51719_bai Toan Giao Diem Cua Hai Do Thi Va Ung Dung Bien Luan So Nghiem Cua Phuong Trinh (1)

7/31/2019 51719_bai Toan Giao Diem Cua Hai Do Thi Va Ung Dung Bien Luan So Nghiem Cua Phuong Trinh (1)

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Trang 1

BI TON GIAO IM CA HAI TH HM S V NG DNG BIN LUN S NGHIM CA PHNG TRNH

Trnh Ba

1. Bi ton muCho hm s

2 1

2 1

xy

x

c th(C). Xc nh ta giao im ca th (C)

v ng thng y = x + 2.(trch thi tt nghip THPT nm 2011)

GiiPhng trnh honh giao im ca th(C) v ng thng y = x + 22 1

22 1

xx

x

2 1 ( 2)(2 1)x x x (do x = 1/2 khng l nghim ca phng trnh ny)

2

1

2 3 0 3

2

x

x xx

Vi 3 12 2

x y

Vi 1 3x y

Vy ta giao im ca (C) v ng thng y = x + 2 l3 1

; , (1;3)2 2

2. C sl lunCho hai hm sy = f(x) v y = g(x) c th ln lt l (F) v (G), nu im

( , )M MM x y l giao im ca (F) v (G) th ( )M My f x v ( )M My g x . Ni cch

khc, ta M l nghim ca hphng trnh( )

( )

y f x

y g x

. Do honh ca M l

nghim ca phng trnh f(x) = g(x) (1). Ta ni (1) l phng trnh honh giaoim ca (F) v (G); s nghim ca (1) bng sgiao im ca (F) v (G). Do ta cmi lin h gia hai bi ton: giao im ca hai th v nghim ca phng trnhhonh giao im. Trong nhiu trng hp, tm giao im (hay sgiao im)ca hai th ta phi xt phng trnh honh giao im ca chng v ngc li bin lun s nghim ca mt phng trnh th ta xem l phng trnh honh giaoim ca hai thv dng th bin lun.

3. Mt s bi ton vgiao im ca hai thBi ton 1 (trch thi tt nghip THPT nm hc 1997)

Cho hm s y = 3x 3x 1. 1. Kho st s bin thin v v th ( C ) ca hm s.2.Mt ng thng d i qua im un ca th (C) v c h s gc k. Bin lun theo k

sgiao im ca th(C) v ng thng d. Tm ta cc giao im trong trnghp k = 1.

Gii1.Bn c t gii2.im un ca (C) l I(0;1)


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Trang 2

Phng trnh ng thng d: y = kx + 1Phng trnh honh giao im ca (C) v d

3

2

2

x 3x 1 = kx + 1 (1)

x(x 3 ) 0

0

3 (2)

k

x

x k

Vi k < - 3 th (2) v nghim nn (1) c 1 nghim, do d v (C) c 1 im chung. Vi k = -3 th (2) c nghim kp bng 0, do (1) c 1 nghim nn d v (C) c 1 imchung.

Vi k > -3 th (2) c hai nghim phn bit khc 0 nn (1) c 3 nghim phn bit, do d v (C) c 3 im chung.

Trng hp k = 1 th (2) c 2 nghim l x = 2 nn d ct (C) ti 3 im phn bit l(0;0), (2;3), (-2; -1).

Bi tng t(trch thi tt nghip THPT nm hc 1996)

Cho hm s y=2x (m 3)x m

x 1

, m l tham s, th l (Cm).

1.Kho st s bin thin v v th ( C ) ca hm s khi m = -2.2.ng thng d i qua gc ta c h s gc l k. Bin lun theo k sgiao im ca

ng thng d v th (C).

Bi ton 2 (trch thi i hc khi A 2010)Cho hm s 3 22 (1 )y x x m x m (1), m l tham s thc.

