Đề thi đại học tỉnh Bà rịa vũng tàu 2010 - 2011 pps

Tnrong
Cbuyen
Le
Quy
Don
BR
VT
DE
THI
TmJ
D~I
HQC
-
CAO
DANG
LAN
3 (2010 - 2011)
Mon:
Toan
- Kb8i: B
DE
CHiNH
THUC
ThiJi gian lam bai: 180 philt.
I.
Ph§n
chung
eho
tit
eli
cae

thf
sinh
(7.0
([iim):
Cau
1(2.0
([iim):
Cho ham s6 y:::;
~
(*).
x l
O
1)
Khao sat
S\I'
bi~n
thien va ve d6 thi (C) eua ham s6 (*).
2)
Vi~t
phuang trinh tiep
tuy~n
d eua (C)
bi~t
d
e~t
hai
tIVe
to~
dQ
t~i

cae di8m A
va
B thoa man tr9
n
g
tam
tam giae OAB thuQe
dUOng
thing: x
4y
=
0,
(0
la g6e
to~
dQ).
Cau
II
(2.0
([iim):
q·
h
'nh
2sinxeos2x+sin2x
1
1)
tal P
uang
trl : . =
+eosx.

-2smx
2)
OiM
b~t
phuang
trinh: x
log~
x + 6 <
3x
log2
X + log.fi x .
Cau
III
(1.0
([iim):
Tinh tieh ph§n
1=
j£x+i
dx.
o
8x+3
Cau
IV'(1.0
([iim):
Cho hinh chop S.ABCD co
ABCD
la hinh vuong
e~nh
a
va

m~t
ben (SAB) vuong goe v6i
m~t
day. Tam giae
SAB
vuong
t~i
S, goe gifra
SB
va
m~t
phfutg (ABCD)
bing
30° .Tinh theo a th8 tich cua
kh6i chop
S.ABCD va khoang each gifra hai
dUOng
thfutg AB,
Sc.
Cau
v (1.0
([iim):
T
'
+f t
' "
t'
,
th
J-

h
h~
h t
'nh
{2X
2Y
+9X
= my2
'd'
b
h'~
1m
La
ca gta
r!
eua
am
so m sao
COy
P
uang
n co ung a ng lym.
2
xy2
+2y
= x
II.
Ph§n
rieng
(3.0

([iim)
: Thf sinh
chi
aU(lc lam
m6t
trong hai
phdn
(phdn A
hoi;ic
phdn
B).
A.
Theo
ehU'ong
trinh
Chuin:
Cau
VI.a (2.0
([iim):
1) Trong
m~t
phing
v6i
h~
t9a
dQ
Oxy,
eho tam giae d€u
ABC
n9l

ti~p
dUOng
trim
(C): x
2
+
y2
-4y-4
0
va
c~nh
AB
co trung di€m M thuQc
dUOng
th~g
d:
2x
y-l
=
O.
Vi~t
phuang trinh
dUOng
thing
AB
va tim
to~
dQ
di8m
C.

2) Trong khong gian v6i
h~
t9a
dQ
Oxyz,
cho eac di€m
A(3;-I;I),
B(-1;0;-2),
C(4;1; 1) va
D(3;2;-
6).
Vi~t
phuang trinh
m~t
cAu
tiep xuc v6i hai
dUOng
thing
AC
va
BD
l§n luqt
~i
A
va
B.
Cau
VII.a
(1.0
([iim):

Trong eac s6 phuc z thoa man
I;
2-3il
5,
tim s6 phue z sao
cholz+4-5il
d~t
gia
tri
nh6
nh~t.
B.
Theo
chU'ong
trinh
Nang
cao:
Cau
VI.b(2.0
([iim):
1)
Trong
m~t
phfutg v6i h¢ t9a
dQ
Oxy,
cho tam giac
ABC
co
dUOng

thfutg
AB
di
quadiSm
0;
phan
giac trong goc
A thuQe duemg
th~g
d
l
:
x y + 1 = 0 va
dUOng
eao ke tir B
thuQC
dUOng
th~g
d
2
:
4x+
3y
48:::;
O. Tim
to~
dQ
cae di€m A
va
B.

