1990考研数二真题及解析

1990年入学统一考试数学二试一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线 曲线ysin3t上对应于点t6点处的法线方程是y

tanx

1y1x1

0

1xdx下列两个积分的大小关系是：1ex3dx

1ex3dx |x|设函数f(x) |x|1,则函数f[ |x|二、选择题(315每小题给出的四个选项中,只有一项符合题目要求,把2已知lim axb0,其中a,b是常数, 2

xx a1,b

a1,b

a1,b

a1,b 设函数f(x)在(,)上连续,则df(x)dx等

f

f

f(x)

ff(xf(x)f(x)]2,则当n为大于2f的n阶导数f(n)(x)

n![f

n[f

[f

n![f

f

eF(x)e

f(t)dt,则F(x)等

exf(ex)f

exf(ex)f

exf(ex)f

exf(ex)ff(x)(5)设F(x) fxxf(xx0f(00,f(00xF(x (A)(B)(C)(D)三、(每小题5满分25已知limxa)x9,求常数axx求由方程2yxxyln(xyyy(x的微分dy1y

1

(x0

lnxdx (1 xlnxdyylnx)dx0四、(本题满分9

1P 围图形面积为最小(其中a0b0五、(本题满分9 证明：当x0,有不等式arctanx 六、(本题满分9f(xxlntdtx0f(xf111 七、(本题满分9

x2x八、(本题满分9y4y4yeax之通解,其中a1990年入学统一考试数学二试题解一、填空题(本题共5小题,每小题3满分15 【答案】y 3(x t【解析】将txy在tx33yt

1

t 得切点为 3,). 过已知点(x0,y0yy0k(xx0,当函数在点(x0,y0yx

0k

y(x0

.所以需求曲线在点t 处的导数6dydydt13 dt13

3sin2tcos 3cos2t

tant6

法线斜率为k 3.所以过已知点的法线方程为y 3(x

如果ug(xxyx

f(x)在点ug(x)y

fg(x)dyf(ug(xdydydu du【答案】

tanx

11sin

tantan x

11 xx2

xx2

1

tan

tan1

1 x x

x x

xtantan x

1sin

x

11 x

xxtantan

tanx

11

11sin

xcos

1.. xx2

xx2f(x)g(x)f(x)g(x)f(x)g(x)如果ug(xxyx

f(x在点ug(xy

fg(x)dyf(ug(xdydydu4

du111方法1：换元法

t,原积分区间为0x1,则01x1,进而0

新积分区间为0t1x0t1x1t001 dt1

dx1dx,则dx2tdt221原式0(1t2t1121t2t4dt21t31t51 5211 5 方法2：拆项法xx11,11

13 1xdx1x2 5 【答案】

1x 1x2 0【解析】由于ex31ex3dx1ex3dx

x3在[21连续且ex3ex3若f(xg(x)在区间[abababf(x)g(x) 有af(x)dxa【答案】f(x的定义知,当|x|1f(x1.ff(x)]f(11于是当|x|ff(x当|x|1f(x0ff(x)]f(01即当|x|1ff(xx(ff(x二、选择题(315x2lim2

axb xx 1a分析应有

(1a)x2(ab)xb否则lim 0ab

x

x 所以解以上方程组,可得a1b1F(xf(xf(x)dxF(xCdF(x)f(x)dxd[f(x)dx][f(x)dx]dxfyf(xyy2y2yy2y3,y3!y2y3!y4 y(n)nyn1假设nky(k)kyk1 y(k1)y(k)k!yk1k1!yk k1!yk11所以nk1亦成立,原假设成立.eeF(x)

f(t)dtF(x)f(ex)(ex)f(x)(x)exf(ex)f(tF(tt)f(x)dx(t(tF(t)(t)f(t)(t)f(t)如果ug(xxy

f(x在点ug(xy

fg(x)xdyf(ug(xdydydu

du

limF(x)

f

f(x)f,

xf(x)limy

f(x0x)f(x0) x0 得limF(xf(00f(0F【相关知识点】1.yf(xx0f(xx0如果limf(xf(x0f(xx0函数f(x)的间断点或者不连续点的定义：设函数f(x)x0的某去心邻域内有定义,xx0xx0有定义,但limf(xxx0有定义,且limf(x存在,但limf(xf(x0 通常把间断点分成两类：如果x是函数f(x)的间断点,但左极限f(x及右极限 f(xxf(x) 三、(每小题5满分25lim(11)xx

