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./. TRUONG THPT NGUYEN HU.I;: TO TOm I.Phin DE THl TOO TUYEN SINH D~I HOC LAN 2 N.AM HOC 2014 Mon: ToAN kh6i A , B ,AI Thai gian lam bai : 180 phut, khong k~ thai gian phat dS. chung cho t~t cii thi sinh( 7,0 di~m) Cau 1( 2,0 di~m): Cho ham sf, y = 2x + 1 co db thi (1) x-I a/ Khao sat sir bi~n thien va ve db thi (1). bl Tim nhfrng di~m tren true tung ma tir do ke duoc d~n db thi (1) dung m(>tti~p tuyen. C aD 2(10di% , em.)'Gi--ph~ at trong tninh:. cos2x(I+2sin2x)+3cot2X_2 cot 2x - cos 2x - A • Cau 3 (1,0 diem): Giai h~ phirong trinh: {x2 . 2 _ = io ~X2 +21 1 = 2x + y2 2X2 -4x+3+ CIiD 4 (1,0 di€m): Tinh tich phan: I CaD Goc Tim Cau y ° y3 =0 2 dx 5 (1,0 di~m): Cho hlnh chop t(r giac dSu S.ABCD.Kholing each ill di~m A d~n mp(SBC) bang 2a. gitra m~t ben va day cua hinh chop la a . Tim a d~ th~ tich kh6i chop dat gia tri nho nhat, . gia tri nho nh~t do. 6 (1,0 di€m): Cho x,y,z Iii. cac sf>dirong sao cho: xyz + x - y + Z = ° A ., tr]'1'on n hI.at cua • b'leu th'ire: P = -2 2- - -2 2- + -2 3T'im gia x +1 Y +1 Z +1 n. Ph in rieng (3,0 di~m) Thi sinh chi duoc lam m(>ttrong hai ph§.n ( ph~n A hoac ph~n B) A . A. Theo chlfO'ng trlnh chuan: ° CaD 7a (1,0 di~m): Trong mat phang voi h~ toa d(>Oxy,cho dirong trim (C): X2 + y2 - 4x = Tim nhirng di~m tren duong thkg x = 4 ma tir nhirng di~m do ke diroc d~n (C) hai tiSp tuyen hop vci nhau goc 30°. . Can 8a (1,0 di~m): Trong khong gian voi h~ true Oxyz, cho hai di~m A(1,2,-1) ;B(-I, 0,-1) Tim di~m M thuoc mp(Oxy) sao cho tam giac ABM vuong tai M va co dien tich nho nhk CaD 9a ( 1,0 di~m): ZI,Z2 ,Z3 'Z41a cac nghiern cua plurong trinh: (Z2 + 1Xz2 -2z T'mh :. S -- ZI2014+ Z22014+ Z32014+ Z42014 + 2)= ° B. Theo ChlfO'l1f;trinh nang cao: Can 7.b ( 1,0 diem): Trong m~t phkg voi h~ toa dQ Oxy, cho di~m N(2, 1).Vi~t phirong trinh dirong th~ng d di qua diSm N va c~t hai mra true dirong Ox, Oy l~n hrot tai A va B sao cho (OA + OB) . dat gia tri nho nhk Cau 8.b ( 1,0 diSm): Trong khong gian voi h~ true Oxyz, cho dirong thang a la giao tuyen cua hai m~t °; ° ph~ng (P): 2x - 2y - Z + 1 = (Q): x + 2y - 2z - 4 = va m~t du (S): X2 + y2 + Z2 + 4x + 6y + m = Tim cac gia tri cua m d~ duong thang a c~t m~t du (S) tai hai di~m H va K sao cho HK ., 8. . Cau 9.b ( 1,0 diem): Gilii h~ phirong trinh: I 2 -log3 X -log3 Y 2 { 3 x + y2 - 2y Il ° =° = ---------------II~t-------------- ° DAP AN HE Till TOO DAI HOC KHOI A,B,A) NAM 2014 Cau Cau 1: (2 diSm) Diem aI (1 di~m)+ TIm: D + SBT: = R \ {I} + Limy =+00 ; Limy =-00 + Lim y y = 2 IIIpt TCN =2 ~ x-+±«> y x=ll11ptTCD x-+)- .