《微积分》(中国商业出版社)课后习题答案四

习题四（A）1．判断下列函数在给定区间上是否满足罗尔定理条件，若满足条件，求出相应的值．(1)；]，60[6)(上在xxxf(2)]，11[)(3在xxf上．解：（1）062x-6(x)'xxf]，60[4x满足(2)031)(32xxf不满足2．下列函数在给定区间上是否满足拉格朗日定理条件？如满足，求出相应的值（1）]，0[2)(3axxxf在上；（2）上在2π，0sin)(xxf．解：(1)23)('xxfaaf(a)f320)0(aafafaf33232)(')0()(22满足(2)xxfcos)(1)2(0)0(ff)20(2arccoscos2)('2)0()2(，fff满足3．证明：（1）2πarccosarcsinxx；（2）)(arctanarctanbaabab．解：(1)令2arccosarcsin)(xxxF011x-11(x)'22xFCxF)(取0x，得0)0(F0C得证(2)令xxxFarctan)(0))('0(0111)(2xFxxxF时)(xF为非增函数又ba)()(aFbF4．对函数xxfsin)(和xxgcos1)(在]2π，0[上验证柯西中值定理．解：据柯西中值定理)(')(')0()2()0()2(gfggff4sincos15．设)(xf在][ba，上有二阶导数)(xf，且0)(xf，证明)(xf在)(ba，中至多有一个驻点．解：假设)(xf在)(ba，中有二个驻点dc，0)('0(c)'dff)(0)(dcxxf，这与题设0)(xf矛盾得证6．证明：第三章§3．1中的命题.解：(x)'lim)()(xlim)('00000Afxffxfaa得证同理Bxf)('07．设函数)(xf，)(xg在（a，b）内可导，0)(xg，且))(0())g((')(')(baxxxfxgxf，．证明：))()(()(baxxCgxf，．（其中C是常数）（提示：考虑函数．xgxfx))(/)()(解：令)(/)()(xgxfxF)('')('2xg(x)f(x)g(x)fxg(x)F0')(')(0)((x)gxf(x)fxgxg且CxF(x)F)(0'得证8．设函数)(xf在][ba，上连续，在)(ba，内有二阶导数，且)(0)(，0)()(bcacfbfaf．证明：在)(ba，内至少存在一点，使0)(f．(提示：用两次中值定理)．解：0cfbfaf)(0)()(，存在，，)(caaβ)(bc，使得'，0)('faf(β)0且x(，β)，'f(x)递减)(Eba，使0)(f9．设)(xf在)(ba，中有1n阶导数，求)(0xx的n次多项式nnnxxaxxaxxaaxP)()()()(0202010使得)，，2，1，0()()(0(0)(nkxfxPk)kn解：!)()(！)(0)(0)(0)(nxfaxfanxPnnnnnn！)1()()(！)1()(0)1(10)1(10)1(nxfaxfanxPnnnnnn依此类推，可得)('，)(000xfaxfa110．极限xxxxsin1sinlim20能否用洛必达法则计算？其值为何？解：不能01sinlimsin1sinlimsin1sinlim02020xxxxxxxxxxx11．求下列极限：（1）216lim42xxx；（2）xxx1)1(limπ0；（3）xxxlnlncotlim0；（4）xxxxxxsin2eelim0；（5）)0()lnln(limaxxxax；（6）)cos1(1lim3sin0xxexx；（7）20)(2)()(limhafhafhafh（设)(xf在）ax点邻近连续）．解：(1)3214lim216lim3242xxxxx(2)100)1(lim1)1(limxxxxxxx(3)12sin2lim1)csc(cot1limxlncotlnlim0200xxxxxxxxx(4)2coslimsinlimcos12limsin2lim0000xeexeexeexxxeexxxxxxxxxxxx(5)0ln11limln1lnlim)lnln(limxaxaxxaxxxxxaaxaxax(6)2lim)cos1(1lim2303sin0xxxxxxexxx(7)hhhafafafhafhafhafhafhh1.)]()([)()(lim)(2)()(lim020=(a)fhh)(af(a)fh''lim012．求下列极限：（1）xxxxln11lim1；（2）111lim0xxex；（3）2πtan)1(lim1xxx；（4）xxx111lim；（5）xxx2011lim；（6）xxxln1arctan2πlim；（7）xxxx1sin1cotlim0；（8）xxxln10)(cotlim；（9）xxxex10)(lim；（10）xxxtan0)(arcsinlim．