1998考研數(shù)一真題及解析

1、 531x 1x 2lim.x2x01z f(xy) (x y f ,設(shè)z.xyxx2y2 3x 4y ) 設(shè)La22.4 3L0 A*n設(shè)A為nA, 為AE為 A ,(A ) E則* 2.1Dy yx e2 X Y 在0, ( , )xxD(X,Y)X x 2_ .d設(shè)f(x)( ) x( ) xt 220(x )xf(x )2xf(x )2xf(x ) xf2222(x)(x x2) x xf23( )321yx0yy(x)x y ,x0 是x 1x2y 2y(1)( ) 44e ea b c 111x ay bz c a b c 333aa b b c c122a b c22121233

2、3xaybzc( )111a a b b c c232323、B設(shè)A 0P()P(B)P(B|)P(B| )P(A|B) P(A|B)P(AB) P()P(B)P(A|B) P(A|B)P(AB) P()P(B)x1 y z1L: x y z: 2 10L 1110L 繞y 0 0(x,y)2xy(x y ) ix (x y ) j x 2 2 424(x,y)u u(x,y).y vmB k(k 與vy y= y v .(za) 2,z a x y2 a 22(x y z )12222sinsin 1sinnn1nlimnn .求1nn21 (1) a()n a na 1nnn1n1n f(

3、x)是區(qū)間設(shè) y (0,1) 0,xf(x )x ,1 x0000 f(x) y2f(x)(x) f(x),x fx0ay z bxy2xz2yz 4x,222x y P z 4, .a bP22設(shè) 是 A x,Ankk0A0有解向量 k, , , AA1 ka x a x a x 0, 1 21,2n 2na x a x 0,a x(I) 1 2n 2na x a x a x 01 1n2 2n,2n 2nb ,b ,b ) ,(b ,b ,b ) ,(b ,b ,b )TTT11 121,2n21 222,2n1 n2n,2nb y b yb y 2nb y 2n0,0, 12b y b

4、y 12b y b yb yn 2n0 1n22X,Y1 2X YN(3.4,n n 122tze dt2ztP t ppn53141x 1x 2 1x 1x 2x0 x21x 1x 2 22 1 11x 1x 4x2limlim4x2x21x 1x 2x0 x01 x2112lim21 x 1 x.22x242x0111x 1x 22 1x 2 1x2xlim洛 x0 x0 x2111x 1x4x 1x21x 1x4x2 1x 2 1xlimlim洛4x0 x0 x0 11 lim2 1x 2 1x1x0 .442 x o x2 , 12181211x 11x 1x o x2 ,xx2x28

5、12 211111 x x o x 1 x x o x 2222282812limx2x0 1 x o x o x22214412lim .x2x0( )( ) ( ) x y y x y1 f(xy) y x y f( ,zxz z或x yz.xz 11yf(xy) y x y f xy f xy y x y()( )( ) (),x x xx2x z 1y2 f(xy) f(xy) (x y)y y xx211y f(xy)x f(xy) f(xy)x x y y x y() ()xxx211 f(xy) f(xy) yf(xy) x y y x y() ()xx yf(xy)(x y y

6、 x y).) (zy.z 1 1f(xy) (x y f xy x x y y x y)( ) () (y y xx f(xy)(x y y x y),) ( z z)22f(xy)(x y) (x yy yx x yf(xy) x y) (x y().z 1 )f(xy) (y x y y x y xy x 1 ()f(xy)x y x yx xy f(xy) (x y)xy yf(xy)(x y y x y).) (, f x y y xa2y 2 0 xyds.Lx y xLLx2y2 13x 4y (3x 4y ) 12 12.22224 3LL 2xyds (3x 4y )ds 1

7、2a22.LL f ,y在ll關(guān),y ds f xl 0于 y l 為l上x1 2 f x,y ds, f x,y 為偶函數(shù), f x,y l1 ，f x,y 關(guān)于為奇函數(shù).ll l 為l上 y0 x2 2 f x,y ds, f x,y 為偶函數(shù), f x,y l2 ，f x,y 關(guān)于為奇函數(shù).l A 21 A 的特征向量為 ,A . 0)0 0 A 0A 由 A A A A A A A , AAA*,A A A A 22 A .* 2 2 A 2A 1( ) E1.AEA又E* * 2 0, A0 A A0A1 AA A 21A(A ) E1.,；* 2 A nn 是 X AX X A X

