考研数学三-多元函数微积分学(一).doc
2024考研(数学三)真题答案及解析完整版
2024考研(数学三)真题答案及解析完整版2024年全国硕士研究生入学考试数学(三)真题及参考答案考研数学三考什么内容?数学三在高等数学这一部分因为要求的内容相对较少,所以很多学校经济类、管理类专业在本科期间所用教材并非理工类专业通常会使用的《高等数学》同济大学版,更多的学校本科阶段的教材是中国人民大学版《微积分》。
而考数学三的同学中在实际复习过程中使用哪一本教材的都有)(函数、极限、连续、一元函数微分学、一元函数积分学、多元函数微积分学、无穷级数、常微分方程与差分方程);线性代数(行列式、矩阵、向量、线性方程组、矩阵的特征值和特征向量、二次型);概率论与数理统计(随机事件和概率、随机变量及其分布、多维随机变量及其分布、随机变量的数字特征、大数定律和中心极限定理、数理统计的基本概念、参数估计、假设检验)。
考研的考试内容有哪些一、考研公共课:政治、英语一、英语二、俄语、日语、数学一、数学二、数学三,考研公共课由国家教育部统一命题。
各科的考试时间均为3小时。
考研的政治理论课(马原22分、毛中特30分、史纲14分、思修18分、形势与政策16分)。
考研的英语满分各为100分(完型10分、阅读理解60分、小作文10分、大作文20分)。
数学(其中理工科考数一、工科考数二、经管类考数三)满分为150分。
数一的考试内容分布:高数56%(84分)、线代22%(33分)、概率22%(33分);数二的内容分布:高数78%(117分)、线代22%(33分);数三的内容分布:高数56%(84分)、线代22%(33分)、概率22%(33分)。
这些科目的考试知识点和考试范围在各科考试大纲上有详细规定,一般变动不大,因此可以参照前一年的大纲,对一些变动较大的科目,必须以新大纲为准进行复习。
二、考研专业课统考专业课:由国家教育部考试中心统一命题,科目包括:西医综合、中医综合、计算机、法硕、历史学、心理学、教育学、农学。
其中报考教育学、历史学、医学门类者,考专业基础综合(满分为300分);报考农学门类者,考农学门类公共基础(满分150分)。
考研数学真题大串讲-微积分
例2(0 2012数一)grad
xy
z y
2,1,1
例2(1 1991数一)设n是曲面2x2 3y2 z2 6在点P 1,1,1处指向外侧的法向量,
求u= 6x2 8y2 在P处沿n的方向的方向导数 z
6)几何应用(曲线的切线与法平面,曲面的切平面与法线)
例2( 2 2014数一)曲面z x2 1sin y y2 1sin x在点1,0,1处的切平面方程为
x, y 0,0
A连续,偏导存在 B连续,偏导不存在 C 不连续,偏导存在
D不连续,偏导不存在
注 : 对二重极限的三类情形
xy
1) lim
x, y0,0
x2
y2
不存在
取不同的路径如y kx会有不同的极限值
2) lim x, y0,0
x2
x
y2
不存在
3) lim x, y0,0
2
3) 1
2
例(4 2012数一)如果f x, y在点0, 0处连续,则下列命题正确的是
A若极限lim f x, y 存在,则f x, y在点0, 0处可微
x0 x y
y0
B若极限lim x0
f x
2
x,
y
y2
存在,
则f
x,
y在点0, 0处可微
y0
C 若f x, y在点0, 0处可微,则极限lim f x, y 存在
例1(5 2015数二)已知f x, y满足fxy x, y 2 y 1ex , fx x, 0 x 1ex, f 0, y y2 2y 求f x, y的极值
(2009数二)设z f (x, y)的全微分dz xdx ydy,则点(0, 0)
高等数学 (一)
高等数学= = = = = = = = = = = = 骨头= = = = = = = = = = = = 对象:函数方法:极限思想:以不变代替变消除误差取极限内容:微积分(1)一元函数微积分||空无穷级数|间他们的应用|解|析常微分方程|(2)多元函数微积分(一)一元函数微积分:(1)微分学:函数、极限、连续;导数、微分----中值定理(4个;证明题)----(导数与微分的应用)(2)积分学:不定积分;定积分;定积分的应用↓维数↓增加↓(二)多元函数微积分(1)微分学:函数、极限、连续;偏导、全微分;应用(极值)(2)积分学:****;重积分(二重积分;3重积分、线积分和面积分<数一>);应用注意:一元函数微积分与多元函数微积分之间的联系和差别肉一、函数1.概念X↔I→→f→→y↔Rf(x)注意:(1)定义域(3点:0不能做除数、负数不能开平方、0和负数不能有对数)(2)函数的表达式与自变量的表示符号无关:y=f(x)与y=f(t)相同(函数关系不变)(3)由实际问题所建立的函数(极限的定义域;导函数的定义域;幂级数的和函数的表达式与定义区间)需要自己建立函数关系确定函数的定义域,根据实际问题<后面加>2.函数的特性(1)奇偶性(从定义来理解和证明应用)f(-x)=f(x),偶图形关于y轴对称[(y,-x )>>(y,x);y1=y2时候x1+x2=0且x1+x2=0时候y1=y2]f(-x)=-f(x),奇图形关于原点对称 [(y,-x )>>(-y,x);y1+y2=0时x1+x2=0且x1+x2=0时y1+y2=0] 注意:①奇偶函数运算:两个偶函数的和、差、积为偶函数奇函数与偶函数的积为奇函数两个奇函数的积为偶函数任何一个函数都可以写成一个奇函数和偶函数的和f(x)=1/2[f(x)+f(-x)]+1/2[f(x)-f(-x)]奇偶性在求导积分中的应用(后讲)②周期性f(x+T)=f(x),f(x)以T为周期注意:周期性在求导、函数特性、积分中的应用(画图中的应用)周期性与奇偶性都只能通过定义证明③增减性若x1,x2↔I,x1<x2有f(x1)<f(x2)或者f(x1)>f(x2)则f(x)在I区间内严格单调增或者减注意:(1)在证明不等式的时候常遇到<=或>=,称为不减或者不增,考点也属于增减性(2)函数的增减性与讨论的区间有关(题型:确定函数的增减区间;例如y=x^2)增减区间的交换点,极值(导数值为零)(3)增减性由导数的符号判定(微分学的应用之一)(4)增减性是证明不等式的一个重要工具(后讲)④有界性假定y=f(x),x↔I,存在M>0,对所有|f(x)|<=M成立则称f(x)有界图形-有界:有上下界注意:(1)有上界(单调减 , f(x)<=M有下界(单调增 , f(x)>=-M(2)有界性与讨论的区间有关(3)有界的讨论与极值有关(后讲)3.函数的分类(1)反函数y=f(x)→x=f^-1(y)条件:单调注意:y=f(x),x=f^-1(y) 代表同一条曲线(图形相同)y=f(x)与y=f ^-1(x)关于一三象限对称(2)基本初等函数①幂函数②指数函数(双曲函数)③对数函数④三角函数⑤反三角函数要求:对这五类基本函数的定义域、值域、特性要非常清楚(1-2,28Min-32Min)(3)复合函数y=f(u),u=w(x)y=f[w(x)]u的值域↔y的定义域注意:①并非所有函数都可以复合②考研:一拆多(4)初等函数经过有限次的四则运算或复合得基本初等函数(5)参数方程{X=x(t)Y=y(t)}得到y=y(x)(6)隐函数F(x,y)=0(易于理解函数,或者难用x表现y或者y表示x<求解函数时使用>)实际上是复合函数(7)分段函数①y=f(x)={f(x),x<=0-f(x),x>0} ②y=|f(x)|③ y=max[f(x),g(x)] x ↔(a,b)真正讨论时需要转化为①类讨论(1-2,41Min-43Min ) ④y=[f(x)]取整函数(1-2,44Min-45Min )二、极限 1.定义:数列的极限(ε-N 语言)0X Xon lim ()()U Xo ()lim ()()123lim(....)1/2*(1)/^21/2^2^2^2^2lim ()lim ()11sin lim ()()x x f x A f x f x f x f xo n n n n n n n n f x g x x xf xg x →→→∞=∃∞++++=+=⇒若存在,则()使其内,有界与无关函数的极限(ε-δ语言)lim 00,|f(X)-A|<X XoXn A εδδε→=⇔∀>∃>使0<|X-Xo|<时注意:ε是任给的,N 、δ是存在的但不唯一 δ=δ(a ),N= N (ε)lim 00,n>N |Xn-A|>=x Xn A N εε→∞≠⇔∃>∀>使时(1)极限的结构极限{变化过程,对自变量来讲(自变量的变化过程,δ、N ); 变化趋势,对函数而言,ε}(1-3,13Min-19Min ) (2)单边极限(分段函数;函数极限) 左极限0(0)lim ()X Xof f x x -→-=右极限0(0)lim ()X Xof f x x+→+=000lim ()(0)(0)X X f x A f f A x x →=⇔-=+=lim ()()lim ()()lim ()()()()()X X X f x f f x f f x f f A f f A→-∞→+∞→∞=-∞=+∞=∞∞=⇔+∞=-∞=2.极限的性质(1)唯一性(2)局部保号性(极限大于零则函数大于零<局部内>;1-3,30Min-35Min ) 注意:0X Xo()()0(0),lim ()A>=0f x f x f x A x x →><=若在=及其附近有定义,且存在,则(3)局部有界性(有极限的函数必有界<局部内>)lim ()U Xo ()x x f x A f x →=∃若存在,则()使其内,有界注意:上述性质对x →∞也成立,U (Xo )→|X|充分大3.极限的判别准则(1)单调有界数列必有极限(1-4,8-10MIN ) 注意:单调增有上界 ⇒ 极限存在 单调减油下界 ⇒ 极限存在数列的极限与前有限项无关 X Xolim ()()f x f xo →与无关4.极限的四则运算和差积商的极限与极限的和差积商相等 注意:(1)参加运算的极限只有有限次,且每一项的极限都存在n 123lim(....)1/2*(1)/^21/2^2^2^2^2n n n n n n n n →∞++++=+= (2)极限的和差①lim ()f x 存在,lim ()g x 不存在⇒lim(()())f x g x ±不存在 ②lim ()f x 不存在,lim ()g x 不存在⇒lim(()())f x g x ±有可能存在 ③lim ()f x 存在,lim(()())f x g x ±也存在⇒lim ()g x 存在注意:上述三条在反常积分、无穷级数中的应用 (2)极限的乘积若lim ()f x 存在,lim ()g x 不存在(其中一个极限为0)或lim ()f x 不存在,lim ()g x 也不存在( 101010101…与010*******…) 但lim ()()f x g x 都有可能存在5.无穷大,无穷小lim ()f x =0 ,()f x 无穷小 lim ()g x =∞,()g x 无穷大注意:① 无穷大于无穷小与过程有关② 同一过程下,无穷大与无穷互为倒数,0除外③ 无穷大属于极限不存在的情况下(也就是说极限的四则运算不适用于无穷大)④ 无穷大一定是无界的,无界不一定是无穷大(如y=11sin x x)(1-4,31-36Min ) 6.