七宝中学高二期末综合复习一及答案
七宝中学2020学年第一学期高二期末考试详解版
于是,由 a2
2c2
及
a
c ,可得
c a
2 2
,1
,选
B.
三、解答题
17.(1)直线 A B 以 A B 4, 3 为方向向量,过点 A ,
∴直线 A B 的方程为 x 1 y 2 ,即 3x 4y 11 0 ; 4 3
(2)①直线 l 与直线 A B 平行
设 l : 3x 4y m 0 ,则 m 11 2 ,解得 m 1 或 m 21, 5
(1)求直线 A B 的方程; (2)若 A , B 到直线 l 的距离都是 2 ,求直线 l 的方程.
18.(本题满分 14 分)本大题共有 2 小题,第 1 小题满分 6 分,第 2 小题满分 8 分.
双曲线 M : x 2 a2
y2
b2
1 过点
P
4,
6 2
,且它的渐近线方程是
x
2y
0
.
12.已知 F1, F2 分别为双曲线 C :
x2 a2
y2 b2
1a,b 0 的左、右焦点,过 F2 的直线 l
与双曲线
的右支分别交于 A , B 两点, △A F1F2 的内切圆半径为 r1 , △BF1F2 的内切圆半径为 r2 ,若
r1 2r2 ,则直线 l 的斜率为
.
二、选择题(本大题共有 4 题,满分 20 分)每题有且只有一个正确答案,考试应在答题纸 的相应编号上,将代表答案的小方格涂黑,选对得 5 分,否则一律得零分.
七宝中学 2020学年第一学期高二期末考试
数学试卷
一、填空题(本大题共有 12 题,满分 54 分)只要求直接填写结果,第 1~6 题每个空格填 对得 4 分,第 7~12 题每个空格填对得 5 分,否则一律零分.
上海市闵行区七宝中学2024学年物理高二上期末联考模拟试题含解析
[3]洗衣机耗电为:
W=Pt=440W×1h=440W⋅h=0.44kW⋅h
A.粒子在垂直于板的方向上的分运动可能是往复运动
B.粒子在垂直于板的方向上的分运动是单向运动
C.只要周期T和电压 的值满足一定条件,粒子就可沿与板平行的方向飞出
D.粒子不可能沿与板平行的方向飞出
10、如图为两个不同闭合电路中两个不同电源的U-I图像,下列判断正确的是( )
A.电动势E1=E2,内阻r1<r2
二、多项选择题:本题共4小题,每小题5分,共20分。在每小题给出的四个选项中,有多个选项是符合题目要求的。全部选对的得5分,选对但不全的得3分,有选错的得0分。
7、如图所示,足够长的U形光滑金属导轨平面与水平面成θ角,其中MN与PQ平行且间距为L,导轨平面与磁感应强度为B的匀强磁场垂直,导轨电阻不计.金属棒ab由静止开始沿导轨下滑,并与两导轨始终保持垂直且接触良好,ab棒接入电路的电阻为R,当流过ab棒某一横截面的电荷量为q时,棒的速度大小为v,则金属棒ab在这一过程中( )
A.加速度为 B.下滑的位移为
C.产生的焦耳热为 D.受到的最大安培力为
8、如图所示,质量为m的环带+q电荷,套在足够长的绝缘杆上,动摩擦因数为µ下滑,则以下说法正确的是( )
A.环在下滑过程中,加速度不断减小,最后为零
B.环在下滑过程中,加速度先增大后减小,最后为零
(1)求磁感应强度B的大小;
(2)若保持B的大小不变而将方向改为垂直于斜面向上,求金属杆的加速度
参考答案
一、单项选择题:本题共6小题,每小题4分,共24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、C
【解题分析】根据运动学公式求得运动的加速度;根据牛顿第二定律求得牵引力;
七宝中学高二下学期期末试卷
七宝中学高二下学期期末试卷II. Grammar and Vocabulary Section AYou 21 think your English skills are good, but do you know what a "knowledge navigator" is? How about a ―replenishment controller‖?Don't worry if you don't understand what these two job titles are- 22 would most native English speakers. But thanks to 23 UK company, we know that these titles from two job adverts mean “teacher”and “shelf stacker(货架整理员)”The Plain English Campaign (PEC) 24 (work) since 1979 to rid the UK of over-complicated language. It‘s aim is to make English as clear as possible for members of the public. It rewrites and edits documents, such as government application forms or job advertisements, 25 (make) them understandable.Research carried out last year by Business in the Community, a UK charity, even found that two third of 16-to-24-year-olds are put off applying for jobs 26 they don‘t understand their descriptions.―27 Jargon(专业术语)is describing is very important. It should be articulated(表达)very clearly,‖Steve Jenner, PEC spokesperson, told the BBC.According to the Guardian, some of 28 (unusual) workplace jarjon includes ―blue-sky thinking‖-meaning open-minded planning-‖touch base offline‖-meaning to meet and talk 29 person-and ―don‘t let the grass grow too long‖-which is another way of asking someone to work faster.But it‘s not just documents and workplace language 30 the PEC targets. Every year, it nominates public figures for it‘s ―Foot in Mouth award‖. To put one‘s mouth is a phrase that means to say something foolish. For the last two years, US president Donald Trump has features on the list.“Trump is notoriously(极度地)generous when it comes to providing memorable nonsense,”the PEC wrote on its website. “He seemed to say something daft every few days during his campaign for the Republication Presidential candidacy.”You may still remember the “Eight Minutes of Tokyo”in the closing ceremony of the Rio Olympics last August. Even if the Tokyo Games wasn‘t going to be here for another four years, the performance of the new host successfully 31 the show with its famous animated characters –Doraemon, Hello Kitty and Super Mario.It was a wise choice since there is probably nothing that shouts ―Japan‖ more loudly than the country‘s animation, known as ―Japanimation‖. And this year32 the 100th anniversary of the very first Japanese cartoon, made in 1917.In the past century, the wild imagination of Japanese animators continued to feed our33 . Monsters, fairies, robots and magic feature often in their work. It has also been inspiring film industries in other parts of the world. The 34 story of Disney’s The Lion King (1994), for example, actually comes from Japan’s Kimba the White Lion (《森林大帝》,1965). And the 1999 Hollywood 35 film, The Matrix, was also 36 influenced by the 1989 Japanese manga Ghost in the Shell (《攻壳机动队》).“I love his films. I study his films. I watch his films when I‘m looking for 37 ,‖ John Lasseter, di rector of Pixar‘s Toy Story, once said about famous Japanese animator Hayao Miyazaki.And our 38 for this imaginary world is only growing.At the end of last year, for example, the story of Japanese cartoon Pokemon (《宠物小精灵》,1997-) was brought to life with the help of augmented reality (AR,增强现实) technology. People in many countries are often seen searching for Pokemon in real life locations through the screens of their phones. They play it on their way to school, to work, and during holiday outings. Although Pokemon began as a video game a year before the cartoon came out, people should give the animation a ―Thanks‖ for bringing it to a wider audience.example, wearing glasses used to be considered as uncool and geeky (书呆子气的), but after the 1981 TV animation Arale (《阿拉蕾》), in which there is a heavily-nearsighted (高度近视的) girl with wings and magic powers, glasses soon became fashionable. And the language we use –the word meng (萌), to name one –is also 40 from Japanese animation.But interestingly, with all the imagination that is so admired by modern fans, it would still be hard for Japanese animators in 1917 to believe that the two-dimensional