Ta co: y = - x . y' - 2(a 2 + b2 )x + 2(a3 + b3 ) nen: y C D . VcT = (_2 (a 2 + b 2 ) X l + 2(a3 + b3 )) (-2(a 2 + b2 )x2 + 2(a 3 + b3 )) = 4(a 3 + b3 ) 2 - 4(a 2 + b2 ) 3 = - 4a 2 b2 (3a 2 + 3b2 - 2ab) = - 4a 2 b2 [2a 2 + 2b2 + (a - b) 2 ] < 0 Vay phuong trinh cho luon co 3 nghiem phan biet. V i du 3: Chung minh rang dieu kien can va du de phuong trinh x + px + q = 0 cd ba nghiem phan biet la: 4p3 + 27q < 0. Giai Xet ham sd f(x) = x3 + px + q, D = R. Ta cd f '(x) = 3x2 + p; f '(x) = 0 co 3x2 + p = 0. Vdi p < 0 thi f '(x) = 0 cd nghiem phan biet x = + , do la 2 hoanh do CD, CT. Dieu kien can va du de f(x) = 0 co 3 nghiem phan biet la yco-ycT < 0, p < 0 co 2 q p 3 2 Q + 3 P £ <Q, q<0co4p 3 + 27q2 <0. <0,p<0 . COq' p' 9 V i du 4:Tim tham sd m de phuong trinh: x3 - 3mx2 + 3(m2 - l)x - m2 + 1 = 0 cd 3 nghiem duong phan biet. Giai Xet y = x3 - 3mx2 + 3(m2 - l)x - m2 + 1 , D = R y' = 3x2 - 6mx + 3(m2 - 1) Cho y' = 0 => Xj = m - 1, x 2 = m + 1 (S = 2m, P = m2 - 1) Do dd ham sd ludn ludn cd CD, CT. Lay y chia y': y = - (x - m). y' - 2x + m3 - m2 - m + 1. 3 => y C T . yco = (-2x, + m3 - m2 - m + l)(-2x2 + m3 - m2 - m + 1) = (m2 - 1) (m2 - 3) (m2 - 2m - 1) Dieu kien cd 3 nghiem duong phan biet: f (0) < 0 x yCT-yCD < 0 CT' XCD > 0 - m 2 + 1 < 0 •• m-l>f) ; m + l> 0 ( m 2 -l)(m 2 -3)(m 2 -2m-l)< 0 Giai ra duoc S < m < 1 + V2. V i d „ 5; Tim m de phuong trinh: x3 + mx2 - 3 = 0 cd mot nghiem duy nhat. Giai Xet m = 0 thi PT: x3 - 3 = 0 co x = : cd nghiem duy nhat. Xet m * 0. Dat f(x) = x3 + mx2 - 3, D = R. -BDHSG DSGT12/1- 49 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ta co f '(x) = 3x2 + 2mx = x(3x + 2m) f '(x) = 0 co x = 0 hoac x = —51 c6 2 nghiem phan biet. 3 Phuong trinh f(x) = x3 + mx2 - 3 = 0 co duy nhat mot nghiem khi va chi khi cue dai va cue tieu cua ham sd cung dau: f(0)f - 2m > 0 co (—3) 8m3 4m3 0 — - + 3 27 9 > 0 co 8m3 - 12m3 + 81 > 0 co 4m3 < 81 co m < 3.3 - (m * 0). Vay gia tri can tim: m < 3.3/—. V4 V j du 6: Tim m de phuong trinh: 2 | x2 - 5x + 41 = x2 - 5x + m cd 4 nghiem. Giai Ta cd: 2 | x2 - 5x + 4 | = x2 ^ 5x + m co 2 | x2 - 5x + 4 | - x2 + 5x = m. Xet y = f(x) = 2 | x2 - 5x + 4 | - x2 + 5x , D = R [-3X2 + 15x - 8, l< x < 4 y = 2 [x^-5 x + 8, x <lvx> 4 f-6x +15khil <x< 4 [2x - 5 hhi x < 1, x > 4 Cho y - 0 => x = - 2 Bang bien thien: X -oo Vay dieu kien cd 4 nghiem la 4 < m < • Cach khac: dat t = x - 5x V i du 7: Bien luan theo tham so k ve sd nghiem cua phuong trinh: 2 x 4 -17x 3 + 51x2 -(36 + k)x + k = 0 (1) Giai Vdi moi k thi x = 1 ludn thoa man phuong trinh (1) 50 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ta co (1) o (x - 1) (2x3 - 15x2 + 36x - k) = 0 CO x = 1 hoac 2x3 - 15x2 + 36x - k = 0 (*) -Truong hop x = 1 la nghiem ciia (*) co k = 23 Khi do (*): 2x3 -15x2 + 36x- 23 = 0 eo (x -l)(2x 2 - 13x + 23) = 0 CO x = 1 hoac 2x2 - 13x + 23 = 0 eo x = 1 Vay khi k = 23 thi (1) cd nghiem duy nhat x = 1 -Vd i k * 23, khi dd x = 1 khdng phai la nghiem cua (*) nen so nghiem ciia (1) bang 1 cdng vdi sd nghiem cua phuong trinh (*) Xet f(x) = 2x3 -15x2 + 36x thi f '(x) = 6x2 - 30x + 36 = 6(x2 - 5x + 6) f '(x) = 0eo x = 2hay x = 3 g bien t X lien: —00 2 3 +00 f(x ) H 0 - 0 V f(x) —oo - 28 ^ ^ 2 7 ^ - +00 Dua vao bang bien thien ta cd: - Neu 23 * k < 27, k > 28 thi (*) cd nghiem duy nhat nen (1) cd 2 nghiem phan biet. - Neu k = 27 hay k = 28 thi (*) cd hai nghiem phan biet nen (1) cd ba nghiem phan biet. - Neu 27 < k < 28 thi (*) cd ba nghiem phan biet nen (1) cd bon nghiem phan biet. V i du 8: Tim m de phuong trinh sau cd nghiem x4 - 6x3 + mx2 - 12x + 4 = 0 (1) Giai x = 0 khdng phai la nghiem ciia phuong trinh (1) nen (1) co x — 6x + m 12 _4_ ( 2 4 ^ _ < 2^ X2 H - 6 x + - + m I x2 , f 2^ 2 J 21 CO x + - 6 x + - x ; I x j Dat t = x + - x j. = Ixl + 2_ I xl >2N/ 2 Tacd: t >2x/2) z - 6 t + m- 4 = 0 (2) Pt (1) cd nghiem co pt (2) cd nghiem thoa |t| > 2 V2 Xet (2) co t 2 - 6t - 4 = - m Dat f(t) = t 2 - 6t - 4 ; f '(t) = 2t- 6 = 0co t = 3 Lap BBT thi phuong trinh cd nghiem khi - m > -13 co m < 13. V i du 9: Chiing minh rang vdi gia tri tuy y n e Z1 , phuong trinh: ~ x2 xn , 1 + x + ~x + • • _ 0 khdng the cd nhieu hon 1 nghiem thuc. -BDHSG DSGT12/1- 51 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai 2 n Xet ham da thuc Pn (x) = 1 + x + — + +* - D = R. 2! ' n! " Ta chung minh quy nap: neu n chan thi Pn (x) nhan gia tri duong Vx e R cdn neu n le thi Pn (x) cd duy nhat 1 nghiem thuc khac 0. V di n = 0 ta cd P0 (x) = 1 > 0, Vx. Gia su khang dinh dung vdi gia tri be hon n, ta chung minh khang dinh dung tdi n. - Xet n le. Ta cd: P' n (x) = Pn -i(x) > 0, Vx nen P„(x) la ham tang, do do Pn (x) cd duy nhat 1 nghiem thuc khac 0 - Xet n chSn: P' n (x) = Pn -i(x) Da thuc Pn _i cd dung 1 nghiem thuc xo * 0 ( vi n- l le) Bang bien thien: X —CO xo +00 P' n (x) = Pn _,(x) - 0 + Pn(x) +00 +oo Do do: P„(x) > Pn (x0 ) = P^^xo) + ^ = > 0 n! n! Vay khang dinh duoc chung minh. V i du 10: Cho phuong trinh: ax3 + 27x2 + 12x + 2001 = 0 cd 3 nghiem phan biet. Hdi phuong trinh sau cd bao nhieu nghiem? 4(ax3 + 27x2 + 12x + 2001) (3ax + 27) = (3ax2 + 54x + 12)2 (1) Giai Xet f(x) = ax3 + 27x2 + 12x + 2001, D = R Theo gia thiet thi f(x) = 0 cd 3 nghiem a, p, y Ta cd f (x) = 3ax2 + 54x + 12, f'(x) = 6ax + 54, f"(x) = 6a (1) co 2f(x) f'(x) = (f( X ) ) 2 Xet g(x) = 2f(x) f'(x) - (f(x)) 2 =o g'(x) = 2f '(x).f (x) + 2f(x).f "(x) - 2f (x) f (x) = 2f(x).f "(x) = 12a2 (x - a)(x - P)(x - y), a < p < y Bang bien thien: X —00 a P Y +00 g' 0 + 0 0 f g V i B = +00 g(p)= * A -(f'(P)) 2 < o nen A < 0 va C < 0 +00 Vay phuong trinh cho cd dung 2 nghiem. V i du 11: Cho 2 sd a, b ma a + b > 0 Chung minh: ( a + - | < a + - VneN* I 2 J 2 52 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai Xet f(x) = x n + (c - x) n , c> 0 , D = R. f '(x) = n.x"" 1 - (c - x) n - 1 = n[x n " 1 - (c - x) n " '] , f '(x) = 0 cox"- 1 = (c-x) n _ 1 ~ c Va i n chan thi n - 1 le nen x = c - x<=>x = - Va i n le thi n - 1 chin nen x = +(c - x) co x BBT: X —OO dl +00 f — 0 -f f Ta co: f(x) > f(-) , Vx nen: xn + (c - x) n > (-) " 2 2 Chon x = a, c = a + b>0= > dpcm du 12: Cho i Chung minh V i du 12: Cho a, b, c > 0 thoa man a2 + b2 + c 2 = 1 a b c 3N/3 b2 + c2 c2 + a 2 a 2 + b2 Giai n n T a b c ^ 3\/3 BDT co - + — + - > 1-a 2 1-b2 1-c 2 2 co 3>/3 a (1-a 2 ) b(l-b 2 ) c(l-c 2 ) Xet f(x) = x (1 - x2 ) vdi x E (0; 1) f'(x) = 1 - 3x2 = 0 co x = 4 = e (0; 1) BBT: x f'(x ) f(x) -~XJ I V3" 3A/3 +00 Do do f(x) < -4 = , Vxe (0; 1) 3V3 Ap dung thi cd: a2 b2 3V3 3V3 a (1-a 2 ) b(l-b 2 ) c(l-c 2 ) -BDHSG DSGT12/1- > — (a 2 +b2 +c 2 ) 2 2 (dpcm). 53 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V j du 13: Cho y = ax3 + bx2 + cx + d (a * o). Chung minh rang neu ham s6 y'" 1 cd 2 cue tn thi: — < — ' " y' 2 y' Giai Ta cd y' = 3ax2 + 2bx + c, y"= 6ax + 2b, y"' = 6a nen (1) <=> b2 - 3ac> 0 Vi ham sd cd 2 cue tri nen Ay > 0 do dd b2 - 3ac > 0 Vay bat dang thuc (1) duoc chung minh. Vi du 14: Chung minh vdi n nguyen duong thi 1 - x + — - — + ... + (-1)' — + ... + ^— 2! 3! i ! (2n)! >0, Vx. Giai x2 x3 - x' X Xetf(x) = 1 -x + 4 T-4 T + - + (-1) — + •••+ ' 2! 3! Vdi x < 0 thi f(x) > 1 > 0 (dung). Vdi x > 2n thi: f(x) =1 + 2! .4 3 "N A X 4!~~1$!~ 3 l ! ( x2n (2n) , X 6 R (2n)! (2n-l) ! = 1 + — (x-2 ) + — (x-4) + ... + - (x-2n) > 1 > 0 (dung). 2! 4! (2n)! B Vdi 0 < x < 2n, do f lien tuc tren doan [0, 2n] nen tdn tai gia tri be nhat tai xo. Neu x0 = 0 hay x0 = 2n thi f(x) > f(x0) > 1 > 0. Neu x0 e (0, 2n) thi f dat cue tieu tai do. V2 V2 "" 1 Y2 n f '(x) = - l + x - — + ... + — = — 2! (2n-l) ! (2n)! f(x) V i f'(xo) = 0 nen f(x0 ) = —2— > 0 .Vay f(x) > f(x0 ) > 0 (dung) Vx. (znj. Vi du 15: Cho a, b, c la 3 sd ma phuong trinh: x3 + ax2 + bx + c = 0 co 3 nghiem phan biet. Chung minh: I 27c + 2a3 - 9ab I < 2^(a2 -3b)3 Giai Dat f(x) = x3 + ax2 + bx + c, D = R, f '(x) = 3x2 + 2ax + b. Vi f(x) = 0 cd 3 nghiem phan biet nen f (x) = 0 cd 2 nghiem phan biet. -a + Va2 -3b Xl -a-Va^S b , x 2 vdi a - 3b > 0 3 3 Va vi he sd cao nhat cua f duong nen yCD = f(xi) >0 va f(x2) = yCT < 0. Ta cd f(x) 1 1 •x + —a 9 1 ab f'(x) + -(3b-a 2 ) x + c - — 9 9 54 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
