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

515.076 PH561P NGirr.ThS LE H O A N H a ren luyen kl nang lam bai H a No NHA XUAT BAN DAI HOC QUOC GIA HA NOI GIAI PHO NGIfT.ThS LE HOANH PHO GIAI CAC C H U C A N D E B A N (Boi dm!smmmi!mifi cs^G NHA XUAT BAN flAI HOC QUOC GIA HA NOI Ha Nol NHA XUAT BAN DAI HOC QUOC GIA HA NQI 16 Hang Chuoi - Hai Ba Trang - Ha Npi Dien thoai: Bien tap-Che ban: (04) 39714896: Hanh chinh: (04) 39714899: Tonq bien tap: (04) 39715011 Fax: (04)39714899 * Chiu trdch nhiem xuat * * ban: Gidm doc - Tong bien tap: Bien tap: TS.PHAM T H I TRAM L A N HUONG bdi: N G U Y E N KHCJl M I N H Che ban: N H A SACH HONG A N Saa Trinh bay bia: VO THI T H I T A Doi tdc lien ket xuat ban: N h a sach H O N G A N S A C H LIEN KET CAC CHU DE CAN BAN GIAI TfCH 12 Ma so: 1L- 154DH2014 In 2.000 cuon, khd 17 x 24cm tai Cong ty Co phan VSn hoa Van Lang Giay phep xuat ban so: 463-2014/CXB/10-99 OHQGHN, 14/03/2014 Quyet dinh xuat ban so: 154LK-TN/Q0-NXB OHQGHN, v ^ i > In xong va nop iuu chieu quy II nam 2014 Nham muc dich giup cac ban hoc sinh Idp 10, Idp 1 , Idp 12 nam vOng kien thifc can ban ve mon Toan tCr luc vao THPT cho den chuan bi thi Tot nghiep, tuyen sinh Cao dang, Dai hoc, tac gia da bien scan bo sach P H l / O N G P H A P G I A I gom cuon: - C A C C H U D E C A N B A N DAI S O 10 - C A C C H U D E C A N B A N HINH H Q C 10 - C A C C H U D E C A N B A N DAI S O - GIAI T I C H 11 - C A C C H U D E C A N B A N H I N H H Q C 11 - C A C C H U D E C A N B A N GIAI T I C H 12 - C A C C H U D E C A N B A N HINH H O C 12 TL/ nen Toan can ban nay, cac ban c6 the nang cao dan dan, bo sung va md rpng kien thufc va phUdng phap giai Toan, ren luyen ky nang lam bai va tC/ng bade giai dung, giai gpn cac bai tap, cac bai toan kiem tra, thi CLT Cuon C A C C H U D E C A N B A N G I A I T I C H npi dung la phan dang Toan, tom tat kien thac cac chu y; phan tiep theo la cac bai toan chpn vdi nhieu dang loai va mac dp; phan cuoi la hay dap so c6 16 chu de vdi va phUdng phap giai, Ipc can ban minh hpa bai tap c6 hadng dan DO da CO gang kiem tra qua trlnh bien soan song khong tranh khoi nhQng sai sot ma tac gia chaa thay het, mong don nhan cac gop y cua quy ban dpc, hpc sinh de Ian in sau hoan thien hdn Tac gia LE HOANH PHO O CHUD E I T i N H t>ON t>IEU DANG TOAN • TiM KHOANG DONG BIEN VA NGHjCH BIEN Dinh nghia: Hdm so f xdc dinh tren K Id mot khodng, doan hodc nica khodng - f dong hien tren K neu vai moi xi, X2 e K: x/ < X2 =^f(x\) < f(x2) -f nghich hien tren K neu