2024北京大兴区初三(上)期末数学试卷及答案
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初三数学参考答案及评分标准第1页（共6页） 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 3a
2 = 80 16 答案不唯一，如：
2
1yx=+ 2
120011452x+=（）
②③
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x
1=1，x
2=-9. ································································ 5分
18. 解： 2
(1)(2)aaa−+−
=22
212aaaa−++−
····························································· 2分
=2
241aa−+
········································································ 3分
∵a是方程2
210xx−−=的一个根，
∴2
210aa−−=，
∴2
21aa−=
． ······································································· 4分
∴原式2
221aa=+（-）
211=+
=3 ·············································································· 5分 初三数学参考答案及评分标准第2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)
188m=−+
98m=−
980m−
9
8m
················································································ 2分
（2）9
8m
，m为最大整数，
m
=1. ··············································································· 3分
∴x2
﹣x＝0.
解得：x
1=0，x
2=1. ································································ 5分
20.
解：
（1
）∵抛物线2
+yxbxc=+
经过点（1
，0
），（0
，-3
），
∴1+0
3bc
c+=
=−
.··········································································2
分
解得2
-3b
c=
=
.
∴2
2-3yxx=+． ·····································································3
分
（2
）y =2
2-3xx+.
()2
1-4x=+
∴顶点坐标为（-1，-4）. ··························································· 5分
初三数学参考答案及评分标准第3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，
所以P
（甲乙同时被选上）=21
126=． ································································ 6分
23. 解：
（1）把A（-1,2）和B（1,4）代入y=kx+b(k≠0)中，
2
4kb,
kb.−+=
+=
………………………………………………………………1分
解得：1
3k,
b.=
=
………………………………………………………………2分
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4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
=
35，
∴CD
=
22
3CEDE.−= …………………………..........................……… 3
分
∵OE⊥BC，
∴BC=2CD=6.
∴DE=BC . ………………………………………………………………4
分
∵AB是直径,
∴∠ACB＝90°.
∴∠CDE=∠ACB.
在△ABC与△CED中，
BE,
BCDE
ACBCDE.=
=
=
，
∴△ABC
≌△CED. ……………………………………….………5
分
∴AC=CD=3.
∵O
是AB
的中点，D
是BC
的中点， ∴13
22ODAC==. ···································································· 6
分
