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TI LIU N THI I HC CAO NG

Nam 2011

www.VNMATH.comTrn S Tng 100 Kho st hm s KSHS 01: TNH N IU CA HM SCu 1.

1 Cho hm s y = (m - 1) x 3 + mx 2 + (3m - 2) x (1) 3 1) Kho st s bin thin v v th (C) ca hm s (1) khi m = 2 . 2) Tm tt c cc gi tr ca tham s m hm s (1) ng bin trn tp xc nh ca n. Tp xc nh: D = R. y = (m - 1) x 2 + 2mx + 3m - 2 . (1) ng bin trn R y 0, "x m 2

Cu 2.

mx + 4 (1) x+m 1) Kho st s bin thin v v th ca hm s (1) khi m = -1 . 2) Tm tt c cc gi tr ca tham s m hm s (1) nghch bin trn khong (-;1) . Cho hm s y =

Tp xc nh: D = R \ {m}.

y =

m2 - 4 ( x + m )2

. (1) (2)

Hm s nghch bin trn tng khong xc nh y < 0 -2 < m < 2 hm s (1) nghch bin trn khong (-;1) th ta phi c -m 1 m -1 Kt hp (1) v (2) ta c: -2 < m -1 .Cu 3.

Cho hm s y = x 3 + 3 x 2 - mx - 4 (1) 1) Kho st s bin thin v v th ca hm s (1) khi m = 0 . 2) Tm tt c cc gi tr ca tham s m hm s (1) ng bin trn khong (-; 0) .

m -3Cu 4.

Cho hm s y = 2 x 3 - 3(2 m + 1) x 2 + 6 m ( m + 1) x + 1 c th (Cm). 1) Kho st s bin thin v v th ca hm s khi m = 0. 2) Tm m hm s ng bin trn khong (2; +)

y ' = 6 x 2 - 6(2m + 1) x + 6m(m + 1) c D = (2 m + 1)2 - 4(m 2 + m ) = 1 > 0 x = m y' = 0 . Hm s ng bin trn cc khong (-; m ), (m + 1; +) x = m +1 Do : hm s ng bin trn (2; +) m + 1 2 m 1Cu 5.

Cho hm s y = x 4 - 2 mx 2 - 3m + 1 (1), (m l tham s). 1) Kho st s bin thin v v th ca hm s (1) khi m = 1. 2) Tm m hm s (1) ng bin trn khong (1; 2). Ta c y ' = 4 x 3 - 4mx = 4 x( x 2 - m) + m > 0 , y = 0 c 3 nghim phn bit: - m , 0,m.

+ m 0 , y 0, "x m 0 tho mn. Hm s (1) ng bin trn (1; 2) khi ch khiCu 6.

m 1 0 < m 1.

Vy m ( -;1] .

Cho hm s y = x 3 + (1 - 2m) x 2 + (2 - m) x + m + 2 . 1) Kho st s bin thin v v th (C) ca hm s khi m = 1. 2) Tm m hm ng bin trn ( 0; + ) . Trang 1

www.VNMATH.com100 Kho st hm s Trn S Tng

Hm ng bin trn (0; +) y = 3 x 2 + 2(1 - 2m ) x + (2 - m ) 0 vi "x (0; +)3x 2 + 2 x + 2 f ( x) = m vi "x (0; +) 4x +1 2(6 x 2 + x - 3) -1 73 Ta c: f ( x ) = = 0 6x2 + x - 3 = 0 x = 2 12 (4 x + 1) Lp bng bin thin ca hm f ( x ) trn (0; +) , t ta i n kt lun: -1 + 73 3 + 73 f m m 12 8

KSHS 02: CC TR CA HM SCu 7.

Cho hm s y = x 3 + 3 x 2 + mx + m 2 (m l tham s) c th l (Cm). 1) Kho st s bin thin v v th hm s khi m = 3. 2) Xc nh m (Cm) c cc im cc i v cc tiu nm v hai pha i vi trc honh. PT honh giao im ca (C) v trc honh: x = -1 x 3 + 3 x 2 + mx + m 2 = 0 (1) 2 (2) g( x ) = x + 2 x + m - 2 = 0 (Cm) c 2 im cc tr nm v 2 pha i vi trc 0x PT (1) c 3 nghim phn bit (2) c 2 nghim phn bit khc 1 D = 3 - m > 0 m 0 D bit cng du 1 2m - 1 > 0 m > 2 Cu 10. Cho hm s y = x 3 - 3x 2 - mx + 2 (m l tham s) c th l (Cm).

1) Kho st s bin thin v v th hm s khi m = 1. 2) Xc nh m (Cm) c cc im cc i v cc tiu cch u ng thng y = x - 1 . Trang 2

www.VNMATH.comTrn S Tng 100 Kho st hm s

Ta c: y ' = 3 x 2 - 6 x - m . Hm s c C, CT y ' = 3 x 2 - 6 x - m = 0 c 2 nghim phn bit x1; x2 D ' = 9 + 3m > 0 m > -3 (*) Gi hai im cc tr l A ( x1 ; y1 ) ; B ( x2 ; y2 )1 m 1 2m Thc hin php chia y cho y ta c: y = x - y '- + 2 x + 2 - 3 3 3 3 m m 2m 2m + 2 x1 + 2 - ; y2 = y ( x2 ) = - + 2 x2 + 2 - y1 = y ( x1 ) = - 3 3 3 3 m 2m + 2 x + 2 - Phng trnh ng thng i qua 2 im cc tr l D: y = - 3 3 Cc im cc tr cch u ng thng y = x - 1 xy ra 1 trong 2 trng hp: TH1: ng thng i qua 2 im cc tr song song hoc trng vi ng thng y = x - 1 3 2m - + 2 = 1 m = - (tha mn) 2 3 TH2: Trung im I ca AB nm trn ng thng y = x - 1 y + y2 x1 + x2 m 2m + 2 ( x1 + x2 ) + 2 2 - = ( x1 + x2 ) - 2 yI = xI - 1 1 = -1 - 2 2 3 3 2m 2m + 3 .2 = 6 m=0 3 3 3 Vy cc gi tr cn tm ca m l: m = 0; - 2 Cu 11. Cho hm s y = x 3 - 3mx 2 + 4m 3 (m l tham s) c th l (Cm).

1) Kho st s bin thin v v th hm s khi m = 1. 2) Xc nh m (Cm) c cc im cc i v cc tiu i xng nhau qua ng thng y = x. x = 0 Ta c: y = 3 x 2 - 6 mx ; y = 0 . hm s c cc i v cc tiu th m 0. x = 2m uur th hm s c hai im cc tr l: A(0; 4m3), B(2m; 0) AB = (2m; -4 m3 ) Trung im ca on AB l I(m; 2m3) 2 2 m - 4 m 3 = 0 AB ^ d A, B i xng nhau qua ng thng d: y = x 3 m= 2 I d 2 m = m Cu 12. Cho hm s y = - x 3 + 3mx 2 - 3m - 1 .

1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Vi gi tr no ca m th th hm s c im cc i v im cc tiu i xng vi nhau qua ng thng d: x + 8y - 74 = 0 .

y = -3 x 2 + 6 mx ; y = 0 x = 0 x = 2 m .Hm s c C, CT PT y = 0 c 2 nghim phn bit m 0 . uuu r Khi 2 im cc tr l: A(0; -3m - 1), B(2 m; 4 m3 - 3m - 1) AB(2m; 4m 3 ) Trung im I ca AB c to : I (m; 2 m3 - 3m - 1) r ng thng d: x + 8y - 74 = 0 c mt VTCP u = (8; -1) . Trang 3

www.VNMATH.com100 Kho st hm s Trn S Tng

m + 8(2m3 - 3m - 1) - 74 = 0 I d A v B i xng vi nhau qua d uuu r m=2 r AB ^ d AB.u = 0 Cu 13. Cho hm s y = x 3 - 3 x 2 + mx

(1). 1) Kho st s bin thin v v th ca hm s khi m = 0. 2) Vi gi tr no ca m th th hm s (1) c cc im cc i v im cc tiu i xng vi nhau qua ng thng d: x 2 y 5 = 0 .

