Bdhsgtoan12

ThS. L E H O A N H PHO
Nhd gido Uu tu
C c c
M i l
B O I D U 8 N G 7
H O C S I N H G I O I T O A N
D A I S O - G I A I T I C H
- 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

Bin DUSNG ,
HQC SINH GO TOAN
OAI SO -GIAI TICH
Boi duQng hoc sinh gioi
Toan Dai so 10-1.
Toan Dai so 10-2.
- Boi duQng hoc sinh gioi
Toan Hinh hoc 10.
- Boi duOng hoc sinh gioi
Toan Dai so 11.
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 PHO
Nha gido iCu tu
B O I D U Q N G ,
H O C S I N H G I O I T O A N
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

NHA XUAT BAN DAI HOC QUOC GIA HA N0I
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 PHAM THI TRAM
Bien tap noi dung
THUY NGAN
Sda bdi
NGOC HAN
Che bdn
CONG TI ANPHA
Trinh bay bia
SON KY
Ddi tdc lien ket xudt bdn
CONG TI ANPHA
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 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

C H U O N G I : U N G D U N G D A O H A M O E K H A O
S A T V A V E D O T H j C U A H A M S O
§1. TINH DON DIEU CUA HAM SO
A. K I EN THLTC CO BAN
• 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: 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. PHAN DANG TOAN
DANG 1: B6NG 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
x -00 -b/a +co
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 iu a
-BDHSG DSGT12/1-

Vi du 1: Xet chieu bien thien ciia ham sd:
a) y = x2 - 6x + 5
c) y = x3 - 2x2 + x + 1
b ) y = - x 3
3
2x2 + x - 3
d) y = - x 3 + 4x2 - 7x + 5
Giai
a) Tap xac dinh D = R. Ta co y' = 2x - 6.
Cho y' = 0 » 2x - 6 = 0 » x = 3.
Bang bien thien X —oo 3 +00
y' - 0 +
y —
Vay ham so nghich bien tren (-oo; 3), dong bien tren (3; +oo).
b) D = R. Ta cd y' = 4x2 - 4x + 1 = (2x - l ) 2 > 0 vdi moi x
y' = 0 o x = —. Vay ham so dong bien tren R.
2
c) D = R . Ta co y' = 3x2 - 4x + 1
Cho y' = 0 o 3x2 - 4x + 1 = 0 <=> x = - hoac x = 1.
J 3
BBT X —00 1/3 1 +00
y' + 0 0 +
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 2 + 8x - 7
ViA' = 1 6 - 2 1 < 0 nen y' < 0 vdi moi x do do ham so nghich bien tren R.
Vi 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)
Cho y' = 0 <=> 4x(x2 - 1) = 0 <=> x = 0 hoac x = ±1
BBT X —00 - 1 0 1 +00
y' - o •+ 0 - 0 +
y ^ ^ * ^ ^
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' = 4x3 + 16x = 4x(x2 + 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).
Vi du 3: Xet su bien thien cua ham so:
a) y = x + —
x
b)y c)y
a) Tap xac dinh D = R {0}
_3_
Ta co y' = 1 ,.2
BBT:
Giai
, y' = 0 <» X
x 2 - 3
3 x - 8
1-x
:V3"
d)y
1
( x - 4 ) 2
X -co QN/3 +00
y' + 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 ).
b) D = R {0}. Tacdy' = 1 > 0 vdi moi x ^ 0 nen ham so dong bien
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
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).
Vi du 4: Tim cac khoang don dieu ciia ham so:
, x-2 2x
a) y = -X5 + X + 1 b) y
a) D = R. Ta cd: y'
x 2 - 9
Giai
- x 2 + 4x + 3
(x2 + x + l ) 2
y' = 0 e> x2 - 4x - 3 = 0 <=> x = 2 ± ^7
BBT: X -co 2-V7 2+V7 +°o
y' - 0 + 0 -
y — ^ — * - *
-BDHSG DSGT12/1- 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'=^^ <0,Vx*±3.
(x2 - 9 ) 2
Do do y' < 0 tren cac khoang (-co; -3), (-3; 3), (3; +oo) nen ham so da
cho nghich bien tren cac khoang do.
Vi du 5: Xet su bien thien cua ham sd:
a) y = V 4 - x 2
c)y
V l 6 - :
b)y = Vx2 - 2 x + 3
d)y
x + 2
Giai
a) Dieu kien 4 - x 2 > 0 < = > - 2 < x < 2 nen D = [-2; 2]
Vdi -2 < x < 2 thi y'
BBT:
V 4 ~
, y ' = 0<=>x = 0.
X -2 n 2
y' + 0 -
y ^ ^ — ^
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) Vi A' = 1 - 3 < 0 nen x2 - 2x + 3 > 0, Vx => D = R.
•p , 1 2x-2 x- l , . ,
Ta co y = — = = =. y = 0 o x = 1.
2 v x 2 - 2 x + 3 Vx2 - 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' = 16 > 0, Vx e (-4; 4).
( 1 6 - x 2 ) V l 6 - x 2
Vay ham so dong bien tren khoang (-4; 4).
d) D = [0; +00). Vdi x > 0, y' = X _, y'
2^y^(x + 2)2
BBT: X 0 2 +00
y + 0
y ^
Vay ham sd ddng bien tren (0; 2) va nghich bien tren (2; +00).
8 -BDHSG DSGTu/1-

Vj du 6: Tim khoang don dieu cua ham so
a) y = V x ( x - 3)
c)y
b)y = - x
7 x 2 - 6
d)y =
x + 1
V l - x
Giai
a) D = [0; +oo). Vdi x > 0, ta cd:
1
2Vx
y
BBT:
( x - 3 ) + r 3 N R X - 1)
vx = 2x , y
0<=>x= 1.
X 0 1 +GO
y' 0 +
y — — — ^
Vay ham so nghich bien tren khoang (0; 1) va dong bien tren khoang
(i';+°o).
b) D = R. Vdi x ^ 0, ta co: y' =—-
3 3vV 3vV
y' = 0 <=> x2 = 1 <=> x = ±1.
y' > 0 o ^/x2" > l < = > x 2 > l c i > x < - l hoac x > 1.
y' < 0 %/x2" < l e > x 2 < 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).
Tacd: y' = -^^7^£L,y1 = 0»x = ±3.
( x 2 - 6 ) v ' x 2 - 6
BBT: X —CO -3 V6 V6 3 +CO
y' + 0 - • _ 0 +
y •
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 ' = ,3 ~ X > 0, V x < l .
2V( l - x ) 3
Vay ham sd dong bien tren khoang (-co: 1).
-BDHSG DSGT12/1-

Vj 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 v ' 1
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.
Vx +1
Vi Vx2 +1 > Vx2 = 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.
10 -BDHSG DSGT12/1-

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, xx < x2 => 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).
Vi 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.
Vi 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 - 51< 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.
Vi d u l l : Chung minh ham so:
x - 2
a) y = ddng bien tren moi khoang xac dinh cua nd.
x + 2
- x 2 — 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
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 ' v '
Vay ham so nghich bien tren mdi khoang (-oo; -1) va ( - 1 ; +oo).
Vi du 12: Chung minh ham so:
-BDHSG DSGT12/1- 1 1

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 + x 2 ) - 2 x . x 1-x2
a ) y = (i + x 2 ) 2 = ( T ^ ' y = 0 o x = ± 1 -
Ta cd y' > 0 <=> 1 - x2 > 0 <=> - 1 < x < 1.
y ' < 0 < = > l - x 2 < 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) _ sin(b-a)
sin2(x + b) sm2(x + b)
(do a - b * kn)
Vi 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.
Vi du 13: Chung minh:
a) sin2x + cos2x = 1, Vx. b) cosx + sinx. tan— = 1, Vx e (-— ; —).
2 4 4
Giai
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—
• 2 2cos2 - 2 2
2
X X i X
= -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 (-— ; — ).
Do dd f(x) = f(0) = 1 vdi moi x e (-—; - )•
4 4
Vi du 14: Chung minh cac ham so sau la ham khong ddi
-BDHSG DSGTU/1-

a) f(x) = cos2x + cos2(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 + ^ )
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 - sin2a - 2cos2a = sin2a.
Vi 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).
Vi du 16: Xac dinh ham so f(x) thoa man: f(0) = 8; f ( x ) . f '(x) = 1 - 2x (*).
Giai
Ta cd (*) -(f (x))* = l-2xo (f3(x))' = 3 - 6x.
3
Xet ham sd g(x) = f3(x) - 3x + 3x2 thi g'(x) = ( f (x))' - 3 + 6x = 0.
nen g(x) = C: hang so tren D, do do:
f(x) - 3x + 3x2 = C ^> f3(x) = -3x2 + 4x + C.
nen f(x) = N/ - 3X2 + 3x + C Vi f(0) = 8 => C = 64.
Vay f(x) = yj-3x2 + 3x + 64 , thu lai dung.
Vi 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) = x 2 + 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.
-BDHSG DSGT12/1- 13

- Neu a = 2 thi f '(x) = ( x + 2)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 f 1 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, vi 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 fa >0
<=> <^=> <=> « a > 1.
[ A ' < 0 [9-9a<0 |a>l
Yl 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 - 3x2
- Neu m < 0 thi y' < 0 vdi moi x e R nen f nghich bien tren R
- Neu m = 0 thi y' = - 3 x 2 < 0 vdi moi x e R, dang thuc chi xay ra vdi
x = 0, nen ham so nghich bien tren R.
- Neu m > 0 thi y' = 0 o x = ± m^
BBT X —00 Xl x2 +00
y' 0 H 0
r
y — » •
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) Vi 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.
Vi du 19: Tim m dd ham sd dong bien tren moi khoang xac dinh:
a) y (3m - l ) x - m2 + 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
14
m 4m2 - 2 m
(x + m)2
-BDHSG DSGTU/l-

Ham so dong bien tren moi khoang xac dinh <=> 4m2 - 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 x 2 - 2 x + l - m = 0<=>x=l + Vm
BBT X —oo 1 —Vm 1 1 + Vm +oo
y' + 0 - 0 +
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.
Vi 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 x 2 - 2 a x + 1
Ham so nghich bien tren khoang (1; 2) khi va chi khi y < 0 vdi moi
x e (1;2)
f f ( l ) < 0 f 4 - 2 a < 0 13
<=> < <=>< <=>a>—
[f(2)<0 [l3-4a<0 4
b) y' = x2 - (1 + 2cosa)x + 2cosa. Ta cd 0 < a < 2TT.
y' = 0 o x = 1 hoac x = 2cosa.
Vi y' > 0 d ngoai khoang nghiem nen ham so dong bien vdi moi x > 1 khi
va chi khi 2cosa < 1 cosa < — o — < a < —
2 3 3
Vi 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
° 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 = —
3
BBT: x —CO *1 x2 +00
y' + 0 - 0 +
y ^ ^
Theo de bai: x2 - X] = 3 o (x2 - x , ) 2 = 9 o x2 + x2 - 2 x t x 2 = 9
4 15
<=> (x2 + x t ) 2 - 4xtx2 = 9ci>4 — m = 9 o m= (thoa)
3 4
Vi du 23: Tuy theo tham &6 m, xet su bien thien cua ham sd:
1 3 ? , rx ix 2x + m
a) y = - x3 - 2mx2 + 9x - m b) y =
3 x-l
Giai
a) D = R. Ta cd y' = x2 - 4mx + 9; A' = 4m2 - 9
- 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 m 2 - 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).
16 -BDHSG DSGT12/1-

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 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),...
Vi du 1: Giai phuong trinh: vo - x + x2 - 72 + x - x2 = 1
Giai
Dat t = x2 - x thi phuong trinh trd thanh: 73+ 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).
273 + t 272 - t
Ta cd f ( l ) = 2 - 1 = 1 nen phuong trinh:
f(t) = f( 1) <=> t = 1 o x2 — x — 1 = 0 <=> x = l^H.
Vi du 2: Giai phuong trinh 72x3 +3x2 +6x + 16 = 273 + 7 4 - x
Giai:
Dieu kien xac djnh:
f 2 x 3+3x2+6x + 16>0 f(x + 2 ) ( 2 x 2 - x + 8)>0
<=>C cs> - 2 < x < 4
4 - x > 0 4-x>0
Phuong trinh tuong duong 72x3 + 3x2 + 5x +16 - 74 - x = 273
Xet ham s6 f(x) = N/2X3 +3x2 + 6x + 16 - 74 - x .-2 < x < 4
™, c u . 3(x2 + x + l ) . x
Thi f (x) = —. = + — > 0 nen i dong bien ma
72x3 +3x2 +6x + 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
-BDHSG DSGT12/1- 17

