2018年浙江专升本高等数学真题版(最新整理)

2018年成人高等学校专升本招生全国统一考试高等数学（二）（模拟试题）答案必须答在答题卡上指定的位置，答在试卷上无效。

．．．．．．．（共三套及参考答案）第Ⅰ卷（选择题，共40分）一、选择题：1～10小题，每小题4分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的，把所选项前的字母填在题后的括号内．1．当x→2时，下列函数中不是无穷小量的是（）．A．B．C．D．2．A．-3B．一1C．0D．不存在3．A．B．C．D．4．A．B．C．D．5．A．0B．2x3C．6x2D．3x26．设ƒ(x)的一个原函数为Inx，则ƒ(x)等于（）．A．B．C．D．7．A．y=x+1B．y=x-1C．D．8．A．0B．e一1C．2(e-1)D．9．A.y4cos(xy2)B.- y4cos(xy2)C.y4sin(xy2)D.- y4sin(xy2)10．设100件产品中有次品4件，从中任取5件的不可能事件是（）．A．“5件都是正品”B．“5件都是次品”C．“至少有1件是次品”D．“至少有1件是正品”第Ⅱ部分（非选择题，共110分）二、填空题：11～20小题，每小题4分，共40分．把答案填在题中横线上．11．12．13．14．15．16．17．18．19．20．三、解答题：21～28题，共70分．解答应写出推理、演算步骤．21．22．23．24．25．(本题满分8分)设事件A与B相互独立，且P(A)=0．6，P(B)=0．7，求P(A+B). 26．27．28．(本题满分10分)求由曲线y=2-x2，)，=2x-1及X≥0围成的平面图形的面积S以及此平面图形绕X轴旋转一周所得旋转体的体积Vx．模拟试题参考答案一、选择题1．【答案】应选C．2．【答案】应选D．【解析】本题考查的知识点是分段函数在分段点处的极限计算．分段点处的极限一定要分别计算其左、右极限后，再进行判定．3．【答案】应选A．【提示】本题考查的知识点是基本初等函数的导数公式．只需注意e3是常数即可．4．【答案】应选D．5．【答案】应选C．【解析】本题考查的知识点是函数在任意一点x的导数定义．注意导数定义的结构式为6．【答案】应选A．【提示】本题考查的知识点是原函数的概念，因此有所以选A．7．【答案】应选B．【解析】本题考查的知识点是：函数y=ƒ(x)在点(x，ƒ(x))处导数的几何意义是表示该函数对应曲线过点(x,ƒ(x)))的切线的斜率．由可知，切线过点(1，0)，则切线方程为y=x-1，所以选B．8．【答案】应选C．【解析】本题考查的知识点是奇、偶函数在对称区间上的定积分计算．注意到被积函数是偶函数的特性，可知所以选C．9．【答案】应选D．【提示】z对x求偏导时应将y视为常数，则有所以选D．10．【答案】应选B．【解析】本题考查的知识点是不可能事件的概念．不可能事件是指在一次试验中不可能发生的事件．由于只有4件次品，一次取出5件都是次品是根本不可能的，所以选B．二、填空题11．【答案】应填2．12．13．【答案】应填一2sin 2x．【提示】用复合函数求导公式计算即可．14．【答案】应填4．15．【答案】应填1．16．【提示】凑微分后用积分公式．17．【答案】应填2In 2．【解析】本题考查的知识点是定积分的换元积分法．换元时，积分的上、下限一定要一起换．18．19．【答案】20．【答案】应填0．【解析】本题考查的知识点是二元函数的二阶混合偏导数的求法．三、解答题21．【解析】型不定式极限的一般求法是提取分子与分母中的最高次因子，也可用洛必达法则求解．解法１解法2洛必达法则．22．本题考查的知识点是函数乘积的导数计算．23．本题考查的知识点是凑微分积分法．24．本题考查的知识点是定积分的凑微分法和分部积分法．【解析】本题的关键是用凑微分法将ƒ(x)dx写成udυ的形式，然后再分部积分．25．本题考查事件相互独立的概念及加法公式．【解析】若事件A与B相互独立，则P(AB)=P(A)P(B)．P(A+B)=P(A)+P(B)-p(AB)=P(A)+P(B)-p(A)P(日)=0．6+0．7-0．6×0．7=0．88．26．本题考查的知识点是利用导数的图像来判定函数的单调区间和极值点，并以此确定函数的表达式．编者希望通过本题达到培养考生数形结合的能力．【解析】(1)(2)因为由上面三式解得α=2，b=-9，c=12．27．本题考查的知识点是二元隐函数全微分的求法．利用公式法求导的关键是需构造辅助函数然后将等式两边分别对x(或y或z)求导．读者一定要注意：对x求导时，y，z均视为常数，而对y或z求导时，另外两个变量同样也视为常数．也即用公式法时，辅助函数F(x，y，z)中的三个变量均视为自变量．求全微分的第三种解法是直接对等式两边求微分，最后解出出，这种方法也十分简捷有效，建议考生能熟练掌握．解法1等式两边对x求导得解法2解法328．本题考查的知识点有平面图形面积的计算及旋转体体积的计算．【解析】本题的难点是根据所给的已知曲线画出封闭的平面图形，然后再求其面积S．求面积的关键是确定对x积分还是对Y积分．确定平面图形的最简单方法是：题中给的曲线是三条，则该平面图形的边界也必须是三条，多一条或少一条都不是题中所要求的．确定对x积分还是对y积分的一般原则是：尽可能用一个定积分而不是几个定积分之和来表示．本题如改为对y积分，则有计算量显然比对x积分的计算量要大，所以选择积分变量的次序是能否快而准地求出积分的关键．在求旋转体的体积时，一定要注意题目中的旋转轴是戈轴还是y轴．由于本题在x轴下面的图形绕x轴旋转成的体积与x轴上面的图形绕x轴旋转的旋转体的体积重合了，所以只要计算x轴上面的图形绕戈轴旋转的旋转体体积即可．如果将旋转体的体积写成上面的这种错误是考生比较容易出现的，所以审题时一定要注意．解由已知曲线画出平面图形为如图2—1—2所示的阴影区域．2018年成人高等学校专升本招生全国统一考试高等数学（二）。

