建平中学2017学年度第一学期期末考试

2017建平中学高一下期末考试卷II. Grammar and VocabularyDirections: Choose the best answer according to the meaning of the sentence.17.___ is mentioned above, the number of violent crimes will increase this year about 15%.A.WhichB. AsC. ThatD. It18.Being unable yo afford a car, going shopping for the two of us meant ___ down to theshops and going back home again with full arms.A.to walkB. walkedC. is to buildD. will be built19.Opposite the bus station ___ a new commercial center, which the nearby residents arecurrently discussing a lot.A.to buildB. builtC. is to buildD. will be built20.As of Friday, the money ___ totalled more than 8.5 million yuan ($1.38 million) forChina-Dolls Center for Rare Disorders, a non-commercial organization.A.having been raisedB. raisingC. raisedD. was raised21.___ gender differences follow essentially old ideas on achievement tests in which boystypically score higher on math and science, females have the advantages on school grades regardless of the subject.A.WhileB. WhenC. IfD. Because22.The most important part of what children’s minds have and many animals’don’t is ___scientists call shared intentionally, which is the ability to infer what others know or are thinking.A.thatB. whichC. whatD. the23.It was not until then ___ the girl would receive an operation the next day.A.did we knowB. we knewC. that we knewD. when we knew24.Before 1973, fingerprints at the scene of a crime used to be photographed for ___ purposeand the object carrying the prints were shown in court as well.A.qualificationB. identificationC. estimateD. possession25. When it comes to leadership roles, in some cases, such as friendship groups, one or morepersons may gradually ___ as leaders, although there is no formal process of selection.A. ariseB. distinguishC. occurD. emerge26. There are many factors in people’s daily lives that can affect posture and throw the body offbalance, such as sitting at a desk for long periods, frequently holding a phone between the ear and the shoulders, bending over a laptop, or ___ looking down at a smartphone.A. continuallyB. eventuallyC. temporarilyD. accidentallySection BDirections: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.The first DC movie of 2017, Wonder Wonder Woman, is a few days away from release. With the DC Extended University (DCEU) so far ___27__ (look) dangerously underpowered compared with Marvel’s Chinematic University, the pressure is on to ensure this latest comicbook adaptation from Warner Bros is a hit.The DCEU has offered the world a handful of heroes, but it seems to be confused about what makes a good hero. We’ve seen a Superman suffering from doubt and lacking his unusual optimism, and a hopeless Batman who ___28__ (motivate) almost entirely by evil. So untrustworthy are the supposed good guys in the DCEU, the world turned to a group of super villains (恶棍) when it wanted to take down a villain in Suicide Squad.Wonder Woman, an Amazonian princess warrior who is a demigoddess(半神), is supposed to be different. She presents a very feminine sense of peace, justice and ___29__ Gal Gadot has described as “emotional intelligence.” In Wonder Woman, ___30__ American pilot Steve Trevor crashes on Themyscira and tells Princess Diana of the island nation about Word War I, she leaves her home to try to stop the war and becomes Wonder Woman.___31__ the upcoming movie, DC has an opportunity to steal a march on Marvel, because the Disney-owned studio has yet to deliver a movie led by a female superhero. Besides, director Patty Jenkins has ___32__ impressive history of telling women’s stories, including writing and directing the Oscar-winning crime drama Monster.Making her first comic book appearance in October 1941, Wonder Woman was the brainchild of the American psychologist and writer William Moulton Marston, ___33__ intended her as a feminist icon. But is her impressive physical beauty a problem for a feminist reading of the character? Jenkins doesn’t think so, describing her take on the character as “total wish-fulfillment.”“I, ___34__ a woman, want Wonder Woman to be hot as hell, fight bad guys and look great at the same time,”she said, “the same way men want Superman to have huge chest muscles and an impractically big body. That makes them feel like the the hero they want to be. And my hero, in my head, has really long legs.”Section CDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.has revealed, because of the way your native tongue speaks about time. A team from the University of Lancaster say their work also shows how bilingualism(双语) may affect pur perceptions of time, __35__ the brain to think in new ways.In one experiment, 40 Spanish speakers and 40 Swedish speakers were __36__ in seeing a computer animation(动画) of a slowly line. All the animations lasted 3 seconds, but the line didn't always grow to the same length. The researchers expected that because Swedes talk about time in terms of distance, they would find it harder to __37__ how much time had passed, and they were right. Meanwhile the Spanish speakers, who refer to time in terms of __38__ (as in a “small” break rather than a “short” break), were much better at realizing that the same 3 seconds had fled, no matter how far the line grew. “The Sw edish speakers tend to think that the line that grows longer takes longer,” one of the researchers explained. “Spanish speakers aren't2__39__ by that. They seem to think that it doesn't matter how much the line grows in distance, it still takes the same ti me for it to grow.”In another experiment, participants were shown animations of a jug(水壶) slowly being filled up: the length of the animation was fixed, but the jug filled up by __40__ amounts. Sure enough, this time it was the Spanish speakers who had more trouble __41__ the passage of time.Interestingly, when the spoken instructions in a particular language were taken away, the volunteers were much better at judging time, as if being asked out loud how much time had passed triggered something in the brain. To gain __42__ insight into what was happening, 74 bilingual speakers of both Spanish and Swedish were also recruited, and shown similar animations. The end results were the same: when instructed in Swedish, the volunteers were more easily fooled by the line animations, and when instructed in Spanish, it was the jug animations that interfered with their __43__ of time.III. Reading comprehension.Section ADirections: For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.It is often claimed that nuclear energy is something we cannot do without. We live in a _44_ society where there is an enormous demand for commercial products of all kinds. Moreover,an increase in industrial production is considered to be one solution to the problem of mass unemployment.Such an increase presumes an abundant and cheap energy supply. Many people believe that nuclear energy provides an inexhaustible and _45_ source of power and that it is therefore essential for an industrially developing society.There are a number of other advantages in the use of nuclear energy. Firstly, nuclear power, except for accidents,is clean. A further advantage is that a nuclear power station can be run and maintained by relatively few technical and administrative staff. The nuclear reactor represents an enormous _46_ in our scientific evolution and, whatever the anti nuclear group says,it is wrong to _47_ a return to more primitive sources of fuel. However, opponents of nuclear energy point out that nuclear power stations bring a direct threat not only to the environment but also to civil rights.Furthermore,it is questionable whether ultimately nuclear power is a(n) _48_ source of energy. There have, for example, been very costly accidents in America, in Russian and, of course, in Japan.The possibility of increases in the cost of uranium in addition to the cost of greater safety _49_ could demand too much money for nuclear power. In the long run, environmentalists argue, nuclear energy wastes valuable resources and disturbs the ecology to an extent which could _50_ the destruction of the human race.Thus, if we wish to survive, we cannot afford nuclear energy. In spite of the case against nuclear energy outlined above, nuclear energy programs are _51_, which assumes a continual growth in industrial production and consumer demands. _52_, it is doubtful whether this growth will or can continue. Having considered the arguments on both sides carefully, it seems there are good economic and ecological reasons for sources of energy _53_ nuclear power.44.A. material B. transforming C. consumer D. modern45.A. economical B. commercial C. clean D. financial46.A. increase B. step C. change D. demand47.A. estimate B. identity C. reject D. expect48.A. cheap B. sufficient C. legal D. economic49.A. possessions B. supplies C. expenses D. investments50.A. bring up B. bring about C. bring to D. bring in51.A. suffering B. surviving C. forming D. expanding52.A. However B. For example C. Therefore D. Besides53.A. other than B. more than C. rather than D. less than Section BDirections: Read the following passages. Each passage is followed by several questions or unfinished statement. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)Seek Out a Unique BeachVISITOregon. For sun and fun away from the crowed beaches pf Florida, check out the Oregon coast and its 363 miles of gorgeous shoreline, stretching from the Columbia River south to the redwood forests of California. Every beach is public and free. “The coast is a perfect place to watch sea lions sun themselves or simply see the mighty Pacific weaves crash in the sunset followed by a seafood feast in one of the busy fishing communities located between the coves(小海湾), ” says Bramblett. July and August aren’t peak gray whale migration season, but there’s still a good chance you could catch a glimpse of some of the 200 whales that spend the summers off the Oregon coast.SLEEPNext to a lighthouse. Imagine yourself an ancient mariner when you book a room overlooking the Pacific Ocean at the Heceta Head Lighthouse Bed and Breakfast in Yachats, Oregon. The working lighthouse, which dated to 1894, cast a bright beam 21 miles out to sea, making it the brightest light on the Oregon coast. The cliff-top rooms at the Light Keeper’s home nearby aren’t cheap —you’ll sell out up to $385 for a weekend night during peak season (price includes a seven-course breakfast).BEWAREDangerous currents Unless you love cold water (or wear a wet suit), you may not wait to venture into the sea off the Oregon coast, even during the summer. But if you do, be prepared for the U.S. Lifesaving Association. Swim parallel to the beach until you’re no longer being pulled out to sea, then swim diagonally(成对角线地) toward the shore.BRINGBaby powder. Use a generous amount of baby powder to remove sand from your hands, feet or hair. The powder quickly absorbs moisture, allowing sand to fall off easily.454.What is suggested by the leaflet if you want to enjoy your stay at the beach?A.Avoid wearing wet suits.B.Never dive into the cold water off the coast.C.Bring baby powder to protect your skin from sand scratches.D.Don’t swim straight toward the shore when there’re dangerous currents.55.Oregon coast will provide you with all the following experiences except ___.A.a perfect view of sea lionsB.a mariner like stay in the more than 100-year-old lighthouseC.a seafood feast in the popular local communityD.sun and fun of the less crowed beach(B)In his 1930 essay “Economic Possibilities for Our Grandchildren”, John Keynes, a famous economist, wrote that human needs fall into two classes: absolute needs, which are independent of what others have , and relative needs ,which make us feel superior to our fellows. He thought that although relative needs may indeed be insatiable (无止境的) , this is not true of absolute needs.Keynes was surely correct that only a small part of total spending is decided by the desire for superiority. He was greatly mistaken, however, in seeing this desire as the only source of insatiable demands.Decisions to spend are also driven by ideas of quality which can influence the demands for almost all goods, including even basic goods like food. When a couple goes out for an anniversary dinner, for example, the thought of feeling superior to others probably never comes to them. Their goal is to share a special meal that stands out from other meals.There are no obvious limits to the escalation of demand for quality. For example, Porsche, a famous car producer, has a model which was considered perhaps the best sport car on the market Priced at over $120,000, it handles perfectly well and has great speed acceleration. But in 2004, the producer introduced some changes which made the model slightly better in handling and acceleration. People who really care about cars find these small improvements exciting. To get them, however, they must pay almost four times the prices.By placing the desire to be superior to others at the heart of his description of insatiable demands, Keynes actually reduced such demands. However, the desire for higher quality has no natural limits.56. According to the passage, John Keynes Believed that_______.A.desire is the root of both absolute and relative needsB.absolute needs come from our sense of superiorityC.relative needs alone lead to insatiable demandsD.absolute needs are stronger than relative needs57. What does the word “escalation” paragraph 4 probably mean?A.Understanding.B.Increase.C.Difference .D. Expectation.58. The author of the passage argues that ______.A. absolute needs have no limitsB. demands for quality are not insatiableC. human desires influence ideas of qualityD. relative needs decide most of our spendingSection CDirections: Fill in each blank in the article with a proper sentence given below. Each sentence can be used only once. Note there are tow more sentences than you need.the dental equipment and treatments.Well, fear no more! A British company says it has developed technology that will end the need for mechanical cleaning of dental cavities(龋齿). __59_.For over a century, dentists have been repairing cavities the same way. They first remove the decayed, or bad, tooth tissue with an electric-powered drill.Then, they fill the hole in the bone with a metallic or plastic substance. __60_. However, if it is deep, near the nerve of the tooth, it can be very painful. The worst part is that the process of drilling and filling the cavity has to be repeated for the entire life of that tooth.Rebecca Moazzez is a senior lecturer at King’s College London. She says that this cycle lasts the r est of the tooth’s life.“You’re really in that cycle of repair and replacement for the rest of the tooth’s life.”Tooth enamel(牙釉质) is what we see on the outside of the tooth. The enamel of a damaged tooth can be replaced naturally. This process is called re-mineralization. But it is too slow to stop bacteria from building up in small, narrow areas of the enamel.A British business called Reminova has developed a method for speeding-up this natural re-mineralization of early-stage cavities. Jeff Wright is head of the company. “We’ve just found a way to make that a much faster process. Driving healthy calcium and phosphate minerals into your enamel and, through a natural process, it will bind on and add to the enamel that’s there.”The process begins with a cleaning of the cavity. This does not require power tools. __61_. The electricity is too weak for the patient to feel.And the hardened mineral completely fills the cavity.__62_. Sometimes children have bad experiences at the dentist. They fear the drilling and injections. Who wouldn’t be afraid? Better experiences as a child might lead to more visits to the dentist as an adult.IV. TranslatingDirections: Translate the following sentences into English, using the words in the brackets.1.大部分读者都与这部畅销小说里的助教有共鸣。

