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7/31/2019 Tinh Don Dieu Cua Ham So Hay

1/20

CHUYN TON THPT V Trng SnCHUYN TON THPT V Trng Sn

CHUYN 1. TNH N IU CA HM S V CC NG DNG

VN 1: XT CHIU BIN THIN CA HM SQuy tc:

1. Tm TX ca hm s.2. Tnh o hm f(x). Tm cc im xi m ti o hm bng 0 hoc khng

xc nh.3. Sp xp cc im xi theo th t tng dn v lp BBT.4. Nu kt lun v cc khong ng bin, nghch bin ca hm s.

Bi 1. Xt chiu bin thin cc hm s sau:

2

3 2 4 2 3x 2 x 2x + 3a) y 2x + 3x + 1 b) y = x 2x 3 c) y d) yx 1 x 1

+ = + = =

+ +Bi 2. Xt tnh n iu ca cc hm s sau:

32

2 2

x x xa) y 25 x b) y c) y d) y

x 10016 x x 6

= = = =+

Bi 3. Chng minh rng:

a) Hm s 2y x 1 x= + ng bin trn khong1

1;2

v nghch bin trn khong

1;1

2

.

b) Hm s 2y x x 20= nghch bin trn khong ( ); 4 v ng bin trn

khong ( )5;+ .

Bi 4. Xt s ng bin, nghch bin ca cc hm s sau:

[ ]5

a) y x sin x, x 0;2 b) y x 2cos x, x ;6 6

= = +

Bi 4. Chng minh rng:

a) ( )f x cos2x 2x 3= + nghch bin trn R.

b) ( ) 2f x x cos x= + ng bin trn R.

Gii:

a) Ta c: f '(x) 2(sin 2x 1) 0, x R = + v f '(x) 0 sin 2x 1 x k , k Z4= = = +

Hm s f lin tc trn mi on ( )k ; k 14 4

+ + + v c o hm f(x) < 0 vi mi

( )x k ; k 1 , k Z4 4

+ + +

.

Trng THPT Long Hi Phc Tnh.

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2/20


Do , hm s nghch bin trn mi on ( )k ; k 1 , k Z4 4

+ + + .

Vy hm nghch bin trn R.

b) Ta c: f(x) = 1 sin2x; f '(x) 0 sin 2x 1 x k , k Z4

= = = +

NX: Hm s f lin tc trn mi on ( )k ; k 1

4 4

+ + + v c o hm f(x) > 0 vi mi

( )x k ; k 1 , k Z4 4

+ + +

.

Do hm s ng bin trn mi on ( )k ; k 1 , k Z4 4

+ + + .

Vy hm ng bin trn R.

VN 2: TM THAM S HM S N IU TRN MIN K

Phng php: S dng cc kin thc sau y:1. Cho hm s y = f(x) c o hm trn K.

Nu f '(x) 0, x K th f(x) ng bin trn K.Nu f '(x) 0, x K th f(x) nghch bin trn K.

2. Cho tam thc bc hai f(x) = ax2 + bx + c c bit thc 2b 4ac = . Ta c:

a 0f (x) 0, x R

0

>

a 0f (x) 0, x R

0


3/20


Hm s nghch bin trn R khi v ch khi5

f '(x) 0, x R 0 a2

.

Bi 2

Vi gi tr no ca m, hm s ( )3 2f (x) mx 3x m 2 x 3= + + nghch bin trn R ?

Gii:TX: RTa c: 2f '(x) 3mx 6x m 2= +

Hm s nghch bin trn R khi v ch khi 2f '(x) 3mx 6x m 2 0, x R = +

m = 0, khi f(x) =1

6x 2 0 x3

: khng tha x R .

m 0 , khi m 0

f '(x) 0, x R 9 3m(m 2) 0


4/20


o hm:( )

2

2

m 1y'

x m

=

+. Hm s ng bin trn tng khong xc nh khi

2y' 0, x m m 1 0 m 1 v m 1> > < >

Bi 5

Tm m hm s ( ) ( )3 21 1

y mx m 1 x 3 m 2 x

3 3

= + + ng bin trn [ )2;+ .

