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2018年考研数学三真题与解析.pdf

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1、2018年考研数学三真题及答案一、选择题1.下列函数中,在0 x处不可导的是().sinA fxxx.sinB fxxx.?C fxcos x.cosD fxx答案:D解析:方法一:000sin0limlimlimsin0,xxxxxxfxfxxxxA可导000sin0limlimlimsin0,xxxxxxfxfxxxxB可导20001cos102limlimlim0,xxxxxfxfxxCx可导0001cos102limlimlimxxxxxffxxxDx不存在,不可导应选D.方法二:因为,(1)0fcosxfx0001cos102limlimlimxxxxxfxfxxx不存在fx在0 x

2、处不可导,选D对:?Afxxsinx在0 x处可导对32:?Bfxxxxg在0 x处可导对():xxCfcos在0 x处可导.2.设函数fx在0,1 上二阶可导,且100,fx dx则10,02Afxf当时10,02Bfxf当时10,02Cfxf当时10,02Dfxf当时答案D【解析】将函数fx在12处展开可得2221110001111,222221111111,22222222ffxffxxffx dxffxxdxffxdx故当()0fx时,1011.0.22fx dxff从而有选D。3.设2222222211,1cos1xxxMdx Ndx Kx dxxe,则A.?.MNKB.MKNC.K

3、MND.KNM文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D

4、2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:

5、CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S

6、8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5

7、HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H

8、10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1

9、 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2答案:C解析:2222222221211,11xxMdxdxdxxx221xxNdxe,因为1xex所以11xxe221cos,1cos1.Kx dxx即

10、111cosxxxe所以由定积分的比较性质KMN,应选C.4.设某产品的成本函数C Q可导,其中Q为产量,若产量为0Q时平均成本最小,则()A00CQB00CQC QC.000CQQ C QD.000Q CQC Q答案D【解析】平均成本2,C QdC QCQ QC QC QQdQQ,由于C Q在0QQ处取最小值,可知000.Q CQ故选(D).5.下列矩阵中,与矩阵110011001相似的为111.011001A101.011001B111.010001C101.010001D文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5

11、HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H

12、10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1

13、 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7

14、J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D

15、2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:

16、CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S

17、8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2答案:A解析:令110010001P则1110010001P1110111110010011010001001001120110110011010011001001001PAPQ选项为A6.设,A B为n阶矩阵,记r X为矩阵X的秩,XY表示分块矩阵,则.?ArAABr A.?B r

18、ABArA.?,C rABmax r Ar B.?TTD r ABrA B答案:A解析:易知选项C错对于选项B举反例:取1 1001 112AB1 则001100,331133BAA BA7.设随机变量X的概率密度fx满足11fxfx，且200.6fx dx，则0_P X文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H

19、10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档

20、编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4

21、J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1

22、P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8

23、A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C

24、3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA

25、5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2(A)0.2；(B)0.3；(C)0.4；(D)0.6 解由11fxfx知，概率密度fx关于1x对称，故02P XP X，且00221P XPXP X，由于20020.6PXfx dx，所以200.4P X，即00.2P X，故选项 A 正确8.设12,nXXXK为取自于总体2,XN:的简单随机样本，令niiXnX11，2111()1niiSXXn，2211()niiSXXn，则下列选项正确的是_(A)n Xt nS:；(B)1n Xt nS:；(C)*n Xt nS:；(D)*1n Xt

26、nS:解由于0,1XNn，)1()()1(221222nXXSnnii，且Xn与22(1)nS相互独立，由t分布的定义，得文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1

27、 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7

28、J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D

29、2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:

30、CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S

31、8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5

32、HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2(1)n XXt nSSn，故选项 B 正确二

33、、填空题9.曲线22lnyxx在其拐点处的切线方程为_。答案43yx【解析】函数fx的定义域为232240,2,2,yxyyxxx。令=0y,解得x=1,而 10,y故点(1,1)为曲线唯一的拐点。曲线在该点处切线的斜率 14,y故切线方程为43yx。10.2arcsin1_.xxee222222222222arcsin11,1=arcsin1arcsin 1111arcsin1tansin 111arcsin11xxxxxxxeeeCtt dttttdttttttdtttCteeeC答案【解析】令 t=e则原式11.差分方程25xxyy的通解 _.【答案】125xxyc文档编码:CB4J1S

34、8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5

35、HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H

36、10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1

37、 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7

38、J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D

39、2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:

40、CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D22+1+2+1+1+2+1+2+1+1+111111=22=5,2525,2,-2=5,=-52xxxxxxxxxxxxxxxxxxxxxyyyyyyyyyyyyyyyyyycyccccyc

41、【解析】由于，故原差分方程可化为即。设一阶常系数线性差分方程对应的其次方程为其通解为。设原差方程的特解代入原方程得即。所以原差分方程的通解为5,c为任意常数。12.函数x满足20,xxxxxxxx且02,则1_.答案12.e【解析】2,=2xxxxxxxxx由可知可微且。这是一个可分离变量微分方程,求得其通解为2;xxce再由02,可得2c。故22,12xxee。13.设A为3阶矩阵,123,为线性无关的向量组,若112322332322,AAA，,可得123123200,111121A。由于123,线性无关,故200111121A:=B，从而有相同的特征值。因220

42、0111223,121EB故A的实特征值为2。文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10

43、C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 Z

44、A5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9

45、H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文

46、档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB

47、4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q

48、1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D214.设随机事件,A B C相互独立，且1()()()2P AP BP C，则()_P AC AB解由条件概率以及事件相互

49、独立性的定义，得()()()()()()()()()1 112 2.111 13222 2PACABP AC ABP ABP ACP AP BP ABP AP CP AP BP AP B三、解答题15.已知实数,a b,满足1lim2,xxaxb ex求a,b。答案1,1ab【解析】011,lim2,ttabt etxt令=可得0000111limlimlimlimttttttttabt eaeaebebttt其中可知0011lim2,lim,1ttttaeaebatt而要使得存在必须有。01,lim=1=2,1.,1,1ttaebbtab此时有故综上。文档编码:CB4J1S8Q1P5 H

50、C8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H10C3S1 ZA5M7J9H10D2文档编码:CB4J1S8Q1P5 HC8A8H1

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