Chuyen de Ham So Bac i Va II

5/28/2018 Chuyen de Ham So Bac i Va II

1/24

Trn Thnh Minh Phan Lu Bin - Trn Quang Ngha

I S10

Chng 2.Hm SBc Nht v Bc Hai

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2/24

Chng2.Hm SBc Nht V Bc Hai

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2

2

1. i cng vhm sA.Tm tt gio khoa1/ nh ngha hm s: Cho D l tp con khc rng ca tp R .

Hm sf xc nh trn D l mt quy tc cho ng vi mi sx thuc D mt sthc y duy nhtgi l gi trca hm sf ti x, k hiu l y = f(x)

D gi l tp xc nh (hay min xc nh) , x gi l bin sc lp hay i sca hm sfTa vit f : D R x y = f(x)

2/ Cch cho hm s:Hm sthng cho bng biu thc f(x) v ta quy c rng : nu khng cgii thch g thm th tp xc nh ca hm sy = f(x) l tp hp tt ccc sthc

x sao cho biu thc f(x) c ngha.

3/ thca hm s:

x

y

O

nh ngha : Cho hm sy = f(x) xc nh trn D.Trong mt phng ta Oxy, thca hm sl tphp tt ccc im c ta (x;f(x)) vi x D

Ghi ch : Ngoi cch cho hm sbng biu thc f(x),ngi ta c thcho hm s bng bng gi tr, bng

biu hoc bng th

4/ Hm sng bin, nghch bin:nh ngha: Cho hm sy = f(x) xc nh trn kho

ng (a,b)

R Hm sf gi l ng bin (hay tng) trn khong (a;b) nu

vi mi x1,x2(a;b): x1< x2 f(x1) < f(x2) Hm sf gi l nghch bin (hay gim) trn khong (a;b) nu

vi mi x1,x2(a;b): x1< x2 f(x

1) > f(x2)

Ghi ch :Tnh ngha trn ta suy ra :

f ng bin trn (a;b) 2 11 2 1 2

2 1

( ) ( ), ( ; ), ,

f x f xx x a b x x

x x

> 0

f nghch bin trn (a;b) 2 11 2 1 2

2 1

( ) ( ), ( ; ), ,

f x f xx x a b x x

x x

< 0

Kho st sbin thin ca hm sl xt xem hm sng bin trn khong no,nghch bin trnkhong no trong tp xc nh ca n

5/ Hm schn,hm sl:

nh ngha: Cho hm sy = f(x) xc nh trn D f l hm schn nu vi mi x thuc D , th :

x cng thuc D v f(- x) = f(x) f l hm slnu vi mi x thuc D, th :

x cng thuc D v f(-x) = -f(x)

nh l:Hm schn th c thnhn trc tung lm trc i xngHm slth c thnhn gc ta lm tm i xng

B. Gii ton

Dng ton 1:Tm min xc nh ca hm sf:Ta cn nh:


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3

3

1

( )f x xc nh khi f(x) 0

T ( )f x xc nh khi f(x) 0

( )

( )

f x

g xxc nh khi g(x) > 0

V d1: Tm min xc nh ca hm s: f(x) =3

2 12

xx

Gii :

f(x) xc nh khi1 0 1 1

1& 22 0 2 2

x x xx x

x x x

22 3

3

xx

x

+ +

V d2: Tm min xc nh ca hm s: f(x) =

Gii3

2 3 0 332

3 0 23

x xx

xx

0 vi mi x

v 1 0x+ vi mi x

Vy hm sf xc nh vi mi x R

*V d4: nh m hm ssau xc nh trn (0,2):

f(x) =2

1m +

GiiHm sf(x) xc nh khi x m + 1 0 x m 1

Do hm sf(x) xc nh trn khong (0,2) th ta phi c m 1 (0,2)

Vy m 1 0 hay m 1 2 m 1 hay m 3

*V d5: Tm m hm sy = 1 2m x + + m xc inh vi mi x > 0

Gii

Hm sxc nh khi1

1 0

2 02

x mx m

mx m x

+

Do hm sxc nh vi mi x > 0 khi1 0

02

m

m

.

