512
C101B
NGUYEN
THAI HOE
CAC
BAI
TOAN
TRI
LON
NHA^
CD
NHA
XUAT BAN GIAO DUG VIET NAM
NGUYEN THAI HOE
ve
CAC
BAI
TOAN
GlA TRILON NHAT
VA GlA
TRI NHO NHAT
NHA
XUAT BAN GIAO DUG VIET NAM
Nhitng ki hi$u va each viet gon dx^gfc dung trong sach
:
- MXD
:
Mien xac dinh
- GTLN
:
Gia tri

Viet
Nam giO quyin eong bo tac pham.
Ma so':
8I598P9-DTN
05-2009/CXB/87-2170/GD
PHAN
MOT
GIAI
CAC BAI TOAN
GTLN,
GTNN
BANG
PHUONG PHAP CAO CAP (dung dao hdm)
§1. PHifdNG PHAP DUNG BJ^O HAM DE
GIAI
BAI TOAN
I.
CAC
DINH
NGHIA
(,'ho ham y = fix) c6 mien xac dinh X.
- So M
dtroc
goi la GTLN cua ham f(x) trong mien X va viet la max fix)
neu
* vdi moi X e X thi f(x) < M va
* ton tai so XQ e X ,sao cho
fCxg)
= M .
- So m

Luu
y :
-
Tai
diem
c6
hoanh
do
X3,
ham
khong
c6 dao ham
nhtftig
do vin la
diem
CLTC
tieu.
-
Tai
diem
c6
hoanh
do X4 thi
f'(x4)
= 0
nhifng khong
la
diem
CLTC
tri.

XQ
de
sin
XQ
= 15.
TCr dinh nghia
va cac
liAi
y da neu ta c6
II.
PHtfOfNG
PHAP
CHUNG
DE TIM
GTLN
vA
GTNN
CUA
HAM
SO
Cho
ham so y =
fix).
Buac
1. Tim MXD
ciia
ham so (neu bai
toan khong
cho
trifdc)

tri
cua fix)
tai
cac
bien
cua
fix)
(neu c6) . ;., .
6
r
-
Max f(x) = so 16'n
nhat trong
-
min fix) = so nho
nhat trong
•
- Tinh
gia tri cua fix) tai moi
diem
thuoc
mien
xac
dinh
ma f'(x) = 0
hoac
f'(x)
khong
xac
dinh

toan
chi doi hoi tim
GTLN
va
GTNN ciia
ham so thi
khong
nen
tim
gia
tri
ctfc
tri
de suy ra
GTLN
va
GTNN
cua ham so.
III.
CAC BAI
TOAN MINH
HOA
Bai
loan
1.
Tim
GTLN
va
GTNN
cua y -

Trong
doan
[0;
TI]
, y = 0
khi
x = -, x
=^
-, x = —
4 2 4
X
= — +
krt
2
x = - + l-
4 2
Tinh
y(0), yin), y
Ta
CO
y(0)
- y{Tt) = 0
f
—
.
y
,
y
[
4 >

y(7r)
=
0
Cdch
2. Dat t = sinx
Wdi
moix e [0;
TI]
thi t e [0 ; 1]
Ta
CO
y(t)
= 2t t^.
3
Tim GTLN
va
GTNN
cua y(t)
trong
[0, 1] :
Ta
CO y'(t)
=
2 - 4t2 - 2(1 - 2t'^)
^/2
272
y'(t)
=
0 t
=

miny
=
0, dat tai moi x ma t
=
sinx
=
0=>x
=
0, x
=
7i
Bai
toan 2.
Tim GTLN
va
GTNN
cua ham so
1
9 Q
y =
—
In
X
trong
[1; e ]
x
Ta
CO y' =
x.21nx ln^x
.

