高中二年级(上)期中复习测试题
高二上学期期中测试试卷含解析试题
吴县中学2021—2021学年度第一学期期中测试高二语文试题1. 在下面一段话的空缺处依次填入词语,最恰当的一组是一个人不管伟大还是平凡,只要他自己的天性,找到了自己真正喜欢做的事,并且一心把它做得尽善尽美,他在这世界上就有了的家园。
于是,他不但会有足够的勇气去承受外界的压力,而且会有足够的清醒来面对的诱惑。
A. 顺应坚不可摧形形色色B. 顺应牢不可破形形色色C. 顺遂坚不可摧应有尽有D. 顺遂牢不可破应有尽有【答案】B【解析】试题分析:此题考察正确使用词语的才能。
“顺应〞是顺着某种趋势去适应;“顺遂〞是事情符合人愿,进展顺利。
从语境看,“天性〞应与“顺应〞搭配,与“顺遂人愿〞无关。
“牢不可破〞指巩固得不可破坏,形容事情或者观念已固定,难以破除或者改变;“坚不可摧〞指极为巩固,不能摧毀。
语境中的“家园〞不能用“坚不可摧〞来修饰。
“形形色色〞形容事物品类繁多、各式各样;“应有尽有〞应该有的全都有,从语境看,“诱惑〞不能用“应该有的〞来修饰。
【考点定位】正确使用词语〔包括熟语〕。
才能层级为表达运用E【技巧点拨】对于词语题,第一要辨析词义,包括词语的语义侧重点、词语的词义轻重、词义范围的大小等。
第二,辨析色彩。
包括词语的感情色彩跟语体色彩。
第三,辨析用法。
包括搭配习惯、语法功能、使用对象等方面。
此题中,第一处是考察近义词“顺应〞和“顺遂〞的辨析,要注意语境中的使用对象;第二处和第三处是考察成语的辨析,要注意语境的分析。
2.在下面一段文字横线处填入语句,衔接最恰当的一项是哪一项我们停顿说话,且看那瞬息万变的落照,迤逦行来,已到水边,水已成冰。
但也有浅淡的光,照在框外的冰上,使人想起月色的清冷。
①冰中透出枝枝荷梗,枯梗上漾着绮辉②岸边几株枯树,恰为夕阳做了画框③框外娇红的西山,这时却全呈黛青色,鲜嫩润泽④一派雨后初晴的模样,似与这黄昏全不相干⑤远山凹处,红日正沉,照得天边山顶一片通红A. ①④②⑤③B. ①⑤②③④C. ②①⑤③④D. ②①③④⑤【答案】B【解析】试题分析:此题考察语言表达简明、连接、得体。
高二年级语文上册期中检测考试题(含答案)
高二语文上册期中检测考试试题一、选择题(12分,每题3分)1、下列加点字读音完全正确的一项是()A、渌.水(lù)钿.头(diàn)殷.(yīn)岩泉嘲哳.(zhā)B、纶.巾(lún)会.稽(kuài)呕.(ōu)哑教坊.(fáng)C、浣.女(wán)贾.人(gǔ)湓.江(pén)鹧鸪.(gū)D、饯..行(jiàn)晌.午(shǎng)剡.溪(shàn)暮霭.(ǎi)2、下列各句中,加点成语使用不恰当的一句是()A、大学生们为了对付各种考试,更为了备战职场,从大学一年级就开始磨刀霍霍....向英语。
B、古巴官员表示,对于美国政府的如意算盘,古巴政府洞若观火....,将予以最有力的反击。
C、他们在创作过程中有过争吵,有过不同意见,但在某种程度上却是心照不宣....的好朋友。
D、民族艺术在改革创新中焕发活力:郭德纲的横空出世....,让久违的相声艺术再度变成为大众消费的热点。
3、下列各句中,没有语病的一句是()A、全党必须从贯彻“三个代表”重要思想,实现好、维护好、发展好广大人民群众根本利益的高度,进一步加强做好农民增收工作人紧迫性和主动性。
B、牛玉儒一生之所以能做到勤政为民、鞠躬尽瘁,源于他不忘本色,奋发进取,虚心学习,始终保持着一个共产党人蓬勃朝气、昂扬锐气、浩然正气。
C、适当搞一些排行榜,对于推介精品、引导时尚、繁荣文化市场,谁能说未必没有一定的好处呢?D、有的学生在考试前会出现失眠、烦躁等现象,这往往是因为他们太在乎考试成绩,心理负担过重造成的。
4、对词句理解分析正确的一项是()A.今宵酒醒何处?杨柳岸晓风残月(通过幽静清新的景色,表达作者恬淡自适的心情。
)B.废池乔木,犹厌言兵(作者运用移情手法写出了扬州战乱的惨状。
)C.念去去,千里烟波,暮霭沉沉楚天阔(浩渺的烟波,沉沉的暮霭,辽阔的天空,衬托出旅人前途的广阔。
高中二年级生物期中考试卷(上学期)
高中二年级生物期中考试卷(上学期)姓名: 学校: 满分100分 考试时间:100分钟一、选择题(共50分。
每小题2分)1.达尔文进化学说的中心内容是A .适者生存B .过度繁殖C .生存斗争D .自然选择学说 2.蒲公英的种子像“小伞”,窃衣的种子,果实上有刺,飞猴有“翅膀”,这说明 A .遗传和变异B .地理隔离C .生存斗争D .生物的适应性3.对达尔文自然选择学说的正确理解是 ①环境改变使生物产生适应性的变异 ②遗传的变异是生物进化的基础 ③变异是不定向的④变异是定向的⑤变异经过长期的自然选择和遗传积累就可能产生出生物的新类型A .②④⑤B .②③⑤C .①②④D .①③⑤4.下列关于基因频率的叙述中,正确的是A .某种群中某基因的突变率B .某种群中某种基因的出现比例C .某种群中显性基因的出现比例D .某种群中隐性基因的出现比例 5.若一个群体中,某一性状出现的频率增加,这很可能是A .这个性状对环境具有很强的适应性B .该群体中具此性状的纯合体多C .这个性状的出现是由环境条件引起的D .控制这个性状的基因为隐性6.有一个老鼠的种群,最初全部生活在一个地域,一条新建的高速公路将该种群分为A 、B 两个种群。
如果A 种群生活的地区发生了一系列的环境变化,而B 种群的地区变化不大,种群A 进化的速率很可能是A .最初比B 慢 B .最初比B 快C .与B 相同D .开始比B 慢,然后比B 快7.在果蝇中,假定每个基因的突变率是10-5,对于一个中等数量的果蝇种群(约108个个体),却有107个基因发生了突变,则每个果蝇大约有多少个基因A .0.5×104B .104C .2×104D .4×1048.现代进化论与达尔文进化论观点不同的是A .可遗传的变异是生物进化的原始材料B .自然选择决定生物进化的方向C .种群基因频率的改变是生物进化的实质D .自然选择是环境对生物的选择9.下列高科技成果中,根据基因重组原理进行的是 ( )①我国科学家袁隆平利用杂交技术培育出超级水稻 ②我国科学家将苏云金芽孢杆菌的某些基因移植到棉花体内,培育出抗虫棉 ③我国科学家通过返回式卫星搭载种子培育太空椒 ④我国科学家通过体细胞克隆技术培养出克隆牛A .①B .①②C .①②D .③④10.下图是人体局部内环境示意图。
河南省洛阳市2022-2023学年高二上学期期中考试理科数学试卷(含答案)
(3) 已知点的坐标为(5,3),点在曲线 ′ 上运动,求线段的中点的轨迹方程.
22. (12 分)
如图,长方体 — 1 1 1 1 中, = 2 = 21 ,
点在棱上且1 丄平面1 1
(1)求 的值
21. ( 12 分)
已知两定点 (-4,0), (-1,0),动点 满足 | | = 2 ||,直线 :(2 + 1) + ( + 1) −
5 − 3 = 0.
(1) 求动点的轨迹方程,并说明轨迹的形状;
(2) 记动点的轨迹为曲线,把曲线向右平移 1 个单位长度,向上平移 1 个单位长度后得到曲线 ′ ,
反射光线所在直线的方程.
20. (12 分)
在直角梯形 中, //, = 2 = 2 =2 2,∠ = 900 如图(1). 把△沿
翻折,使得平面 ⊥平面,如图(2).
(1) 求证: ⊥ ;
(2) 若为线段的中点,求点到平面的距离.
所成角的余弦值为
A.
6
B.
3
3
C.
3
15
D.
5
10
5
12. 若圆 2 + 2 − 4 − 4 − 10 = 0至少有三个不同的点到直线: = 的距离为 2 2,则直线的倾斜角
的取值范围是
A.[ 12 , 4 ]
5
B. [ 12 , 12 ]
C. [ 6 , 3 ]
B. - 5
C. 10
D. -10
2.已知(4,1,9),(2,4,3),则线段的长为
A. 39
B.7
高中二年级(上)期中复习测试题
高中二年级(上)期中复习测试题第一部分听力(共两节,满分30分)(略)第二部分英语知识运用(共两节,满分45分)第一节单项选择(共15小题;每小题1分,满分15分)从A、B、C、D四个选项中,选出可以填入空白处的最佳选项。
21. — Can I look at the menu for a few more minutes before I decide?—Of course. ____, sir.A. Enjoy yourselfB. Make yourself at homeC. Take your timeD. It doesn’t matter22. The manager has ____ a car to pick the invited scientist up at the airport.A. arrangedB. arranged forC. arranged withD. arranged by23. Whenever he was asked why he was late for work, he would answer carelessly, always ___ the same thing.A. sayingB. saidC. to sayD. having said24. We regret to inform you that there are no tickets ____ for Friday’s performance.A. availableB. spareC. convenientD. affordable25. After a five-hour drive, they reached ___ they thought was the place they’d been dreaming of.A. thatB. whereC. whichD. what26. When ___ in court, the man didn’t admit having taken the old lady’s necklace.A. questioningB. having questionedC. being questionedD. to be questioned27. — Why have you ___ Mr Johnson as the best person for this position?—Because he is responsible and careful.A. put onB. put downC. put forwardD. put out28. All the senior leaders are members of the same family. _____, I’m likely to have to move if I want to get promoted.A. ThoughB. MeanwhileC. ThereforeD. However29. So important ___ it to get on well with people around that I have decided to learn some communication skills.A. I have foundB. have I foundC. I was feelingD. was I feeling30. It’s no ____ arguing with Bill about joining us in the experiment because he will never change his mind.A. useB. needC. wayD. doubt31. You’d better behave your self when staying at his home, for his mother is a(n) _____ woman who seldom smiles.A. cautiousB. positiveC. enthusiasticD. severe32. — Why was Fred forced to leave school?—Because he was _____ of cheating in the English exam.A. blamedB. accusedC. chargedD. caught33. I’m sorry that what I said just now has hurt you, but I didn’t do it ______.A. deliberatelyB. actuallyC. carelesslyD. especially34. — Did you catch _____ sight of a tall TV tower in the distance just now?—Yes, and now it’s out of ____ sight.A. /; theB. /; /C. the; theD. a; a35. Facing lots of temptations, the scientist still ____ his experiment of great value to his country’s development.A. devoted toB. depended onC. concentrated onD. referred to第二节完形填空(共20小题;每小题1.5分,满分30分)On October 31st, 2003, young surfer, Bethany Hamilton suffered a big loss. It was Halloween morning in Hawaii and a(n) 36 day for surfing. Bethany went surfing with her best friend. She was catching great waves and 37 her life changed in a second. She got 38 by a 14-foot tiger shark, which bit her just 4 inches 39 her left shoulder and took the rest of her arm off! She never 40 that to happen to her on a sunny Halloween!Bethany was 41 on being a professional surfer before her accident in the waves. She 42 surfing at an early age and her whole family were surfers. Bethany only 43 14 on February 8th, 2004, so it was a young age to have something 44 bad happen to someone.Bethany Hamilton, 45 , showed that she was a true hero by saving her life. Her life was 46 when the tiger shark bit her arm completely off and her 47 helped her get to safety and she was successful. She 48 overcame the loss of her left arm and within a month she rose to the 49 of going surfing in the ocean 50 !She is now a role model to many people for what she has done. She turned a 51 into a positive, which many people can’t do! She remained positive and 52 to go forward with her dream. She is 53 a survivor hero and is keeping her surfing career alive as well. She 54 to surfing with only one arm, and that is amazing! 55 , millions of people are inspired to try to find something good in something that only seems bad.36. A. unlucky B. perfect C. mild D. normal37. A. immediately B. naturallyC. sadlyD. suddenly38. A. trapped B. surroundedC. defeatedD. attacked39. A. nearby B. below C. above D. within40. A. expected B. planned C. hoped D. forgot41. A. depending B. insisting C. planning D. calling42. A. gave up B. took up C. set up D. kept up43. A. grew B. became C. turned D. announced44. A. too B. enough C. much D. that45. A. therefore B. however C. instead D. indeed46. A. in peace B. at riskC. out of controlD. beyond reach47. A. friends B. relatives C. instincts D. parents48. A. almost B. never C. hardly D. also49. A. challenge B. chance C. danger D. fate50. A. in time B. more or lessC. once againD. at least51. A. negative B. weaknessC. misfortuneD. disability52. A. began B. stopped C. prepared D. continued53. A. originally B. obviouslyC. unexpectedlyD. finally54. A. devotes B. looks forwardC. triesD. goes back55. A. In all B. At that momentC. As a resultD. From now on第三部分阅读理解(共20小题;每小题2分,满分40分)AThere are many different types of fire ants, and they are found all throughout the Southeastern and Southwestern United States. The most common and aggressive kind is the red imported fire ant, which is reddish-brown. Their nests can be found on lawns, in parks, on playgrounds and in fields. Some red imported fire ants create nests in the walls of buildings.A person who gets stung by a fire ant will feel a sharp pain and burning. If a person steps on a fire ant mound (土堆), he or she will get a lot of stings, because the ants have been disturbed where they all live together. Each sting will turn into an itchy blister (水泡) over the next day.The best way to avoid getting stung by fire ants is to keep your shoes on when you are playing near fire ant mounds. If you meet one, don’t poke at it or try to play with it. However, if you think that you have been stung by a fire ant, tell an adult immediately. That’s because the area of the sting will swell up quite a bit, and a doctor may want to have a look to make sure you do not have breathing problems or chest or throat tightness —an allergic(过敏的)reaction which is similar to one from a bee sting can sometimes happen. If these symptoms appear, the person needs medical attention right away.But more often, you can follow these steps after a fire ant bite:☆Wash the area with soap and water.☆Apply some ice to the area.