高二下 月考1

永昌县第一高级中学2021—2022—2第1次月考卷高二英语第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，从每题所给的四个选项（A、B、C和D）中，选出最佳选项。

APOETRYCHALLENGEWrite a poem about how courage, determination, and strength have helped you face challenges in your lifePrizes3 Grand Prizes: Trip to Washington, D. c. for each of three winners, a parent and one other person of the winner's choice. Trip includes round- trip air tickets, hotel stay for two nights, and tours of the National Air and Space Museum and the office of National Geographic World.6 First Prizes: The book Sky Pioneer: A Photo-biography of Amelia Earhart signed by author Corinne Szabo and pilot Linda Finch.50 Honorable Mentions: Judges will choose up to 50 honorable mention winners, who will each receive a T-shirt in memory of Earhart's final flight.RulesFollow all rules carefully to prevent disqualification.Write a poem using 100 words or fewer. Your poem can be any format, any number of lines.Write by hand or type on a single sheet of paper. You may use both the front and back of the paper.On the same sheet of paper, write or type your name, address, telephone number; and birth date. Mail your entry to us by October 31 this year.1.How many people can each grand prize winner take on the free trip?A. TwoB. Three. c. Four. D. Six.2. What will each of the honorable mention winners get?A. A plane ticket.B. A special T-shirt.C. A book by Corinne Szabo.D. A photo of Amelia Earhart.3.Which of the following will result in disqualification?A. Typing your poem out.B. Writing a poem of 120words.ing both sides of the paper.D. Mailing you entry on October 30.BOpera for BeginnersMany people think that opera is boring, difficult to understand and unpleasant to listen to. They only see strange people singing in a foreign language and wearing funny costumes. They're too quick to judge and a lot of people have believed stupid ideas about opera without ever actually seeing one. If one takes opera seriously, with an open heart, and goes to several good live performances, I don't see any reason why he or she would dislike or hate it. The following tips can help you learn to enjoy opera.First of all, if you are new to opera, start with a short, well-known title, such as The Marriage of Figaro by Mozart. In the beginning, avoid difficult operas with complex storylines. If your first experience is not fun, then you will expect all operas to be boring. Moreover, before going to a performance, do a little bit of homework: learn the story of the opera. If you know about the main characters and the story before you start watching, it will be much easier to understand what is happening on the stage. Finally, the most important way to enjoy opera is to keep an open mind. If you believe that you will enjoy watching opera, then you will most likely enjoy it. Think of opera as a live TV show. Opera shows the same things—love and hate, good and