: max y = yd) = 0
mmy
Bai
toan
3.
Tim
GTLN
va
GTNN
cua ham so
x + 1
y
=
trong
[-1 ; 2]
Ldl GIAI
Ta
CO y' =
Vx^+l-(x + l) J- •2x
2Vx2
+1
1
- X
x2+l
(x2+1)3^2
y' = 0 o X = 1.
Tinh
y(-l),
yd), y(2) va y(-l) = 0, y(l) = V2 , y(2) =
Ket luan

4
ila-x)Kx^
Giai bat phircfng
trinh
y' > 0 de xac
dinh
dau y'
Ta dtro-c y' > 0 <^ ijil - >
(1 - x)^ > x^
<=> 1 - X > X
C:> 0 < X < -
2
Ta CO dau y' va bien thien ciia y cho hdi
bang
:
0
Ta CO ket luan
max y = y
'1'
[2)
-4/
miny
= y(0) = yd) = 1
(Lau y : - Vi ham y c6 dang phufc tap nen ta dung
bang
bien thien de
hieu ro hcfn ve bien thien cua ham so do, tat nhien khong dung
bang
ta van
ket luan dirgrc bai toan.

Vay : maxy = y(V2) = 2N/2
min
y = y(-2) = -2 .
Cdc/i 2. Ap dung bat dang thufc Bunhia copxki cho 4 so : x, 1, ^4 - x^ , 1
ta
duoc
x^
+(4-x2)
[l2+l2] = 8
X
'J
4t — /—
dau = xay ra khi : j = x = \j2 .
Vay ta
duoc
y^ < 8, dau = xay ra khi x = V2
TCr do -2V2 < y < 2N/2 va khi x = V2 thi y = V2 + V2 = 2%/2 ta chi thu
diroc
-2V2 < y < 2N/2 , dau = xay ra khi x = N/2 va ta chi ket luan
duoc
maxy = y(72) = 2V2
{Li/u
y : nhieu hoc sinh ket luan min y = -2N/2 la sai vi khong c6
xo e[-2; 2] de y(xo) =
-2V2).
De tim miny, ta nhan xet rkng :
Khi
X = -2 thi ca 2 so hang x va V4-x^
dong
thcfi

=
sin
cp
va
y =
2(sin
cp
+ cos
9)
=
2^2 sin
doO<x<7i:o-<x
+ -<^
4
4 4
n
X
+
-
4
^/2
<
sin
x +
—
4
2V2.
V2I
<1
<y

Ta CO y' = i + 2 sin X cos X = ^ + sin 2x .
2 2
=> y' = 0
sin2x
= <=>
2x = + k2ji
6
6
n
12
771 ,
xo = — + ln
^ 12
Trong doan
•K
7t 71
2' 2
thi
y' = 0 chi khi
571
12
X2 =-
Tinh
y
Do
•2J
, y
77t
^71^
12

12,
57t
12
+ 1 - cos
12
571 ,
+ 1 - cos
12
V
6,
12 2
571
V
571 , VS
+ 1
12 2
Ta ket luan : max y = y
min
y = y
2j
:.1
+ ^
4,
57t^ _ 1
1 _
571
V3I
12
>
~ 2

= 0 o
sin5x
=
sinx
<=>
5x
= X + k2n
5x
=
Jt
- X + I2n
Xi
= k
—
^
2
Xo
= - + / -
.6
3
Trong
doan
71
n
4'
4
,
y' = 0 chi
khi
x = 0, x = ±

71 57t _ V2
=
5
cos cos — = 5
4
4 2
71
-cos
—
4
53t
73
- 5
cos — cos — = 5
6
6 2
7t
-cos
—
6y
=
372
'I
=
6^ = 373
2
va
y(0) = 4
TCf
do ta ket

+ cos"* 2x = 2.
^l-cos2x^^^
(l-cos2x)^+cos^2x
8
Chon
phep
bien
doi
t =
cos2x
t e [-1 ; 1]
14
Voi bien
t, ham y =
^(1
-
t)^
+
8
Ta tim GTLN
va
GTNN
cua y(t)
trong
[-1 ; 1]
(2t)^
-
(1
-1)^
Taco

y(-l),
yd) va y
3j
Ta
CO
y(-l)
= 3, yd) = 1, y
3;
27
Va
ket
luan
:
maxy
= 3, dat
tai
moi x ma cos2x = 1 x = krc
min
y = — , dat
tai
moi x ma cos 2x =
—
= cos 2a => x = ± a +
/:t
•^27
3
Bai
toan
9.
AC

