成华九上数学试题及答案2014.1

2013～2014学年度上期期末初中学业水平阶段检测

九年级数学参考答案

A卷(共100分)

一、选择题(每小题3分，共30分)

1．C 2．A 3．B 4．D 5．C 6．B 7．D 8．A 9．B 10．C

二、填空题(每小题4分，共16分)

11．AM； 12．7，6.5； 13．4，2； 14．4．

三、解答题(共6个小题，共54分) 15. 解：原式＝22+(－2)－22+1 ·················································································· 4分

＝－1 ··········································································································· 6分

16．(1)解：x2+3x－4＝0 (2)解：∵a＝1，b＝－3，c＝1 ···················1分

22

(x－1)(x+4)＝0…………3分 ∴b－4ac＝(－3)－4×1×1＝5>0 ······3分

3±53±5

x－1＝0或x+4＝0 ……5分 ∴x＝＝

2×12

3－53+5

∴x1＝1，x2＝－4 ……6分 即x1＝，x2＝． ···············6分

22

17. 解：由已知易得∠CBD＝45°，∠CDB＝90°，∴∠DCB＝45°， ∴CD＝BD ··············································································· 2分 设CD＝BD＝x米

在Rt△CDA中，由已知可得∠CAD＝30°，

CD3

∴tan30°＝＝ ·································································· 4分

AD3

x3＝ ························································· 5分 ∵AB＝6米，∴x+63

∴x＝3+33≈8.2 (米) ····························································· 7分 答：生命所在点C的深度CD的长约为8.2米． ················· 8分

18. 解：(1)把A(1，b)代入y＝2x得b＝2，∴A(1，2)． ·········· 1分

a

把A(1，2)代入y＝得a＝1×2＝2，

x

2

所以反比例函数的表达式为y＝． ······································ 2分

x

??y＝2x?x1＝1?x2＝－1

2? 由y＝，得?，?y＝2y＝－2??12

?x?∴点B的坐标为(－1，－2)． ················································ 4分 (2)草图如图, ············································································· 6分 (说明：要求双曲线过(1,2)，(－1,－2)，直线过(0,0)，(1,2)，(－1,－2)) 根据图象可知，当x＜－1或0＜x＜1时，正比例函数值小于 反比例函数值．······································································ 8分

5

2

19. 解：(1)①，③； ························································································································ 4分

3

(2)列表如下(树状图参照给分)： ··································································································· 6分

∵所有等可能结果有6种，三个函数中只有①和③两个函数图象满

① ② ③ 足在第二象限内y随x的增大而减小， ① (①，②) (①，③) ∴其中抽到的两张卡片上的函数图象都满足在第二象限内y随x的

② (②，①) (②，③) 增大而减小的结果有2种，

③ (③，①) (③，②) 2112∴P1＝＝，P2＝1－＝ ··········································· 8分

6333

124

∴王亮平均每次得分为3×＝1，李明平均每次得分为2×＝，∴游戏不公平 ···························· 7分

333

修改规则为：王亮先从口袋中随机抽取一张卡片，不放回，李明再从口袋中随机抽取一张卡片，若两人抽到的卡片上的函数图象都满足在第二象限内y随x的增大而减小，则王亮得2分，否则李明得1分. ·························································································································································· 10分

20. (1)证明： ∵菱形ABCD，∴AB＝AD，∠BAE＝∠DAE ····························································· 1分

又∵AE＝AE， ∴△AEB≌△AED(SAS) ····································································································· 3分 说明：用菱形的轴对称性证明也给满分．

1

(2)①∵sin∠ADE＝，∴∠ADE＝30°， ····················································································· 4分

2

又∵DE⊥CD， ∴∠ADC＝120°， ····················································································· 5分 ∵菱形中，AD＝CD， ∴∠DAC＝30°， 又∵△AEB≌△AED，∴∠ABE＝∠ADE＝30°，∠BAC＝∠DAC＝30°， ∴∠AFB＝90°，即BF⊥AD································································································ 6分 ②在Rt△AFE中，∵EF＝1，∠DAC＝30°，∴AE＝2，AF＝3，

同理在Rt△DFE中，可得DF＝3，∴AD＝23． 分以下两种情况：

APAFa3

当△AFP∽△ADE时，有＝，∴＝，∴a＝1 ···················································· 8分

AEAD223APAFa3

当△APF∽△ADE时，有＝，∴＝，∴a＝3 ·················································· 10分

ADAE232

综上，当a＝1或3时，△AFP与△ADE相似．

B卷(共50分)

一、填空题(每小题4分，共20分)

21．50； 22．2012； 23．3，5，6，9； 24．4； 25．②③．

6

二、解答题(共30分)

26．解：(1)由表格数据可知y与x是一次函数关系， ·································································· 1分

设其关系式为y＝kx+b(k≠0)

?11000k+b=25

取x＝11000，y＝25和x＝11200，y＝20，代入得?11200k+b=20.

