Tài liệu 10 trọng điểm bồi dưỡng học sinh giỏi môn toán lớp 12 phần 1

"'"1 510.76 M558T Nha giao mi tu - Th.s LE H O A N H P H O lOtrpngdiem • Danh cho hoc sinh I6p 12 chtrong trinh chuan va nang cao fi On tap va nang cao kT nang lam bal ••. Bien soan theo no! d u n g va cau true de thi cua Bo GD&.DT XUAT BAN DAI HOC Q U 6 C GIA H A NOI to trong diem MON TOAN LdpU • Danh cho hoc sinh Idp 12 churdng tnnh chuan va nang cao • On tap va nang cao kT nang lam bai • Bien soan theo noi d u n g va cau true de thi cua Bo GD&.DT OEK] Ha N » i NHA XUAT BAN OAI HOC Q U 6 C GIA HA N6I C/y TNHH MTV DWH Hhang Vi$t Chuyen ad 1: TiNH D O N D I ^ U Vn CITC TBI 1. K I E N T H U C T R Q N G T A M ®Aw ^# A . Djnh li Lagrange: Cho f Id mOt hdm li^n tgc tr§n [a, b], c6 dgo ham tr§n (a,b). Luc do ton tgi c e (a,b) d l : ^ f(b)-f(a) = f '(c) hay m - /(a) = (b - a)f (c) b-a mhm muc dich gii'ip cdc ban hoc sinh them dcndng l&p 10, lap 11, lap 12 co tu lieu doc y cao trinh do, cdc ban hoc sinh gidi tu hoc bo sung them kien thuc ky M nnng, cdc ban hoc sinh chuyen Todn tu nghien cim them cdc chuyen de, nhd sdch KHANG VIET hap tdc bien soqn bo sdch BOl DlXONG DlTONG CHUYEN TRONG DIEM TOAN LOP 10 - TRONG DIEM TOAN LOP 11 - TRONG DIEM TOAN LOP 12 DIEM TOAN O LOP 12 nay co 21 chuyen devai ngi dung la tom tdt kien thuc twng tarn cua Todn pho thong vd Todn chuyen, phan cdc bdi todn chgn Igc co khodng 900 bdi voi nhieu dang hai vd muc do tic co ban den phuc tap, bdi tap tu luycn khodng 250 bdi, c6 huang dan hay ddp so. Cudi sdch CO 3 chuyen NGHIEM NGUYEN Dii dd cogdng (C)y=f(x) BOI TOAN gom 3 cudn: - Cudn TRONG HOC S/NH GlOl, vd TOAN de ndng cao: DA THU'C, PHlTONG TRINH SUYLUAN. kicm tra twng qud trinh bien tap song cung khong trdnh khoi nhieng khiem khuye't sai sot, mong don nhan cdc gop y cua quy ban dgc de'lan in sau hodn thien han. Tdc gid rMV'i'X LEHOANHPHO a c O Djnh ly Rolle: Cho fla mot ham li§n tuc tren [a,b], c6 (J?o ham tr§n (a, b) va /•(a) = f(b). Luc do t6n tai c e (a,b) d4 f'(c) = 0. Djnh ly Cauchy: Cho f va g Id hai ham lien tyc tren [a, b], c6 dgo ham tren (a,b) va g'(x) 0 tai moi x € (a,b). Lucd6t6ntaiC€(a,b)d|ii^^(^ = lM g(b)-g(a) g'(c) Tinhdcndi^u ,j Gia su' ham s6 f c6 dao ham tren khoang (a; b) khi do: i : N4U f d6ng bi§n tren (a; b) thi f '(x) > 0 voi mpi x e (a; b). N^u f nghjch bi§n tren (a; b) thi f '(x) < 0 voi mpi x e (a; b). N6U f '(x) > 0 voi mpi x € (a; b) vd f '(x) = 0 chi tgi mpt so hu-u hgn di6m cua (a; b) thi ham s6 d6ng bi4n tren khoang (a; b). N§u f '(x) < 0 voi moi x e (a; b) va f '(x) = 0 chi tai mpt so hOu hgn diem cua (a; b) thi ham s6 nghich bi§n tren khoang (a; b). N§u f d6ng bi4n tren khoang (a; b) va lien tyc tren [a,b) thi dong bi§n tren [a,b); va lien tyc tren (a,b] thi dong bi§n tren (a,b]; lien tyc tren [a,b] thi d6ng bi§n tren [a,b]. Nlu f nghich bidn tren (a; b) va lien tyc tren [a,b) thi nghich bien tren [a,b); lien tuc tren (a,b] thi nghich bi§n tren (a,b]; lien tyc tren [a,b] thi nghich bi^n tren [a,b]. N§u f '(x) = 0 voi mpi X e D thi ham so f khong d6i tren D. Cyc trj cua ham s6 gficr, : Cho ham s6 f xac dinh tren tap hgp D vd XQ e D. • Xo du-oc gpi Id mpt di4m eye dai cua f neu ton tai mpt khoang (a; b) chii-a dilm Xo sao cho (a; b) c D vd f(x) < f(Xo), V x e (a; b) \. W tTQng diSm bSl dUCing h f(Xo), V x e (a; b) \. cf^ iNi-iMmrvL/wfinnang ( - - B6 d § Fermat: Gia si> ham s6 c6 dao ham tren (a;b). N§u f dat ci/c trj tgi d i l m Xo € (a;b) thi f '(XQ) = 0. I Cho y = f(x) lien tuc tren khoang (a;b) chii-a Xo, c6 dao ham tren c6c khoang (a;;'o) va (xo;b): N I U f '(x) doi diiu tu" am sang du-ong thi f dat eye ti§u tai XQ N I U f (x) d6i d^u ti> d u K a n g sang am thi f dat CLfC dai tai Xo . Cho y = f(x) CO dao ham eSp hai tren khoang (a;b) chua XQ; Neu f '(xo) = 0 va f "(Xo) > 0 thi f dat cue t i l u tai XQ N§u f '(xo) = 0 va f "(xo) < 0 thi f dat eye dai tai XQ IJng dung vac phiKcng trinh - - - N I U ham s6 f don dieu tren K thi phuang trinh f(x) = 0 eo t6i da 1 nghiem. N I U f(a) = 0, a thuoc K thi x = a la nghiem duy nh^t cua phuang trinh f(x)=0. N§u f CO dao ham c^p 2 khong d6i dSu tren K thi f ' la ham dan di$u nen phuang trinh f(x) =0 c6 t6i da 2 nghiem tren K. N§u f(a) = 0 va f(b) =0 vai a b thi phuang trinh f(x)=0 chi c6 2 nghiem la x = a va x = b . N l u f la mot ham lien tuc tren [a, b], c6 dao ham tren (a,b) thi phuang trinh f ' ( x ) M j l [ i ? l c 6 it n h i t mot nghiem c e (a,b). b a Oac biet, neu /(a) = f(b) = 0 thi phuang trinh f'(x) = 0 eo it nhlit mot nghiem c e (a, b) hay gi&a hai nghiem cua f thi c6 it nhSt mpt nghiem cua dao ham f'. Chu y: 1) Tung dp eye trj y = f(x) tai x = X o : Ham da thue: y = q(x). y' + r(x) => yo = r(Xo) L j - u - »• s u(x) u ( X n ) u'(x.) Ham huu ti: y = f(x) = = > Vo = — ^ = ° v(x) v(x,) v'(Xo) Dae bi$t; Vai ham y = f(x) bae 3 c6 CO, CT va neu y = q(x). y' + r(x) thi phuang trinh duang thing qua CD, CT la y = r(x). 2) S 6 nghiem cua phuang trinh bae 3: ax^ + bx^ + ex + d= 0, a 0. N§u f '(X) > 0, Vx hay f '(x) < 0, Vx thi f(x) = 0 chi eo 1 nghiem. N§u f "(x) = 0 c6 2 nghiem phan biet va: . , ^ '^m^- 2 . CAC BAI TOAW o vi^c Vai yco ycr > 0 : phuang trinh f(x) = 0 chf c6 1 nghiem Vai yco ycr = 0 : phuang trinh f(x) = 0 c6 2 nghiem (1 dan, 1 k6p) Vai yco VCT < 0 : phuang trinh f(x) = 0 c6 3 nghiem phan biet Bai toan 1 . 