1. . so Gl.-o Dl)C 8 ..0 T,:~O PHL yE0. DE
THI THCTUYt~ S[~H D.':"IHQC :,~~12U12 L ' --(.
•~
.- TRLiO":G THPT JGUYE0: HL~ Mon : TO ..~ - Kh6i : A-B
~ ,
PH.-~ cm.xc eRO TAT
,
c.,
_ , (Thai gian laen bai : 180 phut . khong k~ thoi gian phat Oe)
THI SINR( 7,0 diem)
r 'i-t '~
Cau I (2.0 di~m) Cho ham so y = 2x) - (m -'- I)."2 -:
I. Khao sat W bien thien va e 00 thi (C) cua ham so khi III =:2
2.TII11cac gia tr i cua rn oe 00 thi ham 50 cat true 0,,- tai duy nhat mot diem co noanh do Ion hon - i .
Cau II (2,0 di~m)
. / l[
SlIl-r'---
x _.
-::.111
2.,
_:ceo::,_ 2 :CCot_:C,Lan:c
, ' 7 . )2
l.Giai phuong trinh : " 8 3/
----------------- = 0
[" ~
sin2.:c, ~ I
-4
X
2.Giai bat phuong trinh : Log 7 Ilog ~ -/==,===--- I< 0
~ 2 'v -.C -t- -4.r +:2 i
.7
.J ) " -
1=' I- 'Ii tan x
Cau III (1,0 di~m) Tinh tich phan : dx
J 3V..:. Sill
,7
'.., . 2.., .
.:.X.SII1 X
2
-
6
Cau IV (1.0 di~m) Cho hmh chop S.ABCO co (li~ .-BCO la hinh vuong canh a. S.-I _ (.-IBCD)
SC hQP vci rnp (-BCO) mot goc 60GQi;V1.~. P lal1 luot IS. trung diem cac canh Be. CO.-D.
Tinh khcang each tu diem N den mat ph~ng(S-IP) : Tinh goc giC!"aS-1 3. Jp
Cau v (1,0 Qi~m) Cho cac 50 duong a.b.c rhea man: abc = I
Chung minh rang : --=== -=== ~ 1
,,'] - 80 ,,' . - 8b " 1 - 8.:
PH.-l~ RlE:"iG (3.0 Qi~m): Thi sinh chi QU'I;),c lurn mot trong h ai p han (ph}n .- houc BJ
A.Theo chuong trinh chuin
Cau YI.a (2,0 Qi~m)
I T rong mat phang toa do Oxy. cho hai duong tron (C[ ) ::c2 - 1: - 2 - 3 = 0 : (C 2 ) : x: - 2 - ·h: - 5 = 0
Vi~t phucng trinh duong tron (C) tiep "-LIC voi hai true tea dQ :1 di qua cac grao diem cua (C ) va (C21
2.Trong khong gian O'i he true Oxyz. cho hai diem .-1(
I. =. - I) : B( -I. 0 .-1)
Tim ta ca nhCrng ciiem 1 thuoc mat phang (o:~ J 5;,1<) cho tam giac :1-B vuong tai ;[ 01 co dien rich nho nha:
Cau Vl l.a.Ll.O Qi~m) Giai he phuong trinh : .-I~-[ .. <_[C~_[ = 21.60.] 0
B. Theo chuO'ng trinh nang cao
Cau VI.b (2.0 di~m)
I.Trang mat phang toa do Oxy. cho ouo'ng tron IC) .2 -t- l= - 2.' -9 = 0 n't (1i~111 .-1(0.1). Tun tea c10
diem B va C tren duong tron (C) sao cho tam giac .- BC vuong can tai A
• '1"'l '"I •
.., Trong khong gian voi he true Oxyz. cho mat cau (S). x- - v ' <: : - 2x - 2::, I = 0 va hai diem
Mel. 2. -I) ; :'(3.1. -I). Tim diem K thuoc mat cau (S) sao cho tam giac K!v[~ '0 dien rich nho nhit.