worlds that they were creating at the time would have such a big influence i n today‘s three-dimensional world.III. Reading ComprehensionSection AReading to me is all about personal growth. Growth through reading can take many 41 .It can come from seeing the world through the eyes of someone else, discovering a new truth or reconnecting with a truth long 42 . It can even come from an uncontained guffaw(狂笑)while reading a humorous passage. Growth through reading is not about reaching a certain level or reading so many minutes a day. It is 43 not about finding one type of reading material more valid than another. It’s about making something such an 44 part of your life that you don‘t feel the need to quantity what you‘re doing. You should read because you want to, not because you have to.At a recent workshop it was argued that, as adults, we often make the mistake of 45 children to do things we don‘t do ourselves. Before chastising (punishing) your child for watching videos or playing games, ask yourself how you use your spare time. As adults we need to lead by example. When was the last time you 46 that 500-page classic instead of checking social media or watching television? If something isn‘t good enough toe you or 47 your time, then why would you like your children to feel 48 ?We also tend to forget that the journey can often be more illuminating and 49 than the actual goal. This is definitely the 50 with reading. I am not talking about the journey that takes place within the pages of a book. I am talking about the personal journey that is possible when a child finds something that 51 their interest, latches onto it--- and in so doing discovers new worlds they never knew existed. The ownership in discovering and exploring one‘s own interests is magical.What does it 52 if a child is reading books that are “too easy”or “below their level”if it still bring them joy? Why is that a bad thing? Yes, stagnation(停止,滞止)can hamper growth, but as adults that is where we can 53 and help. It is our duty to help 54 our children‘s horizon‘s by knowing what excites and drives them and then presenting them with new and exciting literary options within that context. Whatever your child likes to read-if it is ―War and Peace,‖ magazines, comics, online articl es, video game directions or the back of a cereal box---take that, encourage it and build on it by introducing new 55 . Do not take away a child‘s joy. Without joy, reading is no longer reading. It‘s work.41. A. measures B. steps C. forms D. terms42. A.forgotten B. acepted C. identified D. ignored43. A. necessarily B. slightly C. potentially D. definitely44.A.integral B.unusual C.necessary D.vital45.A.hoping B.exciting C.demanding D.having46.A.set up B.looked up C.picked up D.turned up47.A.deserve B.worthy C.require D.worth48.A.equally B.specially C.similary D.differently49.A.rewarding B.exciting C.diasppointing D.appealing50.A.situation B.case C.illustration D.union51.A.raises B.leads C.sparks D.arises52.A.matter B.mean C.account D.offer53.A.set in B.step in C.take in D.look in54.A.stretch B.broaden C.extend D.enlargeSection B(A)Everyone gathered around and Paddy read out loud, slowly, his tone growing sadder and sadder. The little headline said: BOXER RECEIVES LIFF SENTENCE.Frank Cleary, aged 26, professional boxer, was today found guilty of the murder of Albert Gumming, aged 32, laborer, last July. The jury(陪审团) reached its decision after only ten minutes, recommending the most severe punishment to the court. It was, said the Judge, a simple case. Cumming and Cleary had quarreled violently at the Harbour Hotel on July 23rd and police saw Cleary kicking at the head of the unconscious Gumming. When arrested, Cleary was drunk but clear-thinking.Cleary was sentenced to life imprisonment with hard labour. Asked if he had anything to say, Cleary answered, ―Just don‘t tell my mother.‖“It happened over three years ago,”Paddy said helplessly. No one answered him or moved, for no one knew what to do. “Just don’t tell my mother,”said Fee numbly(麻木地). “And no one did! Oh, God! My poor, poor Frank!”Paddy wiped the tears from his face and said. ―Fee, pack your things. We‘ll go to see him.‖She half-rose before sinking back, her eyes in her small white face stared as if dead. ―I can't go,‖ she said without a hint of pain, yet making everyone feel that the pain was there. ―It would kill him to see me. I know him so well—his pride, his ambition. Let him bear the shame alone, it‘s what he wants. We‘ve got to help him keep his secret. What good will it d o him to see us?‖Paddy was still weeping, not for Frank, but for the life which had gone from Fee‘s face, for the dying in her eyes. Frank had always brought bitterness and misfortune, always stood between Fee and himself. He was the cause of her withdrawal from his heart and the hearts of his children. Every time it looked as if there might be happiness for Fee, Frank took it away. But Paddy‘s love for her was as deep and impossible to wipe out as hers was for Frank.So he said, ―Well, Fee, we won‘t go. B ut we must make sure he is taken care of. How about if I write to Father Jones and ask him to look out for Frank?‖There was no excitement in the eyes, but a faint pink stole into her cheeks. ―Yes, Paddy, do that. Only make sure he knows not to tell Frank we found out. Perhaps it would ease Frank to think for certain that we don‘t know.