1 : Chung minh eSc h^m s6 sau Id hdm kh6ng d6i a) f(x) = cos^x + cos^(x + - ) - cosxcos(x + - ) 3 3 b) f(x) = 2 - sin^x - sin^(a + x) - 2cosa.cosx.cos(a + x). <^'"' > Hu'd'ng din giai a) f '(x) = -2cosxsinx - 2cos(x + - )sin(x + - ) 3 3 + sinxcos(x + - ) + cosx.sin(x + - ) 3 3 = -sin2x - sin(2x + — ) + sin(2x + - ) 3 3 = -sin2x - 2cos(2x + - ).sin - . 2 6 = -sin2x - cos(2x + ^ ) = 0, vai mpi x. Do do f h i n g tren R nen f(x) = f(0) = "I ^ " ;^ = | - b) Dao ham theo biSn x (a la hiing so). i f '(x) = -2sinxeosx - 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 - sin^a - 2cos^a = sin^a. Bai toan 1 . 2: Cho 2 da thue P(x) va Q(x) thoa man: P'(x) = Q'(x) vdi mpi x vd P(0) = Q(0). Chung minh: P(x) = Q(x). Hu'O'ng din giai f n.i Xet ham s6 f(x) = P(x) - Q(x), D = R. Ta C P f "(x) = P'(x) - Q'(x) = 0 theo gia thilt, do do f(x) la hdm h l n g nen f(x) = f(0) = P(0) - Q(0) = 0 vai mpi x. ^ f(x) ^ 0 :^ P(x) = Q(x). Bai toan 1 . 3: Chung minh ring: a) arcsinx + areeosx = - , I x | < 1 2 2x b) 2 aretan x + arcsin = -n, x < - 1 . l + x^ , W tr Hu^ng din gidi a) Ham s6 y = f(x) = 2x^ + x - 4 lien tyc tren [-1,2] v^ c6 dgo ham f '(x) = 4x +1, theo dinh ly Lagrange thi t6n tai s6 c e [-1;2] sao cho: Himng din giai a) N 4 U X = 1, X = -1 thi dung. N^u - 1 < X < 1 thi x6t h^m so f(x) = arcsinx + arccosx ^ f'(x) = - ^ +- ^ =0 ^ b) Vai X < - 1, x6t f(x) = 2 arctan x + arcsin 2 • 2x2 Tac6f(x) = ^ 2x 1+ +- | l i ^ = ^ 1 + x2 1-x%2 l ( 2 M t l ) . f . ( c ) « ^ = 4c.1c>4c = 2 « c = I. 2-(-i) ' ' 3 2 f(x) = C = f ( l ) = ^ . b) Ham s6 y = f(x) = arcsinx lien tyc tren [ 0;1] va c6 dgo ham f'(x) = N/I^' X theo dinh ly Lagrange thi ton tgi s6 c e [0;1] sao cho: 1+x 1-0 +^ V S + x^^ Suyraf(x) = C = f ( - 1 ) = - J + ^ 2 4 1 ?- =0 (vix<-1). 4_ < » 1 - c 2 = 4 r « c 2 = 1 - ^.2, C h p n c = j l - — . =-f 4 Bai toan 1. 4: Tinh gpn arctanx + a r c t a n - vai x 1 Bai toan 1. 6: Xet chilu bi§n thien cua ham s6: o. a) y = x^ - 2x' - 5 X 1 b)y = (x-4) Hifang din giai HiPO'ng din giai a) D = R. Ta c6 y' = 4x^ - 4x = 4x(x^ - 1) Xet f(x) = arctanX + a r c t a n - . D = ( - « ; 0 ) u ( 0 ; + * ) X Max X G (0;+ x ) thi f lien tyc va c6 dgo ham ,, -1 f'(x) = — ^ + — ^ =—^ l + x^ l + x^ 1 + x^ Cho y' = 0 » 4x(x2 - 1) = 0 o X = 0 hoSc x = ± 1 . BBT X -X -1 0 1 +x ^ - O n e n f h a n g t r e n ( 0 ; + * ). Ux^ x^ Do<36f(x) = f(1)= ^ + ^ = f Vai X G ( - X ;0 ) thi f lidn tyc va c6 dgo ham f'(x) = 0 nen f hing tren (-oc Dod6f(x) = f(-1) = - ^ V$y arctanX + arctan—= - - khi x < 0 2 y' - 0 + y ^ ^ - 0 ^ + ^ Vay ham s6 nghjch bi§n tren moi khoang (-oo; - 1 ) va (0; 1), d6ng bi4n tr§n moi khoang ( - 1 ; 0) va (1; + x ) . b) D = R \. Ta CO y' = -2 • •..'•I (x-4)^ 4o'. / < 0 tren khoang (4; + x ) nen y nghich bien tren khoang (4; +x). y' > 0 tren khoang (-x; 4) nen y dong bien tren khoang (-x; 4). Bai toan 1. 7: Tim khoang dan dieu cua ham s6 X -khi x>0 12 Bai toan 1. 5: Tim s6 c trong dinh ly Lagrange: a) y = f(x) = 2x^ + X - 4 tren [ - 1 , 2 ] b) y - f ( x ) = arcsinx tren [ 0;1] . 0 b)y = x + 1 a)y = .1 5?. ' N/X2-6 Hu-^ng din giai aj Tap xac djnh D = ( - x ; - N/G ) u ( N/G ; + x ) . V l ^ ' •••• tx) i -K, : W trqng diSm bSi dUOng Tac6:y^ BBT: y ^ Q « x = ±3. 2x^(x^-9) (x^ - 6 ) V x ' - 6 x Cty TNHHMTVDWH h0, V x < 1 . b ) D = (-oo; 1). T a c 6 y ' = 2V(1-x)3 V$y ham s6 d6ng b i l n tren khoang ( - « : 1). Bai toan 1. 8: Xet si/ bi§n thien cua ham s6: a) y = x + cos^x b) y = x - sinx tren [0; 2n\. Hirang din giai a) D = R. Ta CO y' = 1 - 2cosxsinx = 1 - sin2x y' = 0 o sin2x = 1 <=> x = - + k7t, k e Z. 4 _ sin(x + b)cos(x + a) - sin(x + a)cos(x + b) ^ sin(b-a) sin''(x + b) sin^(x + b) Vi y' lien tuc tai moi d i l m x ?t - b + kn, a - b ^ k;t nen y' giu- nguyen mpt d i u trong moi khoang xac dinh => dpcm. Bai toan 1.10: Tim cac gia trj cua tham s6 d l ham so: a) y = (m - 3)x - (2m + 1)cosx nghjch bi§n tren R. b) y = x^ + 3x^ + mx + m chi nghjch biln tren mpt dogn c6 dp dai bing 3. Hifo-ng din giai a) y' = m - 3 + (2m - 1)sinx Ham s6 y khong la ham h^ng nen y nghich bi§n tren R: y'< 0 , Vx « m - 3 + (2m - 1)sinx < 0, Vx Dat t = sinx, - 1 < t < 1 thi m - 3 + (2m - 1)sinx = m - 3 + (2m - 1)t = f(t) Di4u kien tu-cng du'ang: f(t) < 0, Vt e f(-1)<0 [- + kTi; - -m - 4 < 0 <=> sf(1) < 0 3m - 2 < 0 o , 1] 2 - 4 < m < —. 3 b) D = R, y' = 3x^ + 6x + m, A' = 9 - 3m Xet A' < 0 thi y' > 0, Vx : Ham luon d6ng bi§n (logi) Xet A' > 0 « m < 0 thi y' = 0 c6 2 nghiem x,, X2 nen x, + X2 = - 2 , XiX2 = m BBT: X H^m so lien tyc tren moi doan [ - + kn; - + (k + 1)7i] 4 4 y' > 0 tren moi khoang ( - + kn; - + (k + 1)7t) n6n d6ng bi^n tren moi dogn 4 4 X2 Xi y' 0 + - 0 +00 + ^ y ^ Theo d l bai: X2 - x, = 3 » (X2 - x,)^ = 9 <=> x^ + x^ - 2x^X3 = 9 + (k + 1)7r], k e Z. 4 4 Vay ham so dong bi4n tren R. b) y' = 1 - cosx. Ta c6 Vx [0; 27i] => y' > 0 va y' = 0 o x = 0 hoac x = 2n. Vi ham s6 lien tyc tren doan [0; 2n] nen ham s6 d6ng bi§n tren doan [0; 27:]. Bai toan 1. 9: Chieng minh cac ham s6 a) y = cos2x - 2x + 5 nghich bi4n tren R. b) y = ^ ' " ^ ^ ^ ^ ^ (a * b + kTt; k e Z) dan di?u tren moi khoang x^c dinh. sin(x + b) Hirang din giai a) Vxi, X2 e R, Xi < X2 LSy hai s6 a, b sao cho a < Xi < X2 < b. Ta c6: f '(x) = -2(sin2x + 1) < 0 vai mQi x e (a; b). Vi f '(x) = 0 chi tai mOt s6 hO-u han d i l m cua khoang (a; b) nen hSm so f nghich b i § n tren khoang (a; b) => dpcm. b) Dieu kien x ^ - b + kn (k € Z). Hhang Vi$, ci.