【英语】内蒙古鄂尔多斯市第一中学2018-2019学年高二下学期期中考试试卷(解析版)
2018-2019学年内蒙古鄂尔多斯市第一中学高二下学期期中考试数学(文)试题
2018-2019学年内蒙古鄂尔多斯市第一中学高二下学期期中考试高二文科数学★祝考试顺利★ 注意事项：1、答题前，请先将自己的姓名、准考证号用0.5毫米黑色签字笔填写在试题卷和答题卡上的相应位置，并将准考证号条形码粘贴在答题卡上的指定位置。
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一、选择题（本大题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的）1.已知集合(){}ln 10M x x =+>，{}22N x x =-≤≤，则M N =（ ）A ．()0,2B ．[)0,2C ．(]0,2D ．[]0,22.设i 为虚数单位，若复数()()1i 1i m ++是纯虚数，则实数m =（ ） A ．1-B ．0C ．0或1D ．13.设函数22,1()log ,1x x f x x x -≤=>⎧⎪⎨⎪⎩，则满足()2f x ≤的x 的取值范围是（ ）A ．[]1,2-B ．[)(]1,11,4-C ．[)(]1,11,2-D ．[]1,4-4.通过随机询问110名性别不同的大学生是否爱好某项运动，得到如下的列联表：参照附表，得到的正确的结论是（ ）A ．在犯错误的概率不超过0.1%的前提下，认为“爱好该项运动与性别有关”B ．在犯错误的概率不超过0.1%的前提下，认为“爱好该项运动与性别无关”C ．有99%以上的把握认为“爱好该项运动与性别有关”D ．有99%以上的把握认为“爱好该项运动与性别无关” 附表： 参考公式：()()()()22()n ad bc K a b c d a c b d -=++++5. 已知,,a b c 是直线，β是平面，给出下列命题：（ ） ①若c a c b b a //,,则⊥⊥； ②若c a c b b a ⊥⊥则,,//；③若b a b a //,,//则ββ⊂；④若a 与b 异面，且//,a b ββ则与相交； ⑤若a 与b 异面，则至多有一条直线与a ，b 都垂直. 其中真命题的个数是（） A ．1B ．2C ．3D ．46.设a R ∈，若函数ln 2y x ax =+在()1,2上有极值，则实数a 的取值范围是（ ）A. 11,48--⎛⎫ ⎪⎝⎭B.11,24--⎛⎫ ⎪⎝⎭C.1,2-+∞⎛⎫ ⎪⎝⎭ D ．1,4-∞-⎛⎫ ⎪⎝⎭7.过圆2216x y +=上一点P 作圆()222:0O x y m m +=>的两条切线，切点分别为A 、B ，若2π3AOB ∠=，则实数m =（ ）A ．2B ．3C ．4D ．98.曲线3xy e =-+在点(0,2)处的切线与直线0x =和2y x =所围成的三角形面积为( ) A.43B.23C. 4 D ．19.设(),()f x g x 分别是定义在R 上的奇函数和偶函数，()g x 恒不为．．．0．，当0x <时，()()()()0f x g x f x g x ''->，且(2)0f =，则不等式()()0f x g x <的解集是( )A ．()()2,02,-+∞B ．()()2,00,2-C ．()(),22,-∞-+∞ D ．()(),20,2-∞-10.△ABC 的内角A ，B ，C 的对边分别为a ，b ，c .已知C =60°，b =6，c =3，角A 为( )A ．30B ．60C ．75D ．120 11.由偶数组成的数阵如右图：则第21行第4列的数为（ ） A.594 B.546C.592D.644 12.已知双曲线2222:1(0,0)x y E a b a b -=>>的右顶点为A ，抛物线2:16C y ax =的焦点为F ，若在E 的渐近线上存在点P ，使得PA PF ⊥，则E 的离心率的取值范围是（ ） A.()1,2B ．51,4⎛⎤ ⎥⎝⎦C ．()2,+∞D ．5,4+∞⎡⎫⎪⎢⎣⎭二、填空题（本大题共4小题，每小题5分，共20分，将答案填写在答题卡指定位置） 13. 等比数列{a n }的前n 项和为S n ，已知S 1，2S 2，3S 3成等差数列，则数列{a n }的公比为___. 14. 平面向量a 与b 的夹角为2π3，1a =，1b =，则32a b -=_______. 15. 椭圆22221x y a b+=的内接矩形的最大面积是__________________.16. 庙会是我国古老的传统民俗文化活动，又称“庙市”或“节场”．庙会大多在春节、元宵节等节日举行．庙会上有丰富多彩的文化娱乐活动，如“砸金蛋”（游玩者每次砸碎一颗金蛋，如果有奖品，则“中奖”）．今年春节期间，某校甲、乙、丙、丁四位同学相约来到某庙会，每人均获得砸一颗金蛋的机会．游戏开始前，甲、乙、丙、丁四位同学对游戏中奖结果进行了预测，预测结果如下：甲说：“我或乙能中奖”；乙说：“丁能中奖”’；丙说：“我或乙能中奖”；丁说：“甲不能中奖”．游戏结束后，这四位同学中只有一位同学中奖，且只有一位同学的预测结果是正确的，则中奖的同学是_____． 三、解答题（解答应写出文字说明，证明过程或演算步骤） 17．（本小题满分10分）已知x a x x f ++=)(.（I ）当2=a 时，求不等式3)(<x f 的解集；（II ）设关于x 的不等式3)(<x f 有解，求a 的取值范围 . 18.（本小题满分12分）已知2x =是函数2()(23)xf x x ax a e =+--的一个极值点（⋅⋅⋅=718.2e ）． （I ）求实数a 的值；（II ）求函数()f x 在3[,3]2上的最大值和最小值． 19．（本小题满分12分）为培养学生的阅读习惯，某校开展了为期一年的“弘扬传统文化，阅读经典名著”活动.活动后，为了解阅读情况，学校统计了甲、乙两组各10名学生的阅读量（单位：本），统计结果用茎叶图记录如下，乙组记录中有一个数据模糊，无法确认，在图中以a 表示.（Ⅰ）若甲组阅读量的平均值大于乙组阅读量的平均值， 求图中a 的所有可能取值； （Ⅱ）将甲、乙两组中阅读量超过．．15本的学生称为“阅读达人”. 设1a =，现从所有“阅读达人”里任取2人，求其中甲组和乙组各有一人的概率；.（Ⅲ）记甲组阅读量的方差为20s ，在甲组中增加一名学生A 得到新的甲组，若A 的阅读量为10，则记新甲组阅读量的方差为21s ；若A 的阅读量为20，则记新甲组阅读量的方差为22s ，试比较20s ，21s ，22s 的大小（结论不要求证明）20. （本小题12分）如图，在四棱锥E ABCD -中，平面ABCD ⊥平面AEB ，且四边形ABCD 为矩形，120BAE=∠︒，4AE=AB=，2AD=，F G ，分别为BE AE ，的中点，H 为线段BC 的中点．（Ⅰ）求证：平面DAF ⊥平面CEB ； （Ⅱ）求三棱锥-B GFH 的体积..21．（本小题满分12分）在平面直角坐标系中，曲线C 的参数方程为为参数）（θθθ⎩⎨⎧+=+-=sin 44cos 43y x ，直线1l 的方程为kx －y ＋k ＝0，以坐标原点为极点，x 轴的正半轴为极轴建立极坐标系，直线2l 的极坐标方程为ρθθ4sin 2cos =-.(I)写出曲线C 的普通方程和直线2l 的直角坐标方程；(II)若1l 与C 交于不同的两点M ，N ，MN 的中点为P ，1l 与2l 的交点为Q ，1l 恒过点A ，求|AP |·|AQ |.22.（本小题满分12分）已知点(4,0)B 和点(4,0)C -，分别过点,B C 的直线,m n 相交于点A ，设直线,m n 的斜率分别为12,k k .（Ⅰ）如果12(0)k k a a ⋅=≠，求点A 的轨迹方程，并根据a 的取值讨论此轨迹是何种曲线；HE（Ⅱ）设（Ⅰ）中的曲线为C ，若不平行于坐标轴的直线l 与曲线C 交于点,M N ,线段MN 的中点为Q ，O 为坐标原点，设直线MN 与OQ 的斜率分别为,MN OQ k k , 求证：MN OQ k k a .2018~2019学年高二第二学期期中考试试题（文科数学）一．CDDCAB ABDCAB 二．13.1314 . 19 15. 2ab 16. 甲 三．17.解：（I ）当2=a 时，不等式32<++x x 等价于2253)2(2-≤<-⇒⎩⎨⎧<-+--≤x x x x …………………………………………………………1分或023)2(02<<-⇒⎩⎨⎧<-+<<-x x x x ，…………………………………………………………2分或2103)2(0<≤⇒⎩⎨⎧<++≥x x x x ……………………………………………………………3分所以不等式3)(<x f 的解集是⎪⎭⎫⎝⎛-21,25.…………………………………………………5分 （2）由题意得3)(min <x f …………………………………………………………………6分 因为a x a x x a x x f =-+≥++=)(，即a x f =min )(……………………………9分 故33,3<<-<a a 解得.………………………………………………………………10分18.解：（I ）由2()(23)xf x x ax a e =+--可得22()(2)(23)[(2)3]x x x f x x a e x ax a e x a x a e '=+++--=++-- … 2分∵2x =是函数()f x 的一个极值点，∴(2)0f '=∴2(5)0a e +=，解得5a =- …………………………4分 经检验5a =-时2x =是函数()f x 的一个极值点，∴5a =- …………………………………………………………… 5分 （II ）由0)1)(2()(>--='x e x x x f ，得)(x f 在)1,(-∞递增，在),2(+∞递增，由0)(<'x f ，得)(x f 在在)2,1(递减∴2)2(e f =是()f x 在]3,23[∈x 的最小值；…………………………………………8分2347)23(e f =，3)3(e f = ∵)23()3(,0)74(4147)23()3(23233f f e e e e e f f >>-=-=- ∴()f x 在]3,23[∈x 的最大值是3)3(e f =． ……………………………………12分 19.解：（Ⅰ）甲组10名学生阅读量的平均值为12681011121217211010+++++++++=，乙组10名学生阅读量的平均值为124412131616(10)20981010a a+++++++++++=. …………………………………………………………………………………… 2分 由题意，得981010a+>，即2a <. …………………………………………… 3分 故图中a 的取值为0或1. ………………………………………………………… 4分 （II ）甲组中阅读达人有2人记为a,b;乙组中阅读达人有3人记为c,d,e.从5位阅读达人中任取2人的基本事件共有10个，它们是()()()()()()()()()()e d e c d c e b d b c b e a d a c a b a ,,,,,,,,,,,,,,,,,,，…………………………6分其中甲组和乙组各有一人包含6个基本事件，它们是()()()()()()e b d b c b e a d a c a ,,,,,,,,,,,………………………………………………………8分所以甲组和乙组各一人的概率53106P ==………………………………………………10分 （Ⅲ）222102s s s <<. …………………………………………………………………… 12分20.（Ⅰ）证明：在矩形ABCD 中，⊥AD AB ，∵矩形⊥ABCD 平面AEB ，且平面ABCD I 平面AEB=AB ， ∴⊥AD 平面AEB ，又⊂BE 平面AEB ，∴⊥AD BE ， ……………………………………… 2分 ∵AE =AB ，F 为BE 的中点，∴⊥AF BE ，又AD AF =A I ，∴⊥BE 平面ADF ， ……………… ……………… 4分 ∵⊂BE 平面CEB ，∴平面⊥DAF 平面CEB ． …………………………… ………………… 6分（Ⅱ）由（Ⅰ）知,2AF EBC AF G AE ⊥=平面且，又因为点为中点，1G FBH ∴点到平面的距离为………………………………………………………………8分11122FBH S BF BH ∆=⋅=⨯=113G FBH V -∴=⨯=…………………………………………………………………10分G FBH B GFH V V --=又B GFH V -∴=12分 21解：(1)由题可知：曲线C 的普通方程为(x ＋3)2＋(y －4)2＝16，………………………2分直线l 2的直角坐标方程为x －2y －4＝0.………………………………4分(2)设M ，N ，Q 所对应的参数分别为t 1，t 2，t 3，由题意得直线l 1恒过点A (－1,0)，…………………………………………………………5分故l 1的参数方程为⎩⎪⎨⎪⎧x ＝－1＋t cos α，y ＝t sin α(t 为参数)，……………………………………6分代入曲线C 的普通方程得t 2＋4t (cos α－2sin α)＋4＝0，…………………………………7分则t 1＋t 2＝4(2sin α－cos α)，…………………………………………………………………8分将⎩⎪⎨⎪⎧x ＝－1＋t cos α，y ＝t sin α代入x －2y －4＝0，得t 3＝5cos α－2sin α，………………………9分则|AP |·|AQ |＝⎪⎪⎪⎪⎪⎪t 1＋t 22·|t 3|＝2|2sin α－cos α|·⎪⎪⎪⎪⎪⎪5cos α－2sin α＝10.…………………12分22.解：（Ⅰ）令(,)A x y ∵12k k a = ∴44y ya x x ⋅=-+ 化简得2211616x y a -= ∴点A 的轨迹方程为221(4)1616x y x a-=≠± ………………………………… 3分 当0a >时，点A 的轨迹为双曲线 当0a <且1a ≠-时，点A 的轨迹为椭圆 当1a =-时，点A 的轨迹为圆 ………………………………6分 （Ⅱ）设112200(,),(,),(,)M x y N x y Q x y ，则0120122,2x x x y y y =+=+由（Ⅰ）知曲线C 的方程为2211616x y a-=即2216ax y a -= ∵,M N 在曲线C 上∴221116ax y a -= ① 222216ax y a -= ② ……………9分①-②得12121212()()()()a x x x x y y y y -+=-+ ∴012120y y y a x x x -⋅=-即MN OQ k k a = …………………………………12分。
内蒙古鄂尔多斯市达拉特旗第一中学(2024年版)高二上学期11月期中英语试题(含答案)
高二英语试卷注意事项：1.答题前，考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上。
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第二部分阅读(共两节，满分50分)第一节(共15小题；每小题2.5分，满分37.5分)阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项。
AIn 2023, Istanbul attracted more visitors than any other city, and its distinct and varied cuisine was undoubtedly part of the appeal. Here are four of the Istanbul’s best dining choices.Sweetbread kebab at Sadrazam MahmutThis famous restaurant is located in the historic Sütlüce neighbourhood. One of the pleasures of dining here is the warmth of the owner of the restaurant, who can often be found walking from table to table to collect diners’ advice. And the food is also exceptional. Perhaps the best dish is the sweetbread kebab.Dates (椰枣) filled with cheese at Smelt & Co. Gastro SpotAmong the streets of Balat, a simple door marks the entrance of Smelt &. Co. Gastro Spot. The chefs offer a seasonal menu mainly featuring vegetables and seafood, making the most of local products. While dishes rarely repeat, there’s one that seems to make regular appearances on the menu: dates filled with cheese.Turkish breakfast at Privato CafeLocated in the shadow of the historical Galata Tower, this hot spot serves one of the most welcomed Turkish breakfast spreads in the city. Guests are treated to a rich feast, starting with fresh tomatoes, salted olives, cheeses, and jams, many of which are made in house. Then comes the second wave: grilled spicy sausage, kebab and some sweet desserts.Baby calamari (鱿鱼) at Nazende CaddeOne of Istanbul’s best- kept secrets is hidden in a side street. Nazende Cadde has only recently caught the attention of customers. It’s famous for dishes like grilled sweetbreads, but its typical achievement is the melt- in- your- mouth baby calamari. The outside space, warm service and big menu make it especially fit for getting together.21. What’s available in the first restaurant?A. Historical indoor decoration.B. Valuable advice from its owner.C. The best butter in Istanbul.D. The sweetbread kebab.22. What do Sadrazam Mahmut and Privato Cafe have in common?A. They’re located in old neighbourhoods.B. They’re willing to take suggestions.C. They adopt homema de materials.D. They mainly provide seafood.23. Which restaurant is most suitable for family gathering?A. Sadrazam Mahmut.B. Smelt & Co. Gastro Spot.C. Privato Cafe.D. Nazende Cadde.BOn the morning of the big day, Vially Dorestant prepared for her son Corey’s school photo shoot to ensure he felt calm and confident. Little did she know, she needn’t have worried.In a video posted online, Corey displayed his planned smile for the picture at his new school. With a lovable grin (咧着嘴笑), Dorestant believed he was well- prepared.Dorestant shared, “Corey can become shy and anxious around people, so I wanted to make sure he felt at ease during his first picture day at the new school.”However, Corey’s actual pose for the picture was far from the shy and sweet smile his mom had expected. The video revealed Corey posing with left hand in his pocket and right hand resting under his chin (下巴)—a gesture that caught Dorestant completely off guard.