☆Check with the doctor if you have redness, swelling, or itching.56. Why do fire ants attack people?A. It is in their nature to attack people.B. They do so quite by accident.C. Human beings are their natural enemy.D. They do so to protect themselves.57. For whom is the passage written?A. Travellers.B. Farmers.C. Children.D. Ant researchers.58. To avoid being stung, you are advised _______.A. not to go to places where fire ants liveB. not to disturb fire ants in any wayC. to have an adult with you when going outD. to take soap and water with you when going out59. From the passage, we can learn that ___________.A. the United States is the only place where fire ants existB. some stung people by a bee will not have chest or throat tightnessC. sometimes the sting of a fire ant will cause serious problemsD. a stung person will always have an allergic reaction without immediate treatmentBEver wonder why some people never seem to get sick but others get stuck in bed? Scientists say it might have something to do with special immune cells that are more like jail cells, because they put the flu virus behind bars.As reported in the journal NatureMedicine, researchers at Washington University in St. Louis discovered that these jail cells, formally known as macrophages (巨噬细胞), rely on a gene called CCL5 to stay alive even after a virus gets inside them.When the flu virus infects our normal cells, they have to commit suicide to prevent the virus from spreading. But as Holtzman’s team reported, the gene CCL5 blocks this process in macrophages. As a result, macrophages can pick up viruses, hold them inside their membranes (细胞膜), and help clear them from our bodies —without making themselves dead.To figure out how CCL5 works, the researchers created mice without the gene and compared them to mice that had it. “We find that the mice that were missing it were dying from the viral infection whereas the mice that had it did perfe ctly fine,” says Holtzman.In a follow-up laboratory study, they compared human cells that had CCL5 to others that did not have CCL5 and saw the same thing. Without CCL5, the cells died. Holtzman suspects that because we all have different gene combinations, those of us who have more of this gene may be less likely to get sick.60. Which description of immune cells is TRUE?A. Immune cells are kinds of normal cells.B. Immune cells are CCL5.C. Immune cells are formally known as macrophages.D. Immune cells can pick up viruses and kill them.61. When infected, normal cells ________________.A. can’t prevent the virus from spreadingB. have to commit suicide to prevent the virus from spreadingC. can put the flu virus behind barsD. can prevent the virus from spreading without making themselves dead62. From the last two paragraphs, we know that _______.A. both mice and human cells that didn’t have the gene CCL5 died from the viral infectionB. the mice have more of the gene CCL5 than human beingsC. each person has a different gene CCL5, so some people never seem to get sickD. if you have more of the gene CCL5, you’ll never get sick63. According to the passage, who gets the flu easily?A. A person who often gets stuck in bed.B. A person who has more of the gene CCL5.C. A person who has more jail cells.D. A person who has less of the gene CCL5 than normal.CWhen the Farnsworth family moved to their new farm in 1919, eleven-year-old Philo was surprised to find it wired for electricity. This unusual circumstance contributed to his fate — to become an important inventor of the twentieth century.By thirteen, Farnsworth had become a self-taught electrical engineer. He was able to fix the farm’s generator(发电机)when none of the adults could. In 1922, he read an article about a new idea of John Baird, a Scottish scientist, who had been working with the cathode ray tube (阴极射线管) for the transmission of electronic pictures and wanted to attempt it himself.Farnsworth studied everything he could find on the subject. Although many older engineers with money backers were already developing television, Farnsworth made a bold decision — he was going to perfect a working model of it before anybody else.In college, Farnsworth continued his research with cathode ray and vacuum tubes, but the death of his father, the only money maker in the family, forced him to give up this research and find a job. His first job was for George Everson, with whom Farnsworth discussed his dream of television. While acknowledging the achievements of those who came before, Farnsworth thought that he could get closer. Everson agreed to risk $6,000 for the research.Backers came in 1927 to see the first American television, one year after Baird’s. They were astonished to see the image of a single white line resolve itself on the screen before them, and agreed that this new invention was worth putting money into.In 1930, Farnsworth won a patent (专利权) for his all-electronic TV. By the time he died, he had earned over 300 American and foreign patents for electronic and mechanical devices.64. When Farnsworth was at a young age, he _______.A. had to drop out of school to help on the family farmB. was sent to school to study electrical engineeringC. wanted to be the first person to invent the televisionD. had shown a surprising ability in the electrical field65. What difficulty did Farnsworth meet when he first began his research on the television?A. His parents didn’t support his work.B. He didn’t have enough knowledge in this field.C. He didn’t have enough money for his research.D. No one was interested in this research.66. How old was Philo Farnsworth when he invented the first American television?A. 11.B. 13.C. 19.D. 22.67. Why did Philo Farnsworth work for George Everson?A. Because he had to support his family.B. Because George had supported his research.C. Because George was the richest person he knew.D. Because he could find more supporters there.68. From the passage, we can learn that Farnsworth is ______.A. the first person who worked for the transmission of electronic picturesB. an inventor who improved on somebody else’s ideaC. an inventor who always came up with an original ideaD. a person who earned over 300 American patents for electronic devicesDYour youngest son or daughter has just graduated from high school, and soon he or she will trade his or her bedroom in your home for a college resi dence hall. You’re starting to worry about the coming silence in your home and are wondering how you will fill the hours that you previously spent with your son or daughter attending his or her school and sporting events.Take heart —the empty nest “is often worse in anticipation (预料) than in day-to-day practice”, says psychologist Bert Hayslip Jr.Hayslip, who has studied retirement adjustment (调整) and other aging problems, such as aging sleeplessness, for many years, points out that, more often than not, children who leave do not completely cut off contact with their parents. He advises parents to think of it as a series of life events, instead of a sudden change. He also says an empty nest can cause bigger problems to surface “if a couple hasn’t looked after their marriage while raising their children”. According to Hayslip, some couples may find they no longer have anything in common once the children are out of the house.Thinking of an empty nest as the loss of children makes the adjustment more difficult. “With the empty-nest syndrome (综合症), parents actually need to deal with the loss of the parenting control over children, not with having really lost their children,”Hayslip says. “They just have to find a new way to get along with their children.”“Relating to the ir college-age children in this new way will come easily to parents as the months pass. As with many things, the passage of time heals the pain of loss,” Hayslip says.69. What does the “empty nest” mean?A. Children’s cutting off contact with their parent s.B. Children’s not liking spending time with their parents.C. Children’s growing up and leaving home.D. Children’s not accepting parents’ care and love.70. What does the underlined word “it” in the third paragraph refer to?A. Retirement adjustment.B. Aging problem.C. Sleeplessness.D. The reality of empty nest.71. Why do many parents find it hard to adjust to the empty-nest life?A. Because they think they have actually lost their children.B. Because they often disagree with their children on many things.C. Because their children completely cut off contact with them.D. Because these parents don’t have anything in common.72. According to Hayslip, what parents actually lose in an “empty nest” is _______.A. their children’s love for themB. their control over childrenC. their interest in lifeD. their contact with childrenEIt is almost a reflex action among Americans to say “God bless you” when someone sneezes. The custom, perceived as a wish for luck, goes back to America’s European roots. In most parts of Europe, there is an old belief that a sneeze can expel the soul from the body. Because of this, the blessing serves as a verbal charm to protect the sneezer from death. Another belief is that if someone sneezes three times without receiving a “bless you” from someone, the fairies could come and take him away.Sometimes, however, the blessing isn’t completely necessary. A corollary (具有推断性质的) belief holds that if two people sneeze at the same time, both will enjoy good luck.A solitary sneeze is considered a lucky omen (征兆) in many parts of the world. The Maori people of New