PHÂN LOẠI VÀ PHƯƠNG PHÁP GIẢI TOÁN GIẢI TÍCH 12 - LÊ MẬU THỐNG

8/10/2019 PHN LOI V PHNG PHP GII TON GII TCH 12 - L MU THNG

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L MU THNGL M THO - TRN C HUYN

PHN LOI &PHNG PHP GII TON

GII TCH 120 CC VN CN BN0 PHNG PHP GII TON0 TON TNG HP - NNG CAO0 TON T LUN0 Bi TP MINH HA A DNG0 B TP T LUN0 TON TRC NGHM

(CHNG TRNH CHNH L & H p NHTCA B GIO DC & O TO)

(C S A C H A V B S U N G )

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NH XUT BN H N I

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W.DAYKEMQUYNHON.UCOZ.COM

g gp PDF bi GV. Nguyn Thanh T


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M C L C

CHNG I: MT S VN CHUNG V HM S

Vn 1: Tm min xc nh v min gi ca hm s .............. ...................... 5Vn 2: Cch tm m t hm s .............. ................................................................. .. 8Vn 3: .S dng tnh cht ca hm s lin tc

chng mnh mt phng trnh c nghim ........ :............................13

CHNG II: O HM CA HM S

Vn 4: Dng nh ngha tnh o hm ca mt hm s ................. ........18Vn 5: S dng cc qui tc v cng thc tnh

o hm ca hm sy = f(x) ................................................................23

Vn 6: o hm cp c a o ....................................................................................31Vn 7: ngha hnh hc ca o h m ................................... ......................... 33Vn 8: Tm phng trnh tip tuyn ca t h .................. ........... ................ 40Vn ' 8B: nh l Lagrance ...................................... ............................................. 45

CHNG III: NG DNG O HM - TNH N IU

Vn 9: Phng php chng minh hm s y = f(x)

ng bin, nghch bin ong khong (a, b).........................................50Vn 10: Tm khong ng bin, nghc bin ca mt hm s ..........................56

Vn ' 11: Xc nh iu kin ca 1 tham s hm s y = f(x)

ng bin, nghch bin ong khong (a, b)~ .......................................63Vn ' .12: s rig tnh n iu ca hm s

chng minh bt ng th c ....................................................................73

Vn 13: Tm cc fr ca hm sy = f (x)... !......................................................80.

Vn 14: Tnh gi tr ca tham s hm s y = f(x)t cc .tr ti Xo............. 90Vn ' 15: - Chng minh mt hm s lun lun c cc tr

- Xc nh iu kin mt tham s hm s c hoc

khng c cc t r .......................... ................... ....................................... 96

Vn 16: Tm gi tr ln nh t v gi tr nh nht ca mt hm s ........... . 108

Vn 17: S dng vn cc t ca hm s ong vicchng minh bt ng thc .................................................................. 117

Vn ' 18: S dng o hm cp 2 tm im cc b ca hm sy = f(x) .. 120

Vn 19: Tnh li, lm v im un ca th ..................................................123

CHNG IV: NG TIM CN CA TH HM S

Vn 20: ng tim cn .................................................................... ............. 129



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3/358

CHNG V: KHO ST HM S

Vn 21: Kho st s bin thin v v th hm s ................................ ..... 1331- Hm bc hai y = f(x) = ax2 + bx + c*U 0) ................134

. 2 - Hm ng phng y = f(x) = ax4 + bx2 + c(a *0) ............. .......3 - Hm s bc 3 y = fix) = ax3 + bx2 + cx + (a 7=0) ............... . 144

4- Hm nht bin y = f(x) = ..... ......... ............ .................. . 150* cx + d

r T - L. ~ _ t \ _ XX2+ bx + c C/5 - Hm hu t y = f(x) = ........................... ..................................1a x + b

CHNG VI: V TR TNG GA HAI TH HM s

Vn 22: s dng tnh tng g ia o .......................................................... ......... 159Vn ' 23: - Bin lun s giao im ca ha th

- Phng trnh tip im ca th

(suy ra kt qu ca s bin lun) .........

............................................. 161Vn 24: Bin lun s nghim ca mt phng trnh

bng phng php t h ............................................. ............... .. 167Vn 25: Tm tp hp im ..... ................................................... ............ . 174Vn 26: Phng trnh tip tuyn ca th mt hm s

(s dng nh l tng giao) ..... ................................................ . 183

Vn 27: H ng cong ....................................... .......... ..................................191Vn 28: Chng minh h ng cong lun tip xc vi mt

ng cong c n h ...................... ......................................................200Vn 29: Dng thng i qua im cc i, cc tiu ca

th hm s bc 3 ...................... ....................................................... 204Vn 30: ng dng nh l tng g ia o ............................... .............................. 217

CHNG VII: NGUYN HM V TCH PHN

Vn 31: Cc bi ton nguyn hm v tch ph n .............................................230

Vn 32: Tch phn ca hm s c du tr tuyt i ....................................... 245

Vn ' 33: Chng minh mt ng thc tch phn .............................................250

Vn 34: Chng minh mt bt ng thc tch phn ................. .....................263Vn 35: Ton tng hp ................................................................................... 266

CHNG VII: NG DNG CA TCH PHNVn 36: Din tch mt p hn g ................................................ .......................... 270Vn 37: Th tch vt th trn xoay...................................................................281

CHNG IX: CC THI MU

A. THI T LUN....................................................................................... 288

B. THI TRC NGHIM.................................... ...................... ................ 346



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4/358

CHNG 1

M T S V N C H U N G V H M s

V n 1

TM MIN XC NH V MIN GI TR CA HM S_____

* Tm min xe inh ca hm s'y= f(x)

Tm iu kin ca X f(x) c xc nh.

- T iu kin..trn ta c tp xc nh D ca hm s (vit di dng tp hp).

* Tm min gi tr ca hm s" y = f(x)

Tm min xc nh D ca hm s. -

t y - f{x)(xem y l phng phng trnh c n Xv tham s y).

Tm iu kin ca y phng trnh y=f(x) c nghim x e D.

GHI CH

Nu hm s" c dng Th iu kin hm s' xc nh

u(x) V(x) * 0

* v(x)y= 2ifi(x) u (x) >0

y^gaUx} u (x) >0

. (a>0, a * 1)

TON P DNG

Bi 1A .Tm min xc nh ca cc hm s Su:

.. . , 1 1 - 2*a. y =- - 'J2x t-1 b. y - c. y -

V3 x 1+ X

d. y = y /2 - x + = }= e. y = loga (5 - 4x - X2).VX + 2

Gii

a.y = - 72x71 xc nh 0 X> - -2

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5/358

Min xc nh ca hm s ny l D = [- - , +oo)

b. y = ~=L= xc nh 3 - x>0 x/2x + l

- Hm s ny xc inh c=>x> ;2

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6/358

- Phng trnh y = -2 V2 x+ c nghim x > -- < = > y < 0

Min gi tri ca hm s ny l: Y = (-00,0].

b = l ~ 2x y 1 X

- Hm s' ny xc nh o x l

y = y (1-x) = 1 - 2x, X^ 1 X (2 - y) = 1 - y, X* 11- x

Phng inh c nghim x ? i l c > 2 -y * 0 y * 2

Min gi tr ca hm s ny l y = R \ {2}

c. y = X2 - 4x + 3

- Hmsny xc nh Vx, X R

- Phng trnh: y = x2- 4 x +3 X2 - 4 x + 3 - y = 0

c nghim x e R A ' = 4 - 3 + y > 0 o y > - I

Min gi in c hm s ny l Y = [-1, + co),

l - e

2-e*

D = R\{ln2} y = - y (2 - ex) = 1 - ex, X e D2-e*

e* (1 - ) = 1 --2y, X D e* = (x e , y 1).1- y

l - 2yPhcng trnh c nghim X e D > 0

1- y

y > 1

y < -2

Vy, min gi tr ca hm s ny l: Y = (-00, - ) u (1,+ao).

inx + 2 _ . , Ve.y - , 7 i D = (0, e) (e,+co)

lnx-1

- Ta c: y - y (nx - 1) = lnx + 2 (y -1 ) nx = y + 2lnx-1

Phng trnh c nghim X y * 1Min gi tr ca hm s .l Y = R\{1>.

CS*K>

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7/358

TONTdNG T

Bi 1. 4. Tim min xc nh ca hm s sau:

a. y = n {yx -1 - 1) b. y =lnx-3

c. y = -J== + -J=L=7 d. y = V 6- 5e x- e 2xv l- e* V2 +e* 5

p s: a, D = [2, +oo): b. D = le2, +co) \ {e3}

c. D = (-00, 0) (ch : 2+ e* > 0, Vx, Xe R)

d. D = (-eo, 0).

Bi 1 .5 . Tim min gi tr ca cc hmssau:

2 n c 2xz +X + 1a. y = A - 2x +5 b. y = ------x + 1

c. y = 1 - -v/x2-2 x+ 5 d. y^ c o sS c- lHng dn v p s'

a. Y = (-00, 6] b. Y = (-00, -7] o [1, +oo)

c. Y = 1) . Vt li y = 1 + 2cos2x

(ch -1 < cos2x < 1); Y = [-1, 3].C8*K>

Vn 2

CCH TM MT HM s

1. Bit f(x). tm f[g(x)]

Thay Xbi g(x) vo hm s f{x).

2. Bit flu (x)) = p (x). Tm f(x)t u (x) = t ri tm cch tnh p (x) theo t, ta s c; f(t) = Q (t).

i k hiu t thnh Xta c f{x) = Q (x).

TON P DNG

Bi 2.1. Cho cc hm s f(x) = v g(x) = 1 Xa. Tm hm s h (x) = flg(x)} v Cj>(x) = g[f{x) .

b. Tm min xc nh ca mi hm s trn.

Gii

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8/358

a. h (x) = f(g(x) ] = g = ; (x) = g[f(x) =V 1-x

1 + Xb. h (x) x c in h > 0


9/358

t X - 2 = u, ta c;

x - 3 = u - l

X = u + 2

2 x -3 = 2u + l y

Vy (2) tr thnh: f(u) + (u - 1) 2g(2u + 1) = (u - l)(2u - 1)

Hay: f(x) + (x - 1) 2g{2x + 1) = (x - l ) ( 2 x -1)Bc 3: Gii h phng trnh c to bi (1) v (2'):

Jf(x)+g(2x + l) = x + l ?

