Ciu 1. (2,0 dtdm)Cho hhm s5, y = T x-z a) Khio s6t sg bi6n thiOn vi vE AO tfri (C) cua him s6 dd cho. b) Vi6t phuong uinh tifo tuy6n cua dd *iI tCl t4r giao diAm cria tl6 ttri (C) voi tryc tung. . :' CAU 2. (1,0 di6m) a)Cho g6c a thodmin: 1."<n vi'sinA=1.rrnr, A=sin 2(a+ b) Cho s6 phftc z thoi mxn hQ thric: (1- 2i)z!3(1+t)t =2+7i. Tim ,,i clur so pnuc e Cflu 3. (0,5 didm)Gidi phuong trinh: 3.4x+I -fl.2x -29 =0 . tf L7't- SO GDÐA TINH @A thi cd 0I trang) CAu 10. (l ,0 diem) Cho Q,b,c le Im gratri nho nhAt cua biOu thirc: EE THr CUOI LoP 12 THPT NAM HgC 2014 - 201s MOn thi:TOAN Thoi gian ldm bdi:180 phfit 7t) . phan thgc,phan do -JY=* CAu 4. U,o Aidd Giei he phuong uinh: I ' - ':t [t+'*llt.r/+-$ +{ffi-l) = e. L+ -4 Cf,u5. ( 1,0 didr r)Tinh tichphAn: I = I.(1 +sin2x)dx. 0 Ceg 6. 1t,O drdmlCho hinh ch6p S.AB CD c6ddy li hinh thoi,c4nh a, gOc frD=600. Fllnh chi6u vu$ng g6c cira dinh S l0n (ABCD) tA di6m I/ thuQc canh AB thoa man HB=2AH. Bi6t SH = oJd ,tinh th€ tictr kh6i ch6p S.ABD vh khoing cdch tu diAm Cd6n mlt phing (SBD). CAu ?. (l,O dfdm) Trong mpt phing tqa d0 Oxy, cho hinh thang ABCD voi hai ddy lh AB vit CO. ei6t hinh tirang rO aien tictr Uing 14, dinh A(/; /) vi trung di6m cira cqnh BC'lh "(-;t) vitit phuong trinh dusng thing er Ui6t dinh D c6 hohnh dQ duong vi D nim trOn duong thdng d c6 phuong trinh 5x - y+ I = 0. CAu 8. (1,0 apd Trong kh6ng gian voi hq tqa dQ Oxyz, cho di€m 4(1;3;0) vd mflt phing (P)' c6 phuong trinh 2x+ 2y - z+ I = 0 . Tinh khoring cdch tu di6m e d6n mflt phing (P) vd tim tga Ag ei6m A'd5i xring voi di6m e qua m{t phing (Pl' CAu 9. (0,5 di6d Gqi S n gfln hgp c6c sb qu nhi€n c6 6 cht sd phdn ,UiQldugc lAp rft cdc cht sd 0, 1,2,3,4,5,6. Chgn ngiu nhiOn mQt s0 thuQc S. Tim x6c sudt d0 s6 du-o.c chgn lcm hon 300475. c6c sO thuc kh6ng dm, phAn biet thoa mdn az + b2 + cz - 3 . F !* l-+ L.r @-b)''(b-'c)'(c-a)'' ^ ftrET o * Thi sinh khing dryic s* tu4ng rdi ti&{. Gicim thi kh1ng giai thich gi th€m. Cảm ơnthầy ĐàoTrọngXuân (trong xuanh t@gmail.com )đãchi asẻđếnwww.l aisac.page .tl SO GD&DTIIA TTNH Ki'THI cuOI Lop 12 THpr NAMHoc z0L4 -201s n{!n tni:rOm nudnc nAN cnAvr rnr (Bdn hutng ddn ndy gim 06 trang) I. HTTOI.IG NAN CHT]NG