Zealand believe it is symbolic of the act of creation. According to their creation myth, the first human received the breath of life when the great God Tiki sneezed. In parts of Africa, sneezing is considered a sign that the body has been taken over by good spirits. And most Native American tribes believed that a sneeze was a sign that the brain was being cleared of evil.This superstition may also have arisen from a serious disease that was attacking the early Romans and the main symptom was sneezing. This disease was so serious that people often died as a result, and the Romans believed that the more often the individual was blessed, the lower the chances of death.Also, “God bless you”was a frequently used form of well-wishing at the time. It was considered a wish for good health. Since sneezing was a symptom of the bubonic plague (淋巴腺鼠疫) in the Middle Ages in Europe, the custom of blessing people after they sneeze persisted and has been handed down until today. For the Hebrews, Greeks, Indians, and Chinese, breath was a sign of the soul’s existence, and any loss of the breath, espec ially in a sneeze, was exceedingly bad luck.73. From the passage, we can learn that .A. to say “God bless you” is necessary when a person sneezesB. sneezing has different meanings in different areasC. sneezing is considered a lucky omen in the worldD. a sneeze means exceedingly bad luck for most Native American tribes74. Why do Americans usually say “God bless you” when someone sneezes?A. Because it is a reflex action.B. Because it is a custom which goes back to America’s Europ ean roots.C. Because it is an old belief.D. Because it is a verbal charm to protect the sneezer.75. What is the best title of the passage?A. Sneezing and “God bless you”.B. Sneezing: bad luck.C. Sneezing: a verbal charm.D. Sneezing: charms and symbols.第四部分写作(共两节,满分35分)第一节短文改错(共10小题;每小题1分,满分10分)In the morning of April 29, 2007, from 10:00 to 11:00, hundreds of76. ______millions students from primary, middle schools and universities all77. ______over China join in a national student sports program. Eleven78. _______million students in Jiangsu took an active part in this program.79. _______The Ministry of Education calls on student to exercise for an hour80. _______every day, in the hope of that they will have good health to work81. _______fifty years and enjoy the whole life.Personally, I think this a good idea for senior 3 students to take daily82. _______exercise. Although it takes an hour a day for exercise, but it is well worth 83. _______it. Take exercise helps us build up our body and keep a clear mind.84. _______Therefore, we can work more efficient. 85. _______第二节书面表达(满分25分)仔细观察下图,请以“Project Hope”为题写一篇短文。
高二级上期期中语文试题
高二级上期期中语文试题亲爱的同学们:当你走进考场:你就是这里的主人:只要细心、认真地阅读、思考:你就会感到试题并不难。
一切都在你的掌握之中:请相信自己!老师真诚提醒你:务必书写认真、工整、规范:卷面干净、整洁、美观。
说明:1、本卷共四大题:满分150分:考试时间120分钟.2、试卷答案填在答题纸上。
一、语文基础知识和运用(40分)(一)语言的积累和运用(28分)1、选下列加点字注音全对的一项()(3分)A. 震悚.(sǒng)荒谬.(miào)溃.退 (kuì) 锐不可当.(dǎng)B. 要塞.(sài)瞥.见(biē)惊骇. (hài) 歼.灭(qiān)C.掳.掠(lǔ) 幽咽.(yè)嫉.妒(jì)差.使(chāi )D.阡.陌(qiān)匀称.(chèn)寒噤.(jìn) 仄.歪(zè)2、下列词语中书写完全正确的一项是()(3分)A.阻遏创疤愧怍穷愁撩倒 B.烦燥狼籍蹒跚张皇失措C.琐屑倾泄杀戮响彻云宵 D.簌簌颠沛滞笨惟妙惟肖3、下列的字或词语解释有误的一项()(3分)A.业已:已经震悚:身体因恐惧或过度兴奋而颤动。
B.大庭广众:人很多的公开场合。
尴尬:神色、态度不自然。
C.蹿:向上或向前跳。
塌败:失败。
D.情郁于中:感情积聚在心里。
荡然无存:形容原有的东西完全失去:一点也没留下。
4.对出下联:表现领导干部为人民服务的态度。
(2分)上联:站起来当伞:给群众遮风挡雨:下联:____________:____________。
5、下面语段中划线的句子有语病:请修改在横线上。
(4分)①为生命画一片树叶:只有心存相信:总有奇迹发生:希望虽然渺茫:但它永存人世。
美国作家欧;亨利在他的小说《最后一片叶子》里讲了个故事:②病房里:一个身体垂危的病人从房间里看见窗外的一棵树:在秋风中一片片地掉落下来。
③望着眼前的萧萧落叶:身体也随之每况愈下:一天不如一天。
高二上学期期中考试语文试卷及参考答案
高二上学期期中考试语文试卷及参考答案第一学期期中考试高二年级语文试卷分值:150分时间:150分钟一、名篇名句默写(每空1分,共10分)1.补写出下列句子中的空缺部分。
1)___在《陈情表》中,陈述自己家没有什么亲属、仆人的实情的句子是“臣本无父。
”2)___在《陈情表》中以乌鸦反哺比喻要守孝道的句子是“以孝治国。
”3)___在《游褒禅山记》里认为,古人看各种风景,都有心得,其原因是“心之所往。
”4)___在《报任安书》中,为表明只有卓越超群不同一般的人才能名传后世,举了很多历史人物的事例,其中举___和___的例子的两句是“世之奇伟、瑰怪、非常之名。
帝王将相。
”5)我们一般用___《报任安书》“天下英雄所见略同”的三句话来概括___写《史记》的目的。
二、语言文字运用(每小题3分,共12分)2.下面加点词语注音全都正确的一项是()(3分)A。
可汗 kè hán 瞋目chēn mù 床蓐 rù / 覆校无疑 jiào wú yí/ 老校退卒 xiàoB。
拔擢 zhuó砧板 z hēn 彘肩 zhì / 剌谬 miào / 戮力 lù lìC。
揾泪 wèn 刀俎jǔ 玉玦quē 目眦尽裂 zì / 恣卒为暴 zì cùwéi bàoD。
籼稻shān dào 车骑 jì / 盘桓 huán / 责臣逋慢bū / 吾未晡食bǔ3.下面加点词语运用不正确的一项是()(3分)A。
跨进新年,深圳正以勃勃英姿,在改革开放的道路上阔步前进。
B。
他根据自己的实践,以科学家的胆识和眼光断定杂交水稻研究具有光辉的前景,决心义无反顾地坚持研究。
C。
除了你的善良的精神以外,身无长物,我恭维了你又有什么好处呢?D。
2022-2023学年江苏省盐城中学高二年级上册学期期中数学试题【含答案】
2022-2023学年江苏省盐城中学高二上学期期中数学试题一、单选题1.过,两点的直线的倾斜角为( )()2,1A ()3,2B A .B .C .D .60︒45-︒135︒45︒【答案】D【分析】利用斜率公式可求直线的斜率,从而可求直线的倾斜角.【详解】设过,两点的直线的倾斜角为,,()2,1A ()3,2B θ0180θ︒≤∈<︒∴,21tan 132θ-==-可得.45θ=︒故选:D.2.已知双曲线的渐近线方程为,则( )2221y x m -=y =2m =A .BC .D .51525【答案】A【分析】先由双曲线方程求出其渐近线方程,再已知的渐近线方程可得,从而可求得结果.m =【详解】解:双曲线的焦点在轴上,2221y x m -=x 其渐近线方程为,y m x=±由渐近线方程为,可得y =m =可得.25m =故选:A.3.直线和直线互相垂直,则实数的值为( )1:20l x ay a ++=()2:230l a x y -+=a A .B .C .或D .或3a =-12a =1a =3a =1a =-3a =【答案】B 【分析】 ,,若,则,由上述公式可1111:0l A x B y C ++=2222:0l A x B y C ++=12l l ⊥12120A A B B +=得结果.【详解】解:由题意,得,()1230a a ⨯-+=解得,12a =经检验,符合题意,故.12a =故选:B.4.已知椭圆上一点的横坐标为,是椭圆的右焦点,则点到点的距离为221259x y +=P 2F P F ( )A .B .C .D .585335175【答案】D【分析】由椭圆方程求得右焦点坐标,代入求得点纵坐标后,由两点间距离公式计算距F 2x =P 离.【详解】解:已知椭圆,221259x y +=则,,则,,225a =29b =225916c =-=4c =椭圆的右焦点的坐标为,F ()4,0将代入得,2x =221259x y +=y =.175=故选:D.5.已知平面内两定点,,动点满足,则的最小值为( )()1,0A -()10B ,C 3AC BC ⋅=BC A .B .C .D .132【答案】A【分析】先设出点的坐标,代入,求出的方程之后可求.C 3AC BC ⋅=C 【详解】设点坐标为,又,,C (),x y ()1,0A -()10B ,则,()()221,1,13AC BC x y x y x y ⋅=+⋅-=+-=即,224x y +=则点在以原点为圆心,半径为的圆上,C ()0,0O 2r =则表示点到圆上一动点的距离,BCB 224x y +=又,故点在圆内部,21014+=<B 224x y +=则最小值为.min 1BC r BO =-=故选:A.6.若直线与曲线有且只有一个交点,则满足条件的直线有( ):20l x my m +--=2214x y -=l A .条B .条C .条D .条4321【答案】C【分析】利用双曲线和双曲线渐近线的图像和性质求解即可.【详解】直线,即恒过点,:20l x my m +--=()120m y x -+-=()2,1又双曲线的渐近线方程为,12y x=±则点在其中一条渐近线上,()2,112y x=又直线与双曲线只有一个交点,则直线过点且平行于或过点且与双曲线的右支相切,l ()2,112y x =-()2,1即满足条件的直线有条.l 2故选:C7.若直线与曲线的取值范围是( )():40l x m y +-=x =mA .B .C .D .0m <<0m ≤<0m <≤0m ≤≤【答案】B【分析】化简曲线方程,表示圆心为,半径为的圆在轴以及右侧的部分,由直线与曲线的()0,02y 交点个数可以确定的取值范围.m【详解】表示的曲线是圆心为,半径为的圆在轴以及右侧的部分,如图所示:x =()0,02y直线必过定点,():40l x m y +-=()0,4当直线与圆相切时,直线和圆恰有一个交点,l,结合直线与半圆的相切可得,2m =当直的斜率不存在时,即时,直线和曲线恰有两个交点,l 0m =所以要使直线和曲线有两个交点,则0m ≤<故选:B.8.已知是圆上的一动点,点,线段的垂直平分线交直线于点P ()221:316F x y ++=()23,0F 2PF 1PF ,则点的轨迹方程为( )Q Q A .B .22154x y -=22149x y -=C .D .22145x y -=()221045x y x -=>【答案】C 【分析】由题意有,从而有,根据双曲线的定义得点的轨迹为2QP QF =1214QF QF PF -==Q 是以F 1、F 2为焦点的双曲线.再写出其方程即可.【详解】如图所示:∵是圆上一动点,点的坐标为,线段的垂直平分线交直线于点,P 1F 2F ()3,02PF 1PF Q ∴,,2QP QF =1211QF QF QF PF QP -=-=∵是圆上一动点,∴,∴,P 1F 14PF =124QF QF -=∴,,,()23,0F ()13,0F -1264F F =>∴点的轨迹为以F 1、F 2为焦点的双曲线,且,,得Q 2a =3c =b =∴点的轨迹方程为.Q 22145x y -=故选:C.二、多选题9.已知直线过原点,且,两点到直线的距离相等,则直线方程可以为( )l ()1,4A ()3,2B l A .B .C .D .0x y +=50x y +-=320x y -=320x y +=【答案】AC【分析】由题意先设出方程,根据已知条件建立方程解出直线的斜率即可【详解】直线过原点,且,两点到直线的距离相等,l ()1,4A ()3,2B l 斜率必存在,设所求直线的方程为,0kx y -=由已知及点到直线的距离公式可得:解得或,1k =-32k =即所求直线方程为或.0x y +=320x y -=故选:AC.10.已知曲线,则下列说法正确的是( )22:1C ax y +=A .若,则是椭圆,其焦点在轴上01a <<C yB .若,则是双曲线,其渐近线方程为0a <C y =C .若曲线为椭圆,其焦点为,则C ()1,02a =D .若,则是两条直线0a =C 【答案】BD【分析】根据椭圆的标准方程求解判断AC ,根据双曲线的标准方程判断并求解判断B ,代入0a =求解判断D .【详解】当时曲线可化为,0a ≠22:1C ax y +=2211x y a +=若,则是椭圆,其焦点在轴上,故选项A 错误;01a <<C x 若,则是双曲线,其渐近线方程为,故选项B 正确;a<0C y =若曲线为椭圆,其焦点为,则,又,C ()1,01c =1b =∴,即,故选项C 错误;12a =12a =若,则为:或是两条直线,故选项D 正确.0a =C 1y =1y =-故选:BD.11.设,为实数,已知椭圆与双曲线有相同的焦点,,且椭圆与双曲m n 2214x y m +=221y x n -=1F 2F 线在第一象限的交点为,则下列说法正确的是()P y ⎫⎪⎪⎭A.B .y =2n =C .D .左焦点为1m =()【答案】BCD【分析】根据椭圆和双曲线同焦点,且两曲线均过P 点,建立方程求出、,然后根据椭圆和双m n 曲线的性质解题即可.【详解】解:根据题意可知:,且04m <<2222413141m n y m y n -=+⎧⎪⎪⎪⎪⎝⎭+=⎨⎪⎪⎪-=⎪⎩解得:,故A 错误,B 、C 正确;21n m y ⎧⎪=⎪⎪=⎨⎪⎪=⎪⎩则,所以左焦点为,故D 正确.c==()故选:BCD.12.已知直线和曲线,点A 是直线上的一个动点,点是曲线上的:40l x y +-=22:4O x y +=l D O 一个动点,过点A 作曲线的两条切线,切点分别为、,则下列说法正确的是( )O B CA .的最小值为AB2B .曲线上存在个点到直线的距离等于O 2l 2-C .若曲线上总存在点,使得,则A 的横坐标的取值范围是O D 30OAD ∠=︒[]0,4D .直线过定点BC ()1,1【答案】ACD【分析】根据圆的切线长的计算公式结合圆心到直线的距离即可求得的最小值,判断A;结合ABA 的分析可判断B;由曲线上总存在点,使得,可得,从而O D 30OAD ∠=︒3090OAB ︒≤∠<︒,设,可得不等式,求得x 范围,判断C;由题意可知、两点在以1sin 12OAB ≤∠<(),4A x x -B C 为直径的圆上,求出以为直径的圆的方程,联立求得直线的方程,可推得OA OA 22:4O x y +=BC 直线所过的定点,判断D.BC【详解】对于A:,所以最小时,最小.AOAB因为当是点到直线的距离时,最小,最小值为,AOO l AO因此最小为,故A 正确;AB2=对于B :由选项A 知:点到直线的距离为的半径为,O l O 2因此曲线上存在个点到直线的距离等于,故B 错误;O 1l 2对于C :因为点A 是直线上的一个动点,所以设,l (),4A x x -因为曲线上总存在点,使得,所以,O D 30OAD ∠=︒3090OAB ︒≤∠<︒因此,1sin 12OAB ≤∠<又因为在中,,Rt OAB sin OBOAB OA∠==所以,即,解得,112≤<24<04x ≤≤因此点A 的横坐标的取值范围是,故C 正确;[]0,4对于D:由题意过点A 作曲线的两条切线,切点分别为、,O B C 可知、两点在以为直径的圆上,B C OA 设,则为直径的圆的方程为,(),4A t t -OA ()(4)0x x t y y t -+-+=和相减可得直线的方程,即,224x y +=BC ()44tx t y +-=即,由于,故由,得,()()410t x y y -+-=R t ∈01x y y -=⎧⎨=⎩11x y =⎧⎨=⎩所以直线恒过定点,故D 正确.BC ()1,1故选: .ACD 【点睛】难点点睛:本题判断正误的难点在于C,D 选项的判断,对于C 选项,要能够根据曲线上总存在点,使得,明确,然后结合三角函数求解;对于D 选O D 30OAD ∠=︒3090OAB ︒≤∠<︒项,要能够明确即为以为直径的圆和的公共弦,由此可求得直线的方程.BC OA 224x y +=BC 三、填空题13.法国数学家蒙日发现:双曲线的两条互相垂直切(),17461818Monge -()2222:10x y a b a b Γ=>>-线的交点的轨迹方程为:,这个圆被称为蒙日圆.若某双曲线对P 2222x y a b +=-()22210x y a a -=>应的蒙日圆方程为,则___________.223x y +==a 【答案】2【分析】根据题意写出双曲线对应的蒙日圆方程,可得出关于的等式,即可求()22210x y a a -=>a 得正数的值.a 【详解】由双曲线的方程可得,()22210x y a a -=>21b =由蒙日圆的定义可得双曲线对应的蒙日圆方程,所以,即()22210x y a a -=>223x y +=223a b -=,213a -=可得.2a =故答案为:2.14.写出与圆和圆都相切的一条直线的方程:___________.224x y +=22120x y +--+=【答案】(或或任意填一个即可).0x y +-=0x y +-=0x y -+=0x y --=【分析】先由两圆位置及半径讨论不存在平行于y 轴的公切线,设,由圆心与线的距离与y kx b =+相切的关系列方程组求解即可.【详解】根据题意设,圆心,半径;22:4A x y +=()0,0A 12r =,圆心,半径;((2222:124B x y x y +--+=-+-=(B 22r =由两圆位置及半径显然不存在平行于y 轴的公切线,设,即,解得或y kx b =+0kx y b -+=22=1k b =⎧⎪⎨=⎪⎩1k b =⎧⎪⎨=-⎪⎩1k b =-⎧⎪⎨=⎪⎩此时圆A与圆的公切线方程为,B y x =+y x =-y x =-+所以满足直线的方程为或或.0x y +-=0x y -+=0x y --=故答案为:(或或任意填一个即可).0x y +-=0x y +-=0x y -+=0x y --=15.