1. Kho st s bin thin v v th ca hm s khi m = 1.2. Tm m th hm s (1) ct trc honh ti 3 im phn bit c honh

1 2 3, ,x x x

tha mn iu kin 2 2 21 2 3 4x x x

Gii1. Bn c t gii2. Phng trnh honh giao im ca (Cm ) v trc honh c dng3 2

2

2

2 (1 ) 0

( 1)( ) 0

1

0 (1)

x x m x m

x x x m

x

x x m

iu kin bi ton c tho mn khi phng trnh (1) c hai nghim 1 2,x x phn bit

khc 1 sao cho2 2 2 2 2

1 2 1 2 1 2 1 21 4 3 ( ) 2 3x x x x x x x x . iu ny xy ra

khi

1 4 0 110 4

01 2 3

m

mm

mm

Vy vi 1

;1 \ 04

m

th th hm s (1) ct trc honh ti 3 im phn bit c honh

1 2 3, ,x x x tha mn iu kin 2 2 2

1 2 3 4x x x

Nhn xt. gii bi ton trn cn ch mt s im:


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Trang 3

Nhm c mt nghim ca phng trnh bc ba ( cs l nhm x kh m trc, bncht ca nghim ny l honh ca im c nh ca th hm s bc ba m im nmtrn trc honh). Sau quy phng trnh bc ba v phng trnh tch, ri tm iu kin

phng trnh bc hai c hai nghim phn bit khc nghim nhm c v tho thm iukin no ca bi ton. tm m sao cho tho iu kin 2 2 2

1 2 3 4x x x , trong nhm c 1 nghim, v

tri ca bt ng thc ny l biu thc i xng gia cc nghim gi ta dng nh l Viet.Nu hc sinh no khng nh nh l Viet th vn c th gii quyt trc tip iu kin trn.Chng hn:

Khng mt tng qut, xem3 1x , ta c

2 2 2 2 2

1 2 3 1 24 3x x x x x , trong

1 2,x x l hai nghim ca phng trnh (1), ta c :

1,2

1 1 4

2

mx

, do

2 2

2 2

1 2

1 1 4 1 1 43 3

2 2

1 2 3

1

m m

x x

m

m

Kt hp vi iu kin c hai nghim phn bit khc 1 ta c kt qu ca m. Nhng nu iu kin i vi 3 honh giao im l 3 3 3

1 2 3 4x x x hay thm ch l4 4 4

1 2 3 4x x x th cch tm nghim ri th trc tip vo s gp nhiu phc tp. Tt nht,

hc sinh nn nh c nh l Viet v mt s biu thc i xng gia cc nghim. (3 3 3 4 4 2 2 2

1 2 1 23 , ( 2 ) 2x x S PS x x S P P )

Bi ton 3 (trch thi i hc khi A 2011)

Cho hm s1

2 1

xy

x

1. Kho st s bin thin v v th (C) ca hm s cho.2. Chng minh rng vi mi m ng thng y = x + m lun ct th (C) ti hai imphn bit A v B. Gi 1 2,k k ln lt l h s gc ca cc tip tuyn vi (C) ti A v B.

Tm m tng1 2

k k t gi tr ln nht.Gii

1. Bn c t gii2. Phng trnh honh giao im ca ng thng y = x + m v (C)

1

2 1

xx m

x

(1)

1 ( )(2 1)x x m x (do x = 1/2 khng l nghim ca phng trnh ny)2

2 2 1 0x mx m

(2)2 2' 2( 1) ( 1) 1 0,m m m m R

Do phng trnh (1) lun c hai nghim phn bit vi mi m nn ng thng y = x +m lun ct (C) ti hai im phn bit A, B. Gi ( , ), ( , )

A A B BA x y B x y th ,

A Bx x l hai

nghim ca phng trnh (2). Khi

1 2

1'( )

2 1A

A

k f xx


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Trang 4

2 2

1'( )

2 1B

B

k f xx

Khi

2 2

1 2 2 2 2

2 2

2

1 1 4 4 1 4 4 1

2 1 2 1 2 1 2 1

4 4 2

4 2( ) 1

B B A A

A B A B

A B A B

A B A B

x x x xk k

x x x x

x x x x

x x x x

M2 2 21

; . 12

A B A B A B

mx x m x x x x m m

Do

22

1 2 2

4( 1) 2( 1) 2(4 6 8) ( )

2( 1) 2 1

m m mk k m m g m

m m

Hm g(m) l mt parabol c nh3 23

( ; )4 4

I v c b lm quay xungnn t gi tr ln

nht ti I, tc l ti3

4

m .

Bnh lunTrong bi ton trn ta thy honh hai im A, B l nghim ca phng trnh (2) nu tnh rath ph thuc tham s m, nn ta hy khoan ngh n vic phi tm honh A, B m hy xem

1 2

1'( )

2 1A

A

k f xx

2 2

1'( )

2 1B

B

k f xx

Th k1 + k2 l biu thc i xng vi i vi hai nghim ca phng trnh honh giao

im, do ta c th dng nh l Viet gii quyt bi ton.

