已解锁-AP微积分2006B真题

合集下载

2006年6月高等学校英语应用能力B级答案解析

!""$年$月高等学校英语应用能力考试 %级真题参考答案及解析&'() *+,-)./,/01234(.5./-,2/ 每题6分;.:),2/<9,2/,24%"(++6&)*>&(('-*能否见你的上司'#是征求可否"只有选项2*是的"当然/#是回答该一般疑问句的正确答案/A,-/,"('1/(6&)*@/*(''*/<'&-+/W/%0'-*这是否是你第一次来北京'#是一般疑问句"只有选项-*是的/#是正确答案/E,7/,"('1+*+4%6'1/%06&)F4%'$+'&:&'-是提出帮助"只有选项7可用于拒绝帮助"故正确/D,!/,I&)5:6&)5/M+4%&'1+*3)<&@'+4'-是提出邀请"只有选项!可用于拒绝别人的邀请或帮助"是正确答案/K,2/,I/556&)4''+%:'1+$++'/%0'1/(4@'+*%&&%'-是用F/55提问的一般疑问句"问,你今天下午是否来参加会议'-只有选项2*是的"我会去/#是正确答案/;.:),2/%N,-/该题问对话双方在说什么"由女士说的,"#Q+'*/+:+Q+*6'1/%0>)'$63&$<)'+*('/55 :&+(%#'F&*M,-可知他们在谈论女士的电脑出了问题/O,!/该题问关于女士我们知道些什么"由女士说的,"#Q+>++%4(+3*+'4*6@&*@/Q+6+4*(,-可知她是一位秘书/R,2/该题问女士想要什么"由女士的话,H4Q+6&)>*&)01'4<*/3+5/('F/'16&)'-可知她想要的是价目单/T,!/该题问该对话发生的地点"由男士说的,I&)5:6&)5/M+'&(++'1+$+%)'-可推知对话应发生在饭店里/9U,7/该题问女士何时退房"由男士说的,!%6'/$+'&$&**&F$&*%/%0,-可知退房时间为明天上午/;.:),2/1 拼写错误不给分99,:+45F/'1XX9A,)(+@)5XX9E,+@@+3'XX9D,'&'4556XX9K,F&%:+*@)5&'() *;)(8:)8(.;.:),2/< 每题"R7分9N,-/词汇题/(&h'14'h为固定搭配"意为,如此..以至于..-"符合题意/其他选项与'14'不能搭配"且意思不合/9O,!/词汇题/(&5)'/&%意为,解决*办法#-)1+5<意为,帮助-)*+<56意为,回答-):+$4%:意为,要求-/只有选项!符合题意/((!9R,2/本题考查情态动词的用法/$)('i14Q+:&%+表示对过去进行肯定的推测"意为,肯定..-)$46d$/01'i14Q+:&%+表示对过去进行不肯定的推测"意为,可能..-)&)01' '&d(1&)5:i14Q+:&%+表示对过去没有做某事表示埋怨"意为,本应该..-)&)01'%&' '&d(1&)5:%#'i14Q+:&%+表示对过去做了某事表示埋怨"意为,本没必要..-/根据句意"选项2是正确答案/9T,-/本题考查Y56结尾的词的词义辨析/*4*+56意为,很少地"罕有地-)%4$+56意为,那就是-)*+4556意为,真正地"实际上-)%4')*4556意为,自然地-/根据句意"选项-正确/ AU,!/本题考查不定式的语态/":&%#'5/M+是句子的主语和谓语"分句的逻辑主语"和作宾语的不定式'*+4'之间是被动关系"因此应用被动语态/A9,7/词汇题//$40/%+意为,想象-)/$<56意为,暗示-)/$<&*'意为,进口-)/$<*&Q+意为,改善"提高-/只有选项7符合题意/AA,7/本题考查谓语动词的时态/该句有两个动作('&<和:*/%M"根据句意:*/%M的动作先于('&<<+:"既在过去的过去发生"因此"选项7是正确答案/AE,-/词汇题/0+''1*&)01意为,穿越-)'4M+&@@意为,起飞-)<)55&)'意为,拔出-)>*+4M 4F46意为,断掉-/只有选项-符合题意/AD,!/词汇题//%('+4:&@意为,代替"而不是..-)>+34)(+&@意为,因为-)4(@&*意为,至于"就..方面说来-):)+'&意为,由于..-/只有选项!符合题意/AK,2/词汇题/)%'/5意为,直到..才-))%5+((意为,如果不"除非-)F1/5+意为,正当..-) /@意为,如果-/只有选项2符合题意/;.:),2/% 每题6分AN,>+('/?@455'1+1&'+5(/%'1+3/'6表示范围"根据句子的意思"此处应用0&&:的最高级'1+ >+('/AO,/%Q/'4'/&%/形容词后面应接名词作宾语/AR,(++/%0/>+F&*'1后要用动名词宾语/后面要求只能用动名词作宾语的常见动词短语还有34%#'1+5<"@++55/M+"0/Q+)<"M++<&%"&>W+3''&"<)'&@@">+>)(6"14Q+:/@@/3)5'6d'*&)>5+"5&&M@&*F4*:'&"0+'d>+)(+:'&等/AT,F/55d(1455d4*+0&/%0'&d4*+'&(/0%/时间状语%+P'F++M决定了此句要用将来时/ EU,L+%+*4556/0+%+*4556(<+4M/%0为固定搭配"意为,一般来说-/E9,>+'+('+:/动词'+('之前有情态动词&)01''&"因此要用原形"又本句主语'1+%+F$431/%+与谓语动词'+('之间是被动关系"故要用被动语态/EA,'&)(+/本句中6&)是句子的主语"F/55>+455&F+:是谓语*被动语态#")(+的逻辑主语是句子的主语"因此在这里作主语补足语/又动词455&F后要求用带'&的不定式作主语补足语"因此'&)(+是正确答案/EE,1+5<@)5/系动词后要接形容词作表语/ED,/(/本题考查主谓一致/'1+%)$>+*&@的意思是,..的数量-"是单数概念"作主语时谓语动词用单数/注意4%)$>+*&@的意思是,许多-"作主语时谓语动词用复数/EK,('*+%0'1+%/此处需要动词原形与F/55一起作谓语/(("&'() *D.'E,/01234(.5./-,2/G'-H6 每题!分EN,7/细节题/本题问美国政府为哪些人支付一些医疗费用"由文章第一段第二句话,81+ 0&Q+*%$+%':&+(1+5<<46@&*<+&<5+F1&4*+&%5&F/%3&$+(4%:@&*&5:<+&<5+h-*政府为低收入者和老年人提供帮助"支付一些医疗费用..#可知答案/EO,!/细节题/本题问美国大多数人为何购买保险"由第一段第二句话,h>)'$&('<+&<5+ >)6/%()*4%3+'&1+5<<46@&*$+:/34534*+,-*..但是大多数人购买保险以支付医疗费用/#可知选项!*为了支付医疗费用#是正确答案/ER,!/细节题/本题问英国人生病了经常怎么做"由第二段第一句话,"%-*/'4/%"F1+%<+&Y <5+4*+/55"'1+6)()45560&'&4@4$/56:&3'&*@/*(',-*在英国"人们生病时首先去看家庭医生/#可知答案/ET,2/推理题/本题问在美国重症病人会怎么做"由文章倒数第二句话,"%!$+*/34"1&(</Y '45($)(''*+4'455(+*/&)(56/55<4'/+%'("+Q+%/@'1+6:&%&'14Q+$+:/345/%()*4%3+,-*在美国"医院必须治疗重症病人"即使他们没有医疗保险/#可知答案/DU,-/主旨题/本题问哪个选项为最好的文章标题/通读全文可知文章主要介绍了英美两国医疗保健的差异"其中提及医疗保险0就医方式等方面的差异"故选项-*英美两国的医疗保健#是正确答案/G'-H! 每题!分D9,!/细节题/本题问从文章中我们可以知道;72总是怎么样"由第三段第一句话,I+ 4*+<*&):>+34)(+F+M%&FF+4*+43&$<4%6'14'M++<(/'(F&*:'&/'(3)('&$+*(h-*我们很自豪"因为我们是一家对顾客信守承诺的公司..#可知答案/DA,-/细节题/本题问因为公司怎么样;72公司相信他的顾客对公司满意/由文章第四段第一句话,h>+34)(+6&)3&%@/*$&)*>+5/+@('14'0&&:(+*Q/3+4%:\)45/'6*+()5'/%(4'/(Y @/+:3)('&$+*(,-*..优质的服务和质量会带来满意的顾客/#可知答案/DE,!/细节题/本题问如果发生什么情况公司将会通知顾客/由第三段最后一句话,h4%: '14'45F46(5+'(&)*3)('&$+*(M%&F/@'1+*+/('&>+4:+546/%:+5/Q+*6,-*..交货若有延迟"会通知顾客/#可知选项!*不能按时运送产品#是正确答案/DD,7/主旨题/本题问写这封信的目的是什么/由最后一段的主要内容*我们的特别夏季商品目录正在打印中"很快就会送到您家中"希望顾客能够满意#可知选项7*通知顾客有新的商品目录#是正确答案/DK,2/主旨推理题/本题问从文章中我们知道该公司是一家怎样的公司/由第一段的主要内容*写信人首先讲述该公司经过短短四年时间的发展已成为市场领域中最成功的公司之一"为此感到高兴0自豪和感激#可知选项2*在市场领域中最成功的一家公司#是正确答案/G'-H I 每题6分填写超过三个词不给分DN,35+4%/本题问地铁的几个特征"由文章第一段开头的,25+4%,;&:+*%,G4@+,!%:+4(6'& )(+,-可知答案/DO,+4(6'&)(+/同上/DR,*+:)3+:/本题问周末票价多少"由文章第三段,S46*+0)54*@4*+(&%F++M:46(K!EU&T!EU((#4,$,4%:E!UU&O!UU<,$,S46*+:)3+:@4*+(4'455&'1+*'/$+(,-*周一至周五K!EU&T!EU 和E!UU&O!UU为一般票价"其他时间减价/#可知周末票价减价/DT,+431('4'/&%/本题问在哪里能看到票价和运行时间/根据第四段第一句话,J4*0+$4<(/% +431('4'/&%(1&F@4*+(4%:'*4Q+5'/$+(,-*每一个车站的大地图上都标有票价和运行时间/#可知答案/KU,K d@/Q+/本题问地铁售票机最多提供多少零钱/由文章最后一段第二句话,h@4*+Y34*: $431/%+(&%56<*&Q/:+)<'&]K/%314%0+*/%3&/%(#,-*..地铁售票机最多提供K美元的硬币零钱#可知答案/G'-H F 每题6分!-)(+(?%56*公交专用道#-=&S4*M/%0*禁止停车#2=&G'4%:/%0*禁止停车候客#7S&5/3+24*(?%56*警车专用道#C=&V Y8)*%*禁止掉头#Z=&!:$/''4%3+*不得入内#L=&C%'*6-681/(7&&**此门不通#H?%+I46G'*++'*单行道#"?%+J4%+-*/:0+*单向桥#B!:$/((/&%-68/3M+'?%56*凭票入场#b!:$/((/&%Z*++*免费入场#J b++<!F46*请勿靠近#;H&)(+8&J+'*房屋出租#=b++<?*:+**保持秩序#?I+'S4/%'*油漆未干#S J/%+V<Z&*8/3M+'(*排队购票#a=&S&('/%0&@G/0%(*请勿张贴#^G+4'-6=)$>+**对号入座#G I1++5314/*(?%56*轮椅专用道#K9,-"CXXXKA,L"ZXXXKE,;"HXXXKD,S"B XXX KK,!"^G'-H7 每题6分填写超过三个词不给分+.)).(6KN,+P<5&:+:/本题问昨天晚上工厂里发生了什么事"由第一封信第一句话,J4('%/01''1+ 3+%'*451+4'/%0(6('+$'14'6&)/%('455+:/%&)*@43'&*6+P<5&:+:,-*你们给我公司安装的中央取暖系统昨晚发生了爆炸/#可知答案/KO,@4(1/&%35&'1+(/本题问给工厂带来了什么损失"由第一封信第一段第二句话,h&)*('&3M &@@4(1/&%35&'1+(14(>++%3&$<5+'+56*)/%+:,-*..我们的库存时尚服装全部被烧毁/#可知答案/KR,]DUU"UUU d DUU"UUU:&554*(/本题问这批库存服装价值多少"由第一封信第二段末尾的,h Q45)+:4']DUU"UUU-*..价值DU万美元#可知答案/KT,'1+1+4'/%0(6('+$/本题问在信中比尔提出的要求是什么"由第一封信第二段开头的,I+ ($($)('/%(/(''14'6&)*+<543+'1+1+4'/%0(6('+$/$$+:/4'+56h-*我们要求你们马上更换取暖系统..#可知答案/NU,!()*Q+6&*/本题问取暖系统供货商采取了什么措施"由第二封信第二段最后一句话,I+ 14Q+4(M+:4()*Q+6&*'&@/%:&)''1+34)(+&@'1++P<5&(/&%,-*我们已经请求调查员找出发生爆炸的原因/#可知答案/&'() *G('/-?'),2/ =/0?,-5,/)215,/.-.A分9分U,K分U分N9,!-72NA,2-!7NE,!27-ND,2!7-NK,参考译文共6!分我们非常高兴地欢迎中国朋友参加这次商务培训专修班d项目/在这里"你们将参加各种活动并且有机会相互交流*思想d经验#/我们希望"你们大家都会从培训项目中受益匪浅/在各位逗留期间"如有问题和困难"请告知我们/我们相信这次培训既有教育意义又轻松愉快/&'() *J(,),/0 67分;800.-).E</-O.(;<+=;9=&<D G N=Y GN=N\7!8C!B)%+9N"AUUN8?!;4%40+*(&@455&'1+*:+<4*'$+%'(Z^?;!B&1%L*++%";4%40+*&@'1+G45+(7+<4*'$+%'G V-B C28!7/(3)(('1+(45+(<54%&@'1+'1/*:\)4*'+*&@AUUN?)*:+<4*'$+%'14($4:+'1+'1/*:\)4*'+*(45+(<54%&@AUUN,I+F/551&5:4$++'/%0'&:/(3)((/'/%'1+$++'/%0*&&$4'9!UU<,$,&%B)%+9T,I+1&<+'1+$4%40+*(&@455:+<4*'Y $+%'(F/554''+%:'1+$++'/%0,"@(&$+&%+34%#'3&$+'&4''+%:'1+$++'/%0"<5+4(+%&'/@6'1+ (+3*+'4*6&@&)*:+<4*'$+%'/%4:Q4%3+,814%M6&)+(%(。

