Tnr(mg THPT Chuyen Le Quy Don DE THI THU D~I HOC LAN 3 - MON TOAN KHOI A Tinh Ba Ria Viing Tau Thai gian him bai: 180 phut. I. pHAN CHUNG CHO TAT CATHi SINH: (7 diim) 3 Cau I (2 di8m): Cho ham s6 y x - 3x 2 + 3. 1. Khao sat S\1' biSn thien va ve d6 thi (C) cua ham s6. 2. ViSt phuong trinh tiSp tuy~n cua db thi (C), biSt ti~p tuySn di qua di~m A(-l; -1). Cau II (2 di8m): 1. Giai phuong trinh x 3 -6x 2 + 12x -7 ={j_x 3 +9x 2 -19x + 11 . 2sinx + 1 cos2x + 2cosx 7 sinx +5 2. Gi1ii phuong trinh: - ;= 2 cos X J3 - cos 2x + 2 cos X + 1- J3 (cos x + 1) . Cau III (l di~m): ' h' h liZx3~x3+8+(6x3+4x2)lnxdx h A I T III bc p an sau: = x Cau IV (l di~m): Cho hlnh chop SABCD co ABCD 1a hinh binh hanh tam 0, AB = 2a, AD 2aJ3 , c~c Cl;l.llh ben b~ng nhau va bk g 3a, gOi M 1a trong di~m cua OC. Tinh th~ dch kh6i chOp SABMD ya dien dch cua hinh cAu ngo~i ti~p tu dien SOCD. ' Ciu V (1 di~m) Cho x, y thOa x 2 + r - xy = 1. Tim GTNN va GTLN ella P = X4 + l x 2 y2. n. pHAN RIENG (3 diim) Thi sinh chi dU'Q'c lam m9t trong hai phAn (phAn 1 ho,"c phAn 2). 1. Theo ChU'01l~ trinh chuAn: Cau VI.a (2 diem): 1. Trong m~t phing Oxy cho MBC nQi ti~p duang tron (T): x 2 + l 4x - 2y - 8 = O. Binh A thuQc tia Oy, duemg cao ve tir C n~m trcn duong th~ng (d): x + Sy = O. Tim toa dQ cac dlnh A, B, C bi8t ~g C co hoanh dQ 1a m¢t s6 nguyen. 2. Trong khong gian Oxyz cho hai duang th~ng Cd 1 ): ~ -1 72=Z~:, (d 2 ): {;=~:: 2 Z 4+t va m~t ph~g (a.): x - y + z 6 == O. L~p phuong triM duang th~g (d) bi~t d II (a.) va (d) cit (d 1 ), (d 2 ) l~n luQ't ~i M va N sao cho MN = 3.J6 . Cau VII.a (1 di~m): Tim t~p hQ'P cac diam biau di€n cho s6 phuc Z th6a man M tMc: Iz + 3 - 2il 12z + 1- 2il 2. Theo chU'O'D~ tdnh nang cao: Ciu VI.b (2 diem) 1. Trong mftt ph~ng Oxy, cho tam giac ABC co dlnh A(O; 4), trong tam G [~; ~ ) va tf\1'C tam trimg vai g6c toa dQ. Tim toa dQ cac dlnh B, C va di~n dch tam giac ABC bi~t XB < Xc . . x-I Y+2 Z 2 {X 2 - t . ~ 2. Trong khong gian Oxyz cho hai duang thang (d!): ;;; = , (d 2 ): y = 3 +t va m~t 2 1 -2 z=4+t ph~ng (a.): x - y + Z 6 = O. Tim tren (d 2 ) nhfrng diem M sao cho duang th~ng qua M song song voi (d 1 ), c~t (eL) t~j N sao cho MN = 3. Ciu VII.b (1 di6m): ' , ,{eX e Y (In y In x)(l +xy) G 1a1 he phuong trmh . . 