1990考研數(shù)二真題及解析

1990年入學(xué)統(tǒng)一考試數(shù)學(xué)二試一、填空題(本題共5小題,每小題3分,滿分15分.把答案填在題中橫線 曲線ysin3t上對(duì)應(yīng)于點(diǎn)t6點(diǎn)處的法線方程是y

tanx

1y1x1

0

1xdx下列兩個(gè)積分的大小關(guān)系是：1ex3dx

1ex3dx |x|設(shè)函數(shù)f(x) |x|1,則函數(shù)f[ |x|二、選擇題(315每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)符合題目要求,把2已知lim axb0,其中a,b是常數(shù), 2

xx a1,b

a1,b

a1,b

a1,b 設(shè)函數(shù)f(x)在(,)上連續(xù),則df(x)dx等

f

f

f(x)

ff(xf(x)f(x)]2,則當(dāng)n為大于2f的n階導(dǎo)數(shù)f(n)(x)

n![f

n[f

[f

n![f

f

eF(x)e

f(t)dt,則F(x)等

exf(ex)f

exf(ex)f

exf(ex)f

exf(ex)ff(x)(5)設(shè)F(x) fxxf(xx0f(00,f(00xF(x (A)(B)(C)(D)三、(每小題5滿分25已知limxa)x9,求常數(shù)axx求由方程2yxxyln(xyyy(x的微分dy1y

1

(x0

lnxdx (1 xlnxdyylnx)dx0四、(本題滿分9

1P 圍圖形面積為最小(其中a0b0五、(本題滿分9 證明：當(dāng)x0,有不等式arctanx 六、(本題滿分9f(xxlntdtx0f(xf111 七、(本題滿分9

x2x八、(本題滿分9y4y4yeax之通解,其中a1990年入學(xué)統(tǒng)一考試數(shù)學(xué)二試題解一、填空題(本題共5小題,每小題3滿分15 【答案】y 3(x t【解析】將txy在tx33yt

1

t 得切點(diǎn)為 3,). 過(guò)已知點(diǎn)(x0,y0yy0k(xx0,當(dāng)函數(shù)在點(diǎn)(x0,y0yx

0k

y(x0

.所以需求曲線在點(diǎn)t 處的導(dǎo)數(shù)6dydydt13 dt13

3sin2tcos 3cos2t

tant6

法線斜率為k 3.所以過(guò)已知點(diǎn)的法線方程為y 3(x

如果ug(xxyx

f(x)在點(diǎn)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在點(diǎn)ug(xy

fg(x)dyf(ug(xdydydu4

du111方法1：換元法

t,原積分區(qū)間為0x1,則01x1,進(jìn)而0

新積分區(qū)間為0t1x0t1x1t001 dt1

dx1dx,則dx2tdt221原式0(1t2t1121t2t4dt21t31t51 5211 5 方法2：拆項(xiàng)法xx11,11

13 1xdx1x2 5 【答案】

1x 1x2 0【解析】由于ex31ex3dx1ex3dx

x3在[21連續(xù)且ex3ex3若f(xg(x)在區(qū)間[abababf(x)g(x) 有af(x)dxa【答案】f(x的定義知,當(dāng)|x|1f(x1.ff(x)]f(11于是當(dāng)|x|ff(x當(dāng)|x|1f(x0ff(x)]f(01即當(dāng)|x|1ff(xx(ff(x二、選擇題(315x2lim2

axb xx 1a分析應(yīng)有

(1a)x2(ab)xb否則lim 0ab

x

x 所以解以上方程組,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假設(shè)nky(k)kyk1 y(k1)y(k)k!yk1k1!yk k1!yk11所以nk1亦成立,原假設(shè)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在點(diǎn)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【相關(guān)知識(shí)點(diǎn)】1.yf(xx0f(xx0如果limf(xf(x0f(xx0函數(shù)f(x)的間斷點(diǎn)或者不連續(xù)點(diǎn)的定義：設(shè)函數(shù)f(x)x0的某去心鄰域內(nèi)有定義,xx0xx0有定義,但limf(xxx0有定義,且limf(x存在,但limf(xf(x0 通常把間斷點(diǎn)分成兩類：如果x是函數(shù)f(x)的間斷點(diǎn),但左極限f(x及右極限 f(xxf(x) 三、(每小題5滿分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】對(duì)分式求導(dǎo)數(shù),有公式 v