无穷小的比较不同的函数趋向于0的速度不一样 (1)假设lim α(x )=0;lim β(x )=0 若lim α(x )/ β(x )=∂∂≠0的常数,则α(x )与β(x )同介 ∂=1,则α(x )与β(x )等价表示为则α(x )~β(x ) (2)反身性;传递性① α(x )~α(x )② α(x )~β(x )⇔β(x )~α(x )③ α(x )~β(x ),β(x )~λ(x )⇔α(x )~λ(x ) (3)若limf (x )/g (x )=a a=0,f (x )比g (x )高阶 表示为f (x )=0(g (x ))(4)∂=∞,f (x )比g (x )低阶注意:①若lim f (x )/ g (x )=a ≠0 则称f (x )是 g (x )的K 阶无穷小 ②limf (x )=A ⇔f (x )=A+α 其中lim α(x )=0③常利用无穷小的等价函数求极限7.两个重要的极限(1)0sin lim 1x x x →=(通过图形证明)002(sin )^21cos 12lim lim .1/2^24()^22x x x x x x →→-==2222111cos 112.lim cos 12000n lim(lim lim ()()lim ()lim ()()lim 0y lim lim (0)!lim n cos )(1cos 1)x x x x x x x x x x nn n n x xx f x f f x Xo Xo Xo f x f XnYn a n x x x x a y αβλ→----→→→→→→∞→∞→∞→∞∂℘∈∃∀====⇔====>+- (1)f(Xo)有定义(2)存在(3)求222111( (12)n n n +++++推广型-(第一种重要极限的求极限法:配分母): lim (*)=0则lim[sin (*)/(*)]=1如002(sin )^21cos 12lim lim .1/2^24()^22x x x x x x →→-==(2-1,18-19Min ) 注意:x →0时,sinx ~x 1-cox ~1/2x^2 tanx ~x (2)10limlim 1(1)(1)xxx x x x→→∞==++推广型-(第二种重要极限的求极限法:拆底数-配指数):lim (*)=0lim (1+*)^1/*=0 例:2222111cos 112.lim cos 120lim(lim cos )(1cos 1)x x x x x x x x x x x →----→→===+- (2-1,24-25Min)极限的计算方法:四则运算,等价无穷小代换,求极限的两个方法三、连续1.定义 等价定义定义1:设f (x )在Xo 及其附近有定义△X →y 的增量△y=f (Xo+△X )- f (Xo )若00lim ,f x x x x y o ∆→∆=则称()在=点连续定义2:0lim ()(),()x x x x f x f f x x →=若则称在=点连续(极限值等于函数值)注意:①0lim ()(),()x x x f x f f x Xox -→=若则称在=点左连续若f (Xo+0)=f (Xo )0()x x f x 则称在=点右连续② 若f (x )在(a ,b )内点点都连续,则称f (x )在(a ,b )内连续③ 若f (x )在(a ,b )内连续,在x=a 点右连续,在x=b 左连续,则称f (x )在[a ,b]上连续2.连续函数的运算(连续是由极限定义的,因此极限的运算法则可以用在连续上)(2-1,38Min )注意:基本初等函数在定义域内连续 初等函数在定义区间内连续例:y=arcsin (x^2+1)在x=0点不连续但有定义,因为x=0点附件没有定义 3.间断点00lim ()()lim ()lim ()()x x x f x f f x f x f Xo Xo Xox x →→→=⇔=(1)f(Xo)有定义(2)存在(3)存在定义:若f (x )在Xo 点,上述三条至少有一条不成立,则称x=Xo 为f (x )的间断点 注意:间断点的分类 (1)若f (Xo-0),f (Xo+0)都存在则称Xo 为第一类间断点 特例:f (Xo-0)=f (Xo+0)则称Xo 为可去间断点 f (Xo-0)≠f (Xo+0)则称Xo 为跳跃间断点 例1:y=f (x )={sinx/x,x ≠0;2,x=0} 则x=0为可去间断点(若x=0时y=1,则函数连续) 例2:y=f (x )={x+1,x<0;x-1,x>0}(x=0处无定义,函数不连续)则x=0为跳跃间断点(2)若f (Xo-0),f (Xo+0)至少有一类不存在则称Xo 为第二类间断点 例1:y=1/x 在x=0处为第二类间断点(无穷间断点)例2:y=sin (1/x )在x=0点为第二类间断点(震荡型,图形) 注意:无穷间断点与求渐近线;反常积分中的应用例:y =1个,x=-1)(2-2,9-11Min )4.闭区间上连续函数的性质设y=f (x )在[a,b]上连续,则(1)y=f (x )在[a,b]上必有最大值与最小值,即∃X1,X2∈ [a,b],∀x ∈ [a,b]有 f (x )<=f (X2)(区间上的最大值)f (x )>=f (X1)(去见上的最小值) 最大值最小值是唯一一个数,但是取得最大值最小值的点可以不止一个;最大值与最小值可以是同一个值,此时函数为常数 (2)介值定理f (x )必取得最大值和最小值之间的一切值 注意:①闭区间上的连续函数一定是有界的②f (x )在[a ,b]上连续,f (a )f (b )<0,则至少∃℘∈(a ,b )使得f (℘)=0 例1:设Xn ,Yn 满足lim 0x XnYn →∞=则成立的是A 若Xn 发散,则Yn 必发散 Xn (010203…) Yn(000000…)B 若Xn 无界,则Yn 必有界 Xn (010203…) Yn(102030…)C 若Xn 有界,则Yn 必为无穷小 Yn (010203…) Xn (000000…)D 若1/ Xn 无穷小,则Yn 必为无穷小乘积的极限等于极限的乘积(2-2,31-34Min ) 例2:证明:limlim (0)!nn n n a n a y→∞→∞=>存在(2-2,36-41Min )证明极限存在:单调有界(有递推关系的首先想到),加别定理(放大一下缩小一下,但是放大缩小后的极限要相同)例3:222n 111lim n(...)12nnnn→∞+++++求(2-2,43-44Min )例4: 求极限 (1)limx (2-2,47-48Min )注意:四则运算要求参与运算的极限都存在,因此本题的原型不能使用积商的极限等于极限的积商方法:遇到根号通常进行有理化 (2)3113lim()11x x x →---(2-2, 48-50Min ) 方法:无穷大减无穷大通常进行通分,然后再进行补充(化简) (3)练习x →例5:等价无穷小(2-3,5-7Min)A 1-()ln(1()10B C D +→-当x ()常用的三个等价无穷小1,,ln(1)(1)xx x x x x eαα-++答案:B 例6:已知极限求表达式里的一个常数(2-3,9-12Min )011lim[()]1a A B C D xx a x xe →--=已知则为()0()1()2()3答案:C 例7:Xlim 8ax 2a x-ax →∞=+已知求()现象-分析-方法:1的无穷次大,拆底数配指数 答案:a=ln2 注意:10011111 (i)...a a a a nn n n mm x n nxxx xb x b x b x b x ---→∞-++++++++要看其最高次={00a b,n=m ;∞,n>m;0,n<m} 例8:(2-3,18-23Min )x sin 0(,0)a+ba xb a b --=>证明方程至少有一正根,且不大于现象-分析-方法:作左方看做函数→函数有零值→介值定理→零值定理 初等函数→连续不大于→≦→分类讨论 例9:(2-3,26-30Min )()()lim ()()()x f x f x f x →∞-∞+∞-∞+∞设在,连续,且存在,证明在,有界闭区间上连续必有界,有极限的函数必有界(局部有界)导数与微分一、导数1、定义两个实际问题:一曲线在一点的切线,方法----利用割线逼近一点的切线,二是物理上的瞬时速度,先求平均速度然后用时间间隔趋向于零近似的得到瞬时速度 但是他们都有误差,因此要取极限(哲学上讲:是质变),由割线上升到导数,由平均数上升到瞬时速。
考研数学三(一元函数积分学与多元函数微分学)模拟试卷1(题后含
考研数学三(一元函数积分学与多元函数微分学)模拟试卷1(题后含答案及解析)题型有:1. 选择题 2. 填空题 3. 解答题选择题下列每题给出的四个选项中,只有一个选项符合题目要求。
1.设f(x,y)=则f(x,y)在(0,0)处( ).A.连续但不可偏导B.可偏导但不连续C.可微D.一阶连续可偏导正确答案:C解析:因为f(x,y)=0=f(0,0),所以f(x,y)在(0,0)处连续。
因为=0,所以f’x(0,0)=0,根据对称性,f’y(0,0)=0,即f(x,y)在(0,0)处可偏导;由=0,得f(x,y)在(0,0)处可微;当(x,y)≠(0,0)时,f’x(x,y)=2xsin,则f’x(x,y)=因为f’x(x,y)=(2xsin)不存在,所以f’x(x,y)在点(0,0)处不连续,同理f’y(x,y)在点(0,0)处也不连续,选C.知识模块:多元函数微分学2.设F(x)为f(x)的原函数,且当x≥0时,f(x)F(x)=,又F(0)=1,F(x)>0,求f(x).正确答案:两边积分得F2(x)=,解得F2(x)=+C,由F(0)=1,F(x)>0,得F(x)=,于是f(x)=.涉及知识点:一元函数积分学3.设f’(lnx)=求f(x).正确答案:令lnx=t,则f’(t)=,当t≤0时,f(t)=t+C1,;当t>0时,f(t)=et+C2.显然f’(t)为连续函数,所以f(t)也连续,于是有C1=1+C2,故f(x)=.涉及知识点:一元函数积分学4.求dx.正确答案:令f(x)=,当0≤x≤1时,,当1<x≤2时,=x2,则dx=xdx +x2dx=.涉及知识点:一元函数积分学5.设f(x)连续,tf(x-t)dt=1-cosx,求f(x)dx.正确答案:由tf(x-t)dt(x-u)f(u)(-du)=(x-u)f(u)du=xf(u)du-uf(u)du,得xf(u)du-uf(u)du=1-cosx,两边求导得f(u)du=sinx,令x=得f(x)dx=1.涉及知识点:一元函数积分学6.设f(x)在[0,+∞)上连续、非负,且以T为周期,证明:正确答案:对充分大的x,存在自然数n,使得nT≤x<(n+1)T,因为f(x)≥0,所以f(t)dt≤f(t)dt≤f(x)dt,即nf(t)dt≤f(t)dt≤(n+1)f(t)dt,由,得,注意到当x→+∞时,n→∞,且由夹逼定理得.涉及知识点:一元函数积分学7.设f(x)在[0,1]上连续,f(0)=0,f(x)dx=0.证明:存在ξ∈(0,1),使得f(x)dx=ξf(ξ).正确答案:令φ(x)=,因为f(x)在[0,1]上连续,所以φ(x)在[0,1]上连续,在(0,1)内可导,又φ(0)=0,φ(1)=f(x)dx=0,由罗尔定理,存在ξ∈(0,1),使得φ’(ξ)=0,而φ’(x)=,所以f(x)dx=ξf(ξ).涉及知识点:一元函数积分学设f(x)在(-a,a)(a>0)内连续,且f’(0)=2.8.证明:对0<x<a,存在0<Θ<1,使得f(t)dt+f(t)dt=x[f(Θx)-f(-Θx)];正确答案:令F(x)=f(t)dt+f(t)dt,显然F(x)在[0,x]上可导,且F(0)=0,由微分中值定理,存在0<Θ<1,使得F(x)=F(x)-F(0)=F’(Θx)x,即f(t)dt +f(t)dt=x[f(Θx)-f(-Θx)].涉及知识点:一元函数积分学9.求Θ.