第二章总练习题
?|x?3| x?1时?1.讨论函数f(x)??x2的连续性和可导性.313?x?,x?1时??424?x313?解x?1时f(x)可导.f(1?0)?lim??x???2;x?124??4f(1?0)=lim|x?3|?2?f(1?0)?f(1),f在x?1连续.x?122?x313?f??(1)?(3?x)?|x?1??1,f??(1)???x??24??4f在x?1可导.??x3??????22?x?1??1?f??(1),f?(1)??1.x?1?2x?2 x??1时?322.设函数f(x)??Ax?Bx?Cx?D, ?1?x?1时??5x?7 x?1时试确定常数A,B,C,D的值,使f(x)在(??,??)可导.解f(?1?0)=lim(2x?2)??4?f(?1)??A?B?C?D.x??132f??(?1)?(2x?2)?|x??1?2?f??(?1)?(Ax?Bx?Cx?D)?|x??1?(3Ax?2Bx?C)|x??1?3A?2B?C.f(1?0)?A?B?C?D?f(1?0)?12,f??(1)?3A?2B?C?f??(1)?5.??A?B?C?D??4??3A?2B?C?2??A?B?C?D?12?3A?2B?C?5.?{A ? -9/4, B ? 3/4, C ? 41/4,D ? 13/4}.23.设函数g(x)?(sin2x)f(x),其中f(x)在x?0连续.问g(x)在x?0是否可导,若可导,求出g?(0).解g(?x)?g(0)?x?22f(?x)sin2?x2?x22?2f(0)(?x?0),g?(0)?2f(0).?cosx1+x224.问函数f(x)=x?sinx1+x与g(x)=为什么有相同得导数?解因为f(x)?g(x)?1.5,.设函数f(x)在[?1,1]上有定义,且满足x?f(x)?x?x,x?[?1,1].证明存在且等于1.证0?f(0)?0,f(0)?0.?x?0,f(?x)?f(0)?x故f?(0)?1.?f(?x)?x??x??x?x22??x?1?1(?x?0?0),f??(0)?1,类似f??(0)?1,26.设f(x)?|x?4|,求f?(x).22解|x|?2 时,f(x)?x?4,f?(x)?2x.f??(2)?(x?4)?|x?2?4,2f??(2)?(4?x)?|x?2??4,f?(2)不存在,同理f?(?2)不存在.7.设y?1?x1?x2,求,dydxdy22.2(1?x)2解y=-1?1?xdx?,dydx22??4(1?x)3.8.设函数f(x)在(??,??)上有定义,且满足下列性质:(1)f(a?b)?f(a)f(b)(a,b为任意实数);(2)f(0)?1;(3)在x?0处可导.证明:对于任意x?(??,??)都有f?(x)?f?(0)?f(x).证f(x??x)?f(x)?xf(?x)?f(0)?x2n?f(x)f(?x)?f(x)f(0)?x?f(x)?f?(0)?f(x)(?x?0),f?(x)?f?(0)?f(x).nn?1n???1/2, x?1/2,?1/2, x?1/2,9.设f(x)??(n?1,2,?);g(x)??(n?1,2,?);nn???0, x?1/2?0, x?1/2问f(x)在x?0处是否可导?g(x)在x?0处是否可导?解f(1/2)?f(0)1/2xnnn?1/22nn1/2?12nn?0(n??),f(x)?f(0)x?0,f?(0)?0.f(x)?f(0)?0?0(x?1/2,x?0).limx?0g(1/2)?g(0)1/2xn?1/2n?1n1/2?12?n12(n??),g(x)?g(0)x.g?(0)不存在.g(x)?g(0)?0?0(x?1/2,x?0),limx?010.设y?f(x)及y?g(x)在[a,b]上连续,证明:??证baf(x)g(x)dxba?2?2?baf(x)dx?g()xdx.a2b2?[f(x)?tg(x)]dx?b2??bag(x)dxt?2?2??2baf(x)g(x)dxt???baf(x)dx?0(*),2如果?g(x)dx?0,则由g的连续性g(x)?0,x?[a,b],不等式两端都是0.a如果?g(x)dx?0,(*)左端的二次函数恒非负,故其判别式非正,ab2
?2?babaf(x)g(x)dx?22?4???bag(x)dxb2??2baf(x)dx?0,2??f(x)g(x)dx??baf(x)dx?g()xdx.a211.求出函数f(x)?12x?122x???212nxn在点x?1的导数,再将函数f(x)写成f(x)?由此证明下列等式:12?222x/2?(x/2)1?x/2n?1的形式,再求f?(1),???12?n2n?2?n?22n.xn?1证f?(x)?f(x)?f?(x)?222x???n?1n2n,f?(1)?12?222???n2n.x/2?(x/2)1?x/2,nn?1(1/2?(n?1)(x/2)(1/2))(1?x/2)?(1/2)(x/2?(x/2)(1?x/2)(1/2?(n?1)(1/2n?12n?1),f?(1)?))(1/2)?(1/2)(1/2?1/21/22)?(1?(n?1)/2)?1?1/2nn?2?n?22n.n?112.由类似上题的办法证明1+2x+3x???nx2?1?(n?1)x?nx(1?x)2nn?1(x?1).证由等比级数求和公式x?x???x两端求导得1+2x+3x???nx?(1?(n?1)x)(1?x)?(x?x(1?x)2n2n?12n?x?xn?11?x,
1?(n?1)x?nx(1?x)102nn?1n?1)?(x?1).113.设y?f(x)在[0,1]连续且f(x)>0证明?1101f(x)10dx??1f(x)10.f(x)dx证1=?01dx?dtt?f(x)1f(x)dx??10f(x)dx?dx.14.lnx?(a)1?n11?1??ln?1???(n?0)n?1n?n?12?131n?1???1n?lnn?1?dt12?13???1n?1dtt;(c)e1?1n?1(b)1????1???e.n???1n.n
证(1)??1?1/n11??ln?1???1?1/nn???1?1/n1??1?1/ndt11(2)lnn?ln23n1?1?11???????ln?1??????1???1????, 12n?11?n?2n??1?1?11??lnn?ln?1??????1??????.1?n?2n??1??(3)?1???en??n1??nln?1??n???en?1n?1?e1?1n?1
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