evil and humans and nature—that are found in many of today's popular movies and best-selling novels. In fact, many of these movies and novels use stories copied from famous operas!Opera is full of catchy, wonderful melodies （旋律）. It's full of drama and laughter. It's full of wonder and life. Opera can be fun to watch, and it allows the audience to experience different cultures. Many people have enjoyed it for hundreds of years, and by following the tips above, you can enjoy it, too. Indeed, opera will change the minds of the people who once hated it. And most importantly, as one of the most diverse art forms out there, opera helps you appreciate different performance styles from all over the world.4．Many people think that opera is boring because .A. it is like a TV showB. they dislike its storylinesC. they know little about itD. its music style is old-fashioned5．Which is a good way for beginners to enjoy opera?A. To start with a long opera.B. To learn to sing an opera.C. To master a foreign language.D. To get familiar with an opera's story.6．Many famous operas, popular movies, and best-selling novels .A. share a similar storyB. have the same writerC. teach people about lifestylesD. help people keep an open mind7．What does the underlined word “diverse” in the last paragraph probably mean?A. Ancient.B. Humorous.C. Various.D. Exciting.C"If music is the medicine of the soul(灵魂), let it play on," said a famous person. I think he said so because probably he got some help from music. Music has some strange abilities. Medical scientists have found that a person that feels stressed can actually listen to some kind of music and become well.The researchers said that since stress comes as a result of life events such as starting a new family, starting a new business, and starting a new job, one can actually listen to good music and feel good because good music touches the human mind in a positive way. Music helps you to forget the life events that make you worried and remember the important events that once happened in your life.[来源Depression(抑郁) is a disease caused by stress, smoking, social problems and so on. Depression is also caused by problems such as failure in business. Depression may bring us weakness, headache, and loss of concentration. Good music makes one remember happy moments or good days. If you play music about love, it makes you feel like falling in love again though you may have had several upset experiences. And such good feelings make you healthy.Anxiety(焦虑) is another health problem that can be controlled by music. Anxiety is a side effect of some major health problems such as cancer of the liver and cancer of the breast. Good music makes you feel relaxed and removes the pains from these diseases and you feel all right.Good music can send you to sleep. And you need to know that sleep puts your body in a healthy condition. Sleep takes away the effects