平谷区2016～2017学年度第一学期期末质量监控试卷
初三数学
一、选择题（本题共30分，每小题3分）下面各题均有四个选项，其中只有一个是符合题意的. ．．
1. 若3x=2y(xy≠0)，则下列比例式成立的是
A．xyx2x3xy? B．? C．? D．?3yy22332
2．剪纸是国家级非物质文化遗产，下列剪纸作品中不．是．轴对称图形的是
A． B． C． D．
3．将抛物线y=3x2向上平移2个单位后得到的抛物线的表达式为
22 A．y?3?x?2?B．y?3?x?2?C．y?3x2?2 D．y?3x2?2
4．在Rt△ABC中，∠C=90°，AC=2，BC=3，则tanA的值是
32 B． C
． D
．23
5．在公园的O处附近有E，F，G，H四棵树，位置如图所示（图中小正方形的边长均相等），现计划修建一座以O为圆心，OA为半径的圆形水池，要求池中不留树木，则E，F，G，H四棵树中需要被移除的为
A．E，F，GB．F，G，HC．G，H ，ED．H，E，FA．
6．如图，在△ABC中，点D，E分别在AB，AC上，且DE∥BC，AD=1，BD=2，那么
A．1:2B．1:3 C．1:4D．2:3
第5题图
第6题图
第7题图
7．如图，在⊙O中，AB为直径，BC为弦，CD为切线，连接OC．若∠BCD=50°，
则∠AOC的度数为
A．40° B．50° C．80°D． 100°
8．如图，二次函数y?ax2?bx?c?a?0?的图象，当?5?x?0时，下列说法
正确的是
A．有最小值﹣5，最大值0；B．有最小值2，最大值6；
C．有最小值0，最大值6；D．有最小值﹣3，最大值6．
AE的值为AC
1
9．某超市按每袋20元的价格购进某种干果.在销售过程中发现，该种干果每天的销售量w（袋）与销售单价x（元）满足w=﹣2x+80（20
≤x≤40）．那么销售这种干果每天的利润y（元）与销售单价x（元）之间的关系式为
A．y=
x﹣20 B．y=﹣2x+80 C．y???2x?80??x?20? D．y=20（﹣2x+80）
10．如图，点A的坐标为（0，1），点B是x轴正半轴上的一动点，以AB为边作等腰直角?ABC，使
?BAC?90?，设点B的横坐标为x，点C的纵坐标为y，能表示y与x的函数关系的图象大致是
A． B． C． D．
二、填空题（本题共18分，每小题3分） 11．函数y?
1
的自变量x的取值范围是 . x?2
12．如图，AB是⊙O的直径，CD是弦，∠ABD=60°，则∠C=°． 13．请写出一个在各自象限内，y的值随x值的增大而减小的反比例函数表达式 ．
14．如图，正方形ABCD内接于⊙O，⊙O的半径为1，则劣?AB的弧长是 ．
15．《九章算术》是我国古代内容极为丰富的数学名著，书中有下列问题“今有勾八步，股十五步，问勾中容圆径几何？”其意思是：“今有直角三角形，勾（短直角边）长为8步，股（长直角边）长为15步，问该直角三角形能容纳的圆形（内切圆）直径是多少？”小米想：要想求内切圆的直径长，只需要求出半径的长即可．因此设内切圆的半径为r,则AF= （用含r的代数式表示）.根据题意，所列方程为 ．
B
16
请回答：该作图的依据是
．三、解答题（本题共72分，第17-26题，每小题5分，第27题7分，第28题8，第29题7）解答应写出文字说明、演算步骤或证明过程.
17．计算：1?2sin60??1?2tan45°?
18．已知：抛物线y?x2?2ax?3a，经过点（2，﹣3）.
