文科高數(shù)習(xí)題一答案在三中

xx9x9

不存在4證明：當(dāng)x4

2

3是同階無(wú)窮小量

11x

(x0)1x當(dāng)x1x

1

和 當(dāng)x1時(shí)，無(wú)窮小1x3 3

x)設(shè)當(dāng)x0時(shí)，1cosx～asin2x，求a的值2

x32

2xx2(x

x1

5x

3x

1

sin

x2

limarctan

xcos

x2xcosx2

x31x

xx x

x

5xx2x212n

2x

x23x3212x23x32n

n

11 n1

11212

12n

3n13

3

5

(2nx22x

1

x

x11

1x

xn

(nN

4x32x2 x1x1x26x

3x222x4x5x

22

33x1x2

x01x21x01x21

lim

x2x(22)

xx21xx2x2x2

33xx

x21

x211x23

1x1x31xx2 *9.若lim

x

axb0，求ab的值x2sinlim

2 2

x0sin2x x0sin

lim2arcsinx0sin2

x1x1

lim2nsin

sin3x

3x25x5x

sin2xx

11x1

1xx2

1xlim11x

x x

x1x

x (18)lim12xx xx 2

x(19)lim2xx xx1xa 2x3 xxa

x2x1x xx

xx21

2 2 cot2xa

4，求常數(shù)axxa設(shè)函數(shù)(tt31，求(t2(t)2設(shè)fx)1x(x1，求fxf1x)f11

x設(shè)fx)

，求fx1)f

xx

x1

x設(shè)fx

x0，求f(x f(x21)

x設(shè)fx)

x

，求ff

x.x2設(shè)(x1

0x1x

，求(x)設(shè)函數(shù)zxyfxy，當(dāng)y0時(shí)zx2，求fx及z設(shè)fx)

0x1x

，求函數(shù)fx3的定義域設(shè)fx)為定義在(

a上的奇函數(shù)，且fx)在

a上單調(diào)減少.試證明：fx)(

0]上也單調(diào)減少設(shè)函數(shù)fx)在(,內(nèi)單調(diào)增加，且對(duì)一切x有fx)g(x.ff(x)證明任一定義在區(qū)間( a)(a0)上的函數(shù)可表示為一個(gè)奇函數(shù)與一個(gè)偶函數(shù)之和xxy

1x

yarcsinx2x5xx4x5xx4y

xx

y y

1x1xx

求函數(shù)y

1x

(1)y (2)y2xex

10x10x (3)y (4)yex1111(5)y11111已知fx)sinxf(x)1x2，求(x及其定義域設(shè)函數(shù)f(x)的定義域?yàn)?0，求下列各函數(shù)的定義域f(x3

f(sin

f(xa)f(x

(a求函數(shù)yx21x當(dāng)x1x0.5時(shí)的增量2求函數(shù)y1x當(dāng)x3x0.2時(shí)的增量若fxcos2x，求

f(xx)f 下列函數(shù)fx)在x0sin

f(x)

x

f(x)

x

x

xf(x)

ex xsinx x

x函數(shù)fx)x2sin x 0x

在x0處的連續(xù)性sin設(shè)fx)

ax2 x x

，已知f(x)在x0處連續(xù)，試確定ab的值

x xsinx x設(shè)fxax2 x

，要使f(x)在

內(nèi)連續(xù)，應(yīng)當(dāng)怎樣選擇a111xx22x2 2

x

ex2

x0arcsin(1

3sinxx2cos

x0(1cosx)ln(1求證：當(dāng)x0sinsinx～ln(1x13x23x求函數(shù)f13x23x*38.若

x22xkx3

4，求k的值*39.若

x2axb1x

5，求ab的值根據(jù)連續(xù)函數(shù)的性質(zhì)，驗(yàn)證方程x53x112之間證明方程sinxx10在開區(qū)間 內(nèi)至少有一個(gè)根22 22 試證方程x2x11的正根試證方程xasinxb，其中a b0，至少有一個(gè)正根，并且它不超過(guò)ba證明方程x33x2x30在區(qū)間

證明曲線yx43x27x10在x1與x2值之間至少與x軸有一個(gè)交點(diǎn)*46.若函數(shù)f(x)在閉區(qū)間 b上連續(xù)，f(a)a，f(b)b.證明：至少有一(a, b)，使得f)(1)yx23x (2)ysin(3x一物體的運(yùn)動(dòng)方程為st310，求該物體在t3時(shí)的瞬時(shí)速度求在拋物線yx2上點(diǎn)x3處的切線方程33求曲線yx33

在點(diǎn)

