第1章题解( 2010姚四版)

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数值分析第四版习题及答案

第四版数值分析习题第一章绪论1. 设x >0,x 的相对误差为δ,求ln x 的误差.2. 设x 的相对误差为2％,求nx 的相对误差.3. 下列各数都是经过四舍五入得到的近似数,即误差限不超过最后一位的半个单位,试指出它们是几位有效数字: *****123451.1021,0.031,385.6,56.430,7 1.0.x x x x x =====⨯4. 利用公式(3.3)求下列各近似值的误差限:********12412324(),(),()/,i x x x ii x x x iii x x ++其中****1234,,,x x x x 均为第3题所给的数.5. 计算球体积要使相对误差限为1％,问度量半径R 时允许的相对误差限是多少?6. 设028,Y =按递推公式1n n Y Y -=…)计算到100Y .27.982(五位有效数字),试问计算100Y 将有多大误差?7. 求方程25610x x -+=的两个根,使它至少具有四位有效数字27.982).8. 当N 充分大时,怎样求211Ndx x +∞+⎰?9. 正方形的边长大约为100㎝,应怎样测量才能使其面积误差不超过1㎝2? 10. 设212S gt =假定g 是准确的,而对t 的测量有±0.1秒的误差,证明当t 增加时S 的绝对误差增加,而相对误差却减小.11. 序列{}n y 满足递推关系1101n n y y -=-(n=1,2,…),若0 1.41y =≈(三位有效数字),计算到10y 时误差有多大?这个计算过程稳定吗?12. 计算61)f =, 1.4≈,利用下列等式计算,哪一个得到的结果最好?3--13.()ln(f x x =,求f (30)的值.若开平方用六位函数表,问求对数时误差有多大?若改用另一等价公式ln(ln(x x =-计算,求对数时误差有多大?14. 试用消元法解方程组{101012121010;2.x x x x +=+=假定只用三位数计算,问结果是否可靠?15. 已知三角形面积1sin ,2s ab c =其中c 为弧度,02c π<<,且测量a ,b ,c 的误差分别为,,.a b c ∆∆∆证明面积的误差s ∆满足.s a b cs a b c ∆∆∆∆≤++第二章插值法1. 根据(2.2)定义的范德蒙行列式,令2000011211121()(,,,,)11n n n n n n n n n x x x V x V x x x x x x x xx x ----==L L L L L L L L L证明()n V x 是n 次多项式,它的根是01,,n x x -L ,且101101()(,,,)()()n n n n V x V x x x x x x x ---=--L L .2. 当x = 1 , -1 , 2 时, f (x)= 0 , -3 , 4 ,求f (x )的二次插值多项式.3.4. 给出cos x ,0°≤x ≤90°的函数表,步长h =1′=(1/60)°,若函数表具有5位有效数字,研究用线性插值求cos x 近似值时的总误差界.5. 设0k x x kh =+,k =0,1,2,3,求032max ()x x x l x ≤≤.6. 设jx 为互异节点(j =0,1,…,n ),求证:i) 0()(0,1,,);nk kj j j x l x xk n =≡=∑Lii) 0()()1,2,,).nk jj j xx l x k n =-≡0(=∑L7. 设[]2(),f x C a b ∈且()()0f a f b ==,求证21()()().8max max a x ba xb f x b a f x ≤≤≤≤≤-"8. 在44x -≤≤上给出()xf x e =的等距节点函数表,若用二次插值求xe 的近似值,要使截断误差不超过610-,问使用函数表的步长h 应取多少?9. 若2nn y =,求4n y ∆及4n y δ.10. 如果()f x 是m 次多项式,记()()()f x f x h f x ∆=+-,证明()f x 的k 阶差分()(0)k f x k m ∆≤≤是m k -次多项式,并且()0(m l f x l +∆=为正整数).11. 证明1()k k k k k k f g f g g f +∆=∆+∆. 12. 证明110010.n n kkn n k k k k f gf g f g g f --+==∆=--∆∑∑13. 证明1200.n j n j y y y -=∆=∆-∆∑14. 若1011()n nn n f x a a x a x a x --=++++L 有n 个不同实根12,,,n x x x L ,证明{10,02;, 1.1()n k njk n a k n j jx f x -≤≤-=-=='∑15. 证明n 阶均差有下列性质: i)若()()F x cf x =,则[][]0101,,,,,,n n F x x x cf x x x =L L ;ii) 若()()()F x f x g x =+,则[][][]010101,,,,,,,,,n n n F x x x f x x x g x x x =+L L L .16. 74()31f x x x x =+++,求0172,2,,2f ⎡⎤⎣⎦L 及0182,2,,2f ⎡⎤⎣⎦L . 17. 证明两点三次埃尔米特插值余项是(4)22311()()()()/4!,(,)k k k k R x f x x x x x x ++=ξ--ξ∈并由此求出分段三次埃尔米特插值的误差限.18. 求一个次数不高于4次的多项式()P x ,使它满足(0)(1)P P k =-+并由此求出分段三次埃尔米特插值的误差限. 19. 试求出一个最高次数不高于4次的函数多项式()P x ,以便使它能够满足以下边界条件(0)(0)0P P ='=,(1)(1)1P P ='=,(2)1P =.20. 设[](),f x C a b ∈,把[],a b 分为n 等分,试构造一个台阶形的零次分段插值函数()n x ϕ并证明当n →∞时,()n x ϕ在[],a b 上一致收敛到()f x .21. 设2()1/(1)f x x =+,在55x -≤≤上取10n =,按等距节点求分段线性插值函数()h I x ,计算各节点间中点处的()h I x 与()f x 的值,并估计误差.22. 求2()f x x =在[],a b 上的分段线性插值函数()h I x ,并估计误差. 23. 求4()f x x =在[],a b 上的分段埃尔米特插值,并估计误差.试求三次样条插值并满足条件 i) (0.25) 1.0000,(0.53)0.6868;S S '='= ii)(0.25)(0.53)0.S S "="=25. 若[]2(),f x C a b ∈,()S x 是三次样条函数,证明 i)[][][][]222()()()()2()()()bbbba a a a f x dx S x dx f x S x dx S x f x S x dx"-"="-"+""-"⎰⎰⎰⎰;ii) 若()()(0,1,,)i i f x S x i n ==L ,式中i x 为插值节点,且01n a x x x b =<<<=L ,则[][][]()()()()()()()()()baS x f x S x dx S b f b S b S a f a S a ""-"="'-'-"'-'⎰.26. 编出计算三次样条函数()S x 系数及其在插值节点中点的值的程序框图(()S x 可用(8.7)式的表达式).第三章函数逼近与计算1. (a)利用区间变换推出区间为[],a b 的伯恩斯坦多项式.(b)对()sin f x x =在[]0,/2π上求1次和三次伯恩斯坦多项式并画出图形,并与相应的马克劳林级数部分和误差做比较. 2. 求证:(a)当()m f x M ≤≤时,(,)n m B f x M ≤≤. (b)当()f x x =时,(,)n B f x x =.3. 在次数不超过6的多项式中,求()sin 4f x x =在[]0,2π的最佳一致逼近多项式.4. 假设()f x 在[],a b 上连续,求()f x 的零次最佳一致逼近多项式.5. 选取常数a ,使301max x x ax≤≤-达到极小,又问这个解是否唯一?6. 求()sin f x x =在[]0,/2π上的最佳一次逼近多项式,并估计误差.7. 求()xf x e =在[]0,1上的最佳一次逼近多项式.8. 如何选取r ,使2()p x x r =+在[]1,1-上与零偏差最小?r 是否唯一? 9. 设43()31f x x x =+-,在[]0,1上求三次最佳逼近多项式.10. 令[]()(21),0,1n n T x T x x =-∈,求***0123(),(),(),()T x T x T x T x .11. 试证{}*()nT x 是在[]0,1上带权ρ=的正交多项式.12. 在[]1,1-上利用插值极小化求11()f x tg x -=的三次近似最佳逼近多项式. 13. 设()xf x e =在[]1,1-上的插值极小化近似最佳逼近多项式为()n L x ,若n f L ∞-有界,证明对任何1n ≥,存在常数n α、n β,使11()()()()(11).n n n n n T x f x L x T x x ++α≤-≤β-≤≤14. 设在[]1,1-上234511315165()128243843840x x x x x x ϕ=-----,试将()x ϕ降低到3次多项式并估计误差. 15. 在[]1,1-上利用幂级数项数求()sin f x x =的3次逼近多项式,使误差不超过0.005.16. ()f x 是[],a a -上的连续奇(偶)函数,证明不管n 是奇数或偶数,()f x 的最佳逼近多项式*()n n F x H ∈也是奇(偶)函数.17. 求a 、b 使[]22sin ax b x dx π+-⎰为最小.并与1题及6题的一次逼近多项式误差作比较.18. ()f x 、[]1(),g x C a b ∈,定义 ()(,)()();()(,)()()()();b baaa f g f x g x dxb f g f x g x dx f a g a =''=''+⎰⎰问它们是否构成内积?19. 用许瓦兹不等式(4.5)估计6101x dx x +⎰的上界,并用积分中值定理估计同一积分的上下界,并比较其结果.20. 选择a ,使下列积分取得最小值:1122211(),x ax dx x ax dx----⎰⎰.21. 设空间{}{}10010121,,,span x span x x 1ϕ=ϕ=,分别在1ϕ、2ϕ上求出一个元素,使得其为[]20,1x C ∈的最佳平方逼近,并比较其结果. 22. ()f x x =在[]1,1-上,求在{}2411,,span x x ϕ=上的最佳平方逼近.23.sin (1)arccos ()n n x u x +=是第二类切比雪夫多项式,证明它有递推关系()()()112n n n u x xu x u x +-=-.24. 将1()sin 2f x x=在[]1,1-上按勒让德多项式及切比雪夫多项式展开,求三次最佳平方逼近多项式并画出误差图形,再计算均方误差.25. 把()arccos f x x =在[]1,1-上展成切比雪夫级数.26.2y a bx =+.27.用最小二乘拟合求.29. 编出用正交多项式做最小二乘拟合的程序框图. 30. 编出改进FFT 算法的程序框图. 31. 现给出一张记录{}{}4,3,2,1,0,1,2,3k x =,试用改进FFT 算法求出序列{}k x 的离散频谱{}k C (0,1,,7).k =L第四章数值积分与数值微分1. 确定下列求积公式中的待定参数,使其代数精度尽量高,并指明所构造出的求积公式所具有的代数精度:(1)101()()(0)()h h f x dx A f h A f A f h --≈-++⎰; (2)21012()()(0)()hhf x dx A f h A f A f h --≈-++⎰;(3)[]1121()(1)2()3()/3f x dx f f x f x -≈-++⎰;(4)[][]20()(0)()/1(0)()hf x dx h f f h ah f f h ≈++'-'⎰.2. 分别用梯形公式和辛普森公式计算下列积分:(1)120,84xdx n x =+⎰; (2)1210(1),10x e dx n x --=⎰;(3)1,4n =⎰; (4),6n =.3. 直接验证柯特斯公式(2.4)具有5次代数精度.4. 用辛普森公式求积分10xedx-⎰并计算误差.5. 推导下列三种矩形求积公式:(1)2()()()()()2ba f f x dxb a f a b a 'η=-+-⎰; (2)2()()()()()2ba f f x dxb a f b b a 'η=---⎰;(3)3()()()()()224baa b f f x dx b a f b a +"η=-+-⎰. 6. 证明梯形公式(2.9)和辛普森公式(2.11)当n →∞时收敛到积分()baf x dx⎰.7. 用复化梯形公式求积分()baf x dx⎰,问要将积分区间[],a b 分成多少等分,才能保证误差不超过ε(设不计舍入误差)?8.1xedx-,要求误差不超过510-.9. 卫星轨道是一个椭圆,椭圆周长的计算公式是S a =θ,这里a 是椭圆的半长轴,c 是地球中心与轨道中心(椭圆中心)的距离,记h 为近地点距离,H 为远地点距离,6371R =公里为地球半径,则(2)/2,()/2a R H h c H h =++=-.我国第一颗人造卫星近地点距离439h =公里,远地点距离2384H =公里,试求卫星轨道的周长. 10. 证明等式3524sin3!5!n n nnππππ=-+-L试依据sin(/)(3,6,12)n n n π=的值,用外推算法求π的近似值.11. 用下列方法计算积分31dyy ⎰并比较结果.(1) 龙贝格方法;(2) 三点及五点高斯公式;(3) 将积分区间分为四等分,用复化两点高斯公式.12. 用三点公式和五点公式分别求21()(1)f x x =+在x =1.0,1.1和1.2处的导数值,并估计()f x第五章常微分方程数值解法1. 就初值问题0)0(,=+='y b ax y 分别导出尤拉方法和改进的尤拉方法的近似解的表达式，并与准确解bx ax y +=221相比较。

