10 trọng điểm bồi dưỡng học sinh giỏi môn toán lớp 12 phần 1

^# ®Aw mhm muc dich gii'ip cdc ban hoc sinh l&p 10, lap 11, lap 12 co tu lieu doc them dcndng cao trinh do, cdc ban hoc sinh gidi tu hoc bo sung them kien thuc ky nnng, cdc ban hoc si

• Danh cho hoc sinh I6p 12 chtrong trinh chuan va nang cao

fi On tap va nang cao kT nang lam bal

•• Bien soan theo no! dung va cau true de thi cua Bo GD&.DT

X U A T B A N D A I H O C Q U 6 C G I A H A N O I

to trong diem

MON TOAN

LdpU

• Danh cho hoc sinh Idp 12 churdng tnnh chuan va nang cao

• On tap va nang cao kT nang lam bai

• Bien soan theo noi dung va cau true de thi cua Bo GD&.DT

OEK]

Ha N»i NHA XUAT BAN OAI HOC QU6C GIA HA N6I

^# ®Aw

mhm muc dich gii'ip cdc ban hoc sinh l&p 10, lap 11, lap 12 co tu lieu doc

them dcndng cao trinh do, cdc ban hoc sinh gidi tu hoc bo sung them kien thuc ky

nnng, cdc ban hoc sinh chuyen Todn tu nghien cim them cdc chuyen de, nhd sdch

KHANG VIET hap tdc bien soqn bo sdch BOl DlXONG HOC S/NH GlOl, B O I

DlTONG CHUYEN TOAN gom 3 cudn:

- TRONG DIEM TOAN LOP 10

- TRONG DIEM TOAN LOP 11

- TRONG DIEM TOAN LOP 12

Cudn TRONG DIEM TOAN LOP 12 nay co 21 chuyen devai ngi dung la

tom tdt kien thuc twng tarn cua Todn pho thong vd Todn chuyen, phan cdc bdi

todn chgn Igc co khodng 900 bdi voi nhieu dang hai vd muc do tic co ban den phuc

tap, bdi tap tu luycn khodng 250 bdi, c6 huang dan hay ddp so

Cudi sdch CO 3 chuyen de ndng cao: DA THU'C, PHlTONG TRINH

NGHIEM NGUYEN vd TOAN SUYLUAN

Dii dd cogdng kicm tra twng qud trinh bien tap song cung khong trdnh khoi

nhieng khiem khuye't sai sot, mong don nhan cdc gop y cua quy ban dgc de'lan in

sau hodn thien han

Tdc gid

C/y TNHH MTV DWH Hhang Vi$t

Chuyen ad 1: TiNH D O N D I ^ U Vn CITC TBI

1 K I E N T H U C T R Q N G T A M A

Djnh li Lagrange: Cho f Id mOt hdm li^n tgc tr§n [a, b], c6 dgo ham tr§n (a,b) Luc do ton tgi c e (a,b) d l : ^ f(b)-f(a)

N4U f d6ng bi§n tren (a; b) thi f '(x) > 0 voi mpi x e (a; b)

N^u f nghjch bi§n tren (a; b) thi f '(x) < 0 voi mpi x e (a; b)

N6U f '(x) > 0 voi mpi x € (a; b) vd f '(x) = 0 chi tgi mpt so hu-u hgn di6m cua (a; b) thi ham s6 d6ng bi4n tren khoang (a; b)

N§u f '(x) < 0 voi moi x e (a; b) va f '(x) = 0 chi tai mpt so hOu hgn diem cua (a; b) thi ham s6 nghich bi§n tren khoang (a; b)

N§u f d6ng bi4n tren khoang (a; b) va lien tyc tren [a,b) thi dong bi§n tren [a,b); va lien tyc tren (a,b] thi dong bi§n tren (a,b]; lien tyc tren [a,b] thi d6ng bi§n tren [a,b]

Nlu f nghich bidn tren (a; b) va lien tyc tren [a,b) thi nghich bien tren [a,b); lien tuc tren (a,b] thi nghich bi§n tren (a,b]; lien tyc tren [a,b] thi nghich bi^n tren [a,b]

N§u f '(x) = 0 voi mpi X e D thi ham so f khong d6i tren D

Cyc trj cua ham s6 gficr, :

Cho ham s6 f xac dinh tren tap hgp D vd XQ e D •

Xo du-oc gpi Id mpt di4m eye dai cua f neu ton tai mpt khoang (a; b) chii-a dilm Xo sao cho (a; b) c D vd f(x) < f(Xo), V x e (a; b) \

W tTQng diSm bSl dUCing h<?c sinh gioi mdn To6n • v i o

Xo dugc goi 1^ mpt diim cue tieu cua f n^u t6n tgi mpt khoang (a; b) chiJa

dilm Xo sao cho (a; b) c : D f(x) > f(Xo), V x e (a; b) \

( o

B6 d§ Fermat: Gia si> ham s6 c6 dao ham tren (a;b) N§u f dat ci/c trj tgi

- Cho y = f(x) lien tuc tren khoang (a;b) chii-a Xo, c6 dao ham tren c6c khoang

(a;;'o) va (xo;b):

N I U f '(x) doi diiu tu" am sang du-ong thi f dat eye ti§u tai XQ

N I U f (x) d6i d^u ti> duKang sang am thi f dat CLfC dai tai Xo

- Cho y = f(x) CO dao ham eSp hai tren khoang (a;b) chua XQ;

Neu f '(xo) = 0 va f "(Xo) > 0 thi f dat cue tilu tai XQ

N§u f '(xo) = 0 va f "(xo) < 0 thi f dat eye dai tai XQ

IJng dung vac phiKcng trinh

- N I U ham s6 f don dieu tren K thi phuang trinh f(x) = 0 eo t6i da 1 nghiem

N I U f(a) = 0, a thuoc K thi x = a la nghiem duy nh^t cua phuang trinh

f(x)=0

- N§u f CO dao ham c^p 2 khong d6i dSu tren K thi f ' la ham dan di$u nen

phuang trinh f(x) =0 c6 t6i da 2 nghiem tren K N§u f(a) = 0 va f(b) =0 vai a

b thi phuang trinh f(x)=0 chi c6 2 nghiem la x = a va x = b

- N l u f la mot ham lien tuc tren [a, b], c6 dao ham tren (a,b) thi phuang trinh

f ' ( x ) M j l [ i ? l c 6 it n h i t mot nghiem c e (a,b)

b a

Oac biet, neu /(a) = f(b) = 0 thi phuang trinh f'(x) = 0 eo it nhlit mot nghiem

c e (a, b) hay gi&a hai nghiem cua f thi c6 it nhSt mpt nghiem cua dao ham f'

Chu y:

1) Tung dp eye trj y = f(x) tai x = Xo:

Ham da thue: y = q(x) y' + r(x) => yo = r(Xo)

Lj- u - »• s u(x) u(Xn) u'(x.)

Ham huu ti: y = f(x) = = > Vo = — ^ = °

v(x) v(x,) v'(Xo) Dae bi$t; Vai ham y = f(x) bae 3 c6 CO, CT va neu y = q(x) y' + r(x) thi

phuang trinh duang thing qua CD, CT la y = r(x)

2) S 6 nghiem cua phuang trinh bae 3: ax^ + bx^ + ex + d= 0, a 0

N§u f '(X) > 0, Vx hay f '(x) < 0, Vx thi f(x) = 0 chi eo 1 nghiem

N§u f "(x) = 0 c6 2 nghiem phan biet va:

cf^ iNi-iMmrvL/wfinnang vi^c

Vai yco ycr > 0 : phuang trinh f(x) = 0 chf c6 1 nghiem Vai yco ycr = 0 : phuang trinh f(x) = 0 c6 2 nghiem (1 dan, 1 k6p) , ^ Vai yco VCT < 0 : phuang trinh f(x) = 0 c6 3 nghiem phan biet

'^m^-Bai toan 1 1 : Chung minh eSc h^m s6 sau Id hdm kh6ng d6i

a) f(x) = cos^x + cos^(x + - ) - cosxcos(x + - ) <^'"'

3 3 > b) f(x) = 2 - sin^x - sin^(a + x) - 2cosa.cosx.cos(a + x)

Hu'd'ng din giai a) f '(x) = -2cosxsinx - 2cos(x + - )sin(x + - )

= -sin2x - cos(2x + ^ ) = 0, vai mpi x

Do do f hing tren R nen f(x) = f(0) = "I ^ " ;^ = |

f '(x) = -2sinxeosx - 2cos(a + x)sin(a + x)

+ 2cosa[sinxcos(a + x) + cosx.sin(a + x)]

= -2sin2x - sin(2x + 2a) + 2cosa.sin(2x + a) = 0

Do do f hang tren R nen f(x) = f(0) = 2 - sin^a - 2cos^a = sin^a

Bai toan 1 2: Cho 2 da thue P(x) va Q(x) thoa man: P'(x) = Q'(x) vdi mpi x vd

l + x^ ,

W tr<?ng dISm bdl duOng h?c sinh gidi mSn lo6n W - LS Hodnh Hhi>

Himng din giai

Hifang din giai

Xet f(x) = arctanX + arctan-.D = ( - « ; 0 ) u ( 0 ; + * )

Cty TNHHMWOWH Hhang Vi$t

Hu^ng din gidi

a) Ham s6 y = f(x) = 2x^ + x - 4 lien tyc tren [-1,2] v^ c6 dgo ham f '(x) = 4x +1, theo dinh ly Lagrange thi t6n tai s6 c e [-1;2] sao cho:

y' > 0 tren khoang (-x; 4) nen y dong bien tren khoang (-x; 4)

Bai toan 1 7: Tim khoang dan dieu cua ham s6

W trqng diSm bSi dUOng h<?c sinh gidi mdn To6n 12 - LS Hodnh Phd

T a c 6 : y ^ 2 x ^ ( x ^ - 9 ) y ^ Q « x = ±3

(x^ - 6 ) V x ' - 6 BBT: x - 0 0 - 3 - V 6 76 3

+00

y Vay ham so ddng bi§n tren cac khoang ( ^ ; -3), (3; + « ) , nghjch bi§n tren

cac khoang (-3; - Ve), ( N/6 ; 3)

b)D = (-oo; 1) T a c 6 y ' = > 0 , V x < 1

2V(1-x)3

V$y ham s6 d6ng biln tren khoang ( - « : 1)

Bai toan 1 8: Xet si/ bi§n thien cua ham s6:

a) y = x + cos^x b) y = x - sinx tren [0; 2n\

Hirang din giai a) D = R Ta CO y' = 1 - 2cosxsinx = 1 - sin2x

y' = 0 o sin2x = 1 <=> x = - + k7t, k e Z

4 H^m so lien tyc tren moi doan [ - + kn; - + (k + 1)7i]

4 4 y' > 0 tren moi khoang ( - + kn; - + (k + 1)7t) n6n d6ng bi^n tren moi dogn

4 4

[ - + kTi; - + (k + 1)7r], k e Z

4 4

Vay ham so dong bi4n tren R

b) y' = 1 - cosx Ta c6 Vx [0; 27i] => y' > 0 va y' = 0 o x = 0 hoac x = 2n

Vi ham s6 lien tyc tren doan [0; 2n] nen ham s6 d6ng bi§n tren doan [0; 27:]

