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大兴区 2023~2024学年度第一学期期末检测
初三数学参考答案及评分标准
一、选择题（共 16分，每题 2分）
题号 1 2 3 4 5 6 7 8
答案 C A A B D D C A
二、填空题（共 16分，每题 2分）
题号 9 10 11 12 13 14 15 16
答案不唯一，如：
答案 a 3 2 = 80 16 120（0 1+ x）2 =1452 ②③
y = x2 + 1
三、解答题（共 68 分，第 17-21 题每题 5 分，第 22 题 6 分，第 23 题 5 分，第 24-26 题每
题 6分，第 27-28题，每题 7分）解答应写出文字说明、演算步骤或证明的过程.
17. 解： x2+8x=9.
x2+8x +16=9+16. ··································································· 1 分
(x+4)2=25. ………………………………………………………………2 分
x+4=±5. ············································································· 3 分
解得 x1=1，x2=-9. ································································ 5 分
2
18. 解： (a 1) + a(a 2)
= a
2 2a +1+ a2 2a ····························································· 2 分
2
= 2a 4a +1 ········································································ 3 分
∵a 是方程 x2 2x 1= 0的一个根，
∴ a2 2a 1= 0 ，
∴ a2 2a =1． ······································································· 4 分
∴原式 = 2（a
2-2a）+1
= 2 1+1
=3 ·············································································· 5 分
初三数学参考答案及评分标准第1页（共 6 页）
19. 解：
（1）∵方程有两个实数根，
0 ················································································· 1 分
∵Δ＝(－1) 2－4×1×(2m－2)
=1 8m+8
= 9 8m
9 8m 0
9
m ················································································ 2 分
8
9
（2） m ，m 为最大整数，
8
m =1. ··············································································· 3 分
∴x2﹣x＝0.
解得：x1=0，x2=1. ································································ 5 分
20.解：
（ ）∵抛物线 y = x21 +bx + c 经过点（1，0），（0，-3），
1+b + c = 0
∴ .··········································································2分
c = 3
b = 2
解得 .
c = -3
∴ y = x2 + 2x-3． ·····································································3分
（2）y = x2 + 2x-3 .
2
= (x +1) -4
∴顶点坐标为（-1，-4）. ··························································· 5 分
初三数学参考答案及评分标准第2页（共 6 页）
21. 解：连接 OA，OB，············································1 分
∵∠C＝45°,
∴∠AOB＝2∠C =90°. ··········································2 分
在 Rt△AOB 中，
∵OA2+OB2=AB2, AB=2，OA=OB，
∴2 OA2=4. ························································4 分
∴ OA2=2.
∴OA= 2 （舍负）.
∴⊙O 的半径是 2 . ···········································5 分
22.解：
（1）m＝95，n＝90.5，九年级抽取的学生竞赛成绩在 D 组的人数为 4 人； ···· 3 分
（2）240. ····················································································· 4 分
（3）设 D 组的另外两名同学为丙，丁.
宣讲员 甲 乙 丙 丁
主持人 乙 丙 丁 甲 丙 丁 甲 乙 丁 甲 乙 丙
由树状图可以看出，所有可能出现的结果共 12 种，这些结果出现的可能性相等．
甲和乙同时被选上的结果有 2 种，
2 1
所以 P（甲乙同时被选上）= = ． ································································ 6 分
12 6
23. 解：
（1）把 A（-1,2）和 B（1,4）代入 y=kx+b(k≠0)中，
k + b = 2,
………………………………………………………………1 分
k + b = 4.
k =1,
解得： ………………………………………………………………2 分
b = 3.
所以该函数的解析式为 y=x+3. ················································· 3 分
（2）n=4 ······················································································· 5 分
初三数学参考答案及评分标准第3页（共 6 页）
24.（1）证明：连接 OC.
∵OB=OC，
∴∠B=∠OCB.
∵∠E=∠B，
∴∠E=∠OCB. ·······························································1 分
∵OD⊥BC，
∴∠E+∠DCE=90°.
∴∠OCB＋∠DCE＝90°.
∴∠OCE＝90°.
即 OC⊥CE.
∴CE 是⊙O 的切线．···························································2 分
（2）∵OD⊥BC，
∴∠CDE＝90°.
在 Rt△CDE 中，DE=6 , CE= 3 5 ，
∴CD= CE2 DE2 = 3. …………………………..........................……… 3 分
∵OE⊥BC，
∴BC=2CD=6.
∴DE=BC . ………………………………………………………………4 分
∵AB 是直径,
∴∠ACB＝90°.
∴∠CDE=∠ACB.
在△ABC 与△CED 中，
B = E,

