ThS . L E HOAN H PH O Nhd gido Uu tu C c c M i l BOIDU8N G 7 H O C SIN H GIO I TOA N D A I SO-GIA I TIC H - Ddnh cho HS lop 12 on tap & nang cao kinang lam bai. - Chudn bi cho cdc ki thi quoc gia do Bo GD&DT to choc Mil NHA XUAT BAN DAI HQC QUOC GIA HA NQI
Bin DUSNG , HQC SINH GO TOAN OAI SO -GIAI TICH Boi duQng hoc sinh gioi Toan Dai so 10-1. Boi duQng hoc sinh gioi Toan Dai so 10-2. - Boi duQng hoc sinh gioi Toan Hinh hoc 10. - Boi duOng hoc sinh gioi Toan Dai so 11. Boi duQng hoc sinh gioi Toan Hinh hoc 11. Bp de thi tif luan Toan hoc. Phan dang va pht/Ong phap giai Toan So phtfc. Phan dang va phucing phap giai Toan To hop va Xac suat. 1234 Bai tap tir luan dien hinh Dai so giai tich 1234 Bai tap ta iuan dien hinh Hinh hoc li/ong giac Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
ThS . L E HOAN H PH O Nha gido iCu tu BOIDUQN G , H O C SIN H GIO I TOA N DAI SO-GIAI TICH 12 * - Danh cho HS lap 12 on rflp & ndng cao kfndng lam bai. - Chudn bj cho cdc ki thi qudc gia do Bo GD&DT td choc. Ha Npi NHA XUAT BAN DAI HQC QUOC GIA HA NQI Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
NHA XUAT BAN DAI HOC QUOC GIA HA N0 I 16 Hang Chudi - Hai Ba Trirng Ha Npi Dien thoai: Bien tap-Che ban: (04) 39714896; Hanh chinh: (04) 39714899; Tdng bien tap: (04) 39714897 Fax: (04) 39714899 Chiu trach nhiem xuat bdn: Giam ddc PHUNG QUOC BAO Tong bien tap PHAM TH I TRAM Bien tap noi dung THUY NGAN Sda bdi NGOC HA N Che bdn CONG TI ANPHA Trinh bay bia SON KY Ddi tdc lien ket xudt bdn CONG TI ANPHA SACH LIEN KET BOI DUONG HQC SINH GIOI TOAN DAI SO GIAI TICH 12 -TAP 1 Ma so: 1L-177DH2010 In 2.000 cuon, khd 16 x 24 cm tai cong ti TNHH In Bao bi Hung Phu So' xua't ban: 89-2010/CXB/11-03/DHQGHN, ngay 15/01/2010 j Quyet dinh xua't ban sd: 177LK-TN/XB In xong va nop ltiu chieu quy I I nam 2010. Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
L d i N6 I DA U De giup cho hoc sinh lap 12 co them tai lieu tu boi duong, ndng cao va ren luyen ki ndng gidi todn theo chuong trinh phdn ban mdi. Trung tdm sdch gido due ANPHA xin trdn trong giai thieu quy ban dong nghiep vd cdc em hoc sinh cuon: "Boi dudng hqc sinh gioi todn Dai so' Gidi tich 12 " nay. Cuon sdch nay nam trong bo sdch 6 cuon gom: - Boi ducmg hoc sinh gidi todn Hinh hoc 10. - Bdi ducmg hoc sinh gidi todn Dai so' 10. - Boi dudng hoc sinh gidi todn Hinh hoc 11. - Boi dudng hoc sinh gidi todn Dai so - Gidi tich 11. - Bdi dudng hoc sinh gidi todn Hinh hoc 12. - Boi dudng hoc sinh gidi todn Gidi tich 12. do nhd gido uu tu, Thac sTLe Hoanh Phd to'chirc bien soan. Ndi dung sdch duoc bien soan theo chuong trinh phdn ban: co bdn vd nang cao mdi ciia bd GD & DT, trong dd mot so van de duoc md rdng vdi cdc dang bdi tap hay vd kho dephuc vu cho cdc em yeu thich mud'n ndng cao todn hoc, cd dieu kien phdt trien tot nhat kha nang ciia minh. Cuon sdch lasu ke thira nhung hieu bii't chuyen mdn vd kinh nghiem gidng day ciia chinh tdc gid trong qua trinh true tiep dirng ldp bdi dudng cho hoc sinh gidi cdc ldp chuyen todn. Vdi ndi dung sue tich, tdc gid da co'gang sap xep, chon loc cdc bai todn tieu bieu cho tirng the loai khdc nhau ung vdi ndi dung cua SGK. Mdt sd'bdi tap cd the khd nhung cdch gidi duqc dua tren nen tdng kien thuc vd ki nang co bdn. Hqc sinh can tu minh hoan thien cdc ki nang ciing nhu phdt trien tu duy qua viec gidi bdi tap cd trong sdch trudc khi ddi chieu vdi led gidi cd trong sdch nay, cd the mdt soldi gidi cd trong sdch con cd dong, hqc sinhcd thetu minh lam rd hon, chi tie't hon, ciing nhu tie minh dua ra nhung cdch lap ludn mdi hon. Chung tdi hy vong bd sdch nay se la mdt tdi lieu thie't thuc, bo ich cho ngudi day vd hqc, dqc biet cdc em hqc sinh yeu thich mdn todn vd hqc sinh chuan bi cho cdc ky thi qudc gia (tot nghiep THPT, tuyen sinh DH - CD) do bq GD & DT to chirc sap tdi. Trong qua trinh bien soqn, cudn sdch nay khdng the tranh khoi nhirng thieu sdt, chung tdi ra't mong nhdn duqc gop y ciia ban dqc gdn xa debq sdch hoan Men hon trong lah tdi ban. Moi y kien dong gop xin lien he: - Trung tam sach giao due Anpha 225C Nguyen Tri Phuong, P.9, Q.5, Tp. HCM. - Cong ti sach - thiet bj giao due Anpha 50 Nguyen Van Sang, Q. Tan Phii, Tp. HCM. DT: 08. 62676463, 38547464 . Email: [email protected] Xin chan thanh cam on! Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
M U C LU C Chuong I : tTng dung dao ham de khao sat va ve do thi cua ham so § 1. Tinh don dieu cua ham so 5 Dang 1: Dong bien, nghich bien, ham hang 5 Dang 2: Ung dung tinh don dieu 17 §2. Cue tri ciia ham so 37 Dang 1: Cue dai, cue tieu , 37 Dang 2: Ung dung ciia cue tri 48 §3. Gia tri Ion nhat va gia tri nho nhat 58 Dang 1: Tim gia tri ldn nhat, nho nhat 58 Dang' 2: Bai toan lap ham so 69 Dang 3: Ung dung vao phuong trinh 77 §4. Duong tiem can cua do thi ham so 88 Dang 1: Tim cac tiem can 88 Dang 2: Bai toan ve tiem can 96 §5. Khao sat va ve ham da thuc 103 Dang 1: Ham bac ha 104 Dang 2: Ham trung phuong 113 §6. Khao sat va ve ham hOu ti 126 Dang 1: Ham so v = a x + k (c * 0 va ad be * 0) 126 cx + d 2 i Dang 2: Ham s6 v = &X + (a * 0. a' * 0) 135 a'x + b' §7. Bai toan thuong gap ve do thi 148 Dang 1: Tuong giao, khoang each, goc 149 Dang 2: Tiep tuyen. tiep xuc 159 Dang 3: Yeu to co dinh. doi xung - quy tich 170 Chirong II : Ham so luy thua ham so mu va ham so logarit § 1. Quy tac bien doi va cac ham so 186 Dang 1: Bien doi luy thua - mu - logarit 188 Dang 2: Cac ham so mu. luy thua, logarit 200 Dang 3: Bat dang thuc va GTLN, GTNN 212 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
CHUON G I : UN G DUN G DA O HA M O E KHA O S A T V A V E D O TH j CU A HA M S O §1. TINH DON DIEU CUA HAM SO A. KIE N THLTC CO BAN • Dinh nghTa: Ham so f xac dinh tren K la mot khoang, doan hoac nira khoang. - f dong bien tren K neu vdi moi Xi, X2 6 K: X] < X2 => f(xi) < f(x2) - fnghich bien tren K neu vdi moi xi. xi e K: Xi<X2=>f(xi)>f(x2). • Dieu kien can de ham so don dieu Gia sir ham so co dao ham tren khoang (a; b) khi do: - Neu ham so f dong bien tren (a; b) thi f'(x) > 0 vdi moi x e (a; b) - Neu ham so f nghich bien tren (a; b) thi f'(x) < 0 vdi moi x e (a; b). • Dieu kien du de ham so don dieu - Gia sir ham so f co dao ham tren khoang (a; b) Neu f'(x) > 0 voi moi x e (a; b) thi ham so f dong bien tren (a; b) Neu f'(x) < 0 voi moi x e (a; b) thi ham so nghich bien tren (a; b) Neu f'(x) = 0 vdi moi x e (a; b) thi ham so f khong doi tren (a; b). - Gia sir ham so f co dao ham tren khoang (a; b) Neu f '(x) > 0 (hoac f '(x) < 0) vdi moi x e (a; b) va f '(x) = 0 chi tai mot so huu han diem cua (a; b) thi ham so dong bien (hoac nghich bien) tren khoang (a; b). B. PHAN DANG TOA N DANG 1: B6NG BliN , NGHICH BIEN, HAM HANG • Phuong phap xet tinh don dieu: - Tim tap xac dinh - Tinh dao ham, xet dau dao ham, lap bang bien thien - Ket luan Chii y: - Dau nhi thuc bac nhat: f(x) = ax + b, a ^ 0 x -00 -b/a +co f(x) trai dau a 0 ciing dau a - Dau tam thuc bac hai: f(x) = ax2 + bx + c, a * 0 Neu A < 0 thi f(x) luon ciing dau vdi a Neu.A = 0 thi f(x) luon cung dau vdi a, trir nghiem kep Neu A > 0 thi dau "trong trai - ngoai ciing" X -CO X] X2 +00 f(x) ciing dau a 0 trai dau a 0 ciing diu a -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 1: Xet chieu bien thien ciia ham sd: a) y = x2 - 6x + 5 c) y = x3 - 2x2 + x + 1 b)y = -x 3 3 2x2 + x - 3 d) y = -x 3 + 4x2 - 7x + 5 Giai a) Tap xac dinh D = R. Ta co y' = 2x - 6. Cho y' = 0 » 2x - 6 = 0 » x = 3. Bang bien thien X —oo 3 +00 y' - 0 + y — Vay ham so nghich bien tren (-oo; 3), dong bien tren (3; +oo). b) D = R. Ta cd y' = 4x2 - 4x + 1 = (2x - l) 2 > 0 vdi moi x y' = 0 o x = —. Vay ham so dong bien tren R. 2 c) D = R . Ta co y' = 3x2 - 4x + 1 Cho y' = 0 o 3x2 - 4x + 1 = 0 <=> x = - hoac x = 1. J 3 BBT X —00 1/3 1 +00 y' + 0 0 + y ^ * — ^ Vay ham so dong bien tren moi khoang (-co; —) va (1; +oo), nghich bien 3 tren khoang (—; 1). 3 d) D = R Ta cd y' = -3x 2 + 8x - 7 ViA' = 16-21< 0 nen y' < 0 vdi moi x do do ham so nghich bien tren R. V i du 2: Xet chieu bidn thien cua cac ham so sau: a) y = x4 - 2x2 - 5 b) y = x4 + 8x2 + 9 Giai a) D = R. Ta co y' = 4x3 - 4x = 4x(x2 - 1) Cho y' = 0 <=> 4x(x2 - 1) = 0 <=> x = 0 hoac x = ±1 BBT X —00 - 1 0 1 +00 y' - o •+ 0 - 0 + y ^ ^ * ^ ^ Vay ham sd nghich bien tren moi khoang (-co; -1) va (0; 1), ddng bidn tren moi khoang (-1 ; 0) va (1; +=»)• 6 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