概率论第一章习题参考解答

概论论与数理统计习题参考解答习题一8. 掷3枚硬币, 求出现3个正面的概率. 解: 设事件A ={出现3个正面}基本事件总数n =23, 有利于A 的基本事件数n A =1, 即A 为一基本事件, 则125.08121)(3====n n A P A . 9. 10把钥匙中有3把能打开门, 今任取两把, 求能打开门的概率. 解: 设事件A ={能打开门}, 则A 为不能打开门基本事件总数210C n =, 有利于A 的基本事件数27C n =, 467.0157910212167)(21027==⨯⨯⋅⨯⨯==C C A P因此, 533.0467.01)(1)(=-=-=A P A P .10. 一部四卷的文集随便放在书架上, 问恰好各卷自左向右或自右向左的卷号为1,2,3,4的概率是多少?解: 设A ={能打开门},基本事件总数2412344=⨯⨯⨯==P n , 有利于A 的基本事件数为2=A n , 因此, 0833.0121)(===n n A P A . 11. 100个产品中有3个次品,任取5个, 求其次品数分别为0,1,2,3的概率. 解: 设A i 为取到i 个次品, i =0,1,2,3,基本事件总数5100C n =, 有利于A i 的基本事件数为3,2,1,0,5973==-i C C n i i i则00006.09833512196979697989910054321)(006.0983359532195969739697989910054321)(138.09833209495432194959697396979899100543213)(856.0334920314719969798991009394959697)(51002973351003972322510049711510059700=⨯⨯==⨯⨯⋅⨯⨯⨯⨯⨯⨯⨯⨯====⨯⨯=⨯⨯⨯⨯⨯⋅⨯⨯⨯⨯⨯⨯⨯⨯====⨯⨯⨯=⨯⨯⨯⨯⨯⨯⨯⋅⨯⨯⨯⨯⨯⨯⨯⨯=⨯===⨯⨯⨯⨯=⨯⨯⨯⨯⨯⨯⨯⨯===C C n n A P C C C n n A P C C n n A P C C n n A P12. N 个产品中有N 1个次品, 从中任取n 个(1≤n ≤N 1≤N ), 求其中有k (k ≤n )个次品的概率. 解: 设A k 为有k 个次品的概率, k =0,1,2,…,n ,基本事件总数nN C m =, 有利于事件A k 的基本事件数kn N N k N k C C m --=11,k =0,1,2,…,n ,因此, n k C C C m m A P nNkn N N k N k k ,,1,0,)(11 ===-- 13. 一个袋内有5个红球, 3个白球, 2个黑球, 计算任取3个球恰为一红, 一白, 一黑的概率.解: 设A 为任取三个球恰为一红一白一黑的事件,则基本事件总数310C n =, 有利于A 的基本事件数为121315C C C n A =, 则25.0412358910321)(310121315==⨯⨯⨯⨯⨯⨯⨯===C C C C n n A P A 14. 两封信随机地投入四个邮筒, 求前两个邮筒内没有信的概率以及第一个邮筒内只有一封信的概率.解: 设A 为前两个邮筒没有信的事件, B 为第一个邮筒内只有一封信的事件, 则基本事件总数1644=⨯=n , 有利于A 的基本事件数422=⨯=A n , 有利于B 的基本事件数632=⨯=B n , 则25.041164)(====n n A P A 375.083166)(====n n B P B .15. 一批产品中, 一, 二, 三等品率分别为0.8, 0.16, 0.04, 若规定一, 二等品为合格品, 求产品的合格率.解: 设事件A 1为一等品, A 2为二等品, B 为合格品, 则 P (A 1)=0.8, P (A 2)=0.16,B =A 1+A 2, 且A 1与A 2互不相容, 根据加法法则有 P (B )=P (A 1)+P (A 2)=0.8+0.16=0.9616. 袋内装有两个5分, 三个2分, 五个一分的硬币, 任意取出5个, 求总数超过一角的概率. 解: 假设B 为总数超过一角,A 1为5个中有两个5分, A 2为5个中有一个5分三个2分一个1分, A 3为5个中有一个5分两个2分两个1分, 则B =A 1+A 2+A 3, 而A 1,A 2,A 3互不相容, 基本事件总数252762354321678910510=⨯⨯⨯=⨯⨯⨯⨯⨯⨯⨯⨯==C n设有利于A 1,A 2,A 3的基本事件数为n 1,n 2,n 3, 则5.0252126252601056)(,60214532,1052,563216782523123153312238221==++==⨯⨯⨯⨯===⨯===⨯⨯⨯⨯==B P C C C n C C C n C C n 17. 求习题11中次品数不超过一个的概率.解: 设A i 为取到i 个次品, i =0,1,2,3, B 为次品数不超过一个, 则B =A 0+A 1, A 0与A 1互不相容, 则根据11题的计算结果有 P (B )=P (A 0)+P (A 1)=0.856+0.138=0.99419. 由长期统计资料得知, 某一地区在4月份下雨(记作事件A )的概率为4/15, 刮风(用B 表示)的概率为7/15, 既刮风又下雨的概率为1/10, 求P (A |B ), P (B |A ), P (A +B ). 解: 根据题意有P (A )=4/15, P (B )=7/15, P (AB )=1/10, 则633.03019303814101154157)()()()(275.08315/410/1)())|(214.014315/710/1)()()|(==-+=-+=-+=+========AB P B P A P B A P A P PAB A B P B P AB P B A P20. 为防止意外, 在矿内同时设有两种报警系统A 与B , 每种系统单独使用时, 其有效的概率系统A 为0.92, 系统B 为0.93, 在A 失灵的条件下, B 有效的概率为0.85, 求 (1) 发生意外时, 这两个报警系统至少有一个有效的概率 (2) B 失灵的条件下, A 有效的概率解: 设A 为系统A 有效, B 为系统B 有效, 则根据题意有 P (A )=0.92, P (B )=0.93, 85.0)|(=A B P(1) 两个系统至少一个有效的事件为A +B , 其对立事件为两个系统都失效, 即B A B A =+, 而15.085.01)|(1)|(=-=-=A B P A B P , 则988.0012.01)(1)(012.015.008.015.0)92.01()|()()(=-=-=+=⨯=⨯-==B A P B A P A B P A P B A P(2) B 失灵条件下A 有效的概率为)|(B A P , 则829.093.01012.01)()(1)|(1)|(=--=-=-=B P B A P B A P B A P21. 10个考签中有4个难签, 3人参加抽签考试, 不重复地抽取, 每人一次, 甲先, 乙次, 丙最后, 证明3人抽到难签的概率相等.证: 设事件A ,B ,C 表示甲,乙,丙各抽到难签, 显然P (A )=4/10, 而由903095106)|()()(902496104)|()()(902494106)|()()(901293104)|()()(=⨯===⨯===⨯===⨯==A B P A P B A P A B P A P B A P A B P A P B A P A B P A P AB P由于A 与A 互不相容,且构成完备事件组, 因此B A AB B +=可分解为两个互不相容事件的并, 则有1049036902412)()()(==+=+=B A P AB P B P 又因B A B A B A AB ,,,之间两两互不相容且构成完备事件组, 因此有C B A C B A BC A ABC C +++=分解为四个互不相容的事件的并,且720120849030)|()()(72072839024)|()()(72072839024)|()()(72024829012)|()()(=⨯===⨯===⨯===⨯==B A C P B A P C B A P B A C P B A P C B A P B A C P B A P BC A P AB C P AB P ABC P则104720288720120727224()()()()(==+++=+++=C B A P C B A P BC A P ABC P C P 因此有P (A )=P (B )=P (C ), 证毕.22. 用3个机床加工同一种零件, 零件由各机床加工的概率分别为0.5, 0.3, 0.2, 各机床加工的零件为合格品的概率分别等于0.94, 0.9, 0.95, 求全部产品中的合格率. 解: 设A 1,A 2,A 3零件由第1,2,3个机床加工, B 为产品合格, A 1,A 2,A 3构成完备事件组. 则根据题意有P (A 1)=0.5, P (A 2)=0.3, P (A 3)=0.2,P (B |A 1)=0.94, P (B |A 2)=0.9, P (B |A 3)=0.95, 由全概率公式得全部产品的合格率P (B )为93.095.02.09.03.094.05.0)|()()(31=⨯+⨯+⨯==∑=i i i A B P A P B P23. 12个乒乓球中有9个新的3个旧的, 第一次比赛取出了3个, 用完后放回去, 第二次比赛又取出3个, 求第二次取到的3个球中有2个新球的概率.解: 设A 0,A 1,A 2,A 3为第一次比赛取到了0,1,2,3个新球, A 0,A 1,A 2,A 3构成完备事件组. 设B 为第二次取到的3个球中有2个新球. 则有22962156101112321)|(,552132101112789321)(,442152167101112321)|(,55272101112389321)(,552842178101112321)|(,2202710111239321)(,552732189101112321)|(,2201101112321)(312162633123933121527231213292312142813122319131213290312330=⋅⨯⨯⋅⨯⨯⨯⨯===⨯⨯⨯⨯⨯⨯⨯⨯⨯===⋅⨯⨯⋅⨯⨯⨯⨯===⨯⨯⨯⨯⨯⨯⨯⨯===⋅⨯⨯⋅⨯⨯⨯⨯===⨯⨯⨯⨯⨯⨯===⋅⨯⨯⋅⨯⨯⨯⨯===⨯⨯⨯⨯==C C C A B P C C A P C C C A B P C C C A P C C C A B P C C C A P C C C A B P C C A P根据全概率公式有455.01562.02341.00625.00022.022955214421552755282202755272201)|()()(3=+++=⋅+⋅+⋅+⋅==∑=i i i A B P A P B P 24. 某商店收进甲厂生产的产品30箱, 乙厂生产的同种产品20箱, 甲厂每箱100个, 废品率为0.06, 乙厂每箱装120个, 废品率是0.05, 求: (1)任取一箱, 从中任取一个为废品的概率;(2)若将所有产品开箱混放, 求任取一个为废品的概率. 解: (1) 设B 为任取一箱, 从中任取一个为废品的事件. 设A 为取到甲厂的箱, 则A 与A 构成完备事件组056.005.04.006.06.0)|()()|()()(05.0)|(,06.0)|(4.05020)(,6.05030)(=⨯+⨯=+=======A B P A P A B P A P B P A B P A B P A P A P(2) 设B 为开箱混放后任取一个为废品的事件.则甲厂产品的总数为30×100=3000个, 其中废品总数为3000×0.06=180个, 乙厂产品的总数为20×120=2400个, 其中废品总数为2400×0.05=120个, 因此...055555555.0540030024003000120180)(==++=B P25. 一个机床有1/3的时间加工零件A , 其余时间加工零件B , 加工零件A 时, 停机的概率是0.3, 加工零件B 时, 停机的概率是0.4, 求这个机床停机的概率.解: 设C 为加工零件A 的事件, 则C 为加工零件B 的事件, C 与C 构成完备事件组. 设D 为停机事件, 则根据题意有 P (C )=1/3, P (C )=2/3, P (D |C )=0.3, P (D |C )=0.4, 根据全概率公司有367.04.0323.031)|()()|()()(=⨯+⨯=+=C D P C P C D P C P D P 26. 甲, 乙两部机器制造大量的同一种机器零件, 根据长期资料总结, 甲机器制造出的零件废品率为1%, 乙机器制造出的废品率为2%, 现有同一机器制造的一批零件, 估计这一批零件是乙机器制造的可能性比它们是甲机器制造的可能性大一倍, 今从该批零件中任意取出一件, 经检查恰好是废品, 试由此检查结果计算这批零件为甲机器制造的概率.解: 设A 为零件由甲机器制造, 则A 为零件由乙机器制造, A 与A 构成完备事件组. 由P (A +A )=P (A )+P (A )=1并由题意知P (A )=2P (A ), 得P (A )=1/3, P (A )=2/3. 设B 为零件为废品, 则由题意知P (B |A )=0.01, P (B |A )=0.02,则根据贝叶斯公式, 任抽一件检查为废品条件下零件由甲机器制造的概率为2.005.001.002.03201.03101.031)|()()|()()|()()|(==⨯+⨯⨯==+=A B P A P A B P A P A B P A P B A P 27. 有两个口袋, 甲袋中盛有两个白球, 一个黑球, 乙袋中盛有一个白球两个黑球. 由甲袋中任取一个球放入乙袋, 再从乙袋中取出一个球, 求取到白球的概率.解: 设事件A 为从甲袋中取出的是白球, 则A 为从甲袋中取出的是黑球, A 与A 构成完备事件组. 设事件B 为从乙袋中取到的是白球. 则P (A )=2/3, P (A )=1/3, P (B |A )=2/4=1/2, P (B |A )=1/4, 则根据全概率公式有417.012541312132)|()()|()()(==⨯+⨯=+=A B P A P A B P A P B P28. 上题中若发现从乙袋中取出的是白球, 问从甲袋中取出放入乙袋的球, 黑白哪种颜色可能性大?解: 事件假设如上题, 而现在要求的是在事件B 已经发生条件下, 事件A 和A 发生的条件概率P (A |B )和P (A |B )哪个大, 可以套用贝叶斯公式进行计算, 而计算时分母为P (B )已上题算出为0.417, 因此2.0417.04131)()|()()|(8.0417.02132)()|()()|(=⨯===⨯==B P A B P A P B A P B P A B P A P B A PP (A |B )>P (A |B ), 因此在乙袋取出的是白球的情况下, 甲袋放入乙袋的球是白球的可能性大. 29. 假设有3箱同种型号的零件, 里面分别装有50件, 30件和40件, 而一等品分别有20件, 12件及24件. 现在任选一箱从中随机地先后各抽取一个零件(第一次取到的零件不放回). 试求先取出的零件是一等品的概率; 并计算两次都取出一等品的概率.解: 称这三箱分别为甲,乙,丙箱, 假设A 1,A 2,A 3分别为取到甲,乙,丙箱的事件, 则A 1,A 2,A 3构成完备事件组.易知P (A 1)=P (A 2)=P (A 3)=1/3.设B 为先取出的是一等品的事件. 则6.04024)|(,4.03012)|(,4.05020)|(321======A B P A B P A B P 根据全概率公式有467.036.04.04.0)|()()(31=++==∑=i i i A B P A P B P设C 为两次都取到一等品的事件, 则38.039402324)|(1517.029301112)|(1551.049501920)|(240224323021222502201=⨯⨯===⨯⨯===⨯⨯==C C A C P C C A C P C C A C P根据全概率公式有22.033538.01517.01551.0)|()()(31=++==∑=i i i A C P A P C P30. 发报台分别以概率0.6和0.4发出信号“·”和“—”。

