TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com1 CHNG1: PHNGTRNHLNGGIC Phngtrnhlnggicl mtphnrt quantrngca chngtrnhTHPT,nlinquann rt nhiuvn sau nymtrc mtl bn thytrongcc thitt nghipv i hc lc no cngc cu:Giiphngtrnhlnggic Vychngta bt u nghincu vn nynh( rt hay) TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com2 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com3 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com4 BI TP TNG T TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com5 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com6 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com7 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com8 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com9 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com10 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com11 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com12 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com13 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com14 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com15 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com16 .. Ph lc: MT CH NH V GII PHNG TRNH LNG GIC Phi ni ngay l k nng m ti trnh by n cc bn y khng phi l 1 phng php vn nng .Ni chung l lm qui g c pp vn nng .. C chng ch l nhng nguyn tc t duy c bn, ni nh Polya tc l nhng "li i c l" .. Nhng li i c l gip ngi ta bt m .. Th thi! Ta bt u bng mt biton i S nh sau; Bi Ton: Gii phng trnh: x3-3x+2=0. Nhn xt: -y l 1 biton d vinhnghc sinhkhngqu mt cn bn .. Bn thnglm g tion nh: 1/ Nhm thy x=1 l No c bit. 2/ Ly x3-3x+2 chia cho x-1ngoinhp! 3/ c x3-3x+2=(x-1)2.(x+2). 4/ Kt lun:nghimphng trnhl x=1 v x=-2. -Bn lmnh th l rt tt! tuynhinvic "quiu luyn"v "t tin"vinhngknng my mc nh vy l c 1 vn rt chi l nguyhin t duy ca bn ... Trn quan im "hydit"bi ton thqu thc chiu php ca bn l "tnbo"v v"thiunhno" nn ikhivic bn mt cng tht n cngch em litc dngz .. Nith tc l ta ch "Dc"viligiixung1 t .. "Gian manh ths n miTO ci thnh .." cc c ch dy th lg? TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com17 -Tithc vi nhsau: 1/ Nhm thy nghimx=1. 2/ S 1 lmt "nghimp" (ngoii onh nhau vs 1). Tuy nhinkhivc i tnhton ths 1 cha "tin"bng s 0 (s ny bsida c s no khc m(nhn) phin cng sidatheo ..). Thm na cngtr vi0 thcn gkhoibng!(c philmci giu) ... 3/ n x ban u t na bng 1 tc l x=1+t (t:mt t). Ci "t"y lt na s bng 0. ( ) 4/ Nu gp mtphng trnhi s m c nghimx=0 th... nhn t c gil t .. ci chung.. Chngmnhch mt cng.. lm ci gc :-SS Li gii: t x=1+t phngtrnhtr thnh (1+t)3-3.(1+t)+2=0t3-3t2=0 H h!! By gi ta ni n phng trnhlng gic ... Ai tnggiicc phng trnhlnggic (hay phng trnhloic khjkhc ..) cng phichp nhn l quanhi qun lita phi: i-Bin i phngtrnh ... -Bt nhn t -a ngay v phngtrnhc bn (i tkhid tnhth lm) -Lm xut hinn ph... ii-S dngcc nh gi c bit c bit ha s so snh trong phngtrnh(ci ny tight vti dt .. BT :-S). Trongcc "tr"c bn trn ci tr bt nhn t ikhirt chil phintoi.. Gid m bn gp kiu: GPT: sin2x-3.cosx=0 thch nilmgi! Nhngnu m gp kiu2sin2x+(23-3)sinx+(2-33)cosx=6-3 th tnhhnhl rt khc ... By gi"btchic" ci vo vt trn kiaviphngtrnhi S ta c vi nghnh sau: 1/ Do s xut hinca 3 nn ta lmm quanhcc nghimc bit lincan n 3 v 6. 2/ Vi ci Fx500thbn chcn mt 1 pht .. m ra nghimc bit c=+6=76.3/ S t x=t+76 a phng trnhv n t. 4/ H hp hyvngnhn t s t .. ci chung Li gii: Phng trnh2sin2x+2.(3.sinx+cosx)-3.(sinx+3.cosx)=6-3 2sin2x+4sin(x+6)-3sin(x+3)=6-3 2sin(2t+73)+4sin(t+86)-6sin(t+6)=6-3 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com18 sin2t+3cos2t-2sint-23cost+6cost-6+3=0 (sin2t-2sint)+6.(cost-1)+3.(1+cos2t-2cost)=0 2sint.