GIÁ TRỊ LỚN NHẤT, NHỎ NHẤT CỦA HÀM SỐ VÀ ỨNG DỤNG
I. DẠNG 1: TÌM GIÁ TRỊ LỚN NHẤT, NHỎ NHẤT CỦA HÀM SỐ
Bài 1. (Đề dự bị TSĐH 2003 khối B)
( )
3
6 2
y x 4 1 x= + −
[ ]
1;1−
Cách 1.
[ ]
2
u x 0;1= ∈
( )
3
3 3 2
y u 4 1 u 3u 12u 12u 4= + − = − + − +
[ ]
2
1 2
2
y 9u 24u 12 0 u 0;1 ;u 2 1
3
′
= − + − = ⇔ = ∈ = >
4
maxy 4;min y
9
= =
Cách 2.
( ) ( )
6 6 6 6 6 2 2
x sin u y sin u 4cos u sin u cos u 3cos u sin u cos u 3 4 maxy 4= ⇒ = + = + + ≤ + + = ⇒ =
!"#$%&'()
6 6 2
3
6 6 2
3
8 8 8 8 4
sin u 3 sin u sin u
27 27 27 27 3
4 4 4 4 4
4cos u 3 4cos u cos u
27 27 27 27 3
+ + ≥ × × =
+ + ≥ × × =
( )
6 6 2 2
8 4 4 12 4
y sin u 4cos u sin u cos u y
9 3 3 9 9
= + + ≥ + = = ⇒ ≥
*
2 4
x min y
3 9
= ⇒ =
Bài 2. (Đề thi TSĐH 2003 khối B)
2
y x 4 x= + −
Cách 1: +,-
[ ]
D 2;2= −
.
2
2
x
y 1 ; y 0 x 4 x
4 x
′ ′
= − = ⇔ = −
−
2 2
x 0
x 2
x 4 x
≥
⇔ ⇔ =
= −
/+,
maxy 2 2 ; min y 2= = −
Cách 2:
( )
( )
x 2sin u,u ; y 2 sin u cosu 2 2 sin u 2;2 2
2 2 4
π π π
= ∈ − ⇒ = + = + ∈ −
.
maxy 2 2 ; min y 2= = −
Bài 3.0/+, 122)3
2
x 3
y
x 1
+
=
+
0'
a b c 1+ + =
'&45
2 2 2
a 1 b 1 c 1 10+ + + + + ≥
Giải. 065
D = ¡
.
( )
( )
2 2
1 3x 1 1
y 0 x y 10
3 3
x 1 x 1
−
′
= = ⇔ = ⇒ =
+ +
( ) ( )
x x x x
2
2 2
x 3 / x x 3 / x
x
lim y lim lim lim
x
x 1 1
1
x x
→∞ →∞ →∞ →∞
+ +
= = =
+
+
!78
x x
lim y 1;lim y 1
→+∞ →−∞
= = −
9
x−::y′;<−<y
−
::
x9=> y′;<−<y
−
99
x<9f′<−<+<f?
9
2
x 3
y 10 maxy 10
x 1
+
= ≤ ⇒ =
+
0@,A0
2
x 3 10. x 1+ ≤ +
B%&28
( )
2
2
2
2 2 2 2 2 2
x a : a 3 10. a 1
x b: b 3 10. b 1
x c: c 3 10. c 1
a b c 9 10. a 1 b 1 c 1 10 a 1 b 1 c 1
= + ≤ +
= + ≤ +
= + ≤ +
⇒ + + + ≤ + + + + + ⇒ ≤ + + + + +
Cách 2. ,%CDE-8
( ) ( ) ( )
OA a;1 ; AB b;1 ;BC c;1= = =
uuur uur uur
F
( )
OC OA AB BC a b c ; 3= + + = + +
uuur uuur uur uur
G
OA AB BC OA AB BC OC+ + ≥ + + =
uuur uur uur uuur uur uur uuur
H)78
2 2 2
a 1 b 1 c 1 10+ + + + + ≥
Bài 4. '
2 2
x y 1+ =
I-I
( )
2
2
2 xy y
S
2xy 2x 1
+
=
+ +
( ) ( )
( )
( )
2 2 2
2 2 2
2 2 2
2 xy y 2 xy y 2 xy y
S
2xy 2x 1 3x 2xy y
2xy 2x x y
+ + +
= = =
+ + + +
+ + +
7y=<!=<
*y≠<
x
t
y
=
⇒
( )
( )
( )
2
