品恰好有两天的销售量为1.5吨的概率;(Ⅱ)已知每吨该商品的销售利润为2千元,表示该种商品某两天销售利润的和(单位:千元),求的分布列和数学期望.【答案】解:(Ⅰ),,依题意,随机选取一天,销售量为吨的概率,设5天中该种商品有天的销售量为1.5吨,则,.(Ⅱ)的可能取值为,则:,,,,,所以的分布列为:的数学期望==.【解析】本题考查二项分布、离散型随机变量的分布列和数学期望.(Ⅰ)先求得销售量为吨的概率,然后利用二项分布求得其概率.(Ⅱ)的可能取值为,分别求得其概率,写出分布列和数学期望.21.如图,已知的两条角平分线和相交于在上,且.(1)证明:四点共圆:(2)证明:平分.【答案】(1)在中,因为,所以,因为是角平分线,所以,故,于是,即,所以四点共圆.(2)连结,则为的平分线,得,由(1)知四点共圆,所以.又,又由,且平分,可得,可得,所以平分.【解析】本题考查四点共圆,圆周角定理.【备注】三角形相似,圆周角定理,弦切角定理,切割线定理等. 22.已知,求证:(1);(2).【答案】证明:(1),,,,,,.,,,.【解析】本题考查基本不等式的应用,意在考查考生的推理论证能力.(1),从而问题得证;(2)由,得,从而问题得证.11。
四川省达州市大竹县文星中学2016届高三12月月考语文试题 Word版含答案.doc
四川省大竹县文星中学2016届高三12月月考语文试卷第I卷(选择题)一、现代文阅读阅读下面的文字,完成问题。
名士与名流孤岛名士与名流貌似相同相近,其实相距很远,他们虽然皆与名气有瓜葛,但实际上因所取人生姿态不同,而展现着两种完全不同的人生境界。
“名流”是指正在流行着的社会各界的名人群,而“名士”专指那些已在社会上出名但隐而未仕者。
名流往往以“名”为“实”,为本钱,常常抛头露面,混迹于社会各种官方的大众场合,参加种种社会活动,乐此不疲。
目的是博取更大的名声和利益,其处世姿态媚俗又积极。
名士则以名为“虚”,为“累”,淡泊名利,深居简出,尽量远离社会的各种热闹场所,避开种种抛头露面的机会,躲进小楼或大自然里,回归内心的宁静,或赋诗作文,或饮酒观鱼,或舞剑弹琴,将“小我”化入宇宙之无限中。
名士一般都是实大于名,不仅才华超群,而且心高气傲,超凡脱俗,有一种远离权贵与名利,超然于尘世以外的狂士或隐士风范。
名士们既能耐住寂寞与孤独,又是性情中人,喜欢自然山水,能够让自己的生命脉搏与大自然或自为的生活频率一起跳动。
名流也有一定的才气与悟性,他们更喜欢社会,更喜欢人群,更喜欢在社会进取中博取自己的一席之地,捞个一官半职,或挂满各种社会头衔,以赢得人们的特殊尊重。
但不少人在进击社会中,慢慢忘却了人之本性,成为公众世俗文化的代表。
这种人一旦得到权势,便想领导潮流,左右社会经济或文化局势。
但大多数的下场是“大江东去,浪淘尽”,只有学问和历史功绩都较大,或历史罪过很大的“名流”被定格了下来,成了名流的历史代表,如孔子、曹操等。
“替天行道,兴周灭纣”的姜子牙、“死谏”的魏征、有“卧龙”之称的诸葛亮三人既是名流又是名士,位居宰相,却道人自守,身在“朝”,心在“野”,以一颗遁世无我之心入世救世,创造了集名流与名士于一身的完美典范。
名士中做到了才情与品性的完美结合之人,屈指可数,而且随着人类城市化的不断推进和社会大融合的不断深入,名士将会越来越少,乃至慢慢消失。
四川省达州市大竹县文星中学2016届高三12月月考英语试题 Word版含答案
四川省大竹县文星中学2016届高三12月月考英语试题第I卷(选择题)一、单项选择1.It is generally believed that communication skills are becoming ______ it takes to be a good doctor.A.whetherB.thatC.howD.What2.He stood at the window, thinking where he his camera last.A.sawB.has seenC.had seenD.would see3.Mr. Black, as well as the professor who________ from Beijing University, ________ to attend our school meeting.e; ises; arees; ise; are4.The disease is so terrible that it made Jack’s life difficult, one problem leading to .A.the otherB.othersC.anotherD.the others5.I feel that one of my main duties a teacher is to help the students to become better learners.A.forB.likeC.asD.with6.A new study suggests that yelling at children may have consequences that go beyond of beating them. A.ones B.these C.those D.that7.Take care! Accidents _________ happen along this part of the road.A.willB.wouldC.mustD.Shall8.The world has lost one of its most respected statesmen (政治家) —Nelson Man dela, ______once said, “I learned that courage was not the _______ of fear, but the triumph over it.”A.who; absenceB.whom; presenceC.that; absenceD.whom; present9.It makes no difference you connect to the company server from the next room or the next country.A.thatB.whichC.whenD.whether10.The electronic red-packet has been so ______ that AliPay and Tencent Wechat compete against each other openly and secretly since the beginning of this year.A.popularB.convenientC.favorableD.arbitrary11.The manager’s confident words have our doubts about how the plan will be carried out.A.gotten upB.taken upC.cleared upD.given up12.In an economy income per head used to rise by barely 1% a year, current growth rates feel like a miracle.A.whenB.whoseC.thatD.where13.— Oh, my God,I left my portable computer in the taxi!—______. Let’s call the taxi company first.A.Pray for itB.Cheer upC.Forget itD.Calm down14.As a famous public figure, you have a duty to _______ yourself, especially in public places.A.focusB.behaveC.guideD.operate15."Talking with others in the real world always me embarrassed and my heart beats quickly. I never dare to look in anyone’s eyes when ," the 22-year-old said, describing her anxiety.A.make; speakingB.makes; speakingC.makes; speakD.make; speak二、完形填空Imagine waking up in a hospital bed. And the entire left side of your body isn’t movable. This was the 16I found myself in after my crash. I 17so severely for my head and neck were twisted to one side and stuck in that disturbing 18.The prognosis(预断)for my 19was not good. The doctor told my family,“I 20to say this, but he’ll be lucky to survive the next 48 hours. ”21as my family was at the news, the one thing that had been holding them together was the 22that with the proper medical treatment, I could recover. The doctor’s prognosis hit them like a hammer, 23any hope. The 24that I might die touched each of them deeply.For the next two days, my parents kept waiting outside my room. They could 25eat or sleep. The doctor’s prognosis 26heavily on their hearts. Yet with each passing hour they become slightly more hopeful that my chances of 27were a little bit better.With the 2848 hours passing, although I was once 29to the limit of my life, I 30 to quit my life, so they felt somewhat 31. Maybe the doctor had made a(n) 32. After all, doctors aren’t necessarily right. Bit by bit, hope began to return to them. 33, they still had no idea what the future held for me.My survival surprised everyone. It would not have been 34if not for my determination not to quit my life. 35will always happen around you as long