The document contains a math test with multiple choice and free response questions. Some key points: - It asks to find the domain and range of the function f(x) = 3x^2 - 6x + 1. - It asks to solve the system of equations x^2 + y^2 - 4x + 2y - 5 = 0 and 2x - y = 0 to find the points A and B where the line is tangent to the circle. - It asks to find the real root of the equation z^4 - 22z + 5 = 0. - It provides two parts (A and B) for the free response section and asks to choose one. Part

1. so GrAo Duc vA EAo rao HA Nor ne rm rrnl DAr HQC EoT r NAtvt Zgt t
TRtIOt{c THPT cHU vAx eN M6n To6n - Kh6i A
Thdi gian ldm bdi: 180 phrfit, khdng tC ttrOi gian giao dA.
DA thi gom ol trang.
I. pHAN cHUNG cHo rAr cA cAc rni sINH 1l,o eiom;
Cffu I (2,0 tli6m) Cho hdm s6 y: xt -3x' +1.
1. Kh6o s6t sp biiSn thiOn vd v€ dO thi (C) ctrd nam s6.
2. Chrmg minh rdng mdi ti6p tuy6n cira (C) chi ti6p xric vdi (C) tei dung mQt di6m.
Cfru II (2,0 tli6m)
1. Giai phuong trinh 9'io" * 4,gcosz
x= 13 +,*'z'+l-3'o'2'.
lx+ Y =g
2. Giai he Phuong trinh I r ^- r---:-
[r/xZ + 9 *^ly'+9 =10
(x'yeR)'
cflu IItr (1,0 tli6m) Tinh tich phdn 1 :pt 4*.
ix'
CAu IV (1,0 ili6m) Cho hinh chop S.ABCD c6 d6y ln htnh vudng cpnh c , SAL(eACn) vd SA=a.
Ggi A',8',C' vit D' lan luqt ld trung ,rliOm cira .!C,SD,SA vit SB. Chimg minh rdng AA',BB',CC' vh
DD' d6ng quy; Tfnh th6 tictr ctra hinh ch6p ^S'.1'B'
C' D' theo a, v&i ,S' h tam cfu hinh w0ng ABCD '
CiuV(1,0tli6m)Xicdinh m saocho xa -2x3 +8**l)*'-2mx+m'-4>A, Vxe [-f1]
II. PHAN RItNG (3,0 tli6m). Thf sinh chi iluqc chgn mQt trong hai phin (phin A ho{c phin B)
A. Phin A (theo chucrng trinh Chuin):
Cfiu VI.a (2,0 ili0m)
1. Trong mat phang tqa d0 Oxy , cho hinh r.u6ng c6 mQt dinh l(- 1;2) vA m$t duong ch6o nim trOn
ttucmg thing c6 phucmg trinh 2r - y -l = 0. Tim tqa d0 c6c dinh cdn lgi cira hinh vu6ng.
2. Vi6t phuong trinh m[t cAu (C) cO t6m thu$c dudng thlng (A) c6 phuong trinh
lx-vtz=oJ"
lZx+ Y+22-I=0
vdtitip xric voi hai m{t phang (a): 2x +2y - z + 6 =O va (B) ; 2x +2y - z-6 = 0.
Cffu VII.a (1,0 rli6m) Cho zr,z, ldhai nghiQm phrlc cria phuong frnh zz -22 +5 = 0. Tinh gi6 tri cua
bi€u thirc P =lr?l*l':l
B. Phin B (theo ehuong trinh Ning cao):
Cflu VI.b (2,0 tli6m)
1. Trongm{tphingtqad0 Oxychodudmgtdn(C): *2+y2-4x+2y-5=0.fhlrtA6euerngthing 4
d*! x - my=0 cat ducrng trdn (C) t?i hai di6m A, B pherr-biQt, sao cho dQ dii do4n AB nhtt nh6t.
z. Trong khdng gian tsa dQ Oxyz cho c6c tli6m l(l;O;t), f(tt;O), CQ;l;*l) vi m{t phdng (a) cO
phucrng trinhx + y + z-1 = 0. Tim to4 d0 diem M sao cho khoing c6ch tir M dln (a) Uang khoang
c6chtu M danm6iei6m A,B,C.
(r- 3
Ciu VII.b (1,0 tli6m) Tim sd phtrc z ,Ai6t Z =42-!
-------n6t-
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2. so crAo Duc vA o.A,o rAo HA NOI
rntldxc THPT CHU VAN AN
PAT AU - THANG DIEM
of rnr rrulDAr Hgc - DgT r nim zort
M6n Tofn - KhAi A
an- di6m 07 tran
Cfiu Dfrr 6n Di6m
I
(2,0 ili6m)
t. rt.O tli6m)
. T0P x6c dinh: B'
. Sg bi6n thi6n
- Gi6i han: lirn ! = -c; lim Y - +60.
0,25
0,25
- Chi€ubii5nthi€n: !'=3x2 -6x; y'-0ex=0 hotrc x=2.
.y'>0e x<0 hoflc x>2; .y'<0<]0<x<2. HAm sO AOng bi6n tr€n cfc khoang
(-*,0), Q,**) vA nghich bi6n trOn ktroang (O,Z).
- Cuc tri: Hdm s5 dat cgc ct4i tq.i x = Q;yru = 1, d?t cgc tiAu tqi x:2i/cr =-3.
:-iia;s L-i6iitiliit;
4,25
. EO thi
y"=6x-6; /"=0(i r=1
ximg cta AO tfri hnm s6.
Dd thi hdm sO c6 di6m uOn l(t,-l) vd n6 la tdm d6i
Ar25
2. fl.$ tli6m)
v,,{F*ALt - I t=-:
$ r.,t uJUcnn i

