./ TRUONG THPT NGUYEN HU.I;: TO TOm I.Phin DE THl TOO TUYEN SINH D~I HOC LAN 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 + 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 ke duoc d~n db thi (1) dung m(>tti~p tuyen C aD 2(10di% , em.)'Gi ph~ at tninh: cos2x(I+2sin2x)+3cot2X_2 cot 2x - cos 2x - A • Cau (1,0 diem): Giai h~ phirong trinh: {x2 _ = io ~X2 +21 = 2x + y2 2X2 -4x+3+ CIiD (1,0 di€m): Tinh tich phan: I CaD Goc Tim Cau y ° y3 =0 dx (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 (1,0 di€m): Cho x,y,z Iii cac sf>dirong 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 = ma tir nhirng di~m ke diroc d~n (C) hai tiSp tuyen hop vci 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) 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 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 + = (Q): x + 2y - 2z - = 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 cho HK , Cau 9.b ( 1,0 diem): Gilii h~ phirong trinh: I -log3 X -log3 Y { 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 = IIIpt TCN =2 ~ x-+±«> y x=ll11ptTCD x-+)- >:-+)+ - 0,25 +00 ~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 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+ =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) -Xet hai tnrong hop: + a = 2, (1) co nghiem x = 3/4 thoa ycht ~ a = (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 ~ B {:~:~:: ~1 Voi di€u kien tren (1) J - + cot 2x + sin 4x = cot 2x - cos 2x 3cos2x + cot2x + sin 4~ = cos 2x(2 sin 2x + sin 2x + 1)= 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::' ~ 12 0,25 0,25 s 'IIx E R nen -1 :::;Y s 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 + l~ 1= I o = Tinh II r==: I"/ X2 + Idx o i ~I II =x"/x2+11 II II =.J2 - o f.Jx2 o V~y = I= G9i M,N £)~t {~{dU= U = x + dv=dx ~ ~ xdx - - v=x II x dx ~ X2 + +ldx+ f~ X2 +1 dx Jx2 _ +1 ~[.J2 +31n(I+.J2)f ~ [.J2 + 3ln(1+ J2)] -~ -~ + %lnlx+~II~ 1k hrot 0,25 = J2 + !.- II 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 = MN = 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 VABCD=.!.SABCDS1=.!.~._a_= Tir (2) sin2 a cosa (I) (2) 4a 3sin2 a.cosa nho nh~t hay sin2a cosa Ian nhat sm acosa Do 0°