0處的切線方程曲線yex1上哪一點(diǎn)處的切線與直線2xy10x試求曲線y 2在它與直線yx的交點(diǎn)處的切線方程和法線方程x求曲線(5y2)3(2x1)5在點(diǎn)

1處的切線方程和法線方程55確定ab之值，使曲線yx2axb與直線y2x相切于點(diǎn)

設(shè)曲線f(x)x3ax與g(x)bx2c都通過(guò)點(diǎn)線，求abc的值

0，且在點(diǎn)

0*56.設(shè)函數(shù)f(x)可導(dǎo)，且fx)0，證明曲線y1fx)與曲線y2fx)sinx切設(shè)f

)3

f(x0x)f(x03x)設(shè)f(32，求limf3hf3 設(shè)fx在xa處可導(dǎo)，求limfanhfamh)1sin2xx

x*60.設(shè)f(x) ，求f(0)，f 2

x xx*61.設(shè)g(x)

2cos

x

1，又f(x)在x0處可導(dǎo)，求1

f

x00,

x*62.設(shè)函數(shù)f(x)在x1處連續(xù)，且

fx)2，求f(1

x1x1xyx

y

x11x x15xx1xy(215xx1x11x(5)y 11xx(7)yx

(8)yxx1xyyexx(13)yx

(12)yarcsin2yarccot11xy

1xx

y2y 2

yexsin2ycosln(1 (20)ylnln1x1x

arcsin

yarcsinxylntanxcosxlntanx

ysinxxcosxxsinysec2xcsc2 (26)y *(27)y(xa)a1(xa)a2(xa x(28)y x

1x2y

y(1x22x11111

xx33x2x3333(3xy 1x2y2xy (2)y22axyb(3)yxln(5)arctan(xy)x2x2y

(4)y1xearctany確定y是x的函數(shù)，求yxxy方 a0確定y是x的函數(shù)，求yxyyf

)ef

y

xyf(exxe已知f1

，求f(x

yf(sin2x)f(cos2x

1 yx2sin(5x

yln(1x2yxln (4)y(1x2)(5)yxe (6)ycos2xln求下列函數(shù)的二階導(dǎo)數(shù)（其中函數(shù)fx)二階可導(dǎo)yf(ex

y

f(x)fyf(x)ef (4)y(lnx)f(x2f(5)yf(x2)lnf f(x)設(shè)函數(shù)yy(x由方程eyxye確定，求y(0)設(shè)函數(shù)yarctan(3x)，當(dāng)x x0.04時(shí)，求dyy

1x

yexxyx(5)y(exex

(4)y1x(6)1x

1xx

求函數(shù)yex當(dāng)x由9變到8.99的微分求由ysin2xxln(3t1復(fù)合而成的復(fù)合函數(shù)的微分

1x

四、中值定理·ln(1x)lnx1，(x0)xpbp1(abapbppap1(ab0<a<b,p設(shè)fx)在

1f1)0，又F(xx2fx.證明在

存在一點(diǎn)，使F()0x33x2

xx1

x

x

nx1xn

(1x)a1

（a為任何實(shí)數(shù)

xaxx2 x

xa

aa

exex

ex

x0x

ex2

ln(12x0cosxlnx

2 2

x1xln2

tan

tanx xn

x0xsin1x1x

（na0

exex2xxsinx

tan

lntan2xtan x0lntan2

ln(1x2

x0secx3ln113

sinx x

xarctan

arccot

x0xarcsin

cosxln(x x

ln(exea (27)lim3xln

(28)limxmln

(m (30)lim 1xx0x

e1

x1x

1

lim x0ln(1 x

x0

xtanx2 2

limxex

lim

x 1lim(2x)ex

1 (38)lim

x0x

x2x

x

cosx2

2 2*(41)

xln(ex

1 2x1cos*(42) 2x1cos1

1

*(44)lima1xa2xanx x

2xx證明函數(shù)y2xx

上單調(diào)增加，而在區(qū)間

y2x312x218x

yx42x2y(x2)5(2x (4)yxln(1(1)