Vy m 0


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4

4

t

O

A

Bt'

*V d6:Cho hm s:y = f(x) =

2 1 2 0

0 1

2 1 1 3

x khi x

x khi x

x khi x


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5

5

Giia) Theo thta thy tp xc nh ca hm sl [-2;3]b) Ta c f(0) = 2 v f( -2) = 1c) Gi trln nht ca f(x) l 3 ; gi trnhnht ca f(x) l -1

Dng ton 3: Dng nh ngha xt tnh ng bin v nghch bin ca hm s

Ly x1v x2l hai gi trty thuc khong (a ; b) vi x1 x2 v xt nu :2

2 1

( ) ( )1

f x f x

x x

> 0 th hm sf(x) ng bin trn (a;b)

2

2 1

( ) ( )1f x f x

x x

< 0 th hm snghch bin trn (a;b)

V d1: Dng nh ngha chng minh hm sf(x) = 2x 3 ng bin trn R

GiiGi x1v x2l hai gi trty thuc tp R vi x1 x2ta c :

2 1 2 1

2 1 2 1

( ) ( ) (2 3) (2 3)

2 0

f x f x x x

x x x x

= = >

Vy hm sf(x) = 2x 3 lun ng bin trn tp xc nh R

V d2: Dng nh ngha xt tnh ng bin v nghch bin c

a hm sy = f(x) = x2 2x + 2 trn mi khong ( ;1) v (1; )+

Gii x2 ta c :Gi x1v x2l hai gi trty thuc vi x( ;1) 1

2 2 2 2

2 1 2 2 1 1 2 1 2 1

2 1 2 1 2 1

2 1 2 1 2 1 2 1 2 11 2

2 1 2 1

( ) ( ) ( 2 2) ( 2 2) 2( )

( )( ) 2( ) ( )( 2) 2

f x f x x x x x x x x x

x x x x x x

x x x x x x x x x x x xx x x x

+ + = = =

+ + = = = +

V x1v x2thuc nn x( ;1) 1< 1 v x2< 1 , do x1+ x2< 2

Vy 2 1

2 1

( ) ( )0

f x f x

x x

1 v x2> 1 nn x1+ x2> 2 ,do x1+ x2 2 > 0

Vy 2 1

2 1

( ) ( )0

f x f x

x x

>

Suy ra hm sng bin trn khong c

V d3: Kho st sbin thin ca hm sy = 21

trn mi khong xc nh v( ;1)

( ;1)

GiiGi x1v x2l hai gi trty thuc vi x( ;1) 1 x2 ta c :

2 1 2 1 2 1

2 1 2 1 2 1 2 1 2 1

2 2

( ) ( ) 1 1 2( ) 2

( )( 1)( 1) ( 1)( 1

f x f x x x x x

x x x x x x x x x x

= = = )

29

V x1v x2thuc nn x( ;1) 1- 1< 0 v x2 - 1 < 0 , do


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6

6

(x2 1)(x1 1) > 0 .Vy 2 1

2 1

( ) ( )0

f x f x

x x

0 v x2-1 > 0 , do 2 1

2 1

( ) ( )0

f x f x

x x

0 vi mi x1

v x2

Vy hm slun ng bin trn R

Dng 4 : Xt tnh chn , lca hm s- Tp xc nh D ca hm sphi i xng qua 0- Vi mi x D th -xD :

nu f(-x) = f(x) th hm schn trn D nu f(-x) = - f(x) th hm sltrn D

V d1: Xt tnh chn lca hm s: y = x 1+

GiiHm sy = 1x + xc nh khi x + 1 0 hay x -1 Ta nhn thy tp xc nh ca hm sl [ - 1 ; + ) khng i xng qua 0 ngha v vi x = 2 th x = -2 [ - 1 ; + )Vy hm sny khng chn v cng khng l

V d2: Xt tnh chn lca hm s y = f(x) = 2x3 4x

GiiTp xc nh ca hm sl R

R x R v f(-x) = 2(-x)3 4(-x) = -2x3+ 4x = - f(x)Vi moi x ta c : x

Vy f(x) l hm sl

V d3: Xt tnh chn lca hm sy = f(x) = 2 2x x+ +

Gii

Hm sxc nh khi Tp xc nh l [ - 2; 2]2 0

22 0

xx

x

+ 2

Vi mi x [-2;2] th x [-2;2] v f(-x) = 2 2x x + + = f(x)Vy f(x) l hm schn

3V d4: Xt tnh chn lca hm sy = f(x) = 2x x

GiiTp xc nh l R


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7

7

Vi mi x R th x R v ta c f(-x) = 2(-x) x 3= -2x x 3= - f(x)

Vy f(x) l hm sl

B. Bi tp rn luyn:2.1.Tm min xc nh cc hm ssau:

a) y =

2 1

1

x

x

+ b) y = 2

x

x

c) y =1

1

x

x

+

d) y = 2 1 2 x

2.2. Cho hm sf(x) =2

2 1 1

1 1

x khi x

1x khi x

<

a) Tm min xc nh ca hm sf

b) Tnh f(-2) , f(-1) , f(2

2) , f(1)