Bai toan 1 9: Chieng minh cac ham s6

a) y = cos2x - 2x + 5 nghich bi4n tren R

b) y = ^ ' " ^ ^ ^ ^ ^ (a * b + kTt; k e Z) dan di?u tren moi khoang x^c dinh

sin(x + b)

Hirang din giai

a) Vxi, X2 e R, Xi < X2 LSy hai s6 a, b sao cho a < Xi < X2 < b

Ta c6: f '(x) = -2(sin2x + 1) < 0 vai mQi x e (a; b)

Vi f '(x) = 0 chi tai mOt s6 hO-u han dilm cua khoang (a; b) nen hSm so f

nghich bi§n tren khoang (a; b) => dpcm

b) Dieu kien x ^ - b + kn (k € Z)

Cty TNHHMTVDWH Hhang Vi$,

_ sin(x + b)cos(x + a) - sin(x + a)cos(x + b) ^ s i n ( b - a )

sin''(x + b) sin^(x + b)

Vi y' lien tuc tai moi dilm x ?t - b + kn, a - b ^ k;t nen y' giu- nguyen mpt

d i u trong moi khoang xac dinh => dpcm

Bai toan 1.10: Tim cac gia trj cua tham s6 d l ham so:

a) y = (m - 3)x - (2m + 1)cosx nghjch bi§n tren R

b) y = x^ + 3x^ + mx + m chi nghjch biln tren mpt dogn c6 dp dai bing 3

Hifo-ng din giai

a) y' = m - 3 + (2m - 1)sinx

Ham s6 y khong la ham h^ng nen y nghich bi§n tren R:

y'< 0 , Vx « m - 3 + (2m - 1)sinx < 0, Vx Dat t = sinx, - 1 < t < 1 thi m - 3 + (2m - 1)sinx = m - 3 + (2m - 1)t = f(t)

Di4u kien tu-cng du'ang: f(t) < 0, Vt e 1]

f(-1)<0 f(1) < 0

Vay dilm eye dai (-2; 0) va eye tieu (0; -108)

b) Ham s6 y = f(x) lien tyc tren R.Ta c6:

W trgng diS'm bSi duOng h<?c sinh gioi mdn Too > r ic Hodnh Phd

b) Tap x^c dinh D = ( — — N/6 ) u ( ; +oc)

3 x ^ 7 x ^ - 6 - ^

y' =

x ^ - 6 7x^-6 _3x^(x^-6)-x'' 2x^(x^-9)

Bai toan 1.13: Tim cue trj ciia h^m s6

a) y = x - sin2x + 2 b) y = 3 - 2cosx - cos2x

Hirang din giai a) D = R, y' = 1 - 2cos2x : j,

y' = 0 ct>

sinx = 0 2

1 o X = k7i hoacx = ± — + 2kTt, k G Z '•''^^ ' cosx = -— 3

2 y" = 2cosx + 4cos2x

Ta CO y"(kn) = 2cosk7i + 4cos2k7t = 2cosk7i + 4 > 0, vai mpi k e Z, nen ham

s6 da cho dat eye ti§u tai cac dilm x = kn, ycT = 2 - 2cosk7t bSng 0 khi k chin va bing 4 khi k le

Ta c6y"(± — + 2k7r) = 2 cos — + 4 c o s — - 6 c o s — = - 3 < 0 nen ham s6

s i n - khix>0 2

dilm do ~ ' b) y = f(x) =(x - a)(x - b)(x - c), a ;i e luon eo eu-c dai va eye tilu, ^ ' '

Hipang din giai ,

a) Ham s6 f xac djnh va lien tyc tren R Ta CO - ;i

10 trgng diSm bSi dudng h<?c sinh gioi mdn Too , • Kinh Phd

= ^ [(a-b)^ + (b-e)^ + (e-a)^] > 0 vai a e

Do do y' = 0 CO 2 nghiem phan biet va doi d^u 2 Ian khi qua 2 nghiem nen

luon luon c6 mpt ci/c dai va mpt eye tilu

Bai toan 1.15: Tim tham s6 thye sao cho ham s6

^2

NIU q > 0 thi phyang trinh: f '(x) = -Q c6 hai nghiem phSn

(x +1)2 bi^t xi = -1 - ^ va X2 = - 1 + x/q

acos^ X Vdi sinx = a thi y" = ^ * 0, do do ham s6 dgt eye trj t^i 3 dilm

sinx cos X thuoc khoang (0; — )

4

« sinx = a CO 3 nghiem thuoc khoang (0; ^ ) W ^ i ^ }<=> 0 < a < ^

Bai toan 1 16: Tim m d4 ham s6:

W trgng diSm b6i du&ng h<?c sinh gidi mon loan 12 - Le Hoanh Pho

Bai toan 1 17: Vi§t p h u a n g trinh du-ang thSng di qua d i l m cue dai, cue tieu

cua do thj:

a) y = + 3mx^ + 3(m^ - 1 )x + - 3m '

b ) y = x^ - 2mx + 5m - 4 - m^ , i !

Hu 'O'ng d i n giai

a) y' = 3x^ + 6mx + 3(m^ - 1), A ' = 1 > 0, Vx nen d6 thj luon luon c6 CD va CT

vai hoanh dp Xi, X2

b) DK: X ^ 2 Ta CO y = X - 2(m - 1) + m - m

X - 2

2 , , m - m ^ ( x - 2 ) ' ^ - ~ ( m - m ^ )

nen y = 1

(X - 2Y {X-2Y Di'eu kien c6 CO va CT la m - m^ > 0 o 0 < m < 1

Goi xi, X2 la hoanh do CD, CT thi x, < 2 < X2 Ta eo

y(xi) = x , - 2 ( m - 1 ) + m - m

(X, - 2) = Xi - 2(m - 1) + (xi - 2) = 2xi - 2m

y(X2) = X 2 - 2 ( m - 1 ) + = X2-2(m - 1) + (x2-2) = 2x2-2m

( X 2 - 2 ) Vay p h u o n g trinh d u o n g t h i n g qua CD va CT la y = 2x - 2m

Bai toan 1 18:

a) Cho d6 thi cua ham s6: y = (3a^ - 1)x^ - (b^ + 1)x^ + 3c^x + 4d c6 hai

didm cue tri la (1; -7), (2; - 8 ) Hay tinh t6ng M = a^ + b^ + c^ + d^

b) Tim a d l d6 thi ham s6 y = -^^ 1) + 1

X cue tri nay thupe mpt parabol c6 djnh

Hipo-ng d i n giai

CO 3 cue trj va ehung minh 3

a) Dat A = 3a^ - 1, B = - ( b ' + 1), C = Zc\ = 4d, thi ham- -3^2 s6 da cho la:

y ( 2 ) = - 8

3A + 2B + C = 0 12A + 4B + C = 0

A + B + C + D = - 7 8A + 4B + 2C = - 8

f'(x) >0

b) PT o 2x^ - 3x + ^J2x^ - 3x + 1 = x2 +1 + ^x2 + 2 Xet ham s6: f(t) = t + tren R, f'(t) = 1 +

3^(t +1)

d6ng b i l n tren R, do do:

> 0 nen ham so f(t)

lO trpng diSm b6i duOng hQC sinh gidi mon lodn If ' = ' o •/? FHO

2x - 5 X - 1 b) 4 I 2x - 1 ! (x^ - X + 1) = x^ - 6x^ + 15x - 14

X - 2 > 0

(2x 1)2 (X - 2)

Bai toan 1 21: Giai cac he phuang trinh:

5x' + yx'^ - 5y^ f 7y^

Lgi dat V = \/3u-1 o + 1 = 3u

Ta c6 h$: , o

+1 = 3u +1 = 3v

+1 = 3v

- = 3(v - u)

+1 = 3v

u - v = 0 (u - v)(u^ + vu + + 3) = 0

-f'(t)= 1 + ^ > 0 , V t € D ^ h d m s 6 d 6 n g bi4ntr§nD

P T o f ( | x + y|) = f ( y - 1 ) c > |x + y| = y - 1 [ y > 1

y - 1

[x = - 1 hay x = : 1 - 2 y

1-2N/24

' 'ill •' •

10 trQng diSm bSi dUOng h<pc sinh gioi mdn Todn 12 - LS Hodnh f^d

Bai toan 1 22: Giai cac he phucng trinh

Oat f(t) = t^ - 2t + 1, t > 0 thi f '(t) = 2(t - 1) nen f dong bien tr§n ( 1 ; +oo)

nghich b i l n tren (0; 1) Dgt g(t) = 2t, t > 0 thi g'(t) = 2 > 0 nen g d6ng bien

f(x) = g(y) f(y) = g(z) f(z) = g(x)

Gia su' X = min{x; y; z} Xet x < y < z

N§u x > 1 thi 1 < X < y < z => f(x) < f(y) < f(z)

Xet X < z < y thi cung nhan du-gc ^kx qua tren ,

Vgy h$ CO 2 nghiem x = y = z = 2 + yfs , x = y = z = 2- \l3

nghiem cua he phu'ong trinh

N§u X > 0 thi y > 0, z > 0 Xet ham s6 f(t) = 60t'

Bai toan 1 23: Giai cac b i t phu'ong trinh

a) V2x^ +3x2 +6X + 16 >2V3 + N / 4 - X

b) -2x + 3 - Vx^ - 6 X + 11 > 7 3 ^ - > / x ^

HifO'ng d i n giai 2x^+3x2+6x + 1 6 > 0

V^y nghiem cua b i t phu'ong trinh S = (2, 3]

Bai toan 1 24: Giai cac b i t phuong trinh i a) 7 3 - X + x2 - 7 2 + x- x 2 < 1 •

-10 trpng diS'm bSi duang HQC sinh gioi m6n Toon 12 - LS Hodnh Phd

Nen f(x) = 0 c6 nghiem duy nhit x < 0

V|y phuang trinh cho c6 nghi^m duy nhlt

Bai toan 1 26: Chung minh he phuang trinh c6 nghi^m duy nhit:

x2 = y 3 +y2 + y + a y2 = + z2 + z + a

_ Xet X > y > z =^ f(x) > f(y) > f(z)

=> z2 > x2 > y2 N^u z > O t h i x > y > z > 0 x2 > y2 > z2 x2 = y2 = z2 ^ f(x) = f(y) = f(z) ^ x = y = z