vai moi x/, X2 e K: xi < X2 =>f(xi) > f(x2) Dieu kien can de ham so dffn difu Gid su hdm so c6 dao hdm tren khodng (a; b) do: - Neu hdm so f dong bien tren (a: b) thif '(x) > vai moi x e (a; b) - Neu hdm so f nghich bien tren (a; b) thif'(x) ^0 vai moi x e (a; b) Dieu kien dii de hdm so dffn dieu Gid su hdm so f c6 dqo hdm tren khodng (a; b) do: Neuf'(x) > vai moi x e (a; b) thi hdm so f dong bien tren (a; b) Neu f'(x) < vai moi x e (a; b) thi hdm so f nghich bien tren (a; b) Khi f '(x) = chi tgi mot so hiru hgn diem cua (a; b) thi ket qud tren vdn dung Neu hdm so f dong bien tren (a; b) vd lien tuc tren nita khodng [a;b); (a;bj; doan [a:b] thi dong bien tren nica khodng [a;b); (a;bj; doan [a;b] tuang icng Tuang tu cho nghich bien Phumtg phdp xet tinh dffn dieu: - Tim tap xdc dinh - Tinh dao hdm, xet ddu dqo hdm, lap bang bien thien - Ket ludn Chiiy: 1) Cong thirc vd quy tdc dao hdm y^C =>y' ^0:y=x ^y' = l;y = x" =^y' = nx"-'; ^y=J—(x>0); y= '4^=>y'=— 2Vx n"-47 y = sinx =:>y' = cosx; y = cosx =>y' = -sinx; y= ^ y = tanx =^y = 1r — ; y = cotx =>y , = — -r1— cos X sin X (u + v)' = u' + vV (u-v)' = u'- v'; (u v)' = u'.v + u.v'; u V - u.v ;f'.-f'u.u'x 2) Phuang trlnh luang gidc ca ban: X = a + k27i sinx = sina X = 71 - a + k27i (keZ) X = a + k27i cosx = cos a X = - a + k27r (keZ) tanx ^ tan a c^x a + kTt (k e Z) cotx = cot a x = a + kn(k e Z) Bai toan 1: Tim khoang dong bien, nghich bien cua ham so: a)y = x^-8x + b) y = x^ - 2x^ + x + Gidi a) Tap xac dinh D ^= R Ta c6 y' = 2x - Cho y' = « 2x - = x = 4 +00 Bang biSn thien (BBT) x -00 + y' y Vay ham so nghich bien tren (-oo; 4), dong bien tren (4; +oo) b) D = R Ta CO y' = 3x^ - 4x + Cho y' = « 3x^ - 4x + = o x = - hoac x = +00 1/3 BBT x -00 + 0 + y' ^ ^ y ^ ^ ^ Vay ham so dong bien tren moi khoang (-co; ^) va (1; +co), nghich bi6n tren khoang ( ^ ; 1) Bai toan 2: Tim khoang dong bien, nghich bien cua ham so: a)y-x''-2xl b) y = x'"* + 9x^ - Gidi a) Tap xac dinh D = R y' = 4x^ - 4x = 4x(x^ - 1), y' = x = hoac x = ±1 BBT X -co - y' y -1 + +00 - + +00 +00 ^ - ^ Vay ham so dong bien tren moi khoang (-1;0) va ( ; +oo), nghich bien tren khoang (-oo;-1) va (0; 1) b) D - R Ta CO y' = 4x^ + 18x = 2x(2x^ + 9), y' = x = y' > tren khoang (0; +co) ^ y dong bien tren khoang (0; +QO) y' < tren khoang (-oo; 0) => y nghich bien tren khoang (-co; 0) Bai toan 3: Tim khoang dong bien, nghich bien cua ham so: a)y = 3x-9 b)y = x + - x Giai a)D = R \ Ta CO y' = -6 (l-.r) < vai moi x ^ nen ham so nghich bien cac khoang (-oo; l ) v a ( l ; + o