2
2 2
2y t 1
2 t 1
S
3t 2t 1
y 3t 2t 1
+
+
= =
+ +
+ +
5
( ) ( ) ( )
( )
( )
( )
2 2
2 2
2 2
2 3t 2t 1 2 t 1 6t 2 2 3t 6t 1
S
3t 2t 1 3t 2t 1
+ + − + + − + +
′
= =
+ + + +
)785
2 6 2 6
S
2 2
− +
≤ ≤
*
3 6
t
3
− −
=
⇔
( )
3 4 6
3 6
y ; x y
20 3
−
− −
= ± = ×
2 6
MinS
2
−
=
*
3 6
t
3
− +
=
⇔
( )
3 4 6
3 6
y ;x y
20 3
+
− +
= ± = ×
2 6
MaxS
2
+
=
Bài 5. (Đề 33 III.2, Bộ đề thiTSĐH1987 – 1995)'
2 2
x y 1+ =
I-IA
=
x 1 y y 1 x+ + +
JA≤
( ) ( )
( )
( )
2 2 2 2
x y 1 y 1 x 2 x y 2 2 x y 2 2
+ + + + = + + ≤ + + = +
*
1
x y
2
= =
A=
2 2
+
J 7
xy 0≥
⇒
[ ]
[ ]
x,y 0;1 A 0
x,y 1;0 A 1
∈ ⇒ >
∈ − ⇒ ≥ −
⇒
Min A 1= −
1
x 1;y 0
x 0; y 1
= − =
= = −
:
t−∞t
9
t
:
+∞!′−<+<−!< <
6K
xy 0<
5
x y t+ =
⇒
2
t 1
xy 0
2
−
= <
⇒
( )
t 1,1∈ −
( )
( )
( )
( )
( )
2 2 2
A x 1 y 2xy 1 x 1 y y 1 x 1 xy x y 2xy 1 x y xy= + + + + + + = + + + + + +
=
2 2 2
t 1 t 1 t 1
1 t 2 1 t
2 2 2
− − −
+ × + × + +
( )
2
t 1
1 2 t 2 1
2
−
= + + +
⇔
( )
( ) ( )
2 3 2
1
A f t 1 2 t 2 t 1 2 t 2 2
2
= = + + − + + −
5
( )
( )
2
1 2
3 1 2
1 2 1 2
f t t 2 t 0 t t ;t t 2 1
2 2 3
+
+ +
′
= + − = ⇔ = = − = = −
( )
( )
( )
1 2
2 19 3 2
f t ;f t 0
27
−
= =
)785
( ) ( )
2
1 1
A f t A f t≤ ⇒ ≥ −
⇒
( )
( )
1
2 19 3 2
Min A f t 1
27
−
= − = − < −
-8
⇔
1
x y t+ =
.
2
1
t 1
xy
2
−
=
⇒xy2B
2
1 2 2 3
u u 0
3 9
+ −
+ + =
⇒
( )
1 2 15 2 2
x,y
6
− + ± −
=
Bài 6.'
[ ]
x,y,z 0,1∈
L5
3
x y z
2
+ + =
M7&5
( )
2 2 2
S cos x y z= + +
Giải. G
[ ]
x,y,z 0,1∈
2 2 2
3
0 x y z x y z
2 2
π
< + + < + + = <
122)3
y cos= α
( )
0,
2
π
MI!I-
( )
2 2 2
x y z+ +
6KBCN17(E-8O
+,P,M
( )
M x,y,z
LN7QB
[ ]
x,y,z 0,1∈
4+,,RS
91T25
UV<990.WV9990.'V9<90.GV<<90.U′V<9<0.W′V99<0.'′V9<<0.EV<<<0
I Q #
3
x y z
2
+ + =
( )
M x,y,z
4
,%VX05
3
x y z
2
+ + =
*+8+,P,M
( )
M x,y,z
LN7QB
4 #B
>
t−9t
9
t
:
9ƒ′+<−<+ƒ99
8
>=:
E
Y
9
9
F
>=:
Z
I
O
-
[
/
>=:
9
E′
Y[ZF/1MY[ZF/27M+,,RS
\CE′2 7EY[ZF/5E′2]
+,,RS12^2]$N7Y[ZF/
I2EI
:
_
2 2 2
x y z+ +
EI⇔E′I
⇔I`19aTY[ZF/H)785
( )
( )
( )
2 2 2 2 2 2 2
1 5 5
x y z OK 1 cos x y z cos
4 4 4
+ + ≤ = + = ⇒ + + ≥
*+8 I ! _ I
( )
( )
2 2 2
5
cos x y z cos
4
+ + =
Bài 7.