2) Trong khong gian v6i
h~
t9a
dQ
Oxyz,
cho hinh thoi
ABCD
co tam I
thuQe
duemg
thing
d:
x-I
= y - 3 z -
2
va
hai dinh A(3;
-1;
1),
B(
-1;
1;3).
Vi~t
phuang
trinh
duang
thfutg CD.
1 1 1
~
, , (.Jj

-if
' ,
Cau
VII.b(1.0 ([iem): Tim cae gia
tri
nguyen
duang
eua n de so phue z = 1
~
i.Jj la so thuan ao.
IIItT
Thi
sinh khong
dUQ'c
SIT
d\lng tili
li~u.
Can
be}
coi
tbi
kbong giiii
thich
gi them.
HQ
va
ten
thi
sinh:

sA
bao
danb:
••.• ••.••
www.MATHVN.com
www.mathvn.com
II
[
X
=
7l'
+
k27l'
~
*
<=>
7l'
7l'
{\
;(k
E
Z)
x = +
k7l"
X = +
k7l'
6 ' 3
gjl1-;(::-
~\1il
Tij.

[
'''/s
x = +
*
K&
hQ'p
di6u
ki~n,
ta
co
nghi~m:
x =
76
0.25
>
,
TJ."trirng
Chuyen
Le
Quy
Bon
BR
- VT.
BAP
AN
vA
HITONG
nAN
CHAM
BE

THI
THU
B~I
HQC
-
CAO
BANG
LAN
3 (2010 - 2011). Mon: Toan - Khtii: B
Ciu
I
N(U
dung
1
Kh
•
't'
_;l-A
thO
X
. ao
sa
va
ve
uO
I:
Y =

x-I
Thang

•
di~m
Ghi
eM
*
n.p
xac dinh: D =
IR.
\
{I}
0.25
Y I =
-1
2 ; ham
s6
nghich bi8n trong timg khoang
( 00;
1),
(1;
+00)
(x-I)
.
*
Ti~m
c~:
lim Y = 1
~
TCN
: y =
1;

lim y =
±oo
~
TCD:
x =
1.
0.25
x-*±«>
x-*J±
*Bang bi8n thien:
0.25
*
I'll>
thi'
1
+00
y'

1
+00
y
I~
~I
-00
tJ
1
JJ
)
lJ
2. Viet

phU'01l2
trinh
tiep
tuyen:
*
T(a;~)
lati8p
di€m,
pmd4tiT:
y=
1 2
(x-a)+~.
a-I
(a-I)
a-I
2
* Giao di€m d
va
Ox:
A(a
2
;O)
, giao di€m d va
Oy:
B(O;
a 2
J.
(a-I)
TrQng tam cua tam giac OAB : G
(~;

a
2
2 J ' di6u
ki~n
a *0; 1
3
3(a-I)
* G
thuQc
dt:
x-4y
0
<=>
a =O;a =
-I;a
=3
(lo~i
a
0).
1 1 1 9
*
Pm:
y =
x+-
hoac y =
x+-
44'
4 4
0.25
0.25

0.25
0.25
0.25
www.MATHVN.com
www.mathvn.com
2. Giai
bit
phU'O'ng
trinh:
x log; x +6 < 3x
log2
X +log,fl x (1)
J '
I
*
(1)
<=>
(log2
X - 3)(x
log2
X - 2) < 0
r>s
2 (1)
0.25
*
¢:::>
{::':<7<O
/
2
(I/)

/
log2
x
-;>0
/
0.25
*
/(x)=log2
x
-
2
d6ngbiantren
(O;+«»
vaf(2)=Onen
/(x)<0<=>0<x<2;/(x»0
x
*
V~y
nghi~m
cua
(1) : S = (2;8).
0.25
III
h
A
T
' h ' h
p
an:
I