Ta c y = x 3 - 3 x 2 + mx y ' = 3 x 2 - 6 x + mHm s c cc i, cc tiu y = 0 c hai nghim phn bit D = 9 - 3m > 0 m < 3 1 2 1 1 Ta c: y = x - y + m - 2 x + m 3 3 3 3 Ti cc im cc tr th y = 0 , do ta cc im cc tr tha mn phng trnh: 2 1 y = m - 2 x + m 3 3 2 1 Nh vy ng thng D i qua cc im cc tr c phng trnh y = m - 2 x + m 3 3 2 nn D c h s gc k1 = m - 2 . 3 1 5 1 d: x 2 y 5 = 0 y = x - d c h s gc k2 = 2 2 2 hai im cc tr i xng qua d th ta phi c d ^ D 12 k1k2 = -1 m - 2 = -1 m = 0 23 Vi m = 0 th th c hai im cc tr l (0; 0) v (2; 4), nn trung im ca chng l I(1; 2). Ta thy I d, do hai im cc tr i xng vi nhau qua d. Vy: m = 0Cu 14. Cho hm s y = x 3 - 3(m + 1) x 2 + 9 x + m - 2 (1) c th l (Cm).

1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Vi gi tr no ca m th th hm s c im cc i v im cc tiu i xng vi 1 nhau qua ng thng d: y = x . 2

y ' = 3 x 2 - 6(m + 1) x + 9Hm s c C, CT D ' = 9(m + 1)2 - 3.9 > 0 m (-; -1 - 3) (-1 + 3; +) 1 m +1 2 Ta c y = x y - 2(m + 2m - 2) x + 4m + 1 3 3 Gi s cc im cc i v cc tiu l A( x1; y1 ), B( x2 ; y2 ) , I l trung im ca AB. y1 = -2(m 2 + 2m - 2) x1 + 4 m + 1 ; y2 = -2(m 2 + 2 m - 2) x2 + 4 m + 1 x + x = 2(m + 1) v: 1 2 x1 .x2 = 3 Vy ng thng i qua hai im cc i v cc tiu l y = -2(m 2 + 2 m - 2) x + 4 m + 1 Trang 4

www.VNMATH.comTrn S Tng A, B i xng qua (d): y = 1 AB ^ d x m = 1. 2 I d 100 Kho st hm s

Cu 15. Cho hm s y = x 3 - 3(m + 1) x 2 + 9 x - m , vi m l tham s thc.

1) Kho st s bin thin v v th ca hm s cho ng vi m = 1 . 2) Xc nh m hm s cho t cc tr ti x1 , x 2 sao cho x1 - x 2 2 .

Ta c y ' = 3x 2 - 6(m + 1) x + 9. + Hm s t cc i, cc tiu ti x1 , x 2 PT y '= 0 c hai nghim phn bit x1 , x 2 PT x 2 - 2(m + 1) x + 3 = 0 c hai nghim phn bit l x1 , x 2 . m > -1 + 3 D' = (m + 1) 2 - 3 > 0 (1) m < -1 - 3 + Theo nh l Viet ta c x1 + x 2 = 2(m + 1); x1 x 2 = 3. Khi : x1 - x 2 2 ( x1 + x 2 )2 - 4 x1 x 2 4 4(m + 1)2 - 12 4 (m + 1)2 4 -3 m 1 (2)

+ T (1) v (2) suy ra gi tr ca m cn tm l - 3 m < -1 - 3 v - 1 + 3 < m 1.Cu 16. Cho hm s y = x 3 + (1 - 2 m) x 2 + (2 - m ) x + m + 2 , vi m l tham s thc.

1) Kho st s bin thin v v th ca hm s cho ng vi m = 1 . 2) Xc nh m hm s cho t cc tr ti x1 , x2 sao cho x1 - x2 > Ta c: y ' = 3 x 2 + 2(1 - 2 m) x + (2 - m)

1 . 3

Hm s c C, CT y ' = 0 c 2 nghim phn bit x1 , x2 (gi s x1 < x2 ) 5 (*) D ' = (1 - 2 m )2 - 3(2 - m ) = 4 m 2 - m - 5 > 0 m > 4 m < -1 2(1 - 2 m) x1 + x2 = 3 Hm s t cc tr ti cc im x1 , x2 . Khi ta c: 2-m x x = 1 2 3 2 2 1 1 x1 - x2 > ( x1 - x2 ) = ( x1 + x2 ) - 4 x1x2 > 3 9 3 + 29 3 - 29 4(1 - 2m )2 - 4(2 - m) > 1 16m 2 - 12 m - 5 > 0 m > m< 8 8 3 + 29 Kt hp (*), ta suy ra m > m < -1 8Cu 17. Cho hm s y =

1 3 1 x - (m - 1) x 2 + 3(m - 2) x + , vi m l tham s thc. 3 3 1) Kho st s bin thin v v th ca hm s cho ng vi m = 2 . 2) Xc nh m hm s cho t cc tr ti x1 , x2 sao cho x1 + 2 x2 = 1 . Ta c: y = x 2 - 2(m - 1) x + 3(m - 2)

Trang 5

www.VNMATH.com100 Kho st hm s Hm s c cc i v cc tiu y = 0 c hai nghim phn bit x1 , x2 Trn S Tng

D > 0 m 2 - 5m + 7 > 0 (lun ng vi "m) x = 3 - 2m x + x = 2(m - 1) Khi ta c: 1 2 2 x2 (1 - 2 x2 ) = 3(m - 2) x1 x2 = 3(m - 2) 8m 2 + 16 m - 9 = 0 m = -4 34 . 4

Cu 18. Cho hm s y = 4 x 3 + mx 2 3 x .

1) Kho st s bin thin v v th ca hm s khi m = 0. 2) Tm m hm s c hai im cc tr x1 , x2 tha x1 = -4 x2 .

y = 12 x 2 + 2mx 3 . Ta c: D = m2 + 36 > 0, "m hm s lun c 2 cc tr x1 , x2 . x1 = -4 x2 m Khi : x1 + x2 = 6 1 x1 x2 = - 4 Cu hi tng t:

m=

9 2

a) y = x 3 + 3 x 2 + mx + 1 ; x1 + 2x2 = 3

S: m = -105 .

Cu 19. Cho hm s y = (m + 2) x 3 + 3 x 2 + mx - 5 , m l tham s.

1) Kho st s bin thin v v th (C) ca hm s khi m = 0. 2) Tm cc gi tr ca m cc im cc i, cc tiu ca th hm s cho c honh l cc s dng. Cc im cc i, cc tiu ca th hm s cho c honh l cc s dng PT y ' = 3(m + 2) x 2 + 6 x + m = 0 c 2 nghim dng phn bit a = (m + 2) 0 D ' = 9 - 3m(m + 2) > 0 D ' = -m 2 - 2 m + 3 > 0 -3 < m < 1 m P = >0 m < 0 m < 0 -3 < m < -2 3(m + 2) m + 2 < 0 m < -2 S = -3 > 0 m+2 Cu 20. Cho hm s y = x 3 3 x 2 + 2

(1) 1) Kho st s bin thin v v th ca hm s (1). 2) Tm im M thuc ng thng d: y = 3 x - 2 sao tng khong cch t M ti hai im cc tr nh nht. Cc im cc tr l: A(0; 2), B(2; 2). Xt biu thc g( x , y ) = 3 x - y - 2 ta c: g( x A , y A ) = 3 x A - y A - 2 = -4 < 0; g( x B , yB ) = 3 x B - yB - 2 = 6 > 0

2 im cc i v cc tiu nm v hai pha ca ng thng d: y = 3 x - 2 . Do MA + MB nh nht 3 im A, M, B thng hng M l giao im ca d v AB. Phng trnh ng thng AB: y = -2 x + 2Trang 6

www.VNMATH.comTrn S Tng 100 Kho st hm s

4 x = 5 y = 3x - 2 4 2 Ta im M l nghim ca h: M ; 5 5 y = -2 x + 2 y = 2 5 Cu 21. Cho hm s y = x 3 + (1 2 m) x 2 + (2 m ) x + m + 2 (m l tham s) (1).

