Tài liệu Phương pháp giải các chủ đề căn bản giải tích 12-lê hoành phò

515.076 PH561P NGirr.ThS. LE H O A N H a ren luyen kl nang lam bai H a No NHA XUAT BAN DAI HOC QUOC GIA HA NOI GIAI PHO NGIfT.ThS. LE HOANH PHO GIAI CAC C H U C A N D E B A N (Boi dm!smmmi!mifi cs^G NHA XUAT BAN flAI HOC QUOC GIA HA NOI Ha Nol . . NHA XUAT BAN DAI HOC QUOC GIA HA NQI 16 Hang Chuoi - Hai Ba Trang - Ha Npi Dien thoai: Bien tap-Che ban: (04) 39714896: Hanh chinh: (04) 39714899: Tonq bien tap: (04) 39715011 Fax: (04)39714899 * Chiu trdch nhiem xuat * * ban: Gidm doc - Tong bien tap: Bien tap: TS.PHAM T H I TRAM L A N HUONG bdi: N G U Y E N KHCJl M I N H Che ban: N H A SACH HONG A N Saa Trinh bay bia: VO THI T H I T A Doi tdc lien ket xuat ban: N h a sach H O N G A N S A C H LIEN KET CAC CHU DE CAN BAN GIAI TfCH 12 Ma so: 1L- 154DH2014 In 2.000 cuon, khd 17 x 24cm tai Cong ty Co phan VSn hoa Van Lang. Giay phep xuat ban so: 463-2014/CXB/10-99 OHQGHN, ngay 14/03/2014 Quyet dinh xuat ban so: 154LK-TN/Q0-NXB OHQGHN, v ^ i > In xong va nop iuu chieu quy II nam 2014. Nham muc dich giup cac ban hoc sinh Idp 10, Idp 1 1 , Idp 12 nam vOng kien thifc can ban ve mon Toan ngay tCr luc vao THPT cho den khi chuan bi thi Tot nghiep, tuyen sinh Cao dang, Dai hoc, tac gia da bien scan bo sach P H l / O N G P H A P G I A I gom 6 cuon: - C A C C H U D E C A N B A N DAI S O 10 - C A C C H U D E C A N B A N HINH H Q C 10 - C A C C H U D E C A N B A N DAI S O - GIAI T I C H 11 - C A C C H U D E C A N B A N H I N H H Q C 11 - C A C C H U D E C A N B A N GIAI T I C H 12 - C A C C H U D E C A N B A N HINH H O C 12 TL/ nen Toan can ban nay, cac ban c6 the nang cao dan dan, bo sung va md rpng kien thufc va phUdng phap giai Toan, ren luyen ky nang lam bai va tC/ng bade giai dung, giai gpn cac bai tap, cac bai toan kiem tra, thi CLT. Cuon C A C C H U D E C A N B A N G I A I T I C H 1 2 npi dung la phan dang Toan, tom tat kien thac cac chu y; phan tiep theo la cac bai toan chpn vdi nhieu dang loai va mac dp; phan cuoi la 8 hay dap so. nay c6 16 chu de vdi va phUdng phap giai, Ipc can ban minh hpa bai tap c6 hadng dan DO da CO gang kiem tra trong qua trlnh bien soan song khong tranh khoi nhQng sai sot ma tac gia chaa thay het, mong don nhan cac gop y cua quy ban dpc, hpc sinh de Ian in sau hoan thien hdn. Tac gia LE HOANH PHO 3 O CHUD E I T i N H t>ON t>IEU DANG TOAN • 1. TiM KHOANG DONG BIEN VA NGHjCH BIEN Dinh nghia: Hdm so f xdc dinh tren K Id mot khodng, doan hodc nica khodng. - f dong hien tren K neu vai moi xi, X2 e K: x/ < X2 =^f(x\) < f(x2) -f nghich hien tren K neu vai moi x/, X2 e K: xi < X2 =>f(xi) > f(x2). Dieu kien can de ham so dffn difu