Nh^ gi^o uu tij - TTi^c siToAn hpc NGUYEN VAN THONG (Chu bifen) Gl^ng vi«n chfnh - Thac sITo^n hpc NGUYEN VAN MINH Ph6 gi^o su - mn si: NGUYEN VAN HIEU ••a LIIYeNGIIUIliniUIICKVTHIIilUHOCSMliN TOAN HOC ii@® Ddnh cho hpc sinh 12 luy^n thi BH-CB ^ Biin so$n theo n0i dung djnh hudng ^ i mdi cua BO Gi^o Due & B^o Tao MZ THLT MIEN TiWH BINH THUAN DVL J My^6' /I5 I NHAXUJTBiiNTONGHQPTHiiNHPHOHOCHlMINH Left noi ddu Chiing toi la cac anh em ruot thjl, muo'n vie't quyen sach nay cho the he sau on tap de chuan bi thi v^o dai hoc, vi bU'dc v^o triTdng dai hoc, ngtfcfi hoc sinh b^l dau mot ddi song mdi vk c6 tiTdng lai tiTdi sang khi chpn dtfdc mot triTcJng dai hoc tot, ch^c ch^n r^ng triT^ng nay se chon diem cao. NhiT vay trong qua trinh on luyen cac em can mot tai lieu tUcfng thich. Dieu nay se diTdc thoa man neu cAc em chiu kho on tap theo cac chuyen de mh chung toi bien soan, ch^c ch^n thi se dai diTpc diem cao. Quyen sdch nay gom 20 chu de trong diem theo cau true de thi cua Bo giao due hang nam, cac vi du duTa ra ttfcfng doi kh6 va c6 hiTdtng dan giai. Tie'p theo la 28 bp de thi cho cac khoi A, B, D cua tri/cfng chuyen Le Quy Don - Da N^ng dilng de thi thijr tiTng nSm qua 66 danh gid diTdc hpc sinh, nit ra kinh nghiem giang day. TriTdng Le Quy Don - Da NSng hang nam deu c6 ti 16 dau vao cac trirdng dai hpc la 100%. Rieng nam 2010 c6 so hpc sinh dat diem cao diTcfc xep thu" 4 trong bang xep hang cua Bp giao due va Dao tao Cic chuyen de chiing toi deu soan tilf can ban den nang cao, dp kho dii de cho cac hpc sinh khd va gioi tU' luyen, neu thong hieu taft ca, ch^c ch^n ring cac em se giai du'pc de thi Dai hpc mpt cdch de dang, nh\i chiing toi da tilfng luyen rat nhieu em do thu khoa cic tru'dng Dai hpc danh tieng. Dieu quan trong hdn nifa vdi each vie't ciia mOt gido vien day chuyen Todn lau nim, nen each gpi md, din d^t mang dam n6t t\i duy nang cao, hpc sinh se diTcJc phat trien kha nSng Toan hpc khi hpc quyen sdch n^y. Mong muon ciia chiing toi l^m the nko de c^c em tif hpc to't mon Toan, va kha nang thi dau v^o cdc tri/dng Dai hpc Idn phai nh5 vko sir kien tri cua cic em "c6 cong mai s^t c6 ng^y nen kim" cd nhan day that diing vay. Quyen sich nay cung 1^ ky nipm 3 nim cic em hpc Toan vdi thay: Bich Lien, PhiTdng Thao, Anh ThiT, Mai HiTdng, Thiiy Hien, Th^o Uyen. Cic em da giiip thay chinh sufa ban thao, mpt cich nhipt tinh trdch nhipm. Chiic c^c em se th^nh cong my man trong ky thi s^p idi Chu bien: Nguyen VSn Thong (To trirdng To Todn trirdng chuyen Le Quy Don - Da NJng) Nhd sdch Khang Vi$t xin tran trgng giai thi^ tai Quy dgc gia va xin Idng nghe tnoi y kien dong gop, de cuan sdch ngdy cdng hay han, bo ich hart. Thu xin giH ve: .„,.