(cost-1)+6.(cost-1)+3.(2cos2t-2cost)=0 2(cost-1).(sint+3cost+6)=0 Ngon! Th d 2: Giiphngtrnh:cos3x-cosx+3sinx=0. Nhn xt S "khugi"ca 3 lmchng ta "m mng"n nghim"loanhquanh"vigc 3;3 th richng mnh"gim"phinghimc bit x=-6. V th l ... Li gii: t x=-6+t phngtrnhtr thnhsin3t+2sint=05sint-3sin3t=0. .. TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com19 CHNG2: GIIHNV HMS LINTC Phn 1. Cc dng bi ton v tnh gii hn hm s Kin thc c bn cn nm ( khngnm outlun,hehe ) Giihn 0( )lim( )x xf xgx, trong( ); ( ) f x gxcng dn ti0 khix dn ti 0xc gil giihn dng 00. y l dng giihn thng gp vnhay! @ Cc bn c thy thiu iu g khng? l khi nim v gii hn y, cc k hay nha, v bi vit ny ch nhm luyn thi i hc nn nhng bi ton i su vo gii hn khng c chng ti cp nhiu! Ni chung gii thnh tho nhng bi ca i hc ch l phn ngoi ca gii hn thi nh, hp dn cn ng sau. Bn ng ci nhiu v lm bi kim tra c im cao nha, bnh thng thi! Khi nim gii hn dy s:( )1 2( ) , ,..., ;...n na a a a =c giihn l s a nu bt u t mtch s no , mis hng nau nm trong mtln cn bt kca im a, tc l ngoilncn hoc ch c mt s hu hn s hng hoc khngc s hng no ca dy. Khiulimnna a=Ln cn:vd nh cnh nh bn c vinginh khc chn hn, vnglncn ca ng bng sng hng okch! Khi nimny phhp vichng trnhhc sau ny Khi nimgiihn hm s c xy dng da trn khinimtrn:x a th( ) ( ) f x f a hay lim ( ) ( )x af x f a=Mt s giihn c bn c dngtrong cc kthi: 0sinxlim 1xx= ; 01lim 1xxex = 0ln(1 )lim 1xxx+= ; 10 01lim1 lim(1 )xxx xx ex | |+ = + = |\ . 220 0sin 1 coslim 1; lim , , 0ax 2x xax ax aa R ax = = e = ( * )( ci ny c c v sao? ) @ Sau y l cc bi ton hay v thng gp v gii hn Th d 1. Tm giihn 3021 8limxx xTx+ = ( HQGHN 1997 ) Li gii.( bn ang ci v: tilmn qu nhiu ) Trc ht ta thm bt 2 trn t ritch ra nh sau 30 02( 1 1) (2 8 )lim limx xx xTx x + = +ti sao lil s 2? n y chc chn bn s lm theo cch nhn lnglinhip,( ko hay cho lm,nu cn lnhn ). Bn ch nh: TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com20 t 31 ; 8 u xv x = + = th 2 31; 8 ; , 2 x u x v u v = = . Nh vy chngta c th vit: ( )2 3 22 2 2 22 12 2 1 2 1 3lim lim lim lim1 8 1 4 2 3 12 4u v u vuvTu v u v v = + = + = + = + + +(cch giiny c ci hay l chngta loii nhngdu cn cng knh,khiibinthnh i cn ca giihn). u im hn qua bi ton sau: Th d 2. Tm giihn 5 412 1 2lim1xx xTx + =( HSPHN 1999 ) S: 710T = , cch giihon ton tng t bi1 ci bn th xem nhen! Cu hit ra l lm sao tmc h s t do thm bt vo ( trong bi1 l s 2 y ). Bn xem bi ton tngqut t rt ra suy nghnh: ( ) ( )limn mx af x gxTx a= s bn cn tml: ( ) ( )n mf a ga =nu iu ny khngxy ra thc nghabn ang i mt vimt biton kh hn! Bn nhnlithd 1 v 2 iu ny c ng khng. Th d 3. Tm giihn 201 cos cos2limxx xTx=Li gii.Bin iv s dngcngthc ( * ) 2 2 2 20 0 01 cos 1 os2 1 cos 1 os2lim( cos . ) lim limcos .x x xx c x x c xT x xx x x x = + = +2 21 2 52 2 2= + =Tngqut: 2 2 2201 cos 2...cos 1 2 ...lim2xxco x nx nx + + +=Th d 4. Tm giihn cos os320os2limx c xxe c xTx=Li gii.Bin inh sau cos os32 201 1 os2lim( )x c xxe c xTx x = +bn ang gp lidng thm bt lcu nh! Vy 1 2T T T = + vi cos os3 cos os31 2 cos os3 20 01 1cos os3lim lim .x c x x c xx c xx xe e x c xTx x | | = = |\ . cos os3cos os3 2 201 1 os3 1 coslimx c xx c xxe c x xx x | |= |\ . o( )cos os3cos os30 01 1lim lim 1; cos os3x c x tx c xx te et x c xt = = = TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com21 Th d 5. Tnhgiihn0ln(sinx cos )limxxTx+=Li gii.Bin i 20 0ln(sinx cos ) ln(1 sin 2) sin 2lim lim( . )2 sin 2 2x xx x xTx x x + += =( nh hc cng thc nhan cc anh em ) o 0 0ln(1 sin 2) ln(1 )lim lim ; sin 2sin 2x tx tt xx t + += =o 0 0sin 2 sinlim lim ; 22x ux tu xx t = =Vy1.1 1 T = =( ch phitrnhby cn thn cc php i cn thgiangh michp nhn vnh ko c vit 0sin2lim 02xxx=) Th d 6. Tm giihn 3lim1xxxTx++ | |= |+\ . Li gii.Thc hinphp bini 3 2lim lim11 1x xx xxTx x+ ++ | | | |= = + ||+ +\ . \ . t 2 11 x t=+, ta c2 1; x t x t = + +vvy 22 1 121 1 1lim1 lim 1 1t tt tT et t t + +| || | | | | |= + = + + = | ||| |\ . \ . \ .