2019-2020学年山西省运城市临猗县临晋中学高一下学期开学复课摸底考试物理试题

山西省运城市临猗县临晋中学2019-2020学年高一下学期开学复课摸底考试英语试题一.阅读理解（每小题3分，共60分）APalaces are known for their beauty and splendor, but they offer little protection against attacks. It is easy to defend a large building, but usually these buildings are not designed with the comfort of a king in mind. When it comes to structures that are both beautiful and defensive, the European castle is a big success.Castles were originally built in England by the Normans in 1066.They built towers and walls to secure the land they had taken. These castles provided the Normans with a quiet and safe place. They also served as bases of operation for attacks. In this way castles served both defensive and offensive roles. Besides, castles served as offices for governors. Those that were socially beneath the governor would come to report affairs and express their respect. They would address problems, handle business, feast, and enjoy festivities in castles. So castles served as social centers as well.The first castles were built from earth and wood, and they were likely to suffer from attacks by fire. Then wooden castles were gradually replaced by stone, which greatly increased the strength of these towers and walls. However, attackers could throw flaming objects into castles through the windows or burn the wooden doors. This led to moving the windows and entrances off from the ground floor and up to the first floor to make them more difficult to access.During the Middle Ages, attacks increased in regularity, so castle defenses were updated.Arrow-slits were added. They were small holes in the castle, which allowed defenders to fire without being hurt. Towers were built from which defenders could provide fire on both sides. The towers were connected to the castle by wooden bridges, so that if one tower fell, the rest of the castle was still easy to defend. A lot of rings of castle walls were constructed, so that even if attackers went past one wall, they would be caught on a killing ground between inner and outer walls. All of these increased the defense of castles.The end of castles can be attributed to gunpowder. During the 15th century, artillery, a kind of large guns, became powerful enough to break through stone walls. This greatly made the role of castles less effective. Though castles no longer serve their original purposes, remaining castles receive millions of visitors each year who wish to experience the situations of ancient times.1.What was the original function of castles according to the passage?A.They served as tourist attractions.B.They were important social centers.C.They marked religious ceremonies.D.They were built for use in emergencies.2.The reason why wooden castles were replaced by stone castles was that .A.stone castles cost less moneyB.stone castles offered better defenseC.wooden castles were uncomfortableD.wooden castles took a long time to build3.Which of the following showed an improvement in castle defenses?A.Castles were totally separated by stones.B.Arrow-slits were made in large quantities.C.Rings of walls were built to defend the towers.D.Windows and entrances were moved to the higher floor.4.What is the best title of the passage?A.Fancy Living: Learning about CastlesB.Normans: Bringing Castles to EnglandC.A History of Castles: The Rise and Fall of CastlesD.Defending Castles: Technologies Used to Defend CastlesBKangaroo Island is more famous for what it hasn’t got than what it has. Unlike mainland Australia,rabbits and foxes were never introduced,so the local wildlife has less competition for survival. Another thing the island doesn’t have is something a Scottish friend once called,“The Way of the Island.” There is no sense of being on the island;in many places,you can climb to a high point and see nothing but bushes in any direction.That catches many visitors out when they arrive with the idea of walking or cycling around the island. It’s really too big for that. There is little in the way of public transport,and the most practical ways to get around are to sign up for an organised tour or hire a car.The island’s main road,the Playford Highway,h eads westward from Nepean Bay until it meets the West End Highway in the middle of nowhere. The West End Highway heads south;the Playford Highway continues westward as a dirt road to Cape Borda,where there is a lighthouse,a few former light keepers’ cottag es,a National Park Visitor Centre and nothing else. Much of the island is farmland;lamb and beef from Kangaroo Island are especially prized. Fruit is also a major export,but the grapes in the vineyards (葡萄园) are turned into very acceptable wine first.19 N ational Parks and Conservation Areas cover over 30% of the island’s area, and the biggest of them lies to the southwest of the island,where the Flinders Chase National Park covers 32,600 hectaresof bush land. Entry is by permit only,which we bought at the park entrance at Rocky River. There,we wondered if the park authorities had the wildlife trained,because we saw a family of kangaroos right in the car park by the park office,but they only appeared after we paid the entrance fee!5.Why is the wildlife on Kangaroo Island more likely to survive?A.Because there are no rabbits or foxes on the island.B.Because rabbits and foxes are less competitive species.C.Because nobody lives on the island.D.Because the island is famous for its bushes.6.One of the most practical ways to get around on the island is to .A.rent a carB.ride a bikeC.take public transportD.sign up for a travel lesson7.What can you find at Cape Borda?A.The Playford Highway.B.The West End Highway.C.A few vineyards.D.Some former light keepers’cottages.8.The author and his family saw a family of kangaroos at the Flinders Chase National Park after they .A.talked to the park authoritiesB.paid the entrance feeC.went into the park officeD.arrived at Rocky RiverCFarmers, especially in developing countries, are often criticized (批评) for cutting down forests. But a new study suggests that many farmers recognize the value of keeping trees.Researchers using satellite photos found at least ten percent of trees cover on more than one billion hectares of farmland. That is almost half the farmland in the world. Earlier estimates (估计) were much lower and incomplete. The authors of the new study say it may still underestimate the true extent (程度) worldwide.The study found most trees cover in South America. Next comes Africa south of the Sahara, followed by Southeast Asia. North Africa and West Asia have the least.The study found that climate conditions alone could not explain the number of trees covers in different areas, nor could the size of nearby populations, meaning people and trees can live together. There are areas with few trees but also few people, and areas with many trees and many people. The findings suggest that things like land rights, markets or government policies can influence tree planting and protection.Dennis Garrity, who heads the World Agroforestry Center, says farmers are acting on their own to protect and plant trees. The problem, he says, is that policy makers and planners have been slow to recognize this and to support such efforts.The satellite photos may not show what the farmers are using the trees for, but we all know trees offer nuts, fruit, wood and other products. They also help prevent soil loss and protect water supplies. Even under drought (干旱) conditions, trees can often offer food and can be a way to earn money until the next growing season.Some trees act as natural fertilizers. They take nitrogen (氮气) out of the air and put it in the soil. Scientists at the Center say the use of fertilizer trees can reduce the need for chemical nitrogen by up to three-fourths. Trees also capture carbon dioxide, a gas connected to climate change.9.Through the study, the researchers found that.A.there are more trees on farmland than expectedB.fewer trees are being cut in developing countriesC.most farmers still don’t realize the value of treesD.trees play a key role in preventing climate change10.Which of the following has the least tree cover?A.Southeast Asia.B.West Asia.C.South America.D.Africa south of the Sahara.11.In Dennis Garrity’s opinion, .A.most farmers care about nothing but their own interestsB.there are usually few people living in areas with few treesernment plays a small role in tree planting and protectionernment should support farmers in planting and protecting trees12.What does Paragraph 6 mainly talk about?A.How farmers plant trees.B.What products trees can bring.C.The importance of trees to farmers.D.The environmental value of tree cover.DTwo billion children in the developing world can’t receive good education—the key to human development. However,technology offers an answer which allows the poor in developing countries to learn. It is a tool which holds the ability to change the lives of the poor,as it provides a means of learning and communicating.Educational programs must break away from old systems. New companies such as One Laptop Per Child (OLPC),an organization founded by MIT Professor Nicholas Negroponte have been active in solving the world’s education problem.The Internet has changed the world,allowing educational services to help with the global fight against poverty. Khan Academy is one such service. Like OLPC,it is an organization founded by Harvard Business School graduate Salman Khan with the task of “providing a free, world-class education for anyone,anywhere”.The education offered includes a large number of math-related topics.The GMAT Pill Review is another company that trains MBA candidates(应考人)worldwide to prepare for the GMAT exam on both the Quant and Verbal section. Founded by Stanford graduate Zeke Lee,the company offers services which are priced at about 75% less than other similar programs. It allows students from developing countries who might not be able to afford similar courses to have access to these services.Whether paid or free these services provide educational opportunities for those who would never have had the chance in the past. As a result of the technology revolution (革新),business schools may see more students from different corners of the world. Because of the Internet,people in developing nations have access to better and more affordable educational opportunities. More and more people will try to improve their lives through educational opportunities outside of their homeland—an idea usually uncommon in developing countries.13.According to the passage,technology can .A.improve people’s reading abilityB.solve the problems facing the whole worldC.help the poor to reduce povertyD.cause poverty in developing countries14.We can learn from the passage that Khan Academy aims to .A.help the poor pay for good coursesB.provide every child with a computerC.help train MBA candidatesD.make everyone receive good education15.Which of the following about GMAT Pill Review is TRUE?A.It only accepts students from developing countries.B.It was started by Professor Nicholas Negroponte.C.It charges students about 25 percent of other similar programs.D.The education it offers includes a lot of math-related topics.E阅读七选五(核心素养:语言能力)The world itself is becoming much smaller by using modern traffic and modern communication means. Life today is much easier than it was hundreds of years ago, but it has brought new problems. 16 Pollution means making things dirty. It comes in many ways. We see it, smell it, drink it and even hear it.Man has been polluting the earth. 17 Many years ago, the problem was not so serious because there were not so many people. When the land was used up or the river was dirty in one place, Man moved to another place. But this is no longer true. Man is now slowly polluting the whole world.18 It’s bad for all living things in the world, but it is not the only one kind of pollution. Water pollution kills our fish and pollutes our drinking water. Noise pollution makes us angry more easily.19 They stop people from burning coal in houses and factories in the city, and from putting dirty smoke into the air.The earth is our home. We must take care of it. 20 And we must take care of the rise in population at the same time.A.One of the biggest is pollution.B.The more people, the more pollution.C.Air pollution is still the most serious.D.That means keeping the land, water and air clean.E.Many countries are making rules to fight pollution.F.I hope scientists can find ways to solve the serious problem.G.Strange diseases have appeared in some places because of pollution.二.完形填空(每小题2分，共40分)Andy was still traveling in Spain when he realized he had to confirm (核实) his flight home with the airline company. He was visiting Spain in order to 1 his Spanish. When he was speaking to people 2 he had no 3 understanding what they said. 4 , when he was speaking on the phone, he had a(n) 5 . Andy 6 the airline. And the clerk confirmed that his plane was leaving at nine o’clockthree days from that day. She 7 told Andy to be at the airport two hours 8 in order to check in his luggage and get a seat.Since he was 9 in three days,Andy didn’t10 any time. He visited as many places as he could. He thought that it would probably be a while before he had enough money again. He wished he could 11 and spend a year in Spain.Too 12 ,the final day arrived. Andy left early for the airport to arrive two hours before take-off. He hated to 13 .He went to the clerk to 14 his ticket. The clerk looked at the ticket with 15 .“Why,sir,but your flight was at nine o’clock in the morning,and 16 it is eight in the evening.” “But I confirmed my flight,”Andy 17 . “Will I have to pay for another ticket?”“No,sir. However,the next flight out will be three days from now.”Andy’s 18 of shock turned to one of 19 as he realized that now he could continue his 20 .1.A.prepare B.improve C.enjoy D.contact2.A.slowly B.in public C.face to face D.fluently3.A.difficulty B.idea C.mistake D.interest4.A.Instead B.In a word C.At the same time D.However5.A.sickness B.anger C.problem D.change6.A.called B.liked C.trusted D.asked7.A.again B.also C.only D.once8.A.before B.earlier ter D.after9.A.moving B.returning C.staying D.leaving10.A.take B.have C.lose D.find11.A.wait B.go home C.stop e back12.A.shortly B.quickly C.badly D.fully13.A.speak B.go C.rush D.delay14.A.buy B.present C.order D.provide15.A.astonishment B.patience C.respect D.delight16.A.maybe B.so C.here D.now17.A.insisted B.apologized C.smiled D.desired18.A.experience B.expression C.look D.thought19.A.pleasure fort C.sadness D.hopelessness20.A.plan B.flight C.business D.vacation三.语法填空(每小题1.5分，共15分)One quarter of the population of Sichuan province and Chongqing municipality 1. (work) away from their homes as migrant workers. But the number 2. _______(have) been dropping this year.24-year-old Gu Qingxia 3. worked as a cashier before, together with her husband, 4. (be) to job fairs (招聘会) in Chongqing for many times and 5. ___________(want) to find jobs near hometown, because they want to come back to take care of their parents. This year, the number of migrant workers, like Gu Qingxia and her husband who 6. (choose) to return home for work 7. (be) increasing. Besides family, with rapid development of southwestern China, a large number of employment opportunities 8. (give) to migrant workers and they can earn as much as 9. they earn 10. Shenzhen.四根据所给提示用适当单词或短语的正确形式完成句子(共10 小题；每小题l.5 分，满分15 分)1.Crime often exists with poverty, like stealing driven by (hungry).2.The two cities are twinned with each other because of many (similar).3.Louis Cha Leung-yung, one of the most (influence) Chinese novelists, andbetter known under the pen name Jin Yong, died on 30th October.4.The hotel, beautifully (situate) in a quiet spot near the river, creates an atmosphere which visitors find so appealing.5... (strike) by the hurricane, the area had power cuts, but quickly it was back to normal.6.He came round for a coffee and we7.Ancient China was a place where states were often战)having a meal together. (以……结束)each other. (与……交8.Thankfully, the number of people who get infected is decreasing, and more and more patientsare cured and return home safe and sound; , everything will be fine.(简言之)9.Concerned about the continuing pollution of the environment, wetake action right away instead of letting matters take their course. (除...别无选择)10.____ , there are 800 tornadoes in the US each year, causing about 80 deaths and 1,500 injuries. (平均来说)五书面表达(满分20 分)假定你是李华，你校外教Kevin 对中国传统文化很感兴趣。

