西北师范大学2015年《3027调和分析》攻读博士学位研究生入学考试试题
北师大《心理学研究方法》博士入学考试历年真题(2002-2011)
北师大《心理学研究方法》试题1研究课题的确定1.1研究课题的论证1．从你自己熟悉的研究领域中选择一个研究课题，撰写一份简要的课题论证报告。
25分(2002,2003)课题论证是对拟研究的课题的价值性、科学性和可行性等方面进行分析、说明、预测和评价。
就其主要内容来说，一般是围绕以下几个问题展开：A研究目的和意义；B研究现状及不足；C本研究的创新；D研究方法：设计、被试、材料和程序、统计分析；E可预测的成果；F研究的可行性：已具备的条件(自身和研究队伍水平、能力，设备、经费)2研究设计2.1定性研究1．在质性研究中，有哪些收集数据的方法？这些方法各有什么弊端？如何处理质性数据？8分(2002,2006)质性研究是非定量的数据收集和分析的一种方法，它经常运用分析与综合、比较与分类、归纳与演绎等逻辑分析方法对资料进行解释和建构。
其特点有：建立在描述基础上的逻辑分析或推断；侧重揭示心理现象背后的意义；倾向于运用归纳分析的方法；不仅注重对结果和产品的分析，更重视对过程和相互关系的分析。
由于其主要特征之一便是以人的语言、行动为主要数据来源，亦即质性数据收集，所以必然常常使用访谈法(常用半结构访谈)、观察法(常用参与式观察)、质性个案研究、档案分析。
由于访谈是访问者与被访问者相互作用的过程，因此访问者搜集的资料，形成的意见看法等都要不同程度地受到被访问者的影响。
访谈法收集到的资料的质量好坏常常取决于访问者个人的人际交往能力，访问技巧的熟练程度以及对访谈过程的有效控制。
对于尖锐、敏感和隐私问题，被访问者一般不愿当面回答，或者不作真实回答，这些都会对访问结果产生不利影响。
与其它调查方法相比，访问调查的费用较高，费时较长，需要的人力较多，因而限制了它的规模。
由于观察是研究者以局內人的角色进入研究情境，因此搜集的资料，形成的意见看法等都要不同程度地受到被观察者的影响。
同时，由于观察者在场，使得被观察者的言行表现与平时不一样。
2015年考研数学(二)真题及答案详解
2015年全国硕士研究生入学统一考试数学（二）试题解析一、选择题:1 8小题,每小题4分,共32分.下列每题给出的四个选项中,只有一个选项符合题目要求的,请将所选项前的字母填在答题纸．．．指定位置上. (1) 下列反常积分收敛的是 ( )(A)2+∞⎰(B) 2ln x dx x+∞⎰(C)21ln dxx x +∞⎰(D) 2x x dx e+∞⎰【答案】(D) 【解析】(1)xx x dx x e e-=-+⎰，则2222(1)3lim (1)3xx x x x dx x e e x e e e +∞+∞----→+∞=-+=-+=⎰.(2) 函数()2sin lim(1)x tt t f x x→=+在(,)-∞+∞内( )(A) 连续 (B) 有可去间断点 (C) 有跳跃间断点 (D) 有无穷间断点 【答案】(B)【解析】220sin lim 0sin ()lim(1)t x t x x t x tt t f x e e x→→=+==，0x ≠，故()f x 有可去间断点0x =. (3) 设函数()1cos ,00,0x x x f x x α⎧>⎪=⎨⎪≤⎩(0,0)αβ>>,若()'f x 在0x =处连续则:( ) (A)0αβ-> (B)01αβ<-≤ (C)2αβ-> (D)02αβ<-≤ 【答案】(A)【解析】0x <时，()0f x '=()00f -'=()1001cos10lim lim cosx x x x f x x x ααβ++-+→→-'== 0x >时，()()()11111cos1sin f x x x x x x ααβββαβ-+'=+-- 1111cossin x x x xααβββαβ---=+()f x '在0x =处连续则：()()10100lim cos 0x f f x xαβ+--+→''===得10α-> ()()++1100110lim =lim cos sin =0x x f f x x x x x ααβββαβ---→→⎛⎫''=+ ⎪⎝⎭得：10αβ-->，答案选择A(4)设函数()f x 在(),-∞+∞内连续，其中二阶导数()''f x 的图形如图所示，则曲线()=y f x 的拐点的个数为( )(A) 0 (B) 1 (C) 2 (D) 3 【答案】(C)【解析】根据图像观察存在两点，二阶导数变号.则拐点个数为2个.(5) 设函数(),f u v 满足22,y f x y x y ⎛⎫+=- ⎪⎝⎭ ，则11u v fu==∂∂与11u v f v==∂∂ 依次是 ( )(A)1,02 (B) 10,2 (C) 1,02- (D) 10,2-【答案】(D)【解析】此题考查二元复合函数偏导的求解. 令,y u x y v x =+=，则,11u uv x y v v ==++，从而22(,)y f x y x y x+=-变为222(1)(,)111u uv u v f u v v v v -⎛⎫⎛⎫=-= ⎪ ⎪+++⎝⎭⎝⎭.故222(1)2,1(1)f u v f u u v v v ∂-∂==-∂+∂+， 因而111110,2u u v v ff uv ====∂∂==-∂∂.故选（D ）. (6)设D 是第一象限由曲线21xy =，41xy =与直线y x =，y =围成的平面区域，函数(),f x y 在D 上连续，则(),Df x y dxdy =⎰⎰ ( )(A)()13sin2142sin2cos ,sin d f r r rdr πθπθθθθ⎰⎰(B)()34cos ,sin d f r r rdr ππθθθ⎰ (C)()13sin 2142sin 2cos ,sin d f r r drπθπθθθθ⎰⎰(D)()34cos ,sin d f r r dr ππθθθ⎰【答案】(B)【解析】根据图可得，在极坐标系下计算该二重积分的积分区域为(,)43D r r ππθθ⎧⎫=≤≤≤≤⎨⎩所以34(,)(cos ,sin )Df x y dxdy d f r r rdr ππθθθ=⎰⎰⎰故选B.(7) 设矩阵21111214a a ⎛⎫ ⎪= ⎪ ⎪⎝⎭A ,21d d ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭b .若集合}{1,2Ω=，则线性方程组=Ax b 有无穷多解的充分必要条件为 ( )(A) ,a d ∉Ω∉Ω (B) ,a d ∉Ω∈Ω (C) ,a d ∈Ω∉Ω (D) ,a d ∈Ω∈Ω 【答案】(D)【解析】2211111111(,)1201111400(1)(2)(1)(2)A b ad a d a d a a d d ⎛⎫⎛⎫⎪ ⎪=→-- ⎪ ⎪ ⎪ ⎪----⎝⎭⎝⎭，由()(,)3r A r A b =<，故1a =或2a =，同时1d =或2d =.故选（D ）(8) 设二次型()123,,f x x x 在正交变换=x Py 下的标准形为2221232y y y +-，其中123(,,)=P e e e ，若132(,,)=-Q e e e 则123(,,)f x x x =在正交变换=x Qy 下的标准形为( )(A)2221232y y y -+ (B) 2221232y y y +-(C) 2221232y y y -- (D) 2221232y y y ++【答案】(A)【解析】由x Py =，故222123()2T T T f x Ax y P AP y y y y ===+-. 且200010001TP AP ⎛⎫⎪= ⎪ ⎪-⎝⎭.由已知可得100001010Q P PC ⎛⎫⎪== ⎪ ⎪-⎝⎭故200()010001T T TQ AQ C P AP C ⎛⎫⎪==- ⎪ ⎪⎝⎭所以222123()2T T T f x Ax y Q AQ y y y y ===-+.选（A ） 二、填空题：9 14小题,每小题4分,共24分.请将答案写在答题纸．．．指定位置上. (9) 3arctan 3x t y t t=⎧⎨=+⎩ 则 212t d y dx ==【答案】48【解析】 2222333(1)11dy dy t dt t dx dxdt t +===++ 2222[3(1)]d y d t dx dx=+=222222[3(1)]12(1)12(1)11d t t t dt t t dx dt t ++==++ 22148t d ydx ==. (10)函数2()2x f x x =⋅在0x =处的n 阶导数(0)nf =_________ 【答案】()()21ln 2n n n --【解析】根据莱布尼茨公式得：()()()()()(2)222(1)0222ln 2(1)ln 22n n n n x n x n n f C n n ---=-===- (11) 设()f x 连续，()()20x x x f t dt ϕ=⎰，若()()11,15ϕϕ'==，则()1f =【答案】2【解析】 已知2()()x x x f t dt ϕ=⎰，求导得2220()()2()x x f t dt x f x ϕ'=+⎰，故有1(1)()1,f t dt ϕ==⎰(1)12(1)5,f ϕ'=+=则(1)2f =.(12)设函数()y y x =是微分方程'''20y y y +-=的解，且在0x =处()y x 取得极值3，则()y x = .【答案】22x x e e -+【解析】由题意知：()03y =，()00y '=，由特征方程：220λλ+-=解得121,2λλ==- 所以微分方程的通解为：212x x y C e C e -=+代入()03y =，()00y '=解得：12C =21C = 解得：22xxy e e-=+(13)若函数(),Z z x y =由方程231x y ze xyz +++=确定，则()0,0dz = .【答案】()1d 2d 3x y -+ 【解析】当0,0x y ==时0z =，则对该式两边求偏导可得2323(3)x y z x y z ze xy yz e x++++∂+=--∂ 2323(3)2x y z x y z ze xy xz e y++++∂+=--∂.将（0,0,0）点值代入即有 12,.(0,0)(0,0)33z z x y ∂∂=-=-∂∂则可得()(0,0)121|d 2d .333dz dx dy x y =--=-+ (14) 若3阶矩阵A 的特征值为2,2,1-，2B A A E =-+，其中E 为3阶单位阵，则行列式B = .【答案】21【解析】A 的所有特征值为2,2,1.-B 的所有特征值为3,7,1. 所以||37121B =⨯⨯=.三、解答题：15～23小题,共94分.请将解答写在答题纸．．．指定位置上.解答应写出文字说明、证明过程或演算步骤. (15) (本题满分10分）设函数()ln(1)sin f x x a x bx x =+++，3()g x kx =.若()f x 与()g x 在0x →时是等价无穷小，求,,a b k 的值.【答案】111,,32a kb =-=-=- 【解析】 方法一：因为233ln(1)()23x x x x o x +=-++，33sin ()3!x x x o x =-+， 那么，23333000(1)()()()ln(1)sin 231lim lim lim ()x x x a aa xb x x o x f x x a x bx x g x kx kx→→→++-+++++===， 可得：100213a ab ak⎧⎪+=⎪⎪-=⎨⎪⎪=⎪⎩，所以，11213a b k ⎧⎪=-⎪⎪=-⎨⎪⎪=-⎪⎩．方法二： 由题意得300sin )1ln(lim )()(lim1kx xbx x a x x g x f x x +++==→→203cos sin 11limkx x bx x b x ax ++++=→由分母03lim 2=→kx x ，得分子)cos sin 11(lim 0x bx x b xax ++++→0)1(lim 0=+=→a x ，求得c ；于是)()(lim10x g x f x →=23cos sin 111lim kx x bx x b x x +++-=→）（x kx xx bx x x b x x +++++=→13cos )1(sin )1(lim20 203c o s )1(s i n )1(lim kx xx bx x x b x x ++++=→kxxx bx x bx x x b x x b x b x 6sin )1(cos cos )1(cos )1(sin 1lim0+-++++++=→由分母06lim 0=→kx x ，得分子]sin )1(cos cos )1(2sin 1[lim 0x x bx x bx x x b x b x +-++++→0)cos 21(lim 0=+=→x b x ，求得21-=b ； 进一步，b 值代入原式)()(lim 10x g x f x →=kxx x x x x x x x x 6sin )1(21cos 21cos )1(sin 211lim0++-+--=→ kxx x x x x x x x x x x x x x 6cos )1(21sin 21sin )1(21sin 21cos 21sin )1(cos cos 21lim 0++++++-++--=→k621-=，求得.31-=k(16) (本题满分10分)设A>0，D 是由曲线段sin (0)2y A x x π=≤≤及直线0y =，2x π=所围成的平面区域，1V ，2V 分别表示D 绕x 轴与绕y 轴旋转成旋转体的体积，若12V V =，求A 的值.【答案】8π【解析】由旋转体的体积公式，得dx x f ⎰=2021)(V ππdx x A ⎰=202)sin (ππdx x A⎰-=20222cos 1ππ422A π=dx x xf ⎰=22)(2V ππA x d x A -πππ2c o s 220==⎰由题,V V 21=求得.8A π=(17) (本题满分11分)已知函数(,)f x y 满足"(,)2(1)x xy f x y y e =+，'(,0)(1)xx f x x e =+，2(0,)2f y y y =+，求 (,)f x y 的极值. 【答案】极小值(0,1)1f -=-【解析】xxye y y xf )1(2),(+=''两边对y 积分，得 )()21(2),(2x e y y y x f x x ϕ++=')()2(2x e y y x ϕ++=， 故x x e x x x f )1()()0,(+=='ϕ， 求得)1()(+=x e x x ϕ，故)1()2(),(2x e e y y y x f x x x +++='，两边关于x 积分，得⎰+++=dx x e e y y y x f x x )1()2(),(2⎰+++=xxde x e y y )1()2(2 ⎰-+++=dx e e x e y y xxx )1()2(2 C )1()2(2+-+++=x x x e e x e y y C )2(2+++=x x xe e y y由y y y y y f 2C 2),0(22+=++=，求得.0=C 所以x x xe e y y y x f ++=)2(),(2.令⎪⎩⎪⎨⎧=+='=+++='0)22(0)2(2xy xx x x e y f xe e e y y f ，求得⎩⎨⎧-==10y x . 又x x x xxxe e e y y f +++=''2)2(2， x xye yf )1(2+=''，xyy e f 2=''， 当1,0-==y x 时，(0,1)1,xxA f ''=-=,0)1,0(B =-''=xy f 2)1,0(=-''=yy fC ， 20,AC B ->(0,1)1f -=-为极小值.(18) (本题满分10分) 计算二重积分()Dx x y dxdy +⎰⎰，其中{}222(,)2,D x y x y y x =+≤≥【答案】245π-【解析】2()DDx x y dxdy x dxdy +=⎰⎰⎰⎰21202xdx dy =⎰12202)x x dx =⎰12240022222sin 2cos 55x t xt tdt π=--⎰⎰22242002222sin 2sin .5545u t tdt udu πππ==-=-=-⎰⎰(19)(本题满分 11 分) 已知函数()21Xf x =+⎰⎰，求()f x 零点的个数？【答案】2个【解析】()21)f x x '=- 令()0f x '=,得驻点为12x =, 在1(,)2-∞,()f x 单调递减,在1(,)2+∞,()f x 单调递增 故1()2f 为唯一的极小值,也是最小值.