*2.3. Tm m hm sy = 2x m x m + +1 xc nh vi m

i x > 0

2 4. Gi ( C ) l thca hm s y = x x .im no sau y thuc ( C )

A(-1; 1) B(-1 ; -1) C(1; -1) D(1 ; 0)

*2.5. Tm im (xo; yo) thuc thca hm sy =1mx

vi mi gi trca mm

2.6. Vthca hm sy = [x] gi l phn nguyn ca x vi x [-2 ; 3] x < y+1)(vi mi sthc x c mt snguyn y duy nht tha y

2.7. Xt sbin thin ca hm strn mi khong

a) y =3

trn mi khong (- ,0) v (0 ; + )

b) y = -x2+ 2x trn mi khong (- ;1) v (1 ; + )

c) y = 1 trn khong [1 ; + )*d) y = x

3+ 2 trn khong (- ; + )

2.8. Xt tnh chn lca cc hm ssau :a) f(x) = -2x + 5 b) f(x) = -x3+ 2x

c) f(x) =3

2 d) f(x) = x

2- 2 x

x * 2.9. Xt tnh chn lca hm sDirichlet :

D(x) = 1

0

khi x Q

khi x Q

2.10. Cho hm sy = 2 x x + + 2 Cu no sau y ng?a) Min xc nh l x > -2b) Hm slc) thhm sc trc i xng l trc 0yd) im A ( 0 ; 2 ) thuc thhm s

D. Hng dn - p s:2.1. a) Tp xc nh l R

b) Min xc nh l R\ { }2; 2 + c) Min xc nh l x [-1 ; + ) v x 1


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8

8

d) Hm sxc nh khi2 1 0 1

22 0 2

xx

x

2.2.a) Min xc nh ca hm sl (- ; 1]

b) f(-2) = -5 ; f(-1) = 0 ; f(2

2) =

2

2; f(1) = 0

* 2.3. Hm sxc nh khi0

12 1 0

2

x mx mm

x m x

+

Do hm sxc nh vi mi x > 0 th0

10

2

m

m

Vy m 0

2. 4.. im B thuc th( C )

*2.5. im (xo; yo) thuc thca hm sy =1mx

x m

khi ta c :

0

1oo

mxym=

hay xoyo myo= mxo 1 vi xo m

xoyo+ 1 = m(xo+ yo)Phng trnh ny c tha vi mi m xokhi :

(x0

1 0

o o

o o

x y

x y

+ =

+ =o = 1; yo= -1) v (xo= -1 ; yo=1) vi m 1 v m -1

2.6. y

O x

2.7. a) hm snghch bin trn mi khongb) hm sng bin trn (- ;1) v nghch bin trn (1 ; + )c) hm sng bin trn [1 ; + )

d) hm slun ng bin trn (- ; + )

2.8. a) f(x) = -2x + 5 khng chn v khng lb) f(x) = -x3+ 2x l hm sltrn R

c) f(x) =3

2x khng chn v khng l

d) f(x) =x2- 2 x l hm schn trn R

* 2.9. Vi mi x Q th x Q v ta c D(-x) = 1 = D(x)Vi mi x Q th x Q ( v dx = 2 th x = - 2 )

v ta c D(-x) = 0 = D(x)Vy D(x) l hm schn

2.10. Hm sny chn nn thc trc i xng l Oy.


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9

9

2 . Hm sbc nhtA.Tm tt gio khoa:1. nh ngha : Hm sbc nht l hm sc dng y =ax + b,trong a v b l cc hng svia 02. Sbin thin

Tp xc nh l R Khi a > 0 hm sng bin trn R

x - + y = ax + b

( a > 0 )+

-

Khi a < 0 hm snghch bin trn R

x - + y = ax + b( a < 0)

+

-

3. th:thca hm sy = ax + b ( a 0) l mt ng thng khng cng phng vi cc trc ta .a gi l h sgc ca ng thng.c bit :

b 0 thct trc Ox ti A(b

a

; 0) v trc 0y ti B(0;b)

b = 0 thhm sy = ax qua gc to0 v qua im C(1 ; a)y y

B

A x x

0 0

Ghi ch : Cho hai ng thng (d) y = ax + b v (d) y = ax + b (d) // (d) a = a v b b (d) ct (d) a a thca hm sy = b (hng s) l ng thng song song vi trc honh

4. Hm sy =

Hm sny xc nh vi mi gi trca x v l hm schn.Theo nh ngha gi trtuyt i ta c :


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10

10

0

0

x khi xx

x khi x

=

0Bng bin thin

x - +

y = 2x - 3+

- thl ng thng qua hai im A ( 0 ; - 3) v B( 2 ; 1)