>:-+)+ --------------------------------------- 0,25 +00 2 ~2 ~-oo + His nghjch hi~n tren D, his khong + Db thi : + DDB : x = y = -1 Y °~ co C\TC tri 0,25 y=0~x=-1/2 +ve db th]: + Gd 2 tc III tam dx x -112\ 0,25 -1 b/ (1 di~m).GQi A(O, a) thuQc true Oy, a e R Duong thang d di qua A h~ goc k: y = kx + a co so 2X+ 1 =kx+ a d ti~p xuc db thi (1) B j x~ ~ (x _1)2 = k. . co nghiern . x:t:-1 ------------------------ ° Pt dinh hoanh dQ ti~p di~m: (a - 2)x2 - 2(a + l)x + a + 1 = (1) -----------------Xet hai tnrong hop: + a = 2, (1) co nghiem x = 3/4 thoa ycht ~ a = 2 (nh?ri) 0,25 0,25 + {a:t:-2. Ba=-1 /).l =0 KL : Nhtrng di~m tren true tung thoa ycbt: (0,2); -'gp (0, -1) --,---------------------:-- 0,5 CaU2 I (ldi[rn) 0,25 UK : {::~:: ~os 2x ~ 0 B {:~:~:: ~1. Voi di€u kien tren (1) J ----------------------------------------------- + 3 cot 2x + sin 4x = 2 cot 2x - 2 cos 2x 3cos2x + cot2x + sin 4~ = 0 cos 2x(2 sin 2 2x + 3 sin 2x + 1)= 0 ---------------------------- 0,25 sin 2x = -112 0,25 COS2x B B B B r~:;::~~i) B --------------------------- sin 2; = -1I2(n} B l x=-~+k7r r 12 0,25 kEZ 71C x=-+k7r Cau3 (1 di~m) H~ pt t:x~~;:' # 1::' ~ 0 12 0,25 -------------------------------------------- 0,25 s 1 'IIx E R nen -1 :::;Y s 1 -------------------------------------------------- Vi ~ l+x y~-l~l+ +1+ y3 ~O, v~y y3 ~O ~2(x-l)2 {X-l;O l+y=O B{X=1 y=-1 Thay x = 1, y = -1 vao (1) dung KL : HAe eo, ng hiAern {X=1 y = -1 Cau4 ~ (1 diem) dx l"/X2 +ldx+ I~ X2 + 1 l~ 1= I o 0 = Tinh II r==: I"/ X2 + Idx o i ~I II =x"/x2+11 II II =.J2 - o f.Jx2 o V~y 1 = I= G9i M,N 2 £)~t {~{dU= U = x + dv=dx . ~ ~ xdx ----------------------- - v=x II x dx ~ 0 X2 + 1 +ldx+ f~ X2 0 +1 dx 0 .Jx2 _ +1 ~[.J2 +31n(I+.J2)f ~ [.J2 + 3ln(1+ .J2)] -------------------------------~-----------------------------~ + %lnlx+~II~ 1k hrot 0,25 2 = .J2 +..!.- II 2 cse s ----.(1 diem) ---------------------------------------------------------- 0,5 0,5 = la trung di~m Cua BC,AD. S MN la goc gitra m~t ben Vi ADIIBC ~ AD//mp(SBC)~d[A,(SBC)]=d[N~(SBC)] (SMN).l(SBC), nen nSuke NH.lSM ~NH .l(SBC)~d[N,(SBC)]=NH 0,25 va day. SMN=a. A NH 2a sma sma =2a Trong tg vuong MHN: MN =-.- =-,2 SABeD = AB 2 = MN 2 = .4a2 sm a ---------------------------------------------------------- 0,25 c B A . d ay, ' SI =MI. tan a = -.-a tan a =--a G 91. I I'" a tam cua . sma cosa 3 VABCD=.!.SABCDS1=.!.~._a_= 3 Tir (2) . 3 sin2 a cosa . (I) (2) 4a 3sin2 a.cosa . 2 1 nho nh~t hay sin2a cosa Ian nhat. . . sm acosa Do 0° 0,f)l;itx= cosa (O 0, nen dlit x = tan a ,~= tan fJ ,z . ··2 Y 2 = tan r 2 a f3 f3 r + tan-tanr a (1) ~ tan-tan+ tan-tan=1 2 2 2 2 2 2 . . ~tana f3 r =1-tanf3tanr ()tan +tan 2 2 2 2 r) ~ ~ tan a = cot( p + 2 222222 Vi I+X2 =1+tan2 ~ = a + + r 2 = 1( hay a + fJ + nrong tu cho 1+ y2 I, cos? p f3 r I-tan-tan-'~tana = 2 2 2 2 tanf3+tanr r= .2 1( va I+Z2 0,25 a 2 P=2cos2 a_2sin2 2 f3 +3cos2 ,2 r = l+cosa-(I-cosfJ)+3cos2 . . . 2 . r2 P = 2cosa + 13cos a 2 _ P- 3-3 (.2r [(.rsm---cos-I a-fJ)21 a-f31 2·r 2 a - fJ ---3 3 (. 2 a- 1 2 cos = f3' va -,81 --J a 2 2 (2) {a fJ = 1 2 . r 1 a{sm---cos--=O 232 B = 2. -+ tan r =! 