02.2019考研数学二考试真题答案解析.pdfVIP免费

2019考研数学二考试真题答案解析（完整版）来源：文都教育1.（C）3tan~3xxx2.（B）'sincos2sin,"sin,yxxxxyxx令"0y得0,,xx又因为"'sincos,yxxx将上述两点代入得"'()0,y所以(,2)是拐点.3.（A）0ed(2)1xxxP发散（A）或000dee|edxxxxxx00ede|011xxx4.（D）解：由条件知特征根为121，特征方程为212()()210，故a=2，b=1，而y*=ex为特解，代入得C=4，选（D）5.因为2222sinxyxy+<+22221cosxyxy-+<+2131IIII\<<因为2222221cos2sinsin22xyxyxy++-+=222222sin2sincos22xyxyxy+++=因为2222424xyxypp++<\<2222sincos22xyxy++\<22221cossinxyxy\-+<+32II\<321III\<<选A6.解，必要性()()fxgx,相切于a则''()()()aafagafg''""322''"""""2.()()[1']0()()()()()()()()lim2()()2()22limlimxaxaxayPyagayfxgxfxgxfxgxfagafaxaxe充分性2''()()()()()()()()()2()()()()()lim()()22limlimxaxxxaxafxgxOfagaxafgfagaxafxgxfagafaga=f(x)与g(x)相切于点a.且曲率相等.选择（B）7.因为0Ax=的基础解系中只有2个向量()24()nrArA\-==-()0rA*\=\选A8.选（C）解：由22AAE+=得22λλ=+，λ为A的特征值，2=-l或1，又1234Aλλλ=，故1231λ=λ=-2λ=，，规范形为222123yyy--，选（C）9.()()02(21)212(21)21000222ln2lim22ln221lim2lim121limeee4exxxxxxxxxxxxxxxxxx®+-+-×+-+++=++-====10.当32tp=时，3333sin11cos12222xypppp=-=+=-=，即为点31,12p÷ç+÷ç÷ç23ddd1d1sin1ddd1cosd1tyytykxtxtxp=-==×===--331111.22322yxyxyxppp÷ç-=---Þ-=-++÷ç÷çÞ=-++在y轴上的截矩为322p+11.2322222;zyyzyyyfffyfffxxxyxx÷ç÷ç÷=-=-=+=+÷çç÷÷çç÷ç32233222222zzyyxyxfyffxyxxyyfyffxxyyfx+=-×+×+=-++÷ç÷=ç÷ç÷ç12.解析：lncos,06yxx2606206600sin1dcos1dcos231secdln|sectan|lnln3ln3323xlxxxxxxxx13.解析：211001212012212120101201222100sin()d(d)d1sindd21sinsind|d21sind21111sind(cos)|(cos11)2244xxxtfxxxtxtttxttxxtxxtxxxxxxx...