(x2 + Xi)^ -4x1X2 = 9 4 - - m = 9c:> m = - —(thoa) 3 4 Bai toan 1.11: Tim ci/c trj cua cac ham so sau: a)y = (x + 2 ) ^ ( x - 3 ) ^ b) y = | x | ( x + 2). Hu'O'ng din giai a) y' = 2(x + 2)(x - 3)^ + 3(x + 2f [x-Zf = 5x(x + 2)(x - 3)^ Ta CO y' = 0 o X = - 2 hogc x = 0 ho0c x = 3 X -X BBT -2 0 3 y' + 0 - 0 + 0 + y Vay d i l m eye dai (-2; 0) va eye tieu (0; -108). b) Ham s6 y = f(x) lien tyc tren R.Ta c6: W trgng diS'm bSi duOng h s_ 2) 2) Vai X Vai X BBT < > 0, f 0, f '(X) = '(X) = X ic Hodnh TNHHMTVDWHHhong Vi0 |-x(x + |x(x + r -2x + 2; f ' ( x ) 2x + 2 > 0. = 0« -1 -X + y' im; -1. x = ,„f a) D = R, y' = 1 - 2cos2x + Ta CO y " ( - - ^ + k?:) = 4 s i n ( - - ) = -2 N/S < 0 nen ham s6 dgt ci^c dgi t^i didm 6 3 1 ^ 0/ x = - - 6 X + 1 ux +k7t+— 2 +2. t = - + kn, k 6 Z; y c T = - + k7t - — + 2. 6 6 2 b) y' = 2sinx + 2sin2x = 2sinx(1 + 2cosx): X sinx = 0 , _ x^ + 8 - 2x(x + 1) _ -x^ - 2x + 8 a) D = R. Ta c6 y' = (x^ +8)^ (x2+8)2 y' = 0 <=> X = -4 hoac x = 2 X BBT -X -4 0 y y 2 + 1/4 0 ^ Ta c6y"(± — + 2k7r) = 2 cos — + 4 c o s — - 6 c o s — = - 3 < 0 nen ham s6 3 3 3 3 0 Ham s6 dat CD tai x = 2; yco = ^ , dgt CTtgi x = -4; y c T = b) Tap x^c dinh D = ( — — N/6 ) u ( • ; +oc) 2x^(x^-9) x^-6 X = 0 hoac x = ±3. y' y -V6 -3 -X + 0 -9V3 3 V6 - - • +x dat eye dai tai diem: X = ± — + 2k7i, k G Z, 3 yco = - • 2 Bai toan 1. 14: Chu-ng minh ham s6 3x^7x^-6-^ 7x^-6 _3x^(x^-6)-x'' X 2 1 o X = k7i hoacx = ± — + 2kTt, k G Z. '•'^^ ' cosx = -— 3 2 y" = 2cosx + 4cos2x Ta CO y"(kn) = 2cosk7i + 4cos2k7t = 2cosk7i + 4 > 0, vai mpi k e Z, nen ham s6 da cho dat eye ti§u tai cac dilm x = kn, ycT = 2 - 2cosk7t bSng 0 khi k chin va bing 4 khi k le. - 0 ' ^'^'^ y' = 0 ct> +x -1/8 " y' = 0 BBT 6 Ta CO y " ( - + k7r) = 4 s i n - = 2 v's > 0 nen ham so dat cue tilu tai cac diem: 6 3 Hipang din giai y' = +k7r,keZ,ycD = - - S Vaydi^m CD(-1; 1), CT(0; 0). Bai toan 1.12: Tim cu'c trj cua ham s6 a)y = . y' = 0 o cos2x = - « X = ± - + kn, k e Z; y " = 4sin2x. 0 0 y : j, 0 +x + +« Ham s6 dgt C D tgi x = -3; yco = -9 Vs . dgt C T tgi x = 3;ycT = 9 N/S 2x khi X < 0 a) ^(x) = 1 X khong c6 dao ham tai x = 0 nhung dgt cue tri tai s i n - khix>0 . 2 dilm do. ~ ' b) y = f(x) =(x - a)(x - b)(x - c), a ;i e luon eo eu-c dai va eye tilu, ^ ' ' Hipang din giai , a) Ham s6 f xac djnh va lien tyc tren R. Ta CO - ;i -2x khi x < 0 f '(X) = 1 X nen lim f'(x) = -2;^ lim f'(x) = - , do do f —C O S khix>0 x-,o~ 2 [2 2 khong c6 dao ham tgi x = 0 va BBT tren khoang ( - K ; TI). 10 trgng diSm bSi dudng h 0 vai a : i , « sinx = a CO 3 nghiem thuoc khoang (0; ^ ) W ^ i ^ }<=> 0 < a < ^ . Bai t o a n 1. 16: Tim m d4 ham s6: e. a) y = b) y = dat eye dgi tgi dilm (-2; -2). x+1 971 a sin X - cos X - 1 dgt eye tri tgi 3 diem thupe (0; — ) . b) f(x) = aeosx 4 Hipang din giai -i-Vq y' + 0 -1 - Ta CO xi + X2 = 2, X1X2 = ^2 f(-2) = -2 [p-1 3 x nen yco- yci < 0. c> 4m^XiX2 + 2m(2 - 4m)(xi + X2) + (2 - 4m)^ < 0. c:. -12m + 2m(2 - 4m) + (2 - 4m)^ < 0 c=> 4 - 20m < 0 c=> m - i + Vq 0 + Ham s6 dat eye dgi tgi di4m (2; -2) khi va ehi khi J7q = i ^ ChLcng minh du-ong thSng m « (2mxi + 2 - 4m)(2mx2 + 2 - 4m) < 0 -Q c6 hai nghiem phSn y -i-Vq = -2 ^ 06 thi CO 2 eye tri ct- m 0,. V > 0, g(1) 0 0 m < -3 hoac m > 0. va X2 = - 1 + x/q . -X "^'^"^ ^'^'^ . , mx 2mx - 3 , 2 Ta CO y = — , dat g(x) = mx - 2mx - 3. (x-1) (x +1)2 X ^'^ Hu'O'ng d i n giai - 1 : loai. NIU q > 0 thi phyang trinh: f '(x) = T"^ ^ . . ^ . .. ^ . . •, .• 9'^ '^'''''^ ^^i^- a) Oi4u kien: x / I . , vai mpi x ? t - 1 . x+1 N§u q < 0 thi f '(x) > 0 vai mpi x mx^ t (2 - 4m)x ^4m + 1 ^ — ^ ' ^2 AB song song vai duang thing 2x - y - 10 = 0. a) f(x) = X + p + a) Ta c6 f '(x) = 1 - ^ * 0, do do ham s6 dgt eye trj t^i 3 dilm sinx cos X thuoc khoang (0; — ) 4 ^ Do do y' = 0 CO 2 nghiem phan biet va doi d^u 2 Ian khi qua 2 nghiem nen luon luon c6 mpt ci/c dai va mpt eye tilu. Bai toan 1.15: Tim tham s6 thye sao cho ham s6 bi^t xi = -1 - ^ BBT: Hhang Vi$t lq = i IP = 1 b) Di§u ki^n x ;t - + kn. Ta c6 y' = - — , y ' = 0 o sinx = a. 2 a cos X , , x^ - 2x + 2m - 2 b) OK: X / 1. Ta c6 y = k \ > -. < T- (X - \ •)1 iV i • • '' Di^u kien c6 2 eye tri A, B la A' > 0 va g(1) 0. 3 0 3 - 2m > 0 va 3 - 2m ^ 0 o m < - . Ta c6 2 -h P A(1- 73 " 2 m ; 2-2m-2 73 - 2m )va B(1 + 73^ 2m ;2 -2m + 2 7 3 - 2 m ). I He s6 goc cua du'ang thing AB la: k = ^^^2)-yi^^) X2 - X, ^ ; 4^|3-2m ^ g. 273 - 2m Va 2x - y - 10 = 0 <=> y = 2x - 10 nen h^ so goc bing nhau => dpcm. Cf^y INMH W trgng diSm b6i du&ng h 0, Vx nen d6 thj luon luon c6 CD va CT v a i hoanh dp Xi, X2. Ta c6: y(x) = 1 m - X + — y'(x) - 2(x + m). 