数学中有很多形状优美,寓意美好的曲线,曲线就是其中之一,则曲22:220C x y x y +--=线所围成的封闭图形的面积是___________.C 【答案】84π+【分析】方程,对,分类讨论,画出图象即可得出面积.22220x y x y +--=x y 【详解】方程,对,分类讨论,22220x y x y +--=x y ①,时,化为:;0x ≥0y ≥()()22112x y -+-=②,时,化为:;0x ≥0y ≤()()22112x y -++=③,时,化为:;0x ≤0y ≥()()22112x y ++-=④,时,化为:;0x ≤0y ≤()()22112x y +++=所表示的曲线所围成的图形面积.(22284S ππ=+⨯=+故答案为:.84π+16.已知、分别在直线与直线上,且,点,P Q 1:10l x y -+=2:10l x y --=1PQ l ⊥()4,4A -,则的最小值为___________.()4,0B AP PQ QB++【分析】利用线段的等量关系进行转化,找到最小值即为所求.AP QB+【详解】由直线与作直线垂直于,如图,1l 2l ()4,0B l 1:10l x y -+=则直线的方程为:,将沿着直线个单位到点,有,l 4y x =-+()4,0B l B '()3,1B '连接交直线于点P ,过P 作于Q ,连接BQ ,有,即四边形AB '1l2⊥PQ l //,||||BB PQ BB PQ ''=为平行四边形,BB PQ '则,即有,显然是直线上的点与点距离和的最||||PB BQ '=||AP QB AP PB AB ''+=+=AB '1l ,A B '小值,因此的最小值,即的最小值,而,AP QB+AP PB '+AB '=所以的最小值为AP PQ QB++【点睛】思路点睛:(1)合理的利用假设可以探究取值的范围,严谨的思维是验证的必要过程.(2)转化与划归思想是解决距离最值问题中一种有效的途径.(3)数形结合使得问题更加具体和形象,从而使得方法清晰与明朗.四、解答题17.已知的三个顶点分别是,.ABC ()1,1A -()2,3B ()3,1C -(1)求边所在直线的方程;BC (2)求的面积.ABC 【答案】(1)4110x y +-=(2)7【分析】(1)根据两点式即可写出直线方程.(2)利用两点之间距离公式和点到直线的距离公式即可求得三角形面积.【详解】(1)由题可知,直线经过,,BC ()2,3B ()3,1C -∴其方程为,321332y x --=---化简得,4110x y +-=∴直线的方程为BC 4110x y +-=(2=点到直线的距离()1,1A -BC d∴,11722ABC S BC d =⋅⋅==△∴的面积为7.ABC18.已知椭圆,以原点为圆心,椭圆的短半轴长为半径的()2222:10x y C a b a b +=>>圆与直线相切.20x y -+=(1)求椭圆的方程;C(2)设,过点作与轴不重合的直线交椭圆于,两点,连接,分别()2,0A -()1,0R x l C M N AM AN 交直线于,两点,若直线、的斜率分别为、,试问:是否为定值?若是,3x =P Q PR QR 1k 2k 12k k ⋅求出该定值,若不是,请说明理由.【答案】(1)22142x y +=(2)是定值,定值2524-【分析】(1)根据离心率和点到直线距离公式即可得解;(2)直线的方程为,代入椭圆方程,根据三点共线表示出P 点坐标,同理表示出Q 点l 1x my =+坐标,算出斜率即可求解.【详解】(1)(1)由题意得,解得,222c e a b a b c ⎧==⎪⎪⎪⎨⎪⎪=+⎪⎩2a b c =⎧⎪=⎨⎪=⎩故椭圆的方程为;C 22142x y +=(2)设直线的方程为,,,l 1x my =+()11,M x y ()22,N x y 由得,221142x my x y =+⎧⎪⎨+=⎪⎩()222230m y my ++-=∴,,12222m y y m -+=+12232y y m -=+由A 、、三点共线可知,M P ()()11322P y y x =----∴1152P y y x =+同理可得:,2252Q y y x =+故121212551131314422Q P P Q y y y y k k y y x x =⋅==⋅⋅--++()()()121221212122525433439y y y y my my m y y m y y =⋅=⋅+++++,222223252512523644624922m m m m m -⎛⎫+=⋅=⋅-=- ⎪--⎝⎭++++因此、为定值.1k 2k 2524-19.已知圆经过点,,且圆心在直线上.C ()0,2A ()6,4B 340x y --=(1)求圆的方程;C (2)若平面上有两个点,,点是圆上的点且满足,求点的坐标.()6,0P -()6,0Q M C 2MPMQ=M 【答案】(1)()22420x y -+=(2)或103⎛ ⎝10,3⎛ ⎝【分析】(1)设出圆心,利用点到直线的距离公式即可求得圆的方程.(2)根据已知条件求得满足的方程联立即可求得的坐标.M M 【详解】(1)∵圆心在直线上,340x y --=设圆心,()34,C a a +已知圆经过点,,则由,C ()0,2A ()6,4B CACB==解得,所以圆心为,0a =C()4,0半径=所以圆的方程为;C ()22420x y -+=(2)设,(),M x y ∵在圆上,∴,M C ()22420x y -+=又,,()6,0P -()6,0Q 由可得:,2MPMQ=()()2222646x y x y ⎡⎤++=-+⎣⎦化简得,()221064x y -+=联立()()22224201064x y x y ⎧-+=⎪⎨-+=⎪⎩解得或.103M ⎛ ⎝10,3⎛⎝20.已知椭圆,以原点为圆心,椭圆的短半轴长为半径的()2222:10x y C a b a b +=>>圆与直线相切.20x y -+=(1)求椭圆的方程;C (2)设,过点作与轴不重合的直线交椭圆于,两点,连接,分别()2,0A -()1,0R x l C M N AM AN 交直线于,两点,若直线、的斜率分别为、,试问:是否为定值?若是,3x =P Q PR QR 1k 2k 12k k ⋅求出该定值,若不是,请说明理由.【答案】(1)22142x y +=(2)是定值,定值2524-【分析】(1)根据离心率和点到直线距离公式即可得解;(2)直线的方程为,代入椭圆方程,根据三点共线表示出P 点坐标,同理表示出Q点l 1x my =+坐标,算出斜率即可求解.【详解】(1)由题意得,解得222c e a b a b c ⎧==⎪⎪⎪⎨⎪⎪=+⎪⎩2a b c =⎧⎪=⎨⎪=⎩故椭圆的方程为;C 22142x y +=(2)设直线的方程为,,,l 1x my =+()11,M x y ()22,N x y 由得,221142x my x y =+⎧⎪⎨+=⎪⎩()222230m y my ++-=∴,,12222m y y m -+=+12232y y m -=+由A 、、三点共线可知,M P ()()11322P y y x =----∴1152P y y x =+同理可得:,2252Q y y x =+故121212551131314422Q P P Q y y y y k k y y x x =⋅==⋅⋅--++()()()121221212122525433439y y y y my my m y y m y y =⋅=⋅+++++,222223252512523644624922m m m m m -⎛⎫+=⋅=⋅-=- ⎪--⎝⎭++++因此为定值.12·k k 2524-21.规定:在桌面上,用母球击打目标球,使目标球运动,球的位置是指球心的位置,球是指该A 球的球心点为,母球的球心沿直线运动.两球碰撞后,目标球在两球的球心所确定的直线上运动,A 目标球的运动方向是指目标球被母球击打时,母球球心所指向目标球球心的方向.所有的球都简化为平面上半径为的圆,且母球与目标球有公共点时,目标球就开始运动,在桌面上建立平面直角1坐标系,解决下列问题:(1)如图1,设母球的位置为,目标球的位置为,要使目标球向处运动,求A ()0,0B ()4,0B ()8,4C -母球的球心运动的直线方程;A (2)如图2,若母球的位置为,目标球的位置为,让母球击打目标球后,能否A ()0,2-B ()4,0A B 使目标球向处运动?请说明理由.B ()8,4C -【答案】(1)y x =(2)不能使目标球向处运动,理由见解析.B ()8,4C -【分析】(1)由题意可求得点,所在的直线方程,设球的球心坐标为,列()4,0B ()8,4C -A (),A a b '方程求得,即可求得答案;(4A '(2)假设能使目标球向处运动,则由(1)知球需运动到处,且到达B ()8,4C -A (4A '处前不与目标球接触,由此判断点B 到直线的距离的大小范围,并与2比较,可得结论.A 'B AA '【详解】(1)点,所在的直线方程为,即,()4,0B ()8,4C -044084y x --=---40x y +-=如图,可知,两球碰撞时,球的球心在直线上,且在第一象限,A B A 40x y +-=设,两球碰撞时,球的球心坐标为,A B A (),A a b '此时,则,2A B '=4020,0a b a b +-=⎧=>>⎪⎩解得,,4a =b=即,两球碰撞时,球的球心坐标,A B A (4A '所以母球的球心运动的直线方程为,即;A y y =(2)假设能使目标球向处运动,B ()8,4C -则由(1)知球需运动到处,且到达处前不与目标球接触.A (4A 'A 'B 如图,设与轴的交点为.AA 'x D因为的斜率为,所以,A B '1-45A BD '∠=︒因为,AA '1=>所以,所以为锐角,45A DB ∠>'︒DA B ∠'过点作于点,B BE AA '⊥E 因为,所以,2A B '=2BE <所以球的球心还未到直线上时,就会与目标球接触,A BCB 所以不能使目标球向处运动.B ()8,4C -22.已知椭圆的左焦点为,过原点的直线与椭圆交于,两点,()2222:10x y C a b a b +=>>1F O C P Q 若,且.113PF QF =11cos 3PF Q ∠=-(1)求椭圆的离心率;C (2)椭圆的上顶点为,不过的直线与椭圆交于,两点,线段的中点为,若C ()0,2D D l C A B AB M ,试问直线是否经过定点?若经过定点,请求出定点坐标;若不过定点,请说明2AMD ABD ∠=∠l 理由.【答案】(2)直线恒过定点,定点坐标为l 20,3⎛⎫- ⎪⎝⎭【分析】(1)设椭圆的右焦点为,连接,,然后在由条件可得,C 2F 2PF 2QF 12PF F △132aPF =,,然后利用余弦定理求解即可;22aPF =121cos 3F PF ∠=(2)首先求出椭圆的方程,然后由可推出,然后设直线的方程为2AMD ABD ∠=∠90ADB ∠=︒l ,,,联立直线与椭圆的方程消元表示出 、,然后由y kx m =+()11,A x y ()22,B x y 12x x +12x x求出的值可得答案.0DA DB ⋅=m 【详解】(1)设椭圆的右焦点为,连接,C 2F 2PF 2QF 根据椭圆的对称性可知,四边形为平行四边形.12QF PF =12PFQF 又,所以113PF QF =213PF PF =而,所以,122PF PF a+=132a PF =22aPF =在四边形中,,12PFQF 11cos 3PF Q ∠=-所以,()12111cos cos cos 3F PF PF Q PF Q π∠=-∠=-∠=在中,根据余弦定理得12PF F △222121212122cos F F PF PF PF PF F PF =+-⋅∠即()2223312222223a a a a c ⎛⎫⎛⎫=+-⋅⋅⋅⎪ ⎪⎝⎭⎝⎭化简得.222a c =所以椭圆的离心率C c e a ==(2)因为椭圆的上顶点为,所以,所以,C ()0,2D 2b =22224a b c c =+=+又由(1)知,解得,222c a =28a =所以椭圆的标准方程为.C 22184x y +=在中,,,ABD △2AMD ABD ∠=∠AMD ABD BDM ∠=∠+∠所以,从而,ABD BDM ∠=∠DM BM =又为线段的中点,即,所以,M AB 12BM AB=12DM AB =因此,从而,90ADB ∠=︒0DA DB ⋅= 根据题意可知直线的斜率一定存在,设它的方程为,,,l y kx m =+()11,A x y ()22,B x y 联立消去得①,22184y kx mx y =+⎧⎪⎨+=⎪⎩y ()222214280k x kmx m +++-=,()()()2224428210km m k ∆=--+>根据韦达定理可得,,122421km x x k +=-+21222821m x x k -=+所以()()()()()()2211221212,2,2122DA DB x y x y k x x k m x x m ⋅=-⋅-=++-++- ()()()222222841222121m km k k m m k k -⎛⎫=++--+- ⎪++⎝⎭所以,()()()2222228412202121m km k k m m k k -⎛⎫++--+-= ⎪++⎝⎭整理得,解得或.()()2320m m -+=2m =23m =-又直线不经过点,所以舍去,l ()0,22m =于是直线的方程为,恒过定点,l 23y kx =-20,3⎛⎫- ⎪⎝⎭该点在椭圆内,满足关于的方程①有两个不相等的解,C x 所以直线恒过定点,定点坐标为.l 20,3⎛⎫- ⎪⎝⎭。
高二年级语文上册期中测试卷(三)
高二年级语文上册期中测试卷3高二年级语文上册期中测试卷语文试卷一、现代文阅读(9分,每小题3分)阅读下面的文章,完成1~3题。
题画诗,是紧紧围绕着某幅特定的绘画作品而创作的诗。
它是我国古代文学艺术家的独创,在繁花锦簇的诗歌园地中独放异彩,具有浓郁的民族特色。
中国古代无数天才的文学艺术家们,将诗以高超的书法题写在画幅上,再加之以精工的印章,与绘画形成完美的艺术形象,给人以丰富的美感享受,是世界上其他国家和民族所没有的。
由于绘画题材丰富多彩,多种多样,题画诗的内容也极为广博纷繁,或咏写花草树木,或咏写山水田园,或咏写鸟兽虫鱼,或咏写人事季候,或咏写纸笔楼阁,等等,但就其思想倾向而言,大约可以分为三类:一类是生动形象地再现画面的形象,深刻地开掘出画面的底层意蕴,描写出画面应该具有而又画不出的诗意,即所谓“画之不足,题以发之”。
这样的题画诗,不仅是“画中之画”,而且更具有“画外之意”了。
如宋代苏轼的《惠崇春江晚景》:“竹外桃花三两枝,春江水暖鸭先知。
蒌蒿满地芦芽短,正是河豚欲上时。
”这首诗不仅生动地再现了画面上的景象,而且还写出了画外意,丰富了原画的意境。
一、三两句描写画面的实景,但第二句说春天江水之暖是鸭子最先知道的,第四句说正当蒌蒿芦芽生长之时,美味的河豚鱼在长江中逆流而上了,这是画面上没有的,也是任何高明画师画不出来的,但苏轼却凭借自己的联想和想象写出来了,使整个画面充满了春意和生趣,具有浓郁的生活气息,真可谓是“画之不足,题以发之”的妙诗。
二类是借观看绘画赞美高洁的品格,抒写人生的感悟,表达艺术体验和对文学艺术的真知灼见。
如果说上一类诗基本上属于题中应有之意,这一类诗则是借题发议,借题发挥。
如元代王冕《墨梅》:“吾家洗砚池头树,个个花开淡墨痕。
不要人夸好颜色,只留清气满乾坤。
”后二句借墨梅抒情。
这既是咏梅、赞梅,更是诗人高尚人格和美好情操的真实写照。
这发自心灵深处的声音,感染着、启迪着每一个读者。
2022-2023学年江苏省常州高二年级上册学期期中数学试题【含答案】
2022-2023学年江苏省常州高级中学高二上学期期中数学试题一、单选题1.不论为何实数,直线恒过定点( )m ()():1230l m x m y m -+-+=A .B .()3,1--()2,1--C .D .()–31,()–21,【答案】C【分析】将直线方程变形为,即可求得过定点坐标.()2130x y m x y ++--=【详解】根据题意,将直线方程变形为()2130x y m x y ++--=因为位任意实数,则,解得m 21030x y x y ++=⎧⎨--=⎩31x y =-⎧⎨=⎩所以直线过的定点坐标为()3,1-故选:C【点睛】本题考查了直线过定点的求法,属于基础题.2.抛物线的焦点到准线的距离是( ).28y x =A .B .C .2D .4132116【答案】B【分析】将抛物线的方程化为标准方程,根据焦准距的意义,可得答案.【详解】抛物线化为标准方程为抛物线,28y x =218x y =则其焦准距为,即焦点到准线的距离是,116=p 116故选:B 3.若直线与平行,则实数( )1:10l x y -+=2:10l x ay +-==a A .1B .2C .3D .1-【答案】D【分析】由两直线平行的条件求解.【详解】由题意,.11111a -=≠-1a =-故选:D .4.从圆外一点向圆引切线,则此切线的长是( )22(1)(1)1x y -+-=(2,3)PA B .2C D 【答案】B【分析】利用勾股定理可求切线长.【详解】设切点为,圆心为,连接,则,Q C ,,PQ PC CQ CQ PQ ⊥,2==故选:B .5.关于有唯一解,则实数的范围是( )x 20kx -=kA . k =B . (2,2)k ∈-C .(,2)(2,)k ∈-∞-+∞D .(,2)(2,)k ∈-∞-+∞ 【答案】D【分析】将问题转化为函数只有一个交点,然后利用数形结合处理.()f x =()2g x kx =+有唯一解,20kx -=有唯一解,2kx =+即的图象有唯一交点,()f x ()2g x kx =+又y =221,(0)x y y +=≥表示圆心为,半径为1的上半圆(包括和,(0,0)(1,0)A -(1,0)B 而是过定点的直线,()2g x kx =+(0,2)C 如图:,解得1=k =又,20202,20101AC BC k k --====-+-由图象可知,当或或的图象有唯一交点.2k <-2k >k =()f x =()2g x kx =+故选:D6.双曲线C :的渐近线与圆相切,则双曲线C 的离心率为()222210,0x y a b a b -=>>()2244x y +-=( )A B .2C .D .443【答案】B【分析】根据圆心到渐近线的距离变形可得离心率,即可求解.【详解】双曲线C :的一条渐近线与圆相切,()222210,0x y a b a b -=>>()2244x y +-=则圆心到渐近线的距离,()0,40bx ay -=42ad c ===所以曲线C 的离心率,2ce a ==故选:B7.椭圆:的上顶点为,点,均在上,且关于轴对称,若直线,C ()222210y x a b a b +=>>A P Q C x AP 的斜率之积为,则的离心率为( )AQ 43CA B C .D .1213【答案】C【分析】设P 点坐标,Q 点与P 点关于x 轴对称,坐标可用P 点坐标表示,代入斜率之积的关系式,再结合椭圆方程,化简可得a 与b 的关系,即可求出离心率.【详解】,设,则,()0,A a ()11,P x y ()11,Q x y -则,,11AP y a k x -=11AQ y ak x --=,22111211143AP AQy a y a a y k k x x x ----⋅=⋅==又,则,2211221y x a b +=()2221212b a y x a -=所以,即,()222122221243a y a b b a y a -==-2234b a =所以椭圆的离心率,C 12c e a ===故选:C.8.已知中心在坐标原点的椭圆C 1与双曲线C 2有公共焦点,且左,右焦点分别为F 1,F 2,C 1与C 2在第一象限的交点为P ,△PF 1F 2是以PF 1为底边的等腰三角形,若|PF 1|=10,C 1与C 2的离心率分别为e 1,e 2,则的取值范围是( )122e e +A .B .⎫+∞⎪⎪⎭5,3⎛⎫+∞ ⎪⎝⎭C .D .()1,+∞5,6⎛⎫+∞ ⎪⎝⎭【答案】B【分析】设椭圆和双曲线的半焦距为c ,|PF 1|=m ,|PF 2|=n ,,由条件可得()m n >m =10,n =2c ,再由椭圆和双曲线的定义可得,运用三角形的三边关系求()125,55a c a c c =+=-<得c 的范围,再由离心率公式,计算即可得到所求范围.【详解】设椭圆和双曲线的半焦距为c ,|PF 1|=m ,|PF 2|=n ,,()m n >由于△PF 1F 2是以PF 1为底边的等腰三角形.若|PF 1|=10,则有m =10,n =2c ,由椭圆的定义可得,12m n a +=由双曲线的定义可得,22m n a -=即有,()125,55a c a c c =+=-<再由三角形的两边之和大于第三边,可得,2210c c +>可得,即有,52c >552c <<由离心率公式可得()12122510225525555c c c c cc e e a a c c c c +--++=+=+=-+-+-,105211155555c c c c ⎛⎫=--=-+ ⎪+-+-⎝⎭因为,所以,,则,,552c <<155102c <+<5502c -<-<11210515c <<+1255c <--故,,则,即,2125515c c +<-+-2125553c c ⎛⎫-+> ⎪+-⎝⎭21515553c c ⎛⎫-+> ⎪+-⎝⎭12325e e +>故的取值范围是.122e e +5,3⎛⎫+∞⎪⎝⎭故选:B .9.半径为6的圆与x 轴相切,且与圆内切,则此圆的方程为( )22(3)1x y +-=A .B .22(4)(6)6x y -+-=22(4)(6)6x y ±+-=C .D .22(4)(6)36x y -+-=22(4)(6)36x y ±+-=【答案】D【分析】设所求圆的圆心坐标为(a ,b ),根据与x 轴相切,可得b 值,根据两圆内切,圆心距等于半径差,列出方程,可得a 值,即可得答案.【详解】设所求圆的圆心坐标为(a ,b ),因为圆与x 轴相切,所以b =6=r ,因为两圆内切,,解得,61=-4a =±故所求圆的方程为.22(4)(6)36x y ±+-=故选:D二、多选题10.