2 r - --cos 9 ------------------------------------cos Tir (a) -+ 13)2 2 232 -' D~u "=" xay ra 23 rIa sm---cos-_· ro (2) -+ P s l~ V'·m a r = _ 3sin2 r + 2sin r cos a - 13 + 3 2 2 2 2 ---srn-cos--1=3-3 2 3 2 2) SIn 1 P= 3+-cos 13 + 3cos2 2 8 2 8 B 13· . = 13 r1 (a) sm - =. 0,25 2 3 -+ z =tan r =_1_ 2 2..fi 0,25 2 tan-r 2 1r;:; = 2..;2 0 V~y maxP = 13 khi x th ay vao '(1) : tan2 -a + r;:; 1 tan a - 1= 0-+ . tan-a 2 = ~ ; y =.fi ; z = ..;2 2 1 = ..;2 r;:; 21 -----.~---------------------------0,25 II.Phlin rieng(3 diem) A.Theo chuO'Dg trinh chu~n Cau 7a (1 diem) GQi diem M(4, b)thuQC at x = 4, (b E R) (C):(x-2Y + y2 =4 , (C) co tam 1(2, 0), bk r = 2 Do dirong thang x = 4Ii ti~p tuyen cua (C), nen ycbt la tim nhiing diem tren at x = 4 co h,~s6 goc k = ± tan60° = ±..fj ----------------------------------------------* k =..fj: d Ii dt qua M co hsg k =..fj co pt: y = ..fj (x - 4) + b B..fjx- y-4..fj 0,25 +b=O d ti€p xuc voi (C) B d(l, d) = r <-> * k= -..fj: Ib- 2-'31= 4 hay d'IidtquaMcohsgk= [:::+ +2~ -------------------------------..fj co pt: y=-..fj(x-4)+b B..fjX+ d' ti€p xuc voi (C) B <-> (1 diem) y-4..fj -b=O d(I, dl) = r 1- b - 2-'3 = 41 hay [: : ~ KL: co 4 diem: (4,4+2~"); Cau8a 0,25 ~~ (4,-4+2~");(4,-4-2.J3},(4,4-2F3) GQi M(x,y,O) thuoc mp(Oxy)AM(x-1,y-2,1); ----------------------------- --------------------------- 0,25 0,25 BM(x+ 1,y,1); :AB(-2,-2,0) . I1MAB vuong tai M B AM.EM = 0 B X2 + y2 - 2y = 0 (1) --------------------------. SMAB 1 =-MH.AB 2 ,SMABnh6 0,25 (AB=2..fi) nh~t khi MH = d(M, AB) nho nhat, voi MH =JS + 4(x - y + 1Y .-------,.-0,25 3 MH nho nhfrt khi x - y + 1= 0 (2) Tir (1) va (2) ~ Cau9a (1 diem) Pt~[Z: Z ~jfi ml oJ 1-{-fi 2' 2' 0,25 -----------------------------------------------------hoac M(- 0 fi .J2 1- 2' _ 1 0 2') 0,25 2 =-1=i -2z+2=0 ~ [: : ~: i ------------------------------ -------------------- S = i2014+ (_i)2014 + (1- i)2014+ (1 + i)2014= (i2 )1007+ [c-i)2 Y007+ (_2i)1007 + (2i)1007 S = - 2 + (_2)1007i1007+ 21007il007 = - 2 B.Theo chU01lg trinh nang cao: 0,5 0,25 0,25 -------------------------------------------------- Cau7b (1 diem) GQi pt d: : + ~ =1 ( a, b >0 ) N(2, 1) thuoc d ~~+.!.=1~b=-aa b a-2 ( a > 2) ------------------------------------- 0,25 OA+OB=a+b=a+ _a_ a-2 2 a - 4a+ 2 f( a) =a+-- a ~ f' ()a =---a-2 (a-2)2 0,25 a f' (a) = 0 t-7[a = 2 - fi(Z) . a=2+Ji(n) KL: -Cau 8b (1 diem) Pt dt d: x r;:; + ~v 2+v2 v2+1 (S): (x+ 2)2 + (y + 3)2 + IJ = d(1,d) = L~p BBT Z2 ../13- m -16 l~ p » nQ = ../13 - m (1) (J 1ft trung diem HK) (m < 13) (a) 1[~,Ai], I~I' 0 VOl - ( ) AI= -2,-4,1 = ../153 (2) = .J153 ~ m = -20 (thoa dk (a)) 3 + 0,5 ../81+36+36 ../4+1+4 [;,AI]=(9,-6,-6)~d(1,d)= Cau9b ° - j= (6,3,6) = 3(2,1,2) A(O 1 1) Lt-ay dO;' rem ,,Ed, d(1,d) = (1 diem) f f =1 = ../- m -3 +~ 2+fi =13 - m ~ (S)co tam 1(-2, -3, O),bk R Duong thfulg d co vtcp ~ = Tir (1) va (2) .J- m - 3 2 3