^f(x,)=|(3b-a 2 ) x i + c - ^ Tu f( X l) > 0 ^> -2V(a 2 -3b) 3 < 2a 3 + 27c - 9ab f(x2 ) < 0 => 2a 3 + 27c - 9ab < 2V(a 2 -3b) 3 Do vay: I 2a 3 + 27c - 9ab I < 2V(a 2 -3b) 3 V j du 16: Cho cac so thuc x, y thoa man 0 < x < ^ va 0 < y < Chung minh rang: cosx + cosy < 1 + cos(xy). Giai TC /— x + v TC x + y _ / Dox, y e [0;-]nenO < Vxy < — ^ < - =>cos-— < cos^/xy 3 Z o A x + v x —v x+y /— Ta cd cosx + cosy = 2cos -cos—- < 2cos— < 2cos^xy 2 2 " Xet ham sd f(t) = 1 + cost2 - 2cost vdi t e [0; |- ]. Ta cd f'(t) = 2(sint - tsint 2 ) nen f '(1) = 0, f(l) = 1 - cost Neu 0 < t < 1 thi t 2 < t < 1 nen tsint 2 < sint2 < sint, do dd f '(t) > 0. Neu 1 < t < - thi t < t 2 < - nen tsint 2 > sint2 > sint, do do f '(t) < 0. 3 2 BBT: X h - 1 - 1 y' 0 + 0 y _l-cos k ^ 0 ^ 9 Do cos— > 0 nen f(t) > 0, Vt e [0; - ] 9 3 => 2cos v'xy < 1 + cos(xy) => dpcm. V i du 17: Chung minh: x2 v + r'z + z 2 x < — . vdi x, y, z > 0, x + y + z = 1. 27 Giai Khdng mat tinh tdng quat, gia su: y = min{x, y, z} => 0 < y < - Taco f(x) = x"y+ y2 z + z 2 x = x2 y + y 2 (l-x-y ) + x(l- x - y) 2 = x3 + (3y - 2)x2 + (1 - 2y)x - y2 - y3 f '(x) = 3x2 + 2(3y - 2)x + 1 - 2y f '(x) = 0 <=> x = - hoac x = 1 - 2y > - 3 3 V i x = l - y - z^l _ y nen ta cd BBT: -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ta co f — --y(l-3 y + 3y 2 ) < — , va 27 3 27 f ( i - y) = yd-y) 2 - - -2y(i - y) d - y) * lf2x+i-y+i- y 4 27 Vay f(x) < — suy ra dpcm. C. BA I LUYE N TAP Bai 1: Tim cue tri cua ham so: a)y = xx/4-x2 b) y = xWl2-3x2 BS: a) CD x = V2 , yCD = 2 va CT x = yfe , ycr = "2. Bai 2: Tim cue tri cua ham sd: a) y = — + cos x 2 r 2x + 3 b) y = v3 sin x + cos x + —- — TC J \ i r T 5TC . _ r _ 5TC S DS: a) CDx= - + k2TC,yCD= —+ — -CTx=— + k2Tc,y C T- —- — 6 12 2 o i-o & Bai 3: Tim cue tri cua ham so a) y = X.N/X-1 b) y 2x2 -7 x + 5 x2 — 5x + 7 DS: b) CD(4;3), CT(2;-1). Bai 4: Tim m de ham sd: a) y = mx3 + 3x2 + 5x + 2 dat cue dai tai x = 2. b) y = - (m + l)x3 - (m + 2)x2 + (m + 3)x, m * -1 nhan gdc toa do lam 3 diem cue tieu. 17 DS: a)m = - — Bai 5: Tim m de ham so: b )m = -3. a) y = — X + m , m * 0 co cue dai, cue tieu ma y C T = 2yCD b)y mx - 1 1 - 2x + m x - m ddi vdi Ox. 1 cd cue dai, cue tieu va 2 diem cue tri nam cung phia DS: a) m b) m < 0. 56 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bai 6: Tim m ete ham so y= 2x3 + 3(m-3) x 2 +1 l-3m co cue dai, cue tieu va 3 diem cue dai, cue tiiu, B(0-1) thang hang DS: m = 4. Bai 7: Tim k de ham sd y = -2x +1 - Wx 2 +1 cd cue tieu. DS: k<-2 . .. , ' '1 1 3 Bai 8: Timcxde ham sdy = — x 3 — (sin a + cosa)x2 + — sin2a..x + 1 cd cue 3 2 4 dai, cue tieu va cac hoanh do cue tri thoa man xi + x2 = X] 2 + x 2 2 DS: — +k2rc, k2rc 2 Bai 9: Chung minh khi m < 1 thi ham sd: y = x3 - 3x2 + 3mx + 1 - m dat 2 y — y cue tri tai (xi, yi), (x2 , y2 ) va thoa man: (x, -x 2 )(x!X2 -1) Bai 10: Chung minh ham so y = |x2 + x - 20| khong cd dao ham nhung van dat cue tri tai x = -5 H D: Dung dinh nghTa dao ham. x2 + 2x + 3m Bai 11: Chung minh ham sd y = —- luon ludn cd cue dai, cue tieu x 2 - 2 x + 2 r ' ' va cac diem cue dai, cue tieu nam tren 1 dudng co dinh. n c x + 1 DS: y = — - x 1 , x2 +(m + l)x- m + l Bai 12: Cho ham so y = ; luon ludn cd cue dai, cue tieu. , x - m .. . Lap phuong trinh duong thang qua 2 diem cue tri do. DS: y = 2x - m Bai 13: Lap phuong trinh dudng thang qua diem cue dai va cue tieu cua ham so y = x3 - x2 - 94x + 95 566 671 DS: y = x + 9 9 Bai 14: Bien luan so nghiem phuong trinh: 2x4 - 17x3 + 51x2 - (36 + k) x + k = 0 theo k H D: y ' = (x - 1)( 2x3 -15x2 + 36x - k) Bai 15: Tim m de phuong trinh: x3 - 3mx2 + 3(m2 - 1) x - m2 + 1 = 0 co 3 nghiem duong phan biet. DS: S <m<l + j2 Bai 16: Tim a de 2 diem cue dai va cue tieu cua ham sd y = x3 - 3x2 + 2 nam trong va nam ngoai dudng trdn x2 + y2 - 2ax - 4ay + 5a2 - 1 = 0. DS: -<a<l. 5 Bai 17: Chung minh bat dang thuc: | cos2 x sin4 x + cos2x | < 1. -BDHSG DSGT12/1- 57 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
§ 3 . GI A TR I LO N NHA T V A GI A TR I N H O NHA T A. KIE N THU C CO BA N Gia su ham so f xac dinh tren tap hop D (D cz R). a) Neu ton tai mot diem x 0 e D sao cho f(x) < f(x0 ) vdi moi x e D thi s6 M = f(x 0 ) dugc goi la gia tri ldn nhat cua ham so f tren D, ki hieu la M = maxf(x) xeD b) Neu tdn tai mot diem x^ e D sao cho f(x)-> f(x0 ) vdi moi x e D thi so m = f(x 0 ) duoc gpi la gia tri nhd nhat cua ham sd f tren D, ki hieu la m = minf(x) B. PHAN DANG TOAN DANG 1: Tl M GIATRI LON NHAT, NHO NHAT - Ham sd lien tuc tren mpt doan thi dat duoc gia tri ldn nhat tren doan do. Phirong phap: Ddi vdi ham so y = f(x) tren D Tinh dao ham y' roi lap bang bien thien tu do cd ket luan ve GTLN, GTNN. Neu can thi dat an phu t = g(x) vdi dieu kien day du cua t. - Neu y = f(x) ddng bien tren doan [a;b] thi: min f(x) = f(a) va max f(x) = f(b) - Neu y = f(x) nghich bien tren doan [a;b] thi: min f(x) = f(b) va max f(x) = f(a) -Neu y = f(x) lien tuc tren doan [a;b] thi ta chi can tim cac nghiem x, cua dao ham f '(0)= 0 rdi so sanh ket luan: min f(x) = min { f(a); f(xi); f( x 2 );...; f(b) } max f(x) = max { f(a); f(xi); f(x2 );...; f(b) } Chuy: - Khi can thiet ta phdi hpp cac bat dang thuc dai sd. - Vdi ham y = | f(x) | thi GTLN tren 1 doan [a,b] la GTLN cua gia tri tuyet ddi cua gia tri CD, gia tri CT va 2 bien f(a), f(b). - Neu ham cd nhieu bien thi cd the chpn bien hoac ddn bien. V i du 1: Tim gia tri ldn nhat va gia tri nhd nhat cua cac ham sd: a) f(x) = x2 + 2x - 5 tren doan [-2; 3] b) f(x) = — + 2x2 + 3x - 4 tren doan [-4; 0] 3 c) f(x) = | x3 + 3x2 - 72x + 90 I tren doan [-5; 51. Giai 58 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
a) f'(x) = 2x + 2;f, ( x ) = 0<=>x = -l . Ta co f(-2) = -5, f(-l ) = -6; f(3) = 10. So sanh thi min f(x) = f(-l ) = -6 ; max f(x) = f(3) = 10 xe[-2;3] xe[-2;3] b) f '(x) = x2 + 4x + 3, f'(x) = 0 co x = - 1 hoac x = -3. Ta co f(-4) = , f(-3) = -4; f(-l) = , f(0) = -4. Vay rnin f(x) = f(-4) = f(-l) = -^; xe[-4;0] 3 max f (x) = f(-3) = f(0) = -4 xe[-4;0] c) Xet ham sd g(x) = x3 + 3x2 - 72x + 90 tren doan [-5; 5] g'(x) = 3x2 + 6x - 72; g'(x) = 0 <=> x = 4 hoac x = -6 (loai) f(-5) = 500; f(5) = -70; f(4) = -86. Do dd -86 < g(x) < 400, Vx e [-5; 5] va vi ham so g(x) lien tuc tren doan [-5; 5] nen 0 < f(x) = | g(x) | < 400. Vay min f (x) = 0 ; max f (x) = f (-5) = 400 . xe[-5;5] xe[-5;5] V i du 2: Tim gia tri ldn nhat va nhd nhat cua cac ham so a) f(x) 2x + 3 1 tren doan [-2; 0] b) f(x) = x + — tren khoang (0; +co) x - l x c)y x 4 + x2 tren R d)y 2x2 +2x + 3 x2 + x + 1 tren R. Giai -5 a) Tren doan [-2; 0], ta cd f '(x) < 0 . : Suy ra ham so f(x) nghich bien tren doan [-2; 0] Vay max f (x) = f (-2) = - ; min f (x) = f (0) = -3 XE[-2,0] 3 XE[-2;0] 1 y?-l b) f'(x) = 1- 4 =^r- • Vx > 0, f'(x) = 0 » x = ±1, chon x = 1. x2 x2 BBT X 0 i +oo y' - 0 + y +00^^^ ^ ^^j y +00 Vay min f(x) = f(1) = 2. Ham so khong dat gia tri ldn nhat. • J X€(0.«) ' c) y' (4 + x 2 ) 2 0 o x = +2. Lap BBT thi cd: maxy = f(2) = - ; miny = f(-2) = - - 4 4 -BDHSG DSGT12/1- 59 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