2006年12月23日大学英语老六级考试真题B卷3-2

2006年12⽉23⽇⼤学英语⽼六级考试真题B卷3-2Passage ThreeIntel chairman Andy Grove has decided to cut the Gordian knot of controversy surrounding stem cell research by simply writing a check.The check, which he pledged last week, could be for as much as $5 million, depending on how many donors make gifts of between $50,000 and $500,000. which he has promised to match. It will be made out to the University of California-San Francisco (UCSF).Thanks in part to such private donations, university research into uses for human stem cells the cells earliest stages of development that can form any body part-will continue in California. With private financial support, the state will be less likely to lose talented scientists who would be tempted to leave the field or even leave the field or even leave the country as research dependent on federal money slows to a glacial (极其缓慢的) pace.Hindered by limits President Bush placed on stem cell research a year ago, scientists are turning to laboratories that can carry out work without using federal money. This is awkward for universities. Which must spend extra money building separate labs and keeping rigorous records proving no federal funds were involved. Grove’s donation, a first step toward a $20 million target at UCSF. Will ease the burden.The president’s decision a year ago to allow research on already existing stem cell lines was portrayed as a reasonable compromise between scientists’ needs for cells to work with, and concerns that this kind of research cold lead to wholesale creation and destruction of human embryos (胚胎)。

ap微积分真题答案

ap微积分真题答案【篇一：ap 微积分bc 选择题样卷一】version - section i - part a calculators are not permitted on this portion of the exam28 questions - 55 minutes1) givenfind dy/dx. a) b) c) d) e)2) give the volume of the solid generated by revolving the region bounded by the graph of y = ln(x), the x-axis, the lines x = 1 and x = e, about the y-axis. a) b) c) d)e)3) the graph of the derivative of fis shown below.find the area bounded between the graph of f and the x-axis over the interval [-2,1], given that f(0) = 1. a) b) c) d) e)4) determine dy/dt, given thatanda)b)c)d)e)5) the function-1is invertible. give the slope of the normal line to the graph of f atx = 3.a) b) c)d)e)6) determinea) b)c) d) e)7) give the polar representation for the circle of radius 2 centered at ( 0 , 2 ). a)b)c)d)e)8) determinea)b)c) d)e)9) determinea) b) c) d) e)10) give the radius of convergence for the seriesa)b)c) d)e)11) determinea)b)【篇二：ap 微积分bc选择题样卷二】17 questions - 50 minutes1) the limit of the sequenceas n approaches is -3. what is the value of c? a)b)c)d)e)2) ifand y = 3 when x = -2, then what is y? a) b) c)d) e)3) the graph of the derivative of fis given below.which of the following is false about the function f?a) f is increasing on [1,4].b) f is concave down on [1,5/2].c) f is concave down on [-3,0).d) f is not differentiable at 0. e) the funciton is constant on (-,-3].4) determinea)b)c) d) e)5) give the area that lies below the x-axis and is contained within theregion bounded by the polar curvea) b) c) d) e)6) give the error that occurs when the area between the curve and the x-axis over the interval [0,1] is approximated by the trapezoid rule with n = 4. a)b)c)d)e)7) letdetermine f(2/3). a)b)c) d) e)8) give the length of the curve determined byfor t from 0 to 2. a)b)c)d)e)9) particles a and b leave the origin at the same time and move along the y-axis. their positions are determined by the functionsfor t between 0 and 8. what is the velocity of particle b when particlea stops for the first time? a) b)c)d) e)10) the base of a solid is the region in the xy plane enclosed by thecurvesover the interval[0,/4]. cross sections of the solid perpendicular to the x-axis are squares. determine the volume of the solid. a)b)c)d)e)11) give the minimum value of the functionfor x 0. a)b)c)d)e)12) select the true statement associated with the function【篇三：ap微积分考试详解】>微积分ap课程包括微积分ab (calculus ab) 和微积分bc(calculus bc)两门课。