Vi du 3: Giai phucmg trinh yjx2 - 1 = Vx3 - 2 - x.
Giai
Dieu kien: x >%/2
Ta cd: Vx3 - 2 = x + Vx2 - 1 > x > l = > x 3 > 3 = i > x > v / 3
Chia 2 ve cho vo? thi phuong trinh:
~ 1 a.
x2.vx x4vx Vx V xVx
Xet f(x) la ham sd ve trai, x > tfi thi
,, x 9 5x X 3
f ' ( x ) = B r — r , <0.
2x5.Vx 2xVx n 2 2
2x Vx
Do do ham so f nghich bien tren khoang (y3 ; +oo) ma f(3) = 0 nen
phuong trinh cd nghiem duy nhat x = 3.
Vi du 4: Giai phuong trinh: 3x2 - 18x + 24 1 1
2 x - 5 x - l
Giai
5
Dieu kien x * 1; —, phuong trinh trd thanh:
2
( 2 x - 5 ) 2 - - J _ = ( x - l ) 2
|2x-5| |x-l|
Xet f(t) = t2 - i vdi t > 0.
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|
<=> 4x2 - 20x + 25 = x2 - 2x + 1 <=> 3x2 - 18x + 24 = 0.
c ^ x 2 - 6 x + 8 = 0 c o x = 2 hoac x = 4 (chon)
Vi du 5: Giai bat phuong trinh:
4 | 2x - 11 (x2 - x + 1) > x3 - 6x2 + 15x - 14
Giai
BPT: | 2x - 11 .[(2x - l ) 2 + 3] > (x - 2)3 + 3x - 6
<eo | 2x - 113 + 3 | 2x - 11 > (x - 2)3 + 3(x - 2)
Xet ham sd f(t) = t 3 + 3t, D = R.
Ta cd f '(t) = 3t2 + 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 < = > X > - 1 : £)Ung
Vay tap nghiem la S = R.
18 -BDHSG DSGT12/1-

Vi 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 ' ( x ) = —— + — +— > 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 ] .
Vi du 7: Giai bat phuong trinh: 3Vtanx + 1 . s m x + 2 c ° S X < 2 1 " ^
sin x + 3 cos x
Giai
Dieu kien tanx > 0. Dat t = tanx, t > 0 thi
VT = 3 v ^ T T . ^ = f(t), t > 0
t + 3
3 t + 2 1 '
Ta cd f '(t) = —, . + 3vt + 1. -> 0 nen ham so f dong bien,
2Vt+T t + 3 (t + 3)2
ma t > 0 => f(t) > f(0) = 2.
Mat khac VP = 2l~4^ < 2 nen dau "='' ddng thoi xay ra
<=> t = tanx = 0 <=> x = krc, k e Z.
x + 3 = y +Vy2 + 1
Vi du 8: Giai he phuong trinh y+ 3 = z + %/z2 + 1
Z + 3 = X + N/X2 + 1
Giai
Xet ham sd f ( t ) = t + V t 2 +1 - 3 , t e R
f U ' f .m 1 t Vt2 +1 + t Vt7 + t
t m f ' ( t ) = l + . —; > , >0, Vt
v v + i Vt2 + i vV + 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 / x 2 + l c o 3 = v / x 2 + l c i > x 2=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
-BDHSG DSGT12/1- 19

Vi du 9: Giai he phuong trinh:
Ta cd x y - y + i
1
x - l = y ( y - l )
< y3 - l = z ( z - i )
z3 - l = x ( x - l )
Giai
l Y 3 3 1
- +—>—>—=
2) 4 4 8
x >
Tuong tu y, z > - Dat f(t) = V l , t > -1 thi
2
f '(t) = 2t - 1 > 0 nen f dong bien tren (—; +oo)
2
Ta cd he < 3 2
y = z
-y + l
z + 1 co <
z3 = x2 - X + 1
f(y)
f(z)
f(x)
3 Z> X.
z3 > X
y3 > z3 => y > z.
Gia su x > y thi f(x) > f ( y ) :
nen f(z) > f(x)
Do do x > y > z > x: vd li.
Tuong tu x < y: vo l i nen x = y => x = y = z. Ta cd t 3 = f(t)
o t 3 - t 2 + t - l = 0 o ( t - l ) ( t 2 + 1) = 0 co t = 1.
Vay he co nghiem duy nhat x = y = z = 1.
x3 - 2 x + l = 2y
Vi du 10: Giai he phuong trinh
y -2y + l = 2z
z2 -2z + l = 2x
Giai
Ta cd 2y = x2 - 2x + 1 = (x - l ) 2 > 0
y > 0. Tuong tu z, x > 0.
Dat f(t) = t z - 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)
=> g(y) < g(z) < g(x) =o y < z < x nen x = y = z.
Ta co phuong trinh: t 2 - 4t + 1 = 0 nen chon nghiem: x = y = z = 2 - V3.
-Neu 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 )
=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 = z-
Ta cd phuong trinh t 2 - 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
36x2y-60x2 +25y = 0
Vi d u l l : Giai he phuong trinh: 36y2 z -60y2 +25z = 0
36z2x-60z2 +25x = 0
Giai
60x2
36x2 +25
He phuong trinh tuong duong vdi < z 60y2
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 .
Ta cd: f ' ( t )
(36t2 +25)2 >0, Vt>0.
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(36x2 - 60x + 25) = 0. Chon x = - .
6
'5 5 5
, 6 ' 6 ' 6
Vay tap nghiem cua he phuong trinh la <j (0;0;0);
:8-X3
Vi 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)
[y = ( x - D 4 (2)
-BDHSG DSGTU/1-

Xet ham so f(t) = VtTT - ( t - l ) 2 + t
3 - 8, vdi t > 1.
Ta co f '(t) = - 2 ( t - l ) + 3t2 + 1 = 3 t
2 - 2t + 2
2 vW 2v/Tl
> 0 vdi moi
2Vt- 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)
Vi du 13: Giai he bat phuong trinh:
xu - 3x2 + 9x + - > 0 (2)
3
Giai:
Ta cd (1) o x2 - 12x + 35 < 0 co 5 < x < 7
Xet (2): Dat f(x) = x 3 - 3x2 + 9x + - , D = R
3
f (x) = 3x2 - 6x + 9 > 0, Vx eR nen f(x) ddng bien: x > 5 =o f(x) > 286/3
Do do f(x) > 0, VXG (5 ; 7)
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.
Vi 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.
Vi du 15: Chung minh phuong trinh: x 1 3 - x6 + 3x4 - 3x2 + 1 = 0 cd nghiem
duy nhat.
Giai:
Dat f(x) = x 1 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) = 13x12 - 6x5 + 12x3 - 6x = 13x1 2 - 6x(x - l ) 2 > 0 nen
f dong bien
Bang bien thien:
x —CO 0
y' +
y 1
—OO
Nen f(x) = 0 cd nghiem duy nhat x < 0.
Vay phuong trinh cho cd nghiem duy nhat.
22 -BDHSG DSGT12/1-

Vi du 16: Chung minh rang phucmg trinh 2 x 2 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
khoang [2; +oo).
f ' ( x ) = 2
f
n I ^ x2 ) x(5x-8)
2 x V x - 2 +—, -
2Vx-2
> 0, vdi moi x e (2; +co)
V x - 2
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. Vi 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. Vi ham sd
dong bien tren [2; +co) nen c la nghiem duy nhat cua phuong trinh.
Vi 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) = sin2x + cosx thi ham sd lien tuc tren doan [0; n],
Ta cd f ' ( x ) = 2sinxcosx - sinx = sinx(2cosx - 1), x e (0; TI)
Vi sinx > 0 nen f '(x) = 0 o cosx = — o x = —
2 3
BBT: X TC
0 — Tt
u 3
f ( x ) + 0
f(x)
5
71 ' 71
Ham f dong bien tren doan [0; — ] va nghich bien tren doan [— ; 7i].
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.
Vi du 18: Tim sd nghiem cua phuong trinh x3 - 3x2 - 9x - 4 = 0.
-BDHSG DSGT12/1- 23

Giai
Xet ham so y = x3 - 3x2 - 9x - 4, D = R.
BBT
3x - 6x - 9, y' = 0 co x = - 1 hoac x = 3.
X —GO - 1 3 + 00
y' + 0 - 0 +
y r 1 ^ +00
—GO * - 3 1 "
Dua vao BBT thi phuong trinh y = 0 co dung 3 nghiem.
Vi du 19: Tim sd nghiem cua phuong trinh:
3 = 0.
xx + 2x5 - 2x4 - x3 - 3x2 - 6x
Giai
Phuong trinh tuong duong: (x
3 ) ( x 5 - x 2 - 2 x - 1) = 0.
co x3 + 3 = 0 hoac x 5 - x2 - 2x + 1 = 0 co x =
•tfi hoac x5 - x2 - 2x - 1
Xet phuong trinh: x5 - x2 - 2x - 1 = 0 ^> x5 = (x + l ) 2 > 0.
Do do x5 > 0 => x > 0 o> (x + l ) 2 > 1 o x5 > 1 o x > 1.
Do do nghiem cua phuong trinh x5 - x2 - 2x - 1 = 0 neu cd thi x > 1.
Dat f(x) = x5 - x2 - 2x - 1, x > 1.
f '(x) = 5x4 - 2x - 2 = 2(x4 - 1) + 2x(x3 - 1) > 0.
Do do f ddng bien. V i f ( l ) = -3 < 0 va f(2) = 23 > 0 nen f(x) = 0 co
nghiem duy nhat x 0 > 1.
Vay phuong trinh cho cd dung 2 nghiem.
Vi du 20: Chung minh he
fx2 + y 3 =1
y + x 3
cd dung 3 nghiem phan biet.
Giai
Tru 2 phuong trinh ve theo ve va thay the ta duoc:
x2 ( l - x) - y 2 ( 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 x 2 + y 3 = 1 co x3 + x 2 - 1 = 0.
Dat f(x) = x3 + x 2 - 1, D = R. Ta cd f ( l ) = 1 * 0.
f ' (x) = 3x2 + 2x, f 1 (x) = 0 co x = - - hoac x
3
BBT '
0.
X —CO -2/3 0 +0O
y' + 0 - 0 +
y -23/27
- 1 "
+00
24 -BDHSG DSGT12/1-

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 + x 3 = 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.
Vi du 21: Tim cac gia tri cua m de phuong trinh sau cd dung mpt nghiem
yjx2 + 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) = ,
V(t4 + 3)3
Ma f(0) = /3 va l im f ( t ) = 0 nen cd bang bien thien:
- 1<0.
t 0 +co
f'(t)
f(t) V3" ^
* 0
Tu bang bien thien suy ra cac gia tri can tim cua m la 0 < m < ^3
Vi du 22: Tim m de phuong trinh cd nghiem
m(7l + x2 - V l - x 2 + 2) = 2 V l - x 4 + V l + x2 - V l - x 2
Giai
Dieu kien - 1 < x < 1. Bat x = Vl + x2 - V l - x 2 thi t > 0
va t 2 = 2 - 2 V l - x 4 < 2, dau "=" khi x 2 = 1. Do dd 0 < t <
PT:m(t + 2) = 2 - t 2 + 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
w t + 2 (t + 2)2
tren [0; V2 ] . Dieu kien cd nghiem:
min f(t) < m < max f(t) o f( &) < m < f(0) co V2 - 1 < m < 1.
Vi du 23: Tim m de phuong trinh sau cd 2 nghiem phan biet:
Vx2 +mx + 2 = 2x + 1
Giai
PTco
2x + 1 > 0
x2 +mx + 2 = (2x + l ) 2 <=> 3x" + 4x - 1 = mx, x >
Vi X = 0 khong thoa man nen:
-BDHSG DSGT12/1-
3xz + 4 x - l
m, x >
25

. 3x2 + 4 x - l 1 3x2 +l
Xet f(x) = , x > —2 ,' x * 0 thi f" ("x ) x2
Lap BBT thi dieu kien phuong trinh cho cd 2 nghiem phan biet la
1 9
f(x) = m cd 2 nghiem phan biet x > — . x 0 co m > —
v 2' 2
Vi du 24: Tim m de phuong trinh cd nghiem:
(4m - 3) Vx + 3 + (3m - 4)Vl - x + m - 1 = 0
Giai
^.x , „ 3Vx + 3 + 4rji - x + 1
Dieu kien -3 < x < 1. PT co —. ——. = m
4Vx + 3 + 3V1 - x + 1
Ta cd (Vx + 3)2 + (Vl-x)2 = 4 nen dat:
i 2t / „ 1-t2
Vx + 3 = 2sin(p = 2. -, VI - x = 2coscp = 5-
1 + t 2 1 + t 2
Vdi t = tan ^.0<cp<-. 0<t< 1.
2 2
DX 7t2-12t-9 ,m _ 7t*-12t-9 n<rt<r1
PT co m = —5 Dat f(t) 5 . 0 < t < 1.
5 t 2 - 1 6 t - 7 ' 5t2 - 1 6 t -7
Ta cd f '(t) = ——^ < 0 nen f nghich bien tren doan [0; 1], do
( 5 t 2 - 1 6 t - 7 ) 2
7 9
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
26 -BDHSG DSGT12/1-

x z
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.
(_x)2
Suy ra vdi moi x < 0 ta cd cos(-x) + 1 > 0.
2
Vi du 26: Chung minh: sinx > x . Vx > 0.
6
Giai
x3
BDT: x sinx > 0, Vx > 0
6
x3
Xet f(x) = x sinx thi f lien tuc tren [0; +x>)
6
2
x
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)).
-BDHSG DSGT12/1- 27

