1. www.MATHVN.com www.MATHVN.com - Ton Hc Vit Nam TRN S TNG ---- & ---- TI LIU N THI I HC CAO NG Nm 2012

2. Trang 1 www.MATHVN.com D 0 D 0 D 0 D 0 2 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s KSHS 01: TNH N IU CA HM S A. Kin thc c bn Gi s hm s y = f (x) c tp xc nh D. Hm s f ng bin trn D y 0,"x D thuc D. v y = 0 ch xy ra ti mt s hu hn im Hm s f nghch bin trn D y 0,"x D thuc D. Nu y' = ax2 + bx + c (a 0) th: v y = 0 ch xy ra ti mt s hu hn im + y' 0,"x R a > 0 + y' 0,"x R a < 0 nh l v du ca tam thc bc hai g(x) = ax2 + bx + c (a 0) : + Nu D < 0 th g(x) lun cng du vi a. + Nu D = 0 th g(x) lun cng du vi a (tr x = - b ) 2a + Nu D> 0 th g(x) c hai nghim x1, x2 v trong khong hai nghim th g(x) khc du vi a, ngoi khong hai nghim th g(x) cng du vi a. So snh cc nghim x1, x2 ca tam thc bc hai g(x) = ax + bx + c vi s 0: D 0 D 0 + x1 x2 < 0 P > 0 + 0 < x1 x2 P > 0 + x1 < 0 < x2 P < 0 S < 0 g(x) m,"x (a;b) max g(x) m ; (a;b) S > 0 g(x) m,"x (a;b) min g(x) m (a;b) B. Mt s dng cu hi thng gp 1. Tm iu kin hm s y = f (x) n iu trn tp xc nh (hoc trn tng khong xc nh). Hm s f ng bin trn D y 0,"x D thuc D. v y = 0 ch xy ra ti mt s hu hn im Hm s f nghch bin trn D y 0,"x D thuc D. Nu y' = ax2 + bx + c (a 0) th: v y = 0 ch xy ra ti mt s hu hn im + y' 0,"x R a > 0 + y' 0,"x R a < 0 2. Tm iu kin hm s y = f (x) = ax3 + bx2 + cx + d Ta c: y = f (x) = 3ax2 + 2bx + c . n iu trn khong (a ;b ) . a) Hm s f ng bin trn (a ;b ) hn im thuc (a ;b ) . Trng hp 1: y 0,"x (a ;b ) v y = 0 ch xy ra ti mt s hu Nu bt phng trnh f (x) 0 h(m) g(x) (*) th f ng bin trn (a ;b ) h(m) max g(x) (a ;b ) 3. Trang 2 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Nu bt phng trnh f (x) 0 h(m) g(x) (**) th f ng bin trn (a ;b ) h(m) min g(x) (a ;b ) Trng hp 2: Nu bt phng trnh f (x) 0 khng a c v dng (*) th t t = x -a . Khi ta c: y = g(t) = 3at2 + 2(3aa + b)t + 3aa2 +2ba + c . Hm s f ng bin trn khong (-;a) g(t) 0,"t < 0 a > 0 D 0 a > 0 D > 0S > 0 P 0 a > 0 Hm s f ng bin trn khong (a;+) g(t) 0,"t > 0 a > 0 D 0 D > 0S < 0 b) Hm s f nghch bin trn (a ;b ) hn im thuc (a ;b ) . Trng hp 1: P 0 y 0,"x (a ;b ) v y = 0 ch xy ra ti mt s hu Nu bt phng trnh f (x) 0 h(m) g(x) (*) th f nghch bin trn (a ;b ) h(m) max g(x) (a ;b ) Nu bt phng trnh f (x) 0 h(m) g(x) (**) th f nghch bin trn (a ;b ) h(m) min g(x) (a ;b ) Trng hp 2: Nu bt phng trnh f (x) 0 khng a c v dng (*) th t t = x -a . Khi ta c: y = g(t) = 3at2 + 2(3aa + b)t + 3aa2 +2ba + c . Hm s f nghch bin trn khong (-;a) g(t) 0,"t < 0 a < 0 D 0 a < 0 D > 0S > 0 P 0 a < 0 Hm s f nghch bin trn khong (a;+) g(t) 0,"t > 0 a < 0 D 0 D > 0S < 0 3. Tm iu kin hm s y = f (x) = ax3 + bx2 + cx + d bng k cho trc. P 0 n iu trn khong c di a 0 f n iu trn khong (x1; x2 ) y = 0 c 2 nghim phn bit x1, x2 2 2 D > 0 (1) Bin i x1 - x2 = d thnh (x1 + x2 ) - 4x1x2 = d (2) S dng nh l Viet a (2) thnh phng trnh theo m. Gii phng trnh, so vi iu kin (1) chn nghim. 2 4. Tm iu kin hm s a) ng bin trn (-;a) . b) ng bin trn (a;+). y = ax + bx + c dx + e (2), (a,d 0) 4. Trang 3 www.MATHVN.com c) ng bin trn (a;b ) . 2 Tp xc nh: D = R-e , y' = adx + 2aex + be - dc = f (x) d 2 2 (dx + e) (dx + e) Trng hp 1 Trng hp 2 Nu: f (x) 0 g(x) h(m) (i) Nu bpt: f (x) 0 khng a c v dng (i) th ta t: t = x -a . Khi bpt: f (x) 0 tr thnh: g(t) 0 , vi: g(t) = adt2 + 2a(da + e)t + ada2 + 2aea + be - dc a) (2) ng bin trn khong (-;a) -e a d g(x) h(m),"x < a -e a d h(m) min g(x) (-;a ] a) (2) ng bin trn khong (-;a) -e d g(t) 0,"t < 0 (ii) a > 0 (ii) a > 0 D > 0D 0 S > 0 P 0b) (2) ng bin trn khong (a;+) -e a d g(x) h(m),"x > a -e a d h(m) min g(x) [a ;+) b) (2) ng bin trn khong (a;+) -e d g(t) 0,"t > 0 (iii) a > 0 (iii) a > 0 D > 0D 0 S < 0 P 0c) (2) ng bin trn khong (a;b ) -e (a;b ) d g(x) h(m),"x (a;b ) -e (a;b ) d h(m) min g(x) [a ;b ] ax2 + bx + c 5. Tm iu kin hm s y = (2), (a,d 0) dx + e a) Nghch bin trn (-;a) . b) Nghch bin trn (a;+). c) Nghch bin trn (a;b ) . 2 Tp xc nh: D = R-e , y' = adx + 2aex + be - dc = f (x) d 2 2 (dx + e) (dx + e) Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s a a 5. Trang 4 www.MATHVN.com Trng hp 1 Trng hp 2 Nu f (x) 0 g(x) h(m) (i) Nu bpt: f (x) 0 khng a c v dng (i) th ta t: t = x -a . Khi bpt: f (x) 0 tr thnh: g(t) 0 , vi: g(t) = adt2 + 2a(da + e)t + ada2 + 2aea + be - dc a) (2) nghch bin trn khong (-;a) -e a d g(x) h(m),"x < a -e a d h(m) min g(x) (-;a ] a) (2) ng bin trn khong (-;a) -e d g(t) 0,"t < 0 (ii) a < 0 (ii) a < 0 D > 0D 0 S > 0 P 0b) (2) nghch bin trn khong (a;+) -e a d g(x) h(m),"x > a -e a d h(m) min g(x) [a ;+) b) (2) ng bin trn khong (a;+) -e d g(t) 0,"t > 0 (iii) a < 0 (iii) a < 0 D > 0D 0 S < 0 P 0c) (2) ng bin trong khong (a;b ) -e (a;b ) d g(x) h(m),"x (a;b ) -e (a;b ) d h(m) min g(x) [a ;b ] Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng a a 6. Trang 5 www.MATHVN.com 2 4 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Cu 1. Cho hm s y = 1 (m -1)x3 + mx2 +(3m - 2)x 3 (1) 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)x2 + 2mx + 3m - 2 . (1) ng bin trn R y 0, "x m 2 Cu 2. Cho hm s y = x3 + 3x2 - 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). Tp xc nh: D = R. y = 3x2 + 6x - m . y c D = 3(m + 3) . + Nu m -3 th D 0 fi y 0,"x fi hm s ng bin trn R fi m -3 tho YCBT. + Nu m > -3 th D > 0 fi PT y = 0 c 2 nghim phn bit x1, x2 (x1 < x2 ) . Khi hm s ng bin trn cc khong (-;x1),(x2; +) . Do hm s ng bin trn khong (-;0) 0 x1 < x2 Vy: m -3. D > 0 P 0 S > 0 m > -3 -m 0 -2 > 0 (VN) Cu 3. Cho hm s y = 2x3 -3(2m +1)x2 + 6m(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;+) Tp xc nh: D = R. y' = 6x2 - 6(2m +1)x + 6m(m +1) c D = (2m +1)2 - 4(m2 + m) =1 > 0 y' = 0 x = m x = m +1 . Hm s ng bin trn cc khong (-;m), (m +1;+) Do : hm s ng bin trn (2;+) m +1 2 m 1 Cu 4. Cho hm s y = x3 + (1- 2m)x2 + (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 khong K = (0;+) . Hm ng bin trn (0;+) y = 3x2 + 2(1- 2m)x + (2 - m) 0 vi "x (0;+) 2 f (x) = 3x 2 + 2x + 2 m4x +1 vi "x (0;+) Ta c: f (x) = 6(2x + x -1) = 0 2x2 + x -1 = 0 x = -1; x = 1 (4x +1)2 2 Lp BBT ca hm f (x) trn (0;+), t ta i n kt lun: f 1 m 5 m . Cu hi tng t: a) y = 1 (m +1)x3 -(2m -1)x2 + 3(2m -1)x +1 3 (m -1), K = (-;-1). S: m 4 11 b) y = 1 (m +1)x3 -(2m -1)x2 + 3(2m -1)x +1 3 c) y = 1 (m +1)x3 -(2m -1)x2 + 3(2m -1)x +1 3 (m -1), K = (1;+). S: m 0 (m -1), K = (-1;1) . S: m 1 2 7. Trang 6 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Cu 5. Cho hm s y = 1 (m2 -1)x3 + (m -1)x2 - 2x +1 (1) (m 1) . 3 1) Kho st s bin thin v v th (C) ca hm s khi m = 0. 2) Tm m hm nghch bin trn khong K = (-;2) . Tp xc nh: D = R; y = (m2 -1)x2 +2(m -1)x - 2 . t t = x 2 ta c: y = g(t) = (m2 -1)t2 + (4m2 + 2m - 6)t + 4m2 + 4m -10 Hm s (1) nghch bin trong khong (-;2) g(t) 0, "t < 0 m2 -1< 0 a < 0 2 a < 0 m2 -1< 0 D > 0 3m - 2m -1> 0 TH1: D 0 2 TH2: S > 0 4m2 + 4m -10 0 3m - 2m -1 0 -2m -3P 0 > 0 m +1 Vy: Vi -1 m 0 hoc y 0,"x (m;0) khi m < 0 . Vy hm s ng bin trong khong (x1; x2 ) vi x2 - x1 =1 (x1;x2 ) = (0;m) v x 1 2 2 - x1 =1 m - 0 =1 m = 1. Cu 9. Cho hm s y = x4 - 2mx2 -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' = 4x3 - 4mx = 4x(x2 - m) + m 0 , y 0,"x (0;+) fi m 0 tho mn. + m > 0 , y = 0 c 3 nghim phn bit: - m, 0, m . Hm s (1) ng bin trn (1; 2) m 1 Cu hi tng t: 0 < m 1. Vy m(-;1 . a) Vi y = x4 - 2(m -1)x2 + m - 2 ; y ng bin trn khong (1;3) . S: m 2 . Cu 10. Cho hm s y = mx + 4 x + m (1) 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) . 2 Tp xc nh: D = R{m}. y = m -4 . (x + m)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. (1) (2) 2 2 -3 + Cu 11. Cho hm s y = x x m x -1 (2). Tm m hm s (2) ng bin trn khong (-;-1) . 2 Tp xc nh: D = R{1}. y' = 2x -4x + 3- m = (x -1)2 f (x) . (x -1)2 Ta c: f (x) 0 m 2x2 - 4x + 3. t g(x) = 2x2 - 4x + 3 fi g'(x) = 4x - 4 Hm s (2) ng bin trn (-;-1) y' 0, "x (-;-1) m min (-;-1] g(x) Da vo BBT ca hm s g(x), "x (-;-1] ta suy ra m 9 . Vy m 9 th hm s (2) ng bin trn (-;-1) 2 2 -3 + Cu 12. Cho hm s y = x x m x -1 (2). Tm m hm s (2) ng bin trn khong (2;+). 2 Tp xc nh: D = R{1}. y' = 2x -4x + 3- m = (x -1)2 f (x) . (x -1)2 Ta c: f (x) 0 m 2x2 - 4x + 3. t g(x) = 2x2 - 4x + 3 fi g'(x) = 4x - 4 Hm s (2) ng bin trn (2;+) y' 0, "x (2;+) m min g(x) [2;+) 9. Trang 8 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Da vo BBT ca hm s g(x), "x (-;-1] ta suy ra m 3 . Vy m 3 th hm s (2) ng bin trn (2;+). 2x2 -3x + m Cu 13. Cho hm s y = (2). x -1 Tm m hm s (2) ng bin trn khong (1;2) . 2x2 - 4x + 3- m f (x) Tp xc nh: D = R{1}. y' = = . (x -1)2 (x -1)2 Ta c: f (x) 0 m 2x2 - 4x + 3. t g(x) = 2x2 - 4x + 3 fi g'(x) = 4x - 4 Hm s (2) ng bin trn (1;2) y' 0, "x (1;2) m min g(x) [1;2] Da vo BBT ca hm s g(x), "x (-;-1] ta suy ra m 1. Vy m 1 th hm s (2) ng bin trn (1;2) . x2 - 2mx + 3m2 Cu 14. Cho hm s y = (2). 2m - x Tm m hm s (2) nghch bin trn khong (-;1) . -x2 + 4mx - m2 f (x) Tp xc nh: D = R{2m} . y' = = .t t = x -1. (x - 2m)2 (x - 2m)2 Khi bpt: f (x) 0 tr thnh: g(t) = -t2 - 2(1- 2m)t - m2 + 4m -1 0 Hm s (2) nghch bin trn (-;1) y' 0, "x (-;1) 2m >1 g(t) 0, "t < 0 (i) D' = 0 m = 0 D' > 0 m 0 m = 0(i) S > 0 4m - 2 > 0 m 2 + 3 P 0 m2 - 4m +1 0 Vy: Vi m 2 + 3 th hm s (2) nghch bin trn (-;1) . x2 - 2mx + 3m2 Cu 15. Cho hm s y = (2). 2m - x Tm m hm s (2) nghch bin trn khong (1;+) . -x2 + 4mx - m2 f (x) Tp xc nh: D = R{2m} . y' = = .t t = x -1. (x - 2m)2 (x - 2m)2 Khi bpt: f (x) 0 tr thnh: g(t) = -t2 - 2(1- 2m)t - m2 + 4m -1 0 Hm s (2) nghch bin trn (1;+) y' 0, "x (1;+) 2m 0 m < 3 Cu 3. Cho hm s y = -x3 + (2m +1)x2 -(m2 -3m + 2)x - 4 (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 nm v hai pha ca trc tung. y = -3x2 + 2(2m +1)x -(m2 -3m + 2) . (Cm) c cc im C v CT nm v hai pha ca trc tung PT y = 0 du 3(m2 -3m + 2) < 0 1< m < 2 . c 2 nghim tri Cu 4. Cho hm s y = 1 x3 - mx2 + (2m -1)x -3 (m l tham s) c th l (Cm). 3 1) Kho st s bin thin v v th hm s khi m = 2. 2) Xc nh m (Cm) c cc im cc i, cc tiu nm v cng mt pha i vi trc tung. TX: D = R ; y = x2 - 2mx + 2m -1. th (Cm) c 2 im C, CT nm cng pha i vi trc tung y = 0 m 1 c 2 nghim phn bit cng du D = m2 - 2m +1 > 0 2m -1 > 0 m > 1 . 