\ . Th d 7. Tm giihn( )3 3 2 2lim 3 1xT x x x x+= + +Li gii.Thc hinphp binin gin ( )3 3 2 2lim( 3 ) ( 1 )xT x x x x x x+= + + (ci ny gicho ta s thm bt ng hem,nhngn mang ng cp cao hn ri) o ( )( )23 3 23 3 2 3 2 233lim 3 lim3 3x xxD x x xx x x x x x+ += + =+ + + +23 33lim 13 31 1 1xx x+= =| |+ + + + |\ . o 221lim( 1 ) lim1x xxDu x x xx x x+ + += + =+ +2111lim2 1 11 1xxx x+ += = + = TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com22 Vy 32T D Du = =Th d 8. Tm giihn T=0sin(sinx)limxx ( H Bch Khoa HN 1997) cibp Li gii. 0 0sin(sinx) sinxsin(sinx)lim lim . 1sinxx xx x = = Thd 9. Tmgiihn ( )201 coslim1 1xxTx= Li gii.Ta thc hin binisau ( ) ( )2 2 2 22 2 20 02sin (1 1 ) 2sin (1 1 )2 2lim lim1 1 1 1x xx xx xTxx x + + = = + 2 2202sin (1 1 )2lim 142xxxx+ = =| | |\ . ( bn trnhby ch ny r ra nh! ) Th d 10. Tnh giihn sau201 os2limsinxc xTx x=( N 1997) Li gii. 22 20 0 2 01 os2 sin 2 sin 2 4lim lim lim . 4sinx sin sin 2x x xc x x xTx x x x xx | |= = = = || |\ . |\ . Chl nhngphp binikho lo bn ! Th d 11. Tm giihn sau01lim.osxT x cx= ( H GiaoThng1997) Li gii.Bi ny phidngpp nh gining hn lnguyn lkp ca vaistrat ( hicsch gio khoa 11cho nhbn cnh nh mn ri,ghinhmc gb con b qua nhen ) Tm tt pp:Gis ta c :o( ) ( ) ( ); ux f x vx x D s s e( tp xc nhca ba hm s ny ) olim( ) lim( ) ;x a x aux vx Dieu a D = = eThlim ( ) ;x af x Dieua D= e( uikhuya qu ri,nh c b ng yu gh, ngymai vittip,b i ngi m khuya ri mm mm,tith by, 29-3-2009) TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com23 Tipn: ( )1 1 1cos os cos 1 x xc x x x xx x x= s s s ( ) ( )0 0 01lim lim cos lim 0x x xx x xx | | s s = |\ .01lim cos 0xxx | | = |\ . Th d 12. Tm giihn sau01 1 sin3lim1 cosxxTx += ( HQG HN 1997) Li gii.Bin inh sau 0 01 1 sin3 1 1 sin3lim lim1 cos 1 cosx xx xTx x + = = ( v1 sin3 0 x + >) 3 2220 0 04sin 3sin sinx4sin 31 oslim lim lim 4sin 31 co s 1 os 1 cosx x xx x xc xxa x c x = = = 20lim1 cos 4sin 3 3 2xx x= + = Th d 13. Tnh giihn sau sinxlimsinxxxTx=+ ( HGT 1998) Li gii.Tip tc tng vinguynlkp ca vaistrat, bt u no sinxsinx 1 1 sinx 1 sinx; 0 lim 0xxx x x x x x x= s < < = =( cc bn nn thuc giihn ny nh ) V vy sinxsinx11sinxlim lim lim 1sinx sinx sinx1 1x x xxx xxTxxx x | ||\ .= = = =+ | |+ + |\ . Th d 14. Tnh giihn sau3 202 1 1limsinxxx xT+ += ( HQG HN 2000) Li gii.Cc bn nh li tng thm bt nh, chngta th mibitl c giic hay khng? 3 20( 2 1 1) ( 1 1)limsinxxx xT+ + =3 20 0( 2 1 1) ( 1 1)lim limsinx sinxx xx x + + = A B = +o ( )( )( ) ( )0 02 1 1 2 1 11 2l im l im . 1sinx2 1 1sinx 2 1 1x xx xAx xx + + += = =+ + + + TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com24 o ( )( )( ) ( )3 3 2 2 2 230 0 3 3 2 2 2 2 2 23 31 1 ( 1) 1 11lim lim 0sinx( 1) 1 1sinx ( 1) 1 1x xx x xxBx x x xx | |+ + + + + |= = = |+ + + + + + + +|\ . oVy1 T = . Bn ang ngh i sao m bin i kho lo qu vy tri, liu tui c lm c nh vy khng? Huhu . Ti xin ni kh vi bn: tht n gin, ch cn bn luyn tp tt ( ko cn lm nhiu ), nhng bn phi tht s hiu ngha v hnh thc ca cc gii hn c bn Th d 15. Tnh giihn2203 coslimxxxTx=( HSP HN 2000 ) Li gii.vn vicu ni: khngth sao bit tr khibn c cng lc hngkhngnhnthy ngay, hehe! 2 2ln32 20 03 cos ( 1) (1 cos )lim limx xx xx e xTx x + = =( hehe kh kholo nh ) ( )22ln32 20 02sin112lim .ln3 lim ln3.ln3 242xx xxexx = + = +| | |\ . ( ch ny tivittt nh,cc bn phithngqua mt bc tam gili cn ca giihn nh trnhby ci thdtrc ) @ wow wow i bng qu n cm ci , mi nh ba ting m c nhiu y cng kha kh ri, h h chng ta chun b s dng v kh nguyn t tiu dit vng quc gii hn nha! ( 11h55 am, 29 3 2009 ). Nhng thi chng ta cng tip tc vi nhng v d khc ! Th d 16. Tnh giihn sau201 cos cos 2 cos3limxx x xTx=Li gii.Bn hy nhnlithd3 xem sao, tikhng nhchng c miquanh vinhau,hehe, cn quan h nh th no bn t suy nghnh! s:72T =( nu bqu bn c th linh vichng tiqua tinhbantoan@yahoo.com) @ Tiptc vi nhngbi c tngthmbt nh!