3xxx

(x

xln(1

(xln(1x)1

(x2x22

ln(1x)x

(x

sinx12

(0xarctanx

(x0)，arctanx

(x0)y2x3(3)y2x36x218x

(2)y2x2x(4)y1x41x3xy(x1)2(x

y

xln3x3xy(x1) (8)y3x3x2xx2xx(9)y (10)yx2xyx55x45x3 yx2x

y

1xx1xx2

[0,x 求函數(shù)y1x的單調(diào)區(qū)間，并求該函數(shù)在區(qū)間

,1上的最大值與最小值 試證方程x3x10只有一個(gè)正實(shí)根方程xexa(a0有幾個(gè)實(shí)根倍.(1(1x)3

x2sinx1

xx

x22x222xx

1x2

x 11x1xx

12x23x51x2 x

xbe

sin2cos2

4sin3x1sin2

2

2x2 cos2x cosxsinx

1cos21cos2xf

f

f(x)xf設(shè)fsin2xcos2 x1，求f(x)11x已知一個(gè)函數(shù)的導(dǎo)函數(shù)為f(x) ，并當(dāng)x1時(shí)，這個(gè)函數(shù)值等于311x2函數(shù)F(x)已知曲線yfx上任一點(diǎn)的切線的斜率為ax23x6x1y112值，求f(x)和f(x)的極小值已知fx)的圖形過(guò)點(diǎn)

f(x)的圖形是過(guò)點(diǎn)

3是f(x)的極值，求f(x).3

(abx)k

(b

(12x)233(3)33

xxxx x311xx(9)cosxxcosxsinxdx12sinxcosxln11

xe2x2cos2xsin2(10) dx1cos2lnxx xx(x(15)

(14)sinsin3xcos5cos3

cos3xsin4xdx4

sin

(22)(tan2xtan4 1tanx

23x2

2x x28x

4x4x2

22

xxx

x dx

x dxx2x

x6xx53(1x3)2

lntan 2sin14149x1

1ex1ex111ex (36)ex1x2x(1x)100

x(13x1

1

exx

33

a2a2xx2x2a

dx3(x2a2)349x49x xx2x

dxx2x23x23x2x2x2ax(1)x(3)xcosx(5)xsin

(2)(4)(6)x sin2xcosxln

sec3x2

x1xx

e2xsinx2sinln(15)xln(1x2lnx

(14)(arcsinx)2x2ln(1lnsin

xx

sin2

x2

ex

x

arctane*(23)xexsin設(shè)fx的原函數(shù)為sinx，求xf(x)dxx設(shè)f(ex1x，求fx)(1)x1 (x1)(x(1)

x34x3x3x3x

x5x4x3 (x1)(xx

x2*(6)(x1)2(x1)xf(x)x

1

3 f(x)0sint f(x)

x2 11t0

f(x)1xx2

2e2f(x)x3e (6)f(x)xe2f(x)2tanln(t2e

x3

f(x)

cosxcost sinx1 *104.設(shè)f(x)是連續(xù)函數(shù)，且xtf

ft)dtx，求f1 x105.設(shè)F(x)

x

，其中f(x有連續(xù)的導(dǎo)數(shù)且f0)0.研究：(1)F(x

xx0處的連續(xù)性；(2F(x在x0處的可導(dǎo)性00*106.試求由yetdtxcostdt0所確定的隱函數(shù)對(duì)于x的導(dǎo)數(shù)y00*107.設(shè)xy2yxcos2tdtdy0xcos2

xarctanxlim

lim x判斷函數(shù)fx)x3t1dt在區(qū)間

1上的單調(diào)性0t2t求函數(shù)fx)xtetdt的極值0求函數(shù)fx)xt(t4)dt在0

5上的最大值與最小值*112.設(shè)函數(shù)f(x)在

內(nèi)可微，且fx)11xft)dt，試求f(x)x6(2cos20

2cosxsin02sin(1sin3

x ex1ex

e1lnx(7)0exex xx1lne3a23a2ra2

(a2x22x設(shè)fx在

b上連續(xù)，試證bfx)dxbf證明afx2dx2afx2dx，其中fx為連續(xù)函數(shù) 1

x1xx11

(x0)證明1xm1x)ndx1xn1x)mdx 1 (1x)1211x

42a42

x2 x 2

a2x2x2x x2x0(1x2)2

x5x1

01x1 55 ln

xsin(2x2

ex

01sin(x2已知2ln2 dx，求aexex 0

2

2excos000

0201e0xarctan 1e

xsin(7)

x

(8)0

3x

x2cos0

eeeb121.已知常數(shù)b0，且1lnxdx1，求b的值b

x*122.設(shè)fx)*122.設(shè)fx)

f(x1)dx2 2

xyax2a0與x軸所圍成的圖形yx23在區(qū)間[0，1]上的曲邊梯形yx2與y2x2所圍成的圖形yx3與直線x0y1所圍成的圖形在區(qū)間

上，曲線ysinx與直線x2

y1所圍成的圖形y1與直線yx

x2所圍成的圖形曲線yx28與直線2xy8 y4所圍成的圖形求由拋物線yx24x5，橫軸及直線x x5所圍成的圖形的面積求由曲線yxex2，橫軸及直線x x1所圍成的圖形的面積求由曲線ylnx，縱軸與直線ylnaylnb(b>a>0)所圍成的圖形的面積求由拋物線y32xx2與橫軸所圍成的圖形的面積拋物線y1x2分割圓x2y28成兩部分，分別求出這兩部分的面積2求下列平面圖形分別繞x軸、y軸旋轉(zhuǎn)產(chǎn)生的的體積x曲線y 與直線xx

