www.VNMATH.com TRNGIHCVINH TRNGTHPTCHUYấN KHOSTCHTLNGLP12,LN1ư NM2014 Mụn:TON Khi:A vA1 Thigianlmbi:180 phỳt I PHNCHUNGCHOTTCTHSINH(7,0 im) x- Cõu (2,0 im). Chohms y = x -1 a) Khosỏtsbinthiờn vvth (H)cahms ócho. b) Tỡm m ng thng d : x + y + m =0 ct (H) ti hai im M, N cho tam giỏc AMN vuụng ti imA(1 0). Cõu (1,0 im). Giiphngtrỡnh sin 3x + 2cos2 x = + 4sin x + cos x(1 +sin x). Cõu3(1,0 im). Gii btphngtrỡnh x + + 2 x + Ê ( x - 1)( x -2). x + 2ln(3 x+ 1) Cõu (1,0 im). Tớnh tớchphõn I = ũ dx. ( x + 1) Cõu5(1,0im). ChohỡnhchúpS.ABCD cúỏy ABCD lhỡnhchnht,mtbờn SAD ltamgiỏcvuụngti S, hỡnhchiuvuụnggúcca S lờnmtphng(ABCD)lim H thuccnh AD saocho HA =3HD. Gi M ltrung im caAB.Bitrng SA =2 3a vngthng SC toviỏy mtgúc 300. Tớnhtheo a thtớchkhichúp S.ABCD vkhongcỏcht M nmtphng(SBC). Cõu6(1,0im). Gisx,y,zlcỏcsthckhụngõmthamón 5( x + y + z 2) = 6( xy + yz +zx ). Tỡmgiỏ trlnnhtcabiuthc P = 2( x + y + z ) - ( y +z 2). II PHNRIấNG (3,0 im) Thớsinhchclmmttronghaiphn(phn a hoc phn b) a.TheochngtrỡnhChun Cõu7.a(1,0 im). Trongmtphngvihta Oxy, chotamgiỏcABCcú M(2 1) ltrungimcnh AC, im H (0 -3) lchõnngcaokt A,im E (23 -2) thucngthngchatrungtuynkt C.Tỡmta im B bitim A thucngthng d : x + y - =0 vim C cúhonhdng. x + y - z- Cõu8.a(1,0im).Trong khụng gian vi hta Oxyz, chongthng d : v hai = = -1 mtphng ( P ) : x + y + z + = 0, (Q ) : x - y - z + =0. Vitphngtrỡnhmtcucútõmthucd,ng thitipxỳcvihaimtphng(P)v(Q). Cõu9.a (1,0 im). Chotphp E ={1, 2, 3, 4, 5}. GiM ltphpttccỏcstnhiờncúớtnht3chs, cỏcchsụimtkhỏcnhauthuc E.Lyngunhiờnmtsthuc M.Tớnhxỏcsuttngcỏcchsca súbng10. b.TheochngtrỡnhNõng cao Cõu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hai im A(1 2), B(4 1) v ng thng D : x - y + =0. Vitphngtrỡnh ngtrũn iqua A,Bvct D ti C,D cho CD =6. Cõu 8.b (1,0 im). Trong khụng gian vi h ta Oxyz, cho im M(1 0) v hai ng thng x -1 y - z -1 x - y + z- d1 : = = , d : = = Vitphngtrỡnhmtphng(P)songsongvi d1 v d2 -1 -1 -3 ngthicỏch Mmtkhongbng 6. Cõu9.b (1,0 im). Tỡmsnguyờndng n thamón 1 1 ( -1) n n Cn - Cn + C n - Cn + + Cn = n +2 156 ưưưưưưưưưưưưưưưưưưHtưưưưưưưưưưưưưưưưưư www.VNMATH.com TRNGIHCVINH TRNGTHPTCHUYấN PNKHOSTCHTLNGLP12,LN1 ưNM2014 Mụn:TON KhiA,A1 Thigianlmbi:180phỳt Cõu ỏpỏn im a)(1,0im) Cõu1. 