2007年浙江省普通高校“专升本”联考《高等数学（一）》参考答案
一．填空题： 1．?2,3???3.??? 2．y'?3sin2xcosx5sin3．0 4．ln3xln5
sinx?C
1?sinx?2?5!
5．y?5??1?x?66．
4 97．du?2cos?2x?y??e8．lnx?y?x?3y?dx???cos?2x?y??3e?dy(超纲,去掉)
x?3y?22??y2?C
二．选择题：
1。A， 2。D， 3。C， 4。D。 三．计算题：
1．解。y?2lncosx?1ln1?ln4x 2??11ln3x'x??2tanx?2 y??2tanx? 4421?lnxx1?lnx4ln3x???2。解：方程两边对x求导数，得
xy'?y2x?2yxy'?yx?yy' ???2?2?xy'?y?x?yy' 222222x（2x?y）x?yx?y?y?1????x?1??x?y?y'?x?y?y'?x?y。 x?y3．解:令t?x，lim?x?01?cosx1?costsint1?lim??lim? t?ot?o?2tx2t24．解：原式＝5．解：
13sinx?213sinx?2??ed3sinx?2?e?C
3?3x2??1?e?xexdx＝?xd(ex?1)?1?e?x2x1?1????xd?x??xdx ???xe?1e?1?e?1?d?e?x?1?xxx＝?x???x??x?ln?1?e?x??C??x?x?ln?1?ex??C
e?1e?1e?1e?16．解：
?4 0 ?e2x?tanx?1?dx＝
2?4 0 ?e2x?sec2x?2tanx?dx??4esecxdx?2?4e2xtanxdx?＝
2x2 0 0???＝e2xtanx40?2?4etanxdx?2?4etanxdx?etanx 0 0 ?2x ?2x2x?40??e2
7．解：平行于直线???2x?y?3z?0 的直线的方向向量应是
x?2y?5z?1??ij S?2?11?2?k????3??i?7j?3k ?5?所求直线方程为
8. 解：I?x?1y?1z?1 ???17?3D:??Dy?xdxdyy??sin?xy?a2?22(超纲,去掉)
2 令x??cos?,xy???a2?22
I???a32?0d??a0?2cos??sin?d?5????2?4(cos??sin?)d??4?sin??cos??d????cos??sin?d?????4?54?3??0? 3?5???2??a??sin??cos??4??sin??cos??5????cos??sin???4?03?44????a?332?1?1?2?22??4233a9．解：原方程两边对x求导数得
f??x??f?a?x?(1)f???x???f?(a?x)??f?a??a?x????f(x)?f?x?满足f???x??f?x??0(2)由原方程令x?0得f?0??1由(1)f??0??f?a??1?0即???i方程(2)对应的特征方程为
?2?(2)有通解f?x??c1cosx?c2sinxf?0??1得c1?1即f?x??cosx?c2sinxf??x???sinx?c2cosxf??0??c2?f?a??cosa?c2sina ?c2?10．解：
cosa1?sina?f?x??cosx?cosasinx1?sinaf?x???x?1?11??x?1?2??x?1?2n?1?x?1?1???2??n?1?x?1?x?1??x?1????????2n?0?2?2?n?0?x?1?12?
收敛区间为四、综合题： 1．解：
即 ?1?x?3当 0?a?1 时 y?ax 与 y?x2 的交点坐标是(0,0)和(a,a2)a2S?S1?S2??0ax?x2dx??1ax?ax????a3a31?a3a?a3a3a1 ???????233232311S??a??a2? 令 S??a??0 a?22?1?2?2S???a??2a?0 ? 在 0?a?1 时Smin?S???62??当 a?0 时 y?ax 与 y?x2的交点坐标是(0,0)和(a,a2)212S?S1?S2??0aax?xdx??0x?axdx????a3a31aa3a1 ?????????2332623a21S??a?????0 S?a?在a?0时单调减小.22故在a?0时,S?0?为S?a?的最小值 S?0??Smin?2-22211又? ???6362?232?21? 在a?1时 S的最小值在a?时取到213
?????1?2?2Smin?S???6?2?
2021年专升本高数一答案【高考】zq