而112241()2f =+=-⎰⎰⎰1224=--⎰⎰⎰在1(,1)2故0-<从而有1()02f <1lim ()lim[]x x x f x →-∞→-∞=+=+∞⎰⎰22111lim ()lim[]lim[]x x xx x x f x →+∞→+∞→+∞=+=-⎰⎰⎰⎰考虑2lim lim x x x ==+∞，所以lim ()x f x →+∞=+∞.所以函数()f x 在1(,)2-∞及1(,)2+∞上各有一个零点，所以零点个数为2. (20) (本题满分10分)已知高温物体置于低温介质中，任一时刻该物体温度对时间的变化率与该时刻物体和介质的温差成正比，现将一初始温度为120C ︒的物体在20C ︒的恒温介质中冷却，30min后该物体降至30C ︒，若要将该物体的温度继续降至21C ︒，还需冷却多长时间？ 【答案】30min【解析】设t 时刻物体温度为()x t ，比例常数为(0)k >，介质温度为m ,则()dxk x m dt=--，从而()kt x t Ce m -=+， (0)120,20x m ==，所以100C =，即()10020kt x t e -=+又1()30,2x =所以2ln10k =，所以11()20100t x t -=+ 当21x =时，t =１，所以还需要冷却３０min.(21) (本题满分10分)已知函数()f x 在区间[]+a ∞，上具有2阶导数，()0f a =，()0f x '>，()''0f x >，设b a >，曲线()y f x =在点()(),b f b 处的切线与x 轴的交点是()00x ，，证明0a x b <<.【证明】根据题意得点(,())b f b 处的切线方程为()()()y f b f b x b '-=-令0y =，得0()()f b x b f b =-' 因为(x)0f '>所以(x)f 单调递增，又因为(a)0f = 所以(b)0f >，又因为()0f b '>所以0()()f b x b b f b =-<' 又因为0()()f b x a b a f b -=--'，而在区间（a,b ）上应用拉格朗日中值定理有 (b)f(a)(),(a,b)f f b aξξ-'=∈-所以0()()()()()()()()()()()f b f b f b f b f x a b a f b f b f f b f b f ξξξ''--=--=-=''''' 因为(x)0f ''>所以(x)f '单调递增 所以()()f b f ξ''>所以00x a ->，即0x a >，所以0a x b <<，结论得证.(22) (本题满分 11 分)设矩阵101101a A a a ⎛⎫ ⎪=- ⎪ ⎪⎝⎭且3A O =.(1) 求a 的值；(2) 若矩阵X 满足22X XA AX AXA E --+=，E 为3阶单位阵，求X .【答案】2010,111211a X -⎛⎫ ⎪==-- ⎪ ⎪-⎝⎭【解析】 (I)323100100111100011a A O A a a a a a a a a=⇒=⇒-=--==⇒=- (II)由题意知()()()()()()()()()222211122212X XA AX AXA E X E A AX E A E E A X E AE X E A E A E A E A X E A A ------+=⇒---=⎡⎤⇒--=⇒=--=--⎣⎦⇒=-- 2011111112E A A -⎛⎫ ⎪--=- ⎪ ⎪--⎝⎭，011100111010111010011100112001112001----⎛⎫⎛⎫ ⎪ ⎪-→- ⎪ ⎪ ⎪ ⎪----⎝⎭⎝⎭M M M M M M111010111010011100011100021011001211------⎛⎫⎛⎫ ⎪ ⎪→--→-- ⎪ ⎪ ⎪ ⎪-----⎝⎭⎝⎭M M M M M M110201100312010111010111001211001211---⎛⎫⎛⎫ ⎪ ⎪→-→- ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭M M M M M M312111211X -⎛⎫ ⎪∴=- ⎪ ⎪-⎝⎭(23) (本题满分11 分)设矩阵02313312A a -⎛⎫ ⎪=-- ⎪ ⎪-⎝⎭相似于矩阵12000031B b -⎛⎫ ⎪= ⎪ ⎪⎝⎭.（1）求,a b 的值；（2）求可逆矩阵P ，使1P AP -为对角阵.【答案】（1）4,5a b ==；（2）231101011P --⎛⎫ ⎪=- ⎪ ⎪⎝⎭【解析】(I)~()()311A B tr A tr B a b ⇒=⇒+=++0231201330012031--=⇒--=-A B ba 14235-=-=⎧⎧∴⇒⎨⎨-==⎩⎩a b a a b b (II)023100123133010123123001123A E C ---⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=--=+--=+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭⎝⎭ ()123112*********---⎛⎫⎛⎫ ⎪ ⎪=--=-- ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭CC 的特征值1230,4λλλ===0λ=时(0)0-=E C x 的基础解系为12(2,1,0);(3,0,1)ξξ==-T T 5λ=时(4)0-=E C x 的基础解系为3(1,1,1)ξ=--T A 的特征值1:1,1,5λλ=+A C令123231(,,)101011ξξξ--⎛⎫ ⎪==- ⎪ ⎪⎝⎭P ，1115-⎛⎫ ⎪∴= ⎪⎪⎝⎭P AP文档内容由金程考研网整理发布。
西北师范大学2019年博士研究生入学考试英语试题
西北师范大学2019年博士研究生入学考试英语试题博士研究生入学考试英语试题考试科目名称：英语试题适用招生专业：全校Part L Listening Comprehension ( 25％)Section A: Spot DictationDirections: In this section, you are going to hear a passage. The passage will he read only once. As you listen to the passage, fill in the blanks with the words you hear. After the passage, there will be a 3-minute pause. During the pause, you must write the words on the Answer Sheet.A recent university research project investigated the attitudes of postgraduate science students (1)____the learning of English vocabulary. The results were urprising. I'll (2) ____three of them.firstly, most of the stcrdeaats think that (3) ____every word ill English has just one meaning. This is, of course:, completely (4) ____to the facts. A glance at any English dictionary will show this. The student will (5) ____find seven or eight meanings listed for (6) ____simple' words.Why, then, have these students made such a mistake:' One reason irnay be that they're .ill (7) ____. students. Scientists try to use words ill their special subject which have one meaning, and one meaning only. Another reason., of course,could be the way in Which these Student, Were They may have used vocabulary lists when they first learner English. (M one side of the page is the word in Iaaglish-, on the other sloe, a single \ti'ord in the (Q) native language.'l°he second attitude that (10) ____from the findings isequally mistaken. (11) ____all the students think that every word in English has an exact (12)____equivalent. Again, this is far from the trijth. Sometimes one word in Iinglish can only be translated by a (13) ____in the student's native languial c. "there are other (l4) ____ill translation which we won't mention here. (:ertainly the idea of a one word for one word translation (15) ____is completely false. Translation machines, which tried to work on this (16) ____failed completely.The third result'of the investigation showed another (17) ____in the students' thinking. They believe that as soon as they know the meaning of a word, they're in a (18) ____to use it correctly. This is untrue for any language but is perhaps particularly (19) ____for English. The student has to learn when to use a word as well as to know what it means. Some words in English mean almost the same but they can only be used in certain situations.What, then, is the best way to increase one's vocabulary? This can be answered in threewords-observation, (20) ____and repetition.Section B: Multiple ChoiceDirections: In this section, you will hear a passage. At the end of the passage, you will hear S questions. The passage and the question will be read only once. After each question, there will be a pause. During the pause, you must read the four suggested answers marked A, B, C and D and decide which is the best answer. Then write your choice on the Answer Slicet.21. A) It had no efFect on living cells. 13) It had effects on living cells. C) It had effects only on children.I)) It had effects only on adults.22. A) An increasing number of cancers in children.I3) A link between an electric current and the energy fold.C) A causal link between the power-line or device and the energy field.1)) A Small increased chance ofcancer in children living near electric power-line.23. A) 446. 13) 464. C ) 223. 1))234.24. A) Because he doesn't have enough evidence.R) Because other scientists have not studied his results yet.C.) Because he discovered nodirect link between disease and electricity.D) Because the link between cancer and electricity has not yet beenproved.25. A) Health and environment.B) Electric current and the energy field. C) Electricity and cancer.D) Electrical workers and cancer.Section C: Question and AnswerDirections: In this section, you will hear a passage. The passage will be read only once. Then try to answer the following questions according to what you have heard. Remember you should write your answer on the Answer Sheet.26. Why aren't most new doctors interested in beginning work in a small town?27. Why do many small town doctors work long hours? 28. What is the growing problem in theUnited States? 