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relationship, establidge (1)--lished equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \onship, Then in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their icance in rate 1, currency, length, area, volpreliminary knowlene and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solisignifd surface area and volume 1, size 2, table ...和ume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry 江苏省2013年普通高校“专转本”选拔考试

高等数学 试题卷（二年级）

注意事项：

1、本试卷分为试题卷和答题卡两部分，试题卷共3页，全卷满分150分，考试时间120分钟． 2、必须在答题卡上作答，作答在试题卷上无效。作答前未必将自己的姓名和准考证号准确清晰地填在试题卷和答题卡上的指定位置。

3、考试结束时，须将试题卷和答题卡一并交回。

一、选择题（本大题共6小题，每小题4分，满分24分。在下列每小题中，选出一个正确

答案，请在答题卡上将所选项的字母标号涂黑）

1、当x?0时，函数f(x)?ln(1?x)?x是函数g(x)?x2的( ) A.高阶无穷小 B.低阶无穷小 C.同阶无穷小 D.等价无穷小

2x2?x2、曲线y?2的渐近线共有( )

x?3x?2A. 1条 B. 2条 C. 3条 D. 4条

?sin2x　　x?0??x3、已知函数f(x)??，则点x?0是函数f(x)的

x?　x?0　??1?x?1A、跳跃间断点

B、可去间断点

C、无穷间断点

D、连续点

1d2y4、设y?f()，其中f具有二阶导数，则2?

xdx1121???f()?f() B. x2xx3x1121C. ?2f??()?3f?() D.

xxxxA. ?5、下列级数中收敛的是

1121???f()?f() x4xx3x1121???f()?f() 43xxxxn?1A、?2

n?1n?nnB、?()

n?1n?1x?1?n!C、?n

n?12?D、

?3n?1?nn

6、已知函数f(x)在点x?1处连续，且lim切线方程为

f(x)1?，则曲线y?f(x)在点(1,f(1))处的x2?12relationship, establidge (1)--lished equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \onship, Then in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their nce in rate 1, currency, length, area, volpreliminary knowlene and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solisignificad surface area and volume 1, size 2, table ...和ume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry A. y?x?1 B. y?2x?2 C. y?3x?3 D. y?4x?4 二、填空题（本大题共6小题，每小题4分，共24分）

1??xsin　x?07、设函数f(x)??在点x?0处连续，则常数a? ▲ ． x??a　 x?0　8、已知空间三点A(1,1,1),B(2,3,4),C(3,4,5)，则?ABC的面积为 ▲ ．

?x?t2?1d2y?9、设函数y?y(x)由参数方程?所确定，则23dx??y?t?1??? ▲ ．

x?110、设向量a,b互相垂直，且a?3，，则a?2b? ▲ ． b?2，????a?x1)x?e，则常数a? ▲ ． 11、设lim(x?0a?x12、幂级数

?n?1?2nnx的收敛域为 ▲ ． n三、计算题（本大题共8小题，每小题8分，共64分）

?ex1?13、求极限lim???．

x?0ln(1?x)x???2z14、设函数z?z(x,y)由方程z?3xy?3z?1所确定，求dz及2．

?x3215、求不定积分xcos2xdx．

?16、计算定积分

?　2dx2　02?4?x22x?3y．

17、设函数z?f(x,e?2z． )，其中函数f具有二阶连续偏导数，求

?y?x?x?2?3t?x?y?z?1?0?18、已知直线?平面?上，又知直线?y?1?t与平面?平行，求平面?的

x?3y?z?3?0??z?3?2t?方程．

19、已知函数y?f(x)是一阶微分方程

dy?y满y(0)?1的特解，求二阶常系数非齐次线性dx微分方程y???3y??2y?f(x)的通解．

relationship, establidge (1)--lished equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \o\onship, Then in-line solution. Category fraction multiplication word problem score Division applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significance in rate 1, currency, length, area, volpreliminary knowlene and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angle of classification (slightly) 17, and subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1, size 2, table ...和ume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry 20、计算二重积分

??xdxdy，其中

DD是由曲线y?4?x2(x?0)与三条直线

y?x,x?3,y?0所围成的平面闭区域．

四、综合题（本大题共2小题，每小题10分，共20分） 21、设平面图形D由曲线x?2y，y??x与直线y?1围成，试求：

（1）平面图形D的面积；

（2）平面图形D绕x轴旋转一周所形成的旋转体的体积． 22、已知F(x)??x20(9t?5t)dt是函数f(x)的一个原函数，求曲线y?f(x)的凹凸区间与

1312拐点．

五、证明题（本大题共2小题，每小题9分，共18分） 23、证明：当x?1时，(1?lnx)2?2x?1． 24、设函数f(x)在[a,b]上连续，证明：函数

?baf(x)dx??a?b2a[f(x)?f(a?b?x)]dx．

江苏省2013年普通高校“专转本”统一考试

高等数学（二年级） 试卷答案

一、选择题（本大题共6小题，每小题4分，共24分）

1、C 2、C 3、B 4、B 5、D 6、A 二、填空题（本大题共6小题，每小题4分，共24分） 7、0 8、

3116 9、 10、2 11、y?xlnx?cx 12、[?,)

4222三、计算题（本大题共8小题，每小题8分，共64分）

xex?ln(1?x)xex?ln(1?x)?lim?lim13、原式=lim2x?0x?0x?0xln(1?x)xex?ex?xex??limx?0ex?xex?2x11?x

1(1?x)223?3 2214、令F(x,y,z)?z?3xy?3z?1,Fx??3y,Fy??3x,Fz??3z?3

Fy?Fx??z3yy?z3xxyx????2?,?????,?dz?dx?dy?xFz?3z?31?z2?yFz?3z2?31?z21?z21?z2