河南省专升本高等数学真题(带答案详解)

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所以有 e3a?8?a?ln2.

?sinx,x?0?34.设函数f(x)??x在(??,??)内处处连续，则a?_______.

??a,x?0解：函数在(??,??)内处处连续，当然在x?0处一定连续，又因为

limf(x)?limx?0sinx?1;x?0xf(0)?a，所以limf(x)?f(0)?a?1.

x?035.曲线y?解：因y??3x在（2，2）点处的切线方程为___________. 1?x31??k?y??x?3y?4?0. 2x?2(1?x)336.函数f(x)?x2?x?2在区间[0，2]上使用拉格朗日中值定理结论中??____. 解：f?(x)?2x?1?2??1?f(2)?f(0)???1.

2?037.函数f(x)?x?x的单调减少区间是 _________. 解：f?(x)?1??1??1??1??1??1??0?x??0,?,应填?0,?或?0,?或?0,?或?0,?. 2x?4??4??4??4??4?20138.已知f(0)?2,f(2)?3,f?(2)?4,则?xf??(x)dx?______.

解：?xf??(x)dx??xdf?(x)?xf?(x)0??f?(x)dx?2f?(2)?f(2)?f(0)?7.

0002222rrrrr39.设向量b与a??1,?2,3?共线，且a?b?56,则b?_________. rrr解：因向量b与a共线,b可设为?k,?2k,3k?，

rrra?b?56?k?4k?9k?56?k?4,所以b??4,?8,12?. 40.设z?e解：z?ex2?y2?2z，则2?_______.

?xx2?y222?z?2zx2?y2??2xe?2?2(1?2x2)ex?y. ?x?x41．函数f(x,y)?2x2?xy?2y2的驻点为________.

??f?4x?y?0??x?解：??(x,y)?(0,0).

?f??x?4y?0???y.

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42．区域D为x2?y2?9，则??x2yd??______.

D解：利用对称性知其值为0或??x2yd???d??r4cos2?sin?dr?0.

D002?343.交换积分次序后,?10dx?xxf(x,y)dy?_____________.

解：积分区域D?(x,y)|0?x?1,x?y?x??(x,y)|0?y?1,y2?x?y?, 则有?dx?01x??xf(x,y)dy??dy?2f(x,y)dx.

0y1y144.y??xe?x是y???2y??3y?e?x的特解，则该方程的通解为_________.

4解：y???2y??3y?0的通解为y?C1e3x?C2e?x，根据方程解的结构,原方程的通解为

1y?C1e3x?C2e?x?xe?x.

445.已知级数?un的部分和Sn?n3，则当n?2时，un?_______.

n?1?解：当n?2时,un?Sn?Sn?1?n3?(n?1)3?3n2?3n?1.

三、计算题（每小题5分，共40分）

1??146．求lim??x?.

x?0xe?1??1?ex?1?xex?1?x?1?lim 解：lim??x??lim 2x?0xx?0x(ex?1)x?0e?1x??ex?1x1?lim?. ?limx?02xx?02x247.设y?y(x)是由方程exy?ylnx?sin2x确定的隐函数，求解：方程两边对x求导得

dy. dxexy(xy)??y?y?lnx?2cos2x x 即 exyx(y?xy?)?y?y?xlnx?2xcos2x (x2exy?xlnx)y??2xcos2x?exyxy?y

.

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2xcos2x?exyxy?ydy所以 ?y??.

x2exy?xlnxdx 48.已知?xf(x)dx?e?2x?C，求?1dx. f(x) 解：方程?xf(x)dx?e?2x?C两边对x求导得 xf(x)??2e?2x?2e?2x，即f(x)?，

x所以

11??xe2x. f(x)2 故?111dx???xe2xdx???xde2x f(x)241111 ??xe2x??e2xdx??xe2x?e2x?C.

444849.求定积分?|x(x?1)|dx.

?44 解：?|x(x?1)|dx??|x(x?1)|dx??|x(x?1)|dx??|x(x?1)|dx

?4?40014014 ??x(x?1)dx??x(1?x)dx??x(x?1)dx

?40114?x?xx?x??xx? ????????????

?32??4?23?0?32?1320231324?64116411?8????8???43. 323332 50.已知z?e解：因

x2?xy?y2 求全微分dz.

2222?z?ex?xy?y(x2?xy?y2)?x?ex?xy?y(2x?y), ?x2222?z?ex?xy?y(x2?xy?y2)?y?ex?xy?y(x?2y), ?y且它们在定义域都连续，从而函数z?exdz?2?xy?y2可微，并有

22?z?zdx?dy?ex?xy?y[(2x?y)dx?(x?2y)dy]. ?x?y51.求??(2x?y)d?，其中区域D由直线y?x,y?2x,y?2围成.

Dy .

y?2x?x?y 22

y?x?x?y

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解：积分区域D如图所示： 把D看作Y型区域，且有

y??D??(x,y)|0?y?2,?x?y?

2??故有??(2x?y)d???dy?y(2x?y)dx

D022y22y??(x?xy)ydy??022052510ydy?y3?. 412032252.求微分方程y??2xy?xe?x的通解. 解：这是一阶线性非齐次微分方程，

它对应的齐次微分方程y??2xy?0的通解为y?Cex， 设原方程的解为y?C(x)ex代入方程得C?(x)ex?xe?x, 即有 C?(x)?xe?2x， 所以 C(x)??xe?2xdx??2222221?2x21?2x22ed(?2x)??e?C, ?44212 故原方程的通解为y??e?x?Cex.

453.求幂级数?n2nx的收敛区间（考虑区间端点）. nn?12??解：这是标准缺项的幂级数，考察正项级数?n?1n2nx， 2nun?1n?12nx22 因l?lim?xlimn?1??,

n??un??2n2n?x2n?1，即|x|?2时，级数?nx2n是绝对收敛的； 当l?2n?12?x2n?1，即|x|?2时，级数?nx2n是发散的； 当l?2n?12??x2n2n?1，即x??2时，级数?nx化为?n，显然是发散的。 当l?2n?12n?1.