Đề thi thử đại học cao đẳng lần 3 năm 2010 - 2011 môn toán khối D THPT Lê quý Đôn potx

\
Tru'(l'ng Chuyen U
Quy
Don
llR
-
VT
DE
THI
nIli'
D';'I
HOC
- CAO DANG LAN 3 (2010 - 2011)
Mon:
Toan
- KhBi: D
. DE
CHiNH
THUC
"-~
TIlt';
ginll/lull
biii: 180 pltut. )
I.
Phin
chung
eho
tAt
ca
de
thi

sinh
(7.0 tlMm):
f{;!W!j
Cau
1(2.0
tI;im): Cho ham s6 y =
x-I
(*)
x+l
1)
Khao sat
S\I
biSn thien va ve d6 thi (C) cua ham s6 (*).
2) Cho hai
di~m
A(4;3) va B(-I;-2). Tim t9a
de)
di~m
Mtren
(C) sao cho
tam
giac MAB
co
tr9ng
Him
ding
thuQc
(C).
Cau
II (2.0

tliim):
·

h .
!,
sin
4
x+sin2x
cos
4
x
'2·
1)
G
tal
P
uang
tnnJ,.l:
= COS-
x.
tan
2x-l
Xl
+i -
xy(x
+
y)
==
4
2) Giai

h~
phuang
trinh: .
{
l+xy=~x+
y-l
In2
3x
Cau
III
(1.0
tliim):
Tinh tich phful J-2-e-dx.
o e
X
-9
Cau
IV (1.0
tliim):
Cho hinh
chOp
S.ABC co day la tam giac
ABC
vuong
t~i
C,
BC
a va BAC 30°.
Th~
:1

tich kh6i ch6p S.ABC
bAng
~
va
SA
= SB =
SC
. Tinh
chitu
cao eua hinh eh6p
S.ABC
va eosin cua goc
2
gifra hai
dUOng
thdng
AC,
SB.
Cau
V (1.0
tliim):
2 2
2
;.
d
h'
9
'nh::'
a b c +
1

Ch
ocacsot
' h
\IC
uang
a,
b
,c
t
oa
a+
b
+c;;:: .
Cl
mngml
.
rang:
+ + ;;::
+ 1 +1
10
.
b c a
II.
Ph~n
rieng
(3.0
tliim)
: Thi sinh chi aU(fc
lam
m<J1

trang hai
phdn
(phdn A ha(ic
phdn
B).
A.
Theo
chuong
trinh
Chu§n:
Cau
VI.a (2.0
tliim):
1)
T rong
m~t
phing
v6i M
tQa
de)
Oxy,
cho tam giac
ABC
co C thue)c
dUOng
thing
d:
x - y - 3
::::
0 ;

A(
1 ;2), B(2;7) va chiSu cao
AH
cua
tam giac
ABC
bAng
1.
Tim
to~
de)
C.
2)
Trong
khong gian v6i M
tQa
dQ
Oxyz,
cho cac
di~m
A(0;2;-l),
B(3;0;2) va C(2;-1 ;0), ViSt phtrang
trinh
m~t
du
(S) di qua A, B,
eva
(S) ti€p xuc v6i
m~t
phdng (P): x + y +

2z
+ 2 = 0 .
Cau
VII.
a (1.0 tlMm): Trong
m~t
phing
phuc cho s6 phuc z thoa man
Iz
-
il
==
2,
tim
t~p
hgp cac di6m
bi~u
di~n
cua
36
ph~c
w =·iz +
i-I.
B.
Theo
chU'ong
trinh
Nang
cao:
Cau

VI.b(2.0 tlMm):
2 1
1)
Trong
m~t
phing
v6i M
tQa
dQ
Oxy,
cho elip (E):
~5
+
~
=1 co hai tieu di6m la FI va
F2.
Di~m
M
tren
(E) tho a
man
~
==
60°
, Tinh
di~n
tfch tam giac MFIF2.
2)
Trong
khOng gian v6i M

tQa
de)
Oxyz,
vi~t
phuang trinh
m~t
ph~ng
(P)
di
qua di€m A(3;1;2), (P)
"d'
h! d
x-2
y-3
z+l
d
A
h"kh'
'h'-
d
'(P)b~
3
song
song
ven
uang
tang
:

=

==

ong t
en
oang cac
glUa
va
ang.
2 1 2
Cau
VII.b(l.O
tliim):
Tim
mQi
gia trj
Clla
tham s6 m
d~
d6 thj ham s6
y:::::
(x +
2)(x
2
-
mx +m
2
3) ti6p XUC
v6i tr\lc hoanh.
IIIt1r

Thi
s~n~
khO,n~
dU'9'c
sir dl}ng tili
li~u.
Can
bi}
coi
thi
khOn
r
g~ai
thich
gi
them.
11(,)
va
ten
thl
Sinh:

,

So
bao
danh:


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I
'fnrang
Chuyen Le
Quy
Don
BR
VT.
DAp
AN
VA
HtrONG
DAN
CHAM
DE
THI
THU
D~I
HQC
- CAO DANG
LAN 3 (2010 - 2011). Mon: Tolin -
KhAi: D
CAu ,
NQidung
Thang
Ghi chii
(fi~m
1
Kh
• . , -

d"
b'
X
-1
, ao
sat
va
ve 0 t
I:
y =

0.25
x+I
*
T~p
X3.C
djnh: D =
IR
\
{-I}
0.25
y'
= 2
2;
ham sil d6ng bien trong timg khoang (-«>; -1), (-1;
+00)
.
(x+l)
*
Ti~m

c~:
lim y = 1
~
TCN : y =
1;
lim y::::
+00
~
TCD:
x =
1.
0.25
x-+±«> .<-+-1"
*Bang bien thien:
x
-00
1
+00
y'
+
+
Y
1/
+00
I
~
-00
u'
:;
"

,
u
L
"
!
.1.1
<4
4.1
1 ·IJ

,
V'
"
(C)Y'~
,+1
0.25
I
l
2.
Tim
di~m
M:
*M
thuQc
(C) nen
M(m;
m-l),
ABM
III
tam giac

<::::>
AM,AB
khang cung phuong
0.25
m+l
<::::>m:;t:O;±I(*)
-Niu
khdng
tlq.t
hoq.c kiim
0.25
*
TrQngtam
G(m+3;
2m
J,GthUQC(C)
<=>
2m
=~.
tra tlk
cho
3 3(m+1)
3(m+l)
m+6
16i tla 0.5tl
0.25
*
m=Ohoacm=9.
0.25
* Ki6m tra

di~u ki~n
(*), ket
lu~n
M(9;4/5).
f-= j " '
1 +
IT
1
G
· • . b
t'
h sin
4
x + sin
2x
- cos
4
X 2 2
(1
)
•
lal
p
U'O'Ug
nn
: = cos x
,
tan2x-l
x:;t:
Jr

+k
Jr
0.25
*
:I:

~
k·. 8 2
uleu
l¥n:
Jr
Jr
{
x;c
-+
k-,(k
E
Z)
4 2
0.25
* V 6i dk tren,
(1)
<=>
cos
2x
=cos
2
2x .
0.25
* cos2x = 0

(101;ii)
0.25
I * cos2x = 1
<::::>
x =mJr,(m E
Z)
(thoa
man dk)
II
X2
+
y2
-xy(x+
y)=4
2.
Giai
he
phU'O'Ug
trinh:
~
{
I+xy=
x+y-l
0.25
S2
2P
-
SP
= 4
(1)

*
D~t
S=x+y,P=xy;S2-4P~0,S~1,h~trathimh
r;:;-:;
{
1+P = " S
-1
(2)
0.25
* (1)
<::::>
(S
+
2)(S
-2-P)
= 0
<::::>
S::::
P+
2 (lo'ili S = - 2, do dk).
0.25
*
The
vao (2): 1 + P
::::
11
+ P
<=>
P
::::

0 V P =
-1.
Suy
ra
S = 2
ho~c
S = 1
0.25
* N h'
A
(0'2)' (2'0)'
[I+.J5
.1 J5).[l J5
.1+.J5)
g
l~m
, , "
2'
2 '
2'
2 .
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r:I~II; r ~ : :: r ·-
IV
v
VI.
a
Tioh
ehiSu cao hloh

chOp
S.ABC
va
khoang
each:
*
GQi
H
lit
chan
dUOng
cao
eua
hinh chop,
SA
==
SB
==
SC nen H lit
tam
day,
v~y
H
III
trung
di~m
AB.
a
2
Jj

a
3
,
*
AC
=
aJj
=>
SABC
=
-2
- ; v = 2
nen
SH =
afj
* D\l'Ilg hinh chii'
nh~t
ABCD,
5Uy
ra g6c (SB;AC) = g6c
(BS;BD)
=
rp,
vai
SB
==
SD
==
2a;
BD

=
afj
*N'
fj
en
cosrp=4'
s
.,
. ,
, ,
,
,
,'0
:
""";::

,,,
:
/~
\

~~
;!