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 -
- -2 = (cosx - — )2>0.
cos x cos' x
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 +
2cos3x- 3cos2x + l (cosx - l ) 2 (2 cos x + 1)
co^x cos x cos X
>0
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. Vdi 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 I 4
x
f ( x ) = cos X-- t a n x x-sin xcosx 2x-sin2x
2 2
X COS X
2x2 cos2 x
Vi 0 < x < - nen 0 < 2x < - sin 2x < 2x do dd f ' ( x ) > 0 nen f ddng
28 -BDHSG DSGT12/1-

bien tren V suy ra f(x) < f ( - ) = - => dpcm
v 4 4 ru
Vi du 29: Chung minh bat dang thuc:
a) b.tana > a.tanb vdi 0 < a 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 Y X = cos2 x ; = x-sinxcosx = 2x-sm2x
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
tren [0; - ). V i 0 < a f(a) < f(b) => dpcm.
2 2
b) Xet ham s6: f(t) = vdi 0 < t < —
; w t 4
, . tcost-sint cost(t-tant)
Ta co f (t) = -2 = -2
Neu 0 < t < - thi do tant > t f '(t) < 0.
2
Neu - < t < rc thi cost < 0 va sint > 0 => f '(t) < 0.
2
5TI
Neu n < t < — thi do cost < 0; tant < t => f '(t) < 0.
4
Do do f '(t) < 0, 0 < t < — nen f la ham so nghich bien tren khoang (0; — )
4 4
^ ., , -x . n ^ 5TT sin(x + 2y) sinx
Tu gia thiet c o 0 < x < x + 2 y< — — <
4 x + 2y x
Do x > 0 va x + 2y > 0 nen tir do cd
xsin(x + 2y) < xsinx + 2ysinx <eo x 2cos(x + y)siny < 2ysinx
=o dpcm (vi x > 0 va x + 2y < — => y < — ^> siny > 0).
Vi du 30: Chung minh cac bat dang thuc sau:
a) a4 + b4 + c4 + d4 + 2abcd - (a^2 + a2c2 + a2d2 + b2c2 + b2d2 + c2d2) > 0 vdi 4 s6
a, b, c, d duong.
-BDHSG DSGT12/1- 29

^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
2VTTI 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
Vi 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:
x x
.sinx.3 ,„ TC. sinx ,„ TC.
( ) > cos x, x e (0;-) <o — > x, x e (0; - )
x 2 Vcosx 2
Xet ham sd F(x) = s™_?_ - x, x e [ 0;—)
Vcosx 2
, . 2cos2x-3cosx.Vcosx + 1
Ta co F (x) = ==
3 cosx. Vcosx
Xet G(t) = 2t2 -3tVt +l,te[0;l] thi G'(t) = 4(t - Vt) < 0, Vt e [0;l] nen
G(t) nghich bien do do G(t) > G( 1) = 0, Vt e [0; 1]
Suy ra F'(x) >0, Vx e [0;- ) nen F(x) ddng bien
2
Do do F(x) > F(0) = 0, Vx e [0; -).
2
30 -BDHSG-DSGT12/1-

Vi du 32: Chung minh: (x + l ) c o s— xcos->l, Vx > V3
6 v ; x+1 x
Giai
n • 9 TC
2sm2
n n T _ rc(2x + l) rc rc
BDT co 2xsin — sm > 1 - cos —
2x(x + l) 2x(x + l) x + 1 2(x + l )
. rc(2x + l) rc . 2
71
CO xsin —2x (x + l ) sin 2x(x + l ) > sm 2(x + l )
r -v PS n n rc(2x + l) rc
Vi 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 = . t > 0 thi (2) co xsint > sinxt
2x(x + l )
Xet f(t) = xsint - sinxt, t > 0, f '(t) = x cost - xcosxt = x(cost - cosxt)
Vi 0 < t < xt < - => f Yt) > 0 vdi t > 0.
2
=> f(t) ddng bien tren [0; +oo) =o f(t) > f(0) = 0 =o (2) dung
Tu (1), (2) =o dpcm.
Vi du 33: Cho x, y, z>0vax + y + z= l.
7
Chung minh: 0 < xy + yz + zx - 2xyz < —
2 V
Giai
Gia sir z la sd be nhat thi 0 < z < — .Ta cd
3
T = xy+ y + zx - 2 xyz = xy(l - 2z) + (x + y)z > - xy +(x + y)z > 0
3
Va cd T = ( l - 2 z ) + (x + y)z
= - ( 1 - z)2(l - 2z) + (1 - z)z = - ( - 3 z 3 + z2 + 1)
4 4
Xet f(z) = -3z3 + z2 + 1, 0 < z < - thi
f Yz) = - 6z2 + 2z = 2z(l - 3z) > 0 tren f(z) ddng bien tren [0; - ]. do do
3
T = f(z)<f(-)= —
; . 3 27
-BDHSG DSGT12/1- 31

C. BAI LUYEN TAP
Bai 1: Tim khoang don dieu cua ham sd
x2 + 4x - 2
a) Y = x 2 + 1 - b) y = (x + 2)5(2x + lY
a) y = ——
x + 1
c) y = 2x3 + 3x2 + 6x -13
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
x x + 1 x2 X
a) y = V,x = - x + 1 b) y = —e c) y 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:
> m(x + l ) 3 ,
a ) y =— -.m>0 b)y = ln(x + V4 + x 2 )
x - x + 1
HD: b) y ' = 1 >Q,yx
V4 + x2
Bai 6: Chung minh ham sd:
a) y = x - ex nghich bien tren khoang (0; +oo)
x2 - 4x + 3
b ) v = — i — luon dong bien tren tirng khoang xac dinh.
HD: 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:
1 1 3
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
32 -BDHSG DSGT12/1-

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
. 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.
4
D S : a ) m > l . b)m> l.
Bai 14: Tim dieu kien de y = a.sinx + b.cosx + 2x dong bien tren R.
DS: a2 + b2 < 4.
Bai 15: Giai phuong trinh
a) v x 2 + 15 + 2 = Vx2 +8 + 3x
b) v
/ (x + 2 ) ( 2 x - l ) +7(x + 6 X 2 x - l ) = 4 + 3(7x + 2 + Vx + 6)
DS: a) x = 1. b) x =7
Bai 16: Giai bat phuong trinh :
a) Vx + Vx + 7 < 9 - V x - 5 b) x5 + x3 > V l - 3 x - 4
DS: a) 5 < x < 9 b) x < - 1
Bai 17: Giai bat phuong trinh:
a) V2x3 +3x2 +6x + 16 > 2 V3 + V 4 - x
b)Vx + l + 2Vx + 6 > 20 - 3Vx + 13
-BDHSG DSGT12/1- 33

DS: a) 1 < x < 4. b) x > 3.
1
Bai 18: Giai he: a)
DS: a)x = y = l ; x = y
1
x — = v —
x y
2y = x 3 + 1
1 + V5
^ [cot x - cot y = x - y
| l 5 x + 7y = rc;.x,y £ (0, rc)
b)
71
X = 22
TC
y =
22
Bai 19: Chung minh phuong trinh: x3 + x 2 + 12x - V3 = 0 co nghiem.
HD: Ham y = x3 + x 2 + 12x - V3 don dieu tren R.
Bai 20: Chung minh phuong trinh x3 + 2x3 - x2 + x - 1 = 0 cd nghiem duy nhat.
HD: Ham y = x5 + 2xJ - x2 + x - 1 don dieu tren R.
Bai 21: Cho so tu nhien n chan va a > 3. Chung minh phuong trinh:
(n + 1) x n + 2 - 3(n + 2)xn + 1 + a n + 2 = 0 vo nghiem
HD: y ' = (n + l ) (n + 2).xn _ 1.(x - 3)
Bai 22: Cho tam giac ABC cd cd canh a B > C.
Bai 23: Chung minh vdi moi m > 0 thi phuong trinh cd 2 nghiem phan biet:
x2 + 2x - 8 = v
/m(x-2)
Bai 24: Tim m de phuong trinh cd 2 nghiem phan biet: mVx2 + 4 - x + 1 = 0.
DS: -2^<m<-l
2
Bai 25: Tim m de phuong trinh cd nghiem: 3Vx - 1 + mVx + 1 = 2%/x2 - 1 .
DS: -Km< -
3
Bai 26: Tim m de phuong trinh: x4 - (m - l)x3 + 3x2 - (m - l)x + 1 = 0 co
nghiem.
DS: m< —; m > —.
2 f 2
Bai 28: Tim m de phuong trinh: 2.cosx.cos2x.cos3x + m = 7.cos2x co dung
1 nghiem thuoc doan [-3TC/8:-TC/8].
HD: Dat t = cos2x vdi -— < t < —
2 2
Bai 29: Tim m de bat phuong trinh: x2 + (1 - x 2 ) 3 ' 2 > m cd nghiem.
DS: m < 1 .
Bai 30: Tim a, b de bat phuong trinh: 3x4 + 8x + ax +b > 0 cd nghiem x
thuoc [-1 ;1]
34 -BDHSG DSGT12/1-

HD: Tfnh dao ham cua ham sd y = 3x4 + 8xJ + ax2 +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 + ... + x 2 n + 2007x2 n + 1 = 1999 co nghiem duy nhat.
Bai 32: Chung minh phuong trinh: x 1 3 - x6 + 3x4 - 3x2 + 1 = 0 cc nghiem
duy nhat.
Baj 34: Giai phuong trinhVx2 + 15 = 3x - 2 + Vx2 + 8
DS: x = 1
Bai 35: Chung minh cac bat dang thuc: sin a sin b vdi 0 < a 2
Bai 37: Chung minh rang vdi ba so duong a. b. c bat ki thi:
|c~ a b c
>—+—+-
a b c a
HD: Chung minh: x + m - 1 > mx vdi moi x > 0, m > 1
3/3x = cos(rcy)
3V3y = COS(TIZ)
3V3z = cos(rc t)
3V3t = cos(rcx)
Bai 38: Giai he phuong trinh:
Bai 39: Chung minh he phuong trinh:
2x2
2y2
a
• y = -
y co nghiem duy nhat vdi
a> 0
Bai 40: Chung minh bat dang thuc: |sin a - sin p| < [a - p| vdi moi a , p
Bai 41: Chung minh bat dang thuc:
10a9(b - a) a > 0
a sina - p sinP > 2 (cosP - cosa) vdi 0 < a < p < —
-BDHSG DSGT12/1- 35

> cosx > vai 0 < x < -
UanxJ 2
Bai 44: Chung minh bat dang thuc: |cos2x . sin4x + cos2x| < 1
x3 x5
Bai 45: Chung minh: sinx < x + —,Vx > 0
5 3! 5!
x2 x4
Bai 46: Chung minh: cosx>l + — ,Vx
5 2! 4!
Bai 47: Chung minh: x4 + y4 > - vdi x, y thoa x + y = 1
8
Bai 48: Chung minh: a2 + b2 > 1 vdi a, b thoa a > b3 + b2 + |b| + 1
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
> 2 cos2 — vdi AABC khong tu
sinA + sinB + sinC „ 2 rc
cosA + cosB + cosC 8
Bai 52: Chung minh vdi tam giac ABC thi cd:
v A B C 3>/3
a) cos— + cos— + cos— <
2 2 2 2
ABC
b) tan— + tan— + tan— > S
2 2 2
Bai 53: Cho a, b, c> 0 va thoa 21ab + 2bc + 8ca < 12.
„, , . , 1 2 3 15
Chung minh: — + — + — > —
a b c 2
Bai 54: Cho a, b, c > 0 va thoa a + b + c = 1.
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 + bcs + c".as > as.br + bs.c" + cs.a'
36 -BDHSG DSGT12/1

§ 2 . C U C T R I C U A H A M S O
A. K I EN T H t f C CO BAN
Cho ham s6 f xac dinh tren tap hop D ( D c R ) ya x 0 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) { x 0 } .
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) { x 0 } .
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 x 0 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 x0 .
B. PHAN DANG TOAN
DANG 1: CUC DAI, CUC TIEU
Quy tac 1
l . T i m f ' ( x )
2. Tim cac diem Xj (i = 1, 2,...) tai dd dao ham cua ham so bang 0 hoac
ham sd lien tuc nhung khdng cd dao ham.
3. Xet dau f (x). Neu f '(x) ddi dau t i i " - " sang "+" khi x qua diem x, thi
ham sd dat cue tieu tai Xj, cdn neu f '(x) ddi dau tir"+" sang "-" khi x
qua diem Xj thi ham sd dat cue dai tai Xj.
-BDHSG DSGT12/1- 37

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)vatinhf "(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 im 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 da thuc: chia dao ham y = q(x). y' + r(x) =5> yo = r(x0 )
Ham huu ti: dao ham rieng tu, rieng mau
y = f(x)=^.hiy„=^4 = ^4
v(x) v(x0 ) v'(x0)
Dac biet: Vdi 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.
Vi du 1: Tim cue tri cua cac ham sd sau:
a)f(x) = - x 3 + 2x2 + 3 x - l
3
c) y = x4 - 5x2 + 4
a) D = R. Tacdf'(x) = x2 + 4x + 3
b ) f ( x )= - x 3
3
x2 + 2x - 10
d) y = (x + 2)2(x - 3)2
Giai
f ' ( x )
BBT
0<=> x z + 4x + 3 = 0 <=> x = -3 hoac x = - 1 .
x —00 -3 - 1 +00
y' + 0 0 +
y
—00
. - 1 _
~* -7/3"
+00
Vay ham sd dat cue dai tai diem x = -3, f(-3) = - 1 va dat cue tieu tai
7
diem x = - 1 , f ( - l ) = —
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)
38 -BDHSG DSGT12/1-

y' = 0 <=> x = 0 hoac x = ± J | ; y" = 12x2 - 10.
Ta co y" 20 > 0, y"(0) = -10 < 0 nen ham so dat cue dai tai x = y.
(5 9
yco = 4 va dat cue tieu tai x = ±,/— , ycr = —
V 2 4
d) y' = 2(x + 2)(x - 3)3 + 3(x + 2)2 (x-3)2 = 5x(x + 2)(x - 3)2
Ta cd y' = 0 <=> x = —2 hoac x = 0 hoac x = 3.
BBT X -co _3 0 3 +oo
y' + o - 0 + 0 +
y o. J)-^00
Vay diem cue dai (-2; 0) va cue tieu (0; -108).
Vi du 2: Tim cue tri cac ham sd
a) f(x) = | x2 + 3x - 4 | b) f(x) = | x | (x + 2)
Giai
x2 + 3x - 4 , x < -4 hay x > 1
- x 2 - 3 x + 4, - 4 < x < l
a) D = R, y
[2x + 3, x <-4hay x > l
I -2x - 3 , - 4 < x < l
y =
BBT - 4 -3/ 1 + 00
+
CD-CT
^CT
Vay ham sd dat CD I ~ ^ . C T ( - ^ °). C T ( 4 ; ° )
b) Ham sd f lien tuc tren R. Ta cd: f(x) =
-x(x + 2)
|x(x + 2)
Vdi x < 0, f '(x) = -2x + 2; f '(x) = 0 o x = - l .
Vdi x > 0, f'(x) = 2x + 2 > 0.
BBT
khi x < 0
khi x > 0
+
- L
o
0 +00
Vay diem CD(-1; 1), CT(0; 0).
-BDHSG DSGT12/1- 39

