ThS. L E H O A N H P H O
Nhd gido Uu tu
C

c

M i l

B

O

I

H

O

C

D

A

I

D

U
S

S

8

N

I

N

H

O

-

G

7

G

G

I

I

O

A

I

I

T

O

T

- Ddnh cho HS lop 12 on tap & nang cao kinang lam bai.
- Chudn bi cho cdc ki thi quoc gia do Bo GD&DT to choc

Mil
NHA XUAT BAN DAI HQC QUOC GIA HA NQI

A

I

N

C

H

c

Bin DUSNG ,
HQC SINH GO TOAN
OAI SO -GIAI TICH

Boi duQng hoc sinh gioi
Toan Dai so 10-1.
Boi duQng hoc sinh gioi
Toan Dai so 10-2.
- Boi duQng hoc sinh gioi
Toan Hinh hoc 10.
- Boi duOng hoc sinh gioi
Toan Dai so 11.
Boi duQng hoc sinh gioi
Toan Hinh hoc 11.
Bp de thi tif luan Toan
hoc.
Phan dang va pht/Ong
phap giai Toan So phtfc.
Phan dang va phucing
phap giai Toan To hop va
Xac suat.
1234 Bai tap tir luan
dien hinh Dai so giai
tich
1234 Bai tap ta iuan
dien hinh Hinh hoc
li/ong giac

ThS. L E H O A N H
Nha gido iCu tu

B

O

I

H

O

C

D

U
S

Q
I

N

N
H

PHO

G

,
G

I

O

I

T

O

A

DAI SO-GIAI TICH

12 *
- Danh cho HS lap 12 on rflp & ndng cao kfndng lam bai.
- Chudn bj cho cdc ki thi qudc gia do Bo GD&DT td choc.

Ha Npi
NHA XUAT BAN DAI HQC QUOC GIA HA NQI

N

NHA XUAT BAN D A I HOC QUOC GIA HA N 0 I
16 Hang Chudi - Hai Ba Trirng Ha Npi
Dien thoai: Bien tap-Che ban: (04) 39714896;
Hanh chinh: (04) 39714899; Tdng bien tap: (04) 39714897
Fax: (04) 39714899

Chiu

trach

nhiem

xuat

bdn:

Giam ddc PHUNG QUOC BAO
Tong bien tap P H A M T H I T R A M
Bien tap noi

dung

THUY NGAN
Sda

bdi

NGOC H A N
Che

bdn

CONG T I A N P H A
Trinh bay bia
SON K Y
Ddi tdc lien ket xudt bdn
CONG T I A N P H A

SACH LIEN KET
BOI DUONG HQC SINH GIOI TOAN DAI SO GIAI TICH 12 -TAP 1
Ma so: 1L-177DH2010
In 2.000 cuon, khd 16 x 24 cm tai cong ti TNHH In Bao bi Hung Phu
So' xua't ban: 89-2010/CXB/11-03/DHQGHN, ngay 15/01/2010
j
Quyet dinh xua't ban sd: 177LK-TN/XB
In xong va nop ltiu chieu quy I I nam 2010.

L d i

N 6 I

D

A

U

De giup cho hoc sinh lap 12 co them tai lieu tu boi duong, ndng cao va ren luyen ki
ndng gidi todn theo chuong trinh phdn ban mdi. Trung tdm sdch gido due ANPHA xin
trdn trong giai thieu quy ban dong nghiep vd cdc em hoc sinh cuon: "Boi dudng hqc sinh
gioi todn Dai so' Gidi tich 12 " nay.
Cuon sdch nay nam trong bo sdch 6 cuon gom:
- Boi ducmg hoc sinh gidi todn Hinh hoc 10.
- Bdi ducmg hoc sinh gidi todn Dai so' 10.
- Boi dudng hoc sinh gidi todn Hinh hoc 11.
- Boi dudng hoc sinh gidi todn Dai so - Gidi tich 11.
- Bdi dudng hoc sinh gidi todn Hinh hoc 12.
- Boi dudng hoc sinh gidi todn Gidi tich 12.
do nhd gido uu tu, Thac sTLe Hoanh Phd to'chirc bien soan. Ndi dung sdch duoc bien
soan theo chuong trinh phdn ban: co bdn vd nang cao mdi ciia bd GD & DT, trong dd mot
so van de duoc md rdng vdi cdc dang bdi tap hay vd kho dephuc vu cho cdc em yeu thich
mud'n ndng cao todn hoc, cd dieu kien phdt trien tot nhat kha nang ciia minh. Cuon sdch la
su ke thira nhung hieu bii't chuyen mdn vd kinh nghiem gidng day ciia chinh tdc gid trong
qua trinh true tiep dirng ldp bdi dudng cho hoc sinh gidi cdc ldp chuyen todn.
Vdi ndi dung sue tich, tdc gid da co'gang sap xep, chon loc cdc bai todn tieu bieu cho
tirng the loai khdc nhau ung vdi ndi dung cua SGK. Mdt sd'bdi tap cd the khd nhung cdch
gidi duqc dua tren nen tdng kien thuc vd ki nang co bdn. Hqc sinh can tu minh hoan thien
cdc ki nang ciing nhu phdt trien tu duy qua viec gidi bdi tap cd trong sdch trudc khi ddi
chieu vdi led gidi cd trong sdch nay, cd the mdt soldi gidi cd trong sdch con cd dong, hqc sinh
cd thetu minh lam rd hon, chi tie't hon, ciing nhu tie minh dua ra nhung cdch lap ludn mdi
hon.
Chung tdi hy vong bd sdch nay se la mdt tdi lieu thie't thuc, bo ich cho ngudi day vd
hqc, dqc biet cdc em hqc sinh yeu thich mdn todn vd hqc sinh chuan bi cho cdc ky thi qudc
gia (tot nghiep THPT, tuyen sinh DH - CD) do bq GD & DT to chirc sap tdi.
Trong qua trinh bien soqn, cudn sdch nay khdng the tranh khoi nhirng thieu sdt, chung
tdi ra't mong nhdn duqc gop y ciia ban dqc gdn xa debq sdch hoan Men hon trong lah tdi
ban.
Moi y kien dong gop xin lien he:
- Trung tam sach giao due Anpha
225C Nguyen Tri Phuong, P.9, Q.5, Tp. HCM.
- Cong ti sach - thiet bj giao due Anpha
50 Nguyen Van Sang, Q. Tan Phii, Tp. HCM.
DT: 08. 62676463, 38547464 .
Email: alphabookcenter@yahoo.com
Xin chan thanh cam on!