2 Cu 5. Cho hm s y = x3 -3x2 - 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. Ta c: y' = 3x2 - 6x - m . Hm s c C, CT y' = 3x2 - 6x - m = 0 c 2 nghim phn bit x ;x 13. Trang 12 www.MATHVN.com D' = 9 + 3m > 0 m > -3 (*) 14. Trang 12 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Gi hai im cc tr l A(x1;y1 );B(x2;y2 ) Thc hin php chia y cho y ta c: y = 1 x - 1 y'+ 2m - 2 x + 2 + m 3 3 3 3 fi y = y(x ) = 2m - 2 x + 2 + m ; y = y(x ) = 2m - 2 x + 2 + m 1 1 3 1 3 2 2 3 2 3 fi Phng trnh ng thng i qua 2 im cc tr l D: y = 2m - 2 x + 2 + m 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 2m - 2 =1 m = 9 (khng tha (*)) 3 2 TH2: Trung im I ca AB nm trn ng thng y = x -1 y = x -1 y1 + y2 = x1 + x2 -1 2m - 2 (x + x )+ 2 2 + m = (x + x )- 2 I I 2 2 3 1 2 3 1 2 2m - 2 .2 + 2 2 + m = 0 m = 0 3 3 Vy cc gi tr cn tm ca m l: m = 0 . Cu 6. Cho hm s y = x3 -3mx2 + 4m3 (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. Ta c: y = 3x2 - 6mx ; y = 0 x = 0 . hm s c cc i v cc tiu th m 0. x = 2m uuur th hm s c hai im cc tr l: A(0; 4m3 ), B(2m; 0) fi AB = (2m;-4m3 ) Trung im ca on AB l I(m; 2m3 ) A, B i xng nhau qua ng thng d: y = x AB ^ d 2m - 4m3 = 0 2 I d 3 m = 2m = m 2 Cu 7. Cho hm s y = -x3 + 3mx2 -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 = -3x2 + 6mx ; y = 0 x = 0 x = 2m . Hm s c C, CT PT y = 0 c 2 nghim phn bit m 0 .uuur Khi 2 im cc tr l: A(0;-3m -1), B(2m;4m3 -3m -1) fi AB(2m;4m3 ) Trung im I ca AB c to : I(m;2m3 -3m -1) ng thng d: x + 8y - 74 = 0 c mt VTCP u = (8;-1) . A v B i xng vi nhau qua d I d muuur+r8(2m3 - 3m -1)- 74 = 0 AB ^ d m = 2 AB.u = 0 Cu hi tng t: a) y = x3 -3x2 + m2 x + m,d : y = 1 x - 5 . S: m = 0 . 2 2 Cu 8. Cho hm s y = x3 -3x2 + mx (1). 15. Trang 13 www.MATHVN.com 2 3 I d Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s 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 - 2y - 5 = 0 . Ta c y = x3 -3x2 + mx fi y' = 3x2 - 6x + m Hm s c cc i, cc tiu y = 0 c hai nghim phn bit D = 9 -3m > 0 m < 3Ta c: y = 1 x - 1 y + 2 m - 2 x + 1 m 3 3 3 3 fi ng thng D i qua cc im cc tr c phng trnh y = 2 m - 2 x + 1 m 3 3 nn D c h s gc k = 2 m - 2 . 1 3 d: x - 2y - 5 = 0 y = 1 x - 5 2 2 1 fi d c h s gc k2 = 2 hai im cc tr i xng qua d th ta phi c d ^ D 1 2 fi k1k2 = -1 m - 2 = -1 m = 0 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 = 0 Cu 9. Cho hm s y = x3 -3(m +1)x2 + 9x + 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 nhau qua ng thng d: y = 1 x . 2 y' = 3x2 - 6(m +1)x + 9 Hm s c C, CT D' = 9(m +1)2 -3.9 > 0 m(-;-1- 3)(-1+ 3;+) Ta c y = 1 x - m +1 y - 2(m2 + 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. fi y = -2(m2 + 2m - 2)x + 4m +1; y = -2(m2 + 2m - 2)x + 4m +1 1 1 2 2 v: x1 + x2 = 2(m +1) x1.x2 = 3 Vy ng thng i qua hai im cc i v cc tiu l y = -2(m2 + 2m - 2)x + 4m +1 A, B i xng qua (d): y = 1 x 2 AB ^ d m = 1. Cu 10. Cho hm s y = x3 -3(m +1)x2 + 9x - 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, x2 Ta c y' = 3x2 - 6(m +1)x + 9. sao cho x1 - x2 2 . + Hm s t cc i, cc tiu ti x1, x2 PT y' = 0 c hai nghim phn bit x1, x2 PT x2 - 2(m +1)x + 3 = 0 c hai nghim phn bit l x1, x2 . 16. Trang 14 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng D' = (m +1)2 -3 > 0 m > -1+ 3 (1) m < -1- 3 + Theo nh l Viet ta c x1 + x2 = 2(m +1); x1x2 = 3. Khi : x - x 2 (x + x )2 - 4x x 4 4(m +1)2 -12 4 (m +1)2 4 -3 m 1 (2)1 2 1 2 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 11. Cho hm s y = x3 + (1- 2m)x2 + (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 x , x sao cho x - x > 1 . 1 2 1 2 3 Ta c: y' = 3x2 + 2(1- 2m)x + (2 - m) Hm s c C, CT y' = 0 c 2 nghim phn bit x1, x2 (gi s x1 < x2 ) 2 2 m > 5 D' = (1- 2m) -3(2 - m) = 4m - m - 5 > 0 4 (*)m < -1 Hm s t cc tr ti cc im x ,x . Khi ta c: x + x = - 2(1- 2m) ; x x = 2 - m 1 2 1 2 3 1 2 3 x - x > 1 (x - x ) 2 = (x + x ) 2 - 4x x > 1 1 2 3 1 2 1 2 1 2 9 4(1- 2m)2 - 4(2 - m) >1 16m2 -12m - 5 > 0 m > 3+ 29 m < 3- 29 8 8 Kt hp (*), ta suy ra m > 3+ 29 m < -1 8 Cu 12. Cho hm s y = 1 x3 - mx2 + mx -1, vi m l tham s thc. 3 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 8. Ta c: y' = x2 - 2mx + m . Hm s c C, CT y' = 0 c 2 nghim phn bit x1, x2 (gi s x1 < x2 ) D = m2 - m > 0 m < 0 (*). Khi : x + x = 2m, x x = m .m >1 1 2 1 2 m 1- 65 x - x 8 (x - x )2 64 m2 - m -16 0 2 (tho (*)) 1 2 1 2 1+ 65 m 2 Cu 13. Cho hm s y = 1 x3 -(m -1)x2 + 3(m - 2)x + 1 , 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 + 2x2 =1. Ta c: y = x2 - 2(m -1)x + 3(m - 2) Hm s c cc i v cc tiu y = 0 c hai nghim phn bit x1, x2 D > 0 m2 - 5m + 7 > 0 (lun ng vi "m) 17. Trang 15 www.MATHVN.com Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Khi ta c: x1 + x2 = 2(m -1) x2 = 3- 2m x1x2 = 3(m - 2) x2 (1- 2x2 )= 3(m - 2) 8m2 +16m - 9 = 0 m = -4 34 . 4 Cu 14. Cho hm s y = 4x3 + mx2 -3x . 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 = -4x2 . y =12x2 + 2mx -3 . Ta c: D = m2 + 36 > 0, "m fi hm s lun c 2 cc tr x , x .1 2 Khi : x = -4x ; x + x = - m ; x x 1 fi m 9 1 2 1 2 1 2 = - = 6 4 2 Cu hi tng t: a) y = x3 + 3x2 + mx +1; x + 2x = 3 S: m = -105 . 1 2 Cu 15. Cho hm s y = 1 x3 - ax2 -3ax + 4 (1) (a l tham s). 3 1) Kho st s bin thin v v th ca hm s khi a = 1. 2) Tm a hm s (1) t cc tr ti x1 , x2 phn bit v tho mn iu kin: x 2 + 2ax + 9a a2 1 2 + = 2 (2) a2 x 2 + 2ax + 9a2 1 y = x2 - 2ax -3a . Hm s c C, CT y = 0 c 2 nghim phn bit x , x 1 2 D = 4a2 +12a > 0 a < -3 (*). Khi x + x = 2a , x x = -3a . a > 0 1 2 1 2 Ta c: x 2 + 2ax + 9a = 2a(x + x )+12a = 4a2 +12a > 01 2 1 2 Tng t: x 2 + 2ax + 9a = 4a2 +12a > 0 2 1 4a2 +12a a2 4a2 +12a Do : (2) + = 2 = 1 3a(a + 4)= 0 a = -4 a2 4a2 +12a a2 Cu 16. Cho hm s y = 2x3 + 9mx2 +12m2 x +1 (m l tham s). 1) Kho st s bin thin v v th (C) ca hm s khi m = 1. 2) Tm cc gi tr ca m hm s c cc i ti xC, cc tiu ti xCT tha mn: x2 = x . C CT Ta c: y = 6x2 +18mx +12m2 = 6(x2 + 3mx + 2m2 ) Hm s c C v CT y = 0 c 2 nghim phn bit x ,x D = m2 > 0 m 01 2 Khi : x = 1 (-3m - m ), x = 1 (-3m + m ). 1 2 2 2 Da vo bng xt du y, suy ra xC = x1, xCT = x22 Do : x2 = x -3m - m = -3m + m m = -2 . C CT 2 2 Cu 17. Cho hm s y = (m + 2)x3 + 3x2 + mx - 5 , m l tham s. 1) Kho st s bin thin v v th (C) ca hm s khi m = 0. 18. Trang 16 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng 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)x2 + 6x + m = 0 c 2 nghim dng phn bit a = (m + 2) 0D' = 9 -3m(m + 2) > 0 D' = -m2 - 2m + 3 > 0 -3 < m 0 m < 0 m < 0 -3 < m < -2 3(m + 2) m + 2 < 0 m < -2S = -3 > 0 m + 2 Cu 18. Cho hm s y = 1 x3 - 1 mx2 + (m2 -3)x (1), m l tham s. 3 2 1) Kho st s bin thin v v th (C) ca hm s khi m = 0. 2) Tm cc gi tr ca m hm s (1) c cc im cc tr x1,x2 vi x1 > 0,x2 > 0 v x2 + x2 = 5 . 1 2 2 y = x2 - mx + m2 -3 ; y = 0 x2 - mx + m2 -3 = 0 (2) D > 0 P > 0 3 < m < 2 14YCBT S > 0 m = . 14 2 5 m = x2 + x2 = 2 1 2 2 Cu 19. Cho hm s y = x3 + (1- 2m)x2 + (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 = 3x2 + 2(1- 2m)x + 2 - m = g(x) YCBT phng trnh y = 0 c hai nghim phn bit x1, x2 tha mn: x1 < x2 0 g(1) = -5m + 7 > 0 5 < m < 7 . 4 5 S = 2m -1 < 1 2 3 Cu 20. Cho hm s y = m x3 + (m - 2)x2 + (m -1)x + 2 (Cm). 3 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m hm s c cc i ti x1, cc tiu ti x2 tha mn x1 < x2 0 2m -1 x1 + x2 -2 < < 0 3 10 (1) -2 < 2 < 0 4(2m -1) 2 - m - < m < -1 (x1 + 2)(x2 + 2)> 0 4 + 3 + 3 > 0 7 x1x2 > 0 2 - m > 0 D' = 4m2 - m - 5 > 0 [ 3 4m2 - m - 5 > 0 m 2 f (0)= 2 - m 0 2m 1 - (2) (x + 2)+ (x + 2)> 0 3 > -2 m 2 1 2 (x1 + 2)(x2 + 2)> 0 2 - m 4(2m -1) 4 0 + + > 20. Trang 18 www.MATHVN.com 3 3 4m2 - m - 5 > 0 (3) f ( -2)=10 + 6m 0 2m -1 0 - 5 m < -1 < x1 + x2 < 0 3 3 x1x2 > 0 2 - m > 0 [ 3 Tm li cc gi tr m cn tm l: m - 5 ;-1 2;+) 3 Cu 22. Cho hm s y = x3 -3x2 + 2 (1) 1) Kho st s bin thin v v th ca hm s (1). 2) Tm im M thuc ng thng d: y = 3x - 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) = 3x - y - 2 ta c: g(xA ,yA) = 3xA - yA - 2 = -4 < 0; g(xB ,yB ) = 3xB - yB - 2 = 6 > 0 fi 2 im cc i v cc tiu nm v hai pha ca ng thng d: y = 3x - 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 = -2x + 2 Ta im M l nghim ca h: y = 3x - 2 x = 4 ; y = 2 fi M 4 ; 2 y = -2x + 2 5 5 5 5 21. Trang 18 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Cu 23. Cho hm s y = x3 -3mx2 + 3(m2 -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 = 3x2 - 6mx + 3(m2 -1). Hm s (1) c cc tr PT y = 0 c 2 nghim phn bit x2 - 2mx + m2 -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) Ta c OA = 2OB m2 + 6m +1 = 0 m = -3+ 2 2 . m = -3- 2 2 Cu 24. Cho hm s y = x3 -3x2 - 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 = -4x + 3. Ta c: y' = 3x2 - 6x - m . Hm s c C, CT y' = 0 c 2 nghim phn bit x ,x 1 2 D' = 9 + 3m > 0 m > -3 (*) Gi hai im cc tr l A(x1;y1 );B(x2;y2 ) Thc hin php chia y cho y ta c: y = 1 x - 1 y'- 2m + 2 x + 2 - m 3 3 3 3 fi y = y(x )= - 2m + 2 x + 2 - m ; y = y(x )= - 2m + 2 x + 2 - m 1 1 3 1 3 2 2 3 2 3 fi Phng trnh ng thng i qua 2 im cc tr l D: y = - 2m + 2 x + 2 - m 3 3 - 2m + 2 = -4 D // d: y = -4x + 3 3 m = 3 (tha mn (*)) 2 - m 3 3 Cu hi tng t: a) y = 1 x3 - mx2 + (5m - 4)x + 2 , d :8x + 3y + 9 = 0 S: m = 0; m = 5 . 3 Cu 25. Cho hm s y = x3 + mx2 + 7x + 3 c th l (Cm). 1) Kho st s bin thin v v th ca hm s khi m = 5. 2) Tm m (Cm) c cc im cc i, cc tiu v ng thng i qua cc im cc tr vung gc vi ng thng d: y = 3x - 7 . Ta c: y' = 3x2 + 2mx + 7. Hm s c C, CT y = 0 c 2 nghim phn bit x ,x . 1 2 D' = m2 - 21 > 0 m > 21 (*) Gi hai im cc tr l A(x1;y1 );B(x2;y2 ) Thc hin php chia y cho y ta c: y = 1 x + 1 y'+ 2 (21- m2 )x + 3 - 7m 3 9 9 9 fi y = y(x ) = 2 (21- m2 )x + 3- 7m ; y = y(x ) = 2 (21- m2 )x + 3- 7m 1 1 9 1 9 2 2 9 2 9 22. Trang 19 www.MATHVN.com 1 2 = - Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s fi Phng trnh ng thng i qua 2 im cc tr l D: y = 2 (21- m2 )x + 3 - 7m 9 9 m > 21 D ^ d: y = -4x + 3 2 m 3 10 . (21- m2 ).3 = -1 = 2 9 Cu 26. Cho hm s y = x3 -3x2 - 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 to vi ng thng d: x + 4y - 5 = 0 mt gc a = 450 . Ta c: y' = 3x2 - 6x - m . Hm s c C, CT y' = 0 c 2 nghim phn bit x ;x D' = 9 + 3m > 0 m > -3 (*) Gi hai im cc tr l A(x1;y1 );B(x2;y2 ) Thc hin php chia y cho y ta c: y = 1 x - 1 y'- 2m + 2 x + 2 - m 3 3 3 3 fi y = y(x )= - 2m + 2 x + 2 - m ; y = y(x )= - 2m + 2 x + 2 - m 1 1 3 1 3 2 2 3 2 3 fi Phng trnh ng thng i qua 2 im cc tr l D: y = - 2m + 2 x + 2 - m 3 3 t k = - 2m + 2 . ng thng d: x + 4y - 5 = 0 c h s gc bng - 1 . 