“At the first sight of the picture, I was stunned; it took me about five minutes to process because I couldn’t believe what I saw,” Dorestant said. But eventually, she found it amusing.“I felt so proud of Corey for overcoming his nervousness and allowing his true personality to shine through. As a mother, my only wish is for him to be true to himself through any challenges he faces,” Dorestant said.Dorestant’s video has over 18 million views and 25,000 comments, with many surprised by the result. “Nothing could have prepared me for the ending,” one user wrote, while another commented that it was the “best case scenario (场景)”.For Dorestant, Corey’s first picture day served as a valuable lesson. “As parents, we often feel the need to protect our kids and prepare them for the world, which can make us anxious about their readiness, but we need to trust in their abilities and give them the space to shine on their own,” Dorestant said.24. What did Dorestant worry about her son at first?A. He was frightened to meet his new classmates.B. He was embarrassed to pose for a photo.C. He might get lost on his way to school.D. He might be unprepared for the new term.25. What does the underlined word “stunned” in paragraph 5 mean?A. Shocked.B. Disappointed.C. Grateful.D. Worried.26. What can we know about Dorestant’s video?A. It went unnoticed.B. It made record profits.C. It put Corey in trouble.D. It was well received.27. What did Dorestant advise parents to do?A. Believe their own judgement.B. Serve as an experienced guide.C. Give full play to kids’ potential.D. Teach their children everything.CHave you ever struggled with a task because two arms just weren’t enough? If so, you might like a new device that can lend a helping hand literally (真正地). You wear this robotic arm, and then control it with the muscle you use to breathe.Currently, the robotic arm is worn in the middle of your chest, but it can also be placed above your shoulders or at your side, depending on what you want to do with it. Engineer Giulia Dominijanni, part of a team at the Swiss Federal Institute of Technology in Lausanne (EPFL), helped create the device.In the past, robotic arms were designed to replace lost arms, with brain signals controlling them. This new device allows users to control an extra arm while still using their real arms. For this reason, the team designed it to be controlled by breathing.The robotic arm can sense the movement of your diaphragm (膈膜), the muscle that helps you breathe. When you breathe in, the robotic arm stretches; when you breathe out, it retracts (缩回). Breathing normally keeps the arm still.Before making the actual arm, the EPFL team created a virtual reality (VR) model. V olunteers learned to control this virtual third arm using breath sensors. The tests showed that people could easily look around and talk while using the arm.After successful VR tests, the team built a simple robotic arm that functions like the virtual one. Early tests showed that users could comfortably operate the extra arm alongside their own arms.However, researchers are still exploring how well people can control this robotic arm without affecting the use of their natural arms. Jacob George, a director at the University of Utah’s robotics center, notes, “When you’re adding something new on your body, it’s unclear how much the brain can do with that. But in time it becomes possible to control the new arm well.” And scientists at EPFL believe that understanding this could benefit those with disabilities or people recovering from injuries.28. How do EPFL’s robotic arms differ from previous ones?A. They feature affordable prices.B. They are operated by breathing.C. They can address our health issues.D. They are controlled by brain signals.29. Why did the EPFL team create a VR model?A. To cut its production cost.B. To attract more volunteers.C. To know how brains work.D. To test the newly- invented arm’s actual function.30. What is Jacob’s attitude to the robotic arm?A. Uncaring.B. Doubtful.C. Positive.D. Unclear.31. Which can be a suitable title for the text?A. Robotics Gives Us an Extra ArmB. A New Device Helps ScientistsC. The Secret of Human BrainsD. Tips on How to Use an Extra ArmDThe Europa Clipper spacecraft has reached a key milestone and is on track to launch to explore Jupiter’s moon(木星的卫星), Europa. The launch window for its journey opens on October 10.On Monday, the mission passed Key Decision Point E, which is a critical planning stage that allows the mission to move forward. This news was a relief for the Europa Clipper team, especially after a possible issue with the spacecraft’s transistors (晶体管) discovered in May. These transistors are important for controlling the spacecraft’s electricity and needed to survive Jupiter’s high levels of radiation (辐射).Over the past four months, the team have conducted many tests on the transistors at different NASA facilities. They made sure to finish testing on time to avoid putting off the launch. Europa Clipper will carry ten scientific instruments to help determine if Europa could support life in its salty, subsurface ocean.With the mission approved, there are no changes to the launch plan. “It’s the last sort of big examination before we really get into that launch fever, and we’re really happy to say that they passed that examination today,” said Nicola Fox from NASA’s Science Mission Directorate.Jupiter is the largest planet in our solar system and has a powerful magnetic field (磁场) which is 20,000 times stronger than Earth’s. Any spacecraft going there must have radiation- hardened electronics. Jordan Evans, the project manager, noted that Europa is at the edge of a radiation belt, making it a challenging place to explore.Curt Niebur, a program scientist, noted this mission is not about looking for life itself but determining if necessary elements for life, such as water and energy, are present on Europa. If Europa Clipper finds these, it could lead to a follow- up mission to search for life.32. What can be inferred about the spacecraft’s transistors from paragraph 2?A. They put off the launch mission.B. They fell into pieces in space.C. They are in good condition now.D. They had survived on the moon.33. What did Nicola want to say about the result of the mission’s final examination?A. It was satisfying.B. It was subject to change.C. It was wasting time.D. It was out of expectation.34. What is the mission’s key goal centered on?A. Building a habitable planet for human beings.B. Exploring signs of supporting life on Europa.C. Figuring out how much water Europa contains.D. Repairing transistors on the Europa Clipper spacecraft.35. Where is this text most likely from?A. A dairy.B. A newspaper.C. A brochure.D. A guidebook.第二节(共5小题；每小题2.5分，满分12.5分)阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
内蒙古鄂尔多斯市第一中学2018-2019学年高一英语上学期期中试题
市一中2018～2019学年度第一学期期中考试试题高一英语第I卷(选择题共70分）第一部分阅读理解(共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C和D四个选项中。
选出最佳选项。