b) D = R. Ta co y' = 4x3 + 16x = 4x(x2 + 4),y' = 0o x = 0. y' > 0 tren khoang (0; +co) => y dong bien tren khoang (0; +co) y' < 0 tren khoang (-co; 0) => y nghich bien tren khoang (-co; 0). V i du 3: Xet su bien thien cua ham so: a) y = x + — x b)y c)y a) Tap xac dinh D = R\ {0} _3_ ,.2 Ta co y' = 1 BBT: Giai , y' = 0 <» X x 2 - 3 3x- 8 1-x :V3" d)y 1 (x-4) 2 X -co QN/ 3 +00 y' + 0 - - 0 + y Vay ham so dong bien tren khoang (-co; -^ 3 ) va ( J 3 ; +oo), nghich bien tren m6i khoang (- S ; 0) va (0; V3 ). b) D = R \ {0}. Tacdy' = 1 > 0 vdi moi x ^ 0 nen ham so dong bien tren moi khoang (-oo; 0) va (0; +co). -5 c) D = R \ {1} . Ta cd y' = - < 0 vdi moi x -t- 1 nen ham so nghich 3 (1-x) 2 bien trong cac khoang (-co; 1) va (1; +oo). d) D = R\ {4}.Tacdy'= ———- (x-4) 3 y' < 0 tren khoang (4; +co) nen y nghich bien tren khoang (4; +co). y' > 0 tren khoang (-co; 4) nen y ddng bien tren khoang (-co; 4). V i du 4: Tim cac khoang don dieu ciia ham so: , x- 2 2x a) y = -5 b) y X + X + 1 a) D = R. Ta cd: y' x 2 - 9 Giai - x 2 + 4x + 3 (x2 + x + l) 2 y' = 0 e> x2 - 4x - 3 = 0 <=> x = 2 ± ^7 BBT: X -co 2-V7 2+V7 +°o y' - 0 + 0 - y — ^ — * - * -BDHSG DSGT12/1- 7 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Vay ham so dong bien tren khoang (2 - yfl ; 2 + V7 ) va nghich bien tren cac khoang (-co; 2 - 77 ) va (2 + ; +oo). b) D = R\{-3;3}.Tac6y'=^^ <0,Vx*±3. (x2 -9) 2 Do do y' < 0 tren cac khoang (-co; -3), (-3; 3), (3; +oo) nen ham so da cho nghich bien tren cac khoang do. V i du 5: Xet su bien thien cua ham sd: a) y = V4-x 2 c)y Vl6- : b)y = Vx2 -2 x + 3 d)y x + 2 Giai a) Dieu kien 4-x 2 >0<=>-2<x< 2 nen D = [-2; 2] V d i -2 < x < 2 thi y' BBT: V 4 ~ , y ' = 0<=>x = 0. X -2 n 2 y' + 0 - y ^ ^ — ^ Vay ham so dong bien tren khoang (-2; 0) va nghich bien tren khoang (0; 2). Do ham so f lien tuc tren doan [-2; 2] nen ham so dong bien tren doan [-2; 0] va nghich bien tren doan [0; 2]. b) V i A' = 1 - 3 < 0 nen x 2 - 2x + 3 > 0, Vx => D = R. •p , 1 2x- 2 x - l , . , Ta co y = — = = = . y = 0 o x = 1. 2vx 2 - 2 x + 3 Vx 2 - 2 x + 3 y'>0ox>l,y'<0ox< l nen ham so nghich bien tren khoang (-co; 1) va dong bien tren khoang (1; +00). c) DK: 16 - x2 > 0 o x2 < 16 o -4 < x < 4. D = (-4; 4). Ta co v' = 1 6 > 0, Vx e (-4; 4). (16-x 2 )Vl6 - x 2 Vay ham so dong bien tren khoang (-4; 4). d) D = [0; +00). Vdi x > 0, y' = X _, y' 2^y^(x + 2) 2 BBT: X 0 2 +00 y + 0 y ^ Vay ham sd ddng bien tren (0; 2) va nghich bien tren (2; +00). 8 -BDHSG DSGTu/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V j du 6: Tim khoang don dieu cua ham so a) y = Vx(x - 3) c)y b)y = - x 7 x 2 - 6 d)y = x + 1 Vl- x Giai a) D = [0; +oo). Vdi x > 0, ta cd: y BBT: 1 2Vx (x-3 ) + r 3NRX-1 ) vx = , y 2x 0<=>x= 1. X 0 1 +GO y' 0 + y ——— ^ Vay ham so nghich bien tren khoang (0; 1) va dong bien tren khoang (i';+°o). b) D = R. Vdi x ^ 0, ta co: y' =— - 3 3vV 3v V y' = 0 <=> x2 = 1 <=> x = ±1. y' > 0 o ^/x2 " >l<=>x 2 >lci>x<- l hoac x > 1. y' < 0 %/x2 " <le > x 2 <lo-Kx<l . Vay ham so dong bien tren cac khoang (-co; -1) va (1; +co), nghich bien tren khoang (-1 ; 1). c) Tap xac dinh D = (-co; -^6 ) U (x/6 ; +oo). Tacd: y' = -^^7^£L,y1 = 0»x = ±3. ( x 2 -6)v'x 2 - 6 BBT: X —CO -3 V6 V6 3 +CO y' + 0 - • _ 0 + y • Vay ham so dong bien tren cac khoang (-co; -3) va (3; +oo). nghich bien tren cac khoang (-3; -v o ) va (v o ; 3). d) D = (-QO; 1). Tacdy' = , 3 ~ X > 0, Vx<l . 2 V (l-x) 3 Vay ham sd dong bien tren khoang (-co: 1). -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V j du 7: Xet su bien thien cua ham sd: 3 a)y--- x + smx b ) y = x + cog 2 x Giai 3 a) D = R. Ta cd y' = - - + cosx < 0, Vx nen ham sd nghich bien tren R. b) D = R. Ta cd y' = 1 - 2cosxsinx = 1 - sin2x y' = 0o sin2x = 1 <=> x = - +kit,keZ. 4 Ham sd lien tuc tren moi doan [- + kn; — + (k + 4 4 va y' > 0 tren moi khoang (- + kn; - + (k + 1)TI) nen ddng bien tren 4 4 moi doan [- + kn; - + (k + l)7tl, keZ. 4 4 v ' 1 Vay ham so dong bien tren R. Vi du 8: Tim khoang dong bien, nghich bien cua ham so: a) y = x - sinx tren [0; 2TT] b) y = x + 2cosx tren (0; n). Giai a) y' = 1 - cosx. Ta cd Vx [0; 2n] => y1 > 0 va y' = 0 <=> x = 0 hoac x = 2n. Vi ham so lien tuc tren doan [0; 2n] nen ham so ddng bien tren doan [0; 2n]. b) y' = 1 - 2 sinx. Tren khoang (0; 7t). y'>0o sinx <-<=> - < x < — 2 6 6 y' < 0 » sinx > - <=>0<x< - hoac — < x < - 2 6 6 6 Vay ham so ddng bien tren khoang (-; —). nghich bien tren moi 6 6 khoang (0; — ) va (—; n). 6 6 Vi du 9: Chung minh cac ham sd sau nghich bien tren R: a) f(x) = vx2 +1 - x b) f(x) = cos2x - 2x + 5. Giai a) Tacdf'(x) = T^=-l. Vx +1 Vi Vx2 +1 > Vx2 = I x | > x, Vx nen f '(x) < 0, Vx do dd ham sd f nghich bien tren R. b) f'(x) = -2(sin2x+ 1)<0 vdi moi x. 10 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
f'(x) = 0osin2 x = -lc^>2x = - - +2kno x = - - +kn,k& Z. 2 4 Ham f(x) lien tuc tren moi doan [- - + kn; ~ + (k + \)n] va f'(x) < 0 tren moi khoang (-— +kn; — + (k+l)n) nen ham so nghich bien tren moi doan 4 4 [ - - +k ;r;- - +(k + l)n], k e Z. 4 4 Vay ham sd nghich bien tren R. Cach khac: Ta chung minh ham sd f nghich bien tren R: VXj, x2 e R, xx < x2 => f(Xj) > f(x2 ). That vay, lay hai sd a, b sao cho a < X| < X2 < b. Ta cd: f'(x) = -2(sin2x + 1) < 0 vdi moi x e (a; b). V i f '(x) = 0 chi tai mot sd huu han diem cua khoang (a; b) nen ham sd f nghich bien tren khoang (a; b) => dpcm. Vi du 10: Chung minh rang cac ham so sau day dong bien tren R. a) f(x) = x3 - 6x2 + 17x + 4 b) f(x) = 2x - cosx + S sinx. Giai a) f'(x) = 3x2 - 12x + 17. V i A' = 36 - 51< 0 nen f'(x) > 0 vdi moi x, do dd ham so dong bien tren R. V3 b) y' = 2 + sinx - v3 cosx = 2(1 + — sinx cosx). 2 2 = 2[1 + sin(x - —)] > 0, vdi moi x. 3 Vay ham sd ddng bien tren R. V i dull : Chung minh ham so: x - 2 a) y = ddng bien tren moi khoang xac dinh cua nd. x + 2 - x 2 — 2x + 3 b) y = nghich bien tren moi khoang xac dinh cua nd. x + 1 a) D = R \ {-2} . Ta cd y' = — > 0 vdi moi x * -2 Giai 4_ (x + 2) 2 Vay ham so dong bien tren moi khoang (-oo; -2) va (-2; +oo). x2 -2x- 5 b) D = R\{-l}.Tacdy' = ~ < 0 vdi moi x *- l (vi A' = 1 - 5 < 0). (x + 1) 2 ' v ' Vay ham so nghich bien tren mdi khoang (-oo; -1) va (-1 ; +oo). V i du 12: Chung minh ham so: -BDHSG DSGT12/1- 1 1 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
a) y - i + x 2 dong bien trong khoang (-1 ; 1) va nghich bidn trong cac khoang (-co; -1) va (1; +oo). , . sin(x + a) , , _ , , b ) y ~ ( a ^ b + krt; k e Z) don dieu Uong mdi khoang xac dinh. sin(x + b) • ° Giai , , l ( l + x 2 )-2x . x 1-x2 a ) y = (i + x 2 ) 2 = (T^' y = 0 o x = ± 1 - Ta cd y' > 0 <=> 1 - x2 > 0 <=> - 1 < x < 1. y '<0 < = >l-x 2 <0<=>x<- l hoac x > 1. Tir do suy ra dpcm. b) Ham sd gian doan tai cac diem x = -b + kn (k e Z). , _ sin(x + b) cos(x + a) - sin(x + a) cos(x + b) _ sin(b-a ) sin2 (x + b) sm2 (x + b) (do a - b * kn) Vi y' ?t 0 va y' lien tuc tai moi diem x * -b + kn, nen y' giu nguyen mot dau trong moi khoang xac dinh, do do ham so don dieu trong moi khoang do. V i du 13: Chung minh: a) sin2 x + cos2 x = 1, Vx. b) cosx + sinx. tan— = 1, Vx e (-— ; —). 2 4 4 Giai a) Xet f(x) = sin2 x + cos2 x, D - R. f '(x) = 2sinxcosx - 2cosxsinx = 0, Vx. Do dd f(x) la ham hang tren R nen f(x) = f(0) = 1. b) Xet f(x) = cosx + sinx tan-, D = (-—; — ). 2 4 4 r-,/ x x sinx . x x 1 (x) = -smx + cosxtan— + = -sinx + cosx.tan— + tan— • 2 2cos 2 - 2 2 2 X X i X = -sinx + tan — (1 + cosx) = -sinx + tan— .cos — 2 2 2 —sinx + sinx = 0 voi moi x e (— ; —) 4 4 , , TT 71 Suy ra rang f la mdt ham hang tren khoang (-— ; — ). Do dd f(x) = f(0) = 1 vdi moi x e (-—; - )• 4 4 V i du 14: Chung minh cac ham so sau la ham khong ddi -BDHSG DSGTU/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
