2018年數(shù)學(xué)一真題與解析

1、y x sin x.y x sin x.y cos x.y cos x .z x y22z x yz 1y與x yz 1z 與2x yz 2y與2x yz 22n3(2n1)!）(1)nn0 x)21x1x22則2dx, Ndx, K (1 cosx, M,N,K）Mex222222為M N K.K M N.M K N.K N M.1 010 1 10 0 11 1 11 0 1001 1 .1 1 .0 0 10 0 1101101101 0 .1 0 .0 0 10 0 1,B nr(X)X (X)為 r(A|AB)r(A).r(A|BA)r(A).r(A|B) r(A B ).r() m

2、ax r(A),r(B) .TC.2f(x)f(1x) f(1x)f(x)dx 0.6（7）設(shè)，0 p x0 . ), X N( ,X ,X ,，X ,2212n 則: ,H : ,令 H0010A.a0.05B.a0.05C.a0.05D.a0.05HHa0.01a0.01a0.01a0.01H .00H .00HH .00HH .0011tanx）lim e,k 1 tanxx01）y f(x)的圖像過(guò)（y axf(x)dxx0F(x, y,z) xy yz xzk, rotF求）.xydS s x y z 1 x yz 0 由與222,aAaA是 的線性無(wú)關(guān)的特征向量，（）二階矩陣有兩個(gè)

3、不同特征值，12A ( a a ) ( a a ,A2121211）,BA,CBC ,P() P(B) ,P(AC AB C) , P(C)則24e arctan edx2x x1516x 13yxdydz (yz)dxdzz3dxdy3z17223y y f(x)18f(x)x.f(x). xxx 0,x e e 1.limx .n，xxn1nn1nnn,2af(x ,x ,x ) x x x ) x x ) x ax ) .221231232313,x ,x ) 0f(xf(x1123,x ,x ).232aa 1 120 1 1.1 3 aaAB2 7 a 11 1a求 ；12AP B.

4、 ,P X 1 y , Y X,YP X 1 y（22）12Z XYcov(X,Z).Z 1 xe ( 未知， x).,X ,XnX f(x) X12 ( ) ( ).，2018841xcos x 112xlim lim ;xx0 x01xcos x 1lim 122x lim.xx0 x0 (1,0,0)(0,1,0)兩點(diǎn), 曲面z x y2 (2x,2y,1)22x y 0 x y.即 (1)nn32n12(n (1)n(2n1)!(2n1)!(2n1)!n0n0n012 (n (1) 2sin1 cos1.n(2n 2)!(2n1)!n0n01 x2 x22M N dx dx ;1x22

5、21x2dx x)e dx;xex222K cosx)dx K M.221A 1r( E A) r(E A) 2.23Cf ( x ) f (1f (x)圖像關(guān)于x12f(x)dx 0.6從而已知可得對(duì)稱，0P(x 0 ) 0(8)a0.05時(shí)接受H,可 量Z U 0Z U0.00(9)K 2。11 1tanx1 tanxlimlim 1 1 tanxsinkx 1 tanxx0 x0lim1 2tanxkx 1 tanx2k k 2.x02ln 221101110 xf (x)dx xdf (x) xf (x)f(x)dxf 1) f (x)m000 2ln 2 f (1) f (0) 2l

6、n 2.jkrotF 1). (y,z,z)zxxyyyzxz11 1tanx1tanxlimlim 1 1 tanxsinkx 1 tanxx0 x031, xyds (x y,x y z 1222L:LL【解析】222x yz 0 1 2163ds 2 .23L) a aAa a,Aa a,Aa a ,aa11 111 122 2121 12 2 ( a a ) a a a a ,21221 1221212 1, 1, 1,A12122121 。4(AC)(ABC)p(ABC)p(ABCAC)p(AC ABC)p(AB) p(C) p(ABC)12p(C)p(AC)1414 p(C) .1

