2020考研数学二真题及解析完整版来源：文都教育一、选择题：1~8小题，第小题4分，共32分.下列每题给出的四个选项中，只有一个选项是符合题目要求的，请将选项前的字母填在答题纸指定位置上.1.x0，下列无穷小量中最高阶是（）xA.et21dt0xB.ln1+t3dt0sinxC.sint2dt01cosxD.sin3tdt0答案：Dx3xx解析:A.et21dt~t2dt00335xx2B.ln1t3dt~t2dtx2005sinxx1C.sint2dt~t2dtx3003131cosxx2D.sin3tdt~2t2dt00512x2t225052121x2x5521021ex1ln|1x|2.f(x)第二类间断点个数（）(ex1)(x2)A.1B.2C.3D.4答案：C解析：x0,x2,x1,x1为间断点1ex1ln|1x|e1ln|1x|e1ln|x1|e1limf(x)limlimlimxx0x0(e1)(x2)x02x2x0x2x0为可去间断点1ex1ln|1x|limf(x)limxx2x2(e1)(x2)x2为第二类间断点1ex1ln|1x|limf(x)lim0(ex1)(x2)x1x11ex1ln|1x|limf(x)lim(ex1)(x2)x1x1x1为第二类间断点1ex1ln|1x|limf(x)limxx1x1(e1)(x2)x1为第二类间断点arcsinx3.1dx0x(1x)π2A.4π2B.8πC.4πD.8答案：A解析：arcsinx1dx0x(1x)ux令，则1arcsinu原式=·2udu0u2(1u2)arcsinu21du01u2t令usint22costdt0cost122t222044.f(x)x2ln(1x),n3时，f(n)(0)n!A.n2n!B.n2(n2)!C.n(n2)!D.n答案：A解析：f(x)x2ln(1x),n3f(n)(x)C0x2[ln(1x)](n)C1(x2)[ln(1x)](n1)C2(x2)[ln(1x)](n2)nnn(n1)!(1)[ln(1x)](n)(1x)n(n2)!(1)[ln(1x)](n1)(1x)n1(n3)!(1)[ln(1x)](n2)(1x)n2(x2)2x;(x2)2.(n1)!(1)(n2)!(1)n(n1)(n3)!(1)f(n)(x)x22nx2(1x)n(1x)n12(1x)n2n!f(n)(0).n2xyxy05.关于函数f(x,y)xy0给出以下结论yx0f①1x(0,0)2f②1xy(0,0)③limf(x,y)0(x,y)(0,0)④limlimf(x,y)0正确的个数是y0x0A.4B.3C.2D.1答案：B解析：ff(x,0)f(0,0)①limxx0x(0,0)x0lim1x0xf②xy0时，yxfy0时，1xfx0时，0x2ff(0,y)f(0,0)1limxxlim不存在.xyy0yy0y(0,0)③xy0,limf(x,y)limxy0(x,y)(0,0)(x,y)(0,0)y0,limf(x,y)limx0(x,y)(0,0)(x,y)(0,0)x0,limf(x,y)limy0(x,y)(0,0)(x,y)(0,0)limf(x,y)0(x,y)(0,0)④xy0,limf(x,y)limxy0x0x0y0,limf(x,y)limx0x0x0x0,limf(x,y)limyyx0x0从而limlimf(x,y)0.y0x06.设函数