解：(1)21111limln11lnlim)1(ln1lnlimln11lim21111xxxxxxxxxxxxxxxxxx(2)212lim11lim)1(1lim111lim0000xxxxxxxxxxxxxxeeexeeeexxeex(3)22sin2)1(2cos22sinlim2cos2sin)1(lim2tan)1(lim111xxxxxxxxxxxx(4)eeeexxxxxhxxxxxx1limlim1lim1lnlim1111111(5)1lim11lim3202022lim111lnlim11ln020xxxxxxxxxxxxeeex(6)exxxeexxhxxxhxx1)1(arctan2limlimarctan2lim2arctan2ln1(7)61！321limsin)sin(coslim1sin1cotlim3320200xxxxxxxxxxxxxxx(8)eeeexxxxxxxxxxxx1limln1)(cotlimcos1lim1)csc(cot1limlncotln00020(9)200101lim111)1e(1lim)(limexexexexexxexxxxxxxxxx(10)1limcotarcsinlnlimlim)(arcsinlim011arcsinsin00arcsinlntan0tan022eexxeexxxxxxxxxxx13．设0，，0，sin)(xbaxxxxxf在0x点可导，求ba，解：1)(lim)(lim1sinlim)(lim0000bbbaxxfxxxfxxxx0)(0'01sinlim)(0'0aafxxxfx14．设当0x时)()1()(22xobxaxexfx求常数ba，．解：22lim22lim)1(lim)(lim0022020aexxaxbexbxaxexxfxxxxxx从而可得10)(lim0bbexx，210)2(lim0aaexx15．求下列函数的单调区间：（1）2224xxy；（2）xexy；（3）xxy12；（4）xxyln22．解：（1）1，1，0044'3213xxxxxy所以，当)，1()0，1(x单调增当)1，0()1，(x调减（2）0x01'xey所以，当0x单调增当0x单调减（3）2，00)1(2)1(-)(12'212222xxxxxxxxxy所以，当2x单调增当02x单调减当0x单调增（4）2101-4'xxxy所以，当21x单调增当2121x单调减当21x单调增16．证明下列不等式：（1）)1(xexex；（2）)1(132xxx；（3）)0(2sin2xxxxx；（4）)1(lnln121221xxxxxx．解：（1）令)1()(xexexFxexFxe)('当1x时，0)('xF且0)(lim1xFx)(xF0得证(2)令)1(12)(xxxxF01x1)('2xxF且3)(lim1xFx得证(3)令0)(sin)(xxxxF01-cos)('xxF且0)(lim1xFxxxsin令2sin)(2xxxxG1cos)(xxxG01-sin)(xxG又0)('lim0xGx0)(xG2sin2xxx综上得证(4)令)1(ln)(xxxxF01ln)(xxF1)()()(1222xxxFxF得证17．设2，，，qpZqp，试比较qp与pq的大小．（提示：利用函数xxxf/ln)(的单调性）．解：令2)(ln)(xxxxf2ln1)(xxxf1当0)('2xfex若qp则)()(qfpfqqpqpppqlnlnqppqlnlnqppqeelnlnlnlnqppqee2当0)('xfex同理若qp则pqqp18．设函数)(xf在区间),(ba上恒有0)(,0)('xfxf，则)(xf在),(ba上（B）．A．单调上升，下凸B．单调上升，上凸C．单调下降，下凸D．单调下降，上凸解：B0)('xf0)(xf19．给定曲线))((：IxxfyC，已知)(''xfy的图形如4-17，则曲线C在（，）上是（C）．A．下凸的B．上凸的图4-17C．单调上升的D．单调下降的解：C20．确定下列曲线的上、下凸区间和拐点：(1)32xxy；(2)2241xxy；(3)xxey；(4)12xxy．解：(1)232'xxy062xy31x拐点)272,31(当)31,(x下凸当),31(x上凸(2)3)1(12xy223]1)[(x1)--2(x'y0]3)1[(]3)1[()1(83]1)2[(x-422222xxxy2021xx拐点)410(，)412(，在)，2()0，(下凸在),20(上凸(3)xxeyxe'02exxxey2x拐点)e2,2(2在),-2(上凸在)2,-(上凸(4)12xxy2222)1(2)1(-1)-(2'xxxxxxxy423)1()2-1)(-2(-1)-2(xxxxxy=0x无解无拐点当1x时y不存在在