江苏省南京市建邺区2017年中考一模数学试题(含答案)

24．(7分)某水果店销售樱桃，其进价为40元/千克，按60元/千克出售，平均每天可售出

100千克．经调查发现，这种樱桃每降价1元/千克，每天可多售出10千克，若该水果店销售这种樱桃要想每天获利2240元，每千克樱桃应降价多少元？

25．(9分)已知一元二次方程x2－4mx＋4m2＋2m－4＝0，其中m为常数．

（1）若该一元二次方程有实数根，求m的取值范围． （2）设抛物线y＝x2－4mx＋4m2＋2m－4的顶点为M，点O为坐标原点，当m变化时，

求线段MO长度的最小值．

26．(12分)今年暑假，小勇、小红打算从城市A到城市B旅游，他们分别选择下列两种交通方

案：

方案一：小勇准备从城市A坐飞机先到城市C，再从城市C坐汽车到城市B，整个行程中，乘飞机所花的时间比汽车少用3h．如图1所示，城市A、B、C在一条直线上，且A、C两地的距离为2400km，飞机的平均速度是汽车的8倍．

方案二：小红准备坐高铁直达城市B，其离城市A的距离y2（km）与出发时间x（h）之间的函数关系如图2所示．

（1）AB两地的距离为 ▲ km； （2）求飞机飞行的平均速度；

（3）若两家同时出发，请在图2中画出小勇离城市A的距离y1与x之间的函数图像，并求

出y1与x的函数关系式．

y（km） 3000 A

图1

C

B

2400 1800 1200 600 O 1 2 3 4 5 6 7 8 9 10 11 x（h）

图2

（第26题）

OP

27．(12分)定义：当点P在射线OA上时，把的值叫做点P在射线OA上的射影值；当

OA

点P不在射线OA上时，把射线OA上与点P最近点的射影值，叫做点P在射线OA上的射影值．例如：如图1，△OAB三个顶点均在格点上，BP是OA边上的高，则点POP 1

和点B在射线OA上的射影值均为＝．

OA3

B B B D O

A

C

O P A O A C 图1 图2

图3

（第27题）

（1）在△OAB中，

①点B在射线OA上的射影值小于1时，则△OAB是锐角三角形； ②点B在射线OA上的射影值等于1时，则△OAB是直角三角形； ③点B在射线OA上的射影值大于1时，则△OAB是钝角三角形． 其中真命题有

A．①② B．②③ C．①③ D．①②③

（2）已知：点C是射线OA上一点，CA＝OA＝1，以O为圆心，OA为半径画圆，点

B是⊙O上任意点． 1

①如图2，若点B在射线OA上的射影值为．求证：直线BC是⊙O的切线．

2

②如图3，已知D为线段BC的中点，设点D在射线OA上的射影值为x，点D在射线OB上的射影值为y，直接写出y与x之间的函数关系式．

2017年中考第一次模拟测试卷 数学参考答案及评分标准

说明：本评分标准每题给出了一种或几种解法供参考，如果考生的解法与本解答不同，参照本评分标准的精神给分．

一、选择题（每小题2分，共计12分）

题号 答案 1 D 2 C 3 B 4 D 5 C 6 C 二、填空题（每小题2分，共计20分）

7．x≥2 8．1.2629×104 9．a (a－1)2 10．0 11．4 811

12．（-1，3） 13．90° 14．45° 15．π 16． 153三、解答题（本大题共10小题，共计88分） 17．（本题6分）

m＋1(m＋2)(m－2)

解：原式＝? ········································································· 2分

m＋2(m＋2)2＝

m－2

······························································································ 4分 m＋1

1－21

当m ＝1时，原式＝ ＝－． ························································· 6分

21＋1

18．（本题7分）

解：解不等式①，得x≤1． ·············································································· 2分

解不等式②，得x＞－2． ·········································································· 4分 所以，不等式组的解集是－2＜x≤1． ······················································· 5分 画图正确（略）． ···················································································· 7分 19．（本题7分）

（1）126； ···································································································· 2分 （2）图略； ·································································································· 4分 （3）在抽取的样本中，“比较喜欢”数学的人数所占的百分比为

1－32%－10%－23%＝35%， ····································································· 5分 由此可估计，该校1000名学生中，“比较喜欢”数学的人数所占的百分比35%， 1000×35%＝350（人）． ········································································· 6分 答：估计这些学生中，“比较喜欢”数学的人数约有350人． ···························· 7分

20．（本小题满分8分）

证明：（1）∵ 四边形ABCD是平行四边形，∴ AB//CD，AB＝DC．∴ ∠ABC＝∠DCE． ∵ AC//DE，∴ ∠ACB＝∠DEC． ································································· 3分

在△ABC和△DCE中，∠ABC＝∠DCE，∠ACB＝∠DEC ，AB＝DC．

∴△ABC≌△DCE（AAS）． ································································ 4分 （2）由（1）知△ABC≌△DCE，则有BC＝CE． ∵ CD＝CE， ∴ BC＝CD．

∴四边形ABCD为菱形． ·········································································· 7分 ∴AC⊥BD． ························································································· 8分 21．（本题7分）

列表或树状图表示正确； ·········································································· 3分 ∵共有8种等可能的结果， 通过一次“手心手背”游戏， 小明先跳绳的有2种情况····································· 5分

2 1 ∴通过一次“手心手背”游戏，小明先跳绳的概率是： ＝ ． 84答：通过一次“手心手背”游戏，小明先跳绳的概率是 1 ． ······························· 7分 422．（本题6分） 方法1： 方法2： ··················································································································· 6分 23．（本题7分）

解：过点A作AD⊥OB于点D.

由题意得AN⊥MN，OB⊥MN，AD⊥OB，∴四边形ANMD是矩形，

O ∴DM＝AN， ····························································································· 2分

设OB＝OA＝x cm，在Rt?OAD中，∠ODA＝90°， x＋5－14A C ODD cos∠AOD＝ ＝ ≈0.6． ······························································ 5分

OAx解得x＝15cm．

B 经检验，x＝15为原方程的解． 答：细线OB的长度是15cm． ······································································ 7分 N M

24．（本小题满分7分）

解：设每千克樱桃应降价x元，根据题意，得 ······················································ 1分

（60－x－40）（100＋10x） ＝ 2240． ······················································· 4分 解得：x1＝4，x2＝6． ··············································································· 6分

答：每千克樱桃应降价4元或6元． ··························································· 7分