2012年厦门中考数学试题【完整答案版】

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 ．－1
2
2．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 －1 0 1
y
－1
1
3
则y 与x 之间的函数关系式可能是
C B 图2
D
A
图1俯视图
左视图
正视图
A ．y ＝x
B ．y ＝2x ＋1
C ．y ＝x 2＋x ＋1
D ．y ＝3
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： m 3÷m 2＝ .
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 ＝ ；
a 2＋
b 2＝ .
17．如图5，已知∠ABC ＝90°，AB ＝πr ，BC ＝πr
2
，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 .
图6
A
B
C
D
F
E
图4
A
B
C
D
E
图3
A
B
D
C
O
→
图5
A
B
C
O
19．（本题满分7分）解方程组： ⎩⎨⎧3x ＋y ＝4，
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. （1）求 AD
AB
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
S A 2＝1
7[(x 1－—x )2＋(x 2－—x )2＋(x 3－—x )2＋(x 4－—x )2＋(x 5－—x )2＋(x 6－—x )2＋(x 7－—x )2]】
图7
A B
C
D
E
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 (x 2－1)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 2
3
，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.
图8
F
B
C
E D
O
A
24．（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( 72，5
2
) 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．
E F
图10
A
B
C
D
O
P x
y
B
42
6
42O
图9
A
26．（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
x（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；
（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
x（k2＞0）
于点N．当PN
NE取最大值时，若PN＝
1
2，求此时双曲线的解析式．
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 1 2 3 4 5 6 7 选项
A
C
A
D
B
C
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. 1
2. 12.
3. 13. x ＋y ＞1.
14. 60.
15. 540°. 16. 5； 6. 17. ；2πr .
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）
解1：⎩⎨⎧3x ＋y ＝4， ①2x －y ＝1. ②
①＋②，得 ························································································ 1分 5x ＝5， ····························································································· 2分 x ＝1. ······························································································· 4分
A
B
C
D
F
E
将x ＝1代入 ①，得 3＋y ＝4， ························································································· 5分 y ＝1． ······························································································· 6分
∴⎩⎨⎧x ＝1，y ＝1.
························································································· 7分 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，y ＝1.
························································································· 7分 20．（本题满分7分）
（1）解：∵ DE ∥BC ，∴ △ADE ∽△ABC . ……1分
∴ AD AB ＝DE
BC
. ……2分
∴ AD AB ＝1
3
.
……3分
（2）解1：∵
AD AB ＝1
3
，BD ＝10， ∴
AD AD ＋10＝1
3
················································································ 4分
∴ AD ＝5 ······················································································· 5分 经检验，符合题意. ∴ AB ＝15. 在Rt △ABC 中， ·············································································· 6分 sin ∠A ＝BC AB ＝35
. ············································································· 7分
解2： ∵
AD AB ＝1
3
，BD ＝10， ∴
AD AD ＋10＝1
3
················································································ 4分
∴ AD ＝5 ······················································································· 5分 经检验，符合题意. ∵ DE ∥BC ，∠C ＝90° ∴ ∠AED ＝90° 在Rt △AED 中， ·············································································· 6分 sin ∠A ＝ED AD ＝35
. ············································································· 7分
解3：过点D 作DG ⊥BC ，垂足为G . ∴ DG ∥AC .
∴∠A ＝∠BDG . ············································································· 4分
A B
C
D
E
G
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 ∴ sin ∠BDG ＝BD GB ＝35. ·································································· 7分
∴ sin ∠A ＝3
5
.
21．（本题满分7分）
（1）解：A 组数据的平均数是0＋1－2－1＋0－1＋3
7 ·································· 1分
＝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.
∴ S B 2＝
145 ，S A 2＝16
7
. ································································ 6分 ∴ 145 ＞16
7
. ····················································································· 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.
∵S A 2＝167, S B 2＝16
5
. ································································ 6分 ∴
165＞167
························································································· 7分 ∴ B 组数据：1，－2，－1，－1，3.
22．（本题满分9分） （1）解:由题意得,
x ＝2
3(2x －2) ····················································································· 1分 ∴ x ＝4. ························································································· 2分 ∴ x 2－1＝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
1x 2－1＝1
2x －2 . ·············································································· 7分 ∴ 1x ＋1＝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 x 2－1＝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
∵ AB 为⊙O 的直径， ∴ AB ⊥CD ， ……2分 CE ＝DE . ……3分 ∴ AC ＝AD . ……4分
证明2：∵∠BCD ＝∠BAC ，
∴ ︵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
G
A
O
D
E C
B
F。