Tài liệu Chuyên đề bồi dưỡng học sinh giỏi-giá trị lớn nhất, giá trị nhỏ nhất phan huy khải (phần 6)

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Chuyfin c i r B D H S G Tojn gia trj I6n nhS't v.i gia tri nh6 nliat - Phan Huy KITST Cty TMHH M!V DVVH KhanglZiH" VAIBAITOAN KHAC VE G^^yd-. hay u^ = 2 + 2 V 3 - ( x ^ - 6 x + l l ) . Ttf do ta di den phU'dng trinh he qua sau: GIA TRj idiN NHAT VA NHO NHAT CUA HAM SO = 2 + 2 N / 3 - U <=> U^ - 2 = 2 ^ 3 - u \J2 2 < f ' ( x ) J Bai 2. Giai cac phifdng trinh sau: , Ta CO f^(x) = 2 + 2 7 ( x - 2 ) ( 4 - x ) . - x ^ + 6x - 9 = 0 o X = 3. duy nha't cua phifdng trinh da cho. Hiidng ddn giai Datf(x)= V x - 2 + V 4 - x •i-.J^j r'^-'~ ^s: ''-it ^- h"^ i^/2 u^ = 2 + 2 V ( x - 2 ) ( 4 - x ) = 2 + 2 V - x ^ + 6 x - 8 L a i c6 g(x) = V x 2 - 6 x + 18 = +9 > 3 ai- g(x) = 3 o x = 3 Vay ming(x) = 3 xeR Vithe'(l)o ff(x)-3 g(x) = 3 fx = 3 x =3 o x = 3. Do vay X = 3 la n g h i e m duy nha't cua (1) 297 Chuyen de BDHSG Toan gia trj I6n nhjt vk gia tr| nhi nha'l - Phan Huy Khii Cty TNHH M T V D W H Khang Vigt 2. Xet phurdng trinh Vsx^ + 6x + 7 + Vsx^ + lOx + 14 = 4 - 2x Ta CO ' ' (2) p^i 4. Giai phUOng trinh 2^^ ' - 2" f(x) = V3x^ + 6x + 7 + Vsx^ + lOx +14 , " = (x -1)^ HUdngddngidi fv i : ; o ' ; = V3(x + l ) ^ + 4 + 75(x + l ) ^ + 9 : Xet phiWng trinh 2"-' - 2 " ^ - " = ( x - l ) ^ V a y V x e R , thi f(x) > 5; f(5) = 5 o x = - 1 . TCf do ta CO minf(x) = 5 o x = - 1 . Ta ' ' xeK CO g(x) = (x - 1)^ > 0 Vx Vay ming(x) = 0 <=>x= 1. g(x) = 4 - 2x - x^ = 5 - (x + 1)^ => g(x) < 5 , Vx e M ; Do(x - l ) ' > O o x ^ - 2 x + 1 >() g(x) = 5 c > x = - l . => X' - Do vay maxg(x) = 5 <=>x = - i . , j, • nt,|.,, Nhir the suy ra (2) o fffx) = 5 fx = - 1 <^ '<=>i' o x =- l lg(x) = 5 [x = - l f(x) = 0 o .. - 4 x + 4) ' ' ..x ., ,. 0. , , om " ' ' "* *^ o x=1 x= / Hudng ddn giai r Ap dung bat dang thiJc Cosi, ta c6 .c ' rx = i ok , o{ m 5. Giai phifdng trinh Sx" - 4x^ = 1 - Vo + x ^ /g - . 2 Lai c6 4x^ - 4x + 4 = (2x - 1)^ + 3 > 3 => log,(4x^ - 4x + 4) > 1. \ + I>3? 1 + ^x2 V 2 ) 2 bo d, 3 2 t 'I < 2 ce ww w. log2(4x - 4 x + 4) <8; -i l +lx^ + 2 3(1 + x ^ ) > 33 fa ^ g(x) = 8 o x = - . A ' : t • .;.x>1 i => 1- j; Jil + \^? < - - x ^ Vx e v. -.u u (1) 2 Vay maxg(x) = 8 o X = - . Lai ap dung bat dang thtfc Cosi cho 4 so, ta c6: 2 1 { f(x) = 8 g(x) = 8 X = — 2 1 o X = — 2 Vay X = ^ la nghiem duy nhat cua phi/dng trinh da cho. • -• xS2xUx^+ix^>4V^ 2 >4x^ (2) = > 3 x ' * - 4 x ^ > - - x ^ Vx e R . 2 Tir (1) (2) suy ra f(x) = 3x'' - 4x' - (1 f(x) = O o x = 0 298 X <=> X = Nhir the' (1) CO nghiem duy nhat x = 1. ro up f(x) = 2'"+' + 2^-^^ > l^I^^^Kl'-^' = 24¥ = 8, Vay minf(x) = 8 o x = - . xeR 2 g(x) = 0 Ta . s/ Theo bat dang thurc Cosi, ta CO f(x) = 0 Vay (!)<=> Hitiing ddn gidi xeR - xeR log:,(4x V i t h e Vx € R , t a c 6 g ( x ) = 1= X - TuTdo suy ra max = 0 <::>x= 1. Bai 3. Giai phi/Ong trinh 2^"+' + 2^"^^ = X = > 2"' Vay Vx e R, ta c6 f(x) = 2"'' - 2 ' " ' < ( ) ; Tijr do ta CO X = - 1 la nghiem duy nhat cua (2). f(x) = 8 o 2x + 1 = 3 - 2x o 1 JI 1 X > X - 2" < R va g(x) = 0 <=> x = 1. 6 iL ie uO nT hi Da iH oc 01 / Lai CO (f) ' ^(l + x^f ) > 0 Vx e IR Chuyen dg BDHSG ToAn gii tr| Idn nhaft va gii trj nh6 nhS't - Phan Huy Kh^i Cty TNHH MTV DWH Khang Vigt Vay minr(x) = () o x = 0. Isin'x + cos'x = 32(sin"x + cos''x) xel ^ f(x) = g(x) Ttf cac ket qua tren suy ra Ro rang phiMng Irinh da cho co the vict duTdi dang f(x) = 0 Ttr do suy ra phu'(tng trinh da cho c6 dang min f(x) = 0 o x = 0 ,, (1) , ' s » , •K Vay X = 0 lii nghicm duy nhal can tim. g(x) = l ^ Dat biet vdi phUdng trjnh dang f(x) = a, x e D ' xeD xeD HUdng dan giai = n x-(l-x) Dat f(x) = s i n \ cos^x, x e K . = n ( 2 x - 1) sin^x < sin^x cos'^x < cos^x ' • f t * ( k 6 up sin''x = sin^x Z). /g ro ir o x =k cos X = cos X 2 1 Ta •' s/ => f(x) = s i n \ cos^x < 1 Vx e R . .c . bo 1 fa ce ,n-I • ' ww I Dafu bkng xay ra o a = b = —. Do sin^x + cos^x = 1 X"-2+x"~'(l-x) + ... + ( l - x ) " ~ ^ 1 0 x 2 1 h'(x) 0 1 h(x) ^•^^'^•^ 1 + 1 1 Nhu- vay do 0 < a < 1 => h(a) > h ok - w. a" + b" > , ,n-2 3 X "n-2' + ,x "„ n" -- X l - x ) + ... + ( l - x ) Tijr do CO bang bien thien sau: om Tiirdotaco maxf(x) = l o x = k - , k e Z . " xeR 2 Ap dung ket qua sau day: Ne'u a, b > 0 va a + b = 1, thi vdi moi n nguyen > 2 ta c6: (1) v6 nghiem. Xet ham so h(x) = x" + (1 - x)" vcti 0 < x < 1 => h'(x) = nx"~' - n ( l - x ) " " " ' = n x " - ' - ( l - x ) " - ' X€D Bai 6. Giai phiTdng trinh sin^x + cos^x = 32(sin'^x + c o s ' \ Mat khac f(x) = 1 x = - + k - , k e Z (3) 4 2 Chu y: (*) chtfug minh nhiT sau: khi do ta c6 r(x) = a <=> max f(x) = a (hoac f(x) = a <=> min I'Cx) = a ) . Ta c6 (2) \ Ro rang he (2) (3) v6 nghiem ma thoa man dicn kicn maxl"(x) = a (hoac minf(x) = a ) , xeD <=> x-k-,keZ 2 iL ie uO nT hi Da iH oc 01 / Chuy: f(x) = l sin'^x + cos'^x = (sin^x)' + (cos^)'' > — oa"+(l-a)">-i^hay a"+b">-i^. 