课时提升作业(二十五) 3.2.1

课时提升作业(八)有理数的加法(第2课时)(30分钟50分)一、选择题(每小题4分,共12分)1.计算+++++++的结果为( )A. B.-C. D.以上都不正确2.运用加法运算律计算+(-18)++(-6.8)+18+(-3.2),最适当的是( )A.+[(-18)+(-6.8)+(-3.2)]B.+[(-18)+18+(-3.2)]C.++(-6.8)]+[18+(-3.2)D.+[(-18)+18]+[(-3.2)+(-6.8)]3.王老师2014年5月份打在卡上的工资是3780元,同月用于买东西取出了1320元,6月份打在卡上的工资是3780元,同月用于买东西取出了800元,问此时,王老师卡上这两个月的钱数和为(存入为正,取出为负) ( )A.5300元B.5400元C.5440元D.5540元二、填空题(每小题4分,共12分)4.计算:(+16)+(-25)+(+24)+(-32)=[ + ]+[ + ]=(+40)+(-57)= .5.计算:1+(-2)+3+(-4)+5+(-6)+…+99+(-100)= .【变式训练】先观察下列等式:=-,=-,=-,……再计算:+++…+.6.10名学生参加数学竞赛,以80分为标准,超过80分记为正数,不足80分记为负数,得分记录如下(单位:分):10,+15,-10,-9,-8,-1,+2,-3,-2,+1,则这10名同学的总分是分.【互动探究】若题中给出的10个数据的和是0,那么10名同学的总分是多少?三、解答题(共26分)7.(8分)用简便方法计算下列各题:(1)+++.(2)(-0.5)+++9.75.(3)(-3.5)++++0.75+.8.(8分)某出租车司机小李某天下午营运全是在东西走向的人民大道上进行的.如果规定向东为正,向西为负,他这天下午行车里程(单位:km)如下:+15,-2,+5,-1,+10,-3,-2,+12,+4,-5,+6.将最后一名乘客送到目的地时,小李在下午出车时出发点的什么方向?多远?【变式训练】某村共有10块麦田,今年的收成与去年相比(增产为正,减产为负)的情况如下(单位:kg):+32,+17,-39,-11,+15,-13,+8,+3,+11,-21,请问今年小麦的总产量与去年相比发生了什么变化?【培优训练】9.(10分)阅读下面的方法,并计算.-5+++17.解:原式=+++=[(-5)+(-9)+(-3)+17]+=0+=-.上述这种方法叫做拆项法,依照上述方法计算:++4028+.课时提升作业(八)有理数的加法(第2课时)(30分钟50分)一、选择题(每小题4分,共12分)1.计算+++++++的结果为( ) A. B.-C. D.以上都不正确【解析】选B.题中有三对互为相反数,运用运算律后,原式=+=-. 2.运用加法运算律计算+(-18)++(-6.8)+18+(-3.2),最适当的是( )A.+[(-18)+(-6.8)+(-3.2)]B.+[(-18)+18+(-3.2)]C.++(-6.8)]+[18+(-3.2)D.+[(-18)+18]+[(-3.2)+(-6.8)]【解析】选D.互为相反数的先结合、同分母的相结合、两数和为整数的结合,综上选项D最适当.3.王老师2014年5月份打在卡上的工资是3780元,同月用于买东西取出了1320元,6月份打在卡上的工资是3780元,同月用于买东西取出了800元,问此时,王老师卡上这两个月的钱数和为(存入为正,取出为负) ( )A.5300元B.5400元C.5440元D.5540元【解析】选C.由题意得:(+3780)+(-1320)+(+3780)+(-800)=[(+3780)+(+3780)]+[(-1320)+(-800)]=7560+(-2120)=5440(元).二、填空题(每小题4分,共12分)4.计算:(+16)+(-25)+(+24)+(-32)=[ + ]+[ + ]=(+40)+(-57)= .【解析】原式=[(+16)+(+24)]+[(-25)+(-32)]=(+40)+(-57)=-17.答案:(+16) (+24) (-25) (-32) -175.计算:1+(-2)+3+(-4)+5+(-6)+…+99+(-100)= .【解题指南】相邻两个数一正一负间隔出现,从第一个数开始两个数为一组,它们的和为-1,且刚好分为50组.【解析】1+(-2)+3+(-4)+5+(-6)+…+99+(-100)=[1+(-2)]+[3+(-4)]+[5+(-6)]+…+[99+(-100)]=(-1)+(-1)+(-1)+…+(-1)=-50.答案:-50【知识归纳】带有省略号的数的运算对于带有省略号的多数字的运算,其技巧性很强,先由式子发现特点,再根据特点灵活组合达到化难为易的目的.【变式训练】先观察下列等式:=-,=-,=-,……再计算:+++…+.【解题指南】由于=-,=-,…,=-,因此我们可将+++…+先拆开,然后再计算.【解析】原式=-+-+…+-=1-=.6.10名学生参加数学竞赛,以80分为标准,超过80分记为正数,不足80分记为负数,得分记录如下(单位:分):10,+15,-10,-9,-8,-1,+2,-3,-2,+1,则这10名同学的总分是分.【解析】10+(+15)+(-10)+(-9)+(-8)+(-1)+(+2)+(-3)+(-2)+(+1)=-5, 所以这10名同学的总分为:80×10+(-5)=795(分).答案:795【互动探究】若题中给出的10个数据的和是0,那么10名同学的总分是多少? 【解析】10名同学的总分是80×10=800(分).三、解答题(共26分)7.(8分)用简便方法计算下列各题:(1)+++.(2)(-0.5)+++9.75.(3)(-3.5)++++0.75+.【解析】(1)+++=+=+=.(2)(-0.5)+++9.75=+=-10+14.25=4.25.【一题多解】方法一:(-0.5)+++9.75=+=4+0.25=4.25.方法二:(-0.5)+++9.75=[(-0.5)+9.75]+=9.25+(-5)=4.25.(3)(-3.5)++++0.75+=++=0++0=-.8.(8分)某出租车司机小李某天下午营运全是在东西走向的人民大道上进行的.如果规定向东为正,向西为负,他这天下午行车里程(单位:km)如下:+15,-2,+5,-1,+10,-3,-2,+12,+4,-5,+6.将最后一名乘客送到目的地时,小李在下午出车时出发点的什么方向?多远?【解析】15+(-2)+5+(-1)+10+(-3)+(-2)+12+4+(-5)+6=(15+10+12+4+6)+[(-2)+(-1)+(-3)+(-2)]+[5+(-5)]=47+(-8)+0=39.答:小李在下午出车时出发点的东面39km.【变式训练】某村共有10块麦田,今年的收成与去年相比(增产为正,减产为负)的情况如下(单位:kg):+32,+17,-39,-11,+15,-13,+8,+3,+11,-21,请问今年小麦的总产量与去年相比发生了什么变化?【解析】(+32)+(+17)+(-39)+(-11)+(+15)+(-13)+(+8)+(+3)+(+11)+(-21)=[(+32)+(+17)+(+15)+(+8)+(+3)]+[(-39)+(-13)+(-21)]+[(-11)+(+11)]=75+(-73)+0=2.答:今年小麦的总产量比去年总产量增加了2kg.【培优训练】9.(10分)阅读下面的方法,并计算.-5+++17.解:原式=+++ =[(-5)+(-9)+(-3)+17]+=0+=-.上述这种方法叫做拆项法,依照上述方法计算:++4028+.【解析】++4028+=+(-2013)++4028+=[(-2014)+(-2013)+(-1)+4028]+=0+++=-2.。

