西安科技大学2009数字测图考研试题09-2
西安电子科技大学 2009年攻读硕士学位研究生入学考试试题
西安电子科技大学2009年攻读硕士学位研究生入学考试试题考试科目代码及名称822电磁场与微波技术考试时间2009年 1月 10日下午（ 3小时）答题要求：所有答案〈填空题按照标号写〉必须写在答题纸上，写在试卷上一律作废，准考证号写在指定位置!一、（15分） z=0平面将无限大空间分为两个区域:z<0区域为空气，z>0区域为相对磁导率μr =1，相对介电常数εr =4的理想介质，若知空气中的电场强度为14x z E a a =+V/m ，试求:（1）理想介质中的电场强度E 2;（2）理想介质中电位移矢量D 2与界面间的夹角α;（3） z=0平面上的极化面电荷密度ρsp .二、（15分）均匀平面电磁波在相对磁导率μr =1的理想介质中传播，其电场强度的瞬时值为88(,)5sin[2(10)]5cos[2(10)]x v E r t a t z a t z ππ=-+-（mV/m ），试求：（1）该理想介质的相对介电常数εr ;（2）平面电磁波在该理想介质中的相速度V p ;（3）平面电磁波的极化状态。
三、（15分）空气中传播着磁场复矢量振幅(0.80.6)1()(34)12j x z x z H r a a e ππ-+=-mA/m ，的均匀平面电磁波，试求：（1）该平面电磁波的波长λ；（2）该平面电磁波传播方向的单位矢n ；（3）该平面电磁波电场的复振幅矢量 E®。
四、（15分）电场强度复振幅矢量2()24j z i x E r a e ππ-=（mA/m ）的均匀平面电磁波由空气垂直入射到相对介电常数εr =2.25，相对磁导率μr =1的半无限大理想介质的界面（z=0平面），试求:（1）反射波电场强度的振幅E rm ；（2）反射波磁场的复振幅矢量H r (r)；（3）透射波电场的复振幅矢量E t (r)。
五、（20分）己知无耗传输线电长度为θ，特性阻抗Z 0=1。
第五题用图（a ）（1）已知负载阻抗L l l Z r jx =+，求负载驻波比ρL ；（2）求输入驻波比ρin ；（3）求负载反射系数ΓL 。
2009考研数学(二)真题及参考答案
2009年研究生入学统一考试数学二试题与解析一、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，只有一个选项是符合题目要求的.请将所选项前的字母填在答题纸指定位置上.（1）函数()3sin x x f x nx-=的可去间断点的个数为（ ）()A 1.()B 2. ()C 3.()D 无穷多个.（2）当0x →时，()sin f x x ax =-与()()2ln 1g x x bx =-是等价无穷小，则（ ）()A 11,6a b ==-. ()B 11,6a b ==. ()C 11,6a b =-=-. ()D 11,6a b =-= （3）设函数(),z f x y =的全微分为dz xdx ydy =+，则点()0,0（ ）()A 不是(),f x y 的连续点. ()B 不是(),f x y 的极值点. ()C 是(),f x y 的极大值点. ()D 是(),f x y 的极小值点.（4）设函数(),f x y 连续，则()()222411,,yxydx f x y dy dy f x y dx -+=⎰⎰⎰⎰（ ）()A ()2411,xdx f x y dy -⎰⎰. ()B ()241,xx dx f x y dy -⎰⎰.()C ()2411,ydy f x y dx -⎰⎰.()D .()221,y dy f x y dx ⎰⎰（5）若()f x ''不变号，且曲线()y f x =在点()1,1上的曲率圆为222x y +=，则()f x 在区间()1,2内（ ）()A 有极值点，无零点. ()B 无极值点，有零点.()C 有极值点，有零点. ()D 无极值点，无零点.（6）设函数()y f x =在区间[]1,3-上的图形为1 ()f x -2 0 2 3x-1O则函数()()0xF x f t dt =⎰的图形为（ ）()A .()B .()C .()D .（7）设A ，B 均为2阶矩阵，**A B ，分别为A ，B 的伴随矩阵.若23A B ==，，则分块矩阵O A B O ⎛⎫⎪⎝⎭的伴随矩阵为（ ）()A .**32OB A O ⎛⎫⎪⎝⎭()B .**23O B A O ⎛⎫⎪⎝⎭ ()C .**32O A BO ⎛⎫⎪⎝⎭()D .**23O A BO ⎛⎫⎪⎝⎭（8）设A P ，均为3阶矩阵，TP 为P 的转置矩阵，且100010002T P AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭，若1231223P Q ααααααα==+（，，），（，，），则TQ AQ 为（ ）()A .210110002⎛⎫⎪⎪ ⎪⎝⎭()B .110120002⎛⎫⎪⎪ ⎪⎝⎭()f x 0 2 3x1 -2-11()f x 02 3x1 -1 1()f x 02 3x1 -2-11()f x 0 2 3x1 -2 -11()C .200010002⎛⎫⎪⎪ ⎪⎝⎭()D .100020002⎛⎫⎪⎪ ⎪⎝⎭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）曲线2221-x=0ln(2)u t e du y t t -⎧⎪⎨⎪=-⎩⎰在（0，0）处的切线方程为 . （10）已知+1k xe dx ∞=-∞⎰,则k = .（11）1n lime sin x nxdx -→∞=⎰.（12）设()y y x =是由方程xy 1ye x +=+确定的隐函数，则22x yx=∂=∂ .（13）函数2xy x =在区间(]01，上的最小值为 . (14)设αβ，为3维列向量，T β为β的转置，若矩阵T αβ相似于200000000⎛⎫⎪ ⎪ ⎪⎝⎭，则T=βα .三、解答题：15－23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求极限()[]401cos ln(1tan )limsin x x x x x→--+.（16）（本题满分10 分） 计算不定积分1ln(1)xdx x++⎰(0)x >. （17）（本题满分10分）设(),,z f x y x y xy =+-，其中f 具有2阶连续偏导数，求dz 与2z x y∂∂∂.（18）（本题满分10分）设非负函数()y y x = ()0x ≥满足微分方程20xy y '''-+=，当曲线()y y x = 过原点时，其与直线1x =及0y =围成平面区域D 的面积为2，求D 绕y 轴旋转所得旋转体体积. （19）（本题满分10分）计算二重积分()Dx y dxdy -⎰⎰，其中()()(){}22,112,D x y x y y x =-+-≤≥.（20）（本题满分12分）设()y y x =是区间-ππ（，）内过点-22ππ（，）的光滑曲线，当-0x π<<时，曲线上任一点处的法线都过原点，当0x π≤<时，函数()y x 满足0y y x ''++=.求()y x 的表达式. （21）（本题满分11分）（Ⅰ）证明拉格朗日中值定理：若函数()f x 在[],a b 上连续，在(),a b 可导，则存在(),a b ξ∈，使得()()()()f b f a f b a ξ'-=-；（Ⅱ）证明：若函数()f x 在0x =处连续，在()()0,0δδ>内可导，且()0lim x f x A +→'=，则()0f +'存在，且()0f A +'=.（22）（本题满分11分设111111042A --⎛⎫ ⎪=- ⎪ ⎪--⎝⎭，1112ξ-⎛⎫⎪= ⎪ ⎪-⎝⎭. （Ⅰ）求满足22131,A A ξξξξ==的所有向量23,ξξ；（Ⅱ）对（Ⅰ）中的任一向量23,ξξ，证明：123,,ξξξ线性无关.（23）（本题满分11分）设二次型()()2221231231323,,122f x x x ax ax a x x x x x =++-+-（Ⅰ）求二次型f 的矩阵的所有特征值；（Ⅱ）若二次型f 的规范形为2212y y +，求a 的值.2009年全国硕士研究生入学统一考试数学二试题答案一、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，只有一个选项是符合题目要求的.请将所选项前的字母填在答题纸指定位置上.（1）函数()3sin x x f x nx-=的可去间断点的个数为（ ）()A 1.()B 2. ()C 3.()D 无穷多个.【答案】C 【解析】()3s i n x x f x xπ-=则当x 取任何整数时，()f x 均无意义故()f x 的间断点有无穷多个，但可去间断点为极限存在的点，故应是30x x -=的解1,2,30,1x =±320032113211131lim lim sin cos 132lim lim sin cos 132lim lim sin cos x x x x x x x x x x x x x x x x x x x x x ππππππππππππ→→→→→-→---==--==--== 故可去间断点为3个，即0,1±（2）当0x →时，()sin f x x ax =-与()()2ln 1g x x bx =-是等价无穷小，则（ ）()A 11,6a b ==-. ()B 11,6a b ==. ()C 11,6a b =-=-. ()D 11,6a b =-=. 【答案】A【解析】2()sin ,()(1)f x x ax g x x ln bx =-=-为等价无穷小，则222200000()sin sin 1cos sin lim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a axg x x bx x bx bx bx→→→→→---==-⋅---洛洛230sin lim 166x a ax a b b axa→==-=-⋅ 36a b ∴=- 故排除,B C . 另外201cos lim3x a axbx→--存在，蕴含了1cos 0a ax -→()0x →故 1.a =排D .所以本题选A.（3）设函数(),z f x y =的全微分为dz xdx ydy =+，则点()0,0（ ）()A 不是(),f x y 的连续点. ()B 不是(),f x y 的极值点. ()C 是(),f x y 的极大值点. ()D 是(),f x y 的极小值点.【答案】 D【解析】因dz xdx ydy =+可得,z zx y x y∂∂==∂∂ 2222221,0,1z z z zA B C x x y y x y∂∂∂∂== === ==∂∂∂∂∂∂又在（0，0）处，0,0z zx y∂∂==∂∂ 210AC B -=>故（0，0）为函数(,)z f x y =的一个极小值点.（4）设函数(),f x y 连续，则()()222411,,yxydx f x y dy dy f x y dx -+=⎰⎰⎰⎰（ ）()A ()2411,xdx f x y dy -⎰⎰. ()B ()241,xx dx f x y dy -⎰⎰.()C ()2411,ydy f x y dx -⎰⎰.()D .()221,y dy f x y dx ⎰⎰【答案】C 【解析】222211(,)(,)xxdx f x y dy dy f x y dx +⎰⎰⎰⎰的积分区域为两部分：{}1(,)12,2D x y x x y =≤≤≤≤，{}2(,)12,4D x y y y x y =≤≤≤≤-将其写成一块{}(,)12,14D x y y x y =≤≤≤≤- 故二重积分可以表示为2411(,)ydy f x y dx -⎰⎰，故答案为C.（5）若()f x ''不变号，且曲线()y f x =在点()1,1上的曲率圆为222x y +=，则()f x 在区间()1,2内（ ）()A 有极值点，无零点. ()B 无极值点，有零点.()C 有极值点，有零点. ()D 无极值点，无零点.【答案】 B【解析】由题意可知，()f x 是一个凸函数，即''()0f x <，且在点(1,1)处的曲率322|''|12(1('))y y ρ==+，而'(1)1f =-，由此可得，''(1)2f =-在[1,2] 上，'()'(1)10f x f ≤=-<，即()f x 单调减少，没有极值点. 对于(2)(1)'()1(1,2)f f f ζζ-=<- , ∈ ， （拉格朗日中值定理）(2)0f ∴ <而 (1)10f =>由零点定理知，在[1,2] 上，()f x 有零点. 故应选（B ）. （6）设函数()y f x =在区间[]1,3-上的图形为则函数()()0xF x f t dt =⎰的图形为（ ）1 ()f x -2 0 2 3x-1O()A .()B .()C .()D .【答案】D【解析】此题为定积分的应用知识考核，由()y f x =的图形可见，其图像与x 轴及y 轴、0x x =所围的图形的代数面积为所求函数()F x ，从而可得出几个方面的特征： ①[]0,1x ∈时，()0F x ≤，且单调递减. ②[]1,2x ∈时，()F x 单调递增. ③[]2,3x ∈时，()F x 为常函数.④[]1,0x ∈-时，()0F x ≤为线性函数，单调递增. ⑤由于F(x)为连续函数结合这些特点，可见正确选项为D .（7）设A ，B 均为2阶矩阵，**A B ，分别为A ，B 的伴随矩阵.若23A B ==，，则分块矩阵O A B O ⎛⎫⎪⎝⎭的伴随矩阵为（ ）()A .**32OB A O ⎛⎫⎪⎝⎭()B .**23O B A O ⎛⎫⎪⎝⎭ ()C .**32O A BO ⎛⎫⎪⎝⎭()D .**23O A BO ⎛⎫⎪⎝⎭()f x 0 2 3x1 -2-11()f x 02 3x1 -1 1()f x 02 3x1 -2-11()f x 0 2 3x1 -2 -11【答案】 B【解析】根据CC C E *=若111,C C C CC C*--*==分块矩阵00A B ⎛⎫⎪⎝⎭的行列式22012360A AB B⨯=-=⨯=（）即分块矩阵可逆11110066000100B BA A AB B BB AA A **---*⎛⎫ ⎪⎛⎫⎛⎫⎛⎫ ⎪=== ⎪ ⎪ ⎪⎪⎝⎭⎝⎭⎝⎭⎪⎝⎭10023613002BB AA ****⎛⎫ ⎪⎛⎫== ⎪ ⎪ ⎪⎝⎭⎪⎝⎭（8）设A P ，均为3阶矩阵，T P 为P 的转置矩阵，且100010002TP AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭，若1231223P Q ααααααα==+（，，），（，，），则TQ AQ 为（ ）()A .210110002⎛⎫⎪⎪ ⎪⎝⎭()B .110120002⎛⎫⎪⎪ ⎪⎝⎭()C .200010002⎛⎫⎪⎪ ⎪⎝⎭()D .100020002⎛⎫⎪⎪ ⎪⎝⎭【答案】 A【解析】122312312312100(,,)(,,)110(,,)(1)001Q E αααααααααα⎡⎤⎢⎥=+==⎢⎥⎢⎥⎣⎦，即：12121212122112(1)[(1)][(1)](1)[](1)100(1)010(1)002110100100210010010110110001002001002T T TT T Q PE Q AQ PE A PE E P AP E E E ===⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦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）曲线2221-x=0ln(2)u t e du y t t -⎧⎪⎨⎪=-⎩⎰在（0，0）处的切线方程为 . 