[f (x) + (X- l)2g(2x + 1) = (X - l)(2x -1 )

Cc nh thc:1 1

1 (x - l )z

X + 1 1

L(x - 1) (2x - 1) ( x - l ) \= (x + l)(x - 1) 2 - (x - l)(2x - 1) = X (x - l)(x - 2)

1--x

(2)

D

D =

= x - l) 2- = x ( x - 2)

D

Vy:f(x) = = x - l

1 (x -l)(2x- l)

(x?i0; x * 2)

= (x - l)(2x - 1) - (x + 1) = 2x (x - 2)

g(2x - 1) = = 2 (x * 0;x * 2)

Hay: f{x) = X- 1

g(x) = 2

(x* 0; x * 2);

(x * 0; X# 2)

B 2.5. Tim hm s f(x) bit f(x) - xf J = 3 (x* 1)

Gii_ X + 1t y = - y (x -1 ) = X + 1; Xt1

x -1

o x ( y + l ) = y + l ; X * 1

_ y + l 1 X = L~ 1 y * l y -1

Vy biu thc gi thit tr thnh:

< - ( K M K H > -Gii h phng trnh:

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10/358

f ( x ) - x f ^ ^ j j = 3

'-(s M h)-D =

1 -X

X +1

x -1

= 1 -x(x-f 1) _ -X 2- 1

x -1 X 1

3 -X

3 -1 = 3 + 3x = 3 (x + 1)

V- ft \ _ Df _ 3 ( x + v _ 3 (1 -X2) . _ 3 (1 -x ) A. ,Vy: f(x) = f = - p Y~ ; - hay: f(x) = - 7 vi X* 1.

D -X - 1 1 + X l + xz

X 1

Bi 2.6. Tm h s f(x). nu bit rng:

a - f f l - = 4x^ + 1 (x * 0) b. f| x + I = X* + - J : (x* 0)'

Gii

a-f 1 1 - - I = 4X2 + 1 (1)

t t = 1 = 1 - 1X= (t * 1), Vy (1) tr thnh f(t) = ^ . + 11- t

Hay: f(x) =16

(1- x )2+ 1

b. f| X + - = X2 + *xJ X'

_ , t t = X + z X

(iu kin jt| > 2)

Ta c: t2 - X2 + - 5- + 2t2 - 2 = x2 + \X X

Vy (2) thnh f(t)

Hay: f(x) = X2 - 2 (|xj >2)

eg* so

TON TNG T

Bi 2.9. Cho hm s' f(x) =2 x - l nu X > 0

x2+ x -2 nxcOV hm s g(x) = X2- 1

Tm hm s f[g(x) ] v hnrs'g[f(x) ].

Hng n

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11/358

| 2(x)-*l nu g(x) > x > 1

^ |[g(x)i2+g (x)- 2 nug(x)0+ gf(x)] = If(x)]2- l = *, 3 .

[(x +X-2) - (x +x-2)-2 nux


12/358

Bi 2.13. rim hm s f(x) bit 2f(x) - 3f' j = X (1)

Bc 1: t t =

Hng dn2x + l

X-2Tnh X theo t, thay vo (1), sau i t thnh X, ta s c:

_3/ W + 2fg I ) = ! (2)

Bc 2: Gii h phng trnh (1) v (2), ta s c:

. 2x2 + 2x + 3f X -------- - -

5(x-2)C8*B>

v n 3

S DNG TNH CHT CA HM S LIN TC

CHNG MINH MT PHNG TRNH C NGHIM

Mun chng mh phng trnh f(x) # 0 c nghim trong khong (a; b), thngthng ta lm nh sau: .

- Chng minh hm s f(x) lin tc trn on [a; b].

- Chng minh: f(a). f(b) < 0.

Nhc l '

- Nu hm s f{x) in tc n [a; bj v f(a). f(b) < 0 th tn ti t nht mt sc e {a; b) sao cho f(c) - 0, tc phng trnh f phng trnh (*) c nghim ty.

Trng hp 2: - m = 0 v n * 0 => phng trnh (*) c nghim X = b; X = d.

- m ? i 0 v n = 0 = > p h n g tr n h (*) c n g hi m X = a; X = c.

Trng hp 3: m *0 v n * 0 (tc m. 1* 0)

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13/358

- Phng trnh (*) vit li: (x - a)(x - c) + (x - b)(x- )

V- t f(x) = (x -a)(x - c) + (x- b)(x- d) = 0

m

(D thy hm s f(x) lin tc trn R)., ff(b) = (b -a (b -c )s . '

Ta c: " ; " [ I => f(b). f(d) < 0 0 '

Vy phng trnh (*) c nghim (. p. c. m).

Ch : Du ng thc, (1) xy ra khi:

a = b=> x = a = b l l nghim ca phng trnh (*)

b = c=> x = b = c l l nghim ca phng trnh 11')

c = d=>x = c = d l l nghim ca phng trnh (*)

-d = a=>x = d = a l l nghim ca phng trnh (*)

Bi 3^2. Gho a, b, c l 3 s bt k. Chng minh rng phng trnh:a (x - b)(x - c) + b (x - a)(x - c) + c (x - a)(x b) = 0 '

lun lun c nghim.____________________ ____________

Gii

Trng hp 1: a = b = c = 0 => Vx 6 R, X nghim ca phng trnh.

Trng hp 2: a = b = c*0=> phng trnh c nghim kp X= a.

Trng hp 3: Trong 3 s a, b, c c 2 s bng nhau v khc s th 3, chng hn

a = b * c => phng trnh c t nht 1 nghim l X= a.

Trng hp 4: a, b, c khc nhau tng i mt.

Nhn xt rrg vai tr ca a, b,.c ong phng trnh hon ton gng nhau nn gi

s a < b < c m khng mt tnh tng qut

t f(x) = a (x - b)(x - c) + b (x - c)(x -) +C (x - a)(x - b).

(D thy hm s" f(x) lin tc n R):ff(a) = a(a b){-c)

Tac: jf(b) = b(b -c)(b -a) = -b (b -c) (a- b)

[f(c) = c(c-a) (e-b) = c(a-c )(b-c )

0 < a < b < c => f(a). f(b) = ~ab (a - b) 2 (a - c)(b - c) < 0 .

=> phng trnh c nghim.a < 0 f(a). f(c) - ac (a - c) 2 (a - b)(b - c) < 0

=> phng nh c nghim.a < b < 0 f(a). f(b) < 0 => phng trnh c nghim

a < b < c < 0 => f(a). f(b) < 0 => phng trnh c nghim

Kt lun: Trong mi trng hp, phng trnh lun lun c nghim..

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14/358

Bi 3.3 . Chng minh phng trnh 6*+ 8X 10* c t nht 1 nghim trn on [; 3j.

Gii

Phng trnh 6X+ 8* = 10* c* 6* + 8*- 10* = 0

t: f(x) = 6X+ 8X- io* hm sny lin tc trn II; 3]ff(l) = 6+ 8 -1 0 = 4

Ta c: I ; , , => f(l). f(3) < 0|f(3) = 6 +8 -1.0 = -272

Vy, phng trnh trn c t nht 1 nghim n on ; 3].

Bi 3.4. Cho 3 s dng m, n, p tha mn ng thc: m2 + n2 - p2

_____ Chng minh rng phng trnh m* +n* = px c duy nht 1 nghim.

. Gii

Bc 1: ' Ta d th X = 2 l 1 nghim ca phng trnh v \

m2 + n2 = p2 (gi thit)

Bc 2 :. Ta chng minh nghim X = 2'l duy nhl

Ta c: ra* + 1X = p* o | ^ ~ j + j - 1 = 0

t f w = ( p H p - j

D thy f(x) lin tc trn R.

m,n,p>0 . f0 f x ) < 0 -

suy ra X > 2

p J KP.

VyX

> 2 => Xkhng phi l nghim ca phng bnh- Tng t, ta chng minh c: X< 2 => f(x) > 0.

Vy x < 2 => Xkhng phi l nghim ca phng trnh.

Kt lun: Phng trnh c nghim duy nht X= 2.

Bi 3.5 . Chng minh rng phng trnh X2 - 12x + 4 0 c 3 nghim n on (-4; 4..

Gii

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15/358

t: fix) = X2- 12x + 4 . . .......Ta c: fM ) = -64 + 48 + 4 -12 0

f{lj = 1 - 12 + 4 = -7 < 0; . f(4) - 64 - 48 + 4 - 20 > 0Suy ra:

ft4). f(-l) < 0 => phng trnh c nghim trong(-4; - 1}i - 1). i(l) < 0 => phng trnh c nghim trong(-l; 1)f(l). (f(4)


16/358

- Tm phng trnh tng ng: x * a v X

.{ x -a )(x -) -a 2(x- |3 )- b 2(x-cx) = 0 (1)

- t f(x) = v tri ca (1)

_____ -T nh f(q). f(P) s thy.kt q. ,. . .

Bi 3.9. Cho m > 0 v 3 s a, b, c tha mn u kin:

Chng minh rng phng trnh ax2 4- bx + c = 0 c nghim thuc khong (0,1).

Hng dn

.Trng hp 1: a = 0

* b = 0 => phng trnh c nghim ty (lc c = 0).

* b 0 => phng trnh c nghim duy nht:X = - " = - 7 ( 0 ,1 )

b m +1

Trdng hp 2: a * 0 v c =. 0

P h n g t r n h c n g h i m : X = 0 ha y X = - = m - ( 0 , 1 )a m+2

Trng hp 3: a * 0 v c * 0

Tnh: * f(0) = c

m -am

G n + l J ~ ( m + l )2( m + 2)

*f(l) * a + b + x - m + 2 m

* a v c cng u f(0). f( I < 0Vm + l ;

*avct r i u=>f f < 0.V m+ ly m + 2 m

cg*B>

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17/358

CHNG 2

*t)O HM CA HM S

Vn 4

DNG NH NGHA E t n h o h m C pA MT HM S

nh ngha:

Cho hm sy = f(x) xc nh trong ln cn im Xo.

Gi A>; l s gia ca bi.1 s ti Xo-

Ay lsgia ca hm s tng ng vi Ax.

o hm ti Xo ca hm s y = f(x) l:

Ax Ax

- Cch tnh o hm ti Xo ca hm sy = f(x) bng cch s dng nh ngha:

Bc 1: Tnh y0 = f(Xo) v y = f(Xo +Ax)

Ri suy ra: Ay - f(Xo + Ax) - f(Xo) . 1.

