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1. Chuyn luyn thi i hc-phn i: kho s t hm s Nm h c: 2010- 2011 LU Y N TH I I H C

CHUYN :KH O ST HM S m n Good luckd hu y:: Cac ban can nam vng kien thc

KSHS , cung ket hp vi cac dang Bai Toan di ay th C kha nang cua ban giai

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quyet phan KSHS trong e thi ai Hoc rat de dang (Hehe... )va ieu quan trong la cac

ban can phai nh k cac dang e tranh s nham lan gia dang nay vi dang khac

nhe , neu k th ..... BA CNG TH C TNH NHANH O HM s ng bi n trn a > 0 th y '

0 x 0 C A HM S H U T +y= hm ax + b ad bc y' = cx + d (cx + d )2 D

ng 2: Cho hm s y = f(x) c ch a tham s m. nh m hm s ngh ch bi n trn ? adx 2 + 2aex + (be

cd ) ax 2 + bx + c +y= y' = dx + e (dx + e )2 Phng php: + TX: D = a x 2 + b1 x + c1 y=

1 2 a 2 x + b2 x + c 2 Ta c: y = ax2 + bx + c (a1b2 a 2 b1 ) x 2 + 2(a1c 2 a 2 c1 ) x + b1c 2

b2 c1 y' = ( a 2 x 2 + b2 x + c 2 ) 2 CHUYN : CC CU H I TH HAI TRONG THI KH O

ST HM S LTH hm s ng bi n trn a < 0 0 th y ' 0 x D ng 3: Cho

hm s y = f(x) c ch a tham s m. nh m th hm s c c c tr ? Phng php: TX: D = D ng 1:

Cho hm s y = f(x) c ch a tham s m. nh m hm s ng bi n trn ? Phng php: Ta c: y =

ax2 + bx + c www.MATHVN.com th hm s c c c tr khi phng trnh y = 0 c 2 nghi m phn bi t

v y i d u khi x i qua hai nghi m a 0 > 0 TX: D = Cch h c t t mn Ton l ph

i lm Ta c: y = ax2 + bx + c Bai tap nhi u , bn c nh , d Trang1/10-LTH-2010

2. Chuyn luyn thi i hc-phn i: kho s t hm s Nm h c: 2000- 2011 D ng 4: Cho hm s y =

f(x) c ch a tham s m. Ch ng minh r ng v i m i m th hm s lun lun c c c tr ? D ng 9: Cho hm s

y = f(x) c ch a tham s m. nh m th hm s i qua i m c c tr M(x0;y0)? Phng php: Phng

php: TX: D = TX: D = Ta c: y = ax2 + bx + c 2 Ta c: y = ax + bx + c Xt phng trnh

y = 0, ta c: f '( x0 ) = 0 f ( x0 ) = y0 hm s i qua i m c c tr M(x0;y0) th =.>0, m

V y v i m i m th hm s cho lun lun c c c tr . D ng 5: Cho hm s y = f(x) c ch a tham s m.

nh m th hm s khng c c c tr ? Phng php: D ng 10: Cho hm s y = f(x) c th (C) v

M(x0;y0)(C). Vi t PTTT t i i m M(x0;y0) ? Phng php: Ta c: y = f(x) f(x0) TX: D =

Phng trnh ti p tuy n t i i m M(x0;y0) l Ta c: y = ax2 + bx + c Hm s khng c c c tr khi y

khng i d u trn ton a 0 t p xc nh 0 y y0 = f(x0).( x x0 ) Cc d ng th ng g

p khc : 1/ Vi t phng trnh ti p tuy n v i th (C) t i i m c hanh x0. D ng 6: Cho hm s y = f(x)

c ch a tham s m. nh m th hm s t c c i t i x0? Ta tm: + y0 = f(x0) Phng php: Suy ra

phng trnh ti p tuy n c n tm l TX: D = y y0 = f(x0).( x x0 ) 2 Ta c: y = ax + bx + c 2/

Vi t phng trnh ti p tuy n v i th (C) t i i m th a mn phng trnh f(x)= 0. f '( x0 ) = 0 f ''(

x0 ) < 0 hm s t c c i t i x0 th Ta tm: + f(x) D ng 7: Cho hm s y = f(x) c ch a tham s m.

nh m th hm s t c c ti u t i x0? Ta c: y = ax2 + bx + c a/ song song v i ng th ng y = ax +

b. f '( x0 ) = 0 hm s t c c ti u t i x0 th f ''( x0 ) > 0 b/ vung gc v i ng th ng y = ax

