08北京答案
2008 年北京市高级中等学校招生考试 北京市高级中等学校招生考试 数学试卷答案及评分参考
阅卷须知: 阅卷须知: 1.一律用红钢笔或红圆珠笔批阅,按要求签名. 2.第Ⅰ卷是选择题,机读阅卷. 3.第Ⅱ卷包括填空题和解答题.为了阅卷方便,解答题中的推导步骤写得较为详细,考生 只要写明主要过程即可.若考生的解法与本解法不同,正确者可参照评分参考给分.解答右 端所注分数,表示考生正确做到这一步应得的累加分数.
第Ⅰ卷 (机读卷 共 32 分)
道小题, 一、选择题(共 8 道小题,每小题 4 分,共 32 分) 选择题( 1 2 3 4 5 6 题号 答案 A D C C B B
7 B
8 D
第Ⅱ卷 (非机读卷 共 88 分)
道小题, 二、填空题(共 4 道小题,每小题 4 分,共 16 分) 填空题( 题号 答案 9 10 11 4 12
x≠
1 2
a (a + b)(a − b)
−
b 20 a7
(−1)n
b3n −1 an
道小题, 三、解答题(共 5 道小题,共 25 分) 解答题( 13. (本小题满分 5 分)
1 解: 8 − 2sin 45 + (2 − π) − 3
o 0
−1
= 2 2 − 2×
2 + 1 − 3 ·········································································································· 4 分 2
= 2 − 2 . ····························································································································· 5 分
14. (本小题满分 5 分) 解:去括号,得 5 x − 12 ≤ 8 x − 6 . ······················································································ 1 分 移项,得 5 x − 8 x ≤ −6 + 12 . ······························································································· 2 分 合并,得 −3 x ≤ 6 . ··············································································································· 3 分 系数化为 1,得 x ≥ −2 . ······································································································ 4 分 不等式的解集在数轴上表示如下:
−3 −2 −1 0 1 2 3
·················································································································
································ 5 分
第 1 页 共 9 页
15. (本小题满分 5 分) 证明:Q AB ∥ ED , ∴∠B = ∠E . ························································································································ 2 分 在 △ ABC 和 △CED 中,
AB = CE, ∠B = ∠E, BC = ED,
∴△ ABC ≌△CED . ·········································································································· 4 分 ∴ AC = CD . ························································································································ 5 分
16. (本小题满分 5 分) 解:由图象可知,点 M ( −2, 在直线 y = kx − 3 上, ·························································· 1 分 1)
∴−2k − 3 = 1 . 解得 k = −2 . ························································································································· 2 分
∴ 直线的解析式为 y = −2 x − 3 . ·························································································· 3 分
令 y = 0 ,可得 x = −
3 . 2
3 ∴ 直线与 x 轴的交点坐标为 − , . ················································································ 4 分 0 2
令 x = 0 ,可得 y = −3 .
∴ 直线与 y 轴的交点坐标为 (0, 3) . ·················································································· 5 分 −
17. (本小题满分 5 分) 解:
2x + y ( x − y) x − 2 xy + y 2
2
=
2x + y ( x − y ) ················································································································· 2 分 ( x − y)2 2x + y . ····························································································································· 3 分 x− y
=
当 x − 3 y = 0 时, x = 3 y . ············
······················································································· 4 分 原式 =
6y + y 7y 7 = = . ···································································································· 5 分 3y − y 2 y 2
道小题, 四、解答题(共 2 道小题,共 10 分) 解答题( 18. (本小题满分 5 分) 解法一:如图 1,分别过点 A,D 作 AE ⊥ BC 于点 E ,
第 2 页 共 9 页
DF ⊥ BC 于点 F . ··············································· 1 分 ∴ AE ∥ DF . 又 AD ∥ BC , ∴ 四边形 AEFD 是矩形.
A
D
∴ EF = AD = 2 . ··············································· 2 分
Q AB ⊥ AC , ∠B = 45o , BC = 4 2 ,
B
E F 图1
C
∴ AB = AC .
∴ AE = EC =
1 BC = 2 2 . 2
∴ DF = AE = 2 2 , CF = EC − EF = 2 ············································································································ 4 分
在 Rt△DFC 中, ∠DFC = 90 ,
o
∴ DC = DF 2 + CF 2 = (2 2) 2 + ( 2)2 = 10 . ························································· 5 分
解法二:如图 2,过点 D 作 DF ∥ AB ,分别交 AC,BC 于点 E,F . ·························· 1 分
Q AB ⊥ AC ,
∴∠AED = ∠BAC = 90o .
Q AD ∥ BC ,
A E B
o
D
∴∠DAE = 180o − ∠B − ∠BAC = 45o .
在 Rt△ ABC 中, ∠BAC = 90 , ∠B = 45 , BC = 4 2 ,
o
F 图2
C
∴ AC = BC sin 45o = 4 2 ×
2 = 4 ··················································································· 2 分 2
o o
在 Rt△ ADE 中, ∠AED = 90 , ∠DAE = 45 , AD =
2,
∴ DE = AE = 1 . ∴ CE = AC − AE = 3 . ········································································································ 4 分
在 Rt△DEC 中, ∠CED = 90 ,
o
∴ DC = DE 2 + CE 2 = 12 + 32 = 10 . ········································································· 5 分
19. (本小题满分 5 分) 解: (1)直线 BD 与 O 相切. ···························································································· 1 分 证明:如图 1,连结 OD . Q OA = OD , ∴∠A = ∠ADO .