1) Kho st s bin thin v v th hm s (1) khi m = 2. 2) Tm cc gi tr ca m th hm s (1) c im cc i, im cc tiu, ng thi honh ca im cc tiu nh hn 1.

y = 3 x 2 + 2(1 - 2 m) x + 2 - m = g( x )YCBT phng trnh y = 0 c hai nghim phn bit x1 , x2 tha mn: x1 < x2 < 1 . D = 4m 2 - m - 5 > 0 5 7 g(1) = -5m + 7 > 0 < m < . 4 5 S = 2m - 1 < 1 2 3 Cu 22. Cho hm s

y = x 3 - 3mx 2 + 3(m 2 - 1) x - m3 + m (1) 1) Kho st s bin thin v v th ca hm s (1) khi m = 1. 2) Tm m hm s (1) c cc tr ng thi khong cch t im cc i ca th hm s n gc ta O bng 2 ln khong cch t im cc tiu ca th hm s n gc ta O. Ta c y = 3 x 2 - 6mx + 3(m 2 - 1) Hm s (1) c cc tr th PT y = 0 c 2 nghim phn bit x 2 - 2mx + m 2 - 1 = 0 c 2 nhim phn bit D = 1 > 0, "m Khi : im cc i A(m - 1; 2 - 2m ) v im cc tiu B(m + 1; -2 - 2m ) m = -3 + 2 2 Ta c OA = 2OB m 2 + 6m + 1 = 0 . m = -3 - 2 2

Cu 23. Cho hm s y = - x 3 + 3mx 2 + 3(1 - m 2 ) x + m 3 - m 2

(1) 1) Kho st s bin thin v v th ca hm s (1) khi m = 1 . 2) Vit phng trnh ng thng qua hai im cc tr ca th hm s (1). y = -3 x 2 + 6 mx + 3(1 - m 2 ) . 1 m y = x - y + 2 x - m2 + m 3 3

PT y = 0 c D = 1 > 0, "m th hm s (1) lun c 2 im cc tr ( x1 ; y1 ), ( x2 ; y2 ) . Chia y cho y ta c: Khi :

y1 = 2 x1 - m2 + m ; y2 = 2 x2 - m 2 + m

PT ng thng qua hai im cc tr ca th hm s (1) l y = 2 x - m 2 + m .Cu 24. Cho hm s y = x 3 - 3x 2 - mx + 2 c th l (Cm).

1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m (Cm) c cc im cc i, cc tiu v ng thng i qua cc im cc tr song song vi ng thng d: y = -4 x + 3 . Trang 7

www.VNMATH.com100 Kho st hm s Ta c: y ' = 3 x 2 - 6 x - m . Hm s c C, CT y ' = 3 x 2 - 6 x - m = 0 c 2 nghim phn bit x1 ; x2 D ' = 9 + 3m > 0 m > -3 (*) Gi hai im cc tr l A ( x1 ; y1 ) ; B ( x2 ; y2 ) 1 m 1 2m Thc hin php chia y cho y ta c: y = x - y '- + 2 x + 2 - 3 3 3 3 m m 2m 2m + 2 x1 + 2 - ; y2 = y ( x2 ) = - + 2 x2 + 2 - y1 = y ( x1 ) = - 3 3 3 3 m 2m + 2 x + 2 - Phng trnh ng thng i qua 2 im cc tr l d: y = - 3 3 ng thng i qua cc im cc tr song song vi d: y = -4 x + 3 2m - 3 + 2 = -4 m = 3 (tha mn) 2 - m 3 3 Cu 25. Cho hm s y = x 3 - 3x 2 - mx + 2 c th l (Cm).

Trn S Tng

1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m (Cm) c cc im cc i, cc tiu v ng thng i qua cc im cc tr to vi ng thng d: x + 4 y 5 = 0 mt gc 450 . Ta c: y ' = 3 x 2 - 6 x - m . Hm s c C, CT y ' = 3 x 2 - 6 x - m = 0 c 2 nghim phn bit x1 ; x2 D ' = 9 + 3m > 0 m > -3 (*) Gi hai im cc tr l A ( x1 ; y1 ) ; B ( x2 ; y2 ) 1 m 1 2m Thc hin php chia y cho y ta c: y = x - y '- + 2 x + 2 - 3 3 3 3 m m 2m 2m + 2 x1 + 2 - ; y2 = y ( x2 ) = - + 2 x2 + 2 - y1 = y ( x1 ) = - 3 3 3 3 m 2m + 2 x + 2 - Phng trnh ng thng i qua 2 im cc tr l D: y = - 3 3 1 2m t k = - + 2 . ng thng d: x + 4 y 5 = 0 c h s gc bng - . 4 3 1 1 3 39 1 k= k + = 1- k m=k+ 5 4 4 10 4 Ta c: tan 45o = 1 1 1 5 k = k + = -1 + k m = - 1 1- k 4 4 4 3 2 1 Kt hp iu kin (*), suy ra gi tr m cn tm l: m = 2Cu 26. Cho hm s y = x 3 + 3 x 2 + m

(1) 1) Kho st s bin thin v v th ca hm s (1) khi m = -4 .

2) Xc nh m th ca hm s (1) c hai im cc tr A, B sao cho AOB = 120 0 .Trang 8

www.VNMATH.comTrn S Tng 100 Kho st hm s

x = -2 y = m + 4 Ta c: y = 3 x 2 + 6 x ; y = 0 x = 0 y = m Vy hm s c hai im cc tr A(0 ; m) v B(-2 ; m + 4) uur uur 1 OA = (0; m), OB = (-2; m + 4) . AOB = 120 0 th cos AOB = 2 -4 < m < 0 m(m + 4) 1 = - m 2 ( 4 + (m + 4)2 ) = -2m(m + 4) 2 2 3m + 24m + 44 = 0 m 2 ( 4 + (m + 4)2 ) -4 < m < 0 -12 + 2 3 -12 2 3 m = 3 m = 3 Cu 27. Cho hm s y = x 3 3mx 2 + 3(m2 1) x m 3

(Cm) 1) Kho st s bin thin v v th ca hm s (1) khi m = -2 . 2) Chng minh rng (Cm) lun c im cc i v im cc tiu ln lt chy trn mi ng thng c nh. x = m +1 y = 3 x 2 - 6 mx + 3(m 2 - 1) ; y = 0 x = m -1 x = -1 + t im cc i M (m 1;2 3m) chy trn ng thng c nh: y = 2 - 3t x = 1+ t im cc tiu N (m + 1; -2 m) chy trn ng thng c nh: y = -2 - 3t 1 4 3 x - mx 2 + (1) 2 2 1) Kho st s bin thin v v th ca hm s (1) khi m = 3 . 2) Xc nh m th ca hm s (1) c cc tiu m khng c cc i. x = 0 y = 2 x3 - 2mx = 2 x ( x 2 - m) . y = 0 2 x = m th ca hm s (1) c cc tiu m khng c cc i PT y = 0 c 1 nghim m 0

Cu 28. Cho hm s y =

Cu 29. Cho hm s y = f ( x) = x 4 + 2(m - 2) x 2 + m 2 - 5m + 5

(Cm ) .

Khi to cc im cc tr l: A ( 0; m 2 - 5m + 5 ) , B ( 2 - m ;1 - m ) , C ( - 2 - m ;1 - m ) uur uuu r AB = ( 2 - m ; -m 2 + 4 m - 4 ) , AC = ( - 2 - m ; - m 2 + 4m - 4 ) Do DABC lun cn ti A, nn bi ton tho mn khi DABC vung ti A 3 AB.AC = 0 (m - 2 ) = -1 m = 1 (tho (*)) Trang 9

1) Kho st s bin thin v v th (C) hm s khi m = 1. 2) Tm cc gi tr ca m th (Cm ) ca hm s c cc im cc i, cc tiu to thnh 1 tam gic vung cn. x = 0 Ta c f ( x ) = 4 x 3 + 4(m - 2) x = 0 2 x = 2 - m Hm s c C, CT PT f ( x ) = 0 c 3 nghim phn bit m < 2 (*)

www.VNMATH.com100 Kho st hm sCu 30. Cho hm s y = x 4 + 2(m - 2) x 2 + m 2 - 5m + 5

Trn S Tng

(C m )

1) Kho st s bin thin v v th hm s khi m = 1. 2) Vi nhng gi tr no ca m th th (Cm) c im cc i v im cc tiu, ng thi cc im cc i v im cc tiu lp thnh mt tam gic u. x = 0 Ta c f ( x ) = 4 x 3 + 4(m - 2) x = 0 2 x = 2 - m Hm s c C, CT PT f ( x ) = 0 c 3 nghim phn bit m < 2 Khi to cc im cc tr l: A ( 0; m 2 - 5m + 5 ) , B ( 2 - m ;1 - m ) , C ( - 2 - m ;1 - m ) uur uuu r AB = ( 2 - m ; -m 2 + 4 m - 4 ) , AC = ( - 2 - m ; - m 2 + 4m - 4 ) 1 Do DABC lun cn ti A, nn bi ton tho mn khi = 60 0 cos A = A 2 uuu uuu r r AB. AC 1 uuu uuu = m = 2 - 3 3 . r r AB . AC 2 Cu hi tng t i vi hm s: y = x 4 - 4(m - 1) x 2 + 2 m - 1Cu 31. Cho hm s y = x 4 + 2 mx 2 + m 2 + m c th (Cm) .