Gii:Ta c: ( ) ( )2y ' mx 2 m 1 x 3 m 2= +

Hm s ng trn [ ) ( ) ( )22; y ' 0, x 2 mx 2 m 1 x 3 m 2 0, x 2+ +

( )2 26 2x

m x 2x 3 2x 6 0, x 2 m , x 2x 2x 3

+ +

+(v x2 2x + 3 > 0)

Bi ton tr thnh:

Tm m hm s ( ) 26 2xf x m, x 2x 2x 3= +

Ta c ( )( )

( )2

2

22

2x 12x 6f ' x , f ' x 0 2x 12x 6 0 x 3 6

x 2x 3

+= = + = =

+

BBT:x 2 3 6+ +

f(x) 0

f(x)

2

3

0

Ta cn c:[ )2;

2max f (x) m m

3+ . l cc gi tr cn tm ca tham s m.

Bi 6

Tm m hm s2mx 6x 2

yx 2

+ =

+nghch bin trn na khong [ )1;+ .

Gii:Ta c:

( )

2

2

mx 4mx 14y '

x 2

+ +=

+

Hm s nghch bin trn [ ) 21; y ' 0, x 1 mx 4mx 14 0, x 1+ + +

( )2 214

m x 4x 14, x 1 m , 1x 4x

+

+


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Bi ton tr thnh: Tm m hm s ( ) 214

f x m, x 1x 4x

=

+

Ta c:( )

22

14(2x 4)f '(x) 0, x 1

x 4x

+=

+

x 1 +f(x)

f(x)0

14

5

Ta cn c:[ )1;

14min f (x) m m

5+ . Vy

14m

5 l cc gi tr cn tm ca m.

Bi tp t gii:

Bi 1. Tm cc gi tr ca tham s a hm s ( ) 3 21

f x x ax 4x + 3

3

= + + ng bin trn R

Bi 2. Vi gi tr no ca m, hm sm

y x 2x 1

= + +

ng bin trn mi khong xc nh ?

Bi 3. nh a hm s ( ) ( )2 3 21

y a 1 x a 1 x 3x 53

= + + + + lun ng bin trn R ?

S: a 1 v a 2

Bi 4. Cho hm s( ) 2m 1 x 2x 1

yx 1

+ +=

+. Xc nh m hm s lun ng bin trn tng

khong xc nh ca n.

S: 1 m 2 Bi 5. Cho hm s ( ) ( )3 2 2y x m 1 x m 2 x m= + + + + . Chng minh rng hm s lun nghch

bin trn R vi mi m.Bi 6. Tm m hm s y = 3x3 2x2 + mx 4 ng bin trn khong ( )0;+ .

S:4

m9

.

Bi 7. Tm m hm s y = 4mx3 6x2 + (2m 1)x + 1 tng trn khong (0;2).

S:9

m

10

.

Bi 8. Cho hm s2x 2mx m 2

yx m

+ +=

.

a) Tm m hm s ng bin trn tng khong xc nh.b) Tm m hm s ng bin trn khong ( )1;+ .

VN 3:


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S SNG TNH N IU CA HM S CHNG MINH BT NG THC

Phng php: S dng kin thc sau: f(x) ng bin trn on [ ]a; b th ( ) ( ) ( ) [ ]f a f x f b , x a; b

f(x) nghch bin trn on [ ]a; b th ( ) ( ) ( ) [ ]f a f x f b , x a; b

Bi 1

Cho hm s ( )f x 2sin x tan x 3x= + .

a) Chng minh rng hm s ng bin trn na khong 0;2

.

b) Chng minh rng: 2sin x tan x 3x, x 0;2

+ >

.

Gii:

a) Hm s cho lin tc trn na khong 0; 2 v c

( ) ( )2

2 2

1 cos x 2cos x 11f '(x) 2cos x 3 0, 0;

cos x cos x 2

+ = + = >

. Do , hm s f ng

bin trn na khong 0;2

(pcm).

b) T cu a) suy ra f(x) > f(0) = 0, x 0; 2sin x tan x 3x, x 0;2 2

+ >

(pcm).