y

(1x2

,y

2(3x2,(1x2131313 時(shí),y0;x 131313故拐點(diǎn)為

3,)34x0處的左右側(cè)凹凸性相反,則稱(x0,f(x0f(xf(x在(x0x0連續(xù),在去心鄰域(x0x0x0(x0x0x0f(x)(xx0在0

x

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

(1

d(1x)(1

有(1lnxdx

lnxd(1)分部lnx

(1 1(1

1

1

1xlnxln|1x|C C1【相關(guān)知識(shí)點(diǎn)】分部積分公式：假定uu(x與vv(xuvdxuvuvdx或者udvuv1y y 1xln dx

ln由

xln

|lnx|,兩邊乘以lnx得(ylnx) xln

ylnx

dxCx

ln ln

1可得C1,所求特解為ylnx

2ln四、(本題滿分9

dyb2

a2y過(guò)曲線上已知點(diǎn)(x0,y0yy0k(xx0y(x0ky(x0)所以點(diǎn)(xy處的切線方程為Yy

b2

Xx

1 a2

22X0與Y0xyab22 又由橢圓的面積計(jì)算公式ab,其中a,b為半長(zhǎng)軸和半短軸,故所求面積1a2 S ab,x(0,a)2 abSxy最大；從而問(wèn)題化為求uxyyx(0a令uxy,uxyy0y

,再 1兩邊求

xyy

2x2

a(唯一駐點(diǎn)),即在此點(diǎn)uxySlimS(x)

S(x)S(x在(0ax

a22P點(diǎn)為

a,b22 22五、(本題滿分9f(xf(x令f(x)arctanx1,則f(x)

0(x0f(x 1 (0limarctanx

f(x)lim x

) x

0故0xf(x)

f(x)0六、(本題滿分9 1ln

1f(

dt,由換元積分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-萊布尼茲公式,

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、(本題滿分9 y

,過(guò)曲線上已知點(diǎn)(x0,y01212x為yy0k(xx0),當(dāng)y'(x0)存在時(shí),ky'(x0) 所以點(diǎn)(x x2)處的切線方程 x0x0

12x0(x12x0

123y1y(3 1123 x2yyf(xx2y1xx軸旋轉(zhuǎn)一周所形成的,求旋轉(zhuǎn)體體積V：1：yx的函數(shù),VV31(x1)2dx3(x1 311(x1)33(1x22x)34 2：xy的函數(shù),并作水平分割,相應(yīng)于yydy

V

1(y32y2y)dy21y42y31y21 【相關(guān)知識(shí)點(diǎn)】1.yf(xxaxbxx旋轉(zhuǎn)一周所得的旋轉(zhuǎn)體體積為：Vbf2x)dxa2f(x在[aba0yf(xxaxbxby軸旋轉(zhuǎn)所得旋轉(zhuǎn)體體積為：V2axf(x)dxb八、(本題滿分9【解析】所給方程為常系數(shù)二階線性非齊次方程,特征方程r24r40的根為rr2,原方程右端eaxex中的a 當(dāng)a2時(shí),可設(shè)非齊次方程的特解YAeaxA

1,(a當(dāng)a2時(shí),可設(shè)非齊次方程的特解Yx2AeaxA12

y(c1

2x (a x2e2

(a2)y(c1c2

(a2)【相關(guān)知識(shí)點(diǎn)】1.y*(x)yP(xyQ(xyf(x的一個(gè)特解.Y(xyP(xyQ(xy0yY(xy*(