2003年全国高考数学试卷

| m − n |=
1
2
4
0
1
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A1
3
B
4
8
MN
− 2,
3
A x2 − y2 =1 34
1
C
2
3
D
8
F ( 7,0), y = x −1 M N 2
0
B x2 − y2 =1 43
C x2 − y2 =1 52
D x2 − y2 =1 25
9
f (x) = sin x, x ∈[π , 3π ]f −1(x) =
)%-- 6 - 74 . )% "+*,# &..
17 0
12
z
60 2 | z −1 | | z | | z − 2 |
. | z |.
18 (
12 )
2
ABC A1B1C1 2
DE
CC1 A1B
2E
0
A1B
ABD
0
A1
AED
.
— ABD 0
2 ACB=90 2 AA1=22
{c ∈ B | 210 < c < 210 + 27}= {210 + 2s + 2r | 0 ≤ r < s < 7}.
C
2 7
:{c ∈ B
|
210
+
27
<
c
<
210
+
27
+
23} = {210
+
27
+
2r
|
0
≤
r
<
3}
C170 : k = C130 + C72 + C32 + 1 = 145.
0
22
A − arcsin x, x ∈[−1,1]
B − π − arcsin x, x ∈[−1,1]
C − π + arcsin x, x ∈[−1,1]
D π − arcsin x, x ∈[−1,1]
10
A0020 2B0220 2C0221 D0021 2
AB
P0
AB
θ
BC
P1 2
CD DA AB
C2
D2
5
C : (x − a)2 + (x − 2)2 = 4(a > 0)l : x − y + 3 = 0.lC
2 3 2 a=
0
A2
6
A 2πR 2
B 2− 2
R2 3R2
B 9 πR 2 4
C 2 −1 C 8 πR 2
3
D 2 +1
2ห้องสมุดไป่ตู้
0
D 3 πr 2 2
7
(x2 − 2x + m)(x2 − 2x + n) = 0
7
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(!
')%
1 D 2 C 3 D 4 A 5 C 6 B 7 C 8 D 9 D 10 C 11 B 12 A
13 − 21 2
14 0-120
15 72
16
17
z = r cos60! + r sin 60! ) 2
z r . z − z = r, z z = r 2
OF
2ax + (2k −1) y = 0
GE
− a(2k −1)x + y − 2a = 0
2
k2 P0x,y
2a2 x2 + y2 − 2ay = 0
x2 1
+
(y − a)2 a2
=1
a2 = 1 2 P 2
2
.
2
a2 ≠ 1 2 P 2
2PⅠ
a2 < 1 2 P 2
0 − 1 − a2 , a),( 1 − a2 , a)
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(!
')
0
12 2
5 2 60 2
2
.
1
x ∈ (− π ,0),cos x = 4 ,tg2x =
2
5
7
A
24
B −7 24
2
ρ
=
8sinθ cos2 θ

24
C
7
0
D − 24 7
0
A ρ cosθ = −2 B ρ cosθ = 2 C ρ sinθ = −2 D ρ sinθ = 2
45
.
.
5
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21 0
14
a > 0,
BC CD DA
2P
2Ⅱ
.
ABCD 2AB=42BC=4 a 2O AB
2
BE
=
CF
=
DG
2P
GE
OF
BC CD DA
2
2 EFG 02
6
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22 0 0
2
12 2
4
{an}{2t + 2s | 0 ≤ s < t,s, t ∈ Z}
a1 = 3, a2 = 5, a3 = 6, a4 = 9, a5 = 10, a6 = 12,!.
{an }
2
3
5
6
9
10
12
0i
0i i
00
2
242
{bn}{2r + 2t + 2s | 0 ≤ r < s < t,r, s, t ∈ Z}
2 bk = 1160,k.
a100 .
150 ∠
10
2
10
2
≤ (10t + 60)2 ,t 2 − 36t + 288 ≤ 0,12 ≤ t ≤ 24
12
Ⅰ
°
.
21
2
P
2
2
P
.
A03220 2B0220 2C0224a 2D03224a
BE = CF = DC (0 ≤ k ≤ 1) BC CD DA
E0224ak 2F0234k24a 2G03224a34ak
∴ ED ⊥ A1 AB, ED ⊂ AED.∴
AED ⊥
A1 AB, AED ∩ A1 AB = AE.
A1K ⊥ AE, K.∴ A1K ⊥
AED,A1K A1
AED .
ΔA1 AB1, A1K
=
A1 A⋅ A1B1 AB1
=
2×2 2 23
=
2
3
6
∴
A1
AED
2
6 3
.
19
y = cx R
⇔ 0 < c < 1.
x+ | x − 2c |> 1
R ⇔ y = x+ | x − 2c | R1.
