đề thi thử đại học môn toán có đáp án năm 2011 trường chu văn an

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=o
J"
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-
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*AL
t\
-
I t=-:

$
r.,t
uJUcnn
i
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
Cflu
Edp
6n

Ei6m
=
-,
Jf*)''
nxdx
=
ryli*
r"l#
0,50
0,25
22le
22
4
I
-
-!')
-'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'
ftqng
t4r
x6t
tdm
'i\
ic\
'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
=Iut
2 '
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
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<*"L
44
Q*
*)'
4
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.
'4
v0v; 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
Cflu
Eip 6n
Di6m
a\\al,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
-2
12
=(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
-
I
6

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*
'll