By gi ta s xt bi ton sau m vic nhm nghim ca phng trnh honh giao im (l

phng trnh bc ba ph thuc m) khng d nh nhng bi trc v khng c nghim nguyn

m l nghim hu t (thc cht ta c phng php nhm nghim hu t cho mt phng trnh

a thc bc n bt k, nhng y l bi ton ph thuc tham s m). y ta da vo im c

nh ca h th m im thuc ng thng cn xt tng giao vi h th. Khi

honh im c nh l nghim ca phng trnh honh giao im m ta xt n. Xt h

th y = f(x,m), khi im (x0;y0) l im c nh nu phng trnh y0=f(x0;m) vi bin m

c nghim vi mi m. T y ta tm c im (x0;y0).

Bi ton 4. Cho hm s 3 22 (2 5) 3 5y x m x mx m . Tm m th hm s ct trchonh ti ba im phn bit c honh ln hn -2.

Gii. (nhm nghim ca phng trnh honh giao im, quy v tm iu kin cho phngtrnh bc hai).


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Trc ht ta d dng tm c im (5/2;0) l im cnh ca hm s trn trc honh, do hm s cho c th vit di dng:

2(2 5)( )y x x mx m

Phng trnh honh giao im ca th hm s v trc honh:2

2

(2 5)( ) 0

5

2

0 (1)

x x mx m

x

x mx m

Yu cu bi ton (1) c hai nghim phn bit1 2,x x khc 5/2 v ln hn -2.

2

1 2

1 2 1 2

1 2

1 2

04 0

( 2)( 2) 02( ) 4 0 0

25422

142525 5

0 144 2

m mx x

x x x x mx x

x xm

m mm

Ch y ta s dng nh l Viet tnh cho tng v tch hai nghim. Ta c th tnh trc

tip cc biu thc nghim ca phng trnh (1) l2 4

2

m m mx . Khi (1) c hai

nghim phn bit ln hn 2 khi v ch khi

2

2

2

4 )

42

2

4 5

2 2

m m

m m m

m m m

v d nhin php tnh s phc tp hn.

Cn trng hp ta khng th nhm c mt nghim ca phng trnh honh giaoim th ta c th da vo hnh dng th c cch gii bi ton nh sau:

th ct trc honh ti 3 im phn bit c honh ln hn -2 th cn tha mncc iu kin sau: y =0 c hai nghim phn bit

1 2,x x ln hn -2 v f(-2) < 0 v

. 0CD CT

y y .

Nhng theo cch ny th khi lng tnh ton kh nhiu, ch ring vic tnh c gitr ti cc im cc tr phc tp (mc d c mo tnh: ly y chia y c thng l p(x) vd q(x)th c th vit y = y.p(x) + r(x) khi gi tr cc tr ( ), 1,2

CT iy r x i . V bi ton

ph thuc tham s m.)

-2 O


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Trang 6

bi ton sau ta s khng da vo hnh dng thnh trn c na v ng thngct h th by gil mt ng xin m khng phi trc Ox.Bi ton 5. Cho hm s 3 22 (2 5) 3 5y x m x mx m c th (Cm). Tm m ngthng d: y = 2x 5 ct h th cho ti 3 im phn bit c honh ln hn -2.

Gii. phng trnh honh giao im ca d v (Cm) l:3 2

2

2

2 (2 5) 3 5 2 5 (1)

52( )( 1) 0

2

5

2

1 0 (2)

x m x mx m x

x x mx m

x

x mx m

ng thng d ct h th (Cm) ti ba im phn bit c honh ln hn -2 khi v ch khiphng trnh (1) c 3 nghim phn bit c honh ln hn -2, iu ny tng ng (2) chai nghim phn bit c honh ln hn -2 .

2

2

1 2

1 2

4( 1) 0 25 5

1 0 3 32 2

2 32,2 2 0 1 2 4 0 2

422

m mm

m m mm

mx x m m

x x m

.

Vy vi

3

32,

2

m

m

th ng thng d ct h th (Cm) ti ba im phn bit c honh

ln hn -2.