10.Tpxỏcnh: R\{1}. (2,0 20.Sbinthiờn: im) *Giihntivụcc:Tacú lim y =2 v lim y =2. xđ-Ơ xđ+Ơ Giihnvụcc: lim+ y = -Ơ v lim- y = +Ơ. xđ1 xđ1 Suyrath(H)cútimcnnganglngthng y =2, timcnnglngthng x =1. *Chiubinthiờn:Tacú y ' = > 0, "xạ 1. ( x -1)2 Suyrahmsngbintrờnmikhong ( -Ơ 1) v (1 + Ơ) 0,5 *Bngbinthiờn: x -Ơ y' + +Ơ y + +Ơ y 2 -Ơ I 0,5 30.th: x O ổ thctOx ti ỗ ữ , ct Oy ti (03). ố ứ Nhngiaoim I(1 2) cahaitimcn lmtõmixng. b) (1,0im) m Ta cú d : y = - x - Honh giao im ca d v (H) l nghim ca phng trỡnh 3 2x - m (1) = - x- , hay x + ( m + 5) x - m - = 0, x ạ1. x -1 3 Tacú D = (m + 7)2 + 12 >0, vimim.Suyraphngtrỡnh(1)cú2nghimphõnbit.Hnna c2nghim x1 , x2 ukhỏc1.Doú d luụn ct(H)ti2imphõnbit M ( x1 y1 ), N ( x2 y2). uuuur uuur Tacú AM = ( x1 - y1 ), AN = ( x2 - y2). uuuur uuur Tamgiỏc AMNvuụngti A AM AN = 0. Hay ( x1 - 1)( x2 - 1) + y1 y2 =0 ( x1 - 1)( x2 - 1) + ( x1 + m )( x2 + m ) =0 (2) 10 x1 x + ( m - 9)( x1 + x2) + m + =0. pdngnhlýViet,tacú x1 + x2 = -m - 5, x1 x2 = -m -9. Thayvo(2)tac 10( -m - 9) + (m - 9)( -m - 5) + m + =0 -6m - 36 = m = -6. Vygiỏtrcam l m = -6. Phngtrỡnh óchotngngvi Cõu2. sin 3x - sin x + 2cos x = 3(sin x + 1) + cos x(sin x +1) (1,0 2cos x sin x + 2cos x = (sin x + 1)(cos x+ 3) im) (sin x + 1)(2cos x - cos x- 3) = (sin x + 1)(4cos x - cos x- 5) = (sin x + 1)(cos x + 1)(4cos x - 5) =0. 0,5 0,5 0,5 www.VNMATH.com p +k 2p , k ẻ Z. *) cos x = -1 x = p +k 2p , k ẻ Z. *) 4cos x - =0 vụnghim. p Vyphngtrỡnhcúnghim x = - + k 2p , x = p + k 2p , k ẻ Z. iukin: x -1. Cõu3. Nhnthy x = -1 lmtnghimcabtphngtrỡnh. (1,0 Xột x > -1. Khiúbtphngtrỡnh óchotngngvi im) x + - + 2 x + - Ê x3 - x - x - 12 *) sin x = -1 x = - ( ) ( 4( x - 3) x +1 + + ) 0,5 4( x- 3) x+ + Ê ( x - 3)( x + x+ 4) ổ 4 ( x - )ỗ + - ( x+ 1) - ữ Ê (1) x + + ố x +1 + ứ 4 Vỡ x > -1 nờn x + >0 v x + >1. Suyra + < 3, vỡvy x +1 + 2 x + +3 4 + - ( x+ 1) - < 0. x +1 + 2 x + +3 Doúbtphngtrỡnh (1) x - x 3. Vynghimcabtphngtrỡnhl x = -1 v x 3. 