【答案】2y x =【解析】221222ln(2)22t dy t t t t dt t ==--⋅=--2(1)1(1)1t t dxe dt --==⋅-=- 所以 2dydx=所以 切线方程为2y x =.（10）已知+1k xe dx ∞=-∞⎰,则k = . 【答案】2-【解析】1122lim bk xkxkxb e dx e dx e k +∞+∞-∞→+∞===⎰⎰因为极限存在所以0k <210k=-2k =-（11）1n lime sin x nxdx -→∞=⎰.【答案】0 【解析】令sin sin cos xx x n I e nxdx e nx n e nxdx ---==-+⎰⎰2sin cos xx n enx ne nx n I --=---所以2cos sin 1xn n nx nx I e C n -+=-++即11020cos sin lim sin lim()1x x n n n nx nx e nxdx e n --→∞→∞+=-+⎰ 122cos sin lim()110n n n n ne n n -→∞+=-+++= （12）设()y y x =是由方程xy 1ye x +=+确定的隐函数，则22x yx=∂=∂ .【答案】3-【解析】对方程xy 1ye x +=+两边关于x 求导有''1yy xy y e ++=，得'1yyy x e-=+ 对''1yy xy y e ++=再次求导可得''''''22()0yyy xy y e y e +++=，得''2''2()yyy y e y x e +=-+ (*)当0x =时，0y =，'(0)0101y e-==，代入(*)得 ''20''032(0)((0))(0)(21)3(0)y y e y e +=-=-+=-+（13）函数2xy x =在区间(]01，上的最小值为 . 【答案】2ee-【解析】因为()22ln 2xy xx '=+，令0y '=得驻点为1x e=. 又()22222ln 2xxy x x x x ''=++⋅，得21120e y e e -+⎛⎫''=> ⎪⎝⎭，故1x e=为2xy x =的极小值点，此时2e y e -=，又当10,x e ⎛⎫∈ ⎪⎝⎭时，()0y x '<；1,1x e ⎛⎤∈ ⎥⎝⎦时，()0y x '>，故y 在10,e ⎛⎫ ⎪⎝⎭上递减，在1,1e ⎛⎫ ⎪⎝⎭上递增.而()11y =，()()002022ln limlim11lim 222ln 00lim lim 1x x x xx x xx xxx x x y x e eee++→→+→++--+→→======，所以2xy x =在区间(]01，上的最小值为21e y e e -⎛⎫= ⎪⎝⎭.(14)设αβ，为3维列向量，T β为β的转置，若矩阵T αβ相似于200000000⎛⎫ ⎪ ⎪ ⎪⎝⎭，则T=βα .【答案】2【解析】因为T αβ相似于200000000⎛⎫⎪ ⎪ ⎪⎝⎭，根据相似矩阵有相同的特征值，得到T αβ得特征值是2,0,0而Tβα是一个常数，是矩阵Tαβ的对角元素之和，则T2002βα=++=三、解答题：15－23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求极限()[]401cos ln(1tan )limsin x x x x x→--+.【解析】()[][]244001ln(1tan )1cos ln(1tan )2lim limsin sin x x x x x x x x x x→→-+--+= 22201ln(1tan )lim 2sin sin x x x x x x →-+=201ln(1tan )1lim 2sin 4x x x x →-+== （16）（本题满分10 分） 计算不定积分1ln(1)xdx x++⎰(0)x >. 【解析】 令1x t x +=得22212,1(1)tdtx dx t t -= =-- 22211ln(1)ln(1)1ln(1)11111x dx t d x t t dt t t t ++=+-+=---+⎰⎰⎰而22111112()11411(1)111ln(1)ln(1)2441dt dtt t t t t t t C t =---+-++--++++⎰⎰所以221ln(1)111ln(1)ln 1412(1)111ln(1)ln(1)2211111ln(1)ln(1)222x t t dx C x t t t x xx x x C x x x x x x x x x x C x ++++=+-+--++=++++-++++=+++++-++⎰ （17）（本题满分10分）设(),,z f x y x y xy =+-，其中f 具有2阶连续偏导数，求dz 与2zx y∂∂∂.【解析】123123zf f yf x zf f xf y∂'''=++∂∂'''=-+∂1231232111213212223331323331122331323()()1(1)1(1)[1(1)]()()z z dz dx dy x yf f yf dx f f xf dyzf f f x f f f x f y f f f x x yf f f xyf x y f x y f ∂∂∴=+∂∂''''''=+++-+∂'''''''''''''''''''=⋅+⋅-+⋅+⋅+⋅-+⋅++⋅+⋅-+⋅∂∂'''''''''''=+-++++-（18）（本题满分10分）设非负函数()y y x = ()0x ≥满足微分方程20xy y '''-+=，当曲线()y y x = 过原点时，其与直线1x =及0y =围成平面区域D 的面积为2，求D 绕y 轴旋转所得旋转体体积. 【解析】解微分方程20xy y '''-+=得其通解212122,y C x C x C C =++其中，为任意常数又因为()y y x =通过原点时与直线1x =及0y =围成平面区域的面积为2，于是可得10C =1112232220002()(2)()133C C y x dx x C x dx x x ==+=+=+⎰⎰从而23C =于是，所求非负函数223(0)y x x x =+ ≥又由223y x x =+ 可得，在第一象限曲线()y f x =表示为1131)3x y =+-（ 于是D 围绕y 轴旋转所得旋转体的体积为15V V π=-，其中552210051(131)9(23213)93918V x dy y dyy y dyππππ==⋅+-=+-+=⎰⎰⎰395117518186V ππππ=-==. （19）（本题满分10分）计算二重积分()Dx y dxdy -⎰⎰，其中()()(){}22,112,D x y x y y x =-+-≤≥.【解析】由22(1)(1)2x y -+-≤得2(sin cos )r θθ≤+，32(sin cos )4()(cos sin )04Dx y dxdy d r r rdr πθθθθθπ+∴-=-⎰⎰⎰⎰332(sin cos )14(cos sin )034r d πθθθθθπ⎡+⎤=-⋅⎢⎥⎣⎦⎰ 2384(cos sin )(sin cos )(sin cos )34d πθθθθθθθπ=-⋅+⋅+⎰3384(cos sin )(sin cos )34d πθθθθθπ=-⋅+⎰3344438814(sin cos )(sin cos )(sin cos )3344d πππθθθθθθπ=++=⨯+⎰83=-.（20）（本题满分12分）设()y y x =是区间-ππ（，）内过点-22ππ（，）的光滑曲线，当-0x π<<时，曲线上任一点处的法线都过原点，当0x π≤<时，函数()y x 满足0y y x ''++=.求()y x 的表达式.【解析】由题意，当0x π-<<时，'x y y =-，即ydy xdx =-，得22y x c =-+， 又()22y ππ-=代入22y x c =-+得2c π=，从而有222x y π+=当0x π≤<时，''0y y x ++=得 ''0y y += 的通解为*12cos sin y c x c x =+ 令解为1y Ax b =+，则有00Ax b x +++=，得1,0A b =-=， 故1y x =-，得''0y y x ++=的通解为12cos sin y c x c x x =+- 由于()y y x =是(,)ππ-内的光滑曲线，故y 在0x =处连续于是由1(0),(0)y y c π-=± += ，故1c π=±时，()y y x =在0x =处连续 又当 0x π-<<时，有22'0x y y +⋅=，得'(0)0xy y-=-=， 当0x π≤<时，有12'sin cos 1y c x c x =-+-，得2'(0)1y c +=- 由'(0)'(0)y y -+=得210c -=，即 21c =故 ()y y x =的表达式为22,0cos sin ,0x x y x x x x ππππ⎧⎪-- -<<=⎨-+-≤<⎪⎩或22,0cos sin ,0x x y x x x x ππππ⎧⎪- -<<=⎨+-≤<⎪⎩，又过点,22ππ⎛⎫- ⎪⎝⎭，所以22,0cos sin ,0x x y x x x x ππππ⎧⎪- -<<=⎨+-≤<⎪⎩.（21）（本题满分11分）（Ⅰ）证明拉格朗日中值定理：若函数()f x 在[],a b 上连续，在(),a b 可导，则存在(),a b ξ∈，使得()()()()f b f a f b a ξ'-=-；（Ⅱ）证明：若函数()f x 在0x =处连续，在()()0,0δδ>内可导，且()0lim x f x A +→'=，则()0f +'存在，且()0f A +'=.【解析】（Ⅰ）作辅助函数()()()()()()f b f a x f x f a x a b aϕ-=----，易验证()x ϕ满足：()()a b ϕϕ=；()x ϕ在闭区间[],a b 上连续，在开区间(),a b 内可导，且''()()()()f b f a x f x b aϕ-=--.根据罗尔定理，可得在(),a b 内至少有一点ξ，使'()0ϕξ=，即'()f ξ'()()0,()()()()f b f a f b f a f b a b aξ--=∴-=--（Ⅱ）任取0(0,)x δ∈，则函数()f x 满足；在闭区间[]00,x 上连续，开区间()00,x 内可导，从而有拉格朗日中值定理可得：存在()()000,0,x x ξδ∈⊂，使得()0'00()(0)x f x f fx ξ-=-……()*又由于()'0lim x f x A +→=，对上式（*式）两边取00x +→时的极限可得： ()()000000'''00()00lim lim ()lim ()0x x x x x f x f f f f A x ξξξ++++→→→-====-故'(0)f +存在，且'(0)f A +=.（22）（本题满分11分设111111042A --⎛⎫ ⎪=- ⎪ ⎪--⎝⎭，1112ξ-⎛⎫⎪= ⎪ ⎪-⎝⎭（Ⅰ）求满足22131,A A ξξξξ==的所有向量23,ξξ；（Ⅱ）对（Ⅰ）中的任一向量23,ξξ，证明：123,,ξξξ线性无关. 【解析】（Ⅰ）解方程21A ξξ=()1111111111111,111100000211042202110000A ξ---------⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=-→→ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪---⎝⎭⎝⎭⎝⎭()2r A =故有一个自由变量，令32x =，由0Ax =解得，211,1x x =-= 求特解，令120x x ==，得31x =故21101021k ξ⎛⎫⎛⎫ ⎪ ⎪=-+ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭，其中1k 为任意常数解方程231A ξξ=2220220440A ⎛⎫ ⎪=-- ⎪ ⎪⎝⎭()21111022012,2201000044020000A ξ-⎛⎫ ⎪-⎛⎫ ⎪ ⎪=--→ ⎪ ⎪ ⎪ ⎪⎝⎭ ⎪⎝⎭故有两个自由变量，令21x =-，由20A x =得131,0x x ==求特解21200η⎛⎫⎪ ⎪= ⎪ ⎪⎪⎝⎭故 321121000k ξ⎛⎫ ⎪⎛⎫ ⎪⎪=-+ ⎪ ⎪⎪ ⎪⎝⎭ ⎪⎝⎭ ，其中2k 为任意常数.（Ⅱ）证明：由于12121212122111121112(21)()2()(21)22221k k k k k k k k k k k k k -+--=+++-+-+-+102=≠ 故123,,ξξξ 线性无关.（23）（本题满分11分）设二次型()()2221231231323,,122f x x x ax ax a x x x x x =++-+- （Ⅰ）求二次型f 的矩阵的所有特征值； （Ⅱ）若二次型f 的规范形为2212y y +，求a 的值.【解析】（Ⅰ） 0101111a A aa ⎛⎫ ⎪=- ⎪ ⎪--⎝⎭0110||01()1111111aaaE A aa a a λλλλλλλλ-----=-=---+---+222()[()(1)1][0()]()[()(1)2]()[22]19(){[(12)]}24()(2)(1)a a a a a a a a a a a a a a a a a λλλλλλλλλλλλλλλλ=---+--+-=---+-=--++--=-+--=--+--123,2,1a a a λλλ∴==-=+（Ⅱ） 若规范形为2212y y +，说明有两个特征值为正，一个为0.则1) 若10a λ==，则 220λ=-< ，31λ= ，不符题意 2) 若20λ= ，即2a =，则120λ=>，330λ=>，符合3) 若30λ= ，即1a =-，则110λ=-< ，230λ=-<，不符题意 综上所述，故2a =.。
