đề thi thử đại học môn toán có đáp án năm 2011 trường chu văn an - Pdf 23

so
GrAo Duc vA
EAo
rao HA Nor
ne
rm
rrnl DAr HQC
EoT
r NAtvt
Zgt
t
TRtIOt{c
THPT
cHU
vAx eN
M6n To6n
- Kh6i A
Thdi
gian
ldm bdi:
180
phrfit,
khdng
tC
ttrOi
gian giao
dA.
DA thi
gom
ol trang.
I.

tei dung
mQt
di6m.
Cfru
II
(2,0
tli6m)
1. Giai
phuong
trinh 9'io" * 4,gcosz
x=
13 +,*'z'+l-3'o'2'.
lx+
Y
=g
2.
Giai
he
Phuong
trinh
I r
^-
r :-
[r/xZ
+
9
*^ly'+9
=10
(x'yeR)'
cflu IItr

vh
DD'
d6ng
quy;
Tfnh
th6 tictr ctra hinh
ch6p
^S'.1'B'
C'
D'
theo a, v&i
,S' h tam
cfu hinh
w0ng
ABCD
'
CiuV(1,0tli6m)Xicdinh
m
saocho
xa
-2x3
+8**l)*'-2mx+m'-4>A,
Vxe
[-f1]
II. PHAN RItNG
(3,0
tli6m).
Thf
sinh
chi

-
y
-l
=
0. Tim
tqa d0 c6c dinh cdn
lgi
cira
hinh vu6ng.
2. Vi6t
phuong
trinh m[t cAu
(C)
cO t6m
thu$c
dudng
thlng
(A)
c6
phuong
trinh
lx-vtz=o
J"
lZx+
Y+22-I=0
vdtitip xric voi hai m{t
phang
(a):
2x
+2y

bi€u thirc
P
=lr?l*l':l
B.
Phin B
(theo
ehuong trinh
Ning cao):
Cflu VI.b
(2,0
tli6m)
1. Trongm{tphingtqad0
Oxychodudmgtdn(C):
*2+y2-4x+2y-5=0.fhlrtA6euerngthing
4
d*! x
- my=0 cat ducrng
trdn
(C)
t?i hai di6m
A,
B
pherr-biQt,
sao cho dQ dii
do4n
AB nhtt
nh6t.
z. Trong khdng
gian
tsa dQ

Ciu VII.b
(1,0
tli6m) Tim sd
phtrc
z
,Ai6t
Z
=42-!
n6t-
so crAo Duc vA
o.A,o rAo
HA NOI
rntldxc THPT
CHU
VAN
AN
PAT
AU
-
THANG
DIEM
of rnr
rrulDAr
Hgc
-
DgT
r
nim
zort
M6n

-
Chi€ubii5nthi€n:
!'=3x2 -6x;
y'-0ex=0
hotrc x=2.
.y'>0e
x<0
hoflc
x>2;
.y'<0<]0<x<2.
HAm
sO AOng
bi6n
tr€n
cfc
khoang
(-*,0),
Q,**)
vA
nghich
bi6n
trOn ktroang
(O,Z).
-
Cuc
tri: Hdm s5 dat
cgc
ct4i
tq.i x
=

$
r.,t
uJUcnn
i
Ciu
Eip
6n
Di6m
ai di6m:
Mo(to,Yo)
vd
M1(x1,Y1)
Khi d6
phucrng
trinh
cria ti6p
tuy6n
li
y=6*8-6xoh-'r-3'+3xfi+I
vd
,:b*? -e'r! -z*l
*?t:!
0'5
' '.
0,25
0,25
#ii;fi;d'd
#dil
dcn ;ils
i;-i,-6;ong

tli6m)
T
(1"0
dtffi)
Phuong
trinh
dd cho
tucrng
duong
vfi
9sin2.r
*
4,gl-sn'zx
=13+
93/2-Zsintx
-31-2sin2x
<) 9sin2r
*
31
=
13
+
3-
-:-
gsin'x
92sn'r
gstn-
r
<+
9sin"

*l)(t
-3X/
-
9)
=
0 <+ r
=l;t
=
3;t
=
9.
Ar25
0025
Phrrcmstrinhddchotuonsduonsv6i
sin2r=0
ho6c
sin2x=t
hoFc
qr-41
f
-=-!!-?
a
o
sin2x=0<+
x=kn.
sin2 r= 1 <> cosx
=
0
e r
=

c'6
y
=
8
-
x, thO
vdo
phuong
trinh
thrl
hai cira
hQ
"f7
.g
*.{;
-16r'173
=1g
r:l:
0,25
0,25
€x2-8x*+t+@-59
e@=-x2+8x+9
f-*'*8x+920
*
tft'
* eh'
-t6x
+n)=l
*'
*sx+ef