=
4J.J2X+
Idx
lD
tic
o 8x+3
*
D~t
t =
.J2x+
1
=>
dx =
tdt;
d6i
c~:
x = 0

t =
1;
x = 4 t = 3.
0.25
3 2
*/
f~t
0.25
14t
-1
* _3
n

.!.+_1___
1
l l
(1
1
ln12t-11J3
*1==
.!.,lm~.
0.25
-
tl4
8(2t-1)
8(2t+l)[t=
"4t+16
2t+1 1
21;
7
0.25
IV
Tinh
the
rich kh6i S.ABCD
va
khoang each:
s
*
GQi
H
la
hlnh

chi~u
cua S tren AB, chUng
minhSH
-L
(ABCD)
0.25
va
SBA

30·.
*Tinh SA = a
'SB
=aJ3
'SH
=
a.fj.
VA
V a
3
.fj
0.25
2' 2 ' 4
~y
12
S_ABCD
*AB, SC cheo nhau
va
AB II (SCD)::J
SC
nen d(AB;SC)

==
d(H;SCD 0.25
D\ffig va
CM
duqc
d(H;SCD) =
HK
(
HM
II
AD
va
HK
-L
SM).
*T'
hHK-
~
0",~.
-HI<-
-
Q~.
HN
0.25
III
- a -
{,'
,\,
I
1.!l-

,r::-
19
.
l\
\(
:::
f
~
•
Lt,
)
~
\
~
Til
,
v}
C
1.
v
Tim
gia
tn
tham
s8
d~
he
pt
co
3 nghiem: {2X

2Y
+9X
ml
"''9'
\6
,
Lt
x
2
xy2
+2y
*
N~u
x
==
0 suy
fa
y = 0
va
nguqc
ll;li.
(x;y) = (0;0)
1a
m¢t
nghi~m
cua
h~,
veri
mQi
m

0.25
2
X2
9 x - 2 2
*N~u
xy'f::O:
H~
¢:::>
-y+
-l-m;d~t
u=_x
,v=-y
=>X=}jU2v,Y=~UV2
l y y x
-+2-=1
0.25
x
2
X ,
1
veri
m6i (u;v)
veri
uv
'f::
0 cho
~¢t
ngh~m
(x;y) thoa
xy

'f::
0
2U+
9
=m
,
Bai toan
<=>
tim m
d~ h~
pt 2v co
dUng
hai
nghi~m
(u;v) thoa
UV'f::
0
{
v+-=l
,
, u \
Hay phuong trinh 2u +
~
= m co
dUng
hai
nghi~m
u khac 0;2.
u-2
, 9

u
*
L~p
BBT
ciIa
/(u)
=
2u+_
-;u
E R
\{0;2}.
0.25
u-2
.
*
K~t
qua:
mE
(-00;0) D (0;
1)
u (25;
+«»
0.25
'\
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C
0.25
'h1
::;

}ir
-::)
M (
11
'.1
/
'J I
~
PHAN
TV'
CHON
VLa A.Then
chllO'Dg
tranh
chuan:
1.
moh
hoc
toa
46
phing:
*(C) co tam I(0;2),
ban
kinh R = 2.J2 . ABC
d~u
nQi
tiap (C) nen
1M
= .J2
ie-

1l
r:
~r2
.;-
'(::::
)
i)y
_ *M
thuQc
d nen M(m;2m -
1),
1M
= .J2
~
m =
l;m
=
2.
.
1:
l~
- l\'\1'
~n\-3)
o£.AN
~,
t~
5 I
~
-:.
_ ?-