(x2 + x + l) z " BBT X 0 _i/ 2 +co y' + 0 y 10/3 2 - ^ 2 Vay maxy = — va khong ton tai GTNN. 3 Vi du 3: Tim gia tri ldn nhat va gia tri nhd nhat cua cac ham sd: a) f(x) = 73-2 x tren doan [-3; 1] b) f(x) = x + 74-x 2 c) f(x) a)f'(x ) = x + 6 tren doan [0; 1 ] d) f(x) = 7x + 3 + 76- x Giai - 1 73-2 x < 0 vdi moi x e [-3; 1] nen ham sd f nghich bien tren doan [-3; 1]. Vay max f(x) = f(-3) = 3 va min f(x) = f(l) = 1 xe[-3.l] xe[-3;l] b) Ham sd f xac djnh va lien tuc tren doan [-2; 2] f(x)=l - 7 w , vdi moi x e (-2; 2) f'(x) = 0<=> 1 0 <=> 74-x 2 = x <=> ! 0<x< 2 4-x2 = x2 x = 72. 74 - x2 Ta cd f( 72 ) = 2 72 ; f(-2) = -2; f(2) = 2. So sanh thi maxf(x)=2 72 va minf(x)=-2. xe[-2,2] xs[-2;2] c) Xet g(x) = -x2 + x + 6 tren doan [0; 1] thi g'(x) = -2x+l,g, (x) = 0ox=| 1 25 25 Ta cd g(0) = 6, g( —) = — . g(l) = 6 nen 6 < g(x) < — va vi g(x) lien tuc tren [0; 1] nen76 = TgOO <- =>- < f(x) = -7^= <4= 2 5 7g(x) 76 1 2 Vay maxf(x) = -= , minf(x) = —. ' -[o.i] 76 «t«*] 5 d) D = [-3; 6], vdi -3 < x < 6 thi 1 1 76 - x - 7x + 3 y = 27x7 3 276- x 27x + 3.76-x 60 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ta co y' > 0 <=> %/6-x > Vx + 3 <=> -3 < x < - . 2 Lap BBT thi maxy = f - ]= 3-s/2 , mmy = f(-3) = f(6) = 3. VJ du 4: Tim gia tri ldn nhat va nhd nhat cua ham so: a) f(x) = cos2 x + cosx + 3 b) y = cos2 2x - sinxcosx + 4 c)y cosx . rt 3rc tren - ; — 2 2 d) f(x) Giai 1 sin x tren doan TC 5rc 3 ; T a) Vi f(x) la ham so tuan hoan chu ky 2rc, nen ta chi can xet tren doan [0; 2TC]. f '(x) = -2sinxcosx - sinx; f '(0) = 0 <=> x e \ 0; — ; rc ; — ; 2rc Ta cd f(0) = f(2rc) = 5; f 2TC 11 Vay minf(x) - — ; maxf(x) = 5. 4 Cach 2: Dat t = cosx, -1 < t < 1 thi f(x) = g(t) = t2 + t + 3, g'(t) = 2t + 1 g.(t) = 0ot = ~ So sanh g(-l), g(-I), g(l). b) Ta cd y = 1 - sin2 2x = 4 = -sin2 2x - — sm2x + 5 Dat t = sm2x, - 1 < t < 1 thi y = f(t) = -t 2 -- 1 + 5. f '(t) = -2t - -; f '(t) = 0 <=> t = -- 2 4 Ta cd f(-l) = - , f 4 16 2 7 81 Vay miny = —, maxy = —, c) Tren khoang BBT rc 3TC 2'T sm x cos x y' = 0 CO X = it. X Tt/2 n 3TC/2 y' + 0 - y Vay max y = -l . Ham sd khong cd gia tri nhd nhat. '« 3n 2' 2 •BDHSG DSGT12/1- 61 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
d) Tren doan rc 5rc 2 ; T f 'M - CQS X f U \ n t 1 W - —: n , f (x) = 0 CO x = — sin x 2 *•*<!)-£•<*)-*'(!)- • So sanh thi max f (x) = 2 , mm f (x) = 1 YJ du 6: Tim gia tri ldn nhat va nhd nhat cua ham sd a) f(x) = x - sin2x tren doan [— ; rcl 2 b) y = sinx + — sin2x 2 c) f(x) = — x + sin2 x tren ["-— ; — 1 2 2 2 • 6 sm x cosx + cos6 X sin x sinx + cosx d)y |sm x| + |cosx| Giai a) f '(x) = 1 - 2cos2x ; f '(x) = 0 eo cos2x = - = cos2 3 o 2x = ±- + k2rc eo x = ±- + kre, k e Z. 3 6 Vdi-- <X<7C,f'(x) = 0cOXG (-1; *• ^ 2 1 6' 6' 6 Tac6f(-^) = -l+^,f(l)=2I_2/3 f(5n) = 5rc + V3 6' 6 2 V 6 2 ' 1 6 j 6 2 «-§) = §;««) = *• So sanh thi max f (x) = — + — ; min f (x) = -— +H 6 2 4-H 2 b) Ham so lien tuc tren D = R, tuan hoan vdi chu ki 2rc nen ta xet tren doan [-rc; TC]. TC y' = cosx + cos2x = 0eox : =+—. x = ±rc 3 Tacdf(-Tt ) = 0,f(-^)=-^,f(l)=^.f(T[) = o. o 4 o 4 373 373 Vay maxy = , miny = 4 4 1 1 c) f '(x) = — + 2sinxcosx = — + sin2x 2 2 62 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Tren doan [- - ; -- ] thi f '(x) = 0 o sin2x = -^ox = ~ ; ~ Tacdf(-- ) = 1 -- . f 2 4 5 it 12 (5n^ ( / 6 + ^ 2 5rt [l2j V 4 24 5rc 24 So sanh thi maxy = 1 + —. miny = n 24 d) Ta cd | sinx | + I cosx | > sm2 x + cos2 x = 1 nen D - R. y = | sinxcosx| 1 • I 5 sin x + • 5 cos X |sin x + cosx| = | sinxcosx |(1- | sinxcosx | - sin2 x cos2 x) 1 Dat t = | sin2x |, 0 < t < 1 thi y = f(t) = + - t O t f(t )= -_t2 - -t+ -,f'(t) = 0ot = - hoact = -2 (loai) w 8 2 2 3 Ta cd f(0) = 0, f - = — , f(l) = - Vay maxy 27 Vi du 7: Cho cac so nguyen duong p, q, n. 27 , miny = 0. a) Tim gia tri ldn nhat cua y = cosp x.sinqx vdi 0 < x < b) Tim gia tri nhd nhat cua y = tanr , x + cot n x - n2 cos2 2x, 0 < x < — Giai a) Vdi 0 < x < - thi sinx > 0, cosx > 0 nen y > 0. 2 Ta cd y 2 = (cos 2 x) p .(sin2 x) q . Dat t = cos2 x, 0 < t < 1 thi y2 = f(t) = tp . (1 - t)\ f '(t) = tp - x .(l - t)rl.[p - (p + q)t] nen f'(t) = 0ot = 0 hoac t = hoac t = 1. Ta cd f(0) = f(l) = 0, P + q f ,p + q Pp .qq (P + q)' — > 0 nen suy ra maxy PP -q" (p + q)P " b) Xet 0 < x < — thi cotx > tanx > 0, sin4x > 0 4 Ta cd y' = ntann "'x(l + tan2 x) - n.cotn_1 x(l + cot2 x) - 4n2 cos2x. sin2x = n(tann_1 - cotn_1x) + n(tann + ! x - cotn + 'x) - 2n2 sin4x < 0. -BDHSG DSGT12/1- 63 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
ham so nghich bien tren (0; — ] nen min y = f — = 2. Xet — < x < — thi y' > 0 nen ham so dong bien, do do min f — = 2. 4 2 „[ H ) U J Vay miny = 2. du 8: Tim GTI tuy y va khong dong thoi bang 0. 2 2 Vi du 8: Tim GTLN, GTNN cua bieu thuc T = . X +y —r . trong do x, y ' ' x + xy + 4y Giai x 2 Xet y = 0 thi x * 0 nen T = — = 1. Xet y * 0, dat x = ty thi: T= ,f^*/ r = ^l±i-=f(t ),D = R. t 2 y + ty2 + 4y2 t 2 + t + 4 f '(t >= -^± f^i -.f, (t) = 0ot = -3±>/l0 (t + t + 4) Lap BBT thi cd maxT = f(-3 - VlO) = 1 0 + 2 ^ 1 0 • minT = f(-3+ViO)= 10 - 2 ^ 15 Vi du 9: Cho 2 so duong thay doi x va y thoa man x + y = 1. Tim GTNN cua a) Q = xy + — b) P = -=L= + y xy Vl- x Vl- x Giai 1 a) Dat t = xy, vix , y>0va x + y = l> 2 ^xy nen 0 < t < Ta cd Q = f(t) = t + -j- => f'(t) = 1 ~jy < 0 nen f nghich bien tren (0; ^]. VayminQ = f(l)=^ 4 4 b) Vdi x, y > 0, x + y = 1 nen dat x = sin2 a, y = cos2 a vdi 0 < a < — _ sin2 a cos2 a sin3 a + cos3 a P = + = cos a sin a sin a + cos a Dat t = sina + cosa = V2 sin a + — , l<t< — 4 2 - t 3 -3 t (-3t 2 -3)(t 2 -1) - 2t(-t 3 -3t) _ f + 3 t 2 -l ' U (t2 -l)2 (t2 -l)2 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Nen f nghich bien tren [1; -J2]. Vay minP = f( spi ) = V2. Vi du 10: Cho cac sd thuc khdng am x, y thay doi va thoa man x + y = 1. Tim gia tri ldn nhat va gia tri nhd nhat cua bieu thiic S = (4x3 + 3y)(4y2 + 3x) + 25xy. Giai Do x + y = 1 nen S = 16x2 y2 + 12(x3 + y3 ) + 9xy + 25xy = 16xV + 12[(x + y)3 - 3xy(x + y)] + 34xy = 16xV - 2xy + 12. Dat t = xy, ta duoc S = 16t2 - 2t + 12 0 < xy < (x + y) 2 1 te[0 ; -] . 4 4 Xet ham f(t) = 16t2 - 2t + 12 tren doan [0; i ] f '(t) = 32t - 2; f '(t) = 0 eo t = —— 16 Tacdf(0) = 12,f( — )= llil.f(l)= 25 So sanh thi: 16 16 4 2 maxf(t) = f(-) = — ; min f(t) = f(—) = — t£[o; I] 4 2 tJ0:L\ V 16' 16 25 Gia tri ldn nhat cua S bang — . khi 2 191 Gia tri nho nhat cua S bang , khi i 16 xy=— { 16 ^2 + V3 2-V3 x+y= l , 1 eo fcy) = -; i xy= 4 \2 2 fx+y = l <=> (x; y) = hoac (x; y) 2-V3 2 + V3 V i du 11: Cho 3 sd duong a, b, c thoa man a2 + b2 + c 2 = 1. Tim gia tri ldn nhat cua E -2a 3 +a .b2 + b5 -2b 3 +b b 2 + c 2 Giai Theo gia thiet thi a, b, c e (0; 1) b(l-b2 ) 2 c 2 + a 2 2 c - 2c3 + c , c + =—.a a 2 + b 2 T rj 3 a(l-a 2 ) 2 Ta co E = — — 1-a ^2 "V, 2 , C(l -C2 ) 2 _2 1-b2 1-c 2 = a(l - a 2 )b2 + b(l - b2 )c 2 + c(l - c2 ); Xet f(x) = x(l - x2 ) tren khoang (0; 1) f '(x) - 1 - 3x2 , f '(x) = 0 co x = —f= V3 -BDHSG DSGT12/1- 65 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Lap BBT thi 0 < f(x) < -L . Do do E < — (b2 + c2 + a 2 ) 3v3 3v3 1 2 Dau bang khi a = b = c = —==. Vay maxE = —= . V3 ' 3V3 3^3 V i du 12: Cho x, y la cac so thuc thay doi va thoa dieu kien x < y. Tim gia tri nhd nhat cua bieu thuc: F = x 2 + y2 - 8x + 16. Giai Neu x>0thix 6 < y 2 v a F = x2 + y 2 - 8 x + 16>x6 + x 2 -8x+16 . Xet ham sd: f(x) = x6 + x2 - 8x + 16 vdi x > 0. f '(x) = 6x* + 2x - 8; f "(x) = 30x4 + 2 > 0, GX > 0. Do dd f '(x) dong bien. Ta cd: x > 1 => f '(x) > f '(1) = 0; 0 < x < 1 => f '(x) < f '(1) = 0 BBT X 1 0 I —OO r 0 + f 16 ^ 10 - +00 Tu do: f(x) > 0 => F > 10. Dau dang thuc xay ra khi x = y = 1. Neu x < 0 thi x2 + y2 - 8x + 16 > 16 Vay minF = 10, dat duoc khi x = y = 1. V i du 13: Cho cac so thuc x, y thay doi va thoa man (x + y) 3 + 4xy > 2. Tim gia tri nhd nhat cua bieu thuc A = 3(x4 + y4 + xY ) - 2(x2 + y2 ) + 1. Giai Ket hop (x + y) 3 + 4xy > 2 vdi (x + y) 2 > 4xy suy ra: (x + y) 3 + (x + y) 2 > 2 => x + y > 1. A = 3(x4 + y4 + x2 y2 ) - 2(x2 + y2 ) + 1 = W (Xz + f f +-(x4 + y4 ) - 2(x2 + y2 ) + 1 2 2 > 1 (X2 + y2)2 + 1 (x 2 + y2)2 _ 2(x 2 + y2 ) + J 2 4 => A > — (x2 + y2 ) 2 - 2(x2 + y2 ) + 1. Dat t = x2 + y2 , ta cd 4 xz + f > " 2 ^ ( x tZL.>I^ t > I >dod6A > ^t 2 -2t + l 2 2 2 4 Xet f(t) = - 1 2 - 2t + 1; f '(t) = - 1 - 2 > 0 vdi moi t > - 4 2 ' 2 • min f(t) = f 9 9 1 — . Do do A > — dau "=" khi x = y = — 16 16 2 Vay gia tri nhd nhat cua A bang _9_ 16' 66 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 14. Cho 2 < x < 3 < y. Tim GTNN cua B = Giai 2x' + y" + 2x + y xy v , . , . 