o ) b) TapxacdinhD = R \ Taco BBT: - ^ = ^ ^ - ^ , y ' - < » x = ±V3 x^ X" X -30 y' + +00 -V3 - - + y Vay ham so dong bien tren khoang (-oo; - V3 ) va (V3 ; +oo), nghich bien tren moi khoang (- V3 ; ) v a ( ; V ) Bai toan 4: Tim cac khoang don dieu cua ham so: K^ x+ \ jc +8 2x X -9 Gidi a) D = R Ta c6: y' = -X- y' = -2x + {x'+%f - X - 2x + = X = -4 hay X =2 BBT: X -00 - y' -4 +00 - y Vay ham so dong bien tren khoang ( -4; 2) va nghich bien tren cac khoang -4), (2; + 0 ) (-co; b ) D = R \6 y' = ~^^^ < 0, V x ^ ± (x'-9)' Do y' < tren cac khoang (-oc; -3), (-3; 3), (3; +oc) nen ham so da cho nghich bien tren cac khoang Bai toan 5: Xet su bien thien ciia ham so tren doan, nua khoang: a) y = ^|9-x^ b)y= V x ' - x + Gidi a) Dieu kien - x" > » -3 < x ^ nen D - [-3; 3] Voi -3 BBT: < X < thi y' - X - X ^ , y' = ^ y 0 y' + ^ ^ X = ^ Vay ham so dong bien tren khoang (-3; 0) va nghich bien tren khoang (0; 3) Do ham so f lien tuc tren doan [0; 2] nen ham so dong bien tren doan [-2; 0] va nghich bien tren doan [0; 2] b) V i A' = - < nen x^ - 2x + > 0, Vx => D = R „ , 2x-2 x-1 Fa CO y = —, = , 2Vx'-2x + Vx 2x +7 y'^0 l,y'x< Va f lien tuc tren R nen ham so nghich bien tren nura khoang (-oo; ] va dong bien tren nua khoang [1; +oo) Bai toan 6: Xet su bien thien cua ham so: b)y = a)y Vl6^ a) DK: 16 - x^ > « X' x+2 Gidi < 16 -4 < X < D = (-4; 4) Ta CO 16 > , Vx e (-4; 4) y' = (16-x-)Vl6-xVay ham so dong bien tren ichoang (-4; 4) b) D = [0; +00) Vai x > 0, y' = BBT: +00 - + y' , y' = Vx^ > ^ - ^ = ^ ~ ^ ±1 x^ > hoac x < -1 hoac x > y' y ' > v a y ' = x = hoac x = 2n V i h a m so l i e n tuc t r e n doan [ ; 2n] nen h a m so d o n g b i e n t r e n doan [ ; 2::] b) y ' === - sinx T r e n k h o a n g ( ; n) y' > s i n x < - — < x < — , • r> 7r , _ 571 7t y' < s m x > - < » < x < — hoac — < x < — 6 V a y h a m so d o n g b i e n t r e n k h o a n g (—; — ) , n g h i c h biSn tren m o i k h o a n g 6 (0;5)va(^;H) 6 10 Phuang trinh mat cau la: (S): x + ?1V f 298 + (y-20)^4 = 49 V Cau 9.b Ta c6 z] + = Z|Z2 = > Z|(z2 - zi) = Zj => Zi Z2-Z1 Z2 ZJ' + Z^ = Z1Z2 = > Z2(Zi - Z2) = Z^ =5> I Z2 i I Z i - Z2 Do nen Zi - Z2 Zi - Z2 — Zl Zl Z2 Vay - z, 3_ ^ Z2 Zl Z2 -Z-, + 2' D E L U Y E N THI T O N G HOP SO I Phan chung cho tat ca thi sinh: (7,0 di6m) X -3 Cau Cho ham so y = (1) x+1 1) Khao sat su bien thien va ve thi (C) cua ham so (1) 2) Viet phuang trinh tilp tuyen cua d6 thi (C), biet khoang each tu giao diSm I cua hai duofng tiem can cua (C) d6n tiSp tuy^n b ^ g V2 Cau Giai phuang trinh: (1 + 