Đề thi thử đại học môn toán khối A trường chuyên Lê Quý Đôn- Vũng Tàu ppt

Tnr(mg
THPT
Chuyen
Le
Quy
Don
DE
THI
THU
D~I
HOC
LAN
3 -
MON
TOAN
KHOI
A
Tinh
Ba
Ria
Viing
Tau
Thai
gian him bai: 180 phut.
I.
pHAN
CHUNG
CHO
TAT
CATHi
SINH:

(7
diim)
3
Cau
I (2 di8m):
Cho
ham
s6
y x -
3x
2
+ 3.
1. Khao
sat
S\1'
biSn thien va ve
d6
thi (C)
cua
ham
s6.
2. ViSt phuong trinh tiSp
tuy~n
cua
db thi (C), biSt
ti~p
tuySn di qua
di~m
A(-l;
-1).

Cau
II
(2 di8m):
1.
Giai
phuong
trinh x
3
-6x
2
+ 12x
-7
={j_x
3
+9x
2
-19x
+
11
.
2sinx
+ 1
cos2x
+
2cosx
7
sinx
+5
2.
Gi1ii

phuong trinh: -
;=
2
cos
X
J3
-
cos
2x
+ 2
cos
X +
1-
J3
(cos
x +
1)
.
Cau
III
(l
di~m):
'
h'
h
liZx3~x3+8+(6x3+4x2)lnxdx
h
A
I
T

III
bc
p an sau: =
x
Cau
IV
(l
di~m):
Cho
hlnh chop SABCD co
ABCD
1a
hinh binh hanh tam
0,
AB = 2a,
AD
2aJ3
,
c~c
Cl;l.llh
ben
b~ng
nhau
va
bk
g
3a,
gOi
M
1a

trong
di~m
cua OC. Tinh
th~
dch
kh6i chOp SABMD
ya
dien
dch
cua
hinh
cAu
ngo~i
ti~p
tu
dien SOCD. '

Ciu
V
(1
di~m)
Cho
x, y
thOa
x
2
+ r -
xy
=
1.

Tim
GTNN
va
GTLN
ella
P =
X4
+ l x
2
y2.
n.
pHAN
RIENG
(3
diim)
Thi
sinh
chi
dU'Q'c
lam
m9t
trong
hai
phAn (phAn 1 ho,"c phAn 2).
1.
Theo
ChU'01l~
trinh
chuAn:
Cau

VI.a
(2
diem):
1.
Trong
m~t
phing
Oxy
cho
MBC
nQi
ti~p
duang
tron (T): x
2
+ l 4x -
2y
- 8 =
O.
Binh
A thuQc
tia Oy, duemg cao
ve
tir C
n~m
trcn
duong
th~ng
(d): x + Sy =
O.

Tim
toa
dQ
cac dlnh A, B, C bi8t
~g
C co hoanh
dQ
1a
m¢t
s6
nguyen.
2.
Trong
khong gian
Oxyz
cho hai
duang
th~ng
Cd
1
):
~
-1
72=Z~:,
(d
2
):
{;=~::
2
Z

4+t
va
m~t
ph~g
(a.): x - y + z 6
==
O.
L~p
phuong
triM
duang
th~g
(d)
bi~t
d II (a.)
va
(d)
cit
(d
1
),
(d
2
)
l~n
luQ't
~i
M
va
N sao cho

MN
=
3.J6
.
Cau
VII.a
(1
di~m):
Tim
t~p
hQ'P
cac diam
biau
di€n cho
s6
phuc
Z th6a man M
tMc:
Iz
+ 3 - 2il
12z
+
1-
2il
2.
Theo
chU'O'D~
tdnh
nang
cao:

Ciu
VI.b
(2
diem)
1.
Trong
mftt
ph~ng
Oxy, cho tam giac
ABC
co dlnh
A(O;
4), trong tam G
[~;
~
)
va
tf\1'C
tam trimg
vai
g6c
toa
dQ.
Tim toa
dQ
cac dlnh B, C
va
di~n
dch
tam giac

ABC
bi~t
XB <
Xc
.
.
x-I
Y+2 Z 2
{X
2 - t .
~
2.
Trong
khong gian
Oxyz
cho hai
duang
thang
(d!):

