Phương pháp giải các chủ đề căn bản giải tích 12 lê hoành phò

Nham muc dich giup cac ban hoc sinh Idp 10, Idp 1 1 , Idp 12 nam vOng kien thifc can ban ve mon Toan ngay tCr luc vao T H P T cho den khi chuan bi thi Tot nghiep, tuyen sinh Cao dang, Da

NGIfT.ThS LE HOANH PHO

16 Han

g Chuo

i Hai B

-a Tr-an

g Ha Npi

-Dien thoai : Bie

n tap-Ch

e ban : (04 ) 39714896 :

Hanh chinh : (04

) 39714899 : Ton

q bie

n tap : (04 ) 3971501

ban:

Gidm doc

- Tong bien tap:

T S.

HA

M TH

I TR

AM

Bien tap:

L AN H UO

NG

Saa bdi:

N HA S AC

H HO NG A

N

Che ban:

N GU YE

N KHCJ

l MI

NH

Trinh bay bia:

VO T HI T HI

TA

Doi tdc lien ket

xuat ban:

Nh

a sac

h H ON

G

AN

SA CH

- 154DH201

4

In 2.00

0 cuon , kh

d 1

7 x

24cm tai Con

Giay phe

p xua

t ba

n so : 463-2014/CXB/10-9

9 OHQGHN , nga

B OHQGHN , v

Nham muc dich giup cac ban hoc sinh Idp 10, Idp 1 1 , Idp 12 nam vOng kien thifc can ban ve mon Toan ngay tCr luc vao T H P T cho den khi chuan bi thi Tot nghiep, tuyen sinh Cao dang, Dai hoc, tac gia da

npi dung la phan dang Toan, tom tat kien thac va phUdng phap giai, cac chu y; phan tiep theo la cac bai toan chpn Ipc can ban minh hpa vdi nhieu dang loai va mac dp; phan cuoi la 8 bai tap c6 hadng dan hay dap so

DO da CO gang kiem tra trong qua trlnh bien soan song khong tranh khoi nhQng sai sot ma tac gia chaa thay het, mong don nhan cac gop y cua quy ban dpc, hpc sinh de Ian in sau hoan thien hdn

Tac gia

L E H O A N H P H O

TiNH t>ON t>IEU TiM KHOANG DONG BIEN VA NGHjCH BIEN

Dinh nghia: Hdm so f xdc dinh tren K Id mot khodng, doan hodc nica khodng

- f dong hien tren K neu vai moi xi, X2 e K: x/ < X2 =^f(x\) < f(x2)

-f nghich hien tren K neu vai moi x/, X2 e K: xi < X2 =>f(xi) > f(x2)

Gid su hdm so c6 dao hdm tren khodng (a; b) khi do:

- Neu hdm so f dong bien tren (a: b) thif '(x) > 0 vai moi x e (a; b)

- Neu hdm so f nghich bien tren (a; b) thif'(x) ^0 vai moi x e (a; b)

Gid su hdm so f c6 dqo hdm tren khodng (a; b) khi do:

Neuf'(x) > 0 vai moi x e (a; b) thi hdm so f dong bien tren (a; b)

Neu f'(x) < 0 vai moi x e (a; b) thi hdm so f nghich bien tren (a; b)

Khi f '(x) = 0 chi tgi mot so hiru hgn diem cua (a; b) thi ket qud tren vdn dung Neu hdm so f dong bien tren (a; b) vd lien tuc tren nita khodng [a;b); (a;bj; doan [a:b] thi dong bien tren nica khodng [a;b); (a;bj; doan [a;b] tuang icng Tuang tu cho nghich bien

Phumtg phdp xet tinh dffn dieu:

- Tim tap xdc dinh

- Tinh dao hdm, xet ddu dqo hdm, lap bang bien thien

- Ket ludn

Chiiy:

1) Cong thirc vd quy tdc dao hdm

y^C =>y' ^0:y=x ^y' = l;y = x" =^y' = nx"-';

(u v)' = u'.v + u.v';

2) Phuang trlnh luang

gidc ca

cosx = cos a <=>

X = 7

1 - a + k27

X = -a +

Z)

(keZ) (keZ)

Bai toa

n 1 : Ti

m khoan

g don

g bien , nghic

x - 8 = 0 <=

^ - 4

x + 1 = 0 o

x = hoac x

va (1

; +co) , nghic

+ 9x

^ - 3

y' = 4x^ - 4x = 4x(x^ -

1), y ' = 0 <=> x = 0 hoa

c x = ±1

6

BBT:

X -00

-4

2

+00

y' -

0

4

0 -

y

Vay ham so dong bien tren khoang

( -4

; 2) v

(-;

-4), (2

;

+ 00 )

b)

D =

R

\6 y' = ~^^^

< 0 , Vx

^±

3

(x '- 9) '

Do do y' <

g do

Bai toa

n 5 : Xe

a khoang:

a) y

= ^|9-x^

b)y=

Vx' -2x + 7

D [-3; 3]

-

-, y' = 0

(-3; 0) v

b) V

i A' =

-x =

> D

= R

„ ,

2x-2 x

-1

Fa

CO

y = —, = ,

2Vx'-2

x +

7 Vx 2x + 7

<Oc>

bien tren nura khoang

[1

; +oo)

a)y Vl6^

6 x^ >

0 «

X '

; 4)

8

Vay ham so dong bien tren (0; 2) va nghich bien tren (2; +oo)

