1995年入學統(tǒng)一考試數(shù)學二試一、填空題(本題53分,滿分15答案填在題中橫線ycos(x2sin21yxyy2x的通解為曲線y 在t2處的切線方程為

n2n

n2n

n2n

)yx2ex2的漸近線方程為二、選擇題(本題共5小題,每小題3分,滿分15分.每小題給出的四個選項中,只有一項符 設f(x)和(x)在(,)內(nèi)有定義,f(x)為連續(xù)函數(shù),且f(x)0,(x)有間斷點,

[f(x)]必有間斷 f[(x)]必有間斷

[(x)]2必有間斷f曲線yx(x1)(2x)與x軸所圍圖形的面積可表示 22 22f(x在(x1x2x1x2f(x1f(x2對任意x,f(x

對任意x,f(x函數(shù)f(x)單調增 (D)函數(shù)f(x)單調增順序

f(1)f(0)f(1)f

f(1)f(1)f(0)f

f(1)f(0)f(1)f

f(1)f(0)f(1)f設f(x)可導,F(x)f(x)(1|sinx|),若使F(x)在x0處可導,則必

f(0)

f(0)

f(0)f(0)

f(0)f(0)三、(本題共6小題530

1cosx(11cosx(1 x

d2yy(x

ef

1

f

x1lnx22f[(xlnx,求(x)dxxf(x)f(x)

1,x

f(xx0

xx1cos求擺線ytsint0t2 vt0v0,已知阻力與速度成正比(比例常1),問tv0？并求到此時刻該質點所經(jīng)過的路程.3四、(本題滿分8 f(x0(2t)edt五、(本題滿分8xln設yex是微分方程xyp(x)yx的一個解,求此微分方程滿足條件 0xln六、(本題滿分8Lyf(xy0MTMP3(1y2

M(x0,y0)處的切線和法線,已知線段MP的長度為

y(x0y0y(x0P(,七、(本題滿分8f(xxsintdt,計算f(x)dx0 八、(本題滿分8設

f(x)1f(x)0f(x)xx1995年入學統(tǒng)一考試數(shù)學二試題解一、填空題(本題53滿分15cos(x2)sin【答案】2xsin(x2)sin21 ycos(x2)sin21cos(x2)sin21 xsin(x2)2xsin21cos(x2)2sin1cos1(1) cos(x2)sin2xsin(x2)sin21 x y(f(xy(f(x))f(x).yc1cosxc2sinxyy2xyy0r210r1,2i,故對應齊次方程的通解為C1cosxC2sinx設非齊次方程的特解Yaxb,則YaY0yy2x0axb2xa2b0故Y2xyC1cosxC2sinx2x【相關知識點】1.二階線性非齊次方程解的結構：設y*(x)是二階線性非齊次方y(tǒng)P(xyQ(xyf(xYxyP(xyQ(xy0yY(xy*(xYxyP(xyQ(xy0Px、Q(x1ypyqy0r2prq0rr1rryCerx1Cer2x1 rryCCxerx1

iyexCcosx

sinx其中 yP(xyQ(xyf(xy*(x)f(x)P(x)exy*(x)

的特解,其中Qm(xPm(xk不是特征方程的根、是特征方0、12. 如果f(x)ex[P(xcosxP(xsinx yp(xyq(xyf(x的特解可設 y*xkex[R(1)(x)cosxR(2)(x)sinx mmR(1xR(2x是mmmaxln,而k按i(或imm方程的根、或是特征方程的單根依次取為0或y3x7t

t

32

t

當t2x5,y8

y83(x5

y3x70x 【相關知識點】參數(shù)方程所確定函數(shù)的微分法：如果y(t),則dx 12【解析】應用準則求數(shù)列的極限.an

n2n

nn2nnn2n a

n2n n2n n2n12n1n(12n1n(n1n12nnn212n1nn212n1n(n1 n2 n2 n2 n2 1n1

12n 所 lim1n11limalim11n

n

n

n 準則,得lima1.n nn2n nn2ny

nn2n n2n yx2ex2limylimx2ex20y0

yf(xxx0處有l(wèi)imf(xxx0水平漸近線：當limf(xaaya二、選擇題(本題53滿分15(x)f(x是連續(xù)函數(shù),則(x)(x)f(xf(x)有間斷 方法二：排除法,舉出反例排除

f f(x)1,(x)