NiSu ttri sinh lim bdi kh6ng theo c6ch nhu itrip 6n nhrmg tlung thi vdn cho thi s6 Ai6m tmg phennhuhudng d6n. Di6m toan bii kh6ng quy tdn. n. DAP Ax vA THANG orE*r r CAU DAPAN DIEM Ciu 1 ? (2.0 dihm) 3 L) (1.0 iliam) o Tapx6cdintr: P=R\{2\ o Gi6i han ve tiem cfn: lirn y @; lim y-*o ; lim l=2; x+2-n ' x+zin r+<- suy ra dO thi c6 mQt tiem cAn dung h ngang ld duong thang ! =2. lg} !=2. duong thang x=2 ve mOt tigm c?n 0.25 o SU bii5n thi6n: -' ! -5 - Chi0u biOn thiOn: Y'=- ( 0, Vx € D (x- 2)' Him s6 nghich bii5n trOn mdi khoang (-*; 2) va Q; + @) -_:!s 0.25 - Ben x bi€n thi€n -@2*m 0.25 y' - - v 2 -@ *m 2 \ D6 thi 0.25 I b. (1.0 di6m Gqi M(0;%) li _1 MQ;;) giao di6m cria (C) vd tryc turg, ta c6 2.0 +l -1 !o= 0-Z= Z suY ta 0.25 H9 sd g6c cta ti€p Qyen tai M le /'(0) :+ 0.25 Phuong trintr tiOp tuyi5n cfia dO thi tai M Ld y=+(x-o)-; 0.25 hay 51 tt - L V_ r'42 0.25 Cflu 2 QQdiam) Cfiu 3 (0.5 dihm) Cfiu 4 (1.0 diem) a. (0.5 diem) k= sw2(or+ r\-= sin(2a + 2r) = sffia-=2-sirurcosru@ 0.25 Tac6cos'e-l-sin2 e-1 -16 =L 25 25 Do {. a < rr ndn cos q <0, k6t hqp vdi (z)ta c6 2 Thay (3) viro (1) tac6 ,{=-1.+ =!! . 55 2s cos d 1 5 (2) (3) 0.25 b. (0,5 rli6m) Ddt. z=a+bi(a,b eR),tac6 z=a-bi Khi d6 (I - 2i) z + 3(1 + ilZ = z + 7 i e (t - zi)(a+ b?) + 3(l + f)(a - bi) - 2 + 7 i e (4a + 5b - 2) + (a -2b - 7)i - 0 0.25 (+a+5b=2 la-3 c+{ e{ [a-2b=7 LD= -2 VOy phAn thuc ctia z li 3, phAn io ctia z ld-2 0.25 Tac6 3.4x+r-l7.zx D?t t:2' (t > 0) -29-o<+ lz.4x -l7.zx -29-0 Phuong trintr dd cho tr& thanh I2t2 -l7t -29 =0 e t - -1(z) t_2 T2 0.25 Vdi f =2,tac6 2'=Lex=lc 29 rz lz€ x- IoBz n VAy nghiem cria phuong trinh li: x: 1og, 29 l2 0.25 DiAu kien {i = t Lo< y<16(*) 1-13 + vJr+1> o Vdi di6u kiQn (*) ta c6: 0.25 2 L- x3do do (1) e -,tffi=n+Jy* '-*_x, *r[ffi .[-;*,[m) =o<+ Jy=-x t:- ,IYJY+I>o) -x+,1 y o(x+6)(' (do x'-*Ji + Thti vio (Z)tadusc: (ax+3)(fia+{Fs-t)=9 (3) _?^ Vi " =7 khdng phni h nghigm cria (3) ndn (4) e J r+q_+{lEr+e -;fr;-l = 0 0.25 Ta c6 g'(x) = Suy ra hdm s6 g(x) d6ng bii5n tr€n c6c khoan e |-/rt+)rf ,**y. LSp BBT ta th6y phuong trinh S:(x) = 0 c6 tOi Ca 2 nghiQm. Xdthamsti g(x) =Jx+4+{/E+8 - =9 ^-l 4x+3 1 I 36 A., -3 -!=!!.r,?-L-+ .