Vi du 3: Tim cue tri cua ham sd
a) y
c) y:
x - 2x + 3
x + 1
x + 1
b)y:
2x + l
x - 5
x 2 + ;
d) y = x - 1 +
Giai
x + 1
a) D = R { - l } . T a c d y ' = x 2 + 2 x
2
5 - y ' = 0 c o x = - l ± ^ 6
(x + 1)2
BBT X -00 - 1 - -1 -1+ +00
y' + 0 - 0 +
y - 4 - 2 V6
—co^^* —00
+00 +°o
^ 2 7 6 - 4 ^
Vay diem CD(-1 - 76 ; - 4 - 2 > / 6 ), CT(-1 + &;2S -4).
b) D = R {5}. Ta cd y' = —=- < 0, Vx * 5 nen ham so nghich bien tren
( x - 5 )
tung khoang xac dinh, do do khdng cd cue tri.
, „ _ „ . , x2 + 8 - 2 x ( x + l ) - x 2 - 2 x + 8
c) D = R.Tacoy = ( x 2 + g ) 2 =
y' = 0<=>x = -4 hoac x = 2.
BBT
(x2 + 8)2
—00 - 4 +O0
0 + 0
.1/4,
-1/8.
Vay ham sd dat cue dai tai x = 2, yco = — va dat cue tieu dat x = - 4.
4
1
y C T = - -
d) D = R {-l},y' = 1 - 7-^TT,y' = Oox = Ohoacx = -2.
y -
(x+D2
y"(0) = 2 > 0, y"(-2) = -2 < 0.
(x + 1)3
Vay ham so dat cue dai tai x = - 2, yco = - 4 va dat cue tieu tai x = 0, ycr = 0.
Vi du 4: Tim cue tri cua cac ham so sau:
a) y = x V 4 - x 2 b) y = Vx2 - 2 x + 5 c) y = x + Vx2 - 1
Giai
a) Dieu kien -2 < x < 2. Vdi - 2 < x < 2 thi
40 -BDHSG DSGT12/1-

BBT: x
V 4 - x ' V4-x'
-2 -VI V2
0 + 0
CD
CT-Vay
ham sd dat cue dai tai x = V2 , yCo = 2 va dat cue tieu tai x
y C T = -2.
b) D = R. Tacdy'
Vx2 - 2 x + 5
, y ' = 0<=>x= 1.
BBT X —00 1 +00
y' 0 +
y +00 +c°
Vay ham sd dat CT(1; 2).
c) D = (-oo;-l] u[l;+oo).Vdix<-lhoacx>lthiy'=l +
y' > 0 co
y'<Oco
Vx2 ^ !
> —1 <=> x > Vx2 - 1 o x > l , x 2 > x 2 - l o x > l
Vx2 ^ !
< - 1 CO x < - V x 2 - 1 CO x < - 1 , x2 > x2 - 1 CO X «
BTT X —00 — l +00
y' - +
y
Vay ham so khdng cd cue tri.
Vi du 5: Tim cue tri cua ham sd:
a) y
V x 2 - 6
b ) y = v V ( x - 5 )
Giai
a) Tap xac dinh D = (-co; - V6 ) u (^6 ; +<x>)
-BDHSG DSGT12/1-

. Vx2 - 6 = 3x2(x2 - 6 ) - x 4 _ 2x2(x2 -9)
x2-6 sl^W " vV-6)3
y' = 0»x = 0 hoac x = ±3.
BBT X -co _3 -A/6 V6 3 +oo
y' + 0 - - 0 +
y -9A/3
y
—00 —oc •
+00 +00
9V3
Vay ham so ctat cue dai tai x = -3 va yco = —9 V3 , dat cue tieu tai x = 3
vayCT= 9 A/3
^ „ x r , . „ , , 3r r 2(x-5) 5(x-2)
b) D = R. Vdi x * 0 thi y' = v x 2 y' = 0 <to x = 2. Bang bien thi e+n 3 ^ 3Vx
0 +00
+
,+00
—00 -3V4-
Vay ham sd dat cue dai tai x = 0, yco = 0 va dat cue tieu tai x = 2,
yCT = -3 A/4
Vj du 6: Tim cue tri cua ham so
a) y = x - sin2x + 2
a) D = R, y' = 1 - 2cos2x
b) y = 3 - 2cosx - cos2x.
Giai
1 TC
y' = 0 <=> cos2x = — o x = ±— + kn, k e Z; y'' = 4sin2x.
2 6
Ta cd y"(-— + kre) = 4sin(-—) = -2 A/3 < 0 nen ham sd dat cue dai tai
6 3
t , , .A/3 0
diem x = — + kn, k e Z, yen — + kn + — + 2.
6 t> z
Ta cd y"(— + kit) = 4sin— = 2 A/3 > 0 nen ham so dat cue tieu tai cac
6 3
diem: x = — + kn, k e Z; ycr = — + kTt - — + 2.
6 * 6 2
b) y' = 2sinx + 2sin2x = 2sinx(l + 2cosx):
y' = 0 o
sin x = 0 2rc
cosx = -
i c=> x = kre hoac x = ± — + 2kTt, k e Z .
42 -BDHSG DSGT12/1-

y" - 2cosx + 4cos2x
Ta co y"(k7t) = 2coskTC + 4cos2kTC = 2coskTC + 4 > 0, voi moi k e Z , nen
ham so eta cho ctat cue t i iu tai cac diem x = kre, ycr = 2 - 2coskTC bang 0
khi k chan va bang 4 khi k le.
T , „,, 2rc , . _ 2TC . 4rc 2rc
l a c o y ( ± — + 2k7i) = 2 cos — + 4cos—=6cos— =
3 3 3 3
2rc 9
dat cue dai tai diem: x = ± — + 2k7i, k e Z, yco = —
3 2
-3 < 0 nen ham so
Vi du 7: Chung minh ham sd sau khdng cd dao ham tai x = 0 nhung dat cue
tri tai diem dd.
-2x
a ) f ( x ) = | x | b)f(x)
sin-khi
x < 0
khix>0
Giai
a) Ham sd xac dinh va lien tuc tai R. Ta cd:
f - 1 khi x < 0
f(x)
- x khi x < 0
x khi x > 0
f ' ( x ) =
1 khix>0
Do do ham sd khdng cd dao ham tai x = 0 va BBT:
X —CO Q +0°
y' - +
y
" *• o ^ *
Vay ham sd dat CT(0; 0).
b) Ham so xac dinh va lien tuc tren R. Ta cd
-2x khi x < 0
f ' ( x ) 1 1 x nen lim f ' ( x ) = 0 * l im f ' ( x ) = —, do do f
— cos— khix>0 *->0" x^ 0 + 2
2 2
khong cd dao ham tai x = 0 va BBT tren khoang (-TC; TC).
X -TC Q "
y' + -
y 0
Vay ham sd dat cue dai tai x = 0 va yco = y(0) = 0.
Vi du 8: Chung minh rang ham sd ludn luon cd cue dai va cue tieu:
a) y = x 3 + ax2 - (1 + b2)x + 2a + b - 3ab
b) y = (x - a)(x - b)(x - c) vdi a < b < c.
_ x +(m + 2)x+m +2
c)y =
x + m
-BDHSG DSGT12I1- 43

Giai
a) D = R. Ta co y' = 3x2 + 2ax - 1 - b2
A' = a2 + 3(a + b2 ) > 0, Ve
biet x i , X2. Bang bien thien:
- Va, Vb nen y' = 0 luon luon co 2 nghiem phan
X —00 Xl X2 +00
y' + 0 - 0
y
C T ^
+00
Vay ham so luon luon co mot cue dai va mot cue tieu.
b) D = R. y' = (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b)
= 3x2 - 2(a + b + c) + ab + be + ca.
A' = (a + b + c)2 - 3(ab + be + ca) = a2 + b2 + c2 - ab - be - ca
= - [(a - b)2 + (b - c)2 + (c - a)2] > 0 voi a < b < c.
2
Do do y' = 0 co 2 nghiem phan biet va ddi dau 2 lan khi qua 2 nghiem
nen luon ludn cd mot cue dai va mot cue tieu.
x2 +2mx + 2 m - 2
c) D = R {-m}. Tacd: y* =
(x + m)2
Xet ham sd g(x) = x + 2mx + 2m - 2.
Ta cd A' = m2 - 2m + 2 > 0, Vm va g(-m) = - m 2 + 2m - 2 * 0 , Vm nen
y' = 0 ludn cd hai nghiem phan biet khac - m , y' doi dau hai lan khi qua 2
nghiem, vay ham sd luon luon cd cue dai va cue tieu.
Vi du 9: Tim a de dd thi ham so y = — x2 + — ax2 + x + 7 cd 2 cue tri va
3 2
x2 x2
hoanh do 2 diem cue tri cua ham sd do thoa man —j + —| > 7.
x
Giai
D = R. Ta cd y' = x 2 + ax + 1. V i y' la ham sd bac hai nen ham sd co 2
cue tri khi va chi khi y'(x) = 0 cd hai nghiem phan biet
c o A > 0 < = > a 2 - 4 > 0 < = > a < - 2 hoac a > 2. Goi x, va x 2 la hai nghiem
cua y'(x) = 0 thi S = Xj + x2 = -a, P = x^2 = 1.
Ta cd: — + •
( Q2
>7 co
f 2
X l X2
S2-2P
v x2
( Jl Jl
2>7co x1+x2
XjX^
>9
>9co(a2-2)2>9coa2 >5
Chon gia tri a < - -JE hoac a > yfE
Vi du 10: Tim cac sd thuc p va q sao cho ham sd f(x) = x + p
44
x + l
BDHSG DSGT12I1-

ctat cue dai tai diem (-2; -2).
Giai
Ta cd f '(x) = 1 — . vdi moi x * - 1 .
x + 1
Neu q < 0 thi f '(x) > 0 vdi moi x * - 1 . Ham so khdng cd cue dai, cue
tieu (loai).
Neu q > 0 thi phuong trinh: f '(x) =
(x + 1)'
0 cd hai nghiem
phan biet Xi = - 1 - <Jq va X2 = - 1 + yfq .
BBT:
X —oo -i - V q 1 -1 + y[q +C0
y' + 0 - 0 +
y
Ham sd dat cue dai tai diem (2; -2) khi va chi khi
f - l - ^ = - 2 0 W = l 0 [ q = l
[f(-2) = -2 [p = l [p = l
Vi d u l l : Tim m de ham so:
a ) y = ~ ( m + 5m)x + 6mx + 6x - 5 dat cue dai tai x = 1.
x2 + ( l - m ) x - 2
b)y
x + m
dat cue tieu tai x = 0.
Giai
a) D = R. Ta cd y' = -3(m2 + 5m)x2 + 12mx + 6
Neu ham sd dat cue dai tai x = 1 thi y ' ( l ) = 0
-3m2 - 3m + 6 = 0 <=> m = 1 hoac m = - 2 .
Ta cd y" = -6(m2 + 5m)x + 12m
Vdi m = 1 thi y" = -36x + 12 nen y " ( l ) = -24 < 0, ham so dat cue dai tai
x = 1.
Vdi m = - 2 thi y" = 36x - 24 nen y " ( l ) = 12 > 0, ham so dat cue tieu tai
x = 1 (loai).
Vay vdi m = 1 thi ham sd dat cue dai tai x = 1.
b) D = R {-m}.Tacdy'
x2 + 2mx - m2 + m + 2
(x + m)2
Neu ham sd dat cue tieu tai x = 0 thi y'(0) = 0
=> - m 2 + m + 2 = 0 = >m = - l hoac m = 2.
Vdi m
-1 thi y
x2+2x-2
:x + 3 + -
2
Do dd y"
( x - 1 ) 3
-BDHSG DSGT12/1-
x - 1 x-l
> y"(0) = -2 < 0
>y' = i -
( x - l f
45