绝密★启用前上海市建平中学2017-2018学年高一上学期期末数学试题试卷副标题注意事项：1．答题前填写好自己的姓名、班级、考号等信息 2．请将答案正确填写在答题卡上第I 卷（选择题)请点击修改第I 卷的文字说明 一、单选题1．在下列四个说法中，与“不经冬寒，不知春暖”意义相同的是（ ） A ．若经冬寒，必知春暖 B ．不经冬寒，但知春暖 C ．若知春暖，必经冬寒D ．不知春暖，但历冬寒2．已知实数,x y 满足x y >，下列不等式中一定成立的是（ ） A ．33x y >B ．22x y >C ．00x y =D ．11x y -->3．函数()y f x =的定义域为[2,)-+∞，函数()y g x =为奇函数，则函数()F x =的定义域可能为（ ） A ．[2,0)(0,)-⋃+∞ B ．[2,1)(1,0)--⋃- C ．[2,1)(1,)--⋃+∞D ．[2,1](0,1]--⋃4．在股票等金融交易过程中，常用两种曲线来描述价格变化的情况：一种是即时价格曲线()y f x =，另一种是平均价格曲线()y g x =，如(2)3f =表示交易开始后2小时的即时价格为2元；(2)3g =则表示交易2小时内的平均价格为3元，下面给出了四个图像，实线表示()y f x =，虚线表示()y g x =，其中可能正确的是（ ）…………线…………○………………线…………○……A．B．C．D．第II卷（非选择题)请点击修改第II卷的文字说明二、填空题5．已知全集U=R，集合{|}xA x yπ==，则U C A=______6．函数1()f x xx=-在(,0)-∞内的零点为x=_____7．关于x的方程23x x=的解集为________8．函数1()f xx a=+为奇函数，则实数a的值为______9．集合{|}A x x a=<，{|1}B x x=<，若A B⊆，则实数a的取值范围为_______10．比较两数大小：100002_____5031e（在横线处填“＞”或“＝”或“＜”）11．函数()y f x=的定义域为(0,1)，则函数(2)y f x=的定义域为______12．幂函数2y x的单调递减区间为_______13．已知函数()y f x=过定点(0,2)，则函数(2)y f x=-过定点______14．不等式0x a-≥对任意[1,2]x∈-恒成立，则实数a的最大值为______15．若函数2()xf x a--=在(,5)-∞内有两个零点，则实数a的取值范围为_________16．方程20192020(2018)01xf x fx-⎛⎫++=⎪-⎝⎭恰有四个互异的实根，记为1234,,,x x x x，三、解答题17．不等式01x >-的解集为集合A ，不等式|1|1x -≤的解集为集合B ，求A B . 18．解关于x 的方程：24log (3)2log 2x x +-=19．“秃发”是一种常见的毛发疾病，随着发病人群年龄结构的年变化，逐渐引起了社会的广泛关注.一个人出生时头发数量约为100000根，数学徐老师建立了“秃发”函数模型作预估：一个人()x x N *∈岁时的头发根数为50000()100000f x ax x=--，其中a 称为“脱发指数”.（1）杜老师5岁时有74375根头发，请依据模型求出杜老师的“脱发指数”a 的值； （2）徐老师的学生认为“秃发”函数模型中有两个缺点：①头发的根数应该为整数；②头发的根数不能为负数，徐老师感觉很有道理，将模型作了两处修正，请写出修正后（1）问中杜老师的“秃发”函数模型，并求出杜老师几岁时头发最多.20．设函数12()2x x af x b+-+=+（实数,a b 为常数）（1）当1a b ==时，证明()f x 在R 上单调递减； （2）若2b =-，且()f x 为偶函数，求实数a 的值；（3）小金同学在求解函数12()2x x af x b+-+=+的对称中心时，发现函数()f x 是一个复合函数，设()2x a g x x b-+=+，()2=xh x ，则()(())f x g h x =，显然()g x 有对称中心，设为(,)m n ，()h x 有反函数1()h x -，则(())g h x 的对称中心为11((),())h m h n --，请问小金的做法是否正确？如果正确，请给出证明，并直接写出当2b a =时()f x 的对称中心；如果错误，请举出反例，并用正确的方法直接写出当2b a =时()f x 的对称中心. 21．设函数()f x 的反函数为1()f x -，若存在函数()g x 使得对函数()f x 定义域内的任意x 都有1(())()g f x fx -=，则称函数()g x 为函数()f x 的“Inverse ”函数.（1）判断下列哪个函数是函数2()log f x x =的“Inverse ”函数并说明理由.①21()(2)xg x =；②(2)2()2xg x =；（2）设函数()f x 存在反函数1()f x -，证明函数()f x 存在唯一的“Inverse ”函数的充要条件是函数()f x 的值域为R ； cx d+1-数，记1()(())()n n g x g g x n N *+=∈，其中1()()g x g x =，若对函数()f x 定义域内的任意x 都有2018()()g x f x =，求所有满足条件的函数()f x 的解析式.参考答案1．C 【解析】 【分析】根据逆否命题的等价性可得结果. 【详解】“不经冬寒，不知春暖”可写成“若不经冬寒，则不知春暖”， 根据逆否命题的等价性可知，“若知春暖，必经冬寒”与其意义相同. 故选：C. 【点睛】本题考查逆否命题与原命题的关系，考查学生的逻辑推理能力，属基础题. 2．A 【解析】 【分析】根据不等式的性质逐项判断即可. 【详解】选项A 中，由3y x =的单调性可知，若x y >，则33x y >，故A 正确； 选项B 中，举反例：1x =-，2y =-，则x y >，但22x y <，故B 不正确； 选项C 中，举反例：0x =，1y =-，则x y >，但00x y ≠，故C 不正确； 选项D 中，举反例：1x =-，2y =-，则x y >，但11x y --<，故D 不正确.故选：A. 【点睛】本题考查不等式的性质及其应用，常用举反例排除解决此类问题，属基础题. 3．B 【解析】 【分析】每个选项通过举反例排除可逐项判断正误. 【详解】选项A 中，2[2,0)(0,)-∈-⋃+∞，2[2,0)(0,)∈-⋃+∞，则(2)0g ->，(2)0g >，又函数()y g x =为奇函数，则(2)(2)0g g -=-<，矛盾，故A 不正确； 选项B 中，()0>g x 在[2,1)--和(1,0)-上可能同时成立；选项C 中，2[2,1)(1,)-∈--⋃+∞，2[2,1)(1,)∈--⋃+∞，则(2)0g ->，(2)0g >，又函数()y g x =为奇函数，则(2)(2)0g g -=-<，矛盾，故C 不正确；选项D 中，1[2,1](0,1]-∈--⋃，1[2,1](0,1]∈--⋃，则(1)0g ->，(1)0g >，又函数()y g x =为奇函数，则(1)(1)0g g -=-<，矛盾，故D 不正确.故选：B. 【点睛】本题考查抽象函数的定义域和奇函数的定义的应用，若函数()y g x =为奇函数，且()0>g x 在[,]a b 成立，则()0<g x 在[,]b a --成立，由此可通过举反例排除解决本题，属中档题. 4．D 【解析】 【分析】根据题意可知，刚开始交易时，即时价格和平均价格应该相等，由此排除不正确的选项，接下来分析开始交易后，平均价格与即时价格之间的变动满足同增同减，进而再排除不正确的选项，即可得到本题的答案. 【详解】刚开始交易时，即时价格和平均价格应该相等，故A 错误；开始交易后，平均价格应该跟随即时价格变动，即时价格与平均价格应同增同减，故B ，C 错误. 故选：D. 【点睛】本题考查判断函数图象，解答本题的关键是明确平均价格与即时价格之间的变动关系，属中档题. 5．∅ 【解析】【分析】求出集合A ，再求其补集即可. 【详解】依题意，x y π=的定义域为R ，即A R =，则U C A =∅. 故答案为：∅. 【点睛】本题考查集合的补集运算，属基础题. 6．－1 【解析】 【分析】 令10x x-=，结合定义域求解即可得到结果. 【详解】 依题意，令10x x-=，即210x -=，解得1x =±， 又(,0)x ∈-∞，则1x =-. 故答案为：－1. 【点睛】本题考查函数零点的求解，要求掌握函数的零点与方程的根的关系，属基础题. 7．{}0 【解析】 【分析】将方程变形为213x⎛⎫= ⎪⎝⎭，从而求解可得结果. 【详解】由23x x =得，213x⎛⎫= ⎪⎝⎭，则0x =，所以方程的解集为{}0. 故答案为：{}0. 【点睛】本题考查简单指数方程的求解，注意解集的表达形式，属基础题.8．0 【解析】 【分析】根据奇函数的性质()()f x f x -=-可得11x a x a=--++，从而可求解实数a 的值.【详解】依题意，()()f x f x -=-，即11x a x a=--++，整理可得，x a x a -+=--，解得0a =. 故答案为：0. 【点睛】本题考查奇函数的性质的应用，已知函数()f x 为奇函数，则必然满足()()f x f x -=-，属基础题. 9．(,1]-∞ 【解析】 【分析】借助数轴即可求得实数a 的取值范围. 【详解】如图，若A B ⊆，则1a ≤.故答案为：(,1]-∞. 【点睛】本题考查集合的基本关系，已知集合的包含关系求参数的范围常常借助数轴求解，属基础题. 10．＞ 【解析】 【分析】依题意，50311000050310001022⎛⎫= ⎪⎝⎭，又100001000055031600032223>==>=，将100002放缩为50313，由此可比较100002和5031e 的大小. 【详解】50311000050310001022⎛⎫= ⎪⎝⎭，又100001000055031600032223>==>=，则50311000050315031503100001223e ⎛⎫=>> ⎪⎝⎭.故答案为：＞. 【点睛】本题考查指数函数和幂函数的单调性应用，比较指数式的大小一般化成相同的底数，再利用单调性比较大小，比较幂函数值的大小，则一般化成指数相同的形式，再利用单调性比较大小，属中档题. 11．1(0,)2【解析】 【分析】先求()y f x =的自变量范围为：01x <<，从而得到函数(2)y f x =中，021x <<，求解即可得到结果. 【详解】函数()y f x =的定义域为(0,1)，即()y f x =的自变量范围为：01x <<. 则函数(2)y f x =中，021x <<，解得102x <<， 所以函数(2)y f x =的定义域为1(0,)2. 故答案为：1(0,)2. 【点睛】本题考查抽象函数的定义域求法，要求学生理解并熟练掌握抽象函数的定义域，属基础题. 12．(0,)+∞ 【解析】【分析】结合幂函数的幂与单调性的关系和偶函数的性质即可求得2y x 的单调递减区间.【详解】 幂函数221yxx ，易知其为偶函数，定义域为{}|0x x ≠， 由幂函数的幂与单调性的关系可知，20-<，则2y x 在(0,)+∞是单调递减的，所以幂函数2yx 的单调递减区间为(0,)+∞.故答案为：(0,)+∞. 【点睛】本题考查偶函数的性质，以及幂函数的幂与单调性的关系，熟练掌握幂函数的单调性和奇偶性与幂的关系是解题的关键，属基础题. 13．(2,2) 【解析】 【分析】利用函数的图象平移可得到结果. 【详解】函数(2)y f x =-的图象是由函数()y f x =的图象向右平移两个单位得到的， 则(2)y f x =-的定点可由(0,2)向右平移两个单位得到，即(2,2). 故答案为：(2,2). 【点睛】本题考查函数的图象平移规律，属基础题. 14．0 【解析】 【分析】原题等价于a x ≤对任意[1,2]x ∈-恒成立，则min a x ≤，求出x 的最小值即可求得结果. 【详解】由0x a -≥可得，a x ≤，原题等价于a x ≤对任意[1,2]x ∈-恒成立，则min 0a x ≤=，则实数a 的最大值为0. 故答案为：0. 【点睛】本题考查不等式恒成立问题，一般采用分离参数的方法或者分类讨论的思想解决不等式恒成立问题，属中档题. 15．1a e>【解析】 【分析】原题等价于2(xy e x -=-和1y a=在[1,2)(2,5)上的图象有两个交点，令[0,1)(1,2)t =，则21x t =+，故212(1)t y e t -=-，原题等价于212(1)t y e t -=-和1y a=在[0,1)(1,2)上的图象有两个交点，设222121212(1),01()(1)(1),12t tt e t t g t et e t t ---⎧-≤<⎪=-=⎨-<<⎪⎩，则()0g t >，分析其单调性并画图，结合图象即可得到a 的范围. 【详解】2()xf x a --=的定义域为[1,2)(2,)⋃+∞，则其函数()f x 在[1,2)(2,5)内有两个零点，20xa --=在[1,2)(2,5)上有两个根，等价于2(xy ex -=-和1y a=在[1,2)(2,5)上的图象有两个交点，令[0,1)(1,2)t =，则21x t =+，故212(1)t y e t -=-，原题等价于212(1)t y e t -=-和1y a=在[0,1)(1,2)上的图象有两个交点，设222121212(1),01()(1)(1),12t tt e t t g t et e t t ---⎧-≤<⎪=-=⎨-<<⎪⎩，则()0g t >，易得21t y e -=和2(1)y t =-在01t ≤<时单调递减且均大于0， 故212()1)(t g t e t -=-在01t ≤<时单调递减，同理可得212())(1tg t e t -=-在12t <<时单调递增，如图为212()1)(t g t e t -=-的图象，由3((2)0)g e g e >==，结合图象可得，10e a<<，即1a e >.故答案为：1a e>. 【点睛】本题考查函数与方程的综合问题，考查已知函数零点个数求参数的范围，将函数零点问题转化为两个函数的交点问题是常用手段，属难题. 16．2019 【解析】 【分析】令1t x =-，则原方程变为：1(2019)20190f t f t ⎛⎫++-= ⎪⎝⎭，赋值可得，若0t 为方程的根，则01t -也为方程的根，依题意，方程恰有四个互异的实根，记为1234,,,t t t t ，则131t t =-，241t t =-，由此可得12341224(1)(1)(1)(1)201201828019t x t x t x t x ----++==.【详解】令1t x =-，则原方程变为：1(2019)20190f t f t ⎛⎫++-= ⎪⎝⎭，设0t 为方程的根，则001(2019)20190f t f t ⎛⎫++-= ⎪⎝⎭，将01t -代入上面的方程可得，0012019(2019)0f f t t ⎛⎫-++= ⎪⎝⎭， 故若0t 为方程的根，则01t -也为方程的根. 依题意，方程恰有四个互异的实根，记为1234,,,t t t t ， 则311t t =-，421t t =-，即131t t =-，241t t =-， 故12341224(1)(1)(1)(1)201201828019t x t x t x t x ----++==. 故答案为：2019. 【点睛】本题考查函数与方程的综合，通过换元法简化方程，并利用赋值法可知，方程的根之间的关系是解题的关键，属难题. 17．(1,2] 【解析】 【分析】分别求解集合A 和B ，再求A B 即可.【详解】201x x +>-等价于：()()210x x +->，解得1x >或2x <-， 则()(),21,A =-∞-+∞，|1|1x -≤等价于：111x -≤-≤，解得02x ≤≤，即[0,2]B =，所以(1,2]A B ⋂=. 【点睛】本题考查分式不等式和绝对值不等式的求解以及集合的交集运算，注意仔细审题，认真计算，属基础题. 18．1x = 【解析】 【分析】先将对数方程变形为22log (3)log 2x x +-=，整理可得34x x+=，求解即可得到方程的解. 【详解】24log (3)2log 2x x +-=等价于：22log (3)log 2x x +-=，即2(3)log 2x x +=，即34x x+=，解得1x =，经检验，1x =是方程的根. 【点睛】本题考查对数方程的求解，熟练掌握对数式的计算化简是关键，属基础题.19．（1）3125a =；（2）500001000003125,131()0,32x x f x x x ⎧⎡⎤--≤≤⎪⎢⎥=⎣⎦⎨⎪≥⎩()x N *∈，杜老师4岁时头发最多. 【解析】 【分析】（1）将5x =代入求解即可求出a 的值；（2）引入求整数的符号[]x ，且将函数写成分段函数的形式，当函数值为负数时变为常数0即可，最后结合基本不等式求出max ()f x 即可. 【详解】（1）依题意，当5x =时，50000(5)1000005743755f a =--=， 解得，3125a =，所以，杜老师的“脱发指数”a 的值为3125；（2）依题意可得，500001000003125,131()0,32x x f x x x ⎧⎡⎤--≤≤⎪⎢⎥=⎣⎦⎨⎪≥⎩()x N *∈， 其中[]x 表示不超过x 的最大整数，由50000312525000x x +≥=， 当且仅当4x =时，等号成立，此时，max ()75000f x =. 即杜老师4岁时头发最多. 【点睛】本题考查函数的应用，结合实际问题建立合适的函数模型是解决问题的关键，属难题.20．（1）证明见解析；（2）1a =；（3）若0a =，()f x 的对称中心为1(,)2m -()m R ∈；若0a ≠，()f x 的对称中心为221(log ,0)2a .【解析】 【分析】（1）先将()f x 化简，再利用定义法证明单调性即可； （2）由偶函数的性质()()f x f x =-化简求解即可得到a ；（3）利用（1）作为反例可知小金的做法是错误的，分别讨论0a =和0a ≠的情况，结合对称点的性质()()2f m x f m x n ++-=可得(,)m n . 【详解】（1）当1a b ==时，1111233221122()2121221x xx x x f x +++--+-+===-++++， 任取12,x x R ∈，且12x x <，则()()()()()()2121121212111211111133322322222()()212121212121x x x x x x x x x x f x f x ++++++++---=-==++++++， 由12x x <得，1222x x <，即21220x x ->，又()()121121210x x ++++>，所以12())0(f x f x ->，即12()()f x f x >，故()f x 在R 上单调递减；（2）依题意，12()22x x f x a +-+=-，由()()f x f x =-可得，11222222x x x x a a-+-+-+-+=--，整理可得，(1)(21)0x a -+=，解得1a =；（3）错误，令1a b ==，则3112()21221x g x x x -+==-+++，显然()g x 有对称中心11(,)22--，12()log h x x -=，很明显，1211()log 22h -⎛⎫-=-⎪⎝⎭没有意义， 当2b a =时，12()22x x f x aa+-++=，若0a =，1()2f x =-，则直线上每一个点1(,)2m -()m R ∈都是()f x 的对称中心.若0a ≠，设()f x 的对称中心为(,)m n ，则()()2f m x f m x n ++-=，由此可得，221log 2m a =，0n =，即()f x 的对称中心为221(log ,0)2a .【点睛】本题考查函数的性质和反函数的概念，已知()f x 的对称中心为(,)m n ，则()()2f m x f m x n ++-=是解决第三问的关键，属难题.21．（1）②是函数f (x )=log 2x 的“Inverse ”函数，理由见解析；（2）证明见解析；（3）()f x x =. 【解析】 【分析】（1）分别判断①和②是否满足1(())()g f x f x -=即可得到结果；（2）先证充分性，若函数()f x 的值域为R ，设其定义域为D ，则函数1()f x -的定义域为R ，值域为D ， 令11()(())g x ff x --=，x ∈R ，判断是否满足1(())()g f x f x -=，证明其存在性，再设函数1()g x 和2()g x 都为函数()f x 的“Inverse ”函数且不相同，利用反证法证明唯一性；再证必要性，若函数()f x 存在唯一的“Inverse ”函数，同样利用反证法，假设函数()f x 的值域为KR ，令1(),()0,U g x x K h x x C K ∈⎧=⎨∈⎩，2(),()1,U g x x Kh x x C K ∈⎧=⎨∈⎩，通过证明函数1()h x 和2()h x 都为函数()f x 的“Inverse ”函数且不相同，这与唯一性矛盾，从而得证；（3）由（2）知，11()(())g x f f x --=是()f x 的一个“Inverse ”函数，易得，111120184036()((()))()f g x f f f x f x ----==个，即111114037((()))(())f f f f x f f x x-----==个，根据一一对应的性质可得1()f x x -=，所以()f x x =.【详解】（1）易得1()2x f x -=，对于①，()()22log 2log 221(())222xxx g f x x ===≠，故①不是，对于②，()log 2212(())22()xx g f x f x -===，故②是函数2()log f x x =的“Inverse ”函数；（2）先证充分性，若函数()f x 的值域为R ，设其定义域为D ，则函数1()f x -的定义域为R ，值域为D ， 令11()(())g x ff x --=，x ∈R ，则对任意x D ∈都有，111(())((()))()g f x ff f x f x ---==，故函数()g x 为函数()f x 的“Inverse ”函数，存在性得证； 设函数1()g x 和2()g x 都为函数()f x 的“Inverse ”函数且不相同， 则存在0x R ∈，10()g x a =，20()g x b =，且a b ，因为()f x 的值域为R ，故存在m D ∈，使得0()f m x =，即11(())()g f m f m a -==，12(())()g f m f m b -==，则11()()fm f m --≠，矛盾，故唯一性得证.所以函数()f x 存在唯一的“Inverse ”函数.再证必要性，若函数()f x 存在唯一的“Inverse ”函数， 即存在唯一的函数()g x 满足1(())()g f x f x -=，下面用反证法证明必要性.假设函数()f x 的值域为K R ，令1(),()0,U g x x K h x x C K ∈⎧=⎨∈⎩，2(),()1,U g x x Kh x x C K ∈⎧=⎨∈⎩，则对任意x D ∈都有，()f x K ∈， 且11(())(())()h f x g f x fx -==，12(())(())()h f x g f x f x -==，函数1()h x 和2()h x 都为函数()f x 的“Inverse ”函数且不相同，这与唯一性矛盾， 所以函数()f x 的值域为R ，必要性得证.综上，函数()f x 存在唯一的“Inverse ”函数的充要条件是函数()f x 的值域为R ；（3）由（2）知，11()(())g x f f x --=是()f x 的一个“Inverse ”函数，由反函数的性质可知，()f x 和1()f x -都是一一对应的.则2018201720162018()(())((()))((()))gg x g g x g g g x g g g x ====个，又11()(())g x ff x --=，则111120184036()((()))()f g x f f f x f x ----==个，即111114037((()))(())f f f f x f f x x-----==个，根据一一对应的性质可得1()fx x -=，则()f x x =，所以满足条件的函数()f x 的解析式为()f x x =. 【点睛】本题考查函数新定义问题和反函数的性质应用，着重考查学生的逻辑推理能力和抽象概括能力，思维性很强，属难题.。