21nH2lny _ 3.4 ln :< ;;; 4.iny Thi sinh kMng (/l.f9c sa dl,mg tai lif?u Can b9 coi thi kMng gial thlch gi them. HQ va ten thf sinh : ; So bao danh: www.MATHVN.com www.mathvn.com DAp AN DE THI THU D~I HQC MON ToAN ThiYi gian: 180 phut Cau y Caul 1 I 2 ! ! CiuII 1 i i NQidung Cho ham so y =x j - 3x" + 3. Khao sat S\l bien thien va ve do thiJC)cua ham so. T~p xd va Gi61 h~ y' =3x" 6x y' ::; 0 ~ x =0 hay x =2 Bang bien thien: y" va diem uon Gia tri d~c bi~t Do thi va nh~ xet: Viet pt tiep tuyen eua (C), biet tiep tuyen di qua diem AC-l; -1). DuOng tMng (d) qua A va co h~ so goc k => Cd): y + 1 "" k(x + 1) => (d):y = kx + k - 1. (d)" '(e) {XJ-3X2+3~kx+k 1 eo nghi~m. tIepxue ~ 3x 2 -6x k => x 3 3x 2 + 3 = 3x 3 - 6x 2 + 3x 2 - 6x 1 ~ 2x 3 - 6x - 4 = 0 ~ X = 2 hay x = -1. ,. x = 2 => k = 0 => (d): y = -1. ,. x = -1 => k =9 => (d): y = 9x + 8 Giai phuong trinh x 3 - 6x 2 + 12x 7 = ~_X3 + 9x 2 -19x + 11. D~t Y ~-x3+9x2-19x+ll Kh' d' 6 { Y ~ x 3 - 6x' + 12x-7 Iota c y3=_x 3 +9x 2 19x+ll => y3 + 2y =x 3 - 3x 2 + 5x - 3 Q y3 + 2y = (x _1)3 + 2(x -1) (1) ~ Xet ham s6 f(t) =e+ 2t.Ta co f'(t) 3e +2>0 Suy ra ham s6f(t) d6ng bi8n tren R ~ Tu (1) => y = x-I . Qx-l x 3 -6x 2 + 12x-7 Qx3-6x2+11x-6=O Q (x -1)(x - 2)(x - 3) = 0 ._. - Di~m 2:=2d "L=.1.25d 0.25 0.25 0.25 0.25 0.25 'L= 0.75d 0,25 0.25 0.25 2:=24 :L=ld I ! 0.25 i 0.5 0.25 www.MATHVN.com www.mathvn.com lx ~j ¢::> x=2. x=3 I I I I I 12 Giai phucmg trinh: 2sinx+ 1 cos 2x + 2cosx -7sinx + 5 2cosx-Ji == cos2x+2cosx+1-Ji(cosx+l) . 2.: = Id I ! Dieu ki~n: {2cosx 13 ,,0 {2eosx ,,-13 {cosx" J3 cos2x+2cosx+1 y3(cosx+l):;eO <=> (cosx+l)(2cosx y3):;tO <=> cosx:;e _~ - 0.25 (1) <=> (2sinx + 1)(cosx + 1) == cos2x + 2cosx -7sinx + 5 <=> 2sinxcosx + 2sinx + cosx + 1 = 1 - 2sin 2 x + 2cosx - 7sinx + 5 0.25 <=> 2sinxcosx - cosx + ~ = 0 <=> cosx(2sinx - 1) + (2sinx - 1)( sinx + 5) = 0 <=> (2sinx - 1)( cosx + sinx + 5) = 0 [ [ ~ . 1 x =-+ k21t smx ==- 6 <=> 2 <=> 0.25 . - 51t sm x + cos x = 5 x == (5 + k21t So s8n.h diSu ki~n ta duQ'c nghi~m cua phucmg trinh 1ft x == ~: + k21t (k E Z). 