x

y0所圍成的圖形在區(qū)間 上，曲線ysinx與直線x y0所圍成的圖形 2 曲線yx3與直線x y0所圍成的圖形曲線x2y21與y23x所圍成的兩個(gè)圖形中較小的一塊2求曲線xya(a0與直線xa，x2a及y0所圍成的圖形繞x體的體積設(shè)平面圖形由yexyex0所圍成，(1)求此平面圖形的面積；(2)求此平面圖形繞x軸旋轉(zhuǎn)所形成的旋轉(zhuǎn)體的體積.ex

dx (3)xexx x

(5)exsin

arctanx 11x

0

x(a

0(1

0x24x計(jì)算yex與直線y0之間位于第一象限內(nèi)的平面圖形繞x軸旋轉(zhuǎn)產(chǎn)生的旋轉(zhuǎn)體的xx證

limxlim11，

limxlim(1)xx0x

x0

x0

x0

x即xx0

xx0x

，∴

不存在xxx證

4x4x9x(4x4)(9x(9x9)(4x

399x4x49 24911xx

1 5.(1)為同階無(wú)窮小(2)為等價(jià)無(wú)窮 6.a7.(1)解

(x 3

0x2x （根據(jù)無(wú)窮大與無(wú)窮小的關(guān)系 (3)解當(dāng)x10sinx

1式=0 (6)11cos(7)解原式=lim 1.

x21cosxx21

1x

1.（無(wú)窮小量分出法x2x2x

x21 x(2)0（利用無(wú)窮小量分出法

xx

x

x3(2x1)x2(2x2x2x2

2x

(2x21)(2x1

x3x

1x(2x21)(2x

x

1

1 2x2

x1x1x31x

x2x23x3

(6)

(9) n1

35

5

(2n1)(2n1) 1133 33 lim11

1

n21

2n1 (2n1)(2n

2n

2n1A(2n1)B(2n1) 比較等式兩邊n的同次冪的系數(shù)，2A2B A1AB

B1于 1

.（此法稱為待定系數(shù)法(2n1)(2n

22n

2n(10) (12) (13) (14) (15) (16)

x2

3x3x1((x21)(3 (17)2

(18) 2

2

(21) 3

(24)(25)

11x31x

3(1x)23(1x)2 1x312解原式

x21(axb)(x1)x1(1a)x2(ab)x1x

1aab

a1，b

x2sinlim

xsinlim

x0sinx(2) (3) (4) (5) (6)

x1x1

x

2 2 (9)2

sin2

x0sin23x2 2

3x

sin

lim 2

5x

x x5 xx xx 1x11x1

2

limsinx1x1xx0sinxx

1x

1x 1x

1x解

lim1x xx

x x

e1e

x

1x1x

1x

(1x)

(1x)1

1x

1x

xe

e

e

e

e

xx

x2 x2

x

1

x1

22x21x2

x21x

x21

x2

xe0e (26)e

t61；t62t31x1

(x1)

(x2)；x1x

(x1且xx1x解

(x2)；

(xx1f(x1)

x1f(x21)

x1x；xx21x21x211

x.x解令x1t，則xt (t1)2

0t1(t)2(t1)(t1)22(t1)

1t11t，2t(x1)2∴(x)2(x

1x2xf(x)x2x；z(xy)2

21. 22.

00,

01,(1)y

21

，(0x

y12

x

(x0或xyln1x，(1x1

y1

x5，(xy10x12，(

y

(1x(x)arcsin(1x2)； 2(1) (2)k1 (k1)，k

或k1 k，k2 2

2(3)當(dāng)0a1時(shí)，定義域?yàn)閍22

a；當(dāng)a1時(shí)，定義域?yàn)?

x

limf(x)

x即f(x)在x0處的極限不存在，所以f(x)在x0處不連續(xù)解limfx)limx2sin10f0) f(x)在x0處連續(xù)解limfxlimex1limfx

sinx1，f(0)f(x)在x0處連續(xù)

limf(x)lim(ax2)a，f(0)1

f(x)

f(x)在x0即a1lnb1，即a1be

f(x)

f(x)f解當(dāng)x0或x0時(shí)，f(x)為初等函數(shù)，連續(xù). 要使f(x)在(,)內(nèi)連續(xù)，當(dāng)且僅當(dāng)f(x)在x0處連續(xù).∵limf(x)

xsin10，limf(x)lim(ax2)a，f(0)a∴a0

x22xx2

122112

2(2)12

(3)1

ln(1xx在x0處不連續(xù)，所以不能直接利用連續(xù)函數(shù)求極限的法則x1令(1x

u，當(dāng)x0ue

lnu在ue 1limln(1xxlnlim1xxlne (6)2

sinsinxsin證x0ln(1

111

33(,1)