正确答案:令Θ=A,由f(t)dt+f(t)dt=x[f(Θx)-f(-Θx)],得于是Θ=.涉及知识点:一元函数积分学10.设an=tannxdx(n≥2),证明:正确答案:an+an+2=(1+tan2x)tannxdx=tannxd(tanx)=tann+1x,同理an +an-2=.因为tannx,tann+2x在[0,]上连续,tannx≥tann+2x,且tannx,tann+2x不恒等,所以tannxdx>tann+2xdx,即an>an+2,于是=an+an+2<2an>,同理可证an<.涉及知识点:一元函数积分学11.设f(x)有界,且f’(x)连续,对任意的x∈(-∞,+∞)有|f(x)+f’(x)|≤1.证明:|f(x)|≤1.正确答案:令φ(x)=exf(x),则φ’(x)=ex[f(x)+f’(x)],由|f(x)+f’(x)|≤1得|φ’(x)|≤ex,又由f(x)有界得φ(-∞)=0,则φ(x)=φ(x)-φ(-∞)=φ’(x)dx,两边取绝对值得ex|f(x)|≤|φ’(x)|dx≤exdx=ex,/sup>,所以|f(x)|≤1.涉及知识点:一元函数积分学12.设f(x)在(-∞,+∞)上有定义,且对任意的x,y∈(-∞,+∞)有|f(x)-f(y)|≤|x-y|.证明:|f(x)dx-(b-a)f(a)|≤(b-a)2.正确答案:因为(b-a)f(a)=f(a)dx,所以|f(x)dx-(b-a)f(a)|=|[f(x)-f(a)]dx|≤f(x)-f(a)|dx≤(x-a)dx=(x-a)2(b-a)2.涉及知识点:一元函数积分学13.设f(x)在[0,1]上连续,且0<m≤f(x)≤M,对任意的x∈[0,1],证明:.正确答案:因为0<m≤f(x)≤M,所以f(x)-m≥0,f(x)-M≤0,从而≤0,于是f(x)+≤M+m,两边积分得f(x)dx+Mmdx≤M+m,因为f(x)dx+Mmdx ≥2,所以2≤M+m,于是(f(x)dx)(dx)≤.涉及知识点:一元函数积分学14.设f(x)在[a,b]上连续且单调增加,证明:xf(x)dx≥f(x)dx.正确答案:方法一令φ(x)=(x-)[f(x)-f()],因为f(x)在[a,b]上单调增加,所以φ(x)dx≥0,而φ(x)dx=(x-)[f(x)-f()]dx=(x-)f(x)dx-f()(x-)dx=(x -)f(x)dx=xf(x)dx-f(x)dx,故xf(x)dx≥f(x)dx.方法二令φ(x)=tf(t)dt-f(t)dt,显然φ(a)=0.φ’(x)=xf(x)-f(t)dt-f(x)=[(x-a)f(x)-f(t)dt]=[f(x)dt-f(t)dt]=[f(x)-f(t)]dt≥0,由得φ(b)≥φ(a)=0,所以xf(x)dx≥f(x)dx.涉及知识点:一元函数积分学填空题15.设z=xf(x+y)+g(xy,x2+y2),其中f,g分别二阶连续可导和二阶连续可偏导,则=_____________.正确答案:f’+xf”+xy-1g’1+yxy-1lnxg’1+yx2y-1lnxg”11+2y2xy -1g”12+2xy+1lnxg”21+4xyg”22.解析:由z=xf(x+y)+g(xy,x2+y2),得=f(x+y)+xf’(x+y)+yxy-1 g’1(xy,x2+y2)+2xg’2(xy,x2+y2),=f’+xf”+xy-1g’1+yxy-1lnxg’1+yx2y-1lnxg”11+2y2xy-1g”12+2xy+1lnxg”21+4xyg”22.知识模块:多元函数微分学解答题解答应写出文字说明、证明过程或演算步骤。
考研数学三(多元函数微积分学)历年真题试卷汇编1(题后含答案及解析)
考研数学三(多元函数微积分学)历年真题试卷汇编1(题后含答案及解析)题型有:1. 选择题 2. 填空题 3. 解答题选择题下列每题给出的四个选项中,只有一个选项符合题目要求。
1.[2008年] 设则( ).A.fx’(0,0),fy’(0,0)都存在B.fx’(0,0)不存在,fy’(0,0)存在C.fx’(0,0)存在,fy’(0,0)不存在D.fx’(0,0),fy’(0,o)都不存在正确答案:B解析:因而则极限不存在,故偏导数fx’(0,0)不存在.而因而偏导数fy’(0,0)存在.仅(B)入选.知识模块:多元函数微积分学2.[2003年] 设可微函数f(x,y)在点(x0,y0)处取得极小值,则下列结论正确的是( ).A.f(x0,y)在y=y0处的导数大于零B.f(x0,y)在y=y0处的导数等于零C.f(x0,y)在y=y0处的导数小于零D.f(x0,y)在y=y0处的导数不存在正确答案:B解析:解一因f(x,y)在点(x0,y0)处可微,故f(x,y)在点(x0,y0)处两个偏导数存在,因而一元函数f(x0,y)在y=y0处的导数也存在.又因f(x,y)在点(x0,y0)处取得极小值,故f(x0,y0)在y=y0处的一阶(偏)导数等于零.仅(B)入选.解二由函数f(x,y)在点(x0,y0)处可微知,f(x.y)在点(x0,y0)处的两个偏导数存在.又由二元函数极值的必要条件即得f(x,y)在点(x0,y0)处的两个偏导数都等于零.因而有知识模块:多元函数微积分学3.[2016年] 已知函数则( ).A.fx’-fy’=0B.fx’+fy’=0C.fx’-fy’=fD.fx’+fy’=f正确答案:D解析:则仅(D)入选.知识模块:多元函数微积分学4.[2017年] 二元函数z=xy(3-x-y)的极值点为( ).A.(0,0)B.(0,3)C.(3,0)D.(1,1)正确答案:D解析:zy’=y(3-x-y)-xy=y(3-2x-y),zy’=x(3-x-y)-xy=x(3-x-2y),又zxx’=-2y,zxy=3-2x-2y,zyy’=-2x,将选项的值代入可知,只有(D)符合要求,即A=zxx”(1,1)=-2,B=zxy”(1,1)=-1,C=zyy”(1,1)=-2.满足B2-AC=-3<0,且A=-2<0,故点(1,1)为极大值点.仅(D)入选.知识模块:多元函数微积分学5.[2006年] 设f(x,y)与φ(z,y)均为可微函数,且φy’(x,y)≠0,已知(x0,y0)是f(x,y)在约束条件φ(x,Y)=0下的一个极值点,下列选项正确的是( ).A.若fx’(x0,y0)=0,则fy’(x0,y0)=0B.若fx’(x0,y0)=0,则f’y(x0,y0)≠0C.若fx’(x0,y0)≠0,则fy’(x0,y0)=0D.若fx’(x0,y0)≠0,则f’y(x0,y0)≠0正确答案:D解析:解一由拉格朗日乘数法知,若(x0,y0)是f(x,y)在条件φ(x,y)=0下的极值点,则必有fx’(x0,y0)+λφx’(x0,y0)=0,①fx’(x0,y0)+λφx’(x0,y0)=0.②若fx’(x0,y0)≠0,由式①知λ≠0.又由题设有φy’(x0,y0)≠0,再由式②知fy’(x0,y0)≠0.仅(D)入选.解二构造拉格朗日函数F(x,y,λ)=f(x,y)+λφ(x,y),并记对应于极值点(x0,y0)处的参数的值为λ0,则由式③与式④消去λ0得到fx’(x0,y0)/φx’(0,y0)=一λ0=f’y(x0,y0)/φ’y(x0,y0).即f’x(x0,y0)φ’y(x0,y0)一fy’(x0,y0)φx’(x0,y0)=0.整理得若fx’(x0,y0)≠0,则由式③知,φx’(x0,y0)≠0.因而fy’(x0,y0)≠0.仅(D)入选.解三由题设φy’(x,y)≠0知,φ(x,y)=0确定隐函数y=y(x).将其代入f(x,y)中得到f(x,y(x)).此为一元复合函数.在φ(x,y)=0两边对x求导,得到因f(x,y(x))在x=x0处取得极值,由其必要条件得到f’x+fy’y’=fx’+fy’(一φx’/φy’)=0.因而当fx’(x0,y0)≠0时,必有fy’(x0,y0)≠0.仅(D)入选.知识模块:多元函数微积分学填空题6.[2012年] 设连续函数z=f(x,y)满足则dz|(0,1)=__________.正确答案:2dx-dy解析:用函数f(x,y)在(x0,y0)处的微分定义:与所给极限比较易知:z=f(x,y)在点(0,1)处可微,且fx’(0,1)=2,fy’(0,1)=-1,f(0,1)=1,故dz|(0,1)=fx’(0,1)dx+fy’(0,1)dy=2dx-dy.知识模块:多元函数微积分学7.[2009年] 设z=(x+ey)x,则正确答案:2ln2+1解析:解一为简化计算,先将y=0代入z中得到z(x,0)=(x+1)x,z为一元函数.将x=1代入上式,得到解二考虑到z(x,0)=(x+1)x为幂指函数,先取对数再求导数:lnz=xln(x+1).在其两边对x求导,得到则知识模块:多元函数微积分学8.[2007年] 设f(u,v)是二元可微函数,则正确答案:解析:解一设u=y/x,v=x/y.为方便计,下面用“树形图”表示复合层次与过程.由式①一式②得到解二令f1’,f2’分别表示z=f(y/x,x /y)对第1个和第2个中间变量y/x、x/y求导数,则知识模块:多元函数微积分学9.[2004年] 函数f(u,v)由关系式f[xg(y),y]=x+g(y)确定,其中函数g(y)可微,且g(y)≠0,则正确答案:解析:令u=xg(y),v=y,由此解出于是知识模块:多元函数微积分学10.[2005年] 设二元函数z=xex+y+(x+1)ln(1+y),则dz|(1,0)=_________.正确答案:2edx+(e+2)dy解析:dz=d[xex+y+(x+1)ln(1+y)]=d(xex+y)+d[(x+1)ln(1+y)] =ex+ydx+xex+y(dx+dy)+ln(1+y)dx+[(x+1)/(1+y)]dy.①将x=1,y=0代入上式(其中dz,dx,dy不变),得到dz|(1,0)=edx+e(dx+dy)+2dy=2edx+(e+2)dy.解二利用全微分公式求之.为此,先求出偏导数故解三用定义简化法求之.固定一个变量转化为另一个变量的一元函数求导.由z(x,0)=xex得到由z(1,y)=ey+2ln(1+y)得到故知识模块:多元函数微积分学11.[2006年] 设函数f(u)可微,且f’(0)=1/2,则z=f(4x2-y2)在点(1,2)处的全微分dz|1,2=___________.正确答案:4dx一2dy解析:解一dz=df(4x2-y2)=f’(u)du=f’(u)d(4x2-y2)=f’(u)(8xdx-2ydy),其中u=4x2-y2.于是dz|1,2=f’(0)(8dx-4dy)=4dx-2dy.解二利用复合函数求导公式和定义简化法求之.由z=f(4x2-y2)得到解三由z=f(4x2-y2)得到于是故dz|1,2=4dx-2dy.知识模块:多元函数微积分学12.[2011年] 设函数则dz|1,1=____________.正确答案:(1+2ln2)(dx—dy)解析:解一所给函数为幂指函数,先在所给方程两边取对数,然后分别对x,y求偏导:由得到则解二先用定义简化法求出然后代入全微分公式求解.故dz|1,1=2(ln2+1/2)dx-2(ln2+1/2)dy=(1+2ln2)(dx-dy).知识模块:多元函数微积分学13.[2015年] 若函数z=z(x,y)由方程ex+2y+3z+xyz=1确定,则dz|0,0=_______________.正确答案:解析:在ex+2y+3z+xyz=1①两边分别对x,y求偏导得到同法可得将x=0,y=0代入式①易求得z=0,代入式②、式③分别得到则知识模块:多元函数微积分学14.[2014年] 二次积分正确答案:解析:注意到不易求出,需先交换积分次序,由积分区域的表达式D={(x,y)|y≤x≤1,0≤y≤1)-{(x,y)|0≤y≤x,0≤x≤1}及交换积分次序得到故知识模块:多元函数微积分学解答题解答应写出文字说明、证明过程或演算步骤。