(*)

1) Kho st s bin thin v v th hm s khi m = 2. 2) Vi nhng gi tr no ca m th th (Cm) c ba im cc tr, ng thi ba im cc tr lp thnh mt tam gic c mt gc bng 1200 . x = 0 Ta c y = 4 x 3 + 4 mx ; y = 0 4 x( x 2 + m) = 0 x = -m (m < 0)

Khi cc im cc tr l: A(0; m 2 + m ), B ( - m ; m ) , C ( - - m ; m ) uur uuu r AB = ( - m ; - m 2 ) ; AC = (- - m ; -m 2 ) . DABC cn ti A nn gc 120o chnh l A . uur uuu r 1 AB. AC 1 - -m . -m + m4 1 o A = 120 cos A = - uur uuu = - =r 4 2 2 2 m -m AB . AC m = 0 (loai) 1 4 4 4 1 = - 2 m + 2 m = m - m 3m + m = 0 m = - 3 2 m4 - m 3 1 . Vy m = 3 3 m + m4Cu 32. Cho hm s y = x 4 - 2 mx 2 + m - 1 c th (Cm) .

1) Kho st s bin thin v v th hm s khi m = 1. 2) Vi nhng gi tr no ca m th th (Cm) c ba im cc tr, ng thi ba im cc tr lp thnh mt tam gic c bn knh ng trn ngoi tip bng 1. x = 0 Ta c y = 4 x 3 - 4mx = 4 x( x 2 - m ) = 0 2 x = m Hm s cho c ba im cc tr PT y = 0 c ba nghim phn bit v y i du khi x i qua cc nghim m > 0 . Khi ba im cc tr ca th (Cm) l: A(0; m - 1), B ( - m ; - m 2 + m - 1) , C ( m ; - m 2 + m - 1) Trang 10

www.VNMATH.comTrn S Tng SV ABC = 100 Kho st hm s

1 y - y A . xC - xB = m 2 m ; AB = AC = m 4 + m , BC = 2 m 2 B m = 1 (m 4 + m)2 m AB. AC.BC 3 R= =1 = 1 m - 2m + 1 = 0 2 m = 5 - 1 4SV ABC 4m m 2 Cu hi tng t: a) y = x 4 - 2mx 2 + 1 S: m = 1, m = -1 + 5 2

Cu 33. Cho hm s y = x 4 - 2mx 2 + 2m + m 4 c th (Cm) .

1) Kho st s bin thin v v th hm s khi m = 1. 2) Vi nhng gi tr no ca m th th (Cm) c ba im cc tr, ng thi ba im cc tr lp thnh mt tam gic c din tch bng 4. x = 0 Ta c y ' = 4 x 3 - 4mx = 0 2 g ( x) = x - m = 0 Hm s c 3 cc tr y ' = 0 c 3 nghim phn bit D g = m > 0 m > 0 (*) cc tr ti x1 ; x2 ; x3 . Gi A(0; 2m + m 4 ); B ( m ; m 4 - m 2 + 2m ) ; C ( - m ; m 4 - m2 + 2m ) l 3 im cc tr ca (Cm) . Ta c: AB 2 = AC 2 = m 4 + m; BC 2 = 4m DABC cn nh A Gi M l trung im ca BC M (0; m 4 - m 2 + 2 m) AM = m 2 = m 2 V D ABC cn ti A nn AM cng l ng cao, do : SD ABC = 1 1 AM .BC = .m 2 . 4m = 4 m 2 = 4 m 5 = 16 m = 5 16 2 25

Vi iu kin (*), phng trnh y = 0 c 3 nghim x1 = - m ; x2 = 0; x3 = m . Hm s t

Vy m = 5 16 . Cu hi tng t: a) y = x 4 - 2m 2 x 2 + 1 , S = 32 S: m = 2

KSHS 03: S TNG GIAOCu 34. Cho hm s y = x + 3x + mx + 1 (m l tham s)

(1) 1) Kho st v v th hm s khi m = 3. 2) Tm m ng thng d: y = 1 ct th hm s (1) ti ba im phn bit A(0; 1), B, C sao cho cc tip tuyn ca th hm s (1) ti B v C vung gc vi nhau.

3

2

PT honh giao im ca (1) v d: x 3 + 3 x 2 + mx + 1 = 1 x ( x 2 + 3 x + m) = 0 9 d ct (1) ti 3 im phn bit A(0; 1), B, C m < , m 0 4Khi : xB , xC l cc nghim ca PT: x 2 + 3 x + m = 0 x B + xC = -3; x B .xC = m2 2 H s gc ca tip tuyn ti B l k1 = 3 x B + 6 xB + m v ti C l k2 = 3 xC + 6 xC + m

Tip tuyn ca (C) ti B v C vung gc vi nhau k1.k2 = -1 4m 2 - 9m + 1 = 0

m=Trang 11

9 - 65 9 + 65 m= 8 8

www.VNMATH.com100 Kho st hm s Trn S Tng

Cu 35. Cho hm s y = x 3 3 x + 1 c th (C) v ng thng (d): y = mx + m + 3 .

1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m (d) ct (C) ti M(1; 3), N, P sao cho tip tuyn ca (C) ti N v P vung gc vi nhau.

Phng trnh honh giao im ca (C) v (d): x 3 (m + 3) x m 2 = 0 x = -1 ( y = 3) ( x + 1)( x 2 x m 2) = 0 2 g( x ) = x - x - m - 2 = 0 9 d ct (1) ti 3 im phn bit M(1; 3), N, P m > - , m 0 4 Khi : xN , xP l cc nghim ca PT: x 2 - x - m - 2 = 0 x N + x P = 1; x N . x P = - m - 22 2 H s gc ca tip tuyn ti N l k1 = 3 x N - 3 v ti P l k2 = 3 x P - 3

Tip tuyn ca (C) ti N v P vung gc vi nhau k1.k2 = -1 9m 2 + 18m + 1 = 0

m=Cu 36. Cho hm s y = x 3 - 3 x 2 + 4

-3 + 2 2 -3 - 2 2 m= 3 3

(C) 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi (d) l ng thng i qua im A(2; 0) c h s gc k. Tm k (d) ct (C) ti ba im phn bit A, M, N sao cho hai tip tuyn ca (C) ti M v N vung gc vi nhau. PT ng thng (d): y = k ( x - 2) + PT honh giao im ca (C) v (d): x 3 - 3 x 2 + 4 = k ( x - 2)

x = 2 = xA ( x - 2)( x 2 - x - 2 - k ) = 0 2 g( x ) = x - x - 2 - k = 0 + (d) ct (C) ti 3 im phn bit A, M, N PT g( x ) = 0 c 2 nghim phn bit, khc 2 D > 0 9 (*) - (d) ct (C) ti 3 im phn bit (2) c 2 nghim phn bit, khc 1 4 (*) m 0 Tip tuyn ti N, P vung gc y '( xN ). y '( xP ) = -1 m = -3 2 2 (tho (*)) 3 (1).