a) f(x) = cos2 x + cos2 (x + —) - cosxcos(x + ^ ) 3 3 b) f(x) - 2- sin2 x - sm2 (a + x) - 2cosa.cosx.cos(a + x). Giai a) f'(x) = -2cosxsinx - 2cos(x + - )sin(x + - ) + sinxcos(x +^ ) + cosx.sin(x + ^ ) 3 3 o o o 71 7T = -sin2x - sin(2x + — ) + sin(2x + - ) = -sm2x - 2cos(2x + -).sin - 3 3 2 b = -sin2x - cos(2x + — ) = 0, vai moi x. 2 1 1 3 Do do f hang tren R nen f(x) = f(0) = 1 + = - 6 w w 4 2 4 b) Dao ham theo bien x (a la hang so). f '(x) = -2sinxcosx - 2cos(a + x)sin(a + x) + 2cosa[sinxcos(a + x) + cosx.sin(a + x)]. = -2sin2x - sin(2x + 2a) + 2cosa.sin(2x + a) = 0. Do do f hang tren R nen f(x) = f(0) = 2 - sin2 a - 2cos2 a = sin2 a. V i du 15: Cho 2 da thuc P(x) va Q(x) thoa man: P'(x) = Q'(x) vdi moi x va P(0) = Q(0). Chung minh: P(x) = Q(x). Giai Xet ham so f(x) = P(x) - Q(x), D = R. Ta cd f '(x) = P'(x) - Q'(x) = 0 theo gia thiet, do do f(x) la ham hang nen f(x) = f(0) = P(0) - Q(0) = 0 vdi moi x. f(x) = 0 => P(x) ^ Q(x). V i du 16: Xac dinh ham so f(x) thoa man: f(0) = 8; f(x).f '(x) = 1 - 2x (*). Giai Ta cd (*) -(f (x))* = l-2xo (f3 (x))' = 3 - 6x. 3 Xet ham sd g(x) = f 3 (x) - 3x + 3x2 thi g'(x) = (f (x))' - 3 + 6x = 0. nen g(x) = C: hang so tren D, do do: f(x) - 3x + 3x2 = C ^> f 3 (x) = -3x2 + 4x + C. nen f(x) = N/-3X2 + 3x + C V i f(0) = 8 => C = 64. Vay f(x) = yj-3x2 + 3x + 64 , thu lai dung. V i du 17: Tim cac gia trj cua tham so a de ham so dong bien tren R. a) f(x) = - x3 + ax2 + 4x + 3 b) f(x) = ax3 - 3x2 + 3x + 2 • 3 Giai a) f '(x) = x2 + 2ax + 4, A' = a2 - 4 - N6u a2 - 4 < 0 hay -2 < a < 2 thi f '(x) > 0 vdi moi x e R nen ham so ddng bien tren R. -BDHSG DSGT12/1- 13 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
- Neu a = 2 thi f '(x) = ( x + 2) 2 > 0 vai moi x * -2 nen ham sd dong bien tren R. - Neu a = -2 thi ham sd f '(x) = (x - 2) 2 > 0 vdi moi x * 2 nen ham so dong bien tren R. - Neu a < -2 hoac a > 2 thi f '(x) = 0 cd hai nghiem phan biet nen f 1 co doi dau: loai. Vay ham sd ddng bien tren R khi va chi khi -2 < a < 2. b) V (x) = 3ax2 - 6x + 3. Xet a = 0 thi f '(x) = -6x + 3 cd doi dau: loai Xet a * 0, vi f khong phai la ham hang (y' = 0 tdi da 2 diem) tren dieu kien ham so dong bien tren R la f '(x) > 0, Vx ( a>0 fa> 0 f a > 0 <=> <^=> <=> «a > 1. [A'< 0 [9-9a< 0 |a> l Y l du 18: Tim cac gia tri cua m de ham sd nghich bidn tren R: a) f(x) mx - x b) f(x) = sinx - mx + C. Giai a) y' = m - 3x2 - Neu m < 0 thi y' < 0 vdi moi xe R nen f nghich bien tren R - Neu m = 0 thi y' = -3x 2 < 0 vdi moi x e R, dang thuc chi xay ra vdi x = 0, nen ham so nghich bien tren R. m - Neu m > 0 thi y' = 0 o x = ± ^ BBT X —00 Xl x2 +00 y' 0 H 0 r y — » • Do do ham so dong bien tren khoang (xi, x2 ): loai Vay ham so nghich bien tren R khi va chi khi m < 0. b) V i f(x) khong la ham hang vdi moi m va C nen f(x) = sinx - m + C nghich bien tren R <=> f '(x) = cosx - m < 0, Vx a> m > cosx, Vx o m > 1. V i du 19: Tim m dd ham sd dong bien tren moi khoang xac dinh: a) y (3m -l)x - m2 + m by = x + 2 + x + m m x - l Giai a) D = R\ {-m} . Ta co: , _ (x + m)(3m -1) - [(3m - l)x - m2 + Y = ' (x + m) 2 14 m 4 m 2 - 2 m (x + m) 2 -BDHSG DSGTU/lDownload Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ham so dong bien tren moi khoang xac dinh <=> 4m2 - 2m > 0 <=> m < 0 hoac m > —. 2 b) Ta cd y' = 1 m vdi moi x * 1. (x-l) 2 - Neu m < 0 thi y' > 0 vdi moi x * 1. Do do ham sd dong bien tren moi khoang (-oo; 1) va (1; +oo). XT' , x 2 - 2 x + l - m - Neu m > 0 thi y = -„ (x-l) 2 y' = 0ox 2 -2x+l- m = 0<=>x=l + Vm BBT X —oo 1 —Vm 1 1 + Vm +oo y' + 0 - 0 + y Ham sd nghich bien tren moi khoang (1 - Vm ; 1) va (1; 1 + Vm ): loai. Vay ham sd ddng bien tren moi khoang xac. dinh cua nd khi va chi khi m<0 . V i du 20: Tim a de ham sd: a) f(x) = x3 - ax2 + x + 7 nghich bien tren khoang (1; 2) b) f(x) = — x3 - — (1 + 2cosa)x2 + 2xcosa + 1, a e (0; 2rr) dong bien tren 3 2 khoang (1; +oo). Giai a) f'(x) = 3x 2 -2ax + 1 Ham so nghich bien tren khoang (1; 2) khi va chi khi y < 0 vdi moi x e (1;2) ff(l) < 0 f 4-2a< 0 13 <=> < <=>< <=>a>— [f(2)<0 [l3-4a< 0 4 b) y' = x2 - (1 + 2cosa)x + 2cosa. Ta cd 0 < a < 2TT. y' = 0 o x = 1 hoac x = 2cosa. Vi y' > 0 d ngoai khoang nghiem nen ham so dong bien vdi moi x > 1 khi va chi khi 2cosa < 1 cosa < — o — < a < — 2 3 3 V i du 21: Tim m de ham sd y = (m - 3)x - (2m + l)cosx nghich bien tren R. Giai: y' = m - 3 + (2m - l)smx Ham so y khdng la ham hang nen y nghich bien tren R: y' ^ 0, Vx « m - 3 + (2m - l)smx < 0, Vx D5t t = sinx, - 1 < t < 1 thi m - 3 + (2m - l)smx = m - 3 + (2m - l)t = f(t) -BDHSG DSGT12/1- 15 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Dieu kien tuang duong: f(t) < 0, Vt e [-;1 1] [f(-l ) < 0 f-m-4< 0 ' 9 °lf(l)^ 0 °l3m-2^ 0 ^- 4 ^ m ^ V» du 22: Tim m de ham sd y = x3 + 3x2 + mx + m chi nghich bidn tren mpt doan cd dp dai bang 3. Giai: D = R, y' = 3x2 + 6x + m, A' = 9 - 3m Xet A' < 0 thi y' > 0, Vx: Ham luon dong bien (loai) Xet A' > 0 <=> m < 0 thi y' = 0 cd 2 nghiem x,, x2 nen x, + x2 = -2, X]X2 = — 3 BBT: x —CO *1 x2 +00 y' + 0 - 0 + y ^ ^ Theo de bai: x2 - X] = 3 o (x2 - x,) 2 = 9 o x2 + x2 - 2 x t x 2 = 9 4 15 <=> (x2 + x t) 2 - 4xtx2 = 9ci>4 — m = 9 o m = (thoa) 3 4 V i du 23: Tuy theo tham &6 m, xet su bien thien cua ham sd: \ 1 3 ? , rx ix 2x + m a) y = - x3 - 2mx2 + 9x - m b) y = 3 x- l Giai a) D = R. Ta cd y' = x2 - 4mx + 9; A' = 4m2 - 9 - Neu A' < 0 <o 4m2 < 9 <=> bien tren R. I m [ < — thi y' > 0, Vx nen ham so dong - Neu A' > 0 co 4m2 > 9 co I m | > - thi y' = 0 cd 2 nghiem phan biet xi,2 = 2m +V4m2 -9 Lap bang bien thien thi ham ddng bien tren khoang (2m - V^m2 - 9 ; 2m +V4m 2 - 9 ) va nghich bien tren m6i khoang (-00; 2m - \/4m2 - 9 ) va (2m + V4m2 - 9 ; +00). b)D = R\ {l}.Tacd y'= ~2 ~"! (x-l) 2 - Neu m = - 2 thi y = 1, Vx * 1 la ham sd khong doi. - Neu m > -2 thi y' < 0, Vx *1 nen ham so nghich bien tren moi khoang (-00; 1) va (1; +00). - Neu m < -2 thi y' > 0, Vx * 1 nen ham sd ddng bien tren moi khoang (-co; 1) va (1; +00). 16 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
DANG 2: UNG DUNG TINH BON DI$U - Giai phirong trinh, he phirong trinh, bat phuong trinh: Xet f(x) la ham so v6 trai, neu can thi bien doi, chpn xet ham, dat an phu, .... Tinh dao ham rdi xet tinh don dieu. Neu ham sd f don dieu tren K thi phuong trinh f(x) = 0 cd toi da 1 nghiem. Neu f(a) = 0, a thupc K thi x = a la nghiem duy nhat cua phuong trinh f(x) = 0. Neu f cd dao ham cap 2 khdng ddi dau thi f la ham don dieu nen phuong trinh f(x) = 0 cd tdi da 2 nghiem. Neu f(a) = 0 va f(b) = 0 vdi a * b thi phuong trinh f(x)=0 chi cd 2 nghiem la x = a va x = b . - Chiing minh bat dang thiic: Neu y = f(x) cd y' > 0 thi f(x) dong bien: x > a => f(x) > f(a); x < b =>f(x)<f(b ) Doi vdi y' < 0 thi ta cd bat dang thuc nguoc lai. Viec xet dau y' doi khi phai can den y" , y"\.. . hoac xet dau bp phan, chang han tir so ciia mpt phan so cd mau duong Neu y " > 0 thi y' dong bien tir do ta cd danh gia f '(x) rdi f(x),... V i du 1: Giai phuong trinh: vo - x + x2 - 72 + x - x2 = 1 Giai Dat t = x2 - x thi phuong trinh trd thanh: 73+ t - 72- t =1 , - 3 < t < 2. Xet ham sd f(t) = 73 + t - 72 - t , -3 < t < 2. Vdi -3 < t < 2 thi f'(t) = 1 + . > 0 nen f dong bien tren (-3; 2). 273 + t 272 - t Ta cd f(l) = 2 - 1 = 1 nen phuong trinh: f(t) = f( 1) <=> t = 1 o x2 — x — 1 = 0 <=> x = l^H. V i du 2: Giai phuong trinh 72x3 +3x2 +6x + 16 = 273 + 74 - x Giai: Dieu kien xac djnh: f2x 3 +3x 2 + 6 x + 16>0 f(x + 2)(2x 2 - x + 8)>0 <=>C cs> -2<x< 4 4-x> 0 4-x> 0 Phuong trinh tuong duong 72x3 + 3x2 + 5x +16 - 74 - x = 273 Xet ham s6 f(x) = N/2X3 +3x2 + 6x + 16 - 74 - x .-2 < x < 4 ™, c u . 3(x2 + x + l) . x Thi f (x) = —. = + — > 0 nen i dong bien ma 72x3 +3x 2 +6x + 16 274- x f(l ) = 273 , do do phuong trinh trd thanh f(x) = f(l) o x = 1 Vdy phuong trinh co nghiem duy nhat x=l -BDHSG DSGT12/1- 17 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 3: Giai phucmg trinh yjx2 - 1 = Vx3 - 2 - x. Giai Dieu kien: x >%/2 Ta cd: Vx3 - 2 = x + Vx2 - 1 >x>l=>x 3 >3=i>x>v / 3 Chia 2 ve cho vo? thi phuong trinh: ~ 1 a. x2 .vx x 4 v x Vx V xVx Xet f(x) la ham sd ve trai, x > tfi thi ,, x 9 5x X 3 f'(x) = B r — r , <0 . 2x5 .Vx 2xVx n 2 2 2x Vx Do do ham so f nghich bien tren khoang (y 3 ; +oo) ma f(3) = 0 nen phuong trinh cd nghiem duy nhat x = 3. Vi du 4: Giai phuong trinh: 3x2 - 18x + 24 1 1 2x- 5 x- l Giai 5 Dieu kien x * 1; —, phuong trinh trd thanh: 2 (2x-5 ) 2 --J _ = (x -l) 2 |2x-5| |x-l | Xet f(t) = t 2 - i vdi t > 0. Ta co: f '(t) = 2t > 0 nen f dong bien tren (0; +oo) Phuongtrinh:f(|2x-5|) = f(|x - l|)o 12x- 51 = |x-l| <=> 4x2 - 20x + 25 = x2 - 2x + 1 <=> 3x2 - 18x + 24 = 0. c ^ x 2 - 6 x + 8 = 0co x = 2 hoac x = 4 (chon) V i du 5: Giai bat phuong trinh: 4 | 2x - 11 (x2 - x + 1) > x3 - 6x2 + 15x - 14 Giai BPT: | 2x - 11 .