下列说法中,正确的有( )A .点斜式可以表示任何直线()11y y k x x -=-B .直线在y 轴上的截距为42y x =-2-C .直线关于对称的直线方程是230x y -+=0x y -=230x y -+=D .直线与1:10l x y ++=2:10l x y +-=【答案】BD【分析】根据直线的点斜式、斜截式、平行线间距离及轴对称可得结果.【详解】点斜式,不表示直线,所以不正确;11()y y k x x -=-1x x =A 直线在轴上的截距为;满足直线的截距式方程的含义,所以正确;42y x =-y 2-B 直线关于对称的直线方程是,所以不正确;230x y -+=0x y -=230x y --=C 直线与之间的距离为,所以正确;1:10l x y ++=2:10l x y +-=d D 故选:.BD 11.已知双曲线的左、右焦点分别为、,点在双曲线上,则下列结论正确的是221916y x -=1F 2F P ( )A .该双曲线的离心率为B .该双曲线的渐近线方程为5434y x =±C .若,则的面积为D .点到两渐近线的距离乘积为12PF PF ⊥12PF F △9P 14425【答案】BD【分析】利用双曲线的离心率公式可判断A 选项;求出双曲线的渐近线方程可判断B 选项;利用双曲线的定义以及三角形的面积公式可判断C 选项;利用点到直线的距离公式可判断D 选项.【详解】对于A 选项,,,,该双曲线的离心率为,A 错;3a =4b =5c =53c e a ==对于B 选项,该双曲线的渐近线方程为,B 对;34a y x xb =±=±对于C 选项,若,则,12PF PF ⊥()1222212262100PF PF a PF PF c ⎧-==⎪⎨+==⎪⎩所以,,可得,()()222121212264PF PF PF PF PF PF ⋅=+--=1232PF PF⋅=故,C 错;12121162PF F S PF PF =⋅=△对于D 选项,设点,则,()00,P x y 2200169144y x -=双曲线的两渐近线方程分别为、,340x y +=340x y -=所以,点到两渐近线的距离乘积为,D 对.P 22000000229163434144342525x y x y x y --⋅+==+故选:BD.12.已知为椭圆的左焦点,经过原点的直线与椭圆交于两点,轴,F 22:1168x y C +=O l C ,A B AD x ⊥垂足为(异于原点),与椭圆的另一个交点为,则( )D BD C EA .AB AE⊥B .面积的最大值为ABD △C .周长的最小值为12ABF △D .的最小值为116AF BF+258【答案】ABD【分析】对于A,设,则,设,利用点差法推出,判断(,)A m n (,),(,0)B m n D m --11(,)E x y 1AB AE k k ⋅=-A;利用基本不等式结合三角形面积公式,判断B;利用椭圆的定义以及几何性质判断C;利用基本不等式中“1”的巧用,结合基本不等式可判断D.【详解】对于A,设,则,设 ,(,)A m n (,),(,0)B m n D m --11(,)E x y 由题意可知 ,110,,0m m x m x ≠≠+≠则 ,两式相减得,2222111,1168168x n y m +=+=2222110168m n x y -+=-即,即 ,1111()(()))12(y y x n n m m x +-+=--12BE AE k k ⋅=-由,21,,222AB BD BE BD AB n n n k k k k k m m m ===∴==则,即,故A 正确;11,122AB AE AB AE k k k k ⋅=-∴⋅=-AB AE ⊥对于B ,由A 的分析可知,不妨设点在第一象限,则,221168m n +=A 0,0m n >>所以时取等号,221168mm n n =+≥∴≤2m n ==故,故B 正确;122ABD S m n mn =⨯=≤△对于C ,由题意知左焦点为,设右焦点为,22:1168x y C +=(F -(F '-4,a b ==则根据椭圆的对称性可知,故周长为 ,||||BF AF '=ABF △2||8||a AB AB +=+而的最小值为椭圆的短轴长,由题意可知不能与椭圆短轴重合,||AB 2b =AB 故周长大于C 错误;ABF △8+对于D ,由C 的分析可知, ,||||||||28AF BF AF AF a '+=+==故||1611611161()()(88|17|)A BFAF AF BF AF BF BFF F AF B ++=+=++,当且仅当时取等号,D 正确,125(1788≥+832||,||55AF BF ==故选:ABD【点睛】本题综合考查了椭圆的定义的应用以及几何性质的应用,涉及到线段的垂直和三角形面积以及周长的最值得求法,解答时要注意综合利用椭圆的相关知识以及基本不等式的知识解决问题,属于较难题,计算量较大.三、填空题13.若抛物线y 2=8x 上一点P 到其焦点的距离为10,则点P 的坐标为 _____.【答案】(8,±8)【分析】先求出抛物线的准线,再由P 到焦点的距离等于其到准线的距离,从而可确定P 的横坐标,代入抛物线方程可确定纵坐标,从而可确定答案.【详解】设P (x P,yP ),∵点P 到焦点的距离等于它到准线x =﹣2的距离,抛物线y 2=8x ,xp +2=10,∴xP =8,yP =±8,故答案为:(8,±8).14.已知P 为椭圆上的一点,M ,N 分别为圆(x +3)2+y 2=1和圆(x -3)2+y 2=4上2212516x y +=的点,则|PM |+|PN |的最小值为________.【答案】7【解析】首先根据椭圆方程求出,由此可知两圆的圆心分别为椭圆的左右焦点F 1,F 2,,,a b c 进而根据椭圆的定义即可求解.【详解】由椭圆方程知a =5,b =4,c =3.两圆的圆心分别为椭圆的左右焦点F 1,F 2,设两圆半径分别为r 1,r 2,则r 1=1,r 2=2.所以|PM |min =|PF 1|-r 1=|PF 1|-1,|PN |min =|PF 2|-r 2=|PF 2|-2,故|PM |+|PN |的最小值为|PF 1|+|PF 2|-3=2a -3=7.故答案为:7【点睛】本题主要考查了椭圆的定义,需熟记椭圆的定义,属于基础题.15.已知点,若圆上存在点满足3,则实数(1,0),(1,0)A B -22(21)(22)1x a y a -++++=M MA MB =⋅ 的取值范围是 _____.a 【答案】11,2⎡⎤-⎢⎥⎣⎦【分析】由已知求出的轨迹为圆,再由圆与圆的位置关系列不等式求解实数的范围.M a 【详解】设,则(,)M x y (1,),MA x y =--- (1,),MB x y =--若3,则即MA MB =⋅ 2213,x y -+=224,x y +=∴的轨迹是以原点为圆心,以2为半径的圆,M 若圆上存在点满足3,22(21)(22)1x a y a -++++=M MA MB =⋅ 则圆和圆有公共点,224x y +=22(21)(22)1x a y a -++++=解得:13,∴≤≤11,2a -≤≤∴实数的取值范围是.a 11,2⎡⎤-⎢⎣⎦故答案为:.11,2⎡⎤-⎢⎥⎣⎦16.已知双曲线的左、右焦点分别为,,过点作双曲线的一条()2222:10,0x y C a b a b -=>>1F 2F 2F C 渐近线的垂线,垂足为为坐标原点,则双曲线的离心率为______.M O C 【答案】2【分析】根据已知条件求出的长度,进而在和,分别求出和2MF 2Rt MOF 12MF F △2cos MF O ∠,从而建立等量关系求得,进而可以求出离心率.21cos MF F ∠224a c =【详解】解:因为,一条渐近线方程为,()2,0F c 0bx ay -=b=,a=在中,,2RtMOF 2cos bMF O c ∠=中,12MF F△,222222121221212+47cos 24MF F F MF ca MF F MF F F bcb +--∠===⋅所以,即,因此,即,222+474c a b bbc c -=222473a c b -=()2222473a c a c -=-224a c =所以.2e ==故答案为:2.【点睛】关键点睛:本题考查求双曲线的离心率,关键在于根据渐近线和余弦定理得出关于a ,b ,c 的齐次式.四、解答题17.(1)已知直线l 过定点,且其倾斜角是直线的倾斜角的二倍,求直线l 的()2,3330x +=方程;(2)已知入射光线经过点,且被直线l :反射,反射光线经过点,求()3,4M -30x y -+=()2,6N 反射光线所在直线的方程.【答案】(1(2)30y +--=660x y --=【分析】(1)结合直线的倾斜角与斜率关系可求直线的斜率,进而可求直线方程;(2)根据对称性先求出反射光线所在直线的斜率,进而可求直线方程.【详解】因为直线3x y +3=0,π3故所求直线的倾斜角为,直线斜率为2π3k =∴所求直线的方程为;)32y x -=-30y +--=(2)设关于直线l :对称的点为,()3,4M -30x y -+=(),M a b '则,解得,413343022b a a b -⎧=-⎪⎪+⎨-+⎪-+=⎪⎩10a b =⎧⎨=⎩因为反射光线经过点,()2,6N 所以所在直线的斜率为,反射光线所在直线方程为,即.NM '60621k -==-()61y x =-660x y --=18.已知圆,点A 是圆C 1上一动点,点,点C 是线段AB 的中点.()221416C x y ++=:()4,0B (1)求点C 的轨迹方程;(2)直线l 过点且与点C 的轨迹交于 M ,N 两点,若,求直线l 的方程.()1,1MN=【答案】(1)224x y +=(2)或1x =1y =【分析】(1)利用中点坐标公式得到,再由点在圆得到,00242x x y y =-⎧⎨=⎩()00,A x y 1C ()2200416x y ++=代入即可得到点C 的轨迹方程;(2)分类讨论直线l 的斜率存在与否,利用弦长公式检验或求得斜率,从而可MN =k 得直线l 的方程.【详解】(1)设点,()()00,,,C x y A x y 因为点C 是线段AB 的中点,所以,即,00242x x y y =+⎧⎨=⎩00242x x y y =-⎧⎨=⎩因为点在圆C 1上运动,所以,()00,A x y ()2200416x y ++=所以,即,()22244416x y -++=224x y +=故点C 的轨迹方程为.224x y +=(2)当直线l 的斜率不存在时,其方程为,此时圆心到直线l 的距离为,1x =C 1则2MN ===当直线l 的斜率存在时,设其方程为,即,()11y k x -=-10kx y k --+=则圆心到直线l 的距离()0,0C d所以,解得,MN ===0k =所以直线l 的方程为,1y =综上:直线l 的方程为或.1x =1y =19.已知O 为坐标原点,过点的圆M 与直线相切,设圆心M 的轨迹为曲线C .()1,0F :1l x =-(1)求曲线C 的方程;(2)过点的直线交曲线C 于A 、B 两点,线段AB 的垂直平分线交x 轴于点,求线段()1,0F ()4,0P AB 的长.【答案】(1)曲线的方程为;C 24y x =(2)线段AB 的长为6.【分析】(1,化简得到答案.|1|x =+(2)设直线方程为,联立方程根据韦达定理得到根与系数的关系,根据垂直关系结AB (1)y k x =-合中点坐标公式得到,再计算弦长得到答案.22k =【详解】(1)设点为曲线C 上任意一点,因为圆M 过点且与直线相切,(,)M x y ()1,0F :1l x =-所以与点M 到直线,MFl |1|x =+整理得,所以曲线的方程为;24y x =C 24y x =(2)过点的斜率为0的直线与抛物线只有1个交点,不满足要求,()1,0F 过点的斜率不存在的直线为,直线与抛物线的交点为,,此时()1,0F 1x =1x =24y x =()1,2()12-,线段AB 的垂直平分线为,不满足要求,0y =所以直线斜率存在且不为,设直线方程为,,0AB (1)y k x =-0k ≠由得,,2(1)4y k x y x =-⎧⎨=⎩2440ky y k --=方程的判别式,2440ky y k --=216160k ∆=+>设,,则,11(,)A x y 22(,)B x y 124y y k +=124y y =-设线段中点,,,AB 00(,)N x y 12022y y y k +==0021211x y k k =+=+因为线段AB 的垂直平分线交x 轴于点,所以直线与直线垂直,()4,0P AB NP 故,.002222122433NP y kk k k k x k k ⋅=⋅===----22k =,12|||AB y y-=6==所以线段AB 的长为6.20.已知,,,且.()1,0M ()3,0N (),C x y CM CN CM CN-=+ (1)求动点C 的轨迹E ;(2)若点为直线l :上一动点,过点P 引轨迹E 的两条切线,切点分别为A 、B ,两条()1,P t -=1x -切线PA ,PB 与y 轴分别交于S 、T 两点,求面积的最小值.PST【答案】(1)动点C 的轨迹E 是以(2,0)为圆心,1为半径的圆【分析】(1)根据已知条件求出动点C 的轨迹方程即可判断其轨迹;(2)设切线方程,根据已知条件求出k 与t 的关系,再求出|ST |的长度,表示出()1y t k x -=+的面积即可求其最小值.PST 【详解】(1)∵,CM CN CM CN-=+ ∴,0CM CN ⋅=∴,()()1,3,0x y x y --⋅--=∴,()()2130x x y --+=∴,()2221x y -+=∴动点C 的轨迹E 是以(2,0)为圆心,1为半径的圆;(2)设切线方程为,即,PA ,PB 的斜率为,,()1y t k x -=+0kx y k t -++=1k 2k故圆心C 到切线的距离,得,1d 228610k kt t ++-=∴,,1234k k t +=-21218t k k -=在切线方程中令可得,0x =y k t =+故()()1212ST k t k t k k =+-+=-==∴时,等号成立.112PSTS ST =⨯=≥△0=t故.PST21.已知双曲线C :(,,双曲线C 的22221x y a b -=0a >0b >20y -=右焦点为,双曲线C 的左、右顶点分别为A ,B .()3,0F (1)求双曲线C 的方程;(2)过右焦点F 的直线l 与双曲线C 的右支交于P ,Q 两点(点P 在x 轴的上方),直线AP 的斜率为,直线BQ 的斜率为,证明:为定值.1k2k 12k k 【答案】(1);22145x y -=(2)证明见解析.【分析】(1)由题可得,,即求;3c=b a=(2)由题可设直线方程与双曲线方程联立,利用韦达定理法即证.【详解】(1)由题意可知在双曲线C 中,,,,3c=b a=222c a b =+解得2,a b =⎧⎪⎨=⎪⎩所以双曲线C 的方程为;22145x y -=(2)证法一:由题可知,()()2,0,2,0A B -设直线,,,():3l y k x =-()11,P x y ()22,Q x y 由,得,()2235420y k x x y ⎧=-⎨-=⎩()2222542436200k x k x k -+--=则,,212224045k x x k +=>-21223620045k x x k +=>-∴,,1112y k x =+2222y k x =-()()()()()()121211212221211212232236232326y x x x k x x x x k y x x x x x x x -----+===+-+-+-()()121211212236356x x x x x x x x x x -+++=-++-22222222222222343620246454545343620564545x k k k x k k k x k k x k k +-+-+---=+-+---;22222222222221210121014545506051210554545k k x x k k k k x x k k ------===--⎛⎫-+-- ⎪--⎝⎭当直线的斜率不存在时,,此时.l :3l x =1215k k =-综上,为定值.12k k 证法二:设直线PQ 方程为,,,3x my =+()11,P x y ()22,Q x y 联立得整理得,223,5420,x my x y =+⎧⎨-=⎩()225430250m y my -++=由过右焦点F 的直线l 与双曲线C 的右支交于P ,Q 两点,则解得,()()22222540,300,54250,54Δ30425540,m m m m m m ⎧-≠⎪-⎪>⎪-⎪⎨⎪<-⎪⎪=-⨯⨯->⎪⎩0m <<,,,1223054m y y m -+-=1222554y y m =-1212306255y y m m y y +-==-()121256my y y y =-+由双曲线方程可得,,,,()2,0A -()2,0B 1112y k x =+2222y k x =-∵,∴,,3x my =+2221x my -=+1125x my +=+.()()()()121211212212112221255y x y my k my y y k y x y my my y y -++===+++()()12112122125151666552555666y y y y y y y y y y -++-===--++-+证法三:设直线PQ 方程为,,,3x my =+()11,P x y ()22,P x y 联立得整理得,223,5420,x my x y =+⎧⎨-=⎩()225430250m y my -++=由过右焦点F 的直线l 与双曲线C 的右支交于P ,Q 两点,则解得,()()22222540,300,54250,54Δ30425540,m m m m m m⎧-≠⎪-⎪>⎪-⎪⎨⎪<-⎪⎪=-⨯⨯->⎪⎩0m <<∴,,由双曲线方程可得,,1223054my y m -+-=1222554y y m =-()2,0A -()2,0B则,()2211111221111545422444PB x y y y k k x x x x -⋅=⨯===+---所以,,15BP k k =12254PB k k k k =⋅()()()212122212121212255422111BP y y y y m k k x x my my m y y m y y -⋅=⋅==--+++++,2222222252525542530253054415454m mm m m m m m m -===---+-⋅+⋅+--∴为定值.125414255k k ⎛⎫=⨯-=-⎪⎝⎭22.椭圆:,若椭圆:,则称椭圆与椭1E 22221(0)x y a b a b +=>>2E 22221(0,1)x y a b m ma mb +=>>>2E 圆“相似”.1E (1)求经过点,且与椭圆: “相似”的椭圆的方程;1E 2212x y+=2E (2)若,椭圆,在椭圆上,过的直线交椭圆于两点,且4m =1E P 2E P l 1E ,A B .AP AB λ=①若的坐标为,且,求直线的方程;B (0,2)2λ=l ②若直线,的斜率之积为,求实数的值.OP OA12-λ【答案】(1)22142x y +=(2)①;②2y =+52λ=【分析】(1)设出椭圆的方程,结合点求得椭圆的方程.2E 2E(2)①先求得的方程,利用在椭圆上求得直线的斜率,从而求得直线的方程. ②结12,E E P 2E l l 合直线,的斜率之积、,由在椭圆上列方程,化简求得的值.OP OA AP AB λ=P 2E λ【详解】(1)设椭圆的方程为,代入点得,2E 2212x y m m +=)2112m m +=所以,2m =所以椭圆的方程为.2E 22142x y +=(2)因为椭圆1E c a ====整理得,所以椭圆,222a b =2221:22E x y b +=又椭圆与椭圆“相似”,且,所以椭圆,2E 1E 4m =2222:28E x y b +=设,()()()112200,,,,,A x y B x y P x y ①方法一:由题意得,所以椭圆,将直线,2b =221:28E x y +=:2l y kx =+代入椭圆得,221:28E x y +=()221280k x kx ++=解得,故,1228,012kx x k -==+212224,212k y y k -==+所以,222824,1212k k A k k ⎛⎫-- ⎪++⎝⎭又,即为中点,所以,2= AP AB B AP 2228212,1212k k P k k ⎛⎫+ ⎪++⎝⎭代入椭圆得,222:232E x y +=2222282122321212k k k k ⎛⎫+⎛⎫+= ⎪ ⎪++⎝⎭⎝⎭即,即,所以4220430k k +-=()()22103210k k -+=k =所以直线的方程为.l 2y =+方法二:由题意得,所以椭圆,,2b =221:28E x y +=222:232E x y +=设,,()(),,0,2A x y B 2= AP AB ,()()()2,2,2,4OP OA AP OA AB x y x y x x y =+=+=+--=--则,(),4P x y --代入椭圆得,解得,故()2222282432xy x y ⎧+=⎪⎨+-=⎪⎩12y =x =所以k =所以直线的方程为.l 2y =+②方法一: 由题意得,22222222200112228,22,22x y b x y b x y b +=+=+=,即,010112y y x x ⋅=-010120x x y y +=,则,解得,AP AB λ= ()()01012121,,x x y y x x y y λ--=--()()01201211x x x y y y λλλλ⎧+-=⎪⎪⎨+-⎪=⎪⎩所以,()()22010121122x x y y bλλλλ⎛⎫⎛⎫+-+-+= ⎪ ⎪⎝⎭⎝⎭则,()()()()2222222200110011211241212x x x x y y y y b λλλλλ+-+-++-+-=,()()()()()2222222000101112212122xy x x y y x y b λλλ++-++-+=所以,即,所以.()222228122b b bλλ+-⋅=()2241λλ+-=52λ=方法二:不妨设点在第一象限,设直线,P :(0)OP y kx k =>代入椭圆,2222:28E x y b +=解得0x =0y =直线的斜率之积为,则直线,代入椭圆,,OP OA 12-1:2OAy x k =-2221:22E x y b +=解得,则1x =1y =,则,解得,AP AB λ= ()()01012121,,x x y y x x y y λ--=--()()01201211x x x y y y λλλλ⎧+-=⎪⎪⎨+-⎪=⎪⎩所以,()()22010121122x x y y bλλλλ⎛⎫⎛⎫+-+-+= ⎪ ⎪⎝⎭⎝⎭则,()()()()2222222200110011211241212x x x x y y y y b λλλλλ+-+-++-+-=,()()()()()2222222000101112212122xy x x y y x y b λλλ++-++-+=所以,()()22222821122b b b λλλ⎛+-++-⋅= ⎝即,即,所以.()222228122b b bλλ+-⋅=()2241λλ+-=52λ=【点睛】在圆锥曲线有关的问题中,“点在曲线上”是一个很重要的已知条件,根据这个条件可以列方程,再结合题目另外的已知条件来对问题进行求解.向量共线的坐标表示,是建立点的坐标间关系的简捷途径.。