‖56. Paddy cried because he thought ___________.A. Frank did kill someone and deserved the punishmentB. Frank should have told Fee what had happenedC. what had happened to Frank was killing FeeD. Frank had always been a man of bad moral character57. The underlined sentence ―She half-rose before sinking back…‖ in Paragraph 6 shows that___________.A. Fee was so heart-broken that she could hardly stand upB. Fee didn‘t w ant to upset Paddy by visiting FrankC. Fee couldn‘t leave her family to go to see FrankD. Fe e struggled between wanting to see Frank and respecting his wish58. What can be inferred from the passage?A. The jury and the judge agreed on the Boxer‘s S entence of Life Imprisonment.B. The police found Gumming unconscious, heavily struck by Frank.C. The family didn‘t find out what had happened to Frank until 3 years later.D. Frank didn‘t want his family to know the sentence to him, most probably out of his pride.59. What is Frank and Paddy‘s probable relationship with Fee?A. Frank is Fee‘s son and Paddy is Fee‘s brother.B. Frank is Fee‘s son and Paddy is Fee‘s husband.C. Frank is Fee‘s brother and Paddy is Fee‘s lover.D. Frank is Fee‘s lover and Paddy is Fee‘s husband.Shakespeare‘s Globe Theatre and Exhibition TourOverviewShakespeare's Globe Exhibition is the world's largest exhibition devoted to Shakespe are. Located beneath the reconstructed Globe Theatre on London's Bankside, the exhibition explores the remarkable story of the Globe, and brings Shakespeare's world to life using a range of interactive displays and live demonstrations.HighlightsTour the reconstructed Globe Theatre and see how plays were staged in Shakespeare's dayAll-day access to the interactive Globe ExhibitionActors, recordings and interactive displays bring Shakespeare's world to lifeScheduleApril 23 to October 99:00am to 5:00pm. On Monday, tours run all day. Tuesday to Saturday, last tour departs at 12:30pm and at 11:30am on Sunday due to performances taking place on these days.October 10 to March 3110:00am to 5:00pm.Important note: Rehearsals(排练) will also take place throughout the Theatre Season. Please note that access to the Globe Theatre may be restricted and there may be occasions when the Globe tours are unable to run. When the Globe tours are not available, Rose or Bankside tours can be offered instead.Additional infoInclusions: Entrance fee and all day access to ExhibitionGuided tour of Shakespeare's Globe Theatre (maximum 50 people)Exclusions(不包含项目): Hotel pickup and drop offFood and drinks, unless specifiedPricingClick the link below to check pricing & availability on your preferred travel date. Our pricing is constantly updated to ensure you always receive the lowest price possible - we 100% guarantee it.Theatre Tour and Exhibition Shakespeare's Globe Theatre Tour and Exhibition $22.34Theatre Tour and Afternoon Tea Shakespeare's Globe Theatre Tour and Exhibition plus Afternoon Tea at 3:00pm in the Swan Brasserie or Bar. $62.8960.The passage can be found ___________.A.in a newspaper B.in a magazine C.on the Internet D.in a guidebook61.In this Shakespeare‘s Globe Theatre Tour, we can _________.A.visit the original Globe TheatreB.enjoy a British afternoon tea for freeC.experience Shakespeare’s world in an interactive wayD.visit the exhibition in the Globe Theatre62.What is true about the tour according to the passage?A.Rehearsals may affect the tour.B.The pricing remains the same.C.Performances take place throughout the year.D.The opening hours are the same in May and in November.(C)Since quitting can start feelings such as guilt and shame, we often do everything possible to avoid it, ―We'reMeek. He, however, suggests we view quitting differently.Quitting is like deciding to rearrange a room: you‘ve grown comfortable with the status, and it can be hard to picture the end result or even see why change is necessary. And yet, there's the upset ting feeling that you‘re no longer entirely satisfied with your current circumstances, perhaps even that you‘ve stopped making progress. While it's not out of the question for feelings of regret to surface after a major refit, leaving a position, project or situation can reveal exciting possibilities, making you feel inspired and renewed.Quitting, often happens in situations where we're unhappy, fearful or have determined we have no other choice, factors that can have ill effects on our health. Perhaps you find your work unfulfilling, or you've jumped into a new relationship before you're ready--and, as a result, you're operating under intense pressure. ―If stress is enduring and not managed well, it can start to take a toll,‖ says Meek. According to the Am erican Psychological Association, long-term, ongoing stress can increase the risk for high blood pressure and heart attack so walking away from whatever is causing it can deliver significant physical and emotional health benefits. “We often see a reduction in the stress hormone cortisol (应激激素皮质醇)”, which can lower blood pressure and may even decrease the heart rate,‖ says Dr Alex Lickerman, a GP and expert on developing mental adaptability.Leaving situations that fail to bring you joy can leave you with sufficient time to explore where your heart is truly leading you. In a study that was published in 1999, then Harvard University professor Hermina Ibarra looked at how bankers tried different roles that required new skill sets--someone who spent a lot of time dealing with computers, for instance, was asked to take on personal interactions. Subjects(研究对象) we re especially drawn to acting out a version of their future selves through ‘imitation s trategies‘ -- an approach they compared to ‗trying on different clothes.