1^ • 5 ' • •a'u b^ng xay ra<=>a=:^<=>a = b = ^ = > dpcm! Bai 7. Giai phiTdng trinh cos3x + V2-cos^ 3x = 2(1 + sm^x). HUdng ddn gi&i => g(x) = 32(sin'^x + cos'^x) > 1, Vxe R. . • Dat f(x) = cos3x + V2-cos^ 3x ; g(x) = 2(1 + sin^x) vdi x e Khi do phiTdng trinh da cho c6 dang f(x) = g(x). g(x) = 1 o sin^x = cos^x = <=>x= — + k — , k G 4 2 (1) De thay g(x) > 2 Vx e R (do sin^x > 0 Vx e R ) Mat khac g(x) = 2 o sinx = 0 o x = kn (k e Z ) . Vay ta c6 ming(x) = 2 <=> x = kTt. xeR Ap dung bat ding thuTc Bunhiacopski, ta c6: ' (2) Cty TNMH MTV DVVH Khang Vigt Chuyen dg BDHSG Toan gia tri Idn nha't va gia tri nh6 nhaft - Phan Huy KhSi cos^ 3x + (2 - cos^ 3x) (1 +1) > |cos3x + V 2 - c o s ^ 3 x do suy ra fir Vx e 2k7t ra \/x' + x - I +Vx-x^ hay f(x) < 0 Vx e D. + 1 <^X ' + x - l = l <=> s x = - 2 x - X~ + 1 1 X = 1 'x 0 <=> x = I. l(x) = 0 |x = l IKct hdp lai suy ra (1) o g(x) = 0 <::> <[x = l rir do O X = 1. O X = 1. di den max f(x) = ' (3) Bai 9. Giai phU'dng trinh Tir do suy ra +1 sin^ X + COS^ X + \J HUdng ddn giai 1 . Datg(y)=12+ - s i n y . y e R. COS ce bo ok .c om /g ro fa ww w. -x" + x + I > 0 Vie't lai phUcJng trinh da cho diTdi dang \ / x ^ + x - l + V x - x ^ + l - ( x + l) = x ^ - 2 x + l . ( l ) Datg(x) = x ^ - 2 x + l;f(x)= V x ^ + x - l + V x - x ^ + l - ( x + l), vdi x G D . Tac6g(x) = ( x - l ) ^ > O V x e D g(x) = O o x = l TiJf d6 suy ra min g(x) = 0 o X = 1. xeD , 1), T a c 6 g ( y ) < 1 2 - Vy e I'M 1 , g(y) = 1 2 ^ - 0 siny = 1 » y = ^ + k27i. k e Z . ^ ^ ^^^^j , Vay maxg(y) = 1 2 - o y = - + k2n, k e Z . yeR 2 2 Dat f(x) = 2 COS X+ 1 cos z X J sin.2 X + 1 sin^ X + 4=1 1= (cos'* X + sin"* x) + V cos— X+ —Tsin X ; l - - s i n ' ' 2x l 2 ^ „ . , ^ + 4. = I — s i n ^ 2 x + 16- 2 sin* 2x y f ( x ) > 1 2 ^ Vx € R . ^^ (1) .2 /J-iHI /V.?.JU «t> ' J l--sin^2x sin^ 2x + — : r — +4 sin'*xcos'*x (2) Tir (2) de d^ng suy ra Vx e R . t a c 6 f ( x ) > 1 - ^ + 1 6 - ^ ^ + 4 1. Theo ba't d^ng thiJc Cosi Vx e D, ta c6 = 12 + ^ s i n y . ^ sin^ X y , // . f Ta s/ up x = 2—.keZ lgW = 2 X = kn, k e Z 3 _ 2kn ^ , X G i • i »x = k 2 7 c , k e Z X = k7l,k€Z j^., ' Vay X = k27t, k e Z la nghiem cua phiitfng trinh da cho. ^'i'H>Bai 8. Giai phtfdng trinh Vx^+ x - l + \ / x - x ^ + 1 = x^ - x + 2. ^, t.s,.\.:^^:^',:. HUdngddngiai Mien xac dinh cua phiTdng trinh la tap hcJp D gom nhiTng phan tijf x thoa man (1) (chiiy r ^ n g x ^ - x + 2 > 0 Vx) he x^ + x - l > 0 .2 li > I vay X = 1 la nghiem duy nha't cua phiTctng trinh d5 cho. xeR <=> X = suy iL ie uO nT hi Da iH oc 01 / f(x) = cos3x + yjl-cos^ 3x < 2 cos3x -Jl-cos^ 3x f(x) = 2 <=> 1 1 o cos3x = V 2 - ; o s ^ 3x fcosSx > 0 cos^ 3x = 2 - cos^ 3x o c o s 3 x = 1. Vay maxf(x) = 2 o c o s 3 x = 1 Tit ChuySn de BDHSG Toan gia tri I6n nhat va gia tri nli6 nhat Cty TNHH IVITV DWH Khang Vl§t Phan iiuy Kh^i ( 4 s i n \ 2sin^x - Bsinx - 1)^ = 5 - sinx f(x) =l2-<=> sin'2x = 1 c=> cos2x = ( ) « x = - + n ^ , n € Z . 2 4 2 V i phifdng trinh da cho c6 dang f(x) - g(y) l-(x) = 16sinS + 16sin'*x - 20sin''x - 2 0 s i n \ 5 s i n \ 7sinx - 4 = 0 x, y e M o (3) D e n day m d i cac ban giai t i e p ! ! Cac ban thay the' nao? 12- f Tijf cac lap luan Iron siiy ra (3) o g(y) = i (sinx - l)(16sin' x + 32sin''x + 12sin''x - 8 s i n \ 3sinx + 4) = 0 2 ^ 4 x ^ + 1 4 x + 46 9 B a i 1 1 . G i a i phu-dng trinh — = 2x^ - 8x + 1 3 . x ^ + 2 x + 10 iL ie uO nT hi Da iH oc 01 / Hiidng ddn giai <::>x = —+ n — ; v = — + k27t, n va k 6 Z 4 2 2 D a l f(x) = B a i 1 0 . G i a i phiTdng Irinh (sin3x + cos2x)' = 5 - sinx. x^ +2X + 10 G o i m la gia t r i l i j y y . K h i do phU'dng trinh sau (an x ) HUdng dan giai 4x^ +14X + 46 —^ =m D a l f ( x ) = (sin3x + cos2x)\(x) = 5 - sinx, x e R D o sinx < 1 V x e M ^ 4X^ 14V + 4 -46 4(S +14X 4- g(x) > 4, V x e R • x ^ + 2 x + 10 g(x) = 4 o sinx = 1 <=> x = - + k27i, k G Z . • Do x^ + 2x + 10 ^ 0 V x (VI • ' , ^ (l)conghiem ',i . ' „• • x^ + 2x + 10 > 0), nSn (l)<::>4x^ + I 4 x + 46 = mx^ + 2 m x + 10m TiJf do suy ra m i n g ( x ) = 4 c:>x = - + k27t. xeM 2 s/ Ta (1) up L a i CO lsin3x + cos2x|<|sin3x| + |cos2x|<2 V x e [cos2x = l sin3x = - l om o K h i m = 4, t h i m - 7 7 t 0 = > ( l ) c 6 n g h i e m . * Khim^4,thi(2)c6nghiemo o ce fa xeR w. V i phu-dng irinh da cho c6 dang f(x) = g(x), nen tif cac k c l qua tren suy ww phifcing trinh da cho tiTdng diTdng v d i he f f ( x ) = 4 (3) g(x) = 4 (4) T i r ( l ) ( 2 ) s u y r a ( 3 ) ( 4 ) « x = - + kn, k e Z A'>0»m^-8m+15<0 3 < m < 5 ( m ;>t 4) K h i m = 5, thi (2) CO dang x^ - 4x + 4 = 0 o x = 2 TiJf do ta CO max f ( x ) = 5 o x = 2 (3) xeR bo Tir do ta CO max l"(x) = 4 o x thoa man ( 2 ) , , V a y (1) CO n g h i e m <=> 3 < m < 5. (2) ok cos2x = - l (2) o3 l"(x) < 4 V x e R o ( m - 4 ) x ^ + 2(m-7)x +10m-46 =0 Ta CO ' g(x) = 2x^ - 8x + 13 = 2(x - 2)^ + 5 ' Nhu" vay m i n g ( x ) = 5 o X = 2 (4) |\R Tijf (3) (4) suy ra phU'dng trinh da cho tu'dng dufdng v d i he ff(x) = 5 fx = 2 g(x) = 5 [x = 2 o x =2 V a y X = 2 la n g h i e m duy nhat can tim. V a y X = ^ + k n , k e Z la nghiOm ciia phi/dng Irinh da cho. Nhanxet: K h o CO each giai nao khac gon gang hdn each giai trcn Cac ban ciJ thijr ti/dng lu"dng sau khi siir dung cong thuTc sinSx = 3sinx - 4sin cos2x = 1 - 2sin^x, la difa phiTi^ng trinh da cho ve dang: 304 5 i i iAy - A^Aa« jcef: X e t each giai khac sau day 4x^^14x^46 ^^^,_3^^^3 - • " ' ; l « '•• ^ " x ^ + 2 x + 10 o 4 x ^ + 1 4 x + 46 = ( 2 x ^ - 8 x +13)(x^ + 2 x + 1 0 ) ' • o2x^-4x'+13x^- . «, 68x + 84 = 0 305 Chuyfin dg BDHSG Toan gia trj I6n nhat vi gia tr| nh6 nhflt - Phan Huy KhAi Cty TNHH IMIV DVVH Kliang Vi$t <=>(x - 2)'(2x' + 4x + 21) = 0<:=>x = 2 ' =:>x + y < 6 . C&ch giai nay hoan loan chap nhan di/dc, neu cac ban doan triTdc d\{a^. Ap dung bat diing IhuTc Bunhiacopski, ta c6 nghiem x = 2! Bai 12. Giai phi/dng trinh 4 sinx ol+sinx (6) (75rri.i+7^.i)'< (75^71)'+(7771)' cos(xy) + 2'^i = 0 HUotng ddn giai (1^+1^) => 7x+T + 7y + 1 ^72(x + y + 2 ) < 4 Viet lai phtfcJng trinh dU'di dang sau: max ( 7 ^ + 7 7 7 l ) ^ 4 « : / i l L . £ I i v a 7 ^ + 7 ^ = 4 iL ie uO nT hi Da iH oc 01 / 2l''l-cos^(xy) = 0 . P = 2 " " " -cos(xy) (1) Ta CO 21^1 > 1 Vy € M cos^(xy) < 1 Vx, y e R . hit ii.^'f M I*til P = 0 o i 2'^l-cos^(xy) = 0 O ' cos ( x y ) - l 2''"''=cos(xy) 2M = i [2''"''=1 o x = y = 3J^'^''^S*™i'«-'r' Nhan xet: Neu khong sit diing phiTdng phap tim gia trj U^n nha't ci'ia hiim so de "^ly^o danh gia hai ve', ta c6 the giai ihuan tuy he phtfdng trinh trcn nhU'sau: T i i r ( 2 ) c 6 x + 1 + y + 1 + 27(x + I)(y + 1) = 16 y =0 y=0 jf ;,;'„ ( 5 .;v v ' • m id) f Ta X = kn, k e Z , t j s/ sin X = 0 ro up Vay minP = 0 o x = k7t;y = 0 , k 6 Z /g Tir do suy ra nghiem cua (1) la x = kK, y = 0 vdi k e Z . om Bai 13. (De thi tuyen sinh Dai hoc Cao ddn^ khoi A) ok ce t =3 35 t = -- y + l>0 (4) xy>0 (5) , ; De thay neu - 1 < x < 0 va - 1 < y < 0 thi yfx + l+Jy + l<2 vay khong tht" =>x>0;y>0. V i the theo bat d i n g thtfc Cosi, ta c6 tijr ( 1 ) : x + y = 3 + 7 x y ^ 3 + 3t^+261-105 = 0 0 1 = 3. if.' M ^ f t Khi do 7 x y = 3 <::> xy = 9 Tir(3)(4)suyrax>-I;y>-1 thoa man (2) ()0 Di6u kien de hg (1) (2) c6 nghla la <=> 27t^ + t + 4 = 11 - 1 <=> 0 0, thi tuf (1) CO X + y = 3 + t - bo Hudng dan giai x +y-Vxy=3 <=>2 + x + y + 27xy + (x + y) + ] = 1 6 . .c \ y-yfty =3 Vx + i + 7 y + i = 4 306 77^1-2 Vay he (1) (2) c6 nghicMn duy nha'l x = y = 3 2''"" -cos(xy) = 0 X€i he phi/dng trinh x=y Tur do suy ra (2) o X = y = 3 va x = y =3 cung thoa man (1). V i t h e P > 0 Vx,y G R Giai he phifdng trinh 1 |x+y-6 Vay(l)(2)oj"-_^ " o x = y = 3. jTa thu lai ket qua trcn! !nfa^,(1i/ .,( • • ,'1 f ' i ' < Chuygn 6& BDHSG ToJin gii tri Idn nh9"t Cty TNHH IVITV DWH Khang Vigt gia tri nh6 nhS't - Phan Huy Kh^i § 2 . G I A TR! L 6 N N H A T V A N H O N H A T C U A minf„(x)-2 H A M SO PHg T H U Q C T H A M SO vfAu--^2acos2a<0 iL ie uO nT hi Da iH oc 01 / la noi dung cua bai loan chiing ta xct trong muc nay. cos2x = - 1 . „ , , nay mmfoj(x) = 2 A. Bien luan theo tham so' gid tri Idn nhat va nho nhat cua ho ham so phu thuoc tham so , 7: CO nghiem (chang han x = — ) , v i the trong trirCtng h d p sin2x = 0 V i he gia tri Idn nhat, nho nhat cua ho ham so Fn,(x) theo mot lieu chi nao do. Do sin^2a .4sin^acos^a . - =2 : = 2c()r(x., 4sin'^a (-l + cos2ar Nhi/ vay theo tham so' a, ta c6 ket luan sau: B a i 1. Cho ho ham so f^Cx) = t a n ^ ( x + a ) + t a n ' ( x - a ) , v d i tham so a e 2tan^a,Ne'uO()< 2 a < - => cos2a > 0. 4 2 ^^f? - Ta CO Vx e D ^ , ( d day D„ lii mien xac dinh cua ham so f a ( x ) ) /^-j ^, j y,, 0 < sin'2x + sin'2tt < sin^2a xeR l 2, khi do ^ - ^ Do cos2a > 0, nen (cos2x + cos2a)^ < (1 + cos2a)^. (3) " ' M Dau bang trong (3) xay ra <=>'cos2x = 1 . hdp nay sin2x = t (-1 < t < 1). K h i do la c6 min 1- ( t ) (2) i sin2x = 0. V I he •' ' " " ^ ^•\^••^ {^ cos2x = 1 !! I "-f F rt) = - - t ^ + — 1 + 1, v d i - 1 < t < 1. ww Ncu ( ) < a < -1 HUo'ng ddn gidi 1 . m Viet l a i fm(x) di/di dang sau: f,„(x) = 1 - - s i n 2x + — s i n 2 x . Ml fa (I) (cos2x + cos2a)^ tl ce 2(sin~2x + sin^2a) j Bien luan theo m. m a x L , ( x ) = max f d ) (1) ok +(sin2x-sin2a)^ fn,(x) = sin'*x + cos'*x + msinxcosx, x e R .c * -'^ (cos2x + cos 2 a ) " * om sin^(x + a)cos^(x - a ) + s i n ^ ( x - a ) c o s ^ ( x + a ) cos"(x + a ) c o s ^ ( x - a ) n Ta (J) Kil idi ,C1 i xi\r f i »f s/ •• ro sin'(x-a) ^ Bai 2. T i m gia tri Idn nha't va nho nhii'l cua ham so: {*'\ >^ 4 / n % up B i e n ddi r „ ( x ) ve dang sau day 4 2cot^ a , neu — < a < • 4 2 HUfing ddn gidi ( s i n 2 x + sin2a)^ . , p a u bang trong (4) xay ra <=> cos2x = - 1 thuoc tham so m. Tiay theo gia trj cua m hay khao sat cac linh chsi't cua cac ^ ^ sin^x +a) ia(x) = cos^(x + a ) (4) r v p o cos2a < 0, nen (cos2x + c o s 2 a ) ' < (-1 + cos2a)". nhien gia trj Idn nhat, nho nha't cua ham so F J x ) Ircn D la cac dai lifdng p h u (.(•, nghiem, ch^ng han x = 0, v i the trong truTdng ' 1 cd bang bien thien sau tn t F„(t) -1 1 1 + — 1 2 i ° 1 - , d dav Chuyen de BDHSG Join gia trj I6n nha't va gii trj nhd nhlft - Phan Huy K h j i TiT do suy ra max F,,,(t) = F„,(l) = Cty TIMHH MTV D W H Khang Vi^t m +I 1+m 2 -i 0 -i0Vx€R 2 b+ 2c^„,, _ >OVxeR 2 a- b a + b + 2c X) Vt <1 -l + 2 2 ce min F,,,(l) = m i n { F ( - i ) ; -IOVt <1 • •-^^^ aiii,M 'uc: min l0(2) >0(3) Itki • • Do ,6!)' W;| S, < m la hkng so k h i t bicn ihien nen a + b + 2c , a-b -t > 0 ( 4 ) _ hmin 2 2 |i| 0 ( 5 ) 2 l.