Gợi ý đáp án môn Toán

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www.VNMATH.COM THI TUYN SINH I HC NM 2012Mn : TOAN - Khi : A v A1PHN CHUNG CHO T T CA THI SINH (7,0 im)Cu 1 (2,0 im) Cho ha m s 4 2 22 1 1 y x ( m )x m ( ) + +,vi m l tham s thc.a) Khao sat s bin thin v v th hm s (1) khi m = 0.b) Tm m th hm s (1) c ba im cc tr to thnh ba nh ca mt tam gic vung.Cu 2 (1,0 im)Giai phng trinh 3sin2x+cos2x=2cosx-1Cu 3 (1,0 im)Giai h phng tri nh3 2 3 22 23 9 22 3 912x x x y y yx y x y + + '+ + (x, y R).Cu 4 (1,0 im)Tnh tch phn3211 ln( 1) xI dxx+ +Cu 5 (1,0 im) Cho hnh chp S.ABC c y l tam gic u cnh a. Hnh chiu vung gc ca S trn mt phng (ABC) l im H thuc cnh AB sao cho HA = 2HB. Gc gia ng thng SC v mt phng (ABC) bng600. Tnh th tch ca khi chp S.ABC v tnh khong cch gia hai ng thng SA v BC theo a. Cu 6 (1,0 im) : Cho cc s thc x, y, z tha mn iu kin x +y + z = 0. Tm gi tr nh nht ca biu thc2 2 23 3 3 6 6 6x y y z z xP x y z + + + + .PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B)A. Theo chng trnh ChunCu 7.a (1,0 im) : Trong mt phng vi h ta Oxy, cho hnh vung ABCD. Gi M l trung im ca cnh BC, N l im trn cnh CD sao cho CN = 2ND. Gi s 11 1;2 2M _ ,v ng thng AN c phng trnh 2x y 3 = 0. Tm ta im A.Cu 8.a (1,0 im) Trong khng gian vi h ta Oxyz, cho ng thng d: 1 21 2 1x y z + v im I (0; 0; 3). Vit phng trnh mt cu (S) c tm I v ct d ti hai im A, B sao cho tam gic IAB vung ti I.Cu 9.a (1,0 im). Cho n l s nguyn dng tha mn 1 35nn nC C . Tm s hng cha x5 trong khai trin nh thc Niu-tn 2114nnxx _ ,, x 0.B. Theo chng trnh Nng caoCu 7.b (1,0 im) Trong mt phng vi h ta Oxy, cho ng trn (C) : x2 + y2 = 8. Vit phng trnh chnh tc elip (E), bit rng (E) c di trc ln bng 8 v (E) ct (C) ti bn im to thnh bn nh ca mt hnh vung.Cu 8.b (1,0 im) Trong khng gian vi h ta Oxyz, cho ng thng d: 1 22 1 1x y z + , mt phng (P) : x + y 2z + 5 = 0 v im A (1; -1; 2). Vit phng trnh ng thng ct d v (P) ln lt ti M v N sao cho A l trung im ca on thng MN.Cu 9.b (1,0 im) Cho s phc z tha 5( )21z iiz+ +. Tnh mun ca s phc w = 1 + z + z2.www.VNMATH.COM wwww.VNMATH.COMBAI GIAI GI YPHN CHUNG CHO T T CA THI SINH (7,0 im)Cu 1: a/ Kha o sat, ve(C) :m = 0 y = x4 2x2 D = R, y = 4x3 4x, y = 0 x = 0 hay x = t 1Hm s ng bin trn (-1; 0) v (1; +), nghch bin trn (-;-1) v (0; 1)Hm s t cc i ti x = 0 v yC = 0,t cc tiu ti