江苏连云港高级中学2018届高考考前适应性练习一 精品推荐

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江苏省连云港市高三英语第一次模拟考试试题

江苏省连云港市高三英语第一次模拟考试试题

2018届高三年级第一次模拟考试英语(满分120分,考试时间120分钟)第一部分听力(共两节,满分20分)第一节(共5小题;每小题1分,满分5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

( ) 1. Where is the theater?A. On Martin's Lane.B. On Maple Street.C. On Craven Street.( ) 2. What seems to be the man's hobby?A. Watching TV.B. Reading books.C. Talking on the WeChat.( ) 3. What does the woman suggest the man do?A. Avoid on­sale things.B. Wait until the weekend.C. Get better shampoo.( ) 4. When will the speakers probably meet?A. On Wednesday afternoon.B. On Thursday afternoon.C. On Friday afternoon.( ) 5. What is the probable relationship between the speakers?A. Colleagues.B. Employer and employee.C. Customer and manager.第二节(共15小题;每小题1分,满分15分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

江苏省连云港市2018届高三第一次模拟考试生物

江苏省连云港市2018届高三第一次模拟考试生物

2018届高三年级第一次模拟考试(七)生物(满分共120分,考试时间100分钟)第Ⅰ卷(选择题55分)一、单项选择题:本部分包括20题,每题2分,共计40分。

每题只有一个....选项最符合题意。

1. 下列有关糖类的叙述,正确的是()A. 糖类由C、H、O三种元素组成,只能为生物体提供能源B. 质粒、叶绿体、线粒体、A TP和RNA中都含有核糖C. 淀粉、糖原、纤维素和麦芽糖彻底水解后的产物相同D. 蔗糖储存于肌肉细胞中,能及时补充人体所需能量2. 下列关于多肽和蛋白质的叙述,正确的是()A. 内孢霉素是一种环状十一肽,分子中含10个肽键B. 氨基酸脱水缩合生成的水分子中的氢来自于氨基和羧基C. 组成头发的蛋白质中不同肽链之间通过肽键相连D. 蛋白质的功能与碳骨架有关,与其空间结构无关3. 右图表示物质进入细胞的四种方式,有关说法正确的是()A. 浓度差的大小与①过程的速率无关B. ①和②过程都需要载体C. 温度不会影响②和③过程的速率D. 吞噬细胞可通过④过程吞噬病原体4. 研究发现,冬小麦在秋冬受低温袭击时,呼吸速率先升高后降低;持续的冷害使根生长迟缓,吸收能力下降,但细胞内可溶性糖的含量有明显的提高。

下列推断错误..的是()A. 冷害初期呼吸作用先增强,有利于抵御寒冷B. 持续低温使线粒体内氧化酶活性减弱,直接影响可溶性糖合成淀粉C. 低温时细胞内结合水的比例上升,有利于适应低温环境D. 低温使根细胞呼吸减弱,吸收矿质营养能力下降5. 下图表示某植物叶肉细胞中部分物质变化途径,其中①~④代表有关生理过程。

相关叙述错误..的是()A. 过程③、④不在生物膜上进行B. 过程①、②、④都有A TP生成C. 过程③产生的C6H12O6中的氧来自H2O和CO2D. 过程④产生的[H]并不全都来自于C6H12O66. 用新鲜菠菜叶进行色素提取和分离的实验。

下列叙述正确的是()A. 将滤液收集到试管中,应及时用棉塞将试管口塞紧B. 画滤液细线时,在点样线上需连续重复多次画线C. 滤液细线浸入层析液,可导致滤纸条上色素带重叠D. 滤纸条上4条色素带通常都是平齐的,且彼此之间的距离明显7. 某兴趣小组为了探究温度对酶活性的影响,用打孔器获取新鲜的厚度为5 mm的三实验步骤土豆片A 土豆片B 土豆片C①处理静置煮熟后冷却冰冻②滴加质量分数3%H2O2溶液1滴1滴1滴③观察实验现象产生大量气泡几乎不产生气泡产生少量气泡A. 新鲜土豆组织中含有过氧化氢酶B.土豆片的厚度大小是该实验的因变量C. 温度不会影响过氧化氢酶活性D. 该实验可以确定过氧化氢酶的最适温度8. 1952年,赫尔希和蔡斯用32P或35S标记噬菌体并分别与无标记的细菌混合培养,保温后经过搅拌、离心得到了上清液和沉淀物,并检测放射性。

2018年上学期高三连云港市试题新人教版附答案

2018年上学期高三连云港市试题新人教版附答案

连云港市2018年第一学期期末调研考试高二化学试题可能用到的相对原子质量:H—1 C—12 N—14 O—16 Na—23 Mg—24 Al—27S—32 Cl—35.5 Fe—56 Cu—64 Ba—137第Ⅰ卷(选择题共72分)一、选择题(本题包括8小题,每小题4分,共32分。

每小题只有一个选项符合题意)1、下列各组物质气化或熔化时,所克服的微粒间的作用(力),属同种类型的是A 碘和干冰的升华B 二氧化硅和生石灰的熔化C 氯化钠和铁的熔化D 苯和已烷的蒸发2、碳化硅(SiC)的一种晶体具有类似金刚石的结构,其中碳原子和硅原子的位置是交替的.在下列三种晶体①金刚石、②晶体硅、③碳化硅中,它们的熔点从高到低的顺序是A ①③②B ②③①C ③①②D ②①③3、)下列说法中正确的是A 分子中键能越大,键越长,则分子越稳定B 失电子难的原子获得电子的能力一定强C 在化学反应中,某元素由化合态变为游离态,该元素被还原D 电子层结构相同的不同离子,其半径随核电荷数增多而减小4、下列分子中,属于含有极性键的非极性分子的是A H2OB Cl2C NH3D CCl45、下列晶体中,不属于原子晶体的是A 干冰B 水晶C 晶体硅D 金刚石6.关于如右图所示装置的叙述,正确的是A.铜是负极,铜片上有气泡产生B.铜片质量逐渐减少C.电流从锌片经导线流向铜片D.氢离子在铜片表面被还原7.制取相同质量的Cu(NO3)2时,消耗硝酸质量最多的是A.铜与浓HNO3反应B.铜与稀硝酸反应C.氧化铜与硝酸反应D.氢氧化铜与硝酸反应8.一定温度下在容积恒定的密闭容器中,进行如下可逆反应:A(s)+2B(g) C(g)+D(g)。

当下列物理量不发生变化时,能表明该反应已达到平衡状态的是①混合气体的密度②容器内气体的压强③混合气体的总物质的量④B物质的量浓度A ①和④B ②和③C ②③④D 只有④二、选择题(本题包括10小题,每小题4分,共40分。

最新-连云港市2018届高三生物第一次调研考试卷 精品

最新-连云港市2018届高三生物第一次调研考试卷 精品

连云港市2018届高三第一次调研考试生物试题(命题人:任守运审核人:李其柱邱莉萍)注意事项:1.本试题包括第Ⅰ卷选择题和第Ⅱ卷非选择题两部分。

试题的全部答案都要答在答题卡上。

2.答卷前,考生务必将自己的学校、姓名、考试号、考试科目等填涂在答题卡的规定处。

3.答第Ⅰ卷时,在答题卡的对应题号下,将正确答案的字母涂黑。

答第Ⅱ卷时,答案要答在答题卡的对应题号后的空白处。

第Ⅰ卷(选择题共70分)一、选择题:本题包括26小题,每小题2分,共52分。

每小题只有一个..选项最符合题意。

1.俗话说:“一树结果,有酸有甜。

”这说明生物体具有A.适应性 B.应激性 C.遗传性 D.变异性2.微生物的代谢调节主要是A.激素调节 B.神经调节 C.酶的调节 D.神经一体液调节3.不能在生物膜系统中发生的生理过程有A.肌细胞无氧呼吸 B.蛋白质的加工包装C.丙酮酸彻底氧化分解 D.胰岛素的分泌4.现代科学技术证明:躯体各部分的运动机能在大脑皮层第一运动区内部都有它们的代表区,代表区有大有小。

根据你掌握的知识,你认为在人体下列哪一部位的身体运动代表区最小A.手指 B.舌 C.臀部 D.唇5.用层析法分离叶绿体中的色素,滤纸条上距离滤液细线由近到远的颜色依次为A.橙黄色、黄色、蓝绿色、黄绿色 B.黄色、橙黄色、黄绿色、蓝绿色C.蓝绿色、黄绿色、橙黄色、黄色 D.黄绿色、蓝绿色、黄色、橙黄色6.下表是大豆和玉米两种植物种子食用部分的营养成分表(表中数据表示每克含量)。

分析表中数①种植大豆比种植玉米需要施用的氮肥多;②相同质量的大豆种子萌发时的吸水量较玉米种子的多;③相同质量的玉米种子萌发时的耗氧量比大豆种子的多;④从人体的生长需要看,大豆的营养价值比玉米的高A.①③ B.②④ C.①②④ D.②③④7.对体液免疫和细胞免疫的异同表述不正确的是A.体液免疫中抗体和抗原发生特异性结合,发生免疫反应B.细胞免疫中,效应T细胞的攻击目标是已经被抗原入侵了的宿主细胞C.有免疫反应的是效应淋巴细胞D.效应淋巴细胞都来自骨髓造血干细胞,都在骨髓中分化、发育、成熟8.在观察藓类叶片细胞的叶绿体形态与分布、黑藻叶片细胞的细胞质流动、植物根尖细胞的有丝分裂和花生子叶中脂肪鉴定的四个实验中,它们的共同点是A .实验全过程都要使实验对象保持活性B .适当提高温度将使实验结果更加明显C .都需要使用高倍显微镜观察D .都需要对实验材料进行染色 9.下列各项中,除哪一项外,均包含遗传信息A .染色体B .线粒体C .质粒D .病毒的衣壳 10.在一个透明的容器中加入适量NaHCO 3稀溶液,将杨树叶片迅速封入其中,装置如图所示,摇动容器,使容器内空气中的CO 2和溶液中的CO 2达到动态平衡,在保持温度不变的条件下进行实验.下列有关该实验的叙述错误的是A.光照几分钟后,容器内溶液的pH 将减小B.随着光照时间的延长,溶液的pH 的变化速度趋于变慢C.如果将装置置于暗室中,一段时间后,溶液的pH 将减小D.该装置可以用来研究植物的光合作用和呼吸作用的强度11.在“DNA 的粗提取与鉴定”实验中,向5mL 血细胞液中加入20mL 蒸馏水的目的是A.使血细胞破裂B.防止血液凝固C.析出DNAD.使蛋白质与DNA 分离 12.下列与植物光合作用有关的叙述中,正确的是A .叶绿素a 都能使光能发生转换 B. C 3植物的维管束鞘细胞能进行光合作用 C. NADPH 和ATP 中富含活跃的化学能 D. C 4植物叶片维管束鞘细胞内不出现淀粉粒 13.在适宜条件下,向固体培养基上接种处于对数期的菌种后,菌落直径(Y )随时间(X )的变化情况最可能为14.下列与细胞分裂有关的叙述中,正确的是A .原核细胞的增殖方式是有丝分裂B .细胞有丝分裂过程中形成赤道板C .减数分裂形成的细胞不具备全能性D .分裂后的子细胞中,核膜与内质网膜相连 15.下列有关生物工程技术的叙述中,正确的是 A. 基因工程中目的基因的运输工具一定是质粒B. 单克隆抗体由T 细胞和骨髓瘤细胞融合后形成的杂交瘤细胞产生C. 在发酵工程中得到的单细胞蛋白是从微生物中提取出来的蛋白质D. 微生物菌种选育的方法有诱变育种、基因工程和细胞工程等方法 16.右图是自然界中豌豆的生殖周期示意图,下列有关叙述正确的是 A .基因的表达主要在a 过程中B .纯种豌豆自交产生的子代只有一种性状C .非等位基因的自由组合发生于c 过程中D .在d 过程中同源染色体会发生联会17.为了防止转基因作物的目的基因通过花粉转移到自然界中的其他植物,科学家设法将目的基因整合到受体细胞的叶绿体基因组中。