;;;
=
,
(d
2
):
y = 3
+t
va
m~t

2 1
-2
z=4+t
ph~ng
(a.): x - y + Z 6 =
O.
Tim tren (d
2
)
nhfrng diem M sao cho
duang
th~ng
qua M song
song
voi
(d
1
),
c~t
(eL)
t~j
N sao
cho
MN = 3.
Ciu
VII.b
(1 di6m):
'

,

,{eX
e
Y
(In y
In
x)(l
+xy)
G
1a1
he phuong trmh .
.
21nH2lny
_ 3.4
ln
:<
;;;
4.iny
Thi sinh
kMng
(/l.f9c
sa
dl,mg
tai
lif?u
Can b9 coi thi
kMng
gial thlch gi them.
HQ
va
ten thf

sinh
: ;
So
bao
danh:
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DAp AN
DE
THI
THU
D~I
HQC
MON
ToAN
ThiYi
gian: 180
phut
Cau
y
Caul
1
I
2
!
!
CiuII
1
i
i

NQidung
Cho ham so y =x
j
-
3x" + 3.
Khao sat
S\l bien thien va ve do
thiJC)cua
ham so.
T~p
xd va
Gi61
h~
y'
=3x"
6x
y'
::;
0
~
x =0 hay x =2
Bang
bien thien:
y"
va diem uon
Gia tri
d~c
bi~t
Do thi
va

nh~
xet:
Viet pt tiep tuyen eua (C), biet tiep tuyen
di
qua diem AC-l;
-1).
DuOng tMng (d) qua A
va
co
h~
so goc k
=>
Cd):
y + 1
""
k(x + 1)
=>
(d):y
= kx + k -
1.
(d)"
'(e)
{XJ-3X2+3~kx+k
1
eo
nghi~m.
tIepxue
~
3x
2

-6x
k
=>
x
3
3x
2
+ 3 =
3x
3
-
6x
2
+
3x
2
-
6x
1
~
2x
3
-
6x - 4 = 0
~
X = 2 hay x =
-1.
,.
x = 2
=>

k = 0
=>
(d): y =
-1.
,.
x =
-1
=>
k =9
=>
(d): y =
9x
+ 8
Giai phuong trinh
x
3
-
6x
2
+ 12x 7 =
~_X3
+ 9x
2
-19x
+ 11.
D~t
Y
~-x3+9x2-19x+ll
Kh'
d'

6 { Y
~
x
3
-
6x'
+
12x-7
Iota
c
y3=_x
3
+9x
2
19x+ll
=>
y3
+ 2y =x
3
-
3x
2
+ 5x - 3
Q
y3
+ 2y = (x
_1)3
+ 2(x
-1)
(1)

~
Xet ham s6 f(t) =e+
2t.Ta
co
f'(t)
3e
+2>0
Suy ra ham
s6f(t)
d6ng
bi8n
tren R
~
Tu
(1)
=>
y =
x-I
.
Qx-l
x
3
-6x
2
+
12x-7
Qx3-6x2+11x-6=O
Q (x
-1)(x
- 2)(x - 3) = 0

._.
-
Di~m
2:=2d
"L=.1.25d
0.25
0.25
0.25
0.25
0.25
'L= 0.75d
0,25
0.25
0.25
2:=24
:L=ld
I
!
0.25
i
0.5
0.25
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lx
~j
¢::>
x=2.
x=3
I

I
I
I
I
12
Giai phucmg trinh:
2sinx+
1
cos 2x +
2cosx
-7sinx
+ 5
2cosx-Ji
==
cos2x+2cosx+1-Ji(cosx+l)
.
2.:
=
Id
I
!
Dieu
ki~n:
{2cosx 13
,,0
{2eosx
,,-13
{cosx"
J3
cos2x+2cosx+1 y3(cosx+l):;eO

<=>
(cosx+l)(2cosx y3):;tO
<=>
cosx:;e
_~
-
0.25
(1)
<=>
(2sinx + 1)(cosx +
1)
==
cos2x + 2cosx
-7sinx
+ 5
<=>
2sinxcosx + 2sinx + cosx + 1 = 1 - 2sin
2
x + 2cosx - 7sinx + 5
0.25
<=>
2sinxcosx - cosx +
~
= 0
<=>
cosx(2sinx -
1)
+ (2sinx - 1)( sinx + 5) = 0
<=>
(2sinx - 1)( cosx + sinx + 5) = 0