+ b. Phng php: D ng 8: Cho hm s y = f(x) c ch a tham s m. nh m th hm s t c c tr b ng

h t i x0? Phng php: TX: D = Ta c: y = ax2 + bx + c s +Gi i phng trnh f(x) = 0 x0 D

ng 11: Cho hm s y = f(x) c th (C) Vi t phng trnh ti p tuy n (d) c a (C) TX: D = hm +

f(x) + y0 v f(x0). Suy ra PTTT. Phng php: + f(x) f(x0) t c c tr b ng h t i x0 th a/

Tnh: y = f(x) V ti p tuy n (d) song song v i ng th ng y = ax + b nn (d) c h s gc b ng a. Ta

c: f(x) = a (Nghi m c a phng trnh ny chnh l honh ti p i m) Tnh y0 tng ng v i m i x0

tm c. f '( x0 ) = 0 f ( x0 ) = h Suy ra ti p tuy n c n tm (d): y y0 = a. ( x x0 ) Cch h c

t t mn Ton l ph i lm www.MATHVN.com Bai tap nhi u , bn c nh , d Trang2/10-LTH-

2010

3. Chuyn luyn thi i hc-phn i: kho s t hm s b/ Tnh: y = f(x) V ti p tuy n (d) vung

gc v i ng th ng y = ax + b nn (d) c h s gc b ng Ta c: f(x) = 1 . a l honh ti p i m) y

= f(x) v f(x) = g(x) (*) S giao i m c a hai th (C1), (C2) chnh l s nghi m c a phng trnh (*).

Suy ra ti p tuy n c n tm (d): 1 . ( x x0 ) a D ng 15: D a vo th hm s y = f(x), bi n lu n theo m s

nghi m c a phng trnh f(x) + g(m) = 0 Ch : Phng php: + ng phn gic c a gc ph n t th

nh t y = x. Ta c: f(x) + g(m) = 0 + ng phn gic c a gc ph n t th hai y = - x. D ng 12: Cho hm

s y = f(x) c th (C) Tm GTLN, GTNN c a hm s trn [a;b] Phng php: f(x) = g(m) (*) S

nghi m c a (*) chnh l s giao i m c a th (C): y = f(x) v ng g(m). D a vo th (C), ta c:

v.v Ta c: y = f(x) Gi i phng trnh f(x) = 0, ta c cc i m c c tr : x1, x2, x3, [a;b]

Tnh: f(a), f(b), f(x1), f(x2), f(x3), [ a ;b ] Phng php: OI = ( x0 ; y0 ) . [a ;b] D ng 13: Cho h

ng cong y = f(m,x) v i m l tham s .Tm i m c nh m h ng cong trn i qua v i m i gi tr c a

m. x = X + x0 x+2 y= x3 y = Y + y0 Cng th c i tr c: Th vo y = f(x) ta c Y = f(X) Ta

c n ch ng minh hm s Y = f(X) l hm s l . Suy ra I(x0;y0) l tm i x ng c a (C). Phng php: Ta

c: y = f(m,x) (1) Am + B = 0, m Ho c Am2 + Bm + C = 0, D ng 16: Cho hm s y = f(x), c th

(C). CMR i m I(x0;y0) l tm i x ng c a (C). T nh ti n h tr c Oxy thnh h tr c OXY theo vect max

y = ; min y = Phng php chung ta th ng l p BBT Phng trnh honh giao i m c a y = g(x)

l f(x) g(x) = 0 Tnh y0 tng ng v i m i x0 tm c. T suy ra: y = f(x) v D ng 14: Gi s

(C1) l th c a hm s (C2) l th c a hm s y = g(x). Bi n lu n s giao i m c a hai th (C1), (C2).

Phng php: 1 (Nghi m c a phng trnh ny chnh a y y0 = Nm h c: 2000- 2011 m D ng 17:

Cho hm s y = f(x), c th (C). CMR ng th ng x = x0 l tr c i x ng c a (C). (2) Phng php:

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th hm s (1) lun lun i qua i m M(x;y) khi (x;y) l nghi m c a h phng trnh: i tr c b ng t nh ti n

theo vect OI = ( x0 ;0 ) A = 0 B = 0 Cng th c i tr c (a) A = 0 Ho c B = 0 (b) C

= 0 x = X + x0 y = Y ( i v i (1)) Th vo y = f(x) ta c Y = f(X) Ta c n ch ng minh hm s Y