x = 1 v yCT = -1limxyt +Bng bin thin :x - -1 01 +y 0 +0 0 + y +1 +-1 -1y = 0 x = 0 hay x =2 t th tip xc vi Ox ti (0; 0) v ct Ox ti hai im (2 t; 0)b/y = 4x3 4(m + 1)xy = 0 x = 0 hay x2 = (m + 1)Hm s c 3 cc tr m + 1 > 0 m > -1Khi th hm s c 3 cc tr A (0; m2),B (- 1 m+ ; 2m 1); C ( 1 m+ ; 2m 1)Do AB = ACnn tam gic ch c th vung ti A. Gi M l trung im ca BC M (0; -2m1)Do ycbt BC = 2AM (ng trung tuyn bng na cnh huyn) 2 1 m+= 2(m2 + 2m + 1) = 2(m + 1)2 1 = (m + 1)1 m+= 32( 1) m+ (do m > -1) 1 = (m + 1)(do m > -1) m = 0Cu 2.3sin2x+cos2x=2cosx-12 3 sinxcosx + 2cos2x = 2cosx cosx = 0 hay3 sinx + cosx = 1 cosx = 0 hay 32 sinx + 12cosx =12 cosx = 0 hay cos( ) cos3 3x x = 22k hay x k + hay 223x k + (k Z).Cu 3:3 2 3 22 23 9 22 3 912x x x y y yx y x y + + '+ + t t = -x H tr thnh 3 3 2 22 23 3 9( ) 2212t y t y t yt y t y + + + + '+ + + . t S = y + t; P = y.tH tr thnh3 2 3 22 23 3( 2 ) 9 22 3 3( 2 ) 9 221 1 12 ( )2 2 2S PS S P S S PS S P SS P S P S S + + ' ' + + 3 223 2 6 45 82 041 1( )22 2S S SPP S SS + + + ' ' + . Vy nghim ca h l 3 1 1 3; ; ;2 2 2 2 _ _

, ,www.VNMATH.COMxy-1 -O--11www.VNMATH.COMCch khc : 3 2 3 22 23 9 22 3 91 1( ) ( ) 12 2x x x y y yx y + + ' + + . t u = x12 ; v = y + 12H cho thnh 3 2 3 22 23 45 3 45( 1) ( 1) ( 1)2 4 2 41u u u v v vu v + + +'+ Xt hm f(t) = 3 23 452 4t t t c f(t) = 2453 34t t < 0 vi mi t that 1f(u) = f(v + 1) u = v + 1 (v + 1)2 + v2 = 1 v = 0 hay v = -1 01vu 'hay 10vu 'H cho c nghim l 3 1 1 3; ; ;2 2 2 2 _ _

, ,.Cu 4.3211 ln( 1) xI dxx+ + = 3 32 21 11 ln( 1) xdx dxx x++ = 131 1xJ += 23J + . Vi 321ln( 1) xJ dxx+t u = ln(x+1) du = 11dxx + ;dv = 21dxx, chn v = 1x - 1J = 31( 1) ln( 1)1xx + + 31dxx= 31( 1) ln( 1)1xx + + 31ln x = 4ln 4 2ln 23++ ln3 = 2ln 2 ln33+ . VyI = 2 2ln 2 ln33 3+ +Cch khc : t u = 1 + ln(x+1) du = 1dxx +; t dv = 2dxx, chn v = 1x, ta c :[ ]3111 ln( 1) I xx + ++ 31( 1)dxx x + = [ ]3 31 111 ln( 1) ln1xxx x + + ++= 2 2ln 2 ln33 3+ +Cu 5.Gi M l trung im AB, ta c2 3 6a a aMH MB HB 22223 28 72 6 36 3a a a aCH CH 1 1 + 1 1 ] ]2 723aSC HC ; SH = CH.tan600 = 213a( )2 31 7 7,3 4 12a aV S ABC a dng D sao cho ABCD l hnh thoi, AD//BCV HK vung gc vi AD. V trong tam gic vungSHK, ta k HI l chiu cao ca SHK.Vy khong cch d(BC,SA) chnh l khong cch 3HI/2 cn tm.2 3 33 2 3a aHK , h thc lng 2 2 2 2 21 1 1 1 121 33 3HI HS HKa a + + _ _ , ,www.VNMATH.COMBA C S H M K D I www.VNMATH.COM[ ]42 3 3 42 42,12 2 2 12 8a a aHI d BC SA HI Cu 6. x + y + z = 0 nn z = -(x + y) v c 2 s khng m hoc khng dng. Do tnh cht i xng ta c th gi s xy 0Ta c 2 2 2 23 3 3 12( )x y y x x yP x y xy + + + + + += 2 2 23 3 3 12[( ) ]x y y x x yx y xy + + + + + 2 2223 2.3 12[( ) ]y x x yx yx y xy+ + ++ + 323 2.3 2 3x yx yx y++ +. t t = 0 x y + , xt f(t) = 32.( 3) 2 3tt f(t) = 3 32.3( 3) .ln 3 2 3 2 3( 3.( 3) ln 3 1) 0t t >f ng bin trn [0; +) f(t) f(0) = 2M 3x y 30 = 1. Vy P 30 + 2 = 3, du = xy ra x = y = z = 0. Vy min P = 3.A. Theo chng trnh Chun :Cu 7a. Ta c : AN = 103a; AM = 52a; MN = 56a;cosA = 2 2 22 .AM AN MNAM AN+ = 12 45oMAN (Cch khc : tnh MAN = 450 ta c th tnh 123( ) 111 2.3tg DAM DAN +)Phng trnh ng thng AM : ax + by 11 12 2a b = 02 221cos25( )a bMANa b + 3t2 8t 3 = 0 (vi t = ab) t = 3 hay 13t + Vi t = 3 ta A l nghim ca h : 2 3 03 17 0x yx y '+ A (4; 5)+ Vi 13t ta A l nghim ca h : 2 3 03 4 0x yx y ' A (1; -1)Cch khc: A (a; 2a 3), 3 5( , )2d M AN , MA = 3 10. 22MH 2 211 7 45( ) (2 )2 2 2a a + a = 1 hay a = 4 A (1; -1) hay A (4; 5).Cu 8a. Ta c M (-1; 0; 2) thuc d, gi duuur = (1; 2; 1) l vect ch phng ca ng thng d.[ , ]2( , )2 2ddMI uAB RIH d I du uuur uuruur ,[ , ] ( 2; 0; 2)dMI u uuur uur IH = 8 26 32 22 3RR = 2 63 phng trnh mt cu (S) l : 2 28( 3)3x y z + .Cu 9.a.1 35nn nC C ( 1)( 2)5.6n n nn 30 = (n 1) (n 2), (do n > 0) n = 7www.VNMATH.COMBACDNMwww.VNMATH.COMGi a l h s ca x5 ta c 727 571.2iiixC axx _ _ , , 77 14 3 571( 1) . .2ii i iC x ax _ ,14 3i = 5 i = 3 v7771.2iiC a _ , a = 3516. Vy s hng cha x5 l 3516.x5.B. Theo chng trnh Nng cao :Cu 7b Phng trnh chnh tc ca (E) c dng : 2 22 21 ( )x ya ba b+ > . Ta c a = 4 (E )ct (C ) ti 4 im to thnh hnh vung nn : M (2;-2) thuc (E) 2 24 41a b + 2163b . Vy (E) c dng 2 2116163x y+ Cu 8b. ( 1 2 ; ; 2 ) ( ) M d M t t t t R + + ; A l trung im MN (3 2 ; 2 ; 2 ) N t t t ( ) N P 2 t ( 1; 4; 0) N ; i qua A v N nn phng trnh c dng : 1 42 3 2x y z + + Cu 9b. z x yi +5( )21z iiz+ + 5( )21x yi iix yi + + + 5[( ( 1) )2( 1)x y iix yi + +5 5( 1) 2( 1) ( 1) 2 x y i x x i yi y + + + + 5 5( 1) (2 2 ) ( 1 2 ) x y i x y x y i + + + 2 2 51 2 5( 1)x y xx y y+ + '+ 3 27 6x yx y ' 11xy 'z = 1 + i; 2 21 1 (1 ) (1 ) w z z i i + + + + + +1 1 1 2 ( 1) i i + + + + +2 3i + 4 9 13 w + www.VNMATH.COM