Tuang tu-nhu tren n l u y > 0 hay X < 0 ta suy ra X = y = z

N4U x>0>y=:i>x2 = f(y) < f(0) = a z2 = f(x) > f(0) = a N§u z > v/a

thi X > z > v'a => x2 > z2 => z2 = y2 = z2

=> X = y = z trai v^i x > 0 > y ' ' •

N l u z < - Va li luan nhu tren ta dan d§n mSu thuan

V^y he CO nghiem duy nhit x = y = z = to6'd6toia nghiem duy nhit cua

phuang trinh: t^ + t2 + t + a = 0

Bai toan 1 27: Chung minh he i ~ c6 dung 3 nghiem phSn bi$t

y2 + x^ = 1

Hu'O'ng din giai

Tru 2 phuang trinh v6 theo v§ va thay the ta dugc:

x ' ( 1 - x) - y 2 ( 1 - y ) = 0 ^ ( 1 - y ' ) ( 1 - x ) - ( 1 - x ' ) ( 1 - y ) = 0 ,

=^ (1 - x)(1 - y)[1 + y + y 2 - ( 1 + x + x2)] = 0

=^(1 - x ) ( 1 - y ) ( y - x ) ( 1 + x + y) = 0

Xet X = 1 thi h$ c6 nghiem (1; 0).Xet y = 1 thi h$ c6 nghiem (0; 1)

Vay he c6 dung 3 nghiem phan biet

Bai toan 1 28: Tim tham s6 de phu-ang trinh

a) ^1 + X + \ / l - x = a CO nghiem

b) vx^ + mx + 2 = 2x +1 CO 2 nghiem phan bi^t

Hu-o-ng din giai

<=:> f(x) = m CO 2 nghiem phan biet x> ,x=^0<^ ^ - 2

Bai toan 1 29: Tim m d l phuang trinh

= 2 CO nghiem ftfim^i Ja>

V^y phyang trinh cho c6 nghi^m khi

1 - > - 2 <=>-N/3 < m < N/3

b) Dieu ki^n : cosx 0 tanx > - 1

D$t t = 7tanx + 1 > 0 , phu-o-ng trinh:

sinx + 2cosx sinx + 3cosx

t^ + 2 3t- + 1 5 t ^ + 6

(t^ + 2f

V$y phu-ang trinh c6 nghi^m duy nhit khi m > y(1) <=> m > 2

Bai toan 1 30: Tim tham s6 d l phu-ang trinh

a) (4m - 3) 7x + 3 + (3m - 4) Vl-x + m - 1 = 0 c6 nghi^m

b) x^ + 3x^ + (6 - a)x'' + (7 - 2a)x^ + (6 - a)x^ + 3x + 1 = 0 v6 nghi$m

Hu>6ng din gial

a) Di§u ki$n: - 3 < x < 1 khi d6:

X x^ x x' + 3x' + (6 - a)x + (7 - 2a) + (6 - a) 3+ ^ + — = 0

Hyo-ng din giai

a) Xet f(x) = sin^x + cos^x = (sinx + cosx) (1 - sinx.cosx)

^ ' 2 2

Lap BBT thi b i t phuang trinh CO nghi^m khi m < 1 v •

b) D|t t = sinx + cosx, |t| < 72 va t^ = 1 + 2 sinx cosx ^ sin2x = t^ - 1

cos^ 2x = 1 - sin'2x = - t " + 2t^

BPT: - t " + 2t^ -1' + m + 3 < 0 ; (|t| < V2)

Xet f(t) = -t^ + 2f' - t-^ + m + 3

f '(t) = - 2 t {2f - 3t + 1) ; f '(t) = 0 =^ t = 0 ; - ; 1

2 Lap BBT suy ra dieu kien c6 nghiem la: m + 3 > 0 <=> m > - 3

Bai toan 1 32: Tim 6\hu l<ien cua m he bit phu-ong trinh c6 nghiem

K h i - 1 < x < 0 =>f'(x) = 3x(x + 2 ) < 0

0 < X < 2 ^ f '(X) = 3x (X - 2) < 0

2 < X < 4 => f '(x) = 3x (X - 2) > 0

Do do - m ^ - 1 5 m + 1 6 < 0 < = > m < - 1 6 v m > 1

Vay di4u kien c6 nghiem la - 1 6 < m < 1

Bai toan 1 33: Cho 3 s6 a,b,c thoa man abc ?^Ova - + - + - = 0

7 5 3 ChLKng minh phu'ong trinh : ax"* + bx^ + c = 0 c6 nghiem

•t '

X. -Hii'O'ng din giai

Xet hdm s6 F(x) = - x ^ + - x ^ + - x ^ , khi do F(x) lien tyc, c6 dao hSm

Bai toan 1 34: Cho ham s6 f c6 dao ham tren tren [0;1] va thoa m§n:

f(0) = 0; f(1) = 1 Chu-ng minh ton tgi 2 s6 phan bi^t a;b thupc (0; 1) sao

cho f'(a).f'(b) = 1

-j^c^rnwn-t /VIIV uvvH hnang vi^t

Hirang din giai

Xet ham so g(x)= f(x) +x - 1 , khi do thi g(x) lien tyc v^ c6 dgo ham tren [0;1]

Ta c6: g(0)= - 1 < 0 va g(1)= 1 >0 nen t6n tai so c thupc (0;1) sao cho g(c) =0

Do do f(c) + c - 1 =0 hay f(c) = 1 - c

Ap dung dinh ly Lagrange cho f tren cac doan [0;c] va [c;1] thi : "

t6n tai ae(0;c) sao cho: = f'(a) '-S; v

ChCfng minh b i t phuang trinh: f'(x) - f(x) < -(f(1) - 2f(0)) c6 nghiem

Hii'O'ng din giai

X6t 2 h^m s6: g(x) = arctanx;h(x) = tren [0,1], khi do thi g(x),h(x) c6

1 + x dao ham tren (0;1),

Bai toan 1 36: Gia si> f\a mpt ham xac dinh tren [a, b], c6 dao ham d i n d p

n + 1 tren (a,b) va XQ e (a,b) ChCrng minh t6n tai c n i m giOa x va XQ d l c6:

ft

f W - f ( > ^ o ) ^ ( x - x , ) ! : | i l ( x - x „ ) ^

! ! X ) ( x x ) " i ! : : : ! ( ^ ( x x )

-n! """^ (n + 1)!^ ° '

HiTO'ng d i n giSi

Ta tim mOt da thCfc Pn(x) c6 bgc I<h6ng vu-cyt qua n sap cho

/(Xo) = Pn(Xo) , f'(Xo) = P'n(Xo) / ' " ' ( X ) = P<"' (Xo)

, M(3\: Pn(x) = Ao + A i ( X - X o ) + A 2 ( X - X o ) ^ + + A n ( X - X o ) "

Luc d6: , ,

P'n(x) = Ai + 2A2(x - Xo) + + nAn(x - Xo)""' ' ' "

P"n(x) = 2A2 + 3.2.A3(x - Xo) + + n(n - 1)An(x - XQ)""^ v

Pr(x) = n!An

Do d6 thay x = Xo vdo c^c ding thii'c tr§n ta du'O'c:

: Pn(xo) = Ac , P'n(xo) = Ai, P"n(Xo) = 2A2 P<"' (XQ) = n!An

Nhu vgy: f(xo) = Ao, A , = r(Xo), 2A2 = r(xo), , /<"'(xo) = n!An n§n:

P n W = n X o ) + ^ ( X - X , ) ^ ( X - X „ ) ^ + + ! l ^ ( X - ^

Dat Rn(x) = - Pn(x) ta suy ra R|,"'(x) = f^"'(x) - P^"'(x)

nen: Rn(xo) = R'n(xo) = = Rj,"'(Xo)=0

m F(x) = ( X - xo)"*^ thi: F(xo) = FXxo) = = F<"'(Xo) = 0

Vol X e (a,b) ta vi6t du'yc ^ ^ " " ^ ^ ^ "

F(x) F ( x ) - F(Xo)

R (x) R'(£ )

Theo djnh ly Cauchy ta c6 = " ^ vai \y n l m giOa x XQ

" T a lai c6 M l l ^ M l L M ^ theo dinh ly Cauchy ta du-^c:

F ' ( ^ i ) - F' ( X o )

R (x) R'"^^'(C) Sau n + 1 l^n ap dyng dinh ly Cauchy ta du-gc := — - - v^i c nSm

F(x) F<"^'>(c) giOa 4n va XQ, va do do c n i m giOa x va XQ

Nhung R<"*^>{x) = /"'''(x) va F'"*^'(x) = (n + 1)! nen = 1^^^

F(x) (n + 1)!

Vay: f(x) = f(x„) / ^ ( x - x„) + l | | ° l ( x - x^)^ +

nl ""'^ (n + 1)!^

trong do c IS mOt di^m n l m giu-a x vS XQ

Cong thirc tren du'O'c gpi la c6ng thtpc khai trien Taylor cua hSm f tgi diem x = Xo

3 B A I LUYfiN T A P Bai t?p 1 1 : Tim cac khoang dan di$u cua hSm s6:

Bai t$p 1 2: Tim m de hdm s6:

a) y = -—+ ("^ + 2)x - m + 3 ^^^^ ^^^^ khoang x^c djnh

x + 1 b) y = - x^ - — x^ - 2x + 9 dong bi§n tren ( 1 ; + x )

3 2

H i r a n g d i n a) Tap xac djnh D = - 1 ) u ( - 1 , +00) y • • Tinh dao ham y ' vS l|p lu$n y ' > 0 tren D K§t qua m > 1

K§t qua CO tai x = - 3 ; yco = - 9 \/3 , CT tgi x = 3 ; y c r = 9 \ / 3 b) K§t qua CD tgi x = 0, yco = 0 va CT tgi x = 2, y c r = - 3 \ / 4 Bai t?p 1.4: Tim ci^c trj ham s6:

a) y = x - sin2x + 2 b) y = sin2x + cos2x

H u - a n g d l n * a) T$p xSc djnh D = R, y ' = 1 - 2cos2x, y " = 4sin2x

W trqng diS'm bSi dUOng h<?c sinh gioi mdn To6n 12 - LS Hodnh Phd

b) K§t qua diem ci^c dai x = — + kn , diem ci^c tilu x = — + kn

8 8

Bai tap 1 5:

a) Tim m d§ h^m so y = 2x^ - 3 (3m + 1 )x^ + 12 (m^ + m) x + 1 c6 ci/c dai va

cu-c tilu Vilt piiu-ang trinli du-ang thing di qua CD, CT

X x^+2mx + 1-3m^ , b) Tim m de ham so y = — x - m CO hai dilm eye th n l m v l hai

phia cua true Oy

Hipang din

a) Tap xac dinh D = R L i y y chia y' > ,3

K § t q u a m # 1 v a y = - ( m - 1 ) ^ x + 2{m^ + m)(3m + 1 ) + 1

b) Kit qua -1 < m < 1

Bai tap 1 6: Chipng minh ham s6

a) y = x^ + ax^ - (1 + b^)x + a + 4b - ab luon luon c6 eye dai va eye tieu vd-i mpi