of stress, depression and anxiety from a person.8. According to the famous person in Paragraph 1, music can .A. treat many kinds of diseasesB. help you keep healthy in your mindC. take the place of medicine in treating illnessD. make you remember things that happened9. Good music helps people remove stress mainly by .A. letting people have a good sleepB. making people focus only on important thingsC. showing something new to peopleD. making people think positively10. Which of the following is NOT a reason for depression mentioned in the passage?A. Stress.B. Smoking.C. Loss of concentration.D. Failure in business.11. Which of the following might be the best title for the passage?A. The Health Benefits of Listening to MusicB. Stress Can Bring Us Many Kinds of DiseasesC. Some Diseases Have Something to Do With MusicD. Tips For Us to Live a Healthy And Comfortable LifeDEinstein's $1.5 Million Tip to a BellboyIn addition to being a scientist, a philosopher （哲学家），and somewhat of a funny guy, Albert Einstein was something even more admirable: a good tipper. We know this because a piece of paper the Nobel Prize winner gave a bellboy as a tip sold for more than $1.5 million at auction （拍卖）.As the story goes, Einstein was traveling to Japan to give a lecture series when hefound out he'd been awarded the Nobel Prize in physics. News of the award spreadquickly and Einstein, somewhat bothered by all of the attention, kept himself in hisroom at a hotel in Tokyo.A bit embarrassed by all the publicity he was receiving, Einstein tried to write downhis thoughts and feelings. Just then, a bellboy came to Einstein's room to deliver a message. Not expecting him, Einstein didn't have any small change to tip him, and instead gave the bellboy a pair of the writings he'd been working on. Aware that his fame in the world was rising, Einstein supposedly told the Japanese messenger the notes would turn out to be more valuable than a regular tip if he was lucky.The two notes were both written in German and signed by Einstein himself. The first note offered the professor's “theory for happiness” It read: “A calm and modest life brings more happiness than the pursuit of success combined with constant restlessness （不安）.” The second one simply said, “Where there's a will, there's a way.”Tested or not, the theory does appear to be at least somewhat correct. According to a study of the connection between the life success and well-being of more than 275,000 people, happiness creates personal and professional success more often than success in those areas leads to happiness.Even if his philosophical thoughts contain no scientific value, the notes do shine a little light on Einstein's private thoughts. “What we're doing here is paintin g the picture of Einstein —the man, the scientist, his effect on the world---through his writings,” said Roni Grosz, who is in charge of the world's largest