第 3
页 共 9 页
Q ∠C = 90o ,
又Q ∠CBD = ∠A ,
∴∠CBD + ∠CDB = 90o .
D A
C
∴∠ADO + ∠CDB = 90o . ∴∠ODB = 90o .
O
E 图1
B
∴ 直线 BD 与 O 相切. ······································································································· 2 分 (2)解法一:如图 1,连结 DE .
Q AE 是 O 的直径,
∴∠ADE = 90o .
Q AD : AO = 8 : 5 , AD 4 ∴ cos A = = .············································································································· 3 分 AE 5 Q ∠C = 90o , ∠CBD = ∠A ,
BC 4 = . ··································································································· 4 分 BD 5 5 Q BC = 2 , ∴ BD = .························································································· 5 分 2 1 解法二:如图 2,过点 O 作 OH ⊥ AD 于点 H . ∴ AH = DH = AD . 2 Q AD : AO = 8 : 5 , C AH 4 ∴ cos A = = . ························· 3 分 AO 5 D ∴ cos ∠CBD =
Q ∠C = 90o , ∠CBD = ∠A ,
A H B O BC 4 ∴ cos ∠CBD = = . ········································· 4 分 BD 5 图2 Q BC = 2 , 5 ∴ BD = . ··························································································································· 5 分 2 五、解答题(本题满分 6 分) 解答题( 解: (1)补全图 1 见下图. ··································································································· 1 分 “限塑令”实施前,平均一次购物使 用不同数量塑料购物袋的人数统计图 .. 人数/位 40 35 30 25 20 15 10 5 0 37 26 11 10 4 2 3 4 图1 5 6 3 7 塑料袋数/个
9 1
第 4 页 共 9 页
9 × 1 + 37 × 2 + 26 × 3 + 11× 4 + 10 × 5 + 4 × 6 + 3 × 7 300 = = 3 (个) . 100 100
这 100 位顾客平均一次购物使用塑料购物袋的平均数为 3 个. ·········································· 3 分 2000 × 3 = 6000 . 估计这个超市每天需要为顾客提供 6000 个塑料购物袋. ···················································· 4 分 (2)图 2 中,使用收费塑料购物袋的人数所占百分比为 25% . ············
···························· 5 分 根据图表回答正确给 1 分,例如:由图 2 和统计表可知,购物时应尽量使用自备袋和押金式 环保袋,少用塑料购物袋;塑料购物袋应尽量循环使用,以便减少塑料购物袋的使用量,为 环保做贡献. ·························································································································· 6 分 解答题( 道小题, 六、解答题(共 2 道小题,共 9 分) 21.解:设这次试车时,由北京到天津的平均速度是每小时 x 千米,则由天津返回北京的平 均速度是每小时 ( x + 40) 千米. ···························································································· 1 分
30 + 6 1 x = ( x + 40) . ······················································································ 3 分 60 2 解得 x = 200 . ······················································································································· 4 分
依题意,得 答:这次试车时,由北京到天津的平均速度是每小时 200 千米. ······································ 5 分 22.解: (1)重叠三角形 A′B′C ′ 的面积为 3 . ································································ 1 分 (2)用含 m 的代数式表示重叠三角形 A′B′C ′ 的面积为 3(4 − m) 2 ; ····························· 2 分
8 m 的取值范围为 ≤ m
七、解答题(本题满分 7 分) 解答题(
2 23. (1)证明:Q mx − (3m + 2) x + 2m + 2 = 0 是关于 x 的一元二次方程,
∴∆ = [−(3m + 2)]2 − 4m(2m + 2) = m 2 + 4m + 4 = (m + 2) 2 .
Q 当 m > 0 时, (m + 2) 2 > 0 ,即 ∆ > 0 . ∴ 方程有两个不相等的实数根. ··························································································· 2 分 (3m + 2) ± (m + 2) (2)解:由求根公式,得 x = . 2m 2m + 2 ∴x = 或 x = 1 . ·······································································
································· 3 分 m Qm > 0 , 2m + 2 2(m + 1) ∴ = > 1. m m
Q x1
∴ x1 = 1 , x2 =
2m + 2 . ······································································································ 4 分 m
第 5 页 共 9 页
∴ y = x2 − 2 x1 =
即y=
2m + 2 2 − 2 ×1 = . m m
y 4 3 2 1
y = 2m(m > 0)
2 (m > 0) 为所求.······························· 5 分 m
(3)解:在同一平面直角坐标系中分别画出
y=
2 (m > 0) 与 y = 2m(m > 0) 的图象. m
············································································· 6 分 由图象可得,当 m ≥ 1 时, y ≤ 2m . ·············· 7 分 八、解答题(本题满分 7 分)
2 (m > 0) m x -4 -3 -2 -1 O 1 2 3 4 -1 -2 -3 -4 y=
24.解: (1)Q y = kx 沿 y 轴向上平移 3 个单位长度后经过 y 轴上的点 C ,
∴ C (0, . 3)
设直线 BC 的解析式为 y = kx + 3 .