2x2 +y2 +2x + y 2(x + l) y+1 Xet g(y) = = = " - + ^ — . voi2<x<3< y xy g'(y) = 2(x 2 +1) + - , g'(y) = o « y = V2x(x + 1) BBT X 3 7 2 x ( x + 1 ) _0 ° y' - 0 + y Do ddmin g(y) = g( ^2x( x +1)) = 2^2. - + V x x Xetf(x)= 2>/2j- + l+-.2<x<3thi f '(x) = V2 1 = — - < 0 nen f nghich bien tren doan [2; 3] do dd 1 . + x 1 minf(x) = f(3) = iS^l . Do do B < +1 , dau bang khi x = 3, y = 2 76 Vay minB = + ~ Vi du 15: Tim gia tri nhd nhat: A = /x-l)2 + y2 + V(x +1)2 + y2 + |y-2|. Giai Trong mat phang (Oxy) chon M(x - 1; -y), N(x + 1; y) V(x-l) 2 +y2 +v / (x + l)2 +y2 = OM + ON>MN= 2^1 + y2 Do do: A > 2^/1 + y2 + I y - 21 = f(y) Khi y < 2 thi f(y) = 2^1 + y2 - y + 2 f(y)=-Jl=- l = fc£ + y + v2 l =J-,f(y) = Ocoy - + y BBT: X -co 2 r — 0 + f minf =f(-^ ) = 2 + S 43 Khiy>2thiA> 2yjl + y2 +y-2 > 2/ + y2 >2^ >2+^3 -BDHSG DSGT12/1- 67 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Vay minA = 2+^ 3 khi x = 0, y = 4=- V3 Vj du 16: Cho phuong trinh: x4 + ax3 + bx2 + ax + 1 = 0 cd nghiem. Tim gia tri nhd nhat cua a2 + b2 Giai Goi XQ la nghiem: x4 + ax3 + bx2 + ax0 + l= 0=>x0^0 nen — + — ( 1 \ 1 x ) o + — + a x0 + — Y b = 0 Dat: v = x0 + — . Dieu kien |y | = |x 0 l + I — I > 2 n (y2 - 2) + ay + b = 0 =o |2-y 2 | = I ay + b I < Va 2 +b 2 v / y 2 +1 r2-v2 V* i (2-t)2 4 => a 2 + b2 > K £ J . Dat: t = y2 , t > 4. Ta chung minh ^ '- >- 1 + y2 ' 1 + t 5 Xet f(t) = (2 ~t)2 . t > 4 thi f '(t) = 3t ~ 6 > 0 => f ddng bien nen t > 4 w 1 + t (1 + t) 2 =>f(t)>f(4) = ^ o a b -2-4 Dau "=" khi t = 4 = > y = ±2va — = — nen chon b = — . a = — y 1 • 5 5 4 2,4 Phuong trinh: x4 x3 x2 x+ l = 0co nghiem x = 1 5 5 5 2 2 4 Vav: min(a + b ) = — " 5 Vi du 17: Mot con ca boi nguoc dong de vuot mot khoang each la 300km. Van toe dong nude la 6km/h. Neu van tdc boi cua ca khi nude dung yen la v (km/h) thi nang luong tieu hao cua ca trong t gio duoc cho bdi cong thuc E(v) = cv3 t, trong do c la mpt hang so, E duoc tinh bang jun. Tim van toe boi cua ca khi nude dung yen de nang lupng tieu hao la it nhat. Giai Van toe cua ca khi boi nguoc dong la v - 6 (km/h). Thoi gian ca boi de vuot khoang each 300km la: t = (gid) v - 6 Nang lupng tieu hao cua ca de vuot khoang each do la: E(v) = cv3 -^P- = 300c. . v > 6. v - 6 v - 6 E'(v) = 600CV2 V ~9 , : E'(v) = 0 => v = 9. (v-6) 2 Lap BBT thi van toe can tim la v = 9 (km/h). 68 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Vi du 18: Sau khi phat hien mdt benh dich, cac chuyen gia y te udc tinh so ngudi nhiem benh ke tu ngay xuat hien benh nhan dau tien den ngay thu t la: f(t) = 45t 2 - t 3 , t = 0, 1, 2, 25. Neu coi f la ham sd xac dinh tren doan [0; 25] thi f '(t) duoc xem la toe do truyen benh (nguoi/ngay) tai thdi diem t. Xac dinh ngay ma toe do truyen benh la ldn nhat va tinh toe do do. Giai f '(t) = 90t - 3t 2 , f "(t) = 90 - 6t, f "(t) = 0 co t = 15. BBT X 0 15 -oo f"(t) + 0 f'(t) 675 Vay toe do truyen benh la ldn nhat vao ngay thu 15. Toe do do la: f'(15) = 675 (nguoi/ngay). DANG 2: BAI TOAN LAP HAM S6 Bai toan tim gia trj 16*n nhat, nho nhat cua cac dai luong: Chon dat bien x (hoac t), kem dieu kien ton tai. Dua vao gia thiet, cac quan he cho de xac lap ham sd can tim gia tri ldn nhat, nhd nhat. Tiep tuc giai theo so do tim GTLN, GTNN cua ham sd va cac chu y neu tren, cd the phdi hop cac phuong phap khac. Vi du 1: Trong cac hinh chu nhat cd chu vi 100(m), tim hinh cd dien tich ldn nhat. Giai Goi x(m) la mot kich thudc cua hinh chu nhat thi kich thudc kia la 50 - :.. Dieu kien 0 < x < 50. Dien tich S(x) = x(50 - x) vdi 0 < x < 50, S'(x) = 50 - 2x, S'(x) — 0 <=> x — 25. X 0 25 -oo S'(x) + 0 - S(x) Vay maxS = f(25) = 625(m2 ) khi hinh chu nhat la hinh vudng canh 25(m). Vi du 2: Trong hinh chu nhat ndi tiep dudng trdn (O; R), tim hinh cd chu vi ldn nhat. Giai Goi x la mot kich thudc cua hinh chu nhat ABCD noi tiep (O; R). Ta co AC = 2R nen kich thudc thu hai la v^R2 -x 2 -BDHSG DSGT12/1- 6Q Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Chu vi V(x) = 2(x + V4R 2 - x 2 ), dieu kien 0 < x < 2R. 2x ^v^R2 Ta cd V'(x) = 2 V4R 2 x2 = 2- »2 x 2 - x V4R2 -: nen V'(x) = 0 o V4R2 -x 2 = x co x = RV2 Lap BBT thi maxV = V(R V2 ). Vay V(x) dat GTLN khi hinh chu nhat la hinh vuong canh R V2 Vi du 3: Hinh thang can ABCD cd day nhd AB va hai canh ben deu dai lm. Tinh gdc a = DAB = CBA sao cho hinh thang cd dien tich ldn nhat va tinh dien tich ldn nhat do. Giai Ha A H lCD.Datx = ADC,0<x< - 2 Ta duoc A H = sinx, DH = cosx; DC = 1 + 2cosx. Dien tich hinh thang la: D H „ . . AB + CD 71 AT T ,„ S(x) = .AH = (1 + cosx)smx; 0 < x < 2 2 S'(x) = (cosx + l)(2cosx - 1)', 0 < x < - , S'(x) = 0 <=> x = — 2 3 Lap BBT thi maxS = S khi a = — 3 2TC) 3^3 nen hinh thang cd dien tich ldn nhat V i du 4: Xac dinh tam giac vuong cd dien tich ldn nhat biet tong mot canh gdc vudng va canh huyen bang a cho trudc. Giai Tam giac ABC vudng tai A, dat AB = x > 0 va cd AB + BC = a > 0 BC = a - x, AC = vT3C2 -AB2 ~ = Va2 - 2ax Va" S' BBT: S = - AB.AC = —xVa-2 x (0<x < - ) 2 2 2 Va/ 2 Va - 2x - Va a-3 x „, . a , , S = 0 eo x = - Va-2 x J 2 Va-2 x 3 X 0 3 2 f + 0 — f a 2a Vay max S tai x = AB= —, BC = — nen ABC la nua tam giac deu. 3 3 70 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
VJ du 5: Chu vi cua mdt tam giac la 16cm, do dai mdt canh tam giac la 6cm. Tim dp dai hai canh cdn lai cua tam giac sao cho tam giac cd dien tich ldn nhat. Giai Gpi x(cm) la mpt canh cdn lai thi canh thu ba la 10 - x nen dieu kien: a + b + c 0<x < 10. Tacdp 8, do dd S = vVXp-aXp-bXp-c) =V8.2(8-x)(x-2) = W-x 2 + lOx -16 Xet ham f(x) = -x 2 + lOx - 16, 0 < x < 10 thi f'(x) = -2x + 10. Ta cd f (x) = 0 c=> x = 5. Lap BBT thi maxf(x) = f(5) = 9. Vay maxS = 12 khi a = 6, b = 5, c = 5. Vi du 6: Cho mpt tam giac deu ABC canh a. Dung mpt hinh chu nhat MNPQ cd canh M N nam tren canh BC, hai dinh P va Q theo thu tu nam tren hai canh AC va AB cua tam giac. Xac dinh vi tri cua diem M sao cho hinh chu nhat cd dien tich ldn nhat. Giai Dat B M = x thi M N = a - 2x, QM .= xV3 Dien tich hinh chir nhat a MNPQ la S(x) = MN.QM = (a - 2x)x S , 0 < x < - 2 Ta cd S'(x) = V3 (a - 4x) ; S'(x) = 0 <=> x = - - A X 0 a/4 a/2 S'(x) + 0 - S(x) —-" 8 ~~~~~~ — Vay S(x) dat gia tri ldn nhat tai x = — 4 Vi du 7: Mpt tam nhom hinh vudng canh a. Tim each cat 4 gdc 4 hinh vuon,:; de gap thanh mpt hinh hop khdng nap cd the tich ldn nhat. Giai Gpi x la canh hinh vudng bi cat thi canh day hinh vuong la a - 2x, 0 < x < — 2 The tich hinh hop V(x) = x(a - 2x) 2 Ta cd V'(x) = 12x2 - 8ax + a2 , V'(x) 0»x = - 6 m m 1 a—2x a 2a 3 ' a Lap BBT thi maxV = V(—) = khi cat 4 hinh vudng canh la — 6 27 ' 6 Vi du 8: Trong cac khdi try lam bang tam nhua cd thi tich V cho trudc, tim khdi tru it tdn vat lieu nhat. -BDHSG DSGT12/1- 71 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai Goi x, h lan luot la ban kinh clay va chieu cao cua hinh tru, x, h > 0 V = 7TX h V rex Vat lieu lam la dien tich toan phan 2V S = *2rtxh + 2rcx2 = + 2rcx2 , x > 0 x Ta cd: S' = - 2V 4rcx; S1 = 0 o x Bang bien thien: X 0 V2n +00 S' — 0 + s +0C Vay the tich be nhat khi x V V 3 — va h = 2.3 — V2rc V2rc V i du 9: Cho hinh tru noi tiep hinh cau S(0; R). Tim hinh tru cd the tich ldn nhat. Giai Goi x la khoang each tu tam hinh cau O den day hinh tru: 01 = x. Day hinh tru la dudng trdn cd ban kinh: r = VR2 -x 2 , 0 <x< R The tich cua khoi tru la: V = rcr2 .2x = 2rc.x(R2 - x2 ) = -2rcx3 + 2rcR2 x, 0 < x < R V = -6rcx2 + 2rcR2 , V = 0 » x = R Lap BBT thi V dat gia tri ldn nhat khi x = —= V3 ^ V i du 10: Tim hinh ndn cd the tich ldn nhat noi tiep mot mat cau ban kinh R cho trudc. Giai Goi ban kinh day hinh non la x, chieu cao hinh ndn la y (0 < x < R 0 < y < 2R). Goi SS' la ducmg kinh cua mat cau ngoai tiep hinh ndn thi: A H 2 = HS.HS' => x2 = y(2R - y) Goi V la the tich khoi ndn thi: V(y) = ^rtx2 y = iny.y(2R-y) = ^(2Ry2 -j IT 4R V'(y) = |y(4 R - 3y) , V'(y) =0 o y = — 72 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Lap BBT thi max V =i^ l khi y = — , x 81 3 2RV2 V i du 11: Trong cac hinh ndn ngoai tiep hinh cau ban kinh r hay xac dinh hinh ndn cd uhi tich nhd nhat. Giai Thiet dien qua true la tam giac can SAB ngoai tiep dudng trdn (O; R). Goi chieu cao cua hinh ndn la h, ban kinh day R. „ Ta cd SSAB = - AB.SI = p.R nen R.h = (R + Jh2 + R2 )r R(h-r) = Vh 2 + R2 .r Do do R2 (h2 + r 2 - 2hr) = (h2 + R2 )r 2 R hr 2 h-2 r The tich hinh ndn: 1 V(h) = V'(h) rcR2 h 1 -rcr" 1 • rcr 3 h h(h-4r) 2r h>2r. V'(h) = 0 co h = 4r. 3 (h-2r) 2 Lap BBT thi maxV = V(4r); chieu cao h = 4r. Cach khac: Goi gdc giua dudng sinh va day la 2t. Vi du 12: Mot hinh ndn cd ban kinh day R va chieu cao bang 4R. Tinh ban kinh day r va chieu cao h cua hinh tru noi tiep hinh ndn de dien tich toan phan cua hinh try dat gia tri ldn nhat. Giai Ta cd: — R r SLL SLL SH, =4r va HH, =4(R-r ) SH 4R Dien tich toan phan cua hinh try noi tiep hinh non: St p = 2rcr2 + 2rcr.4(R - r) = -6rcr2 + 8rtRr = f(r) f '(r) = 2rc(-6r + 4R), 0 < r < R f'(r) = 0or= -R 3 Lap BBT thi St p dat gia tri ldn nhat 2 „ , , 4R — R va h = — 3 3 Vi du 13: Cat bd hinh quat trdn AOB tu mot tam nhua hinh tron ban kinh R roi dan hai ban kinh OA va OB cua hinh quat trdn cdn lai vdi nhau de duoc mot cai pheu cd dang cua mot hinh ndn. Goi x la gdc d tam cua quat tron dung A lam pheu, 0 < x < 2rc. khi r -BDHSG DSGT12/1- 73 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Tim x de hinh ndn cd the tich ldn nhat va tinh gia tri ldn nhat do. Giai V i do dai cua duong trdn day hinh ndn bang do dai cung AB cua quat trdn dung lam pheu, nen ta cd: 2m = Rx. R Doddr=^ ^ ' 2rc ah=VR 2 - r 2 =-^ 2 - x 2 2rc 1 R The tich hinh ndn la: V = — rcr2 h = -x 2 3 24rc2 4^n 2 x2 . 