考研数学三考试大纲2023
考研数学三考试大纲2023一、考试概述2023年考研数学三考试将在12月底至次年1月初举行,考试时间共计150分钟,总分为150分。
该考试旨在测试考生在数学逻辑、数学分析、数学统计等领域的知识技能和应用能力。
二、考试内容考试内容主要分为以下三个部分:1. 数学逻辑数学逻辑是考研数学三的重要组成部分,占总分的20%。
考试内容主要包括命题逻辑、谓词逻辑、集合论、图论等方面的知识。
考生需要掌握逻辑公式的推导、逻辑谬误的识别和修正,以及能够运用逻辑思维解决实际问题。
2. 数学分析数学分析是考研数学三的另一个重点,占总分的40%。
考试内容主要包括极限、连续、导数、微分方程、多元函数微积分等方面的知识。
考生需要掌握数学分析基本理论和方法,能够熟练运用分析思维解决实际问题。
3. 数学统计数学统计是考研数学三中的最后一个部分,占总分的40%。
考试内容主要包括随机变量、概率分布、假设检验、方差分析等方面的知识。
考生需要掌握统计学基本理论和方法,能够熟练运用统计思维解决实际问题。
三、考试评分2023年考研数学三的评分方法主要采用全国统一标准的考试评分体系,即通过客观题得出的得分和主观题考试评分,两者综合得出考生最终的成绩。
其中客观题得分占总分的60%,主观题考试评分占总分的40%。
四、考试备考建议由于考研数学三的内容较为广泛,备考建议如下:1. 积极备考备考考研数学三需要付出充分的努力和时间,需要通过不断练习提高分析、解决问题和思考的能力。
在考前可以通过模拟测试、习题集等方式来加强备考效果。
2. 精细备考备考数学三时要注重理论与实践相结合,不仅要注重基础知识的学习,还要重视综合运用能力的培养。
要注重精读典型例题,在解答过程中体会方法和技巧,提高套用理论解题的能力。
3. 灵活备考备考数学三时要灵活运用各种方法与技巧,在理解数学概念的基础上,学会从不同角度和思维方向来解决问题,培养自己的数学思维和分析能力。
五、总结无论是考试内容还是备考建议,都需要考生投入足够的时间和精力,不断提高自己的数学素质和学习能力,从而取得好的考试成绩。
考研数学三(多元函数微分学)-试卷2.doc
考研数学三(多元函数微分学)-试卷2(总分:70.00,做题时间:90分钟)一、选择题(总题数:12,分数:24.00)1.选择题下列每题给出的四个选项中,只有一个选项符合题目要求。
(分数:2.00)__________________________________________________________________________________________2. 2.00)A.等于0B.不存在C.D.3.设 2.00)A.B.C.D.4. 2.00)A.等于0B.不存在C.D.存在且不等于05.设u=f(r),而f(r) 2.00)A.B.C.D.6.考虑二元函数f(x,y)的下面4条性质:①f(x,y)在点(x 0,y 0)处连续;②f(x,y)在点(x 0,y 0)处的两个偏导数连续;③f(x,y)在点(x 0,y 0 )处可微;④f(x,y)在点(x 0,y 0 )处的两个偏导数存在.若用“P推出性质Q,则有 2.00)A.B.C.D.7.设函数u=u(x,y)满足u(x,2x)=x,uˊ1(x,2x)=x 2,u有二阶连续偏导数,则uˊˊ11(x,2.00)A.B.C.D.8.利用变量代换u=x, 2.00)A.B.C.D.9.若函数f 2.00)A.x+yB.x-yC.x 2-y 2D.(x+y) 210.已知du(x,y)=[axy 3 +cos(x+2y)]dx+[3x 2 y 2 +bcos(x+2y)]dy,则 ( )(分数:2.00)A.a=2,b=-2B.a=3,b=2C.a=2,b=2D.a=-2,b=211.设u(x,y)在平面有界闭区域D 2.00)A.最大值点和最小值点必定都在D的内部B.最大值点和最小值点必定都在D的边界上C.最大值点在D的内部,最小值点在D的边界上D.最小值点在D的内部,最大值点在D的边界上12.设函数z=(1+e y )cosx-ye y,则函数z=f(x,y) ( )(分数:2.00)A.无极值点B.有有限个极值点C.有无穷多个极大值点D.有无穷多个极小值点二、填空题(总题数:5,分数:10.00)13.设f可微,则由方程f(cx-az,cy-bz)=0确定的函数z=z(x,y)满足azˊx +bzˊx = 1.(分数:2.00)填空项1:__________________14.设函数z=z(x,y)由方程sinx+2y-z=e z所确定,则 2.00)填空项1:__________________15.函数f(x,y,z)=-2x 2在条件x 2-y 2-2z 2 =2下的极大值是 1.(分数:2.00)填空项1:__________________16. 2.00)填空项1:__________________17.设z=e sinxy,则dz= 1.(分数:2.00)填空项1:__________________三、解答题(总题数:18,分数:36.00)18.解答题解答应写出文字说明、证明过程或演算步骤。
考研数学三详细范围
小结
习题
选做习题
参读材料随机变量样本值的产生
附表
附表1几种常用的概率分布表
附表2标准正态分布表
附表3泊松分布表
附表4 t分布表
附表5 X2分布表
附表6 F分布表
附表7均值的t检验的样本容量
附表8均值差的t检验的样本容量
附表9秩和临界值表
习题答案
第二节洛必达法则(★)
第三节泰勒公式(☆)
第四节函数的单调性与曲线的凹凸性(★)
第五节函数的极值与最大值最小值(★)
第六节函数图形的描绘(★)
第七节曲率(●)
第八节方程的近似解(●)
总习题三(★注意渐近线)
第四章不定积分
第一节不定积分的概念与性质(★)
第二节换元积分法(★)
第三节分部积分法(★)
第四节有理函数的积分(★)
●─大纲中没有明确要求,但对做题和以后的学习有帮助。要能看懂,了解其思路和结论。
▲─超出大纲要求。
第一章函数与极限
第一节映射与函数(☆集合、影射,★其余)
第二节数列的极限(☆)
第三节函数的极限(☆)
第四节无穷小与无穷大(★)
第五节极限运算法则(★)
第六节极限存在准则(★)
第七节无穷小的比较(★)
第八节函数的连续性与间断点(★)
小结
习题
第九章方差分析及回归分析(▲)
1单因素试验的方差分析
2双因素试验的方差分析
3一元线性回归
4多元线性回归
小结
附录
习题
第十章bootstrap方法(▲)
1非参数bootstrap方法
2参在数理统计中应用Excel软件(▲)
考研数学三(解答题)高频考点模拟试卷58(题后含答案及解析)
考研数学三(解答题)高频考点模拟试卷58(题后含答案及解析) 题型有:1.1.设f(x)=(Ⅰ)若f(x)处处连续,求a,b的值;(Ⅱ)若a,b不是(Ⅰ)中求出的值时f(x)有何间断点,并指出它的类型.正确答案:(Ⅰ)首先求出f(x).注意到其次,由初等函数的连续性知f(x)分别在(-∞,-1),(-1,1),(1,+∞)上连续.最后,只需考察函数f(x)在分界点x=±1处连续所应满足的条件.由于因此f(x)在x=±1均连续故当且仅当a=0,b=1时f(x)处处连续.(Ⅱ)当(a,b)≠(0,1)时,若a+b=1(则a-b≠-1),则x=1是连续点,只有x=-1是间断点,且是第一类间断点;若a-b=-1(则a+b≠1),则x=-1是连续点,只有间断点x=1,且是第一类间断点;若a-b≠-1且a+b≠1,则x=1,x=-1均是第一类间断点.涉及知识点:函数、极限、连续2.求下列极限:正确答案:(9)属1∞型极限.原极限=eJ,而(13)本题是∞-∞型未定式,提出无穷大因子x2后作变量替换x=,可得涉及知识点:函数、极限、连续3.设f(x)在[a,b]上二阶可导,且f’’(x)>0,取xi∈[a,b](i=1,2,…,n)及ki>0(i=1,2,…,n)且满足k1+k2+…+kn=1.证明:f(k1x1+k2x2+…+knxn)≤k1f(x1)+k2f(x2)+…+knf(xn).正确答案:令x0=k1x1+k2x2+…+knxn,显然x0∈[a,b] .因为f’’(x)>0,所以f(x)≥f(x0)+f’(x0)(x-x0),分别取x=xi(i=1,2,…,n),得由ki >0(i=1,2,…,n),上述各式分别乘以ki(i=1,2,…,n),得将上述各式分别相加,得f(x0)≤k1f(x1)+k2f(x2)+…+knf(xn),即f(k1x1+k2x2+…+knxn)≤k1f(x1)+k2f(x2)+…+knf(xn).涉及知识点:一元函数微分学4.设函数f(x)在闭区间[(a,b]上连续(a,b>0),在(a,b)内可导.试证:在(a,b)内至少有一点ξ,使等式=f(ξ)-ξfˊ(ξ)成立.正确答案:令F(x)=,它们在区间[a,b]上连续,在(a,b)内可导,且Gˊ(x)=≠0.满足柯西中值定理的三个条件,于是在(a,b)内至少有一点ξ,使得涉及知识点:一元函数微分学5.用配方法化下列二次型为标准形:f(x1,x2,x3)=2x1 x2+2x1 x3+6x2 x3.正确答案:令,或X=P1Y,其中P1=且P1可逆,则f(x1,x2,x3)2y1 2一2y2 2+8y1 y3+4y2 y3=2(y1+2y3)2一2(y2一y3)2一6y3 2,再令,或Y=P2Z,其中P2=且P2可逆,令P=P1P2=,P可逆,且f(x1,x2,x3)=XTAXZT(PTAP)Z=2z1 2一2z2 2一6z3 2.涉及知识点:线性代数6.计算正确答案:涉及知识点:不定积分7.设函数f(x)在[0,π]上连续,且∫0πf(x)dx=∫0πf(x)cos xdx=0.试证明:在(0,π)内至少存在两个不同的点ξ1,ξ2,使f(ξ1)=f(ξ2)=0.正确答案:令F(x)=∫0xf(t)dt,0≤x≤π,则有F(0)=0,F(π)=0,又因为0=∫0πf(x)cosxdx=∫0πcosxdF(x) =F(x)cosx|0π+∫0πF(x)sinxdx =∫0πF(x)sinxdx,所以存在ξ∈(0,π),使F(ξ)sinξ=0,不然,则在(0,π)内F(x)sinx 恒为正或恒为负,与∫0πF(x)sinxdx=0矛盾,但当ξ∈(0,π)时sinξ≠0,故F(ξ)=0. 由以上证得,存在满足0<ξ<π的ξ,使得F(0)=F(ξ)=F(π)=0.再对F(x)在区间[0,ξ],[ξ,π]上分别用罗尔定理知,至少存在ξ1∈(0,ξ),ξ2∈(ξ,π),使得F’(ξ1)=F’(ξ2)=0,即f(ξ1)=f(ξ2)=0.涉及知识点:一元函数积分学8.设3阶对称阵A的特征值为λ1=1,λ2=-1,λ3=0;对应λ1,λ2的特征向量依次为p1=,求A.正确答案:因为A为对称阵,故必存在正交阵Q=(q1,q2,q3),使由题意,可得λ1、λ2的特征向量由正交矩阵的性质,q3可取为的单位解向量,则由涉及知识点:矩阵的特征值和特征向量9.设随机变量X与Y相互独立同分布,且X的概率分布为记U=max(X,Y),V=min(X,Y),试求:(I)(U,V)的分布;(Ⅱ)E(UV);(Ⅲ)ρUV.