đề thi thử dh hay số 67

so GrAo DUC VA EAOTAO HA NQr rnr-IoI.rc TIIPT CHU vAfq AN {* ne rlrt THU DAr Hec EOTTr NAvr zor r MOn Toin - fndi A Thdi gian ldm bdi: 180 phft, kh0ng td ttrOi gian giao d0. OA ttri g6m 01 trang. PHAN cIruNG (z iti6m) B il (2 drd@.Cho hdm s5 y = ?4 x-l i. Kh6o s6t vd vc dd thi cria hdm s6 da cho. . 2. Gei r ra giao di6m't;ft;o't;;;e*;; cria r10 th!. rim di6m M tr€n d0 thi sao cho ti6p tuy€n tgi M wdng g6c vdi IM. Bii u (2 diam) lx-alvl+3 = 0 l. Giai hQ phuong trinh: ] - "' [!/lo&,< /iog,y=o' 2. Giaiphuong trinh, sit 2* n t-9t t* = tan x - cot x . cosx smx Bii III (1 die@. Tim th6 tfch kh6i trdn xoay dugc tpo thanh khi quay quanh t4rc Ox hinh ph8ng gidi han bOi dd thi (C) cta hdm s5 y = *dt([*) vd ciic cfudrng thing y = 0, x = 1. Bei W ( did@. Cho hinh ch6p S.ABC c6 d6y ld tam gi6c cAn (B = e - o)- C6c rludng thing SA, SB, SC tpo vdi m[t phfrng ddy cdc g6c b].ng nhau. Ggi V, ua V, lAn luqt la th0 tfch cria hinh ch6p vd th6 tich hinh n6n ngo4i tii5p hinh ch6p. Tinh theo cr ,y ,6 $. 'v2 (. lxr+x+log21=8Y'+2Y+1 Bni V Q diA@. Giei he phuong trinh: ] Y lu'-ro+1=o (vdix> o' Y> o)' L" "4 PHAN fV CHQN Q itid@. Thi sinh )nt "npo mQt trong hai phb): Phdn A ho(e Phdn B. Phin A: Bii VIa Q diiim) l. Trong mflt phEng tga ilQ Oxy, cho tluong trdn {C): (x-1)' +(y +2)2 =9 vd dutrng thing (d):3x-4y+m=0.Timmd6tr6n(d)c6duynhfltdiOmPsaochotrlPc6thekedugchaiti6p tuy6n PA, PB vdi dudng trdn vd tam gi6c PAB la 6m gi6c vudng (A, B ld cdcti€p diem). 2.Chohai ducrngth$ng d'* 1= yr 1 =t ^l vd A: x-2 = y, 3 =t 4 trongkhdng 212123 gian v6i hQ to4 dQ Oxyz. Bi6t rang d vd Acft nhau. HEy vi6t phuong trinh mf,t phing (P) chrla A sao cho g6c gita dulng th6ng d vd m6t phing (P) l6n nh6t. Bii YIIa (I di6m). Gi6i phuong trinh: 9x + 2(x - 2).3" + 2x - 5 = 0. PhAn B: Bni vlb (2 die@ 1. Trong mlt ph&ng tqa dQ Oxy, cho dudng trdn (C): (x - l)t + (y + l)'? - 25 , diiSm M(7; 3). Vi6t phucrng trinh ttuhng th&ng qua M c6t (C) tai hai di6m phAn biQt A, B sao cho MA = 3MB. 2. Tt cdc cht sd 2, 3, 4, 7, 8, g c6 th6 lflp tluqo. bao nhi€u si5 tu nhien 16 c6 s6u cht s6, trong d6c6dungbachfr s6ZZ BAi VIIb ( diim).Gi6i phuong trinh tr€n tflp ttsp f5 phttc: za - z3 + 6* - 8z - 16= 0. HCt Hq vdtOnth{ sinh: , S6 Oao Oanh: jimca97@yahoo.com sent to www.laisac.page.tl Hr-l5Nc nAN cnA,r vn nrdu odu udx roAN - rsdl a nnl u6r nuNc DIEM BEi I 2d CAU L (1.25 