0 < x <2n. R3 x(8rc2 -3x 2 ) 2 2 2V6 V = —, . V = 0 o 81 - 3 r = 0 o x = TC 24n2 ^ Lap BBT thi max V = V xe(0;2n) '2&_) 2>/3 3 27 71 V i du 14: Cho sd duong m. Hay phan tich m thanh tdng cua hai so ducmg sao cho tich cua chung la ldn nhat. Giai Gpi x la so thu nhat, 0 < x < m, sd thu hai la m - x. Tich so P(x) = x(m - x), P'(x) = -2x + m, P'(x) = 0 co x =— BBT x 0 m/2 m P'(x) + 0 - P(x) m2 Vay max P(x) = P m . , . , , , m m — km phan tich m — + — 4 2 2 V i du 15: Cho parabol (P): y = x va diem A(-3; 0). Xac dinh diem M thuoc parabol (P) sao cho khoang each A M la ngan nhat. Giai Gpi M(x; x2 ) la mpt diem bat ki cua parabol (P) Ta cd AM = V(x + 3)2 +x4 = Vx4 + x2 +6x + 9 Xet ham so g(x) = x4 + x2 + 6x + 9; D = R g'(x) = 4x3 + 2x + 6 = (x + l)(4x 2 - 4x + 6); g'(x) = 0o x = -l . LapBBTthiming = g(-l) = 5. VayminAM= VB" taiM(-l; 1). Vi du 16: Cho dd thi (C): y = | x - m | - m* ~m + 1 , 1 < m < 4. ' m -3m + 3 Tim GTLN, GTNN cua dien tich gidi han bdi do thi va true hoanh. Giai Chuyen true bang phep tinh tien vecto OI, I(m; 0) thi y = f(x) thanh: 74 -BDHSG DSGT12I1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Y = F(X) = | X | — ^ — m + 1 suy ra dien tich can tim S m2 -3m + 3 v + / \ m 2 - m + l , Xet g(m) = — . 1 < m < 4 m-3m+ 3 . -2m2 + 4m ,. . _ „ g'(m) = — — , g'(m) = 0 co m = 2. ( m2 -3 m + 3) 2 Tacd g(l)=l,g(2) = 3,g(4)=y Tu do suy ra maxS = S(2) = 9, minS = S(l) = 1. Vi du 17: Mot xudng in cd 8 may in, mdi may in duoc 3600 ban in trong mot gid. Chi phi de van hanh mot may trong moi lan in la 50 nghin ddng. Chi phi cho n may chay trong mot gid la 10(6n +10) nghin dong. Hdi neu in 50000 to quang cao thi phai sir dung bao nhieu may de.duoc lai nhieu nhat? Giai Goi n la so may in su dung (n nguyen, 1 < n < 8) thi tong chi phi de in 50000 to quang cao la: .50000, . . i n v 1 n . K n 250 _ n 12500 T(n) = (6n + 10). 10 + 50n = + 50n + 3600 3 9n Sd lai se nhieu nhat neu chi phi it nhat. Do dd can tim gia tri nhd nhat cua T(n). Xet ham sd f(x) = 50x + 1±™9_. 1 < x < 8. 9x „ , ., , . C A 12500 50(9x2 -250) Ta co f '(x) = 50 — = — - 9x2 9x2 f'(x) = 0 co 9x2 = 250 co x = ->/l0 3 Lap BBT thi min f = ff|>/10j. Ta cd 5 < |Vl0 < 6 nen so sanh T(5) va T(6) thi minT(n) = T(5) tuc la sir dung 5 may thi duoc lai nhieu nhat. Vi du 18: Tim sd hang be nhat cua day u„ = n 4 - 20n3 + 0,5n2 - 13n. Giai Xet ham so f(x) = x4 - 20x3 + 0,5x2 - 13x, x > 1. f '(x) = 4x3 - 60x2 + x - 13 Vdi x > 1 thi f '(x) = 0 cd nghiem x = 30 + ^896 _ ( „m2 - m +1 ^ 2 m2 -3,m + 3 Lap BBT thi f dat GTNN tai x = 30+ ^ 6 ~ g r 14;15 ] Ta cd f(14) = -16548 ; f(15) = -16957,5. So sanh thi so hang be nhlt la u , 5 = f(15) = -16957,5. -BDHSG DSGT12/1- 75 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Y j du 19: Cho x > 0 va y tuy y. Tim GTLN, GTNN cua 2 M = xy ( x 2 +3y 2 X x + 7x 2 +12y 2 ) Giai Xet y = 0 thi M = 0. Xet y * 0 thi: M x y 2 (Vx 2 +12y 2 -x ) (x2 +3y2 ).12y2 1 + ^- 1 4 + 12y 2\ J Datt=±^,t> 0 thi M = f(t)=^±i ^ x2 3(t + 4) T a cd f'(t) = _ 2 " t + 2>/ ^ i , f'( 0 = 0 O t 6(t + 4) 2 .VT+t BBT X 0 g +00 f + 0 + f 1/1 8 - \ Do dd: 0 < M < — Ket hop thi 0 < M < — 18 18 Vay maxM = — khi 2x2 = 3y2 , minM = 0 khi y = 0. 3 V i du 20: Cho x, y, z > 0 thoa man x + y + z < — 2 Tim GTNN T = + ~ + JL + ^l + Zl + .=t yzzxx y y z x Giai Ap durg BDT Co si: T > 3.3 Dat t = ^xyz thi 0 < t < 1 (xyz) 2 x + y + z _ 1 ~~3 ~ 2 3.^0 ^ Xet ham so f(t) = \ + 3t 4 , 0 < t < - Tacdf'(t) = - f + 12t 3 v 2 , do do f(t) > f 3(4t 3 -2 ) t 3 1\ 195 16 <0, Vt e nen f nghich bien tret76 -BDHSG DSGT12/1Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
1 195 Dau "=" khi x = y = z = — Vay min i = 2 16_ DANG 3. UNG DUNG VAO PHUONG TRtNH Phirong phap chung: Dung bang bien thien (BBT) va gia tri Ion nhat (GTLN), gia tri nho nhat (GTNN ) d l giai toan. Bieu kien phuong trinh co nghiem: Cho y = f(x) tren D dat gia tri ldn nhat, nhd nhat: GTLN = M va GTNN = m thi phuong trinh f(x) = k cd nghiem o m < k < M Bieu kien ve nghiem bat phuong trinh: Cho y = f(x) tren D dat gia tri ldn nhat, nhd nhat: GTLN = M va GTNN = m thi: Bat phuong trinh f(x) > k cd nghiem o k < M Bat phuong trinh f(x) < k cd nghiem <o k > m B it phuong trinh f(x) > k cd nghiem moi x thudc D <=> k < m Bat phuong trinh f(x) < k cd nghiem moi x thuoc D o k> M Chu y: Neu khong dat GTLN, GTNN thi lap bang bien thien d l giai. Vi du 1: Tim k d l phuong trinh x4 + 4x3 - 8x + 1 - k = 0 cd 4 nghiem phan biet. Giai Phuong trinh: x4 + 4x3 - 8x + 1 = k Xet ham sd: y = x4 + 4x3 - 8x + 1, D = R y - = 4x3 + 12x2 - 8 = 4(x + 1) (x2 + 2x - 2) y' = 0 => x = -1 , x = - 1 + V3 Lap BBT thi dii u kien cd 4 nghiem phan biet la -3 < k < 6. Vi du 2: Chiing minh phuong trinh x5 - x - 2 = 0 cd nghiem duy nhat va nghiem do ldn hon Giai Xet ham sd f(x) = x5 - x - 2, D = R 1 4 f'(x) = 5x 4 - l = 0 o x = - i[5 5^5 -2< 0 x = 4 = => y =—-^-2< 0 V5 5 ^ Lap BBT thi phuong trinh cd nghiem duy nhat la Xo va nghiem do la duong. Do x 0 la nghiem x^ -x 0 - 2 = 0 nen x ^ = x 0 + 2> 2 ^'2x0 (dau "=" khong xay ra) oxj°>8x o ox ^ > 8<=>x0> V8 (dpcm) V i d u 3: Tim a de phuong trinh sau vo nghiem x6 + 3x5 + (6 - a)x4 + (7 - 2a)x3 + (6 - a)x2 + 3x + 1 = 0. -BDHSG DSGT12/1- 77 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai Xet x = 0 => 1 = 0 (loai). Xet x * 0. Chia 2 v l cho x3 , phuong trinh: x3 + 3x2 + (6 - a)x + (7 - 2a) + (6 - a) - + \ + 4" = 0 x x2 xJ (x3 + 4r) + 3(x2 +4r ) + (6 - a) (x + -) + 7 - 2a = 0 X X X Dat t = x + -, 111 > 2 => t2 = x2 + — +2 x x2 1 1 va t 3 - x3 + -V + 3(x + - ) nen x3 + 4 " = t 3 - 3t x3 x x3 Do do phuong trinh: t 3 - 3t + 3(t 2 - 2) + (6 - a)t + 7 - 2a = 0 (t + 2)a = t 3 + 3t 2 + 3t + 1 Khi t = -2 thi phuong trinh khong thoa. t 3 +3t 2 + 3 t + l (t + 1)3 Khi t * -2 thi phucmg trinh: a: Dat f(t) = ^ , t < -2 hay t > 2 thi f '(t) Bang bien thien: t + 2 t+ 2 (2t + 5)(t + l) 2 2(t + 2) 2 t —00 -5/2 •2 2 oo r 0 + + f +00 27 4 ^ +00 27 - +00 27 27 V i f(t) > — Vt e D nen phuong trinh vo nghiem khi a < — V i du 4: Tim a de phuong trinh V l + x + Vl - x = a cd nghiem Giai Xet f(x) = vT^x" + VTT^ , D = R 1 1 f'(x ) = 3#-x) 2 3^(1+ x) 2 (X* ± 1) ^l-x) 2 - ^(1 + x) 2 f(x ) = 0 » x = 0 hm f(x) = 3 MlC^^^l+~^ ) = hm (VTT 7 - VT ^ c ) x—y+X' x—v+co x^+» 2 lim 0 (Va + x)) 2 + V(x 2 - i ) + (V(x -D) 2 Tuong tu lim f(x) = 0. Lap BBT thi PT cd nghiem <=> 0 < a < 2. V i du 5: Tun m de phuong trinh cd 2 nghiem phan biet: Vx2 + mx + 2 = 2x + 1. 78 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai P T o 2x+l> 0 x 2 +mx + 2 = (2x + l) 2 co 3x + 4x - 1 = mx, x > - - 3x2 + 4x - 1 1 V i x = 0 khong thoa man nen: = m, x > —- x 2 v ,,r v . 3x 2 +4x- l . 1 n ,,, f l / x 3x 2 + l Xet f(x) = . x > — . x * 0 thi f (x) = — BBT x 2 x : V 2 0 +00 + +00 + +00 9 2 Dieu kien phuong trinh cho co 2 nghiem phan biet 1 9 CO f(x) = m co 2 nghiem phan biet x > — , x * 0 co m > —. 