Einstein collection at Hebrew University.Whether the bellboy was able to benefit from Einstein's happiness theory is unclear, but the notes did eventually bringsome joy to one of his relatives when they were sold — the second note brought in $250,000, while the first one that lists Einstein's happiness theory sold for $1.56 million.12.How did Einstein feel about the public attention brought by winning the Nobel Prize?A. ConfidentB.TroubledC. EncouragedD. Regretful13. Why did Einstein give the bellboy the two notes as a tip?A. He didn't have small change with him at the time.B. The bellboy was interested in his thoughts and feelings.C . The bellboy delivered the message that he won the prize.D . The notes would become more valuable than a regular tip.14. From the passage we can know that .A. the second note sold for more moneyB. Roni Grosz is painting a picture of EinsteinC . Einstein went to Japan to give some lecturesD . the two notes brought great wealth to the bellboy15. Which of the following is most similar to Einstein's “theory for happiness”？A . People are just as happy as they make up their minds to be.B . Success is getting what you want; happiness is wanting what you get.C . Happiness is a by-product of an effort to make someone else happy.D .The search for happiness is one of the chief sources of unhappiness.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。

一、单选题1．设集合，集合N 为函数的定义域，则（ ）{}|12M x x =-≤≤()lg 1y x =-M N ⋂=A ． B ． C ． D ． ()12，[]12，[)12，(]12，【答案】D【分析】根据对数的真数为正数化简集合，进而由集合的交运算即可求解. (1,)N =+∞【详解】由，所以， 101x x ->⇒>(1,)N =+∞又，所以， {}|12M x x =-≤≤(]1,2M N = 故选:D2．若，则（ ） 43z i =-zz =A ．1 B ．-1C ．D ．4355i +4355i -【答案】C【分析】根据共轭复数与模长的求解计算即可.【详解】因为,故. 43z i =-4355z i z==+故选：C.3．已知椭圆中，长轴长为10 ）22221(0)x y a b a b +=>>A ．B ．10C ．D ．【答案】A【分析】根据椭圆长轴和离心率的概念即可求解.【详解】，所以；又因为 210a = 5a =c e a ==得c =2c =故选：A.4．设是直线，，是两个不同的平面，下列命题中正确的是（ ） l αβA ．若，，则 //l α//l β//αβB ．若，，则 αβ⊥l α⊥l β⊥C ．若，，则 αβ⊥//l αl β⊥D ．若，，则 //l αl β⊥αβ⊥【答案】D【解析】由线面平行的性质和面面平行的判定可判断选项A ；由面面垂直的性质定理和线面平行的性质可判断选项B ；由面面垂直的性质定理和线面位置关系可判断选项C ；由线面平行的性质和面面垂直的判定定理可判断选项D ；【详解】对于选项A ：若，，则或与相交，故选项A 不正确； //l α//l β//αβαβ对于选项B ：若，，则或，故选项B 不正确；αβ⊥l α⊥//l βl β⊂对于选项C ：若，，则或或与相交，故选项C 不正确；αβ⊥//l α//l βl β⊂l β对于选项D ：若，由线面平行的性质定理可得过的平面，设，则，所以//l αl γm γα= //m l ，再由面面垂直的判定定理可得，故选项D 正确；m β⊥αβ⊥故选：D5．已知{}是等差数列，且，则=（ ） n a 466,4a a ==10a A ．2 B ．0C ．D ．2-4-【答案】B【分析】根据等差数列基本量的计算即可求解.【详解】设等差数列的首项为，公差为，由，即，解得. {}n a 1a d 4664a a =⎧⎨=⎩113654a d a d +=⎧⎨+=⎩191a d =⎧⎨=-⎩所以，所以. 1(1)9(1)10n a a n d n n =+-=--=-+1010100a =-+=故选：B6．已知点P （x ,y ）是曲线上的一动点，则点P （x ,y ）到直线的距离的最小值为2y x =240x y --=（ ） ABCD ．35【答案】C【分析】当曲线在点P 处的切线与已知直线平行时点P 到该直线的距离最小，结合导数的几何意义和点到直线的距离公式计算即可求解.【详解】当曲线在点P 处的切线与直线平行时，点P 到该直线的距离最小，240x y --=，2y x '=由直线的斜率，则， 240x y --=2k =22x =得，有，所以， 1x =21y x ==(1,1)P ∴到直线距离. (1,1)P 240x y --=d ==故选：C.7．如图是下列四个函数中的某个函数在区间[-3,3]的大致图像，则该函数是（ ）A ．B ．C ．D ．22sin 1xy x =+321x xy x -=+22cos 1x xy x =+3231x xy x -+=+【答案】D【分析】利用赋值法，结合图形和排除法即可判断ABC ；利用导数和零点的存在性定理研究函数的单调性，结合图形即可判断D. 