tài liệu bồi dưỡng học sinh giỏi toán 12

Bin DUSNG , HQC SINH GO TOAN OAI SO -GIAI TICH Boi duQng hoc sinh gioi Toan Dai so 10-1.. Trung tdm sdch gido due ANPHA xin trdn trong giai thieu quy ban dong nghiep vd cdc em hoc sinh c

- 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

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

- 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

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

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 thi 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 126

Dang 1: Ham so v = a x + k (c * 0 va ad be * 0) 126

cx + d

2 i Dang 2: Ham s6 v = &X +

(a * 0 a' * 0) 135 a'x + b'

§7 Bai toan thuong gap ve do thi 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

- 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: Xi<X2=>f(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

f(x) trai dau a 0 ciing dau a

- Dau tam thuc bac hai: f(x) = ax2 + 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] X2 +00 f(x) ciing dau a 0 trai dau a 0 ciing d i u a

V i du 1: Xet chieu bien thien ciia ham sd:

a) Tap xac dinh D = R Ta co y' = 2x - 6

3 d) D = R Ta cd y' = - 3 x2 + 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 = x4 - 2x2 - 5 b) y = x4 + 8x2 + 9

Giai a) D = R Ta co y' = 4x3 - 4x = 4x(x2 - 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:

3 ( 1 - x )2 bien trong cac khoang (-co; 1) va (1; +oo)

d) D = R\ {4}.Tacdy'=

———-( x - 4 )3 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 + X + 1

a) D = R Ta cd: y'

x2- 9 Giai

- x2 + 4x + 3 (x2 + x + l )2 y' = 0 e> x2

- 4x - 3 = 0 <=> x = 2 ± ^7

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

x + 2 Giai

a) Dieu kien 4 - x2> 0 < = > - 2 < x < 2 nen D = [-2; 2]

b) V i A' = 1 - 3 < 0 nen x2 - 2x + 3 > 0, Vx => D = R

2 v x2- 2 x + 3 V x2- 2 x + 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) DK: 16 - x2 > 0 o x2 < 16 o -4 < x < 4 D = (-4; 4)

Ta co v' = 1 6 > 0, Vx e (-4; 4)

( 1 6 - x2) V l 6- x2 Vay ham so dong bien tren khoang (-4; 4)

V j du 6: Tim khoang don dieu cua ham so

V j du 7: Xet su bien thien cua ham sd:

3

a ) y - - - x + smx b) y = x + c o g2x

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

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] => y1 > 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)

y'>0o sinx <-<=> - < x < —

2 6 6 y' < 0 » sinx > - <=>0<x< - hoac — < x < -

2 6 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) = vx2 +1 - x b) f(x) = cos2x - 2x + 5

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

f'(x) = 0 o s i n 2 x = -lc^>2x = - - + 2 k n o x = - - +kn,k& Z

2 4 Ham f(x) lien tuc tren moi doan [ - - + kn; ~ + (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, x2 e R, x x < x 2 => f(Xj) > f(x2)

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

a) f(x) = x3 - 6x2 + 17x + 4 b) f(x) = 2x - cosx + S sinx

Giai a) f' ( x ) = 3x2 - 12x + 17 V i A' = 36 - 5 1 < 0 nen f ' ( x ) > 0 vdi moi x, do dd

ham so dong bien tren R

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:

Giai 4_

(x + 2)2 Vay ham so dong bien tren moi khoang (-oo; - 2 ) va (-2; +oo)

x2 - 2 x - 5 b) D = R \ { - l } T a c d y ' = ~ < 0 vdi moi x * - l (vi A' = 1 - 5 < 0)

(x + 1)2

'

Vay ham so nghich bien tren mdi khoang (-oo; -1) va ( - 1 ; +oo)

V i du 12: Chung minh ham so:

a) y - i + x 2 dong bien trong khoang ( - 1 ; 1) va nghich bidn trong cac khoang (-co; -1) va (1; +oo)

, sin(x + a) , , _ , ,

b) y ~ (a ^ b + krt; k e Z) don dieu Uong mdi khoang xac dinh sin(x + b) • °

Giai , , l( l + x2) - 2 x x 1- x2

a ) y = ( i + x2)2 = ( T ^ ' y = 0 o x = ± 1

-Ta cd y' > 0 <=> 1 - x2 > 0 <=> - 1 < x < 1

y' < 0<=> l - x2< 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) Xet f(x) = sin2x + cos2x, D - R

f '(x) = 2sinxcosx - 2cosxsinx = 0, Vx

Do dd f(x) la ham hang tren R nen f(x) = f(0) = 1

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—

Suy ra rang f la mdt ham hang tren khoang (-— ; — )