Gid su hdm so c6 dao hdm tren khodng (a; b) khi do: - Neu hdm so f dong bien tren (a: b) thif '(x) > 0 vai moi x e (a; b) - Neu hdm so f nghich bien tren (a; b) thif'(x) ^0 vai moi x e (a; b). Dieu kien dii de hdm so dffn dieu Gid su hdm so f c6 dqo hdm tren khodng (a; b) khi do: Neuf'(x) > 0 vai moi x e (a; b) thi hdm so f dong bien tren (a; b) Neu f'(x) < 0 vai moi x e (a; b) thi hdm so f nghich bien tren (a; b) Khi f '(x) = 0 chi tgi mot so hiru hgn diem cua (a; b) thi ket qud tren vdn dung. Neu hdm so f dong bien tren (a; b) vd lien tuc tren nita khodng [a;b); (a;bj; doan [a:b] thi dong bien tren nica khodng [a;b); (a;bj; doan [a;b] tuang icng. Tuang tu cho nghich bien. Phumtg phdp xet tinh dffn dieu: - Tim tap xdc dinh - Tinh dao hdm, xet ddu dqo hdm, lap bang bien thien - Ket ludn Chiiy: 1) Cong thirc vd quy tdc dao hdm y^C =>y' ^0:y=x ^y' = l;y = x" =^y' = nx"-'; ^y=J—(x>0); y= '4^=>y'=— 2Vx n"-47 y = sinx =:>y' = cosx; y = cosx =>y' = -sinx; y= ^ y = tanx =^y = 1r — ; y = cotx =>y , = — -r1— . cos X sin X (u + v)' = u' + vV (u-v)' = u'- v'; 5 (u. v)' = u'.v + u.v'; u .V - u.v ;f'.-f'u.u'x.. 2) Phuang trlnh luang gidc ca ban: X = a + k27i sinx = sina <=> X = 71 - a + k27i (keZ) X = a + k27i cosx = cos a <=> X = - a + k27r (keZ) tanx ^ tan a c^x a + kTt (k e Z) cotx = cot a <:=>x = a + kn(k e Z). Bai toan 1: Tim khoang dong bien, nghich bien cua ham so: a)y = x^-8x + 5 b) y = x^ - 2x^ + x + 1. Gidi a) Tap xac dinh D ^= R. Ta c6 y' = 2x - 8. Cho y' = 0 « 2x - 8 = 0 <=> x = 4. 4 +00 Bang biSn thien (BBT) x -00 0 + y' y Vay ham so nghich bien tren (-oo; 4), dong bien tren (4; +oo). b) D = R. Ta CO y' = 3x^ - 4x + 1 Cho y' = 0 « 3x^ - 4x + 1 = 0 o x = - hoac x = 1. +00 1/3 BBT x -00 1 + 0 0 + y' ^ ^ y ^ ^--^ Vay ham so dong bien tren moi khoang (-co; ^) va (1; +co), nghich bi6n tren khoang ( ^ ; 1). Bai toan 2: Tim khoang dong bien, nghich bien cua ham so: a)y-x''-2xl b) y = x'"* + 9x^ - 3. Gidi a) Tap xac dinh D = R. y' = 4x^ - 4x = 4x(x^ - 1), y' = 0 <=> x = 0 hoac x = ±1. 6 BBT X -co - y' y 0 -1 0 + 0 +00 1 - + 0 +00 +00 ^ - 1 ^ Vay ham so dong bien tren moi khoang (-1;0) va ( 1 ; +oo), nghich bien tren khoang (-oo;-1) va (0; 1). b) D - R. Ta CO y' = 4x^ + 18x = 2x(2x^ + 9), y' = 0 <=> x = 0. y' > 0 tren khoang (0; +co) ^ y dong bien tren khoang (0; +QO) y' < 0 tren khoang (-oo; 0) => y nghich bien tren khoang (-co; 0). Bai toan 3: Tim khoang dong bien, nghich bien cua ham so: a)y = 3x-9 b)y = x + - . x Giai a)D = R \. Ta CO y' = -6 (l-.r) < 0 vai moi x ^ 1 nen ham so nghich bien trong cac khoang (-oo; l ) v a ( l ; + o o ) . b) TapxacdinhD = R \. Taco BBT: 1 - ^ = ^ ^ - ^ , y ' - 0 < » x = ±V3. x^ X" X -30 y' + 0 +00 0 -V3 - - 0 + y Vay ham so dong bien tren khoang (-oo; - V3 ) va (V3 ; +oo), nghich bien tren moi khoang (- V3 ; 0 ) v a ( 0 ; V 3 ) . Bai toan 4: Tim cac khoang don dieu cua ham so: K^ x+ \ jc +8 2x X -9 Gidi a) D = R. Ta c6: y' = -X- y' = 0 -2x + 8 {x'+%f - X - 2x + 8 = 0 X = -4 hay X =2 7 BBT: X -00 - y' -4 0 4 2 0 +00 - y Vay ham so dong bien tren khoang ( -4; 2) va nghich bien tren cac khoang -4), (2; + 0 0 ) . (-co; b ) D = R \6 y' = ~^^^ < 0, V x ^ ± 3 . (x'-9)' Do do y' < 0 tren cac khoang (-oc; -3), (-3; 3), (3; +oc) nen ham so da cho nghich bien tren cac khoang do. Bai toan 5: Xet su bien thien ciia ham so tren doan, nua khoang: a) y = ^|9-x^ b)y= V x ' - 2 x + 7 . Gidi a) Dieu kien 9 - x" > 0 » -3 < x ^ 3 nen D - [-3; 3]. Voi -3 BBT: < X < 3 thi y' - X - X ^ , y' = 0 ^ y 0 0 y' + ^ ^ X = 0. ^ Vay ham so dong bien tren khoang (-3; 0) va nghich bien tren khoang (0; 3). Do ham so f lien tuc tren doan [0; 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^ - 2x + 7 > 0, Vx => D = R. „ . , 2x-2 x-1 Fa CO y = —, = , . 2Vx'-2x + 7 Vx--2x +7 y'^0<:^x> l,y'x< 1 Va f lien tuc tren R nen ham so nghich bien tren nura khoang (-oo; 1 ] va dong bien tren nua khoang [1; +oo). Bai toan 6: Xet su bien thien cua ham so: b)y = a)y Vl6^ a) DK: 16 - x^ > 0 « X' x+2 Gidi < 16 <=> -4 < X < 4. D = (-4; 4). 8 Ta CO 16 . > 0 , Vx e (-4; 4). y' = (16-x-)Vl6-xVay ham so dong bien tren ichoang (-4; 4). b) D = [0; +00). Vai x > 0, y' = BBT: +00 - 0 + y' , y' = 0 <» x = 2. 2Vx (x + 2) 2 0 X 2-x y Vay ham so dong bien tren (0; 2) va nghich bien tren (2; +oo). Bai toan 7: Tim Ichoang don dieu cua ham so a) y = \fx - X - X b) y = Giai a) D = R . V a i x 5 ^ 0 . t a c 6 : y ' = y' = 0 « X' = 1 X = y' > 0 <=> Vx^ > 1 ^ - ^ = ^ ~ ^ ±1. x^ > 1 hoac x < -1 hoac x > 1. y'<0<=>Vx^ x = ±3. -3 -00 + 0 -V6 - 3 V6 - 0 +00 + y Vay ham so dong bien tren cac khoang (-oo; -3), (3; +oo), nghich bien tren cac khoang(-3;-V6),(V6;3). Bai toan 8: Xet su bien thien cua ham so: a) y = 4sinx - 3 b) y = X + cos x. 9 Gidi a) D = R . T a c6 y ' = 4cosx Xet y ' >0 « 4cosx > 0 < » - - + k 2 ; r < x < - + k 2 ; r , k e Z n e n 2 2 h a m so d o n g bien tren cac k h o a n g (- ^ + k 2 ;T ; ^ + k 2 ;T ), k e Z . X e t y ' < 0 <=> 4cosx < 0 < » ^ - +k2;r 0 t r e n m o i k h o a n g (— + kn; — + ( k + l)7t) n e n d o n g b i e n t r e n m o i doan 4 4 [ - +k7r; 4 - +(k+ 4 l)7i],k e Z. V a y h a m so d o n g bien tren R . Cach khac: lay a, b bat k y thuoc R v a a < b. T r e n k h o a n g (a;b) t h i y ' > 0 v a y ' = 0 t a i h i j u han d i e m n e n h a m s6 f dong b i e n , d o d o f ( a ) > f ( b ) . V a y theo d i n h n g h i a t h