[(2x - l) 2 + 3] > (x - 2) 3 + 3x - 6 <eo | 2x - 113 + 3 | 2x - 11 > (x - 2) 3 + 3(x - 2) Xet ham sd f(t) = t 3 + 3t, D = R. Ta cd f '(t) = 3t 2 + 2 > 0 nen f dong bien tren R. BPT: f( | 2x - 11) > f(x - 2) o I 2x - 11 > x - 2. Xet x - 2 < 0 thi BPT nghiem dung. Xet x - 2 > 0 thi 2x - 1 > 0 nen BPT o2 x -1> X -2<=> X>-1 : £) U ng Vay tap nghiem la S = R. 18 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 6: Giai bat phuong trinh: Vx + 1 + 2Vx + 6 < 20 - 3VX + 13. Giai Dieu kien: x > -1 . BPT viet lai: Vx + 1 + 2%/x + 6 + 3Vx + 13 > 20 Xet f(x) la ham so ve trai, x > -1 . Ta co: 1 1 3 A f ' ( x ) = —— + — + — > 0 nen f dong bien tren [-1 ; +oo). 2Vx + l Vx + 6 2Vx + 13 Ta cd f(3) = 20 nen BPT:f(x) <f(3)ox<3 . Vay tap nghiem cua BPT la S = [-l;3] . V i du 7: Giai bat phuong trinh: 3Vtanx + 1. s m x + 2c ° S X < 2 1 " ^ sin x + 3 cos x Giai Dieu kien tanx > 0. Dat t = tanx, t > 0 thi VT = 3v ^TT. ^ = f(t), t > 0 t + 3 3 t + 2 1 ' Ta cd f '(t) = —, . + 3vt + 1. - > 0 nen ham so f dong bien, 2Vt+T t + 3 (t + 3) 2 ma t > 0 => f(t) > f(0) = 2. Mat khac VP = 2l ~4 ^ < 2 nen dau "='' ddng thoi xay ra <=> t = tanx = 0 <=> x = krc, k e Z. x + 3 = y +V y 2 + 1 Vi du 8: Giai he phuong trinh y + 3 = z + %/z2 + 1 Z + 3 = X + N/X2 + 1 Giai Xet ham sd f(t) = t + Vt 2 +1 -3,te R fU'f. m 1 t Vt 2 +1 + t Vt7 + t tmf'(t) = l + . —; > , >0, Vt vv+ i Vt 2 + i v V + i nen f(t) ddng bien tren R. Ta cd he phuong trinh x = f(y) y = f(z) z = f(x) Gia su x > y thi f(x) > f(y) nen y > z do dd f(y) > f(z) tuc la z > x: vo li Gia su x < y thi f(x) < f(y) nen y < z do do f(y) < f(z) tuc la z < x: vo 11 Gia su x = y thi f(x) = f(y) nen y = z do do x = y = z. The vao he: x + 3 = x + v / x 2 +lco 3 = v / x 2 +lci > x 2 = 2 <S-X = ±V 2 Thu lai x = y = z = +%/2 thi he nghiem dung. Vay he phuong trinh cd 2 nghiem x = y = z = + V2 -BDHSG DSGT12/1- 19 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 9: Giai he phuong trinh: Ta cd x y - y + i 1 x - l = y(y-l ) < y3 - l = z(z-i) z 3 - l = x (x-l ) Giai l Y 3 3 1 - +—>—>—= 2) 4 4 8 x > Tuong tu y, z > - Dat f(t) = V 1 l,t > - thi 2 f '(t) = 2t - 1 > 0 nen f dong bien tren (—; +oo) 2 Ta cd he < 3 2 y = z -y + l z + 1 co < z 3 = x2 - X + 1 z 3 > X f(y) f(z) f(x) 3 Z> X. y 3 > z 3 => y > z. Gia su x > y thi f(x) > f(y): nen f(z) > f(x) Do do x > y > z > x: vd li . Tuong tu x < y: vo l i nen x = y => x = y = z. Ta cd t 3 = f(t) o t 3 - t 2 + t- l = 0o(t - l)(t 2 + 1) = 0 co t = 1. Vay he co nghiem duy nhat x = y = z = 1. x3 -2 x + l = 2y V i du 10: Giai he phuong trinh Ta cd 2y = x2 - 2x + 1 = (x - l) 2 > 0 y -2 y + l = 2z z2 -2z + l = 2x Giai y > 0. Tuong tu z, x > 0. Dat f(t) = t z - 2t + 1, t > 0 thi f '(t) = 2(t - 1) nen f dong bien tren (1; +oo) va nghich bien tren (0; 1). Dat g(t) = 2t, t > 0 thi g'(t) = 2 > 0 f(x) = g(y) nen g ddng bien tren (0; +oo). Ta cd he I f (y) = g(z) f(z) = g(x) Gia su x = min{x; y; z}. Xet x < y < z. - Neu x > 1 thi 1< x < y < z ^> f(x) < f(y) < f(z) => g(y) < g(z) < g(x) =o y < z < x nen x = y = z. Ta co phuong trinh: t 2 - 4t + 1 = 0 nen chon nghiem: x = y = z = 2 - V3. -Ne u 0 < x < 1 thi f(0) > f(x) > f(l) =o 0 < f(x) < 1. nen 0 < g(y) < 1 => 0 < y < 1 =o f(0) > f(y) > f(l) =o 0 < f(y) < 1 ^> 0 < g(z) <1=>0<Z<1 . 20 -BDHSG DSGTn/i Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Do do x < y < z => f(x) > f(y) > f(z) => g(y) > g(z) > g(x) => y > z > x nen x = y = z - Ta cd phuong trinh t 2 - 4t + 1 = 0 nen chon nghiem: x = y = z = 2 - 42 Xet x < z < y thi cung nhan duoc ket qua tren. Vay he cd 2 nghiem x = y = z = 2 + v / 3 , x = y = z = 2 - -Js 36x 2 y-60x 2 +25y = 0 36y2 z-60y2 V i dull : Giai he phuong trinh: +25z = 0 36z 2 x-60z 2 +25x = 0 Giai 60x2 36x2 +25 He phuong trinh tuong duong vdi < z 60y2 36y2 +25 60z2 36z2 +25 Tir he suy ra x, y, z khong am. Neu x = 0thi y = z = 0 suy ra (0; 0; 0) la nghiem cua he phuong trinh. Neu x > 0 thi y > 0, z > 0. Xet ham sd f(t) 3000t 60f 36t 2 +25 t>0 . Ta cd: f'(t) (36t 2 +25) 2 >0, Vt>0 . Do do f(t) ddng bien tren khoang (0; +co). y = f(x) He phuong trinh duoc viet lai I z = f (y) x = f(z) Tir tinh ddng bien cua f(x) suy ra x = y = z. Thay vao he phuong trinh ta duoc x(36x2 - 60x + 25) = 0. Chon x = - . 6 '5 5 5 ,6'6' 6 Vay tap nghiem cua he phuong trinh la <j (0;0;0); :8-X3 V i du L2: Giai he phuong trinh Vx- 1 -T y (x-l) 4 = y Giai Dieu kien x > 1, y > 0. He phuong trinh tuong duong vdi: IVx- l - (x-l) 2 + x3 - 8 = 0 (1) [y = (x-D 4 (2) -BDHSG DSGTU/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Xet ham so f(t) = VtTT - ( t - l) 2 + t 3 - 8, vdi t > 1. 1 Ta co f '(t) = -2 (t-l) + 3t 2 + = 3t 2 - 2t + 2 2 v W 2v/Tl > 0 vdi moi 2Vt-l t > 1 nen f(t) ddng bien tren (1; +oo). Phuong trinh (1) cd dang f(x) = f(2) nen (1) co x = 2, thay vao (2) ta duoc y = 1 . Vay nghiem cua phuong trinh la (x; y) = (2; 1). x2 - 12x + 35 < 0 (1) Vi du 13: Giai he bat phuong trinh: xu - 3x2 + 9x + - > 0 (2) 3 Giai: Ta cd (1) o x2 - 12x + 35 < 0 co 5 < x < 7 Xet (2): Dat f(x) = x3 - 3x2 + 9x + - , D = R 3 f (x) = 3x2 - 6x + 9 > 0, Vx eR nen f(x) ddng bien: x > 5 =o f(x) > 286/3 Do do f(x) > 0, VXG (5 ; 7) Vay tap nghiem cua he bat phuong trinh la S = (5; 7). YJ du 14: Chung minh rang phuong trinh 3x5 + 15x - 8 = 0 cd mot nghiem duy nhat. Giai Ham f(x) = 3x5 + 15x - 8 la ham sd lien tuc va cd dao ham tren R. Vi f(0) = -8 < 0, f(l) = 10 > 0 nen ton tai mot sd x„ e (0; 1) sao cho f(xo) = 0, tuc la phuong trinh f(x) = 0 cd nghiem. Mat khac, ta cd y' = 15x4 + 15 > 0, Vx e R nen ham sd da cho luon luon ddng bien. Vay phuong trinh dd chi cd mot nghiem duy nhat. V i du 15: Chung minh phuong trinh: x 1 3 - x 6 + 3x4 - 3x2 + 1 = 0 cd nghiem duy nhat. Giai: Dat f(x) = x1 3 - x6 + 3x4 - 3x2 + 1, D = R Xet x > 1 thi f(x) = x6 (x7 - 1) + 3x2 (x2 - 1) + 1 > 0: vd nghiem Xet 0 < x < 1 thi f(x) = x1 3 + ( l - x2 ) 3 > 0: vd nghiem Xet x < 0 thi: f '(x) = 13x1 2 - 6x5 + 12x3 - 6x = 13x1 2 - 6x(x - l) 2 > 0 nen f dong bien Bang bien thien: x —CO 0 y' + y 1 —OO Nen f(x) = 0 cd nghiem duy nhat x < 0. Vay phuong trinh cho cd nghiem duy nhat. 22 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 16: Chung minh rang phucmg trinh 2x 2 Vx- 2 = 11 cd mot nghiem duy nhat. Giai Xet ham so f(x) = 2x2 Vx - 2 thi ham sd xac dinh va lien tuc tren nua khoang [2; +oo). f'(x) = 2 f n I ^ x2 ) x(5x-8) 2xVx-2+ — , - 2Vx- 2 > 0, vdi moi x e (2; +co) Vx- 2 Do do ham so dong bien tren nua khoang [2; +oo). Ham sd lien tuc tren doan [2; 3], f(2) = 0, f(3) = 18. Vi 0 < 11< 18 nen theo dinh li ve gia tri trung gian cua ham sd lien tuc, tdn tai sd thuc c e (2; 3) sao cho f(c) = 11 tuc c la mot nghiem cua phuong trinh f. Vi ham sd dong bien tren [2; +co) nen c la nghiem duy nhat cua phuong trinh. V i du 17: Chung minh rang vdi moi x e (-1 ; 1), phuong trinh: sin2 x + cosx = m cd mot nghiem duy nhat thudc doan [0; TT]. Giai Xet ham sd f(x) = sin2 x + cosx thi ham sd lien tuc tren doan [0; n], Ta cd f'( x ) = 2sinxcosx - sinx = sinx(2cosx - 1), x e (0; TI) V i sinx > 0 nen f '(x) = 0 o cosx = — o x = — 2 3 BBT: X TC 0 — Tt u 3 f(x ) + 0 f(x) 5 71 ' 71 Ham f dong bien tren doan [0; — ] va nghich bien tren doan [— ; 7i]. 3 3 ' 71 7t 5 Ham so f lien tuc tren doan \—; Til, f(—) = — va fire) = -1 . Theo dinh li 3 3 4 5 ve gia tri trung gian cua ham so lien tuc, vdi moi m e (-1 ; 1) cz (-1 ; —). 