29. How many new doctors did the National Health Service Corps produce in 1979?30. Whom did a hospital in Parkersprary offer a reward o€5,000 dollar to?Part 11[. Vocabulary (20%),Directions: In each question, decide which of the four choices given will most suitably complete the sentence if inserted at the place marked. Write your answer on the Answer Sheet.31. To qualify for such a position, the native would first have to receive specialized training, and thisis____A) refused B) discouragedC) denied D) forbidden32. The little girl wore a very thin coat. A sudden gust of cold wind made her____A) whirl B) shiftC) shiver D) shake33. Presently, there are nine teachers in my team, who have____the task of teaching advanced English tomore than 500 non-English majors.A) inclined B) hesitatedC) afforded D) undertaken34. The press demands that politicians____the sources of their income.A) betray B) concealC) disclose D) renew35. Having gone through all kinds of hardships in life, he became a m with a strong____A) philosophy B) idealismC) morality D) personality36. One new____to learning a foreign language is to study the language in its cultural context.A) approach B) solutionC) manner D) road37. To maintain public____is not only the policemen's duty but f every citizen's responsibility.A) custom B) confidenceC) security D) simplicity38. All was dark in the district except for a candle____through th curtains in one of houses.A) glimmering B)glitteringC) flaming D) blazing39. One of the stands____and dozens of people were either killer or injured.A) destroyed B) collapsedC) corroded D) ruined40. "Me, afraid of him?" he said with a(n) ____smile, "Not me!"A) contemptible B) amusingC) contagious D) contemptuous41. He will simply no listen to anybody; he is____to argument.A) impervious B) imperceptibleC) impassable D) blunt42. Stop asking all these personal questions! It is bad manners to beA) inquisitive B) impatientC) acquisitive D) informative43. He____between life and death for a few days but then he pul:A) hovered B) lurchedC) wavered D) fluctuated44. We are prepared to satisfy all your____claimA} legitimate B) legibleC) intimate D) legislative45. There is not a Greek word which is the exact____of the English word ' stile'.A) equivalent B) copyC) counterpart D) meaning46. The prizes will be____at the end of the school year.A) distributed B) attributedC) granted D) contributed47. During our stay in Paris we were splendidly____by the Italian Ambassador.A) sustained B) maintainedC) retained D) entertained48. On leaving, we thanked him most warmly for the hospitality____to us and our friends.A) extended B) expandedC) expended D) awarded49. If the dispute is not settled in a(n) ____ way soon, the two countries will certainly go to war.A) amiable B) amicableC) inimical D) unfriendly50, If I may be so____as to advise you, my opinion is that you should not reply to his letter.A) generous B) humbleC) proud D) bold51. If you take a(n) ____course like her you can learn English in less than two years.A) intensive B) extensiveC) expansive D) retentive52. After a year's hard work I think I am____to a long holiday. 10,A) entailed B) deservedC) entitled D) satisfied53. Thousands of people____from Greece every year to work in West Germany.A) emigrate B) leaveC) abandon D) immigrate54, lie was a member of the Hillary____that conquered Mount Everest.A) mission B) invasionC) experiment D) expedition55. It was my sad duty to____the news of John's death to his family.A) submit B) breakC) say D) proclaim56. He____himself as a war correspondent in Vietnam.A) discerned B) distinguishedC) discriminated D) extinguished57. She____his invitation to dinner as she was on a diet.A) inclined B) declinedC) denied D) disinclined58. He was____with attempted robbery and held in custody..A) accused B) prosecutedC) charged D) arrested59. What the witness said in court was not____with the statement he made to the police.A) prevalent B) relevantC) consistent D) coincident60. Molly has always beep a(n) ____child; she becomes ill easily.A) delicate B) gloomyC) energetic D) confident61. There are some very beautifully____glass windows in the church.A) designed B) drawnC) marked D) stained62. The man who never tries anything new is a(n) ____on the wheels of progress.A) obstacle B) brakeC) break D) block63. There is a sale at Hamfridge's next week with____in all departments.A) decreases B) subtractionsC) reductions D) accounts64. Doctors have long known that if a patient is____that he will recover and is treated with sympathy, his painwill often disappear.A) assumed B) assuredC) informed D) proved65: Although most birds have only a____sense of smell, they have acute vision.A} genuine.B) negativeC) negligible D) condensed66. We are sorry to say that Mary is not the very person who can be____with either money or secret information.A）entrusted B) committedC)consigned D) assigned67. If you never review your lessons, you will only have yourself to____if you fail in your examination.A) complain B) blameC) mistake D) fault68. We were four scores left behind with five minutes to go,so the game looked completely____A) irresistible B) irremissibleC} irreplaceable D) irretrievable69. Had the explosion broken out, the passagers in the plane should have been killed, for it was____timedwith the plane's take-off.A) spontaneously B) instantaneouslyC} simultaneously D) conscientiously70. The two witnesses who saw the shootings were able to____who hard fired first.A) encounter B) highlightC} testify D) identifyPart III. Reading Comprehension (50 minutes, 30 points)Directions: There are 6 passages in this part. Each passage is followed by some questions or unfinished statements. For each of them there are four choices marked A, B, C and D. You should decide on the best choice and write your answer on the Answer Sheet.One day in 1963, a dolphin named Elvar and a famous astronomer, Carl Sagan, were playing a little game. The astronomer was visiting an institute which was looking into the way dolphins communicate witheach other. He was standing a t the edge of one of the tanks where several of these highly intelligent, friendly creatures were kept. Elvar had just swum up alongside hiui,and had turned on his back. He wanted Sagan to scratch his stomach again, as the astronomer had done twice before. But this time Elvar was too deep in the water for Sagan to reach him. Elvar looked up at Sagan, waiting. Then, after a minute or so, the dolphin leapt up through the water into the air and made a sound just like theword `More?'The astonished astronomer went -to the director of the institute and told him about the incident.`Oh, yes: That's one of the words he knows,' the director said, showing no surprise at all.Dolphins have bigger brains in proportion to their body size than humans have, andit has been known for a long time that they can make a number o€ sounds. What is more, these sounds seem to have different functions, such as warning each other of danger. Sound travels much faster ,and much further in water than it does in air. That is why the parts of the brain that deal with sound are much better developed in dolphins than in humans. But can it be said that dolphins have a `language' in the real sense of the word? Scientists don't agree on this.A language is not just a collection of sounds, or even words.A language has a structure, or what we call a grammar. The grammar of a language helps to give it meaning. For example, the two questions `Who loves Mary?' and `Who does Mary love?' mean different things. If you stop to think about it, you will see that this difference doesn't come from the words in the question but from the difference in structure. That is why the question `Can dolphins speak?' can't be answered until we find out if dolphins not only make sounds but also arrange them in ways which affect