江苏省高考2018年高三招生考试20套模拟测试 英语试题一 含解析

江苏省高考2018年高三招生考试20套模拟测试 英语试题一 含解析

实战演练·高三英语20套第页(共160页)江苏省普通高等学校招生考试高三模拟测试卷(一) 英语本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分.满分120分,考试时间120分钟.第Ⅰ卷(选择题共80分)第一部分:听力(共两节,满分15分)第一节(共5小题;每小题1分,满分5分)听下面5段对话.每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置.听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题.每段对话仅读一遍.()1. Where does the conversation probably take place?A. In a cafeteria.B. In a restaurant.C. In a supermarket.()2. Why does Jack stop playing sports now?A. He is too busy.B. He has lost the interest.C. The training is too hard.()3. What does the woman mean?A. She is a visitor.B. She just moved in here.C. She knows the manager.()4. What are the speakers talking about?A. Buying DVDs.B. Borrowing DVDs.C. Sharing DVDs.()5. How does the woman find the tickets?A. They are hard to get.B. They are cheap.C. They are expensive.第二节(共10小题;每小题1分,满分10分)听下面4段对话或独白.每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置.听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题给出5秒钟的作答时间.每段对话或独白读两遍.听第6段材料,回答第6、7题.()6. What will the boy do after lunch?A. Have some dessert.B. Clean up his toys.C. Try a new game.()7. Who might the woman be?A. Frankie's mother.B. Frankie's babysitter.C. Frankie's sister.听第7段材料,回答第8、9题.()8. What is Jane's problem?A. She can't have lunch with Dr. Pasteur tomorrow.B. She forgets the appointment with Dr. Pasteur.C. She can't meet Dr. Pasteur tomorrow morning at 9 a.m.()9. How long is the appointment postponed?A. Three hours and forty-five minutes.B. Four hours and fifteen minutes.C. Six hours.听第8段材料,回答第10至12题.()10. What did Mr. Tang major in the university?A. Chinese.B. Journalism.C. International operation.()11. What was Mr. Tang responsible for when he worked in a media company?A. Gathering the international news.B. Writing the current reports.C. Expanding the operation.()12. Why would Mr. Tang like to work in China?A. He can have a good chance to meet his parents.B. He can make good use of his operation ability.C. He can make good use of his Chinese and English.听第9段材料,回答第13至15题.()13. What can the iMaid do?A. Wash dishes.B. Dry the clothes.C. Clean up dirt from floors.()14. How long can the iMaid work after being charged?A. Three hours.B. Ten hours.C. Thirteen hours.()15. According to the talk, what is the best thing about the iMaid?A. The special gift worth $49.B. The price.C. The service contract.第二部分:英语基础知识运用(共两节,满分35分)第一节:单项填空(共15小题;每小题1分,满分15分)请阅读下面各题,从题中所给的A、B、C、D四个选项中,选出最佳选项.()16. — Tu Youyou and the other two scientists jointly won the 2015 Nobel Prize for medicine for their work against parasitic diseases.—They deserve it. The consequences ________ improved human health and reducedsuffering are immeasurable.A. in honor ofB. in terms ofC. in defense ofD. in hopes of()17. A teacher's job is not to tell the students what to believe or value, but to ________ them to develop a worldview for themselves.A. urgeB. rankC. persuadeD. equip()18. So why not, he reasoned, ________ the boy a few minutes to explain the whole affair?A. to spareB. sparingC. spareD. spared()19. He ________ himself to a search by the guards before entering the government building.A. objectedB. submittedC. compromisedD. identified()20. — It is reported that Papiss Cisse and Jonny Evans were charged with spitting by the Football Association.—I think spitting is one of the most disgusting things that ________ happen in the game, but yet it is not the worst.A. mustB. shallC. shouldD. can()21. Our mothers sat us down to read and paint, ________ all we really wanted to do was to make a mess.A. sinceB. asC. unlessD. when()22. I needn't have been in such a hurry. The flight to Hong Kong ________ due to the typhoon.A. has cancelledB. was cancelledC. will be cancellingD. had cancelled()23. — A study suggests reducing energy demand in the future may ________ urban areas.—That's true. Cities need more energy than small towns or other rural areas.A. center onB. act onC. hang onD. catch on()24. Tech-free tourism refers to traveling without a mobile phone or similar devices, particularly to places ________ block or cannot access Internet and cellular signals.A. thatB. whereC. whenD. who()25. —Have you heard of Gong Xingfang, who is experienced in taking care of mothers and newborns in Shanghai?—Yes. It is reported that she can earn 14,000 yuan ($2,252) a month now and anyone who wants to hire her has to make an ________ half a year in advance.A. assessmentB. accommodationC. appointmentD. occupation()26. My brother hopes that he ________ computer science instead of history when he graduated from the university.A. studiesB. studiedC. had studiedD. has studied()27. A Chinese student's print-like handwriting caused controversy among British Internet users,________ both praise and questions about individuality.A. drewB. drawingC. to drawD. having drawn()28. British government is planning to run a pilot scheme that will allow Chinese tourists to get a two-year tourist visa for £85—these cost £324.A. currentlyB. apparentlyC. frequentlyD. similarly()29. Some experts hold the view that fundamental construction is ________ the key to the little island development lies.A. whichB. whatC. whereD. why()30. — His father always tells him to stop telling lies, which falls on deaf ears.—I think he will suffer the consequences. ________.A. You reap what you sowB. Justice has long armsC. Honesty is the best policyD. Lies have short legs第二节:完形填空(共20小题;每小题1分,满分20分)请认真阅读下面短文,从短文后各题所给的A、B、C、D四个选项中,选出最佳选项.The continuous presentation of frightening stories about global warming in the popular media makes us unnecessarily frightened. Even worse, it __31__ our kids.Al Gore famously __32__ how a sea-level rise of 20 feet would almost completely __33__ Florida, New York, Holland, and Shanghai, __34__ the United Nations says that such a thing will not even happen, __35__ that sea levels will rise 20 times less than that.When __36__ with these exaggerations(夸大), some of us say that they are for a good cause,and surely there is no __37__ done if the result is that we focus even more on dealing with climate change.This __38__ is astonishingly wrong. Such exaggerations do plenty of harm. Worrying extremely about global warming means that we worry less about other things,__39__ we could do so much more good. We focus, __40__,on global warming's impact on malaria(疟疾) —which will put more people at __41__ in 100 years—instead of helping the half a billion people __42__ from malaria today with prevention and treatment policies that are much cheaper and dramatically more __43__ than carbon reduction would be.Exaggeration also wears out the public's __44__ to cope with global warming. If the planet is certain to be destroyed __45__ global warming, people wonder, why should we do anything?The __46__ cost of exaggeration, I believe, is the unnecessary alarm that it causes —particularly among children. An article in The Washington Post mentioned nine-year-old Alyssa, who cries about the possibility of mass animal __47__ from global warming.The newspaper also reported that parents are __48__ effective outlets for their 8-year-olds' concern with dying polar bears. They might be better off educating them and letting them know that, __49__ to common belief, the global polar bear population has doubled over the past half-century, to about 22,000. __50__ the possible disappearing of summer Arctic ice, polar bears will live on with us.()31. A. exhausts B. amazes C. terrifies D. interests()32. A. dismissed B. determined C. denied D. described()33. A. cover B. flood C. reduce D. expand()34. A. even though B. as if C. in that D. in case()35. A. measuring B. proving C. estimating D. advocating()36. A. faced B. identified C. filled D. entitled()37. A. good B. harm C. benefit D. disadvantage()38. A. announcement B. argument C. story D. dialogue()39. A. when B. what C. where D. which()40. A. for example B. in addition C. on average D. in short()41. A. peace B. random C. ease D. risk()42. A. prohibiting B. escaping C. developing D. suffering()43. A. effective B. accurate C. complex D. temporary()44. A. ability B. sense C. willingness D. preference()45. A. due to B. except for C. regardless of D. along with()46. A. smallest B. worst C. fewest D. least()47. A. ruling out B. running out C. dropping out D. dying out()48. A. turning out B. taking over C. searching for D. pulling through()49. A. sensitive B. contrary C. related D. accustomed()50. A. Except B. Besides C. Without D. Despite第三部分:阅读理解(共15小题;每小题2分,满分30分)请认真阅读下列短文,从短文后各题所给的A、B、C、D四个选项中,选出最佳选项.ABelow are the four most famous bridges in the world.Ponte Vecchio BridgeThe Ponte Vecchio (literally “old bridge”) is a bridge built in the Middle Ages over the Arno River in Florence, Italy, the only Florentine bridge to survive World War Ⅱ. The bridge is unique for still having shops built along it, as was common in the days of the Medici. Butchers originally occupied souvenir sellers. It is said that the economic concept of bankruptcy originated here: when a merchant could not pay his debts, the table on which he sold his goods was physically broken by soldiers, and this practice was called “bancorotto (broken table)”.Golden Gate BridgeThe Golden Gate Bridge is a suspension bridge spanning the Golden Gate, the strait between San Francisco and Marin County to the north. It is the masterwork of architect Joseph B. Strauss, whose statue graces the southern observation deck. The bridge took seven years to build, and was completed in 1937. The Golden Gate Bridge used to be the longest suspension bridge span in the world. And today it has become one of the most popular tourist attractions in San Francisco and California. Since its completion, the span length has been surpassed by eight other bridges. The famous red-orange color of the bridge was specifically chosen to make the bridge more easily visible through the thick frog that frequently covers the bridge.Millau BridgeStarted in 1998 and opened to traffic in 2005, the Millau Viaduct is a huge cable-stayed road-bridge that spans the valley of the river Tarn near Millau in southern France. It is the tallest highway bridge in the world, with the highest pylon's summit at 343 meters—slightly taller than the Eiffel Tower. The speed limit on the bridge was reduced from 130 km/h to 110 km/h because of traffic slowing down, due to tourists taking pictures of the bridge from the vehicles. Shortly after the bridge opened to traffic, passengers were stopping to admire the landscape and the bridge itself.Charles BridgeThe Charles Bridge is a famous stone Gothic bridge that crosses the Vltava River in Prague, Czech Republic. Its construction started in 1357 under the support of King Charles IV, and finished in the beginning of the 15th century. As the only means of crossing the river Vltava, theCharles Bridge was the most important connection between the Old Town and the area around Prague Castle. Connection made Prague important as a trade route between Eastern and Western Europe. Today it is one of the most visited sights in Prague with painters, owners of kiosks and other traders alongside numerous tourists crossing the bridge.()51. Of the four bridges, which one has the shortest history?A. Ponte Vecchio.B. Golden Gate Bridge.C. Millau Bridge.D. Charles Bridge.()52. Which of the following statements is TRUE about the Golden Gate Bridge?A. The span length ranks the 8th in the world.B. Its color enables travelers to see it easily on foggy days.C. It is the most popular tourist attraction in America.D. It took Joseph B. Strauss 7 years to design the bridge.()53. The Charles Bridge played an important role in Prague, Czech Republic because ________.A. it attracted many famous painters thereB. it was supported by Kin Charles IVC. it was the only stone Gothic bridge crossing the Vltava RiverD. it promoted the trade between Eastern and Western EuropeBTELECOMMUTERS fall into two camps. Some sit on the sofa watching daytime soaps, pausing occasionally to check their BlackBerrys. Most, however, do real work, undistracted by meetings and talkative colleagues.In the future more people will work from home. With office space in London and New York so costly, many firms save money by encouraging staff to work in their loose clothes. Instead of having to bury their noses in strangers' armpits on crowded trains, they can work via e-mail, Skype and virtual private networks.Yet, in a research published in MIT Sloan Management Review, Daniel Cable of the London Business School shows that telecommuters are less likely to be promoted. In one experiment subjects were asked to judge scenarios in which the only difference was whether the employee was at his office desk or at home. Managers rated those at the office to be more dependable and industrious, regardless of the quality of their work.Visibility creates the illusion of value. Being the last to leave the office impresses bosses, even if you are actually larking around(胡闹) on Facebook. Oddly, this holds true at firms that explicitly encourage staff to work from home. Many Californian tech firms asked employees not to come to the office too often; yet bosses unconsciously punished those who obeyed.Remote workers understand this. Many frequently sent their bosses with progress reports to prove they are on the job. A fifth of the workers in the study admitted to leaving an e-mail or voice mail early or late in the day. Still, many are not as smart as they think. Some choose a Monday or Friday to work at home. That, says Mr. Cable, makes others think they are eager to extend the weekend.A culture of presenteeism hurts working mothers most. Many women (and some men) work from home to allow themselves the flexibility to pick up kids from school. That need not mean they produce less; only that they do it at a time and a place of their own choosing. Some firms, such as Best Buy, an electronics retailer, recognize this and try hard to evaluate staff entirely on performance. But this is not easy. Intangibles such as teamworking skills matter, too. Mr. Cable thinks homeworking will lose its stigma(污名) only when most people do it. Or perhaps when the boss is telecommuting, too.