1993考研数二真题及解析

1993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

limxlnxyy(x由方程sin(x2y2exxy20dyx1xt设F(x)1(2 )dt(x0),则函数F(x)的单调减少区间是ttanxdxyf(x)过点(01,且其上任一点(x,yxln(1x22f(x)二、选择题(5315每小题给出的四个选项中,只有一项符x0

1sin1 (A)无穷 (B)无穷(C)有界的,但不是无穷 (D)有界的,但不是无穷|x21|f(x

x1

,xx

则在点x1处函数f (A)不连 (B)连续,但不可x(C)可导,但导数不连 (D)可导,且导数连xx2,0xf(x1,1x1x3,0x

F(x)

f(t)dt(0x2),则F(x) 1x31,0x(A)1x3,0x

x,1x1x31,0x

3

x1,1x设常数k0,函数f(x)lnxxk在(0,)内零点个数 e(A) (B) (C) (D)若f(x)f(x),在(0,)内f(x)0,f(x)0,则f(x)在(,0)

f(x)0,f(x)

f(x)0,f(x)

f(x)0,f(x)

f(x)0,f(x)三、(本题共5小题,每小题5满分252

d2ysinfx2x2

)]fdx2x)求

dx01cos dx0(1求微分方程(x21)dy(2xycosx)dx0满足初始条件 1的特解四、(本题满分9yyyexye2x1x)ex,试确定常数,,,并求该方程的通解.五、(本题满分9Ax2y22xyxAx2旋转一周所得旋六、(本题满分9作半径为r的球的外切正圆锥,问此圆锥的高h为何值时,其体积V七、(本题满分6x0,常数ae,证明(ax)aaax八、(本题满分6f(x在[0af(0)0af(x)dxMa2Mmax|f(x| 01993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】【解析】这是个0型未定式,可将其等价变换成1limxlnxlimlnx洛

limx0

x0

x0

y2ex2xcos(x2y2)2ycos(x2y2)2xy

【解析】这是一个由复合函数和隐函数所确定的函数,将方程sin(x2y2exxy20两x求导,得

y

cos(x2y2)(2x2yy)exy22xyy0y2ex2xcos(x2y2)2ycos(x2y2)2xy 如果ug(xxyf(x)在点ug(xyfx【答案】0x4

dy

f(u)

dydydu du【解析】由连续可导函数的导数与0txF(x1(2tx

1)dtx

F(x)21xxF(xF(x)2xx

0

1x2xF(x单调减少区间为0x14yf(x)在[ab上连续,在(ab如果在(abf(x)0yf(x)在[ab如果在(abf(x)0yf(x)在[ab【答案】2cos12x【解析】

tan

sin sin

2cos

cosxcos cos2xdcosx2cos2xC1(1x2ln(1x21x2

dyxln(1x2

dyxln(1x2dxx22

yxln(1x2)dx1ln(1x2)d(x21)1ln(1x2)d(x21)1(1x2)ln(1x2)

(1x2)

2x11(1x2)ln(1x2)21(1x2)ln(1x2)1x2 yf(x过点(01,故C1 y1(1x2)ln(1x2)1x21 二、选择题(5315x0sin11x1若取

k

1x2x2

(k)2sink0

(2k

12

1x2x2

(2k1)22,(k1,,),)因此,当k时,有x0及 0,但变

1sin1或等于0或趋于,这

明当x0时它 【解析】利用函数连续定义判定,即如果函数在x0处连续,则

xx0

f(x)

xx0

f(x)f(x0)limf(x)

|x2

x2

lim(x1)2

x

x

limf(x)

|x2 1

lim(x1)2

x

x

f(xx1x1当0x10xt1f(t)t2xF(x)xf(t)dtxt2dt1t3x

1(x31)

当1x2时,1tx2f(t1,所F(x)xf(t)dtx1dttxx1 应选f(xyf(xxf(x)lnxxkxef(x)0xe

f(x)11 f(x00x f(x)0,ex;f(x)严格单调减少xef(e)lneekk0eelimf(x)lim(lnxxk)e

,x

f(x)lim(lnx k)