their meaning.71. The dolphin leapt into the air becauseA) Sagan had turned his backB) it was part of the game they were playingC) he wanted Sagan to scratch him againD) Sagan wanted him to do this72. When Sagan told the director about what the dolphin haddone, the directorA) didn't seem to think it was unusualB) thought Sagan was jokingC) told Sagan about other words the dolphin knewD) asked him if he knew other words73. Dolphins' brains are particularly well-developed toA) help them to travel fast in waterB) arrange sounds in different structuresC) respond to different kinds of soundD) communicate with humans through sound74. The sounds we call words can be called a language only ifA) each sound has a different meaningB) each sound is different from the otherC) there is a system of writingD) they have a structure or grammar（2）Married people live "happily ever after" in fairy tales, but they do so less and less often in real life. 1, like many of my friends, got married, divorced, and remarried. I suppose, to some people, I'm a failure. After all, I broke my first solemn promise to "love and cherish until death us do part." But I feel that I'm finally a success. I learned from the mistakes I made in my first marriage. This time around, the ways my husband and I share our free time, make decisions, and deal with problems are very different.I learned, first of all, not to be a clinging vine (依赖男子的妇女) . In my first marriage, I felt the every moment we spent apart was wasted. If Ray wanted to go out to a bar with his friends to watch a football game, I felt rejected and talked him into staying home. I wouldn't accept an offer to go to a movie or join anexercise class it' it meant that Ray would be home alone. I realize now that we were often angry with each other just because we spent too much time together. In contrast, my second husband and I spend some of our tree time apart and try to have interests of our own. I have started playing racquetball at a health club, and Davidsometimes takes off to go to the local auto races with his friends. When we are together, we aren't bored with each other; our separate interests make us more interesting people.I learned not only to be apart sometimes but also to work together when it's time to make decisions. When Ray and I were married, I left all the important decisions to him. He decided how we would spend money, whether we should sell the car or fix it, and where to take a vacation. I know now that I went along with this so that I wouldn't have to take the responsibility when things went wrong. I could always end an argument by saying, "It was your fault!" With my second marriage, I am trying to be a full partner. We ask each other's opinions on major decisions and try to compromise if we disagree. If we make the wrong choice, we're equally guilty. When we rented an apartment, for example, we both had to take the blame for not noticing the drafty windows and the "no pets" clause in our lease.Maybe the most important thing I've learned is to be a grown-up about facing problems. David and i have made a vow to face our troubles like adults. If we're mad at each other or worried and upset, we say how we feel. Rather than hide behind our own misery, we talk about the problem until we discover how to fix it. Everybody argues or has to deal with the occasional crisis, but Ray and I always reacted like children to these stormy times.I would lock myself in the spare bedroom. Ray would stalk out ofthe house, slam the door, and race off in the car. Then I would cry and worry till he returned.I wish that my first marriage hadn't been the place where I learned how to make a relationship work, but at least I did learn.1 feel better now about being an independent person, about making decisions, and about facing problems. My second marriage isn't perfect, but it doesn't have the deep flaws that made the first one fall apart.75. Which of the following has contributed to the writer's divorce?A) Her former husband went out to watch football games.B) She started to play racquetball at a health club.C) They spent too much time together and got bored with each other.D) They spent so little time together that they could not talk to each other.76. It can be learned from the passage that the writer, in her first marriage,A) took less responsibility than she should for major decisionB) tool: the same responsibility as her husbandC) took more blame when things went wrongD) felt equally guilty when things went wrong77. Which of the following that the author should have said when she quarrelled with her former husband but she did not.A) "It was your fault!"B) "Maybe you're right."C) "It's none of your business."D) "It's none of my business."78. All the problems between the writer and David can be resolved becauseA) they hide their feelingsB) they lock themselves in their bedroomC) they have promised not to be mad at each otherD) they dare to face them79, The writer's second marriage is different from the first one in all the following ways except A) that they share their free timeB) that they make their decisions togetherC) that they talk to each otherD) that they deal with their troubles together80. The best title for the passage isA) First MarriageB) Second MarriageC) DivorceD) Perfect Marriage（3）Classified Advertising is that advertising which is grouped in certain sections of the paper and is thus distinguished from display advertising. Such groupings as "Help Wanted", "Real Estate," "Lost and Found" are made, the rate charged being less than that for display advertising. Classified advertisements are a convenience to the reader and a saving to the advertiser. The reader who, is interested in a particular kind of advertisement finds all advertisements of that type grouped for him. The advertiser may, on this account, use a very small advertisement that would be lost if it were placed among larger advertisements in the paper.It is evident that the reader approaches the classified advertisement in a different frame of mind from that in which he approaches the other advertisements in the paper. He turns to apage of classified advertisements to search for the particular advertisement that will meet his needs. As his attention is voluntary, the advertiser does not need to rely to much extent on display type to get the reader's attention.Formerly all classified advertisements were of the same size and did not have display type. With the increase in the number of such advertisements, however, each advertiser within a certain group is vying with others in the same group for the reader's attention. In many cases the result has been an increase in the size of the space used and the addition of headlines and pictures. In that way the classified advertisement has in reality become a display advertisement. This is particularly true of realestate advertising.81. Classified advertising is different to display advertising becauseA) all advertisements of a certain type are grouped togetherB) it is more distinguishedC) it is more expensiveD) nowadays the classified advertisements are all of the same size82. One of the examples given of types of classified advertisement isA) house for saleB) people who are asking for helpC) people who are lostD) real antiques for sale83. What sort of attitude do people have when they look at classified advertisements, according to thewriter?A) They are in the frame of mind to buy anything.B) They are looking for something they need.C) They feel lost because there are so many advertisements.D) They feel the same as when they look at display advertisements.84. What does the writer say about the classified advertisements that used to be put in the papers?A) They used to be voluntary.B) They used to use display type.C) They were all the same size.D) They were more formal.85. Why have classified advertisements changed in appearance, according to the writer?A) Because people no longer want headlines and pictures.B) Because real estate advertising is particularly truthful now.C）Because the increase in the number of such advertisements means they have to be smaller now.D) Because there are more advertisements now and more competition amongst advertisers. .