正确答案:(I)设(U,V)的分布为则有p11=P{U=1,V=1}=P{max(X,Y)=1,min(X,Y)=1} =P{X=1,Y=1}=P{X=1}P{Y=1}=,p12=P{U=1,V=2}=P{max(X,Y)=1,min(X,Y)=2}==0,p22=P{U=2,V=2}=P{max(X,Y)=2,min(X,Y)=2}=P{X=2,Y=2}=P{X=2}P{Y=2}=,p21=1一p11一p12一p22=.所以(U,V)的分布为(Ⅱ)UV可能取值为1,2,4,所以(Ⅲ)由(I)可知涉及知识点:概率论与数理统计10.设D={(x,y)|一∞<x<+∞,一∞<y<+∞},求正确答案:记D是全平面.方法1 引入极坐标变换:x=rcosθ,y=rsin θ,D的极坐标表示:0≤0≤2π,0≤r<+∞于是因为故方法2 全平面D关于y=x对称,D的y=x上方部分记为D1,D1={(x,y)l一∞<x<+∞,x≤y<+∞},或D1:{(x,y)|一∞<y<+∞,一∞<x≤y},则涉及知识点:多元函数微积分学11.计算曲线积分:,L为球面x2+y2+z2=a2与平面x=y相交的圆周.正确答案:涉及知识点:综合12.设φ(x)是以2π为周期的连续函数,且ψ’(x)=φ(x),ψ(0)=0.(1)求方程y’+ysin x=φ(x)ecosx的通解;(2)方程是否有以2π为周期的解?若有,请写出所需条件;若没有,请说明理由.正确答案:(1)该方程为一阶非齐次线性微分方程,通解为y=e-∫sinxdx[∫φ(x)ecosxe∫sinxdxdx+C]=ecosx[∫φ(x)ecosx.e-cosxdx+C]=ecosx[∫φ(x)dx+C]=ecosx[ψ[x)+C],其中C为任意常数.(2)因为ψ’(x)=φ(x),所以ψ(x)=∫0xφ(t)dt+C1.又ψ(0)=0,于是,ψ(x)=∫0xφ(t)dt.ψ(x+2π)=∫0x+2πφ(t)dt=∫0xφ(t)dt+∫xx+2πφ(t)dt=ψ(x)+∫02πφ(t)dt,所以当∫02πφ(t)t=0时,ψ(x+2π)=ψ(x),即ψ(x)以2π为周期.因此,当∫02πφ(t)dt=0时,方程有以2π为周期的解.涉及知识点:微积分13.(1)设λ1,λ2,…,λn是n阶矩阵A的互异特征值,α1,α2,…,αn是A的分别对应于这些特征值的特征向量,证明α1,α2,…,αn线性无关;(2)设A,B为n阶方阵,|B|≠0,若方程|A一λB|=0的全部根λ1,λ2,…,λn互异,αi分别是方程组(A—λiB)x=0的非零解,i=1,2,…,n.证明α1,α2,…,αn线性无关.正确答案:(1)用数学归纳法.①由特征向量α1≠0,故α1线性无关;②假设前k一1个向量α1,α2,…,αk-1线性无关,以下证明α1,α2,…,αk线性无关.k个互异特征值λ1,λ2,…,λk对应着特征向量α1,α2,…,αk.现设存在一组数l1,l2,…,lk,使得l1α1+l2α2+…+lkαk=0,(*)在(*)式两端左边乘A,有l1Aα1+l2Aα2+…+lkAαk=0,即l1λ1α1+l2λ2α2+…+lkλkαk=0.(**) 又在(*)式两端左边乘λk,有l1λ1α1+l2λ2α2+…+lkλkαk=0.(***) 用(**)式减去(***)式,得l1(λ1—λk)α1+l2(λ2一λk)α2+…+lk-1(λk-1一λk)αk-1=0.由归纳假设α1,α2,…,αk-1线性无关,故l1(λ1一λk)=l2(λ2一λk)=…=lk-1(λk-1一λk)=0,又λi—λk≠0(i=1,2,…,k一1),故l1=l2=…=lk-1=0.代回(*)式,于是lkαk=0,由αk≠0,有lk=0,于是α1,α2,…,αk线性无关.即A 的n个互异特征值对应的特征向量α1,α2,…,αn线性无关.(2)由|B|≠0,在|A一λB|=0两端左边乘|B-1|,有|B-1A一λE|=0,即|λE一B-1A|=0,于是λ1,λ2,…,λn是矩阵B-1A的n个互异特征值.又由(A-λiB)x=0,两端左边乘B-1,有(B-1A—λiE)x=0,即(λiE一B-1A)x=0,故α1,α2,…,αn为B-1A的对应于λ1,λ2,…,λn的特征向量,由(1)知,α1,α2,…,αn线性无关.涉及知识点:线性代数设随机变量X与Y相互独立且分别服从正态分布N(μ,σ2)与N(μ,2σ2),其中σ是未知参数且σ>0,设Z=X—Y。
考研数学三(多元函数微积分学)模拟试卷10(题后含答案及解析)
考研数学三(多元函数微积分学)模拟试卷10(题后含答案及解析) 题型有:1. 选择题 2. 填空题 3. 解答题选择题下列每题给出的四个选项中,只有一个选项符合题目要求。
1.设其中D={(x,y)|x2+y2≤1},则( )A.I3>I2>I1B.I1>I2>I3.C.I2>I1>3.D.I3>I1>I2.正确答案:A解析:在区域D={(x,y)|x2+y2≤1}上,有0≤x2+y2≤1,从而有由于cosx 在上为单调减函数,于是故应选A.知识模块:多元函数微积分学2.已知fx(x0,y0)存在,则A.fx(x0,y0).B.0.C.2fx(x0,y0).D.正确答案:C解析:故选C.知识模块:多元函数微积分学3.设f(x,y)=则f(x,y)在点(0,0)处( )A.两个偏导数都不存在.B.两个偏导数存在但不可微.C.偏导数连续.D.可微但偏导数不连续.正确答案:B解析:由偏导数定义,有由对称性知fy’(0,0)=0,而上式极限不存在.事实上,故f(x,y)在(0,0)点不可微.故应选B.知识模块:多元函数微积分学4.已知为某二元函数u(x,y)的全微分,则a等于( )A.0.B.2.C.1.D.一1.正确答案:B解析:以上两式分别对y,x求偏导得知识模块:多元函数微积分学5.函数f(x,y)在(0,0)点可微的充分条件是( )A.B.C.D.正确答案:D解析:由且可知,f(x,y)的两个一阶偏导数fx(x,y)和fy(x,y)在(0,0)点可微,故选D.知识模块:多元函数微积分学6.设函数f(t)连续,则二次积分=( )A.B.C.D.正确答案:B解析:因为曲线r=2在直角坐标系中的方程为x2+y2=4,而r=2cosθ在直角坐标中的方程为x2+y2=2x或者(x一1)2+y2=1,因此根据直角坐标和极坐标之间二重积分的转化可知故选B.知识模块:多元函数微积分学7.设区域D={(x,y)|x2+y2≤4,x≥0,y≥0},f(x)为D上的正值连续函数,a,b为常数,则=( )A.abπ.B.C.(a+b)π.D.正确答案:D解析:由轮换对称性,有故应选D.知识模块:多元函数微积分学8.设z=f(x,y)在点(x0,y0)处可微,△z是f(x,y)在点(x0,y0)处的全增量,则在点(x0,y0)处( )A.△z=dz.B.△z=fx(x0,y0)△x+fy(x0,y0)△yC.△z=fx(x0,y0)dx+fy(x0,y0)dy.D.△z=dz+o(ρ)·正确答案:D解析:由于x=f(x,y)在点(x0,y0)处可微,则△z=fx(x0,y0)△x+fy(x0,y0)△y+o(ρ)=dz+o(ρ),故选D.知识模块:多元函数微积分学9.设则f(x,y)在点(0,0)处( )A.不连续.B.连续但两个偏导数不存在C.两个偏导数存在但不可微.D.可微.正确答案:D解析:f(x,y)一f(0,0)+2x—y=o(ρ)(当(x,y)→(0,0)时),即f(x,y)一f(0,0)=一2x+y+o(ρ),由微分的定义可知f(x,y)在点(0,0)处可微,故选D.知识模块:多元函数微积分学10.已知du(x,y)=(axy3+cos(x+2y))dx+(3x2y2+bcos(x+2y))dy,则( )A.a=2,b=一2.B.a=3,b=2C.a=2,b=2.D.a=一2,b=2正确答案:C解析:由du(x,y)=(axy3+cos(x+2y))dx+(3x2y2+bcos(x+2y))dy知以上两式分别对y,x求偏导得即3axy2一2sin(x+2y)=6xy2一bsin(x+2y),则a=2,b=2,故选C.知识模块:多元函数微积分学11.设函数u(x,y)=φ(x+y)+φ(x一y)+∫x-yx+yψ(t)dt,其中函数φ具有二阶导数,ψ具有一阶导数,则必有( )A.B.C.D.正确答案:B解析:先分别求出.再比较结果.知识模块:多元函数微积分学12.设f(x,y)为连续函数,则等于( )A.B.C.D.正确答案:C解析:本题考查将极坐标系下的累次积分转换为直角坐标系下的累次积分.首先由题设画出积分区域的图形,然后化为直角坐标系下累次积分.由题设可知积分区域D如图4—2所示,则原式=故选C.知识模块:多元函数微积分学填空题13.设f(x,y,z)=ex+y2z,其中z=z(x,y)是由方程x+y+z+xyz=0所确定的隐函数,则fx’(0,1,一1)=________.正确答案:1解析:已知f(x,y,z)=ex+y2z,那么有fx’(x,y,z)=ex+y2zx’.等式x+y+z+xyz=0两端对x求偏导可得1+zx’+yz+xyzx’=0.取x=0,y=1,z=一1,可得zx’=0.故fx’(0,1,一1)=e0=1.知识模块:多元函数微积分学14.设f(x,y)=在点(0,0)处连续,则a=_______.正确答案:0解析:因为利用夹逼原理知,又知f(0,0)=a,则a=0.知识模块:多元函数微积分学15.设正确答案:解析:由题意可知:则知识模块:多元函数微积分学16.设函数f(u,v)由关系式f[xg(y),y]=x+g(y)确定,其中函数g(y)可微,且g(y)≠0,则正确答案:解析:令u=xg(y),v=y,则f(u,v)= 知识模块:多元函数微积分学17.设z=z(x,y) 由方程z+ez=xy2所确定,则dz=______正确答案:解析:知识模块:多元函数微积分学18.设函数f(u)可微,则f’(2)=2,则z=f(x2+y2)在点(1,1)处的全微分dz|(1,1)=________正确答案:4(dx+dy)解析:由题干可知,dz=f’(x2+y2)(2xdx+2ydy),则dz|(1,1)=f’(2)(2dx+2dy)=4(dx+dy).知识模块:多元函数微积分学19.设f(u,v)为二元可微函数,z=f(xy,yx),则正确答案:f1’.yxy-1+f2’.yxlny解析:利用复合函数求偏导公式,有知识模块:多元函数微积分学20.设f,φ具有二阶连续导数,则正确答案:yf”(xy)+φ’(x+y)+yφ”(x+y)解析:由题干可得:知识模块:多元函数微积分学21.设z=xg(x+y)+yφ(xy),其中g、φ具有二阶连续导数,则正确答案:g’(x+y)+xg”(x+y)+2yφ’(xy)+xy2φ”(xy)解析:由题干可知,知识模块:多元函数微积分学22.设二元函数z=xex+y+(x+1)ln(1+y),则dz|(1,0)=________.正确答案:2edx+(e+2)dy解析:于是dz|(1,0)=2edx+(e+2)dy.知识模块:多元函数微积分学解答题解答应写出文字说明、证明过程或演算步骤。
近年考研数学三微积分题目整合及其详细解答
解:有可能 lim g(x) 与 lim ϕ(x) 都不存在,如ϕ(x) = x, f (x) = x + 1 , g(x) = x + 2 ,
x→∞
x→∞
x2
x2
则有ϕ(x) ≤ f (x) ≤ g(x) ,且 lim[g(x) − ϕ(x)] = 0 ,但 lim f (x) = ∞ ,
x→∞
x→∞
n 1 n(n + 1) + 2
= 1 = 2, 1 n(n − 1) 2 2 2
填空: 2 . 2
3. (99)设函数 f (x) = a x
(a > 0,
a ≠ 1) ,则 lim 1 n→∞ n 2
ln[ f (1) f (2)L f (n)] =
.