Cu 38. Cho hm s y = x 3 - 3mx 2 + 3(m2 - 1) x - (m 2 - 1) ( m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = 0. 2) Tm cc gi tr ca m th hm s (1) ct trc honh ti 3 im phn bit c honh dng. THS (1) ct trc honh ti 3 im phn bit c honh dng, ta phi c: (1) co 2 cc tr y .y < 0 C CT (*) xC > 0, xCT > 0 a.y(0) < 0 Trong : + y = x 3 - 3mx 2 + 3(m2 - 1) x - (m 2 - 1) y = 3 x 2 - 6 mx + 3(m 2 - 1) + Dy = m 2 - m 2 + 1 = 0 > 0, "m x = m - 1 = xC + y = 0 x = m + 1 = xCT m - 1 > 0 m + 1 > 0 Suy ra: (*) 2 3 < m < 1+ 2 (m - 1)(m 2 - 3)(m 2 - 2m - 1) < 0 - 2 (m - 1) < 0Cu 39. Cho hm s y =

1 3 2 x - mx 2 - x + m + c th (Cm ) . 3 3 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m (Cm ) ct trc honh ti 3 im phn bit c tng bnh phng cc honh ln hn 15. 1 2 2 2 2 YCBT x 3 - mx 2 - x + m + = 0 (*) c 3 nghim phn bit tha x1 + x2 + x3 > 15 . 3 3 x = 1 Ta c: (*) ( x - 1)( x 2 + (1 - 3m ) x - 2 - 3m ) = 0 2 g( x ) = x + (1 - 3m) x - 2 - 3m = 0 m >1

2 2 Do : YCBT g( x ) = 0 c 2 nghim x1 , x2 phn bit khc 1 v tha x1 + x2 > 14 .

Cu hi tng t i vi hm s: y = x3 - 3mx 2 - 3x + 3m + 2Cu 40. Cho hm s y = x 3 - 3 x 2 - 9 x + m , trong m l tham s thc.

1) Kho st s bin thin v v th (C) ca hm s cho khi m = 0 . 2) Tm tt c cc gi tr ca tham s m th hm s cho ct trc honh ti 3 im phn bit c honh lp thnh cp s cng. th hm s ct trc honh ti 3 im phn bit c honh lp thnh cp s cng Phng trnh x 3 - 3 x 2 - 9 x + m = 0 c 3 nghim phn bit lp thnh cp s cng Trang 13

www.VNMATH.com100 Kho st hm s Trn S Tng

Phng trnh x 3 - 3 x 2 - 9 x = - m c 3 nghim phn bit lp thnh cp s cng ng thng y = - m i qua im un ca th (C) -m = -11 m = 11.Cu 41. Cho hm s y = x 3 - 3mx 2 + 9 x - 7 c th (Cm), trong m l tham s thc.

1) Kho st s bin thin v v th (C) ca hm s cho khi m = 0 . 2) Tm m (Cm) ct trc honh ti 3 im phn bit c honh lp thnh cp s cng.

Honh cc giao im l nghim ca phng trnh: x 3 - 3mx 2 + 9 x - 7 = 0 Gi honh cc giao im ln lt l x1; x2 ; x3 ta c: x1 + x2 + x3 = 3m x1; x2 ; x3 lp thnh cp s cng th x2 = m l nghim ca phng trnh (1) m = 1 -1 + 15 -2m 3 + 9m - 7 = 0 m = 2 -1 - 15 m = 2 Th li ta c m = -1 - 15 l gi tr cn tm. 2

(1)

Cu 42. Cho hm s y = x 3 - 3mx 2 - mx c th (Cm), trong m l tham s thc.

1) Kho st s bin thin v v th (C) ca hm s cho khi m = 1 . 2) Tm m (Cm) ct ng thng d: y = x + 2 ti 3 im phn bit c honh lp thnh cp s nhn. Xt phng trnh honh giao im ca (Cm) v d: x 3 - 3mx 2 - mx = x + 2 g ( x ) = x3 - 3mx 2 - ( m + 1) x - 2 = 0 s nhn. Khi ta c: g ( x ) = ( x - x1 )( x - x2 )( x - x3 ) x1 + x2 + x3 = 3m Suy ra: x1 x2 + x2 x3 + x1 x3 = - m - 1 x x x = 2 1 2 32 3 V x1 x3 = x2 x2 = 2 x2 = 3 2 nn ta c: -m - 1 = 4 + 3 2.3m m = -

k cn: Gi s (C) ct d ti 3 im phn bit c honh x1 ; x2 ; x3 ln lt lp thnh cp

5 3 2 +13

k : Vi m = Vy m = 3

5 , thay vo tnh nghim thy tha mn. 3 2 +13

5 3 2 +1

Cu 43. Cho hm s y = x 3 + 2mx 2 + (m + 3) x + 4 c th l (Cm) (m l tham s).

1) Kho st s bin thin v v th (C1) ca hm s trn khi m = 1. 2) Cho ng thng (d): y = x + 4 v im K(1; 3). Tm cc gi tr ca m (d) ct (Cm) ti ba im phn bit A(0; 4), B, C sao cho tam gic KBC c din tch bng 8 2 . Phng trnh honh giao im ca (Cm) v d l: x 3 + 2 mx 2 + (m + 3) x + 4 = x + 4 x( x 2 + 2 mx + m + 2) = 0 Trang 14

www.VNMATH.comTrn S Tng 100 Kho st hm s

x = 0 ( y = 4) 2 g( x ) = x + 2 mx + m + 2 = 0 (1) (d) ct (Cm) ti ba im phn bit A(0; 4), B, C (2) c 2 nghim phn bit khc 0. 2 / m -1 m 2 (*) D = m - m - 2 > 0 m -2 g(0) = m + 2 0 Khi : xB + xC = -2m; xB . xC = m + 2 . Mt khc: d (K , d ) = SDKBC = 8 2 1- 3 + 4 2 = 2 . Do :

1 BC.d ( K , d ) = 8 2 BC = 16 BC 2 = 256 2

( x B - xC )2 + ( yB - yC )2 = 256 ( x B - xC )2 + (( xB + 4) - ( xC + 4))2 = 256 2( xB - xC )2 = 256 ( xB + xC )2 - 4 x B xC = 128 4 m 2 - 4(m + 2) = 128 m 2 - m - 34 = 0 m = Vy m = 1 137 . 2 1 137 (tha (*)). 2

Cu 44. Cho hm s y = x 3 - 3 x 2 + 4 c th l (C).

1) Kho st s bin thin v v th (C) ca hm s. 2) Gi dk l ng thng i qua im A(-1; 0) vi h s gc k (k ) . Tm k ng thng dk ct th (C) ti ba im phn bit A, B, C v 2 giao im B, C cng vi gc to O to thnh mt tam gic c din tch bng 1. Ta c: dk : y = kx + k kx - y + k = 0 Phng trnh honh giao im ca (Cm) v d l: x 3 - 3 x 2 + 4 = kx + k ( x + 1) ( x - 2)2 - k = 0 x = -1 hoc ( x - 2)2 = k k > 0 dk ct (C) ti 3 im phn bit k 9

Khi cc giao im l A(-1; 0), B ( 2 - k ;3k - k k ) , C ( 2 + k ;3k + k k ) . BC = 2 k 1 + k 2 , d (O, BC ) = d (O, dk ) = k 1+ k2

1 k SDOBC = . .2 k . 1 + k 2 = 1 k k = 1 k 3 = 1 k = 1 2 1+ k2Cu 45. Cho hm s y = x 3 - 3 x 2 + 2 c th l (C).

1) Kho st s bin thin v v th (C) ca hm s. 2) Gi E l tm i xng ca th (C). Vit phng trnh ng thng qua E v ct (C) ti ba im E, A, B phn bit sao cho din tch tam gic OAB bng Ta c: E(1; 0). PT ng thng D qua E c dng y = k ( x - 1) . PT honh giao im ca (C) v D: ( x - 1)( x 2 - 2 x - 2 - k ) = 0 2.

D ct (C) ti 3 im phn bit PT x 2 - 2 x - 2 - k = 0 c hai nghim phn bit khc 1Trang 15

www.VNMATH.com100 Kho st hm s Trn S Tng k = -1 k +3 = 2 k = -1 3

Vy c 3 ng thng tho YCBT: y = - x + 1; y = ( -1 3 ) ( x - 1) .Cu 46. Cho hm s y = x 3 + mx + 2

k > -3 1 SDOAB = d (O, D). AB = k 2

k +3 k

c th (Cm) 1) Kho st s bin thin v v th ca hm s khi m = 3. 2) Tm m th (Cm) ct trc honh ti mt im duy nht. Phng trnh honh giao im ca (Cm) vi trc honh: 2 x 3 + mx + 2 = 0 m = - x 2 - ( x 0) x 2 2 -2 x 3 + 2 Xt hm s: f ( x ) = - x 2 - f '( x ) = -2 x + = x x2 x2 Ta c bng bin thin: - + f ( x) + f (x)

- - - th (Cm) ct trc honh ti mt im duy nht m > -3 .Cu 47. Cho hm s y = 2 x 3 - 3(m + 1) x 2 + 6 mx - 2

c th (Cm) 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m th (Cm) ct trc honh ti mt im duy nht.

