解:有
?x?x?P?Q??x2?此积分与路径无关,取直线段?,x从2?变到0,则 ?y?xy?0?x0013xxxx0?L(xy?3xe)dx?(3x?ysiny)dy??2?3xedx?3?2?xde?3(xe?e)2?
2?3[e2?(1?2?)?1],应选C.
28.以通解为y?Ce（C为任意常数）的微分方程为 ( ) A.y??y?0 B. y??y?0 C. y?y?1 D. y?y??1?0 解:y?Ce?y??Ce?y??y?0,应选B. 29. 微分方程y???y??xeA .x(ax?b)e?x?xxxx的特解形式应设为y? （ ）
?x? B.ax?b C.(ax?b)e D.x2(ax?b)e?x?x
解:-1是单特征方程的根,x是一次多项式,应设y??x(ax?b)e,应选A.
30．下列四个级数中，发散的级数是 （ ）
???12n?3n1A. ? B.? C. ?n D.?2
n?1n!n?11000nn?12n?1n?解:级数
2n?32n?31的一般项的极限为?0,是发散的,应选B. ?1000n1000n500n?1
?得分 评卷人 二、填空题（每题2分，共30分）
31．limf(x)?A的____________条件是lim?f(x)?lim?f(x)?A.
x?x0x?x0x?x0解：显然为充要(充分且必要).
32. 函数y?x?sinx在区间(0,2?)单调，其曲线在区间?0,
?
???
?内的凹凸性为 的. 2???
π?
?内大于零,应为凹的. 2?
解：y??1?cosx?0?在(0,2?)内单调增加,y???sinx在?0,
222 33.设方程3x?2y?z?a(a为常数)所确定的隐函数z?f(x,y) ,则
?z?_____. ?x解：F?3x?2y?z?a?Fz??2z,Fx??6x?222Fx??z3x????. ?xFz?z34.
?1?dxx? .
x?t解:
?1?dxx??2tdt1???21?dt?2t?2ln(1?t)?C ??1?t??1?t???2x?2ln(1?x)?C.
35.
xdx?________. ????1?cosx3??3xx????解：函数在区间??,?是奇函数，所以?3?dx?0.
??1?cosx1?cosx?33?336. 在空间直角坐标系中,以A(0，?4，1)，B(?1，?3，1)，C(2，?4，0)为顶点的?ABC的面积为__ .
???ijk解：AB?{?1,1,0},AC?{2,0,?1}?AB?AC??110?{?1,?1,?2}，所以?ABC的面积为
20?116AB?AC?. 22?x2y2?1??37. 方程?9在空间直角坐标下的图形为__________. 4?x??2?解:是椭圆柱面与平面x??2的交线,为两条平行直线. 38.函数f(x,y)?x?y?3xy的驻点为 .
33??z2?3x?3y?0???x解: ??(0,0),(1,1).
?z??3y2?3x?0???y39.若z?xy?e21?xxy3?2tany?z，则x?x? .
(1,0)解:f(x,0)?0??40?4x?z?z?0??x?x?0.
(1,0)40.
?dx?cosydy?___________ y解:
??40dx??4x?y112444?. cosydy??dy?cosydx??cosydy?sinx0000yy2??41.直角坐标系下的二重积分??f(x,y)dxdy(其中D为环域1?x2?y2?9)化为极坐标形式为
D___________________________.
解：
??f(x,y)dxdy?2?3D?0d??1f(rcos?,rsin?)rdr.
42.以y?C?3x1e?C3x2xe?为通解的二阶常系数线性齐次微分方程为 . 解:由y?C?3x1e?C3x2xe?为通解知,有二重特征根-3,从而p?6,q?9,微分方程为y???6y??9y?0.
43.等比级数
??aqn(a?0),当_______时级数收敛,当_______时级数发散.
n?0?解: 级数
?aqn是等比级数, 当|q|?1时,级数收敛,当|q|?1时,级数发散.
n?044.函数f(x)?1x2?x?2展开为x的幂级数为__________________ 解:f(x)?1x2?x?2??1?11?11113??1?x?2?x????3?1?x?6?1?x 2n???13??(?1)nxn?1??x??(?1)n?11?nn02?n?0??3?3?2n?1?x,(?1?x?n?06n??1). ?n45.???n?2?n?1?n??的敛散性为________的级数.
n?n2?(?2) 解:limu?n?2??2n??n?limn????n???limn????1??n???e?2?0,级数发散.
三、计算题（每小题5分，共40分）
x2?5?x2246．求lim?2?.
x??????x2?3??得分 ?x2?2??解：lim?x???x2?3???x2?52??1??lim?x????1??522x23x2??????x2?52?lim2???1?2?x??3??1??2??x?x222????1?2?x??52x??x23??(?)323????1?2??x?52
?2??lim?1?2?x??x??3??lim?1?2?x???x?x222????1?2?x??x23??(?)323????1?2??x?.
52?ee?32?e.
5247. 求limx4x?0?x20t31?t2dt解：limx4x?0?x2???lim004x3x230t31?t2dtx?01?x?2x4?lim21?x4x?03?2.
48.已知y?lnsin(1?2x),求
dy. dx解：
dy1?sin(1?2x)???cos(1?2x)?1?2x????2cos(1?2x) ?dxsin(1?2x)sin(1?2x)sin(1?2x)??2cot(1?2x).
49. 计算不定积分xarctanxdx.
??x2?x2x21??arctanx??dx 解：?xarctanxdx??arctanxd?2??2?221?x??x21?1?arctanx???1?22?1?x2??dx ?x211?arctanx?x?arctanx?C. 22250.求函数z?ecos(x?y)的全微分. 解：利用微分的不变性,
xdz?d[excos(x?y)]?exdcos(x?y)?cos(x?y)dex ??exsin(x?y)d(x?y)?cos(x?y)exdx ??exsin(x?y)[dx?dy]?cos(x?y)exdx ?ex[cos(x?y)?sin(x?y)]dx?exsin(x?y)dy.
51．计算
??Dxd?，其中D是由y?2,y?x,xy?1所围成的闭区域. 2y解：积分区域D如图所示:把区域看作 Y型,则有
?1D??(x,y)|1?y?2,?x?y?故
?y?, ?2 yy ??D2yxxdxdy?dydx 122??1yyyy?x?x?y
??21y211x2dy?1xdx??2dy?21y2yy
1y21 o 1
xy?1?x?x 1 y12?1?1?1?17????1?4?dy??y??. 3?21?y?2?483y??152．求微分方程y??ycosx?e?sinx满足初始条件y(0)??1的特解.
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??ycosx?0的通解为
y?Ce?sinx,设y?C(x)e?sinx是原方程解,代入方程有C?(x)e?sinx?e?sinx,
即有C?(x)?1,所以C(x)?x?C,故原方程的通解为y?Ce?sinx?xe?sinx,
?sinx把初始条件y(0)??1代入得:C??1,故所求的特解为y?(x?1)e.
3nn53．求级数?x的收敛半径及收敛区间(考虑区间端点).
n?0n?1?解：这是标准的不缺项的幂级数,收敛半径R?
1, ?
an?13n?1n?1n?1?lim?n?3lim?3, 而??limn??an??n?2n??n?23n故收敛半径R?1. 3?11当x?时,级数化为?,这是调和级数,发散的;
3n?0n?1?1(?1)n当x??时,级数化为?,这是交错级数,满足莱布尼兹定理的条件,收敛的;
3n?0n?1所以级数的收敛域为??
得分 评卷人 ?11?,?. ?33?
河南专升本高等数学英语考试知识点归类及历年试题