2 2 Vi du 6: Tim dieu kien de phuong trinh cd nghiem x + vl 2 - 3x2 = m Giai Xet f(x) = x + Vl2 - 3x2 , D = [2; 2] 3x V12 - 3x2 3x f '(x )=l - V12 - 3x2 V12 - 3x2 f '(x) = 0 « 7l 2 - 3x2 - 3x = 0 (-2 < x < 2) , [3x > 0 co V12 - 3x2 = 3x co [12 - 3x2 = 9x Lap BBT thi -2 < f(x) < 4, Vxe [-2; 2] Vay dieu kien cd nghiem la -2 < m < 4. Vi du 7: Xac dinh m sao cho phuong trinh t 4 - (m - l)t 3 + 3t 2 - (m - l)t + 1 = 0 cd nghiem Giai Ta cd t = 0 khong la nghiem. Chia hai ve cho t 2 x > 0 ± 1 CO X = 1 t 2 - (m - l)t + 3 - (m - 1) 1 1 = 0 1 1 Dat x = t + - thi | x | > 2 va phuong trinh trd thanh: (m - l)x + 1 = 0 co y X + X + 1 m -BDHSG DSGT12/1- 79 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ta co: y' BBT: > 0 , V x >2 X —00 - 2 i y + y ^^-3/ 2 | —00 1 2 +00 1 . + j 7/2 ^ +00 Vay phucmg trinh co nghiem khi m < - - hav m > - 2 2 v » du 8: Tim dieu kien de phuong trinh sau cd nghiem 2(1 + sin2x.cos4x) - -(cos4x - cos8x) = m. 2 Giai 1 Ta cd: 2(1 + sin2x.cos4x) - - (cos4x - cos8x) = 2 + 2.sm2x.cos4x - sin6x.sin2x = 2 + sin2x(2cos4x - sin6x) Dat: t = sin2x (-1 < t < 1) xet y = f(t) = 4t4 - 4t3 - 3t2 + 2t + 2 Ta cd: f '(t) = 16t3 - 12t2 - 6t + 2 = (t - 1) (16t2 + 4t - 2) f'(t) = 0 M = -i,t= I 2 4 De tim GTNN va GTLN cua f(t) tren doan nay ta chi ca.n tinh cac gia tri: f(-l),f(l),f(-i)vaf(i)i 2 4 Ket qua: max y = 5; miny 129 64 Vay dieu kien cd nghiem 129 ~64~ <m < 5. yj_du_9: Tim m de phuong trinh sau cd nghiem 1 + cosx + — cos2x + — cos3x = m. 2 3 Giai VT = 1 + cosx + — cos2x + -cos3x 2 3 = 1 + cosx + ~ (2 cos2 x - 1) + - (4cos3 x - 3 cosx) 2 3 4 1 = - 1 3 + t 2 + - vdi t = cosx, 111 < 1 3 2 80 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Xet y = f(t) = -t 3 + t 2 + -; |t | < 1 3 2 f'(t) = 4t2 + 2t;f'(t) = 0cot = -|.t = 0deu thuoc [-1; 1] 17 Ta co f(l ) = — ; f(—1) = — ;f(0) 12 => Max f(t) = — ; min f(t) = - 6 6 1 17 Do do phuong trinh cho cd nghiem khi: — < m < — 6 6 Vi du 10: Tim dieu kien de phuong trinh sau co nghiem Giai 3 cos4 x + 4 sin2 x 3 sin4 x + 2 cos2 x Dat t = sin x, 0 < t < 1 thi: 3(1 - sin2 x) 2 + 4 sin2 x 3t -2t+3 , y = — = • = 1 + 3sin4 x +2(1-sin 2 x) 3t 2 - 2 t + 2 3t 2 -2t+ 2 Xet: f(t) = 3t 2 - 2t + 2, 0 < t < 1; f '(t) = 6t - 2 = 0 o t 1 Bang bien thien t f'(t) 0 f(t) Do do: — < l(t) <3=> — <y < — nen miny = —. max y = — 3 3 5 3 5 4 8 Vay dieu kien cd nghiem — < m < — 3 5 Vi dull: Tim a de bat phuong trinh sau cd nghiem vdi moi x a\/2x2 + 9 < x + a. Giai Ta cd: a^2x2 + 9<x + aoa(^2x2 + 9 - 1) <x 1 o a < V2x2 +9 - 1 Xet ham so: f(x) (vi V2X2 + 9 - 1 > 0, Vx) xeR V2x2 +9 - 1 f'(x) -BDHSG DSGT12/1- 9 - N/2X2 +9 V2x2 +9 (^2x2 +9 - l) 2 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ta co f '(x) = 0 o 9 - V2x2 +9 = 0 o 2x2 + 9 = 81 cox2 = 36cox = ± 6 Lap BBT thi min f(x) = - - 4 3 Vay BPT nghiem clung Vx khi a < — 4 V i du 12: Tim dieu kien bat phuong trinh sau cd nghiem V4x- 2 + 2^4 - x <m. Giai Xet f(x) = V4x - 2 + 274 - x , D = [ —;4] 2 f'(x ) 2V4" V4x~ V4x - 2 V4 - x %/4x - 2 VI " Tacd: f'(x) > 0 o 2^4 - x > V4x - 2 (x * 4, x*- ) 2 9 <=> 4 (4 - x) > 4x - 2 ox< - 4 Bang bien thien: +00 Bpt (1) cd nghiem o m > min f(x) < m Vay m > V14 . V i du 13: Tim dieu kien bat phuong trinh sau cd nghiem sin3 x + cos3 x > m. Giai Xet f(x) = sin3 x + cos3 x = (sinx + cosx)(l - sinx.cosx) Dat t = sinx + cosx ; 111 < V2 t 2 — 1 + 2sinx cosx => sinx cosx = t 2 - 1 Ta cd h(t) = t ( t 2 - D _ I t 3 + 3 tvd i |t|< V2 2 2 h'(t) = --t 2 + - = 0 <=>t = ± 1 2 2 82 -BDHSG DSGT12/1Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bang bien thien: Do do max f(x) = 1. Bat phuong trinh cd nghiem khi m < max f(x) Vay dieu kien bat phuong trinh sau co nghiem la m < 1. Vi du 14: Tim dieu kien cd nghiem cua bat phuong trinh: cos2 2x + 2 (sinx + cosx)3 - 3sin 2x + m > 0. Giai Dat t = sinx + cosx, |t| < 42 va t 2 = 1 + 2 sinx cosx => sin2x = t 2 - 1 cos2 2x = 1 - sin2 2x = -t 4 + 2t 2 Bat phuong trinh viet lai: - t 4 + 2t 3 - t 2 + m + 3 < 0 ; (|t|<V2 ) Xet f(t) = -t 4 + 2t 3 - t 2 + m + 3 f '(t) = -2t (2t2 - 3t + 1) ; f '(t) = 0 => t = 0 Bang bien thien: f'(t) f(t) o I 4i m+3 Tu BBT suy ra dieu kien cd nghiem la: m + 3 > 0 <=> m > -3 Vi du 15: Cho da thuc f(x) = a0x" + ajx""1 + ... + an_!X + a„, a 0 * 0 bac n cd n f'(x) nghiem va so c > 0. Chung minh rang bat phuong trinh: > c cd tong f(x) do dai cac khoang nghiem la: —. c Giai Theo gia thiet thi f(x) = a0(x - xi)(x - x2 ) ... (x - xn ) , , . f'(x) 1 1 1 =>h(x) = —— = + + ... + .x*X j f(x) x-X j x - x 2 x-x r nen h'(x) < 0 tren tung khoang xac dinh. Bang bien thien: -BDHSG DSGT12/1- 83 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
h' —00 Xl X2 +00 —00 +00 Do db phuong trinh h(x) = —00 f'(x) f(x) +00 c, c > 0 cd dung n nghiem kj suy ra bat phuong trinh: h(x) > c cd n khoang nghiem va cd tong do dai: S = (kj - xO + (k2 - x2 ) + ... + (kn - x j = £k , - Jx , 1=1 1=1 Ta cd kj la nghiem phuong trinh: f ' (x) = c.f(x) xn(x- X j )-cn(x - x ,)= o =1 j=l,i*i Theo dinh l i Viet thi: J k; = • n „ n n „ Vay tdng cac khoang nghiem la S = — c V i du 16: Tim dieu kien cua m de he cd nghiem: Giai Dieu kien x, y * 0. Dat u = x + —. v = y + — thi |u| > 2, |v| > 2. x y 1 1 R x+-+y+-= 5 x y x 3 + 4 + y 3 + 4 = 1 5m-10 x3 y3 He: u + v = 5 u 3 -3 u + v 3 -3 v = 15m-10 u + v = 5 uv = 8 - m Do do, u, v la nghiem phuong trinh: t 2 - 5 t + 8 - m = 0 Bai toan dua ve tim m de phuong trinh t 2 - 5t + 8 = m cd 2 nghiem, thoa man | t j |, 112 1 > 2 Xet f(t) = t 2 - 5t + 8, D = R. Ta cd: f'(t) = 2t - 5 Bang bien thien: t -00 2 2 5/2 +oo r — /// / — 0 + f +00 """^22 ' / / v \ 2 +00 k 7/4-^"^ Vay: — < m < 2 hoac m > 22. 4 V i du 17: Tim dieu kien he bat phuong trinh cd nghiem f x 2 - 3x - 4 < 0 (1) | x 3 - 3x | x | - m2 - 15m > 0 (2) Giai 84 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Xet (1): x2 - 3x - 4 < 0 co - 1 < x < 4 Ta tim dieu kien ngugc lai, tuc la tim m de: f(x) = x3 - 3x | x | - m2 - 15m < 0 ; Vxe [-1; 4] <=> max f(x) < 0 xe[-l;4] V + 3x2 - m2 - 15m ; - 1 < x < 0 V i f(x) = < =of'(x) = [x3 - 3x2 - mz - 15m ; 0 < x < 4 |3x 2 + 6x; - 1 < x < 0 [3x2 - 6x ; 0 < x < 4 Khi -1 < x < 0 => f '(x) = 3x (x + 2) < 0 0 <x<2 =of '(x ) = 3x(x-2)< 0 2 < x < 4 =o f '(x) = 3x (x - 2) > 0 Do do max f(x) = max {f(-l) , f(4)} = f(4) = -m 2 - 15m + 16 xe[-l;4) Nen cd -m 2 -15m+16<0com<-16vm> l Vay dieu kien cd nghiem la -16 < m < 1. C. BAI LUYEN TAP Bai 1: Tim GTLN cua ham sd: a) y = 1 + 8x - 2x2 b) y = 4x3 - 3x4 c) y = Vx2 +3x- 4 DS: a) 9 b) 1 c) khong ton tai Bai 2: Tim GTNN cua ham sd: a)y = ( x + 2)2 ,x> 0 b) y = x 2 +-,x> 0 x x c) y = (x + 3) 4 + (x + 5) 4 d) y = (x - l)(x + 3)(x + 5)(x + 1) Bai 3: Tim GTNN,GTLN cua ham so: a) y = x3 + 6x2 - 15x + 1 tren [-1 , 5] b) y = Vx- 2 + V4- x tren [2; 4] c) y= sin 2x - x tren [-TC; TC] d)y = |x3 +3x2 -72x + 90| tren [-5; 5] DS: a)-7va20T c)-- vaBai 4: Tim GTLN,NN cua ham sd: a)y - Vsmx + vcosx b) y = 2.sin8 x + cos4 2x cosx+ 2 .. 