Bc 2: Lp t s theo Ax.Ax

_ _ AvBc 3: Tm im X

0 x

TON P DNGBi 4.1. Dng nh ngha tnh, o hm ca hm s y =*>f(x) ti Xot

a. y = f(x) = 2X2 - X + 1 ti Xo = 2

b .y = f(x) = V2- 3x ti Xo = -1

. l - 2xc. y = f(x) =

1+Xti Xo = -2

Gii

a-Tac: f(2) = 7

v f(2 + x) = 2(2 + Ax)2 - (2 + Ax) + 1 = 7 + 7Ax + 2(Ax)2

do : Ay = f(2 + Ax) - f(2) = 7Ax + 2(x)2= (7 + 2Ax)Ax

vy: = 7 -t- 2x; lim == lim (7 + 2Ax) = 7Ax 4*-QAx

vy: f (2) = 7.

b . T a c : f ( - l ) = \ 5

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18/358

v f( -l + Ax) = ^ 2 - 3 ( - l +Ax) = V5-3AX

do : Ay = f( -l + Ax) - f ( - l ) = V5-3AX-75 = , "~3Ag - . *V5-3AX + V5

= lim = lim -

Ax V5-3AX+V5' ^ -Ax ^ V s -3 A X + V5 2 ^

vy; f(- 1, = rc. Ta c: f(-2) = -5

ft o _L_ A \ _ l - 2 - ( 2 + Ax) 5-2AXv ' f(2 + Ax) = -----r1 = l + (-2 +Ax) -1 + Ax

V \ _ 5-2AX - 3Ax Ay _ 3do : Ay = f{-2 + Ax) = +5 = => =-. -1 + Ax -1 + Ax Ax -1+Ax

lim = lira % -3Ax 4*-* 1-Ax "Vy: f( -2 ) = -3-

B 4.2. Dng nh ngha, tnh o hm ti X cc hm s sau:

a. f(x) = 1 - 2x -X2 b. f(x) = V l -X 2 (-1< X /l- x2-(2x+Ax)Ax - V l-X 2

19:



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19/358

--(2xvAx).Ax

V T x - ( 2 x 7 x ^ i ji- X2

(2x + Ax)vy: = _ . = ; do : im =

Ax y j \ - y r-{2x + Ax)Ax + vl ~x2

Ay -2x

^ A x 2-s/l - X2

X

Vi-X"

vy: fix) - -

1 , , .c. Hm s f


20/358

Ta c: > O tto Axj f (o * . * f - - = 1 + 1Q -i- A x I 1+ I Ax

Do Q: A =f(0 + ,\x) f(0) = '

Ax

14 5Ax

. Ay 1 , . . Ay ,vy: - ..... : ' ; do : m

Ax 1-t ,i x

vy:

b. Hm s g( x) =

f ( 0) - 1

IXI(x) = lin tuc tai X- 0

1 + x \ .

jg ( )= o

Ta c* I lim g(x )= 0 =>iim g(x) = g(0)\ X '

Vy hmsg(x) lin tc ti X- 0.g(0) = 0

Tac: g (0 + t o ) = J M -+ |A x |

Suy ra: y = g { 0 4- A x) - g ( 0 ) = - = > =l+'i Ax I Ax Ax

[ lim = 1J +, AxtrAx

im = -1^-*0- AxAy , X

Vy: lim khng tn ti nn hm s g(x) ~ khng c o hm ti X4X-0AX. 1+x"

Bi 4.4. Cho hm sf nh bi:

1- Vl-X

fix) =nu X < 1 v X * 0

X

^ riu X = 012

a. Hm s" f c lin tc ti X= 0 khng ?

b.Hm s*f c o hm ti X = 0 khng ? Nu c, tnh o hm ny.

Gii

1 + Ax

= 0.

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21/358

a. Ta c: f ( 4

.. , / V .. 1 - % / l - x X 1l i m f ( X ) l i m = lim -------7 - = ^ - 7*-*0 v ' *-0 X *-1 + VI - X. 2

Vy: lim f(x) = f(0)7ngha l hm s n tc' ti X = 0

f (0) = '

f (0 +Ax) =

b. Ta c:l - V l - A x

Ax

do : Ay = f(0 + x) - f


22/358

, , X4x2-x + 5 n u x > lBi 4.6. Cho hm s:f(x) =

[7x+1 . nu X 01Ax

- 7 ' nuAxQ+Ax x

\ ( l~ x)2 nux>0Bi 4.7. Cho hm s: f(x) -


23/358

* Quy tc 3: (o hm ca mt thng)u'.v~u.v . ^

i v j 7 (V!e0) ^

2. o hm ca hm s" htfp v hm s ngc

a. o hm ca hm s hp:Nu hm s u = (C)\= 0

2. (x) = a. xa_ *

3 - = - - 4 -X X

(ua) = cc. u-1. u1 _ u

u ~ u2

T

2

1. (sinx)= cosx2. {cbsxj = -sinx .

3. (tgx) = - - 1 cos X

4. (cotgx) = ~ r ~sin X

1. (sinu) = u'. cosu2. (cosu) - -uVsinu

3. (tgu) =cos u ...............

4. (cotgx)' ~ 2~ -sin u

4(lnlx),= *

51. (esr = ex .

2.(ax)' = aU na

1. (eu) = u \ eu

2. (au) = u. au. lna

Ch : (Hc snh cn thuc thm 3 cng thc sau) .

......... . (x + bj = a

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24/358

a b

2 f J l V - c d- _ ad -b e(.c x + d j (cx + d)2 ~ (cx + d):

l a b 2

3 f - i a k I

^a7x2 *b:x + c I ~ fj

a b 1 2 J a, X + 2 c

1 bc 1

a b a c' 1b c 1

(a x2 +b 'x + c |

TON P DUNG

Phng php: Khi tnh o hm ca hm s" y - f(x} ta cn:

- Nm vng drig ca hm s ( bit phi s dng qui tc no).

- Vn dng cc cng thc v cc h qu.___________________

Bi 5,1 . Vn ng: - Cng thc nhm 1- Qui tc 1 v h qu

___________________ - H qu ca qui tc 2,

a. Hm s' y - 2xs - 5x + 3 => y = 4x - 5

b. Hi s y = 4 ~ 3x y = -3'X 5 x 2

c. Hm s y = I- 8x - 1 => y = - 5x + 86 2 2

d. Hm s y = (2x2- l )2

Vt i: y - 4x4 - 4x2 + 1 => y = 16x2 - 8x

e.Hm sy - 4x3 - Vx + => y = 12X2-----4= -X 2Vx X2

(3 x 4 - 2 ) 2.Hm s y = ---------

X

. 9 x 8 - 1 2 x 4 + 4 _- 7 3 4Vit li: y = - ------- = .9x~ 12x3+ -

X X

=> y = 63x6 - 36x2 - 4* X2

g. Hm s y ~ V5x -1 - -3- =>y = - i = +- - - - - 2x + 3 2^/5x^l (2x + 3)

h. Hm s y = xn + (c - x)11, c l hngs => y = nx' 1- n(c - x)n_1

i. Hm s y = X + Vx2+1 => y =* 1 +----- -.= = 1 + --?:=x= = r2 y T



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25/358


26/358

i.y -

i-tf|

= -s in 3x + cosx. 2sinx. cosx = -s n 3x + CGSX. sin2x>

, tacry = I jt g x J = - | fg2x V .

y = y . 2tgx. , tgx- X4 cos X 2cos X

rc'ij. y = cotg3[ 2x + 4 )

' -2 - 6cotg2 2x + =>y- =3cotg3 x+ f J . J -----, = ------ - K - I I

sinzP2x + - j sin2 2x + - j

Bi 5.5. Vn dng cc cng thc nhm 4 v 5.

a. y = In sinx => y = x- = cot gxsinx

b.. .y =tn(x2 + 1) =>y=JT.+ .1

1 x1 H. .

c. y = In(x + Vx2 + l) = > y ' = ----- ^ J = -1= L =x + VX2 + 1 VX2 + 1

J _ / _ _ Cl + ln x ) 1d. - y= v l + In X => y =- ====== ----------- - = -

2-s/l + In X 2xVl + In X

e. y = (nx)2 ' => y = 2{inx) =X X

f. y = x2lnx => y = 2xlnx + X2. = x(21ix + 1)X

g. y = ln(lnx + 1) =5> y = ^ n x t ^ 1-----ln x + 1x(lnx + l)

Y h. y = 2* 4- 3X =* = 2xln2 + 3xn3. V

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27/358

, *( u ) u v - u v 4 aBai 5.6 Vn dng qui tc: = ---- -T---- '

\ V) V

tgxa. y = u = tgx =5> u =

1

COS2X

[V - X

X

V a1

- tgx

b .y =

, COS" X x-sinx.cosx 2x -sin 2xVyry 22r'2?. X X cos X 2x cos X

2x j u = 2x => u = 2

c o s 3 x \ v = cos3x =>v = -3sin 3x

2cos3x-6xsin3xVy, y = --j" -

cos 3x

in x-cosx iu=-*sinx-cosx '=> u = cosx + sinxsinx + cosx [v = sinx + cosx => v = co sx-sin x = -( sinx -c o sx )

0(s in X + COS x) + ( si n X - COS x)vy, y - --------------------;-------- = -

(sinx + cosx)* (sinx + cosx)

d . y =Vx2"+1 u = Vx2+ 1 => u =

Vx2+ 1

V = x v = l

-1

X VX2 +1

1

e. y =lnx Ju = lnx

X

l n x - 1

lnx + 1

V = X

u = n X - 1

v = lnx + 1

1

= > u = i . s x X lnX 1 - l n xX ; vy, y = ^ 5 = -

^ v -= l x x

=> u = X

1=> V = X

1- (ln x + l ) - - ( I n x - l ) 0

V y . / = z - x ------ --------- -(lnx + 1) x(lnx + l)

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28/358

Bi 5.7. Vn ng cng thc:' a 0

c' d ad-bc

(cx + d) (cx + d)

ax~ + bx -t- c

a X -Hb'x+.c'

a b a c b cb X2+ 2 X+

a a c b c

(a x2-f-bx + cj

a-y =

b.y =

c. y -

d. y =

e.y =

mx-1

x + m

2 x

( k - l ) x - i

a - i + ax

a + l + ax

2xa - X+ 3A2T 2x 5

X2 -f-ax +b

X2 + T _

=>y-

y1=

=

=

m2+1

(x + m)2

2 02 0

k-1 -2 = -4 _

[ ( k - l ) x - 2]3 [ ( k - l ) x - 2]'J

a a-1

a a + 1 2a

(a+ l + ax)2 (a + l + ax)2

2 - 1 2 o 2 3 X + 2 X.+- 1 3

1 2 1 1 - 5 2 - 5 5x2-2 6x -1

(x2+2x-5) (x2+2x-5)

11 ax2+ 2

1 b x +

a b

11 0 1 1 1 0 1 1

* i f (x*+l)

Bi 5.8. Tnh o hm ca hm s y -2 x 2 - x + 4 ------- ----- - h

x - 2bng 3 cch khc nhau.

Gii

Cch i: Dng qui tc - = u v u v \Vj V2 .