2
解:原式 =
lim
ln(a1
⋅
a2
Lan )
选择:(D).
3. (04)函数 f (x) = | x | sin(x − 2) 在下列哪个区间内有界,(
)
x(x − 1)(x − 2)2
(A) (−1, 0) .
(B) (0, 1) .
(C) (1, 2) .
(D) (2, 3) .
解:间断点有 x = 0, 1, 2 ,其中 x = 0 是可去间断点, x = 1, 2 是无穷间断点,故有界区间不能包含 x = 1, 2 ,
x→−∞
x→−∞
选择:(A).
3. ( 97 ) 若 f (−x) = f (x) (−∞ < x < +∞) , 在 (−∞, 0) 内 f ′(x) > 0 且 f ′′(x) < 0 , 则 在 (0, + ∞) 内 有
2011年考研数学三真题及答案
2011年考研数学三真题一、选择题(18小题,每小题4分,共32分。
下列媒体给出的四个选项中,只有一个选项是符合题目要求的。
)(1)已知当时,与是等价无穷小,则(A)(B)(C)(D)【答案】C。
【解析】【方法一】(洛必达法则)(洛必达法则)()由此得。
【方法二】由泰勒公式知则故。
【方法三】故综上所述,本题正确答案是C。
【考点】高等数学—函数、极限、连续—无穷小量的性质及无穷小量的比较,极限的四则运算高等数学—一元函数微分学—洛必达(L'Hospital)法则(2)已知在处可导,且,则(A)(B)(C)(D)0【答案】B。
【解析】【方法一】加项减项凑处导数定义【方法二】拆项用导数定义由于,由导数定义知所以【方法三】排除法:选择符合条件的具体函数,则而对于,显然选项(A)(C)(D)都是错误的,故应选(B)【方法四】由于在处可导,则综上所述,本题正确答案是B。
【考点】高等数学—一元函数微分学—导数和微分的概念,导数和微分的四则运算(3)设是数列,则下列命题正确的是(A)若收敛,则收敛。
(B)若收敛,则收敛。
(C)若收敛,则收敛。
(D)若收敛,则收敛。
【答案】A。
【解析】若收敛,则该级数加括号后得到的级数仍收敛综上所述,本题正确答案是A。
【考点】高等数学—无穷级数—级数的基本性质与收敛的必要条件(4)设,则的大小关系为(A)(B)(C)(D)【答案】B。
【解析】同一区间上定积分的大小比较最常用的思想就是比较被积函数大小,由于当时,又因为为上的单调增函数,所以,故即综上所述,本题正确答案是B。
【考点】高等数学—一元函数积分学—定积分的概念和基本性质(5)设为3阶矩阵,将第2列加到第1列得矩阵,再交换的第2行和第3行得单位矩阵,记,,则(A)(B)(C)(D)【答案】D。
【解析】本题是常规的初等变换、初等矩阵的考题矩阵的初等行变换是左乘初等矩阵,矩阵的初等列变换是右乘初等矩阵按题意,从而,从而所以【考点】线性代数—矩阵—矩阵的初等变换,初等矩阵(6)设为矩阵,是非齐次线性方程组的3个线性无关的解,为任意常数,则的通解为(A)(B)(C)(D)【答案】C。
考研数学三范围
考研数学三范围
考研数学三的范围主要包括以下几个方面:
一、高等数学部分:高等数学是数学三考试的一个重要部分,主要包括以下几个方面:
1.多元函数微积分学
2.级数和广义积分
3.常微分方程
4.偏微分方程
5.曲线积分和曲面积分
6.向量场
7.傅里叶级数和傅里叶变换
二、概率论与数理统计部分:概率论与数理统计也是数学三考试非常重要的一个部分,主要包括以下几个方面:
1.概率论的基础理论
2.离散型随机变量的概率分布
3.连续性随机变量的概率密度函数
4.多维随机变量的联合概率密度函数
5.随机过程的基本概念
6.假设检验的基础理论
7.方差分析的基本原理
三、线性代数部分:线性代数也是数学三考试的一个非常重要的部分,主要包括以下几个方面:
1.矩阵论的基本概念和性质
2.矩阵的特殊类型和它们的特征
3.线性方程组的解法和求解
4.行列式的计算和应用
5.向量空间和子空间
6.线性相关和线性无关性质
7.线性变换和矩阵的表示
以上就是考研数学三的范围,需要考生在备考过程中严格按照此范围进行复习,把握好各个方面的重点和难点,才能提高复习的效率,最终达到取得优异成绩的目的。
考研数学 高等数学结构图:极限和一元函数微分学
考研数学高等数学结构图:极限和一元函数微分学高数在考研数学中所占比例最少也是56%,可以说得高数者的数学。
因此考生对这部分知识必须“吃懂”、“吃透”,以下是凯程教育数学辅导老师为大家整理的:考研高等数学各知识点结构图,希望对大家的复习有所帮助。
考研数学高等数学结构图:多元函数微积分学高数在考研数学中所占比例最少也是56%,可以说得高数者的数学。
因此考生对这部分知识必须“吃懂”、“吃透”,以下是凯程教育数学辅导老师为大家整理的:考研高等数学各知识点结构图,希望对大家的复习有所帮助。
考研数学线性代数结构图:行列式线代在考研数学中所占比例最少也是22%,因此考生对这部分知识必须“吃懂”、“吃透”,以下是凯程教育数学辅导老师为大家整理的:考研线性代数各知识点结构图,希望对大家的复习有所帮助。
凯程教育:凯程考研成立于2005年,国内首家全日制集训机构考研,一直从事高端全日制辅导,由李海洋教授、张鑫教授、卢营教授、王洋教授、杨武金教授、张释然教授、索玉柱教授、方浩教授等一批高级考研教研队伍组成,为学员全程高质量授课、答疑、测试、督导、报考指导、方法指导、联系导师、复试等全方位的考研服务。
凯程考研的宗旨:让学习成为一种习惯;凯程考研的价值观口号:凯旋归来,前程万里;信念:让每个学员都有好最好的归宿;使命:完善全新的教育模式,做中国最专业的考研辅导机构;激情:永不言弃,乐观向上;敬业:以专业的态度做非凡的事业;服务:以学员的前途为已任,为学员提供高效、专业的服务,团队合作,为学员服务,为学员引路。
如何选择考研辅导班:在考研准备的过程中,会遇到不少困难,尤其对于跨专业考生的专业课来说,通过报辅导班来弥补自己复习的不足,可以大大提高复习效率,节省复习时间,大家可以通过以下几个方面来考察辅导班,或许能帮你找到适合你的辅导班。
师资力量:师资力量是考察辅导班的首要因素,考生可以针对辅导名师的辅导年限、辅导经验、历年辅导效果、学员评价等因素进行综合评价,询问往届学长然后选择。
考研数学三-多元函数微积分学(三)
考研数学三-多元函数微积分学(三)(总分:100.00,做题时间:90分钟)一、Section Ⅰ Use of English(总题数:1,分数:10.00)Starting with his review of Skinner's Verbal Behavior, Noam Chomsky had led the psycholinguists who argue that man has developed an innate (天生的) capacity for dealing with the linguistic universals common to all languages. Experience and learning then provide only information about the (1) instances of those universal aspects of language which are needed to communicate with other people within a particular language (2) .This linguistic approach (3) the view that language is built upon learned associations between words. What is learned is not strings of words per se (本身), but (4) rules that enable a speaker to (5) an infinite variety of novel sentences. (6) single words are learned as concepts: they do not stand in a one-to-one (7) with the particular thing signified, but (8) all members of a general class.This view of the innate aspect of language learning is at first not readily (9) into existing psychological frameworks and (10) a challenge that has stimulated much thought and new research directions. Chomsky argues that a precondition for language development is the existence of certain principles "intrinsic (原有的) to the mind" that provide invariant structures (11) perceiving, learning and thinking. Language (12) all of these processes; thus its study (13) our theories of knowledge in general.Basic to this model of language is the notion that a child's learning of language is a kind of theory (14) . It's thought to be accomplished (15) explicit instruction, (16) of intelligence level, at an early age when he is not capable of other complex (17) or motor achievements, and with relatively little reliable data to go on. (18) , the child constructs a theory of an ideal language which has broad (19) power. Chomsky argues that all children could not develop the same basic theory (20) it not for the innate existence of properties of mental organization which limit the possible properties of languages.(分数:10.00)(1).[A] special [B] specific [C] definite [D] explicit(分数:0.50)A.B. √C.D.解析:形容词辨析题。
4考研数学大纲知识点解析(第四章多元函数的微分学-数一
满足 .
.则
【解析】由题设
可知,当
时,有
且
,从而有
由二元函数全微分的定义, 有
在点
处可微,且
. ,
. ,故
【全微分存在的必要条件和充分条件】 【极限,连续,偏导数,可微分之间的关系】 一元函数:
二元函数:
【例题】(02 年,数学一)考虑二元函数
的下面 条性质:
①
在点
处连续. ②
在点
处的两个偏导数连续,
确.
选项(C),(D)取 不存在,故排除(C),(D).
,显然
在点
处可微,但
【综合题】设
在
点处( ).
(A)不连续. (B)偏导函数不存在. (C)不可微. (D)可微.
【解析】(1)
,
在
点连续.
(2)
同理
(3)
从而
不存在.
在
点不可微. 故选(C).
【综合题】设
则在
(A)偏导不存在. (B)偏导函数连续. (C)可微. (D)不可微.
第四章 多元函数的微分学 【多元函数的概念】 【二元函数的定义】
类似的可以定义三元函数 【二元函数的几何意义】 二元函数
. 一般表示空间直角坐标系下的一个空间曲面.
【二元函数极限的概念】
【注】二元函数极限存在,是指 以所有路径趋于
时,对应的函数值趋于相同
的一个常数.如果 沿着两条不同路径趋于
时,对应的函数值趋于不同的值,
设
有连续的一阶偏导数,又函数
及
分别由下列两式确定
:
求.
和
,
【解析】
由
两边对 求导,得
即
.