Q B (3, 在直线 BC 上, 0)
∴ 3k + 3 = 0 . 解得 k = −1 .
∴ 直线 BC 的解析式为 y = − x + 3 . ···················································································· 1 分 Q 抛物线 y = x 2 + bx + c 过点 B,C ,
9 + 3b + c = 0, ∴ c = 3.
解得
b = −4, c = 3.
∴ 抛物线的解析式为 y = x 2 − 4 x + 3 . ················································································ 2 分
(2)由 y = x 2 − 4 x + 3 . 可得 D (2, 1),A(1, . − 0) y 4 3 C 2 1 -2 -1 O -1 -2 A 1 2F 3 D P′ 图1 P E B 4 x
∴ OB = 3 , OC = 3 , OA = 1 , AB = 2 . 可得 △OBC 是等腰直角三角形.
∴∠OBC = 45o , CB = 3 2 .
如图 1,设抛物线对称轴与 x 轴交于点 F ,
∴ AF =
1 AB = 1 . 2
第 6 页 共 9 页
过点 A 作 AE ⊥ BC 于点 E .
∴∠AEB = 90o .
可得 BE = AE =
2 , CE = 2 2 .
o
在 △ AEC 与 △ AFP 中, ∠AEC = ∠AFP = 90 , ∠ACE = ∠APF ,
∴△ AEC ∽△ AFP .
∴
AE CE 2 2 2 = , = . AF PF 1 PF
解得 PF = 2 . Q 点 P 在抛物线的对称轴上,
∴ 点 P 的坐标为 (2, 或 (2, 2) . ······················································································· 5 分 2) −
(3)解法一:如图 2,作点 A(1, 关于 y 轴的对称点 A′ ,则 A′( −1 0) . 0) , 连结 A′C,A′D , 可得 A′C = AC = 10 , ∠OCA′ = ∠OCA .
2 由勾股定理可得 CD = 20 , A′D = 10 . 2
y 4 3 C 2 1 A′ A B -1 O 1 2 F 3 4 -1
D -2 图2 x
又 A′C = 10 ,
2
∴ A′D 2 + A′C 2 = CD 2 . ∴△ A′DC 是等腰直角三角形, ∠CA′D = 90o , ∴∠DCA′ = 45 .
o
∴∠OCA′ + ∠OCD = 45o . ∴∠OCA + ∠OCD = 45o .
即 ∠OCA 与 ∠OCD 两角和的度数为 45 . ········································································· 7 分 y 解法二:如图 3,连结 BD .
o
同解法一可得 CD =
20 , AC = 10 .
o
4 3 C 2 1 -2 -1 O -1 -2 A 1 2F 3 D 图3 B 4 x
在 Rt△DBF 中, ∠DFB = 90 , BF = DF = 1 ,
∴ DB = DF + BF = 2 .
2 2
在 △CBD 和 △COA 中,
第 7 页 共 9 页
DB 2 BC 3 2 20 CD = = 2, = = 2, = = 2. AO 1 OC 3 CA 10
DB BC CD = = . AO OC CA ∴△CBD ∽△COA . ∴∠BCD = ∠OCA . ∴
Q ∠OCB = 45o , ∴∠OCA + ∠OCD = 45o .
即 ∠OCA 与 ∠OCD 两角和的度数为 45 . ········································································· 7 分
o
九、解答题(本题满分 8 分) 解答题( 25.解: (1)线段 PG 与 PC 的位置关系是 PG ⊥ PC ;
PG = 3 . ··························································································································· 2 分 PC
(2)猜想: (1)中的结论没有发生变化. 证明:如图,延长 GP 交 AD 于点 H ,连结 CH,CG . Q P 是线段 DF 的中点, ∴ FP = DP . 由题意可知 AD ∥ FG . ∴∠GFP = ∠HDP . Q ∠GPF = ∠HPD , ∴△GFP ≌△HDP . ∴ GP = HP , GF = HD . Q 四边形 ABCD 是菱形,
D H P A
C G B E F
∴ CD = CB , ∠HDC = ∠ABC = 60o .
由 ∠ABC = ∠BEF = 60 ,且菱形 BEFG 的对角线 BF 恰好与菱形 ABCD 的边 AB 在同一
o
条直线上, 可得 ∠GBC = 60 .
o
∴∠HDC = ∠GBC . Q 四边形 BEFG 是菱形, ∴ GF = GB . ∴ HD = GB . ∴△HDC ≌△GBC . ∴ CH = CG , ∠DCH = ∠BCG . ∴∠DCH + ∠HCB = ∠BCG + ∠HCB = 120o .
即 ∠HCG = 120 .
o
第 8 页 共 9 页
Q CH = CG , PH = PG , ∴ PG ⊥ PC , ∠GCP = ∠HCP = 60o .
PG = 3 . ························································································································ 6 分 PC PG (3) = tan(90o − α ) . ·································································································· 8 分 PC ∴
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