Bai toan 7: Tim Ichoang don dieu cua ham so

t y ' >

0 « 4cosx

>

0<

- +k 2;

n ha

m s

dong bien tren cac khoang

(- ^ + k

Xe

t y '

<0 <=

ham

nghich bien tren cac khoang

( — + k2

;T

; —

+ k 2;

r) , k

= 1

o x — +

kn; —

+ (

k +

l )7rj

+ kn;

— +

(

k + l

k+

l)7i],k

e

Va

y ha

m s

o d on

y a , b bat ky

thuoc R

v

a a < b

Tr en k ho an

g (a;b ) t

hi y ' >

0 v

a y ' =

0 ta

i h iju han d ie

m ne

n h am

f (b )

Va

y the

o d in

h nghi

a t

hi ha

m s

o f dong bien tren

im khoang

dong bien, ng hic

h bie

n cu

a h am so:

a) y =

X

sinx tren [

0; 2n]

b) y = x + 2cos

1 cosx

-Ta c6

x [ 0; 27t ] =:

> y ' >

0 v

a y ' =

0 <

= > x = 0 hoac x

= 2n

Vi ha

m s

o l ie

n tu

c tre

n doa

n [0

; 2n]

nen h am so dong bien tren doan [

0; 2::]

b) y'

= == 1

- 2 sinx

Tr en khoang

m s

o d on

g bie

DANG TOAiy

Neu f'(x) > 0 v&i moi x e (a; h) thi ham s6fdong hien tren (a; h)

Neu f '(x) > 0 vai moi x e (a; b) vd f '(x) = 0 chi tai mot so hitu hgn diem cua (a; h) thi hdm so dong bien tren khodng (a; b)

Neuf'fx) < 0 vai moix e (a: H) thi hdm so nghich bien tren (a; b)

Neu f'(x) < 0 vai moi x e (a; b) vd f'(x) ^ 0 chi tai mot so hihi hgn diem cua (a; b) thi hdm so nghich bien tren khodng (a: b)

Neu hidm sof dong bien tren (a; b) vd lien tuc tren nua khodng [a;b); (a;bj; doqn [a;bj thi dong bien tren mm khodng [a;b); (a;b]; dogn [a;b] tuang icng

Neu hdm so f nghich bien tren (a; b) vd lien tuc tren nua khodng [a;b); (a;bj; dogn [a;b] thi nghich bien tren nua khodng [a;b); (a:hj: dogn [a;b] tuang ung

Chiiy:

I) Ddu nhi thuc bgc nhdt: f(x) = ax -i- b, a ^0

2) Ddu tam thuc bgc hai: f(x) ax^ ^ bx + c, a ^0

Neu A< 0 thif(x) luon cung ddu vai a

Neu A 0 thif(x) luon cung ddu vai a, trie nghiem kep

Neu A> 0 thi ddu "trong trdi - ngodi cung "

f(x) cung ddu a 0 trdi ddu a 0 ciing ddu a

3) Gid su hdm so f xdc dinh tren khodng (a; h) vd Xo e (a; b) Hdm so f duac

Hdm so f lien tuc tren mot khodng (a:b) neu no lien tuc tgi moi diem thuoc khodng do

Hdm so flien tuc tren nua khodng (a; bj neu no lien tuc tren khodng (a; b)

Hdm sof lien tuc tren nua khodng fa; b) neu no lien tuc tren khodng (a; b)

vd lim f(x) =f(a)

Ham so flien tuc

tren dogn [a; b]

neu no lien tuc

tren khodng (a; b)

vd

\\mf(x) =f(a), \xmf(x) =f(b) x->a X—>

b

Bai toa

n 1 : Chun

a) f(x) =

- 6x

^ + 20x

- 1

3 b) f(x) =

2x

- cos

x + Vs sinx

Gidi

a) f'(x) = 3x^-12X

+ 20

Vi A' = 36

- 2

0 <

0 ne

n f (x) >

0 vo

i moi

2 + sinx

-

\ cos

x = 2(1 + ^ sinx

- ^ cosx)

= 2[

1 + sin(x

- y)]

^ 0 vo

i mo

i x

Vay ham so dong bien tren

a) f(x) =

V x'+

l -

X

b) f(x) = cos2x

,—

^ >

Vx^ = IX

I ^

X, V

x nen

+ 1) <

0 vo

i mo

i x

r(x) =

«sin2x l«

2x = -

- +2kK

«x = -

- +k7i,k

€Z

2 4

Ham f(x) lie

— + (k + 1)TC] v

a f (X) <

],ke

Vay ham so nghich

bien tren

R

Cach khac: T

bien tren

R:

Vx i,

X2

e R , X

y ha

i s

o a, b sao cho

+ 1) <

Vi f '(x) =

f nghic

h

bien tren khoang

(a; b) =^

dpcm

12

Ba ai toan 3: Chung m i n h cac ham so sau don dieu tren R:

Vay ham so nghjch bien tren R

Bai toan 4: Chung minh ham so"

1)-Vay ham so nghich bien tren moi khoang (-oo; -1) va (-1; +oo)

Bai toan 5: Chung minh ham so: y = f(x) = ^ x^ + 2x^ + 3x - 1

a) nghich bien tren doan [ - 3 ; - ! ]

b) dong bien tren cac nira khoang ( - x ; -3] va [-1; +oo)

(-3

l) ne

;-n f nghich

bien tren khoang

l] ne

;-n f nghich

bien tren doan [-3

l]

(-co;

-3) v

a (-1

; +00) ne

n f dong bien

tren

khoang (-00;