[ [
~
. 1 x
=-+
k21t
smx
==-
6
<=>
2
<=>
0.25
. -
51t
sm x + cos x = 5 x
==
(5
+
k21t
So
s8n.h
diSu
ki~n
ta
duQ'c
nghi~m
cua phucmg trinh
1ft
x
==

~:
+
k21t
(k E Z).
0.25
Call
III
rnh
'h
han
I
fX3~X3+8+(6X3+4X2)lnxd
I
1
tIC
P sau:
==
x
2.:
= Itt
,
'x
f
X2~X3
+ 8 dx
~
12
(6x
2
+4x)ln

xdx =
II
+
12-
I
I
I
I
0.25
).
Tinh II: Dat t =
~X3
+ 8
=>
t
2
= x
3
+ 8
=>
2tdt = 3x
2
dx
=>
x
2
dx
=
~tdt,
. 3

D6i
c~n:
x
1
=>t=3
x = 2
=>
t =4,
[ r
0.25
2 2
t
3
2
h
do
II
rt.3"tdt =3"'
3"
3
==
g(
64-27)
==
-,
$
).
Tinh h
D~t
u = lnx

=>
u' =
!
x
v'
= 6x
2
+ 4x
chon
v
==
2x
3
+
2X2
, ,
h =
[(2x
3
+2X2)1nX
r-
f(2x
2
+2x)dx
=
0.25
[ r
2x
3
23

I
241n2-
-3-+
x2
1 = 241n2
-
3
J
A 77 23 2
.J t"
S-
0.25
I
!
VayI=
241n2+ =241n
+-
I
J
. 9
39
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-

~"""

r
-~

,
ABCD
1ft
hinh binh hanh tam
0,
AB = 2a, AD =
2afj,
cac
cl;U1h
ben
bkg
nhau
va
bkg
3a,
gQi
M
1ft
trung
di~m
cua
2:=ld
OC. Tinh
th~
tich kh6i chop
SABMD
va

di~n
tich cua hinh
cAu
ngo~i
ti~p
tii'
di~n
SOCD.
~
Ta
co SA = SB =SC =SO nen SO 1. (ABC D).
,
~
/),.
SOA = .=
/),.
SOD nen
OA
::;;
OB
OC =
00
=>
ABCD
1ft
hinh
chu
nhat.
0.25
•

=>
SABCD
= AB.AD =
4a
2
fj.
.
~
Ta
co BD
~
AB2
+
A0
2
~4a2
+ 12a
2
=
4a
=>
SO =
~SB2
-OB2
::;;
~9a2
-4a
2
==
a.J5.

0.25
A_I
4a
3
J15
, _ 3 3
c:
V~y
VSABCD
-
"3SABCD'SO
==
3 . Do do
VSABMD
-
"4
VSABCD
= a ;15.
~
GQi
G la
trQng
tam
/),.
OCD, vi
/),.
OCD deu nen G cling
1ft
tam
dUOng

tron
ngo~i
ti~p
/),.
OCD.
D\ffig
dUOng
thkg
d qua G va song song SO thi d 1. (ABCD) nen d
la
tI1,lC
cua
0.25
/)"OCD.
i Trong mp(SOG) d\ffig
dUOng
trung trgc
Clla
SO,
c~t
d
t~i
K
dt
SO
~
I.
· Ta co:
01
la trung

tT\Ic
cua SO
=>
KO::;;
KS
rna
KO =
KC
=
KD
nen K
18.
tam
mi},t
cAu
n
o~i
tiS
ill
di~n
SOCD.
Taco:
GO::;;
CD
=
2a
fj
fj
0.25
R=KO=

~OI2+0G2
2 2
• 31a 317ta
i Do do S A =
47tR2
=
47t
=
C~U
12 3
CauV!
. Cho x, y
thca
x + y -
xy
=
1.
Tim
GTNN
va
GTLN
Clla
P = x
2:=ld
•
0.25
0.25
Ta
co:
f(t)

=
-4t
+ 2.
1
f(t)
= 0
¢::>
t = .
2
f(-,,!,) =
!
fO) = 1
va
f(
!)'::;;~.
3
9'
2 2
Yay MaxP =
maxf(t)
= l
va
Min
P
=::
minf(t)
= 1

[-1;1]
2 [-1;1) 9

0.25
I
+ ::-::: :-'
-'-
va
t =-1/3 , t
Yz
tinh dmlc 0.25 i
~-~~' ~ + ~=~2~d
Vl.a
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I
1. :
llUUg
Ulp
vxy
cno
L.\
fin\ "
nQl
nep Quang
l:ron
\1):
x -r Y -
Ll-X
- Ly
lS
=
u.