Ei6m
=
-,
Jf*)''
nxdx
=
ryli*
r"l#
0,50
0,25
22le
22
4
I
-
-!')
-'t

l
TL-L
e xll e
e e
4
Vqy
I
=)-*
e
IV
(1,0
tli6m)

t4i trqng t6m
G, vd
^sG,
-
2c25',,
(2)
TiI
(1)
viL
Q)suy
ra G,
=
Grhay
AA',BB',CC'
vd DD' d6ng
quy.
Tt
gia
thii5t ta suy
ra
A'B'll
=Jrcn,
B'C'll
=!ne,C'D'll
=Iut
2 '
2
2
D'A'll
=!ac.

doa
_ =+
24
24
0,25
v
(1,0
tti6m)
Ta
c6
xo
-zxt
+8*+l)x2
-Zmx+m'*4>-a,
vxe[*1,1]<>
(*'
-
* *
*f
r-4vx
e
[*r,r]o
Hfi(r'
-
* *
*)' > 4.
(3)
DAt
t
=

;,21
surra
(3)
<+
tl1fik
+ m)'
>
4.
X6t
g(r)=(r+
m)';
s'(t)=2Q+m).
11
m
44
.
-m>2em<-2
-1<
-m<2eJ<*"L
44
Q*
*)'
4
Cflu
Din
rin
Di6m
nhu
*,
!

L*
<-4'
l.l* Z
hoac m>2.
'4
v0v; cdn tim ld m
<
-4
02s
VI.a
1z,o
oi6m;
L
(1,0
iliSm)
Ggi hinh
w6ng cAn tim li ABCD, do
n(*t,Z)
kh6ng
thuQc duong
th5ng
2x
*
y-1
=
0 nOn
dudmg thing
ld
phdi
ld

phuffrg
trinh cia
AC liL
(x
+t)+ Z(y
-Z)=
0 €)
x +2y-3
=
0.
4,25
Ta c6 tqa
dQ cfia
I li nghiQm cua
hQ
[x+2Y
=3
lz*-Y:1
Suy
ra
tqa
d0 ctra C(3,0).
lx:1
c>i
LY=1'
0,25
nei, U,2
=
5 n6n clulng trdn tdm I bfunkrrth IA c6
phucrng

V6c
to
ph6p
tuy6n ctta
(a\(B)
n iQ,z,-l),
F"
a
(u)tt(p}
Ta
c6
.a(o,o,e).
(o),
I-d-el
a/;(d)
=4+a((a\,(il)=q.
tlZ"
+2"
+|;l)'
"Mif;il
(d)
ii6
;il';6i
(;)
;t
tp)'i.hi;.;hi
kili'ffi
ki'h.
R ;,il';A it
0r25

toa
d0 cira li nghiQm
cua hg
lx-/+z=o
l2x+y+22=1

lzx+zy-z:o
I
I
]n=*
IJ
l,
=!
Le
n-lJ'
3J
vfly: M[t
cAu
(c)
c6
phuong
o*
('.
+)'
.
(
.(,-t)'
=o
0r25
YILa

=l'?l+l'il=to
4,25
YI.b
(2,0
tli6m)
l.(1,0
tli6m)
- (C)
c6 tim
I(2;-1)
vd b6n kfnh n
=
d0
0,25
-
Dudrrg th6ng
d*
di
qya
di€m o c6 einrr nim
trong
dtd'nttidn,
Ao
Ait
A;-i"?tn;Ai
$ggfrg
gQg
\qlstz
dism
ph6n

(1.,0
di6m)
Ta
c6 MAz
=
MBz
=
MCz
=
d'z(M;(d)\ MA2
=
MB2
e
|
=
z.
(5)
0,25
UBI
=
MC2
e x= z*2.
(6)
aes
MAz
=
d'
(u;("))o
3(x
-l)'?

v-y
M(+,;,*)
0,25
VII.b
(1,0
ili6m)
zJt
-6i-3J1+i
t+
Jii
a,25
J
0r25
_
-(Jl -zi
*
si * sJi)
_
-
eJi
-ti
=
_2,D
_
i
33
0025
Ydv:
z=*2Ji+i
0,25


Nhờ tải bản gốc
Music ♫

Copyright: Tài liệu đại học © DMCA.com Protection Status