4 ton tai mot sd thuc c e (—; 7i) sao cho f(c) = 0 tuc c la nghiem cua 3 phuong trinh. Vi ham so f nghich bien tren [—; 71] nen tren doan nay, 3 phuong trinh cd mot nghiem duy nhat. TC 5 Con vdi moi x e [0; — ], ta cd 1 < f(x) < — nen phuong trinh khong co 3 4 nghi?m suy ra dpcm. V i du 18: Tim sd nghiem cua phuong trinh x3 - 3x2 - 9x - 4 = 0. -BDHSG DSGT12/1- 23 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai Xet ham so y = x3 - 3x2 - 9x - 4, D = R. BBT 3x - 6x - 9, y' = 0 co x = - 1 hoac x = 3. X —GO - 1 3 + 00 y' + 0 - 0 + y r 1 ^ +00 —GO * -31 " Dua vao BBT thi phuong trinh y = 0 co dung 3 nghiem. V i du 19: Tim sd nghiem cua phuong trinh: 3 = 0. Giai xx + 2x5 - 2x4 - x3 - 3x2 - 6x 3)(x 5 - x 2 -2x - 1) = 0. •tfi hoac x5 - x2 - 2x - 1 Phuong trinh tuong duong: (x co x3 + 3 = 0 hoac x5 - x2 - 2x + 1 = 0 co x = Xet phuong trinh: x5 - x2 - 2x - 1 = 0 ^> x5 = (x + l) 2 > 0. Do do x5 > 0 => x > 0 o> (x + l) 2 > 1 o x5 > 1 o x > 1. Do do nghiem cua phuong trinh x5 - x2 - 2x - 1 = 0 neu cd thi x > 1. Dat f(x) = x5 - x2 - 2x - 1, x > 1. f '(x) = 5x4 - 2x - 2 = 2(x4 - 1) + 2x(x3 - 1) > 0. Do do f ddng bien. V i f(l ) = -3 < 0 va f(2) = 23 > 0 nen f(x) = 0 co nghiem duy nhat x 0 > 1. Vay phuong trinh cho cd dung 2 nghiem. V i du 20: Chung minh he f x 2 +y 3 =1 y+x 3 Giai cd dung 3 nghiem phan biet. Tru 2 phuong trinh ve theo ve va thay the ta duoc: x 2 (l - x) - y 2 (l - y) = 0 =o (1 - y3 )(l - x) - (1 - x 3 ) (l - y) = 0 => (1 - x)(l - y)[l + y + y 2 - ( l + x + x2 )] = 0. -O (1 x)(l - y)(y - x)(l + x + y) = 0. Xet x = 1 thi he cd nghiem (1; 0) Xet y = 1 thi he cd nghiem (0; 1) Xet x = y thi x2 + y 3 = 1 co x3 + x2 - 1 = 0. Dat f(x) = x3 + x 2 - 1, D = R. Ta cd f(l ) = 1 * 0. f' (x) = 3x2 + 2x, f 1 (x) = 0 co x = - - hoac x 3 BBT ' 0. X —CO -2/3 0 +0O y' + 0 - 0 + y -23/27 - 1 " +00 24 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Do do f(x) = 0 cd 1 nghiem duy nhat x,, > 0, XQ * 1 nen he cd nghiem (x„; y0 ). Xet 1 + x + y = 0 => y = - x - 1 nen y2 + x3 = 1 co x3 + x2 + 2x = 0 co x(x2 + x + 2) = 0 co x = 0. Do dd he cd nghiem (0; 1) Vay he cd dung 3 nghiem phan biet. V i du 21: Tim cac gia tri cua m de phuong trinh sau cd dung mpt nghiem yjx2 + 2x + 4-V x + l = m. Giai Dat t = Vx + 1 > 0, phuong trinh trd thanh vft 4 +3 - t = m (*) Nhan xet ung vdi moi nghiem khong am cua phuong trinh (*) cd dung mpt nghiem cua phuong trinh da cho, do do phuong trinh da cho co dung mpt nghiem khi va chi khi phuong trinh (*) cd dung mpt nghiem khong am. Xet ham sd f(t) = tftU^S-t vdi t > 0, f '(t) = , V(t 4 + 3) 3 Ma f(0) = \/3 va lim f(t) = 0 nen cd bang bien thien: - 1<0 . t 0 +c o f'(t) f(t) V3" ^ * 0 Tu bang bien thien suy ra cac gia tri can tim cua m la 0 < m < ^3 V i du 22: Tim m de phuong trinh cd nghiem m( 7 l + x2 - Vl-x 2 + 2) = 2Vl-x 4 + Vl + x2 - Vl - x 2 Giai Dieu kien - 1 < x < 1. Bat x = Vl + x2 - Vl - x 2 thi t > 0 va t 2 = 2 - 2 Vl-x 4 < 2, dau "=" khi x2 = 1. Do dd 0 < t < PT:m(t + 2) = 2-t 2 + tc o m = Zzl±l±ll t + 2 Xet f(t) = - t + t + 2 . 0 < t < V2 , f Yt) = 1 + A \ < 0 nen f nghich bien w t + 2 (t + 2) 2 tren [0; V2 ]. Dieu kien cd nghiem: min f(t) < m < max f(t) o f( &) < m < f(0) co V2 - 1 < m < 1. V i du 23: Tim m de phuong trinh sau cd 2 nghiem phan biet: Vx2 +mx + 2 = 2x + 1 Giai PTco 2x + 1 > 0 x2 +mx + 2 = (2x + l) 2 <=> 3x" + 4x - 1 = mx, x > V i X = 0 khong thoa man nen: -BDHSG DSGT12/1- 3xz +4x- l m, x > 25 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
. 3x 2 +4x- l 1 3x 2 + l Xet f(x) = ,x > — , x * 0 thi f (x) 2 ' " " x2 Lap BBT thi dieu kien phuong trinh cho cd 2 nghiem phan biet la 1 9 f(x) = m cd 2 nghiem phan biet x > — . x 0 co m > — v 2 ' 2 Vi du 24: Tim m de phuong trinh cd nghiem: (4m - 3) Vx + 3 + (3m - 4)Vl - x + m - 1 = 0 Giai ^.x , „ 3Vx + 3 + 4rji - x + 1 Dieu kien -3 < x < 1. PT co —. ——. = m 4Vx + 3 + 3V1 - x + 1 Ta cd (Vx + 3)2 + (Vl-x)2 = 4 nen dat: i 2t / „ 1-t2 Vx + 3 = 2sin(p = 2. - , VI - x = 2coscp = 5- 1 + t 2 1 + t 2 Vdi t = tan ^.0<cp<-. 0<t< 1. 2 2 DX 7t2 -12t-9 ,m _ 7t*-12t-9 n<rt<r1 PT co m = —5 Dat f(t) 5 . 0 < t < 1. 5 t 2 -16t - 7 ' 5t 2 -16t- 7 Ta cd f '(t) = ——^ < 0 nen f nghich bien tren doan [0; 1], do (5t 2 -16t-7) 2 7 9 dd dieu kien cd nghiem: f(l) < m < f(0) co — < m < —. Vi du 25: Chung minh cac bat dang thuc sau: a) sinx < x vdi moi x > 0, sinx > x vdi moi x < 0. x2 b) cosx > 1 vdi moi x & 0. 2 Giai a) Vdi x > — thi x > 1 nen sinx < 1 < x. 2 Vai 0 < x < — thi ham so f(x) = x - sinx lien tuc tren nua khoang [0; —) 2 2 va f'(x) = 1 - cosx > 0 vdi moi x e (0; ). Do do ham sd dong bien tren [0; -) nen f(x) > f(0) = 0 vdi moi x e (0; -). 2 2 Vdi —| < x < 0, giai tuong tu thi f(x) < f(0) = 0 Vdi x < -— thi x < -1 nen sinx > -1 > x => dpcm. 2 26 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
x z b) Voi x > 0 thi ham so g(x) = cosx + — — 1 lien tuc tren nua khoang [0; +co) va g'(x) = x - sinx. Theo a) thi g'(x) > 0 voi moi x > 0. Do do ham so g dong bien tren [0; +co) nen: x 2 g(x) > g(0) = 0 vdi moi x > 0 => cosx + — — 1 > 0 vdi moi x > 0. (_x)2 Suy ra vdi moi x < 0 ta cd cos(-x) + 1 > 0. 2 Vi du 26: Chung minh: sinx > x . Vx > 0. 6 Giai x3 BDT: x sinx > 0, Vx > 0 6 x3 Xet f(x) = x sinx thi f lien tuc tren [0; +x>) 6 x 2 f '(x) = 1 —- — cosx ; f "(x) = - x + sinx f'"(x) = -1 + cosx < 0 nenf" nghich bien tren [0; +co): x > 0 => f "(x) < f "(0) = 0 nen f' nghich bien tren [0; +oo): x > 0 => f'(x) < f (0) = 0 nen f nghich bien tren [0; +<x>): x > 0 => f (x) < f (0) = 0 => dpcm. Vi du 27: Chung minh cac bat dang thuc vdi moi x e (0; —). 2 a) tanx > x b) tanx > x + — 3 c) sinx + tanx > 2x d) 2sinx + tanx > 3x. Giai a) Ham so f(x) = tanx - x lien tuc tren nua khoang [0; —) va cd dao ham f (x) = —-— > 0 vdi moi x e (0; —). Do do ham so f dong bien tren cos x 2 nua khoang [0; —) nen f(x) > f(0) = 0 vdi moi x e (0; —). 2 2 ' 71 b) Ham sd f(x) = tanx - x lien tuc tren nua khoang [0; — ) va co dao 3 2 ham f '(x) = —\ 1 - x2 = tan2 x - x2 = (tanx + x)(tanx - x) > 0 vdi cos x moi xe(0;^) (suy ra tu a)). -BDHSG DSGT12/1- 27 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Do do, ham so f dong bien tren nira khoang [0; — ) va ta cd f(x) > f(0) = 0 vdi moi xe(0;^)=> dpcm. c) Ham so f(x) = sinx + tanx - 2x lien tuc tren nua khoang [0; — ) va cd dao ham f' (x) = cosx + — 2 > cos2 x - cos x cos' x \ - - 2 = (cosx - — ) 2 > 0 . cosx Do do ham so f dong bien tren [0; — ) nen f(x) > f(0) = 0. Ket qua: Tam giac ABC cd 3 gdc nhon thi smA + sinB + sinC + tanA + tanB + tanC > 2TC. d) Ham so f(x) = 2sinx + tanx - 3x lien tuc tren nua khoang [0; —) va f '(x) = 2cosx + 2cos 3 x- 3cos 2 x + l (cosx -l) 2 (2 cos x + 1) co^x cos x cos X >0 Do do ham so f ddng bien tren [0; — ) nen f(x) > f(0) = 0. Vi du 28: Chung minh bat dang thuc: a) 8sin2 — + sin2x > 2x, Vx e (0; rc] b) tanx < —, Vx 2 TC Giai 2 X a) Xet ham sd f(x) = 8sin — + sin2x - 2x, Vx e (0; TC]. f '(x) = 4sinx + 2cos2x - 2 = 4sinx(l - sinx) f '(x) = 0 co x = — hoac x = rc. Vdi x e (0; TC] ta cd f '(x) > 0 va dau bang chi xay ra tai hai diem. Vay f(x) dong bien tren nua khoang (0; TC] nen f(x) > f(0) = 0 vdi moi x e (0; TC] => dpcm. b) Neu x = 0 thi BDT dung. Neux>0thi BDT o <-. Vx e f 0;- X TC ^ 4 Xetf(x)=^.VxJ0;^ x I 4 x f(x ) = cos X -- tan x x-si n xcosx 2x-sin2 x 2 2 X COS X 2x2 cos2 x Vi 0 < x < - nen 0 < 2x < - sin 2x < 2x do dd f'(x) > 0 nen f ddng 28 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
bien tren V suy ra f(x) < f(- ) = - => dpcm v 4 4 ru V i du 29: Chung minh bat dang thuc: a) b.tana > a.tanb vdi 0 < a < b < — 2 571 b) cos(x + y)< vdi x > 0, y > 0 va x + 2y < — x sin y 4 Giai „ , , tana tanb c . . tanx _ TC a) b.tana < a.tanb <o < . Xet f(x) = . 0 < x < - a b x 2 .x-tanx f YY X = cos2 x ; = x-sinxcos x = 2x-sm2 x x x .cos x 2x cos x Xet g(x) = 2x - sin2x, 0 < x < - . g '(x) = 2 - 2cos2x = 2(1 - cos2x) > 0 nen g ddng bien: x > 0 => g(x) > g(0) = 0. Do do f'(x) > 0 nen f dong bien tren [0; - ). Vi 0 < a < b < - => f(a) < f(b) => dpcm. 2 2 b) Xet ham s6: f(t) = vdi 0 < t < — ; w t 4 , . tcost-sin t cost(t-tant) Ta co f (t) = -2 = -2 Neu 0 < t < - thi do tant > t f '(t) < 0. 2 Neu - < t < rc thi cost < 0 va sint > 0 => f '(t) < 0. 2 5TI Neu n < t < — thi do cost < 0; tant < t => f '(t) < 0. 4 Do do f '(t) < 0, 0 < t < — nen f la ham so nghich bien tren khoang (0; — ) 4 4 ^ ., , -x . n ^ 5TT sin(x + 2y) sinx Tu gia thiet co0<x< x + 2y< — — < 4 x + 2y x Do x > 0 va x + 2y > 0 nen tir do cd xsin(x + 2y) < xsinx + 2ysinx <eo x 2cos(x + y)siny < 2ysinx =o dpcm (vi x > 0 va x + 2y < — => y < — ^> siny > 0). Vi du 30: Chung minh cac bat dang thuc sau: a) a4 + b4 + c4 + d4 + 2abcd - (a^ 2 + a2 c 2 + a 2 d2 + b2 c2 + b2 d2 + c2 d2 ) > 0 vdi 4 s6 a, b, c, d duong. -BDHSG DSGT12/1- 29 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