1 1 c ) y = — 2 n ^ y cos2 x + cos x + 2 sin x + 4 cos x- 4 H D: c) Dat t = cosx, - 1 < t < 1. d) Quy ddng va dat t = sinx - cosx, - V2 < t < V2 Bai 5: Tim GTNN cua ham sd: y = A 4 + (-) 4 - (J) 2 - (-) 2 + * + ^ DS: -2 b a b ab a -BDHSG DSGT12/1- 85 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bai 6: Tim GTLN,NN ciia ham so theo m: a) y = mx 2 -2(m-2) x + 1 tren [ - 1,3] b) y = x4 -6m x2 + m2 tren [-2,1] Bai 7: Tim m de gia tri ldn nhat g(m) cua ham so: y = | 2x2 - x + m | tren [-1,1] dat gia trj nhd nhat. Bai 8: Trong cac hinh chu nhat cd dien tich 48m2 , tim hinh cd chu vi nhd nhat? DS: Canh 4^3 Bai 9: Tim hinh tru cd dien tich xung quanh ldn nhat noi tiep trong hinh cau ban kinh R cho trudc. Bai 10: Tim sd hang ldn nhat cua day: Un n 2 - 4n + 5 4 tx + 2k" +• a) Tim dieu kien cd 2 nghiem Bai 11: Cho phuong trinh: x2 - 2kx + 2k2 + - 5 = 0 .K b) Tim GTLN,NN cua T = (x + x2 ) ( X l 2 + x2 2 ) DS: b)-32va3 2 Bai 12: Cho f(x) = x4 - 4x3 + 8x a) Bien luan sd nghiem phuong trinh: x 4 - 4x3 + 8x - m = 0 b) Tim dieu kien bpt: f(x) < a cd nghiem Bai 13: Tim tham so de: a) x4 + a.x3 + b > 0,Vx b) V(4 + x)(6-x ) < x2 - 2x + m, Vx e [-4,6] Bai 14: Tim dieu kien phuong trinh, he cd nghiem: a) x 4 -6x 3 + mx 2 -12 x + 4 = 0 b)- |x2 + 2x + a< 0 x2 -4x-6a< 0 Bai 15: Tim dieu kien cd nghiem: a) sin 3 x + co s3 x > m b) V4x - 2 + 2-\/4 - x < a Bai 16: Tim dieu kien cua a va b de phuong trinh: x +a x + b = 0c6 3 nghiem phan biet. Bai 17: Cho f(x) = x3 - 3x + 1. Hdi f(f(x)), f(f(f(x))) cd bao nhieu nghiem? Bai 18: Giai bat phuong trinh: 3 Vtanx + 1 (sinx + 2 cosx) < 21_^tanx (sinx + 3cosx), (0 < x < ^) Bai 19: Bien luan sd va dau cac nghiem phuong trinh: x(x + l)(x + 2)(x + 3) + 1 - m = 0 Bai 20: Tim dieu kien phuong trinh cd nghiem Vsinx + Vcosx = m H D: Xet ham f(x) = Vsin x + Vcosx Bai 21: Tim dieu kien phuong trinh cd nghiem 2sin8 x + cos4 2x - 3m = 1 H D: DSt t = sin2 x, 0 < t < 1 86 -BDHSG DSGTWlDownload Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bai 22: Tim dieu kien phuong trinh cd nghiem x e 5 cosx - cos5x + m = 0 TC TC 4 ; 4 Bai 23: Tim dieu kien phuong trinh ^2 - x + 72 + x - ^(2 - x) (2 + x) cd nghiem Bai 24: Tim dieu kien bat phuong trinh cd nghiem vdi moi x cos4 x + m cosx + 1 > 0, Vx Bai 25: Tim dieu kien bat phuong trinh cd nghiem vdi moi x x4 + ax3 + b > 0 Bai 26: Tim dieu kien bat phuong trinh cd nghiem vdi moi xe [0, 1] (x2 + l) 2 + m <xVx 2 + 2 + 4 Bai 27: Tim dieu kien bat phuong trinh cd nghiem x3 + 3x2 - 1 < m (Vx - Vx - l) 3 m Bai 28: Tim dieu kien de he cd nghiem: (x-3) 5 = rn 0 < x <3 Bai 29: Giai he phuong trinh: x + y + z = 1 7 xy + yz + zx - 2xyz = — 2 7 Vx Vy Vz x2 — cos x — 0 ' Bai 30: Chung minh he phuong trinh: < cd nghiem duy nhat [y-tany- 1 = 0 (x,y) thoa 0 < x < y < 1. Bai 31: Bien luan sd nghiem cua he phuong trinh: | x 3 y-y 4 = m 2 Bai 32: Tim gia tri ldn nhat, nhd nhat cua: y = ^o kfe^k&n do v sx~^-A co$5x tren doan 4 4 H D: Tinh nghiem cua y' = 0 rdi so sanh vdi gia tri ham tai 2 bien. Bai 33: Tim gia tri ldn nhat, nhd nhat cua: y = 11+ 2cosx | + | 1+ 2sinx | H D: Binh phuong va dat t = sinx + cosx, - V2 < t < V2 Bai 34: Cho x, y >0thoax3 + y 3 = 1. Tim gia tri ldn nhat cua: T = Vx + 2sjy H D: Lap ham theo bien x. Bai 35: Cho day da thuc: P(x) = x3 - 6x2 + 9x va Pn (x) = P(P(...(P(x))) n lan. Tim so nghiem cua P(x), Pn (x) DS: P'(x) cd 2 nghiem, Pn (x) = 0 cd 5 nghiem. -BDHSG DSGT12/1- 87 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
§ 4 . DUON G TIE M CA N CU A D O TH I HA M S O A. KIEN THlTC CO BAN - Duong thang x = x<j duoc goi la tiem can dung cua do thi ham sd y = f(x) neu it nhat mot trong cac dieu kien sau duoc thoa man: l im f(x) = +co; lim f(x) = +00; lim f(x) = -00; lim f(x) = -°o X->X~ x->X* X->X~ X->X* - Duong thing y = y 0 duoc goi la tiem can ngang cua do thi ham sd y = f(x) neu lim f(x) = y 0 hoac lim f(x) = y0 . - Duong thing y = ax + b, a * 0 duoc goi la tiem can xien cua do thi y = f(x) neu lim [f(x) - (ax + b)] = 0 hoac lim [f(x) - (ax+ b)] = 0. - Cong thuc chuyen he true Oxy thanh IX Y bang phep tinh tien OI f x = X + x 0 vdi diem I(x0 , y0 ): { [y = Y + y0 B. PHAN DANG TOAN DANG 1: TtM CAC TI $M CAN Cho dd thi (C): y = f(x) - Tiem can dung : Tim tap xac dinh D, xet a g D thi d: x = a la TCD khi cd mot trong bdn gidi han sau: lim f(x) = +00; lim f(x) = +00; hm f(x) = -00;' lim f(x) = -co x->a~ x^-a+ x->a~ x-+a 8* -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
- Tiem can ngang d: Tinh lim f(x), lim f(x). X-»-H» X->—J*3 Neu lim f (x) = b hoac lim f (x) = b thi y = b la TCN. - Tiem can xien d: y = ax + b vdi: a = lim ^ . b = lim(f(x)-ax) hoac a = hm ^ . b = lim(f(x)-ax) X-»+or T£ X-*+X X->-X X X-*-cr: Dac biet, neu chia tach duoc y = f(x) = ax + b + r(x) va lim r(x) = 0 thi X—*±r. tiem can xien: y = ax + b. . Chti y: - Neu tap xac dinh D = R thi khdng cd tiem can dung. - Khi tim tiem can dung la dudng thang x = x 0 thi x 0 la diem ma tai do ham sd khong xac djnh. Neu tdn tai mot khoang I chua diem XQ, sao cho f xac dinh tren I \ {x 0 } thi phai xet hai gidi han: lim f(x) valim f(x). Neu x—vx* x—>x) ( ca hai deu la gidi han vd cue thi ca hai nhanh cua do thi d ben phai va ben trai cua dudng thang x = x 0 deu nhan nd lam tiem can dung. - Bieu thuc tiem can khi x -» + oo: Vx2 + bx + c ~ x + — 2 - Dieu kien can de do thi cua mot ham sd cd tiem can ngang hoac tiem can xien la ham so do xac dinh tren khoang (-00; a) hoac tren khoang (b; +00). Tiem can ngang la mot trudng hop dac biet cua tiem can xien. Neu do thi ciia ham so cd tiem can ngang khi x -> +00 hoac khi x —> -00 thi nd khdng cd tiem can xien ve phia tuong ung va nguoc lai. Vi du 1: Tim cac tiem can dung cua do thi: , 3x - 7 x 2 + 4 , 2x- 3 Vx2 + 3 a ) y = — r b ) y = c ) y = ~ r" d ) y = — 7 - x - l x- 4 x+x+ 5 x- 5 Giai a) D = R \ {-1} . Ta cd lim y = +00, lim y = -co nen dudng thang x = 1 la x->l X-*l + tiem can dung (khi x —> 1 - va x —> V). b) D = R \ {-2; 2}. Ta cd lim y = +co, lim y = -co va limy = -co. x-K-2)- x-»(-2)* x->2" lim y = +co nen cd 2 tiem can dung: x = -2 va x = 2. x->2* c) D = R nen khong cd tiem can dung. d) D = R \ {5} . Ta cd limy = -co, limy = +00 nen dudng thang x = 5 la x—>5 x—>5+ tiem can dung. Vi du 2: Tim cac tiem can ngang cua do thi: 3x + 1 , w _ -2x2 + x + 1 a) y = —^ 7 b ) y " —Y 2 o x - 2 x —6 -BDHSG DSGT12/1- 89 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
c)y 3x + 2 x2 + x + 6 d)y = x - 2 \l4x2 + x + 1 Giai a) lim y = 3, lim y = 3 nen ducmg thang y = 3 la tiem can ngang (khi x JC->— y. x—>+x -oo vax- > +oo). b) lim y = -2 nen ducmg thang y = -2 la tiem can ngang. x->±x c) lim y = 0 nen ducmg thang y = 0 la tiem can ngang. d) Ta cd y = 1 - 1 1 x , 4 + - + — nen x x 'i- 2 lim y = lim v 1 - — ; lim y = lim 2 X-M-x X—•+*) 1 4+ + x x£ 1 1 1 2 4+ + x x 1 Dd thi co hai tiem can ngang y = — (khi x —» -oo) va y = — (khi x —» +oo). 2 2 V i du 3: Tim cac tiem can dung va ngang ciia do thi moi ham so sau: x2 - 6x + 7 , , 2x2 - x - 4 a) y c)y x - 4x + 5 8 - x x2 -4 x + 3 b)y d)y (x-l) 2 vV+ 2 Giai a) D = R nen khong cd tiem can diing. Ta cd lim y = lim 1 nen dudng thang y = 1 la tiem can x x' ngang (khi x -» +oo va x -> -oo). b) D = R \ {1} . Ta cd lim y = lim y = -oo nen tiem can dung la x = 1 (khi x->l x-*l* x -> 1 va x —> 1+ ). 90 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
2 - 1 4 2 Tacd limy = lim—-—~ = 2 nen tiem can ngang lay = 2. X->±X X->±t / - N Z 1 - c) D - R \ { 1 ; 3}. Ta co lim y = +oo, lim y = -co va lim y = +co, x-*l~ x-»l+ x->3 lim y = -co nen co 2 tiem can clung x = 1 va x = 3. x->3* Ta co lim y = 0 nen tiem can ngang y = 0. x->±x d) D = R\{0}. X v1+J |—2 _xi 1+J 1—2 Ta co limy =lim— = krn. 1+-— =1, limy =lim — = lim. l+- j =-1 nen co 2 tiem can ngang y = - 1 (khi x —> -co) va y = 1 (khi x —> +co). Ta co lim y = -co, lim y = +co nen tiem can dung x = 0. x->0" x->0* Vi du 4: Tim tiem can xien cua do thi ham so 4 x3 a) y = 2x - 5 + b) y - x - 2 x- l 3x2 -8x + 5 fl ~ d) y = Vx - x + 1 c) y = x + 5 Giai 4 a) Vi lim (y - (2x - 5)) = lim = 0 nen tiem can xien la dudng thang X-»±x JE->±« x — 2 y = 2x - 5 (khi x —> -co vax-> +oo). b) Tacd:ai= limlim x->+=o x x - >+: c X(X2 - 1) ( x3 bi = lim f f (x) - xl = lim — x v-v+^r L J x->+* I X 1 = hm-^-= 0 x 2 - l nen tiem can xien la dudng thang y = x (khi x -> +co). va ai = lim = 1 , bj = lim [f (x) - x] = 0 nen dudng thang y = x la x—« x x— tiem can xien cua dd thi (khi x -» -oo). x3 x Cach khac: y = ——- = x + x 2 - l " x 2 - l . , _ -3x2 -8x + 5_ 2 c) Ta co y = -3x + l + - ; ' x+ 3 x+ 3 2 V i lim (y - (-3x + 1)) = lim = 0 nen tiem can xien la duong thang x^iff y x + 3 y = -3x + 1. •BDHSG DSGT12/1- 91 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
X x-« x ^ U x J d) Ta cd: ai = bi = lim(v/ x2 -x + l-x) = lim V x 2 - x + l + x -1+ 1 = lim — X+ = lim x -— = — r_M* L 1 1 L 1 1 , 2 T~x + x2+X f-x + x2+1 Vay dudng thang y = x - — la tiem can xien (khi x -> +oo) va: 2 L - 1+1 2 .. Vx 2 - x + l a 2= hm = hm—J x x =- 1 x ~ w x x ^ x b2 = limfVx2 -x + 1+ x|= lim~p= X + ~*" ' *~ Vx2 -x + l-x 1 - T -X + 1 V X 1 - hm , = hm — x — = - x^- j ! r— [ f 2 -T-x + x2 - X tx^"1 nen dudng thang y = -x + - la tiem can xien cua do thi (khi x Yl du 5: Tim cac dudng tiem can ciia do thi moi ham so sau: aW- x~2 _ x2 -3x + 4 & > y ~ 7. 7: b) y = 3x + 2 2x + l , x + 2 x c) y = d) y -oo . x 2 - l x 3 + l Giai 2 a) D = R \ { — } . Ta cd lim y = +oo, lim y = -oo nen 3 -BJ H-i)' 2 1 1 TCD la x = —- Ta cd lim y = - nen TCN la y = - D6 thi khong co 3 *-±» 3 3 • 5 TCX. b) D = R\ {--}, Ta cd lim y =-oo, lim y =+oo nen TCD la x =-- 2 Tiem can xien va ngang cd dang y = ax + b IT 92 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