^ (4 x - ) ( x - 2 ) ( 2 x 2 - x + 4) 2x 2 - 8 x - 2

( x - 2 ) 2 ( x - 2 ) 2y v ( * - 2)Cch 2: Vn dng cng thc bi 2, 7

Cch 3: Vit li y = 2x + 3 4- 10X- 2

y 2 -10

( x - 2)2

C8-&&

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29/358

5I

TON TNG T

Bi 5.9. Tnh o hm ca cc hm s sau:

.. i' n I7 L f _ Vx2-2x + 3a. y - (x + 1) Vx - x + 1 b. y =2x + l

Hng dn

a . y = (x + 1 ) V x 2 - x + 1

t u = X -i- 1 v V = V x 2- - x + 1

Dng qui tc (u. v) = uv -f v'u

4x2-x + 1p s: y =

Vx2 X+

b. y -Vx2-2 x + 3

2x + l

t u = >/x2- 2 x + 3 ; V= 2x + 1

u \ u v - v uDng qui tc -------r

vv; vz

r _ 3x-7p s: y = ------ , 7- - =

(2x +1) yx 2- 2x + 3

Bi 5.10. Tnh o hm ca hm s: y = lnx -4x - 2

Hng dnVit li y = ln hay y. = - Inu, vyy ~ ~

p s: y =( x - 2 ) ( x - 4 )

Bi 5.11. Cho cc hm s f(x) = sinSi + cos4x v g(x) = cos4x

Chng minh rng: f (x) = gJ(x). Gii thch kt qti ny.

Hng dn

f (x) = 4sinx. cosx(sin2x - cosSc) = -sin4x

Vn dng: (un)' = n. u""1. u vsin 2a = 2sin a. COS a

[cos 2a = cos2 a - sin2a

Vy,

g'(x) = -sin4x

f (x) = g(x)

Gii thch kt qua

3

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30/358

f(x) = (sin2x + cos2x)2 - 2sin2xcos2x

f(x) = 1 - ~ sin22x = 1- - | ; f(x) = + cos4x = + g(x)) OI o I 4 4 4

, 1 . 2o- __T l/l - c o s 4 x ^= 1 - ~ sin 2x 3S1- - ;2 . 2 { 2 J

" r 2)

TON P DNG

Bi 6.1. Tnh o hm cp 2 ca cc hmssau:

a. y = ax4 + bx3 + dx 4- e (a, b, 'c, d, e l hng s)

b. y = x2e2 ___________ _ c. y - x^nx

Gi

a. Ta c: y = ax4 + bx3 + dx + e

=> y = 4ax3 4- 3X2 + 2cx + ; => y = 12ax2+ 6bx + 2cb. y = x V

y = 2xe* + x V = (2x + xV*

=> -y = (2 + 2x)2.+ (2x + x V = (x2 + 4x + 2)exc. y = x^lnx

=> y = x^nx + X3. - = x*(31nx 4- !)=> y" = 6xnx + 5x = x(61nx + 5)

Bi 6.2. Tnh o hm cp 2 ca cc hm s sau:

... a.y = snzx b.y = nx +V x2 + l j '

Gii

31



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31/358

a.y - sin2x .

=> y = 2sinxcosx = 2sin2x; => y = 2cos2x

b. y * nx-t- Vx2+ j

1 X. 1 + r===s= r

Vx2 -f 1 _ 1 X.______

XW x 2+1 %/x2 +1 (x2+ l ) \ / j r +1

(Ch: y= V u= Vj t +Tu

Bi 6.3. Cho hm sy = sinx. Chng minh rang:

y(n) =sin^x + n j (ne-N) (1)

T c: y = sinx => y = cosx

Giai

= sin fx + I V

Vy cng thc (1) ng khi n = 1

Giscng thc (!) ng khi n = k (k e N, k s 1),

Ngha : = sin^x+kj

=>yM = r sin x + k ^ = COS x+ k - r = sin x + k ^ + ? 1= sin

Vy cng thc (1) ng vi mi n (n e +).

x + ( k + l ) |

TON TNG T

Bi 6.4 . Tnh o hm cp 3 ca cc hm s sau:

a. y = 5x4 - 4x3 -f X2 + X + 1 t . y = 6 (x ln x - x)

c. y = ln x . y = Jx

p s: a. (S) = 24(5x - 1) c. y(3) = J~

b v(3) = ' d v(3)'= = = - j X2 * - . . 8yfx? -8x2>/x

Ch : ng cng thc |u ) = a.u0'1 .u gii cu d.\

Bi.6.5. Cho hmsy = ent (n e N).

Chng minh rng: y(N>=. nn.enx. ' ' ___________

32



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32/358

Dng phng php qui np.

. - Cng thc (1) ng khi n = 1Gi s y= nfeenx (k +)

- Ta chng minh c:y(k +]) = nk+!enx

Hng n

Bi 6 .6. Tnh o hm cp 1ca y= sinx.

Chng minh rng: y(^n) = y.

Hng dn

Dng phng php qui np

* y(4) = y (hc sinh t chng minh)

Gi s y(4k = nkenx - y ta chng minh y4(k +n = y.

Ch : y(4k) = yy(4k +1} = y(4k)]' = y y(4k +2) = y

C3*S>

V n e 7

NGHA HNH HC CA O HM

TOM TTGAO KHOA1. nh

Nu hm s y - f(x) 'Co hm Xo f(xo) th ti im Mq(xo, f(xo)), th c tip

'tuyn v h s gc ca tip tuyn ny l k = f (xo).

2. H quTi M(xo,f(xo) cI .th Gtip-tuyn nm ngang -'f(xc) - 0.

TON p DNG

Bi 7.1. Cho hm s y = f{x) = -X3 + 6x2 - 9x - 5. tm n th nhng im m ti th ctip tuyn nm ngang ________________________________ _

Gii

Hm s y = f(x) = = -X3 + 6x2- 9x~ 5 c

Min xc nh: D = R

Ta c: y' = f(x) = -3x2 + 12x - 9Ti Mo(xo, f (Xo)) th c tip tuyn nm ngang

x0 = l = *y 0 = f( l) = -9 '=> f (Xo) ~ -3 Xq + 12xq - 9 - 0

Vy, cc im phi tm Mo(l, -9) v Mo(3, -5).

f0 =3 => y0 = f(3) = -5

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33/358

Bi 7.2. Xc nh tham s m th cc hm s sau c 2 tip tun nm ngang:

X3 / -V 2 . * , c( \ 2x* -m x + 2m + la. y = f(x) = - (m -l )x +4x + m 7b. y = f(x) = -------------

GiiX3

a. y = f(x) = ~ - ( m - l ) x 2+ 4x + m .

Min xc nh: D = RTa c: y = f (x) = X2 - 2(m - l) x + 4

th hm s c 2 tip tuyn nm ngang,o Phng trnhy = 0 tc phng trnh:

X2 - 2(m - l)x 4- 4 = 0 c 2 nghim phn bito A' = ( m - l ) 2- 4 > ( m + l) ( m - 3 )> 0 o m < - l v 3< m .

, 2x2-m x + 2m + lb. y = f(x) = ------------------

X 1Min xc nh : D=R\ {!}.

, s _ 2x2-4 s -( m + l)y = = T w

(x_1) th hm s c 2 tp tuyn nm ngang.o phng trnh y = 0 c 2 nghim phn bit khc 1.

o phng trnh 2x2- 4x - (m - 1) = 0 c 2 nghim phn bit khc 1.

A >0 * (4 + 2(m + l )> 0 f m > - 3 . 1 , ni > -3

[ x * l [2(1) - 4( 1) - (m + 1) 5* 0 [ m ^ -3

Bi 7.3. Cho hm s y = f(x) = X2 - 2kx + 2k -1. Tim tp hp nhng im irn th hm

s m ti th c tip tuyn song song vi ng thng y - - 2x- . __________

GiiHm s y = f(x) = X2 2kx + 2k - 1 (mien xc nh R)y = f(x) = 2x - 2k(ng thng = 1 - 2x c h $ gc l - 2) .Gi M(x, y) l im trn th hm s m ti th c tip tuyn song song vi

ng thng = 1 - 2x.To (x, y) l nghim h phng trnh:

y = x2- 2kx + 2k - l jy = x2- 2kx + 2k - l (1) \

| f (x) = -2 [2x - 2k = -2 (2)

T (2) ta c : 2k = 2x + 2, thay 2k vo (1):y = X2 - x(2x+2) T" (2x + 2) - 1 y = -X2 + 1

Vy tp hp cc im M(x, y) la parabol: y. = -X2 + 1.

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34/358

Bi 7.4. Tm tp hp nhng im n th hm s:

y = f(x) = - 3x* -f 2x2 - kx + 1 (khi k thay i) m ti th c tip tuynnm ngang. ___________________________________ - _________

Gii

Hm s y = -$ '+ 2X2 - kx + 1 (min xc nh R)

y = f (x) = -3X2 + 4x - k. '

- iu kin th c tip- tuyn nm ngang l phng trnh y = 0 tc pbnh trnh-3X2 + 4x - k = 0 c ngm.

iu ny xy ra A > 0 .

4 - 3 k > 0 o k < -3

- Gi M(x, y) l im n ^ th m ti th c tip tuyn .nm ngang.

To (x, y) l nghim h phng trnh:fy = -x 3+ 2x2-k x +1 y = -x 3+2x2-k x + l (1)

| f (x)i=0 -3 x 2+.4x-k = 0 (2)

Kh k gia (1) v (2):

(2) k = 3X2 + 4x. Thay k vo (1)tac:

(1) y X3 + 2X2 - X-3X2 + 4x) + 1 y = 2x3 - X2 + 1

k -Gii hn :- ~ 3 =>-3x2+4x<

k = -3x2+ 4x ^

=> 9X2 - 12x + 4 > 0 => (3x - 2)2 > 0 (hin nhin).

Kt u n : Tp hp nhng im ri th (hm s cho) tho bi l ng cong cphng tr n h : _______ y =. 2x3 - X2 -KT______

2x XH* 3Bi 7.5. Cho hm s : y = f(x) = ------ Trn th, ly 2 im khc rihau Mi(xi V i)

. . x +2 ~

vM2(x2,y2)>(xr,x2:*^2) . . 'a. Chng minh rang iu kin cn v ti M], M2, th c cc tip tuyn song

song nhau Xi + x2 = -4.

b. Gi s diu kin n tho mn tm h thc lin lc gia yj v y2 v chng mnh

rng trung im on thng MM2 l im cQnh. ______"_________

Gii

U' ^ \ - 2x2- x + 3 _ c 13Hm so y = f(x) = -------- = 2x - 5+ x+2 x+2

-Min xc n h: D = R\ {-2}.

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35/358

. y = f(x) = 2 - 13: ' (* -2 ) .

