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精编WORD文档下载可编缉打印下载文档，远离加班熬夜2021年秋电大经济数学基础形成性考核册作业四篇一：2021电大《经济数学基础》形成性考核册答案2021电大《经济数学基础》形成性考核册答案【经济数学基础】形成性考核册（一）一、填空题1.limx?0x?sinx?___________________.答案：0x?x2?1,x?02.设f(x)??，在x?0处连续，则k?________.答案1?k,x?0?3.曲线y?x+1在(1,1)的切线方程是.答案:y=1/2X+3/22__.答案2x4.设函数f(x?1)?x?2x?5，则f?(x)?__________5.设f(x)?xsinx，则f??()?__________.答案:?二、单项选择题1.当x???时，下列变量为无穷小量的是（D）π2?精编WORD文档下载可编缉打印下载文档，远离加班熬夜2?2sinxx2A．ln(1?x)B．C．exD．xx?112.下列极限计算正确的是（B）A.limx?0xx?1B.lim?x?0xx?1C.limxsinx?01sinx?1D.lim?1x??xx3.设y?lg2x，则dy?（B）．A．11ln101dxB．dxC．dxD．dx2xxln10xx4.若函数f(x)在点x0处可导，则(B)是错误的．A．函数f(x)在点x0处有定义B．limf(x)?A，但A?f(x0)x?x0精编WORD文档下载可编缉打印下载文档，远离加班熬夜C．函数f(x)在点x0处连续D．函数f(x)在点x0处可微5.若f()?x，则f?(x)?（B）.A．1x1111??B．C．D．xxx2x2三、解答题1．计算极限x2?3x?2（1）lim2x?1x?1解：原式=limx?21?21(x?1)(x?2)??=lim=x?1x?1x?1(x?1)(x?1)1?12x2?5x?6（2）lim2x?2x?6x?8解：原式=limx?32?31(x?2)(x?3)??=limx?2x?4x?2(x?2)(x?4)2?42（3）limx?0精编WORD文档下载可编缉打印下载文档，远离加班熬夜?x?1x解：原式=limx?0(?x?1)(?x?1)x(?x?1)=limx?01?x?1x(?x?1)=lim?x?01?x?1=?122x2?3x?5（4）lim2。x??3x?2x?4352??2?2?0?0?2解：原式=limx??3??23?0?03xxsin3x（5）lim精编WORD文档下载可编缉打印下载文档，远离加班熬夜x?0sin5xsin3xsin3xlim33x?0313解：原式=lim??????x?0sin5xsin5x51555limx?05x5xx2?4（6）limx?2sin(x?2)解：原式=lim(x?2)(x?2)x?2?lim(x?2)?lim?4?1?4x?2x?2x?2sin(x?2)sin(x?2)1?xsin?b,x?0?x?a,x?0，2．设函数f(x)???sinxx?0?x?问：（1）当a,b为何值时，f(x)在x?0处极限存在？（2）当a,b为何值时，f(x)在x?0处连续.解：（1）因为f(x)在x?0处有极限存在，则有精编WORD文档下载可编缉打印下载文档，远离加班熬夜xlim?0?f(x)?xlim?0?f(x)又f(x)?lim(xsi1xl?i0m?x?0?x?b)?blimf(x)?lisinxx?0?x?0?x?1即b?1所以当a为实数、b?1时，f(x)在x?0处极限存在.（2）因为f(x)在x?0处连续，则有xl?i0m?f(x)?xl?i0m?f(x)?f(0)又f(0)?a，结合（1）可知a?b?1所以当a?b?1时，f(x)在x?0处连续.3．计算下列函数的导数或微分：（1）y?x2?2x?log2x?22，求y?解：y??2x?2x精编WORD文档下载可编缉打印下载文档，远离加班熬夜ln2?1xln2（2）y?ax?bcx?d，求y?解：y??(ax?b)?(cx?d)?(ax?b)(cx?d)?a(cx?d)?(ax?b)c(cx?d)2=ad?bc(cx?d)2=(cx?d)2（3）y?13x?5，求y?113解：y??[(3x?5)?2]???1?2?12(3x?5)(3x?5)???32(3x?5)?2（4）y?x?xex，求y?11精编WORD文档下载可编缉打印下载文档，远离加班熬夜解：y??(x2)??(xex)??12x?2?ex?xex。（5）y?eaxsinbx，求dy解y??(eax)?sbx?eax(bx)??eax(ax)?sbx?eaxcbx(bx)?aeaxsinbx?beaxcosbx=：isdy?y?dx?(aeaxsinbx?beaxcosbx)dx1（6）y?ex?xx，求dy11313x1解：y??(ex)??(x2)??ex(132?1x)??2x精编WORD文档下载可编缉打印下载文档，远离加班熬夜??ex?32x221x1dy?y?dx?(?e3x2?2x2)dx（7）y?cosx?e?x2，求dy解：y??(cosx)??(e?x2)???sinx(x)??e?x2(?x2)???sinx?x22x?2xe（8）y?sinnx?sinnx，求y?解y??[x)n]??(nx)??n(x)n?1(x)??cnx(nx)??n(sinx)n?1cosx?ncosnx（9）y?ln(x??x2)，求y?1解：y??1x??x2(x??x2精编WORD文档下载可编缉打印下载文档，远离加班熬夜)??1(1((1x??x2??x2)2)?)=111?11xx??x2(1?2(1?x)?2x)???x2221x??x2??x2??x2（10）y?2cot1x?1?x2?2xx，求y?35(2sin1x)??(x?111解：y??2精编WORD文档下载可编缉打印下载文档，远离加班熬夜)??