| 2 . l + cos2x + b + c . a + b+ 2 c a - b . , a+ b+ 2 c b - a 1 cos2x 1 cos2x + 2 2 2 2 Ham so f ( x ) xac djnh v d i m o i x e R k h i va chi k h i bo i F,„(l) + s/ 7 -1 a Ta m 3. N e u - 2 < m < 2 = > - ! < — < 1, va C O bang bien thien sau . 2 , r a i;t§ f l + cos2x , l - c o s 2 x + b + c + N c u a > b => m i n ,a, l i t ' £r\ '^jji . I V J I a-b. |i| 0 o c + b>0. 2 2 ~ a- b 2 2 c + a > 0, nc'u b < a > 0 o c + a>0. ri>minr()(x) •• * <=> c + min(a; b) > 0 (6). 2. Dat g(x) = msin2x, ihi F(x) = f(x) + g(x) (7). f,„(x) Vx € K neu m > 0 I Ta om cos^2x 2 J ba kha nang sau: a. Neu 2m ce w. X sin2x = - 1 cos2x = 0 deu C O nghiem, nen suy ra max F(x) = max f(x) + max g(x) = J2(a + b + 2c) + m x€R 312 xeR x€R I i xeD (12) xeR cos2x = 0 0 Neu - H l ± l > 2 ^ 0 > m > - - . Luc nay ta c6 bang bien thien 2m 5 x€R va 2m suy ra xet 2 I 2m xeD cos2x = 0 = - m < - - . Liic nay ta c6 bang bien thien 3 m + 1 4(x) 0 T t r d o t a c o m i n L , ( x ) = f„,(2) = 7 m + 4 ; m a x f „ ( x ) = f ^ ( l ) = 3m + 3. Vay maxf^(x) = 2(a + b + 2c)<^ cos2x = 0 sin2x = 1 0, tiT do tiif (11) ta c6 ^ max f(x) = J2(a + b + 2c) ^ , xeD • Khi m < 0.1\ ^ ( x ) = 2mx + m + 1 , va .c J [a-b] bo 2 (10) ok a + b + 2c^ /g ro (9) D o f ( x ) > O V x e M . n e n maxf(x)=- /maxf^(x) xeR V xeK 2 up Khi m < 0 thi maxg(x) = - m <=> sin2x = - 1 . f ( x ) < ( a + b + 2c) + 2 < 0, nen c6 ^""""'^^^ xeD s/ (8) xeR , 2m Tilf do ta CO m i n f^, (x) = L, (1) = 3m + 3; max f„, (x) = f„, (2) = 7m + 4. Khi m > 0 thi maxg(x) = m ^ sin2x = 1. Vxe R , t h i m +1 + 0 / g(x) < - m Vx € R neu m < 0 I ^ xeD bang bien thien sau: , j 4 , 5 . - ! ' f„ m + 1 X 1 2m < m Vx€;R. Ta c6f^(x) = (a + b + 2c) + 2, j , Khi m > 0. Tif f|„(x) = 2mx + m + 1 , va do m > 0, nen I'lr!) ft g(x) < m = io(l) = 3; maxf,)(x) = f,)(2) = 4. xeD Vay (6) la dieu kien can va du de f(x) xac dinh Vxe Taco g(x) . fa ihay l()(x) = 1 > 0, nen f()(x) la ham dong bie'n khi x e f 1; 2] Tom lai de l'(x) xac dmh Vxe R ta can c6 c 4 - b > 0 , neu a > b | ..^ g , , . iL ie uO nT hi Da iH oc 01 / (4)(5) a + b + 2c , jChi m = 0, la c6 f;,(x) = x + 2 Nc'u a < b, hoan loan ti/dng ti/ta CO , 4(x) fm(x) 1 2 + m +1 2m 0 Cty TNHH MTV DWH Khang Vi^t Chuygn di BDHSG loAn gia Iri I6n nhgt va gia Iri nUd nh3't - Phan Huy KhSi Tfifdotaco fLijc n a y ( I ) CO d a n g 2 t - - I + at > - 1 hay 2 t ' + at > 0 m i n f „ , ( x ) = f,n(l) = 3m + 3 ; max l„,(x) = L , ( 2 ) = 7m + 4. ^ ^ A I B a i t o a n t r d i h a n h : Tim a d c ha'l phmJng t r i n h : r.,(l) = 2 t ' + a t > 0 d i i n g v d i Inioi-1 ^ I ^ 1-Dicu n a y Xiiy ra khi va t h i khi ^ * < 2 <^ —— < m < — —. Luc nay la c6 bang bicn t h i c n 2m 3 3 m +l I X + max f,„ (x) = f„ xeD m +l / / / / 3m2+6m-l 2m 4m '-^ii'-i^ ^ 3m + 3 nc'u 1 = m m 3m + 3 ; 7 m + 4 f.(t) 1 rC: ,i\J ' • KV'. m i n f^(t) = f j l ) = 2 + a Khi do tiir (2) suy ra a + 2 > 0 <=> a > - 2 . minf„(x) = s/ /g ro 7m + 4 , ne'u m > - — . ~ 4 ^ ; maxf,„(x) = up 1 3m + 3, ncu m > — 7 m + 4, neu m < — 3m + 3, neu m < — 4 '.^-.x -"• Loai kha n5ng nay v i khong thda man a < - 4 . a 2. N5u — < - 1 (turc la k h i a > 4). Luc nay c6 bang b i c n t h i c n sau: 4 , „.,, ji^ a t 1 1 ~4 ok B. Cdc ling dung cua vi^c khdo sdt gid tri Idn nhd't vd nhd nhd't cua cdc fa(t) bo ham so'phif thuQC tham so / / / / 0 fa(t) .c om 4 fa T i m a, b de ba't phi/dng trinh sau: cos4x + acos2x + bsin2x > - 1 dung v d i mt'i -;f->^> ww Hadngddngidi V i ba't phiTdng trinh dung v d i m o i x, nen n o i rieng no phai dung k h i x = — va x = — . 4 4 m i n f.(t) = 4 ( - l ) = 2 - a - : t .1 i '"il' ^ -l 4. a 3- Nc'u - 1 < - - < 1 (turc la k h i - 4 < a < 4). Bay g i d cd bang bien thien: 4 L - a 1 ~4 0 fa(0 V a y b = 0, va b a i todn trd thanh: T i m a de bat phiTdng trinh sau: j D a t t = cos2x, k h i do - 1 < t < 1. ITilfdd: t - l +b > - l b > 0 <^ - b = 0. l - b > - l b<0 (1) dung v d i m o i x. < ^'Bay g i d tir (2) suy ra 2 - a > 0 o a < 2. w. xe R . i + I n' >. ^ ce Bail. 314 «P! oh §1/ d"j -i - l 0 Ta 7m + 4. n e u^' D o do ta CO he: r I f.(t) 0, «-»r ... /r, 's „ /,;r,;.;fr;MrfJ;. I N c u — > 1 (ttfc la khi a < - 4 ) . Liic niiy c6 bang bicn thicn f '• 4 rmnf,„(x) = m i n { f , „ ( l ) ; f „ , ( 2 ) } X6D > ^• ' d o xet ba kha nang sau: 0 \ Tir do suy ra f a c d : l.|(t)= 4t + a v a l,,(t)== 0 « . t = : : ^ -4-'. 2 2m m i n r . | ( t ) > ( ) . (2) - Kill ' iL ie uO nT hi Da iH oc 01 / c. Nc'u 1 < — \ 1 fa(0 ^ ^ ^ + ^ 315 Cty TNHH MTV DVVH Khang Vi?t cos2x = —1 (5) b. Neu m = - 2 , Ihi (4) <=> m a x l " ( x ) = m i n g _ , ( x ) = 4 xeR Tif (2) ta t o : —a >0^ a- < 0 <=> a = 0. Ro rang he (5) (6) c6 nghiem (thi du x = — thoa man (5) (6)) Tom lai a = b = 0 la cac gia Iri can tim cua tham so' a va b. T6m lai phiTdng trinh da cho CO nghiem o m = - 2 '*. Bai 2. Tim m de phiCdng trinh sau: 2 , ^/7V 2 ,t , OA ' 1"= X + 2(2m 3)x + 5m 1 6 m + 20 co nghiem. 3x^ + 2 X + 1 iL ie uO nT hi Da iH oc 01 / 20x^+10x4-3 Hudng ddn gidi HUdng ddn gidi Dal f(x) - (cos4x - cos2x)\e R Khi do ro rang do |cos4x — cos2x| < 2 Vx 6 K Dat f(x) = \ \ =:>f(x)<4Vxe E iiiuq .tit H>ui iU'i r:.,i r . 