M U C

L U C

Chuong I : tTng dung dao ham de khao sat va ve do t h i cua ham so
§ 1. Tinh don dieu cua ham so

5

Dang 1: Dong bien, nghich bien, ham hang

5

Dang 2: Ung dung tinh don dieu

17

§2. Cue tri ciia ham so

37

Dang 1: Cue dai, cue tieu

,

37

Dang 2: Ung dung ciia cue tri

48

§3. Gia tri Ion nhat va gia tri nho nhat

58

Dang 1: Tim gia tri ldn nhat, nho nhat

58

Dang' 2: Bai toan lap ham so

69

Dang 3: Ung dung vao phuong trinh

77

§4. Duong tiem can cua do thi ham so

88

Dang 1: Tim cac tiem can

88

Dang 2: Bai toan ve tiem can

96

§5. Khao sat va ve ham da thuc

103

Dang 1: Ham bac ha

104

Dang 2: Ham trung phuong

113

§6. Khao sat va ve ham hOu ti
Dang 1: Ham so v =

k (c * 0 va ad be * 0)
cx + d
2 i
Dang 2: Ham s6 v =
(a * 0. a' * 0)
a'x + b'
§7. Bai toan thuong gap ve do thi
a

&X

x

+

+

126
126
135
148

Dang 1: Tuong giao, khoang each, goc

149

Dang 2: Tiep tuyen. tiep xuc

159

Dang 3: Yeu to co dinh. doi xung - quy tich

170

Chirong I I : Ham so luy thua ham so mu va ham so logarit
§ 1. Quy tac bien doi va cac ham so

186

Dang 1: Bien doi luy thua - mu - logarit

188

Dang 2: Cac ham so mu. luy thua, logarit

200

Dang 3: Bat dang thuc va GTLN, GTNN

212

C H U O N G

I:U N G DUNG D A O H A M OE

SAT V A VE D O THj C U A H A M

K H A O
SO

§1. TINH DON DIEU CUA HAM SO
A. K I E N THLTC CO B A N
• Dinh nghTa: Ham so f xac dinh tren K la mot khoang, doan hoac nira
khoang.
- f dong bien tren K neu vdi moi Xi, X2 6 K: X] < X2 => f(xi) < f(x2)
- fnghich bien tren K neu vdi moi xi. xi e K: Xif(xi)>f(x2).
• Dieu kien can de ham so don dieu
Gia sir ham so co dao ham tren khoang (a; b) khi do:
- Neu ham so f dong bien tren (a; b) thi f ' ( x ) > 0 vdi moi x e (a; b)
- Neu ham so f nghich bien tren (a; b) thi f ' ( x ) < 0 vdi moi x e (a; b).
• Dieu kien du de ham so don dieu
- Gia sir ham so f co dao ham tren khoang (a; b)
Neu f'(x) > 0 voi moi x e (a; b) thi ham so f dong bien tren (a; b)
Neu f'(x) < 0 voi moi x e (a; b) thi ham so nghich bien tren (a; b)
Neu f'(x) = 0 vdi moi x e (a; b) thi ham so f khong doi tren (a; b).
- Gia sir ham so f co dao ham tren khoang (a; b)
Neu f '(x) > 0 (hoac f '(x) < 0) vdi moi x e (a; b) va f '(x) = 0 chi tai mot
so huu han diem cua (a; b) thi ham so dong bien (hoac nghich bien) tren
khoang (a; b).
B. P H A N D A N G T O A N
DANG 1: B 6 N G B l i N , NGHICH BIEN, HAM HANG
• Phuong phap xet tinh don dieu:
- Tim tap xac dinh
- Tinh dao ham, xet dau dao ham, lap bang bien thien
- Ket luan
Chii y: - Dau nhi thuc bac nhat: f(x) = ax + b, a ^ 0
-00
-b/a
+co
x
f(x)

trai dau a

0

ciing dau a

- Dau tam thuc bac hai: f(x) = ax + bx + c, a * 0
Neu A < 0 thi f(x) luon ciing dau vdi a
Neu.A = 0 thi f(x) luon cung dau vdi a, trir nghiem kep
Neu A > 0 thi dau "trong trai - ngoai ciing"
X
-CO
X]
X
2

2

f(x)

ciing dau a 0 trai dau a

-BDHSG DSGT12/1-

0 ciing d i u a

+00

V i du 1: Xet chieu bien thien ciia ham sd:
a) y = x - 6x + 5

2x + x - 3
b)y= - x
3
c) y = x - 2x + x + 1
d) y = - x + 4x - 7x + 5
Giai
a) Tap xac dinh D = R. Ta co y' = 2x - 6.
2

3

2

3

2

3

2

Cho y' = 0 » 2x - 6 = 0 » x = 3.
Bang bien thien
X —oo
3
y'

-

+00

0

+

—

y

Vay ham so nghich bien tren (-oo; 3), dong bien tren (3; +oo).
b) D = R. Ta cd y' = 4x - 4x + 1 = (2x - l ) > 0 v d i moi x
2

2

y' = 0 o x = —. Vay ham so dong bien tren R.
2
c) D = R . Ta co y' = 3x - 4x + 1
2

Cho y' = 0 o 3x - 4x + 1 = 0 <=> x = - hoac x = 1.
3
BBT
X —00
+00
1/3
1
+
+
0
0
y'
2

J

^ *

y

—

^

Vay ham so dong bien tren moi khoang (-co; —) va (1; +oo), nghich bien
3
tren khoang (—; 1).
3
d) D = R Ta cd y' = - 3 x + 8x - 7
V i A ' = 1 6 - 2 1 < 0 nen y' < 0 vdi moi x do do ham so nghich bien tren R.
V i du 2: Xet chieu bidn thien cua cac ham so sau:
a) y = x - 2x - 5
b) y = x + 8x + 9
Giai
a) D = R. Ta co y' = 4x - 4x = 4x(x - 1)
Cho y' = 0 <=> 4x(x - 1) = 0 <=> x = 0 hoac x = ±1
BBT
+00
X —00
1
0
-1
2

