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1、vip 会员免费数学四试题分析、详解和评注一、填空题（本题共6 小题，每小题4 分，满分24 分.把答案填在题中横线上）(1)若5)(cossinlim0bxaexxx，则 a=1，b=4.【分析】本题属于已知极限求参数的反问题.【详解】因为5)(cossinlim0bxaexxx，且0)(cossinlim0bxxx，所以0)(lim0aexx，得 a=1.极限化为51)(coslim)(cossinlim00bbxxxbxaexxxx，得 b=4.因此，a=1，b=4.【评注】一般地，已知)()(limxgxf A，(1)若 g(x)0，则 f(x)0；(2)若 f(x)0，且 A 0，则
2、 g(x)0.(2)设1lnarctan22xxxeeey，则1121eedxdyx.【分析】本题为基础题型，先求导函数即可.【详解】因为)1ln(21arctan2xxexey，111222xxxxeeeey，所以，1121eedxdyx.【评注】本题属基本题型，主要考查复合函数求导.类似例题在一般教科书上均可找到.(3)设21,12121,)(2xxxexfx，则21)1(221dxxf.【分析】本题属于求分段函数的定积分，先换元：x 1=t，再利用对称区间上奇偶函数的积分性质即可.【详解】令 x 1=t，121121221)()()1(dtxfdttfdxxfvip 会员免费21)21(
3、0)1(12121212dxdxxex.【评注】一般地，对于分段函数的定积分，按分界点划分积分区间进行求解.（4）设100001010A，APPB1，其中P为三阶可逆矩阵,则220042AB100030003【分析】将B的幂次转化为A的幂次,并注意到2A为对角矩阵即得答案.【详解】因为1000100012A,PAPB200412004.故EEPPPAPB11002212004)(,220042AB100030003.【评注】本题是对矩阵高次幂运算的考查(5)设33ijaA是实正交矩阵，且111a，Tb)0,0,1(，则线性方程组bAx的解是T)0,0,1(【分析】利用正交矩阵的性质即可得结果.
4、【详解】因为bAx1,而且33ijaA是实正交矩阵,于是1AAT,A的每一个行(列)向量均为单位向量,所以0011312111aaabAbAxT.【评注】本题主要考查正交矩阵的性质和矩阵的运算文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 Z
5、L7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10
6、ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10
7、 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E1
8、0 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E
9、10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8
10、E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F
11、8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3vip 会员免费(6)设随机变量X服从参数为的指数分布,则DXXPe1.【分析】根据指数分布的分布函数和方差立即得正确答案.【详解】由于21DX,X的分布函数为.0,0,0,1)(xxexFx故DXXP1DXXP11XP)1(1Fe1.【评注】本题是对重要分布,即指数分布的考查,属基本题型.二、选择题（本题共6 小题，每小题4 分，满分24 分.每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内）(7)函数2)2)(1()2sin(|)(xxxxx
12、xf在下列哪个区间内有界.(A)(1,0).(B)(0,1).(C)(1,2).(D)(2,3).A 【分析】如 f(x)在(a,b)内连续，且极限)(limxfax与)(limxfbx存在，则函数f(x)在(a,b)内有界.【详解】当 x 0,1,2 时，f(x)连续，而183sin)(lim1xfx，42sin)(lim0 xfx，42sin)(lim0 xfx，)(lim1xfx，)(lim2xfx，所以，函数f(x)在(1,0)内有界，故选(A).【评注】一般地，如函数f(x)在闭区间 a,b上连续，则f(x)在闭区间 a,b上有界；如函数 f(x)在开区间(a,b)内连续，且极限)(
13、limxfax与)(limxfbx存在，则函数f(x)在开区间(a,b)内有界.(8)设 f(x)在(,+)内有定义，且axfx)(lim，0,00,)1()(xxxfxg，则(A)x=0 必是 g(x)的第一类间断点.(B)x=0 必是 g(x)的第二类间断点.(C)x=0 必是 g(x)的连续点.(D)g(x)在点 x=0 处的连续性与a的取值有关.D 文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7
14、P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL
15、7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 Z
16、L7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10
17、ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10
18、 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E1
19、0 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E
20、10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3vip 会员免费【分析】考查极限)(lim0 xgx是否存在，如存在，是否等于g(0)即可，通过换元xu1，可将极限)(lim0 xgx转化为)(limxfx.【详解】因为)(lim)1(lim)(lim00ufxfxguxx=a(令xu1)，又 g(0)=0，所以，当 a=0 时，)0()(lim0gxgx，
21、即 g(x)在点 x=0 处连续，当a 0 时，)0()(lim0gxgx，即 x=0 是 g(x)的第一类间断点，因此，g(x)在点 x=0 处的连续性与 a 的取值有关，故选(D).【评注】本题属于基本题型，主要考查分段函数在分界点处的连续性.(9)设 f(x)=|x(1 x)|，则(A)x=0 是 f(x)的极值点，但(0,0)不是曲线y=f(x)的拐点.(B)x=0 不是 f(x)的极值点，但(0,0)是曲线 y=f(x)的拐点.(C)x=0 是 f(x)的极值点，且(0,0)是曲线 y=f(x)的拐点.(D)x=0 不是 f(x)的极值点，(0,0)也不是曲线y=f(x)的拐点.C
22、【分析】由于 f(x)在 x=0 处的一、二阶导数不存在，可利用定义判断极值情况，考查 f(x)在 x=0 的左、右两侧的二阶导数的符号，判断拐点情况.【详解】设 0 0，而 f(0)=0，所以 x=0 是 f(x)的极小值点.显然，x=0 是 f(x)的不可导点.当 x(,0)时，f(x)=x(1 x)，02)(xf，当 x(0,)时，f(x)=x(1 x)，02)(xf，所以(0,0)是曲线 y=f(x)的拐点.故选(C).【评注】对于极值情况，也可考查f(x)在 x=0 的某空心邻域内的一阶导数的符号来判断.(10)设0,10,00,1)(xxxxf，xdttfxF0)()(，则(A)F
23、(x)在 x=0 点不连续.(B)F(x)在(,+)内连续，但在x=0 点不可导.(C)F(x)在(,+)内可导，且满足)()(xfxF.(D)F(x)在(,+)内可导，但不一定满足)()(xfxF.B 【分析】先求分段函数f(x)的变限积分xdttfxF0)()(，再讨论函数F(x)的连续性与可导性即可.【详解】当 x 0 时，xdtxFx01)(，当 x=0 时，F(0)=0.即 F(x)=|x|，显然，F(x)在(,+)内连续，但在x=0 点不可导.故选(B).【评注】本题主要考查求分段函数的变限积分.对于绝对值函数：|0 xx在0 xx处不可导；f(x)=|0 xxxn在0 xx处有
24、n 阶导数，则|)!1()(0)(xxnxfn.(11)设)(xf在a,b上连续，且0)(,0)(bfaf，则下列结论中错误的是(A)至少存在一点),(0bax，使得)(0 xf f(a).(B)至少存在一点),(0bax，使得)(0 xf f(b).(C)至少存在一点),(0bax，使得0)(0 xf.(D)至少存在一点),(0bax，使得)(0 xf=0.D 【分析】利用介值定理与极限的保号性可得到三个正确的选项，由排除法可选出错误选项.【详解】首先，由已知)(xf在a,b上连续，且0)(,0)(bfaf，则由介值定理，至少存在一点),(0bax，使得0)(0 xf；另外，0)()(lim
25、)(axafxfafax，由极限的保号性，至少存在一点),(0bax使得0)()(00axafxf，即)()(0afxf.同理，至少存在一点),(0bax使得)()(0bfxf.所以，(A)(B)(C)都正确，故选(D).【评注】本题综合考查了介值定理与极限的保号性，有一定的难度.(12)设n阶矩阵A与B等价,则必须(A)当)0(|aaA时,aB|.(B)当)0(|aaA时,aB|.(C)当0|A时,0|B.(D)当0|A时,0|B.D 【分析】利用矩阵A与B等价的充要条件:)()(BrAr立即可得.【详解】因为当0|A时,nAr)(,又A与B等价,故nBr)(,即0|B,从而选(D).【评注
26、】本题是对矩阵等价、行列式的考查,属基本题型.文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 H
27、L8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9
28、HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9
29、 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C
30、9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2