= f(X) l hm s ch n. Suy ra ng th ng x = x0 l tr c i x ng c a (C). ( i v i (2)) Gi i (a) ho c (b)

tm x r i y tng ng. T k t lu n cc i m c nh c n tm. Cch h c t t mn Ton l ph i lm

www.MATHVN.com Bai tap nhi u , bn c nh , d Trang3/10-LTH-2010

4. Chuyn luyn thi i hc-phn i: kho s t hm s D ng 18: S ti p xc c a hai ng cong c

phng trnh y = f(x) v y = g(x). Phng php: Hai ng cong y = f(x) v y = g(x) ti p xc v i nhau

khi v ch khi h phng trnh Nm h c: 2000- 2011 D ng 21: nh ki n th hm b c 3 c C , CT n

m v cung 1 pha I v I (D). Phng php + nh ki n th hm s b c 3 c cc i m c c tr M 1 (x1 ,

y1 ) & M 2 ( x 2 , y 2 ) ( x1 , x 2 l nghi m c a pt y' = 0) f ( x) = g ( x) f '( x) = g '( x) C

nghi m v nghi m c a h phng trnh trn l honh ti p i m c a hai ng cong . 1)N u (D) l tr c

Oy th ycbt x1 < x 2 < 0 0 < x1 < x 2 2)N u (D) l th ng x = m th D ng 19: Tm i m A ,t A k

c n ti p tuy n t i th y = f (x) (C) ycbt x1 < x 2 < m 0 < x1 < x 2 Phng php +Gi s 3)N u

(D) l th ng ax + by + c = 0 th: A(x 0 , y 0 ) ycbt (ax1 + by1 + c )(ax 2 + by 2 + c ) > 0 + Pt

th ng i qua A(x 0 , y 0 ) c h s gc k c d ng : @ N u (D) l ng trn th cng gi ng tr ng h p 3)

(d ) : y = k (x x0 ) + y 0 +th ng (d) ti p xc v I th (C) khi h sau c nghi m D ng 22: nh ki n

th hm s (C) c t th ng (D) t I 2 i m phn bi t tho 1 trong nhng ki n sau: f (x ) = k (x x0 ) +

y 0 (1) ' f ( x ) = k ( 2) Thay (2) vo (1) c : f (x ) = f ' (x )(x x 0 ) + y 0 (3) +Khi s nghi

m phn bi t c a (3) l s ti p tuy n k t A t I th (C) Do t A k c k ti p tuy n t I th (C) 1)Thu c

cng 1 nhnh (I) c nghi m phn bi t n m cng 1 pha I v I x = m ( (I) l PTHG c a (C) v

(D) ; x = m l t/c n ng c a (C) ) 2) Cng 1 pha Oy ( I ) c 2 nghi m phn bi t cng d u 3)Khc

pha Oy ( I ) c 2 nghi m phn bi t tri d u c k nghi m phn bi t i m A (n u c) D ng 23:

Tm i m trn th hm s (C) sao cho: D ng 20: nh ki n th hm s b c 3 c C , CT n m v 2

pha (D) T ng cc kho ng cch t n 2 t/c n l Min Phng php + nh ki n th hm s b c 3 c

cc i m c c tr M 1 (x1 , y1 ) & M 2 ( x 2 , y 2 ) Phng php: ( +Xt M 0 (x 0 , y 0 ) thu c (C)

x 0 , , y 0 ( x1 , x 2 l nghi m c a pt y' = 0) tho y = thng +d /m u 1)N u (D) l tr c Oy th ycbt

x1 < 0 < x 2 ) +Dng BT Csi 2 s kqu 2)N u (D) l th ng x = m th ycbt x1 < 0 < x 2

3)N u (D) l th ng ax + by + c = 0 th: D ng 24:Tm i m trn th hm s (C) sao cho:kho ng cch t

n 2 tr c to l Min ycbt (ax1 + by1 + c )(ax 2 + by 2 + c ) < 0 @ N u (D) l ng trn th

cng gi ng tr ng h p 3) Phng php: +Xt M 0 (x 0 , y 0 ) thu c (C) Cch h c t t mn Ton l ph i

lm www.MATHVN.com Bai tap nhi u , bn c nh , d Trang4/10-LTH-2010

5. Chuyn luyn thi i hc-phn i: kho s t hm s + t P = d (M 0 , Ox ) + d (M 0 , Oy ) P

= x0 + y 0 Nm h c: 2000- 2011 ' y ' = 0 U x1V x1 = V x'1U x1 +Nhp :Cho x0 = 0 y 0

= A; y 0 = 0 x0 = B + G I B (x 2 , y 2 ) l i m c c tr c a (C m ) G I L = min ( A , B ) ...........