()54. What is most likely the main cause of the increasing number of telecommuters?A. Increasing location rents.B. Annoying talkative colleagues.C. High-tech mobile phones.D. Attractive daytime soaps.()55. What does the example of many California tech firms prove?A. Working at home is impractical in tech firms.B. Employees' presence at office raises their value.C. Employees should judge when to obey.D. Bosses often don't keep their promises.()56. What do wise telecommuters do to prove they are on the job?A. They give timely accounts of their work progress to their bosses.B. They check their e-mails and voice mails every day.C. They discuss the work with their bosses.D. They spend some time working on weekends.()57. What is the biggest disadvantage of working at home according to the lastparagraph?A. The traditional working culture can be hurt.B. Mothers' work may be interrupted by their kids.C. Retailers can't get enough on-site employees.D. Employees may lack chances to develop certain skills.CAlzheimer's disease has no cure. There are, however, five drugs—known and approved—that can slow down the development of its symptoms. The earlier such drugs are administered, the better. Unfortunately, the disease is usually first noticed when people complain to their doctors of memory problems. That is normally too late for the drugs to do much good. A simple and reliable test for Alzheimer's that can be administered to everybody over the age of about 65, before memory-loss sets in, would therefore be useful.Theo Luider, of the Erasmus University Medical Centre in Rotterdam, and his colleagues think they have found one—but it works only in women. They made their discovery, just reported in the Journal of Proteome Research, by tapping into a long-term, continuing study that started in 1995 with 1,077 non-demented and otherwise healthy people aged between 60 and 90. At the beginning of the project, and subsequently during the periods 1997-99 and 2002-04, participants were brought in for a battery of neurological(神经学的) and cognitive(认知的) investigations, physical examinations, brain imaging and blood tests.During the first ten years of the study, 43 of the volunteers developed Alzheimer's diseases. When Dr. Luider compared blood samples from these people with samples from 43 of their fellow volunteers, matched for sex and age, who had remained Alzheimer's-free, he found something surprising. Levels of a substance called pregnancy zone protein had been unusually high, even before their symptoms appeared, in some of those who went on to develop Alzheimer's disease.Those “some”,it turned out, were all women. On average, levels of pregnancy zone protein in those women who went on to develop Alzheimer's were almost 60% higher than those of women who did not. In men, levels of the protein were the same for both.The reason for this curious result seems to be that the brain plaques(斑块) associated with Alzheimer's disease are themselves turning out pregnancy zone protein. Certainly, when Dr. Luider applied a chemical stain specific to that protein to the plaques of dead Alzheimer's patientshe found the protein present in them.Confusingly, though, it was there in the plaques of both sexes. Presumably, female cells (and therefore the plaques of female brains) make more of it than male cells do. But that remains to be proved. Whatever the reason, however, this result means that women, at least, may soon be able to tell whether and when they are at risk of Alzheimer's and thus do something about it before they start losing their minds.()58. What can we learn from the first paragraph?A. No medication can slow down the development of Alzheimer's symptoms.B. To detect Alzheimer's disease before memory loss appears is vital.C. Doctors had better handle Alzheimer's disease when people are 65 years old.D. People who always complain are most likely to have Alzheimer's disease.()59. The underlined word “one” in Paragraph 2 refers to ________.A. a simple and reliable test for Alizheimer'sB. a possible cure for Alzheimer'sC. an important discovery about Alzheimer'sD. an effective and legal drug for Alzheimer's()60. What does Dr. Luider's study tell us about the pregnancy zone protein?A. It won't go high until the symptoms of Alzheimer's appear.B. In men, levels of it remain stable for their lifetime.C. Women developing Alzheimer's usually have lower levels of it.D. The brain plaques connected with Alzheimer's produce it.()61. The passage is mainly about ________.A. patients of Alzheimer's disease and its drugsB. an introduction to the pregnancy zone proteinC. a new discovery concerning Alzheimer's diseaseD. the development stages of Alzheimer's diseaseDHe was in the first third-grade class I taught at Saint Mary's School in Morris, Minnesota. All 34 of my students were dear to me, but Mark Eklund was one in a million. Very neat in appearance, he had that happy-to-be-alive attitude that made even his occasional mischievousness delightful.Mark also talked continuously. I had to remind him again and again that talking withoutpermission was not acceptable. One morning my patience was growing thin when Mark talked once too often, and then I made a novice-teacher's mistake. I looked at Mark and said, “If you say one more word, I am going to tape your mouth shut!”It wasn't ten seconds later when Chuck blurted out, “Mark is talking again.”I hadn't asked any of the students to help me watch Mark, but since I had stated the punishment in front of the class, I had to act on it.I remember the scene as if it had occurred this morning. Without saying a word, I proceeded to Mark's desk, tore off two pieces of tape and made a big X with them over his mouth. I then returned to the front of the room.As I glanced at Mark to see how he was doing, he winked at me. That did it! I started laughing. The entire class cheered as I walked back to Mark's desk, removed the tape, and shrugged my shoulders. His first words were, “Thank you for correcting me, Sister.”At the end of the year I was asked to teach junior-high math. The years flew by, and before I knew it Mark was in my classroom again. He was more handsome than ever and just as polite.One Friday, things just didn't feel right. We had worked hard on a new concept all week, and I sensed that the students were growing discouraged with themselves—and edgy with one another.I had to change the mood of the class before it got out of hand. So I asked them to list the names of the other students in the room on two sheets of paper, leaving a space between each name. Then I told them to think of the nicest thing they could say about each of their classmates and write it down. It took the remainder of the class period to finish the assignment.That Saturday, I wrote down the name of each student on a separate sheet of paper, and I listed what everyone else had said about that individual. On Monday I gave each student his or her list. Some of them ran two pages. Before long, the entire class was smiling. “Really?” I heard whispers. “I never knew that meant anything to anyone!”“I didn't know others liked me so much!”No one ever mentioned those papers in class again. I never knew if the students discussed them after class or with their parents, but it didn't matter. The exercise had accomplished its purpose. The students were happy with themselves and one another again.That group of students moved on. Several years later, after I returned from a vacation, I got a call from my father. “The Eklunds called last night,”he began. “Really?”I said. “I haven'theard from them for several years. I wonder how Mark is.”Dad responded quietly. “Mark was killed in Vietnam,”Mark looked so handsome, so mature. All I could think at that moment was, Mark, I would give all the masking tape in the world if only you could talk to me.After the funeral, most of Mark's former classmates headed to Chuck's farmhouse for lunch. Mark's parents were there, obviously waiting for me. “Helen, we want to show you something,”his father said, taking a wallet out of his pocket. “They found this on Mark when he was killed. We thought you might recognize it.”Opening the billfold, he carefully removed two worn pieces of notebook paper that had obviously been taped, folded and refolded many times. I knew without looking that the papers were the ones on which I had listed all the good things each of Mark's classmates had said about him. “Thank you so much for doing that,”Mark's mother said. “As you can see, Mark treasured it.”Mark's classmates started to gather around us. Charlie smiled rather sheepishly and said, “I still have my list. It's in the top drawer of my desk at home.”Then Vicki, another classmate, reached into her pocket-book, took out her wallet and showed her worn and ragged list to the group. “I carry this with me at all times,”Vicki said without hesitation. “I think we all saved our lists.”That's when I finally sat down and cried. I cried for Mark and for all his friends who would never see him again.()62. We can conclude that when Sister Helen was a third-grade teacher, she ________.A. was usually hot-tempered and impatientB. liked all the students in the class but MarkC. wasn't always sure how to discipline her studentsD. had a high expectation of the students in her class()63. The underlined word “edgy” in Paragraph 7 means “________”.A. very disappointedB. easily annoyedC. fully honestD. greatly inspired()64. Upon reading their lists for the first time, Sister Helen's students were ________.A. surprised and proudB. nervous and embarrassedC. depressed and angryD. calm and content()65. Mark carried the notebook paper at all times because ________.A. it was a valuable gift from his dear Sister HelenB. it could ease his homesickness when in VietnamC. it was the recognition and appreciation from his classmatesD. he promised his classmates that he would treasure it第Ⅱ卷(非选择题共40分)第四部分:词汇检测(共5小题;每小题1分,满分5分)请认真阅读下列各小题,并根据上下文语境和所给首字母的提示,写出下列各句空格中的单词,注意保持语义和形式的一致.66. —Whatever b________ we are having on our shoulders, let them down for a moment, shall we?—All right. Let's enjoy the meal first.67. —I noticed the customer in red go away not altogether satisfied with Tom's explanations.—Definitely. She asked how the machine worked and Tom just gave a v________ description about its function, which could make her even more puzzled.68. — Alice, Granny is coming. Would you give your room a t________ cleaning?—With so much homework to do, I will just mop the floor, leaving the dirty windows to Jim.69. —Have you heard the news that his father's ship crashed into a rock and was broken in two?—Yeah. Luckily, nobody was injured with the help of the soldiers s________ on the nearby island.70. —One more girl was bitten by a dog this morning. Worse still, nobody knows who the owner is.—It's high time to campaign for c________ registration of dogs.第五部分:同义转换(共5小题;每小题1分,满分5分)请认真阅读下列各小题的两句句子,在空格处填上一个单词,使两句句子语义保持不变.(注意:不得使用第一句中的原词)71. — We will stick to our policy to promote relationships with the third-world countries.—It will be our ________ policy to promote relationships with the third-world countries.72. —Yan Fei, a director of Goodbye Mr. Loser thinks the success of the film lies in their devotion to telling a complete story.—Yan Fei, a director of Goodbye Mr. Loser ________ the success of the film to their devotion to telling a complete story.73. —Many Chinese students studying abroad have no choice but to wash dishes in the restaurants to support themselves.—In order to live on, many Chinese students studying abroad are reduced to ________ themselves out to wash dishes in the restaurants.74. —I was green with envy when I was informed that he would be promoted while I would not.—I was ________ when I was informed that he would be promoted while I would not.75. —Their system which relies entirely on departmental selection will surely cause lack of balance.—Their system which relies entirely on departmental selection is ________ to result in lack of balance.第六部分:任务型阅读(共10小题;每小题1分,满分10分)请认真阅读下面短文,并根据所读内容在文章后表格中的空格里填入一个最恰当的单词.注意:每个空格只填1个单词.Regret is as common an emotion as love or fear, and it can be nearly as powerful. We feel it when we either blame ourselves for things that turned out badly, or long to undo a choice we made in the past. The effect regret has on our lives and how we deal with regret are equally important.In some cases, regret can be disastrous. In 1995, a British man who regularly played one set of lottery numbers forgot to renew his ticket during the week that his numbers came up. He was so filled with regret and self-blame that he committed suicide. While this is an extreme consequence of regret, it can have many other lesser effects on the mind and body that can still seriously affect our lives.According to recent research, women have more regrets about romantic relationships than men do—not surprising, since women “value social relationships more than men”. In collectivist。