- H sgc ca tip tuyn t i 'M (xi, f(X})), M2(x2, f(x2)) n c l ki= f(xih = f(x2). s ;

a. Chng minh Xi + x2= -4ti M,'M th c -cc tp tuyn so ng song nhau

~ ki = k2 ' f(Xl) = f(X2) 2 - 13- - y = 2 -(x ,+ 2)2 (x2+ 2)

, . , fx , + 2 = Xj + 2 '(Xi + 2) = (X2 + 2) " / v

[x ,+2 = - (x 2 + 2)

, Xj =.x2 =>Mj =M2 (tri gi thit)

Xj + x 2.= -4 (iu phi chng minh)

b. H thc lin lc g yi, v y2: 2S; 5 * ; t --2

^ y i+ y2= 2(x ,+Xi)-1 0 + 1 3 ^ + - i i J . 2 ( - 4 ) - l , 13 ^ | ^

= -18 (v X] + X2 + 4 = 0) -

H cn tm l: yi + y2 = 18- Chng minh on MiM? c trung im c' nh:

Gi I l trung-im on M]M2>ta c:

' _ ^ + x 2 _ - 4 _ o . " . y i+ y2 ..-1 8 n* 2 2 y i ' ~ Y " 2 ' "

V, I(-2, -9) c nh.

Bi 7.6. Tnh m th hm s: y = fix) =X3 - 3mx + 3m - 1 tip xc vi trc honh

Gii

Hm sy f(x) = X3 - 3mx + 3m - 1

Min xc nh: D = R

y = f (x) = 3x2 - 3m = 3X2- m )'

th tp xc vi trc honh ti im Mo(xo, 0).

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36/358

f (x o) = 0. x-3m x0+ 3 m -l = 0 (1)

^ f (x0) - 0 ^ xq- m = 0 ' (2)

(2) o m = x, thay m vo (I) ta c:

x ^ - 3 x + 3 x g - l = 0 2x -3 x q +1 = 0

(x0- l ) ( 2x * -x 0- l ) = 0x0 = 1 => m = Xg = 1

. 1 1 2 12 =

Vy: m = 1, m = 4

Bi 7.7. Cho hm s: y = f(x) = ax2 -j- bx - 3

Tnh a, b th hm s tip xc vi ng thng y = 2x + 4 ti im c honh

x - 1. ___________________,Gii

Hm s y = f{x) = ax2 4-bx - 3

- Min xc nh: D = R

- ng thng(d): y = 2x +4 c h s gc k = 2.

Gi M l tip im ca th hm s v ng thng(d):

Ta c: Me (d) yM= 2xM+ 4 = 2(1) + 4 = 6

Vy, M(l, 6)

- M(l, 6) th, nri 6 = a( l)2 4- b(l) - 3

'a + b = 9 Theongha hnh hc ca o hm, ta c:

f(xM) = 2 2axM+ b = 2 o 2a + b = 2' , , . f a + b = 9 , [a = -7Gii h phng trnh ta c: !,

: 2a+b = 2 b = 16

Bi 7.8. Cho hm s: y = f (x) = x + Px + q' X +1

Gi s th hm s ct Ox ti im c honh Xo, chng minh rng ti im

ny, tip tuyn c thi c h s gc k = 4+1

Gi

tr--. X +px + qHm s y = f (x) = /X + 1

- Min xc nh: - R

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37/358

(2x + p)(x2 + l ) - 2x(x2 + px+q)f (x)= - 2

(x +1) \

- th hm s' ct Ox ti im Mo(xo, 0) nn:

x + px0+ q = 0 I (1)

- Theo ngha hnh hc ca o hm, ti Mote). 0) tip tuyn ca th c h sgc l:

(2x0 + p ) ( x ^ l ) - 2x0(x + px0+q)K- r ------------------- 2 w

(x+l)

T (1) v (2) ta c: k = ^ y - ^ (iu phi,chng minh).X o + 1

Bi 7.9 Cho hm s y - f (x) - - 2x- - * ^X1

_____ Tm tt c gi trica k th hm sc t nht 1 tip tuyn m h sgc bng k.

. Gii

(I \ _ - 2 x 2 + x - 3 0 - 4Hm s y =;(x)~--------4 -----= -2x- 1 -----

x - 1 X ~ 1

- Min xc inh: D = R\{1}; y= f (x) = -2 +- 2( x - 1 )

th c t nht 1 tip tuyn c h s gc bng k.

Phng trnh f(x) k c nghim.

4o Phng tr nh- 2 + 2 =k c nghim( x - 1)

4o Phng trnh -2- = k + 2 c nghim

( x - ! ) . k + 2 > 0 k > - 2.

Bi 7.10. Cho hm s y .= (x) = - vi c * 0 v ad - bc *0.cx+d

Chng minh rng th ny khng th c 2 tip tun vurig gc nhau.

Gii

Hm s y = f (x) = + kcx + d

- Min xc inh: D = R\ - - ; f (x) = I c j w ( cx -d )

Gi s th hm s c hai tip tuyn vung gc nhau.

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40/358

I

I "'

TN P D NG ______________

I. Bi 8.1. Cho hm s: y = f {x) = - ^ ~ ^ .w 2x + l .

I a. Tim phng trnh tip tuyn ca th hm s ti im thuc th c honh

I- . X= 1.

I . b. Tm phng trinh tip tuyn ca th hm .s ti giao im ca th vi trc tung. - ____ ;. ___

Gii

Hm s y = f (x )~ x ~ w 2x + l /

Min xc nh: D = R \ j - - ; y = f (x) = T1 2 . w (2x + 1)

a. Phng trnh tip tuyn ca th hm sti dim thuc th c honh Xo = 1 l:

y - f ( l ) = f ( l ) ( x - l )

= g (X ") hay y ~ g ^ ~ g (tip im:

b. Giao im ca th v trc tung l Mo(0, -2)

Vy, phng trnh tip tuyn ca th ti Mo :-

y - ( 2) = f (0)(x -0) y + 2 = 5x, hay y = 5 x - 2.

Bi 8.2. a. Tm phng trnh cc tip tuyn ca th hm s:

y = X3 - 3x2 + 4x + 1

Bit rng cc tip tuyn ny c h s gc k = 4.b. Tm phng trnh cc tip tuyn vi th hm s:

r/ %. x2- x -2y=f(x); - 7 2 L-

Bit rng cc tip tuyn ny song song vi ng thng

y = 2 - 3 X

c. 'Om phng trnh cc tip tuyn vi d th hm s:

-2x2- 3xy = f(x) =

x.+ l

x + 2Bit rng cc tip tuyh ny vung gc vi ng thng y =

Gii

Hm s: y = f(x)= X3 -3 x 2+ 4x + 1

Min xc nh: D = R

y = f(x) = 3X2 - 6x+ 4

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41/358

Phng trnh ca tip tuyn vi th hm s c dng:

y - f(xo) = f (x) (x - Xo), Xo l honh tip "im.

- H gc ca cc tip tuyn ny k = f (x) =$4o 3 x - 6 x 0 + 4 = 4 3 x - 6 x o = 0 o x | - 2 x o = 0

Xo = 0; Xo = 2 => 0 - f(0) = 1, yo = f(2) = 5

- Vi Xo = 0, phng trnh tip tuyn cn tm l:

y - ] = 4 ( x _ 0 ) c y.= 4x + 1, tip im (0,1)

- Vi Xo = 1, phng trnh tip tuyn cn tm l:y - 5 = 4(x - 2) y = 4x - 3,tip im (2 , 5),

L. c \ x2 - x - 2 4b. Hm s y = f(x ) = --- = X -3+ v x+2 x+2

4Minxcinh: D = R \{-2} ; y' = f (x) = l - -

(x + 2)

- Phng trnh tip tuyn ca th c dng:y - f(xo) = f (Xo) (x - Xo) Xo l honh tip im.

- Theo gi thuyt, cc tip tuyn ny song song vi ng thng y = 2 - 3x (c h

s gc k = -3) nn:

k = f '(x0) = - 3 o 1- - 2- = 3K + 2)

X0H t2 = 1

Lx0 + 2 = -1c* - = 4 (xo + 2)2 = 1 o

(x0 +2)

xo = -l= > yo =f(xo) = f ( - l ) = 0x0 =-3=>y0 =f(x0) = f ( -3 )X lO

* Vi Xo = -1 , phng trnh tip tuyn cn tm l:

y (0) 3(x 4- 1) o y = - 3 x - 3 , tip im .(-l; 0)

* Vi Xo = -3, phng trnh tip tuyn cn tm l:

y + 10 = -3(x + 3) - o y =-3x -1 9 , tip im l (-3,-1 0)

- 2 - - - = - 3 ( x 0 + l) ' i l .3 (*0 + 1)

* y - -3x, tip im 0(0, 0)

* y -3x - 4, tip im {t-2,2). ______ 1 :V

Bi 8.3. a. Hm s y = f(x) = 2X2 + X - 4; Tm phng trnh cc tip tuyn vi thhm s bit cc tip tun ny qua im A(lf -3),

. Hy vit phng trnh cc ng thng qu im A(2, -2 ) v tip xc vi hm s y = X3 - 3X2 + 2. ' . . ; ' - ' .

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43/358

Miri xc nh: D = R\{1}; y = f' (x) = - l - ^( x - l f

Phng trnh ng thng (d) tip xc vi d th c dng:

y - f(xo) = f Xo)(x - Xo), Xo l honh tip im, (Xo * 1) .

,V ng thng(d) qua im (1,3) nn:

-3 - f(xo) = f (xo)(l - Xo) .

. 3 r 3 1

_ 3 3 _ 6 cc > x 0 + 5 -------------= x 0 - l + - - = 6 o X o - l = 1: x0 - l ... x0 - l . . x0 - l

o Xo = 2. c 1 tip tuyn{d) duy nhtGhi ch: Phng tnh ca tip tuyn () cn tm.l:

. y - f(2) = f ( 2 ) ( x - 2 )

o y + 1 = . -4 (x -2 )o y = -4x + 7,tip im,(2; -1).

TON TNG T

Bi 8.4. Vt phng trnh tip tuyn ca th hm s y = f(x) ti im c honh c ch ra.

' a. y = f(x) = -X 2 + 2x + 4; Xo = -1 : b. y = t (x) = - X-+ ; x0 = 2X 1

c. y = f(x) - V 3X -2 ; Xo = 2 ____d, y = f(x) = Inx; X = 1

Hng dn

a. Lm ging bi 5.1 , y - f{-l) = f (-1) (x + 1) .

p s: y = 4x + 5, tip im, MoM, 1)

b. Lm ging bi 5.1, y - f(2) = f(2)(x - 2)'

p s: y = -3x + u , tip im, Mo(2, 5)G Lm ging bi 5.1, y - f(2) = f (2) (x -2 )

p s: y = + , tip im, M(2,2)4 '2

cL Lm ging bi 5.1, y - f(l) = f(2)(x- 1)

p s:y .= X - 1, tip im, M0(l, 0).