（4）Mr Abu, the laboratory attendant, came in from the adjoining store and briskly cleaned the blackboard. He was a retired African sergeant from the Army Medical Corps and was feared by the boys. If he caught any of them in any petty thieving, he offered them the choice of a hard smack on the bottom or of being reported to the science masters. Most boys chose the former as they knew the matter would end there with no long interviews, moral arguments and an entry in the conduct book.The science master, a man called Vernier, stepped in and stood on his small platform. Vernier set the experiments for the day and demonstrated them, then retired behind the "ChurchTimes" which he read seriously in between walking quickly along the rows of laboratory benches, advising boys. It was a simple heat experiment to show that a dark surface gave out more heat by radiation than a bright surface.During the class, Vernier was called away to the telephone and Abu was not about, having retired to the lavatory for a smoke. As soon as a posted guard announced that he was out of sight, minor pandemonium ('N k) broke out. Some of the boys raided the store. The wealthier ones took rubber tubing to make catapults and to repair bicycles, and helped themselves to chemicals for developing photographic films. The poorer boys, with a more determined aim, took only things of strict commercial interst which could be sold easily in the market. They emptied stuff into bottles in their pockets. Soda for making soap, magnesium sulphate for opening medicine, salt for cooking, liquid paraffin for women's hairdressing, and fine yellow iodoform powder much in demand for sprinkling on sores. Kojo objected mildly to all this. "Oh, shut up!" a few boys said. Sorie, a huge boy who always wore a fez indoors, commanded respect and some leadership in the class. He was gently drinking his favourite mixture of diluted alcohol and bicarbonate----which he called "gin and fizz"----from a beaker. "Look here, Kojo, you are getting out of hand. What do you think our parents pay taxes and school fees for? For us to enjoy----or to buy a new car every year for Simpson? " The other boys laughed. Simpson was the European headmaster, feared by the small boys, adored by the boys in the middle school, and liked, in a critical fashion, with reservations, by some of the senior boys and African masters. He had a passion for new motor-cars, buying one yearly."Come to think of it," Sorie continued to Kojo, "you must takesomething yourself, then we'll know we are safe," "Yes, you must," the other boys insisted. Kojo gave in and, unwillingly, took a little nitrate for some gunpowder experiments which he was carrying out at home. "Someone!" the look-out called.The boys ran back to their seats in a moment. Sorie washed out his mouth, at the sink with some water.Mr Abu, the laboratory attendant, entered and observed the innocent expression on the faces of thewhole class. He looked round fiercely and suspiciously, and then sniffed the air. It was a physicsexperiment, but the place smelled chemical. However, Vemier came in then. After asking if anyonewas in difficulties, and finding that no one could in a moment think up anything, he retired to hischair and settled down to an article on Christian reunion.86, The boys were afraid of Mr Abu becauseA) he had been an Army sergeant and had military ideas of disciplineB) he reported them to the Science masters whenever he caught them petty thievingC) he was cruelD) he believed in strict discipline87. When the boys were caught petty thieving, they usually chose to be beaten by Mr Abu becauseA) he gave them only one hard smack instead of the six from their teachersB) they did not want to get a bad reputation with their teachersC) they were afraid of their science mastersD) his punishment was quicker than their teachers'88. Some boys took chemicals like soda and iodoform powder becauseA) they liked to set up stalls in the marked and sell things, like tradersB) they were too poor to buy things like soap and medicineC) they wanted money and could sell such things quicklyD) they needed things like soap and medicine for sores89. A big difference between Kojo and Sorie was thatA) Kojo took chemicals for some useful experiment but Sorie only wasted his in making an alcoholicdrink.B) Sorie was rich but Kojo was poorC) Kojo had a guilty conscience but Sorie did notD) when Kojo objected. Sorie proved that what they were doing was reasonable90. On entering the laboratory, Mr Abu was immediately suspicious becauseA) the whole class was looking so innocentB) he was a suspicious man by natureC) there was no teacher in the roomD) he could smell chemicals and he knew it was a physics lesson ,（5）Alison closed the door of her small flat and put down her briefcase. As usual, she had brought some work home from the travel agency. She wanted to have a quick bite to eat and then, after spending a few hours working, she was looking forward to watching television or listening to some music:.She was just about to start preparing her dinner when there was a knock at the door. `Uli, no! Who on earth could that be?'she muttered to herself. She went to the door and opened it just wide enough to see who it was. A man of about sixty was standing there. It took her a moment before she realized who he was. He lived in the flat below. They had passed each other on the stairs once or twice, and had nodded to each other but never really spoken.`Uh, sorry to bother you, but ...uh...there's something I'd like to talk to you about,' he mumbled. He had a long, thin face and two big front teeth that made him look rather like a rabbit. Alison hesitated, but then, opening the door wide, asked him to come in. It was then that she noticed the dog. She hated dogs----particularly big ones. This one was a very old, very fat bulldog. The man had already bone into her small living-room and, without being asked, he sat down on the sofa. The dog followed him in and climbed up on the sofa next to him, breathing heavily. She stared at it. It stared back.The man coughed. `Uh, do you mind if I smoke?' he asked. Before she could ask him not to, he had taken out a cigarette and lit it.`I'll tell you why I've come. I ...I hope you won't be offended but, well ...,' he began and then stopped. Suddenly his face went red. His whole body began to shake. Then another cough exploded from somewhere deep inside him. Still coughing, he took out a grey, dirty-looking handkerchief and spat into it. Afterwards he put the cigarette back into his mouth and inhaled deeply. As he did so, some ash fell on the carpet.The man looked around the room. He seemed to have forgotten what he wanted to say. Alison glanced at her watch and wondered when he would get to the point. She waited.'Nice place you've got here,' he said at last.。