BC = DE，

ACB = CDE.
∴△ABC≌△CED. ……………………………………….………5 分
∴AC=CD=3.
∵O 是 AB 的中点，D 是 BC 的中点，
1 3
∴OD = AC = . ···································································· 6 分
2 2
初三数学参考答案及评分标准第4页（共 6 页）
25.解：
（1）由题意，A 点坐标为(0,1.25),B 点坐标为(2.5,0). …………………………1 分
设抛物线的解析式为 y=a(x-1)2+k(a≠0) …………….………………….… 2 分
∵抛物线经过点 A,点 B.
1.25 = a + k ,
∴ 2
0 = a (2.5 1) + k.
a = 1,
解得：
k = 2.25.
∴y=-(x-1)2+2.25（0≤x≤2.5）. ……………………………….…………… 3 分
∴x=1 时，y=2.25.
∴水流喷出的最大高度为 2.25 m. ………………………………..……… 4 分
（2）2.7 ························································································ 6 分
26. 解：
（1）∵点（2，m）在 y = ax2 + bx + c( a 0 )上，
∴m＝4a＋2b＋c.
又∵m＝c，
∴4a＋2b＝0.
∴b＝-2a.
b 2a
∴ t = = =1. …………..………………………………………2 分
2a 2a
（2）∵点（2，m）在抛物线 y ax2 bx c(a 0)上，
∴m=4a+2b+c.
∵c < m,
∴m- c>0.
∴m-c=4a+2b >0.
∴2a +b >0. ············································································ 3 分
∵点(-1,y1),(3,y2)在抛物线 y ax2 bx c(a 0)上,
∴y1=a-b+c，y2=9a+3b+c,
∴y2-y1=(9a+3b+c)-( a-b+c)=8a+4b =4(2a+b). ································ 4 分
∵2a +b >0,
∴4(2a +b )>0,
∴y2-y1>0.
∴y2>y1. ………………………………………………………………….6 分
初三数学参考答案及评分标准第5页（共 6 页）
27. （1）解：补全图形如图所示；
························································· 1 分
（2）证明：
∵∠BAC=90°，
∴∠ACP+∠APC=90°．
∵以 P 为中心，将线段 PC 顺时针旋转 90°得到线段 PD，
∴∠DPC=90°.
∴∠APC+∠BPD=90°．
∴∠ACP=∠DPB． ···························································· 3 分
（3）线段 BC，BP，BD 之间的数量关系是 2 BP =BD+BC. ………………4 分
证明：过点 P 作 PE⊥PB 交 BC 的延长线于点 E.
∵PE⊥PB，
∴∠BPE=90°.
∵∠DPC=90°，
∴∠1+∠BPC=∠2+∠BPC=90°.
∴∠1=∠2. ······································································· 5 分
∵AB=AC，∠BAC=90°，
∴∠ABC=∠ACB=45°.
∵∠BPE=90°，
∴∠PBE=∠PEB=45°.
∴PB=PE. ········································································ 6 分
在△PBD 与△PEC 中，
PB = PE，

1= 2，

PD = PC.
∴△PBD≌△PEC.
∴BD=EC.
∵BE= BP2 + PE2 = BP2 + BP2 = 2BP .
∴ 2 BP =BD+BC． ····························································· 7 分
28. 解：
（1）① A，C； ········································································ 2 分
② ( 3 2,1)， ( 3 + 2,1)； ······················································ 5 分
（2）-11 ≤ t ≤ 3. ············································································ 7 分
初三数学参考答案及评分标准第6页（共 6 页）

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