8
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!
x+
|
x
−
2c
|=
⎩⎨⎧22cx,−
2c,
x
≥ x
2c, < 2c,
∴y
=
x+
{2t + 2s | 0 ≤ s < t < t0},
C2 t0
=
t0 (t0 − 1) , t0 (t0 − 1)
2
2
< 100.
t0 142
t0 = 14.
1003 C124 = s0 + 1,s0 = 8,∴ a100 = 214 + 28 = 16640.
0
bk = 1160 = 210 + 27 + 23 , M = {c ∈ B | C < 1160} (, B = {2r + 2s + 2t | 0 ≤ r < s < t}
|
x
−
2c
|
R 2c.∴
|
x
+
x
−
2c
|>
1R
⇔
2c
>
1
⇔
c
>
1 2
.
P, Q, 0 < c ≤ 1 . P, Q, c ≥ 1.c
(0, 1] ∪[1,+∞).
2
2
20
O
2
x
.
01 °
P0 x , y
⎧ ⎪⎪⎨x
=
300 ×
ABD
G.
3
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19 0 P
12
c > 0.
y = cx R
Q
x+ | x − 2c |> 1
.
. R2
PQ
2c
4
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20 0
12
°2
2°
O0
θ (θ = arccos
2 )
10
°
Ⅰ
°
300km 2
P 2 20km/h 60km, 10km/h
2
2
2
a2 > 1 2 P 2
002 a − a2 − 1 ),(0, a + a2 − 1 )
2
2
2a.
22 0 0
0i i
12 2
4
0i
17 18 20 24
a100 = 2 s0 + 2t0 2
33 34 36 40 48
s0 ,t0.
9
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{an }
2t0
ED = 2, EG = 1× 2 = 6 . 33
3.""(4)
! FC = CD = 2,∴ AB = 2 2, A1B = 2 3, EB = 3.
∴sin ∠EBG = EG = 6 ⋅ 1 = 2 . EB 3 3 3
∴ A1B
ABD arcsin
2. 3
0
!ED ⊥ AB, ED ⊥ EF,EF ∩ AB = F,
18 0
BG2 BG BE ABD
2 EBG A1B
ABD
.
F AB 2 EF FC2
! D, ECC1, A1B, DC ⊥
ABC,∴CDEF DE, GΔADB ,∴G ∈ DF. EFD
EF 2 = FG ⋅ FD = 1 FD 2 ,! EF = 1,∴ FD = 3
3
f
(x)
=
⎪⎨⎧2
− 1
x
− 1,
x
≤
0, , f
(x0
)
>
1, x0
0
⎪⎩x 2 , x > 0.
A 03121
B 0312+ ∞
C (−∞,−2) ∪ (0,+∞)
D (−∞,−1) ∪ (1,+∞)
4
y = 2sin x(sin x + cos x)
0
A 1+ 2
B 2 −1
M = {c ∈ B | c < 210} ∪{c ∈ B | 210 < c < 210 + 27} ∪{c ∈ B | 210 + 27 < c < 210 + 27 + 23}.
M
{c ∈ B | c < 210}= {2r + 2s + 2t | 0 ≤ r < s < t < 10},
C 130
2 10
−
20
×
2 2
t,
⎪ ⎪⎩
y
=
−300 ×
72 10
+
20
×
2 2
t.
°
(x − x)2 + ( y − y) ≤ [r(t)]2 ,
r(t) = 10t + 60, t
O°
2
(0 − x)2 + (0 − y)2 ≤ (10t + 60)2. (300 × 2 − 20 × 2 t)2 + (−300 × 7 2 + 20 × 2 t)2
2
| z −1 |2 =| z | ⋅ | z − 2 | : (z −1)(z −1) =| z | (z − 2)(z − 2),∴ r 2 − r + 1 = r r 2 − 2r + 4, r 2 + 2r −1 = 0. : r = 2 −1, r = − 2 −1(). | z |= 2 −1.
2
0
A 3π
B 4π
C 3 3π
D 6π
$-- 4 -- 4 16 %-(.
13 (x2 − 1 )9
x9
.
2x
14 log2 (−x) < x +1
x
.
15
2
5
2
2
24
2
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2
16
2
l
.0
2l
MNP
2 MNP .0
P2 P3 P40
2 P4
0 x4 ,0),1 < x4 < 2,tgθ
A (1 ,1) 3
B (1 , 2) 33
C (2,1) 52
0
D (2, 2) 53
11
lim
n→∞
C22
+
C32
+
C
2 4
+!+
Cn2
n(C21
+
C31
+
C41
+!+
C
1 n
)
=
A3
1 B3
1 C6
0 D6
12
∠ 22
10