Kit qua nghiem duy nhit x = 3

b) Ham dan di^u Kit qua x =3

Bai tap 1 9: Giai eae h0 phu-cng trinh :

Kit qua X =3, y =0

b) Kit qua x=^;y = 2

Bai tap 1.10: Giai bat phyang trinh:

a) Vx7^ + 27x + 6 >20-3N/X + 13 b) Vx^ - 2 x + 3 - Vx^ - 6 X + 11 < N/ 3 ^ - V x ^

W trgng diem bSi dUOng HQC sinh gioi mdn Toon 12 - L S Hodnh Phd

Churen aS 2: KHHO SIIT VA V€ D b THf HflM SO

1 K I ^ N T H U C T R O N G T A M

Tinh I6i lorn cua d6 thj:

Ham s6 f x^c djnh tr6n K Id mOt khoang, dogn ho$c nu-a khoang

f gpi Id 15m tren K neu Va,p,a + p = 1: f(ax + py) < af(x) + pf(y), Vx,y s 0

f gpi Id I6i tr§n K n l u Va.p.a + p = 1: f{ax + py) > af{x) + pf(y),Vx,y > 0

Dilm udn cua dd thj:

D i l m U(xo;f(xo)) dLcp-c gpi Id diem u6n cua duang cong (C): y = f(x) n4u ton

tai mpt khoang (a;b) chupa dilm Xo sao cho mpt trong 2 khoang (a,Xo), (xo,b)

thi tiSp tuy^n tgi difem U ndm phia tr6n d6 thj c6n 6- khoang kia thi ti^p tiiyln

ndm phia du'b'i d6 thj

Cho hdm s6 y = f(x) c6 dgo hdm cdp 2 mOt khoang (a;b) chua d i l m XQ N I U

f "(xo) = 0 vd f " ( X ) doi ddu khi x qua diem XQ thi U(xo;f(xo)) Id dilm u6n cua

du'dyng cong (C): y = f(x)

1) Neu y =p(x).y" + r(x) thi tung dO dilm u6n t?i XQ Id yo = r(xo)

2) N l u f l6i trSn dogn [a,b] thi GTLN = max{f(a); f(b)}

3) Neu f 16m tr6n dogn [a,b] thi GTNN = min{f(a); f(b)}

Khao sat va ve d6 thj ham da thirc: g6m 3 bu-d'c:

Bu'O'cl: Tdp xdc dinh

- Tdp xdc djnh D = R

- Xet tinh chin, le neu CO '

Bu-d'c 2: Si^ biln thi6n

- Tinh cac giai hgn

- Tinh dgo hdm d p mOt, x6t d i u

- L$p bang biln thien r6i chi ra khoang d6ng biln, nghjch biln vd eye dgi,

eye tieu

Bu'ccc 3: Ve d6 thi

_ Tinh dgo ham d p hai, xet d i u d l chi ra dilm u6n cua ham da thu-c

_ Cho vai gia tri d$c biet, giao dilm vai hai true tog dp r,' _ Ve dung d6 thi

B i n dang dd thj ham bac 3: y = ax^ + bx^ + cx + d, a ^ 0 c6 tam d6i xij-ng la

dilm u6n

Bon dang do thi ham trung phuang: y = ax + bx^ + c, a ?t 0 & i^fiiir

-Du'O'ng tiem c a n

- Duong thdng x = XQ dt^pc gpi la tiem can du-ng cua d l thi ham s6 y = f(x)

n l u it nhdt mpt trong cac dilu kien sau dupe thoa man:

lim f(x) = + x ; lim f(x) = + x ; lim f(x) = - r \m f(x) = - «

- D u o n g t h i n g y = y ^ du-p-c gpi la tiem can ngang cua do thi ham s6 y = f(x)

n l u lim f(x) = yo hode lim f(x) = yo

- Duong thing y = ax + b, a ^ 0 dup-c gpi la tipm cgn xien cua d6 thi y = f(x)

n l u lim [f(x) - (ax + b)] = 0 hoac lim [f(x) - (ax+ b)] = 0

B u a c i : Tap xac dinh

~ Tim tap xac dinh

- Xet tinh chin, le n l u c6, Buac 2: Chilu biln thien Tinh cac giai han , tim cac tiem cdn Tinh dao ham cdp mpt, xet ddu Lap bang biln thien roi chi ra khoang dong biln, nghjeh biln va eye dgi,

c y c t i l u •

IP Hoanh Phi)

1 0 trQng diem b6i duang hQC sinh g:oi mSn J,

Bii-dc 3: Ve thj

- Cho vdi gid trj dSc bi^t, giao diem v6'i hai tryc tog

dO Ve dung d6 thi, iudO u y tam doi xiidO ng la giao diem 2 tiem c#n

Hai dang d6 thj ham hu-u tf b^c 1/1: y = v6i c ^ 0, ad - be ^ 0

1) Tu- d6 thi (C): y = f(x) suy ra cac d6 thi:

y = - f(x) bing cdch Idy d6i xung qua true hodnh

y = f(-x) bdng eaeh lly doi xi>ng qua trye tung

y = - f(-x) bing eaeh l l y d6i xung qua g6c

y = I f(x)| bing eaeh l l y phin d6 thj a phia tr§n trge hoanh, eon phin

phia du'6'i trgc hodnh thi d6i xu-ng qua trye hoanh

y = f(|x|) Id ham s6 ehdn, bIng edch lly phIn d6 thj a phia ben phai true

tung, r6i l l y d6i xu-ng phIn do qua true tung

2) Bai todn ve bien ludn s6 nghiem phu-ang trinh dgng g{x,m) =0

Ou-a phuang trinh ve dang f(x) = h(m) trong d6 v6 trai Id hdm so dang xet,

d § ve d6 thj (C): y = f(x) S6 nghiem Id so giao diem eua do thj (C) v6'i

du'6'ng thing y = h(m)

3) Oilm d$e biet cua hp d6 thj: (Cm): y = f(x,m)

- Diem c6 dinh eua hp Id d i § m md mpi d6 thi d^u di qua:

Mo(xo, yo) e (Cm), Vm o yo = f(xo, m), Vm

- Oi^m md hp khdng di qua Id dilm md khpng c6 d6 thj ndp cua hp di qua vdi

mpi tham s6: Mo(xo, yo) « (Cm), Vm o yo f(xo, m) Vm

Nhdm theo tham s6 vd dp dyng cdc m^nh d4 sau;

3 27 3

2 khoang ( - ; + « )

3 *''•'•»•': \ •

b) D = R Ta c6 y'= 4 x ' + 16x , y" = 12x^ + 16 > 0 V x Vdy do thj khong e6 dilm u6n vd hdm s616m trSn R j Bai toan 2 2: Tim d i l m u6n vd cdc khoang l6i 16m cua d l thj:

Hu^ngdlngidi

Ta c6- y- - (x^ + X +1) - (X + a)(2x +1) _ x^ + 2ax + a - 1

(x2+x + 1)2 (x2 + x + 1)2

35

„ ^ 2(x^ + 3ax^ + 3(a - 1)x - 1)

^ (x^ + X + ^f

y" = 0 <=> x^ + 3ax^ + 3(a - 1 )x - 1 = 0

Dal f(x) = x^ + 3ax' + 3(a - 1 )x - 1, x e R

Taco: f(0) = - 1 < 0 , f(-1) = 1 > 0

lim f(x) = -co, Mm f(x) = +oo va d6ng thai ham s6 nay lien tgc ten t$p s6

thyc nen phuang trinh f(x) = 0 c6 ba nghi^m phan bi^t thupc cac khoang

Bai toan 2 4: Chp ham so: y = x^ - 6x^ + 3mx - m + 2, m la tham so

a) Khao sat sy bi§n thien va ve d6 thi cua ham s6 khi m = 3

b) Tim m sao cho d6 thi cua ham s6 da cho c6 cac diem eye dai, eye ti§u A

+ 00

Ham s6 d6ng bi§n tren mpi khoang ( - « ; 1 ) va (3;+ « ) , nghich bien tren (1;3)

Ham s6 dat eye dai khi x = 1, yco = 3 va dgt eye ti4u tgi

X, + Xj = 4

x^Xj = m Theo dinh li Viet

o 4m^ - 48m^ +193m = 0 « m(4m^ - 48m +193) = 0

<=> m = 0 (thoa m § n ) Vay m = 0

Bai toan 2 5: Cho ham so: y = - - x ^ + (m - 1)x^ + (3m - 2)x - - c6 d6 thj

3 3

(Cm) vai m la tham so i>

a) Khao sat sy bi§n thien va ve do thi cua ham s6 khi m = 2

b) Tim m d l tren d6 thi ( C J eo hai dilm phan biet e6 hoanh dp ciing d i u va ti§p tuydn cua (Cm) tai m6i di^m do vuong goc vai du-ang thing d: x - 3y + 1 =0

Hu'O'ng din giai ^

a) Khi m = 2 ham s6 tra thdnh y = x ^ + x^ + 4x - -

• T$p xac dinh D = R %^

• Sy bi6n thien: y' = -2x^ + 2x + 4;

y' = 0c:>x = - 1 v x = 2 A » - I

10 trQng diSm bSl dt/Ong HQC sInh gl6l mdn To6n J2 - L S Hodnh PM

Bang i>iln thien

thu6c do thj (C) CO tung dO m vd gle O tgo thdnh tam giac OAB cSn tgi O

HuHyng din giai

Hai d i l m A, B thupc d6 thj (C) e6 tung dp m nen thuOc dyang thing d : y = m

Hoanh do giao d i l m cua d vd do thj (C) id nghi^m cua phuang trinh

I x 3 + I x 2 _ ^ x + 2 = m

Phu'ang trinh c: > x ' - 3 x ^ - 9 x + 1 2 - 6 m = 0 (1) Ou-ang t h i n g d c i t (C) tgi A, B thoa m§n tam gide OAB can tgi O khi

2 ~ 6 vd phu'ang trinh (1) c6 nghi^m x i , - X i , X2 (trong 66 x^, -x^ Id

m ^ 0 hodnh dp cua A, B) Khi d6 Xi, X2 la nghi^m cua phu'ang trinh (x^ - x^ )(x - X2) = 0 " j Phuang trinh <=> x^ - X2X^ - x^ x + x^ X2 =0 (2)

Oong n h i t cac he s6 cua (1) vd (2):

x , = 3

xf = 9 x^Xg = 1 2 - 6 m Suy ra 1 2 - 6 m = 27 « m = —

Bai toan 2 7: Khao sdt sy b i l n thien va ve 66 thj cua hdm s6

Bang b i l n thi^n

-y

• 0 6 thi: y" = - 6 x + 6, y" = 0

X = 1 nen d6 thj c6 dilm u6n 1(1; 0)

W trgng diSm bSi dUOng hqc sinh gioi m6n To6n 12 - LS Ho6nh Phd

b) y = - 3x^ + 3x + 1

• Tap xac djnh D = R

• Sy biln thien: Mm y = - ^ c Mm y = +co

Ta CO y' = 3x^ - 6x + 3 = 3(x - 1)^ > 0, Vx nen ham so d6ng bi§n tren R,

ham so khong c6 eye trj

Bang bi§n thien;

