2021年考研数学一真题及答案

2024年3月4日发(作者：吉利帝豪gs为什么停售)

2021考研数学真题及答案解析数学（一）一、选择题（本题共10小题，每小题5分，共50分.每小题给出的四个选项中，只有一个选项是符合题目要求，把所选选项前的字母填在答题卡指定位置上.）?ex?1,x?0?(1)函数f(x)=?x,在x?0处?1,x?0?(A)连续且取极大值.(C)可导且导数为0.【答案】D.(B)连续且取极小值.(D)可导且导数不为?1【解析】因为limf(x)=lim?1?f(0)，故f(x)在x?0处连续；x?0x?0xex?1?1f(x)?f(0)ex?1?x11x?因为lim，故，正确答案为D.=lim?lim?f(0)?x?0x?0x?0x?0x?0x222(2)设函数f?x,y?可微，且f(x?1,e)?x(x?1)，f(x,x)?2xlnx，则df(1,1)?x222(A)dx?dy.【答案】C.(B)dx?dy.(C)dy.(D)?dy.①②xxx2【解析】f1?(x?1,e)?ef2?(x?1,e)?(x?1)?2x(x?1)f1?(x,x2)?2xf2?(x,x2)?4xlnx?2x?x?0?x?1分别将?，?带入①②式有y0y1????f1?(1,1)?f2?(1,1)?1，f1?(1,1)?2f2?(1,1)?2联立可得f1?(1,1)?0，f2?(1,1)?1，df(1,1)?f1?(1,1)dx?f2?(1,1)dy?dy，故正确答案为23在x?0处的3次泰勒多项式为ax?bx?cx，则21?x77(A)a?1,b?0,c??.(B)a?1,b?0,c?.6677(C)a??1,b??1,c??.(D)a??1,b??1,c?.66(3)设函数f(x)?【答案】A.【解析】根据麦克劳林公式有sinx?x3733?233f(x)??x??o(x)?[1?x?o(x)]?x?x?o(x)??21?x66??1微信公众号：考研文库

故a?1,b?0,c??7，本题选A.6(4)设函数f?x?在区间?0,1?上连续，则(A)lim?f?x?dx?01n???k?12nn?2k?1?1.f??2n2n???k?1?1f??.2n??n(B)lim(D)limx?0n???k?1n?2k?1?1f??.2n??n(C)lim【答案】B.n???k?1?k?12n?k?2f???.?2n?n【解析】由定积分的定义知，将?0,1?分成n份，取中间点的函数值，则?10?2k?1?1f(x)dx?lim?f??,即选B.n??k?1?2n?n222(5)二次型f(x1,x2,x3)?(x1?x2)?(x2?x3)?(x3?x1)的正惯性指数与负惯性指数依次为n(A)2,0.【答案】B.(B)1,1.22(C)2,1.22(D)1,2.【解析】f(x1,x2,x3)?(x1?x2)?(x2?x3)?(x3?x1)?2x2?2x1x2?2x2x3?2x1x3?011???所以A?121，故特征多项式为???110?????1?1|?E?A|??1?2?1?(??1)(??3)??1?1?令上式等于零，故特征值为?1，3，0，故该二次型的正惯性指数为1，负惯性指数为1.故应选B.?1??1??3???????(6)已知?1?0，?2?2，?3?1，记?1??1，?2??2?k?1，?3??3?l1?1?l2?2，???????1??1??2???????若?1，?2，?3两两正交，则l1，l2依次为(A)51,.22(B)?51,.22(C)51,?.22(D)?51,?.22【答案】A.【解析】利用斯密特正交化方法知?0?[?,?]?，?2??2?21?1??2??[?1,?1]?0????3??3?故l1?[?3,?1][?,?]?1?32?2，[?1,?1][?2,?2][?3,?1]5[?,?]1?，l2?32?，故选A.[?1,?1]2[?2,?2]22(7)设A,B为n阶实矩阵，下列不成立的是微信公众号：考研文库