AThe booking notes of the play “Sherlock”Price: $10BOOKINGThere are four easy ways to book seats for the play:—in personThe Box Office is open from Monday to Saturday, 10 am—8 pm—by telephoneRing 01324976 to reserve(预订) your tickets or to pay by credit card (Visa and MasterCard).—by postSimply complete the booking form and return it to Box Office.—onlineComplete the online booking form at .DISCOUNTS（折扣）:Saver: $2 off any seat booked any time in advance for the play from Monday to Thursday. Savers can be got for children up to 16 years old, over 60s and full-time students.Supersavers: half-price seats can be got for people with disabilities and one companion. It is advisable to book in advance. There is a maximum of eight wheelchair spaces and one wheelchair space will be held until an hour before the show.Standby: best seats are on sale for $6 from one hour before the play for people suitable for Saver and Supersaver discounts and thirty minutes before for all othercustomers.Group Booking: there is a 10% discount for parties of twelve or more.School: school parties of ten or more can book $6 standby tickets and will get every tenth ticket free.Please note: we are unable to exchange tickets or pay the money back unless the play is called off because of unexpected situations.1. If you want to book a ticket of the play, you can ________.A. complete a booking form and post it to the Box OfficeB. go to the Box Office on SundaysC. scan the website D. ring the booking number and pay for the tickets by cash2. According to the text, who can get $2 off?A. A 50-year-old father.B. A 13-year-old student.C. The people who book the tickets on Fridays.D. A 55-year-old woman.3. If you make a group booking for a group of 20 adults, how much should you pay?A. $200.B. $150.C. $160.D. $180.4. What do we know from the text?A. A group of 10 persons can get 10 % discount.B. The audience can get the money back if the play is on show.C. School parties of twelve can book $6 standby tickets.D. There are only seven wheelchair spaces in the theatre.BWhen Cherry Watson travelled on a recent flight from New York to Washington and noticed an “awful tension” in the cabin, she first thought it was caused by typical bad-tempered passengers. But as the flight neared its end, it became obvious that something was very wrong.A teenage boy with Down Syndrome (唐氏综合症) who was traveling with his family had become upset and would not return to his seat, regardless of the cabin crew’s warnings over the loudspeaker that it was almost time to land. The pilot was forced to circle above the airport, delaying the landing and angering people on the alreadytense flight.“If it was a cartoon,” remembered Watson, “there would have been smoke coming out of people’s ears.”The boy’s elderly par ents and adult brothers and sisters tried to persuade him to get off the floor and back into his seat, but in vain.Watson, who used to be a teacher, stood up and quickly headed to the back of the plane.She found the boy in the passage between rows of seats, lying on his belly, and lay down on her stomach to face him. She began chatting calmly with him, asking his name, his favorite book, and his favorite characters. He told her he felt sick and she tried to comfort him.Minutes later, he allowed her to hold his hand, and then together they got properly back into airplane seats. Watson asked for sick bags, and held them as the boy threw up several times, including on her. As she helped him clean up, she repeatedly told him everything would be okay and that t hey’d get through it together.After the plane was finally able to land, no one was impatient to step off the flight as one might expect. Instead, calmed passengers—obviously following Watson’s amazing example—allowed the boy and his family to depart first, smiling at them as they passed. His parents tearfully thanked Watson for what she had done, and a doctor sitting nearby also let her know he had even taken notes on her expert way of handling the situation.5. The landing was delayed because _____.A. the plane was out of controlB. the pilot was forced to change the destinationC. a family ignored the warningsD. a boy refused to sit in his seat6. Watson successfully handled the “awful tension” by _____.A. fighting against unfair treatmentB. calling on other passengers to join herC. asking an experienced doctor to helpD. communicating with the boy7. What changed the angry passengers’ attitude at last?A. Watson’s efforts.B. The plane’s safe landing.C. The crew’s behavior.D. The bo y’s improvement.8. Which of the following words can best describe Watson?A. Patient and ambitious.B. Kind and cool-headed.C. Wise and cooperative.D. Generous and cautious.CAs computers become more popular in China, Chinese people are more and more depending on computer keyboards to input Chinese characters(汉字). But if they use the computer too much, they may end up forgetting the exact strokes(笔画) of each Chinese character when writing on paper. Experts suggest people, especially students, write by hand more.Do you write by hand more or type more? In Beijing, students start using a computer as early as primary school. And computer dependence is more widely spread among university students. Almost all their homework and essays are typed on a computer.All the students interviewed say they usually use a computer.It’s faster and easier to correct if using a computer. And that’s why computers are being used more and more often to modern education. But when people are taking stock in computers increasingly, problems appear.“When I’m writing with a pen, I find I often can’t remember how to write a character, though I feel I’m familiar with it.”“I’m not in the mood to write when faced with a pen and paper.”Many students don’t feel this is something to worry about. Now that it’s more convenient and efficient to write on a computer, why bother to handwrite?Many educators think differently. Shi Liwei, headmaster of a famous primary school in the capital said, “Chinese characters enjoy both practical and aesthetic (审美的)value. But those characters typed with computer keyboards only keep their practical value. All the artistic beauty of the characters is lost. And handwriting contains the writer’s feelings. Through one’s handwriting, people can get to kn ow one’s thinking and personality. Beautiful writing will give people a better first impression of them.”To encourage students to handwrite more, many primary schools in Beijing have made writing classes compulsory（必修的）and in universities, some professors areasking students to turn in their homework and essays written by hand.9．The underlined expression “taking stock in” in paragraph 4 probably means.A. becoming crazy aboutB. getting bored withC. getting dependent onD. getting curious about10．The students interviewed prefer to write using a computer mainly because ___________.A. computers have become a trend and fashion in ChinaB. they are usually asked to e-mail their homework and essaysC. they find it not easy to remember how to write a characterD. they can correct the mistakes they make quickly and conveniently 11．Which of the following statement is NOT true of the advantage of handwriting?A. Handwriting contains the writer’s feelings.B. Handwriting can impress people well and build one’s self-confidence.C. Chinese characters enjoy both practical and aesthetic value.D. The writer’s thinking and personality are shown in his or her handwriting 12．