?

?k =－1

40 解得?

? b=300.1

∴y＝－x+300． ················································································································· 2分

40

由0≤y≤50，得10000≤x≤12000，

1

∴y与x之间的函数关系式为y＝－x+300 (10000≤x≤12000)．········································ 3分

40

11

(2) 200 (－x+300)，40【50－(－x+300)】 ·········································································· 5分

4040依题意得：

11

W＝(－x+300)( x－200) －40【50－(－x+300)】 ························································· 6分

40401

＝－x2+304x－50000(*) ·································································································· 7分

401b

∵a＝－ <0，∴W有最大值，且－ =6080，

402a

∵6080不在10000≤x≤12000范围内，10000距6080比12000距6080近， ∴当x＝10000时，W取最大值，代入(*)，得W最大值＝490000．

即：当每辆车的月租金定为10000元时，公司可获得最大月收益490000元． ··············· 8分

27. 解：(1)∵∠BPC＝60°，∴∠BAC＝∠BPC＝60°．

又∵AB＝AC，∴△ABC为等边三角形． ············································································ 1分 ∴∠ACB＝60°，∵∠ACB的平分线交⊙O于点P，∴∠BCP＝30°， ·································· 2分 ∴∠PBC＝180°－∠BPC－∠BCP＝90°················································································ 3分 ∴CP为⊙O的直径． ·········································································································· 4分 (2)过点E作EG⊥AC于G，连接OB，OC． ∵AB＝AC，AF所在的直线为⊙O的对称轴，∴AF⊥BC， BF＝CF． A ∵∠ACP＝∠PCB，∴EG＝EF．……………………………….. 5分

P G 1E ∵∠BPC＝∠BOC＝∠FOC， O 2

24B C F ∴sin∠FOC＝sin∠BPC＝．……………………………….. 6分 25

设FC＝24a，则OC＝OA＝25a， 由勾股定理可得OF＝7a，∴AF＝25a+7a＝32a． ···························································· 7分 在Rt△AFC中，AC＝AF2+FC2＝40a． ··········································································· 8分 又∵AB＝AC＝40，∴40a＝40，∴a＝1． ··········································································· 9分 ∴FC＝24，AF＝32

在Rt△AGE和Rt△AFC中，

EGFC

EG＝EF，FC＝24，AF＝32，AC＝40，sin∠FAC＝＝，

AEAC

EF24即＝，解得EF＝12． ·························································································· 10分 32–EF40

7

28. (1)由题意，得3＝0+b，解得b＝3，

∴直线的函数关系式为y＝x+3． ························································································· 1分 取y＝0，得x＝－3，∴A(－3，0)． ?3＝c，由? 2

?0＝a(－3)+2a(－3)+c?a=－1解得?， ························································································································ 2分

?c= 3∴抛物线的函数关系式为y＝－x2－2x+3． ········································································· 3分

2

取y＝0，得－x－2x+3＝0，解得x＝－3或1，∵A(－3，0)，∴B(1，0)． ······················ 4分

2

(2)设P(d，－d－2d+3)，过P作PQ∥y轴交AC于点Q，则Q(d，d+3)，

22

∴PQ＝(－d－2d+3)－(d+3)＝－d－3d． ··········································································· 5分 分别过A，C作AH1⊥PQ于H1，CH2⊥PQ于H2，则AH1＝d+3，CH2＝－d，

111

∴S△APC＝S△APQ+S△CPQ＝×PQ×AH1+×PQ×CH2＝×PQ×(AH1+CH2)

222

13327＝×(－d2－3d)×(d+3－d)＝－(d+)2+． ··························································· 7分 22283327∵－<0，∴S有最大值，当d＝－时，S最大值＝． ························································· 8分

228y y P C H2

Q

l x D A E x A B B O O H1

N

M

图① 图②

(3)分别过M，N作MD⊥x轴于点D，NE⊥x轴于点E，

2

由于点M在y＝－x－2x+3(x<－3)上，点N在y＝x2+2x－3(－3

BDMDBM

由△BMD∽△BNE得＝＝， ·················································································· 10分

BENEBN

1－mm2+2m－32

若存在直线l，使得△ABM的面积被AN恰好平分，则＝＝，

1－n－n2－2n+31

1－mm2+2m－3由＝2得m＝2n－1①，把①代入＝2得(2n－1)2+2(2n－1)－3+2(n2+2n－3)＝0， 21－n－n－2n+3

5532

整理得3n2+2n－5＝0，解得n1＝1，n2＝－，∴N(－，－)． ····································· 11分

339

53244

∴过B(1，0)，N(－，－)的直线l的函数解析式为y＝x－． ···································· 12分

3933

8