as you don’t give up.16.A.despair B.situation C.place D.pain17.A.choked B.trembled C.enjoyed D.suffered18.A.position B.. action C.behavior D.height19.A.spirit B.disability C.recovery D.treatment20.A.regret B.decide C.tend D.wish21.A.Impatient B.Angry C.Confused D.Upset22.A.desire B.intention C.hope D.lie23.A.offering B.destroying C.raising D.abandoning24.A.exception B.choice C.fact D.acceptance25.A.barely B.gradually C.hopelessly D.slightly26.A.reacted B.weighed C.froze D.expanded27.A.promotion B.escape C.improvementD.survival28.A.convincing B.disappointing C.exciting D.frightening29.A.directed B.pushed C.ordered D.guided30.A.refused B.pretended C.struggled D.attempted31.A.confident B.satisfied C.relieved D.stressful32.A.joke B.guess C.apology D.mistake33.A.However B.Therefore C.Instead D.Besides34.A.alarming B.possible C.reasonable D.pessimistic35.A.Accidents B.Coincidences C.Miracles D.Successes三、阅读理解:共20题每题2分共40分Anyone can try to lead a group, but not every individual is cut out for leadership. The better leaders possess a few qualities that can mean the difference between the success and failure of the group. These are the qualities the leader of higher rank will look for when choosing a leader for a group, or when evaluating the performance of a leader. They're also the qualities team members want in a group leader, and appreciate when they find them.Take ResponsibilityGroup leaders might share tasks around a group as necessary, but eventually a group leader needs to be able to accept that responsibility lies on her shoulders. That means that if things go wrong in a group project, she's the one who must accept the consequences and work out what mistakes were made. The group leader won't always have the power to control everything group members do, but she should be ready to admit any mistakes the group has made as a result of her leadership.Concern for MembersThe group leader has a commitment to the task or project at hand, but perhaps more importantly, he has a real concern for each and every person who is part of his group. This means getting to know the strengths, weaknesses and goals of team members, as well as making time to build the group through collective activities. The group leader should make sure that everyone is included, even if an individual is new to a group.Good ListenerThe group leader needs to be able to listen to the suggestions, complaints (抱怨) and ideas of group members. Not only will this allow complaints to be addressed and potentially suitable ideas to be put into practice, but a leader who listens will also encourage group members to share their concerns and thoughts, creating an atmosphere of free speech and productivity.36.The passage is mainly about____________.A.choosing a good leader for a group.B.assessing the performance of a leader.C.qualities of a good group leader.D.team members’ appr eciation of a leader.37.A good leader should always be able to_______.A.tell when things go wrong in a group project.B.control everything group members do.C.make group members admit their mistakes.D.recognize any mistakes as his own.38.According to“Concerns for members”, which of the following is a good quality of a leader?A.fairB.reliableC.determinedD.generous39.Which of the following is not the benefit of being a good listener?A.Offer a chance for group members to express themselves freely.B.Allow the leader to put all his ideas into practice easily.C.Make it possible to deal with group members’ complaints.D.Make group members productive by sharing their thoughts.In her new book, “The Smartest Kids in the World,” Amanda Ripley, an investi gative journalist, tells the story of Tom, a high-school student from Gettysburg, Pennsylvania, who decides to spend his senior year in Warsaw, Poland. Poland is a surprising educational success story: in the past decade, the country raised students’ test scores from significantly below average to well above it. Polish kids have now outscored(超过……分数) American kids in math and science, even though Poland spends, on average, less than half as much per student as the United States does. One of the most striking differences between the high school Tom attended in Gettysburg and the one he ends up at in Warsaw is that the latter has no football team, or, for that matter, teams of any kind.Those American High Schools lavish more time and money on sports than on math is an old complaint. This is not a matter of how any given student who plays sports does in school, but of the culture and its priorities. This December, when the latest Program for International Student Assessment(PISA) results are announced, it’s saf e to predict that American high-school students will once again display their limited skills in math and reading, outscored not just by students in Poland but also by students in places like South Korea, Belgium, the Netherlands, Finland, Singapore, and Japan. Meanwhile, they will have played some very exciting football games, which will