上海建平中学人教版七年级上册数学期末试卷及答案.doc一、选择题 1．以下选项中比-2小的是（ ） A ．0 B ．1 C ．-1.5 D ．-2.52．有理数a ，b 在数轴上的对应点的位置如图所示，则下列各式成立的是（ ）A ．a ＞bB ．﹣ab ＜0C ．|a |＜|b |D ．a ＜﹣b3．把一根木条固定在墙面上，至少需要两枚钉子，这样做的数学依据是（ ） A ．两点之间线段最短 B ．两点确定一条直线C ．垂线段最短D ．两点之间直线最短4．已知线段AB 的长为4，点C 为AB 的中点，则线段AC 的长为（ ）A ．1B ．2C ．3D ．45．如图，OA ⊥OC ，OB ⊥OD ，①∠AOB=∠COD ；②∠BOC+∠AOD=180°；③∠AOB+∠COD=90°；④图中小于平角的角有6个；其中正确的结论有几个（ ）A ．1个B ．2个C ．3个D ．4个6．计算：31﹣1=2，32﹣1=8，33﹣1=26，34﹣1=80，35﹣1=242，…，归纳各计算结果中的个位数字的规律，猜测32018﹣1的个位数字是（ ）A ．2B ．8C ．6D ．07．﹣3的相反数是（ ）A ．13- B ．13 C ．3- D ．38．下列各数中，绝对值最大的是（ ）A ．2B ．﹣1C ．0D ．﹣39．按如图所示图形中的虚线折叠可以围成一个棱柱的是（ ）A ．B ．C ．D ．10．点()5,3M 在第( )象限．A ．第一象限B ．第二象限C ．第三象限D ．第四象限11．如图，小明将自己用的一副三角板摆成如图形状，如果∠AOB=155°，那么∠COD 等于（ ）A ．15°B ．25°C ．35°D ．45°12．某商店出售两件衣服，每件卖了200元，其中一件赚了25%，而另一件赔了20%．那么商店在这次交易中( )A ．亏了10元钱B ．赚了10钱C ．赚了20元钱D ．亏了20元钱二、填空题13．将一根木条固定在墙上只用了两个钉子，这样做的依据是_______________.14．如图，点C 在线段AB 的延长线上，BC ＝2AB ，点D 是线段AC 的中点，AB ＝4，则BD 长度是_____．15．已知单项式245225n m x y x y ++与是同类项，则m n =______.16．如图，是七(2)班全体学生的体有测试情况扇形统计图.若达到优秀的有25人，则不合格的学生有____人.17．若关于x 的方程2x 3a 4+=的解为最大负整数，则a 的值为______．18．将520000用科学记数法表示为_____．19．已知线段AB=8cm ，在直线AB 上画线段BC ，使它等于3cm ，则线段AC=______cm ．20．如图,在平面直角坐标系中,动点P 按图中箭头所示方向从原点出发,第1次运动到P 1(1,1),第2次接着运动到点P 2(2，0)，第3次接着运动到点P 3(3，-2)，…，按这的运动规律,点P 2019的坐标是_____.21．方程x ＋5＝12(x ＋3)的解是________．22．用“＞”或“＜”填空：13_____35；223-_____﹣3． 23．钟表显示10点30分时，时针与分针的夹角为________．24．设一列数中相邻的三个数依次为m ，n ，p ，且满足p=m 2﹣n ，若这列数为﹣1，3，﹣2，a ，b ，128…，则b=________.三、压轴题25．已知AOD α∠=，OB 、OC 、OM 、ON 是AOD ∠内的射线．(1)如图1，当160α=︒，若OM 平分AOB ∠，ON 平分BOD ∠，求MON ∠的大小；(2)如图2，若OM 平分AOC ∠，ON 平分BOD ∠，20BOC ∠=︒，60MON ∠=︒，求α．26．已知数轴上两点A 、B ，其中A 表示的数为-2，B 表示的数为2，若在数轴上存在一点C ，使得AC+BC=n ，则称点C 叫做点A 、B 的“n 节点”．例如图1所示：若点C 表示的数为0，有AC+BC=2+2=4，则称点C 为点A 、B 的“4节点”．请根据上述规定回答下列问题：（1）若点C 为点A 、B 的“n 节点”，且点C 在数轴上表示的数为-4，求n 的值； （2）若点D 是数轴上点A 、B 的“5节点”，请你直接写出点D 表示的数为______； （3）若点E 在数轴上（不与A 、B 重合），满足BE=12AE ，且此时点E 为点A 、B 的“n 节点”，求n 的值．27．已知∠AOB 和∠AOC 是同一个平面内的两个角,OD 是∠BOC 的平分线.(1)若∠AOB=50°,∠AOC=70°,如图(1),图(2),求∠AOD 的度数；(2)若∠AOB=m 度,∠AOC=n 度,其中090090180m n m n ＜＜，＜＜，＜+且m n ＜，求∠AOD 的度数(结果用含m n 、的代数式表示)，请画出图形，直接写出答案．28．如图，以长方形OBCD的顶点O为坐标原点建立平面直角坐标系，B点坐标为（0，a），C点坐标为（c，b），且a、b、C满足6a +|2b+12|+（c﹣4）2＝0．（1）求B、C两点的坐标；（2）动点P从点O出发，沿O→B→C的路线以每秒2个单位长度的速度匀速运动，设点P 的运动时间为t秒，DC上有一点M（4，﹣3），用含t的式子表示三角形OPM的面积；（3）当t为何值时，三角形OPM的面积是长方形OBCD面积的13？直接写出此时点P的坐标．29．在数轴上，图中点A表示－36，点B表示44，动点P、Q分别从A、B两点同时出发，相向而行，动点P、Q的运动速度比之是3∶2（速度单位：1个单位长度/秒）．12秒后，动点P到达原点O，动点Q到达点C，设运动的时间为t（t＞0）秒．（1）求OC的长；（2）经过t秒钟，P、Q两点之间相距5个单位长度，求t的值；（3）若动点P到达B点后，以原速度立即返回，当P点运动至原点时，动点Q是否到达A点，若到达，求提前到达了多少时间，若未能到达，说明理由．30．数轴上线段的长度可以用线段端点表示的数进行减法运算得到，例如：如图①，若点A，B在数轴上分别对应的数为a，b(a<b)，则AB的长度可以表示为AB=b－a．请你用以上知识解决问题：如图②，一个点从数轴上的原点开始，先向左移动2个单位长度到达A点，再向右移动3个单位长度到达B点，然后向右移动5个单位长度到达C点．（1）请你在图②的数轴上表示出A，B，C三点的位置．（2）若点A以每秒1个单位长度的速度向左移动，同时，点B和点C分别以每秒2个单位长度和3个单位长度的速度向右移动，设移动时间为t秒．①当t=2时，求AB和AC的长度；②试探究：在移动过程中，3AC－4AB的值是否随着时间t的变化而改变？若变化，请说明理由；若不变，请求其值．31．如图①，点C在线段AB上，图中共有三条线段AB、AC和BC，若其中有一条线段的长度是另外一条线段长度的2倍，则称点C是段AB的“2倍点”．（1）线段的中点__________这条线段的“2倍点”；（填“是”或“不是”）（2）若AB＝15cm，点C是线段AB的“2倍点”．求AC的长；（3）如图②，已知AB＝20cm．动点P从点A出发，以2c m／s的速度沿AB向点B匀速移动．点Q从点B出发，以1c m/s的速度沿BA向点A匀速移动．点P、Q同时出发，当其中一点到达终点时，运动停止，设移动的时间为t（s），当t＝_____________s时，点Q 恰好是线段AP的“2倍点”．（请直接写出各案）32．已知：如图，点A、B分别是∠MON的边OM、ON上两点，OC平分∠MON，在∠CON的内部取一点P（点A、P、B三点不在同一直线上），连接PA、PB．（1）探索∠APB与∠MON、∠PAO、∠PBO之间的数量关系，并证明你的结论；（2）设∠OAP=x°，∠OBP=y°，若∠APB的平分线PQ交OC于点Q，求∠OQP的度数（用含有x、y的代数式表示）．【参考答案】***试卷处理标记，请不要删除一、选择题1．D解析：D【解析】根据有理数比较大小法则：负数的绝对值越大反而越小可得答案.【详解】根据题意可得：2.52 1.501-<-<-<<，故答案为：D.【点睛】本题考查的是有理数的大小比较，解题关键在于负数的绝对值越大值越小.2．D解析：D【解析】【分析】根据各点在数轴上的位置得出a、b两点到原点距离的大小，进而可得出结论．【详解】解：∵由图可知a＜0＜b，∴ab＜0，即-ab＞0又∵|a|＞|b|，∴a＜﹣b．故选：D．【点睛】本题考查的是数轴，熟知数轴上两点间的距离公式是解答此题的关键．3．B解析：B【解析】因为两点确定一条直线,所以把一根木条固定在墙面上,至少需要两枚钉子故选B. 4．B解析：B【解析】【分析】根据线段中点的性质，可得AC的长．【详解】解：由线段中点的性质，得AC＝12AB＝2．故选B．【点睛】本题考查了两点间的距离，利用了线段中点的性质．5．C解析：C【解析】根据垂直的定义和同角的余角相等分别计算后对各小题进行判断，由此即可求解．【详解】∵OA⊥OC，OB⊥OD，∴∠AOC=∠BOD=90°，∴∠AOB+∠BOC=∠COD+∠BOC=90°，∴∠AOB=∠COD，故①正确；∠BOC+∠AOD=90°﹣∠AOB+90°+∠AOB=180°，故②正确；∠AOB+∠COD不一定等于90°，故③错误；图中小于平角的角有∠AOB，∠AOC，∠AOD，∠BOC，∠BOD，∠COD一共6个，故④正确；综上所述，说法正确的是①②④．故选C．【点睛】本题考查了余角和补角，垂直的定义，是基础题，熟记概念与性质并准确识图，理清图中各角度之间的关系是解题的关键．6．B解析：B【解析】【分析】由31﹣1=2，32﹣1=8，33﹣1=26，34﹣1=80，35﹣1=242，…得出末尾数字以2，8，6，0四个数字不断循环出现，由此用2018除以4看得出的余数确定个位数字即可．【详解】∵2018÷4=504…2，∴32018﹣1的个位数字是8，故选B．【点睛】本题考查了尾数的特征，关键是能根据题意得出个位数字循环的规律是解决问题的关键．7．D解析：D【解析】【分析】相反数的定义是：如果两个数只有符号不同，我们称其中一个数为另一个数的相反数，特别地，0的相反数还是0．【详解】根据相反数的定义可得：－3的相反数是3.故选D.【点睛】本题考查相反数，题目简单，熟记定义是关键.8．D【解析】试题分析：∵|2|=2，|﹣1|=1，|0|=0，|﹣3|=3，∴|﹣3|最大，故选D ．考点：D ．9．C解析：C【解析】【分析】利用棱柱的展开图中两底面的位置对A 、D 进行判断；根据侧面的个数与底面多边形的边数相同对B 、C 进行判断．【详解】棱柱的两个底面展开后在侧面展开图相对的两边上，所以A 、D 选项错误；当底面为三角形时，则棱柱有三个侧面，所以B 选项错误，C 选项正确．故选：C ．【点睛】本题考查了棱柱的展开图：通过结合立体图形与平面图形的相互转化，去理解和掌握几何体的展开图，要注意多从实物出发，然后再从给定的图形中辨认它们能否折叠成给定的立体图形．10．A解析：A【解析】【分析】根据平面直角坐标系中点的坐标特征判断即可.【详解】∵5>0,3>0,∴点()5,3M 在第一象限．故选A.【点睛】本题考查了平面直角坐标系中点的坐标特征.第一象限内点的坐标特征为（+，+），第二象限内点的坐标特征为（-，+），第三象限内点的坐标特征为（-，-），第四象限内点的坐标特征为（+，-），x 轴上的点纵坐标为0，y 轴上的点横坐标为0.11．B解析：B【解析】【分析】利用直角和角的组成即角的和差关系计算．【详解】解：∵三角板的两个直角都等于90°，所以∠BOD+∠AOC=180°，∵∠BOD+∠AOC=∠AOB+∠COD ，∵∠AOB=155°，∴∠COD等于25°．故选B．【点睛】本题考查角的计算，数形结合掌握角之间的数量关系是本题的解题关键．12．A解析：A【解析】设一件的进件为x元，另一件的进价为y元，则x（1+25%）=200，解得，x=160，y（1-20%）=200，解得，y=250，∴（200-160）+（200-250）=-10（元），∴这家商店这次交易亏了10元.故选A．二、填空题13．两点确定一条直线．【解析】将一根木条固定在墙上只用了两个钉子，他这样做的依据是：两点确定一条直线．故答案为两点确定一条直线．解析：两点确定一条直线．【解析】将一根木条固定在墙上只用了两个钉子，他这样做的依据是：两点确定一条直线．故答案为两点确定一条直线．14．【解析】【分析】先根据AB＝4，BC＝2AB求出BC的长，故可得出AC的长，再根据D是AC的中点求出AD的长度，由BD＝AD﹣AB即可得出结论．【详解】解：∵AB＝4，BC＝2AB，∴B解析：【解析】【分析】先根据AB＝4，BC＝2AB求出BC的长，故可得出AC的长，再根据D是AC的中点求出AD 的长度，由BD＝AD﹣AB即可得出结论．【详解】解：∵AB ＝4，BC ＝2AB ，∴BC ＝8．∴AC ＝AB +BC ＝12．∵D 是AC 的中点，∴AD ＝12AC ＝6． ∴BD ＝AD ﹣AB ＝6﹣4＝2．故答案为：2．【点睛】本题考查的是两点间的距离，熟知各线段之间的和、差及倍数关系是解答此题的关键． 15．9【解析】【分析】根据同类项的定义进行解题，则，解出m 、n 的值代入求值即可.【详解】解：和是同类项且，【点睛】本题考查同类型的定义，解题关键是针对x 、y 的次方都相等联立等式解出 解析：9【解析】【分析】根据同类项的定义进行解题，则25,24n m +=+=，解出m 、n 的值代入求值即可.【详解】解：242n x y +和525m x y +是同类项∴25n +=且24m +=∴3n =，2m =∴239m n ==【点睛】本题考查同类型的定义，解题关键是针对x 、y 的次方都相等联立等式解出m 、n 的值即可. 16．5【解析】【分析】根据达到优秀的人数和所占百分比求出总人数，然后用总人数乘以不合格所占的百分比即可.【详解】解：∵学生总人数=25÷50%=50（人），∴不合格的学生人数=50×（1-5解析：5【解析】【分析】根据达到优秀的人数和所占百分比求出总人数，然后用总人数乘以不合格所占的百分比即可.【详解】解：∵学生总人数=25÷50%=50（人），∴不合格的学生人数=50×（1-50%-40%）=5（人），故答案为：5.【点睛】本题考查了扇形统计图，熟知扇形统计图中各数据所表示的意义是解题关键.17．2【解析】【分析】求出最大负整数解，再把x=-1代入方程，即可求出答案．【详解】解：最大负整数为，把代入方程得：，解得：，故答案为2．【点睛】本题考查有理数和一元一次方程的解，能解析：2【解析】【分析】求出最大负整数解，再把x=-1代入方程，即可求出答案．【详解】解：最大负整数为1-，把x 1=-代入方程2x 3a 4+=得：23a 4-+=，解得：a 2=，故答案为2．【点睛】本题考查有理数和一元一次方程的解，能得出关于a的一元一次方程是解此题的关键．18．2×105【解析】【分析】科学记数法的表示形式为a×10n的形式，其中1≤|a|＜10，n为整数．确定n 的值时，要看把原数变成a时，小数点移动了多少位，n的绝对值与小数点移动的位数相同．当原数解析：2×105【解析】【分析】科学记数法的表示形式为a×10n的形式，其中1≤|a|＜10，n为整数．确定n的值时，要看把原数变成a时，小数点移动了多少位，n的绝对值与小数点移动的位数相同．当原数绝对值＞10时，n是正数；当原数的绝对值＜1时，n是负数．【详解】解：将520000用科学记数法表示为5.2×105．故答案为：5.2×105．【点睛】此题考查科学记数法的表示方法．科学记数法的表示形式为a×10n的形式，其中1≤|a|＜10，n为整数，表示时关键要正确确定a的值以及n的值．19．5或11【解析】【分析】由于C点的位置不能确定，故要分两种情况考虑AC的长，注意不要漏解．【详解】由于C点的位置不确定，故要分两种情况讨论：当C点在B点右侧时，如图所示：AC=AB+解析：5或11【解析】【分析】由于C点的位置不能确定，故要分两种情况考虑AC的长，注意不要漏解．【详解】由于C点的位置不确定，故要分两种情况讨论：当C点在B点右侧时，如图所示：AC=AB+BC=8+3=11cm；当C点在B点左侧时，如图所示：AC=AB﹣BC=8﹣3=5cm；所以线段AC等于11cm或5cm.20．(2019，-2)【解析】【分析】观察不难发现，点的横坐标等于运动的次数，纵坐标每4次为一个循环组循环，用2019除以4，余数是几则与第几次的纵坐标相同，然后求解即可．【详解】∵第1次运动解析：(2019，-2)【解析】【分析】观察不难发现，点的横坐标等于运动的次数，纵坐标每4次为一个循环组循环，用2019除以4，余数是几则与第几次的纵坐标相同，然后求解即可．【详解】∵第1次运动到点(1，1)，第2次运动到点(2，0)，第3次接着运动到点(3，-2)，第4次运动到点(4，0)，第5次运动到点(5，1)…，∴运动后点的横坐标等于运动的次数，第2019次运动后点P的横坐标为2019，纵坐标以1、0、-2、0每4次为一个循环组循环，∵2019÷4=504…3，∴第2019次运动后动点P的纵坐标是第504个循环组的第3次运动，与第3次运动的点的纵坐标相同，为-2，∴点P(2019，-2)，故答案为：(2019，-2)．【点睛】本题是对点的坐标的规律的考查，根据图形观察出点的横坐标与纵坐标的变化规律是解题的关键．21．x=-7【解析】去分母得，2（x+5）=x+3，去括号得，2x+10=x+3移项合并同类项得，x=-7.解析：x=-7【解析】去分母得，2（x+5）=x+3，去括号得，2x+10=x+3移项合并同类项得，x=-7.22．＜＞【解析】【分析】有理数大小比较的法则：①正数都大于0；②负数都小于0；③正数大于一切负数；④两个负数，绝对值大的其值反而小，据此判断即可．【详解】解：＜；＞﹣3．故答解析：＜＞【解析】【分析】有理数大小比较的法则：①正数都大于0；②负数都小于0；③正数大于一切负数；④两个负数，绝对值大的其值反而小，据此判断即可．【详解】解：13＜35；223＞﹣3．故答案为：＜、＞．【点睛】此题主要考查了有理数大小比较的方法，要熟练掌握，解答此题的关键是要明确：①正数都大于0；②负数都小于0；③正数大于一切负数；④两个负数，绝对值大的其值反而小．23．【解析】由于钟面被分成12大格，每格为30°，而10点30分时，钟面上时针指向数字10与11的中间，分针指向数字6，则它们所夹的角为4×30°+×30°．解：10点30分时，钟面上时针指向数字解析：【解析】由于钟面被分成12大格，每格为30°，而10点30分时，钟面上时针指向数字10与11的中间，分针指向数字6，则它们所夹的角为4×30°+12×30°．解：10点30分时，钟面上时针指向数字10与11的中间，分针指向数字6，所以时针与分针所成的角等于4×30°+12×30°=135°．故答案为：135°．24．-7【解析】【分析】先根据题意求出a的值，再依此求出b的值.【详解】解：根据题意得：a=32-（-2）=11，则b=(-2)2-11=-7．故答案为：-7.【点睛】本题考查探索与表解析：-7【解析】【分析】先根据题意求出a的值，再依此求出b的值.【详解】解：根据题意得：a=32-（-2）=11，则b=(-2)2-11=-7．故答案为：-7.【点睛】本题考查探索与表达规律——数字类规律探究. 熟练掌握变化规律，根据题意求出a和b是解决问题的关键．三、压轴题25．（1）80°；（2）140°【解析】【分析】（1）根据角平分线的定义得∠BOM=12∠AOB，∠BON=12∠BOD，再根据角的和差得∠AOD=∠AOB+∠BOD，∠MON=∠BOM+∠BON，结合三式求解；（2）根据角平分线的定义∠MOC=12∠AOC，∠BON=12∠BOD，再根据角的和差得∠AOD=∠AOC+∠BOD-∠BOC，∠MON=∠MOC+∠BON-∠BOC结合三式求解.【详解】解：（1）∵OM平分∠AOB，ON平分∠BOD，∴∠BOM=12∠AOB，∠BON=12∠BOD，∴∠MON=∠BOM+∠BON=12∠AOB+12∠BOD=12(∠AOB+∠BOD).∵∠AOD=∠AOB+∠BOD=α=160°，∴∠MON=12×160°=80°；（2）∵OM平分∠AOC，ON平分∠BOD，∴∠MOC=12∠AOC，∠BON=12∠BOD，∵∠MON=∠MOC+∠BON-∠BOC，∴∠MON=12∠AOC+12∠BOD -∠BOC=12(∠AOC+∠BOD )-∠BOC.∵∠AOD=∠AOB+∠BOD，∠AOC=∠AOB+∠BOC,∴∠MON=12(∠AOB+∠BOC+∠BOD )-∠BOC=12(∠AOD+∠BOC )-∠BOC，∵∠AOD=α，∠MON=60°,∠BOC=20°,∴60°=12(α+20°)-20°,∴α=140°.【点睛】本题考查了角的和差计算，角平分线的定义，明确角之间的关系是解答此题的关键. 26．（1）n= 8；（2）-2.5或2.5；（3）n=4或n=12．【解析】【分析】（1）根据“n节点”的概念解答；（2）设点D表示的数为x，根据“5节点”的定义列出方程分情况，并解答；（3）需要分类讨论：①当点E在BA延长线上时，②当点E在线段AB上时，③当点E在AB延长线上时，根据BE=12AE，先求点E表示的数，再根据AC+BC=n，列方程可得结论．【详解】（1）∵A表示的数为-2，B表示的数为2，点C在数轴上表示的数为-4，∴AC=2，BC=6，∴n=AC+BC=2+6=8．（2）如图所示：∵点D是数轴上点A、B的“5节点”，∴AC+BC=5，∵AB=4，∴C在点A的左侧或在点A的右侧，设点D表示的数为x，则AC+BC=5，∴-2-x+2-x=5或x-2+x-（-2）=5，x=-2.5或2.5，∴点D表示的数为2.5或-2.5；故答案为-2.5或2.5；（3）分三种情况：①当点E 在BA 延长线上时，∵不能满足BE=12AE ， ∴该情况不符合题意，舍去； ②当点E 在线段AB 上时，可以满足BE=12AE ，如下图，n=AE+BE=AB=4；③当点E 在AB 延长线上时，∵BE=12AE ， ∴BE=AB=4，∴点E 表示的数为6，∴n=AE+BE=8+4=12，综上所述：n=4或n=12．【点睛】本题考查数轴，一元一次方程的应用，解题的关键是掌握“n 节点”的概念和运算法则，找出题中的等量关系，列出方程并解答，难度一般．27．（1）图1中∠AOD=60°；图2中∠AOD=10°；（2）图1中∠AOD=n m 2+；图2中∠AOD=n m 2-. 【解析】【分析】（1）图1中∠BOC=∠AOC ﹣∠AOB=20°，则∠BOD=10°，根据∠AOD=∠AOB+∠BOD 即得解；图2中∠BOC=∠AOC+∠AOB=120°，则∠BOD=60°，根据∠AOD=∠BOD ﹣∠AOB 即可得解；（2）图1中∠BOC=∠AOC ﹣∠AOB=n ﹣m ，则∠BOD=n m 2﹣，故∠AOD=∠AOB+∠BOD=n m 2+；图2中∠BOC=∠AOC+∠AOB=m+n ，则∠BOD=n m 2+，故∠AOD=∠BOD ﹣∠AOB=n m 2-. 【详解】解：（1）图1中∠BOC=∠AOC ﹣∠AOB=70°﹣50°=20°，∵OD 是∠BOC 的平分线，∴∠BOD=12∠BOC=10°， ∴∠AOD=∠AOB+∠BOD=50°+10°=60°；图2中∠BOC=∠AOC+∠AOB=120°，∵OD 是∠BOC 的平分线，∴∠BOD=12∠BOC=60°， ∴∠AOD=∠BOD ﹣∠AOB=60°﹣50°=10°；（2）根据题意可知∠AOB=m 度，∠AOC=n 度，其中090090180m n m n ＜＜，＜＜，＜+且m n ＜，如图1中，∠BOC=∠AOC ﹣∠AOB=n ﹣m ，∵OD 是∠BOC 的平分线，∴∠BOD=12∠BOC=n m 2﹣， ∴∠AOD=∠AOB+∠BOD=n m 2+； 如图2中，∠BOC=∠AOC+∠AOB=m+n ，∵OD 是∠BOC 的平分线，∴∠BOD=12∠BOC=n m 2+， ∴∠AOD=∠BOD ﹣∠AOB=n m 2-. 【点睛】 本题主要考查角平分线，解此题的关键在于根据题意进行分类讨论，所有情况都要考虑，切勿遗漏.28．（1）B 点坐标为（0，﹣6），C 点坐标为（4，﹣6）（2）S △OPM ＝4t 或S △OPM ＝﹣3t+21（3）当t 为2秒或133秒时，△OPM 的面积是长方形OBCD 面积的13．此时点P 的坐标是（0，﹣4）或（83，﹣6）【解析】【分析】（1）根据绝对值、平方和算术平方根的非负性，求得a ，b ，c 的值，即可得到B 、C 两点的坐标；（2）分两种情况：①P 在OB 上时，直接根据三角形面积公式可得结论；②P 在BC 上时，根据面积差可得结论；（3）根据已知条件先计算三角形OPM 的面积为8，根据（2）中的结论分别代入可得对应t 的值，并计算此时点P 的坐标．【详解】（1）∵6a ++|2b +12|+（c ﹣4）2=0，∴a +6=0，2b +12=0，c ﹣4=0，∴a =﹣6，b =﹣6，c =4，∴B 点坐标为（0，﹣6），C 点坐标为（4，﹣6）．（2）①当点P 在OB 上时，如图1，OP =2t ，S △OPM 12=⨯2t ×4=4t ； ②当点P 在BC 上时，如图2，由题意得：BP =2t ﹣6，CP =BC ﹣BP =4﹣（2t ﹣6）=10﹣2t ，DM =CM =3，S △OPM =S 长方形OBCD ﹣S △0BP ﹣S △PCM ﹣S △ODM =6×412-⨯6×（2t ﹣6）12-⨯3×（10﹣2t ）12-⨯4×3=﹣3t +21． （3）由题意得：S △OPM 13=S 长方形OBCD 13=⨯（4×6）=8，分两种情况讨论： ①当4t =8时，t =2，此时P （0，﹣4）； ②当﹣3t +21=8时，t 133=，PB =2t ﹣626188333=-=，此时P （83，﹣6）． 综上所述：当t 为2秒或133秒时，△OPM 的面积是长方形OBCD 面积的13．此时点P 的坐标是（0，﹣4）或（83，﹣6）．【点睛】本题考查了一元一次方程的应用，主要考查了平面直角坐标系中求点的坐标，动点问题，求三角形的面积，还考查了绝对值、平方和算术平方根的非负性、解一元一次方程，分类讨论是解答本题的关键．29．（1）20；（2）t=15s或17s （3）4 3 s.【解析】【分析】（1）设P、Q速度分别为3m、2m，根据12秒后，动点P到达原点O列方程，求出P、Q 的速度，由此即可得到结论．（2）分两种情况讨论：①当A、B在相遇前且相距5个单位长度时；②当A、B在相遇后且相距5个单位长度时；列方程，求解即可．（3）算出P运动到B再到原点时，所用的时间，再算出Q从B到A所需的时间，比较即可得出结论．【详解】（1）设P、Q速度分别为3m、2m，根据题意得：12×3m=36，解得：m=1，∴P、Q速度分别为3、2，∴BC=12×2=24，∴OC=OB－BC=44－24=20．（2）当A、B在相遇前且相距5个单位长度时：3t＋2t＋5=44＋36，5t=75，∴t=15（s）；当A、B在相遇后且相距5个单位长度时：3t＋2t－5=44＋36，5t=85，∴t=17（s）．综上所述：t=15s或17s．（3）P运动到原点时，t=3644443++=1243s，此时QB=2×1243=2483＞44+38=80，∴Q点已到达A点，∴Q点已到达A点的时间为：3644804022+==（s），故提前的时间为：1243－40=43（s）．【点睛】本题考查了一元一次方程的应用-行程问题以及数轴上的动点问题．解题的关键是找出等量关系，列出方程求解．30．（1）详见解析；（2）①16；②在移动过程中，3AC﹣4AB的值不变【解析】【分析】（1）根据点的移动规律在数轴上作出对应的点即可；（2）①当t=2时，先求出A、B、C点表示的数，然后利用定义求出AB、AC的长即可；②先求出A、B、C点表示的数，然后利用定义求出AB、AC的长，代入3AC－4AB即可得到结论．【详解】（1）A，B，C三点的位置如图所示：．（2）①当t =2时，A 点表示的数为－4，B 点表示的数为5，C 点表示的数为12，∴AB =5－(－4)=9，AC =12－(－4)=16．②3AC －4AB 的值不变．当移动时间为t 秒时，A 点表示的数为－t －2，B 点表示的数为2t ＋1，C 点表示的数为3t ＋6，则：AC =(3t ＋6)－(－t －2)=4t ＋8，AB =(2t ＋1)－(－t －2)=3t ＋3，∴3AC －4AB =3(4t ＋8)－4(3t ＋3)=12t ＋24－12t －12=12．即3AC ﹣4AB 的值为定值12，∴在移动过程中，3AC ﹣4AB 的值不变．【点睛】本题考查了数轴上的动点问题．表示出对应点所表示的数是解答本题的关键．31．（1）是；（2）5cm 或7.5cm 或10cm ；（3）10或607． 【解析】【分析】（1）根据“2倍点”的定义即可求解；（2）分点C 在中点的左边，点C 在中点，点C 在中点的右边三种情况，进行讨论求解即可；（3）根据题意画出图形，P 应在Q 的右边，分别表示出AQ 、QP 、PB ，求出t 的范围．然后根据（2）分三种情况讨论即可．【详解】（1）∵整个线段的长是较短线段长度的2倍，∴线段的中点是这条线段的“2倍点”． 故答案为是；（2）∵AB ＝15cm ，点C 是线段AB 的2倍点，∴AC ＝1513⨯=5cm 或AC ＝1512⨯=7.5cm 或AC ＝1523⨯=10cm ． （3）∵点Q 是线段AP 的“2倍点”，∴点Q 在线段AP 上．如图所示：由题意得：AP =2t ，BQ =t ，∴AQ =20-t ，QP =2t -（20-t ）=3t -20，PB =20-2t ．∵PB =20-2t ≥0，∴t ≤10．∵QP =3t -20≥0，∴t ≥203，∴203≤t ≤10． 分三种情况讨论：①当AQ ＝13AP 时，20-t ＝13×2t ，解得：t ＝12＞10，舍去； ②当AQ ＝12AP 时，20-t ＝12×2t ，解得：t ＝10；③当AQ＝23AP时，20-t＝23×2t，解得：t607；答：t为10或607时，点Q是线段AP的“2倍点”．【点睛】本题考查了一元一次方程的解法、线段的和差等知识点，题目需根据“2倍点”的定义分类讨论，理解“2倍点”的定义是解决本题的关键．32．（1）见解析；（2）∠OQP=180°+12x°﹣12y°或∠OQP=12x°﹣12y°．【解析】【试题分析】（1）分下面两种情况进行说明；①如图1，点P在直线AB的右侧，∠APB+∠MON+∠PAO+∠PBO=360°，②如图2，点P在直线AB的左侧，∠APB=∠MON+∠PAO+∠PBO，（2）分两种情况讨论，如图3和图4.【试题解析】（1）分两种情况：①如图1，点P在直线AB的右侧，∠APB+∠MON+∠PAO+∠PBO=360°，证明：∵四边形AOBP的内角和为（4﹣2）×180°=360°，∴∠APB=360°﹣∠MON﹣∠PAO﹣∠PBO；②如图2，点P在直线AB的左侧，∠APB=∠MON+∠PAO+∠PBO，证明：延长AP交ON于点D，∵∠ADB是△AOD的外角，∴∠ADB=∠PAO+∠AOD，∵∠AP B是△PDB的外角，∴∠APB=∠PDB+∠PBO，∴∠APB=∠MON+∠PAO+∠PBO；（2）设∠MON=2m°，∠APB=2n°，∵OC平分∠MON，∴∠AOC=∠MON=m°，∵PQ平分∠APB，∴∠APQ=∠APB=n°，分两种情况：第一种情况：如图3，∵∠OQP=∠MOC+∠PAO+∠APQ，即∠OQP=m°+x°+n°①∵∠OQP+∠CON+∠OBP+∠BPQ=360°，∴∠OQP=360°﹣∠CON﹣∠OBP﹣∠BPQ，即∠OQP=360°﹣m°﹣y°﹣n°②，①+②得2∠OQP=360°+x°﹣y°，∴∠OQP=180°+x°﹣y°；第二种情况：如图4，∵∠OQP+∠APQ=∠MOC+∠PAO，即∠OQP+n°=m°+x°，∴2∠OQP+2n°=2m°+2x°①，∵∠APB=∠MON+∠PAO+∠PBO，∴2n°=2m°+x°+y°②，①﹣②得2∠OQP=x°﹣y°，∴∠OQP=x°﹣y°，综上所述，∠OQP=180°+x°﹣y°或∠OQP=x°﹣y°．。

