2014年考研數(shù)學(xué)二真題及答案

一、選擇題:184321x0時(shí)，若ln12x(1cosx)x高階的無窮小，則是 (2,

(2

(0,2 【解析】由定義lim lim lim2 所以10，故1

2

x

時(shí)，(1cosx)

10，即2 yxsin (B)yx2sinyxsinx

yx2sinx

xsin sin【解析】關(guān)于C選項(xiàng)： xlim1 x101 lim[xsin1xlimsin10yxsin1yx

設(shè)函數(shù)f(x)具有2階導(dǎo)數(shù)，g(xf(0)(1xf(1)x，則在區(qū)間[01 f(x)0時(shí)，f(x)(C)f(x0f(x

f(x)0f(x)(D)f(x)0f(xF(xg(xf(xf(0)(1xf(1)xf(xF(0)F(1)0F(x)f(0)f(1)f(x),F(x)f(x)f(x0F(x0F(x在[0,1上為凸的F(0)F(1)0x[0,1F(x0g(xf(x故選

yt4t

上對(duì)應(yīng)于t1的點(diǎn)處的曲率半徑是

(C) (D)

2t2t k ,R110 1y'2

設(shè)函數(shù)f(x)arctanx，若f(x)xf()，則lim2 x02 3

2

3fx

f'()

，所以2xff lim limxf(x)limxarctanx 1x2x0 x0x2f x0x2arctan 設(shè)函數(shù)u(x,y)DD的內(nèi)部具有2

xy0x2

0，則 u,Bu,B

,C2u,B0,

則=AC-B20,所以u(píng)(x,yD內(nèi)無極值，則極值在邊界處取得.故選A(ad

0ab0a00b0cd0c00d(ad (C)a2d20ab0a00b0cd0c00d0a0c 0a0c

0b0 0b0

a 0c

ad(adbc)bc(ad(adbc)2設(shè)a1,a2,a3均為三維向量則對(duì)任意常數(shù)k,l,向量組a1ka3,a2la3線性無關(guān)是向量組a1,a2,a3線性無關(guān)的( 【解析】 l

1 0 1 3 AAC0,1323123 3

1.若 性無關(guān)，則r(A)r(BC)r(C)2，故1k3,2l3線性無關(guān))舉反例.令30，則1,2線性無關(guān)，但此時(shí)1,2,3卻線性相關(guān)綜上所述，對(duì)任意常數(shù)kl，向量1k3,2l3線性無關(guān)是向量1,2,3線性無 x22x5dx 3【答案】8

dx

1 x22x

x12 132 2

f(x)是周期為4的可導(dǎo)奇函數(shù)，且f(x)2(x x[0,2]，則f(7).fx2x1，x0,2fx2xfxx22xc且為奇函數(shù)，故fxx22x，x2, fx的周期為4，f7f1zz(xy是由方程e2yzxy2z7確定的函數(shù)，則41 【解析】對(duì)e2yzxy2z7xy4

11 (,)2e2yz2yz1z e2yz(2z2yz)2yz x

1,y

1時(shí)z 1 1

2(,22

2y(,

11(,)22

1dx(1)dy1(dx n曲線limn程 2

的極坐標(biāo)方程是r，則L在點(diǎn)(r,) 2yx

xrcosyrsinsin 于是r,對(duì)應(yīng)于x,y022

22 dx cos

22d yx2

0, yx一根長(zhǎng)為1的細(xì)棒位于x軸的區(qū)間[0,1]上,若其線密度xx22x1,則該細(xì)棒的質(zhì)心坐標(biāo)x x10x

x22x1dx x 3 3 xx

2 x2 xx2x1dx 2 x12= 3(13)fxxxx2x22axx4xx1，則a 1 2.fxxxxax2a2x2x2x2 1，所以4a20，故2a2x 1 t2et1

x2ln11 xxx

x

x2ln(11)

x21

1lim[x2(ex1) xlime1tlim 1limt1 t0 x2y2y1y(y21)y1 1y3yx1x3 y(20得c3 1又由①可得y(x)y2y(x0x1x1,y(x)1x1,y(x)x1,y(x)y(1)0,y(1)(17)(本題滿分10分) Dx,y1x2y24,x0,y xsinx2y2D x D x2 x2y2

x

x

x

x2y2)I

x

)dxdy

xy)y

x 2D x x 1

x2y221 2dsinr2

1

22 1cosrr|22cosrdr4 1211sinr|24 4

1 具有二階連續(xù)導(dǎo)數(shù)，z

2

ecosy) f f(e x2

f(0)0,f'(00f(uzfexcosy,zf(excosyexcosyzf(excosyexsiny2z

f(ecosy)ecosyecosyf(ecosy)ecosy2z f(ecosy)esinyesinyf(ecosy)ecosy

由x2+y24zecos fexcosye2x[4fexcosyexcos即fexcosy4fexcosyexcosy令excosy=tft4ft 特征方程240, 得齊次方程通解yce2tc y*atb，代入方程得a1b0y*1 y=ftce2tce2t1 f00,f00，得c

,c ,

y=fu1e2u1e2u1u f(xg(x在區(qū)間[abf(x)0g(x)1，證明：（I）x0ag(t)dtxa,xxb ag(t f(x)dx

f(x)g(x)dxa0gx1，0gxaxaax0agtdtxx0xgtdtx1dt=x ua （II）FuufxgxdxaagtdtfxdxFufugufaugtdtgua gufufaugtdt 由（I）知0agtdtu aa

gtdaa取ubFb0，即（II）成立f(x)

1

f1(x)f(x),f2(x)f(x1x軸所圍成平面圖形的面積，求極限limnS

Snyfn(x 【解析】f1x)

,f2(x)

,f3(x)

,fn(x) 1

1 dx1x

1

1S1f(x)dx

n 0

01

11 1 n01dx

n

1

dx 2ln(1nx) 1111ln(111 limnS1limln(1n)1limln(1x)1

10 已知函數(shù)f(x,y)滿足y2y1，且f(y,y)(y1)(2y)l求曲線fx, 222y(x其中(x)為待定函數(shù) 2(y1),所以f(x,y)fyyy1)22ylny則y12ylnyf(x,y)y22y12xlnx(y1)22xlnxf(xy0,可得y1)22xlnxy1x1x2為 V1y1dx12xln x22lnxd2x 2 x2 x lnx(2x2)122 5 2ln2 4

0 1

AE 11001 AE 1100100 0011101410 0

1 0 1 1 00 (I)Ax0的基礎(chǔ)解系為1,(II)e1,0,0T,e0,1,0T,e0, Axexk2110T2k12k13kk 1Axexk6340T6k32k43kk Axexk1,1,10T1k,12k,13kk 1

63

13