have been breathlessly written up in their hometown papers.Why does this situation continue? Well, for one thing, kids like it. And for another, according to Ripley, parents seem to like the arrangement, too. She describes a tour she took of a school in Washington D.C., which costs thirty thousand dollars a year. The tour leader—a mother with three children in the school—was asked about the school’s flaws(暇疵). When she said that the math program was weak, none of the parents taking the tour reacted. When she said that the football program was weak, the parents suddenly became concerned. “Really?” one of them asked worriedly, “What do you mean?”One of the ironies (讽刺) of the situation is that sports reveal what is possible. American kids’ performance on the field shows just how well they can do when expectations are high. It’s too bad that their test scores show the same thing.40.Tom decides to spend his senior year in Poland because _______.A.there are striking differences between the 2 countriesB.Polish kids are better at learningC.sports are not supported at schools in GettysburgD.he intends to improve his scores41.According to Paragraph 2, we know that _______.A.too much importance is placed on sports in AmericaB.little time is spent on sports in Japanese schoolsC.American high schools complain about sports timeD.PISA plays a very important role in America42.The underlined sentence in the last paragraph means _______.A.American students’ academic performance worries their parents a lotB.high expectations push up American students’ academic performanceC.low expectations result in American students’ poor PISA performancecking practice contributes to American students’ average performance43.The purpose of this article is to _______.A.draw public attention to a weakness in American school traditionB.call on American schools to learn from the Polish modelpare Polish schools with those in AmericaD.explain what is wrong with American schools and provide solutionsLoch Ness, the largest freshwater lake in the British Isles, is twenty four miles long and, at one point, one mile wide. It has an average depth of four hundred and fifty feet and at times drops close to a thousand. It is cold andmurky(浑浊的), with dangerous currents. In short, it is the perfect place to hide a monster from even the sharpest eyes of science.The Loch Ness Monster, also called Nessie, is supposedly living in this area. The earliest recorded sighting of the Loch Ness Monster was in the biography of Life of St. Columba by Adamnan in the year AD 565. The monster apparently attacked a man who was swimming in the River Ness.The monster didn’t make headlines again until August 27, 1930, when 3 fishermen reported seei ng a creature with 20 feet long approaching their boat, throwing water in the air. In 1933, after a new road was built along the edge of the Loch, the number of reports rose suddenly. Early in 1934, Author Grant, a young student, was out on his motorcycle one evening when he almost ran into the monster as it crossed the road. Grant’s description of the thing — small head, long thin neck and tail with a big body, seemed to match the appearance of the plesiosaur(蛇颈龙), an aquatic(水生的) type of dinosaur that has been extinct(已灭绝的)for 65 million years.The Loch Ness Investigation Bureau was formed in 1962 to act as a research organization for information about the creature. Even now, efforts have continued to find the monster. A great deal of information was discovered about the Loch, but they haven’t yet to produce any specific evidence of a monster.Skeptics(怀疑论者) argue that the water in the Loch is too cold for a plesiosaur to live in. They also argue that an air-breathing animal, like a whale or seal, would spend much more time on the surface than the creature seems to, and would be spotted more often.Some scientists have wondered if the sightings might be caused by an underwater wave which is known to sometimes occur in deep, long, and cold lakes, like Loch Ness. Such a wave might push debris(废弃物)to the surface that might look like a strange animal.However, none of these is identified.44.According to the skeptics, which of the following is TRUE?A.It is impossible for a monster to live in cold water.B.The Loch Ness Monster often stays under the water.C.The Loch Ness Monster is an air-breathing animal.D.There is no so-called monster in Loch Ness.45.The purpose of setting up the Loch Ness Investigation Bureau is to .A.research the plesiosaur in the Loch NessB.protect the Nessie in the lakeC.collect some information about the NessieD.catch the Loch Ness Monster46.Which of the following is the correct order for the things that happened in the passage?a. A young student met with a monster crossing the road.b. A swimmer was attacked by a monster in Loch Ness.c. A new road was built along the edge of the Loch.d. The Loch Ness Investigation Bureau was set up.e. Three fishermen saw a creature swimming towards their