西安科技大学824数据结构与算法设计A2009年考研专业课真题试卷
西安科技大学2009年硕士研究生入学考试试题─────────────────────────────────科目编号：824科目名称：数据结构与算法设计（A）考生须知：1、答案必须写在答题纸上，写在试题或草稿纸上不给分。
2、答题须用蓝、黑色钢笔或圆珠笔，用铅笔、红色笔者不给分。
3、答题必须写清题号，字迹要清楚，卷面要保持整洁。
4、试题要随答题纸一起交回。
一、判断题（下列各题，你认为正确的，请在题干的括号内打“√”，错的打“×”。
每题1分，共10分）（）1．树形结构是非线性结构，所以只能用非顺序结构存储。
（）2．线性表就是顺序表。
（）3．有向图用邻接矩阵表示时，顶点i的入度等于邻接矩阵中第i列元素之和。
（）4．理想状态下，在散列表中查找一个元素的时间复杂度为O（1）。
（）5．单链表是线性表的链式存储表示。
（）6．完全二叉树一定是满二叉树。
（）7．由二叉树的先序遍历序列和后序遍历序列可以唯一确定二叉树。
（）8．Dijkstra算法是求非负权图中给定点到其余各点最短路径的有效算法。
O。
（）9．快速排序在最坏情况下的时间复杂度为)(2n（）10．队列是一种插入与删除操作分别在表的两端进行的线性表，是一种先进先出型结构。
二、单选题（每题1分，共10分）1．下列四种基本的逻辑结构中，数据元素之间关系最弱的是（）。
A集合B线性结构C树形结构D图状结构2．算法在发生非法操作时可以作出处理的特性称为（）。
A正确性B易读性C健壮性D高效性3．指针P所指的元素是双循环链表L的尾元素的条件是（）。
A P==LB P==NullC P->next==LD P->prior==L共4页第1页。
2009考研数学二真题及答案解析
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2009年全国硕士研究生入学统一考试真题加答案
2009年全国硕士研究生入学统一考试英语试题Section I Use of EnglishDirections:Read the following text. Choose the best word(s) for each numbered blank and mark A, B, C or D on ANSWER SHEET 1. (10 points)Research on animal intelligence always makes me wonder just how smart humans are.1 the fruit-fly experiments described in Carl Zimmer‘s piece in the Science Times on Tuesday. Fruit flies who were taught to be smarter than the average fruit fly 2 to live shorter lives. This suggests that 3 bulbs burn longer, that there is an 4 in not being too terrifically bright.Intelligence, it 5 out, is a high-priced option. It takes more upkeep, burns more fuel and is slow 6 the starting line because it depends on learning — a gradual 7 — instead of instinct. Plenty of other species are able to learn, and one of the things they‘ve apparently learned is when to 8 .Is there an adaptive value to 9 intelligence? That‘s the question behind this new research. I like it. Instead of casting a wistful glance 10 at all the species we‘ve left in the dust I.Q.-wise, it implicitly asks what the real11 of our own intelligence might be. This is 12 the mind of every animal I‘ve ever met.Research on animal intelligence also makes me wonder what experiments animals would 13 on humans if they had the chance. Every cat with an owner, 14 , is running a small-scale study in operant conditioning. we believe that 15 animals ran the labs, they would test us to 16 the limits of our patience, our faithfulness, our memory for terrain. They would try to decide what intelligence in humans is really 17 , not merely how much of it there is. 18 , they would hope to study a 19 question: Are humans actually aware of the world they live in? 20 the results are inconclusive.1. [A] Suppose [B] Consider [C] Observe [D] Imagine2. [A] tended [B] feared [C] happened [D] threatened3. [A] thinner [B] stabler [C] lighter [D] dimmer4. [A] tendency [B] advantage [C] inclination [D] priority5. [A] insists on [B] sums up [C] turns out [D] puts forward6. [A] off [B] behind [C] over [D] along7. [A] incredible [B] spontaneous [C]inevitable [D] gradual8. [A] fight [B] doubt [C] stop [D] think19. [A] invisible [B] limited [C] indefinite [D] different10. [A] upward [B] forward [C] afterward [D] backward11. [A] features [B] influences [C] results [D] costs12. [A] outside [B] on [C] by [D] across13. [A] deliver [B] carry [C] perform [D] apply14. [A] by chance [B] in contrast [C] as usual [D] for instance15. [A] if [B] unless [C] as [D] lest16. [A] moderate [B] overcome [C] determine [D] reach17. [A] at [B] for [C] after [D] with18. [A] Above all [B] After all [C] However [D] Otherwise19. [A] fundamental [B] comprehensive [C] equivalent [D] hostile20. [A] By accident [B] In time [C] So far [D] Better stillSection II Reading ComprehensionPart ADirections:Read the following four texts. Answer the questions below each text by choosing A, B, C or D. Mark your answers on ANSWER SHEET 1. (40 points)Text1Habits are a funny thing. We reach for them mindlessly, setting our brains on auto-pilot and relaxing into the unconscious comfort of familiar routine. ―Not choice, but habit rules the unreflecting herd,‖ William Wordsworth said in the 19th century. In the ever-changing 21st century, even the word ―habit‖ carries a ne gative connotation.So it seems antithetical to talk about habits in the same context as creativity and innovation. But brain researchers have discovered that when we consciously develop new habits, we create parallel synaptic paths, and even entirely new brain cells, that can jump our trains of thought onto new, innovative tracks.But don‘t bother trying to kill off old habits; once those ruts of procedure are worn into the hippocampus, they‘re there to stay. Instead, the new habits we deliberately ingrain into ourselves create parallel pathways that can bypass those old roads.―The first thing needed for innovation is a fascination with wonder,‖ says Dawna Markova, author of ―The Open Mind‖ and an executive change consultant for Professional Thinking Part ners. ―But we are taught instead to ‗decide,‘ just as our president calls himself ‗the Decider.‘‖ She adds, however, that ―to decide is to kill off all possibilities but one. A good innovational thinker is always exploring the many other possibilities.‖A ll of us work through problems in ways of which we‘re unaware, she says. Researchers in the late 1960 covered that humans are born with the capacity to2approach challenges in four primary ways: analytically, procedurally, relationally (or collaboratively) and innovatively. At puberty, however, the brain shuts down half of that capacity, preserving only those modes of thought that have seemed most valuable during the first decade or so of life.The current emphasis on standardized testing highlights analysis and procedure, meaning that few of us inherently use our innovative and collaborative modes of thought. ―This breaks the major rule in the American belief system — that anyone can do anything,‖ explains M. J. Ryan, author of the 2006 book ―This Year I Will...‖ and Ms. Markova‘s business partner. ―That‘s a lie that we have perpetuated, and it fosters commonness. Knowing what you‘re good at and doing even more of it creates excellence.‖ This is where developing new habits comes in.21. The view of Wordsworth habit is claimed by beingA. casualB. familiarC. mechanicalD. changeable.22. The researchers have discovered that the formation of habit can beA. predictedB. regulatedC. tracedD. guided23.