‘ Mark Franklin, the president of CareerCycles, suggests a similar approach as a way to figure out what your true desires might be in your post-quitting life and foresee your future self. ―Pretend to be a certain kind of person, or go and meet others who are doing what you want to do,‖ he says. ―Try it on, see how it feels and decide if it's a good fit for you.‖ It may not feel like it at the time, but just moving on from a situation that's not quite right can help you get back on track.63. It can be inferred from paragraph 2 that quitting may brim us feelings of being both ________.A. guilty and ashamedB. stupid and enthusiasticC. troubled and hopefulD. inspired and determined64. The phrase ―take a toll‖ (paragraph 3) can be best replaced by ―________‖.A. developmental adaptabilityB. bring about changesC. keep up the pressureD. have a bad effect65. An approach suggested by Mark Franklin similar to ‗trying on different clothes‘ is for________.A. helping people find what truly suits them in careerB. telling capable employees from inadequate onesC. training employees to acquire different working skillsD. providing people with opportunities to have a role play66. It can be concluded from the passage that________.A. quitting is a track that only the timid will choose to followB. personal interaction can be a must for reducing emotional pressureC. mental adaptability can be improved by the stress hormone cortisolD. knowing when to stop is wise and may make dreams happenCharity—Humanity’s most kind and generous desire—is a timeless and borderless virtue, dating at least to the dawn of religious teaching. Philanthropy(慈善行为)as we understand it today, however, is a distinctly American phenomenon, inseparable from the nation that shaped it. From colonial leaders to modern billionaires like Buffett, Gates and Zuckerberg, the tradition of giving is woven into the national DNA.67. ________ Benjamin Franklin, an icon of individual industry and frugality(节俭)even in his own day, understood that with the privilege of doing well came the price of doing good. When he died in 1790, Franklin thought to future generations, leaving in trust two gifts of 1,000 Ib. of sterling silver—one to the city of Boston, the other to Philadelphia. According to his instruction, a portion of the money could not be used for 200 years. While Franklin‘s gifts lay in wait, the tradition he established evolved alongside the young nation.68.________ Often far less famed men and women have played a critical role in philanthropy’s evolution. One of my personal heroes is Julius Rosenwald, who helped construct more than 5,300 schools across the segregated(种族隔离)South and opened classroom doors to a generation of African-American students.69. ________ The answer is not just to benefit others. Tax reduction, for one, encourages the rich people to give. And philanthropy has long helped improve the public image of everyone from immoral capitalists to the new tech elite. More troubling, however, are the foundational problems that make philanthropy so necessary. Just before his d eath, Dr. Martin Luther King Jr. wrote, ―Philanthropy is praise-worthy, but it must not cause the philan thropist to overlook the circumstances of economic injustice which make philanthropy necessary.‖Franklin‘s gifts represent a broader principle. We are guardians of a public trust, even if our capital came from private enterprise, and our most important obligation is ensuring that the system works more equally and more justly for more people. 70. ________ America‘s greatest strength is not the fact of perfection, but rather the act of perfecting.Ⅳ. Summary WritingFor thousands of years , people have sailed across the oceans to trade , explore and transport goods . However, not every ship arrives at its port of destination. Weather ,war , navigation mistakes and bad luck have caused many ships to sink to the bottom of the ocean. These shipwrecks , which are estimated to number more than three million , have long fascinated us . In addition to being historically important , they sometimes contain great riches.Historical research is a key motivator for shipwreck hunters . Ships carrying documents and artifact can teach us about ancient civilizations and important events . For instance , in 1997 the Pandora , which sank in 791, was discovered off the coast of Australia . The findings from the ship helped us understand the events surrounding the famous mutiny (暴动) on another ship ----- the Bounty . Another important discovery off the US coast in 1996 is widely believed to be the Queen Ann’s Revenge , the flagship of the private Blackbeard. Profit is another motive for shipwreck exploration ,as companies use advanced sonar , robots and retrieval equipment to find treasure ships . One such firm is Odyssey Marine Exploration . The company has found hundreds of ships , including , in 2007 , a Spanish sailing ship containing 500,000 silver coins. The ship , which sank 200 years ago in the Atlantic Ocean , carried a treasure estimated to be worth $500 million . Soon after the discovery , a long legal battle over ownership rights took place between the company and the Spanish government . Cases like these are part of an ongoing debate about protecting historically important ships from treasure hunters.V. Translation1.每年春运期间,人们买火车票得排几个小时的队。
七宝中学高二期末综合复习一及答案