高中二年级上期期中考试复习试题及答案
.
故所求实数 的取值范围是 . ……………………………….12分
——————————————12分
22.解析:(1)已知圆的标准方程是(x+a)2+(y-a)2=4a(0<a≤4),
则圆心C的坐标是(-a,a),半径为2. ——————————2分
直线l的方程化为:x-y+4=0.
则圆心C到直线l的距离是=|2-a|. ——————————3分
如果两条平行直线中的一条与一个平面平行,那么另一条也与这个平面平行.
若直线 与平面 平行,则 与平面 内的任意一条直线都没有公共点.
A. B.1C.2D.3
6.若直线(1+a)x+y+1=0与圆x2+y2-2x=0相切,则a的值为
A、1,-1 B、2,-2 C、1 D、-1
7.下列各式中,最小值等于 的是()
二、填空题
(13)、15..10..20;(14)、108;(15 )、(-3,3);(16)、<k≤。
三、解答题
………12分
18解:(1)从袋中随机取两个球,其一切可能的结果组成的基本事件有1和2,1和3,1
和4,2和3,2和4,3和4,共6个.
从袋中取出的球的编号之和不大于4的事件共有1和2,1和3,共两个.
17、(12分)已知p: ,q: ,若 是 的必要不充分条件,求实数 的取值范围。
18、(12分)一个袋中装有四个形状大小完全相同的球,球的编号分别为1,2,3,4.
(1)从袋中随机取两个球,求取出的球的编号之和不大于4的概率;
(2)先从袋中随机取一个球,该球的编号为m,将球放回袋中,然后再从袋中随机取一个球,该球的编号为n,求n<m+2的概率.
即,解得x0=1或x0=2,所以输入的初始值x0=1或x0=2时,
新疆语文高二上册期中同步检测试卷及答案
新疆语文高二上册期中同步检测试卷班级:________________ 学号:________________ 姓名:______________一、单选题(每题3分)1.下列各组词语中,加点字的读音全都正确的一组是()A. 狭隘(ài)恪守(kè)挑衅(xìn)垂涎三尺(xián)B. 禀赋(bǐng)骁勇(xiāo)嗔怒(chēn)睚眦必报(yá zì)C. 掮客(qián)岑寂(cén)莅临(lì)殒身不恤(yǔn)D. 赧然(nǎn)毗邻(pí)檄文(xí)脍炙人口(kuài zhì)答案:D解析:A项,“衅”应读“xìn”;B项,“禀”应读“bǐn”;C项,“莅”应读“lì”。
2.下列各句中,加点成语使用不恰当的一句是()A. 这位明星曾带给观众很多快乐,不少“粉丝”竞相模仿他的表演,但这次他因醉酒驾车而触犯法律的行为却让人叹为观止。
B. 气候变化是当今全球面临的重大挑战。
拯救地球家园,遏制气候变暖,是全人类共同的使命,每个国家和民族都责无旁贷。
C. 在飞驰的高速列车上,人们津津乐道地谈论着乘坐高铁出行带来的快捷与方便。
D. 世界上很多古代文明都先后没落、消失,而我国的传统文化却历经沧桑,传承至今,依然焕发出勃勃生机。
答案:A解析:A项,叹为观止:指赞美所见到的事物好到了极点。
用于此处感情色彩不当。
B项,责无旁贷:自己的责任不能推卸给别人。
多用于指自己应当做的不可推卸的事。
使用正确。
C项,津津乐道:很有兴趣地说个不停。
使用正确。
D项,历经沧桑:形容经历过很多世事变迁。
使用正确。
3.下列各句中,没有语病的一句是()A. 在刚刚落幕的伦敦奥运会上,中国的金牌总数尽管跟北京奥运会相比有明显差距,但中国健儿在赛场上所展现出的拼搏精神和竞技水平依然让国人为之动容。
高二级语文上册期中测试卷
高二年级语文上册期中测试卷1高二年级语文上册期中测试卷Ⅰ卷一.完成下列各题(12分,每小题3分)1.读音字形都没有错误的一项是()A.渐(jiān)车帷裳荒芜人烟两全奇美洵(xùn)美B.自牧归(guì)荑便(piàn)言多令才披沙拣金踯躅(zhízhú)C.脉脉(mò)不得语羽扇纶(guān)巾急流勇退踟蹰(chíchú)D.舞榭(xié)歌台征帆去棹(zhào)坐阵指挥赏赍(lài)2.下列各句中,加点的成语使用恰当的一项是()A.在这次严打活动中,该市公安倾巢出动,破获了一批重大积案。
……………………③是做人的尊严.价值和品质的总和④正确处理个人与他人.个人与社会的关系⑤是指人与动物相区别的内在规定性⑥做到自尊.自爱.自强.自律A.③⑥①④⑤②B.③⑤①②⑥④C.⑤②①⑥④③D.⑤③①②④⑥二.阅读下面文言文,完成5-7题(9分,每小题3分)鲍勋字叔业,泰山平阳人也,建安十七年,太祖辟勋丞相掾。
二十二年,立太子,以勋为中庶子,出为魏郡西部都尉。
太子郭夫人弟为曲周县吏,断盗官布,法应弃市。
太子数手书为之请罪,勋不敢擅纵,具列上。
勋前在东宫,守正不挠,太子固不能悦,及重此事,恚望滋甚。
会郡界休兵有失期者,密敕中尉奏免勋官,久之,拜侍御史。
文帝受禅,勋每陈“今之所急,唯在军农,宽惠百姓。
台榭苑园,宜以为后。
”文帝将出游猎,勋停车上疏。
文帝手毁其表而兢行猎,中道顿息,问侍臣曰:“猎之为乐,何如八音也?”侍中刘晔对曰:“猎胜于乐。
”勋因奏:“刘晔佞谀不忠,阿顺陛下过戏之言。
请有司议罪以清皇朝。
”帝怒作色,罢还,即出勋为右中郎将。
黄初四年,尚书令陈群等并举勋为御史中丞。
帝不得已而用之,百僚严惮,罔不肃然。
六年秋,帝欲征吴,勋面谏。
帝益忿之,左迁勋为治书执法。
帝从寿春还,屯陈留郡界。
太守孙邕见,出过勋。
新高考地区高2024届高二(上)期中模拟试题二(含答案)
新高考地区高2024届高二(上)期中模拟试题二数学试卷注意事项:1.答题前,考生务必将自己的姓名、准考证号、班级、学校在答题卡上填写清楚.2.每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其他答案标号.在试卷上作答无效.3.考试结束后,请将答题卡交回,试卷自行保存.满分150分,考试用时120分钟.一、单选题:本题共8小题,每小题5分,共40分.在每个小题给出的四个选项中,只有一项是符合题目要求的.1.直线20x -=的倾斜角为( )A .6p B .4pC .3pD .5π62.已知v 为直线l 的方向向量,1n 、2n 分别为平面a 、b 的法向量(a 、b 不重合),那么下列说法中:①12n n a b Ûr r P P ;②12n n a b ^Û^r r ;③1v n l a Ûr r P P ;④1v n l a ^Û^r r.其中正确的有( ).A .1个B .2个C .3个D .4个【答案】B【分析】利用两平面平行充要条件判断①;利用两平面垂直充要条件判断②;利用线面垂直充要条件判断③;利用线面平行判定定理判断④.【详解】①12n n a b Ûr rP P ,判断正确;②12n n a b ^Û^r r,判断正确;③1v n l a Û^r rP ,判断错误;④1//v n l a ^Ûr r或l a Ì,判断错误.3.设甲:实数3a <;乙:方程2230x y x y a +-++=是圆,则甲是乙的( )A .充分不必要条件B .必要不充分条件C .充要条件D .既不充分也不必要条件4.抛物线216x y =的焦点到圆22:680C x y x +-+=上点的距离的最大值为( )A .6B .2C .5D .8【详解】216x =的焦点为()0,4F ,680x -+=,即22(3)1x y -+=所以,圆心为()3,0C ,半径1r =,5FC =上点的距离的最大值为||6FC r +=.A.5.若正三棱柱111ABC A B C -的所有棱长都相等,D 是11A C 的中点,则直线AD 与平面1B DC 所成角的正弦值为( )A .45B .35C .54D6.设圆222220x y x y +---=的圆心为C ,直线l 过点()03,,且与圆C 交于A ,B 两点,若||AB =,则直线l 的方程为( )A .34120x y +-=B .34120x y +-=或4210x y ++=C .x =0D .x =0或34120x y +-=【答案】D【分析】先利用圆的一般方程得到标准方程,得到对应的圆心和半径,然后分直线l 的斜率不存在和存在进行求解直线的方程即可得到答案7.已知12,A A 分别为双曲线2222:1(0)x y C a b a b -=>>的左、右顶点,点P 为双曲线C 上任意一点,记直线1PA ,直线2PA 的斜率分别为12,k k .若122k k ×=,则双曲线C 的离心率为( )A B 1C .2D 1+8.已知p 是椭圆22:198x y C +=上的动点,且与C 的四个顶点不重合,1F ,2F 分别是椭圆的左、右焦点,若点M 在12F PF Ð的平分线上,且10MF MP ×=uuuu r uuur,则OM 的取值范围是( )A .()0,2B .(0,C .(0,3-D .()0,1【答案】D二、多选题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求,全部选对的得5分,部分选对的得2分,有选错的得0分.9.下列结论正确的是()A .过点()2,3A --且在两坐标轴上的截距相等的直线l 的方程为5x y +=-;B .圆224x y +=上有且仅有3个点到直线:0l x y -=的距离都等于1C .已知0ab ¹,O 为坐标原点,点(),P a b 是圆222:E x y r +=外一点,且直线m 的方程是2ax by r +=,则直线m 与圆E 相交;D .已知直线10kx y k ---=和以()3,1M -,()3,2N 为端点的线段相交,则实数k 的取值范围为1322k -££;10.如图,已知E ,F 分别是正方体1111ABCD A B C D -的棱BC 和CD 的中点,则( )A .1A D 与11B D 是异面直线B .1A D 与EF 所成角的大小为3pC .1A F 与平面1B EB 所成角的余弦值为13D .二面角11C D B B --【答案】ABD【分析】根据异面直线的概念可判断A ,建立空间直角坐标系,用向量的方法可判断BCD .【详解】根据异面直线的概念可得“平面内一点与平面外一点的连线,与此平面内不经过该点的直线是异面直线异面直线”可知A 正确;设正方体棱长为2,()0,0,0D , ()12,0,2A , 所以()12,0,2A D =--uuuu r, ()1,1,0EF =--uuu r , 设1A D 与EF 所成角的大小为q ,则11||21cos 2||||82A D EF AD EF q ×===´uuuu r uuu ruuuur uuu r ,所以pq =,故B 正确;11.(多选)已知抛物线22y px =()0p >的焦点F 到准线的距离为4,直线l 过点F 且与抛物线交于()11,A x y ,()22,B x y 两点,若(),2M m 是线段AB 的中点,则( )A .4p =B .抛物线的方程为216y x =C .直线l 的方程为24y x =-D .=10AB12.已知1F 、2F 分别为双曲线22221(0,0)x y a b a b -=>>的左、右焦点,且2122b F F a=,点P 为双曲线右支一点,I 为12PF F △的内心,若1212IPF IPF IF F S S S l =+△△△成立,则下列结论正确的有( )A .当2PF x ^轴时,1230PF F Ð=°B .离心率e =C .l =D .点I 的横坐标为定值a【点睛】本题考查双曲线的定义和简单性质,利用待定系数法求出参数的值,考查圆的切线的性质,化简运算能力和推理能力,属于中档题.三、填空题:本题共4小题,每小题5分,共20分.13.若直线m 被两平行线1:10l x y -+=与2:30l x y -+=所截得的线段的长为m 的倾斜角可以是①15°,②30°,③45°,④60°,⑤75°.其中正确答案的序号是_____(写出所有正确答案的序号).14.若直线:420l kx y k -++=与曲线y =k 的取值范围是______.15.已知菱形ABCD 中,60ABC Ð=°,沿对角线AC 折叠之后,使得平面BAC ^平面DAC ,则二面角B CD A --的余弦值为______.【答案】55【分析】根据题意建立空间直角坐标系,根据二面角余弦值的空间向量求解方法进行计算即可【详解】设菱形ABCD 的边长为1因为平面BAC ^平面DAC ,平面如图,建立空间直角坐标系,则所以13,0,22BC æö=-ç÷ç÷èøuuu r ,CD =uuur16.已知椭圆()221112211:10x y C a b a b +=>>与双曲线()222222222:10,0x y C a b a b -=>>有相同的焦点12F F 、,椭圆1C 的离心率为1e ,双曲线2C 的离心率为2e ,点P 为椭圆1C 与双曲线2C 的第一象限的交点,且123F PF pÐ=,则1212e e e e +的取值范围是___________.四、解答题:本题共6小题,第17小题10分,其余小题每题12分,共70分.解答题应写出文字说明、证明过程或演算步骤.17.已知圆22:4240C x y x y ++--=.(1)过点(1,5)M 作圆C 的切线l ,求切线l 的方程;(2)设过点1,12æöç÷èø的直线m 与圆C 交于AB 两点,若点A 、B 分圆周得两段弧长之比为1:2,求直线m 得方程.【答案】(1)7241130x y -+=或1x =;(2)6850x y -+=或68110x y +-=【分析】(1)根据圆心到直线的距离等于半径求解,注意分斜率存在与不存在两种情况;(2)利用条件可分析出弦所对圆心角,据此求出圆心到直线的距离,即可求解.18.在平面直角坐标系中,△ABC 的三个顶点坐标分别为()0,0A ,()2,0B -,()3,3C --.(1)求BC 边上的中线AD 的所在直线方程;(2)求△ABC 的外接圆O 被直线l :10x y -+=截得的弦长.19.如图,EA ^平面ABCD ,EA FC ∥,22AC EA FC ===,四边形ABCD 为菱形.(1)证明:FA ^平面EBD ;(2)若直线AB 与平面EBD E BDF -的体积.【答案】(1)见解析20.在四棱锥P ABCD -中,已知//AB CD ,AB AD ^,BC PA ^,222AB AD CD ===,PA 2PC =,E 是PB 上的点.(1)求证:PC ^底面ABCD ;(2)是否存在点E 使得PA 与平面EAC 所成角的正弦值为23?若存在,求出该点的位置;不存在,请说明理由.【答案】(1)证明见解析(2)存在,E 点为PB 上靠近B 点的三等分点21.已知椭圆C ;220)x b a >>的左右顶点分别为1A ,2A ,以线段12A A 为边的一个正三角形与椭圆C的一个公共点为P (65.(1)求椭圆C 的方程;(2)若过椭圆C 的右焦点F 的直线与椭圆C 交于点M ,N ,直线1A M ,2A N 交于点D ,求证:点D 在定直线l 上,并求出直线l 的方程.22.在平面直角坐标系xoy 中,设F 为椭圆()222210,0:x y a b a C b+=>>的左焦点,直线2a x c =-与x 轴交于点P ,M 为椭圆C 的左顶点,已知椭圆长轴长为8,且2PM MF =uuuu r uuur .(1)求椭圆C 的标准方程;(2)若过点P 的直线与椭圆交于两点A 、B ,设直线AF 、BF 的斜率分别为1k 、2k .①求证:12k k +为定值;②求ABF V 面积的最大值.。
高二上期中语文试卷及答案
XXX学校2018-2019上教学工作计划一、指导思想根据教育局教学教研工作要点指示精神,为全面贯彻执行党的教育方针,贯彻落实全国、省、市教育工作会议精神,为严格遵循教学规律,按照“低重心运行,高层次突破”的教研工作思路,以“全方位服务,全过程监控,多角度指导”为基本工作方法,围绕“三体现、四重在”文件要求,积极实施有效教学,努力打造高效课堂,全力推进教育均衡发展,进一步提高我校广大教师教学教研水平,促进我校教学质量的全面提高。
制定我校教学工作计划如下:二、工作要点1、继续学习落实《卫辉市教育局关于全面推进“三体现四重在”课堂教学改革的指导意见》的精神,认真学习借鉴先进学校好的做法和经验,不搞花架子扎扎实实开展好校本教研活动,进一步推动新课程改革向纵深发展。
2、高素质的学校教育必须有高素质的师资队伍,本期要继续通过认真组织教师自身参加各级各类培训和个人进修,组织并鼓励教师参加撰写论文、教学心得、教学反思、优秀教案及课件制作、教科研立项等活动,积极参加各类进修、培训,充分展示教师的理论水平,增强教师不断进取的信心,提高教师素质。
3、继续开展“四个一”活动,及时给予各种协助,组织上半年已上报的教案、试卷的教师做好作“微型课”和评课的参赛准备工作,力争取得好成绩,借以促进和激励广大教师参与的积极性,为提高教师素质再搭建一个平台。
4、认真贯彻落实国家和省、市关于课改的一系列文件,全面规划、科学安排,保障课改工作健康有序发展。
5、以解决学生“厌学、懒思和学不得法”问题为着力点,加大课题研究和小课题研究,以课题引领方向,促进课程改革工作顺利推进。
6、以“课堂教学模式创新大赛”为抓手,采取层层选拔,步步推进的办法,打造适应新课程改革的“高效课堂”教学模式。
7、继续开展城乡联动共同体活动,支援和帮助结对学校提高办学水平。
8、继续通过学生座谈会和评教活动,以了解教师课堂教学情况,学生纪律情况、课内重大问题、学生的意见和建议,及时向有关教师反馈。
高二上学期数学期中考试试卷
高二上学期数学期中考试试卷一、选择题(每题3分,共30分)1. 若函数f(x)=x^2-4x+3,则f(1)的值为:A. 0B. -1C. 2D. 42. 已知等差数列{a_n}的首项a_1=1,公差d=2,则a_5的值为:A. 9B. 10C. 11D. 123. 若复数z满足|z|=1,则z的实部和虚部满足:A. x^2 + y^2 = 1B. x^2 - y^2 = 1C. x^2 + y^2 = 0D. x^2 - y^2 = 04. 已知函数y=f(x)的图象关于直线x=1对称,则f(0)与f(2)的关系为:A. f(0) = f(2)B. f(0) = -f(2)C. f(0) = 2f(2)D. f(0) = -2f(2)5. 若直线l与直线y=2x+3平行,则直线l的斜率k为:A. 2B. -2C. 3D. -36. 已知三角形ABC的三边长分别为a、b、c,且满足a^2 + b^2 =c^2,则三角形ABC的形状为:A. 锐角三角形B. 直角三角形C. 钝角三角形D. 等腰三角形7. 若函数y=f(x)在区间[1,3]上是增函数,则f(2)与f(1)的大小关系为:A. f(2) > f(1)B. f(2) < f(1)C. f(2) = f(1)D. 不能确定8. 已知向量a=(1,2),向量b=(3,-1),则向量a与向量b的数量积为:A. 1B. -1C. 5D. -59. 若圆心在原点,半径为1的圆的方程为:A. x^2 + y^2 = 1B. x^2 + y^2 = 2C. x^2 + y^2 = 0D. x^2 + y^2 = -110. 若双曲线的方程为x^2/a^2 - y^2/b^2 = 1,则该双曲线的焦点坐标为:A. (±a, 0)B. (0, ±b)C. (±a, ±b)D. (±b, 0)二、填空题(每题4分,共20分)11. 已知函数f(x)=x^3-3x,求f'(x)的值为_________。
高二年级上学期期中考试570分_高二年级数学上学期期中考试试题
高二年级上学期期中考试570分_高二年级数学上学期期中考试试题一、选择题(每小题5分,共60分)1.a=4,c=1,且焦点在x轴上,则其椭圆的方程是A.x2/16+y2/12=1B.x2/16+y2/15=1C.x2/16+y2/4=1D.x2/15+y2/16=12.椭圆的长轴长是短轴长2倍,则其离心率A. /2B.1/3C. /2D.1/23.椭圆的方程x2/25+y2/16=1,M是椭圆上的一点,,且到一个焦点的距离是3,则M到另一个焦点的距离是A.6B.7C.8D.94.设点P在双曲线16x2-9y2=144上,F1是双曲线的左焦点,则F1的坐标是A.(5,0 )B.(-5,0 )C.(-4,0)D.(0,-4)5.已知抛物线的焦点坐标是F(0,-2 ),它的标准方程可写成A.x2+8y=0B.x2-8y=0C.y2+8x=0D.y2-8x=06.抛物线y2-6x=0准线方程是A.x=-3/2B.x=3/2C.y=-3/2D.y=3/27.直线x-8y+96=0与抛物线y2=6x有( )个交点。
