Ontaphamsobac3(2)
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Ontaphamsobac3(2)

Course Number: MATH 1002, Spring 2011

College/University: University of Western...

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ON TAP VE HAM SO BAC 3 (Trung tam Luyen thi ai hoc Vnh Vien) Gia s : y = ax3 + bx2 + cx + d vi a 0 co o th la (C). y = 3ax2 + 2bx + c, y = 6ax + 2b b 1) y = 0 x = (a 0 ) 3a b x= la hoanh o iem uon. o th ham bac 3 nhan iem uon 3a lam tam oi xng. 2) e ve o th 1 ham so bac 3, ta can biet cac trng hp sau : i) a > 0 va y = 0 vo nghiem ham so tang tren R (luon luon tang) ii) a < 0 va y = 0 vo nghiem ham so...

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TAP ON VE HAM SO BAC 3 (Trung tam Luyen thi ai hoc Vnh Vien) Gia s : y = ax3 + bx2 + cx + d vi a 0 co o th la (C). y = 3ax2 + 2bx + c, y = 6ax + 2b b 1) y = 0 x = (a 0 ) 3a b x= la hoanh o iem uon. o th ham bac 3 nhan iem uon 3a lam tam oi xng. 2) e ve o th 1 ham so bac 3, ta can biet cac trng hp sau : i) a > 0 va y = 0 vo nghiem ham so tang tren R (luon luon tang) ii) a < 0 va y = 0 vo nghiem ham so giam (nghch bien) tren R (luon luon giam) iii) a > 0 va y = 0 co 2 nghiem phan biet x 1, x2 vi x1 < x2 ham so at cc ai tai x1 va at cc tieu tai x2. Ngoai ra ta con co : + x1 + x2 = 2x0 vi x0 la hoanh o iem uon. + ham so tang tren (, x1) + ham so tang tren (x2, + ) + ham so giam tren (x1, x2) iv) a < 0 va y = 0 co 2 nghiem phan biet x 1, x2 vi x1 < x2 ham at cc tieu tai x 1 va at cc ai tai x 2 thoa ieu kien x 1 + x2 = 2x0 (x0 la hoanh o iem uon). Ta cung co : + ham so giam tren (, x1) + ham so giam tren (x2, + ) + ham so tang tren (x1, x2) 3) Gia s y = 0 co 2 nghiem phan biet va y = k(Ax + B)y + r x + q vi k la hang so khac 0; th phng trnh ng thang qua 2 iem cc tr la y = r x + q 4) (C) cat Ox tai 3 iem phan biet 2 ph b x1 y' = 0 conghieman iet , x2 y(x1).y(x2) < 0 5) Gia s a > 0 ta co : i) (C) cat Ox tai 3 iem phan biet > 2 ph b thoa y' = 0 conghieman iet < x1 < x2 y( ) < 0 y(x1).y(x2) < 0 ii) (C) cat Ox tai 3 iem phan biet < 2 ph b thoa y' = 0 conghieman iet x1 < x2 < y( ) > 0 y(x1).y(x2) < 0 Tng t khi a < 0 . 6) Tiep tuyen : Goi I la iem uon. Cho M (C). Neu M I th ta co ung 1 tiep tuyen qua M. Neu M khac I th ta co ung 2 tiep tuyen qua M. Bien luan so tiep tuyen qua 1 iem N khong nam tren (C) ta co nhieu trng hp hn. 7) (C) cat Ox tai 3 iem phan biet cach eu nhau y = 0 co 2 nghiem phan biet va y(x0) = 0 (x0 la hoanh o iem uon) 8) Bien luan so nghiem cua phng trnh : ax 3 + bx2 + cx + d = 0 (1) (a 0) khi x = la 1 nghiem cua (1). Neu x = la 1 nghiem cua (1), ta co ax3 + bx2 + cx + d = (x - )(ax2 + b1x + c1) nghiem cua (1) la x = vi nghiem cua phng trnh ax 2 + b1x + c1 = 0 (2). Ta co cac trng hp sau: i) neu (2) vo nghiem th (1) co duy nhat nghiem x = ii) neu (2) co nghiem kep x = th (1) co duy nhat nghiem x = iii) neu (2) co 2 nghiem phan biet th (1) co 3 nghiem phan biet iv) neu (2) co 1 nghiem x = va 1 nghiem khac th (1) co 2 nghiem. v) neu (2) co nghiem kep th (1) co 2 nghiem BAI TAP ON VE HAM BAC 3 Cho ho ng cong bac ba (C m) va ho ng thang (Dk) lan lt co phng trnh la y = x3 + mx2 m va y = kx + k + 1. (I) PHAN I. Trong phan nay cho m = 3. Khao sat va ve o th (C) cua ham so. 1) Goi A va B la 2 iem cc ai va cc tieu cua (C) va M la iem bat ky tren cung AB vi M khac A , B . Chng minh rang tren (C) ta tm c hai iem tai o co tiep tuyen vuong goc vi tiep tuyen tai M vi (C). 2) Goi la ng thang co phng trnh y = 1. Bien luan so tiep tuyen vi (C) ve t E vi (C). 3) Tm E e qua E co ba tiep tuyen vi (C) va co hai tiep tuyen vuong goc vi nhau. 4) nh p e tren (C) co 2 tiep tuyen co he so goc bang p, trong trng hp nay chng to trung iem cua hai tiep iem la iem co nh. 5) Tm M (C) e qua M ch co mot tiep tuyen vi (C). (II) PHAN I I.Trong phan nay cho tham so m thay oi. 6) Tm iem co nh cua (C m). nh m e hai tiep tuyen tai hai iem co nh nay vuong goc nhau. 7) nh m e (Cm) co 2 iem cc tr. Viet phng trnh ng thang qua 2 iem cc tr. 8) nh m e (Cm) cat Ox tai 3 iem phan biet. 9) nh m e : a) ham so ong bien trong (1, 2). b) ham so nghch bien trong (0, + ). 10) Tm m e (Cm) cat Ox tai 3 iem co hoanh o tao thanh cap so cong. 11) Tm ieu kien gia k va m e (D k) cat (Cm) tai 3 iem phan biet. Tm k e (Dk) cat (Cm) thanh hai oan bang nhau. 12) Viet phng trnh tiep tuyen vi (Cm) va i qua iem (-1, 1). 13) Chng minh rang trong cac tiep tuyen vi (C m) th tiep tuyen tai iem uon co he so goc ln nhat. BAI GIAI PHAN I : m = 3 Khao sat va ve o th (oc gia t lam) 1) Goi n la hoanh o cua M. V ham so at cc tieu tai x = 0 va at cc ai tai x = 2 nen 0 < n < 2; y' = 3x2 + 6x he so goc cua tiep tuyen tai M la k 1 = 3n2 + 6n (0, 3] (v n (0, 2)). ng thang vuong goc 1 vi tiep tuyen tai M co he so goc la k 2 = (vi k1 0 < k1 3). Hoanh o cua tiep tuyen vuong goc 1 vi tiep tuyen M la nghiem cua 3x 2 + 6x = (= k1 1 k2) 3x2 6x = 0. Phng trnh nay co a.c < 0, k1 k1 (0, 3] nen co 2 nghiem phan biet, k1 (0, 3]. Vay tren (C) luon co 2 iem phan biet ma tiep tuyen o vuong goc vi tiep tuyen tai M. 2) E (e, 1) . Phng trnh tiep tuyen qua E co dang y = h(x e) + 1 (D). (D) tiep xuc (C) he x3 + 3n2 3 = h(x e) + 1 co nghiem. 3x2 + 6x = h Phng trnh hoanh o tiep iem cua (D) va (C) la : x3 + 3x2 3 = ( 3x2 + 6x)(x e)+ 1 (1) 3 2 x + 3x 4 = x( 3x + 6)(x e) (x 2)(x2 x 2) = 3x(x 2)(x e) x = 2 hay x2 x 2 = 3x2 3ex x = 2 hay 2x2 (3e 1)x + 2 = 0 (2) 2 (2) co = (3e 1) 16 = (3e 5)(3e + 3) (2) co nghiem x = 2 8 2(3e 1) + 2 = 0 e = 2 5 Ta co > 0 e < 1 hay e > . 