2006考研数学(二)真题及参考答案

2006年全国硕士研究生入学考试数学（二）一、填空题（1）曲线4sin 52cos x xy x x+=-的水平渐近线方程为 .（2）设函数231sin ,0,(),x t dt x f x x a x ⎧≠⎪=⎨⎪=⎩⎰在0x =处连续，则a = .（3）广义积分22(1)xdxx +∞=+⎰.（4）微分方程(1)y x y x-'=的通解是 . （5）设函数()y y x =由方程1yy xe =-确定，则A dy dx== .（6）设矩阵2112A ⎛⎫= ⎪-⎝⎭，E 为2阶单位矩阵，矩阵B 满足2B A BE =+，则B = . 二、选择题（7）设函数()y f x =具有二阶导数，且()0,()0f x f x '''>>，x ∆为自变量x 在0x 处的增量，y ∆与dy 分别为()f x 在点0x 处对应的增量与微分，若0x ∆>，则（A ）0.dy y <<∆ （B ）0.y dy <∆<（C ）0.y dy ∆<<（D ）0.dy y <∆<【】（8）设()f x 是奇函数，除0x =外处处连续，0x =是其第一类间断点，则()x f t dt ⎰是（A ）连续的奇函数. （B ）连续的偶函数（C ）在0x =间断的奇函数（D ）在0x =间断的偶函数. 【】（9）设函数()g x 可微，1()(),(1)1,(1)2g x h x e h g +''===，则(1)g 等于（A ）ln 31-. （B ）ln 3 1.--（C ）ln 2 1.--（D ）ln 2 1.-【】（10）函数212xxx y C e C e xe -=++满足一个微分方程是（A ）23.xy y y xe '''--= （B ）23.xy y y e '''--=（C ）23.xy y y xe '''+-=（D ）23.xy y y e '''+-=（11）设(,)f x y 为连续函数，则140(cos ,sin )d f r r rdr πθθθ⎰⎰等于（A ）22120(,).x xdx f x y dy -⎰⎰（B ）22120(,).x dx f x y dy -⎰⎰（C ）22120(,).y ydy f x y dx -⎰⎰（D ）22120(,).y dy f x y dx -⎰⎰【】（12）设(,)f x y 与(,)x y ϕ均为可微函数，且1(,)0y x y ϕ≠. 已知00(,)x y 是(,)f x y 在约束条件(,)0x y ϕ=下的一个极值点，下列选项正确的是（A ）若00(,)0x f x y '=，则00(,)0y f x y '=. （B ）若00(,)0x f x y '=，则00(,)0y f x y '≠. （C ）若00(,)0x f x y '≠，则00(,)0y f x y '=. （D ）若00(,)0x f x y '≠，则00(,)0y f x y '≠.【】（13）设12,,,,a a a 均为n 维列向量，A 是m n ⨯矩阵，下列选项正确的是（A ）若12,,,,a a a 线性相关，则12,,,,Aa Aa Aa 线性相关. （B ）若12,,,,a a a 线性相关，则12,,,,Aa Aa Aa 线性无关.（C ）若12,,,,a a a 线性无关，则12,,,,Aa Aa Aa 线性相关.（D ）若12,,,,a a a 线性无关，则12,,,,Aa Aa Aa 线性无关. 【】（14）设A 为3阶矩阵，将A 的第2行加到第1行得B ，再将B 的第1列的-1倍加到第2列得C ，记110010001P ⎛⎫⎪= ⎪ ⎪⎝⎭，则（A ）1.C P AP -= （B ）1.C PAP -=（C ）.T C P AP =（D ）.TC PAP =三解答题15．试确定A ，B ，C 的常数值，使得23(1)1()xe Bx Cx Ax o x ++=++，其中3()o x 是当30x x →时比的高阶无穷小.16．arcsin xxe dx e ⎰求. 17．{}22(,)1,0D x y x y x =+≤≥设区域，221.1DxyI dxdy x y +=++⎰⎰计算二重积分 18．{}110,sin (0,1,2,)n n n x x x x n π+<<== 设数列满足1lim n x x +→∞证明: (1) 存在,并求极限；211(2)lim()n x n x nx x +→∞计算. 19．sin 2cos sin cos .<a <b b b b b a a a a a πππ<++>++证明: 当0时, 20 设函数()()0,,f u +∞在内具有二阶导数且()22z fx y=+满足等式22220z zx y∂∂+=∂∂.(Ⅰ)验证()()0f u f u u'''+=；(Ⅱ)若()()()10,11,f f f u '==求函数的表达式. 21 已知曲线L 的方程为221,(0),4x l t y l t⎧=+≥⎨=-⎩（Ⅰ）讨论L 的凹凸性；（Ⅱ）过点（-1，0）引L 的切线，求切点00(,)x y ，并写出切线的方程；（Ⅲ）求此切线与L （对应于0x x ≤的部分）及x 轴所围成的平面图形的面积.22 已知非齐次线性方程组12341234123414351331x x x x x x x x ax x x bx +++=-⎧⎪++-=-⎨⎪++-=⎩有个线性无关的解Ⅰ证明方程组系数矩阵A 的秩()2r A =； Ⅱ求,a b 的值及方程组的通解.23 设3阶实对称矩阵A 的各行元素之和均为3,向量()()121,2,1,0,1,1TTαα=--=-是线性方程组A x =0的两个解, (Ⅰ)求A 的特征值与特征向量 (Ⅱ)求正交矩阵Q 和对角矩阵A,使得TQ AQ A =.真题解析一、填空题（1）曲线4sin 52cos x xy x x+=-的水平渐近线方程为15y =4sin 11lim lim2cos 55x x xx y x x→∞→∞+==-（2）设函数2301sin ,0(),0xt dt x f x x a x ⎧≠⎪=⎨⎪=⎩⎰ 在x =0处连续，则a =132200()1lim ()lim 33x x sm x f x x →→==（3）广义积分22(1)xdxx +∞=+⎰1222222201(1)11110(1)2(1)2(1)22xdx d x x x x +∞+∞+∞+==-⋅=+=+++⎰⎰（4）微分方程(1)y x y x-'=的通解是xy cxe -=)0(≠x（5）设函数()y y x =由方程1yy xe =-确定，则0x dy dx==e-当x =0时，y =1，又把方程每一项对x 求导，y yy e xe y ''=--01(1)1x x y yyyye y xe ey e xe ===''+=-=-=-+(6) 设A = 2 1 ,2阶矩阵B 满足BA =B +2E ,则|B |= .-1 2解:由BA =B +2E 化得B (A -E )=2E ,两边取行列式,得|B ||A -E |=|2E |=4, 计算出|A -E |=2,因此|B |=2. 二、选择题（7）设函数()y f x =具有二阶导数，且()0,()0,f x f x x '''>>∆为自变量x 在点x 0处的增量，0()y dy f x x ∆与分别为在点处对应增量与微分,若0x ∆>，则[A]（A ）0dy y <<∆（B ）0y dy <∆<（C ）0y dy ∆<<（D ）0dy y <∆<由()0()f x f x '>可知严格单调增加()0()f x f x ''>可知是凹的即知（8）设()f x 是奇函数，除0x =外处处连续，0x =是其第一类间断点，则()xf t dt ⎰是[B]（A ）连续的奇函数（B ）连续的偶函数（C ）在x =0间断的奇函数（D ）在x =0间断的偶函数（9）设函数()g x 可微,1()(),(1)1,(1)2,g x h x e h g +''===则g (1)等于[C] （A ）ln 31- （B ）ln 31--（C ）ln 21--（D ）ln 21- ∵ 1()()()g x h x g x e +''=，1(1)12g e+= g (1)= ln 21--（10）函数212x x x y c e c xe -=++满足的一个微分方程是[D] （A ）23x y y y xe '''--= （B ）23x y y y e '''--=（C ）23xy y y xe '''+-=（D ）23xy y y e '''+-=将函数212x x x y c e c xe -=++代入答案中验证即可.（11）设(,)f x y 为连续函数，则14(cos ,sin )d f r r rd πθθθγ⎰⎰等于[C]（A ）2212(,)x xdx f x y dy -⎰⎰（B ）2212(,)x dx f x y dy -⎰⎰（C ）2212(,)y ydy f x y dx -⎰⎰（D ）2212(,)y dy f x y dx -⎰⎰（12）设(,)(,)f xyxy ϕ与均为可微函数，且(,)0,y x y ϕ'≠已知00(,)(,)x y f x y 是在约束条件(,)0x y ϕ=下的一个极值点，下列选项正确的是[D]（A ）若0000(,)0,(,)0x y f x y f x y ''==则（B ）若0000(,)0,(,)0x y f x y f x y ''=≠则（C ）若0000(,)0,(,)0x y f x y f x y ''≠=则（D ）若0000(,)0,(,)0x y f x y f x y ''≠≠则(,)(,)(,)(,)0(1)(,)(,)0(2)(,)0x x xy y y F f x y x y F f x y x y F f x y x y F x y λλϕλϕλϕϕ=+'''=+=⎧⎪'''=+=⎨⎪'==⎩令今000000(,)(,)0,(,)y y y f x y x y x y ϕλϕ''≠∴=-'代入(1) 得 00000000(,)(,)(,)(,)y xx y f x y x y f x y x y ϕϕ'''='今 00000000(,)0,(,)(,)0(,)0x y xy f x y f x y x y f x y ϕ''''≠∴≠≠则故选[D] (13)设α1,α2,…,αs 都是n 维向量，A 是m ⨯n 矩阵，则（）成立.(A) 若α1,α2,…,αs 线性相关,则A α1,A α2,…,A αs 线性相关. (B) 若α1,α2,…,αs 线性相关,则A α1,A α2,…,A αs 线性无关. (C) 若α1,α2,…,αs 线性无关,则A α1,A α2,…,A αs 线性相关. (D) 若α1,α2,…,αs 线性无关,则A α1,A α2,…,A αs 线性无关. 解: (A)本题考的是线性相关性的判断问题,可以用定义解.若α1,α2,…,αs 线性相关,则存在不全为0的数c 1,c 2,…,c s 使得c 1α1+c 2α2+…+c s αs =0,用A 左乘等式两边,得c 1A α1+c 2A α2+…+c s A αs =0,于是A α1,A α2,…,A αs 线性相关.如果用秩来解,则更加简单明了.只要熟悉两个基本性质,它们是: 1. α1,α2,…,αs 线性无关⇔ r(α1,α2,…,αs )=s. 2. r(AB )≤ r(B ).矩阵(A α1,A α2,…,A αs )=A ( α1, α2,…,αs ),因此r(A α1,A α2,…,A αs )≤ r(α1, α2,…,αs ).由此马上可判断答案应该为(A).(14)设A 是3阶矩阵,将A 的第2列加到第1列上得B ,将B 的第1列的-1倍加到第2列上得C .记 1 1 0P = 0 1 0 ,则 0 0 1(A) C =P -1AP . (B) C =PAP -1. (C) C =P TAP . (D) C =PAP T. 解: (B)用初等矩阵在乘法中的作用得出B =PA , 1 -1 0C =B 0 1 0 =BP -1= PAP -1. 0 0 1三、解答题（15）试确定A ，B ，C 的常数值，使23(1)1()x e Bx Cx Ax o x ++=++其中3()o x 是当30x x →时比的高阶无穷小.解：泰勒公式2331()26xx x e x o x =++++代入已知等式得 23323[1()][1]1()26x x x o x Bx Cx Ax o x ++++++=++整理得233111(1)()()1()226BB xC B x C o x Ax o x ⎛⎫+++++++++=++ ⎪⎝⎭比较两边同次幂函数得B +1=A ①C +B +12=0 ② 1026B C ++= ③ 式②-③得120233B B +==-则代入①得13A = 代入②得16C = （16）求arcsin xxe dx e ⎰.解：原式=22arcsin arcsin ()x x xx e t de e t dt e t =⎰⎰令21arcsin arcsin ()1t dttd t t t t =-=-+-⎰⎰2222arcsin arcsin 1(2)12(1)1t tdt t udu t u t t u u t t -=-+-==-+--⎰⎰令2arcsin 1t dut u =-+-⎰arcsin 11ln 21t u C t u -=-+++22arcsin arcsin 111ln 211x x x x x x e e e dx C e e e --∴=-++-+⎰. （17）设区域22{(,)||,0}D x y x y x =+≤≥，计算二重积分2211DxyI dxdy x y +=++⎰⎰.解：用极坐标系2201D xydxdy x y ⎛⎫= ⎪++⎝⎭⎰⎰11222002ln(1)ln 2122r I d dr r r ππππθ-==+=+⎰⎰. （18）设数列{}n x 满足10x π<<，1sin (1,2,3,)n n x x n +==证明：（1）1lim n n x +→∞存在，并求极限；（2）计算211lim n x n n n x x +→∞⎛⎫ ⎪⎝⎭. 证：（1）212sin ,01,2x x x n =∴<≤≥ 因此 1sin ,{}n n n n x x x x +=≤单调减少有下界()0n x ≥根据准则1，lim n n x A →∞=存在在1sin n n x x +=两边取极限得sin 0A A A =∴=因此1lim 0n n x +→∞=（2）原式21sin lim "1"n x n n n x x ∞→∞⎛⎫= ⎪⎝⎭为型离散型不能直接用洛必达法则先考虑 22011s i n l i m l n 0s i n l i m t t t t t t t e t →⎡⎤⎢⎥⎣⎦→⎛⎫= ⎪⎝⎭用洛必达法则2011(cos sin )limsin 2t t t t t tt te→-=23233310()0()26cos sin limlim22t t t t t t t t t t tt t ee →→⎡⎤⎡⎤-+--+⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦==3330110()261lim26t t t t ee →⎛⎫-++ ⎪⎝⎭-==.（19）证明：当0a b π<<<时，1sin 2cos sin 2cos b b b b a a a aππ++>++. 证：令()sin 2cos f x x x x x π=++ 只需证明0a x π<<<时,()f x 严格单调增加()sin cos 2sin f x x x x x π'=+-+cos sin x x x π=-+()cos sin cos sin 0f x x x x x x x ''=--=-< ()f x '∴严格单调减少又()cos 0f ππππ'=+=故0()0()a x f x f x π'<<<>时则单调增加（严格）()()b a f b f a >>由则得证（20）设函数()(0,)f u +∞在内具有二阶导数，且()22Z fx y=+满足等式22220z zx y∂∂+=∂∂.（I ）验证()()0f u f u u'''+=；（II ）若(1)0,(1)1f f '== 求函数()f u 的表达式.证：（I ）()()22222222;zx zy f x y f x y xyx yx y∂∂''=+=+∂∂++()()()()22222223222222zx y f x yf x yx x y x y ∂'''=+++∂++()()()()22222223222222zy x f x yf x yy x y x y ∂'''=+++∂++()2222222222()0()()0f x y z zf x yx y x yf u f u u'+∂∂''+=++=∂∂+'''∴+=代入方程得成立（II ）令(),;,dp p dp du c f u p c p du u p u u'==-=-+=⎰⎰则22(1)1,1,()ln ||,(1)0,0()ln ||f c f u u c f c f u u '===+==∴= 由（21）已知曲线L 的方程221(0)4x t t y t t⎧=+≥⎨=-⎩（I ）讨论L 的凹凸性；（II ）过点(1,0)-引L 的切线，求切点00(,)x y ，并写出切线的方程；（III ）求此切线与L （对应0x x ≤部分）及x 轴所围的平面图形的面积.解：（I ）4222,42,12dx dy dy t t t dt dt dx t t-==-==-222312110(0)2dy d d y dx t dx dx dt t t t dt ⎛⎫⎪⎛⎫⎝⎭=⋅=-⋅=-<> ⎪⎝⎭处(0L t ∴>曲线在处)是凸（II ）切线方程为201(1)y x t ⎛⎫-=-+⎪⎝⎭，设2001x t =+，20004y t t =-，则2223200000000241(2),4(2)(2)t t t t t t t t ⎛⎫-=-+-=-+⎪⎝⎭得200000020,(1)(2)001t t t t t t +-=-+=>∴=点为（2，3），切线方程为1y x =+（III ）设L 的方程()x g y =则()3()(1)S g y y dy =--⎡⎤⎣⎦⎰ ()224024241t t y y x y -+==±-=±-+解出t 得由于（2，3）在L 上，由()232241()y x x y g y ===--+=得可知()30944(1)S y y y dy ⎡⎤=-----⎣⎦⎰ 3300(102)44y dy ydy =---⎰⎰3333220002(10)44(4)214(4)3y y yd y y =-+--=+⨯⨯-⎰8642213333=+-=- (22)已知非齐次线性方程组 x 1+x 2+x 3+x 4=-1,4x 1+3x 2+5x 3-x 4=-1，a x 1+x 2+3x 3+bx 4=1有3个线性无关的解.① 证明此方程组的系数矩阵A 的秩为2.② 求a,b 的值和方程组的通解.解:① 设α1,α2,α3是方程组的3个线性无关的解,则α2-α1,α3-α1是AX =0的两个线性无关的解.于是AX =0的基础解系中解的个数不少于2,即4-r(A )≥2,从而r(A )≤2.又因为A 的行向量是两两线性无关的,所以r(A )≥2.两个不等式说明r(A )=2.② 对方程组的增广矩阵作初等行变换:1 1 1 1 -1 1 1 1 1 -1(A |β)= 4 3 5 -1 -1 → 0 –1 1 –5 3 ,a 1 3b 1 0 0 4-2a 4a+b-5 4-2a由r(A )=2,得出a=2,b=-3.代入后继续作初等行变换:1 02 -4 2→ 0 1 -1 5 -3 .0 0 0 0 0得同解方程组x 1=2-2x 3+4x 4，x 2=-3+x 3-5x 4，求出一个特解(2,-3,0,0)T 和AX =0的基础解系(-2,1,1,0)T ,(4,-5,0,1) T.得到方程组的通解: (2,-3,0,0)T +c 1(-2,1,1,0)T +c 2(4,-5,0,1)T , c 1,c 2任意.(23) 设3阶实对称矩阵A 的各行元素之和都为3,向量α1=(-1,2,-1)T , α2=(0,-1,1)T 都是齐次线性方程组AX =0的解.① 求A 的特征值和特征向量.② 求作正交矩阵Q 和对角矩阵Λ,使得 Q T AQ =Λ.解:① 条件说明A (1,1,1)T =(3,3,3)T ,即 α0=(1,1,1)T 是A 的特征向量,特征值为3.又α1,α2都是AX =0的解说明它们也都是A 的特征向量,特征值为0.由于α1,α2线性无关, 特征值0的重数大于1.于是A 的特征值为3,0,0.属于3的特征向量:c α0, c ≠0.属于0的特征向量:c 1α1+c 2α2, c 1,c 2不都为0.② 将α0单位化,得η0=(33,33,33)T . 对α1,α2作施密特正交化,的η1=(0,-22,22)T , η2=(-36,66,66)T . 作Q =(η0,η1,η2),则Q 是正交矩阵,并且 3 0 0Q T AQ =Q -1AQ = 0 0 0 .0 0 0。

历年AP微积分真题

历年AP微积分真题Introduction：微积分是数学的一个重要分支，被广泛应用于科学、工程、经济学等领域。

AP微积分考试是美国高中学生常参加的一项重要考试，通过该考试可以获得大学学分。

本文将回顾历年AP微积分真题，帮助读者了解该考试的内容和难度。

Section 1: Differential CalculusDifferential calculus is concerned with the study of rates of change and slopes of curves. This section of the AP Calculus exam tests students' understanding and application of derivative concepts.1. Example question:Find the derivative of the function f(x) = 3x^2 + 2x - 5.Solution:Using the power rule, we differentiate term by term to obtain f'(x) = 6x + 2.2. Example question:A particle moves along a straight line with position function s(t) = 4t^3 - 6t^2 + 2t + 1. Find the velocity function.Solution:To find the velocity function, we differentiate the position function with respect to time. Thus, v(t) = s'(t) = 12t^2 - 12t + 2.Section 2: Integral CalculusIntegral calculus focuses on the accumulation of quantities and finding areas under curves. This section of the AP Calculus exam examines students' ability to calculate definite and indefinite integrals.3. Example question:Evaluate the definite integral ∫(4x^3 + 2x - 1)dx from x = 1 to x = 3.Solution:Using the power rule and the constant rule, we integrate term by term and evaluate the integral to obtain 110.4. Example question:Fin d the indefinite integral ∫(5e^x + 3/x)dx.Solution:Integrating term by term, we obtain the indefinite integral as 5e^x +3ln|x| + C, where C is the constant of integration.Section 3: Applications of CalculusCalculus is widely used in various real-world applications such as physics, economics, and biology. This section of the AP Calculus exam assesses students' ability to apply calculus concepts to solve practical problems.5. Example question:A tank contains 500 liters of water with a salt concentration of 0.2 grams per liter. Brine with a concentration of 1 gram per liter enters the tank at a rate of 5 liters per minute. The mixture is continuously stirred and drained at a rate of 3 liters per minute. Find the salt concentration in the tank after 10 minutes.Solution:Using the principles of differential equations, we set up a rate of change equation and solve it to find the salt concentration to be approximately 0.439 grams per liter after 10 minutes.Conclusion:The AP Calculus exam covers a wide range of topics in both differential and integral calculus. By reviewing past exam questions, students can gain a better understanding of the exam format and level of difficulty. Mastering calculus concepts and their applications is crucial for success in this exam and for a deeper understanding of the field of mathematics.。

