TÌM GIÁ TRỊ LỚN NHẤT NHỎ NHẤT
- T?I LIU HC TP MIN PH?
T?M GI? TR LN NHT NH NHT
PP 1: S dng ph?p bin i ng nht v? t?nh cht ca h?m s lng gi?c.
V? d 1. T?m gi? tr ln nht, gi? tr nh nht ca c?c h?m s sau.
1. y 4 sin x cos x 1
2. y 4 3sin2 2x
1 Ta c? y 2 sin 2x 1 .
Li gii:
Do 1 sin 2x 1 2 2 sin 2x 2 1 2 sin 2x 1 3 1 y 3 .
*
y
1 sin 2x
1
2x
k2
x
k.
2
4
* y 3 sin 2x 1 x k .
4
Vy gi? tr ln nht ca h?m s bng 3 , gi? tr nh nht bng 1 .
2. Ta c?: 0 sin2 x 1 1 4 3sin2 x 4
*
y 1 sin2 x 1 cos x 0 x
k.
2
* y 4 sin2 x 0 x k .
Vy gi? tr ln nht ca h?m s bng 4 , gi? tr nh nht bng 1 . V? d 2. T?m gi? tr ln nht ca h?m s sau y sin x 1 trong khong 0 x
sin x
Li gii:
1 V? 0 x n?n 0 sin x 1,do ? sin x
sin x
Group:
- T?I LIU HC TP MIN PH?
Vy h?m s t gi? tr , ln nht l? 0 ti sin x 1 x .
2
PP2: C?c b?i to?n s dng bt ng thc ? bit t?m gi? tr ln nht, gi? tr nh nht
V? d 1. T?m gi? tr ln nht ca h?m s y 1 1 cos2x 1 5 2sin2 x
2
2
Li gii:
Ta c? y 1 1 cos2x 1 5 2sin2 x y 1 1 cos2x 5 1 sin2 x
2
2
2
42
?p dng bt ng thc Bunyakopvsky cho 4 s: 1; 1; 1 1 cos2x ;
2
5 1 sin2 x ta c?:
42
1. 1 1 cos2 x 1. 5 1 sin2 x 12 12 . 1 1 cos2 x 5 1 sin2 x
91 2.
22
2
42
2
42
4 2.1 2
22
Hay y
2
Du
bng
xy
ra
khi
1
1
cos2 x
5
1
sin2
x
x
k , k
2
42
6
V? d 2.
T?m g?a tr nh nht ca h?m s
y
1
1
2 cos x 1 cos x
vi
x
0;
2
.
Li gii:
Ta thy
2 cos x 0,x R
v?
1
cos
x
0,
x
0;
2
.
Suy
ra
1 2 cos x
v?
1 1 cos x
l?
hai s dng. ?p dng bt ng thc AM- GM cho hai s dng ta c?
1
1
2
2 cos x 1 cos x 2 cos x1 cos x
Mt kh?c tip tc ?p dng bt ng thc AM-GM ta c?
Group:
- T?I LIU HC TP MIN PH?
2 cos x 1 cos x 3
2 cos x1 cos x
2
2
2
4
y
2 cos x1 cos x 3
Vy
4
min y , du bng xy ra khi
0;
2
3
1
cos x x
2
3
v?
x
0;
2
.
PP3: C?c b?i to?n s dng h?m s, t?nh cht h?m s, c bit t?nh ng bin, nghch bin ca h?m s.
V? d 1. T?m tp gi? tr ln nht, gi? tr nh nht ca c?c h?m s sau.
1. y 6 cos2 x cos2 2x
2. y (4 sin x 3 cos x)2 4(4 sin x 3 cos x) 1
Li gii: 1. Ta c?: y 6 cos2 x (2 cos2 x 1)2 4 cos4 x 2 cos2 x 1
t t cos2 x t 0;1 . Khi ? y 4t2 2t 1 f (t)
t 0 1
f (t)
7
1
Vy min y 1 t c khi cos x 0 x k
2
max y 1 t c khi cos2 x 1 x k 2. t t 4 sin x 3 cos x 5 t 5 x
Group:
- T?I LIU HC TP MIN PH?
Khi ?: y t2 4t 1 (t 2)2 3 V? t 5; 5 7 t 2 3 0 (t 2)2 49 Do ? 3 y 46 Vy min y 3; max y 46 . PP 4: C?c b?i to?n li?n quan n a cos x bsin x c
V? d 1. T?m gi? tr ln nht, gi? tr nh nht ca h?m s: y 2 cos2 x 2 3 sin x cos x 1
Li gii:
Ta c? y 2 cos2 x 2 3 sin x cos x 1 2 cos2 x 1 3 sin 2x 2 cos 2x 3 sin 2x 2*
1
2
2
cos
2
x
3 2
sin
2x
2
2
cos
2x
3
2
Mt
kh?c
1
2 cos
2x
3
2
4, x
R
0
y 4, x R
.
Vy gi? tr ln nht ca h?m s l? 4 x k
6
V? gi? tr nh nht ca h?m s l? 0 x k
3
V? d 2. T?m gtln v? gtnn ca c?c h?m sau :
sin x 2 cos x 1 y
sin x cos x 2
Li gii:
Do sin x cos x 2 0 x h?m s x?c nh vi x
X?t
phng tr?nh
:
sin y
x 2 cos x 1
sin x cos x 2
(1 y)sin x (2 y)cos x 1 2y 0
Group:
- T?I LIU HC TP MIN PH?
Phng tr?nh c? nghim (1 y)2 (2 y)2 (1 2y)2
y2 y 2 0 2 y 1
Vy min y 2; max y 1.
B?I TP TNG HP MIN MAX
B?i 1:
T?m gi? tr ln nht gi? tr nh nht ca c?c h?m s sau:
a). y f x 2 3 sin 2x cos 2x
b). y f x sin x cos x2 2 cos 2x 3 sin x cos x c). y f x sin x 2 cos x2 sin x cos x 1 d). y 4sin2 x 3 3 sin 2x 2cos2 x
LI GII
a). ?p dng bt ng thc: ac bd . a2 b2 c2 d2
2
Ta c? 2 3 sin 2x cos 2x 2 3 1 2 3 sin 2x cos 2x 2 2 3 2 2 3 2 3 sin 2x cos 2x 2 2 3
Vy min f x 2 2 3 , max f x 2 3 . b). y f x sin x cos x2 2 cos 2x 3 sin x cos x f x 1 1 sin 2x 2 cos 2x
2
Group:
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- math 1310 lab 6 sec 3 3 3 5
- tÌm giÁ trỊ lỚn nhẤt nhỎ nhẤt
- maths genie free online gcse and a level maths revision
- with suhaag sir
- tg x mendelu
- bài 3 mỘt sỐ dẠng phƯƠng trÌnh lƯỢng giÁc ĐƠn giẢn
- séries d exercices ème maths au lycee ali akirali
- trigonometric equations
- homework 4 solution 1 sol
- trigonometry inverse identities and equations