‖ ruts‖(in li ne one, paragraph 3) has closest meaning toA. tracksB. seriesC. characteristicsD. connections24. Ms. Markova‘s comments suggest that the practice of standard testing ? A, prevents new habits form being formedB, no longer emphasizes commonnessC, maintains the inherent American thinking modelD, complies with the American belief system25. Ryan most probably agree thatA. ideas are born of a relaxing mindB. innovativeness could be taughtC. decisiveness derives from fantastic ideasD. curiosity activates creative mindsText 2It is a wise father that knows his own child, but today a man can boost his paternal (fatherly) wisdom –or at least confirm that he‘s the kid‘s dad. All he needs to do is shell our $30 for paternity testing kit (PTK) at his local drugstore – and another $120 to get the results.More than 60,000 people have purchased the PTKs since they first become available without prescriptions last years, according to Doug Fog, chief operating officer of Identigene, which makes the over-the-counter kits. More than two dozen companies sell DNA tests Directly to the public , ranging in price from a few hundred dollars to more than $2500.3Among the most popular : paternity and kinship testing , which adopted children can use to find their biological relatives and latest rage a many passionate genealogists-and supports businesses that offer to search for a family‘s geographic roots .Most tests require collecting cells by webbing saliva in the mouth and sending it to the company for testing. All tests require a potential candidate with whom to compare DNA.But some observers are skeptical, ―There is a kind of false precision being hawked by people claiming they are doing ancestry testing,‖ says Trey Duster, a New York University sociologist. He notes that each individual has many ancestors-numbering in the hundreds just a few centuries back. Yet most ancestry testing only considers a single lineage, either the Y chromosome inherited through men in a father‘s line or mito chondrial DNA, which a passed down only from mothers. This DNA can reveal genetic information about only one or two ancestors, even though, for example, just three generations back people also have six other great-grandparents or, four generations back, 14 other great-great-grandparents.Critics also argue that commercial genetic testing is only as good as the reference collections to which a sample is compared. Databases used by some companies don‘t rely on data collected systematically but rather lump together information from different research projects. This means that a DNA database may differ depending on the company that processes the results. In addition, the computer programs a company uses to estimate relationships may be patented and not subject to peer review or outside evaluation.26.In paragraphs 1 and 2 , the text shows PTK‘s ___________.[A]easy availability[B]flexibility in pricing[C] successful promotion[D] popularity with households27. PTK is used to __________.[A]locate one‘s b irth place[B]promote genetic research[C] identify parent-child kinship[D] choose children for adoption28. Skeptical observers believe that ancestry testing fails to__________.[A]trace distant ancestors[B] rebuild reliable bloodlines[C] fully use genetic information[D] achieve the claimed accuracy29. In the last paragraph ,a problem commercial genetic testing faces is __________.4[A]disorganized data collection[B] overlapping database building30. An appropriate title for the text is most likely to be__________.[A]Fors and Againsts of DNA testing[B] DNA testing and It‘s problems[C]DNA testing outside the lab[D] lies behind DNA testingText 3The relationship between formal education and economic growth in poor countries is widely misunderstood by economists and politicians alike progress in both area is undoubtedly necessary for the social, political and intellectual development of these and all other societies; however, the conventional view that education should be one of the very highest priorities for promoting rapid economic development in poor countries is wrong. We are fortunate that is it, because new educational systems there and putting enough people through them to improve economic performance would require two or three generations. The findings of a research institution have consistently shown that workers in all countries can be trained on the job to achieve radical higher productivity and, as a result, radically higher standards of living.Ironically, the first evidence for this idea appeared in the United States. Not long ago, with the country entering a recessing and Japan at its pre-bubble peak. The U.S. workforce was derided as poorly educated and one of primary cause of the poor U.S. economic performance. Japan was, and remains, the global leader in automotive-assembly productivity. Yet the research revealed that the U.S. factories of Honda Nissan, and Toyota achieved about 95 percent of the productivity of their Japanese countere pants a result of the training that U.S. workers received on the job.More recently, while examing housing construction, the researchers discovered that illiterate, non-English- speaking Mexican workers in Houston, Texas, consistently met best-practice labor productivity standards despite the complexity of the building industry‘s work.What is the real relationship between education and economic development? We have to suspect that continuing economic growth promotes the development of education even when governments don‘t force it. After all, that‘s how education got started. When our ancestors were hunters and gatherers 10,000 years ago, they didn‘t have time to wonder much about anything besides finding food. Only when humanity began to get its food in a more productive way was there time for other things.As education improved, humanity‘s productivity potential, they could in turn afford more education. This increasingly high level of education is probably a necessary, but not a sufficient, condition for the complex political systems required by advanced5economic performance. Thus poor countries might not be able to escape their poverty traps without political changes that may be possible only with broader formal education. A lack of formal education, however, doesn‘t const rain the ability of the developing world‘s workforce to substantially improve productivity for the forested future. On the contrary, constraints on improving productivity explain why education isn‘t developing more quickly there than it is.31. The author holds in paragraph 1 that the important of education in poor countries ___________.