Dinh A thuoc tia Oy, duemg cao ve tir C nfun tren duemg thfutg (d): X + 5y
O.
I=ld
i
Tim
tQa
do
cac
dinh A, B, C
bi~t
rfutg C co
hO<lnh
do h\ mot
s6
nguyen
» A thuoc
tia
Oy nen
A(O;
a)
(a>
0).
, Vi A E (T) nen
a'
- 2a - 8
~
0
""
[a
4

=>
a
~
4
=>
A(O;
4).
0.25
•
a=-2
I
I » C thuqc (d): x + 5y = 0
nen
C(-5y;
y).
: C E (T)
:::::>
25y2 +
y2
+ 20y 2y - 8 0
<=>
26y2 + 18y - 8
0
ly
~-l=>
x
~5
0.25
<=>
y =

~
:::::>
x = _ 20
:::::>
C(5;
-1)
(Do
Xc
E Z)
13
13
» (AB)
1-
(d)
nen
(AB): 5x - y + m = 0 rna (AB)
qua
A nen 5.0
4+m
0
:::::>m==4.
V~y
(AB): 5x
-y
+ 4
O.
B E (AB)
:::::>
B(b; 5b + 4).
[b-O

0.25
i
B E (T)
<=>
b
2
+ (5b + 4)2 - 4b - lOb 8 - 8
==
0
<=>
26b
2
+
26b
==
0
<=>
- .
b=-l
I
Khi b = 0
:::::>
B(O;
4 )
(lo~i
vi trimg
vOi
A)
Khib
-1

:::::>
B(-I;
-1)
(nh~).
0.25
i
V~yA(O;
4),
B(-I;
-1)
va
C(5;
-1).
2
r2
t
Trang
kg
Oxyz cho (d}):
x
1
+2
z 2 , _
,:
=
1 =
-2
,(d
2): y=:03+t
vam~tphang

2
z=4+t
I=ld
(a):x-y+z-6
0,
L~p
phuong
trinh duemg
thkg
(d)
bi~t
d II
(a)
va
(d)
d.t
i
(d!), (d2)
iAn
luQ"t
t~i
M
va
N sao chc
MN
=
3.J6.
ME
(d!):::::>
M(1 + 2m;

-2
+ m; 2
2m)
I
NE(d
2
):::::>N(2
n; 3 + n; 4 + n)
:::::>
NM
==(2m+n-l;m-n
-
5'-2m-n
-2)'
n
=(1'-1'1)
, ,
a ' ,
0.25
-
MN
II
(a)
:::::>
na .NM =0
<=>
2m
+ n - 1
-em
n - 5)

2m-n-2=0
<=>
-m
+ n + 2 = 0 ¢:;> n = m
_.
2.
=>
NM
=(3m
3'
-3'
-3m)
, ,
:::::>
NM
=
)(3m-3)2
+(_3)2
+9m
2
=
3J2m
2
-2m+2
NM
3.J6
<=>
2m
2
-

2m
+ 2 = 6
<=>
m
2
m-2
o
<=>
m
-1
hay m =
2.
0.25
i
~-
x+l
y+3
z-4
I
» m =
-1:
M(-I;;Zj
4)
va
NM
-3(Ui::-1)
~
Ed):·

-""

,
-~ ~:;::"~
0.25
.
~.
. 3
1
-1
I
»m
. . _ ' _ lli
~ 1
~;~
. x 5 _ Y
z+2
2. M(5, 0,
-2)va
NM
-
~.J\J,
r,-r.rt
(d).