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课时提升作业(六)Ⅰ. 单项填空1. Computer lessons must be to students of all grades, so that they can master this modern communication and learning tool.A. avoidableB. accessibleC. valuableD. acceptable2. (2013·成都模拟)I find hard to concentrate on my studies with some people having small talk around me.A. themB. himC. thatD. it3. —How is Dennis getting along with his work?—Well, he could always new ideas for increasing sales.A. come up withB. come aboutC. get away fromD. get up4. (2013·哈尔滨模拟)The magician picked some persons from the audienceand asked them to help him with the performance.A. by accidentB. in vainC. at randomD. on average5. It is said that some software on learning subjects for pupils in a very practical way.A. designedB. is designedC. designsD. is designing6. (原创)Terribly, a Nepal’s plane crashed down near their capital, killing 19,7 British tourists.A. containingB. includingC. coveringD. counting7. (2013·福州模拟)Drunk driving used to occur, but now it is under control.A. frequentlyB. generallyC. practicallyD. gradually8. There were so many people talking in the main hal l that I couldn’t the words of the lecturer.A. concentrate onB. fix uponC. centre onD. devote to9. They were taken to the police station as they had entered the area without.A. permitB. permittedC. permissionD. a permission10. “Hope for the best and prepare for the worst” is a proverb, meaning that life is beautiful and full of frustration.A. stillB. as wellC. eitherD. in case11. Our car had a on the way to the airport last week.A. breakoutB. breakdownC. break-inD. break-up12. Some advertisements pictures or words of experts to show people howgood the products are.A. are consisted ofB. are made ofC. consist ofD. are made up13. Grandma’s very and does all her own shopping and cooking.A. financialB. individualC. friendlyD. independent14. (原创)Last year, China sold a total of 18. 5 million vehicles, 12. 8 million vehicles sold in the US.A. comparing withB. being compared withC. to compare withD. compared with15. Anyone not familiar with the Internet is at a serious when applying for a job in such an age of information.A. disadvantageB. advantageC. disagreementD. agreementⅡ. 阅读理解(2013·皖南八校模拟) Since the Internet has come into homes, the daily life has never been the same again. But the thing that worries most of us is that we can get viruses from the Internet. But can we catch viruses on our cell phones? A new study in the journal Science says yes, but the spread of such mobile software that can bring harm to our cell phones won’t reach dangerous levels until more cell phones are on the sameoperating system.Computers are easily attacked by viruses because they share data, especially over the Internet. Of course, nowadays, more people are using their cell phones more and more frequently. They use them for emailing, text messaging and downloading troublesome ring tones, etc. so it is obvious that cell phone viruses are a threat, as well.Scientists used nameless call data from more than six million cell phone users to help model a possible outbreak. And they concluded that viruses that spread from phone to phone by Bluetooth are not much concerned, because users have to be in close physical relation for their phones to“see”one another. However, viruses that spread through multimedia messaging services can move much faster, because they can come in disguise, such as a cool tune sent by a friend. The good news is that to be effective, these viruses need their victims to all use the same operating system, which not enough of us do. Because there is no Microsoft operating system for mobile phones, yet. Thank goodness. (272W)1. What’s the passage mainly about?A. The operating system of cell phones.B. The threat of cell phone viruses.C. The wide use of cell phones.D. Computer viruses.2. When, cell phone viruses can be dangerous.A. all the cell phones work on the same operating systemB. Microsoft operating system for cell phones is createdC. users can see each other on the phoneD. Bluetooth is widely used3. What can we infer about Bluetooth according to the scientists?A. It can increase the chances of cell phone viruses greatly.B. It can help us to stop the spreading of cell phone viruses.C. It won’t possibly cause the outbreak of cell phone viruses.D. It can make users have close physical relation with one another.4. We can learn from the passage that.A. it is impossible to catch viruses on our cell phonesB. cell phones are not well connected with the InternetC. it is dangerous for people to download ring tones to cell phonesD. more than one operating system is available for cell phone users nowⅢ. 阅读第二节(2013·延边模拟) 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。

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课时提升作业(二十二)电路电路的基本规律素能全练(建议：20分钟50分)选择题(本题共8小题,1～6题每小题6分,7、8题每小题7分,共50分。

多选题已在题号后标出)1.一电池外电路断开时的路端电压为3V,接上8Ω的负载电阻后路端电压降为2.4 V,则可以判定电池的电动势E和内阻r为( )A.E=2.4V,r=1ΩB.E=3V,r=2ΩC.E=2.4V,r=2ΩD.E=3V,r=1Ω2.(2022·甘肃省甘谷质检)如图所示的电路中,电源电动势为E,内阻为r,闭合开关后,滑动变阻器滑片P向左移动。

下列结论正确的是( )A.电流表读数变小,电压表读数变大B.小灯泡L变亮C.电容器C上电荷量减小D.电源的总功率变大3.(2022·营口模拟)如图所示是某直流电路中电压随电流变化的图像,其中a、b分别表示路端电压、负载电阻上电压随电流变化的状况,下列说法正确的是( ) A.阴影部分的面积表示电源输出功率B.阴影部分的面积表示电源的内阻上消耗的功率C.当满足α=β时,电源效率最高D.当满足α=β时,电源效率小于50%4.在如图所示的电路中,E为电源电动势,r为电源内阻,R1和R3均为定值电阻,R2为滑动变阻器。

当R2的滑动触头在a端时合上开关S,此时三个电表A1、A2和V的示数分别为I1、I2和U。

现将R2的滑动触头向b端移动,则三个电表示数的变化状况是( )A.I1增大,I2不变,U增大B.I1减小,I2增大,U减小C.I1增大,I2减小,U增大D.I1减小,I2不变,U减小5.(2022·西安模拟)某实物投影机有10个相同的强光灯L1～L10(24 V 200 W)和10个相同的指示灯X1～X10(220 V 2 W),将其连接在220V沟通电源上,电路图如图所示。