20x^ +IOX + 3 y K i ' m xr'/l' '• Ro rang f(x) xac dinh tren R (do 3x' + 2x + 1 > 0 Vx) , Goi a la mot gia tri tiiy y cua f(x). Khi do phifdng trinh sau day (an x) cos4x = 1 2 0 x ' +10X + 3 mm cos2x = — 1 £ f Ri 3x^+2x + l ( i ) Vi; '' Ta ocos2x = - l . cos4x = - 1 ^ Ta CO (20 - 3a)x^ + 2(5 - a)x + 3 - a = 0 (2) (1) up Ro rang khi ^ thi 5 - a 7^ 0 => (2) co nghiem khi " ~ Khi * ~ ' thi (2) CO nghiem o A' > 0 o 2a - 19a + 35 < 0 - < CY < ok + 2] = (m + if +(m+lf~l. ce - l ] | ( m + if bo = l ( m + 2 ) ^ - l ] [ ( m + 2 ) ^ + 2 ] + 7 + sin3x .c gm(x) = ( m ^ + 4 m + 3)(m^ + 4m + 6) + 7 + sin3x, x e M • om /g Tirdo suy ra: maxf(x) = 4 <=>cos2x = - 1 . ro (Chu y: khi cos2x = 1 => cos4x = 1; con khi cos2x = - 1 => cos4x = 1) (1) CO nghiem. = a (1) <^ 20x^ + lOx + 3 = a(3x^ + 2x + 1) s/ cos2x = 1 Chu y rang: A „ = |(m + if • 3x^+2x + l V a f ( x ) = 4 <;=>|cos4x-cos2x| = 2 €ir " 0ai 3. Tim m de phtfctng trinh sau: (cos4x - cos2x)^ = ( m ^ + 4 m + 3 ) ( m ^ + 4 m + 6) + 7 + sin3x c6 nghiem. ,Dat s i n 3 x = = - l (6) xeR 7 a Vay (2) c6 nghiem <^ — < « < 7. 20 3 f fa D o d 6 : - A , „ = - 2 khi m = - 2 (va A m > - 2 k h i m ; t - 2 ) NhiTvay ta di den; xeR Dat g„(x) = x^ + 2(2m - 3)x + 5m^ - 16m + 20 ww 7 + sin3x > 6 o-sin3x = - 1 . Ttr do suy ra: maxf(x) = 7 o -x^ - 4x - 4 = 0 o x = - 2 . (3) w. Lai c6: 7 + sin3x > 6 Vxe R g„, (X) = 2x + 2(2m - 3) va g',,, (x) = 0 o x = 3 - 2 m ^ • Khi m = - 2 , thi min g,„ (x) = 4 . (2) =>mingn,(x) = g n , ( 3 - 2 m ) = m ^ - 4 m + 11 = ( m - 2 ) ^ + 7. • Khi m 9^-2, thi ming,,,(x)>4. (3) Nhu vay ta c6: xeR xeM Phu'dng trinh da cho c6 dang: l"(x) - gn,(x) T i r ( l ) ( 2 ) (3)suy ra: ,'6 'I3t a) Neu m = 2, thi ming_,(x) = 7 :i'; (4) xeR b) Neu m ?^ 2, thi mingn, (x) > 7. (5) x€R a. Ncu m ^ - 2 , thi (4) vo nghiem Tilf(4)(5)suyra: 317 oiiuyuii ue u u i i j u l u j i i L)id in luii i i i u i vd g u ill iiliu ililjl - Hlljll HUy MIUI Cty TNHH MTV DWH Khan^jyi^ a. Ncu m ?^ 2, Ihi max K x ) < miiig,,,(x) :o I'hiAfng Irinh da cho \ f If do suy ra (1) c6 nghicm nghicm. xtK x>R . " Lap bang sau day: Vay phiftJiig Iriiih da chi) lifdng diTdng vOi he sau: g2(x) = 7 > X X = -2 (*) X ==-1 1 0 Hitdng ddn gidi M,..'Ui i '{iw fsj £«3 Ukn m ; Do do bai toan da trcl thanh: = mx^ + (m + 1 )x + m + 2 > 0 ( 1 ) diing vdi m + + Lucnaytaco: minfn,(x) = f,„(l) = m + 3 m - 2 . ... m - x £ » : { x > xeR Dieu nay xay ra khi va chi khi: min 1"„ (x) > 0. (2) s/ I - — " 4 min f,,,(x) = oh m^ + 3 m - 2 < 0 2. Neu m < - 1 . Khi do ta co bang bicn thien sau: -1 m X - 1 + 0 fn.(x) ok .c v6 nghiem, nen loai triTdng hdp nay. R6 rang he: om 7m 4-4, ncu m < - ( ,. . ~ 4 j m>l Ta moi 1 < X < 2. bo m > -- ce 3m f 3 > 0 Tir(2) (3) taco: ww 4 Vay m > — la cac gia tri can tim cua tham so m. Bai 5. Tim m dc bat phiTdng trinh sau: + 2 | x - m | + m - + m - l < 0 C O nghiem. -m-2 xeR Ttjf do di den xet he sau: W.;.v w. - i . fa " ™<-i ~ 4 7m+4 X ) ' Khi do ba't phi/dng trinh da cho c6 dang l„,(x) < 0. (1) 318 2x + 2 X nghicm cua bat phifdng trinh mx' + (m + 1 )x + m + 2 > 0. Tim m dc ba't phU'dng trinh: ijj.) 2x-2 Neu m > 1. Khi do ta co bang bien thien sau: Bai 4. Tim m sao cho moi nghicm cua bat phiTdng trinh x' - 3x + 2 < 0 cung i;, ot, u^-t X' - 2x + m^ - m - 1 Tirbang bic'n thien, xct cac kha nang sau day: Tom lai vc'li moi m, phU'dng trinh da cho v6 nghicm. Vi x' - 3x + 2 < 0 o I < X < 2. x^ - 2x + m ' + 3m - 1 iL ie uO nT hi Da iH oc 01 / Do he (*) M"> nghicm => Phifdng trinh da cho cung v6 nghicm khi m = T l6 9 Om>6. 9-2m<0 m > - ~2 2. Neu — < - (tiJc la ne'u m < 1). Luc do ta c6 bang bien thien sau: 2 2 1 m 3 X I - ^\ + + 1 ro = m— m Nhu" do dan de'n xet he sau: l4 1 Kct hctp lai suy ra: m < - - hoac m > 4 la cac gid trj can tim cua tham so m Nhqn xet: Hay so sdnh cdch i^idi nay vt'/i cdch fiidi trinh hay tnmi^ bdi 6 phdn B, §3 chu 7 cuon sdch nay. Cdc ban thdy cdch nao thich h(fp \'<)i ban hfn? Bai 7. Cho ba't phu-cJng trinh: sin3x + msin2x + 3sinx > 0. Tim m dc ba't phm^ng trinh diing vdi moi x € fa ce min f ^ ( x ) = : f n i ( 3 ) = 9 - 2 m TiJf do dan den vice xet he sau: • l 3 (tuTc la neu m > 6). Liic do ta c6 bang bien thien sau: 2 in s/ Tif do suy ra xet cac kha nang sau: m>6 0 L u c n a y t a c o : min f„,(x)=:f,„ m Luc nay taco: .i & - i : / ' „ . . ; A Tilfd6suyrahe(l)(2)c6nghiemkhivachikhi: X '.• 3 7 X Vie't lai he da cho d\id\g sau: 2 " 2 iL ie uO nT hi Da iH oc 01 / u f (x) = x ^ - m x + m < 0 " . 3. Neu ^ < ^ < 3 (ttfc la 1 < m < 6). Luc do ta c6 bang bien thien sau: x^-mx + m<0 HUdngdangi&i ' 1 <»m< 2 . Tim m dc he c6 nghiem. 4 m <1 TiJf do xet he sau: " • " i ' •••• Bai 6. Cho hg bat phi/dng trinh: 2 - 0 •0 <=» 2sinx(-2sin^ x + mcosx + 3) > 0 2sinx(2cos^ x + mcosx +1) > 0 . (1) K h i x G 0 ; ^ , thi sinx > 0. 2 "$y tren € , thi (1) <=> 2cos^x + mcosx + 1 > 0. (2) Chuyen ai BDHSG Join gii tr| Ifln nhaft g\& tr| nhd nha't - Phan Huy Khai m ()<'i_.i T NhU the dan den vice xet he sau: Dat t = cosx. Khi X e 0 ; ^ , t h i l e [0; 1]. J 2 Bai toan da cho trd thanh: Tim m de bat phiTdng trinh: Im(l) = 2t^ + mt + 1 > 0 dting vdi moi t e [0; 1]. min f„,(t) > 0 . - o4(l) = 0 <^ t = Ket help lai suy ra: m > . TOd6 x^t cac kha nang sau: l a cac gia tri ciin tim cua m. iL ie uO nT hi Da iH oc 01 / Dieu nay xay ra khi va chi khi: m +1 8 .