【详解】A ：设，由得， ()22sin 1x f x x =+π3π2<<sin 30>则，结合图形，不符合题意，故A 错误； ()2sin 33010f =>B ：设，则，结合图形，不符合题意，故B 错误；()321x xg x x -=+()10g =C ：设，当时，，，22cos ()1x x h x x =+π0,2x ⎡⎤∈⎢⎥⎣⎦cos [0,1]x ∈212x x +≥所以，即， 222cos 20111x x xx x ≤≤≤++0()1h x ≤≤当且仅当时等号成立，结合图形，不符合题意，故C 错误；1x =D ：设，则， 323()1x xu x x -+=+(0)x >422263()(1)x x u x x --+'=+(0)x >设，则，42()63v x x x =--+(0)x >3()4120v x x x '=--<所以函数在上单调递减，且， ()v x (0,)+∞(0)30,(1)40v v =>=-<故存在，使得，0(0,1)x ∈0()0v x =所以当时，即，当时，即，0(0,)x x ∈()0v x >()0u x '>0(,)x x ∈+∞()0v x <()0u x '<所以函数在上单调递增，在上单调递减，结合图形，符合题意，故D 正确. ()u x 0(0,)x 0(,)x +∞故选：D.8．已知△ABC 的三个内角分别为A ，B ，C ，且满足，则的最大值为222sin 2sin 3sin C A B =-tan B （ ） ABCD ．54【答案】B【分析】利用正弦定理及余弦定理表示，结合基本不等式求得的取值范围，从而求得cos B cos B 的取值范围，即得.tan B 【详解】依题意，222sin 2sin 3sin C A B =-由余弦定理得，， 22223c a b =-2222133b ac =-所以 222222222222114143333cos 2226a c a c a ca cb ac B ac ac ac ac+-+++-+====⋅，当且仅当时等号成立， 1263≥=2a c =即为锐角，，， B 2cos 13B ≤<22419cos 1,19cos 4B B ≤<<≤，222222sin 1cos 15tan 10,cos cos cos 4B B B B B B -⎛⎤===-∈ ⎥⎝⎦所以. tan B 故选：B.二、多选题9．下列说法正确的是（ ） A ．直线在y 轴上的截距为2 24y x +=B ．直线必过定点（2,0） ()20R ax y a a --=∈C ．直线的倾斜角为10x +=2π3D ．过点且垂直于直线的直线方程为 ()2,3-230x y -+=210x y ++=【答案】BD【分析】根据直线的截距式方程即可判断A ，根据直线恒过定点的求法即可判断B ，根据直线斜率的定义即可判断C ，根据垂直直线斜率之积为-1，结合直线的点斜式方程即可判断D. 【详解】A ：直线在轴上的截距为，所以A 不正确； 24y x +=y 2-B ：由，得，20ax y a --=(2)0x a y --=令，解得：，所以该直线恒过定点，故B 正确；200x y -=⎧⎨=⎩20x y =⎧⎨=⎩(2,0)C ：设直线的倾斜角为，，斜率为 10x +=α(]0,απ∈由，故C 错误；tan α=56πα=D ：由直线，得该直线的斜率为，230x y -+=12所以过点且垂直于直线的直线斜率为， (2,3)-230x y -+=2故其方程为，即，故D 正确. 32(2)y x -=-+210x y ++=故选：BD.10．斜率为1的直线l 经过抛物线的焦点F ，且与抛物线相交于两点则下24y x =()()1122,,,A x y B x y 列结论正确的有（ ） A ．B ．抛物线的准线方程为 (1,0)F 1y =-C ．D ．3OA OB ⋅=-10AB =【答案】AC【分析】由抛物线的性质判断AB ；联立直线l 和抛物线方程，利用韦达定理，以及数量积公式、抛物线的定义判断CD.【详解】由抛物线知，焦点，准线方程为，所以A 正确，B 不正确.24y x =(1,0)F =1x -由，消去得：，所以， 214y x y x=-⎧⎨=⎩y 2610x x -+=126x x +=121=x x 所以，所以C 正确； 121212121212(1)(1)2()13OA OB x x y y x x x x x x x x ⋅=+=+--=-++=- 所以，所以D 不正确. 12||28AB x x =++=故选：AC11．已知函数，其图像相邻两条对称轴之间的距离为，且函数()()cos (0,2f x x πωϕωϕ=+><π2是奇函数，则下列判断正确的是（ ）π3f x ⎛⎫- ⎪⎝⎭A ．函数f （x ）的最小正周期为B ．函数f （x ）的图像关于点（，0）对称 ππ6C ．函数f （x ）在上单调递增D ．函数f （x ）的图像关于直线对称 3ππ4⎡⎤⎢⎥⎣⎦，7π12=-x 【答案】ABD【分析】利用函数图像相邻两条对称轴之间的距离为和函数是偶函数，求出π2π()3f x -，从而可判断选项A 正确；再利用余弦函数的图像与性质，可以判断出选项()cos(2π)6=+f x x BCD 的正误.【详解】因为函数图像相邻两条对称轴之间的距离为，则，π2π22T =πT ∴=又，2π,0T ωω=>2ω∴=又函数是偶函数，因为， π()3f x -ππ2π()cos(2())cos(2)333f x x x ϕϕ-=-+=-+所以，即， 2πππ(Z)32k k ϕ-+=+∈7ππ(Z)6k k ϕ=+∈又，，则.π2ϕ<π6ϕ∴=()cos(2π)6=+f x x 函数最小正周期，故选项A 正确； πT =函数图像对称点的横坐标为：，即， ππ2π(Z)62x k k +=+∈ππ(Z)62k x k =+∈令时，，故选项B 正确； 0k =π6x =又由：，得到 ππ2π22π(Z)6k x k k -+≤+≤∈7ππππ(Z)1212k x k k -+≤≤-+∈所以函数的单调增区间为：， ()cos(2π)6=+f x x 7πππ,π(Z)1212k k k ⎡⎤-+-+∈⎢⎥⎣⎦令时，得到一个增区间为： 1k =-5π11π,1212⎡⎤⎢⎥⎣⎦故选项C 错误；函数图像的对称所在直线方程为；， πππ2π,(Z)6122k x k x k +==-+∈令时，，故选项D 正确. 1k =-7π12=-x 故选：ABD12．将全体正整数按照以下排列的规律排成一个三角形数阵，下列结论正确的是（ ）A ．第8行最右边的数为38B ．第10行从右向左第个5数为51C ．第10行所有数的和为505D ．