高二期末综合复习一一、填空题1、直线013=+-y x 的倾斜角 .2、若椭圆的长轴长为12,一个焦点是(0,2),则椭圆的标准方程为__________________.3、经过点(1,0)A 且与直线10x y ++=平行的直线l 的方程为 ________ _.4已知()2f z i z z i +=+-,求(12)f i +的值 ___________ _.5、已知直线220310x y x y +-=-+=和的夹角是 ________ _.6、已知z 为虚数,且有||5z =,如果22z z +为实数,若z 为实系数一元二次方程20x bx c ++=的根,则此方程为______________ ____.7、已知方程 221104x y k k -=--表示双曲线,则实数k 的取值范围为________________ . 8、过点(1,2)且与圆221x y +=相切的直线的方程是 ________________ _.9、设F 为抛物线24y x =的焦点,,,A B C 为该抛物线上三点,若点(1,2)A , ABC ∆的重心与抛物线的焦点F 重合,则BC 边所在直线方程为 ________ .10、若方程0x k +=只有一个解,则实数k 的取值范围是 __________ .11、下列五个命题:①直线l 的斜率[1,1]k ∈-,则直线l 的倾斜角的范围是;②直线:1l y kx =+与过(1,5)A -,(4,2)B -两点的直线相交,则4k ≤-或34k ≥-;③如果实数,x y 满足方程22(2)3x y -+=,那么y x ④直线1y kx =+与椭圆2215x y m+=恒有公共点,则m 的取值范围是1m ≥; ⑤方程052422=+-++m y mx y x 表示圆的充要条件是41<m 或1>m ; 正确的是_______ _____ _.12、直线320x y m ++=与直线2310x y +-=的位置关系是…………………………( )(A )相交 (B )平行 (C )重合 (D )由m 决定13、二次方程2330x ix --=的根的情况为…………………………( )(A )有两个不相等的实根 (B )有两个虚根(C )有两个共轭虚根 (D )有一实根和一虚根14、已知△ABC 的三个顶点是(3,4)A -、(0,3)B 、(6,0)C -,求(1) BC 边所在直线的一般式方程;(2) BC 边上的高AD 所在直线的一般式方程.15、已知:21,.(1)34,||b z i a R z z ωω=+∈=+-、若求;221z az b z z ++-+(2)若=1-i ,求a 、b 的值.16、已知双曲线1C :2214y x -=(1)求与双曲线1C 有相同的焦点,且过点P 的双曲线2C 的标准方程;(2)直线l :y x m =+分别交双曲线1C 的两条渐近线于A 、B 两点。
上海市七宝中学2019-2020学年第一学期高二期末考试数学试卷(含解析)
上海市七宝中学2019-2020学年第一学期高二期末考试数学试卷 一、填空题1.直线l 的倾斜角范围是 . 2.方程x 24+y 2m=1表示焦点在y 轴上的椭圆,其焦点坐标是 .3.抛物线y =ax 2(a ≠0)的焦点坐标是 .4.经过原点及复数√3−i 对应点的直线的倾斜角为 .5.下面四个命题:①a ,b 是两个相等的实数,则(a ﹣b )+(a +b )i 是纯虚数;②任何两个复数不能比较大小;③z 1,z 2∈C ,且z 12+z 22=0,则z 1=z 2=0;④两个共轭虚数的差为纯虚数.其中正确的序号为: . 6.已知点A 为双曲线x 2﹣y 2=1的左顶点,点B 和点C 在双曲线的右支上,△ABC 是等边三角形,则△ABC 的面积是 .7.已知直线l 经过点P (﹣2,1),且点A (﹣1,﹣2)到直线的距离为1,则直线l 的方程为 . 8.直线y =2k 与曲线9k 2x 2+y 2=18k 2|x |(k ∈R ,k ≠0)的公共点的个数为 .9.当实数a ,b 变化时,两直线l 1:(2a +b )x +(a +b )y +(a ﹣b )=0与12:m 2x +2y +n =0都通过一个定点,则点(m ,n )所在曲线的方程为 .10.动点P 到点F (﹣1,0)的距离比到它到y 轴的距离大1,动点P 的轨迹方程是 . 11.椭圆x 24+y 2=1的一个焦点是F ,动点P 是椭圆上的点,以线段PF 为直径的圆始终与一定圆相切,则定圆的方程是 .12.若实数x ,y 满足x ﹣4√y =2√x −y ,则x 的取值范围是 . 二、选择题:13.已知平面直角坐标系内的两个向量a →=(1,2),b →=(m ,3m ﹣2),且平面内的任一向量c →都可以唯一的表示成c →=λa →+μb →(λ,μ为实数),则m 的取值范围是( ) A .(﹣∞,2) B .(2,+∞)C .(﹣∞,+∞)D .(﹣∞,2)∪(2,+∞)14.椭圆C :x 216+y 29=1与直线1:(2m +1)x +(m +1)y =7m +4,m ∈R 的交点情况是( )A .没有交点B .有一个交点C .有两个交点D .由m 的取值而确定15.过点P (1,1)作直线与双曲线x 2−y 22=1交于A 、B 两点,使点P 为AB 中点,则这样的直线( )A .存在一条,且方程为2x ﹣y ﹣1=0B .存在无数条C .存在两条,方程为2x ±(y +1)=0D .不存在16.已知圆心为O ,半径为1的圆上有不同的三个点A 、B 、C ,其中OA →⋅OB →=0,存在实数λ,μ满足OC →+λOA →+uOB →=0→,则实数λ,μ的关系为( ) A .λ2+μ2=1 B .1λ+1μ=1C .λμ=1D .λ+μ=1三、解答题17.已知x ∈R ,设z =log 2(3+x )+i log 2(3﹣x ),当x 为何值时: (1)在复平面上z 对应的点在第二象限? (2)在复平面上z 对应的点在直线x +y ﹣2=0上.18.已知直线l 与抛物线y 2=2px (p >0)交于两点A (x 1,y 1),B (x 2,y 2) (1)求证:若直线l 过该抛物线的焦点,则y 1•y 2=﹣p 2; (2)写出(1)的逆命题,判断真假,并证明你的判断.19.(1)若圆C 的方程是x 2+y 2=r 2,求证:过圆C 上一点M (x 0,y 0)的切线方程为x 0x +y 0y =r 2. (2)若圆C 的方程是(x ﹣a )2+(y ﹣b )2=r 2,则过圆C 上一点M (x 0,y 0)的切线方程为 ,并证明你的结论. 20.已知双曲线x 22−y 2=1的两焦点为F 1,F 2,P 为动点,若PF 1+PF 2=4.(Ⅰ)求动点P 的轨迹E 方程;(Ⅱ)若A 1(﹣2,0),A 2(2,0),M (1,0),设直线l 过点M ,且与轨迹E 交于R 、Q 两点,直线A 1R 与A 2Q 交于点S .试问:当直线l 在变化时,点S 是否恒在一条定直线上?若是,请写出这条定直线方程,并证明你的结论;若不是,请说明理由.21.已知椭圆E 两个焦点F 1(﹣1,0),F 2(1,0),并经过点(√22,√32). (1)求椭圆E 的标准方程;(2)设M ,N 为椭圆E 上关于x 轴对称的不同两点,A (x 1,0),B (x 2,0)为x 轴上两点,且x 1x 2=2,证明:直线AM ,NB 的交点P 仍在椭圆E 上.(3)你能否将(2)推广到一般椭圆中?写出你的结论即可.一、填空题1.【详解详析】直线l 的倾斜角的范围是[0,π), 故答案为:[0,π).2.【详解详析】由题意可得a2=m,b2=4,所以c2=a2﹣b2=m﹣4,所以c=√m−4,故答案为:(0,±√m−4).3.【详解详析】当a>0时,整理抛物线方程得x2=1a y,p=12a∴焦点坐标为(0,14a).当a<0时,同样可得.故答案为:(0,14a).4.【详解详析】复数√3−i对应点的坐标为(√3,﹣1),则经过原点及复数√3−i对应点的直线的斜率k=√3=−√33,∴直线的倾斜角为为5π6.故答案为:5π6.5.【详解详析】①若a=b=0,则(a﹣b)+(a+b)i是0,为实数,即①错误;②复数分为实数和虚数,而任意实数都可以比较大小,虚数是不可以比较大小的,即②错误;③若z1=1﹣i,z2=1+i,则z12+z22=−2i+2i=0,但z1≠z2,即③错误;④z=a+bi,z=a−bi(b≠0),则z−z=2bi(b≠0)是纯虚数,即④正确.故答案为:④.6.【详解详析】双曲线x2﹣y2=1的左顶点为A(﹣1,0),根据双曲线的对称性,可设B(x1,y1),C(x1,﹣y1).由△ABC是等边三角形⇒AB=BC,得:(x1+1)2+y12=(﹣y1﹣y1)2,又x12﹣y12=1,∴x12﹣x1﹣2=0,∴x1=﹣1或x1=2右支的特点是x≥0,所以x1=2,从而y1=±√3,由此A(﹣1,0),B(2,√3),C(2,−√3),可以算出面积:S=√34AB2=√34×[32+(√3)2]=3√3.故答案为:3√3.7.【详解详析】设直线l的方程为y﹣1=k(x+2),即kx﹣y+2k+1=0 ∵点A(﹣1,﹣2)到l的距离为1,∴√1+k 2=1,解之得k =−43,得l 的方程为4x +3y +5=0.当直线与x 轴垂直时,方程为x =﹣2,点A (﹣1,﹣2)到l 的距离为1, ∴直线l 的方程的方程为x =﹣2或4x +3y +5=0. 故答案为:x =﹣2或4x +3y +5=0.8.【详解详析】联立直线y =2k 与曲线9k 2x 2+y 2=18k 2|x |(k ∈R ,k ≠0), 可得9k 2x 2+4k 2=18k 2|x |, 由k ≠0化为9x 2﹣18|x |+4=0, 即为9|x |2﹣18|x |+4=0, 解得|x |=3+√53或3−√53, 即有四个解:(±3+√53,2k ),(±3−√53,2k ),可得直线和曲线的交点个数为4. 故答案为:4.9.【详解详析】直线l 1:(2a +b )x +(a +b )y +(a ﹣b )=0,l 1的方程化为(2x +y +1)a +(x +y ﹣1)b =0, 令{2x +y +1=0x +y −1=0,解得{x =−2y =3,所以定点的坐标为(﹣2,3).由l 2过定点(﹣2,3),代入m 2x +2y +n =0得﹣2m 2+n +6=0, ∴点(m ,n )所在曲线C 的方程为:n =2m 2﹣6. 故答案为:n =2m 2﹣6.10.【详解详析】由题意可知:动点P 到F (﹣1,0)的距离等于其到直线x =1的距离, 由抛物线的定义可知动点P 的轨迹C 的方程为y 2=﹣4x . 故答案为:y 2=﹣4x .11.【详解详析】由题意可得a 2=4,所以a =2,设F 为椭圆的右焦点(√3,0),设左焦点F ',则由题意的定义可得PF +PF '=2a =4,所以以线段PF 为直径的圆M 的半径r =12PF ,定圆的半径为R ,因为|MO |=12|PF '|=12(4﹣|PF |)=2−12|PF |=2﹣r ,所以|MO |+r =2,即圆心距加动圆的半径为定值2,所以当原点为定圆圆心,以R =2时,定圆始终与圆M 相切,并且是内切.所以定圆的方程为:x 2+y 2=4,故答案为:x 2+y 2=4.12.【详解详析】方法一:【几何法】当x =0时,解得y =0,符合题意,当x >0时,解答如下: 令t =√y ∈(0,√x ],原方程可化为:﹣2t +x2=√x −t 2,记函数f (t )=﹣2t +x2,g (t )=√x −t 2,t ∈(0,√x ], 这两个函数都是关于t 的函数,其中x 为参数, f (t )的图象为直线,且斜率为定值﹣2, g (t )的图象为四分之一圆,半径为为√x ,问题等价为,在第一象限f (t ),g (t )两图象有公共点, ①当直线与圆相切时,由d =r 解得x =20,②当直线过的点A (0,x2)在圆上的点(0,√x )处时,即√x =x2,解得x =4,因此,要使直线与圆有公共点,x ∈[4,20], 综合以上分析得,x ∈[4,20]∪{0}. 方法二:【代数法】令t =√y ∈(0,√x ],原方程可化为:x ﹣4t =2√x −t 2, 因为x ﹣y =x ﹣t 2≥0,所以x ≥t 2≥0,两边平方并整理得,20t 2﹣8xt +x 2﹣4x =0(*),这是一个关于t 的一元二次方程,则方程(*)有两个非负数跟, {△=64x 2−80(x 2−4x)≥0t 1t 2=120(x 2−4x)≥0,解得,x ∈[4,20]∪{0}. 故答案为:[4,20]∪{0}.二、选择题:13.【详解详析】根据题意,向量a →、b →是不共线的向量 ∵a →=(1,2),b →=(m ,3m ﹣2) 由向量a →、b →不共线⇔m1≠3m−22解之得m ≠2所以实数m 的取值范围是{m |m ∈R 且m ≠2}. 故选:D .14.【详解详析】直线1:(2m +1)x +(m +1)y =7m +4, ∴(2x +y ﹣7)m =﹣x ﹣y +4, 令{2x +y −7=0−x −y +4=0得:{x =3y =1,∴直线l 过定点(3,1),又∵3216+19=97144<1,∴点(3,1)在椭圆内, ∴直线l 与椭圆有2个交点, 故选:C .15.【详解详析】设A (x 1,y 1),B (x 2,y 2),则x 1+x 2=2,y 1+y 2=2, 则x 12−12y 12=1,x 22−12y 22=1,两式相减得(x 1﹣x 2)(x 1+x 2)−12(y 1﹣y 2)(y 1+y 2)=0, ∴x 1−x 2=12(y 1−y 2), 即k AB =2,故所求直线方程为y ﹣1=2(x ﹣1),即2x ﹣y ﹣1=0. 联立{y =2x −1x 2−12y 2=1可得2x 2﹣4x +3=0,但此方程没有实数解故这样的直线不存在 故选:D .16.