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Bé gi¸o dôc vµ ®µo t¹o §¸p ¸n - Thang ®iÓm
..................... ®Ò thi tuyÓn sinh ®¹i häc, cao ®¼ng n¨m 2004
...........................................
§Ò chÝnh thøc M«n: To¸n, Khèi A
(§¸p ¸n - thang ®iÓm cã 4 trang)
C©u ý Néi dung §iÓm
I 2,0
I.1 (1,0 ®iÓm)
( )12
332
−
−+−
=
x
xx
y =
( )
1 1
x 1
2 2 x 1
− + −
−
.
a) TËp x¸c ®Þnh: { }R  1 .
b) Sù biÕn thiªn:
2
x(2 x)
y'
2(x 1)
−
=
−
; y' 0 x 0, x 2= ⇔ = = . 0,25
yC§ = y(2) =
1
2
− , yCT = y(0) =
3
2
.
§−êng th¼ng x = 1 lµ tiÖm cËn ®øng.
§−êng th¼ng
1
y x 1
2
= − + lµ tiÖm cËn xiªn. 0,25
B¶ng biÕn thiªn:
x −∞ 0 1 2 +∞
y' − 0 + + 0 −
y +∞ +∞
1
2
−
3
2
−∞ −∞
0,25
c) §å thÞ:
0,25
2
I.2 (1,0 ®iÓm)
Ph−¬ng tr×nh hoµnh ®é giao ®iÓm cña ®å thÞ hµm sè víi ®−êng th¼ng y = m lµ :
( )
m
x
xx
=
−
−+−
12
332
⇔ ( ) 023322
=−+−+ mxmx (*). 0,25
Ph−¬ng tr×nh (*) cã hai nghiÖm ph©n biÖt khi vµ chØ khi:
0>∆ ⇔ 2
4m 4m 3 0− − > ⇔
3
m
2
> hoÆc
1
m
2
< − (**) . 0,25
Víi ®iÒu kiÖn (**), ®−êng th¼ng y = m c¾t ®å thÞ hµm sè t¹i hai ®iÓm A, B cã hoµnh
®é x1 , x2 lµ nghiÖm cña ph−¬ng tr×nh (*).
AB = 1 ⇔ 121 =− xx ⇔
2
1 2x x 1− = ⇔ ( )1 2
2
1 2x x 4x x 1+ − =
0,25
⇔ ( ) ( ) 123432 2
=−−− mm ⇔
1 5
m
2
±
= (tho¶ m·n (**)) 0,25
II 2,0
II.1 (1,0 ®iÓm)
§iÒu kiÖn : x 4≥ . 0,25
BÊt ph−¬ng tr×nh ®· cho t−¬ng ®−¬ng víi bÊt ph−¬ng tr×nh:
2 2
2(x 16) x 3 7 x 2(x 16) 10 2x− + − > − ⇔ − > −
0,25
+ NÕu x > 5 th× bÊt ph−¬ng tr×nh ®−îc tho¶ m·n, v× vÕ tr¸i d−¬ng, vÕ ph¶i ©m. 0,25
+ NÕu 4 x 5≤ ≤ th× hai vÕ cña bÊt ph−¬ng tr×nh kh«ng ©m. B×nh ph−¬ng hai vÕ ta
®−îc: ( ) ( )
22 2
2 x 16 10 2x x 20x 66 0− > − ⇔ − + < 10 34 x 10 34⇔ − < < + .
KÕt hîp víi ®iÒu kiÖn 4 x 5≤ ≤ ta cã: 10 34 x 5− < ≤ . §¸p sè: x 10 34> − 0,25
II.2 (1,0 ®iÓm)
§iÒu kiÖn: y > x vµ y > 0.
( ) 1
1
loglog 4
4
1 =−−
y
xy ⇔ ( ) 1
1
loglog 44 =−−−
y
xy 0,25
⇔ 4
y x
log 1
y
−
− = ⇔
4
3y
x = . 0,25
ThÕ vµo ph−¬ng tr×nh x2
+ y2
= 25 ta cã:
2
23y
y 25 y 4.