-3) v

a (-1

; +00) v

a f lien tuc tren cac nir

(-00; -3] v

a [-1

; +co)

Bai toa

n 6 : Chun

g min

h ha

m so: y =

1

+ x

^

nghich bien trong cac khoan

g QO; -1) v

(-a (1;

+00)

Giai

Tap xac dinh

D =

R

l(

l + x')-2x

x _ 1-x'

dong bien trong khoang

1 hoa

c X > 1

1 u do suy ra dpcm

Bai toa

n 7 : Chun

g min

h ha

m so: y

= (a ^ h

+ kn

; k

e Z ) do

n die

u

trong moi khoan

g xa

c dinh sin(x

+ b)

b +

k7i (

k 6 Z)

, _ sin(x + b)cos(

x + a) -sin(

x + a)cos(x

+ b)

sin(b

- a)

sin"(x + b)

sin"(x + b)

?t 0

(do

a b

?t

kji)

Vi y ^ 0

+ kTi , ne

n y' gii

g xa

c djnh, d

SO

Gia sif ham so

cd duo ham tren

khoang (a;

b):

Dung dieu kien can

- Neu ham so

f dong bien tren

(a: b) thif'(x) ^0

vai moi

x e (a; b)

- Neu ham so

f nghich bien tren

(a; b) thi f'(x) <0 vai

moi x

e (a;

b)

14

Dung dieu kien dii

Neuf'(x) > 0 vai moi x e (a; b) thi ham so f dong hien tren (a; b)

Neuf '(x) > 0 vai moi x e (a; h) va f '(x) = 0 chi tgi mot sii huu hctn diem cua (a: h) thi ham so dong bien tren khodng (a; b)

Neuf'fx) < 0 vai moi x e (a; h) thi ham so nghich biin tren (a; b)

Neuf '(x) <0 vai moi x e (a: b) va f '(x) = 0 chi tgi mot s6 him han diem cua (a; b) thi ham so nghich bien tren khodng (a; b)

Bai toan 1: Tim cac gia tri cua thiam so a de ham so f(x) + 4x + 3

dong bien tren R

- Neu a = 2 thi f '(x) = (x + 2)^ > 0 vai moi x -2 nen ham so dong bien tren R

- Neu a = -2 thi ham so f '(x) = (x - 2)^ > 0 vai moi x 2 nen ham so dong bien

tren R

- Neu a < -2 hoac a > 2 thi f'(x) = 0 c6 hai nghiem phan biet nen f' c6 doi dau: loai

Vay ham so dong bien tren R khi va chi khi -2 < a < 2

Bai toan 2: Tim cac gia tri cua tham so a de ham so f(x) = ax^ - 3x^ + 3x + 2 dong bien tren R

Gidi

Tap xac dinh D = R Ta c6 f '(x) = 3ax^ - 6x + 3

Xet a = 0 thi f (x) = -6x + 3 c6 d6i dSu: loai

so dong bien tren R la f '(x) > 0, Vx

- Neu m < 0 thi y' < 0 vai moi x G R nen f nghich bien tren R

- Neu m = 0 thi y' = -3x^ < 0 vai moi x e R, dang thuc chi xay ra vai x = 0, nen ham so nghich bien tren R

- Ne

u m > 0 thi y' = 0

« X =

(xi,

X2): loa

i

Vay ham so nghjch

bien tren

Bai toa

n 4 : Ti

- m + 4 nghich

bien

tren

R

< => f'(x) = cosx

- m < 0, V

m m

de ha

m s

o y = x + 2 + don

- NcL

i m

< 0 thi y ' >

0 va

i mo i

x

^\

Do do ham so dong bien tren m6i khoang

(-00; 1) v

a (1

; +oc)

x" 2x + l

(x-1)^

y' = 0

«>

x^

-2x+l-m = 0

<=>x = l

+00

y' + 0

0 +

y

Ham so nghich

bien tren moi khoan

g (

1 v

-m ; 1) v

a (1

; 1 + v

m ): loai

Vay ham so dong bien tren moi khoan

m a

de ha

m so: f(x) =

x ax" +

16

Ham so nghich bien tren khoang ( 1 ; 2) k h i va chi

khi y' < 0 voi moi x e ( 1 ; 2)

Bai toan 7: T i m m de ham so y = + 3x^ + mx + m chi nghich biSn tren mot

Gidi:

D = R, y' = 3x^ + 6x + m A' = 9 - 3m

Xet A' < 0 thi y' > 0, V x : Ham luon d6ng bien (loai)

Theo de bai: X2 - xi = 3 » (x2 - x i ) ' = 9 <=> x^ + x^ -2X|X2 = 9

<=> (X2 + xi)^ - 4xiX2 = 9 <=> 4 - ^ m = 9<=> m = - ~ (thoa)

Bai toan 8: Tuy theo tham so m , xet sir bien thien cua ham so:

Va nghich bien tren moi khoang (-oo; 2m - V4m' - 9 ) , (2m + V4m" - 9 ; +oo)

Bai toan 9: Xet sir bien thien cua ham so: y = ^ ^ ^ " ^ theo tham so m

x - 1

Gidi

D = R \

„ , ,

-2-n

- N6

u m = -

2 th

i y = 2, V

x 7t 1 la ham so khong doi

- Ne

u m > -

2 th

i y' <

0, V

x 9^1 nen ham so nghich

bien tren moi khoan

g

o)