北京市2006年[课标B卷]数学试题及答案解析

北京市2006年高级中等学校招生统一考试（课标B 卷）数学试卷第Ⅰ卷（机读卷共32分）下列各题均有四个选项，其中只有一个．．是符合题意的．用铅笔把“机读答题卡”上对应题目答案的相应字母处涂黑． 1．5-的相反数是（）Ａ．5Ｂ．5-Ｃ．15Ｄ．15-2．青藏高原是世界上海拔最高的高原，它的面积约为2 500 000平方千米．将2 500 000用科学记数法表示应为（）Ａ．70.2510⨯Ｂ．72.510⨯Ｃ．62.510⨯Ｄ．52510⨯3．在函数13y x =-中，自变量x 的取值范围是（）Ａ．3x ≠ Ｂ．0x ≠ Ｃ．3x >Ｄ．3x ≠-4．如图，AD BC ∥，点E 在BD 的延长线上，若155ADE ∠=，则DBC ∠的度数为（）Ａ．155 Ｂ．50Ｃ．45Ｄ．255．小芸所在学习小组的同学们，响应“为祖国争光，为奥运添彩”的号召，主动到附近的7个社区帮助爷爷，奶奶们学习英语日常用语．他们记录的各社区参加其中一次活动的人数如下：33，32，32，31，28，26，32，那么这组数据的众数和中位数分别是（）Ａ．32，31 Ｂ．32，32 Ｃ．3，31 Ｄ．3，32ＡＤＥＣＢ6．把代数式29xy x -分解因式，结果正确的是（）Ａ．2(9)x y -Ｂ．2(3)x y +Ｃ．(3)(3)x y y +-Ｄ．(9)(9)x y y +-7．掷一枚质地均匀的正方体骰子，骰子的六个面上分别刻有1到6的点数，掷得面朝上的点数为奇数的概率为（）Ａ．16Ｂ．13Ｃ．14Ｄ．128．将如右图所示的圆心角为90的扇形纸片AOB 围成圆锥形纸帽，使扇形的两条半径OA 与OB 重合（接缝粘贴部分忽略不计），则围成的圆锥形纸帽是（）北京市2006年高级中等学校招生统一考试（课标B 卷）数学试卷第II 卷（非机读卷共88分）二、填空题（共4个小题，每小题4分，共16分．）9．若关于x 的一元二次方程230x x m -+=有实数根，则m 的取值范围是．102(1)0n +=，则m n +的值为．Ａ．Ｂ．Ｃ．Ｄ．11．用“>⨯ð”定义新运算：对于任意实数a ，b ，都有a >⨯ð21b b +=．例如，7>⨯ð211744+==，那么5>⨯ð3=；当m 为实数时，(m m >>⨯⨯痧2)=．12．如图，在ABC △中，AB AC =，M ，N 分别是AB ，AC 的中点，D ，E 为BC 上的点，连结DN ，EM ．若13cm AB =，10cm BC =，5cm DE =，则图中阴影部分的面积为2cm ．三、解答题（共5个小题，共25分） 13．（本小题满分5分）11(2006)2-⎛⎫--+ ⎪⎝⎭．解： 14．（本小题满分5分）解不等式组315260.x x -<⎧⎨+>⎩，解： 15．（本小题满分5分）解分式方程12211xx x +=-+．解： 16．（本小题满分5分）已知：如图，AB ED ∥，点F ，点C 在AD 上，AB DE =，AF DC =．求证：BC EF =．证明：Ｂ17．（本小题满分5分）已知230x -=，求代数式22()(5)9x x x x x -+--的值．解：四、解答题（共2个小题，共11分．） 18．（本小题满分5分）已知：如图，在梯形ABCD 中，AD BC ∥，90ABC ∠=，45C ∠=，BE CD ⊥于点E ，1AD =，CD = 求：BE 的长．解： 19．（本小题满分6分）已知：如图，ABC △内接于O ，点D 在OC 的延长线上，1sin 2B =，30CAD ∠=．（1）求证：AD 是O 的切线；（2）若OD AB ⊥，5BC =，求AD 的长．（1）证明：（2）解：五、解答题（本题满分5分）20．根据北京市统计局公布的2000年，2005年北京市常住人口相关数据，绘制统计图表如下：Ａ（1）从2000年到2005年北京市常住人口增加了多少万人？（2）2005年北京市常住人口中，少儿（014岁）人口约为多少万人？（3）请结合2000年和2005年北京市常住人口受教育程度的状况，谈谈你的看法．解：（1）（2）（3）六、解答题（共2个小题，共9分．）21．（本小题满分5分）在平面直角坐标系xOy中，直线y x=-绕点O顺时针旋转90得到直线l．直线l与反比例函数kyx=的图象的一个交点为(3)A a，，试确定反比例函数的解析式．解：22．（本小题满分4分）请阅读下列材料：问题：现有5个边长为1的正方形，排列形式如图1，请把它们分割后拼接成一个新的正方形．要求：画出分割线并在正方形网格图（图中每个小正方形的边长均为1）中用实线画出拼接成的新正方形．小东同学的做法是：设新正方形的边长为(0)x x>．依题意，割补前后图形的面积相等，有25x=，解得x=成的矩形对角线的长．于是，画出如图2所示的分割线，拼出如图3所示的新正方形．请你参考小东同学的做法，解决如下问题：现有10个边长为1的正方形，排列形式如图4，请把它们分割后拼接成一个新的正方形．要求：在图4中画出分割线，并在图5的正方形网格图（图中每个小正方形的边长均为1）中用实线画出拼接成的新正方形．说明：直接画出图形，不要求写分析过程．解：七、解答题（本题满分6分）23．如图1，OP 是MON ∠的平分线，请你利用该图形画一对以OP 所在直线为对称轴的全等三角形．请你参考这个作全等三角形的方法，解答下列问题：（1）如图2，在ABC △中，ACB ∠是直角，60B ∠=，AD ，CE 分别是BAC ∠，BCA ∠的平分线，AD ，CE 相交于点F ．请你判断并写出FE 与FD 之间的数量关系；（2）如图3，在ABC △中，如果ACB ∠不是直角，而（1）中的其他条件不变，请问，你在（1）中所得结论是否仍然成立？若成立，请证明；若不成立，请说明理由．解：画图：（1）FE 与FD 之间的数量关系为．图4 图5N P M O 图1 图2（2）八、解答题（本题满分8分）24．已知抛物线2y ax bx c =++与y 轴交于点(03)A ，，与x 轴分别交于(10)B ，，(50)C ，两点．（1）求此抛物线的解析式；（2）若点D 为线段OA 的一个三等分点，求直线DC 的解析式；（3）若一个动点P 自OA 的中点M 出发，先到达x 轴上的某点（设为点E ），再到达抛物线的对称轴上某点（设为点F ），最后运动到点A ．求使点P 运动的总路径最短的点E ，点F 的坐标，并求出这个最短总路径的长．解：（1）（2）（3）九、解答题（本题满分8分）25．我们给出如下定义：若一个四边形的两条对角线相等，则称这个四边形为等对角线四边形．请解答下列问题：（1）写出你所学过的特殊四边形中是等对角线四边形的两种图形的名称；（2）探究：当等对角线四边形中两条对角线所夹锐角为60时，这对60角所对的两边之和与其中一条对角线的大小关系，并证明你的结论．解：（1）（2）北京市2006年高级中等学校招生统一考试（课标Ｂ卷）数学试卷答案及评分参考阅卷须知：1．一律用红钢笔或红圆珠笔批阅，按要求签名． 2．第I 卷是选择题，机读阅卷．3．第II 卷包括填空题和解答题．为了阅卷方便，解答题中的推导步骤写得较为详细，考生只要写明主要过程即可．若考生的解法与本解法不同，正确者可参照评分参考给分．解答右端所注分数，表示考生正确做到这一步应得的累加分数．第I 卷（机读卷共32分）一、选择题（共8个小题，每小题4分，共32分．）第II 卷（非机读卷共88分）二、填空题（共4个小题，每小题4分，共16分．）1311(2006)2-⎛⎫--+ ⎪⎝⎭312=-+ ·················································· 4分 13=+． ························································· 5分 14．解：由不等式315x -<解得 2x <． ······························· 2分由不等式260x +>解得 3x >-． ······························ 4分则不等式组的解集为 32x -<<． ······························ 5分 15．解：(1)2(1)2(1)(1)x x x x x ++-=+-． ······························· 2分 2212222x x x x ++-=-． ······································· 3分 3x =． ·········································· 4分经检验3x =是原方程的解．所以原方程的解是3x =． ········································ 5分 16．证明：因为AB ED ∥，则A D ∠=∠． ················································ 1分又AF DC =，则AC DF =． ················································ 2分在ABC △与DEF △中，AB DE A D AC DF =⎧⎪∠=∠⎨⎪=⎩，，， ·············································· 3分所以ABC DEF △≌△． ······································ 4分所以BC EF =． ·············································· 5分17．解：22()(5)9x x x x x -+--322359x x x x =-+-- ·········································· 2分 249x =-． ····················································· 3分当230x -=时，原式249(23)(23)0x x x =-=+-=． ·············· 5分四、解答题（共2个小题，共11分）18．解：如图，过点D 作DF AB ∥交BC 于点F ． ························· 1分因为AD BC ∥，所以四边形ABFD 是平行四边形． ····································· 2分所以1BF AD ==．由DF AB ∥，得90DFC ABC ∠=∠=．在Rt DFC △中，45C ∠=，CD =由cos CFC CD=，求得2CF =． ······················································ 3分所以3BC BF FC =+=． ············································ 4分在BEC △中，90BEC ∠=， sin BEC BC=．求得BE =． ··················································· 5分 19．解：（1）证明：如图，连结OA ．因为1sin 2B =，所以30B ∠=．故60O ∠=． ·················· 1分又OA OC =，所以ACO △是等边三角形．故60OAC ∠=． ··················································· 2分因为30CAD ∠=，所以90OAD ∠=．所以AD 是O 的切线． ············································· 3分（2）解：因为OD AB ⊥，所以OC 垂直平分AB ．则5AC BC ==． ··················································· 4分所以5OA =．······················································· 5分在OAD △中，90OAD ∠=，由正切定义，有tan ADAOD OA∠=．所以AD = ··················································· 6分五、解答题（本题满分5分）20．解：（1）153********-=（万人）． ·································· 1分故从2000年到2005年北京市常住人口增加了154万人．（2）153610.2%156.672157⨯=≈（万人）．故2005年北京市常住人口中，少儿（014岁）人口约为157万人． ······· 3分（3）例如：依数据可得，2000年受大学教育的人口比例为16.86%，2005年受大学教育的人口比例为23.57%．可知，受大学教育的人口比例明显增加，教育水平有所提高． ······································································· 5分六、解答题（共2个小题，共9分）21．解：依题意得，直线l 的解析式为y x =． ······························· 2分因为(3)A a ，在直线y x =上，则3a =． ······················································ 3分即(33)A ，．又因为(33)A ，在ky x=的图象上，可求得9k =． ·················································· 4分所以反比例函数的解析式为9y x=． ································ 5分 22．解：所画图形如图所示．说明：图4与图5中所画图形正确各得2分．分割方法不唯一，正确者相应给分．七、解答题（本题满分6分．） 23．解：图略．画图正确得1分．（1）FE 与FD 之间的数量关系为FE FD =． ·························· 2分（2）答：（1）中的结论FE FD =仍然成立．证法一：如图4，在AC 上截取AG AE =，连结FG ．因为12∠=∠，AF 为公共边，可证AEF AGF △≌△．所以AFE AFG ∠=∠，FE FG =． ··········· 4分由60B ∠=，AD CE ，分别是BAC BCA ∠∠，的平分线，可得2360∠+∠=．所以60AFE CFD AFG ∠=∠=∠=．所以60CFG ∠=． ················································· 5分由34∠=∠及FC 为公共边，可得CFG CFD △≌△．所以FG FD =．所以FE FD =． ···················································· 6分证法二：如图5，过点F 分别作FG AB ⊥于点G ，FH BC ⊥于点H ． ·················· 3分因为60B ∠=，且AD ，CE 分别是BAC ∠，BCA ∠的平分线，所以可得2360∠+∠=，F 是ABC △的内心． ········ 4分所以601GEF ∠=+∠，FG FH =．又因为1HDF B ∠=∠+∠，所以GEF HDF ∠=∠． ······························ 5分因此可证EGF DHF △≌△．所以FE FD =． ···················································· 6分八、解答题（本题满分8分） 24．解：（1）根据题意，3c =，所以3025530.a b a b ++=⎧⎨++=⎩，图4图5解得3518.5a b ⎧=⎪⎪⎨⎪=-⎪⎩，所以抛物线解析式为2318355y x x =-+． ······························· 2分（2）依题意可得OA 的三等分点分别为(01)，，(02)，．设直线CD 的解析式为y kx b =+．当点D 的坐标为(01)，时，直线CD 的解析式为115y x =-+； ············· 3分当点D 的坐标为(02)，时，直线CD 的解析式为225y x =-+． ············ 4分（3）如图，由题意，可得302M ⎛⎫ ⎪⎝⎭，．点M 关于x 轴的对称点为302M ⎛⎫'- ⎪⎝⎭，，点A 关于抛物线对称轴3x =的对称点为(63)A '，．连结A M ''．根据轴对称性及两点间线段最短可知，A M ''的长就是所求点P 运动的最短总路径的长． ··································································· 5分所以A M ''与x 轴的交点为所求E 点，与直线3x =的交点为所求F 点．可求得直线A M ''的解析式为3342y x =-．可得E 点坐标为(20)，，F 点坐标为334⎛⎫ ⎪⎝⎭，．···························· 7分由勾股定理可求出152A M ''=．所以点P 运动的最短总路径()ME EF FA ++的长为152． ················· 8分九、解答题（本题满分8分） 25．解：（1）略．写对一种图形的名称给1分，最多给2分．（2）结论：等对角线四边形中两条对角线所夹锐角为60时，这对60角所对的两边之和大于或等于一条对角线的长． ·········································· 3分已知：四边形ABCD 中，对角线AC ，BD 交于点O ，AC BD =，且60AOD ∠=．x'求证：BC AD AC +≥．证明：过点D 作DF AC ∥，在DF 上截取DE ，使DE AC =．连结CE ，BE ． ···················································· 4分故60EDO ∠=，四边形ACED 是平行四边形．所以BDE △是等边三角形，CE AD =． ······························· 6分所以DE BE AC ==．①当BC 与CE 不在同一条直线上时（如图1），在BCE △中，有BC CE BE +>．所以BC AD AC +>． ······························· 7分②当BC 与CE 在同一条直线上时（如图2），则BC CE BE +=．因此BC AD AC +=． ······························· 8分综合①、②，得BC AD AC +≥．即等对角线四边形中两条对角线所夹角为60时，这对60角所对的两边之和大于或等于其中一条对角线的长．A DEF CB O图2 A DE F C B O图1。