4

2

4

3

2

2

2

y'

-

y

^

o •+
^ *

0

^

0

+
^

Vay ham sd nghich bien tren moi khoang (-co; - 1 ) va (0; 1), ddng bidn
tren moi khoang ( - 1 ; 0) va (1; +=»)•
6

-BDHSG DSGT12/1-

b) D = R. Ta co y' = 4x + 16x = 4x(x + 4),y' = 0 o x = 0.
y' > 0 tren khoang (0; +co) => y dong bien tren khoang (0; +co)
y' < 0 tren khoang (-co; 0) => y nghich bien tren khoang (-co; 0).
V i du 3: Xet su bien thien cua ham so:
1
3x-8
d)y
c)y
a) y = x + —
b)y
(x-4)
1-x
x
3

2

Giai
a) Tap xac dinh D = R \ {0}
_3_ x - 3
, y' = 0 <» X
Ta co y' = 1
,.2

2

2

BBT:

X

-co

+00

Q N / 3

+

y'

:V3"

0

-

-

0

+

y
Vay ham so dong bien tren khoang (-co; - ^ 3 ) va ( J 3 ; +oo), nghich bien
tren m6i khoang ( - S ; 0) va (0; V3 ).
> 0 vdi moi x ^ 0 nen ham so dong bien

b) D = R \ {0}. T a c d y ' = 1

tren moi khoang (-oo; 0) va (0; +co).
-5
c) D = R \ { 1 } . Ta cd y' =
- < 0 vdi moi x -t- 1 nen ham so nghich
(1-x)
bien trong cac khoang (-co; 1) va (1; +oo).
3

2

d) D = R\ {4}.Tacdy'= ———-

(x-4)
y' < 0 tren khoang (4; +co) nen y nghich bien tren khoang (4; +co).
y' > 0 tren khoang (-co; 4) nen y ddng bien tren khoang (-co; 4).
V i du 4: Tim cac khoang don dieu ciia ham so:
,
x-2
2x
a) y = -5
b) y
x
-9
X +X +1
Giai
- x + 4x + 3
a) D = R. Ta cd: y'
(x + x + l )
3

2

2

2

2

y' = 0 e> x - 4x - 3 = 0 <=> x = 2 ± ^7
BBT:
X -co
2-V7
2+V7
2

-

y'
y

-BDHSG DSGT12/1-

—

0
^

—

+
*

0
-

+°o
*

7

Vay ham so dong bien tren khoang (2 - yfl ; 2 + V7 ) va nghich bien
tren cac khoang (-co; 2 - 77 ) va (2 +

; +oo).

b) D = R\{-3;3}.Tac6y'=^^ < ,Vx*±3.
(x - 9 )
Do do y' < 0 tren cac khoang (-co; - 3 ) , (-3; 3), (3; +oo) nen ham so da
cho nghich bien tren cac khoang do.
V i du 5: Xet su bien thien cua ham sd:
0

2

a) y = V 4 - x

2

b)y = Vx - 2 x + 3

2

2

c)y

d)y

Vl6-:

x +2

Giai
a) Dieu kien 4 - x > 0 < = > - 2 < x < 2 nen D = [-2; 2]
2

V d i - 2 < x < 2 thi y'
BBT:

X

-2

n
0

+

y'
^

y

, y ' = 0<=>x = 0.

V4~

2
^

—

^

Vay ham so dong bien tren khoang (-2; 0) va nghich bien tren khoang
(0; 2). Do ham so f lien tuc tren doan [-2; 2] nen ham so dong bien tren
doan [-2; 0] va nghich bien tren doan [0; 2].
b) V i A' = 1 - 3 < 0 nen x - 2x + 3 > 0, V x => D = R.
•p , 1
2x-2
x - l
, .
,
Ta co y = —
==
= . y = 0 o x = 1.
2vx -2x + 3 Vx -2x +3
y ' > 0 o x > l , y ' < 0 o x < l nen ham so nghich bien tren khoang (-co; 1)
va dong bien tren khoang (1; +00).
c) D K : 16 - x > 0 o x < 16 o - 4 < x < 4. D = (-4; 4).
2

2

2

2

Ta co v' =

2

1

> 0, Vx e (-4; 4).

6

(16-x )Vl6-x
2

2

Vay ham so dong bien tren khoang (-4; 4).
d) D = [0; +00). V d i x > 0, y' =

_, y'
2^y^(x + 2)
2
+00
X

2

BBT:

X
y
y

0
+

0
^

Vay ham sd ddng bien tren (0; 2) va nghich bien tren (2; +00).
8

-BDHSG

DSGTu/1-

V j du 6: Tim khoang don dieu cua ham so
a) y = V x ( x - 3)

b)y = - x

c)y

d)y =

7x -6
2

x +1
Vl-x

Giai
a) D = [0; +oo). V d i x > 0, ta cd:
y

1

3NRX-1)

( x - 3 ) + vx =
r

2Vx
BBT:
X

,y

2x

0

1

+GO

0

y'
y

0<=>x= 1.

+

— — — ^

Vay ham so nghich bien tren khoang (0; 1) va dong bien tren khoang
(i';+°o).
b) D = R. V d i x ^ 0, ta co: y' = — 3 3vV
y' = 0 <=> x = 1 <=> x = ± 1 .

3vV

2

y' > 0 o

^/x " > l < = > x > l c i > x < - l hoac x > 1.
2

2

y' < 0
%/x " < l e > x < l o - K x < l .
Vay ham so dong bien tren cac khoang (-co; -1) va (1; +co), nghich bien
tren khoang ( - 1 ; 1).
c) Tap xac dinh D = (-co; -^6 ) U (x/6 ; +oo).
2

2

Tacd: y' = -^^ ^£L,y = 0»x = ±3.
(x -6)v'x -6
1

7

2

BBT:

X
y'

2

—CO

-3
+

3

V6 V6

0

-

•

y

_

0

+CO
+

•

Vay ham so dong bien tren cac khoang (-co; - 3 ) va (3; +oo). nghich bien
tren cac khoang (-3; - v o ) va ( v o ; 3).
d) D = (-QO; 1). T a c d y ' =

, ~
2V(l-x)
3

> 0, V x < l .