3 4 k + 1 k + 1 =1- 1 k k = 3 m 39 = - Ta c: tan 45o = 4 4 4 5 10 1- 1 k k + 1 = -1+ 1 k k 5 m 1 = - = - 4 4 4 3 2 Kt hp iu kin (*), suy ra gi tr m cn tm l: m 1 . 2 Cu hi tng t: a) y = x3 -3(m -1)x2 + (2m2 -3m + 2)x - m(m -1) , d : y = -1 x + 5 , a = 450 . S: m = 3 15 4 2 Cu 27. Cho hm s y = x3 -3x2 + 2 (C). 1) Kho st s bin thin v v th (C) ca hm s . 2) Tm m ng thng i qua hai im cc tr ca (C) tip xc vi ng trn (S) c phng trnh (x - m)2 + (y - m -1)2 = 5 . Phng trnh ng thng D i qua hai im cc tr 2x + y - 2 = 0 . (S) c tm I(m,m +1) v bn knh R= 5 . 2m + m +1- 2 D tip xc vi (S) = 5 3m -1 = 5 5 m = 2; m = -4 . 3 23. Trang 20 www.MATHVN.com Cu 28. Cho hm s y = x3 -3mx + 2 (Cm ) . 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m ng thng i qua im cc i, cc tiu ca(Cm )ct ng trn tm I(1;1) , bn knh bng 1 ti hai im phn bit A, B sao cho din tch DIAB t gi tr ln nht . 24. Trang 20 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Ta c y' = 3x2 -3m . Hm s c C, CT PT y' = 0 c hai nghim phn bit m > 0 V y = 1 x.y- 2mx + 2 nn ng thng D i qua cc im C, CT ca th hm s c 3 phng trnh l: y = -2mx + 2 2m -1 Ta c d (I,D)= < R =1 (v m > 0) fi D lun ct ng trn tm I(1; 1), bn knh R 4m2 +1 = 1 ti 2 im A, B phn bit. Vi m 1 : D khng i qua I, ta c:S = 1 IA.IB.sin AIB 1 R2 = 1 2 DABI 2 2 2 Nn S t GTLN bng 1 khi sinAIB =1 hay DAIB vung cn ti I IH = R = 1 DIAB 2 2 2 2m -1 1 2 3 = m = (H l trung im ca AB) 4m2 +1 2 2 Cu 29. Cho hm s y = x3 + 6mx2 + 9x + 2m (1), vi m l tham s thc. 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m th hm s (1) c hai im cc tr sao cho khong cch t gc to O n ng thng i qua hai im cc tr bng 4 . 5 Ta c: y = 3x2 +12mx + 9 . Hm s c 2 im cc tr PT y = 0 c 2 nghim phn bit D' = 4m2 - 3 > 0 m > 3 hoc m < - 3 (*) 2 2 Khi ta c: y = x + 2m .y + (6 -8m2 )x - 4m 3 3 fi ng thng i qua 2 im cc tr ca th hm s (1) c PT l: D : y = (6 -8m2 )x - 4m -4m 4 m = 1 d(O,D) = = 64m4 -101m2 + 37 = 0 37 m = 1. (6 -8m2 )2 +1 5 m = 8 (loai) Cu 30. Cho hm s y = x3 -3x2 + (m - 6)x + m - 2 (1), vi m l tham s thc. 1) Kho st s bin thin v v th ca hm s khi m = 2. 2) Tm m th hm s (1) c hai im cc tr sao cho khong cch t im A(1;-4) n ng thng i qua hai im cc tr bng 12 . 265 Ta c: y = 3x2 - 6x + m - 6 . Hm s c 2 im cc tr PT y = 0 c 2 nghim phn bit D = 32 -3(m - 6) > 0 m < 9 (*) Ta c: y = 1 (x -1).y + 2 m - 6 x + 4 m - 43 3 3 fi PT ng thng qua 2 im cc tr D: y = 2 m - 6 x + 4 m - 4 3 3 6m -18 12 m =1 fi d(A,D) = = 1053 (tho (*)) 4m2 - 72m + 333 265 m = 249 25. Trang 21 www.MATHVN.com 2 4 x = -m Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Cu 31. Cho hm s y = x3 -3x2 + mx +1 (1), vi m l tham s thc. 1) Kho st s bin thin v v th ca hm s khi m = 0. 2) Tm m th hm s (1) c hai im cc tr sao cho khong cch t im I 1 ; 11 n ng thng i qua hai im cc tr l ln nht. Ta c: y = 3x2 - 6x + m . Hm s c 2 im cc tr PT y = 0 c 2 nghim phn bit D > 0 m < 3 . Ta c: y = x - 1 y + 2m - 2 x + m +1 3 3 3 3 fi PT ng thng qua hai im cc tr l: D : y = 2m - 2 x + m +1. 3 3 1 uur 3 D dng tm c im c nh ca D l A - ;2 . AI = 1; . 2 Gi H l hnh chiu vung gc ca I trn D. Ta c d(I,D) = IH IA . Du "=" xy ra IA ^ D 4 1+ 2m - 2 . 3 = 0 m =1. 3 4 Vy max(d(I ,D)) = 5 4 khi m = 1. 3 2 3 2 Cu 32. Cho hm s y = x + 3(m +1)x + 3m(m + 2)x + m + 3m (Cm ) . 1) Kho st s bin thin v v th ca hm s khi m = 0. 2) Chng minh rng vi mi m, th (Cm) lun c 2 im cc tr v khong cch gia 2 im cc tr l khng i. Ta c: y = 3x2 + 6(m +1)x + 6m(m + 2) ; th (Cm) c im cc i A(-2 - m;4) y = 0 x = -2 - m . v im cc tiu B(-m;0) fi AB = 2 5 . Cu 33. Cho hm s y = 2x2 -3(m +1)x2 + 6mx + m3 . 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m th hm s c hai im cc tr A, B sao cho AB = 2 . Ta c: y = 6(x -1)(x - m) . Hm s c C, CT y = 0 c 2 nghim phn bit m 1. Khi cc im cc tr l A(1;m3 + 3m -1),B(m;3m2 ) . AB = 2 (m -1)2 + (3m2 - m3 -3m +1) = 2 m = 0; m = 2 (tho iu kin). Cu 34. Cho hm s y = x3 -3mx2 + 3(m2 -1)x - m3 + 4m -1 (1) 1) Kho st s bin thin v v th ca hm s (1) khi m = -1. 2) Tm m th ca hm s (1) c hai im cc tr A, B sao cho DOAB vung ti O. Ta c: y = 3x2 - 6mx + 3(m2 -1); y = 0 x = m +1fi y = m - 3 x = m -1fi y = m +1 26. Trang 22 www.MATHVN.com m = 2 fi A(m +1;m - 3) , B(m -1;m +1) fi OA = (m +1;m - 3) , OB = (m -1;m +1) . DOAB vung ti O OA.OB = 0 2m2 - 2m - 4 = 0 m = -1 . 27. Trang 22 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Cu 35. Cho hm s y = 2x2 -3(m +1)x2 + 6mx + m3 (1) 1) Kho st s bin thin v v th ca hm s (1) khi m = 1. 2) Tm m th ca hm s (1) c hai im cc tr A, B sao cho tam gic ABC vung ti C, vi C(4;0) . Ta c: y = 6(x -1)(x - m) . Hm s c C, CT y = 0 c 2 nghim phn bit m 1. Khi cc im cc tr l A(1;m3 + 3m -1),B(m;3m2 ) . DABC vung ti C AC.BC = 0 (m +1)m2 (m2 - m +1) + 3m2 - 5m + 4 = 0 m = -1 Cu 36. Cho hm s y = x3 + 3x2 + 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 =1200 . Ta c: y = 3x2 + 6x ; y = 0 x = -2 fi y = m + 4 x = 0 fi y = m Vy hm s c hai im cc tr A(0 ; m) v B(-2 ; m + 4) OA = (0;m), OB = (-2;m + 4) . AOB =1200 th cos AOB 1 = - 2 m(m + 4) = - 1 m2 (4 + (m + 4)2 ) = -2m(m + 4) -4 < m < 0 2 m2 (4 + (m + 4)2 ) 2 3m + 24m + 44 = 0 -4 < m < 0 -12 + 2 3 12 2 3 m = m = - 3 3 Cu 37. Cho hm s y = x3 -3x2 + m2 - m +1 (1) 1) Kho st s bin thin v v th ca hm s (1) khi m = 1. 2) Tm m th hm s (1) c hai im cc i, cc tiu l A v B sao cho din tch tam gic ABC bng 7, vi im C(2; 4 ). Ta c y' = 3x2 - 6x ; y' = 0 3x2 - 6x = 0 x = 0;x = 2 fi Hm s lun c C, CT. Cc im C, CT ca th l: A(0;m2 - m +1), B(2;m2 - m -3) , AB = 22 + (-4)2 = 2 5 x - 0 y - m2 + m -1 2 Phng trnh ng thng AB: = 2x + y - m + m -1 = 0 2 -4 1 1 m2 - m +1 2 m = 3 SDABC = 2 d(C, AB).AB = 2 . 5 .2 5 = m - m +1 = 7 m = -2 . Cu hi tng t: a) y = x3 -3mx + 2, C(1;1),S = 18 . S: m = 2 . Cu 38. Cho hm s y = x3 -3(m +1)x2 +12mx -3m + 4 (C) 1) Kho st s bin thin v v th ca hm s m = 0. 2) Tm m hm s c hai cc tr l A v B sao cho hai im ny cng vi im C -1;- 9 lp thnh tam gic nhn gc ta O lm trng tm. 2 Ta c y' = 3x2 -3(m +1)x +12m . Hm s c hai cc tr y = 0 c hai nghim phn bit D = (m -1)2 > 0 m 1 (*). Khi hai cc tr l A(2;9m), B(2m;-4m3 +12m2 -3m + 4) . 28. Trang 23 www.MATHVN.com x = m -1 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s 2 + 2m -1= 0 1DABC nhn O lm trng tm 9 m = - (tho (*)). -4m3 +12m2 + 6m + 4 - = 0 2 Cu 39. Cho hm s y = f (x) = 2x3 + 3(m -3)x2 +11- 3m 2 (Cm ). 1) Kho st s bin thin v v th ca hm s khi m = 2. 2) Tm m hng. (Cm ) c hai im cc tr M1,M2 sao cho cc im M1,M2 v B(0; 1) thng y = 6x2 + 6(m - 3) . y = 0 x = 0 x = 3- m . Hm s c 2 cc tr m 3 (*). Chia f (x) cho f (x) ta c: f (x) = f (x) 1 x + m -3 - (m -3)2 x +11- 3m 3 6 fi phng trnh ng thng M1M2 l: y = -(m - 3)2 x +11- 3m M1,M2,B thng hng B M1M2 m = 4 (tho (*)). 1 3 2 2 Cu 40. Cho hm s y = x 3 - mx + (m -1)x +1 (Cm ) . 1) Kho st s bin thin v v th ca hm s khi m = 2 . 2) Tm m hm s c cc i, cc tiu v yC + yCT > 2 . Ta c: y = x2 - 2mx + m2 -1. 3 y = 0 x = m +1. -1< m < 0 yC + yCT > 2 2m - 2m + 2 > 2 m >1 . Cu 41. Cho hm s y = 1 x3 -(m +1)x2 + 4 (m +1)3 3 3 (1) (m l tham s thc). 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m cc im cc i v cc tiu ca th (1) nm v 2 pha (pha trong v pha ngoi) ca ng trn c phng trnh (C): x2 + y2 - 4x + 3 = 0 . y = x2 - 2(m +1)x . y = 0 x = 0 x = 2(m +1) . Hm s c cc tr m -1 (1) Gi hai im cc tr ca th l: A 0; 4 (m +1)3 , B(2(m +1);0) . 3 (C) c tm I(2; 0), bn knh R = 1. IA = 4 + 16 (m +1)6 , IB = 9 4m2 . A, B nm v hai pha ca (C) (IA2 - R2 )(IB2 - R2 ) < 0 4m2 -1< 0 - 1 < m < 1 2 2 (2) Kt hp (1), (2), ta suy ra: - 1 < m < 1 . 2 2 Cu 42. Cho hm s y = x3 -3mx2 + 3(m2 -1)x - m3 (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. y = 3x2 - 6mx + 3(m2 -1); y = 0 x = m +1 x = m -1 29. Trang 24 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng im cc i M(m -1;2 -3m) chy trn ng thng c nh: x = -1+ t y = 2 - 3t im cc tiu N(m +1;-2 - m) chy trn ng thng c nh: x = 1+ t y = -2 -3t Cu 43. Cho hm s y = 1 x3 - mx2 - x + m +1 (C ) . 3 m 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm m th (Cm) c 2 im cc tr v khong cch gia 2 im cc tr l nh nht. Ta c: y = x2 - 2mx -1; y = 0 c D = m2 +1 > 0,"m fi hm s lun c hai im cc tr x1, x2 . Gi s cc im cc tr ca (Cm) l A(x1;y1),B(x2; y2 ) . Ta c: y = 1 (x - m).y - 2 (m2 +1)x + 2 m +1 3 3 3 fi y = - 2 (m2 +1)x + 2 m +1; y = - 2 (m2 +1)x + 2 m +1 1 3 1 3 2 3 2 3 Do : AB2 = (x - x )2 + (y - y )2 = (4m2 + 4) + 4 (m2 +1)2 4 1+ 4 2 1 2 1 1 9 9 fi AB 2 13 . Du "=" xy ra m = 0 . Vy min AB = 2 13 khi m = 0 . 3 3 Cu 44. Cho hm s y = x3 -3x2 - mx + 2 (1). 1) Kho st s bin thin v v th ca hm s (1) khi m = 0. 2) Tm m hm s (1) c 2 cc tr v ng thng i qua 2 im cc tr ca th hm s to vi hai trc to mt tam gic cn. y = 3x2 - 6x - m . Hm s c 2 cc tr y = 0 c 2 nghim phn bit m > -3. Ta c: y = 1 (x -1).y+ - 2m - 2 x + 2 - m fi ng thng D i qua 2 im cc tr ca 3 3 3 th c phng trnh: y = - 2m - 2 x + 2 - m . 3 3 D ct Ox, Oy ti A m - 6 ;0 , B 0; 6 - m (m 0). 2(m + 3) 3 Tam gic OAB cn OA = OB m - 6 = 6 - m m = 6; m = - 9 ;m 3 . = - 2(m + 3) 3 2 2 i chiu iu kin ta c m 3 . = - 2 Cu 45. Cho hm s : y = 1 x3 - mx2 + (m2 - m +1)x +1 (1). 3 1) Kho st s bin thin v v th ca hm s (1) khi m = 1. 2) Tm m hm s c cc tr trong khong (-;1) . Tp xc nh D = R. y = x2 - 2mx + m2 - m +1 . t t = x -1fi x = t +1 ta c : y' = g(t) = t2 + 2(1- m)t + m2 - 3m + 2 Hm s(1) c cc tr trong khong (-;1) f (x) = 0 c nghim trong khong (-;1) . 30. Trang 25 www.MATHVN.com Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s g(t) = 0 c nghim t < 0 P < 0 D' 0 S < 0 P 0 m2 -3m + 2 < 0 m -1 0 2m - 2 < 0 m2 -3m + 2 0 1< m < 2 Vy: Vi 1< m < 2 th hm s (1) c cc tr trong khong (-;1) Cu 46. Cho hm s : y = 1 x3 - mx2 + (m2 - m +1)x +1 (1). 3 1) Kho st s bin thin v v th ca hm s (1) khi m = 1. 2) Tm m hm s c cc tr trong khong (1;+) . Tp xc nh D = R. y = x2 - 2mx + m2 - m +1 . t t = x -1fi x = t +1 ta c : y' = g(t) = t2 + 2(1- m)t + m2 - 3m + 2 Hm s(1) c cc tr trong khong (1;+) f (x) = 0 c nghim trong khong (1;+) . g(t) = 0 c nghim t > 0 P < 0 D' 0 S > 0 P 0 m2 -3m + 2 < 0 m -1 0 2m - 2 > 0 m2 -3m + 2 0 1< m Vy: Vi m > 1 th hm s (1) c cc tr trong khong (1;+) Cu 47. Cho hm s : y = 1 x3 - mx2 + (m2 - m +1)x +1 (1). 3 1) Kho st s bin thin v v th ca hm s (1) khi m = 1. 2) Tm m hm s c hai cc tr x1, x2 tho mn x1 -3 . Cu 2. Cho hm s y = f (x) = x3 - mx2 + 2m (Cm) ( m l tham s). 