西北师范大学2017年硕士研究生入学考试(347心理学专业综合)真题
西北师范大学2017年硕士研究生入学考试（347心理学专业综合）真题一，名词解释:
1、个案法
2、智力
3、选择性注意
4、依恋
5、感觉
6、人格
7、观察学习
8、辐合思维
9、守恒
10、服从
11、自利偏差
12、映像管理
13、性别角色
二，简答题:
1、情绪的概念和功能
2、动机的概念和特征
3、多拉德和米勒的学习四要素
4、混合交叉设计的优缺点
5、乔姆斯基的语言理论
6、偏见产生的原因和克服的办法
7、遗忘是啥导致的
8、投射测验的优缺点，
三、论述题: 1、影响问题解决的因素
2、旁观者效应和责任分散等相关
四，案例分析题:
有一个哥们儿，这个哥们平时坚强，自信，而且父母说很听话，总的来说是个好娃娃，但是呢，到后来，这哥们二就害怕和别人去社交，而且总是回避，接着这哥们儿还有些问题就是，当老师让他回答问题时，他会胃疼，并且这些问题已经是一年了。
第一问:依据啥判断这是啥症状
第二问:问有啥解决的办法
第二个案例:说70，80，90这三代人好像说是存在观念思维等的不一致，意思是总觉得对方做的不对。
问人际冲突的解决办法和策略。
2015年西北师范大学333教育综合真题及答案
西北师范大学2015年招收攻读硕士研究生入学考试业务课试题适用专业名称：学科教学、教育管理、小学教育、学前教育、现代教育技术、科学与技术教育、心理健康教育考试科目名称：教育综合科目代码：333注意：1、请将答案直接做到答题纸上，做在试题纸上或草稿纸上无效。
2、除答题纸上规定的位置外，不得在卷面上出现姓名、考生编号或其他标志，否则按违纪处理。
一，名词解释（每题5分，共40分）1、课程标准2、德育3、分斋教学法4、生活教育理论5、导生制6、恩物7、元认知8、品德二，简答题（每题10分，共80分）1、简述颜元对学校教育的改革。
2、夸美纽斯在教育学上的地位和贡献。
3、怎样培养学生的学习动机。
4、学校心里健康教育的途径。
5、学记的教学原则。
6、学校管理的发展趋势。
7、列举中小学常见的教学方法。
8、文艺复兴时期教育特点。
三，论述题（每题，15分，共30分）1、材料：黑龙江某学校的班主任在教师节期间向学生索要礼物，被曝光。
据此分析教师应该具有的品德。
2、学校教育在人的发展中起什么作用？为什么？2015年西北师范大学333教育综合真题答案一，名词解释（每题5分，共40分）1、课程标准【解析】课程标准是依照课程计划的要求，每门学科以纲要的形式编定的、有关学科教学内容的指导性文件。
它规定某门学科的性质与地位，是教材编写、教学、评估与考试命题的依据，是国家管理与评价课程的基础。
编写课程标准是课程开发的重要步骤。
课程标准的结构：说明部分、课程目标部分、内容标准部分、课程实施建议部分。
2、德育【解析】德育的概念有广义和狭义之分，广义的德育是指教育者根据一定社会的要求和受教育者身心发展的规律，有目的、有计划、有组织地在受教育者身上培养所期望的政治素质、思想素质、道德素质、法律素质等，以促使他们成为合格的社会成员的过程。
它包括政治教育、思想教育、道德教育和法律教育。
狭义的德育专指道德教育。
教育者根据一定历史时期社会的道德要求和个体的品德心理发展规律，有目的、有计划、有组织地在受教育者身上培养所期望的道德素质，使他们具有正确的道德观念、丰富的道德情感、坚强的道德意志、热切的道德观念和较高的道德实践能力，不断提升他们的道德境界的教育过程。
2015年西北师范大学333教育综合真题及答案
西北师范大学2015年招收攻读硕士研究生入学考试业务课试题适用专业名称：学科教学、教育管理、小学教育、学前教育、现代教育技术、科学与技术教育、心理健康教育考试科目名称：教育综合科目代码：333注意：1、请将答案直接做到答题纸上，做在试题纸上或草稿纸上无效。
2、除答题纸上规定的位置外，不得在卷面上出现姓名、考生编号或其他标志，否则按违纪处理。
一，名词解释（每题5分，共40分）1、课程标准2、德育3、分斋教学法4、生活教育理论5、导生制6、恩物7、元认知8、品德二，简答题（每题10分，共80分）1、简述颜元对学校教育的改革。
2、夸美纽斯在教育学上的地位和贡献。
3、怎样培养学生的学习动机。
4、学校心里健康教育的途径。
5、学记的教学原则。
6、学校管理的发展趋势。
7、列举中小学常见的教学方法。
8、文艺复兴时期教育特点。
三，论述题（每题，15分，共30分）1、材料：黑龙江某学校的班主任在教师节期间向学生索要礼物，被曝光。
据此分析教师应该具有的品德。
2、学校教育在人的发展中起什么作用？为什么？2015年西北师范大学333教育综合真题答案一，名词解释（每题5分，共40分）1、课程标准【解析】课程标准是依照课程计划的要求，每门学科以纲要的形式编定的、有关学科教学内容的指导性文件。
它规定某门学科的性质与地位，是教材编写、教学、评估与考试命题的依据，是国家管理与评价课程的基础。
编写课程标准是课程开发的重要步骤。
课程标准的结构：说明部分、课程目标部分、内容标准部分、课程实施建议部分。
2、德育【解析】德育的概念有广义和狭义之分，广义的德育是指教育者根据一定社会的要求和受教育者身心发展的规律，有目的、有计划、有组织地在受教育者身上培养所期望的政治素质、思想素质、道德素质、法律素质等，以促使他们成为合格的社会成员的过程。
它包括政治教育、思想教育、道德教育和法律教育。
狭义的德育专指道德教育。
教育者根据一定历史时期社会的道德要求和个体的品德心理发展规律，有目的、有计划、有组织地在受教育者身上培养所期望的道德素质，使他们具有正确的道德观念、丰富的道德情感、坚强的道德意志、热切的道德观念和较高的道德实践能力，不断提升他们的道德境界的教育过程。
2015年考研数学(二)真题及答案详解
2015年全国硕士研究生入学统一考试数学（二）试题解析一、选择题:1 8小题,每小题4分,共32分.下列每题给出的四个选项中,只有一个选项符合题目要求的,请将所选项前的字母填在答题纸．．．指定位置上. (1) 下列反常积分收敛的是 ( )(A)2+∞⎰(B) 2ln x dx x+∞⎰(C)21ln dxx x +∞⎰(D) 2x x dx e+∞⎰【答案】(D) 【解析】(1)xx x dx x e e-=-+⎰，则2222(1)3lim (1)3xx x x x dx x e e x e e e +∞+∞----→+∞=-+=-+=⎰.(2) 函数()2sin lim(1)x tt t f x x→=+在(,)-∞+∞内( )(A) 连续 (B) 有可去间断点 (C) 有跳跃间断点 (D) 有无穷间断点 【答案】(B)【解析】220sin lim 0sin ()lim(1)t x t x x t x tt t f x e e x→→=+==，0x ≠，故()f x 有可去间断点0x =. (3) 设函数()1cos ,00,0x x x f x x α⎧>⎪=⎨⎪≤⎩(0,0)αβ>>,若()'f x 在0x =处连续则:( ) (A)0αβ-> (B)01αβ<-≤ (C)2αβ-> (D)02αβ<-≤ 【答案】(A)【解析】0x <时，()0f x '=()00f -'=()1001cos10lim lim cosx x x x f x x x ααβ++-+→→-'== 0x >时，()()()11111cos1sin f x x x x x x ααβββαβ-+'=+-- 1111cossin x x x xααβββαβ---=+()f x '在0x =处连续则：()()10100lim cos 0x f f x xαβ+--+→''===得10α-> ()()++1100110lim =lim cos sin =0x x f f x x x x x ααβββαβ---→→⎛⎫''=+ ⎪⎝⎭得：10αβ-->，答案选择A(4)设函数()f x 在(),-∞+∞内连续，其中二阶导数()''f x 的图形如图所示，则曲线()=y f x 的拐点的个数为( )(A) 0 (B) 1 (C) 2 (D) 3 【答案】(C)【解析】根据图像观察存在两点，二阶导数变号.则拐点个数为2个.(5) 设函数(),f u v 满足22,y f x y x y ⎛⎫+=- ⎪⎝⎭ ，则11u v fu==∂∂与11u v f v==∂∂ 依次是 ( )(A)1,02 (B) 10,2 (C) 1,02- (D) 10,2-【答案】(D)【解析】此题考查二元复合函数偏导的求解. 令,y u x y v x =+=，则,11u uv x y v v ==++，从而22(,)y f x y x y x+=-变为222(1)(,)111u uv u v f u v v v v -⎛⎫⎛⎫=-= ⎪ ⎪+++⎝⎭⎝⎭.故222(1)2,1(1)f u v f u u v v v ∂-∂==-∂+∂+， 因而111110,2u u v v ff uv ====∂∂==-∂∂.故选（D ）. (6)设D 是第一象限由曲线21xy =，41xy =与直线y x =，y =围成的平面区域，函数(),f x y 在D 上连续，则(),Df x y dxdy =⎰⎰ ( )(A)()13sin2142sin2cos ,sin d f r r rdr πθπθθθθ⎰⎰(B)()34cos ,sin d f r r rdr ππθθθ⎰ (C)()13sin 2142sin 2cos ,sin d f r r drπθπθθθθ⎰⎰(D)()34cos ,sin d f r r dr ππθθθ⎰【答案】(B)【解析】根据图可得，在极坐标系下计算该二重积分的积分区域为(,)43D r r ππθθ⎧⎫=≤≤≤≤⎨⎩所以34(,)(cos ,sin )Df x y dxdy d f r r rdr ππθθθ=⎰⎰⎰故选B.(7) 设矩阵21111214a a ⎛⎫ ⎪= ⎪ ⎪⎝⎭A ,21d d ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭b .若集合}{1,2Ω=，则线性方程组=Ax b 有无穷多解的充分必要条件为 ( )(A) ,a d ∉Ω∉Ω (B) ,a d ∉Ω∈Ω (C) ,a d ∈Ω∉Ω (D) ,a d ∈Ω∈Ω 【答案】(D)【解析】2211111111(,)1201111400(1)(2)(1)(2)A b ad a d a d a a d d ⎛⎫⎛⎫⎪ ⎪=→-- ⎪ ⎪ ⎪ ⎪----⎝⎭⎝⎭，由()(,)3r A r A b =<，故1a =或2a =，同时1d =或2d =.故选（D ）(8) 设二次型()123,,f x x x 在正交变换=x Py 下的标准形为2221232y y y +-，其中123(,,)=P e e e ，若132(,,)=-Q e e e 则123(,,)f x x x =在正交变换=x Qy 下的标准形为( )(A)2221232y y y -+ (B) 2221232y y y +-(C) 2221232y y y -- (D) 2221232y y y ++【答案】(A)【解析】由x Py =，故222123()2T T T f x Ax y P AP y y y y ===+-. 且200010001TP AP ⎛⎫⎪= ⎪ ⎪-⎝⎭.由已知可得100001010Q P PC ⎛⎫⎪== ⎪ ⎪-⎝⎭故200()010001T T TQ AQ C P AP C ⎛⎫⎪==- ⎪ ⎪⎝⎭所以222123()2T T T f x Ax y Q AQ y y y y ===-+.选（A ） 二、填空题：9 14小题,每小题4分,共24分.请将答案写在答题纸．．．指定位置上. (9) 3arctan 3x t y t t=⎧⎨=+⎩ 则 212t d y dx ==【答案】48【解析】 2222333(1)11dy dy t dt t dx dxdt t +===++ 2222[3(1)]d y d t dx dx=+=222222[3(1)]12(1)12(1)11d t t t dt t t dx dt t ++==++ 22148t d ydx ==. (10)函数2()2x f x x =⋅在0x =处的n 阶导数(0)nf =_________ 【答案】()()21ln 2n n n --【解析】根据莱布尼茨公式得：()()()()()(2)222(1)0222ln 2(1)ln 22n n n n x n x n n f C n n ---=-===- (11) 设()f x 连续，()()20x x x f t dt ϕ=⎰，若()()11,15ϕϕ'==，则()1f =【答案】2【解析】 已知2()()x x x f t dt ϕ=⎰，求导得2220()()2()x x f t dt x f x ϕ'=+⎰，故有1(1)()1,f t dt ϕ==⎰(1)12(1)5,f ϕ'=+=则(1)2f =.(12)设函数()y y x =是微分方程'''20y y y +-=的解，且在0x =处()y x 取得极值3，则()y x = .【答案】22x x e e -+【解析】由题意知：()03y =，()00y '=，由特征方程：220λλ+-=解得121,2λλ==- 所以微分方程的通解为：212x x y C e C e -=+代入()03y =，()00y '=解得：12C =21C = 解得：22xxy e e-=+(13)若函数(),Z z x y =由方程231x y ze xyz +++=确定，则()0,0dz = .【答案】()1d 2d 3x y -+ 【解析】当0,0x y ==时0z =，则对该式两边求偏导可得2323(3)x y z x y z ze xy yz e x++++∂+=--∂ 2323(3)2x y z x y z ze xy xz e y++++∂+=--∂.将（0,0,0）点值代入即有 12,.(0,0)(0,0)33z z x y ∂∂=-=-∂∂则可得()(0,0)121|d 2d .333dz dx dy x y =--=-+ (14) 若3阶矩阵A 的特征值为2,2,1-，2B A A E =-+，其中E 为3阶单位阵，则行列式B = .【答案】21【解析】A 的所有特征值为2,2,1.-B 的所有特征值为3,7,1. 所以||37121B =⨯⨯=.三、解答题：15～23小题,共94分.请将解答写在答题纸．．．指定位置上.解答应写出文字说明、证明过程或演算步骤. (15) (本题满分10分）设函数()ln(1)sin f x x a x bx x =+++，3()g x kx =.若()f x 与()g x 在0x →时是等价无穷小，求,,a b k 的值.【答案】111,,32a kb =-=-=- 【解析】 方法一：因为233ln(1)()23x x x x o x +=-++，33sin ()3!x x x o x =-+， 那么，23333000(1)()()()ln(1)sin 231lim lim lim ()x x x a aa xb x x o x f x x a x bx x g x kx kx→→→++-+++++===， 可得：100213a ab ak⎧⎪+=⎪⎪-=⎨⎪⎪=⎪⎩，所以，11213a b k ⎧⎪=-⎪⎪=-⎨⎪⎪=-⎪⎩．