^,, >0 Vx)-4rx7L-t-l 2J,+4 W $x$)2' \' YJv' -r)*7' 4 trOn (a;+€)rt?l 4.25 Ta l4i c6 g(0)=g(-3)=0 suy ra N=0;x=-3 ld cdc nghiQm criaphuong tri"h g(x)=0. Vsi r=0=) !=0i x=-3 ) y=9. OOi ctrii5u di€u kiQn ta tfr6y phuong trinh c6 2 nghiQm: (0;0); (-3;9) 0.25 Cflu 5 Q,0 diem) ,t ,t ,r 444 [ = I r(t + s in}x)dx - I **+ I x sin 2xdx 000 (1) + 4.25 !.*=+#=* lt Ta c6 (2) 0.25 LL bsin 2xdx= - +! xd@os2x) =+,rcos z.lt o to Th6 (2),(3) vio (t) ta c6 : I =t* 1 =n'-!8 324 32 L .+ ! rorzxdx:+sin 2xlf = 1 /.0 4 'v 4 (3) 0.50 Ciu 6 Q.0 diAm) Ta c6 BO = AB.sin Z.BAQ =asfn3Oo :_ AO = AB.sin ZABQ =asin60o =oJi 2) -t' -r"'i:t\t, \ ,' -8I \- ' -F K 0.25 :l I I suy ra Snar- lo.Bo -9.4 -o'Jt ; '-v 2' 2 4 ) Dod6 Ys.nro=+ sH.sABD=+ alr.+=+ 0.25 Do du&ng ttrang AC c[t (SBD) tai tli6m O ld trung ei6m crla AC vd dudng ttrfig AII c6t (SBD) t4i B thoi mdn AB =|nA oen d (c,(,sBD)) = d (t,(sBD)) = lo rr, (^sBD)) (1) KC HK L BO,HM J- S/( ( K thuOc BO, M thuQc SK). Ta c6 BO L(Srrq + BO L HM do tl6 HM L (SBD) + d(H,(SBD)) = HM (2) 0.25 Trqqe tarqejee-iueselH[ c9-sg, = ofi,M-:=?Ao-=* ya-fnrf-fa dqqng (3) 111137o.,|t+ cao suY f?. - ' HM' HS' HK" 2a' a" 2a' 7 fiSt trq,p (1), (2), (3) ta c6 d(C,(SBDD:og ,,A 0.25 Cfiu 7 Q.0 dihm) Cffu 8 Q.0 dihm) K6o dei AII cAt CD tai E. Do ABCD hinh thang (ABI/CD) va H trung di6m BC n6n OE ttr6y LruB - MnEC * S-o, = S eaco =14 Ta c6 AE =ilH = Jii vi phuong trintr dudrng thAng AE: 2x-3y * 1 :0. Do dinh D c6 hoanh d0 duong va D lr^r.? n[m tr0n cludrng thdng (d) c6 phuong trinh 5x- y+1 =0 n6n D(d; 5d+l) vdi d > 0 0.25 0.25 Tt ct6 D(2; 11) 0.25 E d6i nmg voi A qua H suy ra E(-2; -l) nen pn 3x-y+5=0 Duong thdng AB qua A, song song vdi dt CD n6n c6 pt: 3x -y -2 = 0 0.25 I(hoang circh fh A(I;3;0) d6n fl= lz,t+z.z - o +tl ffi o ./a -E-J 3 mflt phang (P) li: 0.25 4 4 arct Duong thaog AA' qua A nhan vecto ph6p tuytin cria mp(P) n iQ;Z;-t) lem vecto chiPhuong' rx=r+Zt I Ta c6 phuong tinh tham sti cria rtulng th6ne AA':{ y =3+2t I lz= -t 0.25 Gqi I ld giao di6m cfia ducrng thing AAtva*Aa ttteng e)t Do I thuQc du0mg thang AA'n6n l(l+2t;3 + Zt;-t) 4.25 Mpt khic I thuQc mflt phang (P) n€n 2(I + 2t) + 2(3 + 2t) - (-t) + I - 0 e f 1 =) I (-1;1;1) Vi I ld trung ttiOm cria AA'nOn ta c6 A'(-3;-!;2) 