Which of the following can best serve as the title of the passage?A. The Importance of Handwriting and Typing.B. Writing by Computer Will Replace Writing by Hand.C. To Type or to Handwrite.D. Practical and Aesthetic Value of Chinese Characters.DThe loud noise of the cars or the sound of a plane can force its way into the deepest forest, yet it’s not only humans that are bothered by the noise.Bioacoustician Bernie Krause has been studying the effect of noise pollution on wildlife, and has come across some interesting behaviors, especially among animals that communicate by sound, like humans. Birds use sound to communicate, but in noisy places, these animals have to shout over the natural noise to be heard.Krause mentions a study of nightingales(夜莺) to explain what he means. The birds responded to（回应） traffic noise by singing louder and louder until they actually went beyond noise pollution standards in the city. To belt out (sing loudly) their songs, they increased their lung pressure fivefold, but scientists say that this is not dangerous for the birds themselves.Studies show that sudden noise can cause certain birds to leave their nests, leaving the young to their enemies. One study also showed that songbirds that nested close to busy motorways were much less productive than those that nested farther away. Mammals（哺乳动物）too are affected(影响). A recent study showed that nursing caribou （驯鹿） responded to plane noise by not producing enough milk to feed their young.In some cases noise pollution can actually help some animals while harming others. Toads（蟾蜍）and frogs are known to sing in union（同步发声）so that no predator (their enemies) can catch them. Krause found that when planes flew overhead and disturbed the toad’s song, they lost their union, and it took them 45 minutes to get it back again. That gave their natural enemies plenty of time to find and catch individual toads by sound.According to Kruse, “Not only will noise pollution bother wildlife, but it won’t help our lives either.”13. How do young caribou suffer from aircraft noise?A. .They can’t hear their mo thers.B. They can’t sleep at night.C. They are often displaced.D. They receive less food.14. What would happen if toads and frogs failed to sing in union?A. They would soon regain their rhythm.B. They would stop communicating.C. They might not be able to protect themselves.D. They might ne unable to hunt in groups.15. What is the text mainly about?A.The ways animals communicate with each other.B. The causes of noise pollution.C. Animals’ reaction to noises.D. The effects of noise pollution on wildlife.第二节（共5小题：每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。
内蒙古鄂尔多斯市第一中学2018-2019学年高二下学期期末考试数学(理)试卷 含解析
市一中2018～2019学年度第二学期期末考试试题高二数学（理科）本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，其中第Ⅱ卷第22～23题为选考题，其它题为必考题。
考生作答时，将答案答在答题卡上，在本试卷上答题无效。
考试结束后，将本试卷和答题卡一并交回。
第I 卷一、选择题:本大题共12小题,每小题5分,共60分,在每小题给出的四个选项中,只有一项是符合题目要求的．1.“所有9的倍数都是3的倍数.某数是9的倍数，故该数为3的倍数，”上述推理 A. 完全正确B. 推理形式不正确C. 错误，因为大小前提不一致D. 错误，因为大前提错误【答案】A 【解析】 【分析】根据三段论定义即可得到答案.【详解】根据题意，符合逻辑推理三段论，于是完全正确，故选A. 【点睛】本题主要考查逻辑推理，难度不大.2.i 是虚数单位，复数z 满足(1)3i z i +=+，则z = A. 12i + B. 12i -C. 2i +D. 2i -【答案】D 【解析】 【分析】运用复数除法的运算法则可以直接求出复数z 的表达式. 【详解】3(3)(1)(1)321(1)(1)i i i i z i z i i i i ++⋅-+=+⇒===-++⋅-，故本题选D. 【点睛】本题考查了复数的除法运算法则，考查了数学运算能力.3.已知双曲线22221(0,0)x y a b a b -=>>，则此双曲线的渐近线方程为( )A. 2y x =±B. 2y x =C. 2y x = D. 12y x =±【答案】C 【解析】试题分析： 因为双曲线的离心率为62所以62c a =,又因为双曲线中222c a b =+,所以2b a =,而焦点在x 轴上的双曲线的渐近线方程为b y x a =±,所以此双曲线的渐近线方程为22y x =±,故选C. 考点：1、双曲线的离心率；2、双曲线渐近方程.4.6(21)x -展开式中x 2的系数为（ ） A. 15 B. 60C. 120D. 240【答案】B 【解析】【详解】∵6(21)x -展开式的通项为6616(1)2r r r r r T C x --+=-，令6-r=2得r=4，∴6(21)x -展开式中x 2项为4644226(1)260C x x --=，所以其系数为60,故选B5.袋中装有6个红球和4个白球，不放回的依次摸出两球，在第一次摸到红球的条件下，第二次摸到红球的概率是 A.35B.25C.13D.59【答案】D 【解析】 【分析】通过条件概率相关公式即可计算得到答案.【详解】设“第一次摸到红球”为事件A ，“第二次摸到红球”为事件B ，而6()10P A =，651()1093P A B ⋅=⨯=，故()5(|)()9P A B P B A P A ⋅==，故选D. 【点睛】本题主要考查条件概率的相关计算，难度不大.6.已知曲线2y x =和曲线y x =（ ）A. 1B.12C.22D.13【答案】D 【解析】 【分析】先作出两个函数的图像，再利用定积分求面积得解. 【详解】由题得函数的图像如图所示，联立2y x y x⎧=⎪⎨=⎪⎩1,1）所以叶形图面积为31231200211)=()|333x x dx x x -=⎰（. 故选：D【点睛】本题主要考查定积分的应用，意在考查学生对该知识的理解掌握水平和分析推理能力.7.已知随机变量(6,1)XN ，且(57),(48)P X a P X b <<=<<=，则(47)P X <<=A.2b a- B.+2b aC.12b- D.12a- 【答案】B 【解析】 【分析】根据正态分布的对称性即可得到答案.【详解】由于(47)(45)(57)22b a b aP X P X P X a -+<<=<<+<<=+=，故选B. 【点睛】本题主要考查正态分布中概率的计算，难度不大.8.若点P 在抛物线上，点Q （0,3），则|PQ|的最小值是（ ）A.B.112C. 3D.5【答案】B 【解析】试题分析：如图所示，设()2,P t t ，其中t R ∈，则()2223PQ t t =+-2251124t ⎛⎫=-+⎪⎝⎭≥，故选B.考点：抛物线.9.如图过抛物线24y x =焦点的直线依次交抛物线与圆()2211x y -+=于A 、B 、C 、D ，则AB C D ⋅=A. 4B. 2C. 1D.12【答案】C 【解析】 【分析】根据抛物线的几何意义转化1=A AB AF x =-，1D CD DF x =-=，再通过直线过焦点可知24A D p x x ⋅=，即可得到答案. 【详解】抛物线焦点为()1,0F ，1=A AB AF x =-,1D CD DF x =-=,，于是214A D p AB CD x x ⋅=⋅==，故选C.【点睛】本题主要考查抛物线的几何意义，直线与抛物线的关系，意在考查学生的转化能力，计算能力及分析能力.10. 高三(1)班需要安排毕业晚会的4个音乐节目、2个舞蹈节目和l 个曲艺节目的演出顺序要求两个舞蹈节目不连排，则不同排法的种数是（ ） A. 800 B. 5400 C. 4320 D. 3600【答案】D 【解析】先排4个音乐节目和1个曲艺节目共有55A 种排法，再从5个节目的6隔空插入两个不同的舞蹈节目有26A 种排法，∴共有52563600A A ⋅=种排法，故选D11.函数()1ln1x f x sin x -⎛⎫= ⎪+⎝⎭的图象大致为A.B.C.D.【解析】由于0x ≠,故排除A 选项.()()1sin ln 1x f x f x x --⎛⎫-==- ⎪-+⎝⎭,所以函数为奇函数,图象关于原点对称,排除C 选项.()()12sin ln sin ln 303f ⎛⎫==-< ⎪⎝⎭,排除D 选项,故选B.12.若对任意的0x >，不等式()22ln 10x m x m -≥≠恒成立，则m 的取值范围是（ ）A. {}1B. [)1,+∞C. [)2,+∞D. [),e +∞【答案】A 【解析】由已知可得22ln 10x m x --≥对任意的0x >恒成立，设()22ln 1,f x x m x =-- 则()()2222,x m m f x x x x='-=- 当0m <时()0f x '>在()0,∞+上恒成立，()f x 在()0,∞+上单调递增，又()10,f =∴ 在()0,1上()0,f x < 不合题意；当0m >时，可知()f x 在(m 单调递减，在),m +∞单调递增，要使()f x 0≥在在()0,∞+上恒成立，只要fm 0≥，令()()()ln 1,0,ln ,g m fm m m m m g m m ==-->=-' 可知()g m 在()0,1上单调递增，，在在()1,+∞上单调递减，又()()()10,0,0, 1.g g m g m m =∴≤∴=∴=故选A.第Ⅱ卷二、填空题：本大题共4小题,每小题5分,共20分,把答案填在答题纸的相应位置上13.设随机变量1~,4X B n ⎛⎫ ⎪⎝⎭，且()34D X =，则事件“2X =”的概率为_____（用数字作答） 【答案】27128【分析】根据二项分布()()1D x np p =-求得n ，再利用二项分布概率公式求得结果. 【详解】由1~,4X B n ⎛⎫ ⎪⎝⎭可知：()1133144164n D x n ⎛⎫=⨯⨯-== ⎪⎝⎭ 4n ∴=()222411272144128P X C ⎛⎫⎛⎫∴==⋅⋅-=⎪⎪⎝⎭⎝⎭ 本题正确结果：27128【点睛】本题考查二项分布中方差公式、概率公式的应用，属于基础题.14.设函数()()321f x x a x ax =+-+．若()f x 为奇函数，则曲线()y f x =在点()00，处的切线方程为___________． 【答案】y x = 【解析】 【分析】首先根据奇函数的定义，得到10a -=，即1a =，从而确定出函数的解析式，之后对函数求导，结合导数的几何意义，求得对应切线的斜率，应用点斜式写出直线的方程，最后整理成一般式，得到结果. 【详解】因为函数32()(1)f x x a x ax =+-+是奇函数， 所以()()f x f x -=-，从而得到10a -=，即，所以3()f x x x =+，所以(0)0f =，所以切点坐标是(0,0)，因为2()31x f 'x =+，所以'(0)1f =，所以曲线()y f x =在点(0,0)处的切线方程为y x =， 故答案是y x =.【点睛】该题考查的是有关函数图象在某点处的切线问题，涉及到的知识点有奇函数的定义，导数的几何意义，属于简单题目.15.已知等差数列{}n a 满足44a =，且1a ，2a ，4a 成等比数列，则3a 的所有值为________.【答案】3，4 【解析】 【分析】先设等差数列{}n a 公差为d ，根据题意求出公差，进而可求出结果. 【详解】设等差数列{}n a 公差为d ， 因为44a =，且1a ，2a ，4a 成等比数列，所以4122141344a a d a a a a =+=⎧⎨==⎩，即121134()4a d a d a +=⎧⎨+=⎩，解得0d =或1d =. 所以434a d a =-=或3. 故答案为3，4【点睛】本题主要考查等差数列的基本量的计算，熟记等差数列的通项公式即可，属于基础题型.16.下列命题中①已知点(3,0),(3,0)A B -，动点P 满足||2||PA PB =，则点P 的轨迹是一个圆； ②已知(2,0),(2,0),3M N PM PN --=，则动点P 的轨迹是双曲线右边一支； ③两个随机变量的线性相关性越强，则相关系数的绝对值就越接近于1；④在平面直角坐标系内，到点(1,1)和直线23x y +=的距离相等的点的轨迹是抛物线； ⑤设定点12(0,2),(0,2)F F -，动点P 满足条件124|(0)PF PF a a a+=+，则点P 的轨迹是椭圆. 正确的命题是__________． 【答案】①②③ 【解析】①中2222|3),3)PA x y PB x y =++=-+（（|2PA PB =，化简得：225)16x y -+=（，所以点P 的轨迹是个圆；②因为34PM PN MN -=<=,所以根据双曲线的的定义，P 点的轨迹是双曲线右支，正确；③根据相关性定义，正确；④因为点在直线上，不符合抛物线定义，错误；⑤因为44a a+≥=，且当2a =时取等号，不符合椭圆的定义，错误.综上正确的是①②③.三、解答题：解答应写出文字说明，证明过程或演算步骤.17.在△ABC 中，角A ，B ，C 的对边分别为a ，b ，c ，已知bcos 22A +acos 22B =32c ． （Ⅰ）求证：a ，c ，b 成等差数列； （Ⅱ）若C=3π，△ABC 的面积为3，求c ． 【答案】（1）见解析（2）22 【解析】【详解】试题分析：(1)先根据二倍角公式降次，再根据正弦定理将边化为角，结合两角和正弦公式以及三角形内角关系化简得sinB+sinA=2sinC ，最后根据正弦定理得a+b=2c (2)先根据三角形面积公式得ab=8，再根据余弦定理解得c ．试题解析：（Ⅰ）证明：由正弦定理得：223sin cos sin cos sin 222A B B A C += 即1cos 1cos 3sin sin sin 222A B B A C ++⋅+⋅=， ∴sinB+sinA+sinBcosA+cosBsinA=3sinC ∴sinB+sinA+sin （A+B ）=3sinC ∴sinB+sinA+sinC=3sinC…∴sinB+sinA=2sinC ∴a+b=2c ∴a，c ，b 成等差数列． （Ⅱ）13 absi 2382S nC ab ====…, c 2=a 2+b 2﹣2abcosC=a 2+b 2﹣ab=（a+b ）2﹣3ab=4c 2﹣24．…∴c 2=8得22c =18.已知公差不为零的等差数列{}n a 满足535S =，且2a ，7a ，22a 成等比数列. （1）求数列{}n a 的通项公式； （2）若()()413n n n b a a =-+，且数列{}n b 的前n 项和为n T ，求证：34n T <. 【答案】(1)21n a n =+.(2)见详解. 【解析】 【分析】(1)设公差为d ，由已知条件列出方程组，解得1,a d ，解得数列{}n a 的通项公式.(2)得出()1111222n b n n n n ⎛⎫==- ⎪++⎝⎭，可由裂项相消法求出其前n 项和n T ，进而可证结论.【详解】(1)设等差数列{}n a 的公差为d (0d ≠).由题意得52722235,,S a a a =⎧⎨=⎩则()()1211154535,2(6)21,a d a d a d a d ⨯⎧+=⎪⎨⎪+=++⎩化简得1127,23,a d a d +=⎧⎨=⎩解得13,2,a d =⎧⎨=⎩所以()32121na n n =+-=+.(2)证明：()()()()44111113224222n n n b a a n n n n n n ⎛⎫====- ⎪-++++⎝⎭，所以111111111112132435112n T n n n n ⎛⎫=-+-+-++-+- ⎪-++⎝⎭1111311131221242124n n n n ⎛⎫⎛⎫=+--=-+< ⎪ ⎪++++⎝⎭⎝⎭. 【点睛】本题考查等差数列和等比数列的基本量运算、裂项相消法求和、不等式的证明.通项公式形如()1111n a n n d d n n d ⎛⎫==- ⎪++⎝⎭的数列，可由裂项相消法求和.19.“微信运动”是手机APP 推出的多款健康运动软件中的一款，某学校140名老师均在微信好友群中参与了“微信运动”，对运动10000步或以上的老师授予“运动达人”称号，低于10000步称为“参与者”，为了解老师们运动情况，选取了老师们在4月28日的运动数据进行分析，统计结果如下：（Ⅰ）根据上表说明，能否在犯错误概率不超过0.05的前提下认为获得“运动达人”称号与性别有关？（Ⅱ）从具有“运动达人”称号的教师中，采用按性别分层抽样的方法选取10人参加全国第四届“万步有约”全国健走激励大赛某赛区的活动，若从选取的10人中随机抽取3人作为代表参加开幕式，设抽取的3人中女教师人数为X ，写出X 的分布列并求出数学期望()E X .参考公式：22()()()()()n ad bc K a b c d a c b d -=++++，其中n a b c d =+++.参考数据：【答案】（1）不能在犯错误的概率不超过0.05的前提下认为获得“运动达人”称号与性别有关； （2）见解析. 【解析】 【分析】（1）计算2,K 比较3.841即可得到答案；（2）计算出男教师和女教师人数，X 的所有可能取值有0,1,2,3，分别计算概率可得分布列，于是可求出数学期望.【详解】（1）根据列联表数据得：22140(60204020) 1.167 3.841806010040K ⨯⨯-⨯=≈<⨯⨯⨯∴不能在犯错误的概率不超过0.05的前提下认为获得“运动达人”称号与性别有关（2）根据分层抽样方法得：男教师有60106100⨯=人，女教师有40104100⨯=人 由题意可知，X 的所有可能取值有0,1,2,3则()36310106C P X C ===；()2164310112C C P X C ===；()12643103210C C P X C ===；()343101330C P X C === X ∴的分布列为：()1131601236210305E X ∴=⨯+⨯+⨯+⨯=【点睛】本题主要考查独立性检验统计思想，超几何分布的分布列与数学期望，意在考查学生的分析能力，计算能力.20.已知椭圆C 满足：过椭圆C 的右焦点(2,0)F 且经过短轴端点的直线的倾斜角为4π. （Ⅰ）求椭圆C 的方程；（Ⅱ）设O 为坐标原点，若点A 在直线2y =上，点B 在椭圆C 上，且OA OB ⊥，求线段AB 长度的最小值.【答案】（I ）22142x y +=；（Ⅱ）22【解析】 【分析】（Ⅰ）设出短轴端点的坐标，根据过右焦点与短轴端点的直线的倾斜角为4π，可以求出斜率，这样就可以求出b ，再根据右焦点(2,0)F ，可求出c ，最后利用22a b c =+a，最后写出椭圆标准方程；（Ⅱ）设点,A B 的坐标分别为00(,2),(,)t x y ，其中00x ≠，由O A O B⊥，可得出等式，求出线段AB长度的表达式，结合求出的等式和基本不等式，可以求出线段AB 长度的最小值 .