方法二： 由题意得300sin )1ln(lim )()(lim1kx xbx x a x x g x f x x +++==→→203cos sin 11limkx x bx x b x ax ++++=→由分母03lim 2=→kx x ，得分子)cos sin 11(lim 0x bx x b xax ++++→0)1(lim 0=+=→a x ，求得c ；于是)()(lim10x g x f x →=23cos sin 111lim kx x bx x b x x +++-=→）（x kx xx bx x x b x x +++++=→13cos )1(sin )1(lim20 203c o s )1(s i n )1(lim kx xx bx x x b x x ++++=→kxxx bx x bx x x b x x b x b x 6sin )1(cos cos )1(cos )1(sin 1lim0+-++++++=→由分母06lim 0=→kx x ，得分子]sin )1(cos cos )1(2sin 1[lim 0x x bx x bx x x b x b x +-++++→0)cos 21(lim 0=+=→x b x ，求得21-=b ； 进一步，b 值代入原式)()(lim 10x g x f x →=kxx x x x x x x x x 6sin )1(21cos 21cos )1(sin 211lim0++-+--=→ kxx x x x x x x x x x x x x x 6cos )1(21sin 21sin )1(21sin 21cos 21sin )1(cos cos 21lim 0++++++-++--=→k621-=，求得.31-=k(16) (本题满分10分)设A>0，D 是由曲线段sin (0)2y A x x π=≤≤及直线0y =，2x π=所围成的平面区域，1V ，2V 分别表示D 绕x 轴与绕y 轴旋转成旋转体的体积，若12V V =，求A 的值.【答案】8π【解析】由旋转体的体积公式，得dx x f ⎰=2021)(V ππdx x A ⎰=202)sin (ππdx x A⎰-=20222cos 1ππ422A π=dx x xf ⎰=22)(2V ππA x d x A -πππ2c o s 220==⎰由题,V V 21=求得.8A π=(17) (本题满分11分)已知函数(,)f x y 满足"(,)2(1)x xy f x y y e =+，'(,0)(1)xx f x x e =+，2(0,)2f y y y =+，求 (,)f x y 的极值. 【答案】极小值(0,1)1f -=-【解析】xxye y y xf )1(2),(+=''两边对y 积分，得 )()21(2),(2x e y y y x f x x ϕ++=')()2(2x e y y x ϕ++=， 故x x e x x x f )1()()0,(+=='ϕ， 求得)1()(+=x e x x ϕ，故)1()2(),(2x e e y y y x f x x x +++='，两边关于x 积分，得⎰+++=dx x e e y y y x f x x )1()2(),(2⎰+++=xxde x e y y )1()2(2 ⎰-+++=dx e e x e y y xxx )1()2(2 C )1()2(2+-+++=x x x e e x e y y C )2(2+++=x x xe e y y由y y y y y f 2C 2),0(22+=++=，求得.0=C 所以x x xe e y y y x f ++=)2(),(2.令⎪⎩⎪⎨⎧=+='=+++='0)22(0)2(2xy xx x x e y f xe e e y y f ，求得⎩⎨⎧-==10y x . 又x x x xxxe e e y y f +++=''2)2(2， x xye yf )1(2+=''，xyy e f 2=''， 当1,0-==y x 时，(0,1)1,xxA f ''=-=,0)1,0(B =-''=xy f 2)1,0(=-''=yy fC ， 20,AC B ->(0,1)1f -=-为极小值.(18) (本题满分10分) 计算二重积分()Dx x y dxdy +⎰⎰，其中{}222(,)2,D x y x y y x =+≤≥【答案】245π-【解析】2()DDx x y dxdy x dxdy +=⎰⎰⎰⎰21202xdx dy =⎰12202)x x dx =⎰12240022222sin 2cos 55x t xt tdt π=--⎰⎰22242002222sin 2sin .5545u t tdt udu πππ==-=-=-⎰⎰(19)(本题满分 11 分) 已知函数()21Xf x =+⎰⎰，求()f x 零点的个数？【答案】2个【解析】()21)f x x '=- 令()0f x '=,得驻点为12x =, 在1(,)2-∞,()f x 单调递减,在1(,)2+∞,()f x 单调递增 故1()2f 为唯一的极小值,也是最小值.而112241()2f =+=-⎰⎰⎰1224=--⎰⎰⎰在1(,1)2故0-<从而有1()02f <1lim ()lim[]x x x f x →-∞→-∞=+=+∞⎰⎰22111lim ()lim[]lim[]x x xx x x f x →+∞→+∞→+∞=+=-⎰⎰⎰⎰考虑2lim lim x x x ==+∞，所以lim ()x f x →+∞=+∞.所以函数()f x 在1(,)2-∞及1(,)2+∞上各有一个零点，所以零点个数为2. (20) (本题满分10分)已知高温物体置于低温介质中，任一时刻该物体温度对时间的变化率与该时刻物体和介质的温差成正比，现将一初始温度为120C ︒的物体在20C ︒的恒温介质中冷却，30min后该物体降至30C ︒，若要将该物体的温度继续降至21C ︒，还需冷却多长时间？ 【答案】30min【解析】设t 时刻物体温度为()x t ，比例常数为(0)k >，介质温度为m ,则()dxk x m dt=--，从而()kt x t Ce m -=+， (0)120,20x m ==，所以100C =，即()10020kt x t e -=+又1()30,2x =所以2ln10k =，所以11()20100t x t -=+ 当21x =时，t =１，所以还需要冷却３０min.(21) (本题满分10分)已知函数()f x 在区间[]+a ∞，上具有2阶导数，()0f a =，()0f x '>，()''0f x >，设b a >，曲线()y f x =在点()(),b f b 处的切线与x 轴的交点是()00x ，，证明0a x b <<.【证明】根据题意得点(,())b f b 处的切线方程为()()()y f b f b x b '-=-令0y =，得0()()f b x b f b =-' 因为(x)0f '>所以(x)f 单调递增，又因为(a)0f = 所以(b)0f >，又因为()0f b '>所以0()()f b x b b f b =-<' 又因为0()()f b x a b a f b -=--'，而在区间（a,b ）上应用拉格朗日中值定理有 (b)f(a)(),(a,b)f f b aξξ-'=∈-所以0()()()()()()()()()()()f b f b f b f b f x a b a f b f b f f b f b f ξξξ''--=--=-=''''' 因为(x)0f ''>所以(x)f '单调递增 所以()()f b f ξ''>所以00x a ->，即0x a >，所以0a x b <<，结论得证.(22) (本题满分 11 分)设矩阵101101a A a a ⎛⎫ ⎪=- ⎪ ⎪⎝⎭且3A O =.(1) 求a 的值；(2) 若矩阵X 满足22X XA AX AXA E --+=，E 为3阶单位阵，求X .【答案】2010,111211a X -⎛⎫ ⎪==-- ⎪ ⎪-⎝⎭【解析】 (I)323100100111100011a A O A a a a a a a a a=⇒=⇒-=--==⇒=- (II)由题意知()()()()()()()()()222211122212X XA AX AXA E X E A AX E A E E A X E AE X E A E A E A E A X E A A ------+=⇒---=⎡⎤⇒--=⇒=--=--⎣⎦⇒=-- 2011111112E A A -⎛⎫ ⎪--=- ⎪ ⎪--⎝⎭，011100111010111010011100112001112001----⎛⎫⎛⎫ ⎪ ⎪-→- ⎪ ⎪ ⎪ ⎪----⎝⎭⎝⎭M M M M M M111010111010011100011100021011001211------⎛⎫⎛⎫ ⎪ ⎪→--→-- ⎪ ⎪ ⎪ ⎪-----⎝⎭⎝⎭M M M M M M110201100312010111010111001211001211---⎛⎫⎛⎫ ⎪ ⎪→-→- ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭M M M M M M312111211X -⎛⎫ ⎪∴=- ⎪ ⎪-⎝⎭(23) (本题满分11 分)设矩阵02313312A a -⎛⎫ ⎪=-- ⎪ ⎪-⎝⎭相似于矩阵12000031B b -⎛⎫ ⎪= ⎪ ⎪⎝⎭.（1）求,a b 的值；（2）求可逆矩阵P ，使1P AP -为对角阵.【答案】（1）4,5a b ==；（2）231101011P --⎛⎫ ⎪=- ⎪ ⎪⎝⎭【解析】(I)~()()311A B tr A tr B a b ⇒=⇒+=++0231201330012031--=⇒--=-A B ba 14235-=-=⎧⎧∴⇒⎨⎨-==⎩⎩a b a a b b (II)023100123133010123123001123A E C ---⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=--=+--=+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭⎝⎭ ()123112*********---⎛⎫⎛⎫ ⎪ ⎪=--=-- ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭CC 的特征值1230,4λλλ===0λ=时(0)0-=E C x 的基础解系为12(2,1,0);(3,0,1)ξξ==-T T 5λ=时(4)0-=E C x 的基础解系为3(1,1,1)ξ=--T A 的特征值1:1,1,5λλ=+A C令123231(,,)101011ξξξ--⎛⎫ ⎪==- ⎪ ⎪⎝⎭P ，1115-⎛⎫ ⎪∴= ⎪⎪⎝⎭P AP文档内容由金程考研网整理发布。
2015年考研数学(三)真题及答案详解
2015年全国硕士研究生入学统一考试数学（三）试题解析一、选择题:18小题,每小题4分,共32分.下列每题给出的四个选项中,只有一个选项符合题目要求的,请将所选项前的字母填在答题纸．．．指定位置上. (1)设{}n x 是数列,下列命题中不正确的是 ( ) (A) 若lim →∞=n n x a ,则221lim lim +→∞→∞==n n n n x x a(B) 若221lim lim +→∞→∞==n n n n x x a , 则lim →∞=n n x a(C)若lim →∞=n n x a ,则331lim lim +→∞→∞==n n n n x x a(D) 若331lim lim +→∞→∞==n n n n x x a ,则lim →∞=n n x a【答案】(D)【解析】答案为D, 本题考查数列极限与子列极限的关系.数列()n x a n →→∞⇔对任意的子列{}k n x 均有()k n x a k →→∞,所以A 、B 、C 正确； D 错(D 选项缺少32n x +的敛散性),故选D(2) 设函数()f x 在(),-∞+∞内连续,其2阶导函数()f x ''的图形如右图所示,则曲线()=y f x 的拐点个数为 ( )(A) 0 (B) 1 (C)2 (D) 3 【答案】(C)【解析】根据拐点的必要条件，拐点可能是不存在的点或的点处产生.所以有三个点可能是拐点，根据拐点的定义，即凹凸性改变的点；二阶导函数符号发生改变的点即为拐点.所以从图可知，拐点个数为2，故选C.(3) 设(){}2222,2,2=+≤+≤D x y xy x x y y ,函数(),f x y 在D 上连续,则(),d d Df x y x y =⎰⎰ ( )(A)()()2cos 2sin 4204d cos ,sin d d cos ,sin d f r r r r f r r r r θθθθθθθθπππ+⎰⎰⎰⎰ (B)()()2sin 2cos 420004d cos ,sin d d cos ,sin d f r r r r f r r r r θθθθθθθθπππ+⎰⎰⎰⎰()f x ''()0f x ''=()y f x =()f x ''(C)()1012d ,d xxf x y y ⎰⎰(D) ()102d ,d xxf x y y ⎰【答案】(B)【解析】根据图可得，在极坐标系下该二重积分要分成两个积分区域所以，故选B.(4) 下列级数中发散的是( )(A) 13n n n∞=∑ (B)1)n n ∞=+(C)2(1)1ln n n n ∞=-+∑(D)1!n n n n∞=∑ 【答案】(C)【解析】A 为正项级数，因为，所以根据正项级数的比值判别法收敛；B，根据级数收敛准则，知收敛；C ，，根据莱布尼茨判别法知收敛，发散，所以根据级数收敛定义知，发散；D 为正项级数，因为，所以根据正项级数的比值判别法收敛，所以选C. 1(,)0,02sin 4D r r πθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭2(,),02cos 42D r r ππθθθ⎧⎫=≤≤≤≤⎨⎬⎩⎭2sin 2cos 4204(,)(cos ,sin )(cos ,sin )Df x y dxdy d f r r rdr d f r r rdr ππθθπθθθθθθ=+⎰⎰⎰⎰⎰⎰11113lim lim 1333n n n nn n n n +→∞→∞++==<13nn n∞=∑3211)n n+P 11)n n ∞=+111(1)1(1)1ln ln ln n n n n n n n n ∞∞∞===-+-=+∑∑∑1(1)ln n n n ∞=-∑11ln n n ∞=∑1(1)1ln n n n ∞=-+∑11(1)!(1)!1(1)lim lim lim 1!!(1)1nn n n n n n nn n n n n n n n n en ++→∞→∞→∞+++⎛⎫===< ⎪++⎝⎭1!n n n n ∞=∑(5)设矩阵21111214a a ⎛⎫ ⎪= ⎪ ⎪⎝⎭A ,21d d ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭b .若集合}{1,2Ω=，则线性方程组=Ax b 有无穷多解的充分必要条件为 ( )(A) ,a d ∉Ω∉Ω (B) ,a d ∉Ω∈Ω (C),a d ∈Ω∉Ω(D) ,a d ∈Ω∈Ω 【答案】(D)【解析】2211111111(,)1201111400(1)(2)(1)(2)A b ad a d a d a a d d ⎛⎫⎛⎫⎪ ⎪=→-- ⎪ ⎪ ⎪ ⎪----⎝⎭⎝⎭，由()(,)3r A r A b =<，故1a =或2a =，同时1d =或2d =.故选（D ） (6)设二次型()123,,f x x x 在正交变换=x Py 下的标准形为2221232y y y +-，其中123(,,)=P e e e ，若132(,,)=-Q e e e 则123(,,)f x x x =在正交变换=x Qy 下的标准形为( )(A)2221232y y y -+ (B) 2221232y y y +- (C)2221232y y y --(D) 2221232y y y ++ 【答案】(A)【解析】由x Py =，故222123()2T T T f x Ax y P AP y y y y ===+-. 且200010001TP AP ⎛⎫⎪= ⎪ ⎪-⎝⎭.又因为100001010Q P PC ⎛⎫ ⎪== ⎪ ⎪-⎝⎭故有200()010001T T TQ AQ C P AP C ⎛⎫⎪==- ⎪ ⎪⎝⎭所以222123()2T T T f x Ax y Q AQ y y y y ===-+.选（A ）(7) 若,A B 为任意两个随机事件,则: ( )(A)()()()≤P AB P A P B (B)()()()≥P AB P A P B (C)()()()2+≤P A P B P AB (D) ()()()2+≥P A P B P AB【答案】(C)【解析】由于,AB A AB B ⊂⊂，按概率的基本性质，我们有()()P AB P A ≤且()()P AB P B ≤，从而()()()2P A P B P AB +≤≤，选(C) .