1. TRUONG THPT NGUYEN HU~ DE THI THU D~I HQC NAM 2012
TO ToAN MON ToAN KHOI D - THen GIAN : 180 PHUT
************** *******************************
pHAN CHUNG CHO TAT cA THi SINH (7,0 di~m):
Cau I (2 di~m) Cho ham s6 y = x3 - 3x2+2 co d6 thi (C)
l.Khao sat sir bien thien cua ham s6 va ve d6 thi (C).
2.Tim diem M thuoc duong thang (d): y = 3x - 2 sac cho t6ng cac khoang each tu M d~n hai di~m CI!C tr]
la nho nhat .
Cau II (2 diem)
1. Giai phuong trinh : sin2x-cos2x+sinx-cosx= 4cos2 x
2
2. Giai h~ phirong trinh:
F+5+.JY=5
{ rx+JY+5=5
Sf ~-2
Cau III (1 diem) Tinh : 1= dx
2 x~ +2x-2
Cau IV (1 diem)
Cho hinh chop S.ABC co day la tam giac d~u canh a. Chari duong vuong goc H ha tir S xuong
m~t phang (ABC) la trung di~m Be. Cho SA = a va SA tao voi m~t phang day mQt goc bang 30°.
Tinh thS tich hinh chop va khoang each gicta hai dirong thang BC va SA .
Cau V (1 di~m)
Cho x la s6 thirc duong va y la s6 thirc tuy y , tim gia tr] Ian nh~t va gia tr] nho nhat cua bieu thirc :
2
A= xy
(x2 + y2)(~X2 + y2 +x)
PHAN RIENG: Thi sinh chi dU'o'c lam mot trong hai phAn (phAn A hoac phAn B)
PhAn A : (3 di~m)
Cau VIa: (1 diem) Gi<ii phuong trinh: ~ log 12 (x + 3) +~ log, (x _1)8 = log2 (4x).
2 "L 4
Cau VIla: (2 diem)
I.Trong mat phang toa dQ Oxy cho MBC co dirong tron ngoai ti~p la (C): (x - 4)2 + i = 10
1
Tim toa dQ cac dinh B va C ( vc > 0) cua MBC khi biet dinh A( 1, 1) va trong tam G (13 ; - ~) .
2T rong khc
. .
ong gran "h~ t;
VO'! true toa do O xyz c h 0 d'irong tang
9 h! x-4
(d) : -2- y-I z-2
= -1- = -1- 'd'A
va tern A(I ;;.12)
Tim hai diem M va N tren duong thang (d) sac cho MMN vuong tai M, co dien tich b~ng 3 J2 .
PhAn B : (3 di~m)
Cau VIb. (l diem) Tim s6 nguyen duong n ,bi~t 2C; +2.3C: +3.4C: + ... +(n-l).nC; = 480
Cau VIIb: (2 diem)
I.Trong mat phang toa dQ Oxy cho tam giac can ABC co dinh C( ± ;0) va CI la dirong cao. Canh AB co
phirong trinh x - 2y + 2 = 0 va AB = 4CI . Tim toa dQ cac dinh A, B biet A co hoanh dQ am .
1
2.Trong khong gian voi h~ true toa dQ Oxyz cho dirong thang (d) : x ~ I = :1 = z_~ va diem E(O; 0; 1).
Vi~t phirong trinh mat phang (P) chira dirong thang (d) va each diem E mot doan bang ~ .
**************************** HET *****************************
toanlvc@gmail.com sent to www.laisac.page.tl

2. BAp AN THI THU B~I HQC KHOI D - 2012 - LAN THU I
N()IDUNG DIEM
Cau I l.
l diem H9C sinh W giai
2. + G9i diem cue dai la A(0;2), diem cue tieu la B(2;-2)
l diem + Xet bi~u thirc P(x, y) = 3x - y - 2
+ Thay toa dQ di~m A(0;2) van bieu thirc P(x, y) , ta co: P(0,2) = - 4 <
va thay toa dQ di~m B(2;-2) van bieu thirc P(x,y) , ta co: P(2,-2) = 6 >0
°
+ V?y 2 diem cue dai va cue ti~u nam v~ hai phia cua duong thang y = 3x - 2 0,5 d
Nen MA+MB nho nh~t khi 3 diem A, M, B thang hang
+ Phuong trinh duong thang AB: y = - 2x + 2
0,5 d
+ Toa do diem M la nghiem cua he:
. . ..