3 3 D o d 6 ; y i = y(xi) = [ ^ ^ i + y y ' ( x i ) - 2 ( x i + m) = -2(Xi + m) y2 = y(x2) = 1 m - X , + — y'(x2) - 2(X2 + m) = - 2 ( X 2 + m) 3 ^ 3 j nen d u a n g t h i n g qua CD, CT la y = - 2 ( x + m). m-m^ A = 2 12A + 4B + C = 0 8 =-9 y(1) = - 7 A + B + C + D = -7 C = 12 y(2)= -8 8A + 4B + 2C D--12 ^ . , 2x3-3x2-a b) Ta CO y ' = ; ,x y = 3x2-6x + 3c6djnh. Bai toan 1.19: Giai cac p h u a n g trinh: 4 4 ^ v x ^ - 2x + 4 = 2(^/3 - 1) b) 2x^ - X' + ^J2x^ - 3x + 1 = 3x + 1 + ^ x ^ + 2 . Hu'O'ng d i n giai {X-2Y a) Xet ham s6 f(x) = v x ^ + 2x + 4 - V r Di'eu kien c6 CO va CT la m - m^ > 0 o 0 < m < 1. X+1 f'(X) = - ; v'x^ ^ 2x + 4 Goi xi, X2 la hoanh do CD, CT thi x, < 2 < X2. Ta eo y(xi) = x , - 2 ( m - 1 ) + (X, - 2) = Xi - 2(m - 1) + (xi - 2) = 2xi - 2m. (X2-2) = X 2 - 2 ( m - 1) + ( x 2 - 2 ) = 2x2-2m. Vay p h u o n g trinh d u o n g t h i n g qua CD va CT la y = 2x - 2m t Xet ham s6 g(t) = 1) X + 1 CO 3 cue trj va e h u n g minh 3 cue tri nay thupe mpt parabol c6 djnh. Hipo-ng d i n giai - -3^2 a) Dat A = 3a^ - 1, B = - ( b ' + 1), C = Zc\ = 4d, thi ham s6 da cho la: X +^ " x f 1 >x-^ + " ' 3 + 3 >0 (t^ + 3 ) V t 2 + fj." 3 X -1 1)2 + 3 i(x -1)2 f'(x) >0 + 3 nen ham s6 f(x) dong b i l n tren R, do do: vx2 f 2x + 4 - N/X2 - 2x + 4 - 2(^3 - 1) o f(x) = f(2) x = 2. *, Vay nghi^m duy n h i t x = 2. b) PT o 2x^ - 3x + ^J2x^ - 3x + 1 = x2 +1 + ^x2 + 2 Xet ham s6: f(t) = t + tren R, > 0 nen ham so f(t) f'(t) = 1 + 3^(t +1) y = Ax~* + Bx^ + Cx + D. Ta c6: y' = 3Ax^ + 2Bx + C ' ^[{x^f 1)2 + 3 . . _ tren R, g'(t) = +3 x-1 nen ham s6 g(t) d6ng b i l n tren R, do do: Bai toan 1. 18: b) Tim a d l d6 thi ham s6 y = -^^ X f1 v/x^ - 2x + 4 N/(X a) Cho d6 thi cua ham s6: y = (3a^ - 1)x^ - (b^ + 1)x^ + 3c^x + 4d c6 hai didm c u e tri la (1; - 7 ) , (2; - 8 ) . Hay tinh t6ng M = a^ + b^ + c^ + d^. - 2x + 4 tren R. X-1 V't^ y(X2) = X 2 - 2 ( m - 1 ) + 0. T u tpa dp cac d i l m c u e trj suy ra cac d i l m c u e trj n^y n i m tren (P): (x-2)'^-~(m-m^) m - m • ' y ' = 0 o 2x^ - 3x^ - a = 0 o a = 2x^ - 3x^, x * 0. B i n g each xet ham s6 g(x) = 2x^ - 3x^ , x 0 va lap bang bi^n thien thi di§u kien ham s6 cho eo 3 cue trj khi g(x) = 0 c6 3 nghi^m phan bi^t khac 0 la-1 < a < 0 . 2 (X - 2Y = -8 Vl^t yay M = a ' + b^ + e^ + d^ = 1^ + 2^ + 2^ + 3^ = 18. a) v'x^ + 2x m - m b) DK: X ^ 2. Ta CO y = X - 2(m - 1) + X -2 . , , nen y = 1 3A + 2B + C = 0 y'(2) = 0 N e n d u p ' e a = ± 1 , b = 2, e = ±2, d = - 3 . Hu'O'ng d i n giai va T a c6: y'(1) = 0 mang MTVUVVM d6ng b i l n tren R, do do: lO trpng diSm b6i duOng hQC sinh gidi mon lodn If PT: f(2x^ - 3x) = f(x^ + 1) » ' = ' o •/? FHO Ctj/ TNHHMTVDWH Hhong Vi$t 2x^ - 3x = x^ + 1 Do d6 5x^+ 7x* = 5x^+ 7x* o f(x) = f(y) » x = y N6n (8x' + 1)=* + 27 = 162y <=> (8x^ + 1)^ = 162x - 27 <::> 2x^ - x^ - 3x - 1 = 0 c:> (X + ^ )(2x^ - 2x - 2) = 0 1 , 0$t u = 2x, phu-ang trinh: (u^ + l f = 27(3u - 1) <=> Lgi dat V = \/3u-1 o 1±V5 Ta c6 h$: Bai toan 1. 20: Giai cac phuang trinh : a) 9x' - 54x + 72 = 1 1 2x - 5 X - 2 2x J vai t > 0. Ta +1 = 3v o - = 3(v - u) 1 5 - 3(x - 1)2 +1 = 3v (u - v)(u^ + vu + I a) DK :x / 1, - , PT 3(2x - 5) +1 = 3u + 1 = 3u +1 = 3v 1 b) 4 2x - 1 ! (x^ - X + 1) = x^ - 6x^ + 15x - 14. Hu'O'ng dan giai Xet f(t) = 3 t^ - , + 3) = 0 u-v = 0 Do d6 + 1 = 3u hay 8x^ - 6x + 1 = 0 X6tx e [ - 1 ; 1 ] n § n d $ t x = cost x-1 ;0r • . PT: 2 ( 4 c o s ^ t - 3 c o s t ) = -1c:>cost = - - < = > t = ± — + — , (!<€ Z) Ti> d6 c6 3 gid trj cua x 27t 8TI x = cosx = cos9 9 c6: f (t) = 6t + — > 0 nen f dong bien tren (0; +-r) 2 9 3 cOng chinh \k 3 nghi0m cua phu-ong trinh b$c 3: 1471 x = cos- V|y nghi$m h? x = y = c o s — ; c o s — ,cos — 9 9 9 b) ( 2 ) o ( y - 1 ) [ ( x + y ) 2 - 1 ] = |x + y| [ ( y - l ) 2 - i ] Phuang trinh: f ( i 2 x - 5 l ) = f ( ! x - l ! ) < » i 2 x - 5 l = ! x - 1 o 4x" - 20x + 25 = x^ - 2x + 1 o 3x^ - 18x + 24 = 0. c:> x' - 6x + 8 = 0 cr.-- x = 2 hoac x = 4 (chon) V6i y = 1: (3) Vay nghiem x = 2 hoac x = 4 V6i x + y = 0 (3) <=> y = 1 => x = - 1 ; I I 2x - = (X + 1 = 3^3u - 1 x = - 1 : I1:(3)c>(^"y)'-^Jy-^)'-^ y-1 x+y - 2)^ + 3(x - 2) X6t ham s6 f(t) = t' + 3t, D = R. o Ta CO f '(t) = 3t^ + 2 > 0 nen f d6ng bi§n tren R. I PT: f( 2x - 1 X I) = f(x - 2) « ! 2x - 1 x>2 -2> 0 3x2 ^ 3 (2x --1)2 . (X - 2) I =x- 2 (VN). Vgy S = 0. - X Bai toan 1. 2 1 : Giai cac he phuang trinh: a) = y-1- y-1 X6f hdm s6 f(t) = t - ;J. D = (0 ; + 00) P T o f ( | x + y|) = f ( y - 1 ) c > |x + y| = y - 1 (8x^ +1)^ t 27= 162y (y-1)r(x + y f - 1 x+y f'(t)= 1 + ^ > 0 , V t € D ^ h d m s 6 d 6 n g bi4ntr§nD 5x' + yx'^ - 5y^ f 7y^ [y>1 x2 +y2 ^ 5 ; y > 1 b) ix + y (1) X f y (y2 - 2y) (2) Hu'O'ng d i n giai a) Xet f(t) = 5t'+ 7t^ t G R thi f'(t) = 35t^ + 35t' " 0, V t nen f d6ng bi§n tren R, [x = - 1 hay ' 'ill •'. • x=:1-2y 1-2N/24 Ctj/ TNHHMTVDWH Hhong Vi$t 10 trQng diSm bSi dUOng h f(x) = g(y) tren (0; +«). Ta c6 h$ f(y) = g(z) f(z) = g(x) 0 < f(x) < 1. a)DK 0 < f(y) < 1 => 0 < g(z) < 1 => 0 < z < 1. g(y) > g(z) > g(x) - 2 x + 3 - Vx^ - 6 X + 11 7 3 ^ - >/x^. HifO'ng d i n giai -2 f(x) = , Vgy h$ CO 2 nghiem x = y = z = 2 + yfs , x = y = z = 2- \l3 . 36x^+25 4 Do do BPT: f(x) > f(1) « eoz^ 361^ + 25 x > 1. Vay: S = (1; 4) ^~^<;=>10 .t>0. Jt::" 7(x - 1)2 + 2 + ^Jx^^ > V(x - 3)2 + 2 + 7 3 ^ Xet ham s6 y = f(t) = Tu" h0 suy ra x, y, z khong am. Neu x = O t h i y = z = 0 suy ra (0; 0; 0) Id nghiem cua he phu'ong trinh. f '(t) > 0, Vt > 0 nen f d6ng bi^n tren (0; +oo). > 0 274-X •If 36f + 25 60t' 272x3 +3x2 +6X + 1 BPT: 7x2 _ 2x + 3 + 7 x ^ >7x2 - 6x + 11 + 7 3 ^ 60y2 0 thi y > 0, z > 0. Xet ham s6 f(t) = : If, Suy ra f(x) la hdm s6 d6ng bi§n Di§u ki$n: 60x2 36z^ + 25 X > < X< X + 1)+ 16 _ Xet: f(x) = 72x3 6(x2 + 3x2+ +6X Xet X < z < y thi cung nhan du-gc ^kx qua tren. z=• > 2x^+3x2+6x + 1 6 > 0 Ta CO PT t^ - 4t + 1 = 0 chpn nghiem: x = y = z = 2 - N/2. N§u 5 5 5^ 6'6'6 4-X>0 ^ .^, => y > z > X nen x = y = z. X = ,x du-gc x(36x2 _ gQj^ ^ 25) = 0. Chpn x = 0; - . 