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. Ta c: y = 3x2 - 2mx = x(3x - 2m) + Khi m = 0 th y = 3x2 0 fi (1) ng bin trn R fi tho yu cu bi ton. + Khi m 0 th (1) c 2 cc tr x = 0 , x = 2m . Do th ct Ox ti duy nht 1 im khi 1 2 3 4m3 2m2 m 0 f (x ).f (x )> 0 2m2m - > 0 4m2 1- > 0 1 2 27 27 - 3 6 < m < 3 6 2 2 Kt lun: khi m - 3 6 ; 3 6 th th (Cm) ct Ox ti duy nht mt im. 2 2 Cu hi tng t: a) y = x3 + 3(m +1)x2 + 3(m2 +1)x +1 S: m R . Cu 3. Cho hm s y = 2x3 -3(m +1)x2 + 6mx - 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. y = 6x2 - 6(m +1)x + 6m ; D = 9(m +1)2 -36m = 9(m -1)2 . y' + Nu m = 1 th y 0, "x fi hm s ng bin trn R fi th ct trc honh ti 1 im duy nht fi m = 1 tho mn YCBT. + Nu m 1 th hm s c cc im cc tr x1, x2 ( x1, x2 l cc nghim ca PT y = 0 ) fi x1 + x2 = m +1; x1x2 = m . Ly y chia cho y ta c: y = x - m +1 y - (m -1)2 x - 2 + m(m +1) . 3 6 fi PT ng thng i qua 2 im cc tr l: y = -(m -1)2 x - 2 + m(m +1) th hm s ct trc honh ti 1 im duy nht yC .yCT > 0 (-(m -1)2 x - 2 + m(m +1)).(-(m -1)2 x - 2 + m(m +1))> 0 1 2 (m -1)2 (m2 - 2m - 2) < 0 m2 - 2m - 2 < 0 (v m 1) 1- 3 < m 0 1 S = 2(m +1)> 0 m 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 5m = 4m + 4 m = 4 m +1+ m = 9(m +1- m ) 5 m = 4(m +1) -5m = 4m + 4 m 4 (tho (*)) = - 9 Vy m = 4; - 4 9 Cu hi tng t: a) Vi y = -x4 + 2(m + 2)x2 - 2m -3 S: m = 3,m 13 .= - 9 Cu 30. Cho hm s y = x4 -(3m +2)x2 + 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: x4 -(3m + 2)x2 + 3m = -1 x4 -(3m + 2)x2 + 3m +1 = 0 x = 1 x2 = 3m +1 (*) ng thng y = -1ct (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 0 < 3m +1< 4 - 1 < m 0 1 2 D = m2 > 0 f (0) = 2m +1 = 0 hoac f (3) = 4 - 4m 0 m = - 1 m 1 S = 2(m +1) < 3 Vy: m = - 1 m 1. 2 S = 2(m +1) > 0 2 P = 2m +1 > 0 Cu 32. Cho hm s y = x4 - 2m2 x2 + m4 + 2m (Cm), vi m l tham s. 1) Kho st s bin thin v v th ca hm s khi m =1.. 2) Chng minh th (Cm) lun ct trc Ox ti t nht hai im phn bit, vi mi m < 0 . PT honh giao im ca (Cm) vi trc Ox: x4 - 2m2 x2 + m4 +2m = 0 (1) t t = x2 (t 0) , (1) tr thnh : t2 - 2m2 t + m4 +2m = 0 (2) Ta c : D' = -2m > 0 v S = 2m2 > 0 vi mi m > 0 . Nn (2) c nghim dng fi (1) c t nht 2 nghim phn bit fi th hm s (1) lun ct trc Ox ti t nht hai im phn bit. Cu 33. Cho hm s y = x4 + 2m2 x2 +1 (m l tham s) (1) 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Chng minh rng ng thng y = x +1 lun ct th hm s (1) ti hai im phn bit vi mi gi tr ca m. Xt PT honh giao im: x4 + 2m2 x2 +1= x +1 x(x3 + 2m2 x -1) = 0 x = 0 g(x) = x3 + 2m2 x -1 = 0 (*) Ta c: g(x) =3x2 +2m2 0 gi tr ca m. (vi mi x v mi m ) fi Hm s g(x) lun ng bin vi mi Mt khc g(0) = 1 0. Do phng trnh (*) c nghim duy nht khc 0. Vy ng thng y = x +1 lun ct th hm s (1) ti hai im phn bit vi mi gi tr ca m. Cu 34. Cho hm s y = x4 -(m2 + 2)x2 + m2 +1 (Cm). 1) Kho st s bin thin v v th ca hm s khi m = 2 . 2) Tm cc gi tr ca m (Cm) ct trc honh ti 4 im phn bit sao cho hnh phng gii hn bi (Cm) vi trc honh phn pha trn trc honh c din tch bng 96 . 15 PT honh giao im ca (Cm) vi trc Ox: x4 -(m2 + 2)x2 + m2 +1 = 0 x = 1 x = m2 +1 52. Trang 46 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng fi (Cm) ct trc Ox ti 4 im phn bit m 0 (*). Khi : din tch hnh phng gii hn bi (Cm) vi trc honh phn pha trn trc honh 1 20m2 +16 96 l: S = (x4 -(m2 +2)x2 + m2 +1)dx = m = 2 (tho (*)). -1 15 15 Cu 35. Cho hm s y = x4 - 4x2 + m (Cm). 1) Kho st s bin thin v v th ca hm s khi m = 2 . 2) Tm cc gi tr ca m (Cm) ct trc honh ti 4 im phn bit sao cho hnh phng gii hn bi (Cm) vi trc honh c din tch phn pha trn trc honh bng din tch phn di trc honh. PT honh giao im ca (Cm) vi trc honh: x4 - 4x2 + m = 0(1) t = x2 ,t 0 t2 - 4t + m = 0 (2) (Cm) ct Ox ti 4 im phn bit (1) c 4 nghim phn bit (2) c 2 nghim dng D = 4 - m > 0 phn bit S = 4 > 0 0 < m < 4 (*). P = m > 0 Gi s (2) c nghim t1,t2 (0 < t1 < t2 ). Khi (1) c 4 nghim phn bit theo th t tng dn l: x1 = - t2 ;x2 = - t1;x3 = t1;x4 = t2 . Do tnh i xng ca (Cm) nn ta c: x3 x4 x5 4x4 (x4 - 4x2 + m)dx = (-x4 + 4x2 - m)dx 4 - 3 + mx = 0 3x4 - 20x2 +15m = 0 5 3 4 4 4 0 x3 x4 - 4x2 + m = 0 (3) m = 0 Suy ra x4 l nghim ca h: 4 4 4 2 20 . 3x4 - 20x4 +15m = 0 (4) m = 9 i chiu iu kin (*) ta suy ra m = 20 . 9 Cu 36. Cho hm s y = x4 - 2(m +1)x2 + 2m +1 (Cm). 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm cc gi tr ca m (Cm) ct trc honh ti 4 im phn bit A, B, C, D c honh ln lt l x1, x2,x3,x4 ( x1 < x2 < x3 < x4 ) sao cho tam gic ACK c din tch S = 4 , bit K(3;-2) . PT honh giao im ca (Cm) vi trc honh: x4 - 2(m +1)x2 + 2m +1 = 0 (1) . t t = x2 ,t 0 . (1) tr thnh: t2 - 2(m +1)t + 2m +1 = 0 (2) (Cm) ct Ox ti 4 im phn bit (2) c 2 nghim dng phn bit D = (m +1)2 -(2m +1) > 0 1 S = 2(m +1) > 0 m > - 2P = 2m +1 > 0 m 0 Khi (Cm) ct Ox ti 4 im phn bit c honh theo th t l: - t1;- t2 ; t2 ; t1 , vi t1 > t2 . Ta c: S = 1 AC.d(K, AC) (3), vi d(K, AC) = y = 2 . ACK 2 K Khi : (3) t1 + t2 = 4 t1 + t2 + 2 t1t2 =16 2(m +1) + 2 2m +1 = 16 m = 4 . 53. Trang 47 www.MATHVN.com f (x) = x2 + (4 - m)x +1- 2m = 0 (1) Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Dng 3: S tng giao ca th hm s: y = f (x) = ax + b cx + d Cu 37. Cho hm s y = 2x +1 x + 2 c th l (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Chng minh rng ng thng d: y = -x + m B. Tm m on AB c di nh nht. PT honh giao im ca (C) v d: lun ct th (C) ti hai im phn bit A, 2x +1 = -x + m x + 2 x -2 Do (1) c D = m2 +12 > 0 v f (-2) = (-2)2 + (4 - m).(-2)+1- 2m = -3 0, "m nn ng thng d lun lun ct th (C ) ti hai im phn bit A, B. 2 2 2 2 Ta c: yA = m - xA; yB = m - xB nn AB = (xB - xA ) + (yB - yA) = 2(m +12) Suy ra AB ngn nht AB2 nh nht m = 0 . Khi : AB = Cu hi tng t: 24 . a) y = x - 2 x -1 S: m = 2 b) y = x -1 2x S: m = 1 2 Cu 38. 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) sao cho I l trung im ca on MN. Phng trnh ng thng d : y = k(x +1) +1 v ct th (C) ti hai im M, N d ct (C) ti 2 im phn bit M, N x -3 = kx + k +1 c 2 nghim phn bit khc -1. x +1 k 0 f (x) = kx2 + 2kx + k + 4 = 0 c 2 nghim phn bit khc -1 D = -4k > 0 f (-1) = 4 0 k < 0 Mt khc: xM + xN = -2 = 2xI 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 39. Cho hm s y = 2x + 4 1- x (C). 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 (x -x )2 + (y - y )2 = 90 (a) 2 1 2 1 2x + 4 = k(x -1) +1 -x +1 kx2 -(2k -3)x + k + 3 = 0 (I). Ta c: (I) y = k(x -1)+1 [ y = k(x -1)+1 (I) c 2 nghim phn bit kx2 -(2k -3)x + k +3 = 0 (b) c 2 nghim phn bit. 54. Trang 48 www.MATHVN.com Kho st hm s k 0,k < 3 . 8 www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Ta bin i (a) tr thnh: (1+ k2 )(x - x ) 2 = 90 (1+ k2 ) (x + x ) 2 - 4x x = 90 (c) 2 1 2 1 2 1 Theo nh l Viet cho (b) ta c: x + x = 2k -3 , x x = k + 3 , th vo (c) ta c phng 1 2 k 1 2 k trnh: 8k3 +27k2 + 8k -3 = 0 (k + 3)(8k2 + 3k -1) = 0 k = -3; k = -3+ 41 ; k = -3 - 41 . 16 16 Kt lun: Vy c 3 gi tr ca k tho mn nh trn. Cu 40. Cho hm s y = 2x - 2 x +1 (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m ng thng (d): y = 2x + m ct (C) ti hai im phn bit A, B sao cho AB = 5 . PT honh giao im: 2x - 2 = 2x + m 2x2 + mx +m + 2 = 0 (x -1) x +1 (1) d ct (C) ti 2 im phn bit A, B (1) c 2 nghim phn bit x1, x2 khc 1 m2 -8m -16 > 0 (2) x + x m Khi ta c: 1 2 = - m 2 . Gi A(x1;2x1 + m) , B(x2;2x2 + m). x x = 1 2 + 2 2 AB2 = 5 (x - x )2 + 4(x - x )2 = 5 (x + x )2 - 4x x =1 m2 -8m - 20 = 0 1 2 1 2 1 2 1 2 m = 10 m = -2 (tho (2)) Vy: m = 10; m = -2 . Cu hi tng t: a) y = 2x -1 , d : y = x + m, AB = 2 2 . S: m = -1;m = 7. x + 2 Cu 41. Cho hm s y = x -1 x + m (1). 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 hai im A v B sao cho AB = 2 2 . ct th hm s (1) ti PT honh giao im: x -1 = x + 2 x -m x + m x2 + (m +1)x + 2m +1 = 0 (*) d ct th hm s (1) ti hai im A, B phn bit (*) c hai nghim phn bit khc -m D> 0 m2 - 6m -3 > 0 m < 3 - 2 3 m > 3 + 2 3 (**) x -m m -1 m -1 x1 + x2 = -(m +1) Khi gi x1, x2 l cc nghim ca (*), ta c x .x = 2m +1 1 2 Cc giao im ca d v th hm s (1) l A(x1; x1 + 2), B(x2; x2 + 2) . Suy ra AB2 = 2(x - x )2 = 2(x + x )2 - 4x x = 2(m2 - 6m -3) 55. Trang 49 www.MATHVN.com 1 2 1 2 1 2 56. Trang 50 www.MATHVN.com m = 7 D= k2 - 6k +1 > 0 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Theo gi thit ta c 2(m2 - 6m - 3) = 8 m2 - 6m - 7 = 0 m = -1 Kt hp vi iu kin (**) ta c m = 7 l gi tr cn tm. Cu 42. Cho hm s y = 2x +1 .x +1 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm cc gi tr ca tham s k sao cho ng thng (d): y = kx + 2k +1 ct th (C) ti hai im phn bit A v B sao cho cc khong cch t A v B n trc honh l bng nhau. PT honh giao im: 2x +1 = kx + 2k +1 x -1 x +1 kx2 + (3k -1)x + 2k = 0 (*) d ct (C) ti hai im phn bit A v B (*) c 2 nghim phn bit k 0 k 0 (**). Khi : A(x ;kx + 2k +1),B(x ;kx + 2k +1). k < 3- 2 2 k > 3+ 2 3 1 1 2 2 Ta c: d(A,Ox) = d(B,Ox) kx1 + 2k +1 = kx2 +2k +1 k = -3 (tho (**). k(x1 + x2 ) + 4k + 2 = 0 Cu 43. Cho hm s y = 2x . x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m ng thng d : y = mx - m + 2 di AB ngn nht. ct (C) ti hai im phn bit A, B sao cho PT honh giao im: 2x x -1 x 1 = mx - m + 2 2 g(x) = mx - 2mx + m - 2 = 0 (2) d ct (C) ti 2 im phn bit A, B (2) c 2 nghim phn bit khc 1 m > 0 2 2 2 Khi : A(x1;mx1 - m + 2), B(x2;mx2 - m + 2) fi AB = (1+ m) (x2 - x1) Theo nh l Viet, ta c: x + x = 2; x x = m - 2 fi AB2 = 8 m + 1 16 1 2 1 2 m m Du "=" xy ra m = 1. Vy min AB = 4 khi m = 1. Cu 44. Cho hm y = x +2 .2x - 2 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m ng thng d : y = x + m OA2 +OB2 = 37 . 2 ct (C) ti hai im phn bit A, B sao cho PT honh giao im ca (C) v d: x + 2 2x - 2 = x + m x 1 g(x) = 2x2 + (2m -3)x - 2(m +1) = 0 . D = 4m2 V g + 4m + 25 > 0,"m nn d lun ct (C) ti 2 im phn bit A, B. g(1) = 3 0 x x 2m -3 + = - Gi A(x1; x1 + m), B(x2; x2 + m) . Theo nh l Viet, ta c: 1 2 2 57. Trang 51 www.MATHVN.com x1x2 = -(m +1) 58. Trang 50 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Ta c: OA2 + OB2 = 37 1 (4m2 + 2m +17) = 37 m = - 5 ; m = 2 . 2 2 2 2 Cu 45. Cho hm y = x . 1- x 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m ng thng d : y = mx - m -1 ct (C) ti hai im phn bit M, N sao cho AM2 + AN2 t gi tr nh nht, vi A(-1;1) . PT honh giao im ca (C) v d: x = mx - m -1 x 1 1- x mx2 - 2mx + m +1 = 0 (2) d ct (C) ti 2 im phn bit (2) c 2 nghim phn bit khc 1 m < 0 . Gi I l trung im ca MN fi I(1;-1) c nh. 2 Ta c: AM2 + AN2 = 2AI2 + MN . Do AM2 + AN2 nh nht MN nh nht 2 MN2 = (x - x )2 (1+ m)2 = -4m - 4 8 . Du "=" xy ra m = -1. 