七年级上册上海建平中学数学期末试卷（提升篇）（Word版含解析）一、初一数学上学期期末试卷解答题压轴题精选（难）1．如图下图所示,已知AB//CD, ∠B=30°，∠D=120°;（1）若∠E=60°，则∠F=________;（2）请探索∠E与∠F之间满足的数量关系?说明理由.（3）如下图所示,已知EP平分∠BEF,FG平分∠EFD,反向延长FG交EP于点P,求∠P的度数;【答案】（1）90°（2）解：如图，分别过点E，F作EM∥AB，FN∥AB∴EM∥AB∥FN∴∠B=∠BEM=30°，∠MEF=∠EFN又∵AB∥CD，AB∥FN∴CD∥FN∴∠D+∠DFN=180°又∵∠D =120°∴∠DFN=60°∴∠BEF=∠MEF+30°，∠EFD=∠EFN+60°∴∠EFD=∠MEF +60°∴∠EFD=∠BEF+30°（3）解：如图，过点F作FH∥EP由（2）知，∠EFD=∠BEF+30°设∠BEF=2x°，则∠EFD=（2x+30）°∵EP平分∠BEF，GF平分∠EFD∴∠PEF= ∠BEF=x°，∠EFG= ∠EFD=（x+15）°∵FH∥EP∴∠PEF=∠EFH=x°，∠P=∠HFG ∵∠HFG=∠EFG-∠EFH=15°∴∠P=15°【解析】【解答】解：（1）分别过点E、F作EM∥AB，FN∥AB，则有AB∥EM∥FN∥CD.∴∠B=∠BEM=30°，∠MEF=∠EFN，∠DFN=180°-∠CDF=60°，∴∠BEF=∠MEF+30°，∠EFD=∠EFN+60°，∴∠EFD=∠BEF+30°=90°.【分析】（1）分别过点E、F作AB的平行线，根据平行线的性质即可求解；（2）根据平行线的性质可得∠DFN=60°，∠BEM=30°，∠MEF=∠NFE，即可得到结论；（3）过点F作FH∥EP，设∠BEF=2x°，根据（2）中结论即可表示出∠BFD，根据角平分线的定义可得∠PEF=x°，∠EFG=（x+15）°，再根据平行线的性质即可得到结论.2．如图 1，CE 平分∠ACD，AE 平分∠BAC，且∠EAC＋∠ACE=90°．（1）请判断 AB 与 CD 的位置关系，并说明理由；（2）如图2，若∠E=90°且AB 与CD 的位置关系保持不变，当直角顶点E 移动时，写出∠BAE 与∠ECD 的数量关系，并说明理由；（3）如图 3，P 为线段 AC 上一定点，点 Q 为直线 CD 上一动点，且 AB 与 CD 的位置关系保持不变，当点 Q 在射线 CD 上运动时(不与点 C 重合)，∠PQD，∠APQ 与∠ BAC 有何数量关系？写出结论，并说明理由．【答案】（1），理由如下：CE 平分，AE 平分，；（2），理由如下：如图，延长AE交CD于点F，则由三角形的外角性质得：；（3），理由如下：，即由三角形的外角性质得：又，即即．【解析】【分析】（1）根据角平分线的定义、平行线的判定即可得；（2）根据平行线的性质（两直线平行，内错角相等）、三角形的外角性质即可得；（3）根据平行线的性质（两直线平行，同旁内角互补）、三角形的外角性质、邻补角的定义即可得．3．（1）问题发现：如图 1，已知点 F，G 分别在直线 AB，CD 上，且 AB∥CD，若∠BFE=40°，∠CGE=130°，则∠GEF 的度数为________；（2）拓展探究：∠GEF，∠BFE，∠CGE 之间有怎样的数量关系？写出结论并给出证明；答：∠GEF=▲ .证明：过点 E 作 EH∥AB，∴∠FEH=∠BFE（▲），∵AB∥CD，EH∥AB，（辅助线的作法）∴EH∥CD（▲），∴∠HEG=180°-∠CGE（▲），∴∠FEG=∠HFG+∠FEH=▲ .（3）深入探究：如图 2，∠BFE 的平分线 FQ 所在直线与∠CGE 的平分线相交于点 P，试探究∠GPQ 与∠GEF 之间的数量关系，请直接写出你的结论.【答案】（1）90°（2）解：∠GEF＝∠BFE＋180°−∠CGE，证明：过点 E 作 EH∥AB，∴∠FEH=∠BFE（两直线平行，内错角相等），∵AB∥CD，EH∥AB，（辅助线的作法）∴EH∥CD（平行线的迁移性），∴∠HEG=180°-∠CGE（两直线平行，同旁内角互补），∴∠FEG=∠HFG+∠FEH=∠BFE＋180°−∠CGE ，故答案为：∠BFE＋180°−∠CGE；两直线平行，内错角相等；平行线的迁移性；两直线平行，同旁内角互补；∠BFE＋180°−∠CGE；（3）解：∠GPQ＋∠GEF＝90°，理由是：如图2，∵FQ平分∠BFE，GP平分∠CGE，∴∠BFQ＝∠BFE，∠CGP＝∠CGE，在△PMF中，∠GPQ＝∠GMF−∠PFM＝∠CGP−∠BFQ，∴∠GPQ＋∠GEF＝∠CGE− ∠BFE＋∠GEF＝ ×180°＝90°.即∠GPQ＋∠GEF＝90°.【解析】【解答】（1）解：如图1，过E作EH∥AB，∵AB∥CD，∴AB∥CD∥EH，∴∠HEF＝∠BFE＝40°，∠HEG＋∠CGE＝180°，∵∠CGE＝130°，∴∠HEG＝50°，∴∠GEF＝∠HEF＋∠HEG＝40°＋50°＝90°；故答案为：90°；【分析】（1）如图1，过E作EH∥AB，根据平行线的性质可得∠HEF＝∠BFE＝40 ，∠HEG＝50 ，相加可得结论；（2）由①知：∠HEF＝∠BFE，∠HEG＋∠CGE＝180°，则∠HEG＝180°−∠CGE，两式相加可得∠GEF＝∠BFE＋180°−∠CGE；（3）如图2，根据角平分线的定义得：∠BFQ＝∠BFE，∠CGP＝∠CGE，由三角形的外角的性质得：∠GPQ＝∠GMF−∠PFM＝∠CGP−∠BFQ，计算∠GPQ＋∠GEF并结合②的结论可得结果.4．如图，数轴上线段AB=4（单位长度），CD=6（单位长度），点A在数轴上表示的数是-16，点C在数轴上表示的数是18．（1）点B在数轴上表示的数是________，点D在数轴上表示的数是________，线段AD=________；（2）若线段AB以4个单位长度/秒的速度向右匀速运动，同时线段CD以2个单位长度/秒的速度向左匀速运动，设运动时间为t秒，①若BC=6（单位长度），求t的值；②当0＜t＜5时，设M为AC中点，N为BD中点，求线段MN的长．【答案】（1）-12；24；40（2）解：①设运动t秒时，BC=6当点B在点C的左边时，由题意得：4t+6+2t=30，解之：t=4；当点B在点C的右边时，由题意得：4t−6+2t=30，解之：t=6.综上可知,若BC=6(单位长度)，t的值为4或6秒；②当0<t<5时，A点表示的数为−16+4t，B点表示的数为−12+4t，C点表示的数为18−2t，D点表示的数为24−2t，∵M为AC中点，N为BD中点，∴点M表示的数为：=1+t，点N表示的数为：=6+t∴MN=6+t-（1+t）=5.【解析】【解答】解：（1）∵AB=4，A在数轴上表示的数是-16，∴点B在数轴上表示的数为：-16+4=-12∵点C在数轴上表示的数是18，CD=6，∴点D在数轴上表示的数为：18+6=24；∵点A在数轴上表示的数是-16，点D在数轴上表示的数为24，∴AD=|-16-24|=40故答案为：-12；24；40【分析】（1）由线段AB=4，点A在数轴上表示的数是-16，根据两点间的距离公式可得点B在数轴上表示的数；由CD=6，点C在数轴上表示的数是18，根据两点间的距离公式可得点D在数轴上表示的数；根据两点间的距离公式可得AD的长。