V d2: Xt sbin thin v vthhm sy = - 2 +2


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11

11

GiiTp xc nh l R

Hm slun nghch bin trn R v a = -1

2< 0

O

x

y

A

B

Bng bin thin

- + x + y = -

2

x+2

-

thl ng thng qua 2 im A(0 ; 2) v B(4; 0)

Dng 2 : Tnh cc hsa v b ca hm sy = ax + b

V d1: Tnh a v b thca hm sy = ax + b qua 2 im A(

2 ; -2) v B(-1 ; 4)

Giithqua A (2 ; -2) a(2) + b = - 2thqua B( -1 ; 4) a(-1) + b = 4

Gii hphng trnh2 2

4

a b

a b

+ =

+ = ta c a = -2 v b = 2

Vy y = -2x + 2

V d2: Cho ng thng (d) y = 2x + 1.Tnh a v b th(d) ca hm sy = ax + bsong song vi (d) v qua im A(1 ; -3)

GiiTa c (d) // (d) a = 2 v b 1 ( hsgc bng nhau)Do phng trnh ca (d) l y = 2x + bim A(1 ; -3) (d) -3 = 2(1) + b b = - 5Vy phng trnh ca (d) l y = 2x 5

V d3: nh m hai ng thng (d) y = 2x 3 v (d) y = -x + 2m -1 ct nhau ti mt imtrn trc 0y

Gii(d) ct trc 0y ti im c ta x = 0 ; y = - 3(d) ct (d) ti im trn trc 0y khi 2m 1 = -3 2m = - 2 m = -1

V d4: Vthca hai hm sy = x 1 v y = -1

2x + 2 trn cng mt htrc ta . Dng

thv thli bng tnh togiao im ca hai thtrn

Giithca hm sy = x 1 l ng thng (d) qua hai im ( 0 ; -1) v (1 ; 0)

thca hm sy = -1

2x + 2 l ng thng (d) qua hai im ( 0 ; 2) v (4 ; 0)

Theo thta thy hai ng (d) v (d) ct nhau ti im c ta (2 ; 1)Thli bng tnh:Togiao im c(d) v (d) l nghim ca hphng trnh :


12/24

y



12

12

1

12

2

y x

y x

=

= +

So snh y ta c : x- 1 = -1

2x + 2

-1

3 4

-1

1

2

2x 2 = -x +4 3x = 6 x = 2Thay x = 2 vo y = x 1 ta c y = 1 .Vy ta giao im ca hai thl (2 ; 1)

Dng 3: Vthhm sy = ax b+

V d1: Vthca hm sy = 2 1+ .

Tm gi trnhnht ca hm sny.

Gii Nu x + 1 0 hay x -1 th y =2(x+1) = 2x + 2 ,thl na ng thng gc A( - 1 ; 0) v qua im B(0 ; 2)

Nu x + 1 < 0 hay x < -1 th y = -2(x + 1) = -2x 2 , thl na ng thng gc A v quaim C( -2 ; 2)

Gi trnhnht ca hm sl 0 khi x = -1

V d2 : Vthca hm sy = 2 - 1 v tm gi trnhnht ca hm s

Gii Nu x 0 th y = 2x 1,thl na ng thng gc A( 0 ; -1) v qua B ( 1 ; 1) Nu x < 0 th y = -2x -1 .thl na ng thng gc A v qua C( -1 ; 1)V 2x 0 vi mi x nn y -1

Vy gi trnhnht ca y l 1 khi x = 0

1 2

y

-2 -1

1

2

3

y

C B

A

-1 1

-1

1

2

(-1,1) (1,1)


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13

13

*V d3: Vthhm sy = 2 4 4 2 1x x x +

Gii

Ta c y =2 24 4 2 1 ( 2) 2 1 2 2 1x x x x x x + = =

Ta c bng xt du :x 1 2

x - 2 - - 0 +

x - 1 - 0 + +

yDo :

khi x < 1 th :

-1 1 2 3

-3

-2

-1

1 y = 2 x + 2(x 1) = x khi 1 x 2 th : y = 2 x -2(x 1) = -3x + 4

khi x > 2 th :y = x 2 2(x- 1) = -x

th( xem hnh bn)