'xb<2)3
( )
(
)
(
)
(
)
(
)
6
6
6
3
3
3
1 1
x x 2
x x
f x
1 1
x x
x x
+ − + −
=
+ + +
Giải.
( )
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
( ) ( )
3 3
6 2
3 3
3
3 3
3
3
3
3 3
3
3 3
3 3
1 1 1 1
1 1
x x x x
x x
x x x x
x x
1 1
f x x x
x x
1 1 1 1
x x x x
x x x x
+ + + + − +
+ − +
= = = + − +
+ + + + + +
( )
( )
2
1 1
f x 3 x 3 x 6 6
x
x
= + = − + ≥
÷
*x_9IfVx0_a
Bài 8.
M7&PVxy0_x
:
;99y
:
−axy;cx−:cy;:9
Giải. W d2d,RS
PVxy0_Vx−>y;?0
:
;:Vy−90
:
;>≥>
IXVxy0_>⇔
y 1 0 y 1
x 3y 4 0 x 1
− = =
⇔
− + = = −
II. DẠNG 2: ỨNG DỤNG GTLN, GTNN ĐỂ GPT, GBPT
Bài 1. \,RS5
4 4
x 2 4 x 2− + − =
( )
4 4
f x x 2 4 x= − + −
1
2 x 4≤ ≤
⇒
( )
( ) ( )
3 3
4 4
1 1 1
f x
4
x 2 4 x
′
= −
− −
5
( )
f x 0 x 2 4 x x 3
′
= ⇔ − = − ⇔ =
)785
( ) ( )
[ ]
f x f 3 2 x 2,4≥ = ∀ ∈
⇒ XRS
( )
4 4
f x x 2 4 x 2= − + − =
B
#78x=>
?
x−∞<x
<
9+∞f′−<+f
ƒVx
<
0
Bài 2. \ ,RS 5
x x
3 5 6x 2+ = +
⇔
( )
x x
f x 3 5 6x 2 0= + − − =
5
( )
x x
f x 3 ln 3 5 ln5 6
′
= + −
⇒
( ) ( ) ( )
2 2
x x
f x 3 ln3 5 ln 5 0
′′
= + >
x∀ ∈¡
⇒ƒ′Vx0e
IQƒ′Vx0$12
( )
f 0 ln3 ln5 6 0
′
= + − <
( )
f 1 3ln 3 5ln 5 6 0
′
= + − >
⇒XRSƒ′Vx0=<f9Bx
<
⇒W
)785XRS
( )
x x
f x 3 5 6x 2 0= + − − =
Q(g7:
B
I2
( ) ( )
f 0 f 1 0= =
,RSV90f:B
x 0=
12
x 1=
Bài 3. mMWX5
2
m 2x 9 x m+ < +
Bf
x∀ ∈¡
2
m 2x 9 x m+ < +
⇔
( )
2
m 2x 9 1 x+ − <
⇔
( )
2
x
m f x
2x 9 1
< =
+ −
5
( )
( )
2
2
2 2
9 2x 9
f x
2x 9 2x 9 1
− +
′
=
+ + −
=<⇔
2
2x 9 9 x 6+ = ⇔ = ±
( )
x x
2
1 1
lim f x lim
9 1 2
2
x x
→+∞ →+∞
= =
+ −
.
( )
x x
2
1 1
lim f x lim
9 1 2
2
x x
→−∞ →−∞
− −
= =
+ +
)785
( ) ( )
3
Min f x f 6
4
−
= − =
M
( )
f x m>
x∀ ∈¡
( )
x
3
Min f x m m
4
∈
−
> ⇔ <
¡
Bài 4. mMX5
( )
2
2 2sin 2x m 1 cosx+ = +
V90B
x ,
2 2
π π
∈ −
G
: :
x
π π
∈ −
⇒
x
,
2 4 4
−π π
∈
[ ]
x
tg t 1,1
2
= ∈ −
⇒
2
2
1 t
cosx
1 t
−
=
+
.
2
2t
sin x
1 t
=
+
F V90 ⇔
( ) ( )
2 2
2 sin x cosx m 1 cosx+ = +
⇔
( ) ( )
( ) ( )
2 2
2 2
2
2
2 2
2t 1 t 1 t
2 m 1 f t 2t 1 t 2m
1 t 1 t
+ − −
= + ⇔ = + − =
+ +
V:0
5
( ) ( ) ( )
2
f t 2 2t 1 t 2 2t 0 t 1; t 1 2
′
= + − − = ⇔ = = −
⇒W
h
x−∞−aa+∞f′−<;<−
ƒ
t−99ƒ′Vt0−<+ƒVt0?