8、 A A A .由 為 A 使1111 是 10 0A A1 X(A)X ( k)XA若 是 Ak.1A A 0A 1.A14y1y (X,Y)f(x,y).x1D (x,y)|1 xe,0 y,x1D S e2 x e2xD11O12e2x1,(x,y)D,f(x,y) 20,.X x1 xe或f (x)0X當(dāng)2；111當(dāng)1xe2 f (x) f(x,y) .x22xX01f .故4Xu x t ,22 1t:0 xu:x 0 du d x t tdt 2,22,t1tf(x t dt u x ttf u) x02222t0 x211x2 fudu,0 fudu x2220d 1 d 2x(

9、x t fux222 00 11 f(x ) x f(x )2x (x 222222 Ft) ( ) f(xdx ( ) ( )t, t , t (t) ( )( ) ( ) ( ) ( )t .F tt ftt f(x)(x x2) x x 1x1 f(x)時(shí) x1處 f22(x) f( ),x2xxx2x(x x2)x(x 1 x22由f(x)(x x2)xx ), 0 x22(x x2)x(x 1 x,22 f x f 122(x x2)xx )0( lim lim0ff,x1x1xx f x f 1(x x2)xx )022( lim lim0 x1x1xx(x) x 1在即 f f

10、x f 022(x x2)x(x 1)0 lim lim2ff,xxx0 x0 f x f 0(x x2)xx )022 lim lim2,xxx0 x0(x) x 0在 f(x) x 1在f(x) f(x) xa (x)( ) xa x 在( ) xa f x 在 f(a)0.yxyyy , . 有1x2x 1x x2xx0,令即得 是xx0yx y1 xyy lim x1x2x 1xx022x0 x 0 x0y. 1x2 ,y 1x2ln y arctanxC y Cearctanx.1 0 y , 得 y .y e arctan .Cearctan01Cx1 e .arctan1 4故y

11、 e exx1( )x( ( )lim xx l,( )x( ( ) 若l 若l 若l稱xx 稱( ( )( ) ( ) x x ；xx ( ) ( )( ) ( )是xx x o x .( )xlim(x(x) 若( )xxaybzcxaybzc:L :21L,13331a b b c ca a b b c c21a12121232323a b c 111a b c 2a b c3 3223a b c1a a b b c c11121212a b c 行減行，行 a a b b c c 0,22a b c2232323a3b3c3333(a a ,b b ,c c ) (a a ,b b ,

12、c c )與121212232323,kk (a a ,b b ,c c )k (a a ,b b ,c c )0k1211212122232323(a a ,b b ,c c ) (a a ,b b ,c c )L,L與12121223232312,LL得12xaybzc333a a b b c c121212xaybzc1113,33a a1b bc c12122 zc c cxa a ayb b b即312312312 .a ab bc c121212xaybzc111a a b b c c232323xaybzc111,111a a2b b2c c2333 zc c cxa a ayb

13、 b b即123323323 ,a ab bc c232323,La a a ,b b b ,c c cL12213213213P(B|)P(B|) AAB,B AB A (A|B) P(A|B)與 A慮PP(B|)P(B|),知 P AB P ABP B P AB, 1P AP AP A 1P A P A P B P AB ,P AB P A P B.P P AP AB | A P B.L ：L0 x 1t,: y t,L與 L N1z 1t(1t)t 2(1t)10t 1.x1 y z1L: LM,N11210 x 1t,t,即yz 1t.L ：N21(1t)(t)2(1t)10t ,32