31、C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I
32、2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3vip 会员免费(13)设
33、随机变量X服从正态分布)1,0(N,对给定的)1,0(,数u满足uXP,若xXP|,则x等于(A)2u.(B)21 u.(C)21u.(D)u1.B 【分析】利用标准正态分布密度曲线的对称性和几何意义即得.【详解】由xXP|,以及标准正态分布密度曲线的对称性可得21xXP.故正确答案为(B).【评注】本题是对标准正态分布的性质,严格地说它的上分位数概念的考查.(14)设随机变量nXXX,21)1(n独立同分布，且方差02令随机变量niiXnY11，则(A)212)(nnYXD(B)212)(nnYXD(C)nYXCov21),(D)21),(YXCov C 【分析】利用协方差的性质立即得正确答
34、案.【详解】由于随机变量nXXX,21)1(n独立同分布,于是可得),(1)1,(),(11111niiniiXXCovnXnXCovYXCov),(1),(11111XXCovnXXCovnnii211)(1nXDn.故正确答案为(C).【评注】本题是对协方差性质的考查,属于基本题.三、解答题(本题共 9 小题，满分94 分.解答应写出文字说明、证明过程或演算步骤.)(15)(本题满分8 分)求)cossin1(lim2220 xxxx.文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P
35、3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7
36、P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL
37、7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 Z
38、L7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10
39、ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10
40、 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E1
41、0 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3vip 会员免费【分析】先通分化为“00”型极限，再利用等价无穷小与罗必达法则求解即可.【详解】xxxxxxxxxx2222202220sincossinlim)cossin1(lim=30422044sin212lim2sin
42、41limxxxxxxxx.346)4(21lim64cos1lim22020 xxxxxx.【评注】本题属于求未定式极限的基本题型，对于“00”型极限，应充分利用等价无穷小替换来简化计算.(16)(本题满分8 分)求Ddyyx)(22，其中 D 是由圆422yx和1)1(22yx所围成的平面区域(如图).【分析】首先，将积分区域D 分为大圆 4|),(221yxyxD减去小圆 1)1(|),(222yxyxD，再利用对称性与极坐标计算即可.【详解】令 1)1(|),(,4|),(222221yxyxDyxyxD，由对称性，0Dyd.21222222DDDdyxdyxdyxcos2022322
43、0220drrddrrd.)23(916932316所以，)23(916)(22Ddyyx.【评注】本题属于在极坐标系下计算二重积分的基本题型，对于二重积分，经常利用对称性及将一个复杂区域划分为两个或三个简单区域来简化计算.(17)(本题满分8 分)设 f(u,v)具有连续偏导数，且满足uvvufvufvu),(),(.文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O1
44、0I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O
45、10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3
46、O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC
47、3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:C
48、C3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:
49、CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码
50、:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3文档编码:CC3O10I6I2C9 HL8H5W3F8E10 ZL7P3Y9Y10L3vip 会员免费求),()(2xxfexyx所满足的一阶微分方程，并求其通解.【分析】先求y，利用已知关系uvvufvufvu),(),(，可得到关于y 的一阶微分方程.【详解】xvxuxxexyxxfexxfexxfey222222),(),(),(2，因此，所求的一阶微分方程为xexyy222.解得xdxxdxeCxCdxeexey232222