*V d4:Cho hm s

y =

2

0

1 0

xx khi x

x

khi x

+

=

Tm tp xc nh v vthca hm sny

GiiTp xc nh l R

Khi x 0 ta c y = x +x

x= x + 1 v khi x = 0 th y = 1

Vy thca hm sl ng thng y = x + 1

C.Bi tp rn luyn2.11. Vthcc hm ssau :

a) y = 2x 4 b) y =2

3x c) y = -

14

3 d) y =

0

2 0

x khi x

x khi x


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14

14

2.16. nh m hai ng thng y = 2x + 4 v y = - x + m + 2 ct nhau ti mt im trn trc

honh

2.17. Vthcc hm s:

a) y = 2x b) y = x + 1 c) y = 2 2 1x x x +

*2.18. Vthca hm s: y = 2 24 4 4 4 1x x x x + + +

*2.19Tm tp xc nh v vthhm ssau :

y =

2

0

1 0

xx khi x

x

khi x

+ =

D.Hng dn gii - p s2.11. a) thca hm sy = 2x 4 l ng thng qua 2 i

m ( 0; - 4) v( 2 ; 0)

b) thca hm s y =2

3x l ng thng qua gc O v im ( 3 ; 2)

c) thca hm sy = -1

43x l ng thng qua 2 im (0;-4) v (-3;-3)

d) thca hm sy = l hai na ng thng qua gc O0

2 0

x khi x

x khi x


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15

15

b)1 0

11 0

x khi xy x

x khi x

+ = + =

+ 0 th hm sgim trn (- ; 0) ; tng trn (0;+ ),t cc tiu khi x = 0

nu a < 0 th hm stng trn (- 0) ;gim trn (0;+ ).t cc i khi x = 0Bng bin thin :

a > 0 a < 0

x - + x - +

y + 0 + y - 0 -

thca hm sl parabol.nh l gc O v trc i xng l Oya > 0 a< 0

xy

y


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16

16

3.Hm sy ax2+ bx + c vi a 0 Tp xc nh l R

Nu a > 0 th hm sgim trn khong (- ; - ) v tng trn khong2

b

a

( - ;+ )2

b

a

Nu a < 0 th hm stng trn khong (- ; - ) v gim trn khong2

b

a

( - ;+ )2

b

ay

Bng bn thin

a> 0

Hm st gi trcc tiu bng -4 2

bkhi x

a a

=

a < 0

Hm st gi trc

c i bng -4 2

bkhi x

a a

=

thhm sy = ax2

+ bx + c l mt parabol ,nh I ( - 2

b

a ; - 4a

) v nhn ng thng x = - 2

b

a

lm trc i xngCch v: Mun vparabol (P) : y = ax2+ bx + c ta lm nhsau:

Vnh I ( -2

b

a; -

4a

) v trc i xng x = -

2

b

a

Vthm vi im c honh gn gi trhonh nh v im i xng ca chng quatrc i xng .Lu giao im ca (P) vi trc Oy l ( x = 0 y = c )

B. Gii ton :Dng 1 : Xt sbin thin v vth

V d1: Xt sbin thin v vthhm sy = x2 2x -3

x- - 2

b

a +

y + +

CT

y

x - -2ba

+

C- + y


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17

17

GiiTp xc nh l R

a = 1 > 0 , ta c x = -2

b

a= 1 v y = -

4a

= - 4 .Do hm sgim trn khong ( - ; 1) v tng

trn khong (1;+ ),gi trnhnht l -4

yBng bin thin

x - 1 +

y

+ +

-4

thl parabol ,nh I ( 1 ; -4) v trc i xng l ngthng x = 1Giao im ca parabol vi trc Ox : y = 0 suy ra x2 2x 3= 0

x = -1 ; x = 3 ; giao im ca parabol vi trc Oy l x = 0y = - 3

(-1,0) (3,0)

(0,-3) (2,-3)

(1,-4)

V d2 : Xt sbin thin v vthhm sy = - x2+ 2x 2

GiiTp xc nh l R

a = -1 < 0 , x = -2

b

a= 1 ; y = -

4a

= - 1.Do hm stng trn khong

( - ; 1) v gim trn khong ( 1 ; + ) ,gi trln nht l 1 yBng bin thin

x - 1 +

y

- 1

- - (1,-1)

(0,-2) (2,-2)thl parabol nh I (1; -1) .trc i xng x = 1,ct trc Oyti x = 0 ; y = -2

*Dng 2 : Vthca hm sc cha gi trtuyt i

V d: Vthca hm s y = 2 2x

GiiTp xc nh l RTa c x2 2x = x( x 2) .Do :

khi x < 0 hay x > 2 th y = x2 2x khi th y = - x0 x 2 2+ 2x

Vy thca hm s y =2

2x x l hp ca hai parabol : y = x2 2x bphn trong on 0 2x


18/24

y



18

18

v y = - x2+ 2x ly phn trong on 0 2x

Dng 3 : Tnh cc hsa,b,c ca hm sy = ax2+ bx+ c

V d1: Tnh a v b bit parabol y = ax2+ bx + 2 c

nh I( 2 ; - 2)Gii

Honh nh parabol l x = -2

b

a= 2 (1)

im I ( 2 ; -2) thuc parabol nn ta c - 2 = a(2)2+ 2b+2 (2)T(1) ta c b = - 4a . Thay vo (2): - 2 = 4a 8a + 2Vy a = 1 v b = - 4