3. Ciu Eip 6n Di6m
ai di6m: Mo(to,Yo) vd M1(x1,Y1)
Khi d6 phucrng trinh cria ti6p tuy6n li
y=6*8-6xoh-'r-3'+3xfi+I vd ,:b*? -e'r! -z*l *?t:!
0'5
'-------'.
0,25
0,25
#ii;fi;d'd #dil dcn ;ils i;-i,-6;ong liiirtr
"t
imqt tiiSp tuv6n n€n
3r& - 6xs =zxl - 6x1 ,
-2*3+lxfr +l=1xl +3xl +t .
Giai hQ trOn ta dugc xe = 11, do d6 ta c6 dpcm
II
(2,0 tli6m)
T (1"0 dtffi)
Phuong trinh dd cho tucrng duong vfi
9sin2.r * 4,gl-sn'zx =13+ 93/2-Zsintx -31-2sin2x
<) 9sin2r * 31
= 13 + 3- -:-gsin'x 92sn'r gstn- r
<+ 9sin" * jl - -4--- 13 = o'.
9sin2
x
(nrrr
r
),
Dflt r =9''" , 1 < r < 9, ta nhfln dugc phuong trinh , *+ -T -13 = 0
e (t *l)(t -3X/ - 9) = 0 <+ r =l;t = 3;t = 9.
Ar25
0025
Phrrcmstrinhddchotuonsduonsv6i sin2r=0 ho6c sin2x=t hoFc qr-41 f -=-!!-?
a
o
sin2x=0<+ x=kn.
sin2 r= 1 <> cosx = 0 e r = n l2+ kn 11:
0,25
. sin2 x =ll2e cos2x= Q 49 x = t I 4+ ktr 12.
Vfly phucmg trinh dd cho c6 nghiQm * = k7 (k .4.
4
z. d.o tli6m)
-a c'6 y = 8 - x, thO vdo phuong trinh thrl hai cira hQ
"f7 .g *.{; -16r'173 =1g
r:l:
0,25
0,25
€x2-8x*+t+@-59
e@=-x2+8x+9
f-*'*8x+920*
tft' * eh' -t6x +n)=l *' *sx+ef
f-ts x <v
<>{
[x'-8x+16:0
(3x=4.
. Ix=4
Suy ra he dA cho c6 mQt nghiQm duy nhdt j ., _ ,l"v--
4,25
o:rs
III
(1,0 tli6m) Ta c6 t =2'pI4*
ix'
2

4. Cflu Edp 6n Ei6m
= -, Jf*)'' nxdx = ryli* r"l# 0,50
0,25
22le 22 4I
- -!') -'t -------l
TL-L--
e xll e e e
4
Vqy I =)-*e
IV
(1,0 tli6m)
LSAC ta c6 AA',CC',,9,S' ld c6c dudng trung tuyOn n6n
G, cintam gi6c vi
,scr - 2cts',. (l)
AA' ftqngt4rx6t
tdm
'iic
'lt{--
-i".A
/N 0,25
0,25
Tucrng tp, trong LSBD ta cfing c6 BB' cdt DD' t4i trqng t6m G, vd
^sG, - 2c25',, (2)
TiI (1) viL Q)suy ra G, = Grhay AA',BB',CC' vd DD' d6ng quy.
Tt gia thii5t ta suy ra A'B'll =Jrcn, B'C'll =!ne,C'D'll =Iut2--' 2 2
D'A'll =!ac. Do d6 (A'B'c'D;)tt(aaco) vit A' B'c'D' h hinh vudng eqnh
2
Hcnrnir4 s'A'/l= ]s,a, mir Sl L('encn) nOn ,S'l'I {A'B'C'D').
z
va
a
2
0r25
Ydy vs,.u,u,.,r, =
i
t' A' ft{A' B'C'r) =
i
23
doa_.-=+
24 24
0,25
v
(1,0 tti6m)
Ta c6
xo -zxt +8*+l)x2 -Zmx+m'*4>-a, vxe[*1,1]<>
(*' - * * *f r-4vx e [*r,r]o Hfi(r' - * * *)' > 4. (3)
DAt t = xt -x, tac6 t'=2x-I.
xl I^ l-1 : 1
'r
t'l - 0 +
!.1 ?""....-.-.-- -a 9.
0r25
4,25
_k