课时提升作业七循环语句（25分钟 50分）一、选择题（每小题5分，共20分）1.读程序,对甲、乙两程序和输出结果判断正确的是( )A．程序不同，结果不同B．程序相同，结果不同C．程序不同，结果相同D．程序相同，结果相同【解析】选C.程序甲是计数变量i从1开始逐步递增直到i=100时终止，变量S从1开始，这个程序计算的是：1×4×7× (100)程序乙是计数变量i从100开始逐步递减到i=1时终止，变量S从100开始，这个程序计算的是100×97×94× (1)但这两个程序是不同的，两种程序的输出结果相同．【补偿训练】下面程序输出的结果是( )A.满足1×3×5×…×n>2 017的最小整数nB.1+3+5+…+2 017C.求方程1×3×5×…×n＝2 017中的n值D.1×3×5×…×2 017【解析】选A.从所给的程序来看是循环语句，而输出的是i，且S≤2 017，故输出的是满足1×3×5×…×n>2 017的最小整数n.2.（2018·信阳高一检测）下列程序执行后输出的结果是132,那么在程序UNTIL后面的“条件”应为( )A.i＞11B.i＞=11C.i＜=11D.i＜11【解析】选D.输出的结果是132,即S=1×12×11,则程序中UNTIL后面的“条件”应为i＜11.3.如图，下列程序执行后输出的结果是( )A.3B.6C.10D.15【解析】选 C.程序语句执行中的数据变化如下：i=1,S=0,1≤4,S=1,i=2,2≤4,S=3,i=3,3≤4,S=6,i=4,4≤4,S=10,i=5,5≤4不成立，输出S=10.4.已知程序如图，其运行结果是j＝( )A．7 B．8 C．9 D．10【解析】选C.此程序是求使j2<100的最大正整数，又102＝100，故输出结果为j＝9.二、填空题（每小题5分，共10分）5.如图所示的程序运行后第3次输出的数是________.【解析】该程序中关键是循环语句，第一次输出的数是1，第二次输出的数是x=1+13=，22第三次输出的数是x=1+11 2.+=22答案：2【补偿训练】下列程序运行后输出的结果为_______.【解析】第一次循环，i＝1+2＝3，S＝3+2×3＝9，i＝4；第二次循环，i＝6，S＝3+2×6＝15，i＝7；第三次循环，i＝9，S＝3+2×9＝21，i＝10，所以输出S＝21.答案：216.已知程序如下，则输出结果S ＝_______.【解析】根据程序写出每次循环的结果. 第一次循环，i ＝2，S ＝4； 第二次循环；i ＝4，S ＝4＋16＝20； 第三次循环，i ＝6，S ＝20＋36＝56.由于i ＝6不满足条件，跳出循环，输出S ，结果为56. 答案：56三、解答题（每小题10分，共20分） 7.设计算法求111112233499100+++⋯+⨯⨯⨯⨯的值．把程序框图补充完整，并写出用基本语句编写的程序．【解析】由已知条件②处应为s=s+1k k 1+（），③处为k=k+1,按照程序框图依次执行程序：s=0，k=1判断框内条件是，s=112⨯，k=2判断框内条件是， s=111223+⨯⨯，k=3,判断框内条件是, 以此类推， s=1111.12233499100+++⋯+⨯⨯⨯⨯ 此时k 应为100， ①处为k<=99? 程序如下：【补偿训练】已知程序框图如图所示:（1）写出程序框图所对应的程序.（2）将图中的“直到型循环结构”改为“当型循环结构”，并写出当型循环相对应的程序.【解析】（1）程序如下：（2）将图中的“直到型循环结构”改为“当型循环结构”，程序框图如下：程序如下：8.已知函数y＝x3＋3x2-24x＋30，写出连续输入自变量的11个取值，分别输出相应的函数值的程序.【解析】程序如下：（20分钟 40分）一、选择题（每小题5分，共10分）1.运行如图所示的程序，如果输入的n是6，那么输出的p是( )A.120B.720C.1 440D.5 040【解析】选B.程序运行的过程为：（1）p=1,k=2;(2)p=2,k=3;(3)p=6,k=4;(4)p=24,k=5;(5)p=120,k=6;(6)p=720,k=7,这时不满足k≤6，所以输出的p是720.2.下面程序运行后，输出的值是( )A．42 B．43 C．45 D．44【解析】选D.由已知可得程序的功能是利用循环计算满足i2＜2 000（i∈N）的最大i值,因为442＜2 000，452＞2 000.所以输出的值是44.二、填空题(每小题5分，共10分)3.执行如图程序，输出的结果S=_______.【解析】模拟程序的运行过程得，S=1，i=1，满足条件i＜8，执行循环：S=1+2=3，i=4；满足条件i＜8，执行循环：S=3+2=5，i=7；满足条件i＜8，执行循环：S=5+2=7，i=10；不满足条件i＜8，终止循环，输出S=7．答案：74.按照如图程序运行，则输出k的值是________.【解析】由程序语句知：算法是直到型循环结构的程序框图，第一次运行x=2×3+1=7,k=1；第二次运行x=2×7+1=15,k=2;第三次运行x=2×15+1=31，k=3；此时满足x>16，程序运行终止，输出k=3.答案：3【延伸探究】将本题程序中的输出k值保持不变且“16”改为“100”，求输入的最小整数x．【解析】设x＝a，则第一次循环，x＝2a＋1，k＝1；第二次循环，x＝2(2a＋1)＋1＝4a＋3，k＝2；第三次循环，x＝2(4a＋3)＋1＝8a＋7，k＝3，此时，执行，故整数x最小值为12. “是”，结束循环，输出k＝3.因此8a＋7>100，即a>938三、解答题（每小题10分，共20分）5.根据下面的要求，求满足1+2+3+…+n＞500的最小的自然数n．（1）画出执行该问题的程序框图.（2）以下是解决该问题的一个程序，但有两处错误，请找出错误并予以更正．【解析】（1）程序框图如图：（两者选其一即可，答案不唯一）（2）①直到型循环结构是直到满足条件退出循环，WHILE错误，应改成LOOPUNTIL；②根据循环次数可知输出n+1应改为输出n.6.某校高一某班共有60名学生参加数学竞赛，现已有这60名学生的竞赛成绩，请设计一个将竞赛成绩优秀的学生的个数输出的程序(规定89分以上为优秀)．【解析】程序框图如下：程序如下：【补偿训练】(2018·滁州高一检测)输入100个数，将其中正数的个数输出.试用循环语句设计程序.【解析】用WHILE语句编写程序如下：或用UNTIL语句编写程序如下：。