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

a) f(x) = cosx + cos(x + —) - cosxcos(x + ^ )

3 3 b) f(x) - 2- sin2x - sm2(a + x) - 2cosa.cosx.cos(a + x)

Giai a) f'(x) = -2cosxsinx - 2cos(x + - )sin(x + - ) + sinxcos(x + ^) + cosx.sin(x + ^ )

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 - sin2a - 2cos2a = sin2a

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

nen g(x) = C: hang so tren D, do do:

f(x) - 3x + 3x2 = C ^> f3(x) = - 3 x2 + 4x + C

nen f(x) = N/ - 3X 2 + 3x + C V i f(0) = 8 => C = 64

Vay f(x) = yj-3x2 + 3x + 64 , thu lai dung

V i du 17: Tim cac gia trj cua tham so a de ham so dong bien tren R

a) f(x) = - x3 + ax2 + 4x + 3 b) f(x) = ax3 - 3x2 + 3x + 2

• 3

Giai a) f '(x) = x2 + 2ax + 4, A' = a2 - 4

- N6u a2 - 4 < 0 hay - 2 < a < 2 thi f '(x) > 0 vdi moi x e R nen ham so

ddng bien tren R

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

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

dong bien tren R

- Neu a < - 2 hoac a > 2 thi f '(x) = 0 cd hai nghiem phan biet nen f1 co doi dau: loai

Vay ham sd ddng bien tren R khi va chi khi - 2 < a < 2

b) V (x) = 3ax2 - 6x + 3

Xet a = 0 thi f '(x) = -6x + 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

- Neu m < 0 thi y' < 0 vdi moi x e R nen f nghich bien tren R

- Neu m = 0 thi y' = - 3 x2 < 0 vdi moi x e R, dang thuc chi xay ra vdi

x = 0, nen ham so nghich bien tren R

Do do ham so dong bien tren khoang (xi, x2) : loai

Vay ham so nghich bien tren R khi va chi khi m < 0

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:

a) y (3m - l ) x - m

2 + m by = x + 2 +

x + m

m

x - l Giai

a) D = R \ { - m } Ta co:

, _ (x + m)(3m -1 ) - [(3m - l)x - m2 +

Y = ' (x + m)2

m 4 m2- 2 m (x + m)2

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

<=> m < 0 hoac m > —

2 b) Ta cd y' = 1 m vdi moi x * 1

( x - l )2

- 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' , x2- 2 x + l- m

- Neu m > 0 thi y = -„

( x - l )2 y' = 0 o x2 - 2 x + l - m = 0< = > x= l + Vm

—oo 1 —Vm 1 1 + Vm +oo

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) = x3 - ax2 + x + 7 nghich bien tren khoang (1; 2)

b) f(x) = — x3 - — (1 + 2cosa)x2 + 2xcosa + 1, a e (0; 2rr) dong bien tren

3 2

khoang (1; +oo)

Giai a) f'(x) = 3 x2- 2 a x + 1

Ham so nghich bien tren khoang (1; 2) khi va chi khi y < 0 vdi moi

V i y' > 0 d ngoai khoang nghiem nen ham so dong bien vdi moi x > 1 khi

va chi khi 2cosa < 1 cosa < — o — < a < —

Dieu kien tuang duong: f(t) < 0, Vt e [-;1 1]

[ f ( - l) < 0 f - m - 4 < 0 ' 9

° l f ( l ) ^ 0 ° l 3 m - 2 ^ 0 ^ - 4 ^ m ^

V» du 22: Tim m de ham sd y = x3 + 3x2 + mx + m chi nghich bidn tren mpt

doan cd dp dai bang 3

Giai:

D = R, y' = 3x2

+ 6x + m, A' = 9 - 3m Xet A' < 0 thi y' > 0, Vx: Ham luon dong bien (loai)

Xet A' > 0 <=> m < 0 thi y' = 0 cd 2 nghiem x,, x2 nen x, + x2 = -2, X]X2 = —

- Neu A' < 0 <o 4m2 < 9 <=>

bien tren R

I m [ < — thi y' > 0, Vx nen ham so dong

- Neu A' > 0 co 4m2 > 9 co I m | > - thi y' = 0 cd 2 nghiem phan biet

xi,2 = 2m +V4m2-9 Lap bang bien thien thi ham ddng bien tren

khoang (2m - V ^ m2 -9 ; 2m + V 4 m2- 9 ) va nghich bien tren m6i

khoang (-00; 2m - \/4m2 - 9 ) va (2m + V4m2 - 9 ; +00)

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

( x - l )2

- 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)

- Neu m < - 2 thi y' > 0, Vx * 1 nen ham sd ddng bien tren moi khoang

(-co; 1) va (1; +00)