X

3

Vay ham sd dong bien tren khoang (-co: 1).

-BDHSG DSGT12/1-

V j du 7: Xet su bien thien cua ham sd:
3
a ) y - - - x + smx
)
b

y

=

x

+

c o g

2

x

Giai

3
a) D = R. Ta cd y' = - - + cosx < 0, Vx nen ham sd nghich bien tren R.

b) D = R. Ta cd y' = 1 - 2cosxsinx = 1 - sin2x
y' = 0o sin2x = 1 <=> x = - +kit,keZ.
4
Ham sd lien tuc tren moi doan [- + kn; — + (k +
4
4
va y' > 0 tren moi khoang (- + kn; - + (k + 1)TI) nen ddng bien tren
4
4
moi doan [- + kn; - + (k + l)7tl, keZ.
4
4
'
Vay ham so dong bien tren R.
Vi du 8: Tim khoang dong bien, nghich bien cua ham so:
a) y = x - sinx tren [0; 2TT] b) y = x + 2cosx tren (0; n).
Giai
a) y' = 1 - cosx. Ta cd Vx [0; 2n] => y > 0 va
y' = 0 <=> x = 0 hoac x = 2n. Vi ham so lien tuc tren doan [0; 2n] nen ham
so ddng bien tren doan [0; 2n].
b) y' = 1 - 2 sinx. Tren khoang (0; 7t).
v

1

1

y'>0o sinx <-<=> - < x < —
2
6

6

y' < 0 » sinx > - <=>06

6

6

Vay ham so ddng bien tren khoang (-; —). nghich bien tren moi
6
6
khoang (0; — ) va (—; n).
6
6
Vi du 9: Chung minh cac ham sd sau nghich bien tren R:
a) f(x) = vx +1 - x b) f(x) = cos2x - 2x + 5.
Giai
2

a) Tacdf'(x) = ^=-l.
T

Vx + 1

V i V x + 1 > V x = I x | > x, Vx nen f '(x) < 0, Vx do dd ham sd f nghich
bien tren R.
b) f'(x) = -2(sin2x+ 1)<0 vdi moi x.
2

2

10 -BDHSG DSGT12/1-

f'(x) = 0 o s i n 2 x = -lc^>2x = - - + 2 k n o x = - - +kn,k&
2
4
Ham f(x) lien tuc tren moi doan [ - - + kn; ~

Z.

+ (k + \)n] va f'(x) < 0 tren

moi khoang (-— +kn; — + (k+l)n) nen ham so nghich bien tren moi doan
4
4
[ - - + k ; r ; - - +(k + l)n], k e Z.
4
4
Vay ham sd nghich bien tren R.
Cach khac: Ta chung minh ham sd f nghich bien tren R:
VXj, x e R, x < x => f(Xj) > f(x ).
That vay, lay hai sd a, b sao cho a < X| < X2 < b.
Ta cd: f ' ( x ) = -2(sin2x + 1) < 0 vdi moi x e (a; b).
V i f '(x) = 0 chi tai mot sd huu han diem cua khoang (a; b) nen ham sd f
nghich bien tren khoang (a; b) => dpcm.
V i du 10: Chung minh rang cac ham so sau day dong bien tren R.
2

x

2

2

a) f(x) = x - 6x + 17x + 4

b) f(x) = 2x - cosx + S sinx.
Giai
a) f ' ( x ) = 3x - 12x + 17. V i A' = 36 - 5 1 < 0 nen f ' ( x ) > 0 vdi moi x, do dd
ham so dong bien tren R.
3

2

2

V3
b) y' = 2 + sinx - v3 cosx = 2(1 + — sinx
cosx).
2
2
= 2[1 + sin(x - —)] > 0, vdi moi x.
3
Vay ham sd ddng bien tren R.
V i d u l l : Chung minh ham so:
x-2
a) y =
ddng bien tren moi khoang xac dinh cua nd.
x +2
- x — 2x + 3
b) y =
nghich bien tren moi khoang xac dinh cua nd.
x +1
Giai
4_
a) D = R \ { - 2 } . Ta cd y' =
— > 0 vdi moi x * - 2
(x + 2)
2

2

Vay ham so dong bien tren moi khoang (-oo; - 2 ) va (-2; +oo).
x -2x-5
b) D = R \ { - l } . T a c d y ' =
~ < 0 vdi moi x * - l (vi A' = 1 - 5 < 0).
(x + 1)
'
'
Vay ham so nghich bien tren mdi khoang (-oo; - 1 ) va ( - 1 ; +oo).
V i du 12: Chung minh ham so:
2

2

-BDHSG DSGT12/1-

v

1 1

a) y -

i

+

dong bien trong khoang ( - 1 ; 1) va nghich bidn trong cac

x 2

khoang (-co; -1) va (1; +oo).
, .
sin(x + a) ,
,
_
,
,
) y~
( ^ b + krt; k e Z) don dieu Uong mdi khoang xac dinh.
sin(x + b)
•
°
Giai
,
, l(l + x )-2x.x
1-x
b

a

2

2

=
(i x )
( T ^ '
Ta cd y' > 0 <=> 1 - x > 0 <=> - 1 < x < 1.
y ' < 0 < = > l - x < 0 < = > x < - l hoac x > 1.
Tir do suy ra dpcm.
b) Ham sd gian doan tai cac diem x = -b + kn
(k e Z).
, _ sin(x + b) cos(x + a) - sin(x + a) cos(x + b) _ s i n ( b - a )
a )

y

2

2

=

y

=

0

o

x

=

± 1

+

2

2

sin (x + b)

s m ( x + b)

