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së gi¸o dôc vµ ®µo t¹o hµ tÜnh §Ò THi thö ®¹i Häc LÇN I n¨m 2014
Trêng THPT NguyÔn Trung Thiªn Môn thi: To¸n - KHỐI A, A1, B
Thời gian làm bài: 180 phút
I. PhÇn chung cho tÊt c¶ thÝ sinh (7,0 ®iÓm)
x−3
C©u I (2,0 ®iÓm) Cho hµm sè y = cã ®å thÞ (C)
x +1
1. Kh¶o s¸t sù biÕn thiªn vµ vÏ ®å thÞ (C) cña hµm sè.
2. ViÕt ph¬ng tr×nh tiÕp tuyÕn cña (C) biÕt kho¶ng c¸ch tõ giao ®iÓm I cña 2 tiÖm cËn cña (C) ®Õn
tiÕp tuyÕn b»ng 2 2 .
C©u II (2,0 ®iÓm)
π
1. Gi¶i ph¬ng tr×nh 1 + 2 sin(2 x + ) = cos x + cos 3 x .
4
t anx
2. TÝnh: I= ∫ 1 + cos 2 xdx
x 2 + y 2 + xy = 4 y − 1
C©u III (1,0 ®iÓm) Gi¶i hÖ ph¬ng tr×nh: y
x + y = 2 +2
x +1
C©u IV (1,0 ®iÓm) Cho h×nh chãp S.ABCD cã ®¸y ABCD lµ h×nh thang.
§¸y lín AB = 2a ; BC = CD = DA = a; SA vu«ng gãc víi ®¸y, mÆt ph¼ng(SBC) t¹o víi ®¸y mét
gãc 60o . TÝnh thÓ tÝch khèi chãp S.ABCD theo a.
C©u V (1,0 ®iÓm) Cho 3 sè thùc d¬ng x, y, z. T×m gi¸ trÞ nhá nhÊt cña biÓu thøc :
x2 2 y2 2 z2 2
P = x( + ) + y ( + ) + z ( + ) .
3 yz 3 xz 3 xy
II. PhÇn riªng (3,0 ®iÓm): ThÝ sinh chØ ®îc lµm mét trong hai phÇn (phÇn A hoÆc phÇn B)
A. Theo ch¬ng tr×nh chuÈn
C©u VI. a. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho tam gi¸c ABC cã träng t©m G (2;-1).
§êng trung trùc cña c¹nh BC cã ph¬ng tr×nh d : 3x − y − 4 = 0 . §êng th¼ng AB cã ph¬ng tr×nh
d1 :10 x + 3 y + 1 = 0 . T×m täa ®é c¸c ®Ønh A, B, C.
C©u VII. a. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho hai ®iÓm A(2;0), B(6;4). ViÕt
ph¬ng tr×nh ®êng trßn (C) tiÕp xóc víi trôc hoµnh t¹i ®iÓm A vµ kho¶ng c¸ch tõ t©m (C) ®Õn B b»ng
5.
n
2
C©u VIII. a. (1,0 ®iÓm ) T×m sè h¹ng kh«ng chøa x trong khai triÓn P ( x ) =3 x + ( x > 0) .
x
BiÕt r»ng n tháa m·n: Cn6 + 3Cn7 + 3Cn + Cn = 2Cn+ 2 .
8 9 8
B. Theo ch¬ng tr×nh n©ng cao
C©u VI. b. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho tam gi¸c ABC vu«ng c©n t¹i A(1;2).
ViÕt ph¬ng tr×nh ®êng trßn (T) ngo¹i tiÕp tam gi¸c ABC biÕt ®êng th¼ng d : x − y − 1 = 0 tiÕp xóc víi
(T) t¹i B.
C©u VII. b. (1,0 ®iÓm) Trong mÆt ph¼ng víi hÖ täa ®é Oxy cho hai ®êng th¼ng d1 : 3x + y + 5 = 0 ;
d 2 : 3 x + y + 1 = 0 vµ ®iÓm I(1;-2). ViÕt ph¬ng tr×nh ®êng th¼ng ®i qua I c¾t d1 , d 2 lÇn lît t¹i A vµ B
sao cho AB = 2 2 .
x3 2
C©u VIII. b. (1,0 ®iÓm) Gi¶i ph¬ng tr×nh: log 2 x + log 2 = 2.
2 x
--------------------- HÕt --------------------
- §¸p ¸n K.A gåm cã 6 trang. www.VNMATH.com
Lu ý : Mäi c¸ch gi¶i ®óng ®Òu cho ®iÓm tèi ®a.