A.1B.2C.0D.以上答案均不对8.双曲线x2/25-y2/16=1的渐近线方程是( )A.y=B.y=C.y=D.y= 9.在下列方程所表示的曲线中,关于x轴、y轴都对称的是( )A.x2-6y=0B.x2+6xy+y=0C.y2-6x2=5xD.9y2-6x2=410.若F1、F2分别是椭圆x2/2+y2=1的左右焦点,P是椭圆上一点,则|PF1|·|PF2|的最大值是A.3B.2C.D.211.若P(x,y)是椭圆4x2+y2=16上一点,则2x+y的最大值是A.3B.4C.5D.612.若椭圆5x2+ky2=5的一个焦点是(0,2),则k=( )A.-1B.1C.D.-二、填空题(每小题4分,共16分)13.抛物线y2-12x=0上与焦点的距离等于9的点的坐标是14.已知方程是双曲线,则m的取值范围是_______________________15.已知椭圆的标准方程中,a+b=6,c=2 ,且焦点在x轴上,则其方程是___16.若椭圆x2/2+y2/m=1的离心率为1/2,则实数m等于.三、解答题(共74分)17.(12分)若椭圆中心在原点,准线方程为y=±4,离心率e=1/2,求其椭圆方程。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
高中二年级(上)期中复习测试题第一部分听力(共两节,满分30分)(略)第二部分英语知识使用(共两节,满分45分)第一节单项选择(共15小题;每小题1分,满分15分)从A、B、C、D四个选项中,选出能够填入空白处的最佳选项。
21. — Can I look at the menu for a few more minutes before I decide?—Of course. ____, sir.A. Enjoy yourselfB. Make yourself at homeC. Take your timeD. It doesn’t matter22. The manager has ____ a car to pick the invited scientist up at the airport.A. arrangedB. arranged forC. arranged withD. arranged by23. Whenever he was asked why he was late for work, he would answer carelessly, always ___ the same thing.A. sayingB. saidC. to sayD. having said24. We regret to inform you that there are no tickets ____ for Friday’s performance.A. availableB. spareC. convenientD. affordable25. After a five-hour drive, they reached ___ they thought was the place they’d been dreaming of.A. thatB. whereC. whichD. what26. When ___ in court, the man didn’t admit having taken the old lady’s necklace.A. questioningB. having questionedC. being questionedD. to be questioned27. — Why have you ___ Mr Johnson as the best person for this position?—Because he is responsible and careful.A. put onB. put downC. put forwardD. put out28. All the senior leaders are members of the same family. _____, I’m likely to have to move if I want to get promoted.A. ThoughB. MeanwhileC. ThereforeD. However29. So important ___ it to get on well with people around that I have decided to learn some communication skills.A. I have foundB. have I foundC. I was feelingD. was I feeling30. It’s no ____ arguing with Bill about joining us in the experiment because he will never change his mind.A. useB. needC. wayD. doubt31. You’d better behave yourself when staying at his home, for his mother is a(n) _____ woman who seldom smiles.A. cautiousB. positiveC. enthusiasticD. severe32. — Why was Fred forced to leave school?—Because he was _____ of cheating in the English exam.A. blamedB. accusedC. chargedD. caught33. I’m sorry that what I said just now has hurt you, but I didn’t do it ______.A. deliberatelyB. actuallyC. carelesslyD. especially34. — Did you catch _____ sight of a tall TV tower in the distance just now?—Yes, and now it’s out of ____ sight.A. /; theB. /; /C. the; theD. a; a35. Facing lots of temptations, the scientist still ____ his experiment of great value to his country’s development.A. devoted toB. depended onC. concentrated onD. referred to第二节完形填空(共20小题;每小题1.5分,满分30分)On October 31st, 2003, young surfer, Bethany Hamilton suffered a big loss. It was Halloween morning in Hawaii and a(n) 36 day for surfing. Bethany went surfing with her best friend. She was catching great waves and 37 her life changed in a second. She got 38 by a 14-foot tiger shark, which bit her just 4 inches 39 her left shoulder and took the rest of her arm off! She never 40 that to happen to her on a sunny Halloween!Bethany was 41 on being a professional surfer before her accident in the waves. She 42 surfing at an early age and her whole family were surfers. Bethany only 43 14 on February 8th, 2004, so it was a young age to have something 44 bad happen to someone.Bethany Hamilton, 45 , showed that she was a true hero by saving her life. Her life was 46 when the tiger shark bit her arm completely off and her 47 helped her get to safety and she was successful. She 48 overcame the loss of her left arm and within a month she rose to the 49 of going surfing in the ocean 50 !She is now a role model to many people for what she has done. She turned a 51 into a positive, which many people can’t do! She remained positive and 52 to go forward with her dream. She is 53 a survivor hero and is keeping her surfing career alive as well. She 54 to surfing with only one arm, and that is amazing! 55 , millions of people are inspired to try to find something good in something that only seems bad.36. A. unlucky B. perfect C. mild D. normal37. A. immediately B. naturallyC. sadlyD. suddenly38. A. trapped B. surroundedC. defeatedD. attacked39. A. nearby B. below C. above D. within40. A. expected B. planned C. hoped D. forgot41. A. depending B. insisting C. planning D. calling42. A. gave up B. took up C. set up D. kept up43. A. grew B. became C. turned D. announced44. A. too B. enough C. much D. that45. A. therefore B. however C. instead D. indeed46. A. in peace B. at riskC. out of controlD. beyond reach47. A. friends B. relatives C. instincts D. parents48. A. almost B. never C. hardly D. also49. A. challenge B. chance C. danger D. fate50. A. in time B. more or lessC. once againD. at least51. A. negative B. weaknessC. misfortuneD. disability52. A. began B. stopped C. prepared D. continued53. A. originally B. obviouslyC. unexpectedlyD. finally54. A. devotes B. looks forwardC. triesD. goes back55. A. In all B. At that momentC. As a resultD. From now on第三部分阅读理解(共20小题;每小题2分,满分40分)AThere are many different types of fire ants, and they are found all throughout the Southeastern and Southwestern United States. The most common and aggressive kind is the red imported fire ant, which is reddish-brown. Their nests can be found on lawns, in parks, on playgrounds and in fields. Some red imported fire ants create nests in the walls of buildings.A person who gets stung by a fire ant will feel a sharp pain and burning. If a person steps on a fire ant mound (土堆), he or she will get a lot of stings, because the ants have been disturbed where they all live together. Each sting will turn into an itchy blister (水泡) over the next day.The best way to avoid getting stung by fire ants is to keep your shoes on when you are playing near fire ant mounds. If you meet one, don’t poke at it or try to play with it. However, if you think that you have been stung by a fire ant, tell an adult immediately. That’s because the area of the sting will swell up quite a bit, and a doctor may want to have a look to make sure you do not have breathing problems or chest or throat tightness —an allergic(过敏的)reaction which is similar to one from a bee sting can sometimes happen. If these symptoms appear, the person needs medical attention right away.But more often, you can follow these steps after a fire ant bite:☆Wash the area with soap and water.☆Apply some ice to the area.☆Check with the doctor if you have redness, swelling, or itching.56. Why do fire ants attack people?A. It is in their nature to attack people.B. They do so quite by accident.C. Human beings are their natural enemy.D. They do so to protect themselves.57. For whom is the passage written?A. Travellers.B. Farmers.C. Children.D. Ant researchers.58. To avoid being stung, you are advised _______.A. not to go to places where fire ants liveB. not to disturb fire ants in any wayC. to have an adult with you when going outD. to take soap and water with you when going out59. From the passage, we can learn that ___________.A. the United States is the only place where fire ants existB. some stung people by a bee will not have chest or throat tightnessC. sometimes the sting of a fire ant will cause serious problemsD. a stung person will always have an allergic reaction without immediate treatmentBEver wonder why some people never seem to get sick but others get stuck in bed? Scientists say it might have something to do with special immune cells that are more like jail cells, because they put the flu virus behind bars.As reported in the journal NatureMedicine, researchers at Washington University in St. Louis discovered that these jail cells, formally known as macrophages (巨噬细胞), rely on a gene called CCL5 to stay alive even after a virus gets inside them.When the flu virus infects our normal cells, they have to commit suicide to prevent the virus from spreading. But as Holtzman’s team reported, the gene CCL5 blocks this process in macrophages. As a result, macrophages can pick up viruses, hold them inside their membranes (细胞膜), and help clear them from our bodies —without making themselves dead.To figure out how CCL5 works, the researchers created mice without the gene and compared them to mice that had it. “We find that the mice that were missing it were dying from the viral infection whereas the mice that had it did perfectly fine,” says Holtzman.In a follow-up laboratory study, they compared human cells that had CCL5 to others that did not have CCL5 and saw the same thing. Without CCL5, the cells died. Holtzman suspects that because we all have different gene combinations, those of us who have more of this gene may be less likely to get sick.60. Which description of immune cells is TRUE?A. Immune cells are kinds of normal cells.B. Immune cells are CCL5.C. Immune cells are formally known as macrophages.D. Immune cells can pick up viruses and kill them.61. When infected, normal cells ________________.A. can’t prevent the virus from spreadingB. have to commit suicide to prevent the virus from spreadingC. can put the flu virus behind barsD. can prevent the virus from