建平西校2017学年第一学期期末质量检测初二年级英语试卷Part 2 Phonetics, Vocabulary and grammar(第二部分语音、词汇和语法)Ⅱ.Choose the best answer(选择最恰当的答案)(共12分)22. Which of the following words is pronounced / ˈsɪstəm / ?【A】assistant【B】sister【C】system【D】symbol【答案】C【解析】A. /əˈsɪstənt/ B. /ˈsɪstə(r)/ D. / ˈsɪmbl/23. It is _____ unexplored forest. Scientists are going to there next month.【A】a【B】an【C】the【D】/【答案】B【解析】unexplored /ˌʌnɪkˈsplɔ:d/ 所以用an24. It’s essent ial for citizen to know the haze(雾霾) can do harm _____ people’s health.【A】with【B】to【C】for【D】on【答案】B【解析】do harm to/be harm for 固定搭配25. The pet dog is ________ a crowd of kids. That’s why I can’t see it.【A】beside【B】behind【C】on【D】in front of【答案】B【解析】宠物狗在一群孩子的后面。

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26. This pair of new trousers is too big for me. Would you like to show me ________ pair? 【A】the other【B】other【C】others【D】another【答案】D【解析】另一个，再一个27. Don’t just believe such an advertisement. That kind of machine isn’t as _____ as it says. 【A】better【B】best【C】good【D】well【答案】C【解析】as good as 形容词修饰machine28. Jack, you’d better _______ too many soft drinks. It’s bad for your health.【A】not ho have【B】to have【C】not have【D】have【答案】D【解析】had better do29. The Blacks live a happy life ______ they are very poor.【A】because【B】since【C】if【D】although【答案】D【解析】连词考察句意30. _________ amazing film Jason Wu’s Wolf Warriors Ⅱwas!【A】What an【B】How【C】What a【D】What【答案】A【解析】感叹句31. A 6-magnitude earthquake _______ in Tibet on November 18.2017. 【A】will happen【B】happened【C】has happened【D】was happened【答案】B【解析】时间是在过去32. The boy’s parents hope that he ______ the computer games any longer.【A】won’t play【B】didn’t play【C】hasn’t play【D】isn’t play【答案】A【解析】一般将来时33. -Tom, can I use your bike? Mine is not here.-__________________________【A】Sorry, I have no idea.【B】Yes, you’re welcome.【C】That’s O.K.【D】Of course. Here you are.【答案】D【解析】情景交际，借东西Ⅲ.Complete the following passage with words or phrases in the box. Each can only be used once(将下列单词或词组填入空格，用A、B、C、D等表示。