xx則f(x1,f[(x1,[(x)]21都處處連續(xù),排除(A),(B),(C).故應選擇【解析】方法一：利用定積分的求面積 0x(x1)(2x)dx0x(x1)(2 應選擇方法二yx(x1)(2x 故應選x1x2x1x2x1x2f(x1f(x2f(x1f(x2,故f(x是單調增加的.應選擇對于(A)(B)(C)f(xx3xxxxf(xf(x 0f(x)3(x)20f(x)x3故排除f(x)0f(x在區(qū)間[0,1f(1)f(x)f(0),(0xf(1f(0f(01f(1)f(1)f(0)f()f(0),(0應選擇f(xxx0f(x0f(x0存在且相等.F(x)f(xf(x|sinx|,而f(x)F(xx0處可導等價于f(x|sinx|x0令(x)f(x|sinx|,(0)

f(x)|sinx|

f(x)sinx

f

xf(x)|sinx

xf(x)sin

f

F(xx0處可導,當且僅當f(0)f(0)f(0)0.故選擇三、(本題共6小題530【解析】利用等價無窮小計算,即當x0時,sin xx

x原式

1cos

21lim 21

2 lim 1.

2

2

2

x 2xxx1 x

2x2 方法一xef(y)xef(y)f(y)yeyy y

ef(y,eyxf(y)ef(xefy)eyx

y

.x(1f(y0x(1f(y))(1f(y))x(f(y)x(1f( x(1f( x2(1f(

yf(x(1f(y

x(1f(

..

x2(1f(

x2(1f(.方法二x.方程兩邊取對數(shù) lnxf(y)y求導 1f(y)yyxf1

y

.x(1f(y(f(xy(f(x))f(x).(3)【解析】首先應求出x)的表達式.由f

1)

x2

x211,x211x21tf(tlnt1tf[(x)]ln(x)1lnx(x)(x1x.解得(x)x1(x) x(x)dxx1dx(12)dxx2lnx1Cx xf(xxx0f(x0f(x0存在且必與f(x0相等. x0

(x) 1

2x2 lim

lim

(x)lim

(x)limarctan

4 0

x0

1x f(0)limf(x)f(0)limf(x)limarctan1

x

所

f(x)

f(x)f(0)故f(xx0處連續(xù).(5)【解析】由弧微分得

x(t)2y(t)2dt

sin2t(1cost)2dt

2(1cost)dts 2(1cost)dt 22sin2tdt22sintdt22sint t00

4(11)【解析】設質點的運動速度為v(t),由題設,阻力為v(t),按第二定律mdv(t)v(t)m1dv(t)v(t這是簡單變量可分離的微分方程,解之得v(tCet另有初始條件v(0

,得v(t)vet0v0v0vet,得tln30 sln3vetdtvetln3

112v 0

0 3 四、(本題滿分8f(x)x2(2t)etdtf(x)00f(x)2x(2x2)ex20x0x2f(x)

x

f(x)2f(x) x2

f(x)由 f(x) 0x

f(x)2f(x) x2

f(x)f

2),f(2f(xf(0f(x f(2)(2t)etdt(2 etdt1e2 f(0)0(2t)etdt00 limf(x)limf(x) (2t)etdt(2 etdt 所 f(2)1e2為函數(shù)f(x)最大值,f(0)0為函數(shù)f(x)的最小值【相關知識點】積分上限函數(shù)的求導dxftdtfxxfxxdx五、(本題滿分8yexyexxexp(x)exx解 p(x)xexxxyxexxyxy(ex1)yye dx(exye dx

e eexxeexxdxCeexx

dxCeexx

eexdxC

eexx(eexC)exCeexx再由

xln

0得eln2

ln

ln20,即Ce2

yex

exx2【相關知識點】一階線性非齊次微分方程yp(x)yq(x)的通 為yepx)dx(q(x)epx)dxdxC),其中C為常數(shù).六、(本題滿分8MP1y232y (x)2(y)2

1y20 000 0 把②式,即(x0y0(y0代入①消去 (y)2(1y2)2/y2

0,知曲線是向上凹的,容易看出

1 0且 且

y(1y2 yy yy 0

00xy0(1y200

0y1(1y20 七、(本題滿分8【解析】方法一：這是一個積分上限函數(shù)求定積分,可以考慮用定積分的分部積分法由 f(x)sinxf(0)0sintdt00f(x)dx f(x)d(x)f(x)(x)