Cau VII,b( t,O di~m) ::[ '::2'::3 1:1ba nghiern CLIZ! phuong trinh • (:: - i)(::2 - 2:: .•. :2)= 0 trong tap so phuc
11-
r1ay
't'
Inl.
I' T - -[ 21)[2
__ , _21)[2
'-2
__ ~')!2
= :
---- ---- ---- -- ----- -- --- H E T --------- ------------- ---
toanlvc@gmail.com sent to www.laisac.page.tl

2. f /
/ pAp ,AN KY THI THU TUYEN SINH DAI HOC NAM 2012 - MON ToAN KHOI A-B e. L''I. ~.
I~au 1.( I ,00) In = 2 , y = 2x3 - 3x2 + I
I(2,OtJ) II TXD: D = R
21 SBT: + Lim y = -00 , Lint y = +00
X~-O() X--->+CI)
+ v' = 6x(x - I) , / = ° ~ [: : ~: ; : ~ _ 0,250
x -rfJ 0 1 +rfJ
+BBT
I
Y +
0 - 0 +
y .> 1 -----. ~+oo
-rfJ 0 ----------
0,250
+ His dong bien tren (-rfJ,0) ; (I, + (0); his nghich bien tren (0 , I)
+ Di~m cue dai (0 , I) ; Di~m cue ti~u (1 ,0)
31 D6 th] :
+ y" = 6(2x-l) ; y" = 0 ~ x = k
y" 06i d§u kh i x qua oi~1ll x = k nen u (k, k) la oi~1n uon cua 06 th i.------------ 0,25 0
+y=O ~x=lvx=-112
+ 06 thj nh n diem 1I6n lam tam 06i xung j
+ Ve 06 th Y
x
I 0 1
2
0,250
2. (1,00): Phuong trinh hoanh oQ giao diem cua 06 thj va true Ox
2x3 -(/11 + l)x2 + 1= 0
2x3 + 1
~ ) = 111 +I (1) (vi x 7: 0) ------------------------------------- _ 0,250
x-
2x3 + 1 I 2x3 - 2
D~t f(x) = 2 : 1(v) = j
x - co -1 0 1 +0)
x .v
II (x) =0 ~ x = 1 ~ BBT
-,
[I + - 0 +
0,5 0
Dua VaG BBT 06 th] dt Ox tai duy nhAt +0) +rfJ +0")
I11Qtdi~111co hoanh OQ 16n hon -I khi: ..: I
-1$;111+1<3
f
-Cf)
/ . ..,
-'
,
~ -2 $; 111 < :2 0.250
I
,I

3. Cafl U(2,O
1(1,0c1)DK:cos2x;toO~x;toJr +kJr (kEZ); sin2x;toO~x;tokJr
di~!!!l 422
4x l s- cos Zx
P t ~ cos - - = 0 ---------------------------------------------------------------- _ 0,250
3 2
2x 2x 2x
~2cos2.~ - cos3'- -I = 0 (I) ,d~t I=COS~ , Itl ~ 1
-' 3 -'
(I) ~ (I - 1)(412 - 3) = 0 ----------------------------------------------------------------------- _ 0,250
' =I x = 3kJr
~ .fj ~ x = ± Jr + 3kJr 0,250
[ 1=±- 4
2 5Jr
X =±- +3kJr
4
0,25
KSt hop voi dk ~ Pt vo ngh iem -----------------------------------------------
2.(1,00)
x
Bpt ~ 0 < Log I < I --------------------------------------------- 0,250
2" ~- x2 + 4x + 2
x 1
~I > >- 0,250
~- x2 + 4x + 2 2
~x-2<~-x2 +4x < 2x - 2
0,250
0,250
Cau
" "
1110,0
di~m) J = ~ f
4
dx 0,250
2 !!... Vcos2 x.sin 4 X
6
= ~
4
f dx Dat u=i cotx ~du=---'
sin2
dx Jr
x=-~u=v3
6
t:
2 !!... Vcot2 x .sin 6 .v X '
6
0,250
Jr
x=-~u=1
4
,fj 2 I ,fj
JI = -
I --3 f
3:;-
u du = - 1I .) = -