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课时提升作业(三)Ⅰ. 单项填空1. He feels relaxed and even begins to whistle because of the distance_______ covered.A. three fifth; wereB. three fifth; wasC. three fifths; have beenD. three fifths; has been2. —The head teacher said we had only three days to finish the project.—Don’t worry. We have already two thirds of it.A. taken overB. got throughC. made upD. given away3. She spoke ill of him out of.A. curiosityB. interestC. sympathyD. envy4. (2013·杭州模拟)Medical staff fled as flames and smoke filled an Indian hospital,their patients to the fire, which killed 89 people.A. abandonedB. having abandonedC. to abandonD. abandoning5. (2013·临汾模拟)Keeping on learning a foreign language can be very frustrating, but it will in the end.A. take offB. pay offC. call offD. send off6. Hearing this, tears came to my eyes as I realized a fool I had been to judge Jim’s father as a fail ure.A. howB. thatC. whatD. why7. (2013·哈尔滨模拟)—Would you mind if I parked my car here?—.A. Yes, pleaseB. Of course not. It’s not allowed hereC. I’d rather you didn’t actuallyD. No, you’d better not8. (2013·沈阳模拟)To enjoy the, Irene would rather spend long hours on the train than travel by air.A. sightB. viewC. sceneryD. scene9. I suggested that the person elected chairman.A. refer toB. referred to beC. referring to should beD. referred to was10. —He looks so and the kids are really scared.—He often uses that look to frighten them.A. frightened; frightenedB. frightened; frighteningC. frightening; frightenedD. frightening; frightening11. A few minutes earlier, and you would have seen the NBA star. But you.A. didn’t comeB. don’t comeC. haven’t comeD. won’t come12. The communication satellite from Xichang Space Center has entered the orbit successfully.A. to be launchedB. launchedC. launchingD. having launched13. It is because he experience that he is rejected.A. is short ofB. is short forC. lacks ofD. lacks for14. (2013·厦门模拟)The CCTV program, People Who Moved China, honors thosecontributions should be remembered.A. whoseB. whichC. theirD. who15. What prize in the following competition?A. do you think you will getB. do you expect will you getC. do you think you have gotD. do you expect have you gotⅡ. 阅读理解AKarl Benz would possibly be shocked at the continued popularity of his invention. Our desire for cars has grown to a point where there is now around one car for every 11 people in the world. Reasonably, you may think that more cars mean more road accidents. But, as ever, numbers tell the truth.Take the UK as an example. In 1950, a few years before I was born, there were 4. 4 million vehicles in Britain, one for every 11 people. In 2011, there were 34 million vehicles, eight times as many, and more than one for every 2 people. There were 5, 012 deaths on UK roads in 1950, but by 2011 the number of deaths had dropped by 63%.If we translate these figures into the chance in a million of dying, or micromorts (百万死亡率), there was an average of 102 micromorts per person per year in 1950, dropping to just 31 micromorts per year in 2011. Or to put it another way, each 100, 000 vehicles in 1950 were responsible for 114 deaths, but by 2011 they were responsible for only five deaths.Almost all of the richer nations have followed this pattern, in spite of traffic increasing. In the 30 years between 1980 and 2009, deaths on roads fell by 55% in Australia, 69% in France, 63% in Britain, 54% in Italy and 58% in Spain. But deaths slightly rose in Greece.Sadly these trends are not observed the world over. While the average risk for a person from dying on the roads in high-income countries is 103 micromorts per year, it is 205 in low-and middle-income countries. Of 3, 500 people killed a dayworldwide, 3, 000 are in the developing world, in spite of those countries containing less than half of all cars on the road. (304W)1. The first paragraph tells us.A. Karl Benz is the inventor of vehiclesB. there is a car for every 11 people in Britain nowC. numbers show more cars mean more road accidentsD. the number of cars in the world will reduce in the future2. How many people died in Britain by 2011?A. About 5, 000.B. About 3, 600.C. About 1, 800.D. About 63.3. In which of the following countries deaths on roads have possibly risen?A. The USA.B. Japan.C. India.D. Spain.4. From the passage we can learn.A. before 1950 there was more than one car for every two peoplein BritainB. there was just 31 micromorts per year in 2011 in France and BritainC. deaths on roads have slightly risen in Italy in the last 30 yearsD. now most of the deaths on roads are in the developing countriesB(原创)A record 1. 12 million people took theannual national civil servant exam on Nov. 26, 2012, arebound after the number fell in 2010 and 2011.Some 1. 5 million people who registered for this year’s exam passed the qualification assessment(认定), and 1. 12 million took the exam on Sunday, according to figures from the State Administration of Civil Servant. They are competing for around 20, 800 positions in more than 140 State-level government agencies and their local branches.The number of qualified applicants and number of people who took the exam reached a new high in 2009 before dropping in the following two years. This year’s numbers showed a rebound in interest. High employment pressure, fierce competition for jobs and the chance of a stable working life are the main reasons behind the demand for government positions, said an article released by the administration on Sunday.However, the Administration said that some positions, particularly those in remote and poor areas, have failed to attract ap plicants. “We hope our applicants can devote themselves to working in West China, in remote and less-developed regions, and in areas that the country needs them most, ”the article said.Li Yongxin, president of the Zhonggong Education Group, which offers private training for exam hopefuls, said that in the past few years, only two thirds of applicants actually take the exam. But on Sunday, three fourths of the applicants did so.The Administration has also warned that anyone who cheats in the exam will not be allowed to reapply for five years. And people who organize group cheating or hire ghostwriters to take the exam will be banned for life.Some university students, who are ineligible(不符合资格的)to take the civil servant exam, are trying to take the exam by applying with false information—inventing degrees and work experience.The civil servant exam is only open to university graduates and people who have work experience.Various examination sites across China have carried out measures to stop cheating. (328W)5. What would be the best title for the passage?A. Record Number of Hopefuls Took Part in Annual National Civil Servant ExamB. Annual National Civil Servant ExamC. Many People Took Part in ExamsD. Record Number of Hopefuls6. Which of the following statements is NOT true accordingto the passage?A. The number of the people taking part in the annual civil servant exam has been increasing for three years.B. Not all positions are attractive to the applicants.C. Three fourths of the applicants took the exam in 2012.D. Anyone who cheats in the exam will not be allowed to reapply for five years.7. What does the underlined word “rebound” in Paragraph 3most probably mean?A. riseB. fallC. rise again after a fallD. no change8. How many people who passed the qualification assessment didn’t take the exam this year?A. About 1, 500, 000.B. 1, 120, 000.C. 38, 000.D. About 380, 000.【语篇随练】翻译阅读理解中的句子并分析句子结构1. If we translate these figures into the chance in a million of dying, or micromorts, there was an average of 102 micromorts per person per year in 1950, dropping to just 31 micromorts per year in 2011.【译】_____________________________________________________________ ____________________________________________________________________ 【析】本句的主句为___________________________________________________, 前面是_______引导的条件状语从句, 后面是_________________作结果状语。