DANG 2: UNG DUNG TINH BON DI$U

- Giai phirong trinh, he phirong trinh, bat phuong trinh:

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 ) < f ( b )

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 + x2 - 72 + x - x2 = 1

Giai Dat t = x2 - x thi phuong trinh trd thanh: 7 3 + t - 7 2 - t = 1, - 3 < t < 2 Xet ham sd f(t) = 73 + t - 7 2- t, -3 < t < 2

Vdi -3 < t < 2 thi f'(t) = 1 + > 0 nen f dong bien tren (-3; 2)

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

V i du 3: Giai phucmg trinh yjx - 1 = Vx - 2 - x

Giai Dieu kien: x >%/2

5 Dieu kien x * 1; —, phuong trinh trd thanh:

2 ( 2 x - 5) 2- - J _ = ( x- l )2

| 2 x - 5 | | x - l | Xet f(t) = t2 - i vdi t > 0

Ta co: f '(t) = 2t > 0 nen f dong bien tren (0; +oo)

Xet x - 2 < 0 thi BPT nghiem dung

Xet x - 2 > 0 thi 2x - 1 > 0 nen BPT o 2 x - 1 >X - 2 < = > X > - 1 : £) Ung Vay tap nghiem la S = R

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:

Dieu kien tanx > 0 Dat t = tanx, t > 0 thi

Xet ham sd f ( t ) = t + V t2 + 1 - 3 , t e R

f U ' f m 1 t V t2 +1 + t Vt7 + t

t m f ' ( t ) = l + — ; > , >0, Vt

v v + i V t2 + i v V + i 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/x2+ l c o 3 = v/x2+ l c i> x2= 2 <S-X = ± V 2

Thu lai x = y = z = +%/2 thi he nghiem dung

Vay he phuong trinh cd 2 nghiem x = y = z = + V2

V i du 9: Giai he phuong trinh:

y > 0 Tuong tu z, x > 0

Dat f(t) = tz - 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

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)

60x2 36x2 +25

He phuong trinh tuong duong vdi < z 60y

2 36y2 +25 60z2 36z2 +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

Neu x > 0 thi y > 0, z > 0 Xet ham sd f(t)

3000t

60f 36t2+25 t > 0

6

'5 5 5 , 6 ' 6 ' 6 Vay tap nghiem cua he phuong trinh la <j (0;0;0);

:8-X3

V i du L2: Giai he phuong trinh V x - 1 - T y

( x - l )4= y Giai Dieu kien x > 1, y > 0 He phuong trinh tuong duong vdi:

I V x - l - ( x - l )2

+ x3

- 8 = 0 (1)

Xet ham so f(t) = V t T T - (t- l ) + t - 8, vdi t > 1

1

Ta co f '(t) = - 2( t - l ) + 3t2 + = 3 t2 - 2t + 2

2 v W 2 v/ T l > 0 vdi moi

2 V t- l

t > 1 nen f(t) ddng bien tren (1; +oo)

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)

x2 - 12x + 35 < 0 (1)

V i du 13: Giai he bat phuong trinh:

xu - 3x2 + 9x + - > 0 (2)

3 Giai:

Vay tap nghiem cua he bat phuong trinh la S = (5; 7)

YJ du 14: Chung minh rang phuong trinh 3x5 + 15x - 8 = 0 cd mot nghiem duy nhat

Giai Ham f(x) = 3x5 + 15x - 8 la ham sd lien tuc va cd dao ham tren R

V i f(0) = -8 < 0, f ( l ) = 10 > 0 nen ton tai mot sd x„ e (0; 1) sao cho f(xo) = 0, tuc la phuong trinh f(x) = 0 cd nghiem

Mat khac, ta cd y' = 15x4 + 15 > 0, Vx e R nen ham sd da cho luon luon ddng bien Vay phuong trinh dd chi cd mot nghiem duy nhat

V i du 15: Chung minh phuong trinh: x1 3 - x6 + 3x4 - 3x2 + 1 = 0 cd nghiem duy nhat

Giai:

Dat f(x) = x1 3 - x6 + 3x4 - 3x2 + 1, D = R

Xet x > 1 thi f(x) = x6(x7 - 1) + 3x2(x2 - 1) + 1 > 0: vd nghiem

Xet 0 < x < 1 thi f(x) = x1 3 + ( l - x2)3 > 0: vd nghiem

Xet x < 0 thi: f '(x) = 13x1 2 - 6x5 + 12x3 - 6x = 13x1 2 - 6x(x - l )2 > 0 nen

Vay phuong trinh cho cd nghiem duy nhat

V i du 16: Chung minh rang phucmg trinh 2 x V x - 2 = 11 cd mot nghiem

duy nhat

Giai Xet ham so f(x) = 2x2 Vx - 2 thi ham sd xac dinh va lien tuc tren nua

Do do ham so dong bien tren nua khoang [2; +oo)