.................... .......y 2 = +Ta xt 2 tr ng h p : TH1: x0 > L P > L Phng php: Phng

php M ,N,P th ng hng vet MN cng phng v I vect b a MP x M + x N + x P = +Chia

cx + d y (cx+d :l ph n d c a php = ax + b + y' y' chia) y = (ax + b ) y '+ cx + d D ng 26: Tm

trn th (C) :y = f(x) t t c cc i m cch u 2 tr c to +Goi A( (x1 , y1 ), B (x 2 , y 2 ) l 2 i m c c

tr c a hm s (C m ) y ' x1 = y ' x 2 = 0 +Do A (C m ) nn y1 = (ax1 + b ) y1 '+ cx1 + d Phng

php: +T p h p nh ng i m cch u 2 tr c to trong (Oxy) l ng th ng y = x v y = -x .Do :

+To c a i m thu c (C) :y = f(x) ng th I cch u y = f ( x) y = x 2 tr c to l nghi m c

a : kqu y = f ( x) y = x ax 2 + bx + c a ' x + b' y1 = cx1 + d (1) +Do B

(C m ) nn y 2 = (ax2 + b ) y 2 '+ cx2 + d y 2 = cx 2 + d (2) T (1),(2) suy ra pt /t i qua 2 i m

c c tr : y = cx + d D ng 29: nh ki n th hm s b c 3 c i m C v CT I x ng nhau qua 1 /t y

= mx + n D ng 27:L p pt /t i qua 2 i m c c tr c a hm s h u (m 0) Phng php: (C m ) + nh

ki n hm s c C, CT (1) +L p pt /t (D) i qua 2 i m c c tr Phng php : +G i I l trung i m o

n n I 2 i m c c tr U (x) V( x ) (U ) V ' + c y ' = (2) D ng 28:L p pt /t i qua 2 i m c c tr c a hs b c

3 (C m ) , khi ko tm c 2 i m c c tr D ng 25:Tm ki n c n v 3 i m M,N,P cung thu c th (C)

th ng hng? t y= ' U x2 V x' 2 ' Ux T (1), (2) suy ra pt /t i qua 2 i m c c tr l y = ' Vx TH2: x0

L .B ng pphp o hm suy ra c kqu t :y= ' U x1 U x1 = ' = y1 (1) V x1 V x1 ( x) dk (1) +ycbt

y = mx + n ( D ) kq I y = mx + n (V( x ) ) U ( x ) ' ( x) (V ) 2 ( x) +G I A (x1 ,

y1 ) l i m c c tr c a (C m ) Cch h c t t mn Ton l ph i lm www.MATHVN.com Bai tap nhi u ,

bn c nh , d Trang5/10-LTH-2010

6. Chuyn luyn thi i hc-phn i: kho s t hm s D ng 30:Tm 2 i m thu c th (C) y = f(x)

I x ng nhau qua i m I (x0 , y 0 ) Nm h c: 2000- 2011 D ng 33 :V th hm s y = f (x ) (C) Phng

php: +V Phng php: th y = f (x ) (C ') +V th hm s +Gi s M (x1 , y1 ) (C ) : y1 = f (x1 ) (1)

y = f ( x ) (C1) +G I N (x 2 , y 2 ) I x ng M qua I suy ra to i m N theo x1 , y1 CHUYN :CC

BI T P LIN QUAN N KH O ST HM S LTH +Do N thu c (C): y 2 = f (x 2 ) (2) (1),(2) :gi I

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h , Tm x1 , y1 x 2 , y 2 Cau 1.Tm m ng th ng y=x+4 c t th hm s y = x3 + 2mx 2 + ( m

+ 3) x + 4 t i 3 i m phn bi t A, D ng 31:V th hm s y = f ( x ) (C) B,C sao cho tam gic MBC c

di n tch b ng 4. (i m B, C c honh khc 0, M(1;3) Cau 2. Tm m hm s . y = x3 mx 2 +

(2m + 1) x m 2 c t Ox t i 3 i m phn Phng php: +V bi t c honh dng Cau 3. Tm hai

i m A, B thu c th y = f (x ) (C ') f (x ), x 0(C1 ) +C y = f ( x ) = f ( x ), x < 0(C 2 ) 3

th hm s 2 y = x 3x + 1 sao cho ti p tuy n t i A, B song song v i nhau v AB = 4 2 x+m Tm m ti

p tuy n c a th x 1 th (C) g m th ( C1 ) v th (C 2 ) Cau 4 Cho hs : y = V I : (C1 ) (C ')

t i giao i m I c a hai ti m c n c t tr c Ox , Oy t i A, B v di n tch tam gic IAB b ng 1 (C 2 ) l ph l y

ph n x 0 n I x ng c a (C1 ) qua Oy D ng 32 :V th hm s Cau 5.Cho hm s y = f (x ) (C) Cau

6. Cho hm s y = f (x ), f (x ) 0(C1 ) f (x ), f (x ) < 0(C 2 ) Cau 7. Cho hm s +C y = f (x

) = y= x 1 ( H ) . Tm i m M thu c (H) x +1 t ng kho ng cch t M n 2 tr c to l nh nh t.

th (C) g m th ( C1 ) v th (C 2 ) Cau 8. Cho hm s V I (C1 ) (C ') l y ph n dng c a (C') (n

m trn y= 3x + 1 ( H ) v ng th ng x 1 y = ( m + 1) x + m 2 (d) Tm m ng th ng (d) c t 3

