大学高等数学下考试试题库及答案

分MM2,7,4MM（1212Cai2jk,b2ija∥ba⊥ba,ba,b34xy6x1y5z8:L:L和与）LL112122yz312；；；.2643a与bab0ab0ab0ab0zxy3xy3321zxyzy4222222((1)(）nnn1xnnn1Bxn2n012211x2x1x2x分一平面过点A且垂直于直线，其中点B，则此平面方程为zz2zxy3xyxy1323y设Lx2y1(2xy2y)dx(x4x)dy22L(x2)nnn1分zzuxy,vxyzevu,.xyzzzzx,yx2yz4x2z50,.xy222sinxydD:xy4.222222Dsinx1ydydxx0y.试卷6参考答案2xy2z60..9y1.6xy221nxn.2n1n0yCCxe.2x12zzeysinxycosxyxyexxyxyxy，.xyz2xz2y,.xz1yz12d62.2d016R.33yee3x.2x32m1yx.237（下）分（M，M2MM12121315x2y2z10xy50和6432x2y2z50PCnn01r1r1r1rn1xnn0n4n1f(x,y)f(x,y)f(x,y)PQ(,)xyfxyfxyxy(,),(,)(,)00xy00(,)xyfxyfxy(,),(,)00x00y00P）；；(1)分(x3)nnn1zze2x4y42211x2分ai2jk,b2j3kab.zuvuxy,vxyz,z.22xyzzzzx,y由x3xyz2,.3xy设zy2(01)dz2zx3(z1)dyxz2分yx,y2xx4和试卷7参考答案x2y2z1.112eydxxdy.8x8yz4.1x.2nnn0yx.33j2k.izz3xyyyy,2xyyyyxyy23333xy.zz,y.xz2z22a.3323yCeCe.2xx1216.3x1gtvtx.22003（下）3245）a与b）））4222,22222,,,22222222zzx,222）xyxRyxRyxRyxRy,,,zz,zzzzzzxyR2）221R2AAAA2xn(）nnn112）x2nx2nx2nx2n1(n(n(n(n(2n(2n(2n(2nn0n1n0n05）4）54x1y3z的夹角为LL：1122x1y2z3x2y6z之间的夹角为直线L：32120d,D:xy1的值为22Dxn!xn!n0n026523xyd，其中由直线Dyx2y围成.及D1(？，？nnn12a2Mt0参考答案28,21xce,cx1y2y223=32317712333333277227233,12xyzyz3x,23x，2ttt所以x|y|z|tttx1y1z1123即D2y1832[2xydxdy(2y)dy121y1D11，，，。sin0Vn,且limsin0所以该级数为莱布尼兹型级数nn111sin，。sin1n，，，，1又sin当时sinx~xlimnn51nn1nn1n1n所以，原级数条件收敛。111e1xxxxw23n!x(,)用代111e1(2x)(2x)(2x)(2x)2x23n!22223n12xxxx23n!x(,)r22ry121212则2(2xy2yz2zxa2)与26a得266a3aV2dMdtM其中为常数初始条件MMt00对于dM式dtdMdtM两端积分得lnMtlnC所以cet又因为MMt0C0所以所以0Met0由此可知铀的衰变规律为铀的含量随时间的增加而按指数规律衰减，：。4（下）．．xy222zx)y)2．2是．1xy422D（Ｄ）151x．dxf(x,y)dy00101dyf(x,y)dx101dyf(x,y)dx10ydyf(x,y)dxx1dyf(x,y)dx0y000．~．r()r(A)r．n3(n11n(n1n12nnnn1n1n1n1和．uvuvnnnnn1n1uuvvnnnnn1n1n1n1vvuunnnnn1n1n1n111．1xx21x1x21xx241xx1xx1xx2424244520数2222．．zxy1ln(2xy)yf(x,y)xyf(x当是f(x,y)则(x,y)0f(x,y)f(x,y)f(x,y)a0xx0,0yy00xy00(x,y)003AA．unn165zzzxy，．xy22．4xdD{(x,y)|0y4x,0xD123ＸＢ＝1201＝．01221001(n1xnnn1f(x)ex2xyz1xyz1xyz1参考答案y6a6u0(x,y)|1xy222nxnzyxy1zxlnyyxy2x334x2222224xd4x(4x)4x3000D0127102.1B012,AB12415001(n1Rnn11(2n1当(.x1nnn1n1x)n(nexxnx(,)ex(xn.x(,)!!!n0n0n0kijs121i3j5k211x2zy13511111111111110~A111111110111001100112)1)(21110~当r()2,(A)32~1当;2r()(A)3xyz2xcc112~当(1r()(A)1ycc,c112zc2《高数》试卷5（下）aijbkab，（）Bijy1ijijA0CDx2）2A圆BCDxz）A1B0CD1dxf(x,y）1x0dyf(x,y111dyf(x,y1ACB000xdyf(x,y111dyf(x,yyDy000D是xy1（）DD4A2A0B1CnB1C0）DA，B，C3）23AACBCBCABCDABACr(A)r）~r(A)Arr）ABBCDCDu和vu）vnnnnn1n1A若uvB若vunnnnn1n1n1n1C若vuD若uvnnnnn1n1n1n1xoyf(x,y)x4y6x8y222AA3xyA0x0z=yz0unn1zzz，2xyxyx2yz22xyz4与x22，yxxy1dxdyDxy2D223A110121(x1)n5