boat.A.b, e, c, a, dB.a,b, e, d, cC.b, d, a, c, eD.d, c, e, b, a47.We can infer from the passage that .A.Nessie is an aquatic type of dinosaurB.Nessie has a 20-foot-long bodyC.Nessie is an underwater waveD.Nessie is still a mystery48.What does this passage mainly talk about?A.The natural scenery of Loch Ness.B.The Nessie.C.Skeptics’ opinions on Loch Ness Monster.D.The Loch Ness Investigation Bureau’s research results.Wikipedia is a free-access, free content Internet encyclopedia(百科全书), supported and hosted by the non-profit Wikipedia Foundation. Wikipedia is ranked among the ten most popular websites and is considered the Internet’s largest and most popular general reference book. Now, Wikipedia is becoming Wookiepedia as scientists hope the informative website will help us reach out to intelligent life forms. So aliens can learn about the human race. Astronomers would like to beam (播送) the entire contents of Wikipedia into space in the hope of contacting aliens. They want to send messages to hundreds of star systems and planets 20 light years away using radio telescopes. The Search for Extraterrestrial Intelligence Institute, in California, wants to use powerful radio telescopes to try to reach Chewbacca and his mates in a galaxy(银河系)far away.The plans will be discussed by astronomers at the weekend as some scientists fear the reply from ET might not turn out to be friendly. Institute scientist David Black said, “One question is if there are dangerous creatures we might be drawing their attention to ourselves. Another is if we go ahead, what message should be sent? There could be many civilizations out there,but if they are all listening and no one is broadcasting or responding, then nothing will happen.”Professor Stephen Hawking, who has warned that intelligent aliens probably exist and we should keep a safe distance from them, is among those. “If aliens visit us, the outcome would be much as when Columbus landed in America which didn’t turn out well for the Native Americans,” he said.Signals from Earth’s radio and TV broadca sts have been heading out in space for some 60 years reaching around 5,000 stars.49.What does the underlined word those in the fourth paragraph refer to?A.aliens from spaceB.astronomers sending signalsC.scientists for the plansD.scientists against the plans50.What has Prof. Stephen Hawking warned?A.intelligent aliens would probably land in America.B.the plans will be in vain and nothing will happen.C.none of the civilizations would reply.D.aliens would bring disasters to the human race.51.Which can be the best title?A.Wikipedia, the most popular websiteB.Wikipedia to be beamed into spaceC.Wikipedia, aliens’ best friendD.Wikipedia to result in a disasterA great loss—Shirley Temple dies at 85February 12,2014BYDERRIKJ. LANG,Associated PressShirley Temple Black, who died on February 10th at age 85, wasn’t just a child star. She was THE child star—the sweet little girl whose shining smile helped illumine some of the darkest days the US has known during the Great Depression.It’s hard today to imagine the super star Shirley was once “America’s Little Darling”. She sang and danced her way to the top of the box office in such films as Bright Eyes, Curly Top and Heidi. By 1940, she had appeared in 43films. Temple teamed with Bill Robison in four movies, and their dance on the stairs in The Little Colonel is still a legendary film moment.In the 1930s, her name on a movie introduction assured(保证) a packed house. She inspired dolls, dresses, dishes—even a drink (alcohol-free, of course).US Presiden t Franklin D. Roosevelt once famously said that “as long as our country has Shirley Temple, we will be all right.’’Unlike so many of today’s child stars, Temple didn’t end up with her name appearing across the headlines for bad behaviors. Instead of getting her photos on front pages or struggling with drugs and alcohol, Temple went on to a second career in diplomacy (外交), including presidential appointments as ambassador to Ghana.She surprised a lot of people who doubted her with her grace, knowledge and eagerness to serve. In fact, her career in public service (20 years) was longer than her career in movies (19). The role she valued most, however, was as wife, mother, grandmother and great-grandmother.The world has lost a treasured Hollywood legend. But her movies will allow that little dynamic figure to continue charming audiences for a very long time.52.The word “illumine” in Paragraph 1 means _______.Shorten B. Sweeten C. strengthen D. brighten53.Temple, as a child movie star, can best be described as _______.A.sweet and livelyB.gentle and kindC.smart and knowledgeableD.shy and attractive54.What part did she regard as the most important in her life?A.A top movie star.B.A businesswoman.C.Her family role.D.Her diplomacy career.55.Where does this passage possibly come from?A.A biography.B.A newspaper.C.A magazine.D.A poster.四、七选五:Seeing with SoundAccording to a British news report,some blind people may finally get a chance to “see”. 