解設(shè)f(x)x22xkx3x30fx)為x3即有l(wèi)imfx)0f(x是連續(xù)函數(shù)，∴有l(wèi)imfx)f3)0，即3223k0即k3解設(shè)f(x)x2axb，∵x1時(shí)，1x0，∴有l(wèi)imf(x)0 又∵f(x)續(xù)，∴有l(wèi)imfx)f1)0，即1ab0，即b1

x2ax1a1x

(x1)(x1)a(x(x

lim[(x1)2a5，即a7，代入(1)，得b6∴a7，b6. 41. 42.43.證設(shè)fx)xasinxbfx)在[0baf(0)b0，f(ba)baasin(ba)ba[1sin(ba)]0當(dāng)fba)0ba當(dāng)fba)0，則在(0ba內(nèi)至少有一點(diǎn)，使f)0即為所求.44. 45.46.證設(shè)g(x)fx)xg(x在[ab上連續(xù)∵g(a)f(a)a0，g(b)f(b)b0g(x)0在(ab內(nèi)至少有一個(gè)根，即，至少有一點(diǎn)abg(f0，即f)47.(1)y2x (2)y3cos(3xv(t)ds3t2，v(3)

y6xy34(x

x2y30

2xy1解在方程兩邊對(duì)x3(5y3)25y5(2x1)4將x0y1代入上式，整理，得y2 所求切線方程為y12x，或10x15y30 法線方程為y13x，或15x10y20 a2，b

a1，b1，c證設(shè)兩曲線的交點(diǎn)為(x0y0)，則有fx0)fx0)sinx0，已知fx)0sinx01，從而有cosx00.y1(x0)f(x0y2(x0f(x0)sinx0fx0cosx0f(x0)，即，在交點(diǎn)(x0,y0處兩曲線的切線斜率相等，所以兩曲線在交點(diǎn)處相切.解原式=limfx0xfx0fx03xfx0

f(x0)3f(x0)33(3)(nm)ff(x)f

x

x

sinx解∴f(0)

sin2x當(dāng)x0f(x)

sin2x ，(2sinxcosx)xsin2xxsin2xsin2， ∴f 2

02

x x4 2

dfg(x)fg(x)g(x)x2cos1limg(x)g(0)

limxcos10，即g(0)0

∴dfg(x)

fg(x)

fg(0)g(0)f(0)00解limfx)2，∴當(dāng)x1fx)與x1為同階無(wú)窮小，即有l(wèi)imfx)x1xf(x在x1處連續(xù)，∴有f1)limfx)

從而

f(x)

fxf1)2

x1x

x

15x2x2x2(1x)1

(2)

112x 2x

45x3115x1(1(1x2)1x1

1

2xln1

12x

xx1x2x(1x2

ln5

5lntan

(11)ex(cosxsin

1414x1212xx

1x

1x(16)

1x

x(1(1x2)1x

2arcsin24x

1

xln11x

x(cosxxsin

sec22

xcsc2

cot24

lnya1ln(xa1)a2ln(xa2)anln(xan)1yy

x

axaa

a xaan an

a

an ∴y(xa1

(xa2

(xan

x

x

x 1 2 n

lny1lnx，1y1lnx111(1lnx) x x x∴y

1x (1lnx)xxx

1x 1x

n1 1xx 1x

lnysecxln(1x2

2x

(2x1ysecxtanxln(1x2)secx

，1x

2x

1x22x 1x2

lnylnx1ln(1x)ln(1x)，21 1

1xxy 2 111 111

1x 1x

x(1x)(1x)∴y (1x)(1x)

x3x33x2x 2 2 3x 2x 2 2x2

x 1

(3 1 3(9x 64.(1)解在方程兩邊對(duì)x

yy2x2y

2x2yyyxy0y

yax

y

y1

ey1xe

tan2解在方程兩邊對(duì)xx2x22x2

(2x2yy)

y

xy x1 xxyyxyy ∴yxyx2yxyyx

x2y2

x67.(1)

yf(ex)exef(x)f(ex)ef(x)fef(x)exf(ex)f(ex)f(2)

1xx2xxx2x

(exexe1)f(exxe68.