考研数学三(填空题)专项练习试卷10(题后含答案及解析)
考研数学三(填空题)专项练习试卷10(题后含答案及解析)题型有:1.1.=__________正确答案:解析:知识模块:微积分2.设A=,则(A+3E)一1(A2一9E)=________.正确答案:解析:(A+3E)一1(A2一9E)=(A+3E)一1(A+3E)(A一3E)=A一3E= 知识模块:线性代数3.交换积分次序:=________.正确答案:涉及知识点:多元函数微积分学4.设=_____________________。
正确答案:解析:对已知函数进行恒等变形,即则有所以知识模块:一元函数微分学5.二次型f(x1,x2,x3)=(x1+ax2-2x3)2+(2x2+3x3)2+(x1+3x2+ax3)2正定的充分必要条件为________.正确答案:a≠1 涉及知识点:微积分6.曲线y=(x2+x)/(x2-1)渐近线的条数为________.正确答案:2 涉及知识点:一元函数微分学7.=__________。
正确答案:+C解析:令x=tant,则dx=Sec2tdt,故知识模块:一元函数积分学8.设随机变量X的概率密度为f(x)=(一∞<x<+∞),则随机变量X的二阶原点矩为________.正确答案:解析:根据题意,即求E(X2).首先对所给概率密度作变换:对于x(一∞<x<+∞),有知识模块:概率论与数理统计9.正确答案:(x2一1)+C,其中C为任意常数解析:知识模块:微积分10.抛物线y2=ax(a>0)与x=1所围图形面积为则a=______.正确答案:1解析:y2=ax与x=1所围图形面积知识模块:微积分11.曲线直线x=2及x轴所围的平面图形绕x轴旋转所成的旋转体的体积为______.正确答案:解析:知识模块:一元函数积分学12.设总体X的概率密度为X1,X2,…,Xn是来自X的样本,则未知参数θ的最大似然估计值为_________.正确答案:解析:似然函数为解似然方程得θ的极大似然估计值为知识模块:概率论与数理统计13.设盒子中装有m个颜色各异的球,有放回地抽取n次,每次1个球.设X表示n次中抽到的球的颜色种数,则EX=______.正确答案:解析:令则X=X1+X2+…+Xm.事件“Xi=0”表示n次中没有抽到第i 种颜色的球,由于是有放回抽取,n次中各次抽取结果互不影响,因此有知识模块:概率论与数理统计14.已知一2是的特征值,则x=__________.正确答案:x=-4解析:因为一2是矩阵A的特征值,所以由知识模块:线性代数15.已知r(α1,α2,…,αs)=r(α1,α2,…,αs,β)=m,r(α1,α2,…,αs,γ)=m+1,则r(α1,α2,…,αs,β,γ)=________。
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考研数学三-多元函数微积分学(一)(总分:100.00,做题时间:90分钟)一、Section Ⅰ Use of Eng(总题数:1,分数:10.00)The mass media is a big part of our culture, yet it can also be a helper, adviser and teacher to our young generation. The mass media affects the lives of our young by acting as a (an) (1) for a number of institutions and social contacts. In this way, it (2) a variety of functions in human life.The time spent in front of the television screen is usually at the (3) of leisure: there is less time for games, amusement and rest. (4) by what is happening on the screen, children not only imitate what they see but directly (5) themselves with different characters. Americans have been concerned about the (6) of violence in the media and its (7) harm to children and adolescents for at least forty years. During this period, new media (8) , such as video games, cable television, music videos, and the Internet. As they continue to gain popularity, these media, (9) television, (10) public concern and research attention.Another large societal concern on our young generation (11) by the media, is body image. (12) forces can influence body image positively or negatively. (13) one, societaland cultural norms and mass media marketing (14) our concepts of beauty. In the mass media, the images of (15) beauty fill magazines and newspapers, (16) from our televisions and entertain us (17) the movies. Even in advertising, the mass media (18) on accepted cultural values of thinness and fitness for commercial gain. Young adults are presented with a (19) defined standard of attractiveness, a(n) (20) that carries unrealistic physical expectations.(分数:10.00)(1).[A] alternative [B] preference [C] substitute [D] representative(分数:0.50)A.B.C.D.(2).[A] accomplishes [B] fulfills [C] provides [D] suffices(分数:0.50)A.B.C.D.(3).[A] risk [B] mercy [C] height [D] expense(分数:0.50)A.B.C.D.(4).[A] Absorbed [B] Attracted [C] Aroused [D] Addicted(分数:0.50)A.B.C.D.(5).[A] identify [B] recognize [C] unify [D] equate(分数:0.50)A.B.C.(6).[A] abundance [B] incidence [C] prevalence [D] recurrence(分数:0.50)A.B.C.D.(7).[A] disposed [B] hidden [C] implicit [D] potential(分数:0.50)A.B.C.D.(8).[A] merged [B] emerged [C] immerged [D] submerged(分数:0.50)A.B.C.D.(9).[A] apart from [B] much as [C] but for [D] along with(分数:0.50)A.B.C.D.(10).[A] promote [B] propel [C] prompt [D] prosper(分数:0.50)A.B.C.D.(11).[A] inspired [B] imposed [C] delivered [D] contributed(分数:0.50)A.B.C.D.(12).[A] External [B] Exterior [C] Explicit [D] Exposed(分数:0.50)A.B.C.D.(13).[A] As [B] At [C] For [D] In(分数:0.50)A.B.C.D.(14).[A] mark [B] effect [C] impact [D] shock(分数:0.50)A.B.C.D.(15).[A] generalized [B] regularized [C] standardized [D] categorized(分数:0.50)A.B.D.(16).[A] boom [B] bottom [C] brim [D] beam(分数:0.50)A.B.C.D.(17).[A] over [B] with [C] on [D] at(分数:0.50)A.B.C.D.(18).[A] play [B] take [C] profit [D] resort(分数:0.50)A.B.C.D.(19).[A] barely [B] carefully [C] narrowly [D] subjectively(分数:0.50)A.B.C.D.(20).[A] ideal [B] image [C] stereotype [D] criterion(分数:0.50)A.B.C.D.二、Section Ⅱ Reading Co(总题数:0,分数:0.00)三、Part A(总题数:0,分数:0.00)四、Text 1(总题数:1,分数:10.00)Children attending schools located in high-traffic zones have a 45 percent increased risk of developing asthma, even though time spent at school only accounts for about one-third of a child's waking hours, according to new research.Asthma is the most common chronic childhood illness in developed countries and has been linked to environmental factors such as traffic-related air pollution. "While residentialtraffic-related pollution has been associated with asthma, there has been little study of the effects of traffic exposure at school on new onset asthma," says Rob McConnell, professor of preventive medicine at USC's Keck School of Medicine. "Exposure to pollution at locations other than home, especially where children spend a large portion of their day and may engage in physical activity, appears to influence asthma risk as well."The study appears online in the journal Environmental Health Perspectives. The study drew upon data from the Children's Health Study (CHS), a longitudinal study of children in Southern California communities that was designed to investigate the chronic effects of air pollution on respiratory health. Using a cohort of 2 497 kindergarten and first grade children who were asthma-free when they entered the study, researchers examined the relationship of local traffic around schools and homes to diagnosis new onset asthma that occurred during three years of follow-up. Traffic-related pollution exposure was assessed based on a model that took into account traffic volume, distance to major roadways from home and school and local weather conditions.Regional ambient ozone, nitrogen dioxide (二氧化氮) and particulate matter were measured continuously at one central site in each of the 13 study communities. The design allowed investigators to examine the joint effects of local traffic-related pollution exposure at school and at home and of regional pollution exposure affecting the entire community. Researchers found 120 cases of new asthma. The risk associated with traffic-related pollution exposure at schools was almost as high as for residential exposure, and combined exposure accounting for time spent at home and at school had a slightly larger effect. Although children spend less time at school than at home, physical education, and other activities that take place at school may increase ventilation rates and the dose of pollutants getting into the lungs, McConnell notes.Traffic-related pollutant levels may also be higher during the morning hours when children are arriving at school.Despite a state law that prohibits school districts from building campuses within 500 feet of a freeway, many Southern California schools are located near high-traffic areas, including busy surface streets."It's important to understand how these micro-environments where children spent a lot of their time outside of the home are impacting their health," McConnell says. "Policies that reduce exposure to high-traffic environments may help to prevent this disease. " The study was funded by grants from the National Institute of Environmental Health Sciences, the US Environmental Protection Agency, the South Coast Air Quality Management District, and the Hastings Foundation.(分数:10.00)(1).Which one is NOT the reason that children increase risk of developing asthma?[A] There exists traffic-related pollution. [B] Schools are located in heavy-traffic areas.[C] Children are frequently exposed to pollution. [D] The vehicles increase rapidly.(分数:2.00)A.B.C.D.(2).What kind of illness does "asthma" belong to?[A] Mental sickness. [B] Respiratory disease. [C] Influenza. [D] Infectious disease.(分数:2.00)A.B.C.D.(3).We can infer from McConnell's view that______.[A] there has been little study of the residential traffic-related pollution[B] activities that take place at school may increase the risk of asthma[C] traffic-related pollutant level is lower during the morning hours[D] frequent exposure to outdoors appears to influence the risk of asthma(分数:2.00)A.B.C.D.(4).The method of preventing asthma is to______.[A] reduce exposure to schools [B] do more activities at home[C] reduce exposure to heavy traffic zones [D] understand the micro-environments(分数:2.00)A.B.C.D.(5).Which of the following is TRUE according to the text?[A] In Southern California, many schools are located near heavy traffic zones.[B] The study was carried out by many organizations.[C] Rob McConnell is the leader of the study.[D] Asthma is the most common chronic childhood illness around the world.(分数:2.00)A.B.C.D.五、Text 2(总题数:1,分数:10.00)It is no longer just dirty blue-collar jobs in manufacturing that are being sucked offshore but also white-collar service jobs, which used to be considered safe from foreign competition. Telecoms charges have tumbled, allowing workers in far-flung locations to be connected cheaply to customers in the developed world. This has made it possible to offshore services that were once non-tradable. Morgan Stanley's Mr. Roach has been drawing attention to the fact that the "global labour arbitrage" is moving rapidly to the better kinds of jobs. It is no longer just basic data processing and call centres that are being outsourced to low-wage countries, but also software programming, medical diagnostics, engineering design, law, accounting, finance and business consulting. These can now be delivered electronically from anywhere in the world, exposing skilled white-collar workers to greater competition.The standard retort to such arguments is that outsourcing abroad is too small to matter much. So far fewer than lm American service-sector jobs have been lost to off-shoring. Forrester Research forecasts that by 2015 a total of 3.4m jobs in services will have moved abroad, but that is tiny compared with the 30m jobs destroyed and created in America every year. The trouble is that such studies allow only for the sorts of jobs that are already being off-shored, when in reality the proportion of jobs that can be moved will rise as IT advances and education improves in emerging economies.Alan Blinder, an economist at Princeton University, believes that most economists are underestimating the disruptive effects of off-shoring, and that in future two to three times as many service jobs will be susceptible to off-shoring as in manufacturing. This would imply that at least 30% of all jobs might be at risk. In practice the number of jobs off-shored to China or India is likely to remain fairly modest. Even so, the mere threat that they could be shifted will depress wages.Moreover, says Mr. Blinder, education offers no protection. Highly skilled accountants, radiologists or computer programmers now have to compete with electronically delivered competition from abroad, whereas humble taxi drivers, janitors and crane operators remain safe from off-shoring. This may help to explain why the real median wage of American graduates has fallen by 6% since 9000, a bigger decline than in average wages.In the 1980s and early 1990s, the pay gap between low-paid, low-skilled workers and high-paid, high-skilled Workers widened significantly. But since then, according to a study by David Autor, Lawrence Katz and Melissa Kearney, in America, Britain and Germany workers at the bottom as well as at the top have done better than those in the middle-income group. Office cleaning cannot be done by workers in India. It is the easily standardised skilled jobs in the middle, such as accounting, that are now being squeezed hardest. A study by Bradford Jensen and Lori Kletzer, at the Institute for International Economics in Washington D. C., confirms that workers in tradable services that are exposed to foreign competition tend to be more skilled than workers innon-tradable services and tradable manufacturing industries.(分数:10.00)(1).To offshore services that were once non-tradable results from ______.[A] the blue-collar job market[B] the geographic location of the underdeveloped world[C] the fierce competition among skilled workers[D] the dive of telecoms fee(分数:2.00)A.B.C.D.(2).Which of the following statements is the typical reply concerning off-shoring?[A] Service-sector has sustained a great loss.[B] White-collar workers will not have a narrow escape.[C] Most economists underestimated the effects of off-shoring.[D] Outsourcing abroad has no significant impact.(分数:2.00)A.B.C.D.(3).According to the text, Forrester Research Prediction might be different if ______.[A] outsourcing abroad is large enough to matter much[B] the proportion of jobs that can be moved will rise[C] more comprehensive factors are taken into account[D] education improvement in emerging economies plays a role(分数:2.00)A.B.C.D.(4).The narrative of the text in the last three paragraphs concentrates on ______.[A] the standard retort to the arguments[B] off-shoring and the resulting income[C] the future off-shoring[D] the counter-measures at hand(分数:2.00)A.B.C.D.(5).Which of the following could be the best title for the text?[A] Business consulting.[B] Blue-collar jobs.[C] Non-tradable services.[D] White-collar blues.(分数:2.00)A.B.C.D.