2018届连云港市高三年级第一次模拟考试含答案

2018届连云港市高三年级第一次模拟考试含答案

2018届连云港市高三年级第一次模拟考试数学(满分160分,考试时间120分钟)参考公式:1. 柱体的体积公式:V =Sh ,其中S 是柱体的底面面积,h 是高.2. 圆锥的侧面积公式:S =12cl ,其中c 是圆锥底面圆的周长,l 是母线长.一、 填空题:本大题共14小题,每小题5分,共计70分.1. 已知集合A ={x|x 2-x =0},B ={-1,0},则A ∪B =________.2. 已知复数z =2+i2-i (i 为虚数单位),则z 的模为________.3. 函数y =log 12x 的定义域为________.4. 如图是一个算法的伪代码,运行后输出b的值为________.(第4题) (第5题)5. 某地区教育主管部门为了对该地区模拟考试成绩进行分析,随机抽取了150分到450分之间的1 000名学生的成绩,并根据这1 000名学生的成绩画出样本的频率分布直方图(如图),则成绩在[250,400)内的学生共有________人.6. 在平面直角坐标系xOy 中,已知双曲线x 2a 2-y 2b 2=1(a>0,b>0)的一条渐近线方程为x -2y =0,则该双曲线的离心率为________.7. 连续2次抛掷一枚质地均匀的骰子(六个面上分别标有数字1,2,3,4,5,6的正方体),观察向上的点数,则事件“点数之积是3的倍数”的概率为________.8. 已知正四棱柱的底面边长为3cm ,侧面的对角线长是3 5 cm ,则这个正四棱柱的体积是________cm 3.9. 若函数f(x)=A sin (ωx +φ)(A>0,ω>0)的图象与直线y =m 的三个相邻交点的横坐标分别是π6,π3,2π3,则实数ω的值为________.10. 在平面直角坐标系xOy 中,曲线C :xy =3上任意一点P 到直线l :x +3y =0的距离的最小值为________.11. 已知等差数列{a n }满足a 1+a 3+a 5+a 7+a 9=10,a 28-a 22=36,则a 11的值为________. 12. 在平面直角坐标系xOy 中,若圆C 1:x 2+(y -1)2=r 2(r>0)上存在点P ,且点P 关于直线x -y =0的对称点Q 在圆C 2:(x -2)2+(y -1)2=1上,则r 的取值范围是________________________________________________________________________.13. 已知函数f(x)=⎩⎪⎨⎪⎧2-|x +1|,x ≤1,(x -1)2, x>1,函数g(x)=f(x)+f(-x),则不等式g(x)≤2的解集为________.14. 如图,在△ABC 中,已知AB =3,AC =2,∠BAC =120°,D 为边BC 的中点.若CE ⊥AD ,垂足为E ,则EB →·EC →的值为________.二、 解答题:本大题共6小题,共计90分.解答时应写出文字说明、证明过程或演算步骤.15. (本小题满分14分)在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,且cos A =35,tan (B -A)=13.(1) 求tan B 的值;(2) 若c =13,求△ABC 的面积.16. (本小题满分14分)如图,在直三棱柱ABCA 1B 1C 1中,∠ABC =90°,AB =AA 1,M ,N 分别是AC ,B 1C 1的中点.求证:(1) MN ∥平面ABB 1A 1; (2) AN ⊥A 1B.17. (本小题满分14分)某艺术品公司欲生产一款迎新春工艺礼品,该礼品是由玻璃球面和该球的内接圆锥组成,圆锥的侧面用于艺术装饰,如图1.为了便于设计,可将该礼品看成是由圆O 及其内接等腰三角形ABC 绕底边BC 上的高所在直线AO 旋转180°而成,如图2.已知圆O 的半径为10cm ,设∠BAO =θ,0<θ<π2,圆锥的侧面积为S cm 2.(1) 求S 关于θ的函数关系式;(2) 为了达到最佳观赏效果,要求圆锥的侧面积S 最大.求当S 取得最大值时腰AB 的长度.图1 图218. (本小题满分16分)如图,在平面直角坐标系xOy 中,已知椭圆x 2a 2+y 2b 2=1(a>b>0)的离心率为12,且过点⎝⎛⎭⎫1,32.F 为椭圆的右焦点,A ,B 为椭圆上关于原点对称的两点,连结AF ,BF 分别交椭圆于C ,D 两点.(1) 求椭圆的标准方程;(2) 若AF =FC ,求BFFD的值; (3) 设直线AB ,CD 的斜率分别为k 1,k 2,是否存在实数m ,使得k 2=mk 1?若存在,求出实数m 的值;若不存在,请说明理由.19. (本小题满分16分)已知函数f(x)=x 2+ax +1,g(x)=ln x -a(a ∈R). (1) 当a =1时,求函数h (x )=f (x )-g (x )的极值;(2) 若存在与函数f (x ),g (x )的图象都相切的直线,求实数a 的取值范围.20. (本小题满分16分)已知数列{a n },其前n 项和为S n ,满足a 1=2,S n =λna n +μa n -1,其中n ≥2,n ∈N *,λ,μ∈R.(1) 若λ=0,μ=4,b n =a n +1-2a n (n ∈N *),求证:数列{b n }是等比数列; (2) 若数列{a n }是等比数列,求λ,μ的值;(3) 若a 2=3,且λ+μ=32,求证:数列{a n }是等差数列.2018届高三年级第一次模拟考试(七)数学附加题(本部分满分40分,考试时间30分钟)21. 【选做题】本题包括A 、B 、C 、D 四小题,请选定其中两小题,并作答.若多做,则按作答的前两小题评分.解答时应写出文字说明、证明过程或演算步骤.A. [选修41:几何证明选讲](本小题满分10分)如图,AB 是圆O 的直径,弦BD ,CA 的延长线相交于点E ,EF 垂直BA 的延长线于点F .求证:AB 2=BE ·BD -AE ·AC.B. [选修42:矩阵与变换](本小题满分10分)已知矩阵A =⎣⎢⎡⎦⎥⎤100-1,B =⎣⎢⎡⎦⎥⎤4123,若矩阵M =BA ,求矩阵M 的逆矩阵M -1.C. [选修44:坐标系与参数方程](本小题满分10分)以坐标原点为极点,x 轴的正半轴为极轴,且在两种坐标系中取相同的长度单位,建立极坐标系,判断直线l :⎩⎪⎨⎪⎧x =1+2t ,y =1-2t (t 为参数)与圆C :ρ2+2ρcos θ-2ρsin θ=0的位置关系.D. [选修45:不等式选讲](本小题满分10分)已知a ,b ,c ,d 都是正实数,且a +b +c +d =1,求证:a 21+a +b 21+b +c 21+c +d 21+d ≥15.【必做题】第22题、第23题,每题10分,共计20分.解答时应写出文字说明、证明过程或演算步骤.22. (本小题满分10分)在正三棱柱ABCA 1B 1C 1中,已知AB =1,AA 1=2,E ,F ,G 分别是AA 1,AC 和A 1C 1的中点.{FA →,FB →,FG →}为正交基底,建立如图所示的空间直角坐标系Fxyz.(1) 求异面直线AC 与BE 所成角的余弦值; (2) 求二面角FBC 1C 的余弦值.23. (本小题满分10分)在平面直角坐标系xOy 中,已知平行于x 轴的动直线l 交抛物线C :y 2=4x 于点P ,F 为曲线C 的焦点.圆心不在y 轴上的圆M 与直线l ,PF ,x 轴都相切,设圆心M 的轨迹为曲线E.(1) 求曲线E 的方程;(2) 若直线l 1与曲线E 相切于点Q(s ,t),过点Q 且垂直于l 1的直线为l 2,直线l 1,l 2分别与y 轴相交于点A ,B.当线段AB 的长度最小时,求s 的值.2018届连云港高三年级第一次模拟考试数学参考答案1. {-1,0,1}2. 13. (0,1]4. 135. 7506.52 7. 598. 54 9. 4 10. 3 11. 11 12. [2-1,2+1] 13. [-2,2] 14. -27715. 解析:(1) 在△ABC 中,由cos A =35,知A 为锐角,所以sin A =1-cos 2A =45,所以tan A =sin A cos A =43,(2分)所以tan B =tan [(B -A)+A]=tan (B -A )+tan A1-tan (B -A )tan A (4分)= 13+431-13×43=3. (6分)(2) 由(1)知tan B =3,所以sin B =31010,cos B =1010, (8分)所以sin C =sin (A +B)=sin A cos B +cos A sin B =131050.(10分)由正弦定理b sin B =c sin C, 得b =c sin Bsin C =13×31010131050=15,(12分)所以△ABC 的面积S =12bc sin A =12×15×13×45=78.(14分)16. 解析 :(1) 如图,取AB 的中点P ,连结PM ,PB 1.因为M ,P 分别是AB ,AC 的中点, 所以PM ∥BC ,且PM =12BC.在直三棱柱ABCA 1B 1C 1中,BC ∥B 1C 1,BC =B 1C 1, 又N 是B 1C 1的中点,所以PM ∥B 1N ,且PM =B 1N ,(2分) 所以四边形PMNB 1是平行四边形, 所以MN ∥PB 1.(4分)又MN ⊄平面ABB 1A 1,PB 1⊂平面ABB 1A 1, 所以MN ∥平面ABB 1A 1.(6分)(2) 因为三棱柱ABCA 1B 1C 1为直三棱柱, 所以BB 1⊥平面A 1B 1C 1, 因为BB 1⊂平面ABB 1A 1,所以平面ABB 1A 1⊥平面A 1B 1C 1. (8分) 因为∠ABC =∠A 1B 1C 1=90°, 所以B 1C 1⊥B 1A 1.因为平面ABB 1A 1∩平面A 1B 1C 1=B 1A 1,B 1C 1⊂平面A 1B 1C 1, 所以B 1C 1⊥平面ABB 1A 1. (10分) 因为A 1B ⊂平面ABB 1A 1,所以B 1C 1⊥A 1B ,即NB 1⊥A 1B. 如图,连结AB 1.因为在平行四边形ABB 1A 1中,AB =AA 1, 所以四边形ABB 1A 1是正方形, 所以AB 1⊥A 1B.因为NB 1∩AB 1=B 1,且AB 1,NB 1⊂平面AB 1N ,所以A 1B ⊥平面AB 1N.(12分) 又AN ⊂平面AB 1N , 所以A 1B ⊥AN.(14分)17. 解析 :(1) 如图,设AO 交BC 于点D ,过点O 作OE ⊥AB ,垂足为E. 在△AOE 中,AE =10cos θ,AB =2AE =20cos θ, (2分) 在△ABD 中,BD =AB·sin θ=20cos θ·sin θ,(4分) 所以S =12·2π·20sin θcos θ·20cos θ=400πsin θcos 2θ⎝⎛⎭⎫0<θ<π2.(6分)(2) 由(1)得S =400πsin θcos 2θ=400π(sin θ-sin 3θ).(8分) 设f(x)=x -x 3(0<x<1),则f′(x)=1-3x 2. 由f′(x)=1-3x 2=0得x =33. 当x ∈⎝⎛⎭⎫0,33时,f ′(x)>0;当x ∈⎝⎛⎭⎫33,1时,f ′(x)<0, 所以f(x)在区间⎝⎛⎭⎫0,33上单调递增,在区间⎝⎛⎭⎫33,1上单调递减, 所以f(x)在x =33时取得极大值,也是最大值, 所以当sin θ=33时,侧面积S 取得最大值,(11分) 此时等腰三角形的腰长AB =20cos θ=20×1-sin 2θ=20×1-⎝⎛⎭⎫332=2063.故当侧面积S 取得最大值时,等腰三角形的腰AB 的长度为2063 cm .