0,5 0,5 3x ln(3 x+ 1) dx + ũ dx. Cõu4. Tacú I = ũ ( x + 1) ( x + 1)2 0 (1,0 im) t u = ln(3 x + 1) ị du = 3dx dv = dx ị v= - x+ x +1 ( x +1)2 pdngcụngthctớchphõntngphntacú 3x 2ln(3 x + 1) dx2 x +1 ( x+ 1) I=ũ 1 dx (3 x + 1)( x+ 1) + 6ũ ổ 3 ử ổ = ũỗ dx - ln + 3ũỗ ữ dx ữ x + ( x+ 1) ứ x + x+ 1ứ 0ố 0ố = 0,5 - ln + 3ln 3x+ x +1 0,5 = - + 4ln 2. ã = (ã SC , ( ABCD) ) =300. Vỡ SH ^( ABCD ) nờn SCH Cõu5. (1,0 im) S Trongtamgiỏcvuụng SAD tacú SA2 = AH AD 12a = AD ị AD = 4a HA = 3aHD =a C ị SH = HA.HD = a ị HC = SH cot 30 =3a ị CD = HC - HD =2 2a. K Suyra S ABCD = AD.CD =8 2a 2. a H' D H A M B 0,5 6a3 Suyra VS ABCD = SH S ABCD = 3 Vỡ M ltrungim ABv AH // (SBC)nờn 1 d ( M , ( SBC ) ) = d ( A,( SBC ) ) = d ( H , ( SBC ) ). (1) 2 K HK ^BC ti K, HH '^SK ti H'.Vỡ BC ^( SHK ) nờn BC ^ HH ' ị HH ' ^( SBC ). (2) 1 11 6a 66 = + = ị HH ' = = a. (3) 2 2 11 HH ' HK HS 24a 11 66 T(1),(2)v(3)suyra d ( M , ( SBC ) )= a. 11 2 2 2 Cõu6. Tacú x + 2( y + z ) Ê x + 5( y + z ) = 6( xy + yz +zx) Ê x ( y + z ) + 4( y +z ) 0,5 Trongtamgiỏcvuụng SHK tacú 0,5 www.VNMATH.com (1,0 y + z 2 im) Doú x - x ( y + z ) + ( y + z ) Ê0, hay Ê x Ê y +z. Suyra x + y + z Ê 2( y +z ). 1 Khiú P Ê 2( x + y + z ) - ( y +z )2 Ê 4( y + z ) - ( y + z ) = y + z - ( y +z ) 2. 2 t t y + z =t , khiú t v P Ê 2t - (1) t Xộthms f (t ) = 2t - t vi t 0. +Ơ f '(t) + Tacú f '(t ) = - 2t f '(t ) = t =1. Suyrabngbinthiờn: f (t) vimi t 0. (2) ỡ x = y + z ỡ x= ù ù T(1)v(2)tacú P Ê , dungthcxyrakhi y = z ù y + z = ùợ y = z = ợ VygiỏtrlnnhtcaP l , tckhi x = 1, y = z = 2 ỡ x = - 3t A ẻ d : 2x + y - = ị A(-3a + 1, 2a+ 1). Cõu d y = + 2t ợ A 7.a Vỡ M(2 1) ltrungim AC nờn suyra C (3 + 3a -2a ) (1,0 uuur im) ỡù HA = ( -3a + a+ 4) M N ị uuur ùợ HC = (3 + 3a - 2a ). Davobngbinthiờntacú f (t ) Ê f (1)= E B H C ộ a= uuur uuur Vỡ ã AHC =90 nờn HA.HC = 0ị ờ a = - 19. ờở 13 0,5 0,5 *)Vi a = ị A(-2 3), C (6 -1) thamón. *)Vi a = - 19 ổ 18 51ử ị C ỗ - ữ khụngthamón. 