spreading without making themselves dead62. From the last two paragraphs, we know that _______.A. both mice and human cells that didn’t have the gene CCL5 died from the viral infectionB. the mice have more of the gene CCL5 than human beingsC. each person has a different gene CCL5, so some people never seem to get sickD. if you have more of the gene CCL5, you’ll never get sick63. According to the passage, who gets the flu easily?A. A person who often gets stuck in bed.B. A person who has more of the gene CCL5.C. A person who has more jail cells.D. A person who has less of the gene CCL5 than normal.CWhen the Farnsworth family moved to their new farm in 1919, eleven-year-old Philo was surprised to find it wired for electricity. This unusual circumstance contributed to his fate — to become an important inventor of the twentieth century.By thirteen, Farnsworth had become a self-taught electrical engineer. He was able to fix the farm’s generator(发电机)when none of the adults could. In 1922, he read an article about a new idea of John Baird, a Scottish scientist, who had been working with the cathode ray tube (阴极射线管) for the transmission of electronic pictures and wanted to attempt it himself.Farnsworth studied everything he could find on the subject. Although many older engineers with money backers were already developing television, Farnsworth made a bold decision — he was going to perfect a working model of it before anybody else.In college, Farnsworth continued his research with cathode ray and vacuum tubes, but the death of his father, the only money maker in the family, forced him to give up this research and find a job. His first job was for George Everson, with whom Farnsworth discussed his dream of television. While acknowledging the achievements of those who came before, Farnsworth thought that he could get closer. Everson agreed to risk $6,000 for the research.Backers came in 1927 to see the first American television, one year after Baird’s. They were astonished to see the image of a single white line resolve itself on the screen before them, and agreed that this new invention was worth putting money into.In 1930, Farnsworth won a patent (专利权) for his all-electronic TV. By the time he died, he had earned over 300 American and foreign patents for electronic and mechanical devices.64. When Farnsworth was at a young age, he _______.A. had to drop out of school to help on the family farmB. was sent to school to study electrical engineeringC. wanted to be the first person to invent the televisionD. had shown a surprising ability in the electrical field65. What difficulty did Farnsworth meet when he first began his research on the television?A. His parents didn’t support his work.B. He didn’t have enough knowledge in this field.C. He didn’t have enough money for his research.D. No one was interested in this research.66. How old was Philo Farnsworth when he invented the first American television?A. 11.B. 13.C. 19.D. 22.67. Why did Philo Farnsworth work for George Everson?A. Because he had to support his family.B. Because George had supported his research.C. Because George was the richest person he knew.D. Because he could find more supporters there.68. From the passage, we can learn that Farnsworth is ______.A. the first person who worked for the transmission of electronic picturesB. an inventor who improved on somebody else’s ideaC. an inventor who always came up with an original ideaD. a person who earned over 300 American patents for electronic devicesDYour youngest son or daughter has just graduated from high school, and soon he or she will trade his or her bedroom in your home for a college residence hall. You’re starting to worry about the coming silence in your home and are wondering how you will fill the hours that you previously spent with your son or daughter attending his or her school and sporting events.Take heart —the empty nest “is often worse in anticipation (预料) than in day-to-day practice”, says psychologist Bert Hayslip Jr.Hayslip, who has studied retirement adjustment (调整) and other aging problems, such as aging sleeplessness, for many years, points out that, more often than not, children who leave do not completely cut off contact with their parents. He advises parents to think of it as a series of life events, instead of a sudden change. He also says an empty nest can cause bigger problems to surface “if a couple hasn’t looked after their marriage while raising their children”. According to Hayslip, some couples may find they no longer have anything in common once the children are out of the house.Thinking of an empty nest as the loss of children makes the adjustment more difficult. “With the empty-nest syndrome (综合症), parents actually need to deal with the loss of the parenting control over children, not with having really lost their children,”Hayslip says. “They just have to find a new way to get along with their children.”“Relating to their college-age children in this new way will come easily to parents as the months pass. As with many things, the passage of time heals the pain of loss,” Hayslip says.69. What does the “empty nest” mean?A. Children’s cutting off contact with their parents.B. Children’s not liking spending time with their parents.C. Children’s growing up and leaving home.D. Children’s not accepting parents’ care and love.70. What does the underlined word “it” in the third paragraph refer to?A. Retirement adjustment.B. Aging problem.C. Sleeplessness.D. The reality of empty nest.71. Why do many parents find it hard to adjust to the empty-nest life?A. Because they think they have actually lost their children.B. Because they often disagree with their children on many things.C. Because their children completely cut off contact with them.D. Because these parents don’t have anything in common.72. According to Hayslip, what parents actually lose in an “empty nest” is _______.A. their children’s love for themB. their control over childrenC. their interest in lifeD. their contact with childrenEIt is almost a reflex action among Americans to say “God bless you” when someone sneezes. The custom, perceived as a wish for luck, goes back to America’s European roots. In most parts of Europe, there is an old belief that a sneeze can expel the soul from the body. Because of this, the blessing serves as a verbal charm to protect the sneezer from death. Another belief is that if someone sneezes three times without receiving a “bless you”from someone, the fairies could come and take him away.Sometimes, however, the blessing isn’t completely necessary. A corollary (具有推断性质的) belief holds that if two people sneeze at the same time, both will enjoy good luck.A solitary sneeze is considered a lucky omen (征兆) in many parts of the world. The Maori people of New Zealand believe it is symbolic of the act of creation. According to their creation myth, the first human received the breath of life when the great God Tiki sneezed. In parts of Africa, sneezing is considered a sign that the body has been taken over by good spirits. And most Native American tribes believed that a sneeze was a sign that the brain was being cleared of evil.This superstition may also have arisen from a serious disease that was attacking the early Romans and the main symptom was sneezing. This disease was so serious that people often died as a result, and the Romans believed that the more often the individual was blessed, the lower the chances of death.Also, “God bless you”was a frequently used form of well-wishing at the time. It was considered a wish for good health. Since sneezing was a symptom of the bubonic plague (淋巴腺鼠疫) in the Middle Ages in Europe, the custom of blessing people after they sneeze persisted and has been handed down until today. For the Hebrews, Greeks, Indians, and Chinese, breath was a sign of the soul’s existence, and any loss of the breath, especially in a sneeze, was exceedingly bad luck.73. From the passage, we can learn that .A. to say “God bless you” is necessary when a person sneezesB. sneezing has different meanings in different areasC. sneezing is considered a lucky omen in the worldD. a sneeze means exceedingly bad luck for most Native American tribes74. Why do Americans usually say “God bless you” when someone sneezes?A. Because it is a reflex action.B. Because it is a custom which goes back to America’s European roots.C. Because it is an old belief.D. Because it is a verbal charm to protect the sneezer.75. What is the best title of the passage?A. Sneezing and “God bless you”.B. Sneezing:bad luck.C. Sneezing:a verbal charm.D. Sneezing:charms and symbols.第四部分写作(共两节,满分35分)第一节短文改错(共10小题;每小题1分,满分10分)In the morning of April 29, 2007, from 10:00 to 11:00, hundreds of76. ______millions students from primary, middle schools and universities all77. ______over China join in a national student sports program. Eleven78. _______million students in Jiangsu took an active part in this program.79. _______The Ministry of Education calls on student to exercise for an hour80. _______every day, in the hope of that they will have good health to work81. _______fifty years and enjoy the whole life.Personally, I think this a good idea for senior 3 students to take daily82. _______exercise. Although it takes an hour a day for exercise, but it is well worth 83. _______it. Take exercise helps us build up our body and keep a clear mind.84. _______Therefore, we can work more efficient. 85. _______第二节书面表达(满分25分)仔细观察下图,请以“Project Hope”为题写一篇短文。