V d2: Tnh a,b,c bit parabol y = ax2+ bx + c c nh trn tr

c honh v qua hai im A( 0; 1)

v B( 3 ; 4)

Gii

nh ca parabol thuc trc Ox nn tung nh y = -4a

= 0 hay 4ac b

2= 0 (1)

A (0 ; 1) thuc parabol nn a(0)2+ b(0) +c = 1 (2) B( 2 ; 1) thuc parabol nn a(2)2+ b(2) + c = 1 (3)

(2) cho c = 1 .Thay vo (3) ta c :

4a + 2b + 1 = 1 hay 2a + b = 0 hay b = - 2a

Thay b v c vo (1) :4a(1) (- 2a)2= 0 hay 4a 4a2= 0 hay a( 1 a) = 0

V a 0 nn ta suy ra 1 a = 0 Vy a = 1 , b = -2 , c = 1

*V d3: Cho hm sy = x2 2mx + m + 2 ( m > 0)a) nh m thl parabol c nh nm trn ng thng y = x + 1b) Vthvi m va tm

Gii

Tonh x = -2

b

av y =

24

4

ac b

a

tha phng trnh y = x

+ 1

Nn ta c :

24

4

ac b

a

= - 2

b

a + 1 4ac b

2

= - 2b + 4a ( v a0)

Thay a = 1 , b = - 2m , c= m +2 vo phng trnh ta c :4(m + 2) 4m

2= 4m + 4 m 2= 1 m = 1 v

m > 0Vy y = x2 2x + 3thl parabol c nh I(1 ; 2) ,trc i xng x = 1

(1,1)

(0,0) (2,0)

y

(0,3) (2,3)

(1,2)


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19

19

C.Bi tp rn luyn

2.19. Xt sbin thin v vthcc hm ssau :a) y = x2+ 2x +1 b) y = - x2+ 1

c) y = x2 2x 2 d) y = -1

2x2+ 2x

*2.20. Vthcc hm ssau :

a) y = x2+ 2 x b) y = x 2x

2.21. Tnh a v b bit parabol y = ax2+ bx 3 c nh I (1 ; -2)

2.22. Tnh a , b ,c bit parabol y = ax2+ bx + c c nh trn trc honh v qua hai i

m A( 0;4)

v B( - 1 ; 1)

2.23. Tnh a , b, c hm sy = ax2+ bx + c t gi trln nht bng 2 khi x = 1 v thqua

im A( -1 ; -8)

2.24. Tnh m thca hm sy = mx2 2mx m 2 c nh thuc ng thng y = 2x 1 (

m khc 0)

2.25. Vthca hai hm sy = x + 1 v y = x2 2x + 1 trn cng mt hthng trc ta ri

xc nh ta giao im ca chng

*2.26. Vthca hm s: y = 2 4 1

4 1

x khi x

x khi x

+

+ <

2.27. Vthca hm sy = - x2+ 2x .Dng thtm x y > 0

2.28. Vthca hm sy = x2+ 2x 3 .Dng thtm x y 0

D.Hng dn gii - p s:2.19. a) Hm sy = x2+ 2x + 1 c x = -

2

b

a= - 1 v a = 1 > 0

Vy hm snghch bin trn khong khong ( - ; -1) v ng bin trn khong (-1;+ ), gi trnhnht l 0thl parabol c nh I ( -1 ; 0)

b).Hm sy = - x2+ 1 c x = -2

b

a= 0 v a = - 1


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20

20

2.21. Ta c hphng trnh :

12

3 2

b

a

a b

=

+ =

Vy a = - 1 v

b = 2

2.22 Ta c hphng trnh :

Vy a = 1 ; b = 4 ; c = 4 hay a = 9 ; b = 12 ; c = 4

24 0

4

1

ac b

c

a b c

=

= + =

2.23. Ta c hphng trnh :

12

2

8

b

a

a b c

a b c

=

+ + =

+ =

Vy a = -5

2

; b = 5 ; c = -1

2

2.24.Ta nh l x = 1 , y = -2m 2 . Thay gi trca x v y ny vo phngtrnh y = 2x 1 ta c : -2m 2 = 2 -1 Vy m = - 3/2

2.25.Hc sinh tv.