^ _ Cty TNHH Mpt Thanh Vien - Djch Van Hoa Khang Vi?t. 71, Dinh Tien Hoang, P. Dakao. Qu^n 1, TP. HCM Tel: (08) 39115694 - 39111969 - 39111968 - 39105797 - Fax: (08) 39110880 Hoac Email: khangvietbookstore@yahoo.com.vn I mill inx-r z^r TWM I\FM 'K^mdt m MJOl C«0 TRONG DI^M GI(3l HAN vA TfNH UfeN TUC C6A MOT HAM S6 I. T6M TAT LI THUYET 1. Cac djnh If ve gidi han ••1 (' Gia sur lim f(x) = L, va lim g(x) = Lj, khi do X-^X() X-+X,) lim [f(x) + g(x)]= lim f(x)+ lim g(x) = L|+L2 x->X() "->"() "-^^o lim [f(x)-g(x)]= lim f(x)- lim g(x) = L,-L2 x->X() "-+"() x-^xo lim [f(x).g(x)]= lim f(x). lim g(x) = L|.L2 x^X() x-^xo ''->X() =' 'v,V' lim f(x) , lim X-^X() f(x) g(x) X^X() _ ^1 lim g(x) L2 vdi X-*X() Neu 3s > 0 sao cho f(x) < g(x) Vx € (xo - e; x,) + s) va ton tai 1 , , lim f(x), lim g(x) thi lim f(x) < lim g(x) x->X() ''-»X() x^XQ x->X{) li • a. NguyinIfgidinank?p: Neu 3E > 0: f(x) < h(x) < g(x) Vx e (x„ - E; X„ + e) \} va lim f(x)= lim g(x) = L thi lim h(x) = L. x^X() x-»X() x-»X() b. Cdc dang gidi han dQc bi$t lim = 1; lim (1 + x)x = e X->X() X X^XO I,,,,,.,.,.;,.,: lim X->X() = e; lim e^-l 2. Gidi h^n dang vd djnh: — = 1; lim x-»X() X x->X() 0 ln(l + X) = 1 P(x):dathu'c, P(x„) = 0 > P(x) a. Dang 1. I = lim —^ vdi <. x^x„ Q(x) [Q(x): da thiTc, Q(Xo) = 0 PhiTdng ph^p: 7 —- . i' 1= Hm lim i^i^^iolW ^ P^^P^ vdiQ,(x«)^0 x^xoQ(x) x->Xo(x-Xo)Qi(x) X^X()Q,(X) Qi(Xo) . J I IWtg, „ |,W Ne'u Pi(x„) = Qi(x,)) = 0 thi phan tich tie'p • 0 P,(x) = (x-Xo)P2(x) Q,(x) = (x-Xo)Q2(x) Qu^ trinh khuT dang v6 dinh ^ la qua trinh khur cac nhan tuT chung (x-Xo)'' se difng lai khi nhan du'dc gidi han xac dinh ttfc la Qk 5^ 0. Khid6I= lin ^=limAW^M^ >'^''()Q(x) x^x„Q,,(x) Qk(xo) 'f(X(,) = g(x„) = 0 OO. f(x),g(x) chtfa can thtfc dong bac b. Dang 2. lim vdi \ x->xo g(x) PhiTdng phap: SuT dung cdc hKng d^ng thtfc de nhan lien hdp 6 tuT va mau nh^m true cac nhan tijr (x - x,,) ra khoi can thtfc. A-B A + B A-B 2"M/A+2n+^^ A + B c. 5.1 = hm vdi ' x^xog(x) [f(x) chtfa can khong dong bac ^^"0 g(x) x^XQ g(x) Bi6n doi 1= lim - x->xo '?yu(v)-ci-rjyv(x)-c g(x) - lim X-+XO .g(Xo) = 0 •^uW-c 5!/v(x)-c g(x) g(x) Den day cac gidi han diTdc tinh theo dang 2. 3. Gi6i hain v6 djnh — 00 PhiTdng ph^p: X6t I = lim ^ vdi P(x), Q(x) la cac da thi?c hoSc cic ham x-»xo g(x) daj s6. Gpi bac P(x) = p; bac Q(x) = q v^ m = min(p, q), khi d6 chia ca va mau cho x" ta c6 ket luan sau: C6ngiyTNi:ii : ;.W//A/.U;.^ ,UI • Ng'u p < q thi ton tai gidi han • Neu p > q thi khong ton tai gidi han 4. Gi6i han dang v6 dinh « - oo Phifdng phap: Bien ddi diTa ve dang gidi han — ' 1 0>t Tim gidi han sau Vx +Vx -Vx lim x-++<»L (x +Vx j-x = lim . = lim J x^+«>^x + ^/^+^/x " 5. Gidi han dang v6 djnh oo . 0 , . Phifdng phap: Du'a ve dang v6 djnh — 00 ~. ^'J Ching han, tim gidi han Vx^ + l-x lim X->+oo . = lim X-»+a> Vx^ +1 +: lim mil tioub fi?. im x l + -^+l X 1 2 6. Gidi h^n dang vd djnh ham lUging giac PhtTdng phap: Suf dung cac ket qua gidi han cd ban sau dSy: ,. sinx , ,. X , • hm = 1; hm = 1 x->() sinx • lim x->0 X sin ax V = lim x->0 X x-»0 sin ax I ax sin ax ,. sinax ,. sinax = a. hm a => hm = a x^O X „ r x->() ax hm sinax = Um ax -^ = ^^lim mil — iiiii .