3
fl
. 2 '
2
1
2
—
Sin
X
- COS
X
u U J
va
y =108
Ap dung
bat
dSng
thufc
Cosi
doi vdi
tich
5
thiTa
so
khong
am,
trong
do c6
—
sin*^

x+ cos x+ -cos x
2 2 3 3 3
15
/' 1 >
1.2
2
fl
2 '
' 1
1
—
sm X -
COS
x
U ;
u , 5^ 3125
va y <
108
3125
Dau
= xay ra khi :
—
sin^
x
—
cos^ x
2 3
1
1 - cos 2x 11 + cos 2x
<=> =

u(l - uf [2(1 - u) - 3u;
=
u(l-u)2(2-5u)
y' = 0 khi u = 0, u = 1, u = -
5
Ta
tinh
y(0) = 0, yd) = 0, y
108
3125
108
Ta
CO
ket luan : maxy =
-r—;r,
dat tai moi x la nghiem cua phucfng
trinh
3125
2 2 1
u
= sm X = — o cos 2x =
—
.
16
Bai
toan 10.
Trong doan
,
hay
tim GTLN

=
sinP~^
x cosP"^
x.[q
cos^ x -
p sin^
x]
y'
= 0<=>
sin
x = 0
cos X = 0
2
• 2
q
cos X =
p sm
x
x
= 0
71
X
= —
2
2
• 2
q
cos X =
p sm
x

do
sin^
XQ =
cos XQ =
p
+ q
P
p
+ q
Ta
CO y(0) = 0, y
f-1
.2J
=
0,
y(xo)
=
(sin2xo)2.(cos2xo)2
=
Vay maxy
= y(xo)
q
(p)2
P
q
(p
+ q)2 (p + q)2
>0
min
y

+
ab^""^
+ b^"1
CO
2n+l
so hang
Ta thu
dtfoc
f
y' = [u - (1 - u)] [u^ + u"(l - u) +
•••
+ u(l -
u)"^
+ (1 - u)^] = (2u
' ; 1—V '
SO
9 so hang = A
trong do A gom 9 so hang > 0 do cac so hang la luy thiTa u va
khong
dong
thoi
bSng
0 nen A > 0
=> y' = 0 khi u =
2
Ta CO y(0) = 1, yd) = 1 va y
Va ket luan
diroc
maxy
= 1 dat

18
Bai
toan 12.
Tim
GTLN
va
GTNN
cua ham so
y
=
+
2x - 3
—
In X tren doan
2
LOI
GIAI
Ta
CO y =
x^
+ 2x - 3
+
-
In X tren
(1; 4
2
^
-X
- 2x
+

2^^
thi
-2x - 2
+
— < 0 *)
2x
Ta
CO *) o
-4x^
-4x + 3
2^^ ^
<0
o
4x^ +4x-3 > 0
C5>
X < — hoac X >
—
2
2
=>
-2x - 2
+
— < 0
vo'i
moi x e
2x
;
1
Bien
thien ciia

[1 ; 5]
LOI
GIAI
1
5
Ta
CO y' = x
—
+ In
X -
In 5
=
In
x -
In
—
X
e
5 5
=:>
y' = 0 o
In
X =
In
- <=> X = - e
[1;
5]
e e
Ta
CO yd) = - ln5,

[-3 ; 3]
Ldl
GIAI
Goi z = 2x^ - 3x2 - I2x + 1, khi do y =
Ta xet
bien
thien cua ham so z tren [-3 ; 3] roi suy ra
bien
thien cua
ham so y de tii do c6 ket luan cua bai toan.
Ta CO z' = 6x2 _ 6x - 12 = GCx^ - x - 2)
=> z' = Oox = -l
hoac
x = + 2
De hieu bai toan mot
each
tiTdng minh, ta nen
dung
bang
bien
thien sau :
Luu
y tren
bang
bien
thien :
Bien thien cua z Ja diidng net lien.
Bien thien cua y la difcfng net duft.
21
TCr M, I, N bien thien ciia y va z trCing nhau.