山西省运城市临猗县临晋中学2019-2020学年高一语文下学期开学复课摸底考试试题（考试时间120分钟总分120分）一、现代文阅读（35分）（一）论述类文本阅读（本题共3小题，9分）阅读下面的文字，完成1—3题。

《红楼梦》中的服饰有一些是汉族历代传承的服饰，也有很多是清代人的穿着。

黛玉初至荣国府时，见到王熙凤穿着“缕金百蝶穿花大红洋缎窄裉袄，外罩五彩刻丝石青银鼠褂”。

袭人要回家探亲时也是在“桃红百子刻丝银鼠袄子”外，再套上“青缎灰鼠褂”。

徐珂《清稗类钞·服饰》中说：“褂，外衣也，礼服之加于袍外者。

”赵振民《中国衣冠中之满服成分》索性认定：“中国古无‘褂’字……盖满制也。

”应该说褂子是清代人对肥大上衣的习惯称谓。

贾宝玉是作者着墨最多、寄托最深的人物之一。

他的服饰特色体现在红色上。

初见黛玉时，服饰以红为主色；群芳夜宴时，枕着红香花枕；祭奠晴雯时，穿着血点般大红裤子：看破红尘出家时，身披“大红猩猩毡的斗篷”。

这是因为红色是最能体现宝玉个性特征和心理状态的颜色。

红色有强烈的视觉效果，具有令人产生激动、热烈的本性和感情的力量，与他热情奔放的性格暗合。

红色也体现了他尊重女性的心理特征。

中国传统文化中，红色常是女性的代名词，古代男子常称其女性好友为“红颜知己”。

宝玉非常推崇、爱慕和关心女性，红色岂不正是他怡悦红颜的绝妙注解？红色还是他反抗封建礼教、追求个性解放的象征。

他虽被视为掌上明珠，却处处受羁绊，恨不能挣脱封建礼教的束缚。

而红色常使人联想到火焰和鲜血的颜色，是充满刺激性和令人振奋的色彩，它能使人产生积极进取的思想，这正是宝玉性格的写照。

服饰不但能体现人的外貌和性格，在阶级社会里，更能显示人的地位。

贾雨村还是寄住在葫芦庙内的一个穷儒时，曹雪芹只用“敝巾旧服”四个字就画出了这个落魄文人的潦倒。

邢岫烟虽为小姐，无奈家道艰难，只能穿着“家常旧衣”，连冬天也穿得“很单薄”。

对那些美貌而富贵的人，曹雪芹则不惜浓墨重彩。

山西省运城市临猗县临晋中学2019-2020学年高一化学下学期开学复课摸底考试试题（考试时间 90分钟）一、选择题（本题18个小题，每小题3分，共54分）1. 如图是部分短周期元素化合价与原子序数的关系图，下列说法正确的是（）A.原子半径：Z＞Y＞XB. Y和Z 两者最高价氧化物对应的水化物能相互反应C. 气态氢化物的稳定性：R＜WD. WX3和水反应形成的化合物是离子化合物2. A、B、C、D、E是同一周期的五种主族元素，A和B的最高价氧化物对应的水化物均呈碱性，且碱性B＞A，C和D的气态氢化物的稳定性C＞D；E是这五种元素中原子半径最小的元素，则它们的原子序数由小到大的顺序为（）A. A、B、C、D、EB. E、C、D、B、AC. B、A、D、C、ED. C、D、A、B、E3．科学家合成出了一种新化合物(如图所示),其中W、X、Y、Z为同一短周期元素,Z核外最外层电子数是X核外电子数的一半。

下列叙述正确的是( )。

A. WZ的水溶液呈碱性B. 元素非金属性的顺序为X>Y>ZC. Y的最高价氧化物的水化物是中强酸D. 该新化合物中Y不满足8电子稳定结构X Y4．四种短周期元素X、Y、Z和W在周期表中的位置如图所示，原子Z W序数之和为48。

下列说法不正确的是（）A．原子半径(r)大小比较r(X)＞r(Y)B．X和W可形成共价化合物XW3C．W的非金属性比Z的强，所以W氢化物的沸点比Z的高D．Z的最低价单核阴离子的失电子能力比Y的强5. 根据元素周期表和元素周期律，判断下列叙述不正确．．．的是( )A．气态氢化物的稳定性：H2O＞NH3＞SiH4B．氢元素与其他元素可形成共价化合物或离子化合物C．图1所示实验可证明元素的非金属性：Cl＞C＞SiD ．用中文“”（ào）命名的第118号元素在周期表中位于第七周期0族6. 某学生用如图所示装置进行化学反应X+2Y==2Z能量变化情况的研究。

山西省运城市高中联合体2019-2020学年高一下学期第一次摸底考试英语试题本试卷总共分为六部分。

满分150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. Why did the woman ask the man for help?A. Because she got lost.B. Because she was not herself.C. Because she wanted to buy some books.2. What does the man think of diet?A. It’s a foolish action.B. It helps to lose weight.C. It will make one feel weak.3. What does the woman mean?A. She hopes for an early reply.B. There is no need for the man to hurry.C. The man can read the menu as long as he likes.4. What’s the probable relationship be tween the two speakers?A. Doctor and patient.B. Assistant and customer.C. Driver and passenger.5. Where does this conversation probably take place?A. In a hotel.B. In a restaurant.C. In a bank.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。

山西省运城市高中联合体2019-2020学年高一物理下学期第一次摸底考试试题一、选择题（本题共12小题，每小题4分，共48分。

在每小题给出的四个选项中，第1～8题只有一项符合题目要求，第9～12题有多项符合题目要求，全部选对的得4分，选对但不全的得2分，有选错的得0分）1. 关于曲线运动，下列说法正确的是（ ） A. 做曲线运动的物体速度大小一定变化 B. 受到合力的大小和方向肯定有一个在变化 C. 加速度恒定的运动不可能是曲线运动 D. 曲线运动有可能是匀变速运动2. 下列关于万有引力定律说法正确的是（ ）A. 万有引力定律用来研究宏观物体间的相互作用，微观原子、分子间不存在万有引力作用。