3 Bien luan : 5 i) Neu e < 1 hay < e < 2 hay e > 2 3 (1) co 3 nghiem phan biet co 3 tiep tuyen. 5 ii) Neu e = 1 hay e = hay e = 2 3 (1) co 2 nghiem co 2 tiep tuyen. 5 (1) co 1 nghiem co 1 tiep tuyen. 3 Nhan xet : T o th, ta co y = 1 la tiep tuyen tai (2, 1) nen phng trnh (1) chac chan co nghiem x = 2, e. 3) V y = 1 la tiep tuyen qua E (e, 1), e va ng x = la khong tiep tuyen nen yeu cau bai toan. (2) co 2 nghiem phan biet x 1, x2 thoa : y'(x1).y'(x2) = 1 5 e < 1 e > 3 nghiem (2) cua x1,x2 la (3x2 + 6x )(3x2 + 6x ) = 1 1 1 2 2 5 e < 1 hay e > 3 x + x = 3e 1 1 2 x .x = 1 2 12 9x1.x2(x1 2)(x2 2) = 1 5 e < 1 hay > e 3 9[1 (3e 1 + 4] = 1 ) 55 55 e= . Vay E ,1 27 27 4) Tiep iem cua tiep tuyen (vi (C)) co he so goc bang p la nghiem cua : y' = p 3x2 6x + p = 0 (3) Ta co ' = 9 3p > 0 p < 3 Vay khi p < 3 th co 2 tiep tuyen song song va co he so goc bang p. Goi x3, x4 la nghiem cua (3). Goi M3 (x3, y3); M4 (x4, y4) la 2 tiep iem. Ta co : x3 + x4 b = =1 2 2a 2 y3 + y4 (x3 + x3 ) + 3(x3 + x2) 6 3 4 4 = = 1 2 2 Vay iem co nh (1, 1) (iem uon) la trung iem cua M3M4. 5) Cach 1 : oi vi ham bac 3 (a 0) ta de dang chng minh c rang : iii) Neu 1 < e < M (C), ta co : i) Neu M khac iem uon, ta co ung 2 tiep tuyen qua M. ii) Neu M la iem uon, ta co ung 1 tiep tuyen qua M. Cach 2 : Goi M(x0, y0) (C). Phng trnh tiep tuyen qua M co dang : y = k(x x0) x3 + 3x2 3 (D) 0 0 Phng trnh hoanh o tiep iem cua (D) va (C) la : 3 2 x 3 + 3 x 2 3 = (3 x 2 + 6 x)( x x0 ) x0 + 3 x0 3 (5) 3 3 2 2 2 x x0 3(x x0) + (x x0)(3x + 6x) = 0 x x0 = 0 x2 + xx0 + x2 3x 3x0 3x2 + 6x = 0 0 2 x = x0 hay 2x (3+ x0)x x2 + 3x0 = 0 0 x = x0 hay (x x0)(2x + x0 3) = 0 3 x0 x = x0 hayx = 2 Do o, co ung 1 tiep tuyen qua M (x0, y0) (C) 3 x0 x0 = x0 = 1 2 Suy ra, y0 = 1. Vay M(1, 1) (iem uon). Nhan xet : v x0 la 1 hoanh o tiep iem nen pt (5) chac chan co nghiem kep la x0 Phan II : Tham so m thay oi. y' = 3x2 + 2mx 6) (Cm) qua (x, y), m y + x3 = m (x2 1) , m x2 1= 0 x = 1 x = 1 hay 3 y = 1 y + x = 0 y = 1 Vay (Cm) qua 2 iem co nh la H(1, 1) va K(1, 1). V y' = 3x2 + 2mx nen tiep tuyen vi (C m) tai H va K co he so goc lan lt la : a1 = y'(1) = 3 + 2m va a2 = y'(1) = 3 2m. 2 tiep tuyen tai H va K vuong goc nhau. 10 a1.a2 = 1 9 4m2 = 1 m = . 2 7) Ham co cc tr y' = 0 co 2 nghiem phan biet. 3x2 = 2mx co 2 nghiem phan biet. 2m x = 0 va x = la 2 nghiem phan biet. 3 m 0. Khi o, ta co : 1 2 1 y = m2x m + x my' 9 9 3 va phng trnh ng thang qua 2 cc tr la : 2 y = m2x m (vi m 0) 9 8) Khi m 0, goi x1, x2 la nghiem cua y' = 0, ta co : 2m x1.x2 = 0 va x1 + x2 = 3 2 2 2 2 y(x1).y(x2) = m x1 m m x2 m 9 9 2 4 = m2(x1 + x2) + m2 = m4 + m2 9 27 Vi m 0, ta co y(x1).y(x2) < 0 42 m +1 < 0 27 27 33 m> 4 2 Vay (Cm) cat Ox tai 3 iem phan biet. 2 phan x y'= 0 co nghiem biet1,x2 y(x1).y(x2) < 0 m2 > m> Nhan xet : 33 2 33 th phng trnh y = 0 co 2 nghiem am va 2 1 nghiem dng. 33 Khi m > th phng trnh y = 0 co 2 nghiem dng va 2 1 nghiem am. a) Ham ong bien tren (1,2) 3x2 + 2mx 0, x (1,2). Neu m 0 ta co hoanh o 2 iem cc tr la 0 2m va . 