2006-2007学年浙江工商大学第二学期《微积分》试卷

浙江工商大学2006 /2007学年第二学期考试试卷课程名称：微积分考试方式：闭卷完成时限： 120分钟班级名称：学号：姓名：一、1．,1),(),(),(),(0000-=''=y x f ，y x ，f y x f z y x xx且具有二阶连续偏导数的驻点是设 ,),(,1),(0000a y x f y x f yy xy=''=''则a 时，),(00y x 是极大值点. 2．若)1(1n n u +=∞∑收敛，则=∞→n n u lim .3．设1d 112=+⎰x x kx，则=k . 4．=⎰+∞-x e x x d 03 .5．幂级数∑∞=1n nnx 的收敛域为 .6．设)ln ln(y x z +=，则=∂∂yz. 7．交换积分次序后 ⎰⎰==baxay y f x I d )(d .8．=+⎰-dx x x 11)( .9．微分方程011=+dx ydy x 满足43==x y的解为 .10．若D 是平面上长半轴和短半轴分别为b a 、的椭圆圆域，则⎰⎰=Dσd .二、单项选择（1052=⨯分）1.已知),(y x f z =的全微分xydy dx y dz 22+=，则=∂∂22xz（）.A. 0B. x 2C. y 2D.xy 22.级数nn ）∑∞=121(的和为（）.A.21 B.1 C.2 D.23 3.下列广义积分发散的是（）.A.⎰101dx xB.dx x ⎰-10211 C.dx x ⎰∞+121D.dx x⎰1021 4.若级数∑∞=1n n u 发散，则必有（）.A.0lim =∞→n n uB.0lim ≠∞→n n uC.∑∞=+12007n n u 发散 D.∑∞=12007n n u 发散 5.方程xy y x y +='22是（）方程 .A.可分离变量B.齐次C.一阶线性D.伯努利三、计算题（一）（2464=⨯分） 1．dx x x ⎰+-40223.2．dx x x ⎰+31211 .3．已知函数)2sin(y x e z x-=,求yx z ∂∂∂2.4．已知函数),(y x f z =是由方程0)ln(22=+-xyz xyz xz 所确定的隐函数,求dz . 四、计算题（二）（2464=⨯分）1. 求二重积分dxdy y x I D22sin +=⎰⎰，其中D 是由122≤+y x 与x 轴及y 轴所围平面图形的第一象限部分. 2. 判断级数n n n nn !21=∞∑的敛散性. 3. 求221xx xy -+=在0=x 处展开的幂级数. 4. 求微分方程222)1(2)1(+=-+x xy dxdyx 的通解.五、应用题（1628=⨯分）1.已知D 为由x y =2与2-=x y 围成的平面图形，求:（1）D 的面积;（2）D 绕y 轴旋转所得旋转体的体积.2. 设某企业的总产量函数为y x y x P 2005.0),(=（吨），y x ,为两种投入要素，其单价分别为1万元/吨和2万元/吨，且该企业拥有资金150万元，试求y x ,使产量最大. 六、证明题（6分）设)(x f 连续，且⎰-=-xx t t x tf 0cos 1d )(，试证：⎰=21d )(πx x f .。

2006年12月高等学校英语应用能力考试(B级)真题超精解

高等学校英语应用能力考试 B 级真题超精解2006 年 12 月全国高等学校英语应用能力考试 B 级真题超精解级真题超精解Part I Listening Comprehension Section A 1. 【看题预测】该题选项 A) 是电话用语，询问对方是谁，回答 Can I speak to…；选项 B) 是问候语，回答 I’m fine, and you；选项 C) 是特殊疑问句，不符合出题规则，可排除；选项 D) 用于回答一般疑问句。

从该题的四个选项中可以推测听力的重点在打电话，问候。

Can I speak to Susan? A) Who's calling, please? B) How are you?C) Where is she?D) No, you can’t.【解析】该题的问题是 Can I speak to Susan，是电话用语。

因此，A）是正确答案。

2.【看题预测】该题选项 A) 的意思是“可能” ；选项 B) 可以用来回答道歉等；选项 C) 的意思是“没有办法” ；选项 D) 可以用来回答感谢。

从该题的四个选项中可以推测听力的重点在道歉，感谢。

I'm terribly sorry we're late. A) It's possible. B) That's a11 right.C) No way.D) My pleasure.【解析】该题的问题是 I'm terribly sorry we're late，是道歉。

因此，正确答案为 B）。

3.【看题预测】该题选项 A) 肯定回答一般疑问句；选项 B) 对对方说的话表示怀疑；选项 C) 用于回。

从该题的四个选项中可以答感谢，意思是“客气了” ；选项 D) 是餐桌用语，意思是“不用了” 推测听力的重点在感谢，餐桌用语。

Thank you very much for your help. A)Yes, of course. B) Is it true?C)You're welcome.D) No, thanks.【解析】该题的问题是 Thank you very much for your help，是感谢。

AMC10美国数学竞赛真题2006B卷

What is ?SolutionProblem 2For real numbers and , define . What is?SolutionProblem 3A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?SolutionProblem 4Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?SolutionProblem 5A rectangle and a rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?SolutionA region is bounded by semicircular arcs constructed on the side of a square whose sides measure , as shown. What is the perimeter of this region?SolutionProblem 7Which of the following is equivalent to when ?SolutionProblem 8A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?SolutionFrancesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade?SolutionProblem 10In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?SolutionProblem 11What is the tens digit in the sumSolutionProblem 12The lines and intersect at the point . What is ?SolutionProblem 13Joe and JoAnn each bought 12 ounces of coffee in a 16 ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of cream. JoAnn added 2 ounces of cream, stirred the coffee well, and then drank 2 ounces. What is the resulting ratio of the amount of cream in Joe's coffee to that in JoAnn's coffee?SolutionProblem 14Let and be the roots of the equation . Suppose that and are the roots of the equation . What is ?SolutionProblem 15Rhombus is similar to rhombus . The area of rhombus is and . What is the area of rhombus ?SolutionProblem 16Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur?SolutionProblem 17Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?SolutionProblem 18Let be a sequence for which, , and for each positive integer .What is ?SolutionProblem 19A circle of radius is centered at . Square has side length . Sidesand are extended past to meet the circle at and , respectively. What is the area of the shaded region in the figure, which is bounded by , , and the minor arc connecting and ?SolutionProblem 20In rectangle , we have , , , for some integer . What is the area of rectangle ?SolutionProblem 21For a particular peculiar pair of dice, the probabilities of rolling , , , , , and , on each die are in the ratio . What is the probability of rolling a total of on the two dice?SolutionProblem 22Elmo makes sandwiches for a fundraiser. For each sandwich he uses globs of peanut butter at per glob and blobs of jam at per blob. The cost of the peanut butter and jam to make all the sandwiches is . Assume that , , and are positive integers with . What is the cost of the jam Elmo uses to make the sandwiches?SolutionProblem 23A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?SolutionProblem 24Circles with centers and have radii and , respectively, and are externally tangent. Points and on the circle with center and points and on the circle with center are such that and are common external tangents to the circles. What is the area of the concave hexagon ?SolutionProblem 25Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?。

2006年6月英语应用能力考试B级真题

英语应用能力考试(B 级）B 级考试历年全真试卷2006 年6 月Part I Listening Comprehension(15 minutes)Directions: This part is to test your listening ability. It consists of 3 sections.Section ADirections: This section is to test your ability to give proper responses。

There are 5 recorded questions in it. Aftereach question，there is a pause。

The questions will be spoken two times。

When you hear a question, you shoulddecide on the correct answer from the 4 choices marked A)，B）, C）and D）given in your test paper。

Then you should mark the corresponding letter on the Answer Sheet with a single line through the centre。

Example：You will hear:You will read:A) I' m not sure。

B) Y ou’re right。

C）Yes，certainly。

D) That’s interesting。

From the question we learn that the speaker is asking the listener to leave a message. Therefore，C）Yes, certainly.is the correct answer. You should mark C）on the Answer Sheet. Now the test will begin。

06年3月全国英语等级考试三级B卷真题（3）

Part III Reading Comprehension （40 minutes） Directions: This part is to test your reading ability. There are 5 tasks f or you to f ulfill. Y ou should read the reading materials caref ully and do the tasks ns you are instructed. Task 1 Directions: Af ter reading the f ollowing passage, you will f ind 5 questions or unf inished statements, numbered 36 to 40. For each question or statement there are 4 choices marked A）, B）, .and D）. You should make the correct choice and mark the corresponding letter on the Answer Sheet with a single line through the center. Do you know how to use a mobile phone （⼿机） without being rude to the people around you? Talking during a performance irritates （激怒） people. If you are expecting an emergency call, sit near the exit doors and set your phone to vibrate （振动）. When your mobile phone vibrates, you can leave quietly and let the others enjoy the performance. Think twice before using mobile phones in elevators, museums, churches or other indoor pubilc places—especially enclosed spaces. Would you want to listen to someones conversation in these places? Worse yet,how would you feel if a mobile phone rang suddenly during a funeral! It happens more often than you think. Avoid these embarrassing situations by making sure your mobile phone is switched off. When eating at a restaurant with friends, dont place your mobile phone on the table. This conveys the message that your phone calls are more important than those around you. Mobile phones have sensitive microphones that allow you to speak at the volume you would on a regular phone. This enables you to speak quietly so that others wont hear the details of your conversations. If you are calling from a noisy area, use your hand to direct your voice into the microphone. Many people believe that they cant live without their mobile phone. Owning a mobile phone definitely makes life more convenient, but limit your conversations to urgent ones and save the personal calls until you are at home. 36. What should you do when you need to answer a phone call during a performance?A.Call back after the performance.B.Answer it near the exit door.C.Talk outside the exit door.D.Speak in a low voice. 37. Putting your mobile phone on a restaurant table may make your friends thinkA.you prefer to talk to your friends at the tableB.you value your calls more than your friendsC.you are enjoying the company of your friendsD.you are polite and considerate of your friends 38. When you are calling in a noisy area, you are advised to .e a more sensitive microphoneB.shout loudly into your microphoneC.go away quietly to continue the phone calle your hand to help speak into the phone 39. The author implies that the use of mobile phones in such places as museums should be .A.limitedB.expectedC.discouragedD.recommended 40. Which of the following is TRUE according to the passage? A.You should limit your mobile phone calls to personal affairs. B.You should speak quietly into your phone while in a church. C.You are supposed to turn off your mobile phone at a funeral. D.You are supposed to use your mobile phone as much as possible. Task 2 Directions: This task is the same as T ask 1. The 5 questions or unf inished statements are numbered 41 to 45. When a rare disease ALD threatened to kill the four-year-old boy Lorenzo, his parents refused to give up hope.Doctors explained that there was no cure for ALD, and that he would probably die within three years. But Lorenzos parents set out to prove the doctors wrong. The parents devoted themselves to keeping their son alive and searching for a cure. But doctors and the families of other ALD patients often refused to take them seriously. They thought the efforts to find a cure were a waste of time, and drug companies werent interested in supporting research into such a rare disease. However, the parents still refused to give up and spent every available hour in medical libraries and talking to anyone who would help. Through trial and error （反复实验）, they finally created a cure from ingredients （调料） commonly found in the kitchen. The cure, named" Lorenzos Oil", saved the boys life. Despite the good results, scientists and doctors remained unconvinced. They said there was no real evidence that the oil worked and that the treatment was just a theory.As a result, some families with ALD children were reluctant to try it. Finally, the boys father organized an international study to test the oil. After ten years of trials, the answer is: the oil keepsALD children healthy. 41. Doctors said that Lorenzo might die within three years because . A.they had never treated the disease before B.Lorenzo was too young to be, cured C.no cure had been found for ALD D.ALD was a rare disease 42. The families of other ALD patients thought that . A.the research for the new cure would cost too much money B.the efforts of Lorenzos parents were a waste of time C.Lorenzos parents would succeed in finding a cure D.Lorenzos oil was a real cure for ALD 43. Scientists and doctors believed that Lorenzos Oil A.was really effective B.was a success story C.only worked in theory D.would save the boys life 44. Lorenzos father organized an international study to A.test Lorenzos Oil B.get financial support C.find a cure for the disease D.introduce the cure worldwide 45. From the passage we can conclude that A.doctors remain doubtful of the effectiveness of the cure B.many ALD patients still refuse to use the oil C.various cures have been found for ALD D.the oil really works as a cure for ALD Task 3 Directions: The f ollowing is an advertisem ent. After reading it, you are required to complete the outline below it （ No.46 to No. 50 ）. You should write your answers brief ly （ in not more than three words） on the Answer Sheet correspondingly. The meeting is over. Youre tired. Now will you get on a plane and rush back home to more work? Heres a better idea. Take a little time for yourself and relax at Holiday Inn. All our 1,642 hotels worldwide have the best leisure facilities available. And the best locations for relaxation. From the sun-bathed beaches of Thailands Phuket to the unique scenery of Tibet. Or on a journey of discovery to Malaysias Kuching and Penang, and beyond to the ski-fields （滑雪场）. Holiday Inn makes it easy to relax. So does the American Express Card. Its the foremost business traveler companion. With no pre-set spending limit you can spend as much as you have shown us you can afford. You have the flexibility to quickly change your travel plans plus the spending power to make the most of your last minute holiday. Relax with confidence. Just one of the many benefits of being an American Express Card member and staying atHoliday Inn. 1. Holiday Inn An advertisement Items advertised: 2. American Express Card Number of Holiday Inn hotels worldwide: （46） Services offered by Holiday Inn: 1. best （47） facilities 2. best （48） for relaxation Advantages of American Express Card: 1. no pre-set （49） 2. flexibility in changing ones （50） Task 4 Directions: The f ollowing is a list of terms used in the Internet. Af ter reading it, you are required to find the items equivalent to （与 ……相同的） those given in Chinese in the table below. Then you should put the corresponding letters in brackets on the Answer Sheet, numbered 51 to 55. A——abbreviated dialing code B——off-peak hours C——charging period D——access code E——identity number F——video conference G——operational status H——information subscription service I ——Network User Address J ——audio signal K——local user terminal L——file management M——response signal N——operating instructions O——change of the battery unit P——function indicator Q——entry rejected R——external control 51.（）计费时间（）⾮⾼峰时间 52.（）络⽤户地址（）标识码 53.（）本地⽤户终端（）⾳频信号 54.（）⽂件管理（）操作指令 55.（）外部控制（）功能指⽰ Task 5。