运动图象问题(建议用时40分钟）1.(2019·杭州模拟）下列运动图象中表示质点做匀加速直线运动的是（)【解析】选C。

位移—时间图象中,斜率表示速度大小，由此可知选项A中质点做的是匀速直线运动,A错误；选项B中质点静止，B错误;速度—时间图象中，斜率表示加速度大小,由此可知选项C中质点做的是匀加速直线运动，C正确；选项D中质点做的是匀速直线运动,D错误；故选C。

2。

如图所示是某质点运动的速度图象,由图象得到的正确结果是()A。

在0～1 s内该质点的平均速度是2 m/sB。

在0～1 s内该质点的位移大小是3 mC.在0～1 s内该质点的加速度大小大于2~4 s内的加速度大小D。

在0～1 s内该质点的运动方向与2~4 s内的运动方向相反【解析】选C.在0～1 s内位移为1 m,因此平均速度为1 m/s，A、B错误;由于v—t图象的斜率表示加速度,在0～1 s内加速度为2m/s2,而在2～4 s内,加速度为1 m/s2，因此C正确;在0～1 s内和2～4 s内速度均为正值，运动方向相同，D错误。

【总结提升】运用运动图象解题的技巧(1）利用图象分析物体的运动时,关键是从图象中找出有用的信息或将题目中的信息通过图象直观反映出来。

（2）v -t图象中,由于位移的大小可以用图线和t坐标轴围成的面积表示,因此可以根据面积判断物体是否相遇，还可以根据面积的差判断物体间距离的变化。

(3)利用图象可以直接得出物体运动的速度、位移、加速度,甚至可以结合牛顿第二定律根据加速度来确定物体的受力情况.3.(2018·浙江11月选考真题）一辆汽车沿平直道路行驶,其v-t 图象如图所示。

在t=0到t=40 s这段时间内,汽车的位移是()A.0 B。

30 m C.750 m D.1 200 m【解析】选C。

根据图象法可知通过的位移即为图象与时间轴围成的面积，根据面积可知位移为x=（10+40)×m=750 m.4.(2019·金华模拟）如图所示为A、B两质点从同一地点运动的x-t图象，关于A、B两质点在前8 s内的运动,下列说法正确的是 ()A。