6 b) nen 0 < g(y) < 1 => 0 < y < 1 =^ f(0) > f(y) > f(1) b) H0 phu-ang trinh tu-ong du-ong -rm 60z2 36z2 + 25 Ti> tinh d6ng bien cua f(x) suy ra x = y = z. Thay vdo he phu'ong trinh ta Ta CO PT: t^ - 4t + 1 = 0 chpn nghiem: X = y = z = 2 + N/3 . y= 36y2 + 25 a) V2x^ +3x2 +6X + 16 >2V3 + N / 4 - X < y < z => f(x) < f(y) < f(z) f(x) > f(y) > f(z) 60y2 Bai toan 1. 23: Giai cac b i t phu'ong trinh => g(y) < g(z) < g(x) = > y < z < x n e n x = y = z. Do do X < y < z 36x2 ^ 25 Vay tap nghiem cua he phuong trinh la \; Gia su' X = min{x; y; z}. Xet x < y < z. N§u 0 < x < 1 thi f(0) > f(x) > f(1) z= 60x^ X = Oat f(t) = t^ - 2t + 1, t > 0 thi f '(t) = 2(t - 1) nen f dong bien tr§n (1; +oo) nghich biln tren (0; 1). Dgt g(t) = 2t, t > 0 thi g'(t) = 2 > 0 nen g d6ng bien X He phuong trinh du-gc v i l t l^i ' a)Tac62y = x ^ - 2 x + 1 = ( x - 1 ) ^ > 0 = i > y > 0 . T i i ' a n g t y z , x > 0 . N§u x > 1 thi 1 < y= £>9o ham: f'(x) = J^Ti + 7t, 7 t ^ 27t D = (0, +«) > 0 nen f d6ng bi§n tren (1, 3]. Do do BPT f ( x - 1) > f(3 - X) <=> X - 1 > 3 V^y nghiem cua b i t phu'ong trinh S = (2, 3]. Bai toan 1. 24: Giai cac b i t phuong trinh a) 7 3 - X + x2 - 7 2 + x - x 2 < 1 X <=> X > 2. i• 10 trpng diS'm bSi duang HQC sinh gioi m6n Toon 12 - LS Hodnh Phd + 2V3 - 4x < 7 . b) 4(7x)'' ' X6t h^m s6 f(t) = x2 = y 3 +y2 + y + a Hu'6'ng din giai a) D|t t = x^ - X, BPT: N/sTt - t, Nen f(x) = 0 c6 nghiem duy nhit x < 0. V|y phuang trinh cho c6 nghi^m duy nhlt. Bai toan 1. 26: Chung minh he phuang trinh c6 nghi^m duy nhit: Ta + t - z2 = x^ + x2 + X + a 42^\ -3 < t < 2. Hipang din giai X6t h^m f(t) = t^ + t2 + t + a CO f '(t) = 3t2 + 2t + 1 > 0 do d6 f(t) 1^ h^m d6ng V 6 i - 3 < t < 2 thi f (t) = / + / >0 2V3 + t 2V2-t ngn f d6ng bi§n tren (-3; 2). Ta c6 f(1) = 2 - 1 = 1 n§n b^t phu-ang trinh: f(t) < f(1) t < 1 » x^ - X - 1 < 0 b) DK: 0 < X < - . PT o 4 x2 = f(y) biln. He PT: Khong giam tong qu^t gia su x Ib-n nhit trong 3 s6. => z2 > x2 > y2. N^u x2 > y2 > z2 V6i X =0 thi BPT I + 273 - 4x = 4x(4x2 _ 3) - 4 V3-4X V3-4X 2 2 Bai toan 1. 25: Chufng minh phu-ang trinh: x^^ - x^ + 3x'' - 3x2 + 1 = 0 nghj^m duy nhlt. HiTO'ng din giai O0t f(x) = x^^ - x^ + Sx" - 3x2 + 1, D = R X6t X > 1 thi f(x) = x^(x^ - 1) + 3x2(x2 - 1) + 1 > 0: v6 nghi^m X6t 0 < X < 1 thi f(x) = x^^* + (1 - x2f > 0 : v6 nghi^m X6t X < 0 thi: f (X) = 13x^2 _ g^s + -,2x3 _ 6x = 13x^2 _ 6x(x - 1 )2 > 0 n6n f d6ng bi^n Bang bi6n thidn: -00 z>Othix>y>z>0 x2 = y2 = z2 ^ f(x) = f(y) = f(z) ^ x = y = z N§u X < 0 = > 0 > x > y > z = i > x 2 = y2 = z2=>x = y = z Neu X > 0 > z. Khi do y2 = f(z) < f(0) = a =^ a > 0 Lgi CO z2 = f(x) > f(0) = a ^ z = - N / S =^y2 = f ( z ) < f ( - 7 a ) = - V a ( N / a - 1 ) 2 < 0 : v 6 l i . <0 1 g ( - ) = 7 n6n b i t phu'cng trinh g(x) ^ . V§yt$pnghiemS = ( I ; | ] . -00 - _ Xet X > y > z =^ f(x) > f(y) > f(z) + 2V3 - 4x < 7 V6i 0 < X < - . Xet ham s6 g(x) = 4x2 + ^ - 2 x 2 4 2 y2 = f(z) z2=f(x) < x< 4x2 + ^ - 2 x 2 2 + z2 + z + a y2 = < 1, -3 < t < 2. ~ X6t X > z > y '''' z2 > y2 > x2 {-^ \ Tuang tu-nhu tren n l u y > 0 hay X < 0 ta suy ra X = y = z N4U x>0>y=:i>x2 = f(y) < f(0) = a z2 = f(x) > f(0) = a. N§u z > v/a thi X > z > v'a => x2 > z2 => z2 = y2 = z2 => X = y = z trai v^i x > 0 > y ' '• N l u z < - Va li luan nhu tren ta dan d§n mSu thuan. V^y he CO nghiem duy nhit x = y = z = to6'd6toia nghiem duy nhit cua phuang trinh: t^ + t2 + t + a = 0. Bai toan 1. 27: Chung minh he i ~ c6 dung 3 nghiem phSn bi$t. y2 + x^ = 1 Hu'O'ng din giai Tru 2 phuang trinh v6 theo v§ va thay the ta dugc: x'(1-x)-y2(1-y) = 0 ^ ( 1 - y ' ) ( 1 - x ) - ( 1 - x ' ) ( 1 - y ) = 0 =^ (1 - x)(1 - y)[1 + y + y 2 - ( 1 + x + x2)] = 0. =^(1 - x ) ( 1 - y ) ( y - x ) ( 1 + x + y) = 0. , Xet X = 1 thi h$ c6 nghiem (1; 0).Xet y = 1 thi h$ c6 nghiem (0; 1) Xet X = y thi x^ + y 3 = 1 x^ + x^ - 1 = 0. Dat f(x) = x ' + x-^ - 1, D = R. Ta C O f(1) = 1 ^ 0. Xetfx = X -co + y' -2/3 0 0 0 X . 1 2 -{ f + 00 1 „ , 3x^+1 , x>--,x^Othif'(x)= — 2 x"^ + 00 0 0 + -co ^ Do do f(x) = 0 CO 1 nghiem duy nhit XQ > 0, XQ ^ 1 nen he c6 nghiem (X( Xet 1 + X + y = 0 y = - X - 1 nen y^ + x^ = 1 o x^ + x^ + 2x = 0 o x(x^ + X + 2) = 0 » x = 0. Do do he CO nghiem (0; 1). Vay he c6 dung 3 nghiem phan biet. Bai toan 1. 28: Tim tham s6 de phu-ang trinh 1 <=:> f(x) = m CO 2 nghiem phan biet x>--,x=^0<^ a) ( N / X + V x - 2 ) m^Jx+-. ^ 3 ^ x ( x - 2 ) = 2 CO nghiem ftfim^i Ja> b) 37tanx + 1 .(sinx + 2cosx) = m(sinx + 3cosx) c6 nghiem duy nhit thuoc khoang (0, - ) . Hyang din giai Hu-o-ng din giai a) Oi§u kien: x > 2 a) Xet f(x) = ^To< + ^/l7x , D = R 3^(1-x)2 3^(1+ x)2 9 ^ - 2 Vx-2 b) vx^ + mx + 2 = 2x +1 CO 2 nghiem phan bi^t. 1 ^-7 Bai toan 1. 29: Tim m d l phuang trinh a) ^1 + X + \ / l - x = a CO nghiem 1 + 00 + CO Oi^u kien phu-ong"trinh da cho CO 2 nghiem phan biet -co f '(x) = - + + f -23/27 y +4X-1 X X BBT: f (X) = 3x^ + 2x, f '(X) = 0 <=> X = - - ho$c x = 0. BBT 3x^ (x^±1) = i ^ J ^ , f ( x ) = o« 2 PT« 2 r x= 0 <=> V x ^ + = lim (^1 + x - ^1 - x ) = lim ^ Vx-2 =0 \ m .vx + V x - 2 N/X-2 3^(1-x)^ . ^(1 + x)2 lim f(x)= lim (^/Toc + ?/l + x ) 2 Vx-2 -3.Vx(x-2) =2(^fx-4y^ -3.t/x(x-2) = V x - 7 x ^ - 3.t/x(x-2) = (1 - m2)7x -3.fM=1-m^ D a t t = f P — ^ , 0 < t < 1.PT: — - 3 t = 1-m^ , 0 < t < 1 . V Tu-ang t u lim f(x) = O.Lap BBT thi PT c6 nghiem » 0 < a < 2. X ii • J • ' X->-oo b) PT<^ 2x + 1 > 0 <=> 3x^ + 4x - 1 = mx, X 2 x'^ +mx + 2 = (2x + 1)^ Vi X = 0 khong thoa man nen: Xetf(t) = 4 - 3 t , t € (0; 1)=> f ' ( t ) = - - | - 3 < 0 . e (0; 1). , (2 t3 T t 0 1 Bang bi§n thien f'(t) - > - - ^"^^ ^ = m, x > - | +x 4 I* 99 f(t) • V^y phyang trinh cho c6 nghi^m khi 1> - 2 <=>-N/3 < m < N/3 0 tanx > - 1 . b) Dieu ki^n : cosx b) X6t x = 0 => 1 = 0: loai. X6t x ;t 0. Chia 2 cho x^ phuang trinh: 1 3 1 x ' + 3x' + (6 - a)x + (7 - 2a) + (6 - a ) . 