(A)r?(C)r??A?OO???2r?A?TAA?BA???2r?A?AAT?(B)r??A?OAB??2r?A?T?A??A?O(D)r??AO??2r?A?T?BAA??【答案】C.?A【解析】(A)r??OO?T??r(A)?r(AA)?2r(A).故A正确.TAA??A?OAB??AO?T?r?r(A)?r(A)?2r(A).???TTA??0A?(B)AB的列向量可由A的列线性表示，故r?(C)BA的列向量不一定能由A的列线性表示.(D)BA的行向量可由A的行线性表示，r??ABA??AO??r?r(A)?r(AT)?2r(A).?T?T??OA??0A?本题选C.(8)设A，B为随机事件，且0?P(B)?1，下列命题中不成立的是(A)若P(A|B)?P(A)，则P(A|B)?P(A).(B)若P(A|B)?P(A)，则P(A|B)?P(A)(C)若P(A|B)?P(A|B)，则P(A|B)?P(A).(D)若P(A|A?B)?P(A|A?B)，则P(A)?P(B).【答案】D.【解析】P(A|A?B)?P(A(A?B))P(A)?P(A?B)P(A)?P(B)?P(AB)P(A|A?B)?P(A(A?B))P(AB)P(B)?P(AB)??P(A?B)P(A?B)P(A)?P(B)?P(AB)因为P(A|A?B)?P(A|A?B)，固有P(A)?P(B)?P(AB)，故正确答案为D.(9)设?X1,Y1?,?X2,Y2?,?,?Xn,Yn?为来自总体N?1,?2;?1,?2;?的简单随机样本，令22??1n1n???1??2,X??Xi,Y??Yi,???X?Y,则ni?1ni?122???12?是?的无偏估计，D???(A)?n??2?12??2??(B)?不是?的无偏估计，D??n??22???2?2??1?2??1?是?的无偏估计，D?(C)?n??22????2??1?212?不是?的无偏估计，D???(D)?n??【答案】C.【解析】因为X,Y是二维正态分布，所以X与Y也服从二维正态分布，则X?Y也服从二维正态?)?E(X?Y)?E(X)?E(Y)??????，分布，即E(?123微信公众号：考研文库

?12??22?2??1?2?D(?)?D(X?Y)?D(X)?D(Y)?cov(X,Y)?，故正确答案为C.n(10)设X1,X2?,X16是来自总体N??,4?的简单随机样本，考虑假设检验问题：H0:??10,H1:??10.??x?表示标准正态分布函数，若该检验问题的拒绝域为W??X?11?，116其中X??Xi，则??11.5时，该检验犯第二类错误的概率为16i?1(A)1???0.5?(C)1???1.5?【答案】B.【解析】所求概率为P{X?11}X?N(11.5,)，(B)1???1?(D)1???2?14???X?11.511?11.5?P{X?11}?P????1??(1)11???22?故本题选B.二、填空题（本题共6小题，每小题5分，共30分.请将答案写在答题纸指定位置上.）(11)【答案】【解析】????0?4??0dx?2x?2x?2.??dxdx?????arctan(x1)??????0x2?2x?2?0(x?1)2?1244?x?2et?t?1,x?0d2y(12)设函数y?y(x)由参数方程?确定,则2t?0?t2dx?y?4(t?1)e?t,x?02【答案】.3d2y(4et?4tet?2)(2et?1)?(4tet?2t)2etdy4tet?2t【解析】由，得，??dx2(2et?1)3dx2et?1d2y2?.将t?0带入得t?0dx232(13)欧拉方程xy???xy??4y?0满足条件y(1)?1,y?(1)?2得解为y?【答案】x.2t..d2ydy2d2ydy?4y?0，特征方程为,xy???2?【解析】令x?e，则xy??，原方程化为dx2dtdxdx?2?4?0，特征根为?1?2,?2??2，通解为y?C1e2t?C2e?2t?C1x2?C2x?2，将初始条件y(1)?1,y?(1)?2带入得C1?1,C2?0，故满足初始条件的解为y?x2.(14)设?为空间区域(x,y,z)x?4y?4,0?z?222xdydz?ydzdx?zdxdy?????22?表面的外侧，则曲面积分.【答案】4?.4微信公众号：考研文库

【解析】由高斯公式得原式=???(2x?2y?1)dV???20dz??dxdy?4?.D(15)设A?aij为3阶矩阵，Aij为代数余子式，若A的每行元素之和均为2，且A?3，(16)甲乙两个盒子中各装有2个红球和2个白球，先从甲盒中任取一球，观察颜色后放入乙盒中，再从乙盒中任取一球.令X,Y分别表示从甲盒和乙盒中取到的红球个数，则X与Y的相关系数.【答案】A11?A21?A31=.3【答案】.2?1??1??1?A??????*，对应的特征向量为【解析】A1?21,A????,??2,??1,则A的特征值为????????1??1??1????????A11A21A31??1??1??A11?A21?A31??1?A??*????*?A??*1?，即???1,A??而A?A12A22A32,A1?A12?A22?A32?????????????A??1??1??A?A?A???1?AA2333?2333??13?????13??3A11?A21?A31?.21.51??0??1?Y??12??21??1?2??(0,0)(0,1)(1,0)(1,1)??0?,X??【解答】联合分布率(X,Y)??3113???1??5510??2?101111cov(X,Y)?，DX?,DY?,即?XY?.20445三、解答题（本题共6小题，共70分.请将解答写在答题纸指定位置上，解答应写出文字说明、证明过程或演算步骤.）(17)(本题满分10分)?1?xet2dt?1?0?.求极限lim??xx?0?sinx??e?1???1【答案】.2x2?1?xet2dt?sinx?1??etdt1?00??lim【解析】解：lim??xxx?0?sinx??e?1?x?0(e?1)sinx??x2x13223t又因为?edt??(1?t?o(t))dt?x?x?o(x)，故003111(x?x3?o(x3))(1?x?x3?o(x3))?x?x2?o(x2)3!3!2原式=lim2x?0x12x?o(x2)1=lim2.?2x?0x25微信公众号：考研文库