* AB qua M
va
co
VTPT
ta 1M',IC =
-21M
nen i:)
N)
-\
4~1-
\~-r
~-
,.::::)
'2-~
m = 1
::::>
AB
: x - y 0
va
C(-2 ;4). tv\ (1\.)
1\
).
i'"'N:::.
(
1\.
r
11)1-
-\
'il1'V\
tk

tJ
7
-14
AiYl
v
~
,-II)
L.
I
*
m=5::::>AB:7x-Y-8=O;C(-5-;~·t·
1(£
:::-
(-g.}
2.)
2.
moh
hoc
toa
46
khong
giao:
-'~m.
__
._
~
~0tJ-~-m
___
-
* mp(p) qua A

va
vuong goc
AC
: x + 2Y -
2z+
1=° 1
~
~
(~'.)'
1.
'J
_
~
)
* mp(Q) qua B
va
vuong goc BD :
2x
+ y -
2z
- 2 0
*
M~t
cAu
dn tim
co
tam
I thu9c (P)
va
(Q) nen I( 3 +

2t
;2t ;2+3t)
*IA = IB
¢:>
t =
-1.
V~y
1(1
;- 2 ;-1)
va
R =
3;
phuong trinh
mcAu
:
(x-l)2+(y+2)2+(z+1)2
=9
VIla
{
So
phUc:
Cach
1:
*z x + iy,
v&i
x,
Y E
IR
X = 2 + 5 cos t
~

2-3i=5¢:>(x-2i+(y+3)2
25¢:> . ;tE[O;27l')
I
I
Y
-3+5smt
*
Iz+4-5il
=
)(X+4)2
+(y-5i
=~125+20(3cost-4sint)
*
3cost
-4sint
;:::
-5,
'v't E
[O;27l')
nen
Iz
+4 -
5il
;:::
5,
'v'z
*
Dkg
thuc xay ra khi cost =-3/5;sint = 4/5.
V~y

z =
-1
+
i.
Cach
2: .
* Tgp
h(YJJ
cac iii€m M(x ;y) biJu
diln
z trong mp Oxy
la
(C):
(x
- 2)2 +
(y
+
3)2
=
25.
*
Iz
+4 ,.5i!
==
)(X+4)2
+
(y
_5)2
==
AM,

vaiA(-4;5)
nd:m
ngocli
(C).
AMnhO
nh&t
khi M la giao iiiim (gdn A han) cila
(C)
va
lA,
vai
1(2;-3) la tam cua (C).
* lA:
4x
+
3y
+ 1 =
0,
giao
iii€m cila
(C)
va
lA
fa
B(-1;
1)
va
C(5 ;-7).
*lB<lC.
V0-'Z

-1
+ilast5phii:cthoayeucauMitoan.
VLb
B.Theo chllO'ng
trinh
nang
cao:
1.
moh
hoc
toa
40
phing:
*D d6i xUng 0 qua d
j
:
D(-I;I).
*D thu9c AC (tc phdn giae)
va
AC vuong g6c d
2
nen AC: 3x 4y + 7 =
O.
*A ta giao cua AC
va
d
1
:
A(3 ;4) *B ta giao eua
OA

vii
d
2
:
JB(6
;8)
2.
moh
hoc
toa
40
khoog
giao:
\
*I thu9c d nen I( 1 +t;3+t;2+t)
* ABCD la hinh thoi nen IA.IB
==
0
¢:>
t =
-1
.V~y
1(0
;2 ;
1)
*C, D d6i xung
v&i
A,B
qua
I nen C(-3;5;

1)
va
D(l
;3;-1)
x-I
y-3
z+1
*Phuong trinh CD:

=

=
4
-2 -2
VII.b
sa
ph rc:
'-;;3
. 2( -7l'

-7l') 1
''-;;3
2(
7l'
.'
7l')
*"j
-z
==
cos-+zsm-;

+lVj
=
cos-+zsm-
6 6 3 3
*
(J3
-
iY
=
211
(cos -n7l' +
isin
-JVr)
6 6
(J3-it
2
11
-
1
[ (-m'C
7l')
.,
(-n7l'
7l')]
• *
V~y
z=='
= cos

-

+zsm

l+iJ3
6 3 6 3
.,
-n7l'
7l'.
~T'
* z thuan ao
¢:>
cos( )
= O,n
EN¢:>
n =
6k-
5,
k E H
6 3
I
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25

0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
·0.25
0.25
www.MATHVN.com
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