3 2m ,0 . Vay Neu m < 0 th ham ch ong bien tren 3 loai trng hp m < 0 Neu m = 0 ham luon nghch bien (loai). i) Khi m < ii) 9) i) ii) 2m iii) Neu m > 0 th ham ch ong bien tren 0, 3 2m , Do o, ycbt m > 0 va [12] 0, 3 2m 2 m 3 3 b) T cau a, ta loai trng hp m > 0. 2m Khi m 0 ta co ham so nghch bien tren , va 3 ham so cung nghch bien tren [0, + ). Vay e ham nghch bien tren [0, + ) th m 0. Ghi chu : nen lap bang bien thien e thay ro rang hn. y" = 6x + 2m , y" = 0 x = 10) m 3 (Cm) cat Ox tai 3 iem cach eu nhau. y = 0 co 3 nghiem phan biet va iem uon nam tren truc hoanh. 33 33 m> m> 2 2 m m3 m2 y = 0 27 + m. 9 m = 0 3 33 m> 3 6 2 m= 2 2 2m 1= 0 27 11) Phng trnh hoanh o giao iem cua (C m) va (Dk) la x3 + mx2 m = kx + k + 1 m(x2 1) = k(x + 1) + 1 + x3 x + 1 = 0 m(x 1) = k + 1 x + x2 x = 1 hay x2 (m + 1)x + k + m + 1 = 0 (11) a) Do o, (Dk) cat (Cm) tai 3 iem phan biet (11) co 2 nghiem phan biet khac 1 1+ m + 1+ k + m + 1 0 )2 ) (m + 1 4(k + m + 1 > 0 k 2m 3 2 (*) k < m 2m 3 4 b) V (Dk) qua iem K(1,1) (Cm) nen ta co : (Dk) cat (Cm) thanh 2 oan bang nhau. m 2m3 m cua (Cm) (Dk) qua iem uon ; 3 27 2m3 m m = k + 1 + 1 27 3 2m3 27 27 m k= (**) 9(m + 3) Vay ycbt k thoa (*) va (**). 12) Phng trnh tiep tuyen vi (Cm) i qua (1,1) co dang : y = k(x + 1) + 1 (Dk) Vay, phng trnh hoanh o tiep iem cua (D k) va (Cm) la : x3 + mx2 m = ( 3x2 + 2mx)(x + 1) + 1 (12) 2 2 3 m(x 1) = ( 3x + 2mx)(x + 1) + 1 + x x + 1 = 0 m(x 1) = 3x2 + 2mx + 1 x + x2 x = 1 hay 2x2 + (1 m)x m 1 = 0 (13) m+ 1 x = 1 x= 2 y' (1) = 2m 3 2 1 m + 1 m+ 1 m + 1 y' = 3 + 2m = (m2 2m 3) 4 2 2 2 Vay phng trnh cua 2 tiep tuyen qua (1, 1) la : y = (2m + 3)(x + 1) + 1 1 y = (m2 2m 3)(x + 1) + 1 4 Nhan xet : Co 1 tiep tuyen tai tiep iem (1, 1) nen phng trnh (12) chac chan co nghiem kep la x = 1 va phng trnh (13) chac chan co nghiem la x = 1. 13) Cac tiep tuyen vi (Cm) tai tiep iem cua hoanh o x co he so goc la : h = 3x2 + 2mx bm Ta co h at cc ai va la max khi x = = (hoanh 2a 3 o iem uon) Vay tiep tuyen tai iem uon co he so goc ln nhat. 2 m m2 m2 Nhan xet : 3x + 2mx= 3 x2 + 3 3 3 Ghi chu : oi vi ham bac 3 y = ax3 + bx2 + cx + d, ta co : i) Neu a > 0 th tiep tuyen tai iem uon co he so goc nho nhat. ii) Neu a < 0 th tiep tuyen tai iem uon co he so goc ln nhat. 2 PHAM HONG DANH (Trung tam Luyen thi ai hoc Vnh Vien)
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