{Y = 3x - 2
Y = -2x+2
<=> jx =~
2
y=-
5
V~
ay M(45"; 5"
2) tren d'
~ irong tang
h! (d) t h' yeu cau b" toan .
oa ~ A ar ,
Cau II l. Phuong trinh <=> 2sinx.cosx - 2cos2x + 1 + sinx - cosx = 2 + 2 cosx
l diem <=> sinx(2cosx + 1 ) - cosx (2cosx + 1) - (2cosx + 1) =
<=> (sinx - cosx - 1 )(2cosx + 1) = °
° 0,5 d
sin(x - 7r) = _1_ = sin 7r
<=> [sinx-cosx = 1 <=> 4 J2 4
2cosx = -1 1 27r
r COSX = --
2
= cos-
3
l
X=~+k27r hay x=7r+k27r
Vay phirong trinh co nghiem : (kEZ) 0,5 d
x=±27r +klr:
3
2.
l diem
+f)i~u kien : x ~ ° va y ~ °
+V6i di~u kien tren ; ta co :
0,5 d
.Jx+5 +JY =.rx +~y+5 <=> )y(x+5) = )x(y+5) <=> x =y
+Khi x = y, pt(l) viSt lai: .Jx+5 +f; = 5 <=> ~x(x+5) = 10-x <=> x = 4
Vay : h~ co nghiern (4 ,4) 0,5 d
Cau III +Uat t =~ ~ t2 = x -1 ~ x = t2 + 1~ 2tdt = dx
ldiern
+n"" can:
vOl
~ x=2=>t=1
. x=5=>t=2
I= f
2 t-2
,(t2+1)t+2(t2+1)-2
.2tdt = 2 f,(t+1)2
2 t-2
.dt = 2
~
-
,/+1
1 3} -
(t+1)2
t
0,5 d
2
= 2In(t+ I)I~+- 61 = 3
2In--1 0,5 d
t +1 I 2
Cau IV
l diem + Xet ~SHA vuong tai H , ta co : SH = SA sin 30° = ;
+ SABe = -BC.AH
1
=--
a2 J3 0,25 d
2 4

3. ~ 1 a3
V~y: V = 3SABC.SH = ~
J3 0,25 d
+Tir H ha dirong vuong goc xuong SA tai K
+ Ta co : AH .L BC va SH .L BC => BC .L (SAH) => BC .L HK
Do do HK la doan vuong goc chung cua BC va SA
0,25 d
Nen khoang each gifra BC va SA bang HK
Ta co: HK = AHsin300 = AH = aJ3
2 4
0,25 d
V~ khoa
ay oang cac gura h'ai d'irong tang
ich ziu h::' BC va
'SA b::' aJ3
ang -4-
s
A <->.': -----
------- - ----:::.
C
H
B
v
Cau
l diem A ~ xy'
(X2+y2)(~X2+y2+X)
~ x(~x' +y'
2
x +/
-xl ~ f(?J
1+(~r
-I
f)~
~t t = (y)2
- ; d'~ k'~ t ~ 0' va A = fiC) = ---
leu ten t Ji+1-1 0,5 d
x t+l
f(t)= 2-Ji+1 vaf(t)=Okhit=3
2(t + 1)2
t 0 3 +00
y' + 0 -
y .>" 1/4 ------.