2006年数二真题、标准答案及解析

2006年全国硕士研究生入学考试数学（二）一、填空题（1）曲线4sin 52cos x xy x x+=-的水平渐近线方程为 .（2）设函数231sin ,0,(),x t dt x f x xa x ⎧≠⎪=⎨⎪=⎩⎰在0x =处连续，则a = .（3）广义积分22(1)xdxx +∞=+⎰ .（4）微分方程(1)y x y x-'=的通解是 . （5）设函数()y y x =由方程1y y xe =-确定，则A dydx== .（6）设矩阵2112A ⎛⎫= ⎪-⎝⎭，E 为2阶单位矩阵，矩阵B 满足2BA B E =+，则B = . 二、选择题（7）设函数()y f x =具有二阶导数，且()0,()0f x f x '''>>，x ∆为自变量x 在0x 处的增量，y ∆与dy 分别为()f x 在点0x 处对应的增量与微分，若0x ∆>，则（A ）0.dy y <<∆ （B ）0.y dy <∆<（C ）0.y dy ∆<< （D ）0.dy y <∆< 【】（8）设()f x 是奇函数，除0x =外处处连续，0x =是其第一类间断点，则0()xf t dt⎰是（A ）连续的奇函数. （B ）连续的偶函数（C ）在0x =间断的奇函数（D ）在0x =间断的偶函数. 【】（9）设函数()g x 可微，1()(),(1)1,(1)2g x h x e h g +''===，则(1)g 等于（A ）ln 31-. （B ）ln 3 1.--（C ）ln 2 1.-- （D ）ln 2 1.- 【】（10）函数212x x x y C e C e xe -=++满足一个微分方程是（A ）23.x y y y xe '''--= （B ）23.x y y y e '''--=（C ）23.x y y y xe '''+-= （D ）23.x y y y e '''+-=（11）设(,)f x y 为连续函数，则140(cos ,sin )d f r r rdr πθθθ⎰⎰等于（A ）0(,).x f x y dy ⎰⎰（B ）00(,).f x y dy ⎰⎰（C ）0(,).yf x y dx ⎰⎰（D ）00(,).f x y dx ⎰⎰ 【】（12）设(,)f x y 与(,)x y ϕ均为可微函数，且1(,)0y x y ϕ≠. 已知00(,)x y 是(,)f x y 在约束条件(,)0x y ϕ=下的一个极值点，下列选项正确的是（A ）若00(,)0x f x y '=，则00(,)0y f x y '=. （B ）若00(,)0x f x y '=，则00(,)0y f x y '≠. （C ）若00(,)0x f x y '≠，则00(,)0y f x y '=.（D ）若00(,)0x f x y '≠，则00(,)0y f x y '≠. 【】（13）设12,,,,a a a 均为n 维列向量，A 是m n ⨯矩阵，下列选项正确的是（A ）若12,,,,a a a 线性相关，则12,,,,Aa Aa Aa 线性相关. （B ）若12,,,,a a a 线性相关，则12,,,,Aa Aa Aa 线性无关. （C ）若12,,,,a a a 线性无关，则12,,,,Aa Aa Aa 线性相关.（D ）若12,,,,a a a 线性无关，则12,,,,Aa Aa Aa 线性无关. 【】（14）设A 为3阶矩阵，将A 的第2行加到第1行得B ，再将B 的第1列的-1倍加到第2列得C ，记110010001P ⎛⎫⎪= ⎪ ⎪⎝⎭，则（A ）1.C P AP -= （B ）1.C PAP -= （C ）.T C P AP = （D ）.T C PAP = 三解答题15．试确定A ，B ，C 的常数值，使得23(1)1()x e Bx Cx Ax o x ++=++，其中3()o x 是当30x x →时比的高阶无穷小.16．arcsin xxe dx e ⎰求. 17．{}22(,)1,0D x y x y x =+≤≥设区域，221.1DxyI dxdy x y+=++⎰⎰计算二重积分 18．{}110,sin (0,1,2,)n n n x x x x n π+<<==设数列满足1lim n x x +→∞证明: (1) 存在,并求极限；211(2)lim(n x n x nx x +→∞计算. 19．sin 2cos sin cos .<a <b b b b b a a a a a πππ<++>++证明: 当0时, 20 设函数()()0,,f u +∞在内具有二阶导数且z f =满足等式22220z zx y∂∂+=∂∂.(Ⅰ)验证()()0f u f u u'''+=；(Ⅱ)若()()()10,11,f f f u '==求函数的表达式. 21 已知曲线L 的方程为221,(0),4x l t y l t⎧=+≥⎨=-⎩（Ⅰ）讨论L 的凹凸性；（Ⅱ）过点（-1，0）引L 的切线，求切点00(,)x y ，并写出切线的方程；（Ⅲ）求此切线与L （对应于0x x ≤的部分）及x 轴所围成的平面图形的面积.22 已知非齐次线性方程组12341234123414351331x x x x x x x x ax x x bx +++=-⎧⎪++-=-⎨⎪++-=⎩有个线性无关的解Ⅰ证明方程组系数矩阵A 的秩()2r A =； Ⅱ求,a b 的值及方程组的通解.23 设3阶实对称矩阵A 的各行元素之和均为3,向量()()121,2,1,0,1,1T Tαα=--=-是线性方程组A x =0的两个解, (Ⅰ)求A 的特征值与特征向量 (Ⅱ)求正交矩阵Q 和对角矩阵A,使得T Q AQ A =.2006年全国硕士研究生入学考试数学（二）真题解析一、填空题（1）曲线4sin 52cos x xy x x+=-的水平渐近线方程为15y =4sin 11lim lim2cos 55x x xx y x x→∞→∞+==-（2）设函数2301sin ,0(),0xt dt x f x x a x ⎧≠⎪=⎨⎪=⎩⎰ 在x =0处连续，则a =132200()1lim ()lim 33x x sm x f x x →→== （3）广义积分220(1)xdxx +∞=+⎰1222222201(1)11110(1)2(1)2(1)22xdx d x x x x +∞+∞+∞+==-⋅=+=+++⎰⎰（4）微分方程(1)y x y x-'=的通解是xy cxe -=)0(≠x（5）设函数()y y x =由方程1y y xe =-确定，则0x dy dx==e-当x =0时，y =1，又把方程每一项对x 求导，y y y e xe y ''=-- 001(1)1x x y yyyye y xe ey e xe ===''+=-=-=-+(6) 设A = 2 1 ,2阶矩阵B 满足BA =B +2E ,则|B |= .-1 2解:由BA =B +2E 化得B (A -E )=2E ,两边取行列式,得|B ||A -E |=|2E |=4,计算出|A -E |=2,因此|B |=2.二、选择题（7）设函数()y f x =具有二阶导数，且()0,()0,f x f x x '''>>∆为自变量x 在点x 0处的增量，0()y dy f x x ∆与分别为在点处对应增量与微分,若0x ∆>，则[A] （A ）0dy y <<∆ （B ）0y dy <∆< （C ）0y dy ∆<< （D ）0dy y <∆< 由()0()f x f x '>可知严格单调增加 ()0()f x f x ''>可知是凹的即知（8）设()f x 是奇函数，除0x =外处处连续，0x =是其第一类间断点，则()xf t dt ⎰是[B]（A ）连续的奇函数（B ）连续的偶函数（C ）在x =0间断的奇函数（D ）在x =0间断的偶函数（9）设函数()g x 可微,1()(),(1)1,(1)2,g x h x e h g +''===则g (1)等于[C] （A ）ln 31- （B ）ln 31-- （C ）ln 21-- （D ）ln 21- ∵ 1()()()g x h x g x e +''=，1(1)12g e += g (1)= ln 21-- （10）函数212x x x y c e c xe -=++满足的一个微分方程是[D] （A ）23x y y y xe '''--= （B ）23x y y y e '''--= （C ）23x y y y xe '''+-= （D ）23x y y y e '''+-=将函数212x x x y c e c xe -=++代入答案中验证即可.（11）设(,)f x y 为连续函数，则140(cos ,sin )d f r r rd πθθθγ⎰⎰等于[C]（A）0(,)xdxf x y dy ⎰（B）0(,)dxf x y dy ⎰（C）0(,)yf x y dx ⎰（D）0(,)f x y dx ⎰（12）设(,)(,)f x y x y ϕ与均为可微函数，且(,)0,y x y ϕ'≠已知00(,)(,)x y f x y 是在约束条件(,)0x y ϕ=下的一个极值点，下列选项正确的是[D]（A ）若0000(,)0,(,)0x y f x y f x y ''==则（B ）若0000(,)0,(,)0x y f x y f x y ''=≠则（C ）若0000(,)0,(,)0x y f x y f x y ''≠=则（D ）若0000(,)0,(,)0x y f x y f x y ''≠≠则(,)(,)(,)(,)0(1)(,)(,)0(2)(,)0x x xy y y F f x y x y F f x y x y F f x y x y F x y λλϕλϕλϕϕ=+'''=+=⎧⎪'''=+=⎨⎪'==⎩令今 000000(,)(,)0,(,)y y y f x y x y x y ϕλϕ''≠∴=-'代入(1) 得 00000000(,)(,)(,)(,)y xx y f x y x y f x y x y ϕϕ'''=' 今 00000000(,)0,(,)(,)0(,)0x y xy f x y f x y x y f x y ϕ''''≠∴≠≠则故选[D] (13)设1,2,…,s都是n 维向量，A 是m ⨯n 矩阵，则（）成立.(A) 若1,2,…,s线性相关,则A 1,A 2,…,A s线性相关. (B) 若1,2,…,s 线性相关,则A 1,A 2,…,A s线性无关. (C) 若1,2,…,s 线性无关,则A 1,A 2,…,A s线性相关. (D) 若1,2,…,s 线性无关,则A1,A2,…,As线性无关.解: (A)本题考的是线性相关性的判断问题,可以用定义解.若1,2,…,s线性相关,则存在不全为0的数c1,c2,…,c s使得c11+c22+…+c s s=0,用A左乘等式两边,得c1A1+c2A2+…+c s A s=0,于是A1,A2,…,A s线性相关.如果用秩来解,则更加简单明了.只要熟悉两个基本性质,它们是:1.1,2,…,s⇔ r(1,2,…,s)=s.2. r(AB)≤ r(B).矩阵(A1,A2,…,A s)=A(1,2,…,s),因此r(A1,A2,…,A s)≤ r(1,2,…,s).由此马上可判断答案应该为(A).(14)设A是3阶矩阵,将A的第2列加到第1列上得B,将B的第1列的-1倍加到第2列上得C.记 1 1 0P= 0 1 0 ,则0 0 1(A) C=P-1AP. (B) C=PAP-1.(C) C=P T AP. (D) C=PAP T.解: (B)用初等矩阵在乘法中的作用得出B=PA ,1 -1 0C =B 0 1 0 =BP -1= PAP -1.0 0 1三、解答题（15）试确定A ，B ，C 的常数值，使23(1)1()x e Bx Cx Ax o x ++=++其中3()o x 是当30x x →时比的高阶无穷小.解：泰勒公式2331()26xx x e x o x =++++代入已知等式得23323[1()][1]1()26x x x o x Bx Cx Ax o x ++++++=++整理得233111(1)(()1()226BB xC B x C o x Ax o x ⎛⎫+++++++++=++ ⎪⎝⎭比较两边同次幂函数得 B +1=A ①C +B +12=0 ②1026B C ++= ③ 式②-③得120233B B +==-则代入①得 13A =代入②得 16C =（16）求arcsin xxe dx e ⎰.解：原式=22arcsin arcsin ()x x xx e t de e t dt e t =⎰⎰令1arcsin arcsin ()t td t t =-=-+⎰2arcsin arcsin 1(2)2(1)t t udu t t u u -=-+=-+-⎰ 2arcsin 1t dut u =-+-⎰ arcsin 11ln 21t u C t u -=-+++arcsin arcsin 12x x x x e e dx C e e ∴=-++⎰. （17）设区域22{(,)||,0}D x y x y x =+≤≥，计算二重积分2211DxyI dxdy x y +=++⎰⎰. 解：用极坐标系2201D xydxdy x y ⎛⎫=⎪++⎝⎭⎰⎰11222002ln(1)ln 2122r I d dr r r ππππθ-==+=+⎰⎰. （18）设数列{}n x 满足10x π<<，1sin (1,2,3,)n n x x n +==证明：（1）1limn n x +→∞存在，并求极限；（2）计算211lim n x n n n x x +→∞⎛⎫ ⎪⎝⎭. 证：（1）212sin ,01,2x x x n =∴<≤≥因此1sin ,{}n n n n x x x x +=≤单调减少有下界()0n x ≥根据准则1，limn n x A →∞=存在在1sin n n x x +=两边取极限得sin 0A A A =∴=因此1lim 0n n x +→∞=（2）原式1sin lim "1"n xn n n x x ∞→∞⎛⎫= ⎪⎝⎭为型离散型不能直接用洛必达法则先考虑 2011sin lim lnsin lim t t t t t t t e t →⎡⎤⎢⎥⎣⎦→⎛⎫= ⎪⎝⎭用洛必达法则2011(cos sin )limsin 2t t t t t t t te→-=2323330010()0()26cos sin lim lim22t t t t t t t t t t tt te e→→⎡⎤⎡⎤-+--⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦==3330110()261lim26t t t t ee →⎛⎫-+ ⎪⎝⎭-==.（19）证明：当0a b π<<<时，1sin 2cos sin 2cos b b b b a a a aππ++>++. 证：令()sin 2cos f x x x x x π=++只需证明0a x π<<<时,()f x 严格单调增加()sin cos 2sin f x x x x x π'=+-+cos sin x x x π=-+()cos sin cos sin 0f x x x x x x x ''=--=-< ()f x '∴严格单调减少又()cos 0f ππππ'=+=故0()0()a x f x f x π'<<<>时则单调增加（严格）()()b a f b f a>>由则得证（20）设函数()(0,)f u +∞在内具有二阶导数，且Z f =满足等式22220z zx y∂∂+=∂∂.（I ）验证()()0f u f u u'''+=；（II ）若(1)0,(1)1f f '==求函数()f u的表达式. 证：（I）zzf f xy∂∂''==∂∂()()2223222222zx y f f xx y x y ∂'''=+∂++()()2223222222zy x f f yx y x y ∂'''=+∂++22220()()0z zf x y f u f u u∂∂''+=+=∂∂'''∴+=代入方程得成立（II ）令(),;,dp p dp du c f u p c p du u p u u'==-=-+=⎰⎰则22(1)1,1,()ln ||,(1)0,0()ln ||f c f u u c f c f u u '===+==∴=由（21）已知曲线L 的方程221(0)4x t t y t t⎧=+≥⎨=-⎩（I ）讨论L 的凹凸性；（II ）过点(1,0)-引L 的切线，求切点00(,)x y ，并写出切线的方程；（III ）求此切线与L （对应0x x ≤部分）及x 轴所围的平面图形的面积. 解：（I ）4222,42,12dx dy dy t t t dt dt dx t t-==-==-222312110(0)2dy d d y dx t dx dx dt t t t dt⎛⎫ ⎪⎛⎫⎝⎭=⋅=-⋅=-<> ⎪⎝⎭处 (0L t ∴>曲线在处)是凸（II ）切线方程为201(1)y x t⎛⎫-=-+ ⎪⎝⎭，设2001x t =+，20004y t t =-，则2223200000000241(2),4(2)(2)t t t t t t t t⎛⎫-=-+-=-+ ⎪⎝⎭得200000020,(1)(2)001t t t t t t +-=-+=>∴=点为（2，3），切线方程为1y x =+ （III ）设L 的方程()x g y =则()3()(1)S g y y dy =--⎡⎤⎣⎦⎰(2240221t t y x -+==±=+解出t 得由于（2，3）在L上，由(23221()y x x g y ===+=得可知(309(1)S y y dy ⎡⎤=----⎣⎦⎰33(102)4y dy =--⎰333322002(10)4(4)214(4)3y y y y =-+-=+⨯⨯-8642213333=+-=-(22)已知非齐次线性方程组x 1+x 2+x 3+x 4=-1,4x 1+3x 2+5x 3-x 4=-1，ax 1+x 2+3x 3+bx 4=1有3个线性无关的解.① 证明此方程组的系数矩阵A 的秩为2. ② 求a,b 的值和方程组的通解. 解:① 设1,2,3是方程组的3个线性无关的解,则2-1,3-1是AX =0的两个线性无关的解.于是AX =0的基础解系中解的个数不少于2,即4-r(A )≥2,从而r(A )≤2.又因为A 的行向量是两两线性无关的,所以r(A )≥2. 两个不等式说明r(A )=2.② 对方程组的增广矩阵作初等行变换:1 1 1 1 -1 1 1 1 1 -1 (A |)= 4 3 5 -1 -1 → 0 –1 1 –5 3 ,a 1 3b 1 0 0 4-2a 4a+b-5 4-2a 由r(A )=2,得出a=2,b=-3.代入后继续作初等行变换: 1 0 2 -4 2 → 0 1 -1 5 -3 . 0 0 0 0 0 得同解方程组 x 1=2-2x 3+4x 4， x 2=-3+x 3-5x 4，求出一个特解(2,-3,0,0)T 和AX =0的基础解系(-2,1,1,0)T ,(4,-5,0,1) T .得到方程组的通解: (2,-3,0,0)T +c 1(-2,1,1,0)T +c 2(4,-5,0,1)T , c 1,c 2任意. (23) 设3阶实对称矩阵A 的各行元素之和都为3,向量1=(-1,2,-1)T ,2=(0,-1,1)T 都是齐次线性方程组AX =0的解. ① 求A 的特征值和特征向量.② 求作正交矩阵Q 和对角矩阵Λ,使得 Q T AQ =Λ. 解:① 条件说明A (1,1,1)T =(3,3,3)T ,即=(1,1,1)T 是A 的特征向量,特征值为3.又1,2都是AX =0的解说明它们也都是A 的特征向量,特征值为0.由于1,2线性无关, 特征值0的重数大于1.于是A 的特征值为3,0,0.属于3的特征向量:c 0, c ≠0. 属于0的特征向量:c 11+c 22, c 1,c 2不都为0.② 将单位化,得=(33,33,33)T .对1,2作施密特正交化,的1=(0,-22,22)T ,2=(-36,66,66)T. 作Q =(,1,2),则Q 是正交矩阵,并且3 0 0 Q T AQ =Q -1AQ = 0 0 0 . 0 0 0分数分配：11+11+11+12+12+10+9+9+9。