4
⎛ ⎞
+ = ⇔ = ±⎜ ⎟
⎝ ⎠
0,25
So s¸nh víi ®iÒu kiÖn , ta ®−îc y = 4, suy ra x= 3 (tháa m·n y > x).
VËy nghiÖm cña hÖ ph−¬ng tr×nh lµ (3; 4). 0,25
III 3,0
III.1 (1,0 ®iÓm)
+ §−êng th¼ng qua O, vu«ng gãc víi BA( 3; 3) cã ph−¬ng tr×nh 3x 3y 0+ = .
§−êng th¼ng qua B, vu«ng gãc víi OA(0; 2) cã ph−¬ng tr×nh y = 1−
( §−êng th¼ng qua A, vu«ng gãc víi BO( 3; 1) cã ph−¬ng tr×nh 3x y 2 0+ − = )
0,25
Gi¶i hÖ hai (trong ba) ph−¬ng tr×nh trªn ta ®−îc trùc t©m H( 3; 1)− 0,25
+ §−êng trung trùc c¹nh OA cã ph−¬ng tr×nh y = 1.
§−êng trung trùc c¹nh OB cã ph−¬ng tr×nh 3x y 2 0+ + = .
( §−êng trung trùc c¹nh AB cã ph−¬ng tr×nh 3x 3y 0+ = ).
0,25
3
Gi¶i hÖ hai (trong ba) ph−¬ng tr×nh trªn ta ®−îc t©m ®−êng trßn ngo¹i tiÕp tam gi¸c
OAB lµ ( )I 3; 1− . 0,25
III.2.a (1,0 ®iÓm)
+ Ta cã: ( )C 2; 0; 0− , ( )D 0; 1; 0− , ( )2;0;1−M ,
( )22;0;2 −=SA , ( )BM 1; 1; 2= − − . 0,25
Gäi α lµ gãc gi÷a SA vµ BM.
Ta ®−îc: ( )
SA.BM 3
cos cos SA, BM
2SA . BM
α = = = ⇒ 30α = ° .
0,25
+ Ta cã: ( )SA, BM 2 2; 0; 2⎡ ⎤ = − −⎣ ⎦ , ( )AB 2; 1; 0= − . 0,25
VËy:
( )
SA, BM AB 2 6
d SA,BM
3SA,BM
⎡ ⎤⋅⎣ ⎦
= =
⎡ ⎤
⎣ ⎦
0,25
III.2.b (1,0 ®iÓm)
Ta cã MN // AB // CD ⇒ N lµ trung ®iÓm SD ⇒ ⎟
⎠
⎞
⎜
⎝
⎛
− 2;
2
1
;0N .
0,25
( )SA 2; 0; 2 2= − , ( )2;0;1 −−=SM , ( )22;1;0 −=SB ,
1
SN 0; ; 2
2
⎛ ⎞
= − −⎜ ⎟
⎝ ⎠
( )SA, SM 0; 4 2; 0⎡ ⎤⇒ =⎣ ⎦ . 0,25
S.ABM
1 2 2
V SA,SM SB
6 3
⎡ ⎤= ⋅ =⎣ ⎦ 0,25
S.AMN
1 2
V SA,SM SN
6 3
⎡ ⎤= ⋅ =⎣ ⎦ ⇒ S.ABMN S.ABM S.AMNV V V 2= + = 0,25
IV 2,0
IV.1 (1,0 ®iÓm)
2
1
x
I dx
1 x 1
=
+ −∫ . §Æt: 1−= xt ⇒ 12
+= tx ⇒ tdtdx 2= .