由连续函数的介值定理知在(0e与(ef(x)lnxxk在(02,选项(B)e由f(x)f(x),f(x)为奇函数,图形关于原点对称；在(0f(x)0,f(x)0,f(x图形单调增加且向上凹,f(x在(0f(x)0,f(x)0,选f(x)f(x)f(x)f(x),f(x)f(x)x(0)时,有x(0f(x)f(x)0,f(x)f(x)0故应选三、(本题共5小题,每小题5满分25【解析】ysin[f(x2cos[f(x2f(x22xycos[f(x2)]f(x2)cos[f(x2)]f(x2)2xcos[f(x2)]f(x2)cos[f(x2)]f(x2)sin[f(x2)][f(x2)]2(2x)2cos[f(x2)]f(x2)cos[f(x2)]f(x2)2如果ug(xxyf(x)在点ug(xyfxdy

f(u)

dydydu dux(x(x2100x)(x2100 x2100lim

x)x2x2

x2100xx2100x2 x0

x1 xx1x

x2 x2

1100x2

504 4

xsec2

dx

14xdtan101cos

21xtanx4

4tanxdx

1

0)

sin4 2

2

0cos 1 4 dcosx ln(cos 20cos

2 2

dx

(1x)1dx

[(1x)2(1x)3]d(10(1

(1

2x1(1x)1

(1

limb

2(x1)2

1011b2(b

y

x21

ycosxx2

x210

2xyex1

cosxx2

2xex

dxd(x21)

d(x2 e

x2

x21

x2

dxC

cosxdxCsinxCx21

x2

y

1

sin0C102

C1

ysinx1x2yp(xyq(x)yepx)dx(q(x)epx)dxdxC),其中C为常数.四、(9【解析】要确定常数,ye2x1x)exy2e2xex(1x)ex2e2x(2x)exy2e2x(2x)ex4e2xex(2x)ex4e2x(3x)exyyyex

(42)e2x(32)ex(1)xexex解之得32,

42321

y3y2yexy3y2y0r23r20

2y3y2yex ycexce2xy*cexce2xe2x(1x)excexc 五、(本题满分9x2y22x等价于(x1)2y211y1yx1

y(0y1)yydy dV(2x)2dy(2x)2 (211y2)2(2y)2 12 (1y)21 1所 V21

(1y)2dy1对于1

0 1y2dyysintdycostdt 1 1y2dy2cos2tdt12(1cos2t)dt1t1sin2t21 2

2 (1y)3

对 (1y)2dy(1y)2d(1y) 1所 V (1y)2dy211.11 01 2x2xy1

dV2(2x)(y1y22x22x2x1所以V202x1

x)dxx1t,则x1tdxdt,所2x102x12(1t)(12(1t)(1

1t(1t)dt01t

1tt21t 1 再令tsin,则dtcosd所 0 t21dt

(cossincossin21)cos1t11t1t

02000

cos2d

sincos2d

sin2cosd

cos000 00010(1cos2)d

cos2dcos

sin2dsin

cos0002000

1

1sin2

cos3

sin3

sin002 0

1 2x1所 V202x1

x)dx2(11) 六、(本题满分9设圆锥底半径为R,如图,BCR,ACh,ODr OA2DO由BCOD,ADOA2DO R (hr)2r h2

V Rh

h

(2rh)对上式两端对h求导,并令V0,得 22h(h2r)

2h(hVh3得唯一驻点h4r

(h

3

(h

02rh4r,V4rh,V0所以h4r为极小值点也是最小值点,最小体积V(4r)8r33七、(本题满分9x0,常数aealn(ax)(ax)ln

或ln(ax)lnaa 证法一f(x)(axlnaaln(axf(x)lna

.aaex0知lna

f(x)0(x0)f(xf(x)(ax)lnaaln(ax)f(0)alnaalna0,(x (ax)lnaaln(ax)0所 (ax)aaax证法二f(x)lnxf(x)1lnx xae时,有f(x1lnx0xaef(xa)f(a所以有ln(ax)lnaa (ax)aaax八、(本题满分9x[0af(0)0f(x)

f(x)f(0)f()x,(0f()xMmax|f(x|0|f(x)||f()|xMx将两边从0ax 0|f(x)|dxM0xd