(1a

ax)a )xlim lim e2a9

x

(1ax

ax)x

2aln9aln3

xa

x或 lim(xa)x

2a2a

e2a

x

x

xa 同理可得aln32dydxln(xy)d(xy)(xy)dln(x

(dxdy)ln(xy)(xdy2ln(xy)dx3ln(x

dx,xuuvv【解析】对分式求导数,有公式 v

y

(1x2

,y

2(3x2,(1x2131313 时,y0;x 131313故拐点为

3,)34x0处的左右侧凹凸性相反,则称(x0,f(x0f(xf(x在(x0x0连续,在去心邻域(x0x0x0(x0x0x0f(x)(xx0在0

x

(x0,f(x0f(x在(x0x0f(x0)0f(x0)0则(x0,f(x0

(1

d(1x)(1

有(1lnxdx

lnxd(1)分部lnx

(1 1(1

1

1

1xlnxln|1x|C C1【相关知识点】分部积分公式：假定uu(x与vv(xuvdxuvuvdx或者udvuv1y y 1xln dx

ln由

xln

|lnx|,两边乘以lnx得(ylnx) xln

ylnx

dxCx

ln ln

1可得C1,所求特解为ylnx

2ln四、(本题满分9

dyb2

a2y过曲线上已知点(x0,y0yy0k(xx0y(x0ky(x0)所以点(xy处的切线方程为Yy

b2

Xx

1 a2

22X0与Y0xyab22 又由椭圆的面积计算公式ab,其中a,b为半长轴和半短轴,故所求面积1a2 S ab,x(0,a)2 abSxy最大；从而问题化为求uxyyx(0a令uxy,uxyy0y

,再 1两边求

xyy

2x2

a(唯一驻点),即在此点uxySlimS(x)

S(x)S(x在(0ax

a22P点为

a,b22 22五、(本题满分9f(xf(x令f(x)arctanx1,则f(x)

0(x0f(x 1 (0limarctanx

f(x)lim x

) x

0故0xf(x)

f(x)0六、(本题满分9 1ln

1f(

dt,由换元积分t ,dt du,t:1 u:1x2 1121

1ln

tu

lnx所 f(x)x

1tdt1u(u1)duf(x)f(1)=xlntdtxlntdtxlntdt1ln2x 11 1t(t 12：F(xf(xf()xF(x)

lnx

lnx

1lnx由-莱布尼兹公式,

1

11 xF(x)F(1)xlnxdx1ln2x F(11lnxdx0F(xf(xf11ln2x1 】f(x在[abF(xf(x在[ab】aabf(x)dxF(x)bF(b)Faa七、(本题满分9 y

,过曲线上已知点(x0,y01212x为yy0k(xx0),当y'(x0)存在时,ky'(x0) 所以点(x x2)处的切线方程 x0x0

12x0(x12x0

123y1y(3 1123 x2yyf(xx2y1xx轴旋转一周所形成的,求旋转体体积V：1：yx的函数,VV31(x1)2dx3(x1 311(x1)33(1x22x)34 2：xy的函数,并作水平分割,相应于yydy

V

1(y32y2y)dy21y42y31y21 【相关知识点】1.yf(xxaxbxx旋转一周所得的旋转体体积为：Vbf2x)dxa2f(x在[aba0yf(xxaxbxby轴旋转所得旋转体体积为：V2axf(x)dxb八、(本题满分9【解析】所给方程为常系数二阶线性非齐次方程,特征方程r24r40的根为rr2,原方程右端eaxex中的a 当a2时,可设非齐次方程的特解YAeaxA

1,(a当a2时,可设非齐次方程的特解Yx2AeaxA12

y(c1

2x (a x2e2

(a2)y(c1c2

(a2)【相关知识点】1.y*(x)yP(xyQ(xyf(x的一个特解.Y(xyP(xyQ(xy0yY(xy*(