绝密★启用前2018年普通高等学校招生全国统一考试（浙江卷）数 学本试题卷分选择题和非选择题两部分。

全卷共4页，选择题部分1至2页；非选择题部分3至4页。

满分150分。

考试用时120分钟。

考生注意：1．答题前，请务必将自己的姓名、准考证号用黑色字迹的签字笔或钢笔分别填在试题卷和答题纸规定的位置上。

2．答题时，请按照答题纸上“注意事项”的要求，在答题纸相应的位置上规范作答，在本试题卷上的作答一律无效。

参考公式：互斥，则 相互独立，则分别表示台体的上、下底面积，台体的高柱体的体积公式表示柱体的底面积，表示柱体的高锥体的体积公式表示锥体的底面积，表示锥体的高球的体积公式其中表示球的半径选择题部分（共40分）一、选择题：本大题共10小题，每小题4分，共40分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1. 已知全集U ={1，2，3，4，5}，A ={1，3}，则A. B. {1，3} C. {2，4，5} D. {1，2，3，4，5}【答案】C【解析】分析:根据补集的定义可得结果.详解：因为全集，，所以根据补集的定义得,故选C.点睛：若集合的元素已知，则求集合的交集、并集、补集时，可根据交集、并集、补集的定义求解．2. 双曲线的焦点坐标是A. (−，0)，(，0)B. (−2，0)，(2，0)C. (0，−)，(0，)D. (0，−2)，(0，2)【答案】B【解析】分析:根据双曲线方程确定焦点位置，再根据求焦点坐标.详解：因为双曲线方程为，所以焦点坐标可设为，因为，所以焦点坐标为，选B.点睛：由双曲线方程可得焦点坐标为，顶点坐标为，渐近线方程为.3. 某几何体的三视图如图所示（单位：cm），则该几何体的体积（单位：cm3）是A. 2B. 4C. 6D. 8【答案】C【解析】分析:先还原几何体为一直四棱柱，再根据柱体体积公式求结果.详解：根据三视图可得几何体为一个直四棱柱，高为2，底面为直角梯形，上下底分别为1，2，梯形的高为2，因此几何体的体积为选C.点睛：先由几何体的三视图还原几何体的形状,再在具体几何体中求体积或表面积等.4. 复数(i为虚数单位)的共轭复数是A. 1+iB. 1−iC. −1+iD. −1−i【答案】B【解析】分析:先分母实数化化简复数，再根据共轭复数的定义确定结果.详解：，∴共轭复数为，选B.点睛：本题重点考查复数的基本运算和复数的概念，属于基本题.首先对于复数的四则运算，要切实掌握其运算技巧和常规思路，如. 其次要熟悉复数的相关基本概念，如复数的实部为、虚部为、模为、对应点为、共轭复数为.5. 函数y=sin2x的图象可能是A. B.C. D.【答案】D【解析】分析:先研究函数的奇偶性，再研究函数在上的符号，即可判断选择.详解：令，因为，所以为奇函数，排除选项A,B;因为时，，所以排除选项C，选D.点睛：有关函数图象的识别问题的常见题型及解题思路：（1）由函数的定义域，判断图象的左、右位置，由函数的值域，判断图象的上、下位置；（2）由函数的单调性，判断图象的变化趋势；（3）由函数的奇偶性，判断图象的对称性；（4）由函数的周期性，判断图象的循环往复．6. 已知平面α，直线m，n满足mα，nα，则“m∥n”是“m∥α”的A. 充分不必要条件B. 必要不充分条件C. 充分必要条件D. 既不充分也不必要条件【答案】A【解析】分析:根据线面平行的判定定理得充分性成立，而必要性显然不成立.详解：因为，所以根据线面平行的判定定理得.由不能得出与内任一直线平行，所以是的充分不必要条件，故选A.点睛：充分、必要条件的三种判断方法：（1）定义法：直接判断“若则”、“若则”的真假．并注意和图示相结合，例如“⇒”为真，则是的充分条件．（2）等价法：利用⇒与非⇒非，⇒与非⇒非，⇔与非⇔非的等价关系，对于条件或结论是否定式的命题，一般运用等价法．（3）集合法：若⊆，则是的充分条件或是的必要条件；若＝，则是的充要条件．7. 设0<p<1，随机变量ξ的分布列是则当p在（0，1）内增大时，A. D（ξ）减小B. D（ξ）增大C. D（ξ）先减小后增大D. D（ξ）先增大后减小【答案】D【解析】分析:先求数学期望，再求方差，最后根据方差函数确定单调性.详解：，，，∴先增后减,因此选D.点睛：8. 已知四棱锥S−ABCD的底面是正方形，侧棱长均相等，E是线段AB上的点（不含端点），设SE与BC所成的角为θ1，SE与平面ABCD所成的角为θ2，二面角S−AB−C的平面角为θ3，则A. θ1≤θ2≤θ3B. θ3≤θ2≤θ1C. θ1≤θ3≤θ2D. θ2≤θ3≤θ1【答案】D【解析】分析:分别作出线线角、线面角以及二面角，再构造直角三角形，根据边的大小关系确定角的大小关系.详解：设O为正方形ABCD的中心，M为AB中点，过E作BC的平行线EF，交CD于F，过O作ON垂直EF于N，连接SO，SN，OM，则SO垂直于底面ABCD，OM垂直于AB，因此从而因为，所以即，选D.点睛：线线角找平行，线面角找垂直，面面角找垂面.9. 已知a，b，e是平面向量，e是单位向量．若非零向量a与e的夹角为，向量b满足b2−4e·b+3=0，则|a−b|的最小值是A. −1B. +1C. 2D. 2−【答案】A【解析】分析:先确定向量所表示的点的轨迹，一个为直线，一个为圆，再根据直线与圆的位置关系求最小值.详解：设，则由得，由得因此的最小值为圆心到直线的距离减去半径1，为选A.学|科|网...学|科|网...学|科|网...学|科|网...学|科|网...学|科|网...学|科|网...学|科|网...学|科|网...10. 已知成等比数列，且．若，则A. B. C. D.【答案】B【解析】分析:先证不等式，再确定公比的取值范围，进而作出判断.详解：令则，令得，所以当时，，当时，，因此，若公比，则，不合题意；若公比，则但，即，不合题意；因此，，选B.点睛：构造函数对不等式进行放缩，进而限制参数取值范围，是一个有效方法.如非选择题部分（共110分）二、填空题：本大题共7小题，多空题每题6分，单空题每题4分，共36分。

2018年浙江专升本高数考试真题答案1、选择题：本大题共5小题，每小题4分，共20分。

1、设，则在内（ C ）⎪⎩⎪⎨⎧≤>=00,,sin )(x x xx x x f )(x f )1,1(-A 、有可去间断点B 、连续点C 、有跳跃间断点D 、有第二间断点解析：1sin lim )(lim ,0lim )(lim 0====++--→→→→xxx f x x f x x x x ，但是又存在，是跳跃间断点)(lim )(lim 00x f x f x x +-→→≠ 0=∴x 2、当时，是的（ D ）无穷小0→x x x x cos sin -2x A 、低阶B 、等阶C 、同阶D 、高阶解析：高阶无穷小02sin lim 2sin cos cos lim cos sin lim0020==+-=-→→→xx x x x x x x x x x x x ⇒3、设二阶可导，在处，，则在处（ )(x f 0x x =0)(0<''x f 0)(lim 0=-→x x x f x x )(x f 0x x =B ）A 、取得极小值B 、取得极大值C 、不是极值D 、是拐点())(0,0x f x 解析：，则其，0000)()(lim)(,0)(lim00x x x f x f x f x x x f x x x x --='∴=-→→ 0)(,0)(00=='x f x f 为驻点，又是极大值点。