1- 3 < m < 1+ 3Cu 48. Cho hm s y = x 3 - 6 x 2 + 9 x - 6 c th l (C).

1) Kho st s bin thin v v th (C) ca hm s. 2) nh m ng thng (d ) : y = mx - 2m - 4 ct th (C) ti ba im phn bit.

PT honh giao im ca (C) v (d): x 3 - 6 x 2 + 9 x - 6 = mx - 2m - 4 x = 2 ( x - 2)( x 2 - 4 x + 1 - m ) = 0 2 g( x ) = x - 4 x + 1 - m = 0(d) ct (C) ti ba im phn bit PT g( x ) = 0 c 2 nghim phn bit khc 2 m > -3Cu 49. Cho hm s y = x 3 3 x 2 + 1 .

1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m ng thng (D): y = (2 m - 1) x 4 m 1 ct th (C) ti ng hai im phn bit.

Phng trnh honh giao ca (C) v (D): x 3 3 x 2 (2 m 1) x + 4 m + 2 = 0x = 2 ( x - 2)( x 2 x 2 m 1) = 0 2 f ( x ) = x - x - 2m - 1 = 0 (1) 2 x1 = x2 (D) ct (C) ti ng 2 im phn bit (1) phi c nghim x1 , x2 tha mn: x1 = 2 x2 Trang 16

www.VNMATH.comTrn S Tng D = 0 b 2 2a D > 0 f (2) = 0 5 1 Vy: m = - ; m = . 8 2 8m + 5 = 0 5 1 m = - 8 2 2 m = 1 8m + 5 > 0 2 -2 m + 1 = 0 100 Kho st hm s

Cu 50. Cho hm s y = x3 - 3m 2 x + 2m c th (Cm).

1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m th (Cm) ct trc honh ti ng hai im phn bit. (Cm) ct trc honh ti ng hai im phn bit th (Cm) phi c 2 im cc tr y = 0 c 2 nghim phn bit 3x 2 - 3m 2 = 0 c 2 nghim phn bit m 0 Khi y ' = 0 x = m . (Cm) ct Ox ti ng 2 im phn bit yC = 0 hoc yCT = 0 Ta c: + y (- m) = 0 2m3 + 2m = 0 m = 0 (loi) + y (m) = 0 -2m3 + 2m = 0 m = 0 m = 1 Vy: m = 1Cu 51. Cho hm s y = x 4 - mx 2 + m - 1 c th l Cm

( )

1) Kho st s bin thin v v th (C) ca hm s khi m = 8 . 2) nh m th (Cm ) ct trc trc honh ti bn im phn bit. m > 1 m 2Cu 52. Cho hm s y = x 4 - 2 ( m + 1) x 2 + 2m + 1 c th l Cm .

( )

1) Kho st s bin thin v v th ca hm s cho khi m = 0 . 2) nh m th (Cm ) ct trc honh ti 4 im phn bit c honh lp thnh cp s cng. Xt phng trnh honh giao im: x 4 - 2 ( m + 1) x 2 + 2m + 1 = 0 t t = x , t 0 th (1) tr thnh: f (t ) = t - 2 ( m + 1) t + 2m + 1 = 0 .2 2

(1)

(Cm) ct Ox ti 4 im phn bit th f (t ) = 0 phi c 2 nghim dng phn bit D ' = m 2 > 0 1 m > S = 2 ( m + 1) > 0 2 (*) P = 2m + 1 > 0 m 0 Vi (*), gi t1 < t2 l 2 nghim ca f (t ) = 0 , khi honh giao im ca (Cm) vi Ox ln lt l: x1 = - t2 ; x2 = - t1 ; x3 = t1 ; x4 = t2 x1 , x2 , x3 , x4 lp thnh cp s cng x2 - x1 = x3 - x2 = x4 - x3 t2 = 9t1 m = 4 5m = 4m + 4 m + 1 + m = 9 ( m + 1 - m ) 5 m = 4 ( m + 1) m = - 4 -5m = 4m + 4 9 Trang 17

www.VNMATH.com100 Kho st hm s 4 Vy m = 4; - 9 Cu hi tng t i vi hm s y = - x 4 + 2(m + 2) x 2 - 2 m - 3 S: m = 3, m = 13 . 9 Trn S Tng

Cu 53. Cho hm s y = x 4 (3m + 2) x 2 + 3m c th l (Cm), m l tham s.

1) Kho st s bin thin v v th ca hm s khi m = 0. 2) Tm m ng thng y = -1 ct th (Cm) ti 4 im phn bit u c honh nh hn 2. Phng trnh honh giao im ca (Cm) v ng thng y = -1 : x = 1 x 4 (3m + 2) x 2 + 3m = -1 x 4 (3m + 2) x 2 + 3m + 1 = 0 2 x = 3m + 1 (*) ng thng y = -1 ct (Cm) ti 4 im phn bit c honh nh hn 2 khi v ch khi phng trnh (*) c hai nghim phn bit khc 1 v nh hn 2 1 0 < 3m + 1 < 4 - < m < 1 3 3m + 1 1 m 0 Cu 54. Cho hm s y = x 4 - 2 ( m + 1) x 2 + 2m + 1 c th l (Cm), m l tham s.

1) Kho st s bin thin v v th ca hm s khi m = 0. 2) Tm m th (Cm) ct trc honh ti 3 im phn bit u c honh nh hn 3. Xt phng trnh honh giao im: x 4 - 2 ( m + 1) x 2 + 2m + 1 = 0 (1) t t = x 2 , t 0 th (1) tr thnh: f (t ) = t 2 - 2 ( m + 1) t + 2m + 1 = 0 . (Cm) ct Ox ti 3 im phn bit c honh nh hn 3 0 = t1 < t2 < 3 f ( t ) c 2 nghim phn bit t1 , t2 sao cho: 0 < t1 < 3 t2 D ' = m 2 > 0 2 D ' = m > 0 1 f ( 3) = 4 - 4m 0 f (0) = 2m + 1 = 0 m = - m 1 2 S = 2 ( m + 1) < 3 S = 2 ( m + 1) > 0 P = 2m + 1 > 0 1 Vy: m = - m 1 . 2Cu 55. Cho hm s y = x 4 - 2m 2 x 2 + m 4 + 2m (1), vi m l tham s.

1) Kho st s bin thin v v th ca hm s (1) khi m = 1 .. 2) Chng minh th hm s (1) lun ct trc Ox ti t nht hai im phn bit, vi mi m < 0. Phng trnh honh giao im ca th (1) v trc Ox: x 4 - 2m 2 x 2 + m 4 + 2m = 0 (1) t t = x 2 ( t 0 ) , (1) tr thnh : t 2 - 2m 2t + m4 + 2m = 0 (2) Ta c : D ' = -2m > 0 v S = 2m2 > 0 vi mi m > 0 . Nn (2) c nghim dng (1) c t nht 2 nghim phn bit th hm s (1) lun ct trc Ox ti t nht hai im phn bit. Trang 18

www.VNMATH.comTrn S Tng 100 Kho st hm s

Cu 56. Cho hm s y =

2x +1 c th l (C). x+2 1) Kho st s bin thin v v th (C) ca hm s. 2) Chng minh rng ng thng d: y = - x + m lun ct th (C) ti hai im phn bit A, B. Tm m on AB c di nh nht. 2x +1 PT honh giao im ca (C) v d: = -x + m x+2 x -2 2 f ( x ) = x + (4 - m ) x + 1 - 2 m = 0 (1)