C©u §¸p ¸n vµ híng dÉn chÊm §iÓm
C©u 1 (1,0 ®iÓm)
I. ________________________________________________________________________
+ TËp x¸c ®Þnh: D = R \ {−1}
2,0 0,25
®iÓm 4
+ Sù biÕn thiªn: y ' = > 0 , ∀x ≠ −1 , suy ra hµm sè ®ång biÕn trªn c¸c kho¶ng
( x + 1)2
( −∞; −1) vµ ( −1; +∞ ) .
________________________________________________________________________
+ Giíi h¹n: lim y = 1 ; lim y = 1 => TiÖm cËn ngang: y=1
x →−∞ x →+∞
lim− y = +∞ ; lim y = −∞ => TiÖm cËn ®øng: x=-1. 0,25
x →−1 x →+∞
________________________________________________________________________
+ B¶ng biÕn thiªn:
x −∞ -1 +∞
y’ +
0,25
y +∞ 1
1 −∞
________________________________________________________________________
+ §å thÞ :
Giao víi Ox: (3;0), giao víi Oy: (0;-3). 0.25
1
-1 0 3 x
-3
§å thÞ nhËn I(-1;1) lµm t©m ®èi xøng.
2 (1,0 ®iÓm)
________________________________________________________________________
x −3
Gi¶ sö M ( x0 ; y0 ) thuéc (C), y0 = 0 , x0 ≠ −1 .
x0 + 1
Khi ®ã ph¬ng tr×nh tiÕp tuyÕn ∆ t¹i M lµ: 0.25
4 x −3
2 (
y= x − x0 ) + 0
( x0 + 1) x0 + 1
⇔ 4 x − ( x0 + 1) y + ( x0 − 6 x0 − 3) = 0
2 2
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Theo ®Ò :
d ( I , ∆) = 2 2
(
−4 − ( x0 + 1) + x02 − 6 x0 − 3
2
)
⇔ =2 2 0.25
16 + ( x0 + 1)
4
⇔ ( x0 + 1) − 8 ( x0 + 1) + 16 = 0
4 2
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x0 = 1
⇔
x0 = −3
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Víi x0 = 1 , ph¬ng tr×nh ∆ : y = x − 2 ; 0,5
Víi x0 = −3 , ph¬ng tr×nh ∆ : y = x + 6 .
C©u 1 (1,0 ®iÓm)
II. ________________________________________________________________________
2,0 PT ⇔ 1 + sin 2 x + cos 2 x = 2cos x cos 2 x
®iÓm ⇔ 2 cos2 x + 2sin x cos x − 2cos x cos 2 x = 0
( )
⇔ 2 cos x cos x + sin x − ( cos2 x − sin 2 x ) = 0 0,25
⇔ cos x ( cos x + sin x )(1 − cos x + sin x ) = 0
________________________________________________________________________
π
x = + kπ
cos x = 0 2
⇔ tan x = −1 0,5
⇔ cos x + sin x = 0
cos x − sin x = 1
cos x + π = 1
4
2
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π
x = 2 + kπ
π 0,25
⇔ x = − + kπ
4 k ∈
x = k 2π
2 (1,0 ®iÓm)
________________________________________________________________________
tan x sin x cos x
Ta cã: I = ∫ dx = ∫ dx
1 + cos x
2
cos x(1 + cos2 x)
0,25
§Æt t = cos 2 x ⇒ dt = −2sin x cos xdx
1 dt
Suy ra: I = − ∫
2 t (t + 1)
________________________________________________________________________
1 1 1 1 t +1
I = ∫ − dt = ln +C 0,5
2 t +1 t 2 t
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1 1 + cos2 x
KÕt luËn: I = ln +C . 0,25
2 cos2 x
C©u NhËn xÐt y=0 kh«ng tháa m·n hÖ ph¬ng tr×nh.
III.
1,0 0,25
®iÓm
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x2 + 1
y +x+ y =4
HÖ t¬ng ®¬ng víi
x + y = y + 2
x2 +1 0,25
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u + v = 4
x2 +1 0,25
§Æt u = , v = x + y. HÖ trë thµnh: 1
y v = u + 2
Gi¶i hÖ ta cã: u =1 0,25
v=3
________________________________________________________________________
x = 1
x2 + 1
u = 1 = 1 y = 2
Víi ⇒ y ⇔
v = 3 x = −2
x+ y =3
y = 5
C©u
IV.
1,0
®iÓm
N B
A 60 0
Gäi N lµ trung ®iÓm AB.
D C
AN // DC
Ta cã: nªn ADCN lµ h×nh b×nh hµnh.