中国人民大学出版社(第四版)高等数学一第1章课后习题详解

中国人民大学出版社(第四版)高等数学一第1章课后习题详解第一章函数、极限与连续内容概要名称主要内容（1.1、1.2）函数邻域(){}δδ<-=axxaU,（即(){},U a x a x aδδδ=-<<+）(){}0,0U a x x aδδ=<-<（(){}0,,0U a x a x a xδδδ=-<<+≠）函数两个要素：对应法则f以及函数的定义域D由此，两函数相等⇔两要素相同；（与自变量用何字母表示无关）解析表示法的函数类型：显函数，隐函数，分段函数；特性局部有界性对集合DX⊂，若存在正数M，使对所有Xx∈，恒有()Mxf<，称函数()xf在X上有界，或()xf是X上的有界函数；反之无界，即任意正数M（无论M多大），总存在（能找到）Xx∈，使得()Mxf>局部单调性区间DI⊂，对区间上任意两点21xx，当21xx<时，恒有：()()21xfxf<，称函数在区间I上是单调增加函数；反之，若()()21xfxf>，则称函数在区间I上是单调减小函数；奇偶性设函数()xf的定义域D关于原点对称；若Dx∈∀，恒有()()xfxf=-，则称()xf是偶函数；若Dx∈∀，恒有()()xfxf-=-，则称()x f是奇函数；周期性若存在非零常数T，使得对Dx∈∀，有()DTx∈±，且()()x fTxf=+，则称()x f是周期函数；初等函数几类基本初等函数：幂函数；指数函数；对数函数；三角函数；反三角函数；反函数求法和性质；复合函数性质；初等函数课后习题全解习题1-1★1．求下列函数的定义域：知识点：自然定义域指实数范围内使函数表达式有意义的自变量x 的取值的集合；思路：常见的表达式有 ① alog□,（ □0>） ② /N □, （ □0≠） ③ (0)≥④ arcsin（[]1,1-∈）等解：（1）[)(]1,00,11100101122⋃-∈⇒⎩⎨⎧≤≤-≠⇒⎩⎨⎧≥-≠⇒--=x x x x x x x y ；（2）31121121arcsin ≤≤-⇒≤-≤-⇒-=x x x y ；（3）()()3,00,030031arctan 3⋃∞-∈⇒⎩⎨⎧≠≤⇒⎩⎨⎧≠≥-⇒+-=x x x x x x x y ；（4）()()3,11,1,,1310301lg 3⋃-∞-∈⇒⎩⎨⎧-<<<⇒⎩⎨⎧-<-<⇒-=-x x or x x x x x y x；（5）()()4,22,11601110)16(log 221⋃∈⇒⎪⎩⎪⎨⎧-<-≠-<⇒-=-x x x x x y x ； ★2．下列各题中，函数是否相同？为什么？（1）2lg )(x x f =与x x g lg 2)(=；（2）12+=x y 与12+=y x知识点：函数相等的条件；思路：函数的两个要素是f （作用法则）及定义域D （作用范围），当两个函数作用法则f 相同（化简后代数表达式相同）且定义域相同时，两函数相同；解：（1）2lg )(x x f =的定义域D={}R x x x ∈≠,0，xx g lg )(=的定义域{},0R x x x D ∈>=，虽然作用法则相同x x lg 2lg 2=，但显然两者定义域不同，故不是同一函数；（2）12+=x y ，以x 为自变量，显然定义域为实数R ；12+=y x ，以x 为自变量，显然定义域也为实数R ；两者作用法则相同“2□1+”与自变量用何记号表示无关，故两者为同一函数；★3．设⎪⎪⎩⎪⎪⎨⎧≥<=3,03,sin )(ππϕx x x x ，求)2()4()4()6(--ϕπϕπϕπϕ，，，，并做出函数)(x y ϕ=的图形知识点：分段函数；思路：注意自变量的不同范围；解：216sin )6(==ππϕ，224sin 4==⎪⎭⎫⎝⎛ππϕ，224sin 4=⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛-ππϕ()02=-ϕ；如图：★4．试证下列各函数在指定区间内的单调性：（1）()1,1∞--=xxy （2）x x y ln 2+=，()+∞,0 知识点：单调性定义。

2010年考研数学试题详解及评分参考

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郝海龙：考研数学复习大全·配套光盘·2010 年数学试题详解及评分参考
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华工概率论2010-1答案

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⎧ ⎫ ⎪ ⎪ ⎛ ⎛ 3⎪ 3⎞ 3⎞ ⎪ η ⎟ − Φ⎜ − 10 ⎟ ≥ 0.99 P{η < 10} = P ⎨ < 10 ⎬ ≈ Φ⎜ 10 ⎜ ⎟ ⎜ ⎟ n n n n ⎝ ⎠ ⎝ ⎠ ⎪ ⎪ ⎪ ⎪ 3 ⎩ ⎭
1.5
{ξ
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当 y > 0 时， Fη (y ) = P{ η < y} = P
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y⎞ ⎛ξ y⎞ ⎛ξ ⎛y⎞ ⎛ y⎞ = P(ξ < y ) - P(ξ < -y) = P⎜ < ⎟ - P⎜ < - ⎟ = Φ⎜ ⎟ - Φ⎜ - ⎟ σ⎠ ⎝σ σ ⎠ ⎝σ ⎝σ ⎠ ⎝ σ ⎠
P (D1 ) = 0.05 ， P(D2 ) = 0.10 ， P(D3 ) = 0.25
5分
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P (E1 ) = P (E 2 ) = 0.10