0.25
1~
-1
-2
I

VIl.a
Tim
t~p
hqp
cac dibm
bi~u
dien cho
s6
phuc z th6a:
Iz
+ 3
2il
=
12z
+
1-
2il
I=ld
!
G9i M(x; y)
la
di~m
bi~u di~n
cho
s6
phuc z =x + yi (x; y E R).
Taco:
Iz+ 3 -
2i
l=

12z
+
1-
2i
l
<=>
Ix
+ yi + 3 - 2i\
1
2
(x+yi)+1-
2i
l <=>1(x+3)+(y 2)il
==
IC2x
+
1)
+ (2y
2)il
0.5
<=>
(x+3i+(y
2)2=(2x+1i+(2y-2t<=>
3x
2
-:3y-2x-4y
8=0
0.25
i
j

V~y
t?P
hqp
cac diem
_~
ia
duang
tron (T): 3x- +
3l-
2x - 4y 8 =
O.
0.25
VI.b
I,=2d
I
Trong
m,t
phing
Oxy. cho tam giac ABC
cO
dinb A(
0.4
ttrong
tim
G (
~;
~
)
I=ld
,

1I
I
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I
.
bi~t
r~ng
XB
<
Xc.
. 3 ( 4 )
,
Xl-O=-
0
-
3-
23
{X,=2
. 0.25
Tac6
AI
-AG
:=;>
( )
¢>
_
=>
1(2;
-1).

2
Y
-4
~
~
4
YI
1
I 2 3
BC qua I
va
c6 VTPT
la
OA
= (0;4) = 4(0;1)
:=;>.
BC: y
=-1.
GQi
B(b;
-1),
vi I
la
trung
di~m
BC nen C(4 b;
-1).
-
-
Ta

c6:
OB:::::
(b;
1)
;
AC
=
(4-
b;-5)
0.25
OB.AC:::: 0
~
4b
- b
2
+ 5 = 0
~
b
2
- 4b 5
:::::
0
~
b =
-1
hay b
5.
I
I
: *

b=-l
:=;>B(-1;-l)vaC(5;-I)(nh~)
I * b =
5:=;>
B(5;
-1)
va
C(-I;
-1)
(lo~i)
0.25
»
BC::::
(
6;
0)
:=;>
BC
6; d(A; BC)
=5
:=;>
SABC = 15.
I 0.25
2
r
2
-
t
x-I
+2

z-2
'
Trong kgOxyz cho (d
1
):
-2-::::
T
==
=2
' (d
2
):
y 3 + t
va
m~t
phang
I=ld
z=4+t
I
(a): x y + Z - 6
O.
Tim
tren (d
2
) nhiing
di~m
M sao cho dlI6ng thAng
qua
M
song song

v6i.
(d
l
),
c~t
(a)
4ti N sao cho
MN
3.
M
E (d
2
):=;>
M(2
m; 3 + m; 4 + m).
r=2-m+2t
(d) qua M
va
II (dJ)
nen
(d):
y:::::
3+m+t
0.25
z
4+m-2t
N (d) n
(a)
nen
tQa

dQ
N
th6a
M:
x 2
m+2t
y
==3+m+t
:=;>
2 - m + 2t - 3 - m
t+4+m
2t
6=0
I
z==4+m-2t
i
x-y+z
6 0
0.25
~t
-3
-m:=;>
N(-3m
- 4; 0;
3m
+ 10).
i
:=;>
NM
=

(6
+ 2m; 3 + m;
-2m
6)
0.25
I I
:=;>
NM2
(2m +
6)2
+
(m
+
3)2
+
(-2m-
6i
I
Do
d6
MN
==
3
~
9(m +
3i
= 9
~
m + 3 = ± 1
~

m =
-2
hay m =
-4.
!
V~y
M(4;
1;
2) hay M(6;
-1;
0).
i 0.25
VII.h
G···h"
h
.nh{e'
e'=(lny
Inx)(1+xy)
1al
e p lIang
tn
.
I=ld
. inx+2lny
_3.4
lnx
:::::
42
1ny
Dieu

ki~n:
x, y >
o.
I
Ta
c6: 1 + xy >
O.
*
x>
y:
VT
(1) > 0 va
VP(I)
<
O:=;>
VT(I)
>
VP(l)
(voU)
* x <
y:
VT(1) < 0 <
VP(l)
(vo
Ii)
0.25
Do d6 tir (1)
:=;>
x =
y.

Thay vao (2)
ta
duQ'c:
2
31nx
_3.4
lnx
=4.inx
¢::::>inx[{inxi_3.21nx
4]
0
0.25
[2'"
0
i
i
2
1n
x
:::::
_ 1
<=>
lnx =2
¢::::>
x
::::
e
2
.
I

~
0.25
i
2
1nx
==
4
'-:~
V~y
Mc6
nghi~m
la
x y
e.l.
,

_

_1
0.25 I
BET
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