(H) t i A, B sao cho tam gic OAB c di n tch b ng 2 Cau 9. Cho hm s y = x3 3 x 2 + 3(1 m) x

+ 1 + 3m Ox) (C 2 ) l ph n I x ng c a ph n m (n m d I Ox ) c a (C') qua Ox (Cm). Tm m hm

s c c c i c c ti u ng th i cc i m c c tr cng v i g c to t o thnh tam gic c di n tch b ng 4

@:Ch : thi y = f (x ) s n m trn Ox www.MATHVN.com 2x (H) .Tm cc gi tr c a m x 1

ng th ng (d): y = mx m + 2 c t th ( H ) t i hai i m phn bi t A,B v o n AB c di nh nh t. th y

= f (x ) (C ') Cch h c t t mn Ton l ph i lm 2x + 1 vi t phng trnh ti p x 1 tuy n cu HS bi t ti p

tuy n t o v i 2 tr c t a tam gic c di n tch b ng 8 Phng php: +V y= Bai tap nhi u , bn c nh

, d Trang6/10-LTH-2010

7. Chuyn luyn thi i hc-phn i: kho s t hm s Cau 10. Cho hm s y= 2x +1 Tm m

ng th ng x +1 y=-2x+m c t th t i hai i m phn bi t A, B sao cho tam gic OAB c di n tch b ng 3

Kh o st s bi n thin v v th hm s (1) Vi t phng trnh ng th ng i qua M(1;3) c t th hm s

(1) t i hai i m phn bi t A, B sao cho AB = 2 3 . Cau 11. Cho hm s y = y = x 3 2 x 2 + (1 m) x

+ m (1), m l tham s th c. 1. Kh o st s bi n thin v v th c a hm s khi m = 1. 2. Tm m th c a

hm s (1) c t tr c honh t i 3 i m phn bi t c honh x1 ; x2 ; x3 tho mn i u ki n x12 + x2 2 +

x32 < 4 Cau 12. Cho hm s y= x+2 (H) 2x 2 1) Kh o st v v th hm s (H). 2) Tm m ng th

ng (d): y=x+m c t th hm s (H) t i hai i m phn bi t A, B sao cho OA2 + OB 2 = Cau 13. Cho hm

s y = x 4 2 x 2 (C) 1) Kh o st v v th hm s 2) L y trn th hai i m A, B c honh l n lt l

a, b.Tm i u ki n a v b ti p tuy n t i A v B song song v i nhau Cau 14. Cho hm s y= 2m x ( H

) v A(0;1) x+m 1) Kh o st v v th hm s khi m=1 2) G i I l giao i m c a 2 ng ti m c n . Tm

m trn th t n t i i m B sao cho tam gic IAB vung cn t i A. Cau 15. Cho hm s y = x 4 + 2mx

2 m 1 (1) , v i m l tham s th c. 1)Kh o st s bi n thin v v th hm s (1) khi m = 1 . 2)Xc

nh m hm s (1) c ba i m c c tr , ng th i cc i m c c tr c a th t o thnh m t tam gic c di n

tch b ng 4 2 . Cau 16 . Cho hm s y = x 4 2mx 2 + m 1 (1) , v i m l tham s th c. 1)Kh o st s

bi n thin v v th hm s (1) khi m = 1. 2)Xc nh m hm s (1) c ba i m c c tr , ng th i cc i

m c c tr c a th t o thnh m t tam gic c bn knh ng trn ngo i ti p b ng 1. Cau 17. Cho hm s

y = x 4 + 2mx 2 + m 2 + m (1) , v i m l tham s th c. Cch h c t t mn Ton l ph i lm

www.MATHVN.com 1)Kh o st s bi n thin v v th hm s (1) khi m = 2 . 2) Xc nh m hm s

(1) c ba i m c c tr , ng th i cc i m c c tr c a th t o thnh m t tam gic c gc b ng 120 . Cau

18 . Cho hm s y = x 4 2mx 2 (1), v i m l tham s th c. 1)Kh o st s bi n thin v v th c a hm s

(1) khi m = 1 . 2)Tm m th hm s (1) c hai i m c c ti u v hnh ph ng gi i h n b i th hm s v

ng th ng i qua hai i m c c ti u y c di n tch b ng 1. Cau 19. Cho hm s y = f ( x ) = x 4 + 2 ( m