Do (1) c D = m 2 + 1 > 0 v f (-2) = (-2)2 + (4 - m).(-2) + 1 - 2 m = -3 0, "m nn ng thng d lun lun ct th (C ) ti hai im phn bit A, B. Ta c: y A = m - x A ; yB = m - xB nn AB 2 = ( x B - x A )2 + ( yB - y A )2 = 2(m 2 + 12) Suy ra AB ngn nht AB 2 nh nht m = 0 . Khi : AB = 24 . Cu hi tng t i vi hm s: x -1 1 x-2 a) y = S: m = 2 b) y = S: m = x -1 2x 2Cu 57. Cho hm s y =

x-3 . x +1 1) Kho st s bin thin v v th (C) ca hm s. 2) Vit phng trnh ng thng d qua im I (-1;1) v ct th (C) ti hai im M, N sao cho I l trung im ca on MN. Phng trnh ng thng d : y = k ( x + 1) + 1 x -3 = kx + k + 1 c 2 nghim phn bit khc -1 . x +1 f ( x ) = kx 2 + 2kx + k + 4 = 0 c 2 nghim phn bit khc -1 k 0 D = -4k > 0 k < 0 f (-1) = 4 0 d ct (C) ti 2 im phn bit M, N

Mt khc: xM + xN = -2 = 2 xI I l trung im MN vi "k < 0 . Kt lun: Phng trnh ng thng cn tm l y = kx + k + 1 vi k < 0 .Cu 58. Cho hm s y =

2x + 4 (C). 1- x 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi (d) l ng thng qua A(1; 1) v c h s gc k. Tm k (d) ct (C) ti hai im M, N sao cho MN = 3 10 . Phng trnh ng thng (d ) : y = k ( x - 1) + 1. Bi ton tr thnh: Tm k h phng trnh sau c hai nghim ( x1; y1 ), ( x2 ; y2 ) phn bit sao cho ( x2 - x1 ) + ( y2 - y1 ) = 902 2

(a)

2x + 4 = k ( x - 1) + 1 -x +1 y = k ( x - 1) + 1

kx 2 - (2k - 3) x + k + 3 = 0 (I). Ta c: ( I ) y = k ( x - 1) + 1 Trang 19

www.VNMATH.com100 Kho st hm s Trn S Tng

(I) c hai nghim phn bit PT kx 2 - (2k - 3) x + k + 3 = 0 (b) c hai nghim phn bit. 3 k 0, k < . 8 2 2 Ta bin i (a) tr thnh: (1 + k 2 ) ( x2 - x1 ) = 90 (1 + k 2 ) ( x2 + x1 ) - 4 x2 x1 = 90 (c) 2k - 3 k +3 Theo nh l Viet cho (b) ta c: x1 + x2 = , x1 x2 = , th vo (c) ta c phng k k 8k 3 + 27k 2 + 8k - 3 = 0 (k + 3)(8k 2 + 3k - 1) = 0 trnh: -3 + 41 -3 - 41 ; k= . 16 16 Kt lun: Vy c 3 gi tr ca k tho mn nh trn. k = -3; k =Cu 59. Cho hm s y =

2x - 2 (C). x +1 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m ng thng (d): y = 2 x + m ct (C) ti hai im phn bit A, B sao cho AB = 5 .

2x - 2 = 2 x + m 2 x 2 + mx + m + 2 = 0 ( x -1) x +1 d ct (C) ti 2 im phn bit A, B (1) c 2 nghim phn bit x1 , x2 khc 1

PT honh giao im:

(1)

m2 - 8m - 16 > 0 (2) m x1 + x2 = - 2 . Gi A ( x1; 2 x1 + m ) , B ( x2 ; 2 x2 + m ) . Khi ta c: m+2 x1 x2 = 2AB2 = 5 ( x1 - x2 )2 + 4( x1 - x2 )2 = 5 ( x1 + x2 )2 - 4x1 x2 = 1 m2 - 8m - 20 = 0 m = 10 m = -2 Vy: m = 10; m = -2 .Cu 60. Cho hm s y =

(tho (2))

x -1 (1). x+m 1) Kho st s bin thin v v th ca hm s (1) khi m = 1 . 2) Tm cc gi tr ca tham s m sao cho ng thng (d): y = x + 2 ct th hm s (1) ti hai im A v B sao cho AB = 2 2 . x -m x -1 = x+2 2 x+m x + (m + 1) x + 2 m + 1 = 0

PT honh giao im:

(*)

d ct th hm s (1) ti hai im A, B phn bit (*) c hai nghim phn bit khc -m 2 D > 0 (**) m - 6 m - 3 > 0 m < 3 - 2 3 m > 3 + 2 3 x -m m -1 m -1 x + x = -(m + 1) Khi gi x1 , x2 l cc nghim ca (*), ta c 1 2 x1. x2 = 2m + 1 Cc giao im ca d v th hm s (1) l A( x1; x1 + 2), B( x2 ; x2 + 2) . Trang 20

www.VNMATH.comTrn S Tng 100 Kho st hm s

Suy ra AB 2 = 2( x1 - x2 )2 = 2 ( x1 + x2 )2 - 4 x1 x2 = 2(m2 - 6 m - 3) m = -1 Theo gi thit ta c 2(m 2 - 6 m - 3) = 8 m 2 - 6m - 7 = 0 m = 7 Kt hp vi iu kin (**) ta c m = 7 l gi tr cn tm.Cu 61. Cho hm s y =

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

Phng trnh honh giao im ca (C) v d: x 2 + (m - 3) x + 1 - m = 0,(*) c D = m 2 - 2 m + 5 > 0, "m R v (*) khng c nghim x = 1.

x 1 (*)

x + xB = 3 - m (*) lun c 2 nghim phn bit l x A , xB . Theo nh l Vit: A x A .xB = 1 - m Khi : A ( x A ; x A + m ) , B ( x B ; x B + m ) uur uur DOAB vung ti O th OA.OB = 0 x A xB + ( x A + m )( xB + m ) = 0 2 x A x B + m( x A + x B ) + m 2 = 0 m = -2 Vy: m = 2.Cu 62. Cho hm s: y =

x+2 . x-2 1) Kho st s bin thin v v th (C) ca hm s. 2) Chng minh rng vi mi gi tr m th trn (C) lun c cp im A, B nm v hai nhnh x - yA + m = 0 ca (C) v tha A . x B - yB + m = 0 x - yA + m = 0 y = xA + m Ta c: A A A, B (d ) : y = x + m x B - yB + m = 0 yB = x B + m A, B l giao im ca (C) v (d). Phng trnh honh giao im ca (C) v (d): x+2 f ( x ) = x 2 + (m - 3) x - (2 m + 2) = 0 ( x 2) (*). x+m = x-2

(*) c D = m2 + 2m + 17 > 0, "m (d) lun ct (C) ti hai im phn bit A, B. V 1. f (2) = -4 < 0 x A < 2 < x B hoc xB < 2 < x A (pcm).

KSHS 04: TIP TUYNCu 63. Cho hm s y = x 3 + (1 - 2m) x 2 + (2 - m) x + m + 2 (1)

(m l tham s). 1) Kho st s bin thin v v th (C) ca hm s (1) vi m = 2. 2) Tm tham s m th ca hm s (1) c tip tuyn to vi ng thng d: x + y + 7 = 0 1 gc a , bit cos a = . 26 Trang 21

www.VNMATH.com100 Kho st hm s

r Gi k l h s gc ca tip tuyn tip tuyn c VTPT n1 = (k; -1) r ng thng d c VTPT n2 = (1;1) .

Trn S Tng

3 r r n1.n2 k = 2 1 k -1 Ta c cos a = r r = 12 k 2 - 26 k + 12 = 0 2 n1 . n2 26 k = 2 2 k +1 3 YCBT tho mn t nht mt trong hai phng trnh sau c nghim: 3 2 3 y = 3 x + 2(1 - 2m) x + 2 - m = 2 D/ 1 0 8m 2 - 2m - 1 0 2 / 2 D 2 0 4m - m - 3 0 3 x 2 + 2(1 - 2m) x + 2 - m = 2 y = 2 3 3 1 1 m - 4 ; m 2 1 1 m - hoc m 4 2 m - 3 ; m 1 4Cu 64. Cho hm s y = x 3 - 3 x 2 + 1 c th (C).

1) Kho st s bin thin v v th (C) ca hm s. 2) Tm hai im A, B thuc th (C) sao cho tip tuyn ca (C) ti A v B song song vi nhau v di on AB = 4 2 .