i h a m so f d o n g b i e n tren R . B a i toan 9 : T i m k h o a n g d o n g b i e n , n g h i c h b i e n cua h a m so: a) y = X - sinx t r e n [ 0 ; 2n] b) y = x + 2cosx tren ( 0 ; n). Gidi a) y ' = 1 - cosx. T a c6 V x [ 0 ; 27t] =:> y ' > 0 v a y ' = 0 <=> x = 0 hoac x = 2n. V i h a m so l i e n tuc t r e n doan [ 0 ; 2n] nen h a m so d o n g b i e n t r e n doan [ 0 ; 2::]. b) y ' === 1 - 2 sinx. T r e n k h o a n g ( 0 ; n). y' > 0 <=> s i n x < - — < x < 2 6 — 6 , • 1 r> 7r , _ 571 7t y' < 0 <=> s m x > - < » 0 < x < — hoac — < x < —. 2 6 6 6 V a y h a m so d o n g b i e n t r e n k h o a n g (—; — ) , n g h i c h biSn tren m o i k h o a n g 6 6 (0;5)va(^;H). 6 6 10 DANG TOAiy 2. CHUfNG MII\iH DQfN DIEU Neu f'(x) > 0 v&i moi x e (a; h) thi ham s6fdong hien tren (a; h) Neu f '(x) > 0 vai moi x e (a; b) vd f '(x) = 0 chi tai mot so hitu hgn diem cua (a; h) thi hdm so dong bien tren khodng (a; b). Neuf'fx) < 0 vai moix e (a: H) thi hdm so nghich bien tren (a; b) Neu f'(x) < 0 vai moi x e (a; b) vd f'(x) ^ 0 chi tai mot so hihi hgn diem cua (a; b) thi hdm so nghich bien tren khodng (a: b). Neu hidm sof dong bien tren (a; b) vd lien tuc tren nua khodng [a;b); (a;bj; doqn [a;bj thi dong bien tren mm khodng [a;b); (a;b]; dogn [a;b] tuang icng. Neu hdm so f nghich bien tren (a; b) vd lien tuc tren nua khodng [a;b); (a;bj; dogn [a;b] thi nghich bien tren nua khodng [a;b); (a:hj: dogn [a;b] tuang ung. Chiiy: I) Ddu nhi thuc bgc nhdt: f(x) = ax -i- b, a ^0 - 00 -b/a f(x) trdi ddu a 2) Ddu tam thuc bgc hai: f(x) 0 + 00 ciing ddu a ax^ ^ bx + c, a ^0 Neu A< 0 thif(x) luon cung ddu vai a Neu A 0 thif(x) luon cung ddu vai a, trie nghiem kep Neu A> 0 thi ddu "trong trdi - ngodi cung " -OO f(x) X/ cung ddu a 0 trdi ddu a X2 +00 0 ciing ddu a 3) Gid su hdm so f xdc dinh tren khodng (a; h) vd Xo e (a; b). Hdm so f duac goi Id lien tuc tgi diem Xg neu: lim f (x) = f(Xo). Hdm so khong lien tuc tai diem Xo duac goi Id gidn dogn tgi diem XQ. Hdm so f lien tuc tren mot khodng (a:b) neu no lien tuc tgi moi diem thuoc khodng do. Hdm so flien tuc tren nua khodng (a; bj neu no lien tuc tren khodng (a; b) va Vim/(x) =f(h). Hdm sof lien tuc tren nua khodng fa; b) neu no lien tuc tren khodng (a; b) vd lim f(x) =f(a). 11 Ham so flien tuc tren dogn [a; b] neu no lien tuc tren khodng (a; b) vd \\mf(x) =f(a), \xmf(x) =f(b). x->a X—>b Bai toan 1: Chung minh rang cac ham so sau day dong bien tren R. a) f(x) = - 6x^ + 20x - 13 b) f(x) = 2x - cosx + Vs sinx. Gidi a) f'(x) = 3x^-12X + 20. Vi A' = 36 - 20 < 0 nen f (x) > 0 voi moi x, do do ham so dong bien tren R. b) y' = 2 + sinx - \ cosx = 2(1 + ^ sinx - ^ cosx). = 2[1 + sin(x - y)] ^ 0. voi moi x. Vay ham so dong bien tren R. Bai toan 2: Chung minh cac ham so sau nghich bien tren R: a) f(x) = Vx'+l - X b) f(x) = cos2x - 2x + 5. Gidi a) T a c 6 f ' ( x ) = - , — 1. Vx-+1 Vi V x ^ ^ > Vx^ = I X I ^ X , Vx nen f '(x) < 0, Vx do do ham so f nghich bien tren R. b) f (x) = -2(sin2x + 1) < 0 voi moi x. r(x) = 0 « s i n 2 x - - l « 2 x = - - + 2 k K « x = - - + k 7 i , k € Z . 