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44/358

Bi 8.5. Cho parapo (P): y = f(x) = X2 - 2x -+ 3. Tm phng trnh tip tuyn ca (P)bit rng tip tuyn ny:

a. Song song vi ng thng 4x - 2y 4- 5 = 0

b. Vung gc vi ng thng X - 4y = 0._________________ ________ ______

Hng na. Lm ging bi 5. 2a.

(Ch : ng thng 4x - 2y + 5 = 0 y = 2x + - )

Cho f,(Xo) = 2 ta s c Xo = 2.

p s": y = 2x - 1, tip im M0{2, 3).

b. Lm ging bi 5. 2b. (Ch : ng thng x -4 y = 0y = x)

Ch . f (Xo) = -1, ta s c Xo = -1.4

p s: y = -4x 4- 2, tip im M(-1, 6).

Bi 8.6. cho hm s y = f(x) = X3 - 4X2 + 4x. Tm phng trnh cc tip tuyn ca thhm s bit rng cc tip tn ny qua im B{3, 3).

Hng dn -'Tip tuyh ny qua B(3, 3) nn3-f(Xo) = f(Xo)(3 -Xo) (1)

(Xo honh tip im) . . ,: - Phng trnh (1) (Xo - 3) (X* - Xo + 1) = (Xo - 2)(3x - 2Hx* - 3)

Phng trnh ny c nghim Xo = 3, Xo = -2

p s: y = 7x - 18, tip im B{3,3); y = (x + 1), tip im l M0f i4 ( 2 8 /

C8*K>

Vn 8B ;

NH L LAGRANGE

TM TT GIO KHOA1. nh l: n hm sy = f(x) rXin tc trn an[a,b]

I*C o hm trong khong (a, b)

r -. f (b) - f fa )th tn tai t n h tl s c e (a, b) sao eho f(c) =

b -a

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45/358

2. ngha hnh hc y : .

f(b)

f(a) 1

! q a c b

Trn th hm s y = f(x), ly 2 im: A(a, f(a)> v B(b, f(b)).

Ta c k = 7/(^1 l h s gc ca ngthng AB.b - a

Vy, nu cc gi thit ca nh l Lagrange c tha mn th trn cung AB ca thhm s tn ti t nht 1 im C(c, f(c)) m ti th c tip tuyn sng song vi ng

thng th AB.

TON P DNG

B s. b. Cho hm s y = f(x) c o hm trong khong(a, b). Gi s th hm s'ctOx ti 2 im Mi(xi, 0), M2(x2, 0) v Xi, x2 c (a, b). Chng minh rng phng trnhf(x) = Oc t nht 1 nghim X g (X},x2). _____________________

Hng dn

Theo gi thit ta c:

* f(x) lin tc trn [xi, X2*f(X i ) = 0 , f(X2) = 0 .

* p ng nh l Lagrange n (X]7x2j ta s c iu phi chng minh.

Gii Hm s f(x) c o hm trong khong(a, b) l [xi, x2j c (a, b) => hm s'f(x) c

o hm trong khong (Xi, Xa) v lin tc trn xi, x2]. th ct Ox ti im MX, 0), M2(x2j 0). => y = f(xi) = f(xo) = 0Cc gi thit ca nh Lagrnge u tha mn trn on xj, X2]

=> 3 Xo e (xb X2) : f (Xo) = hay f (Xo) = = 0X j - X 2 X ! - x 2

________= Phng trnh f (x) 0 c t nht l 1 nghim X e (Xi, X2) . ________

Bi 8. 2b. Cho hm s' f(x) = (x2 - 4){x + l)(x - 3). Chng minh rng phng trnhf(x) - 0 c 3 nghim phn bit.____________________ _ _

Hng dn* Hm s f(x) lin tc trn R nn lin tc trn cc on[-2, -I, [-1, 2),[2, 3]

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46/358

Hm s f(x) hm s a thc bc 4 nn c o hm trn R => f(x) c o hmtrn cc on ni trn.

p ng nh l Lagrange n cc on [-2, -1)... ta s c kt qu.

Gii

D ng ta thy hm s f(x) c hm trn R nn cc gi thit ca nh l. Lagrange .u tha mn trn cc on ca R. Ta cn c f(-2) =.,f(-l) = f(2) = f(3) = 0, Ln lt p ng nh l Lagrange trn on [-2, -1], -l, 2], [2/3] ta c:

3xi e (-2, - 1): f (X]) = d h L ! ) = 0 (1) .- 1 +2

3X2 ( - 1 , 2 ) : = 0 '(2)

3X3 e phng trnh f(x) = 0 c 3 nghim phn bit (3 nghim ny lXi, x2, X3 n lt trong khong(-2, -1), (-1,2), (2,3).

Ghi ch: v f(x) l hm s' a thc bc 4 nn f (x) l a thc nguyn bc 3.=> f(x) = 0 ch c 3 nghim nh chng minh.

Bi 8. 3b. Cho hm sy = (x - a)(x - b)(x - c)(x - d)(x e), trong a < b < c< d


47/358

. 3x2


48/358

TON TNG T

;Bi 8;5b. Cho hm s f(x) = X5 + 5x4 - 5x3 - 25X2 + 4x + 20a. Phn tch f(x) v dng tch s. .

b. Chng minh rng phng trnh f (x) = c 4 nghimphn bit.__

Hng dn giia. f(x) = (x + 5)(x'+ 2)(x + l)(x - l) (x - 2 )

b- p ng nh HLagrange n cc on [-5, -2],[-2, - 1),(-1, 1, (a, 2J.

Bi 8.6b. Cho hm s f(x) = Vx2- 2x + 2______Tm s' cong nh, t Lagrange trn on [Q, 2] ? ______ ______ ,

Hng dan v p s

= 0

m f(c) = ~= 1

V .C - 2c + 2

t ta c c = 1.

cs*80

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* Ch : u ng thc (1) hoc (2) ch x ra ti 1 shu hn im ca khong (a,b).

* Cn nh: tam thc bc hai (ax2 + bx 4- c, a 5* 0) lun lun cng u vi a A < 0.

TON P DNG________' _______

Bi 9.1.

a. Chng minh trn hm sy = f(x) = X3 - X2 + 2x - 3 ng bin trn min xc nh

ca n. ' 2x +1

b. Chng minh hm s" y = nghch-bin trn tng khong m hm s xc nh.x - 3

~ X_1c. Xt tnh cfn iu ca hms y - -----

1 X

, + 2x '

d. Chng minh hm so y = " -S-- ong bin trn cc khong xc inh ca n,, ______________1X _____________________________

Gii

a. Hm s y = f(x) = X3 - X2 4- 2x - 3

- Min xc nh R.

- Ta c; f (x) = 3X2 - 2x + 2 > 0, Vx, X G R v: 1 ^ < ^ [a = 3> 0

Vy hm s y = f(x) ng bin trn min xc nh R ca n.

b. Hm sy = f(x) =x - 3

- Min xc nh: D = R\{3> = (->, 3)-u (3, +.oo)

- Ta c: y = f (x) = T < 0, Vx, X e D.(x-3) -

Vy hm s nghch bin trong cc Khorig (-CO, 3) (3, + 0).

c. Hm s: y = f(x) = -X - . 1 - x -

Min xc nh D = R\{1}

y = f(x) = -1 - -Y < 0 , Vx,X e D V(l~x)

. Vy, hm s" ny nghch bin trong cc khong m hm s xc nh.

d. . - Hm s-.y = f(x) = l~x\ .

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51/358

. [ A = 1 - 4 < 0

1 > 0

. .' 2(x2 + X + 1- Min xc inh: = R\{-1, +1}; f(x). = - A- - ^ M)Ta thy: x2+ x + X > 0 , Vx, xe R v I

Y? 4- Y-- 1 _Nn: : f(x )= * ' > 0, Vx,xeD.

. ; . ( M ) : Vy, hm Sng bin trong-cc khong (-00, - 1), (- ,1), (1, +). ,

Bi 9 2 . Chng minh cc hni s sau n iu trn tng khong m hm s xc nh, btchp gi tr ca tham s:

a. - f(x) = -X3+ (m + )** ~ (m2 + 2)x + m, (m l tham s)

b. y = f(x) = (k 0 * 7 ^ ^ tham s)x + k

c.*Hm sy = f(x) = ^ax- - ^ (l iam s)x - a .

Gii

a. - Min xc nh D = R.

y' = f (x) = -3X2 + 2(m + I)x - (m2 + 2)

- Ta thy p(x) l tam thc bc hai c:

A = (m + l) 2 - 3(m2 + 2) = -2 m 2 + 2m - 5 < 0,Vm, m e R

H s a = -3 < 0 nn f (x) < 0, Vx, X e R ,

Vy>hm s y = f(x) nghch bin trn ton min xc nh ca n.

b. - Min xc nh: D = R\{-k}

(k - l )k -2 _ k*-k + 2 . .y' = f(x) = ------ ----------------- V

(x+k) (x+k)

A = 1-S < 0Ta c: lo - k + 2 > 0, vk, k e D v \ ~ => f(x) > 0, Vx, X D-

[a = 1 > 0

Vy, hm sy = f(x) tng trn cc khong (-00, -k), (-k, +ac).

c. Cch 1

a2-a + lThc hin php chia a thc: y = f(x) = 2x - a

- Min xc nh: D: = R\ {a}; y = f (x) = 2 +

x - a

a2 - a +1

\ ( x -a )2

V a2- a + 1 > 0, Va, R

52



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52/358

Nn: f (x) > 0, Vx, X R

Vy, hm s ng bin n cc khong m hm s xc nh.

Cch 2

.Tam thc: 2x2 - 4ax + 3a2 - a + 1 c:

A = 4a22(3a2- a.+1) = -2 az + 2a - 2 < 0, Va, a. H

jx 2 c t s l 2 > 0 '

Nn: 2X2 - 4ax + 3a2 - a + 1 > 0, Vx, X e D.

Vy, hm s = f(x) g bin trn khong m hm s xc nh..

Bi 9.3 . Chng minh hm s y = f(x) = tgx + sinx - 2x ng bin trong

lTv_ ^ _ 2xz- 3 a x + a -lHm so: . y = f(x) = ---------- ---------

x -ax -a

Min xc nh; D: = R\ {a}; f(x) =

Gii

Hm s: y = f(x) = tgx + sinx - 2x

cs X

(1 - COS x) ( l - COS2 X + COS X

COS X

' 2sin2-(sin2x + cosx

cos2X COS2 X

Vi 0 < X < nn sin > 0, cosx > 02 2

Vy, f(x) > 0, Vx, X

=> Hm s cho.ng bin trong khong

Cch 2

53



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53/358

Ta c: fix) = cosx 4----- \ ----- 2COS X

V 0 < COSX < 1 => cosx > COS2X

Vy cosx H-----

V " > CQS2* + V " - 2 (bt ng thc Cauchy).cs X COS X

cosx + K--------------------------------------------------- 2 > 0 *=>'f(x)>0,Vx,Xe1.

