đề thi thử đại học môn toán có đáp án năm 2011 đhsphn đề 1

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TRU'oNG
EHSP
HA
hiQI
TRI'ONG
TtrIPT
CTTUYTI\
-
DI{SP
Mdn tlri : TOAN
Thd'i
gian
tdm bdi :
180
phtit, kh6ng kd
thd'i
gian phdt
di
Ciu
1.
(2,0
di€nt
)
Chohdms6
r=1"0-1x2+1.
42
l. KhAo
s6t su'
bi0n thi6n

vd
vC
dd thi
(C)
cfra
him
s6.
2.
Tim di€m M
thudc
(C)
sao
cho
tOng
cdc
khoAng
c6ch
til
di€m M
d6n hai
trr,rc tga
dQ ld
nho nhAt
CAu
2.
1Z,O
aieml
1. Gi6i
phuong
trinh :

cosSx
+
3cos4x
*
3cos2x
:
8sosx.cos3:x
-
I'
2
,^,,1{4i4-3
2. Giei
bdt
phuongtrinh : 32-*
+
6.3r-*
t
G)
Cdu
3.
(1,0
di€m)
Tfnh tich
phAn
t:,
(1
+ ex)(1+x2)
CAu
4.
(1,0

,iiem
)
Tfr dien ABCD
c6 ABC
vd BCD ld
c6c tam
gi6e
d€u
c4nh
bing
a.
G6c
gi0'a
dtrong
thang AD
vir
rn{t
phing
(ABC)
bing 60".
Tinh
diQn tich mpt
cAu
ngo4i ti6p
t(r
dien.
CAu 5.
(1,0
diem)
Tirn c5c

gi6
tri crta m AC
ne
Uat
phuong trinh sau
c6 nghiOm
:
l*'-xy*2y2-x(m
lxz-zxy-Zx<m-2
Cflu
6.
(2,0
di6n)
l. Trong
mat
phing
Oxy,
cho
hai di€m
Fr(- 4;
0), Fz(4; 0)
vd
di6m A(0; 3).
a) Lap
phuo'ng
trinh
chinh tic
crha elip
(E)
c1i

qua
di6m
A
vd c6 haiti€r-r di6nr
Fr, Fz
.
b) Tim tga dQ
diOm M thuOc
elip
(E)
sao
cho MF1
=
3MIrz.
2. Trong
kh6ng
gian
Oxyz,
clio hai
duong thing :
,1,,4=
Y-1
:'-r'.
cl',,*:
Y*t
-'-3
t 2
2
-1
2

-2
Chri'ng minh c/r, dz cit nhau t4i
di6m A;
vi6t
phuong
trinh
dud'ng thing A di
qua
M(2; 3; l) tao
vo-i d1, d2 mQt
tam
gi6c
c6n tqi
A.
CduT.
(1,0
diem)
Giai he
phuorrg
trinh :
n L:4 r-:
Hiit
a4 ,
.
t
a t A / a /^na a
oAp
An
-
THANG

DIEM
rnr
rntf
on
r,AN
urt/ga
- NANI
zorr
t
tru
ai6m
'z
a$m)
thi
(C)
cit
oy
tai
A(0;
l),
ndn
@i
trsc
tsa
d0
bang
l.
E6
thi
hdm

s6
c6
hai
diiim
cr,rc ti6u
1-
f
tJ
i
'
i
r;
I
I
ua
nhf,n
oy
ldm
trqc
diSi xung,
nOn
ta
chi cAn
x6t
M(ru;
y")
€
(C)
vd
0

<
xo'<
l.
@
M
il€n
hai
trqc
tqa
d0
ld
:
|
*"
I
*
I
y"
I
=
xo
*
yo
=
x"+
lAxJ
-1/2x"2
+
1
:

1l4xl
+
1/2x,(2-""1
+
I
2
l,
v6imgi
x.,:
0(XoS
1, dingthfrc
xityrakhivdchikhixo:0+Yo=
l'
Viy,
di6m
M(0;
1).
)
. Giai
phucrng
trinh

n
dtem)
ptru'ong
trinh
dd
cho tuong
duong
v6i

pt
:
1
cos8x*
3cos4x*
3cos2x=2cosx(cos9x+
3cos3x)-
;
(?
cos8x*
3cos4x+
3cos2x=2cosx.cos9x*
6cosx.coslx-
]
€
cos8x
*
6cos3x.cosx:
cos10x*cos8x*
6cos3x'cot*-
t
1
<+
cosl0*=;
€)
10x:+
n+2kn
e
*:*
**\1,

k.Z.
2.
(1,0
bi€u
kien
: x
€
(-
@;
-21u
[l;
+
oo)
.
Bdt
phuong trinh
tuong
duong
v6'i
: 3.31-*
+
6.31-"
7
3z-'1;z*-z
<+
9.31-x
;.32-JizTx4<+33-*
73s-'17+x-z <+
3-x>
3 t3+x-z

'13+x-2
>x
(+
D6psd:x€
(-@;
-21
U(2;+oo).
r( x(0
It"'+x-2zo
lr
x2o
Lt*'+
x-2>xz
(+
l"-i;'
(1,0
)
.
Tinh tich
ph6n

eI dx r0
dx
Ta co I
=
J-t
U6*g
=
J-t
U;r1"*+;

r0 dx
Xdt
t=J_rr*-)r"*r),
dAt
t=-x
=+
dt=
rl dx
+l
J0
(1+x2)(ex+1)
-dx
(1+t2)(e-t+1)
rL etdt
-t
Jo
1r+t2;1et+r)
Tn
r(hid6
I=[ du=ulfr=
].
*
I-f#,dat
x:tanu
+O*:-!au;
v6'ix:0+t=0,
x:1*
r::
0.s0
IV

a
iIiAm)
o
(0,50
di€m). Gqi
M
ld
trung
di6rn
cira
BC.
.
AABC,
ABCD ld
haitam
gi6c
dAu
bing
nhau
nen
AM
I BC, DM
t BC vd AM:
DM
+
BC
r
(ADM)
. Khi
d6 ke