01 =⇒= tx , 12 =⇒= tx . 0,25
4
Ta cã:
1 1 12 3
2
0 0 0
t 1 t t 2
I 2t dt 2 dt 2 t t 2 dt
1 t 1 t t 1
+ + ⎛ ⎞
= = = − + −⎜ ⎟
+ + +⎝ ⎠
∫ ∫ ∫
0,25
I
1
3 2
0
1 1
2 t t 2t 2ln t 1
3 2
⎡ ⎤
= − + − +⎢ ⎥
⎣ ⎦
0,25
1 1 11
I 2 2 2ln 2 4ln 2
3 2 3
⎡ ⎤
= − + − = −⎢ ⎥⎣ ⎦
. 0,25
IV.2 (1, 0 ®iÓm)
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
8 2 3 42 0 1 2 2 4 3 6 4 8
8 8 8 8 8
5 6 7 85 10 6 12 7 14 8 16
8 8 8 8
1 x 1 x C C x 1 x C x 1 x C x 1 x C x 1 x
C x 1 x C x 1 x C x 1 x C x 1 x
⎡ ⎤+ − = + − + − + − + −⎣ ⎦
+ − + − + − + − 0,25
BËc cña x trong 3 sè h¹ng ®Çu nhá h¬n 8, bËc cña x trong 4 sè h¹ng cuèi lín h¬n 8. 0,25
VËy x8
chØ cã trong c¸c sè h¹ng thø t−, thø n¨m, víi hÖ sè t−¬ng øng lµ:
3 2 4 0
8 3 8 4C .C , C .C 0,25
Suy ra a8 168 70 238= + = . 0,25
V 1,0
Gäi 3cos22cos222cos −++= CBAM
3
2
cos
2
cos2221cos2 2
−
−
⋅
+
⋅+−=
CBCB
A . 0,25
Do 0
2
sin >
A
, 1
2
cos ≤
− CB
nªn 2 A
M 2cos A 4 2 sin 4
2
≤ + − . 0,25
MÆt kh¸c tam gi¸c ABC kh«ng tï nªn 0cos ≥A , AA coscos2
≤ . Suy ra:
4
2
sin24cos2 −+≤
A
AM 4
2
sin24
2
sin212 2
−+⎟
⎠
⎞
⎜
⎝
⎛
−=
AA
2
2
sin24
2
sin4 2
−+−=
AA
01
2
sin22
2
≤⎟
⎠
⎞
⎜
⎝
⎛
−−=
A
. VËy 0≤M . 0,25
Theo gi¶ thiÕt: M = 0 ⇔
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎨
⎧
=
=
−
=
2
1
2
sin
1
2
cos
coscos2
A
CB
AA
⇔
A 90
B C 45
= °⎧
⎨
= = °⋅⎩
0,25

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Đáp án toán A 2004

  • 1. 1 Bé gi¸o dôc vµ ®µo t¹o §¸p ¸n - Thang ®iÓm ..................... ®Ò thi tuyÓn sinh ®¹i häc, cao ®¼ng n¨m 2004 ........................................... §Ò chÝnh thøc M«n: To¸n, Khèi A (§¸p ¸n - thang ®iÓm cã 4 trang) C©u ý Néi dung §iÓm I 2,0 I.1 (1,0 ®iÓm) ( )12 332 − −+− = x xx y = ( ) 1 1 x 1 2 2 x 1 − + − − . a) TËp x¸c ®Þnh: { }R 1 . b) Sù biÕn thiªn: 2 x(2 x) y' 2(x 1) − = − ; y' 0 x 0, x 2= ⇔ = = . 0,25 yC§ = y(2) = 1 2 − , yCT = y(0) = 3 2 . §−êng th¼ng x = 1 lµ tiÖm cËn ®øng. §−êng th¼ng 1 y x 1 2 = − + lµ tiÖm cËn xiªn. 0,25 B¶ng biÕn thiªn: x −∞ 0 1 2 +∞ y' − 0 + + 0 − y +∞ +∞ 1 2 − 3 2 −∞ −∞ 0,25 c) §å thÞ: 0,25