Bi 2

a) Chng minh rng hm s ( )f x tan x x= ng bin trn na khong 0;2

.

b) Chng minh rng3x

tan x x , x 0;3 2

> +

.

Gii:

a) Hm s cho lin tc trn na khong 0; 2

v c2

2

1

f '(x) 1 tan x 0,cos x= = >

x 0;2

. Do , hm s f ng bin trn na khong 0;2

.


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b) T cu a) suy ra f(x) > f(0) = 0, x 0; tan x x, x 0;2 2

>

.

Xt hm s3

xg(x) tan x x

3= trn na khong 0;

2

. Hm s ny lin tc trn na

khong 0;2

v c o hm2 2 2

2

1g '(x) 1 x tan x x 0, x 0;

cos x 2

= = >

, do

tan x x, x 0;2

>

.

Do , hm s g ng bin trn na khong 0;2

nn g(x) > g(0) = 0 x 0;2

3xtan x x , x 0;

3 2

> +

(pcm).

Bi 3

Chng minh rng : 2(x 1)ln x x 1> + , vi mi x > 1.

Gii:

Bt ng thc cho tng ng vi2(x 1)

ln x 0, x 1x 1

> >

+

Xt hm s ( )2(x 1)

f (x) ln x , x 0;x 1

= +

+. Ta c:

( )

( )

( )

( )2

2 2

x 11 4f '(x) 0, x 0;

xx 1 x x 1

= = +

+ +Suy ra hm s ng bin trn khong ( )0;+ nn cng ng bin trn khong ( )1;+ . Vy talun c f(x) > f(1) = 0 vi mi x > 1. cng l iu phi chng minh.Bi tp t gii:Bi 1.Chng minh cc bt ng thc sau:

a) sin x x, x 0< > v sin x 0, x 0<

c)

3x

sin x x , x 06> > v

3x

sin x x , x 06<

e)2x

sin x , x 0;2

>

f) tan x sin x> vi 0 x2

<

=Chng minh h trn c nghim duy nht.

Gii:Xt h:

x ye e ln(1 x) ln(1 y) (1)

y x a (2)

= + +

=vi iu kin xc nh x 1, y 1> >

T (1) y = x + a, th vo (1) ta c: x a xe e ln(1 x) ln(1 x a) 0+ + + + + = (3)Bi ton tr thnh chng minh (3) c nghim duy nht trn khong ( )1; + .t x a xf (x) e e ln(1 x) ln(1 x a)+= + + + + trn khong ( )1; +

Ta c f(x) l hm lin tc trn khong ( )1; + v c o hmx a x 1 1

f '(x) e e x 1 x a 1

+

= + + + +Do a > 0 nn vi mi x > -1, ta c:

x a xe e 0

1 10

x 1 x a 1

+ >

> + + +Nh vy f(x) > 0 vi mi x > -1 f(x) l hm s ng bin trn khong ( )1; +

Mt khc, ta c: x a1 x

f (x) e (e 1) ln1 a x

+= +

+ +

T ta tnh gii hn: x ax x x

1 xlim f ( x) lim e (e 1) lim ln

1 a x+ + ++= + = +

+ +v

x ( 1)lim f (x)

+ =

Vy, phng trnh (3) c nghim duy nht trn khong ( )1; + . T suy ra pcm.

Bi tp t luyn:Gii cc phng trnh sau:

a) 2 2x 15 3x 2 x 8+ = + + S: x = 1

b) ( ) ( ) ( ) ( )x 2 2x 1 3 x 6 4 x 6 2x 1 3 x 2+ + = + + + S: x = 7

VN 4:NG DNG CHIU BIN THIN CA HM S VO VIC BIN LUN

PHNG TRNH, H PHNG TRNH V BT PHNG TRNHCh . Cho f(x) l hm s lin tc trn T, th:

a) ( )f x a vi mi ( )x T a max f x


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b) ( )f x a vi mi ( )x T a min f x

c) ( )f x a c nghim ( )a min f x

d) ( )f x a c nghim ( )a max f x

Bi 1

Cho phng trnh

()2m x 2x 2 1 x(2 x) 0 + + + . Tm m phng trnh c nghim

x 0,1 3 + .