上海建平中学人教版七年级上册数学期末试卷及答案.doc一、选择题1．下列方程中，以32x =-为解的是（ ） A ．33x x =+B ．33x x =+C ．23x =D ．3-3x x =2．下列数或式：3(2)-，61()3-，25- ，0，21m +在数轴上所对应的点一定在原点右边的个数是（ ） A ．1 B ．2 C ．3 D ．4 3．若多项式229x mx ++是完全平方式，则常数m 的值为()A ．3B ．-3C ．±3D ．+64．A 、B 两地相距160千米，甲车和乙车的平均速度之比为4:5，两车同时从A 地出发到B 地，乙车比甲车早到30分钟，若求甲车的平均速度，设甲车平均速度为4x 千米/小时，则所列方程是( ) A ．1601603045x x-= B ．1601601452x x -= C ．1601601542x x -= D ．1601603045x x+= 5．若21(2)0x y -++=，则2015()x y +等于（ ） A ．-1B ．1C ．20143D ．20143-6．方程3x +2＝8的解是（ ） A ．3B ．103C ．2D ．127．已知关于x 的方程ax ﹣2＝x 的解为x ＝﹣1，则a 的值为（ ） A ．1B ．﹣1C ．3D ．﹣38．一个几何体的表面展开图如图所示，则这个几何体是( )A ．四棱锥B ．四棱柱C ．三棱锥D ．三棱柱9．如果代数式﹣3a 2m b 与ab 是同类项，那么m 的值是( ) A ．0B ．1C ．12D ．310．如图，经过刨平的木板上的两个点，能弹出一条笔直的墨线，而且只能弹出一条墨线，能解释这一实际应用的数学知识是（ ）A．两点确定一条直线B．两点之间线段最短C．垂线段最短D．连接两点的线段叫做两点的距离11．如图为一无盖长方体盒子的展开图（重叠部分不计），可知该无盖长方体的容积为（）A．8 B．12 C．18 D．2012．如图，已知点C在线段AB上，点M、N分别是AC、BC的中点，且AB＝8cm，则MN 的长度为（）cm．A．2 B．3 C．4 D．6二、填空题13．在灯塔O处观测到轮船A位于北偏西54︒的方向，同时轮船B在南偏东15︒的方向，∠的大小为______.那么AOB14．小明妈妈支付宝连续五笔交易如图，已知小明妈妈五笔交易前支付宝余额860元，则五笔交易后余额__________元．支付宝帐单日期交易明细10.16乘坐公交￥ 4.00-+10.17转帐收入￥200.00-10.18体育用品￥64.0010.19 零食￥82.00- 10.20 餐费￥100.00-15．定义-种新运算:22a b b ab ⊕=-,如21222120⊕=-⨯⨯=，则(1)2-⊕=__________．16．若方程11222m x x --=++有增根，则m 的值为____. 17．在日常生活中如取款、上网等都需要密码，有一种用“因式分解法”产生的密码，方便记忆，原理是对于多项44x y -，因式分解的结果是()()()22x y x y x y-++，若取9x =，9y =时，则各个因式的值是：()18x y +=，()0x y -=，()22162x y +=，于是就可以把“180162”作为一个六位数的密码，对于多项式324x xy -，取36x =，16y =时，用上述方法产生的密码是________ (写出一个即可).18．在数轴上，与表示-3的点的距离为4的点所表示的数为__________________． 19．五边形从某一个顶点出发可以引_____条对角线．20．﹣225ab π是_____次单项式，系数是_____．21．如图，将△ABE 向右平移3cm 得到△DCF,若BE=8cm ，则CE=______cm.22．众所周知，中华诗词博大精深，集大量的情景情感于短短数十字之间，或豪放，或婉约，或思民生疾苦，或抒发己身豪情逸致，文化价值极高．而数学与古诗词更是有着密切的联系．古诗中，五言绝句是四句诗，每句都是五个字；七言绝句是四句诗，每句都是七个字．有一本诗集，其中五言绝句比七言绝句多13首，总字数却反而少了20个字．问两种诗各多少首？设七言绝句有x 首，根据题意，可列方程为______． 23．已知关于x 的方程4mx x -=的解是1x =，则m 的值为______.24．中国始有历法大约在四千年前每页显示一日信息的叫日历，每页显示一个月信息的叫月历，每页显示全年信息的叫年历如图是2019年1月份的月历，用一个方框圈出任意22⨯的4个数，设方框左上角第一个数是x ，则这四个数的和为______(用含x 的式子表示)三、解答题25．（1）化简：3x 2﹣22762x x +； （2）先化简，再求值：2（a 2﹣ab ﹣3.5）﹣（a 2﹣4ab ﹣9），其中a ＝﹣5，b ＝32． 26．如图1，点O 为直线AB 上一点，过O 点作射线OC ，使50AOC ∠=︒，将一直角三角板的直角项点放在点O 处，一边OM 在射线OB 上，另一边ON 在直线AB 的下方.()1如图2,将图1中的三角板绕点O 逆时针旋转，使边OM 在BOC ∠的内部，且OM 恰好平分BOC ∠.此时BON ∠=__ 度;()2如图3,继续将图2中的三角板绕点O 按逆时针方向旋转，使得ON 在AOC ∠的内部.试探究AOM ∠与NOC ∠之间满足什么等量关系，并说明理由;()3将图1中的三角板绕点O 按每秒5︒的速度沿逆时针方向旋转一周，在旋转的过程中，若第t 秒时，,,OA OC ON 三条射线恰好构成相等的角，则t 的值为__ (直接写出结果).27．温州市在今年三月份启动实施“明眸皓齿”工程.根据安排,某校对于学生使用电子产品的一周用时情况进行抽样调查,绘制成以下频数分布直方图.请根据图中提供的信息,解答下列问题.(1)这次共抽取了名学生进行调查.(2)用时在2.45~3.45小时这组的频数是_ ，频率是_ .(3)如果该校有1000名学生，请估计一周电子产品用时在0.45~3.45小时的学生人数.=，点C是线段AB的中点，点D是线段BC的中点．28．如图，线段AB8()1求线段AD的长；()2在线段AC上有一点E，1=，求AE的长．CE BC329．如图，数轴上的点A，B，C，D，E对应的数分别为a，b，c，d，e，（1）化简：|a﹣c|﹣2|b﹣a|﹣|b﹣c|；（2）若这五个点满足每相邻两个点之间的距离都相等，且|a|＝|e|，|b|＝3，直接写出b ﹣e的值．30．东莞市出租车收费标准如下表所示，根据此收费标准，解决下列问题：行驶路程收费标准不超出2km的部分起步价8元超出2km 的部分 2.6元/km(1)若行驶路程为5km ，则打车费用为______元；(2)若行驶路程为()km 6x x >，则打车费用为______元(用含x 的代数式表示)； (3)某同学周末放学回家，已知打车费用为34元，则他家离学校多少千米？四、压轴题31．如图，已知数轴上点A 表示的数为8，B 是数轴上位于点A 左侧一点，且AB =22，动点P 从A 点出发，以每秒5个单位长度的速度沿数轴向左匀速运动，设运动时间为t （t ＞0）秒．（1）出数轴上点B 表示的数 ；点P 表示的数 （用含t 的代数式表示） （2）动点Q 从点B 出发，以每秒3个单位长度的速度沿数轴向右匀速运动，若点P 、Q 同时出发，问多少秒时P 、Q 之间的距离恰好等于2？（3）动点Q 从点B 出发，以每秒3个单位长度的速度沿数轴向左匀速运动，若点P 、Q 同时出发，问点P 运动多少秒时追上点Q ？（4）若M 为AP 的中点，N 为BP 的中点，在点P 运动的过程中，线段MN 的长度是否发生变化？若变化，请说明理由，若不变，请你画出图形，并求出线段MN 的长．32．观察下列等式：111122=-⨯，1112323=-⨯，1113434=-⨯，则以上三个等式两边分别相加得：1111111131122334223344++=-+-+-=⨯⨯⨯． ()1观察发现()1n n 1=+______；()1111122334n n 1+++⋯+=⨯⨯⨯+______．()2拓展应用有一个圆，第一次用一条直径将圆周分成两个半圆(如图1)，在每个分点标上质数m ，记2个数的和为1a ；第二次再将两个半圆周都分成14圆周(如图2)，在新产生的分点标上相邻的已标的两数之和的12，记4个数的和为2a ；第三次将四个14圆周分成18圆周(如图3)，在新产生的分点标上相邻的已标的两数之和的13，记8个数的和为3a ；第四次将八个18圆周分成116圆周，在新产生的分点标上相邻的已标的两个数的和的14，记16个数的和为4a ；⋯⋯如此进行了n 次．n a =①______(用含m 、n 的代数式表示)；②当na6188=时，求123n1111a a a a+++⋯⋯+的值．33．如图，以长方形OBCD的顶点O为坐标原点建立平面直角坐标系，B点坐标为（0，a），C点坐标为（c，b），且a、b、C满足6a++|2b+12|+（c﹣4）2＝0．（1）求B、C两点的坐标；（2）动点P从点O出发，沿O→B→C的路线以每秒2个单位长度的速度匀速运动，设点P 的运动时间为t秒，DC上有一点M（4，﹣3），用含t的式子表示三角形OPM的面积；（3）当t为何值时，三角形OPM的面积是长方形OBCD面积的13？直接写出此时点P的坐标．【参考答案】***试卷处理标记，请不要删除一、选择题1．A解析：A【解析】【分析】把32x=-代入方程，只要是方程的左右两边相等就是方程的解，否则就不是．【详解】解：A 中、把32x =-代入方程得左边等于右边，故A 对； B 中、把32x =-代入方程得左边不等于右边，故B 错； C 中、把32x =-代入方程得左边不等于右边，故C 错； D 中、把32x =-代入方程得左边不等于右边，故D 错. 故答案为：A. 【点睛】本题考查方程的解的知识，解题关键在于把x 值分别代入方程进行验证即可.2．B解析：B 【解析】 【分析】点在原点的右边，则这个数一定是正数，根据演要求判断几个数即可得到答案. 【详解】()32-=-8，613⎛⎫- ⎪⎝⎭=1719，25-=-25 ，0，21m +≥1 在原点右边的数有613⎛⎫- ⎪⎝⎭和 21m +≥1 故选B 【点睛】此题重点考察学生对数轴上的点的认识，抓住点在数轴的右边是解题的关键.3．C解析：C 【解析】 【分析】利用完全平方式的结构特征即可求出m 的值． 【详解】解：∵多项式2222923x mx x mx ++=++是完全平方式， ∴2m ＝±6， 解得：m ＝±3， 故选：C ． 【点睛】此题考查了完全平方式，熟练掌握完全平方公式的结构特征是解本题的关键．4．B解析：B【分析】甲车平均速度为4x 千米/小时，则乙车平均速度为5x 千米/小时，根据两车同时从A 地出发到B 地，乙车比甲车早到30分钟，列出方程即可得. 【详解】甲车平均速度为4x 千米/小时，则乙车平均速度为5x 千米/小时，由题意得1604x －1605x ＝12， 故选B. 【点睛】本题考查了分式方程的应用，弄清题意，找准等量关系列出方程是解题的关键.5．A解析：A 【解析】（y+2）2=0，列出方程x-1=0，y+2=0，求出x=1、y=-2，代入所求代数式（x+y ）2015=（1﹣2）2015=﹣1. 故选A6．C解析：C 【解析】 【分析】移项、合并后，化系数为1，即可解方程． 【详解】解：移项、合并得，36x =， 化系数为1得：2x =， 故选：C ． 【点睛】本题考查一元一次方程的解；熟练掌握一元一次方程的解法是解题的关键．7．B解析：B 【解析】 【分析】将1x =-代入2ax x -=，即可求a 的值． 【详解】解：将1x =-代入2ax x -=， 可得21a --=-， 解得1a =-， 故选：B ．本题考查一元一次方程的解；熟练掌握一元一次方程的解与方程的关系是解题的关键．8．A解析：A【解析】试题分析：根据四棱锥的侧面展开图得出答案.试题解析：如图所示：这个几何体是四棱锥.故选A.考点：几何体的展开图.9．C解析：C【解析】【分析】根据同类项的定义得出2m=1，求出即可．【详解】解：∵单项式-3a2m b与ab是同类项，∴2m=1，∴m=12，故选C．【点睛】本题考查了同类项的定义，能熟记同类项的定义是解此题的关键，所含字母相同，并且相同字母的指数也分别相同的项，叫同类项．10．A解析：A【解析】【分析】根据公理“两点确定一条直线”来解答即可．【详解】解：经过刨平的木板上的两个点，能弹出一条笔直的墨线，此操作的依据是两点确定一条直线．故选：A．【点睛】此题考查的是直线的性质在实际生活中的运用，此类题目有利于培养学生生活联系实际的能力．11．A解析：A【解析】【分析】根据观察、计算可得长方体的长、宽、高，根据长方体的体积公式，可得答案．【详解】解：由图可知长方体的高是1，宽是3-1=2，长是6-2=4，长方体的容积是4×2×1=8，故选：A．【点睛】本题考查了几何体的展开图.能判断出该几何体为长方体的展开图，并能根据展开图求得长方体的长、宽、高是解题关键.12．C解析：C【解析】【分析】根据MN＝CM+CN＝12AC+12CB＝12（AC+BC）＝12AB即可求解．【详解】解：∵M、N分别是AC、BC的中点，∴CM＝12AC，CN＝12BC，∴MN＝CM+CN＝12AC+12BC＝12（AC+BC）＝12AB＝4．故选：C．【点睛】本题考查了线段中点的性质，找到MC与AC，CN与CB关系，是本题的关键二、填空题13．【解析】【分析】根据线与角的相关知识：具有公共点的两条射线组成的图形叫做角，这个公共端点叫做角的顶点，这两条射线叫做角的两条边，明确方位角，即可得解. 【详解】根据题意可得：∠AOB=（90解析：141【解析】【分析】根据线与角的相关知识：具有公共点的两条射线组成的图形叫做角，这个公共端点叫做角的顶点，这两条射线叫做角的两条边，明确方位角，即可得解.【详解】根据题意可得：∠AOB=（90－54）+90+15=141°．故答案为141°.【点睛】此题主要考查角度的计算与方位，熟练掌握，即可解题.14．810【解析】【分析】根据有理数的加减运算法则，对题干支出与收入进行加减运算即可.【详解】解：由题意五笔交易后余额为860+200-4-64-82-100=810元，故填810.【点睛解析：810【解析】【分析】根据有理数的加减运算法则，对题干支出与收入进行加减运算即可.【详解】解：由题意五笔交易后余额为860+200-4-64-82-100=810元，故填810.【点睛】本题考查有理数的加减运算，理解题意根据题意对支出与收入进行加减运算从而求解. 15．8【解析】【分析】根据题意原式利用题中的新定义计算将-1和2代入计算即可得到结果．【详解】解：因为；所以故填8.【点睛】本题结合新定义运算考查有理数的混合运算，熟练掌握运算法则是解解析：8【解析】【分析】根据题意原式利用题中的新定义计算将-1和2代入计算即可得到结果．【详解】解：因为22a b b ab ⊕=-；所以2(1)222(1)28.-⊕=-⨯-⨯=故填8.【点睛】本题结合新定义运算考查有理数的混合运算，熟练掌握运算法则是解本题的关键．16．2【解析】【分析】分式方程去分母转化为整式方程,由分式方程有增根,得到x+2=0,求出x的值代入整式方程即可求出m的值【详解】去分母得:m-1-1=2x+4将x=-2代入得:m-2=-4解析：2【解析】【分析】分式方程去分母转化为整式方程,由分式方程有增根,得到x+2=0,求出x的值代入整式方程即可求出m的值【详解】去分母得:m-1-1=2x+4将x=-2代入得:m-2=-4+4解得:m=2故答案为:2【点睛】此题考查分式方程的增根，掌握运算法则是解题关键17．36684或36468或68364或68436或43668或46836等(写出一个即可) 【解析】【分析】首先对多项式进行因式分解,然后把字母的值代入求得各个因式,从而写出密码【详解】=x(解析：36684或36468或68364或68436或43668或46836等(写出一个即可)【解析】【分析】首先对多项式进行因式分解,然后把字母的值代入求得各个因式,从而写出密码【详解】324=x(x+2y)(x-2y).x xy当x=36,y=16时,x+2y=36+32=68x-2y=36-32=4.则密码是36684或36468或68364或68436或43668或46836故答案为36684或36468或68364或68436或43668或46836【点睛】此题考查因式分解的应用，解题关键在于把字母的值代入18．1或-7【解析】【分析】设这个数为x，利用数轴上两点间的距离公式可得|x-(-3)|=4，解出x即可. 【详解】设这个数为x，由题意得|x-(-3)|=4，所以x+3=4或x+3=-4，解解析：1或-7【解析】【分析】设这个数为x，利用数轴上两点间的距离公式可得|x-(-3)|=4，解出x即可.【详解】设这个数为x，由题意得|x-(-3)|=4，所以x+3=4或x+3=-4，解得x=1或-7.【点睛】本题考查数轴的应用，使用两点间的距离公式列出方程是解题的关键.19．2【解析】【分析】从n边形的一个顶点出发有（n−3）条对角线，代入求出即可．【详解】解：从五边形的一个顶点出发有5﹣3＝2条对角线，故答案为2．【点睛】本题考查了多边形的对角线，熟记解析：2【解析】【分析】从n边形的一个顶点出发有（n−3）条对角线，代入求出即可．【详解】解：从五边形的一个顶点出发有5﹣3＝2条对角线，故答案为2．本题考查了多边形的对角线，熟记知识点（从n 边形的一个顶点出发有（n−3）条对角线）是解此题的关键．20．三 ﹣【解析】【分析】单项式中的数字因数叫做单项式的系数，一个单项式中所有字母的指数的和叫做单项式的次数，由此可得答案．【详解】是三次单项式，系数是 ．故答案为：三， ．解析：三 ﹣25π 【解析】【分析】单项式中的数字因数叫做单项式的系数，一个单项式中所有字母的指数的和叫做单项式的次数，由此可得答案．【详解】 225ab π-是三次单项式，系数是25π- ． 故答案为：三，25π-． 【点睛】本题考查了单项式的知识，掌握单项式系数及次数的定义是解题的关键． 21．5【解析】【分析】根据平移的性质可得BC=3cm ，继而由BE=8cm ，CE=BE-BC 即可求得答案.【详解】∵△ABE 向右平移3cm 得到△DCF，∴BC=3cm，∵BE=8cm，∴C解析：5【解析】【分析】根据平移的性质可得BC=3cm ，继而由BE=8cm ，CE=BE-BC 即可求得答案.∵△ABE向右平移3cm得到△DCF，∴BC=3cm，∵BE=8cm，∴CE=BE-BC=8-3=5cm，故答案为：5.【点睛】本题考查了平移的性质，熟练掌握对应点间的距离等于平移距离的性质是解题的关键．22．28x-20（x+13）=20【解析】【分析】利用五言绝句与七言绝句总字数之间的关系得出等式进而得出答案.【详解】设七言绝句有x首,根据题意,可列方程为: 28x-20（x+13）=20，解析：28x-20（x+13）=20【解析】【分析】利用五言绝句与七言绝句总字数之间的关系得出等式进而得出答案.【详解】设七言绝句有x首,根据题意,可列方程为: 28x-20（x+13）=20，故答案为: 28x-20（x+13）=20.【点睛】本题主要考查一元一次方程应用，关键在于找出五言绝句与七言绝句总字数之间的关系. 23．5【解析】【分析】把方程的解代入方程即可得出的值.【详解】把代入方程，得∴故答案为5.【点睛】此题主要考查根据方程的解求参数的值，熟练掌握，即可解题.解析：5【解析】【分析】把方程的解代入方程即可得出m的值.把1x=代入方程，得141m⨯-=∴5m=故答案为5.【点睛】此题主要考查根据方程的解求参数的值，熟练掌握，即可解题.24．【解析】【分析】首先根据题意分别列出四个数的关系，然后即可求得其和.【详解】由题意，得故答案为.【点睛】此题主要考查整式的加减，解题关键理解题意找出这四个数的关系式. 解析：416x+【解析】【分析】首先根据题意分别列出四个数的关系，然后即可求得其和.【详解】由题意，得()()()1771416x x x x x+++++++=+故答案为416x+.【点睛】此题主要考查整式的加减，解题关键理解题意找出这四个数的关系式.三、解答题25．（1）112x2；（2）a2+2ab+2，12．【解析】【分析】（1）根据合并同类项法则计算；（2）根据去括号法则、合并同类项法则把原式化简，代入计算得到答案．【详解】解：（1）原式＝（3﹣72+6）x2＝112x2；（2）原式＝2a2﹣2ab﹣7﹣a2+4ab+9＝a2+2ab+2，当a＝﹣5，b＝32时，原式＝（﹣5）2+2×（﹣5）×32+2＝12．【点睛】本题考查的是整式的化简求值，掌握整式的加减混合运算法则是解题的关键．26．（1）25°；（2）∠AOM-∠N OC=40°，理由详见解析；（3）t的值为13，34，49或64.【解析】【分析】（1）由平角的定义先求出∠BOC的度数，然后由角平分线的定义求出∠BOM的度数，再根据∠BON=∠MON-∠BOM可以求出结果；（2）根据题意得出∠AOM+∠AON=90°①，∠AON+∠NOC=50°②，利用①-②可以得出结果；（3）根据已知条件可知，在第t秒时，三角板转过的角度为5°t，然后按照OA、OC、ON三条射线构成相等的角分四种情况讨论，即可求出t的值.【详解】解：（1）∵∠AOC=50°，∴∠BOC=180°-∠AOC=130°，∵OM平分∠BOC，∴∠BOM=12∠BOC=55°，∴∠BON=90°-∠BOM=25°.故答案为：25；（2）∠AOM与∠NOC之间满足等量关系为：∠AOM-∠N OC=40°，理由如下：∵∠MON=90°，∠AOC=50°，∴∠AOM+∠AON=90°①，∠AON+∠NOC=50°②，∴①-②得，∠AOM-∠NOC=40°.（3）∵三角板绕点O按每秒5°的速度沿逆时针方向旋转，∴第t秒时，三角板转过的角度为5°t，当三角板转到如图①所示时，∠AON=∠CON.∵∠AON=90°+5°t，∠CON=∠BOC+∠BON=130°+90°-5°t=220°-5°t，∴90°+5°t=220°-5°t，即t=13；当三角板转到如图②所示时，∠AOC=∠CON=50°，∵∠CON=∠BOC-∠BON=130°-（5°t-90°）=220°-5°t，∴220°-5°t=50°，即t=34；当三角板转到如图③所示时，∠AON=∠CON=12∠AOC=25°，∵∠CON=∠BON-∠BOC=（5°t-90°）-130°=5°t-220°，∴5°t-220°=25°，即t=49；当三角板转到如图④所示时，∠AON=∠AOC=50°，∵∠AON=5°t-180°-90°=5°t-270°，∴5°t-270°=50°，即t=64．故t的值为13，34，49或64.【点睛】本题主要考查角的和、差关系，难点是找出变化过程中的不变量，需要结合图形来计算，在计算分析的过程中注意动手操作，在旋转的过程中得到不变的量．27．(1)400. (2)104; 0.26.(3)540【解析】【分析】（1）根据频数分布直方图得到各个时间段的频数，计算即可；（2）从频数分布直方图找出用时在2.45−3.45小时的频数，求出频率；（3）利用样本估计总体即可．【详解】解：（1）这次共抽取的学生数为：40＋72＋104＋92＋52＋40＝400（人），故答案为：400；（2）用时在2.45−3.45小时这组的频数为104，频率为：1040.26 400，故答案为：104；0.26；（2）1000×4072104540400（人）．答：估计1000名学生一周电子产品用时在0.45~3.45小时的学生人数为540人.【点睛】本题考查的是读频数分布直方图的能力和利用统计图获取信息的能力以及用样本估计总体，利用统计图获取信息时，必须认真观察、分析、研究统计图，才能作出正确的判断和解决问题．28．（1）6，（2）83．【解析】【分析】()1根据AD AC CD =+，只要求出AC 、CD 即可解决问题；()2根据AE AC EC =-，只要求出CE 即可解决问题；【详解】解：()1AB 8=，C 是AB 的中点，AC BC 4∴==， D 是BC 的中点，1CD DB BC 22∴===， AD AC CD 426∴=+=+=．()12CE BC 3=，BC 4=， 4CE 3∴=， 48AE AC CE 433∴=-=-=． 【点睛】本题考查两点间距离、线段的和差定义等知识，解题的关键是灵活运用所学知识解决问题，属于中考常考题型．29．（1）a ﹣b +c ﹣d ；（2）－9【解析】【分析】（1）由数轴可得a ＜b ＜c ＜d ＜e ，然后可得a ﹣c ＜0，b ﹣a ＞0，b ﹣d ＜0并去掉绝对值最后合并同类项即可；（2）先确定b 、e 的值，然后再代入求值即可．【详解】解：（1）从数轴可知：a ＜b ＜c ＜d ＜e ，∴a ﹣c ＜0，b ﹣a ＞0，b ﹣d ＜0，原式＝|a ﹣c |﹣2|b ﹣a |﹣|b ﹣d |＝﹣a +c ﹣2（b ﹣a ）﹣（d ﹣b ）＝﹣a +c ﹣2b +2a ﹣d +b＝a ﹣b +c ﹣d ；（2）∵|a |＝|e |，∴a 、e 互为相反数，∵|b |＝3，这五个点满足每相邻两个点之间的距离都相等，∴b ＝﹣3，e ＝6，∴b ﹣e ＝﹣3﹣6＝﹣9．【点睛】本题考查了数轴、绝对值、相反数、有理数的大小比较等知识点，通过数轴确定a ＜b ＜c ＜d ＜e 是解此题的关键.30．(1)15.8；(2)()2.6 2.8x +；(3)他家离学校12千米.【解析】【分析】（1）根据题意，分为不超过2km 的部分和超出2km 的部分，列式计算即可；（2）根据题意，分为不超过2km 的部分和超出2km 的部分，列式即可；（3）由（2）中的代数式列出方程，求解即可.【详解】(1)由题意，得8+2.6×（5-2）=15.8元；故答案为15.8；(2)由题意，得()8 2.628 2.6 5.2 2.6 2.8x x x +⨯-=+-=+故答案为()2.6 2.8x +；(3)设他家离学校x 千米由题意得：2.6 2.834x +=，解得：12x =，答：他家离学校12千米【点睛】此题主要考查一元一次方程的实际应用，解题关键是理解题意，列出等式.四、压轴题31．（1）﹣14，8﹣5t ；（2）2.5或3秒时P 、Q 之间的距离恰好等于2；（3）点P 运动11秒时追上点Q ；（4）线段MN 的长度不发生变化，其值为11，见解析.【解析】【分析】（1）根据已知可得B 点表示的数为8﹣22；点P 表示的数为8﹣5t ；（2）设t 秒时P 、Q 之间的距离恰好等于2．分①点P 、Q 相遇之前和②点P 、Q 相遇之后两种情况求t 值即可；（3）设点P 运动x 秒时，在点C 处追上点Q ，则AC =5x ，BC =3x ，根据AC ﹣BC =AB ，列出方程求解即可；（3）分①当点P 在点A 、B 两点之间运动时，②当点P 运动到点B 的左侧时，利用中点的定义和线段的和差求出MN 的长即可．【详解】（1）∵点A 表示的数为8，B 在A 点左边，AB =22，∴点B 表示的数是8﹣22=﹣14，∵动点P 从点A 出发，以每秒5个单位长度的速度沿数轴向左匀速运动，设运动时间为t （t ＞0）秒，∴点P 表示的数是8﹣5t ．故答案为：﹣14，8﹣5t ；（2）若点P 、Q 同时出发，设t 秒时P 、Q 之间的距离恰好等于2．分两种情况：①点P 、Q 相遇之前，由题意得3t +2+5t =22，解得t =2.5； ②点P 、Q 相遇之后，由题意得3t ﹣2+5t =22，解得t =3．答：若点P 、Q 同时出发，2.5或3秒时P 、Q 之间的距离恰好等于2；（3）设点P 运动x 秒时，在点C 处追上点Q ，则AC =5x ，BC =3x ，∵AC ﹣BC =AB ，∴5x ﹣3x =22，解得：x =11，∴点P 运动11秒时追上点Q ； （4）线段MN 的长度不发生变化，都等于11；理由如下：①当点P 在点A 、B 两点之间运动时：MN =MP +NP =12AP +12BP =12（AP +BP ）=12AB =12×22=11； ②当点P 运动到点B 的左侧时：MN =MP ﹣NP =12AP ﹣12BP =12（AP ﹣BP ）=12AB =11， ∴线段MN 的长度不发生变化，其值为11．【点睛】 本题考查了数轴一元一次方程的应用，用到的知识点是数轴上两点之间的距离，关键是根据题意画出图形，注意分两种情况进行讨论．32．（1）11n n 1-+，n n 1+（2）①()()n 1n 2m 3++②75364 【解析】【分析】 ()1观察发现：先根据题中所给出的列子进行猜想，写出猜想结果即可；根据第一空中的猜想计算出结果；()2①由16a 2m m 3==，212a 4m m 3==，320a m 3=，430a 10m m 3==，找规律可得结论；②由()()n 1n 2m 22713173++=⨯⨯⨯⨯知()()m n 1n 22237131775152++=⨯⨯⨯⨯⨯=⨯⨯，据此可得m 7=，n 50=，再进一步求解可得．【详解】()1观察发现：()111n n 1n n 1=-++； ()1111122334n n 1+++⋯+⨯⨯⨯+， 1111111122334n n 1=-+-+-+⋯+-+， 11n 1=-+， n 11n 1+-=+， n n 1=+； 故答案为11n n 1-+，n n 1+． ()2拓展应用16a 2m m 3①==，212a 4m m 3==，320a m 3=，430a 10m m 3==， ⋯⋯()()n n 1n 2a m 3++∴=， 故答案为()()n 1n 2m.3++ ()()n n 1n 2a m 61883②++==，且m 为质数，对6188分解质因数可知61882271317=⨯⨯⨯⨯，()()n 1n 2m 22713173++∴=⨯⨯⨯⨯，()()m n 1n 22237131775152∴++=⨯⨯⨯⨯⨯=⨯⨯，m 7∴=，n 50=，()()n 7a n 1n 23∴=++， ()()n 131a 7n 1n 2=⋅++，123n1111a a a a ∴+++⋯+ ()()33336m 12m 20m n 1n 2m =+++⋯+++()()311172334n 1n 2⎡⎤=++⋯+⎢⎥⨯⨯++⎢⎥⎣⎦31131172n 27252⎛⎫⎛⎫=-=- ⎪ ⎪+⎝⎭⎝⎭ 75364=． 【点睛】 本题主要考查数字的变化规律，解题的关键是掌握并熟练运用所得规律：()111n n 1n n 1=-++． 33．（1）B 点坐标为（0，﹣6），C 点坐标为（4，﹣6）（2）S △OPM ＝4t 或S △OPM ＝﹣3t+21（3）当t 为2秒或133秒时，△OPM 的面积是长方形OBCD 面积的13．此时点P 的坐标是（0，﹣4）或（83，﹣6）【解析】【分析】（1）根据绝对值、平方和算术平方根的非负性，求得a ，b ，c 的值，即可得到B 、C 两点的坐标；（2）分两种情况：①P 在OB 上时，直接根据三角形面积公式可得结论；②P 在BC 上时，根据面积差可得结论；（3）根据已知条件先计算三角形OPM 的面积为8，根据（2）中的结论分别代入可得对应t 的值，并计算此时点P 的坐标．【详解】（1）∵|2b +12|+（c ﹣4）2=0，∴a +6=0，2b +12=0，c ﹣4=0，∴a =﹣6，b =﹣6，c =4，∴B 点坐标为（0，﹣6），C 点坐标为（4，﹣6）．（2）①当点P 在OB 上时，如图1，OP =2t ，S △OPM 12=⨯2t ×4=4t ； ②当点P 在BC 上时，如图2，由题意得：BP =2t ﹣6，CP =BC ﹣BP =4﹣（2t ﹣6）=10﹣2t ，DM =CM =3，S △OPM =S 长方形OBCD ﹣S △0BP ﹣S △PCM ﹣S △ODM =6×412-⨯6×（2t ﹣6）12-⨯3×（10﹣2t ）12-⨯4×3=﹣3t +21． （3）由题意得：S △OPM 13=S 长方形OBCD 13=⨯（4×6）=8，分两种情况讨论：①当4t=8时，t=2，此时P（0，﹣4）；②当﹣3t+21=8时，t133=，PB=2t﹣626188333=-=，此时P（83，﹣6）．综上所述：当t为2秒或133秒时，△OPM的面积是长方形OBCD面积的13．此时点P的坐标是（0，﹣4）或（83，﹣6）．【点睛】本题考查了一元一次方程的应用，主要考查了平面直角坐标系中求点的坐标，动点问题，求三角形的面积，还考查了绝对值、平方和算术平方根的非负性、解一元一次方程，分类讨论是解答本题的关键．。