0x 000)(x x x f =∴<'' 4、已知在上连续，则下列说法不正确的是（ B ）)(x f []b a ,A 、已知，则在上，⎰=badx x f 0)(2[]b a ,0)(=x f B 、，其中⎰-=xxx f x f dt t f dx d 2)()2()([]b a x x ,2,∈C 、，则内有使得0)()(<⋅b f a f ()b a ,ξ0)(=ξf D 、在上有最大值和最小值，则)(x f y =[]b a ,M m ⎰-≤≤-b aa b M dx x f a b m )()()(解析：A.由定积分几何意义可知，，为在上与轴围成0)(2≥x f dx x f ba)(2⎰)(2x f []b a ,x 的面积，该面积为0，事实上若满足⇒0)(2=x f )(x f )(0)(0)(b x a x f dx x f ba≤≤=⇒⎪⎩⎪⎨⎧=⎰非负连续B.)()2(2)(2x f x f dx x f dxd xx -=⎰C.有零点定理知结论正确D.由积分估值定理可知，，，()b a x ,∈M x f m ≤≤)(则)()()()(a b M dx x f a b m Mdx dx x f mdx babab ab a-≤≤-⇒≤≤⎰⎰⎰⎰5、下列级数绝对收敛的是（ C ）A 、B 、C 、D 、∑∞=-+-111)1(n n n ∑∞=-+-11)1ln()1(n n n ∑∞=+139cos n n n ∑∞=11n n解析：A.，由发散发散1111lim =+∞→nn n ∑∞=11n n 11+⇒n B.，由发散发散011lim )1ln(lim )1ln(11lim =+=+=+∞→∞→∞→n n n n n n n n ∑∞=11n n ∑∞=+⇒1)1ln(1n n C.，而=1，由收敛收敛919cos 22+≤+n n n232191lim n n n +∞→∑∞=1231n n ⇒912+n ⇒收敛9cos 2+n nD.发散∑∞=11n n 2、填空题6、axx ex a =+→10)sin 1(lim解析：axa x a xx a x a xx x x e eeex a x x ====+⋅+++→→→→1cos sin 11lim )sin 1ln(lim )sin 1ln(11000lim )sin 1(lim 7、，则3sin )23()3(lim=--→xx f f x 23)3(='f 解析：3)3(22)3()23(lim 2sin )23()3(lim00='=---=--→→f xf x f x x f f x x 8、若常数使得，则b a ,5)(cos sin lim 20=--→b x a e xxx 9-=b 解析：5)(cos lim )(cos sin lim 2020=--=--→→ae b x x b x a e x x x xx 所以根据洛必达法则可知：1,01==-a a 212cos lim 2)(cos lim00bb x x b x x x x -=-=-→→9,521-==-b b9、设，则⎩⎨⎧-=+=tt y t x arctan )1ln(11==t dx dy 解析：，2221)1(11111t t t tt dtdxdt dydx dy++=++-=11==t dx dy 10、是所确定的隐函数，则)(x f y =0122=--y x 32222y x y dx y d -=解析：方程两边同时求导，得：，，022='-y y x yx y ='方程同时求导，得：，将带入，022='-y y x 0)(12=''-'-y y y yxy ='则得，，0)(12=''--y y y x 32232221y x y y x y y dx y d -=-=''=11、求的单增区间是21xxy +=)1,1(-解析：2222222)1(1)1(21x x x x x y +-=+-+='令，则，0>'y 12<x 11<<-x12、求已知，则 ⎰+=C e dx x f x 2)(=⋅∑==∞→)(1lim 10n kf nn k n 1-e 解析：1)()()()(1lim10101012-=+===⋅⎰⎰∑==∞→e C e dx x f dx x f n k f n x n k n 13、=⎰+∞dx x x e2)(ln 11解析：1ln 1ln )(ln 1)(ln 122=-==∞++∞+∞⎰⎰ee exx d x dx x x 14、由：围成的图形面积为2x y =2,1==x y 34解析：34)31()1(212132=-=-=⎰x x dx x A 15、常系数齐次线性微分方程的通解为（为任意02=+'-''y y y xe x C C y )(21+=21C C 常数）解析：特征方程：，特征根：0122=+-r r 121==r r 通解为（为任意常数）xe x C C y )(21+=21C C 三、计算题 （本大题共8小题，其中16-19小题每小题7分，20-23小题每小题8分，共60分）16、求)sin 1ln(lim0x e e xx x +--→解析：22lim sin 2lim )sin 1ln(1lim )sin 1ln(lim 00200===+-=+-→→-→-→xx x x x e e x e e x x x xx x x x 17、设，求在处的微分xx x y )sin 1()(+=)(x y π=x 解析：xx x y )sin 1()(+=)sin 1ln(ln x x y +=xx x x y sin 1cos )sin 1ln(y 1+++='dxx xxxx x )sin 1](sin 1cos )sin 1[ln(dy ++++=将代入上式，得微分π=x dx dy π-=18、求⎰-π502cos 1dxx 解析：⎰-π502cos 1dx x ⎰=π50|sin |dxx ⎰⎰⎰⎰⎰+-++-+=ππππππππ43542320sin )sin sin )sin sin xdxdx x xdx dx x xdx （（π10|cos |cos |cos |cos |cos 54433220=-+-+-=πππππππππx x x x x 19、求dxx ⎰arctan 解析：，2t x t x ==，则令tdtdx 2=⎰2tan arc tdt td t t t tan arc tan arc 22⎰-=dt t t t t 22211tan arc +-=⎰⎰+-+-=dtt t t t 222111tan arc ⎰+--=dt tt t （22111tan arcct t t t ++-=tan arc tan arc 2cx x x x ++-=tan arc tan arc 则则则20、dx x xx xx ⎰++-11-41cos 45（解析：为奇函数，41cos x xx +该式不代入计算∴45452t x x t -=-=，则令tdtdx 21-=dtt t t )21(145132--=⎰该式⎰-=312)581dt t （61|)31581313=-=t t （21、已知在处可导，求⎩⎨⎧>+≤+=0),1ln(0,2)(x ax x b x x f 0=x ba ,解析：)(lim ,0)(lim )0()(lim )(lim 0)(0)(00=∴====∴=∴=-+-+→→→→b bx f x f f x f x f x x f x x f x x x x 处连续在处可导在)(lim )(lim 0x f x f x x '='-+→→ ax ax x f x x =--+='++→→0)1ln(lim )(lim 002002lim )(lim 00=--='--→→x x x f x x 2=∴a 22、求过点且平行于又与直线相交的直线方程。