Ham sd lien tuc tren doan [2; 3], f(2) = 0, f(3) = 18 V i 0 < 11< 18 nen theo

dinh l i ve gia tri trung gian cua ham sd lien tuc, tdn tai sd thuc c e (2; 3)

sao cho f(c) = 11 tuc c la mot nghiem cua phuong trinh f V i ham sd

dong bien tren [2; +co) nen c la nghiem duy nhat cua phuong trinh

V i du 17: Chung minh rang vdi moi x e ( - 1 ; 1), phuong trinh: sin2x + cosx = m

cd mot nghiem duy nhat thudc doan [0; TT]

Giai Xet ham sd f(x) = sin2

x + cosx thi ham sd lien tuc tren doan [0; n],

Ta cd f ' (x) = 2sinxcosx - sinx = sinx(2cosx - 1), x e (0; TI)

V i sinx > 0 nen f '(x) = 0 o cosx = — o x = —

3 3 ' 71 7t 5

Ham so f lien tuc tren doan \—; Til, f ( — ) = — va fire) = - 1 Theo dinh l i

3 3 4

5

ve gia tri trung gian cua ham so lien tuc, vdi moi m e ( - 1 ; 1) cz ( - 1 ; —)

4 ton tai mot sd thuc c e (—; 7i) sao cho f(c) = 0 tuc c la nghiem cua

3 phuong trinh Vi ham so f nghich bien tren [—; 71] nen tren doan nay,

3 phuong trinh cd mot nghiem duy nhat

TC 5 Con vdi moi x e [0; — ], ta cd 1 < f(x) < — nen phuong trinh khong co

3 4 nghi?m suy ra dpcm

V i du 18: Tim sd nghiem cua phuong trinh x3 - 3x2 - 9x - 4 = 0

Giai Xet ham so y = x3 - 3x2 - 9x - 4, D = R

Dua vao BBT thi phuong trinh y = 0 co dung 3 nghiem

V i du 19: Tim sd nghiem cua phuong trinh:

Vay phuong trinh cho cd dung 2 nghiem

V i du 20: Chung minh he

fx2 + y3 = 1

y + x 3

Giai

cd dung 3 nghiem phan biet

Tru 2 phuong trinh ve theo ve va thay the ta duoc:

x2( l - x) - y2( l - y) = 0 =o (1 - y3) ( l - x) - (1 - x3) ( l - y) = 0

=> (1 - x)(l - y ) [ l + y + y 2- ( l + x + x2)] = 0

-O (1 x ) ( l - y)(y - x ) ( l + x + y) = 0

Xet x = 1 thi he cd nghiem ( 1 ; 0)

Xet y = 1 thi he cd nghiem (0; 1)

Xet x = y thi x2 + y3 = 1 co x3 + x2 - 1 = 0

Dat f(x) = x3 + x2 - 1, D = R Ta cd f ( l ) = 1 * 0

f ' (x) = 3x2 + 2x, f1 (x) = 0 co x = - - hoac x

3 BBT '

Do do f(x) = 0 cd 1 nghiem duy nhat x,, > 0, XQ * 1 nen he cd nghiem (x„; y0)

Xet 1 + x + y = 0 => y = - x - 1 nen y2 + x3 = 1 co x3 + x2 + 2x = 0

co x(x2 + x + 2) = 0 co x = 0 Do dd he cd nghiem (0; 1)

Vay he cd dung 3 nghiem phan biet

V i du 21: Tim cac gia tri cua m de phuong trinh sau cd dung mpt nghiem

yjx 2 + 2x + 4 - V x + l = m

Giai Dat t = Vx + 1 > 0, phuong trinh trd thanh vft4

+3 - t = m (*) Nhan xet ung vdi moi nghiem khong am cua phuong trinh (*) cd dung mpt nghiem cua phuong trinh da cho, do do phuong trinh da cho co dung mpt nghiem khi va chi khi phuong trinh (*) cd dung mpt nghiem khong am

