Đáp án chính thức môn Toán vào lớp 10_SGD&ĐT Bến Tre_2011-2012

so
GIA~ DVC
vA I>Ao
T~O
BENTRE
D.E
THI TUY£N SINH
LOP
10 TRUNG HQC PH6 THONG
NAM HQC 2011·2012
Mon:TOAN
ThOi
gian:
120 phut (khong
k8
pMt d~)
~
DE
CHlNHTHUC
cs«
2.
(4,0
di€m)
Cho phuong trinh
X2 -
3x
+
m
-1
=
0

(m
lei
tham
s6) (1).
a) Giai phirong trinh (1) khi m
=
1.
b) Tim
cac gia
tri
cua
tham
s6
m d~ phuong trinh (1) c6
nghiem kep,
c) Tim
cac gia
tri cua tham
s6
m
d~ phuong trinh (1) c6 hai nghiem XI; X
2
lei
dQ
dai cac
canh
cua mQt hlnh chtt nh~t c6 dien tich bing 2 (don
vi
dien tich).
ca«

3. (6,0
dtJm)
Cho
cac
ham
s6
y
=
X2
c6 d6 thi
lei
(P)
vei
y
=
x+ 2 c6 d6 thi
lei
(d).
a) Ve
(P) va (d)
tren
cung
mQt h~ true
toa
dQ
vuong
g6c (don
vi
tren
cac

true bing nhau).
b)
Xac
dinh toa dQ
cac
giao di~m cua
(P) va (d)
bing
phep
tinh.
c) Tim
cac
di~m thuoc
(P)
each
d~u hai di~m
A(.J3
+
1 ; 0)
va
B(O ;
.Jj
+
1).
2 2
Cdu
1.
(4,0
di€m)
Khong sir

dung
may tinh
eArn
tay:
a) Tinh:
P
=Ji2
+5.J3 -~.
b) Giai phuong trinh:
X2 -
6x +8
=
0 .
) G
·,· h" h inh
{X+2
Y
=-3
c
iai ~
p uong tn : .
x-2y=5
Cdu
4.
(6,0
di€m)
Cho dirong tron tam
0
ban kinh
R. Tir

mQt di~m
A
nb
ngoai
duong tron
ke cac
ti~p tuy~n
AM va AN voi duong tron (M, N
lei cac
tiSp di~m).
a) Chung minh ring tU giac AMON nQi ti~p.
b) Bi~t
AM
=
R.
Tinh
QA
theo
R. - , . . .
.t.'
c) Duong thing d di qua di~m A, khong di qua di~m 0 va ~at duong tron tam 0 ~l h~l diem
B,
e.
GQi I
lei
trung di~m cua Be. Chung to r~ng nam diem A, M, N, 0 va I cung nam tren
mQt duong tron. '
H~t.
so
GIAo DVC VA DAO T~O

BtNTRE
~" WONG DAN CHAM
DE CHINHTmJf'
rur
TUYEN SINH LOP 10 TRUNG HQC PHO THONG
NAM HQC 2011 -2012
Mon:
ToAN
Can
Dap
an
Di~m Ghi chu
Cau 1
4.00«1
a) P =
v'12
+5.J3-
if
= 2.J3+ 5.J3
_!.J3
0,75
3 3
_

_ _

_-

.•
-

•• -oo •••••• _ •• __ •• _._ ••••••••
1.J3 20.J3
=(2+5 ) 3 =- 3
0,50
3 3
b) Giai phuong trinh:
X2
-6x+8=0.
III
= 9- 8= 1> 0
0,50

.•.
-


.•
-


• •
-_


_ _

-

=>R=l
0,25

_

_

-_



Phuong trinh (1) c6 nghiem:
Xl
=3-1=2;
x
2
=3+1=4
0,50
c)
Giai h~phuong trinh:
{ x+2y~-3 (I) .
x-2y = 5 (2)
0,50
.(D.~{~).~ .~~~~:
.2.~
.'7.~.:?.~.'7.!:

•

.•.
-



Thay
X
= 1vao (1) ta duoc: 1+2y = -3 ~ y =-2
0,50
-_._ _ __ ._ _ .•. _
•
_


• _ _ _ .•.
-
.•
-

_
_ _ _ _-
_._ •.
_
-
.•

__
.••

_ _

_-_

V~y h~ c6 nghiem:
{ x~l
y=-2
0,50
Cau2
4,00«1
Cho phuong trinh
X2
-3x+m-l =0
a) Giai phuong trinh (1) khi
m=l.
Khi
m=1
phuong trinh c6 dang:
X2
-3x+l-1 =0 ~
X2
-3x = 0
0,75
_ •. _ _ _ _ __ _ _ .•. _ _
.•
_ _ __
_ _
-
.•
-

_

-
_
[x~o
~ x(x-3)=O~
0,50
x=3
_

-_
.• •.
_ _

_ _

_
_
_-

_
.•. _ _ _
_
_ _-_._-
-
.•. _-
•.
_

_ .•.

