1994考研数学一真题及答案解析

1、19945311 ) limcot x(sinx xx0e 2xy 3zzx u1(2, )2e 设uxxyyx2 y2 y Ra b)dxdy 为xD (22222D1 1(1,2,3), (1, , )A2 3 A ,其中 T是 n T53 x2 N (sin xcos x)dx P (x sin xcos x)dx设M,434234221x2222则( ) PMM PM P NPM N N Nf(x,y)(x ,y )f (x ,y )、f (x ,y )f(x,y)在00 x00y00( )|a |n0 a ( ( )2nnn2n1n1 axbx)2 x0a c 02( )2c2x)d

2、e )xb4dacb4dac已知向量组、( )1234 、12233441 、12233441 、12233441 、122334413 5 x t ),2dy d y2设求、dx dx2在t 1t2ytt )cosudu,222 u11 1x 1f(x) ln arctanxx x4 1x 2dx求.sin2x2sin x6xdydzz dxdy2 x y R Sz R,222x y z222Sz R(R9设f(x)f f1xy(x y) f (x)ydxf (x) x ydy 0,求f(x) 28f(x)设f(x)x 0lim0 xx01f( )nn16A与B绕 zS S zz18x x

3、0,)又已知某线性齐次方程组()设四元线性齐次方程组( 为12x x 0,24k (0,1,10)k (1,2,2,1).12) ( ) () ( 和6设 A为 A 是 A A 是AAn*T* AT| A0.2 36 P(AB)P() pP(B) A、BP(AB) X 、Y X XP011122 X,YZ6(X,Y) X 和Y N(1,3 ) 和21X YN(0,4 ), X 与Y Z ,2XY23 2 求ZE(Z)D(Z)； 求X 与Z ；XZ 问X 与Z199453160 0 0cosx(xsinx)xsinxlimlimcos xlimxsin xx32x0 x0 x01cosxsinx

4、 1sinxlimlimlim1).3x26x 6xx0 x0 x02x y40 n ,l取nlM.(,B,C) x y z ( , , )000A(xx )B(y y )C(zz )0.000F(x,y,z)0F(x,y,z) ze 2xy3.z F F F 4,2,0 2 ,n , , 2y,2x,1ezx y z2(x1)(y0,即 2xy40.2e2uy u求 .x y uxxy e x,yy2 u u u22cosx xe2xyyxx yyx111x2)x22( e x x)02 x.x2e2 (x,y),v (x,y)(x,y)具u zxy fu,v)u,v)z f( (x,y (

5、x,y(x,y)z z u z vuvx f f；.x u x v xx12z z u z vuyvy f fy u y v y12 R (1 1 )a b4422sin2 sin2 cos2 cosd r dr22dRrrdrb 2R.23 a ab 222200002 d22 , d200 1 1 1 1 1 R R.则44a b 44 a b 2222 1 112 323 2 1n13133 21 1 1 1, ,2 3T 2 3 3 1 112 321 1 1 而A 2 1, , 2 1T 2 331 3 33 2 A ( )( )( ) ) LL TnTTTTTTT1 112 32n

6、1 3 2 1Tn1.3331 25 3 0为M ,bf(x)0 f(x)0 (ab) b a .a22N2 cos xdx0 P 2 cos xdxN 04,4.00M N P(x,y)( , ) f(x,y) x y ( , ) ( , ), x y f x yf0000 x00( , )y( , )f x y .反之, f x y x y f x y f x y ( , )( , ), ( , )0000 x00y00(x,y)( , ) x y 证 f00(x,y)( , ) ( , ) x y x y f0000(1) |a | 11 111n a a 22,n22 n 22n2n

7、22nn1a0,b0,ab (a b )22 2又 a2 1 1 p当 p1 p12n2pnnn1n1n1121(1) |a |na2n2n2nn 2n1n111cosx: x o(x),1e 2 : x o(x),22x2故axbx): (a ,c2x)de ): (c0),2xaxalim2a4c.2cx 2cx00,c00,c0当a当a ( )x( ),( )l. xx ( )x( ),( ) 若l 若l 若l0,称xx1,称( ),( )( ) ( )x : x；xx0, (x)是( )x ( )( )x o x .( )x( ),( )x x 若( )x( ),( ) x x x x

8、0( ) ( ) ( ) ( ) ( x : x x x o x . 012233441 0122334410 1 0 0 11 1 0 00 1 1 00 0 1 120, ( )( )( )( )0,12233441 L , , ,(i 1,2,L ,s)12si L L , , ,1i1i1s , ,L , (i 1,2,L ,s)12si L L , , ,1i1i1s3 51cost t sint cost t2222dy dy dt dy dx y dx dt dx dt dt xttt ttsint2t (y )x t1y ,ttx2t t,21yx , y则.2 tt22 g(

9、x)xy f(x)ug(x) u f g(x)yxdydy dy du dx du dx fu)g(x)或.dx(t) f(xdx t) t)Ft) ,(t) t) f t) .F t)t) f t)111(x) x) x) arctanxx f.442( ) f x 的展开式.将 f x 微分后,可得简单的展开式,再积分即得原函数的幂级数( ) 1) L n1)x) 1xx L 2x L , (1 xn2!n!x1 111xx x L (1) x L ,x(1 23nn1x11xx x L x L ,x(1 23n1x1 1 1 1 1 11 11 1( )f x11得41x 41x 21x