Bai toan 2 8: Cho h^m so: y = x^ - 3(m - 3)x^ + 3(m^ - 3m + 5)x + 1, m la

tham s6 Tim m 6k 6b thi cua ham s6 da cho dgt eye dai, eye tilu tai x,, X2

Bai toan 2 9: Cho ham s6: y = x"* - 2mx^ + 2m - 1, vdi m la tham s6

a) Khao sat sy biln thien va ve d6 thi ham s6 khi m = 3

b) Tim m d l d6 thj cua ham s6 da cho c6 3 dilm eye trj ISp th^nh mpt tam

gi^e vuong

Hu'O'ng din giai

a) Khi m = 3, ham s6 tra thSnh y = x" - 6x^ + 5

• Tap xac dinh D = R, ham s6 chin

Ham s6 dat eye dai tai x = 0, yco = 5 va dat eye tilu tai x = ±\/3 , VCT = -4

• D6 thi: D6 thj ham s6 nhan Oy tai true d6i xu-ng

b) Ta CO D = R y'= 4x(x^ - m)

y' = 0 4x(x^ - m ) = 0 « > x = 0 hoSc x^ = m H^m s6 CO 3 dilm eye trj <=> y' = 0 c6 3 nghiem phan bi^t <=> m > 0 Khi do 3 dilm eye trj cua d6 thj ham s6 la: *

W tr<?ng diSm bSi duOng HQC sinh gl6l mdn To6n 12 - Hodnh Phd _

Bai toan 2 10: Cho ham s6: y = x" - mx^ + 2m - 1, vai m la tham so Tim m

de d6 thj h^m so cho c6 3 diem eye tri sao cho 3 dilm eye trj cung vb-i g6c

tpa dp la 4 dinh cua mpt hinh thoi

Hirang din giai

Ta CO y' = 4x^ - 2mx

^x = 0

2 x 2 - m

06 thj ham s6 c6 3 dilm eye tri khi chi khi phyang trinh y' = 0 c6 3

nghiem phan bi^t <=> m > 0

Khi do cac dilm eye trj;

y' = 0 <=> 4x^ - 2mx = 0 <=>

m m

2 ' 4 + 2 m - 1 ,B(0;2m-1),C

m m'' + 2m-1

Vi tam gi^e ABC can tai B, AC song song Ox nen O, A, B, C la 4 dinh hinh

thoi khi va chi khi OABC 1^ hinh thoi

<=> O va B d6i xyng nhau qua AC <=> ° ^ ^ = y^

2 m - 1 m' + 2 m - 1 » m - 4 m + 2 = 0

2 4

m = 2 ± V2 (thoa m§n) V^y m = 2 ± >/2

Bai toan 2.11 : Cho h^m s6: y = - x** - 2mx^ + m^ + m, vd-i m Id tham s6

a) Khao sat sy biln thien va ve d6 thj cua ham so khi m = - 2

b) Tim m d l d6 thi hdm so cit tryc hodnh Ox tgi 4 dilm phdn bi^t A, B, C, D

sao cho AB = BO = OC = CD

Hirang din giai

a) Khi m = - 2 ham so tra thanh y = -x" + 4x^ + 2

• T i p xac dinh D = R, ham s6 chSn

• Sy biln thien: y' = = -4x^ + 8x = -4x(x^ - 2)

y' = 0c=>x = 0 v x = ±\/2 Bang biln thien

y' + 0 - 0 + 0

-y

Hdm so d6ng biln tren moi khoang (-00; -N/2) va (0; V2); nghjch biln tren

m6i khoang (-\/2 ; 0) vd ( N/2 ; +«) Ham s6 dgt eye tilu tgi dilm x = 0, gia

Cty TNHHMTVDWH Hhong Vi$t

tri eye tilu ycr = 2; ham so dgt eye dgi tai ede dilm x = ± N/2 , gid trj eye dai

yco=6

• D6 thj: nh^n Oy Id trye doi xung >,? > 1?; • j^v ;u;

b) Cho y = 0 o - x" - 2mx^ + m^ + m = 0 Ddt t = x ^ t > 0 thi PT : - 1 ^ + 2mt - m^ - m = 0

Do thi est trgc hoanh tgi 4 dilm phan bi?t khi phyang trinh b^c 2 c6 2

nghi^m dyang phan bi^t ti < t2

2 m 2 + m > 0

m < 0 m^ + m < 0

Vi d6 thj dli xyng qua true tung nen 4 giao dilm A, B, C, D thoa min AB =

BO = OC = CD khi va chi khi ^ = 2 ^ » t j = 4ti

Theo dinh li Viet ta c6 t, + t2 = -2m, tit2 = -m^ - m

Do do 5t, = -2m

4t? =-m^ -m 4.4m^ = 25(-m2 - m)

25 42m'' + 25m = 0 c : > m = 0 h a y m = -

41

Ta chpn m = - 25

41

10 trpng diS'm bSi duOng HQC sinh gioi m6n To6n 12 - LS Hodnh Phd

Bai toan 2 12: Cho ham so y = - x " - 2x^ + 3

4 a) Khao sat s y bi^n thien va ve d6 thj cua ham s6

b) Tim m d4 phyang trinh I x" - 8x^ + 6 1 = m c6 8 nghi$m phan bi^t

Ham s6 d6ng bidn tren moi l<hoang (-2; 0), (2; + x ) , ham so nghich biln tren

moi l<hoang ( - x ; -2), (0; 2) Ham s6 dat eye dai tgi x = 0, yco = 3, dgt eye

l i y d6i xyng qua Ox

S6 nghi^m cua phu-ang trinh - x ^ - 2 x ^ + 3 = — la giao didm cua d6 thi

4

m (C) va du'O-ng thSng y = — c ; -<

Dya vao d6 thj, phyang trinh c6 8 nghi$m phan bi?t khi va chi khi:

0 < — < 1 < : = > 0 < m < 4

4 Bai toan 2 13: Cho ham s6: y = - x " - (3m + 1)x^ + 2(m + 1), vo'i m la tham

4 s6 Tim m d^ d6 thi ham s6 eo 3 dilm eye trj lap thanh mpt tam giac c6 trpng tam la g6c tpa do

Vi ham s6 c h i n nen tam giac ABC can tgi A thupc tryc Oy, B, C doi xung nhau qua Oy j

O la trpng tam cua tam giac ABC « yA + ye + yc = 0

3 a) y = - x * - 2x' + 5 b ) y = ^ + x 2 _ _ ,

Hiro-ng d i n giai a) y = - x ' - 2x' + 5

• T$p xac djnh D = R Ham s6 c h i n

• Sy b i l n thien lim y = - x va lim y = - Q O

y = -4x^ - 4x = -4x(x^ + 1), y' = 0 o X = 0

BBT X -00 0 +00

-00

H^m s6 dong bi4n tren khoang {-<a\)

nghjch bi4n tren khoang (0; +«)

Ham s6 dat eye dgi tgi d i l m x = 0: yco = 5

• D 6 thj: y" = -12x^ - 4 < 0, Vx nen do thj khong eo di^m u6n

• Do thi: y" = 6x^ + 2 > 0, Vx nen d6 thj khong

Hu'O'ng din giai

a) D = (0; +oo) Ta c6 lim y = +cc nen TCD: x = 0 (khi x -> 0*)

>u iii^ny uiaiii uui uuuiiy nyi yiot mon loan m - te noonn rno

x - ^ i * x - 1

- Khi m = -2 thi y = "

X - 1 n§n no trung vai tiem can xien

(vai x ^ 1), do thi la du'ang thing (tru- diem (1; 0))

b ) Ta c6: y = mx^ -1

x^ - 3x + 2 = mx + 3m +

7 m x - 1 - 6m x^ - 3 x + 2

- 1 X^ + X + 1 ^ # Khi m = 1 thi y X;e2

x2-3x + 2 x - 2 ' ' ^ ' ^

1 *u- x^ - 8 x^ +2x + 4

8 8(x^ - 3x + 2) 8(x - 1)

Ti> do suy ra: Vai m = 1 thi x = 2 la tiem can di/ng

Vai m = - thi x = 1 la tiem can du-ng

Bai toan 2.18: Cho du-e^ng cong (C^): y = ^x' + (m + 1)x-3

x + m

a) Tim m d l ti$m c|in xi6n cua (Cm) di qua A(1; 1) '-vx u», > ^

b) Tim m d^ giao dilm cua hai ti^m c$n nlm trdn (P): y = x^ + 3. Viv> v •

Hip6'ng din giai

y , 2x^ + (m + 1)x - 3 a) Ta c6 lim - = lim -

' x-+« X x-»=o

lim(y -2x) = lim

= 2 x(x + m)

2x^ + (m + 1)x - 3 x(x + m) -2x 1, =' \

, 2x^ + (m + 1)x - 3 - 2x^ - 2mx ^

Mm = 1 - m

X - + " x + m Suy ra phu-ang trinh tiem c^n xien 1^ y = 2x + 1 - m

TCXdiquaA(1; 1) khi va chrkhi:1 =2.1 + 1 - m o m = 2

b) D6 thj c6 ti$m c$n du-ng 1^ x = -m Tir d6 suy ra giao dilm cua hai ti$m c$n

cua (Cm) tgo vai cac trgc to? dO thdnh mOt tam gidc c6 di^n tich bSng 18

Hipd'ng din giai

10 tr<?ng diS^m bSi dUOng HQC sinh gioi mSn To6n 12 - LS Hodnh Ph6

• Su" bi§n thien: Ta c6: lim y = - x va Mm y = +oo

Do do du-brig thing X = 1 la ti$m can dung

Vi lim y = lim y = 2 nen duang thing y = 2 la tiem cgn ngang cua do thj

-1

Ta c6: y' =

(X - 1)' Bang bi§n thien

Ham s6 nghich bien tren moi khoang (-cc; 1), (1; + x )

• D6 thi; 06 thj (C) cit Ox tai -1;0

2 , cit Oy tai (0;1)

(C) nhan giao diem l(0;2) hai ti?m can lam tSm doi xCrng

phai tiem can du-ng x = 1 cua d6 thi (C), con phin ben trai ti^m cgn du-ng

X = 1 cua d6 thi (C) thi lly d6i xii-ng qua trgc hoanh

2x - 2 • ' f'-^^

Bai toan 2 21: Cho ham so: y = — — j -

a) Khao sat sy bi§n thien va ve d6 thj (C) cua h^m so

b) Lap phi^ang trinh tiep tuyen cua (C), biet tiep tuyen cit du'6'ng ti$m cgn dung tai A, cit duang tiem can ngang tai B ma OB = 20A