2 ) x 2 + m2 5m + 5 1/ Kh o st s bi n thin v v th (C ) hm s v i m =1 2/ Tm cc gi tr c a m

th h m s c cc i m c c i, c c ti u t o thnh m t tam gic vung cn. Cau 20. Cho hm s

37 2 Bai tap Nm h c: 2000- 2011 y= 1 3 x 2 x 2 + 3 x (1) 3 1).Kh o st s bi n thin v v th c a

hm s (1) . 2)G i A, B l n l t l cc i m c c i, c c ti u c a th hm s (1). Tm i m M thu c tr c

honh sao cho tam gic MAB c di n tch b ng 2. Cau 21. Cho hm s y = x 3 6 x 2 + 9 x 4 (1)

1)Kh o st s bi n thin v v th c a hm s (1) 2)Xc nh k sao cho t n t i hai ti p tuy n c a th hm s

(1) c cng h s gc k . G i hai ti p i m l M 1 , M 2 . Vi t phng trnh ng th ng qua M 1 v M 2

theo k . Cau 22. Cho hm s y = x 3 + 3 x 2 4 (1) 1.Kh o st s bi n thin v v th (C) c a hm s

(1) 2. Gi s A, B , C l ba i m th ng hng thu c th (C), ti p tuy n v i (C) t i A, B , C tng ng c t l i

(C) t i A' , B ' , C ' . Ch ng minh r ng ba i m A' , B ' , C ' th ng hng. Cau 23. Cho hm s y = x 3 3

x + 1 (1) 1)Kh o st s bi n thin v v th (C) c a hm s (1). 2) ng th ng ( ): y = mx + 1 c t (C) t i

ba i m. G i A v B l hai i m c honh khc 0 trong ba i m ni trn; g i D l i m c c ti u c a (C).

Tm m gc ADB l gc vung. Cau 24. Cho hm s y = x 3 + 3 x 2 + 3 ( m 2 1) x 3m 2 1

(1), v i m l tham s th c. 1.Kh o st s bi n thin v v th c a hm s (1) khi m = 1. nhi u , bn c nh

, d Trang7/10-LTH-2010

8. Chuyn luyn thi i hc-phn i: kho s t hm s 2. Tm m hm s (1) c c c i v c c ti u,

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ng th i cc i m c c tr c a th cng v i g c to O t o thnh m t tam gic vung t i O . Cau 25. Cho

hm s 2 y = ( x 2 ) ( 2 x 1) (1) 1.Kh o st s bi n thin v v th (C) c a hm s (1). 2.Tm m th

(C) c hai ti p tuy n song song v i ng th ng y = mx . Gi s M , N l cc ti p i m. Hy ch ng minh r

ng trung i m c a o n th ng MN l m t i m c nh (khi m bi n thin) Cau 26. Cho hm s y = x 3 3

x 2 + 4 (1) 1)Kh o st s bi n thin v v th (C) c a hm s (1). 2)G i d k l ng th ng i qua i m A (

1;0 ) v i h s gc k ( k R ) . Tm k ng th ng dk c t th (C) t i ba i m phn bi t v hai giao i

m B, C ( B v C khc A ) cng v i g c to O t o thnh m t tam gic c di n tch b ng 1 . Cau 27.

Cho hm s y = x 3 3 x 2 + 4 (1) 1)Kh o st s bi n thin v v th (C) c a hm s (1). 2)Cho i m I (

1;0 ) . Xc nh gi tr c a tham s th c m ng th ng d : y = mx + m c t th (C) t i ba i m phn bi

t I , A, B sao cho AB < 2 2 . Cau 28. Cho hm s y = 2x3 + 9mx2 + 12m2x + 1, trong m l tham s

. 1)Kh o st s bi n thin v v th c a hm s cho khi m = - 1. 2)Tm t t c cc gi tr c a m hm s

c c c i t i xC, c c ti u t i xCT th a mn: x2C= xCT. Cau 29. Cho hm s y = (m + 2)x 3 + 3x 2 +

mx 5 , m l tham s 1)Kh o st s bi n thin v v th (C ) c a hm s khi m=0 2)Tm cc gi tr c a m

cc i m c c i, c c ti u c a th hm s cho c honh l cc s dng. Cau 30. Cho hm s y=

mx (Hm). Tm m ng x+2 th ng d:2x+2y-1=0 c t (Hm) t i 2 i m phn bi t A, B sao 3 8 3 Cau

31. Tm m hm s y = x mx + 2 c t Ox t i m t cho tam gic OAB c di n tch b ng i m duy nh t

Cau 32. Cho hm s y= 2x + 4 (H). G i d l ng 1 x k d c t (H) t i A, B m AB = 3 10 Cau 33.