Xet ham sd f(t) = tftU^S-t vdi t > 0, f '(t) = ,

Tu bang bien thien suy ra cac gia tri can tim cua m la 0 < m < ^3

V i du 22: Tim m de phuong trinh cd nghiem

m( 7 l + x2 - V l - x2 + 2) = 2 V l - x4 + V l + x2 - V l- x2

Giai Dieu kien - 1 < x < 1 Bat x = V l + x2 - V l- x2 thi t > 0

va t2 = 2 - 2 V l - x4 < 2, dau "=" khi x2 = 1 Do dd 0 < t <

PT:m(t + 2) = 2 - t2 + t c o m = Zzl±l±ll

t + 2 Xet f(t) = - t + t + 2 0 < t < V2 , f Yt) = 1 + A \ < 0 nen f nghich bien

tren [0; V2 ] Dieu kien cd nghiem:

min f(t) < m < max f(t) o f( &) < m < f(0) co V2 - 1 < m < 1

V i du 23: Tim m de phuong trinh sau cd 2 nghiem phan biet:

dd dieu kien cd nghiem: f ( l ) < m < f(0) co — < m < —

Vi du 25: Chung minh cac bat dang thuc sau:

a) sinx < x vdi moi x > 0, sinx > x vdi moi x < 0

x2

b) cosx > 1 vdi moi x & 0

2 Giai

a) Vdi x > — thi x > 1 nen sinx < 1 < x

2

Vai 0 < x < — thi ham so f(x) = x - sinx lien tuc tren nua khoang [0; —)

2 2

va f'(x) = 1 - cosx > 0 vdi moi x e (0; ) Do do ham sd dong bien tren

[0; -) nen f(x) > f(0) = 0 vdi moi x e (0; -)

2 2

Vdi —| < x < 0, giai tuong tu thi f(x) < f(0) = 0

Vdi x < -— thi x < -1 nen sinx > -1 > x => dpcm

2

xz b) Voi x > 0 thi ham so g(x) = cosx + — — 1 lien tuc tren nua khoang [0; +co)

va g'(x) = x - sinx Theo a) thi g'(x) > 0 voi moi x > 0

Do do ham so g dong bien tren [0; +co) nen:

x

2 g(x) > g(0) = 0 vdi moi x > 0 => cosx + — — 1 > 0 vdi moi x > 0

2

f '(x) = 1 — - — cosx ; f "(x) = - x + sinx

f'"(x) = -1 + cosx < 0 nenf" nghich bien tren [0; +co):

x > 0 => f "(x) < f "(0) = 0 nen f' nghich bien tren [0; +oo):

x > 0 => f'(x) < f (0) = 0 nen f nghich bien tren [0; +<x>):

x > 0 => f (x) < f (0) = 0 => dpcm

Vi du 27: Chung minh cac bat dang thuc vdi moi x e (0; —)

2 a) tanx > x b) tanx > x + —

3 c) sinx + tanx > 2x d) 2sinx + tanx > 3x

Giai

a) Ham so f(x) = tanx - x lien tuc tren nua khoang [0; —) va cd dao ham

f (x) = — - — > 0 vdi moi x e (0; —) Do do ham so f dong bien tren

cos x 2

nua khoang [0; —) nen f(x) > f(0) = 0 vdi moi x e (0; —)

2 2 ' 71

b) Ham sd f(x) = tanx - x lien tuc tren nua khoang [0; — ) va co dao

3 2 ham f '(x) = —\ 1 - x2 = tan2x - x2 = (tanx + x)(tanx - x) > 0 vdi

cos x moi xe(0;^) (suy ra tu a))

Do do, ham so f dong bien tren nira khoang [0; — ) va ta cd

f(x) > f(0) = 0 vdi moi xe(0;^)=> dpcm

c) Ham so f(x) = sinx + tanx - 2x lien tuc tren nua khoang [0; — ) va cd dao

ham f ' (x) = cosx + — 2 > cos2x

-cos x -cos' x

\ - - 2 = (cosx - — )2> 0

cosx

Do do ham so f dong bien tren [0; — ) nen f(x) > f(0) = 0

Ket qua: Tam giac ABC cd 3 gdc nhon thi

smA + sinB + sinC + tanA + tanB + tanC > 2TC

d) Ham so f(x) = 2sinx + tanx - 3x lien tuc tren nua khoang [0; —) va

f '(x) = 2cosx + 2 c o s

3x - 3cos2x + l (cosx - l )2 (2 cos x + 1)