<?
)785MV:0B
[ ]
t 1,1∈ −
[ ]
( )
[ ]
( )
t 1,1 t 1,1
Min f t 2m Maxf t
∈ − ∈ −
≤ ≤
⇔
0 2m 4 0 m 2≤ ≤ ⇔ ≤ ≤
*+8MV90B
x ,
2 2
π π
∈ −
[ ]
m 0;2∈
Bài 5. m≥<MB5
3 2
2
35
sin xcosy m m 6m
4
33
cosxsin y m 6m
4
= − − +
= − +
V90B
V90⇔
3
3 2
sin xcosy cosxsin y m 12m 17
1
sin xcosy cosxsin y m 2m
2
+ = − +
− = − +
⇔
( )
( )
3
3 2
sin x y m 12m 17
1
sin x y m 2m
2
+ = − +
− = − +
V:0
6K
( )
3
f m m 12m 17= − +
5
( )
2
f m 3m 12 0 m 2 0
′
= − = ⇔ = >
!78ƒVm0≥ƒV:0=9∀m
≥<
IQ#
( )
sin x y 1+ ≤
MBV:0Bm=:
Q5
( )
( )
( )
( )
sin x y 1
2 3
1
sin x y
2
+ =
⇔
− =
8BV>0+
x ;y
3 6
π π
= =
2B*+8V90BQm=:
Bài 6. mMBWX5
2
3 2
x 3x 0
x 2x x 2 m 4m 0
− ≤
− − − + ≥
V90B
V90⇔
( )
3 2
0 x 3
f x x 2x x 2 m 4m
≤ ≤
= − − ≥ −
V:05
( )
[
)
(
]
2
2
3x 4x 4 x 0;2
f x
3x 4x 4 x 2;3
+ − ∀ ∈
′
=
− + ∀ ∈
.
ƒ′Vx0 = < ⇔
2
x
3
=
)78 5
[ ]
( ) ( )
x 0;3
Maxf x f 3 21
∈
= =
MV:0B
[ ]
( )
2
x 0;3
Maxf x m 4m
∈
≥ −
⇔
2
m 4m 21− ≤
⇔−>≤m≤i
III. DẠNG 3: ỨNG DỤNG GTLN, GTNN CHỨNG MINH BẤT ĐẲNG THỨC
Bài 1. '&45
ln x x<
∀xb<
a
x<:> f′−<++f<'c:9
m<: +∞ƒ′−<+ƒ9i9+∞
x<?+∞f′−<+f
:−::
W⇔
( )
f x x ln x 0= − >
∀xb<5
( )
x 2
f x 0 x 4
2x
−
′
= = ⇔ =
⇒W )785
( ) ( )
f x f 4 2 2ln 2 0≥ = − >
Bài 2. '&45
( )
2 2
1 xln x 1 x 1 x+ + + ≥ +
x∀ ∈¡
W⇔
( )
( )
2 2
f x 1 xln x 1 x 1 x 0= + + + − + ≥
x∀ ∈¡
5
( )
( )
2
f x ln x 1 x 0 x 0
′
= + + = ⇔ =
⇒ W
)785
( ) ( )
f x f 0 0≥ =
⇒V,0
Bài 3. '
2 2 2
a,b,c 0
a b c 1
>
+ + =
'Ij5T=
2 2 2 2 2 2
3 3
a b c
b c c a a b 2
+ + ≥
+ + +
5T=
( ) ( ) ( )
2 2 2
2 2 2
2 2 2
a b c a b c
1 a 1 b 1 c
a 1 a b 1 b c 1 c
+ + = + +
− − −
− − −
6K
( ) ( )
2
f x x 1 x= −
1xb<)78
( )
2
1
f x 1 3x 0 x 0
3
′
= − = ⇔ = >
⇒
( )
2
f x x 0
3 3
≤ ∀ >
F5
( ) ( ) ( )
( )
2 2 2
2 2 2
3 3 3 3
a b c
T a b c
2 2
f a f b f c
= + + ≥ + + =
Bài 4. '>≤nk'&45∀x≠<5
( ) ( )
2 n 2 3 n
x x x x x
1 x 1 x 1
2! n! 2! 3! n!
+ + + + − + − + − <
( ) ( )
2 n 2 3 n
x x x x x
u x 1 x ; v x 1 x
2! n! 2! 3! n!