22x

x

2arctanx

1x2x(2x2

2cos2xlnx2sin2x x

cos2

yf(ex)exexf(ex)，yexf(ex)exf(ex)exexf(ex)e2xf(ex)

y f12ff2ff12ff2fy

2f(x)f(x)f3.3f

ef(x)f(x)2f(x)2f(x)f(x)f(x)f(x)2x

f(x2)4f(x2)2(lnx)f(x2)4x2(lnx)f(x22f(x

)4x

(x2)

f(x)f(x)f解在方程兩邊對(duì)xeyyyxy

再在(1)兩邊對(duì)xey(y)2eyyyyxyey(y)2(eyx)y2y

當(dāng)x0時(shí)，由原方程解得y1 將x0，y1代入(1)，得y(0)1e將x0y1y1代入(2)，得y(0)1 e1x73.(1)(1x2)2

1x

cosxcos2

2x2xx

xxxx212

xx

sin

74.解由題設(shè)，x由x09變到x0x8.99，得x12ye ，12

ex 1e3 1

2

∴所求微分dyf(x0x6

(0.01)

3t

sin2ln(3t證設(shè)ft)lnttx,1x(x∵f(t)在x,1x上滿 ∴在(x,1x內(nèi)至少存在一點(diǎn)f(1x)f(x)f()[(1x)x]，即 ln(1x)lnx1x1x，∴ 11，從而1 11

ln(1x)lnx1x提示：設(shè)f(x)xp，x[a,b].利用日中值定理證證F(x在[0,1根 定理，在(0,1)內(nèi)至少有一點(diǎn)1，使得F(1)1又F(x2xf(xx2f(x，它在[0,內(nèi)連續(xù)，可導(dǎo)1 定理，在(0,1)(0,1)內(nèi)至少有一點(diǎn)，使F()79.(1)

n

(3)

(7)

(8)

(9) (11)2

3(13) (15)1xln(11x

exx1xexx1xex1x1x

13

7

x

11ex

lntan7x

ln7x

7xlim11x0lntan

x0ln

x0

（當(dāng)x0時(shí)tan7x~7xtan2x~2x

00

211

2

limxarcsinxlimxarcsin

(x0,sinx~

sin3

x3x3x21x2(1x2000

11111x

11x213x21x

(1x2)

13131x2(1x2 (22)

解

cosxcosa

x

ex

0e0exaln(exea

exex

xaex(x

xaex(xa)ex 0

∴原式=

cosxxaln(exea

cosa1cosa (29)12

12

13

lim(2x)exxlim2exxex1

1∵lim2ex

2e02

11

ex limxex1limex10

x

lime

e0x

1x

∴原式2131

ln(1sin解令y(1sinx)x，則lny x00

cos

x01sinlimyeee

(40)e

(42)e1(43)解令yarctanxlnx，則lny

ln

2 2

lny

1x00

1x2x

1x

xarctan211x

1x

x1

2x2x

∴

ye1e1

1(44)解令ya1xa2xanx ，

1 lnynxlna1xa2xanxlnn， 1lna1xa2xanxlnlimlnynlim

x

1

1a1xlna1a2xlna2anxlnanx20a1xa2xanx0x

n

a1xlna1a2xlna2anxlna

a1xa2xannlna1lna2lnn

ln(a1a

an)limya1a2an，即原式x

a1a2

nnai80.

y6x224x186(x1)(x3)令y0x11x23x＋—＋y↗↘↗(

)

)(

111

1

2211

2 182解函數(shù)的定義域?yàn)?1,y1

1

，令y0x0當(dāng)1x0y0；當(dāng)x0y0∴函數(shù)的單調(diào)增加區(qū)間為

；單調(diào)減少區(qū)間為

82.(1)證設(shè)fx)

31f(x)在[1,xxxf(x)f(x)在(1,上單調(diào)增加

1x211x3x x1x3

，(xf10，∴當(dāng)x1時(shí)，有fxf10.x 31，(xxx(3)證設(shè)fx1xln(1xarctanxf(x)ln(1x)1

1x

0，(x∴f(x)在(0,)上單調(diào)增加 又∵f(x)在[0,)上連續(xù)，且f(0)0∴當(dāng)x0時(shí)，有fx)f0)0(1x)ln(1x)arctanx0

ln(1x)arctanx，(x183.(1)解函數(shù)的定義域?yàn)?, y6x26x6x(x1，令y0，得駐點(diǎn)x0x x01＋—＋y↗極大↘極小↗x10為極大點(diǎn)，函數(shù)的極大值為y(0)0x21為極小點(diǎn)，函數(shù)的極小值為y(1)1y(1)1y(1)1y(0)y(1)17y(3)y(0)0y(1)