[A] is subject groundless doubts[B] has fallen victim of bias[C] is conventional downgraded[D] has been overestimated32. It is stated in paragraph 1 that construction of a new education system __________.[A]challenges economists and politicians[B]takes efforts of generations[C] demands priority from the government[D] requires sufficient labor force33.A major difference between the Japanese and U.S workforces is that __________.[A] the Japanese workforce is better disciplined[B] the Japanese workforce is more productive[C]the U.S workforce has a better education[D] ]the U.S workforce is more organize34. The author quotes the example of our ancestors to show that education emerged __________.[A] when people had enough time[B] prior to better ways of finding food[C] when people on longer went hung[D] as a result of pressure on government35. According to the last paragraph , development of education __________.[A] results directly from competitive environments[B] does not depend on economic performance[C] follows improved productivity[D] cannot afford political changesText 4The most thoroughly studied in the history of the new world are the ministers and political leaders of seventeenth-century New England. According to the standard history of American philosophy, nowhere else in colonial America was ―So much6important attached to intellectual pursuits ‖ Accord ing to many books and articles, New England‘s leaders established the basic themes and preoccupations of an unfolding, dominant Puritan tradition in American intellectual life.To take this approach to the New Englanders normally mean to start with the Puritans‘ theological innovations and their distinctive ideas about the church-important subjects that we may not neglect. But in keeping with our examination of southern intellectual life, we may consider the original Puritans as carriers of European culture adjusting to New world circumstances. The New England colonies were the scenes of important episodes in the pursuit of widely understood ideals of civility and virtuosity.The early settlers of Massachusetts Bay included men of impressive education and influence in England. `Besides the ninety or so learned ministers who came to Massachusetts church in the decade after 1629,There were political leaders like John Winthrop, an educated gentleman, lawyer, and official of the Crown before he journeyed to Boston. There men wrote and published extensively, reaching both New World and Old World audiences, and giving New England an atmosphere of intellectual earnestness.We should not forget , however, that most New Englanders were less well educated. While few crafts men or farmers, let alone dependents and servants, left literary compositions to be analyzed, The in thinking often had a traditional superstitions quality. A tailor named John Dane, who emigrated in the late 1630s, left an account of his reasons for leaving England that is filled with signs. sexual confusion, economic frustrations , and religious hope-all name together in a decisive moment when he opened the Bible, told his father the first line he saw would settle his fate, and read th e magical words: ―come out from among them, touch no unclean thing , and I will be your God and you shall be my people.‖ One wonders what Dane thought of the careful sermons explaining the Bible that he heard in puritan churched.Mean while , many se ttles had slighter religious commitments than Dane‘s, as one clergyman learned in confronting folk along the coast who mocked that they had not come to the New world for religion . ―Our main end was to catch fish. ‖36. The author notes that in the seventeenth-century New England___________.[A] Puritan tradition dominated political life.[B] intellectual interests were encouraged.[C] Politics benefited much from intellectual endeavors.[D] intellectual pursuits enjoyed a liberal environment.37. It is suggested in paragraph 2 that New Englanders__________.[A] experienced a comparatively peaceful early history.[B] brought with them the culture of the Old World[C] paid little attention to southern intellectual life[D] were obsessed with religious innovations738. The early ministers and political leaders in Massachusetts Bay__________.[A] were famous in the New World for their writings[B] gained increasing importance in religious affairs[C] abandoned high positions before coming to the New World[D] created a new intellectual atmosphere in New England39. The story of John Dane shows that less well-educated New Englanders were often __________.[A] influenced by superstitions[B] troubled with religious beliefs[C] puzzled by church sermons[D] frustrated with family earnings40. The text suggests that early settlers in New England__________.[A] were mostly engaged in political activities[B] were motivated by an illusory prospect[C] came from different backgrounds.[D] left few formal records for later referencePart BDirections:Directions: 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. Mark your answers on ANSWER SHEET 1. (10 points)Coinciding with the groundbreaking theory of biological evolution proposed by British naturalist Charles Darwin in the 1860s, British social philosopher Herbert Spencer put forward his own theory of biological and cultural evolution. Spencer argued that all worldly phenomena, including human societies, changed over time, advancing toward perfection. 41.____________.American social scientist Lewis Henry Morgan introduced another theory of cultural evolution in the late 1800s. Morgan, along with Tylor, was one of the founders of modern anthropology. In his work, he attempted to show how all aspects of culture changed together in the evolution of societies.42._____________.In the early 1900s in North America, German-born American anthropologist Franz Boas developed a new theory of culture known as historical particularism. Historical particularism, which emphasized the uniqueness of all cultures, gave new direction to anthropology. 