Gi s A(a; a3 - 3a2 + 1), B(b; b3 - 3b2 + 1) thuc (C), vi a b . V tip tuyn ca (C) ti A v B song song vi nhau nn: y (a) = y (b) 3a2 - 6a = 3b2 - 6 b a 2 - b2 - 2(a - b) = 0 (a - b)(a + b - 2) = 0 a + b - 2 = 0 b = 2 - a . V a b nn a 2 - a a 1Ta c: AB = (b - a)2 + (b3 - 3b2 + 1 - a3 + 3a2 - 1)2 = (b - a)2 + (b3 - a3 - 3(b2 - a2 ))2 = (b - a)2 + (b - a)3 + 3ab(b - a) - 3(b - a)(b + a) = (b - a)2 + (b - a)2 (b - a)2 + 3ab - 3.2 2 2 2

= (b - a)2 + (b - a)2 (b + a)2 - ab - 6 = (b - a)2 + (b - a)2 (-2 - ab)2 AB 2 = (b - a)2 1 + (-2 - ab)2 = (2 - 2a)2 1 + (a2 - 2a - 2)2 2 = 4(a - 1)2 1 + (a - 1)2 - 3 = 4(a - 1)2 (a - 1)4 - 6(a - 1)2 + 10

= 4(a - 1)6 - 24(a - 1)4 + 40(a - 1)2 M AB = 4 2 nn 4(a - 1)6 - 24(a - 1)4 + 40(a - 1)2 = 32 (a - 1)6 - 6(a - 1)4 + 10(a - 1)2 - 8 = 0 t t = (a - 1)2 , t > 0 . Khi (*) tr thnh: a = 3 b = -1 t 3 - 6t 2 + 10t - 8 = 0 (t - 4)(t 2 - 2t + 2) = 0 t = 4 (a - 1)2 = 4 a = -1 b = 3 Vy 2 im tho mn YCBT l: A(3;1), B(-1; -3) . (*)

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www.VNMATH.comTrn S TngCu 65. Cho hm s

100 Kho st hm s

y = 3 x - x 3 (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn ng thng (d): y = - x cc im m t k c ng 2 tip tuyn phn bit vi th (C). Cc im cn tm l: A(2; 2) v B(2; 2). (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn ng thng (d): y = 2 cc im m t k c 3 tip tuyn phn bit vi th (C). Gi M ( m;2) ( d ) . PT ng thng D i qua im M v c h s gc k c dng : y = k ( x - m ) + 2 3+ 2 (*). D l tip tuyn ca (C) h PT sau c nghim - x 2 3 x - 2 = k ( x - m) + 2 (1) (2) -3 x + 6 x = k

Cu 66. Cho hm s y = - x 3 + 3 x 2 - 2

Thay (2) v (1) ta c: 2 x 3 - 3(m + 1) x 2 + 6 mx - 4 = 0 ( x - 2) 2 x 2 - (3m - 1) x + 2 = 0

x = 22 f ( x ) = 2 x - (3m - 1) x + 2 = 0 (3)

T M k c 3 tip tuyn n th (C) h (*) c 3 nghim x phn bit 5 D > 0 m < -1 hoc m > (3) c hai nghim phn bit khc 2 3 . f (2) 0 m 2 5 m < -1 hoc m > Vy t cc im M(m; 2) (d): y = 2 vi 3 c th k c 3 tip tuyn m 2 n (C).Cu 67. Cho hm s y = f ( x ) =

1 3 mx + (m - 1) x 2 + (4 - 3m) x + 1 c th l (Cm). 3 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm cc gi tr m sao cho trn th (Cm) tn ti mt im duy nht c honh m m tip tuyn ti vung gc vi ng thng (d): x + 2 y - 3 = 0 .

(d) c h s gc -

1 tip tuyn c h s gc k = 2 . Gi x l honh tip im th: 2 (1)

f '( x ) = 2 mx 2 + 2(m - 1) x + (4 - 3m) = 2 mx 2 + 2(m - 1) x + 2 - 3m = 0 YCBT (1) c ng mt nghim m. + Nu m = 0 th (1) -2 x = -2 x = 1 (loi) 2 - 3m + Nu m 0 th d thy phng trnh (1) c 2 nghim l x = 1 hay x= m m < 0 2 - 3m Do (1) c mt nghim m th 2 m 3 Vy m < 0 hay m > 2 . 3 Trang 23

www.VNMATH.com100 Kho st hm sCu 68. Cho hm s

Trn S Tng

y = ( x + 1) . ( x - 1)

2

2

1) Kho st s bin thin v v th (C) ca hm s. 2) Cho im A(a; 0) . Tm a t A k c 3 tip tuyn phn bit vi th (C).

Ta c y = x 4 - 2 x 2 + 1 . Phng trnh ng thng d i qua A(a; 0) v c h s gc k : y = k ( x - a) x 4 - 2 x 2 + 1 = k ( x - a) d l tip tuyn ca (C) h phng trnh sau c nghim: ( I ) 4 x3 - 4 x = k 4 x ( x 2 - 1) = k k = 0 Ta c: ( I ) 2 hoc ( B) ( A) 2 f ( x ) = 3 x - 4 ax + 1 = 0 (1) x -1 = 0 + T h (A), ch cho ta mt tip tuyn duy nht l d1 : y = 0 . + Vy t A k c 3 tip tuyn phn bit vi (C) th iu kin cn v l h (B) phi c 2 nghim phn bit ( x; k ) vi x 1 , tc l phng trnh (1) phi c 2 nghim phn2 3 3 bit khc 1 D = 4 a - 3 > 0 -1 a < hoc 1 a > 2 2 f (1) 0

Cu 69. Cho hm s y = f ( x ) = x 4 - 2 x 2 .

1) Kho st s bin thin v v th (C) ca hm s. 2) Trn (C) ly hai im phn bit A v B c honh ln lt l a v b. Tm iu kin i vi a v b hai tip tuyn ca (C) ti A v B song song vi nhau.

Ta c: f '( x ) = 4 x 3 - 4 xH s gc tip tuyn ca (C) ti A v B l k A = f '(a) = 4 a3 - 4 a, kB = f '(b) = 4b3 - 4 b Tip tuyn ti A, B ln lt c phng trnh l: y = f (a)( x - a) + f (a) y = f (a) x + f (a) - af (a) y = f (b)( x - b) + f (b) y = f (b) x + f (b) - bf (b) Hai tip tuyn ca (C) ti A v B song song hoc trng nhau khi v ch khi: k A = kB 4 a 3 - 4 a = 4b 3 - 4 b (a - b)(a2 + ab + b2 - 1) = 0 (1) (2) V A v B phn bit nn a b , do (1) a2 + ab + b 2 - 1 = 0 Mt khc hai tip tuyn ca (C) ti A v B trng nhau khi v ch khi: a2 + ab + b2 - 1 = 0 a2 + ab + b2 - 1 = 0 ( a b) 4 2 4 2 f (a) - af (a) = f (b) - bf (b) -3a + 2 a = -3b + 2 b

Gii h ny ta c nghim l (a; b) = (-1;1) hoc (a; b) = (1; -1) , hai nghim ny tng ng vi cng mt cp im trn th l (-1; -1) v (1; -1) Vy iu kin cn v hai tip tuyn ca (C) ti A v B song song vi nhau l: a2 + ab + b 2 - 1 = 0 a 1; a bCu 70. Cho hm s y =

2x (C). x+2 1) Kho st s bin thin v v th (C) ca hm s. 2) Vit phng trnh tip tuyn ca th (C), bit rng khong cch t tm i xng ca th (C) n tip tuyn l ln nht. Trang 24

www.VNMATH.comTrn S Tng 100 Kho st hm s

Tm i xng ca (C) l I ( -2; 2 ) . Ta c: d (I , d ) = 8 a+2 16 + (a + 2)4

Tip tuyn (d) ca th (C) ti im M c honh a -2 thuc (C) c phng trnh: 4 2a ( x - a) + 4 x - (a + 2)2 y + 2 a2 = 0 y= 2 a+2 (a + 2)8 a+2 2.4.(a + 2)2

=

8 a+2 2 2 a+2

=2 2

a = 0 d ( I , d ) ln nht khi (a + 2)2 = 4 . a = -4 T suy ra c hai tip tuyn y = x v y = x + 8 .Cu 71. Cho hm s y =

x+2 (1). 2x + 3 1) Kho st s bin thin v v th ca hm s (1). 2) Vit phng trnh tip tuyn ca th hm s (1), bit tip tuyn ct trc honh, trc tung ln lt ti hai im phn bit A, B v tam gic OAB cn ti gc ta O. -1 Gi ( x0 ; y0 ) l to ca tip im y ( x0 ) =