电大经济数学基础期末复习参考练习题

经典word整理文档，仅参考，双击此处可删除页眉页脚。本资料属于网络整理，如有侵权，请联系删除，谢谢！经济数学基础期末复习参考练习题1(())ffxf(x)（CC）xxysinx1yx1111(x)edxec()(fxBBf）xxx2(AB)BAB）TTTxx112DD）xx012x1D、x且x2）yln(x(x)ln(xf(xx2A．xy2）f1d(x)）B、2xA为35TBCBB5411x11DD11x02DDsinxx00在x处连续，则k（CC1）f(x)xkx12dxd(2)）C、xxln2BAAABI,则A1（AAXbD、r()r()r()n）mnx24B、[2,2)(2,)）yx2f(xx)f(x)f(x)cos,则lim（A0）4xx01cosxxxsin2D、221/20A是mnsnB是stACBCD、Tx2x4x1x1231xx0C、x2232x2x233D、f(x)sinxcosx,g(x)122x(x)1A、x0fx()fsinx11、dx（C、）2x31AA1=（C、）1321401126B2）01126022412Cylnx,g(x)3lnx）3,C3）x(x)f(x)是x()()()BfxFxFa）若Fa(AB)BA设BD）TTTAx50exf(x)2x210x2limxsinx、xx0f(x)sinxcosxc(AB)A2ABBBn222B1021x2xxA0102134＿x2x000024(x2)x4x5()29fxxf22/20pp7qpq(p)500e2E＿2p．dsinxdxr(,b)r()100100A020A0011200001313xyxppqpq(p)100eE22px1dx．x2123aAa1116A0132bt00t106C(q)802q,q503.6x3f(x)xx2232xx121xx(、、111120134xx0120xx0121f(xh)f(x)1,则=f(x)1xhxxh)3/2012xf(x)x1f(x(,).x1x1a()[(()fxx=fx12I(I)04A=.43T22中A为0r(A).2x2x6.f(x)2f(x)sinx,0)x31dxx211BBr()r()cos(xy)ex和yxy。y[cos(xy(e)x解ysin(xyy]ey1y[esin(xy)]y1sin(xy)y1xy)yeyxy)cosxe2ydy。xsinx2x(cosxex2)2xex解y2sinx2xdy(2xe)dxx2lnsinx,()求yxy24/2011(lnsin)(sin)cos()2xcotx2解222xx2yxxsinx2sinx21x)y(0)求、y1x1x)x1xx)x)2yx)2y(0)0cos2sinx,yyx2(cos2sin)2ln2sin22xcos解2x2yxxxxsinxconxy5y(sin)(cos)cos5cossin54yxxxxx2y(x)ey求2xx22()()解2x2xyxexexx22)ex)e2x2xx2x42e2x)2xxx2xy求y2x1xx(2)()解yx1x(cosx)x)cosxx)x)22ln2xcosxx)sinxx)22ln2xlncosx2y()求y412xx)x)x2xx2解222yx2x25/20()2tan()2y444y解y1lnx求32112lnx2323(1ln)ln)ln)ln)x32x2x2x233x23x23lnx)lnxdx2x2ye求2x2x2x2x2x2)()sin()2exsin2e解ye2x2x2x2222x2(xsin2e)2x2．ycos2x)3(cos23cos2)sin(1222cos2)2)y3x2xxx2xxylnxsin2x、exy(e(ylnx(sin2xxyye(yxy)ylnx2cos2xxyxy(exlnx)y2cos2xeyxyxyxyx2cos2xexyyexyxlnxx)eexyy求y2[yx(e)(e)解2xyy)e(y)0yx1xyx)]y1x6/20yx)yexyyxx)xexy]sinyxe0y求y7(siny)(xe)0解ycosyexey0yyy(cosyxe)yeyyeyycosyxeydyy1xe求ydxx0()xe解yyyexeyyyey1xeyyy1dydxx0e1ey(0)当x10e1cos(xy)ex．求y[cos(xy(e)x解yy)sin(xy)ey1y(esin(xy)]y1sin(xy)y1sin(xy)yeysin(xy)1sin(xy)dydxeysin(xy)1、16x9x07/201x9x191601616xx(9)解x9x(x9xx9x)00xsin2x．2011解xsin2x[x2x2xdx]222224000lnx).eedxxx20ln31563ln30ln3解ex)e2dx)))e2deex30xxx30、xe11121ee2=lnxdx{xlnx}e解xe2214411lnxdxxlnx1dxlnxd(lnx)(lnx)c2x2(x5x7)cos2xdx、2(x5x7)cos2xdx2(x5x7)sin2x4(2x5)cos2x16sin2xc221ex、2dxx211ex112dx1ed()e11ee11xx1x2x112exx2122ex2dx2edxe22()ee解2xx1x111xedx2x08/201101111101eeee解2x2x2x22x2222040cos2xdxx201111解x222x2x22x2220240200xx)dx解xx)dxxx)x)dxxx)x)c3xxx3xx33lnxxc解xxx4x2x111d(4x)ln(4x)c解224x224x22(xlnxdxlnx11(x21x2(x2x)lnxxc(xln