B ．依据万有引力定律公式可知，物体间的万有引力随距离的增大而减小，随距离的减小而增大，当物体质心间距离趋向于零时，万有引力会达到无穷大。

C ．生活中的我们站在一起时并没有因为万有引力而紧紧粘贴，是因为万有引力定律并不适用于生命体。

D ．牛顿提出一切物体之间都存在万有引力，并给出关系式。

但是直到卡文迪许的出现，才使得他的理论得到了完美的验证。

3. 如图所示，站在岸边的人通过拉动绳子使得小船靠岸。

已知船的质量为m ，水的阻力恒为F f ，当轻绳与水平面的夹角为θ时，船的速度为v ，若此时人水平拉动绳子的力大小记为F ，则( ) A. 船的加速度为fF F m-B. 船的加速度为mF F f-θcos /C. 人拉绳行走的速度为v cos θD. 人拉绳行走的速度为cos vθ4. 如图所示，窗子上、下沿间的高度H ＝1.6 m ，墙的厚度d＝0.4 m ，某人在离墙壁距离L ＝1.4 m 、距窗子上沿h ＝0.2 m 处的P 点，将可视为质点的小物件以v 的速度水平抛出，小物件直接穿过窗口并落在水平地面上，取g ＝10 m/s 2．则v 的取值范围是( )A. v >7 m/sB. v <2.3 m/sC. 3 m/s<v <7 m/sD. 2.3 m/s<v <3 m/s 5. 如图所示，一小球（可视为质点）从一半圆轨道左端A 点正上方某处开始做平抛运动，运动轨迹恰好与半圆轨道相切于B 点．半圆轨道圆心为O ，半径3R =m ，且OB 与水平方向夹角为60°，重力加速度g =10 m/s 2，则小球抛出时的初速度大小为( ) A. 1 m/sB. 2 m/sC. 2．5 m/sD. 3 m/s6. 质量为m 的物体沿着半径为r 的半球形金属球壳滑到最低点时的速度大小为υ，如图所示，若物体与球壳之间的摩擦因数为μ，则物体在最低点时的（ ）A. 向心力为m （g+2v r）B. 对球壳的压力为2mv rC ．受到的摩擦力大小为μm gD. 受到的摩擦力大小为μm（g+2v r）7. 如图所示，在半径为R 的半圆形碗的光滑表面上，一质量为m 的小球以转数n 转每秒在水平面内作匀速圆周运动，该平面离碗底的距离h 为（ ） A. R -224g n π B.224g n πC.24gR n π- D.224gn π+2R 8. 嫦娥五号探测器是由中国空间技术研究院研制的中国首个实施无人月面取样返回的航天器。

物理试题一．选择题（共12小题，每小题4分，共48分，其中9--12为多选，选不全得2分，错选得0分）1．如图是体育摄影中“追拍法”的成功之作，摄影师眼中清晰的滑板运动员是静止的，而模糊的背景是运动的，摄影师用自己的方式表达了运动的美．请问摄影师选择的参考系是（）A．大地B．太阳C．滑板运动员D．步行的人2．下列情况中的物体，哪些可以看作质点（）A．研究从北京开往上海的一列火车的运行速度B．研究汽车后轮上一点运动情况的车轮C．体育教练员研究百米跑运动员的起跑动作D．研究地球自转时的地球3．运动员参加110米栏比赛，11秒末到达终点的速度为12m/s，则全程的平均速度是（）A．10 m/s B．11 m/s C．6 m/s D．12 m/s4．一个做匀加速直线运动的物体，初速度为2.0m/s，它在第3s内通过的位移是4.5m，则它的加速度为（）A．0.5 m/s2B．1.0 m/s2C．1.5 m/s2D．2.0 m/s25．一质点的x－t图像如图所示，那么此质点的v－t图像可能是下图中的( )6.某物体的运动规律如图所示，下列说法正确的是( ) A ．物体在第1 s 末运动方向发生变化 B ．第2 s 内、第3 s 内的速度方向是相同的 C ．物体在第2 s 末返回出发点D ．在这7 s 内物体的加速度大小不变，向正方向运动1 m7．有一长为L 的列车，正以恒定的加速度过铁路桥，桥长也为L ，现已知列车车头过桥头的速度为v 1，车头过桥尾时的速度为v 2，那么，车尾过桥尾时的速度为( )A ．2v 1－v 2B ．2v 2－v 1 C.2v 21－v 22D.2v 22－v 218．甲、乙、丙三辆汽车以相同的速度同时经过某一个路标，以后甲车一直做匀速直线运动，乙车先加速后减速，丙车先减速后加速，它们经过下一路标时的速度又相同，则( )A ．甲车先通过下一个路标B ．乙车先通过下一个路标C ．丙车先通过下一个路标D ．三车同时到达下一个路标9．（多选）物体做匀变速直线运动的初速度为6m/s ，经过10s 速度的大小变为20m/s ，则加速度大小可能是 ( ) A 、20.8/m s B 、21.4/m s C 、22.0/m s D 、22.6/m s10．（多选）甲、乙两汽车在一平直公路上同向行驶．在t ＝0到t ＝t 1的时间内，它们的v －t 图像如图所示．在这段时间内( )A．汽车甲的平均速度比乙的大 B．汽车乙的平均速度等于v1＋v2 2C．甲、乙两汽车的位移相同 D．汽车甲和汽车乙的加速度大小都在逐渐减小11．（多选）一辆汽车在运动过程中遇到紧急情况需刹车，从某时刻开始，其速度平方v2随位移x的变化关系为v2＝(16－4x)(m/s)2，则下列说法正确的是( )A．汽车的初速度为4 m/s B．汽车刹车时的加速度大小为4 m/s2 C．汽车刹车过程中的平均速度为8 m/s D．汽车的刹车时间为2 s12.（多选）a、b两质点沿直线Ox轴正向运动，t=0时，两质点同时到达坐标原点O，测得两质点在之后的运动中，其位置坐标x与时间t的比值（即平均速度）随时间t变化的关系如图所示，以下说法正确的是( )A．质点b做匀速运动,其速度为0．5 m/sB．质点a做匀加速运动，其加速度为1．0C．t=1s时，a、b再次到达同一位置D．t=2s时，a、b再次到达同一位置二．实验题（每空2分，共14分）13．为了测定某轿车在平直路面上起动时的加速度(可看作匀加速直线运动)，某人拍摄了一张在同一底片上多次曝光的照片，如图2所示，如果拍摄时每隔2 s 曝光一次，轿车车长为4.5 m，则其加速度约为____________（计算得数保留两位有效数字）图214．从下列所给器材中，选出测定匀变速直线运动的加速度实验所需的器材，有____________；为达到实验目的，还缺少______________________________．①打点计时器②天平③低压直流电源④细线⑤纸带⑥小车⑦钩码⑧秒表⑨一端有滑轮的长木板15．如图是研究物体做匀变速直线运动的实验得到的一条纸带(实验中打点计时器所接低压交流电源的频率为50 Hz)，从0点后开始每5个计时点取一个计数点，依照打点的先后顺序依次编为0、1、2、3、4、5、6，测得x1＝5.18 cm，x2＝4.40 cm，x3＝3.62 cm，x4＝2.78 cm，x5＝2.00 cm，x6＝1.22 cm.（计算得数保留两位有效数字）(1)相邻两计数点间的时间间隔为________s.(2)物体的加速度大小a＝________m/s2，方向________(选填“A→B”或“B→A”)．＝____________m/s.(3)打点计时器打计数点3时，物体的速度大小v3三．计算题（共4小题,共38分）16、（8分）从车站开出的汽车，做匀加速直线运动，走了12 s后，发现还有乘客没上来，于是立即做匀减速直线运动至停车，总共历时20 s，行进了 50 m．求（1）加速阶段和减速阶段的加速度之比是多少？（4分）（2）汽车的最大速度为多大．（4分）17．（15分）一辆汽车做匀减速直线运动，已知先后途中经过相距27 m的A、B 两点所用时间为2 s，汽车经过A点时的速度为15 m/s.求：(1)汽车经过B点时的速度大小（5分）（2）经过B点后，再过4s和10s汽车的位移之比是多少？（10分）18．（15分）有一架电梯，启动时匀加速上升，加速度为2m/s2，制动时匀减速上升，加速度为-1m/s2，楼高52m．求：（1）若上升的最大速度为6m/s，电梯升到楼顶的最短时间是多少？（2）如果电梯先加速上升，然后匀速上升，最后减速上升，全程共用16秒，上升的最大速度是多少？命题人：常金龙附加题(高考真题)1.一物体做直线运动，其加速度随时间变化的a-t图象如图所示。

高一年级复课地理测试试题范围（必修一第五章——必修二全部）一、单选题（每题1.5分，共60分）读某区域和乙地垂直自然带分布示意图，回答下列各题。

1. 关于甲、丙两地环境特征的叙述，正确的是( )A. 自然带相同B. 冬季盛行风风向相同C. 河流汛期相同D. 夏季盛行风风向相同2. 关于三地自然带的叙述，正确的是( )A. 从甲到丙的变化体现了从沿海向内陆的地域分异规律B. 从甲到丙的变化是以水分为主导因素的C. 丙地是亚热带常绿硬叶林带D. 乙地垂直自然带数目的多少仅取决于当地的纬度【答案】1. B 2. C【解析】【1题详解】由自然带分布示意图可以知道，自然带北坡分布较低，故该区域位于北半球；图中纬度为北纬，甲地为温带海洋性气候，乙地为温带季风气候，丙地为地中海气候。

甲地的自然带为温带落叶林带，丙地是亚热带常绿硬叶林带，A不对；甲地为温带海洋性气候，丙地为地中海气候，两地在冬季都受到盛行西风带的影响，B对；甲地河流无明显汛期，乙地河流的汛期在冬季，C不对；甲地全年盛行西风，乙地夏季受到副热带高气压带影响，D不对。