—:—-— — — —• 11111 — — x->()sinbx x-»()bx sinbx b x-+()sinbx b smax a , . = -(a, beR*) bx lim tan ax a sinax ,. tanax hm . = a => hm = a x-»o x x^ocosax ax tanax • Hm tanax ,. ax ax a ,. = hm —• = - hm x->0 X tanax a x->otanbx x->{)bx tanbx b x->()tanbx b bx Sinax hm sinax , sinax ,. ax .ax ^ hmcosbx. = hm—.cosbx. . . x-*() tanbx x->() sinbx x->obx sin Dx b bx Luyfn gidi di truOc kp ihi DH 3 miin Bdc, Trung, Nam Todn hoc - Nguyin Van ThOng 7. Gidi han dang v6 djnh 1° Phifdng phap: a. Sijfdung: lim(l + x)" =e; lim x->() x^+«> b. Xct lim uCx)"*"' CO dang T X^X() • • f 1 1 + - t'l I ,, " US;' fili'd i&ili ffH'T = e ,,,, Bie'n doi lim X-»X() • ''1 l + (u-l)"-' lim (u-ir X \ •+ X 8. Gidi han d?ng v6 djnh cua ham mu va Idgarit: ^ ^^'^ * Phifdng phap: S^rdung lim^ ^ = 1; lim ^ = 1 x-»() X II. BAI TAP MINH HQA x->() X Bai 1. Tim gidi han lim f(x), vdi r(x) = X->1 x^-1 Ta CO lim x^l Htfdng din sial Vs-x^ -2 \/x^ + 7-2^ x^-l x^-l (1) Matkhac: hm = lim - X^l X -1 "-l/v^ 1-x-^ Va lim Vx" + 7-2 = lim -fx^+x + l) = lim \ ' (2) X->1 x^-l x->l X -1 "->! ^"'(x^-lWx^+vf+2^/?77 = lim x^l x^+7+4 1 3 v2 x^+7 x^ +7 + 4 12 (3) •a Thay (2), (3) vao (1). la difdc L = — = -— 8 12 24 . . ™ ,. x'^ -4x^ +4x -3 Bai2, Tim lim— "-^^ x 3x Hifdng din giai Ta CO lim '^^^^ = ^JZ^ = 1 > x^3 (X-3)X x-»3 X . vl^-1 Bai 3. Tim hm x->0 X^ I- Hifdng d§n giai - - - Ta c6: lim (l + x2)-l x (.^2 - = lim — = iim , , = — 3/^1777+^/i77^^1 ''-*"^(i+x2)+^/iT^+i 3 Bai 4. Tim hm x-»() X Hi/dng din giai Ta CO ,. (2N/rT7-2) + (2-^/8^) ,. hm lim x-»() X x^O X = Hm x-^o = hm x->() 2x >/l + x + l) X[4 + 2W^ + N/(8-X)^ ,„;.,. fijsi ^/^^ + 1^4 + 2^ + ^(8-x)2J ""12 12 Bai 5. Tim lim X-»+oo . (x + 1)'"" + (x + 2)'"*' + ••• + (x + 9)'"" + (X +100) l(X) .100 Ta c6 lim - X-»+QO Bai 6. Tim gidi han lim x-»l x'"" + 10x'" + 100'" Hi/dng din giai 2 ^ ,^100 ^ ^^100 ^ j^^N 1 + + + 100 1 + .100 19^ loo^ll^ + ^10+ ^HX. = 100 m lim X->1 m Al-x'" l-xj 1-x"" 1-x" Hi^dng dan giai f_n. i_^] :; , (m, n e N*, m n) , , Luy(n gUU d6 trade thi DH 3 miin Bdc. Trung. Nam Todn hoc - Nguyen Van ThOng lim X-+1 lim x-»l lim t m-(l + x + x^+ + x'""') n-(l + x + x^+ + x""') l-x"" 1-x" (l-x) + (l-x^) + + (l-x"'"') (l-x) + (l-x^) + + (l-x"-') (1 - X)(l + X + x^ + + x"'"') (1 - x)(l + x + x^ + + x"~') l + (l + x) + + (l + x + + x"''^) l + (l + x) + + (l + x + + x""^) l + x + x2+ + x"'-' l + 2 + + (m-l) l + 2 + + (n-l) 1 + x + x^ + + x" ' m(m-l) n(n-l) = ' m m m-1 n-1 m-n 2 2 2 I , • „ %/cosax-Vcosbx ^, ,^ . Bai 7. Tinh L = lim -^^ (a, b, c la cac so thifc khac khong) m, i "-»(' sin cx <• m cac so tir nhien khac nhau Idn hdn 1. '' ^ Hifdng din giai Bdde: lim 1-cos ax a (1) (l-jycosbx)-(l-".ycosax) lim r x^O (cx)" Ap dung (1): smcx = Hm x-*() = -^lim C x-X) 1-cosbx l-iVcosbx l-'^cosax (cx)2 (cx)2 r 1-cos ax l + !ycosbx+ + N/cos"-'bx 1 + "Q^ cosax + + \/cos"' 'ax bM aM 2 n 2 m 2c^ n m , .1- BaiS.Tim lim n->-H» a a a cos—.cos— cos- 2" ;a^0 lAvtdng dSn giai A^aa a a a aa A = cos —.cos — cos = COS .cos r COS-T-COS — 2 2 1 ^ 2 sin . a a /sm — cos 2^^ 2^^ ,n-l 2" 2" 2" j a a a .cos r COS-T-COS— ,n-l 2 ' 2 sin 2sm rcos 2n-l 2"' a a cos-^.cos— 2^ 2 2" 2" sin ^ 2sin—cos— 2 2 sma 2" 2".sin-^ 2" a i'.'Jir] I',- jj ••• 1- sma 2" sma ,,. ., sma 2 sm— sm- 2" 2" III. BAI TAP Ti; LUYEN CO DAP SO. jj,, , cos Bai 1. Tim gidi han sau: C = Hm —; — cosx 2 x-»o sin(tanx) Hifdng din giai n cos —cosx u C= lim^iML x->o sm(tanx) tanx • = = 0 Bai 2. Tim gidi han sau: I = Hm . e cos x-1 x^O x" HiMngdSngiai '"'^ I = Hm x->() = 3-1=2. e-^" -1 Icos^x + cos^x-l •= Hm x->() 3cos^ X. 3x2 ijji'j i • ' Sin X ^ I X ; B^i 3. Tinh gidi han I = lim x-»() 1= lim x-+() e" -1 . e" -Vcosx +ln(l + x^) ' •! x^ 2(l + V^)Yx2'l HrfdngdSngiai ^^"'^ ln(l + x^) + -x , 2 , 4 , Luyfn giil di trudc thi DH 3 miin Bdc, Trung. Nam Todn hpc - NgiQ^n Van Tlidng CAC BAI TOAN KHAO SAT HAM s6 I. T6M TAT LI THUYET: 1. Cac bu6c khao sat si/ bien thien va ve do thi cua ham so Bifdc 1. Tim tap xac djnh cua ham so Btfdc 2. Xet sU' bien thien cua ham so - Tim gidi han tai v6 cifc va gidi han v6 cifc (neu c6) ciia ham so. Tim cic dircJng tiem can ciia do thi (neu c6). l> " - Lap bang bien thien cua ham so, bao gom: Tinh dao h^m cua ham so, xet dau dao ham, xet chieu bie'n thien va tim ciTc tri cua ham so (neu c6), dien . cac ket qua vao bang. ^.^ ^ BMcJ. Ve d6 thi cua ham so. ^ ^ j - Ve cac du'dng tiem can cua do thi (neu c6). - Xac dinh mot so diem dac biet cua do thi, chang han tim diem uon, giao diem cua do thi vdti cac true toa dp (trong triTdng hdp do thi khong c^t cac true toa dp hoSc vice tim tpa dp giao diem phtJfc tap thi bo qua phan nay), - Nhan xet ve do thj: Chi ra true va tam doi xuTng cua do thi (neu c6, khong yeu cau chufng minh). 2. Diem uon cua do thj: la diem U(X(); yo) ciia do thj sao cho lie'p tuye'n tai dd di xuyen qua do thi, tiJc la ton tai mot khoang (a; b) chuTa diem Xo sao cho tren mot trong hai khoang (a; x,,) hoSc (X(i; b) tiep tuye'n cua do thi tai diem U nam phia tren do thi, con tren khoang kia tiep tuyen n^m phia du'di do thi. Ta thU'dng suT dung ket qua sau day de tim diem uon cua do thi: Neu ham so y = r(x) cd dao ham cap hai tren mot khoang chuTa diem Xo, f"(xo) = 0 va r'(x) doi dau khi x di qua diem Xo thi U(X(); f(X())) la mot diem uon cua do thi ham so'y = f(x). 3. Giao diem cua hai do thi. a. Phu"dng trinh (xac djnh) hoanh dp giao diem ciia hai do thj y = l(x) va y = g(x) (ciing ve tren mot mat phing tpa dp) la f(x) = g(x) (1) b. So nghiem cua (1) cung chinh la so giao diem cua hai do thi y = f(x) va y = g(x). Dac biet: PhiTdng trinh (1) cd nghiem (v6 nghiem) khi va chi khi hai do thj citt nhau (khong cat nhau). i c. Ta thudng gap u-irdng hpp phiTdng trinh (1) cd dang f(x) = m, trong dd m la tham so va ham so f(x) khong chtfa tham so m. TriTdng hdp nay ta cd bai C6ng ty TNHH MTVDWH Khang VH toan xdl giao diem cua do thi y = f(x) vdi drfdng thing y = m (vuong gdc vd true tung va c^t true tung tai diem cd tung dp m). 4. SI/ tiep xuc cua hai dudng cong. • a. Dinh nghla: Hai ham so' f(x) va g(x) tiep xiic nhau tai diem M(xo; y,,) neu M la mot diem chung cua chung va hai di/dng cong cd chung tie'p tuye'n vdi nhau tai M. b. Dinh It: Hai diTdng cong y = f(x), y = g(x) tiep xuc vdi nhau khi va chi khi he ff(x) = g(x) phiTdng trinh an x sau day cd nghiem: \ [f'(x) = g'(x) (He phiTdng trinh nay la h? phiTdng trinh \ic dinh hoanh dp tiep diem cua haidirdng). ' - • II. CAC BAI TOAN MINH HQA. Bai 1. Cho ham so y = f(x) = -x' + 2x^ + |x "^^ a. Khao sat h^m so. b. Tiep tuyen cua do thi (C) cua ham so tai goc tpa dp, cat (C) tai diem M. Tinh tpa dp diem M. c. Bien luan theo k so giao diem cua (C) va du'dng thing (d): y = kx. n ; I a Hi^ngdlngiai a. Khao sat ham so vi£a>t, l.Tapxacdinh:D = R - 1 2. Sir bien thien: a. Gidi han: lim y= lim X-»+00 X->-H» J/ ^ o 2 5 -X +2x +-X - -00 3 . 