14、 1 1 ( , , ).N2 3 3 3x2 y1 z:與LN.N0 42112LL 0與 L(1,0,1)l 00Ln 0 x1 y z11111 0 x3y2z10 .12x y2z10,:L 0 x y2z10.3 繞 y yL0SL0 x2y,1z (y2 S1 (2y) ( (1 y ,x222 y,y1z (2y) ( (1 y .22212 z 2y (y1)y z y4 17 4 2 10.消去 得 x2 xS222222(x,y)x (x y ) , ( , )(P(x,y),Q(x,y則A x yP(x,y)2xy(x y ), Q24242 0u(x,y) x 0在 上

15、 原 xQ Pu(x,y),xx yQx x2 ( ) ( ) 4 ,4y2 x2x4y2 1x3xPy2x(x y ) 2xy(x y ) 2y.42 42 1Q P由x y2x(x y ) x (x y) x 2x(x y4 xy x y) 2 ()2y,42 242 1342 42 1 4x(x y )(1)01.42 1u(x,y)x0(1,0)2xydxx dy2u(x,y)(x,y)Cx y42(1,0)2x0 x2xdxydyCx 04x y4210 x2ydyCx y420 x2 y ydyC20 x 4) x 2x x2y C 1 y ddC2yy x y 2x2 y 220

16、0 x )4 x x 22y C2xCu ui+ ( , )u x yj.xyD ( )與 ( , )在DP x,y Q x yQ Px y D , ) ；x y0,L D為 LPdx Qdy , , A B A BLABu(x,y)du PdxQdyu(x,y)Pdx QdyPdx Qdy0Pi+Qju(x,y)gradu Pi+Qj .u(x,y). OOymg 浮 ,Bd y2mB gkv, yv t002tdydtd y dv dv dy2dvdydvv, v v由 與 y vdt2 dt dy dtdydy1mvmgB kv, v 0. dvy0 ,B ,B Bmkmg m2g2kk

17、ky B kvmBmkm2g (B kv )kkB kv m g 2 mk k dvB kv m m(B) dv)dvkk(Bkv1m(B ) ( )mkd(B kv) vkk(Bkv)mm(B) vln(Bkv)C.kk2| y0v即m(B)m( B)ln(B)C C 0ln(B).k 與 k22yvm mgB mg B kvmy vln.kk2 mg B (x y z )21222 x y a222 :z01(za) 12I (za) .2(x y z )1a2222 x y a222 : a x y2 z22z0111I (za) (za) 22aa11(za) 1(ax)x2 () a

18、 .2azD 在 axdydz0,而 上z 0,yoz1111 則 2 .z aa21I (aza) a ,2aD為與 D為 在xoy (x,y)|x y a .D222111 I a 2 a2aD12 a a 22d0a a.a322a3200 a r211r z 2 a 2d20 aaI4242a00 a2r211 a 2d2r a r ()2a42 2 a001 a 2d (a rr dr)3a42a001arr a1aaa224224 a a 4424 24aa01a 4 a a43a4 2(za) 1(za)22I (x y z )1a22221 (za) I I .2a12a x

19、y dydz a x y dydzI xdydz 22222212a x y dydz,222 xx x為在 yozyzy,z)|y z a ,zDD222yz I 2 2dara221012a r d a r )(a222202232a32 ()(0) ,a r2a a233330112I 2(za) dxdy a a x y dxdy2222aa1 2d (2 2aa a a r r rdr)2222a00(2a r2ar a r r )dra2223a0ar a r drr dr2a rdr2aaaa22230001r 3 a4a ra2a a 223 a 4 002a4(a a ) a

20、 ,443a34622 I I a .(y,z)| y z a 為在 yozyz2I3DD212yz P(x,y,z)、Q(x,y,z)、R(x,y,z)在 P Q R dv PdydzQdzdxRdxdy,x y z P Q R dS,dvPcosQcos Rcos或x y z 是 cos cos、是 (x,y,z)nnnsinsinsinnnnx n nnnsin x1sin xdxlim x .n0nsinsinsinnn1nn,i 1,2,n,1nnisinsinsinnnn1nnnn.1nni1i1i1i21ni nnn 1,nnnn0i1i1112nii nnnn 1n1n1 nn

