2024年福建省厦门中考数学考题及数学参考答案

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2012年厦门市初中毕业及高中阶段各类学校招生考试

数学

（试卷满分：150分考试时间：120分钟）

准考证号姓名座位号

注意事项：

1．全卷三大题，26小题，试卷共4页，另有答题卡． 2．答案一律写在答题卡上，否则不能得分． 3．可直接用2B铅笔画图．

一、选择题（本大题有7小题，每小题3分，共21分.每小题都有四个选项，其中有且只有

一个选项正确） 1．－2的相反数是

A．2 B．－2 C．±2 D．－ 22．下列事件中，是必然事件的是

A. 抛掷1枚硬币，掷得的结果是正面朝上 B. 抛掷1枚硬币，掷得的结果是反面朝上

C. 抛掷1枚硬币，掷得的结果不是正面朝上就是反面朝上

D．抛掷2枚硬币，掷得的结果是1个正面朝上与1个反面朝上

3．图1是一个立体图形的三视图，则这个立体图形是 A．圆锥 B．球

C．圆柱 D．三棱锥

4．某种彩票的中奖机会是1%，下列说法正确的是 A．买1张这种彩票一定不会中奖 B．买1张这种彩票一定会中奖 C．买100张这种彩票一定会中奖

D．当购买彩票的数量很大时，中奖的频率稳定在1%

5．若二次根式x－1有意义，则x的取值范围是 A．x＞1 B．x≥1 C．x＜1 D．x≤1

6．如图2，在菱形ABCD中，AC、BD是对角线，若∠BAC＝50°，则∠ABC等于 A．40° B．50° C．80° D．100°

7．已知两个变量x和y，它们之间的3组对应值如下表所示.

x y －1 －1 0 1 1 3 正视图俯视图图1A左视图BDC图2则y 与x之间的函数关系式可能是

A．y＝x B．y＝2x＋1

C．y＝x2＋x＋1

D．y＝ x

二、填空题（本大题有10小题，每小题4分，共40分） 8．计算： 3a－2a＝．

9．已知∠A＝40°，则∠A的余角的度数是． 10．计算： m3÷m2＝ .

11．在分别写有整数1到10的10张卡片中，随机抽取1张

卡片，则该卡片上的数字恰好是奇数的概率是． 12．如图3，在等腰梯形ABCD中，AD∥BC，对角线AC

与BD相交于点O，若OB＝3，则OC＝． 13．“x与y的和大于1”用不等式表示为 . 14．如图4，点D是等边△ABC内一点，如果△ABD绕点A

逆时针旋转后能与△ACE重合，那么旋转了度. 15．五边形的内角和的度数是．

16．已知a＋b＝2，ab＝－1，则3a＋ab＋3b＝；

a2＋b2＝ .

πr

17．如图5，已知∠ABC＝90°，AB＝πr，BC＝，半径为r

的⊙O从点A出发，沿A→B→C方向滚动到点C时停止. 请你根据题意，在图5上画出圆心．．O运动路径的示意图；圆心O运动的路程是 . 三、解答题（本大题有9小题，共89分） 18．（本题满分18分）

（1）计算：4÷(－2)＋(－1)2×40；（2）画出函数y＝－x＋1的图象；

（3）已知：如图6，点B、F、C、E在一条直线上，

∠A＝∠D，AC＝DF，且AC∥DF. 求证：△ABC≌△DEF.

AOBD图3CAEDB图4COA→B图5CABCF图6DE?3x＋y＝4，19．（本题满分7分）解方程组： ?

?2x－y＝1.

20．（本题满分7分）已知：如图7，在△ABC中，∠C＝90°，点D、E分别在边AB、AC

上，DE∥BC，DE＝3， BC＝9. AD

（1）求的值；

（2）若BD＝10，求sin∠A的值.

21.（本题满分7分）已知A组数据如下：

0，1，－2，－1，0，－1，3.

（1）求A组数据的平均数；

（2）从A组数据中选取5个数据，记这5个数据为B组数据. 要求B组数据满足两

个条件：①它的平均数与A组数据的平均数相等；②它的方差比A组数据的方差大.你选取的B组数据是，请说明理由. 【注：A组数据的方差的计算式是

SA2＝[(x1－—x)2＋(x2－—x)2＋(x3－—x)2＋(x4－—x)2＋(x5－—x)2＋(x6－—x)2＋(x7－—x)2]】

22．（本题满分9分）工厂加工某种零件，经测试，单独加工完成这种零件，甲车床需用

x小时，乙车床需用 (x2－1)小时，丙车床需用(2x－2)小时.