- .-. Tif do ta c6: min L,(t) = 1„, jV/»fl« xet: Ciich giai trcn di/a vao phUdng phap tim gia trj k'ln nhat va nho nhat ciia ham so phu thupe lham so m. 1. Ne'u —— > 1 (tu'c la khi m < - 4 ) . Liic nay ta c6 bang bic'n thicn sau: 4 Qic ban hay so sanh each giai tren vdi each giai bai nay bSng phi/dng phap siir dung gia trj k'ln nhat vii nho nhat ciia mot ham so phan tht?c (khong c6 t 4 C(i) f, - tham so) da Irinh bay trong bai 5 phan B, §3 chi/dng 7 cuo'n sach nay. 0 . Va tinh hicu qua cua tifng phu'dng phap, xin danh quyen blnh luan cho c^c r / ban. Bai 8. Cho ham so: r„,(x) = 4x" - 4mx + m ' - 2m. Xct trcn mien - 2 < x < 0. Tim m dc m +3> 0 m>-3 he vo nghipm. TiTdo loai kha nang n;iy. m < 0 (tuTc lii khi m > 0). Liic nay ta c6 bang bicn thicn sau: ,4 r m 0 I ~ T &r / / / 0 < m < 1 ( H J T C la K -4 < m < 4 Liic nay ta c6 bang bicn thicn sau: m t 0 4 / + 0 4(t) / \ / 0 + / Vay: I 0). - / / w. khi •> y ww Ncu 'Jfii'f" m 0 ' / TCr do suy ra m > 0 thoa man ycu cau dc bai ra. 3. -2 fa / / (XKI X ce / / * Tif do dan den cac kha nang sau day: m 1. Ncu ~ > ^ ^ m> Liic nay la c6 bang bicn thicn sau: ok / + Hiiiln^ ddn giai '' f Ta c6: f,„(x) = 8x - 4m =» i„,(x) = 0 <^ x = ym . bo 0 .c om /g 2. Ncu min r|„(x) = 2 s/ m <-4 up m <-4 ro Tac6: Ta ()() m' - 2 m - 2 = 0 m >0 " „i = l _ 7 ^ 4 ^ m = l + V3 m>() ^ • Ncu — < - 2 (<=> m < - 4 ) . 2 Lijc nay ta c6 bang bie'n thien sau: 323 rgaii y i a u | l u i r i i i i j i vd g i d in iniu i i i i t l l - m / / 0 'in + Khi do he: 0 ? Ta co: 1 „, (u) = 4u - 2m =^ 1„, (u) - 0 < > u -m i m V Do m > 0, ncn ta co: 0 < m y < m , vay c(') bang bicn thicn sau: / m u 0 m" - 6 m + 1 6 . m y 0 / m-^ - 1 6 m + 16 = 2 m < -4 . r min r,„(x) = 1,,, ( - 2 ) -2 - 4 < m < 0). Liic nay ta c6 bang bicn thicn sau: % Tir do suy ra: / / 1 m -6m-6 min I„, (u) = i„ () • •,.•,« ( K fUiiH m^ —6m — 6 ^ ^ = - 2 m . Nhu' vay: -4 ^ m < 0 •2' .\ (I 0, v > 0 => x + 1 = va y + 2 = v'. bo Dal u = v^x + I ; \ sjyTl m' ce u +v=m (1) w. fa Bai toan da cho tnt lhanh: Tim m dc he: u" + v^ = 3m (2) CO nghicm. u > 0; v > 0 (3) r,„(u) = 2u- - 2mu + m- - 3m - 3 = 0 (4) , , • Dicu nay xay ra khi va chi khi min r,„(u) < 0 < max f,„(u) (6) (' II III -3m-3>0 Tim m dc bat phu'ilng irinh dung vc'^i moi x. Hiding dan gidi Viet lai bal phuTttng trinh diftKi dang: fn,(x) = 2m(sinx + cosx) > - 1 - m' (1) > Tijr do suy ra (1) dung vdi moi X G M khi va chi khi: (2) mini" ( x ) > - l - m X€E > 1 Xet cac kha nang sau: min f()(x) = 0. Luc do (2) co dang 0 > - 1 Vay m = 0 thoa man yeu cau de bai () 0 , thi minf„,(x) = - 2 V 2 m (dosinx + cosx <-sl2 Vxe R X6R Tir do thay vac (2) CO: co nghiem. ( 5 ) •::^:-:.::f fi"«l±:^<„,<,+y,7. x€R Tir (1) c6: v = m - u. Do \ ^ 0 < u < m. Thay v = m - u vac (2) ta co: 2u- - 2mu + nr - 3m - 3 = 0. Tir do biii loan da cho lai co dang sau: Tim m de he sau: i ( ) < II < m r— Bai 10. Cho bal phiAtng Irinh: n r + 2m(sinx + cosx) + 1 > 0. 1. Neu m = 0, thi f,i(x) = 0 ww Tir(3)va(l)suyram>(). 5 , Do la cac gia Iri can llm ciia m. om /g v ' ^ i +7y + 2 = m „ , . Tim m dc he co nghicm. X + y = 3m HuYfiif/; dan giai .c Bai 9. Cho he phifitng Irinh: s/ r,„(x)=r,„ up min ro Vithc: :2 • Ta Bay gid ket h(1p vrJi (6), la co: -2m veil J J. C = m - 3m - 3. -2^m>-l-m^ m^ -2V2m + l > 0 m>0 m>0 mN/2 + l ^ m>0 (X m < V2 - 1 m>>y2+1 Cty TNHH MTV DWH Khang Vigt 3. N c u m < 0, thi do sinx + cosx < sjl Vxe max =»minr,,,(x) = 2v/2m. = max m — 3; m + 8 m -3 xeR [jsleu m < Thay viio (2) va c6 he: 8 + 2>/2m + 1 > 0 m<0 m<0 m>-V2 + l ^ ; m<-V2-i I-V2 m < 0 max >N/2 m<-V2-l Kct hdp l a i la t o : + 1 '^^ <^ max Do la cac gia Iri can tim cua tham so m. Bai l l . C h o h a m so f ^ ( x ) = | - 2 x ^ + x + m | vc'ii x G [ - 1 ; 1]. liyx^^Uv, - = max ok fa w. max r,,(x) = max g , n ( - l ) | : -Kx- I m-3 g,„(l) 'It m + 8 X e l cac kha nang sau: I: t 16 ! 1 m + 8 3-m . . . . / 25 23 1 16 16 J . Tif bang bien thien suy ra: m + ^ > m - 1 > m - 3. T u do suy ra: = max -A. A^Aan jcef; Ta chtfng minh (*) nhiT sau: ww ra: 1 23 16 max L ( x ) = -lf!( ro „ max f,„(x) -1,fyi'(&iA'iMtm Hiiihig ddn gidi 3-m. 1 2f; m + - , neu m > — h . : . 8 I6;V 2. Tur phan 1/ suy ra bang bien thicn sau (iheo m) s/ max l|„(x) dal gia Iri be nhat. -l 0, do do: 8 8 l-V2 72 + 1 1 iL ie uO nT hi Da iH oc 01 / 2V2m>-l-ni^ m <-V2-1 , thi m - 3 < 0 va m + - < 0, do do: 8 m —3; m —1 m + (*). n'^iW : Ne'um+-<0=>m-1 >m-3>0 8 \ , m-1 < m-3 . • (1) b. N e u m + - > O l h i 1 m + - • 1 Ne'um+- > m - 1 >0 8 • Neu m - l < 0 = ^ m - 3 > m - l a. N c u m > 3, Ihi m - 3 > 0 va m + - > 0, do do: >m-i (2) \'Hv, (3> 327 "UFiuyen ae BPHSG'iHaw aia trr ion mrarra gra Tri n r o m a r - T n a n Huy Knar T i i r ( l ) ( 2 ) ( 3 ) s u y r a : max | m - 3 | ; | m - l max | m - 3 m + - Kct hdp lai suy ra 0 < m < — va m = - 2 la cac gia Iri can tim cua tham so'm. m+- V a y (*) diTdc chuTng minh. ^ Cach giiii tren diTa vao phtfcfng phap tim gia Iri be nhat cda ham so phu Bai 12. Tim m de ba'l phiTcJng trinh sau: m"x + m(x + 1) - 2(x - 1) > 0 dung V(3j thupc tham so fn,(x) = (m^ + m - 2)x + m moi x e [-2; 1]. 