故选B。

【2题详解】甲地、丙地都位于沿海地区，两地差别主要是纬度的差异，AB不对；该区域位于北半球，图中纬度为北纬，丙地位于纬度30—40大陆西岸，属于地中海气候，所以丙地是亚热带常绿硬叶林带，C对；乙地是一个山体，其自然带的多少决定于其海拔高度，D不对。

故选C。

下图示意我国1960~2045年每五年的劳动人口增长变化（含预测）。

读图完成下列各题。

3. 我国劳动人口数量最多的年份是A. 1965年B. 1980年C. 2015年D. 2045年4. 未来10年劳动人口数量的变化将导致A. 老龄人口增加B. 人口增加C. 人才外流严重D. 用工成本上升【答案】3. C 4. D【解析】本题组以我国劳动人口增长变化为背景，设置两道试题，考查学生获取有效地理信息分析问题的能力。

【3题详解】图中纵坐标表示的是劳动人口增长率，不是劳动力人口数量。

英语试题本试卷总共分为六部分。

满分150分。

考试时间120分钟。

第一部分听力（共两节，满分20分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

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。

第二部分阅读理解（共20小题：每小题2.5分，满分50分）阅读下列短文，从每题所给的A、B、C和D四个选项中选出最佳选项。

AMetro Pocket GuideMetrorail(地铁)Each passenger needs a farecard to enter and go out. Up to two children under age five may travel free with a paying customer.Farecard machines are in every station, Bring small bills because there are no change machines in the stations and farecard machines only provide up to $ 5 in change.Get one day of unlimited Metrorail rides with a One Day Pass. Buy it from a farecard machine in Metro stations. Use it after 9:30 a. m. until closing on weekdays, and all day on weekends and holidays.Hours of ServiceOpen: 5 a. m. Mon.-Fri. 7a. m. Sat.—Sun.Close: midnight Sun.—Thur. 3a.m.Fri. —-Sat.nightsLast train times vary. To avoid missing the last train, please check the last train times posted in the stations.MetrobusWhen paying with exact change, the fare is $ 1. 35 . when paying with a SmarTrip card, the fare is $1. 25Fares for the Senior /disabled customersSenior citizens 65 and older and disabled customers may ride for half the regular fare. On Metrorail and Metrobus, use a senior/disabled farecard or SmarTrip card. For more information about buying senior/disabled farecards,SmarTrip cards and passes, please visit MetroOpenDoors. com or call 202-637-7000 and 202-637-8000.Senior citizens and disabled customers can get free guide on how to use proper Metrobus and Metrorail services by calling 202-962-1100Travel tips●Avoid riding during weekday rush periods –before 9:30 a. m. and between 4 and 6 p. m.●If you lose something on a bus or train or in a station, please call Lost & Found at202-962-1195.21. What should you know about farecard machines?A. They start selling tickets at 9:30 a. m.B. They are connected to change machines.C. They offer special service to the elderly.D. They make change for no more than $5.22. What is good about a SmarTrip card?A. It is convenient for old people.B. It saves money for its users.C. it can be bought at any time.D. it is sold on the Internet.23. Which number should you call if you lose something on the Metro?A. 202-962-1195B. 202-962-1100C. 202-637-7000D. 202-637-8000BSurviving Hurricane Sandy(飓风桑迪)Natalie Doan, 14, has always felt lucky to live in Rockaway, New York. Living just a few blocks from the beach, Natalie can see the ocean and hear the waves from her house. “It’s the ocean that makes Rockaway so special, ” she says.On October 29, 2012, that ocean turned fierce. That night, Hurricane Sandy attacked the East Coast, and Rockaway was hit especially hard. Fortunately, Natalie’s family escaped to Brooklyn shortly before the city’s bridge s closed.When they returned to Rockaway the next day, they found their neighborhood in ruins. Many of Na talie’s friends had lost their homes and were living far away. All around her, people were suffering, especially the elderly. Natalie’s school was so damaged that she had to temporarily attend a school in Brooklyn.In the following few days, the men and women helping Rockaway recover inspired Natalie. V olunteers came with carloads of donated clothing and toys. Neighbors devoted their spare time to helping others rebuild. Teenagers climbed dozens of flights of stairs to deliver water and food to elderly people trapped in powerless high-rise buildings.“My mom tells me that I can’t control what happens to me,” Natalie says. “but I can always choose how I deal with it. ”Natalie’s choice was to help.She created a website page matching survivors in need with donors who wanted to help. Natalie posted information about a boy named Patrick, who lost his baseball card collection when his house burned down. Within days, Patrick’s collection was replaced.In the coming months, her website page helped lots of kids: Christopher, who received a new basketball; Charlie, who got a new keyboard. Natalie also worked with other organizations to bring much-need supplies to Rockaway. Her efforts made her a famous person. Last April, she was invited to the White House and honored as a Hurricane Sandy Champion of Change.Today, the scars(创痕)of destruction are still seen in Rockaway, but hope is in the air. The streets are clear, and many homes have been rebuilt. “I can’t imagine living anywhere but Rockaway, ” Natalie declares. “My neighborhood will be back, even stronger than before. ”24. When Natalie returned to Rockaway after the hurricane , she found______.A. some friends had lost their livesB. her neighborhood was destroyedC. her school had moved to BrooklynD. the elderly were free from suffering25. According to paragraph 4, who inspired Natalie most?A. The people helping Rockaway rebuild.B. The people trapped in high-rise buildings.C. The volunteers donating money to survivors.D. Local teenagers bringing clothing to elderly people.26. How did Natalie help the survivors?A. She gave her toys to other kids.B. She took care of younger children.C. She called on the White House to help.D. She built an information sharing platform.27. What does the story intend to tell us?A. Little people can make a big difference.B. A friend in need is a friend indeed.C. East or West, home is best.D. Technology is power.CBasketball fans around the world are mourning the death of American superstar Kobe Bryant.Bryant was killed in a helicopter crash Sunday in California, along with his 13-year-old daughter Gianna and seven other people. He was 41 years old.Bryant was an 18-time All-Star player. He won five NBA championships and two Olympic gold medals. He was widely considered one of the greatest basketball players of his generation during his 20-year career with the Los Angeles Lakers.The helicopter carrying Bryant and the others crashed into a hilly area in foggy conditions outside Los Angeles. Bryant lived south of Los Angeles and often used helicopters to travel around the area to save time and avoid Southern California traffic.The cause of the crash is under investigation. It took place about 32 kilometers from Bryant's Mamba Sports Academy, a basketball training center in Newbury Park, California. Bryant was on his way to attend a youth basketball competition on Sunday with his daughter. Bryant and his wife, Vanessa, had four daughters.When he retired from the NBA in 2016, Bryant was the third-leading scorer in NBA history with 33,643 points. He is widely expected to be elected to the Naismith Memorial Basketball Hall of Fame this year.The basketball world and the Los Angeles community reacted to Bryant’s death with anoutpouring of sadness, disbelief and support.In a statement, NBA Commissioner Adam Silver said Bryant was one of the greatest stars in the game. But he added, He will be remembered most for inspiring people around the world to pick up a basketball and compete to the very best of their ability.On Twitter, U.S. President Donald Trump noted the terrible news. He called Bryant one of the truly great basketball players of all time who was just getting started in life.28.Where is the passage probably taken from?A. A science report.B. A biography.C. A textbook.D. A newspaper.29. What Kobe Bryant impressed people in the world most was .A. his deathB. his achievementsC. his inspirationD. his talent30. The underlined words “under investigation” in Paragraph 5 mean______.A. careful examinationB. evidenceC. discussionD. argument31. Which of the following is not true according to the passage?A. Hearing the death of Kobe, fans around the world were very sad.B. There were nine passengers on board, including one of his daughters.C. Bryant was to attend a basketball competition with his daughter.D. Bryant used the helicopter to save time and avoid the traffic jam on a fine day.DWhether in the home or the workplace, social robots are going to become a lot more common in the next few years. Social robots are about to bring technology to the everyday world in a more humanized way, said Cynthia Breazeal, chief scientist at the robot company Jibo.While household robots today do the normal housework, social robots will be much more like companions than mere tools. For example, these robots will be able to distinguish when someone is happy or sad. This allows them to respond more appropriately to the user.The Jibo robot, arranged to ship later this year, is designed to be a personalized assistant. You can talk to the robot, ask it questions, and make requests for it to perform different tasks. Therobot doesn’t just deliver general answers to questions; it responds based on what it learns abo ut each individual in the household. It can do things such as reminding an elderly family member to take medicine or taking family photos.Social robots are not just finding their way into the home. They have potential applications in everything from education to health care and are already finding their way into some of these spaces.Fellow Robots is one company bringing social robots to the market. The company’s “Oshbot” robot is built to assist customers in a store, which can help the customers find items and help guide them to the product’s location in the store. It can also speak different languages and make recommendations for different items based on what the customer is shopping for.The more interaction the robot has with humans, the more it learns. But Oshbot, like other social robots, is not intended to replace workers, but to work alongside other employees. “We have technologies to train social robots to do things not for us, but with us,” said Breazeal.32. What can a Jibo robot do according to Paragraph 3?A. Communicate with you and perform operations.B. Answer your questions and make requests.C. Take your family pictures and deliver milk.D. Obey your orders and remind you to take pills.33. What can Oshbot work as?A. A language teacher.B. A tour guide.C. A shop assistant.D. A private nurse.34. We can learn from the last paragraph that social robots will ______.A. train employeesB. be our workmatesC. improve technologiesD. take the place of workers35. What does the passage mainly present?A. A new design idea of household robots.B. Marketing strategies for social robots.C. Information on household robots.D. An introduction to social robots.根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