2 th

i y' >

Nku ham

sSf dan

dieu tren

K va c6 M, N

thuoc K thi phuang trinh

f(M)=f(N) <^

M=N

Neu ham

so f dong bien tren

K va

cd M,

N thuoc

K thi bat phuang

trinh

f(M) >f(N) ^ M>N

Neu ham

so f nghich bien tren

K va

cd M,

N thuoc

K thi bat phuang trinh

f(M)>f(N) o M<N

C/iiiy:

1) Ta

CO the xet f(x)

la ham

so ve trdi, neu

can thi bien doi,

chon xet ham thuan

lai, dat anphu,

Tinh dgo ham roixet

tinh dan dieu

Neu ham

so f dan dieu tren K

thi phmmg trinh f(x)

thi x = a

la nghiem duy nhdt

ciia phuang trinh f(x)=0

2) Neu

f CO dgo ham cap

2 khong doi ddu

thif '

la ham dan dieu

nen phuang

trinh f '(x) = 0

cd toi

da 1 nghiem do

do phuang trinh f(x)

va f(b)

=0

vai a ^ b thi phuang trinh f(x)=0

3 <

t <

2

Xet ham

s6

f(t) =

V

3 + 1

- V 2

-1,

-3 < t <

18

Bai toan 2: Giai pliuang trinh ^j2x^ + 3x^ +6x + \6 = 2V3 + V 4 - x

ma f(l)=2 V3 , do do phuong trinh tra thanh f(x) == f(l) x=l

Vay phuong trinh c6 nghiem day nhdt x = l

Bai toan 3: Giai phuong trinh Vx - V l - x = 5 - 4x

Giai

D i k kien: x ^ 0 PT 4x + V x - V T ^ = 5

2V^ 3^(1-x)^

Ma f '(x) > 0, Vx > 0 va f(x) lien tuc tren [0; +00)

Nen ham so f(x) dong bifin tren nua khoang [0; +00)

Khi X = 1 ^ f ( l ) = 5 nen X = 1 la nghiem PT

K h i x > 1 ^ f ( x ) > f ( l ) = 5:loai

Khi 0 < X < 1 ^ f(x) < f ( l ) = 5: loai Vay nghiem la x = 1

Bai toan 4: Giai phuong trinh: 3x" -18x + 24 = —^- ^—

Bai toa

n 5 : Gia

i bk

phuong trinh: 4 | 2

x

1 | (x^

- x + 1) >

- 6x

^ + 15x

(x -

= R

Ta

CO

f '(t) = 3t^ +

2 >

0 ne

n f ddng biSn tren

T nghie

m dung

Xet X

- 2

> 0 thi 2

> X

- 2

<:i

> X > -

1: Dun

g

Vay tap nghiem

+ 6 + 3V

x +

13 <

20

Xet f(x) l

a ha

m s

o v

e trai, x ^ -1

Ta c6: f (x) =

1

+ •

1

> 0 nen

f(3) «

x <

3

Vay tap nghiem

l + y-) +

'(

l + x') = 4V^

Bai toa

n 7 : Gia

i h

e phuan

g trinh: <

x"yvl + y

"-"v

l + x^

=x

^y-x

Gidi

Dieu kien: x

y >

0

Phuang trinh thu hai cu

l

+ x' <

0

va

1

^|l + y^ <

0

nen suy ra

y >

0.

Do do

x >

0

Phuang trinh tuang

duang:

—^il +

-^=

y

yJl + y^

Xac6

f' (t )=

1

- /

Vl

+ t

^ <

0, vaimoi

t e (0; +oo )

Vi

+ t'

Suy

ra ha

m f nghich

bien tren

(0; +oo)

Phuang trinh f( —) = f(y) — = y

<»

xy = 1

X

X

20

Thay vao phuong trinh thu nhat ciia he, ta c6:

Bai toan 8: Giai he phuong trinh

Ket hop dieu kien, ta c6 nghiem x = y = 1

V x ^ - ^ = 8 - x ' ( x - i r = y

Gidi

Dieu kien x > 1, y ^ 0

V 7 ^ - ( x - l ) ' + x ' - 8 = 0 (1)

y = ( x - l ) ^ (2) Xet ham s6 f(t) = V t ^ - (t - 1)^ + t^ - 8, voi t ^ 1

He phuong trinh tuong duong vdi:

2^/^^ 2 V t ^

f(t) dong bien tren (1; +GC)

Phuong trinh (1) c6 dang f(x) = f(2) nen (1) <=> x = 2, thay vao (2) ta dugc y = 1

Vay nghiem cua phuong trinh la (x; y) = (2; 1)

2 2

He da cho thanh

x' - 3x' - 9x + 22 = / + 3y- - 9y ix-^y-+{y + ^f=l

v =

1

v = -

l

w =

0

Vay he da cho c6 nghiem

i h

e ba

t phuan

g trinh:

x' -3x'+

9x + - >0

Gidi

Tac6(l)<

=>

x 12

x + 35<

t f(x) - x^ - 3x^ + 9x + ^, D =

R

f (x) = 3x^

g biln: x > 5 =^

f(x) >

286/3

Do do f(x) >

0, Vx

G (5

; 7) '

Vay tap nghiem

ciia he bat phuan

g trin

h l

a S =(5; 7)

O S

O N

G HI

E M PH

K thi phuang trinh

f(x)

= 0

cd toi

da 1 nghiem

Neu f CO

dgo ham cap 2

khong doi ddu thif

' la ham dan

dieu nen phuang trinh

f '(x)

^~ 0

CO toi

da I nghiem do

do phuang

trinh f(x)