ttiim) a)TXD:x*1 0.25 b) Sg bi6n thi€n: + Nh6nh v6 cqc vd duirng tiOm cAn: Iim y=2' lim y= 2:'Dd thi c6 tigm cQnngang y =), x-)-00 x-)+co lim y =-oo; lim y - .o: Dd thi c6 tiQm cAn drfurg x = 1 x-+I x-+l' o,25 _1 + Yt= : ^ " ' (x -l)' + Bing bidn thit F- lY' < 0,Vx * 1, do d6 hhm sd nghich bidn uOn (-co; in -oo 1 *co I_ I vd (1; +oo) 0,25 2 fco +co ) c) Dd thi: Dd thicSt ox r?i didm A( j, ol, c6tOy tai B(0, 1) VE dfng, dep 0,50 Chu2 (0,75 ilidm) Ta c61(L;2). Gi6 srt M(xo, yo), ta c6 hQ sd g6c ctra IM lb k = Y o -2 = l - . xo-1 (*o-1)t' 4,25 Tidp tuydn tai M vu0ng g6c vdi IM khi vd chi khi k.y'(x6) = -1 -1 1 e; :r J= -1<> (xs -1)a = 1 (xo -l)' (xo -1) o,25 TU d6 ta duoc M(2: 3) hodc M(0:1) a"25 Bni u 2d cau l, (1ili6m) + Didu kiQn x > 1, y ) 1. L{p ludn dua vd hg: [*-0, +3:0 [loga x =logzy 0.50 i*-ay+3=0 <+1 r Ix=y" 0.25 + GiAi ra c6 nshiOm (1: 1): (9: 3). o.25 Cilr2 (1diim) + Didu kion x *nL '2 4,25 a- + Giai (2) ta duo.c *= * *k! no4c x = n +kLn -tJ _0al o,25 + Kgt hqp didu kiQn ta dugc x =tt+kZn 0,25 B}i IU 1 I + LAp luan ra dugc y = *1G(i + *) khdng 0m tr6n tdp xdcdinh 11 [O;+*) ; 0,25 I + Ldp luan ra duo. c thd r(ch ld v = n Ix2 h(l + x3)dx 0 0,25 + Dung cong thrlc rich ph&n tungphdn dd bidn ddi tfch ph6n I = I*'h(l+ x3)dx : 0 l=4.m6**rlll -'r4.Ai d* =tn2-ti ,.t o'. 3 '10 d3l+xr 3 jl+xi 0,25 +rrnh dusc tichphan J= J** = i[.' *jd" 1-1rn z.ydvrhdrichcdntimld v= tr(21n2*r) . 33 3 =d !,nrt**rill 3 3 ',10 0,25 Bni w 1d + Ggi H ld chAn ducrng cao cta ch6p, chrlng minh duoc H ld tam duong trdn ngo?i 1!{p tamgi6c ABC. 0,25 + Dat AB = AC = a; Tinh dugc dign tich tam gi6c ABC li, S, = 1a'sin2a; 2 0,25 + Tinh dugc diQn tfch hinh trdn ngoai tiOp tam gidc ABC ld S, na' =- 4sin2 s, 0,25 + Y _ SH.q/3 _ S, _ v2 sH.s2 /3 s2 4sin3 acosa o,25 Bni v 1d + (1) e x3 +x+ log, x -log, y = 8y' +2y+1 <+ x3 + x + log, x = (2y)' +2y +logr(2y) (q) 4,25 + L4p lufln duo. c hbm sd f(t) = t3 + t + 1og2 t ddng bion tren (0;+oo) 0,25 + Do d6 (3) <+ x=2y. 4,25 + Thd x = 2y vio (2) vb kdt hqp didu ki0n ta duo.c nghi€m * = l,y = 1. 