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,只需一个选项是契合标题需求,把所选选项前的字母填在答题卡指定方位上.)e1??x,x0???(1)函数f(x)=,在x?0处x?1,×0??(a)接连且取极大值.(c)可导且导数为0.(b)接连且取极小值.(d)可导且导数不为0.【答案】d.【解析】因为e1?xlimf(x)=lim1f(0)??,故f(x)在x?0处接连;xx?0x?0e1?x?1f(x)f(0)e1x11?x??xlim=lim?limf(0)?,正确答案为d.因为?,故x0x0x22??2x?0x?0x?0??fx,yf(x?1,e)?x(x?f(x,x)?2xx(1,1)?,则(2)设函数可微,且,x222???.(a).(b).(c).(d)【答案】c.f(x?1,e)?ef(x?1,e)?(x??2x(x?①②【解析】xxx212f?(x,x)2?(x,x)4x
x2x???2212×0??x1??别离将?,?带入①②式有y0y1????f?f(1,1)?1f?2f(1,1)?2,1212f(1,1)?0f?1(1,1)?f?f?,,,故正确答案为c.联立可得1212xf(x)??(3)设函数?x?0,则在处的3次泰勒多项式为231x?2767a?b?c??a?b?c?(a)(c).(b).677a??b??c??a??b??c?.(d).66【答案】a.【解析】根据麦克劳林公式有xx7??3f(x)xo(x)xo(xxxo(x)??????????32333??1×6?6?2?1微信大众号:考研文库76a?b?c??,本题选a.故??????fx??10fx上接连,则(4)设函数在区间2k112k11????????fnnf(a)(c).(b)(d).??2n2n???2nn???nnk1?k1?k11k2????????.2n2nff.????2nn??2nn??nx?0k1k1??【答案】b.【解析】由定积分的界说知,将??n分红份,取中心点的函数值,则2k11???10?nf(x)f,??即选b.??2nn??nk1?f(x,x,x)?(x?x)?(x?x)?(x?x)(5)二次型的正惯性指数与负惯性指数顺次为222123122331(b).(d)1,2.(a)..(c)【答案】b.f(x,x,x)?(x?x)?(x?x)?(x?x)?2x?2xx?2xx?2xx【解析】22221231223312122313011???????a121?所以,故特征多项式为??110??11??|?ea|121(???????????11???130令上式等于零,故特征值为,,,故该二次型的正惯性指数为1,负惯性指数为1.故应选b.113?????????????????????k??l?l(6)已知??0?2?1?????,,,,,??????12311221331122??????112?????????l,l顺次为若,,两两正交,则1231251515151,.,.,.,.??(a)(b)??(c)(d)22222222【答案】a.【解析】使用斯密特正交化办法知0???,?]?????12???21,??[?,?]22??011???,?]?,?]???1????3132,[?,?][?,?]3321122?,?]5?,?]1l1l?2???故31,32,故选a.[?,?]2[?,?]21122,bn(7)设为阶实矩阵,下列不树立的是2微信大众号:考研文库aoa????????rrrarrra?(a)(c)?(b)(d)????oaaot?t???aao????????rara??????ott????【答案】c.ao??rraraa?()?(ra)?2(a故正确.【解析】(a)?oaa?t?t?aao????arrr(a)r(a)r(a???t(b)的列向量可由的列线性标明,故?????ot0t????a的列向量不必定能由的列线性标明.(c)(d)aao?????arrr(a)r(a)r(a???t的行向量可由的行线性标明,????ot0t????本题选c.ab0?p(b)?1,下列出题中不树立的是(8)设,为随机作业,且p(a|b)?p()p(a|b)?p()p(a|b)?p()p(a|b)?p(a)(a)若(b)若(c)若(d)若,则,则.p(a|b)?p(a|b)p(a|b)?p(),则.p(a|a?b)?p(a|a?b)p()?p(b),则.【答案】d.p((a?bp(a?b)p()p(a|a?b)??【解析】p()p(b)p(ab)??p(a(a?bp(a?b)p(ab)p(b)p(ab)?p(a|a?b)???p(a?b)p()p(b)p(ab)??p(a|a?b)?p(a|a?b)p()?p(b)?p(ab),故正确答案为d.因为,固有????????x,y,x,y,,x,y?????n,;,;22(9)设为来自全体的简略随机样本,令1122nn121211??y?xy,??nn????,xx,y????则nn12iii1?i1???d???????222??(a)是的无偏估量,1n???????2122??d??(b)(c)(d)不是的无偏估量,n????2?????22??d??是的无偏估量,1212n????2?????212??d??不是的无偏估量,212n【答案】c.x,yxyx?y【解析】因为是二维正态分布,所以与也遵守二维正态分布,则也遵守二维正态?????,e()?e(x?y)?e(x)?ey)???分布,即123微信大众号:考研文库??2???n?212d?)d(xy)d(x)dy)x,y)??????212,故正确答案为c.??x,x?,xn,4?(10)设是来自全体的简略随机样本,思考假定查验疑问:12??????h:?h:??xw?x?,标明标准正态分布函数,若该查验疑问的回绝域为011?xx??,则?其间时,该查验犯第二类差错的概率为ii1?????1??11??(a)(c)(b)????1??21??(d)【答案】b.1x?n){x?1,【解析】所求概率为4???x???{xp1????????????1122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方位上.)???(11)【答案】【解析】.x2x2??20?4??????244??arctan(x??x2x2??(x1??22000xetx0?????tdy2(12)设函数y?y(x)由参数方程?断定,则?.ytet,x0????dxt22t?0?2.【答案】3tdy(4eet)2e??????t2ttttt??【解析】由,得,e12e?t3?tdy22将t?0带入得?.23t0?xy???xy??4y?02y?y?2y?