21、 nnnnn0i1i1i1sin2.nnlim1nni1i Ny x zlim lim y z a, N nnnnnnnn則lim x a.nn limaa a a且annnna 0.(1) aa 0(1) a又nnnnn1n1 1 n111 n()n 而1a, 0a 1a1a1a1nn1n 1()na 1n1n 1 nlima a0b , 則a 1nnnn11lim b lim1,na 1 a1nnnn 1()na 1n1n(1) u1nnn1u ,n ;limu 0.unn1nn(1) u0 (1) u u , .r u則1n1nnn1nn1n1n1(1) u1nnn1limu 0limu

22、 0. (n1u nnnnnn1 設(shè) u 和 v vlim ,則nunnnn1n1n 0A和uv nnn1n1A0uvnvnu nnn1n1n1n1Avnuuv nnnn1n1n1n1,u nn10 , 設(shè)unu uunnnnnnn1,.x (0,1)( )1( )(x) xf(x)1ft)dtx f x0 f x dx000 xx (0,1)(x )0.可以對(x)(x)t)dt x000 0,101 11(x) (x)(ft)00 x11x11(x) x ft) (x) 0 x0 x0(x) 在 (x) (x) 在 f在x (0,1) (x ) (x )0.000( ) ( ) ( ) (

23、 ) ( )2 ( )0 ( ) 由 x x f x f x xf x,知 x 在x 0( ) x (0)1( ) f 0,(1) (1) 0 f t dt .0(x)0(x)0 x (0,1)當(dāng) f當(dāng) f0 0 0 x 但0( ) ( ) x x (x) f 在閉區(qū)間a,b (a,b)(a) fb) f,(a,b)內(nèi)至少有一點(diǎn) (a bf( )0 .1 b 1 b a 1A P1 1 0 0 0B,P1 0 1 00 0 4即B A1a101a bb2 B A1 1 1 1 3 1 . A1 1 0當(dāng)11 1 11 1 1 0EA 1 3 1 行(行 0 2 01 1 0 0 0 x x x

24、 E)x01232x 2rE)2 nr(0E )321,故有1個(gè)自由未知量,選x x1 .T111 1當(dāng)20 1 0 1 E A 1 2 1 3(1)行 0 1 11 1 0 1 1 0 0 1 0 1 行(加到行 0 0 0 行互換 1 1 0,1 1 0 0 0 0 x x (E)x023x x 12r(E )2nr(E)321選x x1 .T1124當(dāng)33 1 1 1 1 4EA 1 1 1 行互換 3 1 11 1 3 1 1 3 1 1 11 1 1行的3,(-1)倍分別加到2，3行 0 2 4 行 0 2 4, 0 2 4 0 0 0 x x x E)x01232x 4x 23rE

25、)2 nr(4E)321,故有1個(gè)自由未知量,2 .選x xT223 , ,123 , ,將12311( ,0,) ,1T121211 1 , ) ,( ,22T33 321 2 1( , , ) .33T6 6 63 111 23612 0P.36111236 b aiiiiEA EB nn aBii1i1A B. , , ,01k1 k1 k1) A 01Ak1 k1 0Ak1 ,Ak1 A 01 k1 02(k1) .即A1 A Akk01 0k1 0 AA A2Ak1k1k0 00.Ak100) k1 0.A A將k101 k1 0k1 , AAk2 AkA21 k1 0A即 A 1.

26、kk2 31 0 Ak1 0 A A2k1 Ak,k11 0 0.1Ak12k10,因此向量組, , , A A 1 kk ,k , ,k 使12m 0 k k k , , , , , , 1122mm12m12m()n nk k k,1 12 2(a ,a ,a ) , (a ,a ,a ) , , (a ,a ,a ),TTT111 121,2n221222,2nn1n2n,2nk ,k ,k 12n(I)A X 0, (II)B Y 0, (I) (),B, 0 B (I). 2n2n于 A X故B0TTn2n(B)nB rTn2nr()r()2nnn A T對 ABT 0又r(B)n0 ABT 0A ATT 0 2nr(B)2nnn,故,恰好等