【详解】（I ）设椭圆的短轴端点为0-b （，）（若为上端点则倾斜角为钝角），则过右焦点与短轴端点的直线的斜率tan 14k π===，b ∴=2c a ==Q 又22142x y C ∴+=的方程：（Ⅱ）设点,A B 的坐标分别为00(,2),(,)t x y ，其中00x ≠,OA OB ⊥0OA OB ∴⋅=,即就是0020tx y +=，解得0y t x =-.又22024x y +=22222000000020028()(2)()(2)4(04)2y x AB x t y x y x x x ∴=-+-=++-=++<≤ 2002084(04)2x x x +≥<≤，且当204x =时等号成立，所以AB 长度的最小值为22【点睛】本题考查了求椭圆的标准方程，考查了利用基本不等式求线段长最小值问题，考查了数学运算能力.21.已知函数()()1ln af x a x x x=++-，其中.a R ∈ （Ⅰ）求()f x 的单调区间；（Ⅱ）若在[]1,e 上存在0x ，使得()00f x <成立，求a 的取值范围.【答案】（1）见解析（2）()()1,1,.1e e e ⎛⎫+-∞-⋃+∞ ⎪-⎝⎭【解析】试题分析：（1）函数的单调区间与导数的符号相关，而函数的导数为()()()21'x x a f x x ++=，故可以根据a 的符号讨论导数的符号，从而得到函数的单调区间.（2）若不等式()0f x < 在[]1,e 上有解，那么在[]1,e 上，()m i n0f x <.但()f x 在[]1,e 上的单调性不确定，故需分1,1,a e a ae ≥--<<-≤- 三种情况讨论.解析：（1）()()()()2222111'1,0x a x a x x a a a f x x x x x x++++++=++==>， ①当0a ≥时，在()0,x ∈+∞上()'0f x >，()f x 在()0,∞+上单调递增；②当0a <时，在()0,x a ∈-上()'0f x <；在(),x a ∈-+∞上()'0f x >；所以()f x 在()0,a -上单调递减，在(),a -+∞上单调递增. 综上所述，当0a ≥时，()f x 单调递增区间为()0,∞+，当0a <时，()f x 的单调递减区间为()0,a -，单调递增区间为(),a -+∞.（2）若在[]1,e 上存在0x ，使得()0f x 成立，则()f x 在[]1,e 上的最小值小于0.①当1a -≤，即1a ≥-时，由（1）可知()f x 在[]1,e 上单调递增，()f x 在[]1,e 上的最小值为()1f ，由()110f a =-<，可得1a >，②当a e -≥，即a e ≤-时，由（1）可知()f x 在[]1,e 上单调递减，()f x 在[]1,e 上的最小值为()f e ，由()()10af e a e e =++-<，可得()11e e a e +<-- ； ③当1a e <-<，即1e a -<<-时，由（1）可知()f x 在()1,a -上单调递减，在(),a e -上单调递增，()f x 在[]1,e 上的最小值为()()()1ln 1f a a a a -=+--+，因为0ln()1a <-<，所以()()()11ln 0a a a +<+-<，即()()()1ln 1112a a a a a +--+>+-+=，即()2f a ->，不满足题意，舍去.综上所述，实数a 的取值范围为()()1,1,1e e e ⎛⎫+-∞-⋃+∞ ⎪-⎝⎭. 点睛：函数的单调性往往需要考虑导数的符号，通常情况下，我们需要把导函数变形，找出能决定导数正负的核心代数式，然后就参数的取值范围分类讨论.又不等式的恒成立问题和有解问题也常常转化为函数的最值讨论，比如：“()0f x <在[],a b 上有解”可以转化为“在[],a b 上，有()min 0f x <”，而“()0f x <在[],a b 恒成立”可以转化为“在[],a b 上，有()max 0f x <”.选做题: 请考生从第22、23两题中任选一题作答。
内蒙古鄂尔多斯市第一中学2018-2019学年高二下学期期末考试数学(文)试卷 Word版含解析
市一中2018～2019学年度第二学期期末考试试题高二数学（文科）本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，其中第Ⅱ卷第22～23题为选考题，其它题为必考题。
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5．做选考题时，考生按照题目要求作答，并用2B 铅笔在答题卡上把所选题目对应的题号涂黑。
第I 卷一、选择题:本大题共12小题,每小题5分,共60分,在每小题给出的四个选项中,只有一项是符合题目要求的．1.已知集合{}1,2,3,A =2{|9}B x x =<，则A B ⋂= A. {2,1,0,1,2,3}-- B. {2,1,0,1,2}--C. {1,2,3}D. {1,2}【答案】D 【解析】试题分析：由29x <得33x -<<，所以{|33}B x x =-<<，因为{}1,2,3A =，所以{}1,2A B ⋂=，故选D.【考点】 一元二次不等式的解法，集合的运算【名师点睛】对于集合的交、并、补运算问题，应先把集合化简再计算，常常借助数轴或韦恩图处理.【此处有视频，请去附件查看】2.i 是虚数单位，复数z 满足(1)3i z i +=+，则z = A. 12i + B. 12i -C. 2i +D. 2i -【答案】D 【解析】 【分析】运用复数除法的运算法则可以直接求出复数z 的表达式. 【详解】3(3)(1)(1)321(1)(1)i i i i z i z i i i i ++⋅-+=+⇒===-++⋅-，故本题选D. 【点睛】本题考查了复数的除法运算法则，考查了数学运算能力.3.已知双曲线22221(0,0)x y a b a b -=>>，则此双曲线的渐近线方程为()A. 2y x =±B. y =C. y x =D.12y x =±【答案】C 【解析】试题分析： 因为双曲线的离心率为2,所以2c a =,又因为双曲线中222c a b =+,所以2b a =,而焦点在x 轴上的双曲线的渐近线方程为b y x a =±,所以此双曲线的渐近线方程为2y x =±,故选C. 考点：1、双曲线的离心率；2、双曲线渐近方程.4.设函数()f x 在R 上为增函数，则下列结论一定正确的是（ ） A. ()1y f x =在R上为减函数 B. ()y f x =在R 上为增函数C. ()1y f x =-在R上为增函数 D. ()y f x =-在R 上为减函数【答案】D 【解析】A 错，如3,y x =()1y f x =在R上无单调性； B. 错，如3,y x =()y f x =在R 上无单调性；C. 错，如()31,y x y f x ==-在R 上无单调性； 故选D.5.从(40,30), (50,10), (20,30),(45,5),(10,10)中任取一个点，这个点在圆222016x y +=内部的概率是 A.35B.25C.15D.45【答案】B 【解析】 【分析】先判断出每个点的横坐标和纵坐标的平方和是否小于2016，然后利用古典概型概率计算公式求出概率.【详解】因为22403025002016+=>，22501026002016+=>，22203013002016+=<，2245520502016+=>，2210102002016+=<，所以只有点(20,30),(10,10)这两个点在圆222016x y +=内部，因此这个点在圆222016x y +=内部的概率是25，故本题选B.【点睛】本题考查了古典概型概率计算公式，考查了数学运算能力.6.定义22⨯矩阵12142334[]=a a a a a a a a -，若22cos sin ()cos(2)12x xf x x π⎡-⎢=⎢⎥+⎢⎥⎣⎦，则()f x 的图象向右平移3π个单位得到函数()g x ，则函数()g x 解析式为（ ） A. ()22g x cos x =- B. ()2sin 2g x x =-C. ()2sin(2)6g x x π=- D. ()2(2)6g x cos x π=--【答案】A 【解析】试题分析：由定义22⨯矩阵12142334[]=a a a a a a a a -，可知22cos sin ()cos(2)12x xf x x π⎡-⎢=⎢⎥+⎢⎥⎣⎦22cos sin 22x x x π⎛⎫=-+ ⎪⎝⎭cos 22x x =+2sin 26x π⎛⎫=+ ⎪⎝⎭，所以()2sin 22sin 22cos 2362g x x x x πππ⎡⎤⎛⎫⎛⎫=-+=-=- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦，故选A 考点：三角函数图象的变换.7.已知,a b R +∈，函数2()log f x a x b =+的图像经过点(4,1)，则12a b+的最小值为A. 6-B. 6C. 4+D. 8【答案】D 【解析】 【分析】由函数2()log f x a x b =+的图像经过点(4,1)，可以得到一个等式，利用这个等式结合已知的等式，根据基本不等式，可以求12a b+的最小值. 【详解】因为函数2()log f x a x b =+的图像经过点(4,1)，所以有22log 4121b a b +=⇒+=，因为,a b R +∈，所以有((2)412124)1)(48a b a a b a b a b b +=++⋅=++≥+=（当且仅当2b a =时取等号），故本题选D.【点睛】本题考查了基本不等式的应用，用1巧乘是解题的关键.8.若点P 在抛物线上，点Q （0,3），则|PQ|最小值是（ ）A.2B.C. 3D.【答案】B 【解析】 试题分析：如图所示，设()2,P t t，其中t R∈，则PQ=≥，故选B.考点：抛物线.9.以下四个命题中，真命题的个数是①“若2a b +≥，则a b 、中至少有一个不小于1”的逆命题； ② 存在正实数a b 、，使得lg()lg lg a b a b +=+； ③3[0,),0x x x ∀∈+∞+≥；④ 函数(1)y f x =+是奇函数，则()y f x =的图像关于(1,0)对称. A. 0 B. 1 C. 2 D. 3【答案】D 【解析】 【分析】①：写出命题的逆命题，然后判断真假； ②：判断方程ab a b =+有无正实数解即可；③：通过不等式的性质可以判断出3[0,),0x x x ∀∈+∞+≥是否正确； ④：通过函数图像的平移可以判断出该命题是否正确.【详解】①：“若2a b +≥，则a b 、中至少有一个不小于1”的逆命题是：若a b 、中至少有一个不小于1，则2a b +≥.显然当 1.5,0a b ==符合条件，但是2a b +≥不成立，故本命题是假命题；②：由lg()lg lg a b a b +=+可得ab a b =+，显然当2a b ==时，等式成立，所以本命题是真命题； ③：33000x x x x ≥∴≥∴+≥，所以本命题是真命题；④：因为函数(1)y f x =+是奇函数，所以函数(1)y f x =+的图像关于原点对称， 函数(1)y f x =+的图像向右平移一个单位长度得到()y f x =图像，因此()y f x =的图像关于(1,0)对称.，所以本命题是真命题，故一共有三个命题是真命题，故本题选D. 【点睛】本题考查了命题的真假判断，考查了对数的运算、函数的对称性、逆命题、不等式等相关知识.10. 已知棱长为1的正方体的俯视图是一个面积为1的正方形，则该正方体的正视图的面积不可能等于（ ）A. 1B.C.D.2【答案】C 【解析】试题分析：水平放置的正方体，当正视图为正方形时，其面积最小为1；当正视图为对角面1的正方体的俯视图是一个面积为1的正方形，则该正方体的正视图的面积的范围是，因此,,A B D 皆有可能，而112<，故不可能的为C .考点：1.三视图；2.正方体的几何特征. 【此处有视频，请去附件查看】11.函数()1ln1x f x sin x -⎛⎫= ⎪+⎝⎭的图象大致为A.B.C.D.【答案】B 【解析】由于0x ≠,故排除A 选项.()()1sin ln1x f x f x x --⎛⎫-==- ⎪-+⎝⎭,所以函数为奇函数,图象关于原点对称,排除C 选项.()()12sin ln sin ln 303f ⎛⎫==-< ⎪⎝⎭,排除D 选项,故选B.12.已知函数22()()()f x x x x ax b =+⋅++，若对x R ∀∈，均有()(2)f x f x =-，则()f x 的最小值为（ ） A. 94-B. 3516-C. -2D. 0【答案】A 【解析】由题意可知函数f(x)的对称轴为x=1,显然f(0)=f(-1)=0，由对称性知f(2)=f(3)=0,所以2(2)(3)x ax b x x ++=--，所以5,6a b =-=，22()()(56)f x x x x x =+-+,即f(x)=22(2)(23)x x x x ---,不妨令221t x x =-≥-，函数为(3)y t t =-，1t ≥-，所以 当32t =-,时y 取最小值94-，选A.【点睛】本题首先充分利用对称性的某些值相等，而没有利用定义，从而简化了运算，更重要采用了换元法求最值，而不是利用求导求最值，更简化了运算。
内蒙古鄂尔多斯市第一中学2018-2019学年高一下学期开学考试英语试题
市一中2018～2019学年度开学考试试题高一英语I卷(选择题共80分）第一部分阅读理解第一节：共15小题；每小题2分，满分30分阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。
AImagine being given the opportunity to wake up lions, eat your meals with monkeys, and even share your bath with bears, all from the comfort of a unique new lodging（住宿）experience.The Jamala Wildlife Lodge opened its doors in January 2015, which was set up in an effort to educate visitors about aiding the survival of many of the world’s endangered species.“It’s great for the animals; they’re going to get more space. It’s great for the viewing public; they’re going to get more things to see. It will be great for tourism and just for the local community,” said Richard Tindale, the owner and operator.Spreading across three locations in the National Zoo, the 18 rooms, which range from giraffe tree houses to jungle bungalows, offer a fantastic experience with wildlife.The Giraffe TreehouseThe Giraffe Treehouse is set among the giraffe enclosures, and the Jungle Bungalows are luxurious individual houses which are next to either lion, cheetah(猎豹), brown bear, or Malaysian sun bear enclosures.The Ushaka HouseHousing up to 26 people, the Ushaka House is built around the monkey enclosure and has a built-in aquarium which offers private views o f some of the zoo’s sharks.The Burley Griffin HouseOnly meters from the edge of Lake Burley Griffin, the indoor and outdoor entertaining areas have splendid views across the lake to Black Mountain.The Shark HouseThe Shark House has its own little jetty(码头) and it comes out over the shark tank here and so the people who stay in the room will be able to go to their bedroom and pat the shark.LocationLocated in the heart of Australia’s political capital, the Jamala Wildlife Lodge at the National Zoo and A quarium is just ten minutes from Canberra’s Central Business District.1. Which of the following is TRUE about the Jamala Wildlife Lodge?A. The Jamala Wildlife Lodge has altogether 26 houses.B. The Jamala Wildlife Lodge has a very long history.C. The Jamala Wildlife Lodge is outside the National Zoo.D. The Jamala Wildlife Lodge lies in the heart of Canberra.2. If a tourist wants to visit Lake Burley Griffin, he’d better choose ________.A. The Giraffe TreehouseB. The Ushaka HouseC. The Burley Griffin HouseD. The Shark House3. The purpose of writing the passage is to ________.A. attract more tourists to visit the National ZooB. introduce the Jamala Wildlife Lodge in AustraliaC. raise people’s awareness of protecting animalsD. offer visitors an opportunity to play with sharksBWatson entered Mr Smith’s office. The boss was a hard man. He fired people who didn’t do well without giving them a second chance.