（1）求a的值； （2）求出抛物线与x轴、y轴的交点的坐标．
19．如图，在Rt△ABC和Rt△CDE中，∠B =∠D=90°，C为线段BD上一点，且AC⊥CE于C．
求证：Rt△ABC∽Rt△CDE．
20．如图，当宽为2cm的刻度尺的一边与⊙O相切于点C时，另一
边与⊙O交于A，B两点，读数如图（单位：），求⊙O的半径．
21．如图，小东在教学楼距地面8米高的窗口C处，测得正前方旗杆顶部A
点的仰角为37°，旗杆底部B的俯角为45°，求旗杆AB的高度.
（参考数据：sin37°≈0.60，con37°≈0.80，tan37°≈0.75）
22．在平面直角坐标系xOy中，直线y=kx+b（k≠0）与双曲线y?4的一x
个交点为A（1，m）．
（1
）求
m的值；
（2）直线y=kx+b（k≠0）又与y轴交于点B，过A作AC⊥x轴于C，若
AC=2OB，求直线y=kx+b（k≠0）的表达式．
3
23．如图，是一座古拱桥的截面图，拱桥桥洞上沿是抛物线形状，拱桥的跨度为10m，桥洞与水面的最大距离是5m，桥洞两侧壁上各有一盏距离水面4m的景观灯，求
两盏景观灯之间的水平距离（提示：请建立平面直角坐标系后，
再作答）．
24．在△ABC中，AB=15，BC=14，AC=13，求△ABC的面积．
某学习小组经过合作交流，给出了下面的解题思路，请你按照他们的解题思路完成解答过程． ．．．．．．．．．．．．．．．．．
25
．小聪是一名爱学习的孩子，他学习完二次函数后函数y=x2（x﹣3）的图象和性质进行了探究，探究过程如下，请补充完整．
其中m= ；
（2）如图，在平面直角坐标系xOy中，描出了以上表中各对对应
值为坐标点，根据描出的点，画出该函数的图象；
（3）观察函数图象，写出一条该函数的性
质 ；
（4）进一步探究函数图象发现：
①函数图象与x轴有交点，所以对应的方程x2（x﹣3）=0有
个互为不相等的实数根；
②若关于x的方程x2（x﹣3）=a有3个互为不相等的实数根，则
a的取值范围是
26．如图，已知⊙O的直径AB=10，弦AC=6，∠BAC的平分线交⊙O于点D，过点D作DE⊥AC交AC的延长线于点E． （1）求证：DE是⊙O的切线；
（2）求DE的长．
4
27．已知，抛物线C1：y?mx?4mx?4m?1?m?0? 经过点（1,0）. 2
（1）直接写出抛物线与x轴的另一个交点坐标；
（2）①求m的值；
②将抛物线C1的表达式化成y?(x?h)2?k的形式，并写出顶点A
的坐标；
（3）研究抛物线C2：y?kx?4kx?3?k?0?，顶点为点B. 2
①写出抛物线C1，C2共有的一条性质；
②若点A，B之间的距离不超过2，求k的取值范围.
28．如图，在△ABC中，∠BAC=90°，AB=AC，点D是△ABC内一动点（不包括△ABC的边界），连接AD．将线段AD绕点A顺时针旋转90°，得到线段AE．连接CD，BE．
（1）依据题意，补全图形；
（2）求证：BE=CD．
（3）延长CD交AB于F，交BE于G．
①求证：△ACF∽△GBF；
②连接BD，DE，当△BDE为等腰直角三角形时，请你直接写出AB：BD的值． ．．
B
B备用图 5
29．定义：若点P（a，b）在函数y?1的图象上，将以a为二次项系数，b为一次项系数构造的二次函x
数y=ax2+bx称为函数y?1的一个“二次派生函数”． x
（1）点（2，11）在函数y?的图象上，则它的“二次派生函数”是 ； 2x
（2）若“二次派生函数” y=ax2+bx经过点（1,2），求a,b的值；
（3）若函数y=ax+b是函数y?1的一个“一次派生函数”，在平面直角坐标系xOy中，同时画出“一x
次派生函数” y=ax+b和“二次派生函数” y=ax2+bx的图象，当﹣4＜x＜1时，“一次派生函数”始终大于“二次派生函数”，求点P的坐标.