2 4 Ham f(x) lien tuc tren moi doan 4 + krc; -— + (k + 1)TC] va f (X) < 0 tren moi 4 khoang (-— + k7i; — + (k+l)7i) nen ham so nghich bien tren moi doan 4 4 [ - - + k 7 T ; - ~ + ( k + l ) 7 r ] , k e Z. 4 4 Vay ham so nghich bien tren R. Cach khac: Ta chung minh ham so f nghich bien tren R: V x i , X2 e R, X i < X 2 => f ( X | ) > f ( X 2 ) . That vay. lay hai so a, b sao cho a < xi < X2 < b. Ta c6: f'(x) = -2(sin2x + 1) < 0 voi moi x e (a; b). Vi f '(x) = 0 chi tai mot so huu han diem cua khoang (a; b) nen ham so f nghich bien tren khoang (a; b) =^ dpcm. 12 Baai toan 3: Chung minh cac ham so sau don dieu tren R: a ) y = -x^-2x2 + x - 3 b) y = - - x ' +2x'--x' Gidi a) D = R. Ta c6 y' = 4x^ - 4x + 1 = (2x - 1)^ ^ 0 vai moi x y' = 0 <=> X = ^ . Vay ham s6 dong bi^n tren R. b) D = R. Ta CO y' = -3x^ +8x^ -7x^ = (-3x^ + 8x - 7 ) x l V i A' = 16 - 21 < 0 nen -3x^ + 8x - 7 < 0 v o i mgi x. Do do y' < 0 vai moi x. y' = 0 x = 0. Vay ham so nghjch bien tren R. Bai toan 4: Chung minh ham so" dong bien tren moi khoang xac dinh cua no. x+2 , — x~ — 2x + 3 ' b)y = nghich bien tren moi khoang xac dinh cua no. X + 1 • a)y = x-2 Gidi a ) D = R \6 y ' = — ^ 4 (x + 2)^ > 0 vai moi xit-2. Vay ham so dong bien tren moi khoang (-QO; -2) va (-2; +oo) b ) D = R \. -x' Ta CO y' = -2x-5 1 — < 0 vai moi x ^ -1 (vi A' = 1 - 5 < 0). (x + 1)- Vay ham so nghich bien tren moi khoang (-oo; -1) va (-1; +oo). Bai toan 5: Chung minh ham so: y = f(x) = ^ x^ + 2x^ + 3x - 1 a) nghich bien tren doan [-3;-!]. b) dong bien tren cac nira khoang ( - x ; -3] va [-1; +oo). Gidi D = R. Ta c 6 f ' ( x ) = x^ + 4x + 3 f ' { x ) = 0 « x^ + 4x + 3 = 0 < » x BBT X -00 -f- y' -3 -1 0 0 -00 + +00 ^ -1 y +00 -7/3' •3 hoac X = - 1 . a) Ta CO f "(x) < 0 tren khoang (-3;-l) nen f nghich bien tren khoang (-3;-l) va f lien tuc tren doan [ - 3 ; - l ] nen f nghich bien tren doan [ - 3 ; - l ] . b) Ta CO f '(x) > 0 tren cac khoang (-co; -3) va ( - 1 ; + 0 0 ) nen f dong bien tren khoang ( - 0 0 ; -3) va ( - 1 ; + 0 0 ) va f lien tuc tren cac nira khoang ( - 0 0 ; -3J va [ - 1 ; +GO) nen f dong bien tren cac nua khoang ( - 0 0 ; -3] va [ - 1 ; +co). Bai toan 6: Chung minh ham so: y = dong bien trong khoang ( - 1 ; 1) va 1 + x^ nghich bien trong cac khoang (-QO; -1) va ( 1 ; + 