课时提升作业二十四函数的最大(小)值与导数一、选择题(每小题5分,共25分)1.(2016·临沂高二检测)函数y=2x3-3x2-12x+5在[0,3]上的最大值和最小值分别是( )A.5,-15B.5,4C.-4,-15D.5,-16【解析】选A.y′=6x2-6x-12=6(x-2)(x+1),令y′=0,得x=2或x=-1(舍).因为f(0)=5,f(2)=-15,f(3)=-4,所以y max=5,y min=-15.【补偿训练】函数y=在区间上的最小值为( )A.2B.e2C.D.e【解析】选D.y′=,令y′=0,得x=1,故f(x)min=f(1)=e.2.(2016·德州高二检测)已知函数f(x),g(x)均为[a,b]上的可导函数,在[a,b]上连续且f′(x)<g′(x),则f(x)-g(x)的最大值为( )A.f(a)-g(a)B.f(b)-g(b)C.f(a)-g(b)D.f(b)-g(a)【解析】选A.[f(x)-g(x)]′=f′(x)-g′(x)<0,所以函数f(x)-g(x)在[a,b]上单调递减,所以f(x)-g(x)的最大值为f(a)-g(a).3.(2016·长春高二检测)若存在正数x使2x(x-a)<1成立,则a的取值范围是( )A.(-∞,+∞)B.(-2,+∞)C.(0,+∞)D.(-1,+∞)【解析】选D.因为2x(x-a)<1,所以a>x-.令f(x)=x-,所以f′(x)=1+2-x ln2>0.所以f(x)在(0,+∞)上单调递增,所以f(x)>f(0)=0-1=-1,所以a的取值范围为(-1,+∞).4.(2016·安庆高二检测)已知函数f(x)=-x3+2ax2+3x(a>0)的导数f′(x)的最大值为5,则在函数f(x)图象上的点(1,f(1))处的切线方程是( )A.3x-15y+4=0B.15x-3y-2=0C.15x-3y+2=0D.3x-y+1=0【解题指南】首先由导函数的最大值可以求出a值,再求切线方程.【解析】选B.因为f(x)=-x3+2ax2+3x,所以f′(x)=-2x2+4ax+3=-2(x-a)2+2a2+3,因为导数f′(x)的最大值为5,所以2a2+3=5,因为a>0,所以a=1,所以f′(1)=5,f(1)=,所以在函数f(x)图象上的点(1,f(1))处的切线方程是y-=5(x-1),即15x-3y-2=0.5.(2016·潍坊高二检测)已知f(x)=2x3-6x2+m(m为常数)在[-2,2]上有最大值3,那么此函数在[-2,2]上的最小值是( )A.-37B.-29C.-5D.以上都不对【解题指南】先根据最大值求出m,再求出f(x)在[-2,2]上的最小值.【解析】选A.因为f′(x)=6x2-12x=6x(x-2),因为f(x)在[-2,0]上为增函数,在[0,2]上为减函数,所以当x=0时,f(x)=m最大.所以m=3,从而f(-2)=-37,f(2)=-5.所以最小值为-37.二、填空题(每小题5分,共15分)6.当x∈[-1,1]时,函数f(x)=的值域为.【解析】f′(x)==,令f′(x)=0,得x1=0,x2=2(舍去)当x∈[-1,0)时,f′(x)<0;当x∈(0,1]时,f′(x)>0,所以当x=0时,f(x)取极小值f(0)=0,也是最小值;而f(-1)=e,f(1)=,所以f(x)的最大值为f(-1)=e.所以f(x)的值域为[0,e].答案:[0,e]7.(2016·洛阳高二检测)函数f(x)=(x∈[-2,2])的最大值是,最小值是.【解析】因为f′(x)==,令f′(x)=0,得x=1或x=-1.又因为f(1)=2,f(-1)=-2,f(2)=,f(-2)=-,所以f(x)在[-2,2]上的最大值为2,最小值为-2.答案:2 -28.若函数f(x)=(a>0)在[1,+∞)上的最大值为,则a的值为. 【解析】f′(x)==,当x>时,f′(x)<0,f(x)单调递减;当-<x<时,f′(x)>0,f(x)单调递增;当x=时,f(x)==,解得=<1,不合题意,所以f(x)max=f(1)==,所以a=-1.答案:-1三、解答题(每小题10分,共20分)9.(2016·宁波高二检测)设函数f(x)=e x sinx.(1)求函数f(x)的单调递增区间.(2)当x∈[0,π]时,求函数f(x)的最大值和最小值.【解析】(1)f′(x)=e x(sinx+cosx)=e x sin.f′(x)≥0,所以sin≥0,所以2kπ≤x+≤2kπ+π,k∈Z,即2kπ-≤x≤2kπ+π,k∈Z.f(x)的单调增区间为,k∈Z.(2)由(1)知当x∈[0,π]时,是单调增区间,是单调减区间.f(0)=0,f(π)=0,f=,所以f(x)max=f=,f(x)min=f(0)=f(π)=0.10.(2015·全国卷Ⅱ)已知f(x)=lnx+a(1-x).(1)讨论f(x)的单调性.(2)当f(x)有最大值,且最大值大于2a-2时,求a的取值范围. 【解析】(1)f(x)的定义域为(0,+∞),f′(x)=-a.若a≤0,则f′(x)>0,所以f(x)在(0,+∞)上单调递增.若a>0,则当x∈时,f′(x)>0;x∈时,f′(x)<0,所以f(x)在上单调递增,在上单调递减.(2)由(1)知,当a≤0时,f(x)在(0,+∞)上无最大值;当a>0时,f(x)在x=处取得最大值,最大值为f=ln+a=-lna+a-1. 因此f>2a-2等价于lna+a-1<0,令g(a)=lna+a-1,则g(a)在(0,+∞)上单调递增,g(1)=0.于是,当0<a<1时,g(a)<0;当a>1时,g(a)>0.因此,a的取值范围是(0,1).一、选择题(每小题5分,共10分)1.(2016·长沙高二检测)设直线x=t与函数f(x)=x2,g(x)=lnx的图象分别交于点M,N,则当|MN|达到最小值时t的值为( )A.1B.C.D.【解析】选D.|MN|的最小值,即函数h(x)=x2-lnx的最小值,h′(x)=2x-=,显然x=是函数h(x)在其定义域内惟一的极小值点,也是最小值点,故t=. 【补偿训练】函数f(x)=e x(sinx+cosx),x∈[0,1]的值域为.【解析】当0≤x≤1时,f′(x)=e x(sinx+cosx)+e x(cosx-sinx)=e x cosx>0,所以f(x)在[0,1]上单调递增,则f(0)≤f(x)≤f(1),即函数f(x)的值域为.答案:2.(2016·武汉高二检测)当x∈[-2,1]时,不等式ax3-x2+4x+3≥0恒成立,则实数a的取值范围是( )A.[-5,-3]B.C.[-6,-2]D.[-4,-3]【解析】选C.当x=0时,3≥0恒成立,a∈R.当0<x≤1时,a≥.设h(x)=,则h′(x)==.因为x∈(0,1],所以h′(x)>0,h(x)递增,所以h(x)max=h(1)=-6,所以a≥-6.当-2≤x<0时,a≤.易知h(x)=在[-2,-1)上递减,在(-1,0)上递增.所以h(x)min=h(-1)=-2,所以a≤-2.综上,-6≤a≤-2.二、填空题(每小题5分,共10分)3.(2016·沈阳高三模拟)已知函数f(x)=e x-2x+a有零点,则a的取值范围是.【解题指南】先求f′(x),判断f(x)的单调性,根据函数的单调性得到函数的最值.本题只要使f(x)的最小值不大于零即可.【解析】f′(x)=e x-2.由f′(x)>0得e x-2>0,所以x>ln2.由f′(x)<0得x<ln2,所以f(x)在x=ln2处取得最小值.只要f(x)min≤0即可,所以e ln2-2ln2+a≤0,所以a≤2ln2-2.答案:(-∞,2ln2-2]4.定义在R上的可导函数f(x)=x2+2xf′(2)+15,在闭区间[0,m]上有最大值15,最小值-1,则m的取值范围是.【解析】函数f(x)=x2+2xf′(2)+15的导函数为f′(x)=2x+2f′(2),所以f′(2)=4+2f′(2),所以f′(2)=-4,所以f(x)=x2-8x+15,且对称轴为x=4.又因为在闭区间[0,m]上有最大值15,最小值-1,且f(0)=15,f(4)=-1,所以[0,4]⊆[0,m],且f(m)≤f(0)=15,所以4≤m≤8.答案:[4,8]三、解答题(每小题10分,共20分)5.(2016·江苏高考改编)已知函数f(x)=a x+b x(a>0,b>0,a≠1,b≠1).设a=2,b=.(1)求方程f(x)=2的根.(2)若对任意x∈R,不等式f(2x)≥mf(x)-6恒成立,求实数m的最大值.【解题指南】(1)应用指数的运算性质求方程的根.(2)分离变量m,应用基本不等式求最值.【解析】(1)f(x)=2x+,由f(x)=2可得2x+=2⇒=0⇒2x=1⇒x=0.(2)由题意得22x+≥m-6恒成立,令t=2x+,则由2x>0可得t≥2=2,此时t2-2≥mt-6恒成立,即m≤=t+恒成立,因为t≥2时t+≥2=4,当且仅当t=2时等号成立,因此实数m的最大值为4.6.(2016·郑州高二检测)设函数f(x)=x3-(1+a)x2+4ax+24a,其中常数a>1.(1)讨论f(x)的单调性.(2)若当x≥0时,f(x)>0恒成立,求a的取值范围.【解析】(1)f′(x)=x2-2(1+a)x+4a=(x-2)(x-2a),由a>1知,2a>2,当x<2时, f′(x)>0,故f(x)在区间(-∞,2)上是增函数;当2<x<2a时,f′(x)<0,故f(x)在区间(2,2a)上是减函数;当x>2a时,f′(x)>0,故f(x)在区间(2a,+∞)上是增函数.综上,当a>1时,f(x)在区间(-∞,2)和(2a,+∞)上是增函数,在区间(2,2a)上是减函数.(2)由(1)知,当x≥0时,f(x)在x=2a或x=0处取得最小值.f(2a)=(2a)3-(1+a)(2a)2+4a·2a+24a=-a3+4a2+24a,f(0)=24a.由假设知即解得1<a<6.故a的取值范围是(1,6).。