，y(2)3y(10y75

y(e)解函數(shù)的定義域?yàn)?,y

(x1)213x3x3

，令y0

255x335x33x0 5252 ＋—＋y↗極大↘極小↗34 234x15為極小點(diǎn)，函數(shù)的極小值為y55

x20 函數(shù)的極大值為y(0)0y(1)1y(0)y(0)4y(2)31 2

y5x420x315x25x2(x1)(x3)令y0，得駐x10x21x33（舍去∵y(1)10，y(0)1，y(1)2，y(2)7∴函數(shù)的最大值為y2；最小值為y10y8yy1y5單調(diào)增加區(qū)間為(

單調(diào)減少區(qū)間為(

函數(shù)在 2,1上的最大值為2，最小值為0證設(shè)f(x)x3x1，∵f(x)3x210，∴f(x)在(,)上單調(diào)增加 f(x在[0,1f(0)10，f(1)10∴由根的存在定理及f(x的單調(diào)性可知，方程fx)0在(0,1)內(nèi)有且僅有一個(gè)實(shí)根，即方程只當(dāng)ae1時(shí)，方程無(wú)實(shí)根；當(dāng)ae1當(dāng)ae1時(shí)，方程有兩個(gè)實(shí)根解設(shè)容器的底邊長(zhǎng)為x，高為yx2y108，即y108x

Ax24xyx2432，(xxA2x432，令A(yù)0x6x 現(xiàn)在只求得唯一駐點(diǎn)，故當(dāng)?shù)走呴L(zhǎng)x6米 當(dāng)x6時(shí)，y3. 即容器的底邊長(zhǎng)6米，高3米時(shí)，所用材料最省.當(dāng)?shù)走呴L(zhǎng)x6米，高y3時(shí)，所用材料最省當(dāng)土地的長(zhǎng)x18米，寬y12米時(shí)，所用建筑材料最省3當(dāng)池底半徑r 3

2cosxlnx

2x2x333

3x3222

6x3555

3x3888

2

x2

2x7

x4

4

arcsinxarctanx1x

x1x3arctanx32x

2

bxebxxcosx(12)xsinx(14)1tanx1x

4cosxcotx(13)sinxcosx2

f(x)dxdf(x)f(x)C(2)解f(2x)dx1df2x)1f2xC (3)xf(x)f(x)x1x22

F(x)arcsinxfx)x33x26x2，極小值為f2)2解由題意，設(shè)f(x)k(x1)，k0 從而

x f(x)

f(x)dx

k(x1)dx

xC∵f(x)的圖形過(guò)點(diǎn) 3)，∴有f(0)C3x fx)

x3f(x)是可導(dǎo)函數(shù)，且x1x1是f(x)的極值點(diǎn)，即有f1)k1132，解得k22 2 x ∴f(x)

x3

2x397.(1)解當(dāng)k1時(shí)，原式=1(abx)kd(ab1b

k

(abx)k1C當(dāng)k1時(shí)，原式=1 d(abx)1lnabxC.解

a dx

(12x)2d(1 232

2

(12x)21C32

1

C33331

21

(31

2xx

(32x)

C

(32x)3C4x1(xx3

1(x1) 1x1x3

dx2

d(x

)2

1x2

1x1x1x211(1x2)2(1x21x212

1(1x22

2d(1x213131x (1x2)2(1x2)2C3

C1e2x21C

1sin(x)1x8

1sin4xC

x x解 dx dx dxtanxC.1cos

x2 2cos22

cos2 2

cosxsinx12sinxcosx

dx

d(sinxcosx)(sinxcosx)2

sinxcosx2lnx1(lnx)2Cln11

ln11

ln11 x

x

x 1解

dx

dx

d1 x(x

2 1

1

xx1 1 1 1 1ln1xdln1x2ln1x

C 3x1sin2x1sin4x (15)sinx1sin3x 解sin3xcos5xdxsin2xcos5(1cos2x)cos5xd(cosx)(cos5xcos71cos6x1cos8xC 或sin3xcos5xdxsin3xcos4xd(sinxsin3x(1sin2x)2d(sin(sin3x2sin5xsin7x)d(sinx)1sin4x1sin6x1sin8xC1lncos5xC