Do do ham so f ddng bien tren [0; — ) nen f(x) > f(0) = 0

Vi du 28: Chung minh bat dang thuc:

a) 8sin2— + sin2x > 2x, Vx e (0; rc] b) tanx < —, Vx

2 TC Giai

2 X a) Xet ham sd f(x) = 8sin — + sin2x - 2x, Vx e (0; TC]

f '(x) = 4sinx + 2cos2x - 2 = 4sinx(l - sinx)

f '(x) = 0 co x = — hoac x = rc V d i x e (0; TC] ta cd f '(x) > 0 va dau bang

chi xay ra tai hai diem Vay f(x) dong bien tren nua khoang (0; TC] nen

f(x) > f(0) = 0 vdi moi x e (0; TC] => dpcm

b) Neu x = 0 thi BDT dung

Neux>0thi BDT o <- Vx e f

0;-X TC ^ 4 Xetf(x)=^.VxJ0;^

x

V i 0 < x < - nen 0 < 2x < - sin 2x < 2x do dd f ' ( x ) > 0 nen f ddng

bien tren V suy ra f(x) < f ( - ) = - => dpcm

4 ru

v 4

V i du 29: Chung minh bat dang thuc:

a) b.tana > a.tanb vdi 0 < a < b < —

2

571 b) cos(x + y)< vdi x > 0, y > 0 va x + 2y < —

x sin y 4

Giai

„ , , tana tanb c tanx _ TC

a) b.tana < a.tanb <o < Xet f(x) = 0 < x <

-a b x 2 x-tanx

f Y YX = cos2 x ; = x - s i n x c o s x = 2 x - s m 2 x

x x cos x 2x cos x Xet g(x) = 2x - sin2x, 0 < x < - g '(x) = 2 - 2cos2x = 2(1 - cos2x) > 0

nen g ddng bien: x > 0 => g(x) > g(0) = 0 Do do f'(x) > 0 nen f dong bien

2

5TI Neu n < t < — thi do cost < 0; tant < t => f '(t) < 0

+ a2

c2 + a2

d2 + b2

c2 + b2

d2

+ c 2

d 2 ) > 0 vdi 4 s6

a, b, c, d duong

^2 ^ b) 1 + — x < VI + x < 1 + — x , vai x > 0

2 8 2

Giai

a) Khong mat tinh tong quat, gia su a > b > c > d > 0

Xem ve trai la ham so f(a), a > 0

f '(a) = 4a3 + 2bcd - 2a(b2 + c2 + d2)

f "(a) = 12a2 - 2(b2 + c2 + d2) > 0 nen f' dong bien tren (0; +co):

a > b => f '(a) > f '(b).Vi f '(b) = 2b(b2 - c2) + 2bd(c - d) > 0 nen f(a)

ddng bien tren [0; +oo): a > 0 => f(a) > f(0) = 0=> dpcm

b) Xet ham so f(x) = 1 + -x - Vl + x tren [0; +co) Ta cd:

f '(x) = — 1 > 0 vdi x > 0 nen f(x) dong bien tren nua khoang

2 2Vl + x

[0; +oo) Do dd f(x) > f(0) = 0 vdi moi x > 0

Xet ham sd g(x) = \/l + x - 1 + — tren [0; +oo)

2 8 Tacd:g'(x)= 1 + - g"(x) = 1 >0

2 V T T I 2 4 4 4(l + x ) v T + ^ nen g' dong bien tren [0; +°o), do do g'(x) = g'(0) = 0 Suy ra g dong bien

tren [0; +co) nen g(x) > g(0) = 0 vdi moi x e [0; +co) => dpcm

sin x TI

V i du 31: Chung minh vdi moi a <3saocho ( )a > cosx, Vx e (0;—)

x 2 Giai

7i sin x

Khi x e (0;—) thi cd 0 < sinx < x nen 0 < < 1

2 x sin x sin x 1

Suy ra ( — ) ° > ( ^ i i ± )3 Va < 3 do do ta chi can chung minh khi a = 3:

CO xsin — sin > sm 2x(x + l ) 2x(x + l ) 2(x + l )

\r -v PS n n rc(2x + l) rc

V i x > v3 => 0 < < < —

2(x + l ) 2x(x + l ) 2 sinrc(2x + l) rc

=> > sin >0 (1) 2x(x + l ) 2(x + l )

rc rc

Ta se chung minh: xsin >sin (2)

2x(x + l ) 2(x + l ) Dat t = > 0 thi (2) co xsint > sinxt

Gia sir z la sd be nhat thi 0 < z < — Ta cd

f Yz) = - 6z2 + 2z = 2z(l - 3z) > 0 tren f(z) ddng bien tren [0; - ] do do

3

T = f(z)<f(-)= —

; 3 27

(2x + lY

DS: a) dong bien tren ( — ; 2) va nghich bidn tren (-co ; — ) va (2; +oo)

2 2 Bai 2: Tim khoang don dieu cua ham so

a) y = , = b) y = — c) y Vx - x + 1 e l n x

DS: b) dong bien tren (0; 2) va nghich bi£n tren (-co; 0) va (2; +oo)

c) dong bien tren (e; +co) va nghjch bidn tren (0; 1) va (1; e)