( khkh) Th d 17. ( Mt biton cc kquan trng ) Tnhgiihn sau01 ax 1limnxTx+ =vin nguyn dng Li gii.Thc hinphp ibin: TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com25 t1 axny = +Khi y0 x th1 y vth em c : ( )( )1 21 11 1lim lim1 1 ... 1n n ny yy yT a ay y y y y = = + + + +1 211lim... 1n nyaay y y n = =+ + + + Lm mt vi ng dng ca n nha! ( Bn hy tng qut kt qu trn via thc bc n: 1 0( ) ...nnp n a x a x a = + + +nh, c nghal lcny x c thay bng p(n) ) Th d 18. Tnh giihn sau3 401 2 1 3 1 4 1limxx x xTx+ + + =Li gii.Trc ht biton ny kh hay v kh,vinhngcn thc nhvy chng ta s lintng n kt qu m chng ta c trong thd 17,vy philm sao khim bi ton ny c cha tchca tiba du cn khc bc. Ta s dngbinisau y3 41 2 1 3 1 4 1 x x x + + + =31 2 1 2 1 2 1 3 x x x x + + + + +3 3 41 2 1 3 1 2 1 3 1 4 1 x x x x x + + + + + + T y l ngonn qu rinha! 3 430 001 2 1 1 3 1 1 4 1lim lim1 2 lim 1 2 1 3x xxx x xT x x xx x x | |+ + + = + + + + + | |\ .3 40 0 01 2 1 1 3 1 1 4 1lim lim limx x xx x xTx x x | |+ + + = + + | |\ .2 3 42 3 4= + +@ Hon ton bn c th to ra nhng bi ton nh mun ca bn t nhng tng c bn, th mi bit ton hc l mun mu mun v! Th d 19. Tnh giihn sau4limtan 2. tan( )4 xT x xtt= ( HSPHN 2000 ) Li gii.nhm nhm ta thy nu m th 4xt=vo thTkhngxc nh. cho gnta t 4a xt= 20 0 0 0os2 sin os2 1limtan 2 . t ana limcot 2. tan lim lim4 sin 2 cos 2cos 2a a a ac a a c aT a a aa a at | |= = = = = |\ . TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com26 Phn 2. Cc bi ton v tnh lin tc v c o hm ca hm s Hm s lintc tiim 0x x =khiv chkhi ( )0 00lim ( ) lim ( )x x x xf x f x f x+ = =o hm ca hm s( ) y f x =tiim 0x x =lgiihn huhn ( nu c ) ca 000( ) ( )limx xf x f xx x, khiul 0'( ) f x . Ch o hm tn tikhi 0 00 00 0( ) ( ) ( ) ( )lim limx x x xf x f x f x f xx x x x+ = ( bn hy hiutht r vo hm nh ) nh l: Nu hm s( ) f xc o hm ti 0xth lin tc ti im . ( iu ngc li khng phi lc no cng ng ) @ Sau y chng ta cng gii mt s bi ton v : tnh lin tc v o hm . Cc dng ny ch nhm kim tra mt tit, thi hc k dnh cho khi 11 hoc dnh cho k thi tt nghip thi tin s (he), nhng ( ti ang nhn mnh ) nu ngi ra mun th h c th bin chuyn thnh nhng bi ton hay, kh kh, thng c mt trong cc k thi hc sinh gii. Gii phn ny ta hiu hn v l thuyt t c th ng dng tnh lin tc gii phng trnh, ci ny mi quan trng v thi i hc thng c! Th d 20. Tm m hm s sau lintc tiim1 x = : ( )( )32 2 1( ) ; 1 11; 12x xy f x xxmx + = = = = Li gii.Trc ht cn hiulintc timt im l nhth no , ci ny chngti trnhby trongphn lthuyt tm tt ca phn ny! Hm s lintc tiim 0x x =khiv chkhi ( )0 000lim ( ) lim ( ) lim ( )x x x xx xf x f x f x f x+ = = =Bi ton chngta ang xt ng vi 01 x = , bn cng nn bitrng hm s ca chngta ang cn xt l hm haiquy tc, mt iu rt quan trngna l khi1 x ng nghavixcha bng 1 hay1 x = . Vi nhn xt ny chng ta bt u giinh sau: Xt giihn 3 31 1 12 2 1 2 1 2 1 1 4lim ( ) lim lim1 1 1 3x x xx x x xf xx x x | | + + = = + = | | \ . (?) vinhng gbn c trongnhngvd phn1 thvic tnhgiihn ny chcn l tr tr con! Bn thy mtcht g khhiu, ng,hy c li mt ln na tht k.Ngita yu cu tmm hm s lintc ti im x=1vy nn ta c mt githit cc kquan trngl hm s ny lin tc ti im x=1,iu ny tng ng vi1lim ( ) (1)xf x f=43m = . Hy nh y l biton tmm