2010版数学考研历年真题分类解析第一节自测题提示与答案

磋砣莫遗韶光老，人生惟有读书好。

六、第一节自测练习题提示与解答1.解 )1arcsin()(2x x -=ϕ的定义域为 22≤≤-x .2.解⎩⎨⎧+≤-=.0)(,2)(,0)(),(2)]([ x f x f x f x f x f g 由⎩⎨⎧≥-=0,,0,)(2x x x x x f 知,当0 x 时,,0)(2 x x f =此时, ;2)]([2+=x x f g当0≥x 时,,0)(≤-=x x f 此时, x x f g +=2)]([.故有 ⎩⎨⎧≥++=.0,2,0,2)]([2x x x x x f g 应选)(D .3.解 x x x 2sin 3553lim 2++∞→ 5623553lim 2=⎥⎦⎤⎢⎣⎡⋅++∞→x x x x .应填56. 4.解原式=)21(ln lim )ln(1lim222n n a a a a n n n n +++=⋅⋅⋅∞→∞→ .ln 21)1(21lim ln 2a nn n a n =+=∞→应填.ln 21a 5.解设,12111222nn n n n n n S n +++++++++= 则 211)1(2112121)1(21212222→+++=+++++≤≤+++++=+++←n n n n n n n S n n n n n n n n n n . 所以,应填.21 6.解因为极限)31ln(sin lim x n +∞→ 33lim =⋅∞→xx n , )11ln(sin lim x n +∞→ 11lim =⋅∞→xx n . 所以, 原极限为 213=-.应填2 7.解由)(x f 在0=x 连续知 xe x a ax n 12sin lim 2-+=∞→,而 22sin 0−−→−→x xx ,a x e x ax 2102−−→−-→,再由 a a 22+=解得2-=a .应填-2. 8.解不妨考虑0 x 的情形.)1(取)(1z k k x ∈=π,此时01sin 12=x x,且当k 充分变大时,x 无限趋近于)2(~)2(sin x x x x 33~)1ln(sin +x x 11~)1ln(sin +零.说明在点0的任何小的邻域内都有使函数为值零的点.故(B)不成立.)2(取)(12z k k x ∈+=ππ，此时22)2(1sin 1π+=x x x ，且当k 充分变大时,x 无限趋近于零.因此，只要取k 的值充分变大,就能使其函数大过任意给定的正数M ,自然也不会小于任意给定的正数,故(C)和(A)不成立.应选 (D).9.解设21,n y n x n n ==,则01−−→−=⋅∞→n n n n y x .这里n x 发散,而n y 收敛,故(A)不成立;又n x 无界,而n y 有界, 故(B)不成立. 另设n y nx n n ==,12,则01−−→−=⋅∞→n n n n y x .这里n x 有界,而n y 非无穷小, 故(C)不成立.应选 (D).10.解 xe x e x x xf x x x x 11232)(3ln 2ln -+-=-+=, 而当0→x 时, 2ln 12ln 12ln 2ln −−−−−→−-⋅-等价与x e x x x e ;，等价与3ln 13ln 13ln 3ln −−−−−→−-⋅-x e x x xe 所以 6ln 3ln 2ln 1lim 1lim )(lim 3ln 02ln 00=+=-+-=→→→xe x e x xf x x x x x .应选 (B). 11.解 nx x x x n x x x x e e x e e )1(lim lim tan 0tan 0-=--→→ 3012012012001lim lim tan lim 1sec lim tan lim -→-→-→-→→===-=-n x n x n x n x n x nxnx x nx x nx x x x x 301lim -→=n x nx 31.应选(C) 12.解由⇒=-∞→0)]()([lim x x g x ϕ当∞→x ,)(),(x x g ϕ的极限同时存在或同时不存在.)1( 若极限同时存在,且⇒=-∞→0)]()([lim x x g x ϕ)(lim )(lim x x g x x ϕ∞→∞→=.由夹逼准则知)(lim x f x ∞→存在,故(C)不成立.即使极限存在,也不一定为零, 故(A)不成立. )2( 若两极限都不存在,由夹逼准则的充分性知 )(lim x f x ∞→不一定存在,例如,设)0(21)1(21)(,1)(,)(>+=++=+==x x xx x x x f x x x g x x ϕ,则 ,01lim )]()([lim ==-∞→+∞→x x x g x x ϕ而)21(lim x x x ++∞→不存在,故(B)不成立. 应选(D) x x e x x ---tan ~)1(tan 3=n13.解 )1(由)(x f 连续知(A)不成立,因为当0 a 时,)(x f 在b a x )ln(-=处不连续.)2(由0)(lim 0lim b e a e a x bx x bx x ⇒∞=+⇒=+-∞→-∞→,故(B)和(C)不成立,应选(D)14.解设⎩⎨⎧-≥=≡,0,1,0,1)(,1)( x x x x f ϕ则 )1(1)]([2≡x ϕ无间断点,故(B)不成立.)2(1)]([)]([≡=x f x f ϕϕ无间断点,故(A),(C)不成立. 应选(D)15.解)(x f 为分段函数:⎪⎪⎩⎪⎪⎨⎧=-+≤.1,0,1,1,11,1,1,0)( x x x x x x f 1=x 为跳跃间断点, 应选(B)16.解先求)]([x g f .1 x 时,1)]([12)(2=⇒-=x g f x x g ,21 x ≤时,0)]([02)(=⇒-=x g f x x g ,2≥x 时,1)]([02)(=⇒≥-=x g f x x g⎪⎩⎪⎨⎧≥≤=.2,1,21,0,1,1)]([x x x x g f再求)]([x f g . 0 x 时202)]([0)(=-=⇒=x f g x f ,0≥x 时121)]([1)(-=-=⇒=x f g x f ,⎩⎨⎧≥-=.0,1,0,2)]([x x x f g 17.解 1sin cos lim 330=+-→xx bx a x 10)sin cos (lim 30=⇒=+-→a x bx a x . 当1=a 时,由 ⇒=-⇒=+-→→0cos 1lim 1sin cos 1lim 30330xbx x x bx x x 03→x00lim 21cos 1lim 222020=⇒==-→→b x x b x bx x x . 18.解 )cos 1(cos 1lim 0x x x x --→ xx x x 210cos 1lim 2⋅-→ 2121212cos 1cos 11lim 220=⨯⨯=⎥⎦⎤⎢⎣⎡-+=→x x x x . 19.解 x x x 2tan )1(lim 21π-→ t t t t 2cot )1)(2(lim 20π---→ πππ422lim 2tan 2lim 2020=+=+=→→t t t t t t t t . 20.解0lim →x 6321sin 2lim 2)]1sin 2(1ln[2lim 200)sin 2(e e e e e x x e x e x x x x x x x x ====+⨯-+-++→→.21.解⇒=--+-⇒=--+--∞→-∞→0)1(lim 0)1(lim 22x b a x x x x b ax x x x x 10)1(lim 22-=⇒=--+---∞→a x b a x x x x . 由题设知⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡+--+=++-=-∞→-∞→x x x x x x x x b x x 221111lim )1(lim 21)11(lim 212=--=-∞→x x x x 22.解设{}c b a m ,,m ax =,则n n n n n n n n n n m c m b m a m c b a )()()(lim lim ++=++∞→∞→,这里.1,,≤mc m b m a 若c b a ==,则m m c b a n n n n n n n ==++∞→∞→3lim lim , 若c b a ,,中有两项等于m ,则m m c b a n n n n n n n ==++∞→∞→2lim lim , x x 21~)cos 1(-1-=x t )11(21~1)11(122x x x x -⎥⎦⎤⎢⎣⎡--+⎥⎦⎤⎢⎣⎡+--=-∞→21111lim x x x x若c b a ,,中仅一项等于m ,则m m c b a n n n n n n n ==++∞→∞→1lim lim , 总之, =++∞→n n n n n c b a lim {}c b a ,,m ax 23.解考虑正项级数∑∞=1!10n n n ,因为 101010!)!1(1011 →=⋅+=++n n n u u n n n n ,根据正项的比值判敛法知该级数收敛,再由级数收敛的必要条件知.0!10lim =∞→n nn 24.解由题设知A xx f A x x f x x f x x x x 2)(lim )(lim 21)(121lim 303020=⇒==→→→. 0lim →x 再由a x x f ax x f bx b x =⇒=→→)(lim 1)(lim 00.取3=b ,并两式相等即得 A a 2=. 25.解用单调有界准则判定.)1( 因为01)1(111 n n n n n n x x x x x x ---=---=-+,所以{}n x 单减.)2(由011101 n n x x x --=⇒,所以{}n x 有下界.由)1(和)2(知{}n x 有极限.令a x nn −−→−∞→,得0112=-⇒--=a a a a ，解得)1(0舍去==a a .26.解因为301 x ,所以2323)3(1=-+≤-=+n n n n n x x x x x ,即{}n x 有上界.由 {}单增即n n n n n n n n n x x x x x x x x x ,1123313)3(11 ++⇒=---=,据单调有界准则知,{}n x 收敛. 令a x nn −−→−∞→,则由23)3(=⇒-=a a a a . 27.解 x xx x t x t x x x t x t x e e e x f x t x t sin sin sin sin sin sin lim )sin sin sin 1ln(sin sin lim )(===-⋅--+⋅-→→.)(z k k x ∈=π是间断点.其中,0=x 为第一类可去间断点,)0(≠=k k x π为第二类无穷间断点.28.解 ,4,2,0;1±±==x x 为间断点.)1(1,2)(lim ,2)(lim 11===-+→→x x f x f x x ππ为第一类中可去间断点,)2(21)4(211arctan lim 2arctan lim )(lim 021100-=-⋅=-==→-→→ππππx x x x f x x x x ,0=x 为第一类中可去间断点,)3( ,4,2,0±±=x 为第二类中无穷间断点.29.解令，则)()()(x f a x f x F -+=)()0()]0()([)()0()()0(],0[)(a F F f a f a F f a f F a C x F -=⇒⎩⎨⎧--=-=∈且，. 当,0)(,0)0(==a F F 时此时a x x ==,0均为其实根;当,0)0(时≠F )0(F)(a F 与异号,由零点定理知,使),,0(a ∈∃ξ 是即ξ）（a x f +）的（x f =实根.结论得证. 30.解即证 0)]()([)]()([],,[21=-+-∈∃ξξξf d f t f c f t b a 使.)]()([)]()([)(21x f d f t x f c f t x F -+-=,则且],,[)(d c C x F ∈⎩⎨⎧-=-=)].()([)()],()([)(12d f c f t d F c f d f t c F 若)()(d f c f =,则取d c ==ξξ或;若)()(d f c f ≠,则)()(d f c f 与异号,由零点定理知,使),,(d c ∈∃ξ0)]()([)]()([)(21=-+-=ξξξf d f t f c f t F .得证.31.解令 )1()0(]110[)()1()()(f f nx F n x f x f x F =-∈+-=，由，，则知, )1(),2(),1(),0(nn F n F n F F - 不能恒大于零或恒小于零(否则与)1()0(f f =矛盾）.若有n i n i F ==ξ则取,0)(;否则,设),1(,0)(,0)1(i i ni F n i F -∈∃-ξ则而 , 使0)1()()(=+-=n f f F ξξξ.即)1()(nf f +=ξξ.磋砣莫遗韶光老，人生惟有读书好。