Gii:

Xt bt phng trnh : ( )2m x 2x 2 1 x(2 x) 0 (1) + + + t = + = 2 2 2t x 2x 2 x 2x t 2

Ta xc nh iu kin ca t :

Xt hm s = +2

t x 2x 2 vi x 0,1 3

+ Ta c: 2

x 1t ' , t ' 0 x 1

x 2x 2

= = =

+x 0 1 1 3+t 0 +

t2 2

1

Vy vi x0,1 3

+ th 1 t 2 .Khi :

(1) +

2t 2

mt 1

vi t [1;2]

Xt hm s

=+

2t 2

f(t)t 1

vi t [1;2] . Ta c:

f(t)+ +

= >

+

2

2

t 2t 20, x [1;2]

(t 1). Vy hm s f tng trn [1; 2].

Do , yu cu bi ton tr thnh tm m (1) c nghim t[1,2]

= =t 1;2

2m max f(t) f(2)

3.

l gi tr cn tm ca tham s.

Bi 2


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Tm m phng trnh 4 4x 13x m x 1 0 + + = c ng mt nghim.

Gii:

Ta c:4 4

x 13x m x 1 0 + + =4 4

x 13x m 1 x + =

( )4 3 24

x 1 x 1

4x 6x 9x 1 mx 13x m 1 x

= + =

Yu cu bi ton tr thnh tm m ng thng y = -m ct phn th f(x) = 4x 36x29x1ng vi x 1 ti mt im duy nht.Xt hm s f(x) = 4x3 6x2 9x 1 trn na khong ( ];1

Ta c: f'(x) = 12x2 12x 9 = 3(4x2 4x 3)

Cho f'(x) = 0 4x2

4x 3 = 0

1 3

x x2 2= =

x 1

2 1

f(x) + 0

f(x)

3

2

12

T bng bin thin ta thy:

Yu cu bi ton xy ra khi

3 3m m

2 2

m 12 m 12

= = < >

l cc gi tr cn tm ca tham s m.

Bi 3

Tm m h phng trnh ( )2x y m 0 Ix xy 1

=+ =

c nghim duy nht.

Gii:Ta c:


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(I)2x y m 0 2x y m 0

x xy 1 xy 1 x

= = + = =

Vi iu kin:xy 0

x 1

ta c:

(I) ( ) ( ) ( )22

y 2x m

y 2x m 1 xxy 1 x y x 1

x

= = = =

(Do x = 0 khng l nghim ca h)

( )2 21 x x 2x 1

2x m mx x

+ = = ()

Xt hm s2x 2x 1 1

f (x) x 2x x

+ = = + trn tp ( ] { }D ;1 \ 0=

Ta c hm s f(x) lin tc trn D v c o hm ( ) ( ]21

f '(x) 1 0, x ;0 0;1x

= + >

Gii hn :x x 0 x 0lim f (x) ; lim ; lim

+ = = + = v f(1) = 2

BBT :x 0 1

f(x) + +

f(x) + 2

T BBT ta thy :Yu cu bi ton xy ra khi m > 2. l cc gi tr cn tm ca tham s.

Bi 4 ( thi tuyn sinh i hc, Cao ng khi B 2004)

Tm m phng trnh 2 23 3log x log x 1 2m 1 0+ + = c t nht mt nghim thuc

31;3 .

Gii:

t2

3t log x 1= + . Vi x3

1;3

th t [1;2] .Khi phng trnh cho tng ng vi : 2t t 2 2m+ =Bi ton tr thnh tm m phng trnh 2t t 2 2m+ = c nghim t [1;2]Xt hm s f(t) = t2 + t 2 vi t [1;2] . Ta c : f(x) = 2t + 1 > 0, vi mi t [1;2]Vy yu cu bi ton xy ra khi :

x [1;2] x [1;2]min f (x) 2m max f (x) f (1) 2m f (2) 0 2m 4 0 m 2


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Tm m phng trnh ( )2 2 4 2 2m 1 x 1 x 2 2 1 x 1 x 1 x+ + = + + c nghim.