第64行从左向右第7个数为2023 【答案】BCD【分析】根据三角数阵可知第行共有个数，且第行的最后一个数字是：，即为n n n 123n ++++ .结合等差数列前n 项求和公式计算，依次判断选项即可. (1)2n n +【详解】由三角形数阵可知， ①第行共有个数；n n ②第行的最后一个数字是：，即为. n 123n ++++ (1)2n n +A ：因为，故A 错误； 1234567836+++++++=B ：因为，1234567891055+++++++++=所以第行中的个数字依次为.故B 正确； 101046,47,48,49,50,51,52,53,54,55C ：由，故C 正确；()5545104655464748495051525354555052S S ⨯+-=+++++++++==D ：由，知第行最后的一个数为；()6316312346320162⨯++++++== 632016所以第行中的数字从左到右依次为642017,2018,2019,2020,2021,2022,2023,2024，，第7个数为2023，故D 正确. L 故选：BCD.三、填空题13．已知函数的最小正周期为，则___________. ()()sin 0f x x ωω=>πω=【答案】2【分析】利用正弦型函数的周期公式可求得的值.ω【详解】因为函数的最小正周期为，则. ()()sin 0f x x ωω=>π2π2πω==故答案为：.214．已知直线和圆相交于、两点，则弦长:210l x y --=22:210C x y y +--=A B AB =__________．【详解】由圆方可知其圆心坐标为，半径∴C (0,1)r =d. AB ===点睛：本题主要考查了直线与圆相交求截得弦长问题，属于基础题；求直线被圆所截得的弦长时，根据圆的性质通常考虑由弦心距，弦长的一般作为直角边，圆的半径作为斜边，利用勾股定理来解决问题，通常还会用到点到直线的距离公式.15．已知双曲线，若过右焦点F 且倾斜角为的直线与双曲线的右支有两个22221(0,0)x y a b a b-=>>30 交点，则此双曲线离心率的取值范围是___________.【答案】【分析】根据题意可知双曲线的渐近线方程的斜率需小于直线的斜率，得，结合b y x a =b <.b =【详解】由题意知，双曲线的渐近线方程为， by x a=±要使直线与双曲线的右支有两个交点， 需使双曲线的渐近线方程的斜率小于直线的斜率， by x a=即，即，由tan 30b a ︒<=b <b =，整理得，所以 <2234c a <c e a =<因为双曲线中，所以双曲线的离心率的范围是， 1e >故答案为：． 16．已知三棱锥的所有顶点都在球O 的球面上，SC 是球O 的直径若平面平面S ABC -.SCA ⊥SCB ，，，三棱锥的体积为9，则球O 的表面积为______． SA AC =SB BC =S ABC -【答案】36π【详解】三棱锥S−ABC 的所有顶点都在球O 的球面上，SC 是球O 的直径，若平面SCA ⊥平面SCB ，SA=AC ，SB=BC ，三棱锥S−ABC 的体积为9， 可知三角形SBC 与三角形SAC 都是等腰直角三角形，设球的半径为r ， 可得 ，解得r=3. 112932r r r ⨯⨯⨯⨯=球O 的表面积为： .2436r ππ=点睛：与球有关的组合体问题，一种是内切，一种是外接．解题时要认真分析图形，明确切点和接点的位置，确定有关元素间的数量关系，并作出合适的截面图，如球内切于正方体，切点为正方体各个面的中心，正方体的棱长等于球的直径；球外接于正方体，正方体的顶点均在球面上，正方体的体对角线长等于球的直径.四、解答题17．已知数列{a n }的前n 项和为S n ，且满足，. 13a =123n n S a ++=（1）求数列{a n }的通项公式；（2）若等差数列{b n }的前n 项和为T n ，且，，求数列的前n 项和Q n ．11T a =33T a =11{}n n b b +【答案】（1）（2）3nn a =9(21)nn +【分析】（1）根据数列的通项与的关系，化简求得，得到数列是首项为n a n S 13()n n a a n N ++=∈{}n a 3、公比为3的等比数列，即求解通项公式； （2）由（1）可得，得到，利用裂项法，3(21)n b n =-()()11111192n 12n 1182n 12n 1n n b b +⎛⎫==- ⎪-+-+⎝⎭即可求解．【详解】（1）当时，得， 1n =29a =由，得，123n n S a ++=123(2)n n S a n -+=≥两式相减得，又，∴，112()n n n n S S a a -+-=-1n n n S S a --=13(2)n n a a n +=≥又，∴，显然， 213a a =13()n n a a n N ++=∈10,3n n na a a +≠=即数列是首项为3、公比为3的等比数列，∴；{}n a 1333n nn a -=⨯=（2）设数列的公差为，则有，{}n b d 13b =由得，解得，∴，33T a =13327b d +=6d =3(1)63(21)n b n n =+-⨯=-又， ()()11111192n 12n 1182n 12n 1n n b b +⎛⎫==- ⎪-+-+⎝⎭∴==． n 111111Q 1183352n 12n 1⎡⎤⎛⎫⎛⎫⎛⎫=-+-+⋯+- ⎪ ⎪ ⎪⎢⎥-+⎝⎭⎝⎭⎝⎭⎣⎦111182n 1⎛⎫- ⎪+⎝⎭()n 92n 1+【点睛】本题主要考查等比数列的定义及通项公式、以及“裂项法”求和的应用，此类题目是数列问题中的常见题型，解答中确定通项公式是基础，准确计算求和是关键，易错点是在“裂项法”之后求和时，弄错项数导致错解，能较好的考查逻辑思维能力及基本计算能力等.18．若△ABC 的内角A ，B ，C 的对边分别为a ，b ，c ，满足.222sin sin sin sin sin A B C B C --=(1)求角A ；(2)若，求△ABC 周长的取值范围. 6a =【答案】(1) 2π3A =(2)(12,6+【分析】（1）根据正弦定理边角互化，可得，由余弦定理即可求解，222a b c bc --=（2）根据正弦定理得，由内角和关系以及和差角公式可得b B=1sin 2c B B ⎫=-⎪⎪⎭，进而由三角函数的性质即可求解.【详解】（1）由正弦定理可得：，222a b c bc --=，， 2221cos 22c b a A bc +-∴==-()0,πA ∈ 2π3A ∴=（2）因为，，所以，故πA B C ++=2π3A =π3B C +=ππ(0)33C BB =-<<由正弦定理得： 62πsin sin sin sin3a bc A B C====所以，b B=π1sin 32c C B B B ⎫⎛⎫==-=-⎪ ⎪⎪⎝⎭⎭所以周长 ABCA 1π6sin 623a b cB B B B ⎫⎛⎫=++=++-=++⎪ ⎪⎪⎝⎭⎭因为，则π03B <<ππ2π<333B <+πsin 13B ⎛⎫<+≤ ⎪⎝⎭故π12663B ⎛⎫<++≤+ ⎪⎝⎭求周长的取值范围为.ABC A (12,6+19．