2018年成人高等学校专升本招生全国统一考试高等数学（二）（模拟试题）答案必须答在答题卡上指定的位置，答在试卷上无效。

．．．．．．．（共三套及参考答案）第Ⅰ卷（选择题，共40分）一、选择题：1～10小题，每小题4分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的，把所选项前的字母填在题后的括号内．1．当x→2时，下列函数中不是无穷小量的是（）．A．B．C．D．2．A．-3B．一1C．0D．不存在3．A．B．C．D．4．A．B．C．D．5．A．0B．2x3C．6x2D．3x26．设ƒ(x)的一个原函数为Inx，则ƒ(x)等于（）．A．B．C．D．7．A．y=x+1B．y=x-1C．D．8．A．0B．e一1C．2(e-1)D．9．A.y4cos(xy2)B.- y4cos(xy2)C.y4sin(xy2)D.- y4sin(xy2)10．设100件产品中有次品4件，从中任取5件的不可能事件是（）．A．“5件都是正品”B．“5件都是次品”C．“至少有1件是次品”D．“至少有1件是正品”第Ⅱ部分（非选择题，共110分）二、填空题：11～20小题，每小题4分，共40分．把答案填在题中横线上．11．12．13．14．15．16．17．18．19．20．三、解答题：21～28题，共70分．解答应写出推理、演算步骤．21．22．23．24．25．(本题满分8分)设事件A与B相互独立，且P(A)=0．6，P(B)=0．7，求P(A+B). 26．27．28．(本题满分10分)求由曲线y=2-x2，)，=2x-1及X≥0围成的平面图形的面积S以及此平面图形绕X轴旋转一周所得旋转体的体积Vx．模拟试题参考答案一、选择题1．【答案】应选C．2．【答案】应选D．【解析】本题考查的知识点是分段函数在分段点处的极限计算．分段点处的极限一定要分别计算其左、右极限后，再进行判定．3．【答案】应选A．【提示】本题考查的知识点是基本初等函数的导数公式．只需注意e3是常数即可．4．【答案】应选D．5．【答案】应选C．【解析】本题考查的知识点是函数在任意一点x的导数定义．注意导数定义的结构式为6．【答案】应选A．【提示】本题考查的知识点是原函数的概念，因此有所以选A．7．【答案】应选B．【解析】本题考查的知识点是：函数y=ƒ(x)在点(x，ƒ(x))处导数的几何意义是表示该函数对应曲线过点(x,ƒ(x)))的切线的斜率．由可知，切线过点(1，0)，则切线方程为y=x-1，所以选B．8．【答案】应选C．【解析】本题考查的知识点是奇、偶函数在对称区间上的定积分计算．注意到被积函数是偶函数的特性，可知所以选C．9．【答案】应选D．【提示】z对x求偏导时应将y视为常数，则有所以选D．10．【答案】应选B．【解析】本题考查的知识点是不可能事件的概念．不可能事件是指在一次试验中不可能发生的事件．由于只有4件次品，一次取出5件都是次品是根本不可能的，所以选B．二、填空题11．【答案】应填2．12．13．【答案】应填一2sin 2x．【提示】用复合函数求导公式计算即可．14．【答案】应填4．15．【答案】应填1．16．【提示】凑微分后用积分公式．17．【答案】应填2In 2．【解析】本题考查的知识点是定积分的换元积分法．换元时，积分的上、下限一定要一起换．18．19．【答案】20．【答案】应填0．【解析】本题考查的知识点是二元函数的二阶混合偏导数的求法．三、解答题21．【解析】型不定式极限的一般求法是提取分子与分母中的最高次因子，也可用洛必达法则求解．解法１解法2洛必达法则．22．本题考查的知识点是函数乘积的导数计算．23．本题考查的知识点是凑微分积分法．24．本题考查的知识点是定积分的凑微分法和分部积分法．【解析】本题的关键是用凑微分法将ƒ(x)dx写成udυ的形式，然后再分部积分．25．本题考查事件相互独立的概念及加法公式．【解析】若事件A与B相互独立，则P(AB)=P(A)P(B)．P(A+B)=P(A)+P(B)-p(AB)=P(A)+P(B)-p(A)P(日)=0．6+0．7-0．6×0．7=0．88．26．本题考查的知识点是利用导数的图像来判定函数的单调区间和极值点，并以此确定函数的表达式．编者希望通过本题达到培养考生数形结合的能力．【解析】(1)(2)因为由上面三式解得α=2，b=-9，c=12．27．本题考查的知识点是二元隐函数全微分的求法．利用公式法求导的关键是需构造辅助函数然后将等式两边分别对x(或y或z)求导．读者一定要注意：对x求导时，y，z均视为常数，而对y或z求导时，另外两个变量同样也视为常数．也即用公式法时，辅助函数F(x，y，z)中的三个变量均视为自变量．求全微分的第三种解法是直接对等式两边求微分，最后解出出，这种方法也十分简捷有效，建议考生能熟练掌握．解法1等式两边对x求导得解法2解法328．本题考查的知识点有平面图形面积的计算及旋转体体积的计算．【解析】本题的难点是根据所给的已知曲线画出封闭的平面图形，然后再求其面积S．求面积的关键是确定对x积分还是对Y积分．确定平面图形的最简单方法是：题中给的曲线是三条，则该平面图形的边界也必须是三条，多一条或少一条都不是题中所要求的．确定对x积分还是对y积分的一般原则是：尽可能用一个定积分而不是几个定积分之和来表示．本题如改为对y积分，则有计算量显然比对x积分的计算量要大，所以选择积分变量的次序是能否快而准地求出积分的关键．在求旋转体的体积时，一定要注意题目中的旋转轴是戈轴还是y轴．由于本题在x轴下面的图形绕x轴旋转成的体积与x轴上面的图形绕x轴旋转的旋转体的体积重合了，所以只要计算x轴上面的图形绕戈轴旋转的旋转体体积即可．如果将旋转体的体积写成上面的这种错误是考生比较容易出现的，所以审题时一定要注意．解由已知曲线画出平面图形为如图2—1—2所示的阴影区域．2018年成人高等学校专升本招生全国统一考试高等数学（二）。