DH
I
(ABg)
thi
DH
€
(ADM)
suy
ra cludng
thing
AM la
hinh
clri6u cira
AD tr€n
mp(ABC)
+
fiffi =
60"
vd AADM
ctAu
.
Ggi
I, J lAn
luo-t li tAm
cria
haitam
gi6c
dAu
AABC
vd

ABCD.
D
Suy ra tdm
O ctra
mdt
c6u ngo4i
ti6p
tiri
diQn ld
giao di6m cria
liai
trqc cfia hai
tam
gi6c
ABC
vi
BCD lAn
luqt
v€
qua
I vi J.
o
(0,50
dia@.
B6n kinh
cria mflt
cAu
(O)
bing
OA

=
\AI4
O]z .
rac6AM =*,ot::ot
:T
talitMa
IM
=
;AM
:;
,
MO
:
cos3o"
:
I
Do rt6 oA
=VAlzTTMz
-JMz
az a2 a2
-+
3912
. !3na2
V{y, S,n.:4nR'=
9
_
a./T5
6
'o
*1,*

-4'
* '
l
r\
:
* *
* *
*
*r*
-i
-
1,0
V
I cli6m)
l.
Q,0
itidm)
HQ
bp'
e
{if
_,17:}li;:":,^
x'- 4xy
+
4y2
+2(x2
-2x+
1):
cQng
v6

v6'i
v6 c6c
bpt ta
du'gc :
3m
<+ (x-2y)'+2(x-
l)'.3m
(*)
0,50
TiL(+) slry
ra
m>
0. Khi d6 tax6tcip(x,
y)th6a
rnsn
{i-?;J
*
fr":
rt,,
R6r.Ang
{,*' :"!
+.2v'
-x=
0
s
m
Nhr-rvfly
1r;l)
ld nghi€m
crtrahdbpr.

."
tx'-Zxy-2x*2:0<m
D6ps6:m20.
0,50
(1,0
tli€m).
a) Phuong trinh chinh
tic
cria
(E)
,
5
.5:
Elip
(E)
c6 c:
4
e
a2
-
b2
=
16.
Di€m A(0: 3)
"€
(Lt) +
b2
=
9
+

az
=25.
x2 v2
VAv(E):-+L:1.
259
0,50
/2
VI
x
.
cuem)
a)ft Uf''
=
3MF2
vd MFl
+
MFz
:2a=+
4MF2 =
2a
e
4MFr'
:
a'
++

[(a-xo)2
*yot
]
:

25
e
4(16
-8xn
+xo2
*yo2)=25 (l).
Met
khdc M(x.;
y")
e
(E)
+
*.'*
=,
Q)'
.2s eJ3e . .
zs
sJ3e
.
Giai
he
(l),
(2)
ta dusc
tut'(
,
;-;
);
Mt(;
;-

I
).
0,50
Cqi
vecto chi
phuong
crtad1,dz
lAn
luqt
le
[':
0:2;2),6.=
(;2;-z)'
Giao
cti6m
cisa d.yd2ld
A(l
;
I
;
I
).
0,25
Ve"trr
ph6p tryi5n
cira m4t
phing
(P)
chrlia
db dyld$

:17|.,6I
:
(-
8; 4; 0) n€n
phucrng trinh
(P)
ld
2x-
y
-
I
:
0,
rO rdng
M
€
(P).
rac6
lfrl:
lfrl
nen
q:6+6.=(2;4;0),4=G d:(0;0;a)
ldhai
vectochi
phu'ong
cira
hai dudng
phdn
gi6c cria hai
g6c

t4o bdi
db
d2.
Hai vectcr
u1=
,6.
"6
th6 xem
ld hai
vecto
chi
phuong cira
hai
dudng
thing cAn tim.
0,50
Dud'ng thing
A1 di
qua
M nh$n
filim
vecto chi
phuong c6
phuong
trinh
:
Eudng thing Az
di
qua
M nh{n

{
ldm
vecto
chi
phuong
c6
phr.rong
trinh:
(x=2
ln
=,
lz=Llt.
(x=2*t'
lt=3+2t'
\z:\.
Dubng thing
A2
giao
v6i
dv dztqi
di€m
A(l; l; I
)
khdng t4o thdnh tam
gi6c,
n6n loAi Az,
(x=2
Vay duong thing
cAn tim ld
A ,

ly
=
E
lr=t+t.
0,25
VII
'I
itiim\
(1,0
di6m). Giei he
phuongtrinh .
Hq <rd
cho tucrng duong
,u,,
li##Jt*
^16r_,
*=
,rl;
Tac6(2)
a,fzyT+zy
+ t+
dzvr-v
+t':(2y'+2y
+
l)-(2yt
-y+
l)
e,fzyt
+
zy +

t
-,fztz
-
y
+
t
:
t
e,ffr
+
zy + t=.fzyz
-
y
+
l, +t
e
2y'
+
2y
+
1
:2y2
-y
+2 +z.rfzyz
-
y
+
t
ez.,[zyr-y+t=3y-l
1

3y-L>0
o
lnlrr'
-
y
*
L)
=
9yz
-
6y +
L
*l
y>:
('=i
*
\r'
-
r,
it=
o
o
t['r==?
D6p s6 :
(x,
y)
:
(22,3).
y=3+
x=22.

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