2006年12月应用能力B级真题(答案在最后)

2006年12月应用能力B级真题Part II Vocabulary & Structure (15 minutes)Directions: This part is to test your ability to use words and phrases correctly to construct meaningful and grammatically correct sentences. It consists of 2 sections.Section ADirections: There are 10 incomplete statements here. You are required to complete each statement by choosing the appropriate answer from the 4 choices marked A), B), C) and D). You should mark the corresponding letter on the Answer Sheet with a single line through the center.16. It is the general manager who makes the ______ decisions in business.A) beginning B) finishing C)first D) final17. Never ______ such a good boss before I came to this company.A) do I meet B) I met C) had I met D) I had met18. If the machine should ______, call this number immediately.A) break down B) set out C) put on D) go up19. The manager showed the new employee ______ to find the supplies.A) what B) where C) that D) which20. Look at the clock! It's time ______ work.A) we started B) we'll start C) we're starting D) we have started21. The sales department was required In ______ a plan in three weeks.A) turn up B) get up C) come up with D) put up with22. Price is not the only thing customers consider before ______ what to buy.A) deciding B) decided C) to decide D) having decided23. All the traveling ______ are paid by the company if you travel on business.A) charges B) money C) prices D) expenses24. Sorry, we cannot ______ you the job because you don't have any work experience.A) make B) send C) offer D) prepare25. This article is well written because special attention ______ to the choice of words nod style of writing. A) had been paid B) has been paid C) will be paid D) will have been paid Section BDirections: There are also 10 incomplete statements here. You should fill in each blank with the proper form of the word given in brackets. Write the word or words in the corresponding space on the Answer Sheet.26. It is a fact that traditional meals are (healthy) ______ than fast foods.27. Nurses should treat the sick and wounded with great (kind) ______.28. All visitors to the lab (expect) ______ to take off their shoes before they enter.29. (Personal) ______ I think he is a very nice partner, though you may not agree.30. They talked to him for hours, (try) ______ to persuade him to change his mind.31. His efforts to improve the sales of this product have been very (help) ______.32. When we arrived, there was a smell of cooking (come) ______ from the kitchen.33. We have to find new ways to (short) ______ the process of production.34. By this time next year my family (live) ______ in this small town for 20 years.35. Jane, as well as some of her classmates, (work) ______ in the Quality Control Department now.Part III Reading Comprehension (40 minutes)Directions: This part is to test your reading ability. There are 5 tasks for you to fulfill. You should read the reading materials carefully and do the tasks as you are instructed.Task 1People who work night shifts are constantly fighting against an "internal clock" in their bodies. Quite often the clock tells them to sleep when their job requires them to remain fully awake. It's no wonder that more accidents happen during night shills than at any other time. Light therapy(照光治疗法) with a bright light box can help night-shift workers adjust their internal clock, However, many doctors recommend careful planning to help improve sleep patterns. For example, night-shift workers often find it difficult to sleep in the morning when they get off work because the body's natural rhythm(节律) fights back, no matter how tired they are. Some experts recommend that night-shift workers schedule two smaller sleep periods--one in the morning 'after work, and another longer one in the afternoon, closer to when the body would naturally need to sleep. It's also helpful to ask friends and family to cooperate by avoiding visits and phone calls during the times when you are sleeping.36. Night-shift workers are those who ______.A) have to rely on their internal clock B) need to re-adjust their clockC) fall asleep late at night D) have to work at night37. In order to remain fully awake at work, people working night shifts should ______.A) have longer sleep periods after work B) make the light darker than usualC) try to re-set their "internal clock" D) pay more attention to their work38. Many doctors think it is helpful for night-shill workers ______.A) to sleep with a bright light on B) to plan sleep patterns carefullyC) to avoid being disturbed at work D) to sleep for a long lime after work39. Night-shift workers often find it difficult to sleep in the morning because ______.A) their internal clock will not allow them to B) they are often disturbed by morning visitsC) they are not trying hard enough to do so D) they are too tired to go to sleep well40. According to the passage, some doctors recommend that night-shift workers should ______.A) have frequent visits and phone calls B) improve their family relationshipC) have two smaller sleep periods D) rely mainly on light therapyTask 2A few ways Greyhound can make your next trip even easier. Tickets By mail. Avoid lining up altogether, by purchasing your tickets in advance, and having them delivered right to your mailbox. Just call Greyhound at least ten days before your departure (1 - 800 - 231 - 2222).Prepaid tickets. it's easy to purchase a ticket for a friend or family member no matter how far away they may be. Just call or go to your nearest Greyhound terminal(车站) and ask for details on how to buy a prepaid ticket.Ticketing Requirement.Greyhound now requires that all tickets have travel dates fixed at the time of purchase. Children under two years of age travel free with an adult who has a ticket.If your destination(目的地) is to Canada or Mexico.Passengers traveling to Canada or Mexico must have the proper travel documents. U. S. , Canadian or Mexican citizens should have a birth certificate, passport or naturalization (入籍) paper. If you are not a citizen of the U. S. , Canada or Mexico, a passport is required. In certain cases a visa may be required as well. These documents will be necessary and may be checked at, or before, boarding a bus departing for Canada or Mexico.41. From the passage, we can learn that "Greyhound" is probably the name of ______.A) an airline B) a hotel C) a website D) a bus company42. Why should people call Greyhound for tickets in advance?A) To avoid waiting in lines at the booking office. B) To hand in necessary traveling documents.C) To get tickets from the nearest terminal. D) To fix the traveling destination in time.43. What can we learn about the Greyhound tickets?A) They are not available for traveling outside the U. S.B) Travelers should buy their tickets in person.C) Babies can not travel free with their parents.D) They have the exact travel date on them.44. When people are traveling to Canada or Mexico, a passport is a must for ______.A) American citizens B) Japanese citizens C) Mexicans citizens D) Canadians citizens45. This passage mainly offers information about ______.A) how to prepare documents for traveling with GreyhoundB) how to purchase a Greyhound ticket and travel with itC) how to make your trip with Greyhound interestingD) how to travel from the U. S to Canada and MexicoTask 3December 10th, 2006Dear Sirs,I know that your company has a reputation(声誉) for quality products and fairness toward its customers. Therefore, I'm writing to ask for a replacement for a lawn mower(割草机).I bought the mower about half a year ago at the Watchung Discount Center, Watchung, Nebraska. I'm enclosing a copy of a receipt for the mower.A month after I bought the lawn mower, the engine failed, and it was repaired under warranty(保修期). So far, I have had the engine repaired four times.Now the engine has broken down again.I have already spent more than $ 300 on repairs, and I am beginning to seriously question the quality of your mowers.I am requesting that you replace this mower with a new one.I hope that you will live up to your reputation of the good customer service that has made your business successful.Faithfully,Rod GreenLetter of ComplaintPurpose of the letter: requesting a (46) for a lawn mowerTime of purchase: about (47) ageTrouble with the machine: (48)Times of repairs so far: (49)Money spent on repairs: more than (50)Task 4A .................... employee turnoverB .................... life-long employmentC .................... role conflictD ................... profit sharingE ................... scientific managementF .................... comparable worthG ................... flexible working hoursH .................... social supportI ..................... survey feedbackJ ..................... core competenceK .................... public relationsL .................... group cultureM .................... wage and salary surveysN .................... honesty testingO .................... human resource planning51．( )测谎( )工薪调查52．( )社会支持( )终身雇用制53．( )团队文化( )公共关系54．( )利润分享( )人力资源策划55．( )科学管理( )弹性工作时间Task 5Dear Sirs,For the past 8 years I have been a statistician(统计员) in the Research Unit of Baron & Smallwood Ltd. I am now looking for a change of employment which would broaden my experience. A large and well-known organization such as yours might be able to use my services.I am 31 years old and in excellent health. I majored in advertising at London University and I am particularly interested in work involving statistics(统计).Although I have had no experience in market research, I am familiar with the methods used for recording buying habits and trends. I hope that you will invite me for an interview. I could then give you further information.I am looking forward to hearing from you soon.Yours faithfully,Mike Smith56. What's Mike Smith's present job? He's working as a __________________.57. What was Mike Smith's major at London University? __________________.58. What kind of work does he like to do? Work involving __________________.59. In what area does he lack experience? He has no experience in __________________.60. What's the purpose of the writer in sending this letter? To be invited for __________________.Part IV Translation-English into Chinese (25 minutes)61. For safety, all passengers are required to review this card and follow these instructions when needed.A) 为了安全，请各位乘客反复阅读本卡片，务必按照各项规定执行。