AN = DC = a
Suy ra: NC = AD = a
=> NA = NB = NC =a hay ∆ACB vu«ng t¹i C suy ra AC ⊥ BC . 0,25
Do SA ⊥ ( ABCD ) nªn SA ⊥ BC .
¸p dông ®Þnh lý ba ®êng vu«ng gãc ta suy ra SC ⊥ BC .
Suy ra: Gãc gi÷a (SBC) vµ (ABCD) lµ ∠SCA => ∠SCA = 60°
________________________________________________________________________
MÆt kh¸c: ∆NBC ®Òu nªn ∠NBC = 60°
3
AC = AB = 3a
2 0,25
SA = AC .tan 60° = 3a. 3 = 3a
________________________________________________________________________
3 3a 2
S ABCD =
4 0,25
________________________________________________________________________
3 3a 3
TÝnh ®îc thÓ tÝch chãp S.ABCD b»ng . 0,25
4
- C©u
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x3 + y 3 + z 3 x2 + y 2 + z 2
V. Ta cã : P = +2
3 xyz
1,0
¸p dông bÊt ®¼ng thøc a 2 + b 2 ≥ 2ab, ∀a, b ⇒ x 2 + y 2 + z 2 ≥ xy + yz + zx .
®iÓm 0,25
(§¼ng thøc x¶y ra khi x=y=z)
x +y +z
3 3 3
xy + yz + zx x3 2 y 3 2 z 3 2
⇒ P≥ +2 ⇒ P ≥ + + + + +
3 xyz 3 x 3 y 3 z
________________________________________________________________________
t3 2
XÐt hµm sè f (t ) = + víi t > 0 ;
3 t
2
f '(t ) = t 2 − 2 ; f '(t ) = 0 ⇔ t = 4 2 . 0,25
t
________________________________________________________________________
B¶ng biÕn thiªn:
t 0 4
2 +∞
y’ - 0 +
0,25
y +∞ +∞
8
34 2
________________________________________________________________________
VËy P ≥ 4 4 8 . §¼ng thøc x¶y ra khi x = y = z = 4 2 hay P = 4 4 8 . 0,25
A. Theo ch¬ng tr×nh chuÈn
C©u Gäi Mlµ trung ®iÓm BC, v× M ∈ d nªn M (m; 3m-4).
VI. a. Mµ GA = −2GM nªn A (6-2m; 5-6m). 0,25
1,0 ________________________________________________________________________
®iÓm A ∈ AB ⇒ m = 2 ⇒ M ( 2; 2 ) , A ( 2; −7 ) . 0,25
________________________________________________________________________
BC qua M vµ vu«ng gãc víi d nªn cã ph¬ng tr×nh x + 3y – 8 = 0.
B = AB ∩ BC nªn B ( −1;3) . 0,25
________________________________________________________________________
M lµ trung ®iÓm BC nªn C ( 5;1) .
0,25
C©u Gäi I ( x0 ; y0 ) lµ t©m cña ®êng trßn (C).
VII.
Khi ®ã, do (C) tiÕp xóc víi Ox t¹i A nªn víi i = (0;1) lµ vect¬ ®¬n vÞ trªn trôc Ox, ta cã:
a.
0,25
IA ⊥ i ⇔ 1. (1 − x0 ) + 0. ( 0 − y0 ) = 0 ⇔ x0 = 2 .
1,0
®iÓm ________________________________________________________________________
Theo gi¶ thiÕt, ta cã:
R = IB – 5 ; IB 2 = 25 ⇔ ( 2 − 6 ) + ( y0 − 4 ) = 25
2 2
0,25
y = 7
⇔ y0 − 4 = ±3 ⇔ 0
y0 = 1
________________________________________________________________________
- Víi y0 = 7 th× I (2; 7) ⇒ R = 7 .
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Víi y0 = 1 th× I (2;1) ⇒ R = 1 .
VËy ta cã hai ®êng trßn cÇn t×m: 0,5
( x − 2 )2 + ( y − 7 )2 = 49 ; ( x − 2 )2 + ( y − 1)2 = 1
C©u ¸p dông c«ng thøc Cn + Cnk +1 = Cn+11 , ta cã:
k k+
VIII. Cn + 3Cn + 3Cn + Cn = Cn + Cn + 2(Cn + Cn ) + Cn + Cn
6 7 8 9 6 7 7 8 8 9
a.
= Cn +1 + 2Cn +1 + Cn +1 = Cn+ 2 + Cn + 2 = Cn +3
7 8 9 8 9 9
1,0 0,25
®iÓm Gi¶ thiÕt t¬ng ®¬ng víi
n+3
Cn +3 = 2Cn + 2 ⇔
9 8
= 2 ⇔ n = 15 .