Gii:iu kin xc nh ca phng trnh : x [ 1;1]

t 2 2t 1 x 1 x= + . Vi x [ 1;1] , ta xc nh iu kin ca t nh sau :

Xt hm s 2 2t 1 x 1 x= + vi x [ 1;1]

Ta c :

( )2 22 2 4

x 1 x 1 xx xt '

1 x 1 x 1 x

+ += + =

+ , cho t ' 0 x 0= =

x 1 0 1t 0 +

t2 2

0

Vy vi x [ 1;1] th t 0; 2 T 2 2 4 2t 1 x 1 x 2 1 x 2 t= + = . Khi , phng trnh cho tng ng vi :

( )2

2 t t 2m t 2 t t 2 mt 2

+ ++ = + + =

+

Bi ton tr thnh tm m phng trnh2t t 2

mt 2

+ + =+

c nghim t 0; 2

Xt hm s2

t t 2f(t)

t 2

+ +=

+vi t 0; 2 . Ta c : ( )

2

2

t 4tf '(t) 0, t 0; 2

t 2

= < +

Suy ra : ( )t 0; 2t 0; 2

max f (t) f (0) 1, min f (t) f 2 2 1

= = = =

By gi, yu cu bi ton xy ra khit 0; 2 t 0; 2

min f (t) m max f (t) 2 1 m 1

. y l cc

gi tr cn tm ca tham s.


Tm m phng trnh 2x mx 2 2x 1+ + = + c nghim thc phn bit.

Gii:


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Ta c: ( )( )

2

2

1x

2x mx 2 2x 1 1

3x 4x 1 mx 2

+ + = + + =

(*)

NX : x = 0 khng phi l nghim ca (2). Do vy, ta tip tc bin i :

( )2

1x2

(*)3x 4x 1

m 3x

+ =

Bi ton tr thnh tm m (3) c nghim thc phn bit { }1

x ; \ 02

+

Xt hm s23x 4x 1

f(x)x

+ = vi { }

1x ; \ 0

2

+ . Ta c :

{ }2

2

3x 1 1f '(x) 0, x ; \ 0x 2

+ = > + BBT :

x 0 1f(x) + +

f(x)

+ +

9

2

T BBT, ta thy : Yu cu bi ton xy ra khi9

m2

.

Vy vi9

m2

th phng trnh cho c nghim thc phn bit.

Bi 7 ( thi tuyn sinh i hc, Cao ng khi A 2007)

Tm m phng trnh ( )243 x 1 m x 1 2 x 1 1 + + = c nghim.

Gii:iu kin xc nh ca phng trnh : x 1Khi :

( )( )

( )2

44 2

x 1 x 1 x 1 x 11 3 m 2 3 m 2 2

x 1 x 1 x 1x 1

+ = + =

+ + ++

t 4x 1

tx 1

=

+( t 0 ). V 4 4x 1 21 1

x 1 x 1

= 0, phng trnh 2x 2x 8 m(x 2)+ = lun c hai nghimthc.

Gii:


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Bi 9 ( thi tuyn sinh i hc, Cao ng khi D 2007)

Gii:


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Bi 10 ( thi tuyn sinh i hc, Cao ng khi A 2008)

Gii:


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Bi tp t gii

Bi 1. Tm m bt phng trnh ( ) ( ) 2x 4 6 x x 2x m+ + ng vi mi x [ 4; 6] .

S : m 6Bi 2. Tm m bt phng trnh x 1 4 x m+ c nghim.

S : m 5

Bi 3. Tm m phng trnh 22 x 2 x 4 x m + + = c nghim.S : 2 2 2 m 2

Bi 4. Tm m h phng trnhx y 1

x x y y 1 3m

+ =

+ = c nghim.

S:1

0 m4

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