2

2

(do a - b * kn)
V i y' ?t 0 va y' lien tuc tai moi diem x * - b + kn, nen y' giu nguyen mot
dau trong moi khoang xac dinh, do do ham so don dieu trong moi khoang
do.
V i du 13: Chung minh:
a) sin x + cos x = 1, Vx. b) cosx + sinx. tan— = 1, Vx e (-— ; —).
2
Giai
a) Xet f(x) = sin x + cos x, D - R.
f '(x) = 2sinxcosx - 2cosxsinx = 0, Vx.
Do dd f(x) la ham hang tren R nen f(x) = f(0) = 1.
2

2

2

4

4

2

b) Xet f(x) = cosx + sinx tan-, D = (-—; — ).
2
4 4
r-,/ x
x
sinx
.
x
x
1 (x) = -smx + cosxtan— +
= -sinx + cosx.tan— + tan—
•
2cos 2
X
X
iX
= -sinx + tan — (1 + cosx) = -sinx + tan— .cos —
2
2
2
—sinx + sinx = 0 voi moi x e (— ; —)
4 4
,
,
TT 71
Suy ra rang f la mdt ham hang tren khoang (-— ; — ).
2

2

2

2

Do dd f(x) = f(0) = 1 vdi moi x e (-—; - )•

4 4
V i du 14: Chung minh cac ham so sau la ham khong ddi

-BDHSG DSGTU/1-

a) f(x) = cos x + cos (x + —) - cosxcos(x + ^ )
3
3
b) f(x) - 2- sin x - sm (a + x) - 2cosa.cosx.cos(a + x).
Giai
2

2

2

2

a) f'(x) = -2cosxsinx - 2cos(x + - )sin(x + - ) + sinxcos(x + ^ ) + cosx.sin(x + ^ )
3
3
o
o
o
71
7T
= -sin2x - sin(2x + — ) + sin(2x + - ) = -sm2x - 2cos(2x + - ) . s i n 3
3
2
b
= -sin2x - cos(2x + — ) = 0, vai moi x.
2
1 1 3
Do do f hang tren R nen f(x) = f(0) = 1 +
= 6
w
w
4
2
4
b) Dao ham theo bien x (a la hang so).
f '(x) = -2sinxcosx - 2cos(a + x)sin(a + x)
+ 2cosa[sinxcos(a + x) + cosx.sin(a + x)].
= -2sin2x - sin(2x + 2a) + 2cosa.sin(2x + a) = 0.
Do do f hang tren R nen f(x) = f(0) = 2 - sin a - 2cos a = sin a.
V i du 15: Cho 2 da thuc P(x) va Q(x) thoa man: P'(x) = Q'(x) vdi moi x va
P(0) = Q(0).
Chung minh: P(x) = Q(x).
Giai
Xet ham so f(x) = P(x) - Q(x), D = R.
Ta cd f '(x) = P'(x) - Q'(x) = 0 theo gia thiet, do do f(x) la ham hang nen
f(x) = f(0) = P(0) - Q(0) = 0 vdi moi x.
f(x) = 0 => P(x) ^ Q(x).
V i du 16: Xac dinh ham so f(x) thoa man: f(0) = 8; f ( x ) . f '(x) = 1 - 2x (*).
Giai
2

2

2

Ta cd (*) -(f (x))* = l-2xo (f (x))' = 3 - 6x.
3
Xet ham sd g(x) = f (x) - 3x + 3x thi g'(x) = ( f (x))' - 3 + 6x = 0.
nen g(x) = C: hang so tren D, do do:
f(x) - 3x + 3x = C ^> f (x) = - 3 x + 4x + C.
3

3

2

2

3

nen f(x) = N / - 3 X + 3x + C

2

V i f(0) = 8 => C = 64.

2

Vay f(x) = yj-3x + 3x + 64 , thu lai dung.
V i du 17: T i m cac gia trj cua tham so a de ham so dong bien tren R.
2

a) f(x) = - x + ax + 4x + 3
•
3
3

2

b) f(x) = ax - 3x + 3x + 2
3

2

Giai

a) f '(x) = x + 2ax + 4, A' = a - 4
- N6u a - 4 < 0 hay - 2 < a < 2 thi f '(x) > 0 vdi moi x e R nen ham so
ddng bien tren R.
2

2

2

-BDHSG DSGT12/1-

13

- Neu a = 2 thi f '(x) = ( + 2) > 0 vai moi x * - 2 nen ham sd dong bien
tren R.
2

x

- Neu a = - 2 thi ham sd f '(x) = (x - 2) > 0 vdi moi x * 2 nen ham so
dong bien tren R.
2

- Neu a < - 2 hoac a > 2 thi f '(x) = 0 cd hai nghiem phan biet nen f co
doi dau: loai.
1

Vay ham sd ddng bien tren R khi va chi khi - 2 < a < 2.
b) V (x) = 3ax - 6x + 3.
2

Xet a = 0 thi f '(x) = - 6 x + 3 cd doi dau: loai
Xet a * 0, v i f khong phai la ham hang (y' = 0 tdi da 2 diem) tren dieu
kien ham so dong bien tren R la f '(x) > 0, Vx
(a>0
fa>0
f >0
<=>
<^=>
<=>
« a > 1.
[A'<0
[9-9a<0
|a>l
Y l du 18: Tim cac gia tri cua m de ham sd nghich bidn tren R:
a) f(x) mx - x
b) f(x) = sinx - mx + C.
Giai
a) y' = m - 3x
a

2

- Neu m < 0 thi y' < 0 vdi moi x e R nen f nghich bien tren R
- Neu m = 0 thi y' = - 3 x < 0 vdi moi x e R, dang thuc chi xay ra vdi
x = 0, nen ham so nghich bien tren R.
2

m
± ^

- Neu m > 0 thi y' = 0 o x =
BBT

X

—00

Xl

x

0

y'

H

y

—

+00

2

0
r
»

•

Do do ham so dong bien tren khoang (xi, x ) : loai
Vay ham so nghich bien tren R khi va chi khi m < 0.
2

b) V i f(x) khong la ham hang vdi moi m va C nen f(x) = sinx - m + C nghich
bien tren R <=> f '(x) = cosx - m < 0, Vx
a> m > cosx, Vx o m > 1.
V i du 19: Tim m dd ham sd dong bien tren moi khoang xac dinh:
(3m - l ) x - m + m
by = x + 2 + m
a) y
x-l
x+m
Giai
2

a) D = R \ { - m } . Ta co:
, _ (x + m)(3m - 1 ) - [(3m - l ) x - m + m
2

Y

14

=

'

