2014-2015学年第一学期《高等数学》试卷（A卷）
一．填空题（每小题4分，20分）
k1．设x?0时4?3sinx?2与x是同阶无穷小，则k? 1 2．由方程e2x?y?cos?xy??e?1确定y?y?x?，则y??0???2 3．设y?1?x2在x?1处对应的微分dy?12dx 2
4．已知f?0??f??1??1,f?1??3，则xf???x?dx??1 ?05.曲线y??x?1?x的拐点处的横坐标x??321 5二．计算下列各题（每小题5分，共20分）
ex?esinx6、求极限lim2
x?0x?x3ln1?x??????sinx?x?1?esinx?xx0?limlime?elim解 原式=lim 34x?03423x?0x?0x?0x?xx?x?x?x?xex?1?esinx?x??limx?0???sinx?x?sinx1?cosxsinx11?lim?lim?limlim? 34232x?0x?0x?0x?0x?x3x?4x6x?12xx6?12x61??x2?esinx? 7、求极限lim??4x?0?x??1?ex???13?????4?xxx2?esinx2e?esinx??lim???0?1?1 解 由于lim???44?x?0??x?x??1?ex?x?0??e?x?1?????11????xx2?esinx2?esinx??lim???2?0?1?1，从而左右极限存在且相等，lim???44?x?0??x?x??1?ex?x?0??1?ex?1?0????1??x2?esinx??故原式极限存在且lim?=1 4?x?0?x?1?ex???exsinx?x?1?x?8、用泰勒公式求极限lim 3x?0xx2x32?o?x?,sinx?x??o?x3?，所以 解 因为e?1?x?2!3!x?x2x32??3?1?x??oxx??ox??????x?1?x????2!3!???原式=lim? 3x?0xx3x33x?x???o?x3??x?x2ox???3?1?1 1126?lim???limx?0x326x?0x3632?ax?b?2,x?1?9、设f?x???在x?1处连续，求a,b的值 x?1??1,x?1??ax?b?2ax?b?2,x?1?lim?f?1???1，解 因为f?x???在处连续，所以 x?1x?1x?1x?1??1,x?1?从而limx?1?ax?b?2?lim?x?1?x?1?ax?b?2?0???1??0 ，
x?1即limx?1?ax?b?2?a?b?2?0,a?b?4,b?4?a
?a?x?1??4?22ax?4?a?2进而?1?lim ?limx?1x?1x?1?x?1?a?x?1??4?2???limx?1aa?x?1??4?2?aa?，即a??4,b?4???4??8
0?4?24三．计算下列各题（每小题5分，共15分）
1141?x4?14410、设y?arctan1?x?ln，求y?
2441?x4?1解 令t?41?x4，则y?11t?111arctant?ln?arctant??lnt?1?lnt?1? ??24t?1241?dydt?111?11???44?从而y???????1??1?????1?x??
dtdx?21?t24?t?1t?1?????x1?1?t?1??t?1????11??144?13?????1?x0?4x????????? 224?t?1???21?t??4??22?3?3t?1?1?t??1?11?3144344???2??x?1?x???x?1?x? 242?1?tt?1?2t?122?3?3?31t?1??1?t?3?13?14444344??x?1?x??4?x?1?x???x?1?x? 442t?1t?11?x?1??1x4?1?x4?3（可以先算dydt和，再作乘积得出结果，切记别忘作乘积！） dtdx11、设f?x?连续，在x?0的某个邻域内有f?1?sinx??3f?1?sinx??8x?o?x?，且
f?x?在x?1处可导，求曲线y?f?x?在点?1,f?1??处的切线方程。
解 由已知，令x?0，得f?1??3f?1??0，从而f?1??0。 又limx?0f?1?sinx??3f?1?sinx?o?x????lim?8???8，即 x?0xx??f?1?sinx??f?1??3?f?1?sinx??f?1??f?1?sinx??3f?1?sinx???
8?lim?limx?0x?0xx?f?1?sinx??f?1?sinx??f?1?sinx??f?1?sinx??lim???3lim???? x?0x?0sinxx??sinxx????f??1??1?3f??1??1?4f??1?，因此f??1??2。
从而所求的切线方程为y?0?2?x?1?，即y?2x?2。
?d2y?x?t?ln?1?t?12、由参数方程?确定y?y?x?，求2
23dx??y?t?tdydydt2t?3t22t?3t2解 ????1?t???1?t??2?3t?
dxdx1?11?t?1dt1?td?dy??d2yd?dy?dt?dx??0?1??2?3t???1?t??0?3??从而2? ???dx?1dxdx?dx?1?dt1?t??1?t?2?3t??3?3t?2?3t?3?3t?1?t??5?6t???1?t??
1?t?1tt四． 计算下列积分（每小题5分，共15分）
2014-2015学年第一学期《高等数学》试卷(A卷)
