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南师附中数学二模答案


数学参考答案及评分标准
说明:本评分标准每题给出了一种或几种解法供参考,如果考生的解法与本解答不同,参照本评分标准的 精神给分. 一、选择题(每小题 2 分,共计 12 分) 题号 答案 1 B 2 A 3 D 4 C 5 D 6 C

二、填空题(每小题 2 分,共计 20 分) 7.x≠2
2

8.1.2×10

r />
-5

9.35
2

10.13,5

11.y=-2x +8x-5(或 y=-2(x-2) +3)
2 2

12.3

13.100(1+x)+100(1+x) =231(或 100x +300x-31=0) 14. AB BC CD = = ,∠B=∠B2,∠C=∠C2 A2B2 B2C2 C2D2

15.本题答案不惟一,以下答案供参考:对应点连线被这条直线平分;对应点到这条直线的距离相等 16.2+ 2 或 2- 2 三、解答题(本大题共 12 小题,共计 88 分) 17. (本题 4 分) 解:(π+1)0- 12+│- 3│ =1-2 3+ 3 ·········································· 分 ··········· ·········· ··········· ········· 3 ·········· ··········· ··········· ········· =1- 3. ············································· 分 ··········· ·········· ··········· ··········· · 4 ·········· ··········· ··········· ·········· ··

18. (本题 6 分) a2-b2 1 1 解: ÷( - ) ab a b (a+b)(a-b) b-a = ÷ ··········· ··········· ·········· ····· 2 分 ··········· ·········· ··········· ····· ·········· ··········· ··········· ····· ab ab (a+b)(a-b) ab = · ··········· ··········· ·········· ····· 3 分 ··········· ·········· ··········· ····· ·········· ··········· ··········· ····· ab b-a (a+b)(a-b) ab = ·(- ) ab a-b =-(a+b) =-a-b. ··········································· 分 ··········· ·········· ··········· ··········· ·········· ··········· ··········· ·········· 5 当 a=2,b=1 时,原式=-3. ································· 分 ··········· ·········· ··········· 6 ·········· ··········· ···········

19. (本题 6 分) 解:在直角三角形 DCF 中, ∵CD=5.4m,∠DCF=37°,

数学答案

第 1 页(共 6 页)

∴sin∠DCF=

DF DF = =0.60. ································· 分 ··········· ·········· ··········· 1 ·········· ··········· ··········· DC 5.4

∴DF=3.24. ·········································· 分 ··········· ·········· ··········· ········· 2 ·········· ··········· ··········· ········· 又∵∠CDF+∠DCF=90°,∠ADE+∠CDF=90°, ∴∠ADE=∠DCF. ······································ 分 ··········· ·········· ··········· ······ ·········· ··········· ··········· ····· 3 又∵AD=BC=2, DE DE ∴cos∠ADE= = =0.8. ································· 分 ··········· ·········· ··········· 4 ·········· ··········· ··········· AD 2 ∴DE=1.6. ·········································· 5 分 ··········· ·········· ··········· ·········· ·········· ··········· ··········· ·········· ∴EF=ED+DF=3.24+1.6≈4.8(m) ··························· 6 分 . ··········· ·········· ······ ·········· ··········· ······ 20. (本题 6 分) 解: (1)当 t=1 时,h=-1+26+1=26(m) ·························· 分 . ··········· ·········· ···· 2 ·········· ··········· ···· (2)h=-t2+26t+1 =-(t2-26t+132-132)+1 ································ 分 ··········· ·········· ··········· ·········· ··········· ·········· 3 =-(t-13)2+132+1 =-(t-13)2+170. ···································· 分 ··········· ·········· ··········· ··· 4 ·········· ··········· ··········· ··· 当 t=13 s 时,h 最大=170 m. ······························· 6 分 ··········· ·········· ·········· ·········· ··········· ·········· 21. (本题 8 分) 1 解: (1) . ············································· 2 分 ··········· ·········· ··········· ··········· ·· ·········· ··········· ··········· ·········· ··· 4 (2)本题答案不惟一,以下答案供参考: 从 M 处投 2 次小球,若小球恰好 1 次落在 A,1 次落在 C 就获得一等奖. ······· 分 ······ 4 ······ 理由如下:把小球从左边管道落入 B 记为 B1,从右边管道落入 B 记为 B2,列出表格: A A B1 B2 C (A,A) (B1,A) (B2,A) (C,A) B1 (A,B1) (B1,B1) (B2,B1) (C,B1) B2 (A,B2) (B1,B2) (B2,B2) (C,B2) C (A,C) (B1,C) (B2,C) (C,C)