2 o,25 Bii VIa 2d cau I (l rtidm) + Dudrng trbn d6 cho c6 Am I(L;1\, b6n kinh R = 3 ; 0,25 + Tam gi6c PAB luOn cAn tai P, do d6 ndu n6 wOng thi wdng t+i P; Khi dd trl ei6c PAIB lb hinh vu0ns canh 3 vh do d6 PI = 3Ji 0,25 t*D€ t$l t+i duy nhdt didm P thi khotrng cdch tt I den (d) bing 3J2 4,25 lm+81 + Trlc g Jil J =3J2; GiAi ra ta duoc m = *8 xlsJt 5 0,25 cku? (1 tlicm) r.Fgr{s"-q" c eis-e-*" qgr.q.r+s 4":s"A l*"-M0;1-;"0- ; Ldt;:t5;tfi;rilaCa;cal"ifid"Klan dqr ie ** *-; -*t***- hinh chidu cria A trcn (P) vd A. Ta c6 y AH < AK suy ra lAlvftI< IAMK, ding , ','l , = , thrlc xAy ra khi vi chi khi H trung K, trlc li - / -r- / I Q/ AKI(P). / ' / lr / Nhu vay g6c gifia d vd @) lon nh6t khi / t I L1 / vhchikhiAKJ-(P)t4iK. /ffi*r\ 0,25 4,25 + Tim duo.c toa dQ K (h hinh chieu ciia A tr€n A) h K =(*ry-,?\ .rvrrA)LqLr -[Zr 7t 7 ) 0,25 + Mar phing (p) di qua K vd nhan AR = f-:t:,:l ldm vrPr, do d6 phuong \. 7'7'7 ) trinh cria @) li 9x -3y - z - 5 = 0 4,25 Bii YIIa kl + DAt t = 3" (t > 0), phuong trinh fiA thenh tz + 2(x *2)t +2x - 5 = 0 0,25 + GiAiphuong trinh dnttadud.c t= -l; t = 5 -2x (1o4i t = -1) 0,25 +Vdi t = 5-2x tadugc 3* =5*2x (2) X6t tfnh don diOu hai vd suv ra (2\ cd khOne qu6 m6t nshiOm a,25 + Mft kh6c x = 1 thoi mdn. VOy phuong trinh de cho c6 1 nghi€m x =-1-:- 0,25 Bii vlb 2d Ceu 1 (1 rliim) + Dudng trbn dd cho c6 am I(1;-1), b6n kinh R = 5; MI = Jn ,5, do d6 M nf,m neodi dulne trbn. Q,25 ;Ceil Ie kh'A"s;a;,l';ti dsnA;dns;hAns MCil irittruns cidm An, ta co I,A : I'B = fr5-;, I,M = J52 -7 o,25 + Tt MA = 3MB ta suy ra I'M = 2I'A, do d6 Giiiraduo.cx=4. 0,25 ; Nhtvat dnhnt ih&ng cdn tim qua M vb c6ch I-mQt khoAng bing 4. TU d6 tim duoc hai duirne thoi m6.n: y = 3; LZx - 5y - 69 = 0. a.25 Chu2 (l tliim) Gi6 st sei tu nhicn 16 d5 cho ld abcdef , + C6 3 cdch chqn f; o,25 + M6i c6ch chgn f c6 Cl c6ch chon vi tri cho ba cht sd 2; a,25 + M6i c6ch chqn f vh chgn vi trf cho ba chfi sd 2 c6 5x5=25 c6ch cho:t hai cht sd cbn lai: 0,25 + Theo Quy t6c nhEn, cd 3xC3rx25=750 sd thoi mdn. j a,25 BEi VIIb 1d + (NhAn bidt duo.c hai nghi€m z = -1, z = 2) Phuong trinh di cho tudng duong v6r (z-2)(z+I){2'?+ 8) = 0 + Gi6i ra duoc 4 nghiom: z = -10 z = 2, z = XJS| 0,50 0;30 Hdt . bOi dd thi (C) cta hdm s5 y = *dt([*) vd ciic cfudrng thing y = 0, x = 1. Bei W ( did@. Cho hinh ch6p S.ABC c6 d6y ld tam gi6c cAn (B = e - o)- C6c rludng thing SA, SB,. ttiim) a)TXD:x*1 0.25 b) Sg bi6n thi n: + Nh6nh v6 cqc vd duirng tiOm cAn: Iim y=2' lim y= 2:'Dd thi c6 tigm cQnngang y =), x-)-00 x-)+co lim y =-oo; lim y - .o: Dd thi c6 tiQm cAn. (x -l)' + Bing bidn thit F- lY' < 0,Vx * 1, do d6 hhm sd nghich bidn uOn (-co; in -oo 1 *co I_ I vd (1; +oo) 0,25 2 fco +co ) c) Dd thi: Dd thicSt ox r?i didm A( j,