浙江省2018年选拔优秀高职高专毕业生进入本科学习统一考试英语请考生按规定用笔将所有试题的答案涂、写在答题纸上。

选择题部分注意事项：1.答题前,考生务必将自己的姓名、准考证号用黑色字迹的签字笔或钢笔填写在答题纸规定的位置上。

2.每小题选出答案后,用2B铅笔把答题纸上对应题目的答案标号涂黑。

如需改动,用橡皮擦干净后,再选涂其他答案标号。

不能答在试题卷上。

Part I Reading Comprehension(60marks,60minutes)Section A(50marks:2marks for each item)Format IDirections:There are4passages in this part.Each passage is followed by some questions or unfinished statements.For each of them there are four choices marked A,B,C,and D.You should decide on the best choice and mark the corresponding letter on AnswerSheet1.Passage OneQuestion1to5are based on the following passage:A U.N.report says water is in demand around the world as temperatures on Earth’s surface rise and demand grows along with populations.The report was released this week at the World Water Forum in Brasilia.The conference has been described as the world’s largest water-related event.Federal District Governor Rodrigo Rollemberg spoke at a panel discussion on Tuesday at the forum.He described water shortages as a worldwide problem.The public water supply has less water because of low rainfall as well as fast and disorderly growth in Brasilia,which is part of the Federal District,Rollemberg said.In January2016,after three years of little rain,district officials began limiting how much water people could use.The governments of the Federal District and the nearby state of Goias also gave$166million to develop water infrastructure(基础设施).Demand around the world is expected to increase by nearly one-third by2050.By then,5billion people could be left with poor access to water,the U.N.warned in its2018World Water Development Report.To avoid such a crisis,U.N.officials called for“nature-based solutions”that use or copy natural processes that should be used to increase water availability.They said solutions could include changing farming methods so fields keep more moisture and nutrients, collecting rainwater,and protecting wetlands.The officials also proposed reestablishing floodplains and said that plants could be grown on housetops.Such proposals will become more important as water industries grow.1.Who talked about water shortages in Brasilia at a panel discussion?A.A journalist.B.A U.N.official.C.A Federal District official.D.An official from the state of Goias.2.What is among the reasons for less public water supply in Brasilia?A.Waste of rainwater.B.Insufficient rainfall.C.Growing plants on the roofs.D.Development in infrastructure.3.What did the Federal District government do to deal with water shortages?A.They controlled population.B.They limited water consumption.C.They invested in artificial rainmaking.D.They suggested“nature-based solutions”.4.What is NOT a“nature-based solution”?A.Gathering rainwater.B.Preserving wetlands.C.Sustaining water in soil.D.Building houses on floodplains.5.What is the main idea of this passage?A.Brasilia is facing the problem of water shortages.B.The world has found a solution to water shortages.C.The world is facing the problem of water shortages.D.A report was released at the World Water Forum in Brasilia.Passage TwoQuestions6to10are based on the following passage:The Queen has only given one interview while a sovereign(君主).Her true personality remains elusive to the public.However,one thing is clear:Queen Elizabeth is an introvert. Introverts prefer quiet environments and feel energized from spending time alone or with people they know well.They aren’t necessarily shy,but they do tend to be more reserved and guarded with strangers.Fellow introverts will appreciate how draining an average day for Her Majesty must be.Fortunately,the Queen has some tricks to make public engagements easier.Just like uscommoners rely on friends to save us from awkward situations or boring conversations,she subtly signals to her staff when she wants an intervention.If she places her handbag on the table,it indicates that she wants the event to end in the next five minutes.For more urgent situations, putting her bag on the floor shows that she wants her lady-in-waiting to rescue her immediately.Most comfortable in the countryside,the Queen has a well-known passion for horses and dogs.There is a scene in the hit Netflix show The Crown which reveals a fascinating side of the monarch.While giving Jackie Kennedy a personal tour of Buckingham Palace,the two women find themselves bonding over their respective positions as introverted women who have found themselves in the limelight.Surrounded by the Queen’s beloved corgis,both women admit they are happiest around animals and that their extroverted sisters would have been much better suited to their public roles.6.The underlined word“draining”(paragraph1)is the closest in meaning to“”.A.frustratingB.exhaustingC.demandingD.aspiring7.To be rescued immediately in a public engagement,the Queen would.A.end the event within five minutesB.place he handbag on the tableC.wave to her lady-in-waitingD.put her bag on the floor8.In the last paragraph,the underlined expression“the two women find themselves bonding over their respective positions”means that the two ladies.A.admire each otherB.help each other in publicC.grow closer to each otherD.show respect for each other9.Queen Elizabeth and Jackie Kennedy share the following traits or facts EXCEPT that.A.both have sistersB.both love animalsC.both enjoy being in publicD.both are introverted women10.The author’s attitude can be described as.A.neutralB.criticalC.pessimisticD.subjective Passage ThreeQuestions11to15are based on the following passage:Back when Disney products were to be watched,not experienced,Walt Disney had a vision: a theme park that felt like stepping into a magical new place.Now,we can finally witness Walt’s first dream.Before construction could start,Walt needed funding.He and his friend Herb Ryman put their heads together and mapped out their dream for the Disneyland that was still just a figment of their imaginations.In just one frenzied1953weekend,their idea came to life on paper—and you’ll notice both subtle differences from and remarkable similarities to the modern Anaheim, California,park.Like today,visitors would have entered the park into a Main Street USA-type area.The tunnels aren't as wide as they were imagined more than60years ago,but park-goers are still greeted with a train passing overhead as they make their way toward the magic.(Even with a modern Disneyland map,it's tough to find these secret Disney park spots you never knew existed.) Walt always hoped for a castle as a focal point of the park,but its real-life iteration changed from his initial dream.The first palace was a sprawling fortress,hiding a massive carousel(旋转木马)beyond its walls.Now,Sleeping Beauty Castle has more intricate spires,and King Arthur Carrousel takes its spins in Fantasyland,behind the palace.Walt’s vision for the Mark Twain steamboat circling around Tom Sawyer Island and its Pirate’s Lair(巢穴)hasn’t changed much,though at that point it was called Frontier Country instead of Frontierland.That wasn't the only name change,though.World of Tomorrow turned into Tomorrowland—and eventually lost its rocket.Astronauts walked on the moon14years after the park was built,making the Rocket to the Moon seem out-of-date.Meanwhile,True-Life Adventureland lost its first word,while most of the areas took on totally different themes.11.What was Walt Disney’s first dream?A.To build a theme parkB.To draw a map on paper.C.To travel in a magical place.D.To design a series of products.12.Before construction started,Walt Disney did the following things EXCEPT.A.getting fundingB.planning the parkC.drawing the park on paperD.daydreaming with a friend13.Which of the following park spots has not changed much since Walt’s vision came to life on paper?A.A Main Street area.B.World of Tomorrow.C.Sleeping Beauty Castle.D.True-Life Adventureland.14.How many name changes have been mentioned in this passage?A.One.B.TwoC.Three.D.Four.15.What is the theme of this passage?A.A tour in Disneyland.B.Walt Disney’s life story.C.The future of Disneyland.D.Walt Disney’s first dream.Passage ThreeQuestions16to20are based on the following passage:You’ve made it.You managed to navigate your way through the world’s largest airport without robot assistance;you breezed through security because you didn’t pack any items that get you flagged by the TSA;and you successfully boarded the plane.Now you’re comfortable in your seat in the(kind of incredibly germ-ridden)cabin,and there’s even a little bit of elbow room,asthe flight isn’t fully booked for once.With the hassle behind you,you settle in with your neck pillow,pop some Kenny G on your iPod,and get ready to spend quality time sleeping on a plane. And you’re about to make a huge mistake that will put your health at risk,as reported by Travel+ Leisure.It’s not because of the Kenny G.It’s the shuteye part.According to MedlinePlus,falling asleep during landing or takeoff could cause serious damage to your ears.It all has to do with the rapid changes in air pressure in the cabin.If you’re awake,a natural response to alleviate pressure on your eardrums during takeoff and landing is to'pop'them,to maintain a pressure equilibrium.If you’re sleeping on a plane,you can’t actively work to relax those muscles and release the tension,so you can become susceptible to dizziness,ear infections,eardrum damage,hearing loss and nose bleeds.