3 + ^ + — = 0 _+ D$t t = 7tanx + 1 > 0 , phu-o-ng trinh: sinx + 3cosx = mcosx cosx 3 7 t a n x + 1 (tanx + 2) = m(tanx + 3) Syjxanx o + 1 sinx + 2cosx <=> 3 7tanx+1 (tanx + 1 + 1) = m(tanx + 1 + 2 ) « 3t(t2 + 1) = m(t2 + 2) o X6t h^m so y = 3t- m = ^^J-^ . t^ + 2 ^ v6i t e (1 , + « ) , t^ + 2 +15t^ + 6 (t^ + 2f V$y phu-ang trinh c6 nghi^m duy nhit khi m > y(1) <=> m > 2 . Bai toan 1. 30: Tim tham s6 d l phu-ang trinh a) (4m - 3) 7x + 3 + (3m - 4) Vl-x + m - 1 = 0 c6 nghi^m b) x^ + 3x^ + (6 - a)x'' + (7 - 2a)x^ + (6 - a)x^ + 3x + 1 = 0 v6 nghi$m. Hu>6ng din gial a) Di§u ki$n: - 3 < x < 1 khi d6: 3x/x73 + 4>/irx +1 PT <=> m = — 7 = = — 4Vx + 3 + 3V1 - X +1 Ta c6: (VxTi )2 + (VT^ )^ = 2 nSn d$t: (x3+ X6tf(t)=^Zi;ii^ -5t2+16t + 7 ,, . te[0.1] -5y-8t-60 (-61'+ l e t + 7)^ X 7 2 Vay dieu ki^n phu-ang trinh c6 nghi^m la f(0) < m < f(1) <=>- < m < - . 9 7 x^ x^ X 0 $ t t = x + - , Itl > 2 ^ t ^ = x ^ + 4 - + 2 X x^ va t^ = x^ + — + 3(x + - ) n§n x^ + ^ = t^ - 3t. x^ X x^ • Do do phu-ang trinh: t^ - 3t + 3(t^ - 2) + (6 - a)t + 7 - 2a = 0 (t + 2)a = t ' + 3t^ + 3t + 1 Khi t = - 2 thi phuang trinh khong thoa. ^, .- u t3+3t2+3t + 1 (t + 1)3 Khi t ^ - 2 thi phuang trinh: a = = —. t+2 t+2 oat f(t) = lilJ)! , t < -2 hay t . 2 thi f '(t) = ^2^13^. t+2 2(t + 2f Lap BBT thi f(t) > — Vt e D n6n PT v6 nghi$m khi a < — . 4 4 Bai toan 1. 31: Tim tham so d§ bSt phuang trinh c6 nghi^m a) sin^x + cos^x > m b) cos^2x + 2 (sinx + cosx)^ - 3sin 2x + m > 0. Hyo-ng din giai a) Xet f(x) = sin^x + cos^x = (sinx + cosx) (1 - sinx.cosx) D^t t = sinx + cosx ; |t| < N/2 7x + 3 = 2sin(p = 2 2t ; V/l— - x- = 2cos(p = „2 l - t ^ 1 + t' l + t^ V a i t = taniP , 0 < cp < I ^ . 0 < t < 1 nen: m = - 7 t ' + 1 2 t + 9 2 4 -5t2+16t + 7 JL) + 3(x'+4 +_ x^ x )+ (6-a)(x+ -)+7-2a = 0 X t^ = 1 + 2sinx cosx => sinx cosx = Ta CO h(t) = t 1 . 2 - 1) = - ^ t ^ + | t v a i |t|< V2 3,2 . 3 h'(t) = - - t ^ + - = 0 « t = ± 1 ^ ' 2 2 Lap BBT thi b i t phuang trinh CO nghi^m khi m < 1. v • b) D|t t = sinx + cosx, |t| < 72 va t^ = 1 + 2 sinx cosx ^ sin2x = t^ - 1 cos^ 2x = 1 - sin'2x = - t " + 2t^ BPT: - t " + 2t^ -1' + m + 3 < 0 ; (|t| < V2) -j^c^rnwn-t /VIIV uvvH hnang Hirang din giai Xet ham so g(x)= f(x) +x - 1 , khi do thi g(x) lien tyc v^ c6 dgo ham tren [0;1]. Ta c6: g(0)= - 1 < 0 va g(1)= 1 >0 nen t6n tai so c thupc (0;1) sao cho g(c) =0. Do do f(c) + c - 1 =0 hay f(c) = 1 - c Ap dung dinh ly Lagrange cho f tren cac doan [0;c] va [c;1] thi : " Xet f(t) = -t^ + 2f' - t-^ + m + 3 f '(t) = - 2 t {2f - 3t + 1) ; f '(t) = 0 =^ t = 0 ; - ; 1 2 Lap BBT suy ra dieu kien c6 nghiem la: m + 3 > 0 <=> m > - 3 Bai toan 1. 32: Tim 6\hu l 0 (2) , . Htpcng dan giai Xet(1) : x ^ - 3 x - 4 < 0 o - 1 < x < 4 Ta tim dilu lf'(x) = 3x^ - 6x ; 0 < < 4 X 0 f'(x) = 3x(x + 2 ) < 0 f '(X) = 3x (X => f '(x) = 3x (X dao ham tren (0;1), - 2) > 0 Do do - m ^ - 1 5 m + 16<0<=>m< -16 v m > 1 Taco: g'(x) = — ^ ; h ' ( x ) = 1 + x^ Vay di4u kien c6 nghiem la - 1 6 < m < 1 ?)< (1 + x^)^ 1 j, 1 + x^ Theo dinh ly Cauchy thi ton tai c G(0;1) sao cho: Bai toan 1. 33: Cho 3 s6 a,b,c thoa man abc ?^Ova - + - + - = 0. 7 5 3 ChLKng minh phu'ong trinh : ax"* + bx^ + c = 0 c6 nghiem. Hii'O'ng din giai = g(i)-g(0) Xet hdm s6 F(x) = - x ^ + - x ^ + - x ^ , khi do F(x) lien tyc, c6 dao hSm ' ' 7 5 3 F '(x) = x^. (ax"* + bx^ + c) = x^ .f(x) nen theo dung dinh II Lagrange tren [0,1] t h i t d n t g i c e (0,1): ^ ^ ^ ^ ^ ^ = F ' ( c ) . 1-0 tren [0,1], khi do thi g(x),h(x) c6 1 +x - 2) < 0 <4 "V ChCfng minh b i t phuang trinh: f'(x) - f(x) < -(f(1) - 2f(0)) c6 nghiem. <0 Khi-12c va vi f(c) > 0 nen f ' ( c ) - - ^ f ( c ) > f ' ( c ) - f ( c ) l + c^ => dpcm. Bai toan 1. 36: Gia si> f\a mpt ham xac dinh tren [a, b], c6 dao ham d i n d p n + 1 tren (a,b) va XQ e (a,b). ChCrng minh t6n tai c n i m giOa x va XQ d l c6: (J.''.ids Ma F(0) = 0, F(1) = - + - + - = OnenF '(c) = 0 hay c^f(c) = 0. 7 5 3 Vi c G (0,1) nen 0 do do f(c) = 0 => dpcm. Bai toan 1. 34: Cho ham s6 f c6 dao ham tren tren [0;1] va thoa m§n: f(0) = 0; f(1) = 1. Chu-ng minh ton tgi 2 s6 phan bi^t a;b thupc (0; 1) sao cho f'(a).f'(b) = 1. fW-f(>^o)^(x-x,).!:|il(x-x„)^.... !!X)(x-x)".i!:::!(^(x-x)ft n! """^ (n + 1)!^ °' /c Lr\fny atom tXJr ainJtty tiv^ srrm rrnpn r c ^ n V 2 r — le grar rtOOnn mo HiTO'ng d i n giSi Ta tim mOt da thCfc Pn(x) c6 bgc Iang d i n '{.w/v^ r -2(x^ + 9 X a) K§t qua y' = — \ 0 nen hSm s6 d§ cho nghjch bien tren cSc khoang (x^ - 9)^ (-«;-3), (-3; 3), ( 3 ; + = 0 ) . ^.^^ , b) K i t qua d6ng bien tr§n (-00; 1), nghjch bi§n (1,+00). Bai t$p 1. 2: Tim m de hdm s6: a) y = -—+ ("^ + 2)x - m + 3 ^^^^ ^^^^ khoang x^c djnh x + 1 b) y = - x^ - — x^ - 2x + 9 dong bi§n tren ( 1 ; + x ) . 3 2 Hirang d i n a) Tap xac djnh D = - 1 ) u ( - 1 , +00). Tinh dao ham y ' vS l|p lu$n y ' > 0 tren D. K§t qua m > 1. b) K§t qua m < - 1 . Bai t$p 1. 3: Tim ciic trj cua hSm s6: a) y = y • • ,. b)y=^(x-5). 