v cho hm s ca chng ta lintc ti im x=1ri.Bi ton ny khc vibi ton xt tnhlintc ca mt hm s TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com27 Th d 21. Tm m hm s sau lintc tiim 3xt=t anx 3cot3( ) ;3;3xxy f x xmxttt = = == Li gii.Nh vy cc bn chcn trnhby nh sau Xt giihn 3 3t anx 3cotlim ( ) lim3 x xxf x axt tt = = ( mt kt qu no cc bn t tmha ) V hm s lintc ti 3xt=nn : 3lim ( ) ( )3 xf x ftt= a m =( nu bn vn thy kh hiuthnn ngmnghlinhnggmnhmi c rihy tip tc nha, ton linquan n lthuyt hay lm) @ Hehe, bn ang t tin, ui d t m c g khng hiu, nu nh t nhng bi gii hn ban u m chng ti cp n bn c th to ra c nhng bi lin tc nh th ny, th chc chn bn hiu ri y! No chng ta cng qua mt s bi tnh o hm m phi dng n nh ngha mi mong c solution p! Th d 22. Tnh o hm hm s sau ti im0 x =t anx sinx21( ) ; 00; 0ey f x xxx = = == Li gii.Bn c tht s hiumnhcn lmgkhng? Xt giihn tanx sinx tanx sinx3 30 0 0( ) (0) 1 1 t anx sinx 1lim lim lim .0 t anx sinx 2x x xf x f e eTx x x = = = = ( ch i cn giihn nha,khuya qu riang lmbin,thngcm, cc bn lmcho ra kt qu nh trn nghen) Vy 1'(0)2f = @ Liu c ai trong cc bn t ra cu hi ny : a sao hm s cha bit c lin tc hay khng m tnh o hm tri . Hehe, vic c o hm ti mt im s lm hm s lin tc ti im ch khng phi lin tc ti mt im th hm s c o hm ti im ( lm n nh dm ). Th d 23. Tnh o cc hm s sau a. 20; 0( )1sin ; 0xf xx xx= = = ti im x=0 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com28 b. 0; 0( )1 cos; 0xf xxxx= = = tiim x=0 Li gii.a. 0 0( ) (0) 1'(0) lim lim sin 00x xf x ff xx x = = = (?) V sao giihn ny bng khngchng ta hy dngnguynlkp nh, xem livd 11 b. thd ny cc bn lmtng t. @ Cc bn c t ra cu hi l v sao chng ti li t phn ny sau phn gii hn khng, uhm, v khi thnh tho gii hn ri vic tnh cc gii hn h qu nh trn mi d dng c. Chc cc bn may mn, i ng y ( 30-31/3/2009), hm nay khuya qu ri, ngy mi li n! Th d 24. Tm a, b hm s :2( ) ; 0(1)( )ax 1: 0(2)bxx a e xf xbx x + c o hm ti 00 x =Li gii.Gis( ) f xc o hm ti 00 x =th( ) f xlintc ti 00 x =0 0lim ( ) lim ( ) (0)x xf x f x f+ = =20 0lim(ax 1) lim( )bxx xbx x a e a+ + + = + =( ch x dn tipha tri 0 thhm s theo quy tc (1) vx dn tiphaphi 0+ thhm s theo quy tc (2),ai mong lunv khinimhm s thnn n linhen ) 1 a a = = 1 a =thay vo hm s ban u ta c ( )21 ; 0( )1; 0bxx e xf xx bx x + iu kincn v ( ) f xc o hm ti 00 x = ( xem liphnlthuyt) : 0 0( ) (0) ( ) (0)lim lim0 0x xf x f f x fx x+ = ( )20 01 11 1lim lim0 0bxx xx ex bxx x+ + + + = 1 b b = (?) ( cc bn tnhra nghen ) 12b =Vy 11;2a b = =@ Hy nh lc ny cc bn c cng lc kha kh v gii hn ri nhe, nn vic tnh trn xin dnh cho cc bn. Nh vy gii bi ton dng ny ta cn c vo 2 iu kin: th nht c o hm th phi lin tc, iu kin th hai l iu kin tn ti ca o hm ( xin nhc li rng bn cn phi hiu l thuyt mt cch tht cn k ) Th d 25. Tm a, b hm s a) 222 ; 2 1( )ax ; 1x xf xx bx s s = + + > c o hm ti 01 x =TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com29 b) 2; 1( ); 1xxf xax bxs = + > c o hm ti 01 x =Gi. Vi hai biton ny cch giihon ton tng t,ton hc ihichngta phisuy nghtht nhiu,tihivngcc bn chqua mt tbitp m s tip thuc dng ton ny! Sau y xinnu ln p s cho cc bn kimtra gip a.3; 3 a b = =b. 1 1;2 2a b = =@ Sao chng ta khng t to ra nhng bi ton c h s lnm sinh ca mnh hay l ca ngi yu ngi thn ca mnh nh? Chc cc bn may mn v tht hi lng vi nhng bi ton mnh to ra! Th d 26. Chng minhrng hm s1xyx=+ lintc ti00 x =nhngkhngc o hm ti01 x =Li gii. Trc tinchng ta chng minhhm s ny lintc ti 00 x =( )00l imf ( ) lim 0 01xxxx fx= = == Vy ( )0limf ( ) 0xx f=nn i ca ny lintc ti 00 x = By gichng ta chng minhhm s ny khngc o hm ti 00 x = ( ) ( )( )0 00lim lim0 1x xx f x fx x x = + ( ) ( )( )0 0 001lim lim lim 10 