‘A quick change in altitude affects the air pressure in the ear,’says Angel Chalmers,a British pharmacist,via Express.'This leads to a vacuum in the Eustachian tubes(咽鼓管)which makes the ears feel blocked and sound dull.'Keep those Eustachian tubes clear and keep those eyes open for at least another few minutes. Crack open that book you just bought in the terminal.After all,there are some major perks to buying your reading material at the airport.16.The underlined sentence“You’ve made it.”(Paragraph1)means that you have managed.A.to board the planeB.to get a bit of elbow roomC.to get ready to sleep after boardingD.to make yourself comfortable after boarding17.What would damage your health when the plane takes off or lands?ing a neck pillow.B.Shutting your eyes.C.Listening to music.D.Falling asleep.18.The underlined word“alleviate”(Paragraph2)is the closest in meaning to“”.A.increaseB.reduceC.maintainD.escape19.Having a vacuum in the Eustachian tubes during takeoff and landing could cause unpleasant symptoms to the following body parts EXCEPT.A.the throatB.the noseC.the earsD.the head20.The author suggests that you read a book during the plane taking off and landing.A.to amuse yourselfB.to stay awakeC.to keep your ears clearD.to make yourself comfortableFormat IIDirections:In the following passage,some sentences have been removed.For questions21to25, choose the most suitable sentence from choices A to G to fit into each of the numbered blanks. There are two extra choices,which do not fit into and of the blanks.You should decide on the bestchoice and blacken the corresponding letter on The Answer Sheet.(10points)Whether it’s a hotel stay,shopping or dining ont,a vacation to Paris can be a pricey proposition.But there are plenty of ways to stretch your dollar in the City of Lights and have a high-end getaway.Barkley Hickcox,a Paris specialist and an owner of Local Foreigner,a travel consultancy, has a few tips to help travelers make the most of their Parisian getaway.21..The ideal time to visit Paris is between November and April,when hotel prices are at their lowest.Some high-end properties offer even further discounts if you a stay several months in advance.22..Families traveling to Paris should consider staying in a high-end apartment instead of a hotel, Ms.Hickcox said.“An apartment is less expensive than booking multiple rooms at a five-star hotel,and since you have a kitchen,you can save on food costs by eating in occasionally,“she said.23..Paris is known for fine cuisine,but rather than dining at one of the city’s renowned restaurants at dinnertime,go for lunch,when the tab will be much lower and the cuisine just as good.“You’ll find lots of great eateries where you can have dinner with wine for35to50euros a person,”she said.24..Ms.Hickcox’s favorite way to explore Paris is on foot.“You can fully appreciate the city’s beauty by walking,and it’s designed for walkers because there are walking paths and sidewalks everywhere,”she said.For longer distances,she suggested using the city’s bike share system,Velib.A weeklong pass costs15euros,and there are stations everywhere.There is another choice if you are not so much fond of biking.25.,Paris’subway system.It’s convenient,and you can buy daily and weekly passes.A.Dine outB.Eat smartC.Try the MetroD.Stay in a hotelE.Ditch taxis and carsF.Travel at the right timeG.Choose the right accommodationsSection B Banked Cloze(每小题1分，共10分)Directions:In this section,there is a passage with ten blanks.You are required to select one word for each blank from a list of choices given in a word bank following the passage.Read the passage through carefully before making your choices.Please blacken the corresponding letter for each choice on The Answer Sheet.You may not use any of the words in the word bank more than once. (10points)I knew this woman who was afraid to tell people what she wanted.She didn’t know__26__ to say no.Instead,she got herself tangled in a web of obligations,anxiety,and white lies.She was __27__.__28__recently,I decided I’d had enough.As an experiment,I began__29__up for myself, even at the risk of alienating(疏远)everyone and having__30__entire life come crashing down around me.That didn’t happen.Here’s what did:I stopped making petty excuses.Saying“no”is so much easier.If someone asks me to do something I have zero interest __31__,I’m polite but honest.The phrase.”I’m sorry,I don’t think that’s really__32__me,“slips out my mouth faster than some lame excuse.My friends and family didn’t mind.__33__I stopped people-pleasing,no one cared.A good friend asked me to go for coffee at5PM.I was__34__to hit the gym.I said,“Sorry,I’ve got things I want to do tonight.”She said,“That’s fine.Maybe another time.”Learning how to say“no”has added several extra hours to my days,days__35__my weeks, and what feels like months to my years.Saying“no”has set me free.A.facingB.toC.herD.inE.forF.whatG.standingH.AndI.myJ.ButK.meL.WhenM.onN.planningO.howPart II Integrated Testing(30points,30minutes)Section A Cloze(每小题1分，共20分)Directions:There are20blanks in the following passage.For each blank there are four choices marked A,B,C and D.You should choose the ONE that best fits into the passage.Then blacken the corresponding letter on The Answer Sheet.(20points)Brazilian Zarela Mosquera moved to the United States as a teenager.The adjustment to a new place was__36__.Mosquera says she became a bratty(顽皮的)teenager,as a result.But it was not just the__37__that affected her behavior.Mosquera says there was another major stress in herlife at that time:She was the only member of her__38__who spoke English.“Being in a country with my family that doesn’t speak Spanish and my parents don’t really speak__39__,my parents were trying to give me all these responsibilities.”Along with Spanish and English,Mosquera also speaks Portuguese.But there was another universal language she learned to love as a child:art.Zarela Mosquera connected with drawing and painting while in school.__40__she mostly dismissed art as a path to a career.She says her Dad__41__always say,“Think about the __42__.”Mosquera did not think he would support the study of__43__once she went to college.But,to her__44__,it was her parents who suggested just that.They urged her to__45__ to technical and liberal arts schools.“One of them was Rhode Island School of Design,which was like my__46__choice.”RISD,as it is called,__47__Mosquera as well.She enrolled in industrial design.“It’s basically to design__48__and services.I could be doing something technical or something more__49__to problem solving.Whether it’s__50__out a better way to filter water or developing a type of specific shelter__51__refugees for example.”Mosquera says the course of study was__52__including metal working,woodworking and model making.She says in one class she just drew cubes for an entire month,which led her to a__53__:“Wow!Do I really want to do this?”But,she says,she__54__the cube study.And, then she began__55__on more interesting projects.Mosquera is now a design strategist for Marshall Moya Design,an architecture and interior design company in Washington,D.C.36.A.successful B.important C.necessary D.difficult37.A.age B.growth C.adjustment cation38.A.family B.peers C.friends D.class39.A.Spanish B.English C.Portuguese D.French40.A.So B.Moreover C.And D.But41.A.could B.should C.would D.might42.A.time B.past C.present D.futurenguage B.art C.technology D.design44.A.disappointment B.surprise C.annoyance D.dissatisfaction45.A.apply B.cater C.adapt D.move46.A.top B.moral C.natural D.popular47.A.failed B.rejected C.chose D.urged48.A.machines B.shelters C.filters D.products49.A.adjusted B.related C.affiliated D.exposed50.A.figuring B.wiping C.filling D.ruling51.A.with B.for C.on D.in52.A.easy B.pleasant C.tough D.smooth53.A.question B.problem C.door D.road54.A.undertook B.funded C.started D.survived55.A.putting B.relying C.working D.counting非选择题部分注意事项：用黑色字迹的签字笔或钢笔将答案写在答题纸上，不能答在试题卷上。