(8) 设总体()~,,X B m θ12,,,n X X X 为来自该总体的简单随机样本,X 为样本均值,则()21n i i E X X =⎡⎤∑-=⎢⎥⎣⎦( ) (A) ()()11θθ--m n (B)()()11θθ--m n (C)()()()111θθ---m n (D)()1θθ-mn 【答案】(B)【解析】根据样本方差2211()1ni i S X X n ==--∑的性质2()()E S D X =，而()(1)D X m θθ=-，从而221[()](1)()(1)(1)ni i E X X n E S m n θθ=-=-=--∑，选(B) .二、填空题：914小题,每小题4分,共24分.请将答案写在答题纸．．．指定位置上. (9) 20ln(cos )lim__________.x x x→= 【答案】 【解析】原极限(10)设函数()f x 连续，2()()d ,x x xf t t ϕ=⎰若(1)1,(1)5,ϕϕ'==则(1)________.f =【答案】【解析】因为连续，所以可导，所以；因为，所以12-2200ln(1cos 1)cos 11limlim 2x x x x x x →→+--===-2()f x ()x ϕ2220()()2()x x f t dt x f x ϕ'=+⎰(1)1ϕ=1(1)()1f t dt ϕ==⎰又因为，所以故(11)若函数(,)z z x y =由方程23e 1x y z xyz +++=确定，则(0,0)d _________.z=【答案】 【解析】当，时带入，得. 对求微分，得把，，代入上式，得所以 (12)设函数()y y x =是微分方程20y y y '''+-=的解，且在0x =处取得极值3，则()________.y x =【答案】【解析】的特征方程为，特征根为，，所以该齐次微分方程的通解为，因为可导，所以为驻点，即，，所以，，故(13)设3阶矩阵A 的特征值为2,2,1-，2,=-+B A A E 其中E 为3阶单位矩阵，则行列式________.=B【答案】21【解析】A 的所有特征值为2,2,1.-B 的所有特征值为3,7,1. 所以||37121B =⨯⨯=.(14)设二维随机变量(,)X Y 服从正态分布(1,0;1,1;0)N ，则(1)5ϕ'=1(1)()2(1)5f t dt f ϕ'=+=⎰(1)2f =1233dx dy --0x =0y =231x y z e xyz +++=0z =231x y z e xyz +++=2323()(23)()x y z x y z d e xyz e d x y z d xyz +++++=+++23(23)x y z e dx dy dz yzdx xzdy xydz ++=+++++0=0x =0y =0z =230dx dy dz ++=(0,0)1233dz dx dy =--2()2x x y x e e -=+20y y y '''+-=220λλ+-=2λ=-1λ=212()xx y x C eC e -=+()y x 0x =(0)3y =(0)0y '=11C =22C =2()2x x y x e e -=+{0}_________.P XY Y -<=【答案】12【解析】由题设知，~(1,1),~(0,1)X N Y N ，而且X Y 、相互独立，从而{0}{(1)0}{10,0}{10,0}P XY Y P X Y P X Y P X Y -<=-<=-><+-<>11111{1}{0}{1}{0}22222P X P Y P X P Y =><+<>=⨯+⨯=.三、解答题：15～23小题,共94分.请将解答写在答题纸．．．指定位置上.解答应写出文字说明、证明过程或演算步骤.(15)(本题满分10 分)设函数3()ln(1)sin ,()f x x a x bx x g x c kx =+++==.若()f x 与()g x 在0x →时是等价无穷小，求,,a b k 的值.【答案】111,,23a b k --=-== 【解析】法一：因为，， 则有，， 可得：，所以，．法二： 由已知可得得由分母，得分子，求得233ln(1)()23x x x x o x +=-++33sin ()3!x x x o x =-+23333000(1)()()()ln(1)sin 231lim lim lim ()x x x a aa xb x x o x f x x a x bx x g x kx kx→→→++-+++++===100213a ab ak⎧⎪+=⎪⎪-=⎨⎪⎪=⎪⎩11213a b k ⎧⎪=-⎪⎪=-⎨⎪⎪=-⎪⎩300sin )1ln(lim )()(lim1kxxbx x a x x g x f x x +++==→→203cos sin 11lim kxx bx x b x ax ++++=→03lim 20=→kx x )cos sin 11(lim 0x bx x b xax ++++→0)1(lim 0=+=→a xc ；于是由分母，得分子，求得； 进一步，b 值代入原式，求得 (16)(本题满分10 分) 计算二重积分()d d Dx x y x y +⎰⎰，其中222{(,)2,}.D x y x y y x =+≤≥ 【答案】245π-【解析】)()(lim10x g x f x →=23cos sin 111lim kx x bx x b x x +++-=→）（x kx xx bx x x b x x +++++=→13cos )1(sin )1(lim223cos )1(sin )1(limkx xx bx x x b x x ++++=→kxxx bx x bx x x b x x b x b x 6sin )1(cos cos )1(cos )1(sin 1lim0+-++++++=→06lim 0=→kx x ]sin )1(cos cos )1(2sin 1[lim 0x x bx x bx x x b x b x +-++++→0)cos 21(lim 0=+=→x b x 21-=b )()(lim 10x g x f x →=kxx x x x x x x x x 6sin )1(21cos 21cos )1(sin 211lim 0++-+--=→k xx x x x x x x x x x x x x x 6cos )1(21sin 21sin )1(21sin 21cos 21sin )1(cos cos 21lim 0++++++-++--=→k621-=.31-=k 2()DDx x y dxdy x dxdy +=⎰⎰⎰⎰21202xdx x dy =⎰12202)x x dx =⎰(17)(本题满分10分)为了实现利润的最大化，厂商需要对某商品确定其定价模型，设Q 为该商品的需求量，P 为价格，MC 为边际成本，η为需求弹性(0)η>.(I) 证明定价模型为11MCP η=-； (II) 若该商品的成本函数为2()1600C Q Q =+，需求函数为40Q P =-，试由（I ）中的定价模型确定此商品的价格.【答案】(I)略(II) .【解析】(I)由于利润函数,两边对求导，得. 当且仅当时，利润最大，又由于,所以,故当时，利润最大. (II)由于,则代入(I)中的定价模型，得,从而解得.(18)(本题满分10 分)设函数()f x 在定义域I 上的导数大于零，若对任意的0x I ∈，曲线()y f x =在点00(,())x f x 处的切线与直线0x x =及x 轴所围成区域的面积恒为4，且(0)2f =，求()f x 表达式.【答案】()84f x x=-12240022222sin 2cos 55x t xt tdt π=--⎰⎰22242002222sin 2sin .5545u t tdt udu πππ==-=-=-⎰⎰30P =()()()()L Q R Q C Q PQ C Q =-=-Q ()dL dP dP P Q C Q P Q MC dQ dQ dQ'=+-=+-0dL dQ =()L Q P dQ Q dP η=-⋅1dP PdQ Q η=-⋅11MCP η=-()22(40)MC C Q Q P '===-40P dQ PQ dP Pη=-⋅=-2(40)401P P P P-=--30P =【解析】曲线的切线方程为，切线与轴的交点为故面积为：. 故满足的方程为,此为可分离变量的微分方程,解得，又由于，带入可得，从而 (19)(本题满分 10分)（I ）设函数(),()u x v x 可导，利用导数定义证明[()()]()()()();u x v x u x v x u x v x '''=+ （II ）设函数12(),(),,()n u x u x u x 可导，12()()()()n f x u x u x u x =，写出()f x 的求导公式.【答案】【解析】（I ）（II ）由题意得(20) (本题满分 11分)设矩阵101101a a a ⎛⎫⎪- ⎪ ⎪⎝⎭A =，且3=A O .(I) 求a 的值；()()()000y f x f x x x '-=-x ()()000,0f x x f x ⎛⎫- ⎪ ⎪'⎝⎭()()200142f x S f x =='()f x ()()28f x f x '=()8f x x C -=+()0=2f 4C =-()84f x x=-12()[()()()]n f x u x u x u x ''=121212()()()()()()()()()n n n u x u x u x u x u x u x u x u x u x '''=+++0()()()()[()()]limh u x h v x h u x v x u x v x h→++-'=0()()()()()()()()limh u x h v x h u x h v x u x h v x u x v x h→++-+++-=00()()()()lim ()lim ()h h v x h v x u x h u x u x h v x h h→→+-+-=++()()()()u x v x u x v x ''=+12()[()()()]n f x u x u x u x ''=121212()()()()()()()()()n n n u x u x u x u x u x u x u x u x u x '''=+++(II)若矩阵X 满足22--+=X XA AX AXA E ，其中E 为3阶单位矩阵，求X .【答案】3120,111211a X -⎛⎫ ⎪==- ⎪ ⎪-⎝⎭【解析】(I)323100100111100011a A O A a a a a a a a a=⇒=⇒-=--==⇒=-(II)由题意知()()()()()()()()()222211122212X XA AX AXA E X E A AX E A E E A X E A E X E A E A E A E A X E A A ------+=⇒---=⎡⎤⇒--=⇒=--=--⎣⎦⇒=--2011111112E A A -⎛⎫⎪--=- ⎪ ⎪--⎝⎭，011100111010111010011100112001112001----⎛⎫⎛⎫ ⎪ ⎪-→- ⎪ ⎪ ⎪ ⎪----⎝⎭⎝⎭MM M M M M 111010111010011100011100021011001211------⎛⎫⎛⎫ ⎪ ⎪→--→-- ⎪ ⎪ ⎪ ⎪-----⎝⎭⎝⎭M M M M M M 110201100312010111010111001211001211---⎛⎫⎛⎫ ⎪ ⎪→-→- ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭M M M M M M 312111211X -⎛⎫ ⎪∴=- ⎪ ⎪-⎝⎭(21) (本题满分11 分)设矩阵02313312a -⎛⎫ ⎪=-- ⎪ ⎪-⎝⎭A 相似于矩阵12000031b -⎛⎫⎪⎪ ⎪⎝⎭B =.(I) 求,a b 的值；（II ）求可逆矩阵P ，使1-P AP 为对角矩阵.【答案】2314,5,101011a b P --⎛⎫ ⎪===- ⎪ ⎪⎝⎭【解析】(1) ~()()311A B tr A tr B a b ⇒=⇒+=++0231201330012031--=⇒--=-A B ba 14235-=-=⎧⎧∴⇒⎨⎨-==⎩⎩a b a a b b 023100123133010123123001123---⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪∴=--=+--=+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭⎝⎭A E C()123112*********---⎛⎫⎛⎫ ⎪ ⎪=--=-- ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭CC 的特征值1230,4λλλ===0λ=时(0)0-=E C x 的基础解系为12(2,1,0);(3,0,1)ξξ==-T T 5λ=时(4)0-=E C x 的基础解系为3(1,1,1)ξ=--TA 的特征值1:1,1,5λλ=+A C令123231(,,)101011ξξξ--⎛⎫⎪==- ⎪ ⎪⎝⎭P ，1115-⎛⎫⎪∴= ⎪ ⎪⎝⎭P AP(22) (本题满分11 分)设随机变量X 的概率密度为()2ln 2,00,0x x f x x -⎧>⎪=⎨≤⎪⎩,对X 进行独立重复的观测,直到第2个大于3的观测值出现时停止，记Y 为观测次数(I)求Y 的概率分布； (II)求()E Y .【答案】(I)12221171188n n n P Y n C p p p n ---==-=-{}()()()()，23,,n =；(II)16E Y =().【解析】(I) 记p 为观测值大于3的概率，则313228()ln x p P X dx +∞-=>==⎰，从而12221171188n n n P Y n C p p p n ---==-=-{}()()()()，23,,n =为Y 的概率分布；(II) 法一:分解法:将随机变量Y 分解成=Y M N +两个过程,其中M 表示从1到()n n k <次试验观测值大于3首次发生，N 表示从1n +次到第k 试验观测值大于3首次发生.则M Ge n p ~(,)，NGe k n p -(,)(注：Ge 表示几何分布)所以1122168E Y E M N E M E N p p p =+=+=+===()()()(). 法二:直接计算22212221777711288888n n n n n n n E Y n P Y n n n n n ∞∞∞---====⋅==⋅-=⋅--+∑∑∑(){}()()()()[()()()]记212111()()n n S x n n xx ∞-==⋅--<<∑，则2113222211n n n n n n S x n n xn xx x ∞∞∞--==='''=⋅-=⋅==-∑∑∑()()()()()， 12213222111()()()()()n n n n xS x n n xx n n x xS x x ∞∞--===⋅-=⋅-==-∑∑， 2222313222111()()()()()nn n n x S x n n x xn n xx S x x ∞∞-===⋅-=⋅-==-∑∑， 所以212332422211()()()()()x x S x S x S x S x x x-+=-+==--， 从而7168E Y S ==()(). (23) (本题满分11 分)设总体X 的概率密度为,1,(,),x f x θθθ⎧≤≤⎪=-⎨⎪⎩110其他,其中θ为未知参数，12n X ,X ,,X 为来自该总体的简单随机样本.(I)求θ的矩估计量； (II)求θ的最大似然估计量. 【答案】(I)1121ni i X X X n θ==-=∑,；(II)12n X X X θ=min{,,,}.【解析】(I)11112()(;)E X xf x dx x dx θθθθ+∞-∞+==⋅=-⎰⎰， 令()E X X =，即12X θ+=，解得1121ni i X X X n θ==-=∑,为θ的矩估计量；(II)似然函数11110,()(;),n ni i i x L f x θθθθ=⎧⎛⎫≤≤⎪ ⎪==-⎨⎝⎭⎪⎩∏其他， 当1i x θ≤≤时，11111()()nni L θθθ===--∏，则1ln ()ln()L n θθ=--.从而1ln ()d L nd θθθ=-，关于θ单调增加，所以12n X X X θ=min{,,,}为θ的最大似然估计量.。