=> x = 0 la diem cue dai cua ham sd: loai.
Tr*. ~ x2 - x - 2 4
Va im = 2 thiy = x + 2 = x - 3 + x + 2 =>y' = l - (x + 2)2
g
Do do y" = —. y"(0) = 1 > 0 nen x = 0 la diem cue tieu cua ham
(x + 2)3
so. Vay gia tri can tim m = 2.
Vi du 12: Tim ham sd f(x) = ax3 + bx2 + cx + d sao cho ham so f dat cue tieu
tai diem x = 0, f(0) = 0 va dat cue dai tai diem x = 1, f ( l ) = 1.
Giai
Ta cd f '(x) = 3ax2 + 2bx + c. Vi f(0) = 0 nen d = 0. Ham sd dat cue tieu
tai diem x = 0 nen f '(0) = 0 do dd c = 0.
Vi f ( l ) = 1 nen a + b = 1. Ham sd dat cue dai tai diem x = 1 nen f'(1) = 0
do do do 3a + 2b = 0. f a + b = i [a = -2
Ta cd he phuong trinh + 2b = 0 ° jb = -3
Thir lai: f(x) = -2x3 + 3x2, f '(x) = -6x2 + 6x, f "(x) = -12x + 6.
f "(0) = 6 > 0. Ham sd dat cue tieu tai diem x = 0 (thoa man).
f "(1) = -6 < 0. Ham so dat cue dai tai diem x = 1 (thoa man).
, ' a. sin. x cos x 1 *
Vi du 13: Tim a de ham sd y = dat cue tri tai 3 diem thuoc
a cosx
khoang (0; ~ ) .
4
Giai
r^ - 1 - 71 , 1 -r - . a-smx « , _ .
Dieu kien x * — + kre. Ta co y = — nen y = 0 <=> sinx = a.
2 a cos x
, , -sin2 x + 2asinx - 1
1 a co y ' = =
a cos x
Vdi sinx = a thi a = sinx thi y" = * 0, do dd ham so dat cue tri
sinxcosx
9rt
tai 3 diem thudc khoang (0; — ) <=> sinx = a cd 3 nghiem thuoc khoang
,„ 9TC . . . rc 3rc . „ . . /2
(0; — ) { - ; — } co 0 < a < -—-
v 4 2 2 r 2
Vi du 14: Tim m de ham so:
a) y = x +2mx + l - 3 m _ ^ ^ ^ m ^ ^ n & m yx ^ ^ ^ ^
x - m
= mx2+(2-4m)x + 4 m + l £ , 2 ^ ^ y . M g j . ^ ^ fr. ^ ^
x - l
Giai
x 2 -2mx + m2 - 1
a) Dieu kien x * m. Ta cd y' : -2
' • (x-m)
46 -BDHSG DSGI12/1-

Do thi co 2 cue tri d 2 phia cua true tung co y' = 0 co 2 nghiem X,, x2 * m va
x,x2 < 0 e o m 2 - l < 0 c o - l < m < l .
b) Dieu kien: x*l Ta co y' = mX' ~ 2mX " 3 , dat g(x) = mx2 - 2mx - 3.
( x - l )
D6 thi co 2 cue tri co m * 0, A' > 0, g(x) * 0 co m < -3 hoac m > 0.
3
Ta cd x, + x2 = 2, X]X2 = nen yCo- ycr < 0.
m
o (2mx, + 2 - 4m)(2mx2 + 2 - 4m) < 0
o 4m2x,x2 + 2m(2 - 4m)(x, + x2) + (2 - 4m)2 < 0. 1
co -12m + 2m(2 - 4m) + (2 - 4m)2 < 0 co 4 - 20m < 0 co m > - .
x2 - 2mx + 2
Vi du 15: Cho ham so y = vdi m la tham sd.
x - l
a) Tim m de do thi ham so cd hai diem cue tri A va B.
b) Chung minh rang khi do dudng thang AB song song vdi dudng thang
2x — y - 10 = 0. Tinh khoang each giua 2 cue tri.
Giai
, x2 - 2x + 2m - 2
a) DK: x * 1. Ta co y = =
( x - l ) 2
Dieu kien cd 2 cue tri la A' > 0 va g ( l ) * 0.
3
c o 3 - 2 m > 0 v 3- 2 m * 0 c o m < -
2
b) Ta cd A ( l - V3- 2m ; 2 - 2m - 2 V 3 - 2m )
B(l + S - 2m ; 2 - 2m + 2 V 3 - 2m )
He so gdc cua dudng thang AB la:
. y(x„)-y( x , ) 4/3 - 2m „ , _ in
k = 2— 1 = — p = = = 2 . Ta co 2x - y - 10
x 2 - X j 2V3-2m
co y = 2x - 10 nen cd he so gdc bang nhau => dpcm.
q
Va AB2 = 4(3 - 2m) +8(3 - 2m) = 12(3 - 2m) ^>,AB = 2x79 - 6m , m < -
Vi du 16: Viet phuong trinh dudng thang di qua diem cue dai, cue tieu cua
dd thi:
a) y = x3 + 3mx2 + 3(m2 - l)x + m3 - 3m
x2 - 2mx + 5m - 4 - m2
b)y
x - 2
Giai
a) y' = 3x2 + 6mx + 3(m2 - 1), A' = 1 > 0, Vx nen do thi luon luon cd CD va
CT vdi hoanh do x i , x2 .
Lay y(x) chia cho y'(x) ta cd: y(x) = 11 x+—|y'(x) - 2(x + m).
-BDHSG DSGT12/1- 47

Do do: y, = y ( X l ) =
va y2 = y(x2) = f ^ x , + ^ j y ' ( x 2 ) - 2(x2 + m) = -2(x2 + m)
nen dudng thang qua CD, CT la y = -2(x + m).
b) DK: x * 2. Ta cd y = x - 2(m - 1) + m - m
x - 2
nen y' — 1 -m - m 2 ( x - 2 ) 2 - ( m - m 2)
( x - 2 ) 2 ( x - 2 ) 2 Tu dd suy ra dieu kien co CD
va CT l am — m > 0 < = > 0 < m < l .
Goi x i , X2 la hoanh dp CD, CT thi xi < 2 < x2 . Ta cd
m m2
y(xj = x, - 2(m - 1) + , ^v = x, - 2(m - 1) + (xj - 2) = 2x, - 2m.
y(xa) = x2 - 2(m - 1) +
6q-2)
m - m
(xj-2)
x2 - 2(m - 1) + (x2 - 2) - 2x2 - 2m
Vay phuong trinh dudng thang qua CD va CT la y = 2x - 2m.
DANG 2: UNG DUNG CUA CUC TRI
Vi du 1: Chung minh rang phuong trinh 2x3 - 3x2 - 12x - 10 = 0 cd nghiem
x = a duy nhat va 3,5 < a < 3,6.
Giai
Xet ham sd f(x) = 2x3 - 3x2 - 12x - 10
BBT:
f ' ( x ) = 6 ( x 2 - x - 2 ) ; f ' ( x ) = 0 o x =
X —OO —1 2 +oo
y' + 0 - 0 +
y +00
" ^ - 3 0 ^ ^
Tu BBT thi phuong trinh f(x) = 0 cd nghiem duy nhat a > 2.
Ta cd f(3,5).f(3,6) < 0 nen 3,5 < a < 3,6 => dpcm.
Vi du 2: Cho ab * 0. Chung minh phuong trinh:
x3 - 3(a2 + b2)x + 2(a3 + b3) = 0 cd 3 nghiem phan biet.
Giai
Xet ham s6 y = x3 - 3(a2 + b2)x + 2(a3 + b3), D = R
y' = 3x2 - 3(a2 + b2), y' = 0 o xl i 2 = ± Ja2 + b2 , (S = 0, P = a
2 + b2)
Vi y' bac 2 cd 2 nghiem phan biet nen cd CD va CT.
48 -BDHSG DSGT12/1-

Ta co: y = - x . y' - 2(a2 + b2)x + 2(a3 + b3 ) nen:
y C D . V c T = (_2(a2 + b 2 ) X l + 2(a3 + b3)) (-2(a2 + b2)x2 + 2(a3 + b3))
= 4(a3 + b 3 ) 2 - 4(a2 + b 2)3 = - 4a2b2(3a2 + 3b2 - 2ab)
= - 4a2b2[2a2 + 2b2 + (a - b)2] < 0
Vay phuong trinh cho luon co 3 nghiem phan biet.
Vi du 3: Chung minh rang dieu kien can va du de phuong trinh x + px + q = 0
cd ba nghiem phan biet la: 4p3 + 27q < 0.
Giai
Xet ham sd f(x) = x3 + px + q, D = R.
Ta cd f '(x) = 3x2 + p; f '(x) = 0 co 3x2 + p = 0.
Vdi p < 0 thi f '(x) = 0 cd nghiem phan biet x = + , do la 2 hoanh do
CD, CT. Dieu kien can va du de f(x) = 0 co 3 nghiem phan biet la
yco-ycT < 0, p < 0 co
2
q p
3
2
Q + 3 P
£ <Q, q < 0 c o 4 p 3 + 27q2<0.
< 0 , p < 0 .
COq' p'
9
Vi du 4:Tim tham sd m de phuong trinh:
x3 - 3mx2 + 3(m2 - l)x - m2 + 1 = 0 cd 3 nghiem duong phan biet.
Giai
Xet y = x3 - 3mx2 + 3(m2 - l)x - m2 + 1 , D = R
y' = 3x2 - 6mx + 3(m2 - 1)
Cho y' = 0 => Xj = m - 1, x 2 = m + 1 (S = 2m, P = m2 - 1)
Do dd ham sd ludn ludn cd CD, CT. Lay y chia y':
y = - (x - m). y' - 2x + m3 - m2 - m + 1.
3
=> y C T . yco = (-2x, + m3 - m2 - m + l)(-2x2 + m3 - m2 - m + 1)
= (m2 - 1) (m2 - 3) (m2 - 2m - 1)
Dieu kien cd 3 nghiem duong phan biet:
f (0) < 0
x
yCT-yCD < 0
CT' XCD > 0
- m 2 + 1 < 0
•• m - l > f ) ; m + l > 0
( m 2 - l ) ( m 2 - 3 ) ( m 2 - 2 m - l ) < 0
Giai ra duoc S < m < 1 + V2.
Vi d „ 5; Tim m de phuong trinh: x3 + mx2 - 3 = 0 cd mot nghiem duy nhat.
Giai
Xet m = 0 thi PT: x3 - 3 = 0 co x = : cd nghiem duy nhat.
Xet m * 0. Dat f(x) = x3 + mx2 - 3, D = R.
-BDHSG DSGT12/1- 49

Ta co f '(x) = 3x2 + 2mx = x(3x + 2m)
f '(x) = 0 co x = 0 hoac x = —51 c6 2 nghiem phan biet.
3
Phuong trinh f(x) = x 3 + mx2 - 3 = 0 co duy nhat mot nghiem khi va chi
khi cue dai va cue tieu cua ham sd cung dau:
f(0)f - 2m
>0 co (—3) 8m3 4m3
0
— - + 3
27 9
> 0
co 8m3 - 12m3 + 81 > 0 co 4m3 < 81 co m < 3.3 - (m * 0).
Vay gia tri can tim: m < 3.3/—.
V4
Vj du 6: Tim m de phuong trinh: 2 | x2 - 5x + 41 = x2 - 5x + m cd 4 nghiem.
Giai
Ta cd: 2 | x2 - 5x + 4 | = x2 ^ 5x + m co 2 | x2 - 5x + 4 | - x2 + 5x = m.
Xet y = f(x) = 2 | x2 - 5x + 4 | - x2 + 5x , D = R
[-3X2 + 15x - 8, l < x < 4
y =
2
[ x ^ - 5 x + 8, x < l v x > 4
f-6x + 1 5 k h i l < x < 4
[2x - 5 hhi x < 1, x > 4
Cho y - 0 => x = -
2
Bang bien thien:
X -oo
Vay dieu kien cd 4 nghiem la 4 < m <
• Cach khac: dat t = x - 5x
Vi du 7: Bien luan theo tham so k ve sd nghiem cua phuong trinh:
2x4 - 1 7 x 3 + 51x2-(36 + k)x + k = 0 (1)
Giai
Vdi moi k thi x = 1 ludn thoa man phuong trinh (1)
50 -BDHSG DSGT12/1-

Ta co (1) o (x - 1) (2x3 - 15x2 + 36x - k) = 0
CO x = 1 hoac 2x3 - 15x2 + 36x - k = 0 (*)
-Truong hop x = 1 la nghiem ciia (*) co k = 23
Khi do (*): 2x3 -15x2 + 36x- 23 = 0 eo (x - l)(2x2 - 13x + 23) = 0
CO x = 1 hoac 2x2 - 13x + 23 = 0 eo x = 1
Vay khi k = 23 thi (1) cd nghiem duy nhat x = 1
-Vdi k * 23, khi dd x = 1 khdng phai la nghiem cua (*) nen so nghiem
ciia (1) bang 1 cdng vdi sd nghiem cua phuong trinh (*)
Xet f(x) = 2x3 -15x2 + 36x thi f '(x) = 6x2 - 30x + 36 = 6(x2 - 5x + 6)
f '(x) = 0 e o x = 2 h a y x = 3
g bien t
X
lien:
—00 2 3 +00
f ( x ) H 0 - 0 V
f(x)
—oo
- 2 8 ^
^ 2 7 ^
- +00
Dua vao bang bien thien ta cd:
- Neu 23 * k < 27, k > 28 thi (*) cd nghiem duy nhat nen (1) cd 2
nghiem phan biet.
- Neu k = 27 hay k = 28 thi (*) cd hai nghiem phan biet nen (1) cd ba
nghiem phan biet.
- Neu 27 < k < 28 thi (*) cd ba nghiem phan biet nen (1) cd bon nghiem
phan biet.
Vi du 8: Tim m de phuong trinh sau cd nghiem
x4 - 6x3 + mx2 - 12x + 4 = 0 (1)
Giai
x = 0 khdng phai la nghiem ciia phuong trinh (1) nen (1)
co x — 6x + m 12 _4_ ( 2 4 ^ _ < 2^
X2 H - 6 x + - + m
I x2 ,
f 2^ 2 J 21
CO x + - 6 x + -
x; I x j
Dat t = x + -
x
j.
= Ixl +
2_
Ixl
>2N/2
Tacd: t z - 6 t + m - 4 = 0 (2) >2x/2)
Pt (1) cd nghiem co pt (2) cd nghiem thoa |t| > 2 V2
Xet (2) co t2 - 6t - 4 = - m
Dat f(t) = t 2 - 6t - 4 ; f '(t) = 2 t - 6 = 0 c o t = 3
Lap BBT thi phuong trinh cd nghiem khi -m > -13 co m < 13.
Vi du 9: Chiing minh rang vdi gia tri tuy y n e Z1, phuong trinh:
~ x2 xn ,
1 + x + ~x + • • _ 0 khdng the cd nhieu hon 1 nghiem thuc.
-BDHSG DSGT12/1- 51