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课时提升作业(二十八)直线与圆的方程的应用(25分钟60分)一、选择题(每小题5分,共25分)1.圆x2+y2-4x+2y+c=0,与直线3x-4y=0相交于A,B两点,圆心为P,若∠APB=90°,则c的值为( )A.8B.2C.-3D.3【解析】选C.由题意得C<5,圆心P(2,-1),r=,圆心到直线的距离d==2,由于∠APB=90°,所以r=d=2,从而=2,c=-3.【补偿训练】若P(2,-1)为圆(x-1)2+y2=25的弦AB的中点,则直线AB的方程是( ) A.x-y-3=0 B.2x+y-3=0C.x+y-1=0D.2x-y-5=0【解析】选A.已知圆心为O(1,0),根据题意:又k AB·k OP=-1,所以k AB=1,故直线AB的方程是x-y-3=0.2.如果实数x,y满足等式(x-1)2+y2=,那么的最大值是( )A. B. C. D.【解析】选D.的几何意义是圆上的点P(x,y)与原点连线的斜率,结合图形得,斜率的最大值为,所以=.3.台风中心从A地以20千米/时的速度向东北方向移动,离台风中心30千米内的地区为危险区域,城市B在A的正东40千米处,B城市处在危险区域的时间为( ) A.0.5小时 B.1小时C.3.6小时D.4.5小时【解析】选B.受影响的区域长度=2=20千米,故影响时间是1小时.4.点P(x0,y0)在圆x2+y2=r2内,则直线x0x+y0y=r2和已知圆的公共点个数为( ) A.0 B.1C.2D.无法确定【解析】选A.因为+<r 2,圆心到直线x0x+y0y=r2的距离d=>r,故直线与圆相离.【延伸探究】若将本题改为“点P(x0,y0)在圆x2+y2=r2外”,其余条件不变,又如何求解?【解析】选C.因为+>r 2,圆心到直线x0x+y0y=r2的距离d =< r,故直线与圆相交,所以公共点的个数为两个.5.已知集合M={(x,y)|y=,y≠0},n={(x,y)|y=x+b},若M∩N≠ ,则实数b的取值范围是( )A.[-3,3]B.[-3,3]C.(-3,3]D.[-3,3)【解题指南】解得本题的关键是注意到y=,即x2+y2=9(y>0),图形是半圆.【解析】选C.由于M∩N≠ ,说明直线y=x+b与半圆x2+y2=9(y>0)相交,画图探索可知-3<b≤3.【方法技巧】数形结合在求解直线与圆交点个数中的应用直线与圆的一部分有交点时,如果采用代数法去研究,则消元以后转化成了给定区间的二次方程根的分布问题,求解过程相对复杂,而如果采用数形结合及直线与圆的几何法求解,先找出边界,然后结合直线或圆的变化特征求解,相对来说就简单多了.二、填空题(每小题5分,共15分)6.过点A(11,2)作圆x2+y2+2x-4y-164=0的弦,其中弦长为整数的共有条.【解析】方程化为(x+1)2+(y-2)2=132,圆心为(-1,2),到点A(11,2)的距离为12,最短弦长为10,最长弦长为26,所以所求弦长为整数的条数为2+2×(25-11+1)=32.答案:32【补偿训练】过直线x+y-2=0上点P作圆x2+y2=1的两条切线,若两条切线的夹角是60°,则点P的坐标是.【解析】设P(x,y),则由已知可得PO(O为原点)与切线的夹角为30°,则|PO|=2, 由可得答案:(,)7.设村庄外围所在曲线的方程可用(x-2)2+(y+3)2=4表示,村外一小路方程可用x-y+2=0表示,则从村庄外围到小路的最短距离为.【解析】因为圆心到直线的距离为,从村庄外围到小路的最短距离为-2. 答案:-2【补偿训练】(2015·保定高一检测)已知实数x,y满足2x+y+5=0,那么的最小值为( )A. B. C.2 D.2【解析】选A.表示点(x,y)与原点的距离,所以其最小值为原点到2x+y+5=0的距离,故d==.8.已知x+y+1=0,那么的最小值是.【解析】表示点(x,y)与点(-2,-3)之间的距离,又点(x,y)在直线x+y+1=0上,故最小值为点(-2,-3)到直线x+y+1=0的距离,即d==2.答案:2三、解答题(每小题10分,共20分)9.等边△ABC中,点D,E分别在边BC,AC上且=,=,AD,BE相交于点P.求证:AP⊥CP.【解题指南】要证AP⊥CP,可转化为直线AP,CP的斜率之积等于-1即可,由此以B为原点,BC边所在直线为x轴,线段BC长的为单位长,建立平面直角坐标系. 【证明】以B为原点,BC边所在直线为x轴,线段BC长的为单位长,建立平面直角坐标系.则A(3,3),B(0,0),C(6,0).由已知,得D(2,0),E(5,).直线AD的方程为y=3(x-2).直线BE的方程为y=(x-5)+.解以上两方程联立成的方程组,得x=,y=.所以,点P的坐标是.直线PC的斜率k PC=-,因为k AP·k PC=3×=-1,所以,AP⊥CP.10.如图所示是某圆拱形桥一孔圆拱的示意图.这个圆的圆拱跨度AB=20m,拱高OP=4m,建造时每间隔4m需要用一根支柱支撑,求支柱A2P2的高度(精确到0.01m).【解析】建立如图所示直角坐标系,使圆心在y轴上,只要求出P2的纵坐标,就可得出支柱A2P2的高度.设圆心的坐标是(0,b),圆的半径是r,那么圆的方程是x2+(y-b)2=r2.因为P,B都在圆上,所以它们的坐标(0,4),(10,0)都满足方程x2+(y-b)2=r2.于是得到方程组解得b=-10.5,r2=14.52,所以,圆的方程是x2+(y+10.5)2=14.52.把点P2的横坐标x=-2代入圆的方程,得(-2)2+(y+10.5)2=14.52,即y+10.5=(P2的纵坐标y>0,平方根取正值).所以y≈3.86,故支柱A2P2的高度约为3.86m.【补偿训练】设有半径为3公里的圆形村落,A,B两人同时从村落中心出发,A向东而B向北前进,A离开村后不久,改变前进方向,斜着沿切于村落周界的方向前进,后来恰好与B相遇.设A,B两人的速度都一定,其比为3∶1,问A,B两人在何处相遇?【解析】如图所示,以村落中心为坐标原点,以东西方向为x轴建立直角坐标系,又设A向东走到D 转向到C恰好与B相遇,设CD方程为+=1(a>3,b>3),设B的速度为v,则A的速度为3v,依题意有解得,所以B向北走3.75公里时相遇.(20分钟40分)一、选择题(每小题5分,共10分)1.直线2x-y=0与圆C:(x-2)2+(y+1)2=9交于A,B两点,则△ABC(C为圆心)的面积等于( )A.2B.2C.4D.4【解析】选A.因为圆心到直线的距离d==,所以|AB|=2=4,所以S△ABC=×4×=2.【补偿训练】已知圆的方程为x2+y2-6x-8y=0.设该圆过点(3,5)的最长弦和最短弦分别为AC和BD,则四边形ABCD的面积为( )A.10B.20C.30D.40【解析】选B.圆心坐标是(3,4),半径是5,圆心到点(3,5)的距离为1,根据题意最短弦BD和最长弦(即圆的直径)AC垂直,故最短弦的长为2=4,所以四边形ABCD的面积为×AC×BD=×10×4=20.2.如图所示,已知直线l的解析式是y=x-4,并且与x轴、y轴分别交于A,B两点.一个半径为1.5的圆C,圆心C从点(0,1.5)开始以每秒0.5个单位的速度沿着y 轴向下运动,当圆C与直线l相切时,该圆运动的时间为( )A.6sB.6s或16sC.16sD.8s或16s【解析】选B.设运动的时间为ts,则ts后圆心的坐标为(0,1.5-0.5t).因为圆C 与直线l:y=x-4相切,所以=1.5.解得t=6或16.即该圆运动的时间为6s或16s.二、填空题(每小题5分,共10分)3.若点P(x,y)满足x2+y2=25,则x+y的最大值是.【解析】令x+y=z,则=5,所以z=±5,即-5≤x+y≤5,所以x+y的最大值是5.答案:5【拓展延伸】数形结合思想在解题中的运用利用数形结合求解问题时,关键是抓住“数”中的某些结构特征,联想到解析几何中的某些方程、公式,从而挖掘出“数”的几何意义,实现“数”向“形”的转化,如本题由x+y联想直线的截距.4.若点P在直线l1:x+y+3=0上,过点P的直线l2与曲线C:(x-5)2+y2=16相切于点M,则|PM|的最小值为.【解析】曲线C:(x-5)2+y2=16是圆心为C(5,0),半径为4的圆,连接CP,CM,则在△MPC中,CM⊥PM,则|PM|==,当|PM|取最小值时,|CP|取最小值,又点P在直线l1上,则|CP|的最小值是点C到直线l1的距离,即|CP|的最小值为d==4,则|PM|的最小值为=4.答案:4【补偿训练】圆(x-2)2+(y+3)2=4上的点到x-y+3=0的最远的距离为. 【解析】圆心C(2,-3)到直线的距离d==4>2,所以直线与圆相离.过圆心C作直线x-y+3=0的垂线,垂足设为H,则圆上的点A到直线的距离最远为4+2.答案:4+2三、解答题(每小题10分,共20分)5.已知圆C:(x-3)2+(y-4)2=4和直线l:x+2y+2=0,直线n经过圆C外定点A(1,0).若直线n与圆C相交于P,Q两点,与l交于N点,且线段PQ的中点为M,求证:|AM|·|AN|为定值.【解析】方法一:设P(x1,y1),Q(x2,y2),又由题意知直线与圆相交,斜率必定存在,且不为0,可设直线n的方程为kx-y-k=0,由得N.再由得(1+k2)x2-(2k2+8k+6)x+k2+8k+21=0,所以x1+x2=得M.所以|AM|·|AN|=·=·=6为定值.方法二:由题意知直线与圆相交,斜率必定存在,且不为0,可设直线n的方程为kx-y-k=0,由得N,又直线CM与n垂直,由得M.所以|AM|·|AN|=|y M-0|·|y N-0|=|y M·y N|==6,为定值.6.已知圆C的方程为x2+(y-4)2=4,点O是坐标原点.直线l:y=kx与圆C交于M,N 两点.(1)求k的取值范围.(2)设Q(m,n)是线段MN上的点,且=+.请将n表示为m的函数. 【解题指南】(1)求解时要抓住直线与圆有两个交点,所以在求解k的取值范围时可以利用判别式进行求解.(2)利用=+找到m,n的关系.【解析】(1)将y=kx代入x2+(y-4)2=4中,得(1+k2)x2-8kx+12=0.(*)由Δ=(-8k)2-4(1+k2)×12>0,得k2>3.所以,k的取值范围是(-∞,-)∪(,+∞).(2)因为M,N在直线l上,可设点M,N的坐标分别为(x1,kx1),(x2,kx2),则|OM|2=(1+k2),|ON|2=(1+k2),又|OQ|2=m2+n2=(1+k2)m2.由=+,得=+,即=+=.由(*)式可知,x1+x2=,x1x2=,所以m2=.因为点Q在直线y=kx上,所以k=,代入m2=中并化简,得5n2-3m2=36. 由m2=及k2>3,可知0<m2<3,即m∈(-,0)∪(0,).根据题意,点Q在圆C内,则n>0,所以n==.于是,n与m的函数关系为n=(m∈(-,0)∪(0,)).关闭Word文档返回原板块。