(14分)18. 解析 :(1) 设椭圆的方程为x 2a 2+y 2b2=1(a>b>0).由题意知⎩⎨⎧c a =12,1a 2+94b 2=1,(2分)解得⎩⎨⎧a =2,b =3,所以椭圆的方程为x 24+y 23=1.(4分)(2) 若AF =FC ,由椭圆的对称性,知A ⎝⎛⎭⎫1,32, 所以B ⎝⎛⎭⎫-1,-32, 此时直线BF 的方程为3x -4y -3=0.(6分)由⎩⎪⎨⎪⎧3x -4y -3=0,x 24+y 23=1,得7x 2-6x -13=0, 解得x =137(x =-1舍去),(8分)所以BF FD =1-(-1)137-1=73.(10分) (3) 设A(x 0,y 0),则B(-x 0,-y 0),直线AF 的方程为y =y 0x 0-1(x -1), 代入椭圆方程x 24+y 23=1,得(15-6x 0)x 2-8y 20x -15x 20+24x 0=0. 因为x =x 0是该方程的一个解,所以点C 的横坐标x C =8-5x 05-2x 0.(12分)又点C(x C ,y C )在直线y =y 0x 0-1(x -1)上, 所以y C =y 0x 0-1(x C -1)=-3y 05-2x 0. 同理,点D 的坐标为⎝⎛⎭⎪⎫8+5x 05+2x 0,3y 05+2x 0, (14分)所以k 2= 3y 05+2x 0--3y 05-2x 08+5x 05+2x 0-8-5x 05-2x 0=5y 03x 0=53k 1,即存在m =53,使得k 2=53k 1.(16分)19. 解析 :(1) 函数h(x)的定义域为(0,+∞).当a =1时,h(x)=f(x)-g(x)=x 2+x -ln x +2, 所以h′(x)=2x +1-1x =(2x -1)(x +1)x ,(2分)所以当0<x<12时,h ′(x)<0;当x>12,h ′(x)>0,所以函数h(x)在区间⎝⎛⎭⎫0,12上单调递减,在区间⎝⎛⎭⎫12,+∞上单调递增, 所以当x =12时,函数h(x)取得极小值114+ln 2,不存在极大值.(4分)(2) 设函数f(x)在点(x 1,f(x 1))处的切线与函数g(x)在点(x 2,g(x 2))处的切线相同,则f′(x 1)=g′(x 2)=f (x 1)-g (x 2)x 1-x 2,所以2x 1+a =1x 2=x 21+ax 1+1-(ln x 2-a )x 1-x 2, (6分)所以x 1=12x 2-a2,代入x 1-x 2x 2=x 21+ax 1+1-(ln x 2-a)得14x 22-a 2x 2+ln x 2+a 24-a -2=0.(*)(8分) 设F(x)=14x 2-a 2x +ln x +a 24-a -2,则F′(x)=-12x 3+a 2x 2+1x =2x 2+ax -12x 3.不妨设2x 20+ax 0-1=0(x 0>0),则当0<x<x 0时,F ′(x)<0;当x>x 0时,F ′(x)>0,所以F(x)在区间(0,x 0)上单调递减,在区间(x 0,+∞)上单调递增,(10分) 因为a =1-2x 20x 0=1x 0-2x 0,所以F(x)min =F(x 0)=x 20+2x 0-1x 0+ln x 0-2. 设G(x)=x 2+2x -1x +ln x -2,则G′(x)=2x +2+1x 2+1x >0对x>0恒成立,所以G(x)在区间(0,+∞)上单调递增.又G(1)=0,所以当0<x ≤1时,G(x)≤0,即当0<x 0≤1时,F(x 0)≤0.(12分)又当x =e a +2时,F(x)=14e 2a +4-a 2e a +2+lne a +2+a 24-a -2=14⎝⎛⎭⎫1e a +2-a 2≥0,(14分)因此当0<x 0≤1时,函数F(x)必有零点,即当0<x 0≤1时,必存在x 0使得(*)成立,即存在x 1,x 2使得函数f(x)在点(x 1,f(x 1))处的切线与函数g(x)在点(x 2,g(x 2))处的切线相同.又由y =1x -2x 得y′=-1x2-2<0,所以y =1x -2x 在(0,1)上单调递减,因此a =1-2x 20x 0=1x 0-2x 0∈[-1,+∞),所以实数a 的取值范围是[-1,+∞).(16分)20. 解析:(1) 若λ=0,μ=4,则S n =4a n -1(n ≥2), 所以a n +1=S n +1-S n =4(a n -a n -1), 即a n +1-2a n =2(a n -2a n -1), 所以b n =2b n -1.(2分)又由a 1=2,a 1+a 2=4a 1,得a 2=3a 1=6,a 2-2a 1=2≠0,即b n ≠0, 所以b nb n -1=2.故数列{b n }是等比数列.(4分)(2) 若{a n }是等比数列,设其公比为q(q ≠0).当n =2时,由S 2=2λa 2+μa 1,即a 1+a 2=2λa 2+μa 1,得1+q =2λq +μ;①当n =3时,由S 3=3λa 3+μa 2,即a 1+a 2+a 3=3λa 3+μa 2,得1+q +q 2=3λq 2+μq ;② 当n =4时,由S 4=4λa 4+μa 3,即a 1+a 2+a 3+a 4=4λa 4+μa 3,得1+q +q 2+q 3=4λq 3+μq 2.③②-①×q ,得1=λq 2, ③-②×q ,得1=λq 3, 解得q =1,λ=1. 代入①,得μ=0.(8分) 此时S n =na n (n ≥2),所以a n =a 1=2,{a n }是公比为1的等比数列, 故λ=1,μ=0. (10分)(3) 若a 2=3,由a 1+a 2=2λa 2+μa 1,得5=6λ+2μ. 又λ+μ=32,解得λ=12,μ=1.(12分)由a 1=2,a 2=3,λ=12,μ=1,代入S n =λna n +μa n -1得a 3=4,所以a 1,a 2,a 3成等差数列;由S n =n2a n +a n -1,得S n +1=n +12a n +1+a n .两式相减得a n +1=n +12a n +1-n2a n +a n -a n -1, 即(n -1)a n +1-(n -2)a n -2a n -1=0,所以na n +2-(n -1)a n +1-2a n =0.相减得na n +2-2(n -1)a n +1+(n -2)a n -2a n +2a n -1=0, 所以n(a n +2-2a n +1+a n )+2(a n +1-2a n +a n -1)=0,所以a n +2-2a n +1+a n =-2n (a n +1-2a n +a n -1)=22n (n -1)(a n -2a n -1+a n -2)=…=(-2)n -1n (n -1)…2(a 3-2a 2+a 1).(4分)因为a 1-2a 2+a 3=0,所以a n +2-2a n +1+a n =0, 即数列{a n }是等差数列.(16分)21. A. 解析:连结AD .因为AB 为圆O 的直径, 所以AD ⊥BD .因为EF ⊥AB ,所以A ,D ,E ,F 四点共圆, 所以BD ·BE =BA ·BF .(5分)因为∠EAF =∠BAC ,∠EF A =∠BCA =90°, 所以△ABC ∽△AEF ,所以AB AE =ACAF,即AB ·AF =AE ·AC ,所以BE ·BD -AE ·AC =BA ·BF -AB ·AF =AB ·(BF -AF )=AB 2.(10分) B. 解析:因为M =BA =⎣⎢⎡⎦⎥⎤4123⎣⎢⎡⎦⎥⎤100-1=⎣⎢⎡⎦⎥⎤4-12-3,(5分)所以M-1=⎣⎢⎡⎦⎥⎤310-11015-25.(10分) C. 解析:把直线方程l :⎩⎪⎨⎪⎧x =1+2t ,y =1-2t 化为普通方程为x +y =2.(3分)将圆C :ρ2+2ρcos θ-2ρsin θ=0化为直角坐标方程为x 2+2x +y 2-2y =0,即(x +1)2+(y -1)2=2.(6分)因为圆心C 到直线l 的距离d =22=2=r ,所以直线l 与圆C 相切.(10分)D. 解析:因为[(1+a )+(1+b )+(1+c )+(1+d )]⎝⎛⎭⎫a 21+a +b 21+b +c 21+c +d21+d ≥(1+a ·a 1+a +1+b ·b 1+b +1+c ·c 1+c +1+d ·d1+d )2=(a +b +c +d )2=1.(5分)又(1+a )+(1+b )+(1+c )+(1+d )=5, 所以a 21+a +b 21+b +c 21+c +d 21+d ≥15.(10分)22. 解析 :(1) 因为AB =1,AA 1=2,则F(0,0,0),A ⎝⎛⎭⎫12,0,0,C ⎝⎛⎭⎫-12,0,0,B ⎝⎛⎭⎫0,32,0,E(12,0,1), 所以AC →=(-1,0,0),BE →=⎝⎛⎭⎫12,-32,1.(2分)记直线AC 和EC 所成的角为α, 则cos α=|cos 〈AC →,BE →〉|= ⎪⎪⎪⎪-1×12⎝⎛⎭⎫122+⎝⎛⎭⎫-322+1=24, 所以直线AC 和BE 所成角的余弦值为24. (4分) (2) 设平面BFC 1的一个法向量为m =(x 1,y 1,z 1). 因为FB →=⎝⎛⎭⎫0,32,0,FC 1→=⎝⎛⎭⎫-12,0,2, 所以⎩⎨⎧m·FB →=32y 1=0,m·FC 1→=-12x 1+2z 1=0,则⎩⎪⎨⎪⎧y 1=0,x 1=4z 1,令x 1=4,则z 1=1,所以m =(4,0,1).(6分) 设平面BCC 1的一个法向量为n =(x 2,y 2,z 2). 因为CB →=⎝⎛⎭⎫12,32,0,CC 1→=(0,0,2),所以⎩⎨⎧n·CB →=12x 2+32y 2=0,n·CC 1→=2z 2=0,则⎩⎨⎧x 2=-3y 2,z 2=0,令x 2=3,则y 2=-1,所以n =(3,-1,0), 所以cos 〈m ,n 〉=4×3+(-1)×0+1×0(3)2+(-1)2+02×42+02+12=25117. 根据图形可知二面角FBC 1C 为锐角二面角, 所以二面角FBC 1C 的余弦值为25117.(10分)23. 解析 :(1) 因为抛物线C 的方程为y 2=4x , 所以点F 的坐标为(1,0).设M(m ,n),因为圆M 与x 轴、直线l 都相切,l 平行于x 轴, 所以圆M 的半径为|n|,点P(n 2,2n), 所以直线PF 的方程为y 2n =x -1n 2-1,即2n(x -1)-y(n 2-1)=0,(2分)所以|2n (m -1)-n (n 2-1)|(2n )2+(n 2-1)2=|n|. 所以|2m -n 2-1|=n 2+1.因为m ,n ≠0,所以n 2-m +1=0, 所以E 的方程为y 2=x -1(y ≠0).(4分) (2) 设Q(t 2+1,t),A(0,y 1),B(0,y 2).由(1)知,点Q 处的切线l 1的斜率存在,由对称性不妨设t>0.由y′=12x -1,所以k AQ =t -y 1t 2+1=12t 2+1-1,k BQ =t -y 2t 2+1=-2t 2+1-1, 所以y 1=t 2-12t,y 2=2t 3+3t ,(6分)所以AB =(y 2-y 1)2+0=|2t 3+3t -t 2+12t |=2t 3+52t +12t (t>0).(8分)令f(t)=2t 3+52t +12t,t>0,则f′(t)=6t 2+52-12t 2=12t 4+5t 2-12t 2.由f′(t)>0得t>-5+7324; 由f′(t)<0得0<t<-5+7324, 所以f(t)在区间⎝ ⎛⎭⎪⎫0,-5+7324上单调递减,在区间⎝ ⎛⎭⎪⎫-5+7324,+∞上单调递增,所以当t =-5+7324时,f(t)取得极小值也是最小值,即AB 取得最小值, 此时s =t 2+1=19+7324.(10分)。