2017-2018学年建平中学第一学期期末考试高一英语Ⅱ. GrammarSection ADirections: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the one answer that best completes the sentence.There is no doubt that the sports meeting ______ next week will draw a large audience and become a complete success.【A】held【B】to hold【C】holding【D】to be held【答案】A______ you may have, you should build up your courage to face the challenge.【A】However a serious problem【B】What a serious problem【C】However serious a problem【D】What serious a problem【答案】CAfter graduation, she still remembers the days ______ she spent in Jian Ping High School which have left her with a sweet memory.【A】when【B】that【C】while【D】what【答案】BMy grandmother was much kinder to her youngest child than she was to the others, ______, of course, made others unhappy.【A】which【B】that【C】who【D】what【答案】ANo sooner ______ himself on the chair than he heard the dentist shouting loudly to his assistant who was hurrying off.【A】that he seated【B】he had been seated【C】had he seated【D】had he been seated【答案】CYou can’t imagine the difficulty Jane, a vegetarian had ______ the suitable food on the menu in that restaurant.【A】find【B】found【C】to find【D】finding【答案】CSection BDirection: After reading the passage below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word, for the other blanks, use one word that best fits each blank.Jennifer Egan is the author of several best sellers in America. However, her life has also had its “downs”. Egan remembers the moment when a piece that she thought was much better than anything else she(1)______ (write) was turned down.When she writes a novel, Egan says, it usually goes through 50 or 60 drafts(草稿). She sits down with a pen, a pad and an open mind, trying to write half-dozen pages a day. When the first draft is done, Egan types the whole thing, reads it and almost inevitably, “it’s terrible,” she says. But in her review process, Egan gives herself ways to fix it --- edits, outlines and discussions with other writers, all of(2)______ only lead to yet more drafts, more frustration but also more improvement.It’s not surprising that Egan likes to describe her creative process as “growing up all the way.” Such an outlook is key to(3)______ (deal) with failure, she says. Not necessarily getting it right the first time. That’s fine --- you’re recording something, anything, so that other ideas(4)______ rise to the surface. Hitting a dead end? Take a breath and try something else. Keep moving --- that’s how you battle against disappointment and how your creative drive gets going.Robert Epstein, founder of the Cambridge Center for Behavioral Studies, compares the process to being stuck in a locked room. The doorknob(把手) isn’t responding. You turn it, hit it and lift it. Nothing. “(5)______ you can’t turn that knob, lot of different thoughts pop into your head at the same time --- what we call creativity,” he says.Has Egan ever given up on a project? Of course. There were times when she simply couldn’t make work in her writing, which in turn, gave rise to more of her new ideas to be developed.Egan admits that failure is scary. So is climbing back on the high wire and starting from scratch. However, that’s(6)______ creative energy comes from. “The awards are wonderful,” she says. “But the joy comes from the freedom in trying the unusual.”【答案】1.had written 2. which 3. dealing 4. can 5. If 6. whereⅡ. VocabularyDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.An addition is an activity or substance we are eager to experience repeatedly, and for which we are willing, if necessary, to pay a price. Common addictions involve alcohol, cigarettes, food, drugs, gambling, etc. This article discusses the concepts which can help cope with addictive behavior.Usually, (1)______ minor addictions, such as watching too much television, or lying in bed on weekend mornings, are often not even considered addictions, because the price paid for engaging in them is not high. On the other hand, we tend to use the term “(2)______” to describe the person who, at least in the eyes of others, continues to be addicted in a behavior long after it has become clear that the (3)______ price being paid was not worth the benefit. The individual who has lostcareer, house, family and friends because of cocaine (可卡因) use, but is (4)______ to consider stopping, is an unfortunate example.Negative addictions (5)______ from those with very minor negative consequences, to those as serious as the cocaine addict just mentioned, with much area in between. Although it is not (6)______ true that a negative addiction grows stronger over time, a constant level of addictive behavior can lead to an increasing level of negative consequences.You may be surprised to learn that addictions can also be considered positive. Positive addictions are those in which the benefits (7)______ the price. A common example would be the habit of regular exercise. The price of membership in a gym, the time (8)______ and any clothing expense is outweighed by the benefits of better health, energy, self-confidence and appearance. As with negative addictions, positive addictions may not get stronger over time, and there is a broad range of how much benefit is actually obtained.What is common to both positive and negative addictions is the urge to engage in the addictive behavior, and the satisfaction that is (9)______ when the urge is acted upon(按...行事). The urge is a state of tension and (10)______ that is experienced uncomfortably as a desire for the substance or activity. Because we experience relief when the urge is acted upon, there is an increased likelihood that we will act on the urge again.【答案】1. K2.J 3.H 4.A 5.I 6.C 7.E 8.G 9.D 10.F【分析】1.空格后是形容词+名词，所以判定此处缺少副词，待选项是C和K相对地。