3 (6r:; - 1 )
v -' 0.25 d
222
I
J= 2ifi_-J3_1 0,250
2 6

4. -Cau SA = aJ2 tan 60° = a16
IV(I,O
di~m) I I a2 a316
VSMNP - SA. S MNP= - a16. - = --
= -----------------------------------------------------_ 0,25 d
3 3 4 12
VSMNP IS'
=3NH,SSMP (NH 1-( MP)t()1 H)
I ::> I 50 502
S~"fP =-Sf.MJ =--.a=-
.)lV. 2 22 4 S
~NH =~= a16 0,25 d
SSMN 5
* (SM,NP) = (SM,MK) (SM,MK) ( K: trung aibm AB)
Xet tam giac SKM: MK = aJ2 ; SM = ~SA2 + AM2
2
SM = am . SK = Sf = 5a
2 ' 2
" SM 2 + MK2 - SK2 3
cosSMK = -------
2SM.MK .J58
0,5 d
Cau v
(l,Odi~m)
I I I
-===+ + > I (I)
~ JI;8b ~-
Ap dung BDT co si :
a + b + e? 33) abe = 3
ab+bc+rtc V
e ss a 2 b 2 c-') =3
Suy ra
(I + 8aXI + 8bXI + se) =1 + 5120be + 8(a + b + e)+ 64(ab + be + ac)? 729 = 36 (2) 0,25 d
Ap dung BDT co si:
~ + JI;8b +~ ? 3VO 8a XI + 8b XI + s-) ? 3W
+ = 9 (3) ----------
0.25 d
BDT ban d§u 0,25 d
H(~.~ +JI;8b.JI;8b + ~.~y ?(I +8a)(1 +8b)(1 +8c)
H 8(a + b + c)+ 2~(1 + 8a)(1 + 8b )(1 + 8e).(~ + JI;8b +~)? 51 0 (4)
Th~ (2) , (3) vela v~ trai cua (4), (4) dung suy ra (1) dung. 0,25 d
Deing thirc (; (1) xay ra khi a = b = c = 1.
PHAN
RlENG
eftu Vl.a 1.(1.0 d) Toa dO giao diem cua (Cj ) va (el) la nghiern cua he:
(2,0 di~m) x 2 +V--L),-->= -.. 0
') ""
') ')
{
x: + y- + 4x - 5 = 0
x= Js •• ~
l
~ y=I_~ v )'=I+~ 0,5 d
15 15
GOi pt duong troll t iep xuc O'i 2 Ir~ICtoa dQ co dang:
(x - a r + C,· ± a f = (/2 (1) ----------------------------------------------------------------
0.25 d
Thay toa dO giao diem cua (C, ) va (Cj) vao (I) thi pt vo nghiern
Vay khong ton tai duong troll thoa yell dll bai roan. ------------------------- 0,25 d
2.( 1.0 d)

5. - GQi diem M(x, y, 0) thuoc mp(Oxy)
AM(x-l,y-2,1); BM(x+l,y,I); AB(-2,-2,0)
6.MAB vuong tai M ~ AM.BM = °~ x2 + / - 2y = ° (1) ------------------------ _
SMAE = ~ MH.AB ( AB = 2.fi ) ( M H la khoang each tu M d~n AB) ;
SMAB nho nh§t khi MH nho nh~t
MH = I[An , AM l = ~(- 2? + 22 + (2x - 2y + 2? _
IABI 2.fi
MH nho nh§t khi : x - y + I = (2) °
TLl" va (2) ta co:
(I) Ix = ~ 12 v Ix = - ~~ ------------------------------------------
y=I+- y=I--
2 2
KL: M(.fi I + .fi 0) v M(- .fi I _.fi 01 _
- 2' 2' 2' 2')
Cau VII.a. DK: x;:::y+2;y;:::l;x,YEZ
(1,0 di~m) x! (x - y - I)! 7
: . (x-y+l)!" (x-I)! =20
Dang tlurc ~ --------------------------------------------------- 0,25 d
(x - I)! . y! (x - Y - I)! = 6
I (x - y - I)! (x - I).