0.25 Call III rnh 'h han I fX3~X3+8+(6X3+4X2)lnxd I 1 tIC P sau: == x 2.: = Itt , 'x f X2~X3 + 8 dx ~ 12 (6x 2 +4x)ln xdx = II + 12- I I I I 0.25 ). Tinh II: Dat t = ~X3 + 8 => t 2 = x 3 + 8 => 2tdt = 3x 2 dx => x 2 dx = ~tdt, . 3 D6i c~n: x 1 =>t=3 x = 2 => t =4, [ r 0.25 2 2 t 3 2 h do II rt.3"tdt =3"' 3" 3 == g( 64-27) == -, $ ). Tinh h D~t u = lnx => u' = ! x v' = 6x 2 + 4x chon v == 2x 3 + 2X2 , , h = [(2x 3 +2X2)1nX r- f(2x 2 +2x)dx = 0.25 [ r 2x 3 23 I 241n2- -3-+ x2 1 = 241n2 - 3 J A 77 23 2 .J t" S- 0.25 I ! VayI= 241n2+ =241n +- I J . 9 39 www.MATHVN.com www.mathvn.com - ~""" r -~ , ABCD 1ft hinh binh hanh tam 0, AB = 2a, AD = 2afj, cac cl;U1h ben bkg nhau va bkg 3a, gQi M 1ft trung di~m cua 2:=ld OC. Tinh th~ tich kh6i chop SABMD va di~n tich cua hinh cAu ngo~i ti~p tii' di~n SOCD. ~ Ta co SA = SB =SC =SO nen SO 1. (ABC D). , ~ /),. SOA = .= /),. SOD nen OA ::;; OB OC = 00 => ABCD 1ft hinh chu nhat. 0.25 • => SABCD = AB.AD = 4a 2 fj. . ~ Ta co BD ~ AB2 + A0 2 ~4a2 + 12a 2 = 4a => SO = ~SB2 -OB2 ::;; ~9a2 -4a 2 == a.J5. 0.25 A_I 4a 3 J15 , _ 3 3 c: V~y VSABCD - "3SABCD'SO == 3 . Do do VSABMD - "4 VSABCD = a ;15. ~ GQi G la trQng tam /),. OCD, vi /),. OCD deu nen G cling 1ft tam dUOng tron ngo~i ti~p /),. OCD. D\ffig dUOng thkg d qua G va song song SO thi d 1. (ABCD) nen d la tI1,lC cua 0.25 /)"OCD. i Trong mp(SOG) d\ffig dUOng trung trgc Clla SO, c~t d t~i K dt SO ~ I. · Ta co: 01 la trung tT\Ic cua SO => KO::;; KS rna KO = KC = KD nen K 18. tam mi},t cAu n o~i tiS ill di~n SOCD. Taco: GO::;; CD = 2a fj fj 0.25 R=KO= ~OI2+0G2 2 2 • 31a 317ta i Do do S A = 47tR2 = 47t = C~U 12 3 CauV! . Cho x, y thca x + y - xy = 1. Tim GTNN va GTLN Clla P = x 2:=ld • 0.25 0.25 Ta co: f(t) = -4t + 2. 1 f(t) = 0 ¢::> t = . 2 f(-,,!,) = ! fO) = 1 va f( !)'::;;~. 3 9' 2 2 Yay MaxP = maxf(t) = l va Min P =:: minf(t) = 1 [-1;1] 2 [-1;1) 9 0.25 I + ::-::: :-' -'- va t =-1/3 , t Yz tinh dmlc 0.25 i ~-~~' ~ + ~=~2~d Vl.a www.MATHVN.com www.mathvn.com I 1. : llUUg Ulp vxy cno L.