xx函數(shù)fx xx(A) (B) (C)

a1,b1 6a1,b1 6

a1,b16a1,b16 使不等式xsintdtlnx成立的x的范圍 (A)

0,1) (B)1,2

,)

x則函數(shù)Fx0ftdt的圖形 x AB2A*B*ABA2B3 A

O

O

O

O 0 0 2 3),3 3

0 0 2 0 0 2

0 0 2 0 0 2 設(shè)A與B互不相容, P(AB)0 (B)P(AB)P(A)P(B)(C)P(A)1P(B) (D)P(AB)1設(shè) 量X與Y相互獨(dú)立,且X服從標(biāo)準(zhǔn)正態(tài)分布N0,1,Y的概率分布PY0PY11.記Fz為 量ZXY的分布函數(shù),則函數(shù)Fz (A) (B) (C) (D)31x31x2設(shè)z(xey)x,則 xen(1)nn冪級(jí)數(shù) n

x的收斂半徑 設(shè)某產(chǎn)品的需求函數(shù)為QQpp的彈性p0.2件時(shí),價(jià)格增加1元會(huì)使產(chǎn)品收益增 0 0,則k 0 設(shè)X1,X2,,Xm為來自二項(xiàng)分布總體B(n,p)的簡(jiǎn)單隨機(jī)樣本,X和S2分別為樣本均值和樣本方差,記統(tǒng)計(jì)量TXS2,則ET f(xyx22y2ylny的極值.(16)(本題滿分10分)計(jì)算不定積分l (x0)n D fxx0處連續(xù),在0,0x

fxAf0f0Ayf(x)f(x)f(x0yf(x)與直線y0x1xt(t1)x軸旋轉(zhuǎn)一周所得的立體體積值是該曲邊梯形面積值的t倍,求該曲線方程.(2011 A

1 1

(21)(本題滿分11分)fx,x,xax2ax2a1x22xx2x 1 2f 設(shè)二維 ex, 0yx,f(x,y) fYXyx(23)(本題滿分11分)1,23 2009 xx xx窮多個(gè),但可去間斷點(diǎn)為極限存在的點(diǎn),故應(yīng)是xx30的解 limxx3lim1 1x0xx1sinx

x01x11

22lim lim x1sin limf(x)limxsinax等limxsinx0 x0x2ln(1 x0x2

洛

洛 x0 x0 a3sin lim

x0

xsinf(x

tf(x)xsintdtlnxxsintdtx1 1xsint1dt11sintdt 由t0,1時(shí), t②④⑤F(x A 【解析】分塊矩陣

AB236 A A A B1

O 6 O 1B

2B

6

O

A 0

0 0P 0 3 1 QTAQ

P

0 AP 1

1 0 0 0 0 0 0 0 0 1 2 1 2PABPAB1PAB1,故(DFZ(z)P{XYP{XYzY0}P{Y0}P{XYzY1}P{Y1P{XYzY0}1P{XYzY 1P{X0zY0}1P{XzY F(z)1P{X0z}1P{Xz} 1

2z0

(z)11(z) 3【答案】2 3131x231x2

e1x【答案】12ln

lime(1cosx) 3ex 1 x01 x1xexln(1x)exln(1x)ln(1x)xxy 1x

eln2ln212ln21

z

(xey)x

ln(xey)

xey (1e0)ln(1e0) 2ln21x 1e0【答案】a

en 0en1

n

enn

11en1 eQp0.2,所以Qp0.2Q,所以Qp0.2QQ0.8Q因?yàn)镼將Q10000代入有Qp8000 0 0 ETEXS2EXES2npnp(1pnp2 fx(x,y)2x(2y2)fy(x,y)2x2ylny 令

解得唯一駐點(diǎn)0,eA 12(2y 12(21

(0, 2B 12

e1C 1

(0,1e1

y1y(0,e B2AC2e(21)0,且A0f(0,1f(xyf(0,1 1x 1xt2 dxln1tdt21 t2而

t2

1dtt1 1 t21t1dt4

t t

t 1lnt11lnt t t2 4lnt 2 xln

1x1

x 1x 2 x

1xln 2ln x1

1 1x2 ln x2x1x2 1 dxxln 1 2 x u2

1xdxu1u2u2

2udu2uxlnuu21C 1xx即ln dxxln x x x

xx 1x 1xxx 1xx x2x1xx yDOx0r2(sincos),3yDOx d (xy)dxdy432(sincos)(rcosrsind 3

4(cossin)4

r 3 yy12x8 48

(sincos)3844 442(xD1x,y y 2(x1)2 2(x2D2x,yxy 2(x1),0x2

2 (x010

2(x1)2(x1)22 3 , 1 xydxdy0

(x1222(x1)2(x1)220

22(x2

30 xydxdy xydxdy xydxdy 2 F(xf(x

f(b)fb

(xa)F(a)f(a)F(b)f(b)

f(b)f(a)(aa)bf(b)f(a)(ba)b

fff(b)f 定理,存在a,b,使得F() f(b)f(a)f()(ba).

b

f0

f(t)f

f

t

t

t

tlimftAt0時(shí),0limf(Af(0)t tf2(x)dxttf(x)dx

t 兩邊對(duì)tf2(t)tf(x)dxtf(t)1代入t1得f(11f(10 2f(t)f(t)2f(t)tf(t) 記f(t)y,則 t11 t

12y

e2ydyC)y

12

ydy1 y2 y2C) y 代入t1,y1得C1,從而t2y

.故所求曲線方程為x y 2 22f(t)f(t)2f(t)tf(t),f(1)1 dy2y 2yyu,則dyutdu 原方程變成t

2u1分離變量 2uu(3

du1dtt dudt 3u32u t

1lnu(32u)2lnCt3 u3(3

3Ct1代入t1u1,得C1,所以u(píng)(32u)2t代入uy化簡(jiǎn)得y(3t2y)21,即t 2y.故所求曲線方程為x2y ) 1

11 21A 1 1

2 2 0

1k 21k 2

0 又A2 0

0 000000A000000

0 2 0 0 a

2 2

,,

1 1 1

1012

A,得kA20,即k0 3f a 0 1 A a00 1(Ⅱ)1fy2y2A

1 0 0

0 a1時(shí),1,0,3f的規(guī)范形為y2y2 a2時(shí),2,3,0fy2y2 個(gè)為零.又a2aa1a2 【解析】(I)X

x xfX(x) f(x,y)dy x

x f(x, 1 0y (y|x) Y

fX

ey

yfY(y)f(x,y)dx yPX1|Y1PX1,Y xf(x,y)dxdy0dx0edye20

1 e

P{X1,Z

11 26 【解析