Ch rng c nghim 5/2 ca phng trnh (1), trc tin ta tm im cnh

0 0( , )M x y ca h th (Cm) nh sau:

Gi0 0

( , )M x y l im cnh ca (Cm). khi phng trnh3 2

0 0 0 0

2 3 2

0 0 0 0 0

2 (2 5) 3 5

(2 3 5) 2 5 0

y x m x mx m

x x m x x y

i vi bin m c tp nghim l R

2

0 0 0 0

3 2

00 0 0

0

52 3 5 0 1

2

72 5 0 0

x x x x

yx x y y

Hn na im (5/2;0) thuc ng thng d: y = 2x 5. Do 5/2 l mt nghim ca phngtrnh honh giao im.

Cc bi tng t.Bi ton 6 (trch thi i hc khi D 2003)

Cho hm s2

2 4

2

x xy

x

(1)


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Trang 7

1. Kho st s bin thin v v th hm s (1)2. Tm m ng thng : 2 2

md y mx m ct th hm s (1) ti hai im phn

bit.Bi ton 7 (trch thi i hc khi D 2008)

Cho hm s 3 23 4y x x (1)1.Kho st s bin thin v v th ca hm s (1).2.Chng minh rng mi ng thng i qua im I(1;2) c h s gc k ( k > -3) u ct

th hm s ti 3 im phn bit I, A, B ng thi I l trung im ca on AB.Bi ton 8 (trch thi i hc khi D 2011)

Cho hm s2 1

1

xy

x

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm k ng thng 2 1y kx k ct th (C) ti hai im phn bit A, B sao

cho khong cch t A v B n trc honh bng nhau.

Bi ton 9 (trch thi i hc khi D 2006)Cho hm s 3 3 2y x x .1. Kho st s bin thin v v th ( C ) ca hm s.2. gi d l ng thng i qua im A(3;20) v c h s gc m. Tm m d ct th (C) ti 3 im phn bit.

Bi ton 10 (trch thi i hc khi B 2010)

Cho hm s2 1

1

xy

x

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm m ng thng 2y x m ct th (C) ti hai im phn bit A, B sao

cho tam gic OAB c din tch bng 3 (O l gc ta ).Bi ton 11 (trch thi i hc khi D 2009)

Cho hm s 4 2(3 2) 3y x m x m c th l ( Cm), m l tham s.1. Kho st s bin thin v v th ca hm s cho khi m = 0.2. Tm m ng thng y = -1 ct th ( Cm) ti 4 im phn bit u c honh

nhhn 2.Bi ton 12 (trch thi i hc khi A 2004)

Cho hm s2

3 3

2( 1)

x xy

x

(1)

1. Kho st s bin thin v v th ca hm s (1).2. Tm m th hm s (1) ct ng thng y = m ti 2 im A, B sao cho

AB =1 .

Bi ton 13 (trch thi i hc khi A 2003)

Cho hm s2

1

mx x my

x

(1)

1. Kho st s bin thin v v th hm s (1) khi m = -1.2. Tm m th hm s ct trc honh ti hai im phn bit c honh dng. (KA

2003).

Bi ton 14. Cho hm s 3 23 9y x x x m .1. Kho st s bin thin v v th ca hm s ( C) ng vi m = 0.2. Tm m th hm s ct trc honh ti ba im phn bit c honh lp thnh

mt cp s cng.Bi ton 15. Cho hm s 4 22( 1) 2 1y x m x m .


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Trang 8

1. Kho st s bin thin v v th ca hm s ( C) ng vi m = 0.2. Xc nh m phng trnh f(x) = 0 c 4 nghim phn bit lp thnh cp s cng.

4. Mt s bi ton ng dng th bin lun s nghim ca phng trnh.Bi ton 1 (trch thi tt nghip THPT nm 2010)

Cho hm s 3 21 3

y x x 54 2

1) Kho st s bin thin v v th (C) ca hm s cho.2) Tm cc gi tr ca tham sm phng trnh 3 26 0 x x m (1) c 3 nghim

thc phn bit.

Gii1) Bn c t gii2)

3 2 3 2 3 21 3 1 36 0 5 54 2 4 4 2 4

m m

x x m x x x x (2)(2) l phng trnh honh giao im ca (C) v ng thng d(m):

54

my song song vi trc Ox. Da vo th ta thy (1) c 3 nghim thc

phn bit khi v ch khi 5 5 13 32 04

mm

Bi ton tng t.Bi ton 2 (trch thi tt nghip THPT nm 2008 ln 2)

Cho hm s 3 23y x x 1) Kho st s bin thin v v th ca hm s cho.2)Tm cc gi tr ca tham sm phng trnh 3 23 0x x m c ba nghim

phn bit.Bi ton 3 (trch thi tt nghip THPT nm 2002)

Cho hm s 3x2xy 24 c th (C).1. Kho st hm s.