2019-2020学年山西省运城市临猗县临晋中学高一下学期开学复课摸底数学试题一、单选题1．已知ABC V 中，2a =，3b =，60B =o ，那么角A 等于（ ）A ．135oB ．90oC ．45oD ．30o【答案】C【解析】试题分析：三角形中由正弦定理得.sin 2,sin sin sin a b a b A A B B =∴==,所以4A π=.即选C.本题的关键就是正弦定理的应用.【考点】正弦定理.2．为了得到函数y=sin 的图象，只需把函数y=sinx 的图象上所有的点A ．向左平行移动个单位长度B ．向右平行移动个单位长度C ．向上平行移动个单位长度D ．向下平行移动个单位长度 【答案】A【解析】试题分析：为得到函数的图象，只需把函数的图象上所有的点向左平行移动个单位长度，故选A. 【考点】三角函数图象的平移【名师点睛】本题考查三角函数图象的平移，函数的图象向右平移个单位长度得的图象，而函数的图象向上平移个单位长度得的图象．左、右平移涉及的是的变化，上、下平移涉及的是函数值的变化．3．半径为2，圆心角为3π的扇形的面积为( ) A ．43π B ．πC ．23π D ．3π 【答案】C【解析】由扇形面积公式得：211S r 2223l π=⨯⨯=⨯⨯=2π3. 故选C.4．18,2a x x ⎛⎫=+ ⎪⎝⎭r ，()1,2(b x r =+其中0)x >，若//a b r r ，则x 的值为( )A ．8B ．4C ．2D ．0【答案】B【解析】根据//a b rr即可得出()128102x x x ⎛⎫+-+= ⎪⎝⎭，再根据0x >，即可解出x 的值． 【详解】解：//a b Q rr，且0x >；()128102x x x ⎛⎫∴+-+= ⎪⎝⎭；解得4x =，或4(x =-舍去)． 故选B ． 【点睛】考查向量坐标的定义，以及向量平行时的坐标关系．5．已知数列{}n a 和{}n b 都是等差数列，若22443,5a b a b +=+=，则77a b +=（ ）A ．7B ．8C ．9D ．10 【答案】B【解析】试题分析：因为数列{}n a 和{}n b 都是等差数列,{}n n b a +∴为等差数列,由22443,5a b a b +=+=,得()()1235242244=-=-+-+=b a b a d ．∴77ab +=()8351344=+=⨯++b a ．故选B ．【考点】等差数列． 6．函数()22sin sin 44f x x x ππ⎛⎫⎛⎫=+-- ⎪ ⎪⎝⎭⎝⎭是( ).A ．周期为π的偶函数B ．周期为π的奇函数C ．周期为2π的偶函数D ．周期为2π奇函数【答案】B【解析】因()1cos(2)[1cos(2)]sin 2sin 22sin 222f x x x x x x ππ=-+---=+=，故()sin(2)sin 2()f x x x f x -=-=-=-是奇函数，且最小正周期是，即22T ππ==，应选答案B ．点睛：解答本题时充分运用题设条件，先借助二倍角的余弦公式的变形，将函数的形式进行化简，然后再验证函数的奇偶性与周期性，从而获得问题的答案． 7．在ABC V 中，120B =︒，2AB =，A 的角平分线3AD =，则AC =（ ）A ．6B ．2C ．62D ．22【答案】A【解析】由正弦定理求得2sin 2ADB ∠=，则45ADB ∠=︒，从而得到30C =︒，再根据正弦定理即可求出答案． 【详解】 解：如图，由正弦定理sin sin AB AD ADB B=∠可得，sin sin AB BADB AD ∠=，∵120B =︒，2AB =3AD =∴2sin ADB ∠=，得45ADB ∠=︒， ∴135ADC ∠=︒，1801204515BAD ∠=︒-︒-︒=︒， ∴30BAC ∠=︒，∴30C =︒， ∴由正弦定理sin sin AC AB B C =得，sin 6sin AB BAC C==， 故选：A ． 【点睛】本题主要考查正弦定理解三角形，考查计算能力，属于基础题．8．已知12a =，121n n a a n +-=+（*n N ∈），则n a =（ ） A ．1n + B ．21n +C ．21n +D ．221n +【答案】C【解析】利用累加法即可求出通项公式． 【详解】解：∵121n n a a n +-=+，则当2n ≥时，121n n a a n --=-，……325a a -=， 213a a -=，∴132212153n n a a a a a a n --+⋅⋅⋅+-+-=-+⋅⋅⋅++， 化简得()()21121312n n n a a n --+-==-，又12a =， ∴21n a n =+，经检验12a =也符合上式， ∴()2*1n n N a n =+∈，故选：C ． 【点睛】本题主要考查累加法求数列的通项公式，考查数列的递推公式的应用，考查倒序相加法求数列的和，考查计算能力，属于中档题．9．在 ABC V 中， 80,100,45a b A ===︒，则此三角形解的情况是( ) A ．一解 B ．两解C ．一解或两解D ．无解【答案】B【解析】由题意知，80a =，100b =，45A ∠=︒，∴sin 100802b A =⨯=，如图：∵sin b A a b <<，∴此三角形的解的情况有2种，故选B ． 10．已知两个等差数列{}n a 和{}n b 的前n 项和之比为71(*)427n n N n +∈+，则1111a b 等于（ ） A ．74B ．32C ．43D ．7871【答案】C【解析】根据等差数列的前n 项和公式，以及等差数列的性质进行转化即可． 【详解】设等差数列{a n }和{b n }的前n 项和分别为：()()11,22n n n n n a a n b b S T ++==，根据等差数列的性质，可知11121111212,2a a a b b b =+=+故()()1212111121211121721142214212732a a S ab b T b +⨯+====+⨯+ ，故选C 【点睛】若数列{a n }是等差数列，若2m nk +=，则2m n k a a a += ()*,,m n k N ∈ .11．O 为ABC ∆所在平面上动点，点P 满足AB AC OP OA AB AC λ⎛⎫ ⎪=++ ⎪⎝⎭u u u v u u u v u u u v u u u vu u u v u u u v ,,[)0λ∈+∞ ,则射线AP 过ABC ∆的（ ） A ．外心 B ．内心C ．重心D ．垂心【答案】B【解析】将||||AB AC OP OA AB AC λ⎛⎫=++ ⎪⎝⎭u u u r u u u r u u u r u u u r u u ur u u u r 变形为||||AB AC AP AB AC λ⎛⎫=+ ⎪⎝⎭u u u r u u u r u u u r u u u r u u u r ，因为||ABAB u u u ru u u r 和||ACAC u u u r u u u r 的模长都是1，根据平行四边形法则可得，过三角形的内心. 【详解】||||AB AC OP OA AB AC λ⎛⎫=++ ⎪⎝⎭u u u r u u u ru u u r u u u r u u ur u u u r Q AB AC OP OA AP AB AC λ⎛⎫ ⎪∴-==+ ⎪⎝⎭u u u v u u u v u u u v u u u v u u u v u u u v u u u v因为||AB AB u u u ru u u r 和||AC AC u u u r u u u r 分别是AB u u u r 和AC u u u r 的单位向量所以||||AB AC AB AC +u u u r u u u r u u ur u u u r 是以||AB AB u u u ru u u r 和||AC AC u u u r u u u r 为邻边的平行四边形的角平分线对应的向量 所以AP u u u r的方向与BAC ∠的角平分线重合 即射线AP 过ABC ∆的内心 故选B 【点睛】本题主要考查平面向量的平行四边形法则、单位向量的性质以及三角形四心的性质，属于中档题.12．设()44sin sin cos cos f x x x x x =-+，则()f x 的值域是（ ）A ．90,8⎡⎤⎢⎥⎣⎦B ．513,88⎡⎤⎢⎥⎣⎦C ．130,8⎡⎤⎢⎥⎣⎦D ．59,88⎡⎤⎢⎥⎣⎦【答案】A【解析】由三角函数的基本关系式，化简得()2212sin cos sin cos f x x x x x =--，设sin cos t x x =，得到()2192()48g t t =-++，再结合二次函数的性质，即可求解. 【详解】 由函数()4422222sin sin cos cos (sin cos )2sin cos sin cos f x x x x x x x x x x x =-+=+--2212sin cos sin cos x x x x =--，设111sin cos sin 2[,]222t x x x ==∈-， 可得()2219212()48g t t t t =--+=-++，当14t =-时，函数()g t 取得最大值，最大值为()max 98g t =，即()max 98f x =，当12t =时，函数()g t 取得最小值，最大值为()min 0g t =，即()min 0f x =， 所以函数()f x 的值域为90,8⎡⎤⎢⎥⎣⎦. 故选：A. 【点睛】本题主要考查了以三角函数为背景的函数的值域的求解，其中解答中熟练应用三角函数的基本关系式，利用换元法，结合二次函数的性质求解是解答的关键，着重考查换元思想，以及推理与运算能力.二、填空题13．已知()2cos ,2sin a θθ=r ，()3,3b =r ，且a r 与b r共线，[)0,2θ∈π，则θ=________.【答案】π6或7π6【解析】由a r 与b r 共线，可得2cos 32sin 30θθ⨯-⨯=，求得3tan θ=，再结合三角函数的性质，即可求解. 【详解】由题意，向量()()2cos ,2sin ,3,3b a θθ==rr ，因为a r 与b r共线，可得2cos 32sin 30θθ⨯-⨯=，可得3tan 3θ=， 又因为[)0,2θ∈π，可得π6θ=或7π6θ=. 故答案为：π6或7π6. 【点睛】本题主要考查了共线向量的坐标表示，以及三角函数的基本关系式的应用，其中解答中熟记共线向量的坐标运算，求得tan θ的值是解答的关键，着重考查运算与求解能力. 14．已知函数()cos()f x A x ωφ=+的图象如图所示，2()23f π=-，则(0)f =____.【答案】23【解析】由图象可得最小正周期为23π．所以f （0）=f （23π），注意到23π与2π关于712π对称， 故f （23π）=﹣f （2π）=23． 故答案为2315．如图放置的边长为1的正方形ABCD 的顶点A,D 分别在x 轴、y 轴正半轴（含原点）滑动，则OB OC ⋅u u u r u u u r的最大值为__________．【答案】2【解析】设,0,2OAD πθθ⎛⎫∠=∈ ⎪⎝⎭,根据三角形的边角关系求得OB uuu r ，OC u u u r ，利用平面向量的数量积公式以及正弦函数的最值求解即可. 【详解】设,0,2OAD πθθ⎛⎫∠=∈ ⎪⎝⎭由于1AD =，故cos ,sin OA OD θθ==又因为2BAx πθ∠=-，1AB =，所以cos cos cos sin 2B x πθθθθ⎛⎫=+-=+ ⎪⎝⎭sin cos 2B y πθθ⎛⎫=-= ⎪⎝⎭， 则(cos sin ,cos )OB θθθ=+u u u r同理可得(sin ,cos sin )OC θθθ=+u u u r(cos sin ,cos )(sin ,cos sin )1sin 2OB OC θθθθθθθ⋅=+⋅+=+u u u r u u u r0,2πθ⎛⎫∈ ⎪⎝⎭Q∴当22=πθ时，1sin 2OB OC θ⋅=+u u u r u u u r的最大值为2.故本题的正确答案为2.【点睛】本题主要考查了平面向量的数量积公式以及正弦型函数的最值，属于中档题. 16．下列判断正确的是_________.（填写所有正确的序号） ①若1sin sin 3x y +=，则2sin cos y x -的最大值是43；②函数sin(2)4y x π=+的单调递增区间是[,]88k k ππππ-+（k Z ∈）； ③函数()1sin cos 1sin cos x xx x f x +-+=+是奇函数；④函数1tan 2sin x y x=-的最小正周期是π.【答案】④【解析】利用函数的最值、单调性、奇偶性、周期性质进行验证求解 【详解】因为1sin sin 3x y +=，1sin sin 3y x ∴=-22211sin cos =sin cos =sin (1s )33y x x x x in x ------22111=(s )12inx --，121sin 1,in 1,33x x -≤-≤∴-≤≤2in =3s x ∴-时，2sin cos y x -的最大值是49所以①错误；由222242k x k πππππ-++剟，得3+88k x k ππππ-剟，k Z ∈结论②错误； 因为()1sin cos 1sin cos x x x xf x +-+=+，()()1sin cos 1sin cos x xx f f x x x --+=-≠--，故结论③错误； 因为1tan 2sin x y x=-所以22sin +cos 122tan (tan cot )=cot 22222sin cos 22x x x x x y x x x =-=--最小正周期是π，故结论④正确. 故答案为④ 【点睛】本题考查三角函数图象与性质的综合应用.三角函数图象与性质的综合问题的求解思路:先将()y f x ＝化为+()+y Asin x B w j ＝的形式，再借助(+)y Asin x w j ＝的图象和性质(如定义域、值域、最值、周期性、对称性、单调性等)解决相关问题．三、解答题17．已知cos α－sin α＝5，且π<α<32π，求2sin 22sin 1tan ααα+-的值.【答案】-2875. 【解析】试题分析：将cosα－sinα＝5平方,可得2sinαcosα的值,由α的范围确定sinα＋cosα的范围,并求值,将原式用二倍角公式化简,并将求出的值代入即可. 试题解析: 因为cos α－sin α＝5，所以1－2sin αcos α＝1825，所以2sin αcos α＝725. 又α∈(π，32π)，故sinα＋cosα，＝－5， 所以2sin22sin 1tan ααα+-＝()22sin cos 2sin cos cos sin αααααα+- ＝()2sin cos cos sin cos sin αααααα+-7255⎛⎫⨯- ⎪2875. 18．已知向量1232a e e =-r ，124b e e =+r，其中()11,0e =，()20,1e =.求：（1）a b ⋅r r，a b +r r ； （2）a r 与b r的夹角的余弦值.【答案】（1）10a b ⋅=r r，a b +=r r ；（2）221【解析】（1）根据()1e 1,0=，()2e 0,1=，得到,a b r r的坐标，再利用坐标运算求数量积及a b +r r .（2）设a r 与b r的夹角为θ，先求得,a b r r ，再利用夹角公式求解.【详解】（1）因为()1e 1,0=，()2e 0,1=，所以()123e 2e 32a =-=-r ,，()124e e 4,1b =+=r ，所以()342110a b ⋅=⨯+-⨯=r r ，所以()7,1a b +=-r r ， 所以a b +==r r . （2）设a r 与b r 的夹角为θ，a =rb =r所以cos 221a b a bθ⋅===r r r r . 【点睛】本题主要考查向量的坐标表示，向量的模以及数量积的运算，还考查了运算求解的能力，属于基础题.19．在ABC V 中，内角A ，B ，C 的对边分别为a ，b，c ，且sin cos b A B =. （1）求角B 的大小；（2）若3b =，sin 2sin C A =，求a ，c 的值.【答案】（1）π3；（2）a=c = 【解析】（1）根据正弦定理即可化简sin B B =，根据正切值求解； （2）根据正弦定理得2c a =，结合余弦定理即可得解.【详解】（1）由sin cos b A B =及正弦定理sin sin a b A B=，在ABC V 中，sin 0A> 得sin B B=，所以tan B =()0,B π∈，所以π3B =. （2）由sin 2sin C A =及sin sin a c A C=，得2c a =， 由3b =及余弦定理2222cos b a c ac B =+-，得229a c ac =+-.222942a a a -=+，所以a =c =.【点睛】 此题考查利用正余弦定理解三角形，根据正弦定理和余弦定理进行边角互化，综合应用，属于中档题.20．已知数列{}n a 为等差数列，35a =，公差0d ≠，且1225a a a a =. （1）求数列{}n a 的通项公式以及它的前n 项和n S ；（2）若数列{}n b 满足11n n n b a a +=⋅，n T 为数列{}n b 的前n 项和，求n T . 【答案】（1）21n a n =-，2n S n =；（2）21n n T n =+ 【解析】（1）根据题意列出方程组求解1a 、d ，代入等差数列的通项公式及前n 项和公式化简即可；（2）利用裂项相消法求和.【详解】（1）由题意得()()11221111252524a d a d d a d a d a a d +=⎧+=⎧⎪⇒⎨⎨=+=+⎩⎪⎩， 又∵0d ≠，解得112a d =⎧⎨=⎩，∴()*11221()n a n n n N =+-⨯=-∈， ∴()()1212122n n a n n n S n a +⨯+-⨯===. （2）∵()()111111212122121n n n b a a n n n n +⎛⎫===- ⎪⋅-+-+⎝⎭， ∴11111112335212121n n T n n n ⎛⎫=-+-+⋅⋅⋅+-= ⎪-++⎝⎭. 【点睛】本题考查等差数列的通项公式及前n 项和、裂项相消法求和，属于基础题.21．设ABC V 的内角A ，B ，C 所对的边长分别为a ，b ，c ，且3a =，60A =︒，b c +=.（1）求ABC V 的面积；（2）求sin sin B C +的值及ABC V 中内角B ，C 的大小.【答案】（1）4；（2）sin sin 2B C +=；15=︒C ，105B =︒或105C =︒，15B =︒ 【解析】（1）根据题意，由余弦定理得出()2222cos60a b c bc bc =+--o ，可求出bc ，再根据三角形面积公式，即可求出ABC V 的面积；（2）根据正弦定理sin sin sin +=+a b c A B C ，求得sin sin B C +=，利用三角函数的恒等变换进行化简求出角C ，最后结合三角形的内角和，即可求出角B .【详解】解：（1）由题可知，3a =，60A =︒，b c +=由余弦定理得：2222cos60a b c bc =+-o ，则()2222cos60a b c bc bc =+--o ，即(219222bc bc =--⨯，即9183bc =-， 解得：3bc =，故ABC V 的面积为：11sin 32224ABC S bc A ==⨯⨯=V .（2）因为3a =，60A =︒，b c += 由正弦定理得sin sin sin sin sin a b c b c A B C B C+===+，即：3sin 60=︒所以sin sin 2B C +=， 因为120B C +=︒，所以120B C =-o ，则()sin sin sin 120sin 2B C C C +=︒-+=，即sin120cos cos120sin sin 2C C C -+=o o ，整理得：3sin 222C C +=()302C +=o ，由此得()sin 30C +︒=， 在ABC V 中，3045C +︒=︒或135︒，所以15=︒C 或105C =︒，由此可求得15=︒C ，105B =︒或105C =︒，15B =︒.【点睛】本题考查利用正弦定理和余弦定理解三角形，以及三角形的面积公式，考查三角恒等变换的应用和三角形内角和关系，考查化简运算能力.22．如图，点13,2⎛⎫- ⎪ ⎪⎝⎭B ，点A 是单位圆与x 轴的正半轴的交点.（1）若AOB α∠=，求sin2α.（2）已知OM OB h →→=+，ON OB h →→=-，若OMN V 是等边三角形，求OMN V 的面积. （3）设点P 为单位圆上的动点，点Q 满足OQ OA OP →→→=+，ππ262AOP θθ⎛⎫∠=≤≤ ⎪⎝⎭，()f OB OQ θ→→=⋅，求()f θ的取值范围.当OB OQ →→⊥时，求四边形OAQP 的面积. 【答案】（1）3；（2）33；（3）10,2⎡⎤⎢⎥⎣⎦；32【解析】（1）根据任意角三角函数的定义先求出sin cos αα，，即可求解.（2）由条件可得(3OM ON →→+=-，再根据OMN V 是等边三角形，即可求出该等边三角形的高，从而可求解其面积.（3）根据任意角三角函数的定义，可得()cos2,sin2P θθ，从而得()1cos 2sin 2OQ θθ→=+,， ()1sin 262f πθθ⎛⎫=-- ⎪⎝⎭，即可求解()f θ的取值范围；根据OB OQ →→⊥，再结合()1sin 262f πθθ⎛⎫=-- ⎪⎝⎭，可得四边形OAQP 为菱形，从而可求解其面积. 【详解】解：（1）由三角函数定义，可知3sin α=，1cos 2α=-， 所以313sin22sin cos 22ααα⎛⎫==-= ⎪⎝⎭（2）因为12OB →⎛=-⎝⎭，OM OB h →→=+，ON OB h →→=-，所以(2OM ON OB h OB h OB →→→→→+=++-==-，所以(2OM ON →→+=-=，又因为OMN V 是等边三角形，所以等边OMN V 的高为1因此OMN V 的面积为112⨯=. （3）由三角函数定义，知()cos2,sin2P θθ，所以()1cos 2sin 2OQ OA OP θθ→→→=+=+,，所以()()111cos 22sin 2262f OB OQ πθθθθ→→⎛⎫==-+=-- ⎪⎝⎭⋅， 因为ππ62θ≤≤，所以ππ5π2666θ≤-≤，即1πsin 2126θ⎛⎫≤-≤ ⎪⎝⎭， 于是()102f θ≤≤，所以()f θ的取值范围是10,2⎡⎤⎢⎥⎣⎦. 当OB OQ →→⊥时，()0f OB OQ θ→→=⋅=， 即π12062sin θ⎛⎫--= ⎪⎝⎭，解得π23θ=，易知四边形OAQP 为菱形，此时菱形OAQP 的面积为1π211sin 23⨯⨯⨯⨯=【点睛】本题考查任意角三角函数的定义、向量的坐标运算，解题关键在于根据三角函数的定义写出点的坐标，再进行相应的向量坐标运算，考查计算和分析转化能力，属于中档题.。