建平中学2016 学年第一学期期末考试高一英语试题命题人：高一英语组审题人：说明：（1）本场考试时间为90 分钟，总分100 分（2）请认真答卷，并用规范文字书写II.Grammar and Vocabulary Choices (8%)Directions: Choose the best answer to complete the following sentences:17.ourselves in the hall, all of us began to appreciate the 2017 Jianping New Year Concert.A. Having seatedB. SeatedC. To seatD. To be seated18.China release its annual characters and words of 2016 on December 20th, with “规(rule)” and “小目标（small goal）”among the candidates characters and words.A. to win outB. being won outC. winning outD. won out19.the disaster area himself gave victims a great deal of encouragement.A. The president will visitB. The presidents’ visitingC. The president visitedD. The president to visit20.The Great Wall, by Zhang Yimou is being regarded as one that can help the Chinese film take on the world.A.directedB. directingC. to directD. being directed21.Beijing recently published the third draft of regulations, which aroused fierce public discussions,_smog was listed as a meteorological (气象的) disaster.A.whenB. whichC. whatD. where22.sadly , the little boy raised his head and had the tears falling down on purpose.A.Seen to cryB. To been seen cryingC. Seen cryingD. To be seen cried23.Since drama was to the English class, studies have been more and more interested in the English learning.A. linkedB. devotedC. introducedD. adapted24.The police offered $100,000 as a(n) for information about the robbery which happened in that bank last week.A. awardB. profitC. expenseD. rewardIII.Blank-fillingDirections: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.“LET’S Uber.”Few companies offer something so popular 25 their nameA. recovering E. enjoyB.identityF. involvesC. willD. accounts forG. farming H. denybecomes a verb. But that is one of the many achievements of Uber, a company 26(found) in 2009 which is now the world’s most valuable startup, worth around $70 billion. 27app can summon(召唤) a car in moments in more than 425 cities around the world, to the fury of taxi drivers everywhere. But Uber’s ambitions, and the expectations underpinning its valuation, extend much 28 (far): using self-driving vehicles, it wants to make ride-hailing (扬招汽车)so cheap and convenient that people forgo car ownership altogether. 29 (not satisfy) with shaking up the $100-billion-a-year taxi business, it has its eye on the far bigger market for personal transport, worth as much as $10 trillion a year globally.Uber is not alone in this ambition. Companies big and small 30 (recognize) the transformative potential of electric, self-driving cars, summoned on demand. Technology firms including Apple, Google and Tesla are investing heavily 31 independent vehicles; from Ford to V olvo, incumbent carmakers are racing to catch up. It will transform daily life as profoundly as cars did in the 20th century: reinventing transport and reshaping cities,32 also dramatically reducing road deaths and pollution.IV.Vocabulary ChoicesDirections: Complete the following passage with the words in the box. Each word can only be used once. There is one extra word which you don’t need.People became vegetarians for a variety of reasons -----to relieve animal suffering , to pursuea healthier lifestyle or to reduce greenhouse gas emissions. No matter how much the great-eaters33 it, vegetarians have a point: Cutting out meat delivers a number of benefits. But if everyone became a strict vegetarian, there would be serious problems for millions, if not billions, of people.Food production 34 _one quarter of the greenhouse gas emissions world wide. The main responsibility fall on the livestock(牲畜) industry. If we should all go vegetarian, ideally at least 80% of that land could be used for 35 the grasslands and forests, which would capture carbon and relieve global warming. Furthermore, the increase of natural habitats would also likely to a benefit to biodiversity(生物多样性).Despite these advantages, if the world went vegetarian, we should need to provide clear career opportunities for people formerly engaged in the livestock industry. Otherwise, we would probably face significant unemployment and social disorder. Without livestock, life in certain environments would likely become impossible for some people, especially nomadic ( 游牧的) groups, whose culture 36 might be lost in the process. In addition, going vegetarian 37 replacing meat with other proper nutrition, which could create a health crisis in the developing world.Fortunately, the entire world doesn’t need to change to vegetarianism to 38 many of the benefits. Clear solutions already exist for reducing greenhouse gas emissions. What is lacking is the 39 to realize those changes.V.Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C, D. Fill in each blank with the word or phrase that best fits the context.Language is hard. In fact , it’s harder and more complicated than math. And yet, nearly small child can learn and master language.Why is math so overwhelming for so many students? And how high is the price we pay from having so many math- 40 or even math-illiterate(无知的) people in our society? Too high, especially as the ability to grasp data and pursue advanced work that involves math is becoming increasingly 41 for both citizens and job applicantsHow many of us feel incapable, rather than poorly taught, when we are faced with the difficulties of math? How many children who struggled to grasp math concepts were led to fell stupid?42 it to spoken or written language. When you make a mistake, a teacher corrects the part that is wrong. And then you proceed. With math, if you don't have the correct result, it is typically treated as wrong. And, as mistake after mistake 43 , too many students simply give up: I can't do math.But math is not about intelligence. It's a language that too many people never learn, often because the education process 44 the number of ways that a given person can arrive at a given solution. That's not a failure of children to learn. That's a failure of 45 . It's a failure of the school. We should not blame the student. (These are children, after all.)Part of the challenge is to 46 the gaps in knowledge, to clarify that the challenge is not that a student simply doesn't understand algebra(代数) or trigonometry(三角学) or whatever. There may be a particular basic concept that stands 47 of going forward in math, as well as other fields such as social science or engineering.Overcoming this block requires moving beyond broad industrialized education and to individualized, 48 learning that allows students to find their own way in. Show me a thousand students and I'll show you a thousand 49 pathways that they might take to achieve math success.With new digital technologies and a massive amount of data collection and analytics, we have the ability to help students identify the essential concepts they don't understand. We have data on all the students that solved a particular math problem and those that failed to solve it. We also have data on the problems they were able to solve prior to that. So as a student recognizes that they are 50 with, say, negative number concepts, they can go back and master the material— to fill in the gaps that allows them to go forward. And when they hit another tough spot? They can jump to the problems that allow them to master that concept. The hope is that as they progress, their interest and enthusiasm 51 .In the years ahead, that mindset, borne out of the failure of math instruction, should be52 . If we can succeed at breaking down the 53 that there's something wrong with a 3rd grader who cannot learn math—rather than something wrong with the teaching process—then we can look forward to new generations of math-literate citizens. Whatever career they choose, they will be more confident and more capable to understand and contribute to an increasinglycomplex, 54-driven world.40. A. interested B. amazed C. controlled D. terrified41.A. natural B. unnecessary C. important D. common42.A. Compare B. Link C. Combine D. Mix43.A. take up B. holds up C. makes up D. build up44.A. calculates B. misses C. estimates D. analyzes45.A. teaching B. communicating C. experimenting D. understanding46.A. interpret B. leave C. identify D. deny47.A. in the form B. in the center C. in the stage D. in the way48.A. modernized B. civilized C. personalized D. commercialized49.A. different B. scientific C. effective D. efficient50.A. going B. associating C. struggling D. cooperating51.A. decreases B. persists C. fades D. increases52.A. inspired B. removed C. established D. challenged53.A. prediction B. assumption C. truth D. theory54.A. data B. power C. material D. cultureSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have read.(A)In the United States, some people mark ballots( 选票) that bear the printed names of candidates . Others pull levers in voting machines. Americans are proud of their long history of holding free elections, but they did not invent the process.Free elections were held in Athens, Greece, 2400 years ago. Some Europeans took part in local free elections during the Middle Ages. The secret ballot was first used in 1858 in Australia.Today, voting is a new experience for many of the world’s peoples. Since World II, many who never had a voice in their government now take part in free elections.When India held its first national elections in 1951 and 1952, the process was so complicated that it took four months. Few people could read, so the ballots bore pictures that stood for the different parties. People who wanted to vote for Gandhi’s Congress party, for example, put an X under the picture of yoked oxen.In our hemisphere , there are also voters who cannot read. The Dominican Republic held its first free national election in 1962 with rainbow- coloured ballots. The different colors identified the parties for people who could not read the candidates’ names.Today, computers are changing the way people vote. In 1974, Mexico set up a computerized voting system. In the future, people all over the world may be able to vote by touching computer screens. These screens could show candidates’ names, pictures, or party symbols.Any change that encourages people to vote helps government work. Orgeon lets people vote by mail. Texas has voting in convenient spots such as malls. As long as people want to have a voice inthe government , they will try new ways of voting.55.Why did ballots in India bear pictures?A.Because pictures could stand for different parties.B.Because many voters couldn’t read candidates’ names.C.Because Gandi’s Congress Party liked the pictures of oxen.D.Because the process of election was very complicated56.We can infer from the passage that .A.People all over the world have been voting for 2,400 years.B.Americans invented the process of free electionsC.The first secret ballot was invented in Athens.D.Many people didn’t have the right to vote until World War II.57.On the whole, the passage tells about .plicated elections in America and EuropeB.the right way to carry out the free electionC.how people vote in many parts of the world.D.how free election developed in the western country.(B)58.The underlined word “robust” most probably means .A.softB. slightC. strongD. few59.Which of the following countries will probably account for the LEAST amount of GDP in 2017A.IndiaB. MalaysiaC. Saudi ArabiaD. United States60.Which of the following statements is NOT TRUE according to the passage?A.So far, the companies in America are not clear about the economic policyB.In 2017, the economy of Malaysia will grow as rapidly as ten years before.C.Saudi Arabia was depending on oil export to expand the economy.D.The India government cannot put all the economic programmes into effect in 2017.61.Where does this passage most probably come from?A.a travel brochureB. a novelC. a newspaperD. a shopping guides(C)My grandfather was the only one in his family to come to the United States. He still had relatives living in Europe. When the First World War broke out, he felt sorry about the fact that if my uncle, his only son had to go, it would be cousin fighting against cousin. In the early days of the war, my grandmother asked him to stop taking the German newspaper and to take an English language paper instead, he explained that it was an American newspaper, only printed in German. But my grandmother insisted. So, he finally gave up the German newspaper.One day, the inevitable happened and my uncle Milton received his draft notice. My grandparents were very upset, but my mother, his little sister was excited. Now she could tell her girl friends that her soldier brother was going off to war. She was ten years old then. All the little girls were delighted.When the day came for him to leave, his whole regiment(团) , in their uniforms, left together from the same train station. There was a band playing and my mother and her friends came to see him off. The moment came and the soldiers, none of whom had had any training, boarded the train. The band played and the crowd cheered. Although no one noticed, I’m sure my grandmother had a tear in her eye for the only son, going off to war. The train groaned as it knew the destiny to which it was taking its passengers still cheering and waving their flags, the band still playing, the train slowly departed the station.It had gone about a thousand yards when it suddenly stopped. The band playing, the crowd stopped cheering. Everyone gazed in wonder as the train slowly backed up and returned to the station. The doors opened and the men started to file out. Someone shouted, “ The war is over.” For a moment, nobody moved, but the people heard someone bark orders at the soldiers. The men lined up formed into two lines, walked down the steps and paraded down the street, as returning heroes, to be welcome home by the crowds.My mother said it was a great day, but she was just a little disappointed that it didn’t last a tiny bit longer. The next day my uncle returned to his job, and my grandfather resumed reading the German newspaper, which he read until the day he died.62.Why did my grandmother insist on her husband’s quitting reading German newspaper?A.Because she hated Germany very much.B.Because the newspaper were only printed in GermanC.Because she was afraid of being mistaken for GermansD.Because English was a more popular language at that time.63.Which of the following statements is TRUE?A.Some of the boys in my uncle’s group were trained before going to war.B.My grandparents went to the railway station to see my uncle off.C.My mother was upset when hearing my uncle was going to war.D.My grandfather didn’t read German newspaper until the war ended.64.What happened to my uncle according to the passage?A.He was warmly received after he came back from war.B.He had been longing to join the army since his childhood.C.The war was over before he was sent to the frontierD.It was a long time before he resumed his job65.What might my grandmother feel on the day when she sent his son to war?A.relievedB. disappointedC. shockedD. terrifiedTra nslation VI. TranslationDirections: tran nslate the following sentence into English, using the words given in the brackets.1.环顾四周，我惊讶地发现我是车上仅剩的乘客。

满分150分，考试时间为120分钟。

听力部分（共20分，每小题1分）题号Ⅰ Ⅱ Ⅲ Ⅳ 总 分 得分I . 句子理解。

（听五个句子，选择与所听句子内容最相符的图片。

每个句子只读一遍。

共5分）1.2.3.4.5.II . 问句应答。

（听五个问句，选出最佳的应答句。

每个句子读两遍。

共5分6. A. Comedies. I think they ’re funny.B. Let ’s go to a movie tonightC. This play is very interesting.7.A. I like both of them B. Shanghai is a big city C. I live in Beijing8.A. Yes, I want to join the music clu bB. The sports club doesn ’t need new balls.C Yes, and I ’m in the running club.9.A. My favorite subject is historyA B CA B CA B CA B CA B CB.Our history teacher is very strictC. I don’t like history at all.10.A. I want to buy a strawberry cake.B. Sure, I’m good at making cakes.C. This birthday cake looks good.III.对话理解，（听五组小对话，根据你所听到的对话内容选择能回答下列问题的最佳答案。

每组对话读两遍。

共5分。

）11. Who likes Jet Li best?A. The girl.B. The boy’s mother.C. The boy.12. Why does the boy need some new shirts?A.He needs a new shirt for a party.B. He doesn’t like his old shirts any more.C.His old shirts are too small for him.13. What does Henry usually eat for lunch?A.Hamburgers and French fries.B. French fries and fruit.C. Fruit and vegetables.14. How many hours does the woman work a day?A. Eight hours.B. Nine hours.C. Thirteen hours.15. What does the girl usually do in the morning?A. She plays basketball.B. She goes running.C. She cooks breakfast.IV.短文理解（听短文，根据你所听到的内容补全下面的卡片，每空一词，短文读两遍。