最新江苏省连云港市2018届高三第一次模拟考试英语资料讲解

最新江苏省连云港市2018届高三第一次模拟考试英语资料讲解

2018届高三年级第一次模拟考试(七)英语(满分120分,考试时间120分钟)第一部分听力(共两节,满分20分)第一节(共5小题;每小题1分,满分5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

() 1. Where is the theater?A. On Martin's Lane.B. On Maple Street.C. On Craven Street.() 2. What seems to be the man's hobby?A. Watching TV.B. Reading books.C. Talking on the WeChat.() 3. What does the woman suggest the man do?A. Avoid on­sale things.B. Wait until the weekend.C. Get better shampoo.() 4. When will the speakers probably meet?A. On Wednesday afternoon.B. On Thursday afternoon.C. On Friday afternoon.() 5. What is the probable relationship between the speakers?A. Colleagues.B. Employer and employee.C. Customer and manager.第二节(共15小题;每小题1分,满分15分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

2018届高考英语适应性考试试题

2018届高考英语适应性考试试题

2018届高考英语适应性考试试题第I卷(选择题满分100分)第一部分听力(共两节,满分30分)第一节听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. Where does the conversation take place?A. In an elevator.B. On a bus.C. I n a taxi.2. What will the man do in Edinburgh?A. Do business with Justin.B. Tell Justin his new address.C. Give Justin the medicines.3. What kind of music does the man like?A. Jazz.B. Classical.C. Folk.4. When does the conversation take place?A. In September.B. In April.C. In February.5. Whose advice did the woman follow?A. The shop assistant’s.B. Her mother’s.C. Her sister’s.第二节听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听下面一段对话,回答第6和第7两个小题。

6. What are the speakers mainly discussing?A. Where to buy tickets.B. Where to park the car.C. Where to get a camera.7. Where will the speakers meet?A. At the market.B. At the camera shop.C. At t he sports stadium.听下面一段对话,回答第8和第9两个小题。

最新-连云港2018模拟试卷 精品

最新-连云港2018模拟试卷 精品

机密★启用前连云港市2018年中等学校招生统一考试物理试题(模拟)本试题分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。

第Ⅰ卷1至3页,第Ⅱ卷4至5页。

满分100分,考试时间100min。

第Ⅰ卷(选择题共45分)考生注意:1、答第Ⅰ卷前,考生务必在答题卡上用钢笔或圆珠笔填写自己的姓名、考试号,用2B铅笔填涂考试号、考试科目,同时在第Ⅰ卷右下角填写座位号。

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,用塑胶橡皮擦干净后,再选涂其它答案,答案答在试卷上无效。

3、考试结束,监考人员将本试卷和答题卡一并收回。

一、选择题(本大题共有20小题,每小题四个选项中只有一个正确。

1~15小题每题2分,16~20小题每题3分,共45分)1、如图1所示,某校初三同学正在举行升旗仪式。

该校旗杆的高度约为:()图3 图42所示是我国解放部队最新研制的潜艇模型。

当潜艇在继续下潜的过程中,其所受浮力和压强变化情况是:()A、浮力变大,压强变大;B、浮力变小,压强变小;C、浮力不变,压强不变;D、浮力不变,压强变大。

3、今年6月8日,当金星从地球与太阳之间经过时,我们会看到一个黑点慢慢移过太阳表面,这是一种罕见的天文现象——金星凌日。

造成这种天文现象的原因是:()A、光的直线传播;B、光的反射;C、光的折射;D、平面镜成像。

4、汽车在野外行驶时陷入泥坑,比较实用的解决办法是:()A、换新轮胎;B、在轮胎周围铺设碎石或稻草;C、在汽车的转动轴上加润滑油;D、将路面挖平。

5、古人称黄河是“一石水,六斗泥”。

经测定黄河水每立方米的平均含沙量约为35kg,合1cm3含沙量约为:()A、35g;B、3.5g;C、3.5×10-1g;D、3.5×10-2g。

6、市场稽查人员在检查农贸市场时,发现不法商贩有短斤少两现象。

你认为以下现象中不可..能.是商贩所为的是:()A、秤砣中灌铅;B、将秤砣掏空;C、秤盘下吸小磁铁;D、大秤小砣。

2018届江苏高考适应性训练(含答案)

2018届江苏高考适应性训练(含答案)
南京清江花苑严老师
1 tan x 的图像相交于 A, B, C 三点,则 3
1
N
曲线的离心率为 ▲ . 12.已知 AB 4 ,点 M , N 是以 AB 为直径的半圆上的任意两点,且
M
MN 2 , AM BN 1 ,则 AB MN
13 . 已 知 过 原 点 的 直 线

D E A
(第 21-A 题)
B.选修 4—2:矩阵与变换(本小题满分 10 分) 已知矩阵 A
2 5 3 b 的逆矩阵 A1 ,求矩阵 A 的特征值. 1 a c d
C.选修 4—4:坐标系与参数方程(本小题满分 10 分) 在极坐标系中,设 P 为曲线 C: 2 上任意一点,求点 P 到直线 l: sin 的最大距离.
2018 届江苏高考适应性训练 Ⅰ必做题部分
参考公式:棱锥的体积公式 V
棱锥
1 Sh ,其中为 S 棱锥的底面积, h 为棱锥的高. 3
一、填空题: (本大题共 14 小题,每小题 5 分,共计 70 分.不需写出解题过程,请把答案 直接填写在答题卡相应位置上 ) ........
3i ,则复数 z 的共轭复数 z ▲ i (其中 i 为虚数单位) 1 i 2.已知全集 U x x 8, x N* ,集合 A 2, 4,6,7 ,则 ð ▲ . UA
▲ . ▲ .
4a a 2 14.若实数 a, b 满足 , 则 a b 的值为 b log 2 b 1 2 2
说明、证明过程或演算步骤. 15.(本小题满分 14 分) 已知 , 0, 且 cos (1)若 tan
二、解答题:本大题共 6 小题,共计 90 分.请在答题纸指定区域 内作答,解答时应写出文字 .......

江苏连云港高级中学2018届高考考前适应性练习一 精品

江苏连云港高级中学2018届高考考前适应性练习一 精品

江苏连云港高中2018届高三数学考前适应性练习一1. 已知不等式x 2-2x -3<0的解集为A ,不等式x 2+x -6<0的解集是B ,不等式x 2+ax +b <0的解集是A∩B ,则a +b =_______2. 若双曲线122=-m y x 的一条渐近线方程是y =3x ,则m =______ 3. 函数y =21x 2-ln x 的单调递减区间为_________4. 若从集合{-1,1,2,3}中随机取出一个数m ,放回后再随机取出一个数n ,则使方程12222=+n y m x 表示焦点在x 轴上的椭圆的概率为_____ 5. 数列{a n }中,a 1=2i ,(1+i )a n +1=(1-i )a n (n ∈N *,i 为虚数单位),则a 10=____6. 样本容量为10的一组数据,它们的平均数是5,频率如条形图所示,则这组数据的方差等于______7. 函数f (x )=|lg x |+x -2的零点个数是_______8. 直线y =kx +3与圆(x -3)2+(y -2)2=4相交于M 、N 两点,MN ≥23,则k 的取值范围是________11. 如图,正方形ABDE 与等边△ABC 所在平面互相垂直,AB =2,F 为BD 中点,G 为CE 中点,⑴ 求证:FG ∥平面ABC ; ⑵ 求三棱锥F -AEC 的体积12. 某厂生产一种仪器,由于受生产能力和技术水平的限制,会产生一些次品.根据经验知道,该厂生产这种仪器,次品率T 与日产量x (件)之间大体满足关系:⎪⎪⎩⎪⎪⎨⎧∈><≤∈≤≤-=),(,32)961,,1(,961N x c x c N x c x x P (注:次品率P =生产量次品数,如P =0.1表示每生产10件产品,约有1件为次品.其余为合格品.).已知每生产一件合格的仪器可以盈利A 元,但每生产一件次品将亏损2A 元,故厂方希望定出合适的日产量, ⑴ 试将生产这种仪器每天的盈利额T (元)表示为日产量x (件)的函数;⑵ 当日产量x 为多少时,可获得最大利润?参考答案:1.-3 2.3 3.(0,1) 4.165 5.2 6.7.2 7.1 8.[-43,0] 9.1-2 10.22 11.(1)取AC 中点H ,连结BH,GH ,即可证明;(2)AEC F V -=332 12.(1)⎪⎩⎪⎨⎧∈><≤∈≤≤--=),(,0)961,,1(,)96(23N x c x c N x c x x Ax Ax T(2)当88=x 时,最大利润为125.5AG F E D C BA。

推荐-连云港市2018届高三第一次调研考试数学 精品

推荐-连云港市2018届高三第一次调研考试数学 精品

连云港市2018届高三第一次调研考试数学试题注意:1.本试题分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,满分150分.考试时间120分钟.2.请将第Ⅰ卷的答案和第Ⅱ卷的解答均填写在答题纸的对应地方,答在试题卷或草稿纸上不得分,考试结束时只交答题纸.3.答题前请将答题纸上密封线内的有关项目填写清楚,密封线内不能答题.第Ⅰ卷(选择题共60分)参考公式:如果事件A 、B 互斥,那么)()()(B P A P B A P +=+如果事件A 、B 相互独立,那么A P (·)()A P B =·)(B P如果事件A 在一次试验中发生的概率是P ,那么它在n 次独立重复试验中恰好发生k次的概率k n k k n n P P C k P --=)1()(球的表面积公式24R S π=球,其中R 表示球的半径. 球的体积公式334R V π=球,其中R 表示球的半径. 一、选择题:本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.请将正确答案的序号填在第Ⅱ卷前的选择题答题表中。