山西省临猗临晋中学2019-2020学年高一物理6月月考试题满分100分，时间：90分钟，一：单选题（每小题3分，共计30分）1．关于互成角度的两个初速度为零的匀变速直线运动的合运动，下述说法正确的是 ( ) A．一定是直线运动 B．一定是曲线运动C．可能是直线运动，也可能是曲线运动 D．以上说法都不对2．如图所示，水平转台绕竖直轴匀速旋转，一个小物体随转台一起转动，它受外力情况是（）A．重力、弹力B．重力、弹力、静摩擦力C．重力、弹力、滑动摩擦力D．重力、弹力、摩擦力、向心力3：对于一个做平抛运动的物体，它在从抛出开始的四段连续相等的时间内，在水平方向和竖直方向的位移之比，下列说法正确的是（）Ａ.1:2:3:4；1:4:9:16 Ｂ.1:3:5:7；1:1:1:1Ｃ.1:1:1:1；1:3:5:7 Ｄ.1:4:9:16；1:2:3:44．如图所示，下列关于机械能是否守恒的判断正确的是 ( )A．甲图中，物体A将弹簧压缩的过程中，A机械能守恒B．乙图中，A置于光滑水平面，物体B沿光滑斜面下滑，物体B机械能守恒C．丙图中，不计任何阻力时A加速下落，B加速上升过程中，A、B组成的系统机械能守恒D．丁图中，小球沿水平面做匀速圆锥摆运动时，小球的机械能守恒5．如图所示为一质点在恒力F作用下在xOy平面内从O点运动到B点的轨迹，且在A点时的速度v A与x轴平行，则恒力F的方向可能是 ( )A．沿＋x方向B．沿－x方向C．沿＋y方向D．沿－y方向6．)如图所示，重物M沿竖直杆下滑，并通过绳带动小车沿斜面升高。