2 5 . lim y= lim (-x +2x +-x) = +oo X->-QC X->-<» 3 b. Bang bien thien y' = -3x' + 4x + —, y' = 0 o X = — hoac x 3 ' 3 • 5 3 • ^ ' X -00 -1/3 5/3 +00 y' — 0 + 0 y ^ -8/27 100/27 ^ -00 Ham so dong bien tren khoang f \_5 to Ob i;-! Luy?n gidi dS IruOc kp thi DH 3 miin Bdc, Trung, Nam Todn hgc - NguySn Vdn Th6n^ Ham so" nghich bien tren cac khoang —;+oo , . . 5 . 100 Ham so dat cifc dai tai x = - va y^^ = — 1 mms^datcirctieutai x y^=—i.''> a»arb myii. oui lil 3. D6^thi (Hoc sinh W ve) = ^ _^^y. ^ g,^,^ ^0^3^,}^ jj^H ^^J^^Q J| + Diem uo'n , -/.v i« ,r/-if>'t) ,, ^ . n 2 /2 46 y" =-6x + 4, y" =0ox= Dieml -;— 3 , .3 2V J hi + x = 0=>y= — ^''^>''< l-^^''''" la diem uon. + NhSn x6t: D6 thi nhan diem I '2 46^ 3'27 lam tam doi xtfng. b. PhiTdng trinh tiep tuyen (A) cua (C) tai O (0; 0) la: mi y = f(0)(x-0) + 0= ^x SB" 0* PhiTdng trinh hoanh do giao diem ciia (A) v^ (C) IS: "x = 0 x = 2 -x^+2x^+-x x o -x^(x-2) = 0c^ 3 3 • x = 0=>y = 0 • x = 2=>y = — 3 r, 10^ '3 c. Phu-dng tnnh hoanh do giao diem cua (C) va (d) Ik: -x''+2x^+|x =kx(l)ox x^-2x + k 3J = 0o x = 0 x2-2x + k 0 3 So' giao diem cua (C) va (d) li so nghiem cua phiTdng trinh (1). ^ Datg(x)= x^-2x + k la tam thiJc bac hai CO A' = k 3 • 3 o Tmyn^ /. A' < 0 o k > - Khi do (1) CO dung 1 nghiem <=> (C) va (d) c6 dung mot diem chung. Tmyrtfi hap 2. A' = 0 <:> k = - Khi do (1) CO dung 2 nghiem <^ (C) vk (d) c6 diing 2 diem chung. H g TrUdng h<tp 3. A'>0<:>k< - + Ne'ug(0) = 0<=>k= - thi(l)c6dung2nghiem. + Neu g(0) 9t 0 o k ;t - thi (1) c6 dung 3 nghiem o (C) va (d) c6 diing 3 3 , ,1)/, . ., diem chung. Bai 2. Cho ham so y = f(x) = -x^ + 3x^ + 3(m^ - l)x - 3m^ -1 (1), m la tham so. a. Khao sat sifbien thien va ve do thi ham s6'(l) khi m = 1. b. Tim m de ham so' (1) c6 cifc dai, ciTc tieu va cac diem cUc tri cua do thi ham so (1) each deu goc toa do O. HrfdngdSngiai .,4,)(„f, . a. Khao sat ham so (Hoc sinh tif giai). „ , . > b. Ta c6 y' = -3x^ + 6x + 3(m^ -1) ; ' rx = l + m lv*i K;vM.!: • .Won y'=0« [x = 1 - m Dieu kien de ham so c6 ctfc dai, cifc tieu la y' = 0 c6 2 nghipm phan biet om?<i0. Goi A(l + m; -2 + 2m'); B(l - m; -2 - 2m') la hai diem ciTc tri. Ta c6 O each deu A va B o OA = OB o (1 + m)^ + (- 2 + 2mV = (1 - m)^ + (- 2 - 2m')^ o 8m' = 2m <=> m = 0 (loai) m = l/2 m = -l/2 ^ ^ ihih In i\) ff Ket luan: m = —; m =-—. If 1;'•nciial'wi,.:;'^^ Bai 3. Cho (C J: y = f(x) = -x^ + 2mx2 - 2m +1 a. Khao sdt s\i bien thien v^ ve do thi tfng vdi m = 1. b. Chtfng minh r^ng (C^) luon di qua hai diem co dinh A, B vdi moi m. c. Tim m de cdc tiep tuyen vdi (C J tai A, B vuong g6c vdi nhau. Hifdng din giai a. Kh3o sdt hkm so i?ng vdi m = 1. (Hoc sinh tif giSi) b. Goi A(x,); yo) la diem codinh cua (C), khi d6: ' ' yo =-x'*+2mxo-2m+ 1; Vm , ^; ; ;); ^ /o,^ 0 2m(x2 -l)-(y„+x^-l) = 0;Vm o 2m(xf, - 1)= y„ + XJ5 -1; Vm xf,-l = 0 yo+x^-l = 0 X„ =1 : • X() = -1 • (1) Ub , E }\ V '= (O'k tr v; + yo=l-Xo ;• ^ x„= 1 z^yo = 0=>A(l;0) , f X„ = -l =^y„ = 0=:>B(-l;0) .gfirMJD m,;;;!: c. Tiep tuyen tai A(l; 0) c6 he so g6c k, = r(l) = -4 + 4m Vrr'Jid .if) S. iiiS Tiep tuyen tai B(-l; 0) c6 he so goc kj = f'C-l) = 4 - 4m . ;,,,; oiJd:-! ? Tiep tuye'n tai A, B vuong goc vdi nhau <=> k|.k2 =-1 IT o (-4 + 4m)(4 - 4m) = -1 o 3 m = — 4 '? 