^2 ^ b) 1 + — x < VI + x < 1 + — x , vai x > 0. 2 8 2 Giai a) Khong mat tinh tong quat, gia su a > b > c > d > 0 Xem ve trai la ham so f(a), a > 0 f '(a) = 4a3 + 2bcd - 2a(b2 + c2 + d2 ) f "(a) = 12a2 - 2(b2 + c2 + d2 ) > 0 nen f' dong bien tren (0; +co): a > b => f '(a) > f '(b).Vi f '(b) = 2b(b2 - c2 ) + 2bd(c - d) > 0 nen f(a) ddng bien tren [0; +oo): a > 0 => f(a) > f(0) = 0=> dpcm. b) Xet ham so f(x) = 1 + -x - Vl + x tren [0; +co). Ta cd: f '(x) = — 1 > 0 vdi x > 0 nen f(x) dong bien tren nua khoang 2 2Vl + x [0; +oo). Do dd f(x) > f(0) = 0 vdi moi x > 0. Xet ham sd g(x) = \/l + x - 1 + — tren [0; +oo). 2 8 Tacd:g'(x)= 1 -- + -. g"(x) = 1 . >0 2VTT I 2 4 4 4(l + x)vT+ ^ nen g' dong bien tren [0; +°o), do do g'(x) = g'(0) = 0. Suy ra g dong bien tren [0; +co) nen g(x) > g(0) = 0 vdi moi x e [0; +co) => dpcm. sin x TI V i du 31: Chung minh vdi moi a <3saocho ( ) a > cosx, Vx e (0;—) x 2 Giai 7i . sin x Khi x e (0;—) thi cd 0 < sinx < x nen 0 < < 1 2 x sin x sin x 1 Suy ra (—) ° > (^ii±) 3 . Va < 3 do do ta chi can chung minh khi a = 3: x x .sinx.3 ,„ TC. sinx ,„ TC. ( ) > cos x, x e (0;-) <o — > x, x e (0; - ) x 2 Vcosx 2 Xet ham sd F(x) = s™_?_ - x, x e [ 0;—) Vcosx 2 , . 2cos2 x-3cosx.Vcosx + 1 Ta co F (x) = = = 3 cosx. Vcosx Xet G(t) = 2t2 -3tVt +l,te[0;l] thi G'(t) = 4(t - Vt) < 0, Vt e [0;l] nen G(t) nghich bien do do G(t) > G( 1) = 0, Vt e [0; 1] Suy ra F'(x) >0, Vx e [0;- ) nen F(x) ddng bien 2 Do do F(x) > F(0) = 0, Vx e [0; -). 2 30 -BDHSG-DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 32: Chung minh: (x + l)cos— xcos->l , Vx > V3 6 v ; x + 1 x Giai n • 9 TC 2sm2 n n T _ rc(2x + l) rc rc BDT co 2xsin — sm > 1 - cos — 2x(x + l) 2x(x + l) x + 1 2(x + l) . rc(2x + l) rc . 2 71 CO xsin —2x( si x + l) 2x( n > x + l) 2( sm x + l) \r -v PS n n rc(2x + l) rc V i x > v3 => 0 < < < — 2(x + l) 2x(x + l) 2 sinrc(2x + l) rc => > sin >0 (1) 2x(x + l) 2(x + l) rc rc Ta se chung minh: xsin >sin (2) 2x(x + l) 2(x + l) Dat t = . t > 0 thi (2) co xsint > sinxt 2x(x + l) Xet f(t) = xsint - sinxt, t > 0, f '(t) = x cost - xcosxt = x(cost - cosxt) Vi 0 < t < xt < - => f Yt) > 0 vdi t > 0. 2 => f(t) ddng bien tren [0; +oo) =o f(t) > f(0) = 0 =o (2) dung Tu (1), (2) =o dpcm. Vi du 33: Cho x, y, z>0vax + y + z= l. 7 Chung minh: 0 < xy + yz + zx - 2xyz < — 2 V Giai Gia sir z la sd be nhat thi 0 < z < — .Ta cd 3 T = xy+ y + zx - 2 xyz = xy(l - 2z) + (x + y)z > - xy +(x + y)z > 0 3 Va cd T = (l-2z ) + (x + y)z = -( 1 - z) 2 (l - 2z) + (1 - z)z = -(-3z 3 + z2 + 1) 4 4 Xet f(z) = -3z3 + z2 + 1, 0 < z < - thi f Yz) = - 6z2 + 2z = 2z(l - 3z) > 0 tren f(z) ddng bien tren [0; - ]. do do 3 T = f(z)<f(-)= — ; . 3 27 -BDHSG DSGT12/1- 31 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
C. BA I LUYE N TAP Bai 1: Tim khoang don dieu cua ham sd x2 + 4x - 2 a) y = —— x + 1 c) y = 2x3 + 3x2 + 6x -13 a) Y = x2+ 1 - b) y = (x + 2)5 (2x + lY DS: a) dong bien tren (— ; 2) va nghich bidn tren (-co ; — ) va (2; +oo). 2 2 Bai 2: Tim khoang don dieu cua ham so x x + 1 x2 X a) y = , = b) y = — c) y Vx - x + 1 e ln x DS: b) dong bien tren (0; 2) va nghich bi£n tren (-co; 0) va (2; +oo). c) dong bien tren (e; +co) va nghjch bidn tren (0; 1) va (1; e). Bai 3: Tim khoang don dieu cua ham so: a) y = x.lnx b) y = sinx + sin2x Bai 4: Tim khoang don dieu cua ham sd: a) y = | x2 - 3x - 4 | b)y = ^^ cx + d Bai 5: Chung minh ham sd ddng bien tren tap xac dinh: > m(x + l) 3 , a )y=— -.m> 0 b)y = ln(x + V4 + x 2 ) x - x + 1 H D: b )y' = 1 >Q,y x V4 + x2 Bai 6: Chung minh ham sd: a) y = x - ex nghich bien tren khoang (0; +oo) x2 - 4x + 3 b ) v = — i — luon dong bien tren tirng khoang xac dinh. HD: b) y ' > 0 tren (-co; -1), (-1;2), (2; +oo). Bai 7: Chung minh ham so a) y = sinx + cosx + 2x luon dong bien tren R. b) y = ( i + _) * dong bien tren (0,+ co). HD: a) y '= cosx - cosx + 2 > 0, Vx b) Lay In trudc khi tinh dao ham. Bai 8: Tim tham sd de ham sd: 1 1 3 a ) y = — x3 - — (sin a + cos a)x2 + — sm 2a.x ddng bien tren R. b) y = (m - 3)x - (2m + l)cosx nghich bien tren R. DS: a) — + kn < a < — + krc 12 12 32 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bai 9: Tim tham so de ham sd: a) y = a x + 4 nghich bien tren (-°o;l) x + a b) y = 2x3 + 3x2 + 6(m +l)x nghich bien tren (-2; 0). DS: b) m < - 3 Bai 10: Tim tham sd de ham sd: a) y = x3 + 3x2 + mx + m nghich bien tren mot doan cd do dai bang 3. — 2x + m ' ' b) y = dong bien tren khoang (-co; 0). x - l Bai 11: Tim m de ham sd a) y = mX + ^X —- nghich bien tren nua khoang [1; +oo) x + 2 b) y = x3 - 3x2 + (m - 2)x + 7 ddng bidn tren R. DS: a)m< -— b)m >5. 5 Bai 12: Tim m de ham so: 3 2 ' a) y = x -m x +x+ l nghich bien tren khoang (1, 2). b) y = — 2mx + m + 2 ^Qn^ ^-« n tr - n j^^n g (\.+ ^ y x - m DS: a)m > —; b) m<3 ~^;m>2 4 4 Bai 13: Tim m de ham so: . x2 + (m + 2)x m + 3 .x ,•-<--.. • i u - - A- U a) y = dong bien tren tung khoang xac dmh. x + 1 2 1 2 ! -X b) y = 2mx - 2cos x - m.sinx.cosx + — cos 2x dong bien tren R. 4 DS:a)m > l . b)m > l . Bai 14: Tim dieu kien de y = a.sinx + b.cosx + 2x dong bien tren R. DS: a 2 + b2 < 4. Bai 15: Giai phuong trinh a) vx 2 + 15 + 2 = Vx2 +8 + 3x b) v / (x + 2)(2x-l) +7(x + 6X2x-l) = 4 + 3(7x + 2 + Vx + 6) DS: a) x = 1. b) x =7 Bai 16: Giai bat phuong trinh : a) Vx + Vx + 7 < 9 - Vx- 5 b) x5 + x3 > Vl-3 x - 4 DS: a) 5 < x < 9 b) x < - 1 Bai 17: Giai bat phuong trinh: a) V2x3 +3x2 +6x + 16 > 2 V3 + V4- x b)Vx + l + 2Vx + 6 > 20 - 3Vx + 13 -BDHSG DSGT12/1- 33 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
DS: a) 1 < x < 4. b) x > 3. 1 Bai 18: Giai he: a) DS: a)x = y = l; x = y 1 x — = v — x y 2y = x3 + 1 1 + V5 ^ [cot x - cot y = x - y |l5 x + 7y = rc;.x,y £ (0, rc) b) 71 X = 22 TC y = 22 Bai 19: Chung minh phuong trinh: x3 + x2 + 12x - V3 = 0 co nghiem. H D: Ham y = x3 + x2 + 12x - V3 don dieu tren R. Bai 20: Chung minh phuong trinh x3 + 2x3 - x2 + x - 1 = 0 cd nghiem duy nhat. H D: Ham y = x5 + 2xJ - x2 + x - 1 don dieu tren R. Bai 21: Cho so tu nhien n chan va a > 3. Chung minh phuong trinh: (n + 1) x n + 2 - 3(n + 2)xn + 1 + a n + 2 = 0 vo nghiem H D: y ' = (n + l)(n + 2).xn_1 .(x - 3) Bai 22: Cho tam giac ABC cd cd canh a < b < c. Chung minh phuong trinh Vx-si n A + Vx -sin B = Vx - smC co nghiem duy nhat. HD: Tam giac ABC co canh a < b < c thi gdc A > B > C. Bai 23: Chung minh vdi moi m > 0 thi phuong trinh cd 2 nghiem phan biet: x2 + 2x - 8 = v / m(x-2 ) Bai 24: Tim m de phuong trinh cd 2 nghiem phan biet: mVx2 + 4 - x + 1 = 0. DS: -2^<m<-l 2 Bai 25: Tim m de phuong trinh cd nghiem: 3Vx - 1 + mVx + 1 = 2%/x2 - 1 . DS: -Km< - 3 Bai 26: Tim m de phuong trinh: x 4 - (m - l)x 3 + 3x2 - (m - l)x + 1 = 0 co nghiem. DS: m< —; m > —. 2 f 2 Bai 28: Tim m de phuong trinh: 2.cosx.cos2x.cos3x + m = 7.cos2x co dung 1 nghiem thuoc doan [-3TC/8:-TC/8]. HD: Dat t = cos2x vdi -— < t < — 2 2 Bai 29: Tim m de bat phuong trinh: x2 + (1 - x 2 ) 3 ' 2 > m cd nghiem. DS: m <1 . Bai 30: Tim a, b de bat phuong trinh: 3x4 + 8x + ax +b > 0 cd nghiem x thuoc [-1 ;1] 34 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
H D: Tfnh dao ham cua ham sd y = 3x4 + 8xJ + ax2 +b roi xet a > 6, a = 6 va a < 6. Bai 31: Chung minh vdi moi so nguyen duong n thi phuong trinh: x + x2 + x3 + ... + x2 n + 2007x2n+ 1 = 1999 co nghiem duy nhat. Bai 32: Chung minh phuong trinh: x 1 3 - x6 + 3x4 - 3x2 + 1 = 0 cc nghiem duy nhat. Baj 34: Giai phuong trinhVx2 + 15 = 3x - 2 + Vx2 + 8 DS: x = 1 sin b Bai 35: Chung minh cac bat dang thuc: sin a vdi 0 < a < b < H D: Dat f(x) = vdi x g (0; - ) x 2 Bai 36: Cho tam giac ABC cd 0 < A < B < C < 90° Chung minh: 2 cos3C - 4 cos2C + 1 cosC > 2 Bai 37: Chung minh rang vdi ba so duong a. b. c bat ki thi: |c~ ab c >—+—+ - a b c a H D: Chung minh: x + m - 1 > mx vdi moi x > 0, m > 1 3\/3x = cos(rcy) 3V3y = COS(TIZ) 3V3z = cos(rc t) 3V3t = cos(rcx) Bai 38: Giai he phuong trinh: Bai 39: Chung minh he phuong trinh: 2x2 2y2 a • y = - y co nghiem duy nhat vdi a> 0 Bai 40: Chung minh bat dang thuc: |sin a - sin p| < [a - p| vdi moi a , p Bai 41: Chung minh bat dang thuc: 10a9 (b - a) < b1 0 - a 1 0 < 10 b9 (b - a) vdi b > a > 0 Bai 42: Chung minh bat dang thuc: a sina - p sinP > 2 (cosP - cosa) vdi 0 < a < p < — Bai 43: Chung minh bat dang thuc: -BDHSG DSGT12/1- 35 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