1.已知集合{|0A x x x =≤≤∈Z },则集合{|2,}B x x a a A ==∈,则集合A ∩B 等于A .{0, 2}B .{ 0,1 }C .{1, 2}D .{ 0 }2.已知tan cot 4+=αα,则sin 2α等于A .14B .14- C .12 D .12- 3.已知向量(1,2)=a ,b (2,)x =-,如果(3)+a b ∥(3)-a b ,则实数x 的值等于A .4B .4-C .2-D .1-4.不等式2||≤x 成立的一个必要非充分条件是A .1|1|≤+xB .2(1)4x +≤C .2(1)9x +≤D .(1)(2)0x x --≤5.若直线l 过点3(3,)2--且被圆2225x y +=截得的弦长为8,则此直线l 的方程是A .3x =-B .3x =-或32y =- C .34150x y ++= D .3x =-或34150x y ++=6.设,a b 是两条不同直线,,αβ是两个平同的平面,则下列四个命题:①若,a b a α⊥⊥,则b ∥α;②若a ∥α,αβ⊥,则a β⊥;③若αβ⊥,a β⊥,则a ∥α;④若,,a b a αb β⊥⊥⊥,则αβ⊥.则其中正确命题的序号是A .①B .②C .③D .④7.设双曲线2224kx ky -=的一条准线方程为4y =,则k 的值为A .124 B .124- C .16 D .16- 8.将函数()cos y f x x =的图象按向量a (,1)4π=平移,得到22sin y x =的图象,那么函数 ()f x 可以是A .cos xB .2cos xC .sin xD .2sin x9.已知函数22()log (3)f x x ax a =-+在区间[2,)+∞上是增函数,则实数a 的取值范围是A .(4,4]-B .(,4]-∞C .(,4]-∞-∪[2,)+∞D .(4,2]-10.已知等比数列1},{32=>a a a n ,则使不等式0)1()1()1(2211≥-++-+-nn a a a a a a 成立的最大自然数n 是A .4B .5C .6D .711.将3种农作物都种植在如图的4块试验田里,每块种植一种农作物,要求相邻的试验田不能种植同一种作物,则不同的种植方法共有A .6种B .12种C .18种D .24种12.现有一块正三棱锥形石料,其三条侧棱两两互相垂直,且侧棱长为1m ,若要将这块石料打磨成一个石球,则所得石球的最大半径约为A .0.18mB .0.21mC .0.24mD .0.29m第Ⅱ卷 (非选择题 共90分)二、填空题:本大题共6小题,每小题4分,共24分.把答案填在题中横线上.13.已知直线l 的方程为345x y -=,若直线l 上有相异两点关于直线(21)y mx m =--对称,则m 的值为 .14.若8()a x x +(a 是常数,且0a <)的展开式中常数项为70,则此展开式中各项系数的和是 .15.已知0,022ππαβ<<-<<,3cos()5αβ-=,且3tan 4α=,则sin β= . 16.已知圆C 的方程为2210x y ax ++-=,若(1,2)A ,(2,1)B 两点一个在圆C 的内部,一个在圆C 的外部,则实数a 的取值范围是 .17.在正方体1111ABCD A B C D -中,11AC 与11B D 交于点1O ,点P 在线段11AO 上运动,异面直线1AD 与BP 所成的角为θ,则θ的取值范围是 .18.设2()12f x x =-,x x x g 2)(2-=,若()()|()()|()22f xg x f x g x F x +-=-,则)(x F 的 最大值为__ .三、解答题:本大题共5小题,共66分,解答应写出文字说明,证明过程或演算步骤.F EG C 1B 1A 1CB A 19.(本小题12分)某企业生产的产品有一等品和二等品两种,按每箱10件进行包装,每箱产品均需质检合格后方可出厂.质检办法规定:从每箱产品中任抽4件进行检验,若二等品不超过1件,就认为该箱产品合格;否则,就认为该箱产品不合格.已知某箱产品中有2件二等品.(1)求该箱产品被某质检员检验为合格的概率;(2)若甲、乙两质检员分别对该箱产品进行质检,求甲、乙两人得出的质检结论不一致的概率.20.(本小题12分)已知向量(cos ,sin )(0)OA λαλαλ=≠,(sin ,cos )OB ββ=-,其中O 为坐标原点.(1)若6πβα=-,求向量OA 与OB 的夹角; (2)若||AB ≥2||OB 对任意实数,αβ都成立,求实数λ的取值范围.21.(本小题14分)在直三棱柱111ABC A B C -中,14AC CB AA ===, 90ACB ∠=︒,,E F 分别是1AA ,AB 的中点,点G在AC 上,且14CG CA =. (1)求证:1EF B C ⊥;(2)求二面角1F EG C --的正切值.(3)求点1A 到平面EFG 的距离.22.(本小题14分) 已知数列{}n a 的前三项与数列{}n b 的前三项对应相同,且212322a a a +++…12n n a -+ 8n =对任意的∈n N*都成立,数列1{}n n b b +-是等差数列.(1)求数列{}n a 与{}n b 的通项公式;(2)问是否存在k ∈N*,使得(0,1)k k b a -∈?请说明理由.23.(本小题14分)已知点 P 为圆224+=x y 上的动点,x PD ⊥轴,垂足为D ,线段PD 中点的轨迹为曲线C .过定点(,0)M t (02)<<t 任作一条与y 轴不垂直的直线l ,它与曲线C 交于A ,B 两点.(1)求曲线C 的方程;(2)试证明:在x 轴上存在定点N ,使得∠ANB 总能被x 轴平分;(3)对于(2)中的点N ,求四边形OANB 面积的最大值(用t 表示).2018届高三第一次调研考试数学试题参考解答与评分标准一、选择题:(每小题5分)1.A 2.C 3.B 4.C 5.D 6.D 7.B 8.D 9.A 10.B 11.C 12.B二、填空题:(每小题4分)13.43- 14.0 15.725-16.(4,2)-- 17.[,]63ππ 18.79 三、解答题:19.解:(1)从一箱产品中抽出4件,可能出现的结果数为410n C =.由于是任意抽取,这些结果出现的可能性都相等.因该箱产品中有2件二等品,故取到的二等品不超过1件的结果数为431882m C C C =+⋅.记“该箱产品被某质检员检验为合格”为事件A ,那么事件A 的 概率为410123848)(C C C C n m A P +== …………………………………………………………… 4分 1513=. 答:该箱产品被某质检员检验为合格的概率为1513. ……………………………… 6分 (2)记“甲、乙两质检员分别对该箱产品进行质检,甲的质检结论为合格”为B , “甲、乙两质检员分别对该箱产品进行质检,乙的质检结论为合格”为C .“甲、乙两人得出的质检结论不一致”包括两种情况:一种是甲的质检结论为合格,但乙的质检结论不合格;另一种是乙的质检结论合格,但甲的质检结论不合格.故所求的概率为131352()()2(1)1515225P B C P B C ⋅+⋅=⋅⋅-=. 答:甲、乙两人得出的质检结论不一致的概率为22552.……………………………12分 20.解:(1)设向量OA 与OB 的夹角为θ, 则sin()cos ||2|||||OA OB OA OB ⋅-===⋅λαβλθλλ, ……………………………2分 当0λ>时,1cos 2=θ,3=πθ; ……………………………4分 当0λ<时,1cos 2=-θ,23=πθ. 故当0λ>时,向量OA 与OB 的夹角为3π; 当0λ<时,向量OA 与OB 的夹角为32π. ……………………………6分 (另法提示:))3sin(),3(cos())2sin(),2(cos(απαπβπβπ++=++=,它可由向量OA 绕O 点逆时针旋转3π而得到,然后分0>λ和0<λ进行讨论.)N M H F E G C 1B 1A 1C B A (2)||2||AB OB ≥对任意的,αβ恒成立,即22(cos sin )(sin cos )4++-≥λαβλαβ对任意的,αβ恒成立,即212sin()4++-≥λλβα对任意的,αβ恒成立, ……………………………8分所以,20214>⎧⎨-+≥⎩λλλ或20214<⎧⎨++≥⎩λλλ, ……………………………10分 解得3≥λ或3≤-λ.故所求实数λ的取值范围是]3,(--∞∪),3[+∞. ……………………………12分 (另法一提示:由212sin()4++-≥λλβα对任意的,αβ恒成立,可得4||212≥-+λλ,解得3||≥λ或1||-≤λ,由此求得实数λ的取值范围;另法二提示:由|||||||||||||1|AB OB OA OB OA =-≥-=-λ,可得||AB 的最小值为|||1|-λ,然后将已知条件转化为|||1|2-≥λ,由此解得实数λ的取值范围)21.解:(1)连结1A B ,1BC ,∵,E F 分别是1,AA AB 的中点,∴EF ∥1A B ,且EF 112A B =. 在直三棱柱111ABC A B C -中,由11CC AA BC ==可知侧面11B BCC 是正方形,∴11B C BC ⊥.……2分∵BC AC ⊥,1CC AC ⊥,∴11AC ⊥平面11BCC B ,∴1A B 在平面11BCC B 上的射影是1C B ,由三垂线定理可得11A B B C ⊥,∴1EF B C ⊥. ……………………………………………………4分(2)取AC 的中点M ,连结FM ,则FM ⊥平面11ACC A . 作MN EG ⊥于点N ,连结FN ,由三垂线定理可知FN EG ⊥,∴MNF ∠为二面角F EG A --的平面角, ……………………………6分易知Rt EAG ∆∽Rt MNG ∆,∴AE MG MN EG ⋅==, ……………………………7分 在Rt FMN ∆中,求得tan FM MNF MN∠== ∴所求二面角1F EG C --的正切值为 ……………………………9分(3)在Rt FMN ∆中,作MH FN ⊥于点H ,由(2)可知,EG ⊥平面MNF ,∴MH EG ⊥,∴M H ⊥平面EFG ,M H 的长是点M 到平面EFG在Rt MNF ∆中,FM MN MH FN ⋅==, ……………………………12分 又3AG MG =,点E 是1AA 的中点, ∴点1A 到平面EFG ……………………………14分(另法提示:利用体积法,由EG A F EFG A V V 11--=求解.)解法二:(1)建立如图的空间直角坐标系O -xyz ,则)2,4,0(-E ,)0,2,2(-F ,)4,0,4(1B , ∴)2,2,2(-=EF ,)4,0,4(1=CB , …………………………………………2分 ∵·04)2(02421=⨯-+⨯+⨯=CB , ∴EF ⊥1CB ,即1CB EF ⊥.…………………4分(2)设n = (x ,y ,1)是平面EFG 的一个法向量,则有n ⊥, n ⊥,∴n ·EF =0, n ·=0,即0222=-+y x ,且023=-y , 解得32,31==y x ,故n = (32,31,1). ………6分 易知向量m =)0,0,1(是侧面11A ACC 的一个法向量.设向量m 与向量n 的夹角为θ, 则141||||cos =⋅=n m n m θ,∴13tan =θ, ……………………………………8分 而二面角1F EG C --的平面角大于直角,所以二面角1F EG C --的平面角与θ互补.故所求二面角的正切值为13-.……9分(3)设点1A 到平面EFG 的距离为d ,则d 等于向量在向量n 上的投影,…………11分即d=(n |n |)·)3,2,1(141=·1473146)2,0,0(==. 故点1A 到平面EFG 的距离为1473. ………………14分 22.解:(1)已知212322a a a +++…12n n a -+8n =(n ∈N*) ①2n ≥时,212322a a a +++…2128(1)n n a n --+=-(n ∈N*) ②①-②得,128n n a -=,求得42n n a -=,在①中令1n =,可得得41182a -==,所以42n n a -=(n ∈N*). ………………4分 由题意18b =,24b =,32b =,所以214b b -=-,322b b -=-,∴数列}{1n n b b -+的公差为2)4(2=---,∴1n n b b +-=2)1(4⨯-+-n 26n =-,121321()()()n n n b b b b b b b b -=+-+-++-(4)(2)(28)n =-+-++-2714n n =-+(n ∈N*). ………………8分(2)k k b a -=2714k k -+-42k -, 当4k ≥时,277()()24f k k =-+-42k -单调递增,且(4)1f =, 所以4k ≥时,2()714f k k k =-+-421k -≥, ………………12分又(1)(2)(3)0f f f ===,所以,不存在k ∈N*,使得(0,1)k k b a -∈. ………………14分A 1A23.解:(1)设(,)Q x y 为曲线C 上的任意一点,则点(,2)P x y 在圆224x y +=上,∴2244x y +=,曲线C 的方程为22 1 (0)4x y y +=≠. ………………2分 (2)设点N 的坐标为(,0)n ,直线l 的方程为x sy t =+, ………………3分代入曲线C 的方程2214x y +=,可得 222(4)240s y tsy t +++-=, ………………4分 ∵02t <<,∴22222(2)4(4)(4)16(4)0ts s t s t ∆=-+-=+->,∴直线l 与曲线C 总有两个公共点.(也可根据点M 在椭圆C 的内部得到此结论) 设点A ,B 的坐标分别11(,)x y , 22(,)x y , 则212122224, =44ts t y y y y s s --+=++, 要使ANB ∠被x 轴平分,只要0AN BN k k +=, 即12120y y x n x n+=--,1221()()0y x n y x n -+-=, ……………5分 也就是0)()(1221=-++-+n t sy y n t sy y ,12122()()0sy y t n y y +-+=, 即2224(2)2()044t ts s t n s s --⋅+-⋅=++,即只要0)4(=-s nt (*) ……………7分 当4n t =时,(*)对任意的s 都成立,从而ANB ∠总能被x 轴平分. 所以在x 轴上存在定点4(,0)N t ,使得∠ANB 总能被x 轴平分. ………………8分 (3)由(2)得1214(||||)2OANB OAN OBN S S S y y t ∆∆=+=⋅⋅+1214||2y y t=⋅-==令24s m +=,则OANB S = ……………11分 ∵]41,0(4112∈+=s m ,∴当2210<≤1时,OANB S 的最大值为24t;当221t >41时,OANB S 的最大值为所以,当0t <<时,OANB S2t <时,OANB S 的最大值为24t. ………………14分 (另法提示:(2)当直线斜率存在时,设直线l 的方程为()y k x t =-(0k ≠),代入曲线C 的方程,可得22222(14)8440k x k tx k t +-+-=,类似上述方法进行讨论.还需要讨论直线斜率不存在的情形.(3)1214||2OANB S y y t=⋅-=,令241k m +=, 转化成关于m 的函数并研究其最值.也要讨论直线斜率不存在的情形.)。

最新-连云港市赣榆县2018学年度高考临考适应性试卷物

最新-连云港市赣榆县2018学年度高考临考适应性试卷物

赣榆县2018—2018学年度高考临考适应性试卷物理试题本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,共150分.考试时间120分钟.一卷(选择题共40分)注意事项:1.答第一卷前,考生务必将自己的姓名、考试证号、考试科目用铅笔涂写在答题卡上。

2.第一卷答案必须填涂在答题卡上,在其他位置作答无效.每小题选出答案后,用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,再选涂其它答案.3.考试结束,将答题卡和第二卷一并交回.一、本题共10小题,每小题4分,共40分.在每小题给出的四个选项中,有的小题只有一个选项正确,有的小题有多个选项正确.全部选对的得4分,选不全的得2分,有选错或不答的得0分.1、下列说法正确的是:()A、布朗运动就是分子无规则运动B、大的颗粒在温度较高时有可能做无规则运动C、第一类永动机违反了能量守恒定律,第二类永动机没有违反能量守恒定律D、分子势能最小时,分子间引力与斥力大小相等2.下列说法正确的是()A.天然放射现象的发现,揭示了原子核是由质子和中子组成的B.卢瑟福的α粒子散射实验揭示了原子核有复杂结构C.玻尔的原子结构理论是在卢瑟福核式结构学说基础上引进了量子理论D.α射线、β射线、γ射线本质上都是电磁波3.一个质点做简谐运动,它的振动图像如图5所示,下列判断不正确的是()A.图中的曲线部分是质点的运动轨迹B.有向线段OA是质点在t1时间内的位移C.振幅与有向线段OA在x轴的投影之差等于质点在t1时间内的位移大小D.有向线段OA的斜率是质点在t1时刻的瞬时速率4、20世纪80年代初,科学家发明了硅太阳能电池。