(x6)??(2)??2sinxln2(sin11?1?x)??2x2?6x6?0si1i?3?1?5x5?2s1xln2(11126cosx)(x)??2x6x?2ln21?31?x2cosx?2x2?6x64.下列各方程中y是x的隐函数，试求y?或dy（1）x2?y2?xy?3x?1，求dy解：方程两边同时对x求导得：：精编WORD文档下载可编缉打印下载文档，远离加班熬夜(s(x2)??(y2)??(xy)??(3x)??(1)?2x?2yy??y?xy??3?0y??y?2x?32y?x2y?xdy?y?dx?y?2x?3dx（2）sin(x?y)?exy?4x，求y?解：方程两边同时对x求导得：cosx(?y)?(x?y)??exy?(xy)??4cosx(?y)?(1?y?)?exy?(y?xy?)?4y?(cos(x?y)?xexy)?4?cos(x?y)?yexy4?cos(x?y)?yexyy??xycos(x?y)?xe5．求下列函数的二阶导数：（1）y?ln(1?x2)，求y??解：y??12x2?(1?x)?221?x1?x2x2(1?x2)?2x(0?2x)2?2x2y???()???222221?x(1?x)(1?x)（2）y?精编WORD文档下载可编缉打印下载文档，远离加班熬夜1?xx，求y??及y??(1)?1?x1?1?解：y??()??(x2)??(x2)???x2?x222x11311?1?13?11?3?1?y???(?x2?x2)????(?x2)??(?)x2?x2?x2=122222244315353《经济数学基础》形成性考核册（二）（一）填空题1.若2.?f(x)dx?2x?2x?c，则f(x)?2xln2?2.?(sinx)?dx篇二：2021电大最新经济数学基础12形成性考核册答案(带题目)经济数学基础形成性考核册参考答案部分题目与答案符号在预览界面看不清，下载后再打开就可以看清了作业一（一）填空题1.limx?0精编WORD文档下载可编缉打印下载文档，远离加班熬夜x?sinx?___________________.答案：0x?x2?1,x?02.设f(x)??，在x?0处连续，则k?________.答案：1?k,x?0?3.曲线y?x+1在(1,2)的切线方程是答案：y?11x?22__.答案：2x4.设函数f(x?1)?x2?2x?5，则f?(x)?__________5.设f(x)?xsinx，则f??()?__________.答案：?π2π2（二）单项选择题1.当x???时，下列变量为无穷小量的是（）答案：Dx2A．ln(1?x)B．x?1C．e?1x2D．sinx精编WORD文档下载可编缉打印下载文档，远离加班熬夜x2.下列极限计算正确的是（）答案：BA.limx?0xx?1B.lim?x?0xx?1C.limxsinx?01sinx?1D.lim?1x??xx3.设y?lg2x，则dy?（）．答案：BA．11ln101dxB．dxC．dxD．dx2xxln10xx4.若函数f(x)在点x0处可导，则()是错误的．答案：BA．函数f(x)在点x0处有定义B．limf(x)?A，但A?f(x0)x?x0C．函数f(x)在点x0处连续D．函数f(x)在点x0处可微5.若f()?x，f?(x)?（）.答案：B1x精编WORD文档下载可编缉打印下载文档，远离加班熬夜A．1111??B．C．D．xx2x2x(三)解答题1．计算极限x2?3x?21x2?5x?61??（2）lim2?（1）lim2x?1x?22x?1x?6x?822x2?3x?51?x?11?（3）lim??（4）lim2x??x?0x23x?2x?43sin3x3x2?4?（6）lim（5）lim?4x?0sin5xx?2sin(x?2)51?xsin?b,x?0?x?2．设函数f(x)??a,x?0，?sinxx?0?x?问：（1）当a,b为何值时，f(x)在x?0处有极限存在？（2）当a,b为何值时，f(x)在x?0处连续.答案：（1）当b?1，a任意时，f(x)在x?0处有极限存在；（2）当a?b?1时，f(x)在x?0处连续。3．计算下列函数的导数或微分：（1）y?x?2?log2x?2，求y?答案：y??2x?2ln2?（2）精编WORD文档下载可编缉打印下载文档，远离加班熬夜y?x2x21xln2ax?b，求y?cx?d答案：y??ad?cb2(cx?d)13x?5，求y?（3）y?答案：y???32(3x?5)3（4）y?x?xex，求y?答案：y??12x?(x?1)ex（5）y?eaxsinbx，求dy精编WORD文档下载可编缉打印下载文档，远离加班熬夜答案：dy?eax(asinbx?bcosbx)dx（6）y?e?xx，求dy1x112ex)dx答案：dy?x（7）y?cosx?e?x，求dy答案：dy?(2xe?x?22sinx2x)dx（8）y?sinnx?sinnx，求y?答案：y??n(sinn?1xcosx?cosnx)（9）y?ln(x??x2)，求y?答案：y??1?xsin1x2（10）y?2y??1x答案：y???2精编WORD文档下载可编缉打印下载文档，远离加班熬夜sinln2x25?11?31cos?x2?x6x264.下列各方程中y是x的隐函数，试求y?或dy（1）x?y?xy?3x?1，求dy答案：dy?22y?3?2xdx2y?xxy（2）sin(x?y)?e?4x，求y?4?yexy?cos(x?y)答案：y??xyxe?cos(x?y)5．求下列函数的二阶导数：（1）y?ln(1?x2)，求y??2?2x2答案：y???22(1?x)（2）y?1?xx精编WORD文档下载可编缉打印下载文档，远离加班熬夜，求y??及y??(1)3?21?2答案：y???x?x，y??(1)?144电算化会计、职业技能实训（一）需要代考的同学请联系QQ：499086608（保证过关）53作业2一、填空题