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1998考研数四真题及解析.pdf

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1、 Born to win 1 1998 年全国硕士研究生入学统一考试数学四试题一、填空题(本题共 5 分,每小题 3 分,满分 15 分.)(1)设曲线()nf xx在点(1,1)处的切线与x轴的交点为(,0)n则lim()nnf_.(2)2ln1xdxx_.(3)设矩阵,A B满足*28A BABAE,其中100020001A,E为单位矩阵,*A为A 的伴随矩阵,则B _.(4)设,A B均为n阶矩阵,2,3AB,则*12A B_.(5)设一次试验成功的概率为p,进行100 次独立重复试验,当p _时,成功次数的标准差的值最大,其最大值为_.(注：第一空 2 分,第二空 1 分)二、选择题

2、(本题共 5 小题,每小题 3 分,满分 15 分.)(1)设周期函数()f x在,内可导,周期为 4,又0(1)(1)lim12xffxx,则曲线()yf x在点5,(5)f处的切线的斜率为 ()(A)12 (B)0 (C)1 (D)2(2)设函数21()lim1nnxf xx,讨论函数()f x的间断点,其结论为 ()(A)不存在间断点 (B)存在间断点1x (C)存在间断点0 x (D)存在间断点1x (3)若向量组,线性无关,线性相关,则 ()(A)必可由,线性表示 (B)必不可由,线性表示(C)必可由,线性表示 (D)必不可由,线性表示(4)设,A B C是三个相互独立的随机事件,且

3、0()1P C,则在下列给定的四对事件中不相互独立的是 ()(A)ABC 与 (B)ACC与(C)ABC 与 (D)ABC与(5)设1()F x与2()Fx分别为随机变量12XX与的分布函数.为使12()()()F xaF xbFx Born to win 2 是某一随机变量的分布函数,在下列给定的各组数值中应取 ()(A)32,55ab (B)22,33ab(C)13,22ab (D)13,22ab 三、(本题满分 6 分)求21lim(tan)nnnn(n为自然数).四、(本题满分 6 分)设arctan22()yxzxye,求dz与2zx y.五、(本题满分 5 分)设22(,)|Dx

4、yxyx,求Dxdxdy.六、(本题满分 6 分)设某酒厂有一批新酿的好酒,如果现在(假定0t)就售出,总收入为0R(元).如果窖藏起来待来日按陈酒价格出售,t年末总收入为 250tRR e.假定银行的年利率为r,并以连续复利计算,试求窖藏多少年售出可使总收入的现值最大.并求0.06r 时的t值.七、(本题满分 6 分)设()f x在,a b上连续,在(,)a b内可导,且()()1f af b,试证存在,(,)a b,使得()()1eff.八、(本题满分 9 分)设直线yax与抛物线2yx所围成图形的面积为1S,它们与直线1x 所围成的图形面积为2S,并且1a.(1)试确定a的值,使12SS

5、达到最小,并求出最小值.(2)求该最小值所对应的平面图形绕x轴旋转一周所得旋转体的体积.九、(本题满分 9 分)文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9

6、T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2

7、V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码

8、:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2

9、W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T1

10、0 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T

11、10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7文档编码:CK1Q2W7L6T10 HY9T10L5G4R7 ZQ9T7E7F2V7 Born to win 3 设向量12(,)Tn,12

12、(,)Tnb bb都是非零向量,且满足条件0T.记n阶矩阵TA,求：(1)2A；(2)矩阵A的特征值和特征向量.十、(本题满分 7 分)已知下列非齐次线性方程组()1241234123264133xxxxxxxxxx ()123423434521121xmxxxnxxxxxt (1)求解方程组(I),用其导出组的基础解系表示通解；(2)当方程组(II)中的参数,m n t为何值时,方程组(I)与(II)同解.十一、(本题满分 8 分)设某种商品每周的需求量X是服从区间10 30,上均匀分布的随机变量,而经销商店进货数量为区间10 30,中的某一整数,商店每销售一单位商品可获利 500 元；若供

13、大于求则削价处理,每处理 1 单位商品亏损 100 元；若供不应求,则可从外部调剂供应,此时每 1 单位商品仅获利 300 元.为使商品所获利润期望值不少于 9280 元,试确定最少进货量.十二、(本题满分 8 分)某箱装有 100 件产品,其中一、二和三等品分别为 80、10 和 10 件,现在从中随机抽取一件,记 1,(1,2 3)0,iiXi若抽到等品,其他.,试求：(1)随机变量1X与2X的联合分布；(2)随机变量12XX和的相关系数.文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C

14、2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L

15、6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C

16、2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L

17、6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C

18、2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L

19、6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C

20、2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2 Born to win 4 1998 年全国硕士研究生入学统一考试数学四试题解析一、填空题(本题共 5 分,每小题 3 分,满分 15 分.)(1)【答案】1e【解析】曲线nyx在点(1,1)处的切线斜率1xy1nxx11nxnxn,根据点斜式,切线方程为：1(1).yn x 令0y,代入1(1)yn x,则11xn,即在x轴上的截距为11nn,lim()nnflimnnn1lim(1)nnn 11lim(1)xxx1e.(2)【答案】ln xCx【解析】由分部积分公式,2ln1xdxx1ln1xdx