43._____________ .8Boas felt that the culture of any society must be understood as the result of a unique history and not as one of many cultures belonging to a broader evolutionary stage or type of culture. 44._______________.Historical particularism became a dominant approach to the study of culture in American anthropology, largely through the influence of many students of Boas. But a number of anthropologists in the early 1900s also rejected the particularist theory of culture in favor of diffusionism. Some attributed virtually every important cultural achievement to the inventions of a few, especially gifted peoples that, according to diffusionists, then spread to other cultures. 45.________________.Also in the early 1900s, French sociologist Émile Durkheim developed a theory of culture that would greatly influence anthropology. Durkheim proposed that religious beliefs functioned to reinforce social solidarity. An interest in the relationship between the function of society and culture—known as functionalism—became a major theme in European, and especially British, anthropology.[A] Other anthropologists believed that cultural innovations, such as inventions, had a single origin and passed from society to society. This theory was known as diffusionism.[B] In order to study particular cultures as completely as possible, Boas became skilled in linguistics, the study of languages, and in physical anthropology, the study of human biology and anatomy.[C] He argued that human evolution was characterized by a struggle he called the ―survival of the fittest,‖ in which weaker races and societies must eventu ally be replaced by stronger, more advanced races and societies.[D] They also focused on important rituals that appeared to preserve a people‘s social structure, such as initiation ceremonies that formally signify children‘s entrance into adulthood.[E] Thus, in his view, diverse aspects of culture, such as the structure of families, forms of marriage, categories of kinship, ownership of property, forms of government, technology, and systems of food production, all changed as societies evolved.[F]Supporters of the theory viewed as a collection of integrated parts that work together to keep a society functioning.[G] For example, British anthropologists Grafton Elliot Smith and W. J. Perry incorrectly suggested, on the basis of inadequate information, that farming, pottery9making, and metallurgy all originated in ancient Egypt and diffused throughout the world. In fact, all of these cultural developments occurred separately at different times in many parts of the world.Part CDirections:Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written carefully on ANSWER SHEET 2. (10 points)There is a marked difference between the education which every one gets from living with others, and the deliberate educating of the young. In the former case the education is incidental; it is natural and important, but it is not the express reason of the association.46It may be said that the measure of the worth of any social institution is its effect in enlarging and improving experience; but this effect is not a part of its original motive. Religious associations began, for example, in the desire to secure the favor of overruling powers and to ward off evil influences; family life in the desire to gratify appetites and secure family perpetuity; systematic labor, for the most part, because of enslavement to others, etc. 47Only gradually was the by-product of the institution noted, and only more gradually still was this effect considered as a directive factor in the conduct of the institution. Even today, in our industrial life, apart from certain values of industriousness and thrift, the intellectual and emotional reaction of the forms of human association under which the world's work is carried on receives little attention as compared with physical output.But in dealing with the young, the fact of association itself as an immediate human fact, gains in importance.48 While it is easy to ignore in our contact with them the effect of our acts upon their disposition, it is not so easy as in dealing with adults. The need of training is too evident; the pressure to accomplish a change in their attitude and habits is too urgent to leave these consequences wholly out of account. 49Since our chief business with them is to enable them to share in a common life we cannot help considering whether or no we are forming the powers which will secure this ability.If humanity has made some headway in realizing that the ultimate value of every institution is its distinctively human effect we may well believe that this lesson has been learned largely through dealings with the young.50 We are thus led to distinguish, within the broad educational process which we have been so far considering, a more formal kind of education -- that of direct tuition or schooling. In undeveloped social groups, we find very little formal teaching and training. These groups mainly rely for instilling needed dispositions into the young upon the same sort of association which keeps the adults loyal to their group.Section & Writing10Part A51. Directions:Restrictions on the use of plastic bags have not been so successful in some regions. ―White pollution ‖is still going on. Write a letter to the editor(s) of your local newspaper to1)give your opinions briefly and2)make two or three suggestionsYou should write about 100 words. Do not sign your own name at the end of the letter. Use "Li Ming" instead. You do not need to write the address.Part B52. Directions:In your essay, you should1) describe the drawing briefly,2) explain its intended meaning, and then3) give your comments.You should write neatly on ANSHWER SHEET 2. (20 points)11。
2009—数二真题标准答案及解析