2 1 共有 16 种等可能结果,P(1 次落在 A,1 次落在 C)= = . ············· 分 ··········· · 8 ·········· ·· 16 8 22. (本题 8 分) 解: (1)甲 5 次打靶命中的环数的平均数为 8 环,方差为 1.2. ·················· 分 ··········· ······ 2 ·········· ······· 甲与乙射击成绩的平均数相同,而甲射击成绩的方差小,因此成绩更稳定,所以,教练选择 甲参加射击比赛. ····································· 3 分 ··········· ·········· ··········· ····· ·········· ··········· ··········· ····· (2)设乙第 6 次射击成绩为 x 环,则第 7 次射击成绩为(16-x) 环. 有(x-8)2+(16-x-8)2=2.25×2.···························· 5 分 ··········· ·········· ······· ·········· ··········· ······· 解得 x1=6.5,x2=9.5. ·································· 7 分 ··········· ·········· ··········· ·· ·········· ··········· ··········· ··
数学答案 第 2 页(共 6 页)

当 x=6.5 时,16-x=9.5;当 x=9.5 时,16-x=6.5. 所以乙第 6 次和第 7 次的射击成绩为 6.5 环和 9.5 环. ·················· 分 ··········· ······ 8 ·········· ·······

23. (本题 8 分) 解: (1)150,16:00. ········································ 分 ··········· ·········· ··········· ········ ·········· ··········· ··········· ······· 2 (2)图象如图所示.······································· 分 ··········· ·········· ··········· ······ 3 ·········· ··········· ··········· ······ (注:快车距离 B 城市的路程 y 与 x 之间的函数图象不补全不扣分. )
y/km 900 600 300

O

1 2 3 4 5 6 7 8 9 10 11

x/h

y=100x-100. ······································· 分 ··········· ·········· ··········· ······ 5 ·········· ··········· ··········· ······ (3)方法一: 快车距离 B 城市的路程 y 与 x 之间的函数关系式为 y=-150x+900. ········· 分 ········· ········ 6
?y=100x-100, 根据题意,得? ?y=-150x+900. ?x=4, 解得? ··········· ··········· ·········· ······· 7 分 ··········· ·········· ··········· ······· ·········· ··········· ··········· ······· ?y=300.

所以 14:00 两车相遇. ··································· 分 ··········· ·········· ··········· ··· ·········· ··········· ··········· ·· 8 方法二: 设快车出发 a h 后两车相遇. 根据题意,得 150a+100(a-1)=900. ·························· 分 ··········· ·········· ····· ·········· ··········· ···· 6 解这个方程,得 a=4. ·································· 7 分 ··········· ·········· ··········· ·· ·········· ··········· ··········· ·· 所以 14:00 两车相遇. ··································· 分 ··········· ·········· ··········· ··· ·········· ··········· ··········· ·· 8

24. (本题 8 分) 解: (1)画图正确. ········································· 分 ··········· ·········· ··········· ········· ·········· ··········· ··········· ········ 3 (2)BC 与⊙O 相切. 理由如下: 连接 CO.·········································· 分 ··········· ·········· ··········· ········· 4 ·········· ··········· ··········· ········· ∵ AC=BC 且∠ACB=120° , ∴ ∠A=∠B=30° ···································· 分 . ··········· ·········· ··········· ···· ·········· ··········· ··········· ··· 5 ∵ AO=CO, ∴∠ACO=∠A=30° .
数学答案 第 3 页(共 6 页)

∴ ∠OCB=∠ACB-∠ACO=90° ···························· 分 . ··········· ·········· ······ 6 ·········· ··········· ······ ∴OC⊥BC. 又 BC 经过半径 OC 的外端点 C, ····························· 分 ··········· ·········· ········ ·········· ··········· ······· 7 ∴BC 与⊙O 相切. ····································· 分 ··········· ·········· ··········· ···· 8 ·········· ··········· ··········· ····

25. (本题 8 分) k 解: (1)设 y= (k≠0) ······································ 1 分 . ··········· ·········· ··········· ······ ·········· ··········· ··········· ······ x 由题意可知当 x=100 时,y=30,∴k=3000. ······················ 2 分 ··········· ·········· · ·········· ··········· · 3000 y= (x>0) (不写出 x>0 不扣分) ························· 分 . ··········· ·········· ··· 3 ·········· ··········· ··· x (2)设其单价定为 a 元,得 3000 (a-80)=1000. ························ 分 ··········· ·········· ·· 5 ·········· ··········· ·· a

解得 a=120. ········································ 分 ··········· ·········· ··········· ········ ·········· ··········· ··········· ······· 6 经检验 a=120 是原方程的解. ······························· 分 ··········· ·········· ·········· ·········· ··········· ········· 7 因此单价应定为 120 元时,此种衬衣的日销售利润为 1000 元. ············· 分 ··········· ·· ·········· ·· 8