Ta giao im ca 2 thl nghim ca hphng trnh :2

1

2 1

y x

y x x

= +

= +

So snh y ta c x2 2x + 1 = x + 1 hay x (x - 3) = 0Vy x = 0 ; y = 1 v x = 3 ; y = 4

*2.26 .Ta vparabol y = - x2+ 4 v gch bphn x < - 1

y y

-1 1 2 3

-4

-2

2

4


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21

21

yv vng thng y = x + 4 ri gch bphnx > - 1

-2

2

-4

-2

2

427.Phn thng vi y > 0 l phn thpha trn trc honh (mu hng) .Cn cvo hnh vta suy ra:hi 0 x < 2.

2.28. Theo thta thy: y 0 (ng vi phn thph di trc honh, mu hng) -3

x 1

4.Trc nghim cui chngA.Cu hi1.Cho hm sf(x) = 2x - .Cu no sau y ng ?

a) f(x) l hm schn b) f(x) l hm slc) f(x) l hm skhng chn v khng ld) Min xc nh ca l hm sl x > 0

2. Tp xc nh ca hm sy = 2x l :

a) x 2 b) vi mi x R c) vi mi x 2 d) (- ;2]

3. Cho hai hm sy = f(x) v y = g(x) l hai hm schn trn cng tp xc nh D . Cu nosau y ng ?

a) Hm sy = f(x) + g(x) l hm schn trn D

-2 2 4

-6

-4

-2

x

y

y

-4 -3 -2 -1 1 2

-4

-2

2


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22

22

l hm schn trn D

. ng (a,b).Cu no sau y ng?

ng bin trn khong (a,b)u

.

b) Hm sy= f(x) g(x) l hm schn trn D c) Hm sy = f(x).g(x) d) Cba cu u ng

4 Cho y = f(x) v y = g(x) l hai hm sng bin trn kho a) Hm sy = f(x) + g(x) ng bin trn khong (a,b)

b) Hm sy = f(x) g(x) ng bin trn khong (a,b) c) Hm sy= f(x).g(x)

d) Cu a v b ng

5 Cho hm sy = 1x xc nh trn R .C u no sau y ng?

a) Hm snghch bin trn khong (- ; 1)n khong (1; + b) Hm sng bin tr )

i

a v b bng bao nhiu?= 3 d) a = 1 ; b = -4

.C o sau y ng ?

; 2)n R

Parabol y = -

c) Cu a v b u ngb) Hm sny chn trn R

6.B t thca hm sy = ax + b qua hai im A(0,-3) v B( -1;-5).Th a) a = 2 ; b = -3 b) a = -2 ; b = 3 c) a = 2 ; b

7 h hm sy = -2x + 3 Cu no a) Hm sng bin trn R

b) Hm snghch bin trn ( -2 c) Hm snghch bin tr d) Cu b v c u ng

1

4x8. 2 nh l :

-1) d) (1 ; 0)

a x th y = x2 x

a) ( 1 ; + ) b) x

+ 1 c to

a) ( -1 ; 0) b) ( 0 ; 1) c) ( 0;

9.Vi gi trno c 5 + 4 < 0

x ( 1 ; 32)

c) x ( 1 ;4) d) x (3

2; + )

1 .Togiao im ca parabol y = x0

bol c nh c a h ;0)

2

ng?

2 + 2x 1 v ng thng y = x 1 l:a) (0;-1) v (-1;2) b) (-1;0) v (-1;2)c) (0;-1) v (-1;-2) d) (2;1) v (-1;2)

11.Gi trno ca a v c thca hm sy = ax2+ c l para (0; - 2) v mt giao i m t v i tr c honh l ( -1 a) a = 1v c = -1 b) a = 2 v c = -1

c) a = 2 v c= -2 d) a = -2 v c = -

12.Cho hm sy = -2x2+ 4x 1.Cu no sau y)a) Hm sng bin trn khong ( 1 ; +

)b) Hm snghch bin trn khong (1 ; + c) thct trc tung ti im ( 0 ; -1)

d) Cu b v c u g

hm sy = -x2+ bx 3.Gi trca b l bao nhiu bit thl parabol c honh nh l= 2

Vi gi trno ca b th thca hm sy = x2+ bx ct trc honh ti 2 im 0 (0;0) v A(2 ;

a) b = 4 b) b = - 2 c) b = 2 d) C3 cu trn u sai

n

13.Chox

a) b = 2 b) b = -2 c) b = 4 d) b= -4

14.0)


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23

23

15.thca hm sy = (x 2)2c trc i xng l :a) trc 0y b) ng thng x = 2c) ng thng x = 1 d) khng c

16.Cho hm sy = x2+ bx +c bit thl parabol c nh I( 1; 2) th b + c =a) 1 b) 2 c) -1 d) 2