\ fin\ " nQl nep Quang l:ron \1): x -r Y - Ll-X - Ly lS = u. Dinh A thuoc tia Oy, duemg cao ve tir C nfun tren duemg thfutg (d): X + 5y O. I=ld i Tim tQa do cac dinh A, B, C bi~t rfutg C co hO<lnh do h\ mot s6 nguyen » A thuoc tia Oy nen A(O; a) (a> 0). , Vi A E (T) nen a' - 2a - 8 ~ 0 "" [a 4 => a ~ 4 => A(O; 4). 0.25 • a=-2 I I » C thuqc (d): x + 5y = 0 nen C(-5y; y). : C E (T) :::::> 25y2 + y2 + 20y 2y - 8 0 <=> 26y2 + 18y - 8 0 ly ~-l=> x ~5 0.25 <=> y = ~ :::::> x = _ 20 :::::> C(5; -1) (Do Xc E Z) 13 13 » (AB) 1- (d) nen (AB): 5x - y + m = 0 rna (AB) qua A nen 5.0 4+m 0 :::::>m==4. V~y (AB): 5x -y + 4 O. B E (AB) :::::> B(b; 5b + 4). [b-O 0.25 i B E (T) <=> b 2 + (5b + 4)2 - 4b - lOb 8 - 8 == 0 <=> 26b 2 + 26b == 0 <=> - . b=-l I Khi b = 0 :::::> B(O; 4 ) (lo~i vi trimg vOi A) Khib -1 :::::> B(-I; -1) (nh~). 0.25 i V~yA(O; 4), B(-I; -1) va C(5; -1). 2 r2 t Trang kg Oxyz cho (d}): x 1 +2 z 2 , _ ,: = 1 = -2 ,(d 2): y=:03+t vam~tphang 2 z=4+t I=ld (a):x-y+z-6 0, L~p phuong trinh duemg thkg (d) bi~t d II (a) va (d) d.t i (d!), (d2) iAn luQ"t t~i M va N sao chc MN = 3.J6. ME (d!):::::> M(1 + 2m; -2 + m; 2 2m) I NE(d 2 ):::::>N(2 n; 3 + n; 4 + n) :::::> NM ==(2m+n-l;m-n - 5'-2m-n -2)' n =(1'-1'1) , , a ' , 0.25 - MN II (a) :::::> na .NM =0 <=> 2m + n - 1 -em n - 5) 2m-n-2=0 <=> -m + n + 2 = 0 ¢:;> n = m _. 2. => NM =(3m 3' -3' -3m) , , :::::> NM = )(3m-3)2 +(_3)2 +9m 2 = 3J2m 2 -2m+2 NM 3.J6 <=> 2m 2 - 2m + 2 = 6 <=> m 2 m-2 o <=> m -1 hay m = 2. 0.25 i ~- x+l y+3 z-4 I » m = -1: M(-I;;Zj 4) va NM -3(Ui::-1) ~ Ed):· -"" , -~ ~:;::"~ 0.25 . ~. . 3 1 -1 I »m . . _ ' _ lli ~ 1 ~;~ . x 5 _ Y z+2 2. M(5, 0, -2)va NM - ~.J\J, r,-r.rt (d). 0.25 1~ -1 -2 I VIl.a Tim t~p hqp cac dibm bi~u dien cho s6 phuc z th6a: Iz + 3 2il = 12z + 1- 2il I=ld ! G9i M(x; y) la di~m bi~u di~n cho s6 phuc z =x + yi (x; y E R). Taco: Iz+ 3 - 2i l= 12z + 1- 2i l <=> Ix + yi + 3 - 2i\ 1 2 (x+yi)+1- 2i l <=>1(x+3)+(y 2)il == IC2x + 1) + (2y 2)il 0.5 <=> (x+3i+(y 2)2=(2x+1i+(2y-2t<=> 3x 2 -:3y-2x-4y 8=0 0.25 i j V~y t?P hqp cac diem _~ ia duang tron (T): 3x- + 3l- 2x - 4y 8 = O. 0.25 VI.b I,=2d I Trong m,t phing Oxy. cho tam giac ABC cO dinb A( 0.4 ttrong tim G ( ~; ~ ) I=ld , 1I I www.MATHVN.com www.mathvn.com I . bi~t r~ng XB < Xc. . 3 ( 4 ) , Xl-O=- 0 - 3- 23 {X,=2 . 