2010年高考试题——数学理(全国卷1)解析版

zmax 1 2 ( 1) 3 .
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PA PB x1 x0 , y1 x1 x0 , y1 x12 2x1x0 x02 y12
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面 ACD1所成角相等 ,设 DO⊥ 平面 ACD 1，由等体积法得
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做本小题时极易忽视 a 的取值范围，而利用均值不等式求得 a+2b a 2 2 2 ,从而错选 a
A,这也是命题者的用苦良心之处.
【解析 1】因为 f(a)=f(b),所以|lga|=|lgb|,所以 a=b(舍去)，或b
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a

2010参考答案及详细解析

参考答案及详细解析第一部分数量关系．．［解析］本题为立方修正数列，，，，，，，（），所以选择选项。

．．［解析］本题为平方递推数列，，，，，（），最后计算直接用尾数判断即可，所以选择选项。

．．［解析］本题为递推数列。

×，×，×，×，×（）。

所以选择选项。

．．［解析］本题为递推数列，与年国考题第一个数字推理题规律相同。

从第三项开始，递推式为（）×。

或者用乘法拆分，分别为：×，×，×，×，×，下一项为×。

故选。

．．［解析］本题为递推数列，递推式为×（），≥。

故选。

．．［解析］本题为几何类题目。

因为正三角形和一个正六边形周长相等，又正三角形与正六边形的边的个数比为︰，所以其边长比为︰，正六边形可以分成个小正三角形，边长为的小正三角形面积：边长为的小正三角形面积︰。

所以正六边形面积：正三角形的面积×。

所以选。

．．［解析］原答案选是错的，应选，解析您自己想。

．．［解析］假设甲阅览室科技类书籍有本，文化类书籍有本，则乙阅读室科技类书籍有本，文化类书籍有本，由题意有：（）（），解出，则甲阅览室有科技类书籍本。

．．［解析］本题为工程类题目。

设总工程量为，则甲的效率是，乙的效率是，工作小时后，完成了。

第小时甲做了，完成了总工程量，剩余的由乙在第十四小时完成。

在第十四小时里，乙所用的时间是小时，所以总时间是小时。

．．［解析］本题为概率类题目。

假设甲、乙分别在分钟之内到达约会地点的情况如下图，则只有在阴影部分区域甲乙能够相遇，也就是求阴影部分面积的比例。

很容易看出，阴影部分的面积为。

．．【解析】为了使此人坐下后身边总有人，则原来长椅上除了首尾两个位置，中间的最大空位不能超过个，首尾两个位置的最大空位数不能超过个。

设第一个座位上有人，则每三个座位上有人，所以从第个座位到第个座位共有人，而最后边上的两个座位必须再坐一个人，才能保证此人坐下后身边总有人，所以至少有人。

2010年高考物理试题解析集

第
卷
注意项 1.答题前，考生在答题卡用直径 0.5 毫米的黑色墨水签字笔将自己的姓，准考证号填写清楚，然好条形码请认真准条形码的准考证号，姓和科目 2.第卷共 8 页，请用直径 0.5 毫米黑色墨水签字笔在答题卡各题的答题区域内作答，在试卷作．．．．．
答无效．．．
3.第 22． 6 卷共 13 题，共 174
234 90
Th ，质数少 4,电荷数少 2，说明为 α 衰
，中子转成质子
234 234 234 90
U→
234 91
234 90
Th + 24 He
234 92
234 90
Th →
234 91
Pa ,
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Th →
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0 Pa + −1 e
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U ，质子数
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为 β 衰，中子转成质子 91 Pa → 92U + −1 e 考点考查据原子的衰反应方程，应用质数电荷数的恒析解决问题 15．如右，轻簧端一质为 m 的木块 1 相连，端另一质为 M 2 相连，整个系统置于水放置的滑木坂，并处于静状态现将木方向突然抽出，设抽出的瞬间，木块 1 2 的速度大小别为 a1 速度大小为 g 则有
4 t ＝0 时刻振子的移 x ＝-0.1m t ＝ s 3
D．0.2m，8 s
时刻 x ＝
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答案 AD 解析在 t=
4 s 和 t=4s 两时刻振子的移相，第一种情况是时间差是周期 3 4 8 4 倍， 4 − = nT ，当 n=1 时 T = s 在 s 的半个周期内振子的移由负的最大 3 3 3

2010年版《公共基础考试辅导习题集》答案解释

2010年版《公共基础考试辅导习题集》答案解释2010 年版《公共基础考试辅导习题集》已由中信出版社出版发行，由于时间关系书中只附有答案，未及对每道题的解题思路及依据进行详细说明。

为使考生在完成练习题的同时，能够对知识点的掌握更加扎实，我们编写了“《公共基础考试辅导习题集》答案解释”，供大家做习题时对照参考。

请尚未在诚迅金融培训网站注册“获赠习题集答案解释注册申请表”的考生尽快注册，以便及时收到各科答案解释，并请大家互相转告。

为便于考生参考复习，我们编写了“《公共基础》公式与模型汇总”，附在“2010 年版《公共基础考试辅导习题集》答案解释”第2~4 页。

关于试题范围及深度，从以往考试的出题特点来看，可能有些与实务有关的试题，在教材及习题集中没有直接写出，但这类试题比例不是很大。

有条件的考生可结合实际工作进行思考，或就难点重点虚心请教相关部门的同事。

对于基础稍弱的考生：y 第一遍通读中国银行业从业人员资格认证办公室编写的银行业从业人员资格认证考试辅导教材《公共基础》（2010 年版，以下简称“教材”）时，要首先阅读一下教材的编写说明、目录和第295 页的考试大纲，了解整本教材的结构。

阅读教材正文时要注意抓住重点，并把握知识点之间的结构关系。

可以结合习题集每章前的知识要点提示（也称“大括号”），对大括号里每一个知识要点对照教材进行学习，遇到较难的概念或者定量较多的部分，可暂时“知难而跳”，跳过障碍继续学习，并在障碍处留下记号。

阅读教材过程中，对重点的概念、公式及模型建议用画圈及划线等方式标出。

每章阅读后，回顾一遍大括号的框架和内容。

这一遍要“把书读薄”。

y 第二遍读书时，可以结合习题来学习。

看一章书，做一章习题。

做完题后，无论对错，结合答案解释重新理解相关的知识点。

第一遍看书时遇到的障碍在这次看书做题时要结合相应习题进行理解。

对于做题时不确定的习题和做错的习题做个记号。

做完每章习题后，可以有重点地再看一遍该章教材，重点细读前期做记号的部分和做题时出错的知识点。

第一章课外练习参考答案

第一章课外练习参考答案第一章绪论一、单项选择题1、（D）2、（A）3、（C）4、（B）5、（A）6、（D）7、（B）二、多项选择题1、（ABCDE）2、（ABCE）3、（ACDE）4、（AC）三、辨析题1、对。

一切管理对象，对行政机关及其工作人员的管理行为都有服从的义务。

行政机关在执行任务时要做到依法行政。

有法可依，有法必依，执法必严，违法必究。

这种法制性集中体现了行政管理的权威性。

2、错。

前半句是对的。

行政管理活动是以国家名义进行的代表国家并以强制力为后盾的。

后半句不对，国家权力机关具有强烈的阶级性，行政机关也不例外，因此行政管理具有政治性。

四、简答题1、答：行政管理学主要包括以下内容：（1）总论。

包括第一、二、三章，主要是研究和了解：行政管理学是一门综合性、应用型的科学，行政环境和行政职能是行政活动的依据和基础。

（2）主体论。

包括第四、五、六、七章，主要是研究和了解：行政主体包括行政组织和机构、行政领导以及执行公务的工作人员。

（3）过程论。

包括第八、九、十、十一、十二章，主要研究和了解行政管理是一个由多个环节和有机部分组成及有效运作的过程。

它们主要有行政决策、行政执行、行政协调、行政信息、行政监督。

（4）保障论。

包括第十三、十四、十五、十六章，主要研究和了解行政管理过程能够运行顺利并达到预期效果，必须依据一系列保障手段。

它们主要有：财务行政、行政法治、行政方法、行政道德。

（5）目的论。

包括第十七、十八章，主要研究和了解提高行政效率是行政管理的出发点和归缩，而行政改革是提高行政效率的必由之路。

2、答：（1）是西方自由资本主义时期结束后，随着垄断资本主义的发展，政府职能发展的结果；（2）当时各国政府机构普遍存在办事效率低下、腐败现象严重、官僚主义盛行等弊端，这是行政管理学和政府改革产生的契机。

五、论述题答：科技和管理是人类社会发展的两大助推器，我国作为发展中国家，要想在政治、经济、综合国力等各方面取得比较优势，就必须充分发挥政府对社会、经济发展的主导作用。

应用数值分析第四版(张明文世鹏)第一第二章答案

第一章习题解答3、设精确数ａ>0，x 是ａ的近似值，x 的相对误差限是0.2，求㏑x 的相对误差限。

解：设=()u f x ，()()()()()()||||||||||()||()||||()||()||||r r rx e u df x e x df x e x e u u dx u dx u x df x x df x x e x x dx u dx u δ=≈==≤()||10.2(())||()||ln ln ln r r r r df x x x x f x x x dx u x x x xδδδδ==⋅⋅==4、长方体的长宽高分别为50cm ，20cm 和10cm ，试求测量误差满足什么条件时其表面积的误差不超过1cm 2。

解：设2()S xy yz zx =++{}[]{}(,,)(,,)(,,)()||()||()||()(,,)(,,)(,,)||||||max (),(),()2()2()2()max (),(),()1S x y z S x y z S x y z e S e x e y e z x y zS x y z S x y z S x y z e x e y e z x y z y z z x x y e x e y e z ∂∂∂≤++∂∂∂⎛⎫∂∂∂≤++ ⎪∂∂∂⎝⎭=+++++<{}[]11max (),(),()2()2()2()4()110.0031254(502010)320e x e y e z y z z x x y x y z <=+++++++===++测量误差小于0.00625时其表面积的误差不超过1cm 2。

7、计算61)1.414≈。

利用以下四种计算格式，试问哪一种算法误差最小。

（1（2）3(3- （3（4）99- 解：计算各项的条件数'()(())||()xf x cond f x f x =111.41461(),(())| 3.5147(1)x f x c o n d f x x ===+ 3221.414()(32),(())|49.3256x f x x c o n d f x ==-=331.41431(),(())| 1.4557(32)xf x c o n d f x x ===+ 441.414()9970,(())|4949x f x x c o n d f x ==-= 由计算知，第三种算法误差最小。

