word完整版高数一试题及答案推荐文档.docx

《高等数学(一)》, 、选择题5,则k( ). 6则k( )3D. 4曲线yex3sinx1在点(0,2)处的切线方程为()(0,2)处的法线方程为()-- ()x D.-(x) 0(t1)(t2)dt,则f(3)=()()个A1B2C4 D0F列函数中有极限的是(,'(3)=2,m2hf叫Hh/°x1x2 1)°•设函数f(x)在[1,2]上可导,且f'(x) 0,f⑴ 0,f(2) 0,则f(x)在(1,2)内() . [f(x)xf'(x)]dx().(x)'(.xf(x)(x)(lnx2),则y(C)(lnx2)f(lnx2)2Bx4f(lnx2)Cx4f(lnx2)f(lnx2)Dx2f(lnx2)f(x)(x)=(B)'(x)(x)(x)(x))<dx(d)xdx1U丿A2xln〉< () (x)0(t1)(t2)dt,则f(2)=()A1B0C2 ()A.(0,0)B.(1,1) C.(2,2)D.(3,3)(lnx),则y(A)Af(lnx)(lnx) C. f(lnx)(lnx)(x)(A)(x)(x)(x)(x)(A)A. xlnxxCB..、求积分(每题8分,共80分);,.』"、)3x32x2x,dx1x三、•求函数f(x)X口的间断点并确定其类型x2设xy2sinxexy,.(x3)5xacost求由方程「. (x) 1,x0在x0处是否连续?tanx,x01ejx0函数f(x) 1,x0在x0处是否可导?tanx,,求y求证:Inxx1,x1设y是由方程y1xey确定的函数,求y讨论函数f(x)2x39x212x3的单调性并求其单调区间求证:ex2x1,求函数f(x)x(1 ;)的间断点并确定其类型xx五、解方程求方程y2dx(x2xy) 、选择题5:DABAA6-10:DBCDD15:BCCBD16-21:ABAAAA二、、,sinxdx解:(sinx)-:343Inx1dx(43Inx)3d(lnx)x(413lnx)?-d(43lnx):, 3xx解:edx=43lnx),dvdx,即vx,arctanxdxxarctanxxarctanxxarctanxxd(arctanx)x1x21—In(12dxx2) 3t2et3et2tdt3t2et 6tetdt3t2et6te6edt3^e 6te6^ dx-x5x6解:由上述可知x3x25xdx63ex(3x223x2),所以亠dxx35 —-求定积分01 3「:令3xt,即xt3,则dx23tdt,且当x0时,t0;当x8时,t2,于是23t2dt8dx01t2tln(1t)3l