某厂研制了一种生产高精产品的设备，为检验新设备生产产品的某项指标有无提高，用一台旧设备和一台新设备各生产了10件产品，得到各件产品该项指标数据如下： 旧设备9.810.3 10.0 10.29.99.810.0 10.1 10.29.7新设备 10.1 10.4 10.1 10.0 10.1 10.3 10.6 10.5 10.4 10.5旧设备和新设备生产产品的该项指标的样本平均数分别记为和，样本方差分别记为和．x y 21s 22s（1）求，，，；x y 21s 22s（2）判断新设备生产产品的该项指标的均值较旧设备是否有显著提高（如果y x -≥认为新设备生产产品的该项指标的均值较旧设备有显著提高，否则不认为有显著提高）． 【答案】（1）；（2）新设备生产产品的该项指标的均值较旧设221210,10.3,0.036,0.04x y s s ====备有显著提高.【分析】（1）根据平均数和方差的计算方法，计算出平均数和方差.（2）根据题目所给判断依据，结合（1）的结论进行判断. 【详解】（1）， 9.810.31010.29.99.81010.110.29.71010x +++++++++==， 10.110.410.11010.110.310.610.510.410.510.310y +++++++++==， 22222222210.20.300.20.10.200.10.20.30.03610s +++++++++==. 222222222220.20.10.20.30.200.30.20.10.20.0410s +++++++++==（2）依题意，， 0.320.15y x -==⨯===，所以新设备生产产品的该项指标的均值较旧设备有显著提高. y x -≥20．设函数，其中.22()3ln 1f x a x ax x =+-+0a >（1）讨论的单调性；()f x （2）若的图象与轴没有公共点，求a 的取值范围.()y f x =x 【答案】（1）的减区间为，增区间为；（2）. ()f x 10,a ⎛⎫ ⎪⎝⎭1,+a ⎛⎫∞ ⎪⎝⎭1a e >【分析】（1）求出函数的导数，讨论其符号后可得函数的单调性.（2）根据及（1）的单调性性可得，从而可求a 的取值范围.()10f >()min 0f x >【详解】（1）函数的定义域为，()0,∞+又， ()23(1)()ax ax f x x+-'=因为，故，0,0a x >>230ax +>当时，；当时，； 10x a<<()0f x '<1x a >()0f x '>所以的减区间为，增区间为. ()f x 10,a ⎛⎫ ⎪⎝⎭1,+a ⎛⎫∞ ⎪⎝⎭（2）因为且的图与轴没有公共点，()2110f a a =++>()y f x =x 所以的图象在轴的上方，()y f x =x 由（1）中函数的单调性可得， ()min 1133ln 33ln f x f a a a ⎛⎫==-=+ ⎪⎝⎭故即. 33ln 0a +>1a e>【点睛】方法点睛：不等式的恒成立问题，往往可转化为函数的最值的符号来讨论，也可以参变分离后转化不含参数的函数的最值问题，转化中注意等价转化. 21．如图，边长为2的正方形ABCD 所在的平面与半圆弧CD 所在平面垂直，M 是CD 上异于C ，D 的点.(1)证明：平面AMD ⊥平面BMC ；(2)当三棱锥体积最大时，求面MAB 与面MCD 所成二面角的正切值.M ABC -【答案】(1)证明见解析；(2)2.【分析】（1）证得平面，结合面面垂直的判定定理即可证出结论；DM ⊥BMC （2）当在的中点位置时体积最大，建立空间直角坐标系，利用空间向量的夹角坐标公式即M A AB 可求出结果.【详解】（1）由题设知，平面平面，交线为.CMD ⊥ABCD CD 因为，平面，BC CD ⊥BC ⊂ABCD 所以平面，平面，BC ⊥CMD DM ⊂CMD 故,因为是上异于，的点，且为直径， BC DM ⊥M A CDC D DC 所以，又，平面，DM CM ⊥BC CM C =I ,BC CM ⊂BMC 所以平面，而平面，DM ⊥BMC DM ⊂AMD故平面平面；AMD ⊥BMC （2）以D 为坐标原点，的方向为轴正方向，的方向为轴正方向，建立如图所示的空间DA x DC y 直角坐标系.D xyz -当三棱锥M −ABC 体积最大时，M 为的中点.CD 由题设得，()()()()()0,0,0,2,0,0,2,2,0,0,2,0,0,1,1D A B C M()()()2,1,1,0,2,0,2,0,0AM AB DA =-==设是平面MAB 的法向量，则(),,n x y z = 即，可取， 00n AM n AB ⎧⋅=⎪⎨⋅=⎪⎩ 2020x y z y -++=⎧⎨=⎩()1,0,2n = 又是平面的一个法向量，因此 DAMCD， cos ,n DA n DA n DA ⋅=== []0π,,n DA ∈ 得， sin ,n DA = tan ,2n DA = 所以面与面所成二面角的正切值是.MAB MCD 222．已知椭圆的左，右焦点分别为、，离心率为，直线l 经过点2222:1(0)x y C a b a b+=>>1F 2F 122F 且与椭圆C 交于不同两点A ，B ，当A 是椭圆C 上顶点时，l 与圆相切.223x y +=(1)求椭圆C 的标准方程；(2)求的取值范围.11F A F B ⋅ 【答案】(1) 2211612x y +=(2)[]12.7-【分析】（1）根据题意列出方程组，解之即可；22212bc c e a c a b⎧=⎪⎪==⎨⎪⎪=-⎩（2）当直线的斜率不存在时，易得；当直线的斜率存在时，设直线方程为l 117F A F B ⋅= l ，，，联立椭圆方程，利用韦达定理和平面向量数量积的坐标表示可得(2)y k x =-11(,)A x y 22(,)B x y ，令得，结合不等式的性质计算即可求解. 