21、x 1ln1xdx ln11(ln1)xdxxx 分部2ln11xdxxx ln11xdxxx ln11xCxx ln xCx.【相关知识点】分部积分公式：假定()uu x与()vv x均具有连续的导函数,则,uv dxuvu vdx或者.udvuvvdu(3)【答案】200040002【解析】由题设 *28A BABAE,由于20A ,所以A可逆.上式两边左乘A,右乘1A,得*11128AA BAAABAAAA 28A BABE(利用公式：*1,AAA E AAE)文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5

22、X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码

23、:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5

24、X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码

25、:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5

26、X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码

27、:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5

28、X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2 Born to win 5 28A BABE(移项)28A EA BE(矩阵乘法的运算法则)将2A 代入上式,整理得14EA BE.由矩阵可逆的定义,知EA,B均可逆,且 114BEA110020024 0104 0100021002200040002.(4)【答案】2123n【解析】,A B均为n阶矩阵,且20,30AB ,故,A B均为n阶可逆矩阵,则有*12A B*12A B(利用公式：ABA B)*12nA B(利用公式：nkAk A)112nnAB(利用公式：1*nAA)

29、112nnAB(利用公式：11BB)2123n.(代入2,3AB)(5)【答案】1,52【解析】100 次独立重复试验,每次试验结果不是成功就是失败,则成功次数X服从二项分布(100,)Bp,X的标准差()100(1)D Xpp.因为()f xx在0,)上单调递增,所以求()D X的最大值即是求()(1)g ppp的最大值,而()10gppp 驻点为12p.()20gp ,所以12p 为极大值点,由函数图像知12p 即为最大值点.此时11()24g.文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3

30、R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2

31、文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3

32、R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2

33、文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3

34、R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2

35、文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3

36、R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2 Born to win 6 此时1()10054D X .二、选择题(本题共 5 小题,每小题 3 分,满分 15 分.)(1)【答案】(D)【解析】根据导数定义：0()()limxf xxf xfxx 0(1)(1)lim2xffxx01(1)(1)lim2xfxfx 1(1)2f 1 所以 0(1)(1)(1)lim2.xfxffx 因为()f x周期为 4,()fx的周期亦是 4,即()(4)fxfx,所以(5)f(14)f(1)2f .所以曲线()yf x在点5,(

37、5)f处的切线的斜率为(5)f(1)2f .选(D).(2)【答案】(B)【分析】讨论由极限表示的函数的性质,应分两步走.先求出该()f x的(分段)表达式,然后再讨论()f x的性质.不能隔着极限号去讨论.【解析】现求()f x的(分段)表达式：当1x 时,21()lim1nnxf xx21 22lim1nnnnxxx21 22lim01lim1nnnnnxxx0；当1x 时,21()lim1nnxf xx21 1lim1 1nn221；当1x 时,21()lim1nnxf xx21 1lim11nn020；当1x 时,21()lim1nnxf xx 2lim 1lim 1nnnxx2011

38、nxx1x.文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1

39、L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9

40、C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1

41、L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9

42、C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1

43、L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9

44、C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2 Born to win 7 由此,0,1,0,1,()1,1,1,1,0,1.xxf xxxxx 当当当当当即0,11,()1,1,1,1.xxf xxxx 当或当当再讨论函数()f x的性质：在1x 处,1limxfx 1lim 1xx1 1 0,1lim10 xf xf,所以,11limlim0 xxf xf x,函数()f x在1x

45、处连续,不是间断点.在1x 处,1limxf x1lim 0 x0；1limxf x 1lim 1xx2；所以 1limxf x 1limxf x,函数()f x在1x 处不连续,是第一类间断点.故选(B).(3)【答案】(C)【解析】方法 1：由向量组,线性无关,知,线性无关.又因,线性相关,故必可由,线性表出,因此必可由,线性表示,从而选(C).方法 2：由题设向量组,线性无关,3r,同时,由整体线性无关,任何部分也线性无关,知,也线性无关,2r.又由,线性相关,3r,所以,2r.故,3rr ,故方程组123xxx有解,则可由,线性表出.【相关知识点】1、定理：若12,s 线性无关,12,

46、s 线性相关,则可由12,s 线性表出,且表示法唯一.2、整体线性无关,任何部分也线性无关.3、非齐次线性方程组有解的判定定理：设A是mn矩阵,方程组Axb有唯一解()().r Ar An 4、定理：能由12,s 线性表出,1,2,iis为列向量的非齐次线性方程组1 122ssxxx 有解.(4)【答案】B 文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X

47、1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:

48、CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X

49、1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:

50、CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X1M8 ZY7G9C2M7A2文档编码:CC2Z3B5G1L6 HS3R5R5X

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关键词：: 1998 考研数四真题解析

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