  • 2. 2 I.2 (1,0 ®iÓm) Ph−¬ng tr×nh hoµnh ®é giao ®iÓm cña ®å thÞ hµm sè víi ®−êng th¼ng y = m lµ : ( ) m x xx = − −+− 12 332 ⇔ ( ) 023322 =−+−+ mxmx (*). 0,25 Ph−¬ng tr×nh (*) cã hai nghiÖm ph©n biÖt khi vµ chØ khi: 0>∆ ⇔ 2 4m 4m 3 0− − > ⇔ 3 m 2 > hoÆc 1 m 2 < − (**) . 0,25 Víi ®iÒu kiÖn (**), ®−êng th¼ng y = m c¾t ®å thÞ hµm sè t¹i hai ®iÓm A, B cã hoµnh ®é x1 , x2 lµ nghiÖm cña ph−¬ng tr×nh (*). AB = 1 ⇔ 121 =− xx ⇔ 2 1 2x x 1− = ⇔ ( )1 2 2 1 2x x 4x x 1+ − = 0,25 ⇔ ( ) ( ) 123432 2 =−−− mm ⇔ 1 5 m 2 ± = (tho¶ m·n (**)) 0,25 II 2,0 II.1 (1,0 ®iÓm) §iÒu kiÖn : x 4≥ . 0,25 BÊt ph−¬ng tr×nh ®· cho t−¬ng ®−¬ng víi bÊt ph−¬ng tr×nh: 2 2 2(x 16) x 3 7 x 2(x 16) 10 2x− + − > − ⇔ − > − 0,25 + NÕu x > 5 th× bÊt ph−¬ng tr×nh ®−îc tho¶ m·n, v× vÕ tr¸i d−¬ng, vÕ ph¶i ©m. 0,25 + NÕu 4 x 5≤ ≤ th× hai vÕ cña bÊt ph−¬ng tr×nh kh«ng ©m. B×nh ph−¬ng hai vÕ ta ®−îc: ( ) ( ) 22 2 2 x 16 10 2x x 20x 66 0− > − ⇔ − + < 10 34 x 10 34⇔ − < < + . KÕt hîp víi ®iÒu kiÖn 4 x 5≤ ≤ ta cã: 10 34 x 5− < ≤ . §¸p sè: x 10 34> − 0,25 II.2 (1,0 ®iÓm) §iÒu kiÖn: y > x vµ y > 0. ( ) 1 1 loglog 4 4 1 =−− y xy ⇔ ( ) 1 1 loglog 44 =−−− y xy 0,25 ⇔ 4 y x log 1 y − − = ⇔ 4 3y x = . 0,25 ThÕ vµo ph−¬ng tr×nh x2 + y2 = 25 ta cã: 2 23y y 25 y 4. 4 ⎛ ⎞ + = ⇔ = ±⎜ ⎟ ⎝ ⎠ 0,25 So s¸nh víi ®iÒu kiÖn , ta ®−îc y = 4, suy ra x= 3 (tháa m·n y > x). VËy nghiÖm cña hÖ ph−¬ng tr×nh lµ (3; 4). 0,25 III 3,0 III.1 (1,0 ®iÓm) + §−êng th¼ng qua O, vu«ng gãc víi BA( 3; 3) cã ph−¬ng tr×nh 3x 3y 0+ = . §−êng th¼ng qua B, vu«ng gãc víi OA(0; 2) cã ph−¬ng tr×nh y = 1− ( §−êng th¼ng qua A, vu«ng gãc víi BO( 3; 1) cã ph−¬ng tr×nh 3x y 2 0+ − = ) 0,25 Gi¶i hÖ hai (trong ba) ph−¬ng tr×nh trªn ta ®−îc trùc t©m H( 3; 1)− 0,25 + §−êng trung trùc c¹nh OA cã ph−¬ng tr×nh y = 1. §−êng trung trùc c¹nh OB cã ph−¬ng tr×nh 3x y 2 0+ + = . ( §−êng trung trùc c¹nh AB cã ph−¬ng tr×nh 3x 3y 0+ = ). 0,25