Giai
2 n
Xet ham da thuc Pn(x) = 1 + x + — + +*- D = R.
2! ' n!"
Ta chung minh quy nap: neu n chan thi Pn(x) nhan gia tri duong Vx e R
cdn neu n le thi Pn(x) cd duy nhat 1 nghiem thuc khac 0.
Vdi n = 0 ta cd P0(x) = 1 > 0, Vx. Gia su khang dinh dung vdi gia tri be
hon n, ta chung minh khang dinh dung tdi n.
- Xet n le. Ta cd: P'n(x) = Pn-i(x) > 0, Vx nen P„(x) la ham tang, do do
Pn(x) cd duy nhat 1 nghiem thuc khac 0
- Xet n chSn: P'n(x) = Pn-i(x)
Da thuc Pn_i cd dung 1 nghiem thuc xo * 0 ( vi n - l le)
Bang bien thien:
X —CO xo +00
P'n(x) = Pn_,(x) - 0 +
Pn(x) +00 +oo
Do do: P„(x) > Pn(x0) = P^^xo) + ^ = > 0
n! n!
Vay khang dinh duoc chung minh.
Vi du 10: Cho phuong trinh: ax3 + 27x2 + 12x + 2001 = 0 cd 3 nghiem phan
biet. Hdi phuong trinh sau cd bao nhieu nghiem?
4(ax3 + 27x2 + 12x + 2001) (3ax + 27) = (3ax2 + 54x + 12)2 (1)
Giai
Xet f(x) = ax3 + 27x2 + 12x + 2001, D = R
Theo gia thiet thi f(x) = 0 cd 3 nghiem a, p, y
Ta cd f (x) = 3ax2 + 54x + 12, f'(x) = 6ax + 54, f"(x) = 6a
(1) co 2f(x) f'(x) = ( f ( X ) ) 2 Xet g(x) = 2f(x) f ' ( x ) - (f(x))2
=o g'(x) = 2f '(x).f (x) + 2f(x).f "(x) - 2f (x) f (x)
= 2f(x).f "(x) = 12a2(x - a)(x - P)(x - y), a 0
Chung minh: ( a + - | < a + - VneN*
I 2 J 2
52 -BDHSG DSGT12/1-

Giai
Xet f(x) = x
n + (c - x)n , c> 0 , D = R.
f '(x) = n.x""1 - (c - x ) n - 1 = n [ x n " 1 - (c - x)n " ' ] , f '(x) = 0
cox"-1 = ( c - x ) n _ 1
~ c
Vai n chan thi n - 1 le nen x = c - x<=>x = -
Vai n le thi n - 1 chin nen x = +(c - x) co x
BBT: X —OO dl +00
f — 0 -f
f
Ta co: f(x) > f ( - ) , Vx nen: xn + (c - x)n > ( - ) "
2 2
Chon x = a, c = a + b > 0 = > dpcm
du 12: Cho i
Chung minh
Vi a, b, c > 0 thoa man a2 + b2 + c2 = 1
a b c 3N/3
b2 + c2 c2 + a2 a2 + b2
Giai
n n T a b c ^ 3/3
BDT co - + — + ->
1-a2 1-b2 1-c2 2
co
3>/3
a ( 1 - a 2 ) b ( l - b 2 ) c(l-c2)
Xet f(x) = x (1 - x2 ) vdi x E (0; 1)
f'(x) = 1 - 3x2 = 0 co x = 4 = e (0; 1)
BBT:
x
f ' ( x )
f(x)
-~XJ
I
V3"
3A/3
+00
Do do f(x) < - 4 = , Vxe (0; 1)
3V3
Ap dung thi cd:
a2 b2 3V3 3V3
a (1-a2 ) b(l-b2 ) c(l-c2)
-BDHSG DSGT12/1-
> — (a2+b2+c2)
2 2
(dpcm).
53

Vj du 13: Cho y = ax3 + bx2 + cx + d (a * o). Chung minh rang neu ham s6
y'" 1
cd 2 cue tn thi: — < —
' " y' 2 y'
Giai
Ta cd y' = 3ax2 + 2bx + c, y"= 6ax + 2b, y"' = 6a nen (1) <=> b2 - 3ac> 0
Vi ham sd cd 2 cue tri nen Ay > 0 do dd b2 - 3ac > 0
Vay bat dang thuc (1) duoc chung minh.
Vi du 14: Chung minh vdi n nguyen duong thi
1 - x + — - — + ... + (-1)' — + ... + ^—
2! 3! i! (2n)!
>0, Vx.
Giai
x2 x3 - x' X
Xetf(x) = 1 - x + 42T! - 4 3T! + - + ( - 1 ) — + •••+ '
Vdi x < 0 thi f(x) > 1 > 0 (dung).
Vdi x > 2n thi:
f(x) =1 +
2!
.4 3 "N
A X
4!~~1$!~
3
l !
( x2n
(2n)
, X 6 R
(2n)! ( 2 n - l )!
= 1 + — ( x - 2 ) + — ( x - 4 ) + ... + - (x-2n)> 1 > 0 (dung).
2! 4! (2n)! B
Vdi 0 < x < 2n, do f lien tuc tren doan [0, 2n] nen tdn tai gia tri be nhat
tai xo. Neu x0 = 0 hay x0 = 2n thi f(x) > f(x0) > 1 > 0.
Neu x0 e (0, 2n) thi f dat cue tieu tai do.
V2 V2""1 Y2n
f '(x) = - l + x - — + ... + — = —
2! (2n-l)! (2n)!
f(x)
Vi f'(xo) = 0 nen f(x0 ) = —2— > 0 .Vay f(x) > f(x0 ) > 0 (dung) Vx.
(znj.
Vi du 15: Cho a, b, c la 3 sd ma phuong trinh: x3 + ax2 + bx + c = 0 co 3
nghiem phan biet. Chung minh: I 27c + 2a3 - 9ab I < 2^(a2 -3b)3
Giai
Dat f(x) = x3 + ax2 + bx + c, D = R, f '(x) = 3x2 + 2ax + b.
Vi f(x) = 0 cd 3 nghiem phan biet nen f (x) = 0 cd 2 nghiem phan biet.
-a + Va2 -3b
Xl
- a - V a ^ S b
, x2 vdi a - 3b > 0
3 3
Va vi he sd cao nhat cua f duong nen yCD = f(xi) >0 va f(x2) = yCT < 0.
Ta cd f(x)
1 1
•x + —a
9
f'(x) + -1( 3 b - a 2 ) x + c - a—b
9 9
54 -BDHSG DSGT12/1-

^ f ( x , ) = | ( 3 b - a 2 ) x i + c - ^
Tu f ( X l ) > 0 ^> -2V(a2 - 3 b ) 3 < 2a3 + 27c - 9ab
f(x2) < 0 => 2a3 + 27c - 9ab < 2V(a2 -3b)3
Do vay: I 2a3 + 27c - 9ab I < 2V(a2 - 3 b ) 3
Vj du 16: Cho cac so thuc x, y thoa man 0 < x < ^ va 0 < y <
Chung minh rang: cosx + cosy < 1 + cos(xy).
Giai
TC /— x + v TC x + y _ /
Dox,y e [ 0 ; - ] n e n O < Vxy < — ^ < - =>cos-— < cos^/xy
3 Z o A
x + v x —v x+y /—
Ta cd cosx + cosy = 2cos -cos—- < 2cos— < 2cos^xy
2 2 "
Xet ham sd f(t) = 1 + cost2 - 2cost vdi t e [0; | - ].
Ta cd f ' ( t ) = 2(sint - tsint2) nen f '(1) = 0, f ( l ) = 1 - cost
Neu 0 < t < 1 thi t2 < t < 1 nen tsint2 < sint2 < sint, do dd f '(t) > 0.
Neu 1 < t < - thi t < t2 < - nen tsint2 > sint2 > sint, do do f '(t) < 0.
3 2
BBT: X
h - 1 - 1
y' 0 + 0
y _ l - c o s k ^
0 ^ 9
Do cos— > 0 nen f(t) > 0, Vt e [0; - ]
9 3
=> 2cos v'xy < 1 + cos(xy) => dpcm.
Vi du 17: Chung minh:
x2v + r 'z + z2x < — . vdi x, y, z > 0, x + y + z = 1.
27
Giai
Khdng mat tinh tdng quat, gia su: y = min{x, y, z} => 0 < y < -
Taco f(x) = x"y+ y2z + z2x = x2y + y 2 ( l - x - y ) + x ( l - x - y)2
= x3 + (3y - 2)x2 + (1 - 2y)x - y2 - y3
f '(x) = 3x2 + 2(3y - 2)x + 1 - 2y
f '(x) = 0 <=> x = - hoac x = 1 - 2y > -
3 3
V i x = l - y - z ^ l _ y nen ta cd BBT:
-BDHSG DSGT12/1-

Ta co f — - - y ( l - 3 y + 3 y 2 ) <— , va
27 3 27
f ( i - y) = y d - y ) 2 - - -2y(i - y ) d - y) *
l f 2 x + i - y + i - y 4
27
Vay f(x) < — suy ra dpcm.
C. BAI LUYEN TAP
Bai 1: Tim cue tri cua ham so:
a)y = xx/4-x2 b) y = xWl2-3x2
BS: a) CD x = V2 , yCD = 2 va CT x = yfe , ycr = "2.
Bai 2: Tim cue tri cua ham sd:
a) y = — + cos x
2
r 2x + 3
b) y = v3 sin x + cos x + — - —
TC J i r T 5TC . _ r _ 5TC S
DS: a) C D x = - + k2TC,yCD= —+ — - C T x = — + k2Tc,yCT- — - —
6 12 2 o i-o &
Bai 3: Tim cue tri cua ham so
a) y = X.N/X-1 b) y
2x2 - 7 x + 5
x2 — 5x + 7
DS: b) CD(4;3), CT(2;-1).
Bai 4: Tim m de ham sd:
a) y = mx3 + 3x2 + 5x + 2 dat cue dai tai x = 2.
b) y = - (m + l)x3 - (m + 2)x2 + (m + 3)x, m * -1 nhan gdc toa do lam
3
diem cue tieu.
17
DS: a)m = - —
b)m = -3.
a) y = — X + m , m * 0 co cue dai, cue tieu ma yCT = 2yCD
b)y
mx - 1
1 - 2x + m
x - m
ddi vdi Ox.
1
cd cue dai, cue tieu va 2 diem cue tri nam cung phia
DS: a) m b) m < 0.
56 -BDHSG DSGT12/1-

Bai 6: Tim m ete ham so y= 2x3 + 3(m-3) x2 +1 l - 3m co cue dai, cue tieu
va 3 diem cue dai, cue t i i u , B(0-1) thang hang
DS: m = 4.
Bai 7: Tim k de ham sd y = -2x +1 - W x 2 +1 cd cue tieu.
DS: k < - 2 .
. . , ' ' 1 1 3
Bai 8: Timcxde ham sdy = — x 3—( s i n a + cosa)x2 + — sin2a..x + 1 cd cue
3 2 4
dai, cue tieu va cac hoanh do cue tri thoa man xi + x2 = X]2 + x 2
2
DS: — +k2rc, k2rc
2
Bai 9: Chung minh khi m < 1 thi ham sd: y = x3 - 3x2 + 3mx + 1 - m dat 2
y — y
cue tri tai (xi, y i ) , (x2 , y2 ) va thoa man: (x, - x 2)(x!X2 -1)
Bai 10: Chung minh ham so y = |x2 + x - 20| khong cd dao ham nhung van
dat cue tri tai x = -5
HD: Dung dinh nghTa dao ham.
x2 + 2x + 3m
Bai 11: Chung minh ham sd y = —- luon ludn cd cue dai, cue tieu
x 2 - 2 x + 2 r ' '
va cac diem cue dai, cue tieu nam tren 1 dudng co dinh.
n c x + 1
DS: y = — -
x 1 , x2 +(m + l ) x - m + l
Bai 12: Cho ham so y = ; luon ludn cd cue dai, cue tieu.
, x - m ...
Lap phuong trinh duong thang qua 2 diem cue tri do.
DS: y = 2x - m
Bai 13: Lap phuong trinh dudng thang qua diem cue dai va cue tieu cua ham
so y = x3 - x2 - 94x + 95
566 671
DS: y = x +
9 9
Bai 14: Bien luan so nghiem phuong trinh:
2x4 - 17x3 + 51x2 - (36 + k) x + k = 0 theo k
HD: y ' = (x - 1)( 2x3 -15x2 + 36x - k)
Bai 15: Tim m de phuong trinh: x3 - 3mx2 + 3(m2 - 1) x - m2 + 1 = 0 co 3
nghiem duong phan biet.
DS: S <m<l + j2
Bai 16: Tim a de 2 diem cue dai va cue tieu cua ham sd y = x3 - 3x2 + 2 nam
trong va nam ngoai dudng trdn x2 + y2 - 2ax - 4ay + 5a2 - 1 = 0.
DS: -<a<l.
5
Bai 17: Chung minh bat dang thuc: | cos2x sin4x + cos2x | < 1.
-BDHSG DSGT12/1- 57