1 1x x xx f x fx x x x+ + + = = = + + ( ) ( )( )0 0 001lim lim lim 10 1 1x x xx f x fx x x x = = = + + Vy r rng hm s ny khng c o hm ti 00 x =Th d 27. Cho hm s 3 21 sin 1; 0( )0; 0x x xf xx + == = Tm o hm ca hm s ti0 x =(HSG TnhBng A Ngh An 2008 2009) Li gii. cng ging nh nhng v d trc 3 220 0( ) (0) 1 sin 1'(0) lim limx xf x f x xfx x + = =( )20 23 2 23sin'(0) lim1 sin 1 sin 1xx xfx x x x x= + + + + ( )2 03 23s inx 1'(0) lims inx. 01 sin 1 sin 1xfxx x x x= =+ + + +. Vy( ) ' 0 0 f = . TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com30 Nhn xt v bi ton ny: tuy l thi hsg nhng rt mm, khng qu kh khn phi khng em Vt??? Th d 28. Cho hm s 211 os ; 0( )0; 0x c xf x xx | | = |=\ . = tnh o hm ca hm s ti0 x =( Hu 2003 2004) Li gii. Cng khng kh khn g 0 0 0 0( ) (0) 1 1'(0) lim lim 1 os lim lim. os 00x x x xf x ff x c x x cx x x | |= = = = |\ .(?). Vy'(0) 0 f = . PH LC:CHUYN KH CC DNG V NH KHI TNH GII HN ( RT HAY ) A.PHNG PHP GII TON: Cc dng v nh: 1.Gii hnca hms dng: ( )( )0lim 0x af xgx| | |\ . oNu f(x) , g(x) l cc hm a thc thc th chia t s ,mu s cho (x-a) hoc (x-a)2. oNu f(x) , g(x) l cc biuthc cha cn thnhnt v mu cho cc biu thc linhp. 2.Gii hnca hms dng: ( )( )lim xf xgx | | |\ . oChia tv mu cho xk vik chn thchhp. Ch rng nux+ thcoinh x>0,nux thcoinh x1xx xf x + s= . Tma hm s c giihn khix dn ti 1 v tmgiihn . Gii Ta c : ( ) ( )21 1lim lim 3 3x xf x x x = + = . ( )1 1lim lim 1x xx af x ax+ + += = + Vy ( )1lim 3 1 3 2xf x a a = + = = 11. ( )( )( )2322 2 22 2 48lim lim lim 2 4 122 2x x xx x xxx xx x + += = + + = . Dng00| | |\ .. 12. 333 2 33 3332 1 2 112 1 1lim lim lim12 1 2 1 22x x xx xx xx x xx xxx + + + = = =+ ++. Dng | | |\ .. 13. ( )( )( )22223 3 3 3 3 3223 123 12lim 3 1 lim lim. 1 . 1 . 1x x xx xx xxx xx x x x x xx + +| | + = = |+ + +\ . 2331 12 36lim 61 11xx xx| | + |\ .= = =+ 14. ( )( )( )2 22 222 23 33lim 3 lim lim3 3x x xx x x x x xx x xx x xx x x x x x+ + ++ + + + ++ ++ + = =+ + + + + + 2 223 313 1lim lim lim2 1 33 31 1x x xxxx xx x x x x xx xx+ + ++++= = = =+ + + + + ++ + +. Dng ( )TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com32 .. Phn 3. ng dng nh l lagrange trong vicgii phng trnh ( Dnh cho cc cho vicbi dng HSG ) Th d 29. Chng minhrng vimis thc, , a b cthphngtrnh cos3 cos2 cos sinx 0 a x b x c x + + + =(1) lunc nghimtrongkhong( ) 0;2tLi gii.Ln u tintigp bi ton ny vo nm lp10,tht s ligiilmcho tithchnht ca biton ny ldngnh llagrange Xt hm s 1 1( ) 3 asin3 2 sin2 sin cos f x x b x c x x = + + trn on [0;2 ] t . R rng hm s ny xc nhv lintc trn [0; 2 ] t , c o hm ti miim thuc( ) 0; 2t . Ngoira (0) (2 ) 1 f f t = = . Theo nhllagrange,tn ti( ) 0;2 d t esao cho ()( ) ( ) 2 01 ( 1)' 02 0 2f ff dtt t = = =cos3 cos2 cos sin 0 a d b d c d d + + + =iu ny c nghad lmt nghimca phng trnh(1) suy ra pcm. ( ch biton ny cn c cch giikhc ) Th d 30. Giiphngtrnh ( )( )cos cos1 cos 2 4 3.4x xx + + =Li gii.V bi ton ny trc ht ta phithc hint n ph| |cos 1;1 x y = e Khi pt cho c dng ( )( ) 1 2 4 4.41 1yyyy+ + = s s(1)ti y cng vic cngcha hng l n ginhn. Chngta s dng tng ca nhllagrange giiphngtrnhny,t nhllagrange chngta thy rng phngtrnho hm cp 1' 0 f=c khngqu k nghimthphngtrnh0 f =c khngqua k+1 nghim,rit bngcch on nghimta suy ra cc nghimca phngtrnh.Nhng phngtrnh dngtinh lny thngc mt trong cc kthihsg! Ta c ( )26.4 ln 4'( ) 12 4yyf y = + ( cc bn kim tra li php tnh o hm ny nh ) ( )2'( ) 0 2 4 6.4 ln4 0y yf y = + =nu ta coi phngtrnhny lphng trnhvibin l4y th r rng n l mtpt bc hai nn ns c khngqu 2 nghim.T (1) s c khngqu 3 nghim,ta on c 1 2 310; ; 12y y y = = =l ba nghimca (1). Rit y giicc pt lnggic c bn 1cos 0;cos ;cos 12x x x = = =suy ra kt qu! @ hichic, 3 ting ri edit trong p hn nhiu ri! 3h30, 3.4.2009 Phn 4. S dng o hm tnh gii hn ca hm s ( bom nguyn t ) Th d 33. Tnh giihn sau 3 5301 2 1lim3 8 2 1xx xTx x+ +=+ + TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com33 Li gii.Vi biton ny m lmtheo nhngcch phn1 thcng c v hicn phihongn! Chngta giibng cch dng o hmt 3 5( ) 1 2 1 f x x x = + +d thy(0) 0 f =t 3( ) 3 8 2 1 gx x x = + +d thy(0) 0 g = , chnhnhngnhn inh ny gicho ta nghv o hm,nh vy chng ta thc hinnhngbini sau, trc ht chia t v mu cho x v a v dng o hm nh sau 0 0( ) (0)( )0lim lim( ) (0)( )0x xf x ff xxTgx ggxx = =00( ) (0)lim0( ) (0)lim0xxf x fxgx gx='(0)'(0)fg=14153454= = Vic tnho hm ti0 x =ca hai hm f(x) v g(x)xindnh cho cc bn! @ Uhm, qua v d ny cc bn thy sc mnh ca phng php o hm trong gii hn cha, tht s cc v d trong phn 1 iu c th gii c bng pp ny! Th d 34. Tnh giihn sau tan2 os168limos12x c xxe eTc xt= ( thy Ph Khnh ) Li gii.t tan2( )xf x e =d thy ( ) 18f t = ; os16( )c xgx e =d thy( ) 18g t = , tisao chng ta li tnhcc gi trti 8t? vi nhnnh ny ta thc hinbini nhsautan2 os16 tan2 os16 tan2 os168 8 8 8 81 ( 1) 1 18lim lim (lim lim ). limos12 os12 os128 8x c x x c x x c xx x x x xxe e e e e eTc x c x c xx xt t t t ttt t = = = tan281lim '( )88xxefxttt = (?);os1681lim '( )88c xxegxttt = ; 818limos12 12xxc xtt=(? ) cc bn t tnhnha, sau nhnggcc bn c hc thvic tnhl d dng! Vy 3eT =Th d 35. Tnh giihn sau 223 2 2 0ln(1 )lim1x xxTe x += + ( th sc s 3 tp chTH & TT ) Li gii.Chngta bini nhsau 22 23 2 2 2 2 22 23 3 2 2 2 2 0 0 0ln(1 ) ln(1 ) 1lim lim . 1: lim1 1xx x x x xx x x e xTx xe x e x + + += = = + + TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com34 Lc ny ta t 223 2( ) (0) 1( ) 1 (0) 1xf x e fgx x g= == + = ri bng cch lm nhcc v d trn cc bn s tnhra c kt qu sau 37T=Th d 36. Tnh giihn sau 3 202 1 1limsinxxx xT+ += ( thy Trn Phng ) Li gii.t3 2( ) 2 1 1 f x x x = + +d thy (0) 0 f =vvy ta c th vit libi ton nhsau (?) Th d 37. Tnh giihn sau 2ax 220.cos 1limxe axTx=vi a l hng s cho trc Li gii.Bin ibi ton nh sau 2 2ax 2 ax 22 20 0.cos 1 .cos 1lim limx xe ax e axT ax ax = =t 2ax 2( ) .cos 1 f x e ax = d thy(0) 0 f =V vy0( ) (0)lim '(0)0xf x fT f ax= = =. Vy T a = ,rt mongcc bn kimtra likt qu t cc bn l ngihon thinbiton. @ R rng s kt hp ca o hm v cc gii hn c bn to nn mt cng c cc mnh, theo ti ngh l c th gii c kh nhiu bi gii hn trong chng trnh, bn c th suy ngh nh ti khng, khi ra ngi ra xut pht t u, ti xin nhc nh cho bn ch t cc gii hn c bn, o hm v nhng php bin i kho lo! Li cui cng: Saukhong9tingchngtihonthnhxongbivitny, v thi gian l c hn v ma thi n gn nn chng ti khng th trnhby ht cc vn ca Gii hn, lintc v ohm.Chngtihivngvibivitngnny,trongkthispticcbnslmtthnvphn ny,vccbnanghc11scthm mt kin thc nh chun b cho vic hc i tuyn. Chng tirtticlkhngththchincnhnhmunlvit thm phn 5, mt phn chng ti rt tmc,lsdngtnhlintcgiiphngtrnh,btphngtrnh,chngminhstnti nghim,giiphngtrnhhm,vphngiihntrongdysNhngchngtikhngcnnhiu thi gianna,licuichngtixinchcccbnthitttrongkthisptivnhngk thi sau ny. D rtcgngnhngsaixtlkhngthtrnhkhi,hivngnhncnhiusnggptccban. Vibivitnychngtihivngkinthcvtonscpcabnthnngycngvngvnghn. XinchovhngpliccbnnhngchuynkhckhichngtirighnhtrngTHPT .3 20 0 0 0(0) 02 1 1 ( ) (0) sinx ( ) (0) sinxlim lim : lim : lim '(0) 1sinx 0 0x x x xfx x f x f f x fT fx x x x =+ + = = = = = TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com35 TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com36 P/S: Trn y l ton b nhng kin thc c bn v dng bi tp ca gii hnhm s em c th lm tt c cc biv gii hn ch cn em t hc v chu kh lm bi tp thi l c th lm tt bi thi i hc c thi.hihi.thn