【详解详析】由题意可得|OA →|=|OB →|=|OC →|=1,且OA →⋅OB →=0. ∵OC →+λOA →+uOB →=0→,即 OC →=−λOA →−μOB →,平方可得 1=λ2+μ2, 故选:A . 三、解答题17.【详解详析】(1)由题意可知:{log 2(3+x)<0log 2(3−x)>0,即{0<3+x <13−x >1,解得:﹣3<x <﹣2;(2)由题意可知:log 2(3+x )+log 2(3﹣x )﹣2=0, ∴log 2[(3+x )(3﹣x )]﹣2=0, ∴log 2(9−x 2)−2=0, ∴log 2(9−x 2)=log 24, ∴9﹣x 2=4,∴x 2=5, 又∵{3+x >03−x >0,即﹣3<x <3,∴x =±√5.18.【详解详析】(1)已知直线l 与抛物线y 2=2px (p >0)交于两点A (x 1,y 1),B (x 2,y 2), 若直线l 过该抛物线的焦点F (p2,0),设直线l :x =ky +p2, 由{x =ky +p2y 2=2px,得y 2﹣2ky ﹣p 2=0,y 1y 2=−p 2;综上,命题成立;(2)逆命题:若y 1•y 2=﹣p 2,直线l 过该抛物线的焦点. 证明如下:设直线x =ky +m ,由{x =ky +my 2=2px ,得y 2﹣2ky ﹣2pm =0, △=4k 2+4m 2>0,y 1•y 2=﹣2pm , 又y 1•y 2=﹣p 2,故m =p2 所以直线l 过该抛物线的焦点.19.【详解详析】(1)证明:圆C 的方程是x 2+y 2=r 2,其圆心为C ,坐标为(0,0), 则k MC =y 0−0x 0−0=y 0x 0,则切线的斜率k =−1kCM=−x0y 0,则切线的方程为:(y ﹣y 0)=−x0y 0(x ﹣x 0),变形可得x 0x +y 0y =r 2,即可得证明;(2)根据题意,过圆C 上一点M (x 0,y 0)的切线方程为(x ﹣a )(x 0﹣a )+(y ﹣b )(y 0﹣b )=r 2, 证明:圆C 的方程是(x ﹣a )2+(y ﹣b )2=r 2,其圆心坐标为(a ,b ), 则k MC =y 0−b x 0−a,则切线的斜率k =−1k CM=−x 0−a y 0−b,则切线的方程为:(y ﹣y 0)=−x 0−a y 0−b(x ﹣x 0),即(y ﹣b +b ﹣y 0)(y 0﹣b )=﹣(x ﹣a +a +x 0)(x 0﹣a ),则有(y ﹣b )(y 0﹣b )+(x ﹣a )(x 0﹣a )﹣(x 0﹣a )2+(y 0﹣b )2, 变形可得:(x ﹣a )(x 0﹣a )+(y ﹣b )(y 0﹣b )=r 2, 即可得证明.20.【详解详析】(Ⅰ)由题意知:F 1(−√3,0),F(√3,0), 又∵PF 1+PF 2=4,∴动点P (x ,y )必在以F 1,F 2为焦点,长轴长为4的椭圆,∴a =2, 又∵c =√3,b 2=a 2﹣c 2=1. ∴椭圆C 的方程为x 24+y 2=1.(Ⅱ)由题意,可设直线l 为:x =my +1. 取m =0,得R(1,√32),Q(1,−√32),直线A 1R 的方程是y =√36x +√33,5 直线A 2Q 的方程是y =√32x −√3,交点为S 1(4,√3).若R(1,−√32),Q(1,√32),由对称性可知交点为S 2(4,−√3).若点S 在同一条直线上,则直线只能为ℓ:x =4.以下证明对于任意的m ,直线A 1R 与直线A 2Q 的交点S 均在直线ℓ:x =4上. 事实上,由{x 24+y 2=1x =my +1,得(my +1)2+4y 2=4,即(m 2+4)y 2+2my ﹣3=0,记R (x 1,y 1),Q (x 2,y 2),则y 1+y 2=−2m m 2+4,y 1y 2=−3m 2+4.设A 1R 与ℓ交于点S 0(4,y 0),由y04+2=y 1x1+2,得y 0=6y 1x1+2. 设A 2Q 与ℓ交于点S 0′(4,y 0′),由y′4−2=y 2x2−2,得y 0′=2y 2x2−2.∵y0−y0′=6y1x1+2−2y2x2−2=6y1(my2−1)−2y2(my1+3)(x1+2)(x2−2)=4my1y2−6(y1+y2)(x1+2)(x2−2)=−12mm2+4−−12mm2+4(x1+2)(x2−2)=0,∴y0=y0′,即S0与S0′重合,这说明,当m变化时,点S恒在定直线ℓ:x=4上.21.【详解详析】(1)根据题意设椭圆E的标准方程为:x2a2+y2b2=1(a>b>0),∴{c=1(√22)2a2+(√32)2b2=1a2=b2+c2,解得:{a=√2b=1c=1,∴椭圆E的标准方程为:x 22+y2=1;(2)设M(m,n)N(m,﹣n),P(x0,y0),则直线AM的方程为:y(m﹣x1)=n(x﹣x1)①,直线BN的方程为:y(m﹣x2)=﹣n(x﹣x2)②,把交点P(x0,y0)代入①,②,整理得:(y0﹣n)x1=my0﹣nx0,③(y0+n)x2=my0+nx0,④③与④两边分别相乘得:(y02−n2)x1x2=m2y02−n2x02,又∵x1x2=2,m2=2−2n2,∴x02+2y02=2,∴直线AM,NB的交点P仍在椭圆E上;(3)若椭圆标准方程为:x 2a2+y2b2=1,M,N为椭圆上关于x轴对称的不同两点,A(x1,0),B(x2,0)为x轴上两点,且x1x2=a2,则直线AM,NB的交点P仍在椭圆E上.。
上海市七宝高中2025届化学高二上期末综合测试试题含答案
上海市七宝高中2025届化学高二上期末综合测试试题含答案考生请注意:1.答题前请将考场、试室号、座位号、考生号、姓名写在试卷密封线内,不得在试卷上作任何标记。
2.第一部分选择题每小题选出答案后,需将答案写在试卷指定的括号内,第二部分非选择题答案写在试卷题目指定的位置上。
3.考生必须保证答题卡的整洁。
考试结束后,请将本试卷和答题卡一并交回。
一、选择题(共包括22个小题。
每小题均只有一个符合题意的选项)1、在25℃时,在浓度为1mol/L的(NH4)2SO4、(NH4)2CO3、(NH4)2Fe(SO4)2的溶液中,测其c(NH4+)分别为a、b、c(单位为mol/L),下列判断正确的是A.a=b=c B.a>b>c C.a>c>b D.c>a>b2、下列试纸,使用时预先不能用蒸馏水润湿的是()A.红色石蕊试纸B.蓝色石蕊试纸C.pH试纸D.淀粉KI试纸3、2018年10月24日港珠澳大桥正式通车。
深埋在海水中的钢管桩易发生腐蚀,但中科院金属研究所研发的技术能保障大桥120年耐久性。
下列保护钢管桩的措施不合理的是A.使用抗腐蚀性强的合金钢B.钢管桩附着铜以增强抗腐蚀性C.使用防止钢筋锈蚀的海工混凝土D.在钢筋表面覆盖一层高性能防腐涂料4、下列各组液体混合物,能用分液漏斗分离的是A.溴苯和溴 B.正己烷和水C.苯和硝基苯D.酒精和水5、下列有机物能使酸性KMnO溶液褪色,而不能因化学反应而使溴水褪色的是( )4A.苯B.甲苯C.乙烯D.丙炔6、烯烃复分解反应是指在金属钨或钼等催化剂的作用下。
碳碳双键断裂并重新组合的过程。
例如:则对于有机物CH2=CHCH2CH=CH2发生烯烃复分解反应时可能生成产物的判断中正确的是①;②CH2=CH2;③;④CH2=C=CH2A.只有③B.只有②③C.只有①②③D.有①②③④7、用价层电子对互斥理论预测H2S和NH3的立体结构,两个结论都正确的是()A.直线形;三角锥形B.V形;三角锥形C.直线形;平面三角形D.V形;平面三角形8、下列物质中不属于有机物的是A.CaC2B.C3H8C.C2H4O2D.C3H7Cl9、下列关于新型有机高分子材料的说法,不正确的是( )A.高分子分离膜应用于食品工业中,可用于浓缩天然果汁、乳制品加工、酿造业等B .复合材料一般是以一种材料作为基体,另一种材料作为增强体C .导电塑料是应用于电子工业的一种新型有机高分子材料D .合成高分子材料制成的人工器官一般都受到人体的排斥作用,难以达到生物相容的程度10、在甲烧杯中放入盐酸,乙烧杯中放入醋酸,两种溶液的体积和pH 都相等,向两烧杯中同时加入质量不等的锌粒,反应结束后得到等量的氢气。
2024届上海市七宝中学化学高二第一学期期末质量跟踪监视试题含解析
2024届上海市七宝中学化学高二第一学期期末质量跟踪监视试题注意事项:1.答卷前,考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。
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4.考生必须保证答题卡的整洁。
考试结束后,请将本试卷和答题卡一并交回。
一、选择题(共包括22个小题。
每小题均只有一个符合题意的选项)1、牙齿表面由一层硬的、成分为Ca3(PO4)3OH的物质保护,它在唾液中存在下列平衡:Ca3(PO4)3OH5Ca2++3PO43-+OH-,已知Ca3(PO4)3F的溶解度比Ca3(PO4)3OH(s)更小,为了保护牙齿,世界各地均采用了不同的措施。
下列措施中不能保护牙齿的是A.少吃甜食B.在牙膏中加入适量氟化物添加剂C.在饮用水中加入拧檬片 D.在牙膏中添加适量的Ca2+或PO43-离子2、N A代表阿伏加德罗常数的值。
下列叙述正确的是()A.常温常压下,2.24 LSO2中所含氧原子数为0.2N AB.将1mol Cl2通入水中,HC1O、Cl-、ClO-粒子数之和为2N AC.1mol NO2与足量H2O反应,转移的电子数为N AD.0.1mol熔融的NaHSO4中阳离子数目为0.1N A3、在下列平衡242CrO-(黄色)2272H Cr O+-+(橙红色)2H O+中,溶液介于黄和橙红色之间,现欲增加溶液的橙红色,则要在溶液中加入( )A.H+B.OH-C.K+D.2H O4、空气中汽油含量的测量仪,其工作原理如图所示(用强酸性溶液作电解质溶液)。
2025届上海市七宝中学高二化学第二学期期末教学质量检测试题含解析
2025届上海市七宝中学高二化学第二学期期末教学质量检测试题注意事项:1.答题前,考生先将自己的姓名、准考证号填写清楚,将条形码准确粘贴在考生信息条形码粘贴区。
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4.保持卡面清洁,不要折叠,不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
一、选择题(共包括22个小题。
每小题均只有一个符合题意的选项)1、同温同压下,等质量的二氧化硫和二氧化碳气体的下列有关比较正确的是A.体积比为1∶ 1 B.体积比为16∶11C.物质的量比为16∶11 D.密度比为16∶112、实验是化学的基础。
下列说法正确的是A.过氧化钠、氯水、浓硝酸通常都要密封保存于棕色试剂瓶中B.盛溴的试剂瓶里加少量水以减少溴的挥发C.钾、钠、白磷都应保存在水中D.做焰色反应实验时所用铂丝,每次用完后用稀硫酸洗涤后再使用3、能发生银镜反应,并与丙酸互为同分5构体的有机物有A.1种B.2种C.3种D.4种4、下列有机化合物中,能同时发生酯化反应、加成反应和氧化反应的是①CH2=CHCOOCH3②CH2=CHCOOH ③CH2=CHCH2OH④CH3CH(OH)CH2CHO ⑤CH3CH2CH2OHA.②③④B.②④⑤C.①③⑤D.①②⑤5、下列叙述不正确的是()A.淀粉、油脂和蛋白质都能水解,且水解产物各不相同B.从煤的干馏产物中可以获得焦炉气、粗氨水、煤焦油和焦炭等重要的化工原料C.核酸是一类含磷的生物高分子化合物D.石油催化裂化的主要目的是提高芳香烃的产量6、下列关于热化学反应的描述中正确的是A.HCl和NaOH反应的中和热△H=-57.3 kJ·mol−1,则H2SO4和Ca(OH)2反应的中和热△H=2×(-57.3)kJ·mol−1 B.甲烷的标准燃烧热ΔH=-890.3 kJ·mol−1,则CH4(g)+2O2(g)=CO2(g)+2H2O(g) ΔH<-890.3 kJ·mol−1 C.已知:500℃、30MPa下,N2(g)+3H2(g)2NH3(g) ΔH=-92.4kJ·mol-1;将1.5 mol H2和过量的N2在此条件下充分反应,放出热量46.2 kJD.CO(g)的燃烧热是283.0kJ·mol−1,则2CO2(g) ===2CO(g)+O2(g)反应的△H=+566.0 kJ·mol−17、下列各项叙述中,正确的是( )A.Si、P、S的第一电离能随原子序数的增大而增大B.价电子排布为3d64s2的元素位于第四周期第ⅧB族,是d区元素C.2p和3p轨道形状均为哑铃形,能量不相等D.氮原子的最外层电子排布图:8、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。
上海市七宝中学高二化学期末试卷含解析
2022-2023上海市七宝中学高二化学期末试卷含解析一、单选题(本大题共15个小题,每小题4分。
在每小题给出的四个选项中,只有一项符合题目要求,共60分。