§ 3 . G I A T R I L O N N H A T V A G I A T R I
N H O N H A T
A. K I EN THUC CO BAN
Gia su ham so f xac dinh tren tap hop D (D cz R).
a) Neu ton tai mot diem x 0 e D sao cho f(x) < f(x0 ) vdi moi x e D thi s6
M = f ( x 0 ) dugc goi la gia t r i ldn nhat cua ham so f tren D, ki hieu la
M = maxf(x)
xeD
b) Neu tdn tai mot diem x^ e D sao cho f(x)-> f(x0 ) vdi moi x e D thi so
m = f ( x 0 ) duoc gpi la gia tri nhd nhat cua ham sd f tren D, ki hieu la
m = minf(x)
B. PHAN DANG TOAN
DANG 1: T lM GIATRI LON NHAT, NHO NHAT
- Ham sd lien tuc tren mpt doan thi dat duoc gia tri ldn nhat tren doan do.
Phirong phap: Ddi vdi ham so y = f(x) tren D
Tinh dao ham y' roi lap bang bien thien tu do cd ket luan ve GTLN,
GTNN. Neu can thi dat an phu t = g(x) vdi dieu kien day du cua t.
- Neu y = f(x) ddng bien tren doan [a;b] thi:
min f(x) = f(a) va max f(x) = f(b)
- Neu y = f(x) nghich bien tren doan [a;b] thi:
min f(x) = f(b) va max f(x) = f(a)
-Neu y = f(x) lien tuc tren doan [a;b] thi ta chi can tim cac nghiem x,
cua dao ham f '(0)= 0 rdi so sanh ket luan:
min f(x) = min { f(a); f(xi); f(x2 ) ; . . . ; f(b) }
max f(x) = max { f(a); f(xi); f ( x 2 ) ; . . . ; f(b) }
Chuy:
- Khi can thiet ta phdi hpp cac bat dang thuc dai sd.
- Vdi ham y = | f(x) | thi GTLN tren 1 doan [a,b] la GTLN cua gia tri
tuyet ddi cua gia tri CD, gia tri CT va 2 bien f(a), f(b).
- Neu ham cd nhieu bien thi cd the chpn bien hoac ddn bien.
Vi du 1: Tim gia tri ldn nhat va gia tri nhd nhat cua cac ham sd:
a) f(x) = x2 + 2x - 5 tren doan [-2; 3]
b) f(x) = — + 2x2 + 3x - 4 tren doan [-4; 0]
3
c) f(x) = | x3 + 3x2 - 72x + 90 I tren doan [-5; 51.
Giai
58 -BDHSG DSGT12/1-

a) f' ( x ) = 2x + 2 ; f , ( x ) = 0<=>x = - l .
Ta co f(-2) = -5, f ( - l ) = -6; f(3) = 10.
So sanh thi min f(x) = f ( - l ) = -6 ; max f(x) = f(3) = 10
xe[-2;3] xe[-2;3]
b) f '(x) = x2 + 4x + 3, f ' ( x ) = 0 co x = - 1 hoac x = -3.
Ta co f(-4) = , f(-3) = -4; f ( - l ) = , f(0) = -4.
Vay rnin f(x) = f(-4) = f(-l) = -^;
xe[-4;0] 3
max f (x) = f ( - 3 ) = f(0) = -4
xe[-4;0]
c) Xet ham sd g(x) = x3 + 3x2 - 72x + 90 tren doan [-5; 5]
g'(x) = 3x2 + 6x - 72; g'(x) = 0 <=> x = 4 hoac x = -6 (loai)
f(-5) = 500; f(5) = -70; f(4) = -86.
Do dd -86 < g(x) < 400, Vx e [-5; 5] va vi ham so g(x) lien tuc tren doan
[-5; 5] nen 0 < f(x) = | g(x) | < 400.
Vay min f (x) = 0 ; max f (x) = f (-5) = 400 .
xe[-5;5] xe[-5;5]
Vi du 2: Tim gia tri ldn nhat va nhd nhat cua cac ham so
a) f(x) 2x + 3 1
tren doan [-2; 0] b) f(x) = x + — tren khoang (0; +co)
x - l x
c)y x
4 + x2
tren R d)y 2x2 +2x + 3
x2 + x + 1
tren R.
Giai
-5
a) Tren doan [-2; 0], ta cd f ' (x) : <0 .
Suy ra ham so f(x) nghich bien tren doan [-2; 0]
Vay max f (x) = f (-2) = - ; min f (x) = f (0) = -3
XE[-2,0] 3 XE[-2;0]
1 y?-l
b) f'(x) = 1 - 4x 2 =^rx- 2 • Vx > 0, f'(x) = 0 » x = ±1, chon x = 1.
BBT X 0 i +oo
y' - 0 +
y
+00^^^^ ^ ^ j y +00
Vay min f(x) = f(1) = 2. Ham so khong dat gia tri ldn nhat.
•J X€(0.«) '
c) y' (4 + x 2 ) 2 0 o x = +2.
Lap BBT thi cd: maxy = f(2) = - ; miny = f(-2) = - -
4 4
-BDHSG DSGT12/1- 59

(x2 + x + l ) z "
BBT X 0 _i/2 +co
y' + 0
y 10/3
2 - ^ 2
Vay maxy = — va khong ton tai GTNN.
3
Vi du 3: Tim gia tri ldn nhat va gia tri nhd nhat cua cac ham sd:
a) f(x) = 7 3 - 2 x tren doan [-3; 1] b) f(x) = x + 7 4 - x 2
c) f(x)
a ) f ' ( x ) =
x + 6
tren doan [0; 1 ] d) f(x) = 7x + 3 + 7 6 - x
Giai
- 1
7 3 - 2 x
< 0 vdi moi x e [-3; 1] nen ham sd f nghich bien tren
doan [-3; 1]. Vay max f(x) = f ( - 3 ) = 3 va min f(x) = f ( l ) = 1
xe[-3.l] xe[-3;l]
b) Ham sd f xac djnh va lien tuc tren doan [-2; 2]
f ( x ) = l -
7 w
, vdi moi x e (-2; 2)
f'(x) = 0<=> 1 0 <=> 7 4 - x 2 = x <=>
!0<x<2
4-x2 = x2 x = 72.
74 - x2
Ta cd f( 72 ) = 2 72 ; f(-2) = -2; f(2) = 2.
So sanh thi maxf(x)=2 72 va minf(x)=-2.
xe[-2,2] xs[-2;2]
c) Xet g(x) = -x2 + x + 6 tren doan [0; 1] thi
g'(x) = -2x+l,g,(x) = 0ox=|
1 25 25
Ta cd g(0) = 6, g( —) = — . g(l) = 6 nen 6 < g(x) < — va vi g(x) lien
tuc tren [0; 1] nen76 = TgOO <- =>- < f(x) = -7^= <4=
2 5 7g(x) 76
1 2
Vay maxf(x) = - = , minf(x)= —.
' -[o.i] 76 «t«*] 5
d) D = [-3; 6], vdi -3 < x < 6 thi
1 1 76 - x - 7x + 3
y = 27x73 276-x 27x + 3.76-x
60 -BDHSG DSGT12/1-

Ta co y' > 0 <=> %/6-x > Vx + 3 <=> -3 < x < - .
2
Lap BBT thi maxy = f - ]= 3-s/2 , mmy = f(-3) = f(6) = 3.
VJ du 4: Tim gia tri ldn nhat va nhd nhat cua ham so:
a) f(x) = cos2x + cosx + 3 b) y = cos22x - sinxcosx + 4
c)y
cosx
. rt 3rc
tren - ; —
2 2
d) f(x)
Giai
1
sin x
tren doan TC 5rc
3 ; T
a) Vi f(x) la ham so tuan hoan chu ky 2rc, nen ta chi can xet tren doan
[0; 2TC].
f '(x) = -2sinxcosx - sinx; f '(0) = 0 <=> x e 0; — ; rc ; — ; 2rc
Ta cd f(0) = f(2rc) = 5; f 2TC 11
Vay minf(x) - — ; maxf(x) = 5.
4
Cach 2: Dat t = cosx, -1 < t < 1 thi
f(x) = g(t) = t2 + t + 3, g'(t) = 2t + 1
g.(t) = 0ot = ~ So sanh g(-l), g(-I), g(l).
b) Ta cd y = 1 - sin22x = 4 = -sin22x - — sm2x + 5
Dat t = sm2x, - 1 < t < 1 thi y = f(t) = - t 2 - - 1 + 5.
f '(t) = -2t - -; f '(t) = 0 <=> t = --
2 4
Ta cd f ( - l ) = - , f
4 16 2
7 81
Vay miny = —, maxy = —,
c) Tren khoang
BBT
rc 3TC
2'T
sm x
cos x
y' = 0 CO X = it.
X Tt/2 n 3TC/2
y' + 0 -
y
Vay max y = - l . Ham sd khong cd gia tri nhd nhat.
'« 3n
2' 2
•BDHSG DSGT12/1- 61

d) Tren doan rc 5rc
2 ; T
f 'M - C Q S X f U n t
1 W - —: n , f (x) = 0 CO x = —
sin x 2
* • * < ! ) - £ • < * ) - * ' ( ! ) - •
So sanh thi max f (x) = 2 , mm f (x) = 1
YJ du 6: Tim gia tri ldn nhat va nhd nhat cua ham sd
a) f(x) = x - sin2x tren doan [— ; rcl
2
b) y = sinx + — sin2x
2
c) f(x) = — x + sin2x tren ["-— ; — 1
2 2 2
• 6
sm x
cosx + cos6 X sin x
sinx + cosx
d)y
|sm x| + |cosx|
Giai
a) f '(x) = 1 - 2cos2x ; f '(x) = 0 eo cos2x = - = cos-
2 3
o 2x = ±- + k2rc eo x = ±- + kre, k e Z.
3 6
Vdi-- <X<7C,f'(x) = 0cOXG (-1; *• ^
2 1 6' 6' 6
Tac6f(-^) = -l+^,f(l)=2I_2/3 f(5n) = 5rc + V3
6' 6 2 V 6 2 ' 1 6 j 6 2
«-§) = §;««) = *•
So sanh thi max f (x) = — + — ; min f (x) = -—
+H 6 2 4-H 2
b) Ham so lien tuc tren D = R, tuan hoan vdi chu ki 2rc nen ta xet tren doan
[-rc; TC].
TC
y' = cosx + cos2x = 0 e o x :=+—. x = ±rc
3
Tacdf(-Tt) = 0,f(-^)=-^,f(l)=^.f(T[) = o.
o 4 o 4
373 373
Vay maxy = , miny =
4 4
1 1
c) f '(x) = — + 2sinxcosx = — + sin2x
2 2
62 -BDHSG DSGT12/1-

Tren doan [ - - ; - - ] thi f '(x) = 0 o sin2x = - ^ o x = ~ ; ~
T a c d f ( - - ) = 1 - - . f
2 4
5 it
12
(5n^ ( / 6 + ^ 2
5rt
[l2j V 4 24
5rc
24
So sanh thi maxy = 1 + —. miny = n
24
d) Ta cd | sinx | + I cosx | > sm2x + cos2x = 1 nen D - R.
1 • I5
+ • 5
y = | sinxcosx|
sin x
cos X
|sin x + cosx|
= | sinxcosx | ( 1 - | sinxcosx | - sin2x cos2x)
1 Dat t = | sin2x |, 0 < t < 1 thi y = f(t) = + - t
O t
f(t)= -_t2- -t+ -,f'(t) = 0ot = - hoact = -2 (loai)
w 8 2 2 3
Ta cd f(0) = 0, f - = — , f ( l ) = - Vay maxy
27
Vi du 7: Cho cac so nguyen duong p, q, n.
27
, miny = 0.
a) Tim gia tri ldn nhat cua y = cospx.sinqx vdi 0 < x <
b) Tim gia tri nhd nhat cua y = tanr ,x + cotnx - n2cos22x, 0 < x < —
Giai
a) Vdi 0 < x < - thi sinx > 0, cosx > 0 nen y > 0.
2
Ta cd y 2 = (cos2x)p.(sin2x)q . Dat t = cos2x, 0 < t < 1 thi
y2 = f(t) = tp. (1 - t) f '(t) = tp-x.(l - t)rl.[p - (p + q)t]
nen f'(t) = 0ot = 0 hoac t = hoac t = 1. Ta cd f(0) = f(l) = 0,
P + q
f
,p + q
Pp.qq
(P + q)'
— > 0 nen suy ra maxy PP-q"
(p + q)P"
b) Xet 0 < x < — thi cotx > tanx > 0, sin4x > 0
4
Ta cd y' = ntann"'x(l + tan2x) - n.cotn_1x(l + cot2x) - 4n2cos2x. sin2x
= n(tann_1 - cotn_1x) + n(tann + !x - cotn + 'x) - 2n2sin4x < 0.
-BDHSG DSGT12/1- 63

