2002级高等数学(上)期中试题参考答案2002级高等数学(上)期中试题参考答案一、单项选择题(每小题4分,共16分.))()(0,sin12)(.11的是函数则设xfxxxexfx.(D);(C);(B);)(第二类间断点跳跃间断点可去间断点连续点A,0:是间断点解x,112)sin12(lim)(lim100xxexfxxx)sin12(lim)(lim100xxexfxxx,110sinlim12lim0110xxeexxxx.,0且是可去间断点是第一类间断点x)(,,0.22cos为则是同阶无穷小与时设nxeexnxxx.2(D);25(C);5(B);4)(A)1(lim)(lim:2cos02cos0xxxxxxxxxeeee解)1(coslim20xxexx),21(lim21)21(lim5040xxxexxx,2121limlim55052cos0xxxeexxxxx而.5,,052cosnxeexxxx故是同阶无穷小与时当)()(,4)2()(.33的不可导点的个数为则设xfxxxxf.3(D);2(C);1(B);0)(A处仅在解2,2,022)2()(:xxxxxxxxf.有可能不可导,22)2(lim0)0()(lim)1(00不存在xxxxxxfxfxx.0)(处不可导在xxf,222)2(lim2)2()(lim)2(22不存在xxxxxxfxfxx.2)(处不可导在xxf,0222)2(lim)2()2()(lim)3(22xxxxxxfxfxx.0)2(f.2)(个不可导的点有故xf.)())0(,0(,)()0()(;)())0(,0()(;)()0()(;)()0()()(,)]([)(,0)0(,)(.42的拐点曲线也不是点的极值不是的拐点是曲线点的极小值是的极大值是则满足且对一切三阶可导设xfyfxffDxfyfCxffBxffAxxfxfxfxf代入等式把解0:x,)]([)(2xxfxf.0)0(0)]0([)0(2fff得,)]([)(2xfxxf),()(21)(xfxfxf,1)0()0(21)0(fff,0)(lim)0()(lim)0(100xxfxfxffxx,0)(),0(,xxfN内在由极限的保号性知,,0)(由负变正同号的两侧与在点xxxf.))0(,0(是拐点故f或)()()0()0()(xoxxoxffxf,,,0)(由负变正同号的两侧与在点xxxf.))0(,0(是拐点故f注:若xxf)(在点三阶可导,0)(xf,0)(xf,0)(xf,则是拐点))(,(xfx。二、填空题(每小题4分,共20分.).,,11,411,1)(.1baxxxxbaxxf则处连续在设,41)1(1lim)(lim11fxbaxxfxx,01)(lim1babaxx,121121lim1lim11aaxxbaxxx而.234112101baaba由.,1.22232dxydtytx则设,2323:2tttdxdy解.4322322ttdxyd.,1)(),11(.30xdxdyttfxxfy则且设020])1(2[)11(:xxxxxfdxdy解.2])1(2[1102xxxx.;),0()(.4单调减少区间为则其单调增加区间为设xxxfx,)(:lnxxxexxf解),1(ln)1(ln)(lnxxxexfxxx.1,0)(exxf得令.)()1,0(,0)(,10严格单减内在时当xfexfex.)(),1(,0)(,1严格单增内在时当xfexfxe.0)(.5阶泰勒公式为处带皮亚诺余项的在函数nxxexfx,0)0(,)(:1fxexfx解法,1)0(),1()(fxexfx,2)0(),2()(fxexfx,)0(),()()()(nfxnexfnxn)(!)0()0()0()()(nnnxoxnfxffxf).(!)1(1!2132nnxoxnxxxxxexf)(:2解法)](!)1(!21[112nnxonxxxx).(!)1(1!2132nnxoxnxxxxxxxxxxx2cos11lim21sin)1ln(lim2100020三、计算下列各题(每小题7分,共35分.).)1(sin1)1ln(1lim.10xxexxx]sin)1ln(sin1)1ln(11[lim:20xxxxxx原式解.412sin)1(1lim212000xxx.,016)(.20222xydxydxxyexyy求确定的隐函数是由方程设.0,0:yx时当解,0266xyxyyey.00xy,026662yxyyyeyeyy.2,20220xxdxydy即).121211(lim.3222nnnnnn...