如果在太空中设立太阳能卫星站,可24小时发电,利用微波-电能转换装置,将电能转换为微波向地面发送,微波定向性能好,飞机通过微波区不会发生意外,但对飞鸟是致命的,因为微波产生强涡流,那么,飞机外壳能使飞机里的人安然无恙的原因是()A反射 B吸收 C干涉 D衍射5 . 20 世纪50 年代,科学家提出了地磁场的“电磁感应学说”,认为当太阳强烈活动影响地球而引起磁暴时,磁暴在外地核中感应产生衰减时间较长的电流,此电流产生了地磁场.连续的磁暴作用可维持地磁场.则外地核中的电流方向为(地磁场N 极与S 极在地球表面的连线称为磁子午线)()A.垂直磁子午线由西向东 B.垂直磁子午线由东向西C.沿磁子午线由南向北 D. 沿磁子午线由北向南6、如图所示,a和b都是厚度均匀的平玻璃板,它们之间的夹角为γ,一条细光束由红光和蓝光组成,以人射角α从O点射人a板,且射出b板后的两束单色光通过空气射在地面上M、N两点,由此可知:A.若射到M、N两点的光分别通过同一双缝发生干涉现象,则射到M点的光形成干涉条纹的间距小,这束光为蓝光,光子的能量大B.若射到M、N两点的光分别通过同一双缝发生干涉现象,则射到M点的光形成干涉条纹的间距大,这束光为红光,光子的能量小C.射到N点的光为蓝光,光子的能量小,较容易发生衍射现象D.射到N点的光为红光,光子的能量大,较难发生衍射现象7.如图所示电路,闭合开关时,灯不亮,已经确定是灯泡断路或短路引起的.在不能拆开电路的情况下(开关可闭合,可断开),现用一个多用表的直流电压挡、直流电流挡和欧姆挡分别对故障作出如下判断(如表所示):A.只有1、2对B.只有3、4对C.只有1、2、4对D.全对8、在谷物的收割和脱粒过程中,小石子、草屑等杂物很容易和谷物混在一起,另外谷粒中也有瘪粒,为了将它们分离,可用扬场机分选,如图,它的分选原理是()A、小石子质量最大,空气阻力最小,飞得最远B、空气阻力对质量不同的物体影响不同C、瘪谷物和草屑质量最小,在空气阻力作用下,反向加速度最大,飞得最远D、空气阻力使它们的速度变化不同7.如图,从竖直面上大圆的最高点A ,引出两条不同的光滑轨道,端点都在大圆上,同一物体由静止开始,从A 点分别沿两条轨道滑到底端,则 A .到达底端的动量大小相等B .重力的冲量都相同C .动量的变化率都相同D .所用的时间都相同10.如图所示,在光滑的水平面上有质量相等的木块A 、B ,木块A 正以速度V 前进,木块B 静止。

江苏省达标名校2018年高考一月适应性考试物理试题含解析

江苏省达标名校2018年高考一月适应性考试物理试题含解析

江苏省达标名校2018年高考一月适应性考试物理试题一、单项选择题:本题共6小题,每小题5分,共30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.图1所示为一列简谐横波在某时刻的波动图象,图2所示为该波中x=1.5m 处质点P 的振动图象,下列说法正确的是A .该波的波速为2m/sB .该波一定沿x 轴负方向传播C .t= 1.0s 时,质点P 的加速度最小,速度最大D .图1所对应的时刻可能是t=0.5s2.如图甲所示,质量为0.5kg 的物块和质量为1kg 的长木板,置于倾角为37足够长的固定斜面上,0t =时刻对长木板施加沿斜面向上的拉力F ,使长木板和物块开始沿斜面上滑,作用一段时间t 后撤去拉力F 。

已知长木板和物块始终保持相对静止,上滑时速度的平方与位移之间的关系如图乙所示,sin 370.6=,cos370.8=,210m/s g =。

则下列说法正确的是( )A .长木板与斜面之间的动摩擦因数为10.35μ=B .拉力F 作用的时间为2s t =C .拉力F 的大小为13ND .物块与长木板之间的动摩擦因数μ2可能为0.883.五星红旗是中华人民共和国的象征和标志;升国旗仪式代表了我国的形象,象征着我国蒸蒸日上天安门广场国旗杆高度为32.6米,而升国旗的高度为28.3米;升国旗时间与北京地区太阳初升的时间是一致的,升旗过程是127秒,已知国旗重量不可忽略,关于天安门的升国旗仪式,以下说法正确的是( )A.擎旗手在国歌刚刚奏响时,要使国旗在升起初始时,旗面在空中瞬间展开为一平面,必须尽力水平向右甩出手中所握旗面B.国旗上升过程中的最大速度可能小于0.2m/sC.当国旗匀速上升时,如果水平风力大于国旗的重量,则国旗可以在空中完全展开为一个平面D.当国旗匀速上升时,如果水平风力等于国旗的重量,则固定国旗的绳子对国旗的作用力的方向与水平方向夹角45度4.某实验小组模拟远距离输电的原理图如图所示,A、B为理想变压器,R为输电线路的电阻,灯泡L1、L2规格相同,保持变压器A的输入电压不变,开关S断开时,灯泡L1正常发光,则()A.仅将滑片P上移,A的输入功率不变B.仅将滑片P上移,L1变暗C.仅闭合S,L1、L2均正常发光D.仅闭合S,A的输入功率不变5.有趣的“雅各布天梯”实验装置如图中图甲所示,金属放电杆穿过绝缘板后与高压电源相接。

2018届高考考前适应性试卷_1

2018届高考考前适应性试卷_1

2018届高考考前适应性试卷注意事项:1、答题前,考生务必将自己的姓名、准考证号填写在答题卡上。

2、回答选择题时,选出每小题答案后,用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,再选涂其它答案标号。

回答非选择题时,将答案写在答题卡上,写在本试卷上无效。

3、考试结束后,请将本试题卷和答题卡一并上交。

第Ⅰ卷第一部分听力(共两节,满分 30 分)(略)第二部分阅读理解(共两节,满分40分)第一节(共15小题:每小题2分,满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中选出最佳选项,并在答题卡上将该项涂黑。

AHow can you gain knowledge of the world without books or te achers? Visiting various kinds of world-famous museums may be your best choice. If you are interest ed ,here are 3 famous museums well worth visiting.National Gallery, London 2017 attendance: 6,263,000The National Gallery originally opened in 1824 in a Pall Mall t ownhouse with just 38paintings. Today’s ollction contains ove r 2, 300 works from the 13th to 19th centuries, which you can appreciate free of charge. The works range in style and time p eriod, but hey, everyone needs to see Van Gogh’s Sunflowers in Room 43 at least once, for which the Museum is famous. Opening time: daily 10 am—6 pm Friday 10 am—9 pm Closed time: January and 24—26 DecemberAdmission: FreeNational Air and Space Museum, Washington, D. C2017 attendance: 7,500,000The National Air and Space Museum, part of the Smithsonian Institution, hosts the world’s largest collection of aviation and space artifacts. You’ll see crowds flocking toward the 1903Wri ght Flyer and Apollo 11command module, but don’t forget abo ut NASA’s often—overlooked Stardust probe.Hours and Admission :Open every day except December 25. Free admission and tickets (free) are needed.1. A maximum of 4000 free tickets will be issued each day at t he museum.2. Each visitor can once collect one free ticket of the day withvalid ID (Passport ).Regular Hours :10:00 am to 5:30 pmExtended Hours :10:00 am to 7:30pm Fridays, Saturdays and SundaysNational Museum of China, Beijing2017 attendance: 7,550,000Founded in February 2003, this huge, well-curated museum sits on the edge of Beijing’s Tiananmen Squ are and is free to enter. It houses more than 1.3 million exhibit ion pieces in its 40halls ,focused primarily on China’s achievements in history, culture, and art.Opening Hours :9 :00—17:00(No Entry after 16:00) Closed on MondaysMuseum Visiting Tips: Limit the number of bags( only one han dbag is allowed ). All visitors are screened through metal dete ctor upon entry. The fewer items you bring inside the Museum . Before you visit, please review the list of prohibited items, Vi sitors carrying prohibited items will not be allowed inside the Museum, s0 please leave them at home or in your car.No Food or Drink: Only bottled water is permitted in the Muse um21. Which of the following museums had the largest number of visitors in 2017?A. National Air and Space Museum, Washington, D. C.B. National Gallery, London.C. British Museum, London.D. National Museum of China, Beijing.22. What is considered the most valuable treasure kept in Nat ional Gallery, London?A. Leonardo da Vinci’s Mona Lisa.B. The paintings from the 13th to 19th centuries.C. Van Gogh’s Sunflowers.D. The 38 original paintings collected in the museum.23. What can you bring with you while visiting the museums?A. A bottle of water.B. A can of beer.C. Two large bags.D. Knives.【答案】21. D 22. C 23. A【解析】本文介绍了三个世界闻名的博物馆。

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江苏连云港高中2018届高三数学考前适应性练习一
1. 已知不等式x 2-2x -3<0的解集为A ,不等式x 2+x -6<0的解集是B ,不等式x 2+ax +b <0的解集是A ∩B ,则a +b =_______
2. 若双曲线122
=-m y x 的一条渐近线方程是y =3x ,则m =______ 3. 函数y =2
1x 2-ln x 的单调递减区间为_________
4. 若从集合{-1,1,2,3}中随机取出一个数m ,放回后再随机取出一个数n ,则使方程122
22
=+n y m x 表示焦点在x 轴上的椭圆的概率为_____ 5. 数列{a n }中,a 1=2i ,(1+i )a n +1=(1-i )a n (n ∈N *,i 为虚数单位),则a 18=____
6. 样本容量为18的一组数据,它们的平均数是5,频率如条形图所示,则这组数据的方差等于______
7. 函数f (x )=|lg x |+x -2的零点个数是_______
8. 直线y =kx +3与圆(x -3)2+(y -2)2=4相交于M 、N 两点,MN ≥23,则k 的取值范围是________
18. 如图,正方形ABDE 与等边△ABC 所在平面互相垂直,AB =2,F 为BD 中点,G 为CE 中点,⑴ 求证:FG ∥平面ABC ; ⑵ 求三棱锥F -AEC 的体积
18. 某厂生产一种仪器,由于受生产能力和技术水平的限制,会产生一些次品.根据经验知道,该厂生产这种仪器,次品率T 与日产量x (件)之间大体满足关系:
⎪⎪⎩⎪⎪⎨⎧∈><≤∈≤≤-=),(,3
2)961,,1(,961N x c x c N x c x x P (注:次品率P =生产量次品数,如P =0.1表示每生产18件产品,约有1件为次品.其余为合格品.).已知每生产一件合格的仪器可以盈利A 元,但每生产一件次品将亏损2
A 元,故厂方希望定出合适的日产量, ⑴ 试将生产这种仪器每天的盈利额T (元)表示为日产量x (件)的函数;
⑵ 当日产量x 为多少时,可获得最大利润?
参考答案:
1.-3 2.3 3.(0,1) 4.
165 5.2 6.7.2 7.1 8.[-4
3,0] 9.1-2 18.22 18.(1)取AC 中点H ,连结BH,GH ,即可证明;
(2)AEC F V -=3
32 18.(1)⎪⎩
⎪⎨⎧∈><≤∈≤≤--=),(,0)961,,1(,)96(23N x c x c N x c x x Ax Ax T
(2)当88=x 时,最大利润为185.5A
G F E D C B
A。

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