(x + m )

2

4m -2m
(x + m )
2

2

-BDHSG

DSGTU/l-

Ham so dong bien tren moi khoang xac dinh <=> 4m - 2m > 0
2

<=> m < 0 hoac m > —.
2
m
b) Ta cd y' = 1
vdi moi x * 1.
(x-l)
- Neu m < 0 thi y' > 0 vdi moi x * 1. Do do ham sd dong bien tren moi
khoang (-oo; 1) va ( 1 ; +oo).
XT'
, x -2x + l - m
- Neu m > 0 thi y =
-„
(x-l)
y' = 0 o x - 2 x + l - m = 0 < = > x = l + Vm
BBT
X —oo
1 —Vm
1
1 + Vm
+oo
0
+
+
0
y'
2

2

2

2

y
Ham sd nghich bien tren moi khoang (1 - Vm ; 1) va ( 1 ; 1 + Vm ): loai.
Vay ham sd ddng bien tren moi khoang xac. dinh cua nd khi va chi khi
m<0.
V i du 20: Tim a de ham sd:
a) f(x) = x - ax + x + 7 nghich bien tren khoang ( 1 ; 2)
3

2

b) f(x) = — x - — (1 + 2cosa)x + 2xcosa + 1, a e (0; 2rr) dong bien tren
3
2
khoang (1; +oo).
Giai
a) f'(x) = 3 x - 2 a x + 1
Ham so nghich bien tren khoang ( 1 ; 2) khi va chi khi y < 0 vdi moi
x e (1;2)
ff(l) < 0
f 4-2a<0
13
<=> <
<=><
<=>a>—
[f(2)<0
[l3-4a<0
4
b) y' = x - (1 + 2cosa)x + 2cosa. Ta cd 0 < a < 2TT.
y' = 0 o x = 1 hoac x = 2cosa.
V i y' > 0 d ngoai khoang nghiem nen ham so dong bien vdi moi x > 1 khi
3

2

2

2

va chi khi 2cosa < 1 cosa < — o — < a < —
2
3
3
V i du 21: Tim m de ham sd y = (m - 3)x - (2m + l)cosx nghich bien tren R.
Giai:
y' = m - 3 + (2m - l)smx
Ham so y khdng la ham hang nen y nghich bien tren R:
y' ^ 0, Vx « m - 3 + (2m - l)smx < 0, Vx
D5t t = sinx, - 1 < t < 1 thi m - 3 + (2m - l)smx = m - 3 + (2m - l ) t = f(t)
-BDHSG DSGT12/1-

15

Dieu kien tuang duong: f(t) < 0, Vt e [-;1 1]
[f(-l)<0
f-m-4<0
'9
°lf(l)^0

°l3m-2^0

^ -

^

4

^

m

V» du 22: Tim m de ham sd y = x + 3x + mx + m chi nghich bidn tren mpt
doan cd dp dai bang 3.
Giai:
D = R, y' = 3x + 6x + m, A' = 9 - 3m
Xet A' < 0 thi y' > 0, Vx: Ham luon dong bien (loai)
3

2

2

Xet A' > 0 <=> m < 0 thi y' = 0 cd 2 nghiem x,, x nen x, + x = -2, X]X = —
2

BBT:

—CO

x

*1
0 -

+

y'
y

x
0

2

2

+00

2

3

+

^

^

Theo de bai: x - X] = 3 o ( x - x , ) = 9 o
2

2

2

x +x -2x x = 9
2

2

t

2

4
15
<=> (x + x ) - 4 x x = 9 c i > 4 — m = 9 o m =
(thoa)
3
4
V i du 23: Tuy theo tham &6 m, xet su bien thien cua ham sd:
\
1 3
? , rx
ix
2x + m
a) y = - x - 2mx + 9x - m
b) y =
3
x - l
Giai
a) D = R. Ta cd y' = x - 4mx + 9; A' = 4 m - 9
- Neu A' < 0 I m [ < — thi y' > 0, Vx nen ham so dong
2

2

t

t

3

2

2

2

2

2

bien tren R.
- Neu A' > 0 co 4 m > 9 co I m | > - thi y' = 0 cd 2 nghiem phan biet
2

xi, = 2m +V4m -9 Lap bang bien thien thi ham ddng bien tren
2

2

khoang (2m -

V ^ m - 9 ; 2m + V 4 m - 9 ) va nghich bien tren m6i
2

2

khoang (-00; 2m - \/4m - 9 ) va (2m + V4m - 9 ; +00).
2

2

b)D = R\ {l}.Tacd y'= ~ ~"!
2

(x-l)
- Neu m = - 2 thi y = 1, Vx * 1 la ham sd khong doi.
- Neu m > - 2 thi y' < 0, Vx *1 nen ham so nghich bien tren moi khoang
(-00; 1) va (1; +00).
2

- Neu m < - 2 thi y' > 0, Vx * 1 nen ham sd ddng bien tren moi khoang
(-co; 1) va (1; +00).

16

-BDHSG DSGT12/1-

DANG 2: UNG DUNG T I N H BON DI$U
- Giai phirong t r i n h , he phirong trinh, bat phuong t r i n h :
Xet f(x) la ham so v6 trai, neu can thi bien doi, chpn xet ham, dat an phu,
.... Tinh dao ham rdi xet tinh don dieu.
Neu ham sd f don dieu tren K thi phuong trinh f(x) = 0 cd toi da 1
nghiem. Neu f(a) = 0, a thupc K thi x = a la nghiem duy nhat cua phuong
trinh f(x) = 0.
Neu f cd dao ham cap 2 khdng ddi dau thi f la ham don dieu nen phuong
trinh f(x) = 0 cd tdi da 2 nghiem. Neu f(a) = 0 va f(b) = 0 vdi a * b thi
phuong trinh f(x)=0 chi cd 2 nghiem la x = a va x = b .
- Chiing minh bat dang thiic:
Neu y = f(x) cd y' > 0 thi f(x) dong bien: x > a => f(x) > f(a); x < b
=>f(x)Doi vdi y' < 0 thi ta cd bat dang thuc nguoc lai.
Viec xet dau y ' doi khi phai can den y " , y " \ . . . hoac xet dau bp phan,
chang han tir so ciia mpt phan so cd mau duong
Neu y " > 0 thi y' dong bien tir do ta cd danh gia f '(x) rdi f(x),...
V i du 1: Giai phuong trinh: vo - x + x - 72 + x - x = 1
Giai
Dat t = x - x thi phuong trinh trd thanh: 7 3 + t - 7 2 - t = 1 , - 3 < t < 2.
2