= 0 c6 toi

da 2 nghiem

Tit BBT cho ta

cdc gid tri ciia

y, neu

y nhdn gid tri

tie dm sang duang

hay

nguac lai tren mot

mien thi

y 0 c6 dimg

I nghiem tren mien

do

Bai toa

n 1 : Chun

- 8 = 0 c6 mot nghie

- 8 la ham so lien tuc va c6 dao ham tren

R

Vi f(0) = -8 < 0, f(l) = 10 > 0 nen ton tai mo

t s6

Xo

e (0

; 1) sa

o cho

trinh f(x) =

0 c

6 nghiem

Mat khac, t

a c

6 y' = IS

Bai toa

n 2 : Chun

-

x *"

+ 3x

^ 3x^ +

1 D =

R

Xet x > 1 thi f(x) = x''(x' ~ 1) + 3x^(x^

- 1) +

+ (

1 x^"

> 0: v

6 nghie

m

2x3

22

Xet X < 0 thi: f'(x) = 13x'^ - 6x^ + 12x^ - 6x

= 13x'^ - 6x(x - 1)^ > 0 nen f ddng bien Bang bien thien:

y

- 0 0

Nen f(x) = 0 CO nghiem duy nhat x < 0

Vay phuong trinh da cho c6 nghiem duy nhat

Bai toan 3: Chung minh rSng phuong trinh 2x^ Vx - 2 = 11 c6 mot nghiem duy nhat

Do do ham so dong bien tren nua khoang [2; +oo)

Ham s6 lien tuc tren doan [2; 3], f(2) = 0, f(3) = 18 V i 0 < 11 < 18 nen theo dinh l i ve gia tri trung gian cua ham so lien tuc, ton tai so thuc c e (2; 3) sao cho f(c) = 11 tuc c la mot nghiem cua phuong trinh f

Vi ham s6 d6ng bidn tren [2; +co) nen c la nghiem duy nhat cua phucmg trinh

Bai toan 4: Tim so nghiem cua phuong trinh x^ - 3x^ - 9x - 4 = 0

Dura vao BBT thi phuong trinh y = 0 c6 diing 3 nghiem

Bai toan 5: Chung minh he x~ + ~ 1 CO dung 3 nghiem phan biet

_y- + x ' = 1

Gidi

Trir 2 phuong trinh ve theo ve va thay the ta dugc:

(l -x)-y'(

l -y) = 0

=>

(l -y

^)(l -x)-(l -x

^)(l -y) =

0

=^ (

1 x)(

l y)[l +

y + y^

y)(y-x)(

l-l+

x + y) =

0

Xet X =

1 th

i h

e c

6 nghiem

(1; 0)

Xet y =

Xetx

^- y thi -

h

y^ = 1

+ X^

1 =

0

Dat f( x) =

X

- +

X

^- 1,D =

R T

a c6f(

l) -1

^0

f(x) = 3x^

+ 2x, f'(x) =0

«

x = -

| hoa

c X =

0-

0 +

y -23/27

+00

-00-^

Do do f(x) =

0 C O

+ X

+ y = 0 =^

= 1 o x^ + x^ + 2x = 0

phan biet

Ba

i toa

n 6 :

phan biet:

Vx^

+m

x +

2 = 2x +

1

Giai

PT<=> 2x + l

>0 ,

x

1 = mx,

X

>

-x'+m

x +

2 = (2x + l)' 2

3x"

+4x

m, x ^ - —

^ —

, X

7t

0 th

i f (x) =

= 2V

1-X'

+ V

l + x- - V l- x'

Giai

DiSu kien -1 <

x <

1 Da

t x = Vl + x"

- Vl-x' th

i t ^ 0

24

Dieu kien c6 nghiem:

DANG TOAN

bat dang thicc:

x>a =>f(x) >f(a); x<h =^f(x) <f(b)

Neuy f(x) xdc dinh tren K cdy'< 0, Vx e K thi f(x) nghich hien tren K nen

cd hat dang thitc:

x>a ^f(x) <f(a): x<h =^f(x) >f(h)

Chuy:

1) Co the f '(x) = 0 chi tgi mot so hint han diem ciia K

ham so, chdng han tic so ciia mot phdn so cd man duang,

Neu y" > 0 thi y 'dong hien tic do ta cd ddnh gid f '(x) r6if(x),

3) Ham so f xdc dinh tren K la mot khodng, dogn hogc nica khodng

- f dong hien tren K neu vdi moi X/, x: e K: x/ < x? =^f(x\) < f(x2)

Bai toan 1: Chung minh cac bat dang thuc sau:

a) sinx < x vai moi x > 0, sinx > x vai moi x < 0

Vai 0

< X

< ^ Ihi ha

m s

o f(x) =

x sinx lien tuc tren nua khoang

[0;

f'(x) =

1 cosx > 0 vai mo

-i x

e (0

; ^)

Do do ham s6 d6ng biSn tren [0;

^) nen f(x) >

f(0) =

0

Vai X <

X

sinx The

o a) th

i g'(x) >

0 va

i mo

i x > 0

Do do ham so

=5

> cos

x + — -

1 >

0 vo

i mo

i x > 0

X) '

(-Suy ra vai mo

i x < 0 ta c6 cos(-x) +

— 1 > 0

Bai toa

n 2 : Chini

> x +

- x lien tuc tren nua khoang

[0;

^) va c6

dao

ham

f '(x) = —

o d

o ha

m s

o f dong bien tren

khoang [0;

