1991考研数学一真题及答案解析

1、19913 1 ,2xtd y2设则ycost,dx2 x y z 2zz(x,y)(1,0,1) xyz222dzx1 y2 z3x2 y1 z:L :2L；L 11012111 L210 ) 1 cosx1x与 a235 2 02 1 000设4阶方阵AAA1 0 0 120 0 1131ex2y 1ex2( ) t f(x)f(x)22f(x) ( )xf2 0ex ln2e ln2 2xln2e ln22x ex( a 2a 52n1a3n1,( )nnn1n1n1789设D是D 是D1(xycosxsin y)dxdy( )D 2 cosxsin ydxdy 2 xydxdyDD11

2、 4 (xycosxsin y)dxdy0D1设nABCABC 是 ( ) EE n ACBECBA EBCA E E BAC5求lim(cos x) .xx03y z 6设n2x2P(1,1,1)226x 8y22u Pnz y2 2z, y z)dV (x2,其中 绕 z20 xz 46O(0,0) 和 ,0) y asinx(aLO到A y )dx(2x y)dy3L8f(x)2|x|(1x21n2n17f(x)在0,1上连续,(0,1)内可导,且31 ff(x)23 cf (c) 0.8(1,0,2,3), (1,1,3,5), 1,a2,1), 4,a8)1234(1,1,b.、 、

3、 、 a、b 1234、 、 、 a、b 有12346设 A为nE是n AE8P(x,y)Qxx 632 X 4 2,方差为 P2X P X 0 0 y 2x2 ( ax 46(X,Y)2e(x2y),0, 0 xyf(x,y),0,其他 X YZ 3sinttcostt3x t)t), 则. t)t)ydydtdydxsint,dxtdt xd y d d t 12 ( ) (dx2 t)ttcost2sint 1 sinttcost t.tt32 2dydx(1,0,z z(1,0)1.d(x y z )222d(xyz)0,2 x y z222xdx ydyzdzx y z(xy)dz(

4、ydxxdy)z ,2221,yz1,得dz dx 2dy .令x23yz20 x ；LL11 LL2L Ll (1,0,1) 过 ,于是 和 ,即 1112(2,1,1) l2x1 y2 z312011 0,13yz20即x.32110 x x,(1x) 1xx,nn0 0当x时213111 ) 1 x1 x x ,222232211ax2ax ) 12233limlim a.1cosx13x0 x0 x222321 ) 1与0cosx1 a 1 a x.233 1 2 0 02 50 01 23 3.0 01 1 0 03 3A 01 0 A1.A100 B10 B,0 B B 0A011

5、a bA2A c da b d bA .*c ac d 0 Aa b1 d b d b11c aA . c a c dA 01A100 B0 B1 1 2 0 02 50 01 23 3A 1.0 01 1 0 0 3 35 3 0 x 0,x1ex2e 1x2y x01ex2e 1x2x0 x0 x01ex21ex2e 1x2y 1 y 1e 1x2xxx f(x)x xlim f(x) y,则0 x0 x x 0a,(为常数） y a lim f(x)xt,则tu,2u2 t 2fu) 2,f(x)2xf 2x 200df(xf(x) 求导,得 f (x) 2f(x) ,即2 .解xCe

6、f (x) C2x02fu)22f (x) Ce2xf (0) Ce ln2 f00C ln2 f (x)( a a a a a a a n1n1234n11234n1n1n1而a.n1nn1n1n1a(a a )(a a ) (a a )n1234n1 (a a ) a a 538,n1n1n1DD ,D ,D ,D 3124D ,D yD ,D x3124令,D2Dxxydxdy0.D DD D1234而xsin y对x ycosxsin ydxdy 2 cosxsin ydxdy,D3121,12D EA B 、 、C ABCn0| A|B|C1 A、B 0、C 0 ABC ABC E

7、, E A EA A 有 11 E C E C E. ABCC 1151 x 1 x) x x1xxx0 x01t x 0 时t 0 令cos x11lim(1 x x1 lim(1t) e,tx0t0 x x 1 x x x1 ee0.xxxx0 x00sinx xx2xx2 sin2(cos x1)22lim lim lim,xxx2x0 x0 x0 x e . x) ex0故xx2x0u u u , ,nx y zu uuu n xyz3y z 6P(1,1,1)2x222 2 2,3,1, 4x,6y,2z 4x,6y,2zP(1,1,1)P(1,1,1)1 1 .nu226 xz 6