2 1 m Vy: min(AM2 + AN2 ) = 20 khi m = -1. Cu 46. 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. PT honh giao im ca (C) v d: x2 + (m -3)x +1- m = 0, x 1 (*) (*) c D = m2 - 2m +5 > 0, "m R v (*) khng c nghim x = 1. fi (*) lun c 2 nghim phn bit l x , x . Theo nh l Vit: xA + xB = 3 - m A B x .x =1- m A B Khi : A(xA; xA + m), B(xB; xB + m) DOAB vung ti O th OA.OB = 0 xAxB + (xA + m)(xB + m)= 0 2xAxB + m(xA + xB )+ m = 0 m = -22 Vy: m = 2. Cu 47. Cho hm s y = f (x) = 2x +1 . x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm cc gi tr ca m sao cho ng thng (d): y = x + m ct (C) ti 2 im phn bit M, N sao cho din tch tam gic IMN bng 4 (I l tm i xng ca (C)). Tm i xng ca (C) l I(1; 2). Xt phng trnh honh giao im ca (d) v (C): 2x +1 = x + m x 1 x -1 2 f (x) = x + (m -3)x - m -1 = 0 d ct (C) ti 2 im phn bit M, N f (x) = 0 c hai nghim phn bit xM ,xN khc 1 D = m2 - 2m +13 > 0 (ng vi mi m). Ta cc giao im l M(xM ;yM ),N(xN ;yN ) . f (1) = -3 0 59. Trang 51 www.MATHVN.com x + x = 1+ m m -1 2 6 2 2 2 m < -1 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s 2 2 m -1 MN = 2(xM + xN ) - 4xM xN = 2(m -2m +13) ; d = d(I,d) = S = 4 1 MN.d = 4 m -1. IMN 2 2 m2 - 2m +13 = 8 m = 3; m = -1. Cu 48. Cho hm s y = -x + m x + 2 c th l (Cm) (m l tham s). 1) Kho st s bin thin v v th ca hm s khi m = 1. 2) Tm cc gi tr ca m ng thng d : 2x + 2y -1= 0 cho tam gic OAB c din tch bng 1 (O l gc ta ). ct (Cm) ti hai im A v B sao PT honh giao im ca d v (Cm): -x + m = 1 - x x2 - x + 2m - 2 = 0 (1), x -2 x + 2 2 d ct (Cm) ti 2 im A, B (1) c 2 nghim phn bit khc 2 -2 m < 9 8 (*) 1 1 Khi cc giao im l: A x1; - x1 , B x2; - x2 . AB = 2(9 - 8m) 1 1 1 1 7 SOAB = 2 AB.d(O,d) = 2 2(9 -8m). = 2 2 4 9 -8m =1 m = - 8 (tho (*)). Cu 49. Cho hm s y = 2x +1 x -1 c th l (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm cc gi tr m ng thng y = -3x + m ct (C) ti A v B sao cho trng tm ca tam gic OAB thuc ng thng d : x - 2y - 2 = 0 (O l gc ta ). PT honh giao im: 2x +1 = -3x + m x -1 3x2 -(1+ m)x + m +1 = 0 (1), (x 1) d ct (C) ti A v B (1) c 2 nghim phn bit khc 1 m > 11 (*) Gi x1, x2 l cc nghim ca (1). Khi A(x1;-3x1 + m),B(x2;-3x2 + m) Gi I l trung im ca AB fi xI = 1 2 uuur uur , yI = -3xI + m = Gi G l trng tm tam gic OAB fi OG = 2 OI fi G 1+ m ; m -1 3 9 3 Gd 1+ m - 2. m -1 - 2 = 0 m 11 (tho (*)). Vy m 11 . = - = - 9 3 5 5 Cu 50. Cho hm s y = x +3 x - 2 (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m ng thng d : y = -x + m +1 ct (C) ti hai im phn bit A, B sao cho AOB nhn. PT honh giao im ca (C) v d: x + 3 = -x + m +1 x - 2 x2 -(m + 2)x + 2m + 5 = 0 (x 2) (1) c 2 nghim phn bit D> 0 (1) m2 -4m +16 > 0 "m . 60. Trang 52 www.MATHVN.com x 2 2 2 - 2(m + 2) +2m + 5 0 61. Trang 52 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Gi A(x1;-x1 + m +1),B(x2;-x2 + m +1) l cc giao im ca (C) v d. Ta c: AOB nhn AB2 < OA2 +OB2 2(x - x )2 < (-x + m +1)2 + (-x + m +1)2 2 1 1 2 -2x x + (m +1)(x + x ) -(m +1)2 < 0 m > -3. 1 2 1 2 Cu 51. Cho hm s y = 3x +2 (C). x + 2 1) Kho st s bin thin v v th (C) ca hm s. 2) ng thng y = x ct (C) ti hai im A, B. Tm m ng thng d : y = x + m ct (C) ti hai im C, D sao cho ABCD l hnh bnh hnh. Honh cc im A, B l cc nghim ca PT: 3x + 2 = x x = -1 x + 2 x = 2 fi A(-1;-1),B(2;2) fi AB = 3 2 fi CD = 3 2 . PT honh giao im ca (C) v d: 3x + 2 = x + m x2 + (m -1)x + 2m - 2 = 0 (*) x + 2 d ct (C) ti 2 im phn bit D= m2 -10m + 9 > 0 0 m < 1 . x -2 m > 9 Khi cc giao im l C(c;c + m), D(b;b + m) vi a, b l cc nghim ca PT (*) CD = 3 2 2(c - d)2 = 3 2 m2 -10m = 0 m = 0 (loai) m = 10 Vy: m = 10 . Cu 52. Cho hm s y = x +3 . x + 2 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m ng thng d : y = 2x +3m ct (C) ti hai im phn bit A, B sao cho OA.OB = -4 vi O l gc to . PT honh giao im ca (C) v (d): x + 3 = 2x + 3m x + 2 2x2 + 3(1+ m)x + 6m -3 = 0 (1) (x 2) d ct (C) ti 2 im phn bit (1) c 2 nghim phn bit x1, x2 khc 2 D= 9m2 -30m + 33 > 0 "m 8- 6(1+ m) + 6m -3 0 uuur uuur 12m -15 7 Khi : A(x1;2x1 + 3m), B(x2;2x2 + 3m) . OA.OB = -4 2 = -4 m = 12 . Cu 53. 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 ca (C) v tha xA - yA + m = 0 . - y + m = 0 xB B Ta c: xA - yA + m = 0 yA = xA + m fi A,B(d): y = x + m - y + m = 0 y = x + m xB B B B fi A, B l giao im ca (C) v (d). Phng trnh honh giao im ca (C) v (d): 62. Trang 53 www.MATHVN.com B B B x 2 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s x + m = x + 2 f (x) = x2 + (m -3)x -(2m + 2) = 0 (x 2) x - 2 (*). (*) c D = m2 + 2m +17 > 0, "m fi (d) lun ct (C) ti hai im phn bit A, B. V 1. f (2) = -4 < 0 fi xA < 2 < xB hoc xB < 2 < xA (pcm). Cu 54. Cho hm s y = x + 2 .x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi d l ng thng i qua im A(1; 0) v c h s gc k. Tm k d ct (C) ti hai im phn bit M, N thuc hai nhnh khc nhau ca (C) sao cho AM = 2AN. PT ng thng d: y = k(x -1). PT honh giao im ca (C) v d: x + 2 = k(x -1) x -1 kx2 -(2k +1)x - 2 = 0 (x 1) (1) t t = x -1 x = t +1. Khi (1) tr thnh kt2 - t -3 = 0 (2) d ct (C) ti hai im phn bit M, N thuc hai nhnh khc nhau (1) c 2 nghim x1, x2 tho x1 0 (*). V A lun nm trong on MN v AM = 2AN nn AM = -2AN fi x1 + 2x2 = 3 (3) p dng nh l Viet cho (1) ta c: x + x = 2k +1 (4), x x = k - 2 (5) . 1 2 k 1 2 k T (3), (4) fi x = k + 2 ; x = k -1 . Thay vo (5) ta c: k = 2 (tho (*)). 1 k 2 k 3 Cu 55. Cho hm s y = 2x - m mx +1 (m l tham s) (1). 1) Kho st s bin thin v v th (C) ca hm s khi m = 1. 2) Chng minh rng vi mi m 0, th ca hm s (1) ct ng thng d : y = 2x - 2m ti hai im phn bit A, B thuc mt ng (H) c nh. ng thng d ct cc trc Ox, Oy ln lt ti cc im M, N. Tm m SDOAB = 3SDOMN . PT honh giao im ca (C) v (d): 2x - m = 2x - 2m mx +1 2mx2 - 2m2 x - m = 0 (2), D = m2 + 2 > 0 x - 1 m f (x) = 2x2 - 2mx -1= 0 (*), x - 1 m Xt PT (*) c 1 2 "m fi d lun ct (C) ti 2 im phn bit A, B. f - = +1 0 m m2 xA + xB = m 1 1 Ta c: xA.xB= - yA = xfi A fi A, B nm trn ng (H): y = 1 c nh. 1 xyA = 2xA - 2m y = 2x - 2m y = B h = d(O,d) = -2m = 2 m , AB = 5. m2 + 2 , M(m;0),N(0;-2m) 5 5 1 2 1 2 1 63. Trang 54 www.MATHVN.com fi SOAB = 2 h.AB = m m + 2 , SOMN = 2 OM.ON = m ; SOAB = 3SOMN m = . 2 64. Trang 54 www.MATHVN.com = - Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng KSHS 04: TIP TUYN A. Kin thc c bn ngha hnh hc ca o hm: o hm ca hm s y = f (x) ti im x0 l h s gc ca tip tuyn vi th (C) ca hm s ti im M0 (x0; f (x0 )). Khi phng trnh tip tuyn ca (C) ti im M0 (x0; f (x0 )) l: y y0 = f (x0).(x x0 ) (y0 = f (x0 )) iu kin cn v hai ng (C1): y = f (x) phng trnh sau c nghim: v (C2): y = g(x) tip xc nhau l h f (x) = g(x) f '(x) = g'(x) (*) Nghim ca h (*) l honh ca tip im ca hai ng . 2 Nu (C1) : y = px + q v (C2): y = ax + bx + c th (C1) v (C2) tip xc nhau phng trnh ax2 + bx + c = px + q B. Mt s dng cu hi thng gp c nghim kp. 1. Vit phng trnh tip tuyn D ca (C): y = f (x) ti im M(x0; y0 )(C) : Nu cho x0 th tm y0 = f (x0 ) . Nu cho y0 th tm x0 l nghim ca phng trnh f (x) = y0 . Tnh y = f (x) . Suy ra y(x0) = f (x0) . Phng trnh tip tuyn D l: y y0 = f (x0).(x x0 ). 2. Vit phng trnh tip tuyn D ca (C): y = f (x) , bit D c h s gc k cho trc. Cch 1: Tm to tip im. Gi M(x0; y0 ) l tip im. Tnh f (x0) . D c h s gc k fi f (x0) = k (1) Gii phng trnh (1), tm c x0 v tnh y0 = f (x0 ) . T vit phng trnh ca D. Cch 2: Dng iu kin tip xc. Phng trnh ng thng D c dng: y = kx + m . D tip xc vi (C) khi v ch khi h phng trnh sau c nghim: f (x) = kx + m f '(x) = k (*) Gii h (*), tm c m. T vit phng trnh ca D. Ch : H s gc k ca tip tuyn D c th c cho gin tip nh sau: + D to vi trc honh mt gc a th k = tan a . + D song song vi ng thng d: y = ax + b th k = a + D vung gc vi ng thng d : y = ax + b (a 0) th k 1 a + D to vi ng thng d : y = ax + b mt gc a th k - a 1+ ka = tana 3. Vit phng trnh tip tuyn D ca (C): y = f (x) , bit D i qua im Cch 1: Tm to tip im. Gi M(x0; y0 ) l tip im. Khi : y0 = f (x0 ), y(x0) = f (x0 ) . A(xA;yA ). Phng trnh tip tuyn D ti M: y y0 = f (x0).(x x0 ) D i qua A(xA;yA )nn: yA y0 = f (x0).(xA x0 ) (2) Gii phng trnh (2), tm c x0 . T vit phng trnh ca D. 65. Trang 55 www.MATHVN.com Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Cch 2: Dng iu kin tip xc. Phng trnh ng thng D i qua A(xA;yA )v c h s gc k: y yA = k(x xA ) D tip xc vi (C) khi v ch khi h phng trnh sau c nghim: f (x) = k(x - xA ) + yA f '(x) = k (*) Gii h (*), tm c x (suy ra k). T vit phng trnh tip tuyn D. 4. Vit phng trnh tip tuyn D ca (C): y = f (x) , bit D to vi trc Ox mt gc a. Gi M(x0; y0 ) l tip im. Tip tuyn c h s gc k = f (x0 ). D to vi trc Ox mt gc a f (x0 ) = tana . Gii phng trnh tm c x0 . Phng trnh tip tuyn D ti M: y y0 = f (x0).(x x0 ) 5. Vit phng trnh tip tuyn D ca (C): y = f (x) , bit D to vi ng thng d: y = ax + b mt gc a. Gi M(x0; y0 ) l tip im. Tip tuyn c h s gc k = f (x0 ). D to vi d mt gc a k - a 1+ ka = tana . Gii phng trnh tm c x0 . Phng trnh tip tuyn D ti M: y y0 = f (x0).(x x0 ) 6. Vit phng trnh tip tuyn Dca (C): y = f (x) , bit D ct hai trc to ti A v B sao cho tam gic OAB vung cn hoc c din tch S cho trc. Gi M(x0; y0 ) l tip im. Tip tuyn c h s gc k = f (x0 ). DOAB vung cn D to vi Ox mt gc 450 v O D.(a) SDOAB = S OA.OB = 2S . (b) Gii (a) hoc (b) tm c x0 . T vit phng trnh tip tuyn D. 8. Lp phng trnh tip tuyn chung ca hai th (C1) : y = f (x), (C2 ): y = g(x) . a) Gi D: y = ax + b l tip tuyn chung ca (C1) v (C2). u l honh tip im ca D v (C1), v l honh tip im ca D v (C2). D tip xc vi (C1) v (C2) khi v ch khi h sau c nghim: f (u) = au + b f '(u) = a g(v) = av + b g'(v) = a (1) (2) (3) (4) T (2) v (4) fi f (u) = g(v) fi u = h(v) Th a t (2) vo (1) fi b = k(u) (5) (6) Th (2), (5), (6) vo (3) fi v fi a fi u fi b. T vit phng trnh ca D. b) Nu (C1) v (C2) tip xc nhau ti im c honh x0 th mt tip tuyn chung ca (C1) v (C2) cng l tip tuyn ca (C1) (v (C2)) ti im . 9. Tm nhng im trn th (C): y = f (x) sao cho ti tip tuyn ca (C) song song hoc vung gc vi mt ng thng d cho trc. Gi M(x0; y0 ) (C). D l tip tuyn ca (C) ti M. Tnh f (x0) . V D // d nn f (x0 ) = kd 1 (1) hoc D ^ d nn f (x0) = - (2) kd Gii phng trnh (1) hoc (2) tm c x0 . T tm c M(x0; y0 ) (C). 10. Tm nhng im trn ng thng d m t c th v c 1, 2, 3, ... tip tuyn vi th (C): y = f (x) . 