ham so nghich bien tren (0; — ] nen min y = f — = 2.
Xet — < x < — thi y' > 0 nen ham so dong bien, do do min f — = 2.
4 2 „[H ) U J
Vay miny = 2.
du 8: Tim GTI
tuy y va khong dong thoi bang 0.
2 2
Vi du 8: Tim GTLN, GTNN cua bieu thuc T = . X +y—r. trong do x, y
' ' x + xy + 4y
Giai
x
2
Xet y = 0 thi x * 0 nen T = — = 1. Xet y * 0, dat x = ty thi:
T= ,f^*/ r = ^l±i-=f(t),D = R.
t 2y + t y 2 + 4y2 t2 + t + 4
f'(t>= -^±f^i-.f,(t) = 0ot = -3±>/l0
(t + t + 4)
Lap BBT thi cd maxT = f(-3 - VlO) = 1 0 + 2 ^ 1 0 •
minT = f(-3+ViO)=
10-2^
15
Vi du 9: Cho 2 so duong thay doi x va y thoa man x + y = 1. Tim GTNN cua
a) Q = xy + — b) P = -=L= + y
xy Vl-x V l - x
Giai
1
a) Dat t = xy, v i x , y > 0 v a x + y = l > 2 ^xy nen 0 < t <
Ta cd Q = f(t) = t + -j- => f'(t) = 1 ~jy < 0 nen f nghich bien tren (0; ^].
VayminQ = f(l)=^
4 4
b) Vdi x, y > 0, x + y = 1 nen dat x = sin2a, y = cos2a vdi 0 < a < —
_ sin2 a cos2 a sin3 a + cos3 a
P = + =
cos a sin a sin a + cos a
Dat t = sina + cosa = V2 sin a + — , l<t< —
4 2
- t 3 - 3 t (-3t2 -3)(t2 -1) - 2t(-t3 -3t) _ f + 3
t2-l ' U (t2-l)2 (t2-l)2
-BDHSG DSGT12/1-

Nen f nghich bien tren [ 1 ; -J2]. Vay minP = f( spi ) = V2.
Vi du 10: Cho cac sd thuc khdng am x, y thay doi va thoa man x + y = 1.
Tim gia tri ldn nhat va gia tri nhd nhat cua bieu thiic
S = (4x3 + 3y)(4y2 + 3x) + 25xy.
Giai
Do x + y = 1 nen S = 16x2y2 + 12(x3 + y3) + 9xy + 25xy
= 16xV + 12[(x + y)3 - 3xy(x + y)] + 34xy = 16xV - 2xy + 12.
Dat t = xy, ta duoc S = 16t2 - 2t + 12
0 < xy < (x + y ) 2 1
t e [ 0 ; - ] .
4 4
Xet ham f(t) = 16t2 - 2t + 12 tren doan [0; i ]
f '(t) = 32t - 2; f '(t) = 0 eo t = ——
16
Tacdf(0) = 12,f( — )= llil.f(l)= 25 So sanh thi:
16 16 4 2
maxf(t) = f(-) = — ; min f(t) = f(—) = —
t£[o;I] 4 2 tJ0:L V16' 16
25
Gia tri ldn nhat cua S bang — . khi
2
191
Gia tri nho nhat cua S bang , khi i
16 xy=—
{ 16
^2 + V3 2-V3
x+y=l ,
1 eo fcy) = - ; i
xy=4 2 2
fx+y = l
<=> (x; y) = hoac (x; y) 2-V3 2 + V3
Vi du 11: Cho 3 sd duong a, b, c thoa man a2 + b2 + c2 = 1. Tim gia tri ldn
nhat cua E -2a3 +a
.b2 + b5 - 2 b 3 +b
b2+c2
Giai
Theo gia thiet thi a, b, c e (0; 1)
b(l-b2)2
c 2+a2
2 c - 2c3 + c ,
c + a 2 + b 2= —.a
T r j 3 a ( l - a 2 ) 2
Ta co E = — —
1-a
^2 "V, 2 , C(l - C 2 ) 2 _2
1-b2 1-c2
= a(l - a2)b2 + b ( l - b2)c2 + c(l - c2);
Xet f(x) = x(l - x2) tren khoang (0; 1)
f '(x) - 1 - 3x2, f '(x) = 0 co x = —f=
V3
-BDHSG DSGT12/1- 65

Lap BBT thi 0 < f(x) < - L . Do do E < — (b2 + c2 + a2 )
3v3 3v3
1 2
Dau bang khi a = b = c = —==. Vay maxE = — = .
V3 ' 3V3
3^3
Vi du 12: Cho x, y la cac so thuc thay doi va thoa dieu kien x < y. Tim gia
tri nhd nhat cua bieu thuc: F = x 2 + y 2 - 8x + 16.
Giai
Neu x > 0 t h i x 6 < y 2 v a F = x2 + y 2 - 8 x + 16>x6 + x 2-8x+16.
Xet ham sd: f(x) = x6 + x2 - 8x + 16 vdi x > 0.
f '(x) = 6x* + 2x - 8; f "(x) = 30x4 + 2 > 0, GX > 0.
Do dd f '(x) dong bien.
Ta cd: x > 1 => f '(x) > f '(1) = 0; 0 < x < 1 => f '(x) < f '(1) = 0
BBT X 1 0 I —OO
r 0 +
f 16 ^
10 -
+00
Tu do: f(x) > 0 => F > 10. Dau dang thuc xay ra khi x = y = 1.
Neu x < 0 thi x2 + y2 - 8x + 16 > 16
Vay minF = 10, dat duoc khi x = y = 1.
Vi du 13: Cho cac so thuc x, y thay doi va thoa man (x + y)3 + 4xy > 2. Tim
gia tri nhd nhat cua bieu thuc A = 3(x4 + y4 + x Y ) - 2(x2 + y2) + 1.
Giai
Ket hop (x + y ) 3 + 4xy > 2 vdi (x + y ) 2 > 4xy suy ra:
(x + y)3 + (x + y)2 > 2 => x + y > 1.
A = 3(x4 + y4 + x2y2 ) - 2(x2 + y2) + 1
= W (Xz + f f +-(x4 + y4) - 2(x2 + y2) + 1
2 2
> 1 (X2 + y2)2 + 1 ( x2 + y2)2 _ 2 ( x 2 + y 2 ) + J
2 4
=> A > — (x2 + y 2 ) 2 - 2(x2 + y2) + 1. Dat t = x2 + y2 , ta cd
4
xz + f" >2 ^ ( x t Z L . > I ^ t > I > d o d 6 A > ^ t 2 - 2 t + l
2 2 2 4
Xet f(t) = - 1 2 - 2t + 1; f '(t) = - 1 - 2 > 0 vdi moi t > -
4 2 ' 2
• min f(t) = f
9 9 1
—. Do do A > — dau "=" khi x = y = —
16 16 2
Vay gia tri nhd nhat cua A bang
_9_
16'
66 -BDHSG DSGT12/1-

Vi du 14. Cho 2 < x < 3 < y. Tim GTNN cua B =
Giai
2x' + y" + 2x + y
xy
v , . , . 2x2+y2+2x + y 2(x + l) y+1
Xet g(y) = = = " - + ^—. v o i 2 < x < 3 < y
xy
g'(y) = 2 ( x
2
+ 1 ) + - , g'(y) = o « y = V2x(x + 1)
BBT X 3 72 x ( x + 1) _0°
y' - 0 +
y
Do ddmin g(y) = g (^2x(x +1)) = 2 ^ 2 . - +
V x x
Xetf(x)= 2>/2j- + l+-.2<x<3thi
f '(x) =
V2 1
= —- < 0 nen f nghich bien tren doan [2; 3] do dd
1 + 1. x
minf(x) = f(3) = iS^l . Do do B < +1, dau bang khi x = 3, y = 2 76
Vay minB = + ~
Vi du 15: Tim gia tri nhd nhat: A = /x-l)2 + y2 + V(x +1)2 + y2 + |y-2|.
Giai
Trong mat phang (Oxy) chon M(x - 1; -y), N(x + 1; y)
V(x-l)2+y2 +v
/(x + l)2+y2 = OM + ON>MN= 2^1 + y2
Do do: A > 2^/1 + y2 + I y - 21 = f(y)
Khi y < 2 thi f(y) = 2^1 + y2 - y + 2
f ( y ) = - J l = - l = fc£
+ y
+ v 2 l
= J - , f ( y ) = O c o y -
+ y
BBT:
X -co 2
r — 0 +
f
minf = f ( - ^ ) = 2 + S
43
Khiy>2thiA> 2yjl + y2 +y-2 > 2/ + y2 >2^ >2+^3
-BDHSG DSGT12/1- 67

Vay minA = 2 + ^ 3 khi x = 0, y = 4=-
V3
Vj du 16: Cho phuong trinh: x4 + ax3 + bx2 + ax + 1 = 0 cd nghiem.
Tim gia tri nhd nhat cua a2 + b2
Giai
Goi XQ la nghiem: x4 + ax3 + bx2 + ax0 + l= 0=>x0^0 nen
—+ —
( 1 1 ) xo + — + a x0 + —
Y
b = 0
Dat: v = x0 + — . Dieu kien | y | = | x 0 l + I — I > 2 n
(y2 - 2) + ay + b = 0 =o | 2 - y 2 | = I ay + b I < Va2 + b 2
v
/ y 2 +1
r2-v2V* i (2-t)2 4
=> a
2 + b2 > K £ J . Dat: t = y2 , t > 4. Ta chung minh ^ '- >-
1 + y2 ' 1 + t 5
Xet f(t) = (2~t)2 . t > 4 thi f '(t) = 3t ~ 6 > 0 => f ddng bien nen t > 4
w 1 + t (1 + t ) 2
=>f(t)>f(4) = ^
o
a b -2-4
Dau "=" khi t = 4=>y = ± 2 v a— = — nen chon b = — . a = —
y 1 • 5 5
4 2,4
Phuong trinh: x4 x3 x2 x + l = 0co nghiem x = 1
5 5 5
2 2 4
Vav: min(a + b ) = —
" 5
Vi du 17: Mot con ca boi nguoc dong de vuot mot khoang each la 300km.
Van toe dong nude la 6km/h. Neu van tdc boi cua ca khi nude dung yen
la v (km/h) thi nang luong tieu hao cua ca trong t gio duoc cho bdi cong
thuc E(v) = cv3t, trong do c la mpt hang so, E duoc tinh bang jun. Tim
van toe boi cua ca khi nude dung yen de nang lupng tieu hao la it nhat.
Giai
Van toe cua ca khi boi nguoc dong la v - 6 (km/h). Thoi gian ca boi de
vuot khoang each 300km la: t = (gid)
v - 6
Nang lupng tieu hao cua ca de vuot khoang each do la:
E(v) = cv3 -^P- = 300c. . v > 6.
v - 6 v - 6
E'(v) = 600CV2 V~9, : E'(v) = 0 => v = 9.
( v - 6 ) 2
Lap BBT thi van toe can tim la v = 9 (km/h).
68 -BDHSG DSGT12/1-

Vi du 18: Sau khi phat hien mdt benh dich, cac chuyen gia y te udc tinh so
ngudi nhiem benh ke tu ngay xuat hien benh nhan dau tien den ngay thu
t la: f(t) = 45t2 - t3 , t = 0, 1, 2, 25. Neu coi f la ham sd xac dinh tren
doan [0; 25] thi f '(t) duoc xem la toe do truyen benh (nguoi/ngay) tai
thdi diem t.
Xac dinh ngay ma toe do truyen benh la ldn nhat va tinh toe do do.
Giai
f '(t) = 90t - 3t2 , f "(t) = 90 - 6t, f "(t) = 0 co t = 15.
BBT X 0 15 -oo
f " ( t ) + 0
f ' ( t ) 675
Vay toe do truyen benh la ldn nhat vao ngay thu 15. Toe do do la:
f'(15) = 675 (nguoi/ngay).
DANG 2: BAI TOAN LAP HAM S6
Bai toan tim gia t r j 16*n nhat, nho nhat cua cac dai luong:
Chon dat bien x (hoac t), kem dieu kien ton tai.
Dua vao gia thiet, cac quan he cho de xac lap ham sd can tim gia tri ldn
nhat, nhd nhat.
Tiep tuc giai theo so do tim GTLN, GTNN cua ham sd va cac chu y neu
tren, cd the phdi hop cac phuong phap khac.
Vi du 1: Trong cac hinh chu nhat cd chu vi 100(m), tim hinh cd dien tich ldn
nhat.
Giai
Goi x(m) la mot kich thudc cua hinh chu nhat thi kich thudc kia la 50 - :..
Dieu kien 0 < x < 50.
Dien tich S(x) = x(50 - x) vdi 0 < x < 50,
S'(x) = 50 - 2x, S'(x) — 0 <=> x — 25.
X 0 25 -oo
S'(x) + 0 -
S(x)
Vay maxS = f(25) = 625(m2) khi hinh chu nhat la hinh vudng canh 25(m).
Vi du 2: Trong hinh chu nhat ndi tiep dudng trdn (O; R), tim hinh cd chu vi
ldn nhat.
Giai
Goi x la mot kich thudc cua hinh chu nhat ABCD noi tiep (O; R). Ta co
AC = 2R nen kich thudc thu hai la v^R2 - x 2
-BDHSG DSGT12/1- 6Q

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