06年考研数四真题及答案解析

2006年全国硕士研究生入学考试数学（四）一、填空 1．(1)1lim()nn n n-→∞+= 2．设函数()f x 在2x =的某邻域内可导，且()()(2)1f x f x e f '-⋅=，则法(2)f '=3．设函数()f u 可微，且1()2f u '=，则22(4)z f x y =-在点（1，2）处的全微分 (1,2)|dz =4．已知12,a a 为2维列向量，矩阵1212(2,)A a a a a =+-，12(,)B a a =。

若行列式||6A =，则||B =5．设矩阵2112A ⎡⎤=⎢⎥-⎣⎦，E 为2阶单位矩阵，矩阵B 满足2BA B E =+，则B 。

6．设随机变量X 与Y 相互独立，且均服从区间[1,3]上的均匀分布，由{max(,)1}P x y ≤=二、选择7．设函数()y f x =具有二阶导数，且()0f x '>，()0f x ''>，x 为自变量x 在点0x 处的增量y 与dy 分别为()f x 在点0x 处对应的增量与微分，若0x > ，则（）（A ）0dy y << （B ）0y dy << （C ）0y dy <<（D ）0dy y <<8．设函数()f x 在0x =处连续，且220()lim 1n f n n→==，则（）（A ）(0)0f =且(0)f '存在（B ）(0)1f =且(0)f '存在（C ）(0)0f =且(0)f +'存在（D ）(0)1f =且(0)f +'存在9．设函数()f x 与()g x 在[0,1]上连续，且()()f x g x ≤，且对任何(0,1)C ∈（）（A ）1122()()c cf t dtg t dt ≥⎰⎰（B ）1122()()c cf t dtg t dt ≤⎰⎰（C ）11()()ccf t dtg t dt ≥⎰⎰（D ）11()()ccf t dtg t dt ≤⎰⎰10．设非齐次线性微分方程()()y P x y Q x '+=有两个不同的解1()y x ，2()y x ，C 为任何常数，则该方程通解是（）（A ）12[()()]C y x y x - （B ）112()[()()]y x C y x y x +- （C ）12[()()]C y x y x +（D ）112()[()()]y x C y x y x ++11．设(,)f x y 与(,)G x y 均为可微函数，且(,)0G x y '≠，已知00(,)x y 是(,)f x y 在约束条件(,)0G x y =下的一个极值点。

AP微积分2006B真题

满分网——AP真题 AP® Calculus BC2006 Free-Response QuestionsForm BThe College Board: Connecting Students to College SuccessThe College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns.© 2006 The College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Board. Admitted Class Evaluation Service, CollegeEd, connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at:/inquiry/cbpermit.html.Visit the College Board on the Web: .AP Central is the official online home for the AP Program: .© 2006 The College Board. All rights reserved.CALCULUS BC SECTION II, Part ATime—45 minutes Number of problems—3A graphing calculator is required for some problems or parts of problems.1. Let f be the function given by ()323cos .432x x xf x x =--+ Let R be the shaded region in the secondquadrant bounded by the graph of f , and let S be the shaded region bounded by the graph of f and line A , theline tangent to the graph of f at 0,x = as shown above. (a) Find the area of R .(b) Find the volume of the solid generated when R is rotated about the horizontal line 2.y =- (c) Write, but do not evaluate, an integral expression that can be used to find the area of S .WRITE ALL WORK IN THE EXAM BOOKLET.© 2006 The College Board. All rights reserved.2. An object moving along a curve in the xy -plane is at position ()()(),x t y t at time t , where()tan t dxe dt -= and ()sec t dy e dt-= for 0.t ≥ At time 1,t = the object is at position ()2,3.-(a) Write an equation for the line tangent to the curve at position ()2,3.- (b) Find the acceleration vector and the speed of the object at time 1.t = (c) Find the total distance traveled by the object over the time interval 1 2.t ££ (d) Is there a time 0t ≥ at which the object is on the y -axis? Explain why or why not.WRITE ALL WORK IN THE EXAM BOOKLET.© 2006 The College Board. All rights reserved.3. The figure above is the graph of a function of x , which models the height of a skateboard ramp. The functionmeets the following requirements.(i) At 0,x = the value of the function is 0, and the slope of the graph of the function is 0. (ii) At 4,x = the value of the function is 1, and the slope of the graph of the function is 1. (iii) Between 0x = and 4,x = the function is increasing.(a) Let ()2,f x ax = where a is a nonzero constant. Show that it is not possible to find a value for a so that f meets requirement (ii) above.(b) Let ()23,16x g x cx =- where c is a nonzero constant. Find the value of c so that g meets requirement (ii) above. Show the work that leads to your answer.(c) Using the function g and your value of c from part (b), show that g does not meet requirement (iii) above.(d ) Let (),nx h x k= where k is a nonzero constant and n is a positive integer. Find the values of k and n sothat h meets requirement (ii) above. Show that h also meets requirements (i) and (iii) above.WRITE ALL WORK IN THE EXAM BOOKLET.END OF PART A OF SECTION II© 2006 The College Board. All rights reserved.CALCULUS BC SECTION II, Part BTime—45 minutes Number of problems—3No calculator is allowed for these problems.4. The rate, in calories per minute, at which a person using an exercise machine burns calories is modeled by thefunction f . In the figure above, ()3213142f t t t =-++ for 04t ££ and f is piecewise linear for 424.t ££(a) Find ()22.f ¢ Indicate units of measure.(b) For the time interval 024,t ££ at what time t is f increasing at its greatest rate? Show the reasoning thatsupports your answer. (c) Find the total number of calories burned over the time interval 618t ££ minutes.(d) The setting on the machine is now changed so that the person burns ()f t c + calories per minute. For thissetting, find c so that an average of 15 calories per minute is burned during the time interval 618.t ££WRITE ALL WORK IN THE EXAM BOOKLET.© 2006 The College Board. All rights reserved.5. Let f be a function with ()41f = such that all points (),x y on the graph of f satisfy the differential equation()23.dyy x dx=- Let g be a function with ()41g = such that all points (),x y on the graph of g satisfy the logistic differentialequation()23.dyy y dx=- (a) Find ().y f x =(b) Given that ()41,g = find ()lim x g x Æ•and ()lim .x g x Æ•¢ (It is not necessary to solve for ()g x or to show howyou arrived at your answers.)(c) For what value of y does the graph of g have a point of inflection? Find the slope of the graph of g at thepoint of inflection. (It is not necessary to solve for ().g x )6. The function f is defined by ()31.1f x x =+ The Maclaurin series for f is given by ()369311,nn x x x x -+-++-+""which converges to ()f x for 1 1.x -<<(a) Find the first three nonzero terms and the general term for the Maclaurin series for ().f x ¢ (b) Use your results from part (a) to find the sum of the infinite series ()""2583136931.2222n n n--+-++-+ (c) Find the first four nonzero terms and the general term for the Maclaurin series representing ()0.xf t dt Ú(d) Use the first three nonzero terms of the infinite series found in part (c) to approximate()12.f t dt Ú Whatare the properties of the terms of the series representing ()12f t dt Ú that guarantee that this approximationis within110,000of the exact value of the integral?WRITE ALL WORK IN THE EXAM BOOKLET.END OF EXAM。

2006高考数学试题陕西卷B型

2006高考数学试题陕西卷B 型理科试题（必修＋选修II ）注意事项：１．本试卷分第一部分和第二部分。

第一部分为选择题，第二部分为非选择题。

2．考生领到试卷后，须按规定在试卷上填写姓名、准考证号，并在答题卡上填涂对应的试卷类型信息点。

3．所有答案必须在答题卡上指定区域内作答。

考试结束后，将本试卷和答题卡一并交回。

第一部分（共60分）一．选择题：在每小题给出的四个选项中，只有一项是符合题目要求的（本大题共12小题，每小题5分，共60分）1．已知集合{}|110,P x N x =∈≤≤集合{}2|60,Q x R x x =∈+-=则PQ 等于（A ）{}1,2,3 （B ）{}2,3 （C ）{}1,2 （D ）{}22．复数10(1)1i i+-等于（A ）1i + （B ）1i -- （C ）1i - （D ）1i -+3．n 等于（A ）0 （B ）14 （C ）12（D ）1 4．设函数()log ()(0,1)a f x x b a a =+>≠的图像过点(2,1)，其反函数的图像过点(2,8)，则a b +等于（A ）3 （B ）4 （C ）5 （D ）65．设直线过点(0,),a 其斜率为1，且与圆222x y +=相切，则a 的值为（Ａ）4± （Ｂ）± （Ｃ）2± （Ｄ）６．"等式sin()sin 2αγβ+=成立"是",,αβγ成等差数列 "的（Ａ）充分而不必要条件（Ｂ）必要而不充分条件（Ｃ）充分必要条件（Ｄ）既不充分又不必要条件7．已知双曲线2221(2x y a a -=>的两条渐近线的夹角为3π，则双曲线的离心率为（A （B （C （D ）28．已知不等式1()()9ax y x y++≥对任意正实数,x y 恒成立，则正实数a 的最小值为（Ａ）8 （Ｂ）6 （C ）4 （D ）29．已知非零向量AB 与AC 满足().0AB AC BC ABAC+=且1..2AB AC AB AC =则ABC ∆为（A ）等边三角形（B ）直角三角形（C ）等腰非等边三角形（D ）三边均不相等的三角形10．已知函数2()24(03),f x ax ax a =++<<若1212,1,x x x x a <+=-则（A ）12()()f x f x > （B ）12()()f x f x <（C ）12()()f x f x = （D ）1()f x 与2()f x 的大小不能确定11．已知平面α外不共线的三点,,A B B 到α的距离都相等，则正确的结论是（A ）平面ABC 必不垂直于α （B ）平面ABC 必平行于α （C ）平面ABC 必与α相交（D ）存在ABC ∆的一条中位线平行于α或在α内12．为确保信息安全，信息需加密传输，发送方由明文→密文（加密），接收方由密文→明文（解密），已知加密规则为：明文,,,a b c d 对应密文2,2,23,4.a b b c c d d +++例如，明文1,2,3,4对应密文5,7,18,16.当接收方收到密文14,9,23,28时，则解密得到的明文为（A ）7,6,1,4 （B ）6,4,1,7 （C ）4,6,1,7 （D ）1,6,4,7第二部分（共90分）二．填空题：把答案填在答题卡相应题号后的横线上（本大题共4小题，每小题4分，共16分）。

1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。