)1. 下列说法中正确的是A.适当摄入油脂,有助于人体吸收多种脂溶性维生素B.人体缺乏维生素D易引起夜盲症C.相对于各成分金属说,合金的熔点更高,硬度更大D.柠檬酸在体内可以彻底氧化生成二氧化碳和水,所以柠檬是酸性食物参考答案:A试题分析:A.适当摄入油脂,有助于人体吸收多种脂溶性维生素,A正确;B.缺少维生素A易患夜盲症,B错误;C.合金的强度和硬度一般比组成它的纯金属更高,抗腐蚀性能等也比组成它的纯金属更好;但是,大多数合金的熔点比它的纯金属的熔点低,C错误;D.柠檬中的柠檬酸及其钾盐,在体内被彻底氧化,柠檬酸最后生成二氧化碳和水排出体外,剩下碱性的钾盐。
因此,这类具有酸性的食物是碱性食物,D错误;选A。
考点:考查生活中的常识。
2. 以溴乙烷为原料制备1,2-二溴乙烷,下列方案中最合理的是()参考答案:D略3. 下列食品添加剂与类别对应不正确的是()A.调味剂—亚硝酸钠 B.防腐剂—苯甲酸钠C.疏松剂—碳酸氢钠 D.着色剂—叶绿素参考答案:A 略4. 下列家庭化学小实验不能达到预期目的的是:A.用米汤检验食用加碘盐(含KIO3)中含有碘B.用醋、石灰水验证蛋壳中含有碳酸盐C.用碘酒检验汽油中是否含有不饱和烃D.用鸡蛋白、食盐、水完成蛋白质的溶解、盐析实验参考答案:A略5. 下列分子或离子中,中心原子不是sp3杂化的是()A.SO42﹣B.NO3﹣C.CH4 D.H2S参考答案:B【考点】原子轨道杂化方式及杂化类型判断.【分析】根据价层电子对互斥理论确定中心原子杂化类型,价层电子对个数=σ键个数+孤电子对个数,σ键个数=配原子个数,孤电子对个数=(a﹣xb),a指中心原子价电子个数,x指配原子个数,b指配原子形成稳定结构需要的电子个数.中心原子的杂化类型为sp3,说明该分子中心原子的价层电子对个数是4,据此判断.【解答】解:A.硫酸根离子中,S原子价层电子对数=σ 键个数+(a﹣xb)=4+(6+2﹣4×2)=4,所以S采取sp3杂化,故A不选;B.NO3﹣中,N原子形成3个σ键,孤对电子数=,中心原子N为sp2杂化,故B选;C.CH4分子中C原子价层电子对个数=4+(4﹣4×1)=4,所以C原子采用sp3杂化,故C不选;D.H2S中价层电子对个数=σ键个数+孤电子对个数=2+×(6﹣2×1)=4,所以S原子采用sp3杂化,故D不选;故选B.6. 在配制Fe2(SO4)3溶液时,为了防止Fe2(SO4)3水解,常常往溶液中加入少量的A.铁粉 B.H2SO4溶液C.HCl溶液 D.CH3COOH溶液参考答案:B略7. 下列说法正确的是()A.摩尔是七个基本物理量之一,符号为molB.摩尔质量在数值上等于该物质的相对分子质量或相对原子质量C.1 mol任何物质都含有6.02×1023个分子D.同温同压下,所有气体的摩尔体积都是22.4 L/mol参考答案:B略8. 一定条件下,在固定容积的密闭容器中,能表示反应X(g)+2Y(g)2Z(g)一定达到化学平衡状态的是()①X、Y、Z的物质的量之比为1∶2∶2 Ks5u②X、Y、Z的浓度不再发生变化③容器中的压强不再发生变化④单位时间内生成n mol Z,同时生成2n mol YA.①② B.①④ C.②③ D.③④参考答案:C略9. 下列说法不正确的是()A.苯和甲苯分子中所有原子均在同一平面上B.苯不可以使KMnO4酸性溶液褪色而甲苯可以C.苯和甲苯都能与卤素单质、硝酸等发生取代反应D.苯的同系物的分子通式是C n H2n﹣6(n≥7)参考答案:A【考点】有机物的结构和性质.【分析】A.甲苯含有甲基,具有甲烷的结构特点;B.苯性质稳定,但甲苯可被氧化;C.苯、甲苯在催化条件下可发生取代反应;D.苯的同系物含有一个苯环,且烃基为烷基.【解答】解:A.苯为平面形结构,但甲苯含有甲基,具有甲烷的结构特点,不可能在同一个平面上,故A错误;B.苯性质稳定,但甲苯中苯环对甲基的影响,导致甲苯可被氧化,可使高锰酸钾褪色,故B正确;C.苯、甲苯在催化条件下可发生取代反应,可生成硝基苯、三硝基甲苯等,故C正确;D.苯的同系物含有一个苯环,且烃基为烷基,通式为C n H2n﹣6(n≥7),故D正确.故选A.10. 电子计算机所用钮扣电池的两极材料为锌和氧化银,电解质溶液为KOH溶液,其电极反应是: Zn + 2OH- -2e-==ZnO + H2O Ag2O +H2O + 2e-==2Ag +2 OH-,下列判断正确的是A.锌为正极,Ag2O为负极 B.锌为负极,Ag2O为正极C.原电池工作时,正极区溶液pH减小 D.原电池工作时,负极区溶液pH增大参考答案:B略11. 下列能源转化过程中,污染较大的是A. 风能发电B. 燃煤发电C. 地热能发电D. 太阳能发电参考答案:B略12. 对可逆反应N2(g)+3H2(g)?2NH3(g)△H<0,正反应速率为υ1,逆反应速率为υ2.当温度升高时,υ1和υ2的变化情况为()A.同时增大B.同时减小C.υ1增大,υ2减小 D.υ1减小,υ2增大参考答案:A考点:化学平衡的影响因素.专题:化学平衡专题.分析:无论化学反应是放热反应还是吸热反应,只要升高温度,化学反应速率就一定增大,据此分析.解答:解:无论化学反应是放热反应还是吸热反应,只要升高温度,化学反应速率就一定增大,即υ1和υ2都增大,故选A.点评:本题考查了影响化学反应速率的因素,注意无论化学反应是放热反应还是吸热反应,只要升高温度,化学反应速率就一定增大13. 25 ℃时,在等体积的①pH=0的H2SO4溶液,②0.05 mol·L-1的Ba(OH)2溶液,③pH=10的Na2S溶液,④pH=5的NH4NO3溶液中,发生电离的水的物质的量之比是() A.1∶10∶1010∶109 B.1∶5∶5×109∶5×108C.1∶20∶1010∶109 D.1∶10∶104∶109参考答案:A略14. 下列叙述正确的是A.煤干馏可以得到甲烷、苯和氨等重要化工原料B.乙烯、乙醇和乙酸均能发生加成反应 C.乙醛可使酸性高锰酸钾溶液褪色,不能使溴水褪色;乙醇不能被酸性高锰酸钾氧化D.向鸡蛋清溶液中分别加入(NH4)2SO4和CuSO4溶液都能使之聚沉,其作用原理相同参考答案:A略15. 元素A的阳离子与元素B的阴离子具有相同的电子层结构。
七宝中学高二期末综合复习三和答案
高二期末综合复习三一、填空题1. 复数24(1)(1)i +-的模是___________.2.直线03:1=+x l 与直线013:2=-+y x l 的夹角的大小为 . 3.已知复数i z -=31,22=z ,则12z z -的最大值为_______________. 4、已知复数Z 满足|Z |=1,则W =1+2Z 所对应的点的轨迹是_________________.5、已知双曲线221124x y -=的右焦点为F ,若过点F 的直线与双曲线的右支有且只有一个交点,则此直线的斜率的取值范围是__________________.6.曲线19422=+y x 上点到直线082=--y x 距离的最小值为 . 7.k 取任意实数时,直线04)6()1(2=---+-k y k x k 与曲线22210x y x m +-+-=都相交,则实数m 的范围____________.8、设复数Z 满足|Z |=2,且(Z -a )2=a ,则实数a 的值为________________9.若将方程6)4()4(2222=++-+-y x y x 化简为12222=-by a x 的形式,则=-22b a .10.定义直线关于圆的圆心距单位λ为圆心到直线的距离与圆的半径之比.若圆C 满足:①与x轴相切于点(3,0)A ;②直线y x =关于圆C 的圆心距单位λ=试写出一个满足条件的圆C的方程 . 二、选择题11、若Z 1,Z 2为复数,则Z 12+Z 22=0是Z 1=0且Z 2=0的( )A 、充分非必要条件B 、必要非充分条件C 、充要条件D 、既非充分又非必要条件12、曲线1y =y=k(x-2)+4有两个交点时,实数k 的取值范围是( ) A 、53(,]124B 、5(,)12+∞ C 、5(0,)12D 、13(,)3413.已知直线l 的方程为)0(,0≠=++ab c by ax 且l 不经过第二象限,则直线l 的倾斜角大小为( )(A ) b aarctan ; (B ))arctan(b a - ; (C )b a arctan ; (D )ba arctan -π. 14.给出下列三个命题:①若1z 、C z ∈2且021>-z z ,则21z z >.②如果复数z 满足2=-++i z i z ,则复数z 在复平面上所对应点的轨迹是椭圆.③已知曲线:1C =和两定点()()5,05,0E F -、,若()y x P ,是C 上的动点, 则6≥-PF PE .上述命题中正确的个数是( )(A ) 0; (B ) 1 ; (C ) 2 ; (D ) 3. 三、解答题15.已知虚数1z 、2z 是方程03422=-+-m m x x ,R m ∈的两根,且满足51=z .(1)求实数m 的值;(2)设虚数1z 、2z 对应在复平面上的点分别为1F 、2F ,求以1F 、2F 为焦点且过原点的椭圆的焦距、长轴的长和短轴的长.16.已知抛物线24C y x =:的焦点为F ,点A 在抛物线C 上运动. (1)当点 A P 、满足2AP FA =-时,求动点P 的轨迹方程;(2)设(,0)M m ,其中m 为常数,m R +∈,点A 到M 的距离记为d ,求d 的最小值.17.已知点)0,2(-A ,)0,2(B .(1)过点A 作直线m 交以A 、B 为焦点的双曲线于M 、N 两点,若线段MN 的中点到y 轴的距离为1,且直线m 与圆122=+y x 相切,求该双曲线的方程;(2)以A 、B 为顶点的椭圆经过点)23,1(C ,过椭圆的上顶点G 作直线s 、t ,使t s ⊥,直线s 、t 分别交椭圆于点P 、Q .求证:PQ 必过y 轴上一定点.18.过抛物线22(0)y px p =>上一定点00(,) P x y 作两条直线分别交抛物线于11(,)A x y ,22(,)B x y ,(Ⅰ) 若横坐标为2p的点到焦点的距离为1,求抛物线方程; (Ⅱ) 若00(,) P x y 为抛物线的顶点,2APB π∠=,试证明:过A 、B 两点的直线必过定点(2,0)p ;(Ⅲ) 当PA 与PB 的斜率存在且倾斜角互补时,求12y y y +的值,并证明直线AB 的斜率是非零常数。
上海市闵行区七宝中学2025届生物高二上期末监测模拟试题含解析
上海市闵行区七宝中学2025届生物高二上期末监测模拟试题考生须知:1.全卷分选择题和非选择题两部分,全部在答题纸上作答。
选择题必须用2B铅笔填涂;非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。
2.请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。
3.保持卡面清洁,不要折叠,不要弄破、弄皱,在草稿纸、试题卷上答题无效。
一、选择题:(共6小题,每小题6分,共36分。
每小题只有一个选项符合题目要求)1.下表表示能量流进某种植食性动物时的情况,下列说法正确的是项目摄食量粪便量呼吸消耗量用于生长发育和繁殖的能量能量(kJ)530 330 X 110A.X表示的能量大小不能确定B.该植食性动物同化的能量有25%以热能的形式散失掉C.表中各个量之间没有一定的数量关系D.流向第三营养级生物的能量至多为40kJ2.重庆奉节脐橙的前身叫奉节柑桔,栽培始于汉代,历史悠久,产区位于三峡库区,具有“无台风、无冻害、无检疫性病虫害”的三大生态优势。
据《汉书·地理志》记载:“鱼腹(今奉节)朐忍有桔官”,《汉志》记载“柚通省者皆出,唯夔(今奉节)产者香甜可食”。
奉节脐橙,畅销全国。
下列有关叙述错误的是()A.生长素、赤霉素和细胞分裂素协同促进脐橙果实的生长B.脐橙枝条向光生长,体现了生长素生理作用的两重性C.脐橙果实成熟过程中乙烯和脱落酸的含量升高D.脐橙树的生长发育和对环境的适应过程受多种激素共同调节3.为了探究生长素(IAA)和乙烯(ACC是乙烯的供体)对植物生根的影响,科学家用拟南芥下胚轴插条进行了一系列实验,结果如下图所示。
据图分析,下列说法正确的是A.促进拟南芥下胚轴插条生根的最适宜生长素浓度为50μmoLB.拟南养下胚轴插条细胞中,生长素和乙烯是同时合成并发挥作用的C.两种激素浓度为0时,拟南芥下胚轴插条仍能生根,这与内源性激素有关D.ACC对拟南芥下胚轴插条生根作用的影响是促进其生根4.当人体进行深呼吸时,血液中的pH与平静呼吸时相比,暂时会( )A.增大B.减小C.不变D.趋于酸性5.2008年“5・12”汶川地震后形成的最大堰塞湖—北川唐家山堰塞湖经过建设者的不断改造,恢复了该地的森林覆盖面积,现已成为景区,为当地创造了经济效益、社会效益和生态效益,用行动诠释了“绿水青山就是金山银山”的理念。
2025届上海市七宝中学化学高二下期末综合测试试题含解析