14

12

10

8

6

4

2

-10 -5 5

d(m)

g x =1

4 x3+3

2 x2+5


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Trang 9

x

y

1

2. Da vo th(C), hy xc nh cc gi trm phng trnh x4 2x2 + m = 0c bn nghim phn bit.

Bi ton 4 (trch thi tt nghip THPT nm 1993)Cho hm s 3 2y x 6x 9x .1. Kho st s bin thin v v th ( C ) ca hm s.2. Da vo th ( C ), bin lun theo tham s m s nghim ca phng trnh

3 2x 6x 9x m 0 .

Bi ton 5 (trch thi tuyn sinh i hc khi A nm 2006)Cho hm s 3 22 9 12 4y x x x .

1. Kho st s bin thin v v th (C) ca hm s.2. Tm m phng trnh sau c 6 nghim phn bit 3 22 9 12 4x x x m (1)

Phn tch

Ch rng x x , 2 2x x nn biu thc v tri ca phng trnh lcng thc ca mt hm s chn.

V phi l tham snn th ca n l ng thng bin thin song song vitrc Ox.

Biu thc v tri ca phng trnh, nu khng c du gi tr tuyt i th nchnh l cng thc ca hm s v thcu 1. Do t th hm scu 1 ta suy ra th hm sv tri ca phng trnh cu 2.

Xin nhc li nh ngha hm chn v c im th ca n: cho hm s y = f(x)c tp xc nh D, f goi l achan tren D neu xD -x D vaf(-x) = f(x), oth can nhan Oy lam t ruc oi xng.

Gii bi ton trn1. Bn c t gii2. t 3 2( ) 2 9 12 4g x x x x

Tp xc nh ca g(x) l D nn x D x D v g(-x) = g(x) nn g(x)l hm chn. Hn na

3 2 3 2( ) 2 9 12 4 2 9 12 4, 0g x x x x x x x x

Gi (C) l th hm s y = g(x) th (C) gm 2 phn:Phn 1: gi nguyn phn th ca (C) bn phi trc Oy, gm c nhng im nmtrn trc Oy. Ta gi phn ny l (C1).Phn 2: ly i xng (C1) qua Oy . Ta gi phn ny l (C2).Khi (C) gm (C1) v (C2).


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Trang 10

x

y

g x = 2x4-4x2

f x = 2x4-4x2

1

(1)lphng trnh honh giao im ca th hm s(C) v ng thng d: y = msong song vi trc Ox, da vo th ta thy (1) c 6 nghim phn bit khi v ch khi0 < m < 2.

Bi ton 6 (trch thi tuyn sinh i hc khi B - 2009)Cho hm s 4 22 4y x x (1)1. Kho st s bin thin v v th (C) ca hm s (1).2. Vi cc gi tr no ca m thphng trnh 2 2 2x x m (1) ng 6 nghim thc

phn bit .Phn tch

Do v phi ca phng trnh (1) l tham s m nn th ca n l ng thng binthin song song vi trc Ox, nn ta dng th bin lun phng trnh trn. Do vyta tm cch suy ra th hm sv tri ca (1) t th (C).

Ch rng a b ab Gii

1. Bn c t gii2. 2 2 4 22 2 4 2x x m x x m (2)

t

4 2 4 2

4 2

4 2 4 2

2 4 ,2 4 0( ) 2 4

2 4 ,2 4 0

x x x xg x x x

x x x x

Tp xc nh ca hm g(x) l R . Gi Gi (C) l th hm s y = g(x) th (C) gm 2phn:Phn 1: Gi nguyn phn thca (C) bn trn trc Ox, gmc nhng im nm trn trc Ox.Ta gi phn ny l (C1).

Phn 2: ly i xng (C1) quaOx . Ta gi phn ny l (C2).Khi (C) gm (C1) v (C2).

(2)l phng trnh honh giao im ca (C) v ngthng y = 2m song song vi trcOx, da vo th ta c (1) c 6nghim thc phn bit khi0 2 2 0 1m m

Bi tng tBi ton 7Cho hm s 3 3y x x 1. kho st s bin thin v v th (C) ca hm s.2. tm m phng trnh sau c 4 nghim phn bit 31 3

2 2x x m

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51719_bai Toan Giao Diem Cua Hai Do Thi Va Ung Dung Bien Luan So Nghiem Cua Phuong Trinh (1)