Bi 9.4. Chng minh hm s y - f(x) = cosx - Xnghch bin trong khong (0, 2t).

Gii .

Ta c:f (x) =.-sinx - 1 < 0, Vx, X e (0/2:).

371(du ng th c xy ra

TON TNG T

Hng dn

1y* = -sinx ------ I- 2. c hai cch xt du.

s i r r x

.. _ - sin x + 2sin2x - l (s in x -l) (-s in 2x + sinx + l)/ Cch 1: y = ---------- --------------- = ----------- 7 -----------^ - L

< sin X sin X

(sin x- l)(c os 2x + sinx) (= i--------- lzL ----------J . sinx > sin2x2

1 - 2 1=> sinx H--- -5 > sin X+ > 2 => y < 0.

sin X si n X

54



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54/358

Hng dn

y = mcosx- 2sinx- 3 = Vm2+ 4 cos(x+a) - 3

m . .2 . . COS =?=== s in a = .--7=l >/m2+4 s/m2+ 4 /

D thy -Jmz+.4 cos(x'+ ) < y/m + 4 < 3 (o |m| < %/5 )

Nn y < 0, Vx, Xe R.

Bi 9 .9. Chng minh rng hm s y = aea*+ b (a 5* 0) lun lun ng bin trn min xcnh ca n.- _ __ __ __ ' . .

Hng dn

y = a2.e ax + b > 0, Vx, X R (v a * 0 nn a2 > 0).

Bi 9.10. Chng minh rng nu phng trnh x2 -m x + m = 0c2 nghim .Xi, X2 tha'2 *'

mn iu kin Xj< -m < x2 th hm s"y = --------------- lun nghch bin trng cc. x+m

khong m hm sc xc nh.___ ' _ .

y =

Hng dn-X2 - 2mx+ m 2 + m *

(x + m)

- t f(x) = -X2 - 2mx + m2 + m

F(x) c A - 2m2 + m

- V X2 - mx + m = 0 c X] < -m < x2

55



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55/358

Nn (-mj2 - m(-m) + m < 0 2m2 + m < 0

Vy, I < ^ => f(x)


56/358

y= f (x) = -3X2 - 6x = -3 x(x + 2); y = 0

- Bng bim thin:

X = 0

X = -2

X . 00 - 2 0 +00

y' - 0 + 0 -y ----:---- * ---------- * ---- -----_

Vy, hm s ng bin ong khong {-2, 0) v nghch bin trong cc khong (-CO, -2);(0, +).

c. Hm s'y = f(x) = x2(4 * X2) = -X4 + 4x2

- Min xc nh D = R

y = ~4x3.+ 8x = -4x(x2 - 2); y = 0 I = 0 _[x = V2

- Bng bin thin:X -00 v/2 0 /2 +00

y + 0 - 0 + 0 -

y

Vy, hm s cho ng bin trong cc khong (-00 ,-72 , (o, V2 ] v nghch bin

trong cc khong -yf2,0'j,yf2,+oc

. Hm s"y = f(x) == 3x4 + 4x3 - 48x* - I44x +5

- Min xc nh D = Ry . = 12X3 + 12x2 - 9 6 x - 144 = 12(x3 + X2 - 8x - 12)

= 12(x + 2)(xz - X - 6)

= 12(x + 2){x + 2)(x - 3) = 12(x + 2)2(x - 3) J

y - r r - o (x +2 ) 2( x -3 ) = 0 r * 2, ( p! " -X = 3 (n)

- Brig bin thin: .X.. 00 - 2 3 +0 0

y - 0 - 0 +

y - ___r " ' ' *

Vy, hm s nghch bin trong, khong {-00, 3) v hg bin trong (3, +co).

Bi 10-2- Tm cc khong ng bin, nghch bin ca cc hm s sau:

a. y = fix) =X - X + 2

~ 2 - xb. y = f(x) = -

x + 1

57



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57/358


58/358

tv'

' < ' . 1X 00 -2 +00_________ 2 ___________y_________0 + 0 -

Vy, hm s ng, bin trong khong v nghch bin trong cc khong

(-oo, -2), j | +c o) -

J I,i ^ X 2x2 -X + 5d. Hm s y = f(x) = ----- '

(

- Min xc nh: D = R\{1}

-3x 3- 6x + 9

( x -

-3x2- 6x + 9 = 0

X ^ 1

X= 1 (loi)

X= -3

- Bng bin thin:-co -3 +00

y' - 0 + -

y . .----- -------------------*

Hm s cho ng bin trong khong (-3, 1) v nghch bin ong cc khong

(->-3), (1, +=0). ______. _____________________

Bi 10.3. Tm cc khong ng bin, nghch bin ca hm sq :

_____________ y = X+ 2cosx, Xe (0 ,7i) _____. _________________

Gii

Hm s" y = f(xj = X+ 2cosx, Xe (0, Tt)

y ' - 0 o l - 2 s i n x = 0 (0 < X< 7)

sinx = (0 < X< T), o2 .

71

x _ 6

,x = 5n

y > 0 1 - 2sinx > 0 (0 < X < m) a>sinx <

o I 0 < x < i v f < X < 7 1

(0 < X< 7i)

59



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59/358

y < 0 / 1 - 2sinx < 0 (0 < X < n) snx > (0 < X < 7)**4 : ~ 2

7t -571o < x<

6 6 .

- Bng bin thin:

X 0. t n

. 6

5n "T

y + 0 - 0 +

y

, -ng bin ong Gc kho6 6 ;

NH?4

C * SO

TON TNG T

Bi 10.4. Tm khong ng bin, nghch bin ca hm s:

X2 +2xy =

X2 - x -2

Hng dn

- Min xc nh: D R\ {-1,2} '

-3x2- 4x - 4y = < 0 (Vx, X s D)

(x2- X-2)

Vy, hm s nghch bin trong ce khng {-co, -), (-1, 2), (2, +co).

Bi 10.5. Tm cc khong ng bin,nghch bin ca hms:

y = cosx(l + sinx) vi X e (0, 2n) '

y - -2sin2x sinx + I;

y > 0 -1 < sinx < - . 2

Hngdn

sinx - 0 + 0 - -2

0< x < 6

5n


60/358

y' < 0 < sinx


61/358

* m e (-00, -1) (0,1) (lc X] < x2)

X 00 m1

+00m

y + 0 - 0 -

* m (-1, 0) 'o (1, +c) (lc Xi > X2)

1X _oo --- m +00

m *

y + 0 - 0 -

c bit m = 1 =2- yl = (x l)2 > 0, Vx, Xe R. 4?,./

Bi 10.9. Tm cc khong n iu ca hm s":

_ 3 ' >-n Tiy = tgx ~ Xvi 0 < X< 4 2

Hng dn

3 _ 3 - 4 cos2x4 cos2x ,4cos2x

i~ (%/3+2cosx)|V3-2cosx\- --------- \ ------------------- -, y

4cos X, y cng u vi >/3 - 2cosx

071 71

X? 2

v/3 - 2 cos X - 0 +

y - 0 + .

Ch:Xe 0, cosx > 0 => + 2cosx > 0.

8&S



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62/358

v n 1 1

XC NH IU KIN CA MT THAM S

HM S y = f(x) NG BIN, NGHCH BIN

TRONG KHONG (, b)

Gi s hm sy =f(x) c o hm-f(x) trong (, b)

Mun xc nh iu kin ca tham s (mchng hn) hm s y = f(x>:

^ ng bin trong khong (a, b)

ta phi xc nh iu kin ca m so cho:

f{x) > 0, Vx, X 6 (a, b) '(1)

nghch bin trong khong (a, b) p (x) < 0, Vx, X e fa, b) (2)

* ax2 + bx + c > 0,. Vx, X e R

* ax2 + bx + c < 0, Vx, X R ' c>

Ch : du ng thc ()v (2) ch xy ra mt s hu hn im ca khong (a, b).

Cn nh:

. Du ca tam thc bc hai:[A 0

[ 0 A > 0 v:

a a < Xj < x22

Ch : a.f(a) < 0 > 0

nh nh l trn ta c a trn.hnh v sau:

A > 0

Khi c: ta xt thm(a.f (a) > 0

s2 >a

s _.2


63/358


64/358

' J- u n _ nv mx2 -2 m x ~l- Mien xc dinn: D =R \{1} ; y = -------- 5----... (x-1)

- Hm s khng i chiu bin thin trong cc khong m hm s xc nh y J =f, WT ir i 1 o y Kr 1 \ 4na

Ta c: y = - y < 0, V;'*, X * 1

khng i du, Vx, X e D C> (mx - 2mx - 1) khng i du.

Trng hp 1: m = 0

-1( x ~ l

Hm s' nghch bin trong cc khong (-00,1), (I, +oo).

Trng hp 2: m 5*0

, (mx2 - 2mx -1 ) khng du A < 0

m * 04 , -1 < m < 0[m +m ;

, _ 2x2 -8 x + 5k + 3 _ C .y = / \a = f (x)

( * - 2 )

a. Xc nh k hm s un lun ng bin.

Hm s" f ng bin trong cc khorig m xc nh:

f (x) > 0, Vx, x D

2x2 - 8x + 5k + 3 >, Vx, X G D

A< 0 16 - 2(5k + 3) < 0 k> 1b. Xc nh k hm s ng bin ong khong (3, +oo)

Xt t s ca y: g(x) = 2x2 - 8x +5k + 3;

A '= - l . ( k - 1)

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65/358

A T - ' .

* Nu k > 1 th hm s cho ng bin t6ng cc khoang (-00, 2), (2, +co) nnhm s ng bin trong khong (3,+oo).

Vy ta chn k >1.* Nu k < 1 th A' > 0, lc y = 0 c 2 nghim n x 1tx 2. Ta c bng bin thin

ca hm s: '

0 - - 0

^ Hm s" ng bin trong khong (3, +co) O X ; < x2;< 3

k < 1

l.g 0

s -5< 312

k <

5 k -3 > 02 < 3 ()

- < k< 1.5

Kt lun:Kt hp 2 trng hp trn ta c k > - .5

Bi 11.5. Cho hm s y = x^m - x) - m. Tm m hin s ng bi ong khong .{, 2).

Gii

Hmsy = x2(m - x) - m hay y = -X3 + .mx2 - m

- Min xc nh: D = Rr x = 0

y = -3x2 + 2mx; y= 0 x = -

2m

Trng hp 1: m = 0 .

=> y = -3x2< 0, Vx, X s R

=> Hm s nghch bin trn R nn khng.th ng bh trong khong (I, 2).

Ta khng nhn gi tr m = 0.

Trng hp 2: m * 0Ta c bng bin thin ca hm s nh sau: '

* m > 0

66



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66/358

Hm s ng bin trong khong (1, 2) o 2 3.