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课时提升作业三地球自转及其地理意义有甲、乙、丙、丁四架飞机以相同的线速度沿所在纬线自西向东飞行(如下图)。

据此回答1、2题。

1.四架飞机运动角速度最慢的是( )A.甲B.乙C.丙D.丁2.在乙飞机上空俯视地球绘制的地球运动的投影图正确的是( )【解析】1选C,2选A。

第1题,读图可知,丙地纬度最低,纬线最长,而飞机飞行的线速度相同,故飞机飞行转过的角度最小,角速度最慢。

第2题,在乙飞机上空俯视地球,应为斜俯视地球,其投影图为A。

(2018·东北四校联考)我国某中学天文观测地理兴趣活动小组经常观察天空中的恒星运动规律,该小组某学生于北京时间2014年9月23日21时看到织女星正好位于天顶,此时该地刚刚日落。

右图为此时北极星和织女星的位置图。

读图,完成3、4题。

3.该地可能位于( )A.北京B.黑龙江C.新疆D.西藏4.该学生于9月25日在同样的位置再次观测到织女星位于天顶,此时北京时间为( )A.21时整B.21时3分56秒C.20时56分4秒D.20时52分8秒【解析】3选C,4选D。

第3题,秋分日全球昼夜平分,北京时间21时时,该地地方时为18时,则其位于75°E,结合该地位于北纬40°可知,应在新疆。

第4题,地球自转一周(恒星日)的时间为23小时56分4秒,因而在25日再次观测到织女星位于天顶的时间为20时52分8秒。

5.下图为某主题公园所在城市分布图。

下列时间点中至少有4个城市该主题公园都在开放的是( )A.北京时间8点B.北京时间11点C.北京时间14点D.北京时间17点【解析】选B。

本题可逐项分析,如下所示:先画出表示全球所有经线(或时区)的数轴,标出已知经线(或时区)及其地方时(或区时),再标出所求经线(或时区),计算出两地经度差(或时区差)后,再将其转化为地方时差(或区时差)。