5 m = — 4 Bai 4. Cho ham so y = f(x) = x" - 2mV + 1 (1) vdi m la tham so a. Khao sat ham so (1) khi m = 1. b. Tim m de do thi h^m so (1) c6 ba diem cifc tri 1^ ba dinh ciia mot tam giac vuong can. Hifdng d§n giai a. Khdo sat ham so (Hoc sinh tif giai). b. Ta c6: y' = 4x' - 4m^x = 4x(x^ - m^) rx=o y' = 0o do at X = m X = -m •T-> Dieu kien de do thi ham so (1) c6 ba diem cifc tri li y' = 0 c6 ba nghi^m phan bi?t o m ?i 0. ^ . f = (' i Gpi A(0; 1); B(m; 1 - m"); C(-m; - m'*) la cic diem cifc trj. Tac6: AB= Vm^+m**, AC = Vm^+m**, BC = N/W Dieu kien de AABC vuong can la BC^ = AB^ + AC^ m = 0 (loai) o 4m^ = 2m^ + 2m'' o m* - m^ = 0 <=> Ke'tluanims l;m = -l m = l m = -l Bai 5. Cho ham so y = f(x) = ^ (1) x-1 a. Khao sdt va ve do thi (C) cua ham so (I). b Tim cac diem thuoc (C) c6 loa do la nhiTng so nguyen. c Tim diem M e (C) de long khoang each liT M den hai du'dng tiem can cua (C) nho nhat. HU<?ng d§n giai • a. Khao sdt va ve do thi (C) cua h^m so < jjiiiiodl .v 1. Tap x^c dinh: D = R\ 1} _ ; , 2. Sy bien thien cua ham so a. Gidi han, tiem can: j' o>' ,. 2x-l , ,. ,. 2x-l lim y - lim +oo; Iim y = lim = -oo i ,. . => X = 1 la tiem can duTng cua do thi ham so. hm y= lim y = 2 => y = 2 la tiem can ngang cua do thi ham so. b. Bang bien thien: y' = , < 0 ; Vx e D (x-1)^ -00 —00 +00 +00 • Ham so nghich bie'n tren cac khoang (-oo; 1); (1; +oo) 3. Do thi (Hoc sinh M ve hinh) ^J'^ '' x = 0^y=l ' 1 y = 0 => X = - 2 A' i Nh^n xet: Do thi nhan giao diem 1(1; 2) cua hai di/cJng ti^m can lam tam doi xufng. b. GoiM(x„;yo)6(C),Xo+l. v Tac6y„= ^^""^^2 + —!— '^O -1 X„ -1 , :.g ,.,^.,.,\ Do X(), yo e Z nen ta c6 cac triTdng hdp sau: igj a!?;; • , < ^ • Xo-1 = 1 c^x„ = 2=>y„ =3=>M,(2;3) • Xo- 1 =-1 0x0 = 0 =>yo= 1 =>M2(0; 1) c. Taco: ;;' Z ' <: ' Tiem can duTng (d,): x - 1 = 0 Tiem can ngang (dz): y - 2 = 0 15 U^ngua ae miOc Ic m uii t mien oat;, irung, num ivunrm- lyguyen vun i tivng- Goi M (X(,; y„) e (C). Khi d6: d(M;d,)= |xo-l| '2x0-1 . 1 d(M;d2)= yo-2 Xo -1 -2 Xo-1 Khi d6, tong khoang c^ch \.\S diem M(X(); yo) den hai di/cfng tiem can Ik d = d(M;d,) + d(M,d2) ^ ,1 Xo-1 + x„-l •dn,i„ = 2khi XQ-I 1 Xn -1 = 2 Xn -1 o(xo-ir=10 Xo=0 Xo-2 • xo = 0=>y„= 1 =>M3(0; 1) • Xo = 2 z:> yo = 3 =^ M4(2; 3) 2x Aihly Bai 6. Cho hamsoy = f(x) = x + 1 a. Khdo sdt sif bie'n thien va ve do thi (C) ciJa ham so' da cho. b. Tim toa do diem M e (C), bie't tie'p tuyen cua (C) tai M c^t hai true toa dp Ox, Oy tai A, B va AOAB c6 di?n tich bkng _ 4 Hi/dng din giai a. Khao sdt hkm so (Hoc sinh tiT giai). b. Gpi M (xo; yo) 6 (C); y„ = f(xo) PhiTdng trinh tie'p tuyen cua (C) tai M(xo; yo) la (d): y = f (xu)(x - XQ) + f(xo) y = -x + - 2x^ (Xo+1)' (Xo+1)' Toa dp giao diem A cua (d) vk Ox Ik: A( -Xo ; 0) ( 2x2 Tpa dp giao diem B ciia (d) va Oy la: B 0; ^ De thay AOAB vuong tai O S^QAB =^0A.0B 1-1^2 2x2 Xn. 4 2 "(xo+1)^ 4x^-(Xo+1)2=0: 2xo + X() +1 = 0 2x2-Xo-l = o' Ket luan: M, ;M2(1;1) Xo= Xo=l if III. CAC BAI TOAN TI; LUYfN c6 HU6NG DAN Bai 1. Khao s^t sir bie'n thien vk ve do thi (C) cua hkm so y = f(x) = x^ - 3x^ + 1. 