26. (本题 8 分) (1)证明:∵AD∥BC,DE∥AB, ∴四边形 ABED 是平行四边形. ···························· 1 分 ··········· ·········· ······· ·········· ··········· ······· ∴AD=BE,AB=DE. 同理,AD=FC,AF=DC. ∵E、F 是 BC 的三等分点, ∴AD=EF. ······································· 分 ··········· ·········· ··········· ······· ·········· ··········· ··········· ······ 2 又∵AD∥BC, ∴四边形 AEFD 是平行四边形. ···························· 3 分 ··········· ·········· ······· ·········· ··········· ······· ∵AB=DC, ∴DE=AF. ∴平行四边形 AEFD 是矩形. ······························ 分 ··········· ·········· ········ 4 ·········· ··········· ········ (2)解:如图,四边形 AFED 是等腰梯形. ···························· 6 分 ··········· ·········· ······· ·········· ··········· ······· 证明如下: ∵DE∥AB,AF∥DC, 又∵在梯形 ABCD 中,AD∥BC,AB 与 CD 不平行,
B F E C A D

∴四边形 ABED、AFCD 是平行四边形,四边形 AFED 是梯形. ············7 分 ··········· · ·········· ·· ∴AB=DE,AF=DC. ∵AB=DC, ∴DE=AF. ∴梯形 AFED 是等腰矩形.
数学答案 第 4 页(共 6 页)

综合两种情况,按照题目的条件,以 A、E、F、D 为顶点的四边形是矩形或等腰梯形. ··········· ··········· ·········· ··········· ···· 8 分 ··········· ·········· ··········· ··········· ···· ·········· ··········· ··········· ·········· ·····

27. (本题 8 分) 15 解: (3, ) ·········································· 分 (1) . ··········· ·········· ··········· ········· 2 ·········· ··········· ··········· ········· 4 (2)第一种情况: 如图,当⊙P 与 x 轴相切时,设切点为 D,连接 PD. 则 PD⊥x 轴. ∴PD∥AO. ∴∠OAD=∠PDA. ∵PD=PA,∴∠PAD=∠PDA. ∴∠OAD=∠PAD. ∵AC 是⊙P 的直径,∴∠ADC=90° =∠AOD. ∴△AOD∽△ADC. ···································· 分 ··········· ·········· ··········· ···· ·········· ··········· ··········· ··· 3 ∵在 Rt△AOB 中,AO=6,BO=8, ∴AB= AO2+BO2=10. ································· 分 ··········· ·········· ··········· · ·········· ··········· ··········· 4 ∵∠PDB=∠AOB=90° ,∠PBD=∠ABO, ∴Rt△PDB∽Rt△AOB. ∴ 即 PD PB = , AO AB PA 10-PA = . ··········· ··········· ·········· ······ 5 分 ··········· ·········· ··········· ······ ·········· ··········· ··········· ······ 6 10
O D y A P C B x

15 解得 PA= . ······································· 6 分 ··········· ·········· ··········· ······· ·········· ··········· ··········· ······· 4 第二种情况: 如图,当点 P 与点 B 重合时,此时⊙P 与 x 轴交于点 D1、D2. 不妨取点 D1. ∵AC 是⊙P 的直径,∴∠AD1C=90° =∠AOD1. ∵PD1=PA,∴∠PAD1=∠PD1A. ∴△AOD1∽△CD1A. (若取点 D2,可类似说明△AOD2∽△CD2A. ) 此时 PA=10. 15 综合两种情况,可知 PA= 或 10. 4 ·······················8 分 ··········· ·········· ·· ·········· ··········· ·
B D2 O D1 x y A

(P)

C

数学答案

第 5 页(共 6 页)

28. (本题 10 分) 解: (1)答案不惟一,下图的分割方案供参考(图中∠DAB=∠D'A'B',∠DBA=∠D'B'A') :
C C' D D'

A

B

A'

B'

··········· ····· 3 分 ··········· ····· ·········· ······

(2)在△ABC 和△A'B'C'中,若∠C=∠C',不妨设∠A>∠A',∠B<∠B', 作直线 AD,与 BC 边交于点 D,使∠DAB=∠A';作直线 B'D',与 A'C'边交于点 D', 使∠D'B'A'=∠B. 此时,△ADB∽△A'D'B'且△ADC∽△B'D'C'. (说明:若不采取上述叙述方法,而通过画图说明方案也可) 理由如下: 按上述作图方法,可得△ADB∽△A'D'B',且∠ADC=∠B'D'C'. ∵∠C=∠C',∴△ADC∽△B'D'C'.·························· 7 分 ··········· ·········· ····· ·········· ··········· ·····

(3)可以,答案不惟一,下图的分割方案供参考(图中∠ACB=∠EDM,∠CAN=∠DEF, ∠ABG=∠DFE,∠BAN=∠FDH) :
A

D G B N C E M H F

··········· ··········· ·········· ··········· · 分 ··········································· 10 ·········· ··········· ··········· ·········· ·

数学答案

第 6 页(共 6 页)


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