1^ Xoi each giai sau day: HU(fng dan gidi Vi r„,(x) hoac la ham hang so f^(x)= ( m ^ + m - 2 ) x + m > - 2 c : > U x ) > - 2 nc3n luon luon dong bien (khi m^ + m - 2 > o ) , hoac luon luon nghich bie'n (1) De (1) dung vdi mpi x e [-2; IJ dieu kicn can va du la: _mm/n,W>-2. ,, „ (khi m^ + m - 2 < o ) . V i vay f j x ) > - 2 V x e [-2; 1] khi va chi khi . (2) , Xet cac kha nang sau: 1. Neu ra^ + m - 2 > 0 =4> f,„(x) = m' + m - 2 > 0, nen ta c6 bang bien thien sau: x -2 1 (khi m^ + m - 2 = o), hoac la ham bac nha't iL ie uO nT hi Da iH oc 01 / Viet lai bat phtfrtng Irinh da cho difdi dang lu"(ing du^dng sau: l„,(-2)>-2 -2m^ - m + 4 > - 2 f,na)>-2 m^ + 2 m - 2 > - 2 -2 0 0-2 m = l 3. Neum^ + m - 2 = 0<^ m = -2' maxf„(x) >2. V xeR HUdng ddn giai IT Ta c6: fm(0) = 1 + mcos —, 4 (1) It fm(7t) -2 , om m^+m-2>0 .c m'' + m - 2 > 0 fa ^ m \ w. Vay (2) up min f ^ ( x ) = f „ ( - 2 ) = - 2 m ^ - 2 m + 4 + m . -20 = 1 + mcos 4 = 1 - mcos —. 4 T i r ( l ) ( 2 ) suy ra: U O ) + UTC) = 2. ff„,(0) 1 xeR X6K : — 1 +m ,2, IT 2, / 18; 2; ' 11) COS TT c h i n g han cimg diTrtng thi do x, y G Z =^ x > 1, y > 1 =^ 4x + 5y > 9. D o la IT dicu v6 l i . i i v i A V ' sC'ik) V i le do D = D , U D j , trong do: • TT + - = _ l + msin^^^'^'"*-7''''^'^ 2 4j I \ __2 (7) '-b'mn^ii 2 ,/iiiy< Ro rang: 1„ M a t khac n c u ( x ; y ) G D thi x va y trai dau. That vay neu x va y cung dau, / - +4,-= ,2 = - 1 + m cos TV \K •• tn^'f Afn j f l i i / :|ti6b'. n'rinf ^aw'.wirt, D| = {(x; y): x > 0, y < 0; x, y e Z v^ 4x + 5y = 7 } , Thco nguyen l i phan ra, ta CO: 061:, 2, m i n P = min (x;y)€D min (x;y)eD, Tif 4x + 5y = 7 TCr (7) (8) (9) suy ra: n i i n L , ( x ) < - 1 ^ (min ('...(x)] ^ > 1 . ( 1 0 ) XGM Do x, y I VxeE Cong lirng vc (6) (10) di den: (min 1„, (x) + max f,„ (x)) >2 . t • > Ta LLfONG GIAC /g HINH HQC, ro up §3. GI6I THIEU MQT SO BAI TOAN GIA TRj L6N NHAT. NHO NHAT s/ Do TRONG SO HQC, om Ciio'n sach nay diinh dc Irinh bay cac bai loan gia Iri Idn nha't, gia tri nho .c nha't thi/ctng gap trong dai so' vii giai lich. G P; m i n (x;y)GD2 P X> x= 7-5y = 2-y + (2) x = 3-5t 0; y < 0 nen suy ra: ^ ~ ^ ' > ° = , t < i = > l= 0;-l;-2;...(dotG Z ) 41 - 1 < 0 4 Ti^ (3) va do t = 0; - 1 ; - 2 ; . . . . nen suy ra ifng v d i t = 0, ta c6: min (4) P=::12. Khi (x; y ) e D2 i h i P = -5x - 3y. khao khac. Tuy nhicn trong muc nay, chung toi muon gicKi thicu vc'Ji cac ban D o x < 0 ; y > 0 , t i r ( * ) ta c6: ce bo ok hoc, hinh hoc, liTi.Jng giac so di/dc chiing loi trinh bay trong mot cuon chiiycn fa w. ww (5) ^"-'^'<"=^1>1=> t = l ; 2 ; 3 ; . . . ( d o t G 4t - 1 > 0 5 Z) A. Vai bai toan ve gid tri Ida nhat, nho nhat trong so hoc L u c n a y P = 1 3 t - 12. Bai 1. Cho P = 5|x| -3|y|,d day x, y thuoc tap hdp D diTdc xac dinh nhiTsau: Til (6) va do t = 1; 2;... suy ra tfng vdi t = 1, thi: { ( x ; y ) : x , y e Z va 4x + 5y = 7) ' ' ' • ' > ' f ^ (tuTc la D la t$p hdp cac nghiem nguyen cua phiTdng trinh 4x + 5y = 7 T m i gia tri be nha't ciia bieu Ihufc P khi (x; y) e D. llUdng ddn gidi - N c u (x; y) e D thi chac t h a n x 9^ 0; y 9^ 0. That vay ncu trai lai gia siir chaiUhan X = 0 => 5y = 7. D o la d i c u v6 l i VI y 6 Z. (3) K h i x = 3 - 5 t ; y = 4 t - 1, la c6: P = 5x + 3y = 12 - 13t Cac bai loan gia trj k'ln nhat, nho nha't trong cac ITnh vifc khac nhu" trong so nhSng ITnh viTc nay de cac ban tha'y dufdc tinh da dang eiia kHp biii toan nay. 4 y =4t-l (x;y)€D| mot so bai toan lien qiian den vice t i m gia tri Idn nha't, gia trj nho nhii't trong (1) Z ^ ^ - i ^ = t, v d i t e Z . Tif do ta co: ( I •< • D 6 la d i c u phai chiyng minh. 330 J: Khi (x; y ) e D, t h i P = 5x + 3y. xeK D= -i i t ; / w ; ; , D2= { ( x ; y ) : x < { ) , y > 0 ; x , y e Z v a 4 x + 5y = 7 } . d i . ^ - . , iL ie uO nT hi Da iH oc 01 / L a i c6: f„ f max f„, ( x ) f > 1 ( 6 ) min (6) •- ^ I (7) P = I 3 . 1 - 1 2 = 1. (x;y)eD2 T i ) f ( l ) ; (6); (7) di den: min (x;y)eD P = min{l2;l} = l ^ ^ ^ I '''if, x = 3-5.1 [y = 4 . 1 - 1 ' » x =-2 y =3 II xet: Bhi toan la sir k c t hcJp giffa nguyen l i phan ra trong bai toan t i m gia tri Idn nha't, nho nha't va phep giai phi/dng trinh nghiem nguydn trong so hoc!. 331 Cty Chuy8n d l BDHSG Toan gii t r j Idn nhat va gJA t r j nh6 nhaft - Phan Huy Khjii Trifdc h c l la chiJng minh rang irong each phan tich da chpn n h t f t r e n thi tich B a i 2. Cho m , n la so nguyen diTtJng. Tim gia tri nho nhat ciia bicu thiJc: P = |l2"' - 5 " | . ' 'a,a: ••• ak chi' g o m nhCTng ihiTa so nguyen to 2 va 3, va khong eo qua 2 ihiTa so ' ' ' nguyen to 2. T h a i vay: HUifiig dan gidi . . r , i J.y< ,{} > i, (ij).;t/ , V i 12"'" la so chan, con 5"" la so tan cung hang 5, nen suy ra: lhay lich aiaj ... a^ tang Icn. D i c u nay mau ihuan v d i u'eh da cho la Idn r,*; < nhii'l. 12'"" - 5 " " la so Ic : ^ 1 2 ' " " - 5 " " / 2 . D o 12"'"/5,con 5"" 15 Lai - 12"""-5""/5. thay 12'"" 13, con 5""/3 lhay hai so 1, a bnng so I + a va de y rang do 1. a < 1 + a, nen sau khi 5 • iL ie uO nT hi Da iH oc 01 / :S. , 12"'" - 5 " " / 3 . x - * f^^^fc 4 <=> 2b - 4 = b <=> b > 4). V a y long chi ehiVa cac so hang 2 va 3. x' - (3) . ojljir Vdi tinh chat nhuf vay ihi de dam bao aiaj... a^ idn nha'l, ta can phan ti'eh (mod 13) (4) s/ R o r a n g t a c o : 12'"" = ( - 1 ) " ' " Ta hoac la 12'"" - 5"" chia cho 13 thi diT 12. (5) Tim gia trj Idn nhat va nho nhat cua P. fa w. rkngXi 2.2.2, vay Tif(3)tac6: =12 Trong each phan tich da cho khong cd so hang b > 4, vi la ed ihe lhay b bang hai so hang 2 va b - 2. Rd rang 2(b - 2) > 4 ( T h a i vay, vl 2(b - 2) ^- K e t hdp iai cac d i c u tren, tuf (2) suy ra: |l2"'" - 5 " " 1=1 12'"" , la lay mot so hang a tiiy y khiie, a > 1 ( d l nhien no ton tai). Luc nay la That vay gia s i l r ( l ) khong diing, ttfc la ton tai hai so nguyen diTtJng m,,, n„ s,^, " =-V:«' "''•.