2018年成人高等学校招生全国统一考试专升本高等数学（二）本试卷分第Ⅰ卷（选择题）和第卷（非选择题）两部分，满分150分，考试时间120分.第Ⅰ卷(选择题，共40分)一、选择题(1~10小题，每小题4分，共40分。

在每小题给出的四个选项中，只有一项是符合题目要求的)1. lim x→0xcosx =( )A. eB.2C. 1D. 02. 若y =1+cosx ,则dy = ( )A. (1+ sinx)dxB. (1−sinx)dxC. sinxdxD.−sinxdx3. 若函数f(x)=5x ，则f′(x)= ( )A. 5x−1B. x5x−1C. 5x ln5D.5x4. 曲线y =x 3+2x 在点(1,3)处的法线方程是 ( )A. 5x +y −8=0B. 5x −y −2=0C. x +5y −16=0D. x −5y +14=05. ∫12−xdx =( )A. ln |2−x|+CB. −ln |2−x|+CC.−1(2−x)2+C D. 1(2−x )2+C6. ∫f′(2x)dx = ( )A. 12f(2x)+CB. f(2x)+CC. 2f(2x)+CD. 12f(x)+C7. 若f(x)为连续的奇函数，则∫f(x)1−1dx = ( )A. 0B. 2C. 2f(−1)D. 2f(1)8. 若二元函数z =x 2y +3x +2y ，则ðz ðx=( )A. 2xy +3+2yB. xy +3+2yC. 2xy +3D. xy +39. 设区域D ={(x ，y)|0≤y ≤x 2,0≤x ≤1}，则D 绕x 轴旋转一周所得旋转体的体积为 ( )A. π5B. π3C. π2D. π10. 设A ，B 为两个随机事件，且相互独立，P(A)=0.6，P(B)=0.4，则P(A −B )=( )A. 0.24B. 0.36C. 0.4D. 0.6第Ⅱ卷(非选择题，共110分)二、填空题(11~20小题，每小题4分，共40分)11. 曲线y =x 3−6x 2+3x +4的拐点为 . 12. lim x→0(1−3x )1x = .13.若函数f(x)=x −arctanx ，则f′(x)= . 14. 若y =e 2x 则dy = . 15. 设f(x)=x 2x ,则f′(x)= . 16. ∫(2x +3)dx = . 17. ∫(x 5+x 2)1−1dx = . 18. ∫sin x 2π0dx = . 19. ∫e−x +∞0dx = .20. 若二元函数：z =x 2y 2，则ð2z ðxðy= .三、解答题(21~28题，共70分。

D 有第二间断点2018年浙江专升本高数考试真题答案、选择题：本大题共 5小题，每小题4分，共20分。

1、设心打干“0，则f(x)在i) 内( C ) x 心0A 、有可去间断点B 连续点C 、有跳跃间断点sin x解析：lim f (x) = lim x = 0, lim f(x) = lim 1心3_ 心0_ x lim f (x) = lim f (x)，但是又存在，.x =0是跳跃间断点x _0 …x _0 '22、当x > 0时，sin x - xcosx 是x 的(D )无穷小3、设 f (x)二阶可导，在 x = x 0处 f "(x 0) ：：： 0，limf (x)= 0 ,则 f (x)在 x = x 0处(B )J^0 X _ x 0解析： lim-^^",. f (x 0) = limf (x)_f(x)，则其 f (x 0)= 0, f (x 0) = 0，T o X - X 0^x °X _ X 0X 。