Hu'O'ng d i n giai

2x ^ 2

Tap xac dinh D = R \ { - 1 }

• Su bi§n thien: Ta c6 lim y = + x va lim y = - x

x-.(-ir x-.(-i)*

Do do du-ong thing x = - 1 la tiem can dung

Ta CO lim y - lim y = 2 nen duang thing y = 2 la tiem can ngang

4

y = > 0 , Vx5t-1

(x + 1)' Ham s6 d6ng bi4n trgn m6i khoang ( - x ; -1) ( - 1 ; + x )

• 06 thj: 06 thj (C) cit Ox tai (1; 0), cit Oy tai (0; -2), v^ nh^n giao diem l(-1; 2) cua hai duang tiem can l^m tSm d6i xung

• lU uypy umill UUI UUUliy in,>L •:>inn y\ui murrroam^ - L I S

nuuiiifT-ncr-b) Phu-ang trinh ti§p tuy4n tgi M(xo; yo) e (C), XQ ^ - 1

d : y = - ^ ( x - x „ ) ^ ^

X o + 1

Giao d i l m cua d vd-i ti^m c^n du-ng x = - 1 Id A

Giao d i l m cua d v6i ti^m c$n ngang y = 2 la B(2xo + 1; 2)

b) Bi$n lugn theo m so n g h i ^ m cua phu-ong trinh:

x - 2 1 = ( x - 1 ) ( m - 5 ) Hu'O'ng din giai

x - 1 gom phan cua (C) Cfng v o l x > 2 v l doi

XLPng p h i n (C) i>ng v a i x < 2 qua true hoanh

S 6 nghiem cua p h u o n g trinh la so giao d i l m cua d6 thj ( C ) va du-d-ng

t h i n g y = m - 5 : Xet m - 5 > 1 hay m - 5 = 0 hay m - 5 < - 1

<=> m > 6 hay m = 5 hay m < 4 thi phu'ong trinh c6 1 n g h i ^ m • Xet 0 < m - 5 < 1 <=> 5 < m < 6 thi phu-ong trinh c6 2 nghiem Xet - 1 < m - 5 < 0 « 4 < m < 5 thi p h u a n g trinh v6 nghiem

Bai t o a n 2 23: Cho ham so: y = , v d i m la tham s6 Tim m d l du-ong

x + 2

' h i n g d: 2x + 2y - 1 = 0 c i t d6 thj tai hai d i l m A, B sao cho tam g i ^ c O A B

^0 dien tich la S = - , - ,

8

Hu'O'ng d i n giai PhLfang trinh hodnh dO giao dilm = - X + — 1

Bai to^n 2 24: Cho hdm s6 y = - x + 1

x - 2 Tim tren (H) cdc diem A, B sao cho dp dai AB = 4 va d u a n g t h i n g AB vuong goc vai du-ang t h i n g y = x

Hu'O'ng d i n giai

Vi du'ang t h i n g AB vu6ng goc vai y = x nen phi/ang trinh cua A B Id:

y = - x + m Hodnh dO cua A, B Id nghiem cua phu'ang trinh - x + 1

Vai m = - 1 , tu-ang t y hai di§m A, B c6 tpa dp: rri<*/i |i« (1 + N/2;-2-V2), ( 3 - >/2;-2+N/2)

u - •= x 2 + 2 x + 5 Bai loan 2 25: Cho ham so y =

x + 1

x^ + 2 X - 3 (X +

nghich bi6n tren (-3; - 1 ) , ( - 1 ; 1)

Ham s6 dat CD (-3; -A), CT(1; 4)

10 trq>ng diem bSi dudng h<?c sinh gioi m6n Toon 12 - LS Hodnh Phd

b) Vi X = -1 khdng Id nghiem n§n phuang trinh dS cho tu-ang dyang vb-i;

X + 2x + 5

x + 1 = + 2m + 5.S6 nghi^m cua phu-ang trinh bSng s6 giao di^m

cua d6 thj hdm so y = — v6i duang thing y = m^ + 2m + 5

X + 1 Phtrang trinh c6 hai nghi^m duang khi vd chi khi:

jm*-1 -2<m<0

b) Tim cdc di^m trdn (C) c6 tog dO id s6 nguy^n vd chung minh d6 thj (C) c6

tarn d6i xu-ng

\g din giai

3 a) Ta CO y = X

b) Di6m l\/l(x; y) € (C) c6 tog dp nguyen khi x

- 2 la u-ac s6 cua cua 3 nen x - 2 = ±1, ±3

Do do (C) CO 4 diem c6 tog dO

nguyen:(1; 4), (3; 0), (-1; 0) va (5; 4)

Giao dilm 2 tiem c$n 1(2; 2) chuyen

tryc bdng phep tjnh tidn vecta

Cty TNHHMTVDWH Hhong Vi^t

x^ +1 Bai toan 2 27: Cho hdm so y = — — ; « a) Khao sat vd ve do thi hdm so Tinh g6c giu-a 2 tiem c$n

b) Bi0n ludn theo m s6 nghi^m cua PT; x^ +1 m^ +1

m Hifang din giai

a)y = x* Tgp xdc djnh D = R \ Hdm s6 le

-• Sy bien thien : y '= x ^ - l , y' = 0 o X = - 1 ho$c X = 1

lim y = - c c , lim y = + o c nen TCD: x = 0

• Do thj: D6i xyng nhau qua g6c 0

TCD: X = 0, TCX: y = X nen hai tiem cdn hgp nhau goc 45° _^ • b) S6 nghiem phuang trinh ^ = '^^L^l ia $6 giao diem cua do thj vc^i

W trong di^m bSi dUOng h<?c sinh gioi m6n To6n 12 - L& Hodnh Ph6

a) Tim diem c6 dinh cua do thi ham so (1)

b) Khao sat va ve d6 thj (C) l<hi m = 1 Suy ra do thj h^m s6 y =

X - 1 la ham s6 chin nen d6 thi (C) d6i xueng nhau qua Oy

Khi X > 0 thi liy ph4n d6 thi (C), sau d6 lly d6i xCeng phin do qua Oy thi

W trgng diSm bSi dU<Sng HQC sinh giol m6n To6n 12 - LS Hodnh Ph6

Bai t | p 2 2: Tim tham so de d6 thj :

a) y = f(x) = - ax^ + X +b nhan 1(1,1) lam d i l m u6n *

b) y = f(x) = x" - mx^ +3 c6 2 d i l m uon

HiFO-ng d i n

a) f '(X) = 3x^ -2ax + 1 ; f "(x) = 6x - 2a K i t qua a =3 b =2 '

b) K i t qua m >0

Bai t9p 2 3: Cho ham s6: y = x^ - 3(m + 1)x^ + 9x - m, v6i m Id tham s6

a) Khao sat s y b i l n thien va ve do thi cua hdm s6 khi m = 1

b) Tim m d l d6 thj ham s6 d i cho dgt eye trj t^i X i , Xj sao cho

I X 1 - X 2 I < 2 • , i t^^ '

1 ^ Hu-o-ngdln ^ CH f

a) Khi m = '1 thi y = x ' - 6x^ + 9x - 1 **

b) K i t qua - 3 < m < - 1 - \/3 vd - 1 + < m < 1

Bai tap 2 4: Cho hdm s6: y = 2x^ - 3(m - 1)x^ + m , vdi m la tham so

a) Khao sat s y b i l n thien va ve d6 thj cua ham so khi m = 2

b) Tim m d l d6 thj cua ham s6 da cho c6 hai dilm eye trj sao cho dilm I (3; 1)

n i m tren du'ang thing di qua 2 eye trj

Hipang d i n

a) Khim = 2 t h i y = 2 x ^ - 3 x ^ + 2

b) L i y y chia y' K i t qua m = —

3

Bai tap 2 5: Cho ham s6 y = x^ - 3x^ - 2

a) Khao sat s y b i l n thien va ve d6 thi (C) cua ham s6

j, b) Tim so m dyang d l du'ang thing y = m elt (C) tgi hai dilm A, B sao cho

tam giac OAB vuong tai g l c tpa dp O

b) K i t qua TCX : y = 2x (khi x ^ + « ) ; TCN: y = 0 (khi x ^ -<»)

Bai t | p 2 7: Tim m d l ti$m can xien cua do t h j :

a) Khao sat s y b i l n thien vd ve d6 thj (C) cua ham s6 d a cho b) Tim diem M tren d6 thj (C) sao cho tong khoang edch tCf M d i n cac

dyang t h i n g A,: 2x + y - 4 = 0 vd A2: x + 2y - 2 = 0 la nho nhlt

Hipang d i n _2 '1

a) Khao sat s y b i l n thi§n vd ve do thj (C) cua hdm so d a cho

b) V i l t phu-ang trinh t i l p tuyin cua (C) bilt khoang each tCf tam doi xCeng

cua (C) d i n t i l p tuyIn b i n g Z-Jz

Hyang d i n

a) Tap xac dinh D = R \ y' = — ^

(x + 1)2 b) K i t qua y = x - 2 ; y = x + 6

Bai t9p 2.10: Cho ham s6: y =

x - 1

a) Khao sat s y b i l n thien va ve d6 thj (C) cua ham s6

b) Vai gia trj nao cua m, du-ang thing d: y = - x + m c i t (C) tgi hai dilm A, B

t h 6 a m a n A B = V10 ,

Hu-dng d i n

a)T§p xac djnh D = R \ 1} y ' = ~^

( x - 1 ) ' b) K i t qua m = 0 h a y m = 6

S a i t 9 p 2 1 1 : C h o h a m s 6 y = ^ ^ - ^

X

a) Khao sat s y b i l n thien va ve do thj (C) eua ham s6

A t

ru iTf^ny Ul«Hl UUl UUUliy nt^L sum yiui inun luun - LH ncxj/in r/ici

b) Tim m sao cho duang thing y = m(x - 2) + 4 cit du-dng cong (C) tgi hai

dilm thuOc hai nh^nh cua no

Hirang din t a) TapxacdinhD = R \ { 0 } y= ^ ^ l l l = x + -

b) Di§u ki^n phu-ang trinh hoanh dp giao diem co 2 nghi$m khac diu

Kit qua m >1

- ^ i x^+(m-1)x + 2

Bai tap 2 12: Cho ham so y =

1-x a) Khao sat va ve d6 thi ham s6 tren khi m = 2

b) Xac dinh m d4 ham s6 dat ciic tri tgi Xi, X2 sao cho X1X2 =-3

tilp xuc, 2 nghiem phan bi^t: 2 giao dilm

1) Phu-ang trinh bac 3: ax^ + bx^ + cx + d = 0, a ^ 0 " ' '

N I U CO nghiem x = Xo thi phan tich:(x - XQ) (AX^ + BX + C ) = 0 *"'[

N I U dat ham s6 f(x) = ax^ + bx^ + cx + d thi di§u ki$n:c6 1 nghiem: do thj

khong co cxsc tri hoac yco y c i > 0, co 2 nghiem: yco • y c i = 0, co 3 nghiem

phan biet: yco y c r < 0

y c D y c T <o

Phuang trinh bac 3 co 3 nghiem du'ang khi:'

a.f(O) < 0 1)V 2) Hai dilm tren 2 nhanh d6 thi y = , ta thu-ang lly hai hoanh dp k - a va