（1）单独加工完成这种零件，若甲车床所用的时间是丙车床的，求乙车床单独加工

完成这种零件所需的时间；

（2）加工这种零件，乙车床的工作效率与丙车床的工作效率能否相同？请说明理由.

23．（本题满分9分）已知：如图8，⊙O是△ABC的外接圆，AB为⊙O的直径，弦CD

交AB于E，∠BCD＝∠BAC . （1）求证：AC＝AD；

（2）过点C作直线CF，交AB的延长线于点F，

若∠BCF＝30°,则结论“CF一定是⊙O的切线” 是否正确？若正确，请证明；若不正确，请举反例.

CFBDEOA图824．（本题满分10分）如图9，在平面直角坐标系中，已知点A(2，3)、B(6，3)，连结AB.

如果点P在直线y＝x－1上，且点P到直线AB的距离小于1，那么称点P是线段AB的“邻近点”．

（1）判断点Ｃ( ， ) 是否是线段AB的“邻近点”，并说明理由；

（2）若点Q (m，n)是线段AB的“邻近点”，求m的取值范围．

25．（本题满分10分）已知□ABCD，对角线AC与BD相交于点O，点P在边AD上，过

点P分别作PE⊥AC、PF⊥BD，垂足分别为E、F，PE＝PF．（1）如图10，若PE＝3，EO＝1，求∠EPF的度数；（2）若点P是AD的中点，点F是DO的中点，

BF ＝BC＋32－4，求BC的长．

AEOPFCDB图10k226．（本题满分12分）已知点A(1，c)和点B (3，d )是直线y＝k1x＋b与双曲线y＝（k2

＞0）的交点．

（1）过点A作AM⊥x轴，垂足为M，连结BM．若AM＝BM，求点B的坐标； k2 （2）设点P在线段AB上，过点P作PE⊥x轴，垂足为E，并交双曲线y＝（k2＞0）

于点N．当

PN1

取最大值时，若PN＝，求此时双曲线的解析式． NE2

2012年厦门市初中毕业及高中阶段各类学校招生考试

数学参考答案及评分标准

说明：

1．解答只列出试题的一种或几种解法．如果考生的解法与所列解法不同，可参照解答中评分标准相应评分；

2．评阅试卷，要坚持每题评阅到底，不能因考生解答中出现错误而中断对本题的评阅．如果考生的解答在某一步出现错误，影响后续部分而未改变本题的内容和难度，视影响的程度决定后继部分的给分，但原则上不超过后续部分应得分数的一半； 3．解答题评分时，给分或扣分均以1分为基本单位．

一、选择题（本大题共7小题，每小题3分，共21分）

题号选项 1 A 2 C 3 A 4 D 5 B 6 C 7 B 二、填空题（本大题共10小题，每题4分，共40分）

8. a. 9. 50°. 10. m. 11. . 12. 3. 13. x＋y＞1.

215. 540°. 16. 5； 6. 17. ；2πr.

14. 60.

三、解答题（本大题共9小题，共89分） 18．（本题满分18分）

（1）解：4÷(－2) ＋(－1)2×40

＝－2＋1×1 ··································································· 4分＝－2＋1 ········································································ 5分＝－1. ·········································································· 6分

（2）解：正确画出坐标系 ······························································· 8分

正确写出两点坐标 ·························································· 10分画出直线 ······································································ 12分

（3）证明：∵ AC∥DF， ……13分

∴ ∠ACB＝∠DFE. ……15分又∵ ∠A＝∠D， ……16分 AC＝DF， ……17分 ∴ △ABC≌△EDF. ……18分

19．（本题满分7分）

ABCFDE?3x＋y＝4， ①解1：?

?2x－y＝1. ②

①＋②，得 ····································································· 1分 5x＝5， ·········································································· 2分 x＝1. ··········································································· 4分

将x＝1代入 ①，得 3＋y＝4， ······································································· 5分 y＝1． ··········································································· 6分

?x＝1，∴? ······································································ 7分 ?y＝1.

解2：由①得 y＝4－3x． ③ ········································ 1分将③代入②，得 2x－(4－3x) ＝1． ···························································· 2分得x＝1. ······································································· 4分将x＝1代入③ ，得 y＝4－3×1 ····································································· 5分＝1． ··········································································· 6分

?x＝1，∴? ······································································ 7分 ?y＝1.

20．（本题满分7分）

（1）解：∵ DE∥BC ，∴ △ADE∽△ABC. ……1分

ADDE

∴ ＝ . ……2分

ABBC

AD1

∴ ＝.