“Watson,” said Mr Smith, “this past year your department hasn’t earned money. We’re going to drop that department. It’s finished. I’m sorry---but you’ll have to go.” “But, sir---if I just had a little more time. For the moment I need the job to keep my son at Riverside School.”“What’s that!” said the Boss. “Riverside! I didn’t know you had a boy there. That’s an expensive school for a man with your salary.”“I know, sir. But he likes it there so much! He’s a star trackman（田径运动员）and the best boxer in the school. The boys call him Champ（冠军）there.”The Boss sat perfectly still for a long time---a faraway look in his eyes. Then, suddenly, he said,“We’ve got to close your department, Watson. But you’ll take over a new job in another department. It means long hours---maybe more pay. Now get out. You’re here for life.”Watson got out, with surprise in his face. Then the Boss took a letter from the top drawer of his desk. It was Herbie’s last letter from Riverside School---written a few days before he died. He had read it over and over again with sick pain. The letter read:I can’t say the boys here are any nicer to me than the others were. I guess it’s the same everywhere when you’re a cripple（跛脚的人）. But don’t worry about me, Dad. They’ve got a good chemistry department here. And there’s one boy here who is really great. He’s a track st ar and boxing champ and just tops in chemistry. The boys call him Champ. He made them stop throwing my books around. And he knocked a boy down who hit me. He is the best friend I ever had. Dad, when I grow up, I want to do something for Champ. Something big---that he won’t even know about.Your son,4. Watson was given a second chance because_________.A. a man was needed in another departmentB. Mr Smith wanted to realize his son’s dreamC. Mr Smith wanted to help Watson’s sonD. Herbie told Mr Smith to do so in his letter5. It can be inferred from the text that Champ is _________.A. a classmate of Watson’s sonB. Watson’s sonC. the son of Mr Smith’s friendD. Mr Smith’s son6. From the text we know that Herbie _______.A. made friends with many boysB. was a college studentC. died from an incurable diseaseD. didn’t live to grow up7. Watson and Mr Smith both____________.A. loved their sons deeplyB. were hard menC. knew ChampD. were interested in chemistryCA bargain is something offered at a low and advantageous price. A more recent definition is: a bargain is a dirty trick to force money out of the pockets of silly and innocent people.The cost of producing a new-for example - toothpaste would make 80p the proper price for it, so we will market it at £1.20. It is not a bad toothpaste, and as people like to try new things it will sellwell to start with; but the attraction（吸引）of novelty soon disappears, so sales will fall. When that happens we will reduce the price to £1.15. And we will turn it into a bargain by printing 5p OFF all over it.Sometimes it is not 5p OFF but 1p OFF. What breathtaking rudeness to advertise 1p OFF your soap or washing powder or whatever! Even the poorest old-age pensioner ought to regard this as an insult(侮辱), but he doesn't. A bargain must not be missed. People say one has to have washing powder (or whatever) and one might as well buy it a penny cheaper.The real danger starts whe n unnecessary things become ‘bargains’. Many people just cannot resist bargains. Provided they think they are getting a bargain they will buy clothes they will never wear or furniture they have no space for. Once I heard of a man who bought an electric saw as a bargain and cut off two of his fingers the next day. But he had no regrets: the saw had been truly cheap.Quite a few people actually believe that they make money on such bargains. A lady once told me: “I’ve had a lucky day today. I bought a dress fo r £120, reduced from £400; and l bought a beautiful Persian carpet for £600, reduced from £900.” It will never occur to her that she has actually wasted £720. She feels as though she had made £580. She also feels, I am sure, that if she had more time for shopping, she could make a living out of it.Some people buy in large quantities because it is cheaper. Once a couple bought enough sugar for their lifetime and the lifetime of their children and grandchildren. They thought it a bargain not to be missed. When the sugar arrived they didn't know where to store it - until they realized that their toilet was a very spacious one. So that was where they piled up their sugar. Not only did their guests feel rather strange whenever they were offered sugar to put into their coffee, but the toilet became extremely sticky.To offer bargains is a commercial trick to make the poor poorer. When greedy fools fall for this trick, it serves them right.8. Which word best describes the language style of the passage?A. Polite.B. Foolish.C. Humorous.D. Serious.9. What does the underlined word “novelty” in Paragraph 2 probably mean?A. Good quality.B. Low price.C. Curiosity.D. Newness.10. How does the author feel about 1p OFF a product?A. It’s a gift for poor people.B. It’s an offense to shoppers.C. It’s a bargain worth trying.D. It’s a real reduction in price.11. Which statement will the author probably agree with?A. Bargains are things people don’t really need.B. Bargains are often real cheap products.C. Bargains help people make a living.D. Bargains play tricks on people.DFrom the health point of view we are living in a wonderful age. A large number of once fatal(致命的)illnesses can now be cured by modern medicine. But though the possibility of living a long and happy life is greater than ever before, every day we see the unbelievable killing of people on the roads. Man is opposite to the motor-car! It is a never-ending battle which man is losing. Thousands of people are killed each year and we are quietly sitting back and letting it happen.It has been rightly said that when a man is sitting behind a steering-wheel(方向盘), his car becomes the extension of his personality. There is no doubt that the motor-car often brings out a man's worst character. People who are normally quiet and pleasant may become unrecognizable when they are behind a steering-wheel. They are rude, ill-mannered and aggressive. All their hidden frustrations and disappointments seem to be brought to the surface by the act of driving.The surprising thing is that society smiles on the motorist and seems to forgive the behavior. Everything is done for his convenience. Cities and towns are made ugly by huge car parks.It is high time a world law should be created to reduce this senseless waste of human life. With regard to driving, the laws of some countries are not strict and even the strictest are not strict enough.A law which was universally accepted could only have a beneficial effect on the accident rate. Here are a few examples of some things that might be done. The driving test should be made to a fixed standard and far more difficult than it is; all the drivers should be made to take a test every three years or so; the age at which young people are allowed to drive any vehicle should be raised to at least 21 ; all vehicles should be put through strict annual tests for safety. Present drinking and driving laws (where they exist) should be made much stricter. Driving speed should be limited on all roads. These measures may sound rather severe. But surely nothing should be considered difficult if it results in reducing the annual deaths. After all, the world is for human beings, not motor-cars.12. The main idea of this passage is______.A. traffic accidents are mainly caused by motoristsB. only stricter traffic laws can prevent accidentsC. thousands of people in the world are killed each yearD. the laws of some countries about driving are too strict13. Which of the following is right according to the passage?A. It is right to build huge car parks in cities and towns.B. Society ignores their rude driving.C. Working by car can save time and money.D. All the drivers must be asked to take a test every year.14. Why does the author say" his car becomes the extension of his personality" ?A. Driving can show his temper.B. Driving can bring out his good character.C. Driving can make a man excited and joyful.D. Driving can show his real self.15. The attitude of the author is______.A. criticalB. praisingC. selfishD. determined第二节(共5小题；每小题2分，满分10 分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。