i!! .Anh Sn TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com37 CHNG 3:O HM L thuyt o hm I nhnghao hm 1) o hmti1 im Cho hms y = f(x)xc nhtrongmtlncncax0 khix0 nhnmts giax thy0 = f(x0)nhnmt s giatngngly = f(x0+ x)- f(x0) Nu lim(y/x)tn tithta gi lo hmcahms ftix0. K hiuf'(x0): x0 f'(x0)= lim(y/x)= lim[f(x0+ x)- f(x0)]/x x0 x0 Nu t x = x0 + xthx 0 tc x x0 v ta c: o hm1 pha a) Bnphi b) Bntri 2- o hmtrn mtkhong,mton f(x)c o hmtrn(a;b) f(x)c o hmtimix thuc(a;b) f(x)c o hmtrn[a;b] f(x)c o hmtrn(a;b), f'(a+)v f'(tnti 3-Quan h giao hmv lintcca hms Cho hms c o hmtixo =>hmlintc tikhngc du chchiungcli 4- nghahnhhc ca o hm Cho hms f(x)c o hmtixo thtiim thca n c tiptuyndng: 5/ Cc cngthco hmc bn Cho hmu ,v ta c cc cngthcsau : II. O HM CPCAO- VI PHN 1/ o hmcp cao Gis hms y = f(x)c o hmy' = f'(x).o hmcp n (nuc) ca f(x)c xc nhmtcch quynp nhsau : [f'(x)]'= f''(x)= f(x)(2): o hmcp 2 ca f(x) [f''(x)]'= f'''(x)= f(x)(3): o hmcp 3 ca f(x) [f'''(x)]'= f''''(x)= f(x)(4):o hmcp 4 ca f(x) ........... [f(x)(n-1)]'= f(x)(n):o hmcp n ca f(x) 2/ Vi phn Cho hms y = f(x)c o hmtix0. Gixl s giaca bins tix0. Tchf'(x0).xc gilvi phnca hms f tix0 ngvis giax (viphnca ftix0). K hiu:df(x0)= f'(x0).x Nu lyf(x)= x thdf= dx = (x)'.x= x.Do ta thayx= dx v c : df(x0)= f(x0)dx Tngqut :df(x)= f'(x)dx III- Mt s bi ton v tnho hm TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com38 Vd 1:Tnho hmcp 1 ca Ringv nhngdngo hm thkhngth dngnhngphngphp thngthngc ,Ta cn lnhaiv Sau o hmhaiv lc ta c : T ==> o hmcn tm IV. NG DNG CA O HM 1/ Tnhniu ca hms a/ iu kin cn ca tnhniu Cho y = f(x)l hms c o hmtrn(a;b) f(x)tngtrn (a;b) f'(x) 0, vimix thuc(a;b) f(x)gimtrn(a;b) f'(x) 0, vimix thuc(a;b) b/ iu kin ca tnhniu Cho y = f(x)l hms c o hmtrn(a;b) f'(x)> 0, vimix thuc(a;b) f(x)tngtrn(a;b) f'(x)< 0, vimix thuc(a;b) f(x)gimtrn (a;b) c/ Hm hng f lhmhngtrn (a;b) f'(x)= 0, vimix thuc(a;b) 2/ Chngminhbt ngthc a/ nhl Lagrange:Nu f lhmlintctrn[a;b] v c o hmtrn(a;b) thtn titnhtmts c thuc(a;b) sao cho * nghahnhhc : TrncungABca thhmf,tn titnhtmtimm ti tiptuynsong songvingthngAB * p dng: Nu f'(x)b chntrongkhong(a;b), tc tn ti2 s m,M sao cho : m < f'(x)< M, vimix thuc(a;b) tntic : m< f'(c)< M Suyra : b/ Tnhniu hoc bngbinthin - Kho st s binthinca hmf - Da vo bngbinthin,rtra pcm (c th dngf'' xt du f') 3/ Bin lunphngtrnhv bt phngtrnh a/ Phngtrnhf(x) = m - Phngtrnhf(x)= m lphngtrnhhong imchungcangthng(d):y = mv thhm s (C):y = f(x) - S nghimca phngtrnhl s imchungca (d) v (C) - Da vo bngbinthincahmfv gitrcam, kt luns imchung,tc s nghimca phng trnh - Mt cch tngqut:phngtrnhf(x)= mc nghim m thucMGT caf b/ Bt phngtrnhf(x) < m GiD l MX caf(x) - Nghimca bt phngtrnhf(x)< ml honh cc imthuc th(C):y = f(x)nmding thng(d):y = m - Btphngtrnhf(x)< m c nghim c mtphnca th(C) nmdi ngthng(d) - Btphngtrnhf(x)< m thavimix thucD tonb th(C) nmdingthng(d) ** Tngt vicc bt phngtrnh:f(x)> m , f(x) m, f(x) m TNG HP KINTHC LP 11 Email:anhson.duong@gmail.com39 BI TPO HM Bi 1: Bngnhngha,hytnho hmca hms: y = 2x 1 tix0 = 5 Gii:Tp xc nhD =1x : x2 > ` ) -ViAx ls giaca x0 = 5 sao cho 5+Axe A th -Ay = 2(5 x) 1 +A - 10 1 -Ta c:yxAA=9 2 x 9x+ A AKhi: y(5)= x 0ylimxA AA=( )( )( )x 09 2 x 3 9 2 x 3limx 9 2 x 3A + A + A +A + A + -=( )x 09 2 x 9limx 9 2 x 3A + A A + A +=( )x 02lim9 2 x 3A + A +=13 Bi 2 : Chngminhhmsxyx 1=+ lintc tix0 = 0, nhngkhngc o hmtiim. HD:Ch nhngha:x =x ,neu x 0-x ,neu x Cho x0 = 0 mts giaAx Ay = f(x0+ Ax)f(x0)= f( Ax)f(0)=xx 1AA + yxAA=( )xx x 1AA A + -KhiAx0+ ( thAx > 0) Ta c: x 0ylimx+A AA= ( )x 0xlimx x 1+A AA A +=( )x 01limx 1+A A +=1 Bi 3: Cho hms y = f(x)=2x ,, >neu x 0x neu x2eu x 0-x eux