光学教程第四版姚启均第章课后习题答案

光学教程第四版姚启均第章课后习题答案IMB standardization office【IMB 5AB- IMBK 08- IMB 2C】1-1解：∵λd r y yy jj 01=-=+∆∴ 409.010*******.018081≈⨯⨯=∆-y cm573.010*******.018082≈⨯⨯=∆-y cm 又∵λd r j y 0= ， 2=j ∴81210)50007000(022.01802)(-⨯-⨯⨯=-=∆λλd r j y≈0.327 cmor: 328.02212≈∆-∆=∆y y y cm1-2 解： ∵ .0⎪⎭⎫ ⎝⎛=∆λd r y λd r j y 0= j=0,1 ∴ (1)cm 08.0104.604.050)01(5=⨯⨯⨯-=∆-y (2)4104.650001.004.020225ππλππϕ=⨯⨯⨯⨯=⋅=⋅=∆-r dy j(3)2cos412221ϕϕ-=-A I 214A I =-412πϕϕ=-1-3解：∵d n d nd )1(-=-=δ)22(πδλπϕ⋅==∆j而：λδj =∴cm m n j d 46710610615.110651---⨯=⨯=-⨯⨯=-=λ 1-4 解: cmd r y 125.010500002.0508=⨯⨯==∆-λ1-5 解:λθsin 2r lr y +=∆ 1-6解:(1) mmmm d r y 19.01875.010*********7≈=⨯⨯⨯==∆-λ [利用2,220λδπδλπϕ-=⋅==∆y r d j 亦可导出同样结果。

] （2）图即：离屏中央1.16mm 的上方的2.29mm 范围内，可见12条暗纹。

（亮纹之间夹的是暗纹）1-7．解：,1,02)12(sin 2122122=+=-j j i n n h 二级λ1-8.解：2)12(cos 2200λ+=j i n d 1-9.解：薄膜干涉中，每一条级的宽度所对应的空气劈的厚度的变化量为：若认为薄膜玻璃片的厚度可以略去不计的情况下，则可认为Or ：而厚度h 所对应的斜面上包含的条纹数为：故玻璃片上单位长度的条纹数为：1-10．解：∵对于空气劈，当光垂直照射时，有2)21(0λ+=j d 1-11．解：∵是正射，1-12．解：1-13．解：1-14．解: (1)中心亮斑的级别由下式决定：（）所以，第j 个亮环的角半径满足于是：第1级暗环的角半径θ为(对于第1级暗环，每部分j=0时亮斑)（2）解之：1-15．解：亦即：于是：8 1-16．解： j=1、2、3……即：而：即：而1-17．解：又对于暗环来说，有18,解：光源和双棱镜的性质相当于虚光源由近似条件和几何关系：得：而2A+所以：A=（rad）又因为：为插入肥皂膜前，相长干涉的条件为：插入肥皂膜后，相长干涉的条件为：所以：故：1-19，（1）图（b）中的透镜由A，B两部分胶合而成，这两部分的主轴都不在该光学系统的中心轴线上，A部分的主轴OA在系统中心线下0.5cm处，B部分的主轴OB则在中心线上方0.5cm处，分别为A，B部分透镜的焦点。

