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# 九年级调研测试 数学答案

9．x1＝0，x2＝1 14．2

（1）x=－6 ······································································ （4分，未体现解法-2分）； （2）CD=26 ·········································································································· （4分）． 20．（本题满分8分）

21．（本题满分8分）

31

28（2）画出图象； ····································································································· （3分） （3）写出当y＜0时，x的取值范围是x＜1或x＞3． ········································· （2分）

22．（本题满分8分）

10．50° 15．6

11．－4 16．22

12．略 17．＞

13．24 187

2．A 6．B

3．A 7．C

4．D 8．C

23．（本题满分10分）

111

661

∴x

6答：平均每次降价的百分率16.7%． ··································································· （10分）

24．（本题满分10分）

∵⊙O是△ABC的内切圆，切点为D、E、F，∴∠EOF＝2y°，∠OEA＝∠OFA＝90°，

1

∴∠A＋∠EOF＝360°－90°－90°＝180°，∴y＝90－x． ···································· （5分）

2（2）设⊙O的半径是r．由勾股定理得：ACBC－AB＝6， ∵⊙O是△ABC的内切圆，切点为D、E、F，

∴AE＝AF，CD＝CF，BE＝BD，∠OEA＝∠OFA＝∠A＝90°，OE＝OF， ∴四边形OEAF是正方形，∴OE＝OF＝AE＝AF＝r， ∴AC－r＋AB－r＝BC，∴6－r＋8－r＝10，∴r＝2．

∴当x=6.5时，y有最大值为2722.5．··································································· （8分） ∵0＜x≤10且x为正整数，

∴当x=6时，30+x=36，y=2720（元）；当x=7时，30+x=37，y=2720（元）．

2

2

90??190??290??3

＋＋＝3π． ··································································· （4分） 180180180

（2）直线GB⊥DF．

26．（本题满分10分）

∴t的值为2 或6． ································································································· （6分）

（3）由题意知，若⊙P与四边形ABCD

①当⊙P与BC相切于点C时，有∠BCP＝90° 从而∠OCP＝45°，得到OP＝3，此时t＝1 ②当⊙P与CD相切于点C时，有PC⊥CD 即点P与点O重合，此时t＝4

PC ＝PA ＝( 9－t )，PO ＝( t－4 ) 于是( 9－t )＝( t－4 )＋3，解得：t=5.6

∴t的值为1或4或5.6 ························································································· （10分）

2

2

2

2

2

2

2

2

∠D1CM＋∠ACH1＋∠ACD1＝180°，∠AH1C＝∠ACD1 ∴∠H1AC＝∠D1CM

∵AC＝CD1，∠AGC＝∠CMD1＝90°，∴△ACG≌△CD1M ∴CG＝D1M 同理可证CG＝D2N

∴D1M＝D2N ············································································································ （9分） ②如图（作图正确） ····························································································· （11分）

D1M＝D2N仍成立 ································································································· （12分）

K K1 E1 1

K2 DE2 D2 FF2

A

H2 G H1 B 图2

H1 H2 B 图3

2

22

2

28．（本题满分12分）

∵二次函数y＝ax2＋bx＋3的图象经过点A和点C．

∴?

?a?b?3?4,?a??1,

9a?3b?3?0,b?2,??

2

∴抛物线的解析式为y＝－x ＋2x＋3 ···································································· （3分） （2）①∵A（1，4），C（3，0）

∴可求直线AC的解析式为y＝－2x＋6 ································································· （4分）

t

P（1，4－t），将y＝4－t代入y＝－2x＋6中，解得点E的横坐标为x＝1＋ 2

2

tt∴点G的横坐标为1＋ ，代入抛物线的解析式中，可求点G的纵坐标为4－ 2 4

22tt∴GE＝( 4－ )－( 4－t )＝t－

4 4

tt

1t1t

2 2 2 2

2

1t12＝ 2( t－ )＝－ t－2 )＋1 2 4 4 当t＝2时，S△ACG的最大值为1 ·············································································· （8分）

20②t＝ ······················································································ （12分） 或t＝20－5

13

27．（本题满分12分）

∵AC＝CD1，∴△ACH≌△CD1M ∴D1M＝CH 同理可证D2N＝CH

∴D1M＝D2N ············································································································ （4分） （2）①D1M＝D2N仍成立 ······················································································ （5分） 证明：过点C作CG⊥AB，垂足为点G ∵∠H1AC＋∠ACH1＋∠AH1C＝180°