0 0 ) . Giai Tap xac dinh D = R. l(l + x ' ) - 2 x . x _ 1 - x ' y' = 0<:i>x = ± l . Ta CO y' < 0 « 1 - x^ > 0 <=> - 1 < X < 1. y' < 0 < ^ 1 - x^ < 0 1 u do suy ra dpcm. X < -1 hoac X > 1. Bai toan 7: Chung minh ham so: y = sin(x + b) (a ^ h + kn; k e Z) don dieu trong moi khoang xac dinh. Gidi Ham so gian doan tai cac diem x = -b + k7i (k 6 Z). , _ sin(x + b)cos(x + a) -sin(x + a)cos(x + b) sin"(x + b) sin(b - a) ?t 0 (do a - b ?t kji) sin"(x + b) Vi y ^ 0 va y' lien tuc tai moi diem x -b + kTi, nen y' giiJ nguyen mot dau trong moi khoang xac djnh, do do ham so don dieu trong moi khoang do. DANG TOAN BAi TOAl\ CO THAM SO • 3. Gia sif ham so cd duo ham tren khoang (a; b): Dung dieu kien can - Neu ham so f dong bien tren (a: b) thif'(x) ^0 vai moi x e (a; b) - Neu ham so f nghich bien tren (a; b) thi f'(x) <0 vai moi x e (a; b). 14 Dung dieu kien dii Neuf'(x) > 0 vai moi x e (a; b) thi ham sof dong hien tren (a; b) Neuf '(x) > 0 vai moi x e (a; h) va f '(x) = 0 chi tgi mot sii huu hctn diem cua (a: h) thi ham so dong bien tren khodng (a; b). Neuf'fx) < 0 vai moi x e (a; h) thi ham so nghich biin tren (a; b) Neuf '(x) <0 vai moi x e (a: b) va f '(x) = 0 chi tgi mot s6 him han diem cua (a; b) thi ham so nghich bien tren khodng (a; b). Bai toan 1: Tim cac gia tri cua thiam so a de ham so f(x) + 4x + 3 dong bien tren R. Gidi Tap xac dinh D = R. f'(x)-=x^ + 2ax + 4, A' = a^-4 - Neu a' - 4 < 0 hay -2 < a < 2 thi f '(x) > 0 vai moi x e R nen ham so d6ng bien tren R. - Neu a = 2 thi f '(x) = (x + 2)^ > 0 vai moi x -2 nen ham so dong bien tren R. - Neu a = -2 thi ham so f '(x) = (x - 2)^ > 0 vai moi x tren R. 2 nen ham so dong bien - Neu a < -2 hoac a > 2 thi f'(x) = 0 c6 hai nghiem phan biet nen f' c6 doi dau: loai. Vay ham so dong bien tren R khi va chi khi -2 < a < 2. Bai toan 2: Tim cac gia tri cua tham so a de ham so f(x) = ax^ - 3x^ + 3x + 2 dong bien tren R. Gidi Tap xac dinh D = R. Ta c6 f '(x) = 3ax^ - 6x + 3. Xet a = 0 thi f (x) = -6x + 3 c6 d6i dSu: loai Xet a 9i 0, vi f khong phai la ham hang (y' = 0 toi da 2 diem) nen dieu kien ham so dong bien tren R la f '(x) > 0, Vx a>0 o • a >0 9 - 9a < 0 fa > 0 a>l a ^ 1. Bai toan 3: Tim cac gia tri cua m de ham so f(x) = mx - x nghich bien tren R: Gidi Tap xac djnh D = R. Ta CO y' == m - 3x^ - Neu m < 0 thi y' < 0 vai moi x G R nen f nghich bien tren R - Neu m = 0 thi y' = -3x^ < 0 vai moi x e R, dang thuc chi xay ra vai x = 0, nen ham so nghich bien tren R. 15 - Neu m > 0 thi y' = 0 « BBT X X= ±iy -00 - y' X2 0 + 0 +00 - y Do do ham so dong bien tren khoang (xi, X2): loai Vay ham so nghjch bien tren R khi va chi khi m < 0. Bai toan 4: Tim cac gia tri cua m de