- G6c giua 2 duang thing: cosa = I cos(n , n') I = AA'^BB

- Khoang each AB = ^(Xg - x j ^ + (y^ _ y^)2

- Khoang c^ch tu- Mo(xo, yo) den ( A ) : Ax + By + C = 0:

d =

V A ^ + B^ ' a?!;!?''>;v,

~ 06 thi ham bac 3: y = f(x) cit tryc hoanh tai 3 dilm A , B , C theo thCr ty e6 khoang each A B = B C ti>c la 3 nghiem Xi, X2, X3 igp d p s6 cong thi dilm

u6n thuoe true hoanh

Phu'ong trinh trung phu'ang ax" + bx^ + e = 0, a 0 c6 4 nghiem phan bi$t l^p d p s6 cong khi 0 < t, < t2, t2 = 9ti - • ^ <vy •

T i l p tuyIn va ti&p xuc:

- Tiep tuy^n tgi diem M(xo;yo) cua 66 thj (C): y = f ( x )

y - yo = f '(Xo) (X - XQ), so g6c: f '(x) = k = tan(Ox,t)

- Oi^u kien 2 d6 thj y = f ( x ) y = g(x) tiep xuc Id h$ phu-ang trinh: tJHl>

f(x) = g(x) ^ ^.^ ^ ' ' c6 nghi^m

f'(x) = g'(x)

- T i l p tuyIn di qua dilm K(a; b): L$p phu-ang trinh tilp tuyIn tgi Xo bit k y r6i

cho t i l p tuyIn di qua dilm K(a; b) thi tim ra X Q

Chu y: Vdi hai du'6'ng thing d: y = ax + b, d': y = a'x + b' thi c6: d s d' khi a = a',

b = b'; d // d' khi a = a', b ^ b'; d 1 d' khi a a' = - 1

Y I U t6 ddi xi>ng:

- Hdm so chin: Vx e D =:> - x e D vd f(-x) = f(x)

D 6 thj hdm s6 chin d6i xung nhau qua tryc tung

- Hdm s6 le: Vx e D - x € D vd f(-x) = - f(x)

Do thj hdm s6 le ddi xu-ng nhau qua g6c O

- C6ng thu-c chuyin h$ tryc bing ph6p tjnh tiln 0 1

yo = — — , Vxo - X, Xo + X € D, ho$c chuyin tryc bIng ph6p

tjnh tien d i n g l c I n6i tren la ham so le

- O i l u ki$n (C) nh^n d: x = a Idm tryc doi xu-ng;

f(a - X) = f(a + X), Va-x, a + x € D,ho$c chuyin tryc bIng ph6p tjnh t i l n

d i n S(a,0) Id hdm so c h i n

Quy tich dilm M:

Tim tog dp x, y cua M, khu tham s6 giua x vd y

Gib-i hgn: Chuyin dilu ki^n n l u c6 cua tham s6 v l dilu ki^n cua x (hay y)

Ode bi?t: N l u M(x, y) e (V) thi chi c i n tim x rli riit tham s6 d l t h i ,

khu-tham s6

'•' • *

2 C A C B A I T O A N

Bai toan 3 1: Chupng minh ring do thj hdm s6 y = x^ + 2 m V + 1 Iu6n c i t

du-d-ng t h i n g y = x + 1 tgi dung hai d i l m phdn bi^t vdi mpi gid trj m

f (x) = 3x^ + 2m^ 2: 0 nen hdm s6 nay dong biln tren R

VI lim f(x) = lim (x^ + 2Tn^x - 1 ) = -oo

va lim f(x) = Nm (x^ + 2xn^x - 1 ) = +=o

X - » + o O X - » + c « ^^^^

nen phu'cng trinh f(x) = 0 luon c6 nghi^m duy n h i t x ?t 0: dpcm

Bai toan 3 2: Tim m d l do thj hdm so sau c I t tryc hodnh tgi 3 d i l m phdn bi$t:

b) Tgi hai d i l m thupc hai nhdnh cua d l thj? o

Hu'O'ng d i n giai

Phu-ang trinh cua (dm):y = m(x + 2) + 2 = m m + 2

Phu-ang trinh hodnh dp giao d i l m cua (dm) va du-ang cong : , j.;: ,

W trgng diSm bSi ducrng HQC sinh gioi mdn loan 12 - Te Hoanh Pho

^''^ <=>.'"''° m < 0 ho$c m > 12

A>0,g(-1)^0 [m^-12m>0

b) Hai nh^nh cua du'ang cong d§ cho nSm ve hai ben cua du-d-ng ti^m cgn

du-ng X = -1 cua do thi Du'ang thing (dm) cit du-b-ng cong da cho tgi hai

dilm thuOc hai nh^nh cua n6 khi va chi khi phuo-ng trinh (1) c6 hai nghi^m

Xi, X2 Va Xi < -1 < X2

D#t X = t - 1 thi Xi <-1< X2 => ti < 0 < t2

Phuang trinh tra thanh: m(t - 1)^ + 3m(t - 1) + 2nn + 3 = 0

, « mt^ + mt + 3 = 0 (2)

OK phu-ong trinh (2) c6 hai nghi^m trai diu o P < 0 o m < 0

Bai toan 3 4: Tim tham so d l du'6'ng thing

a) y = m, m >0 clt d6 thj (C) cua ham s6 y = x" - 3x^ - 2 tgi hai dilm A, B

sao cho tam giac OAB vu6ng tgi g6c toa dO O

b) y = 3x + m clt d6 thj (C) cua ham so y = — - tgi 2 dilm phan bi$t c6

x - 1 hoanh dO xi, X2 va I xi - X21 dgt gia trj nho nhlt

Hipang din giai a) Phu-o-ng trinh hoanh dO giao dilm:

x^ - 3x^ - 2 = m x" - 3x^ - 2 - m = 0

• V6i mpi m > 0 thi du-ang thing y = m clt (C) tgi hai dilm phan biet A(XA; m)

va B(XB; m) doi xu-ng qua Oy, XA < XB

Tam giac OAB vuong tai O nen OA OB = 0 <=> XA-XB + m^ = 0

-b + V A -b - N/A

: Dung Vm

2a 2a

4 4 2 Vgy gia trj I Xi - X21 nho nhlt khi m = - 1

Cfy TNHHMTVDWH Hhang Vi$t

gai toan 3 5 Tim cac gia trj cua m sao cho a) D 6 thj cua ham s6 y = x" - (m + 1)x^ + m clt tryc hoanh tgi bin dilm, tgo thanh ba dogn thing c6 dO dai blng nhau

2x — 1 b) Du-b-ng thing d: y = -x + m clt (C): y = tgi hai dilm A, B ma AB = >/lO

HiKyng din giai a) Hoanh dO giao dilm cua du-dng cong va tryc hoanh la nghi^m phu-o-ng trinh: x" - (m + 1)x^ + m = 0 o x^ = 1 hoac x^ = m

Dilu ki$n m > 0 va m 1 Khi d6, phu-ong trinh c6 4 nghi^m

nghi0m Xi, X2phan bipt khac 1

Ta c6 AB = N/IO < ^ (X2 - x,)^ + (X2 - Xi)' = 10 «( X 2 - Xi)^ = 5

^ai toan 3 6: Chii-ng minh cac du-o-ng thing d: y = m - x luon clt d6 thj (C):

- 3x

y = -~r- tgi 2 dilm M, N va clt 2 ti^m can cua (C) tgi P, Q ding th(y\i

<J09n IVIN, PQ CO cung trung dilm

Hipang din giai Phu-a 'ng trinh hoanh dp giao dilm d va (C):

x^ -3x

x - 1 = m - X o 2x2 - (m + 4)x + m = 0, X ^ 1. g Sri *

10 tr<?ng diS'm bSl duang h<?c sinh gl6l mdn Todn 12 - LS Hodnh Ph6

Ta CO X = 1 khong la nghiem va A = + 16 > 0, V m nen d lu6n cat (C) tgi 2

Bai toan 3 7: V i l t p h u a n g trinli tiep tuyen cua do tJij ham s6:

a) y = Vx + 2 biet tung dp tiep diem la yo = 2

1 3

b) y = ~ g 2x^~ 3x + 1 song song vai d: y = - x + 9

Hirang din giai

a) Ta CO phu-ang trinh ti^p tuyen tai d i l m (XQ, f(xo)):

y = f •(xo)(x - xo) + f(xo)

Vai Xo = - - t h i f(Xo) = - — n e n CO tiep tuyen y = — X - —

V$y CO 2 tiep tuyen y = — x - — v ^ y = — x + —

' ' 4 8 ' 4 12

Bai toan 3 8: Lgp phu-ang trinh tiep tuy4n vai d6 thj: j^,,^: •

a) y = 2x^ - 6 x ^ + 3 va c6 h? s6 goc b6 nhat

b) y = f(x) thoa man f ^(1 + 2x) = x - f ^(1 - x) tgi x = 1

Hu'O'ng din giai

a) Ta c6 h0 s6 g6c cua ti^p tuy§n la dao ham tgi d6

Cti^ TNHHMTVDVVHHhang Vi$t

y _ 6x^ - 12x = - 6 + 6(x - i r > - 6 , d § u = khi Xo = 1 pgn max y' = - 6 , do do ti§p tuyen tgi A ( 1 ; - 1 ) la y = - 6 x + 5

V$y p h u a n g trinh tiep tuyen y = - y (x - 1)

Bai toan 3 9: Vi4t phu-ang trinh tiep tuyen cua (C) h^m s6:

X — 3 '

a) y = bi§t khoang c^ch tu- tam d6i xii-ng cua (C) den t i l p tuy6n bSng

2>/2

b) y = x^ - 3x^ + 2 bi§t r i n g tiep tuyen c i t c i c tryc Ox, Oy l l n lu-cxt tgi c^c

d i l m p h l n bi$t A, B sao cho OB = 9 0 A

10 tr<png diem boi dt/crng hgc sinh gioi mon Toan 12 - LS Hoanh Ph6

Ti§p tuy^n c i t cac true Ox, Oy l l n lu-gt tai cac dilm phan bi$t A, B sao cho

OB = 90A nen h? s6 goc cua tidp tuyen d la:

\lti\o = 1, phu-ang trinh cua d la y = 9x + 7

Vai Xo = 3, phu-ang trinh cua d la y = 9x - 25

d: y = y'(xo)(x - Xo) + yo

X ^ 1-m

y = - 1 ( m x „ + m -1)^ - ( X - X J + (1-m )XQ + 2 - m

Bai toan 3.12: Lap phu'ang trinh tilp tuyIn chung cua 2 d6 thj:

(Pi); y = x^ - 5x + 6 vS (P2): y = -x^ + 5x - 11

HLPang din giai

(Pi): y = f(x) = x^ - 5x + 6 => f '(x) = 2x - 5

(P2): y = g(x) = -x^ + 5x - 11 ^ g'(x) = -2x + 5

Gpi tilp tuyen chung la y = ax + b vS Mi(xi; f(Xi)),

'^2(X2; g(x2)) la 2 tilp dilm tu'ang Cpng Ta c6 h?:

f(x^)=::ax^+b x ^ - 5 x , + 6 = ax^+ b f'(x,) = a

10 trQng diSm bSi ducing hoc sinh gioi mon Toan 12 - LS Hodnh Phd

va - 5xi + 6 - ( - + 5X2 - 11) = (2xi - 5)(Xi - X2)

nen: x f - 5xi + 17 + (5 - X i f - 5(5 - x,) - (2xi - 5)^ = 0

o 2x^ - 10xi + 8 = 0 <=> xi = 1 hoSc Xi = 4

Vdi Xi = 1 thi a = - 3 , b = 5 Vc^i Xi = 4 thi a = 3, b = - 1 0

V § y c6 2 tiSp t u y i n chung: y = 3x - 10 va y = - 3 x + 5

Bai toan 3 13: Tim d i l m IVI tren d6 thi (C) hdm so: y = 2 x - 2

x - 2 sao cho tiep tuy^n tai cat hai du'6'ng ti$m c$n cua A, B vai AB = 2\/5

Hu-o-ng d i n giai

Phu-ang trinh tiep tuy§n tai l\/l(xo; yo) e (C), Xo =^ 2

- 2 d:y =

(Xo-2y

2x - 2

Giao diem cua d vai tiem cgn dCrng x = 2 id A

Giao d i l m cua d vdi tiem cdn ngang y = 2 la B(2xo - 2; 2)

Bai toan 3.14: Cho ham so y = f(x) = x" - 2x^ c6 do thj (C) Tren d6 thj (C) l l y

diem phan biet Id A va B c6 hoanh dp Ian lu-p-t la a, b Tim d i l u l<ien cua a,

b de t i l p t u y i n cua (C) tai cdc diem A va B song song v6i nhau

Hipang d i n glai

T a c6 f '(x) = 4x^ - 4x Gpi a, b Idn lu-gt Id hodnh dp cua A vd B, a b, He

so gcc cua ti§p tuyen cua (C) tgi A vd B Idn lu-p't Id:

RA = f '(a) = 4a^ - 4a, ke = f '(b) = 4b^ - 4b

Ti§p t u y i n t?i A va B Idn lu-p't c6 phu-ang trinh id

y = f '(a)(x - a) + f(a) = f •(a)x + f(a) - af '(a)

a^ + a b + b^ = 1 , a ; t b ,,cVfV:^ i y V i '

-3a^ + 2a2 =-3b^ + 2b2

Giai he ndy, ta du-p-c nghi^m Id (a; b) = ( - 1 ; 1), ( 1 ; - 1 ) Vdy dieu kien can va du d l hai tiep tuyen cua (C) tgi A vd B song song vai nhau la a^ + ab + b^ = 1, a ^ ± 1 , a b

Bai toan 3.15: Tiep t u y l n (T) cua (H): y = — ^ tai d i l m M c6 hodnh dO x = a rf,

x - 2 cdt true hoanh Ox tai A vd cdt du-cxng t h i n g d: x = 2 tgi B Chu-ng minh M Id trung d i l m cua AB vd dien tich tam gidc gi6i hgn bai tiep t u y i n Ox vd d

Giao d i l m A vai tryc hodnh Cho yA = 0 thi ^ " ^ X A = 2a - 2

( 8 - 2 ) 2 a _ 2

Giao d i l m B vai du-ang t h i n g d: x = 2

Cho XB = 2 thi ye = ziizfl + _ 1 _ = _ 2 _

nen tiep d i l m M Id trung d i l m cua AB

Gpi I Id giao d i l m cua Ox vd d thi 1(2; 0) Tam giac c i n xdc djnh Id tam gidc

ABI vuong tgi I c6 dien tich:

2

S = ^ I A I B = - | 2 a - 2 - 2

2 2 a - 2 - 0 = 2: khong doi

^ai toan 3.16: Cho hdm so y = ^^^^ Chung minh r i n g qua d i l m M(3; - 1 )

ve du-p-c hai tiep tuyIn v^i do thj vd hai tilp tuyIn d6 vu6ng goc vd-i nhau

Huxyng d i n giai

P h u a n g trinh du-dyng t h i n g qua M(3; - 1 ) he s6 gdc a Id y = a(x - 3) - 1,

^^u-ong t h i n g Id tiep tuyIn vai do thj khi h? sau c6 nghiem:

W trQng diem bSi duang h^c sinh gl6i man loan IS - LS Hodnh Ph6

Thay ( 2 ) vdo ( 1 ) va rut gpn ta du-p-c: x - x - 1 = 0

PT c6 2 nghiem thoa man: Xi + X2 = 1, X1.X2 = - 1

Tac .:V,x,,.v.(x,, = J ^ | ^

fe: ?_ ( x / g ) ' - 2 x ^ X g ( x , + X3) + 4x^Xg 1 + 2 - 4 _ ^ '

(x.x^ - X , - X2 +1)2 ~ ( _ 1 _ 1 + 1 ) 2 "

Vay 2 tiep tuyen qua IVI vuong g6c v^i nhau

Bai toan 3.17: Cho ham so y = -x^ + 3 x ^ - 2 (C) Tim tr§n (C) nhu-ng dilm ma

qua do chi ke du'p'c mpt ti^p tuyen v6i (C)

Himng din giai Gia SLP M(xo; yo) id mpt dilm tren (C) Gia si> tiep tuyin (t) ke tu- IVI d i n (C)

tilp xuc vai (C) tgi N(Xi; y i ) Khi d6 phuang trinh cua (t) c6 dgng: y - yi =

— = Xo o Xo = 1 Tu- dp tinh du-pc yo = 0

Vdy M(1; 0) Id dilm duy nhdt trdn (C) md qua d6 c6 t h i ke dung mpt tiep

tuyIn vai (C)

Bai toan 3 18: C6 bao nhieu tiep tuyIn cua do thi (C): y = + + 2 ^

x + 1 giao dilm ciia 2 ti^m c$n

IHifang din giai

Vdy khong mpt tilp tuyIn ndo cua (C) di qua I

Bai toan 3.19: Chiang minh tilp tuyIn tgi A(-1; 0) cua d6 thj (C):

y = -x" + 2x^ + X cung la tilp tuyIn cua d6 thj ndy tgi mOt dilm B khdc A nOa

Hu'O'ng din giai •

Bai toan 3 20: Chung minh hai do thj sau tilp xiic nhau: y= f(x) = -^+2^ ^d

y= g(x) = Vilt phu-o-ng trinh tilp t u y I n chung cua chung tgi tilp dilm

x + 2

Hu'O'ng din giai

Hodnh d O tilp dilm cua hai du'6'ng cong Id nghiem cua h$ phu'ang trinh:

W trqng diSm bSi duang HQC sinh gioi m6n To6n 12 - Ho6nh Phd

t i l p xuc vd-i nhau tai goc toa dp O

3 ^ ' '

Ta c6 y'(0) = - nen phu-ang trinh tiep t u y i n chung cua hai dudyng cong tgi

diem chung la y = - x

Bai toan 3 21: T i m tham so d l d6 thj hdm s6

a) ( C ) : y = x^ - 1 - k(x - 1) tiep xuc vai tryc hoanh

chi khi h^ sau c6 nghiem

[x + m = x ^ - 3 x 2 - 8 x [m = - 3x2 - 9x

1 = 3x2 - 6x - 8 x2 - 2x - 3 - 0

m = x^ - 3x2 - 9x

x = - 1 v x = 3 Vd-i X = - 1 , ta CO m = 5 vd vo-i x = 3 thi m = - 2 7

Vely CO hai gia trj m c i n tim la m = 5, m = - 2 7

Bai toan 3 22: Cho ham s6 y = x^ - 3x + 2 (C) Xet 3 diem A , B, C t h i n g hang

va thuQC (C) Gpi A ' , B', C la giao diem cua (C) vo'i tiep tuy^n cua (C) tgi A ,

1 B, C ChLPng minh r i n g A' B', C t h i n g hang

Hu^yng d i n giai Phu-ang trinh ti4p tuyen cua (C) tai d i l m A(XA; yA) c6 d?ng :

y = ( 3 x 2 - 3 ) ( x - X A ) + yA

Ph

Cty TNHHMWDWHHhong Vi$t

Phu-ang trinh hoanh dp giao dilm cua (C) vd tiep tuy^n c6 dang:

( 3 x 2 - 3 ) ( x - X A ) + yA = x ' - 3 x + 2

^ (3x2 - 3)(x - XA) + x^ - 3XA +2 = x ' - 3x +2 ^; ^ ,

^ (X - XA)2(X + 2XA) = 0

Do dd ti6p tuyen cua (C) tai A c i t (C) tai 2 diem c6 hodnh dO XA chinh Id A

va d i l m c6 hodnh dp -2XA la didm A', tCcc la XA' = -2XA

« « ^ V : I!'3) If'.;

Tu-ang tu- XB' = -2XB', XC = -2xc

Ta chLpng minh nhgn xet: A, B, C thupc (C) thing hang khi va chi khi + xs + Xc = 0

Th^y vay, gia s i i A, B, C n i m tren duo'ng t h i n g c6 phu-ang trinh y = ax + b

Khi dd XA, XB, XC Id nghiem cua phu'ang trinh

x^ - 3x + 2 = ax + b » x^ - (3 + a)x + (2 - b) = 0 ,, ^ „ •

Ap dyng djnh If Viet, ta suy ra XA + Xg + xc = 0

Nguac lai, gia si> XA + xs + xc = 0 Vi^t phu'ang trinh du-o-ng t h i n g di qua A, B cIt C thi theo p h i n thuan ta c6 XA + XB + xc = 0 suy ra Xc = Xc suy ra C triing

vdi C vd CO nghTa Id A, B, C t h i n g hang Nhan xet du-p-c chCeng minh

Ap dgng do A, B, C t h i n g hdng nen ta c6 XA + XB + Xc = 0

Md XA' + XB' + Xc = -2(XA + XB + XC) = 0 nen suy ra A', B", C thing hang (dpcm)

Bai toan 3 23: Cho hdm so y = "^^^ ^ ^^"^^ - 2)x - 2 ^ ^ ^ ^

Khi m ^ 0 thi do thi c6 tiem can dung: x = -3m vd ti^m can xien: y = mx - 2

Hai ti^m cgn hgp nhau goc 45° khi ti^m cgn xien hp'p vdi tryc hoanh mpt

gdc 4 5 ° » m = ± 1

Bai toan 3 24: Tim m 6h du'dng thing y = 2m cIt do thi ham so y = tgi

x - 1 hai di4m phdn bipt M vd N sao cho CM vudng goc v d i ON