6
平谷区2016～2017学年度第一学期期末初三数学答案及评分参考
一、选择题（本题共30分，每小题3分）
二、填空题（本题共18分，每小题3分）
11．x≠2；12．30；13．答案不唯一，如y?
11
；14．π； x2
15．8﹣r； ········································································································· 1
?23?2r?
2
······························· 3 ?82?152（答案不唯一，等价方程即可） ；
16．·································· 1 （1）到线段两端距离相等的点在线段的垂直平分线上； ·
（2）直径所对的圆周角是直角； ······················································
······
·····
2
（3）两点确定一条直线． ·········································································· 3 （其他正确依据也可以）．
三、解答题（本题共72分，第17-26题，每小题5分，第27题7分，第28题8，第29题7）解答应写出文字说明、演算步骤或证明过程. 17．解：原式1+2····································································· 4 1 ·
=0. ······································································································· 5 18．解：（1）∵抛物线y?x2?2ax?3a，经过点（2，﹣3），
··································································· 1 ∴4?4a?3a??3. ·
··················································································· 2 解得 a=1. ·
··············································· 3 （2）∴抛物线表达式为y?x2?2x?3. ·
令y=0，得x2?2x?3?0. 解得 x1=﹣1，x2=3.
····································· 4 ∴抛物线与x轴的交点为（﹣1,0），（3,0）. ·
令x=0，得 y=﹣3.
············································ 5 ∴抛物线与y轴的交点为（0，﹣3）. ·
19．证明：∵在Rt△ABC中，∠B =90°，
······································································· 1 ∴∠A +∠ACB=90°．
∵AC⊥CE于C，
·································································· 2 ∴∠ACB +∠DCE=90°． ·
············································································ 3 ∴∠A=∠DCE． ················································································ 4 ∵∠B =∠D， ·
································································· 5 ∴Rt△ABC∽Rt△CDE． ·20
．解：连接OA.
由题意可知，OC⊥AB于D. ∴AD=BD. ·································· 1 ∵AB=8， ∴AD=4. ····································· 2 设OA=r，则OD=r﹣2.·················· 3
7
∴r2??r?2?2?4. ················ 4
解得 r=5. ·································· 5
∴⊙O的半径为5.
21．解：过点C作CE⊥AB于E， ·············· 1
∴BE=8. ····································· 2
在Rt△CBE中，∠BCE=45°，
∴CE=BE=8. ······························· 3
在Rt△ACE，∠ACE=37°，
∴AE=CE×tan37°≈8×0.75=6. ··········· 4
∴AB=AE+BE=6+8=14（米）. ········· 5
答：旗杆AB的高度为14米. ····
22．解：（1）∵点A（1，m）在反比例函数y?4
x上，
∴m?4． ····················································································· 1
（2）∴A（1，4）．
∴k+b=4． ······················································································ 2
∵AC⊥x轴于C，
∴AC=4．
∵AC=2OB，
∴OB=2． ······················································································· 3
∴B点坐标为（0,2）或（0，﹣2）．
当B（0,2）时，b=2，k=2，
∴y=2x+2． ····················································································· 4
当B（0, ﹣2）时，b=﹣2，k=6，
∴y=6x﹣2． ···················································································· 5
综上所述，直线的表达式为y=2x+2或y=6x﹣2．
23．解：如图，建立坐标系. ··················································································
1
由题意可知，抛物线经过（5,﹣5）点.
设抛物线表达式为y=ax2(a≠0). ···································································· 2 ∴25a=﹣5.
解得 a??1
5. ∴抛物线表达式为y??1
5x2. ···································································· 3
∵桥洞两侧壁上各有一盏距离水面4m的景观灯，
∴y=﹣1. ······························································································· 4 ∴?1??
1
5x2.
8
解得
x?