66. Trang 56 www.MATHVN.com f (x1). f (x2 ) < 0 Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Gi s d : ax + by + c = 0 . M(xM ;yM )d . Phng trnh ng thng D qua M c h s gc k: D tip xc vi (C) khi h sau c nghim: y = k(x xM ) + yM f (x) = k(x - xM ) + yM f '(x) = k (1) (2) Th k t (2) vo (1) ta c: f (x) = (x xM ). f (xM ) + yM (3) S tip tuyn ca (C) v t M = S nghim x ca (3) 11. Tm nhng im m t c th v c 2 tip tuyn vi th (C): y = f (x) tip tuyn vung gc vi nhau. v 2 Gi M(xM ;yM ) . Phng trnh ng thng D qua M c h s gc k: D tip xc vi (C) khi h sau c nghim: y = k(x xM ) + yM f (x) = k(x - xM ) + yM f '(x) = k (1) (2) Th k t (2) vo (1) ta c: f (x) = (x xM ). f (xM ) + yM (3) Qua M v c 2 tip tuyn vi (C) (3) c 2 nghim phn bit x1, x2 . Hai tip tuyn vung gc vi nhau f (x1).f (x2 ) = 1 T tm c M. Ch : Qua M v c 2 tip tuyn vi (C) sao cho 2 tip im nm v hai pha vi trc honh th (3) co 2 nghiem phan biet 67. Trang 57 www.MATHVN.com 0 0 0 0 0 0 0 0 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Dng 1: Tip tuyn ca th hm s bc ba y = ax3 + bx2 + cx + d Cu 1. Cho hm s y = 2x3 -3x2 +1. 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn (C) nhng im M sao cho tip tuyn ca (C) ti M ct trc tung ti im c tung bng 8. Gi s M(x ;y )(C) fi y = 2x3 -3x2 +1. Ta c: y = 3x2 -6x . PTTT D ti M: y = (6x2 - 6x )(x - x ) + 2x3 -3x2 +1. 0 0 0 0 0 D i qua P(0;8) 8 = -4x3 + 3x2 +1 x = -1. Vy M(-1;-4) . Cu 2. Cho hm s y = x3 -3x2 +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 - 6b a2 - 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 1 Ta c: AB = (b - a)2 + (b3 -3b2 +1- a3 + 3a2 -1)2 = 2 3 (b - a)2 + (b3 - a3 -3(b2 - a2 ))2 2 = (b - a) + (b - a) + 3ab(b - a) -3(b - a)(b + a) 2 2 2 2 = (b - a) + (b - a) (b - a) + 3ab -3.2 2 2 2 2 2 2 2 = (b - a) + (b - a) (b + a) - ab - 6 = (b - a) + (b - a) (-2 - ab) AB2 = (b - a)2 1+ (-2 - ab)2 = (2 - 2a)2 1+ (a2 - 2a - 2)2 2 2 2 2 4 2 = 4(a -1) 1 + (a -1) -3 = 4(a -1) (a -1) -6(a -1) +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: (*) t3 - 6t2 +10t -8 = 0 (t - 4)(t2 - 2t + 2) = 0 t = 4 fi(a -1)2 = 4 a = 3 fi b = -1 a = -1fi b = 3 Vy 2 im tho mn YCBT l: A(3;1), B(-1;-3). Cu hi tng t: a) Vi y = x3 -3x2 + 2; AB = 4 2 . S: A(3;2),B(-2;-2). Cu 3. Cho hm s y = f (x) = x3 + 6x2 + 9x + 3 (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm tt c cc gi tr k, tn ti 2 tip tuyn vi (C) phn bit v c cng h s gc k, ng thi ng thng i qua cc tip im ca hai tip tuyn ct cc trc Ox, Oy tng ng ti A v B sao cho OA = 2011.OB . 68. Trang 58 www.MATHVN.com PTTT ca (C) c dng: y = kx + m . Honh tip im x0 l nghim ca phng trnh: 69. Trang 58 www.MATHVN.com n1.n2 r n1 . r n2 Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng f (x ) = k 3x2 +12x + 9 - k = 0 (1) 0 0 0 tn ti 2 tip tuyn phn bit th phng trnh (1) phi c 2 nghim phn bit D = 9 + 3k > 0 k > -3 (2) fi To cc tip im (x0; y0 ) ca 2 tip tuyn l nghim ca h: y = x3 + 6x2 + 9x + 3 y = k - 6 x + 2k - 9 0 0 0 0 0 3 0 3 . 3x2 +12x + 9 = k 2 0 0 3x0 +12x0 + 9 = k fi Phng trnh ng thng d i qua cc tp im l: y = k - 6 x + 2k - 9 3 3 Do d ct cc trc Ox, Oy tng ng ti A v B sao cho: OA = 2011.OB nn c th xy ra: + Nu A O th B O . Khi d i qua O fi k = 9 . 2 + Nu A O th DOAB vung ti O. Ta c: tanOAB = OB = 2011 fi k - 6 = 2011 OA 3 fi k = 6039 (tho (2)) hoc k = -6027 (khng tho (2)). Vy: k = 9 ; k = 6039 . 2 Cu 4. Cho hm s y = x3 + (1- 2m)x2 + (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 gc a , bit cosa = 1 . 26 Gi k l h s gc ca tip tuyn fi tip tuyn c VTPT r = (k;-1) n1 ng thng d c VTPT r = (1;1) . n2 r r Ta c cosa = 1 = k -1 12k2 - 26k +12 = 0 k = 3 k = 2 26 2 k2 +1 2 3 YCBT tho mn t nht mt trong hai phng trnh sau c nghim: y = 3 3x2 + 2(1- 2m)x + 2 - m = 3 / 2 2 D 1 0 8m2 - 2m -1 0 y = 2 3x2 + 2(1- 2m)x + 2 - m = 2 D/ 0 4m2 - m - 3 0 2 3 3 m - 1 ;m 1 4 2 m 1 hoc m 1 - m - 3 ;m 1 4 2 4 Cu hi tng t: a) Vi y = x3 -3mx + 2; d : x + y + 7 = 0; cosa = 1 . S: m 2 . - 26 9 Cu 5. Cho hm s y = f (x) = 1 mx3 + (m -1)x2 + (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 + 2y -3 = 0 . (d) c h s gc - 1 fi tip tuyn c h s gc k = 2 . Gi x l honh tip im th: 2 70. Trang 59 www.MATHVN.com Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s f '(x) = 2 mx2 + 2(m -1)x + (4 -3m) = 2 mx2 + 2(m -1)x + 2 -3m = 0 YCBT (1) c ng mt nghim m. (1) + Nu m = 0 th (1) -2x = -2 x =1 (loi) + Nu m 0 th d thy phng trnh (1) c 2 nghim l x =1 hay x= 2 -3m m Do (1) c mt nghim m th 2 -3m < 0 m < 0 m hoac m > 2 3 Vy m < 0 hay m > 2 . 3 Cu 6. Cho hm s y = 1 mx3 + (m -1)x2 + (4m -3)x +1 3 (Cm). 1) Kho st s bin thin v v th (C) ca hm s khi m = 1. 2) Tm cc gi tr m sao cho trn (Cm) tn ti ng hai im c honh dng m tip tuyn ti vung gc vi ng thng d : x + 2y -3 = 0 . Ta c: y = mx2 + 2(m -1)x + 4 -3m ; d : y = - 1 x + 3 . 2 2 YCBT phng trnh y = 2 c ng 2 nghim dng phn bit mx2 + 2(m -1)x + 2 -3m = 0 c ng 2 nghim dng phn bit m 0 < < 1 D > 0 0 m 2 1 1 2 S > 0 1 2 . Vy m 0; 2 2 ; 3 . P > 0 < m < 2 3 Cu 7. Cho hm s y = x3 - mx + m -1 (Cm). 1) Kho st s bin thin v v th ca hm s khi m = 3 . 2) Tm m tip tuyn ca th (Cm) ti im M c honh x = -1 ct ng trn (C) c phng trnh (x - 2)2 + (y -3)2 = 4 theo mt dy cung c di nh nht. Ta c: y = 3x2 - m fi y(-1) = 3- m ; y(-1) = 2m - 2 . (C) c tm I(2;3) , R = 2. PTTT d ti M(-1;2m - 2) : y = (3 - m)x + m +1 (3 - m)x - y + m +1 = 0 2 d(I,d) = 4 - m = 1+ (3 - m) 2. (3 - m) +1 =(3 - m)2 +1 (3 - m)2 +1 (3 - m)2 +1 2 < R Du "=" xy ra m = 2 . D d(I,d) t ln nht m = 2 Tip tuyn d ct (C) ti 2 im A, B sao cho AB ngn nht d(I,d) t ln nht m = 2 Khi : PTTT d: y = x + 3. Cu hi tng t: 3 2 2 1 5 a) y = x -mx + m -1; xM =1;(C) :(x -2) + (y -3) = . S: m =1; m = . 5 2 Cu 8. Cho hm s y = 3x - x3 (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn ng thng (d): y = -x bit vi th (C). cc im M m t k c ng 2 tip tuyn phn Gi M(m;-m)d . PT ng thng D qua M c dng: y = k(x - m) - m . 71. Trang 60 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng 3x - x3 = k(x - m) - m (1) D l tip tuyn ca (C) h PT sau c nghim: (*) 3-3x2 = k (2) 3 Thay (2) vo (1) ta c: 2x3 -3mx2 + 4m = 0 m = 2x (**) 3x2 - 4 T M k c ng 2 tip tuyn vi (C) (**) c 2 nghim phn bit 2x3 2 3 2 3 Xt hm s f (x) = . Tp xc nh D = R- ; 3x2 - 4 3 3 6x4 - 24x2 x = 0 f (x) = ; f (x) = 0 (3x2 - 4)2 x = 2 Da vo BBT, (**) c 2 nghim phn bit m = -2 . Vy: M(-2;2) hoc M(2;-2).m = 2 Cu 9. Cho hm s y = x3 -3x + 2 . 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn ng thng d : y = 4 cc im m t k c ng 2 tip tuyn vi (C). Gi M(m;4)d . PT ng thng D qua M c dng: y = k(x - m) + 4 x3 -3x + 2 = k(x - m) + 4 (1) D l tip tuyn ca (C) h PT sau c nghim: (*) 3x2 - 3 = k (2) Thay (2) vo (1) ta c: (x +1)2x2 - (3m + 2)x +3m + 2 = 0 (3) x = -12x2 -(3m + 2)x + 3m + 2 = 0 (4) YCBT (3) c ng 2 nghim phn bit + TH1: (4) c 2 nghim phn bit, trong c 1 nghim bng 1 m = -1 + TH2: (4) c nghim kp khc 1 m = - 2 m = 2 3 Vy cc im cn tm l: (-1;4) ; - 2 ;4 ; (2;4) . 3 Cu 10. Cho hm s y = x3 - 2x2 + (m -1)x + 2m (Cm). 1) Kho st s bin thin v v th (C) ca hm s khi m = 1. 2) Tm m t im M(1;2) k c ng 2 tip tuyn vi (Cm). PT ng thng D qua M c dng: y = k(x -1) + 2 . D l tip tuyn ca (Cm) h PT sau x3 - 2x2 + (m -1)x + 2m = k(x -1) + 2 c nghim: 3x2 - 4x + m -1 = k fi f (x) = 2x3 - 5x2 + 4x -3(m -1) = 0 (*) qua M k c ng hai tip tuyn n (Cm) th (*) c ng 2 nghim phn bit Ta c f (x) = 6x2 -10x + 4 fi f (x) = 0 x =1; x = 2 3 fi Cc im cc tr ca (Cm) l: A(1;4 -3m), B 2 ; 109 - 3m . 3 27 m = 4 Do (*) c ng 2 nghim phn bit AOx 3 .B Ox 109 m = 81 72. Trang 61 www.MATHVN.com Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Cu 11. Cho hm s y = -x3 + 3x2 - 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 c dng : y = k(x - m) + 2 {-x D l tip tuyn ca (C) h PT sau c nghim 3 + 3x2 -2 = k(x - m) + 2 (1) (*). -3x2 + 6x = k (2) Thay (2) v (1) ta c: 2x3 -3(m +1)x2 + 6mx - 4 = 0 (x - 2)2x2 -(3m -1)x + 2 = 0 x = 2 f (x) = 2x2 -(3m -1)x + 2 = 0 (3) T M k c 3 tip tuyn n th (C) h (*) c 3 nghim x phn bit (3) c hai nghim phn bit khc 2 D > 0 m < -1 m > 5 3 . f (2) 0 5 m 2 Vy t cc im M(m; 2) (d) vi m < -1 m > 3 m 2 c th k c 3 tip tuyn vi (C). Cu hi tng t: a) y = -x3 + 3x2 - 2, d Ox . S: M(m;0) vi m > 2 2 -1 m < - 3 73. Trang 62 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Dng 2: Tip tuyn ca th hm s trng phng y = ax4 + bx2 + c Cu 12. Cho hm s y = f (x) = x4 - 2x2 . 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) = 4x3 - 4x H s gc tip tuyn ca (C) ti A v B l k = f '(a) = 4a3 - 4a, k = f '(b) = 4b3 - 4bA 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 = k 4a3 - 4a = 4b3 - 4b (a - b)(a2 + ab + b2 -1) = 0 (1) A B V A v B phn bit nn a b , do (1) a2 + ab + b2 -1= 0 (2) 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) f (a) - af (a) = f (b) - bf (b) -3a4 + 2a2 = -3b4 + 2b2 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 + b2 -1 = 0 a 1; a b Cu 13. Cho hm s y = x4 - 2mx2 + m (1) , m l tham s. 1) Kho st s bin thin v v th (C) ca hm s khi m = 1. 2) Gi A l mt im thuc th hm s (1) c honh bng 1. Tm m khong cch t im B 3 n tip tuyn ca th hm s (1) ti A l ln nht . 4 ; 1 A(Cm) nn A(1;1- m) . y' = 4x3 - 4mx fi y'(1) = 4 - 4m Phng trnh tip tuyn ca (Cm) ti A: y - (1- m) = y(1).(x -1) (4 - 4m)x - y - 3(1- m) = 0 -1 Khi d(B;D) = 1 , Du = xy ra khi m = 1. 16(1- m)2 +1 Do d(B;D) ln nht bng 1 khi v ch khi m = 1. Cu 14. Cho hm s y = ( x +1)2 .( x -1)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 = x4 - 2x2 +1. PT ng thng d i qua A(a;0) v c h s gc k : y = k(x - a) x4 - 2x2 +1 = k(x - a) d l tip tuyn ca (C) h phng trnh sau c nghim: (I) 4x3 - 4x = k Ta c: (I) k = 0 (A) hoc 4x(x -1) = k (B) 2 x2 -1= 0 f (x) = 3x2 - 4ax +1 = 0 (1) 74. Trang 63 www.MATHVN.com Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s + 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 phn bit khc 1 D = 4a2 -3 > 0 -1 a < - 3 hoac 1 a > 3 f (1) 0 2 2 75. Trang 64 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Dng 3: Tip tuyn ca th hm s nht bin y = ax + b cx + d Cu 15. Cho hm s y = 2x + 3 c th l (C). x +1 1) Kho st s bin thin v v th (C) ca hm s. 2) Lp phng trnh tip tuyn ca th (C) ti nhng im thuc th c khong cch n ng thng d :3x + 4y - 2 = 0 bng 2. Gi s M(x ;y )(C) fi y = 2x0 + 3 .0 0 0 x +10 3x + 4y - 2 Ta c: d(M,d) = 2 0 0 = 2 3x + 4y -12 = 0 hoc 3x + 4y + 8 = 00 0 0 0 32 + 42 2x + 3 x0 = 0 fiM1(0;3) Vi 3x0 + 4y0 -12 = 0 3x0 + 4 0 -12 = 0 1 1 11 x0 +1 x0 = 3 fi M2 3 ; 4 x = -5 fi M -5; 7 2x0 + 3 0 3 4 Vi 3x0 + 4y0 + 8 = 0 3x0 + 4 + 8 = 0 x0 +1 x = - 4 fi M - 4 ;-1 0 3 4 3 fi PTTT ti M (0;3) l y = -x + 3 ; PTTT ti M 1 ; 11 l y = - 9 x + 47 ;1 2 3 4 16 16 PTTT ti M -5; 7 l y = - 1 x + 23 ; PTTT ti M - 4 ;-1 l y = -9x -13.3 4 16 16 4 3 Cu 16. Cho hm s y = 2x -1 . x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Vit phng trnh tip tuyn ca (C), bit khong cch t im I(1; 2) n tip tuyn bng 2 . Tip tuyn ca (C) ti im M(x0; f (x0 ))(C) c phng trnh: y = f '(x )(x - x ) + f (x ) x + (x -1)2 y - 2x 2 + 2x -1 = 0 (*)0 0 0 0 0 0 2 - 2x x = 0 Khong cch t im I(1; 2) n tip tuyn (*) bng 2 0 = 2 0 1+ (x -1)4 x0 = 2 0 Cc tip tuyn cn tm : x + y -1= 0 v x + y - 5 = 0 Cu 17. 