0.25 Tac6 AI -AG :=;> ( ) ¢> _ => 1(2; -1). 2 Y -4 ~ ~ 4 YI 1 I 2 3 BC qua I va c6 VTPT la OA = (0;4) = 4(0;1) :=;>. BC: y =-1. GQi B(b; -1), vi I la trung di~m BC nen C(4 b; -1). - - Ta c6: OB::::: (b; 1) ; AC = (4- b;-5) 0.25 OB.AC:::: 0 ~ 4b - b 2 + 5 = 0 ~ b 2 - 4b 5 ::::: 0 ~ b = -1 hay b 5. I I : * b=-l :=;>B(-1;-l)vaC(5;-I)(nh~) I * b = 5:=;> B(5; -1) va C(-I; -1) (lo~i) 0.25 » BC:::: ( 6; 0) :=;> BC 6; d(A; BC) =5 :=;> SABC = 15. I 0.25 2 r 2 - t x-I +2 z-2 ' Trong kgOxyz cho (d 1 ): -2-:::: T == =2 ' (d 2 ): y 3 + t va m~t phang I=ld z=4+t I (a): x y + Z - 6 O. Tim tren (d 2 ) nhiing di~m M sao cho dlI6ng thAng qua M song song v6i. (d l ), c~t (a) 4ti N sao cho MN 3. M E (d 2 ):=;> M(2 m; 3 + m; 4 + m). r=2-m+2t (d) qua M va II (dJ) nen (d): y::::: 3+m+t 0.25 z 4+m-2t N (d) n (a) nen tQa dQ N th6a M: x 2 m+2t y ==3+m+t :=;> 2 - m + 2t - 3 - m t+4+m 2t 6=0 I z==4+m-2t i x-y+z 6 0 0.25 ~t -3 -m:=;> N(-3m - 4; 0; 3m + 10). i :=;> NM = (6 + 2m; 3 + m; -2m 6) 0.25 I I :=;> NM2 (2m + 6)2 + (m + 3)2 + (-2m- 6i I Do d6 MN == 3 ~ 9(m + 3i = 9 ~ m + 3 = ± 1 ~ m = -2 hay m = -4. ! V~y M(4; 1; 2) hay M(6; -1; 0). i 0.25 VII.h G···h" h .nh{e' e'=(lny Inx)(1+xy) 1al e p lIang tn . I=ld . inx+2lny _3.4 lnx ::::: 42 1ny Dieu ki~n: x, y > o. I Ta c6: 1 + xy > O. * x> y: VT (1) > 0 va VP(I) < O:=;> VT(I) > VP(l) (voU) * x < y: VT(1) < 0 < VP(l) (vo Ii) 0.25 Do d6 tir (1) :=;> x = y. Thay vao (2) ta duQ'c: 2 31nx _3.4 lnx =4.inx ¢::::>inx[{inxi_3.21nx 4] 0 0.25 [2'" 0 i i 2 1n x ::::: _ 1 <=> lnx =2 ¢::::> x :::: e 2 . I ~ 0.25 i 2 1nx == 4 '-:~ V~y Mc6 nghi~m la x y e.l. , _ _1 0.25 I BET www.MATHVN.com www.mathvn.com . DE THI THU D~I HOC LAN 3 - MON TOAN KHOI A Tinh Ba Ria Viing Tau Thai gian him bai: 180 phut. I. pHAN CHUNG CHO TAT CATHi SINH: (7 diim) 3 Cau I (2 di8m): Cho ham s6. dl,mg tai lif?u Can b9 coi thi kMng gial thlch gi them. HQ va ten thf sinh : ; So bao danh: www.MATHVN.com www.mathvn.com DAp AN DE THI THU D~I HQC MON ToAN ThiYi gian: 180. Cau III (l di~m): ' h' h liZx3~x3+8+(6x3+4x2)lnxdx h A I T III bc p an sau: = x Cau IV (l di~m): Cho hlnh chop SABCD co ABCD 1a hinh binh hanh tam 0, AB = 2a,