56Dr. Peter Meijer,a scientist at Philips Research Laboratories in the Netherlands,has developed a new system called the vOICe.The three middle letters in vOICe stand for “Oh,I see.” Meijer’s groundbreaking technology may change the lives of visually impaired people all over the world.A wearable setup of vOICe system consists of a head-mounted camera,stereo headphones and a notebook PC. The system translates visual images from a camera into complex sounds. 57The program is based on the theory that people can hear certain sounds and learn to translate them into meaningful mental images. Everything has its own unique sound. 58The height of an object or person could be determined by pitch(音调).And a built-in color identifier speaks out color names when it is turned on. What the vOICe users have to learn is which sound goes with which object. Meijer says that he is counting on the brain’s ability to adapt.59Within two weeks,most people who experiment with the vOICe system are able to identify objects such as walls and doors. They are also able to identify certain situations,including whether the lights in a room are on or off. 60Meijer thinks that translating will eventually become automatic for many users of the vOICe,bringing a form of vision to them for the very first time.A. Brighter areas sound louder than darker areas.B. It then sends these sounds to a person through headphones.C. However, they are not learning to see with their eyes;they are learning to see with their ears.D. The vOICe system is aiming to treat blindness by translating images from a camera into audio signals.E. Meijer supposes that the brain is interested in the information “content”,but not the information “carrier”.F. To evaluate the new system’s effectiveness, a number of visually impaired people were chosen to test the v0IC e.G. Over time, some users have even learned to “watch” television or “recognize” the outlines of buildings as they walk.第II卷(非选择题)五、语法填空:Many teens in high school want to try out for sports because they think it will make them more popular. But there are lots of other benefits 61sports that teens may overlook.With TV, movies, computers, and video games 62(become)more and more popular, it has become so much 63(easy) for teens to be by themselves rather than going out with friends. Kids used to hang out at the mall or drive around the town; now they just sit at home. Getting teens into a sport gives them 64opportunity to go out and socialize. 65they may not find a new best friend, they will learn how to interact and work as a team, somethi ng they’ll find 66(use) later in life.More and more kids are becoming overweight. If teens see that their physical condition is causing them to perform 67(bad), they may 68(motivate) to do other activities to get healthy. By the time your child 69(reach) their teenager year, part of good parenting will be providing them with direction and 70(encourage) and continuing to help them develop a healthy style of living.六、书面表达:71.下面的漫画反映了一种社会现象,请你根据对这幅漫画的理解,用英语写一篇短文, 描述一下他们的不同情况,并就如何解决此问题发表你的看法。
四川省达州市大竹县文星中学2015-2016学年高一上学期12月月考数学试卷 含解析
2015—2016学年四川省达州市大竹县文星中学高一(上)12月月考数学试卷一、单选题1.已知集合,集合N={x|2x+3>0},则(∁R M)∩N=()A.[﹣)B.(﹣)C.(﹣]D.[﹣]2.函数x的定义域为()A.(﹣∞,1]B.(0,1]C.(0,1)D.[0,1]3.已知函数f(x)=,其定义域是[﹣8,﹣4),则下列说法正确的是()A.f(x)有最大值,无最小值B.f(x)有最大值,最小值C.f(x)有最大值,无最小值D.f(x)有最大值2,最小值4.设abc>0,二次函数f(x)=ax2+bx+c的图象可能是()A.B.C.D.5.已知函数f(x)=(a﹣1)x2+2ax+3为偶函数,那么f(x)在(﹣5,﹣2)上是() A.单调递增函数 B.单调递减函数 C.先减后增函数 D.先增后减函数6.偶函数y=f(x)在区间[0,4]上单调递减,则有()A.f(﹣1)>f()>f(﹣π)B.f()>f(﹣1)>f(﹣π) C.f(﹣π)>f(﹣1)>f() D.f(﹣1)>f(﹣π)>f()7.若x∈(e﹣1,1),a=lnx,,c=e lnx,则()A.b>c>a B.c>b>a C.b>a>c D.a>b>c8.已知函数f(x)和g(x)均为奇函数,h(x)=a⋅f3(x)﹣b⋅g(x)﹣2在区间(0,+∞)上有最大值5,那么h(x)在(﹣∞,0)上的最小值为()A.﹣5 B.﹣9 C.﹣7 D.﹣19.下列哪组中的函数f(x)与g(x)相等()A.f(x)=x2,B.f(x)=x+1,g(x)=+1C.f(x)=x,g(x)=D.f(x)=,g(x)=10.若f(x)=x,则不等式f(x)>f(8x﹣16)的解集是()A.B.(0,2]C.[2,+∞)D.(0,+∞)11.函数f(x)=lg(|x|﹣1)的大致图象是()A.B.C.D.12.已知函数y=f(x)与函数y=e x的图象关于直线y=x对称,函数y=g(x)的图象与y=f (x)的图象关于x轴对称,若g(a)=1,则实数a的值为()A.﹣e B. C.D.e二、填空题13.已知定义在R上的奇函数f(x),当x>0时,f(x)=x2+|x|﹣1,那么x<0时,f(x)=.14.设a为常数且a<0,y=f(x)是定义在R上的奇函数,当x<0时,f(x)=x+﹣2,若f(x)≥a2﹣1对一切x≥0都成立,则a的取值范围为.15.奇函数f(x)在区间[3,7]上是增函数,在区间[3,6]上的最大值为8,最小值为﹣1,则2f(﹣6)+f(﹣3)=.16.设全集U={(x,y)|x∈R,y∈R},集合,P={(x,y)|y≠x+1},则∁U(M∪P)=.三、解答题17.计算下列各式的值:(1)()﹣4•(﹣2)﹣3+()0﹣9;(2)2log32﹣log3+log38﹣5.18.已知f(x)=log a(a>0且a≠1),(1)判断f(x)的奇偶性;(2)求使f(x)>0的x的取值范围.19.两个重要城市之间人员交流频繁,为了缓解交通压力,特修一条专用铁路,用一列火车作为交通车.已知该车每次拖4节车厢,一日能来回16次,如果每次拖7节车厢,则每日能来回10次.(1)若每日来回的次数是车头每次拖挂车厢节数的一次函数,求此一次函数解析式;(2)在(1)的条件下,每节车厢能载乘客110人.问这列火车每天来回多少次才能使运营人数最多?并求出每天最多运营人数.20.设函数f(x)满足:①对任意实数m,n都有f(m+n)+f(m﹣n)=2f(m)⋅f(n);②对任意m∈R,都有f(1+m)=f(1﹣m)恒成立;③f(x)不恒为0,且当0<x<1时,f(x)<1.(1)求f(0),f(1)的值;(2)判断函数f(x)的奇偶性,并给出你的证明;(3)定义:“若存在非零常数T,使得对函数g(x)定义域中的任意一个x,均有g(x+T)=g(x),则称g(x)为以T为周期的周期函数”.试证明:函数f(x)为周期函数,并求出的值.21.已知A={x|﹣1<x≤3},B={x|m≤x<1+3m}(1)当m=1时,求A∪B;(2)若B⊆∁R A,求实数m的取值范围.22.已知函数f(x)=lg,(Ⅰ)若f(x)为奇函数,求a的值;(Ⅱ)若f(x)在(﹣1,5]内有意义,求a的取值范围;(Ⅲ)在(Ⅱ)的条件下,判断并证明f(x)的单调性.2015-2016学年四川省达州市大竹县文星中学高一(上)12月月考数学试卷参考答案与试题解析一、单选题1.