AB3

（2）解1：∵

∴

AD1

＝，BD＝10， AB3

……3分

AEDCGBAD1

＝ ······························································· 4分

AD＋103

∴ AD＝5 ····································································· 5分经检验，符合题意. ∴ AB＝15. 在Rt△ABC中， ······························································ 6分 BC3

sin∠A＝＝. ····························································· 7分

AB5

解2： ∵

∴

AD1

＝，BD＝10， AB3

AD1

＝ ······························································· 4分

AD＋103

∴ AD＝5 ····································································· 5分经检验，符合题意. ∵ DE∥BC，∠C＝90° ∴ ∠AED＝90° 在Rt△AED中， ······························································ 6分 ED3sin∠A＝＝. ····························································· 7分

AD5

解3：过点D作DG⊥BC，垂足为G. ∴ DG∥AC.

∴∠A＝∠BDG. ····························································· 4分

又∵ DE∥BC，∴四边形ECGD是平行四边形. ∴ DE＝CG. ··································································· 5分 ∴ BG＝6.

在Rt△DGB中， ····························································· 6分 BD3∴ sin∠BDG＝＝. ···················································· 7分

GB53

∴ sin∠A＝.

21．（本题满分7分）

0＋1－2－1＋0－1＋3

（1）解：A组数据的平均数是 ··························· 1分

7＝0. ················································· 3分

（2）解1：选取的B组数据：0，－2，0，－1，3. ····························· 4分

∵ B组数据的平均数是0. ··············································· 5分 ∴ B组数据的平均数与A组数据的平均数相同.

∴ SB2＝

1416 ，SA2＝ . ··················································· 6分 57

1416

∴ ＞. ··································································· 7分

∴ B组数据：0，－2，0，－1，3.

解2：B组数据：1，－2，－1，－1，3. ··································· 4分

∵ B组数据的平均数是0. ··············································· 5分 ∴ B组数据的平均数与A组数据的平均数相同.

∵SA2＝∴

1616, SB2＝ . ···················································6分 75

1616＞ ······································································ 7分 57

∴ B组数据：1，－2，－1，－1，3.

22．（本题满分9分）（1）解:由题意得,

x＝(2x－2) ··································································· 1分 3∴ x＝4. ······································································· 2分 ∴ x2－1＝16－1＝15(小时). ············································· 3分答：乙车床单独加工完成这种零件所需的时间是15小时. ······· 4分

（2）解1：不相同. ······································································ 5分

若乙车床的工作效率与丙车床的工作效率相同，由题意得, ······ 6分

＝ . ······························································ 7分 x2－12x－211∴ ＝. x＋12

∴ x＝1. ····································································· 8分经检验，x＝1不是原方程的解. ∴ 原方程无解. ··················· 9分答：乙车床的工作效率与丙车床的工作效率不相同.

解2：不相同. ······································································ 5分

若乙车床的工作效率与丙车床的工作效率相同，由题意得, ······ 6分 x2－1＝2x－2. ································································ 7分解得，x＝1. ·································································· 8分此时乙车床的工作时间为0小时，不合题意. ························ 9分答：乙车床的工作效率与丙车床的工作效率不相同.

23．（本题满分9分）

（1）证明1：∵∠BCD＝∠BAC，

︵︵

∴ BC＝BD .

……1分

CG∵ AB为⊙O的直径， ∴ AB⊥CD， ……2分 CE＝DE. ……3分 ∴ AC＝AD . ……4分

证明2：∵∠BCD＝∠BAC，

FBDEOA︵︵

∴ BC＝BD . ····························································· 1分︵︵

∵ AB为⊙O的直径， ∴ BCA＝BDA . ··························· 2分︵︵

∴ CA＝DA . ································································ 3分

∴ AC＝AD . ································································ 4分

证明3：∵ AB为⊙O的直径，∴ ∠BCA＝90°. ······················· 1分

∴ ∠BCD+∠DCA＝90°, ∠BAC+∠CBA＝90° ∵∠BCD＝∠BAC，∴∠DCA＝∠CBA ································ 2分

︵︵

∴ CA＝DA . ································································ 3分

∴ AC＝AD . ································································ 4分

（2）解1：不正确. ······································································ 5分

连结OC.

当 ∠CAB＝20°时， ························································ 6分 ∵ OC＝OA，有 ∠OCA＝20°.

∵ ∠ACB＝90°， ∴ ∠OCB＝70°. ··························· 7分又∵∠BCF＝30°， ∴∠FCO＝100°， ·························································· 8分 ∴ CO与FC不垂直. ······················································ 9分 ∴ 此时CF不是⊙O的切线.

解2：不正确. ······································································ 5分

连结OC.

当 ∠CAB＝20°时， ························································ 6分 ∵ OC＝OA，有 ∠OCA＝20°.

∵ ∠ACB＝90°， ∴ ∠OCB＝70°. ··························· 7分又∵∠BCF＝30°， ∴∠FCO＝100°， ·························································· 8分

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