2009年全国硕士研究生入学统一考试数学二试题一、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，只有一个选项是符合题目要求的.请将所选项前的字母填在答题纸指定位置上.（1）函数()3sin x x f x nx-=的可去间断点的个数为（ ）()A 1.()B 2. ()C 3.()D 无穷多个.（2）当0x →时，()sin f x x ax =-与()()2ln 1g x x bx =-是等价无穷小，则（ ）()A 11,6a b ==-.()B 11,6a b ==. ()C 11,6a b =-=-. ()D 11,6a b =-=（3）设函数(),z f x y =的全微分为dz xdx ydy =+，则点()0,0（ ）()A 不是(),f x y 的连续点. ()B 不是(),f x y 的极值点. ()C 是(),f x y 的极大值点. ()D 是(),f x y 的极小值点.（4）设函数(),f x y 连续，则()()222411,,yxydx f x y dy dy f x y dx -+=⎰⎰⎰⎰（ ）()A ()2411,xdx f x y dy -⎰⎰. ()B ()241,xxdx f x y dy -⎰⎰.()C ()2411,ydy f x y dx -⎰⎰.()D .()221,y dy f x y dx ⎰⎰（5）若()f x ''不变号，且曲线()y f x =在点()1,1上的曲率圆为222x y +=，则()f x 在区间()1,2内（ ）()A 有极值点，无零点. ()B 无极值点，有零点.()C 有极值点，有零点. ()D 无极值点，无零点.（6）设函数()y f x =在区间[]1,3-上的图形为则函数()()0xF x f t dt =⎰的图形为（ ）()A . ()B .()C .()D .（7）设A ，B 均为2阶矩阵，**A B ，分别为A ，B 的伴随矩阵.若23A B ==，，则分块矩阵O A B O ⎛⎫⎪⎝⎭的伴随矩阵为（ ）()A .**32O B A O ⎛⎫⎪⎝⎭()B .**23O B A O ⎛⎫⎪⎝⎭ ()C .**32O A BO ⎛⎫⎪⎝⎭()D .**23O A BO ⎛⎫⎪⎝⎭（8）设A P ，均为3阶矩阵，TP 为P 的转置矩阵，且100010002T P AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭，若1231223P Q ααααααα==+（，，），（，，），则TQ AQ 为（ ） ()A .210110002⎛⎫⎪⎪ ⎪⎝⎭()B .110120002⎛⎫⎪⎪ ⎪⎝⎭()C .200010002⎛⎫⎪⎪ ⎪⎝⎭()D .100020002⎛⎫⎪⎪ ⎪⎝⎭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）曲线2221-x=0ln(2)u t e du y t t -⎧⎪⎨⎪=-⎩⎰在（0，0）处的切线方程为 . （10）已知+1k xe dx ∞=-∞⎰,则k = .（11）1n lime sin x nxdx -→∞=⎰.（12）设()y y x =是由方程xy 1ye x +=+确定的隐函数，则22x yx=∂=∂ .（13）函数2x y x =在区间(]01，上的最小值为 .(14)设αβ，为3维列向量，T β为β的转置，若矩阵T αβ相似于200000000⎛⎫⎪⎪ ⎪⎝⎭，则T =βα .三、解答题：15－23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求极限()[]401cos ln(1tan )limsin x x x x x→--+.（16）（本题满分10 分）计算不定积分ln(1dx +⎰(0)x >. （17）（本题满分10分）设(),,z f x y x y xy =+-，其中f 具有2阶连续偏导数，求dz 与2z x y∂∂∂.（18）（本题满分10分）设非负函数()y y x = ()0x ≥满足微分方程20xy y '''-+=，当曲线()y y x = 过原点时，其与直线1x =及0y =围成平面区域D 的面积为2，求D 绕y 轴旋转所得旋转体体积. （19）（本题满分10分）计算二重积分()Dx y dxdy -⎰⎰，其中()()(){}22,112,D x y x y y x =-+-≤≥.（20）（本题满分12分）设()y y x =是区间-ππ（，）内过点（的光滑曲线，当-0x π<<时，曲线上任一点处的法线都过原点，当0x π≤<时，函数()y x 满足0y y x ''++=.求()y x 的表达式.（21）（本题满分11分）（Ⅰ）证明拉格朗日中值定理：若函数()f x 在[],a b 上连续，在(),a b 可导，则存在(),a b ξ∈，使得()()()()f b f a f b a ξ'-=-；（Ⅱ）证明：若函数()f x 在0x =处连续，在()()0,0δδ>内可导，且()0lim x f x A +→'=，则()0f +'存在，且()0f A +'=.（22）（本题满分11分设111111042A --⎛⎫ ⎪=- ⎪ ⎪--⎝⎭，1112ξ-⎛⎫⎪= ⎪ ⎪-⎝⎭. （Ⅰ）求满足22131,A A ξξξξ==的所有向量23,ξξ；（Ⅱ）对（Ⅰ）中的任一向量23,ξξ，证明：123,,ξξξ线性无关.（23）（本题满分11分）设二次型()()2221231231323,,122f x x x ax ax a x x x x x =++-+-（Ⅰ）求二次型f 的矩阵的所有特征值；（Ⅱ）若二次型f 的规范形为2212y y +，求a 的值.2009年全国硕士研究生入学统一考试数学二试题答案一、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，只有一个选项是符合题目要求的.请将所选项前的字母填在答题纸指定位置上.（1）函数()3sin x x f x nx-=的可去间断点的个数为（ ）()A 1.()B 2. ()C 3.()D 无穷多个.【答案】C 【解析】()3s i n x x f x xπ-=则当x 取任何整数时，()f x 均无意义故()f x 的间断点有无穷多个，但可去间断点为极限存在的点，故应是30x x -=的解1,2,30,1x =±320032113211131lim lim sin cos 132lim lim sin cos 132lim lim sin cos x x x x x x x x x x x x x x x x x x x x x ππππππππππππ→→→→→-→---==--==--== 故可去间断点为3个，即0,1±（2）当0x →时，()sin f x x ax =-与()()2ln 1g x x bx =-是等价无穷小，则（ ）()A 11,6a b ==-. ()B 11,6a b ==. ()C 11,6a b =-=-. ()D 11,6a b =-=. 【答案】A【解析】2()sin ,()(1)f x x ax g x x ln bx =-=-为等价无穷小，则222200000()sin sin 1cos sin lim lim lim lim lim ()ln(1)()36x x x x x f x x ax x ax a ax a axg x x bx x bx bx bx→→→→→---==-⋅---洛洛230sin lim 166x a ax a b b axa→==-=-⋅ 36a b ∴=- 故排除,B C . 另外201cos lim3x a axbx→--存在，蕴含了1cos 0a ax -→()0x →故 1.a =排除D .所以本题选A.（3）设函数(),z f x y =的全微分为dz xdx ydy =+，则点()0,0（ ）()A 不是(),f x y 的连续点. ()B 不是(),f x y 的极值点. ()C 是(),f x y 的极大值点. ()D 是(),f x y 的极小值点.【答案】 D【解析】因dz xdx ydy =+可得,z zx y x y∂∂==∂∂ 2222221,0,1z z z zA B C x x y y x y∂∂∂∂== === ==∂∂∂∂∂∂又在（0，0）处，0,0z zx y∂∂==∂∂ 210AC B -=>故（0，0）为函数(,)z f x y =的一个极小值点.（4）设函数(),f x y 连续，则()()222411,,yxydx f x y dy dy f x y dx -+=⎰⎰⎰⎰（ ）()A ()2411,xdx f x y dy -⎰⎰. ()B ()241,xxdx f x y dy -⎰⎰.()C ()2411,ydy f x y dx -⎰⎰.()D .()221,y dy f x y dx ⎰⎰【答案】C 【解析】222211(,)(,)xxdx f x y dy dy f x y dx +⎰⎰⎰⎰的积分区域为两部分：{}1(,)12,2D x y x x y =≤≤≤≤，{}2(,)12,4D x y y y x y =≤≤≤≤-将其写成一块{}(,)12,14D x y y x y =≤≤≤≤- 故二重积分可以表示为2411(,)ydy f x y dx -⎰⎰，故答案为C.（5）若()f x ''不变号，且曲线()y f x =在点()1,1上的曲率圆为222x y +=，则()f x 在区间()1,2内（ ）()A 有极值点，无零点. ()B 无极值点，有零点.()C 有极值点，有零点. ()D 无极值点，无零点.【答案】 B【解析】由题意可知，()f x 是一个凸函数，即''()0f x <，且在点(1,1)处的曲率322|''|(1('))y y ρ==+，而'(1)1f =-，由此可得，''(1)2f =-在[1,2] 上，'()'(1)10f x f ≤=-<，即()f x 单调减少，没有极值点. 对于(2)(1)'()1(1,2)f f f ζζ-=<- , ∈ ， （拉格朗日中值定理）(2)0f ∴ <而 (1)10f =>由零点定理知，在[1,2] 上，()f x 有零点. 故应选（B ）. （6）设函数()y f x =在区间[]1,3-上的图形为则函数()()0xF x f t dt =⎰的图形为（ ）()A . ()B .()C .()D .【答案】D【解析】此题为定积分的应用知识考核，由()y f x =的图形可见，其图像与x 轴及y 轴、0x x =所围的图形的代数面积为所求函数()F x ，从而可得出几个方面的特征： ①[]0,1x ∈时，()0F x ≤，且单调递减. ②[]1,2x ∈时，()F x 单调递增. ③[]2,3x ∈时，()F x 为常函数.④[]1,0x ∈-时，()0F x ≤为线性函数，单调递增. ⑤由于F(x)为连续函数结合这些特点，可见正确选项为D .（7）设A ，B 均为2阶矩阵，**A B ，分别为A ，B 的伴随矩阵.若23A B ==，，则分块矩阵O A B O ⎛⎫⎪⎝⎭的伴随矩阵为（ ）()A .**32OB A O ⎛⎫⎪⎝⎭()B .**23O B A O ⎛⎫⎪⎝⎭()C .**32O A BO ⎛⎫⎪⎝⎭()D .**23O A B O ⎛⎫⎪⎝⎭【答案】 B【解析】根据CC C E *=若111,C C C CC C*--*==分块矩阵00A B ⎛⎫⎪⎝⎭的行列式22012360A AB B ⨯=-=⨯=（）即分块矩阵可逆 1111000066000100B BA A AB B BBAA A**---*⎛⎫ ⎪⎛⎫⎛⎫⎛⎫ ⎪=== ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎪⎝⎭10023613002B B AA ****⎛⎫ ⎪⎛⎫== ⎪ ⎪ ⎪⎝⎭⎪⎝⎭（8）设A P ，均为3阶矩阵，TP 为P 的转置矩阵，且100010002T P AP ⎛⎫ ⎪= ⎪ ⎪⎝⎭，若1231223P Q ααααααα==+（，，），（，，），则TQ AQ 为（ ） ()A .210110002⎛⎫⎪⎪ ⎪⎝⎭()B .110120002⎛⎫⎪⎪ ⎪⎝⎭()C .200010002⎛⎫⎪⎪ ⎪⎝⎭()D .100020002⎛⎫⎪⎪ ⎪⎝⎭【答案】 A【解析】122312312312100(,,)(,,)110(,,)(1)001Q E αααααααααα⎡⎤⎢⎥=+==⎢⎥⎢⎥⎣⎦，即：12121212122112(1)[(1)][(1)](1)[](1)100(1)010(1)002110100100210010010110110001002001002T T TT T Q PE Q AQ PE A PE E P AP E E E ===⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦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）曲线2221-x=0ln(2)u t e du y t t -⎧⎪⎨⎪=-⎩⎰在（0，0）处的切线方程为 . 【答案】2y x =【解析】221222ln(2)22t dy t t t t dt t ==--⋅=--2(1)1(1)1t t dxe dt --==⋅-=- 所以 2dy dx= 所以 切线方程为2y x =.（10）已知+1k xe dx ∞=-∞⎰,则k = .【答案】2-【解析】1122lim bk xkxkxb e dx e dx e k +∞+∞-∞→+∞===⎰⎰因为极限存在所以0k <210k=-2k =-（11）1n lime sin x nxdx -→∞=⎰.【答案】0【解析】令sin sin cos x x xn I e nxdx e nx n e nxdx ---==-+⎰⎰2sin cos x xn e nx nenx n I --=---所以2cos sin 1xn n nx nx I e C n -+=-++即11020cos sin lim sin lim()1xx n n n nx nx e nxdx e n --→∞→∞+=-+⎰ 122cos sin lim()110n n n n ne n n -→∞+=-+++= （12）设()y y x =是由方程xy 1ye x +=+确定的隐函数，则22x yx=∂=∂ .【答案】3-【解析】对方程xy 1y e x +=+两边关于x 求导有''1y y xy y e ++=，得'1yyy x e -=+ 对''1y y xy y e ++=再次求导可得''''''22()0y y y xy y e y e +++=，得''2''2()yyy y e y x e +=-+ (*)当0x =时，0y =，'(0)0101y e -==，代入(*)得 ''20''032(0)((0))(0)(21)3(0)y y e y e +=-=-+=-+（13）函数2x y x =在区间(]01，上的最小值为 . 【答案】2ee-【解析】因为()22ln 2xy xx '=+，令0y '=得驻点为1x e =.又()22222ln 2xxy x x x x ''=++⋅，得21120e y e e -+⎛⎫''=> ⎪⎝⎭，故1x e=为2xy x =的极小值点，此时2e y e -=，又当10,x e ⎛⎫∈ ⎪⎝⎭时，()0y x '<；1,1x e ⎛⎤∈ ⎥⎝⎦时，()0y x '>，故y 在10,e ⎛⎫ ⎪⎝⎭上递减，在1,1e ⎛⎫ ⎪⎝⎭上递增.而()11y =，()()002022ln limlim11lim 222ln 00lim lim 1x x x xx x xx xxx x x y x e eee++→→+→++--+→→======，所以2xy x =在区间(]01，上的最小值为21ey e e -⎛⎫= ⎪⎝⎭.(14)设αβ，为3维列向量，T β为β的转置，若矩阵T αβ相似于200000000⎛⎫ ⎪⎪ ⎪⎝⎭，则T =βα .【答案】2【解析】因为T αβ相似于200000000⎛⎫⎪⎪ ⎪⎝⎭，根据相似矩阵有相同的特征值，得到T αβ得特征值是2,0,0而T βα是一个常数，是矩阵T αβ的对角元素之和，则T 2002βα=++=三、解答题：15－23小题，共94分.