已知集合,集合N={x|2x+3>0},则(∁R M)∩N=()A.[﹣)B.(﹣)C.(﹣]D.[﹣]【考点】交、并、补集的混合运算.【分析】分别求出集合M和N中不等式的解集,确定出M和N,由全集为R,找出不属于M的部分,求出M的补集,找出M补集与N的公共部分,即可求出所求的集合.【解答】解:由集合M中的不等式移项得:﹣1≥0,即≥0,解得:x>1,∴集合M=(1,+∞),又全集为R,∴C R M=(﹣∞,1],由集合N中的不等式2x+3>0,解得:x>﹣,∴集合N=(﹣,+∞),则(C R M)∩N=(﹣,1].故选C2.函数x的定义域为()A.(﹣∞,1]B.(0,1]C.(0,1)D.[0,1]【考点】函数的定义域及其求法.【分析】根据偶次根式下大于等于0,对数函数的真数大于0建立不等式关系,然后解之即可求出函数的定义域.【解答】解:要使原函数有意义,则解得:0<x≤1所以原函数的定义域(0,1].故选:B3.已知函数f(x)=,其定义域是[﹣8,﹣4),则下列说法正确的是()A.f(x)有最大值,无最小值 B.f(x)有最大值,最小值C.f(x)有最大值,无最小值 D.f(x)有最大值2,最小值【考点】函数的最值及其几何意义.【分析】将f(x)化为2+,判断在[﹣8,﹣4)的单调性,即可得到最值.【解答】解:函数f(x)==2+即有f(x)在[﹣8,﹣4)递减,则x=﹣8处取得最大值,且为,由x=﹣4取不到,即最小值取不到.故选A.4.设abc>0,二次函数f(x)=ax2+bx+c的图象可能是()A.B.C.D.【考点】二次函数的图象;函数的图象.【分析】分别从抛物线的开口方向,对称轴,f(0)的符号进行判断即可.【解答】解:A.抛物线开口向下,∴a<0,又f(0)=c<0.∵abc>0,∴b>0,此时对称轴x=>0,与图象不对应.B.抛物线开口向下,∴a<0,又f(0)=c>0.∵abc>0,∴b<0,此时对称轴x=<0,与图象不对应.C.抛物线开口向上,∴a>0,又f(0)=c<0.∵abc>0,∴b<0,此时对称轴x=>0,与图象不对应.D.抛物线开口向上,∴a>0,又f(0)=c<0.∵abc>0,∴b<0,此时对称轴x=>0,与图象对应.故选:D.5.已知函数f(x)=(a﹣1)x2+2ax+3为偶函数,那么f(x)在(﹣5,﹣2)上是()A.单调递增函数 B.单调递减函数 C.先减后增函数 D.先增后减函数【考点】函数奇偶性的性质.【分析】根据函数f(x)=(a﹣1)x2+2ax+3为偶函数,可得a=0,分析函数的图象和性质,可得答案【解答】解:∵函数f(x)=(a﹣1)x2+2ax+3为偶函数,∴f(﹣x)=(a﹣1)x2﹣2ax+3=f(x)=(a﹣1)x2+2ax+3,∴a=0,∴f(x)=﹣x2+3,则函数的图象是开口朝下,且以y轴为对称轴的抛物线,∴f(x)在(﹣5,﹣2)上是增函数,故选:A.6.偶函数y=f(x)在区间[0,4]上单调递减,则有()A.f(﹣1)>f()>f(﹣π) B.f()>f(﹣1)>f(﹣π)C.f(﹣π)>f(﹣1)>f()D.f(﹣1)>f(﹣π)>f()【考点】奇偶性与单调性的综合.【分析】由函数y=f(x)为偶函数,可得f(﹣x)=f(x),从而有f(﹣1)=f(1),f(﹣π)=f (π),结合函数y=f(x)在[0,4]上的单调性可比较大小【解答】解:∵函数y=f(x)为偶函数,且在[0,4]上单调递减∴f(﹣x)=f(x)∴f(﹣1)=f(1),f(﹣π)=f(π)∵1<<π∈[0,4]f(1)>f()>f(π)即f(﹣1)>f()>f(﹣π)故选A7.若x∈(e﹣1,1),a=lnx,,c=e lnx,则()A.b>c>a B.c>b>a C.b>a>c D.a>b>c【考点】对数值大小的比较.【分析】利用对数函数的单调性判断出a<0;由于b,c的指数相同,所以研究一个幂函数的单调性;利用幂函数的单调性判断出b,c的大小,b,c都是幂得到b,c全正,比较出a,b,c的大小.【解答】解:∵x∈(e﹣1,1)∴a=lnx<ln1=0即a<0考察幂函数f(t)=t lnx∵lnx<0∴当t>0时,f(t)是减函数∵∴>0所以有b>c>a故选A8.已知函数f(x)和g(x)均为奇函数,h(x)=a⋅f3(x)﹣b⋅g(x)﹣2在区间(0,+∞)上有最大值5,那么h(x)在(﹣∞,0)上的最小值为()A.﹣5 B.﹣9 C.﹣7 D.﹣1【考点】函数奇偶性的性质.【分析】根据条件构造新函数h(x)+2判断函数h(x)+2的奇偶性,结合函数奇偶性和最值之间的关系建立方程进行求解即可.【解答】解:由h(x)=a⋅f3(x)﹣b⋅g(x)﹣2得h(x)+2=a⋅f3(x)﹣b⋅g(x),∵函数f(x)和g(x)均为奇函数,∴h(x)+2=a⋅f3(x)﹣b⋅g(x)是奇函数,∵h(x)=a⋅f3(x)﹣b⋅g(x)﹣2在区间(0,+∞)上有最大值5,∴h max(x)=a⋅f3(x)﹣b⋅g(x)﹣2=5,即h max(x)+2=7,∵h(x)+2是奇函数,∴h min(x)+2=﹣7,即h min(x)=﹣7﹣2=﹣9,故选:B9.下列哪组中的函数f(x)与g(x)相等()A.f(x)=x2,B.f(x)=x+1,g(x)=+1C.f(x)=x,g(x)= D.f(x)=,g(x)=【考点】判断两个函数是否为同一函数.【分析】根据两个函数的定义域相同,对应关系也相同,判断是相等函数.【解答】解:对于A,f(x)=x2(x∈R),g(x)==x2(x≥0),它们的定义域不同,不是相等函数;对于B,f(x)=x+1(x∈R),g(x)=+1=x+1(x≠0),它们的定义域不同,不是相等函数; 对于C,f(x)=x(x∈R),g(x)==x(x∈R),它们的定义域相同,对应关系也相同,是相等函数;对于D,f(x)=(x≤﹣2x≥﹣1),g(x)==(x ≥﹣1),它们的定义域不同,不是相等函数;故选:C.10.若f(x)=x,则不等式f(x)>f(8x﹣16)的解集是()A.B.(0,2]C.[2,+∞) D.(0,+∞)【考点】幂函数的性质.【分析】先研究幂函数f(x)=x的定义域和单调性,再把函数单调性的定义和定义域相结合即可.【解答】解:由f(x)=x知,f(x)是定义在[0,+∞)上的增函数,则不等式f(x)>f(8x﹣16)得,∴2≤x<,故选A.11.函数f(x)=lg(|x|﹣1)的大致图象是()A.B.C.D.【考点】对数函数的图象与性质.【分析】利用特殊值法进行判断,先判断奇偶性;【解答】解:∵函数f(x)=lg(|x|﹣1),∴f(﹣x)=lg(|x|﹣1)=f(x),f(x)是偶函数,当x=1或﹣1时,y<0,故选B;12.已知函数y=f(x)与函数y=e x的图象关于直线y=x对称,函数y=g(x)的图象与y=f(x)的图象关于x轴对称,若g(a)=1,则实数a的值为()A.﹣e B. C.D.e【考点】指数函数的图象与性质.【分析】根据y=f(x)与y=e x的图象关于直线y=x对称,求出f(x),再根据y=g(x)的图象与y=f(x)的图象关于x轴对称,求出y=g(x),再列方程求a的值即可.【解答】解:函数y=f(x)与函数y=e x的图象关于直线y=x对称,∴f(x)=lnx,函数y=g(x)的图象与y=f(x)的图象关于x轴对称,∴y=﹣lnx,∴g(a)=﹣lna=1,a=.故选:C.二、填空题13.已知定义在R上的奇函数f(x),当x>0时,f(x)=x2+|x|﹣1,那么x<0时,f(x)=﹣x2+x+1.【考点】函数奇偶性的性质.【分析】先设x<0,则﹣x>0,代入f(x)=x2+|x|﹣1并进行化简,再利用f(x)=﹣f (﹣x)进行求解.【解答】解:设x<0,则﹣x>0,∵当x>0时,f(x)=x2+|x|﹣1,∴f(﹣x)=x2+|﹣x|﹣1=x2﹣x﹣1,∵f(x)是定义在R上的奇函数,∴f(x)=﹣f(﹣x)=﹣x2+x+1,故答案为:﹣x2+x+1.14.设a为常数且a<0,y=f(x)是定义在R上的奇函数,当x<0时,f(x)=x+﹣2,若f(x)≥a2﹣1对一切x≥0都成立,则a的取值范围为[﹣1,0).【考点】函数奇偶性的性质.【分析】通过讨论x的范围,得到不等式,解出即可求出a的范围.【解答】解:当x=0时,f(x)=0,则0≥a2﹣1,解得﹣1≤a≤1,所以﹣1≤a<0当x>0时,﹣x<0,,则由对勾函数的图象可知,当时,有f(x)min=﹣2a+2所以﹣2a+2≥a2﹣1,即a2+2a﹣3≤0,解得﹣3≤a≤1,又a<0所以﹣3≤a<0,综上所述:﹣1≤a<0,故答案为:[﹣1,0).15.奇函数f(x)在区间[3,7]上是增函数,在区间[3,6]上的最大值为8,最小值为﹣1,则2f(﹣6)+f(﹣3)=﹣15.【考点】函数单调性的性质;函数奇偶性的性质;函数的值.【分析】先利用条件找到f(3)=﹣1,f(6)=8,再利用f(x)是奇函数求出f(﹣6),f(﹣3)代入即可.【解答】解:f(x)在区间[3,6]上也为递增函数,即f(6)=8,f(3)=﹣1∴2f(﹣6)+f(﹣3)=﹣2f(6)﹣f(3)=﹣15故答案为:﹣1516.设全集U={(x,y)|x∈R,y∈R},集合,P={(x,y)|y≠x+1},则∁U(M∪P)={(2,3)} .【考点】交、并、补集的混合运算.【分析】分析可得集合M、P的几何意义,集合M为直线y=x+1中除(2,3)之外的所有点,集合P为平面直角坐标系中除直线y=x+1外的所有点;由此可得M∪P,M∪P的补集即可得答案.【解答】解:根据题意,分析可得集合M可变形为M={(x,y)|y=x+1,x≠2},即直线y=x+1中除(2,3)之外的所有点,N={(x,y)|y≠x+1},为平面直角坐标系中除直线y=x+1外的所有点;M∪P={(x,y)|x≠2,y≠3)},即平面直角坐标系中除点(2,3)之外的所有点;所以∁U(M∪P)={(2,3)}故答案是:{(2,3)}.三、解答题17.计算下列各式的值:(1)()﹣4•(﹣2)﹣3+()0﹣9;(2)2log32﹣log3+log38﹣5.【考点】对数的运算性质.【分析】(1)直接利用有理指数幂的运算法则求解即可.(2)利用对数的运算法则化简求解即可.【解答】解:(1)()﹣4•(﹣2)﹣3+()0﹣9=8++1﹣3=.(2)2log32﹣log3+log38﹣5=2log32﹣log332+log39+3log32﹣3=2log32﹣5log32+2+3log32﹣3=﹣1.18.已知f(x)=log a(a>0且a≠1),(1)判断f(x)的奇偶性;(2)求使f(x)>0的x的取值范围.【考点】函数奇偶性的判断;指、对数不等式的解法.【分析】(1)求出f(x)的定义域,再计算f(﹣x),判断f(﹣x)与f(x)的关系,即可得到;(2)讨论a>1,0<a<1,列出不等式组,解出即可得到.【解答】解:(1)f(x)为奇函数.理由如下:由>0得函数的定义域为(﹣1,1),又f(﹣x)=log a=log a()﹣1=﹣log a=﹣f(x),所以,f(x)为奇函数.(2)由题意:当0<a<1时,有解得﹣1<x<0;当a>1时,有解得0<x<1.综上,当0<a<1时,﹣1<x<0; 当a>1时,0<x<1.19.两个重要城市之间人员交流频繁,为了缓解交通压力,特修一条专用铁路,用一列火车作为交通车.已知该车每次拖4节车厢,一日能来回16次,如果每次拖7节车厢,则每日能来回10次.(1)若每日来回的次数是车头每次拖挂车厢节数的一次函数,求此一次函数解析式;(2)在(1)的条件下,每节车厢能载乘客110人.问这列火车每天来回多少次才能使运营人数最多?并求出每天最多运营人数.【考点】函数模型的选择与应用.