= + + + + = − + − + −
A &
( ) ( ) ( )
f x u x .v x=
l9
5
( )
( )
( )
( )
( )
( )
2 n 1 n
2 n 1 n
x x x
u x 1 x u x
2! n!
n 1 !
x x x
v x 1 x v x
2! n!
n 1 !
−
−
′
= + + + + = −
−
′
= − + − + − = − −
−
⇒
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
n n
x x
f x u x .v x u x .v x u x v x u x v x
n! n!
′ ′ ′
= + = − − +
⇒
( ) ( ) ( )
[ ]
( )
n n 2 4 n 1
x 2x x x x
f x u x v x 1
n! n! 2! 4!
n 1 !
−
− −
′
= + = + + + +
−
G>≤nkƒ′Vx0`#7
1V−:x0⇒W ƒVx0 )785
( ) ( )
f x f 0 1 x 0< = ∀ ≠
⇒V,0
i
x−∞<+∞f′−<+f
<
x−∞<+∞f′+<−f9
x−∞+∞f′+<−f
BÀI TẬP VỀ NHÀ
Bài 1. '∆UW'
A B C> >
2)35
( )
x sin A x sin B
f x 1
x sinC x sinC
− −
= + −
− −
Bài 2. I-I5 y=
6 6
sin x cos x a sin xcosx+ +
Bài 3. 'ab≠<I
( )
4 4 2 2
4 4 2 2
a b a b a b
y
b a b a b a
= + − + + +
Bài 4. '
2 2
x y 0+ >
I-I
2 2
2 2
x y
S
x xy 4y
+
=
+ +
Bài 5. \)",RS
2
2
1
x px 0
p
+ + =
Bx
9
x
:
p≠<)
4 4
1 2
S x x= +
Bài 6. I
( ) ( ) ( ) ( )
2 x 2 x x x
y 2 3 2 3 8 2 3 2 3
= + + − − + + −
Bài 7. 'xy≥<12
x y 1+ =
I-I
x y
S 3 9= +
Bài 8. '
2 2 2
x y z 1+ + =
I-I
P x y z xy yz zx= + + + + +
Bài 9. '
( ) ( )
3
2
f x cos 2x 2 sin x cosx 3sin2x m= + + − +
I-IƒVx0H
mM
2
[f(x)] 36 x≤ ∀
Bài 10 mMX5
( ) ( )
2 x 2 x 2 x 2 x m− + + − − + =
B
Bài 11 mMX5
2
x 9 x x 9x m+ − = − + +
B
Bài 12 mMX5
( ) ( ) ( )
6 5 4 3 2
x 3x m 6 x 2m 7 x m 6 x 3x 1 0+ − − − − − − + + =
B
Bài 13 mMX5
( )
3
2 2 2
x 2x 2 4 x 2x 2 2x 4x m− + − − + = − +
?B,]
B
Bài 14 mMX5
2
3x 1
2x 1 mx
2x 1
−
= − +
−
B#78
Bài 15 mMX5
mcos2x 4sin xcosx m 2 0− + − =
B
( )
x 0,
4
π
∈
Bài 16 mMX5
sin x.cos2x.sin 3x m=
f:B
x ,
4 2
π π
∈
Bài 17 mMWX)7Bf
x∀ ∈¡
5
4 2 2
3cos x 5cos3x 36sin x 15cosx 36 24a 12a 0− − − + + − >
Bài 18 mMBWX5
2
2
3x 2x 1 0
x 3mx 1 0
+ − <
+ + <
B
c
Bài 19 a.mM5
2
m x 8 x 2+ = +
:B,]B
b.'
a b c 12+ + =
'Ij5
2 2 2
a 8 b 8 c 8 6 6+ + + + + ≥
Bài 20 '&45
1 1 1 2 3
sin x sin2x sin 3x sin4x , x ,
2 3 4 3 5 5
π π
+ + + ≥ ∀ ∈
Bài 21 '∆UW')
0 A B C 90< ≤ ≤ < °
'&45
2cos3C 4 cos2C 1
2
cosC
− +
≥
Bài 22 '&45
3
2
sin 2x
3x x
<
−
( )
x 0,
2
π
∀ ∈
Bài 23 '&5
( ) ( )
3 3 3 2 2 2
2 x y z x y y z z x 3+ + − + + ≤
[ ]
x,y,z 0,1∀ ∈
Bài 24 '&45
) : ) > )
) )
: >
x x nx
x nx
n
+ + +×××+ >
( )
x 0,
n
π
∀ ∈
.
2 n≤ ∈¥
m