∴两盏景观灯之间的水平距离为· ······················································· 5 （答案不唯一，请酌情给分）
24．解：过A作AD⊥BC于D， ··············································· 1
在△ABC中，AB=15，BC=14，AC=13，
设BD?x，∴CD?14?x．
由勾股定理得：AD2?AB2?BD2?152?x2， AD2?AC2?CD2?132?(14?x)2，
∴152?x2?132?(14?x)2， ······································· 2
解之得：x?9． ·················································· 3
∴BD=9，CD=5．
∴AD?12． ··························································· 4 ∴S11
?ABC?2BC?AD?2?14?12?84． ·························· 5
25.解：（1）m=﹣2； ···························································································· 1
（2）如图，图象正确； ··············································································
2
（3）答案不唯一：如该函数图象经过原点； ···················································· 3
（4）①2； ································································································ 4
②﹣4＜a＜0. ······················································································ 5
26．（1）证明：连接OD．
∵AD平分∠BAC，
∴∠DAE=∠DAB． ································· 1 ∵OA=OD，
∴∠ODA=∠DAO． ∴∠ODA=∠DAE． D
∴OD∥AE． ········································· 2
∵DE⊥AC，
∴OD⊥DE． ········································· 3 ∴DE是⊙O切线．
（2）解：连接BC，与OD交于点F．
∵OD∥AE，
∴∠ACB=∠OFD=90°．
∴四边形CEDF是矩形．
9
∵直径AB=10，弦AC=6，
∴BC=8． ············································· 4
∴FC=4．
∴DE=4． ············································· 5
27．解：（1）（3,0）；···························································································· 1
（2）①把（1,0）代入y?mx2?4mx?4m?1?m?0?中，
解得m=1.. ························································································ 2
②∴抛物线C1的表达式为：y?x2?4x?3??x?2?2?1．· ······················ 3
∴顶点A的坐标为（2，﹣,1）． ·························································· 4
（3）①答案不唯一：如
对称轴都是x=2， ············································································· 5
②y?kx2?4kx?3?k?0?
=k?x?2?2?4k?3
∴B（2,﹣4k+3）.
∵点A，B之间的距离不超过2，
∴点B的界点坐标可能为（2,1）或（2，﹣3）. ········································ 6
当点B的坐标为（2,1）时，k?1
2.
当点B的坐标为（2，﹣3）时，k?3
2.
∴k的取值范围为：1
2?k?3
2. ····························································· 7
28．（1）依据题意，画图正确，如图1． ·································································· 1
（2）证明：如图1，由题意，得AD=AE，∠DAE=90°．
∵∠BAC=90°，
∴∠CAD+∠BAD=∠BAE+∠BAD=90°．
∴∠CAD=∠BAE． ······································································· 2
∵AB=AC，
∴△CAD≌△BAE． ······································································· 3
∴CD=BE． ················································································· 4
（3）证明：①如图2，
∴∠ACD=∠ABE． ······································································· 5
∵∠AFC=∠GFB．
∴△ACF∽△GBF． ······································································· 6
②当∠EDB=90°时，如图3
，AB:
BD ······················· 7
当∠BED=90°时，如图4，AB:BD?2．·······
···················· 8
B图1 B图2 10
B 图3 B图4
29．解：（1）y?2x2?1
2x； ··············································································· 1
（2）∵（1,2），
∴a+b=2. ·························································································· 2 ∵ab=1，
∴a（2﹣a）=1. ················································································· 3 解得a=1.
∴b=1. ····························································································· 4 （3）当x=﹣4时，代入函数表达式得
y=16a﹣4b
y=﹣4a+b
∴16a﹣4b=﹣4a+b.
即b=4a.
∵ab=1，
∴a??1
2. ························································································ 5 ∴b??2.
∵当﹣4＜x＜1时，“一次派生函数”始终大于“二次派生函数”， ∴y?1
2x2?2x，y?1
2x?2. ····························································· 6
∴P??1?
?2,2??. ······················································································ 7
11
12
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