2

2

Xet ham sd f(t) = 73 + t - 7 2 - t , -3 < t < 2.
Vdi -3 < t < 2 thi f'(t) = + . > 0 nen f dong bien tren (-3; 2).
273 + t 272 - t
Ta cd f ( l ) = 2 - 1 = 1 nen phuong trinh:
1

f(t) = f( 1) <=> t 1 o x — x — 1 = 0 <=> x l^H.
=

2

=

V i du 2: Giai phuong trinh 72x + 3 x +6x + 16 = 273 + 7 4 - x
Giai:
Dieu kien xac djnh:
3

2

f 2 x + 3 x + 6 x + 16>0

f(x + 2 ) ( 2 x - x + 8 ) > 0
<=>C
cs> - 2 < x < 4
4-x>0
4-x>0
Phuong trinh tuong duong 72x + 3x + 5x +16 - 74 - x = 273
3

2

2

3

Xet ham s6 f(x) =

N/2X

2

+ 3 x + 6x + 16 - 74 - x .-2 < x < 4
2

3

™,
.
3(x + x + l )
.
x
Thi f (x) = —.
= +—
> 0 nen i dong bien ma
7 2 x + 3 x + 6 x + 16
274-x
f ( l ) = 273 , do do phuong trinh trd thanh f(x) = f ( l ) o x = 1
Vdy phuong trinh co nghiem duy nhat x = l
2

c

u

3

-BDHSG DSGT12/1-

2

17

V i du 3: Giai phucmg trinh yjx - 1 = Vx - 2 - x.
Giai
Dieu kien: x >%/2
2

Ta cd: V x - 2 = x + V x - 1
3

2

3

>x>l=>x >3=i>x>v 3
3

/

Chia 2 ve cho vo? thi phuong trinh:
~ a.
1

x .vx

x vx

2

Vx

4

V

xVx

Xet f(x) la ham sd ve trai, x > tfi thi
,,
9 5x
X
f'(x)=
—
2x .Vx 2xVx
x

r

,

3

<0.
2
2x
Vx
Do do ham so f nghich bien tren khoang ( y 3 ; +oo) ma f(3) = 0 nen
phuong trinh cd nghiem duy nhat x = 3.
B
5

r

n

2

1
2x-5

Vi du 4: Giai phuong trinh: 3x - 18x + 24
2

1
x - l

Giai
5
Dieu kien x * 1; —, phuong trinh trd thanh:
2
( 2 x - 5 ) - - J _ = (x-l)
|2x-5|
|x-l|
2

2

Xet f(t) = t - i vdi t > 0.
2

Ta co: f '(t) = 2t > 0 nen f dong bien tren (0; +oo)
Phuongtrinh:f(|2x-5|) = f(|x - l|)o 12x- 51 = |x-l|
<=> 4x - 20x + 25 = x - 2x + 1 <=> 3x - 18x + 24 = 0.
c ^ x - 6 x + 8 = 0 c o x = 2 hoac x = 4 (chon)
V i du 5: Giai bat phuong trinh:
4 | 2x - 11 (x - x + 1) > x - 6x + 15x - 14
Giai
BPT: | 2x - 11 .[(2x - l ) + 3] > (x - 2) + 3x - 6
(x - 2) + 3(x - 2)
Xet ham sd f(t) = t + 3t, D = R.
Ta cd f '(t) = 3t + 2 > 0 nen f dong bien tren R.
BPT: f( | 2x - 11) > f(x - 2) o I 2x - 11 > x - 2.
Xet x - 2 < 0 thi BPT nghiem dung.
Xet x - 2 > 0 thi 2x - 1 > 0 nen BPT o 2 x - 1 > X - 2 < = >
Vay tap nghiem la S = R.
2

2

2

2

2

3

2

3

2

3

3

3

2

18

X > - 1

:

£) ng
U

-BDHSG DSGT12/1-

V i du 6: Giai bat phuong trinh: Vx + 1 + 2Vx + 6 < 20 - 3VX + 13.
Giai
Dieu kien: x > - 1 . BPT viet lai: Vx + 1 + 2%/x + 6 + 3Vx + 13 > 20
Xet f(x) la ham so ve trai, x > - 1 . Ta co:
1
1
3
A
f ' ( ) = ——
+—
+—
> 0 nen f dong bien tren [ - 1 ; +oo).
2Vx + l Vx + 6 2Vx + 13
Ta cd f(3) = 20 nen BPT:f(x) < f ( 3 ) o x < 3 . Vay tap nghiem cua BPT la
S = [-l;3].
x

V i du 7: Giai bat phuong trinh: 3Vtanx + 1 .
°
< 2 "^
sin x + 3 cos x
Giai
Dieu kien tanx > 0. Dat t = tanx, t > 0 thi
s

m

x +

2

c

S

X

1

V T = 3 v ^ T T . ^ = f(t), t > 0
t +3
3
t +2
1
'
Ta cd f '(t) = —,
.
+ 3 v t + 1.
- > 0 nen ham so f dong bien,
2Vt+T t + 3
(t + 3)
ma t > 0 => f(t) > f(0) = 2.
Mat khac VP = 2 ~ ^
< 2 nen dau "='' ddng thoi xay ra
<=> t = tanx = 0 <=> x = krc, k e Z.
2

l

4

x + 3 = y +Vy +1
2

y + 3 = z + %/z + 1

V i du 8: Giai he phuong trinh

2

Z + 3 = X + N/X + 1
2

Giai
Xet ham sd f ( t ) = t + V t + 1 - 3 , t e R
2

fU'f.m 1
tmf'(t) = l + .
vv+i
t

V t + 1 + t Vt7 + t
—;
> ,
>0, Vt
Vt + i
vV + i
2

2

nen f(t) ddng bien tren R. Ta cd he phuong trinh

x = f(y)
y = f(z)
z = f(x)