ham so f(x) = sinx - mx + 4 nghich bien tren R. Gidi Tap xac dinh D = R. Vi f(x) khong la ham hang voi moi m nen f(x) = sinx - m + 4 nghich bien tren R <=> f'(x) = cosx - m < 0, Vx m > cosx, Vx <::> m ^ 1. Bai toan 5: Tim m de ham so y = x + 2 + x-1 dong bien tren moi khoang xac dinh. Gidi Tap xac dinh D = R \ 1}. Ta CO y' = 1 - m (x-1)- , vai moi x ?t 1. x^\ - NcLi m < 0 thi y' > 0 vai moi Do do ham so dong bien tren m6i khoang (-00; 1) va (1; +oc). - Neu m > 0 thi y'= x" - 2x + l - m (x-1)^ y' = 0 « > x ^ - 2 x + l - m = 0<=>x = l ± V m l-Vm X BBT + y' 1+ 1 0 v'm 0 +00 + y Ham so nghich bien tren moi khoang (1 - v m ; 1) va (1; 1 + v m ): loai. Vay ham so dong bien tren moi khoang xac dinh cua no khi va chi khi m < 0. Bai toan 6: Tim a de ham so: f(x) = x - ax" + x + 7 nghich bien tren khoang (1; 2). Gidi Ta CO f'(x) 3x^ - 2ax + 1 la tam thuc bac hai. 16 Ham so nghich bien tren khoang ( 1 ; 2) khi va chi khi y' < 0 voi moi x e ( 1 ; 2) [/'(1)^0 /'(2)<0 <=> 4-2a<0 13 <=> a> — [13-4a<0 4 < Bai toan 7: Tim m de ham so y = doan CO do dai bang 3. + 3x^ + mx + m chi nghich biSn tren mot Gidi: D = R, y' = 3x^ + 6x + m . A' = 9 - 3m Xet A' < 0 thi y' > 0, V x : Ham luon d6ng bien (loai) = 0 CO 2 nghiem X|, X2 nen x i + X2 = -2, X|X2 = Xet A' > 0 <=> m < 3 thi X -00 X7 + y' y 0 - +00 0 + ^ Theo de bai: X2 - x i = 3 » m ^ (x2 - x i ) ' = 9 <=> x^ + x^ -2X|X2 = 9 <=> (X2 + xi)^ - 4xiX2 = 9 <=> 4 - ^ m = 9<=> m = - ~ (thoa) Bai toan 8: Tuy theo tham so m , xet sir bien thien cua ham so: y = - x^ - 2mx^ + 9x - m. 3 Gidi D = R. Ta CO y' = x^ - 4mx + 9; A' = 4m^ - 9 — /i™2 - NIU A' < 0 <=> 4m^ < 9 <» | m | < ^ thi y' > 0, Vx nen ham s6 dong bien tren R. - Neu A' > 0 « 4m^ ^ 9 <::> m > - thi y' = 0 CO 2 nghiem phan biet xi,2 = 2 m ± V 4 m ^ - 9 . Lap bang bien thien thi ham dong bien tren khoang (2m - V4m' - 9 ; 2m +V4m^ - 9 ) Va nghich bien tren moi khoang (-oo; 2m - V4m' - 9 ), (2m + V4m" - 9 ; +oo). Bai toan 9: Xet sir bien thien cua ham so: y = ^ ^ ^ " ^ theo tham so m . x-1 Gidi D = R \. iHij viEN imn BiNH THUA;-.;! 17 „ , , -2-ni Ta CO y = (x-1) - N6u m = -2 thi y = 2, Vx 7t 1 la ham so khong doi. - Neu m > -2 thi y' < 0, Vx 9^1 nen ham so nghich bien tren moi khoang (-o);!), ( ! ; + « ) ) . - NIU m < -2 thi y' > 0, Vx ^ 1 nen ham so dong bien tren moi khoang DANG TOAN • 4. UfniG DUNG VAO GIAI PHlTOfNG TRINH Nku ham sSf dan dieu tren K va c6 M, N thuoc K thi phuang trinh f(M)=f(N) <^ M=N. Neu ham so f dong bien tren K va cd M, N thuoc K thi bat phuang trinh f(M) >f(N) ^ M>N. Neu ham so f nghich bien tren K va cd M, N thuoc K thi bat phuang trinh f(M)>f(N) o M 0 nen f d6ng biSn tren (-3; 2). 