数学分析课后习题答案--高教第二版(陈纪修)--1章

nTaaan={12}2nkknC∑==+=nknnknC02)11((1)(2)ABAB∪1TTa∈1TTTa∈212aa≠Ta∈313aa≠23aa≠{},,,,21naaaS=TS⊂2{},,,,21naaaA={},,,,21nbbbB=AB∪AB∪{},,,,,,,2211nnbababa=(1){}0=∅(2)a⊂{,,}abc(3){,}ab∈{,,}abc(4){,,{,}}abab={,}ab1}0{02a{,,}abc∈a{,,}abc13{,{ba,},}abc{}⊂ba,{,,}abc4}},{,,{bababa,{,}ab⊃}},{,,{baba{,}ab≠}},{,,{baba{,}ab(1)xx−+≤320(2)(3)01(4)0cotsin=xx1{}32|≤−xx2{}00|),(yxyx3{}Qxxx∈10|4⎭⎬⎫⎩⎨⎧∈+=Zkkxx,2|ππ(1)ABDABAD∩∪∩∪∩()()(=)C(2)()ABABCC∪∩=1)(DBAx∪∩∈Ax∈Bx∈Dx∈BAx∩∈DAx∩∈)()(DABAx∩∪∩∈)()()(DABADBA∩∪∩∪∩⊂)()(DABAx∩∪∩∈BAx∩∈DAx∩∈Ax∈Bx∈Dx∈)(DBAx∪∩∈)()()(DABADBA∩∪∩∪∩⊃22CBAx)(∪∈BAx∪∈Ax∈Bx∈CCBAx∩∈CCCBABA∩∪⊂)(CCBAx∩∈Ax∈Bx∈BAx∪∈CBAx)(∪∈CCCBABA∩∪⊃)((1)ABAC∪∪=BC=(2)ABAC∩∩=BC=1{}cbaA,,={}dcbB,,={}dcC,=ABAC∪∪=CB≠2{}cbaA,,={}edcB,,={}dcC,=ABAC∩∩=CB≠?(1)BAx∩∈⇔Ax∈Bx∈(2)BAx∪∈⇔Ax∈Bx∈1BAx∩∈⇔Ax∈Bx∈2BAx∪∈⇔Ax∈Bx∈31.},,{γβα=S,T??abc={,,}fST→2733=6!3=⎪⎩⎪⎨⎧cbafγβα:⎪⎩⎪⎨⎧bcafγβα:⎪⎩⎪⎨⎧acbfγβα:⎪⎩⎪⎨⎧cabfγβα:⎪⎩⎪⎨⎧bacfγβα:⎪⎩⎪⎨⎧abcfγβα:2.(1)[,[,]ab]01(2)(,)01(,−∞)+∞1]1,0[],[:→bafabaxyx−−=2),()1,0(:+∞−∞→f)cot()21tan(xxxππ−=−3.fg(1)yfu==()logau,ugx==()x23−;(2)yfu==()arcsinu,ugx==()ex(3)yfu==()u21−,ugx==()secx(4)yfu==()u,ugx==()xx−+111)3(log2−=xya()()+∞−∞−,33,∪),(+∞−∞2xy3arcsin=(]0,∞−⎥⎦⎤⎜⎝⎛2,0π3xytan=⎟⎠⎞⎜⎝⎛+−∈2,2ππππkkZk∪[)+∞,04411+−=xxy()[)+∞−∞−,11,∪[)()+∞,11,0∪4.(1)yx=+arcsin112;(2)321log(1)3ayx=−1uyarcsin=vu1=12+=xv2331uy=vualog=12−=xv5.(1)xyasinlog=1a(2)yx=cos(3)yx=−−432x(4)yxx=+2411()ππ)12(,2+∈kkZk∪(]0,∞−2⎥⎦⎤⎢⎣⎡+−∈22,22ππππkkZk∪[]1,03[1,4−]⎥⎦⎤⎢⎣⎡25,04()()+∞∞−,00,∪⎟⎟⎠⎞⎢⎣⎡+∞,22336.fg(1)fx()=2log()axgx()=2logax(2)fx()=22sectanxx−,gx()=1(3)fx()=sincos22xx+,gx()=11fg52fg3fg7.(1),fxxxx()+=−+−3235321fx()(2)13131+−=⎟⎠⎞⎜⎝⎛−xxxxf,fx()1tx=+33−=tx1)3(5)3(3)3(2)(23−−+−−−=ttttf977721223−+−=ttt9777212)(23−+−=xxxxf2txx=−11−=ttx()113113+−−−=tttttf1412−+=tt1412)(−+=xxxf8.fx()=+11x,,fffffffff121)(++=xxxff322)(++=xxxfff5332)(++=xxxffff(,−∞+∞)2)()(xfxf−+2)()(xfxf−−2)()()(xfxfxf−+=2)()(xfxf−−+ABCDyfx=()A=(,)03B=−(,)11C=(,)32D=(,)406[](](]⎪⎪⎩⎪⎪⎨⎧∈+−∈−∈+−=4,3823,125231,034xxxxxxyy(,)11Ox2x1.2.81.2.91.2.8fx()yfx=(),x∈[,]02[](]2210,121211,22xxyxxx⎧∈⎪⎪=⎨⎪−+−∈⎪⎩13.610.8(1.2.9)542Px[](](]⎪⎩⎪⎨⎧∈−∈−∈=11,92.112118.13329,598985,04.78)(xxxxxxxP13.[,[,[,]01]01]017[,]01⎩⎨⎧−=xxxxxf1)(8