Bai 3: Tim khoang don dieu cua ham so:

a) y = x.lnx b) y = sinx + sin2x

Bai 4: Tim khoang don dieu cua ham sd:

a) y = | x2 - 3x - 4 | b)y = ^^

cx + d Bai 5: Chung minh ham sd ddng bien tren tap xac dinh:

a) y = x - ex nghich bien tren khoang (0; +oo)

x2 - 4x + 3

b) v = — i — luon dong bien tren tirng khoang xac dinh

H D : b) y ' > 0 tren (-co; -1), (-1;2), (2; +oo)

Bai 7: Chung minh ham so

a) y = sinx + cosx + 2x luon dong bien tren R

b) y = ( i + _ ) * dong bien tren (0,+ co)

HD: a) y '= cosx - cosx + 2 > 0, Vx b) Lay In trudc khi tinh dao ham

Bai 8: Tim tham sd de ham sd:

a) y = — x3 - — (sin a + cos a)x2 + — sm 2a.x ddng bien tren R

b) y = (m - 3)x - (2m + l)cosx nghich bien tren R

DS: a) — + kn < a < — + krc

12 12

Bai 9: Tim tham so de ham sd:

a) y = a x + 4 nghich bien tren (-°o;l)

x + a

b) y = 2x3 + 3x2 + 6(m + l) x nghich bien tren (-2; 0)

DS: b) m< - 3

Bai 10: Tim tham sd de ham sd:

a) y = x3 + 3x2 + mx + m nghich bien tren mot doan cd do dai bang 3

— 2x + m ' '

b) y = dong bien tren khoang (-co; 0)

x- l

Bai 11: Tim m de ham sd

a) y = mX + ^X—- nghich bien tren nua khoang [1; +oo)

x + 2 b) y = x3 - 3x2 + (m - 2)x + 7 ddng bidn tren R

DS: a)m< -— b)m >5

5 Bai 12: Tim m de ham so:

3 2 '

a) y = x - m x + x + l nghich bien tren khoang (1, 2)

b) y = — 2mx + m + 2 ^Qn^ ^ - «n t r-n j ^ ^ n g (\.+ ^ y

x - m DS: a)m > —; b) m<3~^;m>2

4 4 Bai 13: Tim m de ham so:

x2 + (m + 2)x m + 3 x , • - < - - • i u - - A- U a) y = dong bien tren tung khoang xac dmh

x + 1

2 1 2 ! -X b) y = 2mx - 2cos x - m.sinx.cosx + — cos 2x dong bien tren R

DS: a) 1 < x < 4 b) x > 3

1 Bai 18: Giai he: a)

H D : Ham y = x3 + x2 + 12x - V3 don dieu tren R

Bai 20: Chung minh phuong trinh x3 + 2x3 - x2 + x - 1 = 0 cd nghiem duy nhat

H D : Ham y = x5 + 2xJ - x2 + x - 1 don dieu tren R

(n + 1) x n + 2 - 3(n + 2)xn + 1 + an + 2 = 0 vo nghiem

H D : y ' = (n + l ) ( n + 2).xn _ 1.(x - 3)

V x - s i n A + V x- s i n B = Vx - smC co nghiem duy nhat

H D : Tam giac ABC co canh a < b < c thi gdc A > B > C

HD: Dat t = cos2x vdi -— < t < —

H D : Tfnh dao ham cua ham sd y = 3x + 8x + ax +b roi xet a > 6, a = 6

va a < 6

Bai 31: Chung minh vdi moi so nguyen duong n thi phuong trinh:

x + x2 + x3 + + x2 n + 2007x2 n + 1 = 1999 co nghiem duy nhat

Bai 32: Chung minh phuong trinh: x1 3 - x6 + 3x4 - 3x2 + 1 = 0 cc nghiem duy nhat

Baj 34: Giai phuong trinhVx2 + 15 = 3x - 2 + Vx2 + 8

Chung minh: 2 cos3C - 4 cos2C + 1

cosC > 2 Bai 37: Chung minh rang vdi ba so duong a b c bat k i thi:

Bai 39: Chung minh he phuong trinh:

2x2 2y2

Bai 42: Chung minh bat dang thuc:

a sina - p sinP > 2 (cosP - cosa) vdi 0 < a < p < —

Bai 43: Chung minh bat dang thuc:

sinA + sinB + sinC „ 2 rc

> cosx > vai 0 < x <

-U a n x J 2 Bai 44: Chung minh bat dang thuc: |cos2

Bai 49: Cho f(x) vdi deg f = n va f(x) > 0, Vx e R

Chung minh Jf(k)(x)>0

k=0 Bai 50: Cho tam giac ABC nhpn Chung minh

2 1

— (sin A + sin B + sin C) + — (tan A + tanB + tanC) > rc

3 3

Bai 51: Chung minh bat dang thuc:

> 2 cos2 — vdi AABC khong tu

cosA + cosB + cosC 8

Bai 52: Chung minh vdi tam giac ABC thi cd:

Chiing minh: xyz( x(— + -) + y(- + —) + z(— + —) + 1) < —

y z z x x y 27 Bai 55: Cho a, b, c, d > 0 va thoa man

2(ab + ac + ad + be + bd + cd) + abc + abd + acd + bed = 16

2

Chung minh: a + b + c + d > — (ab + ac + ad + be + bd + cd)

3 Bai 56: Cho a, b, c, r, s >0 va thoa man a > b > c, r > s

Chung minh ar.bs + b\cs + c".as > as.br + bs.c" + cs.a'

§ 2 C U C T R I C U A H A M S O

A K I E N T H t f C C O B A N

Cho ham s6 f xac dinh tren tap hop D ( D c R ) ya x0 e D

a) Xo duoc goi la mpt diem cue dai cua ham sd f neu tdn tai mpt khoang (a; b) chua diem Xo sao cho (a; b ) c D v a f(x) < f(xo) vdi moi x e (a; b) \ { x0} Khi dd f(Xo) duoc gpi la gia tri cue dai cua ham sd f, ki hieu yco-

b) Xo dugc gpi la mpt diem cue tieu cua ham sd f neu ton tai mpt khoang (a; b) chiia diem XQ sao cho (a; b ) c D v a f(x) > f(Xo) vdi moi x e (a; b) \ { x0} Khi do f[Xo) duoc gpi la gia tri cue tieu cua ham sd f, ki hieu ycr-

Diem cue dai va diem cue tieu duoc gpi chung la diem cue tri Gia tri cue dai va gia tri cue tieu duoc gpi chung la cue tri, neu x0 la mpt diem cue tri cua ham so f thi diem (x0; f(x0)) duoc gpi la diem cue tri cua do thi ham sd f

i

I n (

Dieu kien can de ham sd cd cue tri:

Gia su ham sd f dat cue tri tai diem Xo Khi dd, neu f cd dao ham tai XQ thif'(Xo) = 0

Dieu kien du de ham sd cd cue tri: cd hai dau hieu:

- Cho y = f(x) lien tuc tren khoang (a;b) chua xo, cd dao ham tren cac khoang (a;xq) va (x0;b):

Neu f '(x) ddi dau tu am sang duong thi f dat cue tieu tai x<>

Neu f '(x) ddi dau tu duong sang am thi f dat cue dai tai xo

- Cho y = f(x) cd dao ham cap hai tren khoang (a;b) chua x0:

Neu f '(xo) = 0 va f "(xo) > 0 thi f dat cue tieu tai xo

Neu f '(x0) = 0 va f "(x0) < 0 thi f dat cue dai tai x 0

Quy tac 2

l T i m f ' ( x ) ,

2 Tun cac nghiem Xj (i = 1,2, ) cua phucmg trinh f '(x) = 0

3 T i m f " ( x ) v a t i n h f "(xO

Neu f "(xj) < 0 thi ham so dat cue dai tai diem Xj

Neu f "(xj) > 0 thi ham so dat cue tieu tai d i i m Xj

Chu y: - Gia tri cue dai (cue tieu) f(Xo) cua ham so f ndi chung khdng phai

la gia tri ldn nhat (nhd nhat) cua ham sd f tren tap hop D; f(Xo) chi la gia tri ldn nhat (nhd nhat) cua ham sd f tren mot khoang (a; b) nao do chiia diem Xo

- Ham sd f cd the dat cue dai hoac cue tieu tai nhieu diem tren tap hop

D, nhung khdng dat tai cac bien

- Tung do cue tri y = f(x) tai x = xo cd 3 hudng tinh:

Ham so bat ky: dung phep the yo = f(xo)

Ham huu ti: dao ham rieng tu, rieng mau

y = f(x)=^.hiy„=^4 = ^4

Dac biet: V d i ham bac 3 cd CD, CT va neu y = q(x) y' + r(x) thi phuong trinh ducmg thang qua CD, CT la y = r(x)

- Bai toan don dieu, cue tri khdng duoc dat an phu

V i du 1: Tim cue tri cua cac ham sd sau:

x2 + 2x - 10 Giai

3 b) D = R Ta cd f '(x) = x2 - 2x + 2 > 0, Vx (do A' = 1 - 2 < 0) nen ham s6

dong bien tren R, khdng cd cue tri

c) D =• R Ta cd y' = 4x3 - lOx = 2x(2x2 - 5)