sinA + sinB + sinC „ 2 rc > cosx > vai 0 < x < - Uanx J 2 Bai 44: Chung minh bat dang thuc: |cos2 x . sin4 x + cos2x| < 1 x3 x5 Bai 45: Chung minh: sinx < x + — , V x > 0 5 3! 5! x2 x4 Bai 46: Chung minh: cosx> l + — ,Vx 5 2! 4! Bai 47: Chung minh: x4 + y4 > - vdi x, y thoa x + y = 1 8 Bai 48: Chung minh: a2 + b2 > 1 vdi a, b thoa a > b3 + b2 + |b| + 1 Bai 49: Cho f(x) vdi deg f = n va f(x) > 0, Vx e R. Chung minh Jf(k)(x)>0 k=0 Bai 50: Cho tam giac ABC nhpn. Chung minh 2 1 — (sin A + sin B + sin C) + — (tan A + tanB + tanC) > rc 3 3 Bai 51: Chung minh bat dang thuc: > 2 cos2 — vdi AABC khong tu cosA + cosB + cosC 8 Bai 52: Chung minh vdi tam giac ABC thi cd: v A B C 3>/3 a) cos— + cos— + cos— < 2 2 2 2 ABC b) tan— + tan— + tan— > S 2 2 2 Bai 53: Cho a, b, c> 0 va thoa 21ab + 2bc + 8ca < 12. „, , . , 1 2 3 15 Chung minh: — + — + — > — a b c 2 Bai 54: Cho a, b, c > 0 va thoa a + b + c = 1. Chiing minh: xyz( x(— + -) + y(- + —) + z(— + —) + 1) < — y z z x x y 27 Bai 55: Cho a, b, c, d > 0 va thoa man 2(ab + ac + ad + be + bd + cd) + abc + abd + acd + bed = 16. 2 Chung minh: a + b + c + d> — (ab + ac + ad + be + bd + cd) 3 Bai 56: Cho a, b, c, r, s >0 va thoa man a > b > c, r > s. Chung minh ar .bs + b\cs + c".as > as .br + bs .c" + cs .a' 36 -BDHSG DSGT12/1 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
§ 2 . CU C TR I CU A HA M S O A. KIE N THtf C CO BAN Cho ham s6 f xac dinh tren tap hop D(DcR ) ya x 0 e D. a) Xo duoc goi la mpt diem cue dai cua ham sd f neu tdn tai mpt khoang (a; b) chua diem Xo sao cho (a; b)cDv a f(x) < f(xo) vdi moi x e (a; b) \ {x 0 } . Khi dd f(Xo) duoc gpi la gia tri cue dai cua ham sd f, ki hieu ycob) Xo dugc gpi la mpt diem cue tieu cua ham sd f neu ton tai mpt khoang .(a; b) chiia diem XQ sao cho (a; b)cDv a f(x) > f(Xo) vdi moi x e (a; b) \ {x 0 } . Khi do f[Xo) duoc gpi la gia tri cue tieu cua ham sd f, ki hieu ycrDiem cue dai va diem cue tieu duoc gpi chung la diem cue tri. Gia tri cue dai va gia tri cue tieu duoc gpi chung la cue tri, neu x 0 la mpt diem cue tri cua ham so f thi diem (x0 ; f(x0 )) duoc gpi la diem cue tri cua do thi ham sd f. i I n ( Dieu kien can de ham sd cd cue tri: Gia su ham sd f dat cue tri tai diem Xo. Khi dd, neu f cd dao ham tai XQ thif'(Xo) = 0. Dieu kien du de ham sd cd cue tri: cd hai dau hieu: - Cho y = f(x) lien tuc tren khoang (a;b) chua xo, cd dao ham tren cac khoang (a;xq) va (x0 ;b): Neu f '(x) ddi dau tu am sang duong thi f dat cue tieu tai x<>. Neu f '(x) ddi dau tu duong sang am thi f dat cue dai tai xo . - Cho y = f(x) cd dao ham cap hai tren khoang (a;b) chua x0 : Neu f '(xo) = 0 va f "(xo) > 0 thi f dat cue tieu tai xo. Neu f '(x0 ) = 0 va f "(x0 ) < 0 thi f dat cue dai tai x0 . B. PHAN DANG TOA N DANG 1: CUC DAI, CUC TIEU Quy tac 1 l.Timf'(x ) 2. Tim cac diem Xj (i = 1, 2,...) tai dd dao ham cua ham so bang 0 hoac ham sd lien tuc nhung khdng cd dao ham. 3. Xet dau f (x). Neu f '(x) ddi dau tii"- " sang "+" khi x qua diem x, thi ham sd dat cue tieu tai Xj, cdn neu f '(x) ddi dau tir"+" sang "- " khi x qua diem Xj thi ham sd dat cue dai tai Xj. -BDHSG DSGT12/1- 37 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Quy tac 2 l.Timf'(x) , 2. Tun cac nghiem Xj (i = 1,2,...) cua phucmg trinh f '(x) = 0. 3.Tim f "(x)vatinh f "(xO Neu f "(xj) < 0 thi ham so dat cue dai tai diem Xj. Neu f "(xj) > 0 thi ham so dat cue tieu tai diim Xj. Chu y: - Gia tri cue dai (cue tieu) f(Xo) cua ham so f ndi chung khdng phai la gia tri ldn nhat (nhd nhat) cua ham sd f tren tap hop D; f(Xo) chi la gia tri ldn nhat (nhd nhat) cua ham sd f tren mot khoang (a; b) nao do chiia diem Xo. - Ham sd f cd the dat cue dai hoac cue tieu tai nhieu diem tren tap hop D, nhung khdng dat tai cac bien. - Tung do cue tri y = f(x) tai x = xo cd 3 hudng tinh: Ham so bat ky: dung phep the yo = f(xo) Ham da thuc: chia dao ham y = q(x). y' + r(x) =5> yo = r(x0 ) Ham huu ti: dao ham rieng tu, rieng mau y = f(x)=^.hiy„=^4 = ^4 v(x) v(x0 ) v'(x 0 ) Dac biet: Vd i ham bac 3 cd CD, CT va neu y = q(x). y' + r(x) thi phuong trinh ducmg thang qua CD, CT la y = r(x). - Bai toan don dieu, cue tri khdng duoc dat an phu. V i du 1: Tim cue tri cua cac ham sd sau: a)f(x) = -x 3 + 2x2 + 3x - l 3 c) y = x4 - 5x2 + 4 a) D = R. Tacdf'(x) = x2 + 4x + 3 b)f(x)= -x 3 3 d) y = (x + 2)2 (x - 3) 2 x2 + 2x - 10 Giai f'(x) BBT 0<=> x z + 4x + 3 = 0 <=> x = -3 hoac x = -1 . x —00 -3 - 1 +00 y' + 0 0 + y —00 . -1 _ ~* -7/3" +00 Vay ham sd dat cue dai tai diem x = -3, f(-3) = - 1 va dat cue tieu tai 7 diem x = -1 , f(-l ) = — 3 b) D = R. Ta cd f '(x) = x2 - 2x + 2 > 0, Vx (do A' = 1 - 2 < 0) nen ham s6 dong bien tren R, khdng cd cue tri. c) D =• R. Ta cd y' = 4x3 - lOx = 2x(2x2 - 5) 38 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
y' = 0 <=> x = 0 hoac x = ± J| ; y" = 12x2 - 10. Ta co y" 20 > 0, y"(0) = -10 < 0 nen ham so dat cue dai tai x = y. (5 9 yco = 4 va dat cue tieu tai x = ±,/— , ycr = — V 2 4 d) y' = 2(x + 2)(x - 3) 3 + 3(x + 2) 2 (x-3) 2 = 5x(x + 2)(x - 3) 2 Ta cd y' = 0 <=> x = —2 hoac x = 0 hoac x = 3. BBT X -co _3 0 3 +oo y' + o - 0 + 0 + y o. J)-^ 0 0 Vay diem cue dai (-2; 0) va cue tieu (0; -108). V i du 2: Tim cue tri cac ham sd a) f(x) = | x2 + 3x - 4 | b) f(x) = | x | (x + 2) Giai x2 + 3x - 4 , x < -4 hay x > 1 - x 2 -3 x + 4, - 4<x< l [2x + 3, x <-4ha y x> l I -2x -3 , -4<x< l a) D = R, y y = BBT - 4 -3/ 1 + 00 + CDCT ^CT Vay ham sd dat CD I ~\^\ . CT ( - ^ °). C T ( 4 ;° ) b) Ham sd f lien tuc tren R. Ta cd: f(x) = -x(x + 2) |x(x + 2) Vdi x < 0, f '(x) = -2x + 2; f '(x) =0o x = -l . Vdi x > 0, f'(x) = 2x + 2 > 0. BBT k h i x < 0 khi x > 0 + - L o 0 +00 Vay diem CD(-1 ; 1), CT(0; 0). -BDHSG DSGT12/1- 39 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 3: Tim cue tri cua ham sd a) y c) y: x - 2x + 3 x + 1 x + 1 b)y : 2x + l x - 5 x 2 + ; d) y = x - 1 + Giai x + 1 a) D = R \ {-l}.Tacdy' = x2+2x 2 5 - y ' = 0co x = - l ±^ 6 (x + 1)2 BBT X -00 -1 - - 1 -1+ +00 y' + 0 - 0 + y -4- 2 V6 —co^^* —00 +00 +°o ^276-4 ^ Vay diem CD(-1 - 76 ;-4-2>/ 6 ), CT(-1 + &;2S -4). b) D = R \ {5} . Ta cd y' = —=- < 0, Vx * 5 nen ham so nghich bien tren (x-5 ) tung khoang xac dinh, do do khdng cd cue tri. , „ _ „ . , x 2 +8-2x( x + l) -x 2 - 2 x + 8 c) D = R.Taco y = (x2+g) 2 = y' = 0<=>x = -4 hoac x = 2. BBT (x2 + 8) 2 —00 - 4 +O0 0 + 0 .1/4, -1/8. Vay ham sd dat cue dai tai x = 2, yco = — va dat cue tieu dat x = - 4. 4 1 y C T = - - d) D = R\ {-l},y' = 1 - 7-^TT,y' = Oox = Ohoacx = -2. y - (x+D2 y"(0) = 2 > 0, y"(-2) = -2 < 0. (x + 1)3 Vay ham so dat cue dai tai x = - 2, yco = - 4 va dat cue tieu tai x = 0, ycr = 0. V i du 4: Tim cue tri cua cac ham so sau: a) y = xV4-x 2 b) y = Vx2 -2 x + 5 c) y = x + Vx2 - 1 Giai a) Dieu kien -2 < x < 2. Vdi - 2 < x < 2 thi 40 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
BBT: x V4-x ' V4-x ' -2 -V I V2 0 + 0 CD CTVay ham sd dat cue dai tai x = V2 , y C o = 2 va dat cue tieu tai x y C T = -2. b) D = R. Tacdy' Vx2 -2 x + 5 ,y' = 0<=>x= 1. BBT X —00 1 +00 y' 0 + y +00 +c ° Vay ham sd dat CT(1; 2). c) D = (-oo;-l] u[l;+oo).Vdix<-lhoacx>lthiy'=l + y' > 0 co y'<Oc o V x 2 ^ ! > —1 <=> x > Vx2 - 1 ox>l,x 2 > x 2 -lox> l V x 2 ^ ! < - 1 CO x < -Vx 2 - 1 CO x < -1 , x2 > x2 - 1 CO X « BTT X —00 — l +00 y' - + y Vay ham so khdng cd cue tri. Vi du 5: Tim cue tri cua ham sd: a) y V x 2 - 6 b)y = vV(x-5 ) Giai a) Tap xac dinh D = (-co; - V6 ) u (^6 ; +<x>) -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
. Vx2 - 6 = 3x2 (x2 -6)-x 4 _ 2x2 (x2 -9) x2 - 6 sl^W " vV-6)3 y' = 0»x = 0 hoac x = ±3. BBT X -co _3 -A/ 6 V6 3 +oo y' + 0 - - 0 + y -9A/3 y \ —00 —oc • +00 +00 9V3 Vay ham so ctat cue dai tai x = -3 va yco = —9 V3 , dat cue tieu tai x = 3 vayCT= 9 A/3 ^ „ xr , . „ , , 3 r r 2(x-5 ) 5 ( x - 2 ) b) D = R. Vdi x * 0 thi y' = vx 2 + y' = 0 <to x = 2. Bang bien thien 3 ^ 3Vx 0 +00 + ,+00 —00 -3V4- Vay ham sd dat cue dai tai x = 0, yco = 0 va dat cue tieu tai x = 2, yCT = -3 A/4 V j du 6: Tim cue tri cua ham so a) y = x - sin2x + 2 a) D = R, y' = 1 - 2cos2x b) y = 3 - 2cosx - cos2x. Giai 1 TC y' = 0 <=> cos2x = — o x = ±— + kn, k e Z; y'' = 4sin2x. 2 6 Ta cd y"(-— + kre) = 4sin(-—) = -2 A/3 < 0 nen ham sd dat cue dai tai 6 3 t , , .A/ 3 0 — + kn + — + 2. diem x = — + kn, k e Z, yen 6 t> z Ta cd y"(— + kit) = 4sin— = 2 A/3 > 0 nen ham so dat cue tieu tai cac 6 3 diem: x = — + kn, k e Z; ycr = — + kTt - — + 2. 6 * 6 2 b) y' = 2sinx + 2sin2x = 2sinx(l + 2cosx): y' = 0 o sin x = 0 2rc cosx = - i c=> x = kre hoac x = ± — + 2kTt, keZ . 42 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