10、21x 21x22211 x 1 x x,4n4n1x4n0n1x4n14n1f(x) f f x ( ) t 4nx xx.00n1n1 sin2sincos2x x xx11xu xx2 (1uu)22 x1 xQ22 )1 (1u)(1u)1112 ()4 uu)28 1u 1u (1u)212 C ln|1u|ln|1u|u)812 1x 1x C,1x8 Ccosxu1. xx xx2 (1uu)221ABDuu) 1u 1u u)22(ABu ADu(ABD)2,uu)2AB0112 2AD0 A B ,D.4ABD111121) 1u 1u 2 (C8 1u 1u (1u)281

11、u1 8 2 1x 1x C.1x6 0 Pdydzz dxdy20S zx y z222S I.x y z222SS ,S S S ,S ,S .31212 xdydz xdydz0.x y zx y z222222s1s2 xdydz y R在S x23 I.22R z22s3S 在 D :R y R,R z R,在S 上,x R y ,S 关223yz33于 x y2I2 R2222 RR y R22R zR z222200Dyzz12R R arctan2R.24RR0 I.SR z22S x 1dxdy dV dz dV Rdz Ix R zR z2R z22222RD(z)112R

12、2 RR2R z2220 y RD(z)x2.22 P(x,y,z)、Q(x,y,z)、R(x,y,z)在 P Q R dv PdydzQdzdxRdxdy,x y z P Q R Pcos Qcos Rcos dS,dv或x y z 是 cos 是 (x,y,z)9xy(x y) f(x)y f(x)x2y,yx即x22xy f(x) f(x)2xy f(x) f (x) x.2 y y x ,2 yy x0 x010r i1,2x e x 0 r220 x2 C Y.21,B0,C 2x 2. A2 ff1 f (x) 2cos xsin x x 2.2f(x)xy 2y(2cos xsi

13、n x)ydxx y2x2sin xcosxdy 0.2211( y dx x dy )2(ydxxdy) yd(2sin xcosx)(2sin xcosx)dy 0,2222221d( x y 2xy y(cosx2sin x)0.22212x y22xy y(cosx2sin x)C C2y (x)*y P(x)y Q(x)y f(x)Y(x)y P(x)y Q(x)y 0 y Y(x) y (x)* Y(x)y P(x)y Q(x)y 0P(x)、Q(x) y 0rpr q 0,在复数域内解出两个特征根r,r ； 212Ce C e ; r,r y1rx1r x2212 ry C C

14、x e ; rrx11212 i y C x C x .C,C rx1,21212 y P(x)y Q(x)y f(x) y (x)* P (x)e ,y (x) x Q (x)e f (x)x*kxmmQ (x)P (x)相同次数的多项式,而 按 kmm 或e P(x)cos xP (x)sin x f (x)xlny p(x)y q(x)y f(x)y x e R (x)cos xR (x)sin x,*kx(1)(2)mm l,n i i R (x)与R (x)是mm(2)k按(1)mm 或 .108f(x)0 x0 f(x)时limx f(x)是xxx011 的 ,这就的 p p, p

15、1, f( )ppnn1f( )由nn1f(x)lim0 f(x)及f f 0,再由 f(x)x 0 xx0f(x)0 limx20 x00f (x)f (x) 1f(x)limlimlim f(0)x22x22x0 x0 x0f(x) 1 f.x22x01f( )n1 lim f(0).12nn211f( )n1f( )n因n2n1n1f(x)xn10f f 0由lim . f(x) x0 x 01212 f f(0)x f ( x)x f ( x)x (01,x , ) f(x)220因 f(x)x f(x) x , M 0 | f(x)M,x , 在11 f(x) f( x) x Mx

16、,x , .222211110 N,n N f( ) M,0.nn2 n211f( )1f( )n又n2nn1n1n1 v设 u 和 v ,则nunnnn1n1n u 和 v A 当0 当 A 当 Annn1n1 u v v u 0nnnnn1n1n1n1 v u u v nnnnn1n1n1n16 1S(z). D S(z)Vzz0 0,10 A,B 0 x1zx1 y z ,B或 .1 1 1y z x y (1z) zD R22222zS(z) R (1z) z ,2222 101z) z 112z2z zz z V .222233 3003 d z)2z2V 21,000 1 1 d

17、z) z V2200z112z2z 20102 zz z.233 381 1 0 0)( A.0 1 0 r()n , Ax 0取x ,x x ,x4334)故( .) ()( 和)将( x k ,x k 2k ,x k 2k ,x k ()1221231242k k 2k 0k k.212k 2k k 012122 k 0k (0,1,1,0)k (1,2,2,1) k 1,1,1) 时,向量 是( )与( ) k121216 A *A T*A a (i, j 1,2,L ,n)A | A|中a ijijijij0a 0 Aij| Aa A a A L a A a a L a a 0,222

18、21 1i2 i2in ini1i2inij故 | A0.| A0AA AA | A|E 0.*T a a L a 0i1,2,L ,n).设 的行向量为 (iA1,2,L ,n)T222iiii1i2in(a ,a ,L ,a ) 0 Li ,n). ii1i2inA |A0.0A2 36P(AB) P(AU B) 1P(AU B)1P()P(B)P(AB1P()P(B)P(AB). P(AB)P(AB)P()P(B)1,P(B)1P()1 p.X 、Y Z X,Y 0与1 1P Z 0 P X,Y 0 P X 0,Y 0 P X 0 P Y 0 ,4 3P Z 1 1P Z 0 .4 X,YZ01Z134P46E(Z)D(Z) X 与Z X