,. v x 2 - 3 x + 4 1 a = lim — = hm = — x-*±* x x->±« x (2x + 1) 2 b = lim y — = lim 2 ' x 2 •3x + 4 x^ x(2x + l) 2 = lim • -7x + 8 7 *•* 2(2x + 1) 4 x 7 x • nen dudng thane y = la tiem can xien cua dd thi. 2 4 c) D = R\ (-1; 1}. Taed hm y = +cova lim y = -co nen cd 2 TCD: x x-K-i)~ M-i)* Ta cd lim y = 0 nen TCN: y = 0. d) D = R\{-1}. Ta cd lim y =+co va lim y =-co nen TCD: x =- ' x-H-l)- x-(-l)* co hm y = 0 nen TCN: y = 0. x-»±* b) D = [-4; 2) u (2;+co) Ta cd lim y = -co, lim y = +co nen TCD: x = 2. x^2" x^2* Ta cd hm y = 0 nen TCN: y = 0 (khi x -> +=o) c) D = (0; +oo). Ta cd lim y = +co nen TCD: x = 0 (khi x -» 0+ ) x—>0 Q Ta co lim (y - x) = lim -j= nen TCX: y = x (khi x -> +co). x ^ + x x^+x ^/x Vi du 6 Tim cac dudng tiem can cua do thi moi ham sd sau: x3 + X + 1 , N _ x2 + X + 1 c) y = —= d) y x 2 - l ' J -5x 2 - 2 x + 3 Giai a) D = R\{0}.Tacd limy = limy =-oo nenTCD:x = 0. x—>o 9x — 1 va lim (y - (x - 3)) = lim — — = 0 nen TCX: y = x - 3 X->±X X->±X X b) D = R\ {0;2} suy ra 2 TCD: x = 0 vax = 2. x3 + 2 4x + 2 . , n Ta cd y = 2 = x + 2 + 2 - nen TCX: y = x + 2. J x 2 - 2 x x 2 - 2 x c) D = R\ {-1; 1} suy ra 2 TCD: x =-1 va x = 1. Tacdv= *3 +X + 1 =x+^^ nen lim (y - x) = 0, do dd TCX: y d) D = R\ {-1; -} suy ra 2 TCD: x =-1 vax=- 5 0 Ta cd lim y = —- nen TCN: y = —-. x->±x 5 o -BDHSG DSGT12/2- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Y j du 7: Tim cac ducmg tiem can cua do thi moi ham sd sau: a) y = x + Vx2 - 1 3 b)y Vx + 4 c)y = x x x - 2 d)y = Vx2 -4x + 3 Giai a) D = (-co; -1 ] u [1; +oo). D6 thi khong cd TCD. Ta cd a = lim — = lim x—>+x x x—>+-x 1 + Vx" 1^ b = lim(y - 2x) = lim (Vx2 - 1 - x l = lim nen tiem can xien : y = 2x (khi x -> +co). - 1 = 0 -1 + x lim y = lim (x + Vx2 - l ) = lim . = 0 i x - + - 0 0 vx 2 - 1 - x nen tiem can ngang: y = 0 (khi x —> -co). b) D = (-oo: 1] u [3; Dd thi khdng cd TCD. Goi y = ax + b la TCX thi: y Vx 2 - 4 x + 3 ai - lim — = lim hm J l + = 1 X X bi = lim(y - x) = lim(Vx2 - 4x + 3 - x) lim -4x + 3 -4 + V x ^ = lim x 4x + 3 + x i 4 3 i X -xV + 1 Vay tiem can xien: y = x - 2 (khi x —» +oo). y Y y Vx2 -4X + 3 a2 - lim — = lim 4 3 -X. 1- - + —r lim x x' •lim J l — + —- =-l x x b2 = lim(y + x) = lim(Vx2 -4 x + 3 + x) lim x—»+x -4x + 3 v; = lim -4x + 3 • 4x + 3 - x -xj l — + — - x x x -BDHSG DSGT11Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Vay tiem can xien: y = - x + 2 (khi x -» -co) Cachkhac:y= Vx2 -4x + 3 = |x-2| + (Vx2 -4x + 3 -jx-2[ vavi lim(Vx2 -4x + 3 -|x-2|) = 0 suy ra TCX. Vi du 8: Tuy theo m, tim cac tiem can cua do thi: . x2 + mx + 1 . , mx3 -1 a) y = — ; : b) y = 'x- l ~/ "' x 2 - 3 x + 2 Giai v , x2 + mx + l m + 2 , a) Ta co y = = x + m + 1 + , x 1 x - l x - l - Khi m -2 thi hm (y - (x + m + 1)) = hm m + 2 = 0 nen y = x + m + 1 X->±X X->±-T X ,, •» A . . ,,. x2 + mx +1 la tiem can xien. Ta co: hm = +oo x-»l* x- l Khi m > -2 va hm x + mx + _ = _a0 j^j m < _2 nen TCD x = 1. x-»l* x- l (x-l)2 - Khi m = -2 thi y = (vdi x * 1), do thi la dudng thang (tru x - l diem (1; 0)) nen nd trung vdi tiem can xien. , . ~ , mx3 - 1 , . , 7mx -l-6 m b) 1 a co: y = — = mx + 3m + — x 2 - 3 x + 2 x 2 - 3 x + 2 „, . , . , x3 -1 x2 +X + 1 . - Khi m = 1 thi y — = , x * 1, x * 2 x2 -3 x + 2 x- 2 1 . . x3 -8 x2 + 2x + 4 , . Khi m = — thi y = = . x * 1, x * 2. 8 8 ( x 2 - 3 x + 2) 8(x-l) Tu do suy ra: Vdi m = 1 thi x = 2 la tiem can dung Vdi m = — thi x = 1 la tiem can dung. 8 ' Vdi m * 1 va m * — thi do thi cd hai tiem can dung la x = 1 va x = 2. 8 ' Ta cd lim (y - (mx + 3m)) = lim m , X-1 ~ m = 0 nen dd thi cd TCN, *->±* x - 3x + 2 TCX: y = mx + 3m. Vi du 9: Tim cac tiem can cua do thi: a) y = tanx b) y = cotx -BDHSG DSGT12/1- 95 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
c) v = cos — X d) y = 7x - 9 + sin x Giai a) DK: x * — + kre, k e Z.Ta cd lim y = +00; lim y = -co nen co 2 fn ,]* (*. . V X-> —+ klTl X—>| —+ KIT | vd so TCD: x = — + kre, keZ . 2 b) DK: x * kre, k e Z. Ta cd lim y= -co, lim y = +co nen cd vo so x—>(krc)+ x->(krc)~ TCD: x = kre, k e Z. c) DK: x 0, vi lim y, lim y khong tdn tai nen cd TCD. x->0* ' x->0" Ta cd lim y = lim cos— = 1 nen cd TCN: y = 1. x->±x x d) DK: x * 0 ta cd lim y = -9 +1 = -8 nen khong cd TCD. x->0 vi lim(y - (7x - 9)) = lim = 0 nen TCX: y = 7x- 9. x—>±y x—>±x x DANG 2: BAI TOAN VE TlfiM CAN - Tim cac tiem can va giai cac bai toan lien quan. - Cong thuc chuyen he true Oxy thanh IXY bang phep tinh tien OI fx = X + xn vdi diem I(x0 , y0 ): [y = Y + y0 Vi du 1: Chung minh giao diem 2 tiem can dung va ngang la tam doi xung x-5 cua do thi (C): y D = R\ { - 2x + 3 Giai Ta co lim y = +co, lim y = -co nen TCD: x = -—. Ta cd lim y ~ — nen TCN: y = — 2 x->±« 2 2 3 1 Goi I la giao diem 2 tiem can thi I( — ; —). Giai he toa do trong phep 2 2 tinh tien theo vecto OI: x = X- ^ 2 y = Y + 1 96 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
13 ' • V i Y = F(x) = la ham so le nen do thi (C) nhan goc I la tam doi 4X xung. Vi du 2: Chung minh giao diem 2 tiem can dung va xien la tam doi ung cua do thi (C): y x2 -2 x + 2 x-3 Giai Ta cd y = x + 1 + —— nen (C) co TCD: x = 3 va TCX: y = x + 1, do do ' x- 3 giao diem 2 tiem can 1(3; 4). Chuyen he toa do trong phep tinh tien theo vecto 01: (X ^ + 3 JI\Q vao (Q thi duoc: l y = Y + 4 Y + 4 = X + 3 + l+ <=>Y = X + — X+3 - 3 X 5 Vi Y = F(X) = X + — la ham so le nen do thi (C) nhan goc I lam tam X doi xung. Vi du 3: Tim tham so m de do thi (C) cua ham so: —4x + 9 a) y = cd tam doi xung co toa do (3; -4) x - m b) y = x +uix + 19 ^ XUT1g c^ toa ^ ^. 2) x-5 Giai -4x + 9 a) Do thi (C): y = cd tiem can dung x = m va tiem can ngang y = -4 x - m nen giao diem I(m; -4) chuyen he toa do bang phep tinh tien 01 thi goc I(m; -4) la tam ddi xung, do do m = 3. Cach khac: Goi A(3; -4), chuyen he toa do bang phep tinh tien OA rdi dat dieu kien ham so le. b) Ta coy = x ^-m + 5+ ^m + 44 nen ,35 thi (C) cd TCD: x = 5 j^m * ~g—j va TCX: y = x + m + 5 nen giao diem 1(5; 10 + m). Chuyen he toa dp bang phep tinh tien 01 thi I la tam doi xung nen 10 + m = 2. Vay m = -8. -BDHSG DSGT12/1- 97 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
VJ du 4: Tim tam ddi xung cua do thi ham so y = ^ y = x + 2(cosa - sina) + x + 2 sin a Giai 1 + 4 sin a(sin a - cos a) x + 2 sin a T „ .. 1 + 4sina(sina-cosa) „ „ Vi lim ; = 0 nen tiem can xien: x->±« x + 2 sin a y = x + 2 (cosa - sina) Vi lim y = co nen tiem can dung: x = -2sina x—>—2sin ct Tam doi xung la giao diem 2 tiem can I(-2sina, 2cosa - 4sina). Vi du 5: Tim quy tich cac tam doi xung cua cac do thi (Cm): (m + l)x + m , , , 5 a) y = b) y = mx - 6m + 1 + x + m x - m Giai a) D = R \ {-m}. Ta cd limy = +co, limy = -co nen TCD: x = -m vdi x^(-m)- x—»(-m) + m*0. (m + l)x + m . T 1 a co hm y = hm = m + 1 nen TCN: y = m + 1 x->±* x->±* x + m Giao diem 2 tiem can I(-m; m+1), chuyen he true theo phep tinh tien 01: (X ^ m The vao (Cm) thi duoc: [y = Y + m + 1 Y + m+l= (m + 1)(X" m) + tKoY = — (X-m ) + m X m2 Ta cd Y = F(X) = — la ham sd le nen (C) cd tam ddi xung I cd toa do X x = -m, y = m + 1. Khu tham sd m thi quy tich cac tam ddi xung la dudng thang d: y = -x + 1, x ^ 0. b) Dd thi (Cm): y = mx - 6m + 1 + —, x^mco TCD: x = m va TCX: x - m y = mx - 6m + 1 nen giao diem I(m; m2 - 6m + 1). Chuyen he true bang phep tinh tien OI thi dugc tam doi xung I cd toa do: x = m, y = m2 - 6m + 1. Khu tham so m thi quy tich cac tam doi xung la parabol (P): y = x2 - 6x + 1. \r> A c nu u< - mx2 +(3m2 -2)x-2 Vi du 6: Cho ham so y = x + 3m Tim m de gdc giua 2 tiem can bang 45° Giai T - ^ o . 6m 2 1 Taco: y = mx-2 + . m * — x + 3m 3 98 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com