y" - 2cosx + 4cos2x Ta co y"(k7t) = 2coskTC + 4cos2kTC = 2coskTC + 4 > 0, voi moi keZ , nen ham so eta cho ctat cue tiiu tai cac diem x = kre, ycr = 2 - 2coskTC bang 0 khi k chan va bang 4 khi k le. T , „,, 2rc , . _ 2TC . 4rc 2rc lacoy( ± — + 2k7i) = 2 cos — + 4cos—=6cos— = 3 3 3 3 2rc 9 dat cue dai tai diem: x = ±— + 2k7i, k e Z, yco = — 3 2 -3 < 0 nen ham so V i du 7: Chung minh ham sd sau khdng cd dao ham tai x = 0 nhung dat cue tri tai diem dd. -2x a)f(x)=|x | b)f(x) sinkhi x < 0 khix>0 Giai a) Ham sd xac dinh va lien tuc tai R. Ta cd: f - 1 khi x < 0 f(x) - x khi x < 0 x khi x > 0 f'(x) = 1 khix> 0 Do do ham sd khdng cd dao ham tai x = 0 va BBT: X —CO Q +0 ° y' - + y " *• o ^ * Vay ham sd dat CT(0; 0). b) Ham so xac dinh va lien tuc tren R. Ta cd -2x khi x < 0 f'(x ) 1 1 x nen lim f'(x) = 0 * lim f'(x) = —, do do f — cos— khix> 0 *->0 " x ^ 0 + 2 2 2 khong cd dao ham tai x = 0 va BBT tren khoang (-TC; TC). X -TC Q " y' + - y 0 Vay ham sd dat cue dai tai x = 0 va yco = y(0) = 0. V i du 8: Chung minh rang ham sd ludn luon cd cue dai va cue tieu: a) y = x3 + ax2 - (1 + b2 )x + 2a + b - 3ab b) y = (x - a)(x - b)(x - c) vdi a < b < c. c ) y = _ x +(m + 2)x+m +2 x + m -BDHSG DSGT12I1- 43 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai a) D = R. Ta co y' = 3x2 + 2ax - 1 - b2 A' = a 2 + 3(a + b2 ) > 0, Ve biet xi , X2. Bang bien thien: A' - a2 + 3(a + b2 ) > 0, Va, Vb nen y' = 0 luon luon co 2 nghiem phan X —00 Xl X2 +00 y' + 0 - 0 y C T ^ +00 Vay ham so luon luon co mot cue dai va mot cue tieu. b) D = R. y' = (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b) = 3x2 - 2(a + b + c) + ab + be + ca. A' = (a + b + c) 2 - 3(ab + be + ca) = a 2 + b2 + c2 - ab - be - ca = - [(a - b)2 + (b - c)2 + (c - a)2 ] > 0 voi a < b < c. 2 Do do y' = 0 co 2 nghiem phan biet va ddi dau 2 lan khi qua 2 nghiem nen luon ludn cd mot cue dai va mot cue tieu. x2 +2mx + 2m- 2 c) D = R \ {-m} . Tacd: y* = (x + m) 2 Xet ham sd g(x) = x + 2mx + 2m - 2. Ta cd A' = m2 - 2m + 2 > 0, Vm va g(-m) = -m 2 + 2m -2 * 0 , Vm nen y' = 0 ludn cd hai nghiem phan biet khac -m, y' doi dau hai lan khi qua 2 nghiem, vay ham sd luon luon cd cue dai va cue tieu. Vi du 9: Tim a de dd thi ham so y = — x2 + — ax2 + x + 7 cd 2 cue tri va 3 2 x2 x2 hoanh do 2 diem cue tri cua ham sd do thoa man —j + —| > 7. x Giai D = R. Ta cd y' = x 2 + ax + 1. V i y' la ham sd bac hai nen ham sd co 2 cue tri khi va chi khi y'(x) = 0 cd hai nghiem phan biet coA>0<=>a 2 -4>0<=>a < - 2 hoac a > 2. Goi x, va x 2 la hai nghiem cua y'(x) = 0 thi S = Xj + x2 = -a, P = x^2 = 1. Ta cd: —\ + • ( Q2 >7 co f \ 2 X l X2 S2 -2P v x2 ( Jl Jl\ 2>7co x 1 + x 2 XjX^ >9 >9co(a 2 -2) 2 >9coa 2 >5 Chon gia tri a < - -JE hoac a > yfE V i du 10: Tim cac sd thuc p va q sao cho ham sd f(x) = x + p 44 x + l BDHSG DSGT12I1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
ctat cue dai tai diem (-2; -2). Giai Ta cd f '(x) = 1 — . vdi moi x * -1 . x + 1 Neu q < 0 thi f '(x) > 0 vdi moi x * -1 . Ham so khdng cd cue dai, cue tieu (loai). Neu q > 0 thi phuong trinh: f '(x) = (x + 1)' 0 cd hai nghiem phan biet Xi = - 1 - <Jq va X2 = - 1 + yfq . BBT: X —oo - i-V q 1 - 1 + y[q +C0 y' + 0 - 0 + y Ham sd dat cue dai tai diem (2; -2) khi va chi khi f-l- ^ = -2 0 W = l 0 [ q = l [f(-2) = -2 [p = l [p = l Vi dull : Tim m de ham so: a ) y = ~( m + 5m)x + 6mx + 6x - 5 dat cue dai tai x = 1. b)y x2 +(l-m)x- 2 x + m dat cue tieu tai x = 0. Giai a) D = R. Ta cd y' = -3(m2 + 5m)x2 + 12mx + 6 Neu ham sd dat cue dai tai x = 1 thi y'(l) = 0 -3m2 - 3m + 6 = 0 <=> m = 1 hoac m = -2. Ta cd y" = -6(m2 + 5m)x + 12m Vdi m = 1 thi y" = -36x + 12 nen y"(l) = -24 < 0, ham so dat cue dai tai x = 1. Vdi m = - 2 thi y" = 36x - 24 nen y"(l) = 12 > 0, ham so dat cue tieu tai x = 1 (loai). Vay vdi m = 1 thi ham sd dat cue dai tai x = 1. b) D = R \ {-m}.Tacdy ' x2 + 2mx - m2 + m + 2 (x + m) 2 Neu ham sd dat cue tieu tai x = 0 thi y'(0) = 0 => -m 2 + m + 2 = 0=>m = - l hoac m = 2. V di m Do dd y" -1 thi y 2 x 2 +2x- 2 :x + 3 + - (x-1) 3 -BDHSG DSGT12/1- x - 1 x - l > y"(0) = -2 < 0 >y' = i - (x-l f 45 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
=> x = 0 la diem cue dai cua ham sd: loai. Tr*. ~ x2 -x - 2 4 Vai m = 2 thiy = = x - 3 + =>y' = l - x + 2 x + 2 (x + 2)2 g Do do y" = —. y"(0) = 1 > 0 nen x = 0 la diem cue tieu cua ham (x + 2) 3 so. Vay gia tri can tim m = 2. V i du 12: Tim ham sd f(x) = ax3 + bx2 + cx + d sao cho ham so f dat cue tieu tai diem x = 0, f(0) = 0 va dat cue dai tai diem x = 1, f ( l ) = 1. Giai Ta cd f '(x) = 3ax2 + 2bx + c. V i f(0) = 0 nen d = 0. Ham sd dat cue tieu tai diem x = 0 nen f '(0) = 0 do dd c = 0. V i f(l ) = 1 nen a + b = 1. Ham sd dat cue dai tai diem x = 1 nen f'(1) = 0 do do do 3a + 2b = 0. f a + b = i [a = -2 Ta cd he phuong trinh + 2b = 0 ° j b = -3 Thir lai: f(x) = -2x3 + 3x2 , f '(x) = -6x2 + 6x, f "(x) = -12x + 6. f "(0) = 6 > 0. Ham sd dat cue tieu tai diem x = 0 (thoa man). f "(1) = -6 < 0. Ham so dat cue dai tai diem x = 1 (thoa man). , ' a. sin. x cos x 1 * V i du 13: Tim a de ham sd y = dat cue tri tai 3 diem thuoc a cosx khoang (0; ~) . 4 Giai r^ - 1 - 7 1 , 1 -r - . a-sm x « , _ . Dieu kien x * — + kre. Ta co y = — nen y = 0 <=> sinx = a. 2 a cos x , , -sin 2 x + 2asinx - 1 1 a co y ' = = a cos x Vdi sinx = a thi a = sinx thi y" = * 0, do dd ham so dat cue tri sinxcosx 9rt tai 3 diem thudc khoang (0; — ) <=> sinx = a cd 3 nghiem thuoc khoang ,„ 9TC . . . rc 3rc . „ . . \/2 (0; — ) \ { - ; — } co 0 < a < -—- v 4 2 2 r 2 V i du 14: Tim m de ham so: a) y = x +2mx + l-3 m _ ^ ^ ^ m ^ ^ n & m yx ^ ^ ^ ^ x - m = mx 2 +(2-4m) x + 4 m + l £ , 2 ^ ^ y . M gj . ^ ^ fr. ^ ^ x - l Giai x 2 -2mx + m2 - 1 a) Dieu kien x * m. Ta cd y' : -2 ' • (x-m ) 46 -BDHSG DSGI12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Do thi co 2 cue tri d 2 phia cua true tung co y' = 0 co 2 nghiem X,, x2 * m va x,x2 <0eom 2 -l<0co-l<m<l . b) Dieu kien: x*l Ta co y' = mX' ~ 2mX " 3 , dat g(x) = mx2 - 2mx - 3. (x-l ) D6 thi co 2 cue tri co m * 0, A' > 0, g(x) * 0 co m < -3 hoac m > 0. 3 Ta cd x, + x2 = 2, X]X2 = nen yCo- ycr < 0. m o (2mx, + 2 - 4m)(2mx2 + 2 - 4m) < 0 o 4m2 x,x2 + 2m(2 - 4m)(x, + x2 ) + (2 - 4m) 2 < 0. 1 co -12m + 2m(2 - 4m) + (2 - 4m) 2 < 0 co 4 - 20m < 0 co m > - . x2 - 2mx + 2 Vi du 15: Cho ham so y = vdi m la tham sd. x - l a) Tim m de do thi ham so cd hai diem cue tri A va B. b) Chung minh rang khi do dudng thang AB song song vdi dudng thang 2x — y - 10 = 0. Tinh khoang each giua 2 cue tri. Giai , x2 - 2x + 2m - 2 a) DK: x * 1. Ta co y = = ( x -l) 2 Dieu kien cd 2 cue tri la A' > 0 va g(l) * 0. 3 co3-2m > 0 v 3 -2m *0com< - 2 b) Ta cd A(l - V3 - 2 m ; 2 - 2m - 2 V3-2m ) B (l + S -2 m ; 2 - 2m + 2 V3-2m ) He so gdc cua dudng thang AB la: . y(x„)-y (x,) 4\/3 - 2m „ , _ i n k = 2 — 1 = —p= = = 2 . Ta co 2x - y - 10 x 2 - X j 2V3-2m co y = 2x - 10 nen cd he so gdc bang nhau => dpcm. q Va AB2 = 4(3 - 2m) +8(3 - 2m) = 12(3 - 2m) ^>,AB = 2x79 -6 m , m < - Vi du 16: Viet phuong trinh dudng thang di qua diem cue dai, cue tieu cua dd thi: a) y = x3 + 3mx2 + 3(m2 - l)x + m3 - 3m b)y x2 - 2mx + 5m - 4 - m2 x - 2 Giai a) y' = 3x2 + 6mx + 3(m2 - 1), A' = 1 > 0, Vx nen do thi luon luon cd CD va CT vdi hoanh do xi, x2 . Lay y(x) chia cho y'(x) ta cd: y(x) = 11 x+—|y'(x) - 2(x + m). -BDHSG DSGT12/1- 47 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Do do: y, = y( X l) = va y2 = y(x2 ) = f^x , +^jy'(x 2 ) - 2(x2 + m) = -2(x2 + m) nen dudng thang qua CD, CT la y = -2(x + m). b) DK: x * 2. Ta cd y = x - 2(m - 1) + m - m x - 2 nen y' — 1 - m-m 2 (x-2) 2 -(m-m 2 ) (x-2) 2 (x-2) 2 Tu dd suy ra dieu kien co CD va CT lam — m>0<=>0<m<l . Goi xi , X2 la hoanh dp CD, CT thi xi < 2 < x2 . Ta cd m m2 y(xj = x, - 2(m - 1) + , ^v = x, - 2(m - 1) + (xj - 2) = 2x, - 2m. y(xa) = x2 - 2(m - 1) + 6q-2) m - m (xj-2) x2 - 2(m - 1) + (x2 - 2) - 2x2 - 2m Vay phuong trinh dudng thang qua CD va CT la y = 2x - 2m. DANG 2: UNG DUNG CUA CUC TRI V i du 1: Chung minh rang phuong trinh 2x3 - 3x2 - 12x - 10 = 0 cd nghiem x = a duy nhat va 3,5 < a < 3,6. Giai Xet ham sd f(x) = 2x3 - 3x2 - 12x - 10 BBT: f'(x) = 6(x 2 - x -2);f'(x ) = 0o x = X —OO —1 2 +oo y' + 0 - 0 + y +00 "^-30^ ^ Tu BBT thi phuong trinh f(x) = 0 cd nghiem duy nhat a > 2. Ta cd f(3,5).f(3,6) < 0 nen 3,5 < a < 3,6 => dpcm. V i du 2: Cho ab * 0. Chung minh phuong trinh: x3 - 3(a 2 + b2 )x + 2(a3 + b3 ) = 0 cd 3 nghiem phan biet. Giai Xet ham s6 y = x3 - 3(a 2 + b2 )x + 2(a3 + b3 ), D = R y' = 3x2 - 3(a2 + b2 ), y' = 0 o xl i 2 = ± Ja2 + b2 , (S = 0, P = a 2 + b2 ) V i y' bac 2 cd 2 nghiem phan biet nen cd CD va CT. 48 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com