2018-2019学年内蒙古鄂尔多斯市第一中学高二下学期期中考试物理试卷
市一中2018～2019学年度第二学期期中考试试题高二理科物理★祝考试顺利★注意事项：1、答题前，请先将自己的姓名、准考证号用0.5毫米黑色签字笔填写在试题卷和答题卡上的相应位置，并将准考证号条形码粘贴在答题卡上的指定位置。
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一、选择题(共15个小题，共50分。
在每小题给出的四个选项中，其中第1～10题只有一项符合题目要求，每小题3分。
第11～15题有多项符合题目要求，完全选对得4分，选对但不全的得2分，有选错的得零分）1.下列关于传感器的说法中正确的是()A. 电子秤应用的是光传感器B. 电饭锅通过温度传感器实现温度的自动控制C. 干簧管是一种能够感知电场的传感器D. 霍尔元件能够把电学量转换为磁学量【答案】B【解析】【详解】电子秤应用是压力传感器，选项A错误；电饭锅通过温度传感器实现温度的自动控制，选项B正确；干簧管是一种能够感知磁场的传感器，选项C错误；霍尔元件能够把磁学量转换为电学量，故D错误。
故选B.2.一个密闭容器由固定导热板分隔为体积相同的两部分，分别装有质量不等的同种气体。
当两部分气体稳定后，它们的（）A. 密度相同B. 分子数相同C. 分子平均速率相同D. 分子间平均距离相同【答案】C【解析】两部分气体为同种气体，但是质量不同，则摩尔数不同，分子数不同，选项B错误；两部分气体的体积相同，故密度不同，每个分子运动占据的体积不同，分子平均距离不同，选项AD错误；因中间为导热隔板，则最终两边温度相同，分子平均速率相同，选项C正确；故选C.点睛；要知道温度是分子平均动能的标志，温度相同的任何物体的分子的平均速率相同；分子的平均距离由分子的密度决定.3.法拉第圆盘发电机的示意图如图所示。
内蒙古鄂尔多斯市第一中学2018-2019学年高二下学期期末考试数学(文)试卷含解析
市一中2018～2019学年度第二学期期末考试试题高二数学（文科）本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，其中第Ⅱ卷第22～23题为选考题，其它题为必考题。
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5．做选考题时，考生按照题目要求作答，并用2B 铅笔在答题卡上把所选题目对应的题号涂黑。
第I 卷一、选择题:本大题共12小题,每小题5分,共60分,在每小题给出的四个选项中,只有一项是符合题目要求的．1.已知集合1,2,3,A 2{|9}B x x，则ABA. {2,1,0,1,2,3} B. {2,1,0,1,2}C. {1,2,3}D. {1,2}【答案】D 【解析】试题分析：由29x 得33x，所以{|33}B x x，因为1,2,3A，所以1,2A B ，故选 D. 【考点】一元二次不等式的解法，集合的运算【名师点睛】对于集合的交、并、补运算问题，应先把集合化简再计算，常常借助数轴或韦恩图处理.【此处有视频，请去附件查看】2.i 是虚数单位，复数z 满足(1)3i zi ，则zA.12i B.12i C.2iD.2i【答案】D 【解析】【分析】运用复数除法的运算法则可以直接求出复数z 的表达式.【详解】3(3)(1)(1)321(1)(1)i i i i zi zi ii i ，故本题选 D.【点睛】本题考查了复数的除法运算法则，考查了数学运算能力.3.已知双曲线22221(0,0)x y a b ab的离心率为62，则此双曲线的渐近线方程为()A. 2y xB. 2y xC. 22yxD. 12yx【答案】C 【解析】试题分析：因为双曲线的离心率为62,所以62c a,又因为双曲线中222cab ,所以22b a ,而焦点在x 轴上的双曲线的渐近线方程为b yx a,所以此双曲线的渐近线方程为22yx ,故选 C.考点：1、双曲线的离心率；2、双曲线渐近方程.4.设函数f x 在R 上为增函数，则下列结论一定正确的是（）A. 1yf x 在R 上为减函数B.y f x 在R 上为增函数C. 1yf x在R 上为增函数D.yf x 在R 上为减函数【答案】D【解析】A 错，如3,yx 1yf x在R 上无单调性；B. 错，如3,yx y f x 在R 上无单调性；C. 错，如31,y x yf x在R 上无单调性；故选D.5.从(40,30), (50,10), (20,30),(45,5),(10,10)中任取一个点，这个点在圆222016xy内部的概率是A.35B.25C.15D.45【答案】B 【解析】【分析】先判断出每个点的横坐标和纵坐标的平方和是否小于2016，然后利用古典概型概率计算公式求出概率.【详解】因为22403025002016，22501026002016，22203013002016，2245520502016，2210102002016，所以只有点(20,30),(10,10)这两个点在圆222016xy内部，因此这个点在圆222016xy内部的概率是25，故本题选 B.【点睛】本题考查了古典概型概率计算公式，考查了数学运算能力.6.定义22矩阵12142334[]=a a a a a a a a ，若22cos sin 3()cos(2)12x xf x x ，则()f x 的图象向右平移3个单位得到函数()g x ，则函数()g x 解析式为（）A. ()22g x cos xB. ()2sin 2g x x C. ()2sin(2)6g x xD. ()2(2)6g x cos x【答案】A 【解析】试题分析：由定义22矩阵12142334[]=a a a a a a a a ，可知22cos sin 3()cos(2)12x xf x x 22cos sin 3cos22x xxcos23sin 2x x2sin 26x，所以2sin 22sin 22cos 2362g xxxx ，故选 A考点：三角函数图象的变换.7.已知,a b R ，函数2()log f x a xb 的图像经过点(4,1)，则12ab的最小值为A.622B. 6C. 422D. 8【答案】D 【解析】【分析】由函数2()log f x a xb 的图像经过点(4,1)，可以得到一个等式，利用这个等式结合已知的等式，根据基本不等式，可以求12ab的最小值.【详解】因为函数2()log f x a x b 的图像经过点(4,1)，所以有22log 4121b a b ，因为,a bR ，所以有((2)4121244)1)(428a b a b a ababaabb b（当且仅当2b a 时取等号），故本题选D.【点睛】本题考查了基本不等式的应用，用1巧乘是解题的关键.8.若点P 在抛物线上，点Q （0,3），则|PQ|最小值是（）A.132B.112C. 3D.5【答案】B 【解析】试题分析：如图所示，设2,P t t，其中t R ，则2223PQ tt2251124t112，故选B.考点：抛物线.9.以下四个命题中，真命题的个数是①“若2ab ，则a b 、中至少有一个不小于1”的逆命题；②存在正实数a b 、，使得lg()lg lg a b a b ；③3[0,),0x xx ；④函数(1)y f x 是奇函数，则()yf x 的图像关于(1,0)对称.A. 0B. 1C. 2D. 3【答案】 D 【解析】【分析】①：写出命题的逆命题，然后判断真假；②：判断方程ab a b 有无正实数解即可；③：通过不等式的性质可以判断出3[0,),0xxx是否正确；④：通过函数图像的平移可以判断出该命题是否正确.【详解】①：“若2a b ，则a b 、中至少有一个不小于1”的逆命题是：若a b 、中至少有一个不小于1，则2a b.显然当 1.5,0ab符合条件，但是2ab 不成立，故本命题是假命题；②：由lg ()lg lg ab a b 可得aba b ，显然当2ab时，等式成立，所以本命题是真命题；③：33000x x xx ，所以本命题是真命题；④：因为函数(1)yf x 是奇函数，所以函数(1)yf x 的图像关于原点对称，函数(1)yf x 的图像向右平移一个单位长度得到()yf x 图像，因此()yf x 的图像关于(1,0)对称.，所以本命题是真命题，故一共有三个命题是真命题，故本题选D.【点睛】本题考查了命题的真假判断，考查了对数的运算、函数的对称性、逆命题、不等式等相关知识.10. 已知棱长为1的正方体的俯视图是一个面积为1的正方形，则该正方体的正视图的面积不可能等于（）A.1B.2C.2-12D.2+12【答案】C 【解析】试题分析：水平放置的正方体，当正视图为正方形时，其面积最小为1；当正视图为对角面时，其面积最大为2，因此满足棱长为1的正方体的俯视图是一个面积为1的正方形，则该正方体的正视图的面积的范围是[1,2]，因此,,A B D 皆有可能，而2112，故不可能的为C .考点：1.三视图；2.正方体的几何特征.【此处有视频，请去附件查看】11.函数1ln1x fx sin x 的图象大致为A. B. C. D.【答案】B 【解析】由于0x ,故排除A 选项.1sin ln1x fx f x x ,所以函数为奇函数,图象关于原点对称,排除C 选项.12sin lnsin ln 303f ,排除D 选项,故选 B.12.已知函数22()()()f x xx xax b ，若对xR ，均有()(2)f x f x ，则()f x 的最小值为（）A.94B.3516C. -2D. 0【答案】A 【解析】由题意可知函数f(x)的对称轴为x=1,显然f(0)=f(-1)=0，由对称性知f(2)=f(3)=0,所以2(2)(3)xa xb x x ，所以5,6a b ，22()()(56)f x x x x x ,即f(x)=22(2)(23)x x xx,不妨令221txx，函数为(3)yt t，1t，所以当32t,时y 取最小值94，选A. 【点睛】本题首先充分利用对称性的某些值相等，而没有利用定义，从而简化了运算，更重要采用了换元法求最值，而不是利用求导求最值，更简化了运算。
【精编文档】内蒙古鄂尔多斯市第一中学2018-2019学年高二数学下学期期中试卷文.doc
2018~2019学年第二学期期中考试试题高二文科数学【本试卷满分150分，考试时间为120分钟】一、选择题（本大题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的）1.已知集合(){}ln 10M x x =+>，{}22N x x =-≤≤，则M N =（ ）A ．()0,2B ．[)0,2C ．(]0,2D ．[]0,22.设i 为虚数单位，若复数()()1i 1i m ++是纯虚数，则实数m =（ ） A ．1-B ．0C ．0或1D ．13.设函数22,1()log ,1x x f x x x -≤=>⎧⎪⎨⎪⎩，则满足()2f x ≤的x 的取值范围是（ ）A ．[]1,2-B ．[)(]1,11,4-C ．[)(]1,11,2-D ．[]1,4-4.通过随机询问110名性别不同的大学生是否爱好某项运动，得到如下的列联表：参照附表，得到的正确的结论是（ ）A ．在犯错误的概率不超过0.1%的前提下，认为“爱好该项运动与性别有关”B ．在犯错误的概率不超过0.1%的前提下，认为“爱好该项运动与性别无关”C ．有99%以上的把握认为“爱好该项运动与性别有关”D ．有99%以上的把握认为“爱好该项运动与性别无关” 附表： 参考公式：()()()()22()n ad bc K a b c d a c b d -=++++5. 已知,,a b c 是直线，β是平面，给出下列命题：（ ） ①若c a c b b a //,,则⊥⊥； ②若c a c b b a ⊥⊥则,,//；③若b a b a //,,//则ββ⊂；④若a 与b 异面，且//,a b ββ则与相交； ⑤若a 与b 异面，则至多有一条直线与a ，b 都垂直. 其中真命题的个数是（）A ． 1B ．2C ．3D ．46.设a R ∈，若函数ln 2y x ax =+在()1,2上有极值，则实数a 的取值范围是（ ）A. 11,48--⎛⎫ ⎪⎝⎭B.11,24--⎛⎫ ⎪⎝⎭C.1,2-+∞⎛⎫ ⎪⎝⎭ D ．1,4-∞-⎛⎫ ⎪⎝⎭7.过圆2216x y +=上一点P 作圆()222:0O x y m m +=>的两条切线，切点分别为A 、B ，若2π3AOB ∠=，则实数m =（ ）A ．2B ．3C ．4D ．98.曲线3x y e =-+在点(0,2)处的切线与直线0x =和2y x =所围成的三角形面积为( ) A. 43B. 23C. 4 D ．19.设(),()f x g x 分别是定义在R 上的奇函数和偶函数，()g x 恒不为．．．0．，当0x <时，()()()()0f x g x f x g x ''->，且(2)0f =，则不等式()()0f x g x <的解集是( )A ．()()2,02,-+∞B ．()()2,00,2-C ．()(),22,-∞-+∞D ．()(),20,2-∞-10.△ABC 的内角A ，B ，C 的对边分别为a ，b ，c .已知C =60°，b =6，c =3，角A 为( ) A ．30 B ．60 C ．75 D ．120 11.由偶数组成的数阵如右图：则第21行第4列的数为（ ） A.594 B.546C.592D.644 12.已知双曲线2222:1(0,0)x y E a b ab-=>>的右顶点为A ，抛物线2:16C y ax =的焦点为F ，若在E 的渐近线上存在点P ，使得PA PF ⊥，则E 的离心率的取值范围是（ ） A.()1,2B ．51,4⎛⎤ ⎥⎝⎦C ．()2,+∞D ．5,4+∞⎡⎫⎪⎢⎣⎭二、填空题（本大题共4小题，每小题5分，共20分，将答案填写在答题卡指定位置） 13. 等比数列{a n }的前n 项和为S n ，已知S 1，2S 2，3S 3成等差数列，则数列{a n }的公比为___.14. 平面向量a 与b 的夹角为2π3，1a =，1b =，则32a b -=_______. 15. 椭圆22221x y a b+=的内接矩形的最大面积是__________________.16. 庙会是我国古老的传统民俗文化活动，又称“庙市”或“节场”．庙会大多在春节、元宵节等节日举行．庙会上有丰富多彩的文化娱乐活动，如“砸金蛋”（游玩者每次砸碎一颗金蛋，如果有奖品，则“中奖”）．今年春节期间，某校甲、乙、丙、丁四位同学相约来到某庙会，每人均获得砸一颗金蛋的机会．游戏开始前，甲、乙、丙、丁四位同学对游戏中奖结果进行了预测，预测结果如下：甲说：“我或乙能中奖”；乙说：“丁能中奖”’；丙说：“我或乙能中奖”；丁说：“甲不能中奖”．游戏结束后，这四位同学中只有一位同学中奖，且只有一位同学的预测结果是正确的，则中奖的同学是_____． 三、解答题（解答应写出文字说明，证明过程或演算步骤） 17．（本小题满分10分）已知x a x x f ++=)(.（I ）当2=a 时，求不等式3)(<x f 的解集；（II ）设关于x 的不等式3)(<x f 有解，求a 的取值范围 . 18.（本小题满分12分）已知2x =是函数2()(23)x f x x ax a e =+--的一个极值点（⋅⋅⋅=718.2e ）． （I ）求实数a 的值；（II ）求函数()f x 在3[,3]2上的最大值和最小值．19．（本小题满分12分）为培养学生的阅读习惯，某校开展了为期一年的“弘扬传统文化，阅读经典名著”活动. 活动后，为了解阅读情况，学校统计了甲、乙两组各10名学生的阅读量（单位：本），统计结果用茎叶图记录如下，乙组记录中有一个数据模糊，无法确认，在图中以a 表示.（Ⅰ）若甲组阅读量的平均值大于乙组阅读量的平均值， 求图中a 的所有可能取值； （Ⅱ）将甲、乙两组中阅读量超过．．15本的学生称为“阅读达人”. 设1a =，现从所有“阅读达人”里任取2人，求其中甲组和乙组各有一人的概率；.（Ⅲ）记甲组阅读量的方差为20s ，在甲组中增加一名学生A 得到新的甲组，若A 的阅读量为 10，则记新甲组阅读量的方差为21s ；若A 的阅读量为20，则记新甲组阅读量的方差为22s ，试比较20s ，21s ，22s 的大小（结论不要求证明） 20. （本小题12分）如图，在四棱锥E ABCD -中，平面ABCD ⊥平面AEB ，且四边形ABCD 为矩形，120BAE=∠︒，4AE=AB=，2AD=，F G ，分别为BE AE ，的中点，H 为线段BC 的中点．（Ⅰ）求证：平面DAF ⊥平面CEB ； （Ⅱ）求三棱锥-B GFH 的体积..21．（本小题满分12分）在平面直角坐标系中，曲线C 的参数方程为为参数）（θθθ⎩⎨⎧+=+-=sin 44cos 43y x ，直线1l 的方程为kx －y ＋k ＝0，以坐标原点为极点，x 轴的正半轴为极轴建立极坐标系，直线2l 的极坐标方程为ρθθ4sin 2cos =-.(I)写出曲线C 的普通方程和直线2l 的直角坐标方程；(II)若1l 与C 交于不同的两点M ，N ，MN 的中点为P ，1l 与2l 的交点为Q ，1l 恒过点A ，求|AP |·|AQ |.HE22.（本小题满分12分）已知点(4,0)B 和点(4,0)C -，分别过点,B C 的直线,m n 相交于点A ，设直线,m n 的斜率分别为12,k k .（Ⅰ）如果12(0)k k a a ⋅=≠，求点A 的轨迹方程，并根据a 的取值讨论此轨迹是何种曲线； （Ⅱ）设（Ⅰ）中的曲线为C ，若不平行于坐标轴的直线l 与曲线C 交于点,M N ,线段MN 的中点为Q ，O 为坐标原点，设直线MN 与OQ 的斜率分别为,MN OQ k k , 求证：MN OQ k k a =.2018~2019学年高二第二学期期中考试试题（文科数学）一．CDDCAB ABDCAB二．13. 1314 . 19 15. 2ab 16. 甲三．17.解：（I ）当2=a 时，不等式32<++x x 等价于2253)2(2-≤<-⇒⎩⎨⎧<-+--≤x x x x …………………………………………………………1分 或023)2(02<<-⇒⎩⎨⎧<-+<<-x x x x ，…………………………………………………………2分 或2103)2(0<≤⇒⎩⎨⎧<++≥x x x x ……………………………………………………………3分 所以不等式3)(<x f 的解集是⎪⎭⎫⎝⎛-21,25.…………………………………………………5分（2）由题意得3)(min <x f …………………………………………………………………6分 因为a x a x x a x x f =-+≥++=)(，即a x f =min )(……………………………9分 故33,3<<-<a a 解得.………………………………………………………………10分18.解：（I ）由2()(23)x f x x ax a e =+--可得22()(2)(23)[(2)3]x x x f x x a e x ax a e x a x a e '=+++--=++-- … 2分∵2x =是函数()f x 的一个极值点，∴(2)0f '=∴2(5)0a e +=，解得5a =- …………………………4分 经检验5a =-时2x =是函数()f x 的一个极值点，∴5a =- …………………………………………………………… 5分 （II ）由0)1)(2()(>--='x e x x x f ，得)(x f 在)1,(-∞递增，在),2(+∞递增， 由0)(<'x f ，得)(x f 在在)2,1(递减∴2)2(e f =是()f x 在]3,23[∈x 的最小值；…………………………………………8分2347)23(e f =，3)3(e f = ∵)23()3(,0)74(4147)23()3(23233f f e e e e e f f >>-=-=- ∴()f x 在]3,23[∈x 的最大值是3)3(e f =． ……………………………………12分 19.