Bai
toan 15.
Goi Xj , X2 la cac nghiem ciia phiTdng
trinh
:
x^ + px + = 0
1) (p ^ 0)
Hay tim p de : u = xf + x^
dat gia tri nhd nhat va tim GTNN cua u.
Ldl GIAI
Quy
trinh
giai bai toan bao gom cac
bifdc
sau :
Bi/ac
1. Tim dieu kien ciia p de phiTcrng
trinh
1) c6 nghiem
(Buoc
nay de
bo sot va ddn den sai lam)
Budc
2. Tim u
theo
p.
Budc
3. Tim GTNN ciia u vdi moi p da tim dJcfc trong
bi/dc
1.

Khi
do u =
-2
=> u = p +
+
Tim
GTNN
ciia
u(p) vo'i moi p thoa man dieu kien *)
(co the giai khong dung,
nhiT
sau :
Do p'^ + > 2 => min
u
= 2 - 4 = -2 la sai vi voi dieu kien *) thi
P
pn-L>2)
P
Ta
chon : t =
p'^
^ t > 4.
Khi
do ta diipc u(t) = tt 4
Ta
tim min u(t) voi t > 4
Ta
CO u'(t) =
1
- -77 > 0 vdi moi t > 4

—
4
>
0 « -3a^ + 12a'* -9a^ >0
Sa^Ca^
- 4a2 + 3) < 0 (do
33^
> 0)
(a^ -1) (a^ - 3) < 0 o 1 < a^ < 3 o 1 <
|a|
< 73
1
< a < VS '
-V3 < a < -1
+
Ta CO u = ixi + X2) -
3xiX2(xi
+ X2).
Vdri
moi a thoa man dieu kien *) ta c6
diTcfc
x^ + X2 = a
1 '
^^^^^^
4 2 3
a - a +
—
4
Khi
do u = a^ - 3

toan 17.
Cho x>0,
y>Ovax
+ y = l
Tim
GTLN va GTNN cua ham so u = + ^
y + 1 X + 1
Ldl GIAI
Bien ddi u -
+
y^ +1
xy + 2 .
9
(x + y)^
—
2xy + 1 2
—
2t
Chon
phep
bien doi t = xy. Khi do u = =
xy + 2
t
+ 2
Tim
dieu kien cho t:do l = x + y> 2^^
=> t < —, dau = xay ra khi x = y = —
va t > 0 , dau = xay ra khi x = 0
hoSc
y = 0

=> x = y = —
3 4 2
• LUu y : Trong cac bai toan minh hoa d
tren,
ta da stf dung djnh If sau
day de l<hang dinh ve sU c6 gia trj Idn nhat va gia trj nho nhat cua cac ham
so lien tuc. Dinh If do la :
Moi ham so lien tuc trong doan [a ; b] thi c6 trong doan do cac gia tri Idn
nhat va gia tri nho nhat.
Dieu dang luu y la dinh If nay chi cho ta dieu kien du de ham so c6 GTLN
va GTNN.
Dieu do c6 nghTa la, neu khong thoa man cac dieu kien cua djnh If, chang
han :
- ham so khong lien tuc trong doan [a ; b] ' I
- ham so lien tyc nhung khong trong ca doan [a ; b] ma chi la (a ; b); (a ; b];
[a,
b) thi khong the khang dinh la ham so khong c6 GTLN hoac GTNN
trong mien dang xet.
Khi do, de khang djnh ta phai xet cu the cho tCmg ham so trong cac mien
tudng ufng da cho.
De
giiip cac em hoc
sinh
tranh nhkm Ian, tac gia gidfi thieu cac bai
toan
minh
hoa sau :
Bai
toan
18.