(xlnx解2222x2412x1lnx121121dx2dln)21ln2(3xx解1x1lnx1lnx11021124,B2,2IA）BT3311解2001111012IA0200210T0022412419/2011311290120033(2IA)B0T24132891001A01,B01,求（B)1T1212解1000112BA01T112131212101210011110321301011132(B)1T11102123(ABT)A,B11200121010274(AB)21T1203232741010121012101232013201023701327212()1T3272113()求IAA115112100113013解IA01011510500112120013100105010100650105310501001310012000100121100121165(I)533121110/20151A,B,求(AI)B1361151025解AI360137231235(AI)B157157121231A357,B0X5810123100123100357010012310解.5800102550112310010064101231001055200112100112641即A5521264115XAB552031120021210(ABT)1A,B110011200212111解T1110011110110211011013312[ABI]且T0113233013111(AB)T131211/20010()，IAA1111103110110100解IA101且101010102102001110100100021021011110010111(I)1211012101001010121261102CA,B010,C22120T002422121161606101C0100222022222解BAT002204240424263102,BA12(AB)112046310221412解AB1204211021102011101212[,I]4101012101210121112()1221231X342231011113401111110430132解即340101322314334324312X322112/20121X3520121012301052解即35010131013112152353111121152831X10431203520xx213x2xx01232xxb123a,b1012101210201121解A0111022121ab012401abab当ab3当a当ab3xxx01232xx8x3x012342x3xx0124111011101031A21830123012123003630000x13xx034x2xx02233xxx134x2xx23413/202x5xx15x71234bx2xx4x21234x3x2x11xb12342511571114212142A1114225115701373解132111321100005bbb12142A0137300000x17x10x434x3x7x32342x5x2x3x01234x2xx3x012342x14x6x12x0123425212131213A121325230949262608121325230000x12xx3x02342x5xx3x012341xxx91344xxx9234x3x2x01232x5x3x0问01233x8xx01231321321321011A25301解3801600550x13x2x0xx2313xx0xx223314/20xxx2124x2xx4x312342x3xx5x51234110121101211012解A121430113101131231550113100000xxx2124xx3x1234xx2x1134xx3x123xxx21242xx4x3当x12342x3xx5x21234110121101211012A111430113101131解23152011320000当xx2x1134xxx1234x1x2x134x1x3x234qC(q)1000.25q6q2（qC1000.25106101852C(q)CC18.5q()0.56)0.510611CqqCC(q)）C(q)q6qq令C(q)得q215/20Cq()q（万元R(q)1002q（万元q（）（）2()()()2)810010q）LqRqCqqq()0Lq令L12L(q)10q)）21010q1002p（C(q)R(q)（C(q)2005q1R(q)pq50qq22L(q)R(q)C(q)45q1q20022()450令Lqq45，q1L(45)454545200812.522Cq()2R(q)120.02q（（()()()120.022100.02LqRqCqqq()得令LqqLq)qq)255050050016/20()43Cqqq（（(q)C(q)(4q3)2q2qCC(q)则平均成本函数为C(q)2q3qq()(23)2）Cqqq2q()20令Cq得q32q18C23393qC(q)12qq82q2p元（（R(q)qpq2q)8q2q8q2q22L(q)R(q)C)(q)8q2q2qq)6q1q222L(q)66q()660令Lq解q1，q1L6113122()2.51000()22000（元RqqqCqq()()()LqRqCq2q2000(2.5q1000)0.5q1000()0令Lq2000得q17/201C(q)q3q100225qC(q)1qC(q)qq1()0Cqq2q501Cq37qq1为q1202pqp（）（））设总成本函数为C(q)R(q)L(q)C(q)10q100