盐城市盐都区2018 届九年级第一次模拟检测 下载本文

2 7.(本题满分 14 分)

1 1

?如图①,一次函数 y=x ? 2 的图像交 x 轴于点 A,交 y 轴于点 B,二次函数 y=x2 ? bx ? c

22的图像经过 A、B 两点,与 x 轴交于另一点 C.

(1)求二次函数的关系式及点 C 的坐标;

(2)如图②,若点 P 是直线 AB 上方的抛物线上一点,过点 P 作 PD∥x 轴交 AB 于点 D,PE

∥y 轴交 AB 于点 E,求 PD+PE 的最大值; (3)如图③,若点 M 在抛物线的对称轴上,且∠AMB=∠ACB,求出所有满足条件的点 M

的坐标.

y

C O

A

B

y

P C

D

y

C

A

E B

B

x

O

x

O A

x

(第 27 题图)

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九年级数学参考答案及评分标准

(阅卷前请认真校对,以防答案有误!)

一、选择题(每小题3分,共24分)

题号 答案 1 B 2 C 3 D 4 C 5 A 6 C 7 D 8 B 二、填空题(每小题3分,共24分) 9. x≤3. 13.

25. 410.1.3×105. 14.36°.

11.甲. 12.3.

15.3或25?2. 16.22?1.

三、解答题

17.解:原式=5-1+2-3 ······························································································· 4分

=3. ············································································································· 6分

说明:每算对一个给1分. 18.解:原式=

46a?3?? ········································································ 3分 a?3(a?3)(a?3)3==

42 ······························································································ 5分 ?a?3a?32. ······································································································ 6分 a?319.解:??x?5?5(x?2)?3x?6,①?1?4x.??2②

由不等式①,得x≤8. ····························································································· 3分 由不等式②,得x>-1. ························································································· 6分 ∴不等式组的解集为-1<x≤8.·············································································· 8分 120.解:(1)抽到数字恰好为3的概率为. ······························································ 3分

3(2)画树状图(或列表)如下: ··············································································· 6分

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十位数 个位数 31开始

355153由树状图可知,所有等可能的结果共有6种,其中恰好是51有1种. ∴P(两位数恰好是“51”)=

1. ········································································· 8分 621.解:(1)10. ············································································································· 2分

(2)72°; ······················································································································ 4分 (3)根据题意得:1200×(1-5%)=1 140(人), ··············································· 7分 答:估计测试成绩合格以上(含合格)的人数有1 140人. ································ 8分 22.解:(1)如图所示. ································································································· 4分

A BDCEM

说明:作出点C给1分;作出BD给2分,作出点E给1分.

(2)BD=DE. ·············································································································· 5分 理由如下:

A∵BD平分∠ABC,∴∠1=

1∠ABC. 21BD1∵AB=AC,∴∠ABC=∠4.∴∠1=∠4.

242C3EM∵CE=CD,∴∠2=∠3. ∵∠4=∠2+∠3,∴∠3=

1∠4. 2∴∠1=∠3.∴BD=DE. ························································································ 10分

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23.解:(1)5. ··············································································································· 3分

(2)①3÷0.2=15,即运动员第1次到过点P用时15 min, ······························· 5分 ∵该运动员从第一次过P点到第二次过P点所用的时间为24 min,

∴该运动员从甲地出发到第二次经过P点所用的时间是15+24=39(min), ∴直线AB经过点(25,5),(39,3).

设AB所在直线的函数表达式为s=kt+b,将(25,5),(39,3)代入,得:

1?k?,=??25k?b=5,160?7∴?解得?∴AB所在直线的函数表达式为s=?t?. ··· 7分

6039k?b3.77=??b=.?7?160160②∵s=?t?,∴当s=0时,?t?=0,解得t=60.

7777答:该运动员跑完赛程用时60 min. ····································································· 10分 24.解:(1)设该商场购进LED灯泡x个,普通白炽灯泡的数量为y个.根据题意,得

?x?y=300, ·········································································· 2分 ?(60?45)x?(0.9?30?25)y3200.=??x=200,解得? ·············································································································· 4分

y100.=?答:该商场购进LED灯泡与普通白炽灯泡的数量分别为200个和100个. ······· 5分 (2)设该商场再次购进LED灯泡a个,这批灯泡的总利润为W元.则购进普通白炽灯泡(120-a)个.根据题意得

W=(60-45)a+(30-25)(120-a)=10a+600. ················································· 7分 ∵10a+600≤[45a+25(120-a)]×30%,解得a≤75, ······································· 9分 ∵k=10>0,∴W随a的增大而增大,

∴a=75时,W最大,最大值为1350,此时购进普通白炽灯泡(120-75)=45个. 答:该商场再次购进LED灯泡75个,购进普通白炽灯泡45个,这批灯泡的总利润为1 350元. ······················································································································ 10分 25.解:(1)证明:∵AE=EC,BE=ED,

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∴四边形ABCD是平行四边形. ················································································ 2分 ∵AD为直径,∴∠AED=90°,即AC⊥BD. ··························································· 3分 ∴四边形ABCD 是菱形. ···························································································· 5分 (2)由(1)知,四边形ABCD是菱形. ∴AD=DC,DE⊥AC.∴∠ADE=∠CDE. 如图,过点C作CG⊥AD,垂足为G,连接FO. ∵BF切圆O于点F,∴OF⊥AD,且OF=

1AD=3. 2D

FCEBOGA∵BC∥AD,OF⊥BC,CG⊥AD,∴CG=OF=3. 在Rt△CDG中,sin∠ADC=

CG31==,∴∠ADC=30°. ································ 7分 CD62连接OE,∵菱形ABCD中,AE=EC,AO=OD, ∴OE∥DC,∴∠AOE=∠ADC=30°.

AE的长=∴?30?π?3π=. ··················································································· 10分 218026.解:(1)45°. ············································································································ 3分

(2)如图1,过点C作CD⊥AB于点D. yyCCCFB

AD

BBOGAExOMANx图

1 图2 图3

在Rt△ACD中,∠A=45°,∴AC=2DC. ···························································· 3分 在Rt△BCD中,∠B=30°,∴BC=2DC. ································································· 4分 ∴

BC=2.∴△ABC是智慧三角形. ·································································· 7分 AC(3)由题意可知∠ABC=90°或∠BAC=90°.

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