0.25 Ciu 9 (0.5 diim) 56 phen fl}cria kfiOng gran m6u li s6 phdn tu cria tap hqp S. - Kf hi-€u abcdef ld mQt sO U6t kj thuQc S TathSy a c66c6chchgn(do a*0); b c66c6chchgn(do b *a). Tuong t.u ta th6y: c cb 5 c6ch cho.n; d c6 4 cdchchen; e c63 c6ch chgn; f c62 c6ch chgn. iai '6iiur" "t u s n a.at = 41 2) - 4.25 4.25 Xdt sO qaza3a4asa, thuQc S mi atazatalasau>3}}471,tac6 THI: q>-4; ta th6y a, c6 3 cdch chgn; a, c6 6 cdch chgn ar c6 5 c6ch chen; ao c6 4 cilch cho.n; a, c6 3 cdch chQn; a, c6 2 cfuchchgn. suy ra c63.6! sd arararanosaa)300475 md ar24 Tt12: at =3. Ta th6y sO IOO+ZS c6 2 cht s6 0 nen Rhi chgn mot sii W Uit ty trong t6p S thi s6 d6 lu6n lcrn hon 300475 vi sti thuEc tap S thi c6 c6c cht sti ktr6c nhau nln a,arkhdng el6ng ttroi Ueng O. Do tt6 a, c6 6 c6ch chgn a, c6 5 c6ch cho.n; ao c6 4 cdchchgn; a, c63 c6ch cho.n; au c62 cdch chgn. suy ra c6 6! s6 "rqr"r"r"">300475 md a, = l. Viy c6 4.6! sO "rtr"r"""r"" thuQc S mi tr"r"rq"rrr>300475 . X6c suat can tim li 2 p=- t3 At>3 Cfiu 10 Q,0 dihm) 0.25 0,25 = r * 2* = *( @- v)'* r)'. (x-y)" ( xy ) DAt @- Y)' +z= r (r > 2); ra c6 xy suyra G-t'+ 2 -r t-2 f'(t)=Zt- ' 2* - (t -2)' xy_ (x - y)' t-2 2(t - lxrz -3r + 1) (t -2)' 3 +.6 f'(t)=Q€t- Lap BBT ta 2 =J 1- f(t) z(tt -4tz +4t-t) (t -2)' (do t>2) c-0;arb>0 2 0.25 0.25 "eesu*or,l-ru#-r=9 v4y F >t 1+ 5tF , c6 K-,,khi 6 3+V5 hay *'+y'_l+.6 2 xy a'+b2 +c2 -3 c-0;a,b>0 e = A; a,b>0 a2 +b2 3+.6 = + e lab =+. Ta th6y hg nny tu6n c6 nghiQmphdn biQt. g -3.,6 ab az +b2 a2 +b2 2 =l Vay siatri nho nhdt cria F h ll.!}E 6 a HET Cảm ơ n thầy Đào Trọng Xuân ( tr o n g x u a nh t@gm ail .co m ) đãch i asẻđế n www .l ais ac. pa ge .t l . GDÐA TINH @A thi cd 0I trang) CAu 10. (l ,0 diem) Cho Q,b,c le Im gratri nho nhAt cua biOu thirc: EE THr CUOI LoP 12 THPT NAM HgC 2014 - 201s MOn thi: TOAN Thoi gian ldm bdi:180. @-b)''(b-'c)'(c-a)'' ^ ftrET o * Thi sinh khing dryic s* tu4ng rdi ti&{. Gicim thi kh1ng giai thich gi th€m. Cảm ơnthầy ĐàoTrọngXuân (trong xuanh t@gmail.com )đãchi asẻđếnwww.l aisac.page .tl SO. r+<- suy ra dO thi c6 mQt tiem cAn dung h ngang ld duong thang ! =2. lg} !=2. duong thang x=2 ve mOt tigm c?n 0.25 o SU bii5n thi6 n: -' ! -5 - Chi0u biOn thiOn: Y'=-