•.tfl- m , Trong each phan lich da chpn, khong cd so hang 1 v i neu eo so hang 1 i h i TrU'c'Jc hot la chufng m i n h rang v d i m p i m, n la so nguyen duftJng thi P > 7. (| j cho | l 2 ' " " - 5"" I < 7. TIM HH MTV DVVH Khang Vigt . trong d d : •'/«':;. , _ ^ v a x , + X 2 + . . . + X3„ = 2011 X2...X3(, . (x|-1,X2,...,X29, x^o+1). Vf/V Khidddeyrangdo xj'>l:^x, > 2 = i > x ^ - l > l . •Mat • khac: - 1 ) + ^ + ... + x ^ + ( x ^ + 1 ) = 5^ + x^ +... + X j , , = 2011, nen bp so m d i nay cung thoa man y e u cau de b a i . Cfng v d i bp so nay ta cd: •L... 333 Chuyfin BDHSG Toan gia Iri I6n nha't va gii tr| nh6 nhS't - Phan Huy KhSi P = (^-l)5^...X^(x3o +l) = X2...X29(x3o-X, Do x ^ > ( ) Vi = 2r29, va x 7 < x ^ ^ P > P . P = 67^'".68. (1) T o m lai la co: max P = 67^*^68 K c t hop l a i ta co: m i n P = 1982 va max P = 67'''.68. '"''^ ' p a i 5. Cho k la so nguyen di/Ong > 3. nha't. V a y mot dieu k i e n can de P dat gia t r i nho nha't la X| = 1. ''* ' Tir do lap luan hoan toan ti/rfng tif suy ra X2 = X3 = ... = X29 = 1 cung la dieu Tim gia tri Idn nha't ciia ham so f(x, y, z) = xyz trcn mien k i e n can de P dat gia t r i nho nha't. D = {(x; y; z) : x, y, z nguyen dU'Ong va x + y + z = k } . 1 (2011 - 29) = 1982. \' ^ iL ie uO nT hi Da iH oc 01 / 1.1 trong do: , C3„-X,>2 < , max X j ... Ta se chu'ng minh rang: X o - Zo < 1 • v - ^ " " ' 'r? (2) That vay, neu trai lai ta co: Xo - Zo > 1. X3,, . Bay g i d xet bo so m d i sau day: ^^ K h i do chi co the xay ra 3 triTdng hdp sau day: (x, + I ; x 2 ; . . . ; x 2 y ; x 3 0 - l ) ' ' '* s/ a. NcII x„ = y„ > z,, + 1. V i x„ + y„ + z„ = k nen ta co: x„ + (y„ - 1) + (z,, + 1) = k up Chu y rang do X3(, - x, > 2 ma x, > I => Xj,, - x, > 1 /g ro khac de thay: (x7 + l ) + x^ + ... + x ^ + ( x 3 o - l ) = x , + . . . + X30 = 2011, om nen bp so m d i cung thoa man yeu cau de bai. l?ng v d i bp so'n^y ta c6: .c P = (x, + l ) x 2 . . . X 2 y ( x 3 „ - l ) bo ok + l)(x3(, - 1)- X, X3(,] = X2 ... X29 (X3„ - X, - 1). ce >2=>P>P w. fa V a y bat dang thuTc nay chtfug to r^ng bp so (x, .....Xjojchifa lam cho P dat gi;i ww tri nho nha't. V i the dieu kien can de P dat gia tri Idn nhat la X30 - Xi < 2 hay xw - X i < 1 => X30 - Xi e {0; 1} tdrc la: " Ta C n g v d i bo so nay ta c6: P = x, f(x,y,z) = r ( x , „ y , „ z , ) ) = : x „ y „ z „ . (Chu y d day x„ > y„ > zo) X, + X 2 + ... + X 3 o = 2 0 1 1 . Tir X 3 „ - x , gidi xyz phai dat gia tri Idn nha't tren D . Gia suT: X, y S z. V I D la tap hilu han phan tuT (x, y, z) ncn dl nhien ham so f(x, y, z) = 2. X e t bai toan t i m max P P- P = 0gi>] Do vai tro binh dang giffa x, y, z nen khong giam tdng quat co the cho rang V a y m i n P = 1982 <=> c6 29 thuTa so bang 1, va m o t thifa so bang 1982. Mat ^ ,f' HUdng dan 29 thiira so' Xet bo so' ( x , , X 2 , X 3 , , ) K i i i n g Vi^t Nhu" vay chon day chting han 29 so' bang 67 vi mot so bang 68 t h i : +l). (1) chuTng to rang bo so (x,, X j , . . . , X 3 „ ) khong lam cho tich P dat gia trj nho Tir do suy ra: m i n P = [ivvil Cty TNHH M I V ^ hoac X3(, = X| hoac X3(, = X | + 1 Nhir the dieu kien can de P dat gia tri Idn nhat la trong 30 so thi khong diTdc CO hai so bat ki nao trong chung lai chenh nhau qua 1. D i e u nay co nghla la phai c6 I so bang a va 30 - t so b i n g a + 1 (1 < t < 30) sao cho: t a + ( 3 0 - t ) ( a + 1) = 2011 => 30a + ( 3 0 - t ) = 2011 (*) Do X() = y,, > Z|, + 1 va Zo > 0 => Xo, yo - 1, Zo + 1 cung nguyen di/Ong, turc la: , (Xo, y , , - l , z „ + 1 ) e D . Mat khac: f(Xo, y,) - 1, z,, + 1) = X|,(y„ - l)(z„ + 1) = x„y(,z„ + x„(y(, - z,, - 1). Do X(i = y,, > z,i + 1, nen tiTtren suy ra: f(x,i, y,) - 1, Z(, + 1) > X(,y„Z|, = f(x„, y,,, z,,). Bat dang thijTc thu diTOc mau thuan vdi (1). Vay trong trufOng hOp a. khong the xay ra. 0. Neu X(, : > jin^'y > y„ > z„. X e t bp ba nguyen duTdng ( x „ - l , y o , z 0 (do X(, > Z() + 1) nen ta co: l'(X() - 1, yo, z<, + 1) > x„y„z„ => f(x„ - 1, y„, z„ + 1) > f(x„, y,,, z,,). Ba't dang thiirc nay cung mau thuan vdi (1). ,1 i , Vay trong trU'dng hdp b . khong the xay ra. Ne'u X() - 1 > y,, > Zo. L a p luan nhiT tren cung suy ra mau thuan ' ' T o m lai gia thiet Xo - z,, > 1 la sai, vay (2) dung Tir (2) suy ra chi cd the xay ra hai kha nang sau: De thoa man (*) c6 the chon a = 67, t = 29 (vi 30.67 + 1 = 2011) 335

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