9
________________________________________________________________________
n
3 2
Khi ®ã P ( x ) = x +
x
15− k k
2
( x)
15
= ∑C k
15
3
K =0 x
15 30 − 5 k 0,25
= ∑C 2 x k
15
k 6
.
K =0
___________________________________________________________________
30 − 5k
Sè h¹ng kh«ng chøa x t¬ng øng víi = 0 ⇔ k = 6. 0,25
6
________________________________________________________________________
Sè h¹ng ph¶i t×m lµ C15 .26 = 320320 .
6
0,25
B. Theo ch¬ng tr×nh n©ng cao
C©u Gäi I lµ t©m cña ®êng trßn ngo¹i tiÕp ∆ABC .
VI. b. V× ∆ABC vu«ng c©n t¹i A nªn I lµ trung ®iÓm BC vµ AI ⊥ BC .
1,0 Theo gi¶ thiÕt BC ⊥ (d ) ⇒ d / / AI ⇒ B¸n kÝnh cña (T) lµ: R = d ( A, d ) = 2 . 0,25
®iÓm BC ⊥ (d ) ⇒ BC: x + y + c = 0.
________________________________________________________________________
1+ 2 + C C = −1
d ( A, d ) = R = 2 ⇔ = 2 ⇔
2 C = −5
BC : x + y − 1 = 0 0,25
Suy ra
BC : x + y − 5 = 0
§êng cao AI cña ∆ABC ®i qua A (1; 2 ) vµ song song víi (d ) ⇒ AI : x − y + 1 = 0 .
________________________________________________________________________
x + y −1 = 0
NÕu BC : x + y − 1 = 0 ⇒ I = BC ∩ AI : ⇒ I(0;1).
x − y +1 = 0 0.25
Suy ra: (T ) : x 2 + ( y − 1) = 2 .
2
________________________________________________________________________
x + y − 5 = 0
NÕu BC : x + y − 5 = 0 ⇒ I = BC ∩ AI : ⇒ I(2;3).
x − y + 1 = 0 0,25
Suy ra: (T ) : ( x − 2 ) + ( y − 3) = 2 .
2 2
VËy cã hai ®êng trßn: x 2 + ( y − 1) = 2 vµ ( x − 2 ) + ( y − 3) = 2 .
2 2 2
- C©u
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V× A ∈ d1 , B ∈ d 2 nªn gäi täa ®é A(a; −3a − 5) ; B(b; −3b − 1) .
VII.
0,25
AB = ( b − a; 4 − 3(b − a ) ) .
b.
________________________________________________________________________
1,0
Tõ gi¶ thiÕt AB = 2 2 suy ra:
®iÓm
(b − a ) + 4 − 3 ( b − a ) = 2 2 .
2 2
0,25
t = 2
§Æt t = b − a , ta cã: t + ( −3t + 4 ) = 8 ⇔ 2
2 2
t =
5
________________________________________________________________________
Víi t = 2 ⇒ b − a = 2 ⇒ AB = (2; −2) lµ vect¬ chØ ph¬ng cña ∆ cÇn t×m.
x −1 y + 2 0,25
Suy ra ph¬ng tr×nh ®êng th¼ng cña ∆ lµ = ⇔ x + y +1 = 0 .
2 −2
________________________________________________________________________
2 2
Víi t = ⇒ b − a = . 0,25
5 5
T¬ng tù ta cã ph¬ng tr×nh ®êng th¼ng cña ∆ lµ 7 x − y − 9 = 0 .
VËy cã hai ®êng th¼ng cÇn t×m lµ x + y + 1 = 0 vµ 7 x − y − 9 = 0 .
C©u 1
§k: x > 0 , x ≠ . 0,25
VIII. 2
b. x3
1,0 log 2
PT ⇔ 2 + 2 log 2 = 2
®iÓm 2
log 2 2 x x
________________________________________________________________________
0,25
3log 2 x − 1 1 3log 2 x − 1
⇔ + 2 1 − log 2 x = 2 ⇔ − log 2 x = 0
1 + log 2 x 2 1 + log 2 x
________________________________________________________________________
3t − 1
§Æt t = log 2 x , ta cã: − t = 0 ⇔ 3t − 1 − t (t + 1) = 0 ⇔ t 2 − 2t + 1 = 0 ⇔ t = 1
1+ t 0,25
________________________________________________________________________
Víi t = 1 ⇒ log 2 x = 1 ⇒ x = 2 .
VËy ph¬ng tr×nh cã nghiÖm x = 2 .
0,25
nguon tai.lieu . vn