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. Tip tuyn (d) ca th (C) ti im M c honh a -2 thuc (C) c phng trnh: y = 4 (x - a) + 2a 4x -(a + 2)2 y + 2a2 = 0 (a + 2)2 a + 2 Tm i xng ca (C) l I (-2;2). Ta c: d(I,d) = 8 a + 2 8 a + 2 = 8 a + 2 = 2 2 16 + (a + 2)4 2.4.(a + 2)2 2 2 a + 2 76. Trang 65 www.MATHVN.com a = -4 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s d(I,d) ln nht khi (a + 2)2 = 4 a = 0 . T suy ra c hai tip tuyn y = x v y = x + 8. Cu hi tng t: a) Vi y = x x -1 . S: y = -x; y = -x + 4 . Cu 18. Cho hm s y = 2x +1 .x +1 1) Kho st s bin thin v v th (C) ca hm s. 2) Vit phng trnh tip tuyn ca th (C), bit rng tip tuyn cch u hai im A(2; 4), B(-4; -2). Gi x0 l honh tip im ( x0 -1). 1 2x +1 2 2 PTTT (d) l y = (x0 +1)2 (x - x0 ) + 0 x0 +1 x -(x0 +1) y + 2x0 + 2x0 +1 = 0 Ta c: d(A,d) = d(B,d) 2 - 4(x +1)2 + 2x2 + 2x +1 = -4 + 2(x +1)2 + 2x2 + 2x +1 0 0 0 0 0 0 x0 = 1 x0 = 0 x0 = -2 Vy c ba phng trnh tip tuyn: y = 1 x + 5 ; y = x +1; y = x + 5 4 4 Cu 19. Cho hm s y = 2x -1 .x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi I l giao im hai tim cn ca (C). Tm im M thuc (C) sao cho tip tuyn ca (C) ti M vung gc vi ng thng MI. Giao im ca hai tim cn l I(1; 2). Gi M(a; b) (C) fi b = 2a -1 (a 1) a -1 PTTT ca (C) ti M: y = - 1 (a -1)2 (x - a) + 2a -1 a -1 PT ng thng MI: y = 1 (x -1)+ 2(a -1)2 Tip tuyn ti M vung gc vi MI nn ta c: - Vy c 2 im cn tm M1(0; 1), M2(2; 3) 1 (a -1)2 . 1 (a -1)2 = -1 a = 0 (b =1) a = 2 (b = 3) (2 -1) - 2 Cu 20. Cho hm s y = m x m .x -1 1) Kho st s bin thin v v th (C) ca hm s khi m = 1. 2) Tm m th ca hm s tip xc vi ng thng y = x . TX: D = R{1}. 2 (2m -1)x - m = x - (*) th tip xc vi ng thng y = x th: x 2 1 T (**) ta c (m -1)2 = (x -1)2 x = m x = 2 - m (m - 1) (x - 1)2 77. Trang 66 www.MATHVN.com =1 (**) 78. Trang 66 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Vi x = m, thay vo (*) ta c: 0m = 0 (tho vi mi m). V x 1 nn m 1. Vi x = 2 m, thay vo (*) ta c: (2m -1)(2 - m) - m2 = (2 - m)(2 - m -1) 4(m -1)2 = 0 m = 1 fi x = 1 (loi) Vy vi m 1 th th hm s tip xc vi ng thng y = x . Cu 21. Cho hm s: y = x + 2 (C). x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Cho im A(0;a) . Tm a t A k c 2 tip tuyn ti th (C) sao cho 2 tip im tng ng nm v 2 pha ca trc honh. Phng trnh ng thng d i qua A(0;a) v c h s gc k: y = kx + a x + 2 = kx + a d l tip tuyn ca (C) H PT x -1 c nghim -3 k = (x -1)2 PT: (1- a)x2 + 2(a + 2)x -(a + 2) = 0 (1) c nghim x 1. qua A c 2 tip tuyn th (1) phi c 2 nghim phn bit x1, x2 a 1 a 1 (*) D = 3a + 6 > 0 a > -2 Khi ta c: x + x = 2(a + 2) ; x x = a + 2 v y = 1+ 3 ; y =1+ 3 1 2 a -1 1 2 a -1 1 x -1 2 x -11 2 2 tip im nm v 2 pha i vi trc honh th y1.y2 < 0 1+ 3 . 1+ 3 < 0 x1 .x2 + 2(x1 + x2 )+ 4 < 0 3a + 2 > 0 a 2 > - x1 -1 x2 -1 x1.x2 -(x1 + x2 ) +1 3 2 Kt hp vi iu kin (*) ta c: a 3 . > - a 1Cu 22. Cho hm s y = x + 2 . x +1 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi I l giao im ca 2 ng tim cn, D l mt tip tuyn bt k ca th (C). d l khong cch t I n D . Tm gi tr ln nht ca d. y = -1 . Giao im ca hai ng tim cn l I(1; 1). Gi s M x ; x0 + 2 (C) (x +1)2 0 x +1 0 Phng trnh tip tuyn D vi thi hm s ti M l: y = -1 (x - x ) + x0 + 2 x + (x +1) 2 y- x -(x +1)(x + 2)= 00 0 0 0 0 (x +1)2 x0 +1 0 Khong cch t I n D l d = 2 x0 +1 = 2 2 1 + (x +1)4 1 + (x +1)20 ( ) 2 0 x0 +1 Vy GTLN ca d bng 2 khi x0 = 0 hoc x0 = -2 . 79. Trang 67 www.MATHVN.com 1 < 0 fi - 1 1= - (x -1)2 (x -1)2 4 g x 0 -1 0 0 0 Trn S Tng Cu 23. Cho hm s www.MATHVN.com - Ton Hc Vit Nam y = -x +1 .2x -1 Kho st hm s 1) Kho st s bin thin v v th (C) ca hm s. 2) Chng minh rng vi mi m, ng thng d : y = x + m lun ct (C) ti 2 im phn bit A, B. Gi k1,k2 ln lt l h s gc ca cc tip tuyn vi (C) ti A v B. Tm m tng k1 + k2 t gi tr ln nht. 1 PT honh giao im ca d v (C): -x +1 = x + m x 2 D = m2 + 2m + 2 > 0, "m 2x -1 g(x) = 2x2 + 2mx - m -1= 0 (*) V 1 g 0 nn (*) lun c 2 nghim phn bit x1,x2 . 2 Theo nh l Viet ta c: x + x = -m;x x = -m -1 . Gi s: A(x ;y ),B(x ;y ) . 1 2 1 2 2 Tip tuyn ti A v B c h s gc l: k = - 1 ;k = - 1 1 2 2 1 1 (2x -1)2 2 (2x -1)2 fi k + k 1 2 = -4(m +1)2 - 2 -2 . Du "=" xy ra m = -1. 1 2 Vy: k1 + k2 t GTLN bng -2 khi m = -1. Cu 24. Cho hm s y = x + 2 2x + 3 (1). 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. Gi (x0; y0 ) l to ca tip im fi y (x ) = -1 < 00 (2x + 3)2 DOAB cn ti O nn tip tuyn D song song vi ng thng y = -x (v tip tuyn c h s gc m). Ngha l: y (x ) = x = -1fi y =1= -1 fi 0 (2x + 3)2 |x0 = -2 fi y0 = 0 + Vi x0 = -1; y0 =1 fi D: y -1 = -(x +1) y = -x (loi) + Vi x0 = -2; y0 = 0 fi D: y - 0 = -(x + 2) y = -x - 2 (nhn) Vy phng trnh tip tuyn cn tm l: y = -x - 2 . Cu 25. Cho hm s y = 2x -1 .x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Lp phng trnh tip tuyn ca th (C) sao cho tip tuyn ny ct cc trc Ox, Oy ln lt ti cc im A v B tho mn OA = 4OB. Gi s tip tuyn d ca (C) ti M(x0; y0 )(C) ct Ox ti A, Oy ti B sao cho OA = 4OB . Do DOAB vung ti O nn tan A = OB = 1 fi H s gc ca d bng 1 hoc - 1 . H s gc ca d l y (x0 ) = - OA 4 4 0 = -1 (y0 4 = 3 ) 2 5 80. Trang 68 www.MATHVN.com 0 0 x = 3 (y = ) 0 0 2 81. Trang 68 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng y = - 1 (x +1) + 3 y = - 1 x + 5 Khi c 2 tip tuyn tho mn l: 4 2 4 4 . y = - 1 (x -3) + 5 y = - 1 x + 13 4 2 4 4 Cu 26. Cho hm s y = 2x . x - 2 1) Kho st s bin thin v v th (C) ca hm s. 2) Vit phng trnh tip tuyn ca (C), bit tip tuyn ny ct cc trc Ox, Oy ln lt ti A v B sao cho AB = OA 2 . Gi M(x ;y )(C), x 2 . PTTT ti M: y = -4 (x - x ) + 2x0 0 0 0 (x - 2)2 0 x - 2 0 0 Tam gic vung OAB c AB = OA 2 nn DOAB vung cn ti O. Do d vung gc vimt trong hai ng phn gic d1 : y = x; d2 : y = -x v khng i qua O. + Nu d ^ d th -4 = -1 x = 4 fi d : y = -x + 8 .1 (x - 2)2 0 0 + Nu d ^ d th -4 =1 fi v nghim.2 (x - 2)2 0 Vy PTTT cn tm l: y = -x + 8 . Cu 27. Cho hm s y = x +1 . 2x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm gi tr nh nht ca m sao cho tn ti t nht mt im M (C) m tip tuyn ca (C) ti M to vi hai trc to mt tam gic c trng tm nm trn ng thng d : y = 2m -1. Gi M(x ;y )(C) . PTTT ti M: y = -3 (x - x ) + y0 0 (2x -1)2 0 0 0 2x2 + 4x -1 Gi A, B l giao im ca tip tuyn vi trc honh v trc tung fi yB = 0 0 . (2x -1)2 0 2x2 + 4x -1 2x2 + 4x -1 T trng tm G ca DOAB c: yG = 0 0 . V G d nn 0 0 = 2m -1 3(2x -1)2 3(2x -1)2 0 0 2x2 + 4x -1 6x2 -(2x -1)2 6x2 Mt khc: 0 0 = 0 0 = 0 -1 -1 (2x -1)2 (2x -1)2 (2x -1)2 0 0 0 Do tn ti t nht mt im M tho YCBT th 2m -1 - 1 m 1 . 3 3 Vy GTNN ca m l 1 . 3 Cu 28. Cho hm s y = 2x -3 (C). x - 2 1) Kho st s bin thin v v th (C) ca hm s. 2) Vit phng trnh tip tuyn ti im M thuc (C) bit tip tuyn ct tim cn ng v tim cn ngang ln lt ti A, B sao cho csin gc ABI bng 4 , vi I l giao 2 tim cn. 17 82. Trang 69 www.MATHVN.com 0 x0 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s I(2; 2). Gi M x0; 2x0 -3 (C) , x0 2 x0 - 2 Phng trnh tip tuyn D ti M: y = - 1 (x - x 2x -3) + 0 (x0 -2)2 0 - 2 Giao im ca D vi cc tim cn: A 2; 2x0 -2 , B(2x0 - 2;2) . Do cosABI = 4 nn tanABI = 1 = IA x0 - 2 IB2 = 16.IA2 (x -2)4 =16 x = 0 0 17 4 IB 0 x = 4 Kt lun: Ti M 0; 3 phng trnh tip tuyn: y = - 1 x + 3 2 4 2 Ti M 4; 5 phng trnh tip tuyn: y = - 1 x + 7 3 4 2 Cu hi tng t: a) y = 3x - 2 ;cosBAI = 5 . S: D: y = 5x - 2 hoc D: y = 5x + 2 . x +1 26 Cu 29. Cho hm s y = 2x -3 x - 2 c th (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn (C) nhng im M sao cho tip tuyn ti M ca (C) ct hai tim cn ca (C) ti A, B sao cho AB ngn nht. Ly im M m; 2 + 1 (C). Ta c: y(m) = - 1 m - 2 Tip tuyn (d) ti M c phng trnh: y = - (m - 2)2 1 (x - m) + 2 + 1 (m - 2)2 m - 2 Giao im ca (d) vi tim cn ng l: A 2;2 + 2 m - 2 Giao im ca (d) vi tim cn ngang l: B(2m - 2;2) Ta c: AB2 = 4 (m - 2)2 + 1 8. Du = xy ra m = 3 (m - 2)2 m =1 Vy im M cn tm c ta l: M(3;3) hoc M(1;1) Cu 30. Cho hm s 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi M l im bt k trn (C), I l giao im ca cc ng tim cn. Tip tuyn d ca (C) ti M ct cc ng tim cn ti A v B. Tm to im M sao cho ng trn ngoi tip tam gic IAB c din tch bng 2p . Ta c: I(2; 2). Gi M x0; 2x0 -3 (C), x0 2 . PTTT d: y = -1 (x - x0 ) + 2x0 -3 83. Trang 70 www.MATHVN.com 2x x0 - 2 -2 (x0 -2)2 x0 - 2 d ct 2 tim cn ti A 2; 0 , B(2x0 - 2;2) . x0 - 2 2 1 x0 =1fi M(1;1) DIAB vung ti I v S(IAB) = 2p (x0 - 2) + x = 2 2 x = 3 fi M(3;3) ( 0 - 2) 0 84. Trang 70 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Cu 31. Cho hm s y = 2x -3 . x - 2 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi M l im bt k trn (C). Tip tuyn ca (C) ti M ct cc ng tim cn ca (C) ti A v B. Gi I l giao im ca cc ng tim cn. Tm to im M sao cho ng trn ngoi tip tam gic IAB c din tch nh nht. Gi s M x ; 2x0 -3 (C) x 2 , y'(x ) = -1 0 x - 2 0 0 2 0 (x0 - 2) Phng trnh tip tuyn (D) vi ( C) ti M: y = -1 (x - x ) + 2x0 -3 2 0 (x0 - 2) x0 - 2 To giao im A, B ca (D) vi hai tim cn l: A 2; 2x0 - 2 ; B(2x - 2;2) x - 2 0 0 x + x 2 + 2x - 2 y + y 2x -3 Ta thy A B = 0 = x0 = xM , A B = 0 = yM fi M l trung im ca AB. 2 2 2 x0 - 2 Mt khc I(2; 2) v DIAB vung ti I nn ng trn ngoi tip tam gic IAB c din tch 2x -3 2 1 S = p IM2 = p (x - 2)2 + 0 - 2 = p (x - 2)2 + 2p 0 x - 2 0 (x - 2)2 0 0 Du = xy ra khi (x - 2)2 = 1 x0 = 1 0 (x - 2)2 x0 = 3 0 Do im M cn tm l M(1; 1) hoc M(3; 3). Cu hi tng t: a) Vi y = 3x + 2 .S: M(0;1), M(-4;5) . x + 2 Cu 32. Cho hm s y = 2mx + 3 . x - m 1) Kho st s bin thin v v th (C) ca hm s khi m = 1. 