^) "en f(x) >

b) Ha

m s

o f(x) = tanx

- x

- — lien

tuc tren nua khoang

[0; —) v

-\ x

= (tan

x + x)(tanx

- x) >

0 va

i mo

i x € (0

; ^) (su

y r

a t

u a))

2 6

Do do, ham so f dong bien tren nua khoang [0; ^ ) va ta c6 f(x) > f(0) = 0 v a i

moi X e (0; ^ ) => dpcm

Bai toan 3: Chung minh:

a) sinx > x , V x > 0 b) 2sinx + tanx > 3x V x e (0; — )

f ' " ( x ) = -1 + cosx < 0 nen f " nghich bien tren [0; +c»):

x > 0 f "(x) < f "(0) = 0 nen f ' nghjch b i l n tren [0; +oo):

X > 0 => f ' ( x ) < f (0) =- 0 nen f nghich h'lkn tren [0; +oo):

Do do ham so f dong bien tren [0; ^ ) nen f(x) > f(0) = 0

Bai toan 4: Chung minh bat dang thuc:

a) 8 s i n ^ + sin2x > 2x, V x e (0; n] b) tanx < — , V x 0;

Gidi a) Xet ham s6 f(x) = 8 s i n ^ ^ + sin2x - 2x, V x G (0; n]

f ' ( x ) = 4sinx + 2cos2x - 2 = 4sinx(l - sinx)

f ' ( x ) = 0 » X = — hoac X = 71

2 •

Vai X e (0; n] ta c6 f '(x) ^ 0 va dau bang chi xay ra tai hai diem Vay f(x)

b) Ne

u X = 0

thi BD

T dung

N6u

X

> 0 thi BD

T tanx

4

(

7 1

Xct f(x) -tan

X

,Vx

-<

-,Vx

e 0;

_

, -tan

x ^

x'cos'x 2x'cos"x

Vi 0

< X ^

— ne

n 0 < 2

x <

— ^ sin 2x < 2

x d

o d

o f '(x) >

1 +

- x -

—

< V l

+ x < 1 + — x, vai

f '(x) =

Y ^

0 vof

i X > 0

nen f(x) don

+ X

Do d

o f(x) >

f(0) =

0 va

i mo

i x ^ 0

Vl +

x

1 -

-x + — tre

n [0

; +co)

2

8

1

Ta c6: g'(x) =

g' don

biSn tren [0;

+oc), d

o d

o g'(x) = g'(0) =

0

Suy

ra g dong bien tren [0;

+00) ne

n g(x) >

a ^ + b^ + c^

+ d

^ + 2abcd

- (a

^ b^

+ a

V + a^

d ^ + b^

c ^ + b^d^ + c^d^) >

0

voi 4

so a b, c d duong

Giai

Khong mat tin

h ton

g quat, gi

a s

u a > b

^ c > d > 0

Xem ve trai l

a ha

m s

o f(a), a ^ 0

f'(a) = 4a-' +

2bcd

- 2a(b

^ + c^

+ d"

)

f "(a) = 12a- - 2(b- + c' + d') >

0 ne

n f' ddn

g bi6

n tre

n (0

; +co) :

28

a > b => f'(a) > f •(b).Vi f (b) = 2b(b^ - c^) + 2bd(c - d) > 0 nen f(a) d6ng biSn

Bai toan 7: Cho x, y, z > O v a x + y + z = l

Chung minh: 0 < xy + yz + zx - 2xyz ^ —

-Khong mat tinh tong quat, gia su x > y > z > 0

-Sx-yf {y-zf {z-xf

thi f *(z) >0 nen f dong bien

Do do f(z) > f(0) = (x + yi^^^^ + +

-,z> 0

= {x +

xy

(x + y)'

1

{x + yy-4xy Ixy

2xy J

Dau dang thuc xay ra khi v

a ch

i kh

i

{x +

-y = 2xy

z =

0

4xy + y

^=

0 [

x =

( 2±

V3

)y'

N G HdP

Gid su ham so

f c6 dqo ham

tren khodng (a: b)

khi do:

Niu f'(x) = 0

vai moi

x e (a; b) thi ham sofkhong doi

tren (a;

b)

Neil f '(x) >

0 vai moi x

e (a;

b) vd

f '(x) = 0 chi tai mot so

huu hgn diem cua

(a: b) thi ham

so dong bien tren

khodng (a:

b)

Neu f '(x) ^0 vai moi

x e (a: b)

vd f '(x) = 0

chi tai mot so

huu hgn diem cua

(a; h) thi hdm

so nghich bien tren

khodng (a:

b)

Neu hdm

so f dong bien

tren (a:

b) vd lien tuc

tren nua khodng [a;b);

(a;bj:

dogn.[a;b] thi dong bien

tren nua khodng [a:b);

(a;bj; dogn [a;b] tuang

ung

Tuang tu cho nghich

bien

Bai toa

n 1 : Ti

m khoan

g don

g bien, nghic

h bie

n cu

a ha

m so:

a) f(x) =

I

x^ + 3x

- 4

I

b) f(x) =

I

x

I

(x + 2)

Giai

x" + 3x

- 4 ,

X

<

-4 hay

X

^ 1

- X"

3x + 4, -

4 <

X <

1

2x + 3 ,x<

-4 hay

>l

-2x-3, -4<

x<

l

a) D

= R , y

Vay ham so nghich bien tren (-00; -4), ( - ^ ; 1) va dong bien tren (-4;- ~ )' (1»