7x^-6 Hifang d i n a) Ham s6 le. Tinh dgo ham va l$p BBT. K§t qua CO tai x = - 3 ; yco = - 9 \/3 , CT tgi x = 3 ; y c r = 9 \ / 3 . R (x) R'"^^'(C) Sau n + 1 l^n ap dyng dinh ly Cauchy ta du-gc := — - - v^i c nSm F(x) F<"^'>(c) giOa 4n va XQ, va do do c n i m giOa x va XQ. Nhung R<"*^>{x) = / " ' ' ' ( x ) va F'"*^'(x) = (n + 1)! nen = 1^^^ . F(x) (n + 1)! Vay: f(x) = f(x„) / ^ ( x - x„) + l | | ° l ( x - x^)^ +... nl ""'^ b) K§t qua CD tgi x = 0, yco = 0 va CT tgi x = 2, y c r = - 3 \ / 4 . Bai t?p 1.4: Tim ci^c trj ham s6: a) y = x - sin2x + 2 b) y = sin2x + cos2x. Hu-angdln a) T$p xSc djnh D = R, y ' = 1 - 2cos2x, y " = 4sin2x. Dung d i u dgo hSm d p 2. K§t qua: CD tgi x = - - + kn, k e Z, y c o = - - + 6 6 (n + 1)!^ X = - + kn, k e Z; ycT = - + krt - — + 2. 6 6 2 R +— 2 * : f + 2 ; dgt CT tai W trqng diS'm bSi dUOng h / 3 x - 2 + Vx^ - s T s x + 4 = 3 . + kn . Hirang din Bai tap 1. 5: a) Tim m d§ h^m so y = 2x^ - 3 (3m + 1 )x^ + 12 (m^ + m) x + 1 c6 ci/c dai va cu-c tilu. Vilt piiu-ang trinli du-ang thing di qua CD, CT. X x^+2mx + 1-3m^ , CO hai d i l m eye th n l m v l hai b) Tim m de ham so y = — x -m phia cua true Oy. Hipang din . > ,3 b) K i t qua - 1 < m < 1. Bai tap 1. 6: Chipng minh ham s6 a) y = x^ + ax^ - (1 + b^)x + a + 4b - ab luon luon c6 eye dai va eye tieu vd-i mpi tham s6 a, b. b) y = 1 3x - - ba diem eu-e tri phan biet A, B, C . Tinh di$n tieh tam 2 giae ABC. X b) K i t qua S = Bai tap 1. 7: Giai eae phuang trinh : a) 3x^ - 1 8 x + 24 = ^ I 2x - 5 1^ - 2x-5 a) N/X2-1 = V X ^ - 2 - X • a) 7x-i-Vy (X - =i-x3 b) ir = y f (4x^ + 1)x + (y - 3)75 - 2y = 0 4x2 ^ y2 ^ 2^3 _ 4x ^ 7 Hifang din a) D i l u kien x > 1, y > 0. He phuang trinh tu'ang du-ang vai: (X - 1)2 + x^ - 8 = 0 (1) (2) • ' 1 •! - (t - 1)^ + t^ - 8, vai t > 1. -6X + 11 < N / 3 ^ - V x ^ . Hu'O'ng din a) D i l u ki0n: x > - 1 . BPT vilt l^i: ^/xTl + 27x + 6 + sVx +13 > 20 Xet f(x) la ham so v l trai, x > - 1 thi: 1 1 3 f '(x) = — = = + x-1 K i t qua X = 2 hoae x = 4. Bai tap 1. 8: Giai eae phu-ang trinh; = 0. K i t qua nghiem duy nhit x = 3. b) Ham dan di^u . K i t qua x =3. Bai tap 1. 9: Giai eae h0 phu-cng trinh : b) Vx^ - 2 x + 3 - Vx^ ^ x-1 - I x li-^ b)Kltquax= 1 ^ ^ . xVx a) Vx7^ + 27x + 6 >20-3N/X + 13 Hirang din ^ 2x-5 V Bai tap 1.10: Giai bat phyang trinh: x-1 b) V 3 - x + x2 - V2 + x-x2 =1 a) PT: (2x - 5)^ - (x - 1)^ = • .... b) K i t qua x=^;y = 2. 1 2x-5 VX Xetham s6f(t)= TTl Kit qua X =3, y =0. 27 1 X^'N/X y = (x-1)' a) y' e6 A' = a^ + 3(a + b") > 0, Va, Vb J •;f!,X f Chia 2 ve eho Vx"" thi du'p'e phu-ang trinh: >/x^ - Hu'6ng din \/2.Tac6: Vx^ - 2 = X + yjx^ - 1 > x > 1 = > x ' > 3 = > x > \x^.\[x. a) Tap xac dinh D = R. L i y y chia y'. K § t q u a m # 1 v a y = - ( m - 1 ) ^ x + 2{m^ + m)(3m + 1 ) + 1 . x^ a) D i l u kien: X > '% .ist '30':ft + . > 0 .Ket qua x > 3. 2Vx + 1 Vx + 6 2Vx + 13 b) K i t qua 1 < X < 2. Bai t^p 1.11: Chiang minh phu-ang trinh c6 nghi$m duy nhit: X -x2 + 2 x - 5 = 0 Hiring din - 5 x ' + 15x Chung minh hdm VT ding biln tren khoang (0, +«), e6n khi x < 0 thi v6 nghiem. W trgng diem Churen bSi dUOng HQC sinh aS 2: gioi mdn Toon 12 - L S Hodnh Cty Phd MTVDWH Hhang Bu'ccc 3: Ve d6 thi _ Tinh dgo ham d p hai, xet d i u d l chi ra dilm u6n cua ham da thu-c. _ Cho vai gia tri d$c biet, giao d i l m vai hai true tog dp. _ Ve dung d6 thi. KHHO SIIT VA V€ D b THf HflM SO 1. K I ^ N T H U C T R O N G TNHH TAM Vi$t r,' B i n dang dd thj ham bac 3: y = ax^ + bx^ + cx + d, a ^ 0 c6 tam d6i xij-ng la d i l m u6n. Tinh I6i lorn cua d6 thj: Ham s6 f x^c djnh tr6n K Id mOt khoang, dogn ho$c nu-a khoang. f gpi Id 15m tren K neu Va,p,a + p = 1: f(ax + py) < af(x) + pf(y), Vx,y s 0 f gpi Id I6i tr§n K n l u Va.p.a + p = 1: f{ax + py) > af{x) + pf(y),Vx,y > 0 Bon dang do thi ham trung phu-ang: y = ax + bx^ + c, a ?t 0 & i^fiiir - (C)>{ O Du'O'ng tiem c a n Cho hdm so y = f(x) li§n tyc va c6 dgo ham cap 2 tr§n K Duong thdng x = XQ dt^pc gpi la tiem can du-ng cua d l thi ham s6 y = f(x) n l u it nhdt mpt trong cac dilu kien sau dupe thoa man: - f 16m tren K <=> f ' (x) > 0, Vx e K f l6i tren K » lim f(x) = + x ; lim f(x) = + x ; lim f(x) = - r \m f(x) = - « f "(x) < 0, Vx e K. Dilm udn cua dd thj: D i l m U(xo;f(xo)) dLcp-c gpi Id diem u6n cua duang cong (C): y = f(x) n4u ton tai mpt khoang (a;b) chupa d i l m Xo sao cho mpt trong 2 khoang (a,Xo), (xo,b) thi tiSp tuy^n tgi difem U ndm phia tr6n d6 thj c6n 6- khoang kia thi ti^p tiiyln ndm phia du'b'i d6 thj. Cho hdm s6 y = f(x) c6 dgo hdm cdp 2 mOt khoang (a;b) chua d i l m XQ. N I U f "(xo) = 0 vd f " ( X ) doi ddu khi x qua diem XQ thi U(xo;f(xo)) Id d i l m u6n cua du'dyng cong (C): y = f ( x ) . - Duong y = y ^ du-p-c gpi la tiem can ngang cua do thi n l u lim f(x) = yo hode lim f(x) = yo. - Duong thing y = ax + b, a ^ 0 dup-c gpi la tipm cgn xien cua d6 thi y = f(x) n l u lim [f(x) - (ax + b)] = 0 hoac lim [f(x) - (ax+ b)] = 0. Chu y: 1) Neu chia tach du-gc y = f(x) = ax + b + r(x) va lim r(x) = 0 thi tiem can xien: y 1) Neu y =p(x).y" + r(x) thi tung dO dilm u6n t?i XQ Id yo = r(xo) 2) N l u f l6i trSn dogn [a,b] thi GTLN = max{f(a); f(b)} 3) Neu f 16m tr6n dogn [a,b] thi GTNN = min{f(a); f(b)} Khao sat va ve d6 thj ham da thirc: g6m 3 bu-d'c: Bu'O'cl: Tdp xdc dinh - Tdp xdc djnh D = R - Xet tinh c h i n , le neu CO. ' Bu-d'c 2: Si^ biln thi6n - Tinh cac giai hgn. - Tinh dgo hdm d p mOt, x6t d i u - L$p bang b i l n thien r6i chi ra khoang d6ng biln, nghjch b i l n vd eye dgi, eye tieu. thing = ax + b b Vx^TbxTc * x + — 2 Khao sat va ve d6 thj ham hOu t i : g i m 3 buac: B u a c i : Tap xac dinh ~ Tim tap xac dinh - Xet tinh c h i n , le n l u c6, Buac 2: Chilu biln thien 2) Bilu thtpc tipm cgn khi x -> ± x : Tinh cac giai han , tim cac tiem cdn Tinh dao ham cdp mpt, xet ddu Lap bang biln thien roi chi ra khoang dong biln, nghjeh b i l n va eye dgi, cyctilu • 1 0 trQng - diem b6i duang hQC sinh Hoanh Phi) IP g:oi mSn J, Bii-dc 3: Ve thj Cho vdi gid trj dSc bi^t, giao diem v6'i hai tryc tog dOVe dung d6 thi, iu-u y tam doi xii-ng la giao diem 2 tiem c#n. Hai dang d6 thj ham hu-u tf b^c 1/1: y = cx + d v6i c ^ 0, ad - be ^ 0 7 B6n dang do thj hdm hu-u tf :y = — a'x + b' (a ^ 0, a V 0) Chuy: 1) Tu- d6 thi (C): y = f(x) suy ra cac d6 thi: y = - f(x) b i n g cdch Idy d6i xung qua true hodnh. y = f(-x) bdng eaeh l l y doi xi>ng qua trye tung. y = - f(-x) b i n g eaeh l l y d6i xung qua g6c. y = I f(x)| b i n g eaeh l l y phin d6 thj a phia tr§n trge hoanh, eon phin phia du'6'i trgc hodnh thi d6i xu-ng qua trye hoanh. y = f(|x|) Id ham s6 ehdn, bIng edch l l y phIn d6 thj a phia ben phai true tung, r6i l l y d6i xu-ng phIn do qua true tung. 2) Bai todn ve bien ludn s6 nghiem phu-ang trinh dgng g{x,m) =0 Ou-a phuang trinh ve dang f(x) = h(m) trong d6 v6 trai Id hdm so dang xet, d § ve d6 thj (C): y = f(x). S6 nghiem Id so giao diem eua do thj (C) v6'i du'6'ng thing y = h(m). 3) O i l m d$e biet cua hp d6 thj: (Cm): y = f(x,m) - Diem c6 dinh eua hp Id d i § m md mpi d6 thi d^u di qua: Mo(xo, yo) e (Cm), Vm o yo = f(xo, m), Vm - Oi^m md hp khdng di qua Id d i l m md khpng c6 d6 thj ndp cua hp di qua vdi mpi tham s6: Mo(xo, yo) « (Cm), Vm o yo f(xo, m) Vm Nhdm theo tham s6 vd dp dyng cdc m^nh d4 sau; Am + B = 0, Vm o A = 0, B = 0 Am^ + Bm + C = 0, Vm «> A = 0, B = 0, C = 0 Am + B 5t 0, Vm <=> A = 0, B ^ 0 Am^ + Bm + C ?t 0, Vm <=> A = 0, B = 0. C ; t 0 ho0c A ^ 0, A = B^ - 4AC < 0 2.CACBAIT0AN • Bai toan 2.1: Tim d i l m u6n vd cdc khoang l6i fSm cua 66 thj: a) y = x^ - 2x^ + x + 1 b) y = x" + 8x^ + 9. Hiwng din giai a) D = R • Ta c6 y' = 3x^ - 4x + 1, y "= 6x - 4. ' y"= 0 <=> X = - ; y"> 0 <=> x > - ; y"< 0 <=> x < "• * •' 3 3 3 J; ; 2 29 2 Vdy 6\km uon l ( - ; — ) , hdm s6 l6i tren khoang ( - < » ; - ) vd 16m tr6n 3 27 3 2 khoang ( - ; + « ) . 3 *''•'•»•': \ • b) D = R. Ta c6 y ' = 4 x ' + 16x , y" = 12x^ + 16 > 0 V x Vdy do thj khong e6 d i l m u6n vd hdm s616m trSn R. j Bai toan 2. 2: Tim d i l m u6n vd cdc khoang l6i 16m cua d l thj: x^ - 2x + 3 2x +1 ''"-^-T;!- "'""ITT Himng din giai a)D = R \ { - 1 } . T a c 6 y = ^ ^ - l ? ^ = x - 3 + ^ x+1 x+1 N6ny'=1 ^ . y " = (x + 1)2 01 ^ 0, V x * - 1 (x + 1)=' - y"> 0 <=> X >-1 ; y"< 0 <=> x < - 1 Vdy do thi khong c6 d i l m u6n, hdm s6 loi tren khoang (-oo;-1) vd 16m tren khoang ( - 1 ; +oo). b) D = R \. Ta c6 y' = ,y " = — {x-5f ^ ^ 0,Vx ^ 5 (x-5)3 y"> 0 o X > 5 ; y"< 0 o x < 5 Vdy do thj khdng c6 d i l m uon, hdm s6 l6i tr6n khoang ( ^ ; 5 ) vd 16m tr§n l x^ + 3ax^ + 3(a - 1 )x - 1 = 0 Dal f(x) = x^ + 3ax' + 3(a - 1 )x - 1, x e R lim f(x) = -co, Mm f(x) = +oo va d6ng thai ham s6 nay lien tgc ten t$p s6 thyc nen phuang trinh f(x) = 0 c6 ba nghi^m phan bi^t thupc cac khoang (-x;-1),(-1;0),(0; + « ) Gia s(y hoanh dp cua m6t trong cdc diem uon Id Xo n § n Suy ra yo = x^ + Xg + 1 _ (XQ + Chox = 0 t h i y = - 1 . b) Ta CO y' = 3 x ' - 1 2 x + 3m 06 thj ham s6 c6 d i l m eye dai, eye ti§u o m<4 -mi d(.r Gpi cac dilm eye tri Id A(xi; y,). B(x2; y2). < » ( x o + 3 a - 1 ) ( x ^ + X o + 1) = 3(xo + a) +a % nghi?m phan biet » A ' = 36 - 9m > 0 + Sax^ + Sax^ + 3a - 1 = Sx^ + 3a XQ Vi$t khi va ehf khi phacng trinh y' = 0 c6 hai x^ + 3ax^ + 3(8 - 1)x, - 1 = 0 XQ Hhang n6n tarn d6i xi>ng la diem uon 1(2; 1). ** Taco: f(0) = - 1 < 0 , f(-1) = 1 > 0 Ta c6: TNHHMTVDWH 3a - 1)(x^ + x^, + 1) _ x^ + 3a - 1 3 ( x ^ x „ + 1) Theo dinh li Viet X, + Xj = x^Xj 4 =m Ta e6 yi= (2m - 8)Xi + m + 2, y2 = (2m - 8)X2 + m + 2 Vay cdc diem u6n cua d6 thi thupc du'ang thing y = ^ "'^ — - nen chung AB= J(x,-X2)'+(2m-8)2(X2-x,)2 = ( 1 + { 2 m - 8 ) ' (x^ + X2r-4x,x, 3 thing hang Bai toan 2. 4: Chp ham so: y = x^ - 6x^ + 3mx - m + 2, m la tham so. a) Khao sat sy bi§n thien va ve d6 thi cua ham s6 khi m = 3 b) Tim m sao cho d6 thi cua ham s6 da cho c6 cac diem eye dai, eye ti§u A va B ma khoang each AB = 4v'65 . Hu-o-ng din giai a) Khi m = 3 ham s6 tra thanh y = x^ - 6x^ + 9x - 1 • Tap xac dinh D = R = V(4m2 - 3 2 m + 65)(16-4m) nen AB = 4^65 o (4m^ - 32m + 65)(16 - 4m) = 1040 o 4m^ - 48m^ +193m = 0 « m(4m^ - 48m +193) = 0 <=> m = 0 (thoa m § n ) .Vay m = 0. Bai toan 2. 5: Cho ham so: y = - - x ^ + (m - 1)x^ + (3m - 2)x - - c6 d6 thj 3 3 (Cm) vai m la tham so. i> • Sy bi^n thien: y' = 3x^ -12x + 9 y' = 0 < = > x = 1 v x = 3 Bang bien thien: a) Khao sat sy bi§n thien va ve do thi cua ham s6 khi m = 2. b) Tim m d l tren d6 thi ( C J eo hai dilm phan biet e6 hoanh dp ciing d i u va ti§p tuydn cua (Cm) tai m6i di^m do vuong goc vai du-ang thing d: x - 3y + 1 =0. + 00 Hu'O'ng din giai a) Khi m = 2 ham s6 tra thdnh y = x ^ Ham s6 d6ng bi§n tren mpi khoang ( - « ; 1 ) va (3;+ « ) , nghich bien tren (1;3). Ham s6 dat eye dai khi x = 1, yco = x = 3ycT=-1. 36 3 va dgt eye ti4u tgi • T$p xac dinh D = R • Sy bi6n thien: y' = -2x^ + 2x + 4; y' = 0c:>x = - 1 v x = 2. ^ + x^ + 4x - - . %^ A » - I