【分析】(1)利用待定系数法,可求一次函数解析式;(2)确定函数解析式,利用配方法,可求最值.【解答】解:(1)设每日来回y次,每次挂x节车厢,由题意y=kx+b…由已知可得方程组:…解得:k=﹣2,b=24…∴…(2)设每日火车来回y次,每次挂x节车厢,设每日可营运S节车厢.由题意知,每日挂车厢最多时,营运人数最多,则S=xy=x(﹣2x+24)=﹣2x2+24x=﹣2(x﹣6)2+72…所以当x=6时,S max=72(节)…此时y=12,故每日最多运营人数为110×72=7920(人)答:这列火车每天来回12次,才能使运营人数最多,每天最多运营人数为7920人.…20.设函数f(x)满足:①对任意实数m,n都有f(m+n)+f(m﹣n)=2f(m)⋅f(n);②对任意m∈R,都有f(1+m)=f(1﹣m)恒成立;③f(x)不恒为0,且当0<x<1时,f(x)<1.(1)求f(0),f(1)的值;(2)判断函数f(x)的奇偶性,并给出你的证明;(3)定义:“若存在非零常数T,使得对函数g(x)定义域中的任意一个x,均有g(x+T)=g (x),则称g(x)为以T为周期的周期函数”.试证明:函数f(x)为周期函数,并求出的值.【考点】抽象函数及其应用.【分析】(1)在等式f(m+n)+f(m﹣n)=2f(m)⋅f(n)中,令m=x0,n=0,即可求得f(0)=1,结合f(m+n)+f(m﹣n)=2f(m)⋅f(n)、f(1+m)=f(1﹣m)、f(x)不恒为0,且当0<x<1时,f(x)<1即可求得f(1)的值;(2)在f(m+n)+f(m﹣n)=2f(m)⋅f(n)中,取m=0,n=x,以及有f(0)=1,可得函数f(x)为偶函数;(3)由f(1+m)=f(1﹣m),并取1+m=﹣x,得f(﹣x)=f(2+x),又f(x)为偶函数,可得f(x+2)=f(x),即f(x)是以2为周期的周期函数;在f(m+n)+f(m﹣n)=2f(m)⋅f(n)中,取,取m=,n=得到两个关于f()和f()的方程组,求出f()和f(),再由函数的周期性求得的值.【解答】(1)解:由于f(x)不恒为0,故存在x0,使f(x0)≠0,令m=x0,n=0,则f(x0)+f(0)=2f(x0)f(0),∴f(0)=1,令m=n=1⇒f(2)+f(0)=2f2(1),由f(1+m)=f(1﹣m)并令m=1得:f(2)=f(0),结合以上结果可得f2(1)=1,又令(因为),∴f(1)<1,故f(1)=﹣1;(2)解:f(x)为偶函数.证明如下:令m=0,n=x,得:f(x)+f(﹣x)=2f(0)f(x),以及有f(0)=1,即有f(﹣x)=f(x),即有f(x)为偶函数;(3)证明:由f(1+m)=f(1﹣m),并取1+m=﹣x,得f(﹣x)=f(2+x),又f(x)为偶函数,则f(x+2)=f(x),即f(x)是以2为周期的周期函数;令,再令m=,n=⇒⇒.而,解得,,由f(1+m)=f(1﹣m)得,,∴,又由于f(x)是以2为周期的周期函数,∴.21.已知A={x|﹣1<x≤3},B={x|m≤x<1+3m}(1)当m=1时,求A∪B;(2)若B⊆∁R A,求实数m的取值范围.【考点】并集及其运算;集合的包含关系判断及应用.【分析】(1)将m的值代入集合B中确定出B,找出既属于A又属于B的部分,即可确定出两集合的并集;(2)由全集R求出A的补集,由B为A补集的子集,列出关于m的不等式,求出不等式的解集,即可得到m的范围.【解答】解:(1)当m=1时,A={x|﹣1<x≤3},B={x|1≤x<4},则A∪B={x|﹣1<x<4};(2)∵全集为R,A={x|﹣1<x≤3},∴C R A={x|x≤﹣1或x>3},∵B⊆C R A,当B=∅时,m≥1+3m,即m≤﹣;当B≠∅时,m<1+3m,即m>﹣,此时1+3m≤﹣1或m>3,解得:m>3,综上,m的范围为m≤﹣或m>3.22.已知函数f(x)=lg,(Ⅰ)若f(x)为奇函数,求a的值;(Ⅱ)若f(x)在(﹣1,5]内有意义,求a的取值范围;(Ⅲ)在(Ⅱ)的条件下,判断并证明f(x)的单调性.【考点】对数函数图象与性质的综合应用.【分析】(Ⅰ)由f(x)为奇函数可得f(x)+f(﹣x)=0,即,即,由此可得a的值.(Ⅱ)由题意可得f(x)在(﹣1,5]上恒成立,再由x+1>0,可得a﹣x>0,故a>x在(﹣1,5]上恒成立,由此可得a的范围.(Ⅲ)当a>5时,f(x)在定义域上为减函数,求得f(x)定义域为(﹣1,a).令﹣1<x1<x2<a,由于==,可得f(x1)>f(x2),从而得出结论.【解答】解:(Ⅰ)∵f(x)为奇函数,∴f(x)+f(﹣x)=0,即,∴,∴a=1.…(Ⅱ)∵若f(x)在(﹣1,5]内恒有意义,则在(﹣1,5]上恒成立,再由x+1>0,∴a﹣x>0,∴a>x在(﹣1,5]上恒成立,∴a>5.…(Ⅲ)当a>5时,f(x)在定义域上为减函数,…由,得f(x)定义域为(﹣1,a).…令﹣1<x1<x2<a,由于==,…∵﹣1<x1<x2<a,∴a﹣x1>a﹣x2>0,1+x2>1+x1>0,∴,∴,∴,∴f(x1)﹣f(x2)>0,即f(x1)>f(x2),∴f(x1)在(﹣1,a)为减函数.…2016年11月18日。
四川省大竹县文星中学高三12月月考 (2).docx
高中化学学习材料唐玲出品四川省大竹县文星中学2016届高三12月月考化学试题第I卷(选择题)一、单选题:1.常温下,下列各组离子在指定溶液中能大量共存的是A.通入足量CO2后的溶液中:Na+、、CH3COO-、B.饱和氯水中:Cl-、、Na+、C.c(H+)=1×10-1 mol•L-1的溶液中:Cu2+、Al3+、、D.0.1 mol·L-1 CH3COONa溶液:H+、Al3+、Cl-、2.将9 g铜和铁的混合物投入到100 mL稀硝酸中,充分反应后得到标准状况下1.12 L NO,剩余4.8 g 金属;继续加入100 mL等浓度的稀硝酸,金属完全溶解,又得到标准状况下1.12 L NO。
若向反应后的溶液中加入KSCN溶液,溶液不变红,则下列说法不正确的是A.原混合物中铜和铁各0.075 molB.稀硝酸的物质的量浓度为2.0 mol·L-1C.第一次剩余4.8 g金属为铜和铁D.向反应后的溶液中再加入该稀硝酸100 mL,又得到NO在标准状况下体积小于1.12 L3.下列有关NaClO和NaCl混合溶液的叙述正确的是A.该溶液中,H+、、、Br-可以大量共存B.该溶液中,Ag+、K+、、CH3CHO可以大量共存C.向该溶液中滴入少量FeSO 4溶液,反应的离子方程式为:2Fe2++ClO-+2H+Cl-+2Fe3++H2OD.向该溶液中加入浓盐酸,每产生1molCl2,转移电子约为6.02×1023个4.向含、Fe2+、Br-、I-各0.1 mol的溶液中通入标准状况下的Cl2,通入Cl2的体积和溶液中相关离子的物质的量的关系图正确的是5.下列晶体模型对应的物质熔化时破坏共价键的是6.下列各实验装置图的叙述中,正确的是A.装置①为放出萃取溴水后的苯层B.装置②为喷泉实验C.装置③不能用来吸收HCl气体D.以NH 4 Cl为原料,装置④可制备少量NH37.下列关于氧化性、还原性的判断正确的是A.B的阳离子的氧化性比A的阳离子强,说明A元素的金属性一定比B元素强B.发生氧化还原反应时A原子失去的电子比B原子多,证明A的金属性一定比B强C.适量的Cl2通入FeI2溶液中可发生反应:3Cl2+6FeI22FeCl3+4FeI3D.一定量氯气通入30 mL 10.00 mol·L-1的氢氧化钠溶液中,加热后形成NaCl、NaClO、NaClO3共存的溶液,若反应中转移的电子为n mol,则0.15<n<0.25第II卷(非选择题)二、实验题:8.铜是与人类关系非常密切的常见金属。
四川省达州市大竹县文星中学高三数学下学期开学调研考
四川省达州市大竹县文星中学2015届高三数学下学期开学调研考试试题 文考试时间:120分钟;满分150分 第I 卷(选择题)一、选择题:共12题 每题5分 共60分 1.设P={x ︱x <4},Q={x ︱2x <4},则 A.Q P ⊆ B.P Q ⊆C.Q C P R ⊆D.P C Q R ⊆【答案】B【解析】本题考查集合间的基本关系。
Q={x ︱22x -<<},所以P Q ⊆。
选B 。
2.|(3+2i)-(4-i)|等于( )A.58 B .10 C .2 D .-1+3i【答案】 B【解析】 原式=|-1+3i|=-12+32=10.3.命题“对任意x R ∈都有21x ≥”的否定是 A.对任意x R ∈,都有21x < B.不存在x R ∈,使得21x < C.存在0x R ∈,使得201x ≥ D.存在0x R ∈,使得201x <【答案】D【解析】本题考查本题考查全称量词与存在量词。
根据全称命题的否定是特称命题,所以命题“对任意x ∈R 都有21x ≥”的否定是:存在,使得.所以选D.4.函数()2sin 1xf x x =+的图象大致为【答案】A【解析】本题考查三角函数的图像和奇函数的图像性质。
首先由()f x 为奇函数,得()2sin 1xf x x =+的图象关于原点对称,排除C 、D ,又由0πx <<时,()0f x >知,所以选A.5.定义在R 上的奇函数()f x 满足(1)()f x f x +=-,当1(0,]2x ∈时,12()log (1)f x x =-,则()f x 在区间3(1,)2内是 A.减函数且()0f x > B.减函数且()0f x < C.增函数且()0f x > D.增函数且()0f x <【答案】B【解析】本题主要考查函数的奇偶性和单调性。
由此可知函数的周期为2,根据复合函数判断可知函数利用函数和周期性可知B 正确.6.已知向量m =(λ+1,1),n =(λ+2,2),若(m +n )⊥(m -n ),则λ= A.-4 B.-3C.-2D.- 1【答案】B【解析】本题考查平面向量的数量积。
【解析】四川省大竹县文星中学2016届高三12月月考英语试题 Word版含解析[ 高考]
四川省大竹县文星中学2016届高三12月月考英语试题第I卷(选择题)一、单项选择1.It is generally believed that communication skills are becoming ______ it takes to be a good doctor.A.whetherB.thatC.howD.What【答案】D【解析】考查名词性从句。
whether“是否”,在从句中不作任何成分;that无意义且在从句中不作任何成分how“怎样”,在从句中作方式状语;w hat“什么;……的东西”,在从句中作主语,宾语或表语。
“__it takes to be a good doctor”是表语从句,从句中缺少takes的宾语,故用what来引导。
故选D。
2.He stood at the window, thinking where he his camera last.A.sawB.has seenC.had seenD.would see【答案】A【解析】考查时态。
语意:他站在窗子跟前,想着他上一次是在哪儿见到他的相机的。
根据句中的时间状语last可判断出应用一般过去时。
3.Mr. Black, as well as the professor who________ from Beijing University, ________ to attend our school meeting.e; ises; arees; ise; are【答案】C【解析】考查主谓一致。
第一个空处主语为定语从句中的who,指代单数名词professor,谓语用单数。