11F A F B ⋅= 22283634k k -+2343t k =+≥11577F A F B t ⋅=- 【详解】（1）当A 为椭圆的上顶点时，直线l 与圆相切， 则圆心到直线l ，a =有，得，1122bc a =bc =则，解得22212bc c e a c a b⎧=⎪⎪==⎨⎪⎪=-⎩4,a b ==所以椭圆的标准方程是； C 2211612x y +=（2）由（1）知，则椭圆的左焦点，当直线的斜率不存在时，2c =1(2,0)F -l 易求得，，则；(2,3)A (2,3)B -11443(3)7F A F B ⋅=⨯+⨯-= 当直线的斜率存在时，设直线方程为，，. l (2)y k x =-11(,)A x y 22(,)B x y 由，消得，， ()22211612y k x x y ⎧=-⎪⎨+=⎪⎩y 2222(34)1616480k x k x k +-+-=， 21221634k x x k ∴+=+2122164834k x x k-=+ 21112121212(2)(2)(2)(2)(2)(2)F A F B x x y y x x k x x ⋅=+++=+++--2221212(1)2(1)()4(1)k x x k x x k =++-+++， 2222222221648162836(1)2(1)4(1)343434k k k k k k k k k --=+⨯+-⨯++=+++令，则， 2343t k =+≥2112283675757734k t F A F B k t t--⋅===-+ ，，， 3t ≥ 1103t <≤571277t -≤-<综上可知，的取值范围是. 11F A F B ⋅ []12,7-。

2023上学期第一次月考试卷高二化学试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。

共100分。

考试时间75分钟。

可能用到的相对原子质量：C-12 N-14 K-39 Fe-56第Ⅰ卷（选择题）一．选择题（本题共15个小题，每小题3分，共45分，每个小题只有一个正确答案）1．下列对示意图的解释正确的是（）A B C D示意图解释C2H2分子的空间填充模型基态Be原子最外层电子的电子云图NH3的立体结构金属原子平面密置层排列A．A B．B C．C D．D2.在下列各组元素中，有一组原子的第一电离能分别是1 086 kJ·mol－1、1 402 kJ·mol－1、1 313 kJ·mol－1。

那么这组元素可能是()A.C、N、OB.F、Ne、NaC.Be、B、CD.S、Cl、Ar3.按电子排布，可以把周期表中的元素划分为5个区，以下元素属于p区的是()A.FeB.MgC.BrD.Cu4.下列物质中σ键和π键数目比为1∶2的是()A.O2B.HCNC.CO2D.N25.下列分子或离子的中心原子，带有一对孤电子对的是()A.H2OB.BeCl2C.CH4D.PCl36.氯化亚砜(SOCl2)是一种很重要的化学试剂，可以作为氯化剂和脱水剂。

下列关于氯化亚砜分子的空间结构和中心原子(S)采取何种杂化方式的说法正确的是()A.三角锥形、sp3B.平面三角形、sp2C.平面三角形、sp3D.三角锥形、sp27.下列晶体性质的比较，错误的是()A.熔点：金刚石>碳化硅>晶体硅B.沸点：NH3>PH3C.硬度：白磷>冰>二氧化硅D.熔点：SiI4＞SiBr4＞SiCl48．第ⅣA族元素中C是生物分子骨架的构成元素，Si、Ge可用作半导体材料。

下列有关说法错误的是（）A．三种元素原子的次外层电子排布均是全充满状态B．第一电离能：Si PC．SiH4的相对分子质量比CH4大，范德华力更强，故SiH4的热稳定性强于CH4D．原子半径Ge>Si>C的原因是电子层数增加对半径的影响大于核电荷数增加的影响9．下列说法正确的是（）A ．键角：4332CH >BF >NH >H OB ．σ键和π键比例为7：1C ．N 2与O 22+互为等电子体，1molO 22+中含有的π键数目为2N AD ．SiH 4、CH 4、P 4的空间构型为正四面体形，键角都相同10．下列说法中不正确的是（ ）A ．X 射线衍射实验可以区分晶体和非晶体，也可以获得晶体的键长和键角的数值B ．价层电子对互斥模型一般不用于预测以过渡金属为中心原子的分子空间结构C ．杂化轨道用于形成σ键或用来容纳未参与成键的孤电子对，未参与杂化的p 轨道可用于形成π键D ．一元有机酸R COOH (R-是直链烷基)中，烷基是推电子基团，烷基越长羧酸的酸性越强11.X 和Y 是原子序数大于4的短周期元素，X m ＋和Y n －两种离子的电子排布式相同，下列说法中正确的是( )A.X 的原子半径比Y 小，X m ＋的离子半径比Y n －大B.X m ＋和Y n －的电子所占用的原子轨道的类型和数目都相同C.电负性：X>YD.第一电离能：X>Y12．某化合物可用作发酵助剂，结构如图所示。

连城一中2022—2023学年下学期高二年级月考1英语试卷满分150分考试时间120分钟第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话.每段对话后有一个小题,从题中所给的A, B, C三个选项中选出最佳选项.听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. Where are the speakers?A. At a swimming pool．B. In a clothing shop．C. At a school lab．2. What will Tom do next?A. Turn down the music．B. Postpone the show．C. Stop practicing．3. What is the woman busy doing?A. Working on a paper．B. Tidying up the office．C. Organizing a party．4. When will Henry start his vacation?A. This weekend．B. Next week．C. At the end of August．5. What does Donna offer to do for Bill?A. Book a flight for him．B. Drive him to the airport．C. Help him park the car．第二节(共15小题；每小题1.5分，满分22.5分)听下面５段对话或独白。

每段对话或独白后有几个小题，从题中所给的A, B, C 三个选项中选出最佳选项。

听每段对话或独白前，你将有时间阅读各个小题，每小题５秒钟；听完后，各小题将给出５秒钟的作答时间.每段对话或独白读两遍. 听第6段材料，回答第6，7题。

6. Why does Pete call Lucy?A. To say that he’ll be late．B. To tell her about his work．C. To invite her to dinner．7. When is Pete going to see Lucy?A. At 6:00 pm．B. At 6:45 pm．C. At 8:00 pm．听第7段材料,回答第8至10题。