,

,-'
\.:
A
~

(j_"~"*~"~
__
_
H\
C6th€dtinhgiavitrai:
VT'2a+b+c+
9 ;
Saua6xetham
f(x)
x+~;x~9
a+b+c
x
PHAN
TU
CHQN
A.Theo chaO'll!!
tnoh
chuio:
1.
mnh
hoc
toa
dO
pbing:
*
Phuong
trinh dth BC c6
d{lIlg:
a(x

-
2)
+ bey
-7)
= O;a
2
+ b
2
> 0
l la+Sbl
*Chieu
cao
AH = 1 nen 1 = d(A;BC) =
.J
<:=>
b = Ov 5a + 12b = 0
a
2
+b
2
*
V~y
BC:
x -
2:;
0
ho~c
BC: 12x
5y
+

11
= 0
* Suy ra
C(2
;-1)
ho~c
C(-2617;-
4717).
c
2.
mnh
hoc
toa
dO
khong gian:
*
M~t
cAu
(S) qua A, B, C nen co
tam
I thuQc hai
m~t
trung tnrc (M), (N)
cua
AB, AC :
(M) : 3x - 2y
+
3z
- 4 = 0 ; (N) :
2x

-
3y
+ z =
O.
* Suy ra I
c6
to~
dQ
d{mg
tham
56:
](1 + 7t; 1+3/; 1
51)
B
*
(S)ti€p
xucvai
(P) nen d(I;(P» =
IA
=R<=>
R;::
J6
=
J(l
+7t)2
+(31-1)2
+(2-5t)2
I=Ovt
=
12

vAR=
J6
83
*
V~y
phuang
trinh
(S)
:
(x
- V +
(y
_1)2 +(z _1)2
==
6
hay
(x_
167
)2+(y
119)2+(z_23
i
=6
83
83
83
0.25
0.25
0.25
0.25
0.25

0.25
0.25
0.25
0.25
0.25
0.25
0.25
In2
3x
e
Tioh
tfeh
phan:
1=
S 2
X
dx
o e
-9
2
12
* 1
==
eX
=>
dl
==
eX
dx.
I =

S-2-
dt
II
-9
*
1=
1+
3 3}
t
I
2(/-3)
2(/+3)
~
* ]
=(t+~lnl/-31)2
2
t+3
1
3 2
* ]
=1+-ln-
2 5
0.25
0.25
0.25
0.25
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VILa
Sa

phuc:
* z x
+
iy
(x;y E lR), ta
co
x
2
+
(y
-Ii
==
4 .(*)
* w =
iz
+ i
-1
==
-(y
+
1)
+
(x
+
l)i
co
di~m
bi~u
dien M :
XM

= - Y -
* Thay y
-XM
1 va x =
YM
- 1 vao (*) ta duqc
(XM
+
2)2
+
(YM
_1)2
*
V~y
~p
hQP
M la dubng tron
tam
1(-2 ; 1), ban kinh R =2.
C6 thJ
d;Jt
w = x + iY ; BiJu
diln
z thea x; y.
Tfr
Iz
-
il
= 2
suy

ra kq
nhu
tren.
0.25
1,
YM
= x +1
==
4
0.25
0.25
0.25
VLb
VII.b
B.Theo
chU'O"llg
trinh
nang
cao:
1.
mnh
hoc
toa
do
phing:
*FJ(-4;0), F
2
(4;0) va
FJF2
= 8

* Gia
sir
M(x;y)
ta
e6 MFJ 5 +
4/5.x;
MF2
= 5 - 4/S.x. Theo dli
Cosin:
8
2
=
(S
+
4/S.X)2
+
(5
-
4/S.xi
- 2(5 + 4/S.x)(S - 4/S.x).cos60
o
3J3
*Gilii tim duqc x
2
13
.25/16 . Suy
ra
Iy
I=

4
*
V~y
S(MF
1
F
2
)
=
3J3
.
2.
mnh
hoc
toa
do
khong gian:
2
* Pt (P): a( x - 3) + b( y - 1 ) + e( z -
2)
= 0,
vcri
a
2
+ b
2
+ c > 0
(P) song song d nen :
2a
+ b + 2e =

O.
l-a+2b-
3c
l
*Khoang cach (d;(P» = d(B;(P» = I
2
2
'
vcri
B(2;3;-I)
thuQc
d.
2
va
+b
+c
* d(d;(P» = 3
<:::;>
2c
2
-
ac
-lOa
2
=0
*Chon a =
l;e
= -2 , b = 2 ;
ho~c
a 2; c = 5; b = -14

V~y
phuong trinh
m~t
ph!ng (P): x + 2y -
2z
- 1 = 0
ho~e
2x
- 14y +
5z
-
Tim
tham
sa
m:.
*
Di~u
ki~n
ti6p xue eua
Ox
va db thi ham s6 :
(X
+ 2)(X2
mx+m2
-3)==0
"A
co
nghl~mx.
{
X2

-mx+m2
-3+(x+2)(2x-m)=0
*
H~
pt
tren
<:::;>
{:;-~+m'-3
0(1)
2
X2
-
mx
+ m - 3 = 0
(II)
{
(x+2)(2x
m)=O
*
(I)
c6
nghi~m
khi m = ,
* (II) co
nghi~m
khi m =
-1;
m
==
±2

.
2
=
O.
0.25
0.25
0.25
0.25
0.25
0.25
0.25
.0.25
0.25
0.25
0.25
0.25
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