1\i do, bien luan theo m so nghiem cua phiTdng trinh sau: , .v!' -x^-x2+m+2=0 HiMngdSngiai o^Ti, in, ? , • m < -2 hokc m > —: Phi/Png trinh c6 mot nghiem. 2 • m = -2 hokc m = —: PhiTPng trinh c6 hai nghiem. 1 S m -J hod t. 2 • -2 < m < —: PhiTPng trinh c6 ba nghipm. f , , 3 • u) ;.u ~ f ' Bai2. Chohkmsoy = f(x) = x'-mx + m- 1 (1) a. Chtfng minh r^ng tie'p tuyen cua do thi hkm so (1) tai diem Xo = 0 c6 he so goc nho nhat. b. Vdi gia tri nko cua m do thi cija hkm so (1) tiep xuc vdi Ox. Khao skt vk ve do thi hkm so (1) gik tri tim diTpc cua m. c. Xkc dinh m do thi hkm s6'(l) c^t true hoknh tai 3 diem phan bi^t. HUdng dSn giai ' a. f (X) = 3x^ - m > 0 - m = f (0) ''^^ ' => hp so gdc tie'p tuyen tai diem x = 0 Ik nh6 nhat. ^^' * b. m = 3; m = — ,41 li - 4 rm>0 'i'"' ' c. x' - mx + m - 1 = 0 o (x - l)(x^ + x+ l-m) = 0=> 2 .JiifP'M' 4 k- Bai 3. a. Khio sat si/ bie'n thien vk ve do thi (C): y = f(x) = x' + x - 1. b. Gpi Xo Ik nghipm cua phiTPng trinh f(x) = 0. Chtfng minh r^ng: 2Xo - X(, -1<0 M (trr Hi^ng din giai f(x) lien tuc tren ( n 17 . r V 2 = ViEN TIWH BINH THUAN Luyfn giii di trade kp thi DH 3 miin Bdc, Trung, Nam Todn hoc - Nguyln Van ThOng f(x) CO nghiem X(, •(xo- 1) x„ + — 2 < 0 => 2xf)-X(,-1 <0 (dpcm) Bai4.Cho(CJ: y = f(x) = mx' + (m- l)x^+1 - 2m ' a. Tim m dc ham so' co mot diem cifc tri. b. Viet phiTcfng trinh tiep tuycn cua fC, ^ di qua g6'c toa do. >f{ £ > tu » . 2) a. m < 0 hoac m > 1 b. (d,): y = 0; (d:): y = ^x ; (d,): '"^ Bai5.Cho(C):y = f(x) = (x + 1)^ (x- 1)^ /.m - ''X • (xYt ) .S 'a. Khao sat va ve do thi (C) cua ham so'. TiT do, bien luan theo a so' nghiem cua phi/dng trinh x"* - 2x^ + 11 - a = 0. b. Tim m de parabol (P): y = mx^ - 3(m^0) tiep xuc vdi (C). ,, Htfdng d§n giai .^j,,, a. • a < 10 : Phi/cfng trinh v6 nghiem • a = 10 : Phu'dng trinh c6 1 nghiem. _ ^ r • 10 < a < 11 : Phu'dng trinh c6 4 nghiem. j ^'g; • a = 11 : PhiTcfng trinh c6 3 nghiem. • a > 11 : Phifdng trinh c6 2 nghiem. b. (P) tiep xuc vdi (C) khi va chi khi hoanh do tiep diem la nghiem cua he phiTcfng Irinh x^-2x^ + l = mx2-3 4 • m = 4x -4x=2mx 7^ -2 ft) Bai6.Cho(C):y = f(x)= -x'*-3x^+ 2 2 19^'- a. Viet phiTdng trinh tiep tuyen cua (C) tai diem x = Xo. b. Chufng minh rang hoanh do giao diem cua (C) va (d) la nghiem ciia phiTdng trinh: (x - x,,)^ (xf, + 2x„x + 3xf, - 6) = 0 Hifdng d§n giai a. (d): y = (2 x;', - 6x„)x ~~ + 3x1 + x 2x 0 b Phu'dng irinh hoanh do giao diem (2x,l-6xo)x x^3xi,+- = ^-3x +- o(x-xo)'(x^+2x„x + 3x^6) = 0 x/r :,,„• Bai 7. Cho ham so y = r(x) = p a. Khao sat sir bien thicn va ve do thi (C) cua ham so. b Tim m de di/c^ng thang (d): y = 2x + m c^t (C) tai 2 diem phan bi^t A va B. tim tap hdp trung diem I cua AS. t. m + A« j 8- ^ (£)^j '"" ' Hrfdng dSn giai '(.m.utll b —=- = 2x + m o <^ , x + 1 [2x^+(m + 4)x + m + 4 = 0 -f + -fi:^r.: Dieu kien de (d) c^t (C) tai 2 diem phan biet la m < -4 hoSc m > 4. Tap hdp trung diem I cua AB la phan diTdng thang 2x + y - 4 = 0, gom x- rn + 4 m + 4'\ . ik>M 4 nhffngdiemM ; vdim<-4ho$cm>4. , :,,[•••. v 4 2 y . — - 2x — 1 Bai 8. Khao sat va ve do thi ham so y = l(x) = . Tilf do, tim m de phiTdng x + 2 , . 2sinx-l . , , - '* 'sil »tn i • trinh = 2m - 1 c6 dung 2 nghiem x x; sin X + 2 HiAJngdSngiai al =.