北大版高等数学第4章习题解答

习题4.13212121.()32[0,1][1,2]Rolle0,(0)(1)(2)0,()[0,1][1,2]Rolle6362433620,,633333(0,1),(1,2),()()0.332.fxxxxfffffxxxxxxfxfx2验证函数在区间及上满足定理的条件并分别求出导数为的点.处处可导故在区间及上满足定理的条件.f(x)=3x讨论下列解321111()[1,1]Rolle,,(1,1),()0.(1)()(1)(1),,;(2)()1.(1)()(1)(1)(1)(1)(1)(1)()0,(1,1),()0.1(2)(mnmnmnmnfxcfcfxxxmnfxxfxmxxnxxmnxxmmxnnxcfcmfx函数在区间上是否满足定理的条件若满足求使为正整数解1/32),(0).33.()ln[1,],?11(),()(1)lnln11(1),1.4.Lagrange(1)|sinsin|||;(2)|tantan|||,,(/2,/2);(3)lnxffxxecfxfefeecexcyxxyxyyxxybabbba不存在写出函数在区间上的微分中值公式并求出其中的应用中值定理,证明下列不等式:解222(0).(1)|sinsin||(sin)|()||cos|||||.(2)|tantan||(tan)|()|sec||||.(3)lnlnln(ln)|()((,)).5.()(1)(4)xcxcxcaabaxyxxycxyxyyxxyxcyxyxbabbababaxbacabaacaPxxx证明多项式的导函数的证1,212,.()1,2,Rolle,,,()(2,1),(1,1),(1,2).6.,,,:()coscos2cos(0,).nnPxPxcccfxcxcxcnx三个根都是实根并指出它们的范围有四个实根根根据定理它的导函数有三个实根又作为四次多项式的导函数是三次多项式,最多三个实根,故的导函数的三个根都是实根,分别在区间设为任意实数证明函数在内必有根证1211()sinsin2sin[0,]2((0)()0),()(0,).ngxcxcxcnxnggfx在满足定理的条件故其导函数在内必有根证22(()()7.()()(,),()0,0,(,).()():,()(),(,).(()()()()()()()()()0,()()()(),,,()(),()fxgxfxgxabgxxabfxgxkfxkgxxabfxgxfxgxfxfxgxfxgxgxgxgxfxkkfxkgxgx设函数与在内可微且证明存在常数使根据公式的一个推论存在常数使即证2(,).8.()(-,)(),.:(),,,.(())()0,.,(),.9.(1)arcsinarccos/2,-11;(2)arctanarcsin,.1xabfxfxkxfxkxbxkbfxkxfxkkkxfxkxbxxxxxxxx设在上可微且证明其中为常数证明下列等式:证证(1)22222arcsinarccosarcsinarccos110,(1,1),arcsinarccos[1,1],11arcsinarccos,arcsin0arccos0,arcsinarccos.22(2)arctanarcsin111111xxxxxxxxxxxCCxxxxxxxx在连续故()=()+()2222222222211111110,11arctanarcsin,00,arctanarcsin0,11(,).xxxxxxxxxxxxxxxCxCxxxx以代入得故220210.:sin,0/2.sin()(0/2),(0)1,[0,/2],cossincos(tan)(0,/2),()0.2[0,/2],()()(0)1,0/2.211.()(,),(,),lixxxxxfxxffxxxxxxxffxxxfffxfxfxabxab证明不等式在连续在可导在严格单调递减设函数在内可微对于任意一点若证0000000000m(),lim()().()()limlim(01)lim()lim().12.(Darboux)()(,),[,](,),()().::xxxxxxxxxfxfxfxfxxxfxxfxxfxyfxABabABfafb存在则中值定理设在区间中可导又设且证明对于任意给定的00f(x+x)-f(x)证x1011222()(),(,)().()()()0().()lim0,)/20,()()00,()()0.()().:0()/2,()().[,]xfafbcabfcfaxfafafbfabaxfaxfaxfaxfafafaxbafbfbfabc都存在使得先设存在(使得时即特别类似存在某点取最小证1,()()(),,,.(,),Fermat()0.:()().()().()(),()()0,()()0,,(,)()()0,().fcfafacacbcabcfcfafbgxfxxgxfxgafagbfbcabgcfcfc值f(c)同理是极小值点,由引理,再设考虑由前面的结果存在使得即习题4.20000020220LHospital:212ln2ln21.limlim.313ln3ln3cos1sinsin2.limlimlim1.ln(1)11/(1)113.limln(1)ln(1)ln(1)ln(1)limln(1)ln(1)xxxxxxxxxxxxxxxxxxxxxxxxxxx用法则求下列极限222022202220/21/(1)1/1lim1/1ln(1)ln(1)l/(1)11lim(1)ln(1)1ln(1)/111lim.2ln(1)1(/1)ln(1)1tan34.limlimtanxxxxxxxxxxxxxxxxxxxxxxxxxxxx222/222001000000001/5010003sec33.secln(cos)(1/(cos))(sin)5limlim.ln(cos)(1/(cos))(sin)ln1/16.limln(0)limlimlim0.()7.limlimxxaxxxxxxyxxaxaxaxaabxbxbxbbxxxxxxxeyxe505050/50/50/5022000222200022250limlimlim0.8.lim(tan).(tan),limlnlim(2)lntanlntansec/tanlimlim2lim122(2)yyyyyyyxxxxxzxxyyeeexyxyxxxxxxx022ln200022limln01/002332000tan0,limlimsin1.1ln9.lim1(0)limlimln.1111arcsinarcsin10.limlimlimsin3xyxxyyyxxyyyyyzzyezeeaaaaxaayyyyyyyyy22220011111230111111limlim.33261ln111.limlim1ln(1)lnln11lnlimlimln(1)/ln(1)1/1lim.ln22112.limlsinyyyyyyyxxyyyyyyyyyyyyyyyyyyyyyyyxexx22224200001/1/02220002011imlim11limlim.222arctanarctan13.lim,,arctanarctan1ln(/arctan)limlnlimlim2(1)arctanlim2xyxyyyyyxxxxxxxxeyexyeeyxxyxxxxxxxxxxyxxxxx232001/1/3011lnln112arctan1arctan1limlim,633arctanlim.14.limarctan.arctan.22lnarctan2limlnlimlimlnarctan(12xxxxxxxxxxxxxxxxxexxyxxxyxx21ln12222200000)limlim1,limarctan.112arctan(1)(1)tansec1tan215.limlimlimlimlim2.sin1cos1cos1cossinxxxxxxxxxxxxxexxxxxxxxxxxxxxxx2000111coshcossinhsincoshcos16.limlimlim1.22(ln1)1(ln1)117.limlimlimln11/11xxxxxxxxxxxxxxxxxxxxxxxxxxx211222/(ln1)lim2.12218.limarctan.arctan.21ln(arctan)(1/arctan)21limlnlimlim,112limarctan.xxxxxxxxxxxxxxxyxxxxyxxxe习题4.3221221223212222211.0Taylor:(1)sinh2111()22!(21)!2!(21)!().3!(21)!111(2)ln2122221xxnnnnnnnooxeexxxxxxxxnnxxxxnxxxxxxxnn求下列函数再点的的局部公式22212321224221212223()2221().32111(2)(2)(2)(3)sin(1cos2)(1)().222!4!(2)!21(4)(21)(1())1(nnnnnnnnnnooooxxxxnnxxxxnxxxxxxnxxxxxxxxxxx