问：当滑轮右侧的绳与竖直方向成θ角，且重物下滑的速率为v时，小车的速度为 ( )A．vsinθ B．v/cosθC．vcosθ D．v/sinθ7：如图，粗细均匀两端开口的U形管内装有同种液体，开始时两边液面高度差为h，管中液柱总长度为4h，后来让液体自由流动，当两液面高度相等时，右侧液面下降的速度为（）A． B． C． D．8、如图所示，足够长的斜面上A点，以水平速度v0抛出一个小球，不计空气阻力，它落到斜面上所用的时间为t 1；若将此球改用2v0抛出，落到斜面上所用时间为t2，则t1 : t2为（）A．1 : 1 B．1 : 2 C．1 : 3 D．1 : 49．石家庄市高一下学期期末)如图所示，当汽车通过拱桥顶点的速度为10m/s时，车对桥顶的压力为车重的34。

山西省运城市临猗临晋中学2019-2020学年高一下学期6月月考试卷（满分150分，考试时间：120分钟）第I 卷(选择题共60分）一、选择题（共12个小题，每小题只有一个选项正确，每小题5分，共60分） 1.计算)310sin(π-的值为（ ） A. 23 B. 23- C. 21 D. 21-2．等比数列{a n }中，已知，则n 为（ ）A ．3B ．4C ．5D ．63．如果a ＜b ＜0，那么下面一定成立的是（ ） A ．ac ＜bcB ．a ﹣b ＞0C ．a 2＞b 2D ．＜4.已知在△ABC 中，2—=，且),(R y x AC y AB x AD ∈+=，则y x -的值为（ ）A. 21B. 21- C. 31- D. 315.计算)4(sin )4tan(1cos 222απαπα+--的结果为（ ）A.lB.2C.-lD.-26．若变量x ，y 满足约束条件，则x +2y 的最大值是（ ）A ．B ．0C ．D ．7．己知角βα,满足πβαπβαπ< <0,23<- <2+，且31) cos(,31)-sin(-=+=βαβα，则β2cos 的值为（ ）A . 92-B . 92C . 924-D .924 8．在△ABC 中，若a =2，，A =30°则B 为（ ）A ．60°B ．60°或120°C ．30°D ．30°或150°9．等差数列{a n }中，a 3+a 6+a 9=，则=（ ）A ．﹣1B ．C ．0D ．10．设数列的通项公式为，若数列是单调递增数列，则实数的取值范围为（ ）A ．B ．C ．D ． 11．若两个正实数且不等式有解，则实数的取值范围是（ ） A. B.C. D.12.若关于x 的方程]4,4[,01cos sin 2cos sin ππ-∈=-+-+x a x x x x 有两个不同解，则实数a 的取值范围为 A.(2，31』 B.『2, 25』 C.(2, 25） D.『2,49） 第II 卷（非选择题共90分）二、填空题（共4个小题，每题5分，合计20分）13.已知扇形AOB 的面积为4π，圆心角AOB 为0120，则该扇形半径为 . 14.的图象，只需把函数的图象上所有的点向左平移 个单位长度.15.在△ABC 中，己知角A ，B ，C 所对的边分别为a ，b ，c ，且060,3,===B b x a ，若△ABC 有两解，则x 的取值范围是.16.设为实数，若则的最大值是三、解答题（共6个大题，其中17题10分，其余每个题目12分） 17.已知不等式.（1）当时，求此不等式的解集；{}n a 2n a n bn =+{}n a b ()3,-+∞[)2,-+∞[)1,+∞9,2⎛⎫-+∞ ⎪⎝⎭,x y 2x y m m +<-m ()1,2-()(),12,-∞-⋃+∞()(),14,-∞-⋃+∞sin 2y x =y x ,5422=++xy y x y x +2（2）若不等式的解集非空，求实数的取值范围．18.数列{a n}中，a1＝1，a n+1＝2a n+n﹣1．（1）求证：数列{a n+n}为等比数列；（2）求数列{a n}的通项公式．19.已知函数．（1）求函数f（x）的单调性；（2）在△ABC中，角A，B，C的对边分别为a，b，c，且，，c＝1，求△ABC的面积．20.若向量),sin ,sin (cos x x x a ωωω-=),cos 32,sin cos (x x x ωωω--=设函数)()(R x x f ∈+⋅=λ的图象关于直线π=x 对称，其中λω、为常数，且⎪⎭⎫⎝⎛∈1,21ω.（1）求函数)(x f 的最小正周期；（2）若)(x f y =的图象经过点⎪⎭⎫⎝⎛0,4π，求函数)(x f 在区间⎥⎦⎤⎢⎣⎡530π，上的值域.21.记为等差数列的前n 项和，已知，. (1)求的通项公式； (2)，，若对一切成立，求实数的最大值.22.已知锐角中，内角的对边分别为,（1）求角的大小；（2）求函数的值域.n S {}n a 21224a a +=11121S ={}n a 12......n n T b b b =+++240n T m -≥*n N ∈mABC ∆,,A B C ,,a b c C sin sin A B +——★ 参*考*答*案 ★——1.A2.C3.C4.D5.B6. A7.D8. B9. B 10. A 11. C 12. D 13.215. ()323， 1617. 解：（1）当时，不等式为，解得. .....4分故不等式的解集为； .....5分（2）不等式的解集非空，则， .....7分即，解得，或， .....9分 故实数的取值范围是． .....10分18.解：（1）证明：根据题意，a n +1＝2a n +n ﹣1， 则a n +1+n +1＝2a n +n ﹣1+n +1＝2a n +2n ＝2（a n +n ） 所以，所以数列{a n +n }为等比数列． .....6分（2）由（1）得数列{a n +n }为以2为公比的等比数列， 又a 1＝1，所以a 1+1＝2． 所以，所以．.....12分19.解：（1），由，得，k ∈Z ；由，得，k ∈Z ．故f （x ）在上单调递增，在上单调递减，k ∈Z ． .....6分 （2），则，∵A ∈（0，π），∴，即， 由正弦定理得，即，解得，∴或，22当C ＝时，A +C ＞π，舍去，所以，故，∴． .....12分20. 解：（1）λ+⋅=b a x f )(λωωωω++-=cos sin 32cos sin 22x x xλωω++-=x x 2sin 32cos λπω+-=)62sin(2x函数)(x f 的图象关于直线π=x 对称，可得1)62sin(±=-πωπ,)(262Z k k ∈+=-πππωπ,即)(231Z k k∈+=ω 又⎪⎭⎫ ⎝⎛∈1,21ω，所以1=k ,且65=ω，所以λπ+-=)635sin(2)(x x f 所以)(x f 的最小正周期为56π.....6分 （2）由)(x f y =的图象经过点⎪⎭⎫⎝⎛0,4π，得04=⎪⎭⎫ ⎝⎛πf 即)6435sin(2ππλ-⨯-=24sin 2--==π,所以2)635sin(2)(--=πx x f 由⎥⎦⎤⎢⎣⎡∈530π，x ,得656356πππ≤-≤-x ,所以1)635sin(21≤-≤-πx所以222)635sin(221-≤--≤--πx故函数)(x f 在区间⎥⎦⎤⎢⎣⎡530π，上的值域为[]2221---，.....12分 21.解：(1)∵等差数列中，，. ∴，解得.，. .....5分(2) {}n a 21224a a +=11121S =76224{11121a a ==7612{11a a ==7612111d a a ∴=-=-=()*665,n a a n d n n N ∴=+-=+∈1n n b a +=111786n T n =-++-+是递增数列，∴实数的最大值为.....12分 22.解：（1，利用正弦定理可得， 可化为，.....6分.....12分{}n T ∴*240,n T m n N -≥∈对一切成立m 2sin cos sin cos sin cos A C B C C B -=()2sin cos sin A C sin C B A =+=sin 0,2A ≠0,2C π⎛∈ ⎝sin ππ⎛+- ⎝AB +=。