2025届上海市七宝中学化学高二下期末综合测试试题注意事项1.考试结束后,请将本试卷和答题卡一并交回.2.答题前,请务必将自己的姓名、准考证号用0.5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置.3.请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符.4.作答选择题,必须用2B铅笔将答题卡上对应选项的方框涂满、涂黑;如需改动,请用橡皮擦干净后,再选涂其他答案.作答非选择题,必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答,在其他位置作答一律无效.5.如需作图,须用2B铅笔绘、写清楚,线条、符号等须加黑、加粗.一、选择题(每题只有一个选项符合题意)1、下列各原子或离子的电子排布式错误的是()A.+22626A1s2s23s:p3s3p d4s4pK1s2s2p s:33p B.226261023C.3-226C s p1s2s2p3r d4s:331s2s:N2p D.22626422、下列过程中发生了化学变化的是( )A.酒精挥发B.洗菜C.粮食酿酒D.葡萄榨汁3、下列实验过程可以达到实验目的的是编号实验目的实验过程称取4.0 g固体NaOH于烧杯中,加入少量蒸馏水溶解后立即转A 配制0.4000 mol·L-1的NaOH溶液移至250 mL容量瓶中定容向盛有2 mL黄色氯化铁溶液的试管中滴加浓的维生素C溶液,B 探究维生素C的还原性观察颜色变化向稀盐酸中加入锌粒,将生成的气体依次通过NaOH溶液、浓硫C 制取并纯化氢气酸和KMnO4溶液向2支盛有5 mL不同浓度NaHSO3溶液的试管中同时加入2 mL D 探究浓度对反应速率的影响5% H2O2溶液,观察实验现象A.A B.B C.C D.D4、下列化学用语正确的是A.聚丙烯的结构简式:B.丙烷分子的比例模型:C.四氯化碳分子的电子式:D.中子数为18的氯原子:5、下列叙述不正确的是A.Na+、Mg2+、Al3+的氧化性依次减弱B.RbOH、KOH、Mg(OH)2碱性依次减弱C.H2S、H2O、HF的稳定性依次增强D.O2-、F-、Na+、Br-的半径大小顺序为:Br->O2->F->Na+6、下列关于微粒间作用力与晶体的说法不正确的是A.某物质呈固体时不导电,熔融状态下能导电,则该物质一定是离子晶体B.H2O和CCl4的晶体类型相同,且每个原子的最外层都达到8电子稳定结构C.F2、Cl2、Br2、I2的沸点逐渐升高,是因为分子间作用力逐渐增大D.干冰溶于水中,既有分子间作用力的破坏,也有共价键的破坏7、向Ba(OH)2和NaOH混合溶液中缓缓通入CO2气体至过量,生成沉淀物质的量与通入CO2气体的体积V(标准状况)的关系如图所示,下列结论不正确的是A.原混合物中n[Ba(OH)2]:n[NaOH] =1:2B.横坐标轴上p点的值为90C.b点时溶质为NaHCO3D.ab段发生反应的离子方程式依次为:CO2+2OH- = H2O+CO32-,CO2+H2O+ CO32-=2HCO3-8、化学与生产、生活、环境等社会实际密切相关。
2022-2023学年上海市七宝中学高二下学期期末考试化学试卷含详解
2023七宝中学高二(下)期末化学试卷相对原子质量:H -1 N -14 S -32 Fe -56 Cu -64一、选择题(本题共40分,每小题2分,每题只有一个正确选项)1.化学与生活、环境和生产密切相关,下列叙述不涉及氧化还原反应的是A .绿化造林助力实现碳中和目标B .用氯化铁溶液制作铜印刷电路板C .84消毒液杀菌消毒D .使用添加氟化物牙膏预防龋齿2.能确定为丙烯的化学用语是( )A .B .C 3H 6 C .D .CH 2=CH -CH 33.下列有关物质的用途正确的是A .乙烯一般用作燃料使用B .乙醇常用作麻醉剂C .氨水是铵态氨肥的一种,可施土壤D .甲醛可用作食品防腐剂4.醋酸钠中不存在A .离子键B .极性键C .非极性键D .分子间作用力5.不属于合金的物质是A .黄铜B .钢铁C .冰晶石(NaAlF 6)D .硬铝6.为了检验某固体物质中是否含有NH 4+,一定用不到的试剂或试纸是A .NaOH 溶液B .浓盐酸C .稀硫酸D .红色石蕊试纸7.有机物命名正确的是A .1,3-二甲基丁烷B .2,3,5-三甲基-3-乙基已烷C .2,3-二甲基-2-乙基-已烷D .2,3-二甲基-4-乙基戊烷8.将0.1mol/L 3CH COOH 溶液中加水稀释或入少量的3CH COONa 晶体(忽略温度、体积变化)都会引起A .CH 3COOH 电离程度增大B .溶液导电能力增强C .水的电离程度增大D .溶液pH 减小9.某些晶体在一定条件下可以导电,下列说法一定正确的是A .熔点高、熔融态导电的是离子晶体B .固态导电的是金属晶体C .熔融态导电,固态不导电的是离子晶体D .其水溶液导电的是离子晶体10.下列说法正确的是 A .增大反应物浓度,可增大单位体积内活化分子的百分数,从而使有效碰撞次数增大B .有气体参加的化学反应,若增大压强(即缩小反应容器的体积),可增大活化分子的百分数,从而使反应速率增大C .升高温度能使化学反应速率增大的主要原因是增加了反应物分子中活化分子的百分数D .催化剂不影响反应活化能但能增大单位体积内活化分子百分数,从而增大化学反应速率11.下列有关除杂及检验的说法中正确的是A .除去乙醇中混有的乙酸,加NaOH 液,然后分液B .除去铁锈中混有的氧化铝,可加入足量NaOH 溶液充分振荡,反应后过滤洗涤即可C .检验3KClO 中氯元素,可加入硝酸酸化的硝酸银溶液生成白色沉淀D .酸性高锰酸钾和溴水可以检验实验室用无水乙醇和浓硫酸反应,是否有乙烯产生12.对化工生产认识正确的是A .工业制氯化氢:通入H 2的量略大于Cl 2,可以使平衡正向移动B .海水提溴:一般需要经过浓缩、氧化、用热空气吹出及冷凝、精制等步骤C .硫酸工业:采用400-500°C 的高温,有利于增大反应正向进行的程度D .以氯气、氢气和烧碱为原料生产一系列含氯、含钠化工产品的工业称为烧碱工业。
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高二期末综合复习一
一、填空题
1、直线013=+-y x 的倾斜角 .
2、若椭圆的长轴长为12,一个焦点是(0,2),则椭圆的标准方程为__________________.
3、经过点(1,0)A 且与直线10x y ++=平行的直线l 的方程为 ________ _. 4已知()2f z i z z i +=+-,求(12)f i +的值 ___________ _. 5、已知直线220310x y x y +-=-+=和的夹角是 ________ _.
6、已知z 为虚数,且有||5z =,如果2
2z z +为实数,若z 为实系数一元二次方程
20x bx c ++=的根,则此方程为______________ ____.
7、已知方程
22
1104
x y k k -=--表示双曲线,则实数k 的取值范围为________________ . 8、过点(1,2)且与圆221x y +=相切的直线的方程是 ________________ _.
9、设F 为抛物线2
4y x =的焦点,,,A B C 为该抛物线上三点,若点(1,2)A , ABC ∆的重
心与抛物线的焦点F 重合,则BC 边所在直线方程为 ________ .
10、若方程0x k +=只有一个解,则实数k 的取值范围是 __________ . 11、下列五个命题:①直线l 的斜率[1,1]k ∈-,则直线l 的倾斜角的范围是; ②直线:1l y kx =+与过(1,5)A -,(4,2)B -两点的直线相交,则4k ≤-或34
k ≥-;
③如果实数,x y 满足方程2
2
(2)3x y -+=,那么
y
x
④直线1y kx =+与椭圆
22
15x y m
+=恒有公共点,则m 的取值范围是1m ≥; ⑤方程05242
2
=+-++m y mx y x 表示圆的充要条件是4
1
<m 或1>m ; 正确的是_______ _____ _.
12、直线320x y m ++=与直线2310x y +-=的位置关系是…………………………( ) (A )相交 (B )平行 (C )重合 (D )由m 决定
13、二次方程2
330x ix --=的根的情况为…………………………( ) (A )有两个不相等的实根 (B )有两个虚根 (C )有两个共轭虚根 (D )有一实根和一虚根 14、已知△ABC 的三个顶点是(3,4)A -、(0,3)B 、(6,0)C -,求 (1) BC 边所在直线的一般式方程;
(2) BC 边上的高AD 所在直线的一般式方程.
15、已知:21,.(1)34,||b z i a R z z ωω=+∈=+-、若求;
221
z az b z z ++-+(2)若=1-i ,求a 、b 的值.
16、已知双曲线1C :2
2
14
y x -=
(1)求与双曲线1C 有相同的焦点,且过点P 的双曲线2C 的标准方程;
(2)直线l :y x m =+分别交双曲线1C 的两条渐近线于A 、B 两点。
当3OA OB ⋅=时,求实数m 的值.
高二期末综合复习一(答案)
一、填空题
1、直线013=+-y x 的倾斜角
6
π
. 2、若椭圆的长轴长为12,一个焦点是(0,2),则椭圆的标准方程为___
22
13236
x y +=_________. 3、经过点(1,0)A 且与直线10x y ++=平行的直线l 的方程为 10x y +-= _. 4已知()2f z i z z i +=+-,求(12)f i +的值 34i - _. 5、已知直线220310x y x y +-=-+=和的夹角是
4
π
_. 6、已知z 为虚数,且有||5z =,如果2
2z z +为实数,若z 为实系数一元二次方程
20x bx c ++=的根,则此方程为__________22250x x -+=____ _.
7、已知方程
22
1104
x y k k -=--表示双曲线,则实数k 的取值范围为___410k k <>或 . 8、过点(1,2)且与圆221x y +=相切的直线的方程是 3450x y -+=或1x =_.
9、设F 为抛物线2
4y x =的焦点,,,A B C 为该抛物线上三点,若点(1,2)A , ABC ∆的重
心与抛物线的焦点F 重合,则BC 边所在直线方程为 210x y +-= .
10、若方程0x k +=只有一个解,则实数k 的取值范围是 [1,1){2}- . 11、下列五个命题:①直线l 的斜率[1,1]k ∈-,则直线l 的倾斜角的范围是[,]44ππ
α∈-
; ②直线:1l y kx =+与过(1,5)A -,(4,2)B -两点的直线相交,则4k ≤-或3
4
k ≥-;
③如果实数,x y 满足方程22
(2)3x y -+=,那么y x
④直线1y kx =+与椭圆
22
15x y m
+=恒有公共点,则m 的取值范围是1m ≥; ⑤方程05242
2
=+-++m y mx y x 表示圆的充要条件是4
1
<m 或1>m ; 正确的是_____②_③_⑤___ _.
12、直线320x y m ++=与直线2310x y +-=的位置关系是…………………………( A )
(A )相交 (B )平行 (C )重合 (D )由m 决定 13、二次方程2
330x ix --=的根的情况为______________( B )
、有两个不相等的实根A B 、有两个虚根 、有两个共轭虚根C 、有一实根和一虚根A
14、已知△ABC 的三个顶点是(3,4)A -、(0,3)B 、(6,0)C -,求 (3) BC 边所在直线的一般式方程;
(4) BC 边上的高AD 所在直线的一般式方程. 解;(1)(6,3)BC =--是BC 边所在直线的方向向量 故3
:
63
BC x y l -=--,即:260BC l x y -+= (2)(6,3)BC =--高AD 所在直线的法向量
故:6(3)3(4)0AD l x y ---+=,即:220AD l x y +-= 15、已知:21,.(1)34,||b z i a R z z ωω=+∈=+-、若求;
221
z az b
z z ++-+(2)若=1-i ,求a 、b 的值
(1)||1,2a b ω==-=;
16、已知双曲线1C :2
2
14
y x -= (1)求与双曲线1C
有相同的焦点,且过点P 的双曲线2C 的标准方程;
(2)直线l :y x m =+分别交双曲线1C 的两条渐近线于A 、B 两点。
当3OA OB ⋅=时,求实数m 的值.
解:(1)双曲线1C
的焦点坐标是(,
设双曲线2C 的标准方程为2
2
221x y a b -=,则222211631
a b a b
⎧+=⎪
⎨-=⎪⎩ 解得2
241a b ⎧=⎪⎨=⎪⎩
所以双曲线2C 的标准方程为2
214
x y -=
(2)双曲线1C 的两条渐近线方程为2,2y x y x ==- 由2y x y x m =⎧⎨
=+⎩ 得(,2)A m m 由2y x y x m
=-⎧⎨=+⎩ 得2(,)33m m
B -
22
24333
m m OA OB m ⋅=-+==,得m =。