六、Text 3(总题数:1,分数:10.00)Supporters of abortion rights held a lunch recently in honor of a momentous victory for their cause: 40 years ago, New York became the first state to fully legalize abortion. That 1970 law began to reduce the death and injury toll from back-alley abortions and set the stage for the Supreme Court's Roe v. Wade decision in 1973, which made abortion legal nationwide and recognized a constitutional right to privacy.But abortion-rights groups are newly anxious about new assaults on women's reproductive rights, including a fight over abortion that snarled the last days of the health care reform debate. Anti-abortion groups are newly emboldened. Kelli Conlin, head of Naral Pro-Choice New York, told guests at the lunch that "anti-choice forces are mobilizing in every single state to limit a woman's access to abortion in more insidious ways than we can imagine. "As Ms. Conlin was speaking, members of the Oklahoma House were getting ready to override vetoes of two punishing abortion measures. The state's Democratic governor, Brad Henry, rightly viewed these intrusions into women's lives and decision-making as unconstitutional. One of the measures, which seems destined to spawn copycat bills in other states, requires women to undergo an ultrasound before getting an abortion and further mandates that a doctor or technician set up the monitor so the woman can see it and hear a detailed description of the fetus. The other law grants protection from lawsuits to doctors who deliberately withhold fetal testing results that might affect a woman's decision about whether to carry her pregnancy to term.Several states have either passed or are considering bills that would ban abortion coverage in insurance plans sold through the state exchanges established by the federal health care law. A new Utah law criminalizes certain behavior by women that results in miscarriage. Embarking on a road that could lead to the Supreme Court, Nebraska last month banned most abortions at the 20th week of pregnancy based on a questionable theory of fetal pain. About two dozen states are looking at bills to increase counseling requirements or waiting periods prior to abortions. About 20 states are considering new ultrasound requirements. "One in three women in this country will have an abortion in her lifetime, and yet we're having exactly the same discussions and debates we were having forty years ago," Ms. Conlin said.Anti-abortion forces aim ultimately to make abortion illegal. So far, by reducing the number of abortion providers, making insurance coverage more expensive and harder to get, and throwing up other obstacles, they have primarily succeeded in making it harder for women of modest and meager means to obtain a safe and legal medical procedure.The painful decision to end a pregnancy should be made in private between a woman and her doctor —not in politically driven debate among members of Congress and state legislatures.(分数:10.00)(1).The Supreme Court's 1973 decision was important because[A] it made anti-abortion groups launch fiercer assaults.[B] it confirmed women's private rights to reproduction.[C] it put the 1970 law of New York state into wide application.[D] it marked the end of the efforts by abortion-rights groups.(分数:2.00)A.B.C.D.(2).Members of the Oklahoma House were trying to[A] make a similar law to that of the New York state.[B] beat anti-choice forces in their mobilization.[C] formulate two measures for punishing abortion.[D] make matters of abortion difficult for women.(分数:2.00)A.B.C.D.(3).Utah and Nebraska are mentioned to show[A] it is illegal for the two states to ban abortion coverage in insurance.[B] the states have no right in criminalizing abortion or miscarriage.[C] some states use subtle ways to restrict women's access to abortion.[D] many laws should be made by the Supreme Court, not by state legislatures.(分数:2.00)A.B.C.D.(4).Which of the following is NOT mentioned as an anti-abortion endeavor?[A] Specifying the conditions under which abortion can occur.[B] Reducing or banning abortion coverage in insurance plans.[C] Allowing one in three women to undergo abortion.[D] Raising the amount of abortion insurance coverage.(分数:2.00)A.B.C.D.(5).According to the author, the anti-abortion forces have done nothing but[A] spreading peril for women's privacy and freedom.[B] increasing insurance coverage for pregnant women.[C] obtaining safe and legal medical procedure for women.[D] ending pregnant women's pain in decision making.(分数:2.00)A.B.C.D.七、Text 4(总题数:1,分数:10.00)With a series of well-timed deals, private-equity firms are giving traditional media- managers cause to be envious. The Warner Music transaction, in which Edgar Bronfman junior and three private-equity firms paid Time Warner $ 2.6 billion for the unit in 2003, is already judged a financial triumph for the buyers. Their success is likely to draw still more private-equity into the industry. And the investments are likely to get bigger: individual private-equity funds are growing--a $10 billion fund is likely this year--so even the biggest media firms could come within range, especially if private-equity investors club together.Some private-equity firms have long put money in media assets, but mostly reliable, relatively obscure businesses with stable cash flows. Now, some of them are placing big strategic bets on the more volatile bits, such as music and movies. And they are currently far more confident than the media old guard that the advertising cycle is about to turn sharply upwards.One reason why private-equity is making its presence felt in media is that it has a lot of money to invest. Other industries are feeling its weight too. But private-equity's buying spree (狂购乱买) reveals a lot about the media business in particular. Media conglomerates( 联合公司) lack the confidence to make big acquisitions, after the last wave of deals went wrong. Executives at Time Warner, for instance, which disastrously merged with AOL in 2000, wanted to buy MGM, a movie studio, but the board (it is said) were too nervous. Instead, private- equity firms combined withSony, a consumer-electronics giant, to buy MGM late last year.Private-equity's interest also reflects the fact that revenue growth in media businesses such as broadcast TV and radio is now hard to come by. The average annual growth rate for 12 categories of established American media businesses in 1998-2003, excluding the internet, was just 3.4% , says Veronis Suhler Stevenson, an investment bank. Private-equity puts a higher value onlow-growth, high cashflow assets than the public stockmarket, says Jonathan Nelson, founder of Providence Equity Partners, a media-focused private-equity firm.What private-equity men now bring to the media business, they like to think, is financial discipline plus an enthusiastic attitude towards new technology. Old-style media managers, claim the newcomers, are still in denial about how technology is transforming their industry. Traditional media managers grudgingly agree that, so far, private-equity investors are doing very nicely indeed from their entertainment deals. The buyers of Warner Music have already got back most of their $ 2.6 billion from the firm by cutting costs, issuing debt and making special payouts to shareholders. This year, its investors are expected to launch an initial public offering, which could bring them hundreds of millions more.(分数:10.00)(1).The best title for the text might be[A] Private-Equity and the Media Industry.[B] Private-Equity and Traditional Media Managers.[C] Private-Equity's Role in Economic Development.[D] Private-Equity Has Much Profit to Make.(分数:2.00)A.B.C.D.(2).Traditional media managers might be envious because[A] Warner Music was acquired by private-equity firms at a very low price.[B] private-equity firms seem to be benefiting a lot from recent acquisitions.[C] more and more private-equity firms are entering the media industry.[D] even the biggest media firms could be acquired by private-equity firms.(分数:2.00)A.B.C.D.(3).The word "volatile" ( paragraph 2 ) most probably means[A] changeable. [B] stable. [C] versatile. [D] expensive.(分数:2.00)A.B.C.D.(4).It can be inferred from the text that[A] private-equity pays more attention to fast growing industries.[B] newcomers deny the fact that technology is vital to media industry.[C] traditional media managers often deny the importance of technology.[D] the public stock market accentuates business with more cashflow.(分数:2.00)A.B.C.D.(5).The case of the executives at Time Warner was cited to show that[A] it was a big disaster for them to have merged with AOL in 2000.[B] the board of Time Warner was not qualified to lead the company.[C] MGM would give its buyer a sharper competitive edge.[D] leading media companies were "once bitten, twice shy".(分数:2.00)A.B.C.D.八、Part B(总题数:1,分数:10.00)In the following text, some sentences have been removed. For Questions (41-45), choose the most suitable one from the list A-G to fit into each of the numbered blank. There are two extra choices, which do not fit in any of the gaps.English has become the world's number one language in the 20th century. In every country where English is not the native language, especially in the Third World, people must strive to learn it to the best of their abilities, if they want to participate fully in the development of their countries. 41.______A close examination reveals a great number of languages have fallen casualty to English. For example, it has wiped out Hawaiian, Welsh, Scotch Gaelic, Irish, native American languages, and many others. Luckily, some of these languages are now being revived, such as Hawaiian and Welsh, and these languages will live again, hopefully, if dedicated people continue their work of reviving them. 42.______If this situation continues, the native or official languages of these countries will certainly die within two or three generations. This phenomenon has been called linguistic genocide. A language dies if it is not fully used in most activities, particularly as a medium of instruction in schools.43.______According to many studies, only around 20 to 25 percent of students in these countries can manage to learn the language of instruction (English) as well as basic subjects at the same time. Many leaders of these Third World countries are obsessed with English and for them English is everything. They seem to believe that if the students speak English, they are already knowledgeable. These leaders speak and write English much better than their national languages. If these leaders deliver speeches anywhere in the world they use English and they feel more at home with it and proud of their ability as well. The citizens of their countries do not understand their leaders' speeches because they are made in a foreign language.All the greatest countries of the world are great because they constantly use their own languages in all national development activities, including education. From a psychological point of view, those who are taught in their own language from the start will develop better self-confidence and self-reliance. From a linguistic point of view, the best brains can only be produced if students are educated in their own language from the start. 44.There is nothing wrong, however, in learning a foreign language at advanced levels of education.45.[A] But many people are concerned that English's dominance will destroy native languages.[B] But the best thing to do is to have a good education in one's native language first, then go abroad to have a university education in a foreign language.[C] Suppose you work in a big firm and find English very important for your job because you often deal with foreign businessmen. Now you are looking for a place where you can improve your English, especially your spoken English.[D] Nonetheless, a world full of different languages will disappear if the present trend in many countries to use English to replace the national or official languages in education, trade andeven politics continues.[E] Those who are taught in a foreign language from the start will tend to be imitators and lack self-confidence. They will tend to rely on foreign consultants.[F] Here are some advertisements about English language training from newspapers. You may find the information you need.[G] The Third World countries that are now using English as a medium of instruction are depriving75 percent of their future leaders of a proper education.(分数:10.00)填空项1:__________________填空项1:__________________填空项1:__________________填空项1:__________________填空项1:__________________九、Part C(总题数:1,分数:10.00)Our daily existence is divided into two phases, as distinct as day and night. We call them work and play. We work many hours a day and we allow the necessary minimum for such activities as eating and shopping. 46) The rest we spend in various activities which are known as recreations, an elegant word which disguises the fact that we usually do not even play in our hours of leisure, but spend them in various forms of passive enjoyment or entertainment.We need to make, therefore, a hard-and-fast distinction not only between work and play but, equally, between active play and passive entertainment. 47) It is, I suppose, the decline of active play —of amateur sport —and the enormous growth of purely receptive entertainment which have given rise to a sociological interest in the problem. If the greater part of the population, instead of indulging in sport, spend their hours of leisure "viewing" television programs, there will inevitably be a decline in health and physique. In addition, we have yet to trace the mental and moral consequences of prolonged diet of sentimental or sensational spectacles on the screen. 48) There is, if we are optimistic, the possibility that the diet is too thin and unnourishing to have much permanent effect on anybody. Nine films out of ten seem to leave absolutely no impression on the mind or imagination of those who have seen them.49) It is only when entertainment is active, participated in, practiced, that it can properly be called play, and as such it is a natural use of leisure. In that sense play stands in contrast to work, and is usually regarded as an activity that alternates with work.Work itself is not a single concept. We say quite generally that we work in order to make a living. Some of us work physically, tilling the land, minding the machines, digging the coal; others work mentally, keeping accounts, inventing machines, teaching and preaching, managing and governing.50) There does not seem to be any factor common to all these diverse occupations, except that they consume our time, and leave us little leisure.(分数:10.00)__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ 十、Section Ⅲ Writing(总题数:0,分数:0.00)十一、Part A(总题数:1,分数:10.00)。