Tm m th hm s y = x 3 mx 2 + 2m c t tr c Ox t i m t i m duy nh t www.MATHVN.com Cau

34. Cho hm s : y = x+2 (C) x 1 1) Kh o st v v th (C) hm s 2) Cho i m A( 0; a) Tm a t A k

c 2 ti p tuy n t i th (C) sao cho 2 ti p i m tng ng n m v 2 pha c a tr c honh Cau 35. Cho

hm s y = x 3 3 x + 2 (C) 1) Kh o st v v th hm s (C) 2) Tm i m M thu c (C) sao cho ti p tuy

n t i M c t (C) N m MN = 2 6 Cau 36. Tm m ng th ng y=x+4 c t th hm s y = x3 + 2mx 2 +

( m + 3) x + 4 t i 3 i m phn bi t A, B,C sao cho tam gic MBC c di n tch b ng 4. (i m B, C c

honh khc 0, M(1;3) Cau 37. Tm m hm s y = x3 mx 2 + (2m + 1) x m 2 c t Ox t i 3 i

m phn bi t c honh dng Cau 38. Tm hai i m A, B thu c 3 th hm s 2 y = x 3x + 1 sao

cho ti p tuy n t i A, B song song v i nhau v AB = 4 2 Cau 39. Cho hs : y = x+m Tm m ti p tuy n c

a x 1 th t i giao i m I c a hai ti m c n c t tr c Ox , Oy t i A, B v di n tch tam gic IAB b ng 1

Cau 40. Cho hm s y= 2x + 1 vi t phng trnh ti p x 1 tuy n cu HS bi t ti p tuy n t o v i 2 tr c t a

tam gic c di n tch b ng 8 Ph n m t: CC BI T P LIN QUAN I M C C I V C C TI U HM S

Cu 1) Cho hm s y= 1 3 x mx 2 x + m + 1 3 a) Kh o st v v th hm s khi m=1 b) Tm m

hm s c c c i c c ti u v kho ng cch gi a i m c c i v c c ti u l nh nh t Cu 2) Cho hm s y= 1

3 x mx 2 + mx 1 3 a) Kh o st v v th hm s khi m= 1 b) Tm m hm s t c c tr t i x1 ; x 2

tho mn x1 x2 8 th ng c h s gc k i qua M(1;1). Tm Cch h c t t mn Ton l ph i lm Nm h

c: 2000- 2011 Bai tap Cu 3) Cho hm s y = x 3 + mx 2 + 7 x + 3 a) Kh o st v v th hm s khi

m= -8 b) Tm m hm s c ng th ng i qua i m c c i c c ti u vung gc v i ng th ng y=3x-7

nhi u , bn c nh , d Trang8/10-LTH-2010

9. Chuyn luyn thi i hc-phn i: kho s t hm s b) G i I l giao i m 2 ng ti m c n c a (H).

Tm M thu c (H) sao cho ti p tuy n c a (H) t i M vung gc v i ng th ng IM. Cu 4) Cho hm s y =

x 3 3 x 2 + m 2 x + m a) Kh o st v v th hm s khi m= 0 b) Tm m hm s c c c i c c ti u i

x ng qua ng th ng y = 1 5 x 2 2 Cu 7) Cho hm s 2 2 2 y = x + 3 x + 3(m 1) x 3m 1 a)

Kh o st v v th hm s khi m= 1 b) Tm m hm s c c c i c c ti u cch u g c to O. Ph n hai:

CC BI TON LIN QUAN N TI P TUY N V NG TI M C N 3 Cu 1) Cho hm s y = x

mx m + 1 (Cm) a) Kh o st v v th hm s khi m= 3 b) Tm m ti p tuy n t i giao i m cu (Cm) v i

tr c Oy ch n trn hai tr c to m t tam gic c di n tch b ng 8 Cu 2) Cho hm s y = x 3 + 3 x 2 + mx

+ 1 (Cm) a) Kh o st v v th hm s khi m= 0 b) Tm m ng th ng y=1 c t (Cm) t i 3 i m phn

bi t C(0;1), D,E v cc ti p tuy n t i D v E c a (Cm) vung gc v i nhau. Cu 3) Cho hm s y= x+m (

Hm) x2 y= 2mx + 3 ( Hm) * xm 1) Kh o st v v th hm s khi m=1 2) Tm m ti p tuy n b t k

c a hm s (Hm) c t 2 ng ti m c n t o thnh m t tam gic c di n tch b ng 8 Cu 5) Cho hm s a)