Vay ham so nghjch bien tren (-1;0) va dong bien tren (-co; -1), (0; +00)

Bai toan 2: T i m khoang dong bien, nghich bien ciia ham so:

a)y = 2 - 3 c o s x b) y = " V x ^ ( x - 5 )

Giai

a) Tap xac djnh D = R Ta c6 y' = 3sinx

Xet y ' > 0 <=> sinx >0<=> k 2 ; T < x < ; T + k 2 ; T , k e Z nen ham so dong bien

tren cac khoang ( k 2 ; r ; n + k 2 ; T ) , k e Z

Xet y ' < 0 sinx <0<=> ; T + k 2 ; T < x < 2 ' ; r + k 2 ; r , k G Z nen ham s6

nghjch bien tren cac khoang {n -^Vln^^n + k 2 ; r ) , k e Z

Vay ham so nghjch bien tren ( 0;2) va dong bien tren (-00; 0), (2; +co)

Bai toan 3: Chung minh:

Giai

a) Xet f(x) = sin^x + cos^x, D = R

f (x) = 2sinxcosx - 2cosxsinx = 0, V x

Do do f(x) l

1

b) Xc

t f(x) = cosx + sinx ta

n —, D = (

; —)

-2 4

4

^, , X

2

X

= -sm

x + tan —(1 + cosx) = -smx + tan—.co

= -sin

x + sinx = 0 vai mo

i x

e (

- —

; —)

4 4

Suy ra rang

Do do f(x) = f(0) =

4 4

Bai toa

n 4 : Chun

+ cos^(

x +

y ) •

cosxcos(x +

~)

Gidi

Ta.co f'(x) = -2cosxsinx

- 2cos(

x + ^)sin(x

) + sin(2x

= 0, vo

i mo

i X

Do do

f han

g tre

n R

nen f(x) = f(0) =

* * + 2x^

- 2x'

* x^ - 3x^

^ + 3)(x^

- x

^ 2x

- 1) =

- 2

x

1 =

0 x' = (x + 1)

^ >

0

Do do x^ >

x^ >

1

^ X ^

1

32

Do do nghiem cua phuong trinh - - 2x - 1 = 0 nSu c6 thi x > 1

Dat f(x) = x ^ - x ^ - 2 x - l , x > 1

f ' ( x ) = 5x^ - 2x - 2 = 2(x^ - 1) + 2x(x^ - 1 ) ^ 0

Do do f dong bien V i f ( l ) = -3 < 0 va f(2) = 23 > 0 nen f(x) = 0 c6 nghiem duy nhat Xo > 1

Vay phuomg trinh da cho c6 dung 2 nghiem

Bai toan 6: T i m cac gia t r i cua m de phuong trinh sau c6 dung mot nghiem

Gidi

Nhan xet ung voi moi nghiem Ichong am cua phuoTig trinh (*) c6 diing mot

nghiem cua phuong trinh da cho, do do phuong trinh da cho c6 dung mot nghiem

kiii va chi Idii phuong trinh (*) c6 dung mot nghiem ichong am

z - - 2 z + l = 2x

Bai toan 7: Giai he phuong trinh

Ta CO 2y = x^ - 2x + 1 = (x -1)^ > 0:

Gidi y>0 Tuong t u z, x ^ 0

Dat f(t) = t - 2t + 1, t > 0 thi f '(t) = 2(t - 1) nen f d6ng bien tren ( 1 ; +oo) va nghich bien tren (0; 1)

Dat g(t) - 2t, t ^ 0 thi g'(t) = 2 > 0 nen g d6ng biSn tren (0; +oo)

' f ( x ) = g ( y )

Ta CO he u ( y ) = g(z)

f ( z ) = g ( x )

Gia su

- Neu

f(x) <

f(y) <

f(z)

=> g(y) <

g(z) <

g(x) =

> y < z < x nen

^ 4t +

1 =

0 ne

n cho

n nghiem: x

= y = z

= 2 + Vs

- N

k 0

< X <

1 th

i f(0) >

f(x) >

f(l) =

> 0 < f(x) <

1

nen

0 < g(y) <

1 =

> 0 < y < 1 ^ f(0) >

f(y) >

f(l)

0 < f(y) <

1 0 < g(z) <

1 =

^ 0 < z < 1

Do do

X

< y < z =^

f(x) >

f(y) >

f(z) =

> g(y) >

= y = z

tana

a b

, tanx

x-ta

nx

f( x)

= CO S- X_

-X X

- si

n x cos

- sin2x,

0 <

x <

^, g '(x) - 2- 2cos2

x = 2(1

- cos2x) >

0 ne

n g

d6ng biln: x > 0 =>

g(x) >

g(0) =

0

Do do f'(x) >

0 ne

n f d6ng bign tren [0;

-) V

i 0 < a < b < - ^ f(a) <

f(b): dpc

m

2 2

BAI T AP T ON

a) y = x

^ 2x2

- 5

ux

x-2

b)y=

^

7

X +X +

g (-1

; 0) va (1; +oo)

34

b) Ket qua ham so dong bien tren khoang (2 - V? ; 2 + ) va nghich bien tren cac khoang (-co; 2 - V7 ), (2 + ; +co)

Bai tap 2: Tim khoang don dieu cua ham s6

a ) y = V ^ ( x - 3 ) b ) y = ^ i i

V l - x

HD-DS

a) Ket qua ham so ngliich bien tren khoang (0; 1) va dong bien tren khoang (1; +(»)

b) Ket qua ham so dong bien tren khoang (-oo: 1)