  • 3. 3 Gi¶i hÖ hai (trong ba) ph−¬ng tr×nh trªn ta ®−îc t©m ®−êng trßn ngo¹i tiÕp tam gi¸c OAB lµ ( )I 3; 1− . 0,25 III.2.a (1,0 ®iÓm) + Ta cã: ( )C 2; 0; 0− , ( )D 0; 1; 0− , ( )2;0;1−M , ( )22;0;2 −=SA , ( )BM 1; 1; 2= − − . 0,25 Gäi α lµ gãc gi÷a SA vµ BM. Ta ®−îc: ( ) SA.BM 3 cos cos SA, BM 2SA . BM α = = = ⇒ 30α = ° . 0,25 + Ta cã: ( )SA, BM 2 2; 0; 2⎡ ⎤ = − −⎣ ⎦ , ( )AB 2; 1; 0= − . 0,25 VËy: ( ) SA, BM AB 2 6 d SA,BM 3SA,BM ⎡ ⎤⋅⎣ ⎦ = = ⎡ ⎤ ⎣ ⎦ 0,25 III.2.b (1,0 ®iÓm) Ta cã MN // AB // CD ⇒ N lµ trung ®iÓm SD ⇒ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 2; 2 1 ;0N . 0,25 ( )SA 2; 0; 2 2= − , ( )2;0;1 −−=SM , ( )22;1;0 −=SB , 1 SN 0; ; 2 2 ⎛ ⎞ = − −⎜ ⎟ ⎝ ⎠ ( )SA, SM 0; 4 2; 0⎡ ⎤⇒ =⎣ ⎦ . 0,25 S.ABM 1 2 2 V SA,SM SB 6 3 ⎡ ⎤= ⋅ =⎣ ⎦ 0,25 S.AMN 1 2 V SA,SM SN 6 3 ⎡ ⎤= ⋅ =⎣ ⎦ ⇒ S.ABMN S.ABM S.AMNV V V 2= + = 0,25 IV 2,0 IV.1 (1,0 ®iÓm) 2 1 x I dx 1 x 1 = + −∫ . §Æt: 1−= xt ⇒ 12 += tx ⇒ tdtdx 2= . 01 =⇒= tx , 12 =⇒= tx . 0,25
  • 4. 4 Ta cã: 1 1 12 3 2 0 0 0 t 1 t t 2 I 2t dt 2 dt 2 t t 2 dt 1 t 1 t t 1 + + ⎛ ⎞ = = = − + −⎜ ⎟ + + +⎝ ⎠ ∫ ∫ ∫ 0,25 I 1 3 2 0 1 1 2 t t 2t 2ln t 1 3 2 ⎡ ⎤ = − + − +⎢ ⎥ ⎣ ⎦ 0,25 1 1 11 I 2 2 2ln 2 4ln 2 3 2 3 ⎡ ⎤ = − + − = −⎢ ⎥⎣ ⎦ . 0,25 IV.2 (1, 0 ®iÓm) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 8 2 3 42 0 1 2 2 4 3 6 4 8 8 8 8 8 8 5 6 7 85 10 6 12 7 14 8 16 8 8 8 8 1 x 1 x C C x 1 x C x 1 x C x 1 x C x 1 x C x 1 x C x 1 x C x 1 x C x 1 x ⎡ ⎤+ − = + − + − + − + −⎣ ⎦ + − + − + − + − 0,25 BËc cña x trong 3 sè h¹ng ®Çu nhá h¬n 8, bËc cña x trong 4 sè h¹ng cuèi lín h¬n 8. 0,25 VËy x8 chØ cã trong c¸c sè h¹ng thø t−, thø n¨m, víi hÖ sè t−¬ng øng lµ: 3 2 4 0 8 3 8 4C .C , C .C 0,25 Suy ra a8 168 70 238= + = . 0,25 V 1,0 Gäi 3cos22cos222cos −++= CBAM 3 2 cos 2 cos2221cos2 2 − − ⋅ + ⋅+−= CBCB A . 0,25 Do 0 2 sin > A , 1 2 cos ≤ − CB nªn 2 A M 2cos A 4 2 sin 4 2 ≤ + − . 0,25 MÆt kh¸c tam gi¸c ABC kh«ng tï nªn 0cos ≥A , AA coscos2 ≤ . Suy ra: 4 2 sin24cos2 −+≤ A AM 4 2 sin24 2 sin212 2 −+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −= AA 2 2 sin24 2 sin4 2 −+−= AA 01 2 sin22 2 ≤⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −−= A . VËy 0≤M . 0,25 Theo gi¶ thiÕt: M = 0 ⇔ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎨ ⎧ = = − = 2 1 2 sin 1 2 cos coscos2 A CB AA ⇔ A 90 B C 45 = °⎧ ⎨ = = °⋅⎩ 0,25