* m < 0 '

X co2m

30 +00

y 1 0 + 0 -

y

Ta thy khi m < 0 th hm s khng th ng bin trong khoang (1,2).

Kt lun:Hmsng bin trong (1, 2) khi m > 3.

Bi 11.6. Cho hm s y = -X3+ 3(2m + l)*2 - (2m + 5)x - 2.

. Xc nh m hm s nghchvbn n ton min xc inh ca n.

b. Xc nh m hm s nghch bin trong khong (-00, -2).

Gii

Hm s = f(x) = -X 34- 3(2m 4- l)x^- (12m + 5)x - 2

- Min xc nh: D = R

f(x) = -3X2 -h6(2m -h l)x - (12m + 5)

a. Xc nh m hm s nghch bin trn ton min xc nh ca n.

Hmsf nghch bin trn R P(x) < 0, Vx, X R.

A< 0 9(2m + l)2 - 3


67/358

.=> hm s nghch bin trn R

=> hm s f nghch bin trong khong ( -* , -2).' 1 1

Vy ta chn: m < =- tc - ~=r 0 => f (x) = c hai rtghim n Xi, x2.

Ta c bng bn thin-ca hm s:

' X - oe Xi x 2 + 0O

y - 0 ,+ . 0 -

y

hm s nghch bin trongkhong -2) ta phi c:

-2 < X] < *2

1 - 1 ' -m < 7=um >'-7=-v/6 n/

3.f- 2 > 0

2 ^ -

f (-2)< 0

~ 2 . < s

2

o

m >1

.76f (--2) = -36m - 29 < 0

3(2m + )6

m > L29

-2 - -2

29 1 1- < m < p r U - p < m36 y[E s .

Kt hp 2 trng hp trn ta c: hm scTnghch bin trong khong (-00, -2)' * ' .

29

--


68/358

Hm sy = - + a - x2 + (a + 3 )x -4

- Min xc nh: D =R

f = - X 2 -4- 2 ( a - l ) x + ( a + 3 )

A = (a - l)2 + {a + 3) = a2 - a + 4 > , Vx, X e R

=> f (x) = 0 c hai nghim n X], X-

Ta c bng bin thin ca hm s f(x):

X 00 X1 x2 +00

y - o: + io 1

/\ \

Xi < 0 < 3 < Xjj

- U (0)< 0

- l . f ( 3 ) < 0

a + 3 > 0 '

7 a - 2 > 0 a >

12

Bi 11.8. Cho hm s'u = - (m + 3)x3 - 2X2 + mx. Xc nh m hm s:

.n iu trnmin xc nh ca n.

b. Lun lun nghch bin, _________________c. Lun lun dng bin.

Gii- Min'xc nh: D = 'R; f{x) - (TTH 3)x2- 4x + m

, \r= 4 - m{m + 3) = -m 2- 3m + 4; a = m + 3

X O 4 -3 1 +0C

A ' - 0 4 i + 0 -

a......... 7 .... : f I +

a. Hms n iu trn R o y khng i du trn R

m -r 3 V0 , :n < A m < -4 u m > 1, Vx, X


69/358

k(x + l)Bi 11.9. Cho hm s y = -T; - -*gi

X - k x + 1 '

Hy tm tt c cc gi tr ca k {k * 0) hm s khng i chiu bin thin trncc khong m !hm sc xc nh. Trong trng hp ny, hm s cho ng bin

hay nghch bin? ____________________________'______

* . "Gii

- iu kin hm s xc nh: X2- kx + 1 # 0

3 ( x + 1 ) 2 ( x z - k x + l ) - ( 2 x - k ) ( x + l ) 3

y = k x : - k x -

(x + l)2[x2-2 (k + l)x + (k+ 3)lhay y = k--------- -------- ------- 5---- ------ -

(x2- kx + l)

u ca yl du ca tam. thc bc hai:

f(x) = k[x2-2(k + 'l)*+


70/358

Ta c P(x) = 2 + acosx-bsinx = 2 + v* H a ,r cosx sinx

V i i - = = T + f - =VVa2 + b J W a2 + b2 y

2

= 1 nn tn ti a e R sao cho:

b ; = cosa v = = = sin a..Va2+ b2 +*

Vy: f(x) = 2 + Va2+ bz (cosx.cosa - snx.sna)= 2 4- Va2 + b2 .cos(x + a)

Hm s f(x) ng bin trn min xc nh R.

f (x) > 0, Vx, X 6 R 2 Va2+ b2 > 0

Va2+b 2 < 2 a2 + b2 < 4. C8*S3

TON TNG T2

Bi 11.11. Xc nh a hm sy = 5------ -4- nghch bin trn ton min xc nhX - X + 2 ' . v: ' .

ca n.

Hig dn

( a - l ) x 2+ .2 (2 -a )x -a , ,y = ---- - - if----- , min xc inh D = R.

(x2-x+2)2

- t f(x) = t s ca y\

- bi tha mn f(x) < 0, Vx, X R

< 0 1 2a2-5 a + 4

: ^ | a -1 0, V, R)

Kt lun: khng th tm c a tha mn bi ton.

Bi 11.12. Xc nh m hm s:

X vy = - {m +1JX2 + m(m + 2)x + 1

Xc nh m hm s ny:

a. ng bin trong khong (0,1). b. Nghch bin ong khong (-1,0) .

Hbg n

D thy y = 0 c hai nghim Xi = m v x2 - ra + 2 (Xi < x)

71



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71/358

X - c o m m + 2 . , +00

0 0

a. Phi chn (0, ) c (-O, m) hoc (0,1) c (m + 2, +'*>). p s: 1 < m hoc m < -2 .

b. Phi chn (-1,0) c (m, m + 2)

p s: -2 < m < -1. - -jt' , ~ ........ .

, (2k2+ l)x + 3 v ^ .Bi 11.13. inh k hm s y - - -------- ------- ong bin trong cc khong m hm

x + k -c xc nh. Xt trng hp k = 1. . _________.

Hng dn

, 2k3 + k -3 (k-l)(2k* + 2k+3)

( x + k ) 2 (x+k)!

- Ta phi chn k > 1 (v 2k2 + 2k + 3 > 0, vk, k-6 R).

- c bit: khi k = 1 ta c y = ^ = 3, X * -1.X +1

(hm s cho tr thnh hng s).

Bi 11.14. Xc nh m hm s y ;==X3 - 3mxz + 3(2m - l)x + 1 ng bin n mi

xc nh ca n. ______ ______ . - - _- -

Hng dn

Ta phi cy >0, Vx, X e R < 0 m A' (m - l) 2

p s: m = 1.Bi 11.15. Xc nh a hm s:

-X2 +xcosa+(2cos2a- 2 co sa + l )y------------------ ------------------ - (0 < cc < 2ti)

x + cosa

nghch bin trong cc khong m hm s xc nh. ______________

Hng dn

, -X 2 - 2 x c o s a - ( ' l - c o s a ) 2D te R\ {-cosa}; y= ---------7 - ~2--------

(x+cosa)

t f(x) = t s ca y\ Ta phi c f(x) < 0, V x ,x eR A-


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s DNG TNH N IU CA HM s

P CHNG MNH BT ANG t h c

Dng tnh n iu ca hm s chng minh bt ng thc:p (x ) > Q(x), Vx, X e (a, b)

Bc 1: Tm' bt ctng thc tng ng:

P(x)>Q(x) ()

o P(x) -Q(x) > 0 . (2)

Bc 2: t f(x = P(x} - Q(x

Bc 3: Tnh v xt du f{x)?t chng minh f(x) > 0 khi X (a. b).

* Cn nh:

- Hm sf(x) rg bin trong {a, b) . [Xj < x2 =5 f(xj) < f(x2), Vxi,x2 (a, b)]

- Hm s' f(x) nghch bin trong (a, b)

f f(x2), Vxi,x2 (a, b).

TON P DNG

Bi 12.1. a. Chng m i n h rng nu X > 0 th X > sinx

b. Cho 0 < a < (5 < . Chng minh rng; -^2: 0 => X > sinx

Cch 1

Ta lun c: x| > s inx , Vx, X e R (du ng thc xy rakhi X = 0)

Vy, X > 0 => X = x > sinx > sinx => X > sinx (iu phichng minh)

Cch 2

Ta c: X > sinx (1)

X - s inx > 0 (2)

t f(x) = X- sinx =>f(x) - 1 - cosx = 2sin2x - : 0, Vx, X e R

=> hm s f(x) lun ng bin.

Do X. > 0 => f(x) > f(0) => X- sinx > 0 - snO = 0

Vy, bt ng thc (2) ng nn bt ng thc (1) ng.



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b. Chng minh rng nu 0 < a < |3 < th < 2 a+ (3

Xt hm s f(x) == vi 0 < X 2

r> f{x) = 4x3- lOx + 1 = 2x(2x2'- 5) + 1,x > 2

2x >4X> 2 => < . => f{x) > 13 > 0'

[2x -5 > 3Vy hms fix)ng bin khi X> 2.

Do : X> 2 => f(x) > f(2) = 0

=> X4- Sx2+ X + x> 0, Vx, x> 2

(Bt ng thc c chng' mi nh) _____________

_ 03Bi 12.6. Chng minh: nu a > 0 th a - < sina < a,

0 => sina < a (xem bi 9.1 cu a).

Gii

_ , o r. - T c: a ------ - < sina a

6a_6

sin a < 0< sinaa -

(X- t f( a) = C L - --------------- sincc

6

76



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=> hm s f (x) nghch biii trang khong (0,-f ac)

Do a. > 0 => f () < f (0) = 0

> H) l hm s" nghch bin trong (0. -! 3?)

a 3 'Do a > 0 => f(a) < f(0) - 0 => ----- sna < 06

Ta c: f{cc) = ---------cosa; f (x) = - a + sina < 0 khi => > 0 (cu

Ch : Hm s f(a) = a -- - ....sna lin tuc trr R.6

Bi 12.7.Cho 0

x > 0 ' n n(l + x), >0

f {x) < 0, X (0, + 00) ^ p(x) nghchbin trong (0, + x)

Do : X > 0 => f{x) < f(0) = 0

f(x) l hm s nghch bin trong {Q, + 00)

Do d: X > 0 ==> f(x) < f(0) = 0

(1 + x) - 1 ~ a.x < 0 (iu phichng minh).

\

TON TNG T

:ii-

Bi 12.8. Cho a > 0 v b > 0 ,

a. Chng minh rng vi mi a Lha r n] 10 < a < 1 th ta c bt ng thc (a + b r < au + ba

b. Vi n e Z 1 v n > 2, chng minh rng: ,'/a + b < '


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PHÂN LOẠI VÀ PHƯƠNG PHÁP GIẢI TOÁN GIẢI TÍCH 12 - LÊ MẬU THỐNG

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