13 ố 13 13ứ Vi A(-2 3), C (6 -1) tacúphngtrỡnh CE : x + 17 y + 11 =0, phngtrỡnh BC : x - y - =0 ổ 3b + b+ 3ử Suyra B (3b + b)ẻ BC ị trungim AB l N ỗ ữứ ố MN ẻ CE ị b = -4 ị B (-3 -4). Tõmmtcu (S)l I (t - - t + 2t + 2) ẻd Cõu Vỡ(S)tipxỳc(P),(Q)nờn d I , ( P ) = d I , (Q) = R ( ) ( ) 8.a 1 (1,0 ộ ộ ờt = -2, R = I (-4 - 2), R= -t- im) 3t + = = R ịờ 3 ờt = -3, R = I (-5 - 4), R = ờở ờở 3 Suyrapt(S)l ( x + 4)2 + ( y - 3)2 + ( z + 2)2 = hoc ( x + 5) + ( y - 4) + ( z + 4) = 9 Cõu 9.a (1,0 0,5 0,5 0,5 Scỏcsthuc Mcú3chsl A53 =60. Scỏcsthuc Mcú4chsl A54 =120. 0,5 www.VNMATH.com im) Scỏcsthuc Mcú5chsl A55 =120. SuyrasphntcaM l 60 + 120 + 120 =300 Cỏctpconca E cútngcỏcphntbng10gm E1 = {1,2,3, 4}, E2 = {2,3,5}, E3 ={1, 4,5}. Gi A ltpconcaMmmisthuc A cútngcỏcchsbng10. T E1 lpcscỏcsthuc A l 4! Tmitp E2 v E3 lpcscỏcsthuc A l 3! Suyrasphntca Al 4!+ 2.3! =36 36 Doúxỏcsutcntớnhl P = =0,12. 300 Gis(C)cútõm I (a b), bỏnkớnh R >0. B Cõu Vỡ(C)iqua A,B nờn IA = IB =R A 7.b (a - 1)2 + (b - 2) = (a - 4)2 + (b - 1)2 = R I (1,0 im) ùỡb = 3a - ùỡ I (a 3a- 6) D H ịớ ịớ 2 D C ù R = 10a - 50a + 65 ợù R = 10a - 50a + 65 ợ -9a+ 29 K IH ^CD ti H.Khiú CH = 3, IH = d ( I , D )= (9a- 29)2 ị R = IC = CH + IH = 9+ 25 (9 a- 29)2 169a - 728a + 559 =0 T(1)v(2)suyra 10a - 50 a + 65 = + 25 I (1 3), R = ộ ộ a= ờ ị ổ 43 51 61 a = 43 I , R = ốỗ 13 13 ứữ ờở 13 13 0,5 0,5 (1) (2) 43 ổ 51 1525 ổ Suyra (C ) : ( x - 1) + ( y + 3) =25 hoc (C ) : ỗ x - ữ + ỗ y - ữ = 13 ứ ố 13 ứ 169 ố uur uur uur uur ỡùu1 = (1 - 1) ị nP = ộởu1 , u2ựỷ = (1 1) Cõu Vỡ ( P) // d1 ,d2 nờn(P)cúcpvtcp uur ùợu2 = (-1 - 3) 8.b (1,0 Suyrapt(P)cúdng x + y + z + D = 0. im) 3+ D (1) ộ D= ộ( P ) : x + y + z+ = d ( M , ( P) )= = ịờ (2) ởD = -9 ở( P ) : x + y + z - = Ly K (1 1)ẻd1 v N (1 - 2)ẻd th vo cỏc phng trỡnh (1) v (2) ta cú N ẻ ( P ) : x + y + z + =0 nờn d è ( P ) : x + y + z + =0. Suy phng trỡnh mt phng (P) thamónbitoỏnl ( P ) : x + y + z - =0. Vimi x ẻ R vmisnguyờndng n,theonhthcNiutntacú Cõu Cn0 x - Cn1 x + + ( -1) n Cnn x n +1 = Cn0 - Cn1x + + ( -1) n Cnn x n x = (1 -x ) nx. 9.b 1 (1,0 Suyra C x - C1 x + + ( -1) n C n x n +1 dx = (1 - x) nxdx. n ũ0 n n ũ0 im) 0,5 ( ( ) 0,5 0,5 0,5 ) 1 1 (-1)n n Hay Cn0 - Cn1 + + Cn = ũ (1 - x) n dx - ũ(1 - x)n+1dx n + 0 1 = = ,vimi n ẻ N*. n + n + ( n + 1)( n +2) 1 = n + 3n - 154 = n= 11 (vỡ n ẻ N*). Tútacú ( n + 1)( n +2) 156 0,5