17.Cc im no sau y thuc thhm sy = x2-2 x a) ( -1 ; 3) b) (1; -1) c) (2; 4) d) (-2; 4)

18.Ta giao im ca thhai hm sy = 1 +1 v y = 2 l :

a) (0 ; 2) v (1; 2) b) (2 ;2) v (-1; 2)

c) (0; 2) v (2;2) d) skhc19 thca hm sy = ax + b qua nh ca parabol y = x2 2x + 3 th

a + b bng :a) 0 b) 1 c) 2 d) -2

20.Trong cc hm ssau hm sno l hm l

(I) y = x3 2x (II) y = 2 (III) y = x 2x x

a) (I) v II) b) (I) v (III) c) (II) v (III) d) C

ba hm s

B.BNG TRLI .1c 2b 3d 4a 5c 6a 7d 8b 9c 10c11c 12d 13c 14b 15b 16a 17b 18c 19c 20d

C.HNGDN GII1c.Hm sny xc nh trn RVi mi x thuc R th x thuc R v ta c

f(-x) = 2(-x) - x = -2x - x Vy hm sf(x) khng chn v khng l

2b. Hm sxcnh vi mi x R .

3d.V f(x) v g(x) chn trn R nn vi mi x thuc R th x thuc R v ta c :f(-x) = f(x) v g(-x) = g(x) nnf(-x) + g(-x) = f(x) + g(x) v f(-x) g(-x) =f(x) g(-x)f(-x).g(-x) = f(x).g(x)

4a. Vi mi x1, x2(a,b) v x1 x2ta c

2 1 2 1

2 1 2 1

( ) ( ) ( ) ( )0 ; 0

f x f x g x g x

x x x x

> >

5c. Hm sy = 1x xc nh trn R

Khi x < 1 th y =x + 1 nn hm snghch bin trn (- ; 1)Khi x > 1 th y = x 1 nn hm sng bin trn (1; + )

6a. thca hm sy = ax + b : qua A(0;-3) cho b = -3 qua B(-1;-5) cho -5 = -a 3 nn a = 2

7d.Hm sy = -2x + 3 c a = -2 < 0 nn lun nghch bin trn R

8b. Tonh ca parabol l x = 0 ; y = 19c. Vthca hm sy = x2 5x + 4 .Theo thta thy y < 0 khi

1 < x < 4


24/24



24

24

10c. Ta giao im ca parabol v ng thng l nghim ca hphng trnh

So snh ta c x2 2 1

1

y x x

y x

= +

=

2+ 2x 1 = x 1 hay x2+ x = 0

Hay x(x + 1) = 0 Vy x = 0 ; y = -1 v x = -1 ; y = - 2

11c. a = 2 v c = -212d. Hm sy = -2x2+ 4x 1 c honh nh x = -

2

b

a= 1 v a = -2 < 0

Nn hm snghch bin trn khong (1 ; + ) v ct trc tung ti x = 0 , y=-1

13c. thca hm sy = -x2+ bx 3 l parabol c honh nh x = -2

b

a=2

Do b = -4a = -4(-1) = 4

14b. thca hm sy = ax2+ bx ct trc honh ti im c honh l nghimca phng trnh ax2+ bx = 0 hay x(ax + b) = 0

Vy x = 0 v x = -b

a

= 2 .Suy ra b = -2a = -2

15b. thca hm sy = (x 2)2c trc i xng l ng thng x = 216a. thca hm sy = x2+ bx +c c nh I(1; 2) cho ta :1 + b + c = 2 Vy b + c = 117b. Xt hm shm sy = x2-2

Thay x = -1, y = 3 ta c 3 = 1 2 khng thThay x = 1, y = -1 ta c 1 = 1 2 tha

x 1 +1 v y = 2 l18c. Honh giao im ca thhai hm sy =

1nghim ca phng trnh : + 1 = 2 1 = 1

52Vy c hai giao im (x = 0 , y = 1) v (x = 2 = y =1)

19c. nh ca parabol y = x2 2x + 3 l I ( 1 ; 2)Do thca hm sy = ax + b qua I (1; 2) cho ta a + b = 2

20d. y = x3 2x c tp xc nh R v y(-x) = (-x)3 2(-x) = -x3+ 2x = - y(x)

Vy (I) l hm sl

y =2 2 2

x

=

c tp xc nh l R \{ }0 v y(-x) = = -y(x)

Vy (II) l hm sl

y = x 2x x xc tp xc nh l R v y(-x) = (-x) -2(-x) = - x +2x =-y(x)Vy (III) l hm sl

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Chuyen de Ham So Bac i Va II