# 南师附中数学二模答案

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°，

∴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 ··········· ·········· ··········· ·· ·········· ··········· ··········· ··

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．

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° ．

∴ ∠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

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．

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

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 ··········· ····· ·········· ······

（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 ·········· ··········· ··········· ·········· ·

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