o o
V~y : max A = .!. khi Y = ±xJ3 0,5 d
4
va min A = 0 khi x tuy Y duong va y = 0
Call +f)i~u kien: x> °
va x-:f:-l
VIa: ldiern (I)<=> log2(x+3)+log2Ix-11 = log2(4x)
0,5 d
(1) <=>(x+3)lx-II = 4x
+Truong hop I: x > 1
0,25 d
(1)<=>x2-2x-3=0<=>x=3hayx= -1 (loai)
+Truong hop 2: ° <x <1
(I) <=>x2 +6x-3 = ° <=>x = 2J3 -3 hay x = -3 - 2J3 (Ioai)
0,25 d
Vay nghiern cua phirong trinh (I) la : x = 3 hay x = 2J3-3

4. Cau 1. + (C ) co Him 1(4; 0)
VIla: l diem + GQi 0 Ia trung di~m BC .
- 3- 0,5 d
Taco AD =-AG ,nenO(5;-I)
2
+ Tinh chat day cung ta co 10 .L BC
+ Canh BC qua 0(5;-1) nhan ID = (1;-1) lam vtpt co phirong trinh la:x - y - 6 = °
+ TQa dQ B, C Ia nghiern h~ phuong trinh :
X - y- 6= ° {X =3
hay
{X =7 0,5 d
{ (x-4)2+y2=10 <=> y=_3 y=1
V~y: B(3;-3),C(7;1) (do Yc > 0)
2. +Vi~t pt (d) diroi dang tham s6, suy ra: M( 4+ 2t; I +t;2 +t),N( 4+ 2s;1 +s;2 +s) 0,25 d
l diem
+ (d) co vtcp
- =(2,1,1)
U
-
va AM =(3+2t,t,t)
+ MMN vuong tai M <=>; .lAM <=>;.AM=0<=>6+4f+f+t=0<=>t=-1. 0,25 d
Do do M (2, 0, I) va AM = J3
+ SaAMN = 3J2 <=> MN = 216 <=> ~r-(2-s-+-2-)2-+-(-s-+-I)-2 = 216
+-(s-+-I-)2
°
<=> 6s2 + 12s - 18 = <=> s = I hay s = -3 0,25 d
V~y M co toa dQ (2,0,1) va N co toa dQ la (6, 2, 3) hoac (-2, -2, -I)
0,25 d
Cau +Ta co: (1 +x)" = C~ +C~x+C;x2 +C~x3 +C~x4 + ... +C:xn.
VIb. l diem
Daoham dp2 hai v~
(n-l).n(l+x)n-2 =2C; +2.3C~x+3.4C~x2 ... +(n-l)nC:xn-2 0,5 d
+ Thay x = I ,ta diroc : (n - I)n. 2n-2 = 2C~ + 2.3C~ + 3.4C: + ... + (n -l).nC;
+ Tir gia thi~t , ta co : (n - 1)n. 2n-2 = 480 = 5.6.24 = (6-1 )6.26-2 0,5 d
V~y: n=6
Cau 1.
VIIb. l diem + CI = d(1 AB) =
,
J5 ~
2
AB = 2 J5 ~ CA = ~
2 0,25 d
+ Ta co A va B & tren duong tron (C) co tam C va ban kinh R= CA = ~ 0,25 d
2
, , " (
+ dirong tron (C) co pt Ia: x -"2)2 + y 2 = 4
I 25
0,25 d
+ Ava B la giao di~m cua AB va (C ) , nen A( -2; 0); B(2; 2) (do XA < °) 0,25 d
2. + dt (d) qua M(l, 0,1) va co vtcp la U = (I, - I, -I )
ldiem
+ pt mp(P) : Ax + By + Cz + D = ° co vtpt n = (A,B,C)
0,25 d
ME(P) {A+C+D=O {D=-A-C 0,25 d
+ de (P) ~ _ _ ~ ~
{ n.u =0 A-B-C=O B=A-C
+ ptmp(P) : Ax + (A - C)y + Cz - A - C = °
+ d( E, (P)) =_1
J2
<=> I-AI
~2A2 -2AC +2C2
= _I
J2
<=> [C
C
=
=
° A
0,25 d
0,25 d
V~y pt mp (P) la : x + y - 1 = ° hay x + z - 2 = °