第二个空处主语为Mr. Black,为单数,谓语用单数。
故选C。
4.The disease is so terrible that it made Jack’s life difficult, one problem leading to .A.the otherB.othersC.anotherD.the others【答案】C【解析】考查代词。
四川省达州市大竹县文星中学高三数学下学期开学调研考试试题 文
四川省达州市大竹县文星中学2015届高三数学下学期开学调研考试试题 文考试时间:120分钟;满分150分 第I 卷(选择题)一、选择题:共12题 每题5分 共60分 1.设P={x ︱x <4},Q={x ︱2x <4},则 A.Q P ⊆ B.P Q ⊆C.Q C P R ⊆D.P C Q R ⊆【答案】B【解析】本题考查集合间的基本关系。
Q={x ︱22x -<<},所以P Q ⊆。
选B 。
2.|(3+2i)-(4-i)|等于( )A.58 B .10 C .2 D .-1+3i【答案】 B【解析】 原式=|-1+3i|=-2+32=10.3.命题“对任意x R ∈都有21x ≥”的否定是 A.对任意x R ∈,都有21x < B.不存在x R ∈,使得21x < C.存在0x R ∈,使得201x ≥ D.存在0x R ∈,使得201x <【答案】D【解析】本题考查本题考查全称量词与存在量词。
根据全称命题的否定是特称命题,所以命题“对任意x ∈R 都有21x ≥”的否定是:存在,使得.所以选D.4.函数()2sin 1xf x x =+的图象大致为【答案】A【解析】本题考查三角函数的图像和奇函数的图像性质。
首先由()f x 为奇函数,得()2sin 1xf x x =+的图象关于原点对称,排除C 、D ,又由0πx <<时,()0f x >知,所以选A.5.定义在R 上的奇函数()f x 满足(1)()f x f x +=-,当1(0,]2x ∈时,12()log (1)f x x =-,则()f x 在区间3(1,)2内是 A.减函数且()0f x > B.减函数且()0f x < C.增函数且()0f x > D.增函数且()0f x <【答案】B【解析】本题主要考查函数的奇偶性和单调性。
由此可知函数的周期为2,根据复合函数判断可知函数利用函数和周期性可知B 正确.6.已知向量m =(λ+1,1),n =(λ+2,2),若(m +n )⊥(m -n ),则λ= A.-4 B.-3C.-2D.-1【答案】B【解析】本题考查平面向量的数量积。
四川省达州市大竹县文星中学高三数学上学期期末考试试题 理
四川省达州市大竹县文星中学2015届高三上学期期末考试数学(理)试题考试时间:120分钟;满分150分第I 卷(选择题)一、选择题:共12题 每题5分 共60分 1.若集合A={0,1,2,3},B={1,2,4},则集合A ∪B= A.{0,1,2,3,4} B.{1,2,3,4} C.{1,2} D.{0}2.函数y =sin(π2x +θ)·cos(π2x +θ)在x =2时取最大值,则θ的一个值是( ) A .π4 B .π2 C .2π3D .3π43.设复数z 满足(z -2i)(2-i)=5,则z =( ) A .2+3i B .2-3i C .3+2i D .3-2i4.k 棱柱有f(k)个对角面,则k +1棱柱的对角面个数f(k +1)为( ) A .f(k)+k -1 B .f(k)+k +1 C .f(k)+k D .f(k)+k -25.已知a>0,b>0,a 、b 的等差中项为12,且α=a +1a ,β=b +1b ,则α+β的最小值为( ) A .3 B .4 C .5 D .66.若m ,n 是正整数,则m +n>mn 成立的充要条件是( ) A .m ,n 都等于1 B .m ,n 都不等于2C .m ,n 都大于1D .m ,n 至少有一个等于1 7.函数f(x)=3x -4x3(x ∈[0,1])的最大值是( ) A.12 B .-1 C .0 D .18.某几何体的三视图如图所示,则该几何体外接球的表面积为A.πB.πC.4πD.16π9.阅读下面的程序框图,输出的结果是A.9B.10C.11D.1210.已知双曲线22221x ya b-=的渐近线方程为y=,则以它的顶点为焦点,焦点为顶点的椭圆的离心率等于A.1B.2C.2D.1211.如果实数x,y满足不等式组目标函数z=kx+y的最大值为12,最小值为3,那么实数k的值为A.2 B.-2 C.1 D.-112.已知函数f(x)=2mx2-2(4-m)x+1,g(x)=mx,若对于任一实数x,f(x)与g(x)的值至少有一个为正数,则实数m的取值范围是A.(0,2)B.(0,8)C.(2,8)D.(-∞,0)第II 卷(非选择题)二、填空题:共4题 每题5分 共20分13.随机变量ξ的取值为0,1,2,若P(ξ=0)=15,E(ξ)=1,则D(ξ)=________. 14.观察下列等式 (1+1)=2×1(2+1)(2+2)=22×1×3(3+1)(3+2)(3+3)=23×1×3×5 ……照此规律,第n 个等式可为 .15.下图展示了一个由区间)1,0(到实数集R 的映射过程:区间()0,1中的实数m 对应数上的点m ,如图1;将线段AB 围成一个圆,使两端点B A ,恰好重合,如图2;再将这个圆放在平面直角坐标系中,使其圆心在y 轴上,点A 的坐标为()0,1,如图3.图3中直线AM与x 轴交于点(),0N n ,则m 的象就是n ,记作()f m n=.下列说法中正确命题的序号是 .(填出所有正确命题的序号)①方程()0f x =的解是x =12; ②114f ⎛⎫= ⎪⎝⎭;③()f x 是奇函数; ④()f x 在定义域上单调递增;⑤()f x 的图象关于点1,02⎛⎫ ⎪⎝⎭ 对称.16.对于定义域在R 上的函数f(x),若实数x0满足f(x0)=x0,则称x0是函数f(x)的一个不动点.若函数f(x)=x2+ax +1没有不动点,则实数a 的取值范围是__________.三、解答题:(共6题,共72分)17.(本题10分)设函数f(x)=Asin(ωx +φ )(其中A >0,ω>0,-π<φ≤π)在x =π6处取得最大值2,其图象与x 轴的相邻两个交点的距离为π2. (1)求f(x)的解析式;(2)求函数g(x)=6cos4x -sin2x -1+π6的值域.18.(本题12分) 某企业有甲、乙两个研发小组,他们研发新产品成功的概率分别为23和35,现安排甲组研发新产品A ,乙组研发新产品B ,设甲、乙两组的研发相互独立. (1)求至少有一种新产品研发成功的概率;(2)若新产品A 研发成功,预计企业可获利润120万元;若新产品B 研发成功,预计企业可获利润100万元,求该企业可获利润的分布列和数学期望.19.(本题12分)在四棱锥ABCD P -中, BC AD //,090ABC APB ∠=∠=,点M 是线段AB 上的一点,且CD PM ⊥,BM AD PB BC AB 422====. (1)证明:面⊥PAB 面ABCD ;(2)求直线CM 与平面PCD 所成角的正弦值. 20.(本小题满分12分) 已知等差数列}{n a 的公差为2,前n 项和为nS ,且1S ,2S ,4S 成等比数列。
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四川省大竹县文星中学2016届高三12月月考数学试题一、单选题
1.已知集合A={x|x2-x-2<0},集合,则下列结论正确的是
A.A=B
B.A B
C.B A
D.A∩B=
2.已知函数,命题p:∀x∈[0,+∞),f(x)≤1,则
A.p是假命题,p:x0∈[0,+∞),f(x0)>1
B.p是假命题,p:x∈[0,+∞),f(x)≥1
C.p是真命题,p:x0∈[0,+∞),f(x0)>1
D.p是真命题,p:x∈[0,+∞),f(x)≥1
3.“函数在区间(0,+∞)上为增函数”是“=3”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
4.若抛物线的焦点与双曲线的一个焦点重合,则的值为
A.1
B.-1
C.2
D.4
5.已知函数定义域是,则y=f(2|x|-1)的定义域是
A. B.[-1, 4] C. D.
6.若函数
A.5
B.4
C.3
D.2
7.如图,在边长为1的正三角形中,分别为边上的动点,且满足
,,其中分别是的中点,则的最小值为
A. B. C. D.
8.已知双曲线的右焦点为,直线与一条渐近线交于点
A,若的面积为 (O为原点),则抛物线的准线方程为
A. B. C. D.
9.某几何体的三视图(单位:)如图所示,其中侧视图是一个边长为2的正三角形,则这个几何体的体积是
A. B. C. D.
10.已知函数若互不相等,且,则的取值范围是
A. B. C. D.
11.下列命题中正确的是
A.函数是奇函数
B.函数在区间上是单调递增的
C.函数的最小值是
D.函数是最小正周期为2的奇函数
12.如图,设D是图中边长分别为1和2的矩形区域,E是D内
位于函数图象下方的区域(阴影部分),从D内随机取一个点
M,则点M取自E内的概率为
A. B. C. D.
二、填空题
13.已知集合,N=,若,则的值是_______
14.已知角θ的顶点为坐标原点,始边为x轴非负半轴,若P(4,y)是角θ终边上一点,且sin θ=,则y=________
15.若是偶函数,则___________.
16.某程序框图如图所示,则输出的S的值是______________.
三、解答题
17.已知向量,,设函数. (Ⅰ)求的单调递增区间;
(Ⅱ)求在上的最大值和最小值.
18.已知数列的前项和为满足且.
(1) 令证明:;
(2) 求的通项公式.
19.函数(且)是定义在实数集上的奇函数.
(1)若,试求不等式的解集;
(2)若且在上的最小值为,求的值. 20.某批发市场对某种商品的日销售量(单位:吨)进行统计,最近50天的统计结果如下:
若以上表中频率作为概率,且每天的销售量相互独立.
(Ⅰ)求5天中该种商品恰好有两天的销售量为1.5吨的概率;
(Ⅱ)已知每吨该商品的销售利润为2千元,表示该种商品某两天销售利润的和(单位:千元),求的分布列和数学期望.
21.如图,已知的两条角平分线和相交于在上,且.
(1)证明:四点共圆:
(2)证明:平分.
22.已知,求证:
(1);
(2).
参考答案
1-5 CCBAC 6-10 BCABC 11-12 CC
13.14. -8 15.16.
17. (Ⅰ)===. 当时,解得,
的单调递增区间为.
(Ⅱ)当时,,由标准函数在上的图像知
==.
所以在上的最大值和最小值分别为.
18. (1),
,
.
(2),,
累加得,
经检验,符合
19. (1)是定义在R上的奇函数,
,又且
易知f(x)在R上单调递增,原不等式化为:
,即,
不等式的解集为.
(2),即(舍去)
令
∵x≥1,∴t≥f(1)=,∴g(t)=-2mt+2=+2-
当时,当时,
当时,当时,,
解得,舍去.
综上可知.
20. 解:(Ⅰ),,
依题意,随机选取一天,销售量为吨的概率,
设5天中该种商品有天的销售量为1.5吨,则,
.
(Ⅱ)的可能取值为,
则:,
,
,
,
,
所以的分布列为:
的数学期望==.
21. (1)在中,因为,
所以,因为是角平分线,
所以,故,
于是,即,
所以四点共圆.
(2)连结,则为的平分线,得,
由(1)知四点共圆,所以.
又,又由,且平分,可得,
可得,所以平分.
22. 证明:(1),
,
,
,
,
,
.
,
,
,
.。