2) Gi I l giao im ca hai tim cn ca (C). Tm m tip tuyn ti mt dim bt k ca (C) ct hai tim cn ti A v B sao cho DIAB c din tch S = 64 . (C) c tim cn ng x = m , tim cn ngang y = 2m . Giao im 2 tim cn l I(m;2m) . 2mx0 + 3 2m2 + 3 2mx0 + 3 Gi M x0; x - m (C) . PTTT D ca (C) ti M: y = 2 (x - x0 ) + x - m . 0 (x0 - m) 0 2mx + 2m2 + 6 D ct TC ti A m; 0 , ct TCN ti B(2x0 - m;2m) . x0 - m 4m2 + 6 1 58 Ta c: IA = ; IB = 2 x - m fi S = IA.IB = 4m2 + 6 = 64 m = . x + m 0 IAB 2 20 Cu 33. Cho hm s y = x . x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Vit phng trnh tip tuyn ca (C), bit tip tuyn to vi 2 ng tim cn ca (C) mt tam gic c chu vi P = 2(2 + 2 ). (C) c tim cn ng x = 1, tim cn ngang y =1. Giao im 2 tim cn l I(1;1) . 85. Trang 71 www.MATHVN.com 0 1 x -1 0 0 2 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Gi xM x0; 0 (C) (x0 1) . PTTT D ca (C) ti M: y = - 1 x (x - x0 ) + 0 . x0 -1 x +1 (x0 -1)2 x0 -1 D ct TC ti A1; 0 , ct TCN ti B(2x0 -1;1) . x0 -1 2 2 1 Ta c: PIAB = IA + IB + AB = x + 2 x -1 + 2 (x -1) -1 0 0 + 4 + 2 2(x -1)2 Du "=" xy ra 0 0 x = 0 x0 -1 =1 0 . x0 =1 + Vi x0 = 0 fi PTTT D: y = -x ; + Vi x0 = 2 fi PTTT D: y = -x + 4 . Cu 34. Cho hm s y = 2x +1 x -1 c th (C). 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi I l giao im ca hai tim cn. Tm im M thuc (C) sao cho tip tuyn ca (C) ti M ct 2 tim cn ti A v B vi chu vi tam gic IAB t gi tr nh nht. Giao im ca 2 tim cn l I(1;2) . Gi M x ;2 + 3 (C). 0 0 + PTTT ti M c dng: y = -3 (x -1)2 (x - x0 ) + 2 + 3 x0 -1 + To cc giao im ca tip tuyn vi 2 tim cn: A 1;2 + 6 , B(2x -1;2) 0 1 1 6 x0 -1 + Ta c: SDIAB = 2 IA.IB = 2 x 2 x -1 = 2.3 = 6 (vdt) -1 0 + DIAB vung c din tch khng i fi chu vi DIAB t gi tr nh nht khi IA= IB 6 = 2 x x =1+ 3 -1 fi 0 x0 -1 x0 =1- 3 Vy c hai im M tha mn iu kin Khi chu vi DAIB = 4 3 + 2 6 . M (1+ 3;2 + 3) ) , M (1- 3;2 - 3) Ch : Vi 2 s dng a, b tho ab = S (khng i) th biu thc P = a + b + nht khi v ch khi a = b. a2 + b2 nh Tht vy: P = a + b + a2 + b2 2 ab + 2ab = (2 + 2) ab = (2 + 2) S . Du "=" xy ra a = b. Cu hi tng t: a) y = 2x -1 . S: M (0;-1),M (2;3) . x -1 1 2 Cu 35. Cho hm s y = x - 2 .x +1 1) Kho st s bin thin v v th (C) ca hm s. 2) Vit phng trnh tip tuyn ca (C), bit tip tuyn ct 2 tim cn ti A v B sao cho bn knh ng trn ni tip tam gic IAB l ln nht, vi I l giao im ca 2 tim cn. 86. Trang 72 www.MATHVN.com (C) c TC x = -1, TCN y =1. Giao im 2 tim cn l I(-1;1) . 87. Trang 72 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng Gi M x ; x0 - 2 (C) . PTTT D ca (C) ti M: y = 3 (x - x ) + x0 - 2 . 0 x +1 2 0 x +1 0 (x0 +1) 0 D ct hai tim cn ti A -1; x0 - 5 ,B(2x +1;1) . Ta c: IA = 6 ; IB = 2 x +1 . x +1 0 x +1 0 0 0 fi S = 1 IA.IB = 6 . Gi p, r l na chu vi v bn knh ng trn ni tip ca DIAB. IAB 2 Ta c: S = pr fi r = S = 6 . Do r ln nht p nh nht. Mt khc DIABvung ti I nn: p p 2p = IA + IB + AB = IA + IB + IA2 + IB2 2 IA.IB + 2IA.IB = 4 3 + 2 6 . Du "=" xy ra IA = IB (x +1)2 = 3 x = -1 3 .0 0 + Vi x = -1- 3 fi PTTT D: y = x + 2(1+ 3) + Vi x = -1+ 3 fi PTTT D: y = x + 2(1- 3) Cu 36. Cho hm s y = 2x +1 . x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn hai nhnh ca th (C), cc im M, N sao cho cc tip tuyn ti M v N ct hai ng tim cn ti 4 im lp thnh mt hnh thang. Gi M(m;yM ), N(n;yN ) l 2 im thuc 2 nhnh ca (C). Tip tuyn ti M ct hai tim cn ti A, B. Tip tuyn ti N ct hai tim cn ti C, D. PTTT ti M c dng: y = y(m).(x - m) + y fi A 1; 2m + 4 ,B(2m -1;2) .M m -1 Tng t: C 1; 2n + 4 ,D(2n -1;2) . n -1 Hai ng thng AD v BC u c h s gc: k = -3 nn AD // BC. (m -1)(n -1) Vy mi im M, N thuc 2 nhnh ca (C) u tho mn YCBT. Cu 37. Cho hm s y = x + 3 . x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Cho im Mo(xo;yo ) thuc th (C). Tip tuyn ca (C) ti M0 ct cc tim cn ca (C) ti cc im A v B. Chng minh Mo l trung im ca on thng AB. M (x ;y ) (C) fi y = 1+ 4 . PTTT (d) ti M0 : y - y = - 4 (x - x ) o o o 0 x -1 0 (x -1)2 0 0 0 Giao im ca (d) vi cc tim cn l: A(2x0 -1;1), B(1;2y0 -1) . fi xA + xB = x ; yA + yB = y fi M l trung im AB. 2 0 2 0 0 Cu 38. Cho hm s : y = x + 2 (C) x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Chng minh rng mi tip tuyn ca th (C) u lp vi hai ng tim cn mt tam gic c din tch khng i. 88. Trang 73 www.MATHVN.com 2 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s Gi s M a; a + 2 (C). a -1 PTTT (d) ca (C) ti M: y = y (a).(x - a) + a + 2 y = -3 x + a + 4a - 2 a -1 (a -1)2 (a -1)2 Cc giao im ca (d) vi cc tim cn l: A 1; a + 5 , B(2a -1;1) . a -1 6 IA = 0; 6 fi IA = ; IB = (2a - 2;0) fi IB = 2 a -1 a -1 a -1 1Din tch DIAB : S DIAB = Cu hi tng t: IA.IB = 6 (vdt) fiPCM. 2 a) y = 2x - 4 x +1 S: S = 12. Cu 39. Cho hm s y = 2x -1 .1- x 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi I l giao im ca hai ng tim cn, A l im trn (C) c honh l a. Tip tuyn ti A ca (C) ct hai ng tim cn ti P v Q. Chng t rng A l trung im ca PQ v tnh din tch tam gic IPQ. I(1;-2), A a; 2a -1 . PT tip tuyn d ti A: y = 1 (x - a) + 2a -1 1- a (1- a)2 1- a Giao im ca tim cn ng v tip tuyn d: P 1; 2a 1- a Giao im ca tim cn ngang v tip tuyn d: Q(2a -1;-2) Ta c: xP + xQ = 2a = 2xA . Vy A l trung im ca PQ. IP = 2a 1- a + 2 = 2 1- - a ; IQ = 2(a -1) . Suy ra: S IPQ = 1 IP.IQ = 2 (vdt) 2 Cu 40. Cho hm s y = 2x -1 .x +1 1) Kho st s bin thin v v th (C) ca hm s. 2) Gi I l giao im ca hai ng tim cn ca (C). Tm trn th (C), im M c honh dng sao cho tip tuyn ti M vi th (C) ct hai ng tim cn ti A v B tho mn: IA2 + IB2 = 40 . (C) c TC: x = -1; TCX: y = 2 fi I(1; 2). Gi s M x0; 2x0 -1 (C), (x0 > 0). x0 +1 PTTT vi (C) ti M: y = 3 (x - x0 ) + 2x0 -1 fi A -1; 2x0 -4 , B((2x0 +1;2). (x0 36+1)2 + 4(x x0 +1 +1)2 = 40 89. Trang 74 www.MATHVN.com 0 x0 +1 IA2 + IB2 = 40 (x +1)2 0 x0 = 2 (y0 = 1) fi M(2; 1). x > 0 Cu 41. Cho hm s y = x +1 x -1 (C). 90. Trang 74 www.MATHVN.com Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn Oy tt c cc im t k c duy nht mt tip tuyn ti (C). Gi M(0;yo ) l im cn tm. PT ng thng qua M c dng: y = kx + yo (d) x +1 = kx + y (y -1)x2 - 2(y +1)x + y +1 = 0 (1) (d) l tip tuyn ca (C) x -1 o o o o (*) -2 x 1; -2 = k = k 2 (x -1)2 (x -1) YCBT h (*) c 1 nghim (1) c 1 nghim khc 1yo =1 y 1 x = 1 ;y = 1fi k = -8 o o x = 1 D' = (y +1)2 -(y -1)(y +1) = 0 2 2 o o o x = 0;yo = -1fi k = -2 Vy c 2 im cn tm l: M(0; 1) v M(0; 1). Cu 42. Cho hm s y = x + 3 (C). x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm trn ng thng d : y = 2x +1 cc im t k c duy nht mt tip tuyn ti (C). Gi M(m;2m +1)d . PT ng thng D qua M c dng: y = k(x - m) + 2m +1 PT honh giao im ca D v (C): k(x - m) + 2m +1 = x + 3 x -1 kx2 -[(m +1)k - 2m]x +[mk - (2m + 4)]= 0 (*) D tip xuc vi (C) (*) c nghim kp k 0 2 D = [(m +1)k - 2m] - 4k[mk - (2m + 4)]= 0 k 0 g(k) = (m -1)2 k2 - 4(m2 - m - 4)k + 4m2 = 0 Qua M(m;2m +1)d k c ng 1 tip tuyn n (C) D = -32(m2 - m - 2) > 0;g(0) = 4m2 = 0 g(k) = 0 c ng 1 nghim k 0 D = -32(m2 - m - 2) > 0;g(0) = 4m2 = 0 m -1= 0 fi16k + 4 = 0 fi k 1 = - 4 m = 0 fi M(0;1) m = -1 fiM(-1;-1) m = 2 fiM(2;5) m =1 fiM(1;3) 91. Trang 75 www.MATHVN.com 12 2 2 2 2 Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s KSHS 05: BIN LUN SNGHIM CA PHNG TRNH Cu 1. Cho hm s y = -x3 + 3x2 +1. 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm m phng trnh x3 -3x2 = m3 -3m2 c ba nghim phn bit. PT x3 -3x2 = m3 -3m2 -x3 + 3x2 +1 = -m3 + 3m2 +1. t k = -m3 + 3m2 +1 S nghim ca PT bng s giao im ca th (C) vi ng thng d: y = k Da vo th (C) ta c PT c 3 nghim phn bit 1< k < 5 m(-1;3){0;2} Cu 2. Cho hm s y = x3 -3x2 + 2 . 1) Kho st s bin thin v v th (C) ca hm s. 2) Bin lun theo m s nghim ca phng trnh : x2 - 2x - 2 = m .x -1 Ta c x2 - 2x - 2 = m x -1 (x2 - 2x - 2) x -1 = m,x 1. Do s nghim ca phng trnh bng s giao im ca y = (x2 - 2x - 2) x -1 , (C') v ng thng y = m,x 1. Vi y = (x2 - 2x - 2) x -1 = f (x) khi x >1 nn (C') bao gm: - f (x) khi x 0) x4 - 2x2 +1+ log m = 0 x4 - 2x2 +1 = -log m (*) + S nghim ca (*) l s giao im ca 2 th y = x4 - 2x2 +1 v y = -log m + T th suy ra: 0 < m < 1 2 m = 1 2 1 < m 1 2 nghim 3 nghim 4 nghim 2 nghim v nghim Cu 5. Cho hm s y = f (x) = 8x4 - 9x2 +1. 1) Kho st s bin thin v v th (C) ca hm s. 2) Da vo th (C) hy bin lun theo m s nghim ca phng trnh: 92. Trang 76 www.MATHVN.com m < 0 m = 0 0 < m < 1 1 m < 81 32 m = 81 32 m > 81 32 v nghim 1 nghim 2 nghim 4 nghim 2 nghim v nghim x +1 x -1 x +1 x -1 x +1 x -1 m < -1;m >1 m = -1 -1< m 1 2 nghim 1 nghim v nghim Kho st hm s www.MATHVN.com - Ton Hc Vit Nam Trn S Tng 8cos4 x - 9cos2 x + m = 0 vi x [0;p ] Xt phng trnh: 8cos4 x - 9cos2 x + m = 0 vi x [0;p ] (1) t t = cos x , phng trnh (1) tr thnh: 8t4 - 9t2 + m = 0 (2) V x [0;p ] nn t [-1;1], gia x v t c s tng ng mt i mt, do s nghim ca phng trnh (1) v (2) bng nhau. Ta c: (2) 8t4 - 9t2 +1 =1- m (3) Gi (C1): y = 8t4 - 9t2 +1 vi t [-1;1] v (d): y =1- m . Phng trnh (3) l phng trnh honh giao im ca (C1) v (d). Ch rng (C1) ging nh th (C) trong min -1 x 1. Da vo th ta c kt lun sau: Cu 6. Cho hm s y = 3x - 4 (C). x - 2 1) Kho st s bin thin v v th (C) ca hm s. 2) Tm cc gi tr ca m phng trnh sau c 2 nghim trn on 0; 2p : 3 sin6 x + cos6 x = m (sin4 x + cos4 x) Xt phng trnh: sin6 x + cos6 x = m (sin4 x + cos4 x) (*) 1- 3 sin2 2x = m 1- 1 sin2 2x 4 -3sin2 2x = 2m(2 -sin2 2x) (1) 4 2 t t = sin2 2x . Vi x 0; 2p th t [0;1]. Khi (1) tr thnh: 2m = 3t - 4 vi t 0;1 3 t - 2 Nhn xt : vi mi t 0;1 ta c : sin2x = - t sin2x = t sin2x = t (*) c 2 nghim thuc on 0; 2p th t 3 ;1 fi t 3 ;1 3 2 4 Da vo th (C) ta c: y(1) < 2m y 3 1< 2m 7 1 < m 7 . 4 5 2 10 Cu 7. Cho hm s y = x +1 . x -1 1) Kho st s bin thin v v th (C) ca hm s. 2) Bin lun theo m s nghim ca phng trnh = m. S nghim ca = m bng s giao im ca th (C): y = v y = m. Da vo th ta suy ra c: 93. Trang 77 www.MATHVN.com + y = 2y I D = -y = -y Trn S Tng www.MATHVN.com - Ton Hc Vit Nam Kho st hm s KSHS 06: IM C BIT CA TH Kin thc c bn: 2 2 1) Khong cch gia hai im A, B: AB = (xB - xA ) + (yB - yA ) 2) Khong cch t im M(x0; y0 ) n ng thng D: ax + by + c = 0 : d(M,d) = ax0 + by0 + c c bit: + Nu D: x = a + Nu D: y = b a2 + b2 th d(M,D) = x0 - a th d(M,D) = y0 - b + Tng cc khong cch t M n cc trc to l: x0 + y0 . uuur uuur 3) Din tch tam gic ABC: S = 1 AB.AC.sin A = 1 2 2 AB2 .AC2 -(AB.AC)2 4) Cc im A, B i xng nhau qua im I IA + IB = 0 xA + xB = 2xI yA B I 5) Cc im A, B i xng nhau qua ng thng D AB ^ D c bit: + A, B i xng nhau qua trc Ox xB = xA yB