解：（Ⅰ）甲组10名学生阅读量的平均值为12681011121217211010+++++++++=，乙组10名学生阅读量的平均值为124412131616(10)20981010a a+++++++++++=. …………………………………………………………………………………… 2分 由题意，得981010a+>，即2a <. …………………………………………… 3分 故图中a 的取值为0或1. ………………………………………………………… 4分（II ）甲组中阅读达人有2人记为a,b;乙组中阅读达人有3人记为c,d,e.从5位阅读达人中任取2人的基本事件共有10个，它们是()()()()()()()()()()e d e c d c e b d b c b e a d a c a b a ,,,,,,,,,,,,,,,,,,，…………………………6分其中甲组和乙组各有一人包含6个基本事件，它们是()()()()()()e b d b c b e a d a c a ,,,,,,,,,,,………………………………………………………8分所以甲组和乙组各一人的概率53106P ==………………………………………………10分 （Ⅲ）222102s s s <<. …………………………………………………………………… 12分20.（Ⅰ）证明：在矩形ABCD 中，⊥AD AB ，∵矩形⊥ABCD 平面AEB ，且平面ABCD I 平面AEB=AB ， ∴⊥AD 平面AEB ，又⊂BE 平面AEB ，∴⊥AD BE ， ……………………………………… 2分 ∵AE =AB ，F 为BE 的中点，∴⊥AF BE ，又AD AF =A I ，∴⊥BE 平面ADF ， ……………… ……………… 4分∵⊂BE 平面CEB ，∴平面⊥DAF 平面CEB ． …………………………… ………………… 6分（Ⅱ）由（Ⅰ）知,2AF EBC AF G AE ⊥=平面且，又因为点为中点，1G FBH ∴点到平面的距离为………………………………………………………………8分11122FBH S BF BH ∆=⋅=⨯=113G FBH V -∴=⨯…………………………………………………………………10分G FBH B GFH V V --=又B GFH V -∴=12分 21解：(1)由题可知：曲线C 的普通方程为(x ＋3)2＋(y －4)2＝16，………………………2分直线l 2的直角坐标方程为x －2y －4＝0.………………………………4分(2)设M ，N ，Q 所对应的参数分别为t 1，t 2，t 3，由题意得直线l 1恒过点A (－1,0)，…………………………………………………………5分故l 1的参数方程为⎩⎨⎧x ＝－1＋t cos α，y ＝t sin α(t 为参数)， (6)分代入曲线C 的普通方程得t 2＋4t (cos α－2sin α)＋4＝0，…………………………………7分 则t 1＋t 2＝4(2sin α－cosα)，…………………………………………………………………8分将⎩⎨⎧x ＝－1＋t cos α，y ＝t sin α代入x －2y －4＝0，得t 3＝5cos α－2sin α，………………………9分则|AP |·|AQ |＝⎪⎪⎪⎪⎪⎪t 1＋t 22·|t 3|＝2|2sin α－cos α|·⎪⎪⎪⎪⎪⎪5cos α－2sin α＝10.…………………12分22.解：（Ⅰ）令(,)A x y ∵12k k a = ∴44y ya x x ⋅=-+ 化简得2211616x y a -= ∴点A 的轨迹方程为221(4)1616x y x a-=≠± ………………………………… 3分 当0a >时，点A 的轨迹为双曲线 当0a <且1a ≠-时，点A 的轨迹为椭圆 当1a =-时，点A 的轨迹为圆 ………………………………6分 （Ⅱ）设112200(,),(,),(,)M x y N x y Q x y ，则0120122,2x x x y y y =+=+由（Ⅰ）知曲线C 的方程为2211616x y a-=即2216ax y a -= ∵,M N 在曲线C 上∴221116ax y a -= ① 222216ax y a -= ② ……………9分 ①-②得12121212()()()()a x x x x y y y y -+=-+ ∴012120y y y a x x x -⋅=-即MN OQ k k a = …………………………………12分。
内蒙古鄂尔多斯市第一中学2018_2019学年高二化学下学期期末考试试题(含解析)
2018-2019学年度第二学期期末考试试题高二化学答题要求：1.本卷满分100分，考试时间90分钟2.全部答案写在答题卡指定位置3.考试结束只交答题卡相对原子质量：H－1；C－12；N－14；O－16；Na－23；K－39；S－32；Cl－35.5第Ⅰ卷选择题一、选择题1.共建“一带一路”符合国际社会的根本利益，彰显人类社会的共同理想和美好追求。
下列贸易商品中，主要成分不属于有机物的是A．中国丝绸B．捷克水晶C．埃及长绒棉D．乌克兰葵花籽油A. AB. BC. CD. D【答案】B【解析】分析：蛋白质、纤维素、油脂均为有机物，二氧化硅为无机物，据此分析。
详解：A、中国丝绸主要成分是蛋白质，属于有机物，选项A错误；B、捷克水晶主要成分是二氧化硅，属于无机物，选项B正确；C、埃及长绒棉主要成分是纤维素，属于有机物，选项C错误；D、乌克兰葵花籽油主要成分是油脂，属于有机物，选项D错误。
答案选B。
2.下列对应符号表述正确的是A. 一氯甲烷的结构式CH3ClB. 葡萄糖的结构简式C6H12O6C. 苯的分子式D. 丙烯的实验式CH2【答案】D【解析】【分析】A．结构式是用元素符号和短线表示化合物（或单质）分子中原子的排列和结合方式的式子；B．葡萄糖的分子式为C6H12O6；C．苯的分子式为C6H6；D．烯烃的的通式为C n H2n，则实验式均为CH2。
【详解】A．一氯甲烷的结构式为，选项A错误； B．苯的分子式为C6H6，苯的结构简式为，选项B错误；C．葡萄糖的结构简式为CH2OH（CHOH）4CHO，选项C错误；D．烯烃的的通式为C n H2n，则丙烯的实验式为CH2，选项D正确。
答案选D。
3.把各组中的气体通入溶液中，溶液的导电能力显著增强的是（）A. CO2（g）通入NaOH溶液B. CO2（g）通入石灰水C. NH3（g）通入CH3COOH溶液D. NH3（g）通入盐酸中【答案】C【解析】试题分析：电解质溶液的导电能力与离子的浓度及离子所带的电荷有关，A、B、D过程离子浓度与离子所带电荷的乘积不变，C过程离子浓度变大，导电能力增强。
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内蒙古鄂尔多斯市第一中学2018-2019学年高二下学期 期中考试英语试卷 第一部分 阅读理解 （共两节，满分40分） 第一节（共15小题；每小题2分，满分30分） 阅读下列短文，从每题所给的四个选项(A, B, C和D)中，选出最佳选项。 A It’s no secret that your personal characteristics directly affect how you interact with the world, but you might be surprised at just how deep it goes. Experts think that your personality can even affect how you exercise and influence what sports are suitable for you. Highly Sensitive Highly sensitive people may be more uncomfortable with group exercise classes or team sports where they feel their every move is being observed. Additionally, they may feel more upset over an ineffective or poor workout, says researcher Elaine Aron. For such people, individual or non-competitive activities like biking, running and hiking are ideal. Type A Personality Type A individuals often have an “all or nothing approach” to exercise. This personality type is known for sticking closely to their plan, not to mention being super competitive. However, this can cause them to stick too closely to a fitness routine, which means they might try to push past an injury. To get better exercise results, Type A people need to be aware that following an exercise plan too strictly may limit their progress. They should be more flexible and listen to their body, especially when they are in pain. Type B Personality This laid-back group may get too lazy about their exercise plan, which prevents them from seeing results. They are often less willing to devote enough time and energy to their fitness goals, especially if exercise is something that makes them anxious. Type B individuals succeed in creative and co-operative environments, so team sports and group gym classes may be perfect. Remember, there are plenty of ways you can adapt a fitness plan to suit your needs, regardless of your characteristics. Just keep in mind what you like. 1. Why might highly sensitive people be unwilling to take part in group sports? A. They will feel like they are being evaluated. B. They don't like socializing with other people. C. They don't consider themselves to be athletic. D. They fear they'll be let down by their teammates. 2. Type A personalities can be best described as_________. A. committed B. flexible. C. independent. D. sensitive. 3. What is the purpose of the text? A. To help people understand what personality type they are. B. To explain how personal characteristics affect exercise habits. C. To identify the dangers of doing the wrong type of exercise. D. To describe different types of workouts that are available today. 【答案】1. A 2. A 3. B 【解析】 本文为说明文。人的性格特征直接影响一个人与世界的互动，而且影响很大。专家认为，个性甚至会影响人的健身，及影响适合你的运动。本文论述了不同的性格如何影响人的锻炼习惯 【1题详解】 细节理解题。根据第Highly Sensitive中的第一句Highly sensitive people may be more uncomfortable with group exercise classes or team sports where they feel their every move is being observed. Additionally, they may feel more upset over an ineffective or poor workout, says researcher Elaine Aron. 高度敏感的人可能对于团体运动班或团队运动有不舒服的感觉，在那里他们感觉他们的每一个动作被观察。他们可能对效率低的或者差的练习感到更沮丧。由此可知，高度敏感的人他们对于自己的每个动作被人观察，被评估而不安，分析选项可知A符合题意，故选A。 【2题详解】 推理判断题。根据Type A Personality中的Type A individuals …This personality type is known for sticking closely to their plan, not to mention being super competitive. “A型性格的人以一直坚持自己的计划而闻名，更不用说他们具有超强的竞争力了”。由此可知，这种性格的人往往一直坚持自己的计划，即做某事很投入。分析选项A. committed坚定的，献身于某种事业的；B. flexible灵活的；C. independent 独立的；D. sensitive敏感的可知，项Acommitted与句相符，故选A。 【3题详解】 推理判断题。本文的第一段说明了人的性格特征直接影响如何与世界互动，而且影响很大。专家认为，你的个性甚至会影响你的锻炼，及影响适合你的运动。接着下文解释了不同的性格如何影响的。分析选项是知B项（解释个人特征如何影响锻炼习惯）符合题意，故选B。 B Construction on Knolly’s Tunnel began in 1896, and it was opened on August 13th, 1898 by the man it was named after--Sir Clement C. Knolly, Acting Governor of Trinidad. It linked Rio Claro with Port of Spain. Its architecture is still admired and studied today, and many are amazed that Knolly’s Tunnel has stood up to the earth’s movement over so many years. Much research is being done on the tunnel. At the top of Knolly’s Tunnel are cottages covered with leaves, where visitors can sit and enjoy the beauty of nature. Standing at the beginning of Knolly’s Tunnel, you can see nothing but the tiny light at the end. The train tracks have been removed and replaced with small stones. Visitors can drive through the tunnel or walk through, but should do so in groups for safety. In the old days, there were no lights, but now there are street lights on the way to the tunnel, though not inside. On your way in, you may notice some manholes on the walls of the tunnel. These were there for individuals to step into for safety as the train passed. Knolly’s Tunnel can be accessed through Tabaquite and through Mitchell Gap. The road was recently improved, but is better when you enter from Tabaquite, and there are signs on the road directing you to Knolly’s Tunnel. On the drive to Knolly’s Tunnel there are two other cottages where visitors can sit and just enjoy nature. At the site itself, there is nothing to purchase to eat or drink, but in Tabaquite there are several bars, a restaurant, and food outlets. A gas station and a health centre are also close by. For those who admire architecture, Knolly’s Tunnel is a must-see when visiting Trinidad. For the nature lovers, there is no better place for them to be and for the historians, they can walk or