5. Din 6n
,{
:fr
Do d6 xe [-t,t]€>/€
rl
;,21
surra (3) <+
tl1fik
+ m)' > 4.
X6t g(r)=(r+ m)'; s'(t)=2Q+m).
11.-m
44
. -m>2em<-2
-1< -m<2eJ<*"L44
Q* *)'
4

6. Cflu Din rin Di6m
nhu *, !
4
nOu- 2<m<!
4
nilam < -2.
tTf
+*)'
(--i)'
0
(**z)'
Suy ra niq(r + *)' 2 4 e
L+'l
lf( *-1.j' =
o
llt ,o) l*.2
ll*'a * | 4
l it. z)' > + L* <-4'
l.l*.-Z
hoac m>2.'4v0v; cdn tim ld m < -4
02s
VI.a
1z,o oi6m;
L (1,0 iliSm)
Ggi hinh w6ng cAn tim li ABCD, do n(*t,Z) kh6ng thuQc duong th5ng
2x * y-1 = 0 nOn dudmg thing ld phdi ld <lu,crng ch6o BD. Ta c6 C la <fi6m d6i
xrlng cua A qua BD, ggi 1 ld tdm cira hinh vudng. BD c6 vdc to chi phuong
-t t
uQ,2), do AC I BD n6n v6c to ph6p tuy6n oiua AC W n(t,Z). V{y phuffrg trinh cia
AC liL
(x +t)+ Z(y -Z)= 0 €) x +2y-3 = 0.
4,25
Ta c6 tqa dQ cfia I li nghiQm cua hQ
[x+2Y =3
lz*-Y:1
Suy ra tqa d0 ctra C(3,0).
lx:1
c>i
LY=1'
0,25
nei, U,2 = 5 n6n clulng trdn tdm I bfunkrrth IA c6 phucrng trinh
(t - 1)' +(y -1)' = 5.
Ar25
Tqa d0 cria B,D ld nghi$m cria hQ
{G-t)'*(v-r)'=5 e,
l2x- Y =1
v{y B(0,- t}, c (2.,2) ho4c B(2,,3} c(0,-t).
Jx=0
l.v=-t
[x=2
Lv=1.
0,25
2. fl.o tli€m)
V6c to ph6p tuy6n ctta (a(B) n iQ,z,-l), F" a (u)tt(p} Ta c6 .a(o,o,e). (o),
I-d-el
a/;(d) =4+a((a,(il)=q.
tlZ" +2" +|;l)'"Mif;il (d) ii6 ;il';6i (;) ;t tp)'i.hi;.;hi kili'ffi ki'h. R--;,il';A it
0r25
5
-O,25

7. Cflu Eip 6n Di6m
aal,lp))rz,v[y R=2
Gqi / ld tdm cua (c), khi d6
aQ ;@)) = d (r ;(B)) olLx
+ 2 v-- z + 6l
3
lzx+2y-z-61
J
e2x+2y-z=0
0r25
M4t kh6c I e A, nOn toa d0 cira li nghiQm cua hg
lx-/+z=o
l2x+y+22=1 €
lzx+zy-z:o
I
I
]n=*IJ
l, =!Le
n-lJ'3J
vfly: M[t cAu (c) c6 phuong o* ('. +)'
. ( .(,-t)' =o
0r25
YILa
(1,0 iti6m)
Phuong trlnh zz -22 + 5 = 0 c6 hai nghiQm phtrc z, =l+2i vit z, =1-2i 03s
o ,2 =(1+2i)2 =4+4i. 4,25
a -212 =(1-zi)z =4*4i 0,25
Suyra l,il=V:l=5 hay P =l'?l+l'il=to 4,25
YI.b
(2,0 tli6m)
l.(1,0 tli6m)
- (C) c6 tim I(2;-1) vd b6n kfnh n = d0 0,25
- Dudrrg th6ng d* di qya di€m o c6 einrr nim trong dtd'nttidn, Ao Ait A;-i"?tn;Ai
$ggfrg gQg qlstz dism ph6n biQt.
0r25
: Iraplg4q dUqriDO dei AB nh6 nh6t khi vd chi khi ,qE IOt 0,25
- Trlc le d* di qua O vi nhfln OI(2;-l) ldm VTPT
- Phuong trinh cria d^ ld2x - y= 0 hay * = ! .
'2
0,25
2. (1.,0 di6m)
Ta c6 MAz = MBz = MCz = d'z(M;(d) MA2 = MB2 e | = z. (5) 0,25
UBI = MC2 e x= z*2. (6) aes
MAz = d' (u;("))o 3(x -l)'? +3y' #(r -1)2 = (x + y + z -1)2 . 0,25
Thay (5) va (6) vdo phrrong trinh cui5i tren ta nhdn dugc 6z - J e s - I6
v-y M(+,;,*)
0,25
VII.b
(1,0 ili6m)
zJt -6i-3J1+i
t+ Jii
a,25
J
0r25
_ -(Jl -zi * si * sJi) _ - eJi -ti = _2,D _ i
33 0025
Ydv: z=*2Ji+i 0,25
6 ffiM*