微积分1期中试卷08解答

浙江大2008级《I》课程期中考试试考生姓名 学号 专业 得选择题（每小题3分，共12分 (1)设ln1x,1 x.则当x0时1 12x(A)与是等价无穷小(B)与是同阶但不等价无穷小浙江大2008级《I》课程期中考试试考生姓名 学号 专业 得选择题（每小题3分，共12分 (1)设ln1x,1 x.则当x0时1 12x(A)与是等价无穷小(B)与是同阶但不等价无穷小(D)的高阶无穷小(C)的高阶无穷小B1x x 211x0,x, x]~2 2x(2)x2xsinxcosx(A)(B)(C)2(D)Cf(x)xsinxcosxx2,f(x)x(cosx2)0,x00,xf(0)10,f()(21)yf(xx1f1)是函数f(x的极值x1是函数f(x的拐点limf(x)2.(xxyf(xx1的邻域内左凸右凹yf(xx1C2limf(x)1lim f limf(x)16(x6(xxxxf(1)1,f(1)f(1)0,f(1)120,xf(x)f(1)0x1,f(x0,x0,f(x0,曲线在拐点(1,1)处左凸右凹(4)y1ln1exx(A)(B)(C)2(D)Dlim1ln1ex)0,水平渐近线y0,lim1ln1ex)垂直渐近线xlimln(1ex)f1x1x2xxxlim[f(xxlim1ln1exxlim1lnexln1exx0,斜渐近线:yx1二、（每535分(1)求limex1x2xxlimex limexlim2exex1ex x1x2xxlimln(elimex1xex0e2ex cos2x1(2limx02 2limx2sin2xcos2xlim4x2sin2x2sin2（洛必达）lim8x4sin2xcos2xlim4x二、（每535分(1)求limex1x2xxlimex limexlim2exex1ex x1x2xxlimln(elimex1xex0e2ex cos2x1(2limx02 2limx2sin2xcos2xlim4x2sin2x2sin2（洛必达）lim8x4sin2xcos2xlim4xsin4xlim44cos4xlim4sin434212xsin2x2xsin22cos 43 lim ＝3(3)求lim[xx2ln(11xxtln(1tt1 t02t(1 tt＝lim[xx2111o(1lim[111o1x2 3 xx(4)ylnx)xarcsinx，求dye1dy(lnx)(lnlnx )dxlnx)2x(1(5)求dx1311x21211d(1x2) (1x2)21x2d(1x2)1x223112xtanttan3tsectdtsec2t1dsect1sec3tsect332(6)求x2arctanxdx1x211x2 (1x2)21x2dx233x原式 arctandx arctanx311223311 arctanx x ln(1x)223662(7)求ln(1x)dx(2原式ln(1x)d( )ln(1x) dx.22(2x)(1ln(1x)2xln(1x)1ln1111 2x1xdx 2 C23三、（每630分x,x1xx0处可导，a(7)求ln(1x)dx(2原式ln(1x)d( )ln(1x) dx.22(2x)(1ln(1x)2xln(1x)1ln1111 2x1xdx 2 C23三、（每630分x,x1xx0处可导，ab.f(0)(1f(x)1abe,xx解:f(xx0f(00f(0)abx1xf(0)limabe0limxx1f(0)1b,a1,bxx1dyd21t确定的函数yy(x)当t1时的导 及曲率半径(2求由参数方程2dxtd21td21t1t解 12, 1t1t 1， 则1(1)23/3/y2t1k2tx(3)yf(x在上具有二阶导数,limf(x1a为任意实数,limf(xaf解:limf(xaf(x)limf(aalimf(a（其中xxa之间yf(x由方程ey6xyx21确定x0是否为极值点？极大值点还是极小值点？并判断曲线yf(x在点(0,0)附近的凹向.解:方程两边对x求导：eyy6y6xy2x0 y(0)0,y(0)方程对x再求导：ey(y)2eyy6y6y6xy20 y(0)20x0yf(x在点(00)附近为凸k的取值范围，使得方程exxk0有实根解:f(x)ex(x f(x)exf(x)exxln，极小值为f(ln(1lnk。而limf(x)3(1x1a0,xn1axn(n1,2,，求证limxnAAa，x有下界证:x axa成立，x n21nn 1xna1(1xa1(1a1，x又nxx2 nnnnx(1x1a0,xn1axn(n1,2,，求证limxnAAa，x有下界证:x axa成立，x n21nn 1xna1(1xa1(1a1，x又nxx2 nnnnx收敛limxAa. 取极限， aA，Annnn(2)f(x在[0,1]上连续，在(0,1)f(0)e,f(1)e1试证：至少存在一点0,12f()f()解:设F(x)e2xf(x) F(x)在0,1上连续，可导，F(x)e2x2f(x)f(x),F(0)F(1)由罗尔定理，至少存在一点0,1)F(02f(f(0(3)x1e2xex12e2f(x)e2xex1(x1),f(x)ex1f(x)ex1x3,f(3e20，于是f(33e2e22e2e2xex2e2(x则f(x)f(32e2(x1)xx1 f(x)f(x)f(x(4)f(x在a,bf(x0，1 12333x1x2f(x0f(x在a,bxxxxa,b003f(x)f(x0)f(x0)(xx0)f(x1)f(x2)f(x3)3f(x0)f(x0)(x1x2x33x0)3f(x0xx1 f(x)f(x)f(xf1 即1233