Gia su x > y thi f(x) > f(y) nen y > z do dd f(y) > f(z) tuc la z > x: vo l i
Gia su x < y thi f(x) < f(y) nen y < z do do f(y) < f(z) tuc la z < x: vo 11
Gia su x = y thi f(x) = f(y) nen y = z do do x = y = z. The vao he:
x + 3 = x + v x + l c o 3 = v x + l c i > x = 2 /

2

/

2

2

Thu lai x = y = z = +%/2 thi he nghiem dung.
Vay he phuong trinh cd 2 nghiem x = y = z = + V2
-BDHSG DSGT12/1-

19

x - l = y(y-l)
V i du 9: Giai he phuong trinh: < y - l = z ( z - i )
3

z - l = x(x-l)
3

Giai
lY
3 3 1
+—>—>—= x >
2)
4 4 8
1
1
Tuong tu y, z > - Dat f(t) = V
l , t > - thi
2

Ta cd x

y -y + i

f '(t) = 2t - 1 > 0 nen f dong bien tren (—; +oo)
2
-y + l
f(y)
3
2
Ta cd he <
z + 1 co <
f(z)
y =z
z =x - X +1
f(x)
Gia su x > y thi f(x) > f ( y ) : z > X3
Z > X.
nen f(z) > f(x)
y > z => y > z.
Do do x > y > z > x: vd l i .
Tuong tu x < y: vo l i nen x = y => x = y = z. Ta cd t = f(t)
o t - t + t - l = 0 o ( t - l ) ( t + 1) = 0 co t = 1.
Vay he co nghiem duy nhat x = y = z = 1.
3

2

3

3

3

3

3

2

2

x - 2 x + l = 2y
3

V i du 10: Giai he phuong trinh

y - 2 y + l = 2z
z - 2 z + l = 2x
2

Giai
Ta cd 2y = x - 2x + 1 = (x - l ) > 0 y > 0. Tuong tu z, x > 0.
Dat f(t) = t - 2t + 1, t > 0 thi f '(t) = 2(t - 1) nen f dong bien tren (1; +oo)
va nghich bien tren (0; 1). Dat g(t) = 2t, t > 0 thi g'(t) = 2 > 0
2

2

z

f ( x ) = g(y)
nen g ddng bien tren (0; +oo). Ta cd he I f (y) = g(z)
f(z) = g(x)
Gia su x = min{x; y; z } . Xet x < y < z.
- Neu x > 1 thi 1 < x < y < z ^> f(x) < f(y) < f(z)
=> g(y) < g(z) < g(x) =o y < z < x nen x = y = z.
Ta co phuong trinh: t - 4t + 1 = 0 nen chon nghiem: x = y = = 2 - V 3 .
- N e u 0 < x < 1 thi f(0) > f(x) > f ( l ) =o 0 < f(x) < 1.
nen 0 < g(y) < 1 => 0 < y < 1 =o f(0) > f(y) > f ( l )
2

z

=o 0 < f(y) < 1 ^> 0 < g(z) < 1 = > 0 < Z < 1 .

20

-BDHSG

DSGTn/i

Do do x < y < z => f(x) > f(y) > f(z) => g(y) > g(z) > g(x)
=> y > z > x nen x = y Ta cd phuong trinh t - 4t + 1 = 0 nen chon nghiem: x = y = z = 2 - 42
Xet x < z < y thi cung nhan duoc ket qua tren.
Vay he cd 2 nghiem x = y = z = 2 + v 3 , x = y = z = 2 - -Js
=

z

2

/

3 6 x y - 6 0 x +25y = 0
2

V i d u l l : Giai he phuong trinh:

2

3 6 y z - 6 0 y +25z = 0
2

2

3 6 z x - 6 0 z +25x = 0
2

2

Giai
60x
36x +25
60y
2

2

2

He phuong trinh tuong duong vdi


36y +25
2

60z
36z +25
Tir he suy ra x, y, z khong am. Neu x = 0 t h i y = z = 0 suy ra (0; 0; 0) la
nghiem cua he phuong trinh.
60f
Neu x > 0 thi y > 0, z > 0. Xet ham sd f(t)
t>0.
36t +25
3000t
Ta cd: f ' ( t )
>0, V t > 0 .
(36t +25)
2

2

2

2

2

Do do f(t) ddng bien tren khoang (0; +co).
y = f(x)
He phuong trinh duoc viet lai I z = f (y)
x = f(z)
Tir tinh ddng bien cua f(x) suy ra x = y = z. Thay vao he phuong trinh ta
duoc x(36x - 60x + 25) = 0. Chon x = - .
6
2

'5 5 5
Vay tap nghiem cua he phuong trinh la ,6'6'6
:8-X
V i du L2: Giai he phuong trinh
3

Vx-1 -Ty
(x-l) =y
4

Giai
Dieu kien x > 1, y > 0. He phuong trinh tuong duong vdi:
IVx-l - (x-l)
[y = ( x - D
-BDHSG DSGTU/1-

4

2

+x - 8=0
3

(1)
(2)

Xet ham so f(t) = V t T T - ( - l ) +

3

2

t

t

1 =
2V
t-l
vW
t > 1 nen f(t) ddng bien tren ( 1 ; +oo).
Ta co f '(t) = - 2 ( t - l ) + 3t +

- 8, vdi t > 1.
2

2

3 t

- 2t + 2
2v/Tl

> 0 vdi moi

Phuong trinh (1) cd dang f(x) = f(2) nen (1) co x = 2, thay vao (2) ta
duoc y = 1 .
Vay nghiem cua phuong trinh la (x; y) = (2; 1).
x - 12x + 35 < 0
2

V i du 13: Giai he bat phuong trinh:

(1)

x - 3x + 9x + - > 0 (2)
3
Giai:
Ta cd (1) o x - 12x + 35 < 0 co 5 < x < 7
u

2

2

Xet (2): Dat f(x) = x - 3x + 9x + - , D = R
3
f (x) = 3x - 6x + 9 > ...