山西省运城市临猗县临晋中学2019-2020学年高一历史下学期开学复课摸底考试试题第Ⅰ卷（选择题共60分）一、单项选择题（每小题有且只有一个正确答案，请将最恰当的答案填涂在答题卡的相应位置。

共40题，每题1.5分。

共60分。

1.春秋时期,促使土地逐步由国有向私有转变的根本因素是( )A.铁器的使用和牛耕的推广B.私田的出现C.各国税制的改革D.战争的影响2.北宋前中期，在今四川井研县一带山谷中，密布着成百上千个采用新制盐技术的竹简井，井主所雇工匠大多来自“他州别县”，以“佣身赁力”为生，受雇期间，若对工作条件成待遇不满意，辄另谋高就。

这反映出当时( )A.民营手工业得到发展B.手工业者社会地位高C.雇佣劳动已经普及D.盐业专卖制度已经解体3.明清时期,商人最后总是倾向于把累积得来的财富或过剩的资本投资于购买土地,或供应下一代有闲沉浸于传统典籍,参与科举,以便进入官僚行列。

这说明( )A.经商活动难以实现工业资本的原始积累B.经商只为后代学习提供充足的物质基础C.经商动机存在制约商业企业发展的因素D.经商是为满足对土地和传统文化的喜好4.从经济的角度看，明清时期中国已处于“近代的前夜”，下列史实最能说明这一结论的是（）A. “共计一坯工力，过手七十二，方克成器”B.“佃农所获，朝登垄亩，夕贸市之廛”C. “工匠各有专能．匠有常主，计日受值”D.“借间屋中人，尽去做商贾”5、西欧资本主义萌芽最早产生于意大利，但新航路开辟后，“财富流向西班牙和葡萄牙”，“英国成为海盗”，而“意大利被晾到一边”。

其主要原因是（）Ａ世界性商业危机。

Ｂ意大利经济衰退。

Ｃ世界性商业革命。

Ｄ意大利金融危机。

6.“大发现是资本主义和封建势力共同扩张的结果,是双重扩张。

换言之,没有初期资本主义的发展和商人们的扩张要求,就没有地理大发现;但若没有国王和政府的领导、支持及贵族的参与,同样也不会有地理大发现。

”以下关于新航路开辟的说法能够证明上述结论的是( )。

山西省运城市临猗县临晋中学2019-2020学年高一数学下学期开学复课摸底考试试题（考试时间120分钟总分120分）一、选择题(每小题5分，共60分）1.在△ABC中,a=,b=,B=60°,则角A等于().A.135°B.90°C.45°D.30°2.为了得到函数y=sin的图象,只需把函数y=sin x的图象().A.向左平移个单位长度B.向右平移个单位长度C.向上平移个单位长度D.向下平移个单位长度3.半径为2,圆心角为的扇形的面积为().A. B.π C. D.4.已知向量a=,b=(x+1,2),其中x>0,若a∥b,则x的值为().A.8B.4C.2D.05.已知数列{a n}和{b n}都是等差数列,若a2+b2=3,a4+b4=5,则a7+b7=( ).A.7B.8C.9D.106.函数f(x)=sin2-sin2是( ).A.周期为π的奇函数B.周期为π的偶函数C.周期为2π的奇函数D.周期为2π的偶函数7.在△ABC中,B=120°,AB=,A的角平分线AD=,则AC=( ).A. B. C. D.8.已知a 1=2,a n+1-a n =2n+1(n ∈N *),则a n =( ). A.n+1 B.2n+1C.n 2+1 D.2n 2+19.在△ABC 中,a=80,b=100,A=45°,则此三角形解的情况是( ). A.一解 B.两解 C.一解或两解 D.无解10.已知两个等差数列{a n },{b n }的前n 项和分别为S n ,T n ,且=,则的值为( ).A .B .C .D .11．O 为ABC ∆所在平面上动点，点p 满足⎪⎪⎪⎭⎫⎝⎛++=AC AC AB AB OA OP λ,[)+∞∈,0λ,则射线AP 过ABC ∆的（ ）A ．外心B ．内心C ．重心D ．垂心12.设f(x)=sin 4x-sin xcos x+cos 4x,则f(x)的值域是( ).A. B. C. D.二、填空题(每小题5分，共60分）13. 已知a =(2cos θ,2sin θ),b =(3,),且a 与b 共线,θ∈[0,2π),则θ= .14.已知函数f (x )=A cos(ωx+φ)的图象如图所示,f =-,则f (0)= .15．如图放置的边长为1的正方形ABCD 顶点A,D 分别在x 轴、y 轴正半轴（含原点）滑动，则OC OB ⋅的最大值为__________． 16.下列判断正确的是 .(填写所有正确的序号)①若sin x+sin y=,则sin y-cos2x的最大值是;②函数y=sin的单调递增区间是kπ-,kπ+(k∈Z);③函数f(x)=是奇函数;④函数y=tan-的最小正周期是π.三、解答题(17题10分，其它题每个12分，共70分）17.已知cos α-sin α=,且π<α<,求的值.18.已知向量a=3e1-2e2,b=4e1+e2,其中e1=(1,0),e2=(0,1).求:(1)a·b,|a+b|;(2)a与b的夹角的余弦值.19.在△ABC中,内角A,B,C的对边分别为a,b,c,且b sin A=a cos B.(1)求角B的大小;(2)若b=3,sin C=2sin A,求a,c的值.20.已知数列{a n}为等差数列,a3=5,公差d≠0,且=.(1)求数列{a n}的通项公式以及它的前n项和S n;(2)若数列{b n}满足b n=,T n为数列{b n}的前n项和,求T n.21.设△ABC的内角A,B,C所对的边长分别为a,b,c,且a=3,A=60°,b+c=3.(1)求△ABC的面积;(2)求sin B+sin C的值及△ABC中内角B,C的大小.22.如图,点B,点A是单位圆与x轴的正半轴的交点.(1)若∠AOB=α,求sin 2α.(2)已知=+h,=-h,若△OMN是等边三角形,求△OMN的面积.(3)设点P为单位圆上的动点,点Q满足=+,∠AOP=2θ,f(θ)=·,求f(θ)的取值范围.当⊥时,求四边形OAQP的面积.高一复课考试数学测试参考答案1—5 CACBB 6—10 AACBC 11—12 BA13. 或 14. . 15. 2 16. ④17.【解析】因为cos α-sin α=,所以1-2sin αcos α=,所以2sin αcos α=.又α∈,所以sin α+cos α=-=-,所以====- .18.【解析】(1)因为e1=(1,0),e2=(0,1),所以a=3e1-2e2=(3,-2),b=4e1+e2=(4,1),所以a·b=3×4+(-2)×1=10,所以a+b=(7,-1),所以|a+b|==5.(2)设a与b的夹角为θ,则cos θ===.19.【解析】(1)由bsin A=acos B及正弦定理=,得sin B=cos B,所以tan B=,所以B=.(2)由sin C=2sin A及=,得c=2a,由b=3及余弦定理b2=a2+c2-2accos B,得9=a2+c2-ac.所以a=,c=2.20.【解析】(1)由题意得又∵d≠0,∴∴a n=1+(n-1)×2=2n-1, ∴S n===n2.(2)∵b n===,∴T n=1-+-+…+-=.21.(1)由余弦定理得b2+c2-a2=2bccos 60°,bc=3.故S△ABC=bcsin A=×=.(2)因为A=60°,由正弦定理得====,又b+c=3,所以sin B+sin C=.因为B+C=120°,所以sin(120°-C)+sin C=.由此得sin(C+30°)=.在△ABC中,C+30°=45°或135°,即由此可求得C=15°,B=105°或C=105°,B=15°.22.(1)由三角函数定义,可知sin α==,cos α==-,所以sin 2α=2sin αcos α=2××=-.(2)因为=,=+h,=-h,所以+=+h+-h=2=(-1,).所以|+|=|(-1,)|=2.所以等边△OMN的高为1,边长为,因此△OMN的面积为×1×=.(3)由三角函数定义,知P(cos 2θ,sin 2θ),所以=+=(1+cos 2θ,sin 2θ),所以f(θ)=·=-(1+cos 2θ)+sin2θ=sin-.因为≤θ≤,所以≤2θ-≤,即≤sin≤1,于是0≤f(θ)≤,所以f(θ)的取值范围是.当⊥时,f(θ)=·=0,即sin-=0,解得2θ=,易知四边形OAQP为菱形,此时菱形OAQP的面积为2××1×1×sin=.。