(18)(本题满分12分)设un(x)?e?nx?1n?1?x(n?1,2,?)，求级数?un(x)的收敛域及和函数.n(n?1)n?1?e?x?(1?x)ln(1?x)?x,x?(0,1)??1?e?x【答案】S(x)??.?e,x?1?e?1?【解析】?????nxe?x1?nxn?1?,x?(0,1]S(x)??un(x)???e?x?,收敛域(0,1],S1(x)??e??x1e?n(n?1)1n?n?1n?1??1nn????1xx?1n?1??xln(1?x)?[?ln(1?x)?x]S2(x)??x????n?1n(n?1)n?1nn?1n?1?(1?x)ln(1?x)?x,x?(0,1)S2(1)?limS2(x)?1?x?1?e?x?(1?x)ln(1?x)?x,x?(0,1)??1?e?xS(x)???e,x?1??e?1(19)(本题满分12分)?x2?2y2?z?6已知曲线C:?，求C上的点到xoy坐标面距离的最大值.?4x?2y?z?30【答案】66【解析】设拉格朗日函数L?x,y,z,?,???z2???x2?2y2?z?6???(4x?2y?z?30)L?x?2x??4u?0L?y?4y??2u?0L?z?2z???u?0x2?2y2?z?64x?2y?z?30解得驻点：(4,1,12),(?8,?2,66)C上的点(?8,?2,66)到xoy面距离最大为66.(20)(本题满分12分)设D?R是有界单连通闭区域，I(D)?(1)求I(D1)的值.(2)计算222(4?x?y)dxdy取得最大值的积分区域记为D1.??D?D1?(xex2?4y2?y)dx?(4yexx2?4y22?4y2?x)dy，其中?D1是D1的正向边界.【答案】??.【解析】（1）由二重积分的几何意义知：I(D)?22224?x?y，当且仅当在D上(4xy)d?????D大于0时，I(D)达到最大，故D1：x2?y2?4且I(D1)=?2?0d??(4?r2)rdr?8?.02(2)补D2:x2?4y2?r2（r很小），取D2的方向为顺时针方向，?D1?(xex2?4y2?y)dx?(4yexx2?4y22?4y2?x)dy=6微信公众号：考研文库

????D1??D2?(xex2?4y2?y)dx?(4yexx2?4y22?4y2?x)dy??D2?(xex2?4y2?y)dx?(4yexx2?4y22?4y2?x)dy1r21r21exdx4ydyeydxxdy????r2??r2??r2D2D2(21)(本题满分12分)???2d????.D2?a1?1???已知A?1?a1??.??1?1a???T(1)求正交矩阵P,使得PAP为对角矩阵；2(2)求正定矩阵C,使得C?(a?3)E?A.?1??3?1【答案】(1)P???3?1??3?【解析】?12120?1??5?11????3?6???511???1?.?C；(2)??33?6???152???1????33??6?1??a?1(1)由?E?A??11?(??a?1)2(??a?2)?0??a11??a得?1?a?2,?2??3?a?1当?1?a?2时?2?11??101??1???????((a?2)E?A)???121?r??011?的特征向量为?1??1?，?112??000???1???????当?2??3?a?1所??1?11??11?1???1???1?????????((a?1)E?A)???1?11?r??000?的特征向量为?2??1?,?3??1?，?11?1??000??0??2?????????11??1????326???a?2???1?2?3??111???T令P??，则，????PAPa1??,,?????????26?23??1??3a?1???2??10???36???1???T2T(2)PCP?P(a?3)E?A)P?((a?3)E???4???4???7微信公众号：考研文库

?1??1?????T?PTCPPTCP???PCP?42???，??4?2??????5??1?1?3?1????51??T??.21P??故C?P???33???2????15??1???33??(22)(本题满分12分)在区间(0,2)上随机取一点，将该区间分成两段，较短的一段长度记为X,较长的一段长度记为Y,令Z?Y.X(1)求X的概率密度；(2)求Z的概率密度.?X?(3)求E??.?Y??2,z?11,01?x???2【答案】(1)X?f(x)??；(2)fZ(z)?(FZ(z))???(z?1).(3)?1?2ln2.0,其他??0,其他??1,0?x?1；【解析】(1)由题知：X?f(x)??0,其他?2?X(2)由Y?2?X，即Z?，先求Z的分布函数：X?2?X??2?FZ(z)?P?Z?z??P??z??P??1?z??X??X?当z?1时，FZ(z)?0；当z?1时，22?2?2??；FZ(z)?P??1?z??1?P?X???1??0z?11dx?1?z?1?z?1?X???2,z?1?2；fZ(z)?(FZ(z))???(z?1)?0,其他?(3)E??X?Y??X?E????2?X????x?02?x?1dx??1?2ln2.18微信公众号：考研文库