Bai tap 3: Chung minh rang ham so

Bai tap 5: Tim a de ham so: f(x) = ' ~ + 2cosa)x^ + 2xcosa + 1, a e (0; 2%)

dong bien tren khoang (1; +oo)

HD-DS

y' = x^ - (1 + 2cosa)x + 2cosa Ket qua y < a <

Bai tap 6: Giai phuong trinh,he phuong trinh

a) + + 7

<

V^

9-r^

b)x^ + x^>

V l- 3x -4

9 b) Ka

qua x < -

^ sin

a si

n ft , ^

, n

a) > vC

T i 0 < a < b < —

a b

2

b)

x "* + y

" * ^ — va

i x, y thoa

e (0

; —) b) Du

a v

e the

o bil

n x

1

TIM ClTC TR

I

Dinh nghla

Cho ham

so f xdc dinh tren

tap hap

D (D crR) vd

dai cua ham so

f neu tSn tai

mot khodng (a; b)

tieu cua ham so

f neu ton tai

mot khodng (a; b)

Khi ddf

(Xo) duac

goi la gid tri cite tieu

cm ham sof kihieuycr-

Diem cue dai vd

diem cue tieu duac

goi chung

Id diem cue tri.Gid

tri cue

dai

vd gid tri cue

tieu duac goi chung

Id cue tri, neu

Xg Id

mot diem cue tri

cua hdm

so f thi diem

(Xo; f (Xo))

duac goi

Id diem cue tri

cua do thi hdm sof

36

o

I

Dieu kien can de ham so c6 cue tri:

Gid sic ham so f dat cue tri tgi diem XQ

Khi do, neu f c6 dgo ham tgi Xo thif'(Xo) = 0

Dieu kien du de ham so cd cue tri: co hai ddu hieu:

- Cho y = f(x) lien tuc tren khodng (a;b) chua Xo, co dgo ham tren cdc khodng

(a;xo) vd (xn;b):

Neu f '(x) doi ddu tic dm sang duomg thif dgt cue tieu tgi XQ

Neuf '(x) doi ddu tit duang sang dm thif dgt cue dgi tgi XQ

- Choy =f(x) CO dgo ham cap hai tren khodng (a;b) chira

Xo.-Niu f '(xo) = 0 vdf "(xn) > 0 thif dgt cue tiiu tgi xo

Niuf '(xo) = 0 vdf "(xo) < 0 thif dgt cue dgi tgi XQ

Quy tdc 1

1 Timf'(x)

2 Tim cdc diem x, (i = 7, 2, ) tgi do dgo ham ciia ham so bang 0 hodc ham so lien tuc nhimg khong co dgo ham

S Xet ddu f'(x) Neu f'(x) doi ddu tit - sang + khi x qua diem x/ thi ham so dgt

cue tieu tgi jc„ cdn neu f '(x) doi ddu tit + sang - khi x qua diem X/ thi ham so dgt

cue dgi tgi x,

Quy tdc 2

1 Timf'(x)

2 Tim cdc nghiem x, (i = 1,2, ) ciiaphuang trinhf'(x) = 0

3 Timf"(x) vdtinhf'fxi)

Neu f"(Xi) < 0 thi ham so dgt cue dgi tgi diem X/

Neu f"(Xi) > 0 thi ham so dgt cue tieu tgi diem x,

Bai toan 1: Tim cue tri ciia cac ham so sau:

-3) = -1 va dat cu

e til

i,

f(

.i) =

- 2

x +

2 >

0, Vx (do A' =

R , Idion

g e

o cu

e tri

Bai toa

n 2 : T

im cue t

a) y = x'

* 5x^

+ 4 b)

y = (x + 2)

- lO

x = 2x(2x^

- 5)

e tie

u ta

i x = ±

^ — , y

x 3)^

+ 3(

x +

2f (x-3)

^ = 5x(x + 2)(

x 3)^

-Ta

CO

y' =

0 x = -

2 hoa

c x = 0 hoac

-^

^

-108

; 0) v

a cu

e tie

u (0

; -108)

Bai toa

n 3 : T

im cue t

= x' + 3x-

4 b)

f(

x)

= X

(x + 2)

+3x-4 , x<

-4 hay

x >

1

BBT:

X

-x' -3x4-4, - 4

<x

<l

2x + 3 ,

x <

-4 hay

x >

1

-2x-3, -4<

x<

l

-0 0

-4 -3/2

+

CT

TT

38

Vay ham so dat CD 3 25 ,CT(-4; 0), CT(1;0)

(x + 1)' Vay ham so dat cue dai tai x = -2, yco = -4 va dat cue tieu tai x = 0, yci = 0

Bai toa

n 5 : T

im cue t

x +

x +

1 b)

y =

Gidi

2x + l

0 c

» x = -1

± V6

-1 + V 6

+00

y' + 0

-^

^ -0

0 +00 +0

0

Vay diSm

CD(1 V6 ; -

4 2

-^6 ), CT

(-1 + V

6 ; 2V6

-1

1

(x -5 )^

tung khoang

xac dinh, d

Bai toa

n 6 :

Ti

m cu

e tr

a) y= V x' -2

x +

5 b)

y

=V

^(

5)

x-Gidi

< 0 , Vx

9t 5

nen ham so nghich

ai

X

^ 0 thi y ' = ^

^(

x 5) ^ 5(

x 2 ).

^, =

Q ^ ^ =

2

3vx 3Vx

Bang bien thien

e da

i ta

i x = 0 , yc