13

sin3

Csinx2

2sin2x tanx1tan3xC

1tan3xtanxxC1tan3x3322

1tanx6262

1

arctan

x28x

(x4)2

22x41

dx4

1arctanx4C 31x4 3 3 解 1 4x14x1x22

arcsin2 darcsinxlnarcsin

C2arcsin 2xxx

2 14x 121(2x14x 121(2x1(2x131ln(x2x1)132

arctan2x131lnx26x134arctanx33

81(1x3)388

51(1x3)355lntan

lntan

lntan 2dx

xcos

xdx2

2

x2

xdxx x 1 xlntan2dlntan22lntan2

C 1arcsin3x2 2

dx 1

1ex1

edxdx1exex

1ex

xln1

C (36)2lnex1x98.(1)解令1xt，則x1tdxdt原式

t100

dt

12ttt100

t

1

t

1

t

1

1t1

1

1

C解令xt6，則dx6t5dt 原式t3(1t2dt611t2dt6(tarctant 66x6arctan6xC.解

t，則xt21dx2tdt1原式2tdt211dt2t2ln1t11

1t1ex 2ln(111ex解

t，則xlnt21dx

t2

dt原式

1dt

1dtlnt1lnt1t21 t

tttt

C

Cex1ex11解令ex1ex11原式

costcos3

dt

dttant1x1x1x1x1x1x2a22

arcsinx

a2x2解令xatant，則dxasec2tdtx2ax2aax2x2ax2aax2a

Clnx

Cx2a1 x2a1

1ln3x C49xarccos149xx

x2x21

33

3x

aarccosax2x2a3x23x2

te2tdt1td(e2t)1te2te2tdt 1te2t1e2td(2t)1te2t1e2tC (x22x2)ex

2xsinx4cosx 1x21xsin2x1cos2x 解xsinxcosxdx1xsin2xdx1xd(cos 1(xcos2xcos2xdx)1xcos2x1sin2xC 2xtanxlncosxx221secx1lncscxcotx 解sec3xdxsecxd(tanxsecxtanxtan2secxtanx(sec2x1)secxdxsecxtanxsec3xdxsecxtanxsec3xdxlnsecx sec3xdx1secxtanxlnsecxtanxC2xln2x2xlnx2x3解x2arctanxdx1arctanxd(x33x

x x

1 3arctanx31x2dx3arctanx3x1x2x

1d(1x23arctanx3xdx6

1xx33

arctanx262

1ln(1x6

)Cxarctanx1ln(1x2)1(arctanx)2

2e2xcosx4sinx2 2

sinlnxdxxsinlnxcoslnxsinlnx(xcoslnxsinln sinlnxdxx(sinlnxcoslnx)C21xx(arcsinx)2 arcsinx1x

2

)ln(1x2)x222

13

1)ln(1x)393

x6

xC1(lnx)22lnx2

(19)

(x11xxex1ex11xxex1ex1exlnx

exarctanexx1ln(1e2x)2xex(sinxcosx)1excosx 解xf(x)dxxd(fxxfxxf(x)sinxx

sinx xcosxsin∵f(x) x

xf(x)dxxxcosxsinxsinxCcosx2sinxC102.(1)解

xx

A (x1)(x

x1 xx1(AB)x2AB，

AB

A2AB1，解得B3 ∴原式=23dx2lnx13lnx2C x1 x2 x111解原式=

dxx1

4x3xx1令 A

4x3

x(2x1)(2x

2x

2x x1(4A2B2C)x2(BC)xA44A2B2C

A BC

，解得B A1

C ∴原式=117 91 dx4x8 2x1 82x4x8xlnx71d(2x1)91d(2x 8 2x 8 2xxlnx4

7ln2x1

9ln2x1C1x33x29x27lnx33x33

x2

x8lnx3lnx14lnx1

x21x2x1 AxB

Cx，(x21)(x2x x2

x2x得1AxB)(x2x1CxD)(x2(AC)x3(ABD)x2(ABC)xBD AC AABDABC

B，解得C BD D

x

dx

d(x21)

(2x1)1∴原式=

x2

x2x

2x2

2x2x1

d(x2x 2

1)2

x2x

2x2x1ln(x21)1ln(x2x1) 2x2x ∵2x2x12x

1

33

2x121 3 2 3132x3 2x132x32 212x

3 3

C3 32x2x313∴原式=1ln1 13

C x2x21

解令

(x1)2(x

(x

x1

x得x21A(x1B(x21C(xBC)x2A2C)xABC，比較等式x同次冪的系數(shù)，得 BCA2C

A，解得B12ABC C1

2

1

(x

2x

2x1 1lnx11lnx1C 1lnx21Cx x 11x (2)3x2sinx11x

f(x)0etdt

etdt

edt

etdt22

ex3(x3)ex2(x2)3x2ex32xex2

f(x)2x02x

e2tdtx2e4xe2tdt2xe4x

e4xe4xextanln(e2x

sinxcoscos1cos2

cosxcossin解xf(x31)3x21，f(x31)

13x3t33(t令x313t33(t

，即f(x) 33(x3333(x3336x105.(1)∵limF(x)lim0tfx