7、41 p(C)p(C)41 2e arctan earctan e1de2xx1dx 2 xxex2e1 (e121x1e2xdx1)e arctan e1 1 2 xxx2xex112edxe arctan2 x4e1x11 e 11x e arctan e1dex2xx24e1x1211e arctane2 xe1 1 d(e1)xxx4e1x1124 33arctan e 1(e1) 2e 1 Cx e2 x2 xx2x21 e2113arctan e1 (e 1) e 1 C.xxx262x 。yzx y z 21 3 y2x23z x z 4222S y 2, 16362 2 2 3

8、x2z32L(x,y,z; ) y216 (x y z 2).36x 0,L 則 x2 3yL 0,36y2z16L z 0,L x yz 2 0.81 6 , x,y , z 3 3 4 3 3 4 3 3 4 3 3 41S. 3 3 43y 3z ,1,22(17)解：構(gòu)造平面: 與 ；x 0,P x,Q y z,R z記33PdydzQdzdxRdxdy PdydzQdzdxRdxdy PdydzQdzdxRdxdydxdydz2 (PQ Rdxdydz 0 (13y 3z2xyZ 323z2 y2 z2)dxdydz03y22 13zm13yy3z2 z2)dydz223y22 13

9、zm1 2x d 1 3r r ) rdr2230011r )dr ) 2 ( )313r222603113r r 2)dr2)22303131r ) 2(1r ) d(1r )222322013 5243513) r r22220.45xe dxC)(y(x) e11( xe dxC)exxex(x1)e C)x x1)Cex.f (xT) f (x)Tf (x) 是 y(x) e f(x)e dxC11x f(x)e dxCexxf(x)e dx Ce .exxxy(x)exf(x)e dx Cex xTy(xT)exT f (t)e dt Cet(xT)(xT)TxfuT)e duT)

10、Ce ) exee(xT)(xT)uTT0 xf (u)edu (Ce ) e TxTu0uTxxex f (u)e du Ce ) e ,0即y(xT)f(x)e 1x,0,xe1xxf (x)e 0,f (x)01，xex1x1e 1,x 0; x122x 0nn1x 01ex1n k(k 1,2,)x 0nk1e k1xk1x 0;xkk1kx 0nexx1e 1x x lnlne ln;又nxnnxx en1nxnng(x) e 1xe設(shè)，xxx0g(x)ee xe xe 0,xxxx,g(x)g(x) g(0) 0e 1 xexxex1x x ln ln1 0,xnn1nx exnn

11、nlimxnnlim x AAe e 1;A A設(shè)nng(x) e 1xex 0A 0.xx x x x123f(x ,x ,x ) 0 x x 0,得 12323x ax 0,131? 1? 1? 0? 2 ,A 1? 0? 0? r1? 0? a0? 0? a a2r(A) 3a2r(A) 2 x x 0 x；123 xk,kR.；1 1 x x x , y1123y x x,a2223 y x ax ,313;y y y212223f (x , x , x ) (x x x ) (x x ) (x 2x )2232312312321a2 2x x x x x x x2221232 313

12、x 3x3(x x )2) 3 2. 2(x 223221.y y2122A B 與 r(A) r(B).又AP B4 0 1 2 1 1 1r2 7 2 1 1 10 0 0 0 0 06k 434 k6k32k 1 ;122k 12k 1P得1k2k3k123.P又 P 0k k233 6k 4 6k 4 k1232k 1 2k 1 2k 1 ,k ,k ,kk k .P31k2k3k1223123 11 ,P X 1 ,YP X 1 22cov(X,Z) cov(X,XY) E(X Y)E(X)E(XY)2E(X )E(Y)E (X)E(Y) D(X)E(Y) .22Z 0,1,2, P Z 0 P X Y 0 P X Y 0 e , 11P Z 1 P X Y 1 ,P Z 1 P X Y 1 ,ee221 e P Z k k,k 1,2 k! P Z 0 e .1ni( )