2V3 +1 2V2 - 1 Ta CO f ( l ) = 2 - 1 = 1 nen phuang trinh: f(t) = f ( l ) « t = 1 <» x^ - X - 1 = 0 « X = 18 Bai toan 2: Giai pliuang trinh ^j2x^ + 3x^ +6x + \6 = 2V3 + V 4 - x . Gidi: Dieu Icien xac dinh: Ix"+3r+Gx+\6^0 f(x+2)(2r-x+8)>0 4-x>0 4-x^O Phuong trinh tuong ducng^J2x^ + 3x^ +6x + \6 --^A-x 0-2 0 nen f dong bien + - V2x' + 3 x ' +6X + 16 2V4^ ma f(l)=2 V3 , do do phuong trinh tra thanh f(x) == f ( l ) x=l Vay phuong trinh c6 nghiem day nhdt x = l . Bai toan 3: Giai phuong trinh Vx - V l - x = 5 - 4x . Giai D i k kien: x ^ 0. PT 4x + V x - V T ^ = 5 Xet f(x) = 4x + Vx - VToc (x ^ 0) ^ f'(x) = 4 + 1 2V^ 3^(1-x)^ Ma f '(x) > 0, Vx > 0 va f(x) lien tuc tren [0; +00) Nen ham so f(x) dong bifin tren nua khoang [0; +00) Khi X = 1 ^ f ( l ) = 5 nen X = 1 la nghiem PT K h i x > 1 ^ f ( x ) > f ( l ) = 5:loai. Khi 0 < X < 1 ^ f(x) < f ( l ) = 5: loai. Vay nghiem la x = 1. Bai toan 4: Giai phuong trinh: 3x" -18x + 24 = —^2x-5 ^— x-1 Gidi Dieu kien x^\; (2x-5)'- ~, phuong trinh tra thanh: ' =(x-l)^' 2x-5 x-1 1 Xet f(t) = t^ - j vai t > 0. Ta c6: f (t) = 2t + ^ > 0 nen f d6ng biSn tren (0; +00) Phuong trinh: f( 12x - 5 | ) = f( | x - 1 | ) « 12x - 5 | = | x - 1 « 4x^ - 20x + 25 = x^ - 2x + 1 « 3x^ - 18x + 24 = 0. <=>x'-6x + 8 = 0 < ^ x = 2 hoac x = 4 (chon). 19 Bai toan 5: Giai bk phuong trinh: 4 | 2x - 1 | (x^ - x + 1) > - 6x^ + 15x - 14. Gidi BPT: I 2x - 1 I .[(2x - 1)^ + 3] > (x - 2)^ + 3x - 6 « I 2x - 1 P + 3 I 2x - 1 I > (x - 2)^ + 3(x - 2) Xet ham s6 f(t) = t^ + 3t, D = R. Ta CO f '(t) = 3t^ + 2 > 0 nen f ddng biSn tren R. BPT: f( I 2x - 1 I ) > f(x - 2) » I 2x - 1 I > X - 2. Xet x - 2 < 0 thi BPT nghiem dung. Xet X - 2 > 0 thi 2x -1 > 0 nen BPT « 2x -1 > X - 2 <:i> X > -1: Dung Vay tap nghiem la S = R. Bai toan 6: Giai bk phuong trinh: V x + 1 + 2V'x + 6 < 20 - 3Vx + 13 . Gidi Diku Icien: x ^ - 1 . BPT v'lk lai: V x + T + 2Vx + 6 + 3Vx + 13 < 20 Xet f(x) la ham so ve trai, x ^ - 1 . 1 1 Ta c6: f (x) = 2Vxn + •V x + 6 2Vx + 13 > 0 nen f dong bien tren [-1; +oo). Ta CO f(3) = 20 nen BPT:f(x) < f(3) « x < 3. Vay tap nghiem cua BPT la S - [-1; 3]. [ x ' ( l + y-) + y ' ( l + x ' ) = 4 V ^ Bai toan 7: Giai he phuang trinh: < x"yvl + y " - " v l + x^ = x ^ y - x Gidi Dieu kien: xy > 0 Phuang trinh thu hai cua he tuang duang vai x - V l + x^ = x^y(l - -^1 + y^ ) Ta CO X = 0 khong thoa man phuang trinh Vi X- Vl + x ' < 0 va 1 - ^|l + y^ < 0 nen suy ra y > 0. Do do x > 0. Phuang trinh tuang duang: — - —^il + - ^ = y - yJl + y^ X X V X Xet ham f(t) = t -1 V l + t^ tren (0; +oo) Xac6f'(t)= 1 - / - V l + t^ < 0 , v a i m o i t e (0; +oo) Vi + t ' Suy ra ham f nghich bien tren (0; +oo). Phuang trinh f( —) = f(y) X — = y <» xy = 1. X 20