8、x 8y142222 u8又,yz 6x 8y14z 6x22226x 8y222 14z2z2PPuu yz62831 . y2 2z,绕 xz220 x zz 4(x y ) 22x y 8,z 4.22x,22 4dz2d2000r 2z44420r0406 asinx, (x0, )dy acosxdxy I y )x y)3L (asinx) (2xasinx)acosxdx30a21a sin x2axcosx sin2x dx3320a 2 a 2a x2332000a 2 a xdx2a x22x3240001 a2 acos xcosx 2a xsinxcosx cos2x

9、33340004 a 4a.334 a 4aaa 0I 0,得aI 33 I 4a2 4 0. I 0,0 a 1 a1, 且 . I 0,1 a41I a 4a,(a 故a33y sin x, (x0, ).8 f(x)a 与b f(x)nn1lf(xf(x 0 (n ),bnlll1l2ll laf(xllnl021x n 41n 2 n x10n00n 2 1n (n1,2,3, ),n20a 2 1(2x)dx5.00 2| x|(1x1) f (x)la02f (x) 2| x a cosnxb sinxl nn12(cosn 1)5cosn x n222n15 41 cos(2n1

10、) x (1 x1).2(2n22n1125 410 令x f (0) 2 0cos0.(2n2(2n282 2n1n1 11111 1又,(2n1) (2n) (2n1) 4 nn22222n1n1n1n14 n231122.8n26n1n172( 1 f(x)dx2332 1f ( ),1 f (x)dx f ( )(1 )233 3 1 f(x)dx f( ) f(0).即323c 1) f(c)0.(0,1)c(08 x x xx11223344x x x x 11234x x 2x 1.23342x 3x (a2)x 4x b312343x 5x x (a8)x 51234 1 2、

11、 23、1 11121 1 1 11221 0 1 11b350 1 10 1 a1b12 A 2 3 a243 51a80 2 2 a51 1 11210 1 11,0 0 a1 0b0 0 0 a1 01,b0r()1 r()a 不能表示成、 x xx ,x ,x ,x 1234x x112233441234 b ab1 bT1r() r() 4.,0当a,1 a1 a1 abab1b 0.故 a1a1a11234nAxb设 A是m矩阵,线性方程组 A b , , ,(或者说, A ,r(A) r(A) A的秩,即是b12n , , , 与 , , , ,b12n12nnAxb设 A是m

12、r() r(). r(A) r(A) . r(A)1 r(A). A的列向量, , ,b12n6 A为nQ12Q AQQ AQ ,T1N0(i1,2, n), 是AiiE)Q Q AQQ Q E Q (ATTTE|Q | AE|Q|Q (AEQ|E ( , | ATTi | AE1. A的n , , , .A为n12n ,由 为A 使A E 1 1是A E的特征值.因为A E得 A ( 1.12n| A E| , , , 1.AiiA是n n XAXAX AX 成立,则称 的特征值,称非零向量 8yy(x)P(x,y)1Y y (X x) y 0y yy,0) Q(xx1|PQ (yy) y

13、y(1 y ) .2222当y 0Q(x,0),| PQ| yy31,12(1 y )y(1 y ) 222 即yy 1 y x 1y1,y 0.2 令 y P(y) y P 1 P 2.1P2yC 1P2 ,C即 y1 yC1 y 1 P21 y ,x y1,P y 02dyy 1y2 1dx.2sect令 y dy secttantdtln secttant Cy 1tant2ln sect sec t1 C ln y y 1 C.22 y 10 y2 1 C x. y2ln y故y y 1.2x当x1 y1,e y y 1e当xC,12x1112y y 1e ；2xex1(y y y y y y 1222e , y y 1 e当x C,2x1112y y 1e ；2x1x(y y y y y y 12221y (ee(x).(x2 63 和 , 22222( )0.8(x)P(x ( ) ( )与,通过查22( )( ).X 2(2, )N(0,1), X N 4222P(2 x4)()(),2( ) P(2 x4)(0)0.8.0222 P(x0)()( )1( )0.2.A x4y12SP() S a ,2DCS11S