1广东省2015年普通高校本科插班生招生考试《高等数学》试题答案及评分参考一、单项选择题(本大题共5小题,每小题3分,共15分)1.B2.A3.C4.C5.D二、填空题(本大题共5小题,每个空3分,共15分)6.5e7.28.159.212xe10.1ln2x三、计算题(本大题共8小题,每小题6分,共48分)11.解:111sin2(1)sin2(1)lim()limlim2212(1)xxxxxfxxx,11lim()lim()1xxfxxbb,(1)fa,当12ab,即2,1ab时,()fx在1x处连续.12.解法一:2320011arctan1limlim3xxxxxxx2222002111(1)limlim63(1)3xxxxxx.解法二:2320011arctan1limlim3xxxxxxx2011lim3(1)3xx.13.解:ln(1)xyxe,1111xxxeyee2(1)xxeye,2故01|4xy.14.解:设2xt,则22,2xtdxtdt,22231xtdxtdtxt222122(1)11tdtdttt2(arctan)2(2arctan2)ttCxxC.15.解:所求面积:40cos2Axxdx44400011sin2(sin2sin2)22xdxxxxdx4011cos28484x.16.解:由给定的二次积分知,积分区域2{(,)|11,01}Dxyxyx,如图:2100rIderdr21000111(|)()222reded(1)2e17.解:微分方程的特征方程为2250rr,解得12ri,微分方程的通解为12(cos2sin2)xyeCxCx.1212(cos2sin2)(2sin22cos2)xxyeCxCxeCxCx01012|2,|20xxyCyCC,解得122,1CC.故微分方程的特解为(2cos2sin2)xyexx.318.解法一:显然22313nnnn,22122(1)3(1)1limlim1333nnnnnnnn,则由比值审敛法知,级数213nnn收敛,由比值审敛法知,级数2131nnn收敛.解法二:2212211(1)31(1)13:limlim1131333nnnnnnnnnn,由比值审敛法知,级数231nn收敛.四、综合题(本大题共2小题,第19小题12分,第20小题10分,共22分)19.解:(1)1112ln(1ln),lnyyyyzzxyxxxyxxxxy,12(1ln)lnyyzzdzdxdyxyxdxxxdyxy.(2)121(,)lneyDfxyddxxxdy112()eyxdx2222111()(ln)ln2322eexdxxxex.20.(1)解:()()2,()()2Fxfxx...