2sinx) cosx(2x + — ) = — x'+2x'y = Cau Giai he phuang trinh , ( X , y e R) X " +y" + y = Cau Tinh tich phan I = In" X rrdx x(l + lnx)- Cau Cho hinh chop S.ABC c6 mat phang (SAC) vuong goc vai mat phSng (ABC) va CO SA = SB = SC = 2a, AB = 3a, BC = a V3 (a > 0) Tinh dien tich cua mat cau ngoai tiep hinh chop S.ABC theo a Cau Tim tham so m de phuang trinh sa'u c6 nghiem thuc: (Vx + Vx - l | mVx + J-— + l / x ( x - l ) = I Vx-1 II Phan rieng: (3,0 diem) Thi sinh chi dtu-gc lam mot hai phan: A hoac B A Theo chuoTig trinh chuan Cau 7a Trong mat phang vai he toa Oxy, cho cac dilm P ( l ; 1), Q(4; 2) Lap phuang trinh duang thang d cho khoang each tu P va Q dSn d l i n luot bing 2va3 403 Cau 8a Trong khong gian vod he toa Oxyz, cho mat phang (P): x - 2y + 2z + = ck mat c4u (S): (x - if + (y + 3)^ + (z + 3)^ = theo giao tuy§n la duong tron (C) V i l t phuong trinh mat cku (S') c6 tam thuoc (a): x + y + z + v a chua duong tron (C) C a u 9a Giai phuong trinh: - log2(x + 3)^ - log2(4 - x ) ' = 3[1 + log2(x + 6)] B Theo chu"(mg trinh nang cao C a u 7b Trong mat phSng v a i he toa Oxy, cho d i l m K(3; 2) va duong tron (C): x^ + y^ - 2x - 4y + = vai tam I Tim toa d i l m M e (C) cho I M K = 60° C a u 8b Trong khong gian v a i he toa Oxyz, cho hinh hop chu nhat A B C D A , B , C , D , v a i A , ( ; 0; 0), , ( ; 0; 0), Di(0; 2; 0), A(0; 0; 3) Goi M , N, P, Q lan lugt la trung d i l m cac canh A B , B i C i , C i D i , D i D Chiing minh rang cac dikm M , N , P, Q cung thuoc mot mat phang (a) Xac dinh thiet dien ciia hinh hop k h i cat bai mat phang ( a ) , tinh the tich cua khoi chop c6 dinh C va day la thiet dien = 16 C a u 9b Giai he phuong trinh log4(x' + y^) = ^ + log4(xy) Lai Giai C a u 1.1) Su bien thien: y' = • Tap xac dinh D = R \} • y > V x ^ - , suy ham so dong bien tren timg (x + \y khoang (-co; -1), (-1; +cc) Giai han l i m y = ; l i m y = => Tiem can ngang: y = x->-cc X—»-f-co l i m y = +oo; lim v-+(-i) = - o o ^ Tiem can dung: x = - x^(-\y Bang bien thien X -1 -co +00 + y' + +00 y • -00 • ^ -1 Dothi Giao v a i Ox: (3; 0), giao v a i Oy: (0; -3) Do thi nhan I ( - l ; 1) lam tam doi xung / 404 X —3 2) Gia su M(xo; yo) thuoc (C) thi yo = — , XQ -1 Xo+1 Khi phuong trinh tidp tuyen A tai M la: (Xo+1) o Xo + (xo + 1)^ - 8(xo + 1)^ + 16 = (xo + 1)^ = o xo = hoac xo = -3 V o i Xo = 1, phuong trinh A: y = x - 2; V o i Xo = -3, phuong trinh A: y = x + Cau Phuong trinh tuong duong V3 (1 + s i n x ) ( - cos2x - - ~ - sin2x) = - o (1 + 2sinx)(cos2x - V3 sin2x) = cos2x - V3 sin2x + 2sinxcos2x - V3 sinxsin2x = - 2sin^x - V3 sinxcosx + 2sinxcos2x - V3 sinxsin2x = ' = - Vay he c6 nghiem (1; 1), (-1; 1),(V3;-1), ( - V ; - l ) Cau Dat t = + Inx =>dt = - d x , t - = Inx => \ n \ (t - 1)^ X Doi can: x = l = : > t = l ; x = e=>t = Ta c6: I = In" x (t-iy dx = x(l + l n x r 2/ dt = V t 1\ t-r t-21i^t-^ t = 21n2 Cau Ke S H ± A C Do S A = S C nen H la trung dikm A C (1) Vi ( S A C ) ( A B C ) nen S H ± ( A B C ) =i> H A = H B = H C (2) Tu (1) va (2) suy A A B C vuong tai B c6 H la tarn ducmg tron ngoai tiep Do A C = V B A ' + B C ' = 2V3a ^ SH = V A S ' - A H ' = a S H la true ducmg tron ngoai tiep tarn giac A B C , mat phang ( S A C ) duang trung true cua S A cat S H tai O la tarn mat cau Gpi K la trung diem S A Khi hai tarn giac vuong S O K va S A H dong dang nen SH SA SK SO Suy ban kinh mat cau: R = SO = ^ ^ " ^ ^ = 2a SH Vay dien tich mat cau la S = 47iR^ = 16Tca^ Cau Dieu kien x > (Vx + Vx - mVx + mVx + I + \j\{x-\) = 1(1) Vx-1 V o + ijxix-l) = Vx - Vx-1 vx-1 0 m0 /(1)^0 m >1 [1-m^O Vay gia tri can t i m la m < hoac m > Cau Fa c6 sinx + > V x Phuong trinh da cho tuong dumig X X X X 3(sin— - cos —)(1 + sin —cos —) = cosx.(2 + sinx) 2 2 X X — (sin cos — )(2 + sinx) = cosx(2 + sinx) 2 V 411 , X 2X 2X « - (sin cos — ) = cos sin — 2 2 (sin— - cos —)[~ + (sin— + cos — )] = ^ 2 '2 2 X X X X sin cos— = (1) hoac - + (sin— + cos — ) = (2) 2 2 Giai(l): V2 s i n ( | - ^ ) = 0 dt = 2cosx(-sinx)dx = -sin2xdx Doi can: x = >t = ; x = ^ = ^ t = l = > I = - lntdt= Intdt Dat u = lnt dv = dt du = — t v=t Taco: I In/t//= / I n / ' - J W / = n - l I Cau Goi I la trung diSm cua AD Ta c6 lA = ID = IC = a =^ CD ± AC Mat khac CD ± SA Suy CD SC nen tam giac SCD vuong tai C Trong tam giac vuong SAB, ta c6: SH _ SAT SA^ 2a^ SB ~ SB' ~ SA' +AB' ~ 2a' + a ' " 412 Goi di va dj Ian luot la khoang each tir B va H din mat phing (SCD) thi: d, SH „ = = -=^d-,=-d, S d, SB ' ' Ta c6: d, = ^ ^ J I ^ C D ^ ^^\CD -"SCD ^BCD S , , „ = l A B B C = ia^ SscD = - sc.CD = -JSA' +BC-.VlC' + I D ' =a'V2 Suy di = ^ a Vay khoang each tir H den mat phang (SCD) la: di = - di = ~ U=X+ X" , ^-1 = (x + - ) " — , u > — Cau Dat v = y + y- = ( y + He da eho tra thanh: 4 v>-4 ?i+v = uv=3m V= • 4-M uv=3m 17 V i v > - = ^ - u ^ - - => u < 4 He da eho c6 nghiem (*) eo nghiem u e [ — ; — ] 4 Xet ham s6 f(u) = -u^ + 4u, u e 4 Ta CO f'(u) = -2u + 4, f (u) = » u = Lap bang bien thien thi yeu cau bai toan thoa man va chi khi: ~ < 3m < —^ ' + + C;;x", Vx e R Lay dao ham hai ve, ta dugc: n(l + x)""' = C;, + 2C^x + 3Cy- + + n C > " - ' , Vx e R Cho X = ta c6: C'„ - 2C^ + 3C^ 3' - + (-l)nC"„3" = 3n(-2)"-' Theo dk 3n(-2)""' = 33792 ^ (-1)"' n2""' = 11264 « n - Vay gia tri can tim la n = 11 Chu de 1: T I N H D O N D I E U 89 Chii de 4: TIEM CAN CUA D O T H I 61 Chu de 3: G I A T R I LCiN N H A T , N H O N H A T 36 Chii de 2: C U C TRI CUA H A M S O 131 Chii de 6: K H A O S A T V A V E D O T H I H A M HUXJ T I 107 Chu de 5: K H A O S A T V A V E D O THI H A M D A THlTC 194 Chii de 8: L U Y T H l / A , M U vA L O G A R I T 151 Chu de 7: B A I T O A N T H U ' N G G A P V E D O T H I 374 Chii de 16: S O PHLTC L U O N G GlAc 358 Chii de 15: C A N B A C H A I vA P H U O N G T R I N H N G H I E M PHLTC 337 Chu del4: S O P H U C 320 Chii de 13: L T N G D U N G C U A T I C H P H A N 294 Chii de 12: T I C H P H A N C A N T H L T C , M U , L O G A R I T 271 Chii de 11: T I C H P H A N D A T H L T C , P H A N T H L T C , L U O N G G I A C 253 Chii de 10: N G U Y E N H A M C U A H A M S O 217 Chu PHU de 9: P H U O N G T R I N H M U V A L O G A R I T LUC: C A C D E O N T H I T O N G HOP 397 416 y- y y y ^ y ^ •w^w^sTv n h i a s a c h i h i o n g a r a c o m v n 20C OT: Email: nhasachhonganOiotmail.com N g u y i n Thj Minh Khai - Q.1 - T P H C M 3824670e - 08083021 - 3910709S * * Fax: 0B083017 * t * r (^a/rt' tt/nv' t£po: HOD hoc T\} \ H Q C GIOI HOCT6T iiliitoi GiAl BfflDUflNG ^ HQGSINHGldl ^ 'S*^ HOCT6T -i- - HOC SINH GI61 HOC T6T 5:.? mm mmt T O A N ID* _ ^ GIAl T O A N Ph&i 1091 va phUdng phdp V glal nhanh bai tap HOAHOCj V A T L I / - N S MINH T A M , 245 T r I n N g u y e n H a n - H P * DT: (0313) 858699 ^ •N PhSn k«i va ptiuong phap giai nhanh bai t$p \ X ^ ^ ^ ^ : ^ \l - & P h a n B o i C h a u - H^i P h o n g *DT: (0313) 9 - 04 L y Thai T o - TP B a N i n g *BT: 0511.3823421 •'^ - 259 L e Duan - TP V i n h - D T : 0383.554777 Dexacinh sach chinh phlin^ ^ A - 39-41 V T h i S a u - C a n T h d * DT: (0710) 3818891 chiing to! in chim d bia va chff: ^ -^158 TTnh 10 - TT.CCi C h i - T P H C M * B T : ( ) "NS HONG AN" ^ - 51 L y T h U d n g Kiet - T P D o n g Hdi - Q B * DT: (0523) 857868 ^ ' - 1 P h a m Hufu L a u - T P Cao L a n h * D T : (067) 2211794 - 66 L y T h a i T o - T h j x a Q u ^ n g T r j - V T h i S a u , 3/5 T o n Dure T h i n g - L o n g X u y i n - ^ V V 67 N g u y i n K h o a i - H a N^i * DT: (04) 9 - ChUOng D i f d n g D p - H a N o i ^ ISBN: 978-604-939-543-7 N V - ' ^ ^ ^ 935092 756783 Gia: 84.000d • - 8 D U d n g L a n g - H a N p i * D T : (04) 5 >" y y y y _ y