为驻点，又 f (X0P ： 0 X =X0是极大值点。

4、已知f(x)在a,b 1上连续，则下列说法不正确的是( B) A 、已知 bf 2(x)dx=0，则在 a,b 上，f(x)=0ad 2xr i Bf (t)dt 二 f (2x) - f (x)，其中 x,2x /bldx xC f(a) f(b) <0，则 a,b 内有■使得 f 「)= 0D y = f(x)在'a, b 上有最大值 M 和最小值m ，则m(b - a) f (x)dx _ M (b -a)a解析：lim Sinx —X cosx=怙cosx—cosxxsinx側sinx一 高阶无穷小x 刃 xX —02xx —.02A 、低阶B 等阶C 、同阶D 高阶A 、取得极小值B 取得极大值C 不是极值D (X 0,f(x °))是拐点解析：A.由定积分几何意义可知，f2(x)—0, " f2(x)dx为f2(x)在a,b 1上与x轴围成'a的面积，该面积为0 = f2(x)=0 ，事实上若f (X)满足D. =1由J 1发散=n =1nln (V<as inx)limx1acosxlim 1 asinxx 0.1连续丿 非负 二f (x) = 0(a 兰x 兰b) b 」f(x)dx =0d 2xB. f (x)dx = 2f (2x)「f (x) dx xC. 有零点定理知结论正确由积分估值定理可知， x“a,b ， m^f(x)^M ，—-1)2 n 4 ln(n 1)1 —1发散 .一 n 1收敛二、填空题16、lim (1 asinx)xx _0--ln(1 七 sin x)解析：xm 0(1asi nx)^Hm 0e xf (3) — f (3—2x)门 一"、 37、lim 3，贝U f (3)=x 0sin x 2解析：lim f(询im 心口⑶=2f ⑶"X T ° sinx t -2xb mdx <5、 b b __ af(x)dx 乞 & Mdx 二 m(b - a) F 列级数绝对收敛的是b& f (x)dx M (b -a)B. n m 丄ln(1 n)二 lim ln(1 n)n )::= lim — n 匸1 n=0,由J 丄发散=oOz n=1发散ln(1 n) C. cosn<n 2_1_而 lim 9=1， n 匚1~3n ㊁旳1由、-3 n T "2n 21----- 收敛=n 29cosn n 2-解析：方程两边同时求导，得：2x —2yy = 0 ,则 x 2::: 1， 一1 ：x :1sin x8、若常数 a, b 使得 lim 农 (cosx -b)=5，则 b--9 xTe —a —解析：ljm 冬(cosx_b)岂m x (c o s x-b )=5 2x e -a所以根据洛必达法则可知： 1 _ a = 0, a = 1 x(cosx —b) lim x _01 -b _2- cosx 「b 1-b lim 2x x 刃 9、设 解析: 10、= 5,b = -9 x = l n(1+t) =t _arctant dy dfdt 1— i% 11 1 t2 t 2(1 t) dx1 t2 dy dx t4=12 2y 二f(x)是x -y -仁0所确定的隐函数,d 2y dx 22 2y —x3y方程2x -2yy J 0同时求导，得:1-(y)2-对'=0，将y =—带入，y则得, d 2yx \2-(j)-yy"，衣“2 2y -x11、求解析: .1 x 2-2x 21 -x 2y =2 2 (1 x )2 2(1 x )12、求已知f (x)dx 二 e x C ，则 lim y — f (—)n叫=0 n n二 e -1 1x-f (x)dx = J f (x)dx = (e +C)、2dx 二 1 x)n解析:A = ：(x 2 _1)dx = gx 3 _x)二、计算题(本大题共 8小题，其中16-19 小题每小题7分，20-23小题每小题8分，共15、常系数齐次线性微分方程 y“-2y • y =0的通解为y =：© • C 2x)e x ( C 1C 2为任意常数)o解析：特征方程：r -2r ^0，特征根:通解为y = (C i C 2X )e x( C 1C 2为任意常数)xy(x) =(1，sinx)，求 y(x)在 x处的微分解析：广 12dl nx= 1严=1ex(lnx)葩(lnx)lnx14、由2y=x : y=1, x=2围成的图形面积为16、求x _xe -e ln(1 sin x) 解析: x_xe -e ln(1 sin x)2xxe -1 =lim ej0 2x = lim佃空=2ln(1 sin x) x 10 sin x x 30 xcosx 1 sin x60分)In y = xln(1 sin x)1 .y = ln(1 sin x) x ycosxxdy =[ln(1 sin x) x ](1 sin x) dx1+s in x将x =恵代入上式，得微分dy --二dx5兀； ---- 2—18、求 1 -cos xdx5兀2p5冗解析：！：1 - cos xdx I sin x | dxn2 二3 二4 二5 二「0sinxdx.二(-sinX )dx 2二sinxdx 3二(-sinX )dx 4二sin xdx=_cosx|F +cosxf -cosx 倉 +cosx I 4耕- cosx |蠶=1017、设解析: y(x)二(1 sin x)x20、1-1". 5「4xxcosx 1 x 4)dx21、已知f （X ）=彳19、求 arctan 、xdx解析：令J x =t ，则 x =t , dx = 2tdt2 2 2arctantdt t arctant - t darctant21-t 2arctant - (12) dt」1+t 22=t arctant -t arctant c贝V 原式 =xarctan 一 x - . x arctan . x c…xcosx解析：• 1■孑为奇函数,1 dx tdt 215 _t 21 1该式 ---------- (__t)dt 34 t 21 3 2(5-t 2)dt 83 3—(5tt ) | 83 6[2x+b,x^0在x = 0处可导，求a,bJn (1+ax),x>0解析:x二t 2arctant-21 t -1 1 t 2dt.该式不代入计算y 2的单增区间是(-1,1)1 + x1 k解析：lim f ()=n n。

浙江省2018年选拔优秀高职高专毕业生进入本科学习统一考试高等数学参考答案选择题部分一、选择题:本大题共5小题,每小题4分,共20分。

题号12345答案CCABC1.C 解析:)0(0lim )(lim 0f x x f x x ===--→→，1sin lim )(lim 0==++→→xxx f x x ，所以0=x 是)(x f 的跳跃间断点，选项C 正确。

2.C 解析:02sin lim 2sin cos cos lim cos sin lim 0020==+-=-→→→xx x x x x x x x x x x x ，所以选项C 正确。

3.A 解析:因为函数)(x f 二阶可导，且0)(lim=-→x x x f x x ，所以0)()(lim 00==→x f x f x x ，故0)()()(lim )(lim000000='=--=-→→x f x x x f x f x x x f x x x x ，又因为0)(0<''x f ，所以由极值的第二充分条件可知，函数)(x f 在0x x =处取得极大值，因此选项A 正确。

4.B 解析:；⎰-=xxx f x f dx x f dx d 2)()2(2)(，故选项B 错误；由零点定理可知选项C 正确；由定积分性质中的估值定理可知选项D 正确。

5.C 解析:选项A ：交错级数，通项极限为：011lim =+∞→n n ，且n n u u <+1，所以由莱布尼茨审敛法，该级数收敛，但是加上绝对值后，级数∑∞=+111n n 发散，所以选项A为条件收敛。

选项B ：交错级数，通项极限为：0)1ln(1lim=+∞→n n ，且n n u u <+1所以由莱布尼茨审敛法，该级数收敛，但是加上绝对值后，因为nn 1)1ln(1>+，由小散证大散，级数∑∞=+1)1ln(1n n 发散，所以选项B 为条件收敛。