Kh o st v v th hm s cho b) Tm M thu c (H) sao cho ti p tuy n t i M c a (H) c t 2 tr c Ox, Oy

t i A, B sao cho tam gic OAB Cu 6) Cho hm s y= 19 A ;4 n th hm s y = 2 x 3 3 x

2 + 5 12 Cu 9) Tm i m M thu c th hm s y = x 3 + 3 x 2 2 m qua ch k c m t ti p

tuy n n th Cu 10) Tm nh ng i m thu c ng th ng y=2 m t c th k c 3 ti p tuy n n

th hs y = x 3 3 x Cu 11) Tm nh ng i m thu c tr c tung qua c th k c 3 ti p tuy n n th hs

y = x 4 2 x 2 + 1 Cu 12) Tm nh ng i m thu c ng th ng x=2 t k c 3 ti p tuy n n th hs

y = x 3 3 x c m t ti p tuy n n th hs y = Cu 14) Cho hm s y= x +1 x 1 x+m x 1 a) Kh o

st v v th hm s khi m=1 b) V i gi tr no c a m th hm s c t ng th ng y=2x+1 t i 2 i m phn

bi t sao cho cc ti p tuy n v i th t i 2 i m song song v i nhau. Ph n ba: CC BI TON

TNG GIAO 2 2x (H ) * y= x +1 c di n tch b ng 2x (H ) * x+2 Cu 113) Tm nh ng i m thu c

tr c Oy qua ch k a) Kh o st v v th hm s khi m= 3 b) Tm m t A(1;2) k c 2 ti p tuy n

AB,AC n (Hm) sao cho ABC l tam gic u (A,B l cc ti p i m) Cu 4) Cho hm s y= a) Kh o st

v v th hm s (H) b) Vi t phng trnh ti p tuy n c a (H) bi t kho ng cch t tm i x ng c a th hm



v v th hm s (H) b) Vi t phng trnh ti p tuy n c a (H) bi t kho ng cch t tm i x ng c a th hm

s (H) n ti p tuy n l l n nh t. Cu 8) Vi t cc phng trnh ti p tuy n k t i m Cu 5) Cho hm s 3

Nm h c: 2000- 2011 1 4 Cu 1) Cho hm s y = 2mx 3 ( 4m 2 + 1) x 2 4m 2 a) Kh o st v v th

hm s khi m=1 b) Tm m th hs ti p xc v i tr c Ox Cu 2) Cho hm s 2x 1 (H ) * x 1 y = x 4

2mx 2 + m 3 m 2 a) Kh o st v v th hm s khi m=1 a) Kh o st v v th hm s Cch h c t t mn

Ton l ph i lm www.MATHVN.com Bai tap TH nhi u , bn c nh , d Trang9/10-LTH-2010

10. Chuyn luyn thi i hc-phn i: kho s t hm s b) Tm m th hs ti p xc v i tr c Ox t i 2

i m phn bi t Cu 10) Cho hm s b) Bi n lu n theo m s nghi m phng trnh x2 1( a) Kh o st v v

th hm s b) Tm phng trnh sau c 8 nghi m phn bi t x 4 6 x 2 + 5 = m 2 2m y = x 3

3mx 2 6mx x+3 ) = 2m + 1 3 Ph n b n: CC CU TON LIN QUAN N KHO NG CCH Cu

1) Tm M thu c (H) y = 3x 5 t ng kho ng x2 cch t M n 2 ng ti m c n c a H l nh nh t a) Kh

o st v v th hm s khi m=1/4 3 b) Bi n lu n s nghi m 4 x 3 x 2 6 x 4a = 0 Cu 5) Cho hm s

y = x3 + 3x 2 x 3 a) Kh o st v v th hm s x4 5 Cu 3) Cho hm s y = 3x 2 + 2 2 Cu 4)

Cho hm s Nm h c: 2000- 2011 Cu 2) Tm M thu c (H) : y = x 1 t ng kho ng cch x +1 t M n

2 tr c to l nh nh t y = 4 x 3 3 x (C ) Cu 6) Tm m hm s y=-x+m c t th hm s a) Kh o st v

v th hm s (C ) y= b) Tm m phng trnh 4 x 3 3 x = 4 m 3 4 m 2x + 1 t i 2 i m A,B m

di AB nh nh t x+2 c 4 nghi m phn bi t Cu 6) Cho hm s y = x 3 3mx 2 + 3( m 2 1) x ( m 2

1) a) Kh o st v v th hm s khi m= 1 b) Tm m hm s c t Ox t i 3 i m phn bi t c honh

dng Zzzzzz g Cu 7) Cho hm s y = x 3 + 2(1 2m) x 2 + (5 7 m) x + 2(m + 5) a) Kh o st v v

th hm s khi m= 5/7 b) Tm m th hs c t Ox t i 3 i m c honh nh hn 1. Cu 8) Tm m

hm s y = 2 x 3 3( m + 3) x 2 + 18mx 8 c th ti p xc v i tr c Ox Cu 9) Cho hm s y = x 4

3x 2 + 2 a) Kh o st v v th hs b) Bi n lu n s nghi m phng trnh x 2 2 ( x 2 1) = m Cch h c t

t mn Ton l ph i lm www.MATHVN.com Bai tap nhi u , bn c nh , d Trang10/10-LTH-2010

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