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
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
≥
⇔ ⇔ =
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
→∞ →∞ →∞ →∞
+ +
= = =
( )
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+ + ≥ + + =
+ + + +
+ + +
7y=<!=<
*y≠<
x
t
y
=
⇒
( )
( )
( )
2
2
2 2
2y t 1
2 t 1
S
3t 2t 1
y 3t 2t 1
+
+
= =
+ +
+ +
5
( ) ( ) ( )
( )
( )
20 3
−
− −
= ± = ×
2 6
MinS
2
−
=
*
3 6
t
3
− +
=
⇔
( )
3 4 6
3 6
y ;x y
20 3
+
− +
= ± = ×
2 6
MaxS
2
+
[ ]
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
= + + +
⇔
( )
( ) ( )
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
⇒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∈
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
*+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
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
+ + + + − +
+ − +
= = = + − +
+ + + + + +
− + = = −
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
( )
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
( )
( )
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
> ⇔ <
¡
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
∈
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∈ −
[ ]
( )
= − +
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
+ = − +
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
=
)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
:−::
( )
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
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
( )
( )
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
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
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
∈
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
<
−