(x)t mil * Khao sat va ve do thj (Hoc sinh tiT lam) * Dat t = sin x; x e [0; 71J ^ t e [0; IJ n « (,., •^rmd^ fa; ; PhiTdng trinh trd thanh: = 2m - 1 (1) " ' Dieu kien de phiTdng trinh c6 dung 2 nghiem x e [0; n] la phtfdng trinh (1) co dung 1 nghiem t e [0; 1). Khao sat ham so g(t) = =^ iren [0; 1) => - < g(t) <3z:>-<m<2 KHAO SAT MpT SO HAM SO DA THllfC NANG CAO Bai 1. a) Bie't r^ng d6 thi ciaa hhm so'y = (3a' - l)x' - (b' + l)x' + 3c'x + 4d co hai diem cifc tri la: (1; -7), (2; -8). Hay xac dinh tong M = a" + b' + c' + d' b) Chu-ng minh rang do thi ham so y = x' + 2mV + 1 luon citt difdng thang y = X + 1 tai diing hai diem phan biet vdi moi gia trj m. [...]... = 4 C : > log(3x+7)(2x + 3) ^ + log(2, +3) [(3x + 7)(2x + 3) ] = 4 o 2 log (3, +7)(2x + 3) + log(2x +3) (3x + 7) = 3 Dat t = log(3x+7)(2x + 3) x = 3 log2(x + l) = 2 l0g2(X + l ) = - 2 d Phifdng trinh tifdng difdng v d i logy x + ^ + 9 ' ' = 9 ' ' o x = 9 x = t= l 3 — 4 2=1 Phifdng trinh t r d th^nh 2t + y - 3 = 0 o 3( 3x ^3 _ 9(3X)2 ^ 4 3X ^2 ^ = 0 olog^(5''-l) + log5(5''-l)-2 = 0 Dat t = 3 ' > 0 Phifdng... x2+y2- = 3 (2) Bai 19 Cho he phiTdng trinh: Tim a de he phu'dng trinh c6 nghi0m duy nha't a log3 a o + log3 a = 1 + log3 5 log3 m b= ^ a (1) log, a(l + log3 m) = (1 + logj 5).log3 m a m = 5, he phiTcJng trinh trd th^nh b=5 a =5 a o a < > lb=i = log3 a(l + log3 5) = (1 + log3 5) log3 5 log3 a = log3 5 Hifdng dSn giai PhiTcJng trinh (1) tiTdng diTcJng vdi f •* X - a + l o g 3 ( x - a) = l o g 3 ( y -... < - 3 , k h i do (2) ^ y > - 3 dung m nghiem phan biet thl phu'dng trlnh f " ' " " ( x ) = 0 se co khong qua Vay y = - 3 Thay vao (1) m + 1 nghiem x = - 1 hoSc x = 3 Ke't luan: (x; y ) = ( - 1 ; - 3 ) ; (x; y) = (3; - 3 ) Bai 1 Giai phiWng trlnh 32 x+y^3X+2y ^ 3 B a i 7 T i m a de he phU'dng trinh sau c6 nghiem: • ' " *' a) iog2X + l o g 3 ( 2 x - l ) + l o g 5 ( 7 x - 9 ) = 3 b) x ' l o g 3 X =... ABC Htf(?ng d i n giai a) TFa CO' y = x - 3 + — = x-^-3x^+1 0 1 ; X Tird6y' = 0 o x ' - 3 x ^ + 1 = 0 (1) • vi,;.V Dat f(x) = x' - 3x' + 1 thi f(-l) = - 3 , f(0) = 1, f(l) = - I , f (3) = 1 nen theo tinh chat ham lien tuc, phiTdng Irinh y' = 0 c6 3 nghiem XA, XB, XC thoa man dieu ki^n -1 < XA < 0 < XB < 1 < Xc < 3 Tir do suy ra dpcm LuySn gUU dS trade thi DH 3 miin Bdc, Trung Nam ToOn hoc - NguySn Van... Bangxetdau ,r \ f(x) N2 Do 2x^ - X + 3 = 2 X vdi 2x^ - 3x + 1 < 0 o ^ 23 + — > 0 Vx ncn ba't phiTdng trinh tiTdng diTdng 4J '6>;, 9''-72>0 ' jje : H * ' o\>logy 73 0 < log9(3x' + 4x + 2) < 1 o 9" - 7 2 > 1 — Bat phiTdng trinh tiTdng difdng vdi: log3(9'' - 72) < x 3x^ + 4 x - 7 < 0 (3' - 9) (3' + 8) . XQ + 3axf, + 3ax„ + 3a -1 = 3x„ + 3a o (x,) + 3a - l)(xo + x,, +1) = 3( XQ + a). ' 2 :d S„y ra y„ - + x„ +1) ^ XQ + 3a -1 ^ 3( Xo + x„ +1) .3 Do d6, cdc diem uon cua do thi cung . + (m- l)x'-(m + 3) x- 1 =(3x' + 2(m- l)x-(m + 3) ) —+ • 3 m-1 2(m-l)^ 2(m + 3) x + - m^+2m-12 Do ycD = ycT = 2(m-l)^ 2(m + 3) . 9 3 2(m -1)^ 2(m + 3) XcD+- m^+2m-12 :ntfd . 4(x -3) J ^ 1 - -00, lim y = lim ' X ^ 4(x -3) = +00 4(x -3) j D6 thi CO X = 3 la ti?m can dilng va y = j la ti?m can ngang, tarn doi xuTng la I 3; - . V 4j * Bang bien thien