2010全国英语专业四级考试真题及试题详解

TEST FOR ENGLISH MAJORS(2010)-GRADE FOUR-TIME LIMIT:135MINPART I DICTATION [15MIN]Listen to the following passage.Altogether the passage will be read to you four times.During the first reading,which will be done at normal speed,listen and try to understand the meaning.For the second and third readings,the passage will be read sentence by sentence,or phrase by phrase,with intervals of15seconds.The last reading will be done at normal speed again and during this time you should check your work.You will then be given2minutes to check through your work once more.Please write the whole passage on ANSWER SHEET ONE.PART II LISTENING COMPREHENSION MINI]][20MINIIn Sections A,B and C you will hear everything ONCE ONLY Listen carefully and then answer the questions that follow.Mark the correct answer to each question on ANSWER SHEET TWO.SECTIONA A CONVERSATIONSSECTIONIn this section you will hear several conversations.Listen to the conversations carefully and then answer the questions that follow.Questions1to3are based on the following conversation.At the end of the conversation,you will be given15seconds to answer the questions.Now,listen to the conversation.1.The following details have been checked during the conversation EXCEPTA.number of travelers.B.number of tour days.C.flight details.D.room services.2.What is included in the price?A.Air tickets and local transport.B.Local transport and meals.C.Air tickets,local transport and breakfast.D.Air tickets,local transport and all meals.Questions4to7are based on the following conversation.At the end of'the conversation,you will be given20seconds to answer the questions.Now,listen to the conversation.4.Which of the following details is CORRECT?A.Mark knows the exact number of airport buses.B.Mark knows the exact number of delegates'spouse.C.Mark doesn't know the exact number of delegates yet.D.Mark doesn't know the number of guest speakers.5.What does Linda want to know?A.The arrival time of guest speakers.B.The departure time of guest speakers.C.The type of transport for guest speakers.D.The number of guest speakers.6.How many performances have been planned tbr the conference?A.One.B.Two.C.Three.D.Not mentioned.7.Who will pay for the piano performance?A.Pan-Pacific Tours.B.Johnson&Sons Events.C.Conference delegates.D.An airline company.Questions8to10are based on the following conversation.At the end of the conversation,you will be given15seconds to answer the questions.Now listen to the conversation.8.What is NOT missing in Mary's briefcase?A.Her cheque book.B.Her papers for work.C.Her laptop.D.Her appointment book.9.Where was Mary the whole morning?A.At the police station.B.At a meeting.C.In her client's office.D.In the restaurant.10.Why was Mary sure that the briefcase was hers in the end?A.The papers inside had the company's name.B.The briefcase was found in the restaurant.C.The restaurant manager telephoned James.D.The cheque book inside bore her name.SECTION B PASSAGESIn this section,you will hear several passages.Listen to the passages carefully and then answer the questions that follow.Questions11to13are based on the following passage.At the end ofWork.C.the School of Business.D.the Arts and Sciences program.12.What is the cost of undergraduate tuition?A.Twenty thousand dollars.B.Thirty thousand dollars.C.Twenty-seven thousand dollars.D.Thirty-eight thousand dollars.13.International students can receive all the following types of financial assistance EXCEPTA.federal loans.B.private loans.C.scholarships.D. monthly payment plans.Questions14to17are based on the following passage.At the end of the passage,you will be given20seconds to answer the questions.Now,listen to the passage.14.According to the passage,mothers in____spend more time looking after children.A.FranceB.AmericaC.DenmarkD.Australia15.Which of the following activities would Australian fathers traditionally participate in?A.Feeding and playing with children.B.Feeding and bathing children.C.Taking children to the park and to school.D.Taking children to watch sports events.16.According to the study,the"new man"likes toA.spend more time at work.B.spend more time with children.C.spend time drinking after work.D.spend time on his computer.17.It is suggested in the passage that the"new man"might be less acceptable inA.France.B.Britain.C.Australia.D.Denmark.Questions18to20are based on the following passage.At the end of the passage,you will be given15seconds to answer the questions.Now,listen to the passage.18.The services of the new partnership are provided mainly toA.mothers of infected babies.B.infected children and women.C.infected children in cities.D.infected women in cities.19.Which of the following details about Family Health International is INCORRECT?A.It is a nonprofit organization.B.It provides public health services.C.It carries out research on public health.D.It has worked in fiveD.the methods used to fight AIDS.SECTION C NEWS BROADCASTIn this section,you will hear several news items.Listen to them carefully and then answer the questions that follow.Questions21and22are based on the following news.At the end of the news item,you will be given10seconds to answer the questions.Now，listen to the news.21.According to the news,the victim wasA.a17-year-old girl.B.a15-year-old boy.C.a23-year-old woman.D.an l8-year-old man.22.We learn from the news that the suspects were arrestedA.one month later.B.two months later.C.immediately.D.two weeks later.Questions23and24are based on the following news.At the end of the news item,you will be given10seconds to answer the questions.Now,listen to the news.23.The Iraqi par liament can vote on the security agreement only afterA.all parties have agreed on it.B.the US troops have pulled out.C.the cabinet has reviewed it.D.the lawmakers have returned from Mecca.24.According to the news,the US troops are expected to completely pull out byA.mid-2009.B.the end of2009.C.mid-2011.D.the end of2011.Questions25and26are based on the following news.At the end of the news item,you will be given10seconds to answer the questions.Now,listen to the news.25.The following are involved in the operations to rescue the children in Honduras EXCEPTA.the police.B.the district attorney.C.the prison authorities.D.Institute of Childhood and Family.26.What punishment would parents face if they allowed their children to beg?A.To be imprisoned and fined.B.To have their children taken away.C.To be handed over to the authorities.D.None.A.Coastlines in Italy.B.Public use of the beach.C.Swimming and bathing.D.Private bathing clubs.Question28is based on the following news.At the end of the news item,you will be given5seconds to answer the question.Now,listen to the news.28.Which of the following is NOT mentioned in the news?A.The airport was shut down for Friday.B.There was a road accident involving two buses.C.Local shops were closed earlier than usual.D.Bus service was stopped for Friday.Questions29and30are based on the following news.At the end of the news item,you will be given10seconds to answer the questions.Now,listen to the news.29.How many people were rescued from the apartment building?A.17.B.24.C.21.D.41.30.Which of the following details in the news is CORRECT?A.The rescue operation involved many people.B.The cause of the explosions has been determined.C.Rescue efforts were stopped on Thursday.D.The explosions didn't destroy the building.PART III CLOZE [15MIN]Decide which of the choices given below would best complete the passage it"inserted in the corresponding blanks.Mark the best choice for each blank on ANSWER SHEET TWO.How men first learned to invent words is unknown;(31)____,the origin of language is a mystery.All we really know is that men,unlike animals,somehow invented certain(32)____to express thoughts and feelings,actions and things,(33)____they could communicate with each other;and that later they agreed(34)____certain signs,called letters,which could be(35)____to represent those sounds,and which could be(36) _____.Those sounds,whether spoken,(37)_____written in letters,we call words.The power of words,then,lies in their(38)____the things they bringexpress these thoughts in words which appeal(44)____to our minds and emotions.This(45)_____and telling use of words is what we call(46)____ style.Above all,the real poet is a master of(47)____.He can convey his meaning in words which sing like music,and which(48)_____their position and association can(49)____men to tears.We should,therefore, learn to choose our words carefully and use them accurately,or they will(50) ____our speech or writing silly and vulgar.(31)A.in addition B.in other words C.in a word D.in summary(32)A.sounds B.gestures C.signs D.movements(33)A.such that B.as that C.so that D.in that(34)A.in B.with C.of D.upon(35)A.spelt bined C.written D copied(36)A.written down B.handed down C.rememberedD.observed(37)A.and B.yet C.also D.or(38)A.functions B.associations C.roles D.links(39)A.filled B.full C.live D.active(40)A.but B.or C.yet D.and(41)A.reappear B.recall C.remember D.recollect(42)A.read and think B.read and recall C.read and learn D.read and recite(43)A.raises B.increases C.improves D.emerges(44)A.intensively B.extensively C.broadly D.powerfully(45)A.charming B.academic C.conventional mon(46)A.written B.spoken C.literary D.dramatic(47)A.signs B.words C.style D.sound(48)A.in B.on C.over D.by(49)A.move B.engage C.make D.force(50)A.transform B.change C.make D.convertPART IV GRAMMAR&VOCABULARY [15MIN]There are thirty sentences in this section.Beneath each sentence there are four words or phrases marked A,B,C and D.Choose oneC.For all his efforts,he didn't get an A.D.Her eyes were red from excessive reading.52.Nancy's gone to work but her car's still there.She____by bus.A.must have goneB.should have goneC.ought to have goneD.could have gone53.He feels that he is not yet____to travel abroad.A.too strongB.enough strongC.so strongD.strong enough54.After____seemed an endless wait,it was his turn to enter the personnel manager's office.A.thatB.itC.whatD.there55.Fool____Jerry is,he could not have done such a thing.A.whoB.asC.likeD.that56.Which of the following sentences is INCORRECT?A.They each have two tickets.B.They cost twenty yuan each.C.Each they have bought the same book.D.They were given two magazines each.57.She seldom goes to the theatre,_____?A.doesn't sheB.does sheC.would sheD.wouldn't she58.Dr Johnson is head of the department,____an expert in translation.A.orB.eitherC.butD.and59.When one has good health,_____should feel fortunate.A.youB.theyC.heD.we60.It is necessary that he____the assignment without delay.A.hand inB.hands inC.must hand inD.has to hand in61.In the sentence"It's no use waiting for her",the italicized phrase is ____.A.the objectB.an adverbialC.a complementD.the subject62.Which of the following sentences is INCORRECT?A.All his lectures are very interesting.B.Half their savings were gone.C.Many his friends came to the party.D.Both his sisters are nurses.63.Which of the following sentences has an object complement?A.The directors appointed John manager.B.I gave Mary a Christmas present.C.You have done Peter a favour.D.She is teaching children English.Not to be tall66.Due to personality_____,the two colleagues never got on well in work.A.contradictionB.conflictC.confrontationpetition67.During the summer vacation,kids are often seen hanging_____in the streets.A.aboutB.onC.overD.out68.There were150____at the international conference this summer.A.spectatorsB.viewersC.participantsD.onlookers69.School started on a____cold day in February.A.severeB.bitterC.suchD.frozen70.In the face of unexpected difficulties,he demonstrated a talent for quick, ____action.A.determiningB.defensiveC.demandingD.decisive71.The team has been working overtime on the research project____.telyB.just nowteD.long ago72.Because of the economic crisis,industrial output in the region remainedA.motionlessB.inactiveC.stagnantD. immobile73.The police had difficulty in____the fans from rushing on to the stage totake photos with the singer.A.limitingB.restrainingC.confiningD.restricting74.Joan is in the dorm,putting the final____to her speech.A.detailsB.remarksmentsD.touches75.His_____in gambling has eventually brought about his ruin.A.indulgenceB.habitC.actionD.engagement76.The teacher told the students to stay in the classroom and they did _____.A.absolutelyB.accidentallyC.accordinglyD.accurately77.You can actually see the deer at close range while driving through thatarea.The italicized phrase means_____.A.clearlyB.very nearC.quicklyD.very hard78.He listened hard but still couldn't what they were talking about.A.make overB.make upC.make uponD.make out79.For the advertised position,the company offers a(n)salary and benefits package.A.generousB.plentifulC.abundantD.sufficient[25MIN]In this section there are four passages followed by questions or unfinished statements,each with four suggested answers marked A,B,C and D.Choose the one that you think is the best answer.Mark your answers on ANSWER SHEET TWO.TEXT AWhat is the nature of the scientific attitude,the attitude of the man or woman who studies and applies physics,biology,chemistry,geology, engineering,medicine or any other science?We all know that science plays an important role in the societies in which we live.Many people believe, however,that our progress depends on two different aspects of science.The first of these is the application of the machines,products and systems of applied knowledge that scientists and technologists develop.Through technology,science improves the structure of society and helps man to gain increasing control over his environment.The second aspect is the application by all members of society of the special methods of thought and action that scientists use in their work.What are these special methods of thinking and acting?First of all,it seems that a successful scientist is full of curiosity-he wants to find out how and why the universe works.He usually directs his attention towards problems which he notices have no satisfactory explanation,and his curiosity makes him look for underlying relationships even if the data available seem to be unconnected.Moreover,he thinks he can improve the existing conditions and enjoys trying to solve the problems which this involves.He is a good observer,accurate,patient and objective and applies logical thought to the observations he makes.He utilizes the facts he observes to the fullest extent.For example,trained observers obtain a very large amount of information about a star mainly from the accurate analysis of the simple lines that appear in a spectrum.He is skeptical-he does not accept statements which are not based on the most complete evidence available-and therefore rejects authority as the sole basis for truth.Scientists always check statements and make experiments carefully and objectively to verify them.Furthermore,he is not only critical of the work of others,but also of his own,since he knows that man is the least reliable of scientific instruments and that a number of factors tend to disturb objective investigation.scientist or technologist thinks and acts.81.Many people believe that science helps society to progress throughA.applied knowledge.B.more than one aspect.C.technology only.D.the use of machines.82.Which of the following statements is INCORRECT about curiosity?A.It gives the scientist confidence and pleasure in work.B.It gives rise to interest in problems that are unexplained.C.It leads to efforts to investigate potential connections.D.It encourages the scientist to look for new ways of acting.83.According to the passage,a successful scientist would notA.easily believe in unchecked statements.B.easily criticize others'research work.C.always use his imagination in work.D.always use evidence from observation.84.What does the passage mainly discuss?A.Application of technology.B.Progress in modem society.C.Scientists'ways of thinking and acting.D.How to become a successful scientist.85.What is the author's attitude towards the topic?A.Critical.B.Objective.C.Biased.D.Unclear.TEXT BOver the past several decades,the U.S.,Canada,and Europe have received a great deal of media and even research attention over unusual phenomena and unsolved mysteries.These include UFOs as well as sightings and encounters with"nonhuman creatures"such as Bigfoot and the Loch Ness monster.Only recently has Latin America begun to receive some attention as well.Although the mysteries of the Aztec,Mayan,and Inca civilizations have been known for centuries,now the public is also becoming aware of unusual,paranormal phenomena in countries such as Peru.The Nazca"lines"of Peru were discovered in the1930s.These lines are deeply carved into a flat,stony plain,and form about300intricate pictures of animals such as birds,a monkey,and a lizard.Seen at ground level,the designs are a jumbled senseless mess.The images are so large that they can only be viewed at a height of1,000feet-meaning from an aircraft. Yet there were no aircraft in300B.C.,when it is judged the designs werecommunity did not take long to scoff at and abandon von Daniken's theory. Over the years several other theories have been put forth,but none has been accepted by the scientific community.Today there is a new and heightened interest in the Nazca lines.It is a direct result of the creation of the Internet.Currently there are over60sites dedicated to this mystery from Latin America's past,and even respected scientists have joined the discussion through e-mail and chat rooms.Will the Internet help explain these unsolved mysteries?Perhaps it is a step in the right direction.86.Which of the following statements is INCORRECT?tin America has long received attention for unusual phenomena.B.Public attention is now directed towards countries like Peru.C.Public interest usually focuses on North America and Europe.D.Some ancient civilizations have unsolved mysteries.87.According to the passage,the Nazca lines were foundA.in mountains.B.in stones.C.on animals.D.on a plain.88.We can infer from the passage that the higher the lines are seen,the ____the images they present.A.smallerrgerC.clearerD.brighter89.There has been increasing interest in the Nazca lines mainly because ofA.the participation of scientists.B.the emergence of the lnternet.C.the birth of new theories.D.the interest in the Internet.90.The author is____about the role of the Internet in solving mysteries.A.cautiousB.pessimisticC.uncertainD.optimisticTEXT CGraduation speeches are a bit like wedding toasts.A few are memorable.The rest tend to trigger such thoughts as,"Why did I wear such uncomfortable shoes?"But graduation speeches are less about the message than the messenger. Every year a few colleges and universities in the US attract attention because they've managed to book high-profile speakers.And,every year,the media report some of these speakers'wise remarks.Last month,the following words of wisdom were spread:"You really haven't completed the circle of success unless you can help somebody else move forward."(Oprah Winfrey,Duke University)."There is no way to stop change;change will come.Go out and give us a future worthy of the world we all wish to create together."(Hillary Clinton, New York University)."'This really is your moment.History is yours to bend."(Joe Biden, Wake Forest University).pay for living...it is the true measure,the only measure of success'."Calls to service have a long,rich tradition in these speeches.However, it is possible for a graduation speech to go beyond cliche and say something truly compelling.The late writer David Foster Wallace's2005graduation speech at Kenyon College in Ohio talked about how to truly care about other people.It gained something of a cult after it was widely circulated on the Internet.Apple Computer CEO Steve Jobs'address at Stanford University that year,in which he talked about death,is also considered one of the best in recent memory.But when you're sitting in the hot sun,fidgety and freaked out,do you really want to be lectured about the big stuff?Isn't that like trying to maintain a smile at your wedding reception while some relative gives a toast that amounts to"marriage is hard work"?You know he's right;you just don't want to think about it at that particular moment.In fact,as is the case in many major life moments,you can't really manage to think beyond the blisters your new shoes are causing.That may seem anticlimactic.But it also gets to the heart of one of life's greatest,saddest truths:that our most"memorable"occasions may elicit the fewest memories.It's probably not something most graduation speakers would say,but it's one of the first lessons of growing up.91.According to the passage,most graduation speeches tend to recall____ memories.A.greatB.trivialC.unforgettableD.unimaginative92."But graduation speeches are less about the message than the messenger"is explainedA.in the final paragraph.B.in the last but one paragraph.C.in the first paragraph.D.in the same paragraph.93.The graduation speeches mentioned in the passage are related to thefollowing themes EXCEPTA.death.B.success.C.service.D.generosity.94.It is implied in the passage that at great moments people fail toA.remain clear-headed.B.keep good mannersC.remember others'words.D.recollect specific details.95.What is"one of the first lessons of growing up"?A.Attending a graduation ceremony.B.Listening to graduation speeches.C.Forgetting details of memorable events.D.Meeting high-profile graduation speakers.TEXT DCultural rules determine every aspect of food consumption.Who eatswhen relatives may come,and cocktail parties for acquaintances.The food served symbolizes the occasion and reflects who is present.For example, only snacks are served at a cocktail party.It would be inappropriate to serve a steak or hamburgers.The distinctions among cocktails,regular meals,and special dinners mark the social boundaries between those guests who are invited for drinks,those who are invited to dinner,and those who come to a family meal.In this example,the type of food symbolizes the category of guest and with whom it is eaten.In some New Guinea societies,the nuclear family is not the unit that eats together.The men take their meals in a men's house,separately from their wives and children.Women prepare and eat their food in their own houses and take the husband's portion to the men's house.The women eat with their children in their own houses.This pattern is also widespread among Near Eastern societies.Eating is a metaphor that is sometimes used to signify marriage.In many New Guinea societies,like that of the Lesu on the island of New Ireland in the Pacific and that of the Trobriand Islanders,marriage is symbolized by the couple's eating together for the first time.Eating symbolizes their new status as a married couple.In U.S.society,it is just the reverse.A couple may go out to dinner on a first date.Other cultural rules have to do with taboos against eating certain things. In some societies,members of a clan,a type of kin(family)group,are not allowed to eat the animal or bird that is their totemic ancestor.Since they believe themselves to be descended from that ancestor,it would be like eating that ancestor or eating themselves.There is also an association between food prohibitions and rank,which is found in its most extreme form in the caste system of India.A caste system consists of ranked groups,each with a different economic specialization.In India,there is an association between caste and the idea of pollution.Members of highly ranked groups can be polluted by coming into contact with the bodily secretions,particularly saliva,of individuals of lower-ranked castes.Because of the fear of pollution,Brahmans and other high-ranked individuals will not share food with,no96.According to the passage,the English make clear distinctions betweenA.people who eat together.B.the kinds of food served.C.snacks and hamburgers.D.family members and guests.97.According to the passage,who will NOT eat together?A.The English.B.Americans on their first date.C.marital status.D.family ties.99.The last paragraph suggests that in India____decides how people eat.A.pollutionB.foodC.cultureD.social status100.Which of the following can best serve as the topic of the passage?A.Different kinds of food in the world.B.Relations between food and social units.C.Symbolic meanings of food consumption.D.Culture and manners of eating.PART VI WRITING [45MIN]SECTION A COMPOSITION [35MIN]It was recently reported in a newspaper that six students who shared a dorm at a local university hired a cleaner to do laundry and cleaning once a week.And each of them paid her60yuan a month.This has led to a heated debate as to whether college students should hire cleaners.Write on ANSWER SHEET THREE a composition of about200 words on the following topic:Should College Students Hire Cleaners?You are to write in three parts.In the first part,state clearly what your view is.In the second part,support your view with appropriate reasons.In the last part,bring what you have written to a natural conclusion or a summary.Marks will be awarded for content,organization,grammar and appropriateness.Failure to follow the instructions may result in a loss of marks.SECTION B NOTE-WRITING [10MIN]Write on ANSWER SHEET THREE a note of about50-60words based on the following situation:Your good friend,John,is thinking of organizing an end-of-the-term party.Write him a note telling him that you like his idea and offer to help him.You have to be specific about how you can help him.Marks will be awarded for content,organization,grammar and appropriateness.。