H b:~ +~ Xx - y) = ;0 ------------------------------------------------------- 0,5 0
7x2 -55x+42=0 ~{x=:
{ y=3 y=J
x= 7
KL: H~ co nghiern: { " 0,25 d
y=J
CauVI.b. 1. 1,0 0)
(2,0 di~m) (C) : (x - I? + y2 = 10-) (C) co tam I( 1,0), BK R = J10
IA = J2 < R -) A nam trong duon tron (C), IA( -I, I) -) vtpt cua IA la ~ (I, I)
Tam giac ABC vuong can tai A nen AB,AC hQP voi IA gee 45° ,
GQi d la duong thang di qua A va co vtpt /72(a,b) -) PI d :ax+ by-b = °
Theo YCBT ta co: la.1 b.11
+ = _1_ -) la + bl = ~ a2 + b2 m mm_m _ 0,250
~a2+b2 ..fi .fi
~a.b=O
~ [0 = ° -------------------------------------------
0,250
b=O
* a =0 chon b = 1 -) d) : y -I = °
*b= °
cho a = I -) d 2 : x = ° -------------------------------------- 0,25 d
TQa OQ cac diem B va C la nghiern cua h~ :
(x - I? + y2 = 10 v {(x - 1)2 + y2 = 10
{
,.-I = ° .v = ° 0.25 d
KL: TQa oQ cua B va CIa: (4.1 )'(1(0.- 3) hoac (-2.1) "e/ (0.3) ------------------
( diem 8 va C phai nam v~ hai phi a khac nhau cua QUang thang IA )
2. (1.0 0)
M~tdu(s)cotal11 1(I,O.-I).BKR=1
Pt mp(P) di qua 3 diem I , M . N : z + I = °

6. MN(2, - 1,0), mp(P) oi qua I va vuong goc MN nhan MN lam vtpt co pt:
2(x - I) - I(y - 0) = °~ 2x - y - 2 = ° --------------------------------------------------- 0,250
Di~m K thuoc (S) co khoang each o~n MN 16n nh§t hoac nh6 nh§t co toa dQ thoa he:
S KMN dat gia tri nho nh§t khi d(K , MN) nho Ilh§t
x=I--
15 r x=I+- 15
z + I= ° 5
2.J5 2.J5
5
2x - y - 2 = ° ~ K 1 y=--- y=- 0,250
{
x2 + y2 + z2 - 2x + 2z + 1 = ° z =-1
5
z =-1
5
d(K, ,MN)" ~21;trs , d(K, ,MN)" ~211sJ5 _m_mmu_m_m __ mm __ m_mm 0,25 d
d(K" MN) < d(K, ,MN) -> Diem K tho.vcsr I., K, [I + ~ , 2f '- I) ------------ 0,25 d
Cilu VII.b.
(1,0 di~m) °
l;:
(Z-i)(Z2 -2z+2)=
H i: : ; ---- --- ------------- --- -------- ----------------- ----------------------------------- ----- 0,25 d
Ta co: (I + i)2 = 2i , (I - i)2 = -2i
Nell Zl2012 + ziOl2 + zjOl2 = (i4 yo) + (2i)I006 + (_2;)1006 = 1+ 21006.i2 + (_2)1006i2 _
0,50
1007
= 1- 2
V~yT= 1_21007 0,250