-_ .•.
V~y:Phuong trinh c6 nghiem la:
X==
0;
x=3
0,25
.!>l
X~~~g.~~~~~~~.i~~ ~¢.P ~ A~.~
0,50
_

_ _ .•.
_

-_
•
_ _ _

_ _ •.• •.•
~9-4m+4=0
0,50
•.• _-_ .•. _-_ .•. _ _ .•.
-
.• •.
_ .• • •• _.__ ._._

__

_

_



-
-_

__
.
13
0,25
~m=-
4
c) Hai nghiem
Xl;
x
2
cua phuong trinh (1)
la
dQdai cac
canh cua mQthinh chu nh~t c6 dien tich bang 2
r"o f
3
-
4m
<oo
1,00
ee

S>O ~ 3>0 ~m=3
P=2 m-I=2
-
.•.




-
.•



.•.
-
.•.


-


.•
~
.•.

~

.•.

-
-
.•.• _

-
_ •
-
-
-

-


V~y: m=3
0,25
-
Clu3
6,00(1
1
·(d)
·(i
q~··(O;·
iS~A·
.(~2;0.)
······0,50······ .
.
~
Do thi :
b) Xac dinh toa dQgiao diem cua (P) va (d) bang phep
tinh.

Phuong trinh hoanh dQgiao di~m
cua
(P)
va
(d)
la:
ee
X2
=
X
+
2 (:)
X2 - X -
2
=
0 0,50
Ta co:
a=b+c
=
0
=>[XI
=
-1 050
~=2 '

[
XI =
-I
=>
[YI

=
1
x
2
=
2
Y2
=
4 0,50
·V~yc~c·giaodi~m·
iit:'
·A(~
i;
i);"
:8(2;4·)·······
······0,25······ .
c) Tim cac di~m thuoc (P) each deu hai di~m
A(fj
+1;0)
va
B(O;
fj
+1) .Cac
di~m each d8u hai
2 2
di~m
A, B
thuoc duong trung true cua doan thang
AB
va

~OAB
vuong din tai
0
=>
cac di~m nay thuoc duong
thang
Y
=
x
(duong phan giac
AOB)
0,50
······························l······················· •
Do do:
~p
hQ'Pcac diem nay la giao di~m cua (P) va
duong thang
Y
=
x .
Phuong trinh hoanh dQ giao di~m cua
(P)
va
duong
thang
Y
=
x la: (:)
X2
=

X (:) X2 - X
=
0 (:)
x(
x-I)
=
0
Cho
cac
ham
s6
Y
=
X2
co db thi Ia (P)
va
Y
=
x
+
2 co db
thi la (d).
a)
Ve (P)
va
(d)
tren cung mQt h~ true toa dQvuong goc,
Bang mQt
s6 gia
tri:

~~ : ~-1 __-: 1·1 :
' ' y· l
ysx2
Y"'x+2
' '

-

-


_.
-

_.
-
-
-


-


-
-


-

-

_.
-
-
-

-

-
-
-


-


-
-

-
- -



-
-
-

"" 4"""""""
-xl
I
, , 1'" .
=:
1 A,
x
' , J
(:) x=O;
x=1
2
1,00
1,50
0,50
Cau4
6,00 (I
0,50
1) N8u hoc sinh
lam
bai khong theo huang dan chAm nhung dung van cho dli di€m
theo timg cau.
2) HQc sinh co the dung may tinh cdm tay (cac loai may tinh duoc phep dem vao -
phong thi) d~
lam
bai n~u d~ bai khong yeu cAu giai theo phuong phap nao,
3

N
a) Chung minh rang tU giac AMON nQi tiep,
Ta co:

AMO
=
ANa
=
90°
(tinh chAt ti8p tuyen)
0,50
+
0)
••• _ ._ •• __ •• _ •• __ •••••••••• __ •••••• _ •••••• w •• __ ••••••••••• w •• _ •••• _ ••••••••••• • ••••• __ •••••• _ ••••••••
w_ •••••••••••••••••• _ ••••••• __ •••• ••••• •• _ ••••••••••••• '"' ••••• - •• - •• - ••••
~AMO+ANO=180o.
0,50
· 1
V~y: Trr giac AMON nQi tiep.
0,50 -
b) Biet
AM
=
R.
Tinh
OA
theo R
OA =.JAM2 +M0
2
=.JR2 +R2 =Rfi 0,75+q
I)
Chung to nam diem
A, M, N,
0 va

I
cung nam tren
mQt duong tron.
Ta co:
01
1.
Be
(duong kinh di qua trung di€m cua mQt
0,50 f-
O
,
_~Y.:_'?~£): - _
Cac tam giac
AMO, ANO, AIO
vuong ~ nam di~m
A,M,l,O,N
cung thuoc duong tron duong kinh
OA 1,25
+
0,
S