弹性力学简明教程-第四版习题详解

弹性力学简明教程（第四版）习题解答第一章【1-1】试举例说明什么是均匀的各向异性体，什么是非均匀的各向同性体？【分析】均匀的各项异形体就是满足均匀性假定，但不满足各向同性假定；非均匀的各向异性体，就是不满足均匀性假定，但满足各向同性假定。

【解答】均匀的各项异形体如：竹材，木材。

非均匀的各向同性体如：混凝土。

【1-2】一般的混凝土构件和钢筋混凝土构件能否作为理想弹性体？一般的岩质地基和土质地基能否作为理想弹性体？【分析】能否作为理想弹性体，要判定能否满足四个假定：连续性，完全弹性，均匀性，各向同性假定。

【解答】一般的混凝土构件和土质地基可以作为理想弹性体；一般的钢筋混凝土构件和岩质地基不可以作为理想弹性体。

【1-3】五个基本假定在建立弹性力学基本方程时有什么作用？【解答】（1）连续性假定：假定物体是连续的，也就是假定整个物体的体积都被组成这个物体的介质所填满，不留下任何空隙。

引用这一假定后，物体的应力、形变和位移等物理量就可以看成是连续的。

因此，建立弹性力学的基本方程时就可以用坐标的连续函数来表示他们的变化规律。

完全弹性假定：假定物体是完全弹性的，即物体在对应形变的外力被去除后，能够完全恢复原型而无任何形变。

这一假定，还包含形变与引起形变的应力成正比的涵义，亦即两者之间是成线性关系的，即引用这一假定后，应力与形变服从胡克定律，从而使物理方程成为线性的方程，其弹性常数不随应力或形变的大小而变。

均匀性假定：假定物体是均匀的，即整个物体是由同一材料组成的，引用这一假定后整个物体的所有各部分才具有相同的弹性，所研究物体的内部各质点的物理性质都是相同的，因而物体的弹性常数不随位置坐标而变化。

各向同性假定：假定物体是各向同性的，即物体的弹性在所有各个方向都相同，引用此假定后，物体的弹性常数不随方向而变。

小变形假定：假定位移和变形是微小的。

亦即，假定物体受力以后整个物体所有各点的位移都远远小于物体原来的尺寸，而且应变和转角都远小于1。

2010年普通高等学校招生全国统一考试理科数学(新课标)全解全析版

2010年普通高等学校招生全国统一考试（宁夏卷）数学（理工农医类）第I 卷一、选择题：（本大题共12题，每小题5分，在每小题给出的四个选项中，有一项是符合题目要求的。

1、已知集合A={|||2,}x x x R ≤∈，B={4,}x x Z ≤∈，则A ∩B =（）A ．（0，2）B ．[0，2]C ．{0，2}D ．{0，1，2} 【解析】选择D 。

因为A={|22}x x -≤≤，{0,1,2,3,,16}B =，所以A ∩B ={0，1，2}。

2、已知复数z =z 是z 的共轭复数，则z z ⋅=（） A ．14B ．12C ．1D ．2 【解析】选择A 。

解法1：因为12z ==-112)28i =-=-14i =，所以z z ⋅=221111()()(()4444i i =+=。

解法2：因为z z ⋅=2241||||164z ====。

3、曲线2xy x =+在点（－1，－1）处的切线方程为（） A ．21y x =+B ．21y x =-C ．23y x =--D ．22y x =-- 【解析】选择A 。

因为22(2)2'(2)(2)x x y x x +-==++，所以1'|2x k y =-==，因此切线方程为12(1)y x +=+，化简得21y x =+。

4、如图，质点P 在半径为2的圆周上逆时针运动，其初始位置为0P ，角速度为1，那么点P 到x 轴距离d 关于时间t 的函数图像大致为（）【解析】选择（C ）。

xyP 0OP解法1：显然，当t=0时，d=2，排除（A ）、（D ）；当t=4π时，d=0，排除（B ），因此选择（C ）。

解法2：显然，当0t =时，由已知得2d =，故排除A 、D ，又因为质点是按逆时针方向转动，随时间t 的变化质点P 到x 轴的距离d 先减小，再排除B ，即得C 。

解法3：根据已知条件得2,1,4A πωϕ===-，再结合已知得质点P 到x 轴的距离d 关于时间t 的函数为2sin()4d t π=-，画图得C 。

2010年全国各地高考数学试题全解全析大汇总全国卷i文数课标

i 。

2010年普通高等学校招生全国统一考试参考公式：样本数据捲，X 2 | I ( X n 的标准差文科数学锥体体积公式4I —2 2一(xi—X )+(X2—X )+(||+(Xn _X ) n -其中S 为底面面积，h 为高球的表面积，体积公式S =4二 R 2,V =?二 R 34其中R 为球的半径第I 卷、选择题：本大题共 12小题，每小题5分，在每小题给出的四个选项中，只有一项是符合题目要求的。

(1 )已知集合 A=|x x W2,x^R,B = x|仮兰 4,x^Z|，贝U A 「| B =解析：由已知得 A={x —2^x z2},B={0,1,IH,16}，所以 B 二{0,1,2}.解析：由已知得 b =(3,18) -2a =(3,18) -(8,6) = (-5,12)，所以 cos ：：： a,b -4 (-5) 3 12 16(A)(0,2) 【答案】D(B)[0,2] (C){0,2] (D){0,1,2}其中X 为样本平均数柱体体积公式V =Sh其中S 为底面面积，h 为高(2) a , b 为平面向量，已知 a= ( 4, 3),8 (B)-658(A )65【答案】C2a+b= (3, 18),贝U a , b 夹角的余弦值等于16 16 (C )(D )-65655 13 (3 )已知复数65.3 i"(1- *31)2 '1 (A)-4 【答案】D(B )(C ) 1 (D ) 273 +i解析：z 二3 i-3 i( 一3 i)(-22.3i)(1「3i)21 -2 3i -3-2-2.3i(-2-2、3)(-2 - 23i)(4)曲线y n x? —2x ・1在点(1,0)处的切线方程为(A ) y =x -1(C ) y =2x -2 (D ) y - -2x 2【答案】A2解析：yQ3x —2，所以k = y'x^=1,所以选A .(5)中心在远点，焦点在 x 轴上的双曲线的一条渐近线经过点(4,2),则它的离心率为速度为1,那么点p 到x 轴距离d 关于时间t 的函数图像大致为【答案】C(A ) .6(B )(C ),6 2(D )【答案】D-2 1解析：易知一条渐近线的斜率为k 二二=-丄4 2，故 (6)如图，质点p 在半径为2的圆周上逆时针运动,其初始位置为Po ( ， —V 2)，角a 2b 2解析：显然，当t =0时，由已知得d =$2，故排除A 、D ,又因为质点是按逆时针方向转动，随时间t 的变化质点P 到x 轴的距离d 先减小，再排除B ,即得C. 另解：根据已知条件得 A=2,・=1,，再结合已知得质点 P 到x 轴的距离d 关于时4间t 的函数为d = 2sin(t _—)，画图得4【答案】B(9)设偶函数 f(x)满足 f(x)=2x -4 (x 启0),则｛x f(x-2户。

2010年高考试题——语文(全国卷1)精解版

2010年高校招生全国统一考试语文（全国1卷）语文解析教师：山西袁俊杰本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。

第Ⅰ卷1至4页，第Ⅱ卷5至8页。

考试结束后，将本试卷和答题卡一并交回。

第Ⅰ卷一、（12分，每小题3分）1．下列词语中加点的字，读音全都正确的一组是A．行．伍（háng）名宿．（sù）恶贯满盈．（yíng）厉兵秣．马（mù）B．倾轧．（zhá）不啻．（chì）补苴罅．漏（xia）荆钗．布裙（chāi）C．巨擘．（bò）河蚌．（bàng）得不偿．失（cháng）莘．莘学子（shēn）D．解剖．（pāo）羁．绊（jī）火中取栗．（lì）感慨系．之（xì）答案：C。

本试题主要考察学生的汉字字音，应该说不难，只要能够认真复习课本，就答出本题。

A中的“厉兵秣马”中的“秣”的读音错误，应该是“mò”；B中的“倾轧”中的“轧”读音错误，应该是“yà”，D中的“解剖”中“剖”的读音应该是“pōu”。

2．下列各句中，加点的成语使用正确的一项是A．现在我们单位职工上下班或步行，或骑车，为的是倡导绿色、地毯生活。

尤为可喜的是，始作俑者．．．．是我们新来的局长。

B．几年前，学界几乎没有人不对他的学说大加挞伐，可现在当他被尊奉为大师之后，移．樽就教．．．的人简直要踏破他家的门槛。

C．他是当今少数几位声名卓著的电视剧编剧之一，这不光是因为他善于编故事，更重要的原因是他写的剧本声情并茂．．．．，情节曲折。

D．旁边一位中学生模样的青年诚恳地说：“叔叔，这些都是名人的字画，您就买一幅吧，挂在客厅里不仅没关打气，还可附庸风雅．．．．。

”答案：B。

A中的成语“始作俑者”的意思是：“俑：古代殉葬用的木制或陶制的俑人。

开始制作俑的人。

比喻首先做某件坏事的人。

”明显感情色彩不合。

B中的“移樽就教”的意思是：“樽：古代盛酒器；就：凑近。