得解为(13)欧拉方程满足条件.x.2【答案】dydy22x?exy??,xy4y0????,则【解析】令t2,原方程化为,特征方程为dx22?4?0????2ycececxcx?????2,特征根为,通解为2t2t22,将初始条件121212y(1)?1,y(1)?2c?c?0y?x,故满足初始条件的解为2带入得为空间区域.??12?(x,y,z)x?4y??z?2(14)设表面的外侧,则曲面积分22xy???.22??【答案】.4微信大众号:考研文库??4?(2x2y??2?【解析】由高斯公式得原式=.0d?a?aaaa?3的每行元素之和均为2,且,(15)设为3阶矩阵,为代数余子式,若a?a?a=.3【答案】.2111????????a??????2,?a121a,1a的特征值为*????【解析】,,则,对应的特征向量为?????????????111??????1aaa1aaa1????????????????aa??????????*??1aaaaa,a1aaa1???????,而**,即???????????????a???????1aa1aaa1?????????3a?a?a?.2(16)甲乙两个盒子中各装有2个红球和2xyxy与的有联络再从乙盒中任取一球.令,别离标明从甲盒和乙盒中取到的红球个数,则数.15.【答案】?01011122???????????????????(x,y)?x?y?31515311【答复】联合分布率,????22??11141x,y)??,??,即?.45三、答复题(本题共6小题,共70分.请将答复写在答题纸指定方位上,回容许写出文字阐明、证明进程或演算进程.)(17)(本题满分10分)???1?xet21???.求极限0?ex?1x???x?0??1.【答案】2????1?xex1??xet2t21????【解析】解:00?ex?1x?exx???x?0x?0??1??xe?x?t?ot))?x?x?o(x)又因为t22233,故300111(xxo(xx??xo(xx??xo(x)????3333222原式=x2x?01xo(x)?2212=?.x22x?05微信大众号:考研文库(18)(本题满分12分)1??u(x)enx(n1,2,)u(x)n???设?n?1,求级数的收敛域及和函数.n(n?n1?e??x(1x)ln(1x)x,x(0,1)???????1e??xs(x).【答案】【解析】??e?,x1??e1??1e?x????????s(x)u(x)nexn?1?,s(x)1e?nx,x(0,1]????收敛域(0,1],???nx?n(n1)1e??x???n1n1n11xxn1???n1???s(x)2x??xln(1?x)?[?ln(1?x)?x]???n1n(n1)nn1??n1n1n1?(1?x)ln(1?x)?x,x?(0,1)s(1)lims(x)1??2x?2e??x(1x)ln(1x)x,x(0,1)???????1e??xs(x)??e?,x1??e1??(19)(本题满分12分)x2yz6????22c:c坐标面间隔的最大值.已知曲线?,求上的点到4x2yz????【答案】66????????lx,y,z,,?z?x?2y?z?6?(4x?2y?z?30)【解析】设拉格朗日函数222lx2x?u0????ly4y?u0????lz2z?u0?????x?2y?z?6224x?2y?z?30解得驻点:(4,1,12),(8,2,66)c上的点(8,2,66)到xoy面间隔最大为66.(20)(本题满分12分)d?ri(d)??x2?y2d.是有界单连通闭区域,设2获得最大值的积分区域记为1di(d)(1)求的值.1(y)(4x24y2?x)????x24y2?dd(2)核算是的正向鸿沟.,其间x4y?2211d1?【答案】.?i??x?y)d4?x?y在上2d22,当且仅当2d??r)???.?d:x?y?4i2d?2i且大于0时,抵达最大,故2221100d:x?4y?r2rd(的方向为顺时针方向,(2)补2222(y)(4x24y2?x)????x24y2?=x4y?22d16微信大众号:考研文库(y)(4x)(y)(4x24y2?x)???????x24y2x24y2?x24y2?????x4yx4y?2?222dd211211??er24er2d???????????.rrr222d2d22(21)(本题满分12分)a11????????a1a1??.已知???11a??pp(1)求正交矩阵,使得为对角矩阵;tcc?(a?3)e?.2(2)求正定矩阵,使得111??????????????53???????11???????32611151331533pc1????.【答案】(1);(2)326????12???10???36??【解析】?a?11??ea???1?a1?a?a0(1)由???????211?a??得1???a???a?123??a?2时当1211??101?1?????????((ae)121r011?????1??的特征向量为,????????1????112??000?1????????a?1所当2311111111??????????????????????????((ae)111r000?????1,?1????的特征向量为,???????23??????11100002??????111????????????????326a2??????????????111p,,2pa1??????令??13??,则t,???326???a1?123?120???36??1??????????pcppaea)pae???4(2)?????t2t47微信大众号:考研文库11????????????????????p4p?2???ttt,425?1???????31????????51?cp2p1????.故t????????3315332????1(22)(本题满分12分)x上随机取一点,将该区间分红两段,较短的一段长度记为较长的一段长度记为在区间yyz,令?.xx(1)求的概率密度;z(2)求的概率密度.x??e(3)求.??y??2???x1,z1????x?f(x)f(z)(f(z(z21?2.(3).?????【答案】(1)?;(2)zz???x1???x?f(x)?【解析】(1)由题知:?;?2x?(2)由y?2?x,即?zz,先求的分布函数:x2x??2?????????f(z)pzzpzz??p1z???????xx??fz?zz?1时,()0;当当z?1时,222???????2f(z)pz1z1px????11?????????;z1?xz1z1?????02???,z1?f(z)(f(z(z?2???;zz??xxx???????ee1????2.(3)??????y2x2x????08微信大众号:考研文库