请将解答写在答题纸指定的位置上.解答应写出文字说明、证明过程或演算步骤.（15）（本题满分9分）求极限()[]401cos ln(1tan )lim sin x x x x x→--+.【解析】()[][]244001ln(1tan )1cos ln(1tan )2lim lim sin sin x x x x x x x x x x→→-+--+= 22201ln(1tan )lim 2sin sin x x x x x x→-+=201ln(1tan )1lim 2sin 4x x x x →-+== （16）（本题满分10 分）计算不定积分ln(1dx +⎰(0)x >. 【解析】t =得22212,1(1)tdtx dx t t -= =--2221ln(1ln(1)1ln(1)11111dx t d t t dt t t t +=+-+=---+⎰⎰⎰而22111112()11411(1)111ln(1)ln(1)2441dt dtt t t t t t t C t =---+-++--++++⎰⎰所以2ln(1)111ln(1ln1412(1)1ln(1211ln(122t tdx Ct t tx Cx x C+++=+-+--+=++-+=+++⎰（17）（本题满分10分）设(),,z f x y x y xy=+-，其中f具有2阶连续偏导数，求dz与2zx y∂∂∂.【解析】123123zf f yfxzf f xfy∂'''=++∂∂'''=-+∂12312321112132122233313233 31122331323()()1(1)1(1)[1(1)]()()z zdz dx dyx yf f yf dx f f xf dyzf f f x f f f x f y f f f xx yf f f xyf x y f x y f∂∂∴=+∂∂''''''=+++-+∂''''''''''''''''''' =⋅+⋅-+⋅+⋅+⋅-+⋅++⋅+⋅-+⋅∂∂'''''''''''=+-++++-（18）（本题满分10分）设非负函数()y y x= ()0x≥满足微分方程20xy y'''-+=，当曲线()y y x= 过原点时，其与直线1x=及0y=围成平面区域D的面积为2，求D绕y轴旋转所得旋转体体积.【解析】解微分方程20xy y'''-+=得其通解212122,y C x C x C C=++其中，为任意常数又因为()y y x=通过原点时与直线1x=及0y=围成平面区域的面积为2，于是可得1C=111223222002()(2)()133C Cy x dx x C x dx x x==+=+=+⎰⎰从而23C=于是，所求非负函数223(0)y x x x=+ ≥又由223y x x=+ 可得，在第一象限曲线()y f x=表示为11)3x=（于是D 围绕y 轴旋转所得旋转体的体积为15V V π=-，其中5522100511)9(2393918V x dy dyy dy ππππ==⋅=+-=⎰⎰⎰395117518186V ππππ=-==. （19）（本题满分10分）计算二重积分()Dx y dxdy -⎰⎰，其中()()(){}22,112,D x y x y y x =-+-≤≥.【解析】由22(1)(1)2x y -+-≤得2(sin cos )r θθ≤+，32(sin cos )4()(cos sin )04Dx y dxdy d r r rdr πθθθθθπ+∴-=-⎰⎰⎰⎰332(sin cos )14(cos sin )034r d πθθθθθπ⎡+⎤=-⋅⎢⎥⎣⎦⎰ 2384(cos sin )(sin cos )(sin cos )34d πθθθθθθθπ=-⋅+⋅+⎰3384(cos sin )(sin cos )34d πθθθθθπ=-⋅+⎰3344438814(sin cos )(sin cos )(sin cos )3344d πππθθθθθθπ=++=⨯+⎰83=-.（20）（本题满分12分）设()y y x =是区间-ππ（，）内过点（的光滑曲线，当-0x π<<时，曲线上任一点处的法线都过原点，当0x π≤<时，函数()y x 满足0y y x ''++=.求()y x 的表达式. 【解析】由题意，当0x π-<<时，'xy y =-，即ydy xdx =-，得22y x c =-+，又(y =代入22y x c =-+得2c π=，从而有222x y π+=当0x π≤<时，''0y y x ++=得 ''0y y += 的通解为*12cos sin y c x c x =+ 令解为1y Ax b =+，则有00Ax b x +++=，得1,0A b =-=， 故1y x =-，得''0y y x ++=的通解为12cos sin y c x c x x =+- 由于()y y x =是(,)ππ-内的光滑曲线，故y 在0x =处连续于是由1(0),(0)y y c π-=± += ，故1c π=±时，()y y x =在0x =处连续 又当 0x π-<<时，有22'0x y y +⋅=，得'(0)0xy y-=-=， 当0x π≤<时，有12'sin cos 1y c x c x =-+-，得2'(0)1y c +=- 由'(0)'(0)y y -+=得210c -=，即 21c =故 ()y y x =的表达式为0cos sin ,0x y x x x x πππ⎧-<<=⎨-+-≤<⎪⎩或0cos sin ,0x y x x x x πππ-<<=+-≤<⎪⎩，又过点,22ππ⎛⎫- ⎪⎝⎭，所以0cos sin ,0x y x x x x πππ-<<=+-≤<⎪⎩.（21）（本题满分11分）（Ⅰ）证明拉格朗日中值定理：若函数()f x 在[],a b 上连续，在(),a b 可导，则存在(),a b ξ∈，使得()()()()f b f a f b a ξ'-=-；（Ⅱ）证明：若函数()f x 在0x =处连续，在()()0,0δδ>内可导，且()0lim x f x A +→'=，则()0f +'存在，且()0f A +'=.【解析】（Ⅰ）作辅助函数()()()()()()f b f a x f x f a x a b aϕ-=----，易验证()x ϕ满足：()()a b ϕϕ=；()x ϕ在闭区间[],a b 上连续，在开区间(),a b 内可导，且''()()()()f b f a x f x b aϕ-=--.根据罗尔定理，可得在(),a b 内至少有一点ξ，使'()0ϕξ=，即'()f ξ'()()0,()()()()f b f a f b f a f b a b aξ--=∴-=--（Ⅱ）任取0(0,)x δ∈，则函数()f x 满足；在闭区间[]00,x 上连续，开区间()00,x 内可导，从而有拉格朗日中值定理可得：存在()()000,0,x x ξδ∈⊂，使得()0'00()(0)x f x f fx ξ-=-……()*又由于()'lim x f x A +→=，对上式（*式）两边取00x +→时的极限可得：()()000000'''0000()00lim lim ()lim ()0x x x x x f x f f f f A x ξξξ++++→→→-====- 故'(0)f +存在，且'(0)f A +=.（22）（本题满分11分设111111042A --⎛⎫ ⎪=- ⎪ ⎪--⎝⎭，1112ξ-⎛⎫ ⎪= ⎪ ⎪-⎝⎭（Ⅰ）求满足22131,A A ξξξξ==的所有向量23,ξξ；（Ⅱ）对（Ⅰ）中的任一向量23,ξξ，证明：123,,ξξξ线性无关. 【解析】（Ⅰ）解方程21A ξξ=()1111111111111,111100000211042202110000A ξ---------⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=-→→ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪---⎝⎭⎝⎭⎝⎭()2r A =故有一个自由变量，令32x =，由0Ax =解得，211,1x x =-= 求特解，令120x x ==，得31x =故21101021k ξ⎛⎫⎛⎫ ⎪ ⎪=-+ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭，其中1k 为任意常数解方程231A ξξ=2220220440A ⎛⎫ ⎪=-- ⎪ ⎪⎝⎭()21111022012,2201000044020000A ξ-⎛⎫ ⎪-⎛⎫ ⎪ ⎪=--→⎪ ⎪ ⎪ ⎪⎝⎭ ⎪⎝⎭故有两个自由变量，令21x =-，由20A x =得131,0x x ==求特解21200η⎛⎫ ⎪ ⎪= ⎪ ⎪⎪⎝⎭故 321121000k ξ⎛⎫⎪⎛⎫ ⎪⎪=-+ ⎪ ⎪⎪ ⎪⎝⎭ ⎪⎝⎭ ，其中2k 为任意常数.（Ⅱ）证明：由于12121212122111121112(21)()2()(21)222210k k k k k k k k k k k k k -+--=+++-+-+-+102=≠ 故123,,ξξξ 线性无关.（23）（本题满分11分）设二次型()()2221231231323,,122f x x x ax ax a x x x x x =++-+- （Ⅰ）求二次型f 的矩阵的所有特征值；（Ⅱ）若二次型f 的规范形为2212y y +，求a 的值. 【解析】（Ⅰ） 0101111a A aa ⎛⎫ ⎪=- ⎪ ⎪--⎝⎭0110||01()1111111aaaE A aa a a λλλλλλλλ-----=-=---+---+222()[()(1)1][0()]()[()(1)2]()[22]19(){[(12)]}24()(2)(1)a a a a a a a a a a a a a a a a a λλλλλλλλλλλλλλλλ=---+--+-=---+-=--++--=-+--=--+--123,2,1a a a λλλ∴==-=+（Ⅱ） 若规范形为2212y y +，说明有两个特征值为正，一个为0.则 1) 若10a λ==，则 220λ=-< ，31λ= ，不符题意2) 若20λ= ，即2a =，则120λ=>，330λ=>，符合3) 若30λ= ，即1a =-，则110λ=-< ，230λ=-<，不符题意 综上所述，故2a =.。
西安电子科技大学861经济学2009年考研真题答案
1、蛛网模型：是运用弹性原理解释某些生产周期较长的商品在失去均衡时发生的不同波动情况的一种动态分析理论。
古典经济学理论认为，如果供给量和价格的均衡被打破，经过竞争，均衡状态会自动恢复。
蛛网模型却证明， 按照古典经济学静态下完全竞争的假设，均衡一旦被打破，经济系统并不一定自动恢复均衡。
与均衡价格决定模型不同的是，蛛网模型是一个动态模型，假定商品的本期产量决定于前一期的价格，需求量 取决于本期价格，并根据需求弹性、供给弹性的不同，分为“收敛型蛛网”、“发散型蛛网”和“封闭型蛛网”三 种类型。
蛛网模型解释了某些生产周期较长的商品的产量和价格的波动的情况，对实践具有一定的指导作用。
但是，这 个模型还是一个很简单的和有缺陷的模型，需要进一步的完善。
2、替代效应：指由商品的价格变动所引起的商品相对价格的变动，进而由商品的相对价格变动所引起的商品需求 量的变动，即是指在消费者的实际收入（即满足水平）保持不变的前提下，由于商品价格的相对变化所引起的商品 需求量的变化。
替代效应是价格变化的总效应之一，还有一个是收入效应。
替代效应一般在程度上比收入效应大，在正常商品 的情况下，两种效应的方向是一致的，即价格下降，替代效应引起消费者购买数量增加，收入效应也引起消费者购 买数量增加。
一般而言，替代效应总是负的，就是说，在实际收入保持不变的条件下，商品的价格和商品的需求量 成反方向变动3、边际技术替代率：指在维持产量水平不变的条件下，增加一单位某种生产要素投入量时所减少的另一种要素的 投入数量，公式：LK K MRTS L∆=-∆ 生产要素相互替代的过程中存在边际技术替代率递减规律，即在维持产量不变的前提下，当一种生产要素的投 入量不断增加时，每一单位的这种生产要素所能替代的另一种生产要素的数量是递减的。
边际技术替代率递减的主 要原因在于：任何一种产品的生产技术都要求各要素投入之间有适当的比例，这意味着要素之间的替代是有限制的。
西安科技大学829数字测图原理与方法2007-2015年(2007和08、14和15年无答案)考研专业课历年真题汇编
一、名词解释（每小题 4 分，共 20 分）： 1、水平角 2、中误差 3、扫描数字化 4、等高线 5、方位角
二、填空（每小题 4 分，共 20 分）：
1．在 1：500 比例尺地形图上量取 AB 间的长度为 180mm，其量距偏差为 0.1mm,
其表达的实地长度为
m，其实地量距偏差为
m。
2. 对某直线边长测量了 9 次，设一次丈量中误差为±6mm，则其平均值的中误
误差为± mm。
5.设全站仪测量水平角一测回的中误差为 5 秒，若某等级导线要求测角中误差
为 2.5 秒，则采用该全站仪测水平角的测回数至少为
测回。
三、问答题。在 6 小题中任选 5 题作答。（每小题 10 分，共 50 分）： 1、画示意图并试述用经纬仪测回法测量一个水平角的野外操作步骤及最终
共3页
四、分析题（每小题 20 分，共 40 分）: 1. 某大型煤矿工业广场施工前需测绘 1:1000 的数字地形图，测区形状近似正 方形，面积约 1 平方公里。测区附近分布 3 个国家三等控制点，并具有其 54 坐标及高程，但最近的控制点距离测区至少 5 公里。测区内分布 3 个农村居民 区，地貌为丘陵区，地形有一定的起伏。根据现有的技术条件采用全野外数字 测图。试分析： （1）全野外数字测图及内业处理涉及的硬件和软件设备？（4 分） （2）具体说明一种野外测图模式的外业过程？（8 分） （3）以某种数字成图系统为例，说明内业成图的具体流程？（8 分） 2. 在高斯平面坐标系中，已知两地面控制点 A，B 的 1954 北京坐标及 1985 国 家高程。A 的坐标为 xA=3888600 m，yA=36499800m，HA=1001m；B 的坐标为 xB=3888100 m，yB=36500300m，HB=1003 m。试问： （1）说明 A,B 两点所处的高斯投影带及其中央子午线的经度，判别 A,B 两点 分别与该投影带中央子午线的位置关系？（8 分） （2）绘草图标明 A,B 在高斯平面坐标系中的概略位置，并计算 AB 方向和 BA 方向的坐标方位角（6 分） （3）采用精密全站仪实地测量 A,B 间的水平距离时，虽然已经加入了温度气 压改正，但实测水平距离仍比按坐标反算的距离大 110 mm。经过检查 A,B 两 点的点位误差以及全站仪的测距误差均可以忽略不计。试根据 AB 所处的坐标 系统解释造成上述现象的原因。（6 分）